MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley [2nd Custom ed.] 0137114036, 9780137114030

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MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Cover Art: Courtesy of Photodisc/Getty Images Copyright © 2021 by Pearson Learning Solutions All rights reserved. Permission in writing must be obtained from the publisher before any part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying and recording, or by any information storage or retrieval system. Additional copyright information is included, where applicable, as a footnote at the beginning of each chapter.

Attention bookstores: For permission to return any unsold stock, contact us at [email protected] Pearson Learning Solutions, 501 Boylston Street, Suite 900, Boston, MA 02116 A Pearson Education Company www.pearsoned.com ISBN 10: 0137114036 ISBN 13: 9780137114030 Printed in the USA

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

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MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Preface Our Goal Fundamentals of Differential Equations is designed to serve the needs of a one-semester course in basic theory as well as applications of differential equations. The flexibility of the text provides the instructor substantial latitude in designing a syllabus to match the emphasis of the course. Sample syllabi are provided in this preface that illustrate the inherent flexibility of this text to balance theory, methodology, applications, and numerical methods, as well as the incorporation of commercially available computer software for this course.

New to This Edition • This text now features a MyMathLab course with approximately 750 algorithmic online homework exercises, tutorial videos, and the complete eText. Please see the “Technology and Supplements” section below for more details. • In the Laplace Transforms chapter (7), the treatments of discontinuous and periodic functions are now divided into two sections that are more appropriate for 50 minute lectures: Section 7.6 “Transforms of Discontinuous Functions” (page 383) and Section 7.7 “Transforms of Periodic and Power Functions” (page 392). • New examples have been added dealing with variation of parameters, Laplace transforms, the Gamma function, and eigenvectors (among others). • New problems added to exercise sets deal with such topics as axon gating variables and oscillations of a helium-filled balloon on a cord. Additionally, novel problems accompany the new projects, focusing on economic models, disease control, synchronization, signal propagation, and phase plane analyses of neural responses. We have also added a set of Review Problems for Chapter 1 (page 29). • Several pedagogical changes were made including amplification of the distinction between phase plane solutions and actual trajectories in Chapter 5 and incorporation of matrix and Jacobian formulations for autonomous systems. • A new appendix lists commercial software and freeware for direction fields, phase portraits, and numerical methods for solving differential equations. (Appendix G, page A-17.) • “The 2014–2015 Ebola Epidemic” is a new Project in Chapter 5 that describes a system of differential equations for modelling for the spread of the disease in West Africa. The model incorporates such features as contact tracing, number of contacts, likelihood of infection, and efficacy of isolation. See Project F, page 314. • A new project in Chapter 1 called “Applications to Economics” deals with models for an agrarian economy as well as the growth of capital. See Project C, page 35. xiii

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

xiv

Preface

• A new project in Chapter 4 called “Gravity Train” invites to reader to utilize differential equations in the design of an underground tunnel from Moscow to St. Petersburg, Russia, using gravity for propulsion. See Project H, page 240. • Phase-locked loops constitute the theme of a new project in Chapter 5 that utilizes differential equations to analyze a technique for measuring or matching high frequency radio oscillations. See Project G, page 317. • A new Project in Chapter 10 broadens the analysis of the wave and heat equations to explore the telegrapher’s and cable equations. See Project E, page 637.

Prerequisites While some universities make linear algebra a prerequisite for differential equations, many schools (especially engineering) only require calculus. With this in mind, we have designed the text so that only Chapter 6 (Theory of Higher-Order Linear Differential Equations) and Chapter 9 (Matrix Methods for Linear Systems) require more than high school level linear algebra. Moreover, Chapter 9 contains review sections on matrices and vectors as well as specific references for the deeper results used from the theory of linear algebra. We have also written Chapter 5 so as to give an introduction to systems of differential equations—including methods of solving, phase plane analysis, applications, numerical procedures, and Poincaré maps—that does not require a background in linear algebra.

Sample Syllabi As a rough guide in designing a one-semester syllabus related to this text, we provide three samples that can be used for a 15-week course that meets three hours per week. The first emphasizes applications and computations including phase plane analysis; the second is designed for courses that place more emphasis on theory; and the third stresses methodology and partial differential equations. Chapters 1, 2, and 4 provide the core for any first course. The rest of the chapters are, for the most part, independent of each other. For students with a background in linear algebra, the instructor may prefer to replace Chapter 7 (Laplace Transforms) or Chapter 8 (Series Solutions of Differential Equations) with sections from Chapter 9 (Matrix Methods for Linear Systems).

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Preface

Week

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Methods, Computations, and Applications

Theory and Methods (linear algebra prerequisite)

Methods and Partial Differential Equations

Sections

Sections

Sections

1.1, 1.2, 1.3 1.4, 2.2 2.3, 2.4, 3.2 3.4, 3.5, 3.6 3.7, 4.1 4.2, 4.3, 4.4 4.5, 4.6, 4.7 4.8, 4.9 4.10, 5.1, 5.2 5.3, 5.4, 5.5 5.6, 5.7, 7.2 7.3, 7.4, 7.5 7.6, 7.7, 7.8 8.1, 8.2, 8.3 8.4, 8.6

1.1, 1.2, 1.3 1.4, 2.2, 2.3 2.4, 3.2, 4.1 4.2, 4.3, 4.4 4.5, 4.6 4.7, 5.2, 5.3 5.4, 6.1 6.2, 6.3, 7.2 7.3, 7.4, 7.5 7.6, 7.7, 7.8 8.2, 8.3 8.4, 8.6, 9.1 9.2, 9.3 9.4, 9.5, 9.6 9.7, 9.8

1.1, 1.2, 1.3 1.4, 2.2 2.3, 2.4 3.2, 3.4 4.2, 4.3 4.4, 4.5, 4.6 4.7, 5.1, 5.2 7.1, 7.2, 7.3 7.4, 7.5 7.6, 7.7 7.8, 8.2 8.3, 8.5, 8.6 10.2, 10.3 10.4, 10.5 10.6, 10.7

xv

Retained Features Flexible Organization

Most of the material is modular in nature to allow for various course configurations and emphasis (theory, applications and techniques, and concepts).

Optional Use of Computer Software

The availability of computer packages such as Mathcad®, Mathematica®, MATLAB®, and Maple™ provides an opportunity for the student to conduct numerical experiments and tackle realistic applications that give additional insights into the subject. Consequently, we have inserted several exercises and projects throughout the text that are designed for the student to employ available software in phase plane analysis, eigenvalue computations, and the numerical solutions of various equations.

Review of Integration

In response to the perception that many of today’s students’ skills in integration have gotten rusty by the time they enter a differential equations course, we have included an appendix offering a quick review of the basic methods for integrating functions analytically.

Choice of Applications

Because of syllabus constraints, some courses will have little or no time for sections (such as those in Chapters 3 and 5) that exclusively deal with applications. Therefore, we have made the sections in these chapters independent of each other. To afford the instructor even greater flexibility, we have built in a variety of applications in the exercises for the theoretical sections. In addition, we have included many projects that deal with such applications.

Projects

At the end of each chapter are projects relating to the material covered in the chapter. Several of them have been contributed by distinguished researchers. A project might involve a more challenging application, delve deeper into the theory, or introduce more advanced topics in differential equations. Although these projects can be tackled by an individual student, classroom testing has shown that working in groups lends a valuable added dimension to the learning experience. Indeed, it simulates the interactions that take place in the professional arena.

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Preface

Technical Writing Exercises

Communication skills are, of course, an essential aspect of professional activities. Yet few texts provide opportunities for the reader to develop these skills. Thus, we have added at the end of most chapters a set of clearly marked technical writing exercises that invite students to make documented responses to questions dealing with the concepts in the chapter. In so doing, students are encouraged to make comparisons between various methods and to present examples that support their analysis.

Historical Footnotes

Throughout the text historical footnotes are set off by colored daggers (†). These footnotes typically provide the name of the person who developed the technique, the date, and the context of the original research.

Motivating Problem

Most chapters begin with a discussion of a problem from physics or engineering that motivates the topic presented and illustrates the methodology.

Chapter Summary and Review Problems

All of the main chapters contain a set of review problems along with a synopsis of the major concepts presented.

Computer Graphics

Most of the figures in the text were generated via computer. Computer graphics not only ensure greater accuracy in the illustrations, they demonstrate the use of numerical experimentation in studying the behavior of solutions.

Proofs

While more pragmatic students may balk at proofs, most instructors regard these justifications as an essential ingredient in a textbook on differential equations. As with any text at this level, certain details in the proofs must be omitted. When this occurs, we flag the instance and refer readers either to a problem in the exercises or to another text. For convenience, the end of a proof is marked by the symbol ◆.

Linear Theory

We have developed the theory of linear differential equations in a gradual manner. In Chapter 4 (Linear Second-Order Equations) we first present the basic theory for linear second-order equations with constant coefficients and discuss various techniques for solving these equations. Section 4.7 surveys the extension of these ideas to variable-coefficient second-order equations. A more general and detailed discussion of linear differential equations is given in Chapter 6 (Theory of Higher-Order Linear Differential Equations). For a beginning course emphasizing methods of solution, the presentation in Chapter 4 may be sufficient and Chapter 6 can be skipped.

Numerical Algorithms

Several numerical methods for approximating solutions to differential equations are presented along with program outlines that are easily implemented on a computer. These methods are introduced early in the text so that teachers and/or students can use them for numerical experimentation and for tackling complicated applications. Where appropriate we direct the student to software packages or web-based applets for implementation of these algorithms.

Exercises

An abundance of exercises is graduated in difficulty from straightforward, routine problems to more challenging ones. Deeper theoretical questions, along with applications, usually occur toward the end of the exercise sets. Throughout the text we have included problems and projects that require the use of a calculator or computer. These exercises are denoted by the symbol .

Laplace Transforms

We provide a detailed chapter on Laplace transforms (Chapter 7), since this is a recurring topic for engineers. Our treatment emphasizes discontinuous forcing terms and includes a section on the Dirac delta function.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Preface

xvii

Power Series

Power series solutions is a topic that occasionally causes student anxiety. Possibly, this is due to inadequate preparation in calculus where the more subtle subject of convergent series is (frequently) covered at a rapid pace. Our solution has been to provide a graceful initiation into the theory of power series solutions with an exposition of Taylor polynomial approximants to solutions, deferring the sophisticated issues of convergence to later sections. Unlike many texts, ours provides an extensive section on the method of Frobenius (Section 8.6) as well as a section on finding a second linearly independent solution. While we have given considerable space to power series solutions, we have also taken great care to accommodate the instructor who only wishes to give a basic introduction to the topic. An introduction to solving differential equations using power series and the method of Frobenius can be accomplished by covering the materials in Sections 8.1, 8.2, 8.3, and 8.6.

Partial Differential Equations

An introduction to this subject is provided in Chapter 10, which covers the method of separation of variables, Fourier series, the heat equation, the wave equation, and Laplace’s equation. Examples in two and three dimensions are included.

Phase Plane

Chapter 5 describes how qualitative information for two-dimensional systems can be gleaned about the solutions to intractable autonomous equations by observing their direction fields and critical points on the phase plane. With the assistance of suitable software, this approach provides a refreshing, almost recreational alternative to the traditional analytic methodology as we discuss applications in nonlinear mechanics, ecosystems, and epidemiology.

Vibrations

Motivation for Chapter 4 on linear differential equations is provided in an introductory section describing the mass–spring oscillator. We exploit the reader’s familiarity with common vibratory motions to anticipate the exposition of the theoretical and analytical aspects of linear equations. Not only does this model provide an anchor for the discourse on constant-coefficient equations, but a liberal interpretation of its features enables us to predict the qualitative behavior of variable-coefficient and nonlinear equations as well.

Review of Algebraic Equations and Matrices

The chapter on matrix methods for linear systems (Chapter 9) begins with two (optional) introductory sections reviewing the theory of linear algebraic systems and matrix algebra.

Technology and Supplements MyMathLab® Online Course (access code required) Built around Pearson’s best-selling content, MyMathLab is an online homework, tutorial, and assessment program designed to work with this text to engage students and improve results. MyMathLab can be successfully implemented in any classroom environment—lab-based, hybrid, fully online, or traditional. MyMathLab’s online homework offers students immediate feedback and tutorial assistance that motivates them to do more, which means they retain more knowledge and improve their test scores. Used by more than 37 million students worldwide, MyMathLab delivers consistent, measurable gains in student learning outcomes, retention, and subsequent course success. Visit www.mymathlab.com/results to learn more.

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xviii

Preface

Learning and Teaching Tools

• Exercises with immediate feedback—Nearly 750 assignable exercises are based on the textbook exercises and regenerate algorithmically to give students unlimited opportunity for practice and mastery. MyMathLab provides helpful feedback when students enter incorrect answers and includes optional learning aids including Help Me Solve This, View an Example, videos, and an eText.

• Learning Catalyticstm is a student response tool that uses students’ smartphones, tablets, or laptops to engage them in more interactive tasks and thinking. Learning Catalytics fosters student engagement and peer-to-peer learning with real-time analytics.

• Instructional videos are available as learning aids within exercises and for self-study within the Multimedia Library. The Guide to Video-Based Assignments makes it easy to assign videos for homework by showing which MyMathLab exercises correspond to each video. • The complete eText is available to students through their MyMathLab courses for the lifetime of the edition, giving students unlimited access to the eText within any course using that edition of the textbook.

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Preface

xix

• Accessibility and achievement go hand in hand. MyMathLab is compatible with the JAWS screen reader, and enables multiple-choice and free-response problem types to be read and interacted with via keyboard controls and math notation input. MyMathLab also works with screen enlargers, including ZoomText, MAGic, and SuperNova. And, all MyMathLab videos have closed-captioning. More information is available at mymathlab.com/accessibility. • A comprehensive gradebook with enhanced reporting functionality allows you to efficiently manage your course. The Reporting Dashboard provides insight to view, analyze, and report learning outcomes. Student performance data is presented at the class, section, and program levels in an accessible, visual manner so you’ll have the information you need to keep your students on track.

Item Analysis tracks class-wide understanding of particular exercises so you can refine your class lectures or adjust the course/department syllabus. Just-in-time teaching has never been easier! MyMathLab comes from an experienced partner with educational expertise and an eye on the future. Whether you are just getting started with MyMathLab, or have a question along the way, we’re here to help you learn about our technologies and how to incorporate them into your course. To learn more about how MyMathLab helps students succeed, visit mymathlab.com or contact your Pearson rep. MathXL® is the homework and assessment engine that runs MyMathLab. (MyMathLab is MathXL plus a learning management system.) MathXL access codes are also an option. Student’s Solutions Manual

ISBN-10: 0321977211 | ISBN-13: 9780321977212 Contains complete worked-out solutions to odd-numbered exercises, providing students with an excellent study tool. Available in print and for download within MyMathLab.

Instructor’s Solutions Manual (downloadable)

ISBN-10: 0134659244 | ISBN-13: 9780134659244 Contains answers to all even-numbered exercises, detailed solutions to the even-numbered problems in several of the main chapters, and additional projects. Available for download in the Pearson Instructor Resource Center www.pearsonhighered.com/irc as well as within MyMathLab.

MATLAB, Maple, and Mathematica Manuals (downloadable)

By Thomas W. Polaski (Winthrop University), Bruno Welfert (Arizona State University), and Maurino Bautista (Rochester Institute of Technology), respectively. These manuals contain a collection of instructor tips, worksheets, and projects to aid instructors in integrating computer algebra systems into their courses. Complete manuals are available for instructor download within MyMathLab. Student worksheets and projects available for download within MyMathLab.

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xx

Preface

Acknowledgments The staging of this text involved considerable behind-the-scenes activity. We thank, first of all, Philip Crooke, Glenn Webb (Vanderbilt University), and Joanna Wares (University of Richmond) who have continued to provide novel biomathematical projects, as well as Greg Huffman (Vanderbilt University) for his project on economics. We also want to thank Frank Glaser (California State Polytechnic University, Pomona) for many of the historical footnotes. We are indebted to Herbert E. Rauch (Lockheed Research Laboratory) for help with Section 3.3 on heating and cooling of buildings, Project B in Chapter 3 on aquaculture, and other application problems. Our appreciation goes to Richard H. Elderkin (Pomona College), Jerrold Marsden (California Institute of Technology), T. G. Proctor (Clemson University), and Philip W. Schaefer (University of Tennessee), who read and reread the manuscript for the original text, making numerous suggestions that greatly improved our work. Thanks also to the following reviewers of this and previous editions: *Miklos Bona, University of Florida Amin Boumenir, University of West Georgia Mark Brittenham, University of Nebraska *Jennifer Bryan, Oklahoma Christian University Weiming Cao, University of Texas at San Antonio Richard Carmichael, Wake Forest University *Kwai-lee Chui, University of Florida Karen Clark, The College of New Jersey *Shaozhong Deng, University of North Carolina Charlotte Patrick Dowling, Miami University *Chris Fuller, Cumberland University Sanford Geraci, Northern Virginia David S. Gilliam, Texas Tech University at Lubbock Scott Gordon, State University of West Georgia *Irvin Hentzel, Iowa State University *Mimi Rasky, Southwestern College Richard Rubin, Florida International University John Sylvester, University of Washington at Seattle Steven Taliaferro, Texas A&M University at College Station Michael M. Tom, Louisiana State University Shu-Yi Tu, University of Michigan, Flint Klaus Volpert, Villanova University *Glenn F. Webb, Vanderbilt University E. B. Saff, A. D. Snider

*Denotes reviewers of the current edition

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

CHAPTER

4

Linear Second-Order Equations

4.1 Introduction: The Mass–Spring Oscillator A damped mass–spring oscillator consists of a mass m attached to a spring fixed at one end, as shown in Figure 4.1. Devise a differential equation that governs the motion of this oscillator, taking into account the forces acting on it due to the spring elasticity, damping friction, and possible external influences.

m

k

b Equilibrium point

y

Figure 4.1 Damped mass–spring oscillator

Newton’s second law—force equals mass times acceleration 1F = ma2—is without a doubt the most commonly encountered differential equation in practice. It is an ordinary differential equation of the second order since acceleration is the second derivative of position 1y2 with respect to time 1 a = d 2y>dt 2 2 . When the second law is applied to a mass–spring oscillator, the resulting motions are common experiences of everyday life, and we can exploit our familiarity with these vibrations to obtain a qualitative description of the solutions of more general second-order equations. We begin by referring to Figure 4.1, which depicts the mass–spring oscillator. When the spring is unstretched and the inertial mass m is still, the system is at equilibrium; we measure the coordinate y of the mass by its displacement from the equilibrium position. When the mass m is displaced from equilibrium, the spring is stretched or compressed and it exerts a force that resists the displacement. For most springs this force is directly proportional to the displacement y and is thus given by (1)

Fspring = -ky ,

where the positive constant k is known as the stiffness and the negative sign reflects the opposing nature of the force. Hooke’s law, as equation (1) is commonly known, is only valid for sufficiently small displacements; if the spring is compressed so strongly that the coils press against each other, the opposing force obviously becomes much stronger. 152

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 4.1

y

153

Introduction: The Mass–Spring Oscillator

y

y

t

(a)

t

(b)

t

(c)

Figure 4.2 (a) Sinusoidal oscillation, (b) stiffer spring, and (c) heavier mass

Practically all mechanical systems also experience friction, and for vibrational motion this force is usually modeled accurately by a term proportional to velocity: (2)

Ffriction = -b

dy = -by′ , dt

where b 1 Ú 02 is the damping coefficient and the negative sign has the same significance as in equation (1). The other forces on the oscillator are usually regarded as external to the system. Although they may be gravitational, electrical, or magnetic, commonly the most important external forces are transmitted to the mass by shaking the supports holding the system. For the moment we lump all the external forces into a single, known function Fext 1t2. Newton’s law then provides the differential equation for the mass–spring oscillator: my″ = -ky - by′ + Fext 1t2

or (3)

my″ + by′ + ky = Fext 1t2 .

What do mass–spring motions look like? From our everyday experience with weak auto suspensions, musical gongs, and bowls of jelly, we expect that when there is no friction 1b = 02 or external force, the (idealized) motions would be perpetual vibrations like the ones depicted in Figure 4.2. These vibrations resemble sinusoidal functions, with their amplitude depending on the initial displacement and velocity. The frequency of the oscillations increases for stiffer springs but decreases for heavier masses. In Section 4.3 we will show how to find these solutions. Example 1 demonstrates a quick calculation that confirms our intuitive predictions. Example 1 Solution

Verify that if b = 0 and Fext 1t2 = 0, equation (3) has a solution of the form y1t2 = cos vt and that the angular frequency v increases with k and decreases with m. Under the conditions stated, equation (3) simplifies to (4)

my″ + ky = 0 .

The second derivative of y1t2 is -v2 cos vt, and if we insert it into (4), we find my″ + ky = -mv2 cos vt + k cos vt , which is indeed zero if v = 2k>m. This v increases with k and decreases with m, as predicted. ◆

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Chapter 4

Linear Second-Order Equations

y

y

t

t

(a)

(b)

Figure 4.3 (a) Low damping and (b) high damping

When damping is present, the oscillations die out, and the motions resemble Figure 4.3. In Figure 4.3(a) the graph displays a damped oscillation; damping has slowed the frequency, and the amplitude appears to diminish exponentially with time. In Figure 4.3(b) the damping is so dominant that it has prevented the system from oscillating at all. Devices that are supposed to vibrate, like tuning forks or crystal oscillators, behave like Figure 4.3(a), and the damping effect is usually regarded as an undesirable loss mechanism. Good automotive suspension systems, on the other hand, behave like Figure 4.3(b); they exploit damping to suppress the oscillations. The procedures for solving (unforced) mass–spring systems with damping are also described in Section 4.3, but as Examples 2 and 3 below show, the calculations are more complex. Example 2 has a relatively low damping coefficient 1b = 62 and illustrates the solutions for the “underdamped” case in Figure 4.3(a). In Example 3 the damping is more severe 1b = 102, and the solution is “overdamped” as in Figure 4.3(b). Example 2 Solution

Verify that the exponentially damped sinusoid given by y1t2 = e-3t cos 4t is a solution to equation (3) if Fext = 0, m = 1, k = 25, and b = 6. The derivatives of y are y′1t2 = -3e-3t cos 4t - 4e-3t sin 4t , y″1t2 = 9e-3t cos 4t + 12e-3t sin 4t + 12e-3t sin 4t - 16e-3t cos 4t = -7e-3t cos 4t + 24e-3t sin 4t , and insertion into (3) gives

my″ + by′ + ky = 112y″ + 6y′ + 25y = -7e-3t cos 4t + 24e-3t sin 4t + 61 -3e-3t cos 4t - 4e-3t sin t2 + 25e-3t cos 4t = 0. ◆

Example 3 Solution

Verify that the simple exponential function y1t2 = e-5t is a solution to equation (3) if Fext = 0, m = 1, k = 25, and b = 10. The derivatives of y are y′1t2 = -5e-5t, y″1t2 = 25e-5t and insertion into (3) produces

my″ + by′ + ky = 112y″ + 10y′ + 25y = 25e-5t + 101 -5e-5t 2 + 25e-5t = 0 . ◆

Now if a mass–spring system is driven by an external force that is sinusoidal at the angular frequency v, our experiences indicate that although the initial response of the system may be

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 4.1

Introduction: The Mass–Spring Oscillator

Fext

155

y

t

(a)

t

(b)

Figure 4.4 (a) Driving force and (b) response

somewhat erratic, eventually it will respond in “sync” with the driver and oscillate at the same frequency, as illustrated in Figure 4.4. Common examples of systems vibrating in synchronization with their drivers are sound system speakers, cyclists bicycling over railroad tracks, electronic amplifier circuits, and ocean tides (driven by the periodic pull of the moon). However, there is more to the story than is revealed above. Systems can be enormously sensitive to the particular frequency v at which they are driven. Thus, accurately tuned musical notes can shatter fine crystal, wind-induced vibrations at the right (wrong?) frequency can bring down a bridge, and a dripping faucet can cause inordinate headaches. These “resonance” responses (for which the responses have maximum amplitudes) may be quite destructive, and structural engineers have to be very careful to ensure that their products will not resonate with any of the vibrations likely to occur in the operating environment. Radio engineers, on the other hand, do want their receivers to resonate selectively to the desired broadcasting channel. The calculation of these forced solutions is the subject of Sections 4.4 and 4.5. The next example illustrates some of the features of synchronous response and resonance. Example 4 Solution

Find the synchronous response of the mass–spring oscillator with m = 1, b = 1, k = 25 to the force sin Ωt. We seek solutions of the differential equation (5)

y″ + y′ + 25y = sin Ωt

that are sinusoids in sync with sin Ωt; so let’s try the form y1t2 = A cos Ωt + B sin Ωt. Since y′ = - ΩA sin Ωt + ΩB cos Ωt , y″ = - Ω 2A cos Ωt - Ω 2B sin Ωt , we can simply insert these forms into equation (5), collect terms, and match coefficients to obtain a solution: sin Ωt = y″ + y′ + 25y = - Ω 2A cos Ωt - Ω 2B sin Ωt + 3 - ΩA sin Ωt + ΩB cos Ωt4 +253A cos Ωt + B sin Ωt4 = 3 - Ω 2B - ΩA + 25B4 sin Ωt + 3 - Ω 2A + ΩB + 25A4 cos Ωt , so

- ΩA + 1 - Ω 2 + 252B = 1 1 - Ω 2 + 252A + ΩB = 0 .

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Chapter 4

Linear Second-Order Equations

0.15 B

0.1 0.05

5

0

10

15

20

V

–0.05 –0.1 A

–0.15 –0.2

Figure 4.5 Vibration amplitudes around resonance

We find A =

-Ω , 2 Ω + 1Ω 2 - 252 2

B =

- Ω 2 + 25 . Ω 2 + 1Ω 2 - 252 2

Figure 4.5 displays A and B as functions of the driving frequency Ω. A resonance clearly occurs around Ω ≈ 5. ◆ In most of this chapter, we are going to restrict our attention to differential equations of the form (6)

ay″ + by′ + cy = f1t2 ,

where y1t2 [or y1x2, or x1t2, etc.] is the unknown function that we seek; a, b, and c are constants; and f1t2 [or f1x2] is a known function. The proper nomenclature for (6) is the linear, second-order ordinary differential equation with constant coefficients. In Sections 4.7 and 4.8, we will generalize our focus to equations with nonconstant coefficients, as well as to nonlinear equations. However, (6) is an excellent starting point because we are able to obtain explicit solutions and observe, in concrete form, the theoretical properties that are predicted for more general equations. For motivation of the mathematical procedures and theory for solving (6), we will consistently compare it with the mass–spring paradigm: 3inertia4 * y″ + 3damping4 * y′ + 3stiffness4 * y = Fext .

4.1

EXERCISES

1. Verify that for b = 0 and Fext 1t2 = 0, equation (3) has a solution of the form y1t2 = cos vt, where v = 2k>m . 2. If Fext 1t2 = 0, equation (3) becomes my″ + by′ + ky = 0 . For this equation, verify the following:

(a) If y1t2 is a solution, so is cy1t2, for any constant c. (b) If y1 1t2 and y2 1t2 are solutions, so is their sum y1 1t2 + y2 1t2. 3. Show that if Fext 1t2 = 0, m = 1, k = 9, and b = 6, then equation (3) has the “critically damped” solutions y1 1t2 = e-3t and y2 1t2 = te-3t. What is the limit of these solutions as t S ∞ ?

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 4.2

Homogeneous Linear Equations: The General Solution

4. Verify that y = sin 3t + 2 cos 3t is a solution to the initial value problem 2y″ + 18y = 0 ;

y102 = 2 ,

157

(a) To investigate this, find the synchronous solution A cos Ωt + B sin Ωt to the generic forced oscillator equation

y′102 = 3 .

(7)

Find the maximum of ƒy1t2ƒ for - ∞ 6 t 6 ∞ . 5. Verify that the exponentially damped sinusoid y1t2 = e-3t sin 1 23 t2 is a solution to equation (3) if Fext 1t2 = 0, m = 1, b = 6, and k = 12. What is the limit of this solution as t S ∞ ? 6. An external force F1t2 = 2 cos 2t is applied to a mass– spring system with m = 1, b = 0, and k = 4, which is initially at rest; i.e., y102 = 0, y′102 = 0. Verify that y1t2 = 12 t sin 2t gives the motion of this spring. What will eventually (as t increases) happen to the spring?

my″ + by′ + ky = cos Ωt .

(b) Sketch graphs of the coefficients A and B, as functions of Ω, for m = 1, b = 0.1, and k = 25. (c) Now set b = 0 in your formulas for A and B and resketch the graphs in part (b), with m = 1, and k = 25. What happens at Ω = 5? Notice that the amplitudes of the synchronous solutions grow without bound as Ω approaches 5. (d) Show directly, by substituting the form A cos Ωt + B sin Ωt into equation (7), that when b = 0 there are no synchronous solutions if Ω = 2k>m. (e) Verify that 12mΩ2 -1t sin Ωt solves equation (7) when b = 0 and Ω = 2k>m. Notice that this nonsynchronous solution grows in time, without bound.

In Problems 7–9, find a synchronous solution of the form A cos Ωt + B sin Ωt to the given forced oscillator equation using the method of Example 4 to solve for A and B. 7. y″ + 2y′ + 4y = 5 sin 3t, Ω = 3 8. y″ + 2y′ + 5y = -50 sin 5t, Ω = 5 9. y″ + 2y′ + 4y = 6 cos 2t + 8 sin 2t, Ω = 2

Clearly one cannot neglect damping in analyzing an oscillator forced at resonance, because otherwise the solutions, as shown in part (e), are nonphysical. This behavior will be studied later in this chapter.

10. Undamped oscillators that are driven at resonance have unusual (and nonphysical) solutions.

4.2 Homogeneous Linear Equations: The General Solution We begin our study of the linear second-order constant-coefficient differential equation (1)

ay″ + by′ + cy = f1t2

1a ≠ 02

with the special case where the function f1t2 is zero: (2)

ay″ + by′ + cy = 0 .

This case arises when we consider mass–spring oscillators vibrating freely—that is, without external forces applied. Equation (2) is called the homogeneous form of equation (1); f1t2 is the “nonhomogeneity” in (1). (This nomenclature is not related to the way we used the term for first-order equations in Section 2.6.) A look at equation (2) tells us that a solution of (2) must have the property that its second derivative is expressible as a linear combination of its first and zeroth derivatives.† This suggests that we try to find a solution of the form y = ert, since derivatives of ert are just constants times ert. If we substitute y = ert into (2), we obtain ar 2 ert + brert + cert = 0 , ert 1ar 2 + br + c2 = 0 .

†

The zeroth derivative of a function is the function itself.

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158

Chapter 4

Linear Second-Order Equations

Because ert is never zero, we can divide by it to obtain (3)

ar 2 + br + c = 0 .

Consequently, y = ert is a solution to (2) if and only if r satisfies equation (3). Equation (3) is called the auxiliary equation (also known as the characteristic equation) associated with the homogeneous equation (2). Now the auxiliary equation is just a quadratic, and its roots are r1 =

-b + 2b2 - 4ac 2a

and

r2 =

-b - 2b2 - 4ac . 2a

When the discriminant, b2 - 4ac, is positive, the roots r1 and r2 are real and distinct. If b2 - 4ac = 0, the roots are real and equal. And when b2 - 4ac 6 0, the roots are complex conjugate numbers. We consider the first two cases in this section; the complex case is deferred to Section 4.3. Example 1

Find a pair of solutions to (4)

Solution

y″ + 5y′ - 6y = 0 .

The auxiliary equation associated with (4) is

r 2 + 5r - 6 = 1r - 121r + 62 = 0 ,

which has the roots r1 = 1, r2 = -6. Thus, et and e-6t are solutions. ◆ Notice that the identically zero function, y1t2 K 0, is always a solution to (2). Furthermore, when we have a pair of solutions y1 1t2 and y2 1t2 to this equation, as in Example 1, we can construct an infinite number of other solutions by forming linear combinations: (5)

y1t2 = c1 y1 1t2 + c2 y2 1t2

for any choice of the constants c1 and c2. The fact that (5) is a solution to (2) can be seen by direct substitution and rearrangement: ay″ + by′ + cy = = = =

a1c1y1 + c2y2 2″ + b1c1y1 + c2y2 2′ + c1c1y1 + c2y2 2 a1c1y1> + c2y2> 2 + b1c1y1= + c2y2= 2 + c1c1y1 + c2y2 2 c1 1ay1> + by1= + cy1 2 + c2 1ay2> + by2= + cy2 2 0+0.

The two “degrees of freedom” c1 and c2 in the combination (5) suggest that solutions to the differential equation (2) can be found meeting additional conditions, such as the initial conditions for the first-order equations in Chapter 1. But the presence of c1 and c2 leads one to anticipate that two such conditions, rather than just one, can be imposed. This is consistent with the mass–spring interpretation of equation (2), since predicting the motion of a mechanical system requires knowledge not only of the forces but also of the initial position y102 and velocity y′102 of the mass. A typical initial value problem for these second-order equations is given in the following example.

Example 2

Solve the initial value problem (6)

y″ + 2y′ - y = 0 ;

y102 = 0 ,

y′102 = -1 .

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Section 4.2

Solution

Homogeneous Linear Equations: The General Solution

159

We will first find a pair of solutions as in the previous example. Then we will adjust the constants c1 and c2 in (5) to obtain a solution that matches the initial conditions on y102 and y′102. The auxiliary equation is r 2 + 2r - 1 = 0 . Using the quadratic formula, we find that the roots of this equation are r1 = -1 + 22

and

r2 = -1 - 22 .

Consequently, the given differential equation has solutions of the form (7)

y1t2 = c1e1-1 + 222t + c2e1-1 - 222t .

To find the specific solution that satisfies the initial conditions given in (6), we first differentiate y as given in (7), then plug y and y′ into the initial conditions of (6). This gives y102 = c1e0 + c2e0 ,

y′102 = 1 -1 + 222c1e0 + 1 -1 - 222c2e0 , or 0 = c1 + c2 ,

-1 = 1 -1 + 222c1 + 1 -1 - 222c2 . Solving this system yields c1 = - 22>4 and c2 = 22>4. Thus, y1t2 = -

22 1-1 + 222t 22 1-1 - 222t e + e 4 4

is the desired solution. ◆ To gain more insight into the significance of the two-parameter solution form (5), we need to look at some of the properties of the second-order equation (2). First of all, there is an existence-and-uniqueness theorem for solutions to (2); it is somewhat like the corresponding Theorem 1 in Section 1.2 for first-order equations but updated to reflect the fact that two initial conditions are appropriate for second-order equations. As motivation for the theorem, suppose the differential equation (2) were really easy, with b = 0 and c = 0. Then y″ = 0 would merely say that the graph of y1t2 is simply a straight line, so it is uniquely determined by specifying a point on the line, (8)

y1t0 2 = Y0 ,

and the slope of the line, (9)

y′1t0 2 = Y1 .

Theorem 1 states that conditions (8) and (9) suffice to determine the solution uniquely for the more general equation (2).

Existence and Uniqueness: Homogeneous Case Theorem 1. For any real numbers a 1 ≠02, b, c, t0, Y0, and Y1, there exists a unique solution to the initial value problem (10) ay″ + by′ + cy = 0 ;

y1t0 2 = Y0 ,

The solution is valid for all t in 1 - ∞, + ∞ 2.

y′1t0 2 = Y1 .

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160

Chapter 4

Linear Second-Order Equations

Note in particular that if a solution y1t2 and its derivative vanish simultaneously at a point t0 1i.e., Y0 = Y1 = 02, then y1t2 must be the identically zero solution. In this section and the next, we will construct explicit solutions to (10), so the question of existence of a solution is not really an issue. It is extremely valuable to know, however, that the solution is unique. The proof of uniqueness is rather different from anything else in this chapter, so we defer it to Chapter 13.† Now we want to use this theorem to show that, given two solutions y1 1t2 and y2 1t2 to equation (2), we can always find values of c1 and c2 so that c1y1 1t2 + c2y2 1t2 meets specified initial conditions in (10) and therefore is the (unique) solution to the initial value problem. But we need to be a little more precise; if, for example, y2 1t2 is simply the identically zero solution, then c1y1 1t2 + c2y2 1t2 = c1y1 1t2 actually has only one constant and cannot be expected to satisfy two conditions. Furthermore, if y2 1t2 is simply a constant multiple of y1 1t2—say, y2 1t2 = ky1 1t2—then again c1y1 1t2 + c2y2 1t2 = 1c1 + kc2 2y1 1t2 = Cy1 1t2 actually has only one constant. The condition we need is linear independence.

Linear Independence of Two Functions Definition 1. A pair of functions y1 1t2 and y2 1t2 is said to be linearly independent on the interval I if and only if neither of them is a constant multiple of the other on all of I.†† We say that y1 and y2 are linearly dependent on I if one of them is a constant multiple of the other on all of I.

Representation of Solutions to Initial Value Problem Theorem 2. If y1 1t2 and y2 1t2 are any two solutions to the differential equation (2) that are linearly independent on 1 - ∞, ∞ 2, then unique constants c1 and c2 can always be found so that c1y1 1t2 + c2y2 1t2 satisfies the initial value problem (10) on 1 - ∞, ∞ 2.

The proof of Theorem 2 will be easy once we establish the following technical lemma.

A Condition for Linear Dependence of Solutions Lemma 1. For any real numbers a 1 ≠02, b, and c, if y1 1t2 and y2 1t2 are any two solutions to the differential equation (2) on 1 - ∞, ∞ 2 and if the equality (11)

y1 1t2y2= 1t2 - y1= 1t2y2 1t2 = 0

holds at any point t, then y1 and y2 are linearly dependent on 1 - ∞, ∞ 2. (The expression on the left-hand side of (11) is called the Wronskian of y1 and y2 at the point t; see Problem 34 on page 164.)

†

All references to Chapters 11–13 refer to the expanded text, Fundamentals of Differential Equations and Boundary Value Problems, 7th ed. †† This definition will be generalized to three or more functions in Problem 35 and Chapter 6.

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Section 4.2

Homogeneous Linear Equations: The General Solution

161

Proof of Lemma 1. Case 1. If y1 1t2 ≠ 0, then let k equal y2 1t2 >y1 1t2 and consider the solution to (2) given by y1t2 = ky1 1t2. It satisfies the same “initial conditions” at t = t as does y2 1t2: y1t2 =

y2 1t2 y1 1t2

y1 1t2 = y2 1t2 ;

y′1t2 =

y2 1t2

y1 1t2

y1= 1t2 = y2= 1t2 ,

where the last equality follows from (11). By uniqueness, y2 1t2 must be the same function as ky1 1t2 on I. Case 2. If y1 1t2 = 0 but y1= 1t2 ≠ 0, then (11) implies y2 1t2 = 0. Let k = y2= 1t2 >y1= 1t2. Then the solution to (2) given by y1t2 = ky1 1t2 (again) satisfies the same “initial conditions” at t = t as does y2 1t2: y1t2 =

y2= 1t2 y 1t2 = 0 = y2 1t2 ; y1= 1t2 1

y′1t2 =

y2= 1t2

y1= 1t2

y1= 1t2 = y2= 1t2 .

By uniqueness, then, y2 1t2 = ky1 1t2 on I. Case 3. If y1 1t2 = y1= 1t2 = 0, then y1 1t2 is a solution to the differential equation (2) satisfying the initial conditions y1 1t2 = y1= 1t2 = 0; but y1t2 K 0 is the unique solution to this initial value problem. Thus, y1 1t2 K 0 3and is a constant multiple of y2 1t24. ◆ Proof of Theorem 2. We already know that y1t2 = c1y1 1t2 + c2y2 1t2 is a solution to (2); we must show that c1 and c2 can be chosen so that y1t0 2 = c1y1 1t0 2 + c2y2 1t0 2 = Y0

and

y′1t0 2 = c1y1= 1t0 2 + c2y2= 1t0 2 = Y1 .

But simple algebra shows these equations have the solution† c1 =

Y0y2= 1t0 2 - Y1y2 1t0 2 y1 1t0 2y2= 1t0 2 - y1= 1t0 2y2 1t0 2

and c2 =

Y1y1 1t0 2 - Y0y1= 1t0 2 y1 1t0 2y2= 1t0 2 - y1= 1t0 2y2 1t0 2

as long as the denominator is nonzero, and the technical lemma assures us that this condition is met. ◆ Now we can honestly say that if y1 and y2 are linearly independent solutions to (2) on 1 - ∞, + ∞ 2, then (5) is a general solution, since any solution yg 1t2 of (2) can be expressed in this form; simply pick c1 and c2 so that c1y1 + c2y2 matches the value and the derivative of yg at any point. By uniqueness, c1y1 + c2y2 and yg have to be the same function. See Figure 4.6 on page 162. How do we find a general solution for the differential equation (2)? We already know the answer if the roots of the auxiliary equation (3) are real and distinct because clearly y1 1t2 = er1t is not a constant multiple of y2 1t2 = er2t if r1 ≠ r2.

Distinct Real Roots

If the auxiliary equation (3) has distinct real roots r1 and r2, then both y1 1t2 = er1t and y2 1t2 = er2t are solutions to (2) and y1t2 = c1er1t + c2er2t is a general solution.

To solve for c1, for example, multiply the first equation by y2= 1t0 2 and the second by y2 1t0 2 and subtract.

†

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162

Chapter 4

Linear Second-Order Equations

y

(Can’t both be solutions)

y(t0)

Slope y9(t0) t

t0 Figure 4.6 y1t0 2, y′1t0 2 determine a unique solution.

When the roots of the auxiliary equation are equal, we only get one nontrivial solution, y1 = ert. To satisfy two initial conditions, y1t0 2 and y′1t0 2, then we will need a second, linearly independent solution. The following rule is the key to finding a second solution.

Repeated Root

If the auxiliary equation (3) has a repeated root r, then both y1 1t2 = ert and y2 1t2 = tert are solutions to (2), and y1t2 = c1ert + c2tert is a general solution.

We illustrate this result before giving its proof. Example 3

Find a solution to the initial value problem (12)

Solution

y″ + 4y′ + 4y = 0 ;

y102 = 1 ,

y′102 = 3 .

The auxiliary equation for (12) is

r 2 + 4r + 4 = 1r + 22 2 = 0 .

Because r = -2 is a double root, the rule says that (12) has solutions y1 = e-2t and y2 = te-2t. Let’s confirm that y2 1t2 is a solution: y2 1t2 = te-2t , y′2 1t2 = e-2t - 2te-2t , y″2 1t2 = -2e-2t - 2e-2t + 4te-2t = -4e-2t + 4te-2t , y″2 + 4y′2 + 4y2 = -4e-2t + 4te-2t + 41e-2t - 2te-2t 2 + 4te-2t = 0 .

Further observe that e-2t and te-2t are linearly independent since neither is a constant multiple of the other on 1 - ∞, ∞ 2. Finally, we insert the general solution y1t2 = c1e-2t + c2te-2t into the initial conditions, y102 = c1e0 + c2 102e0 = 1 , y′102 = -2c1e0 + c2e0 - 2c2 102e0 = 3 ,

and solve to find c1 = 1, c2 = 5. Thus y = e-2t + 5te-2t is the desired solution. ◆

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Section 4.2

Homogeneous Linear Equations: The General Solution

163

Why is it that y2 1t2 = tert is a solution to the differential equation (2) when r is a double root (and not otherwise)? In later chapters we will see a theoretical justification of this rule in very general circumstances; for present purposes, though, simply note what happens if we substitute y2 into the differential equation (2): y2 1t2 = tert ,

y2= 1t2 = ert + rtert ,

y″2 1t2 = rert + rert + r 2tert = 2rert + r 2tert ,

ay2> + by2= + cy2 = 32ar + b4ert + 3ar 2 + br + c4tert . Now if r is a root of the auxiliary equation (3), the expression in the second brackets is zero. However, if r is a double root, the expression in the first brackets is zero also: (13)

r =

-b { 102 -b { 2b2 - 4ac = ; 2a 2a

hence, 2ar + b = 0 for a double root. In such a case, then, y2 is a solution. The method we have described for solving homogeneous linear second-order equations with constant coefficients applies to any order (even first-order) homogeneous linear equations with constant coefficients. We give a detailed treatment of such higher-order equations in Chapter 6. For now, we will be content to illustrate the method by means of an example. We remark briefly that a homogeneous linear nth-order equation has a general solution of the form y1t2 = c1y1 1t2 + c2y2 1t2 + g + cnyn 1t2 ,

where the individual solutions yi 1t2 are “linearly independent.” By this we mean that no yi is expressible as a linear combination of the others; see Problem 35 on page 164.

Example 4

Find a general solution to (14)

Solution

y‴ + 3y″ - y′ - 3y = 0 .

If we try to find solutions of the form y = ert, then, as with second-order equations, we are led to finding roots of the auxiliary equation (15)

r 3 + 3r 2 - r - 3 = 0 .

We observe that r = 1 is a root of the above equation, and dividing the polynomial on the lefthand side of (15) by r - 1 leads to the factorization 1r - 121r 2 + 4r + 32 = 1r - 121r + 121r + 32 = 0 .

Hence, the roots of the auxiliary equation are 1, -1, and -3, and so three solutions of (14) are et, e-t, and e-3t. The linear independence of these three exponential functions is proved in Problem 36. A general solution to (14) is then (16)

y1t2 = c1et + c2e-t + c3e-3t . ◆

So far we have seen only exponential solutions to the linear second-order constant coefficient equation. You may wonder where the vibratory solutions that govern mass–spring oscillators are. In the next section, it will be seen that they arise when the solutions to the auxiliary equation are complex.

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164

Chapter 4

4.2

Linear Second-Order Equations

EXERCISES

In Problems 1–12, find a general solution to the given differential equation. 1. 2y″ + 7y′ - 4y = 0 2. y″ + 6y′ + 9y = 0 3. y″ + 5y′ + 6y = 0 4. y″ - y′ - 2y = 0 5. y″ + 8y′ + 16y = 0 6. y″ - 5y′ + 6y = 0 7. 6y″ + y′ - 2y = 0 8. z″ + z′ - z = 0 9. 4y″ - 4y′ + y = 0 10. y″ - y′ - 11y = 0 11. 4w″ + 20w′ + 25w = 0 12. 3y″ + 11y′ - 7y = 0 In Problems 13–20, solve the given initial value problem. 13. y″ + 2y′ - 8y = 0 ; y102 = 3 , y′102 = - 12 14. y″ + y′ = 0 ; y102 = 2 , y′102 = 1 15. y″ - 4y′ + 3y = 0 ; y102 = 1 , y′102 = 1>3 16. y″ - 4y′ - 5y = 0 ; y1 - 12 = 3 , y′1 - 12 = 9 17. y″ - 6y′ + 9y = 0 ; y102 = 2 , y′102 = 25>3 18. z″ - 2z′ - 2z = 0 ; z102 = 0 , z′102 = 3 19. y″ + 2y′ + y = 0 ; y102 = 1 , y′102 = -3 20. y″ - 4y′ + 4y = 0 ; y112 = 1 , y′112 = 1 21. First-Order Constant-Coefficient Equations. (a) Substituting y = ert, find the auxiliary equation for the first-order linear equation ay′ + by = 0 , where a and b are constants with a ≠ 0. (b) Use the result of part (a) to find the general solution. In Problems 22–25, use the method described in Problem 21 to find a general solution to the given equation. 22. 3y′ - 7y = 0 23. 5y′ + 4y = 0 24. 3z′ + 11z = 0 25. 6w′ - 13w = 0 26. Boundary Value Problems. When the values of a solution to a differential equation are specified at two different points, these conditions are called boundary conditions. (In contrast, initial conditions specify the values of a function and its derivative at the same point.) The purpose of this exercise is to show that for boundary value problems there is no existence–uniqueness theorem that is analogous to Theorem 1. Given that every solution to (17)

y″ + y = 0

is of the form y1t2 = c1 cos t + c2 sin t , where c1 and c2 are arbitrary constants, show that (a) There is a unique solution to (17) that satisfies the boundary conditions y102 = 2 and y1p>22 = 0.

(b) There is no solution to (17) that satisfies y102 = 2 and y1p2 = 0. (c) There are infinitely many solutions to (17) that satisfy y102 = 2 and y1p2 = -2. In Problems 27–32, use Definition 1 to determine whether the functions y1 and y2 are linearly dependent on the interval 10, 12. 27. y1 1t2 = cos t sin t , y2 1t2 = sin 2t 28. y1 1t2 = e3t , y2 1t2 = e-4t 29. y1 1t2 = te2t , y2 1t2 = e2t 30. y1 1t2 = t 2 cos 1ln t2 , y2 1t2 = t 2 sin 1ln t2 31. y1 1t2 = tan2 t - sec2 t , y2 1t2 K 3 32. y1 1t2 K 0 , y2 1t2 = et 33. Explain why two functions are linearly dependent on an interval I if and only if there exist constants c1 and c2, not both zero, such that c1y1 1t2 + c2y2 1t2 = 0

for all t in I .

34. Wronskian. For any two differentiable functions y1 and y2, the function (18)

W3y1, y2 41t2 = y1 1t2y′2 1t2 - y′1 1t2y2 1t2

is called the Wronskian† of y1 and y2. This function plays a crucial role in the proof of Theorem 2. (a) Show that W3y1, y2 4 can be conveniently expressed as the 2 * 2 determinant W3y1, y2 41t2 = `

y1 1t2 y1= 1t2

y2 1t2 ` . y2= 1t2

(b) Let y1 1t2, y2 1t2 be a pair of solutions to the homogeneous equation ay″ + by′ + cy = 0 (with a ≠ 0) on an open interval I. Prove that y1 1t2 and y2 1t2 are linearly independent on I if and only if their Wronskian is never zero on I. [Hint: This is just a reformulation of Lemma 1.] (c) Show that if y1 1t2 and y2 1t2 are any two differentiable functions that are linearly dependent on I, then their Wronskian is identically zero on I. 35. Linear Dependence of Three Functions. Three functions y1 1t2, y2 1t2, and y3 1t2 are said to be linearly dependent on an interval I if, on I, at least one of these functions is a linear combination of the remaining two [e.g., if y1 1t2 = c1y2 1t2 + c2y3 1t2]. Equivalently (compare Problem 33), y1, y2, and y3 are linearly dependent on I if there exist constants C1, C2, and C3, not all zero, such that C1y1 1t2 + C2y2 1t2 + C3y3 1t2 = 0

for all t in I.

Otherwise, we say that these functions are linearly independent on I.

†

Historical Footnote: The Wronskian was named after the Polish mathematician H. Wronski (1778–1863).

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 4.3

For each of the following, determine whether the given three functions are linearly dependent or linearly independent on 1 - ∞ , ∞ 2:

(a) y1 1t2 = 1 , y2 1t2 = t , y3 1t2 = t 2 . (b) y1 1t2 = - 3 , y2 1t2 = 5 sin2 t , y3 1t2 = cos2 t . (c) y1 1t2 = et , y2 1t2 = tet , y3 1t2 = t 2et . (d) y1 1t2 = et , y2 1t2 = e-t , y3 1t2 = cosh t . 36. Using the definition in Problem 35, prove that if r1, r2, and r3 are distinct real numbers, then the functions er1t, er2t, and er3t are linearly independent on 1 - ∞ , ∞ 2. [Hint: Assume to the contrary that, say, er1t = c1er2t + c2er3t for all t. Divide by er2t to get e1r1 - r22t = c1 + c2e1r3 - r22t and then differentiate to deduce that e1r1 - r22t and e1r3 - r22t are linearly dependent, which is a contradiction. (Why?)] In Problems 37–41, find three linearly independent solutions (see Problem 35) of the given third-order differential equation and write a general solution as an arbitrary linear combination of these. 37. y‴ + y″ - 6y′ + 4y = 0 38. y‴ - 6y″ - y′ + 6y = 0 39. z‴ + 2z″ - 4z′ - 8z = 0 40. y‴ - 7y″ + 7y′ + 15y = 0 41. y‴ + 3y″ - 4y′ - 12y = 0 42. (True or False): If f1, f2, f3 are three functions defined on 1 - ∞ , ∞ 2 that are pairwise linearly independent on 1 - ∞ , ∞ 2, then f1, f2, f3 form a linearly independent set on 1 - ∞ , ∞ 2. Justify your answer. 43. Solve the initial value problem: y‴ - y′ = 0 ;

y102 = 2 ,

y′102 = 3 ,

Auxiliary Equations with Complex Roots

165

44. Solve the initial value problem: y‴ - 2y″ - y′ + 2y = 0 ; y102 = 2 , y′102 = 3 , y″102 = 5 . 45. By using Newton’s method or some other numerical procedure to approximate the roots of the auxiliary equation, find general solutions to the following equations: (a) 3y‴ + 18y″ + 13y′ - 19y = 0 . (b) yiv - 5y″ + 5y = 0 . (c) yv - 3yiv - 5y‴ + 15y″ + 4y′ - 12y = 0 . 46. One way to define hyperbolic functions is by means of differential equations. Consider the equation y″ - y = 0. The hyperbolic cosine, cosh t, is defined as the solution of this equation subject to the initial values: y 102 = 1 and y′102 = 0. The hyperbolic sine, sinh t, is defined as the solution of this equation subject to the initial values: y102 = 0 and y′102 = 1. (a) Solve these initial value problems to derive explicit formulas for cosh t and sinh t. Also show that d d cosh t = sinh t and sinh t = cosh t. dt dt (b) Prove that a general solution of the equation y″ - y = 0 is given by y = c1 cosh t + c2 sinh t. (c) Suppose a, b, and c are given constants for which ar 2 + br + c = 0 has two distinct real roots. If the two roots are expressed in the form a - b and a + b, show that a general solution of the equation ay″ + by′ + cy = 0 is y = c1eat cosh1bt2 + c2eat sinh1bt2. (d) Use the result of part (c) to solve the initial value problem: y″ + y′ - 6y = 0, y102 = 2, y′102 = - 17>2.

y″102 = - 1 .

4.3 Auxiliary Equations with Complex Roots The simple harmonic equation y″ + y = 0, so called because of its relation to the fundamental vibration of a musical tone, has as solutions y1 1t2 = cos t and y2 1t2 = sin t. Notice, however, that the auxiliary equation associated with the harmonic equation is r 2 + 1 = 0, which has imaginary roots r = {i, where i denotes 2 -1.† In the previous section, we expressed the solutions to a linear second-order equation with constant coefficients in terms of exponential functions. It would appear, then, that one might be able to attribute a meaning to the forms eit and e-it and that these “functions” should be related to cos t and sin t. This matchup is accomplished by Euler’s formula, which is discussed in this section. When b2 - 4ac 6 0, the roots of the auxiliary equation (1)

ar 2 + br + c = 0

Electrical engineers frequently use the symbol j to denote 2 - 1.

†

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166

Chapter 4

Linear Second-Order Equations

associated with the homogeneous equation (2)

ay″ + by′ + cy = 0

are the complex conjugate numbers r1 = a + ib and r2 = a - ib

1i = 2 -12 ,

where a, b are the real numbers (3)

a = -

b 2a

and b =

24ac - b2 . 2a

As in the previous section, we would like to assert that the functions er1t and er2t are solutions to the equation (2). This is in fact the case, but before we can proceed, we need to address some fundamental questions. For example, if r1 = a + ib is a complex number, what do we mean by the expression e1a+ib2t? If we assume that the law of exponents applies to complex numbers, then (4)

e1a+ib2t = eat+ibt = eateibt .

We now need only clarify the meaning of eibt. For this purpose, let’s assume that the Maclaurin series for ez is the same for complex numbers z as it is for real numbers. Observing that i 2 = -1, then for u real we have eiu = 1 + 1iu2 + = 1 + iu = a1 -

1iu2 n 1iu2 2 + g+ + g 2! n!

u 2 iu 3 u 4 iu 5 + + + g 2! 3! 4! 5!

u2 u4 u3 u5 + + g b + ia u + + gb . 2! 4! 3! 5!

Now recall the Maclaurin series for cos u and sin u: cos u = 1 -

u2 u4 + + g, 2! 4!

u3 u5 + + g. 3! 5! Recognizing these expansions in the proposed series for eiu, we make the identification sin u = u -

(5)

eiU = cos U + i sin U ,

which is known as Euler’s formula.† When Euler’s formula (with u = bt) is used in equation (4), we find (6)

e1a+ib2t = eat 1cos bt + i sin bt2 ,

which expresses the complex function e1a+ib2t in terms of familiar real functions. Having made sense out of e1a+ib2t, we can now show (see Problem 30 on page 172) that (7)

d 1a+ib2t e = 1a + ib2e1a + ib2t , dt

†

Historical Footnote: This formula first appeared in Leonhard Euler’s monumental two-volume Introductio in Analysin Infinitorum (1748).

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Section 4.3

Auxiliary Equations with Complex Roots

167

and, with the choices of a and b as given in (3), the complex function e1a+ib2t is indeed a solution to equation (2), as is e1a - ib2t, and a general solution is given by (8)

y1t2 = c1e1a + ib2t + c2e1a - ib2t

= c1eat 1cos bt + i sin bt2 + c2eat 1cos bt - i sin bt2 .

Example 1 shows that in general the constants c1 and c2 that go into (8), for a specific initial value problem, are complex. Example 1

Use the general solution (8) to solve the initial value problem y″ + 2y′ + 2y = 0 ;

Solution

y102 = 0,

y′102 = 2 .

The auxiliary equation is r 2 + 2r + 2 = 0, which has roots -224 - 8 = -1 { i . 2 Hence, with a = -1, b = 1, a general solution is given by r =

y1t2 = c1e-t 1 cos t + i sin t2 + c2e-t 1 cos t - i sin t2 .

For initial conditions we have

y102 = c1e0 1 cos 0 + i sin 02 + c2e0 1 cos 0 - i sin 02 = c1 + c2 = 0 ,

y′102 = -c1e0 1 cos 0 + i sin 02 + c1e0 1 - sin 0 + i cos 02 -c2e0 1 cos 0 - i sin 02 + c2e0 1 - sin 0 - i cos 02 = 1 -1 + i2c1 + 1 -1 - i2c2 = 2.

As a result, c1 = -i, c2 = i, and y1t2 = -ie-t 1 cos t + i sin t2 + ie-t 1 cos t - i sin t2, or simply 2e-t sin t. ◆ The final form of the answer to Example 1 suggests that we should seek an alternative pair of solutions to the differential equation (2) that don’t require complex arithmetic, and we now turn to that task. In general, if z1t2 is a complex-valued function of the real variable t, we can write z1t2 = u1t2 + iv1t2, where u1t2 and v1t2 are real-valued functions. The derivatives of z1t2 are then given by dz du dv = +i , dt dt dt

d 2z d 2u d 2v = 2 +i 2 . 2 dt dt dt

With the following lemma, we show that the complex-valued solution e1a+ib2t gives rise to two linearly independent real-valued solutions.

Real Solutions Derived from Complex Solutions Lemma 2. Let z1t2 = u1t2 + iv1t2 be a solution to equation (2), where a, b, and c are real numbers. Then, the real part u1t2 and the imaginary part v1t2 are real-valued solutions of (2).†

†

It will be clear from the proof that this property holds for any linear homogeneous differential equation having realvalued coefficients.

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168

Chapter 4

Linear Second-Order Equations

Proof. By assumption, az″ + bz′ + cz = 0, and hence a1u″ + iv″2 + b1u′ + iv′2 + c1u + iv2 = 0 , 1au″ + bu′ + cu2 + i1av″ + bv′ + cv2 = 0 . But a complex number is zero if and only if its real and imaginary parts are both zero. Thus, we must have au″ + bu′ + cu = 0

and

av″ + bv′ + cv = 0 ,

which means that both u1t2and v1t2 are real-valued solutions of (2). ◆ When we apply Lemma 2 to the solution e1a+ib2t = eat cos bt + ieat sin bt , we obtain the following.

Complex Conjugate Roots If the auxiliary equation has complex conjugate roots a { ib, then two linearly independent solutions to (2) are eat cos bt

and

eat sin bt ,

and a general solution is (9)

y1t2 = c1eAt cos Bt + c2eAt sin Bt ,

where c1 and c2 are arbitrary constants.

In the preceding discussion, we glossed over some important details concerning complex numbers and complex-valued functions. In particular, further analysis is required to justify the use of the law of exponents, Euler’s formula, and even the fact that the derivative of ert is rert when r is a complex constant.† If you feel uneasy about our conclusions, we encourage you to substitute the expression in (9) into equation (2) to verify that it is, indeed, a solution. You may also be wondering what would have happened if we had worked with the function e1a-ib2t instead of e1a+ib2t. We leave it as an exercise to verify that e1a-ib2t gives rise to the same general solution (9). Indeed, the sum of these two complex solutions, divided by two, gives the first real-valued solution, while their difference, divided by 2i, gives the second. Example 2

Find a general solution to (10)

Solution

y″ + 2y′ + 4y = 0 .

The auxiliary equation is r 2 + 2r + 4 = 0 , which has roots r =

-2 { 2 -12 -2 { 24 - 16 = = -1 { i23 . 2 2

†

For a detailed treatment of these topics see, for example, Fundamentals of Complex Analysis, 3rd ed., by E. B. Saff and A. D. Snider (Prentice Hall, Upper Saddle River, New Jersey, 2003).

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Section 4.3

Auxiliary Equations with Complex Roots

169

Hence, with a = -1, b = 23, a general solution for (10) is y1t2 = c1e-t cos1 23 t2 + c2e-t sin1 23 t2 . ◆ When the auxiliary equation has complex conjugate roots, the (real) solutions oscillate between positive and negative values. This type of behavior is observed in vibrating springs. Example 3

In Section 4.1 we discussed the mechanics of the mass–spring oscillator (Figure 4.1, page 152), and we saw how Newton’s second law implies that the position y1t2 of the mass m is governed by the second-order differential equation (11)

my″1t2 + by′1t2 + ky1t2 = 0 ,

where the terms are physically identified as

3inertia4y″ + 3damping4y′ + 3stiffness4y = 0 .

Determine the equation of motion for a spring system when m = 36 kg, b = 12 kg/sec (which is equivalent to 12 N-sec/m), k = 37 kg/sec2, y102 = 0.7 m, and y′102 = 0.1 m/sec. After how many seconds will the mass first cross the equilibrium point? Solution

The equation of motion is given by y1t2, the solution of the initial value problem for the specified values of m, b, k, y102, and y′102. That is, we seek the solution to (12)

36y″ + 12y′ + 37y = 0 ;

y102 = 0.7 , y′102 = 0.1 .

The auxiliary equation for (12) is 36r 2 + 12r + 37 = 0 , which has roots r =

-12 { 2144 - 413621372 72

=

-12 { 1221 - 37 1 = - {i. 72 6

Hence, with a = -1>6, b = 1, the displacement y1t2 can be expressed in the form (13)

y1t2 = c1e-t>6 cos t + c2e-t>6 sin t .

We can find c1 and c2 by substituting y1t2 and y′1t2 into the initial conditions given in (12). Differentiating (13), we get a formula for y′1t2: y′1t2 = a -

c1 c2 + c2 b e-t>6cos t + a -c1 - b e-t>6sin t . 6 6

Substituting into the initial conditions now results in the system c1 = 0.7 , -

c1 + c2 = 0.1 . 6

Upon solving, we find c1 = 0.7 and c2 = 1.3>6. With these values, the equation of motion becomes y1t2 = 0.7e-t>6 cos t +

1.3 -t>6 e sin t . 6

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170

Chapter 4

Linear Second-Order Equations

To determine the times when the mass will cross the equilibrium point, we set y1t2 = 0 and solve for t: -t>6 -t>6 0 = y1t2 = 0.7e-t>6 cos t + 1.3 sin t = 1cos t210.7e-t>6 + 1.3 tan t2 . 6e 6e

But y1t2 is not zero for cos t = 0, so only the zeros of the second factor are pertinent; that is, tan t = - 4.2 1.3 . A quick glance at the graph of tan t reveals that the first positive t for which this is true lies between p>2 and p. Since the arctangent function takes only values between -p>2 and p>2, the appropriate adjustment is 4.2 2 + p ≈ 1.87 seconds . ◆ t = arctan 1 -1.3

From Example 3 we see that any second-order constant-coefficient differential equation ay″ + by′ + cy = 0 can be interpreted as describing a mass–spring system with mass a, damping coefficient b, spring stiffness c, and displacement y, if these constants make sense physically; that is, if a is positive and b and c are nonnegative. From the discussion in Section 4.1, then, we expect on physical grounds to see damped oscillatory solutions in such a case. This is consistent with the display in equation (9). With a = m and c = k, the exponential decay rate a equals -b> 12m2, and the angular frequency b equals 24mk - b2 > 12m2, by equation (3). It is a little surprising, then, that the solutions to the equation y″ + 4y′ + 4y = 0 do not oscillate; the general solution was shown in Example 3 of Section 4.2 (page 162) to be c1e-2t + c2te-2t. The physical significance of this is simply that when the damping coefficient b is too high, the resulting friction prevents the mass from oscillating. Rather than overshoot the spring’s equilibrium point, it merely settles in lazily. This could happen if a light mass on a weak spring were submerged in a viscous fluid. From the above formula for the oscillation frequency b, we can see that the oscillations will not occur for b 7 24mk. This overdamping phenomenon is discussed in more detail in Section 4.9. It is extremely enlightening to contemplate the predictions of the mass–spring analogy when the coefficients b and c in the equation ay″ + by′ + cy = 0 are negative. Example 4

Interpret the equation (14)

36y″ - 12y′ + 37y = 0

in terms of the mass–spring system. Solution

Equation (14) is a minor alteration of equation (12) in Example 3; the auxiliary equation 36r 2 - 12r + 37 has roots r = 1 + 2 16 { i. Thus, its general solution becomes (15)

y1t2 = c1e+t>6 cos t + c2e+t>6 sin t .

Comparing equation (14) with the mass–spring model (16)

3inertia4y″ + 3damping4y′ + 3stiffness4y = 0 ,

we have to envision a negative damping coefficient b = -12, giving rise to a friction force Ffriction = -by′ that imparts energy to the system instead of draining it. The increase in energy over time must then reveal itself in oscillations of ever-greater amplitude–precisely in accordance with formula (15), for which a typical graph is drawn in Figure 4.7. ◆

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Section 4.3

Auxiliary Equations with Complex Roots

171

y

et/6

t 2et/6

Figure 4.7 Solution graph for Example 4

Example 5

Interpret the equation (17)

y″ + 5y′ - 6y = 0

in terms of the mass–spring system. Solution

Comparing the given equation with (16), we have to envision a spring with a negative stiffness k = -6. What does this mean? As the mass is moved away from the spring’s equilibrium point, the spring repels the mass farther with a force Fspring = -ky that intensifies as the displacement increases. Clearly the spring must “exile” the mass to (plus or minus) infinity, and we expect all solutions y1t2 to approach {∞ as t increases (except for the equilibrium solution y1t2 K 0). In fact, in Example 1 of Section 4.2, we showed the general solution to equation (17) to be (18)

c1et + c2e-6t .

Indeed, if we examine the solutions y1t2 that start with a unit displacement y102 = 1 and velocity y′102 = v0, we find (19)

y1t2 =

6 + v0 t 1 - v0 -6t e + e , 7 7

and the plots in Figure 4.8 on page 172 confirm our prediction that all (nonequilibrium) solutions diverge—except for the one with v0 = -6. What is the physical significance of this isolated bounded solution? Evidently, if the mass is given an initial inwardly directed velocity of -6, it has barely enough energy to overcome the effect of the spring banishing it to + ∞ but not enough energy to cross the equilibrium point (and get pushed to - ∞ ). So it asymptotically approaches the (extremely delicate) equilibrium position y = 0. ◆ In Section 4.8, we will see that taking further liberties with the mass–spring interpretation enables us to predict qualitative features of more complicated equations. Throughout this section we have assumed that the coefficients a, b, and c in the differential equation were real numbers. If we now allow them to be complex constants, then the roots

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Chapter 4

Linear Second-Order Equations

y

y0 5 0

y0 5 6

1

y0 5 26

t

y0 5 218

y0 5 212

Figure 4.8 Solution graphs for Example 5

r1, r2 of the auxiliary equation (1) are, in general, also complex but not necessarily conjugates of each other. When r1 ≠ r2, a general solution to equation (2) still has the form y1t2 = c1er1t + c2er2t , but c1 and c2 are now arbitrary complex-valued constants, and we have to resort to the clumsy calculations of Example 1. We also remark that a complex differential equation can be regarded as a system of two real differential equations since we can always work separately with its real and imaginary parts. Systems are discussed in Chapters 5 and 9.

4.3

EXERCISES

In Problems 1–8, the auxiliary equation for the given differential equation has complex roots. Find a general solution. 1. y″ + 9y = 0 2. y″ + y = 0 3. z″ - 6z′ + 10z = 0 4. y″ - 10y′ + 26y = 0 5. w″ + 4w′ + 6w = 0 6. y″ - 4y′ + 7y = 0 7. 4y″ + 4y′ + 6y = 0 8. 4y″ - 4y′ + 26y = 0 In Problems 9–20, find a general solution. 9. y″ - 8y′ + 7y = 0 10. y″ + 4y′ + 8y = 0 11. z″ + 10z′ + 25z = 0 12. u″ + 7u = 0 13. y″ - 2y′ + 26y = 0 14. y″ + 2y′ + 5y = 0 15. y″ - 3y′ - 11y = 0 16. y″ + 10y′ + 41y = 0 17. y″ - y′ + 7y = 0 18. 2y″ + 13y′ - 7y = 0 19. y‴ + y″ + 3y′ - 5y = 0 20. y‴ - y″ + 2y = 0 In Problems 21–27, solve the given initial value problem. 21. y″ + 2y′ + 2y = 0 ; y102 = 2 , y′102 = 1 22. y″ + 2y′ + 17y = 0 ; y102 = 1 , y′102 = - 1 23. w″ - 4w′ + 2w = 0 ; w102 = 0 , w′102 = 1

24. 25. 26. 27.

y″ + 9y = 0 ; y102 = 1 , y′102 = 1 y″ - 2y′ + 2y = 0 ; y1p2 = ep , y′1p2 = 0 y″ - 2y′ + y = 0 ; y102 = 1 , y′102 = - 2 y‴ - 4y″ + 7y′ - 6y = 0 ; y102 = 1 , y′102 = 0, y″102 = 0

28. To see the effect of changing the parameter b in the initial value problem y″ + by′ + 4y = 0 ; y102 = 1 , y′102 = 0 , solve the problem for b = 5, 4, and 2 and sketch the solutions. 29. Find a general solution to the following higher-order equations. (a) y‴ - y″ + y′ + 3y = 0 (b) y‴ + 2y″ + 5y′ - 26y = 0 (c) yiv + 13y″ + 36y = 0 30. Using the representation for e1a+ib2t in (6), verify the differentiation formula (7).

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Section 4.3

31. Using the mass–spring analogy, predict the behavior as t S + ∞ of the solution to the given initial value problem. Then confirm your prediction by actually solving the problem. (a) y″ + 16y = 0 ; y102 = 2 , y′102 = 0 (b) y″ + 100y′ + y = 0 ; y102 = 1 , y′102 = 0 (c) y″ - 6y′ + 8y = 0 ; y102 = 1 , y′102 = 0 (d) y″ + 2y′ - 3y = 0 ; y102 = - 2 , y′102 = 0 (e) y″ - y′ - 6y = 0 ; y102 = 1 , y′102 = 1 32. Vibrating Spring without Damping. A vibrating spring without damping can be modeled by the initial value problem (11) in Example 3 by taking b = 0. (a) If m = 10 kg, k = 250 kg/sec2, y102 = 0.3 m, and y′102 = -0.1 m/sec, find the equation of motion for this undamped vibrating spring. (b) After how many seconds will the mass in part (a) first cross the equilibrium point? (c) When the equation of motion is of the form displayed in (9), the motion is said to be oscillatory with frequency b>2p. Find the frequency of oscillation for the spring system of part (a). 33. Vibrating Spring with Damping. Using the model for a vibrating spring with damping discussed in Example 3: (a) Find the equation of motion for the vibrating spring with damping if m = 10 kg, b = 60 kg/sec, k = 250 kg/sec2, y102 = 0.3 m, and y′102 = -0.1 m/sec. (b) After how many seconds will the mass in part (a) first cross the equilibrium point? (c) Find the frequency of oscillation for the spring system of part (a). [Hint: See the definition of frequency given in Problem 32(c).] (d) Compare the results of Problems 32 and 33 and determine what effect the damping has on the frequency of oscillation. What other effects does it have on the solution? 34. RLC Series Circuit. In the study of an electrical circuit consisting of a resistor, capacitor, inductor, and an electromotive force (see Figure 4.9), we are led to an initial value problem of the form (20)

L

q dI + RI + = E1t2 ; dt C

q102 = q0 , I102 = I0 , where L is the inductance in henrys, R is the resistance in ohms, C is the capacitance in farads, E1t2 is the electromotive force in volts, q1t2 is the charge in coulombs on the capacitor at time t, and I = dq>dt is the current in amperes. Find the current at time t if the charge on the capacitor is initially zero, the initial current is zero, L = 10 H, R = 20 Ω, C = 162602 -1 F, and E1t2 = 100 V. [Hint: Differentiate both sides of the

Auxiliary Equations with Complex Roots

173

differential equation in (20) to obtain a homogeneous linear second-order equation for I1t2. Then use (20) to determine dI>dt at t = 0.] R

E(t)

q(t)

C

I(t) L Figure 4.9 RLC series circuit

35. Swinging Door. The motion of a swinging door with an adjustment screw that controls the amount of friction on the hinges is governed by the initial value problem Iu″ + bu′ + ku = 0 ; u102 = u0 , u′102 = v0 , where u is the angle that the door is open, I is the moment of inertia of the door about its hinges, b 7 0 is a damping constant that varies with the amount of friction on the door, k 7 0 is the spring constant associated with the swinging door, u0 is the initial angle that the door is opened, and y0 is the initial angular velocity imparted to the door (see Figure 4.10). If I and k are fixed, determine for which values of b the door will not continually swing back and forth when closing.

Figure 4.10 Top view of swinging door

36. Although the real general solution form (9) is convenient, it is also possible to use the form (21)

d1e1a+ib2t + d2e1a-ib2t

to solve initial value problems, as illustrated in Example 1. The coefficients d1 and d2 are complex constants. (a) Use the form (21) to solve Problem 21. Verify that your form is equivalent to the one derived using (9). (b) Show that, in general, d1 and d2 in (21) must be complex conjugates in order that the solution be real. 37. The auxiliary equations for the following differential equations have repeated complex roots. Adapt the “repeated root” procedure of Section 4.2 to find their general solutions: (a) yiv + 2y″ + y = 0 . (b) yiv + 4y‴ + 12y″ + 16y′ + 16y = 0 . [Hint: The auxiliary equation is 1r 2 + 2r + 42 2 = 0.]

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174

Chapter 4

Linear Second-Order Equations

38. Prove the sum of angles formula for the sine function by following these steps. Fix x. (a) Let f1t2 J sin1x + t2. Show that f ″1t2 + f1t2 = 0, f102 = sin x, and f′102 = cos x. (b) Use the auxiliary equation technique to solve the initial value problem y″ + y = 0, y102 = sin x, and y′102 = cos x.

(c) By uniqueness, the solution in part (b) is the same as f1t2 from part (a). Write this equality; this should be the standard sum of angles formula for sin 1x + t2.

4.4 Nonhomogeneous Equations:

the Method of Undetermined Coefficients In this section we employ “judicious guessing” to derive a simple procedure for finding a solution to a nonhomogeneous linear equation with constant coefficients (1)

ay″ + by′ + cy = f1t2,

when the nonhomogeneity f1t2 is a single term of a special type. Our experience in Section 4.3 indicates that (1) will have an infinite number of solutions. For the moment we are content to find one particular solution. To motivate the procedure, let’s first look at a few instructive examples. Example 1

Find a particular solution to (2)

Solution

y″ + 3y′ + 2y = 3t .

We need to find a function y1t2 such that the combination y″ + 3y′ + 2y is a linear function of t—namely, 3t. Now what kind of function y “ends up” as a linear function after having its zeroth, first, and second derivatives combined? One immediate answer is: another linear function. So we might try y1 1t2 = At and attempt to match up y1> + 3y1= + 2y1 with 3t. Perhaps you can see that this won’t work: y1 = At, y1= = A and y1> = 0 gives us y1> + 3y1= + 2y1 = 3A + 2At , and for this to equal 3t, we require both that A = 0 and A = 3>2. We’ll have better luck if we append a constant term to the trial function: y2 1t2 = At + B. Then y2= = A, y2> = 0, and y2> + 3y2= + 2y2 = 3A + 21At + B2 = 2At + 13A + 2B2 ,

which successfully matches up with 3t if 2A = 3 and 3A + 2B = 0. Solving this system gives A = 3>2 and B = -9>4. Thus, the function y2 1t2 =

3 9 t2 4 is a solution to (2). ◆ Example 1 suggests the following method for finding a particular solution to the equation ay″ + by′ + cy = Ct m, m = 0, 1, 2, c; namely, we guess a solution of the form

yp 1t2 = Amt m + g + A1t + A0 ,

with undetermined coefficients Aj, and match the corresponding powers of t in ay″ + by′ + cy with Ct m.† This procedure involves solving m + 1 linear equations in the m + 1 unknowns In this case the coefficient of t k in ay″ + by′ + cy will be zero for k ≠ m and C for k = m.

†

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Section 4.4

Nonhomogeneous Equations: the Method of Undetermined Coefficients

175

A0, A1, c, Am, and hopefully they have a solution. The technique is called the method of undetermined coefficients. Note that, as Example 1 demonstrates, we must retain all the powers t m, t m - 1, c, t 1, t 0 in the trial solution even though they are not present in the nonhomogeneity f1t2. Example 2

Find a particular solution to (3)

Solution

y″ + 3y′ + 2y = 10e3t .

We guess yp 1t2 = Ae3t because then yp= and yp> will retain the same exponential form: yp> + 3yp= + 2yp = 9Ae3t + 313Ae3t 2 + 21Ae3t 2 = 20Ae3t .

Setting 20Ae3t = 10e3t and solving for A gives A = 1>2; hence, yp 1t2 =

e3t 2

is a solution to (3). ◆ Example 3

Find a particular solution to (4)

Solution

y″ + 3y′ + 2y = sin t .

Our initial action might be to guess y1 1t2 = A sin t, but this will fail because the derivatives introduce cosine terms: y1> + 3y1= + 2y1 = -A sin t + 3A cos t + 2A sin t = A sin t + 3A cos t , and matching this with sin t would require that A equal both 1 and 0. So we include the cosine term in the trial solution: yp 1t2 = A sin t + B cos t , yp= 1t2 = A cos t - B sin t , yp> 1t2 = -A sin t - B cos t ,

and (4) becomes

yp> 1t2 + 3yp= 1t2 + 2yp 1t2 = -A sin t - B cos t + 3A cos t - 3B sin t + 2A sin t + 2B cos t = 1A - 3B2 sin t + 1B + 3A2 cos t = sin t .

The equations A - 3B = 1, B + 3A = 0 have the solution A = 0.1, B = -0.3. Thus, the function yp 1t2 = 0.1 sin t - 0.3 cos t

is a particular solution to (4). ◆ More generally, for an equation of the form (5)

ay″ + by′ + cy = C sin bt 1or C cos bt2 ,

the method of undetermined coefficients suggests that we guess (6)

yp 1t2 = A cos bt + B sin bt

and solve (5) for the unknowns A and B.

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176

Chapter 4

Linear Second-Order Equations

If we compare equation (5) with the mass–spring system equation (7)

3inertia4 * y″ + 3damping4 * y′ + 3stiffness4 * y = Fext ,

we can interpret (5) as describing a damped oscillator, shaken with a sinusoidal force. According to our discussion in Section 4.1, then, we would expect the mass ultimately to respond by moving in synchronization with the forcing sinusoid. In other words, the form (6) is suggested by physical, as well as mathematical, experience. A complete description of forced oscillators will be given in Section 4.10. Example 4

Find a particular solution to (8)

Solution

y″ + 4y = 5t 2et .

Our experience with Example 1 suggests that we take a trial solution of the form yp 1t2 = 1At 2 + Bt + C2et, to match the nonhomogeneity in (8). We find yp = 1At 2 + Bt + C2et ,

yp= = 12At + B2et + 1At 2 + Bt + C2et ,

yp> = 2Aet + 212At + B2et + 1At 2 + Bt + C2et ,

yp> + 4yp = et 12A + 2B + C + 4C2 + tet 14A + B + 4B2 + t 2et 1A + 4A2 = 5t 2et . Matching like terms yields A = 1, B = -4>5, and C = -2>25. A solution is given by yp 1t2 = a t 2 -

4t 2 - b et . ◆ 5 25

As our examples illustrate, when the nonhomogeneous term f1t2 is an exponential, a sine, a cosine function, or a nonnegative integer power of t times any of these, the function f1t2 itself suggests the form of a particular solution. However, certain situations thwart the straightforward application of the method of undetermined coefficients. Consider, for example, the equation (9)

y″ + y′ = 5 .

Example 1 suggests that we guess y1 1t2 = A, a zero-degree polynomial. But substitution into (9) proves futile: 1A2 ″ + 1A2′ = 0 ≠ 5 .

The problem arises because any constant function, such as y1 1t2 = A, is a solution to the corresponding homogeneous equation y″ + y′ = 0, and the undetermined coefficient A gets lost upon substitution into the equation. We would encounter the same situation if we tried to find a solution to (10)

y″ - 6y′ + 9y = e3t

of the form y1 = Ae3t, because e3t solves the associated homogeneous equation and

3 Ae3t 4 ″ - 6 3 Ae3t 4 ′ + 9 3 Ae3t 4

= 0 ≠ e3t .

The “trick” for refining the method of undetermined coefficients in these situations smacks of the same logic as in Section 4.2, when a method was prescribed for finding second solutions to homogeneous equations with double roots. Basically, we append an extra factor of t to the trial

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Section 4.4

Nonhomogeneous Equations: the Method of Undetermined Coefficients

177

solution suggested by the basic procedure. In other words, to solve (9) we try yp 1t2 = At instead of A: (9′)

yp = At, yp= = A, yp> = 0, yp> + yp= = 0 + A = 5 , A = 5, yp 1t2 = 5t .

Similarly, to solve (10) we try yp = Ate3t instead of Ae3t. The trick won’t work this time, because the characteristic equation of (10) has a double root and, consequently, Ate3t also solves the homogeneous equation:

3 Ate3t 4 ″ - 6 3 Ate3t 4 ′ + 9 3 Ate3t 4

= 0 ≠ e3t.

But if we append another factor of t, yp = At 2e3t, we succeed in finding a particular solution:† yp = At 2e3t, yp= = 2Ate3t + 3At 2e3t, yp> = 2Ae3t + 12Ate3t + 9At 2e3t ,

yp> - 6yp= + 9yp = 12Ae3t + 12Ate3t + 9At 2e3t 2 - 612Ate3t + 3At 2e3t 2 + 91At 2e3t 2 = 2Ae3t = e3t ,

so A = 1>2 and yp 1t2 = t 2e3t >2. To see why this strategy resolves the problem and to generalize it, recall the form of the original differential equation (1), ay″ + by′ + cy = f1t2. Its associated auxiliary equation is (11)

ar 2 + br + c = 0 ,

and if r is a double root, then (12)

2ar + b = 0

holds also [equation (13), Section 4.2, page 163]. Now suppose the nonhomogeneity f1t2 has the form Ct mert, and we seek to match this f1t2 by substituting yp 1t2 = 1Ant n + An - 1t n - 1 + g + A1t + A0 2ert into (1), with the power n to be determined. For simplicity we merely list the leading terms in yp, yp= , and yp>: yp = Ant nert + An - 1t n - 1ert + An - 2t n - 2ert + (lower-order terms)

y′p = Anrt nert + Annt n - 1ert + An - 1rt n - 1ert + An - 1 1n - 12t n - 2ert + An - 2rt n - 2ert + 1l.o.t.2 ,

y″p = Anr 2t nert + 2Annrt n - 1ert + Ann1n - 12t n - 2ert

+ An - 1r 2t n - 1ert + 2An - 1r1n - 12t n - 2ert + An - 2r 2t n - 2ert + 1l.o.t2 .

Then the left-hand member of (1) becomes (13)

ayp> + byp= + cyp = An 1ar 2 + br + c2t nert + 3Ann12ar + b2 + An - 1 1ar 2 + br + c24t n - 1ert

+ 3Ann1n - 12a + An - 1 1n - 1212ar + b2 + An - 2 1ar 2 + br + c24t n - 2ert + 1l.o.t.2 ,

Indeed, the solution t 2 to the equation y″ = 2, computed by simple integration, can also be derived by appending two factors of t to the solution y K 1 of the associated homogeneous equation.

†

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178

Chapter 4

Linear Second-Order Equations

and we observe the following: Case 1. If r is not a root of the auxiliary equation, the leading term in (13) is An 1ar 2 + br + c2t nert, and to match f1t2 = Ct mert we must take n = m: yp 1t2 = 1Amt m + g + A1t + A0 2ert . Case 2. If r is a simple root of the auxiliary equation, (11) holds and the leading term in (13) is Ann12ar + b2t n - 1ert, and to match f1t2 = Ct mert we must take n = m + 1: yp 1t2 = 1Am + 1t m + 1 + Amt m + g + A1t + A0 2ert . However, now the final term A0ert can be dropped, since it solves the associated homogeneous equation, so we can factor out t and for simplicity renumber the coefficients to write yp 1t2 = t1Amt m + g + A1t + A0 2ert . Case 3. If r is a double root of the auxiliary equation, (11) and (12) hold and the leading term in (13) is Ann1n - 12at n - 2ert, and to match f1t2 = Ct mert we must take n = m + 2: yp 1t2 = 1Am + 2t m + 2 + Am + 1t m + 1 + g + A2t 2 + A1t + A0 2ert, but again we drop the solutions to the associated homogeneous equation and renumber to write yp 1t2 = t 2 1Amt m + g + A1t + A0 2ert . We summarize with the following rule.

Method of Undetermined Coefficients To find a particular solution to the differential equation ay″ + by′ + cy = Ct mert, where m is a nonnegative integer, use the form (14)

yp 1t2 = t s 1Amt m + g + A1t + A0 2ert,

with (i) s = 0 if r is not a root of the associated auxiliary equation; (ii) s = 1 if r is a simple root of the associated auxiliary equation; and (iii) s = 2 if r is a double root of the associated auxiliary equation. To find a particular solution to the differential equation Ct meatcos bt ay″ + by′ + cy = c or Ct meatsin bt for b ≠ 0, use the form (15)

yp 1t2 = t s 1Amt m + g + A1t + A0 2eat cos bt + t s 1Bmt m + g + B1t + B0 2eat sin bt ,

with (iv) s = 0 if a + ib is not a root of the associated auxiliary equation; and (v) s = 1 if a + ib is a root of the associated auxiliary equation.

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Section 4.4

Nonhomogeneous Equations: the Method of Undetermined Coefficients

179

[The (cos, sin) formulation (15) is easily derived from the exponential formulation (14) by putting r = a + ib and employing Euler’s formula, as in Section 4.3.] Remark 1. The nonhomogeneity Ct m corresponds to the case when r = 0. Remark 2. The rigorous justification of the method of undetermined coefficients [including the analysis of the terms we dropped in (13)] will be presented in a more general context in Chapter 6. Example 5

Find the form for a particular solution to (16)

y″ + 2y′ - 3y = f1t2 ,

where f1t2 equals (a) 7 cos 3t Solution

(b) 2tet sin t

(c) t 2 cos pt

(d) 5e-3t

(e) 3tet

(f) t 2et

The auxiliary equation for the homogeneous equation corresponding to (16), r 2 + 2r - 3 = 0, has roots r1 = 1 and r2 = -3. Notice that the functions in (a), (b), and (c) are associated with complex roots (because of the trigonometric factors). These are clearly different from r1 and r2, so the solution forms correspond to (15) with s = 0: (a) yp 1t2 = A cos 3t + B sin 3t

(b) yp 1t2 = 1A1t + A0 2et cos t + 1B1t + B0 2et sin t

(c) yp 1t2 = 1A2t 2 + A1t + A0 2 cos pt + 1B2t 2 + B1t + B0 2 sin pt

For the nonhomogeneity in (d) we appeal to (ii) and take yp 1t2 = Ate-3t since -3 is a simple root of the auxiliary equation. Similarly, for (e) we take yp 1t2 = t1A1t + A0 2et and for (f) we take yp 1t2 = t1A2t 2 + A1t + A0 2e t. ◆ Example 6

Find the form of a particular solution to y″ - 2y′ + y = f1t2, for the same set of nonhomogeneities f1t2 as in Example 5.

Solution

Now the auxiliary equation for the corresponding homogeneous equation is r 2 - 2r + 1 = 1r - 12 2 = 0, with the double root r = 1. This root is not linked with any of the nonhomogeneities (a) through (d), so the same trial forms should be used for (a), (b), and (c) as in the previous example, and y1t2 = Ae-3t will work for (d). Since r = 1 is a double root, we have s = 2 in (14) and the trial forms for (e) and (f) have to be changed to (e) yp 1t2 = t 2 1A1t + A0 2et

(f) yp 1t2 = t 2 1A2t 2 + A1t + A0 2et

respectively, in accordance with (iii). ◆ Example 7

Find the form of a particular solution to y″ - 2y′ + 2y = 5tet cos t .

Solution

Now the auxiliary equation for the corresponding homogeneous equation is r 2 - 2r + 2 = 0, and it has complex roots r1 = 1 + i, r2 = 1 - i. Since the nonhomogeneity involves eat cos bt with a = b = 1; that is, a + ib = 1 + i = r1, the solution takes the form yp 1t2 = t1A1t + A0 2et cos t + t1B1t + B0 2et sin t . ◆

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180

Chapter 4

Linear Second-Order Equations

The nonhomogeneity tan t in an equation like y″ + y′ + y = tan t is not one of the forms for which the method of undetermined coefficients can be used; the derivatives of the “trial solution” y1t2 = A tan t, for example, get complicated, and it is not clear what additional terms need to be added to obtain a true solution. In Section 4.6 we discuss a different procedure that can handle such nonhomogeneous terms. Keep in mind that the method of undetermined coefficients applies only to nonhomogeneities that are polynomials, exponentials, sines or cosines, or products of these functions. The superposition principle in Section 4.5 shows how the method can be extended to the sums of such nonhomogeneities. Also, it provides the key to assembling a general solution to (1) that can accommodate initial value problems, which we have avoided so far in our examples.

4.4

EXERCISES

In Problems 1– 8, decide whether or not the method of undetermined coefficients can be applied to find a particular solution of the given equation. 1. y″ + 2y′ - y = t -1et 2. 5y″ - 3y′ + 2y = t 3 cos 4t 3. 2y″1x2 - 6y′1x2 + y1x2 = 1 sin x2 >e4x 4. x″ + 5x′ - 3x = 3t 5. y″1u2 + 3y′1u2 - y1u2 = sec u 6. 2v″1x2 - 3v1x2 = 4x sin2 x + 4x cos2 x 7. 8z′1x2 - 2z1x2 = 3x100e4x cos 25x 8. ty″ - y′ + 2y = sin 3t In Problems 9–26, find a particular solution to the differential equation. 9. y″ + 3y = -9 10. y″ + 2y′ - y = 10 12. 2x′ + x = 3t 2 11. y″1x2 + y1x2 = 2x 13. y″ - y′ + 9y = 3 sin 3t 14. 2z″ + z = 9e2t 2 dy dy 15. - 5 + 6y = xex 16. u″1t2 - u1t2 = t sin t 2 dx dx 17. y″ + 4y = 8 sin 2t 18. y″ - 2y′ + y = 8et -3t 19. 4y″ + 11y′ - 3y = - 2te 20. y″ + 4y = 16t sin 2t 21. x″1t2 - 4x′1t2 + 4x1t2 = te2t

22. 23. 24. 25. 26.

x″1t2 - 2x′1t2 + x1t2 = 24t 2et y″1u2 - 7y′1u2 = u 2 y″1x2 + y1x2 = 4x cos x y″ + 2y′ + 4y = 111e2t cos 3t y″ + 2y′ + 2y = 4te-t cos t

In Problems 27–32, determine the form of a particular solution for the differential equation. (Do not evaluate coefficients.) 27. y″ + 9y = 4t 3 sin 3t 28. y″ - 6y′ + 9y = 5t 6e3t 29. y″ + 3y′ - 7y = t 4et 30. y″ - 2y′ + y = 7et cos t 31. y″ + 2y′ + 2y = 8t 3e-t sin t 32. y″ - y′ - 12y = 2t 6e-3t In Problems 33–36, use the method of undetermined coefficients to find a particular solution to the given higher-order equation. 33. y‴ - y″ + y = sin t 34. 2y‴ + 3y″ + y′ - 4y = e-t 35. y‴ + y″ - 2y = tet 36. y 142 - 3y″ - 8y = sin t

4.5 The Superposition Principle and

Undetermined Coefficients Revisited The next theorem describes the superposition principle, a very simple observation which nonetheless endows the solution set for our equations with a powerful structure. It extends the applicability of the method of undetermined coefficients and enables us to solve initial value problems for nonhomogeneous differential equations.

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Section 4.5

The Superposition Principle and Undetermined Coefficients Revisited

181

Superposition Principle Theorem 3.

If y1 is a solution to the differential equation

ay″ + by′ + cy = f1 1t2 ,

and y2 is a solution to

ay″ + by′ + cy = f2 1t2 ,

then for any constants k1 and k2, the function k1y1 + k2y2 is a solution to the differential equation ay″ + by′ + cy = k1 f1 1t2 + k2 f2 1t2 .

Proof. This is straightforward; by substituting and rearranging we find a1k1y1 + k2y2 2″ + b1k1y1 + k2y2 2′ + c1k1y1 + k2y2 2 = k1 1ay1> + by1= + cy1 2 + k2 1ay2> + by2= + cy2 2 = k1 f1 1t2 + k2 f2 1t2 . ◆

Example 1

Solution

Find a particular solution to (1)

y″ + 3y′ + 2y = 3t + 10e3t

(2)

y″ + 3y′ + 2y = -9t + 20e3t .

and

In Example 1, Section 4.4, we found that y1 1t2 = 3t>2 - 9>4 was a solution to y″ + 3y′+ 2y = 3t, and in Example 2 we found that y2 1t2 = e3t >2 solved y″ + 3y′ + 2y = 10e3t. By superposition, then, y1 + y2 = 3t>2 - 9/4 + e3t >2 solves equation (1). The right-hand member of (2) equals minus three times 13t2 plus two times 110e3t 2. Therefore, this same combination of y1 and y2 will solve (2): y1t2 = -3y1 + 2y2 = -313t>2 - 9>42 + 21e3t >22 = -9t>2 + 27>4 + e3t . ◆ If we take a particular solution yp to a nonhomogeneous equation like (3)

ay″ + by′ + cy = f1t2

and add it to a general solution c1y1 + c2y2 of the homogeneous equation associated with (3), (4)

ay″ + by′ + cy = 0 ,

the sum (5)

y1t2 = yp 1t2 + c1y1 1t2 + c2y2 1t2

is again, according to the superposition principle, a solution to (3):

a1yp + c1y1 + c2y2 2″ + b1yp + c1y1 + c2y2 2′ + c1yp + c1y1 + c2y2 2 = f1t2 + 0 + 0 = f1t2 .

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182

Chapter 4

Linear Second-Order Equations

Since (5) contains two parameters, one would suspect that c1 and c2 can be chosen to make it satisfy arbitrary initial conditions. It is easy to verify that this is indeed the case.

Existence and Uniqueness: Nonhomogeneous Case Theorem 4. For any real numbers a1 ≠02, b, c, t0, Y0, and Y1, suppose yp 1t2 is a particular solution to (3) in an interval I containing t0 and that y1 1t2 and y2 1t2 are linearly independent solutions to the associated homogeneous equation (4) in I. Then there exists a unique solution in I to the initial value problem ay″ + by′ + cy = f1t2,

(6)

y1t0 2 = Y0, y′1t0 2 = Y1 ,

and it is given by (5), for the appropriate choice of the constants c1, c2.

Proof. We have already seen that the superposition principle implies that (5) solves the differential equation. To satisfy the initial conditions in (6) we need to choose the constants so that (7)

b

yp 1t0 2 + c1y1 1t0 2 + c2y2 1t0 2 = Y0 , yp= 1t0 2 + c1y1= 1t0 2 + c2y2= 1t0 2 = Y1 .

But as in the proof of Theorem 2 in Section 4.2, simple algebra shows that the choice c1 = c2 =

3 Y0 - yp 1t0 2 4 y2= 1t0 2 - 3 Y1 - yp= 1t0 2 4 y2 1t0 2 y1 1t0 2y2= 1t0 2 - y1= 1t0 2y2 1t0 2

3 Y1 - yp= 1t0 2 4 y1 1t0 2 - 3 Y0 - yp 1t0 2 4 y1= 1t0 2

and

y1 1t0 2y2= 1t0 2 - y1= 1t0 2y2 1t0 2

solves (7) unless the denominator is zero; Lemma 1, Section 4.2, assures us that it is not. Why is the solution unique? If yI 1t2 were another solution to (6), then the difference yII 1t2 := yp 1t2 + c1y1 1t2 + c2y2 1t2 - yI 1t2 would satisfy (8)

b

ayII> + byII= + cyII = f1t2 - f1t2 = 0 , yII 1t0 2 = Y0 - Y0 = 0 , yII= 1t0 2 = Y1 - Y1 = 0 .

But the initial value problem (8) admits the identically zero solution, and Theorem 1 in Section 4.2 applies since the differential equation in (8) is homogeneous. Consequently, (8) has only the identically zero solution. Thus, yII K 0 and yI = yp + c1y1 + c2y2. ◆ These deliberations entitle us to say that y = yp + c1y1 + c2y2 is a general solution to the nonhomogeneous equation (3), since any solution yg 1t2 can be expressed in this form. (Proof: As in Section 4.2, we simply pick c1 and c2 so that yp + c1y1 + c2y2 matches the value and the derivative of yg at any single point; by uniqueness, yp + c1y1 + c2y2 and yg have to be the same function.)

Example 2

Given that yp 1t2 = t 2 is a particular solution to y″ - y = 2 - t 2 , find a general solution and a solution satisfying y102 = 1, y′102 = 0.

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Section 4.5

Solution

The Superposition Principle and Undetermined Coefficients Revisited

183

The corresponding homogeneous equation, y″ - y = 0 , has the associated auxiliary equation r 2 - 1 = 0. Because r = {1 are the roots of this equation, a general solution to the homogeneous equation is c1et + c2e-t. Combining this with the particular solution yp 1t2 = t 2 of the nonhomogeneous equation, we find that a general solution is y1t2 = t 2 + c1et + c2e-t . To meet the initial conditions, set y102 = 02 + c1e0 + c2e-0 = 1 , y′102 = 2 * 0 + c1e0 - c2e-0 = 0 , which yields c1 = c2 = 12 . The answer is 1 y1t2 = t 2 + 1et + e-t 2 = t 2 + cosh t . ◆ 2

Example 3

A mass–spring system is driven by a sinusoidal external force 15 sin t + 5 cos t2. The mass equals 1, the spring constant equals 2, and the damping coefficient equals 2 (in appropriate units), so the deliberations of Section 4.1 imply that the motion is governed by the differential equation (9)

y″ + 2y′ + 2y = 5 sin t + 5 cos t .

If the mass is initially located at y102 = 1, with a velocity y′102 = 2, find its equation of motion. Solution

The associated homogeneous equation y″ + 2y′ + 2y = 0 was studied in Example 1, Section 4.3; the roots of the auxiliary equation were found to be -1 { i, leading to a general solution c1e-t cos t + c2e-t sin t. The method of undetermined coefficients dictates that we try to find a particular solution of the form A sin t + B cos t for the first nonhomogeneity 5 sin t: (10)

yp = A sin t + B cos t ,

yp= = A cos t - B sin t ,

yp> = -A sin t - B cos t ;

yp> + 2yp= + 2yp = 1 -A - 2B + 2A2 sin t + 1 -B + 2A + 2B2 cos t = 5 sin t .

Matching coefficients requires A = 1, B = -2 and so yp = sin t - 2 cos t. The second nonhomogeneity 5 cos t calls for the identical form for yp and leads to 1 -A - 2B + 2A2 sin t + 1 -B + 2A + 2B2 cos t = 5 cos t, or A = 2, B = 1. Hence yp = 2 sin t + cos t. By the superposition principle, a general solution to (9) is given by the sum y = c1e-t cos t + c2e-t sin t + sin t - 2 cos t + 2 sin t + cos t = c1e-t cos t + c2e-t sin t + 3 sin t - cos t . The initial conditions are y102 = 1 = c1e-0 cos 0 + c2e-0 sin 0 + 3 sin 0 - cos 0 = c1 - 1 , y′102 = 2 = c1 3 -e-t cos t - e-t sin t 4 t = 0 + c2 3 -e-t sin t + e-t cos t 4 t = 0 + 3 cos 0 + sin 0 = -c1 + c2 + 3 , requiring c1 = 2, c2 = 1, and thus (11)

y1t2 = 2e-t cos t + e-t sin t + 3 sin t - cos t . ◆

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184

Chapter 4

Linear Second-Order Equations

The solution (11) exemplifies the features of forced, damped oscillations that we anticipated in Section 4.1. There is a sinusoidal component 13 sin t - cos t2 that is synchronous with the driving force 15 sin t + 5 cos t2, and a component 12e-t cos t + e-t sin t2 that dies out. When the system is “pumped” sinusoidally, the response is a synchronous sinusoidal oscillation, after an initial transient that depends on the initial conditions; the synchronous response is the particular solution supplied by the method of undetermined coefficients, and the transient is the solution to the associated homogeneous equation. This interpretation will be discussed in detail in Sections 4.9 and 4.10. You may have observed that, since the two undetermined-coefficient forms in the last example were identical and were destined to be added together, we could have used the form (10) to match both nonhomogeneities at the same time, deriving the condition yp> + 2yp= + 2yp = 1 -A - 2B + 2A2 sin t + 1 -B + 2A + 2B2 cos t = 5 sin t + 5 cos t ,

with solution yp = 3 sin t - cos t. The next example illustrates this “streamlined” procedure. Example 4

Find a particular solution to (12)

Solution

y″ - y = 8tet + 2et .

A general solution to the associated homogeneous equation is easily seen to be c1et + c2e-t. Thus, a particular solution for matching the nonhomogeneity 8tet has the form t1A1t + A0 2et, whereas matching 2et requires the form A0tet. Therefore, we can match both with the first form: yp = t 1 A1t + A0 2 et =

yp= = yp> = = Thus

1 A1t2 + A0t 2 et , 1 A1t2 + A0t 2 et + 1 2A1t + A0 2 et = 3 A1t2 + 1 2A1 + A0 2 t + A0 4 et , 3 2A1t + 1 2A1 + A0 2 4 et + 3 A1t2 + 1 2A1 + A0 2 t + A0 4 et 3 A1t2 + 1 4A1 + A0 2 t + 1 2A1 + 2A0 2 4 et .

y″p - yp = 3 4A1t + 1 2A1 + 2A0 2 4 et = 8tet + 2et ,

which yields A1 = 2, A0 = -1, and so yp = 12t 2 - t2et. ◆ We generalize this procedure by modifying the method of undetermined coefficients as follows.

Method of Undetermined Coefficients (Revisited) To find a particular solution to the differential equation ay″ + by′ + cy = Pm 1t2ert,

where Pm 1t2 is a polynomial of degree m, use the form (13)

yp 1t2 = t s 1Amt m + g + A1t + A0 2ert ;

if r is not a root of the associated auxiliary equation, take s = 0; if r is a simple root of the associated auxiliary equation, take s = 1; and if r is a double root of the associated auxiliary equation, take s = 2.

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Section 4.5

The Superposition Principle and Undetermined Coefficients Revisited

185

To find a particular solution to the differential equation

ay″ + by′ + cy = Pm 1t2eat cos bt + Qn 1t2eat sin bt ,

b≠0,

where Pm 1t2 is a polynomial of degree m and Qn 1t2 is a polynomial of degree n, use the form yp 1t2 = t s 1Akt k + g + A1t + A0 2eat cos bt + t s 1Bkt k + g + B1t + B0 2eat sin bt ,

(14)

where k is the larger of m and n. If a + ib is not a root of the associated auxiliary equation, take s = 0; if a + ib is a root of the associated auxiliary equation, take s = 1.

Example 5

Write down the form of a particular solution to the equation y″ + 2y′ + 2y = 5e-t sin t + 5t 3e-t cos t .

Solution

The roots of the associated homogeneous equation y″ + 2y′ + 2y = 0 were identified in Example 3 as -1 { i. Application of (14) dictates the form yp 1t2 = t1A3t 3 + A2t 2 + A1t + A0 2e-t cos t + t1B3t 3 + B2t 2 + B1t + B0 2e-t sin t . ◆

The method of undetermined coefficients applies to higher-order linear differential equations with constant coefficients. Details will be provided in Chapter 6, but the following example should be clear. Example 6

Write down the form of a particular solution to the equation y‴ + 2y″ + y′ = 5e-t sin t + 3 + 7te-t .

Solution

The auxiliary equation for the associated homogeneous is r 3 + 2r 2 + r = r1r + 12 2 = 0, with a double root r = -1 and a single root r = 0. Term by term, the nonhomogeneities call for the forms A0 e-t cos t + B0 e-t sin t 1for 5e-t sin t2 , t A0 1for 32 , 2 t 1A1t + A0 2e-t 1for 7te-t 2 .

(If -1 were a triple root, we would need t 3 1A1t + A0 2e-t for 7te-t.) Of course, we have to rename the coefficients, so the general form is yp 1t2 = Ae-t cos t + Be-t sin t + tC + t 2 1Dt + E2e-t . ◆

4.5

EXERCISES

1. Given that y1 1t2 = cos t is a solution to

y″ - y′ + y = sin t

and y2 1t2 = e2t >3 is a solution to

y″ - y′ + y = e2t,

use the superposition principle to find solutions to the following differential equations: (a) y″ - y′ + y = 5 sin t . (b) y″ - y′ + y = sin t - 3e2t . (c) y″ - y′ + y = 4 sin t + 18e2t .

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186

Chapter 4

Linear Second-Order Equations

2. Given that y1 1t2 = 11>42 sin 2t is a solution to y″ + 2y′ + 4y = cos 2t and that y2 1t2 = t>4 - 1>8 is a solution to y″ + 2y′ + 4y = t, use the superposition principle to find solutions to the following: (a) y″ + 2y′ + 4y = t + cos 2t . (b) y″ + 2y′ + 4y = 2t - 3 cos 2t . (c) y″ + 2y′ + 4y = 11t - 12 cos 2t . In Problems 3–8, a nonhomogeneous equation and a particular solution are given. Find a general solution for the equation. 3. y″ - y = t , yp 1t2 = - t 4. y″ + y′ = 1 , yp 1t2 = t 5. u″ - u′ - 2u = 1 - 2t , up 1t2 = t - 1 6. y″ + 5y′ + 6y = 6x2 + 10x + 2 + 12ex, yp 1x2 = ex + x2 7. y″ = 2y + 2 tan3 x , yp 1x2 = tan x 8. y″ = 2y′ - y + 2ex , yp 1x2 = x2ex In Problems 9–16 decide whether the method of undetermined coefficients together with superposition can be applied to find a particular solution of the given equation. Do not solve the equation. 9. 3y″ + 2y′ + 8y = t 2 + 4t - t 2et sin t 10. y″ - y′ + y = 1et + t2 2 11. y″ - 6y′ - 4y = 4 sin 3t - e3tt 2 + 1>t 12. y″ + y′ + ty = et + 7 13. y″ - 2y′ + 3y = cosh t + sin3 t 14. 2y″ + 3y′ - 4y = 2t + sin2 t + 3 15. y″ + ety′ + y = 7 + 3t 16. 2y″ - y′ + 6y = t 2e-t sin t - 8t cos 3t + 10t In Problems 17–22, find a general solution to the differential equation. 17. y″ - 2y′ - 3y = 3t 2 - 5 18. y″ - y = - 11t + 1 19. y″1x2 - 3y′1x2 + 2y1x2 = ex sin x 20. y″1u2 + 4y1u2 = sin u - cos u 21. y″1u2 + 2y′1u2 + 2y1u2 = e-u cos u 22. y″1x2 + 6y′1x2 + 10y1x2 = 10x4 + 24x3 + 2x2 - 12x + 18 In Problems 23–30, find the solution to the initial value problem. 23. y′ - y = 1 , y102 = 0 24. y″ = 6t ; y102 = 3 , y′102 = -1 25. z″1x2 + z1x2 = 2e-x ; z102 = 0 , z′102 = 0 26. y″ + 9y = 27 ; y102 = 4 , y′102 = 6 27. y″1x2 - y′1x2 - 2y1x2 = cos x - sin 2x ; y102 = - 7>20 , y′102 = 1>5 28. y″ + y′ - 12y = et + e2t - 1 ; y102 = 1 , y′102 = 3

29. y″1u2 - y1u2 = sin u - e2u ; y102 = 1 , y′102 = - 1 30. y″ + 2y′ + y = t 2 + 1 - et ; y102 = 0 ,

y′102 = 2

In Problems 31–36, determine the form of a particular solution for the differential equation. Do not solve. 31. y″ + y = sin t + t cos t + 10t 32. y″ - y = e2t + te2t + t 2e2t 33. x″ - x′ - 2x = et cos t - t 2 + cos3 t 34. y″ + 5y′ + 6y = sin t - cos 2t 35. y″ - 4y′ + 5y = e5t + t sin 3t - cos 3t 36. y″ - 4y′ + 4y = t 2e2t - e2t In Problems 37– 40, find a particular solution to the given higher-order equation. 37. y‴ - 2y″ - y′ + 2y = 2t 2 + 4t - 9 38. y 142 - 5y″ + 4y = 10 cos t - 20 sin t 39. y‴ + y″ - 2y = tet + 1 40. y 142 - 3y‴ + 3y″ - y′ = 6t - 20 41. Discontinuous Forcing Term. In certain physical models, the nonhomogeneous term, or forcing term, g1t2 in the equation ay″ + by′ + cy = g1t2 may not be continuous but may have a jump discontinuity. If this occurs, we can still obtain a reasonable solution using the following procedure. Consider the initial value problem y″ + 2y′ + 5y = g1t2 ;

y102 = 0 ,

y′102 = 0,

where g1t2 = b

10 if 0 … t … 3p>2 . 0 if t 7 3p>2

(a) Find a solution to the initial value problem for 0 … t … 3p>2. (b) Find a general solution for t 7 3p>2. (c) Now choose the constants in the general solution from part (b) so that the solution from part (a) and the solution from part (b) agree, together with their first derivatives, at t = 3p>2. This gives us a continuously differentiable function that satisfies the differential equation except at t = 3p>2. 42. Forced Vibrations. As discussed in Section 4.1, a vibrating spring with damping that is under external force can be modeled by (15)

my″ + by′ + ky = g1t2 ,

where m 7 0 is the mass of the spring system, b 7 0 is the damping constant, k 7 0 is the spring constant, g1t2 is the force on the system at time t, and y1t2 is the displacement from the equilibrium of the spring system at time t. Assume b2 6 4mk.

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Section 4.6

(a) Determine the form of the equation of motion for the spring system when g1t2 = sin bt by finding a general solution to equation (15). (b) Discuss the long-term behavior of this system. [Hint: Consider what happens to the general solution obtained in part (a) as t S + ∞ .] 43. A mass–spring system is driven by a sinusoidal external force g1t2 = 5 sin t. The mass equals 1, the spring constant equals 3, and the damping coefficient equals 4. If the mass is initially located at y102 = 1>2 and at rest, i.e., y′102 = 0, find its equation of motion. 44. A mass–spring system is driven by the external force g1t2 = 2 sin 3t + 10 cos 3t. The mass equals 1, the spring constant equals 5, and the damping coefficient equals 2. If the mass is initially located at y102 = - 1, with initial velocity y′102 = 5, find its equation of motion. y(t)

m

Speed

k

x 5 2L/2 x 5 L/2 Figure 4.11 Speed bump

45. Speed Bumps. Often bumps like the one depicted in Figure 4.11 are built into roads to discourage speeding. The figure suggests that a crude model of the vertical motion y1t2 of a car encountering the speed bump with the speed V is given by

my″ + ky = b

187

(The absence of a damping term indicates that the car’s shock absorbers are not functioning.) (a) Taking m = k = 1, L = p, and F0 = 1 in appropriate units, solve this initial value problem. Thereby show that the formula for the oscillatory motion after the car has traversed the speed bump is y1t2 = A sin t, where the constant A depends on the speed V.

(b) Plot the amplitude 0 A 0 of the solution y1t2 found in part (a) versus the car’s speed V. From the graph, estimate the speed that produces the most violent shaking of the vehicle. 46. Show that the boundary value problem y″ + l2y = sin t ;

y102 = 0 ,

y1p2 = 1 ,

has a solution if and only if l ≠ { 1, {2, {3, c. 47. Find the solution(s) to y″ + 9y = 27 cos 6t

V

cos(px/L)

y1t2 = 0

Variation of Parameters

for t … -L> 12V2 ,

F0 cos 1pVt>L2 for 0 t 0 6 L> 12V2 r for t Ú L> 12V2. 0

(if it exists) satisfying the boundary conditions (a) y102 = - 1 , y1p>62 = 3 . (b) y102 = -1 , y1p>32 = 5 . (c) y102 = -1 , y1p>32 = -1 . 48. All that is known concerning a mysterious second-order constant-coefficient differential equation y″ + py′ + qy = g1t2 is that t 2 + 1 + et cos t, t 2 + 1 + et sin t, and t 2 + 1 + et cos t + et sin t are solutions. (a) Determine two linearly independent solutions to the corresponding homogeneous equation. (b) Find a suitable choice of p, q, and g1t2 that enables these solutions.

4.6 Variation of Parameters We have seen that the method of undetermined coefficients is a simple procedure for determining a particular solution when the equation has constant coefficients and the nonhomogeneous term is of a special type. Here we present a more general method, called variation of parameters,† for finding a particular solution. Consider the nonhomogeneous linear second-order equation (1)

ay″ + by′ + cy = ƒ1t2

and let {y1 1t2, y2 1t2} be two linearly independent solutions for the corresponding homogeneous equation ay″ + by′ + cy = 0 . †

Historical Footnote: The method of variation of parameters was invented by Joseph Lagrange in 1774.

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188

Chapter 4

Linear Second-Order Equations

Then we know that a general solution to this homogeneous equation is given by (2)

yh 1t2 = c1y1 1t2 + c2y2 1t2 ,

where c1 and c2 are constants. To find a particular solution to the nonhomogeneous equation, the strategy of variation of parameters is to replace the constants in (2) by functions of t. That is, we seek a solution of (1) of the form† (3)

yp 1t2 = V1 1t2y1 1t2 + V2 1t2y2 1t2 .

Because we have introduced two unknown functions, v1 1t2 and v2 1t2, it is reasonable to expect that we can impose two equations (requirements) on these functions. Naturally, one of these equations should come from (1). Let’s therefore plug yp 1t2 given by (3) into (1). To accomplish this, we must first compute yp= 1t2 and yp> 1t2. From (3) we obtain yp= = 1v1= y1 + v2= y2 2 + 1v1y1= + v2y2= 2 .

To simplify the computation and to avoid second-order derivatives for the unknowns v1, v2 in the expression for yp>, we impose the requirement (4)

v1= y1 + v2= y2 = 0 .

Thus, the formula for yp= becomes (5)

yp= = v1y1= + v2y2= ,

and so (6)

yp> = v1= y1= + v1y″1 + v2= y2= + v2y2> .

Now, substituting yp, yp= , and yp>, as given in (3), (5), and (6), into (1), we find (7)

f = ayp> + byp= + cyp

= a1v1= y1= + v1y1> + v2= y2= + v2y2> 2 + b1v1y1= + v2y2= 2 + c1v1y1 + v2y2 2 = a1v1= y1= + v2= y2= 2 + v1 1ay1> + by1= + cy1 2 + v2 1ay2> + by2= + cy2 2

= a1v1= y1= + v2= y2= 2 + 0 + 0

since y1 and y2 are solutions to the homogeneous equation. Thus, (7) reduces to (8)

y1= y1= + y2= y2= =

f . a

To summarize, if we can find v1 and v2 that satisfy both (4) and (8), that is, y1V′1 + y2V′2 = 0 , f y′1V′1 + y′2V′2 = , a then yp given by (3) will be a particular solution to (1). To determine v1 and v2, we first solve the linear system (9) for v1= and v2= . Algebraic manipulation or Cramer’s rule (see Appendix D) immediately gives (9)

v1= 1t2 =

a3

-f 1t2y2 1t2

y1 1t2y2= 1t2

-

y1= 1t2y2 1t2

4

and

v2= 1t2 =

a3

f 1t2y1 1t2

y1 1t2y2= 1t2

- y1= 1t2y2 1t2 4

,

†

In Exercises 2.3, Problem 36, we developed this approach for first-order linear equations. Because of the similarity of equations (2) and (3), this technique is sometimes known as “variation of constants.”

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Section 4.6

Variation of Parameters

189

where the bracketed expression in the denominator (the Wronskian) is never zero because of Lemma 1, Section 4.2. Upon integrating these equations, we finally obtain (10)

v1 1t2 =

L a3

-f 1t2y2 1t2

y1 1t2y2= 1t2

-

y1= 1t2y2 1t2

4

dt

and

v2 1t2 =

L a3

f 1t2y1 1t2

y1 1t2y2= 1t2

- y1= 1t2y2 1t2 4

dt .

Let’s review this procedure.

Method of Variation of Parameters To determine a particular solution to ay″ + by′ + cy = f : (a) Find two linearly independent solutions {y1 1t2, y2 1t2} to the corresponding homogeneous equation and take yp 1t2 = v1 1t2y1 1t2 + v2 1t2y2 1t2 .

(b) Determine v1 1t2 and v2 1t2 by solving the system in (9) for v1= 1t2 and v2= 1t2 and integrating. (c) Substitute v1 1t2 and v2 1t2 into the expression for yp 1t2 to obtain a particular solution. Of course, in step (b) one could use the formulas in (10), but v1 1t2 and v2 1t2 are so easy to derive that you are advised not to memorize them. Example 1

Find a general solution on 1 -p>2, p>22 to (11)

Solution

d 2y dt 2

+ y = tan t .

Observe that two independent solutions to the homogeneous equation y″ + y = 0 are cos t and sin t. We now set (12)

yp 1t2 = v1 1t2 cos t + v2 1t2 sin t

and, referring to (9), solve the system

1 cos t2v1= 1t2 + 1 sin t2v2= 1t2 = 0 , 1 - sin t2v1= 1t2 + 1 cos t2v2= 1t2 = tan t ,

for v1= 1t2 and v2= 1t2. This gives

v1= 1t2 = - tan t sin t , v2= 1t2 = tan t cos t = sin t .

Integrating, we obtain (13)

v1 1t2 = = -

L

tan t sin t dt = -

sin2 t dt L cos t

1 - cos2 t dt = 1 cos t - sec t2 dt L cos t L

= sin t - ln 0 sec t + tan t 0 + C1 , (14)

v2 1t2 =

L

sin t dt = - cos t + C2 .

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190

Chapter 4

Linear Second-Order Equations

We need only one particular solution, so we take both C1 and C2 to be zero for simplicity. Then, substituting v1 1t2 and v2 1t2 in (12), we obtain yp 1t2 = 1sin t - ln 0 sec t + tan t 0 2 cos t - cos t sin t ,

which simplifies to

yp 1t2 = - 1cos t2 ln 0 sec t + tan t 0 .

We may drop the absolute value symbols because sec t + tan t = 11 + sin t2 > cos t 7 0 for -p>2 6 t 6 p>2. Recall that a general solution to a nonhomogeneous equation is given by the sum of a general solution to the homogeneous equation and a particular solution. Consequently, a general solution to equation (11) on the interval 1 -p>2, p>22 is (15)

y1t2 = c1 cos t + c2 sin t - 1 cos t2 ln1sec t + tan t2 . ◆

Note that in the above example the constants C1 and C2 appearing in (13) and (14) were chosen to be zero. If we had retained these arbitrary constants, the ultimate effect would be just to add C1 cos t + C2 sin t to (15), which is clearly redundant. Example 2

Find a particular solution on 1 -p>2, p>22 to (16)

Solution

d 2y dt 2

+ y = tan t + 3t - 1 .

With f 1t2 = tan t + 3t - 1, the variation of parameters procedure will lead to a solution. But it is simpler in this case to consider separately the equations (17)

(18)

d 2y dt 2 d 2y dt 2

+ y = tan t , + y = 3t - 1

and then use the superposition principle (Theorem 3, page 181). In Example 1 we found that yq 1t2 = - 1cos t2 ln1sec t + tan t2

is a particular solution for equation (17). For equation (18) the method of undetermined coefficients can be applied. On seeking a solution to (18) of the form yr 1t2 = At + B, we quickly obtain yr 1t2 = 3t - 1 .

Finally, we apply the superposition principle to get

yp 1t2 = yq 1t2 + yr 1t2 = - 1cos t2 ln1sec t + tan t2 + 3t - 1

as a particular solution for equation (16). ◆ Note that we could not have solved Example 1 by the method of undetermined coefficients; the nonhomogeneity tan t is unsuitable. Another important advantage of the method of variation of parameters is its applicability to linear equations whose coefficients a, b, c are functions of t. Indeed, on reviewing the derivation of the system (9) and the formulas (10), one

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Section 4.6

Variation of Parameters

191

can check that we did not make any use of the constant coefficient property; i.e., the method works provided we know a pair of linearly independent solutions to the corresponding homogeneous equation. We illustrate the method in the next example. Example 3

Find a particular solution of the variable coefficient linear equation (19)

t 2y″ - 4ty′ + 6y = 4t 3 , t 7 0 ,

given that y1 1t2 = t 2 and y2 1t2 = t 3 are solutions to the corresponding homogeneous equation.

Solution

The functions t2 and t3 are linearly independent solutions to the corresponding homogeneous equation on 10, ∞ 2 (verify this !) and so (19) has a particular solution of the form yp 1t2 = v1 1t2t 2 + v2 1t2t 3 .

To determine the unknown functions v1 and v2, we solve the system (9) with f 1t2 = 4t 3 and a = a1t2 = t 2 : t 2v1= 1t2 + t 3v2= 1t2 = 0

2tv1= 1t2 + 3t 2v2= 1t2 = f>a = 4t .

The solutions are readily found to be v1= 1t2 = -4 and v2= 1t2 = 4>t, which gives v1 1t2 = -4t and v2 1t2 = 4 ln t. Consequently, yp 1t2 = 1 -4t2t 2 + 14 ln t2t 3 = 4t 3 1 -1 + ln t2

is a solution to (19). ◆ Variable coefficient linear equations will be discussed in more detail in the next section.

4.6

EXERCISES

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters. 1. y″ + 4y = tan 2t 2. y″ + y = sec t 3. y″ - 2y′ + y = t -1et 4. y″ + 2y′ + y = e-t 5. y″1u2 + 16y1u2 = sec 4u 6. y″ + 9y = sec2 13t2 7. y″ + 4y′ + 4y = e-2t ln t 8. y″ + 4y = csc2 12t2 In Problems 9 and 10, find a particular solution first by undetermined coefficients, and then by variation of parameters. Which method was quicker? 9. y″ - y = 2t + 4 10. 2x″1t2 - 2x′1t2 - 4x1t2 = 2e2t In Problems 11–18, find a general solution to the differential equation. 11. y″ + y = tan t + e3t - 1 12. y″ + y = tan2 t 13. v″ + 4v = sec4 12t2

14. y″1u2 + y1u2 = sec3 u 15. y″ + y = 3 sec t - t 2 + 1 16. y″ + 5y′ + 6y = 18t 2 1 1 17. y″ + 2y = tan 2t - et 2 2 18. y″ - 6y′ + 9y = t -3e3t 19. Express the solution to the initial value problem 1 y″ - y = , y112 = 0 , y′112 = - 2 , t using definite integrals. Using numerical integration (Appendix C) to approximate the integrals, find an approximation for y122 to two decimal places. 20. Use the method of variation of parameters to show that t

f1s2 sin 1t - s2 ds L0 is a general solution to the differential equation y1t2 = c1 cos t + c2 sin t +

y″ + y = f1t2 ,

where f1t2 is a continuous function on 1 - ∞ , ∞ 2. [Hint: Use the trigonometric identity sin 1t - s2 = sin t cos s - sin s cos t .]

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192

Chapter 4

Linear Second-Order Equations

3

21. Suppose y satisfies the equation y″ + 10y′ + 25y = et subject to y102 = 1 and y′102 = - 5. Estimate y10.22 to within {0.0001 by numerically approximating the integrals in the variation of parameters formula.

In Problems 22 through 25, use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linearly independent solutions to the corresponding homogeneous equation for t 7 0. 22. t 2y″ - 4ty′ + 6y = t 3 + 1 ; y1 = t 2 , y2 = t 3

23. ty″ - 1t + 12y′ + y = t 2 ; y1 = et , y2 = t + 1 24. ty″ + 11 - 2t2y′ + 1t - 12y = tet ; y1 = et , y2 = et ln t 25. ty″ + 15t - 12y′ - 5y = t 2e-5t ; y1 = 5t - 1 , y2 = e-5t

4.7 Variable-Coefficient Equations The techniques of Sections 4.2 and 4.3 have explicitly demonstrated that solutions to a linear homogeneous constant-coefficient differential equation, (1)

ay″ + by′ + cy = 0 ,

(2)

ay″ + by′ + cy = f 1t2 ,

are defined and satisfy the equation over the whole interval 1 - ∞, + ∞ 2. After all, such solutions are combinations of exponentials, sinusoids, and polynomials. The variation of parameters formula of Section 4.6 extended this to nonhomogeneous constant-coefficient problems,

yielding solutions valid over all intervals where f1t2 is continuous (ensuring that the integrals in (10) of Section 4.6 containing f1t2 exist and are differentiable). We could hardly hope for more; indeed, it is debatable what meaning the differential equation (2) would have at a point where f1t2 is undefined, or discontinuous. Therefore, when we move to the realm of equations with variable coefficients of the form (3)

a2 1t2y″ + a1 1t2y′ + a0 1t2y = f1t2 ,

the most we can expect is that there are solutions that are valid over intervals where all four “governing” functions—a2 1t2, a1 1t2, a0 1t2, and f1t2—are continuous. Fortunately, this expectation is fulfilled except for an important technical requirement—namely, that the coefficient function a2 1t2 must be nonzero over the interval.† Typically, one divides by the nonzero coefficient a2 1t2 and expresses the theorem for the equation in standard form [see (4), below] as follows.

Existence and Uniqueness of Solutions Theorem 5. If p1t2, q1t2, and g(t) are continuous on an interval 1a, b2 that contains the point t0, then for any choice of the initial values Y0 and Y1, there exists a unique solution y1t2 on the same interval (a, b) to the initial value problem (4)

y″1t2 + p1t2y′1t2 + q1t2y1t2 = g1t2 ; y1t0 2 = Y0 , y′1t0 2 = Y1 .

† Indeed, the whole nature of the equation—reduction from second-order to first-order—changes at points where a2 1t2 is zero.

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Section 4.7

Example 1

193

Determine the largest interval for which Theorem 5 ensures the existence and uniqueness of a solution to the initial value problem 1t - 32

(5) Solution

Variable-Coefficient Equations

d 2y dt 2

+

dy + 2t y = ln t ; y112 = 3 , y′112 = -5 . dt

The data p1t2, q1t2, and g1t2 in the standard form of the equation, y″ + py′ + qy =

d 2y dt

2

+

dy 1 2t ln t + y = = g, 1t - 32 dt 1t - 32 1t - 32

are simultaneously continuous in the intervals 0 6 t 6 3 and 3 6 t 6 ∞ . The former contains the point t0 = 1, where the initial conditions are specified, so Theorem 5 guarantees (5) has a unique solution in 0 6 t 6 3. ◆ Theorem 5, embracing existence and uniqueness for the variable-coefficient case, is difficult to prove because we can’t construct explicit solutions in the general case. So the proof is deferred to Chapter 13.† However, it is instructive to examine a special case that we can solve explicitly.

Cauchy–Euler, or Equidimensional, Equations Definition 2. A linear second-order equation that can be expressed in the form (6)

at 2y″1t2 + bty′1t2 + cy = f1t2 ,

where a, b, and c are constants, is called a Cauchy–Euler, or equidimensional, equation.

For example, the differential equation 3t 2y″ + 11ty′ - 3y = sin t is a Cauchy–Euler equation, whereas 2y″ - 3ty′ + 11y = 3t - 1 is not because the coefficient of y″ is 2, which is not a constant times t 2. The nomenclature equidimensional comes about because if y has the dimensions of, say, meters and t has dimensions of time, then each term t 2y″, ty′, and y has the same dimensions (meters). The coefficient of y″1t2 in (6) is at 2, and it is zero at t = 0; equivalently, the standard form y″ +

f1t2 b c y′ + 2 y = at at at 2

has discontinuous coefficients at t = 0. Therefore, we can expect the solutions to be valid only for t 7 0 or t 6 0. Discontinuities in ƒ, of course, will impose further restrictions.

†

All references to Chapters 11–13 refer to the expanded text, Fundamentals of Differential Equations and Boundary Value Problems, 7th ed.

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194

Chapter 4

Linear Second-Order Equations

To solve a homogeneous Cauchy–Euler equation, for t 7 0, we exploit the equidimensional feature by looking for solutions of the form y = t r , because then t 2y″, ty′, and y each have the form (constant) * tr: y = t r , ty′ = trt r - 1 = rt r , t 2y″ = t 2r1r - 12t r - 2 = r1r - 12t r , and substitution into the homogeneous form of (6) (that is, with g = 0) yields a simple quadratic equation for r: ar1r - 12t r + brt r + ct r = [ar 2 + 1b - a2r + c]t r = 0 , or

(7)

ar 2 + 1b - a2r + c = 0 ,

which we call the associated characteristic equation. Example 2

Find two linearly independent solutions to the equation 3t 2y″ + 11ty′ - 3y = 0 , t 7 0 .

Solution

Inserting y = t r yields, according to (7),

3r 2 + 111 - 32r - 3 = 3r 2 + 8r - 3 = 0 ,

whose roots r = 1>3 and r = -3 produce the independent solutions y1 1t2 = t 1>3 , y2 1t2 = t - 3

1for t 7 02 . ◆

Clearly, the substitution y = t r into a homogeneous equidimensional equation has the same simplifying effect as the insertion of y = ert into the homogeneous constant-coefficient equation in Section 4.2. That means we will have to deal with the same encumbrances: 1. What to do when the roots of (7) are complex 2. What to do when the roots of (7) are equal If r is complex, r = a + ib, we can interpret t a + ib by using the identity t = eln t and invoking Euler’s formula [equation (5), Section 4.3]: t a + ib = t at ib = t aeib ln t = t a 3cos1b ln t2 + i sin1b ln t24 .

Then we simplify as in Section 4.3 by taking the real and imaginary parts to form independent solutions: (8)

y1 = t a cos1b ln t2 , y2 = t a sin1b ln t2 .

If r is a double root of the characteristic equation (7), then independent solutions of the Cauchy–Euler equation on 10, ∞ 2 are given by (9)

y1 = t r , y2 = t r ln t .

This can be verified by direct substitution into the differential equation. Alternatively, the second, linearly independent, solution can be obtained by reduction of order, a procedure to be discussed shortly in Theorem 8. Furthermore, Problem 23 demonstrates that the substitution t = ex changes the homogeneous Cauchy–Euler equation into a homogeneous constant-coefficient equation, and the formats (8) and (9) then follow from our earlier deliberations. We remark that if a homogeneous Cauchy–Euler equation is to be solved for t 6 0, then one simply introduces the change of variable t = -t, where t 7 0. The reader should verify via the chain rule that the identical characteristic equation (7) arises when tr = 1 -t2 r is substituted in the equation. Thus the solutions take the same form as (8), (9), but with t replaced

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Section 4.7

Variable-Coefficient Equations

195

by -t; for example, if r is a double root of (7), we get 1 -t2 r and 1 -t2 r ln 1 -t2 as two linearly independent solutions on 1 - ∞, 02. Example 3

Find a pair of linearly independent solutions to the following Cauchy–Euler equations for t 7 0. (a) t 2y″ + 5ty′ + 5y = 0

Solution

(b) t 2y″ + ty′ = 0

For part (a), the characteristic equation becomes r 2 + 4r + 5 = 0 , with the roots r = -2 { i , and (8) produces the real solutions t -2 cos1ln t2 and t -2 sin1ln t2. For part (b), the characteristic equation becomes simply r 2 = 0 with the double root r = 0, and (9) yields the solutions t 0 = 1 and ln t. ◆ In Chapter 8 we will see how one can obtain power series expansions for solutions to variable-coefficient equations when the coefficients are analytic functions. But, as we said, there is no procedure for explicitly solving the general case. Nonetheless, thanks to the existence/ uniqueness result of Theorem 5, most of the other theorems and concepts of the preceding sections are easily extended to the variable-coefficient case, with the proviso that they apply only over intervals in which the governing functions p1t2, q1t2, g1t2 are continuous. Thus we have the following analog of Lemma 1, page 160.

A Condition for Linear Dependence of Solutions Lemma 3. equation (10)

If y1 1t2 and y2 1t2 are any two solutions to the homogeneous differential

y″1t2 + p1t2y′1t2 + q1t2y1t2 = 0

on an interval I where the functions p1t2 and q1t2 are continuous and if the Wronskian† W3y1, y2 4 1t2 J y1 1t2y2= 1t2 - y1= 1t2y2 1t2 = `

y1 1t2 y1= 1t2

y2 1t2 ` y2= 1t2

is zero at any point t of I, then y1 and y2 are linearly dependent on I.

As in the constant-coefficient case, the Wronskian of two solutions is either identically zero or never zero on I, with the latter implying linear independence on I. Precisely as in the proof for the constant-coefficient case, it can be verified that any linear combination c1y1 + c2y2 of solutions y1 and y2 to (10) is also a solution. In fact, these are the only solutions to (10) as stated in the following result.

Representation of Solutions to Initial Value Problems Theorem 6. If y1 1t2 and y2 1t2 are any two solutions to the homogeneous differential equation (10) that are linearly independent on an interval I, then every solution to (10) on I is expressible as a linear combination of y1 and y2. Moreover, the initial value problem consisting of equation (10) and the initial conditions y1t0 2 = Y0, y′1t0 2 = Y1 has a unique solution on I for any point t0 in I and any constants Y0, Y1.

†

The determinant representation of the Wronskian was introduced in Problem 34, Section 4.2.

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196

Chapter 4

Linear Second-Order Equations

As in the constant-coefficient case, the linear combination yh = c1y1 + c2y2 is called a general solution to (10) on I if y1, y2 are linearly independent solutions on I. For the nonhomogeneous equation (11)

y″1t2 + p1t2y′1t2 + q1t2y1t2 = g1t2 ,

a general solution on I is given by y = yp + yh, where yh = c1y1 + c2y2 is a general solution to the corresponding homogeneous equation (10) on I and yp is a particular solution to (11) on I. In other words, the solution to the initial value problem stated in Theorem 5 must be of this form for a suitable choice of the constants c1, c2. This follows, just as before, from a straightforward extension of the superposition principle for variable-coefficient equations described in Problem 30. As illustrated at the end of the Section 4.6, if linearly independent solutions to the homogeneous equation (10) are known, then yp can be determined for (11) by the variation of parameters method.

Variation of Parameters Theorem 7. If y1 and y2 are two linearly independent solutions to the homogeneous equation (10) on an interval I where p1t2, q1t2, and g1t2 are continuous, then a particular solution to (11) is given by yp = v1y1 + v2y2, where v1 and v2 are determined up to a constant by the pair of equations y1V′1 + y2V′2 = 0 , y1′V1′ + y2′V2′ = g , which have the solution (12)

v1 1t2 =

-g1t2 y2 1t2 g1t2y1 1t2 dt , v2 1t2 = dt . L W3y1, y2 4 1t2 L W3y1, y2 4 1t2

Note the formulation (12) presumes that the differential equation has been put into standard form [that is, divided by a2 1t2].

The proofs of the constant-coefficient versions of these theorems in Sections 4.2 and 4.5 did not make use of the constant-coefficient property, so one can prove them in the general case by literally copying those proofs but interpreting the coefficients as variables. Unfortunately, however, there is no construction analogous to the method of undetermined coefficients for the variable-coefficient case. What does all this mean? The only stumbling block for our completely solving nonhomogeneous initial value problems for equations with variable coefficients, y″ + p1t2y′ + q1t2y = g1t2 ; y1t0 2 = Y0 , y′1t0 2 = Y1 ,

is the lack of an explicit procedure for constructing independent solutions to the associated homogeneous equation (10). If we had y1 and y2 as described in the variation of parameters formula, we could implement (12) to find yp, formulate the general solution of (11) as yp + c1y1 + c2y2, and (with the assurance that the Wronskian is nonzero) fit the constants to the initial conditions. But with the exception of the Cauchy–Euler equation and the ponderous power series machinery of Chapter 8, we are stymied at the outset; there is no general procedure for finding y1 and y2.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 4.7

Variable-Coefficient Equations

197

Ironically, we only need one nontrivial solution to the associated homogeneous equation, thanks to a procedure known as reduction of order that constructs a second, linearly independent solution y2 from a known one y1. So one might well feel that the following theorem rubs salt into the wound.

Reduction of Order Theorem 8. If y1 1t2 is a solution, not identically zero, to the homogeneous differential equation (10) in an interval I (see page 195), then (13)

y2 1t2 = y1 1t2

e- 1p1t2dt dt 2 L y1 1t2

is a second, linearly independent solution.

This remarkable formula can be confirmed directly, but the following derivation shows how the procedure got its name. Proof of Theorem 8. Our strategy is similar to that used in the derivation of the variation of parameters formula, Section 4.6. Bearing in mind that cy1 is a solution of (10) for any constant c, we replace c by a function v1t2 and propose the trial solution y2 1t2 = v1t2y1 1t2, spawning the formulas y2= = vy1= + v=y1 , y2> = vy1> + 2v=y1= + v>y1 . Substituting these expressions into the differential equation (10) yields 1vy1> + 2v′y1= + v″y1 2 + p1vy1= + v′y1 2 + qvy1 = 0 ,

or, on regrouping, (14)

1 y1> + py1= + qy1 2v + y1v″ + 12y1= + py1 2v′ = 0 .

The group in front of the undifferentiated v1t2 is simply a copy of the left-hand member of the original differential equation (10), so it is zero.† Thus (14) reduces to (15)

y1v″ + 12y1= + py1 2v′ = 0 ,

which is actually a first-order equation in the variable w K v′: (16)

y1w′ + 12y1= + py1 2w = 0 .

Indeed, (16) is separable and can be solved immediately using the procedure of Section 2.2. Problem 48 on page 201 requests the reader to carry out the details of this procedure to complete the derivation of (13). ◆ Example 4

Given that y1 1t2 = t is a solution to (17)

y″ -

1 1 y′ + 2 y = 0 , t t

use the reduction of order procedure to determine a second linearly independent solution for t 7 0. This is hardly a surprise; if v were constant, vy would be a solution with v′ = v″ = 0 in (14).

†

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198

Chapter 4

Solution

Linear Second-Order Equations

Rather than implementing the formula (13), let’s apply the strategy used to derive it. We set y2 1t2 = v1t2y1 1t2 = v1t2t and substitute y2= = v′t + v, y2> = v″t + 2v′ into (17) to find (18)

1 1 v″t + 2v′ - 1v′t + v2 + 2 vt = v″t + 12v′ - v′2 = v″t + v′ = 0 . t t

As promised, (18) is a separable first-order equation in v′, simplifying to 1v′2′> 1v′2 = -1>t with a solution v′ = 1>t, or v = ln t (taking integration constants to be zero). Therefore, a second solution to (17) is y2 = vt = t ln t. Of course (17) is a Cauchy–Euler equation for which (7) has equal roots: ar 2 + 1b - a2r + c = r 2 - 2r + 1 = 1r - 12 2 = 0 , and y2 is precisely the form for the independent solution predicted by (9). ◆ Example 5

The following equation arises in the mathematical modeling of reverse osmosis.† (19)

1sin t2y″ - 21cos t2y′ - 1sin t2y = 0 , 0 6 t 6 p .

Find a general solution. Solution

As we indicated above, the tricky part is to find a single nontrivial solution. Inspection of (19) suggests that y = sin t or y = cos t, combined with a little luck with trigonometric identities, might be solutions. In fact, trial and error shows that the cosine function works: y1 = cos t , y1= = -sin t , y1> = -cos t ,

1sin t2y1> - 21cos t2y1= - 1sin t2y1 = 1sin t2 1 - cos t2 - 21cos t2 1 - sin t2 - 1sin t2 1cos t2 = 0 .

Unfortunately, the sine function fails (try it). So we use reduction of order to construct a second, independent solution. Setting y2 1t2 = v1t2y1 1t2 = v1t2cos t and computing y2= = v′cos t - v sin t, y2> = v″cos t - 2v′sin t - v cos t, we substitute into (19) to derive 1sin t23v″cos t - 2v′sin t - v cos t4 - 21cos t23v′cos t - v sin t4 - 1sin t23v cos t4 = v″1sin t21cos t2 - 2v′1sin2 t + cos2 t2 = 0, which is equivalent to the separated first-order equation 1v′2 ′ sec2 t 2 = = 2 . 1v′2 1sin t21cos t2 tan t Taking integration constants to be zero yields ln v′ = 2 ln1 tan t2 or v′ = tan2 t, and v = tan t - t. Therefore, a second solution to (19) is y2 = 1 tan t - t2 cos t = sin t - t cos t. We conclude that a general solution is c1 cos t + c2 1sin t - t cos t2. ◆ In this section we have seen that the theory for variable-coefficient equations differs only slightly from the constant-coefficient case (in that solution domains are restricted to intervals), but explicit solutions can be hard to come by. In the next section, we will supplement our exposition by describing some nonrigorous procedures that sometimes can be used to predict qualitative features of the solutions.

†

Reverse osmosis is a process used to fortify the alcoholic content of wine, among other applications.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 4.7

4.7

22. 1t + 12 2y″1t2 + 101t + 12y′1t2 + 14y1t2 = 0 , t 7 -1 23. To justify the solution formulas (8) and (9), perform the following analysis. (a) Show that if the substitution t = ex is made in the function y1t2 and x is regarded as the new independent variable in Y1x2 J y1ex 2, the chain rule implies the following relationships: dy d 2y dY d 2Y dY = , t2 2 = . dt dx dt dx 2 dx (b) Using part (a), show that if the substitution t = e x is made in the Cauchy–Euler differential equation (6), the result is a constant-coefficient equation for Y1x2 = y1e x 2, namely, t

In Problems 5 through 8, determine whether Theorem 5 applies. If it does, then discuss what conclusions can be drawn. If it does not, explain why. 5. t 2z″ + tz′ + z = cos t ; z102 = 1 , z′102 = 0 6. y″ + yy′ = t 2 - 1 ; y102 = 1 , y′102 = -1 7. y″ + ty′ - t 2y = 0 ; y102 = 0 , y112 = 0 8. 11 - t2y″ + ty′ - 2y = sin t ; y102 = 1 , y′102 = 1

12.

d 2y dt 2

+ 2t

dy - 6y = 0 dt

d 2z dz + 5t + 4z = 0 dt dt 2

d 2w 6 dw 4 + w = 0 + t dt t 2 dt 2

24.

13. 9t 2y″1t2 + 15ty′1t2 + y1t2 = 0 14. t 2y″1t2 - 3ty′1t2 + 4y1t2 = 0 In Problems 15 through 18, find a general solution for t 6 0. 1 5 15. y″1t2 - y′1t2 + 2 y 1t2 = 0 t t 16. t 2y″1t2 - 3ty′1t2 + 6y1t2 = 0 17. t 2y″1t2 + 9ty′1t2 + 17y1t2 = 0 18. t 2y″1t2 + 3ty′1t2 + 5y1t2 = 0 In Problems 19 and 20, solve the given initial value problem for the Cauchy–Euler equation. 19. t 2y″1t2 - 4ty′1t2 + 4y1t2 = 0 ; y112 = - 2 , y′112 = -11

25.

26.

20. t y″1t2 + 7ty′1t2 + 5y1t2 = 0 ; y112 = - 1 , y′112 = 13 2

In Problems 21 and 22, devise a modification of the method for Cauchy–Euler equations to find a general solution to the given equation. 21. 1t - 22 2y″1t2 - 71t - 22y′1t2 + 7y1t2 = 0 , t 7 2

dY d 2Y + 1b - a2 + cY = ƒ1e x 2 . 2 dx dx (c) Observe that the auxiliary equation (recall Section 4.2) for the homogeneous form of (20) is the same as (7) in this section. If the roots of the former are complex, linearly independent solutions of (20) have the form eax cos bx and eax sin bx; if they are equal, linearly independent solutions of (20) have the form erx and xerx. Express x in terms of t to derive the corresponding solution forms (8) and (9). Solve the following Cauchy–Euler equations by using the substitution described in Problem 23 to change them to constant coefficient equations, finding their general solutions by the methods of the preceding sections, and restoring the original independent variable t. (a) t 2y″ + ty′ - 9y = 0 (b) t 2y″ + 3ty′ + 10y = 0 (c) t 2y″ + 3ty′ + y = t + t -1 (d) t 2y″ + ty′ + 9y = -tan13 ln t2 Let y1 and y2 be two functions defined on 1 - ∞ , ∞ 2. (a) True or False: If y1 and y2 are linearly dependent on the interval 3a, b4, then y1 and y2 are linearly dependent on the smaller interval 3c, d4 ⊂ 3a, b4. (b) True or False: If y1 and y2 are linearly dependent on the interval 3a, b4, then y1 and y2 are linearly dependent on the larger interval 3C, D4 ⊃ 3a, b4. Let y1 1t2 = t 3 and y2 1t2 = 0 t 3 0 . Are y1 and y2 linearly independent on the following intervals? (a) 30, ∞ 2 (b) 1 - ∞ , 04 (c) 1 - ∞ , ∞ 2 (d) Compute the Wronskian W3y1, y2 41t2 on the interval 1 - ∞ , ∞ 2. Consider the linear equation (20)

In Problems 9 through 14, find a general solution to the given Cauchy–Euler equation for t 7 0. 9. t 2y″1t2 + 7ty′1t2 - 7y1t2 = 0

11. t 2

199

EXERCISES

In Problems 1 through 4, use Theorem 5 to discuss the existence and uniqueness of a solution to the differential equation that satisfies the initial conditions y112 = Y0 , y′112 = Y1, where Y0 and Y1 are real constants. 1. 11 + t 2 2y″ + ty′ - y = tan t 2. t1t - 32y″ + 2ty′ - y = t 2 3. t 2y″ + y = cos t y′ 4. ety″ + y = ln t t-3

10. t 2

Variable-Coefficient Equations

27.

(21)

a

t 2y″ - 3ty′ + 3y = 0 , for - ∞ 6 t 6 ∞ .

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200

Chapter 4

Linear Second-Order Equations

(a) Verify that y1 1t2 J t and y2 1t2 J t 3 are two solutions to (21) on 1 - ∞ , ∞ 2. Furthermore, show that y1 1t0 2y2= 1t0 2 - y1= 1t0 2y2 1t0 2 ≠ 0 for t0 = 1. (b) Prove that y1 1t2 and y2 1t2 are linearly independent on 1 - ∞ , ∞ 2. (c) Verify that the function y3 1t2 J 0 t 0 3 is also a solution to (21) on 1 - ∞ , ∞ 2. (d) Prove that there is no choice of constants c1, c2 such that y3 1t2 = c1y1 1t2 + c2y2 1t2 for all t in 1 - ∞ , ∞ 2. [Hint: Argue that the contrary assumption leads to a contradiction.] (e) From parts (c) and (d), we see that there is at least one solution to (21) on 1 - ∞ , ∞ 2 that is not expressible as a linear combination of the solutions y1 1t2, y2 1t2. Does this provide a counterexample to the theory in this section? Explain. 28. Let y1 1t2 = t 2 and y2 1t2 = 2t 0 t 0 . Are y1 and y2 linearly independent on the interval: (a) 30, ∞ 2? (b) 1 - ∞ , 04? (c) 1 - ∞ , ∞ 2? (d) Compute the Wronskian W3y1, y2 41t2 on the interval 1 - ∞ , ∞ 2. 29. Prove that if y1 and y2 are linearly independent solutions of y″ + py′ + qy = 0 on 1a, b2, then they cannot both be zero at the same point t0 in 1a, b2. 30. Superposition Principle. Let y1 be a solution to y″1t2 + p1t2y′1t2 + q1t2y1t2 = g1 1t2

on the interval I and let y2 be a solution to

y″1t2 + p1t2y′1t2 + q1t2y1t2 = g2 1t2

on the same interval. Show that for any constants k1 and k2, the function k1y1 + k2y2 is a solution on I to

y″1t2 + p1t2y′1t2 + q1t2y1t2 = k1g1 1t2 + k2g2 1t2 .

31. Determine whether the following functions can be Wronskians on -1 6 t 6 1 for a pair of solutions to some equation y″ + py′ + qy = 0 (with p and q continuous). (a) w1t2 = 6e4t (b) w1t2 = t 3 -1 (c) w1t2 = 1t + 12 (d) w1t2 K 0 32. By completing the following steps, prove that the Wronskian of any two solutions y1, y2 to the equation y″ + py′ + qy = 0 on 1a, b2 is given by Abel’s formula† W3y1, y2 41t2 = C exp 5− t0 and t in 1a, b2 ,

t

Lt 0

p1T2 dT 6 ,

where the constant C depends on y1 and y2. (a) Show that the Wronskian W satisfies the equation W′ + pW = 0. (b) Solve the separable equation in part (a). (c) How does Abel’s formula clarify the fact that the Wronskian is either identically zero or never zero on 1a, b2?

33. Use Abel’s formula (Problem 32) to determine (up to a constant multiple) the Wronskian of two solutions on 10, ∞ 2 to ty″ + 1t - 12y′ + 3y = 0 .

34. All that is known concerning a mysterious differential equation y″ + p1t2y′ + q1t2y = g1t2 is that the functions t, t 2, and t 3 are solutions. (a) Determine two linearly independent solutions to the corresponding homogeneous differential equation. (b) Find the solution to the original equation satisfying the initial conditions y122 = 2, y′122 = 5. (c) What is p1t2? [Hint: Use Abel’s formula for the Wronskian, Problem 32.] 35. Given that 1 + t, 1 + 2t, and 1 + 3t 2 are solutions to the differential equation y″ + p1t2y′ + q1t2y = g1t2, find the solution to this equation that satisfies y112 = 2, y′112 = 0. 36. Verify that the given functions y1 and y2 are linearly independent solutions of the following differential equation and find the solution that satisfies the given initial conditions. ty″ - 1t + 22y′ + 2y = 0 ; y1 1t2 = et , y2 1t2 = t 2 + 2t + 2 ; y′112 = 1 y112 = 0 , In Problems 37 through 39, find general solutions to the nonhomogeneous Cauchy–Euler equations using variation of parameters. 37. t 2z″ + tz′ + 9z = - tan13 ln t2 38. t 2y″ + 3ty′ + y = t -1 3 ≤ 39. t 2z″ - tz′ + z = t ¢1 + ln t 40. The Bessel equation of order one-half 1 t 2y″ + ty′ + at 2 − by = 0 , t + 0 4 has two linearly independent solutions, y1 1t2 = t -1>2 cos t, y2 1t2 = t -1>2 sin t . Find a general solution to the nonhomogeneous equation 1 t 2y″ + ty′ + ¢t 2 - ≤ y = t 5>2 , t 7 0 . 4 In Problems 41 through 44, a differential equation and a nontrivial solution f are given. Find a second linearly independent solution using reduction of order. 41. t 2y″ - 2ty′ - 4y = 0 , t 7 0 ; f1t2 = t -1 42. t 2y″ + 6ty′ + 6y = 0 , t 7 0 ; f1t2 = t -2 43. tx″ - 1t + 12x′ + x = 0 , t 7 0 ; f 1t2 = et 44. ty″ + 11 - 2t2y′ + 1t - 12y = 0 , t 7 0 ; f1t2 = et

†

Historical Footnote: Niels Abel derived this identity in 1827.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 4.8

Qualitative Considerations for Variable-Coefficient and Nonlinear Equations

45. Find a particular solution to the nonhomogeneous equation ty″ - 1t + 12y′ + y = t e , 2 2t

given that f1t2 = e is a solution to the corresponding homogeneous equation. 46. Find a particular solution to the nonhomogeneous equation t

11 - t2y″ + ty′ - y = 11 - t2 2 , given that f1t2 = t is a solution to the corresponding homogeneous equation. 47. In quantum mechanics, the study of the Schrödinger equation for the case of a harmonic oscillator leads to a consideration of Hermite’s equation, y″ − 2ty′ + Ly = 0 , where l is a parameter. Use the reduction of order formula to obtain an integral representation of a second linearly independent solution to Hermite’s equation for the given value of l and corresponding solution f 1t2. (a) l = 4 , f 1t2 = 1 - 2t 2 (b) l = 6 , f 1t2 = 3t - 2t 3 48. Complete the proof of Theorem 8 by solving equation (16). 49. The reduction of order procedure can be used more generally to reduce a homogeneous linear nth-order equation to a homogeneous linear 1n - 12th-order equation. For the equation ty″′ - ty″ + y′ - y = 0,

which has f 1t2 = et as a solution, use the substitution y 1t2 = v 1t2 f 1t2 to reduce this third-order equation to a homogeneous linear second-order equation in the variable w = v′.

201

50. The equation ty″′ + 11 - t2y″ + ty′ - y = 0

has f 1t2 = t as a solution. Use the substitution y 1t2 = v 1t2 f 1t2 to reduce this third-order equation to a homogeneous linear second-order equation in the variable w = v′. 51. Isolated Zeros. Let f1t2 be a solution to y″ + py′ + qy = 0 on (a, b), where p, q are continuous on (a, b). By completing the following steps, prove that if f is not identically zero, then its zeros in (a, b) are isolated, i.e., if f1t 0 2 = 0, then there exists a d 7 0 such that f1t2 ≠ 0 for 0 6 0 t - t0 0 6 d. (a) Suppose f1t 0 2 = 0 and assume to the contrary that for each n = 1, 2, c, the function f has a zero at tn, where 0 6 0 t0 - tn 0 6 1>n. Show that this implies f′1t0 2 = 0. [Hint: Consider the difference quotient for f at t0.] (b) With the assumptions of part (a), we have f1t0 2 = f′1t0 2 = 0. Conclude from this that f must be identically zero, which is a contradiction. Hence, there is some integer n0 such that f1t2 is not zero for 0 6 0 t - t0 0 6 1>n0. 52. The reduction of order formula (13) can also be derived from Abels’ identity (Problem 32). Let f 1t2 be a nontrivial solution to (10) and y1t2 a second linearly independent solution. Show that W3f, y4 y ′ ¢ ≤ = f f2 and then use Abel’s identity for the Wronskian W3 f, y4 to obtain the reduction of order formula.

4.8 Qualitative Considerations for Variable-Coefficient and Nonlinear Equations

There are no techniques for obtaining explicit, closed-form solutions to second-order linear differential equations with variable coefficients (with certain exceptions) or for nonlinear equations. In general, we will have to settle for numerical solutions or power series expansions. So it would be helpful to be able to derive, with simple calculations, some nonrigorous, qualitative conclusions about the behavior of the solutions before we launch the heavy computational machinery. In this section we first display a few examples that illustrate the profound differences that can occur when the equations have variable coefficients or are nonlinear. Then we show how the mass–spring analogy, discussed in Section 4.1, can be exploited to predict some of the attributes of solutions of these more complicated equations. To begin our discussion we display a linear constant-coefficient, a linear variable-coefficient, and two nonlinear equations.

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202

Chapter 4

Linear Second-Order Equations

(a) The equation 3y″ + 2y′ + 4y = 0

(1)

is linear, homogeneous with constant coefficients. We know everything about such equations; the solutions are, at worst, polynomials times exponentials times sinusoids in t, and unique solutions can be found to match any prescribed data y1a2, y′1a2 at any instant t = a. It has the superposition property: If y1 1t2 and y2 1t2 are solutions, so is y1t2 = c1y1 1t2 + c2y2 1t2. (b) The equation

11 − t 2 2y″ − 2ty′ + 2y = 0

(2)

also has the superposition property (Problem 30, Exercises 4.7). It is a linear variable-coefficient equation and is a special case of Legendre’s equation 11 - t 2 2y″ - 2ty′ + ly = 0, which arises in the analysis of wave and diffusion phenomena in spherical coordinates. (c) The equations (3)

y″ - 6y2 = 0 ,

(4)

y″ - 24y1>3 = 0

do not share the superposition property because of the square and the cube root of y terms (e.g., the quadratic term y2 does not reduce to y21 + y22). They are nonlinear†equations.

The Legendre equation (2) has one simple solution, y1 1t2 = t, as can easily be verified by mental calculation. A second, linearly independent, solution for -1 6 t 6 1 can be derived by the reduction of order procedure of Section 4.7. Traditionally, the second solution is taken to be (5)

y2 1t2 =

t 1+t lna b -1. 2 1-t

Notice in particular the behavior near t = {1; none of the solutions of our constant-coefficient equations ever diverged at a finite point! We would have anticipated troublesome behavior for (2) at t = {1 if we had rewritten it in standard form as (6)

y″ -

2t 2 y′ + y = 0, 2 1-t 1 - t2

since Theorem 5, page 192, only promises existence and uniqueness between these points. As we have noted, there are no general solution procedures for solving nonlinear equations. However, the following lemma is very useful in some situations such as equations (3), (4). It has an extremely significant physical interpretation, which we will explore in Project D of Chapter 5; for now we will merely tantalize the reader by giving it a suggestive name.

†

Although the quadratic y2 renders equation (3) nonlinear, the occurrence of t 2 in (2) does not spoil its linearity (in y).

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 4.8

Qualitative Considerations for Variable-Coefficient and Nonlinear Equations

203

The Energy Integral Lemma Lemma 4. (7)

Let y1t2 be a solution to the differential equation

y″ = f1y2 ,

where f1y2 is a continuous function that does not depend on y′ or the independent variable t. Let F1y2 be an indefinite integral of f1y2, that is, f1y2 =

d F1y2 . dy

Then the quantity (8)

E1t2 J

1 y′1t2 2 - F 1 y1t2 2 2

is constant; i.e., (9)

d E1t2 = 0 . dt

Proof. This is immediate; we insert (8), differentiate, and apply the differential equation (7): d 1 d E1t2 = c y′1t2 2 - F 1 y1t2 2 d dt dt 2 dF 1 = 2y′y″ y′ 2 dy = y′3y″ - f1y24 = 0. ◆ As a result, an equation of the form (7) can be reduced to (10)

1 1y′2 2 - F1y2 = K , 2

for some constant K, which is equivalent to the separable first-order equation dy = { 223F1y2 + K4 dt having the implicit two-parameter solution (Section 2.2) (11)

t = {

dy L 223F1y2 + K4

+c.

We will use formula (11) to illustrate some startling features of nonlinear equations.

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204

Chapter 4

Example 1

Solution

Linear Second-Order Equations

Apply the energy integral lemma to explore the solutions of the nonlinear equation y″ = 6y2, given in (3). Since 6y2 =

d 12y3 2, the solution form (11) becomes dy

t = {

dy L 223y3 + K4

+c.

For simplicity we take the plus sign and focus attention on solutions with K = 0. Then we find t = 1 12 y -3>2 dy = -y -1>2 + c, or (12)

y1t2 = 1c - t2 -2 ,

for any value of the constant c. Clearly, this equation is an enigma; solutions can blow up at t = 1, t = 2, t = p, or anywhere—and there is no clue in equation (3) as to why this should happen! Moreover, we have found an infinite family of pairwise linearly independent solutions (rather than the expected two). Yet we still cannot assemble, out of these, a solution matching (say) y102 = 1, y′102 = 3; all our solutions (12) have y′102 = 2y102 3>2, and the absence of a superposition principle voids the use of linear combinations c1y1 1t2 + c2y2 1t2. ◆ Example 2

Solution

Apply the energy integral lemma to explore the solutions of the nonlinear equation y″ = 24y1>3, given in (4). Since 24y1>3 = t = {

d 118y4>3 2, formula (11) gives dy dy L 22318y4>3 + K4

+c.

Again we take the plus sign and focus attention on solutions with K = 0. Then we find t = y1>3 >2 + c. In particular, y1 1t2 = 8t 3 is a solution, and it satisfies the initial conditions y102 = 0, y′102 = 0. But note that y2 1t2 K 0 is also a solution to (4), and it satisfies the same initial conditions! Hence, the uniqueness feature, guaranteed for linear equations by Theorem 5 on page 192 of Section 4.7, can fail in the nonlinear case. ◆ So these examples have demonstrated violations of the existence and uniqueness properties (as well as “finiteness”) that we have come to expect from the constant-coefficient case. It should not be surprising, then, that the solution techniques for variable-coefficient and nonlinear second-order equations are more complicated—when, indeed, they exist. Recall, however, that in Section 4.3 we saw that our familiarity with the mass–spring oscillator equation (13)

Fext = 3inertia4y″ + 3damping4y′ + 3stiffness4y = my″ + by′ + ky

was helpful in picturing the qualitative features of the solutions of other constant-coefficient equations. (See Figure 4.1, page 152.) By pushing these analogies further, we can also anticipate some of the features of the solutions in the variable-coefficient and nonlinear cases. One of the simplest linear second-order differential equations with variable coefficients is (14)

y″ + ty = 0 .

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Section 4.8

Qualitative Considerations for Variable-Coefficient and Nonlinear Equations

205

y

t

Figure 4.12 Solution to equation (14)

Example 3 Solution

Using the mass–spring analogy, predict the nature of the solutions to equation (14) for t 7 0. Comparing (13) with (14), we see that the latter equation describes a mass–spring oscillator where the spring stiffness “k” varies in time—in fact, it stiffens as time passes [“k” = t in equation (14)]. Physically, then, we would expect to see oscillations whose frequency increases with time, while the amplitude of the oscillations diminishes (because the spring gets harder to stretch). The numerically computed solution in Figure 4.12 displays precisely this behavior. ◆ Remark Schemes for numerically computing solutions to second-order equations will be discussed in Section 5.3. If such schemes are available, then why is there a need for the qualitative analysis discussed in this section? The answer is that numerical methods provide only approximations to solutions of initial value problems, and their accuracy is sometimes difficult to predict (especially for nonlinear equations). For example, numerical methods are often ineffective near points of discontinuity or over the long time intervals needed to study the asymptotic behavior. And this is precisely when qualitative arguments can lend insight into the reasonableness of the computed solution. It is easy to verify (Problem 1) that if y1t2 is a solution of the Airy equation (15)

y″ - ty = 0 ,

then y1 -t2 solves y″ + ty = 0, so “Airy functions” exhibit the behavior shown in Figure 4.12 for negative time. For positive t the Airy equation has a negative stiffness “k” = -t, with magnitude increasing in time. As we observed in Example 5 of Section 4.3, negative stiffness tends to reinforce, rather than oppose, displacements, and the solutions y1t2 grow rapidly with (positive) time. The solution known as the Airy function of the second kind Bi1t2, depicted in Figure 4.13, behaves exactly as expected.

Bi(t)

Ai(t) t

Figure 4.13 Airy functions

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Chapter 4

Linear Second-Order Equations

In Section 4.3 we also pointed out that mass–spring systems with negative spring stiffness can have isolated bounded solutions if the initial displacement y102 and velocity y′102 are precisely selected to counteract the repulsive spring force. The Airy function of the first kind Ai1t2, also depicted in Figure 4.13, is such a solution for the “Airy spring”; the initial inwardly directed velocity is just adequate to overcome the outward push of the stiffening spring, and the mass approaches a delicate equilibrium state y1t2 K 0. Now let’s look at Bessel’s equation. It arises in the analysis of wave or diffusion phenomena in cylindrical coordinates. The Bessel equation of order n is written (16)

y″ +

1 n2 y′ + a 1 − 2 b y = 0 . t t

Clearly, there are irregularities at t = 0, analogous to those at t = {1 for the Legendre equation (2); we will explore these in depth in Chapter 8.

Example 4 Solution

Apply the mass–spring analogy to predict qualitative features of solutions to Bessel’s equation for t 7 0. Comparing (16) with the paradigm (13), we observe that • the inertia “m” = 1 is fixed at unity; • there is positive damping 1“b” = 1>t2, although it weakens with time; and • the stiffness 1“k” = 1 - n2 >t 2 2 is positive when t 7 n and tends to 1 as t S + ∞. Solutions, then, should be expected to oscillate with amplitudes that diminish slowly (due to the damping), and the frequency of the oscillations should settle at a constant value (given, according to the procedures of Section 4.3, by 2k>m = 1 radian per unit time). The graphs of the Bessel functions Jn 1t2 and Yn 1t2 of the first and second kind of order n = 12 exemplify these qualitative predictions; see Figure 4.14. The effect of the singularities in the coefficients at t = 0 is manifested in the graph of Y1>2 1t2. ◆ Although most Bessel functions have to be computed by power series methods, if the order n is a half-integer, then Jn 1t2 and Yn 1t2 have closed-form expressions. In fact, J1>2 1t2 = 22> 1pt2 sin t and Y1>2 1t2 = - 22> 1pt2 cos t. You can verify directly that these functions solve equation (16).

J1/2(t) t

Y1/2(t)

Figure 4.14 Bessel functions

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Section 4.8

207

Qualitative Considerations for Variable-Coefficient and Nonlinear Equations

I2(t)

K2(t)

t Figure 4.15 Modified Bessel functions

Example 5

Give a qualitative analysis of the modified Bessel equation of order n: (17)

y″ +

n2 1 y′ − a 1 − 2 b y = 0 . t t

Solution

This equation also exhibits unit mass and positive, diminishing, damping. However, the stiffness now converges to negative 1. Accordingly, we expect typical solutions to diverge as t S + ∞. The modified Bessel function of the first kind, In 1t2 of order n = 2 in Figure 4.15 follows this prediction, whereas the modified Bessel function of the second kind, Kn 1t2 of order n = 2 in Figure 4.15, exhibits the same sort of balance of initial position and velocity as we saw for the Airy function Ai1t2. Again, the effect of the singularity at t = 0 is evident. ◆

Example 6

Use the mass–spring model to predict qualitative features of the solutions to the nonlinear Duffing equation (18)

Solution

y″ + y + y3 = y″ + 11 + y2 2y = 0 .

Although equation (18) is nonlinear, it can be matched with the paradigm (13) if we envision unit mass, no damping, and a (positive) stiffness “k” = 1 + y2, which increases as the displacement y gets larger. (This increasing-stiffness effect is built into some popular mattresses for therapeutic reasons.)† Such a spring grows stiffer as the mass moves farther away, but it restores to its original value when the mass returns. Thus, high-amplitude excursions should oscillate faster than low-amplitude ones, and the sinusoidal shapes in the graphs of y1t2 should be “pinched in” somewhat at their peaks. These qualitative predictions are demonstrated by the numerically computed solutions plotted in Figure 4.16. ◆ y

t

Figure 4.16 Solution graphs for the Duffing equation †

Graphic depictions of oscillations on a therapeutic mattress are best left to one's imagination, the editors say.

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Chapter 4

Linear Second-Order Equations

y O ,

m

u

t

mg sin u mg

Figure 4.17 Solutions to the van der Pol equation

Figure 4.18 A pendulum

The fascinating van der Pol equation (19)

y″ − 11 - y2 2y′ + y = 0

originated in the study of the electrical oscillations observed in vacuum tubes. Example 7 Solution

Predict the behavior of the solutions to equation (19) using the mass–spring model. By comparison to the paradigm (13), we observe unit mass and stiffness, positive damping 3“b” = - 11 - y2 24 when 0 y1t2 0 7 1, and negative damping when 0 y1t2 0 6 1. Friction thus dampens large-amplitude motions but energizes small oscillations. The result, then, is that all (nonzero) solutions tend to a limit cycle whose friction penalty incurred while 0 y1t2 0 7 1 is balanced by the negative-friction boost received while 0 y1t2 0 6 1. The computer-generated Figure 4.17 illustrates the convergence to the limit cycle for some solutions to the van der Pol equation†. ◆ Finally, we consider the motion of the pendulum depicted in Figure 4.18. This motion is measured by the angle u1t2 that the pendulum makes with the vertical line through O at time t. As the diagram shows, the component of gravity, which exerts a torque on the pendulum and thus accelerates the angular velocity du>dt, is given by -mg sin u. Consequently, the rotational analog of Newton’s second law, torque equals rate of change of angular momentum, dictates (see Problem 7) (20)

m/2u″ = -/mg sin u ,

or (21) Example 8 Solution

g mU″ + m sin U = 0 . O

Give a qualitative analysis of the motion of the pendulum. If we rewrite (21) as mu″ +

mg sin u u = 0 / u

†

The vdP equation also provides a model for the world's most accurate pendulum clock, a detail of which appears on the cover.

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Section 4.8

Qualitative Considerations for Variable-Coefficient and Nonlinear Equations

209

(mg, ) sin mg/,

22p

2p

2p

p

Figure 4.19 Pendulum “stiffness”

and compare with the paradigm (13), we see fixed mass m, no damping, and a stiffness given by “k” =

mg sin u . / u

This stiffness is plotted in Figure 4.19, where we see that small-amplitude motions are driven by a nearly constant spring stiffness of value mg>/, and the considerations of Section 4.3 dictate the familiar formula v =

g k = Am A/

for the angular frequency of the nearly sinusoidal oscillations. See Figure 4.20, which compares a computer-generated solution to equation (21) to the solution of the constant-stiffness equation with the same initial conditions.† For larger motions, however, the diminishing stiffness distorts the sinusoidal nature of the graph of u1t2, and lowers the frequency. This is evident in Figure 4.21 on page 210.

True stiffness p rad 6

t

Constant stiffness Figure 4.20 Small-amplitude pendulum motion

†

The latter is identified as the linearized equation in Project D on page 236.

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Chapter 4

Linear Second-Order Equations

3p rad 8

t

Figure 4.21 Large-amplitude pendulum motion

Finally, if the motion is so energetic that u reaches the value p, the stiffness changes sign and abets the displacement; the pendulum passes the apex and gains speed as it falls, and this spinning motion repeats continuously. See Figure 4.22. ◆

p

2p

3p

4p

5p

6p

t

Figure 4.22 Very-large-amplitude pendulum motion

The computation of the solutions of the Legendre, Bessel, and Airy equations and the analysis of the nonlinear equations of Duffing, van der Pol, and the pendulum have challenged many of the great mathematicians of the past. It is gratifying, then, to note that so many of their salient features are susceptible to the qualitative reasoning we have used herein.

4.8

EXERCISES

1. Show that if y1t2 satisfies y″ - ty = 0, then y1 -t2 satisfies y″ + ty = 0. 2. Using the paradigm (13), what are the inertia, damping, and stiffness for the equation y″ - 6y2 = 0? If y 7 0, what is the sign of the “stiffness constant”? Does your answer help explain the runaway behavior of the solutions y1t2 = 1> 1c - t2 2? 3. Try to predict the qualitative features of the solution to y″ - 6y2 = 0 that satisfies the initial conditions y102 = -1, y′102 = - 1. Compare with the computergenerated Figure 4.23. [Hint: Consider the sign of the spring stiffness.]

y

t

21

Figure 4.23 Solution for Problem 3

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Section 4.8

Qualitative Considerations for Variable-Coefficient and Nonlinear Equations

4. Show that the three solutions 1> 11 - t2 2, 1> 12 - t2 2, and 1> 13 - t2 2 to y″ - 6y2 = 0 are linearly independent on 1 - 1, 12. (See Problem 35, Exercises 4.2, page 164.) 5. (a) Use the energy integral lemma to derive the family of solutions y1t2 = 1> 1t - c2 to the equation y″ = 2y3. (b) For c ≠ 0 show that these solutions are pairwise linearly independent for different values of c in an appropriate interval around t = 0. (c) Show that none of these solutions satisfies the initial conditions y102 = 1, y′102 = 2. 6. Use the energy integral lemma to show that motions of the free undamped mass–spring oscillator my″ + ky = 0 obey

211

10. Use the result of Problem 8 to prove that if the pendulum in Figure 4.18 on page 208 is released from rest at the angle a, 0 6 a 6 p, then 0 u1t2 0 … a for all t. [Hint: The initial conditions are u102 = a, u′102 = 0; argue that the constant in Problem 8 equals - 1g>/2cos a.] y

t

m1y′2 2 + ky2 = constant . 7. Pendulum Equation. To derive the pendulum equation (21), complete the following steps. (a) The angular momentum of the pendulum mass m measured about the support O in Figure 4.18 on page 208 is given by the product of the “lever arm” length / and the component of the vector momentum mv perpendicular to the lever arm. Show that this gives angular momentum = m/2

Figure 4.24 Solution to the Rayleigh equation y

du . dt

(b) The torque produced by gravity equals the product of the lever arm length / and the component of gravitational (vector) force mg perpendicular to the lever arm. Show that this gives

t

torque = - /mg sin u .

Figure 4.25 Solution to the Rayleigh equation

(c) Now use Newton’s law of rotational motion to deduce the pendulum equation (20). 8. Use the energy integral lemma to show that pendulum motions obey 1u′2 2

11. Use the mass–spring analogy to explain the qualitative nature of the solutions to the Rayleigh equation y″ − 31 − 1y′2 2 4y′ + y = 0

(22)

g cos u = constant . 2 / 9. Use the result of Problem 8 to find the value of u′102, the initial velocity, that must be imparted to a pendulum at rest to make it approach (but not cross over) the apex of its motion. Take / = g for simplicity.

depicted in Figures 4.24 and 4.25. 12. Use reduction of order to derive the solution y2 1t2 in equation (5) for Legendre’s equation. 13. Figure 4.26 contains graphs of solutions to the Duffing, Airy, and van der Pol equations. Try to match the solution to the equation.

-

y

y

y

t

t

t (a)

(b)

(c)

Figure 4.26 Solution graphs for Problem 13

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Chapter 4

Linear Second-Order Equations

14. Verify that the formulas for the Bessel functions J1>2 1t2, Y1>2 1t2 do indeed solve equation (16). 15. Use the mass–spring oscillator analogy to decide whether all solutions to each of the following differential equations are bounded as t S + ∞ . (a) y″ + t 2y = 0 (b) y″ - t 2y = 0 (c) y″ + y5 = 0

(d) y″ + y6 = 0

(e) y″ + 14 + 2 cos t2y = 0 (Mathieu’s equation)

17. Armageddon. Earth revolves around the sun in an approximately circular orbit with radius r = a, completing a revolution in the time T = 2p1a3 >GM2 1>2, which is one Earth year; here M is the mass of the sun and G is the universal gravitational constant. The gravitational force of the sun on Earth is given by GMm>r 2, where m is the mass of Earth. Therefore, if Earth “stood still,” losing its orbital velocity, it would fall on a straight line into the sun in accordance with Newton’s second law:

(f) y″ + ty′ + y = 0 (g) y″ - ty′ - y = 0 16. Use the energy integral lemma to show that every solution to the Duffing equation (18) is bounded; that is, 0 y1t2 0 … M for some M. [Hint: First argue that y2 >2 + y4 >4 … K for some K.]

m

d 2r GMm = - 2 . 2 dt r

If this calamity occurred, what fraction of the normal year T would it take for Earth to splash into the sun (i.e., achieve r = 0)? [Hint: Use the energy integral lemma and the initial conditions r102 = a, r′102 = 0.]

4.9 A Closer Look at Free Mechanical Vibrations In this section we return to the mass–spring system depicted in Figure 4.1 (page 152) and analyze its motion in more detail. The governing equation is (1)

Fext = 3inertia4

d 2y dt

2

+ 3damping4

dy + 3stiffness4y dt

= my″ + by′ + ky .

Let’s focus on the simple case in which b = 0 and Fext = 0, the so-called undamped, free case. Then equation (1) reduces to (2)

m

d 2y dt 2

+ ky = 0

and, when divided by m, becomes (3)

d 2y dt 2

+ V2y = 0 ,

where v = 2k>m. The auxiliary equation associated with (3) is r 2 + v2 = 0, which has complex conjugate roots {vi. Hence, a general solution to (3) is (4)

y1t2 = C1 cos Vt + C2 sin Vt .

We can express y1t2 in the more convenient form (5)

y1t2 = A sin1Vt + F2 ,

with A Ú 0, by letting C1 = A sin f and C2 = A cos f. That is, A sin1vt + f2 = A cos vt sin f + A sin vt cos f = C1 cos vt + C2 sin vt .

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Section 4.9

A Closer Look at Free Mechanical Vibrations

213

y

A

2f/v

Period 2p/v

Period 2p/v t 2p f –– v 2 –– v

Figure 4.27 Simple harmonic motion of undamped, free vibrations

Solving for A and f in terms of C1 and C2, we find (6)

A = 2C 21 + C 22

and

tan F =

C1 , C2

where the quadrant in which f lies is determined by the signs of C1 and C2. This is because sin f has the same sign as C1 1sin f = C1 >A2 and cos f has the same sign as C2 1cos f = C2 >A2. For example, if C1 7 0 and C2 6 0, then f is in Quadrant II. (Note, in particular, that f is not simply the arctangent of C1 >C2, which would lie in Quadrant IV.) It is evident from (5) that, as we predicted in Section 4.1, the motion of a mass in an undamped, free system is simply a sine wave, or what is called simple harmonic motion. (See Figure 4.27.) The constant A is the amplitude of the motion and f is the phase angle. The motion is periodic with period 2p>v and natural frequency v>2p, where v = 2k>m. The period is measured in units of time, and the natural frequency has the dimensions of periods (or cycles) per unit time. The constant v is the angular frequency for the sine function in (5) and has dimensions of radians per unit time. To summarize: angular frequency = v = 2k>m 1rad/sec2 , natural frequency = v>2p 1cycles/sec2 , period = 2p>v 1sec2 .

Observe that the amplitude and phase angle depend on the constants C1 and C2, which, in turn, are determined by the initial position and initial velocity of the mass. However, the period and frequency depend only on k and m and not on the initial conditions. Example 1

A 1>8-kg mass is attached to a spring with stiffness k = 16 N/m, as depicted in Figure 4.1. The mass is displaced 1>2 m to the right of the equilibrium point and given an outward velocity (to the right) of 22 m/sec. Neglecting any damping or external forces that may be present, determine the equation of motion of the mass along with its amplitude, period, and natural frequency. How long after release does the mass pass through the equilibrium position?

Solution

Because we have a case of undamped, free vibration, the equation governing the motion is (3). Thus, we find the angular frequency to be v =

k 16 = = 822 rad /sec . Am A 1>8

Substituting this value for v into (4) gives (7)

y1t2 = C1 cos1822 t2 + C2 sin1822 t2 .

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Linear Second-Order Equations

Now we use the initial conditions, y102 = 1>2 m and y′102 = 22 m/sec, to solve for C1 and C2 in (7). That is, 1>2 = y102 = C1 , 22 = y′102 = 822C2 , and so C1 = 1>2 and C2 = 1>8. Hence, the equation of motion of the mass is (8)

y1t2 =

1 1 cos1822 t2 + sin1822 t2 . 2 8

To express y1t2 in the alternative form (5), we set A = 2C 21 + C 22 = 211>22 2 + 11>82 2 = tan f =

217 , 8

1>2 C1 = 4. = C2 1>8

Since both C1 and C2 are positive, f is in Quadrant I, so f = arctan 4 ≈ 1.326. Hence, (9)

y1t2 =

217 sin1822 t + f2 . 8

Thus, the amplitude A is 217>8 m, and the phase angle f is approximately 1.326 rad. The period is P = 2p>v = 2p> 18222 = 22p>8 sec, and the natural frequency is 1>P = 8> 1 22p2 cycles per sec. Finally, to determine when the mass will pass through the equilibrium position, y = 0, we must solve the trigonometric equation (10)

y1t2 =

217 sin1822 t + f2 = 0 8

for t. Equation (10) will be satisfied whenever (11)

822 t + f = np

t =

np - f

≈

np - 1.326

, 822 822 n an integer. Putting n = 1 in (11) determines the first time t when the mass crosses its equilibrium position: t =

p-f 822

or

≈ 0.16 sec . ◆

In most applications of vibrational analysis, of course, there is some type of frictional or damping force affecting the vibrations. This force may be due to a component in the system, such as a shock absorber in a car, or to the medium that surrounds the system, such as air or some liquid. So we turn to a study of the effects of damping on free vibrations, and equation (2) generalizes to (12)

m

d 2y dt

2

+b

dy + ky = 0 . dt

The auxiliary equation associated with (12) is (13)

mr 2 + br + k = 0 ,

and its roots are

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Section 4.9

(14)

A Closer Look at Free Mechanical Vibrations

215

1 b -b { 2b2 - 4mk 2b2 - 4mk . = { 2m 2m 2m

As we found in Sections 4.2 and 4.3, the form of the solution to (12) depends on the nature of these roots and, in particular, on the discriminant b2 - 4mk.

Underdamped or Oscillatory Motion (b2 * 4mk) When b2 6 4mk, the discriminant b2 - 4mk is negative, and there are two complex conjugate roots to the auxiliary equation (13). These roots are a { ib, where b 1 (15) a J , b J 24mk - b2 . 2m 2m Hence, a general solution to (12) is (16)

y1t2 = eat 1C1 cos bt + C2 sin bt2 .

As we did with simple harmonic motion, we can express y1t2 in the alternate form (17)

y1t2 = Aeat sin1bt + f2 ,

where A = 2C 21 + C 22 and tan f = C1 >C2. It is now evident that y1t2 is the product of an exponential damping factor, Aeat = Ae-1b>2m2t , and a sine factor, sin1bt + f2, which accounts for the oscillatory motion. Because the sine factor varies between –1 and 1 with period 2p>b, the solution y1t2 varies between -Aeat and Aeat with quasiperiod P = 2p>b = 4mp> 24mk - b2 and quasifrequency 1>P. Moreover, since b and m are positive, a = -b>2m is negative, and thus the exponential factor tends to zero as t S + ∞. A graph of a typical solution y1t2 is given in Figure 4.28. The system is called underdamped because there is not enough damping present (b is too small) to prevent the system from oscillating. It is easily seen that as b S 0 the damping factor approaches the constant A and the quasifrequency approaches the natural frequency of the corresponding undamped harmonic motion. Figure 4.28 demonstrates that the values of t where the graph of y1t2 touches the exponential curves {Aeat are close to (but not exactly) the same values of t at which y1t2 attains its relative maximum and minimum values (see Problem 13 on page 221). y

A

Aeat Aeatsin(bt 1 f)

t

2A

2Aeat

Quasiperiod 5 2p/b

Figure 4.28 Damped oscillatory motion

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Chapter 4

Linear Second-Order Equations

y

y

y

no local max or min

one local max

one local min

t

t

(a)

t

(b)

(c)

Figure 4.29 Overdamped vibrations

Overdamped Motion (b2 + 4mk) When b2 7 4mk, the discriminant b2 - 4mk is positive, and there are two distinct real roots to the auxiliary equation (13): (18)

r1 = -

1 1 b b 2b2 - 4mk , r2 = 2b2 - 4mk . + 2m 2m 2m 2m

Hence, a general solution to (12) in this case is (19)

y1t2 = C1er1t + C2er2t .

Obviously, r2 is negative. And since b2 7 b2 - 4mk (that is, b 7 2b2 - 4mk ), it follows that r1 is also negative. Therefore, as t S + ∞, both of the exponentials in (19) decay and y1t2 S 0. Moreover, since y′1t2 = C1r1er1t + C2r2er2t = er1t 1C1r1 + C2r2e1r2-r12t 2 ,

we see that the derivative is either identically zero (when C1 = C2 = 0) or vanishes for at most one value of t (when the factor in parentheses is zero). If the trivial solution y1t2 K 0 is ignored, it follows that y1t2 has at most one local maximum or minimum for t 7 0. Therefore, y1t2 does not oscillate. This leaves, qualitatively, only three possibilities for the motion of y1t2, depending on the initial conditions. These are illustrated in Figure 4.29. This case where b2 7 4mk is called overdamped motion.

Critically Damped Motion (b2 = 4mk) When b2 = 4mk, the discriminant b2 - 4mk is zero, and the auxiliary equation has the repeated root -b>2m. Hence, a general solution to (12) is now (20)

y1t2 = 1C1 + C2t2e-1b>2m2t .

To understand the motion described by y1t2 in (20), we first consider the behavior of y1t2 as t S + ∞. By L’Hôpital’s rule, (21)

lim lim y1t2 = S

t S +∞

t

+∞

C1 + C2t e

1b>2m2t

= S lim t

C2

+∞ 1b>2m2e1b>2m2t

= 0

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Section 4.9

A Closer Look at Free Mechanical Vibrations

217

(recall that b>2m 7 0). Hence, y1t2 dies off to zero as t S + ∞. Next, since y′1t2 = a C2 -

b b C1 C tb e-1b>2m2t , 2m 2m 2

we see again that a nontrivial solution can have at most one local maximum or minimum for t 7 0, so motion is nonoscillatory. If b were any smaller, oscillation would occur. Thus, the special case where b2 = 4mk is called critically damped motion. Qualitatively, critically damped motions are similar to overdamped motions (see Figure 4.29 again). Example 2

Assume that the motion of a mass–spring system with damping is governed by (22)

d 2y dt

2

+b

dy + 25y = 0 ; y102 = 1 , y′102 = 0 . dt

Find the equation of motion and sketch its graph for the three cases where b = 6, 10, and 12. Solution

The auxiliary equation for (22) is (23)

r 2 + br + 25 = 0 ,

whose roots are (24)

r = -

b 1 { 2b2 - 100 . 2 2

Case 1. When b = 6, the roots (24) are -3 { 4i. This is thus a case of underdamping, and the equation of motion has the form (25)

y1t2 = C1e-3t cos 4t + C2e-3t sin 4t .

Setting y102 = 1 and y′102 = 0 gives the system C1 = 1 ,

-3C1 + 4C2 = 0 ,

whose solution is C1 = 1, C2 = 3>4. To express y1t2 as the product of a damping factor and a sine factor [recall equation (17)], we set A = 2C 12 + C 22 =

C1 5 4 , tan f = = , 4 C2 3

where f is a Quadrant I angle, since C1 and C2 are both positive. Then (26)

y1t2 =

5 -3t e sin14t + f2 , 4

where f = arctan14>32 ≈ 0.9273. The underdamped spring motion is shown in Figure 4.30(a) on page 218. Case 2. When b = 10, there is only one (repeated) root to the auxiliary equation (23), namely, r = -5. This is a case of critical damping, and the equation of motion has the form (27)

y1t2 = 1C1 + C2t2e-5t .

Setting y102 = 1 and y′102 = 0 now gives C1 = 1 , C2 - 5C1 = 0 , and so C1 = 1, C2 = 5. Thus, (28)

y1t2 = 11 + 5t2e-5t .

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218

Chapter 4

Linear Second-Order Equations

y

y 5 23t 4e 11 + 6 11 e(26 1 22

11 )t

1

11 2 6 11 e(26 2 22

11 )t

5 23t sin(4t 1 0.9273) 4e

t

(1 1 5t)e25t

t

b 5 10 b 5 12

b56 2 5 e23t 4

(a)

(b) Figure 4.30 Solutions for various values of b

The graph of y1t2 given in (28) is represented by the lower curve in Figure 4.30(b). Notice that y1t2 is zero only for t = -1>5 and hence does not cross the t-axis for t 7 0. Case 3. When b = 12, the roots to the auxiliary equation are -6 { 211. This is a case of overdamping, and the equation of motion has the form (29)

y1t2 = C1e1-6+2112t + C2e1-6 - 2112t .

Setting y102 = 1 and y′102 = 0 gives C1 + C2 = 1 ,

1 -6 + 2112C1 + 1 -6 - 2112C2 = 0 ,

from which we find C1 = 111 + 62112 >22 and C2 = 111 - 62112 >22. Hence, (30)

y1t2 = =

11 + 6211 1-6+2112t 11 - 6211 1-6 - 2112t e + e 22 22 e1-6+2112t e 11 + 6211 + 111 - 62112e-2211t f . 22

The graph of this overdamped motion is represented by the upper curve in Figure 4.30(b). ◆ It is interesting to observe in Example 2 that when the system is underdamped 1b = 62, the solution goes to zero like e-3t; when the system is critically damped 1b = 102, the solution tends to zero roughly like e-5t; and when the system is overdamped 1b = 122, the solution goes to zero like e1-6+2112t ≈ e-2.68t. This means that if the system is underdamped, it not only oscillates but also dies off slower than if it were critically damped. Moreover, if the system is overdamped, it again dies off more slowly than if it were critically damped (in agreement with our physical intuition that the damping forces hinder the return to equilibrium).

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Section 4.9

A Closer Look at Free Mechanical Vibrations

219

y 1m 2 k 5 4 N/m

1 kg 4 b 5 1 N-sec/m

t

Equilibrium

2.5 maximum displacement is y(0.096) < 20.55 m

(a)

(b)

Figure 4.31 Mass–spring system and graph of motion for Example 3

Example 3

A 1>4-kg mass is attached to a spring with a stiffness 4 N/m as shown in Figure 4.31(a). The damping constant b for the system is 1 N-sec/m. If the mass is displaced 1>2 m to the left and given an initial velocity of 1 m/sec to the left, find the equation of motion. What is the maximum displacement that the mass will attain?

Solution

Substituting the values for m, b, and k into equation (12) and enforcing the initial conditions, we obtain the initial value problem (31)

1 d 2y dy 1 + + 4y = 0 ; y102 = - , y′102 = -1 . 2 4 dt dt 2

The negative signs for the initial conditions reflect the facts that the initial displacement and push are to the left. It can readily be verified that the solution to (31) is (32)

1 1 -2t y1t2 = - e-2t cos 1 223 t 2 e sin 1 223 t 2 , 2 23

or (33)

y1t2 =

7 -2t e sin 1 223 t + f 2 , A 12

where tan f = 23>2 and f lies in Quadrant III because C1 = -1>2 and C2 = -1> 23 are both negative. [See Figure 4.31(b) for a sketch of y1t2.] To determine the maximum displacement from equilibrium, we must determine the maximum value of 0 y1t2 0 on the graph in Figure 4.31(b). Because y1t2 dies off exponentially, this will occur at the first critical point of y1t2. Computing y′1t2 from (32), setting it equal to zero, and solving gives y′1t2 = e-2t e 5 23

5 23

sin 1 223 t 2 - cos 1 223 t 2 f = 0 ,

sin1223 t2 = cos1223 t2 ,

tan1223 t2 =

23 . 5

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220

Chapter 4

Linear Second-Order Equations

Thus, the first positive root is t =

1 223

arctan

23 ≈ 0.096 . 5

Substituting this value for t back into equation (32) or (33) gives y10.0962 ≈ -0.55. Hence, the maximum displacement, which occurs to the left of equilibrium, is approximately 0.55 m. ◆

4.9

EXERCISES

All problems refer to the mass–spring configuration depicted in Figure 4.1, page 152. 1. A 2-kg mass is attached to a spring with stiffness k = 50 N/m. The mass is displaced 1>4 m to the left of the equilibrium point and given a velocity of 1 m/sec to the left. Neglecting damping, find the equation of motion of the mass along with the amplitude, period, and frequency. How long after release does the mass pass through the equilibrium position? 2. A 3-kg mass is attached to a spring with stiffness k = 48 N/m. The mass is displaced 1>2 m to the left of the equilibrium point and given a velocity of 2 m/sec to the right. The damping force is negligible. Find the equation of motion of the mass along with the amplitude, period, and frequency. How long after release does the mass pass through the equilibrium position? 3. The motion of a mass–spring system with damping is governed by y″1t2 + by′1t2 + 16y1t2 = 0 ; y102 = 1 ,

6. The motion of a mass–spring system with damping is governed by

y″1t2 + 4y′1t2 + ky1t2 = 0 ; y102 = 1 , y′102 = 0 .

7.

8.

9.

y′102 = 0 .

Find the equation of motion and sketch its graph for b = 0, 6, 8, and 10. 4. The motion of a mass–spring system with damping is governed by

10.

y″1t2 + by′1t2 + 64y1t2 = 0 ; y102 = 1 ,

y′102 = 0 .

Find the equation of motion and sketch its graph for b = 0, 10, 16, and 20. 5. The motion of a mass–spring system with damping is governed by y″1t2 + 10y′1t2 + ky1t2 = 0 ; y102 = 1 ,

y′102 = 0 .

Find the equation of motion and sketch its graph for k = 20, 25, and 30.

11.

12.

Find the equation of motion and sketch its graph for k = 2, 4, and 6. A 1>8-kg mass is attached to a spring with stiffness 16 N/m. The damping constant for the system is 2 N-sec/m. If the mass is moved 3>4 m to the left of equilibrium and given an initial leftward velocity of 2 m/sec, determine the equation of motion of the mass and give its damping factor, quasiperiod, and quasifrequency. A 20-kg mass is attached to a spring with stiffness 200 N/m. The damping constant for the system is 140 N-sec/m. If the mass is pulled 25 cm to the right of equilibrium and given an initial leftward velocity of 1 m/sec, when will it first return to its equilibrium position? A 2-kg mass is attached to a spring with stiffness 40 N/m. The damping constant for the system is 815 N-sec/m. If the mass is pulled 10 cm to the right of equilibrium and given an initial rightward velocity of 2 m/sec, what is the maximum displacement from equilibrium that it will attain? A 1>4-kg mass is attached to a spring with stiffness 8 N/m. The damping constant for the system is 1>4 N-sec/m. If the mass is moved 1 m to the left of equilibrium and released, what is the maximum displacement to the right that it will attain? A 1-kg mass is attached to a spring with stiffness 100 N/m. The damping constant for the system is 0.2 N-sec/m. If the mass is pushed rightward from the equilibrium position with a velocity of 1 m/sec, when will it attain its maximum displacement to the right? A 1>4-kg mass is attached to a spring with stiffness 8 N/m. The damping constant for the system is 2 N-sec/m. If the mass is pushed 50 cm to the left of equilibrium and given a leftward velocity of 2 m/sec, when will the mass attain its maximum displacement to the left?

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Section 4.10

13. Show that for the underdamped system of Example 3, the times when the solution curve y1t2 in (33) touches the exponential curves { 27>12e-2t are not the same values of t for which the function y1t2 attains its relative extrema. 14. For an underdamped system, verify that as b S 0 the damping factor approaches the constant A and the quasifrequency approaches the natural frequency 2k>m> 12p2. 15. How can one deduce the value of the damping constant b by observing the motion of an underdamped system? Assume that the mass m is known. 16. A mass attached to a spring oscillates with a period of 3 sec. After 2 kg are added, the period becomes 4 sec. Assuming that we can neglect any damping or external forces, determine how much mass was originally attached to the spring.

A Closer Look at Forced Mechanical Vibrations

221

17. Consider the equation for free mechanical vibration, my″ + by′ + ky = 0, and assume the motion is critically damped. Let y102 = y0, y′102 = v0 and assume y0 ≠ 0. (a) Prove that the mass will pass through its equilibrium at exactly one positive time if and only if - 2my0 7 0. 2mv0 + by0 (b) Use computer software to illustrate part (a) for a specific choice of m, b, k, y0, and v0. Be sure to include an appropriate graph in your illustration. 18. Consider the equation for free mechanical vibration, my″ + by′ + ky = 0, and assume the motion is overdamped. Suppose y102 7 0 and y′102 7 0. Prove that the mass will never pass through its equilibrium at any positive time.

4.10 A Closer Look at Forced Mechanical Vibrations We now consider the vibrations of a mass–spring system when an external force is applied. Of particular interest is the response of the system to a sinusoidal forcing term. As a paradigm, let’s investigate the effect of a cosine forcing function on the system governed by the differential equation (1)

m

d 2y dt

2

+b

dy + ky = F0 cos Gt , dt

where F0 and g are nonnegative constants and 0 6 b2 6 4mk (so the system is underdamped). A solution to (1) has the form y = yh + yp, where yp is a particular solution and yh is a general solution to the corresponding homogeneous equation. We found in equation (17) of Section 4.9 that (2)

yh 1t2 = Ae-1b>2m2t sina

24mk - b2 t + fb , 2m

where A and f are constants. To determine yp, we can use the method of undetermined coefficients (Section 4.4). From the form of the nonhomogeneous term, we know that (3)

yp 1t2 = A1 cos gt + A2 sin gt ,

where A1 and A2 are constants to be determined. Substituting this expression into equation (1) and simplifying gives (4)

3 1 k - mg2 2 A1 + bgA2 4 cos gt + 3 1 k - mg2 2 A2 - bgA1 4 sin gt

= F0 cos gt .

Setting the corresponding coefficients on both sides equal, we have 1k - mg 2 2A1 + bgA2 = F0 , -bgA1 + 1k - mg 2 2A2 = 0 .

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222

Chapter 4

Linear Second-Order Equations

Solving, we obtain (5)

F0 1k - mg 2 2

A1 =

1k - mg 2 + b g 2 2

2 2

, A2 =

F0bg

1k - mg 2 2 2 + b2g 2

.

Hence, a particular solution to (1) is (6)

yp 1t2 =

F0

1k - mg 2 2 2 + b2g 2

3 1 k - mg2 2 cos gt + bg sin gt 4 .

The expression in brackets can also be written as 21k - mg 2 2 2 + b 2g 2 sin1gt + u2 ,

so we can express yp in the alternative form (7)

yp 1t2 =

F0

21k - mg 2 2 2 + b 2g 2

sin1gt + u2 ,

where tan u = A1 >A2 = 1k - mg 2 2 > 1bg2 and the quadrant in which u lies is determined by the signs of A1 and A2. Combining equations (2) and (7), we have the following representation of a general solution to (1) in the case 0 6 b2 6 4mk: (8)

y1t2 = Ae-1b>2m2t sina

F0 24mk - b 2 t + fb + sin1gt + u2 . 2m 21k - mg 2 2 2 + b 2g 2

The solution (8) is the sum of two terms. The first term, yh, represents damped oscillation and depends only on the parameters of the system and the initial conditions. Because of the damping factor Ae-1b>2m2t, this term tends to zero as t S + ∞. Consequently, it is referred to as the transient part of the solution. The second term, yp, in (8) is the offspring of the external forcing function f1t2 = F0 cos gt. Like the forcing function, yp is a sinusoid with angular frequency g. It is the synchronous solution that we anticipated in Section 4.1. However, yp is out of phase with f1t2 (by the angle u - p>2), and its magnitude is different by the factor (9)

1

21k - mg 2 2 2 + b2g 2

.

As the transient term dies off, the motion of the mass–spring system becomes essentially that of the second term yp (see Figure 4.32, page 223). Hence, this term is called the steady-state solution. The factor appearing in (9) is referred to as the frequency gain, or gain factor, since it represents the ratio of the magnitude of the sinusoidal response to that of the input force. Note that this factor depends on the frequency g and has units of length/force.

Example 1

Solution

A 10-kg mass is attached to a spring with stiffness k = 49 N/m. At time t = 0, an external force f1t2 = 20 cos 4t N is applied to the system. The damping constant for the system is 3 N-sec/m. Determine the steady-state solution for the system. Substituting the given parameters into equation (1), we obtain (10)

10

d 2y dt

2

+3

dy + 49y = 20 cos 4t , dt

where y1t2 is the displacement (from equilibrium) of the mass at time t.

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Section 4.10

223

A Closer Look at Forced Mechanical Vibrations

y

0.2

y(t)

0.15

0.1

0.05

yp(t) t 2

4

6

8

10

Figure 4.32 Convergence of y1t2 to the steady-state solution yp 1t2 when m = 4, b = 6, k = 3, F0 = 2, g = 4

To find the steady-state response, we must produce a particular solution to (10) that is a sinusoid. We can do this using the method of undetermined coefficients, guessing a solution of the form A1 cos 4t + A2 sin 4t. But this is precisely how we derived equation (7). Thus, we substitute directly into (7) and find (11)

yp 1t2 =

20

2149 - 1602 2 + 1921162

sin14t + u2 ≈ 10.182 sin14t + u2 ,

where tan u = 149 - 1602 >12 ≈ -9.25. Since the numerator, 149 - 1602, is negative and the denominator, 12, is positive, u is a Quadrant IV angle. Thus, u ≈ arctan1 -9.252 ≈ -1.46 , and the steady-state solution is given (approximately) by (12)

yp 1t2 = 10.182 sin14t - 1.462 . ◆

The above example illustrates an important point made earlier: The steady-state response (12) to the sinusoidal forcing function 20 cos 4t is a sinusoid of the same frequency but different amplitude. The gain factor [see (9)] in this case is 10.182 >20 = 0.009 m/N. In general, the amplitude of the steady-state solution to equation (1) depends on the angular frequency g of the forcing function and is given by A1g2 = F0M1g2, where (13)

M1g2 J

1

21k - mg 2 2 2 + b2g 2

is the frequency gain [see (9)]. This formula is valid even when b2 Ú 4mk. For a given system (m, b, and k fixed), it is often of interest to know how this system reacts to sinusoidal inputs of various frequencies (g is a variable). For this purpose the graph of the gain M1g2, called the frequency response curve, or resonance curve, for the system, is enlightening.

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224

Chapter 4

Linear Second-Order Equations

To sketch the frequency response curve, we first observe that for g = 0 we find M102 = 1>k. Of course, g = 0 implies the force F0 cos gt is static; there is no motion in the steady state, so this value of M102 is appropriate. Also note that as g S ∞ the gain M1g2 S 0; the inertia of the system limits the extent to which it can respond to extremely rapid vibrations. As a further aid in describing the graph, we compute from (13) k b2 bd m 2m2 M′1g2 = . 3 1k - mg2 2 2 + b2g2 4 3>2 2m2g c g 2 - a

(14)

It follows from (14) that M′1g2 = 0 if and only if (15)

g = 0 or g = gr J

k b2 . B m 2m2

Now when the system is overdamped or critically damped, so b2 Ú 4mk 7 2mk, the term inside the radical in (15) is negative, and hence M′1g2 = 0 only when g = 0. In this case, as g increases from 0 to infinity, M1g2 decreases from M102 = 1>k to a limiting value of zero. When b2 6 2mk (which implies the system is underdamped), then gr is real and positive, and it is easy to verify that M1g2 has a maximum at gr . Substituting gr into (13) gives (16)

M1gr 2 =

(17)

M1g2 =

1>b

. k b2 B m 4m2 The value gr >2p is called the resonance frequency for the system. When the system is stimulated by an external force at this frequency, it is said to be at resonance. To illustrate the effect of the damping constant b on the resonance curve, we consider a system in which m = k = 1. In this case the frequency response curves are given by 1

211 - g 2 2 2 + b2g 2

,

and, for b 6 22, the resonance frequency is gr >2p = 11>2p2 21 - b2 >2. Figure 4.33 displays the graphs of these frequency response curves for b = 1>4, 1>2, 1, 3>2, and 2. Observe that as b S 0 the maximum magnitude of the frequency gain increases and the resonance frequency gr >2p for the damped system approaches 2k>m>2p = 1>2p, the natural frequency for the undamped system. To understand what is occurring, consider the undamped system 1b = 02 with forcing term F0 cos gt. This system is governed by (18)

m

d 2y dt 2

+ ky = F0 cos gt .

A general solution to (18) is the sum of a particular solution and a general solution to the homogeneous equation. In Section 4.9 we showed that the latter describes simple harmonic motion: (19)

yh 1t2 = A sin1vt + f2 ,

v J 2k>m .

The formula for the particular solution given in (7) is valid for b = 0, provided g ≠ v = 2k>m. However, when b = 0 and g = v, then the form we used with undetermined coefficients

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Section 4.10

A Closer Look at Forced Mechanical Vibrations

225

M(g) 4

1

b 5 4– 3

1

b 5 2–

2

b51 3

b 5 –2

1

b52

g

0

1

2

Figure 4.33 Frequency response curves for various values of b

to derive (7) does not work because cos vt and sin vt are solutions to the corresponding homogeneous equation. The correct form is (20)

yp 1t2 = A1t cos vt + A2t sin vt ,

which leads to the solution (21)

yp 1t2 =

F0 t sin Vt . 2mV

[The verification of (21) is straightforward.] Hence, in the undamped resonant case (when g = v), a general solution to (18) is (22)

y1t2 = A sin1vt + f2 +

F0 t sin vt . 2mv

Returning to the question of resonance, observe that the particular solution in (21) oscillates between { 1F0t2 > 12mv2. Hence, as t S + ∞ the maximum magnitude of (21) approaches ∞ (see Figure 4.34 on page 226). It is obvious from the above discussion that if the damping constant b is very small, the system is subject to large oscillations when the forcing function has a frequency near the resonance frequency for the system. It is these large vibrations at resonance that concern engineers. Indeed, resonance vibrations have been known to cause airplane wings to snap, bridges to collapse,† and (less catastrophically) wine glasses to shatter. †

An interesting discussion of one such disaster, involving the Tacoma Narrows bridge in Washington State, can be found in Differential Equations and Their Applications, 4th ed., by M. Braun (Springer-Verlag, New York, 1993). See also the articles, “Large-Amplitude Periodic Oscillations in Suspension Bridges: Some New Connections with Nonlinear Analysis,” by A. C. Lazer and P. J. McKenna, SIAM Review, Vol. 32 (1990): 537–578; or “Still Twisting,” by Henry Petroski, American Scientist, Vol. 19 (1991): 398–401.

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226

Chapter 4

Linear Second-Order Equations

yp F0 –––– t 2m v

t

F0 2 –––– t 2m v Figure 4.34 Undamped oscillation of the particular solution in (21)

When the mass–spring system is hung vertically as in Figure 4.35, the force of gravity must be taken into account. This is accomplished very easily. With y measured downward from the unstretched spring position, the governing equation is my″ + by′ + ky = mg , and if the right-hand side is recognized as a sinusoidal forcing term with frequency zero 1mg cos 0t2, then the synchronous steady-state response is a constant, which is easily seen to be yp 1t2 = mg>k. Now if we redefine y1t2 to be measured from this (true) equilibrium level, as indicated in Figure 4.35(c), ynew 1t2 J y1t2 - mg>k ,

then the governing equation

my″ + by′ + ky = mg + Fext 1t2

simplifies; we find > = mynew + bynew + kynew = m1y - mg>k2″ + b1y - mg>k2′ + k1y - mg>k2 = my″ + by′ + ky - mg = mg + Fext 1t2 - mg ,

L

y

(a)

L

L

yp 5 mg/k

mg/k m

ynew

(b)

m (c)

Figure 4.35 Spring (a) in natural position, (b) in equilibrium, and (c) in motion

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Section 4.10

or (23)

A Closer Look at Forced Mechanical Vibrations

227

> = mynew + bynew + kynew = Fext 1t2 .

Thus, the gravitational force can be ignored if y1t2 is measured from the equilibrium position. Adopting this convention, we drop the “new” subscript hereafter. Example 2 Solution

Suppose the mass–spring system in Example 1 is hung vertically. Find the steady-state solution. This is trivial; the steady-state solution is identical to what we derived before, yp 1t2 =

20

2149 - 1602 2 + 1921162

sin14t + u2

[equation (11)], but now yp is measured from the equilibrium position, which is mg>k ≈ 10 * 9.8>49 = 2 m below the unstretched spring position. ◆ Example 3

A 64-lb weight is attached to a vertical spring, causing it to stretch 3 in. upon coming to rest at equilibrium. The damping constant for the system is 3 lb-sec/ft. An external force F1t2 = 3 cos 12t lb is applied to the weight. Find the steady-state solution for the system.

Solution

If a weight of 64 lb stretches a spring by 3 in. (0.25 ft), then the spring stiffness must be 64>0.25 = 256 lb/ft. Thus, if we measured the displacement from the (true) equilibrium level, equation (23) becomes (24)

my″ + by′ + ky = 3 cos 12t ,

with b = 3 and k = 256. But recall that the unit of mass in the U.S. Customary System is the slug, which equals the weight divided by the gravitational acceleration constant g ≈ 32 ft/sec2 (Table 3.2, page 110). Therefore, m in equation (24) is 64>32 = 2 slugs, and we have 2y″ + 3y′ + 256y = 3 cos 12t . The steady-state solution is given by equation (6) with F0 = 3 and g = 12: yp 1t2 = =

4.10

3 3 1256 - 2 # 122 2 cos 12t + 3 # 12 sin 12t 4 1256 - 2 # 122 2 2 + 32 # 122 3 1 -8 cos 12t + 9 sin 12t2 . ◆ 580

EXERCISES

In the following problems, take g = 32 ft/sec2 for the U.S. Customary System and g = 9.8 m/sec2 for the MKS system. 1. Sketch the frequency response curve (13) for the system in which m = 4, k = 1, b = 2. 2. Sketch the frequency response curve (13) for the system in which m = 2, k = 3, b = 3.

3. Determine the equation of motion for an undamped system at resonance governed by d 2y

+ 9y = 2 cos 3t ; dt 2 y102 = 1 , y′102 = 0 . Sketch the solution.

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228

Chapter 4

Linear Second-Order Equations

4. Determine the equation of motion for an undamped system at resonance governed by d 2y 2

11.

+ y = 5 cos t ;

dt y102 = 0 ,

y′102 = 1 .

Sketch the solution. 5. An undamped system is governed by d 2y

+ ky = F0 cos gt ; dt 2 y102 = y′102 = 0 ,

m

12.

where g ≠ v J 2k>m. (a) Find the equation of motion of the system. (b) Use trigonometric identities to show that the solution can be written in the form y1t2 =

6. 7.

8.

9.

10.

2F0

m1v - g 2 2

2

sina

v+g v-g tb sina tb . 2 2

(c) When g is near v, then v - g is small, while v + g is relatively large compared with v - g. Hence, y1t2 can be viewed as the product of a slowly varying sine function, sin31v - g2t>24, and a rapidly varying sine function, sin31v + g2t>24. The net effect is a sine function y1t2 with frequency 1v - g2 >4p, which serves as the time-varying amplitude of a sine function with frequency 1v + g2 >4p. This vibration phenomenon is referred to as beats and is used in tuning stringed instruments. This same phenomenon in electronics is called amplitude modulation. To illustrate this phenomenon, sketch the curve y1t2 for F0 = 32, m = 2, v = 9, and g = 7. Derive the formula for yp 1t2 given in (21). Shock absorbers in automobiles and aircraft can be described as forced overdamped mass–spring systems. Derive an expression analogous to equation (8) for the general solution to the differential equation (1) when b2 7 4mk. The response of an overdamped system to a constant force is governed by equation (1) with m = 2, b = 8, k = 6, F0 = 18, and g = 0. If the system starts from rest 3y102 = y′102 = 04, compute and sketch the displacement y1t2. What is the limiting value of y1t2 as t S + ∞ ? Interpret this physically. An 8-kg mass is attached to a spring hanging from the ceiling, thereby causing the spring to stretch 1.96 m upon coming to rest at equilibrium. At time t = 0, an external force F1t2 = cos 2t N is applied to the system. The damping constant for the system is 3 N-sec/m. Determine the steady-state solution for the system. Show that the period of the simple harmonic motion of a mass hanging from a spring is 2p2l>g, where l denotes

13.

14.

15.

16.

the amount (beyond its natural length) that the spring is stretched when the mass is at equilibrium. A mass weighing 8 lb is attached to a spring hanging from the ceiling and comes to rest at its equilibrium position. At t = 0, an external force F1t2 = 2 cos 2t lb is applied to the system. If the spring constant is 10 lb/ft and the damping constant is 1 lb@sec/ft, find the equation of motion of the mass. What is the resonance frequency for the system? A 2-kg mass is attached to a spring hanging from the ceiling, thereby causing the spring to stretch 20 cm upon coming to rest at equilibrium. At time t = 0, the mass is displaced 5 cm below the equilibrium position and released. At this same instant, an external force F1t2 = 0.3 cos t N is applied to the system. If the damping constant for the system is 5 N@sec/m, determine the equation of motion for the mass. What is the resonance frequency for the system? A mass weighing 32 lb is attached to a spring hanging from the ceiling and comes to rest at its equilibrium position. At time t = 0, an external force F1t2 = 3 cos 4t lb is applied to the system. If the spring constant is 5 lb/ft and the damping constant is 2 lb-sec/ft, find the steadystate solution for the system. An 8-kg mass is attached to a spring hanging from the ceiling and allowed to come to rest. Assume that the spring constant is 40 N/m and the damping constant is 3 N-sec/m. At time t = 0, an external force of 2 sin 12t + p>42 N is applied to the system. Determine the amplitude and frequency of the steady-state solution. An 8-kg mass is attached to a spring hanging from the ceiling and allowed to come to rest. Assume that the spring constant is 40 N/m and the damping constant is 3 N-sec/m. At time t = 0, an external force of 2 sin 2t cos 2t N is applied to the system. Determine the amplitude and frequency of the steady-state solution. A helium-filled balloon on a cord, hanging y km above a level surface, is subjected to three forces: (i) the (constant) buoyant force B exerted by the external air pressure; (ii) the weight of the balloon mg, where m is the mass of the balloon; (iii) the weight of the part of the cord that has been lifted off the surface, dgy, where d is the linear density (kg /m) of the cord. (We assume that the cord is longer than the elevation of the balloon.) Express Newton’s second law for the balloon and show that it is exactly analogous to the governing equation for a mass vertically hung from a spring (page 226). Find the equation of motion. What is the frequency of the balloon’s oscillations?

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Chapter 4 Summary

229

Chapter 4 Summary In this chapter we discussed the theory of second-order linear differential equations and presented explicit solution methods for equations with constant coefficients. Much of the methodology can also be applied to the more general case of variable coefficients. We also studied the mathematical description of vibrating mechanical systems, and we saw how the mass–spring analogy could be used to predict qualitative features of solutions to some variable-coefficient and nonlinear equations. The important features and solution techniques for the constant-coefficient case are listed below.

Homogeneous Linear Equations (Constant Coefficients) ay″ + by′ + cy = 0,

a13 02, b, c constants.

Linearly Independent Solutions: y1, y2. Two solutions y1 and y2 to the homogeneous equation on the interval I are said to be linearly independent on I if neither function is a constant times the other on I. This will be true provided their Wronskian, W3y1, y2 4 1t2 J y1 1t2y2= 1t2 - y1= 1t2y2 1t2 ,

is different from zero for some (and hence all) t in I. General Solution to Homogeneous Equation: c1y1 + c2y2. If y1 and y2 are linearly independent solutions to the homogeneous equation, then a general solution is y1t2 = c1y1 1t2 + c2y2 1t2 ,

where c1 and c2 are arbitrary constants. Form of General Solution. The form of a general solution for a homogeneous equation with constant coefficients depends on the roots r1 =

-b + 2b2 - 4ac , 2a

r2 =

-b - 2b2 - 4ac 2a

of the auxiliary equation ar 2 + br + c = 0 ,

a≠0.

(a) When b2 - 4ac 7 0, the auxiliary equation has two distinct real roots r1 and r2 and a general solution is y1t2 = c1er1t + c2er2t . (b) When b2 - 4ac = 0, the auxiliary equation has a repeated real root r = r1 = r2 and a general solution is y1t2 = c1ert + c2tert . (c) When b2 - 4ac 6 0, the auxiliary equation has complex conjugate roots r = a { ib and a general solution is y1t2 = c1eat cos bt + c2eat sin bt .

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230

Chapter 4

Linear Second-Order Equations

Nonhomogeneous Linear Equations (Constant Coefficients) ay″ + by′ + cy = f1t2 General Solution to Nonhomogeneous Equation: yp + c1y1 + c2y2. If yp is any particular solution to the nonhomogeneous equation and y1 and y2 are linearly independent solutions to the corresponding homogeneous equation, then a general solution is y1t2 = yp 1t2 + c1y1 1t2 + c2y2 1t2 , where c1 and c2 are arbitrary constants. Two methods for finding a particular solution yp are those of undetermined coefficients and variation of parameters. cos Bt f . If the right-hand side f1t2 of a Undetermined Coefficients: f1t2 = pn 1t2eAt e sin Bt nonhomogeneous equation with constant coefficients is a polynomial pn 1t2, an exponential of the form eat, a trigonometric function of the form cos bt or sin bt, or any product of these special types of functions, then a particular solution of an appropriate form can be found. The form of the particular solution involves unknown coefficients and depends on whether a + ib is a root of the corresponding auxiliary equation. See the summary box on page 184. The unknown coefficients are found by substituting the form into the differential equation and equating coefficients of like terms.

Variation of Parameters: y1t2 = V1 1t2y1 1t2 + V2 1t2y2 1t2. If y1 and y2 are two linearly independent solutions to the corresponding homogeneous equation, then a particular solution to the nonhomogeneous equation is y1t2 = v1 1t2y1 1t2 + v2 1t2y2 1t2 ,

where

v1=

and v2= are determined by the equations v1= y1 + v2= y2 = 0 v1= y1= + v2= y2= = f 1t2 >a .

Superposition Principle.

If y1 and y2 are solutions to the equations

ay″ + by′ + cy = f1 and ay″ + by′ + cy = f2 , respectively, then k1y1 + k2y2 is a solution to the equation ay″ + by′ + cy = k1 f1 + k2 f2 . The superposition principle facilitates finding a particular solution when the nonhomogeneous term is the sum of nonhomogeneities for which particular solutions can be determined.

Cauchy–Euler (Equidimensional) Equations at 2y″ + bty′ + cy = f1t2 Substituting y = t r yields the associated characteristic equation ar 2 + 1b - a2r + c = 0

for the corresponding homogeneous Cauchy–Euler equation. A general solution to the homogeneous equation for t 7 0 is given by (i) c1t r1 + c2 t r2, if r1 and r2 are distinct real roots; (ii) c1t r + c2 t r ln t, if r is a repeated root; (iii) c1t a cos1b ln t2 + c2 t a sin1b ln t2, if a + ib is a complex root.

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Review Problems for Chapter 4

231

A general solution to the nonhomogeneous equation is y = yp + yh, where yp is a particular solution and yh is a general solution to the corresponding homogeneous equation. The method of variation of parameters (but not the method of undetermined coefficients) can be used to find a particular solution.

REVIEW PROBLEMS FOR CHAPTER 4 In Problems 1–28, find a general solution to the given differential equation. 1. y″ + 8y′ - 9y = 0 2. 49y″ + 14y′ + y = 0 3. 4y″ - 4y′ + 10y = 0 4. 9y″ - 30y′ + 25y = 0 5. 6y″ - 11y′ + 3y = 0 6. y″ + 8y′ - 14y = 0 7. 36y″ + 24y′ + 5y = 0 8. 25y″ + 20y′ + 4y = 0 9. 16z″ - 56z′ + 49z = 0 10. u″ + 11u = 0 2 11. t x″1t2 + 5x1t2 = 0 , t70 12. 2y‴ - 3y″ - 12y′ + 20y = 0 13. y″ + 16y = tet 14. v″ - 4v′ + 7v = 0 15. 3y‴ + 10y″ + 9y′ + 2y = 0 16. y‴ + 3y″ + 5y′ + 3y = 0 17. y‴ + 10y′ - 11y = 0 18. y 142 = 120t 19. 4y‴ + 8y″ - 11y′ + 3y = 0 20. 2y″ - y = t sin t 21. y″ - 3y′ + 7y = 7t 2 - et 22. y″ - 8y′ - 33y = 546 sin t 23. y″1u2 + 16y1u2 = tan 4u 24. 10y″ + y′ - 3y = t - et>2 25. 4y″ - 12y′ + 9y = e5t + e3t 26. y″ + 6y′ + 15y = e2t + 75 27. x2y″ + 2xy′ - 2y = 6x -2 + 3x , x70 28. y″ = 5x -1y′ - 13x -2y , x70 In Problems 29–36, find the solution to the given initial value problem. 29. y″ + 4y′ + 7y = 0 ; y′102 = -2 y102 = 1 , 30. y″1u2 + 2y′1u2 + y1u2 = 2 cos u ; y′102 = 0 y102 = 3 , 31. y″ - 2y′ + 10y = 6 cos 3t - sin 3t ; y′102 = -8 y102 = 2 ,

32. 4y″ - 4y′ + 5y = 0 ; y102 = 1 , y′102 = - 11>2 33. y‴ - 12y″ + 27y′ + 40y = 0 ; y102 = -3 , y′102 = - 6 , y″102 = - 12 34. y″ + 5y′ - 14y = 0 ; y102 = 5 , y′102 = 1 35. y″1u2 + y1u2 = sec u ; y102 = 1 , y′102 = 2 36. 9y″ + 12y′ + 4y = 0 ; y102 = -3 , y′102 = 3 37. Use the mass–spring oscillator analogy to decide whether all solutions to each of the following differential equations are bounded as t S + ∞ . (a) y″ + t 4y = 0 (b) y″ - t 4y = 0 (c) y″ + y7 = 0 (d) y″ + y8 = 0 (e) y″ + 13 + sin t2y = 0 (f) y″ + t 2y′ + y = 0 (g) y″ - t 2y′ - y = 0 38. A 3-kg mass is attached to a spring with stiffness k = 75 N/m, as in Figure 4.1, page 152. The mass is displaced 1>4 m to the left and given a velocity of 1 m/sec to the right. The damping force is negligible. Find the equation of motion of the mass along with the amplitude, period, and frequency. How long after release does the mass pass through the equilibrium position? 39. A 32-lb weight is attached to a vertical spring, causing it to stretch 6 in. upon coming to rest at equilibrium. The damping constant for the system is 2 lb-sec/ft. An external force F1t2 = 4 cos 8t lb is applied to the weight. Find the steady-state solution for the system. What is its resonant frequency?

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232

Chapter 4

Linear Second-Order Equations

TECHNICAL WRITING EXERCISES FOR CHAPTER 4 1. Compare the two methods—undetermined coefficients and variation of parameters—for determining a particular solution to a nonhomogeneous equation. What are the advantages and disadvantages of each? 2. Consider the differential equation d 2y dx2

+ 2b

dy +y = 0, dx

where b is a constant. Describe how the behavior of solutions to this equation changes as b varies. 3. Consider the differential equation d 2y dx2

where c is a constant. Describe how the behavior of solutions to this equation changes as c varies. 4. For students with a background in linear algebra: Compare the theory for linear second-order equations with that for systems of n linear equations in n unknowns whose coefficient matrix has rank n − 2. Use the terminology from linear algebra; for example, subspace, basis, dimension, linear transformation, and kernel. Discuss both homogeneous and nonhomogeneous equations.

+ cy = 0 ,

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Projects for Chapter 4 A Nonlinear Equations Solvable by First-Order Techniques Certain nonlinear second-order equations—namely, those with dependent or independent variables missing—can be solved by reducing them to a pair of first-order equations. This is accomplished by making the substitution w = dy>dx, where x is the independent variable. (a) To solve an equation of the form y″ = F1x, y′2 in which the dependent variable y is missing, setting w = y′ (so that w′ = y″) yields the pair of equations w′ = F1x, w2 , y′ = w .

Because w′ = F1x, w2 is a first-order equation, we have available the techniques of Chapter 2 to solve it for w1x2. Once w1x2 is determined, we integrate it to obtain y1x2. Using this method, solve 2xy″ - y′ +

1 = 0, y′

x70.

(b) To solve an equation of the form y″ = F1y, y′2 in which the independent variable x is missing, setting w = dy>dx yields, via the chain rule, d 2y dx

2

=

dw dw dy dw = = w . dx dy dx dy

Thus, y″ = F1y, y′2 is equivalent to the pair of equations dw = F1y, w2 , dy dy = w. dx

(1)

w

(2)

In equation (1) notice that y plays the role of the independent variable; hence, solving it yields w1y2. Then substituting w1y2 into (2), we obtain a separable equation that determines y1x2. Using this method, solve the following equations: (i)

2y

d 2y dx

2

= 1+ a

dy 2 b . dx

(ii)

d 2y dx

2

+y

dy = 0. dx

(c) Suspended Cable. In the study of a cable suspended between two fixed points (see Figure 4.36 on page 234), one encounters the initial value problem d 2y dx

2

=

dy 2 1 1+ a b ; aB dx

y102 = a ,

y′102 = 0 ,

where a 1 ≠ 02 is a constant. Solve this initial value problem for y. The resulting curve is called a catenary.

233

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234

Chapter 4

Linear Second-Order Equations

y

a x Figure 4.36 Suspended cable

B Apollo Reentry Courtesy of Alar Toomre, Massachusetts Institute of Technology

Each time the Apollo astronauts returned from the moon circa 1970, they took great care to reenter Earth’s atmosphere along a path that was only a small angle a from the horizontal. (See Figure 4.37.) This was necessary in order to avoid intolerably large “g” forces during their reentry. To appreciate their grounds for concern, consider the idealized problem 2 d 2s s >H ds Ke = a , b dt dt 2

where K and H are constants and distance s is measured downrange from some reference point on the trajectory, as shown in the figure. This approximate equation pretends that the only force on the capsule during reentry is air drag. For a bluff body such as the Apollo, drag is proportional to the square of the speed and to the local atmospheric density, which falls off exponentially with height. Intuitively, one might expect that the deceleration predicted by this model would depend heavily on the constant K (which takes into account the vehicle’s mass, area, etc.); but, remarkably, for capsules entering the atmosphere (at “s = - ∞ ”) with a common speed V0, the maximum deceleration turns out to be independent of K. (a) Verify this last assertion by demonstrating that this maximum deceleration is just V 20 > 12eH2. [Hint: The independent variable t does not appear in the differential equation, so it is helpful to make the substitution v = ds>dt; see Project A, part (b).] (b) Also verify that any such spacecraft at the instant when it is decelerating most fiercely will be traveling exactly with speed V0 > 1e, having by then lost almost 40% of its original velocity. (c) Using the plausible data V0 = 11 km/sec and H = 10/1sin a2 km, estimate how small a had to be chosen so as to inconvenience the returning travelers with no more than 10 g’s. s=

o

Di

sta

nce

s

a

Figure 4.37 Reentry path

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Projects for Chapter 4

235

C Simple Pendulum In Section 4.8, we discussed the simple pendulum consisting of a mass m suspended by a rod of length ℓ having negligible mass and derived the nonlinear initial value problem (3)

g d 2U + sin U = 0 ; U102 = A , U′102 = 0 , 2 O dt

where g is the acceleration due to gravity and u1t2 is the angle the rod makes with the vertical at time t (see Figure 4.18, page 208). Here it is assumed that the mass is released with zero velocity at an initial angle a, 0 6 a 6 p. We would like to determine the equation of motion for the pendulum and its period of oscillation. (a) Use equation (3) and the energy integral lemma discussed in Section 4.8 to show that a

2g du 2 b = 1cos u - cos a2 dt /

and hence dt = -

/ A 2g

du

3 cos u - cos a

.

(b) Use the trigonometric identity cos x = 1 - 2 sin2 1x>22 to express dt by dt = -

1 / 2Ag

du

2 3 sin 1a>22 - sin 1u>22 2

.

(c) Make the change of variables sin 1u>22 = sin 1a>22 sin f and show that the elapsed time, T, for the pendulum to fall from the angle u = a (corresponding to f = p>2) to the angle u = b (corresponding to f = Φ), when a Ú b Ú 0, is given by T =

L0

Φ df 1 / du / = , 2 2 B g Lp/2 21 - k 2 sin2f La 2 B g 2 sin 1a>22 - sin 1u>22 b

T

(4)

dt = -

where k J sin 1a>22.

(d) The period P of the pendulum is defined to be the time required for it to swing from one extreme to the other and back—that is, from a to - a and back to a. Show that the period is given by (5)

P = 4

/ Bg L 0

p>2

df 21 - k 2 sin 2f

.

The integral in (5) is called an elliptic integral of the first kind and is denoted by F1k, p>22. As you might expect, the period of the simple pendulum depends on the length / of the rod and the initial displacement a. In fact, a check of an elliptic integral table will show that the period nearly doubles as the initial displacement increases from p>8 to 15p>16 (for fixed /). What happens as a approaches p? (e) From equation (5) show that

(6)

- T + P>4 =

Φ df / F1k, Φ2, where F1k, Φ2 J . Bg Lf = 0 21 - k 2 sin2f

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236

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Linear Second-Order Equations

For fixed k, F1k, Φ2 has an “inverse,” denoted by sn1k, u2, that satisfies u = F1k, Φ2 if and only if sn1k, u2 = sin Φ. The function sn1k, u2 is called a Jacobi elliptic function and has many properties that resemble those of the sine function. Using the Jacobi elliptic function sn1k, u2, express the equation of motion for the pendulum in the form

(7)

b = 2 arcsin e k sn c k,

g 1 -T + P>42 d f, 0 … T … P>4 . A/

(f) Take / = 1 m, g = 9.8 m/sec, a = p>4 radians. Use Runge–Kutta algorithms or tabulated values of the Jacobi elliptic function to determine the period of the pendulum.

D Linearization of Nonlinear Problems A useful approach to analyzing a nonlinear equation is to study its linearized equation, which is obtained by replacing the nonlinear terms by linear approximations. For example, the nonlinear equation

(8)

d 2u + sin u = 0 , dt 2

which governs the motion of a simple pendulum, has

(9)

d 2u +u = 0 dt 2

as a linearization for small u. (The nonlinear term sin u has been replaced by the linear approximation u.) A general solution to equation (8) involves Jacobi elliptic functions (see Project C), which are rather complicated, so let’s try to approximate the solutions. For this purpose we consider two methods: Taylor series and linearization. (a) Derive the first six terms of the Taylor series about t = 0 of the solution to equation (8) with initial conditions u102 = p>12, u′102 = 0. (The Taylor series method is discussed in Project D of Chapter 1 and Section 8.1.) (b) Solve equation (9) subject to the same initial conditions u102 = p>12, u′102 = 0. (c) On the same coordinate axes, graph the two approximations found in parts (a) and (b). (d) Discuss the advantages and disadvantages of the Taylor series method and the linearization method. (e) Give a linearization for the initial value problem. x″1t2 + 0.131 - x2 1t24x′1t2 + x1t2 = 0

x102 = 0.4 , x′102 = 0 ,

for x small. Solve this linearized problem to obtain an approximation for the nonlinear problem.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Projects for Chapter 4

237

E Convolution Method The convolution of two functions g and f is the function g * f defined by 1g * f21t2 J

t

L0

g1t - V2f1V2 dV .

The aim of this project is to show how convolutions can be used to obtain a particular solution to a nonhomogeneous equation of the form (10)

ay″ + by′ + cy = f1t2 , where a, b, and c are constants, a ≠ 0. (a) Use Leibniz’s rule, t

t

d 0h h1t, v2 dv = 1t, v2 dv + h1t, t2 , dt La La 0t to show the following: 1y * f2′1t2 = 1y′ * f21t2 + y102f1t2 1 y * f2″1t2 = 1y″* f21t2 + y′102f1t2 + y102f′1t2 , assuming y and f are sufficiently differentiable. (b) Let ys 1t2 be the solution to the homogeneous equation ay″ + by′ + cy = 0 that satisfies ys 102 = 0, ys= 102 = 1>a. Show that ys * f is the particular solution to equation (10) satisfying y102 = y′102 = 0. (c) Let yk 1t2 be the solution to the homogeneous equation ay″ + by′ + cy = 0 that satisfies y102 = Y0, y′102 = Y1, and let ys be as defined in part (b). Show that 1ys * f21t2 + yk 1t2

is the unique solution to the initial value problem (11)

ay″ + by′ + cy = f1t2 ; y102 = Y0 , y′102 = Y1 .

(d) Use the result of part (c) to determine the solution to each of the following initial value problems. Carry out all integrations and express your answers in terms of elementary functions. (i) y″ + y = tan t ; y102 = 0 , y′102 = - 1 (ii) 2y″ + y′ - y = e-t sin t ; y102 = 1 , y′102 = 1 (iii) y″ - 2y′ + y = 1t et ; y102 = 2 , y′102 = 0

F Undetermined Coefficients Using Complex Arithmetic The technique of undetermined coefficients described in Section 4.5 can be streamlined with the aid of complex arithmetic and the properties of the complex exponential function. The essential formulas are d 1a+ib2 t e = 1a + ib2e1a+ib2 t , e1a+ib2 t = eat 1cos bt + i sin bt2 , dt Re e1a+ib2 t = eat cos bt ,

Im e1a+ib2 t = eat sin bt .

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238

Chapter 4

Linear Second-Order Equations

(a) From the preceding formulas derive the equations (12)

Re31a + ib2e1a+ib2t 4 = eat 1a cos bt - b sin bt2 ,

(13)

Im31a + ib2e1a+ib2t 4 = eat 1b cos bt + a sin bt2 . Now consider a second-order equation of the form L3y4 J ay″ + by′ + cy = g ,

(14)

where a, b, and c are real numbers and g is of the special form g1t2 = eAt 31ant n + P + a1t + a0 2 cos Bt + 1bnt n + P + b1t + b0 2 sin Bt4 ,

(15)

with the aj’s, bj’s, a, and b real numbers. Such a function can always be expressed as the real or imaginary part of a function involving the complex exponential. For example, using equation (12), one can quickly check that g1t2 = Re3G1t24 ,

(16) where (17)

G1t2 = e1A − iB2t 31an + ibn 2t n + P + 1a1 + ib1 2t + 1a0 + ib0 24 .

Now suppose for the moment that we can find a complex-valued solution Y to the equation (18)

L3Y4 = aY″ + bY′ + cY = G .

Then, since a, b, and c are real numbers, we get a real-valued solution y to (14) by simply taking the real part of Y; that is, y = Re Y solves (14). (Recall that in Lemma 2, page 167, we proved this fact for homogeneous equations.) Thus, we need focus only on finding a solution to (18). The method of undetermined coefficients implies that any differential equation of the form (19)

L3Y4 = e1a { ib2t 31an + ibn 2t n + g + 1a1 + ib1 2t + 1a0 + ib0 24

has a solution of the form (20)

Yp 1t2 = t se1a { ib2t 3Ant n + g + A1t + A0 4 ,

where An, . . ., A0 are complex constants and s is the multiplicity of a + ib as a root of the auxiliary equation for the corresponding homogeneous equation L3Y4 = 0. We can solve for the unknown constants Aj by substituting (20) into (19) and equating coefficients of like terms. With these facts in mind, we can (for the small price of using complex arithmetic) dispense with the methods of Section 4.5 and avoid the unpleasant task of computing derivatives of a function like e3t 12 + 3t + t 2 2sin12t2, which involves both exponential and trigonometric factors. Carry out this procedure to determine particular solutions to the following equations: (b) y″ - y′ - 2y = cos t - sin 2t . (c) y″ + y = e-t 1cos 2t - 3 sin 2t2 . (d) y″ - 2y′ + 10y = tet sin 3t . The use of complex arithmetic not only streamlines the computations but also proves very useful in analyzing the response of a linear system to a sinusoidal input. Electrical engineers make good use of this in their study of RLC circuits by introducing the concept of impedance.

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Projects for Chapter 4

239

G Asymptotic Behavior of Solutions In the application of linear systems theory to mechanical problems, we have encountered the equation (21)

y″ + py′ + qy = f1t2 ,

where p and q are positive constants with p2 6 4q and f1t2 is a forcing function for the system. In many cases it is important for the design engineer to know that a bounded forcing function gives rise only to bounded solutions. More specifically, how does the behavior of f1t2 for large values of t affect the asymptotic behavior of the solution? To answer this question, do the following: (a) Show that the homogeneous equation associated with equation (21) has two linearly independent solutions given by eat cos bt , eat sin bt , where a = - p>2 6 0 and b = 12 24q - p2 .

(b) Let f1t2 be a continuous function defined on the interval 30, ∞ 2. Use the variation of parameters formula to show that any solution to (21) on 30, ∞ 2 can be expressed in the form (22) y1t2 = c1eat cos bt + c2eat sin bt t

1 at e cos bt f1v2e-av sin bv dv b L0 t 1 at + e sin bt f1v2e-av cos bv dv . b L0 -

(c) Assuming that f is bounded on 30, ∞ 2 (that is, there exists a constant K such that 0 f1v 2 0 … K for all v Ú 0), use the triangle inequality and other properties of the absolute value to show that y1t2 given in (22) satisfies 2K 0 y1t2 0 … 1 0 c1 0 + 0 c2 0 2eat + 11 - eat 2 0a0b for all t 7 0.

(d) In a similar fashion, show that if f1 1t2 and f2 1t2 are two bounded continuous functions on 30, ∞ 2 such that 0 f1 1t2 - f2 1t2 0 … e for all t 7 t0, and if f1 is a solution to (21) with f = f1 and f2 is a solution to (21) with f = f2, then 2e 0 f1 1t2 - f2 1t2 0 … Meat + 11 - ea1t-t02 2 0a0b for all t 7 t0, where M is a constant that depends on f1 and f2 but not on t.

(e) Now assume f1t2 S F0 as t S + ∞ , where F0 is a constant. Use the result of part (d) to prove that any solution f to (21) must satisfy f1t2 S F0 >q as t S + ∞ . [Hint: Choose f1 = f, f2 K F0, f1 = f, f2 K F0 >q.]

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240

Chapter 4

Linear Second-Order Equations

H GRAVITY TRAIN† Legend has it that Robert Hooke proposed to Isaac Newton that a straight tunnel be drilled from Moscow to St. Petersburg to accommodate a “gravity train” serving the two cities. Upon departure gravity would accelerate the train toward the center of the tunnel, then decelerate it as it continued toward the terminal. It is amusing to investigate the mechanics of such a conveyance, under idealized conditions. As depicted in Figure 4.38, x denotes the distance of the train from the center of the tunnel, R denotes its distance from the center of the earth, and mg denotes the pull of gravity. (a) Show that Newton’s law yields the differential equation

(23)

mx″ = - mgx>R

if friction is neglected. (b) Assume that the tunnel lies so close to the surface of the earth that it is reasonable to regard g and R as constants—in which case (23) is analogous to the undamped mass– spring equation (2) on page 212 of Section 4.9, with mg/R as the effective “spring constant” k. Take g = 9.8 km/s2 and R = 6400 km. The overland (surface) distance from Moscow to St. Petersburg is about 650 km. Find the equation of motion for the train departing at rest from Moscow at time t = 0. (c) How long will it take the train to get to St. Petersburg? (d) The American mathematician/entertainer Tom Lehrer composed a humorous song‡ about Nicolai Ivanovich Lobachevsky, in which a plagiarized dissertation travels from Dnepropetrovsk to Petropavlovsk (16,000 km) to Iliysk (8,500 km) to Novorossiysk (17,000 km) to Alexandrovsk (15,000 km) to Akmolinsk (1,000 km) to Tomsk (1,300 km) to Omsk (900 km) to Pinsk (21,000 km) to Minsk (3,000 km) to Moscow (5,000 km). If gravity trains were to connect all these cities, what would be unusual about their timetables? y

x(t)

Санкт-Петербург

Москва m∙g LR x

Figure 4.38 Force diagram of gravity train

†

Suggested by Valery Ochkov, National Research University Moscow; and Katarina Pisacic, Sveucilisni Centar Varazdin. ‡

As we go to press the song is archived at https://en.wikipedia.org/wiki/Lobachevsky_%28song%29, and Lehrer's performance can be viewed at https://www.youtube.com/watch?v=UQHaGhC7C2E.

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CHAPTER

Introduction to Systems and Phase Plane Analysis

5

5.1 Interconnected Fluid Tanks Two large tanks, each holding 24 liters of a brine solution, are interconnected by pipes as shown in Figure 5.1. Fresh water flows into tank A at a rate of 6 L/min, and fluid is drained out of tank B at the same rate; also 8 L/min of fluid are pumped from tank A to tank B, and 2 L/min from tank B to tank A. The liquids inside each tank are kept well stirred so that each mixture is homogeneous. If, initially, the brine solution in tank A contains x0 kg of salt and that in tank B initially contains y0 kg of salt, determine the mass of salt in each tank at time t + 0.†

6 L/min

A

B

8 L/min

x(t)

y(t)

24 L

24 L

x (0) 5 x0 kg

y (0) 5 y0 kg

6 L/min

2 L/min Figure 5.1 Interconnected fluid tanks

Note that the volume of liquid in each tank remains constant at 24 L because of the balance between the inflow and outflow volume rates. Hence, we have two unknown functions of t: the mass of salt x1t2 in tank A and the mass of salt y1t2 in tank B. By focusing attention on one tank at a time, we can derive two equations relating these unknowns. Since the system is being flushed with freshwater, we expect that the salt content of each tank will diminish to zero as t S + ∞. To formulate the equations for this system, we equate the rate of change of salt in each tank with the net rate at which salt is transferred to that tank. The salt concentration in tank A is x1t2 >24 kg/L, so the upper interconnecting pipe carries salt out of tank A at a rate of 8x>24 kg/min; similarly, the lower interconnecting pipe brings salt into tank A at the rate 2y>24 kg/min (the concentration of salt in tank B is y>24 kg/L). The freshwater inlet, of course, transfers no salt (it simply maintains the volume in tank A at 24 L). From our premise, dx = input rate − output rate , dt †

For this application we simplify the analysis by assuming the lengths and volumes of the pipes are sufficiently small that we can ignore the diffusive and advective dynamics taking place therein.

241

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so the rate of change of the mass of salt in tank A is 8 1 1 dx 2 = yx = y- x. dt 24 24 12 3 The rate of change of salt in tank B is determined by the same interconnecting pipes and by the drain pipe, carrying away 6y>24 kg/min: dy 2 6 1 1 8 = xyy = x- y. dt 24 24 24 3 3 The interconnected tanks are thus governed by a system of differential equations:

(1)

1 1 x′ = - x + y, 3 12 1 1 y′ = x - y . 3 3

Although both unknowns x1t2 and y1t2 appear in each of equations (1) (they are “coupled”), the structure is so transparent that we can obtain an equation for y alone by solving the second equation for x, (2)

x = 3y′ + y ,

and substituting (2) in the first equation to eliminate x: 1 1 13y′ + y2′ = - 13y′ + y2 + y, 3 12 1 1 3y″ + y′ = -y′ - y + y, 3 12 or 3y″ + 2y′ +

1 y = 0. 4

This last equation, which is linear with constant coefficients, is readily solved by the methods of Section 4.2. Since the auxiliary equation 3r 2 + 2r +

1 = 0 4

has roots -1>2, -1>6, a general solution is given by (3)

y1t2 = c1e-t>2 + c2e-t>6 .

Having determined y, we use equation (2) to deduce a formula for x: (4)

x1t2 = 3a -

c1 -t>2 c2 -t>6 1 1 e - e b + c1e-t>2 + c2e-t>6 = - c1e-t>2 + c2e-t>6 . 2 6 2 2

Formulas (3) and (4) contain two undetermined parameters, c1 and c2, which can be adjusted to meet the specified initial conditions: 1 1 x102 = - c1 + c2 = x0 , 2 2

y102 = c1 + c2 = y0 ,

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Section 5.2

Differential Operators and the Elimination Method* for Systems

243

or

y0 - 2x0 y0 + 2x0 , c2 = . 2 2 Thus, the mass of salt in tanks A and B at time t are, respectively, c1 =

x1t2 = - a (5)

y1t2 = a

y0 - 2x0 -t>2 y0 + 2x0 -t>6 be + a be , 4 4

y0 - 2x0 -t>2 y0 + 2x0 -t>6 be + a be . 2 2

The ad hoc elimination procedure that we used to solve this example will be generalized and formalized in the next section, to find solutions of all linear systems with constant coefficients. Furthermore, in later sections we will show how to extend our numerical algorithms for first-order equations to general systems and will consider applications to coupled oscillators and electrical systems. It is interesting to note from (5) that all solutions of the interconnected-tanks problem tend to the constant solution x1t2 K 0, y1t2 K 0 as t S + ∞ . (This is of course consistent with our physical expectations.) This constant solution will be identified as a stable equilibrium solution in Section 5.4, in which we introduce phase plane analysis. It turns out that, for a general class of systems, equilibria can be identified and classified so as to give qualitative information about the other solutions even when we cannot solve the system explicitly.

5.2 Differential Operators and

the Elimination Method* for Systems dy d = y was devised to suggest that the derivative of a function y is the dt dt d result of operating on the function y with the differentiation operator . Indeed, second deriva2 dt dy d d tives are formed by iterating the operation: y″(t) = 2 = y. Commonly, the symbol D dt dt dt d is used instead of , and the second-order differential equation dt y″ + 4y′ + 3y = 0 The notation y′(t) =

is represented† by

D2y + 4Dy + 3y = 1D2 + 4D + 323y4 = 0 .

So, we have implicitly adopted the convention that the operator “product,” D times D, is interpreted as the composition of D with itself when it operates on functions: D2y means D1D3y4 2; i.e., the second derivative. Similarly, the product 1D + 321D + 12 operates on a function via 1D + 321D + 123y4 = 1D + 32 3 1D + 123y4 4 = 1D + 323y′ + y4 = D3y′ + y4 + 33y′ + y4 = 1y″ + y′2 + 13y′ + 3y2 = y″ + 4y′ + 3y = 1D2 + 4D + 323y4 .

*An alternative procedure to the methodology of this section will be described in Chapter 9. Although it involves the machinery of matrix analysis, it is preferable for large systems. † Some authors utilize the identity operator I, defined by I3y4 = y, and write more formally D2 + 4D + 3I instead of D2 + 4D + 3.

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244

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Introduction to Systems and Phase Plane Analysis

Thus, 1D + 321D + 12 is the same operator as D2 + 4D + 3; when they are applied to twice-differentiable functions, the results are identical. Example 1 Solution

Show that the operator 1D + 121D + 32 is also the same as D2 + 4D + 3. For any twice-differentiable function y1t2, we have

1D + 121D + 323y4 = 1D + 12 3 1D + 323y4 4 = 1D + 123y′ + 3y4 = D3y′ + 3y4 + 13y′ + 3y4 = 1y″ + 3y′2 + 1y′ + 3y2 = y″ + 4y′ + 3y = 1D2 + 4D + 323y4 .

Hence, 1D + 121D + 32 = D2 + 4D + 3. ◆

Since 1D + 121D + 32 = 1D + 321D + 12 = D2 + 4D + 3, it is tempting to generalize and propose that one can treat expressions like aD2 + bD + c as if they were ordinary polynomials in D. This is true, as long as we restrict the coefficients a, b, c to be constants. The following example, which has variable coefficients, is instructive. Example 2 Solution

Show that 1D + 3t2D is not the same as D1D + 3t2. With y1t2 as before,

1D + 3t2D3y4 = 1D + 3t23y′4 = y″ + 3ty′ ;

D1D + 3t23y4 = D3y′ + 3ty4 = y″ + 3y + 3ty′ . They are not the same! ◆ Because the coefficient 3t is not a constant, it “interrupts” the interaction of the differentiation operator D with the function y1t2. As long as we only deal with expressions like aD2 + bD + c with constant coefficients a, b, and c, the “algebra” of differential operators follows the same rules as the algebra of polynomials. (See Problem 39 for elaboration on this point.) This means that the familiar elimination method, used for solving algebraic systems like 3x - 2y + z = 4 , x+y-z = 0, 2x - y + 3z = 6 , can be adapted to solve any system of linear differential equations with constant coefficients. In fact, we used this approach in solving the system that arose in the interconnected tanks problem of Section 5.1. Our goal in this section is to formalize this elimination method so that we can tackle more general linear constant coefficient systems. We first demonstrate how the method applies to a linear system of two first-order differential equations of the form a1x′1t2 + a2x1t2 + a3y′1t2 + a4y1t2 = f1 1t2 , a5x′1t2 + a6x1t2 + a7y′1t2 + a8y1t2 = f2 1t2 ,

where a1, a2, . . . , a8 are constants and x1t2, y1t2 is the function pair to be determined. In operator notation this becomes 1a1D + a2 23x4 + 1a3D + a4 23y4 = f1 , 1a5D + a6 23x4 + 1a7D + a8 23y4 = f2 .

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Section 5.2

Example 3

245

Solve the system (1)

Solution

Differential Operators and the Elimination Method* for Systems

x′1t2 = 3x1t2 - 4y1t2 + 1 , y′1t2 = 4x1t2 - 7y1t2 + 10t .

The alert reader may observe that since y′ is absent from the first equation, we could use the latter to express y in terms of x and x′ and substitute into the second equation to derive an “uncoupled” equation containing only x and its derivatives. However, this simple trick will not work on more general systems (Problem 18 is an example). To utilize the elimination method, we first write the system using the operator notation: (2)

1D - 323x4 + 4y = 1 , -4x + 1D + 723y4 = 10t .

Imitating the elimination procedure for algebraic systems, we can eliminate x from this system by adding 4 times the first equation to 1D - 32 applied to the second equation. This gives

1 16 + 1D - 321D + 72 2 3y4

= 4 # 1 + 1D - 32310t4 = 4 + 10 - 30t ,

which simplifies to (3)

1D2 + 4D - 523y4 = 14 - 30t .

Now equation (3) is just a second-order linear equation in y with constant coefficients that has the general solution (4)

y1t2 = C1e-5t + C2et + 6t + 2 ,

which can be found using undetermined coefficients. To find x1t2, we have two options. Method 1. We return to system (2) and eliminate y. This is accomplished by “multiplying” the first equation in (2) by 1D + 72 and the second equation by -4 and then adding to obtain 1D2 + 4D - 523x4 = 7 - 40t .

This equation can likewise be solved using undetermined coefficients to yield (5)

x1t2 = K1e-5t + K2et + 8t + 5 ,

where we have taken K1 and K2 to be the arbitrary constants, which are not necessarily the same as C1 and C2 used in formula (4). It is reasonable to expect that system (1) will involve only two arbitrary constants, since it consists of two first-order equations. Thus, the four constants C1, C2, K1, and K2 are not independent. To determine the relationships, we substitute the expressions for x1t2 and y1t2 given in (4) and (5) into one of the equations in (1)—say, the first one. This yields -5K1e-5t + K2et + 8 = 3K1e-5t + 3K2et + 24t + 15 - 4C1e-5t - 4C2et - 24t - 8 + 1 , which simplifies to

14C1 - 8K1 2e-5t + 14C2 - 2K2 2et = 0 .

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246

Chapter 5

Introduction to Systems and Phase Plane Analysis

Because et and e-5t are linearly independent functions on any interval, this last equation holds for all t only if 4C1 - 8K1 = 0

and

4C2 - 2K2 = 0 .

Therefore, K1 = C1 >2 and K2 = 2C2. A solution to system (1) is then given by the pair 1 (6) x1t2 = C1e-5t + 2C2et + 8t + 5 , y1t2 = C1e-5t + C2et + 6t + 2 . 2 As you might expect, this pair is a general solution to (1) in the sense that any solution to (1) can be expressed in this fashion. Method 2. A simpler method for determining x1t2 once y1t2 is known is to use the system to obtain an equation for x1t2 in terms of y1t2 and y′1t2. In this example we can directly solve the second equation in (1) for x1t2: x1t2 =

1 7 5 y′1t2 + y1t2 - t . 4 4 2

Substituting y1t2 as given in (4) yields 5 1 7 3 -5C1e-5t + C2et + 64 + 3C1e-5t + C2et + 6t + 24 - t 4 4 2 1 = C1e-5t + 2C2et + 8t + 5 , 2

x1t2 =

which agrees with (6). ◆ The above procedure works, more generally, for any linear system of two equations and two unknowns with constant coefficients regardless of the order of the equations. For example, if we let L 1, L 2, L 3, and L 4 denote linear differential operators with constant coefficients (i.e., polynomials in D), then the method can be applied to the linear system L 1 3x4 + L 2 3y4 = f1 , L 3 3x4 + L 4 3y4 = f2 .

Because the system has constant coefficients, the operators commute (e.g., L 2L 4 = L 4L 2) and we can eliminate variables in the usual algebraic fashion. Eliminating the variable y gives (7)

1L 1L 4 - L 2L 3 23x4 = g1 ,

where g1 J L 4[f1] - L 2[f2]. Similarly, eliminating the variable x yields (8)

1L 1L 4 - L 2L 3 23y4 = g2 ,

where g2 J L 1[ f2] - L 3[ f1]. Now if L 1L 4 - L 2L 3 is a differential operator of order n, then a general solution for (7) contains n arbitrary constants, and a general solution for (8) also contains n arbitrary constants. Thus, a total of 2n constants arise. However, as we saw in Example 3, there are only n of these that are independent for the system; the remaining constants can be expressed in terms of these.† The pair of general solutions to (7) and (8) written in terms of the n independent constants is called a general solution for the system.

†

For a proof of this fact, see Ordinary Differential Equations, by M. Tenenbaum and H. Pollard (Dover, New York, 1985), Chapter 7.

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Section 5.2

Differential Operators and the Elimination Method* for Systems

247

If it turns out that L 1L 4 - L 2L 3 is the zero operator, the system is said to be degenerate. As with the anomalous problem of solving for the points of intersection of two parallel or coincident lines, a degenerate system may have no solutions, or if it does possess solutions, they may involve any number of arbitrary constants (see Problems 23 and 24).

Elimination Procedure for 2 : 2 Systems To find a general solution for the system L 1 3x4 + L 2 3y4 = f1 , L 3 3x4 + L 4 3y4 = f2 ,

where L 1, L 2, L 3, and L 4 are polynomials in D = d>dt: (a) Make sure that the system is written in operator form. (b) Eliminate one of the variables, say, y, and solve the resulting equation for x1t2. If the system is degenerate, stop! A separate analysis is required to determine whether or not there are solutions. (c) (Shortcut) If possible, use the system to derive an equation that involves y1t2 but not its derivatives. [Otherwise, go to step (d).] Substitute the found expression for x1t2 into this equation to get a formula for y1t2. The expressions for x1t2, y1t2 give the desired general solution. ◆ (d) Eliminate x from the system and solve for y1t2. [Solving for y1t2 gives more constants—in fact, twice as many as needed.] (e) Remove the extra constants by substituting the expressions for x1t2 and y1t2 into one or both of the equations in the system. Write the expressions for x1t2 and y1t2 in terms of the remaining constants. ◆

Example 4

Find a general solution for (9)

Solution

x″1t2 + y′1t2 - x1t2 + y1t2 = -1 , x′1t2 + y′1t2 - x1t2 = t2 .

We begin by expressing the system in operator notation: (10)

1D2 - 123x4 + 1D + 123y4 = -1 , 1D - 123x4 + D3y4 = t 2 .

Here L 1 J D2 - 1, L 2 J D + 1, L 3 J D - 1, and L 4 J D. Eliminating y gives [see (7)]:

1 1D2 - 12D - 1D + 121D - 12 2 3x4

= D[-1] - 1D + 123t 2 4 ,

which reduces to (11)

1D2 - 121D - 123x4 = -2t - t 2 , or 1D - 12 2 1D + 123x4 = -2t - t 2 .

Since 1D - 12 2 1D + 12 is third order, we should expect three arbitrary constants in a general solution to system (9). Although the methods of Chapter 4 focused on solving second-order equations, we have seen several examples of how they extend in a natural way to higher-order

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248

Chapter 5

Introduction to Systems and Phase Plane Analysis

equations.† Applying this strategy to the third-order equation (11), we observe that the corresponding homogeneous equation has the auxiliary equation 1r - 12 2 1r + 12 = 0 with roots r = 1, 1, -1. Hence, a general solution for the homogeneous equation is xh 1t2 = C1et + C2tet + C3e-t .

To find a particular solution to (11), we use the method of undetermined coefficients with xp 1t2 = At 2 + Bt + C. Substituting into (11) and solving for A, B, and C yields (after a little algebra) xp 1t2 = -t 2 - 4t - 6 .

Thus, a general solution to equation (11) is (12)

x1t2 = xh 1t2 + xp 1t2 = C1et + C2tet + C3e-t - t 2 - 4t - 6 .

To find y1t2, we take the shortcut described in step (c) of the elimination procedure box. Subtracting the second equation in (10) from the first, we find 1D2 - D23x4 + y = -1 - t 2 ,

so that

y = 1D - D2 23x4 - 1 - t 2 .

Inserting the expression for x1t2, given in (12), we obtain

y1t2 = C1et + C2 1tet + et 2 - C3e-t - 2t - 4 - 3C1et + C2 1tet + 2et 2 + C3e-t - 24 - 1 - t 2 ,

(13)

or

y1t2 = -C2et - 2C3e-t - t 2 - 2t - 3 .

The formulas for x1t2 in (12) and y1t2 in (13) give the desired general solution to (9). ◆ The elimination method also applies to linear systems with three or more equations and unknowns; however, the process becomes more cumbersome as the number of equations and unknowns increases. The matrix methods presented in Chapter 9 are better suited for handling larger systems. Here we illustrate the elimination technique for a 3 * 3 system. Example 5

Find a general solution to (14)

Solution

x′1t2 = x1t2 + 2y1t2 - z1t2 , y′1t2 = x1t2 + z1t2 , z′1t2 = 4x1t2 - 4y1t2 + 5z1t2 .

We begin by expressing the system in operator notation: (15)

1D - 123x4 - 2y + z = 0 , -x + D3y4 - z = 0 , -4x + 4y + 1D - 523z4 = 0 .

Eliminating z from the first two equations (by adding them) and then from the last two equations yields (after some algebra, which we omit) (16)

1D - 223x4 + 1D - 223y4 = 0 , - 1D - 123x4 + 1D - 121D - 423y4 = 0 .

†

More detailed treatment of higher-order equations is given in Chapter 6.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 5.2

Differential Operators and the Elimination Method* for Systems

249

On eliminating x from this 2 * 2 system, we eventually obtain 1D - 121D - 221D - 323y4 = 0 ,

which has the general solution (17)

y1t2 = C1et + C2e2t + C3e3t .

Taking the shortcut approach, we add the two equations in (16) to get an expression for x in terms of y and its derivatives, which simplifies to x = 1D2 - 4D + 223y4 = y″ - 4y′ + 2y .

When we substitute the expression (17) for y1t2 into this equation, we find (18)

x1t2 = -C1et - 2C2e2t - C3e3t . Finally, using the second equation in (14) to solve for z1t2, we get z1t2 = y′1t2 - x1t2 ,

and substituting in for y1t2 and x1t2 yields (19)

z1t2 = 2C1et + 4C2e2t + 4C3e3t .

The expressions for x1t2 in (18), y1t2 in (17), and z1t2 in (19) give a general solution with C1, C2, and C3 as arbitrary constants. ◆

5.2

EXERCISES

1. Let A = D - 1, B = D + 2, C = D2 + D - 2, where D = d>dt. For y = t 3 - 8, compute (a) A3y4 (b) B3A3y44 (c) B3y4 (d) A3B3y44 (e) C3y4 2. Show that the operator 1D - 121D + 22 is the same as the operator D2 + D - 2. In Problems 3–18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t. 3. x′ + 2y = 0 , 4. x′ = x - y , x′ - y′ = 0 y′ = y - 4x 5. x′ + y′ - x = 5 , x′ + y′ + y = 1

6. x′ = 3x - 2y + sin t , y′ = 4x - y - cos t

7. 1D + 123u4 - 1D + 123y4 = et , 1D - 123u4 + 12D + 123y4 = 5

8. 1D - 323x4 + 1D - 123y4 = t , 1D + 123x4 + 1D + 423y4 = 1 9. x′ + y′ + 2x = 0 , x′ + y′ - x - y = sin t

10. 2x′ + y′ - x - y = e-t , x′ + y′ + 2x + y = et

11. 1D2 - 123u4 + 5y = et , 12. D2 3u4 + D3y4 = 2 , 2u + 1D2 + 223y4 = 0 4u + D3y4 = 6

dx = x - 4y , dt dy = x+y dt dw = 5w + 2z + 5t , 15. dt dz = 3w + 4z + 17t dt

13.

dx + y = t2 , dt dy -x + = 1 dt dy dx 16. +x+ = e4t , dt dt d 2y 2x + 2 = 0 dt

14.

17. x″ + 5x - 4y = 0 , -x + y″ + 2y = 0 18. x″ + y″ - x′ = 2t , x″ + y′ - x + y = -1 In Problems 19–21, solve the given initial value problem. dx 19. = 4x + y ; x102 = 1 , dt dy = - 2x + y ; y102 = 0 dt 20.

dx = 2x + y - e2t ; x102 = 1 , dt dy = x + 2y ; y102 = - 1 dt

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250

21.

Chapter 5

Introduction to Systems and Phase Plane Analysis

d 2x = y ; x102 = 3 , dt 2 2 dy = x ; y102 = 1 , dt 2

x′102 = 1 , y′102 = - 1

22. Verify that the solution to the initial value problem x′ = 5x - 3y - 2 ; x102 = 2 , y′ = 4x - 3y - 1 ; y102 = 0

satisfies 0 x1t2 0 + 0 y1t2 0 S + ∞ as t S + ∞ . In Problems 23 and 24, show that the given linear system is degenerate. In attempting to solve the system, determine whether it has no solutions or infinitely many solutions. 23. 1D - 123x4 + 1D - 123y4 = - 3e-2t , 1D + 223x4 + 1D + 223y4 = 3et

24. D3x4 + 1D + 123y4 = et , D2 3x4 + 1D2 + D23y4 = 0

In Problems 25–28, use the elimination method to find a general solution for the given system of three equations in the three unknown functions x1t2, y1t2, z1t2. 25. x′ = x + 2y - z , y′ = x + z , z′ = 4x - 4y + 5z

26. x′ = 3x + y - z , y′ = x + 2y - z , z′ = 3x + 3y - z

27. x′ = 4x - 4z , y′ = 4y - 2z , z′ = - 2x - 4y + 4z

28. x′ = x + 2y + z , y′ = 6x - y , z′ = -x - 2y - z

In Problems 29 and 30, determine the range of values (if any) of the parameter l that will ensure all solutions x1t2, y1t2 of the given system remain bounded as t S + ∞ . dx dx 29. = lx - y , = - x + ly , 30. dt dt dy dy = 3x + y = x-y dt dt

tank A into tank B at a rate of 3 L/min and from B into A at a rate of 1 L/min (see Figure 5.2). The liquid inside each tank is kept well stirred. A brine solution with a concentration of 0.2 kg/L of salt flows into tank A at a rate of 6 L/min. The (diluted) solution flows out of the system from tank A at 4 L/min and from tank B at 2 L/min. If, initially, tank A contains pure water and tank B contains 20 kg of salt, determine the mass of salt in each tank at time t Ú 0. 32. In Problem 31, 3 L/min of liquid flowed from tank A into tank B and 1 L/min from B into A. Determine the mass of salt in each tank at time t Ú 0 if, instead, 5 L/min flows from A into B and 3 L/min flows from B into A, with all other data the same. 33. In Problem 31, assume that no solution flows out of the system from tank B, only 1 L/min flows from A into B, and only 4 L/min of brine flows into the system at tank A, other data being the same. Determine the mass of salt in each tank at time t Ú 0. 34. Feedback System with Pooling Delay. Many physical and biological systems involve time delays. A pure time delay has its output the same as its input but shifted in time. A more common type of delay is pooling delay. An example of such a feedback system is shown in Figure 5.3 on page 251. Here the level of fluid in tank B determines the rate at which fluid enters tank A. Suppose this rate is given by R1 1t2 = a3V - V2 1t24, where a and V are positive constants and V2 1t2 is the volume of fluid in tank B at time t. (a) If the outflow rate R3 from tank B is constant and the flow rate R2 from tank A into B is R2 1t2 = KV1 1t2, where K is a positive constant and V1 1t2 is the volume of fluid in tank A at time t, then show that this feedback system is governed by the system dV1 = a 1 V - V2 1t2 2 - KV1 1t2 , dt dV2 = KV1 1t2 - R3 . dt

31. Two large tanks, each holding 100 L of liquid, are interconnected by pipes, with the liquid flowing from

6 L/min 0.2 kg/L

4 L/min

A

3 L/min

B

x(t)

y(t)

100 L

100 L

x (0) 5 0 kg

y (0) 5 20 kg

2 L/min

1 L/min Figure 5.2 Mixing problem for interconnected tanks

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Section 5.2

251

Differential Operators and the Elimination Method* for Systems

A

B

R1 Tank A Pump

4 hr

x(t)

2 hr

y(t)

5 hr

V1 (t) Power R2

Tank B

Figure 5.5 Two-zone building with one zone heated

V2 (t)

R3

Figure 5.3 Feedback system with pooling delay

(b) Find a general solution for the system in part (a) when a = 5 (min) -1, V = 20 L, K = 2 (min) -1, and R3 = 10 L/min. (c) Using the general solution obtained in part (b), what can be said about the volume of fluid in each of the tanks as t S + ∞ ? 35. A house, for cooling purposes, consists of two zones: the attic area zone A and the living area zone B (see Figure 5.4). The living area is cooled by a 2-ton air conditioning unit that removes 24,000 Btu/hr. The heat capacity of zone B is 1>2°F per thousand Btu. The time constant for heat transfer between zone A and the outside is 2 hr, between zone B and the outside is 4 hr, and between the two zones is 4 hr. If the outside temperature stays at 100°F, how warm does it eventually get in the attic zone A? (Heating and cooling of buildings was treated in Section 3.3 on page 102.) 36. A building consists of two zones A and B (see Figure 5.5). Only zone A is heated by a furnace, which generates 80,000 Btu/hr. The heat capacity of zone A is 1>4°F per thousand Btu. The time constant for heat transfer between

zone A and the outside is 4 hr, between the unheated zone B and the outside is 5 hr, and between the two zones is 2 hr. If the outside temperature stays at 0°F, how cold does it eventually get in the unheated zone B? 37. In Problem 36, if a small furnace that generates 1000 Btu/hr is placed in zone B, determine the coldest it would eventually get in zone B if zone B has a heat capacity of 2°F per thousand Btu. 38. Arms Race. A simplified mathematical model for an arms race between two countries whose expenditures for defense are expressed by the variables x1t2 and y1t2 is given by the linear system dx = 2y - x + a ; x102 = 1 , dt dy = 4x - 3y + b ; y102 = 4 , dt where a and b are constants that measure the trust (or distrust) each country has for the other. Determine whether there is going to be disarmament (x and y approach 0 as t increases), a stabilized arms race (x and y approach a constant as t S + ∞ ), or a runaway arms race (x and y approach + ∞ as t S + ∞ ). 39. Let A, B, and C represent three linear differential operators with constant coefficients; for example, A J a2D2 + a1D + a0 , B J b2D2 + b1D + b0 , C J c2D2 + c1D + c0 , where the a’s, b’s, and c’s are constants. Verify the following properties:†

2 hr

(a) Commutative laws:

A

A+B = B+A, AB = BA .

4 hr

4 hr B

24,000 Btu/hr

Figure 5.4 Air-conditioned house with attic

(b) Associative laws:

1A + B2 + C = A + 1B + C2 , 1AB2C = A1BC2 .

(c) Distributive law: A1B + C2 = AB + AC .

We say that two operators A and B are equal if A3y4 = B3y4 for all functions y with the necessary derivatives.

†

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252

Chapter 5

Introduction to Systems and Phase Plane Analysis

5.3 Solving Systems and Higher-Order Equations Numerically

Although we studied a half-dozen analytic methods for obtaining solutions to first-order ordinary differential equations in Chapter 2, the techniques for higher-order equations, or systems of equations, are much more limited. Chapter 4 focused on solving the linear constant-coefficient second-order equation. The elimination method of the previous section is also restricted to constant-coefficient systems. And, indeed, higher-order linear constantcoefficient equations and systems can be solved analytically by extensions of these methods, as we will see in Chapters 6, 7, and 9. However, if the equations—even a single second-order linear equation—have variable coefficients, the solution process is much less satisfactory. As will be seen in Chapter 8, the solutions are expressed as infinite series, and their computation can be very laborious (with the notable exception of the Cauchy–Euler, or equidimensional, equation). And we know virtually nothing about how to obtain exact solutions to nonlinear second-order equations. Fortunately, all the cases that arise (constant or variable coefficients, nonlinear, higherorder equations or systems) can be addressed by a single formulation that lends itself to a multitude of numerical approaches. In this section we’ll see how to express differential equations as a system in normal form and then show how the basic Euler method for computer solution can be easily “vectorized” to apply to such systems. Although subsequent chapters will return to analytic solution methods, the vectorized version of the Euler technique or the more efficient Runge–Kutta technique will hereafter be available as fallback methods for numerical exploration of intractable problems.

Normal Form

A system of m differential equations in the m unknown functions x1 1t2, x2 1t2, . . . , xm 1t2 expressed as (1)

x1= 1t2 = f1 1t, x1, x2, . . . , xm 2 , x2= 1t2 = f2 1t, x1, x2, . . . , xm 2 , O = xm 1t2 = fm 1t, x1, x2, . . . , xm 2

is said to be in normal form. Notice that (1) consists of m first-order equations that collectively look like a vectorized version of the single generic first-order equation (2)

x′ = f1t, x2 ,

and that the system expressed in equation (1) of Section 5.1 takes this form, as do equations (1) and (14) in Section 5.2. An initial value problem for (1) entails finding a solution to this system that satisfies the initial conditions x1 1t0 2 = a1, x2 1t0 2 = a2, . . . , xm 1t0 2 = am

for prescribed values t0, a1, a2, . . . , am. The importance of the normal form is underscored by the fact that most professional codes for initial value problems presume that the system is written in this form. Furthermore, for a linear system in normal form, the powerful machinery of linear algebra can be readily applied. [Indeed, in Chapter 9 we will show how the solutions x1t2 = ceat of the simple equation x′ = ax can be generalized to constant-coefficient systems in normal form.]

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Section 5.3

Solving Systems and Higher-Order Equations Numerically

253

For these reasons it is gratifying to note that a (single) higher-order equation can always be converted to an equivalent system of first-order equations. To convert an mth-order differential equation (3)

y 1m2 1t2 = f 1 t, y, y′, c, y 1m-12 2

into a first-order system, we introduce, as additional unknowns, the sequence of derivatives of y: x1 1t2 J y1t2, x2 1t2 J y′1t2, . . . , xm 1t2 J y 1m-12 1t2 .

With this scheme, we obtain m - 1 first-order equations quite trivially:

(4)

x′1 1t2 = y′1t2 = x2 1t2 , x′2 1t2 = y″1t2 = x3 1t2 , O x′m − 1 1t2 = y 1m − 12 1t2 = xm 1t2 .

The mth and final equation then constitutes a restatement of the original equation (3) in terms of the new unknowns: (5)

x′m 1t2 = y 1m2 1t2 = f1t, x1, x2, . . . , xm 2 .

If equation (3) has initial conditions y1t0 2 = a1, y′1t0 2 = a2, . . . , y1m-12 1t0 2 = am, then the system (4)–(5) has initial conditions x1 1t0 2 = a1, x2 1t0 2 = a2, . . . , xm 1t0 2 = am. Example 1

Convert the initial value problem (6)

y″1t2 + 3ty′1t2 + y1t2 2 = sin t ; y102 = 1 , y′102 = 5

into an initial value problem for a system in normal form. Solution

We first express the differential equation in (6) as y″1t2 = - 3ty′1t2 - y1t2 2 + sin t .

Setting x1 1t2 J y1t2 and x2 1t2 J y′1t2, we obtain x1= 1t2 = x2 1t2 , x2= 1t2 = -3tx2 1t2 - x1 1t2 2 + sin t .

The initial conditions transform to x1 102 = 1, x2 102 = 5 . ◆

Euler’s Method for Systems in Normal Form Recall from Section 1.4 that Euler’s method for solving a single first-order equation (2) is based on estimating the solution x at time 1t0 + h2 using the approximation (7)

x1t0 + h2 ≈ x1t0 2 + hx′1t0 2 = x1t0 2 + h f 1 t0, x1t0 2 2 ,

and that as a consequence the algorithm can be summarized by the recursive formulas (8)

tn + 1 = tn + h ,

(9)

xn + 1 = xn + h f1tn, xn 2,

n = 0, 1, 2, . . .

[compare equations (2) and (3), Section 1.4]. Now we can apply the approximation (7) to each of the equations in the system (1): (10)

xk 1t0 + h2 ≈ xk 1t0 2 + hxk= 1t0 2 = xk 1t0 2 + h fk 1 t0, x1 1t0 2, x2 1t0 2, . . . , xm 1t0 2 2 ,

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254

Chapter 5

Introduction to Systems and Phase Plane Analysis

and for k = 1, 2, . . . m, we are led to the recursive formulas (11)

(12)

tn + 1 = tn + h ,

x1;n + 1 = x1;n + h f1 1tn, x1;n, x2;n, . . . , xm;n 2 , x2;n + 1 = x2;n + h f2 1tn, x1;n, x2;n, . . . , xm;n 2 , O xm;n + 1 = xm;n + h fm 1tn, x1;n, x2;n, . . . , xm;n 2

1n = 0, 1, 2, c2 .

Here we are burdened with the ungainly notation xp;n for the approximation to the value of the pth-function xp at time t = t0 + nh; i.e., xp;n ≈ xp 1t0 + nh2. However, if we treat the unknowns and right-hand members of (1) as components of vectors x1t2 J 3x1 1t2, x2 1t2, . . . , xm 1t24 , f1t, x2 J 3f1 1t, x1, x2, . . . , xm 2, f2 1t, x1, x2, . . . , xm 2, . . . , fm 1t, x1, x2, . . . , xm 2 4 , then (12) can be expressed in the much neater form (13)

Example 2

xn + 1 = xn + h f1tn, xn 2 .

Use the vectorized Euler method with step size h = 0.1 to find an approximation for the solution to the initial value problem (14)

y″1t2 + 4y′1t2 + 3y1t2 = 0; y102 = 1.5 , y′102 = -2.5 ,

on the interval 30, 14. Solution

For the given step size, the method will yield approximations for y10.12, y10.22, . . . , y11.02. To apply the vectorized Euler method to (14), we first convert it to normal form. Setting x1 = y and x2 = y′, we obtain the system (15)

x1= = x2; x2= = -4x2 - 3x1;

x1 102 = 1.5 , x2 102 = -2.5 .

Comparing (15) with (1) we see that f1 1t, x1, x2 2 = x2 and f2 1t, x1, x2 2 = -4x2 - 3x1. With the starting values of t0 = 0, x1;0 = 1.5, and x2;0 = -2.5, we compute

c

c

x1 10.12 ≈ x1;1 = x1;0 + hx2;0 = 1.5 + 0.11 -2.52 = 1.25 , x2 10.1) ≈ x2;1 = x2;0 + h1 -4x2;0 - 3x1;0 2 = -2.5 + 0.13 -41 -2.52 - 3 # 1.54 = -1.95 ; x1 10.22 ≈ x1;2 = x1;1 + hx2;1 = 1.25 + 0.11 -1.952 = 1.055 , x2 10.2) ≈ x2;2 = x2;1 + h1 -4x2;1 - 3x1;1 2 = -1.95 + 0.13 -41 -1.952 - 3 # 1.254 = -1.545 .

Continuing the algorithm we compute the remaining values. These are listed in Table 5.1 on page 255, along with the exact values calculated via the methods of Chapter 4. Note that the x2;n column gives approximations to y′1t2, since x2 1t2 K y′1t2. ◆

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 5.3

TABLE 5.1

Solving Systems and Higher-Order Equations Numerically

255

Approximations of the Solution to (14) in Example 2

t = n10.12

x1;n

y Exact

x2;n

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1.5 1.25 1.055 0.9005 0.77615 0.674525 0.5902655 0.51947405 0.459291215 0.407597293 0.362802203

1.5 1.275246528 1.093136571 0.944103051 0.820917152 0.71809574 0.63146108 0.557813518 0.494687941 0.440172416 0.392772975

- 2.5 - 1.95 - 1.545 - 1.2435 - 1.01625 - 0.842595 - 0.7079145 - 0.60182835 - 0.516939225 - 0.4479509 - 0.391049727

y′ Exact

- 2.5 - 2.016064749 - 1.641948207 - 1.35067271 - 1.122111364 - 0.9412259 - 0.796759968 - 0.680269946 - 0.585405894 - 0.507377929 - 0.442560044

Euler’s method is modestly accurate for this problem with a step size of h = 0.1. The next example demonstrates the effects of using a sequence of smaller values of h to improve the accuracy.

Example 3

For the initial value problem of Example 2, use Euler’s method to estimate y112 for successively halved step sizes h = 0.1, 0.05, 0.025, 0.0125, 0.00625.

Solution

Using the same scheme as in Example 2, we find the following approximations, denoted by y11;h2 (obtained with step size h): 0.1 0.05 0.025 0.0125 0.00625 y1 1;h 2 0.36280 0.37787 0.38535 0.38907 0.39092 h

[Recall that the exact value, rounded to 5 decimal places, is y112 = 0.39277.] ◆ The Runge–Kutta scheme described in Section 3.7 is easy to vectorize also; details are given on the following page. As would be expected, its performance is considerably more accurate, yielding five-decimal agreement with the exact solution for a step size of 0.05: h

0.1 0.05 0.025 0.0125 0.00625 0.39278 0.39277 0.39277 0.39277 0.39277 y1 1;h 2 As in Section 3.7, both algorithms can be coded so as to repeat the calculation of y112 with a sequence of smaller step sizes until two consecutive estimates agree to within some prespecified tolerance e. Here one should interpret “two estimates agree to within e” to mean that each component of the successive vector approximants [i.e., approximants to y112 and y′112] should agree to within e.

An Application to Population Dynamics A mathematical model for the population dynamics of competing species, one a predator with population x2 1t2 and the other its prey with population x1 1t2, was developed independently in

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256

Chapter 5

Introduction to Systems and Phase Plane Analysis

the early 1900s by A. J. Lotka and V. Volterra. It assumes that there is plenty of food available for the prey to eat, so the birthrate of the prey should follow the Malthusian or exponential law (see Section 3.2); that is, the birthrate of the prey is Ax1, where A is a positive constant. The death rate of the prey depends on the number of interactions between the predators and the prey. This is modeled by the expression Bx1x2, where B is a positive constant. Therefore, the rate of change in the population of the prey per unit time is dx1 >dt = Ax1 - Bx1x2. Assuming that the predators depend entirely on the prey for their food, it is argued that the birthrate of the predators depends on the number of interactions with the prey; that is, the birthrate of predators is Dx1x2, where D is a positive constant. The death rate of the predators is assumed to be Cx2 because without food the population would die off at a rate proportional to the population present. Hence, the rate of change in the population of predators per unit time is dx2 >dt = -Cx2 + Dx1x2. Combining these two equations, we obtain the Volterra–Lotka system for the population dynamics of two competing species: (16)

x′1 = Ax1 − Bx1x2 , x′2 = −Cx2 + Dx1x2 .

Such systems are in general not explicitly solvable. In the following example, we obtain an approximate solution for such a system by utilizing the vectorized form of the Runge–Kutta algorithm. For the system of two equations x1= = f1 1t, x1, x2 2 , x2= = f2 1t, x1, x2 2 , with initial conditions x1 1t0 2 = x1;0, x2 1t0 2 = x2;0, the vectorized form of the Runge–Kutta recursive equations (cf. (14), page 254) becomes tn + 1 J tn + h (17)

1n = 0, 1, 2, . . .2 ,

e x1;n + 1 J x1;n + 16 1k1,1 + 2k1,2 + 2k1,3 + k1,4 2 , x2;n + 1 J x2;n + 16 1k2,1 + 2k2,2 + 2k2,3 + k2,4 2 ,

where h is the step size and, for i = 1 and 2, ki,1 J h fi 1tn, x1;n, x2;n 2 ,

(18)

f

ki,2 J h fi 1 tn + h2, x1;n + 12k1,1, x2;n + 12k2,1 2 ,

ki,3 J h fi 1 tn + h2, x1;n + 12k1,2, x2;n + 12k2,2 2 , ki,4 J h fi 1tn + h, x1;n + k1,3, x2;n + k2,3 2 .

It is important to note that both k1,1 and k2,1 must be computed before either k1,2 or k2,2. Similarly, both k1,2 and k2,2 are needed to compute k1,3 and k2,3, etc. In Appendix F, program outlines are given for applying the method to graph approximate solutions over a specified interval 3t0, t1 4 or to obtain approximations of the solutions at a specified point to within a desired tolerance.

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Section 5.3

Example 4

Solving Systems and Higher-Order Equations Numerically

257

Use the classical fourth-order Runge–Kutta algorithm for systems to approximate the solution of the initial value problem (19)

x1= = 2x1 - 2x1x2 ; x2= = x1x2 - x2 ;

x1 102 = 1 , x2 102 = 3

at t = 1. Starting with h = 1, continue halving the step size until two successive approximations of x1 112 and of x2 112 differ by at most 0.0001.

Solution

Here f1 1t, x1, x2 2 = 2x1 - 2x1x2 and f2 1t, x1, x2 2 = x1x2 - x2. With the inputs t0 = 0, x1;0 = 1, x2;0 = 3, we proceed with the algorithm to compute x1 11; 12 and x2 11; 12, the approximations to x1 112, x2 112 using h = 1. We find from the formulas in (18) that k1,1 = h12x1;0 - 2x1;0 x2;0 2 = 2112 - 2112132 = -4 , k2,1 = h1x1;0 x2;0 - x2;0 2 = 112132 - 3 = 0 ,

k1,2 = h 3 2 1 x1;0 + 12k1,1 2 - 2 1 x1;0 + 12k1,1 2 1 x2;0 + 12k2,1 2 4 = 2 3 1 + 12 1 -42 4 - 2 3 1 + 12 1 -42 4 3 3 + 12 102 4 = -2 + 2132 = 4 ,

k2,2 = h 3 1 x1;0 + 12k1,1 2 1 x2;0 + 12k2,1 2 - 1 x2;0 + 12k2,1 2 4 =

3 1 + 12 1 -42 4 3 3 + 12 102 4

= 1 -12132 - 3 = -6 ,

- 3 3 + 12 102 4

and similarly we compute k1,3 = h 3 2 1 x1;0 + 12k1,2 2 - 2 1 x1;0 + 12k1,2 2 1 x2;0 + 12k2,2 2 4 = 6 ,

k2,3 = h 3 1 x1;0 + 12k1,2 2 1 x2;0 + 12k2,2 2 - 1 x2;0 + 12k2,2 2 4 = 0 ,

k1,4 = h321x1;0 + k1,3 2 - 21x1;0 + k1,3 21x2;0 + k2,3 24 = -28 ,

k2,4 = h3 1x1;0 + k1,3 21x2;0 + k2,3 2 - 1x2;0 + k2,3 24 = -18 . Inserting these values into formula (17), we get 1 x1;1 = x1;0 + 1k1,1 + 2k1,2 + 2k1,3 + k1,4 2 6 1 = 1 + 1 -4 + 8 + 12 - 282 = -1 , 6 1 x2;1 = x2;0 + 1k2,1 + 2k2,2 + 2k2,3 + k2,4 2 , 6 1 = 3 + 10 - 12 + 0 + 182 = 4 , 6

as the respective approximations to x1 112 and x2 112.

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258

Chapter 5

Introduction to Systems and Phase Plane Analysis

Repeating the algorithm with h = 1>2 1N = 22 we obtain the approximations x1 11; 2-1 2 and x2 11; 2-1 2 for x1 112 and x2 112. In Table 5.2, we list the approximations x1 11; 2-m 2 and x2 11; 2-m 2 for x1 112 and x2 112 using step size h = 2-m for m = 0, 1, 2, 3, and 4. We stopped at m = 4, since both

0 x1 11; 2-3 2 - x1 11; 2-4 2 0 = 0.00006 6 0.0001 and

0 x2 11; 2-3 2 - x2 11; 2-4 2 0 = 0.00001 6 0.0001 . Hence, x1 112 ≈ 0.07735 and x2 112 ≈ 1.46445, with tolerance 0.0001. ◆

TABLE 5.2

Approximations of the Solution to System (19) in Example 4

m

h

x1 11; h2

x2 11; h2

0 1 2 3 4

1.0 0.5 0.25 0.125 0.0625

- 1.0 0.14662 0.07885 0.07741 0.07735

4.0 1.47356 1.46469 1.46446 1.46445

To get a better feel for the solution to system (19), we have graphed in Figure 5.6 an approximation of the solution for 0 … t … 12, using linear interpolation to connect the vectorized Runge– Kutta approximants for the points t = 0, 0.125, 0.25, . . . , 12.0 (i.e., with h = 0.125). From the graph it appears that the components x1 and x2 are periodic in the variable t. Phase plane analysis is used in Section 5.5 to show that, indeed, Volterra–Lotka equations have periodic solutions.

x1

4 3 x2

2 1

t 0

1

2

3

4

5

6

7

8

9

10 11 12

Figure 5.6 Graphs of the components of an approximate solution to the Volterra–Lotka system (17)

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Section 5.3

5.3

Solving Systems and Higher-Order Equations Numerically

259

EXERCISES

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form. 1. y″1t2 + ty′1t2 - 3y1t2 = t 2 ; y102 = 3 ,

y′102 = -6

y102 = 1 ,

y′102 = 0

2. y″1t2 = cos1t - y2 + y2 1t2 ; 3. y 142 1t2 - y 132 1t2 + 7y1t2 = cos t ; y102 = y′102 = 1 ,

y 132 102 = 2

y″102 = 0 ,

4. y 162 1t2 = 3y′1t24 3 - sin 1 y1t2 2 + e2t ; y102 = y′102 = g = y 152 102 = 0

5. x″ + y - x′ = 2t ; y″ - x + y = -1 ; [Hint: Set x1 = x , 6. 3x″ + 5x - 2y = 0 ; 4y″ + 2y - 6x = 0 ; 7. x‴ - y = t ;

x132 = 5 ,

x′132 = 2 ,

y132 = 1 ,

y′132 = - 1

x2 = x′ ,

x3 = y ,

x4 = y′.]

x102 = - 1 ,

x′102 = 0 ,

y102 = 1 ,

y′102 = 2

x102 = x′102 = x″102 = 4 ,

2x″ + 5y″ - 2y = 1 ;

y102 = y′102 = 1

8. Sturm–Liouville Form. A second-order equation is said to be in Sturm–Liouville form if it is expressed as 3p1t2y′1t24′ + q1t2y1t2 = 0 . Show that the substitutions x1 = y, x2 = py′ result in the normal form

12. t 2y″ + y = t + 2 ; y112 = 1 , y′112 = - 1 on 31, 24 13. y″ = t 2 - y2 ; y102 = 0 , y′102 = 1 on 30, 14 (Can you guess the solution?) In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)† 14. Using the vectorized Runge–Kutta algorithm with h = 0.5, approximate the solution to the initial value problem 3t 2y″ - 5ty′ + 5y = 0 ; 2 3 at t = 8. Compare this approximation to the actual solution y1t2 = t 5>3 - t. 15. Using the vectorized Runge–Kutta algorithm, approximate the solution to the initial value problem y112 = 0 ,

y″ = t 2 + y2 ;

y′112 =

y102 = 1 ,

at t = 1. Starting with h = 1, continue halving the step size until two successive approximations 3of both y112 and y′1124 differ by at most 0.01. 16. Using the vectorized Runge–Kutta algorithm for systems with h = 0.125, approximate the solution to the initial value problem

x1= = x2 >p ,

x′ = 2x - y ;

x102 = 0 ,

x2= = - qx1 .

y′ = 3x + 6y ;

y102 = - 2

If y102 = a and y′102 = b are the initial values for the Sturm–Liouville problem, what are x1 102 and x2 102? 9. In Section 3.6, we discussed the improved Euler’s method for approximating the solution to a first-order equation. Extend this method to normal systems and give the recursive formulas for solving the initial value problem. In Problems 10–13, use the vectorized Euler method with h = 0.25 to find an approximation for the solution to the given initial value problem on the specified interval. 10. y″ + ty′ + y = 0 ; y′102 = 0 on 30, 14 y102 = 1 , 11. 11 + t 2 2y″ + y′ - y = 0 ; y102 = 1 , y′102 = -1 on 30, 14

y′102 = 0

at t = 1. Compare this approximation to the actual solution

x1t2 = e5t - e3t ,

y1t2 = e3t - 3e5t .

17. Using the vectorized Runge–Kutta algorithm, approximate the solution to the initial value problem du = 3u - 4y ; u102 = 1 , dx dy = 2u - 3y ; y102 = 1 dx at x = 1. Starting with h = 1, continue halving the step size until two successive approximations of u112 and y112 differ by at most 0.001.

†

Appendix G describes various websites and commercial software that sketch direction fields and automate most of the differential equation algorithms discussed in this book.

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260

Chapter 5

Introduction to Systems and Phase Plane Analysis

18. Combat Model. A simplified mathematical model for conventional versus guerrilla combat is given by the system x1= = - 10.12x1x2 ; x2=

= -x1 ;

x1 102 = 10 , x2 102 = 15 ,

where x1 and x2 are the strengths of guerrilla and conventional troops, respectively, and 0.1 and 1 are the combat effectiveness coefficients. Who will win the conflict: the conventional troops or the guerrillas? [Hint: Use the vectorized Runge–Kutta algorithm for systems with h = 0.1 to approximate the solutions.] 19. Predator–Prey Model. The Volterra–Lotka predator– prey model predicts some rather interesting behavior that is evident in certain biological systems. For example, suppose you fix the initial population of prey but increase the initial population of predators. Then the population cycle for the prey becomes more severe in the sense that there is a long period of time with a reduced population of prey followed by a short period when the population of prey is very large. To demonstrate this behavior, use the vectorized Runge– Kutta algorithm for systems with h = 0.5 to approximate the populations of prey x and of predators y over the period 30, 54 that satisfy the Volterra–Lotka system x′ = x13 - y2 , y′ = y1x - 32 under each of the following initial conditions: (a) x102 = 2 , y102 = 4 . (b) x102 = 2 , y102 = 5 . (c) x102 = 2 , y102 = 7 . 20. In Project C of Chapter 4, it was shown that the simple pendulum equation u″1t2 + sin u1t2 = 0 has periodic solutions when the initial displacement and velocity are small. Show that the period of the solution may depend on the initial conditions by using the vectorized Runge–Kutta algorithm with h = 0.02 to approximate the solutions to the simple pendulum problem on 30, 44 for the initial conditions: (a) u102 = 0.1 , u′102 = 0 . (b) u102 = 0.5 , u′102 = 0 . (c) u102 = 1.0 , u′102 = 0 . [Hint: Approximate the length of time it takes to reach -u102.] 21. Fluid Ejection. In the design of a sewage treatment plant, the following equation arises:†

where H is the level of the fluid in an ejection chamber and t is the time in seconds. Use the vectorized Runge–Kutta algorithm with h = 0.5 to approximate H1t2 over the interval 30, 54. 22. Oscillations and Nonlinear Equations. For the initial value problem x″ + 10.1211 - x2 2x′ + x = 0 ; x102 = x0 ,

x′102 = 0 ,

use the vectorized Runge–Kutta algorithm with h = 0.02 to illustrate that as t increases from 0 to 20, the solution x exhibits damped oscillations when x0 = 1, whereas x exhibits expanding oscillations when x0 = 2.1. 23. Nonlinear Spring. The Duffing equation y″ + y + ry3 = 0 , where r is a constant, is a model for the vibrations of a mass attached to a nonlinear spring. For this model, does the period of vibration vary as the parameter r is varied? Does the period vary as the initial conditions are varied? [Hint: Use the vectorized Runge–Kutta algorithm with h = 0.1 to approximate the solutions for r = 1 and 2, with initial conditions y102 = a, y′102 = 0 for a = 1, 2, and 3.] 24. Pendulum with Varying Length. A pendulum is formed by a mass m attached to the end of a wire that is attached to the ceiling. Assume that the length l1t2 of the wire varies with time in some predetermined fashion. If u1t2 is the angle in radians between the pendulum and the vertical, then the motion of the pendulum is governed for small angles by the initial value problem l 2 1t2u″1t2 + 2l1t2l′1t2u′1t2 + gl1t2sin 1u1t22 = 0 ;

u102 = u0 ,

u′102 = u1 ,

where g is the acceleration due to gravity. Assume that l1t2 = l0 + l1 cos1vt - f2 , where l1 is much smaller than l0. (This might be a model for a person on a swing, where the pumping action changes the distance from the center of mass of the swing to the point where the swing is attached.) To simplify the computations, take g = 1. Using the Runge– Kutta algorithm with h = 0.1, study the motion of the pendulum when u0 = 0.05, u1 = 0, l0 = 1, l1 = 0.1, v = 1, and f = 0.02. In particular, does the pendulum ever attain an angle greater in absolute value than the initial angle u0?

60 - H = 177.72H″ + 119.4221H′2 2 ; H102 = H′102 = 0 , †

See Numerical Solution of Differential Equations, by William Milne (Dover, New York, 1970), p. 82.

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Section 5.3

Solving Systems and Higher-Order Equations Numerically

In Problems 25–30, use a software package or the SUBROUTINE in Appendix F. 25. Using the Runge–Kutta algorithm for systems with h = 0.05, approximate the solution to the initial value problem y‴ + y″ + y2 = t ; y102 = 1 ,

y′102 = 0 ,

y″102 = 1

Suppose these species have the same growth rates, and the maximum population that the habitat can support is the same for each species. (We assume it to be one unit.) Also suppose the competitive advantage that S1 has over S2 is the same as that of S2 over S3 and S3 over S1. This situation is modeled by the system p1= = p1 11 - p1 - ap2 - bp3 2 ,

p2= = p2 11 - bp1 - p2 - ap3 2 , p3= = p3 11 - ap1 - bp2 - p3 2 ,

at t = 1. 26. Use the Runge–Kutta algorithm for systems with h = 0.1 to approximate the solution to the initial value problem x′ = yz ;

x102 = 0 ,

y′ = - xz ;

y102 = 1 ,

z′ = -xy>2 ;

z102 = 1 ,

27. Generalized Blasius Equation. H. Blasius, in his study of laminar flow of a fluid, encountered an equation of the form y‴ + yy″ = 1y′2 2 - 1 . Use the Runge–Kutta algorithm for systems with h = 0.1 to approximate the solution that satisfies the initial conditions y102 = 0, y′102 = 0, and y″102 = 1.32824. Sketch this solution on the interval 30, 24. 28. Lunar Orbit. The motion of a moon moving in a planar orbit about a planet is governed by the equations d 2y dt

2

= -G

where a and b are positive constants. To demonstrate the population dynamics of this system when a = b = 0.5, use the Runge–Kutta algorithm for systems with h = 0.1 to approximate the populations pi over the time interval 30, 104 under each of the following initial conditions:

(a) p1 102 = 1.0 , p2 102 = 0.1 , p3 102 = 0.1 . (b) p1 102 = 0.1 , p2 102 = 1.0 , p3 102 = 0.1 . (c) p1 102 = 0.1 , p2 102 = 0.1 , p3 102 = 1.0 .

at t = 1.

d 2x mx = -G 3 , 2 dt r

261

my r3

,

where r J 1x2 + y2 2 1>2, G is the gravitational constant, and m is the mass of the planet. Assume Gm = 1. When x102 = 1, x′102 = y102 = 0, and y′102 = 1, the motion is a circular orbit of radius 1 and period 2p. (a) Setting x1 = x, x2 = x′, x3 = y, x4 = y′, express the governing equations as a first-order system in normal form. (b) Using h = 2p>100 ≈ 0.0628318, compute one orbit of this moon (i.e., do N = 100 steps.). Do your approximations agree with the fact that the orbit is a circle of radius 1?

On the basis of the results of parts (a)–(c), decide what you think will happen to these populations as t S + ∞ . 30. Spring Pendulum. Let a mass be attached to one end of a spring with spring constant k and the other end attached to the ceiling. Let l0 be the natural length of the spring and let l1t2 be its length at time t. If u1t2 is the angle between the pendulum and the vertical, then the motion of the spring pendulum is governed by the system k 1l - l0 2 = 0 , m l 2 1t2u″1t2 + 2l1t2l′1t2u′1t2 + gl1t2 sin u1t2 = 0 . l″1t2 - l1t2u′1t2 - g cos u1t2 +

Assume g = 1, k = m = 1, and l0 = 4. When the system is at rest, l = l0 + mg>k = 5. (a) Describe the motion of the pendulum when l102 = 5.5, l′102 = 0, u102 = 0, and u′102 = 0. (b) When the pendulum is both stretched and given an angular displacement, the motion of the pendulum is more complicated. Using the Runge–Kutta algorithm for systems with h = 0.1 to approximate the solution, sketch the graphs of the length l and the angular displacement u on the interval 30, 104 if l102 = 5.5, l′102 = 0, u102 = 0.5, and u′102 = 0.

29. Competing Species. Let pi 1t2 denote, respectively, the populations of three competing species Si, i = 1, 2, 3.

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262

Chapter 5

Introduction to Systems and Phase Plane Analysis

5.4 Introduction to the Phase Plane In this section, we study systems of two first-order equations of the form

(1)

dx = f1x, y2 , dt dy = g1x, y2 . dt

Note that the independent variable t does not appear in the right-hand terms f1x, y2 and g1x, y2; such systems are called autonomous. For example, the system that modeled the interconnected tanks problem in Section 5.1, 1 1 x′ = - x + y, 3 12 1 1 y′ = x - y , 3 3 is autonomous. So is the Volterra–Lotka system, x′ = Ax - Bxy , y′ = -Cy + Dxy , (with A, B, C, D constants), which was discussed in Example 4 of Section 5.3 as a model for population dynamics. For future reference, we note that the solutions to autonomous systems have a “timeshift immunity,” in the sense that if the pair x1t2, y1t2 solves (1), so does the time-shifted pair x1t + c2, y1t + c2 for any constant c. Specifically, if we let X1t2 J x1t + c2 and Y1t2 J y1t + c2, then by the chain rule dX dx 1t2 = 1t + c2 = f 1 x1t + c2, y1t + c2 2 = f 1 X1t2, Y1t2 2 , dt dt dy dY 1t2 = 1t + c2 = g 1 x1t + c2, y1t + c2 2 = g 1 X1t2, Y1t2 2 , dt dt

proving that X1t2, Y1t2 is also a solution to (1). Since a solution to the autonomous system is a pair of functions x1t2, y1t2 that satisfies (1) for all t in some interval I, it can be visualized as a pair of graphs, as shown in Figure 5.7. For purposes of comparing one of the solution pair with its mate, we can compress these two graphs into one by charting the path in the plane traced out by the ordered pair 1 x1t2, y1t2 2 as t increases.

Trajectories and the Phase Plane Definition 1. If x1t2, y1t2 is a solution pair to (1) for t in the interval I, then a plot in the xy-plane of the parametrized curve x = x1t2, y = y1t2 for t in I, together with arrows indicating its direction with increasing t, is said to be a trajectory of the system (1). In such a context we call the xy-plane the phase plane and refer to a representative set of trajectories in this plane as a phase portrait of the system.

A trajectory provides a visualization of the motion of a particle in the plane that starts at some point 1x0, y0 2 and follows a path as time increases that is determined by the solution to the autonomous system with initial conditions x102 = x0, y102 = y0. The trajectory for the

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Section 5.4

x(t)

263

Introduction to the Phase Plane

y(t)

t1

t2

t

t3

t1

t2

t3

t

Figure 5.7 Solution pair for system (1)

solution pair of Figure 5.7 is shown in Figure 5.8. Note, however, that the trajectory contains less information than the individual graphs for x1t2 and y1t2, because the t-dependence has been suppressed. (We can’t tell if the particle’s speed is fast or slow.) In principle we can construct, point by point, the trajectory from the solution graphs, but we cannot reconstruct the solution graphs from the phase plane trajectory alone (because we would not know what value of t to assign to each point). Nonetheless, trajectories provide valuable information about the qualitative behavior of solutions to the autonomous system. Can we find trajectories without knowing a solution pair to the system? Yes; the key is to observe that since t does not appear explicitly on the right-hand side of system (1), if we divide the two equations and invoke the chain rule dy>dt dy , = dx dx>dt we get the “related phase plane differential equation” (without t) g1x, y2 dy (2) . = dx f1x, y2 We mastered equations like (2) in Chapter 2; their solution curves provide highways along which the trajectories travel. What is missing from these highways is the orientation of a trajectory, which can be gleaned from the autonomous system (1) by determining the sign of f or g at a point and recalling that 1dx>dt, dy>dt2 = 1f, g2 is the velocity of the moving particle. Furthermore, as illustrated in Example 1 below, a single curve arising from the related phase plane differential equation might well accommodate several different trajectories. As we shall see in the examples to follow as well as in Figure 5.12, trajectories are typically of the following types: (i) a single point, indicating a stationary particle; (ii) a non-closed arc, without self-intersections; or (iii) a closed curve (cycle), indicating periodic motion. Moreover, thanks to the Existence-Uniqueness Theorem for first-order equations in Chapter 1 (page 11), as long as the functions f and g in (1) are continuously differentiable in the plane, the trajectories of a phase portrait will not intersect each other (unless they coincide). y

Ax(t 2), y(t 2)B

Ax(t 3), y(t 3)B

Ax(t 1), y(t 1)B

x

Figure 5.8 Phase plane trajectory of the solution pair for system (1)

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264

Chapter 5

Introduction to Systems and Phase Plane Analysis

Except for the very special case of linear systems with constant coefficients that was discussed in Section 5.2, finding all solutions to the system (1) is generally an impossible task. But it is relatively easy to find constant solutions; if f1x0, y0 2 = 0 and g1x0, y0 2 = 0, then the constant functions x1t2 K x0, y1t2 K y0 solve (1). For such solutions the following terminology is used.

Critical Points and Equilibrium Solutions Definition 2. A point 1x0, y0 2 where f1x0, y0 2 = 0 and g1x0, y0 2 = 0 is called a critical point, or equilibrium point, of the system dx>dt = f1x, y2, dy>dt = g1x, y2, and the corresponding constant solution x1t2 K x0, y1t2 K y0 is called an equilibrium solution. Notice that trajectories of equilibrium solutions consist of just single points (the equili-brium points). But what can be said about the other trajectories? Can we predict any of their features from closer examination of the equilibrium points? To explore this we focus on the related phase plane differential equation (2) and exploit its direction field (recall Section 1.3, page 15). However, we’ll augment the direction field plot by attaching arrowheads to the line segments, indicating the direction of the “flow” of solutions as t increases. This is easy: When dx>dt is positive, x1t2 increases so the trajectory flows to the right. Therefore, according to (1), all direction field segments drawn in a region where f1x, y2 is positive should point to the right [and, of course, they point to the left if f1x, y2 is negative]. If f1x, y2 is zero, we can use g1x, y2 to decide if the flow is upward [y1t2 increases] or downward 3y1t2 decreases]. [What if both f1x, y2 and g1x, y2 are zero?] In the examples that follow, one can use computers or calculators for generating these direction fields. Example 1

Sketch the direction field in the phase plane for the system

(3)

dx = -x , dt dy = -2y dt

and identify its critical point. Solution

Here f1x, y2 = -x and g1x, y2 = -2y are both zero when x = y = 0, so 10, 02 is the critical point. The direction field for the related phase plane differential equation 2y dy -2y = (4) = -x x dx is given in Figure 5.9 on page 265. Since dx>dt = -x in (3), trajectories in the right half-plane (where x 7 02 flow to the left, and vice versa. From the figure we can see that all solutions “flow into” the critical point 10, 02. Such a critical point is called asymptotically stable.† ◆ Remark. For this simple example, we can actually solve the system (3) explicitly; indeed, (3) constitutes an uncoupled pair of linear equations whose solutions are x1t2 = c1e-t and y1t2 = c2e-2t. By elimination of t, we obtain the equation y = c2e-2t = c2[x1t2 >c1]2 = cx2. So the trajectories lie along the parabolas y = cx2. [Alternatively, we could have solved (4) by separating variables and obtained these same parabolas as the “highways” on which the trajectories travel.] Notice that each such parabola is made up of three trajectories: an incoming trajectory approaching the origin in the right half-plane; its mirror-image trajectory approaching †

See Section 12.3 for a rigorous exposition of stability and critical points. All references to Chapters 11–13 refer to the expanded text, Fundamentals of Differential Equations and Boundary Value Problems, 7th ed.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 5.4

265

Introduction to the Phase Plane

y y

x x

Figure 5.9 Direction field for Example 1

Figure 5.10 Trajectories for Example 1

the origin in the left half-plane; and the origin itself, an equilibrium point. Sample trajectories are indicated in Figure 5.10. Example 2

Sketch the direction field in the phase plane for the system

(5)

dx = x, dt dy = 2y dt

and describe the behavior of solutions near the critical point 10, 02. Solution

This example is almost identical to the previous one; in fact, one could say we have merely “reversed time” in (3). The direction field segments for (6)

dy 2y = x dx

are the same as those of (4), but the direction arrows are reversed. Now all solutions flow away from the critical point 10, 02; the equilibrium is unstable. ◆ Example 3

For the system (7) below, find the critical points, sketch the direction field in the phase plane, and predict the asymptotic nature (i.e., behavior as t S + ∞ ) of the solution starting at x = 2, y = 0 when t = 0.

(7)

Solution

dx = 5x - 3y - 2 , dt dy = 4x - 3y - 1 . dt

The only critical point is the solution of the simultaneous equations f1x, y2 = g1x, y2 = 0: (8)

5x0 - 3y0 - 2 = 0 , 4x0 - 3y0 - 1 = 0 ,

from which we find x0 = y0 = 1. The direction field for the related phase plane differential equation

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266

Chapter 5

Introduction to Systems and Phase Plane Analysis

y

y y = 2x 2 1

x(t)

3

y(t) 2 1 22

21

1

2

x 3

21 22

Figure 5.11 Direction field, trajectories, and a typical solution pair for a trajectory (Example 3)

(9)

dy 4x - 3y - 1 = dx 5x - 3y - 2

is shown in Figure 5.11, along with some trajectories and a typical solution pair.† Note that solutions flow to the right for 5x - 3y - 2 7 0, i.e., for all points below the line 5x - 3y - 2 = 0, and that they slow down near the critical point. The solution curve to (9) passing through 12, 02 in Figure 5.11 apparently is a “highway” extending to infinity. Does this imply the corresponding system solution x1t2, y1t2 also approaches infinity in the sense that 0 x1t2 0 + 0 y1t2 0 S + ∞ as t S + ∞, or could its trajectory “stall” at some point along the highway, or possibly even “backtrack?” It cannot backtrack, because the direction arrows along the trajectory point, unambiguously, to the right. And if 1 x1t2, y1t2 2 stalls at some point 1x1, y1 2, then intuitively we would conclude that 1x1, y1 2 was an equilibrium point (since the “speeds” dx>dt and dy>dt would approach zero there). But we have already found the only critical point. So we conclude, with a high degree of confidence,‡ that the system solution does indeed go to infinity. The critical point 11, 12 is unstable because, although many solutions get arbitrarily close to 11, 12, most of them eventually flow away. Solutions that lie on the line y = 2x - 1, however, do converge to 11, 12. Such an equilibrium is an example of a saddle point. ◆ In the preceding example, we informally argued that if a trajectory “stalls”—that is, if it has an endpoint—then this endpoint would have to be a critical point. This is more carefully stated in the following theorem, whose proof is outlined in Problem 30.

Endpoints Are Critical Points Theorem 1. Let the pair x1t2, y1t2 be a solution on [0, + ∞ 2 to the autonomous system dx>dt = f1x, y2, dy>dt = g1x, y2, where f and g are continuous in the plane. If the limits lim x1t2 x* J S t

+∞

and

lim y1t2 y* J S t

+∞

exist and are finite, then the point 1x*, y*2 is a critical point for the system. †

Solutions to equation (9) can be obtained analytically using the methods of Section 2.6. These informal arguments are made more rigorous in Chapter 12. All references to Chapters 11–13 refer to the expanded text, Fundamentals of Differential Equations and Boundary Value Problems, 7th edition. ‡

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 5.4

y

y

y

x

Node (asymptotically stable)

x

Spiral (unstable)

x

Center (stable)

x

Saddle (unstable)

y

y

267

Introduction to the Phase Plane

y

x

Spiral (asymptotically stable)

x

Node (unstable)

Figure 5.12 Examples of different trajectory behaviors near critical point at origin

Some typical trajectory configurations near critical points are displayed and classified in Figure 5.12. These phase plane portraits were generated by software packages having trajectorysketching options†. A more complete discussion of the nature of various types of equilibrium solutions and their stability is deferred to Chapter 12.‡ For the moment, however, notice that unstable critical points are distinguished by “runaway” trajectories emanating from arbitrarily nearby points, while stable equilibria “trap” all neighboring trajectories. The asymptotically stable critical points attract their neighboring trajectories as t S + ∞. Historically, the phase plane was introduced to facilitate the analysis of mechanical systems governed by Newton’s second law, force equals mass times acceleration. An autonomous mechanical system arises when this force is independent of time and can be modeled by a second-order equation of the form (10)

y″ = ƒ1y, y′2 .

As we have seen in Section 5.3, this equation can be converted to a normal first-order system by introducing the velocity y = dy>dt and writing dy = y, dt (11) dy = ƒ1y, y2 . dt

†

Appendix G describes various websites and commercial software that sketch direction fields and automate most of the differential equation algorithms discussed in this book. ‡ All references to Chapters 11–13 refer to the expanded text, Fundamentals of Differential Equations and Boundary Value Problems, 7th edition.

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268

Chapter 5

Introduction to Systems and Phase Plane Analysis

Thus, we can analyze the behavior of an autonomous mechanical system by studying its phase portrait in the yy-plane. Notice that with y as the vertical axis, trajectories 1y1t2, y1t22 flow to the right in the upper half-plane (where y 7 0), and to the left in the lower half-plane. Example 4

Sketch the direction field in the phase plane for the first-order system corresponding to the unforced, undamped mass–spring oscillator described in Section 4.1 (Figure 4.1, page 152). Sketch several trajectories and interpret them physically.

Solution

The equation derived in Section 4.1 for this oscillator is my″ + ky = 0 or, equivalently, y″ = -ky>m. Hence, the system (11) takes the form y′ = y , ky y′ = - . m The critical point is at the origin y = y = 0. The direction field in Figure 5.13 indicates that the trajectories appear to be either closed curves (ellipses?) or spirals that encircle the critical point. We saw in Section 4.9 that the undamped oscillator motions are periodic; they cycle repeatedly through the same sets of points, with the same velocities. Their trajectories in the phase plane, then, must be closed curves.† Let’s confirm this mathematically by solving the related phase plane differential equation

(12)

(13)

ky dy = . my dy

Equation (13) is separable, and we find y dy = -

ky dy m

or

da

y2 y2 k b = - da b , m 2 2 y

y

Figure 5.13 Direction field for Example 4 †

By the same reasoning, underdamped oscillations would correspond to spiral trajectories asymptotically approaching the origin as t S + ∞.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 5.4

Introduction to the Phase Plane

269

y

y

Figure 5.14 Trajectories for Example 4

so its solutions are the ellipses y2 >2 + ky2 >2m = C as shown in Figure 5.14. The trajectories of (12) are confined to these ellipses, and hence neither come arbitrarily close to nor wander arbitrarily far from the equilibrium solution. The critical point is thus identified as a center in Figure 5.12 on page 267. Furthermore, the system solutions must continually circulate around the ellipses, since there are no critical points to stop them. This confirms that all solutions are periodic. ◆ Remark. More generally, we argue that if a solution to an autonomous system like (1) passes through a point in the phase plane twice and if it is sufficiently well behaved to satisfy a uniqueness theorem, then the second “tour” satisfies the same initial conditions as the first tour and so must replicate it. In other words, closed trajectories containing no critical points correspond to periodic solutions. Through these examples we have seen how, by studying the phase plane, one can often anticipate some of the features (boundedness, periodicity, etc.) of solutions of autonomous systems without solving them explicitly. Much of this information can be predicted simply from the critical points and the direction field (oriented by arrowheads), which are obtainable through standard software packages. The final example ties together several of these ideas. Example 5

Find the critical points and solve the related phase plane equation for

(14)

dx = -y1y - 22 , dt dy = 1x - 221y - 22 . dt

What is the asymptotic behavior of the solutions starting from 13, 02, 15, 02, and 12, 32? Solution

To find the critical points, we solve the system -y1y - 22 = 0,

1x - 221y - 22 = 0 .

One family of solutions to this system is given by y = 2 with x arbitrary; that is, the line y = 2. If y ≠ 2, then the system simplifies to -y = 0, and x - 2 = 0, which has the solution x = 2, y = 0. Hence, the critical point set consists of the isolated point 12, 02 and the horizontal line y = 2. The corresponding equilibrium solutions are x1t2 K 2, y1t2 K 0, and the family x1t2 K c, y1t2 K 2, where c is an arbitrary constant.

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Introduction to Systems and Phase Plane Analysis

y

(2, 3) (2 2 5 , 2)

y52

0

1

2

3

4

5

6

x

Figure 5.15 Phase portrait for Example 5

The related phase plane differential equation for the system (14) is (15)

1x - 221y - 22 dy x-2 = . = y dx -y1y - 22

Solving this equation by separating variables, y dy = - 1x - 22 dx

or

y2 + 1x - 22 2 = C ,

demonstrates that the trajectories lie on concentric circles centered at 12, 02. See Figure 5.15. Next we analyze the flow along each trajectory. From the equation dx>dt = -y1y - 22, we see that x is decreasing when y 7 2. This means the flow is from right to left along the arc of a circle that lies above the line y = 2. For 0 6 y 6 2, we have dx>dt 7 0, so in this region the flow is from left to right. Furthermore, for y 6 0, we have dx>dt 6 0, and again the flow is from right to left. We now observe in Figure 5.15 that there are four types of trajectories associated with system (14): (a) those that begin above the line y = 2 and follow the arc of a circle counterclockwise back to that line; (b) those that begin below the line y = 2 and follow the arc of a circle clockwise back to that line; (c) those that continually move clockwise around a circle centered at 12, 02 with radius less than 2 (i.e., they do not intersect the line y = 2); and finally, (d) the critical points 12, 02 and y = 2, x arbitrary. The solution starting at 13, 02 lies on a circle with no critical points; therefore, it is a periodic solution, and the critical point 12, 02 is a center. But the circle containing the solutions starting at 15, 02 and at 12, 32 has critical points at 12 - 25, 22 and 12 + 25, 22. The direction arrows indicate that both solutions approach 12 - 25, 22 asymptotically (as t S + ∞ 2. They lie on the same solution curve for the related phase plane differential equation (a circle), but they are quite different trajectories. ◆ Note that for the system (14) the critical points on the line y = 2 are not isolated, so they do not fit into any of the categories depicted in Figure 5.12 on page 267. Observe also that all solutions of this system are bounded, since they are confined to circles.

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Section 5.4

5.4

Introduction to the Phase Plane

271

EXERCISES

In Problems 1 and 2, verify that the pair x1t2, y1t2 is a solution to the given system. Sketch the trajectory of the given solution in the phase plane. dy dx 1. = 3y3 , = y; dt dt y1t2 = et x1t2 = e3t , dy dx 2. = 1, = 3x2 ; dt dt x1t2 = t + 1 , y1t2 = t 3 + 3t 2 + 3t In Problems 3–6, find the critical point set for the given system. dx dx 4. 3. = x-y, = y-1, dt dt dy dy = x2 + y2 - 1 = x+y+5 dt dt dx dx = x2 - 2xy , = y2 - 3y + 2 , 6. 5. dt dt dy dy = 3xy - y2 = 1x - 121y - 22 dt dt In Problems 7–9, solve the related phase plane differential equation (2), page 263, for the given system. dx dx = y-1, = x2 - 2y -3 , 7. 8. dt dt dy dy = ex+y = 3x2 - 2xy dt dt dx 9. = 2y - x , dt dy = ex + y dt 10. Find all the critical points of the system dx = x2 - 1 , dt dy = xy , dt and the solution curves for the related phase plane differential equation. Thereby prove that there are two trajectories that lie on semicircles. What are the endpoints of the semicircles? In Problems 11–14, solve the related phase plane differential equation for the given system. Then sketch by hand several representative trajectories (with their flow arrows). dx dx 11. = 2y , = - 8y , 12. dt dt dy dy = 2x = 18x dt dt

13.

dx = 1y - x21y - 12 , dt dy = 1x - y21x - 12 dt

14.

3 dx = , y dt dy 2 = x dt

In Problems 15–18, find all critical points for the given system. Then use a software package to sketch the direction field in the phase plane and from this describe the stability of the critical points (i.e., compare with Figure 5.12). dx dx = 2x + y + 3 , = - 5x + 2y , 15. 16. dt dt dy dy = - 3x - 2y - 4 = x - 4y dt dt dx dx = 2x + 13y , = x17 - x - 2y2 , 17. 18. dt dt dy dy = - x - 2y = y15 - x - y2 dt dt In Problems 19–24, convert the given second-order equation into a first-order system by setting y = y′. Then find all the critical points in the yy-plane. Finally, sketch (by hand or software) the direction fields, and describe the stability of the critical points (i.e., compare with Figure 5.12). d 2y d 2y 19. -y = 0 +y = 0 20. 2 dt dt 2 21.

d 2y dt 2

+ y + y5 = 0

22.

d 2y dt 2

+ y3 = 0

23. y″1t2 + y1t2 - y1t2 4 = 0 24. y″1t2 + y1t2 - y1t2 3 = 0 25. Using software, sketch the direction field in the phase plane for the system dx>dt = y , dy>dt = - x + x3 . From the sketch, conjecture whether the solution passing through each given point is periodic:

(a) 10.25, 0.252 (b) 12, 22 (c) 11, 02 26. Using software, sketch the direction field in the phase plane for the system dx>dt = y , dy>dt = - x - x3 . From the sketch, conjecture whether all solutions of this system are bounded. Solve the related phase plane differential equation and confirm your conjecture.

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27. Using software, sketch the direction field in the phase plane for the system dx>dt = -2x + y , dy>dt = -5x - 4y . From the sketch, predict the asymptotic limit (as t S + ∞ ) of the solution starting at 11, 12. 28. Figure 5.16 displays some trajectories for the system dx>dt = y , dy>dt = -x + x2 .

(d) Argue similarly that the supposition that f1x*, y*2 6 0 also leads to a contradiction; hence, f1x*, y*2 must be zero. (e) In the same manner, argue that g1x*, y*2 must be zero. Therefore, f1x*, y*2 = g1x*, y*2 = 0 , and 1x*, y*2 is a critical point. 31. Phase plane analysis provides a quick derivation of the energy integral lemma of Section 4.8 (page 201). By completing the following steps, prove that solutions of equations of the special form y″ = f1y2 satisfy 1 1y′2 2 - F1y2 = constant , 2

y

1

x

where F1y2 is an antiderivative of f1y2. (a) Set y = y′ and write y″ = f1y2 as an equivalent first-order system. (b) Show that the solutions to the yy-phase plane equation for the system in part (a) satisfy y2 >2 = F1y2 + K. Replacing y by y′ then completes the proof. 32. Use the result of Problem 31 to prove that all solutions to the equation

y″ + y3 = 0

Figure 5.16 Phase plane for Problem 28

What types of critical points (compare Figure 5.12 on page 267) occur at 10, 02 and 11, 02? 29. Find the critical points and solve the related phase plane differential equation for the system dx = 1x - 121y - 12 dt dy = y1y - 12 . dt Describe (without using computer software) the asymptotic behavior of trajectories (as t S ∞ ) that start at (a) 13, 22, (b) 12, 1>22, (c) 1 -2, 1>22, (d) 13, - 22. 30. A proof of Theorem 1, page 266, is outlined below. The goal is to show that f1x*, y*2 = g1x*, y*2 = 0. Justify each step. (a) From the given hypotheses, deduce that limt S +∞ x′1t2 = f1x*, y*2 and limt S +∞ y′1t2 = g1x*, y*2. (b) Suppose f1x*, y*2 7 0. Then, by continuity, x′1t2 7 f1x*, y*2 >2 for all large t (say, for t Ú T). Deduce from this that x1t2 7 tf1x*, y*2 >2 + C for t 7 T, where C is some constant. (c) Conclude from part (b) that limt S +∞ x1t2 = + ∞ , contradicting the fact that this limit is the finite number x*. Thus, f1x*, y*2 cannot be positive.

remain bounded. [Hint: Argue that y4 >4 is bounded above by the constant appearing in Problem 31.] 33. A Problem of Current Interest. The motion of an iron bar attracted by the magnetic field produced by a parallel current wire and restrained by springs (see Figure 5.17) is governed by the equation 1 d 2x = -x + , for l-x dt 2

-x0 6 x 6 l ,

wire

l

x0 x

Figure 5.17 Bar restrained by springs and attracted by a parallel current

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Section 5.4

Introduction to the Phase Plane

where the constants x0 and l are, respectively, the distances from the bar to the wall and to the wire when the bar is at equilibrium (rest) with the current off. (a) Setting y = dx>dt, convert the second-order equation to an equivalent first-order system. (b) Solve the related phase plane differential equation for the system in part (a) and thereby show that its solutions are given by

273

y50 y k m

y = { 2C - x2 - 2 ln1l - x2 , where C is a constant. (c) Show that if l 6 2 there are no critical points in the xy-phase plane, whereas if l 7 2 there are two critical points. For the latter case, determine these critical points. (d) Physically, the case l 6 2 corresponds to a current so high that the magnetic attraction completely overpowers the spring. To gain insight into this, use software to plot the phase plane diagrams for the system when l = 1 and when l = 3. (e) From your phase plane diagrams in part (d), describe the possible motions of the bar when l = 1 and when l = 3, under various initial conditions. 34. Falling Object. The motion of an object moving vertically through the air is governed by the equation d 2y 2

= -g -

dt where y is the upward vertical displacement and V is a constant called the terminal speed. Take g = 32 ft/sec2 and V = 50 ft/sec. Sketch trajectories in the yy-phase plane for - 100 … y … 100, -100 … y … 100, starting from y = 0 and y = - 75, -50, -25, 0, 25, 50, and 75 ft/sec. Interpret the trajectories physically; why is V called the terminal speed? 35. Sticky Friction. An alternative for the damping friction model F = -by′ discussed in Section 4.1 is the “sticky friction” model. For a mass sliding on a surface as depicted in Figure 5.18, the contact friction is more complicated than simply - by′. We observe, for example, that even if the mass is displaced slightly off the equilibrium location y = 0, it may nonetheless remain stationary due to the fact that the spring force - ky is insufficient to break the static friction’s grip. If the maximum force that the friction can exert is denoted by m, then a feasible model is given by

Ffriction = d m sign1y2 , - m sign1y′2 ,

Figure 5.18 Mass–spring system with friction

(16)

d 2y

= -ky + Ffriction . dt 2 Thus, if the mass is at rest, friction balances the spring force if 0 y 0 6 m>k but simply opposes it with intensity m if 0 y 0 Ú m>k. If the mass is moving, friction opposes the velocity with the same intensity m. (a) Taking m = m = k = 1, convert (16) into the firstorder system m

y′ = y ,

if 0 y 0 6 1 and y = 0 , y′ = d -y + sign1y2 , if 0 y 0 Ú 1 and y = 0 , -y - sign1y2 , if y ≠ 0 . (b) Form the phase plane equation for (17) when y ≠ 0 and solve it to derive the solutions 0,

(17)

g dy dy ` ` , V 2 dt dt

ky ,

Friction

if 0 ky 0 6 m and y′ = 0 , if 0 ky 0 Ú m and y′ = 0 , if y′ ≠ 0 .

(The function sign (s) is + 1 when s 7 0, - 1 when s 6 0, and 0 when s = 0.) The motion is governed by the equation

y2 + 1y { 12 2 = c ,

where the plus sign prevails for y 7 0 and the minus sign for y 6 0. (c) Identify the trajectories in the phase plane as two families of concentric semicircles. What is the center of the semicircles in the upper half-plane? The lower half-plane? (d) What are the critical points for (17)? (e) Sketch the trajectory in the phase plane of the mass released from rest at y = 7.5. At what value for y does the mass come to rest? 36. Rigid Body Nutation. Euler’s equations describe the motion of the principal-axis components of the angular velocity of a freely rotating rigid body (such as a space station), as seen by an observer rotating with the body (the astronauts, for example). This motion is called nutation. If the angular velocity components are denoted by x, y, and z, then an example of Euler’s equations is the three-dimensional autonomous system dx>dt = yz , dy>dt = -2xz , dz>dt = xy .

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The trajectory of a solution x1t2, y1t2, z1t2 to these equations is the curve generated by the points 1 x1t2, y1t2, z1t2 2 in xyz-phase space as t varies over an interval I. (a) Show that each trajectory of this system lies on the surface of a (possibly degenerate) sphere centered at the origin 10, 0, 02. [Hint: Compute d 2 2 2 dt 1x + y + z 2.] What does this say about the magnitude of the angular velocity vector? (b) Find all the critical points of the system, i.e., all points 1x0, y0, z0 2 such that x1t2 K x0, y1t2 K y0, z1t2 K z0 is a solution. For such solutions, the angular velocity vector remains constant in the body system. (c) Show that the trajectories of the system lie along the intersection of a sphere and an elliptic cylinder of the form y2 + 2x2 = C, for some constant C. [Hint: Consider the expression for dy>dx implied by Euler’s equations.] (d) Using the results of parts (b) and (c), argue that the trajectories of this system are closed curves. What does this say about the corresponding solutions?

(e) Figure 5.19 displays some typical trajectories for this system. Discuss the stability of the three critical points indicated on the positive axes.

Figure 5.19 Trajectories for Euler’s system

5.5 Applications to Biomathematics:

Epidemic and Tumor Growth Models In this section we are going to survey some issues in biological systems that have been successfully modeled by differential equations. We begin by reviewing the population models described in Sections 3.2 and 5.3. In the Malthusian model, the rate of growth of a population p1t2 is proportional to the size of the existing population: (1)

dp = kp 1k 7 02 . dt

Cells that reproduce by splitting, such as amoebae and bacteria, are obvious biological examples of this type of growth. Equation (1) implies that a Malthusian population grows exponentially; there is no mechanism for constraining the growth. In Section 3.2 we saw that certain populations exhibit Malthusian growth over limited periods of time (as does compound interest).† Inserting a negative growth rate, (2)

dp = -kp , dt

results in solutions that decay exponentially. Their average lifetime is 1>k, and their half-life is 1ln 22 >k (Problems 6 and 8). In animals, certain organs such as the kidney serve to cleanse

†

Gordon E. Moore (1929–) has observed that the number of transistors on new integrated circuits produced by the electronics industry doubles every 24 months. “Moore’s law” is commonly cited by industrialists.

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Section 5.5

Applications to Biomathematics: Epidemic and Tumor Growth Models

275

the bloodstream of unwanted components (creatinine clearance, renal clearance), and their concentrations diminish exponentially. As a general rule, the body tends to dissipate ingested drugs in such a manner. (Of course, the most familiar physical instance of Malthusian disintegration is radioactive decay.) Note that if there are both growth and extinction processes, dp>dt = k + p - k- p and the equation in (1) still holds with k = k + - k- . When there are two-party interactions occurring in the population that decrease the growth rate, such as competition for resources or violent crime, the logistic model might be applicable; it assumes that the extinction rate is proportional to the number of possible pairs in the population, p1p - 12 >2: (3)

dp p ( p - 1) = k1p - k2 dt 2

or, equivalently,

dp = -Ap ( p - p1) .† dt

Rodent, bird, and plant populations exhibit logistic growth rates due to social structure, territoriality, and competition for light and space, respectively. The logistic function p1t2 =

p0p1

p0 + 1p1 - p0 2e-Ap1t

,

p0 J p102

was shown in Section 3.2 to be the solution of (3), and typical graphs of p1t2 were displayed there. In Section 5.3 we observed that the Volterra–Lotka model for two different populations, a predator x2 1t2 and a prey x1 1t2, postulates a Malthusian growth rate for the prey and an extinction rate governed by x1x2, the number of possible pairings of one from each population, (4)

dx1 = Ax1 - Bx1x2 , dt

while predators follow a Malthusian extinction rate and pairwise growth rate (5)

dx2 = -Cx2 + Dx1x2 . dt

Volterra–Lotka dynamics have been observed in blood vessel growth (predator = new capillary tips; prey = chemoattractant), fish populations, and several animal–plant interactions. Systems like (4)–(5) were studied in Section 5.3 with the aid of the Runge–Kutta algorithm. Now, armed with the insights of Section 5.4, we can further explore this model theoretically. First, we perform a “reality check” by proving that the populations x1 1t2, x2 1t2 in the Volterra–Lotka model never change sign. Separating (4) leads to d ln x1 1 dx1 = = A - Bx2 , x1 dt dt while integrating from 0 to t results in (6)

x1 1t2 = x1 102e 10{A - Bx2(t)} dt , t

and the exponential factor is always positive. Thus x1 1t2 [and similarly x2 1t2] retains its initial sign (negative populations never arise).

†

One might propose that the growth rate in animal populations is due to two–party interactions as well. (Wink, wink.) However, in monogamous societies, the number of pairs participating in procreation is proportional to p>2, leading to (1). A growth rate determined by all possible pairings p1p - 12 >2 would indicate a highly hedonistic social order.

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276

Chapter 5

Example 1 Solution

Introduction to Systems and Phase Plane Analysis

Find and interpret the critical points for the Volterra–Lotka model (4)–(5). The system dx1 A = Ax1 - Bx1x2 = -Bx1 a x2 - b = 0 , dt B (7) dx2 C = -Cx2 + Dx1x2 = Dx2 a x1 - b = 0 dt D has the trivial solution x1 1t2 K x2 1t2 K 0, with an obvious interpretation in terms of populations. If all four coefficients A, B, C, and D are positive, there is also the more interesting solution A C , x1 1t2 K . B D At these population levels, the growth and extinction rates for each species cancel. The direction field diagram in Figure 5.20 for the phase plane equation dx2 -Cx2 + Dx1x2 x2 -C + Dx1 (9) = = x1 A - Bx2 dx1 Ax1 - Bx1x2 (8)

x2 1t2 K

suggests that this equilibrium is a center (compare Figure 5.12 on page 267) with closed (periodic) neighboring trajectories, in accordance with the simulations in Section 5.3. However it is conceivable that some spiral trajectories might snake through the field pattern and approach the critical point asymptotically. A rather tricky argument in Problem 4 demonstrates that this is not the case. ◆ The SIR Epidemic Model. The SIR† model for an epidemic addresses the spread of diseases that are only contracted by contact with an infected individual; its victims, once recovered, are immune to further infection and are themselves noninfectious. So the members of a population of size N fall into three classes: S1t2 = the number of susceptible individuals—that is, those who have not been infected; s J S/N is the fraction of susceptibles. I1t2 = the number of individuals who are currently infected, comprising a fraction i J I/N of the population. R1t2 = the number of individuals who have recovered from infection, comprising the fraction r J R/N. x1

x2 (A/B, C/D) Figure 5.20 Typical direction field diagram for the Volterra–Lotka system †

Introduced by W. O. Kermack and A. G. McKendrick in “A Contribution to the Mathematical Theory of Epidemics,” Proc. Royal Soc. London, Vol. A115 (1927): 700–721.

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Section 5.5

Applications to Biomathematics: Epidemic and Tumor Growth Models

277

The classic SIR epidemic model assumes that on the average, an infectious individual encounters a people per unit time (usually per week). Thus, a total of aI people per week are contacted by infectees, but only a fraction s = S/N of them are susceptible. So the susceptible population diminishes at a rate (10)

dS = -saI or (dividing by N) dt

(11)

ds = -asi . dt

The parameter a is crucial in disease control. Crowded conditions, or high a, make it difficult to combat the spread of infection. Ideally, we would quarantine the infectees (low a) to fight the epidemic. The infected population is (obviously) increased whenever a susceptible individual is infected. Additionally, infectees recover in a Malthusian-disintegrative manner over an average time of, say, 1>k weeks [recall (1)], so the infected population changes at a rate (12)

k dI = saI - kI = aa s - b I or a dt

(13)

di k = aa s - b i . a dt

And, of course, the population of recovered individuals increases whenever an infectee is healed: (14)

dR dr = kI or = ki . dt dt With the SIR model, the total population count remains unchanged: d(S + I + R) = -saI + saI - kI + kI = 0 . dt

Thus, any fatalities are tallied in the “recovered/noninfectious” population R. Interestingly, equations (11) and (13) do not contain R or r, so they are suitable for phase plane analysis. In fact they constitute a Volterra–Lotka system with A = 0, B = D = a, and C = k. Because the coefficient A is zero, the critical point structure is different from that discussed in Example 1. Specifically, if -asi in (11) is zero, then only -ki remains on the right in (13), so I1t2 = i1t2 K 0 is necessary and sufficient for a critical point, with S unrestricted. (Physically, this means the populations remain stable only if there are no carriers of the infection.) Our earlier argument has shown that if s1t2 and i1t2 are initially positive, they remain so. As a result we conclude from (11) immediately that s1t2 decreases monotonically; as such, it has a limiting value s1 ∞ 2 as t S ∞. Does i1t2 have a limiting value also? If so, 5s1 ∞ 2, i1 ∞ 26 would be a critical point by Theorem 1 of Section 5.4, page 266, and thus i1 ∞ 2 = 0. To analyze i1t2 consider the phase plane equation for (11) and (13): (15)

di k asi - ki = = -1 + , as ds -asi

which has solutions (16)

i = -s +

k ln s + K . a

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200 180 160

Deaths due to flu

140 120 100 80 60 40 20 0

1

2

3

4

5

6

7

8

9

10

11

12

13

Week Figure 5.21 Mortality data for Hong Kong flu, New York City

From (16) we see that s1 ∞ 2 cannot be zero; otherwise the right-hand side would eventually be negative, contradicting i1t2 7 0. Therefore, (16) demonstrates that i1t2 does have a limiting value i1 ∞ 2 = -s1 ∞ 2 + 1k>a2 ln s1 ∞ 2 + K. As noted, i1 ∞ 2 must then be zero. From (13) we further conclude that if s102 exceeds the “threshold value” k>a, the infected fraction i1t2 will initially increase 1di>dt 7 0 at t = 02 before eventually dying out. The peak value of i1t2 occurs when di>dt = 0 = a3s - 1k>a24i, i.e., when s1t2 passes through the value k>a. In the jargon of epidemiology, this phenomenon defines an “epidemic.” You will be directed in Problem 10 to show that if s102 … k>a, the infected population diminishes monotonically, and no epidemic develops. Example 2 †

According to data issued by the Centers for Disease Control and Prevention (CDC) in Atlanta, Georgia, the Hong Kong flu epidemic during the winter of 1968–1969 was responsible for 1035 deaths in New York City (population 7,900,000), according to the time chart in Figure 5.21. Analyze this data with the SIR model.

Solution

Of course, we need to make some assumptions about the parameters. First of all, only a small percentage of people who contract Hong Kong flu perish, so let’s assume that the chart reflects a scaled version of the infected population fraction i1t2. It is known that the recovery period for this flu is around 5 days, or 5>7 week, so we try k = 7>5 = 1.4. And since the infectees spend much of their convalescence in bed, the average contact rate a is probably less than 1 person per day or 7 per week. The CDC estimated that the initial infected population I102 was about 10, so the initial data for (11), (13), and (14) are s102 =

7,900,000 - 10 10 ≈ 0.9999987, i102 = ≈ 1.2658 * 10-6, r102 = 0 . 7,900,000 7,900,000

†

We borrow liberally from “The SIR Model for Spread of Disease” by Duke University’s David Smith and Lang Moore, Journal of Online Mathematics and Its Applications, The MAA Mathematical Sciences Digital Library, http://mathdl.maa.org/mathDL/4/?pa=content&sa=viewDocument&nodeId=479&bodyId=612, copyright 2000, CCP and the authors, published December, 2001. The article contains much interesting information about this epidemic.

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Section 5.5

1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

R

I t

0 10

15

20

S

1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

S

5

279

Applications to Biomathematics: Epidemic and Tumor Growth Models

25

30

S

1.0

R 0.8

R

0.6 0.4 I

0.2 0

I

t

0 5

10

(a)

15

20

25

t

–0.2

30

(b)

0

5

10

15

20

25

30

(c)

Figure 5.22 SIR simulations (a) k = 1.4, a = 2.0; (b) k = 1.4, a = 3.5; (c) k = 1.4, a = 6.0

Numerical simulations of this system are displayed in Figure 5.22.† The contact rate a = 3.5 per week generates an infection fraction curve that closely matches the mortality data’s characteristics: time of peak and duration of epidemic. ◆ A Tumor Growth Model.‡ The observed growth of certain tumors can be explained by a model that is mathematically similar to the epidemic model. The total number of cells N in the tumor subdivides into a population P that proliferates by splitting (Malthusian growth) and a population Q that remains quiescent. However, the proliferating cells also can make a transition to the quiescent state, and this occurrence is modeled as a Malthusian-like decay with a “rate” r1N2 that increases with the overall size of the tumor: (17)

dP = cP − r1N2P , dt

(18)

dQ = r1N2P . dt

Thus the total population N increases only when the proliferating cells split, as can be seen by adding the equations (17) and (18): (19)

d(P + Q) dN = = cP . dt dt We take (17) and (19) as the system for our analysis. The phase plane equation

(20)

cP - r1N2P r1N2 dP = = 1c dN cP

can be integrated, leading to a formula for P in terms of N (21)

P = N-

1 r1N2 dN + K . cL

†

Appendix G describes various websites and commercial software that sketch direction fields and automate most of the differential equation algorithms discussed in this book. ‡

The authors wish to thank Dr. Glenn Webb of Vanderbilt University for this application. See M. Gyllenberg and G. F. Webb, “Quiescence as an Explanation of Gompertz Tumor Growth,” Growth, Development, and Aging, Vol. 53 (1989): 25–55.

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Introduction to Systems and Phase Plane Analysis

Suppose the initial conditions are P102 = 1, Q102 = 0, and N102 = 1 (a single proliferating cell). Then we can eliminate the nuisance constant K by taking the indefinite integral in (21) to run from 1 to N and evaluating at t = 0: 1

1 = 1-

1 r(N) dN + K 1 K = 0 . c L1

Insertion of (21) with K = 0 into (19) produces a differential equation for N alone: N

(22) Example 3

dN = cN r1u2 du . dt L1

The Gompertz law (23)

N1t2 = ec11 - e

-bt

2/b

has been observed experimentally for the growth of some tumors. Show that a transition rate r1N2 of the form b11 + ln N2 predicts Gompertzian growth. Solution

If the indicated integral of the rate r(N) is carried out, (22) becomes (24)

dN = cN - b1N - 12 - b1N ln N - N + 12 = 1c - b ln N2 N . dt

Dividing by N we obtain a linear differential equation for the function ln N d ln N = -b ln N + c dt whose solution, for the initial condition N102 = 1, is found by the methods of Section 2.3 to be c ln N1t2 = 11 - e-bt 2 , b confirming (23). ◆ Problem 9 invites the reader to show that if the growth rate is modeled as r1N2 = s12N - 12, then the solution of (22) describes logistic growth. Typical curves for the Gompertz and logistic models are displayed in Figure 5.23. See also Figure 3.4 on page 98. Other applications of differential equations to biomathematics appear in the discussions of artificial respiration (Project B, page 81) in Chapter 2, HIV infection (Project A, page 141) and aquaculture (Project B, page 144) in Chapter 3, and spread of staph infections (Project B) and the Ebola epidemic (Project F, page 314) in this chapter. N

N

t

t (b)

(a) Figure 5.23 (a) Gompertz and (b) logistic curves

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Section 5.5

5.5

281

Applications to Biomathematics: Epidemic and Tumor Growth Models

EXERCISES

1. Logistic Model. tic equation

In Section 3.2 we discussed the logis-

dp = Ap1p - Ap2 , dt

p102 = p0 ,

and its use in modeling population growth. A more general model might involve the equation (25)

dp = Ap1p - Apr , dt

is secreted on a 24-h cycle, then the rate of change of the level of the hormone in the blood may be represented by the initial value problem

p102 = p0 ,

where r 7 1. To see the effect of changing the parameter r in (25), take p1 = 3, A = 1, and p0 = 1. Then use a numerical scheme such as Runge–Kutta with h = 0.25 to approximate the solution to (25) on the interval 0 … t … 5 for r = 1.5, 2, and 3. What is the limiting population in each case? For r 7 1, determine a general formula for the limiting population. 2. Radioisotopes and Cancer Detection. A radioisotope commonly used in the detection of breast cancer is technetium-99m. This radionuclide is attached to a chemical that upon injection into a patient accumulates at cancer sites. The isotope’s radiation is then detected and the site located, using gamma cameras or other tomographic devices. Technetium-99m decays radioactively in accordance with the equation dy>dt = - ky, with k = 0.1155>h. The short half-life of technetium-99m has the advantage that its radioactivity does not endanger the patient. A disadvantage is that the isotope must be manufactured in a cyclotron. Since hospitals are not equipped with cyclotrons, doses of technetium-99m have to be ordered in advance from medical suppliers. Suppose a dosage of 5 millicuries (mCi) of technetium-99m is to be administered to a patient. Estimate the delivery time from production at the manufacturer to arrival at the hospital treatment room to be 24 hours and calculate the amount of the radionuclide that the hospital must order, to be able to administer the proper dosage. 3. Secretion of Hormones. The secretion of hormones into the blood is often a periodic activity. If a hormone

dx pt = a - b cos - kx , x102 = x0 , dt 12 where x1t2 is the amount of the hormone in the blood at time t, a is the average secretion rate, b is the amount of daily variation in the secretion, and k is a positive constant reflecting the rate at which the body removes the hormone from the blood. If a = b = 1, k = 2, and x0 = 10, solve for x1t2. 4. Prove that the critical point (8) of the Volterra–Lotka system is a center; that is, the neighboring trajectories are periodic. [Hint: Observe that (9) is separable and show that its solutions can be expressed as [xA2 e-Bx2] # [xC1 e-Dx1] = K .4

(26)

Prove that the maximum of the function xpe-qx is 1p>qe2 p, occurring at the unique value x = p>q (see Figure 5.24), so the critical values (8) maximize the factors on the left in (26). Argue that if K takes the corresponding maximum value 1A>Be2 A 1C>De2 C, the critical point (8) is the (unique) solution of (26), and it cannot be an endpoint of any trajectory for (26) with a lower value of K.†

1

e

xe2x

x 1 Figure 5.24 Graph of xe-x

5. Suppose for a certain disease described by the SIR model it is determined that a = 0.003 and b = 0.5. (a) In the SI-phase plane, sketch the trajectory corresponding to the initial condition that one person is infected and 700 persons are susceptible.

†

In fact, the periodic fluctuations predicted by the Volterra–Lotka model were observed in fish populations by Lotka’s son-in-law, Humberto D’Ancona.

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9. Show that with the transition rate formula r1N2 = s12N - 12, equation (22) takes the form of the equation for the logistic model (Section 3.2, equation (14)). Solve (22) for this case. 10. Prove that the infected population I1t2 in the SIR model does not increase if S102 is less than or equal to k>a. 11. An epidemic reported by the British Communicable Disease Surveillance Center in the British Medical Journal (March 4, 1978, p. 587) took place in a boarding school with 763 residents.† The statistics for the infected population are shown in the graph in Figure 5.25. Assuming that the average duration of the infection is 2 days, use a numerical differential equation solver (see Appendix G) to try to reproduce the data. Take S102 = 762, I102 = 1, R102 = 0 as initial conditions. Experiment with reasonable estimates for the average number of contacts per day by the infected students, who were confined to bed after the infection was detected. What value of this parameter seems to fit the curve best?

(b) From your graph in part (a), estimate the peak number of infected persons. Compare this with the theoretical prediction S = k>a ≈ 167 persons when the epidemic is at its peak. 6. Show that the half-life of solutions to (2)—that is, the time required for the solution to decay to one-half of its value—equals 1ln 22 >k. 7. Complete the solution of the tumor growth model for Example 3 on page 280 by finding P1t2 and Q1t2. 8. If p1t2 is a Malthusian population that diminishes according to (2), then p1t2 2 - p1t1 2 is the number of individuals in the population whose lifetime lies between t1 and t2. Argue that the average lifetime of the population is given by the formula ∞

L0

t ` dt ` dt dp1t2

p102 and show that this equals 1>k.

300

Infected Students

250

200

150

100

50

0

1

2

3

4

5

6

7 Day

8

9

10

11

12

13

14

15

Figure 5.25 Flu data for Problem 11

†

See also the discussion of this epidemic in Mathematical Biology I, An Introduction, by J. D. Murray (SpringerVerlag, New York, 2002), 325–326.

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Section 5.6

283

Coupled Mass–Spring Systems

5.6 Coupled Mass–Spring Systems In this section we extend the mass–spring model of Chapter 4 to include situations in which coupled springs connect two masses, both of which are free to move. The resulting motions can be very intriguing. For simplicity we’ll neglect the effects of friction, gravity, and external forces. Let’s analyze the following experiment. Example 1

On a smooth horizontal surface, a mass m1 = 2 kg is attached to a fixed wall by a spring with spring constant k1 = 4 N/m. Another mass m2 = 1 kg is attached to the first object by a spring with spring constant k2 = 2 N/m. The objects are aligned horizontally so that the springs are their natural lengths (Figure 5.26). If both objects are displaced 3 m to the right of their equilibrium positions (Figure 5.27) and then released, what are the equations of motion for the two objects?

k1 5 4

k1 5 4

k2 5 2 2 kg

x>0

y>0

x 50

y 50

Figure 5.26 Coupled system at equilibrium

Solution

k 2 52

1 kg

2 kg

1 kg

x 53

y 53

x 50

y 50

Figure 5.27 Coupled system at initial displacement

From our assumptions, the only forces we need to consider are those due to the springs themselves. Recall that Hooke’s law asserts that the force acting on an object due to a spring has magnitude proportional to the displacement of the spring from its natural length and has direction opposite to its displacement. That is, if the spring is either stretched or compressed, then it tries to return to its natural length. Because each mass is free to move, we apply Newton’s second law to each object. Let x1t2 denote the displacement (to the right) of the 2-kg mass from its equilibrium position and similarly, let y1t2 denote the corresponding displacement for the 1-kg mass. The 2-kg mass has a force F1 acting on its left side due to one spring and a force F2 acting on its right side due to the second spring. Referring to Figure 5.27 and applying Hooke’s law, we see that F1 = -k1x ,

F2 = +k2 1y - x2 ,

since 1y - x2 is the net displacement of the second spring from its natural length. There is only one force acting on the 1-kg mass: the force due to the second spring, which is F3 = -k2 1y - x2 .

Applying Newton’s second law to these objects, we obtain the system d 2x = F1 + F2 = -k1x + k2 1y - x2 , dt 2 d 2y m2 2 = F3 = -k2 1y - x2 , dt

m1 (1) or

d 2x + 1k1 + k2 2x − k2y = 0 , dt 2 d 2y m2 2 + k2y − k2x = 0 . dt m1

(2)

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In this problem, we know that m1 = 2, m2 = 1, k1 = 4, and k2 = 2. Substituting these values into system (2) yields (3) (4)

d 2x + 6x - 2y = 0 , dt 2 d 2y + 2y - 2x = 0 . dt 2

2

We’ll use the elimination method of Section 5.2 to solve (3)–(4). With D J d>dt we rewrite the system as (5) (6)

12D2 + 623x4 - 2y = 0 ,

-2x + 1D2 + 223y4 = 0 .

Adding 1D2 + 22 applied to equation (5) to 2 times equation (6) eliminates y: 3 1D2 + 2212D2 + 62 - 44 3x4 = 0 ,

which simplifies to (7)

2

d 4x d 2x + 10 + 8x = 0 . dt 4 dt 2

Notice that equation (7) is linear with constant coefficients. To solve it let’s proceed as we did with linear second-order equations and try to find solutions of the form x = ert. Substituting ert in equation (7) gives 21r 4 + 5r 2 + 42ert = 0 . Thus, we get a solution to (7) when r satisfies the auxiliary equation r 4 + 5r 2 + 4 = 0 .

From the factorization r 4 + 5r 2 + 4 = 1r 2 + 121r 2 + 42, we see that the roots of the auxiliary equation are complex numbers i, -i, 2i, -2i. Using Euler’s formula, it follows that z1 1t2 = eit = cos t + i sin t

and

z2 1t2 = e2it = cos 2t + i sin 2t

are complex-valued solutions to equation (7). To obtain real-valued solutions, we take the real and imaginary parts of z1 1t2 and z2 1t2. Thus, four real-valued solutions are x1 1t2 = cos t ,

x2 1t2 = sin t ,

x3 1t2 = cos 2t ,

x4 1t2 = sin 2t ,

and a general solution is (8)

x1t2 = a1 cos t + a2 sin t + a3 cos 2t + a4 sin 2t ,

where a1, a2, a3, and a4 are arbitrary constants.† To obtain a formula for y1t2, we use equation (3) to express y in terms of x: d 2x + 3x dt 2 = -a1 cos t - a2 sin t - 4a3 cos 2t - 4a4 sin 2t

y1t2 =

+ 3a1 cos t + 3a2 sin t + 3a3 cos 2t + 3a4 sin 2t ,

†

A more detailed discussion of general solutions is given in Chapter 6.

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Section 5.6

Coupled Mass–Spring Systems

285

and so y1t2 = 2a1 cos t + 2a2 sin t − a3 cos 2t − a4 sin 2t .

(9)

To determine the constants a1, a2, a3, and a4, let’s return to the original problem. We were told that the objects were originally displaced 3 m to the right and then released. Hence, dx 102 = 0 ; dt

x102 = 3 ,

(10)

y102 = 3 ,

dy 102 = 0 . dt

On differentiating equations (8) and (9), we find dx = -a1 sin t + a2 cos t - 2a3 sin 2t + 2a4 cos 2t , dt dy = -2a1 sin t + 2a2 cos t + 2a3 sin 2t - 2a4 cos 2t . dt Now, if we put t = 0 in the formulas for x, dx>dt, y, and dy>dt, the initial conditions (10) give the four equations dx 102 = a2 + 2a4 = 0 , dt dy 102 = 2a2 - 2a4 = 0 . dt

x102 = a1 + a3 = 3 , y102 = 2a1 - a3 = 3 ,

From this system, we find a1 = 2, a2 = 0, a3 = 1, and a4 = 0. Hence, the equations of motion for the two objects are x1t2 = 2 cos t + cos 2t , y1t2 = 4 cos t - cos 2t , which are depicted in Figure 5.28. ◆

24

22

2

4

x

24

22

2

5

5

10

10

15

15

4

y

20

20 t

t

Figure 5.28 Graphs of the motion of the two masses in the coupled mass–spring system

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The general solution pair (8), (9) that we have obtained is a combination of sinusoids oscillating at two different angular frequencies: 1 rad/sec and 2 rad/sec. These frequencies extend the notion of the natural frequency of the single (free, undamped) mass–spring oscillator (page 212, Section 4.9) and are called the natural (or normal) angular frequencies† of the system. A complex system consisting of more masses and springs would have many normal frequencies. Notice that if the initial conditions were altered so that the constants a3 and a4 in (8) and (9) were zero, the motion would be a pure sinusoid oscillating at the single frequency 1 rad/sec. Similarly, if a1 and a2 were zero, only the 2 rad/sec oscillation would be “excited.” Such solutions, wherein the entire motion is described by a single sinusoid, are called the normal modes of the system.‡ The normal modes in the following example are particularly easy to visualize because we will take all the masses and all the spring constants to be equal. Example 2

Three identical springs with spring constant k and two identical masses m are attached in a straight line with the ends of the outside springs fixed (see Figure 5.29). Determine and interpret the normal modes of the system.

Solution

We define the displacements from equilibrium, x and y, as in Example 1 on page 283. The equations expressing Newton’s second law for the masses are quite analogous to (1), except for the effect of the third spring on the second mass: (11)

mx″ = -kx + k1y - x2 ,

(12)

my″ = -k1y - x2 - ky ,

or

1mD2 + 2k23x4 - ky = 0 ,

-kx + 1mD2 + 2k23y4 = 0 .

Eliminating y in the usual manner results in (13)

3 1mD2 + 2k2 2 - k 2 4 3x4

= 0.

This has the auxiliary equation

1mr 2 + 2k2 2 - k 2 = 1mr 2 + k21mr 2 + 3k2 = 0 ,

with roots { i2k>m, { i23k>m. Setting v J 2k>m, we get the following general solution to (13): (14)

x1t2 = C1 cos vt + C2 sin vt + C3 cos 1 23vt 2 + C4 sin 1 23vt 2 .

k

k

m

x.0 x50

k

m

y.0 y50

Figure 5.29 Coupled mass–spring system with fixed ends

†

The study of the natural frequencies of oscillations of complex systems is known in engineering as modal analysis. The normal modes are more naturally characterized in terms of eigenvalues (see Section 9.5).

‡

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Section 5.6

y

x

1

x

1

y

21

1

5

5

5

5

10

10

10

10

15

15

15

15

20

20

20

20

t

t

287

Coupled Mass–Spring Systems

t

t

(a)

(b) Figure 5.30 Normal modes for Example 2

To obtain y1t2, we solve for y1t2 in (11) and substitute x1t2 as given in (14). Upon simplifying, we get (15)

y1t2 = C1 cos vt + C2 sin vt - C3 cos 1 23vt 2 - C4 sin 1 23vt 2 .

From the formulas (14) and (15), we see that the normal angular frequencies are v and 23 v. Indeed, if C3 = C4 = 0, we have a solution where y1t2 K x1t2, oscillating at the angular frequency v = 2k>m rad/sec (equivalent to a frequency 2k>m>2p periods/sec). Now if x1t2 K y1t2 in Figure 5.29, the two masses are moving as if they were a single rigid body of mass 2m, forced by a “double spring” with a spring constant given by 2k. And indeed, according to equation (4) of Section 4.9 (page 212), we would expect such a system to oscillate at the angular frequency 22k>2m = 2k>m (!). This motion is depicted in Figure 5.30(a). Similarly, if C1 = C2 = 0, we find the second normal mode where y1t2 = -x1t2, so that in Figure 5.29 there are two mirror-image systems, each with mass m and a “spring and a half” with spring constant k + 2k = 3k. (The half-spring will be twice as stiff.) Section 4.9’s equation (4) then predicts an angular oscillation frequency for each system of 23k>m = 23 v, which again is consistent with (14) and (15). This motion is shown in Figure 5.30(b). ◆

5.6

EXERCISES

1. Two springs and two masses are attached in a straight line on a horizontal frictionless surface as illustrated in Figure 5.31. The system is set in motion by holding the mass m2 at its equilibrium position and pulling the mass m1 to the left of its equilibrium position a distance 1 m and then releasing both masses. Express Newton’s law for the system and determine the equations of motion for the two masses if m1 = 1 kg, m2 = 2 kg, k1 = 4 N/m, and k2 = 10>3 N/m.

m1 x.0 x50

k1

k2

m2

y.0 y50

Figure 5.31 Coupled mass–spring system with one end free

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2. Determine the equations of motion for the two masses described in Problem 1 if m1 = 1 kg, m2 = 1 kg, k1 = 3 N/m, and k2 = 2 N/m. 3. Four springs with the same spring constant and three equal masses are attached in a straight line on a horizontal frictionless surface as illustrated in Figure 5.32. Determine the normal frequencies for the system and describe the three normal modes of vibration. k

k

m x. 0 x 50

k

m

k

m

y.0

z.0

y 50

z50

Figure 5.32 Coupled mass–spring system with three degrees of freedom

4. Two springs, two masses, and a dashpot are attached in a straight line on a horizontal frictionless surface as shown in Figure 5.33. The dashpot provides a damping force on mass m2, given by F = - by′. Derive the system of differential equations for the displacements x and y.

is applied to the second object of mass 1 kg. The displacement functions x1t2 and y1t2 now satisfy the system (16)

2x″1t2 + 6x1t2 − 2y1t2 = 0 ,

(17)

y″1t2 + 2y1t2 − 2x1t2 = 37 cos 3t .

(a) Show that x1t2 satisfies the equation (18)

x 142 1t2 + 5x″1t2 + 4x1t2 = 37 cos 3t .

(b) Find a general solution x1t2 to equation (18). [Hint: Use undetermined coefficients with xp = A cos 3t + B sin 3t.] (c) Substitute x1t2 back into (16) to obtain a formula for y1t2. (d) If both masses are displaced 2 m to the right of their equilibrium positions and then released, find the displacement functions x1t2 and y1t2. 7. Suppose the displacement functions x1t2 and y1t2 for a coupled mass–spring system (similar to the one discussed in Problem 6) satisfy the initial value problem x″1t2 + 5x1t2 − 2y1t2 = 0 , y″1t2 + 2y1t2 − 2x1t2 = 3 sin 2t ; x102 = x′102 = 0 ,

k1

m1

k2

b

m2

x 50

y 50

Figure 5.33 Coupled mass–spring system with one end damped

5. Two springs, two masses, and a dashpot are attached in a straight line on a horizontal frictionless surface as shown in Figure 5.34. The system is set in motion by holding the mass m2 at its equilibrium position and pushing the mass m1 to the left of its equilibrium position a distance 2 m and then releasing both masses. Determine the equations of motion for the two masses if m1 = m2 = 1 kg, k1 = k2 = 1 N/m, and b = 1 N-sec/m. [Hint: The dashpot damps both m1 and m2 with a force whose magnitude equals b 0 y′ - x′ 0 .] k1

m1 x.0 x 50

b

m2

y′102 = 0 .

Solve for x1t2 and y1t2. 8. A double pendulum swinging in a vertical plane under the influence of gravity (see Figure 5.35) satisfies the system

y.0

x.0

y102 = 1 ,

1m1 + m2 2l 12u 1> + m2l1l2u 2> + 1m1 + m2 2l1gu1 = 0 , m2l 22u 2> + m2l1l2u 1> + m2l2gu2 = 0 ,

l1

m1 l2

k2

y.0

m2

y 50

Figure 5.34 Coupled mass–spring system with damping between the masses

6. Referring to the coupled mass–spring system discussed in Example 1, suppose an external force E1t2 = 37 cos 3t

Figure 5.35 Double pendulum

when u1 and u2 are small angles. Solve the system when m1 = 3 kg, m2 = 2 kg, l1 = l2 = 5 m, u1 102 = p>6, u2 102 = u 1= 102 = u 2= 102 = 0.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 5.7

9. The motion of a pair of identical pendulums coupled by a spring is modeled by the system mg mx1> = x - k1x1 - x2 2 , l 1 mg mx2> = x + k1x1 - x2 2 l 2 for small displacements (see Figure 5.36). Determine the two normal frequencies for the system. 10. Suppose the coupled mass–spring system of Problem 1 (Figure 5.31) is hung vertically from a support (with mass m2 above m1), as in Section 4.10, page 226. (a) Argue that at equilibrium, the lower spring is stretched a distance l1 from its natural length L1, given by l1 = m1g>k1. (b) Argue that at equilibrium, the upper spring is stretched a distance l2 = 1m1 + m2 2g>k2. (c) Show that if x1 and x2 are redefined to be displacements from the equilibrium positions of the masses m1 and m2, then the equations of motion are identical with those derived in Problem 1.

Electrical Systems

l

289

l

m x1

k

m x2

Figure 5.36 Coupled pendulums

5.7 Electrical Systems The equations governing the voltage–current relations for the resistor, inductor, and capacitor were given in Section 3.5, together with Kirchhoff’s laws that constrain how these quantities behave when the elements are electrically connected into a circuit. Now that we have the tools for solving linear equations of higher-order systems, we are in a position to analyze more complex electrical circuits. Example 1

The series RLC circuit in Figure 5.37 on page 290 has a voltage source given by E1t2 = sin 100t volts (V), a resistor of 0.02 ohms 1Ω 2, an inductor of 0.001 henrys (H), and a capacitor of 2 farads (F). (These values are selected for numerical convenience; typical capacitance values are much smaller.) If the initial current and the initial charge on the capacitor are both zero, determine the current in the circuit for t 7 0.

Solution

Using the notation of Section 3.5, we have L = 0.001 H, R = 0.02 Ω, C = 2 F, and E1t2 = sin 100t. According to Kirchhoff’s current law, the same current I passes through each circuit element. The current through the capacitor equals the instantaneous rate of change of its charge q: (1)

I = dq>dt .

From the physics equations in Section 3.5, we observe that the voltage drops across the capacitor (EC), the resistor (ER), and the inductor (EL) are expressed as (2)

EC =

q , C

ER = RI ,

EL = L

dI . dt

Therefore, Kirchhoff’s voltage law, which implies EL + ER + EC = E, can be expressed as (3)

L

1 dI + RI + q = E1t2 . dt C

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Resistance R

Voltage source

E

Inductance L

Capacitance C Figure 5.37 Schematic representation of an RLC series circuit

In most applications we will be interested in determining the current I1t2. If we differentiate (3) with respect to t and substitute I for dq>dt, we obtain dE d 2I dI 1 +R + I= . 2 dt C dt dt After substitution of the given values this becomes

(4)

L

10.0012

dI d 2I + 10.022 + 10.52I = 100 cos 100t , 2 dt dt

or, equivalently, (5)

d 2I dI + 20 + 500I = 100,000 cos 100t . 2 dt dt The homogeneous equation associated with (5) has the auxiliary equation r 2 + 20r + 500 = 1r + 102 2 + 1202 2 = 0 ,

whose roots are -10 { 20i. Hence, the solution to the homogeneous equation is (6)

Ih 1t2 = C1e-10t cos 20t + C2e-10t sin 20t .

To find a particular solution for (5), we can use the method of undetermined coefficients. Setting Ip 1t2 = A cos 100t + B sin 100t

and carrying out the procedure in Section 4.5, we ultimately find, to three decimals, A = -10.080 ,

B = 2.122 .

Hence, a particular solution to (5) is given by (7)

Ip 1t2 = -10.080 cos 100t + 2.122 sin 100t .

Since I = Ih + Ip, we find from (6) and (7) that (8)

I1t2 = e-10t 1C1 cos 20t + C2 sin 20t2 - 10.080 cos 100t + 2.122 sin 100t .

To determine the constants C1 and C2, we need the values I102 and I′102. We were given I102 = q102 = 0. To find I′102, we substitute the values for L, R, and C into equation (3) and equate the two sides at t = 0. This gives 10.0012I′102 + 10.022I102 + 10.52q102 = sin 0 .

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Section 5.7

Electrical Systems

291

Because I102 = q102 = 0, we find I′102 = 0. Finally, using I1t2 in (8) and the initial conditions I102 = I′102 = 0, we obtain the system I102 = C1 - 10.080 = 0 , I′102 = -10C1 + 20C2 + 212.2 = 0 . Solving this system yields C1 = 10.080 and C2 = -5.570. Hence, the current in the RLC series circuit is (9)

I1t2 = e-10t 110.080 cos 20t - 5.570 sin 20t2 - 10.080 cos 100t + 2.122 sin 100t . ◆

Observe that, as was the case with forced mechanical vibrations, the current in (9) is made up of two components. The first, Ih, is a transient current that tends to zero as t S + ∞. The second, Ip 1t2 = -10.080 cos 100t + 2.122 sin 100t ,

is a sinusoidal steady-state current that remains. It is straightforward to verify that the steady-state solution Ip 1t2 that arises from the more general voltage source E1t2 = E0 sin gt is (10)

Ip 1t2 =

E0 sin1gt + u2

2R + 3gL - 1> 1gC24 2 2

,

where tan u = 11>C - Lg 2 2 > 1gR2 (compare Section 4.10, page 221). Example 2

At time t = 0, the charge on the capacitor in the electrical network shown in Figure 5.38 is 2 coulombs (C), while the current through the capacitor is zero. Determine the charge on the capacitor and the currents in the various branches of the network at any time t 7 0.

Solution

To determine the charge and currents in the electrical network, we begin by observing that the network consists of three closed circuits: loop 1 through the battery, resistor, and inductor; loop 2 through the battery, resistor, and capacitor; and loop 3 containing the capacitor and inductor. Taking advantage of Kirchhoff’s current law, we denote the current passing through the battery and the resistor by I1, the current through the inductor by I2, and the current through the capacitor by I3. For consistency of notation, we denote the charge on the capacitor by q3. Hence, I3 = dq3 >dt.

20 ohms (V)

I3

A I2

I1

5 volts (V)

I1

1q3 1 160

1 henry (H)

loop 1

farads (F)

2q3 loop 3 I1

B

loop 2

I3

Figure 5.38 Schematic of an electrical network

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Introduction to Systems and Phase Plane Analysis

As discussed at the beginning of this section, the voltage drop at a resistor is RI, at an inductor LdI>dt, and at a capacitor q>C. So, applying Kirchhoff’s voltage law to the electrical network in Figure 5.38, we find for loop 1, dI2 dt

(11)

+

20I1

(inductor)

=

(resistor)

5; (battery)

for loop 2, (12)

20I1 (resistor)

+

160q3 (capacitor)

=

5; (battery)

and for loop 3, (13)

−

dI2 dt

+

(inductor)

160q3 = 0 . (capacitor)

[The minus sign in (13) arises from taking a clockwise path around loop 3 so that the current passing through the inductor is –I2.] Notice that these three equations are not independent: We can obtain equation (13) by subtracting (11) from (12). Hence, we have only two equations from which to determine the three unknowns I1, I2, and q3. If we now apply Kirchhoff’s current law to the two junction points in the network, we find at point A that I1 - I2 - I3 = 0 and at point B that I2 + I3 - I1 = 0. In both cases, we get (14)

I1 - I2 -

dq3 = 0, dt

since I3 = dq3 >dt. Assembling equations (11), (12), and (14) into a system, we have (with D = d>dt) (15) (16) (17)

DI2 + 20I1 = 5 , 20I1 + 160q3 = 5 , -I2 + I1 - Dq3 = 0 .

We solve these by the elimination method of Section 5.2. Using equation (16) to eliminate I1 from the others, we are left with (18) (19)

DI2 - 160q3 = 0 , 20I2 + 120D + 1602q3 = 5 .

Elimination of I2 then leads to (20)

20D2q3 + 160Dq3 + 3200q3 = 0 .

To obtain the initial conditions for the second-order equation (20), recall that at time t = 0, the charge on the capacitor is 2 coulombs and the current is zero. Hence, (21)

q3 102 = 2 ,

dq3 102 = 0 . dt

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Section 5.7

Electrical Systems

293

We can now solve the initial value problem (20)–(21) using the techniques of Chapter 4. Ultimately, we find q3 1t2 = 2e-4t cos 12t +

2 -4t e sin 12t , 3 dq3 80 I3 1t2 = 1t2 = - e-4t sin 12t . dt 3

Next, to determine I2, we substitute these expressions into (19) and obtain I2 1t2 =

1 dq3 1t2 - 8q3 1t2 4 dt 1 64 -4t = - 16e-4t cos 12t + e sin 12t . 4 3

Finally, from I1 = I2 + I3, we get I1 1t2 =

1 16 -4t - 16e-4t cos 12t e sin 12t . ◆ 4 3

Note that the differential equations that describe mechanical vibrations and RLC series circuits are essentially the same. And, in fact, there is a natural identification of the parameters m, b, and k for a mass–spring system with the parameters L, R, and C that describe circuits. This is illustrated in Table 5.3. Moreover, the terms transient, steady-state, overdamped, critically damped, underdamped, and resonant frequency described in Sections 4.9 and 4.10 apply to electrical circuits as well. This analogy between a mechanical system and an electrical circuit extends to large-scale systems and circuits. An interesting consequence of this is the use of analog simulation and, in particular, analog computers to analyze mechanical systems. Large-scale mechanical systems are modeled by building a corresponding electrical system and then measuring the charges q1t2 and currents I1t2. Although such analog simulations are important, both large-scale mechanical and electrical systems are currently modeled using digital computer simulation. This involves the numerical solution of the initial value problem governing the system. Still, the analogy between mechanical and electrical systems means that basically the same computer software can be used to analyze both systems.

TABLE 5.3

Analogy Between Mechanical and Electrical Systems

Mechanical Mass–Spring System with Damping

 

mx″ + bx′ + kx = f1t2 Displacement Velocity Mass Damping constant Spring constant External force

  x x′ m b k f1t2

Electrical RLC Series Circuit

 

  Lq″ + Rq′ + 11>C2q = E1t2 Charge q Current q′ = I Inductance L R Resistance (Capacitance) -1 1>C Voltage source E1t2

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294

Chapter 5

5.7

Introduction to Systems and Phase Plane Analysis

EXERCISES

1. An RLC series circuit has a voltage source given by E1t2 = 20 V, a resistor of 100 Ω, an inductor of 4 H, and a capacitor of 0.01 F. If the initial current is zero and the initial charge on the capacitor is 4 C, determine the current in the circuit for t 7 0. 2. An RLC series circuit has a voltage source given by E1t2 = 40 cos 2t V, a resistor of 2 Ω, an inductor of 1>4 H, and a capacitor of 1>13 F. If the initial current is zero and the initial charge on the capacitor is 3.5 C, determine the charge on the capacitor for t 7 0. 3. An RLC series circuit has a voltage source given by E1t2 = 10 cos 20t V, a resistor of 120 Ω, an inductor of 4 H, and a capacitor of 122002 -1 F. Find the steady-state current (solution) for this circuit. What is the resonance frequency of the circuit? 4. An LC series circuit has a voltage source given by E1t2 = 30 sin 50t V, an inductor of 2 H, and a capacitor of 0.02 F (but no resistor). What is the current in this circuit for t 7 0 if at t = 0, I102 = q102 = 0? 5. An RLC series circuit has a voltage source of the form E1t2 = E0 cos gt V, a resistor of 10 Ω, an inductor of 4 H, and a capacitor of 0.01 F. Sketch the frequency response curve for this circuit. 6. Show that when the voltage source in (4) is of the form E1t2 = E0 sin gt, then the steady-state solution Ip is as given in equation (10). 7. A mass–spring system with damping consists of a 7-kg mass, a spring with spring constant 3 N/m, a frictional component with damping constant 2 N-sec/m, and an external force given by f1t2 = 10 cos 10t N. Using a 10-Ω resistor, construct an RLC series circuit that is the analog of this mechanical system in the sense that the two systems are governed by the same differential equation. 8. A mass–spring system with damping consists of a 16-lb weight, a spring with spring constant 64 lb/ft, a frictional component with damping constant 10 lb-sec/ft, and an external force given by f1t2 = 20 cos 8t lb. Using an inductor of 0.01 H, construct an RLC series circuit that is the analog of this mechanical system. 9. Because of Euler’s formula, eiu = cos u + i sin u, it is often convenient to treat the voltage sources E0 cos gt and E0 sin gt simultaneously, using E1t2 = E0eigt. In this case, equation (3) becomes 2

(22)

L

dq dt

2

+R

dq 1 + q = E0eiGt , dt C

where q is now complex (recall I = q′, I′ = q″).

(a) Use the method of undetermined coefficients to show that the steady-state solution to (22) is E0 eigt . qp 1t2 = 1>C - g 2L + igR The technique is discussed in detail in Project F of Chapter 4, page 237. (b) Now show that the steady-state current is E0 Ip 1t2 = eigt . R + i3gL - 1> 1gC24 (c) Use the relation a + ib = 2a2 + b2eiu, where tan u = b>a, to show that Ip can be expressed in the form E0 Ip 1t2 = ei1gt + u2 , 2 2R + 3gL - 1> 1gC24 2 where tan u = 11>C - Lg 2 2 > 1gR2. (d) The imaginary part of eigt is sin gt, so the imaginary part of the solution to (22) must be the solution to equation (3) for E1t2 = E0 sin gt. Verify that this is also the case for the current by showing that the imaginary part of Ip in part (c) is the same as that given in equation (10).

In Problems 10–13, find a system of differential equations and initial conditions for the currents in the networks given in the schematic diagrams (Figures 5.39–5.42 on pages 294–295). Assume that all initial currents are zero. Solve for the currents in each branch of the network. 10. 10 H 30 H

10 V

20 V

40 W I3

I1

I2

Figure 5.39 RL network for Problem 10 10 V

11.

10 V

5V

1 F 30

20 H I1

I2

I3

Figure 5.40 RLC network for Problem 11

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Section 5.8

Dynamical Systems, Poincaré Maps, and Chaos

10 V

12.

10 V

13.

0.02 H

0.025 H I2

I1

0.5 F

1V

0.5 H I1

I3

Figure 5.41 RL network for Problem 12

295

cos 3t V I2

I3

Figure 5.42 RLC network for Problem 13

5.8 Dynamical Systems, Poincaré Maps, and Chaos In this section we take an excursion through an area of mathematics that has received a lot of attention both for the interesting mathematical phenomena being observed and for its application to fields such as meteorology, heat conduction, fluid mechanics, lasers, chemical reactions, and nonlinear circuits, among others. The area is that of nonlinear dynamical systems.† A dynamical system is any system that allows one to determine (at least theoretically) the future states of the system given its present or past state. For example, the recursive formula (difference equation) xn+1 = 11.052xn ,

n = 0, 1, 2, c

is a dynamical system, since we can determine the next state, xn+1, given the previous state, xn. If we know x0, then we can compute any future state 3indeed, xn+1 = x0 11.052 n+1 4. Another example of a dynamical system is provided by the differential equation dx = -2x , dt

where the solution x1t2 specifies the state of the system at “time” t. If we know x1t0 2 = x0, then we can determine the state of the system at any future time t 7 t0 by solving the initial value problem dx = -2x , dt

x1t0 2 = x0 .

Indeed, a simple calculation yields x1t2 = x0e-21t-t02 for t Ú t0. For a dynamical system defined by a differential equation, it is often helpful to work with a related dynamical system defined by a difference equation. For example, when we cannot express the solution to a differential equation using elementary functions, we can use a numerical technique such as the improved Euler’s method or Runge–Kutta to approximate the solution to an initial value problem. This numerical scheme defines a new (but related) dynamical system that is often easier to study. In Section 5.4, we used phase plane diagrams to study autonomous systems in the plane. Many important features of the system can be detected just by looking at these diagrams. For example, a closed trajectory corresponds to a periodic solution. The trajectories for nonautonomous systems in †

For a more detailed study of dynamical systems, see An Introduction to Chaotic Dynamical Systems, by R. L. Devaney (Westview Press, 2003) and Nonlinear Oscillations, Dynamical Systems and Bifurcations of Vector Fields, by J. Guckenheimer and P. J. Holmes (Springer-Verlag, New York, 1983).

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the phase plane are much more complicated to decipher. One technique that is helpful in this regard is the so-called Poincaré map. As we will see, these maps replace the study of a nonautonomous system with the study of a dynamical system defined by the location in the xy-plane 1y = dx>dt2 of the solution at regularly spaced moments in time such as t = 2pn, where n = 0, 1, 2, . . . . The advantage in using the Poincaré map will become clear when the method is applied to a nonlinear problem for which no explicit solution is known. In such a case, the trajectories are computed using a numerical scheme such as Runge–Kutta. Several software packages have options that will construct Poincaré maps for a given system. To illustrate the Poincaré map, consider the equation (1)

x″1t2 + v2x1t2 = F cos t ,

where F and v are positive constants. We studied similar equations in Section 4.10 and found that a general solution for v ≠ 1 is given by (2)

x1t2 = A sin1vt + f2 +

F cos t , v2 - 1

where the amplitude A and the phase angle f are arbitrary constants. Since y = x′, y1t2 = vA cos1vt + f2 -

F sin t . v -1 2

Because the forcing function F cos t is 2p-periodic, it is natural to seek 2p-periodic solutions to (1). For this purpose, we define the Poincaré map as (3)

xn J x12pn2 = A sin12pvn + f2 + F> 1v2 - 12 , yn J y12pn2 = vA cos12pvn + f2 ,

for n = 0, 1, 2, . . . . In Figure 5.43 on page 297, we plotted the first 100 values of 1xn, yn 2 in the xy-plane for different choices of v. For simplicity, we have taken A = F = 1 and f = 0. These graphs are called Poincaré sections. We will interpret them shortly. Now let’s play the following game. We agree to ignore the fact that we already know the formula for x1t2 for all t Ú 0. We want to see what information about the solution we can glean just from the Poincaré section and the form of the differential equation. Notice that the first two Poincaré sections in Figure 5.43, corresponding to v = 2 and 3, consist of a single point. This tells us that, starting with t = 0, every increment 2p of t returns us to the same point in the phase plane. This in turn implies that equation (1) has a 2p-periodic solution, which can be proved as follows: For v = 2, let x1t2 be the solution to (1) with 1 x102, y102 2 = 11>3, 22 and let X1t2 J x1t + 2p2. Since the Poincaré section is just the point 11>3, 22, we have X102 = x12p2 = 1>3 and X′102 = x′12p2 = 2. Thus, x1t2 and X1t2 have the same initial values at t = 0. Further, because cos t is 2p-periodic, we also have X″1t2 + v2X1t2 = x″1t + 2p2 + v2x1t + 2p2 = cos1t + 2p2 = cos t . Consequently, x1t2 and X1t2 satisfy the same initial value problem. By the uniqueness theorems of Sections 4.2 and 4.5, these functions must agree on the interval [0, ∞ 2. Hence, x1t2 = X1t2 = x1t + 2p2 for all t Ú 0; that is, x1t2 is 2p-periodic. (The same reasoning works for v = 3.) With a similar argument, it follows from the Poincaré section for v = 1>2 that there is a solution of period 4p that alternates between the two points displayed in Figure 5.43(c) as t is incremented by 2p. For the case v = 1>3, we deduce that there is a solution of period 6p rotating among three points, and for v = 1>4, there is an 8p-periodic solution rotating among four points. We call these last three solutions subharmonics. The case v = 12 is different. So far, in Figure 5.43(f), none of the points has repeated. Did we stop too soon? Will the points ever repeat? Here, the fact that 12 is irrational plays

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Section 5.8

n 5 0, 1, 2, . . . ( –1– , 2)

2

n 5 0, 1, 2, ...

3

297

Dynamical Systems, Poincaré Maps, and Chaos

2

3

2

1

x

21

1

n50 22

1

2

21 21

x

21

1

x

1

1 21

1

2

=2

=3

=

(a)

(b)

(c)

1 –– 2

2 n 50

2

2

1

1

5 1

n50 22 2

21

2

n50 x

1

1

3 22

3 x

1

21

21

1 21 4

2 21 =

21 1 –– 3

=

(d)

x

1

1 1 –– 4

(e)

= Î2 (f)

Figure 5.43 Poincaré sections for equation (1) for various values of v

a crucial role. It turns out that every integer n yields a distinct point in the Poincaré section (see Problem 8). However, there is a pattern developing. The points all appear to lie on a simple curve, possibly an ellipse. To see that this is indeed the case, notice that when v = 22, A = F = 1, and f = 0, we have xn = sin 1 222 pn 2 + 1 ,

yn = 22 cos 1 222 pn 2 ,

n = 0, 1, 2, . . . .

It is then an easy computation to show that each 1xn, yn 2 lies on the ellipse 1x - 12 2 +

y2 = 1. 2

In our investigation of equation (1), we concentrated on 2p-periodic solutions because the forcing term Fcos t has period 2p. [We observed subharmonics when v = 1>2, 1>3, and 1>4—that is, solutions with periods 212p2, 312p2, and 412p2.] When a damping term is introduced into the differential equation, the Poincaré map displays a different behavior. Recall

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Chapter 5

Introduction to Systems and Phase Plane Analysis

that the solution will now be the sum of a transient and a steady-state term. For example, let’s analyze the equation (4)

x″1t2 + bx′1t2 + v2x1t2 = F cos t ,

where b, F, and v are positive constants. When b2 6 4v2, the solution to (4) can be expressed as (5)

x1t2 = Ae-1b>22t sin a

F 24v2 - b2 t + fb + sin1t + u2 , 2 2 21v - 12 2 + b2

where tan u = 1v2 - 12 >b and A and f are arbitrary constants [compare equations (7) and (8) in Section 4.10]. The first term on the right-hand side of (5) is the transient and the second term, the steady-state solution. Let’s construct the Poincaré map using t = 2pn, n = 0, 1, 2, . . . . We will take b = 0.22, v = A = F = 1, and f = 0 to simplify the computations. Because tan u = 1v2 - 12 >b = 0, we will take u = 0 as well. Then we have x12pn2 = xn = e-0.22pnsin 1 20.9879 2pn 2 ,

x′12pn2 = yn = -0.11e-0.22pnsin 1 20.9879 2pn 2

+ 20.9879 e-0.22pncos 1 20.9879 2pn 2 +

1 . 10.222

The Poincaré section for n = 0, 1, 2, . . . , 10 is shown in Figure 5.44 (black points). After just a few iterations, we observe that xn ≈ 0 and yn ≈ 1> 10.222 ≈ 4.545; that is, the points of the Poincaré section are approaching a single point in the xy-plane (colored point). Thus, we might expect that there is a 2p-periodic solution corresponding to a particular choice of A and f. [In this example, where we can explicitly represent the solution, we see that indeed a 2p-periodic solution arises when we take A = 0 in (5).] There is an important difference between the Poincaré sections for equation (1) and those for equation (4). In Figure 5.43, the location of all of the points in (a)–(e) depends on the initial value selected (here A = 1 and f = 0). (See Problem 10.) However, in Figure 5.44, the first few points (black points) depend on the initial conditions, while the limit point (colored point) does not (see Problem 6). The latter behavior is typical for equations that have a “damping”

5.4

5.2

20.015

20.01

x

20.005 4.8

4.6 Figure 5.44 Poincaré section for equation (4) with F = 1, b = 0.22, and v = 1

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Section 5.8

Dynamical Systems, Poincaré Maps, and Chaos

7

7

6

6

5

5

4

4

3

22

299

3

21

1

(a) x0 = 2,

0

=4

2

x

22

21

1

(b) x0 = 2,

0

2

x

=6

Figure 5.45 Poincaré section for equation (6) with initial values x0, y0

term (i.e., b 7 0); namely, the Poincaré section has a limit set† that is essentially independent of the initial conditions. For equations with damping, the limit set may be more complicated than just a point. For example, the Poincaré map for the equation (6)

x″1t2 + 10.222x′1t2 + x1t2 = cos t + cos 1 22t 2

has a limit set consisting of an ellipse (see Problem 11). This is illustrated in Figure 5.45 for the initial values x0 = 2, y0 = 4 and x0 = 2, y0 = 6. So far we have seen limit sets for the Poincaré map that were either a single point or an ellipse—independent of the initial values. These particular limit sets are attractors. In general, an attractor is a set A with the property that there exists an open set‡ B containing A such that whenever the Poincaré map enters B, its points remain in B and the limit set of the Poincaré map is a subset of A. Further, we assume A has the invariance property: Whenever the Poincaré map starts at a point in A, it remains in A. In the previous examples, the attractors of the dynamical system (Poincaré map) were easy to describe. In recent years, however, many investigators, working on a variety of applications, have encountered dynamical systems that do not behave orderly—their attractor sets are very complicated (not just isolated points or familiar geometric objects such as ellipses). The behavior of such systems is called chaotic, and the corresponding limit sets are referred to as strange attractors.

The limit set for a map 1xn, yn 2, n = 1, 2, 3, . . . , is the set of points 1p, q2 such that limk S ∞ 1xnk, ynk 2 = 1p, q2, where n1 6 n2 6 n3 6 cis some subsequence of the positive integers. ‡ A set B ⊂ R2 is an open set if for each point p ∈ B there is an open disk V containing p such that V ⊂ B. †

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Introduction to Systems and Phase Plane Analysis

1

1

21

1

x

21 (b) F = 0.28

(a) F = 0.2

1

1

x

21

x

1

1

x

21

1

21

21

(c) F = 0.29

(d) F = 0.37

Figure 5.46 Poincaré sections for the Duffing equation (7) with b = 0.3 and g = 1.2

To illustrate chaotic behavior and what is meant by a strange attractor, we discuss two nonlinear differential equations and a simple difference equation. First, let’s consider the forced Duffing equation (7)

x″1t2 + bx′1t2 − x1t2 + x3 1t2 = F sin Gt .

We cannot express the solution to (7) in any explicit form, so we must obtain the Poincaré map by numerically approximating the solution to (7) for fixed initial values and then plot the approximations for x12pn>g2 and y12pn>g2 = x′12pn>g2. (Because the forcing term F sin gt has period 2p>g, we seek 2p>g-periodic solutions and subharmonics.) In Figure 5.46, we display the limit sets (attractors) when b = 0.3 and g = 1.2 in the cases (a) F = 0.2, (b) F = 0.28, (c) F = 0.29, and (d) F = 0.37. Notice that as the constant F increases, the Poincaré map changes character. When F = 0.2, the Poincaré section tells us that there is a 2p>g-periodic solution. For F = 0.28, there is a subharmonic of period 4p>g, and for F = 0.29 and 0.37, there are subharmonics with periods 8p>g and 10p>g, respectively. Things are dramatically different when F = 0.5: The solution is neither 2p>g-periodic nor subharmonic. The Poincaré section for F = 0.5 is illustrated in Figure 5.47 on page 301. This section was generated by numerically approximating the solution to (7) when g = 1.2, b = 0.3, and F = 0.5, for fixed initial values.† Not all of the approximations x12pn>g2 and y12pn>g2 that were calculated are graphed; because of the presence of a transient solution, the first few points were omitted. It turns out that the plotted set is essentially independent of the initial values and has the property that once a point is in the set, all subsequent points lie in the set. †

Historical Footnote: When researchers first encountered these strange-looking Poincaré sections, they would check their computations using different computers and different numerical schemes [see Hénon and Heiles, “The Applicability of the Third Integral of Motion: Some Numerical Experiments,” Astronomical Journal, Vol. 69 (1964): 75]. For special types of dynamical systems, such as the Hénon map, it can be shown that there exists a true trajectory that shadows the numerical trajectory [see M. Hammel, J. A. Yorke, and C. Grebogi, “Numerical Orbits of Chaotic Processes Represent True Orbits,” Bulletin American Mathematical Society, Vol. 19 (1988): 466–469].

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Section 5.8

301

Dynamical Systems, Poincaré Maps, and Chaos

1 x

21

1 21

Figure 5.47 Poincaré section for the Duffing equation (7) with b = 0.3, g = 1.2, and F = 0.5

Because of the complicated shape of the set, it is indeed a strange attractor. While the shape of the strange attractor does not depend on the initial values, the picture does change if we consider different sections; for example, t = 12pn + p>22 >g, n = 0, 1, 2, . . . yields a different configuration. Another example of a strange attractor occurs when we consider the forced pendulum equation (8)

x″1t2 + bx′1t2 + sin 1 x1t2 2 = F cos t ,

where the x1t2 term in (4) has been replaced by sin 1 x1t2 2 . Here x1t2 is the angle between the pendulum and the vertical rest position, b is related to damping, and F represents the strength of the forcing function (see Figure 5.48). For F = 2.7 and b = 0.22, we have graphed in Figure 5.49 approximately 90,000 points in the Poincaré map. Since we cannot express the solution to (8) in any explicit form, the Poincaré map was obtained by numerically approximating the solution to (8) for fixed initial values and plotting the approximations for x12pn2 and y12pn2 = x′12pn2. The Poincaré maps for the forced Duffing equation and for the forced pendulum equation not only illustrate the idea of a strange attractor; they also exhibit another peculiar behavior called chaos. Chaos occurs when small changes in the initial conditions lead to major changes in the behavior of the solution. Henri Poincaré described the situation as follows: It may happen that small differences in the initial conditions will produce very large ones in the final phenomena. A small error in the former produces an enormous error in the latter. Prediction becomes impossible . . . .

F cos t

4 3 2 1 x(t)

0 2bx9(t)

Figure 5.48 Forced damped pendulum

21 23

22

21

x 0

1

2

3

Figure 5.49 Poincaré section for the forced pendulum equation (8) with b = 0.22 and F = 2.7

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Chapter 5

Introduction to Systems and Phase Plane Analysis

In a physical experiment, we can never exactly (with infinite accuracy) reproduce the same initial conditions. Consequently, if the behavior is chaotic, even a slight difference in the initial conditions may lead to quite different values for the corresponding Poincaré map when n is large. Such behavior does not occur for solutions to either equation (4) or equation (1) (see Problems 6 and 7). However, two solutions to the Duffing equation (7) with F = 0.5 that correspond to two different but close initial values have Poincaré maps that do not remain close together. Although they both are attracted to the same set, their locations with respect to this set may be relatively far apart. The phenomenon of chaos can also be illustrated by the following simple map. Let x0 lie in 30, 12 and define xn+1 = 2xn (mod 1) ,

(9)

where by (mod 1) we mean the decimal part of the number if it is greater than or equal to 1; that is xn + 1 = e

When x0 = 1>3, we find x1 = x2 = x3 = x4 =

2# 2# 2# 2#

for 0 … xn 6 1>2 , for 1>2 … xn 6 1 .

2xn , 2xn - 1 ,

11>32 12>32 11>32 12>32

1mod 12 = 1mod 12 = 1mod 12 = 1mod 12 =

2>3 , 4>3 1mod 12 = 1>3 , 2>3 , 1>3 , etc.

Written as a sequence, we get 51>3, 2>3, 1>3, 2>3, . . .6, where the overbar denotes the repeated pattern. What happens when we pick a starting value x0 near 1>3? Does the sequence cluster about 1>3 and 2>3 as does the mapping when x0 = 1>3? For example, when x0 = 0.3, we get the sequence 50.3, 0.6, 0.2, 0.4, 0.8, 0.6, 0.2, 0.4, 0.8, . . .6 .

In Figure 5.50, we have plotted the values of xn for x0 = 0.3, 0.33, and 0.333. We have not plotted the first few terms, but only those that repeat. (This omission of the first few terms parallels the situation depicted in Figure 5.47 on page 301, where transient solutions arise.) It is clear from Figure 5.50 that while the values for x0 are getting closer to 1>3, the corresponding maps are spreading out over the whole interval 30, 14 and not clustering near 1>3 and 2>3. This behavior is chaotic, since the Poincaré maps for initial values near 1>3 behave quite differently from the map for x0 = 1>3. If we had selected x0 to be irrational (which we can’t do with a calculator), the sequence would not repeat and would be dense in 30, 14. x0 5 0.3 0

1 3

x0 5 0.33 2 3

1

1 3

0

(a)

2 3

1

(b)

x0 5 0.333 0

1 3

2 3

1

(c) Figure 5.50 Plots of the map xn + 1 = 2xn (mod 1) for x0 = 0.3, 0.33, and 0.333

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 5.8

Dynamical Systems, Poincaré Maps, and Chaos

303

Systems that exhibit chaotic behavior arise in many applications. The challenge to engineers is to design systems that avoid this chaos and, instead, enjoy the property of stability. The topic of stable systems is discussed at length in Chapter 12.†

5.8

EXERCISES

A software package that supports the construction of Poincaré maps is required for the problems in this section. 1. Compute and graph the points of the Poincaré map with t = 2pn, n = 0, 1, . . . , 20 for equation (1), taking A = F = 1, f = 0, and v = 3>2. Repeat, taking v = 3>5. Do you think the equation has a 2p-periodic solution for either choice of v? A subharmonic solution? 2. Compute and graph the points of the Poincaré map with t = 2pn, n = 0, 1, . . . , 20 for equation (1), taking A = F = 1, f = 0, and v = 1> 13. Describe the limit set for this system. 3. Compute and graph the points of the Poincaré map with t = 2pn, n = 0, 1, . . . , 20 for equation (4), taking A = F = 1, f = 0, v = 1, and b = - 0.1. What is happening to these points as n S ∞ ? 4. Compute and graph the Poincaré map with t = 2pn, n = 0, 1, . . . , 20 for equation (4), taking A = F = 1, f = 0, v = 1, and b = 0.1. Describe the attractor for this system. 5. Compute and graph the Poincaré map with t = 2pn, n = 0, 1, . . . , 20 for equation (4), taking A = F = 1, f = 0, v = 1>3, and b = 0.22. Describe the attractor for this system. 6. Show that for b 7 0, the Poincaré map for equation (4) is not chaotic by showing that as t gets large xn = x12pn2 ≈

F

sin12pn + u2 ,

21v - 12 2 + b2

yn = x′12pn2 ≈

2

F 21v - 12 + b 2

2

2

cos12pn + u2

independent of the initial values x0 = x102 and y0 = x′102. 7. Show that the Poincaré map for equation (1) is not chaotic by showing that if 1x0, y0 2 and 1x0*, y*0 2 are two initial values that define the Poincaré maps {1xn, yn 2} and 51xn*, y*n 26, respectively, using the recursive formulas in (3), then one can make the distance between

1xn, yn 2 and 1x*n, y*n 2 small by making the distance between 1x0, y0 2 and 1x0*, y*0 2 small. [Hint: Let 1A, f2 and 1A*, f*2 be the polar coordinates of two points in the plane. From the law of cosines, it follows that the distance d between them is given by d 2 = 1A - A*2 2 + 2AA*31 - cos1f - f*24.] 8. Consider the Poincaré maps defined in (3) with v = 12, A = F = 1, and f = 0. If this map were ever to repeat, then for two distinct positive integers n and m, sin 1 212pn 2 = sin 1 212pm 2 . Using basic properties of the sine function, show that this would imply that 12 is rational. It follows from this contradiction that the points of the Poincaré map do not repeat. 9. The doubling modulo 1 map defined by equation (9) exhibits some fascinating behavior. Compute the sequence obtained when (a) x0 = k>7 for k = 1, 2, . . . , 6 . (b) x0 = k>15 for k = 1, 2, . . . , 14 . (c) x0 = k>2j, where j is a positive integer and k = 1, 2, . . . , 2j - 1 . Numbers of the form k>2j are called dyadic numbers and are dense in 30, 14. That is, there is a dyadic number arbitrarily close to any real number (rational or irrational). 10. To show that the limit set of the Poincaré map given in (3) depends on the initial values, do the following: (a) Show that when v = 2 or 3, the Poincaré map consists of the single point 1x, y2 = aA sin f +

F , vA cos f b . v2 - 1 (b) Show that when v = 1>2, the Poincaré map alternates between the two points a

F { A sin f , { vA cos f b . v -1 (c) Use the results of parts (a) and (b) to show that when v = 2, 3, or 1>2, the Poincaré map (3) depends on the initial values 1x0, y0 2. 2

†

All references to Chapters 11–13 refer to the expanded text, Fundamentals of Differential Equations and Boundary Value Problems, 7th ed.

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304

Chapter 5

Introduction to Systems and Phase Plane Analysis

11. To show that the limit set for the Poincaré map xn J x12pn2, yn J x′12pn2, where x1t2 is a solution to equation (6), is an ellipse and that this ellipse is the same for any initial values x0, y0, do the following: (a) Argue that since the initial values affect only the transient solution to (6), the limit set for the Poincaré map is independent of the initial values. (b) Now show that for n large,

side of table

pin

long ruler

xn ≈ a sin1212pn + c2 ,

tie tack (pivot)

yn ≈ c + 12a cos1212pn + c2 ,

where a = 1 1 + 210.222 2 2 -1>2, c = 10.222 -1, and c = arctan 5 - 3 10.222 12 4 -1 6 . (c) Use the result of part (b) to conclude that the ellipse x2 +

12.

13. 14. 15.

1y - c2

short ruler (a) double pendulum

2

= a2 2 contains the limit set of the Poincaré map. Using a numerical scheme such as Runge–Kutta or a software package, calculate the Poincaré map for equation (7) when b = 0.3, g = 1.2, and F = 0.2. (Notice that the closer you start to the limiting point, the sooner the transient part will die out.) Compare your map with Figure 5.46(a) on page 300. Redo for F = 0.28. Redo Problem 12 with F = 0.31. What kind of behavior does the solution exhibit? Redo Problem 12 with F = 0.65. What kind of behavior does the solution exhibit? Chaos Machine. Chaos can be illustrated using a long ruler, a short ruler, a pin, and a tie tack (pivot). Construct the double pendulum as shown in Figure 5.51(a). The pendulum is set in motion by releasing it from a position such as the one shown in Figure 5.51(b). Repeatedly

(b) release position Figure 5.51 Double pendulum as a chaos machine

set the pendulum in motion, each time trying to release it from the same position. Record the number of times the short ruler flips over and the direction in which it was moving. If the pendulum was released in exactly the same position each time, then the motion would be the same. However, from your experiments you will observe that even beginning close to the same position leads to very different motions. This double pendulum exhibits chaotic behavior.

Chapter 5 Summary Systems of differential equations arise in the study of biological systems, coupled mass–spring oscillators, electrical circuits, ecological models, and many other areas. Linear systems with constant coefficients can be solved explicitly using a variant of the Gauss elimination process. For this purpose we begin by writing the system with operator notation, using D J d>dt, D2 J d 2 >dt 2, and so on. A system of two equations in two unknown functions then takes the form L 1 3x4 + L 2 3y4 = f1 ,

L 3 3x4 + L 4 3y4 = f2 ,

where L1, L2, L3, and L4 are polynomial expressions in D. Applying L4 to the first equation, L2 to the second, and subtracting, we get a single (typically higher-order) equation in x1t2, namely, 1L 4L 1 - L 2L 3 23x4 = L 4 3 f1 4 - L 2 3 f2 4 .

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Chapter 5 Summary

305

We then solve this constant coefficient equation for x1t2. Similarly, we can eliminate x from the system to obtain a single equation for y1t2, which we can also solve. This procedure introduces some extraneous constants, but by substituting the expressions for x and y back into one of the original equations, we can determine the relationships among these constants. A preliminary step for the application of numerical algorithms for solving systems or single equations of higher order is to rewrite them as an equivalent system of first-order equations in normal form:

(1)

x1= 1t2 = f1 1t, x1, x2, . . . , xm 2 , x2= 1t2 = f2 1t, x1, x2, . . . , xm 2 , f

xm= 1t2 = fm 1t, x1, x2, . . . , xm 2 . For example, by setting y = y′, we can rewrite the second-order equation y″ = ƒ1t, y, y′2 as the normal system y′ = y , y′ = ƒ1t, y, y2 . The normal system (1) has the outward appearance of a vectorized version of a single firstorder equation, and as such it suggests how to generalize numerical algorithms such as those of Euler and Runge–Kutta. A technique for studying the qualitative behavior of solutions to the autonomous system (2)

dx = ƒ1x, y2 , dt

dy = g1x, y2 dt

is phase plane analysis. We begin by finding the critical points of (2)—namely, points 1x0, y0 2 where ƒ1x0, y0 2 = 0

and

g1x0, y0 2 = 0 .

The corresponding constant solution pairs x1t2 K x0, y1t2 K y0 are called equilibrium solutions to (2). We then sketch the direction field for the related phase plane differential equation (3)

g1x, y2 dy = dx ƒ1x, y2

with appropriate direction arrows (oriented by the sign of dx>dt or dy>dt). From this we can usually conjecture qualitative features of the solutions and of the critical points, such as stability and asymptotic behavior. Software is typically used to visualize the solution curves to (3), which provide the highways along which the trajectories of the system (2) travel. Nonautonomous systems can be studied by considering a Poincaré map for the system. A Poincaré map can be used to detect periodic and subharmonic solutions and to study systems whose solutions exhibit chaotic behavior.

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306

Chapter 5

Introduction to Systems and Phase Plane Analysis

REVIEW PROBLEMS FOR CHAPTER 5 In Problems 1–4, find a general solution x1t2, y1t2 for the given system. 1. x′ + y″ + y = 0 , 2. x′ = x + 2y , x″ + y′ = 0 y′ = - 4x - 3y 3. 2x′ - y′ = y + 3x + et , 3y′ - 4x′ = y - 15x + e-t 4. x″ + x - y″ = 2e-t , x″ - x + y″ = 0 In Problems 5 and 6, solve the given initial value problem. 5. x′ = z - y ; x102 = 0 , y′ = z ; y102 = 0 , z′ = z - x ; z102 = 2 6. x′ = y + z ; y′ = x + z ; z′ = x + y ;

x102 = 2 , y102 = 2 , z102 = - 1

7. For the interconnected tanks problem of Section 5.1, page 241, suppose that instead of pure water being fed into tank A, a brine solution with concentration 0.2 kg/L is used; all other data remain the same. Determine the mass of salt in each tank at time t if the initial masses are x0 = 0.1 kg and y0 = 0.3 kg. In Problems 8–11, write the given higher-order equation or system in an equivalent normal form (compare Section 5.3). 8. 2y″ - ty′ + 8y = sin t 9. 3y‴ + 2y′ - ety = 5 10. x″ - x + y = 0 , x′ - y + y″ = 0 11. x‴ + y′ + y″ = t , x″ - x′ + y‴ = 0 In Problems 12 and 13, solve the related phase plane equation for the given system. Then sketch by hand several representative trajectories (with their flow arrows) and describe the

stability of the critical points (i.e., compare with Figure 5.12, page 267). 12. x′ = y - 2 , 13. x′ = 4 - 4y , y′ = 2 - x y′ = -4x 14. Find all the critical points and determine the phase plane solution curves for the system dx = sin x cos y , dt dy = cos x sin y . dt In Problems 15 and 16, sketch some typical trajectories for the given system, and by comparing with Figure 5.12, page 267, identify the type of critical point at the origin. 15. x′ = -2x - y , y′ = 3x - y

16. x′ = -x + 2y , y′ = x + y

17. In the electrical circuit of Figure 5.52, take R1 = R2 = 1 Ω, C = 1 F, and L = 1 H. Derive three equations for the unknown currents I1, I2, and I3 by writing Kirchhoff’s voltage law for loops 1 and 2, and Kirchhoff’s current law for the top juncture. Find the general solution. I1

I3

I2

L

R2

C loop 1

loop 2

R1

Figure 5.52 Electrical circuit for Problem 17

18. In the coupled mass–spring system depicted in Figure 5.26, page 283, take each mass to be 1 kg and let k1 = 8 N/m, while k 2 = 3 N/m. What are the natural angular frequencies of the system? What is the general solution?

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Projects for Chapter 5 A Designing a Landing System for Interplanetary Travel Courtesy of Alfred Clark, Jr., Professor Emeritus, University of Rochester, Rochester, NY

You are a second-year Starfleet Academy Cadet aboard the U.S.S. Enterprise on a continuing study of the star system Glia. The object of study on the present expedition is the large airless planet Glia-4. A class 1 sensor probe of mass m is to be sent to the planet’s surface to collect data. The probe has a modifiable landing system so that it can be used on planets of different gravity. The system consists of a linear spring (force = - kx, where x is displacement), a nonlinear # spring (force = - ax3), and a shock-damper (force = - bx),† all in parallel. Figure 5.53 shows a schematic of the system. During the landing process, the probe’s thrusters are used to create a constant rate of descent. The velocity at impact varies; the symbol VL is used to denote the largest velocity likely to happen in practice. At the instant of impact, (1) the thrust is turned off, and (2) the suspension springs are at their unstretched natural length. (a) Let the displacement x be measured from the unstretched length of the springs and be taken negative downward (i.e., compression gives a negative x). Show that the equation governing the oscillations after impact is $ # mx + bx + kx + ax3 = - mg . (b) The probe has a mass m = 1220 kg. The linear spring is permanently installed and has a stiffness k = 35,600 N/m. The gravity on the surface of Glia-4 is g = 17.5 m/sec2. The nonlinear spring is removable; an appropriate spring must be chosen for each mission. These nonlinear springs are made of coralidium, a rare and difficult-to-fabricate alloy. Therefore, the Enterprise stocks only four different strengths: a = 150,000, 300,000, 450,000, and 600,000 N/m3. Determine which springs give a compression as close as

Probe body

VL

m k

Probe body a

b

k Landing pod

V50

m

surface of Glia- 4

b

a

Landing pod

(a)

(b)

Figure 5.53 Schematic of the probe landing system. (a) The system at the instant of impact. The springs are not stretched or compressed, the thrusters have been turned off, and the velocity is VL downward. (b) The probe has reached a state of rest on the surface, and the springs are compressed enough to support the weight. Between states (a) and (b), the probe oscillates relative to the landing pod. # The symbol x denotes dx>dt.

†

307

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

308

Chapter 5

Introduction to Systems and Phase Plane Analysis

possible to 0.3 m without exceeding 0.3 m, when the ship is resting on the surface of Glia-4. (The limit of 0.3 m is imposed by unloading clearance requirements.) (c) The other adjustable component on the landing system is the linear shock-damper, which may be adjusted in increments of ∆b = 500 N-sec/m, from a low value of 1000 N-sec/m to a high value of 10,000 N-sec/m. It is desirable to make b as small as possible because a large b produces large forces at impact. However, if b is too small, there is some danger that the probe will rebound after impact. To minimize the chance of this, find the smallest value of b such that the springs are always in compression during the oscillations after impact. Use a minimum impact velocity VL = 5 m/sec downward. To find this value of b, you will need to use a software package to integrate the differential equation.

B Spread of Staph Infections in Hospitals—Part I Courtesy of Joanna Wares, University of Richmond, and Glenn Webb, Vanderbilt University

Methicillin-resistant Staphylococcus aureus (MRSA), commonly referred to as staph, is a bacterium that causes serious infections in humans and is resistant to treatment with the widely used antibiotic methicillin. MRSA has traditionally been a problem inside hospitals, where elderly patients or patients with compromised immune systems could more easily contract the bacteria and develop bloodstream infections. MRSA is implicated in a large percentage of hospital fatalities, causing more deaths per year than AIDS. Recently, a genetically different strain of MRSA has been found in the community at large. The new strain (CA-MRSA) is able to infect healthy and young people, which the traditional strain (HA-MRSA) rarely does. As CA-MRSA appears in the community, it is inevitably being spread into hospitals. Some studies suggest that CA-MRSA will overtake HA-MRSA in the hospital, which would increase the severity of the problem and likely cause more deaths per year. To predict whether or not CA-MRSA will overtake HA-MRSA, a compartmental model has been developed by mathematicians in collaboration with medical professionals (see references [1], [2] on page 310). This model classifies all patients in the hospital into three groups: • H1t2 = patients colonized with the traditional hospital strain, HA-MRSA. • C1t2 = patients colonized with the community strain, CA-MRSA. • S1t2 = susceptible patients, those not colonized with either strain. The parameters of the model are • bC = the rate (per day) at which CA-MRSA is transmitted between patients. • bH = the rate (per day) at which HA-MRSA is transmitted between patients. • dC = the rate (per day) at which patients who are colonized with CA-MRSA exit the hospital by death or discharge. • dH = the rate (per day) at which patients who are colonized with HA-MRSA exit the hospital by death or discharge. • dS = the rate (per day) at which susceptible patients exit the hospital by death or discharge. • aC = the rate (per day) at which patients who are colonized with CA-MRSA successfully undergo decolonization measures. • aH = the rate (per day) at which patients who are colonized with HA-MRSA successfully undergo decolonization measures. • N = the total number of patients in the hospital. • Λ = the rate (per day) at which patients enter the hospital.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Projects for Chapter 5

dS

L

S(t)

aC

aH

bC dC

309

bH

C(t)

dH

H(t)

Figure 5.54 A diagram of how patients transit between the compartments

Patients move between compartments as they become colonized or decolonized (see Figure 5.54). This type of model is typically known as a SIS (susceptible-infected-susceptible) model, in which patients who become colonized can become susceptible again and colonized again (there is no immunity). The transition between states is described by the following system of differential equations: bHS1t2H1t2 dS = Λ " dt N entrance rate (1++)++1*

-

acquire HA-MRSA

+

a H1t2

H (+)+* HA-MRSA decolonized

+

bCS1t2C1t2 N (1++)++1* acquire CA-MRSA a C1t2

-

C (+)+* CA-MRSA decolonized

d(1)1* SS1t2 exit hospital

bHS1t2H1t2 dH = - aHH1t2 - dHH1t2 (+)+* (+)+* dt N (++)++* decolonized exit hospital from S

bCS1t2C1t2 dC = - aCC1t2 - dCC1t2. (+)+* (+)+* dt N (++)++* decolonized from S

exit hospital

If we assume that the hospital is always full, we can conserve the system by letting Λ = dSS1t2 + dHH1t2 + dCC1t2. In this case S1t2 + C1t2 + H1t2 = N for all t (assuming you start with a population of size N). (a) Show that this assumption simplifies the above system of equations to dH = 1bH >N21N - C - H2H - 1dH + aH 2H , dt (1) dC = 1bC >N21N - C - H2C - 1dC + aC 2C . dt S is then determined by the equation S1t2 = N - H1t2 - C1t2. Parameter values obtained from the Beth Israel Deaconess Medical Center are given in Table 5.4 on page 310. Plug these values into the model and then complete the following problems. (b) Find the three equilibria (critical points) of the system (1). (c) Using a computer, sketch the direction field for the system (1). (d) Which trajectory configuration exists near each critical point (node, spiral, saddle, or center)? What do they represent in terms of how many patients are susceptible, colonized with HA-MRSA, and colonized with CA-MRSA over time? (e) Examining the direction field, do you think CA-MRSA will overtake HA-MRSA in the hospital? Further discussion of this model appears in Project E of Chapter 12.† †

All references to Chapters 11–13 refer to the expanded text, Fundamentals of Differential Equations and Boundary Value Problems, 7th ed.

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310

Chapter 5

Introduction to Systems and Phase Plane Analysis

TABLE 5.4

Parameter Values for the Transmission Dynamics of Community-Acquired and Hospital-Acquired Methicillin-Resistant Staphylococcus aureus Colonization (CA-MRSA and HA-MRSA)

Parameter

Symbol

Baseline Value

Total number of patients

N

400

Length of stay Susceptible Colonized CA-MRSA Colonized HA-MRSA

1>dS 1>dC 1>dH

5 days 7 days 5 days

bC bH

0.45 per day 0.4 per day

aC aH

0.1 per day 0.1 per day

Transmission rate per susceptible patient to Colonized CA-MRSA per colonized CA-MRSA Colonized HA-MRSA per colonized HA-MRSA Decolonization rate per colonized patient per day per length of stay CA-MRSA HA-MRSA References

1. D’Agata, E. M. C., Webb, G. F., Pressley, J. 2010. “Rapid emergence of co-colonization with communityacquired and hospital-acquired methicillin-resistant Staphylococcus aureus strains in the hospital setting”. Mathematical Modelling of Natural Phenomena 5(3): 76–93. 2. Pressley, J., D’Agata, E. M. C., Webb, G. F. 2010. “The effect of co-colonization with community-acquired and hospital-acquired methicillin-resistant Staphylococcus aureus strains on competitive exclusion”. Journal of Theoretical Biology 265(3): 645–656.

C Things That Bob Courtesy of Richard Bernatz, Department of Mathematics, Luther College

The motion of various-shaped objects that bob in a pool of water can be modeled by a secondorder differential equation derived from Newton’s second law of motion, F = ma. The forces acting on the object include the force due to gravity, a frictional force due to the motion of the object in the water, and a buoyant force based on Archimedes’ principle: An object that is completely or partially submerged in a fluid is acted on by an upward (buoyant) force equal to the weight of the water it displaces. (a) The first step is to write down the governing differential equation. The dependent variable is the depth z of the object’s lowest point in the water. Take z to be negative downward so that z = - 1 means 1 ft of the object has submerged. Let V1z2 be the submerged volume of the object, m be the mass of the object, r be the density of water (in pounds per cubic foot), g be the acceleration due to gravity, and gw be the coefficient of friction for water. Assuming that the frictional force is proportional to the vertical velocity of the object, write down the governing second-order ODE. (b) For the time being, neglect the effect of friction and assume the object is a cube measuring L feet on a side. Write down the governing differential equation for this case. Next, designate z = l (a negative number) to be the depth of submersion such that the buoyant force is equal and opposite the gravitational force. Introduce a new variable, z, that gives

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Projects for Chapter 5

(c)

(d)

(e)

(f)

311

the displacement of the object from its equilibrium position l (that is, z = z + l). You can now write the ODE in a more familiar form. [Hint: Recall the mass–spring system and the equilibrium case.] Now you should recognize the type of solution for this problem. What is the natural frequency? In this task you consider the effect of friction. The bobbing object is a cube, 1 ft on a side, that weighs 32 lb. Let gw = 3 lb-sec/ft, r = 62.57 lb/ft3, and suppose the object is initially placed on the surface of the water. Solve the governing ODE by hand to find the general solution. Next, find the particular solution for the case in which the cube is initially placed on the surface of the water and is given no initial velocity. Provide a plot of the position of the object as a function of time t. In this step of the project, you develop a numerical solution to the same problem presented in part (c). The numerical solution will be useful (indeed necessary) for subsequent parts of the project. This case provides a trial to verify that your numerical solution is correct. Go back to the initial ODE you developed in part (a). Using parameter values given in part (c), solve the initial value problem for the cube starting on the surface with no initial velocity. To solve this problem numerically, you will have to write the secondorder ODE as a system of two first-order ODEs, one for vertical position z and one for vertical velocity w. Plot your results for vertical position as a function of time t for the first 3 or 4 sec and compare with the analytical solution you found in part (c). Are they in close agreement? What might you have to do in order to compare these solutions? Provide a plot of both your analytical and numerical solutions on the same graph. Suppose a sphere of radius R is allowed to bob in the water. Derive the governing second-order equation for the sphere using Archimedes’ principle and allowing for friction due to its motion in the water. Suppose a sphere weighs 32 lb, has a radius of 1>2 ft, and gw = 3.0 lb-sec/ft. Determine the limiting value of the position of the sphere without solving the ODE. Next, solve numerically the governing ODE for the velocity and position of the sphere as a function of time for a sphere placed on the surface of the water. You will need to write the governing second-order ODE as a system of two ODEs, one for velocity and one for position. What is the limiting position of the sphere for your solution? Does it agree with the equilibrium solution you found above? How does it compare with the equilibrium position of the cube? If it is different, explain why. Suppose the sphere in part (d) is a volleyball. Calculate the position of the sphere as a function of time t for the first 3 sec if the ball is submerged so that its lowest point is 5 ft under water. Will the ball leave the water? How high will it go? Next, calculate the ball’s trajectory for initial depths lower than 5 ft and higher than 5 ft. Provide plots of velocity and position for each case and comment on what you see. Specifically, comment on the relationship between the initial depth of the ball and the maximum height the ball eventually attains. You might consider taking a volleyball into a swimming pool to gather real data in order to verify and improve on your model. If you do so, report the data you found and explain how you used it for verification and improvement of your model.

D Hamiltonian Systems The problems in this project explore the Hamiltonian† formulation of the laws of motion of a system and its phase plane implications. This formulation replaces Newton’s second law F = ma = my″ and is based on three mathematical manipulations: †

Historical Footnote: Sir William Rowan Hamilton (1805–1865) was an Irish mathematical physicist. Besides his work in mechanics, he invented quaternions and discovered the anticommutative law for vector products.

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(i) It is presumed that the force F1t, y, y′2 depends only on y and has an antiderivative - V1y2, that is, F = F1y2 = - dV1y2 >dy. (ii) The velocity variable y′ is replaced throughout by the momentum p = my′ (so y′ = p>m). (iii) The Hamiltonian of the system is defined as p2 + V1y2 . H = H1y, p2 = 2m (a) Express Newton’s law F = my″ as an equivalent first-order system in the manner prescribed in Section 5.3. (b) Show that this system is equivalent to Hamilton’s equations (2) (3)

dy 0H = dt 0p dp 0H = dt 0y

a=

p b, m

a= -

dV b. dy

(c) Using Hamilton’s equations and the chain rule, show that the Hamiltonian remains constant along the solution curves: d H1y, p2 = 0 . dt In the formula for the Hamiltonian function H1y, p2, the first term, p2 > 12m2 = m1y′2 2 >2, is the kinetic energy of the mass. By analogy, then, the second term V1y2 is known as the potential energy of the mass, and the Hamiltonian is the total energy. The total (mechanical)† energy is constant—hence “conserved”—when the forces F1y2 do not depend on time t or velocity y′; such forces are called conservative. The energy integral lemma of Section 4.8 (page 203) is simply an alternate statement of the conservation of energy. Hamilton’s formulation for mechanical systems and the conservation of energy principle imply that the phase plane trajectories of conservative systems lie on the curves where the Hamiltonian H1y, p2 is constant, and plotting these curves may be considerably easier than solving for the trajectories directly (which, in turn, is easier than solving the original system!). (d) For the mass–spring oscillator of Section 4.1, the spring force is given by F = - ky (where k is the spring constant). Find the Hamiltonian, express Hamilton’s equations, and show that the phase plane trajectories H1y, p2 = constant for this system are the ellipses given by p2 > 12m2 + ky2 >2 = constant. See Figure 5.14, page 269. The damping force - by′ considered in Section 4.1 is not conservative, of course. Physically speaking, we know that damping drains the energy from a system until it grinds to a halt at an equilibrium point. In the phase plane, we can qualitatively describe the trajectory as continuously migrating to successively lower constant-energy orbits; stable centers become asymptotically stable spiral points when damping is taken into consideration. (e) The second Hamiltonian equation (3), which effectively states p′ = my″ = F, has to be changed to 0H 0H bp - by′ = m 0y 0y when damping is present. Show that the Hamiltonian decreases along trajectories in this case (for b 7 0): p′ = -

p 2 d H1y, p2 = - ba b = - b1y′2 2 . m dt †

Physics states that when all forms of energy, such as heat and radiation, are taken into account, energy is conserved even when the forces are not conservative.

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313

(f) The force on a mass–spring system suspended vertically in a gravitational field was shown in Section 4.10 (page 226) to be F = - ky + mg. Derive the Hamiltonian and sketch the phase plane trajectories. Sketch the trajectories when damping is present. (g) As indicated in Section 4.8 (page 207), the Duffing spring force is modeled by F = - y - y3. Derive the Hamiltonian and sketch the phase plane trajectories. Sketch the trajectories when damping is present. (h) For the pendulum system studied in Section 4.8, Example 8, the force is given by (cf. Figure 4.18, page 208) F = - /mg sin u = -

0 0 1 -/mg cos u2 = - V1u2 0u 0u

(where / is the length of the pendulum). For angular variables, the Hamiltonian formulation dictates expressing the angular velocity variable u′ in terms of the angular momentum p = m/2u′; the kinetic energy, mass * velocity2 >2, is expressed as m1/u′2 2 >2 = p2 > 12m/2 2. Derive the Hamiltonian for the pendulum and sketch the phase plane trajectories. Sketch the trajectories when damping is present. (i) The Coulomb force field is a force that varies as the reciprocal square of the distance from the origin: F = k>y2. The force is attractive if k 6 0 and repulsive if k 7 0. Sketch the phase plane trajectories for this motion. Sketch the trajectories when damping is present. (j) For an attractive Coulomb force field, what is the escape velocity for a particle situated at a position y? That is, what is the minimal (outward-directed) velocity required for the trajectory to reach y = ∞ ?

E Cleaning Up the Great Lakes A simple mathematical model that can be used to determine the time it would take to clean up the Great Lakes can be developed using a multiple compartmental analysis approach.† In particular, we can view each lake as a tank that contains a liquid in which is dissolved a particular pollutant (DDT, phosphorus, mercury). Schematically, we view the lakes as consisting of five tanks connected as indicated in Figure 5.55. 15 Superior 2900 mi3

15

15

99

Michiga

n 1180 3 mi

38

38

850 mi3 Huron 14 17

i3

93 m

io 3 ntar

O

85

68 116

mi3

Erie Figure 5.55 Compartmental model of the Great Lakes with flow rates (mi3/yr) and volumes (mi3) †

For a detailed discussion of this model, see An Introduction to Mathematical Modeling by Edward A. Bender (Dover Publications, New York, 2000), Chapter 8.

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For our model, we make the following assumptions: 1. The volume of each lake remains constant. 2. The flow rates are constant throughout the year. 3. When a liquid enters the lake, perfect mixing occurs and the pollutants are uniformly distributed. 4. Pollutants are dissolved in the water and enter or leave by inflow or outflow of solution. Before using this model to obtain estimates on the cleanup times for the lakes, we consider some simpler models: (a) Use the outflow rates given in Figure 5.55 to determine the time it would take to “drain” each lake. This gives a lower bound on how long it would take to remove all the pollutants. (b) A better estimate is obtained by assuming that each lake is a separate tank with only clean water flowing in. Use this approach to determine how long it would take the pollution level in each lake to be reduced to 50% of its original level. How long would it take to reduce the pollution to 5% of its original level? (c) Finally, to take into account the fact that pollution from one lake flows into the next lake in the chain, use the entire multiple compartment model given in Figure 5.55 to determine when the pollution level in each lake has been reduced to 50% of its original level, assuming pollution has ceased (that is, inflows not from a lake are clean water). Assume that all the lakes initially have the same pollution concentration p. How long would it take for the pollution to be reduced to 5% of its original level?

F The 2014-2015 Ebola Epidemic Courtesy of Glenn Webb, Vanderbilt University

We develop a model for the 2014–2015 Ebola epidemic in West Africa, which began in the spring of 2014 and spread though the countries of Sierra Leone, Liberia, and Guinea. Initially there was great concern that the epidemic might develop with great severity, and even spread globally.† Instead, by early 2015 the epidemic had subsided in these West African countries, and in January 2016, the World Health Organization (WHO) declared that the epidemic was contained in all three countries. The reasons that Ebola subsided are complex, but a key role was increased identification and isolation of infected cases, and contact tracing of these cases to identify additional cases. Our model incorporates the principal features of contact tracing, namely, the number of contacts per identified infectious case, the likelihood that a traced contact is infectious, and the efficiency of the isolation of contact-traced individuals. The model consists of a system of ordinary differential equations for the compartments of the epidemic population, and is based on an earlier version of the model.‡ The model incorporates the unique features of the Ebola outbreaks in this region. These features include the rates of transmission to susceptibles from both infectious cases and improperly handled deceased cases, the rates of reporting cases, and the rates of recovery and mortality of unreported cases.

†

H. Nishiura and G. Chowell, “Early transmission dynamics of Ebola virus disease (EVD), West Africa, March to August 2014.” Eurosurveil., Vol. 19 no. 36 (Sept. 11, 2014). ‡

C. Browne, X. Huo, P. Magal. M. Seydi, O. Seydi, G. Webb, “A model of the 2014 Ebola epidemic in West Africa with contact tracing,” PLOS Currents Outbreaks, published online, January 30, 2015.

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Projects for Chapter 5

315

Reported Cases 16000 14000 12000 10000 8000 6000 4000 2000

Phase 1

12/27/2015

11/27/2015

10/27/2015

9/27/2015

8/27/2015

7/27/2015

6/27/2015

5/27/2015

4/27/2015

3/27/2015

2/27/2015

1/27/2015

12/27/2014

11/27/2014

10/27/2014

9/27/2014

8/27/2014

7/27/2014

6/27/2014

5/27/2014

0

Phase 2

October 16, 2014 Figure 5.56 WHO Ebola Situation Data: reported cases. (Center for Disease Control and Prevention, Previous Case Counts: http://www.cdc.gov/vhf/ebola/outbreaks/2014-west-africa/cumulative-cases-graphs.html)

The model extends the SIR paradigm of Section 5.5, by refining the population classes into susceptibles S1t2 (those who have not been affected), exposed E1t2 (incubating cases), I1t2 (infectious cases), and contaminated deceased C1t2 (improperly handled corpses of cases). The successful containment of the epidemic was due to measures isolating infected individuals and contaminated corpses, implemented around October 16, 2014. The WHO data in Figure 5.56 traces the chronology of the number of reported cases R1t2; the inflection point clearly highlights the initiation of isolation and contact tracing measures. Figure 5.57 on page 316 schematically depicts the transitions between the classifications, as quantified by the subsequent system of differential equations: TABLE 5.5

Ebola Model Parameters in Sierra Leone

Parameter

Description

N = 6 * 106 b = 0.321 e = 0.002 k = 0 (1st phase) k = 8 (2nd phase) a = 0.08 (1st phase) a = 0.145 (2nd phase) p = 0.5

Population of Sierra Leone (assumed to be constant) Transmission rate exclusive of improper handling of deceased cases Transmission rate due to improper handling of deceased cases Average number of contacts traced per reported and hospitalized infectious case Rate of reporting and hospitalization of infectious cases not resulting from contact tracing Probability a contact-traced infected individual is isolated without causing a new case Probability a contact-traced individual is infected Rate of recovery of unreported infectious cases Rate of mortality of unreported infectious cases Average incubation period (denominator) Average number of days before proper handling of deceased unreported cases (denominator)

v g V s c c

= = = = = =

0.1 0.0333 0.125 1>9 1>5 (1st phase) 1>2 (2nd phase)

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E

S Susceptible

b

Exposed

e

C

I s

n

Infectious

kpva

a

Contaminated

c

g Unreported Recovered or Deceased

Hospitalized

Figure 5.57 Schematic of Ebola transitions

dS1t2 = dt

-bS1t2

I1t2

-eS1t2

C1t2

N N c c susceptible- susceptibleinfected contaminated interaction interaction T T I1t2 C1t2 dE1t2 = bS1t2 + eS1t2 -sE1t2 dt N N c incubation T dI1t2 sE1t2 = dt

-vI1t2

c mortality, unreported infected T

dC1t2 = dt dR1t2 = dt

- 1a + g + akpv2I1t2 proper c handling of hospitalization, recovery, unreported or isolation by corpses T contact tracing -cC1t2

vI1t2 aI1t2

+cC1t2

By trial and error and numerical experimentation, the following parameter values were found to give a good fit, to the WHO data, of the solution of this system. The initial values were taken to be S102 = N, E102 = 20, I102 = 20, C102 = 10. Note the change in the values of k, a, and c on the 142nd day, reflecting the initiation of contact tracing. (a) Use a software package to solve the system for the time interval indicated in Figure 5.56 on page 315. Plot the number of reported cases R1t2. Note the high quality of the fit of the simulations to the WHO data. (b) Plot the number of persons affected by the epidemic, 6 * 106 - S1t2. You should see that about 42% of all cases were unreported on October 16, 2014, but this fell to 10% by January, 2016.

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317

G Phase-Locked Loops Today’s high-speed communications technology requires circuitry that will measure or match incoming radio-frequency oscillations in the gigahertz range. Mathematically, that means one must determine the frequency v in a sinusoidal signal A sin1vt + a2 when v is in the neighborhood of 2p * 109 rad>s. In theory this can be accomplished by taking three measurements of the sinusoid within a quarter-period. However, obtaining three accurate measurements of a signal in an interval of 10 -10 seconds is impractical. The phase-locked loop accomplishes the task in hardware, using a voltage-controlled oscillator (VCO) to synthesize another sinusoid B cos u1t2 whose phase, u1t2, matches that of the incoming signal 1vt + a2. The frequency du1t2 >dt of the VCO output is linearly related to the voltage applied to the VCO (say, du1t2 >dt = E + Fv1t22, and it matches v once synchronization is achieved. Thus one can determine the incoming frequency by reading off the VCO voltage. A diagram of the phase-locked loop circuit is given in Figure 5.58. In brief: the mixer comAB sin3vt + a - u1t24 bines the signals A sin 1vt + a2 and B cos u1t2 to create a control signal 2 whose phase equals the difference between the phase of the incoming signal and the phase of the VCO output.† This control signal provides a measure of the lack of synchronization; it is zero when u1t2 = vt + a1{ 2pn2. Thus to help u1t2 catch up to vt + a, the circuit adjusts the VCO voltage v1t2 so as to increase the output frequency, du1t2 >dt, in proportion to a weighted combination of the control signal and its derivative: d1E + Fv2 d 2u d du1t2 = 2 = dt dt dt dt AB d AB = K1 sin3vt + a - u1t24 + K2 e sin3vt + a - u1t24 f . 2 dt 2 We are going to employ the phase plane to analyze how the phase-locked loop approaches synchronicity when it is controlled by equation (4).

(4)

(a) Convert (4) into an autonomous equation by introducing the error e1t2 J vt + a - u1t2 and deriving (5)

d 2u AB de AB d 2e = = - K1 sin e - K2 1cos e2 . 2 2 2 2 dt dt dt (b) Put equation (5) into normal form (two first-order differential equations in x1 J e1t2 and x2 = de>dt) and find all the critical points. v

Input A sin (vt1a)

Mixer

Control signal (AB/2) sin [vt1a2u(t)]

Voltage generator

(t)

VCO

B cos u(t)

(K1.K2.∫.S)

Figure 5.58 Phase-locked loop †

A classic trigonometric identity shows that multiplication of the incoming and VCO signals yields the control signal plus a high-frequency oscillation, which is easily filtered out.

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AB = 4, (c) Use software to sketch the direction field in this system’s phase plane, with K1 2 AB = 1. (The duplicitous nomenclature is unintended: phase plane does not refer K2 2 to the phase of the sinusoid.) Trace a few trajectories to get a feel for how the phaselocked loop performs. Which critical points are stable, and which are unstable? Are all the stable critical points suitable for synchronization? (d) Apply the mass–spring analogy (Section 4.8) to equation (5). The mass m is one, the de AB AB sin e instead of - ke, and the damping force is - K2 1cos e2 spring force is - K1 2 2 dt de instead of - b . If e is small, the mass–spring parameters and the phase-locked loop dt parameters are approximately the same (recall Project D, Chapter 4). Sketch the direction field for the (normal form of the) damped mass–spring system d 2e de = - b - ke 2 dt dt with b = 1, k = 4, and compare with that of the phase-locked loop. Explain why the number of equilibria is different, and why the nature of the phase-locked loop’s “forces” renders certain of its equilibria to be unstable.

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CHAPTER

Matrix Methods for Linear Systems

9

9.1 Introduction In this chapter we return to the analysis of systems of differential equations. When the equations in the system are linear, matrix algebra provides a compact notation for expressing the system. In fact, the notation itself suggests new and elegant ways of characterizing the solution properties, as well as novel, efficient techniques for explicitly obtaining solutions. In Chapter 5 we analyzed physical situations wherein two fluid tanks containing brine solutions were interconnected and pumped so as ultimately to deplete the salt content in each tank. By accounting for the various influxes and outfluxes of brine, a system of differential equations for the salt contents 1x1t2 and y1t22 of each tank was derived; a typical model is dx>dt = −4x + 2y , dy>dt = 4x − 4y .

(1)

Express this system in matrix notation as a single equation.

The right-hand side of the first member of (1) possesses a mathematical structure that is familiar from vector calculus; namely, it is the dot product† of two vectors: (2)

-4x + 2y = 3 -4 24 # 3x y4  .

Similarly, the second right-hand side in (1) is the dot product 4x - 4y = 34

-44 # 3x y4 .

The frequent occurrence in mathematics of arrays of dot products, such as evidenced in the system (1), led to the development of matrix algebra, a mathematical discipline whose basic operation—the matrix product—is the arrangement of a set of dot products according to the following plan: c

-4 4

2 x 3 -4 24 # 3x y4 -4x + 2y d. dc d = c d = c 4x - 4y -4 y 34 -44 # 3x y4

In general, the product of a matrix—i.e., an m by n rectangular array of numbers—and a column vector is defined to be the collection of dot products of the rows of the matrix with the

†

Recall that the dot product of two vectors u and v equals the length of u times the length of v times the cosine of the angle between u and v. However, it is more conveniently computed from the components of u and v by the “inner product” indicated in equation (2).

496

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Section 9.1

Introduction

497

vector, arranged as a column vector: 3row # 14 # v row # 1 3row # 24 # v row # 2 T  , D T  DvT = D f f 3row # m4 # v row # m where the vector v has n components; the dot product of two n-dimensional vectors is computed in the obvious way: 3a1 a2 gan 4 # 3x1 x2 g xn 4 = a1x1 + a2x2 + g + anxn .

Using the notation for the matrix product, we can write the system (1) for the interconnected tanks as c

-4 x′ d = c 4 y′

2 x dc d . -4 y

The following example demonstrates a four-dimensional implementation of this notation. Note that the coefficients in the linear system need not be constants. Example 1

Express the system x1= = 2x1 + t 2x2 + 14t + et 2x4 ,

(3)

x2= = 1sin t2x2 + 1cos t2x3 , x3= = x1 + x2 + x3 + x4 , x4= = 0

as a matrix equation. Solution

We express the right-hand side of the first member of (3) as the dot product 2x1 + t 2x2 + 14t + et 2x4 = 32 t 2 0

14t + et 24 # 3x1 x2 x3 x4 4  .

The other dot products are similarly identified, and the matrix form is given by 2 t2 x1= 0 = x2 0 sin t cos t D =T = D x3 1 1 1 = x4 0 0 0

14t + et 2 0 T  1 0

x1 x D 2T . ◆ x3 x4

In general, if a system of differential equations is expressed as x1= = a11 1t2x1 + a12 1t2x2 + g + a1n 1t2xn x2= = a21 1t2x1 + a22 1t2x2 + g + a2n 1t2xn f

xn= = an1 1t2x1 + an2 1t2x2 + g + ann 1t2xn  ,

it is said to be a linear homogeneous system in normal form.† The matrix formulation of such

†

The normal form was defined for general systems in Section 5.3, page 256.

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Matrix Methods for Linear Systems

a system is then x′ = Ax , where A is the coefficient matrix a11 1t2 a21 1t2 A = A1t2 = D f an1 1t2

a12 1t2 a22 1t2 f an2 1t2

g g g

a1n 1t2 a2n 1t2 T f ann 1t2

and x is the solution vector x1 x x = D 2T  . f xn Note that we have used x′ to denote the vector of derivatives x1 ′ x1= x x2= x′ = D 2 T = D T . f f xn xn= Example 2

Express the differential equation for the undamped, unforced mass–spring oscillator (recall Section 4.1, page 152) (4)

my″ + ky = 0

as an equivalent system of first-order equations in normal form, expressed in matrix notation. Solution

We have to express the second derivative, y″, as a first derivative in order to formulate (4) as a first-order system. This is easy; the acceleration y″ is the derivative of the velocity y = y′, so (4) becomes (5)

my′ + ky = 0 .

The first-order system is then assembled by identifying y with y′, and appending it to (5): y′ = y my′ = -ky . To put this system in normal form and express it as a matrix equation, we need to divide the second equation by the mass m: y ′ 0 c d = c y -k>m

1 y dc d . ◆ 0 y

In general, the customary way to write an nth-order linear homogeneous differential equation an 1t2y 1n2 + an - 1 1t2y 1n - 12 + g + a1 1t2y′ + a0 1t2y = 0

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Section 9.1

Introduction

499

as an equivalent system in normal form is to define the first 1n - 12 derivatives of y (including y, the zeroth derivative, itself) to be new unknowns: x1 1t2 = y1t2 , x2 1t2 = y′1t2 , f

xn 1t2 = y 1n - 12 1t2 .

Then the system consists of the identification of xj 1t2 as the derivative of xj - 1 1t2, together with the original differential equation expressed in these variables (and divided by an 1t2): x1= x2= f

= x2 , = x3 ,

xn= - 1 = xn , a0 1t2 a1 1t2 an - 1 1t2 xn= = x1 x2 - g x  . an 1t2 an 1t2 an 1t2 n For systems of two or more higher-order differential equations, the same procedure is applied to each unknown function in turn; an example will make this clear. Example 3

The coupled mass–spring oscillator depicted in Figure 5.26 on page 283 was shown to be governed by the system 2

(6)

d 2x + 6x - 2y = 0 , dt 2

d 2y dt 2

+ 2y - 2x = 0 .

Write (6) in matrix notation. Solution

We introduce notation for the lower-order derivatives: (7)

x1 = x ,

x2 = x′ ,

x3 = y ,

x4 = y′ .

In these variables, the system (6) states (8)

2x2= + 6x1 - 2x3 = 0 ,

x4= + 2x3 - 2x1 = 0 .

The normal form is then x1= x2= x3= x4=

= = = =

x2 , -3x1 + x3 , x4 , 2x1 - 2x3

or in matrix notation x1 ′ 0 1 x2 -3 0 D T = D x3 0 0 x4 2 0

0 1 0 -2

0 x1 0 x2 T D T. ◆ 1 x3 0 x4

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500

9.1

Chapter 9

Matrix Methods for Linear Systems

EXERCISES

In Problems 1–6, express the given system of differential equations in matrix notation. 1. x′ = 7x + 2y , y′ = 3x - 2y 2. x′ = y , y′ = - x 3. x′ = x + y + z , y′ = 2z - x , z′ = 4y 4. x1= = x1 - x2 + x3 - x4 , x2= = x1 + x4 , x3= = 1px1 - x3 , x4= = 0 5. x′ = 1sin t2x + ety , y′ = 1cos t2x + 1a + bt 3 2y 6. x1= = 1cos 2t2x1 , x2= = 1sin 2t2x2 , x3= = x1 - x2

In Problems 7–10, express the given higher-order differential equation as a matrix system in normal form. 7. The damped mass–spring oscillator equation my″ + by′ + ky = 0 8. Legendre’s equation 11 - t 2 2y″ - 2ty′ + 2y = 0 9. The Airy equation y″ - ty = 0 1 n2 10. Bessel’s equation y″ + y′ + a1 - 2 by = 0 t t In Problems 11–13, express the given system of higherorder differential equations as a matrix system in normal form. 11. x″ + 3x + 2y = 0 , y″ - 2x = 0 12. x″ + 3x′ - y′ + 2y = 0 , y″ + x′ + 3y′ + y = 0 13. x″ - 3x′ + t 2y - 1cos t2x = 0 , y‴ + y″ - tx′ + y′ + etx = 0

9.2 Review 1: Linear Algebraic Equations Here and in the next section we review some basic facts concerning linear algebraic systems and matrix algebra that will be useful in solving linear systems of differential equations in normal form. Readers competent in these areas may proceed to Section 9.4. A set of equations of the form a11x1 + a12x2 + g + a1nxn = b1 , a21x1 + a22x2 + g + a2nxn = b2 , f an1x1 + an2x2 + g + annxn = bn (where the aij’s and bi’s are given constants) is called a linear system of n algebraic equations in the n unknowns x1, x2, c, xn. The procedure for solving the system using elimination methods is well known. Herein we describe a particularly convenient implementation of the method called the Gauss–Jordan elimination algorithm.† The basic idea of this formulation is to use the first equation to eliminate x1 in all the other equations; then use the second equation to eliminate x2 in all the others; and so on. If all goes well, the resulting system will be “uncoupled,” and the values of the unknowns x1, x2, c, xn will be apparent. A short example will make this clear. †

The Gauss–Jordan algorithm is neither the fastest nor the most accurate computer algorithm for solving a linear system of algebraic equations, but for solutions executed by hand it has many pedagogical advantages. Usually it is much faster than Cramer’s rule, described in Appendix D.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 9.2

Example 1

Review 1: Linear Algebraic Equations

501

Solve the system 2x1 + 6x2 + 8x3 = 16 , 4x1 + 15x2 + 19x3 = 38 , 2x1 + 3x3 = 6 .

Solution

By subtracting 2 times the first equation from the second, we eliminate x1 from the latter. Similarly, x1 is eliminated from the third equation by subtracting 1 times the first equation from it: 2x1 + 6x2 + 8x3 = 16 , 3x2 + 3x3 = 6 , -6x2 - 5x3 = -10 . Next we subtract multiples of the second equation from the first and third to eliminate x2 in them; the appropriate multiples are 2 and -2, respectively: 2x1

+ 2x3 = 4 , 3x2 + 3x3 = 6 , x3 = 2 .

Finally, we eliminate x3 from the first two equations by subtracting multiples (2 and 3, respectively) of the third equation: 2x1

= 0, 3x2 = 0 , x3 = 2 .

The system is now uncoupled; i.e., we can solve each equation separately: x1 = 0 , x2 = 0 , x3 = 2 . ◆ Two complications can disrupt the straightforward execution of the Gauss–Jordan algorithm. The first occurs when the impending variable to be eliminated (say, xj ) does not occur in the jth equation. The solution is usually obvious; we employ one of the subsequent equations to eliminate xj. Example 2 illustrates this maneuver. Example 2

Solve the system x1 -x1 -2x1 x1

Solution

+ 2x2 + 4x3 + x4 - 2x2 - 2x3 - 4x2 - 8x3 + 2x4 + 4x2 + 2x3

= = = =

0, 1, 4, -3 .

The first unknown x1 is eliminated from the last three equations by subtracting multiples of the first equation: x1 + 2x2 + 4x3 + x4 2x3 + x4 4x4 2x2 - 2x3 - x4

= = = =

0, 1, 4, -3 .

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502

Chapter 9

Matrix Methods for Linear Systems

Now, we cannot use the second equation to eliminate the second unknown because x2 is not present. The next equation that does contain x2 is the fourth, so we switch the second and fourth equation: x1 + 2x2 + 4x3 + x4 = 0 , 2x2 - 2x3 - x4 = -3 , 4x4 = 4 , 2x3 + x4 = 1 , and proceed to eliminate x2: + 6x3 + 2x4 = 3 ,

x1

2x2 - 2x3 - x4 = -3 , 4x4 = 4 , 2x3 + x4 = 1 . To eliminate x3, we have to switch again, x1

+ 6x3 + 2x4 = 2x2 - 2x3 - x4 = 2x3 + x4 = 4x4 =

3, -3 , 1, 4,

and eliminate, in turn, x3 and x4. This gives x1 2x2

- x4 = = 2x3 + x4 = 4x4 =

0, -2 , 1, 4,

and

x1 2x2

= = 2x3 = 4x4 =

1, -2 , 0, 4.

The solution to the uncoupled equations is x1 = 1 , x2 = -1 , x3 = 0 , x4 = 1 . ◆ The other complication that can disrupt the Gauss–Jordan algorithm is much more profound. What if, when we are “scheduled” to eliminate the unknown xj, it is absent from all of the subsequent equations? The first thing to do is to move on to the elimination of the next unknown xj + 1, as demonstrated in Example 3.

Example 3

Apply the Gauss–Jordan algorithm to the system (1)

Solution

2x1 + 4x2 + x3 = 8 , 2x1 + 4x2 = 6, - 4x1 - 8x2 + x3 = - 10 .

Elimination of x1 proceeds as usual: 2x1 + 4x2 + x3 = 8 , - x3 = -2 , 3x3 = 6 .

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Section 9.2

Review 1: Linear Algebraic Equations

503

Now since x2 is absent from the second and third equations, we use the second equation to eliminate x3 : 2x1 + 4x2 (2)

= 6, - x3 = -2 , 0 = 0.

How do we interpret the system (2)? The final equation contains no information, of course, and we ignore it.† The second equation implies that x3 = 2. The first equation implies that x1 = 3 - 2x2, but there is no equation for x2. Evidently, x2 is a “free” variable, and we can assign any value to it—as long as we take x1 to be 3 - 2x2. Thus (1) has an infinite number of solutions, and a convenient way of characterizing them is x1 = 3 - 2s , x2 = s , x3 = 2 ;

-∞ 6 s 6 ∞ .

We remark that an equivalent solution can be obtained by treating x1 as the free variable, say x1 = s, and taking x2 = 13 - s2 >2, x3 = 2. ◆ The final example is contrived to demonstrate all the features that we have encountered. Example 4

Find all solutions to the system x1 - x2 2x1 - 2x2 3x1 - 3x2 4x1 - 4x2

Solution

+ 2x3 + 4x3 + 6x3 + 8x3

+ 2x4 + 3x4 + 9x4 + 8x4

= = = =

0, 1, -3 , 0.

We use the first equation to eliminate x1: x1 - x2 + 2x3 + 2x4 = 0 , - x4 = 1 , 3x4 = -3 , 0 = 0. Now, both x2 and x3 are absent from all subsequent equations, so we use the second equation to eliminate x4. x1 - x2 + 2x3

= -x4 = 0 = 0 =

2, 1, 0, 0.

There are no constraints on either x2 or x3; thus we take them to be free variables and characterize the solutions by x1 = 2 + s - 2t , x2 = s , x3 = t , x4 = -1,

- ∞ 6 s, t 6 ∞ . ◆

In closing, we note that if the execution of the Gauss–Jordan algorithm results in a display of the form 0 = 1 (or 0 = k, where k ≠ 0), the original system has no solutions; it is inconsistent. This is explored in Problem 12. The occurrence of the identity 0 = 0 in the Gauss–Jordan algorithm implies that one of the original equations was redundant. In this case you may observe that the final equation in (1) can be derived by subtracting 3 times the second equation from the first.

†

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504

Chapter 9

9.2

Matrix Methods for Linear Systems

EXERCISES

In Problems 1–11, find all solutions to the system using the Gauss–Jordan elimination algorithm. 1. x1 + 2x2 + 2x3 = 6 , 2x1 + x2 + x3 = 6 , x1 + x2 + 3x3 = 6 2.

x1 + x2 + x3 + x1 + 2x1 + 2x2 - x3 + x1 + 2x2 - x3 +

x4 x4 x4 x4

3.

x1 + x2 - x3 = 0 , - x1 - x2 + x3 = 0 , x1 + x2 - x3 = 0

4.

x3 + x1 + x2 + x3 + 2x1 - x2 + x3 + 2x1 - x2 + x3 +

x4 x4 2x4 x4

= = = =

1, 0, 0, 0

= = = =

0, 1, 0, 0

8.

x1 + 2x2 + x3 = - 3 , 2x1 + 4x2 - x3 = 0 , x1 + 3x2 - 2x3 = 3

9.

11.

+ x3 = -1 , 2x1 -3x1 + x2 + 4x3 = 1 , - x1 + x2 + 5x3 = 0

13. Use the Gauss–Jordan elimination algorithm to show that the following system of equations has a unique solution for r = 2, but an infinite number of solutions for r = 1.

6. - 2x1 + 2x2 - x3 = 0 , x1 - 3x2 + x3 = 0 , 4x1 - 4x2 + 2x3 = 0 -x1 + 3x2 = 0 , - 3x1 + 9x2 = 0

x1 + x2 + x3 = i , 2x1 + 3x2 - ix3 = 0 , x1 + 2x2 + x3 = i

12. Use the Gauss–Jordan elimination algorithm to show that the following systems of equations are inconsistent. That is, demonstrate that the existence of a solution would imply a mathematical contradiction. (a) 2x1 - x2 = 2 , -6x1 + 3x2 = 4 (b) 2x1 + x3 = -1 , -3x1 + x2 + 4x3 = 1 , - x1 + x2 + 5x3 = 1

5. - x1 + 2x2 = 0 , 2x1 + 3x2 = 0

7.

10.

2x1 - 3x2 = rx1 , x1 - 2x2 = rx2 14. Use the Gauss–Jordan elimination algorithm to show that the following system of equations has a unique solution for r = -1, but an infinite number of solutions for r = 2. x1 + 2x2 - x3 = rx1 , x1

11 - i2x1 + 2x2 = 0 , - x1 - 11 + i2x2 = 0

+ x3 = rx2 ,

4x1 - 4x2 + 5x3 = rx3

9.3 Review 2: Matrices and Vectors A matrix is a rectangular array of numbers arranged in rows and columns. An m * n matrix— that is, a matrix with m rows and n columns—is usually denoted by a11 a A J D 21 f am1

a12 a22 f am2

a13 a23 f am3

g g g g

a1n a2n T, f amn

where the element in the ith row and jth column is aij. The notation 3aij 4 is also used to designate A. The matrices we will work with usually consist of real numbers, but in certain instances we allow complex-number entries.

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Section 9.3

Review 2: Matrices and Vectors

505

Some matrices of special interest are square matrices, which have the same number of rows and columns; diagonal matrices, which are square matrices with only zero entries off the main diagonal (that is, aij = 0 if i ≠ j); and (column) vectors, which are n * 1 matrices. For example, if 3 A = £2 0

4 6 1

-1 5§ , 4

3 0 0 B = £0 0 0§ , 0 0 7

4 x = £2§ , 1

then A is a square matrix, B is a diagonal matrix, and x is a vector. An m * n matrix whose entries are all zero is called a zero matrix and is denoted by 0. For consistency, we denote matrices by boldfaced capitals, such as A, B, C, I, X, and Y, and reserve boldfaced lowercase letters, such as c, x, y, and z, for vectors.

Algebra of Matrices Matrix Addition and Scalar Multiplication. The operations of matrix addition and scalar multiplication are very straightforward. Addition is performed by adding corresponding elements: c

1 1 1 2 3 4 1 2 3 d = c d . d +c 1 1 1 5 6 7 4 5 6

Formally, the sum of two m * n matrices is given by A + B = 3aij 4 + 3bij 4 = 3aij + bij 4 .

(The sole novelty here is that addition is not defined for two matrices whose dimensions m, n differ.) To multiply a matrix by a scalar (number), we simply multiply each element in the matrix by the number: 3c

3 6 9 1 2 3 d . d = c 12 15 18 4 5 6

In other words, rA = r3aij 4 = 3raij 4. The notation -A stands for 1 -12A. Properties of Matrix Addition and Scalar Multiplication. Matrix addition and scalar multiplication are nothing more than mere bookkeeping, and the usual algebraic properties hold. If A, B, and C are m * n matrices and r, s are scalars, then A + 1B + C2 = 1A + B2 + C , A+0 = A, r1A + B2 = rA + rB , r1sA2 = 1rs2A = s1rA2 .

A+B = B+A, A + 1 -A2 = 0 , 1r + s2A = rA + sA ,

Matrix Multiplication. The matrix product is what makes matrix algebra interesting and useful. We indicated in Section 9.1 that the product of a matrix A and a column vector x is the column vector composed of dot products of the rows of A with x: 1 1 2 3 1#1+2#0+3#2 7 c d £0§ = c # d = c d. # # 4 5 6 16 4 1+5 0+6 2 2

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506

Chapter 9

Matrix Methods for Linear Systems

More generally, the product of two matrices A and B is formed by taking the array of dot products of the rows of the first “factor” A with the columns of the second factor B; the dot product of the ith row of A with the jth column of B is written as the ijth entry of the product AB: c

1 3

0 -1

1 1 d £ -1 2 4

x 1+0+4 2+0+1 x+0+z d y§ = c 3 + 1 + 8 6 + 1 + 2 3x - y + 2z z

2 −1 1

= c

5 3 x+z d. 12 9 3x - y + 2z

Note that AB is only defined when the number of columns of A matches the number of rows of B. A useful formula for the product of an m * n matrix A and an n * p matrix B is AB J 3cij 4 ,

n

where

cij J a aikbkj . k=1

The dot product of the ith row of A and the jth column of B is seen in the “sum of products” expression for cij. Since AB is computed in terms of the rows of the first factor and the columns of the second factor, it should not be surprising that, in general, AB does not equal BA (matrix multiplication does not commute): c

2 1 1 2 0 1 d , dc d = c 4 3 3 4 1 0

but

c

3 4 0 1 1 2 d . dc d = c 1 2 1 0 3 4

In fact, the dimensions of A and B may render one or the other of these products undefined: c

2 1 2 0 dc d = c d ; 4 3 4 1

0 1 2 c dc d not defined . 1 3 4

By the same token, one might not expect (AB)C to equal A(BC), since in (AB)C we take dot products with the columns of B, whereas in A(BC) we employ the rows of B. So it is a pleasant surprise that this complication does not arise, and the “parenthesis grouping” rules are the customary ones:

Properties of Matrix Multiplication 1AB2C = A1BC2 1A + B2C = AC + BC A1B + C2 = AB + AC 1rA)B = r1AB2 = A1rB2

(Associativity) (Distributivity) (Distributivity) (Associativity)

To summarize, the algebra of matrices proceeds much like the standard algebra of numbers, except that we should never presume that we can switch the order of matrix factors. (If you think 1A + B2 2 = A2 + B2 + 2AB, what error have you made?) Matrices as Linear Operators. Let A be an m * n matrix and let x and y be n * 1 vectors. Then Ax is an m * 1 vector, and so we can think of multiplication by A as defining an operator that maps n * 1 vectors into m * 1 vectors. A consequence of the distributivity and associativity properties is that multiplication by A defines a linear operator, since A(x + y) = Ax + Ay and A(rx) = rAx. Moreover, if A is an m * n matrix and B is an n * p matrix, then the m * p matrix AB defines a linear operator that is the composition of the linear operator defined by B

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Section 9.3

Review 2: Matrices and Vectors

507

with the linear operator defined by A. That is, (AB)x = A(Bx), where x is a p * 1 vector. Examples of linear operations are (i) stretching or contracting the components of a vector by constant factors; (ii) rotating a vector through some angle about a fixed axis; (iii) reflecting a vector in a plane mirror. The Matrix Formulation of Linear Algebraic Systems. Matrix algebra was developed to provide a convenient tool for expressing and analyzing linear algebraic systems. Note that the set of equations x1 + 2x2 + x3 = 1 , x1 + 3x2 + 2x3 = -1 , x1 + x3 = 0 can be written using the matrix product

(1)

1 2 1 x1 1 £ 1 3 2 § £ x2 § = £ -1 § . 0 1 0 1 x3

In general, we express the linear system a11x1 + a12x2 + g + a1nxn = b1 , a21x1 + a22x2 + g + a2nxn = b2 , f an1x1 + an2x2 + g + annxn = bn in matrix notation as Ax = b, where A is the coefficient matrix, x is the vector of unknowns, and b is the vector of constants occurring on the right-hand side: a11 a A = D 21 f an1

g g

a12 a22 f an2

g

a1n a2n T, f ann

x1 x x = D 2T , f xn

b1 b b = D 2T . f bn

If b = 0, the system Ax = b is said to be homogeneous (analogous to the nomenclature of Section 4.2). Matrix Transpose. The matrix obtained from A by interchanging its rows and columns is called the transpose of A and is denoted by AT. For example, if A = c

1 2 -1 2

1 A = £2 6 T

6 d ,    then -1

-1 2§ . -1

In general, we have 3aij 4 T = 3bij 4, where bij = aji. Properties of the transpose are explored in Problem 7.

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508

Chapter 9

Matrix Methods for Linear Systems

Matrix Identity. There is a “multiplicative identity” in matrix algebra, namely, a square diagonal matrix I with ones down the main diagonal. Multiplying I on the right or left by any other matrix (with compatible dimensions) reproduces the latter matrix: 1 2 1 1 2 1 1 0 0 1 0 0 1 2 1 £0 1 0§ £1 3 2§ = £1 3 2§ = £1 3 2§ £0 1 0§ . 1 0 1 1 0 1 0 0 1 0 0 1 1 0 1 (The notation In is used if it is convenient to specify the dimensions, n * n, of the identity matrix.) Matrix Inverse. Some square matrices A can be paired with other (square) matrices B having the property that BA = I:

(2)

E

3 2

-1

1 2

0

- 32

1

1 2

1 2 1 1 0 0 - 12 U £ 1 3 2 § = £ 0 1 0 § . 1 0 1 0 0 1 1 2

When this happens, it can be shown that (i) B is the unique matrix satisfying BA = I, and (ii) B also satisfies AB = I. In such a case, we say that B is the inverse of A and write B = A -1. Not every matrix possesses an inverse; the zero matrix 0, for example, can never satisfy the equation 0B = I. A matrix that has no inverse is said to be singular. If we know an inverse for the coefficient matrix A in a system of linear equations Ax = b, the solution can be calculated directly by computing A -1b, as the following derivation shows: Ax = b implies A -1Ax = A -1b implies x = A -1b . Using (2), for example, we can solve equation (1) quite efficiently: x1 x £ 2§ = E x3

3 2

-1

1 2

0

- 32

1

1 2

1 - 12 U £ -1 § = E 0 1 2

5 2 1U . 2

- 52

On the other hand, the coefficient matrix for any inconsistent system has no inverse. For example, the coefficient matrices 2 c -6

-1 d 3

and

2 £ -3 -1

0 1 1

1 4§ 5

for the inconsistent systems of Problem 12, Exercises 9.2 (page 504), are necessarily singular. When A-1 is known, solving Ax = b by multiplying b by A-1 is certainly easier than applying the Gauss–Jordan algorithm of the previous section†. So it appears advantageous to be able to find matrix inverses. Some inverses can be obtained directly from the interpretation of the matrix as a linear operator. For example, the inverse of a matrix that rotates a vector is †

(When the effort to compute the inverse is accounted for, Gauss–Jordan emerges as the winner.)

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Section 9.3

Review 2: Matrices and Vectors

509

the matrix that rotates it in the opposite direction. A matrix that performs a mirror reflection is its own inverse (what do you get if you reflect twice?). But in general one must employ an algorithm to compute a matrix inverse. The underlying strategy for this algorithm is based on the observation that if X denotes the inverse of A, then X must satisfy the equation AX = I; finding X amounts to solving n linear systems of equations for the columns 5x1, x2, . . . , xn 6 of X: 1 0 0 Ax1 = F V , f 0 0

0 1 0 Ax2 = F V , f 0 0

0 0 0 g , Axn = F V . f 0 1

The implementation of this operation is neatly executed by the following variation of the Gauss–Jordan algorithm. Finding the Inverse of a Matrix. By a row operation, we mean any one of the following: (a) Interchanging two rows of the matrix (b) Multiplying a row of the matrix by a nonzero scalar (c) Adding a scalar multiple of one row of the matrix to another row. If the n * n matrix A has an inverse, then A -1 can be determined by performing row operations on the n * 2n matrix 3A?I4 obtained by writing A and I side by side. In particular, we perform row operations on the matrix 3A?I4 until the first n rows and columns form the identity matrix; that is, the new matrix is 3I?B4. Then A-1 = B. We remark that if this procedure fails to produce a matrix of the form 3I?B4, then A has no inverse.

Example 1

Solution

1 2 1 Find the inverse of A = £ 1 3 2 § . 1 0 1 We first form the matrix 3A?I4 and row-reduce the matrix to 3I?A-1 4. Computing, we find the following: 1 2 1 ? 1 0 0 £1 3 2 ? 0 1 0§ . 1 0 1 ? 0 0 1

The matrix 3A?I4 Subtract the first row from the second and third to obtain

1 £0 0

1 ? 1 ? 0 ?

1 2 1 ? £0 1 1 ? 0 0 2 ?

Add 2 times the second row to the third row to obtain Subtract 2 times the second row from the first to obtain

2 1 -2

1 £0 0

0 1 0

-1 ? 1 ? 2 ?

1 -1 -1

0 1 0

0 0§ . 1

1 0 0 -1 1 0 § . -3 2 1 3 -1 -3

-2 1 2

0 0§ . 1

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510

Chapter 9

Matrix Methods for Linear Systems

Multiply the third row by 1>2 to obtain

1

0

-1

?

3

-2

≥0

1

1

?

-1

1

0¥ .

?

- 32

1

1 2

1 0 0 ?

3 2

−1

1 2

0

− 12 ¥ .

− 32

1

1 2

0 Add the third row to the first and then subtract the third row from the second to obtain

0

1

≥0 1 0 ? 0 0 1 ?

0

1 2

The matrix shown in color is A-1. [Compare equation (2).] ◆ It is convenient to have an expression for the inverse of a generic 2 * 2 matrix. The following formula is easily verified by mental arithmetic: (3)

c

a c

d b -1 1 c d = d ad - bc -c

-b d if ad - bc ≠ 0 . a

The denominator in (3), whose nonvanishing is the crucial condition for the existence of the inverse, is known as the determinant. Determinants. The determinant of a 2 * 2 matrix A, denoted det A or  A  , is defined by det A = `

a11 a21

a12 ` = a11a22 - a12a21. a22

The determinants of higher-order square matrices can be defined recursively in terms of lower-order determinants, using the concept of the minor; the minor of a particular entry is the determinant of the submatrix formed when that entry’s row and column are deleted. Then the determinant of an n * n matrix equals the alternating-sign sum of the products of the entries of the first row with their minors. For a 3 * 3 matrix A this looks like a11 a12 a13 a22 a23 a21 a23 a21 a22 det A J † a21 a22 a23 † = a11 ` a a a + ` ` ` ` 12 13 a31 a33 a31 a32 ` . 32 a33 a31 a32 a33 For example, 1 †0 2

2 3 1

1 3 5† = 1 ` 1 -1

0 5 ` -2 ` 2 -1

0 3 5 ` ` +1 ` 2 1 -1

= 11 -3 - 52 - 210 - 102 + 110 - 62 = 6 . For a 4 * 4 matrix, we have 4 -2 ∞ 3 5

3 1 0 2

2 2 3 1

-6 1 1 ∞ = 4†0 5 2 -1

2 3 1

-2 + 2† 3 5

-2 1 5† - 3† 3 5 -1 1 0 2

2 3 1

1 5† -1

1 -2 5 † - 1 -62 † 3 5 -1

1 0 2

2 3† . 1

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Section 9.3

Review 2: Matrices and Vectors

511

As we have shown, the first minor is 6; and the others are computed the same way, resulting in 4 -2 ∞ 3 5

3 1 0 2

2 2 3 1

-6 1 ∞ = 4162 - 31602 + 21542 - 1 -621362 = 168 . 5 -1

Although higher-order determinants can be calculated similarly, a more practical way to evaluate them involves the row-reduction of the matrix to upper triangular form. Here we will deal mainly with low-order determinants, and direct the reader to a linear algebra text for further discussion.† Determinants have a geometric interpretation: det A is the volume (in n-dimensional space) of the parallelepiped whose edges are given by the column vectors of A. But their chief value lies in the role they play in the following theorem, which summarizes many of the results from linear algebra that we shall need, and in Cramer’s rule, described in Appendix D.

Matrices and Systems of Equations Theorem 1. Let A be an n * n matrix. The following statements are equivalent: (a) A is singular (does not have an inverse). (b) The determinant of A is zero. (c) Ax = 0 has nontrivial solutions (x ≠ 0). (d) The columns (rows) of A form a linearly dependent set. In part (d), the statement that the n columns of A are linearly dependent means that there exist scalars c1, c, cn, not all zero, such that c1a1 + c2a2 + g + cnan = 0 , where aj is the vector forming the jth column of A. If A is a singular square matrix (so det A = 0), then Ax = 0 has infinitely many solutions. Indeed, Theorem 1 asserts that there is a vector x0 ≠ 0 such that Ax0 = 0, and we can get infinitely many other solutions by multiplying x0 by any scalar, i.e., taking x = cx0. Furthermore, Ax = b either has no solutions or it has infinitely many of them of the form x = xp + xh , where xp is a particular solution to Ax = b and xh is any of the infinity of solutions to Ax = 0 (see Problem 15). The resemblance of this situation to that of solving nonhomogeneous linear differential equations should be quite apparent. To illustrate, in Example 3 of Section 9.2 (page 502) we saw that the system 2 £ 2 -4

4 4 -8

1 x1 8 0 § £ x2 § = £ 6 § 1 x3 -10

has solutions x1 = 3 - 2s ,

x2 = s ,

x3 = 2 ;

-∞ 6 s 6 ∞ .

†

Your authors’ personal favorite is Fundamentals of Matrix Analysis with Applications, by Edward Barry Saff and Arthur David Snider (John Wiley & Sons, Hoboken, New Jersey, 2016).

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512

Chapter 9

Matrix Methods for Linear Systems

Writing these in matrix notation, we can identify the vectors xp and xh mentioned above: x = £

3 -2 3 - 2s s § = £ 0 § + s £ 1 § = xp + xh . 2 2 0

Note further that the determinant of A is indeed zero, det A = 2 `

2 2 0 4 0 ` + 1` ` - 4` -4 -4 1 -8 1

4 ` = 2#4-4#2+1#0 = 0, -8

and that the linear dependence of the columns of A is exhibited by the identity 0 1 4 2 -2£ 2 § + 1£ 4 § + 0£ 0 § = £ 0 § . 0 1 -8 -4 If A is a nonsingular square matrix (i.e., A has an inverse and det A ≠ 02, then the homogeneous system Ax = 0 has x = 0 as its only solution. More generally, when det A ≠ 0, the system Ax = b has a unique solution (namely, x = A-1b2.

Calculus of Matrices

If we allow the entries aij 1t2 in a matrix A1t2 to be functions of the variable t, then A1t2 is a matrix function of t. Similarly, if the entries xi 1t2 of a vector x1t2 are functions of t, then x1t2 is a vector function of t. These matrix and vector functions have a calculus much like that of real-valued functions. A matrix A1t2 is said to be continuous at t0 if each entry aij 1t2 is continuous at t0. Moreover, A1t2 is differentiable at t0 if each entry aij 1t2 is differentiable at t0, and we write dA 1t 2 = A′1t0 2 J 3aij= 1t0 24 . dt 0

(4)

Similarly, we define b

(5)

Example 2

La

A1t2 dt J c

Let A1t2 = c

b

La

aij 1t2 dt d .

t 2 + 1 cos t d . et 1 1

Find: Solution

L0

A1t2dt .

Using formulas (4) and (5), we compute (a) A′1t2 = c

Example 3

(b)

(a) A′1t2 .

2t et

Show that x1t2 = c A = c

0 v

-sin t d. 0

1

(b)

4

A1t2dt = c 3 e-1 L0

sin 1 d. ◆ 1

cos vt d is a solution of the matrix differential equation x′ = Ax, where sin vt -v d . 0

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 9.3

Solution

Review 2: Matrices and Vectors

513

We simply verify that x′1t2 and Ax1t2 are the same vector function: x′1t2 = c

-v sin vt d ; v cos vt

Ax = c

0 v

-v sin vt -v cos vt d . ◆ d = c dc v cos vt 0 sin vt

The basic properties of differentiation are valid for matrix functions.

Differentiation Formulas for Matrix Functions d dA 1CA2 = C (C a constant matrix) . dt dt d dA d B 1A + B2 = + . dt dt dt d d B dA 1AB2 = A + B. dt dt dt

In the last formula, the order in which the matrices are written is very important because, as we have emphasized, matrix multiplication does not always commute.

9.3

EXERCISES

2 1 -1 d and B J c 3 5 2 Find: (a) A + B . (b) 3A - B .

1. Let A J c

2. Let A J c

2 0 5 d 2 1 1 Find: (a) A + B .

3. Let A J c

2 4 d 1 1 Find: (a) AB .

2 1 4. Let A J £ 0 4 § -1 3 Find: (a) AB .

B J c

1 0 (b) 7A - 4B . and

0 d . -3 -1 3

2 d . -2

-1 3 d . 5 2 (b) A2 = AA . (c) B2 = BB .

and

and

B J c

B J c

1 0

1 3

-1 d . 1

(b) BA .

1 -2 1 0 5. Let A J c d , B J c d , and 2 -3 1 1 -1 1 C J c d . 2 1 Find: (a) AB . (b) AC . (c) A1B + C2 . 6. Let A J C J c

1 1 Find: (a)

1 2 0 3 c d , B J c d , 1 1 1 2 -4 d . 1 AB . (b) 1AB2C .

and

(c) 1A + B2C .

7. (a) Show that if u and v are each n * 1 column vectors, then the matrix product uTv is the same as the dot product u # v. (b) Let v be a 3 * 1 column vector with vT = 32 3 54. Show that, for A as given in Example 1 (page 509), 1Av2 T = vTAT. (c) Does 1Av2 T = vTAT hold for every m * n matrix A and n * 1 vector v? (d) Does 1AB2 T = BTAT hold for every pair of matrices A, B such that both matrix products are defined? Justify your answer. 2 -1 1 2 8. Let A J c d and B J c d . -3 4 3 2 Verify that AB ≠ BA. In Problems 9–14, use the method of Example 1 to compute the inverse of the given matrix, if it exists. For Problems 9 and 10, confirm your answer by comparison with formula (3). 9. c

2 1 d -1 4

10. c

1 1 1 11. £ 1 2 1 § 2 3 2 -2 13. £ 2 3

-1 1 1

4 1 d 5 9

1 1 1 12. £ 1 2 3 § 0 1 1 1 0§ -1

1 14. £ 1 1

1 -1 1

1 2§ 4

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514

Chapter 9

Matrix Methods for Linear Systems

15. Prove that if xp satisfies Axp = b, then every solution to the nonhomogeneous system Ax = b is of the form x = xp + xh, where xh is a solution to the corresponding homogeneous system Ax = 0. 2 -1 1 16. Let A = £ -1 2 1§ . 1 1 2 (a) Show that A is singular. 3 (b) Show that Ax = £ 1 § has no solutions. 3 3 (c) Show that Ax = £ 0 § has infinitely many solutions. 3

In Problems 17–20, find the matrix function X-1 1t2 whose value at t is the inverse of the given matrix X1t2. et e4t 17. X1t2 = c t d e 4e4t 18. X1t2 = c

sin 2t 2 cos 2t

et 19. X1t2 = £ et et

cos 2t d -2 sin 2t

1 25. † -1 4

4 -1 5

0 2† -2

1 24. † 0 -1 3 2† 2

In Problems 33 and 34, find dX>dt for the given matrix functions. e5t 3e2t d 33. X1t2 = c -2e5t -e2t cos 2t e-2t sin 2t 34. X1t2 = £ -sin 2t 2 cos 2t 3e-2t § 3 sin 2t cos 2t e-2t

1 26. † 3 1

27. A = c

0 3 2 4 0 6

2 -1 † 1 4 -3 † 2

1 1 3 3 d 28. A = c d -2 4 2 4 0 0 0 29. A = £ 0 1 0 § 1 0 1 30. Illustrate the equivalence of the assertions (a)–(d) in Theorem 1 (page 511) for the matrix

-2 4 2

0 x1t2 = £ et § -3et

In Problems 37 and 38, verify that the given matrix function satisfies the given matrix differential equation. e3t e2t 1 -1 37. X′ = c X1t2 = c 2t d dX , -e -2e3t 2 4 1 38. X′ = £ 0 0

2 2§ 4

0 3 -2

et X1t2 = £ 0 0

In Problems 27–29, determine the values of r for which det1A - rI2 = 0.

4 A = £ -2 2

e-t sin 3t 0 § 32. x1t2 = £ -e-t sin 3t

e3t 31. x1t2 = £ 2e3t § -e3t

0 0 0 36. x′ = £ 0 1 0 § x , 1 0 1

In Problems 21–26, evaluate the given determinant. 12 8 4 3 ` ` 22. ` 21. ` 3 2 -1 2 0 1 5

In Problems 31 and 32, find dx>dt for the given vector functions.

In Problems 35 and 36, verify that the given vector function satisfies the given system. e3t 1 1 35. x′ = c x1t2 = c 3t d dx , 2e -2 4

e-t e2t - e-t 2e2t § e-t 4e2t

e3t 1 t 20. X1t2 = £ 3e3t 0 1 § 9e3t 0 0

1 23. † 3 1

(a) Show that the row-reduction procedure applied to 3A?I4 fails to produce the inverse of A. (b) Calculate det A. (c) Determine a nontrivial solution x to Ax = 0. (d) Find scalars c1, c2, and c3, not all zero, so that c1a1 + c2a2 + c3a3 = 0, where a1, a2, and a3 are the columns of A.

0 -2 § X , 3 0 0 et e5t § et -e5t

In Problems 39 and 40, the matrices A1t2 and B1t2 are given. Find 1

(a)

L

A1t2dt .

(b)

L0

39. A1t2 = c

t et d , 1 et

40. A1t2 = c

1 e-2t d , 3 e-2t

B1t2dt .

(c)

B1t2 = c

cos t sin t

B1t2 = c

as follows.

d 3A1t2B1t24 . dt - sin t d cos t

e-t e-t d -e-t 3e-t

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Section 9.4

41. An n * n matrix A is called symmetric if AT = A; that is, if aij = aji, for all i, j = 1, … , n. Show that if A is an n * n matrix, then A + AT is a symmetric matrix. 42. Let A be an m * n matrix. Show that ATA is a symmetric n * n matrix and AAT is a symmetric m * m matrix (see Problem 41). 43. The inner product of two vectors is a generalization of the dot product, for vectors with complex entries. It is defined by 1x, y2 J a xiyi ,

515

x = col 1x1, x2, . . . , xn 2, y = col 1y1, y2, . . . , yn 2 are complex vectors and the overbar denotes complex conjugation. (a) Show that 1x, y2 = xTy, where y = col 1y1, y2, c, yn 2 . (b) Prove that for any n * 1 vectors x, y, z and any complex number l, we have 1x, y) = 1y, x2 , 1x, y + z2 = 1x, y2 + 1x, z2 , 1lx, y2 = l1x, y) , 1x, ly2 = l1x, y2 .

n

i=1

Linear Systems in Normal Form

where

9.4 Linear Systems in Normal Form In keeping with the introduction presented in Section 9.1, we say that a system of n linear differential equations is in normal form if it is expressed as (1)

x′1t2 = A1t2x1t2 + f1t2 ,

(2)

y 1n2 1t2 + pn - 1 1t2y 1n-12 1t2 + g + p0 1t2y1t2 = g1t2

where x1t2 = col 1 x1 1t2, c, xn 1t2 2 , f1t2 = col 1 f1 1t2, c, fn 1t2 2 , and A1t2 = 3aij 1t24 is an n * n matrix. As with a scalar linear differential equation, a system is called homogeneous when f1t2 K 0; otherwise, it is called nonhomogeneous. When the elements of A are all constants, the system is said to have constant coefficients. Recall that an nth-order linear differential equation

can be rewritten as a first-order system in normal form using the substitution x1 1t2 J y1t2, x2 1t2 J y′1t2, c, xn 1t2 J y 1n-12 1t2; indeed, equation (2) is equivalent to x′1t2 = A1t2x1t2 + f1t2, where x1t2 = col 1 x1 1t2, c, xn 1t22, f1t 2 J col 1 0, c, 0, g1t2 2 , and 0 0 A1t2 J E f 0 -p0 1t2

1 0 f 0 -p1 1t2

0 1 f 0 -p2 1t2

g

g g

0 0 f 0

-pn - 2 1t2

0 0 f 1

U.

-pn - 1 1t2

The theory for systems in normal form parallels very closely the theory of linear differential equations presented in Chapters 4 and 6. In many cases the proofs for scalar linear differential equations carry over to normal systems with appropriate modifications. Conversely, results for normal systems apply to scalar linear equations since, as we showed, any scalar linear equation can be expressed as a normal system. This is the case with the existence and uniqueness theorems for linear differential equations. The initial value problem for the normal system (1) is the problem of finding a differentiable vector function x1t2 that satisfies the system on an interval I and also satisfies the initial condition x1t0 2 = x0, where t0 is a given point of I and x0 = col1x1,0, c, xn,0 2 is a given vector.

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516

Chapter 9

Matrix Methods for Linear Systems

Existence and Uniqueness Theorem 2. If A1t2 and f1t2 are continuous on an open interval I that contains the point t0, then for any choice of the initial vector x0, there exists a unique solution x1t2 on the whole interval I to the initial value problem x′1t2 = A1t2x1t2 + f1t2 ,

x1t0 2 = x0 .

We give a proof of this result in Chapter 13† and obtain as corollaries the existence and uniqueness theorems for second-order equations (Theorem 4, Section 4.5, page 182) and higher-order linear equations (Theorem 1, Section 6.1, page 319). If we rewrite system (1) as x′ - Ax = f and define the operator L3x4 J x′ - Ax, then we can express system (1) in the operator form L3x4 = f. Here the operator L maps vector functions into vector functions. Moreover, L is a linear operator in the sense that for any scalars a, b and differentiable vector functions x, y, we have L3ax + by4 = aL3x4 + bL3y4 . The proof of this linearity follows from the properties of matrix multiplication (see Problem 27). As a consequence of the linearity of L, if x1, c, xn are solutions to the homogeneous system x′ = Ax, or L3x4 = 0 in operator notation, then any linear combination of these vectors, c1x1 + g + cnxn, is also a solution. Moreover, we will see that if the solutions x1, c, xn are linearly independent, then every solution to L3x4 = 0 can be expressed as c1x1 + g + cnxn for an appropriate choice of the constants c1, c, cn.

Linear Dependence of Vector Functions Definition 1. The m vector functions x1, c, xm are said to be linearly dependent on an interval I if there exist constants c1, c, cm, not all zero, such that c1x1 1t2 + g + cmxm 1t2 = 0

(3)

for all t in I. If the vectors are not linearly dependent, they are said to be linearly independent on I.

Example 1 Solution

Example 2

Show that the vector functions x1 1t2 = col1et, 0, et 2, x2 1t2 = col13et, 0, 3et 2, and x3 1t2 = col1t, 1, 02 are linearly dependent on 1 - ∞, ∞ 2.

Notice that x2 is just 3 times x1 and therefore 3x1 1t2 - x2 1t2 + 0 # x3 1t2 = 0 for all t. Hence, x1, x2, and x3 are linearly dependent on 1 - ∞, ∞ 2. ◆ Show that t x1 1t2 = c  t  d

,

x2 1t2 = c

are linearly independent on 1 - ∞, ∞ 2.

t d t

†

All references to Chapters 11–13 refer to the expanded text, Fundamentals of Differential Equations and Boundary Value Problems, 7th ed.

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Section 9.4

Solution

Example 3 Solution

Linear Systems in Normal Form

517

Note that at every instant t0, the column vector x1 1t0 2 is a multiple of x2 1t0 2; indeed, x1 1t0 2 = x2 1t0 2 for t0 Ú 0, and x1 1t0 2 = -x2 1t0 2 for t0 … 0. Nonetheless, the vector functions are not dependent, because the c’s in condition (3) are not allowed to change with t; for t 6 0, the equation c1x1 1t2 + c2x2 1t2 = 0 implies c1 - c2 = 0, but for t 7 0 it implies c1 + c2 = 0. Thus c1 = c2 = 0 and the functions are independent. ◆ Show that the vector functions x1 1t2 = col1e2t, 0, e2t 2, x2 1t2 = col1e2t, e2t, -e2t 2, and x3 1t2 = col1et, 2et, et 2 are linearly independent on 1 - ∞, ∞ 2. To prove independence, we assume c1, c2, and c3 are constants for which c1x1 1t2 + c2x2 1t2 + c3x3 1t2 = 0

holds at every t in 1 - ∞, ∞ 2 and show that this forces c1 = c2 = c3 = 0. In particular, when t = 0 we obtain 1 1 1 c1 £ 0 § + c2 £ 1 § + c3 £ 2 § = 0 , 1 -1 1 which is equivalent to the system of linear equations (4)

c1 + c2 + c3 = 0 , c2 + 2c3 = 0 , c1 - c2 + c3 = 0 .

Either by solving (4) or by checking that the determinant of its coefficients is nonzero (recall Theorem 1 on page 511), we can verify that (4) has only the trivial solution c1 = c2 = c3 = 0. Therefore the vector functions x1, x2, and x3 are linearly independent on 1 - ∞, ∞ 2 (in fact, on any interval containing t = 0). ◆ As Example 3 illustrates, if x1 1t2, x2 1t2, c, xn 1t2 are n vector functions, each having n components, we can establish their linear independence on an interval I if we can find one point t0 in I where the determinant det3x1 1t0 2 cxn 1t0 24

is not zero. Because of the analogy with scalar equations, we call this determinant the Wronskian.

Wronskian Definition 2. The Wronskian of n vector functions x1 1t2 = col1x1,1, c, xn,1 2, c, xn 1t2 = col1x1,n, c, xn,n 2 is defined to be the function x1,1 1t2 x2,1 1t2 W3x1, c, xn 4 1t2 J ∞ f xn,1 1t2

x1,2 1t2 x2,2 1t2 f xn,2 1t2

g g g

x1,n 1t2 x2,n 1t2 ∞ . f xn,n 1t2

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518

Chapter 9

Matrix Methods for Linear Systems

We see that n vector functions are linearly independent on an interval if their Wronskian is nonzero at any point in the interval. But now we show that if these functions happen to be independent solutions to a homogeneous system x′ = Ax, where A is an n * n matrix of continuous functions, then the Wronskian is never zero on I. For suppose to the contrary that W1t0 2 = 0. Then by Theorem 1 the vanishing of the determinant implies that the column vectors x1 1t0 2, x2 1t0 2, c, xn 1t0 2 are linearly dependent. Thus there exist scalars c1, c, cn not all zero, such that at t0 c1x1 1t0 2 + g + cnxn 1t0 2 = 0 .

However, c1x1 1t2 + g + cnxn 1t2 and the vector function z1t2 K 0 are both solutions to x′ = Ax on I, and they agree at the point t0. So these solutions must be identical on I according to the existence-uniqueness theorem (Theorem 2, page 516). That is, c1x1 1t2 + g + cnxn 1t2 = 0

for all t in I. But this contradicts the given information that x1, c, xn are linearly independent on I. We have shown that W1t0 2 ≠ 0, and since t0 is an arbitrary point, it follows that W1t2 ≠ 0 for all t ∈ I. The preceding argument has two important implications that parallel the scalar case. First, the Wronskian of solutions to x′ = Ax is either identically zero or never zero on I (see also Problem 33). Second, a set of n solutions x1, c, xn to x′ = Ax on I is linearly independent on I if and only if their Wronskian is never zero on I. With these facts in hand, we can imitate the proof given for the scalar case in Section 6.1 (Theorem 2, page 322) to obtain the following representation theorem for the solutions to x′ = Ax.

Representation of Solutions (Homogeneous Case) Theorem 3. (5)

Let x1, c, xn be n linearly independent solutions to the homogeneous system

x′1t2 = A1t2x1t2

on the interval I, where A1t2 is an n * n matrix function continuous on I. Then every solution to (5) on I can be expressed in the form (6)

x1t2 = c1x1 1t2 + g + cnxn 1t2 ,

where c1, c, cn are constants. A set of solutions 5x1, c, xn 6 that are linearly independent on I or, equivalently, whose Wronskian does not vanish on I, is called a fundamental solution set for (5) on I. The linear combination in (6), written with arbitrary constants, is referred to as a general solution to (5). If we take the vectors in a fundamental solution set and let them form the columns of a matrix X1t2, that is, x1,1 1t2 x 1t2 X1t2 = 3x1 1t2 x2 1t2 cxn 1t24 = D 2,1 f xn,1 1t2

x1,2 1t2 x2,2 1t2 f xn,2 1t2

g g g

x1,n 1t2 x2,n 1t2 T, f

xn,n 1t2

then the matrix X1t2 is called a fundamental matrix for (5). We can use it to express the general solution (6) as x1t2 = X1t2c ,

where c = col1c1, c, cn 2 is an arbitrary constant vector. Since det X = W3x1, c, xn 4 is never zero on I, it follows from Theorem 1 on page 511 that X1t2 is invertible for every t in I.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 9.4

Example 4

Linear Systems in Normal Form

519

Verify that the set -e-t 0 e2t 2t S = • £ e § , £ 0 § , £ e-t § ¶ e2t e-t -e-t is a fundamental solution set for the system (7)

0 1 1 x′1t2 = £ 1 0 1 § x1t2 1 1 0

on the interval 1 - ∞, ∞ 2 and find a fundamental matrix for (7). Also determine a general solution for (7). Solution

Substituting the first vector in the set S into the right-hand side of (7) gives e2t 0 1 1 2e2t 2t Ax = £ 1 0 1 § £ e § = £ 2e2t § = x′1t2 . 1 1 0 e2t 2e2t Hence this vector satisfies system (7) for all t. Similar computations verify that the remaining vectors in S are also solutions to (7) on 1 - ∞, ∞ 2. For us to show that S is a fundamental solution set, it is enough to observe that the Wronskian e2t W1t2 = † e2t e2t

-e-t 0 e-t

0 0 e-t † = e2t ` -t e -e-t

2t e-t -t e e + ` ` -e-t e2t

e-t ` = -3 -e-t

is never zero. A fundamental matrix X1t2 for (7) is just the matrix we used to compute the Wronskian; that is, (8)

e2t X1t2 J £ e2t e2t

-e-t 0 e-t

0 e-t § . -e-t

A general solution to (7) can now be expressed as e2t -e-t 0 2t x1t2 = X1t2c = c1 £ e § + c2 £ 0 § + c3 £ e-t § . ◆ e2t e-t -e-t It is easy to check that the fundamental matrix in (8) satisfies the equation 0 1 1 X′1t2 = £ 1 0 1 § X1t2 ; 1 1 0 indeed, this is equivalent to showing that x′ = Ax for each column x in S. In general, a fundamental matrix for a system x′ = Ax satisfies the corresponding matrix differential equation X′ = AX. Another consequence of the linearity of the operator L defined by L3x4 J x′ - Ax is the superposition principle for linear systems. It states that if x1 and x2 are solutions, respectively, to the nonhomogeneous systems L3x4 = g1 and L3x4 = g2 ,

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520

Chapter 9

Matrix Methods for Linear Systems

then c1x1 + c2x2 is a solution to L3x4 = c1g1 + c2g2 . Using the superposition principle and the representation theorem for homogeneous systems, we can prove the following theorem.

Representation of Solutions (Nonhomogeneous Case) Theorem 4.

If xp is a particular solution to the nonhomogeneous system

(9)

x′1t2 = A1t2x1t2 + f1t2

(10)

x1t2 = xp 1t2 + c1x1 1t2 + g + cnxn 1t2 ,

on the interval I and 5x1, c, xn 6 is a fundamental solution set on I for the corresponding homogeneous system x1t2 = A1t2x1t2, then every solution to (9) on I can be expressed in the form where c1, c, cn are constants.

The proof of this theorem is almost identical to the proofs of Theorem 4 in Section 4.5 (page 182) and Theorem 4 in Section 6.1 (page 325). We leave the proof as an exercise. The linear combination of xp, x1, c, xn in (10) written with arbitrary constants c1, c, cn is called a general solution of (9). This general solution can also be expressed as x = xp + Xc, where X is a fundamental matrix for the homogeneous system and c is an arbitrary constant vector. We now summarize the results of this section as they apply to the problem of finding a general solution to a system of n linear first-order differential equations in normal form.

Approach to Solving Normal Systems 1. To determine a general solution to the n * n homogeneous system x′ = Ax:

(a) Find a fundamental solution set 5x1, c, xn 6 that consists of n linearly independent solutions to the homogeneous system. (b) Form the linear combination x = Xc = c1x1 + g + cnxn ,

where c = col1c1, c, cn 2 is any constant vector and X = 3x1 cxn 4 is the fundamental matrix, to obtain a general solution. 2. To determine a general solution to the nonhomogeneous system x′ = Ax + f: (a) Find a particular solution xp to the nonhomogeneous system. (b) Form the sum of the particular solution and the general solution Xc = c1x1 + g + cnxn to the corresponding homogeneous system in part 1, x = xp + Xc = xp + c1x1 + g + cnxn , to obtain a general solution to the given system.

We devote the rest of this chapter to methods for finding fundamental solution sets for homogeneous systems and particular solutions for nonhomogeneous systems.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 9.4

9.4

2. r′1t2 = 2r1t2 + sin t , u′1t2 = r1t2 - u1t2 + 1 dx = t 2x - y - z + t , dt dy = etz + 5 , dt dz = tx - y + 3z - et dt

4.

dx = x+y+z, dt dy = 2x - y + 3z , dt dz = x + 5z dt

In Problems 5–8, rewrite the given scalar equation as a firstorder system in normal form. Express the system in the matrix form x′ = Ax + f. 5. y″1t2 - 3y′1t2 - 10y1t2 = sin t d 4w 6. x″1t2 + x1t2 = t 2 7. + w = t2 dt 4 8.

d 3y dt 3

521

EXERCISES

In Problems 1–4, write the given system in the matrix form x′ = Ax + f. 1. x′1t2 = 3x1t2 - y1t2 + t 2 , y′1t2 = - x1t2 + 2y1t2 + et

3.

Linear Systems in Normal Form

-

dy + y = cos t dt

In Problems 9–12, write the given system as a set of scalar equations. 2 5 0 9. x′ = c d d x + e-2t c -3 -2 4 t 2 1 10. x′ = c d x + et c d 1 -1 3 1 0 1 0 1 11. x′ = £ - 1 2 5 § x + et £ 0 § + t £ 1 § 0 0 0 5 1

4 c d 1

1 -3 15. et c d , et c d 5 - 15

14. c

te d , e-t

sin t d , 16. c cos t

1 1 0 17. e2t £ 0 § , e2t £ 1 § , e3t £ 1 § 5 -1 0 sin t sin t cos t 18. c d ,c d ,c d cos t sin t cos t

t2 £0§ t2

20. Let cos t sin t cos t x1 = £ 0 § , x2 = £ cos t § , x3 = £ sin t § . 0 cos t cos t (a) Compute the Wronskian. (b) Are these vector functions linearly independent on ( - ∞ , ∞ )? (c) Is there a first-order homogeneous linear system for which these functions are solutions? In Problems 21–24, the given vector functions are solutions to a system x′1t2 = Ax1t2. Determine whether they form a fundamental solution set. If they do, find a fundamental matrix for the system and give a general solution. 1 -2 x2 = e2t c d d , 21. x1 = e2t c -2 4 3 22. x1 = e-t c d , 2

x2 = e4t c

e-t 23. x1 = £ 2e-t § , e-t

et x2 = £ 0 § , et

et 24. x1 = £ et § , et

et x1 = c t d e

1 d -1

sin t x2 = £ cos t § , - sin t

e3t x3 = £ -e3t § 2e3t - cos t x3 = £ sin t § cos t

and

x2 = c

e-t d 3e-t

are solutions to the homogeneous system x′ = Ax = c

In Problems 13–19, determine whether the given vector functions are linearly dependent 1LD2 or linearly independent 1LI2 on the interval 1 - ∞ , ∞ 2. t 13. c d , 3

t £0§ , t

25. Verify that the vector functions

0 1 0 1 3 12. x′ = £ 0 0 1 § x + t £ - 1 § + £ 1 § -1 1 2 2 0

-t

1 19. £ 0 § , 1

c

-t

e d e-t sin 2t c d cos 2t

2 3

on 1 - ∞ , ∞ 2, and that xp =

-1 d x, -2

3 tet 1 et t 0 c td - c td + c d - c d 2 te 4 3e 2t 1

is a particular solution to the nonhomogeneous system x′ = Ax + f1t2, where f1t2 = col1et, t2. Find a general solution to x′ = Ax + f1t2. 26. Verify that the vector functions e3t x1 = £ 0 § , e3t

- e3t -e-3t 3t x2 = £ e § , x3 = £ -e-3t § 0 e-3t

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522

Chapter 9

Matrix Methods for Linear Systems

are solutions to the homogeneous system 1 -2 2 x′ = Ax = £ - 2 1 2§ x 2 2 1 on 1 - ∞ , ∞ 2, and that 5t + 1 xp = £ 2t § 4t + 2 is a particular solution to x′ = Ax + f1t2, where f1t2 = col1 - 9t, 0, -18t2. Find a general solution to x′ = Ax + f1t2. 27. Prove that the operator defined by L3x4 J x′ - Ax, where A is an n * n matrix function and x is an n * 1 differentiable vector function, is a linear operator. 28. Let X1t2 be a fundamental matrix for the system x′ = Ax. Show that x1t2 = X1t2X-1 1t0 2x0 is the solution to the initial value problem x′ = Ax, x1t0 2 = x0. In Problems 29–30, verify that X1t2 is a fundamental matrix for the given system and compute X-1 1t2. Use the result of Problem 28 to find the solution to the given initial value problem. -1 0 6 0 29. x′ = £ 1 0 1 § x , x102 = £ 0 § ; 1 1 1 0 6e-t - 3e-2t 2e3t X1t2 = £ - e-t e-2t e3t § -t - 5e e-2t e3t 2 3 3 30. x′ = c d x, x102 = c d ; 3 2 -1 e-t e5t X1t2 = c -t 5t d -e e 31. Show that t2 t  t  ` ` K 0 2t 2  t  on 1 - ∞ , ∞ 2, but that the two vector functions c

t2 d , 2t

c

tt d 2t

are linearly independent on 1 - ∞ , ∞ 2. 32. Abel’s Formula. If x1, c, xn are any n solutions to the n * n system x′1t2 = A1t2x1t2, then Abel’s formula gives a representation for the Wronskian W1t2 J W3x1, c, xn 41t2. Namely, W1t2 = W1t0 2expa

t

Lt0

5a11 1s2 + P + ann 1s26 ds b,

where a11 1s2, c, ann 1s2 are the main diagonal elements of A1s2. Prove this formula in the special case when n = 3. [Hint: Follow the outline in Problem 30 of Exercises 6.1, page 327.]

33. Using Abel’s formula (Problem 32), confirm that the Wronskian of n solutions to x′ = Ax on the interval I is either identically zero on I or never zero on I. 34. Prove that a fundamental solution set for the homogeneous system x′1t2 = A1t2x1t2 always exists on an interval I, provided A1t2 is continuous on I. 3Hint: Use the existence and uniqueness theorem (Theorem 2) and make judicious choices for x0.4 35. Prove Theorem 3 on the representation of solutions of the homogeneous system. 36. Prove Theorem 4 on the representation of solutions of the nonhomogeneous system. 37. To illustrate the connection between a higher-order equation and the equivalent first-order system, consider the equation y‴1t2 − 6y″1t2 + 11y′1t2 − 6y1t2 = 0 .

(11)

(a) Show that {et, e2t, e3t} is a fundamental solution set for (11). (b) Using the definition in Section 6.1, compute the Wronskian of {et, e2t, e3t}. (c) Setting x1 = y, x2 = y′, x3 = y″, show that equation (11) is equivalent to the first-order system x′ = Ax ,

(12) where

0 1 0 A J £0 0 1§ . 6 -11 6 (d) The substitution used in part (c) suggests that e2t e3t et t 2t S J • £ e § , £ 2e § , £ 3e3t § ¶ et 4e2t 9e3t is a fundamental solution set for system (12). Verify that this is the case. (e) Compute the Wronskian of S. How does it compare with the Wronskian computed in part (b)? 38. Define x1 1t2, x2 1t2, and x3 1t2, for - ∞ 6 t 6 ∞ , by sin t sin t 0 x1 1t2 = £ sin t § , x2 1t2 = £ 0 § , x3 1t2 = £ sin t § . sin t 0 sin t (a) Show that for the three scalar functions in each individual row there are nontrivial linear combinations that sum to zero for all t. (b) Show that, nonetheless, the three vector functions are linearly independent. (No single nontrivial combination works for each row, for all t.) (c) Calculate the Wronskian W3x1, x2, x3 41t2. (d) Is there a linear third-order homogeneous differential equation system having x1 1t2, x2 1t2, and x3 1t2 as solutions?

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Section 9.5

Homogeneous Linear Systems with Constant Coefficients

523

9.5 Homogeneous Linear Systems with Constant Coefficients

In this section we discuss a procedure for obtaining a general solution for the homogeneous system (1)

x′1t2 = Ax1t2 ,

where A is a (real) constant n * n matrix. The general solution we seek will be defined for all t because the elements of A are just constant functions, which are continuous on 1 - ∞, ∞ 2 (recall Theorem 2, page 516). In Section 9.4 we showed that a general solution to (1) can be constructed from a fundamental solution set consisting of n linearly independent solutions to (1). Thus our goal is to find n such vector solutions. In Chapter 4 we were successful in solving homogeneous linear equations with constant coefficients by guessing that the equation had a solution of the form ert. Because any scalar linear equation can be expressed as a system, it is reasonable to expect system (1) to have solutions of the form x1t2 = ertu , where r is a constant and u is a constant vector, both of which must be determined. Substituting ertu for x1t2 in (1) gives rertu = Aertu = ertAu . Canceling the factor ert and rearranging terms, we find that (2)

1A - rI2u = 0 ,

where rI denotes the diagonal matrix with r’s along its main diagonal. The preceding calculation shows that x1t2 = ertu is a solution to (1) if and only if r and u satisfy equation (2). Since the trivial case, u = 0, is of no help in finding linearly independent solutions to (1), we require that u ≠ 0. Such vectors are given a special name, as follows.

Eigenvalues and Eigenvectors Definition 3. Let A = 3aij 4 be an n * n constant matrix. The eigenvalues of A are those (real or complex) numbers r for which 1A - rI2u = 0 has at least one nontrivial (real or complex) solution u. The corresponding nontrivial solutions u are called the eigenvectors of A associated with r.

As stated in Theorem 1 of Section 9.3, a linear homogeneous system of n algebraic equations in n unknowns has a nontrivial solution if and only if the determinant of its coefficients is zero. Hence, a necessary and sufficient condition for (2) to have a nontrivial solution is that (3)

0 A − rI 0 = 0 .

Expanding the determinant of A - rI in terms of its cofactors, we find that it is an nth-degree polynomial in r; that is, (4)

0 A − rI 0 = p1r2 .

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524

Chapter 9

Matrix Methods for Linear Systems

Therefore, finding the eigenvalues of a matrix A is equivalent to finding the zeros of the polynomial p1r2. Equation (3) is called the characteristic equation of A, and p1r2 in (4) is the characteristic polynomial of A. The characteristic equation plays a role for systems similar to the role played by the auxiliary equation for scalar equations. Many commercially available software packages can be used to compute the eigenvalues and eigenvectors for a given matrix. Three such packages are MATLAB®, available from The MathWorks, Inc.; MATHEMATICA®, available from Wolfram Research; and MAPLESOFT®, available from Waterloo Maple Inc. Although you are encouraged to make use of such packages, the examples and most exercises in this text can be easily carried out without them. Those exercises for which a computer package is desirable are flagged with the icon . Example 1

Find the eigenvalues and eigenvectors of the matrix A J c

Solution

2 1

-3 d . -2

The characteristic equation for A is

0 A - rI 0 = `

2-r 1

-3 ` = 12 - r21 -2 - r2 + 3 = r 2 - 1 = 0 . -2 - r

Hence the eigenvalues of A are r1 = 1, r2 = -1. To find the eigenvectors corresponding to r1 = 1, we must solve 1A - r1I2u = 0. Substituting for A and r1 gives (5)

c

1 1

0 -3 u1 d c d = c d . -3 u2 0

Notice that this matrix equation is equivalent to the single scalar equation u1 - 3u2 = 0. Therefore, the solutions to (5) are obtained by assigning an arbitrary value for u2 (say, u2 = s) and setting u1 = 3u2 = 3s. Consequently, the eigenvectors associated with r1 = 1 can be expressed as (6)

3 u1 = s c d . 1 For r2 = -1, the equation 1A - r2I2u = 0 becomes c

3 1

-3 u1 0 dc d = c d . -1 u2 0

Solving, we obtain u1 = s and u2 = s, with s arbitrary. Therefore, the eigenvectors associated with the eigenvalue r2 = -1 are (7)

1 u2 = s c d . ◆ 1

We remark that in the above example the collection (6) of all eigenvectors associated with r1 = 1 forms a one-dimensional subspace when the zero vector is adjoined. The same is true for r2 = -1. These subspaces are called eigenspaces.

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Section 9.5

Example 2

525

Find the eigenvalues and eigenvectors of the matrix 1 A J £1 4

Solution

Homogeneous Linear Systems with Constant Coefficients

2 0 -4

-1 1§ . 5

The characteristic equation for A is 1-r 0 A - rI 0 = † 1 4

-1 2 -r 1 † = 0, -4 5 - r

which simplifies to 1r - 121r - 221r - 32 = 0. Hence, the eigenvalues of A are r1 = 1, r2 = 2, and r3 = 3. To find the eigenvectors corresponding to r1 = 1, we set r = 1 in 1A - rI2u = 0. This gives (8)

0 £1 4

2 -1 -4

u1 -1 0 1 § £ u2 § = £ 0 § . 4 0 u3

Using elementary row operations (Gaussian elimination), we see that (8) is equivalent to the two equations u1 - u2 + u3 = 0 , 2u2 - u3 = 0 . Thus, we can obtain the solutions to (8) by assigning an arbitrary value to u2 (say, u2 = s), solving 2u2 - u3 = 0 for u3 to get u3 = 2s, and then solving u1 - u2 + u3 = 0 for u1 to get u1 = -s. Hence, the eigenvectors associated with r1 = 1 are (9)

-1 u1 = s £ 1 § . 2 For r2 = 2, we solve -1 £ 1 4

2 -2 -4

-1 u1 0 1 § £ u2 § = £ 0 § 3 u3 0

in a similar fashion to obtain the eigenvectors (10)

-2 u2 = s £ 1 § . 4 Finally, for r3 = 3, we solve -2 £ 1 4

2 -3 -4

-1 u1 0 1 § £ u2 § = £ 0 § 2 u3 0

and get the eigenvectors (11)

-1 u3 = s £ 1 § . ◆ 4

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526

Chapter 9

Matrix Methods for Linear Systems

Let’s return to the problem of finding a general solution to a homogeneous system of differential equations. We have already shown that ertu is a solution to (1) if r is an eigenvalue and u a corresponding eigenvector. The question is: Can we obtain n linearly independent solutions to the homogeneous system by finding all the eigenvalues and eigenvectors of A? The answer is yes, if A has n linearly independent eigenvectors.

n Linearly Independent Eigenvectors Theorem 5. Suppose the n * n constant matrix A has n linearly independent eigenvectors u1, u2, c, un. Let ri be the eigenvalue† corresponding to ui. Then (12)

{er1tu1, er2tu2, c, erntun}

(13)

x1t2 = c1er1tu1 + c2er2tu2 + g + cnerntun ,

is a fundamental solution set (and X1t2 = 3er1tu1 er2tu2 g erntun 4 is a fundamental matrix) on 1 - ∞, ∞ 2 for the homogeneous system x′ = Ax. Consequently, a general solution of x′ = Ax is where c1, c, cn are arbitrary constants. Proof. As we have seen, the vector functions listed in (12) are solutions to the homogeneous system. Moreover, their Wronskian is W1t2 = det3er1tu1, c, erntun 4 = e1r1 + g + rn2t det3u1, c, un 4 .

Since the eigenvectors are assumed to be linearly independent, it follows from Theorem 1 in Section 9.3 that det3u1, c, un 4 is not zero. Hence the Wronskian W1t2 is never zero. This shows that (12) is a fundamental solution set, and consequently a general solution is given by (13). ◆

An application of Theorem 5 is given in the next example. Example 3

Find a general solution of (14)

Solution

x′1t2 = Ax1t2 , where A = c

2 1

-3 d. -2

In Example 1 we showed that the matrix A has eigenvalues r1 = 1 and r2 = -1. Taking, say, s = 1 in equations (6) and (7), we get the corresponding eigenvectors 3 u1 = c d 1

and

1 u2 = c d . 1

Because u1 and u2 are linearly independent, it follows from Theorem 5 that a general solution to (14) is (15)

3 1 x1t2 = c1et c d + c2e-t c d . ◆ 1 1

The eigenvalues r1, c, rn may be real or complex and need not be distinct. In this section the cases we discuss have real eigenvalues. We consider complex eigenvalues in Section 9.6.

†

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Section 9.5

Homogeneous Linear Systems with Constant Coefficients

527

x2 2

1.5

x(0) =

1

u2

1 2

(u1 + u2)

u1

0.5

0

1

2

3

4

5

x1

Figure 9.1 Trajectories of solutions for Example 3

If we sum the vectors on the right-hand side of equation (15) and then write out the expressions for the components of x1t2 = col 1 x1 1t2, x2 1t2 2 , we get x1 1t2 = 3c1et + c2e-t , x2 1t2 = c1et + c2e-t .

This is the familiar form of a general solution for a system, as discussed in Section 5.2. Example 3 nicely illustrates the geometric role played by the eigenvectors u1 and u2. If the initial vector x102 is a scalar multiple of u1 1i.e., x102 = c1u1 2, then the vector solution to the system, x1t2 = c1etu1, will always have the same or opposite direction as u1. That is, it will lie along the straight line determined by u1 (see Figure 9.1). Furthermore, the trajectory of this solution, as t increases, will tend to infinity, since the corresponding eigenvalue r1 = 1 is positive (observe the et term). A similar assertion holds if the initial vector is a scalar multiple of u2, except that since r2 = -1 is negative, the trajectory x1t2 = c2e-tu2 will approach the origin as t increases (because of e-t). For an initial vector that involves both u1 and u2, such as x102 = 12 1u1 + u2 2, the resulting trajectory is a blend of the above motions, with the contribution due to the larger eigenvalue r1 = 1 dominating as t increases; see Figure 9.1. The straight-line trajectories in the x1x2-plane (the phase plane), then, point along the directions of the eigenvectors of the matrix A. (See Section 5.4, Figure 5.11, page 266, for example.) A useful property of eigenvectors that concerns their linear independence is stated in the next theorem.

Linear Independence of Eigenvectors Theorem 6. If r1, c, rm are distinct eigenvalues for the matrix A and ui is an eigenvector associated with ri, then u1, c, um are linearly independent. Proof. Let’s first treat the case m = 2. Suppose, to the contrary, that u1 and u2 are linearly dependent so that (16)

u1 = cu2

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528

Chapter 9

Matrix Methods for Linear Systems

for some constant c. Multiplying both sides of (16) by A and using the fact that u1 and u2 are eigenvectors with corresponding eigenvalues r1 and r2, we obtain (17)

r1u1 = cr2u2 .

Next we multiply (16) by r2 and then subtract from (17) to get 1r1 - r2 2u1 = 0 .

Since u1 is not the zero vector, we must have r1 = r2. But this violates the assumption that the eigenvalues are distinct! Hence u1 and u2 are linearly independent. The cases m 7 2 follow by induction. The details of the proof are left as Problem 48. ◆ Combining Theorems 5 and 6, we get the following corollary.

n Distinct Eigenvalues Corollary 1. If the n * n constant matrix A has n distinct eigenvalues r1, c, rn and ui is an eigenvector associated with ri, then {er1tu1, c, erntun} is a fundamental solution set for the homogeneous system x′ = Ax.

Example 4

Solve the initial value problem (18)

Solution

1 x′1t2 = £ 1 4

-1 1 § x1t2 , 5

2 0 -4

-1 x102 = £ 0 § . 0

In Example 2 we showed that the 3 * 3 coefficient matrix A has the three distinct eigenvalues r1 = 1, r2 = 2, and r3 = 3. If we set s = 1 in equations (9), (10), and (11), we obtain the corresponding eigenvectors -1 u1 = £ 1 § , 2

-2 u2 = £ 1 § , 4

-1 u3 = £ 1 § , 4

whose linear independence is guaranteed by Theorem 6. Hence, a general solution to (18) is (19)

-1 -2 -1 2t 3t x1t2 = c1e £ 1 § + c2e £ 1 § + c3e £ 1 § 2 4 4 t

-et = £ et 2et

-2e2t e2t 4e2t

-e3t c1 e3t § £ c2 § . 4e3t c3

To satisfy the initial condition in (18), we solve -1 x102 = £ 1 2

-2 1 4

-1 c1 -1 1 § £ c2 § = £ 0 § 4 c3 0

and find that c1 = 0, c2 = 1, and c3 = -1. Inserting these values into (19) gives the desired solution. ◆

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Section 9.5

Homogeneous Linear Systems with Constant Coefficients

529

There is a special class of n * n matrices that always have n linearly independent eigenvectors. These are the real symmetric matrices.

Real Symmetric Matrices Definition 4. A real symmetric matrix A is a matrix with real entries that satisfies AT = A.

Taking the transpose of a matrix interchanges its rows and columns. Doing this is equivalent to “flipping” the matrix about its main diagonal. Consequently, AT = A if and only if A is symmetric about its main diagonal. If A is an n * n real symmetric matrix, it is known† that all its eigenvalues are real and that there always exist n linearly independent eigenvectors. In such a case, Theorem 5 applies and a general solution to x′ = Ax is given by (13). Example 5

Find a general solution of (20)

Solution

1 x′1t2 = Ax1t2 , where A = £ -2 2

-2 1 2

2 2§ . 1

A is symmetric, so we are assured that A has three linearly independent eigenvectors. To find them, we first compute the characteristic equation for A: 1-r 0 A - rI 0 = † -2 2

-2 1-r 2

2 2 † = - 1r - 32 2 1r + 32 = 0 . 1-r

Thus the eigenvalues of A are r1 = r2 = 3 and r3 = -3. Notice that the eigenvalue r = 3 has multiplicity 2 when considered as a root of the characteristic equation. Therefore, we must find two linearly independent eigenvectors associated with r = 3. Substituting r = 3 in 1A - rI2u = 0 gives -2 £ -2 2

-2 -2 2

2 u1 0 2 § £ u2 § = £ 0 § . -2 u3 0

This system is equivalent to the single equation -u1 - u2 + u3 = 0, so we can obtain its solutions by assigning an arbitrary value to u2, say u2 = y, and an arbitrary value to u3, say u3 = s. Solving for u1, we find u1 = u3 - u2 = s - y. Therefore, the eigenvectors associated with r1 = r2 = 3 can be expressed as -1 1 s-y u = £ y § = s £ 0 § + y£ 1 § . 0 s 1

†

See Fundamentals of Matrix Analysis with Applications, by Edward Barry Saff and Arthur David Snider (John Wiley & Sons, Hoboken, New Jersey, 2016).

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530

Chapter 9

Matrix Methods for Linear Systems

By first taking s = 1, y = 0 and then taking s = 0, y = 1, we get the two linearly independent eigenvectors -1 u2 = £ 1 § . 0

1 u1 = £ 0 § , 1

(21)

For r3 = -3, we solve 4 1A + 3I2u = £ -2 2

-2 4 2

2 u1 0 2 § £ u2 § = £ 0 § 4 u3 0

to obtain the eigenvectors col1 -s, -s, s2. Taking s = 1 gives -1 u3 = £ -1 § . 1 Since the eigenvectors u1, u2, and u3 are linearly independent, a general solution to (20) is 1 -1 -1 x1t2 = c1e3t £ 0 § + c2e3t £ 1 § + c3e-3t £ -1 § . ◆ 1 0 1 If a matrix A is not symmetric, it is possible for A to have a repeated eigenvalue but not to have two linearly independent corresponding eigenvectors. In particular, the matrix (22)

A = c

1 4

-1 d -3

has the repeated eigenvalue r1 = r2 = -1, but Problem 35 shows that all the eigenvectors associated with r = -1 are of the form u = s col11, 22. Consequently, no two eigenvectors are linearly independent. A procedure for finding a general solution in such a case is illustrated in Problems 35–40, but the underlying theory is deferred to Section 9.8, where we discuss the matrix exponential. A final note. If an n * n matrix A has n linearly independent eigenvectors ui with eigenvalues ri, a little inspection reveals that property (2) is expressed columnwise by the equation f (23) ≥

A

¥ ≥

u1

f u2

f

f

g g f g

f un f

f ¥ = ≥

u1

f u2

f

f

g g f g

f un

¥ ≥

f

r1 0

0 r2

0

0

g g f g

0 0

¥

rn

or AU = UD, where U is the matrix whose column vectors are eigenvectors and D is a diagonal matrix whose diagonal entries are the eigenvalues. Since U’s columns are independent, U is invertible and we can write (24)

A = UDU -1 or D = U-1AU ,

and we say that A is diagonalizable. [In this context equation (24) expresses a similarity transformation.] Because the argument that leads from (2) to (23) to (24) can be reversed, we have a new characterization: An n * n matrix has n linearly independent eigenvectors if, and only if, it is diagonalizable.

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Section 9.5

9.5

Homogeneous Linear Systems with Constant Coefficients

EXERCISES

In Problems 1–8, find the eigenvalues and eigenvectors of the given matrix. -4 2 6 -3 1. c d 2. c d 2 -1 2 1 3. c

1 2

531

-1 d 4

4. c

1 1

5 d -3

1 0 0 5. £ 0 0 2 § 0 2 0

0 1 1 6. £ 1 0 1 § 1 1 0

1 0 0 7. £ 2 3 1 § 0 2 4

-3 8. £ 0 4

1 -3 -8

In Problems 19–24, find a fundamental matrix for the system x′1t2 = Ax1t2 for the given matrix A. -1 1 5 4 19. A = c d 20. A = c d 8 1 -1 0

0 1§ 2

In Problems 9 and 10, some of the eigenvalues of the given matrix are complex. Find all the eigenvalues and eigenvectors. 0 -1 1 2 -1 9. c d 1 0 10. £ 0 1 1§ 0 -1 1 In Problems 11–16, find a general solution of the system x′1t2 = Ax1t2 for the given matrix A. 11. A = £

-1

3 4

-5

3

1 3 12. A = c d 12 1

§

1 2 2 13. A = £ 2 0 3 § 2 3 0

-1 14. A = £ 1 0

1 2 3 15. A = £ 0 1 0 § 2 1 2

-7 0 6 16. A = £ 0 5 0 § 6 0 2

1 2 3

0 1§ -1

17. Consider the system x′1t2 = Ax1t2, t Ú 0, with A = C

1

23

23

-1

S.

(a) Show that the matrix A has eigenvalues r1 = 2 and r2 = - 2 with corresponding eigenvectors u1 = col1 13, 12 and u2 = col11, - 132. (b) Sketch the trajectory of the solution having initial vector x102 = -u1. (c) Sketch the trajectory of the solution having initial vector x102 = u2. (d) Sketch the trajectory of the solution having initial vector x102 = u2 - u1. 18. Consider the system x′1t2 = Ax1t2, t Ú 0, with A = c

-2 1

1 d . -2

(a) Show that the matrix A has eigenvalues r1 = - 1 and r2 = -3 with corresponding eigenvectors u1 = col11, 12 and u2 = col11, - 12. (b) Sketch the trajectory of the solution having initial vector x102 = u1. (c) Sketch the trajectory of the solution having initial vector x102 = - u2. (d) Sketch the trajectory of the solution having initial vector x102 = u1 - u2.

0 1 21. A = £ 0 0 8 - 14

0 1§ 7

2 0 23. A = D 0 0

1 -1 0 0

1 0 3 0

-1 1 T 1 7

4 0 24. A = D 0 0

-1 0 0 0

0 0 2 1

0 0 T -3 -2

3 22. A = £ 1 3

1 -1 3 -1 § 3 -1

25. Using matrix algebra techniques, find a general solution of the system x′ = x + 2y - z , y′ = x + z , z′ = 4x - 4y + 5z . 26. Using matrix algebra techniques, find a general solution of the system x′ = 3x - 4y , y′ = 4x - 7y . In Problems 27–30, use a linear algebra software package such as MATLAB®, MAPLESOFT®, or MATHEMATICA® to compute the required eigenvalues and eigenvectors and then give a fundamental matrix for the system x′1t2 = Ax1t2 for the given matrix A. 0 1.1 0 2 1 1 27. A = £ 0 0 1.3 § 28. A = £ -1 1 0 § 0.9 1.1 - 6.9 3 3 3 0 0 29. A = D 0 2

1 0 0 -6

0 1 0 3

0 0 T 1 3

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532

Chapter 9

0 1 30. A = D 0 0

1 -1 0 0

Matrix Methods for Linear Systems

0 0 0 -2

(c) To obtain a second linearly independent solution to x′ = Ax, try x2 1t2 = te2tu1 + e2tu2. 3Hint: Show that u1 and u2 must satisfy

0 0 T 1 4

In Problems 31–34, solve the given initial value problem. 1 3 3 31. x′1t2 = c d x1t2 , x102 = c d 3 1 1 32. x′1t2 = c

6 2

-3 d x1t2 , 1

1 33. x′1t2 = £ -2 2

-2 1 -2

x102 = c

2 - 2 § x1t2 , 1

0 1 1 34. x′1t2 = £ 1 0 1 § x1t2 , 1 1 0

- 10 d -6

-2 x102 = £ - 3 § 2

-1 x102 = £ 4 § 0

35. (a) Show that the matrix 1 -1 d 4 -3 has the repeated eigenvalue r = -1 and that all the eigenvectors are of the form u = s col11, 22. (b) Use the result of part (a) to obtain a nontrivial solution x1 1t2 to the system x′ = Ax. (c) To obtain a second linearly independent solution to x′ = Ax, try x2 1t2 = te-tu1 + e-tu2. 3Hint: Substitute x2 into the system x′ = Ax and derive the relations A = c

1A + I2u1 = 0 ,

1A + I2u2 = u1 .

Since u1 must be an eigenvector, set u1 = col11, 22 and solve for u2.4

(d) What is 1A + I2 2 u2? (In Section 9.8, u2 will be identified as a generalized eigenvector.) 36. Use the method discussed in Problem 35 to find a general solution to the system x′1t2 = c

5 3

1A - 2I2u2 = u1.4

1A - 2I2u1 = 0 ,

-3 d x1t2 . -1

37. (a) Show that the matrix 2 1 6 A = £0 2 5§ 0 0 2 has the repeated eigenvalue r = 2 with multiplicity 3 and that all the eigenvectors of A are of the form u = s col11, 0, 02. (b) Use the result of part (a) to obtain a solution to the system x′ = Ax of the form x1 1t2 = e2tu1.

(d) To obtain a third linearly independent solution to x′ = Ax, try t 2 2t e u1 + te2tu2 + e2tu3 . 2 3Hint: Show that u1, u2, and u3 must satisfy x3 1t2 =

1A - 2I2u2 = u1 ,

1A - 2I2u1 = 0 ,

1A - 2I2u3 = u2.4

(e) Show that 1A - 2I2 2 u2 = 1A - 2I2 3 u3 = 0. 38. Use the method discussed in Problem 37 to find a general solution to the system 3 -2 x′1t2 = £ 2 -1 -4 4 39. (a) Show that the matrix 2 A = £ 1 -2

1 2 -2

1 1 § x1t2. 1

1 1§ -1

has the repeated eigenvalue r = 1 of multiplicity 3 and that all the eigenvectors of A are of the form u = s col1 -1, 1, 02 + y col1 -1, 0, 12. (b) Use the result of part (a) to obtain two linearly independent solutions to the system x′ = Ax of the form x1 1t2 = etu1 and x2 1t2 = etu2. (c) To obtain a third linearly independent solution to x′ = Ax, try x3 1t2 = tetu3 + etu4. 3Hint: Show that u3 and u4 must satisfy 1A - I2u3 = 0 ,

1A - I2u4 = u3 .

Choose u3, an eigenvector of A, so that you can solve for u4.4 (d) What is 1A - I2 2 u4? 40. Use the method discussed in Problem 39 to find a general solution to the system 1 x′1t2 = £ 0 0

3 7 9

-2 -4 § x1t2 . -5

41. Use the substitution x1 = y, x2 = y′ to convert the linear equation ay″ + by′ + cy = 0, where a, b, and c are constants, into a normal system. Show that the characteristic equation for this system is the same as the auxiliary equation for the original equation.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 9.5

Homogeneous Linear Systems with Constant Coefficients

42. (a) Show that the Cauchy–Euler equation at 2y″ + bty′ + cy = 0 can be written as a Cauchy–Euler system (25)

t x′ = Ax

with a constant coefficient matrix A, by setting x1 = y>t and x2 = y′. (b) Show that for t 7 0 any system of the form (25) with A an n * n constant matrix has nontrivial solutions of the form x1t2 = t ru if and only if r is an eigenvalue of A and u is a corresponding eigenvector. In Problems 43 and 44, use the result of Problem 42 to find a general solution of the given system. 1 3 43. tx′1t2 = c d x1t2 , t 7 0 -1 5 -4 2 44. tx′1t2 = c d x1t2 , t 7 0 2 -1 45. Mixing Between Interconnected Tanks. Two tanks, each holding 50 L of liquid, are interconnected by pipes with liquid flowing from tank A into tank B at a rate of 4 L/min and from tank B into tank A at 1 L/min (see Figure 9.2). The liquid inside each tank is kept well stirred. Pure water flows into tank A at a rate of 3 L/min, and the solution flows out of tank B at 3 L/min. If, initially, tank A contains 2.5 kg of salt and tank B contains no salt (only water), determine the mass of salt in each tank at time t Ú 0. Graph on the same axes the two quantities x1 1t2 and x2 1t2, where x1 1t2 is the mass of salt in tank A and x2 1t2 is the mass in tank B. 3 L/min Pure water

A

4 L/min

B

x1(t)

x 2 (t)

50 L

50 L

x1(0) = 2.5 kg

x 2 (0) = 0 kg

3 L/min

1 L/min Figure 9.2 Mixing problem for interconnected tanks

46. Mixing with a Common Drain. Two tanks, each holding 1 L of liquid, are connected by a pipe through which liquid flows from tank A into tank B at a rate of 3 - a L/min 10 6 a 6 32. The liquid inside each tank is kept well stirred. Pure water flows into tank A at a rate of 3 L/min. Solution flows out of tank A at a L/min and out of tank B at 3 - a L/min. If, initially, tank B contains no salt (only water) and tank A contains 0.1 kg of salt, determine the mass of salt in each tank at time t Ú 0. How does the mass of salt in tank A depend on the choice of a? What is the maximum mass of salt in tank B? (See Figure 9.3.)

3 L/min Pure water

A

533

B L/min

x1(t)

x 2 (t)

1L

1L

x1(0) = 0.1 kg

x 2 (0) = 0 kg

L/min

) L/min

3 L/min Figure 9.3 Mixing problem for a common drain, 0 6 a 6 3

47. To find a general solution to the system 1 3 -1 x′ = Ax = £ 3 0 1§ x , -1 1 2 proceed as follows: (a) Use a numerical root-finding procedure to approximate the eigenvalues. (b) If r is an eigenvalue, then let u = col 1u1, u2, u3 2 be an eigenvector associated with r. To solve for u, assume u1 = 1. (If not u1, then either u2 or u3 may be chosen to be 1. Why?) Now solve the system 1 0 1A - rI2 £ u2 § = £ 0 § u3 0 for u2 and u3. Use this procedure to find approximations for three linearly independent eigenvectors for A. (c) Use these approximations to give a general solution to the system. 48. To complete the proof of Theorem 6, page 527, assume the induction hypothesis that u1, . . . , uk, 2 … k, are linearly independent. (a) Show that if c1u1 + g + ckuk + ck + 1uk + 1 = 0 , then

c1 1r1 - rk + 1 2u1 + g + ck 1rk - rk + 1 2uk = 0 . (b) Use the result of part (a) and the induction hypothesis to conclude that u1, c, uk + 1 are linearly independent. The theorem follows by induction. 49. Stability. A homogeneous system x′ = Ax with constant coefficients is stable if it has a fundamental matrix whose entries all remain bounded as t S + ∞ . (It will follow from Theorem 9 in Section 9.8 that if one fundamental matrix of the system has this property, then all fundamental matrices for the system do.) Otherwise, the system is unstable. A stable system is asymptotically stable if all solutions approach the zero solution as t S + ∞ . Stability is discussed in more detail in Chapter 12.†

†

All references to Chapters 11–13 refer to the expanded text, Fundamentals of Differential Equations and Boundary Value Problems, 7th ed.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

534

Chapter 9

Matrix Methods for Linear Systems

(a) Show that if A has all distinct real eigenvalues, then x′1t2 = Ax1t2 is stable if and only if all eigenvalues are nonpositive. (b) Show that if A has all distinct real eigenvalues, then x′1t2 = Ax1t2 is asymptotically stable if and only if all eigenvalues are negative. (c) Argue that in parts (a) and (b), we can replace “has distinct real eigenvalues” by “is symmetric” and the statements are still true. 50. In an ice tray, the water level in any particular ice cube cell will change at a rate proportional to the difference between that cell’s water level and the level in the adjacent cells.

(a) Argue that a reasonable differentiable equation model for the water levels x, y, and z in the simplified three-cell tray depicted in Figure 9.4 is given by x′ = y - x , y′ = x + z - 2y ,

z′ = y - z .

(b) Use eigenvectors to solve this system for the initial conditions x102 = 3, y102 = z102 = 0.

x

y

z

Figure 9.4 Ice tray

9.6 Complex Eigenvalues In the previous section, we showed that the homogeneous system (1)

x′1t2 = Ax1t2 ,

where A is a constant n * n matrix, has a solution of the form x1t2 = ertu if and only if r is an eigenvalue of A and u is a corresponding eigenvector. In this section we show how to obtain two real vector solutions to system (1) when A is real and has a pair† of complex conjugate eigenvalues a + ib and a - ib. Suppose r1 = a + ib1a and b real numbers) is an eigenvalue of A with corresponding eigenvector z = a + ib, where a and b are real constant vectors. We first observe that the complex conjugate of z, namely z J a - ib, is an eigenvector associated with the eigenvalue r2 = a - ib. To see this, note that taking the complex conjugate of 1A - r1I2z = 0 yields 1A - r1I2z = 0 because the conjugate of the product is the product of the conjugates and A and I have real entries 1A = A, I = I2. Since r2 = r1, we see that z is an eigenvector associated with r2. Therefore, two linearly independent complex vector solutions to (1) are (2) (3)

w1 1t2 = er1t z = e1a + ib2t 1a + ib2 , w2 1t2 = er2t z = e1a-ib2t 1a - ib2 .

As in Section 4.3, where we handled complex roots to the auxiliary equation, let’s use one of these complex solutions and Euler’s formula to obtain two real vector solutions. With the aid of Euler’s formula, we rewrite w1 1t2 as w1 1t2 = eat 1cos bt + i sin bt21a + ib2 = eat 5 1cos bt a - sin bt b2 + i1sin bt a + cos bt b26 .

We have thereby expressed w1 1t2 in the form w1 1t2 = x1 1t2 + ix2 1t2, where x1 1t2 and x2 1t2 are the two real vector functions (4) (5)

x1 1t2 J eat cos bt a - eat sin bt b , x2 1t2 J eat sin bt a + eat cos bt b .

†

Recall that the complex roots of a polynomial equation with real coefficients must occur in complex conjugate pairs.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 9.6

Complex Eigenvalues

535

Since w1 1t2 is a solution to 112, then w1= 1t2 = Aw1 1t2 ,

x1= + ix2= = Ax1 + iAx2 . Equating the real and imaginary parts yields x1= 1t2 = Ax1 1t2

x2= 1t2 = Ax2 1t2 .

and

Hence, x1 1t2 and x2 1t2 are real vector solutions to (1) associated with the complex conjugate eigenvalues a { ib. Because a and b are not both the zero vector, it can be shown that x1 1t2 and x2 1t2 are linearly independent vector functions on 1 - ∞, ∞ 2 (see Problem 15). Let’s summarize our findings.

Complex Eigenvalues If the real matrix A has complex conjugate eigenvalues a { ib with corresponding eigenvectors a { ib, then two linearly independent real vector solutions to x′1t2 = Ax1t2 are

Example 1

(6)

eat cos bt a - eat sin bt b ,

(7)

eat sin bt a + eat cos bt b .

Find a general solution of (8)

Solution

x′1t2 = c

-1 -1

2 d x1t2 . -3

The characteristic equation for A is

0 A - rI 0 = `

-1 - r -1

2 ` = r 2 + 4r + 5 = 0 . -3 - r

Hence, A has eigenvalues r = -2 { i. To find a general solution, we need only find an eigenvector associated with the eigenvalue r = -2 + i. Substituting r = -2 + i into 1A - rI2z = 0 gives c

1-i -1

2 z 0 d c 1d = c d . -1 - i z2 0

The solutions can be expressed as z1 = 2s and z2 = 1 -1 + i2s, with s arbitrary. Hence, the eigenvectors associated with r = -2 + i are z = s col12, -1 + i2. Taking s = 1 gives the eigenvector z = c

0 2 2 d = c d +ic d . 1 -1 -1 + i

We have found that a = -2, b = 1, and z = a + ib with a = col12, -12, and b = col10, 12, so a general solution to (8) is x1t2 = c1 e e-2t cos t c

2 0 d - e-2t sin t c d f -1 1

= + c2 e e-2t sin t c (9)

x1t2 = c1 c

2 0 d + e-2t cos t c d f -1 1

2e-2t cos t 2e-2t sin t d + c2 c -2t d. ◆ -2t e 1cos t - sin t2 -e 1cos t + sin t2

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536

Chapter 9

Matrix Methods for Linear Systems

k1

k2

k3

m1

m2

x1 > 0

x2 > 0

x1 = 0

x2 = 0

Figure 9.5 Coupled mass–spring system with fixed ends

Complex eigenvalues occur in modeling coupled mass–spring systems. For example, the motion of the mass–spring system illustrated in Figure 9.5 is governed by the second-order system (10)

m1x1> = -k1x1 + k2 1x2 - x1 2 ,

m2x2> = -k2 1x2 - x1 2 - k3x2 ,

where x1 and x2 represent the displacements of the masses m1 and m2 to the right of their equilibrium positions and k1, k2, k3 are the spring constants of the three springs (see the discussion in Section 5.6). If we introduce the new variables y1 J x1, y2 J x1= , y3 J x2, y4 J x2= , then we can rewrite the system in the normal form

(11)

0 - 1k1 + k2 2 >m1 y′1t2 = Ay1t2 = D 0 k2 >m2

1 0 0 0

0 k2 >m1 0 - 1k2 + k3 2 >m2

0 0 T y1t2 . 1 0

For such a system, it turns out that A has only imaginary eigenvalues and they occur in complex conjugate pairs: {ib1, {ib2. Hence, any solution will consist of sums of sine and cosine functions. The frequencies of these functions f1 J

b1 2p

and

f2 J

b2 2p

are called the normal or natural frequencies of the system (b1 and b2 are the angular frequencies of the system). In some engineering applications, the only information that is required about a particular device is a knowledge of its normal frequencies; one must ensure that they are far from the frequencies that occur naturally in the device’s operating environment (so that no resonances will be excited). Example 2 Solution

Determine the normal frequencies for the coupled mass–spring system governed by system (11) when m1 = m2 = 1 kg, k1 = 1 N/m, k2 = 2 N/m, and k3 = 3 N/m. To find the eigenvalues of A, we must solve the characteristic equation -r -3 0 A - rI 0 = ∞ 0 2

1 -r 0 0

0 2 -r -5

0 0 ∞ = r 4 + 8r 2 + 11 = 0 . 1 -r

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 9.6

537

Complex Eigenvalues

From the quadratic formula we find r 2 = -4 { 15, so the four eigenvalues of A are {i24 - 15 and {i24 + 15. Hence, the two normal frequencies for this system are 24 - 15 ≈ 0.211 2p

9.6

and

24 + 15 ≈ 0.397 cycles per second. ◆ 2p

EXERCISES

In Problems 1– 4, find a general solution of the system x′1t2 = Ax1t2 for the given matrix A. 2 -4 -2 -5 1. A = c d 2. A = c d 2 -2 1 2 1 3. A = £ 0 0 5 4. A = £ - 1 3

-1 1§ 1

2 1 -1 -5 4 -5

-5 2§ -3

In Problems 5–9, find a fundamental matrix for the system x′1t2 = Ax1t2 for the given matrix A. -1 - 2 -2 -2 5. A = c d 6. A = c d 8 -1 4 2 0 7. A = £ 0 0

0 0 1

1 -1 § 0

0 1 8. A = D 0 0

1 0 0 0

0 0 0 - 13

0 0 T 1 4

0 0 0 -1

0 0 T 1 0

0 -1 9. A = D 0 0

1 0 0 0

In Problems 10–12, use a linear algebra software package to compute the required eigenvalues and eigenvectors for the given matrix A and then give a fundamental matrix for the system x′1t2 = Ax1t2. 0 0 10. A = D 0 13

1 0 0 -4

0 1 0 - 12

0 0 T 1 4

0 0 11. A = D 0 -2

1 0 0 2

0 1 0 -3

0 0 T 1 2

1 0 12. A = E 0 0 0

0 0 1 0 0

0 1 0 0 0

0 0 0 0 -29

0 0 0U 1 -4

In Problems 13 and 14, find the solution to the given system that satisfies the given initial condition. -3 -1 13. x′1t2 = c d x1t2 , 2 -1 -1 1 (a) x102 = c d (b) x1p2 = c d 0 -1 2 (c) x1 - 2p2 = c d 1

0 (d) x1p>22 = c d 1

0 -1 2 0 § x1t2 , 0 1 -2 0 (a) x102 = £ 2 § (b) x1 -p2 = £ 1 § -1 1

1 14. x′1t2 = £ 0 1

15. Show that x1 1t2 and x2 1t2 given by equations (4) and (5) are linearly independent on 1 - ∞ , ∞ 2, provided b ≠ 0 and a and b are not both the zero vector. 16. Show that x1 1t2 and x2 1t2 given by equations (4) and (5) can be obtained as linear combinations of w1 1t2 and w2 1t2 given by equations (2) and (3). [Hint: Show that x1 1t2 =

w1 1t2 + w2 1t2 2

,

x2 1t2 =

w1 1t2 - w2 1t2 2i

.d

In Problems 17 and 18, use the results of Problem 42 in Exercises 9.5 to find a general solution to the given Cauchy– Euler system for t 7 0. -1 -1 0 17. tx′1t2 = £ 2 -1 1 § x1t2 0 1 -1 18. tx′1t2 = c

-1 9

-1 d x1t2 -1

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

538

Chapter 9

Matrix Methods for Linear Systems

19. For the coupled mass–spring system governed by system (10), assume m1 = m2 = 1 kg, k1 = k2 = 2 N/m, and k3 = 3 N/m. Determine the normal frequencies for this coupled mass–spring system. 20. For the coupled mass–spring system governed by system (10), assume m1 = m2 = 1 kg, k1 = k2 = k3 = 1 N/m, and assume initially that x1 102 = 0 m, x1= 102 = 0 m/sec, x2 102 = 2 m, and x2= 102 = 0 m/sec. Using matrix algebra techniques, solve this initial value problem. 21. RLC Network. The currents in the RLC network given by the schematic diagram in Figure 9.6 are governed by the following equations: 4I 2= 1t2 + 52q1 1t2 = 10 ,

13I3 1t2 + 52q1 1t2 = 10 , I1 1t2 = I2 1t2 + I3 1t2 , 1 —– 52

I1 10 volts

farads

I1 I2

I3 13 ohms

4 henries

I1 Figure 9.6 RLC network for Problem 21

I1

50 henries I2 80 ohms

160 volts

I3 1 800

farads

I1 Figure 9.7 RLC network for Problem 22

where q3 1t2 is the charge on the capacitor, I3 1t2 = q3′ 1t2, and initially q3 102 = 0.5 coulombs and I3 102 = 0 amps. Solve for the currents I1, I2, and I3. 3Hint: Differentiate the first two equations, use the third equation to eliminate I3, and form a normal system with x1 = I1, x2 = I 1′ , and x3 = I2.4 23. Stability. In Problem 49 of Exercises 9.5, (page 542), we discussed the notion of stability and asymptotic stability for a linear system of the form x′1t2 = Ax1t2. Assume that A has all distinct eigenvalues (real or complex). (a) Show that the system is stable if and only if all the eigenvalues of A have nonpositive real part. (b) Show that the system is asymptotically stable if and only if all the eigenvalues of A have negative real part. 24. (a) For Example 1, page 535, verify that

where q1 1t2 is the charge on the capacitor, I1 1t2 = q1= 1t2, and initially q1 102 = 0 coulombs and I1 102 = 0 amps. Solve for the currents I1, I2, and I3. 3Hint: Differentiate the first two equations, eliminate I1, and form a normal system with x1 = I2, x2 = I 2= , and x3 = I3.4 22. RLC Network. The currents in the RLC network given by the schematic diagram in Figure 9.7 are governed by the following equations:

-e-2t cos t + e-2t sin t d e-2t cos t -e-2t sin t - e-2t cos t d + c2 c e-2t sin t is another general solution to equation (8). (b) How can the general solution of part (a) be directly obtained from the general solution derived in (9) on page 535? x1t2 = c1 c

50I 1= 1t2 + 80I2 1t2 = 160 ,

50I 1= 1t2 + 800q3 1t2 = 160 , I1 1t2 = I2 1t2 + I3 1t2 ,

9.7 Nonhomogeneous Linear Systems The techniques discussed in Chapters 4 and 6 for finding a particular solution to the nonhomogeneous equation y″ + p1x2y′ + q1x2y = g1x2 have natural extensions to nonhomogeneous linear systems.

Undetermined Coefficients The method of undetermined coefficients can be used to find a particular solution to the nonhomogeneous linear system x′1t2 = Ax1t2 + f1t2

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 9.7

Nonhomogeneous Linear Systems

539

when A is an n * n constant matrix and the entries of f1t2 are polynomials, exponential functions, sines and cosines, or finite sums and products of these functions. We can use the procedure box in Section 4.5 (page 184) and reproduced at the back of the book as a guide in choosing the form of a particular solution xp 1t2. Some exceptions are discussed in the exercises (see Problems 25–28). Example 1

Find a general solution of (1)

Solution

x′1t2 = Ax1t2 + tg ,

where

1 A = £ -2 2

-2 1 2

2 2§ 1

and

g = £

-9 0§ . -18

In Example 5 in Section 9.5, page 529, we found that a general solution to the corresponding homogeneous system x′ = Ax is (2)

1 -1 -1 3t -3t xh 1t2 = c1e £ 0 § + c2e £ 1 § + c3e £ -1 § . 1 0 1 3t

Since the entries in f1t2 J tg are just linear functions of t, we are inclined to seek a particular solution of the form a1 b1 xp 1t2 = ta + b = t £ a2 § + £ b2 § , a3 b3

where the constant vectors a and b are to be determined. Substituting this expression for xp 1t2 into system (1) yields a = A1ta + b2 + tg , which can be written as

t1Aa + g2 + 1Ab - a2 = 0 .

Setting the “coefficients” of this vector polynomial equal to zero yields the two systems (3)

Aa = -g ,

(4)

Ab = a .

By Gaussian elimination or by using a linear algebra software package, we can solve (3) for a and we find a = col15, 2, 42. Next we substitute for a in (4) and solve for b to obtain b = col11, 0, 22. Hence a particular solution for (1) is (5)

5 1 5t + 1 xp 1t2 = ta + b = t £ 2 § + £ 0 § = £ 2t § . 4 2 4t + 2

A general solution for (1) is x1t2 = xh 1t2 + xp 1t2, where xh 1t2 is given in (2) and xp 1t2 in (5). ◆ In the preceding example, the nonhomogeneous term f1t2 was a vector polynomial. If, instead, f1t2 has the form f1t2 = col11, t, sin t2 , then, using the superposition principle, we would seek a particular solution of the form xp 1t2 = ta + b + 1sin t2c + 1cos t2d .

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540

Chapter 9

Matrix Methods for Linear Systems

Similarly, if

f1t2 = col1t, et, t 2 2 ,

we would take

xp 1t2 = t 2a + tb + c + etd .

Of course, we must modify our guess, should one of the terms be a solution to the corresponding homogeneous system. If this is the case, the annihilator method [equations (15) and (16) of Section 6.3, page 337] would appear to suggest that for a nonhomogeneity f1t2 of the form ertt mg, where r is an eigenvalue of A, m is a nonnegative integer, and g is a constant vector, a particular solution of x′ = Ax + f can be found in the form xp 1t2 = ert 5t m + sam + s + t m + s - 1am + s - 1 + g + ta1 + a0 6 ,

for a suitable choice of s. We omit the details.

Variation of Parameters In Section 4.6 we discussed the method of variation of parameters for a general constantcoefficient second-order linear equation. Simply put, the idea is that if a general solution to the homogeneous equation has the form xh 1t2 = c1x1 1t2 + c2x2 1t2, where x1 1t2 and x2 1t2 are linearly independent solutions to the homogeneous equation, then a particular solution to the nonhomogeneous equation would have the form xp 1t2 = y1 1t2x1 1t2 + y2 1t2x2 1t2, where y1 1t2 and y2 1t2 are certain functions of t. A similar idea can be used for systems. Let X1t2 be a fundamental matrix for the homogeneous system (6)

x′1t2 = A1t2x1t2 ,

where now the entries of A may be any continuous functions of t. Because a general solution to (6) is given by X1t2c, where c is a constant n * 1 vector, we seek a particular solution to the nonhomogeneous system (7)

x′1t2 = A1t2x1t2 + f1t2

of the form (8)

xp 1t2 = X1t2Y1t2 ,

where Y1t2 = col 1 y1 1t2, c, yn 1t2 2 is a vector function of t to be determined. To derive a formula for Y1t2, we first differentiate (8) using the matrix version of the product rule to obtain xp= 1t2 = X1t2Y′1t2 + X′1t2Y1t2 .

Substituting the expressions for xp 1t2 and xp= 1t2 into (7) yields (9)

X1t2Y′1t2 + X′1t2Y1t2 = A1t2X1t2Y1t2 + f1t2 .

Since X1t2 satisfies the matrix equation X′1t2 = A1t2X1t2, equation (9) becomes XY′ + AXY = AXY + f , XY′ = f .

Multiplying both sides of the last equation by X-1 1t2 [which exists since the columns of X1t2 are linearly independent] gives Y′1t2 = X-1 1t2 f1t2 .

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 9.7

Nonhomogeneous Linear Systems

541

Integrating, we obtain Y1t2 =

L

X-1 1t2 f1t2dt .

Hence, a particular solution to (7) is (10)

xp 1t2 = X1t2Y1t2 = X1t2

L

X−1 1t2 f1t2dt .

Combining (10) with the solution X1t2c to the homogeneous system yields the following general solution to (7): (11)

x1t2 = X1t2c + X1t2

L

X−1 1t2 f1t2dt .

The elegance of the derivation of the variation of parameters formula (10) for systems becomes evident when one compares it with the more lengthy derivations for the scalar case in Sections 4.6 and 6.4. Given an initial value problem of the form (12)

x1t0 2 = x0 ,

x′1t2 = A1t2x1t2 + f1t2 ,

we can use the initial condition x1t0 2 = x0 to solve for c in (11). Expressing x1t2 using a definite integral, we have t

x1t2 = X1t2c + X1t2

Lt0

X-1 1s2 f1s2ds .

From the initial condition x1t0 2 = x0, we find x0 = x1t0 2 = X1t0 2c + X1t0 2

t0

Lt0

X-1 1s2 f1s2ds = X1t0 2c .

Solving for c, we have c = X-1 1t0 2x0. Thus, the solution to (12) is given by the formula (13)

x1t2 = X1t2X−1 1t0 2x0 + X1t2

t

Lt0

X −1 1s2 f1s2 ds .

To apply the variation of parameters formulas, we first must determine a fundamental matrix X1t2 for the homogeneous system. In the case when the coefficient matrix A is constant, we have discussed methods for finding X1t2. However, if the entries of A depend on t, the determination of X1t2 may be extremely difficult (entailing, perhaps, a matrix power series!).

Example 2

Find the solution to the initial value problem (14)

x′1t2 = c

2 1

-3 e2t d x1t2 + c d , -2 1

x102 = c

-1 d. 0

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542

Chapter 9

Solution

Matrix Methods for Linear Systems

In Example 3 in Section 9.5, we found two linearly independent solutions to the corresponding homogeneous system; namely, 3et e-t x1 1t2 = c t d and x2 1t2 = c -t d . e e Hence a fundamental matrix for the homogeneous system is 3et e-t X1t2 = c t -t d . e e Although the solution to (14) can be found via the method of undetermined coefficients, we shall find it directly from formula (13). For this purpose, we need X-1 1t2. By formula (3) of Section 9.3 (page 510): X 1t2 = C -1

-2

1 e-t

- 12 e-t

- 12 et

- 32 et

S.

Substituting into formula (13), we obtain the solution x1t2 = c

1

3et e-t + c t -t d e e

#

= £

= £

9.7

- 32 et + 12 e-t - 12 et + 12 e-t - 32 et + 12 e-t - 12 et + 12 e-t

3 2

t

0

= £

- 12

3et e-t 2 d£ et e-t - 1 2 £

§c

1 e-s 2 1 - 2 es

§ +c

t

-1 d 0 - 12 e-s 3 es 2 -t

3e e d et e-t

t

#

£

0

§ +c

§c

e2s d ds 1

1 es - 1 e-s 2 2 § ds 1 3s - 2 e + 32 es

1 t 1 -t 3et e-t 2 e + 2 e - 1 d £ § et e-t 3 et - 1 e3t - 4 3 2 6

- 92 et - 56 e-t + 43 e2t + 3 - 32 et - 56 e-t + 13 e2t + 2

§ . ◆

EXERCISES

In Problems 1–6, use the method of undetermined coefficients to find a general solution to the system x′1t2 = Ax1t2 + f1t2, where A and f1t2 are given. 6 1 - 11 1. A = c d , f1t2 = c d 4 3 -5 2. A = c

1 1 d , 4 1

f1t2 = c

-t - 1 d - 4t - 2

1 3. A = £ -2 2 4. A = c

-2 1 2

2 2 d , 2 2

2 2§ , 1 f1t2 = c

2et f1t2 = £ 4et § -2et - 4 cos t d - sin t

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Section 9.7

0 5. A = £ - 1 0 6. A = c

-1 0 0 0§ , 0 1

e2t f1t2 = £ sin t § t

t f1t2 = e-2t c d 3

1 1 d , 0 2

In Problems 7–10, use the method of undetermined coefficients to determine only the form of a particular solution for the system x′1t2 = Ax1t2 + f1t2, where A and f1t2 are given. 0 1 sin 3t 7. A = c d , f1t2 = c d 2 0 t 8. A = c

-1 0 d , 2 2

0 9. A = £ -1 0 10. A = c

2 1

f1t2 = c

-1 0 0 1§ , 0 1 -1 d , 5

t2 d t+1

e2t f1t2 = £ sin t § t

f1t2 = c

te-t d 3e-t

In Problems 11–16, use the variation of parameters formula (11) to find a general solution of the system x′1t2 = Ax1t2 + f1t2, where A and f1t2 are given. 1 f1t2 = c d 0

11. A = c

0 1 d , -1 0

12. A = c

1 2 d , 3 2

13. A = c

8 4

-4 d , -2

f1t2 = c

0 1

-1 d , 0

t2 f1t2 = c d 1

14. A = c 15. A = c

-4 2

16. A = c

0 1 d , -1 0

f1t2 = c

1 d -1

t -2 >2 d t -2

f1t2 = c

2 d , -1

f1t2 = c

-1

t d 4 + 2t -1

8 sin t d 0

In Problems 17–20, use the variation of parameters formulas (11) and possibly a linear algebra software package to find a general solution of the system x′1t2 = Ax1t2 + f1t2, where A and f1t2 are given. 3et f1t2 = £ - et § - et

0 1 1 17. A = £ 1 0 1 § , 1 1 0 1 18. A = £ 0 0 0 -1 19. A = D 0 0

-1 0 -1 1 0 0 0

1 1§ , 2 0 0 0 1

0 0 T, 1 0

0 f1t2 = £ et § et t 0 f1t2 = D -t T e t

543

Nonhomogeneous Linear Systems

0 0 20. A = D 0 8

1 0 0 -4

0 1 0 -2

0 0 T, 1 -1

et 0 f1t2 = D T 1 0

In Problems 21 and 22, find the solution to the given system that satisfies the given initial condition. 0 2 et 21. x′1t2 = c d x1t2 + c t d , -1 3 -e 5 0 (a) x102 = c d (b) x112 = c d 4 1 1 (c) x152 = c d 0 22. x′1t2 = c

(d) x1 -12 = c

-4 d 5

2 4t d x1t2 + c d , -2 - 4t - 2 4 1 (a) x102 = c d (b) x122 = c d -5 1 0 4

23. Using matrix algebra techniques and the method of undetermined coefficients, find a general solution for x″1t2 + y′1t2 - x1t2 + y1t2 = - 1 , x′1t2 + y′1t2 - x1t2 = t 2 . Compare your solution with the solution in Example 4 in Section 5.2, page 247. 24. Using matrix algebra techniques and the method of undetermined coefficients, solve the initial value problem x′1t2 - 2y1t2 = 4t ,

x102 = 4 ;

y′1t2 + 2y1t2 - 4x1t2 = -4t - 2 ,

y102 = -5 .

Compare your solution with the solution in Example 1 in Section 7.10, page 412. 25. To find a general solution to the system et 0 1 d x1t2 + f1t2 , where f1t2 = c d , 0 −2 3 proceed as follows: (a) Find a fundamental solution set for the corresponding homogeneous system. (b) The obvious choice for a particular solution would be a vector function of the form xp 1t2 = eta; however, the homogeneous system has a solution of this form. The next choice would be xp 1t2 = teta. Show that this choice does not work. (c) For systems, multiplying by t is not always sufficient. The proper guess is

x′1t2 = c

xp 1t2 = teta + etb .

Use this guess to find a particular solution of the given system. (d) Use the results of parts (a) and (c) to find a general solution of the given system.

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544

Chapter 9

Matrix Methods for Linear Systems

26. For the system of Problem 25, we found that a proper guess for a particular solution is xp 1t2 = teta + etb. In some cases a or b may be zero. (a) Find a particular solution for the system of Problem 25 if f1t2 = col13et, 6et 2. (b) Find a particular solution for the system of Problem 25 if f1t2 = col1et, et 2. 27. Find a general solution of the system 0 1 1 -1 x′1t2 = £ 1 0 1 § x1t2 + £ - 1 - e-t § . 1 1 0 - 2e-t

33. RL Network. The currents in the RL network given by the schematic diagram in Figure 9.8 are governed by the following equations: 2I 1= 1t2 + 90I2 1t2 = 9 ,

I 3= 1t2 + 30I4 1t2 - 90I2 1t2 = 0 ,

60I5 1t2 - 30I4 1t2 = 0 , I1 1t2 = I2 1t2 + I3 1t2 ,

I3 1t2 = I4 1t2 + I5 1t2 .

3Hint: Try xp 1t2 = e-ta + te-tb + c.4 28. Find a particular solution for the system 1 x′1t2 = c -1

-1 -3 d x1t2 + c d . 1 1

3Hint: Try xp 1t2 = ta + b.4

In Problems 29 and 30, find a general solution to the given Cauchy–Euler system for t 7 0. (See Problem 42 in Exercises 9.5, page 533.) Remember to express the system in the form x′1t2 = A1t2x1t2 + f1t2 before using the variation of parameters formula. 2 -1 t -1 29. tx′1t2 = c d x1t2 + c d 3 -2 1 4 -3 t 30. tx′1t2 = c d x1t2 + c d 8 -6 2t 31. Use the variation of parameters formula (10) to derive a formula for a particular solution yp to the scalar equation y″ + p1t2y′ + q1t2y = g1t2 in terms of two linearly independent solutions y1 1t2, y2 1t2 of the corresponding homogeneous equation. Show that your answer agrees with the formulas derived in Section 4.6. [Hint: First write the scalar equation in system form.] 32. Conventional Combat Model. A simplistic model of a pair of conventional forces in combat yields the following system: x′ = c

-a -c

-b p d x+ c d , -d q

where x = col1x1, x2 2. The variables x1 1t2 and x2 1t2 represent the strengths of opposing forces at time t. The terms - ax1 and -dx2 represent the operational loss rates, and the terms -bx2 and -cx1 represent the combat loss rates for the troops x1 and x2, respectively. The constants p and q represent the respective rates of reinforcement. Let a = 1, b = 4, c = 3, d = 2, and p = q = 5. By solving the appropriate initial value problem, determine which forces will win if (a) x1 102 = 20 , x2 102 = 20 . (b) x1 102 = 21 , x2 102 = 20 . (c) x1 102 = 20 , x2 102 = 21 .

2 henries

I1

I3

1 henry

I2 9 volts

90 ohms

I3

I4

I5

30 ohms

60 ohms

I2 I1

I3

Figure 9.8 RL network for Problem 33

Assume the currents are initially zero. Solve for the five currents I1, . . . , I5. 3Hint: Eliminate all unknowns except I2 and I5, and form a normal system with x1 = I2 and x2 = I5.4 34. Mixing Problem. Two tanks A and B, each holding 50 L of liquid, are interconnected by pipes. The liquid flows from tank A into tank B at a rate of 4 L/min and from B into A at a rate of 1 L/min (see Figure 9.9). The liquid inside each tank is kept well stirred. A brine solution that has a concentration of 0.2 kg/L of salt flows into tank A at a rate of 4 L/min. A brine solution that has a concentration of 0.1 kg/L of salt flows into tank B at a rate of 1 L/min. The solutions flow out of the system from both tanks—from tank A at 1 L/min and from tank B at 4 L/min. If, initially, tank A contains pure water and tank B contains 0.5 kg of salt, determine the mass of salt in each tank at time t Ú 0. After several minutes have elapsed, which tank has the higher concentration of salt? What is its limiting concentration? 4 L/min 0.2 kg/L

1 L/min

A

4 L/min

B

x1(t)

x 2(t)

50 L

50 L

x1(0) = 0 kg

x2(0) = 0.5 kg

1 L/min 0.1 kg/L

4 L/min

1 L/min Figure 9.9 Mixing problem for interconnected tanks

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 9.8

The Matrix Exponential Function

545

9.8 The Matrix Exponential Function In this chapter we have developed various ways to extend techniques for scalar differential equations to systems. In this section we take a substantial step further by showing that with the right notation, the formulas for solving normal systems with constant coefficients are identical to the formulas for solving first-order equations with constant coefficients. For example, we know that a general solution to the equation x′1t2 = ax1t2, where a is a constant, is x1t2 = ceat. Analogously, we show that a general solution to the normal system (1)

x′1t2 = Ax1t2 ,

where A is a constant n * n matrix, is x1t2 = eAtc. Our first task is to define the matrix exponential eAt. If A is a constant n * n matrix, we define eAt by taking the series expansion for eat and replacing a by A; that is, (2)

eAt J I + At + A2

t2 tn + P + An + P. 2! n!

(Note that we also replace 1 by I.) By the right-hand side of (2), we mean the n * n matrix whose elements are power series with coefficients given by the corresponding entries in the matrices I, A, A2 >2!, . . . . If A is a diagonal matrix, then the computation of eAt is straightforward. For example, if A = c

-1 0 d , 0 2

then A2 = AA = c

1 0 d , 0 4

A3 = c

-1 0 d , 0 8

c , An = c

1 -12 n 0

0 d, 2n

and so ∞

At

e

∞

n a 1 -12

tn n=0 = D = aA n! n=0 n

tn n!

0 ∞ nt a 2 n! n=0 n

0

T = c

e-t 0 d . 0 e2t

More generally, if A is an n * n diagonal matrix with r1, r2, c, rn down its main diagonal, then eAt is the diagonal matrix with er1t, er2t, c, ernt down its main diagonal (see Problem 26). If A is not a diagonal matrix, the computation of eAt is more involved. We deal with this important problem later in this section. It can be shown that the series (2) converges for all t and has many of the same properties† as the scalar exponential eat. †

For proofs of these and other properties of the matrix exponential function, see Fundamentals of Matrix Analysis with Applications, by Edward Barry Saff and Arthur David Snider (John Wiley & Sons, Hoboken, New Jersey, 2016), Chapter 7. See also the amusing articles, “Nineteen Dubious Ways to Compute the Exponential of a Matrix,” by Cleve Moler and Charles van Loan, SIAM Review, Vol. 20, No. 4 (Oct. 1978), and “Nineteen . . . Matrix, Twenty-Five Years Later,” ibid., Vol. 45, No. 1 (Jan. 2003).

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546

Chapter 9

Matrix Methods for Linear Systems

Properties of the Matrix Exponential Function Theorem 7. Let A and B be n * n constant matrices and r, s, and t be real (or complex) numbers. Then, (a) eA0 = e0 = I . (b) (c) (d) (e)

eA1t+s2 = eAteAs . 1eAt 2 -1 = e-At . e1A+B2t = eAteBt, provided that AB = BA . erIt = ertI .

Property (c) has profound implications. First, it asserts that for any matrix A, the matrix eAt has an inverse for all t. Moreover, this inverse is obtained by simply replacing t by -t. (Note that (c) follows from (a) and (b) with s = -t.) In applying property (d) (the law of exponents), one must exercise care because of the stipulation that the matrices A and B commute (see Problem 25). Another important property of the matrix exponential arises from the fact that we can differentiate the series in (2) term by term. This gives d At d t2 tn 1e 2 = a I + At + A2 + g + An + g b dt dt 2 n! = A + A2t + A3

t2 tn - 1 + g + g + An 2 1n - 12!

= A c I + At + A2

t2 tn - 1 + gd . + g + An - 1 2 1n - 12!

Hence, d At 1e 2 = AeAt , dt and so eAt is a solution to the matrix differential equation X′ = AX. Since eAt is invertible [property (c)], it follows that the columns of eAt are linearly independent solutions to system (1). Combining these facts we have the following.

eAt Is a Fundamental Matrix Theorem 8. If A is an n * n constant matrix, then the columns of the matrix exponential eAt form a fundamental solution set for the system x′1t2 = Ax1t2. Therefore, eAt is a fundamental matrix for the system, and a general solution is x1t2 = eAtc. If a fundamental matrix X1t2 for the system x′ = Ax has somehow been determined, it is easy to compute eAt, as the next theorem describes.

Relationship Between Fundamental Matrices Theorem 9. Let X1t2 and Y1t2 be two fundamental matrices for the same system x′ = Ax. Then there exists a constant matrix C such that Y1t2 = X1t2C for all t. In particular, (3)

eAt = X1t2X102 -1 .

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Section 9.8

The Matrix Exponential Function

547

Proof. Since X1t2 is a fundamental matrix, every column of Y1t2 can be expressed as X1t2c for a suitable constant vector c, so column-by-column we have £

Y1t2

§ = £

f § £ c1 f

X1t2

f c2 f

g g g

f cn § = X1t2C . f

If we choose Y1t2 = eAt = X1t2C, then (3) follows by setting t = 0. ◆ If the n * n matrix A has n linearly independent eigenvectors ui, then Theorem 5 of Section 9.5 (page 526) provides us with X(t) and (3) gives us (4)

eAt = 3er1tu1 er2tu2 g erntun 4 3u1 u2 g un 4 -1 .

Are there any other ways that we can compute eAt? As we observed, if A is a diagonal matrix, then we simply exponentiate the diagonal elements of At to obtain eAt. Also, if A is a nilpotent matrix, that is, Ak = 0 for some positive integer k, then the series for eAt has only a finite number of terms—it “truncates”—since Ak = Ak + 1 = g = 0. In such cases, eAt reduces to eAt = I + At + g + Ak - 1

tk - 1 t k-1 + 0 + 0 + g = I + At + g + Ak - 1 . 1k - 12! 1k - 12!

Thus eAt can be calculated in finite terms if A is diagonal or nilpotent. Can we take this any further? Yes; a consequence of the Cayley–Hamilton theorem† is that when the characteristic polynomial for A has the form p1r2 = 1r1 - r2 n, then 1r1I - A2 n = 0 = 1 -12 n 1A - r1I2 n. So if A has only one (multiple) eigenvalue r1, then A - r1I is nilpotent, and we exploit that by writing A = r1I + A - r1I: eAt = er1Ite1A - r1I2t = er1t c I + 1A - r1I2t + g + 1A - r1I2 n - 1

Example 1

Find the fundamental matrix eAt for the system x′ = Ax ,

Solution

tn - 1 d . 1n - 12!

where

2 A = £ 1 -2

1 2 -2

1 1§ . -1

We begin by computing the characteristic polynomial for A: 2-r p1r2 = 0 A - rI 0 = † 1 -2

1 2-r -2

1 1 † = -r 3 + 3r 2 - 3r + 1 = - 1r - 12 3 . -1 - r

Thus, r = 1 is an eigenvalue of A with multiplicity 3. By the Cayley–Hamilton theorem, 1A - I2 3 = 0, and so (5)

eAt = ete1A-I2t = et e I + 1A - I2t + 1A - I2 2

t2 f. 2

The Cayley–Hamilton theorem states that a matrix satisfies its own characteristic equation, that is, p1A2 = 0. For a discussion of this theorem, see Fundamentals of Matrix Analysis with Applications, by Edward Barry Saff and Arthur David Snider (John Wiley & Sons, Hoboken, New Jersey, 2016), Section 6.3. †

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548

Chapter 9

Matrix Methods for Linear Systems

Computing, we find 1 A-I = £ 1 -2

1 1 -2

1 1§ -2

and

0 0 0 1A - I2 = £ 0 0 0 § . 0 0 0 2

Substituting into (5) yields eAt

(6)

1 0 0 1 t t = e £ 0 1 0 § + te £ 1 0 0 1 -2

1 1 -2

et + tet 1 tet tet 1 § = £ tet et + tet tet § . ◆ t t t -2 -2te -2te e - 2tet

We are not through with nilpotency yet. What if we have a nonzero vector u, an exponent m, and a scalar r satisfying 1A - r I2 m u = 0, so that A - r I is “nilpotent when restricted to u”? Such vectors are given a (predictable) name.

Generalized Eigenvectors Definition 5. Let A be a square matrix. A nonzero vector u satisfying (7)

1A - rI2 m u = 0

for some scalar r and some positive integer m is called a generalized eigenvector associated with r. [Note that r must be an eigenvalue of A, since the final nonzero vector in the list u, 1A - rI2u, 1A - rI2 2 u, c, 1A - rI2 m - 1 u is a “regular” eigenvector.]

A valuable feature of generalized eigenvectors u is that we can compute eAtu in finite terms without knowing eAt, because eAt u = erIte1A - rI2t u (8)

= ert c Iu + t1A - rI2u + g + = ert c u + t1A - rI2u + g +

tm - 1 tm 1A - rI2 m - 1 u + 1A - rI2 m u + g d 1m - 12! m!

tm - 1 1A - rI2 m - 1 u + 1m - 12!

0

+ gd .

Moreover, eAtu is a solution to the system x′ = Ax (recall Theorem 8). Hence if we can find n generalized eigenvectors ui for the n * n matrix A that are linearly independent, the corresponding solutions xi 1t2 = eAtui will form a fundamental solution set and can be assembled into a fundamental matrix X1t2. (Since the xi’s are solutions that reduce to the linearly independent ui’s at t = 0, they are always linearly independent.) Finally, we get the matrix exponential by applying (3) from Theorem 9: (9)

eAt = X1t2X102 -1 = 3eAtu1 eAtu2 g eAtun 4 3u1 u2 g un 4 -1 ,

computed in a finite number of steps. Of course, any (regular) eigenvector is a generalized eigenvector (corresponding to m = 1), and if A has a full set of n linearly independent eigenvectors, then (9) is simply the representation (4). But what if A is defective, that is, possesses fewer than n linearly

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 9.8

The Matrix Exponential Function

549

independent eigenvectors? Luckily, the primary decomposition theorem in advanced linear algebra* guarantees that when the characteristic polynomial of A is (10)

p1r2 = 1r1 - r2 m1 1r2 - r2 m2 g 1rk - r2 mk ,

where the ri ’s are the distinct eigenvalues of A and mi is the multiplicity of the eigenvalue ri, then for each i there exist mi linearly independent generalized eigenvectors satisfying (11)

1A - riI2 mi u = 0 .

Furthermore, the conglomeration of these n = m1 + m2 + g + mk generalized eigenvectors is linearly independent.† We’re home! The following scheme will always yield a fundamental solution set, for any square matrix A.

Solving x ′ = Ax To obtain a fundamental solution set for x′ = Ax for any constant square matrix A:

(a) Compute the characteristic polynomial p1r2 = 0 A - rI 0 . (b) Find the zeros of p1r2 and express it as p1r2 = 1r1 - r2 m1 1r2 - r2 m2 g 1rk - r2 mk, where r1, r2, c, rk are the distinct zeros (i.e., eigenvalues) and m1, m2, c, mk are their multiplicities. (c) For each eigenvalue ri find mi linearly independent generalized eigenvectors by applying the Gauss–Jordan algorithm to the system 1A - riI2 mi u = 0. (d) Form n = m1 + m2 + g + mk linearly independent solutions to x′ = Ax by computing x1t2 J eAtu = ert c u + t1A - rI2u +

t2 1A - rI2 2 u + g d 2!

for each generalized eigenvector u found in part (c) and corresponding eigenvalue r. If r has multiplicity m, this series terminates after m or fewer terms. We can then, if desired, assemble the fundamental matrix X1t2 from the n solutions and obtain the matrix exponential eAt using (9).‡

Example 2

Find the fundamental matrix eAt for the system (12)

Solution

x′ = Ax ,

where

1 0 0 A = £1 3 0§ . 0 1 1

We begin by finding the characteristic polynomial for A: 1-r

p1r2 = 0 A - rI 0 = † 1 0

0 3-r 1

0 0 † = - 1r - 12 2 1r - 32 . 1-r

*

See Fundamentals of Matrix Analysis with Applications, by Edward Barry Saff and Arthur David Snider (John Wiley & Sons, Hoboken, New Jersey, 2016), Section 7.3. † Some of the generalized eigenvectors satisfy (11) with a lower exponent, but that will not affect the calculation. ‡ With defective matrices this algorithm may prove ineffective for computer implementation; rounding effects inevitably foil the machine’s ability to detect multiple eigenvalues.

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550

Chapter 9

Matrix Methods for Linear Systems

Hence, the eigenvalues of A are r = 1 with multiplicity 2 and r = 3 with multiplicity 1. Since r = 1 has multiplicity 2, we must determine two linearly independent associated generalized eigenvectors satisfying 1A - I2 2 u = 0. From 0 0 0 u1 0 1A - I2 u = £ 2 4 0 § £ u2 § = £ 0 § , 1 2 0 u3 0 2

we find u2 = s, u1 = -2u2 = -2s, and u3 = y, where s and y are arbitrary. Taking s = 0 and y = 1, we obtain the generalized eigenvector u1 = col10, 0, 12. The corresponding solution to (12) is

(13)

0 0 0 0 0 0 t x1 1t2 = e 5u1 + t1A - l2u1 6 = e £ 0 § + te £ 1 2 0 § £ 0 § = £ 0 § et 1 0 1 0 1 t

t

(u1 is, in fact, a regular eigenvector). Next we take s = 1 and y = 0 to derive the second linearly independent generalized eigenvector u2 = col1 -2, 1, 02 and (linearly independent) solution (14)

x2 1t2 = eAtu2 = et 5u2 + t1A - I2u2 6 -2 0 0 0 -2 t = e £ 1 § + te £ 1 2 0 § £ 1 § 0 0 0 1 0 t

-2et -2 0 t = e £ 1 § + te £ 0 § = £ et § .† 0 1 tet t

For the eigenvalue r = 3, we solve 1A - 3I2u = 0, that is, -2 £ 1 0

0 0 1

0 u1 0 0 § £ u2 § = £ 0 § , -2 u3 0

to obtain the eigenvector u3 = col10, 2, 12. Hence, a third linearly independent solution to (12) is

(15)

0 0 x3 1t2 = e3tu3 = e3t £ 2 § = £ 2e3t § . 1 e3t

The matrix X1t2 whose columns are the vectors x1 1t2, x2 1t2, and x3 1t2,  0 X1t2 = £ 0 et

-2et 0 et 2e3t § , tet e3t

Note that x2 1t2 has been expressed in the format discussed in Problem 35 of Exercises 9.5, page 532.

†

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Section 9.8

The Matrix Exponential Function

551

is a fundamental matrix for (12). Setting t = 0 and then computing X-1 102, we find 0 X102 = £ 0 1

-2 1 0

0 2§ 1

and

- 14

- 12

X-1 102 = E - 12

0

0U .

1 4

1 2

0

1

It now follows from formula (3) that

eAt

0 = X1t2X-1 102 = £ 0 et

= E

-2e et tet

t

- 14

0 2e3t § E - 12 e3t 1 4

- 12

1

0

0U

1 2

0

et

0

0

- 12 et + 12 e3t

e3t

0U . ◆

- 14 et - 12 tet + 14 e3t

- 12 et + 12 e3t

et

Use of the fundamental matrix eAt simplifies many computations. For example, the properties eAte-As = eA1t-s2 and 1eAt0 2 -1 = e-At0 enable us to rewrite the variation of parameters formula (13) in Section 9.7 (page 541) in a simpler form. Namely, the solution to the initial value problem x′ = Ax + f1t2, x1t0 2 = x0 is given by x1t2 = eA1t-t02x0 +

t

eA1t-s2f1s2ds , Lt0 which is a system version of the formula for the solution to the scalar initial value problem x′ = ax + f1t2, x1t0 2 = x0. In closing, we remark that the software packages for eigenvalue computation listed in Section 9.5 (page 524) also contain subroutines for computing the matrix exponential. (16)

9.8

EXERCISES

In Problems 1–6, (a) show that the given matrix A satisfies 1A - rI2 k = 0 for some number r and some positive integer k and (b) use this fact to determine the matrix eAt. [Hint: Compute the characteristic polynomial and use the Cayley–Hamilton theorem.] 3 -2 1 -1 1. A = c d 2. A = c d 0 3 1 3 2 3. A = £ - 3 9

1 -1 3

-1 1§ -4

2 4. A = £ 0 0

1 2 0

3 -1 § 2

-2 5. A = £ 4 1

0 -2 0

0 0§ -2

0 6. A = £ 0 -1

1 0 -3

0 1§ -3

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552

Chapter 9

Matrix Methods for Linear Systems

In Problems 7–10, determine eAt by first finding a fundamental matrix X1t2 for x′ = Ax and then using formula (3). 0 1 1 1 7. A = c d 8. A = c d -1 0 4 1 0 9. A = £ 0 1

1 0 -1

0 1§ 1

0 2 2 10. A = £ 2 0 2 § 2 2 0

In Problems 13–16 determine eAt, using a linear software package to find the eigenvectors. 0 1 0 0 0 0 0 1 0 0 13. A = E 1 - 3 3 0 0U 0 0 0 0 1 0 0 0 -1 0 0 0 -1 0 0

0 1 -2 0 0

0 0 0 0 -1

0 0 0U 1 0

0 0 15. A = E -1 0 0

1 0 -3 0 0

0 1 -3 0 0

0 0 0 0 -4

0 0 0U 1 -4

-1 0 16. A = E 0 0 0

0 0 -1 0 0

0 1 -2 0 0

0 0 0 0 -4

0 0 0U 1 -4

1 1 19. A = D 0 0

1 0 -5 0 1 0 0

1 2 2 1

0 -2 0

0 0 § x1t2 , -2

1 x′1t2 = £ 2 0

1 1 -1

1 -1 § x1t2 , 1

1 x102 = £ 1 § . -1

-1 x102 = £ 0 § . 3

23. Use the results of Problem 3 and the variation of parameters formula (16) to find the solution to the initial value problem 2 1 x′1t2 = £ -3 -1 9 3 0 x102 = £ 3 § . 0

-1 0 1 § x1t2 + £ t § , -4 0

24. Use your answer to Problem 9 and the variation of parameters formula (16) to find the solution to the initial value problem 0 x′1t2 = £ 0 1

1 0 -1

0 0 1 § x1t2 + £ 0 § , 1 t

1 x102 = £ -1 § . 0 25. Let A = c

In Problems 17–20, use the generalized eigenvectors of A to find a general solution to the system x′1t2 = Ax1t2, where A is given. 0 17. A = £ 0 -2

-2 x′1t2 = £ 4 1

22. Use your answer to Problem 12 to find the solution to the initial value problem

In Problems 11 and 12, determine eAt by using generalized eigenvectors to find a fundamental matrix and then using formula (3). 5 -4 0 1 1 1 11. A = £ 1 0 2§ 12. A = £ 2 1 -1 § 0 2 5 0 -1 1

1 0 14. A = E 0 0 0

21. Use the results of Problem 5 to find the solution to the initial value problem

0 1§ -4

0 0 1 18. A = £ 0 1 2 § 0 0 1

2 1 T 0 1

-1 20. A = £ - 1 -4

-8 -3 - 16

1 2§ 7

1 2 d -1 3

and

B = c

2 1 d . 0 1

(a) Show that AB ≠ BA. (b) Show that property (d) in Theorem 7 does not hold for these matrices. That is, show that e1A + B2t ≠ eAteBt. 26. Let A be a diagonal n * n matrix with entries r1, c, rn down its main diagonal. To compute eAt, proceed as follows: (a) Show that Ak is the diagonal matrix with entries r k1, c, r kn down its main diagonal. (b) Use the result of part (a) and the defining equation (2) to show that eAt is the diagonal matrix with entries er1t, c, ernt down its main diagonal. 27. In Problems 35–40 of Exercises 9.5, page 532, some ad hoc formulas were invoked to find general solutions to the system x′ = Ax when A had repeated eigenvalues.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Chapter 9 Summary

Using the generalized eigenvector procedure outlined on page 548, justify the ad hoc formulas proposed in (a) Problem 35, Exercises 9.5. (b) Problem 37, Exercises 9.5. (c) Problem 39, Exercises 9.5. 28. Let 5 2 -4 A = £0 3 0§ 4 -5 - 5

553

29. For the matrix A in Problem 28, solve the initial value problem 2 x′1t2 = Ax1t2 + sin12t2 £ 0 § , 4 0 x102 = £ 1 § . -1

(a) Find a general solution to x′ = Ax. (b) Determine which initial conditions x102 = x0 yield a solution x1t2 = col 1 x1 1t2, x2 1t2, x3 1t2 2 that remains bounded for all t Ú 0; that is, satisfies

0 0 x1t2 0 0 J 2x12 1t2 + x22 1t2 + x23 1t2 … M

for some constant M and all t Ú 0.

Chapter 9 Summary In this chapter we discussed the theory of linear systems in normal form and presented methods for solving such systems. The theory and methods are natural extensions of the development for second-order and higher-order linear equations. The important properties and techniques are as follows.

Homogeneous Normal Systems x′1t2 = A1t2x1t2 The n * n matrix function A1t2 is assumed to be continuous on an interval I. Fundamental Solution Set: 5x1, c, xn 6. A set of n vector solutions x1 1t2, c, xn 1t2 of the homogeneous system on the interval I form a fundamental solution set provided they are linearly independent on I or, equivalently, their Wronskian x1,1 1t2 x2,1 1t2 W3x1, c, xn 4 1t2 J det3x1, c, xn 4 = ∞ f xn,1 1t2

x1,2 1t2 x2,2 1t2 f xn,2 1t2

g g g

x1,n 1t2 x2,n 1t2 ∞ f xn,n 1t2

is never zero on I. Fundamental Matrix: X1t2. An n * n matrix function X1t2 whose column vectors form a fundamental solution set for the homogeneous system is called a fundamental matrix. The determinant of X1t2 is the Wronskian of the fundamental solution set. Since the Wronskian is never zero on the interval I, then X-1 1t2 exists for t in I.

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554

Chapter 9

Matrix Methods for Linear Systems

General Solution to Homogeneous System: Xc = c1x1 + g + cnxn. If X1t2 is a fundamental matrix whose column vectors are x1, . . . , xn, then a general solution to the homogeneous system is x1t2 = X1t2c = c1x1 1t2 + c2x2 1t2 + g + cnxn 1t2 ,

where c = col1c1, c, cn 2 is an arbitrary constant vector.

Homogeneous Systems with Constant Coefficients. The form of a general solution for a homogeneous system with constant coefficients depends on the eigenvalues and eigenvectors of the n * n constant matrix A. An eigenvalue of A is a number r such that the system Au = ru has a nontrivial solution u, called an eigenvector of A associated with the eigenvalue r. Finding the eigenvalues of A is equivalent to finding the roots of the characteristic equation

0 A - rI 0 = 0 .

The corresponding eigenvectors are found by solving the system 1A - rI2u = 0. If the matrix A has n linearly independent eigenvectors u1, . . . , un and ri is the eigenvalue corresponding to the eigenvector ui, then 5er1tu1, er2tu2, c, erntun 6

is a fundamental solution set for the homogeneous system. A class of matrices that always has n linearly independent eigenvectors is the set of symmetric matrices—that is, matrices that satisfy A = AT. If A has complex conjugate eigenvalues a { ib and associated eigenvectors z = a { ib, where a and b are real vectors, then two linearly independent real vector solutions to the homogeneous system are eat cos bt a - eat sin bt b ,

eat sin bt a + eat cos bt b .

When A has a repeated eigenvalue r of multiplicity m, then it is possible that A does not have n linearly independent eigenvectors. However, associated with r are m linearly independent generalized eigenvectors that can be used to generate m linearly independent solutions to the homogeneous system (see page 548 under “Generalized Eigenvectors”).

Nonhomogeneous Normal Systems x′1t2 = A1t2x1t2 + f1t2 The n * n matrix function A1t2 and the vector function f1t2 are assumed continuous on an interval I. General Solution to Nonhomogeneous System: xp + Xc. If xp 1t2 is any particular solution for the nonhomogeneous system and X1t2 is a fundamental matrix for the associated homogeneous system, then a general solution for the nonhomogeneous system is x1t2 = xp 1t2 + X1t2c = xp 1t2 + c1 x1 1t2 + g + cn xn 1t2 ,

where x1 1t2, c, xn 1t2 are the column vectors of X1t2 and c = col1c1, c, cn 2 is an arbitrary constant vector. Undetermined Coefficients. If the nonhomogeneous term f1t2 is a vector whose components are polynomials, exponential or sinusoidal functions, and A is a constant matrix, then one can use an extension of the method of undetermined coefficients to decide the form of a particular solution to the nonhomogeneous system.

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Review Problems for Chapter 9

555

Variation of Parameters: X1t2Y1t2. Let X1t2 be a fundamental matrix for the homogeneous system. A particular solution to the nonhomogeneous system is given by the variation of parameters formula xp 1t2 = X1t2Y1t2 = X1t2

L

X-1 1t2f1t2 dt .

Matrix Exponential Function If A is a constant n * n matrix, then the matrix exponential function eAt J I + At + A2

t2 tn + g + An + g 2! n!

is a fundamental matrix for the homogeneous system x′1t2 = Ax1t2. The matrix exponential has some of the same properties as the scalar exponential eat. In particular, e0 = I ,

eA1t + s2 = eAteAs ,

1eAt 2 -1 = e-At .

If 1A - rI2 k = 0 for some r and k, then the series for eAt has only a finite number of terms: eAt = ert e I + 1A - rI2t + g + 1A - rI2 k - 1

tk - 1 f. 1k - 12!

The matrix exponential function eAt can also be computed from any fundamental matrix X1t2 via the formula eAt = X1t2X-1 102 .

Generalized Eigenvectors If an eigenvalue ri of a constant n * n matrix A has multiplicity mi there exist mi linearly independent generalized eigenvectors u satisfying 1A - riI2 mi u = 0. Each such u determines a solution to the system x′ = Ax of the form x1t2 = eri t c I + 1A - ri I2t + g + 1A - riI2 mi - 1

t mi - 1 du 1mi - 12!

and the totality of all such solutions is linearly independent on 1 - ∞, ∞ 2 and forms a fundamental solution set for the system.

REVIEW PROBLEMS FOR CHAPTER 9 In Problems 1–4, find a general solution for the system x′1t2 = Ax1t2, where A is given. 6 -3 3 2 1. A = c d 2. A = c d 2 1 -5 1 1 2 3. A = D 0 0

2 1 0 0

0 0 1 2

0 0 T 2 1

In Problems 5 and 6, find a fundamental matrix for the system x′1t2 = Ax1t2, where A is given. 5. A = c

1 2

-1 d 4

1 1 0 4. A = £ 0 1 0 § 0 0 2

5 6. A = £ 0 0

0 -4 3

0 3§ 4

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556

Chapter 9

Matrix Methods for Linear Systems

In Problems 7–10, find a general solution for the system x′1t2 = Ax1t2 + f1t2, where A and f1t2 are given. 5 1 1 7. A = c f1t2 = c d d , 6 4 1 8. A = c

-4 2

2 9. A = £ -3 9 2 10. A = £ 0 0

f1t2 = c

2 d , -1

-1 1§ , -4

1 -1 3 -2 3 -1

-3 2§ , 2

e4t d 3e4t

t f1t2 = £ 0 § 1 e-t f1t2 = £ 2 § 1

In Problems 11 and 12, solve the given initial value problem. 0 1 1 11. x′1t2 = c d x1t2 , x102 = c d -2 3 -1 12. x′1t2 = c

2 1 te2t d x1t2 + c 2t d , e -4 2

2 x102 = c d 2

In Problems 13 and 14, find a general solution for the Cauchy– Euler system tx′1t2 = Ax1t2, where A is given. 0 3 1 13. A = £ 1 2 1 § 1 3 0

1 14. A = £ 2 -1

2 1 1

-1 1§ 0

17. For each of the following, determine if the statement made is True or False. (a) Every n * n matrix of real numbers has n linearly independent eigenvectors. (b) If X1t2 is a fundamental matrix for x′ = Ax, where 1 3 2 3 d , then the product X1t2 c d is also A = c 8 4 4 6 a fundamental matrix for this system. 0 0 0 (c) The vector functions £ 1 § , £ t § , £ 1 § are linearly et tet 2et dependent on 1 - ∞ , ∞ 2. (d) If A is an n * n matrix of real numbers and x*1t2 satisfies the initial value problem x′ = Ax, x152 = 0, then x*1t2 = 0 for all t. (e) If A is a 3 * 3 matrix of real numbers whose eigenvalues are 2, 2, and 1, then A has two linearly independent generalized eigenvectors that correspond to the eigenvalue 2. (f) The Wronskian W[x1 , x2, . . . , xn]1t2 of n linearly independent n * 1 vector functions xi on an open interval I is never zero on I. (g) If a 3 * 3 matrix of real numbers has eigenvalues 2 + 3i, 2 - 3i, and 5, then the matrix has three linearly independent eigenvectors.

In Problems 15 and 16, find the fundamental matrix eAt for the system x′1t2 = Ax1t2, where A is given. 4 2 3 0 1 4 15. A = £ 2 1 2 § 16. A = £ 0 0 2 § -1 2 0 0 0 0

TECHNICAL WRITING EXERCISES FOR CHAPTER 9 1. Explain how the theory of homogeneous linear differential equations (as described in Section 6.1) follows from the theory of linear systems in normal form (as described in Section 9.4). 2. Discuss the similarities and differences between the method for finding solutions to a linear constant-coefficient differential equation (see Chapters 4 and 6) and the method for finding solutions to a linear system in normal form that has constant coefficients (see Sections 9.5 and 9.6). 3. Explain how the variation of parameters formulas for linear second-order equations derived in Section 4.6 follow

from the formulas derived in Section 9.7 for linear systems in normal form. 4. Explain how you would define the matrix functions sin At and cos At, where A is a constant n * n matrix. How are these functions related to the matrix exponential and how are they connected to the solutions of the system x″ + A2 x = 0? Your discussion should include the cases when A is c

0 0 0 1 0 0 d ,c d ,c d , 0 0 0 0 1 0

and

c

1 1

-1 d . -1

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Projects for Chapter 9 A Uncoupling Normal Systems The easiest normal systems to solve are systems of the form (1)

x′1t2 = Dx1t2 ,

where D is an n * n diagonal matrix. Such a system actually consists of n uncoupled equations (2)

xi′ 1t2 = diixi 1t2 ,

i = 1, c, n ,

whose solution is xi 1t2 = ciediit , where the ci’s are arbitrary constants. This raises the following question: When can we uncouple a normal system? To answer this question, we need the following result from linear algebra. An n * n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors p1, c, pn. Moreover, if P is the matrix whose columns are p1, c, pn, then (3)

P -1AP = D ,

where D is the diagonal matrix whose entry dii is the eigenvalue associated with the vector pi. (a) Use the above result to show that the system (4)

x′1t2 = Ax1t2 ,

where A is an n * n diagonalizable matrix, is equivalent to an uncoupled system (5)

y′1t2 = Dy1t2 ,

where y = P -1x and D = P -1AP. (b) Solve system (5). (c) Use the results of parts (a) and (b) to show that a general solution to (4) is given by x1t2 = c1ed11t p1 + c2ed22t p2 + g + cnednnt pn . (d) Use the procedure discussed in parts (a)–(c) to obtain a general solution for the system 5 x′1t2 = £ 1 -1

-4 0 2

4 1 § x1t2 . 1

Specify P, D, P -1, and y.

557

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558

Chapter 9

Matrix Methods for Linear Systems

B Matrix Laplace Transform Method The Laplace transform method for solving systems of linear differential equations with constant coefficients was discussed in Chapter 7. To apply the procedure for equations given in matrix form, we first extend the definition of the Laplace operator ℒ to a column vector of functions x = col1x1 1t2, c, xn 1t22 by taking the transform of each component: ℒ5x6 1s2 J col 1 ℒ5x1 6 1s2, c, ℒ5xn 6 1s2 2 .

With this notation, the vector analogue of the important property relating the Laplace transform of the derivative of a function (see Theorem 4, Chapter 7, page 362) becomes (6)

ℒ5x′6 1s2 = sℒ5x6 1s2 - x102 . Now suppose we are given the initial value problem

(7)

x′ = Ax + f1t2  ,

x102 = x0 ,

where A is a constant n * n matrix. Let xn 1s2 denote the Laplace transform of x1t2 and fn 1s2 denote the transform of f1t2. Then, taking the transform of the system and using the relation (6), we get ℒ5x′6 = ℒ5Ax + f6 , sxn - x0 = Axn + fn .

Next we collect the xn terms and solve for xn by premultiplying by 1sI - A2 -1: 1sI - A2xn = fn + x0  ,

xn = 1sI - A2 -1 1 fn + x0 2 .

Finally, we obtain the solution x1t2 by taking the inverse Laplace transform: (8)

x = ℒ -1 5xn 6 = ℒ -1 5 1sI - A2 -1 1 fn + x0 26 .

In applying the matrix Laplace transform method, it is straightforward (but possibly tedious) to compute 1sI - A2 -1, but the computation of the inverse transform may require some of the special techniques (such as partial fractions) discussed in Chapter 7. (a) In the above procedure, we used the property that ℒ5Ax6 = Aℒ5x6 for any constant n * n matrix A. Show that this property follows from the linearity of the transform in the scalar case. (b) Use the matrix Laplace transform method to solve the following initial value problems: -1 0 2 d. x102 = c d x1t2 , (i) x′1t2 = c 3 -1 3 3 -2 sin t 0 d, (ii) x′1t2 = c d x1t2 + c x102 = c d . - cos t 4 -1 0 (c) By comparing the Laplace transform solution formula (8) with the matrix exponential solution formula given in Section 9.8 [relation (16), page 551] for the homogeneous case f1t2 K 0 and t0 = 0, derive the Laplace transform formula for the matrix exponential (9)

eAt = ℒ -1 5 1sI - A2 -1 6 1t2 .

(d) What is eAt for the coefficient matrices in part (b) above? (e) Use (9) to rework Problems 1, 2, 7, and 8 in Exercises 9.8, page 551.

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Projects for Chapter 9

559

C Undamped Second-Order Systems We have seen that the coupled mass–spring system depicted in Figure 9.5, page 536, is governed by equations (10) of Section 9.6, which we reproduce here: m1x1> = - k1x1 + k2 1x2 - x1 2 ,

m2x2> = - k2 1x2 - x1 2 - k3x2 . This system was rewritten in normal form as equation (11) in Section 9.6; however, there is some advantage in expressing it as a second-order system in the form c

- k2 m1 0 x1 ″ x1 k1 + k2 dc d = -c dc d . 0 m2 x2 - k2 k2 + k3 x2

This structure, (10)

Mx″ = - Kx ,

with a diagonal mass matrix M and a symmetric stiffness matrix K, is typical for most undamped vibrating systems. Our experience with other mass–spring systems (Section 5.6) suggests that we seek solutions to (10) of the form (11)

x = 1cos vt2 v

or

x = 1sin vt2 v ,

where v is a constant vector and v is a positive constant. (a) Show that the system (10) has a nontrivial solution of the form (11) if and only if v and v satisfy the “generalized eigenvalue problem” Kv = v2 Mv. (b) By employing the inverse of the mass matrix, one can rewrite (10) as x″ = - M -1Kx = : Bx . Show that - v2 must be an eigenvalue of B if (11) is a nontrivial solution. (c) If B is an n * n constant matrix, then x″ = Bx can be written as a system of 2n firstorder equations in normal form. Thus, a general solution can be formed from 2n linearly independent solutions. Use the observation in part (b) to find a general solution to the following second-order systems: (i) x″ = c

0 -4

1 d x. -5

(ii) x″ = c

-2 2

2 d x. -5

-5 (iii) x″ = £ - 1 1

4 0 -2

-4 -1 § x . -1

2 (iv) x″ = £ - 1 -1

-1 2 0

-1 0§ x . 2

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CHAPTER

10

Partial Differential Equations

10.1 Introduction: A Model for Heat Flow Develop a model for the flow of heat through a thin, insulated wire whose ends are kept at a constant temperature of 0°C and whose initial temperature distribution is to be specified. Suppose the wire is placed along the x-axis with x = 0 at the left end of the wire and x = L at the right end (see Figure 10.1). If we let u denote the temperature of the wire, then u depends on the time t and on the position x within the wire. (We will assume the wire is thin and hence u is constant throughout a cross section of the wire corresponding to a fixed value of x.) Because the wire is insulated, we assume no heat enters or leaves through the sides of the wire. To develop a model for heat flow through the thin wire, let’s consider the small volume element V of wire between the two cross-sectional planes A and B that are perpendicular to the x-axis, with plane A located at x and plane B located at x + ∆x (see Figure 10.1). The temperature on plane A at time t is u1x, t2 and on plane B is u1x + ∆x, t2. We will need the following principles of physics that describe heat flow.† 1. Heat Conduction: The rate of heat flow (the amount of heat per unit time flowing through a unit of cross-sectional area at A) is proportional to 0u>0x, the temperature gradient at A (Fick’s law). The proportionality constant k is called the thermal conductivity of the material. In general, the thermal conductivity can vary from point to point: k = k1x2. 2. Direction of Heat Flow: The direction of heat flow is always from points of higher temperature to points of lower temperature.

V

0

A

B

x

x 1 Dx

x

L

Figure 10.1 Heat flow through a thin piece of wire

†

For a discussion of heat transfer, see University Physics with Modern Physics™, 13th ed., by H. D. Young and R. A. Freedman (Addison-Wesley, Reading, Mass., 2011).

560

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Section 10.1

Introduction: A Model for Heat Flow

561

3. Specific Heat Capacity: The amount of heat necessary to raise the temperature of an object of mass m by an amount ∆u is cm ∆u, where the constant c is the specific heat capacity of the material. The specific heat capacity, like the thermal conductivity, can vary with position: c = c1x2. If we let H represent the amount of heat flowing from left to right through the surface A during an interval of time ∆t, then the formula for heat conduction becomes H1x2 = -k1x2 a ∆t

0u 1x, t2 , 0x

where a is the cross-sectional area of the wire. The negative sign follows from the second principle—if 0u>0x is positive, then heat flows from right to left (hotter to colder). Similarly, the amount of heat flowing from left to right across plane B during an interval of time ∆t is H1x + ∆x2 = -k1x + ∆x2 a ∆t

0u 1x + ∆x, t2  . 0x

The net change in the heat ∆E in volume V is the amount entering at end A minus the amount leaving at end B, plus any heat generated by sources (such as electric currents, chemical reactions, heaters, etc.). The latter is modeled by a term Q1x, t2∆x a ∆t, where Q is the energy rate (power) density. Therefore, (1)

∆E = H1x2 - H1x + ∆x2 + Q1x, t2∆x a ∆t = a ∆t c k1x + ∆x2

0u 0u 1x + ∆x, t2 - k1x2 1x, t2 d + Q1x, t2∆x a ∆t . 0x 0x

Now by the third principle, the net change is given by ∆E = cm ∆u, where ∆u is the change in temperature and c is the specific heat capacity. If we assume that the change in temperature in the volume V is essentially equal to the change in temperature at x—that is, ∆u = u1x, t + ∆t2 - u1x, t2, and that the mass of the volume V of wire is a r ∆x, where r = r1x2 is the density of the wire, then (2)

∆E = c1x2ar1x2 ∆x 3u1x, t + ∆t2 - u1x, t24 .

Equating the two expressions for ∆E given in equations (1) and (2) yields a ∆t c k1x + ∆x2

0u 0u 1x + ∆x, t2 - k1x2 1x, t2 d + Q1x, t2 ∆x a ∆t 0x 0x

= c1x2a r1x2 ∆x 3u1x, t + ∆t2 - u1x, t24 .

Now dividing both sides by a ∆x ∆t and then taking the limits as ∆x and ∆t approach zero, we obtain (3)

0 0u 0u c k1x2 1x, t2 d + Q1x, t2 = c1x2r1x2 1x, t2 . 0x 0x 0t

If the physical parameters k, c, and r are uniform along the length of the wire, then (3) reduces to the one-dimensional heat flow equation (4)

0u 02u 1x, t2 = b 2 1x, t2 + P1x, t2 , 0t 0x

where the positive constant b J k> 1cr2 is the diffusivity of the material and P1x, t2 J Q1x, t2 > 1cr2.

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562

Chapter 10

Partial Differential Equations

Equation (4) governs the flow of heat in the wire. We have two other constraints in our original problem. First, we are keeping the ends of the wire at 0°C. Thus, we require that (5)

u10, t2 = u1L, t2 = 0

for all t. These are called boundary conditions. Second, we must be given the initial temperature distribution f1x2. That is, we require (6)

u1x, 02 = f1x2 ,

06x6L.

Equation (6) is referred to as the initial condition on u. Combining equations (4), (5), and (6), we have the following mathematical model for the heat flow in a uniform wire without internal sources 1P = 02 whose ends are kept at the constant temperature 0°C: (7) (8) (9)

eu e2u 1x, t2 = B 2 1x, t2 , et ex u10, t2 = u1L, t2 = 0 , u1x, 02 = f1x2 ,

06x6L,

t70,

t70, 06x6L.

This model is an example of an initial-boundary value problem. Intuitively we expect that equations (7)–(9) completely and unambiguously specify the temperature in the wire. Once we have found a function u1x, t2 that meets all three of these conditions, we can be assured that u is the temperature. (Theorem 7 in Section 10.5, page 602, will bear this out.) In higher dimensions, the heat flow equation (or just heat equation) is adjusted to account for the additional heat flow contribution along the other axes by the simple modification 0u = b ∆u + P1x, y, z, t2 , 0t where ∆u, known as the Laplacian in this context,† is defined in two and three dimensions, respectively, as 02u 02u 02u 02u 02u ∆u J 2 + 2 and ∆u J 2 + 2 + 2 . 0x 0y 0x 0y 0z When the temperature reaches a steady state—that is, when u does not depend on time, and there are no sources—then 0u>0t = 0 and the temperature satisfies Laplace’s equation ∆u = 0 . One classical technique for solving the initial-boundary value problem for the heat equation (7)–(9) is the method of separation of variables, which effectively allows us to replace the partial differential equation by ordinary differential equations. This technique is discussed in the next section. In using separation of variables, one is often required to express a given function as a trigonometric series. Such series are called Fourier series; their properties are discussed in Sections 10.3 and 10.4. We devote the remaining three sections to the three basic partial differential equations that arise most commonly in applications: the heat equation, the wave equation, and Laplace’s equation. Many computer-based algorithms for solving partial differential equations have been developed. These are based on finite differences, finite elements, variational principles, and projection methods that include the method of moments and its wavelet-based implementations. Like the Runge–Kutta or Euler methods for ordinary differential equations, such techniques are more universally applicable than the analytic procedures of separation of variables, but their accuracy is often difficult to assess. Indeed, it is customary practice to gauge any newly proposed numerical procedure by comparing its predictions with those of separation of variables. † Regrettably, it is the same notation as we just used for u1x, t + ∆t2 - u1x, t2. Some authors prefer the symbol ∇ 2 for the Laplacian.

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Section 10.2

Method of Separation of Variables

563

10.2 Method of Separation of Variables The method of separation of variables is a classical technique that is effective in solving several types of partial differential equations. The idea is roughly the following. We think of a solution, say u1x, t2, to a partial differential equation as being a linear combination of simple component functions un 1x, t2, n = 0, 1, 2, c, which also satisfy the equation and certain boundary conditions. (This is a reasonable assumption provided the partial differential equation and the boundary conditions are linear.) To determine a component solution, un 1x, t2, we assume it can be written with its variables separated; that is, as un 1x, t2 = Xn 1x2Tn 1t2 .

Substituting this form for a solution into the partial differential equation and using the boundary conditions leads, in many circumstances, to two ordinary differential equations for the unknown functions Xn 1x2 and Tn 1t2. In this way we have reduced the problem of solving a partial differential equation to the more familiar problem of solving a differential equation that involves only one variable. In this section we will illustrate this technique for the heat equation and the wave equation. In the previous section we derived the following initial-boundary value problem as a mathematical model for the sourceless heat flow in a uniform wire whose ends are kept at the constant temperature zero: (1)

eu e2u 1x, t2 = B 2 1x, t2 , et ex

(2)

u10, t2 = u1L, t2 = 0 ,

t70,

(3)

u1x, 02 = f1x2 ,

06x6L.

06x6L,

t70,

To solve this problem by the method of separation of variables, we begin by addressing equation (1). We propose that it has solutions of the form u1x, t2 = X1x2T1t2 , where X is a function of x alone and T is a function of t alone. To determine X and T, we first compute the partial derivatives of u to obtain 0u = X1x2T ′1t2 0t

and

02u = X″1x2T1t2 . 0x2

Substituting these expressions into (1) gives X1x2T ′1t2 = bX″1x2T1t2 , and separating variables yields (4)

T ′1t2 bT1t2

=

X″1x2 X1x2

.

We now observe that the functions on the left-hand side of (4) depend only on t, while those on the right-hand side depend only on x. If we fix t and vary x, the ratio on the right cannot change; it must be constant. Most authors define this constant with a minus sign, so we adhere to the convention: X″1x2 = -l X1x2

and

T ′1t2 = -l , bT1t2

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564

Chapter 10

Partial Differential Equations

or (5)

X″1x2 = −LX1x2

and

T ′1t2 = −LBT1t2 .

Consequently, for separable solutions, we have reduced the problem of solving the partial differential equation (1) to solving the two ordinary differential equations in (5). Next we consider the boundary conditions in (2). Since u1x, t2 = X1x2T1t2, these conditions are X102T1t2 = 0

and

X1L2T1t2 = 0 ,

t70.

Hence, either T1t2 = 0 for all t 7 0, which implies that u1x, t2 K 0, or (6)

X102 = X1L2 = 0 .

Ignoring the trivial solution u1x, t2 K 0, we combine the boundary conditions in (6) with the differential equation for X in (5) and obtain the boundary value problem (7)

X″1x2 + lX1x2 = 0 ,

X102 = X1L2 = 0 ,

where l can be any constant. Notice that the function X1x2 K 0 is a solution of (7) for every l. Depending on the choice of l, this may be the only solution to the boundary value problem (7). Thus, if we seek a nontrivial solution u1x, t2 = X1x2T1t2 to (1)–(2), we must first determine those values of l for which the boundary value problem (7) has nontrivial solutions. These solutions are called the eigenfunctions of the problem; the eigenvalues are the special values of l. If we write the first equation in (7) as -D2 3X4 = lX and compare with the matrix eigenvector equation Au = r u (Section 9.5, page 523), the designation of l as an eigenvalue (of -D2) becomes clear. To solve the (constant-coefficient) equation in (7), we try X1x2 = erx, derive the auxiliary equation r 2 + l = 0, and consider three cases. Case 1. l 6 0. In this case, the roots of the auxiliary equation are { 1-l, so a general solution to the differential equation in (7) is X1x2 = C1e1-l x + C2e-1-l x . To determine C1 and C2, we appeal to the boundary conditions: X102 = C1 + C2 = 0 , X1L2 = C1e1-l L + C2e-1-l L = 0 . From the first equation, we see that C2 = -C1. The second equation can then be written as C1 1e1-l L - e-1-l L 2 = 0 or C1 1e21-l L - 12 = 0. Since -l 7 0, it follows that 1e21-l L - 12 7 0. Therefore C1, and hence C2, is zero. Consequently, there is no nontrivial solution to (7) for l 6 0. Case 2. l = 0. Here r = 0 is a repeated root to the auxiliary equation, and a general solution to the differential equation is X1x2 = C1 + C2 x . The boundary conditions in (7) yield C1 = 0 and C1 + C2 L = 0, which imply that C1 = C2 = 0. Thus, for l = 0, there is no nontrivial solution to (7). Case 3. l 7 0. In this case the roots of the auxiliary equation are {i1l. Thus a general solution to X″ + l X = 0 is (8)

X1x2 = C1 cos 1l x + C2 sin 1l x .

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Section 10.2

Method of Separation of Variables

565

This time the boundary conditions X102 = X1L2 = 0 give the system C1 = 0 , C1 cos 1l L + C2 sin 1l L = 0 . Because C1 = 0, the system reduces to solving C2 sin 1lL = 0. Hence, either sin 1lL = 0 or C2 = 0. Now sin 1lL = 0 only when 1lL = np, where n is an integer. Therefore, (7) has a nontrivial solution 1C2 ≠ 02 when 1lL = np or l = 1np>L2 2, n = 1, 2, 3, c (we exclude n = 0, since it makes l = 0). Furthermore, the nontrivial solutions (eigenfunctions) Xn corresponding to the eigenvalue l = 1np>L2 2 are given by [cf. (8)] npx b, L where the an’s are arbitrary nonzero constants. The eigenvalues and eigenfunctions for this example have many features that are common to all of the separation of variables solutions that we will study. Take note of these features of Figure 10.2. (9)

Eigenvalue

Xn 1x2 = an sina

Eigenfunction

2

L

4

2

L

2

L

2

L

2

L

2

L

2

16 L

x

(i) The eigenfunctions are solutions to a second-order ordinary differential equation (7) containing a parameter l called the eigenvalue. (ii) Each eigenfunction satisfies a homogeneous boundary condition at each end (7).

9 L

x

SEPARATION OF VARIABLES: EIGENFUNCTION PROPERTIES

2

x

x

(iii) The only values of l that admit nontrivial solutions, i.e., the eigenvalues, form an infinite set accumulating at ∞ . (iv) The eigenfunctions oscillate faster as l increases; the first contains no interior zeros, and each subsequent eigenfunction contains one more zero than its predecessor. (v) If any two distinct eigenfunctions are multiplied together, the resulting function is zero on the average; specifically, sin

l

1m - n2px 1m + n2px mpx npx 1 sin = c cos - cos d L L 2 L L

and both cosines integrate to 0 over 30, L4.

Figure 10.2 Eigenfunctions

Having determined that l = 1np>L2 2 for any positive integer n, let’s consider the second equation in (5): T ′1t2 + b a

np 2 b T1t2 = 0 . L

For each n = 1, 2, 3, c, a general solution to this linear first-order equation is Tn 1t2 = bn e-b1np>L2 t . 2

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566

Chapter 10

Partial Differential Equations

(Note that the time factor has none of the eigenfunction properties.) Combining this with equation (9) we obtain, for each n = 1, 2, 3, c, the functions (10)

un 1x, t2 J Xn 1x2Tn 1t2 = an sin1npx>L2 bn e-b1np>L2 t 2

= cn e-b1np>L2 t sin1npx>L2 , 2

where cn is also an arbitrary constant. It is easy to see that each un 1x, t2 is a solution to (1)–(2). A simple computation also shows that if un and um are solutions to (1)–(2), then so is any linear combination aun + bum. (This is a consequence of the facts that the operator L J 0>0t - b02 >0x2 is a linear operator and the boundary conditions in (2) are linear homogeneous.) This enables us to solve the following example. Example 1

Find the solution to the heat flow problem

(12)

0u 02u = 7 2, 0t 0x u10, t2 = u1p, t2 = 0 ,

t70,

(13)

u1x, 02 = 3 sin 2x - 6 sin 5x ,

06x6p.

(11)

Solution

06x6p,

t70,

Comparing equation (11) with (1), we see that b = 7 and L = p. Hence, we need only find a combination of terms like (10) that satisfies the initial condition (13): u1x, 02 = a cn e0 sin nx = 3 sin 2x - 6 sin 5x. For these data the task is simple; c2 = 3 and c5 = -6. The solution to the heat flow problem (11)–(13) is u1x, t2 = c2e-b12p>L2 t sin12px>L2 + c5e-b15p>L2 t sin15px>L2 2

2

= 3e-28t sin 2x - 6e-175t sin 5x . ◆ What would we do if the initial condition (13) had been an arbitrary function, rather than a simple combination of a few of the eigenfunctions we found? Possibly the most beautiful property of the eigenfunctions—one that we did not list on page 565—is completeness; virtually any function f likely to arise in applications can be expressed as a convergent series of eigenfunctions! For the sines we have been working with, the Fourier sine series looks like (14)

∞ npx f1x2 = a cn sin a b L n=1

for 0 6 x 6 L .

This enables the complete solution to the generic problem given by (1)–(3): (15)

H npx 2 u1x, t2 = a cn e-B1np/L2 t sin a b, L n=1

provided this expansion and its first two derivatives converge. The convergence of the sine series (among others), as well as the procedure for finding the coefficients cn, will be discussed in the next two sections (and generalized in Chapter 11). For now, let us turn to the mathematical description of a vibrating string, another situation in which the separation of variables approach applies. This concerns the transverse vibrations of a string stretched between two points, such as a guitar string or piano wire. The goal is to find a function u1x, t2 that gives the displacement (deflection) of the string at any point

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Section 10.2

Method of Separation of Variables

567

u(x, t)

L

0

x

Figure 10.3 Displacement of string at time t

x 10 … x … L2 and any time t Ú 0 (see Figure 10.3). In developing the mathematical model, we assume that the string is perfectly flexible and has constant linear density, the tension on the string is constant, gravity is negligible, and no other forces are acting on the string. Under these conditions and the additional assumption that the displacements u1x, t2 are small in comparison to the length of the string, it turns out that the motion of the string is governed by the following initial-boundary value problem.†

(17)

e2u e2u = A2 2 , 2 et ex u10, t2 = u1L, t2 = 0 ,

(18)

u1x, 02 = f1x2 ,

0"x"L,

(19)

eu 1x, 02 = g1x2 , et

0"x"L.

(16)

0*x*L,

t+0,

t#0,

Equation (16) essentially states Newton’s third law, F = ma, for the string. The second (time) derivative on the left is the acceleration, and the second (spatial) derivative on the right arises because the restoring force is produced by the string’s curvature. The constant a2 is strictly positive and equals the ratio of the tension to the (linear) density of the string. The physical significance of a, which has units of velocity, will be revealed in Section 10.6. The boundary conditions in (17) reflect the fact that the string is held fixed at the two endpoints x = 0 and x = L. Equations (18) and (19), respectively, specify the initial displacement of the string and the initial velocity of each point on the string. Recall that these are the typical initial data for all mechanical systems. For the initial and boundary conditions to be consistent, we assume f102 = f1L2 = 0 and g102 = g1L2 = 0. Let’s apply the method of separation of variables to the initial-boundary value problem for the vibrating string (16)–(19). Thus, we begin by assuming equation (16) has a solution of the form u1x, t2 = X1x2T1t2 , where X is a function of x alone and T is a function of t alone. Differentiating u, we obtain 02u = X1x2T ″1t2 , 0 t2

02u = X″1x2T1t2 . 0 x2

Substituting these expressions into (16), we have X1x2T″1t2 = a2X″1x2T1t2 , †

For a derivation of this mathematical model, see Partial Differential Equations: Sources and Solutions, by Arthur D. Snider (Dover Publications, N.Y., 2006).

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568

Chapter 10

Partial Differential Equations

and separating variables gives T ″1t2

=

2

a T1t2

X″1x2 X1x2

.

Just as before, these ratios must equal some constant -l: (20)

X″1x2 X1x2

= -L

T ″1t2

and

A2T1t2

= -L .

Furthermore, with u1x, t2 = X1x2T1t2, the boundary conditions in (17) give X102T1t2 = 0 ,

X1L2T1t2 = 0 ,

tÚ0.

In order for these equations to hold for all t Ú 0, either T1t2 K 0, which implies that u1x, t2 K 0, or X102 = X1L2 = 0 . Ignoring the trivial solution, we combine these boundary conditions with the differential equation for X in (20) and obtain the boundary value problem (21)

X″1x2 + lX1x2 = 0 ,

X102 = X1L2 = 0 ,

where l can be any constant. This is the same boundary value problem that we encountered earlier while solving the heat equation. There we found that the suitable values for l are l = a

np 2 b , L

n = 1, 2, 3, c,

with corresponding eigenfunctions (nontrivial solutions) (22)

Xn 1x2 = cn sina

npx b, L

where the cn’s are arbitrary nonzero constants. Recall Figure 10.2, page 565. Having determined that l = 1np>L2 2 for some positive integer n, let’s consider the second equation in (20) for such l: T ″1t2 +

a2n2p2 T1t2 = 0 . L2

For each n = 1, 2, 3, c, a general solution is Tn 1t2 = cn,1 cos

npa npa t + cn,2 sin t. L L

Combining this with equation (22), we obtain, for each n = 1, 2, 3, c, the function un 1x, t2 = Xn 1x2Tn 1t2 = a cn sin

npx npa npa b acn,1 cos t + cn,2 sin tb , L L L

or, reassembling the constants, (23)

un 1x, t2 = a an cos

npa npa npx t + bn sin tb sin . L L L

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Section 10.2

Method of Separation of Variables

569

Using the fact that linear combinations of solutions to (16)–(17) are again solutions, we consider a superposition of the functions in (23): (24)

H nPA nPA nPx t + bn sin t d sin . u1x, t2 = a c an cos L L L n=1

For a solution of the form (24), substitution into the initial conditions (18)–(19) gives (25)

∞ npx u1x, 02 = a an sin = f1x2 , L n=1

(26)

∞ npa 0u npx 1x, 02 = a bn sin = g1x2 , 0t L L n=1

0…x…L, 0…x…L.

We have now reduced the vibrating string problem (16)–(19) to the problem of determining the Fourier sine series expansions for f1x2 and g1x2: (27)

∞ npx f1x2 = a an sin , L n=1

∞ npx g1x2 = a Bn sin , L n=1

where Bn = 1npa>L2bn. If we choose the an’s and bn’s so that the equations in (25) and (26) hold, then the expansion for u1x, t2 in (24) is a formal solution to the vibrating string problem (16)–(19). If this expansion is finite, or converges to a function with continuous second partial derivatives, then the formal solution is an actual (genuine) solution. Example 2

Find the solution to the vibrating string problem (28)

06x6p,

(29)

u10, t2 = u1p, t2 = 0 ,

(30)

u1x, 02 = sin 3x - 4 sin 10 x ,

(31) Solution

02u 02u = 4 2, 2 0t 0x

t70,

tÚ0, 0…x…p,

0u 1x, 02 = 2 sin 4x + sin 6x , 0t

0…x…p.

Comparing equation (28) with equation (16), we see that a = 2 and L = p. Hence, we need only determine the values of the coefficients an and bn in formula (24). The an’s are chosen so that equation (25) holds; that is, ∞

u1x, 02 = sin 3x - 4 sin 10 x = a an sin nx . n=1

Equating coefficients of like terms, we see that a3 = 1 ,

a10 = -4 ,

and the remaining an’s are zero. Similarly, referring to equation (26), we must choose the bn’s so that ∞ 0u 1x, 02 = 2 sin 4x + sin 6x = a n2bn sin nx . 0t n=1

Comparing coefficients, we find 2 = 142122b4

or

1 = 162122b6

or

1 , 4 1 b6 = , 12 b4 =

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570

Chapter 10

Partial Differential Equations

and the remaining bn’s are zero. Hence, from formula (24), the solution to the vibrating string problem (28)–(31) is (32)

u1x, t2 = cos 6t sin 3x +

1 1 sin 8t sin 4x + sin 12 t sin 6x - 4 cos 20t sin 10 x . ◆ 4 12

In later sections the method of separation of variables is used to study a wide variety of problems for the heat, wave, and Laplace’s equations. However, to use the method effectively, one must be able to compute trigonometric series (or, more generally, eigenfunction expansions) such as the Fourier sine series that we encountered here. These expansions are discussed in the next two sections.

10.2

EXERCISES

In Problems 1– 8, determine all the solutions, if any, to the given boundary value problem by first finding a general solution to the differential equation. 1. y″ - y = 0 ; 06x61, y102 = 0 , y112 = - 4 2. y″ - 6y′ + 5y = 0 ; 06x62, y102 = 1 , y122 = 1 3. y″ + 4y = 0 ; 06x6p, y102 = 0 , y′1p2 = 0 4. y″ + 9y = 0 ; 06x6p, y102 = 0 , y′1p2 = -6 5. y″ - y = 1 - 2x ; 06x61, y102 = 0 , y112 = 1 + e 6. y″ + y = 0 ; 0 6 x 6 2p , y102 = 0 , y12p2 = 1 7. y″ + y = 0 ; 0 6 x 6 2p , y102 = 1 , y12p2 = 1 8. y″ - 2y′ + y = 0 ; -1 6 x 6 1 , y1 - 12 = 0 , y112 = 2 In Problems 9–14, find the values of l (eigenvalues) for which the given problem has a nontrivial solution. Also determine the corresponding nontrivial solutions (eigenfunctions). 9. y″ + ly = 0 ; 06x6p, y102 = 0 , y′1p2 = 0 10. y″ + ly = 0 ; 06x6p, y′102 = 0 , y1p2 = 0 11. y″ + ly = 0 ; 0 6 x 6 2p , y102 = y12p2 , y′102 = y′12p2 12. y″ + ly = 0 ; 0 6 x 6 p>2 , y′102 = 0 , y′1p>22 = 0 13. y″ + ly = 0 ; 06x6p, y102 - y′102 = 0 , y1p2 = 0 14. y″ - 2y′ + ly = 0 ; 06x6p, y102 = 0 , y1p2 = 0

In Problems 15–18, solve the heat flow problem (1)–(3) with b = 3, L = p, and the given function f1x2. 15. f1x2 = sin x - 6 sin 4x 16. f1x2 = sin 3x + 5 sin 7x - 2 sin 13 x 17. f1x2 = sin x - 7 sin 3x + sin 5x 18. f1x2 = sin 4x + 3 sin 6x - sin 10 x In Problems 19–22, solve the vibrating string problem (16)– (19) with a = 3, L = p, and the given initial functions f1x2 and g1x2. 19. f1x2 = 3 sin 2x + 12 sin 13 x , g1x2 K 0 20. f1x2 K 0 , g1x2 = -2 sin 3x + 9 sin 7x - sin 10 x 21. f1x2 = 6 sin 2x + 2 sin 6x , g1x2 = 11 sin 9x - 14 sin 15 x 22. f1x2 = sin x - sin 2x + sin 3x , g1x2 = 6 sin 3x - 7 sin 5x 23. Find the formal solution to the heat flow problem (1)–(3) with b = 2 and L = 1 if ∞ 1 f1x2 = a 2 sin npx . n=1 n

24. Find the formal solution to the vibrating string problem (16)–(19) with a = 4, L = p, and ∞ 1 f1x2 = a 2 sin nx , n=1 n ∞

1 - 12 n + 1

n=1

n

g1x2 = a

sin nx .

25. By considering the behavior of the solutions of the equation T′1t2 = - lbT1t2 ,

t70,

give an argument that is based on physical grounds to rule out the case where l 6 0 in equation (5). 26. Verify that un 1x, t2 given in equation (10) satisfies equation (1) and the boundary conditions in (2) by substituting un 1x, t2 directly into the equations involved.

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Section 10.3

In Problems 27–30, a partial differential equation (PDE) is given along with the form of a solution having separated variables. Show that such a solution must satisfy the indicated set of ordinary differential equations. 02u 1 0u 1 02u 27. + = 0 + 0 r 2 r 0r r 2 0 u 2 with u1r, u2 = R1r2ϴ1u2 yields r 2R″1r2 + rR′1r2 - lR1r2 = 0 , ϴ″1u2 + lϴ1u2 = 0 , where l is a constant. 02u 0u 02u 28. 2 + + u = a2 2 0t 0t 0x with u1x, t2 = X1x2T1t2 yields X″1x2 + lX1x2 = 0 ,

T″1t2 + T′1t2 + 11 + la 2T1t2 = 0 , where l is a constant. 0u 02u 02u 29. = be 2 + 2 f 0t 0x 0y with u1x, y, t2 = X1x2Y1y2T1t2 yields 2

T′1t2 + blT1t2 = 0 , X″1x2 + mX1x2 = 0 ,

Y″1y2 + 1l - m2Y1y2 = 0 , where l, m are constants. 02u 1 0u 1 02u 02u + + + = 0 30. 0 r 2 r 0r r 2 0 u 2 0z2 with u1r, u, z2 = R1r2ϴ1u2Z1z2 yields ϴ″1u2 + mϴ1u2 = 0 , Z″1z2 + lZ1z2 = 0 ,

Fourier Series

571

31. For the PDE in Problem 27, assume that the following boundary conditions are imposed: u1r, 02 = u1r, p2 = 0 , u1r, u2 remains bounded as r S 0+ . Show that a nontrivial solution of the form u1r, u2 = R1r2ϴ1u2 must satisfy the boundary conditions ϴ102 = ϴ1p2 = 0 , R1r2 remains bounded as r S 0+ . 32. For the PDE in Problem 29, assume that the following boundary conditions are imposed: u10, y, t2 = u1a, y, t2 = 0 ; 0 … y … b , t Ú 0 , 0u 0u 1x, 0, t2 = 1x, b, t2 = 0 ; 0 … x … a , t Ú 0 . 0y 0y Show that a nontrivial solution of the form u1x, y, t2 = X1x2Y1y2T1t2 must satisfy the boundary conditions X102 = X1a2 = 0 , Y′102 = Y′1b2 = 0 . 33. When the temperature in a wire reaches a steady state, that is, when u depends only on x, then u1x2 satisfies Laplace’s equation 02u>0x2 = 0. (a) Find the steady-state solution when the ends of the wire are kept at a constant temperature of 50°C, that is, when u102 = u1L2 = 50. (b) Find the steady-state solution when one end of the wire is kept at 10°C, while the other is kept at 40°C, that is, when u102 = 10 and u1L2 = 40.

r 2R″1r2 + rR′1r2 - 1r 2l + m2R1r2 = 0 , where m, l are constants.

10.3 Fourier Series While analyzing heat flow and vibrating strings in the previous section, we encountered the problem of expressing a function in a trigonometric series [compare equations (12) and (25) in Section 10.2]. In the next two sections, we discuss the theory of Fourier series, which deals with trigonometric series expansions. First, however, we review some function properties that are particularly relevant to this study: piecewise continuity, periodicity, and even and odd symmetry. In Section 7.2 we defined a piecewise continuous function on 3a, b4 as a function f that is continuous at every point in 3a, b4, except possibly for a finite number of points at which f has a jump discontinuity. Such functions are necessarily integrable over any finite interval on which they are piecewise continuous.

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572

Chapter 10

Partial Differential Equations

Recall also that a function is periodic of period T if f1x + T2 = f1x2 for all x in the domain of f. The smallest positive value of T is called the fundamental period. The trigonometric functions sin x and cos x are examples of periodic functions with fundamental period 2p and tan x is periodic with fundamental period p. A constant function is a periodic function with arbitrary period T. Two symmetry properties of functions will be useful in the study of Fourier series. A function f that satisfies f1 -x2 = f1x2 for all x in the domain of f has a graph that is symmetric with respect to the y-axis [see Figure 10.4(a)]. We say that such a function is even. A function f that satisfies f1 -x2 = -f1x2 for all x in the domain of f has a graph that is symmetric with respect to the origin [see Figure 10.4(b)]. It is said to be an odd function. The functions 1, x2, x4, . . . are examples of even functions, while the functions x, x3, x5, . . . are odd. The trigonometric functions sin x and tan x are odd functions and cos x is an even function. Example 1

Determine whether the given function is even, odd, or neither. (a) f1x2 = 21 + x2

Solution

(c) h1x2 = ex

(b) g1x2 = x1>3 - sin x

(a) Since f1 -x2 = 21 + 1 -x2 2 = 21 + x2 = f1x2, then f1x2 is an even function. (b) Because g1 -x2 = 1 -x2 1>3 - sin 1 -x2 = -x1>3 + sin x = -1x1>3 - sin x2 = -g1x2, it follows that g1x2 is an odd function. (c) Here h1 -x2 = e-x. Since e-x = ex only when x = 0 and e-x is never equal to -ex, then h1x2 is neither an even nor an odd function. ◆ Knowing that a function is even or odd can be useful in evaluating definite integrals. The following result, illustrated in Figure 10.4, is a straightforward consequence of the definition of the definite integral.

Properties of Symmetric Functions Theorem 1.

If f is an even piecewise continuous function on 3 -a, a4, then

a

a

f1x2 dx = 2

(1)

f1x2 dx . L-a L0 If f is an odd piecewise continuous function on 3 -a, a4, then a

(2)

L-a

f1x2 dx = 0 .

y

y f(x)

f(x) A

-a

A

A

a

-a

x

a

x

-A

(a) Figure 10.4 (a) Even function

(b) a 1-a

f = A+A =

a 2 10

a

f (b) Odd function 1-a f = A - A = 0

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Section 10.3

573

Fourier Series

The next example deals with certain integrals that are crucial in Fourier series. Example 2

Evaluate the following integrals when m and n are nonnegative integers: L

(a)

L-L

sin

L

(c) Solution

L-L

L

mpx npx cos dx . L L

cos

(b)

L-L

sin

mpx npx sin dx . L L

mpx npx cos dx . L L

The even and odd functions occurring in the integrands are sketched in Figure 10.5. The given integrals are easily evaluated by invoking the trigonometric formula for products of sines and cosines: 1m - n2px 1 1m + n2px mpx npx 1 cos = sin + sin , L L 2 L 2 L 1m - n2px 1 1m + n2px mpx npx 1 sin sin = cos - cos , L L 2 L 2 L 1m - n2px 1 1m + n2px mpx npx 1 cos cos = cos + cos . L L 2 L 2 L sin

1

1 -L

L

x

cos 0 x L

sin 0 x L

-L

L

-1

-1

1

1 x

cos 1 x L

sin 1 x L

x

-1

-1

1

1 x

cos 2 x L

sin 2 x L

-1

x

x -1

1

1 x

cos 3 x L

sin 3 x L

-1

x -1

Figure 10.5 The sinusoids

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574

Chapter 10

Partial Differential Equations

Thus if m ≠ n, each of the integrals calls for the (signed) area under an oscillating sinusoid over a whole number of periods, and is therefore zero. In fact, the only way to avoid these “oscillators” is to take m = n; whence cos3 1m - n2px>L4 = cos 0 = 1 and subtends an area of 2L, while sin3 1m - n2px>L4 = sin 0 = 0 subtending zero area (again). But note that if m and n are both zero, then cos3 1m + n2px>L4 = cos 0 also subtends area 2L. Restoring the factors of 1>2, we summarize with L

(3)

L-L

sin

mpx npx cos dx = 0 , L L

sin

0, mpx npx sin dx = e L, L L

L

(4)

L-L

0, mpx npx cos cos dx = • L , L L L-L 2L , L

(5)

m≠n, m = n, m≠n, m = n≠0, ◆ m = n = 0.

Equations (3)–(5) express an orthogonality condition† satisfied by the set of trigonometric functions 51 = cos 0x, cos x, sin x, cos 2x, sin 2x, c6, where L = p. We will say more about this later in this section. It is easy to verify that if each of the functions f1, c, fn is periodic of period T, then so is any linear combination c1 f1 1x2 + g + cn fn 1x2 .

For example, the sum 7 + 3 cos px - 8 sin px + 4 cos 2px - 6 sin 2px has period 2, since each term has period 2. Furthermore, if the infinite series ∞ a0 npx npx + a a an cos + bn sin b 2 n=1 L L

consisting of 2L-periodic functions converges for all x, then the function to which it converges will be periodic of period 2L. Just as we can associate a Taylor series with a function that has derivatives of all orders at a fixed point, we can identify a particular trigonometric series with a piecewise continuous function. To illustrate this, let’s assume that f1x2 has the series expansion‡ (6)

f1x2 =

∞ a0 npx npx + a e an cos + bn sin f, 2 n=1 L L

where the an’s and bn’s are constants. (Necessarily, f has period 2L.) To determine the coefficients a0, a1, b1, a2, b2, . . . , we proceed as follows. Let’s integrate f1x2 from -L to L, assuming that we can integrate term by term: L

L-L

L

f1x2 dx =

L L ∞ ∞ a0 npx npx dx + a an cos dx + a bn sin dx . 2 L L n=1 n=1 L-L L-L L-L

†

This nomenclature is suggested by the fact that the formulas for the Riemann sums approximating the integrals (3)–(5) look like dot products (Section 9.1, page 496) of higher-dimensional vectors. In most calculus texts, it is shown that the dot product of orthogonal vectors in two dimensions is zero. The choice of constant a0 >2 instead of just a0 will be motivated shortly.

‡

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.3

Fourier Series

575

The (signed) area under the “oscillators” is zero. Hence, L

L-L

L

f1x2 dx =

a0 dx = a0 L , L-L 2

and so L

a0 =

1 f1x2 dx . L L-L

(Notice that a0 >2 is the average value of f over one period 2L.) Next, to find the coefficient am when m Ú 1, we multiply (6) by cos1mpx>L2 and integrate: L

(7)

L-L

f1x2 cos

L L ∞ a0 mpx mpx npx mpx dx = dx + a an cos dx cos cos L 2 L-L L L L n=1 L-L

L

∞

+ a bn n=1

L-L

sin

npx mpx cos dx . L L

The orthogonality conditions (3)–(5) render the integrals on the right-hand side quite immediately. We have already observed that L

L-L

cos

mpx dx = 0 , L

mÚ1,

and, by formulas (3) and (5), we have L

L-L

sin

npx mpx cos dx = 0 , L L

L

L-L

cos

npx mpx 0, cos dx = e L, L L

n≠m, n = m.

Hence, in (7) we see that only one term on the right-hand side survives the integration: L

L-L

f1x2 cos

mpx dx = amL . L

Thus, we have a formula for the coefficient am: L

am =

1 mpx f1x2 cos dx . L L-L L

Similarly, multiplying (6) by sin1mpx>L2 and integrating yields L

L-L

f1x2 sin

mpx dx = bmL L

so that the formula for bm is L

bm =

1 mpx f1x2 sin dx . L L-L L

Motivated by the above computations, we now make the following definition.

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576

Chapter 10

Partial Differential Equations

Fourier Series Definition 1. Let f be a piecewise continuous function on the interval 3 -L, L4. The Fourier series† of f is the trigonometric series (8)

f1x2 ∼

∞ a0 npx npx + a e an cos + bn sin f, 2 n=1 L L

where the an’s and bn’s are given by the formulas‡ L

(9)

an =

1 npx f1x2 cos dx , L L-L L

(10)

bn =

1 npx f1x2 sin dx , L L-L L

n = 0, 1, 2, c,

L

n = 1, 2, 3, c.

Formulas (9) and (10) are called the Euler-Fourier formulas. We use the symbol ∼ in (8) to remind us that this series is associated with f1x2 but may not converge to f1x2. We will return to the question of convergence later in this section. Let’s first consider a few examples of Fourier series. Example 3

Compute the Fourier series for f1x2 = e

Solution

-p 6 x 6 0 , 06x6p.

0, x,

Here L = p. Using formulas (9) and (10), we have p

p

p

a0 =

1 x2 1 2 = p, f1x2 dx = x dx = p L-p p L0 2p 0 2

an =

1 1 f1x2 cos nx dx = x cos nx dx p L-p p L0

p

p

pn

pn

1 3cos u + u sin u4 2 pn2 0

=

1 pn2 L0

=

1 1 1cos np - 12 = 3 1 -12 n - 14 , pn2 pn2

u cos u du =

p

bn = = =

n = 1, 2, 3, c ,

p

1 1 f1x2 sin nx dx = x sin nx dx p L-p p L0 1 pn2 L0

pn

pn

u sin u du =

1 -12 n + 1 -cos np = , n n

1 3sin u - u cos u4 2 pn2 0 n = 1, 2, 3, c .

†

Historical Footnote: Joseph B. J. Fourier (1768–1830) developed his series for solving heat flow problems. Lagrange expressed doubts about the validity of the representation, but Dirichlet devised conditions that ensured its convergence. Note that the notational choice of a0 >2 in equation (6) made it unnecessary to insert an extra formula for a0 in (9). Notice that f1x2 need not be defined for every x in 3 - L, L4; we need only that the integrals in (9) and (10) exist.

‡

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.3

Fourier Series

577

4 f(x)

3 4 terms

9 terms 2 1

-8

-6

-4

-2

0 -1

x 4

2

6

8

Figure 10.6 Partial sums of Fourier series in Example 3

Therefore, f1x2 ∼

(11)

=

∞ 1 -12 n + 1 p 1 + a e 2 3 1 -12 n - 14cos nx + sin nx f n 4 n = 1 pn

1 p 2 1 - e cos x + cos 3x + cos 5x + g f 4 p 9 25 + e sin x -

1 1 sin 2x + sin 3x + g f . 2 3

Some partial sums of this series are displayed in Figure 10.6. ◆ Example 4

Compute the Fourier series for f1x2 = e

Solution

-1 , 1,

-p 6 x 6 0 , 06x6p.

Again, L = p. Notice that f is an odd function. Since the product of an odd function and an even function is odd (see Problem 7), f1x2cos nx is also an odd function. Thus, p

1 f1x2cos nx dx = 0 , n = 0, 1, 2, c . p L-p Furthermore, f1x2 sin nx is the product of two odd functions and therefore is even, so an =

p

bn = =

p 2 -cos nx 2 2 1 1 -12 n c d = c d , p p n n n 0

0, = • 4 , pn Thus, (12)

p

1 2 f1x2sin nx dx = sin nx dx p L-p p L0

f1x2 ∼

n = 1, 2, 3, c ,

n even , n odd .

4 2 ∞ 31 - 1 -12 n 4 1 1 sin nx = c sin x + sin 3x + sin 5x + g d . a pn=1 p n 3 5

Some partial sums of (12) are sketched in Figure 10.7 on page 578. ◆

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578

Chapter 10

Partial Differential Equations

2

2 terms

9 terms f(x)

1

-8

-6

-4

x

0

-2

2

4

6

8

-2 Figure 10.7 Partial sums of Fourier series in Example 4

In Example 4 the odd function f has a Fourier series consisting only of sine functions. It is easy to see that, in general, if f is any odd function, then its Fourier series consists only of sine terms. Example 5 Solution

Compute the Fourier series for f1x2 = 0 x 0 , -1 6 x 6 1. Here L = 1. Since f is an even function, f1x2sin1npx2 is an odd function. Therefore, 1

f1x2 sin1npx2 dx = 0 , n = 1, 2, 3, c . L-1 Since f1x2 cos1npx2 is an even function, we have bn =

1

a0 =

L-1

1

f1x2 dx = 2

L0

x dx = x2 2

L-1

= 1, 0

1

an =

1

1

f1x2 cos1npx2 dx = 2 pn

L0

x cos1npx2 dx pn

=

2 p2n2 L0

=

2 3 1 -12 n - 14 , p2n2

u cos u du =

2 2 3cos u + u sin u4 2 = 2 2 1cos np - 12 2 2 pn pn 0 n = 1, 2, 3, c .

Therefore, (13)

f1x2 ∼ =

∞ 1 2 + a 2 2 3 1 -12 n - 14cos1npx2 2 n=1 p n

1 1 1 4 e cos1px2 + cos13px2 + cos15px2 + g f . 2 p2 9 25

Partial sums for (13) are displayed in Figure 10.8. ◆ 9 terms

2 terms

-2

1

-1

0

x 1

2

Figure 10.8 Partial sums of Fourier series in Example 5

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Section 10.3

Fourier Series

579

Notice that the even function f of Example 5 has a Fourier series consisting only of cosine functions and the constant function 1 = cos10px2. In general, if f is an even function, then its Fourier series consists only of cosine functions [including cos10px2].

Orthogonal Expansions

Fourier series are examples of orthogonal expansions.† A set of functions 5fn 1x26n∞= 1 is said to be an orthogonal system or just orthogonal with respect to the nonnegative weight function w1x2 on the interval 3a, b4 if b

(14)

La

fm 1x2 fn 1x2w1x2 dx = 0 ,

whenever

m≠n.

As we have seen, the set of trigonometric functions (15)

51, cos x, sin x, cos 2x, sin 2x, c6

(16)

‘f‘ J c

is orthogonal on 3 -p, p4 with respect to the weight function w1x2 K 1. If we define the norm of f as b

La

f 2 1x2 w1x2 dx d

1>2

,

then we say that a set of functions 5fn 1x26n∞= 1 1or 5fn 1x26 nN= 1 2 is an orthonormal system with respect to w1x2 if (14) holds and also ‘ fn ‘ = 1 for each n. Equivalently, we say the set is an orthonormal system if b

(17)

fm 1x2 fn 1x2w1x2 dx = e

0, 1,

m≠n, m = n.

La We can always obtain an orthonormal system from an orthogonal system just by dividing each function by its norm. In particular, since p

L-p

p

cos2nx dx =

L-p

sin2nx dx = p ,

n = 1, 2, 3, c

and p

L-p

1 dx = 2p ,

then the orthogonal system (15) gives rise on 3 -p, p4 to the orthonormal system 5 12p2 -1>2, p -1>2 cos x, p -1>2 sin x, p -1>2 cos 2x, p -1>2 sin 2x, c6 .

If 5fn 1x26n∞= 1 is an orthogonal system with respect to w1x2 on 3a, b4, we might ask if we can expand a function f1x2 in terms of these functions; that is, can we express f in the form (18)

f1x2 = c1 f1 1x2 + c2 f2 1x2 + c3 f3 1x2 + g

for a suitable choice of constants c1, c2, c? Such an expansion is called an orthogonal expansion, or a generalized Fourier series.

†

Orthogonality is also discussed in Section 8.8 on page 474.

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580

Chapter 10

Partial Differential Equations

To determine the constants in (18), we can proceed as we did in deriving Euler’s formulas for the coefficients of a Fourier series, this time using the orthogonality of the system. Presuming the representation (18) is valid, we multiply by fm 1x2w1x2 and integrate to obtain b

(19)

La

f1x2 fm 1x2 w1x2 dx = c1

b

La

f1 1x2 fm 1x2 w1x2 dx + c2 b

∞

= a cn n=1

La

b

La

f2 1x2 fm 1x2 w1x2 dx + g

fn 1x2 fm 1x2 w1x2 dx .

(Here we have also assumed that we can integrate term by term.) Because the system is orthogonal with respect to w1x2, every integral on the right-hand side of (19) is zero except when n = m. Solving for cm gives b

(20)

cm =

La

b

f1x2 fm 1x2 w1x2 dx b

La

2 fm 1x2 w1x2 dx

=

Lc

f1x2 fm 1x2 w1x2 dx ‘ fm ‘ 2

,

n = 1, 2, 3, c.

The derivation of the formula for cm was only formal, since the question of the con∞ vergence of the expansion in (18) was not answered. If the series g n = 1 cn fn 1x2 converges uniformly to f1x2 on 3a, b4, then each step can be justified, and indeed, the coefficients are given by formula (20). The notion of uniform convergence is discussed in the next subsection and in Section 13.2.†

Convergence of Fourier Series Let’s turn to the question of the convergence of a Fourier series. For Example 5 it is possible to use a comparison or limit comparison test to show that the series is ∞ absolutely dominated by a p-series of the form g n = 1 1>n2, which converges. However, this is much harder to do in Example 4, since the terms go to zero like 1>n. Matters can be even worse, since there exist Fourier series that diverge.‡ We state two theorems that deal with the convergence of a Fourier series and two dealing with the properties of termwise differentiation and integration. For proofs of these results, see Partial Differential Equations of Mathematical Physics, 2nd ed., by Tyn Myint-U (Elsevier North Holland, Inc., New York, 1983), Chapter 5; Advanced Calculus with Applications, by N. J. DeLillo (Macmillan, New York, 1982), Chapter 9; or an advanced text on the theory of Fourier series. Before proceeding, we need a notation for the left- and right-hand limits of a function. Let f1x + 2J lim+ f1x + h2 hS0

and

f1x - 2 J lim+ f1x - h2 . hS0

We now present the fundamental pointwise convergence theorem for Fourier series. While reading it, keep in mind Figures 10.6-10.8 . †

All references to Chapters 11–13 refer to the expanded text, Fundamentals of Differential Equations and Boundary Value Problems, 7th ed. ∞ sin nx ‡ In fact, there are trigonometric series that converge but are not Fourier series; an example is a . ln 1n + 12 n=1

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.3

Fourier Series

581

Pointwise Convergence of Fourier Series Theorem 2. 1 -L, L2

If f and f ′ are piecewise continuous on 3 -L, L4, then for any x in

∞ a0 npx npx 1 + a e an cos + bn sin f = 3f1x + 2 + f1x - 24 , 2 n=1 L L 2

(21)

where the an’s and bn’s are given by the Euler-Fourier formulas (9) and (10). For 1 x = {L, the series converges to 3f1 -L+ 2 + f1L- 24 .† 2 In other words, when f and f′ are piecewise continuous on 3 -L, L4, the Fourier series converges to f1x2 whenever f is continuous at x and converges to the average of the left- and right-hand limits at points where f is discontinuous. Observe that the left-hand side of (21) is periodic of period 2L. This means that if we extend f1x2 from the interval 1 -L, L2 to the entire real line using 2L-periodicity, then equation (21) holds for all x for the 2L-periodic extension of f1x2. Example 6

To which function does the Fourier series for f1x2 = e

-1 , 1,

-p 6 x 6 0 , 06x6p,

converge? Solution

In Example 4 we found that the Fourier series for f1x2 is given by (12), and in Figure 10.7, we sketched the graphs of two of its partial sums. Now f1x2 and f ′1x2 are piecewise continuous in 3 -p, p4. Moreover, f is continuous except at x = 0. Thus, by Theorem 2, the Fourier series of f in (12) converges to the 2p-periodic function g1x2, where g1x2 = f1x2 = -1 for -p 6 x 6 0, g1x2 = f1x2 = 1 for 0 6 x 6 p, g102 = 3f10+ 2 + f10- 24 >2 = 0, and at {p we have g1 {p2 = 3f1 -p + 2 + f1p - 24 >2 = 1 -1 + 12 >2 = 0. The graph of g1x2 is given in Figure 10.9. ◆ y

1

-

x 0

2 -1

Figure 10.9 The limit function of the Fourier series for f1x2 = e

-1 , 1,

-p 6 x 6 0 , 06x6p

†

From formula (21), we see that it doesn’t matter how we define f1x2 at its points of discontinuity, since only the left-and right-hand limits are involved. The derivative f ′1x2, of course, is undefined at such points.

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582

Chapter 10

Partial Differential Equations

When f is a 2L-periodic function that is continuous on 1 - ∞, ∞ 2 and has a piecewise continuous derivative, its Fourier series not only converges at each point—it converges uniformly on 1 - ∞, ∞ 2. This means that for any prescribed tolerance e 7 0, the graph of the partial sum sN 1x2 J

N a0 npx npx + a e an cos + bn sin f 2 n=1 L L

will, for all N sufficiently large, lie in an e-corridor about the graph of f on 1 - ∞, ∞ 2 (see Figure 10.10). The property of uniform convergence of Fourier series is particularly helpful when one needs to verify that a formal solution to a partial differential equation is an actual (genuine) solution.

f (x) + sN (x)

f (x)

f (x) – x Figure 10.10 An e-corridor about f

Uniform Convergence of Fourier Series Theorem 3. Let f be a continuous function on 1 - ∞, ∞ 2 and periodic of period 2L. If f ′ is piecewise continuous on 3 -L, L4, then the Fourier series for f converges uniformly to f on 3 -L, L4 and hence on any interval. That is, for each e 7 0, there exists an integer N0 (that depends on e) such that ` f1x2 - c

N a0 npx npx + a e an cos + bn sin fd ` 6e, 2 n=1 L L

for all N Ú N0, and all x ∈ 1 - ∞, ∞ 2.

In Example 5 we obtained the Fourier series expansion given in (13) for f1x2 = 0 x 0 , -1 6 x 6 1. Since g1x2, the periodic extension of f1x2 (see Figure 10.11 on page 583) is continuous on 1 - ∞, ∞ 2 and f ′1x2 = e

-1 ,

-1 6 x 6 0 ,

1,

06x61,

is piecewise continuous on 3 -1, 14, the Fourier series expansion (13) converges uniformly to 0 x 0 on 3 -1, 14. Compare Figure 10.8 on page 578. The term-by-term differentiation of a Fourier series is not always permissible. For example, the Fourier series for f1x2 = x, -p 6 x 6 p (see Problem 9), is ∞

(22)

f1x2 ∼ 2 a 1 -12 n + 1 n=1

sin nx , n

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.3

Fourier Series

583

which converges for all x, whereas its derived series ∞

2 a 1 -12 n+1 cos nx n=1

diverges for every x. The following theorem gives sufficient conditions for using termwise differentiation. g(x)

1

-4

-2

0

2

4

x

Figure 10.11 Periodic extension of f1x2 = 0 x 0 , - 1 6 x 6 1

Differentiation of Fourier Series Theorem 4. Let f1x2 be continuous on 1 - ∞, ∞ 2 and 2L-periodic. Let f ′1x2 and f ″1x2 be piecewise continuous on 3 -L, L4. Then, the Fourier series of f ′1x2 can be obtained from the Fourier series for f 1x2 by termwise differentiation. In particular, if f 1x2 =

∞ a0 npx npx + a e an cos + bn sin f, 2 n=1 L L

then ∞ pn npx npx f ′1x2 ∼ a e -an sin + bn cos f. L L n=1 L

Notice that Theorem 4 does not apply to the example function f1x2 and its Fourier series expansion shown in (22), since the 2p-periodic extension of this f1x2 fails to be continuous on 1 - ∞, ∞ 2. Termwise integration of a Fourier series is permissible under much weaker conditions.

Integration of Fourier Series Theorem 5. f1x2 ∼

Let f1x2 be piecewise continuous on 3 -L, L4 with Fourier series ∞ a0 npx npx + a e an cos + bn sin f. 2 n=1 L L

Then, for any x in 3 -L, L4, x

x

x ∞ a0 npt npt f1t2 dt = dt + a e an cos + bn sin f dt . 2 L L n = 1 L-L L-L L-L

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584

Chapter 10

Partial Differential Equations

A final note on the convergence issue: The question as to under what conditions the Fourier series converges has led to a tremendous amount of beautiful mathematics. As a practical matter, however, time and economics will often dictate that you can afford to compute only a few terms of an eigenfunction series for your particular partial differential equation. Convergence fades into the background, and you need to know how to “do the best with what you’ve got.” So it is extremely gratifying to know that even if you’re going to use only part of an eigenfunction expansion, the Fourier coefficients (9)–(10) are still the best choice; Problem 37 demonstrates that every partial sum of a Fourier series outperforms any comparable superposition of the trigometric functions, in terms of mean-square approximation.

10.3

EXERCISES

In Problems 1– 6, determine whether the given function is even, odd, or neither. 1. f1x2 = x3 + sin 2x 2. f1x2 = sin2x 2 -1>2 4. f1x2 = sin1x + 12 3. f1x2 = 11 - x 2 5. f1x2 = e-x cos 3x 6. f1x2 = x1>5 cos x2 7. Prove the following properties: (a) If f and g are even functions, then so is the product fg. (b) If f and g are odd functions, then fg is an even function. (c) If f is an even function and g is an odd function, then fg is an odd function. 8. Verify formula (5). 3Hint: Use the identity 2 cos A cos B = cos1A + B2 + cos1A - B2.4 In Problems 9–16, compute the Fourier series for the given function f on the specified interval. Use a computer or graphing calculator to plot a few partial sums of the Fourier series. 9. f1x2 = x , - p 6 x 6 p 10. f1x2 = 0 x 0 ,

-p 6 x 6 p

11. f1x2 = e

1,

-2 6 x 6 0 ,

x,

06x62

12. f1x2 = e

0,

06x6p

x ,

13. f1x2 = x2 , 14. f1x2 = e

-p 6 x 6 0 ,

2

-1 6 x 6 1 06x6p,

x, x+p,

15. f1x2 = e , x

-p 6 x 6 0

-p 6 x 6 p

0, -1 , 16. f1x2 = μ 1, 0,

- p 6 x 6 -p>2 , - p>2 6 x 6 0 , 0 6 x 6 p>2 , p>2 6 x 6 p

In Problems 17–24, determine the function to which the Fourier series for f1x2, given in the indicated problem, converges. 17. Problem 9 18. Problem 10 19. Problem 11 20. Problem 12 21. Problem 13 22. Problem 14 23. Problem 15 24. Problem 16 25. Find the functions represented by the series obtained by the termwise integration of the given series from -p to x. ∞

(a) 2 a

n=1

(b)

1 -12 n+1 n

sin nx ∼ x ,

-p 6 x 6 p

4 ∞ sin12n + 12x ∼ f1x2 , p na 12n + 12 =0 f1x2 = e

-1 ,

-p 6 x 6 0 ,

1, 06x6p 26. Show that the set of functions p p 3p 3p e cos x, sin x, cos x, sin x, c , 2 2 2 2 12n - 12p 12n - 12p x, sin x, c f cos 2 2 is an orthonormal system on 3 - 1, 14 with respect to the weight function w1x2 K 1. 27. Find the orthogonal expansion (generalized Fourier series) for 0, -1 6 x 6 0 , f1x2 = e 1, 06x61, in terms of the orthonormal system of Problem 26. 28. (a) Show that the function f1x2 = x2 has the Fourier series, on -p 6 x 6 p, ∞ 1 - 12 n p2 f1x2 ∼ +4 a cos nx . 3 n2 n=1

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Section 10.3

Fourier Series

585

(b) Use the result of part (a) and Theorem 2 to show that ∞ 1 -12 n+1 p2 = . a 2 12 n n=1

particular, show that the inner product of two functions defined by

(c) Use the result of part (a) and Theorem 2 to show that

(23)

∞

8f, g9 J

2

1 p a n2 = 6 . n=1 29. In Section 8.8, it was shown that the Legendre polynomials Pn 1x2 are orthogonal on the interval 3 -1, 14 with respect to the weight function w1x2 K 1. Using the fact that the first three Legendre polynomials are P0 1x2 K 1 , P1 1x2 = x , P2 1x2 = 13>22x2 - 11>22 ,

find the first three coefficients in the expansion

f1x2 = c0P0 1x2 + c1P1 1x2 + c2P2 1x2 + g,

where f1x2 is the function -1 , -1 6 x 6 0 , f1x2 J e 1, 06x61. 30. As in Problem 29, find the first three coefficients in the expansion f1x2 = c0P0 1x2 + c1P1 1x2 + c2P2 1x2 + g,

when f1x2 = 0 x 0 , - 1 6 x 6 1. 31. The Hermite polynomials Hn 1x2 are orthogonal on the interval 1 - ∞ , ∞ 2 with respect to the weight function 2 W1x2 = e-x . Verify this fact for the first three Hermite polynomials: H0 1x2 K 1 , H1 1x2 = 2x ,

H2 1x2 = 4x2 - 2 .

32. The Chebyshev (Tchebichef) polynomials Tn 1x2 are orthogonal on the interval 3 - 1, 14 with respect to the weight function w1x2 = 11 - x2 2 -1>2. Verify this fact for the first three Chebyshev polynomials: T0 1x2 K 1 , T1 1x2 = x ,

T2 1x2 = 2x2 - 1 .

33. Let 5fn 1x26 be an orthogonal set of functions on the interval 3a, b4 with respect to the weight function w1x2. Show that they satisfy the Pythagorean property ‘ fm + fn ‘ 2 = ‘ fm ‘ 2 + ‘ fn ‘ 2 if m ≠ n. 34. Norm. The norm of a function ‘ f ‘ is like the length of a vector in Rn. In particular, show that the norm defined in (16) satisfies the following properties associated with length 1assume f and g are continuous and w1x2 7 0 on 3a, b42: (a) ‘ f ‘ Ú 0, and ‘ f ‘ = 0 if and only if f K 0.

(b) ‘cf ‘ = 0 c 0 ‘ f ‘, where c is any real number. (c) ‘ f + g‘ … ‘ f ‘ + ‘g‘ . 35. Inner Product. The integral in the orthogonality condition (14) is like the dot product of two vectors in Rn. In

b

La

f1x2g1x2w1x2 dx ,

where w1x2 is a positive weight function, satisfies the following properties associated with the dot product 1assume f, g, and h are continuous on 3a, b42: (a) 8f + g, h9 = 8f, h9 + 8g, h9 . (b) 8cf, h9 = c8f, h9 , where c is any real number. (c) 8f, g9 = 8g, f9 . 36. Complex Form of the Fourier Series. (a) Using the Euler formula eiu = cos u + i sin u, i = 1- 1, prove that einx + e-inx 2

cos nx =

and

sin nx =

einx - e-inx . 2i

(b) Show that the Fourier series f1x2 ∼

∞ a0 + a 5an cos nx + bn sin nx6 2 n=1 ∞

= c0 + a 5cneinx + c-ne-inx 6 , n=1

where c0 =

a0 an - ibn an + ibn , cn = , c-n = . 2 2 2

(c) Finally, use the results of part (b) to show that ∞

f1x2 ∼ a cneinx , n = -∞

-p 6 x 6 p ,

where p

cn =

1 f1x2 e-inx dx . 2p L-p

37. Least-Squares Approximation Property. The Nth partial sum of the Fourier series f1x2 ∼

∞ a0 + a 5an cos nx + bn sin nx6 2 n=1

gives the best mean-square approximation of f by a trigonometric polynomial. To prove this, let FN 1x2 denote an arbitrary trigonometric polynomial of degree N: FN 1x2 =

N a0 + a {an cos nx + bn sin nx} , 2 n=1

and define p

E J

L-p

3f1x2 - FN 1x24 2 dx ,

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586

Chapter 10

Partial Differential Equations

which is the total square error. Expanding the integrand, we get p

E =

L-p

p

+

p

f 2 1x2 dx - 2

L-p

F 2N 1x2

L-p

f1x2FN 1x2 dx

E - E* = p e

dx .

L-p

F 2N 1x2 dx = p a

a20 2

+ a21 + g + a2N

+ b21 + g + b2N b

L-p

f1x2FN 1x2 dx = p a

a0a0 + a1a1 + g + aNaN 2 + b1b1 + g + bN bN b .

(b) Let E* be the error when we approximate f by the Nth partial sum of its Fourier series, that is, when we choose an = an and bn = bn. Show that p

E* =

L-p

f 2 1x2dx - p a

2

Hence, the Nth partial sum of the Fourier series gives the least total square error, since E Ú E*. 38. Bessel’s Inequality. Use the fact that E*, defined in part (b) of Problem 37, is nonnegative to prove Bessel’s inequality

a02 + a21 + g + a2N 2

p ∞ a20 1 + a 5a2n + b2n 6 … f 2 1x2 dx . p L-p 2 n=1

(If f is piecewise continuous on 3 -p, p4, then we have equality in (24). This result is called Parseval’s identity.) 39. Gibbs Phenomenon.† Josiah Willard Gibbs, who was awarded the first American doctorate in engineering (Yale, 1863), observed that near points of discontinuity of f, the partial sums of the Fourier series for f may overshoot by approximately 9% of the jump, regardless of the number of terms. This is illustrated in Figure 10.12 for the function

+ b21 + g + b2N b .

f1x2 = e

Overshoot

-1 ,

-p 6 x 6 0 ,

1,

06x6p,

Overshoot 1

-3

-2

+ 1a1 - a1 2 2

+ g + 1bN - bN 2 2 f .

(24)

and p

1a0 - a0 2 2

+ g + 1aN - aN 2 2 + 1b1 - b1 2 2

(a) Use the orthogonality of the functions 51, cos x, sin x, cos 2x, c6 to show that p

(c) Using the results of parts (a) and (b), show that E - E* Ú 0, that is, E Ú E*, by proving that

-1

1

0

1

2

3

x

-3

-2

-1

-1

0

1

2

3

x

-1

(a) Graph of f11(x)

(b) Graph of f51(x)

Figure 10.12 Gibbs phenomenon for partial sums of Fourier series

†

Historical Footnote: Actually, H. Wilbraham discovered this phenomenon some 50 years earlier than Gibbs did. It is more appropriately called the Gibbs–Wilbraham phenomenon.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.4

Fourier Cosine and Sine Series

each interval as the place to evaluate the integrand, then p sin1p>2n2 p sin x dx ≈ + g x n p>2n L0

whose Fourier series has the partial sums 1 4 f2n-1 1x2 = c sin x + sin 3x p 3 sin 12n - 12x d . + g+ 12n - 12

+

To verify this for f1x2, proceed as follows: (a) Show that

=

= p1sin x2f 2n-1 1x2 = 4 sin x 3cos x + cos 3x

= + g + cos12n - 12x4 = 2 sin 2nx .

sin312n - 12p>2n4 p n 12n - 12p>2n

p p f2n-1 a b . 2 2n

(d) Use the result of part (c) to show that the overshoot satisfies p

lim f2n-1 a S

p sin x 2 dx . b = n ∞ p L0 x 2n (e) Using the result of part (d) and a numerical integration algorithm (such as Simpson’s rule, Appendix C) for the sine integral function

(b) Infer from part (a) and the figure that the maximum occurs at x = p> 12n2 and has the value f2n-1 a

587

p p 1 3p 4 b = c sin + sin p 2n 2n 3 2n 12n - 12p 1 sin d . + g+ 2n - 1 2n

z

sin x dx , L0 x show that limn S ∞ f2n-1 1p> 12n22 ≈ 1.18. Thus, the approximations overshoot the true value of f10+ 2 = 1 by 0.18, or 9% of the jump from f10- 2 to f10+ 2. Si1z2 J

(c) Show that if one approximates p

sin x dx x L0

using the partition xk J 12k - 121p>2n2, k = 1, 2, . . . , n, ∆xk = p>n and choosing the midpoint of

10.4 Fourier Cosine and Sine Series A typical problem encountered in using separation of variables to solve a partial differential equation is the problem of representing a function defined on some finite interval by a trigonometric series consisting of only sine functions or only cosine functions. For example, in Section 10.2, equation (25), page 569, we needed to express the initial values u1x, 02 = f1x2, 0 6 x 6 L, of the solution to the initial-boundary value problem associated with the problem of a vibrating string as a trigonometric series of the form (1)

∞ npx f1x2 = a an sina b. L n=1

Recalling that the Fourier series for an odd function defined on 3 -L, L4 consists entirely of sine terms, we might try to achieve (1) by artificially extending the function f1x2, 0 6 x 6 L, to the interval 1 -L, L2 in such a way that the extended function is odd. This is accomplished by defining the function fo 1x2 J e

f1x2 , -f1 -x2 ,

06x6L, -L 6 x 6 0 ,

and extending fo 1x2 to all x using 2L-periodicity.† Since fo 1x2 is an odd function, it has a Fourier series consisting entirely of sine terms. Moreover, fo 1x2 is an extension of f1x2, Strictly speaking, we have extended fo 1x2 to all x other than the integer multiples of L. Continuity considerations often suggest appropriate values for the extended functions at some of these points, as well. Figure 10.13 on page 589 illustrates this.

†

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588

Chapter 10

Partial Differential Equations

since fo 1x2 = f1x2 on 10, L2. This extension is called the odd 2L-periodic extension of f1x2. The resulting Fourier series expansion is called a half-range expansion for f1x2, since it represents the function f1x2 on 10, L2, which is half of the interval 1 -L, L2 where it represents fo 1x2. In a similar fashion, we can define the even 2L-periodic extension of f1x2 as the function fe 1x2 J e

06x6L,

f1x2 , f1 -x2 ,

-L 6 x 6 0 ,

x, ∼ f 1x2 = e x+p,

06x6p,

with fe 1x + 2L2 = fe 1x2. To illustrate the various extensions, let’s consider the function f1x2 = x, 0 6 x 6 p. ∼ If we extend f1x2 to the interval 1 -p, p2 using p-periodicity, then the extension f is given by -p 6 x 6 0 ,

∼ ∼ ∼ with f 1x + 2p2 = f 1x2. In Problem 14 of Exercises 10.3, the Fourier series for f 1x2 was found to be ∞ 1 p ∼ f 1x2 ∼ - a sin 2nx , 2 n=1 n

which consists of both odd functions (the sine terms) and even functions (the constant ∼ term), since the p-periodic extension f 1x2 is neither an even nor an odd function. The odd 2p-periodic extension of f1x2 is just fo 1x2 = x, -p 6 x 6 p, which has the Fourier series expansion ∞

(2)

fo 1x2 ∼ 2 a

n=1

1 -12 n+1 sin nx n

(see Problem 9 in Exercises 10.3). Because fo 1x2 = f1x2 on the interval 10, p2, the expansion in (2) is a half-range expansion for f1x2. The even 2p-periodic extension of f1x2 is the function fe 1x2 = 0 x 0 , -p 6 x 6 p, which has the Fourier series expansion (3)

fe 1x2 =

p 4 ∞ 1 cos 12n + 12x a 2 p n = 0 12n + 12 2

(see Problem 10 in Exercises 10.3). ∼ The preceding three extensions, the p-periodic function f 1x2, the odd 2p-periodic function fo 1x2, and the even 2p-periodic function fe 1x2, are natural extensions of f1x2. There are many other ways of extending f1x2. For example, the function g1x2 = e

x,

06x6p,

0,

-p 6 x 6 0 ,

g1x + 2p2 = g1x2 ,

which we studied in Example 3 of Section 10.3, is also an extension of f1x2. However, its Fourier series contains both sine and cosine terms and hence is not as useful as previous extensions. The graphs of these extensions of f1x2 are given in Figure 10.13 on page 589.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.4

~ f (x)

-3

-2

fo(x)

x

-

2

3

-3

-2

-2

x

-

fe(x)

-3

589

Fourier Cosine and Sine Series

2

3

g (x)

x

-

2

3

-3

-2

-

x 2

3

Figure 10.13 Extensions of f1x2 = x, 0 6 x 6 p

The Fourier series expansions for fo 1x2 and fe 1x2 given in (2) and (3) represent f1x2 on the interval 10, p2 1 actually, they equal f1x2 on 10, p2 2 . This motivates the following definitions.

Fourier Cosine and Sine Series Definition 2. Let f1x2 be piecewise continuous on the interval 30, L4. The Fourier cosine series of f1x2 on 30, L4 is (4)

∞ a0 npx + a an cos , 2 n=1 L

where L

2 npx f1x2 cos dx , n = 0, 1, c . L L0 L The Fourier sine series of f1x2 on 30, L4 is (5)

an =

(6)

∞ npx a bn sin L , n=1

where L

(7)

bn =

2 npx f1x2 sin dx , L L0 L

n = 1, 2, c .

The trigonometric series in (4) is just the Fourier series for fe 1x2, the even 2L-periodic extension of f1x2, and that in (6) is the Fourier series for fo 1x2, the odd 2L-periodic extension of f1x2. These are called half-range expansions for f1x2.

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590

Chapter 10

Example 1

Partial Differential Equations

Compute the Fourier sine series for f1x2 = e

Solution

0 … x … p>2 ,

x,

p>2 … x … p .

p-x,

Using formula (7) with L = p, we find p

2 2 bn = f1x2 sin nx dx = p L0 p L0 2 = pn2 L0

pn>2

p>2

p

2 x sin nx dx + 1p - x2 sin nx dx p Lp>2

p

pn

2 u sin u du + 2 sin nx dx u sin u du 2 pn Lp>2 Lpn>2 pn>2

2 = 3sin u - u cos u4 ` pn2 0

-

2 np c cos pn - cos d n 2

pn

-

=

2 3sin u - u cos u4 2 pn2 pn>2

0, 4 np 41 -12 1n-12>2 • sin = 2 , pn2 pn2

n even , n odd .

So on letting n = 2k + 1, we find the Fourier sine series for f1x2 to be (8)

1 -12 k 4 ∞ 4 1 1 sin12k + 12x = e sin x - sin 3x + sin 5x + g f . ◆ a 2 p k = 0 12k + 12 p 9 25

Since, in Example 1, the function f1x2 is continuous and f ′1x2 is piecewise continuous on 10, p2, it follows from Theorem 2 on pointwise convergence of Fourier series that f1x2 =

4 1 1 1 e sin x - sin 3x + sin 5x sin 7x + g f p 9 25 49

for all x in 30, p4 . Let’s return to the problem of heat flow in one dimension.

Example 2

Find the solution to the heat flow problem 0u 02u = 2 2, 0t 0x

(9)

Solution

06x6p,

(10)

u10, t2 = u1p, t2 = 0 ,

(11)

u1x, 02 = e

x, p-x,

t70,

t70,

0 6 x … p>2 , p>2 … x 6 p .

Comparing equation (9) with equation (1) in Section 10.2 (page 563), we see that b = 2 and L = p. Hence, we need only represent u1x, 02 = f1x2 in a Fourier sine series of the form ∞

a cn sin nx .

n=1

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.4

Fourier Cosine and Sine Series

591

In Example 1 we obtained this expansion and showed that cn =

n even ,

0, 4 np 41 -12 1n-12>2 sin = • , 2 pn2 pn2

n odd .

Hence, from equation (15) on page 566, the solution to the heat flow problem (9)–(11) is ∞

u1x, t2 = a cne-2n t sin nx 2

n=1

=

1 -12 k -212k + 122t 4 ∞ e sin12k + 12x 2 p ka = 0 12k + 12

=

4 1 -50t 1 e e-2t sin x - e-18t sin 3x + e sin 5x + g f . p 9 25

A sketch of a partial sum for u1x, t2 is displayed in Figure 10.14. ◆ u

p x

t

Figure 10.14 Partial sum for u1x, t2 in Example 2

10.4

EXERCISES

In Problems 1–4, determine (a) the p-periodic extension ∼ f , (b) the odd 2p-periodic extension fo, and (c) the even 2p-periodic extension fe for the given function f and sketch their graphs. 1. f1x2 = x2 , 0 6 x 6 p 2. f1x2 = sin 2x , 3. f1x2 = e

06x6p

0 , 0 6 x 6 p>2 , 1 , p>2 6 x 6 p

4. f1x2 = p - x ,

06x6p

In Problems 5–10, compute the Fourier sine series for the given function. 5. f1x2 = - 1 , 0 6 x 6 1 6. f1x2 = cos x , 0 6 x 6 p

7. f1x2 = x2 , 0 6 x 6 p 8. f1x2 = p - x , 0 6 x 6 p 9. f1x2 = x - x2 , 0 6 x 6 1 10. f1x2 = ex , 0 6 x 6 1 In Problems 11–16, compute the Fourier cosine series for the given function. 11. f1x2 = p - x , 0 6 x 6 p 12. f1x2 = 1 + x , 0 6 x 6 p 13. f1x2 = ex , 0 6 x 6 1 14. f1x2 = e-x , 0 6 x 6 1 15. f1x2 = sin x , 0 6 x 6 p 16. f1x2 = x - x2 , 0 6 x 6 1

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592

Chapter 10

Partial Differential Equations

In Problems 17–19, for the given f1x2, find the solution to the heat flow problem. 02u 0u 06x6p, t70, = 5 2, 0t 0x u10, t2 = u1p, t2 = 0 , t70, u1x, 02 = f1x2 ,

17. f1x2 = 1 - cos 2x 18. f1x2 = x1p - x2 19. f1x2 = e

-x , x-p,

0 6 x … p>2 , p>2 … x 6 p

06x6p,

10.5 The Heat Equation In Section 10.1 we developed a model for heat flow in an insulated uniform wire whose ends are kept at the constant temperature 0°C. In particular, we found that the temperature u1x, t2 in the wire is governed by the initial-boundary value problem (1) (2) (3)

0u 02u = b 2, 06x6L, 0t 0x u10, t2 = u1L, t2 = 0 , t70, u1x, 02 = f1x2 , 06x6L

t70,

[see equations (7)–(9) in Section 10.1, page 562]. Here equation (2) specifies that the temperature at the ends of the wire is zero, whereas equation (3) specifies the initial temperature distribution. In Section 10.2 we also derived a formal solution to (1) –(3) using separation of variables. There we found the solution to (1)–(3) to have the form (4)

∞ npx 2 u1x, t2 = a cne-b1np>L2 t sin , L n=1

where the cn’s are the coefficients in the Fourier sine series for f1x2: (5)

∞ npx f1x2 = a cn sin . L n=1

In other words, solving (1)–(3) reduces to computing the Fourier sine series for the initial value function f1x2. In this section we discuss heat flow problems where the ends of the wire are insulated or kept at a constant, but nonzero, temperature. (The latter involves nonhomogeneous boundary conditions.) We will also discuss the problem in which a heat source is adding heat to the wire. (This results in a nonhomogeneous partial differential equation.) The problem of heat flow in a rectangular plate is also discussed and leads to the topic of double Fourier series. We conclude this section with a discussion of the existence and uniqueness of solutions to the heat flow problem. In the model of heat flow in a uniform wire, let’s replace the assumption that the ends of the wire are kept at a constant temperature zero and instead assume that the ends of the wire are insulated—that is, no heat flows out (or in) at the ends of the wire. It follows from the principle of heat conduction (see Section 10.1) that the temperature gradient must be zero at these endpoints, that is, 0u 0u 10, t2 = 1L, t2 = 0 , 0x 0x

t70.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.5

The Heat Equation

593

In the next example we obtain the formal solution to the heat flow problem with these boundary conditions. Example 1

Find a formal solution to the heat flow problem governed by the initial-boundary value problem (6) (7) (8)

Solution

02u 0u = b 2, 06x6L, 0t 0x 0u 0u 10, t2 = 1L, t2 = 0 , t 7 0 , 0x 0x u1x, 02 = f1x2 , 0 6 x 6 L .

t70,

Using the method of separation of variables, we first assume that u1x, t2 = X1x2T1t2 . Substituting into equation (6) and separating variables, as was done in Section 10.2 [compare equation (5) on page 564], we get the two equations (9)

X″1x2 + lX1x2 = 0 ,

(10)

T′1t2 + blT1t2 = 0 ,

where l is some constant. The boundary conditions in (7) become X′102T1t2 = 0

and

X′1L2T1t2 = 0 .

For these equations to hold for all t 7 0, either T1t2 K 0, which implies that u1x, t2 K 0, or (11)

X′102 = X′1L2 = 0 .

Combining the boundary conditions in (11) with equation (9) gives the boundary value problem (12)

X″1x2 + lX1x2 = 0 ;

X′102 = X′1L2 = 0 ,

where l can be any constant. To solve for the nontrivial solutions to (12), we propose X1x2 = e rx and form the auxiliary equation r 2 + l = 0. When l 6 0, arguments similar to those used in Section 10.2 show that there are no nontrivial solutions to (12). When l = 0, the auxiliary equation has the repeated root 0 and a general solution to the differential equation is X1x2 = A + Bx . The boundary conditions in (12) reduce to B = 0 with A arbitrary. Thus, for l = 0, the nontrivial solutions to (12) are of the form X1x2 = c0 , where c0 = A is an arbitrary nonzero constant. When l 7 0, the auxiliary equation has the roots r = {i1l. Thus, a general solution to the differential equation in (12) is X1x2 = C1 cos 1l x + C2 sin 1l x . The boundary conditions in (12) lead to the system 1lC2 = 0 , - 1lC1 sin 1l L + 1lC2 cos 1l L = 0 .

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594

Chapter 10

Partial Differential Equations

Hence, C2 = 0 and the system reduces to solving C1 sin 1lL = 0. Since sin1lL = 0 only when 1lL = np, where n is an integer, we obtain a nontrivial solution only when 1l = np>L or l = 1np>L2 2, n = 1, 2, 3, c . Furthermore, the nontrivial solutions (eigenfunctions) Xn corresponding to the eigenvalue l = 1np>L2 2 are given by npx Xn 1x2 = cn cos (13) , L where the cn’s are arbitrary nonzero constants. In fact, formula (13) also holds for n = 0, since l = 0 has the eigenfunctions X0 1x2 = c0. Note that these eigenvalues and eigenfunctions (see Figure 10.15) have the same properties as those on page 565 (which we now recognize as the components of the sine series): homogeneous boundary conditions, eigenvalues clustering at infinity, increasingly oscillatory, orthogonal. Having determined that l = 1np>L2 2, n = 0, 1, 2, c, let’s consider equation (10) for such l: T′1t2 + b1np>L2 2T1t2 = 0 .

Eigenvalue

For n = 0, 1, 2, c, the general solution is

Eigenfunction

Tn 1t2 = bn e-b1np>L2 t , 2

L

0

L

2

L

x x

2

2

4 L

un 1x, t2 = Xn 1x2Tn 1t2 = c cn cos

x

L

2

where the bn’s are arbitrary constants. Combining this with equation (13), we obtain the functions

un 1x, t2 = an e-b1np>L2 t cos 2

L

2

9 L

x

2

L

2

x

L

2

l

npx , L

where an = bncn is, again, an arbitrary constant. If we take an infinite series of these functions, we obtain (14)

16

npx 2 d 3bn e-b1np>L2 t 4 , L

u1x, t2 =

∞ a0 npx 2 + a an e-b1np>L2 t cos , 2 n=1 L

which will be a solution to (6)–(7) provided the series has the proper convergence behavior. Notice that in (14) we have altered the constant term and written it as a0 >2, thus producing the standard form for cosine expansions.

Figure 10.15 Cosine eigenfunctions

Assuming a solution to (6)–(7) is given by the series in (14) and substituting into the initial condition (8), we get (15)

u1x, 02 =

∞ a0 npx + a an cos = f1x2 , 0 6 x 6 L . 2 n=1 L

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Section 10.5

The Heat Equation

595

This means that if we choose the an’s as the coefficients in the Fourier cosine series for f, L

2 npx f1x2 cos dx , n = 0, 1, 2, c , L L0 L then u1x, t2 given in (14) will be a formal solution to the heat flow problem (6)–(8). Again, if this expansion converges to a continuous function with continuous second partial derivatives, then the formal solution is an actual solution. ◆ an =

Note that the individual terms in the series (4) and (14) are genuine solutions to the heat equation and the associated boundary conditions (but not the initial condition, of course). 2 These solutions are called the modes of the system. Observe that the time factor e-b1np>L2 t decays faster for the more oscillatory (higher n) modes. This makes sense physically; a temperature pattern of many alternating hot and cold strips arranged closely together will equilibrate faster than a more uniform pattern. See Figure 10.16. A good way of expressing how separation of variables “works” is as follows: The initial temperature profile f1x2 is decomposed into modes through Fourier analysis—equations (5) or (15). Then each mode is matched with a time decay factor—equations (4) or (14)—and it evolves in time accordingly. When the ends of the wire are kept at 0°C or when the ends are insulated, the boundary conditions are said to be homogeneous. But when the ends of the wire are kept at constant temperatures different from zero, that is, (16)

u10, t2 = U1

and

t70,

u1L, t2 = U2 ,

then the boundary conditions are called nonhomogeneous. u

u

x t t

x (a) n = 0

(b) n = 1 u

u

x x t (c) n = 2

t (d) n = 3

Figure 10.16 Modes for equation (14)

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596

Chapter 10

Partial Differential Equations

From our experience with vibration problems in Section 4.9 we expect that the solution to the heat flow problem with nonhomogeneous boundary conditions will consist of a steadystate solution y1x2 that satisfies the nonhomogeneous boundary conditions in (16) plus a transient solution w1x, t2. That is, u1x, t2 = Y1x2 + w1x, t2 , where w1x, t2 and its partial derivatives tend to zero as t S ∞. The function w1x, t2 will then satisfy homogeneous boundary conditions, as illustrated in the next example.

Example 2

Find a formal solution to the heat flow problem governed by the initial-boundary value problem

(18)

0u 02u = b 2, 06x6L, 0t 0x u10, t2 = U1 , u1L, t2 = U2 ,

(19)

u1x, 02 = f1x2 , 0 6 x 6 L .

(17)

Solution

t70, t70,

Let’s assume that the solution u1x, t2 consists of a steady-state solution y1x2 and a transient solution w1x, t2—that is, (20)

u1x, t2 = y1x2 + w1x, t2 .

Substituting for u1x, t2 in equations (17)–(19) leads to

(22)

0u 0w 02w = = by″1x2 + b 2 , 0 6 x 6 L , t70, 0t 0t 0x y102 + w10, t2 = U1 , y1L2 + w1L, t2 = U2 , t70,

(23)

y1x2 + w1x, 02 = f1x2 , 0 6 x 6 L .

(21)

If we allow t S ∞ in (21)–(22), assuming that w1x, t2 is a transient solution, we obtain the steady-state boundary value problem y″1x2 = 0 , y102 = U1 ,

06x6L, y1L2 = U2 .

Solving for y, we obtain y1x2 = Ax + B, and choosing A and B so that the boundary conditions are satisfied yields (24)

y1x2 = U1 +

1U2 - U1 2x L

as the steady-state solution. With this choice for y1x2, the initial-boundary value problem (21)–(23) reduces to the following initial-boundary value problem for w1x, t2:

(26)

0w 02w = b 2 , 06x6L, t70, 0t 0x t70, w10, t2 = w1L, t2 = 0 ,

(27)

w1x, 02 = f1x2 - U1 -

(25)

1U2 - U1 2x , L

06x6L.

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Section 10.5

The Heat Equation

597

Recall that a formal solution to (25)–(27) is given by equation (4). Hence, ∞ npx 2 , w1x, t2 = a cn e-b1np>L2 t sin L n=1

where the cn’s are the coefficients of the Fourier sine series expansion f1x2 - U1 -

∞ 1U2 - U1 2x npx = a cn sin . L L n=1

Therefore, the formal solution to (17)–(19) is (28)

u1x, t2 = U1 +

1U2 - U1 2x L

∞ npx 2 + a cn e-b1np>L2 t sin , L n=1

with cn =

L 1U2 - U1 2x npx 2 c f1x2 - U1 d sin dx . ◆ L L0 L L

In the next example, we consider the heat flow problem when a heat source is present but is independent of time. (Recall the derivation in Section 10.1.)

Example 3

Find a formal solution to the heat flow problem governed by the initial-boundary value problem (29)

Solution

0u 02u = b 2 + P1x2 , 0t 0x

(30)

u10, t2 = U1 ,

(31)

u1x, 02 = f1x2 ,

06x6L,

u1L, t2 = U2 ,

t70,

t70,

06x6L.

We begin by assuming that the solution consists of a steady-state solution y1x2 and a transient solution w1x, t2, namely, u1x, t2 = y1x2 + w1x, t2 , where w1x, t2 and its partial derivatives tend to zero as t S ∞. Substituting for u1x, t2 in (29)–(31) yields (32)

0u 0w 02w = = by″1x2 + b 2 + P1x2 , 0t 0t 0x

(33)

y102 + w10, t2 = U1 ,

(34)

y1x2 + w1x, 02 = f1x2 ,

06x6L,

y1L2 + w1L, t2 = U2 ,

t70,

t70,

06x6L.

Letting t S ∞ in (32)–(33), we obtain the steady-state boundary value problem y″1x2 = -

1 P1x2 , b

y102 = U1 ,

06x6L,

y1L2 = U2 .

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598

Chapter 10

Partial Differential Equations

The solution to this boundary value problem can be obtained by two integrations using the boundary conditions to determine the constants of integration. The reader can verify that the solution is given by the formula (35)

y1x2 = c U2 - U1 +

L

L0

a

z

x

z

1 x 1 P1s2 ds b dz d + U1 a P1s2 ds b dz . L L0 b L0 L0 b

With this choice for y1x2, we find that the initial-boundary value problem (32)–(34) reduces to the following initial-boundary value problem for w1x, t2: (36)

0w 02w = b 2 , 0t 0x

06x6L,

t70,

(37)

w10, t2 = w1L, t2 = 0 ,

t70,

(38)

w1x, 02 = f1x2 - y1x2 ,

06x6L,

where y1x2 is given by formula (35). As before, the solution to this initial-boundary value problem is (39)

∞ npx 2 w1x, t2 = a cn e-b1np>L2 t sin , L n=1

where the cn’s are determined from the Fourier sine series expansion of f1x2 - y1x2: (40)

∞ npx f1x2 - y1x2 = a cn sin . L n=1

Thus the formal solution to (29)–(31) is given by u1x, t2 = y1x2 + w1x, t2 , where y1x2 is given in (35) and w1x, t2 is prescribed by (39)–(40). ◆ The method of separation of variables is also applicable to problems in higher dimensions. For example, consider the problem of heat flow in a rectangular plate with sides x = 0, x = L, y = 0, and y = W. If the two sides y = 0, y = W are kept at a constant temperature of 0°C and the two sides x = 0, x = L are perfectly insulated, then heat flow is governed by the initial-boundary value problem in the following example (see Figure 10.17). y

W

08

=0

=0

08

L

x

Figure 10.17 Plate with insulated sides

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Section 10.5

Example 4

Solution

The Heat Equation

599

Find a formal solution u1x, y, t2 to the initial-boundary value problem (41)

0u 02u 02u = be 2 + 2f , 0t 0x 0y

(42)

0u 0u 10, y, t2 = 1L, y, t2 = 0 , 0x 0x

(43)

u1x, 0, t2 = u1x, W, t2 = 0 ,

(44)

u1x, y, 02 = f1x, y2 ,

06x6L,

06y6W,

06y6W, 06x6L,

06x6L,

t70,

t70, t70,

06y6W.

If we assume a solution of the form u1x, y, t2 = V1x, y2T1t2, then equation (41) separates into the two equations (45)

T′1t2 + blT1t2 = 0 ,

(46)

02V 02V 1x, y2 + 2 1x, y2 + lV1x, y2 = 0 , 2 0x 0y

where l can be any constant. To solve equation (46), we again use separation of variables. Here we assume V1x, y2 = X1x2Y1y2. This allows us to separate equation (46) into the two equations (47) (48)

X″1x2 + mX1x2 = 0 ,

Y ″1y2 + 1l - m2Y1y2 = 0 ,

where μ can be any constant (see Problem 29 in Exercises 10.2 on page 571). To solve for X1x2, we observe that the boundary conditions in (42), in terms of the separated variables, become X′102Y1y2T1t2 = X′1L2Y1y2T1t2 = 0 ;

06y6W,

t70.

Hence, in order to get a nontrivial solution, we must have (49)

X′102 = X′1L2 = 0 .

The boundary value problem for X given in equations (47) and (49) was solved in Example 1 [compare equations (12) and (13)]. Here m = 1mp>L2 2, m = 0, 1, 2, . . . , and mpx Xm 1x2 = cm cos , L where the cm’s are arbitrary. (We forgo the “c0 >2” convention here.) To solve for Y1y2, we first observe that the boundary conditions in (43) become (50)

Y102 = Y1W2 = 0 .

Next, substituting m = 1mp>L2 2 into equation (48) yields Y ″1y2 + 1 l - 1mp>L2 2 2 Y1y2 = 0 ,

which we can rewrite as (51)

Y ″1y2 + EY1y2 = 0 ,

where E = l - 1mp>L2 2. The boundary value problem for Y consisting of (50)–(51) has also been solved before. In Section 10.2 [compare equations (7) and (9), pages 564 and 565] we showed that E = 1np>W2 2, n = 1, 2, 3, c , and the nontrivial solutions are given by npy Yn 1y2 = an sin , W

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600

Chapter 10

Partial Differential Equations

where the an’s are arbitrary. Since l = E + 1mp>L2 2, we have l = 1np>W2 2 + 1mp>L2 2 ;

m = 0, 1, 2, . . . ,

n = 1, 2, 3, c .

Substituting l into equation (45), we can solve for T1t2 and obtain Tmn 1t2 = bmn e-1m >L +n >W 2bp t . 2

2

2

2

2

Substituting in for Xm, Yn, and Tmn, we get umn 1x, y, t2 = a cm cos

npy mpx 2 2 2 2 2 b aan sin b 1bmn e-1m >L +n >W 2bp t 2 , L W npy mpx 2 2 2 2 2 umn 1x, y, t2 = amn e-1m >L +n >W 2bp t cos sin , L W where amn J anbmncm 1m = 0, 1, 2, c, n = 1, 2, 3, c2 are arbitrary constants. If we now take a doubly infinite series of such functions, then we obtain the formal series ∞ ∞ npy mpx 2 2 2 2 2 u1x, y, t2 = a a amn e-1m >L +n >W 2bp t cos sin . L W m=0 n=1

(52)

We are now ready to apply the initial conditions (44). Setting t = 0, we obtain (53)

∞ ∞ npy mpx u1x, y, 02 = f1x, y2 = a a amn cos sin . L W m=0 n=1

This is a double Fourier series.† The formulas for the coefficients amn are obtained by exploiting the orthogonality conditions twice. Presuming (53) is valid and permits term-by-term integration, we multiply each side by cos1ppx>L2 sin1qpy>W2 and integrate over x and y: L

W

f1x, y2 cos

L0 L0

∞

qpy ppx sin dy dx L W L

∞

= a a amn m=0 n=1

L0 L0

W

cos

qpy npy ppx mpx sin cos sin dy dx . L W L W

According to the orthogonality conditions, each integral on the right is zero, except when m = p and n = q; thus, L

W

L0 L0

f1x, y2 cos

qpy ppx sin dy dx L W LW

apq , 4 qpy ppx 2 cos dx sin dy = μ L W LW L0 L0 apq , 2 L

= apq

W

p≠0,

2

p = 0.

Hence, L

(54)

a0q =

W

qpy 2 f1x, y2sin dy dx , LW L0 L0 W

q = 1, 2, 3, c ,

†

For a discussion of double Fourier series, see Partial Differential Equations of Mathematical Physics, 2nd ed. by Tyn Myint-U (Elsevier North Holland, New York, 1983), Sec. 5.14.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.5

The Heat Equation

601

and for p Ú 1, q Ú 1, L

(55)

apq =

W

qpy ppx 4 f1x, y2 cos sin dy dx . LW L0 L0 L W

Finally, the solution to the initial-boundary value problem (41)–(44) is given by equation (52), where the coefficients are prescribed by equations (54) and (55). ◆ We derived formulas (54) and (55) under the assumption that the double series (53) truly converged (uniformly). How can we justify the assumption that a double Fourier series converges? A rough argument goes as follows. If we fix y in f1x, y2, then f1x, y2 presumably has ∞ a convergent cosine series in x: f1x, y2 = a m = 0 dm cos 1mpx>L2. But the coefficients dm depend on the value of the y that we fixed: dm = dm 1y2. So presumably each function dm 1y2 ∞ has a convergent sine series in y: dm 1y2 = a n = 1 amn sin 1npy/W2. Assembling all this, we ∞ ∞ get f1x, y2 = a m = 0 a n = 1 amn cos 1mpx>L2 sin 1npy>W2.

Existence and Uniqueness of Solutions In the examples that we have studied in this section and in Section 10.2, we were able to obtain formal solutions in the sense that we could express the solution in terms of a series expansion consisting of exponentials, sines, and cosines. To prove that these series converge to actual solutions requires results on the convergence of Fourier series and results from real analysis on uniform convergence. We will not go into these details here, but refer the reader to Section 6.5 of the text by Tyn Myint-U (see footnote on page 600) for a proof of the existence of a solution to the heat flow problem discussed in Sections 10.1 and 10.2. (A proof of uniqueness is also given there.) As might be expected, by using Fourier series and the method of separation of variables one can also obtain “solutions” when the initial data are discontinuous, since the formal solutions require only the existence of a convergent Fourier series. This allows one to study idealized problems in which the initial conditions do not agree with the boundary conditions or the initial conditions involve a jump discontinuity. For example, we may assume that initially one half of the wire is at one temperature, whereas the other half is at a different temperature, that is, f1x2 = e

U1 , U2 ,

0 6 x 6 L>2 , L>2 6 x 6 L .

The formal solution that we obtain will make sense for 0 6 x 6 L, t 7 0, but near the points of discontinuity x = 0, L>2, and L, we have to expect behavior like that exhibited in Figure 10.7 of Section 10.3, page 578. The question of the uniqueness of the solution to the heat flow problem can be answered in various ways. One is tempted to argue that the method of separation of variables yields formulas for the solutions and therefore a unique solution. However, this does not exclude the possibility of solutions existing that cannot be obtained by the method of separation of variables. From physical considerations, we know that the peak temperature along a wire does not increase spontaneously if there are no heat sources to drive it up. Indeed, if hot objects could draw heat from colder objects without external intervention, we would have violations of the second law of thermodynamics. Thus no point on an unheated wire will ever reach a

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602

Chapter 10

Partial Differential Equations

higher temperature than the initial peak temperature. Such statements are called maximum principles and can be proved mathematically for the heat equation and Laplace’s equation. One such result is the following.†

Maximum Principle for the Heat Equation Theorem 6. equation (56)

Let u1x, t2 be a continuously differentiable function that satisfies the heat

0u 02u = b 2, 0t 0x

06x6L,

t70,

and the boundary conditions (57)

u10, t2 = u1L, t2 = 0 ,

tÚ0.

Then u1x, t2 attains its maximum value at t = 0, for some x in 30, L4—that is, max u1x, t2 = max u1x, 02 .

tÚ0 0…x…L

0…x…L

We can use the maximum principle to show that the heat flow problem has a unique solution.

Uniqueness of Solution Theorem 7. (58) (59) (60)

The initial-boundary value problem

0u 02u = b 2, 06x6L, t70, 0t 0x u10, t2 = u1L, t2 = 0 , tÚ0, u1x, 02 = f1x2 , 06x6L,

has at most one continuously differentiable solution.

Proof. Assume u1x, t2 and y1x, t2 are continuously differentiable functions that satisfy the initial-boundary value problem (58)–(60). Let w = u - y. Now w is a continuously differentiable solution to the boundary value problem (56)–(57). By the maximum principle, w must attain its maximum at t = 0, and since w1x, 02 = u1x, 02 - y1x, 02 = f1x2 - f1x2 = 0 , we have w1x, t2 … 0. Hence, u1x, t2 … y1x, t2 for all 0 … x … L, t Ú 0. A similar argun = y - u yields y1x, t2 … u1x, t2. Therefore we have u1x, t2 = y1x, t2 for ment using w all 0 … x … L, t Ú 0. Thus, there is at most one continuously differentiable solution to the problem (58)–(60). ◆

†

For a discussion of maximum principles and their applications, see Maximum Principles in Differential Equations, by M. H. Protter and H. F. Weinberger (Springer-Verlag, New York, 1984).

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Section 10.5

10.5

02u 0u = 2, 06x6p, t70, 2. 0t 0x u10, t2 = u1p, t2 = 0 , t 7 0 , u1x, 02 = x2 , 0 6 x 6 p

4.

02u 0u = 3 2, 06x6p, t70, 0t 0x 0u 0u 10, t2 = 1p, t2 = 0 , t 7 0 , 0x 0x u1x, 02 = x , 0 6 x 6 p 02u 0u = 2 2, 06x61, t70, 0t 0x 0u 0u 10, t2 = 11, t2 = 0 , t 7 0 , 0x 0x u1x, 02 = x11 - x2 , 0 6 x 6 1

02u 0u = 2, 06x6p, t70, 5. 0t 0x 0u 0u 10, t2 = 1p, t2 = 0 , t 7 0 , 0x 0x x u1x, 02 = e , 0 6 x 6 p 6.

7.

8.

9.

603

EXERCISES

In Problems 1–10, find a formal solution to the given initialboundary value problem. 02u 0u 1. = 5 2, 06x61, t70, 0t 0x u10, t2 = u11, t2 = 0 , t 7 0 , u1x, 02 = 11 - x2x2 , 0 6 x 6 1

3.

The Heat Equation

02u 0u = 7 2, 06x6p, t70, 0t 0x 0u 0u 10, t2 = 1p, t2 = 0 , t 7 0 , 0x 0x u1x, 02 = 1 - sin x , 0 6 x 6 p 02u 0u = 2 2, 06x6p, t70, 0t 0x u10, t2 = 5 , u1p, t2 = 10 , t 7 0 , u1x, 02 = sin 3x - sin 5x , 0 6 x 6 p 02u 0u = 2, 06x6p, t70, 0t 0x u10, t2 = 0 , u1p, t2 = 3p , t 7 0 , u1x, 02 = 0 , 0 6 x 6 p 02u 0u = 2 + e-x , 0 6 x 6 p , t 7 0 , 0t 0x u10, t2 = u1p, t2 = 0 , t 7 0 , u1x, 02 = sin 2x , 0 6 x 6 p

10.

0u 02u = 3 2 +x, 06x6p, t70, 0t 0x t70, u10, t2 = u1p, t2 = 0 , u1x, 02 = sin x , 0 6 x 6 p

11. Find a formal solution to the initial-boundary value problem 0u 02u = 4 2, 06x6p, t70, 0t 0x 0u 10, t2 = 0 , u1p, t2 = 0 , t 7 0 , 0x u1x, 02 = f1x2 , 0 6 x 6 p . 12. Find a formal solution to the initial-boundary value problem 0u 02u = 2, 06x6p, t70, 0t 0x 0u u10, t2 = 0 , u1p, t2 + 1p, t2 = 0 , t 7 0 , 0x u1x, 02 = f1x2 , 0 6 x 6 p . 13. Find a formal solution to the initial-boundary value problem 0u 02u = 2 2 + 4x , 0 6 x 6 p , t 7 0 , 0t 0x u10, t2 = u1p, t2 = 0 , t 7 0 , u1x, 02 = sin x , 0 6 x 6 p . 14. Find a formal solution to the initial-boundary value problem 0u 02u = 3 2 +5, 06x6p, t70, 0t 0x u10, t2 = u1p, t2 = 1 , t 7 0 , u1x, 02 = 1 , 0 6 x 6 p . In Problems 15–18, find a formal solution to the initialboundary value problem. 02u 02u 0u = 2 + 2, 06x6p, 06y6p, t70, 0t 0x 0y 0u 0u 10, y, t2 = 1p, y, t2 = 0 , 0 6 y 6 p , t 7 0 , 0x 0x u1x, 0, t2 = u1x, p, t2 = 0 , 0 6 x 6 p , t 7 0 , u1x, y, 02 = f1x, y2 , 0 6 x 6 p , 0 6 y 6 p , for the given function f1x, y2. 15. f1x, y2 = cos 6x sin 4y - 3 cos x sin 11y

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604

Chapter 10

Partial Differential Equations

16. f1x, y2 = cos x sin y + 4 cos 2x sin y - 3 cos 3x sin 4y 17. f1x, y2 = y 18. f1x, y2 = x sin y

L 7 0 is a consumption rate with units sec -1. Assume the boundary conditions are

19. Chemical Diffusion. Chemical diffusion through a thin layer is governed by the equation

and the initial concentration is given by

C10, t2 = C1a, t2 = 0 ,

C1x, 02 = f1x2 ,

02C 0C = k 2 - LC , 0t 0x

t70,

06x6a.

Use the method of separation of variables to solve formally for the concentration C1x, t2. What happens to the concentration as t S + ∞ ?

where C1x, t2 is the concentration in moles/cm3, the diffusivity k is a positive constant with units cm2/sec, and

10.6 The Wave Equation In Section 10.2 we presented a model for the motion of a vibrating string. If u1x, t2 represents the displacement (deflection) of the string and the ends of the string are held fixed, then the motion of the string is governed by the initial-boundary value problem (1) (2) (3) (4)

02u 0 t2 u10, t2 u1x, 02 0u 1x, 02 0t

02u , 06x6L, 0x2 = u1L, t2 = 0 , t70, = f1x2 , 06x6L, = a2

= g1x2 ,

t70,

06x6L.

Equation (1) is called the wave equation. The constant a2 appearing in (1) is strictly positive and depends on the linear density and tension of the string. The boundary conditions in (2) reflect the fact that the string is held fixed at the two endpoints x = 0 and x = L. Equations (3) and (4) specify, respectively, the initial displacement and the initial velocity of each point on the string. For the initial and boundary conditions to be consistent, we assume f102 = f1L2 = 0 and g102 = g1L2 = 0. Using the method of separation of variables, we found in Section 10.2 that a formal solution to (1)–(4) is given by [compare equations (24)–(26) on page 569] (5)

∞ npa npa npx u1x, t2 = a c an cos t + bn sin t d sin , L L L n=1

where the an’s and bn’s are determined from the Fourier sine series (6) (7)

∞ npx f1x2 = a an sin , L n=1 ∞ npa npx b sin . g1x2 = a bn a L L n=1

See Figure 10.18 on page 605 for a sketch of partial sums for (5).

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.6

The Wave Equation

605

u

u=

x

0 t

Figure 10.18 Partial sums for equation (5) with f1x2 = e

x, p - x,

0 6 x … p>2, and g1x2 = 0 p>2 … x 6 p,

Each term (or mode) in expansion (5) can be viewed as a standing wave (a wave that vibrates in place without lateral motion along the string). For example, the first term, a a1 cos

pa pa px t + b1 sin tb sin , L L L

consists of a sinusoidal shape function sin1px>L2 multiplied by a time-varying amplitude. The second term is also a sinusoid sin12px>L2 with a time-varying amplitude. In the latter case, there is a node in the middle at x = L>2 that never moves. For the nth term, we have a sinusoid sin1npx>L2 with a time-varying amplitude and 1n - 12 nodes. This is illustrated in Figure 10.19. Thus, separation of variables decomposes the initial data into sinusoids or modes [equations (6)–(7)], assigns each mode a frequency at which to vibrate [equation (5)], and represents the solution as the superposition of infinitely many standing waves. Modes with more nodes vibrate at higher frequencies. By choosing the initial configuration of the string to have the same shape as one of the individual terms in the solution, we can activate only that mode. Node

L

0

n=1

Nodes

L

0

n=2

L

0

n=3

Figure 10.19 Standing waves. Time-varying amplitudes are shown by dashed curves

The different modes of vibration of a guitar string are distinguishable to the human ear by the frequency of the sound they generate; different frequencies are discerned as different “pitches.” The fundamental is the frequency of the lowest mode, and integer multiples of the fundamental frequency are called “harmonics.” A cello and a trombone sound different even if they are playing the same fundamental, say F#, because of the difference in the intensities of the harmonics they generate.

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606

Chapter 10

Partial Differential Equations

Many engineering devices operate better in some modes than in others. For example, certain modes propagate down a waveguide or an optical fiber with less attenuation than the others. In such cases, engineers design the systems to suppress the undesirable modes, either by shaping the initial excitation or by introducing devices (like resistor cards in a waveguide) which damp out certain modes preferentially. The fundamental mode on a guitar string can be suppressed by holding the finger lightly in contact with the midpoint, thereby creating a stationary point or node there. According to equation (5), the pitch of the note then doubles, thus producing an “octave.” This style of fingering, called “playing harmonics,” is used frequently in musical performance. An important characteristic of an engineering device is the set of angular frequencies supported by its eigenmodes—its “eigenfrequencies.” According to equation (5), the eigenfrequencies of the vibrating string are the harmonics vn = npa>L, n = 1, 2, 3, . . . . These blend together pleasantly to the human ear, and we enjoy melodies played on string instruments. The eigenfrequencies of a drum are not harmonics (see Problem 21). Therefore, drums are used for rhythm, not for melody. As we have seen in the preceding section, the method of separation of variables can be used to solve problems with nonhomogeneous boundary conditions and nonhomogeneous equations where the forcing term is time independent. In the next example, we will consider a problem with a time-dependent forcing term h1x, t2. Example 1

For given functions f, g, and h, find a formal solution to the initial-boundary value problem (8)

2 02u 20 u a = + h1x, t2 , 0t 2 0x2

(9)

u10, t2 = u1L, t2 = 0 , t 7 0 ,

(10)

06x6L,

t70,

u1x, 02 = f 1x2 , 0 6 x 6 L ,

0u 1x, 02 = g1x2 , 0 6 x 6 L . 0t The boundary conditions in (9) certainly require that the solution be zero for x = 0 and x = L. Motivated by the fact that the solution to the corresponding homogeneous system (1) – (4) consists of a superposition of standing waves, let’s try to find a solution to (8) – (11) of the form (11)

Solution

(12)

∞ npx u1x, t2 = a un 1t2 sin , L n=1

where the un 1t2’s are functions of t to be determined. For each fixed t, we can compute a Fourier sine series for h1x, t2. If we assume that the series is convergent to h1x, t2, then (13)

∞ npx h1x, t2 = a hn 1t2 sin , L n=1

where the coefficient hn 1t2 is given by [recall equation (7) on page 589] hn 1t2 =

L

2 npx h1x, t2 sin dx , n = 1, 2, . . . . L L0 L If the series in (13) has the proper convergence properties, then we can substitute (12) and (13) into equation (8) and obtain ∞

∞ npa 2 npx npx > c u + a d 1t2 b u 1t2 sin = hn 1t2 sin . n n a a L L L n=1 n=1

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.6

The Wave Equation

607

Equating the coefficients in each series (why?), we have un> 1t2 + a

npa 2 b un 1t2 = hn 1t2 . L

This is a nonhomogeneous, constant-coefficient equation that can be solved using variation of parameters. You should verify that un 1t2 = an cos

t

npa npa L npa t + bn sin t+ h 1s2 sin c 1t - s2 d ds npa L0 n L L L

[compare Problem 20 of Exercises 4.6]. Hence, with this choice of un 1t2, the series in (12) is a formal solution to the partial differential equation (8). Since† un 102 = an

and

un= 102 = bn a

npa b, L

substituting (12) into the initial conditions (10)–(11) yields (14)

∞ npx u1x, 02 = f 1x2 = a an sin , L n=1

(15)

∞ 0u npa npx 1x, 02 = g1x2 = a bn a b sin . 0t L L n=1

Thus, if we choose the an’s and bn’s so that equations (14) and (15) are satisfied, a formal solution to (8)–(11) is given by (16)

∞ npa npa u1x, t2 = a e an cos t + bn sin t L L n=1 t

+

L npa npx h 1s2sin c 1t - s2 d ds fsin . ◆ npa L0 n L L

The method of separation of variables can also be used to solve initial-boundary value problems for the wave equation in higher dimensions. For example, a vibrating rectangular membrane of length L and width W (see Figure 10.20 on page 608) is governed by the following initial-boundary value problem for u1x, y, t2: (17)

02 u 02u 02u = a2 a 2 + 2 b , 0 6 x 6 L , 2 0t 0x 0y

06y6W,

(18)

u10, y, t2 = u1L, y, t2 = 0 ,

06y6W,

t70,

(19)

u1x, 0, t2 = u1x, W, t2 = 0 ,

06x6L,

t70,

(20) (21)

u1x, y, 02 = f 1x, y2 , 0u 1x, y, 02 = g1x, y2 , 0t

To compute un= 102, we use the fact that

†

06x6L,

06y6W,

06x6L,

06y6W.

t

t70,

t

d 0G G1s, t2 ds = G1t, t2 + 1s, t2 ds. dt L0 L0 0t

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608

Chapter 10

Partial Differential Equations

y

u=0

W

(

=

u=0

+

(

u=0

L

u=0

0

x

Figure 10.20 Vibrating membrane

Using an argument similar to the one given for the problem of heat flow in a rectangular plate (Example 4 in Section 10.5), we find that the initial-boundary value problem (17)–(21) has a formal solution (22)

∞ ∞ m2 n2 u1x, y, t2 = a a • amn cos ° + apt¢ B L2 W 2 m=1 n=1

+ bmn sin °

m2 n2 mpx npy + apt¢ ¶ sin sin , 2 2 BL L W W

where the constants amn and bmn are determined from the double Fourier series ∞ ∞ npy mpx f1x, y2 = a a amn sin sin , L W m=1 n=1 ∞ ∞ npy m2 n2 mpx g1x, y2 = a a ap + bmn sin sin . 2 2 BL L W W m=1 n=1

In particular, L

(23)

amn =

(24)

bmn =

W

npy 4 mpx f1x, y2 sin sin dy dx , LW L0 L0 L W L

4

W

g1x, y2 sin

npy mpx sin dy dx . L W

m n L0 L0 + 2 2 BL W We leave the derivation of this solution as an exercise (see Problem 19). We mentioned earlier that the solution to the vibrating string problem (1)–(4) consisted of a superposition of standing waves. There are also “traveling waves” associated with the wave equation. Traveling waves arise naturally out of d’Alembert’s solution to the wave equation for an “infinite” string. To obtain d’Alembert’s solution to the wave equation 2

LWpa

2

2 02 u 20 u = a , 0 t2 0x2

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Section 10.6

The Wave Equation

609

we use the change of variables c = x + at ,

h = x - at .

If u has continuous second partial derivatives, then 0u>0x = 0u>0c + 0u>0h and 0u>0t = a10u>0c - 0u>0h2, from which we obtain 02u 02u 02u 02u = + 2 + , 0c0h 0h2 0x2 0c2 02u 02u 02u 02u + 2f . = a2 e 2 - 2 2 0c0h 0h 0t 0c Substituting these expressions into the wave equation and simplifying yields 02u = 0. 0c 0h We can solve this equation directly by first integrating with respect to c to obtain 0u = b1h2 , 0h where b1h2 is an arbitrary function of h, and then integrating with respect to h to find u1c, h2 = A1c2 + B1h2 , where A1c2 and B1h2 = 1 b1h2 dh are arbitrary functions. Substituting the original variables x and t gives d’Alembert’s solution (25)

u1x, t2 = A1x + At2 + B1x − At2 .

It is easy to check by direct substitution that u1x, t2, defined by formula (25), is indeed a solution to the wave equation, provided A and B are twice-differentiable functions. Example 2

Using d’Alembert’s formula (25), find a solution to the initial value problem (26) (27) (28)

Solution

2 02 u 20 u a = 0 t2 0x2 u1x, 02 = f1x2 0u 1x, 02 = g1x2 0t

-∞ 6 x 6 ∞ ,

t70,

-∞ 6 x 6 ∞ , -∞ 6 x 6 ∞ .

A solution to (26) is given by formula (25), so we need only choose the functions A and B so that the initial conditions (27)–(28) are satisfied. For this we need (29)

u1x, 02 = A1x2 + B1x2 = f1x2 ,

(30)

0u 1x, 02 = aA′1x2 - aB′1x2 = g1x2 . 0t

Integrating equation (30) from x0 to x 1x0 arbitrary2 and dividing by a gives x

(31)

A1x2 - B1x2 =

1 g1s2 ds + C , a Lx0

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610

Chapter 10

Partial Differential Equations

where C is also arbitrary. Solving the system (29) and (31), we obtain x

A1x2 =

1 1 C f1x2 + g1s2 ds + , 2 2a Lx0 2

B1x2 =

1 1 C f1x2 g1s2 ds - . 2 2a Lx0 2

x

Using these functions in formula (25) gives x+at

u1x, t2 =

1 1 3f1x + at2 + f1x - at24 + c 2 2a Lx0

x-at

g1s2ds -

Lx0

g1s2ds d ,

which simplifies to x+at

(32)

Example 3

Solution

u1x, t2 =

1 1 3f1x + at2 + f1x - at24 + g1s2 ds . ◆ 2 2a Lx-at

Find the solution to the initial value problem (33)

02 u 02 u = 4 , 0 t2 0 x2

(34)

u1x, 02 = sin x ,

(35)

0u 1x, 02 = 1 , 0t

-∞ 6 x 6 ∞ ,

t70,

-∞ 6 x 6 ∞ , -∞ 6 x 6 ∞ .

This is just a special case of the preceding example where a = 2, f1x2 = sin x, and g1x2 = 1. Substituting into (32), we obtain the solution x+2t

(36)

1 1 3sin1x + 2t2 + sin1x - 2t24 + ds 2 4 Lx-2t = sin x cos 2t + t . ◆

u1x, t2 =

We now use d’Alembert’s formula to show that the solution to the “infinite” string problem consists of traveling waves. Let h1x2 be a function defined on 1 - ∞, ∞ 2. The function h1x + a2, where a 7 0, is a translation of the function h1x2 in the sense that its “shape” is the same as h1x2, but its position has been shifted to the left by an amount a. This is illustrated in Figure 10.21 for a function h1x2 whose graph consists of a triangular “bump.” If we let t Ú 0 be a parameter (say, time), then the functions h1x + at2 represent a family of functions with the same shape but shifted farther and farther to the left as t S ∞ (Figure 10.22 on page 611). We say that h1x + at2 is a traveling wave moving to the left with speed a. In a similar fashion, h1x - at2 is a traveling wave moving to the right with speed a.

h(x + a) x0 - a

h(x)

x1 - a

x0

x1

x

Figure 10.21 Graphs of h1x2 and h1x + a2

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Section 10.6

The Wave Equation

611

h(x)

x 0 Figure 10.22 Traveling wave

If we refer to formula (25), we find that the solution to 02u>0t 2 = a202u>0x2 consists of traveling waves A1x + at2 moving to the left with speed a and B1x - at2 moving to the right at the same speed. In the special case when the initial velocity g1x2 K 0, we have 1 u1x, t2 = 3f1x + at2 + f1x - at24 . 2 Hence, u1x, t2 is the sum of the traveling waves 1 1 f1x + at2 and f1x - at2 . 2 2 These waves are initially superimposed, since 1 1 u1x, 02 = f1x2 + f1x2 = f1x2 . 2 2 As t increases, the two waves move away from each other with speed 2a. This is illustrated in Figure 10.23 for a triangular wave.

0 t=0

0 t=

0 1 t=

0 t=

1

2

Figure 10.23 Decomposition of initial displacement into traveling waves

Example 4 Solution

Express the standing wave cos a

pa px tb sin a b as a superposition of traveling waves. L L

By a familiar trigonometric identity, 1 pa px 1 p p cos a tb sina b = sin 1x - at2 + sin 1x + at2 L L 2 L 2 L = h1x - at2 + h1x + at2 , where 1 px sin . 2 L This standing wave results from adding a wave traveling to the right to the same waveform traveling to the left. ◆ h1x2 =

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612

Chapter 10

Partial Differential Equations

Existence and Uniqueness of Solutions In Example 1 the method of separation of variables was used to derive a formal solution to the given initial-boundary value problem. To show that these series converge to an actual solution requires results from real analysis, just as was the case for the formal solutions to the heat equation in Section 10.5. In Examples 2 and 3, we can establish the validity of d’Alembert’s solution by direct substitution into the initial value problem, assuming sufficient differentiability of the initial functions. We leave it as an exercise for you to show that if f has a continuous second derivative and g has a continuous first derivative, then d’Alembert’s solution is a true solution (see Problem 12). The question of the uniqueness of the solution to the initial-boundary value problem (1)–(4) can be answered using an energy argument.

Uniqueness of the Solution to the Vibrating String Problem Theorem 8. (37) (38) (39) (40)

The initial-boundary value problem

2 02 u 20 u a = , 06x6L, t70, 0 t2 0 x2 u10, t2 = u1L, t2 = 0 , t70, u1x, 02 = f1x2 , 06x6L, 0u 1x, 02 = g1x2 , 06x6L, 0t

has at most one twice continuously differentiable solution.

Proof. Assume both u1x, t2 and y1x, t2 are twice continuously differentiable solutions to (37)–(40) and let w1x, t2 J u1x, t2 - y1x, t2. It is easy to check that w1x, t2 satisfies the initial-boundary value problem (37)–(40) with zero initial data; that is, for 0 … x … L, (41)

w1x, 02 = 0

and

0w 1x, 02 = 0 . 0t

We now show that w1x, t2 K 0 for 0 … x … L, t Ú 0. If w1x, t2 is the displacement of the vibrating string at location x for time t, then with the appropriate units, the total energy E1t2 of the vibrating string at time t equals the integral L

(42)

E1t2 J

1 0w 2 0w 2 c a2 a b + a b d dx . 2 L0 0x 0t

(The first term in the integrand relates to the stretching of the string at location x and represents the potential energy. The second term is the square of the velocity of the vibrating string at x and represents the kinetic energy.) We now consider the derivative of E1t2: L

d 1 0w 2 0w 2 dE = e c a2 a b + a b d dx f . dt dt 2 L0 0x 0t

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Section 10.6

The Wave Equation

613

Since w has continuous second partial derivatives (because u and y do), we can interchange the order of integration and differentiation. This gives L

(43)

0w 02w 0w 02w dE + = c a2 d dx . dt 0x 0t 0x 0t 0t 2 L0

Again the continuity of the second partials of w guarantees that the mixed partials are equal; that is, 02w 02w = . 0t 0x 0x 0t Combining this fact with integration by parts, we obtain L

(44)

L

a2

L0

0w 02w 0w 02w dx = a2 dx 0x 0t 0x 0x 0x 0t L0 0w 0w 0w 0w = a2 1L, t2 1L, t2 - a2 10, t2 10, t2 0x 0t 0x 0t L 2 0 w 0w = a2 2 dx . 0x 0t L0

The boundary conditions w10, t2 = w1L, t2 = 0, t Ú 0, imply that 10w>0t210, t2 = 10w>0t21L, t2 = 0, t Ú 0. This reduces equation (44) to L

L0

L

a2

0w 02w 0w 02w dx = a2 dx . 0x 0t 0x 0t 0x2 L0

Substituting this in for the first integrand in (43), we find L

dE 0w 02w 02w = c 2 - a2 2 d dx . dt 0x L0 0t 0t Since w satisfies equation (37), the integrand is zero for all x. Thus, dE>dt = 0, and so E1t2 K C, where C is a constant. This means that the total energy is conserved within the vibrating string. The first boundary condition in (41) states that w1x, 02 = 0 for 0 … x … L. Hence, 10w>0x21x, 02 = 0 for 0 6 x 6 L. Combining this with the second boundary condition in (41), we find that when t = 0, the integrand in (42) is zero for 0 6 x 6 L. Therefore, E102 = 0. Since E1t2 K C, we must have C = 0. Hence, L

(45)

E1t2 =

1 0w 2 0w 2 c a2 a b + a b d dx K 0 . 2 L0 0x 0t

That is, the total energy of this system is zero. Because the integrand in (45) is nonnegative and continuous and the integral is zero, the integrand must be zero for 0 … x … L. Moreover, the integrand is the sum of two squares and so each term must be zero. Hence, 0w 1x, t2 = 0 0x

and

0w 1x, t2 = 0 0t

for all 0 … x … L, t Ú 0. Thus w1x, t2 = K, where K is a constant. Physically, this says that

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614

Chapter 10

Partial Differential Equations

there is no motion in the string. Finally, since w is constant and w is zero when t = 0, then w1x, t2 K 0. Consequently, u1x, t2 = y1x, t2 and the initial-boundary value problem has at most one solution. ◆

10.6

EXERCISES

In Problems 1 – 4, find a formal solution to the vibrating string problem governed by the given initial-boundary value problem. 02u 02u 1. = , 06x61, t70, 0t 2 0x2 u10, t2 = u11, t2 = 0 , t 7 0 ,

6. A vibrating string is governed by the initial-boundary value problem (1)–(4). The initial conditions are given by f1x2 K 0 and g1x2 = e

u1x, 02 = x11 - x2 , 0 6 x 6 1 , 0u 1x, 02 = sin 7px , 0 6 x 6 1 0t 2.

02u 02u = 4 2, 06x6p, t70, 2 0t 0x u10, t2 = u1p, t2 = 0 , t 7 0 , u1x, 02 = x 1p - x2 , 0 6 x 6 p , 0u 1x, 02 = 0 , 0 6 x 6 p 0t

8.

2

5. The Plucked String. A vibrating string is governed by the initial-boundary value problem (1)–(4). If the string is lifted to a height h0 at x = a and released, then the initial conditions are

and g1x2 K 0. Find a formal solution.

t70,

0u 1x, 02 = 0 , 0 6 x 6 p 0t

06x6p, u1x, 02 = sin 4x + 7 sin 5x , x , 0 6 x 6 p>2 , 0u 1x, 02 = e 0t p - x , p>2 6 x 6 p

h0 1L - x2 > 1L - a2 ,

02u 02u = 2 + x sin t , 0 6 x 6 p , 2 0t 0x u10, t2 = u1p, t2 = 0 , t 7 0 , u1x, 02 = 0 , 0 6 x 6 p ,

02u 02u = 9 2, 06x6p, t70, 4. 2 0t 0x u10, t2 = u1p, t2 = 0 , t 7 0 ,

f1x2 = e

a6x6L,

In Problems 7 and 8, find a formal solution to the vibrating string problem governed by the given nonhomogeneous initial-boundary value problem. 02u 02u 7. = 2 + tx , 0 6 x 6 p , t70, 2 0t 0x u10, t2 = u1p, t2 = 0 , t 7 0 , u1x, 02 = sin x , 0 6 x 6 p , 0u 1x, 02 = 5 sin 2x - 3 sin 5x , 0 6 x 6 p 0t

02u 02u = , 06x6p, t70, 16 0t 2 0x2 u10, t2 = u1p, t2 = 0 , t 7 0 ,

h0x>a ,

y0 1L - x2 > 1L - a2

where y0 is a constant. Find a formal solution.

06x6p, u1x, 02 = sin2x , 0u 1x, 02 = 1 - cos x , 0 6 x 6 p 0t 3.

06x…a,

y0 x>a ,

06x…a, a6x6L,

9. If one end of a string is held fixed while the other is free, then the motion of the string is governed by the initialboundary value problem 02u 02u = a2 2 , 2 0t 0x

06x6L,

u10, t2 = 0 and u1x, 02 = f1x2 , 0u 1x, 02 = g1x2 , 0t

t70,

0u 1L, t2 = 0, 0x 06x6L,

t70,

06x6L.

Derive a formula for a formal solution.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.6

10. Derive a formula for the solution to the following initialboundary value problem involving nonhomogeneous boundary conditions 02u 02u = a2 2 , 2 0t 0x u10, t2 = U1 ,

06x6L,

u1x, 02 = f1x2 , 0u 1x, 02 = g1x2 , 0t

(46) ut + Aux + Buxxx = 0 , where u1x, t2 is the displacement of the water from its equilibrium depth at location x and at time t, and a and b are positive constants. (a) Show that equation (46) has a solution of the form

t70,

06x6L, 06x6L,

(47)

where U1 and U2 are constants. 11. The Telegraph Problem.† Use the method of separation of variables to derive a formal solution to the telegraph problem

0u 1x, 02 = 0 , 0t

(48)

06x6L.

12. Verify d’Alembert’s solution (32) to the initial value problem (26)–(28) when f1x2 has a continuous second derivative and g1x2 has a continuous first derivative by substituting it directly into the equations.

dV dV d 3V + ak + bk 3 3 = 0 . dz dz dz

V1x, t2 = A sin 3lkx - 1alk - bl3k 3 2t + B4 ,

In Problems 13–18, find the solution to the initial value problem. 02u 02u = a2 2 , -∞ 6 x 6 ∞ , t70, 2 0t 0x u1x, 02 = f1x2 , -∞ 6 x 6 ∞ , 0u 1x, 02 = g1x2 , -∞ 6 x 6 ∞ , 0t

19. Derive the formal solution given in equations (22)–(24) to the vibrating membrane problem governed by the initial-boundary value problem (17)–(21).

-w

These solutions, defined by (47), are called uniform waves. (b) Physically, we are interested only in solutions V1z2 that are bounded and nonconstant on the infinite interval 1 - ∞ , ∞ 2. Show that such solutions exist only if ak - w 7 0. (c) Let l2 = 1ak - w2 > 1bk 3 2. Show that the solutions from part (b) can be expressed in the form

06x6L,

for the given functions f1x2 and g1x2. 13. f1x2 K 0 , g1x2 = cos x 14. f1x2 = x2 , g1x2 K 0 15. f1x2 = x , g1x2 = x 16. f1x2 = sin 3x , g1x2 K 1 2 17. f1x2 = e-x , g1x2 = sin x 18. f1x2 = cos 2x , g1x2 = 1 - x

u1x, t2 = V1z2 , z = kx - w1k2t ,

where k is a fixed constant and w1k2 is a function of k, provided V satisfies

02u 02u 0 u + + u = a2 2 , 0 6 x 6 L , t 7 0 , 2 0t 0t 0x u10, t2 = u1L, t2 = 0 , t70, u1x, 02 = f1x2 ,

615

20. Long Water Waves. The motion of long water waves in a channel of constant depth is governed by the linearized Korteweg and de Vries (KdV) equation

t70,

u1L, t2 = U2 ,

The Wave Equation

where A and B are arbitrary constants. [Hint: Solve for w in terms of l and k and use (47).] (d) Since both l and k can be chosen arbitrarily and they always appear together as the product lk, we can set l = 1 without loss of generality. Hence, we have V1x, t2 = A sin 3kx - 1ak - bk 3 2t + B4 as a uniform wave solution to (46). The defining relation w1k2 = ak - bk 3 is called the dispersion relation, the ratio w1k2 >k = a - bk 2 is called the phase velocity, and the derivative dw>dk = a - 3bk 2 is called the group velocity. When the group velocity is not constant, the waves are called dispersive. Show that the standard wave equation utt = a2uxx has only nondispersive waves. 21. Vibrating Drum. A vibrating circular membrane of unit radius whose edges are held fixed in a plane and

†

For a discussion of the telegraph problem, see Project E at the end of this chapter.

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616

Chapter 10

Partial Differential Equations

Show that there is a family of solutions of the form

whose displacement u1r, t2 depends only on the radial distance r from the center and on the time t is governed by the initial-boundary value problem.

un 1r, t2 = 3an cos1knat2 + bn sin1knat24J0 1knr2 , where J0 is the Bessel function of the first kind of order zero (see page 478) and 0 6 k1 6 k2 6 g 6 kn 6 g are the positive zeros of J0. See Figure 10.24.

2 1 0u 02u 2 0 u = a a + b, 0t 2 0r 2 r 0r 06r61, t70,

u11, t2 = 0 , u1r, t2

t70,

remains finite as r S 0+ .

(a)

(b)

(c)

Figure 10.24 Mode shapes for vibrating drum. (a) J0 12.405r2, (b) J0 15.520r2, (c) J0 18.654r2

10.7 Laplace’s Equation In Section 10.1 we showed how Laplace’s equation, ∆u =

02u 02u + = 0, 0x2 0y2

arises in the study of steady-state or time-independent solutions to the heat equation. Because these solutions do not depend on time, initial conditions are irrelevant and only boundary conditions are specified. Other applications include the static displacement u1x, y2 of a stretched membrane fastened in space along the boundary of a region (here u must satisfy Laplace’s equation inside the region); the electrostatic and gravitational potentials in certain force fields (here u must satisfy Laplace’s equation in any region that is free of electrical charges or mass); and, in fluid mechanics for an idealized fluid, the stream function u1x, y2 whose level curves (stream lines), u1x, y2 = constant, represent the path of particles in the fluid (again u satisfies Laplace’s equation in the flow region). There are two basic types of boundary conditions that are usually associated with Laplace’s equation: Dirichlet boundary conditions, where the solution u1x, y2 to Laplace’s equation in a domain D is required to satisfy u1x, y2 = f1x, y2

on 0D ,

with f1x, y2 a specified function defined on the boundary 0D of D; and Neumann boundary conditions, where the directional derivative 0u>0n along the outward normal to the boundary is required to satisfy 0u 1x, y2 = g1x, y2 0n

on 0D ,

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Section 10.7

Laplace’s Equation

617

with g1x, y2 a specified function defined on 0D.† We say that the boundary conditions are mixed if the solution is required to satisfy u1x, y2 = f1x, y2 on part of the boundary and 10u>0n21x, y2 = g1x, y2 on the remaining portion of the boundary. In this section we use the method of separation of variables to find solutions to Laplace’s equation with various boundary conditions for rectangular, circular, and cylindrical domains. We also discuss the existence and uniqueness of such solutions. Example 1

Find a solution to the following mixed boundary value problem for a rectangle (see Figure 10.25):

(3)

02u 02u + = 0, 06x6a, 06y6b, 0x2 0y2 0u 0u 10, y2 = 1a, y2 = 0 , 0…y…b, 0x 0x u1x, b2 = 0 , 0…x…a,

(4)

u1x, 02 = f1x2 ,

(1) (2)

0…x…a. y b

u(x, b) = 0

−u (a, y) = 0 −x

−u (0, y) = 0 −x

0

u(x, 0) = f (x)

a

x

Figure 10.25 Mixed boundary value problem

Solution

Separating variables, we first let u1x, y2 = X1x2Y1y2. Substituting into equation (1), we have X″1x2Y1y2 + X1x2Y ″1y2 = 0 , which separates into X″1x2 Y ″1y2 = = -l , X1x2 Y1y2 where l is some constant. This leads to the two ordinary differential equations (5)

X″1x2 + lX1x2 = 0 ,

(6)

Y ″1y2 - lY1y2 = 0 . From the boundary condition (2), we observe that

(7)

X′102 = X′1a2 = 0 .

†

These generalize the boundary conditions (5) in Section 10.1 and (42) in Section 10.5.

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618

Chapter 10

Partial Differential Equations

We have encountered the eigenvalue problem in (5) and (7) previously (see Example 1 in Section 10.5). The eigenvalues are l = ln = 1np>a2 2, n = 0, 1, 2, c, with corresponding solutions (8)

Xn 1x2 = an cos a

npx b, a

where the an’s are arbitrary constants. Setting l = ln = 1np>a2 2 in equation (6) and solving for Y gives† (9)

Y0 1y2 = A0 + B0y , Yn 1y2 = An cosh a

npy npy b + Bn sinh a b, a a

n = 1, 2, . . .

Now the boundary condition u1x, b2 = 0 in (3) will be satisfied if Y1b2 = 0. Setting y = b in (9), we see that we want A0 = -bB0 and An = -Bn tanh a Yn 1y2 = Cn sinh c

npb b, a

or equivalently,

np 1y - b2 d , a

n = 1, 2, c

(see Problem 18). Thus we find that there are solutions to (1)–(3) of the form u0 1x, y2 = X0 1x2Y0 1y2 = a0B0 1y - b2 = E0 1y - b2 ,

un 1x, y2 = Xn 1x2Yn 1y2 = an cos a = En cos a

npx np 1y - b2 d b Cn sinh c a a

npx np 1y - b2 d , b sinh c a a

n = 1, 2, c ,

where the En’s are constants. In fact, by the superposition principle, (10)

∞ npx np u1x, y2 = E0 1y - b2 + a En cos a 1y - b2 d b sinh c a a n=1

is a formal solution to (1)–(3). Applying the remaining nonhomogeneous boundary condition in (4), we have ∞ npb npx u1x, 02 = f1x2 = -E0b + a En sinh a b cos a b. a a n=1

This is a Fourier cosine series for f1x2 and hence the coefficients are given by the formulas a

1 f1x2 dx , 1 -ba2 L0 a 2 npx b dx , En = f1x2 cos a a npb L0 b a sinh a a E0 =

(11)

n = 1, 2, c .

Thus a formal solution is given by (10) with the constants En given by (11). See Figure 10.26 on page 619 for a sketch of partial sums for (10). ◆ We usually express Yn 1y2 = an enpy>a + bn e-npy>a. However, computation is simplified in this case by using the hyperbolic functions cosh z = 1ez + e-z 2 >2 and sinh z = 1ez - e-z 2 >2 .

†

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.7

Laplace’s Equation

619

u

x

y

Figure 10.26 Partial sums for Example 1 with f1x2 = e

x,

0 6 x … p>2,

p - x,

p>2 … x 6 p

In Example 1 the boundary conditions were homogeneous on three sides of the rectangle and nonhomogeneous on the fourth side, 5 1x, y2: y = 0, 0 … x … a6. It is important to note that the method used in Example 1 can also be used to solve problems for which the boundary conditions are nonhomogeneous on all sides. This is accomplished by solving four separate boundary value problems in which three sides have homogeneous boundary conditions and only one side is nonhomogeneous. The solution is then obtained by summing these four solutions (see Problem 5). For problems involving circular domains, it is usually more convenient to use polar coordinates. In rectangular coordinates the Laplacian has the form ∆u =

02u 02u + . 0x2 0y2

In polar coordinates 1r, u2, we let x = r cos u ,

y = r sin u

so that r = 2x2 + y2 ,

tan u = y>x .

With patience and a little care in applying the chain rule, one can show that the Laplacian in polar coordinates is (12)

𝚫u =

e2u 1 eu 1 e2u + + 2 2 r er er r eU2

(see Problem 6). In the next example we obtain a solution to the Dirichlet problem in a disk of radius a. Example 2

A circular metal disk of radius a has its top and bottom insulated. The edge 1r = a2 of the disk is kept at a specified temperature that depends on its location (varies with u). The steadystate temperature inside the disk satisfies Laplace’s equation. Determine the temperature distribution u1r, u2 inside the disk by finding the solution to the following Dirichlet boundary value problem, depicted in Figure 10.27 on page 620: (13) (14)

02u 1 0u 1 02u + + = 0, 0…r6a, 0r 2 r 0r r 2 0u 2 u1a, u2 = f1u2 , -p … u … p .

-p … u … p ,

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620

Chapter 10

Partial Differential Equations

u(a, ) = f ( ) (r, ) r a

Figure 10.27 Steady-state temperature distribution in a disk

Solution

To use the method of separation of variables, we first set u1r, u2 = R1r2ϴ1u2 , where 0 … r 6 a and -p … u … p. Substituting into (13) and separating variables give r 2R″1r2 + rR′1r2 R1r2

= -

ϴ″1u2 = l, ϴ1u2

where l is any constant. This leads to the two ordinary differential equations (15)

r 2R″1r2 + rR′1r2 - lR1r2 = 0 ,

(16)

ϴ″1u2 + lϴ1u2 = 0 .

For u1r, u2 to be continuous in the disk 0 … r 6 a, we need ϴ1u2 to be 2p-periodic; in particular, we require (17)

ϴ1 -p2 = ϴ1p2

and

ϴ′1 -p2 = ϴ′1p2 .

Therefore, we seek nontrivial solutions to the eigenvalue problem (16)–(17). When l 6 0, the general solution to (16) is the sum of two exponentials. Hence we have only trivial 2p-periodic solutions. When l = 0, we find ϴ1u2 = Au + B to be the solution to (16). This linear function is periodic only when A = 0, that is, ϴ0 1u2 = B is the only 2p-periodic solution corresponding to l = 0. When l 7 0, the general solution to (16) is ϴ1u2 = A cos 1l u + B sin 1l u . Here we get a nontrivial 2p-periodic solution only when 1l = n, n = 1, 2, c. [You can check this using (17).] Hence, we obtain the nontrivial 2p-periodic solutions (18)

ϴn 1u2 = An cos nu + Bn sin nu

corresponding to 1l = n, n = 1, 2, c. Notice that the eigenfunctions are the terms of the Fourier series discussed in Section 10.3 (recall Figure 10.5, page 573).

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Section 10.7

Laplace’s Equation

621

Now for l = n2, n = 0, 1, 2, c, equation (15) is the Cauchy–Euler equation (19)

r 2R″1r2 + rR′1r2 - n2R1r2 = 0

(see Section 4.7, page 192). When n = 0, the general solution is R0 1r2 = C + D ln r . Since ln r S - ∞ as r S 0+ , this solution is unbounded near r = 0 when D ≠ 0. Therefore, we must choose D = 0 if u1r, u2 is to be continuous at r = 0. We now have R0 1r2 = C and so u0 1r, u2 = R0 1r2ϴ1u2 = CB, which for convenience we write in the form (20)

u0 1r, u2 =

A0 , 2

where A0 is an arbitrary constant. When l = n2, n = 1, 2, c, you should verify that equation (19) has the general solution Rn 1r2 = Cnr n + Dnr -n . Since r -n S ∞ as r S 0+ , we must set Dn = 0 in order for u1r, u2 to be bounded at r = 0. Thus, Rn 1r2 = Cnr n . Now for each n = 1, 2, . . . , we have the solutions (21)

un 1r, u2 = Rn 1r2ϴn 1u2 = Cnr n 1An cos nu + Bn sin nu2 ,

and by forming an infinite series from the solutions in (20) and (21), we get the following formal solution to (13): u1r, u2 =

∞ A0 + a Cnr n 1An cos nu + Bn sin nu2 . 2 n=1

It is more convenient to write this series in the equivalent form (22)

u1r, u2 =

∞ a0 r n + a a b 1an cos nu + bn sin nu2 , 2 n=1 a

where the an’s and bn’s are constants. These constants can be determined from the boundary condition; indeed, with r = a in (22), condition (14) becomes f1u2 =

∞ a0 + a 1an cos nu + bn sin nu2 . 2 n=1

Hence, we recognize that an, bn are Fourier coefficients for f1u2. Thus, p

(23)

an =

1 f1u2cos nu du , p L-p

(24)

bn =

1 f1u2sin nu du , p L-p

n = 0, 1, c ,

p

n = 1, 2, c .

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622

Chapter 10

Partial Differential Equations

To summarize, if an and bn are defined by formulas (23) and (24), then u1r, u2 given in (22) is a formal solution to the Dirichlet problem (13)–(14). ◆ The procedure in Example 2 can also be used to study the Neumann problem in a disk: (25)

∆u = 0 ,

(26)

0u 1a, u2 = f1u2 , 0r

-p … u … p ,

0…r6a,

-p … u … p .

For this problem there is no unique solution, since if u is a solution, then the function u plus a constant is also a solution. Moreover, f must satisfy the consistency condition p

f1u2 du = 0 . L-p If we interpret the solution u1r, u2 of equation (25) as the steady-state temperature distribution inside a circular disk that does not contain either a heat source or heat sink, then equation (26) specifies the flow of heat across the boundary of the disk. Here the consistency condition (27) is simply the requirement that the net flow of heat across the boundary is zero. (If we keep pumping heat in, the temperature won’t reach equilibrium!) We leave the solution of the Neumann problem and the derivation of the consistency condition for the exercises. The technique used in Example 2 also applies to annular domains, 5 1r, u2: 0 6 a 6 r 6 b6, and to exterior domains, 5 1r, u2: a 6 r6. We also leave these applications as exercises. Laplace’s equation in cylindrical coordinates arises in the study of steady-state temperature distributions in a solid cylinder and in determining the electric potential inside a cylinder. In cylindrical coordinates,

(27)

x = r cos u ,

y = r sin u ,

z = z,

Laplace’s equation becomes (28)

𝚫u =

e2u 1 eu 1 e2u e2u + + 2 2 + 2 =0. 2 r er er r eU ez

The Dirichlet problem for the cylinder 5 1r, u, z2: 0 … r … a, 0 … z … b6 has the boundary conditions 0…z…b,

(29)

u1a, u, z2 = f1u, z2 ,

-p … u … p ,

(30)

u1r, u, 02 = g1r, u2 ,

0…r…a,

-p … u … p ,

(31)

u1r, u, b2 = h1r, u2 ,

0…r…a,

-p … u … p .

To solve the Dirichlet boundary value problem (28)–(31), we first solve the three boundary value problems corresponding to: (i) g K 0 and h K 0; (ii) f K 0 and h K 0; and (iii) f K 0 and g K 0. Then by the superposition principle, the solution to (28)–(31) will be the sum of these three solutions. This is the same method that was discussed in dealing with Dirichlet problems on rectangular domains. (See the remarks following Example 1.) In the next example we solve the Dirichlet problem when g K 0 and h K 0. Example 3

The base 1z = 02 and the top 1z = b2 of a charge-free cylinder are grounded and therefore are at zero potential. The potential on the lateral surface 1r = a2 of the cylinder is given by u1a, u, z2 = f1u, z2 where f1u, 02 = f1u, b2 = 0. Inside the cylinder, the potential u1r, u, z2

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Section 10.7

Laplace’s Equation

623

satisfies Laplace’s equation. Determine the potential u inside the cylinder by finding a solution to the Dirichlet boundary value problem (32) (33) (34) Solution

02u 1 0u 1 02u 02u + + + = 0, 0…r6a, -p … u … p , 0r 2 r 0r r 2 0u 2 0z2 u1a, u, z2 = f1u, z2 , -p … u … p , 0…z…b, u1r, u, 02 = u1r, u, b2 = 0 , 0…r6a, -p … u … p .

06z6b,

Using the method of separation of variables, we first assume that u1r, u, z2 = R1r2ϴ1u2Z1z2 . Substituting into equation (32) and separating out the Z’s, we find R″1r2 + 11>r2R′1r2 R1r2

+

Z″1z2 1 ϴ″1u2 = = l, 2 ϴ1u2 Z1z2 r

where l can be any constant. Separating further the R’s and ϴ ’s gives r 2R″1r2 + rR′1r2 R1r2

- r 2l = -

ϴ″1u2 = m, ϴ1u2

where m can also be any constant. We now have the three ordinary differential equations: (35) (36) (37)

r 2R″1r2 + rR′1r2 - 1r 2l + m2R1r2 = 0 , ϴ″1u2 + mϴ1u2 = 0 , Z″1z2 + lZ1z2 = 0 .

For u to be continuous in the cylinder, ϴ1u2 must be 2p-periodic. Thus, let’s begin with the eigenvalue problem ϴ″1u2 + mϴ1u2 = 0 , ϴ1 -p2 = ϴ1p2

and

-p 6 u 6 p , ϴ′1 -p2 = ϴ′1p2 .

In Example 2 we showed that this problem has nontrivial solutions for m = n2, n = 0, 1, 2, c, that are given by the Fourier series components (38)

ϴn 1u2 = An cos nu + Bn sin nu ,

where the An’s and Bn’s are arbitrary constants. The boundary conditions in (34) imply that Z102 = Z1b2 = 0. Therefore, Z must satisfy the eigenvalue problem Z″1z2 + lZ1z2 = 0 ,

06z6b,

Z102 = Z1b2 = 0 .

We have seen this eigenvalue problem several times before. Nontrivial solutions exist for l = 1mp>b2 2, m = 1, 2, 3, . . . and are given by the sine series components (39)

Zm 1z2 = Cm sin a

mpz b, b

where the Cm’s are arbitrary constants. Substituting for m and l in equation (35) gives r 2R″1r2 + rR′1r2 - a r 2 a

mp 2 b + n2 b R1r2 = 0 , b

0…r6a.

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624

Chapter 10

Partial Differential Equations

You should verify that the change of variables s = 1mpr>b2 transforms this equation into the modified Bessel’s equation of order n (page 483)† (40)

s2R″1s2 + sR′1s2 - 1s2 + n2 2R1s2 = 0 ,

0…s6

mpa . b

The modified Bessel’s equation of order n has two linearly independent solutions—the modified Bessel function of the first kind: ∞ 1s>22 2k+n , In 1s2 = a k = 0 k! Γ1k + n + 12

which remains bounded near zero, and the modified Bessel function of the second kind: Kn 1s2 = lim S y

n

p I-y 1s2 - Iy 1s2 , 2 sin yp

which becomes unbounded as s S 0. (Recall that Γ is the gamma function discussed in Section 7.6.) A general solution to (40) has the form CKn + DIn, where C and D are constants. Since u must remain bounded near s = 0, we must take C = 0. Thus, the desired solutions to (40) have the form (41)

Rmn 1r2 = Dmn In a

mpr b; b

n = 0, 1, c ,

m = 1, 2, c ,

where the Dmn’s are arbitrary constants. If we multiply the functions in (38), (39), and (41) and then sum over m and n, we obtain the following series solution to (32) and (34): (42)

∞ mpr mpz u1r, u, z2 = a am0 I0 a b sin a b b b m=1 ∞ ∞ mpr mpz + a a 1amn cos nu + bmn sin nu2 In a b sin a b, b b n=1 m=1

where the amn’s and bmn’s are arbitrary constants. The constants in (42) can be obtained by imposing the boundary condition (33). Setting r = a and rearranging terms, we have (43)

∞ mpa mpz f1u, z2 = a am0 I0 a b sin a b b b m=1 ∞ ∞ mpa mpz + a c a amn In a b sin a b d cos n u b b n=1 m=1 ∞ ∞ mpa mpz + a c a bmn In a b sin a b d sin nu . b b n=1 m=1

†

The modified Bessel’s equation of order n arises in many applications and has been studied extensively. We refer the reader to the text Special Functions by E. D. Rainville (Chelsea Publishing, New York, 1972), for details about its solution.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.7

Laplace’s Equation

625

Treating this double Fourier series like the one in Section 10.5 (page 600), we find p

b

(44)

am0 =

(45)

amn = 2

(46)

bmn = 2

L0 L-p b

f1u, z2 sin a p

L0 L-p b

p

L0 L-p

mpz mpa b du dz n pbI0 a b, b b

f1u, z2 sin a

mpz mpa b cos nu du dz n pbIn a b, b b

f1u, z2 sin a

mpz mpa b sin nu du dz n pbIn a b , nÚ1. b b

nÚ1,

Consequently, a formal solution to (32)–(34) is given by equation (42) with the constants amn and bmn determined by equations (44)–(46). ◆

Existence and Uniqueness of Solutions The existence of solutions to the boundary value problems for Laplace’s equation can be established by studying the convergence of the formal solutions that we obtained using the method of separation of variables. To answer the question of the uniqueness of the solution to a Dirichlet boundary value problem for Laplace’s equation, recall that Laplace’s equation arises in the search for steady-state solutions to the heat equation. Just as there are maximum principles for the heat equation, there are also maximum principles for Laplace’s equation. We state one such result here.†

Maximum Principle for Laplace’s Equation Theorem 9. Let u1x, y2 be a solution to Laplace’s equation in a bounded domain D with u1x, y2 continuous in D, the closure of D. 1 D = D ∪ 0D, where 0D denotes the boundary of D.) Then u1x, y2 attains its maximum value on 0D.

The uniqueness of the solution to the Dirichlet boundary value problem follows from the maximum principle. We state this result in the next theorem but leave its proof as an exercise (see Problem 19).

Uniqueness of Solution Theorem 10. Let D be a bounded domain. If there is a continuous solution to the Dirichlet boundary value problem ∆u1x, y2 = 0 in D , u1x, y2 = f1x, y2 on 0D , then the solution is unique.

†

A proof can be found in Section 8.2 of the text, Partial Differential Equations of Mathematical Physics, 2nd ed., by Tyn Myint-U (Elsevier North Holland, New York, 1983).

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626

Chapter 10

Partial Differential Equations

Solutions to Laplace’s equation in two variables are called harmonic functions. These functions arise naturally in the study of analytic functions of a single complex variable. Moreover, complex analysis provides many useful results about harmonic functions. For a discussion of this interaction, we refer the reader to an introductory text on complex analysis such as Fundamentals of Complex Analysis, 3rd ed., by E. B. Saff and A. D. Snider (Prentice Hall, Englewood Cliffs, N.J., 2003).

10.7

EXERCISES

In Problems 1–5, find a formal solution to the given boundary value problem. 02u 02u 1. + = 0, 06x6p, 06y61, 0x2 0y2 0u 0u 10, y2 = 1p, y2 = 0 , 0 … y … 1 , 0x 0x u1x, 02 = 4 cos 6x + cos 7x , 0 … x … p , u1x, 12 = 0 , 2.

0…x…p

0u 0u + = 0, 06x6p, 06y6p, 0x2 0y2 0u 0u 10, y2 = 1p, y2 = 0 , 0…y…p, 0x 0x u1x, 02 = cos x - 2 cos 4x , 0 … x … p , 2

2

u1x, p2 = 0 , 0 … x … p 3.

4.

5.

02u 02u + = 0, 06x6p, 06y6p, 0x2 0y2 u10, y2 = u1p, y2 = 0 , 0 … y … p , u1x, 02 = f1x2 , 0 … x … p , u1x, p2 = 0 , 0…x…p 02u 02u + = 0, 06x6p, 06y6p, 0x2 0y2 u10, y2 = u1p, y2 = 0 , 0 … y … p , u1x, 02 = sin x + sin 4x , 0 … x … p , u1x, p2 = 0 , 0 … x … p 02u 02u + = 0, 06x6p, 06y61, 0x2 0y2 0u 0u 10, y2 = 1p, y2 = 0 , 0 … y … 1 , 0x 0x u1x, 02 = cos x - cos 3x , 0 … x … p , u1x, 12 = cos 2x ,

In Problems 7 and 8, find a solution to the Dirichlet boundary value problem for a disk: 02u 1 0u 1 02u + = 0, 0…r62, + 0r 2 r 0r r 2 0u 2 -p … u … p u12, u2 = f1u2 ,

-p … u … p ,

for the given function f1u2. -p … u … p 7. f1u2 = 0 u 0 , 8. f1u2 = cos2 u , -p … u … p 9. Find a solution to the Neumann boundary value problem for a disk: 02u 1 0u 1 02u + = 0, 0…r6a, + -p … u … p , 0r 2 r 0r r 2 0u 2 0u 1a, u2 = f1u2 , -p … u … p . 0r 10. A solution to the Neumann problem (25)–(26) must also satisfy the consistency condition in (27). To show this, use Green’s second formula L LD

1Y∆u - u∆Y2 dx dy =

LeD

aY

eu eY -u b ds , en en

where 0>0n is the outward normal derivative and ds is the differential of arc length. 3Hint: Take y K 1 and observe that 0u>0n = 0u>0r.4 11. Find a solution to the following Dirichlet problem for an annulus: 02u 1 0u 1 02u + = 0, + 0r 2 r 0r r 2 0u 2 u11, u2 = sin 4u - cos u , u12, u2 = sin u ,

16r62,

-p … u … p ,

-p … u … p ,

-p … u … p .

0…x…p

6. Derive the polar coordinate form of the Laplacian given in equation (12).

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Section 10.7

12. Find a solution to the following Dirichlet problem for an annulus: 02u 1 0u 1 02u + = 0, 16r63, + 0r 2 r 0r r 2 0u 2 u11, u2 = 0 , -p … u … p ,

-p … u … p ,

u13, u2 = cos 3u + sin 5u , - p … u … p . 13. Find a solution to the following Dirichlet problem for an exterior domain: 02u 1 0u 1 02u + + = 0, 16r, 0r 2 r 0r r 2 0u 2 -p … u … p , u11, u2 = f1u2 u1r, u2 remains bounded as r S ∞ .

-p … u … p ,

02u 1 0u 1 02u + = 0, 16r, -p … u … p , + 0r 2 r 0r r 2 0u 2 0u 11, u2 = f1u2 , -p … u … p , 0r u1r, u2 remains bounded as r S ∞ . 15. Find a solution to the following Dirichlet problem for a half disk: 02u 1 0u 1 02u + = 0, 06r61, + 0r 2 r 0r r 2 0u 2 0…r…1, u1r, 02 = 0 ,

06u6p,

0…r…1,

u1r, p2 = 0 , u11, u2 = sin 3u ,

0…u…p,

u10, u2 bounded. 16. Find a solution to the following Dirichlet problem for a half annulus: 02u 1 0u 1 02u + = 0, + p 6 r 6 2p , 0r 2 r 0r r 2 0u 2 p … r … 2p , u1r, 02 = sin r , u1r, p2 = 0 ,

06u6p,

p … r … 2p ,

0…u…p. u1p, u2 = u12p, u2 = 0 , 17. Find a solution to the mixed boundary value problem 02u 1 0u 1 02u + + = 0, 16r63, 0r 2 r 0r r 2 0u 2 -p … u … p , u11, u2 = f1u2 , 0u 13, u2 = g1u2 , -p … u … p . 0r 18. Show that - Bn tanha

-p … u … p ,

627

where Cn = Bn > cosha

npb b. a

19. Prove Theorem 10 on the uniqueness of the solution to the Dirichlet problem. 20. Stability. Use the maximum principle to prove the following theorem on the continuous dependence of the solution on the boundary conditions: Theorem. Let f1 and f2 be continuous functions on 0D, where D is a bounded domain. For i = 1 and 2, let ui be the solution to the Dirichlet problem ∆u = 0 ,

14. Find a solution to the following Neumann problem for an exterior domain:

Laplace’s Equation

u = fi ,

in D , on 0D .

If the boundary values satisfy

0 f1 1x, y2 - f2 1x, y2 0 … e

for all 1x, y2 on 0D ,

where P 7 0 is some constant, then

0 u1 1x, y2 - u2 1x, y2 0 … e

for all 1x, y2 in D .

21. For the Dirichlet problem described in Example 3, let a = b = p and assume the potential on the lateral side 1r = p2 of the cylinder is f1u, z2 = sin z. Use equations (44)–(46) to compute the solution given by equation (42). 22. Invariance of Laplace’s Equation. A complex-valued function f1z2 of the complex variable z = x + iy can be written in the form f1z2 = u1x, y2 + iy1x, y2, where u and y are real-valued functions. If f1z2 is analytic in a planar region D, then its real and imaginary parts satisfy the Cauchy–Riemann equations in D; that is, in D 0u 0y = , 0x 0y

0u 0y = - . 0y 0x

Let f be analytic and one-to-one in D and assume its inverse f -1 is analytic in D′, where D′ is the image of D under f. Then 0y 0x = , 0u 0y

0y 0x = 0y 0u

are just the Cauchy–Riemann equations for f -1. Show that if f1x, y2 satisfies Laplace’s equation for 1x, y2 in D, then c1u, y2 J f 1 x1u, y2, y1u, y2 2 satisfies Laplace’s equation for 1u, y2 in D′.

npy npy npb b cosha b + Bn sinha b a a a

= Cn sinhc

np 1y - b2 d , a

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

628

Chapter 10

Partial Differential Equations

23. Fluid Flow Around a Corner. The stream lines that describe the fluid flow around a corner (see Figure 10.28) are given by f1x, y2 = k, where k is a constant and f, the stream function, satisfies the boundary value problem 02f

+

02f

= 0, x70, 0x 0y2 f1x, 02 = 0 , 0…x, 2

y70,

0…y.

f10, y2 = 0 ,

(u, ) = constant

=0

u

y Figure 10.29 Flow above a flat surface (x, y) = constant =0

x =0 Figure 10.28 Flow around a corner

(a) Using the results of Problem 22, show that this problem can be reduced to finding the flow above a flat plate (see Figure 10.29). That is, show that the problem reduces to finding the solution to 02c

+

02c

= 0, 0u2 0y2 c1u, 02 = 0 ,

y70,

where c and f are related as follows: c1u, y2 = f 1 x1u, y2, y1u, y2 2 with the mapping between 1u, y2 and 1x, y2 given by the analytic function f1z2 = z2. (b) Verify that a nonconstant solution to the problem in part (a) is given by c1u, y2 = y. (c) Using the result of part (b), find a stream function f1x, y2 for the original problem. 24. Unbounded Domain. Using separation of variables, find a solution of Laplace’s equation in the infinite rectangle 0 6 x 6 p, 0 6 y 6 ∞ that is zero on the sides x = 0 and x = p, approaches zero as y S ∞ , and equals f1x2 for y = 0.

-∞ 6 u 6 ∞ ,

-∞ 6 u 6 ∞ ,

Chapter 10 Summary Separation of Variables. A classical technique that is effective in solving boundary value problems for partial differential equations is the method of separation of variables. Briefly, the idea is to assume first that there exists a solution that can be written with the variables separated: e.g., u1x, t2 = X1x2T1t2. Substituting u = XT into the partial differential equation and then imposing the boundary conditions leads to the problem of finding the eigenvalues and eigenfunctions for a boundary value problem for an ordinary differential equation. Solving for the eigenvalues and eigenfunctions, one eventually obtains solutions un 1x, t2 = Xn 1x2Tn 1t2, n = 1, 2, 3, . . . that solve the partial differential equation and the homogeneous boundary conditions. Taking infinite linear combinations of the un 1x, t2 yields complete solutions of the form ∞

u1x, t2 = a cnun 1x, t2 . n=1

The coefficients cn are then obtained using the initial conditions or nonhomogeneous conditions.

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Chapter 10 Summary

629

Fourier Series. Let f1x2 be a piecewise continuous function on the interval 3 -L, L4. The Fourier series of f is the trigonometric series f1x2 ∼

∞ a0 npx npx + a e an cos a b + bn sin a bf , 2 n=1 L L

where the an’s and bn’s are given by Euler’s formulas: L

an =

1 npx f1x2 cos a b dx , L L-L L

bn =

1 npx f1x2 sin a b dx , L L-L L

n = 0, 1, 2, . . . ,

L

n = 1, 2, 3, c .

Let f be piecewise continuous on the interval 30, L4. The Fourier cosine series of f1x2 on 30, L4 is f1x2 ∼

∞ a0 npx + a an cos a b, 2 n=1 L

where L

an =

2 npx f1x2 cos a b dx . L L0 L

The Fourier sine series of f1x2 on 30, L4 is ∞ npx f1x2 ∼ a bn sin a b, L n=1

where L

bn =

2 npx f1x2 sin a b dx . L L0 L

Fourier series and the method of separation of variables are used to solve boundary value problems and initial-boundary value problems for the three classical equations: Heat equation

0u 02u = b 2. 0t 0x

Wave equation

2 02u 20 u a = . 0t 2 0x2

Laplace’s equation

02u 02u + = 0. 0x2 0y2

For the heat equation, there is a unique solution u1x, t2 when the boundary values at the ends of a conducting wire are specified along with the initial temperature distribution. The wave equation yields the displacement u1x, t2 of a vibrating string when the ends are held fixed and the initial displacement and initial velocity at each point of the string are given. In such a case, separation of variables yields a solution that is the sum of standing waves. For an infinite string with specified initial displacement and velocity, d’Alembert’s solution yields traveling waves. Laplace’s equation arises in the study of steady-state solutions to the heat and wave equations.

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630

Chapter 10

Partial Differential Equations

TECHNICAL WRITING EXERCISES FOR CHAPTER 10 1. The method of separation of variables is an important technique in solving initial-boundary value problems and boundary value problems for linear partial differential equations. Explain where the linearity of the differential equation plays a crucial role in the method of separation of variables. 2. In applying the method of separation of variables, we have encountered a variety of special functions, such as sines, cosines, Bessel functions, and modified Bessel functions. Describe three or four examples of partial differential equations that involve other special functions, such as Legendre polynomials, Hermite polynomials, and Laguerre polynomials. (Some exploring in the library may be needed; start with the table on page 483.) 3. A constant-coefficient second-order partial differential equation of the form a

can be classified using the discriminant D J b2 - 4ac. In particular, the equation is called hyperbolic

if

D 7 0,

elliptic

if

D 6 0.

and

Verify that the wave equation is hyperbolic and Laplace’s equation is elliptic. It can be shown that such hyperbolic (elliptic) equations can be transformed by a linear change of variables into the wave (Laplace’s) equation. Based on your knowledge of the latter equations, describe which types of problems (initial value, boundary value, etc.) are appropriate for hyperbolic equations and elliptic equations.

02u 02u 02u +b +c 2 = 0 2 0x 0y 0x 0y

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Projects for Chapter 10 A Steady-State Temperature Distribution in a Circular Cylinder When the temperature u inside a circular cylinder reaches a steady state, it satisfies Laplace’s equation ∆u = 0. If the temperature on the lateral surface 1r = a2 is kept at zero, the temperature on the top 1z = b2 is kept at zero, and the temperature on the bottom 1z = 02 is given by u1r, u, 02 = f1r, u2, then the steady-state temperature satisfies the boundary value problem 02u 1 0u 1 02u 02u + + + = 0, 0…r6a, -p … u … p , 0r 2 r 0r r 2 0u 2 0z2 -p … u … p , 0…z…b, u1a, u, z2 = 0 , u1r, u, b2 = 0 ,

0…r6a,

u1r, u, 02 = f1r, u2 ,

06z6b,

-p … u … p ,

0…r6a,

-p … u … p ,

where f1a, u2 = 0 for - p … u … p (see Figure 10.30). To find a solution to this boundary value problem, proceed as follows: (a) Let u1r, u, z2 = R1r2ϴ1u2Z1z2. Show that R, ϴ, and Z must satisfy the three ordinary differential equations r 2R″ + rR′ - 1r 2l + m2R = 0 , ϴ″ + mϴ = 0 , Z″ + lZ = 0 . (b) Show that ϴ1u2 has the form ϴ1u2 = A cos nu + B sin nu for m = n2, n = 0, 1, 2, c. (c) Show that Z1z2 has the form Z1z2 = C sinh 3b1b - z24 for l = - b2, where b 7 0. z u(r, , b) = 0

u(a, , z) = 0 u50

u(r, , 0) = f (r, ) Figure 10.30 Dirichlet problem for cylinder

631

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

632

Chapter 10

Partial Differential Equations

(d) Show that R1r2 has the form, for each n, Rn 1r2 = DJn 1br2 , where Jn is the Bessel function of the first kind. (e) Show that the boundary conditions require R1a2 = 0, and so Jn 1ba2 = 0 .

Hence, for each n, if 0 6 an1 6 an2 6 g 6 anm 6 g are the zeros of Jn, then bnm = anm >a .

Moreover, Rn 1r2 = DJn 1anmr>a2 . Some typical eigenfunctions are displayed in Figure 10.31. (f) Use the preceding results to show that u1r, u, z2 has the form ∞ ∞ anm 1b - z2 anmr b 1anm cos nu + bnm sin nu2sinha b, u1r, u, z2 = a a Jn a a a n=0 m=1

where anm and bnm are constants.

Eigenvalue

Eigenfunction J0(2.405r /a) a

r

J0(5.520r /a) a

r

J0(8.654r /a)

a

r

J0(11.792r /a) a

r

Figure 10.31 Bessel eigenfunctions. See also Figure 10.24, page 616.

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Projects for Chapter 10

633

(g) Use the final boundary condition and a double orthogonal expansion involving Bessel functions and trigonometric functions to derive the formulas a

a0m =

2p

a0mr 1 b r dr du , f1r, u2 J0 a 2 2 a pa sinh1a0mb>a23J1 1a0m 24 L0 L0

for m = 1, 2, 3, c, and for n, m = 1, 2, 3, c, a

2p

a

2p

anm =

anmr 2 b cos1nu2 r dr du , f1r, u2 Jn a 2 2 a pa sinh1anmb>a23Jn+1 1anm 24 L0 L0

bnm =

anmr 2 b sin1nu2 r dr du . f1r, u2 Jn a 2 2 a pa sinh1anmb>a23Jn+1 1anm 24 L0 L0

(The orthogonality relations for Bessel function expansions are described by equations (21), (22) in Section 11.7, page 689.)*

B Laplace Transform Solution of the Wave Equation Laplace transforms can be used to solve certain partial differential equations. To illustrate this technique, consider the initial-boundary value problem (1) (2) (3) (4) (5)

02u 02u = a2 2 , 06x6 ∞, 2 0t 0x t70, u10, t2 = h1t2 , u1x, 02 = 0 , 0u 1x, 02 = 0 , 0t lim u1x, t2 = 0 ,

xS ∞

t70,

06x6 ∞, 06x6 ∞, tÚ0.

This problem arises in studying a semi-infinite string that is initially horizontal and at rest and where one end is being moved vertically. Let u1x, t2 be the solution to (1)–(5). For each x, let U1x, s2 J ℒ5u1x, t2 6 1x, s2 =

∞

L0

e-stu1x, t2 dt .

(a) Using the fact that ℒe

02u 02 f = ℒ5u6 , 0x2 0x2

show that U1x, s2 satisfies the equation 02U , 06x6 ∞. 0x2 (b) Show that the general solution to (6) is (6)

s2U1x, s2 = a2

U1x, s2 = A1s2e-sx>a + B1s2esx>a , where A1s2 and B1s2 are arbitrary functions of s. †

All references to Chapters 11–13 refer to the expanded text, Fundamentals of Differential Equations and Boundary Value Problems, 7th ed.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

634

Chapter 10

Partial Differential Equations

(c) Since u1x, t2 S 0 as x S ∞ for all 0 … t 6 ∞ , we have U1x, s2 S 0 as x S ∞ . Use this fact to show that the B1s2 in part (b) must be zero. (d) Using equation (2), show that A1s2 = H1s2 = ℒ5h61s2 , where A1s2 is given in part (b). (e) Use the results of parts (b), (c), and (d) to obtain a formal solution to (1)–(5).

C Green’s Function Let Ω be a region in the xy-plane having a smooth boundary 0Ω. Associated with Ω is a Green’s function G1x, y; j, h2 defined for pairs of distinct points 1x, y2, 1j, h2 in Ω. The function G1x, y; j, h2 has the following property. Let ∆u J 02u>0x2 + 02u>0y2 denote the Laplacian operator on Ω; let h1x, y2 be a given continuous function on Ω; and let f1x, y2 be a given continuous function on 0Ω. Then, a continuous solution to the Dirichlet boundary value problem

∆u1x, y2 = h1x, y2 ,

in Ω ,

u1x, y2 = f1x, y2 ,

on 0Ω ,

is given by (7)

u1x, y2 =

G1x, y; j, h2 h1j, h2 dj dh LL Ω

+

L0Ω

f1j, h2

0G 1x, y; j, h2 ds1j, h2 , 0n

where n is the outward normal to the boundary 0Ω of Ω and the second integral is the line integral, with respect to arc length, around the boundary of Ω with the interior of Ω on the left as the boundary is traversed. In (7) we assume that 0Ω is sufficiently smooth so that the integrands and integrals exist. When Ω is the upper half-plane, the Green’s function is† G1x, y; j, h2 =

1x - j2 2 + 1y - h2 2 1 ln c d . 4p 1x - j2 2 + 1y + h2 2

(a) Using (7), show that a solution to (8)

∆u1x, y2 = h1x, y2 ,

(9)

u1x, 02 = f1x2 ,

-∞ 6 x 6 ∞ ,

06y,

-∞ 6 x 6 ∞ ,

is given by u1x, y2 =

∞ f1j2 y dj p L-∞ 1x - j2 2 + y2 ∞ ∞ 1x - j2 2 + 1y - h2 2 1 ln c d h1j, h2 dj dh . + 4p L0 L-∞ 1x - j2 2 + 1y + h2 2

†

Techniques for determining Green’s functions can be found in texts on partial differential equations such as Partial Differential Equations of Mathematical Physics, 2nd ed., by Tyn Myint-U (Elsevier North Holland, New York, 1983), Chapter 10.

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Projects for Chapter 10

635

(b) Use the result of part (a) to determine a solution to (8)–(9) when h K 0 and f K 1. (c) When Ω is the interior of the unit circle x2 + y2 = 1, the Green’s function, in polar coordinates 1r, u2 and 1r, f2, is given by G1r, u; r, f2 =

1 ln 3r 2 + r2 - 2rr cos1f - u24 4p 1 1 ln3r 2 + r-2 - 2rr-1 cos1f - u24 ln r . 4p 4p

Using (7), show that the solution to 02u 1 0u 1 02u + = 0, 0…r61, + 0r 2 r 0r r 2 0u 2 0 … u … 2p , u11, u2 = f1u2 ,

0 … u … 2p ,

is given by u1r, u2 =

1 2p L0

2p

1 - r2 f1f2 df . 1 + r 2 - 2r cos1f - u2

This is known as Poisson’s integral formula. (d) Use Poisson’s integral formula to derive the following mean value property for solutions to Laplace’s equation.

Mean Value Property Theorem 11. If u1x, y2 satisfies ∆u = 0 in a bounded domain Ω in R2 and 1x0, y0 2 lies in Ω, then, (10)

u1x0, y0 2 =

1 2p L0

2p

u1x0 + r cos u, y0 + r sin u2 du

for all r 7 0 for which the disk B1x0, y0; r2 = 5 1x, y2: 1x - x0 2 2 + 1y - y0 2 2 … r 2 6 lies entirely in Ω.

[Hint: Use a change of variables that maps the disk B1x0, y0; r2 to the unit disk B10, 0; 12.4

D Numerical Method for ∆u = f on α Rectangle

Let R denote the open rectangle R = 5 1x, y2: a 6 x 6 b, c 6 y 6 d6 and 0R be its boundary. Here we describe a numerical technique for solving the generalized Dirichlet problem (11)

02u 02u + = f1x, y2 , 0x2 0y2 u1x, y2 = g1x, y2 ,

for 1x, y2 in R , for 1x, y2 on 0R .

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

636

Chapter 10

Partial Differential Equations

y yn = d k

R y0 = c 0

x0 = a

x1

xm - 1 x m = b

h

x

Figure 10.32 Rectangular grid

The method is similar to the finite-difference technique discussed in Chapter 4. We begin by selecting positive integers m and n and step sizes h and k so that b - a = hm and d - c = kn. The interval 3a, b4 is now partitioned into m equal subintervals and 3c, d4 into n equal subintervals. The partition points are

xi = a + ih , yj = c + jk ,

0…i…m, 0…j…n

(see Figure 10.32). The (dashed) lines x = xi and y = yj are called grid lines and their intersections 1xi, yj 2 are the mesh points of the partition. Our goal is to obtain approximations to the solution u1x, y2 of problem (11) at each interior mesh point, i.e., at 1xi, yj 2 where 1 … i … m - 1, 1 … j … n - 1. 3Of course, from (11), we are given the values of u1x, y2 at the boundary mesh points, e.g., u1x0, yj 2 = g1x0, yj 2, 0 … j … n.4 The next step is to approximate the partial derivatives 02u>0x2 and 02u>0y2 using the centered-difference approximations (12)

02u 1 1xi, yj 2 ≈ 2 3u1xi+1, yj 2 - 2u1xi, yj 2 + u1xi-1, yj 24 2 0x h

1for 1 … i … m - 12 ,

1 02u 1xi, yj 2 ≈ 2 3u1xi, yj+1 2 - 2u1xi, yj 2 + u1xi, yj-1 24 2 0y k

1for 1 … i … n - 12 .

and (13)

These approximations are based on the generic formula y″1x2 = 3y1x + h2 - 2y1x2 + y1x - h24 >h2 + 3terms involving h2, h3, c4 , which follows from the Taylor series expansion for y1x2. (a) Show that substituting the approximations (12) and (13) into the Dirichlet problem (11) yields the following system: (14)

h 2 h 2 2c a b + 1 d ui, j - 1ui+1, j + ui - 1, j 2 - a b 1ui, j+1 + ui, j-1 2 = - h2f1xi, yj 2 k k for i = 1, 2, . . . , m - 1, and j = 1, 2, c, n - 1; u0, j = g1a, yj 2 ,

um, j = g1b, yj 2 ,

ui,0 = g1xi, c2 ,

ui,n = g1xi, d2 ,

j = 0, 1, c, n , i = 1, 2, c, m - 1 ,

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637

(xi , yj + 1)

(xi - 1 , yj )

(xi , yj)

(xi + 1 , yj )

(xi , yj -1) Figure 10.33 Approximate solution ui,j at 1xi, yj 2 is obtained from values at the ends of the cross

where ui,j approximates u1xi, yj 2. Notice that each equation in (14) involves approximations to the solution that appear in a cross centered at a mesh point (see Figure 10.33). (b) Show that the system in part (a) is a linear system of 1m - 121n - 12 unknowns in 1m - 121n - 12 equations. (c) For Laplace’s equation where f1x, y2 K 0, show that when h = k, equation (14) yields ui, j =

1 1u + ui+1, j + ui, j-1 + ui, j+1 2 . 4 i-1, j

Compare this averaging formula with the mean value property of Theorem 11 (Project C). (d) For the square plate R = 5 1x, y2: 0 … x … 0.4, 0 … y … 0.46 , the boundary is maintained at the following temperatures: u10, y2 = 0 ,

u10.4, y2 = 10y ,

0 … y … 0.4 .

u1x, 02 = 0 ,

u1x, 0.42 = 10x ,

0 … x … 0.4 .

Using the system in part (a) with f1x, y2 K 0, m = n = 4, and h = k = 0.1, find approximations to the steady-state temperatures at the mesh points of the plate. [Hint: It is helpful to label these grid points with a single index, say p1, p2, . . . , pq, choosing the ordering in a book-reading sequence.]

E The Telegrapher’s Equation and the Cable Equation Electrical engineers often employ transmission lines to guide electromagnetic waves from a source to a device. The transmission line may take any of several forms, such as depicted in Figure 10.34(a) on page 638. They have common characteristics: uniformity in one direction (z) and uniform capacitance, inductance, longitudinal resistance, and transversal resistance per unit length in the z direction (if they are operated at low frequency). Thus any short length dz of the line can be modeled as an RLC circuit as depicted in Figure 10.34(b). In the figure, the circuit parameters are interpreted per unit length; thus the incremental inductance is L dz, the incremental capacitance is C dz, the (very low) incremental longitudinal

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

638

Chapter 10

Partial Differential Equations

Wire pair

Coax

Microstrip

Coplanar

(a) R dz

L dz G dz

I(z, t)

R dz

L dz

+ V(z, t) -

C dz

R dz

I(z+dz, t) C dz

G dz

z

L dz

+ V(z+dz, t) -

G dz

C dz z

z+dz dz (b)

Figure 10.34 Transmission lines (a) common forms (b) circuit model

resistance is R dz, and the (very low) incremental transversal conductance (reciprocal of resistance) is G dz. (a) Apply the circuit analysis of Section 5.7 to the length dz of the transmission line to derive the equations (15)

V1z + dz, t2 = V1z, t2 - 1R dz2I1z, t2 - 1L dz2

0I1z, t2

(16)

I1z + dz, t2 = I1z, t2 - 1G dz2V1z, t2 - 1C dz2

0V1z, t2

0t 0t

, ,

(b) Divide these equations by dz and take limits to obtain the system (17)

0V1z, t2 0z 0I1z, t2 0z

= - RI1z, t2 - L

0I1z, t2

= - GV1z, t2 - C

, 0t 0V1z, t2 0t

.

(c) Eliminate I1z, t2 from the system (17) to derive (18)

02V1z, t2 0z

2

= LC

02V1z, t2 0t

2

+ 3RC + LG4

0V1z, t2 0t

+ RGV1z, t2 .

[Hint: Differentiate the first equation with respect to z and the second with respect to t, and observe the equality of the mixed partials of I1z, t2.] Equation (18) is known as the telegrapher’s equation for the transmission line; (17) is also known as the system of telegrapher’s equations. (d) Make the substitution V1z, t2 = T1t2v1z, t2 in the telegrapher’s equation. For what choices of T1t2 is the term proportional to 0v>0t eliminated? For what choices of T1t2 is the term proportional to v eliminated? (e) Show that the telegrapher’s equation for a lossless line, where the longitudinal resistance R and the transversal conductance G of the line are zero, reduces to the wave equation of Section 10.6. What is the speed of the voltage waves on a lossless line?

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Projects for Chapter 10

639

(f ) Naturally engineers are interested in oscillatory waves on transmission lines, and thanks to Euler’s formula (Section 4.3, page 166), solutions to the system (17) can be constructed from the real and imaginary parts of solutions of the form V1z, t2 = V1z2eivt and I1z, t2 = I1z2eivt. (Elaboration of this procedure is given in Project B, Chapter 7, page 418.) Derive a first-order matrix system of the form (19)

c

V1z2 ′ V1z2 d = Ac d I1z2 I1z2

for such solutions, where A is a constant matrix. (g) For the lossless case 1R = G = 02, construct the matrix exponential eA t for the matrix in (19) by direct calculation of the series expansion in equation (2) of Section 9.8 (page 545). Express a general solution to the oscillatory transmission line system (19) for the lossless line. Ranvier node

Axon

Neuron Figure 10.35 Axon model

(h) In the neurons of an invertebrate’s central nervous system, electric signals are propagated between Ranvier nodes along axons, which are fibers containing an axoplasm liquid and insulated by a myelin sheaf; see Figure 10.35. Their electrical properties are quite similar to those of the coaxial cable in Figure 10.34 on page 638. However, the inductance is negligible. Therefore the voltages in a myelinated axon propagate in accordance with the cable equation, which is the transmission line equation (18) with L omitted: (20)

02V1z, t2 0z

2

= RC

0V1z, t2

+ RGV1z, t2 .

0t

(i) Use the strategy of part (d) to find T1t2 such that the substitution V1z, t2 = T1t2v1z, t2 results in an analog to the heat equation (Section 10.5) (21)

02v1z, t2 0z

2

= RC

0v1z, t2 0t

.

(Therefore propagation along an axon proceeds not like a wavefront, but rather like a warm front.)

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Appendices

APPENDIX

A

Review of Integration Techniques In solving differential equations, we cannot overemphasize the importance of integration techniques. Indeed, we sometimes refer to the problem of solving a differential equation by asking, “How can we integrate this equation?” This section is devoted to reviewing three of the standard techniques that will be useful for integrating many of the functions encountered in this text. Of course, one can always search in a table of integrals (a brief table appears on the inside back cover of this text), but familiarity with tricks of integration is a worthwhile investment of time for several reasons: (i) we may not find the particular integral we need in the table; (ii) using one of the techniques may be quicker; and (iii) we might be curious to know how the various integrals in the table were obtained. So what techniques can we use for the following integrals? (a) (d)

L

x 22x + 1 dx

7x2 + 10x - 1 dx 2 L x + 3x - x - 3 3

(b) (e)

x2 L 29 - x2

dx

(c)

L

x ln x dx

2

L

ex dx

For (a) we can use algebraic substitution, while (b) can be handled with trigonometric substitution. For (c) we have available integration by parts, and the integral in (d) will yield to partial fractions. But try as we may, no trick is sufficient to give (e) in a finite number of terms. By this we mean that if we have available the elementary functions (rational functions, trigonometric functions, inverse trigonometric functions, exponential functions, and logarithmic functions) and permit use of the standard operations—addition, subtraction, multiplication, division, taking roots, and composite functions—a finite number of times, we cannot find a for2 mula for the antiderivative of ex . A similar conundrum arises with integration of 21 + cos 2x. So we can add a fourth reason for honing our skills with integration techniques: They will help us to recognize those integrals that can be expressed in a finite number of terms. In this context, we are speaking of integration in closed form—analytically—as opposed to numerical integration. Thus integration is the inverse of differentiation; it requires finding an antiderivative. As a warmup, then, we begin this appendix by jogging your recollection of rules for derivatives. We hope you will find our approach a little different and refreshing.

Differentiation Of course you remember that the derivative of a constant equals zero, and the derivative of a sum equals the sum of the derivatives. So let’s jump in with the basic power rule. d p Derivative of a power function x = pxp - 1. dx A-1

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A-2

Appendices

This holds for any constant exponent p, be it integer, fraction, or negative. Examples:

d 100 x = 100x99, dx

Derivative of the exponential

d 7>2 7 x = x5>2, dx 2

d 1>3 1 x = x -2>3, dx 3

d -e x = -ex -e - 1 dx

d x e = ex. dx

This is the essential feature of the exponential function. d d Derivative of trig functions sin x = cos x, cos x = -sin x. dx dx Remember which one gets the minus sign. d d d Derivative of a product 1fg2 = f g + g f . dx dx dx d 2 2 Example: 1x sin x2 = x cos x + 1sin x212x2 dx d d d g = 1fg2 - g f, we begin to see how to build an dx dx dx d d antiderivative (fg) for f g, by compensating with - g f . This is the essence of integration dx dx by parts.) d d g f-f g d f dx dx Derivative of a quotient a b = . 2 dx g g (Bonus: If we rewrite this rule as f

(If you forget this, you can just apply the product rule to f g-1) Examples:

1cos x21cos x2 - 1sin x21 -sin x2 d d sin x cos2x + sin2x = tan x = = dx dx cos x cos2x cos2x 1 = = sec2x = 1 + tan2x cos2x

(Bonus: if you reverse the roles, so tan x = y and x = arctan y, and invert, you get a new rule: d 1 arctan y = .) dy 1 + y2 1cos x2102 - 11 -sin x2 sin x d d 1 = = = 1sec x21tan x2 sec x = 2 dx dx cos x cos x cos2x

Chain rule

Examples:

df1u2 du d du d f1u2 = c f1u2 d c d . aThink .b dx du dx du dx d 1 d 17x2 + x - 22 - 1>4 = c - 17x2 + x - 22 - 5>4 d c 17x2 + x - 22 d dx 4 dx 1 14x + 1 = . 2 4 17x + x - 22 5>4 d x2 d 2 2 e = 3e1x 2 4 c 1x2 2 d = 2xe x . dx dx

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Appendix A

Review of Integration Techniques

A-3

Now we begin our review of integration with the method of substitution (also called change of variable), which is derived from the chain rule.

Method of Substitution d f1g1x22 = f ′1g1x22g′1x2 we obtain dx

From the formula

L

f ′ 1 g1x2 2 g′1x2dx = f 1 g1x2 2 + C .

Thus if we can recognize an integrand as being of the form f ′1g1x22g′1x2, then integration is immediate via the substitution u = g1x2. Here are some simple illustrations: •

L

sin3 2x cos 2x dx

Since d1sin 2x2 = 2 cos 2x dx, the substitution u = sin 2x reduces the integration to that of u3 >2 du. •

4

L

x3e5x dx

Observe that x3 is essentially (except for a constant factor) the derivative of 5x4, so we set u = 5x4 and integrate eu >20 du. •

•

6 ln 1x + 52

dx x+5 Recognizing that 1> 1x + 52 is the derivative of ln 1x + 52, we put u = ln 1x + 52 and integrate 6u du. L

L

tan3 x sec x dx

Here it is convenient to use the identity tan2 x = sec2 x - 1 and write the integrand as 1sec2 x - 12sec x tan x. Recalling that 1sec x2′ = sec x tan x, the substitution u = sec x reduces the problem to integrating 1u2 - 12 du. When a portion of the integrand involves some fractional power of an expression, an algebraic substitution can sometimes simplify the integration process. The technique is amply illu-strated in the following example.

Example 1

Find (1)

Solution

I1 J

L

x22x + 1 dx .

The difficulty here arises from the square root factor, so let’s try to rid ourselves of this problem term by making the substitution u = 22x + 1. Here are the details: (2)

u2 = 2x + 1,

x =

u2 - 1 , 2

dx = u du .

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A-4

Appendices

If we use these expressions in (1), we find that in terms of the new variable u, (3)

L

1 u2 - 1 b 1u2 1u du2 = 1u4 - u2 2 du 2 2 L L 1 u5 u3 = a - b +C 2 5 3 u3 = 13u2 - 52 + C . 30 a

x22x + 1 dx =

Returning to the original variable x, via u = 22x + 1, we deduce that I1 =

12x + 12 3>2 30

3312x + 12 - 54 + C =

12x + 12 3>2 15

13x - 12 + C . ◆

We remark that the substitution u = 22x + 1 is not the only substitution that will allow us to compute the integral in (1). Another substitution that works just as well is u = 2x + 1. In a similar manner we can compute the integrals x3

L 11 + x2 2 3>2

dx

ex

and

1>3

+1

dx

L x2>3

by making the substitution u = 1 + x2 in the first and u = x1>3 + 1 in the second. The integration of certain algebraic expressions can sometimes be simplified by utilizing trigonometric substitutions. The effectiveness of these substitutions in handling quadratic expressions under a radical derives from the familiar identities 1 - sin2 u = cos2 u , 1 + tan2 u = sec2 u, sec2 u - 1 = tan2 u . Accordingly, if the integrand involves: (I)

2a2 - x2,

set x = a sin u,

(II)

2a + x ,

set x = a tan u,

(III)

2x - a ,

set x = a sec u.

2

2

2 2

These substitutions are easy to remember if we associate with each of the cases a right triangle whose sides are a, x, and a suitable radical.† The labeling of the triangle depends on the particular case at hand. For example, in (II) it is clear that the hypotenuse must be 2a2 + x2, while in (I) the hypotenuse must be a, and in (III) the hypotenuse is x. Example 2 Solution

Find the indefinite integral I2 J

x2 dx L 29 - x2

.

This integral is of type (I) with a = 3. We first construct a triangle as shown in Figure A.1. x = 3 sin u, 3 u

x

√9 2 x 2 = 3 cos u, dx = 3 cos u du.

√9 2 x 2 Figure A.1 Triangle for Example 2 The radical sign may not be visible. For example, if the integrand is 1> 1a2 + x2 2 5, this comes under case (II), because 1> 1a2 + x2 2 5 = 1> 1 2a2 + x2 2 10.

†

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Appendix A

Review of Integration Techniques

A-5

From this triangle we see immediately that I2 =

x2 dx L 29 - x2

13 sin u2 2 3 cos u du = 9 sin2 u du . 3 cos u L L

=

To integrate sin2 u we use the trigonometric identity sin2 u =

1 - cos 2u 2

and obtain I2 = =

9 9 9 11 - cos 2u2du = u - sin 2u + C 2L 2 4 9 9 u - sin u cos u + C . 2 2

Returning to the original variable x, we can derive the expressions for u, sin u, and cos u directly from the triangle in Figure A.1. We find that x u = arcsin , 3

sin u =

x , 3

cos u =

29 - x2 . 3

Hence I2 =

9 9 x 9 x 29 - x2 x 1 arcsin + C = arcsin - x29 - x2 + C . ◆ 2 3 23 3 2 3 2

Integration by Parts The method of integration by parts is of both practical and theoretical importance. It is based on the product rule for differentiation which we now write in the format d dy du 1uy2 = u +y = uy′ + yu′ . dx dx dx Integrating both sides of this equation and transposing, we obtain (4)

L

uy′dx = uy -

L

yu′dx .

Since dy = y′dx and du = u′dx, equation (4) can be put in the more compact form (5)

L

u dY = uY -

L

Y du .

Equation (5) is the basic equation for integration by parts. The idea is that the integrand on the left side may be very complicated, but if we select the part for u and the part for dy quite carefully, the integral on the right side may be much easier. (Some experimentation may be necessary to find the right selection.)

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

A-6

Appendices

Example 3 Solution

Find the indefinite integral I3 J

L

x ln x dx.

There are two natural ways of selecting u and dy so that the product is x ln x dx: First possibility: Second possibility:

let u = x, let u = ln x,

dy = ln x dx. dy = x dx.

With the first selection we cannot easily find y, the integral of ln x, so we come to a dead end. In the second case we can find y. We arrange the work as follows. Let u = ln x, dy = x dx . Then du =

1 x2 dx, y = . x 2

Making these substitutions in (5), we have for the left side x ln x dx . 1ln x21x dx2 = u dy = L L L and for the right side (6)

uy -

L

y du =

x2 x2 x2 1 1 dx = ln x ln x x dx . x 2 2 2L L2

Therefore, by (5), (7)

I3 =

L

x ln x dx =

x2 1 ln x - x2 + C . ◆ 2 4

Note that we used y = x2 >2 rather than y = x2 >2 + C. It is clear that equation (5) is correct for every value of C, and in most cases it is simpler to set C = 0. The arbitrary constant that we need can be added in the last step, as we have done in going from equation (6) to (7). It is sometimes necessary to apply integration by parts more than once in order to evaluate a particular integral, such as I4 J

x2ex dx . L In the first application, we set u = x2, dy = ex dx to deduce that (8)

I4 = x2ex - 2

xex dx

L and to evaluate the last integral in (8) we perform a second application of integration by parts with u = x, dy = ex dx. There are some integrals that may appear at first not to be split into two parts, but we need to keep in mind that “1” can be considered as a factor of any expression. For example, to compute the integrals I5 J

arctan13x2 dx and I6 J ln x dx . L L an effective splitting for evaluating I5 is u = arctan13x2, dy = 1 dx, while for I6 an appropriate choice is u = ln x, dy = 1 dx. The reader should verify that integration by parts then reduces these integrals to routine problems.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Appendix A

Review of Integration Techniques

A-7

Partial Fractions

Any rational function, that is, any function of the form R1x2 = P1x2 >Q1x2, where P1x2 and Q1x2 are polynomials, can be integrated in finite terms. The first step in the process is to compare the degree of P1x2 with that of Q1x2. If deg P Ú deg Q, then division gives P1x2 p1x2 (9) = s1x2 + , Q1x2 Q1x2 where s1x2 and p1x2 are polynomials and deg p 6 deg Q. Since the integration of the polynomial s1x2 is a trivial matter, we need only focus on the integration of the remainder term p1x2 >Q1x2. The idea of the method of partial fractions is to express p1x2 >Q1x2 as the sum of simpler fractions that are easy to integrate. For example, knowing that (10)

R1x2 =

7x2 + 10x - 1 x + 3x2 - x - 3 3

can be written as 1 4 2 + + x-1 x+1 x+3 leads immediately to the evaluation R1x2 dx = 2 ln 0 x - 1 0 + ln 0 x + 1 0 + 4 ln 0 x + 3 0 + C . L The key to finding such a simple fraction representation is to first factor the denominator polynomial Q1x2. Indeed, in (10) the denominator is just the product 1x - 121x + 121x + 32. In general, we recall that any polynomial Q1x2 with real coefficients can be factored into a product of linear and irreducible quadratic factors with real coefficients. There are four cases that may arise in the factorization: (i) nonrepeated linear factors; (ii) repeated linear factors; (iii) nonrepeated quadratic factors; and (iv) repeated quadratic factors. In the case when a nonrepeated linear factor 1x - a2 occurs in the factorization of Q1x2, we associate the partial fraction A> 1x - a2, where A is a constant to be determined. If a linear factor 1x - a2 is repeated m times in the factorization of Q1x2, we associate the sum of m fractions m Ak a 1x - a2 k ,

k=1

where the m constants Ak are to be determined. In the case of a nonrepeated quadratic factor ax2 + bx + c (where b2 - 4ac 6 0) we associate the single fraction Ax + B , ax2 + bx + c where we now have two constants in the numerator to be determined, while if this quadratic factor is repeated m times, then we associate the sum Akx + Bk a 1ax2 + bx + c2 k , m

k=1

which involves the determination of 2m constants Ak and Bk. We illustrate the approach in the following example. Further discussion of partial fractions appears in Section 7.4, page 370.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

A-8

Appendices

Example 4 Solution

Find the indefinite integral I7 =

3x3 + 3x2 + 3x + 2 dx. x3 1x + 12 L

Since the factor x is repeated three times in the denominator, while 1x + 12 is a nonrepeated factor, our partial fraction decomposition must be (11)

A B C D 3x3 + 3x2 + 3x + 2 = + + + , x + 1 x x2 x3 x3 1x + 12

where A, B, C, and D are unknowns to be determined. Multiplying both sides of (11) by x3 1x + 12, we have (12)

3x3 + 3x2 + 3x + 2 = Ax3 + Bx2 1x + 12 + Cx1x + 12 + D1x + 12 . In order for (12) to be true for all x, the corresponding coefficients must be the same: A+B B+C C+D D

The coefficients of x3 yield The coefficients of x2 yield The coefficients of x1 yield The coefficients of x0 yield

= = = =

3. 3. 3. 2.

Solving this system of four linear equations in four unknowns gives A = 1, B = 2, C = 1, and D = 2. Hence I7 =

2 1 1 2 1 1 + + 2 + 3 b dx = ln 0 1x + 12x2 0 - - 2 + C . ◆ x x x 1 + x x x L a

We remark that the coefficients A and D in the representation (11) can be determined directly. Indeed, if we set x = 0 in (12), we obtain 2 = D, and if we set x = -1 in (12), we obtain -3 + 3 - 3 + 2 = A1 -12 or A = 1. But B and C cannot be obtained so readily. If we are fortunate to have a denominator with only nonrepeated linear factors, then this simple substitution method will work to quickly determine all the unknown coefficients. In dealing with the integration of rational functions that have a quadratic factor in the denominator, the familiar formula 1 dx = arctan x + C 1 + x2 L plays a central role. For example, by completing the square and making a simple change of variables, we find 3 3 dx = dx = 3 arctan1x + 22 + C . 2 2 + 5 22 x + 4x 1x + +1 L L

A

EXERCISES

In Problems 1–26, find the indicated indefinite integral. 1. 3.

L

x3 11 + x2 2 3/2 dx 5

4 6t

L

4t e dt

2. 4.

L

x cos 12x2 dx 3 3x

L

x e dx

5. 7.

L

cos 2 1pu2 du

t+1 dt 2 Lt + 4

6. 8.

x1>3 L x1>3 + 1 L

dx

10 sec2 15x2 dx

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Appendix A

9. 11. 12. 14. 16. 18. 20. 22.

1 L 21 - 2x2

dx

10.

1 dx 2 L 2x + 4x + 4

4 dx 1x 121x 221x - 32 L L

sec12x2 tan 12x2 dx

2x + 41 dx L x + 5x - 14 2

x3 + 4x + 1 dx 2 L x + 3x + 4x + 2 3

x2 dx Lx - 1 2

L L

sec13u + 12 du 36x5exp12x3 2 dx

13.

24. 26.

29x2 - 1 dx x L

29.

19.

31.

21. 23.

L

y sinh y dy 1

L 1x + 42 3>2 2

L

dx

t ln 1t + 32 dt

sin 2x dx

25.

A-9

sin t dt 2 L 1 + cos t

e3y dy 6y L1 + e

In Problems 27–34, use an appropriate trigonometric identity to help find the indicated integral.

x3 + 2x2 + 8 dx 2 2 L x 1x + 42 1 17. dx L x ln x

15.

L

Review of Integration Techniques

27.

33.

L L L L

sin 2/5x cos3x dx

28.

cos 13x2 cos 17x2 dx

30.

tan4u sec6u du

32.

sin2 13x2 cos2 13x2 dx

34.

L L L L

tan3x2sec x dx cot3u du cos1>3x sin 3 x dx cot2x csc4 x dx

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Appendix B

APPENDIX

B

Newton’s Method

A-9

Newton’s Method To solve an equation g1x2 = 0, we must find the point or points where the graph of y = g1x2 meets the x-axis. One procedure for approximating a solution is Newton’s method. x be a root of g1x2 = 0 and let x1 be To motivate Newton’s method geometrically, we let ∼ our guess at the value of ∼ x . If g1x1 2 = 0, we are done. If g1x1 2 ≠ 0, then we are off by some amount that we call dy (see Figure B.1). Then dy = g′1x1 2 , dx and so (1)

dx =

dy . g′1x1 2 y = g(x)

( x1, g(x1)) dy = g(x1) ~ x

x2

dx

x1

x

Figure B.1 Tangent line approximation of root

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

A-10 Appendices

Now dx = x1 - x2 or x2 = x1 - dx, where x2 is at the point where the tangent line through 1 x1, g1x1 2 2 intersects the x-axis (Figure B.1). Using equation (1) and the fact that dy = g1x1 2, we obtain x2 = x1 -

g1x1 2 dy = x1 , g′1x1 2 g′1x1 2

x. which we use as the next approximation to the root ∼ Repeating this process with x2 in place of x1, we obtain the next approximation x3 to the root ∼ x . In general, we find the next approximation xn + 1 by the formula (2)

xn + 1 = xn −

g1xn 2 , g′1xn 2

n = 1, 2, . . . .

The process is illustrated in Figure B.2. x , then the sequence of iterations If the initial guess x1 is sufficiently close to a root ∼ 5xn 6∞n = 1 usually converges to the root ∼ x . However, if we make a bad guess for x1, then the x. process may lead away from ∼ y = g(x)

~ x x4 x3

x2

x1

x

Figure B.2 Sequence of iterations converging to root

Example 1

Find a root to four decimal places of the equation (3)

Solution

x3 + 2x - 4 = 0 .

Setting g1x2 = x3 + 2x - 4, we find that g′1x2 = 3x2 + 2 is positive for all x. Hence, g is increasing and has at most one zero. Furthermore, since g112 = -1 and g122 = 8, this zero must lie between 1 and 2. Thus, we begin the procedure with the initial guess x1 = 1.5. For g1x2 = x3 + 2x - 4, equation (2) becomes (4)

xn + 1 = xn -

xn3 + 2xn - 4 3x2n + 2

,

n = 1, 2, . . . .

With x1 = 1.5, equation (4) gives x2 = 1.5 -

11.52 3 + 211.52 - 4 311.52 + 2 2

= 1.5 -

2.375 ≈ 1.22857 . 8.75

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Appendix B

Newton’s Method

A-11

Using x2 to compute x3 and so on, we find x3 = 1.18085 , x4 = 1.17951 , x5 = 1.17951 , where we have rounded off the computations to five decimal places. Since x4 and x5 agree to four decimal places and we are uncertain of the fifth decimal place because of roundoff, we surmise that the root ∼ x of (3) agrees with 1.1795 to four decimal places. Indeed, g11.17952 = -0.00005 . . .

and

g11.17962 = 0.00056 . . . ,

x = 1.1795 . . . . ◆ so 1.1795 6 ∼ x 6 1.1796. Consequently, ∼ Observe that Newton’s method transforms the problem of finding a root to the equation g1x2 = 0 into the problem of finding a fixed point for the function h1x2 = x - g1x2 >g′1x2; that is, finding a number x such that x = h1x2. [See equation (2).] Several theorems give conditions that guarantee that the sequence of iterations 5xn 6∞n = 1 defined by (2) will converge to a zero of g1x2. We mention one such result.

Convergence of Newton’s Method Theorem 1. Suppose a zero ∼ x of g1x2 lies in the interval 1a, b2 and that in this interval g′1x2 7 0

and

g″1x2 7 0 .

If we select x1 so that ∼ x 6 x1 6 b, then the sequence of iterations defined by (2) will decrease to ∼ x.

We do not give a proof of this theorem, but refer the reader to an introductory numerical analysis text such as Numerical Analysis, 10th ed., by R. Burden, J. Faires, and A. Burden (Cengage Learning, 2016).

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Appendix C

APPENDIX

C

Simpson’s Rule

A-11

Simpson’s Rule A useful procedure for approximating the value of a definite integral is Simpson’s rule. Let the interval [a, b] be divided into 2n equal parts and let x0, x1, . . . , x2n be the points of the partition, that is, xk J a + kh ,

k = 0, 1, . . . , 2n ,

where h J 1b - a2 > 12n2. If yk J f1xk 2 ,

k = 0, 1, . . . , 2n ,

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

A-12 Appendices

then the Simpson’s rule approximation IS for the value of the definite integral b

f1x2 dx

La is given by (1)

h 3y + 4y1 + 2y2 + 4y3 + P + 2y2n − 2 + 4y2n − 1 + y2n 4 3 0 h n = a 1y2k - 2 + 4y2k - 1 + y2k 2 . 3 k=1

IS =

If b

E J

La

f1x2 dx - IS

is the error that results from using Simpson’s rule to approximate the value of the definite integral, then (2)

0E0 …

1b - a2 4 h M, 180

where M J max 0 f 142 1x2 0 for all x in 3a, b4.

Example 1

Use Simpson’s rule with n = 4 to approximate the value of the definite integral 1

(3) Solution

1 dx . 2 L0 1 + x

Here h = 1>8, xk = k>8, k = 0, 1, . . . , 8, and yk = 11 + x2k 2 -1 =

1 k2 1+ 64

=

64 . 64 + k 2

By Simpson’s rule (1), we find IS =

1 18 2 3

c1 + 4 a

64 64 64 b + 2a b + 4a b 64 + 1 64 + 4 64 + 9

+ 2a

64 64 64 b + 4a b + 2a b 64 + 16 64 + 25 64 + 36

+ 4a

64 64 b +a b d = 0.7854 . 64 + 49 64 + 64

Hence, the value of the definite integral in (3) is approximately IS = 0.7854. ◆ For a more detailed discussion of Simpson’s rule, we refer the reader to a numerical analysis book such as Numerical Analysis, 10th ed., by R. Burden, J. Faires, and A. Burden (Cengage Learning, 2016).

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Appendix D

APPENDIX

D

Cramer’s Rule

A-13

Cramer’s Rule When a system of n linear equations in n unknowns has a unique solution, determinants can be used to obtain a formula for the unknowns. This procedure is called Cramer’s rule. When n is small, these formulas provide a simple procedure for solving the system. Suppose that for a system of n linear equations in n unknowns, a11x1 + a12x2 + g + a1nxn = b1 , a21x1 + a22x2 + g + a2nxn = b2 , (1) f f f f an1x1 + an2x2 + g + annxn = bn , the coefficient matrix a11 a12 g a1n a a g a2n (2) A J D 21 22 T f f f an1 an2 g ann has a nonzero determinant. Then Cramer’s rule gives the solutions det Ai , i = 1, 2, . . . , n , (3) xi = det A where Ai is the matrix obtained from A by replacing the ith column of A by the column vector b1 b D 2T f bn consisting of the constants on the right-hand side of system (1).

Example 1

Use Cramer’s rule to solve the system x1 + 2x2 - x3 = 0 , 2x1 + x2 + x3 = 9 , x1 - x2 - 2x3 = 1 .

Solution

We first compute the determinant of the coefficient matrix: 1 det £ 2 1

2 1 -1

-1 1 § = 12 . -2

Using formula (3), we find 0 1 det £ 9 12 1 1 1 x2 = det £ 2 12 1

2 1 -1 0 9 1

1 1 det £ 2 12 1

2 1 -1

x1 =

-1 48 1§ = = 4, 12 -2 -1 -12 1§ = = -1 , 12 -2

0 24 9§ = = 2. ◆ 12 1 MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition. x3 =

A-14 Appendices

For a more detailed discussion of Cramer’s rule, we refer the reader to Fundamentals of Matrix Analysis with Applications, by Edward Barry Saff and Arthur David Snider (John Wiley & Sons, Hoboken, New Jersey, 2016).

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

A-14 Appendices

APPENDIX

E

Method of Least Squares The method of least squares is a procedure for fitting a straight line to a set of measured data. Consider the scatter diagram in Figure E.1, consisting of an x, y-plot of some data points 5 1xi, yi 2: i = 1, 2, . . . , N6. It is desired to construct the straight line y = a + bx that best fits the data points, in the sense that the sum of the squares of the vertical deviations from the points to the line is minimized. y

3

P4

2

P5

P3

1

P1

P2 x

0

1 2 3 4 5 Figure E.1 Scatter diagram and least-squares linear fit

If the y-axis intercept a and the slope b of the line are known, then the y-value on the line corresponding to the measured x-value xi is given by a + bxi. The corresponding measured y-value, yi, thus deviates from the line by 3yi - 1a + bxi 24, and the total sum of squares of deviations is S J a 3yi - 1a + bxi 24 2 . N

i=1

The interesting feature of the function S is that the symbols xi and yi are constants, while a and b are the (unknown) variables. The values of a and b that minimize S will force its partial derivatives to be zero: 0 =

N 0S = a 23yi - 1a + bxi 2 4 1 -12 , 0a i=1

0 =

N 0S = a 23yi - 1a + bxi 2 4 1 -xi 2 . 0b i=1

Displaying these conditions in a form that emphasizes the roles of a and b, we rewrite them after a little algebra as N

N

aN + b a xi = a yi , i=1

i=1

N

N

N

i=1

i=1

i=1

a a xi + b a x2i = a xiyi

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Appendix E

Method of Least Squares

A-15

and obtain the formulas for the optimal values of intercept and slope: a a x2i b a a yi b - a a xi b a a xiyi b a =

N

N

N

N

i=1

i=1 N

i=1 N

i=1

- a a xi b

N a x2i i=1

2

,

i=1

N a xiyi - a a xi b a a yi b N

b =

N

i=1

i=1

N

N a x2i i=1 Example 1 Solution

N

i=1 2

- a a xi b N

.

i=1

Find the least-squares linear fit to the data points P1 11, 12, P2 12, 12, P3 13, 22, P4 14, 2.52, and P5 15, 3.12, which are plotted in Figure E.1. Arranging the data as in Table E.1 yields a = b =

5519.62 - 15134.52 51552 - 1152

2

5134.52 - 1519.62 51552 - 1152 2

= =

10.5 = 0.21 , 50 28.5 = 0.57 . 50

TABLE E.1

i

xi

yi

xiyi

x2i

1 2 3 4 5

1 2 3 4 5

1 1 2 2.5 3.1

1 2 6 10 15.5

1 4 9 16 25

Sums

15

9.6

34.5

55

Thus, y = 0.21 + 0.57x is the equation for the best-fitting line, which is graphed in Figure E.1. ◆

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

A-16 Appendices

APPENDIX

F

Runge–Kutta Procedure for n Equations Below are program outlines for the classical fourth-order Runge–Kutta algorithm for systems discussed in Section 5.3. The first of these, called the “subroutine,” provides approximations to the solution functions of a system of n first-order ordinary differential equations over an interval. The second outline on page A-17 utilizes the method of halving the step size in order to obtain approximations (to within a prescribed tolerance) to the solution functions at a single given point.

Classical Fourth-Order Runge–Kutta Subroutine (n Equations) Purpose

To approximate the solution to the initial value problem xi= = fi 1t, x1, . . . , xn 2 ; i = 1, 2, . . . , n xi 1t0 2 = ai  , for t0 … t … c.

input

n, t0, a1, . . . , an, c, N (number of steps), prntr ( = 1 to print table).

Step 1

Set step size h = 1c - t0 2 >N ,

Step 2 Step 3

t = t0 ,

x1 = a1 , . . . ,

xn = an .

For j = 1 to N do Steps 3–5. Set ki,1 = hfi 1t, x1, . . . , xn 2 , i = 1, . . . , n ; h 1 1 ki,2 = hfi a t + , x1 + k1, 1, . . . , xn + kn,1 b , 2 2 2 i = 1, . . . , n ; h 1 1 ki,3 = hfi a t + , x1 + k1, 2, . . . , xn + kn,2 b , 2 2 2 i = 1, . . . , n ; ki,4 = hfi 1t + h, x1 + k1,3, . . . , xn + kn,3 2 , i = 1, . . . , n .

Step 4

Set t = t+h; xi = xi +

Step 5

1 1k + 2ki,2 + 2ki,3 + ki,4 2 , i = 1, . . . , n . 6 i,1

If prntr = 1, print t, x1, x2, . . . , xn.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Appendix G

Software for Analyzing Differential Equations

A-17

Classical Fourth-Order Runge–Kutta Algorithm with Tolerance (n Equations) Purpose

To approximate the solution to the initial value problem

input

at t = c, with tolerance e. n, t0, a1, . . . , an, c e (tolerance) M (maximum number of iterations)

Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8 Step 9 Step 10 Step 11 output

xi= = fi 1t, x1, . . . , xn 2 ; xi 1t0 2 = ai , i = 1, 2, . . . , n

Set zi = ai, i = 1, 2, . . . , n; set prntr = 0. For m = 0 to M do Steps 3–7 (or, to save time, start with m 7 0). Set N = 2m. Call FOURTH-ORDER RUNGE–KUTTA SUBROUTINE (n EQUATIONS). Print h, x1, x2, . . . , xn. If 0 zi - xi 0 6 e for i = 1, . . . , n, go to Step 10. Set zi = xi , i = 1, . . . , n . Print “xi 1c2 is approximately”; xi 1for i = 1, . . . , n2; “but may not be within the tolerance”; e. Go to Step 11. Print “xi 1c2 is approximately”; xi 1for i = 1, . . . , n2; “with tolerance”; e. STOP. Approximations of the solution to the initial value problem at t = c using 2m steps.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Appendix G

APPENDIX

G

Software for Analyzing Differential Equations

A-17

Software for Analyzing Differential Equations In this section we shall list some commercial software and some freeware that we have found to be useful in the various aspects of analyzing systems of differential equations numerically: direction field plotting, Euler and Runge–Kutta codes for solutions, and phase plane experimentation. These items are available at the time of printing this edition, but the accessibility of freeware, in particular, is notoriously tenuous. Each of the web sites mentioned was successfully accessed on July 6, 2016. Some commercial products that are helpful: MATLAB® (Mathworks, 1 Apple Hill Drive, Natick, MA 01760-2098 USA) (http://www.mathworks.com/products/matlab/): “The MATLAB ODE Suite” at http://www.mathworks.com/videos/solving-odes-in-matlab-9-the-matlab-ode-suite117653.html?s_tid=srchtitle “Choose an ODE Solver” at http://www.mathworks.com/help/matlab/math/choose-an-ode-solver.html “Vector Fields” at http://www.mathworks.com/help/matlab/vector-fields.html .

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

A-18 Appendices

MATHEMATICA® (Wolfram Research, 100 Trade Center Dr, Champaign, IL 61820 USA) (https://www.wolfram.com/mathematica/): “How to Solve a Differential Equation” at https://reference.wolfram.com/language/howto/SolveADifferentialEquation.html “How to Plot a Direction Field” at https://reference.wolfram.com/language/howto/PlotAVectorField.html . MAPLE® (Maplesoft, 615 Kumpf Drive, Waterloo, ON N2V 1K8 CAN) (http://www.maplesoft.com/). “dsolve” at https://www.maplesoft.com/support/help/maple/view.aspx?path=dsolve “dfieldplot” at https://www.maplesoft.com/support/help/maple/view.aspx?path=DEtools%2Fdfieldplot . PTC Mathcad® (Parametric Technology Corporation, 140 Kendrick Street, Needham, MA 02494 USA) (http://www.ptc.com/engineering-math-software/mathcad). “Differential Equations worksheet” in PTC Mathcad Worksheet Library - Applied Math. Some freeware sources that are helpful: Casio Computer Co. Ltd® (6-2, Hon-machi 1-chome, Shibuya-ku, Tokyo 151-8543, Japan) (http://world.casio.com/) “Euler’s Method Calculator” at http://keisan.casio.com/exec/system/1392171850 “Runge–Kutta method (2nd-order) Calculator” at http://keisan.casio.com/exec/system/1392171606 “Runge–Kutta method (4th-order) Calculator” at http://keisan.casio.com/exec/system/1222997077 The codes dfield and pplane are graciously provided by our differential equations colleague John C. Polking of the Department of Mathematics at Rice University. He holds a copyright and they are not in the public domain; however, they are being made available free for use in educational institutions. “dfield and pplane: The Java Versions” (Polking, J. C. and Castellanos, J.) at http://math.rice.edu/~dfield/dfpp.html The Dynamic Web Tools site provides codes dedicated to the FitzHugh-Nagumo equation: “FitzHugh-Nagumo System of Differential Equations” (Martin, M.) at http://math.jccc.edu:8180/webMathematica/JSP/mmartin/fitznag.jsp The MIT Interactive Mathematics Site provides the following dedicated codes: “Vector Fields” (Miller, H., MIT Mathlets) at http://mathlets.org/mathlets/vector-fields/ “Linear Phase Portraits: Matrix Entry” (Miller, H., MIT Mathlets) at http://mathlets.org/mathlets/linear-phase-portraits-matrix-entry/

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Answers to Odd-Numbered Problems 5. (a)

CHAPTER 1 Exercises 1.1, page 5 1. ODE, 2nd-order, ind. var. t, dep. var. x, linear 3. ODE, 1st-order, ind. var. x, dep. var. y, nonlinear 5. ODE, 1st-order, ind. var. x, dep. var. y, nonlinear 7. ODE, 1st-order, ind. var. t, dep. var. p, nonlinear 9. ODE, 2nd-order, ind. var. x, dep. var. y, linear 11. PDE, 2nd-order, ind. var. t, r, dep. var. N 13. dp>dt = kp, where k is the proportionality constant 15. dT>dt = k1M - T2, where k is the proportionality constant 17. Kevin wins by 623 - 426 ≈ 0.594 sec.

p

1 t

0

1

Figure B.4 Direction field for Problem 5(a)

(b) limt S ∞ p1t2 = 3>2 (d) No

Exercises 1.2, page 13 5. No 7. Yes 9. No 11. Yes 3. Yes 13. Yes 19. The left-hand side is always Ú 4. 21. (a) 1>3, -3 (b) 1 { 26 23. Yes 25. Yes 27. No 29. (b) Yes 31. (a) No (c) y = 2x and y = - 2x

7. (a)

(c) limt S ∞ p1t2 = 3>2

p

2 1

Exercises 1.3, page 21 1. (b) y

(c)

y

t

0

1

Figure B.5 Direction field for Problem 7(a) 1 0

1

x

1 0 1

x

(b) 2

(c) 2

(d) 0

(e) No

9. (d) It increases and asymptotically approaches the line y = x - 1. (f ) and (g) Figure B.1 Solution to Problem 1(b)

Figure B.2 Solution to Problem 1(c)

(d) The solutions in parts (b) and (c) both become infinite and have the line y = 2x as an asymptote as x S ∞ . As x S - ∞ , the solution in part (b) becomes infinite and has y = - 2x as an asymptote, but the solution in part (c) does not even exist for x negative. 3. All solutions have limiting value 8 as t S + ∞ .

y

f (x)

0

x

y

y5x–1 Figure B.6 Direction field and sketch of f1x2 for Problems 9 1f2, 1g2

8

1 0 1

t

Figure B.3 Solutions to Problem 3 satisfying y102 = 5, y102 = 8, and y102 = 15

B-1

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-2

Answers to Odd-Numbered Problems

11.

13.

y

y

(0, 4) y 5 f (x)

y 5 f(x) x

x Figure B.8 Solution to Problem 13 Figure B.7 Solution to Problem 11

15.

c50 c52

c 5 –1 y

11. x112 stabilizes at 1.56 { 0.01 for h below 2-10; 3tan112 = 1.5574 c.4 x1t0 2 = 1 for some t0 in 10.78, 0.802; 3arctan 112 = 0.785 c4. 15. (a) T1302 = 311.7 (b) T1602 = 298.2 Review Problems, page 29 1. ind. var. t, dep. var. x, nonlinear 3. ind. var. y, dep. var. x, linear 5. ind. var. t, dep. var. v, linear 7. True 9. False 11. True 13. True 1 n 15. yn = a1 - b , y112 = e-1 ≈ 0.3679 n 17. (a) 500 (b) No (c) 500

CHAPTER 2 1 0

x

1

Figure B.9 Solution to Problem 15

17. It approaches 3. Exercises 1.4, page 28 1. (rounded to three decimal places) xn 0.1 0.2 0.3 0.4 yn

0.5

4.000 3.998 3.992 3.985

3. (rounded to three decimal places) xn 0.1 0.2 0.3 0.4 yn

0.5

1.100 1.220 1.362 1.528 1.721

5. (rounded to three decimal places) xn 1.1 1.2 1.3 1.4 yn

3.975

0.100

0.209

0.325

Exercises 2.2, page 46 1. No 3. Yes 5. Yes 4 7. y = { 2 ln1x4 2 + C 9. e2x 12x - 12 + 4e-t 1t + 12 = C 21 - Cx -8>3 11. y = { 2 y - x3 = C 13. 21 + y2

15. y = { 2C exp 3exp1 - x2 24 - 1 p 17. y = tana -ln  cos x  + b 3 19. y = 1sin x + 12 2 - 1 21. y2 >2 + ln y = sin u - u cos u + 1>2 - p 23. y = arctan1t 2 + 12 3 25. y = 4ex >3 - 1 x

27. (a) y1x2 =

0.564

h

yN

1

p

3.142

2

p/2

1.571

4

p/4

1.207

8

p/8

1.148

9.

xn 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0

x 2

L0

1>3

et dtb

x

21 + sin t dt + p>4b L0 (d) y10.52 ≈ 1.381 (d) 0f>0y is not continuous at 10, 02 . (a) x2 + y -2 = C 1 1 1 ; ; (b) 2 2 21 - x 24 - x 21>4 - x2 1 1 (c) -1 6 x 6 1 ; -2 6 x 6 2 ; - 6 x 6 2 2 1 (d) 0 x 0 6 is domain . a 28.1 kg (a) 82.2 min (b) 31.8 min (c) Never attains desired temperature (c) y1x2 = tana

7. (rounded to three decimal places) N

L0

(b) y1x2 = a1 + 3

1.5

0.444

2

et dt

yn - 0.9 - 0.81654 - 0.74572 - 0.68480 - 0.63176 - 0.58511 - 0.54371 - 0.50669 - 0.47335 - 0.44314

29. 31.

33. 35.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Answers to Odd-Numbered Problems

37. (a) $1105.17 (c) $4427.59 39. A wins.

(b) 27.73 years

Exercises 2.3, page 54 3. Both 5. Neither 1. Linear 7. y = 11>22e3x + Cex 9. r = sin u + C cos u 11. y = - t - 2 + Cet 13. x = y3 + Cy -2 15. y = 1 + C1x2 + 12 -1>2 17. y = xex - x t3 1 t3 17 19. x = ln t + 6 36 2t 36t 3 21. y = x2cos x - p2cos x 23. y1t2 = 138>32e-5t - 18>32e-20t 25. (b) y132 ≈ 0.183 27. (b) 0.9960 (c) 0.9486 , 0.9729 29. x = e4y >2 + Ce2y 31. (a) y = x - 1 + Ce-x (b) y = x - 1 + 2e-x (c) y = x>3 - 1>9 + Ce-3x x - 1 + 2e-x , if 0 … x … 2 (d) y = e x>3 - 1>9 + 14e6 >9 + 2e4 2e-3x , if 2 6 x (e) y 2 1 x 1 2 3 4 5 Figure B.10 Solution to Problem 31(e)

33. (a) y = x is only solution in a neighborhood of x = 0. (b) y = -3x + Cx2 satisfies y102 = 0 for any C. 35. (a) 0.0281 kg>L (b) 0.0598 kg>L 2 cos1pt>122 1p>122sin1pt>122 1 37. x1t2 = 2 2 4 + 1p>122 4 + 1p>122 2 2 19 +a + be-2t 2 4 + 1p>122 2 39. 71.8°F at noon; 26.9°F at 5 p.m. Exercises 2.4, page 64 1. Linear with y as dep. var. 3. Separable 5. Separable, also linear with x as dep. var. 7. Exact 9. y = 1C - 3x2 > 1x2 - 12 3 11. ex sin y - x3 + 2y = C t 1t - 12e + C 13. y = 1 + et 15. r = 1C - eu 2sec u 17. Not exact 19. x2 - y2 + arctan1xy2 = C

B-3

21. ln x + x2y2 - sin y = p2 23. y = -2> 1tet + 22 25. sin x - x cos x = ln y + 1>y + p - 1 (equation is separable, not exact) 27. (a) -ln 0 y 0 + f1x2 (b) cos x sin y - y2 >2 + f1x2, where f is a function of x only 29. (c) y = x2 > 1C - x2 (d) Yes, y K 0 33. (a) x = cy2 ; x K 0 ; y K 0 (b) x2 + 4y2 = c (c) 2y2ln y - y2 + 2x2 = c (d) 2x2 + y2 = c Exercises 2.5, page 69 1. Integrating factor depending on x alone 3. Exact 5. Linear with x as dep. var., has integrating factor depending on x alone 7. m = y -4 ; x2 y -3 - y -1 = C and y K 0 9. m = x -2 ; y = x4 >3 - x ln 0 x 0 + Cx and x K 0 11. m = y -2 ; x2y -1 + x = C and y K 0 13. m = xy ; x2y3 - 2x3y2 = C 15. (a) m1z2 = exp 3 1 H1z2 dz 4 ; z = xy (b) ln1x3y2 + 2xy = 2 0N/0x - 0M/0y depends only on x2y . 17. (a) 2 x M - 2xyN (b) exp1x2y2[x2 + 2xy] = C 19. (b) m = ey ; x = y - 1 + Ce-y Exercises 2.6, page 76 1. Homogeneous and Bernoulli 3. Bernoulli 5. Homogeneous and Bernoulli 7. y′ = G1ax + by2 9. y = - x> 1ln 0 x 0 + C2 and y K 0 11. y = x> 1ln 0 x 0 + C2 and y K 0, x K 0 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 45. 47.

21 + x2 >t 2 = ln 0 t 0 + C 1x2 - 4y2 2 3x2 = C y = 1x + C2 2 >4 - x and y = -x y = x + 16 + 4Ce2x 2 > 11 + Ce2x 2 and y = x + 4 y = 2> 1Cx - x3 2 and y K 0 y = 5x2 > 1x5 + C2 and y K 0 x -2 = 2t 2ln 0 t 0 + Ct 2 and x K 0 r = u 2 > 1C - u2 and r K 0 y-3 2 arctana b - ln 31x + 22 2 + 1y - 32 2 4 = C. x+2 12x + 2y - 32 3 = C12x + y - 22 2 x2 + t 2 - Ct = 0 and t K 0 y = - 3> 1x4 + Cx2 and y K 0 y = 1u>421ln 0 u 0 + C2 2 and u K 0 sin1x + y2 + cos1x + y2 = Cex - y 1x - y + 22 2 = Ce2x + 1 1y - 4x2 2 1y + x2 3 = C (a) v′ + [2 Pu + Q]v = -P (b) y = x + 5x> 1C - x5 2

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-4

Answers to Odd-Numbered Problems

Review Problems, page 79 1. ex + ye-y = C 3. x2y - x3 + y -2 = C 5. y + x sin1xy2 = C 7. t = Cexp11> 17y7 22

9. 1x2 + 4y2 2 3x2 = C 11. tan1t - x2 + t = C 13. y = - 1x2 >22cos12x2 + 1x>42sin12x2 + Cx

15. y = 2x + 3 - 1x + C2 2 >4 17. 19. 21. 23. 25.

y = 2> 11 + Ce2u 2 and y K 0 y2 = x2 + Cx3 and x K 0 xy - x2 - x + y2 >2 - 4y = C y2 + 2xy - x2 = C x2y -2 - 2xy -1 - 4xy -2 = C and y K 0

27. 31y - 42 2 - 31x - 32 2 4 ` 3 231x - 32 + 1y - 42 4

, 3 231x - 32 - 1y - 42 4 `

29. x4y3 - 3x3y2 + x4y2 = C

1>23

= C

31. y = - x3 >2 + 7x>2

-t

35. y = - 2x22x2 - 1 33. x = - t - 2 + 3e 2 37. ln31y - 22 + 21x - 12 2 4 y-2 d = ln 2 + 22 arctan c 221x - 12

39. y = 2119x4 - 12 >2 t er 41. y1t2 = e-t dr + 3e-1t - 22, y132 ≈ 1.1883 c 2 L2 1 + r

CHAPTER 3 Exercises 3.2, page 100 1. 5 - 4.5e-2t>25 kg; 5.07 min 3. 10.421100 - t2 - 13.9 * 10-7 21100 - t2 4 L; 19.96 min 5. 0.0097%; 73.24 h 7. 20 min later; 1>e times as salty 9. 110,868 13. 5970; 6000 15. 6572; 6693 17. (a) 1 dp Logistic Year p dt (Least Squares) 1790 1800 1810 1820 1830 1840 1850 1860 1870 1880 1890

0.0351 0.0363 0.0331 0.0335 0.0326 0.0359 0.0356 0.0267 0.0260 0.0255 0.0210

3.29 4.51 6.16 8.41 11.45 15.53 20.97 28.14 37.45 49.31 64.07

Year

1 dp p dt

1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 2010

0.0210 0.0150 0.0162 0.0073 0.0145 0.0185 0.0134 0.0114 0.0098 0.0132 0.0097  

Logistic (Least Squares) 81.89 102.65 125.87 150.64 175.80 200.10 222.47 242.16 258.81 272.44 283.28 291.73

(b) p1 = 316.920, A = 0.00010050 (c) p0 = 3.28780 (using all data, including 2010) (d) See table in part (a). 19. 11>22ln1152 ≈ 1.354 yr; 14 million tons per yr 21. 1 hr; 2 h 23. 11.7% 25. 31,606 yr 27. e-2t kg of Hh, 2e-t - 2e-2t kg of It, and 1 - 2e-t + e-2t kg of Bu Exercises 3.3, page 107 1. 20.7 min 3. 22.6 min 5. 9:08 a.m. 7. 28.3°C; 32.5°C; 1:16 p.m. 9. 16.3°C; 19.1°C; 31.7°C; 28.9°C 11. 39.5 min 13. 148.6°F 15. T - M = C1T + M2exp32 arctan1T>M2 - 4M 3kt4; for T near M, M 4 - T 4 ≈ 4M 3 1M - T2, and so dT>dt ≈ k1 1M - T2, where k1 = 4M 3k Exercises 3.4, page 115 1. 10.9812t + 10.09812e-10t - 0.0981 m; 1019 sec 3. 18.6 sec 5. 4.91t + 22.55 - 22.55e-2t m; 97.3 sec 7. 241 sec 9. 95.65t + 956.5e-t>10 - 956.5 m; 13.2 sec 11. 13. 17. 19. 21. 25.

eby 0 by - mg 0 mg = ey0b 0 by0 - mg 0 mge-b x>m 2.69 sec; 101.19 m 15. 1v0 - T>k2e-kt>I + T>k 300 sec 2636e-t>20 + 131.8t - 2636 m; 1.768 sec 5e-2t >2 + 6t - 5>2; 6 m/sec 23. Sailboat B (e) 11.18 km/sec (f ) 2.38 km/sec 2

Exercises 3.5, page 121 1. I = 51.44e-100t + cos 120t + 1.2 sin 120t6 >2.44; EL = 1 -7.2e-100t - 6 sin 120t + 7.2 cos 120t2 >2.44 3. - 1ln .42 * 10-10 ≈ 9.2 * 10-11 sec dI d LI 2 5. VI = IRI = I 2R; VI = L I = ; dt dt 2 2 dCEC d CE C VI = EC = dt dt 2 7. - 110 ln .12 >3 ≈ 7.68 sec

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Answers to Odd-Numbered Problems

Exercises 3.6, page 129 h “e” 3. 7. 1 3 0.1 2.72055 0.01 2.71830 0.001 2.71828 0.0001 2.71828 9.

11. 13. 15. 17.

xn

yn

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0.61784 1.23864 1.73653 1.98111 1.99705 1.88461 1.72447 1.56184 1.41732 1.29779

xn 1.1 1.2 1.3 1.4 1.5

yn 0.10450 0.21668 0.33382 0.45300 0.57135

  -3   9   - 27   81   - 243

5. Order 2, f112 ≈ 1.3725; order 4, f112 ≈ 1.3679 7. -11.7679 9. 1.36789 11. x = 1.41 13. x = 0.50 yn 15. xn 0.5 1.0 1.5 2.0 2.5 3.0

0.21462 0.13890 - 0.02668 - 0.81879 - 1.69491 - 2.99510

19. y132 ≈ 0.24193 with h = 0.0625 21. z112 ≈ 2.87083 with h = 0.03125

CHAPTER 4

f112 ≈ x11; 2-3 2 = 1.25494 f112 ≈ y11; 2-3 2 = 0.71698 x = 1.27 xn yn 1h = 0.22 yn 1h = 0.12 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

B-5

-1 1 -1 1 -1 1 -1 1 -1 1

Exercises 4.1, page 156 3. Both approach zero. 5. 0 7. y1t2 = - 130>612 cos 3t - 125>612 sin 3t 9. y1t2 = -2 cos 2t + 13>22 sin 2t yn 1h = 0.0252 0.06250 0.00391 0.00024 0.00002 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000

We conclude that step size can dramatically affect convergence. Tn 19. Time

K = 0.2 K = 0.4

K = 0.6

Midnight 4 a.m. 8 a.m. Noon 4 p.m. 8 p.m. Midnight

65.0000 69.1639 71.4836 72.9089 72.0714 69.8095 68.3852

65.0000 68.1300 73.6678 76.9783 74.7853 69.2831 65.9740

65.0000 68.5644 72.6669 75.1605 73.5977 69.5425 67.0500

Exercises 3.7, page 139 1. yn+1 = yn + h cos1xn + yn 2 h2 - sin1xn + yn 231 + cos1xn + yn 24 2 h2 3. yn+1 = yn + h1xn - yn 2 + 11 - xn + yn 2 2 h3 h4 - 11 - xn + yn 2 + 11 - xn + yn 2 6 24

Exercises 4.2, page 164 1. c1et>2 + c2e-4t 3. c1e-3t + c2e-2t 5. c1e-4t + c2te-4t 7. c1et> 2 + c2e-2t> 3 9. c1et> 2 + c2tet> 2 11. c1e-5t> 2 + c2te-5t> 2 4 1 13. 3e-4t 15. et - e3t 3 3 7t 17. a + 2be3t. 19. e-t - 2te-t 3 21. (a) ar + b = 0 (b) ce-bt>a -4t>5 13t>6 23. ce 25. ce 27. Lin. dep. 29. Lin. indep. 31. Lin. dep. 33. If c1 ≠ 0, then y1 = -(c2/c1)y2 . 35. (a) Lin. indep. (b) Lin. dep. (c) Lin. indep. (d) Lin. dep. 37. 39. 41. 43. 45.

c1et + c2e1-1 - 252t + c3e1-1 + 252t c1e-2t + c2te-2t + c3e2t c1e-3t + c2e-2t + c3e2t 3 + et - 2e-t (a) c1er1t + c2er2t + c3er3t (where r1 = -4.832, r2 = -1.869, and r3 = 0.701) (b) c1er1t + c2e-r1t + c3er2t + c4e-r2t (where r1 = 1.176, r2 = 1.902) (c) c1e-t + c2et + c3e-2t + c4e2t + c5e3t

Exercises 4.3, page 172 1. c1 cos 3t + c2 sin 3t 3. c1e3t cos t + c2e3t sin t 5. c1e-2t cos 22 t + c2e-2t sin 22 t 7. c1e-t> 2 cos 1 25t>22 + c2e-t> 2 sin 1 25t>22 9. c1et + c2e7t

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-6

Answers to Odd-Numbered Problems

11. c1e-5t + c2te-5t 13. c1et cos 5t + c2et sin 5t 15. 17. 19. 21. 23.

c1e13 + 2532t>2 + c2e13 - 2532t>2. c1et>2 cos 1323t>22 + c2et>2 sin 1323t>22 c1et + c2e-t cos 2t + c3e-t sin 2t 2e-t cos t + 3e-t sin t 1 22>423e12 + 222t - e12 - 222t 4

25. et sin t - et cos t 27. e2t - 22 et sin 22t 29. (a) c1e-t + c2et cos 22t + c3et sin 22t (b) c1e2t + c2e-2t cos 3t + c3e-2t sin 3t (c) c1 cos 2t + c2 sin 2t + c3 cos 3t + c4 sin 3t 31. (a) Oscillatory (b) Tends to zero (c) Tends to - ∞ (d) Tends to - ∞ (e) Tends to + ∞ 33. (a) y1t2 = 0.3e-3t cos 4t + 0.2e-3t sin 4t (b) 3p + arctan1 - 1.524 >4 (c) 2>p (d) Decreases the frequency of oscillation, introduces the factor e-3t, causing the solution to decay to zero 35. b Ú 22Ik 37. (a) c1 cos t + c2 sin t + c3t cos t + c4t sin t (b) 1c1 + c2t2e-t cos 1 23t2 + 1c3 + c4t2e-t sin 1 23t2 Exercises 4.4, page 180 1. No 3. Yes 5. No 7. Yes 9. yp K -3 11. yp 1x2 = 31ln 22 2 + 14 -12x 13. cos 3t 15. xex >2 + 3ex >4 17. - 2t cos 2t t 8 19. ¢ + ≤te-3t 21. t 3e2t >6 13 169 1 1 2 2 23. - u 3 u u 21 49 343 25. e2t 1cos 3t + 6 sin 3t2 27. 1A3t 4 + A2t 3 + A1t 2 + A0t2 cos 3t + 1B3t 4 + B2t 3 + B1t 2 + B0t2 sin 3t t 29. e 1A4t 4 + A3t 3 + A2t 2 + A1t + A0 2 31. 1A3t 4 + A2t 3 + A1t 2 + A0t2e-t cos t + 1B3t 4 + B2t 3 + B1t 2 + B0t2e-t sin t 33. 11>52 cos t + 12>52 sin t 1 4 35. ¢ t 2 - t≤ et 10 25 Exercises 4.5, page 185 1. (a) 5 cos t (b) cos t - e2t (c) 4 cos t + 6e2t t -t 3. y = - t + c1e + c2e 5. u = t - 1 + c1e2t + c2e-t 7. y = tan x + c1e22x + c2e-22x 9. Yes 11. No 13. Yes 15. No 4t 1 17. y(t) = -t 2 + + + c1e3t + c2e-t. 3 9

19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41.

43. 45.

47.

1 cos x - sin x2ex >2 + c1ex + c2e2x 11>22u e-usin u + 1c1 cos u + c2 sin u2e-u et - 1 e-x - cos x + sin x - 13>102 cos x - 11>102 sin x - 11>202 cos 2x + 13>202 sin 2x y = - 11>22 sin u - 11>32e2u + 13>42eu + 17>122e-u yp = 1A1t + A0 2t cos t + 1B1t + B0 2t sin t + C # 10t xp 1t2 = 1A cos t + B sin t2et + C2t 2 + C1t + C0 + D1 cos 3t + D2 sin 3t + E1 cos t + E2 sin t yp = 1A1t + A0 2 cos 3t + 1B1t + B0 2 sin 3t + Ce5t yp = t 2 + 3t - 1 yp = 1t>10 - 4>252tet - 1>2 (a) y1 = - 12 cos 2t + sin 2t2e-t + 2 for 0 … t … 3p>2 (b) y2 = yh = 1c1cos 2t + c2sin 2t2e-t for t 7 3p>2 (c) c1 = -21e3p>2 + 12 , c2 = - 1e3p>2 + 12 y = - cos t + 11>22 sin t - 11>22e-3t + 2e-t 2V cos 1p>2V2 sin t for V ≠ 1; (a) y1t2 = V2 - 1 p y1t2 = sin t for V = 1 2 (b) V ≈ 0.73 (a) 2 sin 3t - cos 6t (b) No solution (c) c sin 3t - cos 6t , where c is any constant y y y z y

= = = = =

Exercises 4.6, page 191 1. c1 cos 2t + c2 sin 2t - 11>421 cos 2t2 ln  sec 2t + tan 2t  3. c1et + c2tet + tet ln  t  5. c1 cos 4u + c2 sin 4u + 1u>42 sin 4u + 11>1621cos 4u2 ln  cos 4u 

7. 12 ln t - 32t 2e-2t >4 + c1e-2t + c2te-2t 9. yp = -2t - 4 11. y = - 1cos t2 ln  sec t + tan t  + 11>102e3t -1 + c1 cos t + c2 sin t 13. c1 cos 2t + c2 sin 2t + 11>242sec2 2t - 1>8 + 11>821sin 2t2ln ƒ sec 2t + tan 2t ƒ 15. c1 cos t + c2 sin t - t 2 + 3 + 3t sin t + 31cos t2ln ƒ cos t ƒ 17. c1 cos 2t + c2 sin 2t - et >5 - 11>221 cos 2t2ln ƒ sec 2t + tan 2t ƒ et e-u eu e-t du du 2 L1 u 2 L1 u 3y122 ≈ -1.934 21. 0.3785 23. c1et + c2 1t + 12 - t 2 25. c1(5t - 1) + c2e-5t - t 2e-5t >10 t

t

19. y = e1 - t - et - 1 +

Exercises 4.7, page 199 1. Unique solution on 1 - p>2, p>22 3. Unique solution on 10, ∞ 2 5. Does not apply; t = 0 is a point of discontinuity

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Answers to Odd-Numbered Problems

7. 9. 11. 13.

Does not apply; not an initial value problem c1t + c2t -7 c1t -2 + c2t -2 ln t c1t -1>3 + c2t -1> 3ln t

19. 21. 23. 25. 27.

t - 3t 4 c1 1t - 22 + c2 1t - 22 7 (c) t acos1b ln 0 t 0 2, t a sin 1b ln 0 t 0 2; t r, t r ln 0 t 0 (a) True (b) False (e) No, because the coefficient of y″ vanishes at t = 0 and the equation cannot be written in standard form. Otherwise their Wronskian would be zero at t0, contradicting linear independence. (a) Yes (b) No (c) Yes (d) Yes Cte-t 1 + 2t - t 2 c1 cos 13 ln t2 + c2 sin 13 ln t2 + (1/9) cos13 ln t2ln 0 sec13 ln t2 + tan13 ln t2 0

15. c1 t cos 3 2 ln 1 - t2 4 + c2t sin 3 2 ln 1 - t2 4 17. t -4 5 c1 cos 3 ln 1 - t2 4 + c2 sin 3 ln 1 - t2 4 6

29. 31. 33. 35. 37.

39. c 1t + c2t ln t + 11>22 t 1ln t2 2 + 3t1ln t23ln 0 ln t 0 4 41. t 4 43. t + 1 45. 1t - 12e2t >2 47. (a) 11 - 2t 2 2

2

L

(b) 13t - 2t 2 3

11 - 2t 2 2 -2 et dt 13t - 2t 3 2 -2 et dt 2

L 49. tw″ + 2tw′ + 1t + 12w = 0 f1tn 2 - f1t0 2 0-0 51. (a) f′1t0 2 = lim = lim = 0 nS ∞ n S ∞ tn - t0 tn - t0 Exercises 4.8, page 210 1. Let Y1t2 = y1 -t2. Then Y′1t2 = - y′1 - t2, Y″1t2 = y″1 -t2. But y″1s2 - sy1s2 = 0, so y″1 - t2 + ty1 -t2 = 0 or Y″1t2 + tY1t2 = 0. 3. The spring stiffness is 1 - 6y2, so it opposes negative displacements 1y 6 02 and reinforces positive displacements 1y 7 02. Initially y 6 0 and y′ 6 0, so the (positive) stiffness reverses the negative velocity and restores y to 0. Thereafter, y 7 0 and the negative stiffness drives y to + ∞ . d 4 5. (a) y″ = 2y3 = 1y >22. Thus, by setting K = 0 and dy choosing the ( - ) sign in equation (11), we get dy 1 t = + c = + c, or y = 1> 1t - c2 . y L 22y4 >2 (b) Linear dependence would imply 1> 1t - c1 2 y1 1t2 t - c2 = = K constant y2 1t2 1> 1t - c2 2 t - c1 in a neighborhood of 0, which is false if c1 ≠ c2. (c) If y1t2 = 1> 1t - c2, then y102 = - 1>c, y′102 = - 1>c2 = - y102 2, which is false for the given data.

B-7

7. (a) The velocity, which is always perpendicular to the lever arm, is / du>dt. Thus, (lever arm) times (perpendicular momentum) = / m/ du>dt = m/2du>dt. (b) The component of the gravitational force perpendicular to lever arm is mg sin u, and is directed toward decreasing u. Thus, torque = -/mg sin u. d (c) Torque = (angular momentum) or dt -/mg sin u = 1m/2u′2′ = m/2u″ . 9. 2 or - 2 11. The sign of the damping coefficient 1y′2 2 - 1 indicates that low velocities are boosted by negative damping but that high velocities are slowed. Hence, one expects a limit cycle. 13. (a) Airy (b) Duffing (c) van der Pol 15. (a) Yes (t 2 = positive stiffness) (b) No ( -t 2 = negative stiffness) (c) Yes (y4 = positive stiffness) (d) No (y5 = negative stiffness for y 6 0) (e) Yes (4 + 2 cos t = positive stiffness) (f) Yes (positive stiffness and damping) (g) No (negative stiffness and damping) 17. 1>422 Exercises 4.9, page 220 1. y1t2 = - 11>42 cos 5t - 11>52 sin 5t amplitude = 241>20; period = 2p>5; frequency = 5>2p; 3p - arctan 15>424 >5 sec 3. b = 0: y1t2 = cos 4t y

b50

1

p

t

2p

21 Figure B.11 b = 0

b = 6: y1t2 = e-3t cos 27t + 13> 272e-3t sin 27t

= 14> 272e - 3t sin 1 27t + f2 , where

f = arctan27>3 ≈ 0.723 y

b56

1

4 e –3t Ë7

–1

4 e –3t Ë7

t

Figure B.12 b = 6

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-8

Answers to Odd-Numbered Problems

b = 8: y1t2 = 11 + 4t2e - 4t y

7. y1t2 = 1 - 3>42e - 8t cos 8t - e-8t sin 8t = 15>42e-8t sin 18t + f2, where f = p + arctan13>42 ≈ 3.785 ; damp. factor = 15>42e-8t ; quasiperiod = p>4; quasifreq. = 4>p 9. 0.242 m

b58

1 t

11. 110> 299992 arctan1 299992 ≈ 0.156 sec 13. Relative extrema at t = 3p>3 + np - arctan 1 23>224 > 12232

1 4

Figure B.13 b = 8

b = 10: y1t2 = 14>32e-2t - 11>32e-8t y

for n = 0, 1, 2, c; but touches curves { 27>12e-2t at t = 3p>2 + mp - arctan1 23>224 > 12232 for m = 0, 1, 2, c. 15. First measure half the quasiperiod P as the time between two successive zero crossings. Then compute the ratio y1t + P2 >y1t2 = e-1b>2m2P.

b 5 10

1 t 1 6

ln 4

Exercises 4.10, page 227

1. M1g2 = 1> 211 - 4g 2 2 2 + 4g 2

Figure B.14 b = 10

M

5. k = 20: y1t2 = 311 + 252 >24e1-5 + 252t + 311 - 252 >24e1-5 - 252t y

1

k 5 20 g

1

1

k = 25: y1t2 = 11 + 5t2e-5t

y p 1 3

k 5 25

Figure B.16 k = 25

k = 30: y1t2 = e-5t cos 25t + 25e-5t sin 25t f = arctan11> 252 ≈ 0.421

1

Ë6e –5t

–1

–Ë6e –5t

–Ë6 Figure B.17 k = 30

4p 3

t 2p

Figure B.19

t 1 5

= 26e-5t sin 1 25t + f2 ,

5p 3

p 2p 3

1

k 5 30

4

3. y1t2 = cos 3t + 11>32t sin 3t

Figure B.15 k = 20

Ë6

3

Figure B.18

t

y

2

where

5. (a) y1t2 = - 3F0 > 1k - mg 2 24 cos 1 2k>m t2 + 3F0 > 1k - mg 2 24 cos gt = 1F0 > 3m1v2 - g 2 2421cos gt - cos vt2 (c) y1t2 = sin 8t sin t 1

–1

sin t p

2p

t

–sin t Figure B.20

7. y1t2 = c1er1t + c2er2t +

F0 sin 1gt + u2

21k - mg 2 2 2 + b2g 2

,

where r1, r2 = - 1b>2m2 { 11>2m2 2b2 - 4mk and tan u = 1k - mg 2 2 > 1bg2 as in equation (7) 9. yp 1t2 = 10.082 cos 2t + 10.062 sin 2t = 10.12 sin 12t + u2 , where u = arctan14>32 ≈ 0.927

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Answers to Odd-Numbered Problems

11. y1t2 = - 118>852e-2t cos 6t - 122>2552e-2t sin 6t + 12> 2852 sin12t + u2 , where u = arctan19>22 ≈ 1.352 ; res. freq. = 222>p cycles>sec 13. yp 1t2 = 13>185218 sin 4t - 11 cos 4t2 15. Amp =

B

a-

11 2 3 2 b + ab ≈ 0.011m2 , 986 1972

freq. = 2/p Review Problems, page 231 1. c1e-9t + c2et 3. c1et>2 cos13t>22 + c2et>2 sin13t>22 5. c1e3t>2 + c2et>3 7. c1e-t>3 cos1t>62 + c2e-t>3 sin1t>62 9. c1e7t>4 + c2te7t>4 11. t 1>2 5 c1 cos 3 1 219>22ln t 4 + c2 sin 3 1 219>22ln t 4 6 13. c1 cos 4t + c2 sin 4t + 11>172tet - 12>2892et 15. c1e-2t + c2e-t + c3e-t>3 17. c1et + c2e-t>2 cos1 243t>22 + c3e-t>2 sin1 243t>22 19. c1e-3t + c2et>2 + c3tet>2 21. c1e3t>2 cos1 219t>22 + c2e3t>2 sin1 219t>22 - et >5 + t 2 + 6t>7 + 4>49 23. c1 cos 4u + c2 sin 4u - 11>1621 cos 4u2ln ƒ sec 4u + tan 4u ƒ c1e3t>2 + c2te3t>2 + e3t >9 + e5t >49 c1x + c2x -2 - 2x -2 ln x + x ln x e-2t cos1 23t2 2et cos 3t - 17>32etsin 3t - sin 3t - e-t - 3e5t + e8t cos u + 2 sin u + u sin u + 1cos u2ln ƒ cos u ƒ (a), (c), (e), and (f) have all solutions bounded as tS +∞ 39. yp 1t2 = 11>42 sin 8t ; 262>2p

25. 27. 29. 31. 33. 35. 37.

CHAPTER 5 Exercises 5.2, page 249 1. (a) - t 3 + 3t 2 + 8 (b) - 2t 3 + 3t 2 + 6t + 16 (c) 2t 3 + 3t 2 - 16 (d) - 2t 3 + 3t 2 + 6t + 16 (e) - 2t 3 + 3t 2 + 6t + 16 3. x = c1 + c2e-2t ; y = c2e-2t 5. x K - 5 ; y K 1 7. u = c1 - 11>22c2e-t + 11>22et + 15>32t ; y = c1 + c2e-t + 15>32t 9. x = c1et + 11>42cos t - 11>42sin t ; y = -3c1et - 13>42cos t - 11>42sin t 11. u = c1 cos 2t + c2 sin 2t + c3e23t + c4e-23t - 13>102et ; y = c1 cos 2t + c2 sin 2t - 12>52c3e23t - 12>52c4e-23t + 11>52et

B-9

13. x = 2c2et cos 2t - 2c1et sin 2t ; y = c1et cos 2t + c2et sin 2t 15. w = - 12>32c1e2t + c2e7t + t + 1 ; z = c1e2t + c2e7t - 5t - 2 17. x = c1 cos t + c2 sin t - 4c3 cos 1 26 t 2 - 4c4 sin 1 26 t 2 ; y = c1 cos t + c2 sin t + c3 cos 1 26 t 2 + c4 sin 1 26 t 2 19. x = 2e3t - e2t ; y = -2e3t + 2e2t 21. x = et + e-t + cos t + sin t ; y = et + e-t - cos t - sin t 23. Infinitely many solutions satisfying x + y = et + e-2t 25. x1t2 = - c1et - 2c2e2t - c3e3t , y1t2 = c1et + c2e2t + c3e3t , z1t2 = 2c1et + 4c2e2t + 4c3e3t 27. x1t2 = c1e8t + c2e4t + c3 , 1 y1t2 = 1c1e8t - c2e4t + c3 2 , 2 z1t2 = -c1e8t + c3 29. -3 … l … - 1 20 20 31. x1t2 = - a10 + ber1t - a10 ber2t 27 27 + 20 kg , 30 r1t 30 r2t y1t2 = e e + 20 kg, where 27 27 - 5 - 27 r1 = ≈ - 0.0765 , 100 - 5 + 27 ≈ - 0.0235 r2 = 100 20 - 1025 1 -3 + 252 t>100 33. x = c de 25 20 + 1025 1 -3 - 252 t>100 - c de + 20 ; 25 10 10 y = -a be1 -3 + 252 t>100 + a be1 -3 - 252 t>100 + 20 25 25 35. 90.4°F 37. 460>11 ≈ 41.8°F Exercises 5.3, page 259 1. x1= = x2 , x2= = 3x1 - tx2 + t 2 ; x1 102 = 3 , x2 102 = - 6 3. x1= = x2 , x2= = x3 , x3= = x4 , x4= = x4 - 7x1 + cos t ; x1 102 = x2 102 = 1 , x3 102 = 0 , x4 102 = 2 5. x1= = x2 , x2= = x2 - x3 + 2t , x3= = x4 , x4= = x1 - x3 - 1 ; x1 132 = 5 , x2 132 = 2 , x3 132 = 1 , x4 132 = - 1 7. x1= = x2 , x2= = x3 , x3= = x4 + t , x4= = x5 , x5= = 12x4 - 2x3 + 12>5 ; x1 102 = x2 102 = x3 102 = 4 , x4 102 = x5 102 = 1

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-10 Answers to Odd-Numbered Problems

5. 10, 02 3. 1 1> 22, 1> 22 2 , 1 -1> 22, -1> 22 2 7. ex + ye-y = c 9. ex + xy - y2 = c 11. y2 - x2 = c 13. 1x - 12 2 + 1y - 12 2 = c

9. tn + 1 = tn + h, n = 0, 1, 2, c; h xi, n + 1 = xi,n + 3fi 1tn, x1,n, c, xm,n 2 + 2 fi 1tn + h, x1,n + hf1 1tn, x1,n, c, xm,n 2, c, xm,n + hfm 1tn, x1,n, c, xm,n 24 i = 1, 2, c, m ti 11. i 13. i ti y1ti 2 y1ti 2 1 2 3 4

0.250 0.500 0.750 1.000

0.750000 0.625000 0.573529 0.563603

1 2 3 4

0.250 0.500 0.750 1.000

15. y112 ≈ x1 11, 2-2 2 = 1.69, y′112 ≈ 1.82 17. u11; 2-2 2 = y11; 2-2 2 = 0.36789 19. Part (a) Part (b) i

ti

1 2 3 4 5 6 7 8 9 10 21.

0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 i 1 2 3 4 5 6 7 8 9 10

23. 25. 27. 29.

x1ti 2

y1ti 2

x1ti 2

1.95247 2.25065 1.48118 3.34588 1.83601 2.66294 4.53662 3.36527 5.19629 2.47788 4.32906 3.10706 1.96093 2.71900 1.92574 2.86412 1.96166 2.34143 4.28449 2.77457 3.90106 3.00965 4.11886 3.83241 2.18643 3.14344 2.32171 2.63187 2.25824 2.21926 ti x1 1ti 2 ≈ H1ti 2 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

y

y

y5x x

0.25000 0.50000 0.75000 1.00000

x –1 –2 –3

4

Figure B.23

Figure B.22

15. 1 - 2, 12 is a saddle point (unstable). Part (c)

y1ti 2

x1ti 2

y1ti 2

2.42311 1.45358 2.40348 4.64923 3.32426 2.05910 2.18977 3.89043 3.79362 2.49307

0.91390 1.63657 4.49334 5.96115 1.51830 0.95601 2.06006 5.62642 5.10594 1.74187

2.79704 1.13415 1.07811 5.47788 5.93110 2.18079 0.98131 1.38072 5.10462 5.02491

0.09573 0.37389 0.81045 1.37361 2.03111 2.75497 3.52322 4.31970 5.13307 5.95554

y 3

2

1

24

23

22

x

21

21 Figure B.24

17. 10, 02 is a center (stable). y

x

Yes, yes y112 ≈ x1 11; 2-3 2 = 1.25958 y10.12 ≈ 0.00647, c, y12.02 ≈ 1.60009 (a) P1 1102 ≈ 0.567, P2 1102 ≈ 0.463, P3 1102 ≈ 0.463 (b) P1 1102 ≈ 0.463, P2 1102 ≈ 0.567, P3 1102 ≈ 0.463 (c) P1 1102 ≈ 0.463, P2 1102 ≈ 0.463, P3 1102 ≈ 0.567 All populations approach 0.5.

Figure B.25

19. 10, 02 is a saddle point (unstable). y

Exercises 5.4, page 271 1. x = y3 , y 7 0 y

y

x Figure B.21

Figure B.26

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-11

Answers to Odd-Numbered Problems

21. 10, 02 is a center (stable).

27.

1 x1t2, y1t2 2 approaches 10, 02.

y

y 2

1

y

22 Figure B.27

22

y

Figure B.30

3

29. Critical points are 11, 02 and 1x, 12 with x arbitrary. Phase plane solutions are x = 1, and y = C1x - 12. Solution starting at 13, 22 goes to infinity. The others converge to 11, 02. 31. (a) y′ = y , y′ = f1y2

2 1 y

21

2

1 21

f1y2 dy implies y dy = f1y2 dy + constant, or = y dy L L y2 >2 = F1y2 + K 33. (a) x′ = y, y′ = - x + 1> 1l - x2 -x + 1> 1l - x2 dy implies (b) = y dx 1 y dy = bdx + constant, or a -x + l -x L L 2 2 y >2 = -x >2 - ln 1l - x2 + constant. The solution equation follows with the choice C>2 for the constant. (c) At a critical point y = 0 and -x + 1> 1l - x2 = 0. (b)

22 23 Figure B.28

25. (a) Periodic (b) Not periodic (c) Critical point (periodic) 3

2

1 21

23. 10, 02 is a center (stable); 11, 02 is a saddle point (unstable).

22

x

21

y

2

l { 2l2 - 4 , and 2

Solutions for the latter are x = are real only for l Ú 2.

1

y

(d) 3 x

21

1

2

3

2

1

21 Figure B.29 21

x

20.5

0.5

1

21 Figure B.31 l = 1

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-12 Answers to Odd-Numbered Problems

(e) See Figures B.31 and B.32. When l = 3, one critical point is a center and the other is a saddle. For l = 1, the bar is attracted to the magnet. For l = 3, the bar may oscillate periodically, or (rarely) come to rest at the saddle critical point.

Exercises 5.5, page 281 1. Runge–Kutta approximations tn

y 1

0.5 x 2

1

3

20.5

21 Figure B.32 l = 3

35. (c) For upper half-plane, center is at y = 0, y = - 1; for lower half-plane, center is at y = 0, y = 1. (d) All points on the segment y = 0, - 1 … y … 1 (e) y = -0.5; see Figure B.33.

pn 1r = 1.52

pn 1r = 22

0.25 1.59600 1.54249 0.5 2.37945 2.07410 0.75 3.30824 2.47727 1.0 4.30243 2.72769 1.25 5.27054 2.86458 1.5 6.13869 2.93427 1.75 6.86600 2.96848 2.0 7.44350 2.98498 2.25 7.88372 2.99286 2.5 8.20933 2.99661 2.75 8.44497 2.99839 3.0 8.61286 2.99924 3.25 8.73117 2.99964 3.5 8.81392 2.99983 3.75 8.87147 2.99992 4.0 8.91136 2.99996 4.25 8.93893 2.99998 4.5 8.95796 2.99999 4.75 8.97107 3.00000 5.0 8.98010 3.00000 Limiting population is 31/1r - 12. 3. x1t2 =

y

pn 1r = 32 1.43911 1.64557 1.70801 1.72545 1.73024 1.73156 1.73192 1.73201 1.73204 1.73205 1.73205 1.73205 1.73205 1.73205 1.73205 1.73205 1.73205 1.73205 1.73205 1.73205

1 2 cos1pt/122 1p/122sin1pt/122 2 4 + 1p/122 2 4 + 1p/122 2 2 19 + a + be-2t 2 4 + (p/12)2 I

5. (a) 300 21.5

20.5

3.5 7.5

21 0 1 25.5

150 y 167

700

S

Figure B.34

Figure B.33

(b) 295 persons c[1 - exp1 -bt2] 7. P1t2 = expe - bt f , b c[1 - exp1 -bt2] Q1t2 = expe - bt f[exp1bt2 - 1] b N0 1c + s2 9. N1t2 = , N0 J N102 N0s + 1c + s - N0s2e-1c + s2t 11. Roughly, 2

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-13

Answers to Odd-Numbered Problems

Exercises 5.6, page 287 1. m1x″ = k1 1y - x2 , m2y″ = - k1 1y - x2 - k2y ; x102 = -1 , x′102 = 0 , y102 = 0 , y′102 = 0 x1t2 = - 18>172cos t - 19>172cos 1 220>3 t 2 , y1t2 = - 16>172cos t + 16>172cos 1 220>3 t 2 3. mx″ = - kx + k1y - x2 , my″ = - k1y - x2 + k1z - y2 , mz″ = - k1z - y2 - kz ;

The normal frequency 11>2p2 3 1 2 + 22 2 1k>m2 has

the mode x1t2 = z1t2 = - 1 1> 22 2 y1t2; the normal

frequency 11>2p2 3 1 2 - 22 2 1k>m2 has the mode x1t2 = z1t2 =

1 1> 22 2 y1t2; and the normal frequency

11>2p2 22k>m has the mode x1t2 = - z1t2, y1t2 K 0 . 5. x1t2 = -e-t - te-t - cos t ; y1t2 = e-t + te-t - cos t 7. x1t2 = 12>52cos t + 14>52sin t - 12>52cos 26t +

1 26>5 2 sin 26t - sin 2t ;

-

1 26>10 2 sin 26t - 11>22sin 2t

y1t2 = 14>52cos t + 18>52sin t + 11>52cos 26t 9. 11>2p2 2g>l ;

11>2p2 21g>l2 + 12k>m2

Exercises 5.7, page 294 1. I1t2 = 1 19> 221 2 3 e1 -25 - 52212 t>2 - e1 -25 + 52212 t>2 4 3. Ip 1t2 = 14>512cos 20t - 11>512sin 20t ; resonance frequency is 5>p. 5. M1g2 = 1> 21100 - 4g 2 2 2 + 100g 2 M 0.02 0.01 Ë175/8 Figure B.35

7. L = 35 H, R = 10 Ω, C = 1>15 F, and E1t2 = 50 cos 10t V 11. I1 = 13>52e-3t>2 - 18>52e-2t>3 + 1 , I2 = 11>52e-3t>2 - 16>52e-2t>3 + 1 , I3 = 12>52e-3t>2 - 12>52e-2t>3 13. 11>22I 1= + 2q3 = cos 3t 1where I3 = q3= 2 , 11>22I 1= + I2 = 0, I1 = I2 + I3 : I1 102 = I2 102 = I3 102 = 0 ;

I1 = - 136>612e-t cos 23t - 1 4223>61 2 e-t sin23t + 136>612cos 3t + 130>612sin 3t , I2 = 145>612e-t cos 23t - 1 3923>61 2 e-t sin23t - 145>612cos 3t + 154>612sin 3t , I3 = - 181>612e-t cos 23t - 1 323>61 2 e-t sin23t + 181>612cos 3t - 124>612sin 3t

Exercises 5.8, page 303 1. For v = 3>2: The Poincaré map alternates between the points 10.8, 1.52 and 10.8, -1.52. There is a subharmonic solution of period 4p. For v = 3>5: The Poincaré map cycles through the points 1 -1.5625, 0.62, 1 - 2.1503, - 0.48542, 1 -0.6114, 0.18542, 1 -2.5136, 0.18542, and 1 - 0.9747, -0.4854). There is a subharmonic solution of period 10p. 3. The points become unbounded. 5. The attractor is the point 1 - 1.0601, 0.26242. 9. (a) 5 1>7, 2>7, 4>7, 1>7, c 6 , 5 3>7, 6>7, 5>7, 3>7, c 6 (b) 5 1>15, 2>15, 4>15, 8>15, 1>15, c 6 , 5 1>5, 2>5, 4>5, 3>5, 1>5, c 6 , 5 1>3, 2>3, 1>3, c 6 , 5 7>15, 14>15, 13>15, 11>15, 7>15, c 6 (c) xn = 0 for n Ú j 13. y 2

0

–2 – 2.2

0

x 2.2

Figure B.36

Review Problems, page 306 1. x1t2 = - 1c1 >32t 3 - 1c2 >22t 2 - 1c3 + 2c1 2t + c4 , y1t2 = c1t 2 + c2t + c3 3. x1t2 = c1 cos 3t + c2 sin 3t + et >10 , y1t2 = 13>221c1 + c2 2cos 3t - 13>221c1 - c2 2sin 3t - 111>202et - 11>42e-t 5. x1t2 = 2 sin t, y1t2 = et - cos t + sin t , z1t2 = et + cos t + sin t 7. x1t2 = - 113.9>42e-t>6 - 14.9>42e-t>2 + 4.8 , y1t2 = - 113.9>22e-t>6 + 14.9>22e-t>2 + 4.8 9. x1= = x2 , x2= = x3 , x3= = 11>3215 + etx1 - 2x2 2 11. x1= = x2 , x2= = x3 , x3= = t - x5 - x6 , x4= = x5 , x5= = x6 , x6= = x2 - x3 13. x2 - 1y - 12 2 = c ; critical point 10, 12 is saddle (unstable). 15. Asymptotically stable spiral point 17. I1 + I2 + I3 = 0 , q>C = R2I2 , R2I2 = R1I3 + L dI3 >dt , where q is the charge on the capacitor 1I1 = dq>dt2; I3 = e-t 1A cos t + B sin t2 , I2 = e-t 1B cos t - A sin t2 , I1 = e-t 31A - B2sin t - 1A + B2cos t4 .

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-14 Answers to Odd-Numbered Problems

CHAPTER 6 Exercises 6.1, page 326 1. 1 - ∞ , 02 3. 13p>2, 5p>22 5. 10, ∞ 2 7. Lin. indep. 9. Lin. dep. 11. Lin. indep. 13. Lin. indep. 15. c1e3x + c2e-x + c3e-4x 17. c1x + c2x2 + c3x3 19. (a) c1ex + c2e-x cos 2x + c3e-x sin 2x + x2 (b) - ex + e-x sin 2x + x2 21. (a) c1x + c2x ln x + c3x1ln x2 2 + ln x (b) 3x - x ln x + x1ln x2 2 + ln x 23. (a) 2 sin x - x (b) 4x - 6 sin x 29. (b) Let f1 1x2 = 0 x - 1 0 and f2 1x2 = x - 1 33. e2x, 1sin x - 2 cos x2 >5 , - 12 sin x + cos x2 >5 35. xy‴ - y″ + xy′ - y = 0 Exercises 6.2, page 332 1. c1 + c2e2x + c3e-4x 3. c1e-x + c2e-2x>3 + c3ex>2 5. c1e-x + c2e-x cos 5x + c3e-x sin 5x 7. c1e-x + c2e1 3 + 2652 x>4 + c3e1 3 - 2652 x>4 9. c1e3x + c2xe3x + c3x2e3x 11. c1e-x + c2xe-x + c3x2e-x + c4x3e-x 13. c1 cos 22x + c2x cos 22x + c3 sin 22x + c4x sin 22x 15. c1ex + c2xex + c3e-3x + 1c4 + c5x2e-x cos 2x + 1c6 + c7x2e-x sin 2x -4x 17. c1e + c2e3x + 1c3 + c4x + c5x2 2e-2x + 1c6 + c7x2e-2x cos x + 1c8 + c9x2e-2x sin x + c10 + c11x + c12x2 + c13x3 + c14x4 19. ex - 2e-2x - 3e2x 21. e2x - 22ex sin 22x 23. x1t2 = c1 + c2t + c3et, y1t2 = c1 - c2 + c2t 27. c1e1.120x + c2e0.296x + c3e-0.520x + c4e-2.896x 29. c1e-0.5x cos10.866 x2 + c2 e-0.5x sin10.866 x2 + c3e-0.5x cos11.323 x2 + c4e-0.5x sin11.323 x2 31. (a) 5 x, x -1, x2 6 (b) 5 x, x2, x -1, x -2 6 (c) 5 x, x2 cos13 ln x2, x2 sin13 ln x2 6 33. (b) x1t2 = c1 cos t + c2 sin t + c3 cos 26 t + c4 sin 26 t (c) y1t2 = 2c1 cos t + 2c2 sin t - 1c3 >22 cos 26 t - 1c4 >22sin 26 t (d) x1t2 = 13>52cos t + 12>52cos 26 t , y1t2 = 16>52cos t - 11>52cos 26 t 35. c1 cosh rx + c2 sinh rx + c3 cos rx + c4 sin rx, where r 4 = k> 1EI2 Exercises 6.3, page 337 1. c1xex + c2 + c3x + c4x2 3. c1x2e-2x 5. c1ex + c2e3x + c3e-2x - 11>62xex + 11>62x2 + 15>182x + 37>108

7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 39.

c1ex + c2e-2x + c3xe-2x - 11>62x2e-2x c1ex + c2xex + c3x2ex + 11>62x3ex D5 D+7 1D - 221D - 12 31D + 12 2 + 44 3 1D + 22 2 31D + 52 2 + 94 2 c3 cos 2x + c4 sin 2x + c5 c3xe3x + c4x2 + c5x + c6 c3 + c4x + c5 cos 2x + c6 sin 2x c3xe-x cos x + c4xe-x sin x + c5x2 + c6x + c7 c2x + c3x2 + c6x2ex -2e3x + e-2x + x2 - 1 x2e-2x - x2 + 3 x1t2 = - 11>632e3t + c1 + c2t - c3e22t - c4e-22t , y1t2 = 18>632e3t + c1 + c2t + c3e22t + c4e-22t

Exercises 6.4, page 341 1. 11>62x2e2x 3. e2x >16 5. ln1sec x2 - 1sin x2ln1sec x + tan x2 7. c1x + c2x2 + c3x3 - 11>242x -1 9. - 11>22ex

L

e-xg1x2 dx + 11>62e-x

+ 11>32e2x

L

exg1x2 dx

e-2xg1x2 dx L 11. c1x + c2x -1 + c3x3 - x sin x - 3 cos x + 3x -1 sin x Review Problems, page 343 1. (a) 10, ∞ 2 (b) 1 - 4, -12 , 1 -1, 12 , 11, ∞ 2 5. (a) e-5x 1c1 + c2x2 + e2x 1c3 + c4x + c5x2 2 + 1cos x21c6 + c7x2 + 1sin x21c8 + c9x2 (b) c1 + c2x + c3x2 + c4x3 + ex 1c5 + c6x2 + 1e-x cos 23 x21c7 + c8x2 + 1e-x sin 23 x21c9 + c10x2 7. (a) D3 (b) D2 1D - 32 (c) 3D2 + 44 2 2 3 (d) 31D + 22 + 94 (e) D3 1D + 12 2 1D2 + 421D2 + 92 9. c1x + c2x5 + c3x -1 - 11>212x -2

CHAPTER 7 Exercises 7.2, page 360 1 1 1. 2 , s 7 0 , s76 3. s-6 s s , s70 5. 2 s +4 s-2 , s72 7. 1s - 22 2 + 9 2s + 1 e-ps + 1 , all s 9. e-2s a 2 b , s 7 0 11. 2 s s +1

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Answers to Odd-Numbered Problems

13. 15. 17. 19. 21. 23. 25. 27. 29.

2 8 2 6 - 3+ 2- , s70 s s+3 s s 6 1 s-4 + , s74 4 2 s 1s - 12 1s - 42 2 + 1 6 1 6 , s73 - 4+ 2 s 1 1s - 32 + 36 s 24 s-1 , s75 5 1s - 12 2 + 7 1s - 52 Continuous (hence piecewise continuous) Piecewise continuous Continuous (hence piecewise continuous) Neither All but functions (c), (e), and (h)

Exercises 7.3, page 365 2 2 1. 3 + s 1s - 12 2 + 4 1 1 s+1 + 3. 1s + 12 2 + 9 s - 6 s 1 s 4 - + 5. 1s + 12 3 s2 s2 + 16 24 24 12 4 1 7. 5 - 4 + 3 - 2 + s s s s s 41s + 12 s 11. 2 9. 31s + 12 2 + 44 2 s - b2 1 s 3s s 13. 15. + 2s 21s2 + 42 41s2 + 12 41s2 + 92 s s 17. 2 2 21s + 92 21s + 492 m-n n+m + 19. 2 2 2 23s + 1n + m2 4 23s + 1m - n2 2 4 s-a 21. 1s - a2 2 + b2 2s3 - 6sb2 s2 - b2 (b) 2 25. (a) 2 2 2 1s + b 2 1s + b2 2 3 1 29. 2 33. e-s >s2 s + 6s + 10 35. e-1p>22s > 1s2 + 12 Exercises 7.4, page 374 1. ett 3 3. e-tcos 3t 5. 11>22e-2tsin 2t -2t -2t 7. 2e cos 3t + 4e sin 3t 9. 13>22etcos 2t - 3etsin 2t 6 1 4 2 1 11. 13. s+5 s+2 s-1 1s + 12 2 s 1s - 12 + 2 3 + 15. s + 1 1s - 12 2 + 4 5 11 4 + 17. 6s 101s - 22 151s + 32 1 4 1 s+1 19. d c 17 s - 3 1s + 12 2 + 1 1s + 12 2 + 1

21. 23. 25. 27. 29. 31. 33. 35. 39.

41. 43.

B-15

11>32 + et + 114>32e6t -e-3t + 2te-3t + 6e-t 8e2t - e-tcos 2t + 3e-tsin 2t - 15>32e-t + 15>122e2t + 15>42e-2t 3e-2t + 7etcos t + 11etsin t F1 1s2 = F2 1s2 = F3 1s2 = 1>s2 , ℒ-1 5 1>s2 6 1t2 = f3 1t2 = t e5t >t - e-2t >t 21cos t - cos 3t2 >t A C B + , where + s s-1 s+2 2s + 1 -1 A = = , ` 1s - 121s + 22 s = 0 2 2s + 1 B = = 1, ` s1s + 22 s = 1 2s + 1 -1 C = = ` s1s - 12 s = -2 2 2e-t - 4e3t + 5e2t 21s - 12 + 2132 4 + s+2 1s - 12 2 + 22

Exercises 7.5, page 382 1. 2etcos 2t + etsin 2t 3. -e-3t + 3te-3t 5. t 2 + cos t - sin t 7. cos t - 4e5t + 8e2t 9. 31t - 12et-1 - e-61t - 12 11. 2 - t + e2 - t + 2et - 2 13. 17>52sin t + 111>52cos t + 13>52e2t - p - e-1t - p>22 - s2 + s - 1 15. 2 1s + 121s - 121s - 22 s3 + 5s2 - 6s + 1 s5 + s4 + 6 19. 17. 4 2 s 1s + s - 12 s1s - 121s2 + 5s - 12 3 2 s + s + 2s -s3 + 1 + 3se-2s - e-2s 21. 2 23. 1s + 121s - 12 2 s2 1s2 + 42 t 2 -t 27. 1t - 42e 25. 2e - cos t - sin t 29. 13a - b2et >2 + 1b - a2e3t >2 31. 5>2 + 1a - 5>22e-tcos t + 1a + b - 5>22e-tsin t 35. t 2 >2 37. cos t + t sin t + c1sin t - t cos t2, 1c arbitrary2 39. e1t2 = -a cos 1 2k>I t 2

41. e1t2 = 1 -2aI>24Ik - m2 2 e-mt>12I2sin 124Ik - m2 t>12I22

Exercises 7.6, page 390 1. 2e-s >s3 3. e-2s 14s2 + 4s + 22 >s3 5. 12e-s - e-2s + 2e-3s 2 >s 7. 31e-s - e-2s 21s + 124 >s2 9. 1e-s - 2e-2s + e-3s 2 >s2 11. et - 2u1t - 22 13. e-21t - 22u1t - 22 - 3e-21t - 42u1t - 42 15. e-21t - 32 3cos1t - 32 - 2 sin1t - 324u1t - 32 17. 17e6 - 2t - 6e3 - t 2u1t - 32 19. 10 - 10u1t - 3p231 + e-1t - 3p2 1cos t + sin t24 + 10u1t - 4p231 - e-1t - 4p2 1cos t + sin t24

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-16 Answers to Odd-Numbered Problems 21. sin t + 31 - cos1t - 324u1t - 32 2 1 t

0 1 2 3 4 5 6

–1

Figure B.37

23. t + 34 - t + sin1t - 22 - 2 cos1t - 224u1t - 22 y 6 4

19. (a) 2p>s

2

(b) 1052p> 116s9>2 2 27. (a) and (b), but not (c)

t 2

4

6

8 10

Figure B.38

25. e-tcos t + 2e-tsin t + 11>2231 - e2p-t 1cos t + sin t24u1t - 2p2 - 11>2231 - e4p-t 1cos t + sin t24u1t - 4p2 -t 27. e + e-2t + 11>223e-3t - 2e-21t + 12 + e-1t + 42 4u1t - 22 29. cos 2t + 11>3231 - u1t - 2p24sin t + 11>6238 + u1t - 2p24sin 2t -2t 31. 2e - 2e-3t + 31>36 + 11>621t - 12 - 11>42e-21t - 12 + 12>92e-31t - 12 4u1t - 12 - 319>36 + 11>621t - 52 - 17>42e-21t - 52 + 111>92e-31t - 52 4u1t - 52 33. 0.04 - 0.02 e-3t>125 + 0.02 u1t - 10231 - e-31t - 102>125 4 35. The resulting differential equation has polynomial coefficients, so the Laplace transform method will result in a differential equation for the transform.

f(t) 2 t 1

2

3

4

5

6

1 1 - e-s-1 e-s - e-2s c + d s s+1 1 - e-2s f(t) 1 t 2

L0

t

et - y 1t - y2g1y2dy

g1y2e2y-2tsin1t - y2dy + e-2tcos t + 3e-2tsin t L0 5. 1 - cos t 7. 2e5t - 2e-2t 9. 1t>22sin t 11. 12>32e-2t + 11>32et 13. s-2 1s - 32 -1 15. t>4 + 13>82sin 2t 19. 3 17. cos t 3.

21. e-t>2cos 1 23 t>2 2 - 1 1> 23 2 e-t>2sin 1 23 t>2 2 23. H1s2 = 1s2 + 92 -1 ; h1t2 = 11>32sin 3t ; yk 1t2 = 2 cos 3t - sin 3t ;

3 sin 31t - y2 4 g1y2dy + 2 cos 3t - sin 3t L0 2 25. H1s2 = 1s - s - 62 -1 ; h1t2 = 1e3t - e-2t 2 >5 ; yk 1t2 = 2e3t - e-2t ; y1t2 = 11>32

t

y1t2 = 11>52

t

y1t2 = 11>22

t

3

4

Figure B.40

5

L0

t

L0 + e-2tsin 6t 31. t 2 >2

Figure B.39

1

t

1. 2tet - et +

29. 11>302

1 0

Exercises 7.8, page 404

3e31t - y2 - e-21t - y2 4g1y2dy + 2e3t - e-2t L0 27. H1s2 = 1s2 - 2s + 52 -1 ; h1t2 = 11>22etsin 2t ; yk 1t2 = etsin 2t ;

Exercises 7.7, page 396 1 - 2se-2s - e-2s 1. s2 11 - e-2s 2

3.

1 - e-as 1 7. 2 -as s11 + e 2 as 11 + e-as 2 t-n -t e e 13. 11 + e - en + 1 2 6 2 e-2t 1 + e + e2 - e2n + 2 + c d , 3 e+1 for n 6 t 6 n + 1 ∞ 1 1 15. a n = s-1 n = 1s ∞ 1 -12 n + 1 1 1 17. a = lna1 + 2 b 2n 2 s n = 1 2ns 5.

y

e1t - y2 3 sin 21t - y2 4 g1y2dy + etsin 2t

e-21t - y2 3 sin 61t - y2 4 e1y2dy - e-2tcos 6t

Exercises 7.9, page 410 1. -1 3. - 1 5. e-2 7. e-s - e-3s -s 9. e 11. 0 13. - 1sin t2u1t - p2 15. et + e-3t + 11>421et - 1 - e3 - 3t 2u1t - 12 - 11>421et - 2 - e6 - 3t 2u1t - 22 t-2 17. 21e - e-1t - 22 2u1t - 22 + 2et - t 2 - 2 19. e-t - e-5t + 1e>421e1 - t - e5 - 5t 2u1t - 12

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Answers to Odd-Numbered Problems

21. sin t + 1sin t2u1t - 2p2 ; see Fig. B.41.

Review Problems, page 415 3 1 1 2 1. + e-2s c - 2 d 3. s s s 1s + 92 3 s2 - 36 1 6 2 5 7. 2 5. - 4+ 3- 2 s-2 s s s + 25 1s + 362 2 1 4 8 9. 2e-4s c 3 + 2 + d 11. 17>22t 2e-3t s s s 13. 2et + 2e-2tcos 3t + e-2tsin 3t 15. e-2t + e-t - 2te-t 17. 32et - 2 + 2e4 - 2t 4u1t - 22 19. e2t - e5t 21. - 13>22 + t + t 2 >2 + 13>22e-tcos t - 11>22e-tsin t 23. 1 2> 27 2 e-3t>2sin 1 27 t>2 2

y 3

2 p

1 –1

3p t 4p

2p

–2

–3 Figure B.41

23. sin t + 1sin t2u1t - p2 + 1sin t2u1t - 2p2 y 3 1 –1

2 p

3p 2p

–2

t 4p

–3 Figure B.42

25. 27. 29. 31.

25. 11>22e-2tsin 2t 27. 11>22etsin 2t 29. The mass remains stopped at x1t2 K 0 , t 7 p>2 L 35. 33lx2 - x3 + 1x - l2 3u1x - l24 6EI Exercises 7.10, page 413 1. x = et ; y = et 3. z = et ; w K 0 t 5. x = 17>42e + 17>42e-t - 13>22cos t ; y = 17>42et - 17>42e-t + 11>22sin t 7. x = - 1150>172e5t>2cos 1 215 t>2 2 - 1 334215>85 2 e

sin 1 215 t>2 2 - 13>172e

5t>2

y = 146>172e5t>2cos 1 215 t>2 2

9. 11.

13.

15. 17. 19. 21.

23.

B-17

+ 5 11>42 - 3 3> 1 427 2 4 e-31t - 12>2sin 3 27 1t - 12 >2 4 - 11>42e-31t - 12>2cos 3 27 1t - 12 >2 4 6 u1t - 12 c3t + te-2t + e-2t - 14 19>102e-3t + 11>102cos t - 13>102sin t 1s2 - 5s + 62 -1 ; e3t - e2t x = 1 - 1et + e-t 2 >2 - 31 - 1et - 2 + e-1t - 22 2 >24u1t - 22 , t y = 1e - e-t 2 >2 - 11>223et - 2 - e-1t - 22 4u1t - 22

CHAPTER 8 -3t

;

- 1 146215>85 2 e5t>2sin 1 215 t>2 2 + 122>172e-3t x = 4e-2t - e-t - cos t ; y = 5e-2t - e-t x = 1et - e-t 2 >2 - 11>223et - 2 - e - 1t - 22 4u1t - 22 ; y = 1 - 1et + e-t 2 >2 - 31 - 1et - 2 + e-1t - 22 2 >24u1t - 22 x = e-t - 11>223e-1t - p2 + cos t - sin t4u1t - p2 ; y = e-t + 31 - 11>22e-1t - p2 + 11>22cos t + 11>22sin t4u1t - p2 x = t2 ; y = t - 1 x = 1t - 22et - 2 ; y = et - 2 x = -7e-t + et ; y = 2e-t ; z = - 13e-t + et x = 0.011 - e-t>2 + e-t>6 2 + 30.48 - 0.36e-1t - 52>6 - 0.12e-1t - 52>2 4 u1t - 52 ; y = 0.021e-t>2 + e-t>6 2 + 30.48 - 0.72e-1t - 52>6 + 0.24e-1t - 52>2 4 u1t - 52 2I1 + 10.12I 3= + 10.22I 1= = 6 , 10.12I 3= - I2 = 0 , I1 = I2 + I3 ; I1 102 = I2 102 = I3 102 = 0 ; I1 = - e-20t - 2e-5t + 3 , I2 = - 2e-20t + 2e-5t , I3 = e-20t - 4e-5t + 3

Exercises 8.1, page 425 1. 1 + x + x2 + g 3. x + x2 + 11>22x3 + g 5. 1 - 11>62t 3 + 11>1802t 6 + g 7. 11>62u 3 - 11>1202u 5 + 11>50402u 7 + g 1 1 9. (a) p3 1x2 = 1x - 12 - 1x - 12 2 + 1x - 12 3 2 3 (b) 0 e3 11.52 0 = 0 f 142 1j211.5 - 12 4 >4! 0 … 610.52 4 >4! (c) 0 ln11.52 - p3 11.52 0 = 0.011202… (d) y p3(x)

2

x 22 24

1

2

ln x Figure B.43

13. t + 11>22t - 11>62t 3 + g 2

15. 1 - x2 >6

Exercises 8.2, page 433 1. 3 - 1, 32 3. 1 -4, 02 5. 31, 34 ∞ 1 9. a c + 2-n - 1 d xn n=0 n + 1 11. x + x2 + 11>32x3 + g

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-18 Answers to Odd-Numbered Problems 13. 1 - 2x + 15>22x2 + g 15. (c) 1 - 11>22x + 11>42x2 - 11>242x3 + g ∞

∞

17. a 1 -12 nnxn - 1

19. a ak 2kx2k - 1 n=1 k=1 ∞ 1 - 12 n 21. ln1x + 12 = a xn + 1 n=0 n + 1 ∞

23. a 1k + 12ak + 1x k=0 ∞ k=1

29. a

∞

+ a1 ax + a

25. 27. 29. 31. 35.

1 - 12 n + 1 1x - p2 2n

n=0

∞

k=1

k ∞

25. a ak - 1xk

12n2!

31. 1 + a 2xn n=1

33. 61x - 12 + 31x - 12 + 1x - 12 2

∞

35. (a) a 1 -12 n 1x - 12 n n=0

3 ∞

(b) a

n=1

1 - 12 n - 1 n

1x - 12 n

Exercises 8.3, page 443 1. - 1 5. - 1, 2 3. { 22 7. x = np , n an integer 9. u … 0 and u = np , n = 1, 2, 3, … 11. y = a0 11 - 2x + 13>22x2 - x3 >3 + g 2 13. a0 11 + x4 >12 + g 2 + a1 1x + x5 >20 + g 2 15. a0 11 - x2 >2 - x3 >6 + g 2 + a1 1x + x2 >2 - x3 >6 + g 2 17. a0 11 - x2 >2 + g 2 + a1 1x - x3 >6 + g 2 ∞ 1 2 19. a0 a x2n = a0e x n! n=0 21. a0 11 - 2x2 + x4 >32 ∞ 1 - 321 - 12 g 12k - 52 + a1 c x + a x2k + 1 d 12k + 12! k=1

∞

29. (a) a0 c 1 + a 1 - 12 k

23. a3k + 2 = 0 , k = 0, 1, … ∞ 31 # 4 # 7 g 13k - 224 2 a0 a1 + a x3k b 13k2!

32 # 5 # 8 g 13k - 124 2

x3k + 1 b 13k + 12! 2 + x2 - 15>122x + 111>722x6 + g x + 11>62x3 - 11>122x4 + 17>1202x5 + g 1 - 2x + x2 - 11>62x3 1 + 2x - 13>42x2 - 15>62x3 10 - 25t 2 + 1250>32t 3 - 1775>42t 4 + g k=1 4

Exercises 8.4, page 449 1. 2 3. 23 5. p>2 7. a0 31 - 1x - 12 2 + 11>221x - 12 4 - 11>621x - 12 6 + g 4 9. a0 31 + 1x - 12 2 + g 4 + a1 31x - 12 + 11>321x - 12 3 + g 4 11. a0 31 - 11>821x - 22 2 + 11>3221x - 22 3 + g 4 + a1 31x - 22 + 11>821x - 22 2 - 17>9621x - 22 3 + g 4 2 4 13. 1 - 11>22t + 11>62t - 131>7202t 6 + g 15. 1 + x + 11>242x4 + 11>602x5 + g 17. 1 - 11>621x - p2 3 + 11>12021x - p2 5 + 11>18021x - p2 6 + g 19. -1 + x + x2 + 11>22x3 + g 21. a0 31 + 11>22x2 + 11>82x4 + 11>482x6 + g 4 + 311>22x2 + 11>122x4 + 111>7202x6 + g 4 23. a0 31 - 11>22x2 + g 4 + a1 3x - 11>32x3 + g 4 + 311>22x2 + 11>32x3 + g 4 25. a0 31 - 11>22x2 + g 4 + a1 3x + g 4 + 311>22x2 - 11>62x3 + g 4 27. a0 31 - 11>22x2 - 11>62x3 + g 4 + a1 3x + 11>22x2 + g 4 + 311>62x3 + g 4

n1n - 221n - 42 g 1n - 2k + 221n + 121n + 32 g 1n + 2k - 12

x2k d 12k2! ∞ 1n - 121n - 32 g 1n - 2k + 121n + 221n + 42 g 1n + 2k2 2k + 1 + a1 c x + a 1 - 12 k x d 12k + 12! k=1 2 (c) P0 1x2 = 1 , P1 1x2 = x , P2 1x2 = 11>2213x - 12 31. (a) 1 - 11>22t 2 + 1h>62t 3 + 311 - h2 2 >244t 4 + g (b) Yes k=1

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Answers to Odd-Numbered Problems

Exercises 8.5, page 453 1. c1x -2 + c2x -3 3. c1x cos14 ln x2 + c2x sin14 ln x2 5. c1x3cos12 ln x2 + c2x3sin12 ln x2 7. c1x + c2x -1cos13 ln x2 + c3x -1sin13 ln x2 9. c1x + c2x -1>2cos 3 1 219 >2 2 ln x 4 11. 13. 15. 17.

+ c3x -1>2sin 3 1 219 >2 2 ln x 4 c1 1x - 32 -2 + c2 1x - 32 1>2 c1x + c2x2 + 14>152x -1>2 2t 4 + t -3 131>172x + 13>172x -2cos15 ln x2 - 176>852x -2sin15 ln x2

Exercises 8.6, page 464 1. {1 are regular. 3. {i are regular. 5. 1 is regular, - 1 is irregular. 7. 2 is regular, - 1 is irregular. 9. - 4 is regular and 2 is irregular. 11. r 2 - 3r - 10 = 0 ; r1 = 5 , r2 = - 2 13. r 2 - 5r>9 - 4>3 = 0 ; r1 = 1 5 + 2457 2 >18 ,

r2 = 1 5 - 2457 2 >18 r 2 - 1 = 0 ; r1 = 1 , r2 = - 1 r 2 + r - 12 = 0 ; r1 = 3 , r2 = - 4 a0 3x2>3 - 11>22x5>3 + 15>282x8>3 - 11>212x11>3 + g4 a0 31 - 11>42x2 + 11>642x4 - 11>23042x6 + g 4 a0 3x - 11>32x2 + 11>122x3 - 11>602x4 + g 4 ∞ 1 - 12 nx n + 13>22 25. a0 a n - 1 2 1n + 22!

15. 17. 19. 21. 23.

n=0 ∞

x2n + 2 27. a0 a 2n n = 0 2 1n + 12! n! 29. a0 x2 ∞ xn 31. a0 a = a0ex ; yes, a0 6 0 n = 0 n! ∞ 2n - 2 13n + 42x n + 11>32 33. a0 c x1>3 + a d ; yes, a0 6 0 3nn! n=1

35. a0 3x5>6 - 11>112x11>6 + 11>3742x17>6 - 11>25,8062x23>6 + g 4 4>3 37. a0 3x + 11>172x7>3 + 11>7822x10>3 + 11>68,0342x13>3 + g 4 ∞

39. The expansion a n!xn diverges for x ≠ 0. n=0

41. The transformed equation is z d 2y>dz2 + 3 dy>dz - y = 0. Also zp1z2 = 3 and z2q1z2 = - z are analytic at z = 0; hence, z = 0 is a regular singular point. y1 1x2 = a0 31 + 11>32x -1 + 11>242x -2 + 11>3602x -3 + g 4 -3 -1 45. a0 3x - 11>22x - 11>82x - 11>1442x3 + g 4 + a3 31 + 11>102x2 + 11>2802x4 + 11>15,1202x6 + g 4

B-19

Exercises 8.7, page 473 1. c1y1 1x2 + c2y2 1x2, where y1 1x2 = x2>3 - 11>22x5>3 + 15>282x8>3 + g and y2 1x2 = x1>3 - 11>22x4>3 + 11>52x7>3 + g 3. c1y1 1x2 + c2y2 1x2 , where y1 1x2 = 1 - 11>42x2 + 11>642x4 + g and y2 1x2 = y1 1x2 ln x + 11>42x2 - 13>1282x4 + 111>13,8242x6 + g 5. c1 3x - 11>32x2 + 11>122x3 + g 4 + c2 3x -1 - 14 7. c1y1 1x2 + c2y2 1x2 , where y1 1x2 = x3>2 - 11>62x5>2 + 11>482x7>2 + g and y2 1x2 = x -1>2 - 11>22x1>2 9. c1w1 + c2w2 , where w1 1x2 = x2 + 11>82x4 + 11>1922x6 + g and w2 1x2 = w1 1x2ln x + 2 - 13>322x4 - 17>11522x6 + g 11. c1y1 1x2 + c2y2 1x2 , where y1 1x2 = x2 and y2 1x2 = x2 ln x - 1 + 2x - 11>32x3 + 11>242x4 + g 13. c1y1 1x2 + c2y2 1x2 , where y1 1x2 = 1 + x + 11>22x2 + g and y2 1x2 = y1 1x2ln x - 3x + 13>42x2 + 111>362x3 + g4 15. c1y1 1x2 + c2y2 1x2 , where y1 1x2 = x1>3 + 17>62x4>3 + 15>92x7>3 + g and y2 1x2 = 1 + 2x + 16>52x2 + g ; all solutions are bounded near the origin. 17. c1y1 1x2 + c2y2 1x2 + c3y3 1x2 , where y1 1x2 = x5>6 - 11>112x11>6 + 11>3742x17>6 + g , y2 1x2 = 1 - x + 11>142x2 + g , and y3 1x2 = y2 1x2ln x + 7x - 1117>1962x2 + 14997>2981162x3 + g 19. c1y1 1x2 + c2y2 1x2 + c3y3 1x2 , where y1 1x2 = x4>3 + 11>172x7>3 + 11>7822x10>3 + g , y2 1x2 = 1 - 11>32x - 11>302x2 + g , and y3 1x2 = x -1>2 - 11>52x1>2 - 11>102x3>2 + g 21. (b) c1y1 1t2 + c2y2 1t2 , where y1 1t2 = 1 - 11>621at2 2 + 11>12021at2 4 + g , and y2 1t2 = t -1 31 - 11>221at2 2 + 11>2421at2 4 + g4 (c) c1y1 1x2 + c2y2 1x2 , where y1 1x2 = 1 - 11>621a>x2 2 + 11>12021a>x2 4 + g, and y2 1x2 = x31 - 11>221a>x2 2 + 11>2421a>x2 4 + g4 23. c1y1 1x2 + c2y2 1x2 , where y1 1x2 = 1 + 2x + 2x2 and y2 1x2 = - 11>62x -1 - 11>242x -2 - 11>1202x -3 + g 25. For n = 0, c1 + c2 3ln x + x + 11>42x2 + 11>182x3 + g 4 ; and for n = 1, c1 11 - x2 + c2 311 - x2 ln x + 3x - 11>42x2 - 11>362x3 + g 4 Exercises 8.8, page 485 1 3 5 3 1. c1F a1, 2; ; xb + c2x1>2F a , ; ; xb 2 2 2 2 1 3 3 3 3. c1F a1, 1; ; xb + c2x1>2F a , ; ; xb 2 2 2 2

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-20 Answers to Odd-Numbered Problems

∞ 1 xn = - x -1 ln11 - x2 5. F11, 1; 2; x2 = a n = 0 1n + 12 ∞ 1 3 7. F a , 1; ; x2 b = a 11>22n x 2n > 13>22n 2 2 n=0 ∞

= a x2n > 12n + 12 n=0

9. 13. 17. 19. 21. 27. 29. 37. 39.

41.

1 1+x = x -1 lna b 2 1-x -1 -1 11 - x2 , 11 - x2 ln x c1J1>2 1x2 + c2J-1>2 1x2 15. c1J1 1x2 + c2Y1 1x2 c1J2>3 1x2 + c2J-2>3 1x2 J1 1x2 ln x - x -1 + 13>642x3 - 17>23042x5 + g x3>2J3>2 1x2 ∞ 1 - 12 n + 1 1 1 J0 1x2 ln x + a 2n c 1 + + g + d x2n 2 n 2 n = 1 2 1n!2 1 , x , 13x2 - 12 >2 , 15x3 - 3x2 >2 , 135x4 - 30x2 + 32 >8 1 , 2x , 4x2 - 2 , 8x3 - 12x 1 , 1 - x , 12 - 4x + x2 2 >2 , 16 - 18x + 9x2 - x3 2 >6 (b) z = c1 cos x + c2 sin x , so for x large, y1x2 ≈ c1x -1>2 cos x + c2x -1>2 sin x

Review Problems, page 489 1. (a) 1 - x + 13>22x2 - 15>32x3 + g (b) -1 + x - 11>62x3 + 11>62x4 + g 3. (a) a0 31 + x2 + g 4 + a1 3x + 11>32x3 + g 4 (b) a0 31 + 11>22x2 - 11>62x3 + g 4 + a1 3x - 11>22x2 + 11>22x3 + g 4 5. a0 31 + 11>221x - 22 2 - 11>2421x - 22 4 + g 4 + a1 1x - 22 7. (a) a0 3x3 + x4 + 11>42x5 + 11>362x6 + g 4 (b) a0 3x - 11>42x2 + 11>202x3 - 11>1202x4 + g 4 9. (a) c1y1 1x2 + c2y2 1x2 , where y1 1x2 = x + x2 + 11>22x3 + g = xex and y2 1x2 = y1 1x2ln x - x2 - 13>42x3 - 111>362x4 + g (b) c1y1 1x2 + c2y2 1x2 , where y1 1x2 = 1 + 2x + x2 + g and y2 1x2 = y1 1x2ln x - 34x + 3x2 + 122>272x3 + g 4 (c) c1y1 1x2 + c2y2 1x2 , where y1 1x2 = 1 - 11>62x + 11>962x2 + g and y2 1x2 = y1 1x2ln x - 8x -2 - 4x -1 + 129>362 + g (d) c1y1 1x2 + c2y2 1x2 , where y1 1x2 = x2 - 11>42x3 + 11>402x4 + g and y2 1x2 = x -1 + 11>22 + 11>42x + g

CHAPTER 9 Exercises 9.1, page 500 x ′ 7 2 x 1. c d = c d c d y 3 -2 y x ′ 1 1 1 x 3. £ y § = £ -1 0 2 § £ y § z 0 4 0 z

x ′ sin t 5. c d = c cos t y

et x d c d a + bt 3 y

y ′ 0 7. c d = c y -k>m 9. c

0 x1 ′ d = c t x2 ′ x1

1 y d c d -b>m y

1 x1 d c d ; x1 = y , x2 = y′ 0 x2

0 1 0 0 x1 - 3 0 -2 0 x2 x T D 2T ; 11. D T = D 0 0 0 1 x3 x3 2 0 0 0 x4 x4 x1 = x , x2 = x′ , x3 = y , x4 = y′ 0 1 0 0 0 x1 x1 ′ x2 0 0 x2 cos t 3 -t 2 0 0 1 0 U E x3 U ; 13. E x3 U = E 0 x4 0 0 0 0 1 x4 t x5 -e t x5 0 -1 -1 x1 = x , x2 = x′ , x3 = y , x4 = y′ , x5 = y″ Exercises 9.2, page 504 1. x1 = 2 , x2 = 1 , x3 = 1 3. x1 = -s + t , x2 = s , x3 = t 5. x1 = 0 , x2 = 0 7. x1 = 3s , x2 = s 1 - ∞ 6 s 6 ∞ 2 9. x1 = 2s , x2 = 1 -1 + i2s , with s any complex number 11. x1 = - 1s + 12 >2 , x2 = - 111s + 12 >2 , x3 = s 1- ∞ 6 s 6 ∞2 13. For r = 2, unique solution is x1 = x2 = 0. For r = 1, solutions are x1 = 3s, x2 = s, - ∞ 6 s 6 ∞ . Exercises 9.3, page 513 1 1 7 1. (a) c d (b) c 5 2 7 3. (a) c

18 14 d 4 5 -1 -2 5. (a) c d - 1 -3 7. (c) Yes 1 13. £ - 2 1

(d) Yes 0 1 1

1 -2 § 0

3 d 18

(b) c

8 12 16 3 d (c) c d 3 5 5 19 -5 -1 -6 -3 (b) c d (c) c d - 8 -1 -9 -4 4>9 - 1>9 9. c d 11. Doesn’t exist 1>9 2>9 17. c

14>32e-t - 11>32e-4t

- 11>32e-t d 11>32e-4t

e-t 11>22e-t - 11>22e-t t 19. £ 11>32e - 11>22et 11>62et § -2t - 11>32e 0 11>32e-2t 21. 11 23. -12 25. 25 27. 2, 3 3e3t 5e5t 6e2t 29. 0, 1, 1 31. £ 6e3t § 33. c d 5t -10e -2e2t -3e3t

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-21

Answers to Odd-Numbered Problems

11>22t 2 + c1 et + c2 d t + c3 et + c4 - 1 + cos 1 sin 1 d (b) c 1 - cos 1 sin 1

39. (a) c

(c) J

11 + et 2 cos t + 1et - t2 sin t 1et - 12 sin t + et cos t

1et - t2 cos t - 1et + 12 sin t R 1et - 12 cos t - et sin t

Exercises 9.4, page 521 x′1t2 3 - 1 x1t2 t2 1. c d = c dc d + c td y′1t2 -1 2 y1t2 e dx/dt t2 3. £ dy/dt § = £ 0 t dz/dt

-1 0 -1

-1 x t t e § £y§ + £ 5 § 3 z -et

0 1 x1 1t2 0 x1= 1t2 d; d = c d + c dc 10 3 x2 1t2 sin t x2= 1t2 x1 = y, x2 = y′

5. c

7.

9. 11.

13. 21. 23.

25.

0 1 0 0 0 x1 1t2 x1= 1t2 x2= 1t2 x2 1t2 0 0 1 0 0 T = D T + ≥ ¥; T D D = 0 0 0 1 0 x3 1t2 x3 1t2 = x4 1t2 x4 1t2 -1 0 0 0 t2 x1 = w, x2 = w′, x3 = w″, x4 = w‴ x1= 1t2 = 5x1 1t2 + 2e-2t ; x2= 1t2 = - 2x1 1t2 + 4x2 1t2 - 3e-2t x1= 1t2 = x1 1t2 + x3 1t2 + et ; x2= 1t2 = - x1 1t2 + 2x2 1t2 + 5x3 1t2 + t ; x3= 1t2 = 5x2 1t2 + x3 1t2 19. LI LI 15. LD 17. LI Not a fundamental solution set e-t et e3t -t Yes; £ 2e 0 - e3t § ; -t e et 2e3t -t e et e3t -t c1 £ 2e § + c2 £ 0 § + c3 £ - e3t § e-t et 2e3t 1 et t et e-t 0 3 tet c t d - c t d + c d - c d + c1 c t d + c2 c -t d 2 te 4 3e 2t e 3e 1

5. Eigenvalues are r1 = 1, r2 = 2, and r3 = -2 with 1 0 associated eigenvectors u1 = s £ 0 § , u2 = s £ 1 § , 0 0 1 and u3 = s £ - 1 § . 1 7. Eigenvalues are r1 = 1, r2 = 2, and r3 = 5 with 2 0 associated eigenvectors u1 = s £ - 3 § , u2 = s £ - 1 § , 2 1 0 and u3 = s £ 1 § . 2 9. Eigenvalues are r1 = i and r2 = -i, with associated i -i eigenvectors u1 = s c d , and u2 = s c d , 1 1 11. c1e3t/2 c

3 1 d + c2et/2 c d 10 2

-2 0 1 13. c1e-t £ 1 § + c2e-3t £ - 1 § + c3e-5t £ 1 § 1 1 1 3 1 1 15. c1e-t £ 0 § + c2et £ - 6 § + c3e4t £ 0 § -2 4 1 (c) 17. (b) x2

1. Eigenvalues are r1 = 0 and r2 = - 5 with associated s 2s eigenvectors u1 = c d and u2 = c d . 2s -s

x1

x1 (–Ë3, –1)

(1, –Ë3)

Figure B.43

(d)

Figure B.44

x2

- 11>42et 0 11>42et 14>52e2t - 12>52e2t § ; 29. X-1 1t2 = £ - 11>52e2t -3t -3t 11>52e 19>202e 13>202e-3t -t -2t - 13/22e + 13/52e - 11/102e3t x1t2 = £ 11/42e-t - 11/52e-2t - 11/202e3t § 15/42e-t - 11/52e-2t - 11/202e3t 6t 37. (b) 2e (e) 2e6t ; same Exercises 9.5, page 531

x2

x1 (1 – Ë3, –Ë3 – 1)

Figure B.45

e3t 19. c 3t 4e

e-3t d -2e-3t

e2t 0 23. ≥ 0 0

e-t - 3e-t 0 0

e3t 0 e3t 0

et 21. £ et et 7t -e e7t ¥ 2e7t 8e7t

e2t 2e2t 4e2t

e4t 4e4t § 16e4t

3. Eigenvalues are r1 = 2 and r2 = 3 with associated 1 1 eigenvectors u1 = s c d and u2 = s c d. -and 1 - 2 Equations, Second Custom Edition for University of California Berkeley. MATH 54 Linear Algebra Differential

Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-22 Answers to Odd-Numbered Problems

25. x = - c1et - 2c2e2t - c3e3t , y = c1et + c2e2t + c3e3t , z = 2c1et + 4c2e2t + 4c3e3t e-0.3473t 27. £ - 0.3157e-0.3473t 0.0844e-0.3473t

29.

0.0251e3.4142t 0.0858e3.4142t D 0.2929e3.4142t e3.4142t

2e4t + e-2t 31. c 4t d 2e - e-2t 35. (b) x1 1t2 = (c) x2 1t2 = 37. (b) x1 1t2 = (c) x2 1t2 =

(d) x3 1t2 =

e0.5237t 0.4761e0.5237t 0.1918e0.5237t

- 0.2361e-1.6180t 0.3820e-1.6180t - 0.6180e-1.6180t e-1.6180t

0.0286e-7.0764t - 0.1837e-7.0764t § e-7.0764t

e0.6180t 0.6180e0.6180t 0.3820e0.6180t 0.2361e0.6180t

e0.5858t 0.5858e0.5858t T 0.3431e0.5858t 0.2010e0.5858t

- 3e-t + e5t 33. £ - 2e-t - e5t § e-t + e5t 1 e-t c d 2 1 1 te-t c d + e-t c d 2 1 1 e2t £ 0 § 0 1 0 te2t £ 0 § + e2t £ 1 § 0 0

1 0 0 t 2 2t e £ 0 § + te2t £ 1 § + e2t £ - 6/5 § 2 0 0 1/5

-1 -1 39. (b) x1 1t2 = et £ 1 § ; x2 1t2 = et £ 0 § 0 1 1 1 (c) x3 1t2 = tet £ 1 § + et £ 0 § (d) 0 -2 0 3t 2 t4 43. c1 c 2 d + c2 c 4 d t t 45. x1 1t2 = 12.5>221e-3t>25 + e-t>25 2 , x2 1t2 = 12.521e-t>25 - e-3t>25 2 47. (a) r1 = 2.39091 , r2 = -2.94338 , r3 = 3.55247 1 1 (b) u1 = £ - 2.64178 § ; u2 = £ - 1.16825 § ; - 9.31625 0.43862 1 u3 = £ 0.81004 § -0.12236 (c) c1er1tu1 + c2er2tu2 + c3er3tu3, where the ri’s and the ui’s are given in parts (a) and (b).

Exercises 9.6, page 537 2 cos 2t 2 sin 2t 1. c1 c d + c2 c d cos 2t + sin 2t sin 2t - cos 2t 1 -1 -2 -1 -2 3. c1et cos t£ 1 § - c1et sin t£ 0 § + c2et sin t£ 1 § + c2et cos t£ 0 § + c3et £ 0 § 0 0 1 0 1 Note: Equivalent answer is obtained if col 1 - 1, 1, 02 and col 1 -2, 0, 12 are replaced by col 1 - 5, 1, 22 and col 10, - 2, 12, respectively. sin t 1 cos t e-t cos 4t e-t sin 4t 5. c -t d 7. £ 0 - cos t -sin t § -t 2e sin 4t - 2e cos 4t 0 - sin t cos t cos t -sin t 9. ≥ 0 0

sin t cos t 0 0

0 0 cos t - sin t

0 0 ¥ sin t cos t

et sin t - et cos t 2et sin t 11. D t 2e sin t + 2et cos t 4et cos t

- e-21t - p2 cos t e-2t 1sin t - cos t2 d (b) c -21t - p2 d -2t e 1cos t - sin t2 - 2e sin t e-21t + 2p2 12 cos t - 3 sin t2 ep - 2t cos t (c) c -21t + 2p2 d (d) c p - 2t d e 1cos t + 5 sin t2 e 1sin t - cos t2

- et sin t - et cos t - 2et cos t t 2e sin t - 2et cos t 4et sin t

-cos t sin t cos t -sin t

- sin t -cos t T sin t cos t

13. (a) c

t -1 t -1 cos1ln t2 t -1 sin1ln t2 -1 17. c1 £ 0 § + c2 £ t sin1ln t2 § + c3 £ -t -1 cos1ln t2 § -2t -1 - t -1 cos1ln t2 - t -1 sin1ln t2 19.

39 - 217 222p

≈ 0.249 ;

39 + 217 222p

≈ 0.408

21. I1 = 15>62e-2t sin 3t , I2 = - 110>132e-2t cos 3t + 125>782e-2t sin 3t , I3 = 110>132e-2t cos 3t + 120>392e-2t sin 3t

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Answers to Odd-Numbered Problems

11. c1 c

(b) c (c) c

23.

25.

27.

1 2 t -1 >2 13. c1 c d + c2e6t c d - c -1 d 2 1 t 1 -2 ln 0 t 0 + 18>52t - 8>25 15. c1 c d + c2e-5t c d +c d 2 1 2 ln 0 t 0 + 116>52t + 4>25 1 -1 -1 1 -t -t t 17. c1e £ 1 § + c2e £ 0 § + c3e £ 1 § + e £ - 1 § 1 -1 1 0

29.

2t

c1 cos t + c2 sin t + 1 - c sin t + c2 cos t - t 19. D t1 T c3e + c4e-t - 11>42e-t - t + 11>22te-t c3et - c4e-t - 11>42e-t - 1 - 11>22te-t

33. I1 = I2 + 3I5, I3 = 3I5, I4 = 2I5, where I2 = 1 1> 1 202817 2 2 3 1 13 - 2817 2 e- 1 31 + 2817 2 5t>2 - 1 13 + 2817 2 e1 -31 + 2817 2 5t>2 4 + 1>10 ,

Exercises 9.8, page 551

5. (a) r = - 2 ; k = 2

2 sin t 2 cos t -2 sin t

- 4 + 29e5t - 20te5t - 5 + 5e5t 2 - 2e5t + 10te5t

11 - t + t 2/22et 1t 2/22et 13. E 1t + t 2/22et 0 0

1 3> 1 402817 2 2 3 1 31 - 2817 2 e- 131 + 2817 25t>2 - 1 31 + 2817 2 e1 -31 + 2817 2 5t>2 4 + 3>20

- 2t d 1 1 + 3t - 13/22t 2 t -t + 11/22t 2 -t - 3t § (b) e £ 1 t 3t 1 - 3t + 13/22t 2 9t - 19/22t 2 1 0 0 cos t sin t d (b) e-2t £ 4t 1 0 § 7. c - sin t cos t t 0 1

(b) e3t c

3. (a) r = -1 ; k = 3

1 £ 11. 25

1 -20 + 2e-5 2et + 1 -3e-5 - e-10 2e2t + 14t + 32et d 1 -10 + e-5 2et + 1 -3e-5 - e-10 2e2t + 12t + 32et

- 18et + 1 + 14e2t + 2 - 3e2t + 1 + 14t + 72et d -9et + 1 + 14e2t + 2 - 3e2t + 1 + 12t + 52et x = 1c1 + c2t2et + c3e-t - t 2 - 4t - 6 ; y = -c2et - 2c3e-t - t 2 - 2t - 3 et 2 0 e2t (a) e c t d , c 2t d f (c) tet c d + et c d e 2e 2 1 1 1 2 0 (d) c1et c d + c2e2t c d + tet c d + et c d 1 2 2 1 1 -1 -1 c1e2t £ 1 § + c2e-t £ 0 § + c3e-t £ 1 § 1 1 0 1 1 0 + te-t £ 0 § + e-t £ 0 § + £ 0 § -1 0 1 1 1 c1tc d + c2t -1 c d 3 1 - 13>42t -1 - 11>22t -1 ln t + 1 +c d - 13>42t -1 - 13>22t -1 ln t + 2

I5 =

et + cos t - sin t 1 t 9. £ e - sin t - cos t 2 t e - cos t + sin t

- 2et - 1 + 2e21t - 12 - 3e2t - 1 + 14t - 12et d -et - 1 + 2e21t - 12 - 3e2t - 1 + 12t + 12et

(d) c

sin t cos t 0 d + c2 c d +c d cos t -sin t -1

1. (a) r = 3 ; k = 2

4tet + 5et d 2tet + 4et

21. (a) c

Exercises 9.7, page 542 1 1 2 1. c1e7t c d + c2e2t c d +c d 1 -4 -1 1 1 -1 1 3. c1e-3t £ 1 § + c2e3t £ 0 § + c3e3t £ 1 § + et £ 0 § -1 1 -1 0 0 0 2 1 0 2t e sin t cos t 5. t£ 0 § + £ 0 § + £ - 1 § + £0§ + £ -1 § 3 2 2 -1 -1 0 0 0 1 -1 0 + c1e - t £ 1 § + c2et £ 1 § + c3et £ 0 § 0 1 0 7. xp = ta + b + sin 3t c + cos 3t d 9. x1t2 = ta + b + e2tc + 1 sin t2 d + 1 cos t2 e

B-23

1 0

et - cos t - sin t et + sin t - cos t § et + cos t + sin t

20 - 20e5t 25 - 10 + 10e5t

1t - t 2 2et 11 - t + t 2 2et 1 - 3t - t 2 2et 0 0

- 8 + 8e5t - 40te5t - 10 + 10e5t § 4 + 21e5t + 20te5t

1t 2/22et 1t + t 2/22et 11 + 2t + t 2/22et 0 0

0 0 0 cos t -sin t

0 0 0 U sin t cos t

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-24 Answers to Odd-Numbered Problems 11 + t + t 2/22e-t 1 - t 2/22e-t 15. E 1 - t + t 2/22e-t 0 0

1t + t 2 2e-t 11 + t - t 2 2e-t 1 - 3t + t 2 2e-t 0 0

1t 2/22e-t 1t - t 2/22e-t 11 - 2t + t 2/22e-t 0 0

1 -t 1 17. c1e-t £ -1 § + c2e-t £ - 1 + t § + c3e-2t £ - 2 § 1 4 2-t e-2t 21. £ 4te + e-2t § . te-2t - e-2t -2t

0 0 0 11 + 2t2e-2t -4te-2t

0 0 0 U -2t te 11 - 2t2e-2t

3 0 1 2t 6 1 t t + t2 19. c1e2t D T + c2et D T + c3et D T + c4et D T 1 0 0 0 1 0 0 1

3t 2 - et 1t 2 - 2t + 22 At 1 + et 1t - 12 23. e £ 3 § + e £ § , where eAt is the matrix in the answer to Problem 3. t 2 9t 6 - 3e 1t - 2t + 22 -t

14>132e-3t + 6te3t + 12>132 13 sin 2t - 2 cos 2t2 e3t § 29. x1t2 = £ -3t 3t 18>132e + 13t - 12e + 14>132 13 sin 2t - 2cos 2t2

.

Review Problems, page 555 1 3 1. c1e3t c d + c2e4t c d 1 2

1 0 1 0 1 0 1 0 3. c1e-t D T + c2e-t D T + c3e3t D T + c4e3t D T 0 1 0 1 0 -1 0 1

5. c

e2t -e2t

e3t d -2e3t

1 t -t + 11/22t 2 2+t t § + £ 7 - 3t § 9. c1e-t £ 0 § + c2e-t £ 1 § + c3e-t £ 3 3t 1 - 3t + 13/22t 2 10 -1 1 + c2t -1 £ 0 § + c3t 4 £ 1 § 1 1 - et + 12/32e5t + 11/32e-t - 11/22et + 12/32e5t - 11/62e-t 11/32e5t + 12/32e-t 11/32e5t - 11/32e-t § t -t e -e 11/22et + 11/22e-t (d) true (e) true (f ) false (g) true

1 1 - 1/3 d + c2e3t c d + c d 7. c1e-t c -2 2 -14/3 -3 13. c1t -1 £ 1 § 0 11/22et + 12/32e5t - 11/62e-t 11/32e5t - 11/32e-t 15. £ - 11/22et + 11/22e-t 17. (a) false (b) false (c) false

11. c

3et - 2e2t d 3et - 4e2t

CHAPTER 10 Exercises 10.2, page 570 1. y = 34> 1e - e-1 241e-x - ex 2 3. y K 0 5. y = ex + 2x - 1 7. y = cos x + c sin x; c arbitrary 9. ln = 12n - 12 2 >4 and yn = cn sin312n - 12x>24, where n = 1, 2, 3, … and the cn’s are arbitrary. 11. ln = n2, n = 0, 1, 2, …; y0 = a0 and yn = an cos nx + bn sin nx, n = 1, 2, 3, …, where a0, an, and bn are arbitrary. 13. The eigenvalues are the roots of tan1 2ln p2 + 2ln = 0, where ln 7 0. For n large, ln ≈ 12n - 12 2 >4, n is a positive integer. The eigenfunctions are yn = cn 3sin1 2ln x2 + 2ln cos1 2ln x24 , where the cn’s are arbitrary.

15. 17. 19. 21.

u1x, t2 u1x, t2 u1x, t2 u1x, t2

= = = =

e-3t sin x - 6e-48t sin 4x e-3t sin x - 7e-27t sin 3x + e-75t sin 5x 3 cos 6t sin 2x + 12 cos 39t sin 13x 6 cos 6t sin 2x + 2 cos 18t sin 6x + 111>272 sin 27t sin 9x - 114>452 sin 45t sin 15x ∞

23. u1x, t2 = a n-2e-2p n t sin npx 2 2

n=1

25. If K 7 0, then T1t2 becomes unbounded as t S ∞ , and so the temperature u1x, t2 = X1x2T1t2 becomes unbounded at each position x. Since the temperature must remain bounded for all time, K w 0. 33. (a) u1x2 K 50 (b) u1x2 = 30x>L + 10

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Answers to Odd-Numbered Problems

Exercises 10.3, page 584 1. Odd 3. Even 5. Neither ∞ 21 - 12 n + 1 9. f1x2 ∼ a sin nx n n=1 ∞ npx 2 11. f1x2 ∼ 1 + a c 2 2 1 - 1 + 1 -12 n 2 cos 2 n=1 p n 1 npx + 1 1 - 12 n + 1 - 1 2 sin d pn 2 ∞ 41 -12 n 1 13. f1x2 ∼ + a cos npx 3 n2p2 n=1

∞ 1 - 12 n 15. f1x2 ∼ 31sinh p2 >p4 a1 + 2 a c cos nx 2 n=1 1 + n 1 - 12 n + 1n + sin nx d b 1 + n2 17. The 2p-periodic function g1x2, where x 1 -p 6 x 6 p2 , g1x2 = e 0 1x = {p2 19. The 4-periodic function g1x2, where

1 x g1x2 = e 1>2 3>2

1 -2 6 x 6 02 , 10 6 x 6 22 , 1x = 02 , 1x = {22

21. The 2-periodic function g1x2, where g1x2 = x2 1 - 1 … x … 12 23. The 2p-periodic function g1x2, where g1x2 = e

ex 1ep + e-p 2 >2

1 - p 6 x 6 p2 , 1x = {p2

(b) F1x2 = 0 x 0 - p 12n - 12px 2 c 1 - 12 n + 1 cos a b 27. f1x2 ∼ a 12n 2 12p n=1 12n - 12px bd + sin a 2 29. c0 = 0 ; c1 = 3>2 ; c2 = 0 25. (a) F1x2 = 1x2 - p2 2 >2 ∞

Exercises 10.4, page 591 ∼ 1. (a) The p-periodic function f 1x2, where ∼ 2 f 1x2 = x 10 6 x 6 p2 (b) The 2p-periodic function fo 1x2, where x2 10 6 x 6 p2 , fo 1x2 = e 2 1 - p 6 x 6 02 -x (c) The 2p-periodic function fe 1x2, where x2 10 6 x 6 p2 , fe 1x2 = e 2 1 - p 6 x 6 02 x ∼ 3. (a) The p-periodic function f 1x2, where 0 10 6 x 6 p>22 , ∼ f 1x2 = e 1 1p>2 6 x 6 p2

B-25

(b) The 2p-periodic function fo 1x2, where -1 fo 1x2 = e

0 0 1

1 -p 6 x 6 - p>22 , 1 -p>2 6 x 6 02 , 10 6 x 6 p>22 ,

1p>2 6 x 6 p2

(c) The 2p-periodic function fe 1x2, where 1 0 fe 1x2 = e 0 1

1 - p 6 x 6 - p>22 , 1 - p>2 6 x 6 02 , 10 6 x 6 p>22 , 1p>2 6 x 6 p2

4 ∞ 1 sin12k - 12px p ka 2k -1 =1 ∞ 2p1 -12 n + 1 4 7. f1x2 ∼ a c + 3 1 1 -12 n - 1 2 d sin nx n pn

5. f1x2 ∼ -

n=1 ∞

8 9. f1x2 ∼ a sin12k + 12px 12k + 12 3p3 k=0 p 4 ∞ 1 11. f1x2 ∼ + cos12k - 12x 2 2 p ka = 1 12k - 12 ∞ 1 - 12 ne - 1 13. f1x2 ∼ e - 1 + 2 a cos npx 2 2 n=1 1 + p n 1 1 2 2 ∞ 15. f1x2 ∼ + a b cos 2kx p p ka 2k + 1 2k -1 =1 17. u1x, t2 = 2 ∞ 2 1 1 2 c d e-512k - 12 t sin12k - 12x p ka 2k 2k + 2k 1 1 3 =1 k+1 4 ∞ 1 -12 2 19. u1x, t2 = e-512k + 12 t sin12k + 12x p a 12k + 12 2 k=0

Exercises 10.5, page 603 ∞ 81 -12 n + 1 - 4 2 2 1. u1x, t2 = a e-5p n t sin npx p3n3 n=1

∞ 4 p 2 e-312k + 12 t cos12k + 12x 2 2 ka = 0 p12k + 12 ∞ 2ep 1 -12 n - 2 21ep - 12 2 e-n t cos nx + a 5. u1x, t2 = p p11 + n2 2 n=1

3. u1x, t2 =

30 -2t 5 5 xe sin x + e-8t sin 2x p p p 10 -18t + a1 - be sin 3x p 5 -32t 6 + e sin 4x - a1 + be-50t sin 5x p 2p ∞ 10 2 + a 321 -12 n - 14e-2n t sin nx pn n=6

7. u1x, t2 = 5 +

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

B-26 Answers to Odd-Numbered Problems

∞ e-p - 1 2 x - e-x + 1 + a cne-n t sin nx , p n=1

9. u1x, t2 =

∞ npx 11. u1x, t2 = a an Tn 1t2 sin , where L n=1 L

where 2e

cn = g

-p

2n -2 31 - 12 n + 1e-p + 14 1 - 12 n + pn p11 + n2 2 +

2 31 - 12 n - 14 pn

1n ≠ 22 ,

e-p - 1 4 11 - e-p 2 + 1 + p 5p ∞

1n = 22

11. u1x, t2 = a ane-41n + 1>22 t cos1n + 1>22x, where 2

n=0 ∞

f1x2 = a an cos1n + 1>22x n=0

13. u1x, t2 =

p2 1 x - x3 - 3e-2t sin x 3 3 ∞ 41 -12 n 2 + a e-2n t sin nx 3 n

∞

n=2

15. u1x, y, t2 = e-52t cos 6x sin 4y - 3e-122t cos x sin 11y ∞ 1 - 12 n + 1 2 17. u1x, y, t2 = 2 a e-n t sin ny n n=1

∞ npx 2 2 2 19. C1x, t2 = a cne-3L + kn p >a 4t sina b , where a n=1 a 2 npx cn = f1x2 sina b dx a L0 a Concentration goes to zero as t S + ∞ .

Exercises 10.6, page 614 1 1. u1x, t2 = sin 7pt sin 7px 7p ∞ 8 + a cos12k + 12pt sin12k + 12px 3 k = 0 1 12k + 12p 2 ∞ 4 3. u1x, t2 = a 3 321 -12 n + 1 - 14cos 2nt sin nx n=1 n 2

2h0L

∞

1 npa npx npat 5. u1x, t2 = 2 sin sin cos a 2 L L L p a1L - a2 n = 1 n 5 3 sin 2t sin 2x - sin 5t sin 5x 2 5 ∞ 1 - 12 n + 1 sin nt +2 a ct d sin nx n n3 n=1 ∞ 12n + 12pat 9. u1x, t2 = a c an cos 2L n=0 12n + 12pat 12n + 12px + bn sin d sin , where 2L 2L ∞ 12n + 12px f1x2 = a an sin and 2L

7. u1x, t2 = cos t sin x +

n=0 ∞

12n + 12pa

n=0

2L

g1x2 = a bn

sin

2 npx f1x2 sin dx , and L L0 L 1 Tn 1t2 = e-t>2 acos bnt + sin bntb , where 2bn 1 bn = 23L2 + 4a2p2n2 2L 1 13. u1x, t2 = 3sin1x + at2 - sin1x - at24 2a 1 = sin at cos x a 15. u1x, t2 = x + tx 1 2 2 17. u1x, t2 = c e-1x + at2 + e-1x - at2 2 cos1x - at2 - cos1x + at2 + d a an =

12n + 12px 2L

21. u1r, t2 = a 3an cos1knat2 + bn sin1knat24J0 1knr2 , n=1

where 1

1 f1r2J0 1knr2r dr , and cn L0 1 1 g1r2J0 1knr2r dr , with bn = akncn L0 an =

1

cn =

L0

J 20 1knr2r dr

Exercises 10.7, page 626 4 cos 6x sinh361y - 124 1. u1x, y2 = sinh1 -62 cos 7x sinh371y - 124 + sinh1 -72 ∞

3. u1x, y2 = a An sin nx sinh1ny - np2, where n=1

An =

p

2 f1x2 sin nx dx p sinh1 -np2 L0

cos 3x sinh13y - 32 sinh1 -12 sinh1 -32 cos 2x sinh 2y + sinh122 ∞ p r 2k + 1 7. u1r, u2 = - a cos12k + 12u 2 k = 0 12k + 12 2p22k-1 ∞ a0 r n 9. u1r, u2 = + a a b 1an cos nu + bn sin nu2, where 2 n=1 a a0 is arbitrary, and for n = 1, 2, 3, c p a an = f1u2 cos nu du , pn L-p p a bn = f1u2 sin nu du np L-p 5. u1x, y2 =

cos x sinh1y - 12

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Answers to Odd-Numbered Problems

4 2 2 1 11. u1r, u2 = a r - r -1 bcos u + a r - r -1 bsin u 3 3 3 3 1 4 256 -4 + ar + r bsin 4u 255 255 ∞ a0 13. u1r, u2 = + a r -n 1an cos nu + bn sin nu2, 2 n=1 where p 1 an = f1u2 cos nu du 1n = 0, 1, 2, c2 , p L-p and p 1 bn = f1u2 sin nu du 1n = 1, 2, 3, c2 p L-p 15. u1r, u2 = r 3 sin 3u

∞

17. u1r, u2 = a0 + b0 ln r + a 31anr n + bnr -n 2cos nu n=1

+ 1cnr n + dnr -n 2sin nu4 ,

where p

p

1 3 f1u2 du , g1u2 du , b0 = 2p L-p 2p L-p and for n = 1, 2, 3, c p 1 an + bn = f1u2 cos nu du , p L-p p 1 n3n - 1an - n3-n - 1 bn = g1u2 cos nu du , p L-p and p 1 cn + dn = f1u2 sin nu du , p L-p p 1 n3n - 1cn - n3-n - 1 dn = g1u2 sin nu du p L-p 21. u1r, u, z2 = 3I0 1r2 >I0 1p24sin z 23. (c) f1x, y2 = 2xy a0 =

B-27

APPENDIX A Exercises A, page A-1 1. 1x2 + 12 5/2 15x2 - 22 >35 + C

3.

2 6t5 e +C 15

1 1 u+ sin12pu2 + C 2 4p 1 1 7. ln1t 2 + 42 + arctan1t>22 + C 2 2 9. 1 1> 22 2 arcsin 22 x + C 5.

11. 2 ln 0 x - 1 0 - 4 ln 0 x - 2 0 + 2 ln 0 x - 3 0 + C 1 13. 29x2 - 1 - arccos a b + C 3x 1 2 15. ln1x2 + 42 - + C 17. ln 0 ln x 0 + C x 2 x 19. y cosh y - sinh y + C 21. +C 42x2 + 4 1 t 2 3t 23. 1t 2 - 92 ln1t + 32 - + + C 2 4 2 25. -arctan1cos t2 + C 5 5 27. sin7/ 5 x sin17/ 5 x + C 7 17 sin110x2 sin14x2 + +C 29. 20 8 1 2 1 31. tan9 u + tan7 u + tan5 u + C 9 7 5 1 1 33. x sin112x2 + C 8 96

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Index

Note: n indicates footnote. Abel, Niels, 200 Abel’s identity formula, 200, 324, 327, 522 Absolute stability, 148 Accuracy, 124, 128, 137 Addition, matrix, 505 Aging springs in mass-spring systems, 450, 495 Agrarian economy model, 35 Air conditioning, 107 Air resistance models, 110–111 Aircraft guidance in crosswind, model for, 146–147 Airy equation explanation of, 205 initial value problem for, 493 mass-spring analogy using, 205–206 Airy functions, 205 Algebra, of matrices, 496, 505–512 Algebraic equations, linear, 500–503 Algorithms alphabetization, 491–492 classical fourth-order Runge-Kutta, 134–138, 255–258, A-16–A-17 Euler’s, 121–129 Gauss-Jordan elimination, 500–503, 509 Quicksort, 491–492 with tolerance, 128, 136 Alligator problem, 101 Allometric equation, 144 Alphabetization algorithms, 491–492 Amplitude modulation, 228 Analog simulation, 293 Analytic coefficients, equations with, 445–449 Analytic functions, 432–433 Angular frequency, 213, 286, 536 Angular momentum, 211 Annihilator method examples using, 335–336 explanation of, 334 method of undetermined coefficients for, 337 undetermined coefficients and, 334–337, 346 Apollo reentry problem, 234 Approximation methods centered-difference, 99, 636 Euler’s, 23–27

least-squares property, 585 Newton’s, A-9–A-11 Simpson’s rule, A-11–A-12 Taylor polynomials, 36–37, 421–425 Aquaculture model, 144 Archimedes principle, 310 Armageddon problem, 212 Arms race problem, 251 Asymptotic behavior of linear equation solutions, 88–89, 239 Asymptotically stable critical points, 264 explanation of, 264 homogeneous linear systems, 533 linear systems, 411 Atmospheric pressure, 48 Attractors, 299, 301 Automatic pilot error, 381–382 Autonomous equations, 19, 33 Autonomous systems critical points of, 264–270 endpoints, 266 equilibrium solutions, 264–266 explanation of, 262 Auxiliary equations complex roots of, 165–172, 330 distinct real roots of, 161, 328–329 explanation of, 158, 328 homogeneous equations and,158 repeated roots of, 162, 330–331 roots of, 158, 161–172, 328–331 Axons, 639 Barometric equation, 48 Beam problems deflection under axial force, 341 vibrating, 333, 347 Bernoulli, James, 73n Bernoulli, John, 5n, 41n, 73n, 328n Bernoulli equations, 55, 73–74 Bessel, Frederic Wilhelm, 478n Bessel functions aging springs in spring-mass systems and, 495 explanation of, 477–480 of the first kind, 478 modified, 624 of the second kind, 478

second-order, 207 spherical, 485 Bessel’s equation of order explanation of, 206, 477–480 mass-spring analogy using, 206–207 modified, 207, 624 one-half order, 200 reduction to, 487 Bessel’s inequality, 586 Bifurcation, 34 Biomathematics Ebola epidemic model, 314–316 growth rate models, 274–276 MRSA (staph) infection model, 308–310 SIR epidemic model, 276–279 tumor growth model, 279–280 virus (HIV) infection models, 141 Blasius equation, 261 Block diagrams, 92 Bobbing object model, 310–311 Bode plots, 419 Boundary conditions circular domains, 619–622 Dirichlet, 616, 622 explanation of, 164, 562 homogeneous, 595 Laplace’s equation associations, 616–622 mixed, 617–619 nonhomogeneous, 596 Neumann, 616, 622 Boundary value problems (see also Dirichlet boundary value problem; Initial–boundary value problems)164, 619–625 Bounded solutions, 206 Boxcar function, 384n Brachistochrone problem, 5n Buckling columns, 474–475 towers, 493–494 Building temperature, 102–103, 131 Building time constant, 104, 107 Buoyancy, 311 Cable equation, 639 Calculus of matrices, 512–513 Cancer detection dosage, 281 Capacitance, 119

I-1

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

I-2

Index

Capital, growth of, 36 Carbon dating problem, 101 Casoratian, 348 Catenary curve, 233 Cauchy, Augustin, 329n Cauchy-Euler equations higher-order, 333 homogeneous, 194 linear second-order equations and, 193–195, 450–453 series expansions using, 450–453 singular equation points and, 454–459 Cauchy-Euler system, 533 Cauchy product, 429 Cayley-Hamilton theorem, 547 Center (critical point), 267, 269 Centered-difference approximations, 99, 636 Chain rule, A-2 Chaos, model for, 150–151 Chaos machine problem, 304 Chaotic systems, 299–303 forced Duffing equation, 300–301 forced pendulum equation, 301 Poincaré maps for, 301–302 strange attractors, 299, 301 Characteristic equations, 158, 194, 451, 524 Characteristic polynomials, 524 Chebyshev polynomials, 482, 487 Chemical diffusion problem, 604 Chemical reactions, numerical model for, 140 Circular cylinder, steady-state temperature distribution in, 631–633 Circular domain, Laplace’s equation for, 619–622 Clairaut equations, 85 Clairaut, Alexis, 66n, 85n Classical fourth-order Runge-Kutta method algorithm with tolerance, 136, A-17 mathematical models using, 134–138, 257–258 procedures for n equations, A-16–A-17 subroutine, 135, A-16 Classical mechanics. See Newtonian mechanics Classical orthogonal polynomials, 482 Clinical medicine, equations for, 80–83 Coefficient matrix, 498, 507 Coefficients analytic, 445–449 constant, 156, 194, 328–331 damping, 153, 170 discontinuous, 55 of equations, 4 of kinetic friction, 116 linear, 74–76 undetermined, 174–185, 237–238, 334–337, 346, 538–540 variable, 192–198 Coffee problems, 107, 108 Columns, buckling, 474–475 Combat models, 260, 544

Comparison test for improper integrals, 359–360 Compartmental analysis explanation of, 92 mixing problems, 92–95, 533 population models and, 95–100 Competing species, Runge-Kutta subroutine and, 261 Complex exponential function, 237–238 Complex residue computation, 376 Complex roots of auxiliary equations, 165–172, 330 complex conjugate roots, 168 of polynomial equations, 534n real solutions derived from, 167–168 Compound interest, 3, 48 Computer algebra system, 345–346 Concave function, 35 Confluent hypergeometric equation, 485 Conservative energy system, 312 Consistency condition, 622 Constant coefficients auxilliary equations and, 328–331 differential operators with, 156, 244–246 explanation of, 156, 328, 515 higher-order differential equations, 319, 327–331 homogeneous equations with, 157–163, 327–331 homogeneous linear systems with, 523–530 linear systems with, 515 qualitative considerations for, 201–210 Contact heat transfer, 28–29 Contact tracing, 314–316 Contour integral, 388n Control signal, 317 Control theory, 380–382 Convergence Euler’s method and, 124–126 of Fourier series, 580–584 method of order p, 124–125, 132–133 of Newton’s method, A-11 pointwise, 581 power series and, 426–430 radius of, 427–428, 445–445 Taylor method, 132–134 uniform, 582 Convolution explanation of, 237, 398 impulse response function for, 402–403 integro-differential equations, 401 Laplace transforms and, 397–403 properties of, 398 theorem, 398 transfer function for, 401–402 Cooling air conditioning, 107 of buildings, 102–107 Newton’s law of, 29, 47, 102

Coulomb force field, 313 Coupled equations, 77 Coupled mass-spring systems (see also Mass spring systems) 283–287 Cramer’s rule, 500n, A-13–A-14 Critical load, 494 Critical points asymptotically stable, 264 autonomous systems, 264–270 endpoints as, 266 examples of, 267 explanation of, 264 phase plane analysis and, 264–270 stability of, 264–266 Critically damped motion, 216–217 Crosswinds model, 146–147 Current (magnetic field) problem, 272–273 Curve of pursuit model, 145–146 Cylindrical domain, Laplace’s equation for, 622–625 d’ Alembert, Jean le Rond, 3n d’ Alembert’s solution, 609–611 Damped oscillations, 154 Damping Hamilton’s equations and, 312 mass-spring oscillators and, 152–156 vibrating spring with, 173 vibrating spring without, 173 vibration desire or supression, 154 Damping coefficient, 153, 170 Damping factor, 215 Decay constant, 102, 225 Defective matrix, 548 Deflection of beams, 341 Degenerate system, 247 Demand function, 147 Dependent variables, 4 Derivatives, Laplace transforms of, 362–365 Desired accuracy, 128 Determinants, 510–511 Diagonal matrices, 505 Diagonalizable matrices, 530 Difference equations, 149, 347–349 Differential equations (see also Equations; specific types of equations) autonomous, 19, 33, 262 construction of, 327 direction fields for, 15–20 Euler’s approximation method for, 23–27 exact, 57–63 examples of, 3 existance and uniqueness solutions of, 11–13, 17–19 first-order. See First-order equations higher-order linear. See Higher-order differential equations initial value problem, 10–13 integral equations, 32 linear. See Linear second-order equations

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Index

nonlinear, 4, 201–210, 233–234, 236 order of, 4, 6 ordinary, 4, 6 overview of, 1–5 partial (see also Partial differential equations) 3, 4 second-order. See Linear second-order equations separable, 41–45 series solutions of. See Series solutions software for analyzing, A-17–A-18 solutions of, 2, 6–13 special functions for, 483–484 variables in, 4 Differential operators, 243–249, 320–321 Differentiation chain rule, A-2 derivatives for, A-2 of Fourier series, 582–583 integration using, A-1–A-3 Diffusivity, 561 Dirac delta function impulses and, 405–410 Laplace transform and, 406–410 properties of, 405 Direction fields autonomous equations, 19, 33 differential equations, 15–20 existance-uniqueness solution from, 17–19 explanation of, 15–20 method of isoclines and, 19–20 phase lines for, 33–34 system stability and, 264–266 Dirichlet, Peter Gustav Lejeune, 576n Dirichlet boundary conditions, 616, 622 Dirichlet boundary value problems circular domain, 619–622 cylindrical domain, 622–625 existance-uniqueness solution and, 625 numerical method for solution of, 635–637 Discontinuous coefficients, 55 Discontinuous forcing term, 55, 186 Discontinuous functions jump discontinuity and, 357, 571 Laplace transforms and, 387, 383–389 rectangular window function, 384–386 translation and, 386–389 unit step function, 384 Dispersion relation, 615 Distinct real roots, 161, 328–329 Distribution, 405 Doetsch, G., 351n Dot products, 496–497 Double Fourier series, 600 Drag force, 114–115 Drag race problem, 5 Duffing equation, 207, 260, 300–301, 425 Duhamel, J.M.C., 418 Duhamel’s formula, 417–418

Dynamical systems attractors, 299, 301 chaotic systems, 299–303 explanation of, 295–296 Poincaré maps, 296–299, 301–302 Ebola epidemic model, 314–316 Economics agrarian economy, 35 compound interest, 3, 48 demand and supply functions, 147 growth of capital, 36 marginal benefit, 35 market equilibrium, 147–148 Eigenfunctions, 564–565 Eigenspaces, 524 Eigenvalues complex, 534–537 distinct, 528 explanation of, 523, 535, 564 homogeneous linear systems and, 523–525 of a matrix, 524–525 normal modes as, 286n real-valued, 529–530 Eigenvectors explanation of, 523 generalized, 548–549 homogeneous linear systems and, 523–525 linear independence of, 526–528 of a matrix, 524–525 Electrical circuits analogy between mechanical and, 293 differential equations for, 3 explanation of, 118–119 first-order differential equations for, 118–121 Kirchhoff’s laws for, 118–119 phase-locked loop, 317–318 units and symbols for, 119 Electrical systems, 289–293, 317–318 Electromagnetic wave transmission (see also Transmission lines) 637–638 Elimination method, 243–249 Elliptical integral of first kind, 235 Emden’s equation, 444 Endpoints, 266 Energy argument, 612–614 Energy integral lemma, 203–205, 234, 272 Epidemic models boarding school problem, 282 SIR model, 276–280 Equations (see also Differential equations; specific types of equations) adjustments, 147 with analytic coefficients, 445–449 auxiliary, 158, 328–331 characteristic, 158, 194 coupled, 77 exact, 57–63 of form dy/dx = G(ax + by), 72–73 homogeneous, 70–72

I-3

indicial, 451, 456–457 integral, 32 linear, 48–54 linear algebraic, 412–413 with linear coefficients, 74–76 matrices and systems of, 511–512 separable, 41–45 Equidimensional equations. See Cauchy-Euler equations Equilibrium direction fields for, 264–266 explanation of, 33–34 economic market, 147–148 mass-spring systems, 226–227 saddle point, 266 stable, 243, 264 system stability and, 264–266 unstable, 265–266 Equilibrium points (see also Critical points) 264 Equilibrium solutions, 264 Equipotential lines, 23 Escape velocity, 117 Euler, Leonhard, 60n, 66n, 166n, 328n, 394n Euler-Fourier Formula, 576 Euler’s formula, 166 Euler’s methods approximation, 23–27 convergence and, 124–126 improved, 121–129 initial value problems, 23–27 in normal form, 253–255 recursive formulas for, 24–25 step size, 23 system solutions using, 253–255 Exact differential equations explanation of, 57–59 method for solving, 62–63 test for, 59–61 Existence and uniqueness of solutions direction fields for, 17–19 energy argument for, 612–614 explanation of, 11–13, 319 first-order equations, 2, 53, 56 Fourier series, 601–602 heat flow model, 601–602 homogeneous equations, 159–160 initial value problems, 11–13, 46, 53 Laplace’s equation and, 625–626 limitations of, 13 linear differential equations, 53, 56, 159– 160, 182, 192–193, 319 second-order equations, 159–160, 182, 192–193 linear systems and, 516 nonhomogeneous equations, 182 theorem of, 11 variable-coefficient equations, 192–193 wave equation and, 612–614

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

I-4

Index

Explicit solutions, 6–7 Exponential, derivative of, A-2 Exponential law, 96 Exponential of a matrix, 545–551 Exponential order a, 359, 379 Exponential shift property, 345–346 Exponents of a regular singular point, 456 Factorial function, 475 Factors damping, 215 integrating, 66–69 inverse Laplace transforms and, 370–374 nonrepeated, 370–371 quadratic, 373–374 repeated linear, 372–373 Falling bodies differential equations for, 1–2, 38–40 drag force and, 114–115 limiting or terminal velocity of, 40 motion of, 38–40, 273 velocity of, 40, 115, 131 Feedback system problem, 250–251 Fibonacci rule, 349 Fibonacci sequence, 349 First-order equations Bernoulli equations, 73–74 compartmental analysis and, 92–100 electrical circuits, 118–121 Euler’s method and, 121–129 exact equations, 57–63 general solution to, 39, 50 heating and cooling of buildings and, 102–107 higher-order numerical methods and, 132–138 homogeneous equations, 70–72 initial value problems, 39, 52–53, 86 with linear coefficients, 74–76 linear equations, 48–54, 88–89 mathematical modeling and, 90–92 motion of falling body and, 38–40 multiple solutions of, 86 Newtonian mechanics and, 109–115 Runge-Kutta method and, 133–138 separable equations, 41–45 special integrating factors for, 66–69 substitutions and transformations and, 70–76 Taylor method and, 132–134 Fluid ejection, 260 Fluid flow around a corner, 628 nonisothermal numerical model, 140 Torricelli’s law of, 83–84 velocity potentials, 70 Fluid tanks, interconnected, 241–243, 533, 544 Forced Duffing equation, 300–301 Forced mechanical vibrations, 221–227

Forced pendulum equation, 301 Forced vibration problem, 186–187 Forcing function, 418 Forcing terms, 55, 186 Formal solutions, 569, 595 Fourier, Joseph B.J., 576n Fourier cosine series, 587–591 Fourier series complex forms of, 585 computation of, 576–579 convergence of, 580–584 differentiation of, 582–583 double, 600 Euler formulas for, 576 explanation of, 576 functions in, 571–572 generalized, 579 insulated wire heat flow model, 560–562, 592–602 integrals crucial in, 573–574 integration of, 583 orthogonal expansions and, 579–580 orthogonality condition, 574–575 Fourier sine series, 566, 587–591 Free fall differential equations for, 1–2 examples of, 38–40 problem in, 48 Free mechanical vibrations (see also Vibrations) 212–220 Frequency angular, 213, 286, 536 complex eigenvalues for, 536–537 fundamental, 605 natural, 213, 286, 536 resonance, 224 Frequency gain, 222 Frequency response curve, 223–225 Frequency response modeling, 418–420 Friction coefficient of kinetic, 116 explanation of, 116 static, 117 sticky, 273 Frobenius, George, 455n Frobenius’s method explanation of, 454–463 incidial equations and, 456–457 linearly independent solutions and, 465–466 singular equation points and, 454–459 Frobenius’s theorem, 460 Functions analytic, 432–433 Bessel, 477–480 complex exponential, 237–238 demand, 147 differential equation solutions, 2 Dirac delta, 405–410 discontinuous, 357, 383–389, 581

even, 572 gamma, 394–395 forcing (input), 418 Fourier series, 571–572 generalized, 406 harmonic, 626 homogeneous, 77 impulse response, 402–403, 410 jump discontinuity, 357 Laplace transforms and, 355–360, 383–389, 392–395, 405–410 linear dependence of, 160, 164, 323–324 linear independence of, 160, 165, 323–324 matrix, 513 matrix exponential, 545–551 odd, 572 periodic, 392–393 piecewise continuous (see also Piecewise continuity) 357, 359, 571–572 power, 392, 393–394 special, 474–484 spherically symmetric, 492–493 steady-state (output) solution, 418 supply, 147 symmetric, 572 transfer, 366, 401–402, 410 twice-differentiable, 244 utility, 86–87 vector, 516–517 Fundamental frequency, 605–606 Fundamental matrices, 518, 546–548 Fundamental period, 392, 572 Fundamental solution sets, 324, 326, 518 Gain factor, 222 Galileo, 5n, 392n Gamma function, 394–395 Gas law, 48 Gating variables, 57 Gauss, Carl Friedrich, 476n Gauss elimination process, 304–305 Gauss-Jordan elimination algorithm, 500–503, 509 Gaussian hypergeometric equation, 476 Gegenbauer polynomials, 482 General solutions explanation of, 39, 50, 161–163, 322 first-order differential equations, 39, 50 higher-order differential equations, 322, 325 to homogeneous equations, 157–163, 322 to linear differential equations, 50, 161–163, 322 to linear systems in normal form, 518 matrix systems and, 518 to nonhomogeneous equations, 182, 196, 325 power series for, 439–443 for systems, 246 to variable-coefficient equations, 196

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Index

Generalized Blasius equation, 261 Generalized eigenvectors, 548–549 Generalized Fourier series, 579 Generalized function, 406 Generating function, 482, 486 Geometric series, 428 Gibbs, Josiah Willard, 586 Gibbs phenomenon, 586–587 Gibbs-Wilbraham phenomenon, 586n Global truncation error, 131 Golden ratio, 349 Gompertz law, 280 Grand Prix race problem, 48 Gravitational force, 1, 38, 211 Gravity train problem, 240 Great Lakes clean up model, 313–314 Green’s function, 634–635 Grid lines, 636 Group velocity, 615 Growth of capital, 36 Growth rate models logistics model, 97–100, 275 Malthusian model, 96–97, 274–275 population, 95–96 tumors, 274–276 Volterra-Lotka system, 275–276 Guitar strings, 606 Half life, 282 Half-range expansions, 588, 589 Half-rectified sine wave, 396 Hamilton, Sir William Rowan, 311n Hamilton’s equations, 312 Hamiltonian systems, 311–313 Harmonic equation, 165 Harmonic functions, 626 Harmonic motion, 213 Harmonic series, 428 Harmonics, 605 Heat capacity, specific, 561 Heat conduction, 560, 561 Heat equation, 561–562, 590–591, 601–602 Heat exchange, 28–29 Heat flow model direction of, 560 existence and uniqueness of solutions, 601–602 heat equation for, 561–562 initial-boundary values for, 596–601 insulated wire, 560–562, 592–602 maximum principles for, 601–602 physical principles of, 560–561 separation of variables for, 562, 592–595 Heating and cooling of buildings, 102–107 Heaviside, Oliver, 351n, 375n, 383 Heaviside’s Expansion formula, 375–376 Henon map, 300n Hermite polynomials, 482, 487 Hermites equation, 201

Higher-order differential equations annihilator method, 334–337, 346 basic theory of, 319–325 constant coefficients with homogeneous, 327–331 conversion to normal form, 252–253, 515–520 difference equations, 149, 347–349 exponential shift property, 345–346 homogeneous, 322–324, 327–331 nonhomogeneous, 319, 325, 334–337 numerical methods for, 252–258 undetermined coefficients and, 334–337, 346 variation of parameters method and, 338–340 Wronskian of, 321–324 HIV infection models, 141–143 Hoare, C.A.R., 491 Hodgkin-Huxley theory of the neuron, 57 Homogeneous boundary conditions, 595 Homogeneous equations auxiliary equations and, 158, 162, 328–331 characteristic equation, 158 with constant coefficients, 157–163, 327–331 existance and uniqueness solution of, 159–160 explanation of, 70–72, 157–158, 319 fundamental solutions of, 324 general solutions to, 157–163, 322 higher-order differential equation solutions, 322 homogenous functions in, 77 linear dependence/independence, 322–324 substitutions and transformations and, 70–72 Wronskian functions in, 322–324 Homogeneous linear systems complex eigenvalues for, 534–537 with constant coefficients, 523–530 eigenvalues and eigenvectors for, 523–525 explanation of, 507, 515 fundamental solutions to, 518–520 general solutions to, 518–520 linear dependence/independence, 518 normal form, 515, 518–520 representation of solutions, 518–520 superposition principle and, 519 Homogeneous of order n, 77 Hong Kong flu epidemic model, 278–279 Hooke’s law, 152 Hormone secretion problem, 57, 281 Hyperbolic sine and cosine, 165 Hypergeometric equation confluent, 485 explanation of, 475–477 Identity matrix, 508 Impedance, 238 Implicit function theorem, 8, 15 Implicit numerical method, 125 Implicit solutions, 8–9 Improper integrals, 353, 359

I-5

Improved Euler’s method explanation of, 121–129 predictor-corrector method, 125–126 subroutine, 127 with tolerance, 128 Impulse response function, 402–403, 410 Impulses change in momentum as, 406 Dirac delta function and, 405–410 Inconsistent system, 508 Independent variables, 4 Indicial admittance, 417 Indicial equations, 451, 456–457 Inductance, 118 Inertial reference frame, 109 Initial-boundary value problems example of, 562 heat flow and, 562, 596–601 separation of variables and, 567–569, 607 wave equation for, 606–610 Initial conditions, 10, 158–160, 515, 562 Initial value problems convolution theorem to solve, 399–400 Euler’s approximation method for, 23–27 example of, 39, 52–53, 158–159 existance and uniqueness solution of, 11–13, 46, 53, 159 explanation of, 10–13 impulse response function to solve, 403 interval of definition, 47 Laplace transforms and, 366, 376–382 linear dependence and, 160 normal form of linear systems, 515–516 multiple solutions of, 86 Picard’s method for, 32–33 Taylor polynomial and, 423–424 variable-coefficient equations solution, 195 Inner products, 515, 585 Integral curves, 77 Integral equations comparison test for improper, 359–360 Fourier series, 573–574 Laplace transforms, 355–353, 359 Picard’s method, 32 Integrating factors explanation of, 66–68 finding, 68–69 linear differential equation solution using, 49–50 special, 66–69 Integration techniques differential equations and, 2 differentiation, A-1–A-3 explanation of, A-1 Fourier series, 583 integration by parts, A-5–A-6 method of substitution, A-3–A-5 partial fractions, A-7–A-8 power series differentiation and, 429–430 separable equations, 42

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

I-6

Index

Integro-differential equations, 401 Interconnected fluid tanks, 241–243, 533, 544 Interplanetary travel systems, 307–308 Interval of definition problem, 47 Invariance of Laplace’s equation, 627 Inverse, matrix, 508–510 Inverse Laplace transforms contour integral for, 388n explanation of, 366–367 linearity of, 368–370 nonrepeating linear factors and, 370–371 quadratic factors and, 373–374 repeated linear factors and, 372–373 translation property, 386 Inverse mapping, 366 Irregular singular points, 455 Isoclines, method of, 19–20 Isolated zeros problem, 201 Jacobi elliptic function, 236 Jacobi polynomials, 482 Jump discontinuity, 357, 571 Kinetic energy, 312 Kinetic friction, 116 Kirchhoff, Gustav Robert, 118n Kirchhoff’s laws current law, 118 differential equations using, 3, 118–119 voltage law, 118 Korteweg-de Vries (KdV) equation, 615 Kutta, W., 133n Lagrange, Joseph, 187n, 576n Laguerre polynomials, 474, 482, 487 Laguerre’s differential equation, 474, 487 Landing system design, 307–308 Laplace, Pierre, 351n Laplace transforms convolution and, 397–403 definition of, 351, 353–360 of derivatives, 362–365 discontinuous functions and, 357, 383–389 elementary functions, 356 existence of, 357–360 exponential order a, 359, 379 gamma function and, 394–395 impulses and Dirac delta function and, 405–410 initial value problems and, 366, 376–382 integral equations, 353–355, 359 inverse, 366–374 linear systems solved with, 412–413 linearity property, 355–356 matrix method, 558 method of, 376 mixing problem and, 350–352 piecewise continuity and, 357, 359, 379 periodic functions and, 392–393 power functions and, 392, 393–394

properties of, 361–365 translation property, 361–362 wave equation solution from, 633–634 Laplace’s equation boundary conditions associated with, 616–622 circular domain, 619–622 cylindrical domain, 622–625 Dirichlet boundary conditions, 616, 622 Dirichlet boundary value problems, 619–625 existence and uniqueness of solutions and, 625–626 explanation of, 616–617 heat equation and, 562 invariance of, 627 mean value property, 635 modified Bessel functions for, 624 Neumann boundary conditions, 616, 622 Poisson’s integral formula for, 635 rectangular domain, 617–619 separation of variables using, 617–625 Laplacian, 562 Least-squares approximation property, 585 Least-squares linear fit, 99 Least-squares method, A-14–A-15 Legendre, Adrien Marie, 480n Legendre polynomials, 481–482, 486 Legendre’s equation explanation of, 202, 450, 480 special functions and, 480–483 variable-coefficient equation solution using, 202 Lehrer, Tom, 240 Leibniz, Gottfried, 41n, 73n Leibniz’s rule, 364, 417 Length variable representation, 3 Leonardo da Pisa, 349 Level curves, 57 Lifetime, average, 282 Limit set, 299n Limiting (terminal) velocity, 40, 112 Linear algebraic equations, 412–413, 500–503 Linear coefficients, 74–76 Linear dependence of functions, 160, 195, 323–324 of homogeneous equations, 322–324 of three functions, 164–165 of vector functions, 516–517 of Wronskian, 324 Linear differential equations (see also Higher-order differential equations) asymptotic behavior of solutions to, 88–89 basic theory of, 319–325 definition of, 4 existence and uniqueness of solution, 53, 159–160, 182, 192–193, 319 first-order, 48–54, 88–89 method for solving, 50–53 ordinary points, 436–439

power series solutions to, 435–443 second-order, see Linear second-order equations singular points, 436 standard form, 49, 319 Linear differential operators, 244–246, 320–321 Linear factors, 370–373 Linear independence of eigenvectors, 526–528 explanation of, 160 of functions, 160, 194, 323–324 of homogeneous equations, 322–324 fundamental solution sets and, 324, 518 second linearly independent solutions, 465–473 of vector functions, 516–517 Linear operator explanation of, 506 higher-order linear differential equations, 320 inverse Laplace transform as, 368 Laplace transform as, 355 matrices as, 506–507 Linear second-order equations asymptotic behavior of solutions, 239 auxiliary equations for, 158, 161–162, 165–172 Cauchy-Euler, 193–195, 450–453 with complex roots, 165–172 with constant coefficients, 156, 157–163, 201–210 convolution method for, 237 existance and uniqueness of solution, 159–160, 182, 192–193 forced mechanical vibrations and, 221–227 free mechanical vibrations and, 212–220 homogeneous, 157–163 mass-spring oscillator and, 152–156, 169–172, 205–220 nonhomogeneous, 174–191 nonlinear equations and, 201–210, 233–234, 236 power series method and, 435–443 qualitative considerations for variable coefficient and nonlinear, 201–210 reduction of order, 194, 197–198, 327 superposition principle for, 180–185, 200 undetermined coefficients method for, 174–185, 237–238 variable-coefficient, 192–210 variation of parameters and, 187–191, 196 vibrations and, 154, 186–187, 212–220 Wronskian of, 160, 164, 195 Linear systems Cauchy-Euler, 533 elimination method for, 243–249 frequency response modeling of, 418–420 homogeneous, 518–520, 523–530 Laplace transforms for solution of, 412–413

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Index

matrix formulation of, 507 matrix methods for (see also Matrix methods), 496–499 nonhomogeneous, 520, 538–542 in normal form, 515–520 stability of, 411, 533, 538 Linearized equation, 615 Lobachevsky, Nicolai Ivanovich, 240 Local truncation error, 125, 131 Logistic function, 98, 275 Logistic model biomathematics applications, 275 Malthusian vs., 97, 275 phase line for, 34 for population growth, 18–19, 97–100, 275, 281 Long water waves, 615 Lossless line, 638 Lunar orbit equations, 261 Lungs, mechanical ventilation model for, 81–83 Lymphocytes, 141 Maclaurin series, 433n Magnetic field problems, 23, 77, 272–273 Malthus, Thomas R., 96n Malthusian law, 96 Malthusian model of population growth, 96–97 Marginal benefit, 35 Market equilibrium, 147–148 Mass matrix, 559 Mass-spring systems Airy equation, 205–206 auxiliary equations with complex roots, 169–172 Bessel’s equation, 206–207 coupled, 283–287 Duffing equation, 207 equilibrium of, 226–227 examples with, 169–172, 268–269, 283–287 explanation of, 152–156 forced mechanical vibrations and, 221–227 free mechanical vibrations and, 212–220 harmonic motion of, 213 matrix methods and, 498–499 nonlinear equations for, 205–210 periodic motion of, 213 spring stiffness and, 152, 205, 207–210 vibrations of, 212–227 Mathematical models compartmental analysis, 92–100 development of, 91 electrical circuits, 118–121 heating and cooling of buildings and, 102–107 higher-order numerical methods, 132–138 historical background of, 90 idealization of, 90

improved Euler’s method and, 121–129 insulated wire heat flow, 560–562, 592–602 mechanical ventilation, 81–83 Newtonian mechanics and, 109–115 numerical methods, 121–129, 132–138 oil spill, 80–81 parameters, 92n population dynamics and, 255–258 risk aversion, 86–87 Runge-Kutta method, 133–138, 257–258 Taylor method, 132–134 tests of, 91–92 utility functions, risk aversion using, 86–87 Mathieu’s equation, 212 Matrices algebra of, 496, 505–512 calculus of, 512–513 coefficient, 498 defective, 548 diagonal, 505 explanation of, 504–505 functions, 513, 545–551 fundamental, 518, 546–548 inverse of, 508–510 nilpotent, 547–548 normal form, 497–498 product of, 496–499 symmetric, 529–530 square, 505 transpose, 507 zero, 505 Matrix exponential function fundamental matrices and, 546–548 generalized eigenvector and, 548–549 nilpotent matrices and, 547–548 properties of, 545–546 solving x′ = Ax using, 549–551 Matrix methods complex eigenvalues and, 534–537 constant coefficients and, 523–530 homogeneous linear systems and, 518–520, 523–530 introduction to, 496–499 Laplace transform method, 558 linear algebraic equations and, 500–503 linear systems in normal form and, 515–520 matrices and vectors and, 504–513 matrix exponential function and, 545–551 nonhomogeneous linear systems and, 520, 538–542 uncoupling normal systems, 557 undamped second-order systems, 559 Maximum principles, 601–602 Mean value property, 635 Mechanical ventilation, model for, 81–83 Mechanical vibrations forced, 221–227 free, 212–220

I-7

Mechanics explanation of, 109 Newtonian, 109–115 procedure for models, 110 units for, 110 Mesh points, 636 Method of Frobenius, 455–465 Method of isoclines, 19–20 Method of least squares, A-14–A-15 Method of order p, 124–125, 132–133 Method of substitution, A-3–A-5 Midpoint method, 133–134 Minor of a matrix, 510 Mixing problems common drains and, 533 compartmental analysis for, 92–95 first-order equations and, 47, 56 interconnected tanks of, 241–243, 533, 544 Laplace transforms and, 350–352 Modal analysis, 286n Models. See Mathematical models Modes of a system, 595 Modified Bessel equation, 207, 624 Modified Bessel functions, 624 Moore, Gordon E., 274n Moore’s law, 274n Motion critically damped, 216–217 of falling bodies, 1–2, 38–40, 273 force of gravity and, 1, 38 harmonic, 213 nutation, 273–274 of objects that bob, 310–311 overdamped, 170, 216 pendulum analysis, 208–210 periodic, 213 underdamped (oscillating), 215 vibration of mass-spring system, 212–220 MRSA (staph) infection model, 308–310 Multiplication, matrix, 505–506 Multistep methods, 149 Natural frequency, 213, 286 Neumann boundary conditions, 616, 622 Neumann function, 478 Neuron synapses firing, 57 Newton’s law of cooling, 29, 47, 102 Newton’s law of motion, 109 Newton’s method approximating solutions using, A-9–A-11 convergence of, A-11 Newton’s second law explanation of, 1, 38, 267 mass-spring systems and, 152, 169, 283, 286 Newtonian mechanics inertial reference frame for, 109 model examples, 110–115 procedure for models, 110 terminal (limiting) velocity and, 112 units for, 110

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

I-8

Index

Nilpotent matrix, 547–548 Nodes, 33, 267, 639 Nonhomogeneous boundary conditions, 596 Nonhomogeneous equations annihilator method for, 334–337 existence and uniqueness of solutions, 182 explanation of, 157, 174, 319 general solution, 196, 325 higher-order differential equation solutions, 325, 334–337 method of undetermined coefficients for, 184–185, 337 power series and, 449 superposition and, 180–185 undetermined coefficients and, 174–185, 237–238, 334–337 variation of parameters and, 187–191, 196, 338–340 Nonhomogeneous linear systems explanation of, 515 general solution to, 520 representation of solutions for, 520 undetermined coefficients and, 538–540 variation of parameters and, 540–542 Nonlinear equations Duffing equation, 207 energy integral lemma for, 203–205 explanation of, 4 linearization of, 236 mass-spring oscillator analogy, 205–208 nonlinear solutions using, 233–234 numerical solutions, 201–202 oscillations and, 260 pendulum motion analysis, 208–210 qualitative considerations for, 201–210 solvable by first-order techniques, 233–234 van der Pol equation, 208 Nonlinear spring model, 260 Norm of Fourier series, 579, 585 Normal form Euler’s method in, 253–255 explanation of, 252–253, 497 higher-order equations in, 252–255 linear systems in, 515–520 matrix algebra and, 497–498 system solutions in, 252–255 Normal frequencies, angular, 286, 536 Normal modes, 286 Normal systems, 520, 557 Numerical methods Airy’s equation, 205–206 Bessel’s equation, 206–207 convergence and, 124–126 Dirichlet problem general solution using, 635–637 Euler’s method, 23–27, 121–129, 253–255 for higher-order equations and systems, 252–258 higher-order, 132–138 implicit method, 125

improved Euler’s method, 121–129 Legendre’s equation, 202 mathematical models using, 121–129, 132–138 multistep, 149 predictor-corrector method, 125–126 relative error, 128 Runge-Kutta method, 133–138, 255–258, A-16–A-17 stability of, 125, 148–149 system analysis using, 252–258 Taylor method, 132–134 truncation error, 125, 131 Nutation, 273–274 Oil spill model, 80–81 One-dimensional heat flow equation, 561 One-parameter family of solutions, 8 Operators differential, 243–249, 251, 320–321 integral, 353 linear, 320, 355, 368, 506–507 Ordinary differential equations, 4, 6 Ordinary points, 436–439 Orthogonal expansions, 579–580 Orthogonal system, 579 Orthogonal trajectories, 23, 65 Orthogonality classical polynomials, 482 Fourier series and, 574–575 Legendre polynomials and, 481 Orthonormal system, 579 Oscillation explanation of, 215 nonlinear equations and, 260 Oscillators (see also Mass-spring systems) 152–156, 169–172, 212–220, 285–289 Overdamped motion, 170, 216 Parachutist example, 113–114 Parameters (see also Variation of parameters) 92n Parseval’s identity, 586 Partial differential equations eigenfunctions for, 564–565 electromagnetic wave transmission, 637–638 explanation of, 3, 4 Fourier cosine and sine series and, 587–591 Fourier series and, 571–584 Green’s function for, 634–635 heat equation and, 592–602 heat flow model for, 560–562 Laplace transform solution for, 633–634 Laplace’s equation and, 616–626 numerical method for Dirichlet problem, 635–637 separation of variables method and, 563–570 steady-state temperature distribution using, 631–633

transmission line models, 637–639 wave equation and, 603–614, 633–634 Partial fractions method, 370, A-7–A-8 Parts, integration by, A-5–A-6 Pendulums angular momentum, 211 chaotic behavior of, 301 motion analysis, 208–210, 211, 235–236, 260, 261, 288–289 nonlinear equations for, 208–210 spring, 261 torque, 211 with varying length, 260 Period doubling, model for, 150–151 Periodic functions, 392–393 Periodic motion, 213, 268–269 Periodic solutions, 269 Phase angle, 419 Phase line, 33–34 Phase-locked loops, 317–318 Phase plane analysis, 262–270 Phase plane equation, 262 Phase portrait, 262, 270 Phase velocity, 615 Picard-Lindelöf existence theorem, 32n Picard’s method, 32–33 Piecewise continuity explanation of, 357, 571 Laplace transform and, 357, 359, 379 symmetric functions and, 572 Plucked string problem, 614 Poincaré maps, 296–299, 301–302 Poincaré sections, 296 Pointwise convergence, 581 Poisson’s integral formula, 635 Polynomials characteristic, 524 Chebyshev, 482, 487 classical orthogonal, 482 Gegenbauer, 482 Hermite, 482, 487 Jacobi, 482 Laguerre, 482, 487 Legendre, 481–482, 486 spherical, 481 Taylor, 421–425 ultraspherical, 482 Pooling delay, 250–251 Population models dynamics model, 255–258 growth rate, 95–100, 274–276 integro-differential equations, 401 logistics model, 97–100, 275 Malthusian model, 96–97, 274–275 tumor growth, 274–276 Potential energy, 312 Power functions derivative of, A-2 explanation of, 392 transforms of, 393–394

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Index

Power series analytic coefficients for, 446–449 analytic functions and, 432–433 convergence and, 726–730 differentiation and integration of, 429–430 explanation of, 426–430 index of summation, 430–431 linear differential equations and, 435–443 vanishing on an interval, 428–429 Predator-prey models, 255–258, 260, 275 Predictor-corrector method, 125–126 Product, derivative of, A-2 Pythagorean property, 585 Quadratic factors, 373–374 Quasifrequency, 215 Quasiperiod, 215 Quicksort algorithm, 491–492 Quotient, derivative of, A-2 Radiation, Sefan’s law of, 29 Radioactive decay models, 2, 38, 52, 54, 102, 274–275 Radioisotope dosage, 281 Radius of convergence analytic coefficients and, 445–446 explanation of, 427 minimum value for, 446 ratio test for, 427–428 Ranvier nodes, 639 Rate of change, 2, 241–243 Ratio test, 427–428, 434 Rayleigh equation, 211 RC circuit, 118–121 Rectangular domain, Laplace’s equation for, 617–619 Rectangular window function, 384–386 Recurrence relation, 438, 482 Reduction of order, 194, 197–198, 327 Regular singular points, 455 Relative error, 128 Removable singularities, 436n Repeated roots, 162, 330–331 Residue computation, 375 Resistance, 118 Resonance curve, 223–225 Resonance frequencies, 224 Reverse osmosis model, 198 Riccati, Jacopo, 77n Riccati equation, 77 Rigid body nutation, 273–274 Risk aversion model, 86–87 RL circuit, 118–121, 544 RLC circuit, 289–291, 538, 638 Rocket flight, 117 Rodrigues’s formula, 482, 486 Roots auxiliary equations, 158, 161–162, 165–172, 328–331 complex, 165–172, 330

complex conjugate, 168, 534n distinct real, 161, 328–329 double, 162–163 real solution derived from complex, 167–168 repeated, 162, 330–331 Row operation, 509–510 Runge, C., 133n Runge-Kutta method classical fourth-order algorithms, 134–138, 255–258, A-16–A-17 competing species, subroutine for, 261 midpoint method, 133–134 procedures for n equations, A-16–A-17 for systems, 255–258 Saddle point, 266–267 Sawtooth wave, 396 Scalar multiplication, 505 Scatter diagrams, 99 Schrödinger’s equation, 492–493 Second linearly independent soutions, 465–473 Second-order linear equations. See Linear second-order equations Secretion of hormones problem, 57 Selfadjoint form, 481 Separable equations explanation of, 41 justification of method, 45 method of solving, 42–45 Separation of variables eigenfunction properties, 565 explanation of, 38–39, 562 heat flow model and, 562, 592–595 Laplace’s equation and, 617–625 method of, 563–570 partial differential equations and, 562, 563–570 initial-boundary problem solution using, 567–569, 607 wave equation and, 604–605, 607 Series solutions Airy’s equations, 493 alphabetization algorithms, 491–492 analytical functions and, 432–433 Bessel functions for, 495 buckling of a tower, 493–494 Cauchy-Euler equations and, 450–453 equations with analytic coefficients and, 445–449 linear differential equations and, 435–443 method of Frobenius and, 454–463 power series, 426–433, 435–443 second linearly independent, 465–473 Shrödinger’s equations for, 492–493 special functions and, 474–484 spherically symmetric, 492 spring-mass system with aging spring, 495 Taylor polynomial approximation and, 421–425

I-9

Similarity transformation, 530 Simple harmonic equation, 165 Simple harmonic motion, 213 Simpson’s rule, A-11–A-12 Singular points Cauchy-Euler equations and, 454–459 explanation of, 436 exponents of, 456 first-order, 55–56 irregular, 455 regular, 455 Singular solutions, 85 Sink, 33 Sinusoidal response, 221 Sinusoidal signal inputs, 317, 419–420 SIR epidemic model, 276–279 Snowplow problems, 84–85 Software for analyzing differential equations, A-17–A-18 Solar collector design, 87–88 Solar water heater problem, 108 Solution sets, fundamental, 324, 518 Solutions asymptotic behavior of, 88–89, 239 Clairaut equations and, 85 constant multiples of, 55 differential equations, 6–13, 88–89 equilibrium, 264 explicit, 6–7 formal, 569, 595 fundamental sets, 324, 326 general. See General solutions homogeneous equations, 322–323 implicit, 8–9 initial value problem, 10–13, 86, 158–159, 195, 516 linear combinations of, 157–158 linear dependence of, 160, 195, 324 linear independence of, 160, 194, 324 not expressed using elementary functions, 46 real derived from complex, 167–168 series. See Series solutions singular, 85 spherically symmetric, 492–493 steady-state, 222, 418–419, 596 transient, 222, 596 unique. See Existence and uniqueness of solutions Wronskian of, 321–324 Special functions Bessel’s equation, 477–480 for differential equations, 483–484 explanation of, 474–475 hypergeometric equation, 475–477 Legendre’s equation, 480–483 Special integrating factors, 66–69 Specific heat capacity, 561 Speed bump problem, 187 Spherical Bessel functions, 485

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

I-10

Index

Spherical polynomials, 481 Spherically symmetric solutions, 492–493 Spiral (trajectory) point, 267 Spring stiffness, 152, 205, 207–210 Springs (see also Mass-spring systems) aging, 450, 495 Bessel functions and, 495 nonlinear, 260 pendulum motion analysis, 261 soft vs. hard, 425 vibrating, 173 Square matrix, 505, 545–551 Square pulse, 384 Square wave, 396 Stability absolute, 148 asymptotic, 264, 411, 533 homogeneous linear systems, 533, 538 linear systems, 411 market equilibrium and, 147–148 numerical methods, 125, 148–149 trapezoid scheme, 125 Stable equilibrium, 243 Stable systems, 264, 267, 411, 533 Standard form, linear differential equations in, 49, 319 Standing wave, 603–604 Staph (MRSA) infection model, 308–310 Static friction, 117 Steady-state current, 291 Steady-state response, 223 Steady-state solutions circuit analysis, 291 explanation of, 222, 418 frequency response modeling, 418–419 heat flow model, 596 to sinusoidal inputs, 419 vibrations, 222–223 Steady-state temperature distribution in circular cylinder, 631–633 Stefan’s law of radiation, 29, 108 Step size, 23, 122 Sticky friction, 273 Stiffness matrix, 559 Stopping procedure subroutine, 128 Strange attractors, 299, 301 Struck string problem, 614 Sturm-Liouville equations, 259 Subharmonics, 296 Substitutions Bernoulli equations, 73–74 equations of form dy/dx = G(ax + by), 72–73 equations with linear coefficients, 74–76 homogeneous equations, 70–72 method of, A-3–A-5 procedure for, 70 transformations and, 70–76 trigonometric, A-4–A-5 Summation index, 430–431

Superposition principle explanation of, 181, 519 linear systems, 519 undetermined coefficients and, 180–185 variable coefficients and, 200 Supply function, 147 Suspended cable, 233–234 Swinging door problem, 173 Symmetric functions, 572 Symmetric matrices, 515, 529–530 System analysis biomathematics applications, 274–280 bobbing motion model, 310–311 chaotic systems, 299–303 coupled mass-spring systems, 283–287 differential operators for, 243–249 dynamical systems, 295–303 Ebola epidemic model, 314–316 electrical systems, 289–293 elimination method for, 243–249 general solution for, 246 Great Lakes clean up model, 313–314 Hamiltonian systems, 311–313 interconnected fluid tanks, 241–243 landing system design and, 307–308 Laplace transforms for, 412–413 linear systems, 243–249, 412–413 numerical methods for, 252–258 phase-locked loop circuits, 317–318 phase plane analysis and, 262–270 Poincaré maps, 296–299, 301–302 population dynamics model, 255–258 staph (MRSA) infection model, 308–310 undamped second-order, 559 System parameters, 420 Taylor method of order p, 132–134 Taylor polynomial approximation explanation of, 36–37, 421–422 series solutions and, 421–425 Taylor series (see also Power series) 36–37, 422, 433 Tchebichef (Chebyshev) polynomials, 482, 487 Technetium-99m decay rate, 281 Telegraph problem, 615 Telegrapher’s equations, 638 Temperature distribution in a circular cylinder, 631–633 Temperature of buildings, 102–103, 131 Terminal velocity, 40, 112, 273 Thermal radiation, 29 Threshold value, 278 Time constant, 104–107 Time paths, market equilibrium and, 147–148 Time-shifted pair, 262 Time variable representation, 3, 10 Tolerance classical fourth-order Runge-Kutta algorithm with, 136, A-17 improved Euler’s method with, 128

Torque, 211 Torricelli’s law of fluid flow, 83–84 Total differential, 58 Towers, buckling of, 493–494 Trajectories behaviors of, 267 explanation of, 262 orthogonal, 23, 65 phase plane equations, 262–270 phase portraits of, 262, 270 Transfer function convolution and, 401–402 Duhamel’s formula and, 417–418 impulses and, 410 Laplace tranforms and, 366, 401–402, 410 Transformations. See Substitutions Transient current, 291 Transient solution heat flow model, 596 Poincaré maps and, 300 vibrations, 222 Translation discontinuous functions and, 386–389 inverse Laplace transform as, 386 Laplace transform as, 361–362, 386 Transmission lines cable equation, 639 numerical model for, 140 partial differential equations for, 637–639 telegrapher’s equations, 638 Transpose a matrix, 507 Transverse vibrations of a beam, 347 Trapezoid scheme, 125 Travelling waves, 610–611 Triangular wave, 396 Trig functions, derivative of, A-2 Trigonometric substitutions, A-4–A-5 Truncation error, 125, 131 Tumor growth model, 279–280 Tunnel, 240 U.S. census data problem, 101 Ultraspherical polynomials, 482 Uncoupling normal systems, 557 Undamped, free vibrations, 212–215 Undamped second-order systems, 559 Underdamped motion, 215 Undetermined coefficients method annihilator method and, 334–337, 346 complex arithmetic for, 237–238 explanation of, 174–180, 183–185, 337 nonhomogeneous linear equations and, 334–337 nonhomogeneous linear systems and, 174–185, 538–540 superposition principle and, 180–185 Uniform convergence, 582 Uniform waves, 615 Unique solutions (see also Existence and uniqueness of solutions), 2, 11–13, 602

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

Index

Unit step function, 384 Unit triangular pulse, 391 Unstable systems, 533 Utility functions, risk aversion model using, 86–87 van der Pol equation, 208, 426 Variable-coefficient equations Cauchy-Euler equations, 193–195 energy integral lemma for, 203–205 existance and uniqueness of solutions, 192–193 of higher order, 319 initial value solution using, 195 Legendre’s equation, 202 qualitative considerations for, 201–210 reduction of order, 197–198 of second order, 192–198, 201–210 variation of parameters, 196 Variable resistor, 444 Variable spring constant, 445 Variables differential equations using, 4 model development, 91 separation of, 38–39 (see also Separation of variables) Variation of parameters explanation of, 56, 187–191, 196 higher-order equations and, 338–340 nonhomogeneous linear equations, 187–191 nonhomogeneous linear systems and, 540–542

variable-coefficient equations, 196 Vector functions, 516–518 Vectors, 322, 505 Velocity escape, 117 group, 615 limiting or terminal, 40, 112, 273 phase, 615 Ventilation, 81 Verhulst, P.F., 98n Vibrating beam problem, 333 Vibrating drums, 615–616 Vibrating membrane, 607–608 Vibrating spring problems, 173 Vibrating strings, 3, 612–614 Vibrations critically damped motion, 216–217 desire or supression of, 154 forced, 186–187, 221–227 free, 212–220 frequency and, 213 mass-spring system and, 212–220 overdamped motion, 216 transverse, 333, 347 undamped motion, 212–215 underdamped motion, 215 Virus (HIV) infection models, 141 Volterra, V., 401 Volterra-Lotka system, 255–258, 260, 262, 275–276

I-11

Wave equation d’Alembert’s solution for, 609–611 existance and uniqueness of solutions, 612–614 explanation of, 603 Fourier series and, 603–614, 633–634 fundamental frequency and, 605–606 initial-boundary value problems, 606–610 Laplace transform solution of, 633–634 separation of variables and, 604–605, 607 standing waves, 603–604 travelling waves, 610–611 vibrating string problem and, 612–614 Weber function, 478 Window functions, 384–386 Wire, heat flow in insulated, 560–562, 592–602 Wronskian definition of, 160, 164, 195, 321, 517 fundamental sets, linear dependence and, 324, 517–518 linear combination functions and, 321–323 linear second-order equations and, 164, 195 variable-coefficient equations, 195 vector functions and, 322, 517–518 Young’s modulus, 333, 347 Zero matrix, 505 Zeros, isolated, 201 Zeroth derivatives, 157n

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

A BRIEF TABLE OF INTEGRALS*

1 1 f1u2 + g1u22du = 1 f1u2du + 1 g1u2du.

1 cf1u2du = c 1 f1u2du.

un+1 n 1 u du = n + 1 ,

1 u dv = uv - 1 v du.

u u 1 ue du = 1u - 12e .

u u 1 e du = e .

au u 1 a du = ln a ,

du 1 u = ln 0 u 0 .

n ≠ - 1.

n u n u n-1 u 1 u e du = u e - n 1 u e du.

du 1 u ln u = ln 0 ln u 0 .

1 ln u du = u ln u - u.

a 7 0, a ≠ 1.

ln u 1 n n+1 a b, 1 u ln u du = u n + 1 1n + 12 2

du 1 u 1 a2 + u2 = a arctan a .

n ≠ - 1.

u-a du 1 1 u2 - a2 = 2a ln ` u + a ` .

a+u du 1 1 a2 - u2 = 2a ln ` a - u ` .

a + bu 1 du 1 1a + bu21a + bu2 = ab - ab ln ` a + bu ` 1 a u du a 1 1a + bu21a + bu2 = ab - ab c b ln 0 a + bu 0 - b ln 0 a + bu 0 d .

1

1

du 2u + a 2

2

= ln @ u + 2u2 + a2 @ .

u = arcsin , a 2a - u du 2

2

1 sin u du = -cos u.

1

a2 Ú u2.

du 2u2 - a2

1

= ln @ u + 2u2 - a2 @ ,

du u2u - a 2

1 cos u du = sin u.

1 cot u du = ln 0 sin u 0 . 1 csc u du = -ln 0 csc u + cot u 0 = ln 0 csc u - cot u 0 . 2 1 sec u du = tan u. 2 1 1 1 sin u du = 2 u - 4 sin 2u. n 1 sin u du = -

2 1 csc u du = -cot u. 2 1 1 1 cos u du = 2 u + 4 sin 2u.

2

=

1 a arccos , a u

u2 Ú a2.

u 7 a 7 0.

1 tan u du = - ln 0 cos u 0 . 1 sec u du = ln 0 sec u + tan u 0 .

1 sec u tan u du = sec u. 2 1 tan u du = tan u - u.

sinn - 1u cos u n - 1 + sinn - 2u du. n n 1

n 1 cos u du =

cosn - 1u sin u n - 1 + cosn - 2u du. n n 1

*Note: An arbitrary constant is to be added to each indefinite integral.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

A BRIEF TABLE OF INTEGRALS* (continued)

1 u sin u du = sin u - u cos u.

n n n-1 1 u sin u du = -u cos u + n 1 u cos u du.

1 u cos u du = cos u + u sin u.

n n n-1 1 u cos u du = u sin u - n 1 u sin u du.

au 1 e sin nu du =

eau 1a sin nu - n cos nu2 a2 + n2

au 1 e cos nu du =

.

sin1a + b2u sin1a - b2u 1 sin au sin bu du = - 21a + b2 + 21a - b2 ,

1 cos au cos bu du =

sin1a + b2u sin1a - b2u + , 21a + b2 21a - b2

L0

.

a2 ≠ b2. a2 ≠ b2.

1 sinh u du = cosh u. Γ1t2 =

a2 + n2

a2 ≠ b2.

cos1a + b2u cos1a - b2u 1 sin au cos bu du = - 21a + b2 - 21a - b2 ,

∞

eau 1a cos nu + n sin nu2

1 cosh u du = sinh u.

e-uut - 1du, t 7 0;

Γ 1 12 2 = 2p; and Γ1n + 12 = n!, if n is a positive integer.

SOME POWER SERIES EXPANSIONS f1x2 = f1a2 + f ′1a21x - a2 +

f ″1a2 2!

1x - a2 2 + g +

f 1n2 1a2 1x - a2 n + g 1Taylor series2 n!

∞ 1 -12 nx 2n + 1 sin x = a n = 0 12n + 12!

∞ xn ex = a n = 0 n! ∞

11 - x2 -1 = a x n n=0

∞

n=0

x3 1#3 5 1#3#5 7 + x + x +g 2#3 2#4#5 2#4#6#7

∞ 1 -12 kx 2k J0 1x2 = a 2 2k k = 0 1k!2 2

∞ xn ln11 - x2 = - a n=1 n

11 - x2 -2 = a 1n + 12x n

2 x5 + 17 x7 + 62 x9 + tan x = x + 13 x3 + 15 g 315 2835

arcsin x = x +

1 -12 nx2n n = 0 12n2! ∞

cos x = a

∞ 1 -12 kx2k + 1 J1 1x2 = a 2k + 1 k = 0 k!1k + 12!2

∞ x2n + 1 arctan x = a 1 -12 n 2n + 1 n=0

∞ 1 -12 kx2k + n Jn 1x2 = a 2k + n k = 0 k!Γ1n + k + 122

*Note: An arbitrary constant is to be added to each indefinite integral.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

LINEAR FIRST-ORDER EQUATIONS

The solution to y′ + P1x2y = Q1x2 taking the value y1x0 2 at x = x0 is y1x2 = e- Lx0 P1j2dj c x

x

Lx0

e Lx0 P1z2dz Q1j2dj + y1x0 2 d . j

METHOD OF UNDETERMINED COEFFICIENTS To find a particular solution to the constant-coefficient differential equation ay″ + by′ + cy = Pm 1t2ert ,

where Pm 1t2 is a polynomial of degree m, use the form yp 1t2 = t s 1Amt m + g + A1t + A0 2ert ;

if r is not a root of the associated auxiliary equation, take s = 0; if r is a simple root of the associated auxiliary equation, take s = 1; and if r is a double root of the associated auxiliary equation, take s = 2. To find a particular solution to the differential equation ay″ + by′ + cy = Pm 1t2eat cos bt + Qn 1t2eat sin bt , b ≠ 0 ,

where Pm 1t2 is a polynomial of degree m and Qn 1t2 is a polynomial of degree n, use the form yp 1t2 = t s 1Akt k + g + A1t + A0 2eat cos bt

+ t s 1Bkt k + g + B1t + B0 2eat sin bt ,

where k is the larger of m and n. If a + ib is not a root of the associated auxiliary equation, take s = 0; if a + ib is a root of the associated auxiliary equation, take s = 1.

VARIATION OF PARAMETERS FORMULA If y1 and y2 are two linearly independent solutions to ay″ + by ′ + cy = 0, then a particular solution to ay″ + by ′ + cy = f is y = y1y1 + y2y2, where y1 1t2 =

-f1t2y2 1t2 dt, L aW3y1, y2 4 1t2

y2 1t2 =

f1t2y1 1t2 dt, L aW3y1, y2 4 1t2

and W3y1, y2 4 1t2 = y1 1t2y2= 1t2 - y1= 1t2y2 1t2.

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.

A TABLE OF LAPLACE TRANSFORMS

f 1t2

F1s2 = ℒ5 f 6 1s2

f 1t2 1

2p

2t

2s

F1s - a2

21. 2t

2p 2s3/2

sF1s2 - f102

22. t n - 11>22,

1. f1at2

s 1 Fa b a a

20.

2. eat f1t2 3. f ′1t2 4. f 1n2 1t2

snF1s2 - sn - 1f102 - sn - 2f ′102

5. t n f1t2 6.

1 f1t2 t

1 -12 nF 1n2 1s2

25. cos bt

F1s2 s

n = 1, 2, c

r7 -1

24. sin bt

∞

t

23. t r,

- g - sf 1n - 22 102 - f 1n - 12 102

1s F1u2du

7. 10 f1y2dy

F1s2 = ℒ5 f 6 1s2

26. eat sin bt 27. eat cos bt

1 # 3 # 5 g 12n - 12 2p

2ns n + 11/22 Γ1r + 12 sr+1 b 2 s + b2 s s2 + b2 b 1s - a2 2 + b2 s-a 1s - a2 2 + b2

8. 1 f * g21t2

F1s2 G1s2

28. sinh bt

9. f1t + T2 = f1t2

-st 10 e f1t2dt 1 - e-sT

29. cosh bt

10. f1t - a2 u1t - a2, a Ú 0

e-asF1s2

30. sin bt - bt cos bt

2b3 1s + b2 2 2

11. g1t2 u1t - a2, a Ú 0

e-as ℒ5g1t + a261s2

31. t sin bt

2bs 1s + b2 2 2

12. u1t - a2, a Ú 0

e-as s

32. sin bt + bt cos bt

2bs2 1s + b2 2 2

13. q a,b 1t2,

06a6b

e-sa - e-sb s

33. t cos bt

s2 - b2 1s2 + b2 2 2

14. d1t - a2,

aÚ0

e-as

34. sin bt cosh bt - cos bt sinh bt

1 s-a

35. sin bt sinh bt

T

15. eat 16. t n, n = 1, 2, c 17. eatt n, n = 1, 2, c

n! sn+1 n! 1s - a2 n + 1

1a - b2

18. eat - ebt

1s - a21s - b2

19. aeat - bebt

1s - a21s - b2

b s - b2 s s2 - b2 2

2

2

2

4b3 s + 4b4 2b2s s4 + 4b4 4

36. sinh bt - sin bt

2b3 s4 - b4

37. cosh bt - cos bt

2b2s s - b4

38. Jv 1bt2, v 7 - 1

1a - b2s

4

1 2s2 + b2 - s 2 v bv 2s2 + b2

MATH 54 Linear Algebra and Differential Equations, Second Custom Edition for University of California Berkeley. Copyright © 2021 by Pearson Education, Inc. All Rights Reserved. Pearson Custom Edition.