Advanced engineering mathematics [6 ed.] 9781352010251, 9781352010268

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K.A. STROUD

ADVANCED ENGINEERING MATHEMATICS

ADVANCED ENGI NEERIN G MATHEMATICS K. A. Stroud Formerly Principal Lecturer Department of Mathematics, Coventry University

with

Dexter J. Booth Formerly Principal Lecturer School of Computing and Engineering, University of Huddersfield

SIXTH EDITION REVIEW BOARD FOR THE SIXTH EDITION: Declan Bates, Professor of Bioengineering, School of Engineering, University of Warwick, UK Stewart Chidlow, Senior Lecturer in Applied Mathematics, School of Applied Mathematics, Liverpool John Moores University, UK Patrick Johnson, Senior Lecturer in Mathematics, Department of Engineering and Mathematics, Sheffield Hallam University, UK Colin Steele, Director of Service Teaching, Department of Mathematics, University of Manchester, UK

# K.A. Stroud 1986, 1990, 1996 # K.A Stroud and Dexter J. Booth under exclusive licence to Macmillan Education Limited 2003, 2011, 2020 All rights reserved. No reproduction, copy or transmission of this publication may be made without written permission. No portion of this publication may be reproduced, copied or transmitted save with written permission or in accordance with the provisions of the Copyright, Designs and Patents Act 1988, or under the terms of any licence permitting limited copying issued by the Copyright Licensing Agency, Saffron House, 6–10 Kirby Street, London EC1N 8TS. Any person who does any unauthorized act in relation to this publication may be liable to criminal prosecution and civil claims for damages. The authors have asserted their rights to be identified as the authors of this work in accordance with the Copyright, Designs and Patents Act 1988. This edition published 2020 by RED GLOBE PRESS Previous editions published under the imprint PALGRAVE

Red Globe Press in the UK is an imprint of Macmillan Education Limited, registered in England, company number 01755588, of 4 Crinan Street, London, N1 9XW. Red Globe Press is a registered trademark in the United States, the United Kingdom, Europe and other countries. ISBN 978–1–352–01025–1 paperback ISBN 978–1–352–01026–8 ebook This book is printed on paper suitable for recycling and made from fully managed and sustained forest sources. Logging, pulping and manufacturing processes are expected to conform to the environmental regulations of the country of origin. A catalogue record for this book is available from the British Library. A catalog record for this book is available from the Library of Congress.

Summary of contents Preface to the first edition Preface to the sixth edition Hints on using the book Useful background information

xx xxi xxiii xxiv

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

1 47 93 124 156 194 239 270 300 337 359 377 396 437 481 512 538 581 603 640 683 720 767 817 861 897 944 995 1021 1061 1109 1140

Numerical solutions of equations and interpolation Laplace transforms 1 Laplace transforms 2 Laplace transforms 3 Difference equations and the Z transform Introduction to invariant linear systems Fourier series 1 Fourier series 2 Introduction to the Fourier transform Power series solutions of ordinary differential equations 1 Power series solutions of ordinary differential equations 2 Power series solutions of ordinary differential equations 3 Numerical solutions of ordinary differential equations Matrix algebra Systems of ordinary differential equations Direction fields Phase plane analysis Nonlinear systems Dynamical systems Partial differentiation Partial differential equations Numerical solutions of partial differential equations Multiple integration 1 Multiple integration 2 Integral functions Vector analysis 1 Vector analysis 2 Vector analysis 3 Complex analysis 1 Complex analysis 2 Complex analysis 3 Optimization and linear programming

1165 1174 1213

Appendix Answers Index

iii

Contents xx

Preface to the first edition

xxi

Preface to the sixth edition

xxiii

Hints on using the book

xiv

Useful background information

Programme 1

Numerical solutions of equations and interpolation

1

Learning outcomes Introduction The Fundamental Theorem of Algebra Relations between the coefficients and the roots of a polynomial equation Cubic equations Transforming a cubic to reduced form Tartaglia’s solution for a real root Numerical methods Bisection Numerical solution of equations by iteration Using a spreadsheet Relative addresses Newton-Raphson iterative method Tabular display of results Modified Newton-Raphson method Interpolation Linear interpolation Graphical interpolation Gregory-Newton interpolation formula using forward finite differences Central differences Gregory-Newton backward differences Lagrange interpolation Review summary 1 Can you? checklist 1 Test exercise 1 Further problems 1

1 2 2

25 31 33 34 38 41 42 43

Programme 2

47

Laplace transforms 1

4 7 8 8 10 10 12 12 14 15 16 21 24 24 25

47 48 48

Learning outcomes Introduction Laplace transforms

v

vi

Contents

Differentiating and integrating a transform Theorem 1 The first shift theorem Theorem 2 Multiplying by t and t n Theorem 3 Dividing by t Inverse transforms Rules of partial fractions The ‘cover up’ rule Table of inverse transformations Solution of differential equations by Laplace transforms Transforms of derivatives Solution of first-order differential equations Solution of second-order differential equations Simultaneous differential equations Review summary 2 Can you? checklist 2 Test exercise 2 Further problems 2

55 55 56 58 61 62 68 69 70 71 73 75 80 87 90 90 91

Programme 3

93

Laplace transforms 2

Learning outcomes Introduction Heaviside unit step function Unit step at the origin Effect of the unit step function Laplace transform of uðt  cÞ Laplace transform of uðt  cÞ:f ðt  cÞ (the second shift theorem) Differential equations involving the unit step function Convolution The convolution theorem Review summary 3 Can you? checklist 3 Test exercise 3 Further problems 3

93 94 94 95 95 98 99 109 113 118 119 121 121 122

Programme 4

124

Laplace transforms 3

Learning outcomes Laplace transform of periodic functions Periodic functions Inverse transformations The Dirac delta – the unit impulse Graphical representation Laplace transform of ðt  aÞ The derivative of the unit step function Differential equations involving the unit impulse Harmonic oscillators Damped motion

124 125 125 131 135 136 137 140 141 143 145

Contents

vii

Forced harmonic motion with damping Resonance Review summary 4 Can you? checklist 4 Test exercise 4 Further problems 4

147 150 152 153 154 155

Programme 5

156

Difference equations and the Z transform

Learning outcomes Introduction Sequences Difference equations Solving difference equations Solution by inspection The particular solution The Z transform Table of Z transforms Properties of Z transforms Linearity First shift theorem (shifting to the left) Second shift theorem (shifting to the right) Scaling Final value theorem The initial value theorem The derivative of the transform Inverse transforms Solving difference equations Sampling Review summary 5 Can you? checklist 5 Test exercise 5 Further problems 5

156 157 157 159 161 161 164 167 172 172 172 173 174 175 176 177 177 178 181 184 187 190 191 191

Programme 6

194

Introduction to invariant linear systems

Learning outcomes Invariant linear systems Systems Input-response relationships Linear systems Time-invariance of a continuous system Shift-invariance of a discrete system Differential equations The general nth-order equation Zero-input response and zero-state response

194 195 195 196 197 200 202 203 203 204

viii

Contents

Zero-input, zero response Time-invariance Responses of a continuous system Impulse response Arbitrary input Convolution Exponential response The transfer function HðsÞ Differential equations Responses of a discrete system The discrete unit impulse Arbitrary input Exponential response Transfer function Difference equations Review summary 6 Can you? checklist 6 Test exercise 6 Further problems 6

209 211 212 212 212 213 216 218 220 223 223 224 226 227 228 232 235 236 237

Programme 7

239

Fourier series 1

Learning outcomes Introduction Periodic functions Graphs of y ¼ A sin nx Harmonics Non-sinusoidal periodic functions Analytic description of a periodic function Integrals of periodic functions Orthogonal functions Fourier series Dirichlet conditions Effect of harmonics Gibbs’ phenomenon Sum of a Fourier series at a point of discontinuity Review summary 7 Can you? checklist 7 Test exercise 7 Further problems 7

239 240 240 240 241 242 242 246 249 250 253 259 261 261 264 265 266 267

Programme 8

270

Fourier series 2

Learning outcomes Odd and even functions and half-range series Odd and even functions

270 271 271

Contents

ix

Products of odd and even functions Half-range series Series containing only odd harmonics or only even harmonics Significance of the constant term 12 a0 Functions with periods other than 2 Functions with period T Fourier coefficients Half-range series with arbitrary period Review summary 8 Can you? checklist 8 Test exercise 8 Further problems 8

274 281 285 287 288 288 289 292 295 297 297 298

Programme 9

300

Introduction to the Fourier transform

Learning outcomes Complex Fourier series Introduction Complex exponentials Complex spectra The two domains Continuous spectra Fourier’s integral theorem Some special functions and their transforms Even functions Odd functions Top-hat function The Dirac delta The triangle function Alternative forms Properties of the Fourier transform Linearity Time shifting Frequency shifting Time scaling Symmetry Differentiation The Heaviside unit step function Convolution The convolution theorem Fourier cosine and sine transformations Table of transforms Review summary 9 Can you? checklist 9 Test exercise 9 Further problems 9

300 301 301 301 306 307 308 310 313 313 313 315 317 319 319 320 320 321 321 321 322 323 324 325 326 328 330 330 333 334 335

x

Contents

Programme 10

Power series solutions of ordinary differential equations 1

337

Learning outcomes Higher derivatives Leibnitz theorem – nth derivative of a product of functions Choice of function for u and v Power series solutions Leibnitz-Maclaurin method Cauchy-Euler equi-dimensional equations Review summary 10 Can you? checklist 10 Test exercise 10 Further problems 10

337 338 341 343 344 345 352 356 357 357 357

Programme 11

Power series solutions of ordinary differential equations 2

359

Learning outcomes Introduction Solution of differential equations by the method of Frobenius The indicial equation Review summary 11 Can you? checklist 11 Test exercise 11 Further problems 11

359 360 360 369 375 375 376 376

Programme 12

377

Power series solutions of ordinary differential equations 3

Learning outcomes Introduction Bessel’s equation Gamma and Bessel functions Graphs of Bessel functions J0 ðxÞ and J1 ðxÞ Legendre’s equation Legendre polynomials Rodrigue’s formula and the generating function Sturm-Liouville systems Orthogonality Legendre’s equation revisited Polynomials as a finite series of Legendre polynomials Review summary 12 Can you? checklist 12 Test exercise 12 Further problems 12

377 378 378 379 384 384 385 385 387 389 390 391 392 394 394 395

Contents

Programme 13

xi

Numerical solutions of ordinary differential equations

Learning outcomes Introduction Taylor’s series Function increment First-order differential equations Euler’s method The exact value and the errors Graphical interpretation of Euler’s method The Euler-Cauchy method – or the improved Euler method Euler-Cauchy calculations Runge-Kutta method Second-order differential equations Euler second-order method Runge-Kutta method for second-order differential equations Predictor-corrector methods Review summary 13 Can you? checklist 13 Test exercise 13 Further problems 13

Programme 14

Matrix algebra

Learning outcomes Singular and non-singular matrices Rank of a matrix Elementary operations and equivalent matrices Consistency of a set of linear equations Uniqueness of solutions Solution of sets of linear equations Inverse method Row transformation method Gaussian elimination method Triangular decomposition method Using an electronic spreadsheet Comparison of methods Matrix transformation Rotation of axes Review summary 14 Can you? checklist 14 Test exercise 14 Further problems 14

396 396 397 397 398 399 399 408 412 414 415 420 423 423 425 430 432 434 434 435

437 437 438 439 440 444 445 449 449 453 457 460 466 470 471 473 475 477 478 479

xii

Contents

Programme 15

Systems of ordinary differential equations

481

Learning outcomes Eigenvalues of 2  2 matrices Characteristic equation Sum and product of eigenvalues Eigenvectors Systems of linear, first-order ordinary differential equations Repeated eigenvalues Diagonalization of a matrix Modal matrix Spectral matrix Systems of linear, second-order differential equations Review summary 15 Can you? checklist 15 Test exercise 15 Further problems 15

481 482 482 483 484 487 492 498 498 498 503 508 509 510 510

Programme 16

512

Direction fields

Learning outcomes Differential equations Introduction Family of solutions Direction fields DFIELD Introduction A specific solution Family of solutions Autonomous differential equations Equilibrium solutions The phase line Non-autonomous equations Introduction Review summary 16 Can you? checklist 16 Test exercise 16 Further problems 16

512 513 513 513 516 518 518 519 520 521 521 524 530 530 534 535 536 536

Programme 17

538

Phase plane analysis

Learning outcomes Phase plane analysis Introduction Mass-spring system PPLANE Phase plane analysis

538 539 539 539 542 544

Contents

xiii

Eigenvalues and the phase plane Imaginary eigenvalues Two complex eigenvalues Behaviour around the critical point Two real and negative eigenvalues Behaviour around the critical point Two real and positive eigenvalues Two real eigenvalues of different signs Two identical eigenvalues Star node Singular coefficient matrix The inhomogeneous case Critical point moved to the origin Review summary 17 Can you? checklist 17 Test exercise 17 Further problems 17

545 545 551 554 556 556 560 563 567 568 568 570 571 574 576 577 578

Programme 18

581

Nonlinear systems

Learning outcomes Multiple critical points Introduction Linearization Problems with linearization Review summary 18 Can you? checklist 18 Test exercise 18 Further problems 18

581 582 582 585 594 599 600 600 601

Programme 19

603

Dynamical systems

Learning outcomes Dynamical systems Introduction Predator-prey problems Competition within a single population Two non-interacting populations Two interacting populations Second-order differential equations Undamped pendulum: small oscillations Undamped pendulum: no approximation Damped pendulum Bifurcation First-order equations Second-order equations Limit cycles The Van der Pol equation

603 604 604 604 607 608 610 616 616 617 620 624 624 628 630 632

xiv

Contents

Review summary 19 Can you? checklist 19 Test exercise 19 Further problems 19

634 634 635 635

Programme 20

640

Partial differentiation

Learning outcomes Small increments Taylor’s theorem for one independent variable Taylor’s theorem for two independent variables Small increments Rates of change Implicit functions Change of variables Inverse functions General case Stationary values of a function Maximum and minimum values Saddle point Lagrange undetermined multipliers Functions with three independent variables Review summary 20 Can you? checklist 20 Test exercise 20 Further problems 20

640 641 641 641 643 645 646 647 651 653 659 660 666 671 674 677 679 680 680

Programme 21

683

Partial differential equations

Learning outcomes Introduction Partial differential equations Solution by direct integration Initial conditions and boundary conditions The wave equation Solution of the wave equation Solution by separating the variables The heat conduction equation for a uniform finite bar Solutions of the heat conduction equation Laplace’s equation Solution of the Laplace equation Laplace’s equation in plane polar coordinates The problem Separating the variables The n ¼ 0 case Review summary 21 Can you? checklist 21 Test exercise 21 Further problems 21

683 684 685 685 686 687 688 688 697 698 703 704 708 709 709 713 715 716 717 718

Contents

Programme 22

xv

Numerical solutions of partial differential equations

720

Learning outcomes Introduction Numerical approximation to derivatives Functions of two real variables Grid values Computational molecules Summary of procedures Derivative boundary conditions Second-order partial differential equations Elliptic equations Hyperbolic equations Parabolic equations Second partial derivatives Time-dependent equations The Crank-Nicholson procedure Dimensional analysis Review summary 22 Can you? checklist 22 Test exercise 22 Further problems 22

720 721 721 723 724 727 731 734 738 738 739 739 741 745 750 756 757 760 761 762

Programme 23

767

Multiple integration 1

Learning outcomes Introduction Differentials Exact differential Integration of exact differentials Area enclosed by a closed curve Line integrals Alternative form of a line integral Properties of line integrals Regions enclosed by closed curves Line integrals round a closed curve Line integral with respect to arc length Parametric equations Dependence of the line integral on the path of integration Exact differentials in three independent variables Green’s theorem Review summary 23 Can you? checklist 23 Test exercise 23 Further problems 23

767 768 776 778 780 782 786 787 790 792 793 797 798 799 804 805 812 814 815 816

xvi

Contents

Programme 24

Multiple integration 2

817

Learning outcomes Double integrals Surface integrals Three dimensional coordinate systems Cartesian coordinates Cylindrical coordinates Spherical coordinates Element of volume in the three coordinate systems Volume integrals Change of variables in multiple integrals Curvilinear coordinates Transformation in three dimensions Review summary 24 Can you? checklist 24 Test exercise 24 Further problems 24

817 818 823 829 829 830 831 833 834 843 845 853 855 857 858 858

Programme 25

861

Integral functions

Learning outcomes Gamma and beta functions The gamma function The beta function Reduction formulas Relation between the gamma and beta functions Application of gamma and beta functions Duplication formula for gamma functions The error function The graph of erfðxÞ The complementary error function erfcðxÞ Elliptic functions Standard forms of elliptic functions Complete elliptic functions Alternative forms of elliptic functions Review summary 25 Can you? checklist 25 Test exercise 25 Further problems 25

861 862 862 871 872 875 876 880 881 882 882 884 885 885 889 892 894 895 895

Programme 26

897

Vector analysis 1

Learning outcomes Introduction Triple products Scalar triple products of three vectors Properties of scalar triple products

897 898 903 903 904

Contents

xvii

Coplanar vectors Vector triple products of three vectors Differentiation of vectors Differentiation of sums and products of vectors Unit tangent vectors Partial differentiation of vectors Integration of vector functions Scalar and vector fields grad (gradient of a scalar field) Directional derivatives Unit normal vectors grad of sums and products of scalars div (divergence of a vector function) curl (curl of a vector function) Summary of grad, div and curl Multiple operations Review summary 26 Can you? checklist 26 Test exercise 26 Further problems 26

905 907 910 915 915 918 918 921 921 924 927 929 931 932 933 935 938 940 941 941

Programme 27

944

Vector analysis 2

Learning outcomes Line integrals Scalar field Vector field Volume integrals Surface integrals Scalar fields Vector fields Conservative vector fields Divergence theorem (Gauss’ theorem) Stokes’ theorem Direction of unit normal vectors to a surface S Green’s theorem Review summary 27 Can you? checklist 27 Test exercise 27 Further problems 27

944 945 945 948 952 956 957 960 965 970 976 979 985 988 990 991 992

Programme 28

995

Vector analysis 3

Learning outcomes Curvilinear coordinates Orthogonal curvilinear coordinates Orthogonal coordinate systems in space

995 996 1000 1001

xviii

Contents

Scale factors Scale factors for coordinate systems General curvilinear coordinate system (u, v, w) Transformation equations Element of arc ds and element of volume dV in orthogonal curvilinear coordinates grad, div and curl in orthogonal curvilinear coordinates Particular orthogonal systems Review summary 28 Can you? checklist 28 Test exercise 28 Further problems 28

1005 1006 1008 1009

Programme 29

1021

Complex analysis 1

1010 1011 1014 1016 1018 1019 1019

Learning outcomes 1021 Functions of a complex variable 1022 Complex mapping 1023 Mapping of a straight line in the z-plane onto the w-plane under the transformation w ¼ f ðzÞ 1025 Types of transformation of the form w ¼ az þ b 1029 Nonlinear transformations 1038 Mapping of regions 1043 Review summary 29 1057 Can you? checklist 29 1058 Test exercise 29 1058 Further problems 29 1059

Programme 30

Complex analysis 2

Learning outcomes Differentiation of a complex function Regular function Cauchy-Riemann equations Harmonic functions Complex integration Contour integration – line integrals in the z-plane Cauchy’s theorem Deformation of contours at singularities Conformal transformation (conformal mapping) Conditions for conformal transformation Critical points Schwarz-Christoffel transformation Open polygons Review summary 30 Can you? checklist 30 Test exercise 30 Further problems 30

1061 1061 1062 1063 1065 1067 1072 1072 1075 1080 1089 1089 1090 1093 1098 1104 1105 1106 1107

Contents

Programme 31

xix

Complex analysis 3

Learning outcomes Maclaurin series Radius of convergence Singular points Poles Removable singularities Circle of convergence Taylor’s series Laurent’s series Residues Calculating residues Integrals of real functions Ð 2 Integrals of the form 0 Fðcos , sin Þ d Ð1 Integrals of the form 1 FðxÞ dx  ð1 sin x FðxÞ Integrals of the form dx cos x 1

1109 1109 1110 1114 1115 1115 1116 1116 1117 1119 1123 1125 1126 1126 1128 1131

Review summary 31 Can you? checklist 31 Test exercise 31 Further problems 31

1133 1135 1136 1137

Programme 32

1140

Optimization and linear programming

Learning outcomes Optimization Linear programming (or linear optimization) Linear inequalities Graphical representation of linear inequalities Solver Solver parameters Applications Nonlinear programming Review summary 32 Can you? checklist 32 Test exercise 32 Further problems 32 Appendix Answers Index

1140 1141 1141 1142 1142 1148 1149 1154 1156 1158 1159 1160 1161 1165 1174 1213

Preface to the first edition The purpose of this book is essentially to provide a sound second year course in Mathematics appropriate to studies leading to B.Sc. Engineering Degrees and other qualifications of a comparable level. The emphasis throughout is on techniques and applications, supported by sufficient formal proofs to warrant the methods being employed. The structure of the text and the techniques used follow closely those of the author’s first year book, Engineering Mathematics – Programmes and Problems, to which this further book is a companion volume and a continuation of the highly successful learning strategies devised. As with the previous work, the text is based on a series of self-instructional programmes arising from extensive research and rigid evaluation in a variety of relevant courses and, once again, the individualized nature of the development makes the book eminently suitable both for general class use and for personal study. Each of the course programmes guides the student through the development of a particular topic, with numerous worked examples to demonstrate the techniques and with increased responsibility passing to the student as mastery is achieved. Revision exercises are provided where appropriate and each programme terminates with a Revision Summary of the main points covered, a Test Exercise based directly on the work of the programme and a set of Further Problems which provides opportunity for the additional practice that is essential for ensured success. The ability to work at one’s own pace throughout is of utmost importance in maintaining motivation and in achieving mastery. In several instances, the topic of a programme is a direct extension of basic work covered in Engineering Mathematics and where this is so, the title page of the programme carries a brief reference to the relevant programme in the first year treatment. This clearly directs the student to worthwhile revision of the prerequisites assumed in the further development of the subject matter. A complete set of Answers to all problems and a detailed Index are provided at the end of the book. Grateful acknowledgement is made of the constructive suggestions and cooperation received from many quarters both in the development of the original programmes and in the final preparation of the text. Recognition must also be made of the many sources from which examples have been gleaned over the years and which contribute in no small measure to the success of the work. Finally my sincere appreciation is due to the publishers for their patience, advice and ready cooperation in the preparation of the text for publication. K.A. Stroud

xx

Preface to the sixth edition It is 50 years since Ken Stroud first published his approach to personalized learning with his classic text Engineering Mathematics, now in its eighth edition and having sold over half a million copies. Some 15 years later he followed this with Further Engineering Mathematics which, for its fourth edition, was restyled as Advanced Engineering Mathematics. As in all earlier editions his unique and hugely successful programmed learning style is continued in this sixth edition. As with previous editions I have endeavoured to retain the very essence of the style, particularly the time-tested Stroud format with its close attention to technique development throughout. This methodology has contributed to the mathematical abilities of so many students all over the world

New to this edition To cater for continual changes in engineering mathematics the work of this edition builds upon material that was present in previous editions. In Introduction to invariant linear systems the presentation of various elements of the subject have been revisited and improved and in Power series solutions of ordinary differential equations the method of Frobenius has been restructured and the presentation refined. In Systems of ordinary differential equations there is a new section dealing with a simpler method of obtaining the eigenvalues of a 2  2 matrix and the Cayley-Hamilton theorem has been omitted for inclusion in the sister text Engineering Mathematics. In addition, the notation in the later part of the Programme is changed to ensure a smooth fit with four new Programmes: Direction fields, Phase plane analysis, Nonlinear systems and Dynamical systems. These four Programmes are a major addition to this new edition and address a significant omission from earlier editions. Finally, Optimization and linear programming is given a whole facelift making full use of computer software to optimize objective functions. Lecturers will also find a complete set of PowerpointTM lecture slides to accompany each Programme on the book’s companion website at www.macmillanihe.com/stroud.

Acknowledgements This is a further opportunity that I have had to work on the Stroud books. It is ever a challenge and an honour to be able to deal with Ken Stroud’s material. Ken had an understanding of his students and their learning and thinking processes that was second to none and this is reflected in every page of this book. As always, my thanks go to the Stroud family for their continuing support and encouragement of new projects and ideas which are allowing Ken’s teaching methodology to be offered to a whole new range of students. I should also like to express my thanks and appreciation for the valuable feedback that has been provided by all the reviewers and students during the writing of this new edition and of previous editions upon which this one builds. In particular I should like to thank Professor John C Polking of

xxi

xxii

Preface to the sixth edition

Rice University in Houston, Texas for his permission to make extensive use of his suite of programs, DFIELD and PPLANE; they have proved invaluable in permitting me to display the phase plane so straightforwardly. I should also like to thank Professor Mike Hagerty of Boston College, Massachusetts for his close reading of an earlier edition and providing a number of useful amendments. Engineering is not a static universe and it is always a challenge to best determine how a new edition is to be developed. All the encouraging comments and sympathetic treatment of the new material has been greatly appreciated. Finally, I should like to thank the entire production team at Red Globe Press for all their care, not least Ann Edmondson who has assiduously converted my many documents into the professional looking book you now see. And how could I not mention the one person who has overseen all my efforts for the last twenty years, Helen Bugler my editor in whom I have the utmost admiration for her continued enthusiasm and professionalism. Dexter J Booth Huddersfield January 2020

Hints on using the book This book contains 32 Programmes, each of which has been written in such a way as to make learning more effective and more interesting. It is almost like having a personal tutor, for you proceed at your own rate of learning and any difficulties you may have are cleared before you have the chance to practise incorrect ideas or techniques. You will find that each Programme is divided into sections called frames. When you start a Programme, begin at Frame 1. Read each frame carefully and carry out any instructions or exercise which you are asked to do. In almost every frame, you are required to make a response of some kind, testing your understanding of the information in the frame, and you can immediately compare your answer with the correct answer given in the next frame. To obtain the greatest benefit, you are strongly advised to cover up the following frame, where necessary, until you have made your response. When a series of dots occurs, you are expected to supply the missing word, phrase, or number. At every stage, you will be guided along the right path. There is no need to hurry: read the frames carefully and follow the directions exactly. In this way, you will learn. At the end of each Programme, you will find a Review summary and a Can you? checklist that matches the Learning outcomes given at the beginning of the Programme. Read these carefully to make sure you have not missed anything. Next you will find a short Test exercise. This is set directly on what you have learned in the Programme: the questions are straightforward and contain no tricks. When you have completed these, return to the Can you? checklist as a final reminder of the contents of the Programme. To provide you with the necessary practice, a set of Further problems is also included. Remember that in mathematics, as in many other situations, practice makes perfect or more nearly so. Even if you feel you have done some of the topics before, work steadily through each Programme: it will serve as useful revision and fill in any gaps in your knowledge that you may have.

xxiii

Useful background information 1 Algebraic identities ða þ bÞ2 ¼ a2 þ 2ab þ b2

ða þ bÞ3 ¼ a3 þ 3a2 b þ 3ab2 þ b3

ða  bÞ2 ¼ a2  2ab þ b2

ða  bÞ3 ¼ a3  3a2 b þ 3ab2  b3

ða þ bÞ4 ¼ a4 þ 4a3 b þ 6a2 b2 þ 4ab3 þ b4 ða  bÞ4 ¼ a4  4a3 b þ 6a2 b2  4ab3 þ b4 a2  b2 ¼ ða  bÞða þ bÞ a3 þ b3 ¼ ða þ bÞða2  ab þ b2 Þ a3  b3 ¼ ða  bÞða2 þ ab þ b2 Þ 2 Trigonometrical identities (1) sin2  þ cos2  ¼ 1;

sec2  ¼ 1 þ tan2 ;

cosec2  ¼ 1 þ cot2  (2) sinðA þ BÞ ¼ sin A cos B þ cos A sin B sinðA  BÞ ¼ sin A cos B  cos A sin B cosðA þ BÞ ¼ cos A cos B  sin A sin B cosðA  BÞ ¼ cos A cos B þ sin A sin B tan A þ tan B 1  tan A tan B tan A  tan B tanðA  BÞ ¼ 1 þ tan A tan B (3) Let A ¼ B ¼  ; sin 2 ¼ 2 sin  cos  tanðA þ BÞ ¼

cos 2 ¼ cos2   sin2  ¼ 1  2 sin2  ¼ 2 cos2   1 2 tan  1  tan2    ; sin  ¼ 2 sin cos 2 2   cos  ¼ cos2  sin2 2 2   ¼ 1  2 sin2 ¼ 2 cos2  1 2 2 tan 2 ¼

(4) Let  ¼

 2

xxiv

Useful background information

tan  ¼

2 tan

xxv  2

 2 CþD CD cos (5) sin C þ sin D ¼ 2 sin 2 2 CþD CD sin C  sin D ¼ 2 cos sin 2 2 CþD CD cos C þ cos D ¼ 2 cos cos 2 2 CþD CD cos D  cos C ¼ 2 sin sin 2 2 (6) 2 sin A cos B ¼ sinðA þ BÞ þ sinðA  BÞ 1  tan2

2 cos A sin B ¼ sinðA þ BÞ  sinðA  BÞ 2 cos A cos B ¼ cosðA þ BÞ þ cosðA  BÞ 2 sin A sin B ¼ cosðA  BÞ  cosðA þ BÞ (7) Negative angles:

sinðÞ ¼  sin  cosðÞ ¼ cos  tanðÞ ¼  tan 

(8) Angles having the same trigonometrical ratios: (a) Same sine:  and ð1808  Þ (b) Same cosine:  and ð3608  Þ, i.e. ðÞ (c) Same tangent:  and ð1808 þ Þ (9) a sin  þ b cos  ¼ A sinð þ Þ a sin   b cos  ¼ A sinð  Þ a cos  þ b sin  ¼ A cosð  Þ a cos   b sin  ¼ A cosð þ Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 8 2 2 >

:  ¼ tan1 b ð08 <  < 908Þ a 3 Standard curves (a) Straight line dy y2  y1 Slope, m ¼ ¼ dx x2  x1 Angle between two lines, tan  ¼

m2  m1 1 þ m1 m2

For parallel lines, m2 ¼ m1 For perpendicular lines, m1 m2 ¼ 1

xxvi

Useful background information

Equation of a straight line (slope ¼ m) (1) Intercept c on real y-axis: y ¼ mx þ c (2) Passing through ðx1 , y1 Þ: y  y1 ¼ mðx  x1 Þ y  y1 x  x1 ¼ (3) Joining ðx1 , y1 Þ and ðx2 ; y2 Þ: y2  y1 x2  x1 (b) Circle Centre at origin, radius r: x2 þ y2 ¼ r 2 Centre ðh; kÞ, radius r: ðx  hÞ2 þ ðy  kÞ2 ¼ r 2 x2 þ y2 þ 2gx þ 2fy þ c ¼ 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi with centre ðg; f Þ: radius ¼ g 2 þ f 2  c

General equation:

Parametric equations: x ¼ r cos , y ¼ r sin  (c) Parabola Vertex at origin, focus ða, 0Þ: y2 ¼ 4ax Parametric equations: x ¼ at 2 , y ¼ 2at (d) Ellipse

 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  x2 y2 Centre at origin, foci  a2 þ b2 ; 0 : 2 þ 2 ¼ 1 a b where a ¼ semi-major axis, b ¼ semi-minor axis Parametric equations: x ¼ a cos , y ¼ b sin 

(e) Hyperbola

 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  x2 y2 Centre at origin, foci  a2 þ b2 ; 0 : 2  2 ¼ 1 a b Parametric equations: x ¼ a sec , y ¼ b tan  Rectangular hyperbola:

  a a a2 ¼ c2 Centre at origin, vertex  pffiffiffi ; pffiffiffi : xy ¼ 2 2 2 a where c ¼ pffiffiffi i.e. xy ¼ c2 2 Parametric equations: x ¼ ct, y ¼ c=t 4 Laws of mathematics (a) Associative laws – for addition and multiplication a þ ðb þ cÞ ¼ ða þ bÞ þ c aðbcÞ ¼ ðabÞc (b) Commutative laws – for addition and multiplication aþb¼bþa ab ¼ ba (c) Distributive laws – for multiplication and division aðb þ cÞ ¼ ab þ ac bþc b c ¼ þ (provided a 6¼ 0Þ a a a

Frames 1 to 83

Programme 1

Numerical solutions of equations and interpolation Learning outcomes When you have completed this Programme you will be able to: Appreciate the Fundamental Theorem of Algebra Find the two roots of a quadratic equation and recognize that for polynomial equations with real coefficients complex roots exist in complex conjugate pairs Use the relationships between the coefficients and the roots of a polynomial equation to find the roots of the polynomial Transform a cubic equation to its reduced form Use Tartaglia’s solution to find the roots of a cubic equation Find the solution of the equation f ðxÞ ¼ 0 by the method of bisection Solve equations involving a single real variable by iteration and use a spreadsheet for efficiency Solve equations using the Newton–Raphson iterative method Use the modified Newton–Raphson method to find the first approximation when the derivative is small Understand the meaning of interpolation and use simple linear and graphical interpolation Use the Gregory–Newton interpolation formula with forward and backward differences for equally spaced domain points Use the Gauss interpolation formulas using central differences for equally spaced domain points Use Lagrange interpolation when the domain points are not equally spaced

1

2

Programme 1

Introduction 1

In this Programme we shall be looking at analytic and numerical methods of solving the general equation in a single variable, f ðxÞ ¼ 0. In addition, a functional relationship can be exhibited in the form of a collection of ordered pairs rather than in the form of an algebraic expression. We shall be looking at interpolation methods of estimating values of f ðxÞ for intermediate values of x between those listed among the ordered pairs. First we shall look at the Fundamental Theorem of Algebra, which deals with the factorization of polynomials.

The Fundamental Theorem of Algebra 2

The Fundamental Theorem of Algebra can be stated as follows: Every polynomial expression f (x) = an xn + an1 xn1 +    + a1 x + a0 can be written as a product of n linear factors in the form f (x) = an (x  r 1 )(x  r 2 )(    )(x  rn ) As an immediate consequence of this we can see that there are n values of x that satisfy the polynomial equation f ðxÞ ¼ 0, namely x ¼ r1 , x ¼ r2 , . . . , x ¼ rn . We call these values the roots of the polynomial, but be aware that they may not all be distinct. Furthermore, the polynomial coefficients ai and the polynomial roots ri may be real, imaginary or complex. For example the quadratic equation x2 þ 5x þ 6 ¼ 0 can be written ðx þ 2Þðx þ 3Þ ¼ 0 so it has the two distinct roots x ¼ 2 and x ¼ 3 x2  4x þ 4 ¼ 0 can be written as ðx  2Þðx  2Þ ¼ 0 so it has the two coincident roots x ¼ 2 and x ¼ 2 x2 þ x þ 1 ¼ 0 can be written as ðx þ aÞðx þ bÞ ¼ 0 so it has the two roots x ¼ a and x ¼ b To find the numerical values of a and b we need to use the formula for finding the roots of a general quadratic equation. Can you recall what it is? If not, then refer to Frame 14 of Programme F.6 in Engineering Mathematics, Eighth Edition. The solution to the quadratic equation ax2 þ bx þ c ¼ 0 is . . . . . . . . . . . . The answer is in the next frame

Numerical solutions of equations and interpolation

x¼

b 

3

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b2  4ac 2a

3

So the roots of x2 þ x þ 1 ¼ 0 are . . . . . . . . . . . . Next frame pffiffiffi 1 3 x¼ j 2 2

4

Because

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi b2  4ac 1  1  4 ¼ a ¼ b ¼ c ¼ 1 and so x ¼ 2 2a pffiffiffi 3 1 ¼ j 2 2 b 

This quadratic equation has two distinct complex roots. Notice that the two roots form a complex conjugate pair – each is the complex conjugate of the other. Whenever a polynomial with real coefficients ai has a complex root it also has the complex conjugate as another root. pffiffiffi So given that x ¼ 2 þ j 5 is one root of a quadratic equation with real coefficients then the other root is . . . . . . . . . . . . pffiffiffi x ¼ 2  j 5 Because

pffiffiffi pffiffiffi The complex conjugate of x ¼ 2 þ j 5 is x ¼ 2  j 5 and complex roots of a polynomial equation with real coefficients always appear as conjugate pairs. The quadratic equation with these two roots is . . . . . . . . . . . .

5

4

Programme 1

6

x2 þ 4x þ 9 ¼ 0 Because If x ¼ a and x ¼ b are the roots of a quadratic equation then ðx  aÞðx  bÞ ¼ 0 gives the quadratic equation. That is ðx  aÞðx  bÞ ¼ x2  ða þ bÞx þ ab ¼ 0. pffiffiffi pffiffiffi Here, the two roots are x ¼ 2 þ j 5 and x ¼ 2  j 5 so that  h h pffiffiffii pffiffiffii x  2 þ j 5 x  2  j 5 ¼ 0 h pffiffiffi pffiffiffii h pffiffiffiih pffiffiffii That is x2  x 2 þ j 5  2  j 5 þ 2 þ j 5 2  j 5 ¼ 0. So x2 þ 4x þ 9 ¼ 0. Notice that the coefficients are . . . . . . . . . . . .

7

Real

Relations between the coefficients and the roots of a polynomial equation Let , ,  be the roots of px3 þ qx2 þ rx þ s ¼ 0. Then, writing the expression px3 þ qx2 þ rx þ s in terms of , ,  gives   q 2 r s 3 2 3 px þ qx þ rx þ s ¼ p x þ x þ x þ p p p ¼ ............

8

pðx  Þðx  Þðx  Þ Therefore

  q r s px3 þ qx2 þ rx þ s ¼ p x3 þ x2 þ x þ p p p ¼ pðx  Þðx  Þðx  Þ  ¼ p x2  ½ þ  x þ  ðx  Þ  ¼ p x3  ½ þ x2 þ x  x2 þ ½ þ x    ¼ p x3  ½ þ  þ x2 þ ½ þ  þ x  Þ

Therefore, equating coefficients (a)  þ  þ  ¼ . . . . . . . . . . . . (b)  þ  þ  ¼ . . . . . . . . . . . . (c)  ¼ . . . . . . . . . . . .

Numerical solutions of equations and interpolation

q (a)  ; p

(b)

r ; p

5

(c) 

9

s p

This, of course, applies to a cubic equation. Let us extend this to a more general equation. So on to the next frame In general, if 1 , 2 , 3 . . . n are roots of the equation p0 x þ p1 x n

then sum of sum of sum of sum of

þ p2 x

n2

þ . . . þ pn1 x þ pn

10

ðp0 ¼ 6 0Þ p1 the roots ¼ p0 p2 products of the roots, two at a time ¼ p0 p3 products of the roots, three at a time ¼  p0 pn products of the roots, n at a time ¼ ð1Þn : p0 n1

¼0

So for the equation 3x4 þ 2x3 þ 5x2 þ 7x  4 ¼ 0, if , , ,  are the four roots, then (a)  þ  þ  þ  ¼ . . . . . . . . . . . . (b)  þ  þ  þ  þ  þ  ¼ . . . . . . . . . . . . (c)  þ  þ  þ  ¼ . . . . . . . . . . . . (d)  ¼ . . . . . . . . . . . . 2 (a)  ; 3

(b)

5 ; 3

7 (c)  ; 3

(d) 

4 3

Now for a problem or two on the same topic. Example 1 Solve the equation x3  8x2 þ 9x þ 18 ¼ 0 given that the sum of two of the roots is 5. Using the same approach as before, if , ,  are the roots, then (a)  þ  þ  ¼ . . . . . . . . . . . . (b)  þ  þ  ¼ . . . . . . . . . . . . (c)  ¼ . . . . . . . . . . . .

11

6

Programme 1

12

(a) 8; (b) 9; (c) 18 So we have  þ  þ  ¼ 8 ; 5þ ¼8 Also

Let  þ  ¼ 5 ; ¼3

 ¼ 18 ð3Þ ¼ 18 ;  ¼ 6  þ  ¼ 5 ;  ¼ 5   ; ð5  Þ ¼ 6

2  5  6 ¼ 0

; ð  6Þð þ 1Þ ¼ 0

;  ¼ 1 or 6 ;  ¼ 6 or  1

Roots are x ¼ 1, 3, 6

13

Example 2 Solve the equation 2x3 þ 3x2  11x  6 ¼ 0 given that the three roots form an arithmetic sequence. Let us represent the roots by ða  kÞ, a, ða þ kÞ Then the sum of the roots ¼ 3a ¼ . . . . . . . . . . . . and the product of the roots ¼ aða  kÞða þ kÞ ¼ . . . . . . . . . . . .

14

3 3a ¼  ; 2

; a¼ If k ¼

1 2

5 2

If k ¼ 

5 2



  1 1  k2 ¼ 3 2 4

1 a¼ ; 2 1 a¼ ; 2

aða þ kÞða  kÞ ¼

; k¼

a  k ¼ 3;

aþk¼

ak¼

a þ k ¼ 3

2;

6 ¼3 2

5 2

2

1 ; required roots are 3,  , 2 2 Here is a similar one. Example 3 Solve the equation x3 þ 3x2  6x  8 ¼ 0 given that the three roots are in geometric sequence. a This time, let the roots be , a, ak k a a Then þ a þ ak ¼ . . . . . . . . . . . . and ðaÞðakÞ ¼ . . . . . . . . . . . . k k

Numerical solutions of equations and interpolation

sum of roots ¼ 3;

7

product of roots ¼ 8

15

It then follows that the roots are . . . . . . . . . . . ., . . . . . . . . . . . ., . . . . . . . . . . . . 4, 2,

16

1

The working rests on the relationships between the roots and the coefficients, i.e. if , ,  are the roots of the cubic equation ax3 þ bx2 þ cx þ d ¼ 0 then (a)

 þ  þ  ¼ ............

(b)

 þ  þ  ¼ . . . . . . . . . . . .

(c)

 ¼ . . . . . . . . . . . . b (a)  ; a

(b)

c ; a

(c) 

d a

17

In each of the three examples reconstruct the cubic to confirm that they are correct. Now on to the next stage

Cubic equations The Fundamental Theorem of Algebra tells us that every cubic expression f ðxÞ ¼ ax þ bx þ cx þ d 3

2

can be written as a product of three linear factors f ðxÞ ¼ aðx  r1 Þðx  r2 Þðx  r3 Þ Consequently, every cubic equation f ðxÞ ¼ aðx  r1 Þðx  r2 Þðx  r3 Þ ¼ 0 has three roots which may be distinct or coincident and which may be real or complex. However, because complex roots of a polynomial with real coefficients always appear in complex conjugate pairs we can say that every such cubic equation has at least one . . . . . . . . . . . .

18

8

Programme 1

19

at least one real root To find the value of this real root we can employ a formula equivalent to the formula used to find the two roots of the general quadratic. This is called Tartaglia’s method but before we can proceed to look at that we must first consider how to transform the general cubic to its reduced form. Next frame

20

Transforming a cubic to reduced form In every case, an equation of the form y3 þ py2 þ qy þ r ¼ 0 can be converted into the reduced form x3 þ ax2 þ b ¼ 0 by the substitution p y¼x . 3 The example will demonstrate the method. Example 4 Express y3 þ 3y2 þ 5y þ 8 ¼ 0 in reduced form. p 3 Substitute y ¼ x  ¼ x  ¼ x  1. The equation then becomes 3 3 ðx  1Þ3 þ 3ðx  1Þ2 þ 5ðx  1Þ þ 8 ¼ 0 ðx3  3x2 þ 3x  1Þ þ 3ðx2  2x þ 1Þ þ 5ðx  1Þ þ 8 ¼ 0 which simplifies to . . . . . . . . . . . .

21

x3 þ 2x þ 5 ¼ 0

Tartaglia’s solution for a real root In the sixteenth century, Tartaglia discovered that a root of the cubic equation x3 þ ax þ b ¼ 0, where a > 0, is given by ( rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi)1=3 ( rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi)1=3 b a3 b2 b a3 b2 þ þ þ   x¼  þ 2 2 27 4 27 4 That looks pretty formidable, but it is a good deal easier than it appears. Notice rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b a3 b2 þ that and occur twice and it is convenient to evaluate these first and 2 27 4 then substitute the results in the main expression for x.

Numerical solutions of equations and interpolation

9

Example 5 Find a real root of x3 þ 2x þ 5 ¼ 0. b ¼ 2:5 Here, a ¼ 2, b ¼ 5 ; 2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi a3 b2 8 25 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 6:5463 ¼ 2:5586 þ þ ¼ 27 4 27 4 Then x ¼ ð2:5 þ 2:5586Þ1=3 þ ð2:5  2:5586Þ1=3 ¼ 0:3884  1:7166 ¼ 1:3282

x ¼ 1:328

Once we have a real root, the equation can be reduced to a quadratic and the remaining two roots determined: x ¼ 0:664 þ j 1:823 and x ¼ 0:664  j 1:823 (see Engineering Mathematics, Eighth Edition, Programme F.6). Example 6 Determine a real root of 2x3 þ 3x  4 ¼ 0. This is first written x3 þ 1:5x  2 ¼ 0 ; a ¼ 1:5, b ¼ 2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b a3 b2 Now you can evaluate and þ and so determine 2 27 4 x ¼ ............

22

0.8796 Because ( rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi)1=3 b a3 b2 þ ¼ f2:06066g1=3 ¼ 1:2725 and  þ 2 27 4 ( rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi)1=3 b a3 b2 þ ¼ f0:6066g1=3 ¼ 0:3929,   2 27 4 therefore x ¼ 1:2725  0:3929 ¼ 0:8796 Note: If you wish to find the real root of a cubic of the form x3 þ ax þ b ¼ 0 where a < 0 then it is best that you resort to numerical methods. Read on. Next frame

10

Programme 1

Numerical methods 23

The methods that we have used so far to solve quadratic equations and to find the real root of a cubic equation are called analytic methods. These analytic methods used straightforward algebraic techniques to develop a formula for the answer. The numerical value of the answer can then be found by simple substitution of numbers for the variables in the formula. Unfortunately, general polynomial equations of order five or higher cannot by solved by analytic methods. Instead, we must resort to what are termed numerical methods. The simplest method of finding the solution to the equation f ðxÞ ¼ 0 is the bisection method.

Bisection The bisection method of finding a solution to the equation f ðxÞ ¼ 0 consists of Finding a value of x, say x ¼ a, such that f ðaÞ < 0 Finding a value of x, say x ¼ b, such that f ðbÞ > 0 The solution to the equation f ðxÞ ¼ 0 must then lie between a and b. Furthermore, it must lie either in the first half of the interval between a and b or in the second half. f (x) f (b) > 0 f ([a + b]/2) a

f (a) < 0

a+b 2

b

x

Find the value of f ð½a þ b=2Þ – that is halfway between a and b. If f ð½a þ b=2Þ > 0 then the solution lies in the first half and if f ð½a þ b=2Þ < 0 then it lies in the second half. This procedure is repeated, narrowing down the width of the interval by a half each time. An example should clarify all this. Example 7 Find the positive value of x that satisfies the equation x2  2 ¼ 0. Firstly we note that if x ¼ 1 then x2  2 < 0, and that if x ¼ 2 then x2  2 > 0, so the solution that we seek must lie between 1 and 2. We look for the . . . . . . . . . . . .

Numerical solutions of equations and interpolation

11

24

The mid-point between 1 and 2 which is 1.5 Now, when x ¼ 1:5, x2  2 ¼ 0:25 > 0 so the solution must lie between . . . . . . . . . . . .

25

1 and 1.5 The mid-point between 1 and 1.5 is 1.25. When x ¼ 1:25, x2 2 ¼ 0:4375 < 0 so the solution must lie between . . . . . . . . . . . .

26

1.25 and 1.5 The mid-point between 1.25 and 1.5 is 1.375. We now evaluate x2  2 at this point and determine in which half interval the solution lies. This process is repeated and the following table displays the results. In each block of six numbers the first column lists the end points of the interval and the mid-point. The second column contains the respective values f ðxÞ ¼ x2  2. Construct the table as follows. (a) For each block of six numbers copy the last number in the first column into the second place of the first column of the following block. This represents the centre point of the previous interval. (b) For each block of six numbers copy the number that represents the other end point of the new interval from the first column into the first place of the first column of the following block. Look at the signs in the second column of the first block to decide which is the appropriate number. a b ða þ bÞ=2 a b ða þ bÞ=2 a b ða þ bÞ=2 a b ða þ bÞ=2

1.0000 1.0000 2.0000 0.2500

1.5000 0.2500 1.2500 0.4375

1.5000 0.2500 1.2500 0.4375 1.3750 0.1094

1.5000 0.2500 1.3750 0.1094 1.4375 0.0664

1.3750 0.1094 1.4375 0.0664 1.4063 0.0225

1.4375 0.0664 1.4063 0.0225 1.4219 0.0217

1.4063 0.0225 1.4219 0.0217 1.4141 0.0004

1.4219 0.0217 1.4141 0.0004 1.4180 0.0106

1.4141 0.0004 1.4180 0.0106 1.4160 0.0051

1.4141 0.0004 1.4160 0.0051 1.4150 0.0023

1.4141 0.0004 1.4150 0.0023 1.4146 0.0010

1.4141 0.0004 1.4146 0.0010 1.4143 0.0003

1.4141 0.0004 1.4143 0.0003 1.4142 0.0001

1.4143 0.0003 1.4142 0.0001 1.4142 0.0001

1.4142 0.0001 1.4142 0.0001 1.4142 0.0000

2.0000 1.5000

1.0000 1.0000

The final result to four decimal places is x ¼ 1:4142 which is the correct answer to that level of accuracy – but it has taken a lot of activity to produce it. A much faster way of solving this equation is to use an iteration formula that was first devised by Newton. Next frame

12

Programme 1

Numerical solution of equations by iteration 27

The process of finding the numerical solution to the equation f ðxÞ ¼ 0 by iteration is performed by first finding an approximate solution and then using this approximate solution to find a more accurate solution. This process is repeated until a solution is found to the required level of accuracy. For example, Newton showed that the square root of 2 can be found by a process called iteration. that is, if x2 ¼ 2 then 2x2 ¼ x2 þ 2 and so, dividing through by 2x gives x 1 x¼ þ 2 x pffiffiffi If an approximate value of 2 is then used to evaluate the right-hand side of this pffiffiffi equation this then becomes a better approximation to 2. This better value is then used to evaluate pffiffiffi anew the right-hand side to produce an even better approximation to 2. This procedure is repeated until the required level of accuracy is obtained. This is the process of iteration and it is expressed by the formula: xi 1 i ¼ 0, 1, 2, . . . xiþ1 ¼ þ 2 xi where x0 is the approximation pffiffiffi that starts the iteration off. So, to find a succession of approximate values of 2, each of increasing accuracy, we proceed as follows. Let x0 ¼ 1:5 – found by the first stage of the bisection method. Then   1 2 x0 þ ¼ 0:5ð1:5 þ 2=1:5Þ ¼ 1:4166 . . . x1 ¼ 2 x0 This value is then used to find x2 . By rounding x1 to 1.4167, the value of x2 is found to be . . . . . . . . . . . .

28

x2 ¼ 1:4142 Because

  1 2 x1 þ x2 ¼ ¼ 0:5ð1:4167 þ 2=1:4167Þ ¼ 1:4142 . . . 2 x1

This has achieved the same level of accuracy as the bisection method in just two steps.

Using a spreadsheet This simple iteration procedure is more efficiently performed using a spreadsheet. If the use of a spreadsheet is a totally new experience for you then you are referred to Programme F.4 of Engineering Mathematics, Eighth Edition where the spreadsheet is introduced as a tool for constructing graphs of functions. If you have a limited knowledge then you will be able to follow the text from here. The spreadsheet we shall be using here is Microsoft Excel, though all commercial spreadsheets possess the equivalent functionality.

Numerical solutions of equations and interpolation

13

Open your spreadsheet and in cell A1 enter n and press Enter. In this first column we are going to enter the iteration numbers. In cell A2 enter the number 0 and press Enter. Place the cell highlight in cell A2 and highlight the block of cells A2 to A7 by holding down the mouse button and wiping the highlight down to cell A7. Click the Edit command on the Command bar and point at Fill from the drop-down menu. Select Series from the next drop-down menu and accept the default Step value of 1 by clicking OK in the Series window. The cells A3 to A7 fill with . . . . . . . . . . . . the numbers 1 to 5

29

In cell B1 enter the letter x – this column is going to contain the successive xvalues obtained by iteration. In cell B2 enter the value of x0 , namely 1.5. In cell B3 enter the formula ¼ 0.5*(B2+2/B2) The number that appears in cell B3 is then . . . . . . . . . . . . 1.41666667

30

Place the cell highlight in cell B3, click the command Edit on the Command bar and select Copy from the drop-down menu. You have now copied the formula in cell B3 onto the Clipboard. Highlight the cells B3 to B7 and then click the Edit command again but this time select Paste from the drop-down menu. The cells B4 to B7 fill with numbers to provide the display ............ n 0 1 2 3 4 5

x 1 .5 1.41666667 1.41421569 1.41421356 1.41421356 1.41421356

By using the various formatting facilities provided by the spreadsheet the display can be amended to provide the following n 0 1 2 3 4 5

x 1.500000000000000 1.416666666666670 1.414215686274510 1.414213562374690 1.414213562373090 1.414213562373090

The number of decimal places here is 15, which is far greater than is normally required but it does demonstrate how effective a spreadsheet can be.

31

14

Programme 1

Notice that to find a value accurate to a given number of decimal places or significant figures it is sufficient to repeat the iterations until there is no change in the result from one iteration to the next. Save your spreadsheet under some suitable name such as Newton because you may wish to use it again. Now we shall look at this spreadsheet a little more closely

32

Relative addresses Place the cell highlight in cell B3 and the formula that it contains is ¼ 0.5*(B2+2/B2). Now place the cell highlight in cell B4 and the formula there is ¼ 0.5*(B3+2/B3). Why the difference? When you enter the cell address B2 in the formula in B3 the spreadsheet understands that to mean the contents of the cell immediately above. It is this meaning that is copied into cell B4 where the cell immediately above is B3. If you wish to refer to a specific cell in a formula then you must use an absolute address. Place the cell highlight in cell C1 and enter the number 2. Now place the cell highlight in cell B3 and re-enter the formula ¼ 0.5*(B2+$C$1/B2) and copy this into cells B4 to B7. The numbers in the second column have not changed but the formulas have because in cells B3 to B7 the same reference is made to cell C1. The use of the dollar signs has indicated an absolute address. So why would we do this? Change the number in cell C1 to 3 to obtain the display . . . . . . . . . . . .

33

n 0 1 2 3 4 5

x 1.500000000000000 1.750000000000000 1.732142857142860 1.732050810014730 1.732050807568880 1.732050807568880

pffiffiffi These are the iterated values of 3 – the square root of the contents of cell C1. We can now use the same spreadsheet to find the square root of any positive number. Newton’s iterative procedure to find the square root of a positive number is a special case of the Newton–Raphson procedure to find the solution of the general equation f ðxÞ ¼ 0, and we shall look at this in the next frame.

Numerical solutions of equations and interpolation

15

Newton–Raphson iterative method Consider the graph of y ¼ f ðxÞ as shown. Then the x-value at the point A, where the graph crosses the x-axis, gives a solution of the equation f ðxÞ ¼ 0.

34

y

If P is a point on the curve near to A, then x ¼ x0 is an approximate value of the root of f ðxÞ ¼ 0, the error of the approximation being given by AB. x

Let PQ be the tangent to the curve as P, crossing the x-axis at Q ðx1 ; 0Þ. Then x ¼ x1 is a better approximation to the required root.

 PB dy From the diagram, ¼ i.e. the value of the derivative of y at the QB dx P point P, x ¼ x0 . PB ¼ f 0 ðx0 Þ and PB ¼ f ðx0 Þ QB PB f ðx0 Þ ¼ ¼ h (say) ; QB ¼ 0 f ðx0 Þ f 0 ðx0 Þ

;

f ðx0 Þ f 0 ðx0 Þ If we begin, therefore, with an approximate value (x0 ) of the root, we can determine a better approximation (x1 ). Naturally, the process can be repeated to improve the result still further. Let us see this in operation. x1 ¼ x0  h

; x1 ¼ x0 

On to the next frame

35

Example 1 The equation x3  3x  4 ¼ 0 is of the form f ðxÞ ¼ 0 where f ð1Þ < 0 and f ð3Þ > 0 so there is a solution to the equation between 1 and 3. We shall take this to be 2, by bisection. Find a better approximation to the root. We have f ðxÞ ¼ x3  3x  4 ; f 0 ðxÞ ¼ 3x2  3 If the first approximation is x0 ¼ 2, then f ðx0 Þ ¼ f ð2Þ ¼ 2

and

f 0 ðx0 Þ ¼ f 0 ð2Þ ¼ 9

A better approximation x1 is given by f ðx0 Þ x0 3  3x0  4 ¼ x  0 f 0 ðx0 Þ 3x0 2  3 ð2Þ x1 ¼ 2  ¼ 2:22 9 ; x0 ¼ 2; x1 ¼ 2:22 x1 ¼ x0 

16

Programme 1

If we now start from x1 we can get a better approximation still by repeating the process. x2 ¼ x1 

f ðx1 Þ x1 3  3x1  4 ¼ x1  0 f ðx1 Þ 3x1 2  3

Here x1 ¼ 2:22

36

f ðx1 Þ ¼ . . . . . . . . . . . . ; f ðx1 Þ ¼ 0:281;

f 0 ðx1 Þ ¼ . . . . . . . . . . . . f 0 ðx1 Þ ¼ 11:785

Then x2 ¼ . . . . . . . . . . . .

37

x2 ¼ 2:196 Because 0:281 x2 ¼ 2:22  : ¼ 2:196 11 79 Using x2 ¼ 2:196 as a starter value, we can continue the process until successive results agree to the desired degree of accuracy. x3 ¼ . . . . . . . . . . . .

38

x3 ¼ 2:196 Because f 0 ðx2 Þ ¼ f 0 ð2:196Þ ¼ 11:467 f ðx2 Þ ¼ f ð2:196Þ ¼ 0:002026; f ðx2 Þ 0:00203 ¼ 2:196 (to 4 sig. fig.) ; x3 ¼ x2  0 ¼ 2:196  f ðx2 Þ 11:467 The process is simple but effective and can be repeated again and again. Each repetition, or iteration, usually gives a result nearer to the required root. In general xnþ1 ¼ . . . . . . . . . . . .

39

xnþ1 ¼ xn 

f ðxn Þ f 0 ðxn Þ

Tabular display of results Open your spreadsheet and in cells A1 to D1 enter the headings n, x, f ðxÞ and f 0 ðxÞ Fill cells A2 to A6 with the numbers 0 to 4 In cell B2 enter the value for x0 , namely 2 In cell C2 enter the formula for f ðx0 Þ, namely ¼ B2^3 – 3*B2 – 4 and copy into cells C3 to C6

Numerical solutions of equations and interpolation

17

In cell D2 enter the formula for f 0 ðx0 Þ, namely ¼ 3*B2^2 – 3 and copy into cells D3 to D6 In cell B3 enter the formula for x1 , namely ¼ B2 – C2/D2 and copy into cells B4 to B6. The final display is . . . . . . . . . . . . to 6 dp n 0 1 2 3 4

x

f ðxÞ

f 0 ðxÞ

2 2.222222 2.196215 2.195823 2.195823

2  0.307270 0.004492 0.000001 0.000000

9 11.814815 11.470081 11.464922 11.464920

40

As soon as the number in the second column is repeated then we know that we have arrived at that particular level of accuracy. The required root is therefore x ¼ 2:195823 to 6 dp. Save the spreadsheet so that it can be used as a template for other such problems. Now let us have another example. Next frame

41

Example 2 The equation x þ 2x  5x  1 ¼ 0 is of the form f ðxÞ ¼ 0 where f ð1Þ < 0 and f ð2Þ > 0 so there is a solution to the equation between 1 and 2. We shall take this to be x ¼ 1:5. Use the Newton–Raphson method to find the root to six decimal places. Use the previous spreadsheet as a template and make the following amendments 3

2

In cell B2 enter the number . . . . . . . . . . . . 1 .5

42

Because That is the value of x0 that is used to start the iteration In cell C2 enter the formula . . . . . . . . . . . . ¼ B2^3 + 2*B2^2 – 5*B2 – 1 Because That is the value of f ðx0 Þ ¼ x0 3 þ 2x0 2  5x0  1. Copy the contents of cell C2 into cells C3 to C5. In cell D2 enter the formula . . . . . . . . . . . .

43

18

Programme 1

44

¼ 3*B2^2 + 4*B2 – 5 Because That is the value of f 0 ðx0 Þ ¼ 3x0 2 þ 4x0  5. Copy the contents of cell D2 into cells D3 to D5. In cell B2 the formula remains the same as . . . . . . . . . . . .

45

¼ B2 – C2/D2 The final display is then . . . . . . . . . . . .

46

n 0 1 2 3 4

x 1 .5 1.580645 1.575792 1.575773 1.575773

f ðxÞ 0.625 0.042798 0.000159 0.000000 0.000000

f 0 ðxÞ 7.75 8.817898 8.752524 8.752280 8.752280

The repetition of the x-value ensures that the solution x ¼ 1:575773 is accurate to 6 dp. Now do one completely on your own. Next frame

47

Example 3 The equation 2x3  7x2  x þ 12 ¼ 0 has a root near to x ¼ 1:5. Use the Newton– Raphson method to find the root to six decimal places. The spreadsheet solution produces . . . . . . . . . . . .

48

x ¼ 1:686141 to 6 dp Because Fill cells A2 to A6 with the numbers 0 to 4 In cell B2 enter the value for x0 , namely 1.5 In cell C2 enter the formula for f ðx0 Þ, namely ¼ 2*B2^3 – 7*B2^2 – B2 + 12 and copy into cells C3 to C6 In cell D2 enter the formula for f 0 ðx0 Þ, namely ¼ 6*B2^2 – 14*B2 – 1 and copy into cells D3 to D6 In cell B3 enter the formula for x1 , namely ¼ B2 – C2/D2 and copy into cells B4 to B6. The final display is . . . . . . . . . . . . (formatted to 6 dp)

Numerical solutions of equations and interpolation

n 0 1 2 3 4

f ðxÞ 1.5 0.073275 0.000286 4.46E-09

x 1 .5 1.676471 1.686103 1.686141 1.686141

0

19

49

f 0 ðxÞ 8.5 7.60727 7.54778 7.54755 7.54755

As soon as the number in the second column is repeated then we know that we have arrived at that particular level of accuracy. The required root is therefore x ¼ 1:686141 to 6 dp. First approximations The whole process hinges on knowing a ‘starter’ value as first approximation. If we are not given a hint, this information can be found by either (a) applying the remainder theorem if the function is a polynomial (b) drawing a sketch graph of the function. Example 4 Find the real root of the equation x3 þ 5x2  3x  4 ¼ 0 correct to six significant figures. Application of the remainder theorem involves substituting x ¼ 0, x ¼ 1, x ¼ 2, etc. until two adjacent values give a change in sign. f ðxÞ ¼ x3 þ 5x2  3x  4 f ð0Þ ¼ 4; f ð1Þ ¼ 1; f ð1Þ ¼ 3

f(x)

–1

The sign changes between f ð0Þ and f ð1Þ. There is thus a root between x ¼ 0 and x ¼ 1. Therefore choose x ¼ 0:5 as the first approximation and then proceed as before.

3

0

x

–4

Complete the table and obtain the root x ¼ ............

50

x ¼ 0:675527 The final spreadsheet display is n 0 1 2 3 4

x 0.500000 0.689655 0.675597 0.675527 0.675527

f ðxÞ 1.375000 0.119070 0.000582 0.000000 0.000000

f 0 ðxÞ 7.250000 8.469679 8.386675 8.386262 8.386262

20

51

Programme 1

Example 5 Solve the equation ex þ x  2 ¼ 0 giving the root to 6 significant figures. It is sometimes more convenient to obtain a first approximation to the required root from a sketch graph of the function, or by some other graphical means. In this case, the equation can be rewritten as ex ¼ 2  x and we therefore sketch graphs of y ¼ ex and y ¼ 2  x. x

0.2

0 .4

0.6

0.8

1

x

1.22

1.49

1.82

2.23

2.72

2x

1.8

1 .6

1.4

1.2

1

e

y y = ex

y=2–x

x

It can be seen that the two curves cross over between x ¼ 0:4 and x ¼ 0:6. Approximate root x ¼ 0:4 f ðxÞ ¼ ex þ x  2

f 0 ðxÞ ¼ ex þ 1

x ¼ ............ Finish it off

52

x ¼ 0:442854 The final spreadsheet display is n 0 1 2 3

x 0.400000 0.443412 0.442854 0.442854

f ðxÞ 0.108175 0.001426 0.000000 0.000000

f 0 ðxÞ . 2 491825 2.558014 2.557146 2.557146

Note: There are times when the normal application of the Newton–Raphson method fails to converge to the required root. This is particularly so when f 0 ðx0 Þ is very small, so before we leave this section let us consider this difficulty.

Numerical solutions of equations and interpolation

21

53

Modified Newton–Raphson method If the slope of the curve at x ¼ x0 is small, the value of the second approximation x ¼ x1 may be further from the exact root at A than the first approximation. f (x)

x0

x1

x

If x ¼ x0 is an approximate solution of f ðxÞ ¼ 0 and x ¼ x0  h is the exact solution then f ðx0  hÞ ¼ 0. By Taylor’s series f ðx0  hÞ ¼ f ðx0 Þ  hf 0 ðx0 Þ þ

h2 00 f ðx0 Þ  . . . ¼ 0 2!

(a) If we assume that h is small enough to neglect terms of the order h2 and higher then this equation can be written as f ðx0  hÞ  f ðx0 Þ  hf 0 ðx0 Þ, that is f ðx0 Þ  hf 0 ðx0 Þ  0 and so f ðx0 Þ f ðx0 Þ giving x1 ¼ x0  0 as a better approximation h 0 f ðx0 Þ f ðx0 Þ to the solution of f ðxÞ ¼ 0. This is, of course, the relationship we have been using and which may fail when f 0 ðxÞ is small. Notice: h is positive unless the sign of f ðx0 Þ is the opposite of the sign of f 0 ðx0 Þ. (b) If we consider the first three terms then h2 00 f ðx0 Þ  0, that is 2! 2f ðx0 Þ  2hf 0 ðx0 Þ þ h2 f 00 ðx0 Þ  0

f ðx0  hÞ  f ðx0 Þ  hf 0 ðx0 Þ þ

Since f 0 ðx0 Þ is small we shall assume that we can neglect it so sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ h¼ f 00 ðx0 Þ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ That is h ¼ unless the signs of f ðx0 Þ and f 0 ðx0 Þ are different when it f 00 ðx0 Þ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ is h ¼  . We use this result only when f 0 ðx0 Þ is found to be very f 00 ðx0 Þ small. Having found x1 from x0 we then revert to the normal relationship f ðx0 Þ for subsequent iterations. xnþ1 ¼ xn  0 f ðx0 Þ Note this

22

54

Programme 1

Example 6 The equation x3  1:3x2 þ 0:4x  0:03 ¼ 0 is known to have a root near x ¼ 0:7. Determine the root to 6 significant figures. We start off in the usual way. f ðxÞ ¼ x3  1:3x2 þ 0:4x  0:03 f 0 ðxÞ ¼ 3x2  2:6x þ 0:4 and complete the first line of the normal table. n

xn

0

0.7

f ðxn Þ

f 0 ðxn Þ

h¼

f ðxn Þ f 0 ðxn Þ

xnþ1 ¼ xn  h

Complete just the first line of values.

55

We have n

xn

f ðxn Þ

f 0 ðxn Þ

0

0.7

0.044

0.05

h¼

f ðxn Þ f 0 ðxn Þ

0.88

xnþ1 ¼ xn  h 1.58

We notice at once that (a) The value of x1 is well away from the approximate value (0.7) of the root. (b) The value of f 0 ðx0 Þ is small, i.e. 0.05. To obtain x1 we therefore make a fresh start, using the modified relationship x1 ¼ . . . . . . . . . . . . sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ x1 ¼ x0  f 00 ðx0 Þ

56

f ðxÞ ¼ x3  1:3x2 þ 0:4x  0:03 ¼ ½ðx  1:3Þx þ 0:4x  0:03 ¼ ð3x  2:6Þx þ 0:4 f 0 ðxÞ ¼ 3x2  2:6x þ 0:4 00 f ðxÞ ¼ 6x  2:6

n

x0

f ðx0 Þ

0

0.7

0.044

Complete the line.

f 00 ðx0 Þ

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ h¼ f 00 ðx0 Þ

x1 ¼ x0  h

Numerical solutions of equations and interpolation

00

n

x0

f ðx0 Þ

f ðx0 Þ

0

0.7

0.044

1 .6

23

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ h¼ f 00 ðx0 Þ

x1 ¼ x0  h

0.2345

57

0.9345

Note that in the expression x1 ¼ x0  h, we chose the positive sign since at x0 ¼ 0:7, f ðx0 Þ is negative and the slope f 0 ðx0 Þ is positive. 0.7

x1

x

– 0.044

Having established that x1 ¼ 0:9345, we now revert to the usual xnþ1 ¼ xn 

f ðxn Þ f 0 ðxn Þ

for the rest of the calculation. Complete the table therefore and obtain the required root.

58

The final spreadsheet display is n 0 1 2 3 4 5

x 0.7 0.934521 0.892801 0.887387 0.887298 0.887298

f ðxÞ 0.044 0.024625 0.002544 4.02E-05 1.06E-08 9.16E-16

f 0 ðxÞ . 0 05 0.590233 0.469997 0.45516 0.454919 0.454919

f 00 ðxÞ 1 .6

Therefore to six decimal places the required root is x ¼ 0:887298. Note that we only used the modified method to find x1 . After that the normal relationship is used.

And now . . . To date our task has been to find a value of x that satisfies an explicit equation f ðxÞ ¼ 0. This is quite general because any equation in x can be written in this form. For example, the equation sin x ¼ x  e3x can always be written as sin x  x þ e3x ¼ 0 and then approached by one of the methods that we have discussed so far. What we want to do now is to work the other way – given a value of x, to find the corresponding value of f ðxÞ. If f ðxÞ is given explicitly then this is no problem, it is just a matter of substituting the value of x in the formula and working it out. However, many times a function exists but it is not given explicitly, as in the case of a set of readings compiled as a result of an experiment or practical test. We shall consider this problem in the following frames. Next frame

24

Programme 1

Interpolation 59

When a function is defined by a well-understood expression such as f ðxÞ ¼ 4x3  3x2 þ 7 or f ðxÞ ¼ 5 sinðexp½xÞ the values of the dependent variable f ðxÞ corresponding to given values of the independent variable x can be found by direct substitution. Sometimes, however, a function is not defined in this way but by a collection of ordered pairs of numbers. Example 1 A function can be defined by the following set of data: x

f ðxÞ

1

4

2

14

3

40

4

88

5

164

6

274

Intermediate values, for example, x ¼ 2:5, can be estimated by a process called interpolation. The value of f ð2:5Þ will clearly lie between 14 and 40, the function values for x ¼ 2 and x ¼ 3.

Purely as an estimate,

f ð2:5Þ ¼ . . . . . . . . . . . . What do you suggest?

60

27

Linear interpolation If you gave the result as 27, you no doubt agreed that x ¼ 2:5 is midway between x ¼ 2 and x ¼ 3, and that therefore f ð2:5Þ would be midway between 14 and 40, i.e. 27. This is the simplest form of interpolation, but there is no evidence that there is a linear relationship between x and f ðxÞ, and the result is therefore suspect. Of course, we could have estimated the function value at x ¼ 2:5 by other means, such as ............

Numerical solutions of equations and interpolation

25

61

by drawing the graph of f ðxÞ against x

Graphical interpolation We could, indeed, plot the graph of f ðxÞ against x and, from it, estimate the value of f ðxÞ at x ¼ 2:5. f(x)

This method is also approximate and can be time consuming. f ð2:5Þ  26

x

In what follows we shall look at interpolation using finite differences, which work well and quickly when the values of x are equally spaced. When the values of x are not equally spaced we need to resort to the more involved algebraic method called Lagrangian interpolation (which could also be used for equally spaced points). Next frame

Gregory–Newton interpolation formula using forward finite differences x .. .

f ðxÞ .. .

x0 x1 .. .

f ðx0 Þ f ðx1 Þ .. .

f0 ¼ f ðx1 Þ  f ðx0 Þ

We assume that x0 , x1 , . . . are distinct, equally spaced apart, and x0 < x1 < . . .

For each pair of consecutive function values, f ðx0 Þ and f ðx1 Þ, in the table, the forward difference f0 is calculated by subtracting f ðx0 Þ from f ðx1 Þ. This difference is written in a third column of the table, midway between the lines carrying f ðx0 Þ and f ðx1 Þ. x

f ðxÞ

1

4

2

14

3 .. .

40 .. .

f 10 26

Complete the table for the data given in Frame 59 which then becomes . . . . . . . . . . . .

62

26

Programme 1

63

x

f ðxÞ

1

4

2

14

3

40

4

88

5

164

6

274

f 10 26 48 76 110

We now form a fourth column, the forward differences of the values of f , denoted by 2 f , and again written midway between the lines of f . These are the second forward differences of f ðxÞ. So the table then becomes . . . . . . . . . . . .

64

x

f ðxÞ

1

4

2

14

3

40

4

88

5

164

6

274

f

2 f

10 16

26

22

48

28

76

34

110

A further column can now be added in like manner, giving the third differences, denoted by 3 f , so that we then have . . . . . . . . . . . .

65

x

f ðxÞ

1

4

2

14

3

40

4

88

5

164

6

274

f

2 f

3 f

10 26 48 76 110

16 22 28

6 6 6

34

Notice that the table has now been completed, for the third differences are constant and all subsequent differences would be zero. Now we shall see how to use the table. So move on

Numerical solutions of equations and interpolation

27

To find f ð2:5Þ

x0 h

x1

66

xp

x

f ðxÞ

1

4

2

14

3

40

4

88

5

164

6

274

f

2 f

10

3 f

16

26

22

48

28

76

34

110

6 6 6

We have to find f ð2:5Þ. Therefore denote x ¼ 2 as x0 x ¼ 2:5 as xp x ¼ 3 as x1 Let h ¼ the constant range between successive values of x, i.e. h ¼ x1  x0 xp  x0 Express ðxp  x0 Þ as a fraction of h, i.e. p ¼ , 0 0, say x ¼ b. The solution to the equation f ðxÞ ¼ 0 must then lie between a and b. Furthermore, it must lie either in the first half of the interval between a and b or in the second half.

Numerical solutions of equations and interpolation

5 Numerical solution of equations by iteration The process of finding the numerical solution to the equation f ðxÞ ¼ 0 by iteration is performed by first finding an approximate solution and then using this approximate solution to find a more accurate solution. This process is repeated until a solution is found to the required level of accuracy. 6 Using a spreadsheet Iteration procedures are more efficiently performed using a spreadsheet. 7 Newton–Raphson iteration method If x ¼ x0 is an approximate solution to the equation f ðxÞ ¼ 0, a better approximation x ¼ x1 is given by x1 ¼ x0 

f ðx0 Þ f ðxn Þ , and in general xnþ1 ¼ xn  0 f 0 ðx0 Þ f ðxn Þ

8 Modified Newton–Raphson iteration method If, in the Newton–Raphson procedure f 0 ðx0 Þ is sufficiently small enough to cause the value of x1 to be a worse approximation to the solution than x0 , then x1 is obtained from the relationship sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ x1 ¼ x0  f 00 ðx0 Þ Subsequent iterations then use xnþ1 ¼ xn 

f ðxn Þ . f 0 ðxn Þ

9 Interpolation Linear Graphical 10 Gregory–Newton interpolation formulas using forward finite differences fp ¼ f0 þ pf0 þ

pðp  1Þ 2 pðp  1Þðp  2Þ 3  f0 þ  fo þ    2! 3!

11 Gauss interpolation formulas using central finite differences Gauss forward formula pðp  1Þ 2 ðp þ 1Þpðp  1Þ 3  f0 þ  f0þ1 2 2! 3! ðp þ 1Þpðp  1Þðp  2Þ 4 þ  f0 þ    4! Gauss backward formula fp ¼ f0 þ pf0þ1 þ 2

ðp þ 1Þp 2 ðp þ 1Þpðp  1Þ 3  f0 þ  f01 2 2! 3! ðp þ 2Þðp þ 1Þpðp  1Þ 4  f0 þ    þ 4!

fp ¼ f0 þ pf01 þ 2

39

40

Programme 1

12 Gregory–Newton interpolation formula using backward finite differences fp ¼ f0 þ pf1 þ

pðp þ 1Þ 2 pðp þ 1Þðp þ 2Þ 3  f2 þ  f3 þ    2! 3!

13 Lagrange interpolation If the straight line pðxÞ ¼ a0 þ a1 x passes through the two points ðx0 , f ðx0 ÞÞ and ðx1 , f ðx1 ÞÞ, where a0 and a1 are constants, then the interpolation polynomial (straight line) for this line can be written as x  x1 x  x0 f ðx0 Þ þ f ðx1 Þ pðxÞ ¼ x0  x1 x1  x0 The quadratic interpolating polynomial that passes through the three points ðx0 , f ðx0 ÞÞ, ðx1 , f ðx1 ÞÞ and ðx2 , f ðx2 ÞÞ can be written as pðxÞ ¼

ðx  x1 Þðx  x2 Þ ðx  x0 Þðx  x2 Þ f ðx0 Þ þ f ðx1 Þ ðx0  x1 Þðx0  x2 Þ ðx1  x0 Þðx1  x2 Þ ðx  x0 Þðx  x1 Þ þ f ðx2 Þ ðx2  x0 Þðx2  x1 Þ

The cubic interpolating polynomial that passes through the four data points ðx0 , f ðx0 ÞÞ, ðx1 , f ðx1 ÞÞ, ðx2 , f ðx2 ÞÞ and ðx3 , f ðx3 ÞÞ can be written as pðxÞ ¼

ðx  x1 Þðx  x2 Þðx  x3 Þ f ðx0 Þ ðx0  x1 Þðx0  x2 Þðx0  x3 Þ ðx  x0 Þðx  x2 Þðx  x3 Þ f ðx1 Þ þ ðx1  x0 Þðx1  x2 Þðx1  x3 Þ ðx  x0 Þðx  x1 Þðx  x3 Þ f ðx2 Þ þ ðx2  x0 Þðx2  x1 Þðx2  x3 Þ ðx  x0 Þðx  x1 Þðx  x2 Þ f ðx3 Þ þ ðx3  x0 Þðx3  x1 Þðx3  x2 Þ

The interpolating polynomial that passes through n þ 1 data points is pðxÞ ¼

ðx  x1 Þðx  x2 Þð  Þðx  xn Þ f ðx0 Þ ðx0  x1 Þðx0  x2 Þð  Þðx0  xn Þ ðx  x0 Þðx  x2 Þð  Þðx  xn Þ þ f ðx1 Þ þ    ðx1  x0 Þðx1  x2 Þð  Þðx1  xn Þ ðx  x0 Þðx  x1 Þð  Þðx  xn1 Þ f ðxn Þ  þ ðxn  x0 Þðxn  x1 Þð  Þðxn  xn1 Þ

Numerical solutions of equations and interpolation

41

Can you? Checklist 1 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can:

Frames

. Appreciate the Fundamental Theorem of Algebra?

1

to

3

4

to

6

7

to

17

18

to

20

Yes

No

. Find the two roots of a quadratic equation and recognize that for polynomial equations with real coefficients complex roots exist in complex conjugate pairs?

Yes

No

. Use the relationships between the coefficients and the roots of a polynomial equation to find the roots of the polynomial?

Yes

No

. Transform a cubic equation to reduced form?

Yes

No

. Use Tartaglia’s solution to find the real root of a cubic equation?

Yes

21 and 22

No

. Find the solution of the equation f ðxÞ ¼ 0 by the method of bisection?

Yes

to

33

34

to

52

53

to

58

59

to

61

No

. Understand the meaning of interpolation and use simple linear and graphical interpolation?

Yes

27

No

. Use the modified Newton–Raphson method to find the first approximation when the derivative is small?

Yes

26

No

. Solve equations using the Newton–Raphson iterative method?

Yes

to

No

. Solve equations involving a single real variable by iteration and use a spreadsheet for efficiency?

Yes

23

No

42

Programme 1

. Use the Gregory–Newton interpolation formula using forward and backward differences for equally spaced domain points?

Yes

62

to

73

74

to

77

78

to

83

No

. Use the Gauss interpolation formulas using central differences for equally spaced domain points?

Yes

No

. Use Lagrange interpolation when the domain points are not equally spaced?

Yes

No

Test exercise 1 pffiffiffi 1 Given that x ¼ 1 þ j 3 is one root of a quadratic equation with real coefficients, find the other root and hence the quadratic equation. 2 Solve the cubic equation 2x3  7x2  42x þ 72 ¼ 0. 3 Write the cubic 3x3 þ 5x2 þ 3x þ 5 in reduced form and use Tartaglia’s method to find the real root. 4 Use the method of bisection to find a solution to x3  5 ¼ 0 correct to 4 significant figures. 5 Use the Newton–Raphson method to find a positive solution of the following equation, correct to 6 decimal places: cos 3x ¼ x2 . 6 Use the modified Newton–Raphson method to find the solution correct to 6 decimal places near to x ¼ 2 of the equation x3  6x2 þ 13x  9 ¼ 0. 7 Given the table of values x

f ðxÞ

1 2 3 4 5 6

0 19 70 171 340 595

estimate (a) f ð2:5Þ using the Gregory–Newton forward difference formula (b) f ð3:4Þ using the Gauss central difference formula (c) f ð5:6Þ using the Gregory–Newton backward difference formula.

Numerical solutions of equations and interpolation

8 Given the table of values x

f ðxÞ

1

4

2

9

5

108

use Lagrangian interpolation to estimate the value of f ð2:2Þ.

Further problems 1

pffiffiffi 1  j 3 1 þ j and x ¼ pffiffiffi are two roots of a quartic equation 1 Given that x ¼ 2 2 with real coefficients, find the other two roots and hence the quartic equation.

2 Solve the equation x3  5x2  8x þ 12 ¼ 0, given that the sum of two of the roots is 7. 3 Find the values of the constants p and q such that the function f ðxÞ ¼ 2x3 þ px2 þ qx þ 6 may be exactly divisible by ðx  2Þðx þ 1Þ. 4 If f ðxÞ ¼ 4x4 þ px3  23x2 þ qx þ 11 and when f ðxÞ is divided by 2x2 þ 7x þ 3 the remainder is 3x þ 2, determine the values of p and q. 5 If one root of the equation x3  2x2  9x þ 18 ¼ 0 is the negative of another, determine the three roots. 6 Solve the equation x3  7x2  21x þ 27 ¼ 0, given that the roots form a geometric sequence. 7 Form the equation whose roots are those of the equation x3 þ x2 þ 9x þ 9 ¼ 0 each increased by 2. 8 Form the equation whose roots exceed by 3 the roots of the equation x3  4x2 þ x þ 6 ¼ 0. 9 If the equation 4x3  4x2  5x þ 3 ¼ 0 is known to have two roots whose sum is 2, solve the equation. 10 Solve the equation x3  10x2 þ 8x þ 64 ¼ 0, given that the product of two of the roots is the negative of the third. 11 Form the equation whose roots exceed by 2 those of the equation 2x3  3x2  11x þ 6 ¼ 0. 12 If , ,  are the roots of the equation x3 þ px2 þ qx þ r ¼ 0, prove that 2 þ  2 þ  2 ¼ p2  2q. 13 Using Tartaglia’s solution, find the real root of the equation 2x3 þ 4x  5 ¼ 0 giving the result to 4 significant figures. 14 Solve the equation x3  6x  4 ¼ 0.

43

44

Programme 1

15 Rewrite the equation x3 þ 6x2 þ 9x þ 4 ¼ 0 in reduced form and hence determine the three roots. 16 Show that the equation x3 þ 3x2  4x  6 ¼ 0 has a root between x ¼ 1 and x ¼ 2, and use the Newton–Raphson iterative method to evaluate this root to 4 significant figures. 17 Find the real root of the equations: (a) x3 þ 4x þ 3 ¼ 0

(b) 5x3 þ 2x  1 ¼ 0.

18 Solve the following equations: (a) x3  5x þ 1 ¼ 0

(b) x3 þ 2x  3 ¼ 0

(c) x3  4x þ 1 ¼ 0. 19 Express the following in reduced form and determine the roots: (a) x3 þ 6x2 þ 9x þ 5 ¼ 0 (b) 8x3 þ 20x2 þ 6x  9 ¼ 0 (c) 4x3  9x2 þ 42x  10 ¼ 0. 20 Use the Newton–Raphson iterative method to solve the following. (a) Show that a root of the equation x3 þ 3x2 þ 5x þ 9 ¼ 0 occurs between x ¼ 2 and x ¼ 3. Evaluate the root to four significant figures. (b) Show graphically that the equation e2x ¼ 25x  10 has two real roots and find the larger root correct to four significant figures. (c) Verify that the equation x  cos x ¼ 0 has a root near to x ¼ 0:8 and determine the root correct to three significant figures. (d) Obtain graphically an approximate root of the equation 2 ln x ¼ 3  x. Evaluate the root correct to four significant figures. (e) Verify that the equation x4 þ 5x  20 ¼ 0 has a root at approximately x ¼ 1:8. Determine the root correct to five significant figures. (f) Show that the equation x þ 3 sin x ¼ 2 has a root between x ¼ 0:4 and x ¼ 0:6. Evaluate the root correct to five significant figures. (g) The equation 2 cos x ¼ ex  1 has a real root between x ¼ 0:8 and x ¼ 0:9. Evaluate the root correct to four significant figures. (h) The equation 20x3  22x2 þ 5x  1 ¼ 0 has a root at approximately x ¼ 0:6. Determine the value of the root correct to four significant figures. 21 A polynomial function is defined by the following set of function values x

2

4

6

8

10

y ¼ f ðxÞ

7.00

9.00

97.0

305

681

Find (a) f ð4:8Þ using the Gregory–Newton forward difference formula (b) f ð7:2Þ using the Gauss central difference formula (c) f ð8:5Þ using the Gregory–Newton backward difference formula.

Numerical solutions of equations and interpolation

45

22 For the function f ðxÞ x

4

5

6

7

8

9

10

f ðxÞ

10

12

56

128

234

380

572

Find (a) f ð4:5Þ and f ð6:4Þ using the Gregory–Newton forward difference formula (b) f ð7:1Þ and f ð8:9Þ using the Gregory–Newton backward difference formula. 23

x

2

4

6

8

10

12

f ðxÞ

9

35

231

675

1463

2691

For the function defined in the table above, evaluate (a) f ð2:6Þ and (b) f ð7:2Þ. 24 A function f ðxÞ is defined by the following table x

4

2

0

2

4

6

8

f ðxÞ

277

51

1

17

147

533

1319

Find (a) f ð3Þ and f ð1:6Þ using the Gregory–Newton forward difference formula (b) f ð0:2Þ and f ð3:1Þ using the Gauss central difference formula (c) f ð4:4Þ and f ð7Þ using the Gregory–Newton backward difference formula. 25 Given the table of values x

f ðxÞ

1 3

2.71828 0.04979

5

0.00674

use Lagrangian interpolation to find the value of f ð3:4Þ. 26 Given the table of values x

f ðxÞ

6 7 .2

0.801153 0.82236

9

0.73922 0.994808

13

use Lagrangian interpolation to find the value of f ð8Þ.

46

Programme 1

27 Given the table of values x

f ðxÞ

2

2.63906 2.48491

0 5 6

1.94591 1.79176

use Lagrangian interpolation to find the values of (a) f ð0:8Þ (b) f ð0:8Þ (c) f ð5:5Þ.

Frames 1 to 90

Programme 2

Laplace transforms 1 Learning outcomes When you have completed this Programme you will be able to: Obtain the Laplace transforms of simple standard expressions Use the first shift theorem to find the Laplace transform of a simple expression multiplied by an exponential Find the Laplace transform of a simple expression multiplied or divided by a variable Use partial fractions to find the inverse Laplace transform Use the ‘cover up’ rule Use the Laplace transforms of derivatives to solve differential equations Use the Laplace transform to solve simultaneous differential equations

Prerequisite: Engineering Mathematics (Eighth Edition) Programme 27 Introduction to Laplace transforms 47

48

Programme 2

Introduction 1

The solution of a linear, ordinary differential equation with constant coefficients such as the second-order equation af 00 ðtÞ þ bf 0 ðtÞ þ cf ðtÞ ¼ gðtÞ can be solved by first obtaining the general form for the expression f ðtÞ. This general form will contain a number of integration constants whose values can be found by applying the appropriate boundary conditions (see Engineering Mathematics, Eighth Edition, Programme 26). A more systematic way of solving such equations is to use the Laplace transform which converts the differential equation into an algebraic equation and has the added advantage of incorporating the boundary conditions from the beginning. Furthermore, in situations where f ðtÞ represents a function with discontinuities, the Laplace transform method can succeed where other methods fail. Laplace transform techniques also provide powerful tools in numerous fields of technology such as Control Theory where a knowledge of the system transfer function is essential and where the Laplace transform comes into its own. Let us see what it is all about. (For a more detailed introduction see Engineering Mathematics, Eighth Edition, Programme 27.)

Laplace transforms The Laplace transform of an expression f ðtÞ is denoted by Lff ðtÞg and is defined as the semi-infinite integral ð1 f ðtÞest dt ð1Þ Lff ðtÞg ¼ t¼0

The parameter s is assumed to be positive and large enough to ensure that the integral converges. In more advanced applications s may be complex and in such cases the real part of s must be positive and large enough to ensure convergence. In determining the transform of an expression, you will appreciate that the limits of the integral are substituted for t, so that the result will be an expression in s. Therefore ð1 f ðtÞest dt ¼ FðsÞ Lff ðtÞg ¼ t¼0

Make a note of this general definition: then we can apply it

Laplace transforms 1

So we have

49

Lff ðtÞg ¼

ð1

2

f ðtÞest dt ¼ FðsÞ

0

Example 1 To find the Laplace transform of f ðtÞ ¼ a (constant).

st 1 ð1 1 e a st Lfag ¼ ae dt ¼ a ¼  est 0 s s 0 0 a a ¼  f0  1g ¼ s s a ðs > 0Þ ; Lfag ¼ s

ð2Þ

Example 2 To find the Laplace transform of f ðtÞ ¼ eat (a constant). As with all cases, we multiply f ðtÞ by est and integrate between t ¼ 0 and t ¼ 1. ð1 ð1 eat est dt ¼ eðsaÞt dt ; Lfeat g ¼ 0

0

¼ ............ Finish it off. Lfeat g ¼

Because Lfeat g ¼

ð1

eat est dt ¼

0

ð1

eðsaÞt dt ¼



0

1 eðsaÞt ðs  aÞ 0

1 1 f0  1g ¼ ¼ sa sa 1 ðs > aÞ ; Lfeat g ¼ sa

ð3Þ

So we already have two standard transforms Lfag ¼

a s

and ;

1 sa Lf4g ¼ . . . . . . . . . . . . ;

Lfeat g ¼

Lf5g ¼ . . . . . . . . . . . . ;

3

1 sa

Lfe4t g ¼ . . . . . . . . . . . . Lfe2t g ¼ . . . . . . . . . . . .

50

Programme 2

4

4 ; s 5 Lf5g ¼  ; s Lf4g ¼

1 s4 1 Lfe2t g ¼ sþ2 Lfe4t g ¼

Note that, as we said earlier, the Laplace transform is always an expression in s. Now for some more examples

5

Example 3 To find the Laplace transform of f ðtÞ ¼ sin at. We could, of course, apply the definition and evaluate ð1 sin at  est dt Lfsin atg ¼ 0

using integration by parts. However, it is much shorter if we use the fact that e j ¼ cos  þ j sin  so that sin  is the imaginary part of e j , written iðe j Þ. The function sin at can therefore be written iðe jat Þ so that ð1 ð1 ejat est dt ¼ i eðsjaÞt dt Lfsin atg ¼ Lfiðe jat Þg ¼ i 0 0 (

 1 ) eðsjaÞt 1 ½0  1 ¼i  ¼i ðs  jaÞ ðs  jaÞ 0  1 ¼i s  ja We can rationalize the denominator by multiplying top and bottom by ............

6

s þ ja  s þ ja a ; Lfsin atg ¼ i 2 ¼ 2 s þ a2 s þ a2 a ; Lfsin atg ¼ 2 s þ a2

ð4Þ

We can use the same method to determine Lfcos atg since cos at is the real part of e jat , written > ¼ ; sy  x = s1 1 > > ; sx þ y ¼ sþ1

ð1Þ

 and y by the normal algebraic method. Eliminating (c) We now solve these for x y we have 1 s1 s ¼ sy þ s2 x sþ1  2 1 s2  2s  1 ¼ ; s2 þ 1 x  ¼ s þ 1 s  1 ðs þ 1Þðs  1Þ ¼ sy  x

¼ ; x

s2  2s  1 ðs  1Þðs þ 1Þðs2 þ 1Þ

Representing this in partial fractions gives . . . . . . . . . . . .

Laplace transforms 1

83

1 1 1 1 s 1 ¼ :  : þ þ x 2 s  1 2 s þ 1 s2 þ 1 s2 þ 1

81

Because s2  2s  1 A B Cs þ D þ þ  ðs  1Þðs þ 1Þðs2 þ 1Þ s  1 s þ 1 s2 þ 1   ; s2  2s  1 ¼ Aðs þ 1Þ s2 þ 1 þ Bðs  1Þ s2 þ 1

¼ x

þ ðs  1Þðs þ 1ÞðCs þ DÞ Putting s ¼ 1 and s ¼ 1 gives A ¼  12 and B ¼  12. Comparing coefficients of s3 and the constant terms gives C ¼ 1 and D ¼ 1. 1 1 1 1 sþ1 :  : þ 2 s  1 2 s þ 1 s2 þ 1 ; x ¼ ............ ¼ ; x

x ¼  12 et  12 et þ cos t þ sin t

82

 to obtain y and hence y, in the We now revert to equations (1) and eliminate x same way. Do this on your own. y ¼ ............ y ¼ 12 et þ 12 et  cos t þ sin t Here is the working. 9 s > > s2 y  s x¼ s  1= 1 > > ; y þ sx ¼ sþ1  s 1 s2 þ 2s  1 þ ¼ ; s2 þ 1 y ¼ s  1 s þ 1 ðs  1Þðs þ 1Þ s2 þ 2s  1 A B Cs þ D þ þ  ðs  1Þðs þ 1Þðs2 þ 1Þ s  1 s þ 1 s2 þ 1   ; s2 þ 2s  1 ¼ Aðs þ 1Þ s2 þ 1 þ Bðs  1Þ s2 þ 1

; y ¼

þ ðs  1Þðs þ 1ÞðCs þ DÞ Putting s ¼ 1 and s ¼ 1 gives A ¼ 12 and B ¼ 12. Equating coefficients of s3 and the constant terms gives C ¼ 1 and D ¼ 1. 1 1 1 1 s 1 : þ :  þ 2 s  1 2 s þ 1 s2 þ 1 s2 þ 1 1 1 ; y ¼ et þ et  cos t þ sin t 2 2 ; y ¼

83

84

Programme 2

So the results are 1 t e þ et þ sin t þ cos t ¼ sin t þ cos t  cosh t 2 1 t y ¼ e þ et þ sin t  cos t ¼ sin t  cos t þ cosh t 2 ; x ¼ sin t þ cos t  cosh t; y ¼ sin t  cos t þ cosh t x¼

Simultaneous equations are all solved in much the same way. Here is another. Example 2 Solve the equations 2y_  6y þ 3x ¼ 0 3x_  3x  2y ¼ 0 given that at t ¼ 0, x ¼ 1 and y ¼ 3. Expressing these in Laplace transforms, we have ............ ............

84

¼0 2ðsy  y0 Þ  6y þ 3x   2y ¼ 0   x0 Þ  3x 3ðsx Then we insert the initial conditions and simplify, obtaining ............ ............

85

 þ ð2s  6Þy ¼ 6 3x   2y ¼ 3 ð3s  3Þx

(1) (2)

 (a) To find x  þ ð2s  6Þy ¼ 6 3x   ð2s  6Þy ¼ 3ðs  3Þ ðs  3Þð3s  3Þx  ¼ 3s  9 þ 6 ½ðs  3Þð3s  3Þ þ 3x  2 ; 3s  12s þ 12  x ¼ 3s  3  2 s  4s þ 4  x¼s1

(1) (2)  ðs  3Þ Adding,

¼ ; x

s1 2

ðs  2Þ



A B Aðs  2Þ þ B þ ¼ 2 s  2 ðs  2Þ ðs  2Þ2

; s  1 ¼ Aðs  2Þ þ B ¼ ; x

1 1 þ s  2 ðs  2Þ2

giving

A ¼ 1 and

B¼1

; x ¼ e2t þ te2t

(b) Going back to equations (1) and (2), we can find y. y ¼ ............

Laplace transforms 1

85

  y ¼ 12 6e2t þ 3te2t

86

 we get Because, eliminating x ( ) ( ) 6s  9 1 A B 1 Aðs  2Þ þ B y ¼ þ  ¼ 2 2ðs  2Þ2 2 s  2 ðs  2Þ2 ðs  2Þ2 ; 6s  9 ¼ Aðs  2Þ þ B ( ) 1 6 3 þ ; y ¼ 2 s  2 ðs  2Þ2

; A ¼ 6;

B¼3

  ; y ¼ 12 6e2t þ 3te2t

Simultaneous second-order equations are solved in like manner. Again, with all these solutions it is a worthwhile exercise to substitute the solution back into the differential equation to verify that the solution is correct.

87

Example 3 If x and y are functions of t, solve the equations € þ 2x  y ¼ 0 x €y þ 2y  x ¼ 0 given that at t ¼ 0, x0 ¼ 4; y0 ¼ 2; x1 ¼ 0; y1 ¼ 0.  2   y ¼ 0   sx0  x1 þ 2x s x We start off as usual with  2 ¼0 s y  sy0  y1 þ 2y  x and Inserting the initial conditions, we have   4s þ 2x   y ¼ 0 s2 x ¼0 s2 y  2s þ 2y  x  and hence x. Simplifying these we can eliminate y to obtain x x ¼ ............ pffiffiffi  x ¼ 3 cos t þ cos 3t Because  2   y ¼ 4s s þ2 x  2  þ s þ 2 y ¼ 2s x Eliminating y and simplifying gives ¼ x

4s3 þ 10s ðs2 þ 1Þðs2 þ 3Þ

4s3 þ 10s As þ B Cs þ D  þ 2 s þ3 ðs2 þ 1Þðs2 þ 3Þ s2 þ 1   3 2 2 ; 4s þ 10s ¼ s þ 3 ðAs þ BÞ þ s þ 1 ðCs þ DÞ ¼ ; x

88 ð1Þ ð2Þ

86

Programme 2

Equating coefficients of like powers of s

3 4¼AþC ; AþC¼4 s ½CT 0 ¼ 3B þ D

; 3B þ D ¼ 0

Putting s ¼ 1, Putting s ¼ 1

14 ¼ 4A þ 4B þ 2C ¼ 2D

; 2A þ 2B þ C þ D ¼ 7

14 ¼ 4A þ 4B  2C þ 2D ; 2A  2B þ C  D ¼ 7

Putting C ¼ 4  A and D ¼ 3B in the last two leads to A ¼ ............; C ¼ ............;

89

B ¼ ............; D ¼ ............ A ¼ 3;

B ¼ 0; C ¼ 1; D ¼ 0

3s s þ 2 þ1 s þ3 ; x ¼ ............

¼ ; x

s2

pffiffiffi  x ¼ 3 cos t þ cos 3t

90

To find y we could return to equations (1) and (2) and repeat the process,  so as to obtain y and hence y. eliminating x But always keep an eye on the original equations, the first of which is € þ 2x  y ¼ 0 x € þ 2x. Therefore, in this particular case, y ¼ x So all we have to do is to differentiate x twice and substitute pffiffiffi  x ¼ 3 cos t þ cos 3t pffiffiffi  pffiffiffi x_ ¼ 3 sin t  3 sin 3t pffiffiffi  € ¼ 3 cos t  3 cos 3t x pffiffiffi  pffiffiffi  ; y ¼ 3 cos t  3 cos 3t þ 6 cos t þ 2 cos 3t pffiffiffi  ; y ¼ 3 cos t  cos 3t which is a good deal quicker. So, as we have seen, the method of solving differential equations by Laplace transforms follows a general routine. (a) Express the equation in Laplace transforms (b) Insert the initial conditions (c) Simplify to obtain the transform of the solution (d) Rewrite the final transform in partial fractions (e) Determine the inverse transforms and, by now, you are fully aware of the importance of partial fractions!

Laplace transforms 1

87

That brings us to the end of this particular Programme. We shall continue our study of Laplace transforms in the next Programme. Meanwhile, be sure you are familiar with the items listed in the Review summary that follows, and respond to the questions in the Can you? checklist. You will then have no difficulty with the Test exercise and the Further problems provide additional practice.

Review summary 2 1 Laplace transform Lff ðtÞg ¼

ð1

f ðtÞest dt ¼ FðsÞ.

0

2 Table of transforms f ðtÞ a eat sin at cos at sinh at cosh at tn

Lff ðtÞg ¼ FðsÞ a s 1 sa a s2 þ a2 s s2 þ a2 a s2  a2 s s2  a2 n! snþ1

(n a positive integer)

3 Linearity of the Laplace transform (a) The transform of a sum (or difference) of expressions is the sum (or difference) of the individual transforms. That is Lff ðtÞ  gðtÞg ¼ Lff ðtÞg  LfgðtÞg. (b) The transform of an expression that is multiplied by a constant is the constant multiplied by the transform of the expression. That is Lfkf ðtÞg ¼ kLff ðtÞg First shift theorem   If Lff ðtÞg ¼ FðsÞ, then L eat f ðtÞ ¼ Fðs þ aÞ.

4 Theorem 1

5 Theorem 2

Multiplying by t

If Lff ðtÞg ¼ FðsÞ, then Lftf ðtÞg ¼ 

d fFðsÞg. ds

88

Programme 2

6 Theorem 3

Dividing by t

 ð1 f ðtÞ ¼ Fð Þ d If Lff ðtÞg ¼ FðsÞ, then L t s  f ðtÞ provided that Lim exists. t t!0

7 Inverse transform If Lff ðtÞg ¼ FðsÞ, then L1 fFðsÞg ¼ f ðtÞ. 8 Rules of partial fractions (a) The numerator must be of lower degree than the denominator. If not, divide out. (b) Factorize the denominator into its prime factors. (c) A linear factor ðs þ aÞ gives a partial fraction

A where A is a constant sþa

to be determined. (d) A repeated factor ðs þ aÞ2 gives (e) Similarly ðs þ aÞ3 gives

A B þ . s þ a ðs þ aÞ2

A B C þ þ . 2 s þ a ðs þ aÞ ðs þ aÞ3

 A quadratic factor s2 þ ps þ q gives

Ps þ Q . s2 þ ps þ q  2 (g) A repeated quadratic factor s2 þ ps þ q gives (f)

Ps þ Q Rs þ T . þ s2 þ ps þ q ðs2 þ ps þ qÞ2 9 The ‘cover up’ rule The ‘cover up’ rule often enables the values of the constant coefficients to be written down almost on sight. However, this method only works when the denominator of the original fraction has non-repeated, linear factors. 10 Table of inverse transforms FðsÞ a s 1 sþa n! snþ1 1 sn

f ðtÞ a eat tn t n1 ðn  1Þ!

(n a positive integer)

Laplace transforms 1

FðsÞ a þ a2 s s2 þ a2 a 2 s  a2 s s2  a2 s2

89

f ðtÞ sin at cos at sinh at cosh at

By the first shift theorem If FðsÞ is the Laplace transform of f ðtÞ then Fðs þ aÞ is the Laplace transform of eat f ðtÞ. 11 Laplace transforms of derivatives  Lfxg ¼ x  dx _ ¼ sx_  x0 L ¼ Lfxg dt ( ) d2 x € g ¼ sx   sx0  x1 etc. ¼ Lfx L dt 2 where x0 ¼ value of x at t ¼ 0 x1 ¼ value of

dx at t ¼ 0, etc. dt

12 Solution of differential equations (a) Rewrite the equation in terms of Laplace transforms. (b) Insert the given initial conditions. (c) Rearrange the equation algebraically to give the transform of the solution. (d) Express the transform in standard forms by partial fractions. (e) Determine the inverse transforms to obtain the particular solution. 13 Simultaneous differential equations Convert the simultaneous differential equations into simultaneous algebraic equations by taking the Laplace transform of each equation in turn. Insert the initial values. Solve the simultaneous algebraic equations in the usual manner and take the inverse Laplace transform of the algebraic solutions to find the solutions to the simultaneous differential equations.

90

Programme 2

Can you? Checklist 2 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can:

Frames

. Obtain the Laplace transforms of simple standard expressions?

1

to

14

15

to

17

18

to

26

27

to

42

43

to

46

47

to

79

80

to

90

Yes

No

. Use the first shift theorem to find the Laplace transform of a simple expression multiplied by an exponential?

Yes

No

. Find the Laplace transform of a simple expression multiplied or divided by a variable?

Yes

No

. Use partial fractions to find the inverse Laplace transform?

Yes

No

. Use the ‘cover up’ rule?

Yes

No

. Use the Laplace transforms of derivatives to solve differential equations?

Yes

No

. Use the Laplace transform to solve simultaneous differential equations?

Yes

No

Test exercise 2 1 Determine the Laplace transforms of the following functions. (a) 3e4t  5e4t (b) sin 4t þ cos 4t (c) t 3 þ 2t 2  t þ 4 et  e2t . (d) e2t cos 5t (e) t sin 3t (f) t 2 Determine the inverse transforms of the following. s5 s2 þ 3s  7 (b) (a) ðs  3Þðs  4Þ ðs  1Þðs2 þ 2Þ (c)

s2  3s  4 2

ðs  3Þðs  1Þ

(d)

2s2  6s  1 . ðs  3Þðs2  2s þ 5Þ

Laplace transforms 1

91

3 Solve the following equations by Laplace transforms. dx þ 3x ¼ e2t given that x ¼ 2 when t ¼ 0 (a) dt (b) 3x_  6x ¼ sin 2t given that x ¼ 1 when t ¼ 0 €  7x_ þ 12x ¼ 2 (c) x

given that at t ¼ 0, x ¼ 1 and x_ ¼ 5

€  2x_ þ x ¼ tet (d) x

given that at t ¼ 0, x ¼ 1 and x_ ¼ 0.

4 Solve the following pair of simultaneous equations where x and y are functions of t and given that at t ¼ 0, x ¼ 4 and y ¼ 1. x_ þ y_ þ x þ 2y ¼ e3t x_ þ 3x þ 5y ¼ 5e2t .

Further problems 2 1 Determine the Laplace transforms of the following functions. (a) e4t cos 2t (b) t sin 2t  3 2 (c) t þ 4t þ 5 (d) e3t t 2 þ 4 (e) t 2 cos t

sinh 2t . t

(f)

2 Determine the inverse transforms of the following. 2s  6 5s  8 (a) (b) ðs  2Þðs  4Þ sðs  4Þ (c)

s2  2s þ 3 3

(d)

2  11s ðs  2Þðs2 þ 2s þ 2Þ

ðs  2Þ s s5 (e) 2 (f) 2 . s þ 4s þ 20 ðs þ 1Þðs2 þ 4Þ

In Questions 3 to 11, solve the equations by Laplace transforms. 3 x_  4x ¼ 8 4 3x_  4x ¼ sin 2t €  2x_ þ x ¼ 2ðt þ sin tÞ 5 x €  6x_ þ 8x ¼ e3t 6 x € þ 9x ¼ cos 2t 7 x €  2x_ þ 5x ¼ e2t 8 x € þ 4x_ þ 4x ¼ t 2 þ e2t 9 x

at t ¼ 0, x ¼ 2. at t ¼ 0, x ¼ 13 .

at t ¼ 0, x ¼ 6, x_ ¼ 5. at t ¼ 0, x ¼ 0, x_ ¼ 2.

at t ¼ 0, x ¼ 1, x_ ¼ 3. at t ¼ 0, x ¼ 0, x_ ¼ 1. at t ¼ 0, x ¼ 12 , x_ ¼ 0. at t ¼ 0, x ¼ x_ ¼ 0.

€ þ 8x_ þ 32x ¼ 32 sin 4t 10 x € þ 25x ¼ 10ðcos 5t  2 sin 5tÞ at t ¼ 0, x ¼ 1, x_ ¼ 2. 11 x

92

Programme 2

In Questions 12 to 17, solve the pairs of simultaneous equations by Laplace transforms. ) 12 y_ þ 3x ¼ e2t at t ¼ 0, x ¼ y ¼ 0. x_  3y ¼ e2t 13 4x_  2y_ þ 10x  5y ¼ 0 at t ¼ 0, y ¼ 4, x ¼ 2. y_  18x þ 15y ¼ 10 14 x_  2y_  3x þ 6y ¼ 12 at t ¼ 0, x ¼ 12, y ¼ 8. 3y_ þ 5x þ 2y ¼ 16 15 2x_ þ 3y_ þ 7x ¼ 14t þ 7 at t ¼ 0, x ¼ y ¼ 0. 5x_  3y_ þ 4x þ 6y ¼ 14t  14 16 2x_ þ 2x þ 3y_ þ 6y ¼ 56et  3et at t ¼ 0, x ¼ 8, y ¼ 3. x_  2x  y_  3y ¼ 21et  7et ) €  €y þ x  y ¼ 5e2t 17 x at t ¼ 0, x ¼ 1, y ¼ 2, x_ ¼ 0, y_ ¼ 2. 2x_  y_ þ y ¼ 0 18 Find an expression for x in terms of t, given that €y  x_ þ 2x ¼ 10 sin 2t y_ þ 2y þ x ¼ 0 19

and when t ¼ 0, x ¼ y ¼ 0.

€ þ 8x þ 2y ¼ 24 cos 4t If x and 4€y þ 2x þ 5y ¼ 0 and at t ¼ 0, x ¼ y ¼ 0, x_ ¼ 1, y_ ¼ 2, determine an expression for y in terms of t.

20 Solve completely, the pair of simultaneous equations € þ 12€y þ 6x ¼ 0 5x € þ 16€y þ 6y ¼ 0 5x given that, at t ¼ 0, x ¼ 74, y ¼ 1, x_ ¼ 0, y_ ¼ 0.

Frames 1 to 46

Programme 3

Laplace transforms 2 Learning outcomes When you have completed this Programme you will be able to: Use the Heaviside unit step function to ‘switch’ expressions on and off Obtain the Laplace transform of expressions involving the Heaviside unit step function Solve linear, constant coefficient ordinary differential equations with piecewise continuous right-hand sides Understand what is meant by the convolution of two functions and use the convolution theorem to find the inverse transform of a product of transforms

93

94

Programme 3

Introduction 1

In the previous Programme, we dealt with the Laplace transforms of continuous functions of t. In practical applications, it is convenient to have a function which, in effect, ‘switches on’ or ‘switches off’ a given term at pre-described values of t. This we can do with the Heaviside unit step function.

Heaviside unit step function Consider a function that maintains a zero value for t < c and a unit value for t c.

f (t)

c

f ðtÞ ¼ 0

for t < c

f ðtÞ ¼ 1

for t c

t

This function is the Heaviside unit step function and is denoted by f ðtÞ ¼ uðt  cÞ where the c indicates the value of t at which the function changes from a value of 0 to a value of 1. Thus, the function f (t)

t

is denoted by f ðtÞ ¼ . . . . . . . . . . . .

2

f ðtÞ ¼ uðt  4Þ Similarly, the graph of f ðtÞ ¼ 2uðt  3Þ is ............

Laplace transforms 2

95

3 f (t)

t

So uðt  cÞ has just two values for t < c, uðt  cÞ ¼ . . . . . . . . . . . . for t c, uðt  cÞ ¼ . . . . . . . . . . . . t < c, uðt  cÞ ¼ 0;

t c, uðt  cÞ ¼ 1

Unit step at the origin u(t)

If the unit step occurs at the origin, then c ¼ 0 and f ðtÞ ¼ uðt  cÞ becomes f ðtÞ ¼ uðtÞ

t

i.e.

uðtÞ ¼ 0 for t < 0 uðtÞ ¼ 1 for t 0.

Effect of the unit step function The graph of f ðtÞ ¼ t 2 is, of course, as shown. f (t)

t

Remembering the definition of uðt  cÞ, the graph of f ðtÞ ¼ uðt  2Þ  t 2 is ............

4

96

Programme 3

5

f(t)

t

For

t < 2, uðt  2Þ ¼ 0

; uðt  2Þ  t 2 ¼ 0  t 2 ¼ 0

t 2, uðt  2Þ ¼ 1

; uðt  2Þ  t 2 ¼ 1  t 2 ¼ t 2

So the function uðt  2Þ suppresses the function t 2 for all values of t up to t ¼ 2 and ‘switches on’ the function t 2 at t ¼ 2. Now we can sketch the graphs of the following functions. (a) f ðtÞ ¼ sin t

for 0 < t < 2

(b) f ðtÞ ¼ uðt  =4Þ  sin t These give . . . . . . . . . . . .

for 0 < t < 2.

and . . . . . . . . . . . .

6 f (t) t –

f (t)

–

t

4

That is, the graph of f ðtÞ ¼ uðt  =4Þ  sin t is the graph of f ðtÞ ¼ sin t but suppressed for all values prior to t ¼ =4. If we sketch the graph of f ðtÞ ¼ sinðt  =4Þ we have

f(t) t 4

Since uðt  cÞ has the effect of suppressing a function for t < c, then the graph of f ðtÞ ¼ uðt  =4Þ  sinðt  =4Þ is ............

Laplace transforms 2

97

7 f (t) t

4

That is, the graph of f ðtÞ ¼ uðt  =4Þ  sinðt  =4Þ is the graph of f ðtÞ ¼ sin t ðt > 0Þ, shifted =4 units along the t-axis. In general, the graph of f ðtÞ ¼ uðt  cÞ  sinðt  cÞ is the graph of f ðtÞ ¼ sin t ðt > 0Þ, shifted along the t-axis through an interval of c units. Similarly, for t > 0, sketch the graphs of (a) f ðtÞ ¼ et (b) f ðtÞ ¼ uðt  cÞ  et (c) f ðtÞ ¼ uðt  cÞ  eðtcÞ (d) f ðtÞ ¼ et fuðt  1Þ  uðt  2Þg. Arrange the graphs under each other to show the important differences.

(a)

8 1 f (t)

f(t) = e –t

0 (b)

t 1 f (t) 0

(c)

f(t) = u (t – c).e –t c

t

1 f (t) = u (t – c).e –(t – c)

f (t) 0 (d)

c

t

1 f (t) = e –t {u (t – 1) – u(t – 2)}

f (t) 0

1

2

t

In (a), we have the graph of f ðtÞ ¼ et In (b), the same graph is suppressed prior to t ¼ c In (c), the graph of f ðtÞ ¼ et is shifted c units along the t-axis In (d), the graph of f ðtÞ ¼ et is turned on at t ¼ 1 and off at t ¼ 2 because when t 2, uðt  1Þ  uðt  2Þ ¼ 1  1 ¼ 0.

98

Programme 3

Laplace transform of u (t – c ) Lfuðt  cÞg ¼ Because Lfuðt  cÞg ¼

ecs s ð1

est uðt  cÞ dt

0

but est uðt  cÞ ¼ so that Lfuðt  cÞg ¼



0 for 0 < t < c est for t c

ð1

est uðt  cÞ dt ¼

0

¼

est s

ð1

est dt

c

1 ¼ c

esc s

Therefore, the Laplace transform of the unit step at the origin is LfuðtÞg ¼ . . . . . . . . . . . .

9

1 s Because c ¼ 0. ecs s 1 LfuðtÞg ¼ . s

Lfuðt  cÞg ¼

So and

Also from the definition of uðtÞ: Lð1Þ ¼ Lf1  uðtÞg LðtÞ ¼ Lft  uðtÞg Lff ðtÞg ¼ Lff ðtÞ  uðtÞg Make a note of these results: we shall be using them

10

As we have seen, the unit step function uðt  cÞ is often combined with other functions of t, so we now consider the Laplace transform of uðt  cÞ  f ðt  cÞ.

Laplace transforms 2

99

Laplace transform of u ( t – c ) . f ( t – c ) (the second shift theorem) Lfuðt  cÞ  ðf ðt  cÞg ¼ ecs Lff ðtÞg ¼ ecs FðsÞ Because

ð1 Lfuðt  cÞ  f ðt  cÞg ¼ est uðt  cÞ  f ðt  cÞ dt 0  0 for 0 < t < c st but e uðt  cÞ ¼ st for t c e

so that Lfuðt  cÞ  f ðt  cÞg ¼

ð1

est f ðt  cÞ dt

c

We now make the substitution t  c ¼ v so that t ¼ c þ v and dt ¼ dv. Also for the limits, when t ¼ c, v ¼ 0 and when t ! 1, v ! 1. Therefore ð1 esðcþvÞ f ðvÞ dv Lfuðt  cÞ  f ðt  cÞg ¼ 0 ð1 esv f ðvÞ dv ¼ ecs ð1 Now

0

esv f ðvÞ dv has exactly the same value as

0

ð1

est f ðtÞ dt which is, of course,

0

the Laplace transform of f ðtÞ. Therefore Lfuðt  cÞ  f ðt  cÞg ¼ ecs Lff ðtÞg ¼ ecs FðsÞ Lfuðt  cÞ  f ðt  cÞg ¼ ecs  FðsÞ

11

where FðsÞ ¼ Lff ðtÞg

n o   So L uðt  4Þ  ðt  4Þ2 ¼ e4s  FðsÞ where FðsÞ ¼ L t 2   2e4s 4s 2! ¼e ¼ s3 s3 Note that FðsÞ is the transform of t 2 and not of ðt  4Þ2 . In the same way: Lfuðt  3Þ  sinðt  3Þg ¼ . . . . . . . . . . . .

12

e3s þ1

s2

Because Lfuðt  3Þ  sinðt  3Þg ¼ e3s  FðsÞ   1 ; Lfuðt  3Þ  sinðt  3Þg ¼ e3s 2 s þ1

where FðsÞ ¼ Lfsin tg ¼

1 s2 þ 1

100

Programme 3

So now do these in the same way. n o ¼ ............ (a) L uðt  2Þ  ðt  2Þ3 (b) Lfuðt  1Þ  sin 3ðt  1Þg   (c) L uðt  5Þ  eðt5Þ

¼ ............ ¼ ............

(d) Lfuðt  =2Þ  cos 2ðt  =2Þg ¼ . . . . . . . . . . . .

13

Here they are n o   (a) L uðt  2Þ  ðt  2Þ3 ¼ e2s  FðsÞ where FðsÞ ¼ L t 3   3! 6e2s ¼ e2s 4 ¼ 4 s s (b) Lfuðt  1Þ  sin 3ðt  1Þg ¼ es  FðsÞ where FðsÞ ¼ Lfsin 3tg   3 3es ¼ es 2 ¼ 2 s þ9 s þ9 n o   ðt5Þ 5s (c) L uðt  5Þ  e  FðsÞ where FðsÞ ¼ L et ¼e   1 e5s ¼ ¼ e5s s1 s1 (d) Lfuðt  =2Þ  cos 2ðt  =2Þg ¼ es=2  FðsÞ where FðsÞ ¼ Lfcos 2tg   s s  es=2 ¼ 2 ¼ es=2 2 s þ4 s þ4 So Lfuðt  cÞ  f ðt  cÞg ¼ ecs  FðsÞ where FðsÞ ¼ Lff ðtÞg. Written in reverse, this becomes If FðsÞ ¼ Lff ðtÞg, then ecs  FðsÞ ¼ Lfuðt  cÞ  f ðt  cÞg where c is real and positive. This is known as the second shift theorem. Make a note of it: then we will use it

14

If FðsÞ ¼ Lff ðtÞg, then ecs  FðsÞ ¼ Lfuðt  cÞ  f ðt  cÞg This is useful in finding inverse transforms, as we shall now see.

Laplace transforms 2

101

Example 1

e4s . s2 where c ¼ 4 and therefore indicates uðt  4Þ.

Find the function whose transform is The numerator corresponds to ecs 1 Then 2 ¼ FðsÞ ¼ Lftg ; f ðtÞ ¼ t. s  4s 1 e ; L ¼ uðt  4Þ  ðt  4Þ s2

Remember that in writing the final result, f ðtÞ is replaced by ............ f ðt  cÞ Example 2 Determine L1

15



6e2s . s2 þ 4

The numerator contains e2s and therefore indicates . . . . . . . . . . . . uðt  2Þ The remainder of the transform, i.e. ;

  6 2 , can be written as 3 s2 þ 4 s2 þ 4

6 ¼ FðsÞ ¼ Lf. . . . . . . . . . . .g s2 þ 4 Lf3 sin 2tg

; L1

6e2s ¼ ............ s2 þ 4

Because  2s  6e 6 L1 2 ¼ uðt  2Þ  f ðt  2Þ where f ðtÞ ¼ L1 2 s þ4 s þ4 ¼ uðt  2Þ  3 sin 2ðt  2Þ

Determine L1

17



3uðt  2Þ  sin 2ðt  2Þ

Example 3

16



s  es . s2 þ 9 This, in similar manner, is . . . . . . . . . . . .

18

102

Programme 3

19

uðt  1Þ  cos 3ðt  1Þ Because the numerator contains es which indicates uðt  1Þ. s ¼ FðsÞ ¼ Lfcos 3tg Also 2 s þ9 ; f ðtÞ ¼ cos 3t ; f ðt  1Þ ¼ cos 3ðt  1Þ.  s  es ; L1 2 ¼ uðt  1Þ  cos 3ðt  1Þ s þ9 Remember that, having obtained f ðtÞ, the result contains f ðt  cÞ. Here is a short exercise by way of practice. Exercise Determine the inverse transforms of the following. 2e5s s3 3e2s (b) 2 s 1 8e4s (c) 2 s þ4

2s  e3s s2  16 5es (e) s s  es=2 (f) s2 þ 2 (d)

(a)

20

Results – all very straightforward. (a) uðt  5Þ  ðt  5Þ2 (b) 3uðt  2Þ  sinhðt  2Þ (c) 4uðt  4Þ  sin 2ðt  4Þ (d) 2uðt  3Þ  cosh 4ðt  3Þ (e) 5uðt  1Þ pffiffiffi (f) uðt  1=2Þ  cos 2ðt  1=2Þ. Before looking at a more interesting example, let us collect our results together as far as we have gone.

21

The main points are (a) uðt  cÞ ¼ 0

0 (b) Lfuðt  cÞg ¼ = s 1 > > ; LfuðtÞg ¼ s (c) Lfuðt  cÞ  f ðt  cÞg ¼ ecs  FðsÞ where FðsÞ ¼ Lff ðtÞg (d) If FðsÞ ¼ Lff ðtÞg, then ecs  FðsÞ ¼ Lfuðt  cÞg  f ðt  cÞg Now let us apply these to some further examples.

ð1Þ

ð2Þ ð3Þ ð4Þ

Laplace transforms 2

103

Example 1 Determine the expression f ðtÞ for which Lff ðtÞg ¼

3 4es 5e2s  2 þ 2 s s s

We take each term in turn and find its inverse transform.   3 1 ¼ 3L1 ¼ 3 i.e. 3uðtÞ (a) L1 s s  s 4e (b) L1 ¼ uðt  1Þ  4ðt  1Þ s2  2s 5e ¼ ............ (c) L1 s2 uðt  2Þ  5ðt  2Þ

22

 3 L1 ¼ 3uðtÞ s  s 4e ¼ uðt  1Þ  4ðt  1Þ L1 s2  2s 5e ¼ uðt  2Þ  5ðt  2Þ L1 s2

So we have

; FðtÞ ¼ 3uðtÞ  uðt  1Þ  4ðt  1Þ þ uðt  2Þ  5ðt  2Þ To sketch the graph of f ðtÞ we consider the values of the function within the three sections 0 < t < 1, 1 < t < 2, and 2 < t. Between t ¼ 0 and t ¼ 1, f ðtÞ ¼ . . . . . . . . . . . . f ðtÞ ¼ 3

23

Because in this interval, uðtÞ ¼ 1, but uðt  1Þ ¼ 0 and uðt  2Þ ¼ 0. In the same way, between t ¼ 1 and t ¼ 2, f ðtÞ ¼ . . . . . . . . . . . . f ðtÞ ¼ 7  4t Because between t ¼ 1 and t ¼ 2, uðtÞ ¼ 1, uðt  1Þ ¼ 1, but uðt  2Þ ¼ 0. ; f ðtÞ ¼ 3  4ðt  1Þ þ 0 ¼ 3  4t þ 4 ¼ 7  4t Similarly, for t > 2, f ðtÞ ¼ . . . . . . . . . . . .

24

104

Programme 3

25

f ðtÞ ¼ t  3 Because for t > 2, uðtÞ ¼ 1, uðt  1Þ ¼ 1 and uðt  2Þ ¼ 1 ; f ðtÞ ¼ 3  4ðt  1Þ þ 5ðt  2Þ ¼ 3  4t þ 4 þ 5t  10 ¼ t  3 So, collecting the results together, we have for

0 < t < 1, f ðtÞ ¼ 3 1 < t < 2, f ðtÞ ¼ 7  4t

ðt ¼ 1, f ðtÞ ¼ 3; t ¼ 2, f ðtÞ ¼ 1Þ

2 < t,

ðt ¼ 2, f ðtÞ ¼ 1; t ¼ 3, f ðtÞ ¼ 0Þ

f ðtÞ ¼ t  3

Using these facts we can sketch the graph of f ðtÞ, which is . . . . . . . . . . . .

26 f(t)

t

Here is another. Example 2

 2 3es 3e3s Determine the expression f ðtÞ ¼ L þ 2  2 and sketch the graph s s s of f ðtÞ. 1

First we express the inverse transform of each term in terms of the unit step function. This gives . . . . . . . . . . . .

27

L1

 2 ¼ 2uðtÞ; s

 s 3e ¼ uðt  1Þ  3ðt  1Þ L1 s2  3s 3e L1 ¼ uðt  3Þ  3ðt  3Þ s2

; f ðtÞ ¼ 2uðtÞ þ uðt  1Þ  3ðt  1Þ  uðt  3Þ  3ðt  3Þ So there are ‘break points’, i.e. changes of function, at t ¼ 1 and t ¼ 3, and we investigate f ðtÞ within the three intervals. 0 1 ¼ z ð1  1=zÞ2 z

FðzÞ ¼

So, multiplying numerator and denominator by z2 z FðzÞ ¼ provided jzj > 1 ðz  1Þ2 And another example. The Z transform of the discrete unit impulse:  1 n¼0 ðnÞ ¼ is FðzÞ ¼ . . . . . . . . . . . . 0 otherwise

22

1 Because FðzÞ ¼ Z fðnÞg 1 X ðnÞ ¼ zn n¼0 1 0 0 þ þ þ ... z0 z1 z2 ¼1 ¼

Next frame

172

Programme 5

Table of Z transforms 23

We list the results that we have obtained so far as well as some additional ones for future reference. Sequence

Transform F(z)

Permitted values of z

ðnÞ ¼ f1, 0, 0, . . .g

1

All values of z

uðnÞ ¼ f1, 1, 1, . . .g

z z1 z

jzj > 1

n uðnÞ ¼ f0, 1, 2, 3, . . .g n2 uðnÞ ¼ f0, 1, 4, 9, . . .g n3 uðnÞ ¼ f0, 1, 8, 27, . . .g   an uðnÞ ¼ 1, a, a2 , a3 , . . .   n an uðnÞ ¼ 0, a, 2a2 , 3a3 , . . .

ðz  1Þ2 zðz þ 1Þ ðz  1Þ3  z z2 þ 4z þ 1 ðz  1Þ4 z ðz  aÞ az ðz  aÞ2

jzj > 1 jzj > 1 jzj > 1 jzj > jaj jzj > jaj Next frame

Properties of Z transforms 24

Linearity The Z transform is a linear transform. That is, if a and b are constants then Z ðaf ðnÞ þ bgðnÞÞ ¼ aZ ff ðnÞg þ bZfgðnÞg For example, the Z transform of the sequence n uðnÞ is Z fn uðnÞg ¼ . . . . . . . . . . . . and the Z transform of the sequence e2n uðnÞ is Zfe2n uðnÞg ¼ . . . . . . . . . . . .

25

Zfn uðnÞg ¼

z ðz  1Þ

2

  and Z e2n uðnÞ ¼

z z  e2

Because Zfn uðnÞg ¼

z

from the table and, also from the table, ðz  1Þ2 z so when a ¼ e2 , Z fan uðnÞg ¼ za   z Z e2n uðnÞ ¼ z  e2

Consequently, the Z transform of ð3n  5e2n ÞuðnÞ is . . . . . . . . . . . .

Difference equations and the Z transform

173

 5z3 þ 13z2  z 3e2 þ 5 2

ðz  1Þ ðz 

e2 Þ

Because     Z ð3n  5e2n ÞuðnÞ ¼ 3Zfn uðnÞg  5Z e2n uðnÞ 3z 5z  ¼ 2 z  e2 Þ ð ðz  1Þ  3z z  e2  5zðz  1Þ2 ¼ ðz  1Þ2 ðz  e2 Þ ¼ ¼

3z2  3ze2  5z3 þ 10z2  5z ðz  1Þ2 ðz  e2 Þ  5z þ 13z2  z 3e2 þ 5 3

ðz  1Þ2 ðz  e2 Þ

First shift theorem (shifting to the left) If Z ff ðnÞg ¼ FðzÞ then

 Z ff ðn þ mÞg ¼ zm FðzÞ  zm f ð0Þ þ zm1 f ð1Þ þ . . . þ zf ðm  1Þ

is the Z transform of the sequence that has been shifted by m places to the left. For example Z ff ðn þ 1Þg ¼ zFðzÞ  zf ð0Þ Z ff ðn þ 2Þg ¼ z2 FðzÞ  z2 f ð0Þ  zf ð1Þ These will be used later when solving difference equations. Note the similarity between these results and the Laplace transforms for the first and second derivatives for continuous functions. z then For example, given that Z f4n uðnÞg ¼ z4   Z 4nþ3 uðnÞ ¼ . . . . . . . . . . . .

26

174

Programme 5

27

64z z4 Because

 Z ff ðn þ mÞg ¼ zm FðzÞ  zm f ð0Þ þ zm1 f ð1Þ þ . . . þ zf ðm  1Þ so  

 Z 4nþ3 uðnÞ ¼ z3 Z f4n uðnÞg  z3 40 þ z2 41 þ z42 where Z f4n uðnÞg ¼

z z4

 z  z3 þ 4z2 þ 16z z4

 z4  z3 þ 4z2 þ 16z ¼ z4  z4  z3 þ 4z2 þ 16z ðz  4Þ ¼ z4  4 4 z  z  64z ¼ z4 64z ¼ z4 In this way we have derived the Z transform of the sequence f64, 256, 1024, . . .g by shifting the sequence f1, 4, 16, 64, 256, . . .g three places to the left and losing the first three terms. ¼ z3

Try another. Given that Zfn uðnÞg ¼

z ðz  1Þ2

then

Z fðn þ 1ÞuðnÞg ¼ . . . . . . . . . . . .

28

z2 ðz  1Þ2 Because

 Z ff ðn þ mÞg ¼ zm FðzÞ  zm f ð0Þ þ zm1 f ð1Þ þ . . . þ z f ðm  1Þ so z  ½z  0 Z ff ðn þ 1Þg ¼ z ðz  1Þ2 ¼

z2 ðz  1Þ2

Second shift theorem (shifting to the right) If Z ff ðnÞg ¼ FðzÞ then Z ff ðn  mÞg ¼ zm FðzÞ the Z transform of the sequence that has been shifted by m places to the right. z then For example, given that Z fuðnÞg ¼ z1 Z fuðn  3Þg ¼ . . . . . . . . . . . .

Difference equations and the Z transform

1 z2 ðz  1Þ

175

29

Because Zff ðn  mÞg ¼ zm FðzÞ so z Z fuðn  3Þg ¼ z3 z1 1 ¼ 2 z ðz  1Þ In this way we have derived the Z transform of the sequence f0, 0, 0, 1, 1, 1, . . .g by shifting the sequence f1, 1, 1, 1, . . .g three places to the right and defining the first three terms as zeros. Try this one. The sequence ff ðnÞuðnÞg with Z transform Zff ðnÞuðnÞg ¼

1 , where a is a constant, is Zf. . . . . . . . . . . .g ðz  aÞ f ðnÞ ¼ an1 uðn  1Þ

Because From the table of transforms the nearest transform to the one in question is z which is the Z transform of fan uðnÞg. Now ðz  aÞ 1 1 z ¼  ðz  aÞ z ðz  aÞ ¼ z1 FðzÞ

where FðzÞ ¼ Z fan uðnÞg

and so   1 ¼ Z an1 uðn  1Þ ðz  aÞ which is the Z transform of an uðnÞ, shifted one place to the right.

Scaling If the sequence f ðnÞ has the Z transform Z ff ðnÞg ¼ FðzÞ then the sequence an f ðnÞ  has the Z transform Z fan f ðnÞg ¼ F a1 z . For example, Zfn uðnÞg ¼

z ðz  1Þ2

so that Z f2n n uðnÞg ¼ . . . . . . . . . . . .

30

176

Programme 5

31

2z ðz  2Þ2 Because Since Zfn uðnÞg ¼

z

ðz  1Þ2  Z f2n n uðnÞg ¼ F 21 z ¼ ¼

¼ FðzÞ then by the translation property

21 z ð21 z

2

 1Þ

2z ðz  2Þ2

Final value theorem For the sequence f ðnÞ with Z transform FðzÞ   z1 FðzÞ provided that Lim f ðnÞ exists. Lim f ðnÞ ¼ Lim z n!1 z!1 n!1 1 n For example, the sequence f ðnÞ ¼ 2 uðnÞ has the Z transform z 2z . ¼ FðzÞ ¼ 1 z  2 2z  1 Now Lim z!1



  z1 2ðz  1Þ FðzÞ ¼ Lim ¼0 z 2z  1 z!1

and

 n 1 uðnÞ ¼ 0 which confirms the final value theorem. Lim 2 n!1

Using the final value theorem the final value of the sequence with the Z transform FðzÞ ¼

32

10z2 þ 2z ðz  1Þð5z  1Þ2

is . . . . . . . . . . . .

0.75 Because ( )    z1 z1 10z2 þ 2z FðzÞ ¼ Lim Lim z z z!1 z!1 ðz  1Þð5z  1Þ2 ( ) 10z þ 2 ¼ Lim z!1 ð5z  1Þ2 12 ¼ 16 ¼ 0:75

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177

The initial value theorem For the sequence f ðnÞ with Z transform FðzÞ f ð0Þ ¼ Lim fFðzÞg z!1

For example, the sequence f ðnÞ ¼ an uðnÞ has the Z transform FðzÞ ¼ Lim FðzÞ ¼ Lim z!1

z!1

z 1 ¼ Lim ¼ 1 z  a z!1 1

z and za

ˆ pital’s rule. Furthermore f ð0Þ ¼ a0 ¼ 1, so demonstrating the validity of by L’Ho the theorem.

The derivative of the transform If Z ff ðnÞg ¼ FðzÞ then zF 0 ðzÞ ¼ Z fn f ðnÞg This is easily proved. FðzÞ ¼

1 X

f ðnÞzn and so F 0 ðzÞ ¼

n¼0

1 X

f ðnÞðnÞzn1 ¼ 

n¼0

1 1X f ðnÞn zn z n¼0

1 ¼  Z fn f ðnÞg z

and so  zF 0 ðzÞ ¼ Z fn f ðnÞg

For example, the sequence f ðnÞ ¼ an uðnÞ has the Z transform FðzÞ ¼

z and so za

the sequence n an uðnÞ has Z transform Z fn an uðnÞg ¼ zF 0 ðzÞ ¼ . . . . . . . . . . . . Zfn an uðnÞg ¼

33

az 2

ðz  aÞ

Because

!  z 0 zaz az ¼ ¼ z zF ðzÞ ¼ z za ðz  aÞ2 ðz  aÞ2 0

Notice that this is in agreement with the Table of transforms in Frame 23.

178

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Programme 5

Summary We now summarize the properties that we have just discussed. Linearity Z faf ðnÞ þ bgðnÞg ¼ aZff ðnÞg þ bZfgðnÞg If Z ff ðnÞg ¼ FðzÞ then: Shifting to the left

 Z ff ðn þ mÞg ¼ zm FðzÞ  zm f ð0Þ þ zm1 f ð1Þ þ . . . þ zf ðm  1Þ Shifting to the right Z ff ðn  mÞg ¼ zm FðzÞ Scaling

 Z fan f ðnÞg ¼ F a1 z Final value theorem   z1 FðzÞ provided Lim f ðnÞ exists Lim f ðnÞ ¼ Lim z n!1 z!1 n!1 Initial value theorem f ð0Þ ¼ Lim fFðzÞg z!1

Derivative of the transform zF 0 ðzÞ ¼ Z fnf ðnÞg

Inverse transforms 35

If the sequence f ðnÞ has Z transform Z ff ðnÞg ¼ FðzÞ, the inverse transform is defined as Z 1 FðzÞ ¼ f ðnÞ There are many times when, given the Z transform of a sequence, it is not possible to immediately read off the sequence from the Table of transforms. Instead some manipulation may be required and, as with Laplace transforms, very often this involves using partial fractions.

Difference equations and the Z transform

179

Example z . To find the inverse  5z þ 6 transform, and hence the sequence, we recognize that the denominator can be factorized and separated into partial fractions as The sequence f ðnÞ has Z transform FðzÞ ¼

z2

FðzÞ ¼ . . . . . . . . . . . .

FðzÞ ¼

3 2  z3 z2

36

Because z z2  5z þ 6 z ¼ ðz  2Þðz  3Þ A B þ ¼ z2 z3 Aðz  3Þ þ Bðz  2Þ ¼ ðz  2Þðz  3Þ

FðzÞ ¼

Equating numerators gives z ¼ Aðz  3Þ þ Bðz  2Þ, giving A þ B ¼ 1 and 3A  2B ¼ 0. From these two equations we find that A ¼ 2 and B ¼ 3. So FðzÞ ¼

3 2  z3 z2

The nearest Z transform in the table to either of these two partial fractions z . Therefore if we write is Zfan uðnÞg ¼ za 3 2 FðzÞ ¼  z3 z2 3 z 2 z   ¼  z z3 z z2 so Z 1 FðzÞ ¼ . . . . . . . . . . . . Z 1 FðzÞ ¼ ð3n  2n ÞuðnÞ Because 3 z 2 z    z z3 z z2 ¼ 3  z1 Zf3n uðnÞg  2  z1 Z f2n uðnÞg

FðzÞ ¼

and so Z1 FðzÞ ¼ 3  3n1 uðn  1Þ  2  2n1 uðn  1Þ by the second shift theorem ¼ 3n uðnÞ  2n uðnÞ So f ðnÞ ¼ ð3n  2n ÞuðnÞ.

37

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Programme 5

There is a simpler way of doing this without employing the second shift theorem. Recognizing that z appears in the numerator of FðzÞ, we consider instead the FðzÞ partial fraction breakdown of z FðzÞ ¼ ............ z

38

1 1  z3 z2 Because FðzÞ 1 z ¼  2 z z z  5z þ 6 1 ¼ 2 z  5z þ 6 1 ¼ ðz  2Þðz  3Þ A B þ ¼ z2 z3 Aðz  3Þ þ Bðz  2Þ ¼ ðz  2Þðz  3Þ Equating numerators gives 1 ¼ Aðz  3Þ þ Bðz  2Þ, giving AþB¼0

[z]:

3A  2B ¼ 1 with solution A ¼ 1 and B ¼ 1. So that

[CT]: FðzÞ z FðzÞ

1 1  that is z3 z2 z z  ¼ z3 z2 ¼ Z f3n uðnÞg  Z f2n uðnÞg and so ¼

Z1 FðzÞ ¼ 3n uðnÞ  2n uðnÞ ¼ ð3n  2n ÞuðnÞ Thus the use of the second shift theorem is avoided. So try one yourself. The sequence f ðnÞ has Z transform FðzÞ ¼

ðz2

5z  4z þ 4Þðz þ 2Þ

therefore f ðnÞ ¼ . . . . . . . . . . . .

Difference equations and the Z transform

 5 ð2n  1Þ2n þ ð2Þn uðnÞ 16

181

39

Because FðzÞ 1 5z ¼  2 z z ðz  4z þ 4Þðz þ 2Þ 5 ¼ ðz  2Þ2 ðz þ 2Þ A B C þ þ ¼ 2 z  2 z þ 2 ðz  2Þ ¼

Aðz þ 2Þ þ Bðz  2Þðz þ 2Þ þ Cðz  2Þ2 ðz  2Þ2 ðz þ 2Þ

  Equating numerators gives 5 ¼ Aðz þ 2Þ þ B z2  4 þ C z2  4z þ 4 , giving BþC¼0 A  4C ¼ 0

[z2 ]: [z]: [CT]:

2A  4B þ 4C ¼ 5

with solution A ¼ 5=4, B ¼ 5=16 and C ¼ 5=16, so FðzÞ 5=4 5=16 5=16  ¼ þ giving 2 z z 2 zþ2 ðz  2Þ FðzÞ ¼

5 2z 5 z 5 z   þ  and so  8 ðz  2Þ2 16 z  2 16 z þ 2

5 5 5  n2n uðnÞ   2n uðnÞ þ  ð2Þn uðnÞ 8 16 16  5 ð2n  1Þ2n þ ð2Þn uðnÞ ¼ 16 Move on to the next frame

Z 1 FðzÞ ¼

Solving difference equations If a sequence satisfies a difference equation with given initial terms then the general term of the sequence can be found by using the Z transform. For example, to solve the difference equation f ðn þ 2Þ  5f ðn þ 1Þ þ 6f ðnÞ ¼ 1 where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 we begin by taking the Z transform of both sides of the equation to give: Z ff ðn þ 2Þ  5f ðn þ 1Þ þ 6f ðnÞg ¼ Zf1g that is Z ff ðn þ 2g  5Zff ðn þ 1Þg þ 6Z ff ðnÞg ¼ Zf1g Using the first shift theorem where Z ff ðnÞg ¼ FðzÞ this then becomes  2 z z FðzÞ  z2 f ð0Þ  zf ð1Þ  ð5zFðzÞ  zf ð0ÞÞ þ 6FðzÞ ¼ z1

40

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Programme 5

Collecting like terms and substituting for the initial terms f ð0Þ ¼ 0 and f ð1Þ ¼ 1 gives  2 z  5z þ 6 FðzÞ  z ¼

 z z z2 so z2  5z þ 6 FðzÞ ¼ z þ ¼ that is z1 z1 z1 z2 z2 FðzÞ ¼ ¼ and so 2 ðz  1Þðz  5z þ 6Þ ðz  1Þðz  2Þðz  3Þ FðzÞ z ¼ z ðz  1Þðz  2Þðz  3Þ

This has the partial fraction breakdown FðzÞ . . . . . . . . . . . . . . . . . . ¼ þ þ z z1 z2 z3

41

FðzÞ 1=2 2 3=2 ¼  þ z z1 z2 z3 Because Letting

z A B C ¼ þ þ ðz  1Þðz  2Þðz  3Þ z  1 z  2 z  3 ¼

Aðz  2Þðz  3Þ þ Bðz  1Þðz  3Þ þ Cðz  1Þðz  2Þ ðz  1Þðz  2Þðz  3Þ

and so z ¼ Aðz  2Þðz  3Þ þ Bðz  1Þðz  3Þ þ Cðz  1Þðz  2Þ. Taking z ¼ 1, 2 and 3 in turn yields A ¼ 1=2, B ¼ 2 and C ¼ 3=2. Consequently, FðzÞ ¼

 z  3 z  1 z  2 þ and so f ðnÞ ¼ . . . . . . . . . . . . 2 z1 z2 2 z3 

42

f ðnÞ ¼

 1 3nþ1  2nþ1 þ uðnÞ 2 2

Because f ðnÞ ¼ Z 1 fFðzÞg    z  3  z  z  1 1 2 þ ¼Z 2 z1 z2 2 z3 n o n o n z o 1 z z 3 ¼ Z 1  2Z 1 þ Z 1 2 z1 z2 2 z3 1 3 n n ¼ uðnÞ  2  2 uðnÞ þ  3 uðnÞ 2 2   1 3nþ1 nþ1 2 uðnÞ þ ¼ 2 2

Difference equations and the Z transform

183

Try one yourself. The solution of the second order difference equation f ðn þ 2Þ  f ðnÞ ¼ 1 where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 is f ðnÞ ¼ . . . . . . . . . . . . See the following frames for the answer and working  f ðnÞ ¼

 1 3 ð2n  3Þ þ ð1Þn uðnÞ 4 4

43

Because Taking the Z transform of the difference equation gives Zff ðn þ 2Þ  f ðnÞg ¼ Z f1g. That is Z ff ðn þ 2Þg  Zff ðnÞg ¼ Z f1g so that  2 z . z FðzÞ  z2 f ð0Þ  zf ð1Þ  FðzÞ ¼ z1 Substituting f ð0Þ ¼ 0 and f ð1Þ ¼ 1 gives FðzÞ ¼ . . . . . . . . . . . .

FðzÞ ¼

44

z2 þ 2z 2

ðz þ 1Þðz  1Þ

Because  2 z becomes z FðzÞ  z2 f ð0Þ  zf ð1Þ  FðzÞ ¼ z1  2 z z FðzÞ þ z  FðzÞ ¼ so that z1 z z2 þ 2z ¼ and so ðz2  1ÞFðzÞ ¼ z þ z1 z1 z2 þ 2z z2 þ 2z ¼ FðzÞ ¼ 2 ðz  1Þðz  1Þ ðz þ 1Þðz  1Þ2 Therefore FðzÞ ...... ...... ...... ¼ þ þ z ðz  1Þ2 z  1 z þ 1

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Programme 5

45

FðzÞ 1=2 3=4 3=4 ¼ þ  2 z z  1 z þ1 ðz  1Þ Because FðzÞ z þ 2 ¼ z ðz þ 1Þðz  1Þ2 ¼ ¼

A ðz  1Þ2

þ

B C þ z1 zþ1

Aðz þ 1Þ þ Bðz þ 1Þðz  1Þ þ Cðz  1Þ2 ðz þ 1Þðz  1Þ2

giving

 z þ 2 ¼ Aðz þ 1Þ þ Bðz þ 1Þðz  1Þ þ Cðz  1Þ2 1 3 3 and hence A ¼ , B ¼  and C ¼ 2 4 4 From this we conclude that: f ðnÞ ¼ Z 1 fFðzÞg ( )  1 1 z 3 1 n z o 3 1 z Z Z þ  ¼ Z 2 4 z1 4 zþ1 ðz  1Þ2 1 3 3 n uðnÞ  uðnÞ þ ð1Þn uðnÞ 2 4 4   1 3 ð2n  3Þ þ ð1Þn uðnÞ ¼ 4 4

¼

Move on to the next frame

Sampling 46

If a continuous function f ðtÞ of time t progresses from t ¼ 0 onwards and is measured at every time interval T then what will result is the sequence of values ff ðkTÞg ¼ ff ð0Þ, f ðTÞ, f ð2TÞ, f ð3TÞ; . . .g A new, piecewise continuous function f ðtÞ can then be created from the sequence of sampled values such that  f ðkTÞ if t ¼ kT f ðtÞ ¼ 0 otherwise

Difference equations and the Z transform

185

The graph of this new function consists of a series of spikes at the regular intervals t ¼ kT f *(t)

T

2T

3T

nT

This function can alternatively be described in terms of the delta function ðtÞ as f ðtÞ ¼ f ð0ÞðtÞ þ f ðTÞðt  TÞ þ f ð2TÞðt  2TÞ þ f ð3TÞðt  3TÞ þ . . . 1 X f ðkTÞðt  kTÞ ¼ k¼0

The Laplace transform of f ðtÞ is then given as F ðsÞ ¼ Lff ðtÞg ð1 ¼ ff ð0ÞðtÞ þ f ðTÞðt  TÞ þ f ð2TÞðt  2TÞ þ . . .gest dt 0

¼ f ð0Þ þ f ðTÞesT þ f ð2TÞe2sT þ f ð3TÞe3sT þ . . . 1 X f ðkTÞeksT ¼ k¼0

Define a new variable z ¼ esT and we see that Lff ðtÞg ¼

1 X

f ðkTÞzk ¼

k¼0

1 X f ðkTÞ k¼0

zk

which is the Z transform of the sequenceff ðkTÞg. Example 1 The function f ðtÞ ¼ eat is sampled every interval of T. The Z transform of the sampled function is then . . . . . . . . . . . . FðzÞ ¼ Because Defining f ðtÞ ¼

P1 k¼0



f ðkTÞðt  kTÞ ¼

transform of f ðtÞ is given as F ðsÞ ¼

1 X k¼0

ekaT eksT

47

z z  eaT P1 k¼0

eakT ðt  kTÞ then the Laplace

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Programme 5

This means that the Z transform of ff ðkTÞg is FðzÞ ¼

1 kaT X e k¼0

zk

¼

1 z ¼ eaT z  eaT 1 z

Notice that this agrees with the Z transform of the sequence bn uðnÞ  z  when b is replaced by eaT . which is zb Try another. Example 2 The function f ðtÞ ¼ t is sampled every interval of T. The Z transform of the sampled function is then . . . . . . . . . . . .

48

FðzÞ ¼ Because The Z transform of ff ðkTÞg is FðzÞ ¼ FðzÞ ¼

1 X kT k¼0



¼T

Tz ðz  1Þ2 1 X f ðkTÞ . Here f ðkTÞ ¼ kT and so zk k¼0

zk

 1 2 3 þ 2 þ 3 þ ... z z z

T 1 þ 2z1 þ 3z2 þ 4z3 þ . . . z  d  1 þ z1 þ z2 þ z3 þ . . . ¼ Tz dz     d 1 1 T 1 2 Tz 1 1 ¼ Tz ¼ ¼ dz z z z ðz  1Þ2 ¼

Example 3 The function f ðtÞ ¼ cos t is sampled every interval of T. The Z transform of the sampled function is then . . . . . . . . . . . .

Difference equations and the Z transform

FðzÞ ¼

zðz  cos TÞ z2  2 cos T þ 1

Because f ðkTÞ ¼ cos kT ¼ FðzÞ ¼

z . z  eaT

  e jkT þ ejkT and the Z transform of ekaT is 2

e jkT þ ejkT is Therefore the Z transform of 2     1 z z  1 z z  e jT þ z z  ejT ¼ þ 2 z  ejT z  e jT 2 ðz  ejT Þðz  e jT Þ    2 1 2z  z e jT þ ejT ¼ 2 z2  ½e jT þ ejT z þ 1 zðz  cos TÞ ¼ 2 z  2z cos T þ 1 And that is the end of the Programme on Z transforms. All that remain are the Review summary and the Can you? checklist. Read through these closely and make sure that you understand all the workings of this Programme. Then try the Test exercise; there is no need to hurry, take your time and work through the questions carefully. The Further problems then provide a valuable collection of additional exercises for you to try.

Review summary 5 1 Sequences Any function f whose input is restricted to integer values n has an output f ðnÞ in the form of a discrete sequence of numbers. A sequence can be defined by a prescription for the nth term. Alternatively, it can be defined recursively where terms are defined by the values of previous terms. A recursively defined sequence requires one or more initial terms to start the process of evaluating successive terms. 2 Difference equations The equation that recursively defines a sequence is called a difference equation. A linear, constant coefficient difference equation consists of a sum of general terms of the sequence, each multiplied by a constant. The order of a difference equation is the maximum number of terms between any pair of terms in the equation.

187

49

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Programme 5

3 Solving difference equations In analogy with linear constant coefficient inhomogeneous differential equations, a linear constant coefficient inhomogeneous difference equation can be solved by first finding the inhomogeneous solution in terms of unknown constants, adding this to the particular solution and then applying the initial terms to find the values of the unknown constants. 4 Z transform The Z transform of the sequence f ðnÞ is 1 X f ðnÞ ¼ FðzÞ where the value of z is chosen to zn n¼1 ensure that the sum converges. f ðnÞ and Z ff ðnÞg form a Z transform pair.

Z ff ðnÞg ¼

5 Table of Z transforms Sequence

Transform F(z)

Permitted values of z

ðnÞ ¼ f1, 0, 0, . . .g

1

All values of z

uðnÞ ¼ f1, 1, 1, . . .g

z z1 z

jzj > 1

n uðnÞ ¼ f0, 1, 2, 3, . . .g n2 uðnÞ ¼ f0, 1, 4, 9, . . .g n3 uðnÞ ¼ f0, 1, 8, 27, . . .g   an uðnÞ ¼ 1, a, a2 , a3 , . . .   n an uðnÞ ¼ 0, a, 2a2 , 3a3 , . . .

ðz  1Þ2 zðz þ 1Þ ðz  1Þ3  z z2 þ 4z þ 1 ðz  1Þ4 z ðz  aÞ az ðz  aÞ2

jzj > 1 jzj > 1 jzj > 1 jzj > jaj jzj > jaj.

6 Linearity The Z transform is a linear transform. That is, if a and b are constants then Z faf ðnÞ þ bgðnÞg ¼ aZ ff ðnÞg þ bZfgðnÞg. 7 First shift theorem (shifting to the left) If Z ff ðnÞg ¼ FðzÞ then

 Z ff ðn þ mÞg ¼ zm FðzÞ  zm f ð0Þ þ zm1 f ð1Þ þ . . . þ zf ðm  1Þ

the Z transform of the sequence that has been shifted by m places to the left.

Difference equations and the Z transform

189

8 Second shift theorem (shifting to the right) If Z ff ðnÞg ¼ FðzÞ then Z fn  mg ¼ zm FðzÞ the Z transform of the sequence that has been shifted by m places to the right. 9 Scaling If the sequence f ðnÞ has the Z transform Z ff ðnÞ g ¼ FðzÞ then the sequence an f ðnÞ has the Z transform Z fan f ðnÞg ¼ F a1 z . 10 Final value theorem For the sequence f ðnÞ with Z transform FðzÞ   z1 Lim f ðnÞ ¼ Lim FðzÞ provided that Lim f ðnÞ exists. z n!1 z!1 n!1 11 The initial value theorem For the sequence f ðnÞ with Z transform FðzÞ f ð0Þ ¼ Lim fFðzÞg. z!1

12 The derivative of the transform If Z ff ðnÞg ¼ FðzÞ then zF 0 ðzÞ ¼ Z fnf ðnÞg. 13 Inverse transformations If the sequence f ðnÞ has Z transform Z ff ðnÞg ¼ FðzÞ, the inverse transform is defined as Z 1 FðzÞ ¼ f ðnÞ. 14 Solving difference equations If a sequence f ðnÞ satisfies a difference equation with given initial conditions then the general term of the sequence can be found by using the Z transform where Z ff ðnÞg ¼ FðzÞ. This is referred to as solving the difference equation. 15 Sampling If a continuous function f ðtÞ is sampled at equal intervals, the resulting sequence has a Z transform that is related to the Laplace transform of the piecewise function created from the sequence of sample values. Lff ðtÞg ¼

1 X k¼0

f ðkTÞzk ¼

1 X f ðkTÞ k¼0

zk

¼ Zff ðkTÞg

where ff ðkTÞg ¼ ff ð0Þ, f ðTÞ, f ð2TÞ, f ð3TÞ, . . .g,  f ðkTÞ if t ¼ k

f ðtÞ ¼ 0 otherwise and z ¼ esT .

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Programme 5

Can you? Checklist 5 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can:

Frames

. Convert the descriptive prescription of the output form of a sequence into a recursive description and recognize the importance of initial terms?

1

to

3

4

to

8

9

to

16

17

to

22

Yes

No

. Recognize a difference equation, determine its order and generate its terms from a recursive description?

Yes

No

. Obtain the solution to a difference equation as a sum of the homogeneous solution and the particular solution?

Yes

No

. Define the Z transform of a sequence and derive transforms of specified sequences?

Yes

No

. Make reference to a table of standard Z transforms?

Yes

23

No

. Recognize the Z transform as being a linear transform and so obtain the transform of linear combinations of standard sequences?

Yes

to

34

35

to

39

40

to

45

46

to

49

No

. Create a sequence by sampling a continuous function and demonstrate the relationship between the Laplace and the Z transform?

Yes

26

No

. Use the Z transform to solve linear, constant coefficient difference equations?

Yes

26

No

. Use partial fractions to derive the inverse transforms

Yes

to

No

. Apply the first and second shift theorems, the scaling theorem, the initial and final value theorems and the derivative theorem?

Yes

24

No

Difference equations and the Z transform

Test exercise 5 1 Find a recursive description corresponding to each of the following prescriptions for the output of a sequence: (a) f ðnÞ ¼ 5n  9 where n is an integer 1 (b) f ðnÞ ¼ 23  4n where n is an integer 0 (c) f ðnÞ ¼ 3n where n is an integer 2. 2 Determine the order and find the first six terms of each of the following sequences: (a) f ðn þ 3Þ  f ðnÞ ¼ 5n where f ð0Þ ¼ 1, f ð1Þ ¼ 1 and f ð2Þ ¼ 3 (b) f ðn þ 1Þ  5f ðnÞ þ 6f ðn  1Þ ¼ 2n where f ð1Þ ¼ 0 and f ð0Þ ¼ 1 (c) f ðn þ 2Þ  f ðn þ 1Þ þ 12f ðnÞ ¼ 3uðnÞ where f ð0Þ ¼ 2 and f ð1Þ ¼ 5. 3 Obtain the solution to the following difference equation in the form of a sum of homogeneous and particular solutions: f ðn þ 1Þ  5f ðnÞ þ 6f ðn  1Þ ¼ 2n where f ð1Þ ¼ 0 and f ð0Þ ¼ 1. Check that your answer is in agreement with the answer to 2(b). 4 Find the Z transform of each of the sequences with output: (a) f ðnÞ ¼ ð1Þn uðnÞ (b) f ðnÞ ¼ ð4n  2an ÞuðnÞ (c) f ðnÞ ¼ ðn  3ÞuðnÞ  (d) f ðnÞ ¼ 5nþ2 uðnÞ. 5 Find the inverse Z transform of z2 ðz  3Þ . FðzÞ ¼ 2 ðz  2z þ 1Þðz  2Þ 6 Solve the difference equation f ðn þ 2Þ  4f ðn þ 1Þ þ 4f ðnÞ ¼ 3 where f ð0Þ ¼ 1 and f ð1Þ ¼ 0. 7 The function f ðtÞ ¼ sin t is sampled at equal intervals of t ¼ T. Find the Z transform of the resulting sequence of values.

Further problems 5 1 Find the Z transform of f ðnÞ ¼ ðaÞn where a > 0. 2 Solve each of the following difference equations in the form of the homogeneous solution plus the particular solution: (a) f ðn þ 2Þ þ 5f ðn þ 1Þ þ 6f ðnÞ ¼ 1 where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 (b) 3f ðn þ 2Þ  7f ðn þ 1Þ þ 2f ðnÞ ¼ n where f ð0Þ ¼ 1 and f ð1Þ ¼ 0 (c) f ðn þ 2Þ  9f ðnÞ ¼ 2n2 where f ð0Þ ¼ 1 and f ð1Þ ¼ 1.

191

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Programme 5

3 Given that aðn þ 1Þ ¼ bðnÞ and that bðn þ 1Þ ¼ cðnÞ where cðnÞ ¼ f ðnÞ  gðnÞ, show that f ðn þ 2Þ þ f ðnÞ ¼ gðnÞ and solve for f ðnÞ when gðnÞ ¼ ðnÞ, the unit impulse sequence where f ð0Þ ¼ 0 and f ð1Þ ¼ 1. 4 If pðn þ 1Þ ¼ qðnÞ qðn þ 1Þ ¼ rðnÞ rðnÞ ¼ f ðnÞ  qðnÞ  pðnÞ where  and  are constants, show that pðn þ 2Þ þ pðn þ 1Þ þ pðnÞ ¼ f ðnÞ. Solve this recurrence relation when f ð0Þ ¼ 1, f ð1Þ ¼ 0 for (a)  ¼ 4,  ¼ 4 and f ðnÞ ¼ ðnÞ, the unit impulse sequence (b)  ¼ 4,  ¼ 4 and f ðnÞ ¼ uðnÞ the unit step sequence. 5 Find the Z transform of each of the following sequences. (a) f1, 0, 1, 0, 1, 0, . . .g (b) f0, 1, 0, 1, 0, 1, . . .g (c) f1, 0, 1, 1, 0, 0, 0, 1g (d) f1, 1, 1, 0, 0, 0, 1, 1g (e) f0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1g (f) f1, 1, 0, 0, 0, 1, 1g Note that the last four of these are finite sequences. 6 Find the inverse transform of z (a) FðzÞ ¼ ðz þ 1Þðz þ 2Þðz þ 3Þ z2 ðz þ 1Þðz þ 2Þðz þ 3Þ zð3z þ 1Þ (c) FðzÞ ¼ ðz  2Þðz  3Þ (b) FðzÞ ¼

(d) FðzÞ ¼

z2 . 2  3z þ z2

7 Given FðzÞ ¼

z2

3z2 zþ1

show that Z 1 FðzÞ ¼ f3, 3,  3,  3, . . .g. Hint: Use long division on FðzÞ. 8 Given



FðzÞ ¼

1þ

2 z

3

show that Z 1 FðzÞ ¼ f1,  6, 24,  48, . . .g. Hint: Use the binomial theorem on FðzÞ.

Difference equations and the Z transform

9 Find the final value of the sequence f ðnÞ with Z transform 4z2  z . FðzÞ ¼ 2 2z  3z þ 1 10 What is the initial value of the sequence whose Z transform is given by 2z2  z þ 1 ? FðzÞ ¼ 5  3z  7z2 11 Given the sequence of n terms f ðkÞ for 0  k  n  1 with Z transform Fn ðzÞ, show that the Z transform of the sequence formed by continually repeating the terms f ðkÞ is given as Fn ðzÞ . FðzÞ ¼ 1  zn 12 Using the result of Question 11, show that the Z transform of the sequence obtained by continually repeating the three term sequence f1, 0,  1g is z2 . FðzÞ ¼ 2 z þ1 13 Find the Z transforms of the sequence of values obtained when f ðtÞ is sampled at regular intervals of t ¼ T where (a) f ðtÞ ¼ sinh t (b) f ðtÞ ¼ cosh at (c) f ðtÞ ¼ eat cosh bt. 14 Solve each of the following difference equations using the Z transform (a) f ðn þ 2Þ þ 5f ðn þ 1Þ þ 6f ðnÞ ¼ 1 where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 (b) f ðn þ 2Þ  7f ðn þ 1Þ þ 2f ðnÞ ¼ n where f ð0Þ ¼ 1 and f ð1Þ ¼ 0 (c) 3f ðn þ 2Þ  9f ðnÞ ¼ 2 where f ð0Þ ¼ 1 and f ð1Þ ¼ 1 (d) f ðn þ 2Þ þ 2f ðn þ 1Þ  15f ðnÞ ¼ 4n where f ð0Þ ¼ 0 and f ð1Þ ¼ 1. 15 If f ðn þ 1Þ ¼ 3ðn þ 1Þf ðnÞ show that f ðn þ 1Þ ¼ 3nþ1 ðn þ 1Þ!f ð0Þ. 16 Show that the difference equation gðn þ 2Þ  gðn þ 1Þ  6gðnÞ ¼ 0 can be derived from the coupled difference equation f ðn þ 1Þ ¼ gðnÞ gðn þ 1Þ ¼ gðnÞ þ 6f ðnÞ. Find f ðnÞ and gðnÞ given that f ð1Þ ¼ 0 and gð1Þ ¼ 1. 17 Show that f ðnÞ ¼ n!uðnÞ satisfies the difference equation f ðn þ 1Þ  ðn þ 1Þf ðnÞ ¼ ðn þ 1Þ. 18 Use the derivative property to find the Z transform of f ðnÞ ¼ 3n n uðn  3Þ. 19 Solve the equation for the Fibonacci sequence: f ðn þ 2Þ ¼ f ðn þ 1Þ þ f ðnÞ where f ð0Þ ¼ 0, f ð1Þ ¼ 1 and n 0.

193

Frames 1 to 55

Programme 6

Introduction to invariant linear systems Learning outcomes When you have completed this Programme you will be able to: Recognize a system as a process whereby an input (either continuous or discrete) is converted to an output, also called the response of the system Distinguish between linear and nonlinear systems and recognize time-invariant and shift-invariant systems Determine the zero-input response and the zero-state response Appreciate why zero valued boundary conditions give rise to a time-invariant system Demonstrate that the response of a continuous, linear, time-invariant system to an arbitrary input is the convolution of the input with response of the system to a unit impulse Understand the role of the exponential function with respect to a linear, time-invariant system Use the convolution theorem to find the response of a continuous, linear, time-invariant system to an arbitrary input Derive the system transfer function of a constant coefficient linear differential equation and use it to solve the equation Demonstrate that the response of a discrete, linear, shift-invariant system to an arbitrary input is the convolution sum of the input with response of the system to a unit impulse Understand the role of the exponential function with respect to a discrete linear, shift-invariant system Derive the system transfer function of a constant coefficient linear difference equation and use it to solve the equation Derive the constant coefficient difference equation from knowledge of its unit impulse response. Prerequisites: Advanced Engineering Mathematics Programme 3 Laplace transforms 2, Programme 4 Laplace transforms 3 and Programme 5 Difference equations and the Z transform 194

Introduction to invariant linear systems

195

Invariant linear systems 1

Systems A system is a process that is capable of accepting an input, processing the input and producing an output, also called the response of the system. In Engineering Mathematics, Eighth Edition a function was described as an example of a system where the input was a number x which was processed by the function f to produce a number output f ðxÞ as shown in the box diagram: x

f

f(x)

For example the function f with input x and output f ðxÞ ¼ sin x can be represented as: x

f (x) = sin x

f

This system description simply links the input number to the output number via the function. How the function performs the process of evaluating the sine of the input number is not accounted for in this description it is just accepted that the function f can do it. In this Programme we are going to extend this application of a system to one that will accept an expression as input, process the expression and produce another expression as output. For continuous inputs the box diagram for this system will be: x(t)

L

y(t)

that is yðtÞ ¼ LfxðtÞg or LfxðtÞg ¼ yðtÞ Here the input and output expressions involve the parameter t. In what follows we shall take this to represent the variable time but it can represent whatever variable is appropriate to the problem in hand. For a discrete system the box diagram is: x[n]

L

y[n]

that is y½n ¼ Lfx½ng or Lfx½ng ¼ y½n Here the input and output expressions involve the discrete integer parameter n. For the purposes of this Programme the integer parameter is placed within square braces to indicate the discrete nature as opposed to round braces used to indicate a continuous nature. That is: x1 , x2 , x3 , . . . ;xn , . . . or x½1, x½2, x½3, . . . , x½n, . . .

196

Programme 6

Just as a system can be used to describe the input–output relationship linking two numbers so a system can be used to describe the input–output relationship linking two expressions. What we need to look for now are input–output relationships linking two expressions that can be described by a system. Further, just as there are many different types of relationship there are many different types of system. The specific type of system we shall be interested in is an invariant linear system (but we get ahead of ourselves). Move to the next frame

2

Input-response relationships Many physical situations in science and engineering can be described by a linear, constant coefficient, ordinary differential equation of the type met in the previous Programmes. Their method of solution may differ depending on the structure of the differential equation but the desire to obtain the solution is common to all. Take for instance the particularly simple first order differential equation dy1 ðtÞ ¼ 2t where y1 ð0Þ ¼ 0 dt By integrating this equation: ð ð ð dy1 ðtÞ t2 dt ¼ 2t dt that is dy1 ðtÞ ¼ y1 ðtÞ ¼ 2 þ C ¼ t 2 þ C dt 2 and applying the boundary condition y1 ð0Þ ¼ 0 ¼ 0 þ C we arrive at the solution y1 ðtÞ ¼ t 2 . The same equation, but with a different right-hand side, dy2 ðtÞ ¼ 4t 3 where y2 ð0Þ ¼ 0 has solution . . . . . . . . . . . . dt

3

t4 Because By integrating this equation: ð ð ð dy2 ðtÞ t4 3 dt ¼ 4t dt that is dy2 ðtÞ ¼ y2 ðtÞ ¼ 4 þ C0 ¼ t 4 þ C0 dt 4 and applying the boundary condition y2 ð0Þ ¼ 0 ¼ 0 þ C0 we arrive at the solution y2 ðtÞ ¼ t 4 . The general form of this simple equation can be given as: dyðtÞ ¼ xðtÞ where yð0Þ ¼ 0 dt and in both cases we insert the specific expression xðtÞ in the right-hand side and then manipulate the equation to obtain the solution yðtÞ. It is this commonality of procedure that merits further study.

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197

In each case, the method used to find the solution can be represented by a system where the differential equation specifies the relationship between the input and the output. The process is that of integration and evaluating the integration constant, the input is the term on the right-hand side and the output or the system response is what we are trying to find, the solution to the differential equation. We can use a box diagram to represent the system: 2t

L

t2

4t 3

L

t4

In the first box diagram 2t is input and t 2 is the response and in the second box diagram 4t 3 is input and t 4 is the response The process L is the same for each differential equation; what differs are the respective inputs and their corresponding responses. The response of the differential equation where yð0Þ ¼ 0 is

dyðtÞ ¼ xðtÞ to the input xðtÞ ¼ sin t dt

yðtÞ ¼ . . . . . . . . . . . .  cos t þ 1

4

Because In the differential equation

dyðtÞ ¼ xðtÞ, yðtÞ is the response to the input dt

xðtÞ ¼ sin t so that ð ð dyðtÞ dt ¼ sin t dt. dt That is ð dyðtÞ ¼ yðtÞ ¼  cos t þ A where A is the integration constant. and applying the boundary condition yð0Þ ¼ 0 ¼ 1 þ A we arrive at the response yðtÞ ¼  cos t þ 1. Move to the next frame

Linear systems Systems that are linear are of particular interest because many problems in science and engineering can be posed as linear systems. A system yðtÞ ¼ LfxðtÞg is linear if sums and scalar multiples are preserved, that is if Lfx1 ðtÞ þ x2 ðtÞg ¼ Lfx1 ðtÞg þ Lfx2 ðtÞg and LfxðtÞg ¼ LfxðtÞg where  is a constant. In particular, by choosing  ¼ 0 then Lf0g ¼ 0 which shows that if nothing is put into a linear system nothing will come out – zero input yields zero output.

5

198

Programme 6

These two properties can be combined. yðtÞ ¼ LfxðtÞg is a linear system if: Lfax1 ðtÞ þ bx2 ðtÞg ¼ aLfx1 ðtÞg þ bLfx2 ðtÞg where a and b are constants. For the discrete case, the system is linear if: Lfax1 ½n þ bx2 ½ng ¼ aLfx1 ½ng þ bLfx2 ½ng where a and b are constants. For example, consider the system in which the output is 5 times the input. That is: yðtÞ ¼ LfxðtÞg ¼ 5xðtÞ To show that this is a linear system we consider two distinct inputs x1 ðtÞ and x2 ðtÞ and their respective responses y1 ðtÞ ¼ 5x1 ðtÞ and y2 ðtÞ ¼ 5x2 ðtÞ. We also consider the linear combination of the inputs xðtÞ ¼ ax1 ðtÞ þ bx2 ðtÞ where a and b are constants and where yðtÞ is the corresponding response. Then: yðtÞ ¼ LfxðtÞg ¼ Lfax1 ðtÞ þ bx2 ðtÞg ¼ 5½ax1 ðtÞ þ bx2 ðtÞ ¼ 5ax1 ðtÞ þ 5bx2 ðtÞ

the response is 5 times the input

¼ ay1 ðtÞ þ by2 ðtÞ ¼ aLfx1 ðtÞg þ bLfx2 ðtÞg that is Lfax1 ðtÞ þ bx2 ðtÞg ¼ aLfx1 ðtÞg þ bLfx2 ðtÞg Therefore the system is linear. A system that is not linear is called a nonlinear system. Try one for yourself: The system L with input xðtÞ and response yðtÞ ¼ LfxðtÞg ¼ xðtÞ sin pt where p is a real number is . . . . . . . . . . . . (linear/nonlinear)

6

linear Because If xðtÞ ¼ ax1 ðtÞ þ bx2 ðtÞ then yðtÞ ¼ Lfax1 ðtÞ þ bx2 ðtÞg ¼ ½ax1 ðtÞ þ bx2 ðtÞ sin pt ¼ ax1 ðtÞ sin pt þ bx2 ðtÞ sin pt ¼ aLfx1 ðtÞg þ bLfx2 ðtÞg Now, how about the system L with input xðtÞ and response yðtÞ ¼ LfxðtÞg ¼ exðtÞ ? This system is . . . . . . . . . . . . (linear/nonlinear)

Introduction to invariant linear systems

199

nonlinear

7

Because If xðtÞ ¼ ax1 ðtÞ þ bx2 ðtÞ then yðtÞ ¼ Lfax1 ðtÞ þ bx2 ðtÞg ¼ eax1 ðtÞþbx2 ðtÞ ¼ eax1 ðtÞ ebx2 ðtÞ whereas aLfx1 ðtÞg þ bLfx2 ðtÞg ¼ aex1 ðtÞ þ bex2 ðtÞ 6¼ eax1 ðtÞ ebx2 ðtÞ Therefore Lfax1 ðtÞ þ bx2 ðtÞg 6¼ aLfx1 ðtÞg þ bLfx2 ðtÞg Similar considerations work for discrete systems. For example, the system L with discrete input x½n and response y½n ¼ Lfx½ng ¼ 2x½n is linear because if x½n ¼ ax½ n þ bx2 ½n then y½n ¼ Lfax1 ½n þ bx2 ½ng ¼ 2½ax1 ½n þ bx2 ½n ¼ 2ax1 ½n  2bx2 ½n ¼ aLfx1 ½ng þ bLfx2 ½ng Again, try one for yourself. The discrete system L with input x½n and response y½n ¼ Lfx½ng ¼ x½n þ 4x½n  1 is . . . . . . . . . . . . (linear/nonlinear) linear Because If xðtÞ ¼ ax1 ½n þ bx2 ½n then y½n ¼ Lfax1 ½n þ bx2 ½ng



 ¼ ax1 ½n þ bx2 ½n þ 4 ax1 ½n  1 þ bx2 ½n  1 i h  i h ¼ a x1 ½n þ 4x1 ½n  1 þ bðx2 ½n þ 4x2 ½n  1Þ ¼ aLfx1 ½ng þ bLfx2 ½ng Now, how about the system L with input x½n and response y½n ¼ Lfx½ng ¼ cos x½n ? This system is . . . . . . . . . . . . (linear/nonlinear)

8

200

Programme 6

9

nonlinear Because If x½n ¼ ax1 ½n þ bx2 ½n then y½n ¼ Lfax1 ½n þ bx2 ½ng ¼ cosfax1 ½n þ bx2 ½ng whereas aLfx1 ½ng þ bLfx2 ½ng ¼ a cos x1 ½n þ b cos x2 ½n 6¼ cosfax1 ½n þ bx2 ½ng Therefore Lfax1 ½n þ bx2 ½ng 6¼ aLfx1 ½ng þ bLfx2 ½ng Move to the next frame

10

Time-invariance of a continuous system Consider the plot of the input to and the corresponding response of an arbitrary continuous system x (t)

y(t)

t

t

If this response pattern is retained but shifted wholesale through t0 when the input is similarly shifted through t0 then the system is said to be timeinvariant. In other words it does not matter when we activate the system, we always get the same response for the same input; the response will be the same on Tuesday as it was on Monday. x (t)

t0

x(t – t0)

t

y(t)

t0

y(t – t0)

t

That is, a system is said to be time-invariant if: yðtÞ ¼ LfxðtÞg and yðt  t0 Þ ¼ Lfxðt  t0 Þg where t0 is a constant. To see this analysis more clearly, consider the output in two different ways. Firstly consider it as the response of the system acting on the input y1 ðtÞ ¼ LfxðtÞg and secondly consider it as a dependent variable equated to an expression involving the input as an independent variable. That is: y2 ðtÞ ¼ f fxðtÞg where y1 ðtÞ ¼ y2 ðtÞ

Introduction to invariant linear systems

201

The time delay through t0 then results in y1 ðt  t0 Þ ¼ Lfxðt  t0 Þg and y2 ðt  t0 Þ ¼ f fxðt  t0 Þg If the delayed response is the same as a delayed version of the original response, that is y1 ðt  t0 Þ ¼ y2 ðt  t0 Þ, the system L is time-invariant but if y1 ðt  t0 Þ 6¼ y2 ðt  t0 Þ the system L is time-variant. For example, if yðtÞ ¼ LfxðtÞg ¼ exðtÞ then Lfxðt  t0 Þg ¼ exðtt0 Þ ¼ yðt  t0 Þ (the delayed response is the same as a delayed version of the original response) therefore yðtÞ ¼ LfxðtÞg and yðt  t0 Þ ¼ Lfxðt  t0 Þg so the system is time-invariant. You try one. The system defined by yðtÞ ¼ LfxðtÞg ¼ x2 ðtÞ . . . . . . . . . . . . (is/is not) time-invariant is time-invariant

11

Because Since yðtÞ ¼ LfxðtÞg ¼ x2 ðtÞ then Lfxðt  t0 Þg ¼ x2 ðt  t0 Þ ¼ yðt  t0 Þ therefore yðtÞ ¼ LfxðtÞg and yðt  t0 Þ ¼ Lðt  t0 Þg (the delayed response is the same as a delayed version of the original response) so the system is time-invariant. Quite straightforward. Now try this one. The system defined by yðtÞ ¼ LfxðtÞg ¼ txðtÞ . . . . . . . . . . . . (is/is not) time-invariant is not time-invariant Because Since yðtÞ ¼ LfxðtÞg ¼ txðtÞ (that is, the input is multiplied by t) then Lfxðt  t0 Þg ¼ txðt  t0 Þ ¼ yðt  t0 Þ, the input is still multiplied by t. However, a delayed version of the original response is given as yðt  t0 Þ ¼ ðt  t0 Þxðt  t0 Þ so that yðt  t0 Þ 6¼ Lfxðt  t0 g and so the system is not time-invariant. Try another. The system defined by yðtÞ ¼ LfxðtÞg ¼ xðtÞ sin kt (ka real constant) . . . . . . . . . . . . (is/is not) time-invariant

12

202

Programme 6

13

is not time-invariant Because Since yðtÞ ¼ LfxðtÞg ¼ xðtÞ sin kt (that is, the input is multiplied by sin kt) then Lfxðt  t0 Þg ¼ xðt  t0 Þ sin kt, the input is still multiplied by sin kt. However, a delayed version of the original response is given as yðt  t0 Þ ¼ xðt  t0 Þ sin kðt  t0 Þ so that yðt  t0 Þ 6¼ Lfxðt  t0 g and the system is not time-invariant. Similar considerations apply when dealing with a discrete system. Move to the next frame

14

Shift-invariance of a discrete system If, when a discrete system is such that If y½n ¼ Lfx½ng and y½n  m ¼ Lfx½n  mg the system is said to be shift-invariant. This means that if term n of the input sequence corresponds to term n of the response sequence then moving up or down the input sequence a given number of terms corresponds to moving up or down the same number of terms of the response sequence. For example, the system defined by the response: y½n ¼ Lfx½ng ¼ x3 ½n is shift-invariant because the response to x½n  m is y½n  m ¼ x3 ½n  m which is the same as the shifted version of the original response y½n. However, the system defined by y½n ¼ Lfx½ng ¼ x½n þ x½n is not shift-invariant because, for example, the response to x½n þ m þ x½n þ m (m terms further down the input sequence) is not the same as a shifted version of the original response which is y½n þ m ¼ x½n þ m þ x½ðn þ mÞ ¼ x½n þ m þ x½n  m and where x½n þ m 6¼ x½n  m. Try a couple yourself. The system defined by y½n ¼ 5x½n . . . . . . . . . . . . (is/is not) shift-invariant

Introduction to invariant linear systems

is shift-invariant

203

15

Because For example, if x1 ½n ¼ x½n  mthen Lfx½n  mg ¼ Lfx1 ½ng ¼ 5x1 ½n ¼ 5x½n  m ¼ y½n  m And another. The system defined by y½n ¼ nx½n . . . . . . . . . . . . (is/is not) shift-invariant is not shift-invariant

16

Because For example, if x1 ½n ¼ x½n  m then Lfnx½n  mg ¼ Lfnx1 ½ng ¼ nx1 ½n ¼ nx½n  m 6¼ y½n  m since y½n  m ¼ ðn  mÞx½n  m

Differential equations The general nth-order equation Linear, constant coefficient differential equations define a linear system so let us refresh our memory. The general nth-order, linear, constant coefficient, inhomogeneous differential equation: an

dn yðtÞ dn1 yðtÞ þ a þ . . . þ a0 yðtÞ n1 dt n dt n1 m m1 d xðtÞ d xðtÞ ¼ bm þ bm1 þ . . . þ b0 xðtÞ m m1 dt dt

coupled with the values of the n boundary conditions:   dn yðtÞ dn1 yðtÞ ,  , . . . , yðt0 Þ dt n t¼t0 dt n1  t¼t0

describes the input-response relationship of a continuous linear system with input xðtÞ and response yðtÞ.

17

204

Programme 6

Such an equation has a solution in the form yðtÞ ¼ yh ðtÞ þ yp ðtÞ where yh ðtÞ is the complementary function solution to the homogeneous equation an

dn yðtÞ dn1 yðtÞ þ an1 þ . . . þ a0 yðtÞ ¼ 0 n dt dt n1

and yp ðtÞ is a particular integral or particular solution to the inhomogeneous equation. [Refer to Programme 26 of Engineering Mathematics, Eighth Edition.] The procedure for solving such an equation is: (a) Find the homogeneous solution yh ðtÞ in terms of unknown integration constants (b) Find the particular solution yp ðtÞ and form the complete solution yðtÞ ¼ yh ðtÞ þ yp ðtÞ (c) Apply the boundary conditions to find the values of the unknown integration constants in yh ðtÞ. An alternative method of finding the complete solution requires the computation of the zer-input and zero-state responses. Move to the next frame

18

Zero-input response and zero-state response The zero-input response is the solution to the differential equation with zero input, this being the homogeneous solution with the boundary conditions applied. The zero-state response is the complete solution (the sum of the homogeneous solution and the particular integral) with all the boundary conditions having been set to zero. The final complete solution is the sum of the zero-input response and the zero-state response. Example 1 Solve the equation: dyðtÞ þ 4yðtÞ ¼ tuðtÞ where yð0Þ ¼ 2 dt Zero-input response Yzi ðtÞ The zero-input response is the homogeneous solution to the differential equation with the boundary condition applied. That is, we require the solution to: dyzi ðtÞ þ 4yzi ðtÞ ¼ 0 where yzi ð0Þ ¼ 2. That is dt yzi ðtÞ ¼ . . . . . . . . . . . .

Introduction to invariant linear systems

205

yzi ðtÞ ¼ 2e4t

19

Because The auxiliary equation is m þ 4 ¼ 0 therefore m ¼ 4 so yzi ðtÞ ¼ Ae4t Applying the boundary condition yzi ð0Þ ¼ 2 results in yzi ðtÞ ¼ 2e4t Zero-state response yzs ðtÞ The zero-state response is the complete solution to the differential equation with the boundary condition set to zero. That is, we require the solution to: dyzs ðtÞ þ 4yzs ðtÞ ¼ 0 where yzs ð0Þ ¼ 0. That is dt yzs ðtÞ ¼ . . . . . . . . . . . .

yzs ðtÞ ¼

t 1 e4t  þ 4 16 16

20

Because Taking Laplace transforms we see that 1 1 that is ðs þ 4ÞYzs ðsÞ ¼ 2 s2 s 1 1 1 1 þ yielding ¼ Therefore Yzs ðsÞ ¼ 2  s ðs þ 4Þ 4s2 16s 16ðs þ 4Þ fsYzs ðsÞ  Yzs ð0Þg þ 4Yzs ðsÞ ¼

yzs ðtÞ ¼

t 1 e4t  þ 4 16 16

The final complete solution is then: yðtÞ ¼ . . . . . . . . . . . .

yðtÞ ¼

 1 4t  1 þ 33e4t 16

Because yðtÞ ¼ yzi ðtÞ þ yzs ðtÞ t 1 e4t  þ 4 16 16  1 4t  1 þ 33e4t ¼ 16 ¼ 2e4t þ

The advantage of this process is that it is easier to rework the same equation for different boundary conditions or different inputs because we only need consider one of the responses.

21

206

Programme 6

Example 2 To find the solution for: dyðtÞ þ 4yðtÞ ¼ tuðtÞ where yð0Þ ¼ 6 dt (same input but different boundary condtion) The final complete solution is then: yðtÞ ¼ . . . . . . . . . . . .

22

yðtÞ ¼

 1 4t  1 þ 97e4t 16

Because We only need to rework the zero-input response because the zero-state response is the same as in Example 1. The zero-input response is yzi ðtÞ ¼ Ae4t . Applying the boundary condition yzi ð0Þ ¼ 6 results in yzi ðtÞ ¼ 6e4t . Therefore the complete solution is: yðtÞ ¼ yzi ðtÞ þ yzs ðtÞ t 1 e4t  þ 4 16 16  1 4t  1 þ 97e4t ¼ 16

¼ 6e4t þ

Example 3 To find the solution to: dyðtÞ þ 4yðtÞ ¼ 3uðtÞ where yð0Þ ¼ 2 dt (same boundary condtion but different input) The final complete solution is then: yðtÞ ¼ . . . . . . . . . . . .

Introduction to invariant linear systems

yðtÞ ¼

207

 1 3 þ 5e4t 4

23

Because We only need rework the zero-state response because the zero-input response is the same as in Example 1. dyzs ðtÞ þ 4yzs ðtÞ ¼ 3uðtÞ where yzs ð0Þ ¼ 0, dt 3 3 we see that fsYzs ðsÞ  yzs ð0Þg þ 4Yzs ðsÞ ¼ that is ðs þ 4ÞYzs ðsÞ ¼ s s

Taking Laplace transforms of

Therefore Yzs ðsÞ ¼

3 3 3 3 3e4t ¼  yielding yzs ðtÞ ¼  . sðs þ 4Þ 4s 4ðs þ 4Þ 4 4

Therefore the complete solution is: yðtÞ ¼ yzi ðtÞ þ yzs ðtÞ 3 3e4t  4 4  1 4t ¼ 3 þ 5e 4 ¼ 2e4t þ

So the zero-input and zero-state responses for the differential equation: dyðtÞ  yðtÞ ¼ et uðtÞ where yð0Þ ¼ 1 and yðtÞ ¼ 0 for t < 0 are . . . . . . . . . . . . dt yzi ¼ et uðtÞ yzs ðtÞ ¼ uðtÞ sinh t Because dyðtÞ  yðtÞ ¼ et uðtÞ where yð0Þ ¼ 1 and yðtÞ ¼ 0 for t < 0 dt Zero-input response dyzi ðtÞ  yzi ðtÞ ¼ 0 where yzi ð0Þ ¼ 1. yzi ðtÞ ¼ Aet uðtÞ, yzi ð0Þ ¼ 1 so yzi ðtÞ ¼ et uðtÞ dt Zero-state response dyzs ðtÞ  yzs ðtÞ ¼ et uðtÞ where yzs ð0Þ ¼ 0. dt Taking Laplace transforms ðs  1ÞYzs ðsÞ ¼

  1 1 1 1 1 so Yzs ðsÞ ¼ ¼ þ sþ1 ðs  1Þðs þ 1Þ 2 s þ 1 s  1

Therefore yzs ðtÞ ¼

1  t e þ et ¼ sinh t for t 0. That is, yzs ðtÞ ¼ uðtÞ sinh t 2

24

208

Programme 6

And another. The zero-input and zero-state responses for the differential equation: d2 yðtÞ dyðtÞ 5 þ 6yðtÞ ¼ tuðtÞ dt 2 dt  dyðtÞ where yð0Þ ¼ 1 and ¼ 1 and yðtÞ ¼ 0 for t < 0 dt t¼0 are . . . . . . . . . . . .  yzi ðtÞ ¼ 4e2t  3e3t uðtÞ  uðtÞ yzs ðtÞ ¼ 9e2t þ 4e3t þ 6t þ 5 36

25

Because d2 y dyðtÞ þ 6yðtÞ ¼ tuðtÞ 5 dt 2 dt  dyðtÞ where yð0Þ ¼ 1 and ¼ 1 and yðtÞ ¼ 0 for t < 0 dt  t¼0

Zero-input response

 d2 yzi ðtÞ dyzi ðtÞ dyzi ðtÞ þ 6y  5 ðtÞ ¼ 0 where y ð0Þ ¼ 1 and ¼ 1. zi zi dt 2 dt dt t¼0

The auxiliary equation is m2  5m þ 6 ¼ ðm  2Þðm  3Þ ¼ 0 therefore yzi ðtÞ ¼ Ae2t þ Be3t and

dyzi ðtÞ ¼ 2Ae2t þ 3Be3t . dt

Therefore A þ B ¼ 1 and 2A þ 3B ¼ 1 so A ¼ 4, B ¼ 3 thus  yzi ðtÞ ¼ 4e2t  3e3t uðtÞ. Zero-state response

 d2 yzs ðtÞ dyzs ðtÞ dyzs ðtÞ þ 6yzs ðtÞ ¼ tuðtÞ where yzs ð0Þ ¼ 0 and 5 ¼ 0. dt 2 dt dt t¼0 Taking Laplace transforms ðs2  5s þ 6ÞYzs ðsÞ ¼

1 so s2

1 1 5 1 1 þ þ so ¼ þ s2 ðs  2Þðs  3Þ 6s2 36s 4ðs  2Þ 9ðs  3Þ  uðtÞ yzs ðtÞ 9e2t þ 4e3t þ 6t þ 5 36

Yzs ðsÞ ¼

Move to the next frame

Introduction to invariant linear systems

209

26

Zero-input, zero-response We have already seen that a system yðtÞ ¼ LfxðtÞg is linear if sums and scalar multiples are preserved, that is if Lfx1 ðtÞ þ x2 ðtÞg ¼ Lfx1 ðtÞg þ Lfx2 ðtÞg and LfxðtÞg ¼ LfxðtÞg where  is a constant. In particular, by choosing  ¼ 0 then Lf0g ¼ 0 which shows that if nothing is put into a linear system nothing will come out – zero input yields zero output. In the previous frame we considered two systems described by the differential equations: dyðtÞ  yðtÞ ¼ et uðtÞ where yð0Þ ¼ 1 dt  d2 yðtÞ dyðtÞ dyðtÞ þ 6yðtÞ ¼ tuðtÞ where yð0Þ ¼ 1 and (b)  5 ¼ 1 dt 2 dt dt t¼0 (a)

Do these equations give rise to linear or nonlinear systems? (a) . . . . . . . . . . . . (b) . . . . . . . . . . . . (a) Nonlinear

(b) Nonlinear

Because (a) yzi ðtÞ ¼ et uðtÞ so that the zero-input response is not zero. Therefore this differential equation does not give rise to a linear system but to a nonlinear system.  (b) yzi ðtÞ ¼ 4e2t  3e3t uðtÞ so that the zero-input response is not zero. Therefore this differential equation does not give rise to a linear system but to a nonlinear system. Is it possible for a differential equation to give rise to a linear system? To answer this question we now re-cast these two differential equations with general boundary conditions. Firstly, dyðtÞ  yðtÞ ¼ et uðtÞ where yð0Þ ¼ A dt Here, for t < 0 the equation becomes, with zero-input: dyðtÞ  yðtÞ ¼ 0 where yð0Þ ¼ A dt With solution being the zero-input response yzi ðtÞ ¼ Aet and this can only be zero if A ¼ 0. The differential equation only gives rise to a linear system if the value of the boundary condition is zero.

27

210

Programme 6

Now you try. The conditions that the differential equation:  d2 yðtÞ dy dyðtÞ  5 þ 6yðtÞ ¼ tuðtÞ where yð0Þ ¼ K1 and ¼ K2 dt 2 dt dt t¼0 gives rise to a linear system are . . . . . . . . . . . .

28

K1 ¼ 0 and K2 ¼ 0 Because Here, for t < 0 the equation becomes, with zero-input:

 d2 yðtÞ dyðtÞ dyðtÞ þ 6yðtÞ ¼ 0 where yð0Þ ¼ K1 and 5 ¼ K2 dt 2 dt dt t¼0

with solution being the zero-input response yzi ðtÞ ¼ Ae2t þ Be3t . Applying the boundary conditions: yð0Þ ¼ K1 :  dyðtÞ ¼ K2 : dt t¼0

A þ B ¼ K1 2A þ 3B ¼ K2

with solution A ¼ 3K1  K2 and B ¼ K2  2K1 giving yzi ðtÞ ¼ ð3K1  K2 Þe2t þ ðK2  2K1 Þe3t This can only be zero if: 3K1  K2 ¼ 0 2K1 þ K2 ¼ 0 that is if K1 ¼ 0 and K2 ¼ 0 The differential equation only gives rise to a linear system if the values of the boundary conditions are zero. This is a general property – a constant coefficient, linear differential equation only gives rise to a linear system if the values of all the boundary conditions are zero. This is an important fact to be remembered. Next we look at these differential equations and time-invariance. Move to the next frame

Introduction to invariant linear systems

Time-invariance Consider the differential equation in Frame 17, namely: dyðtÞ þ 4yðtÞ ¼ tuðtÞ where yð0Þ ¼ y0 and yðtÞ ¼ 0 for t < 0 dt with solution yðtÞ ¼

1  ½16y0 þ 1e4t þ 4t  1 uðtÞ. 16

If we re-visit this equation but this time change the boundary condition to yð0Þ ¼ 0 the solution is the zero-state solution: yzs ðtÞ ¼

1  4t e þ 4t  1 uðtÞ 16

If we now delay the input by 3 units so that the input becomes ðt  3Þuðt  3Þ the differential equation becomes: dyðtÞ þ 4yðtÞ ¼ ðt  3Þuðt  3Þ where yð3Þ ¼ 0 and yðtÞ ¼ 0 for t < 3 dt The homogeneous solution is again yh ðtÞ ¼ Ae4t uðtÞ and the particular solution has the form yp ðtÞ ¼ Ct þ D. However, substituting into the differential equation we now find that C þ 4Ct þ 4D ¼ t  3 for t 3 from which we find that 4C ¼ 1 and C þ 4D ¼ 3. Therefore: 13 12 1 3 1 ¼  ¼  and the particular solution is C ¼ 1=4, D ¼  16 16 16 4 16     t 3 1 t 3 1   yp ðtÞ ¼ uðt  3Þ so that yðtÞ ¼ Ae4t þ uðt  3Þ. 4 16 4 16 Applying the boundary condition yð3Þ ¼ 0: yð3Þ ¼ Ae12 

1 e12 ¼ 0 that is A ¼ giving the solution to 16 16

dyðtÞ þ 4yðtÞ ¼ ðt  3Þuðt  3Þ where yð3Þ ¼ 0 as dt  1  4ðt3Þ e yðtÞ ¼ þ 4ðt  3Þ  1 uðt  3Þ 16 This is the same solution but delayed by the same amount as the input. Consequently, the system is not only linear but it is also time-invariant. Indeed, the zero values of the boundary conditions ensure that the general constant coefficient, linear differential equation gives rise to a system that is not only linear but also time-invariant. Move to the next frame

211

29

212

Programme 6

Responses of a continuous system 30

Impulse response We shall soon see that any continuous, linear, time-invariant system has the important property that its response to any input can be found from knowing its response to the unit impulse ðtÞ – a property that can be exploited to solve such differential equations as considered here [refer to Programme 4]. When the input to a linear, time-invariant system is the unit impulse ðtÞ the response is denoted by hðtÞ and is referred to as the impulse response. That is: hðtÞ ¼ LfðtÞg and, because the system is time-invariant hðt  t0 Þ ¼ Lfðt  t0 Þg

Arbitrary input From the properties of the unit impulse ðtÞ we can express an arbitrary input xðtÞ in terms of the unit impulse ðtÞ as: ð1 xð Þðt  Þ d

xðtÞ ¼ 1

so that if the response to this arbitrary input is yðtÞ then: yðtÞ ¼ LfxðtÞg ð 1 xð Þðt  Þ d

¼L 1

Because the variable inside the integral is the variable of integration and the operator L is acting on t and not on the operator can be moved inside the integral. (Recall that an integral is a limit of a sum and, for linear systems, sums are preserved.) Therefore: ð 1 ð1 xð Þðt  Þ d ¼ xð ÞLfðt  Þgd

L 1 ð1 1 ¼ xð Þhðt  Þd

1

because the system is given as time invariant. This is a remarkable result so we shall look at it very closely. We have just found that: ð1 xð Þhðt  Þd . LfxðtÞg ¼ 1

You have seen integrals like this one before, can you recall? This integral is the . . . . . . . . . . . . between the input to the system and the impulse response of the system.

Introduction to invariant linear systems

213

31

convolution Because The convolution between xðtÞ and hðtÞ is obtained by first reversing hðtÞ to form hðtÞ, changing the variable to the dummy variable of the integral to form xð Þ and hð Þ, advancing hð Þ by t to form hð þ tÞ ¼ hðt  Þ, taking the product of this with xð Þ and integrating with respect to to form [refer to Programme 3, Frame 43]: ð1 xð Þhðt  Þd ¼ xðtÞ hðtÞ yðtÞ ¼ 1

Aside: It is also worth remembering that convolution is a commutative operation. That is: ð1 xð Þhðt  Þd and yðtÞ ¼ xðtÞ hðtÞ ¼ ð1 1 hð Þxðt  Þd

yðtÞ ¼ hðtÞ xðtÞ ¼ 1

Bear this fact in mind as we shall make use of it a little later on. This is a most important result because it tells us that if we know the impulse response of a continuous, linear, time-invariant system then we can find the response of the system to any input simply by evaluating the convolution of the input with the system impulse response. To see convolution in action move to the next frame

32

Convolution As an example of convolution consider the response to a unit step input uðtÞ at time t ¼ 0 of a system that has an impulse response: hðtÞ ¼ uðtÞekt

k>0

The graphs of the input and the impulse response are: h(t) = u (t)e–kt

x(t) = u(t) 1

1

t

t

214

Programme 6

To evaluate the convolution yðtÞ ¼ xðtÞ hðtÞ we first need to change t to the variable of integration to form xð Þ and hð Þ. We then require hð Þ to be advanced by t to form hð  tÞ where t > 0. Next we flip hð  tÞ about the vertical to form hð½  tÞ ¼ hðt  Þ to overlap with the unit step. h (t – τ) 1

1

t

τ

t

τ

The only non-zero overlap inside the convolution integral is then between the values ¼ 0 and ¼ t provided t > 0. If t < 0 then there is no overlap at all with the unit step function so that: yðtÞ ¼ xðtÞ hðtÞ ð1 ¼ xð Þhðt  Þd

1

ðt ð1 xð Þhðt  Þd þ xð Þhðt  Þd þ xð Þhðt  Þd

1 0 t ðt xð Þ ¼ 0 for < 0 and hðt  Þ ¼ 0 for > t ¼ 0 þ xð Þhðt  Þd þ 0

¼

ð0

0

Now xð Þ ¼ uð Þ and hðt  Þ ¼ uðt  Þekðt Þ so that ðt uð Þuðt  Þekðt Þ d

yðtÞ ¼

¼0 ðt uðt  Þekðt Þ d

¼

¼0

¼

ð0

uðpÞekðpÞ dðpÞ

p¼t

where p ¼ t  so p ¼ t when ¼ 0 and p ¼ 0 when ¼ t

Furthermore, d ¼ dðpÞ ¼ dp so that ð0 yðtÞ ¼  uðpÞekp dp ¼ ¼

ðt

p¼t

p¼0 ðt

uðpÞekp dp ekp dp

p¼0

kp t e 1  ekt ¼ ¼ k 0 k You try one. The response to the input xðtÞ ¼ uðt  aÞ of a system with impulse response hðtÞ ¼ uðtÞebt where a > 0 and b > 0 is . . . . . . . . . . . .

Introduction to invariant linear systems

yðtÞ ¼

215

33

 1 1  ebðtaÞ uðt  aÞ b

Because The graphs of the input and the impulse response are: h (t) = u (t)e–bt

x(t) = u(t – a) 1

1

t

t

To evaluate the convolution yðtÞ ¼ xðtÞ hðtÞ we first need to change t to the variable of integration to form xð Þ and hð Þ. We then require hð Þ to be advanced by t to form hðt  Þ where t > 0. Next we flip hðt  Þ about the vertical to form hð½  tÞ to overlap with the unit step. If t < a there is no overlap with the unit step function. h(t – τ) 1

1

τ

t

t

a τ

Provided t > a the only non-zero overlap inside the convolution integral is over the interval a to t so that: ð1 xð Þhðt  Þ d

yðtÞ ¼ xðtÞ hðtÞ ¼ 1 ðt ð1 ða xð Þhðt  Þ d þ xð Þhðt  Þ d þ xð Þhðt  Þ d

¼ 1 a t ðt xð Þ ¼ 0 for < a and ¼ 0 þ xð Þhðt  Þ d þ 0 a hðt  Þ ¼ 0 for > t ðt bðt Þ d

¼ uð  aÞuðt  Þe a ðt ¼ uðt  Þebðt Þ d

a

¼ ¼

ð0 ta ð ta

uðxÞebðxÞ dðxÞ uðxÞebx dx

0



where x ¼ t 

ta

 1 1  ebðtaÞ for t > a b 0  1 1  ebðtaÞ uðt  aÞ The response can then be written as yðtÞ ¼ b ebx ¼ b

¼

Next frame

216

34

Programme 6

Exponential response When the input to a continuous, linear, time-invariant system with impulse response hðtÞ is the exponential xðtÞ ¼ Aest (A and s being constants) the system response is given as: yðtÞ ¼ Aest hðtÞ or, because convolution is commutative, yðtÞ ¼ hðtÞ Aest That is: yðtÞ ¼ A

ð1 1

es hðt  Þ d or yðtÞ ¼ A

ð1 1

hð Þesðt Þ d

Taking the latter form of the convolution of the input with the impulse response we see that: ð1 hð Þesðt Þ d

yðtÞ ¼ A 1 ð1 st hð Þes d

¼ Ae 1

¼ Aest HðsÞ ð1 where HðsÞ ¼ hð Þes d . 1

Notice that if hðtÞ ¼ 0 for t < 0 then: ð1 HðsÞ ¼ hð Þes d so that HðsÞ is the . . . . . . . . . . . . transform of hðtÞ 0

35

Laplace Because Recalling from Frame 1 of Programme 2 the Laplace transform FðsÞ of function f ðtÞ is defined as: ð1 FðsÞ ¼ f ðtÞest dt 0

The expression HðsÞ is called the system’s transfer function. Furthermore, because the system response is simply a scaled version of the input, the scaling factor being HðsÞ we can say that:  LfAest g ¼ HðsÞ Aest which tells us that Aest is an eigenfunction of the operator L with corresponding eigenvalue HðsÞ (refer to Engineering Mathematics, Eighth Edition). The converse is also true. The eigenfunctions of a linear, time-invariant system are exponential functions, which places the exponential function in a special position with respect to linear time-invariant systems as we shall appreciate later.

Introduction to invariant linear systems

217

For example, the transfer function corresponding to the input 5est of the continuous, linear, time-invariant system with impulse response: ( t 0 e6t hðtÞ ¼ 0 t 6 ¼ sþ6 ¼5

0

If s  6 then s þ 6  0 and the integral diverges. So the transfer function corresponding to the input 3est uðt  2Þ of the continuous, linear, time-invariant system with impulse response hðtÞ ¼ et uðtÞ is . . . . . . . . . . . .

36

3e2ð1þsÞ provided s > 1 sþ1 Because HðsÞ ¼ 3

ð1

est uðt  2ÞuðtÞe d

ð01

eð1þsÞt d

ð1þsÞt 1 e ¼3 ð1 þ sÞ 2   eð1þsÞ2 ¼3 0 ð1 þ sÞ

¼3

2

¼

3e2ð1þsÞ sþ1

provided s > 1

If s  1 then s þ 1  0 and the integral diverges. Move to the next frame

218

37

Programme 6

The transfer function HðsÞ We have seen that the response yðtÞ of a continuous, linear, time-invariant system to an input xðtÞ is given in terms of the system’s unit impulse response hðtÞ as the convolution: yðtÞ ¼ xðtÞ hðtÞ We have also seen that provided hðtÞ ¼ 0 for t < 0 then the system’s transfer function HðsÞ is the Laplace transform of hðtÞ. That is: ð1 hðtÞest dt HðsÞ ¼ 0

Referring now to Frame 45 of Programme 3 we see that the convolution theorem states that the Laplace transform of a convolution of two functions is equal to the product of their respective Laplace transforms. Therefore, if ð1 ð1 yðtÞest dt and XðsÞ ¼ xðtÞest dt then YðsÞ ¼ XðsÞHðsÞ YðsÞ ¼ 0

0

For example to find the response of a time-invariant linear system with impulse response hðtÞ ¼ uðtÞet to an input xðtÞ ¼ uðtÞ  uðt  1Þ all we need do is: (a) Find the Laplace transforms of hðtÞ ¼ uðtÞet and xðtÞ ¼ uðtÞ  uðt  1Þ These are HðsÞ ¼

es 1 es and XðsÞ ¼  s sþ1 s

(b) Obtain YðsÞ ¼ XðsÞHðsÞ es e2s  sðs þ 1Þ sðs þ 1Þ   1 1  ¼ es  e2s s sþ1

This is YðsÞ ¼

(c) Take the inverse Laplace transform  s  2s e es e e2s   YðsÞ ¼  s sþ1 s sþ1     ðt1Þ so yðtÞ ¼ uðt  1Þ  uðt  1Þe  uðt  2Þ  uðt  2Þeðt2Þ     ¼ uðt  1Þ 1  eðt1Þ  uðt  2Þ 1  eðt2Þ You try one. The response of a time-invariant linear system with impulse response hðtÞ ¼ uðt  1Þ to an input xðtÞ ¼ uðtÞ sin t  uðt  1Þ sinðt  1Þ is yðtÞ ¼ . . . . . . . . . . . .

Introduction to invariant linear systems

uðt  1Þð1  cosðt  1ÞÞ  uðt  2Þð1  cosðt  2ÞÞ Because (a) The Laplace transforms of hðtÞ ¼ uðt  1Þ and xðtÞ ¼ uðtÞ sin t  uðt  1Þ sinðt  1Þ are: HðsÞ ¼

es 1 es and XðsÞ ¼ 2  2 s þ1 s þ1 s

(b) The Laplace transform of yðtÞ is: YðsÞ ¼ XðsÞHðsÞ es e2s  2 þ 1Þ sðs þ 1Þ   s 1 s 2s  ¼ e e s s2 þ 1

¼

sðs2

(c) Then the inverse Laplace transform is  s  2s e ses e se2s  2  2  l1 l1 fYðsÞg ¼ l1 s s þ1 s s þ1 so yðtÞ ¼ ðuðt  1Þ  uðt  1Þ cosðt  1ÞÞ  ðuðt  2Þ  uðt  2Þ cosðt  2ÞÞ ¼ uðt  1Þð1  cosðt  1ÞÞ  uðt  2Þð1  cosðt  2ÞÞ In summary, to find the response of a continuous, linear, time-invariant system all we need to do is take the inverse Laplace transform of the product of the Laplace transforms of the input and the transfer function – the transfer function being the Laplace transform of the unit impulse response. yðtÞ ¼ l1 fXðsÞHðsÞg where HðsÞ ¼ lfhðtÞg So, given the input all we need do to proceed is to determine the impulse response. However, for those differential equations that give rise to continuous, linear, time-invariant systems we have a much simpler way of determining the transfer function. Next frame

219

38

220

39

Programme 6

Differential equations To solve the equation y00 ðtÞ  5y0 ðtÞ þ 6yðtÞ ¼ xðtÞ where yð0Þ ¼ 0, y0 ð0Þ ¼ 0 and y00 ð0Þ ¼ 0 we note that the differential equation gives rise to a continuous, linear, timeinvariant system. That is yðtÞ ¼ LfxðtÞg Accordingly, the exponential function Aest is an eigenfunction of the system whose corresponding eigenvalue is the system’s transfer function HðsÞ. That is: if xðtÞ ¼ Aest then yðtÞ ¼ Aest HðsÞ Substituting these into the differential equation, we then see that:  st 00  0  Ae HðsÞ 5 Aest HðsÞ þ6 Aest HðsÞ ¼ Aest  2  s  5s þ 6 HðsÞAest ¼ Aest 1 HðsÞ ¼ 2 s  5s þ 6 HðsÞ is the Laplace transform of the left-hand side of the differential equation and now you see the importance of all the boundary conditions having a value of zero. If they did not then their non-zero values would be automatically incorporated into the Laplace transform so giving a different expression to this one here. Now if, for example, the input is xðtÞ ¼ e5t uðtÞ then its Laplace transform is 1 giving the Laplace transform of the system’s response yðtÞ as: XðsÞ ¼ sþ5 YðsÞ ¼ XðsÞHðsÞ 1 ðs þ 5Þðs  2Þðs  3Þ P Q R ¼ þ þ sþ5 s2 s3

¼

Taking inverse Laplace transforms we see that yðtÞ ¼ Pe5t þ Qe2t þ Re3t where the values of P, Q and R can be found – the usual partial fractions procedure giving the solution to the differential equation as: yðtÞ ¼

e5t e2t e3t  þ 56 7 8

Therefore, the solution to y00 ðtÞ þ 3y0 ðtÞ  28yðtÞ ¼ et uðtÞ where yð0Þ ¼ 0, y0 ð0Þ ¼ 0 and y00 ð0Þ ¼ 0 is yðtÞ ¼ . . . . . . . . . . . .

Introduction to invariant linear systems

yðtÞ 

221

40

et e7t e4t þ þ 30 66 55

Because 1 s2 þ 3s  28 1 The Laplace transform of the input xðtÞ ¼ et uðtÞ is XðsÞ ¼ sþ1 The Laplace transform of the response yðtÞ is then 1 1  2 YðsÞ ¼ s þ 1 s þ 3s  28 1 ¼ ðs þ 1Þðs þ 7Þðs  4Þ The auxiliary equation is s2 þ 3s  28 ¼ 0 so that HðsÞ ¼

¼

P Q R þ þ sþ1 sþ7 s4

which gives the response as yðtÞ ¼ Pet þ Qe7t þ Re4t . Now 1 P Q R ¼ þ þ ðs þ 1Þðs þ 7Þðs  4Þ s þ 1 s þ 7 s  4 ¼

Pðs þ 7Þðs  4Þ þ Qðs þ 1Þðs  4Þ þ Rðs þ 1Þðs þ 7Þ ðs þ 1Þðs þ 7Þðs  4Þ

¼

ðP þ Q þ RÞs2 þ ð3P  3Q þ 8RÞs þ ð28P  4Q þ 7RÞ ðs þ 1Þðs þ 7Þðs  4Þ

Therefore

0

PþQþR¼0

1

1

B 3P  3Q þ 8R ¼ 0 that is @ 3 3 28P  4Q þ 7R ¼ 1 28 4 0 1 0 1 1=30 P B C B C giving @ Q A ¼ @ 1=66 A R

10

1

0 1 0 CB C B C 8 A@ Q A ¼ @ 0 A 7 R 1 1

P

1=55

et e7t e4t þ þ Therefore yðtÞ ¼  30 66 55 And just one more. To find the solution to y00 ðtÞ  4yðtÞ ¼ ½uðtÞ  uðt  1Þt where yð0Þ ¼ 0, y0 ð0Þ ¼ 0 and y00 ð0Þ ¼ 0 we first need to arrange the input into a form that is Laplace transformable. In other words, we need to convert uðt  1Þt into a form involving uðt  1Þðt  1Þ that is: ½uðtÞ  uðt  1Þt ¼ . . . . . . . . . . . .

222

Programme 6

41

½uðtÞ  uðt  1Þt ¼ uðtÞt  uðt  1Þðt  1Þ  uðt  1Þ Because uðt  1Þðt  1Þ ¼ uðt  1Þt  uðt  1Þ therefore ½uðtÞ  uðt  1Þt ¼ uðtÞt  uðt  1Þðt  1Þ  uðt  1Þ The differential equation then becomes: y00 ðtÞ  4yðtÞ ¼ uðtÞt  uðt  1Þðt  1Þ  uðt  1Þ where yð0Þ ¼ 0, y0 ð0Þ ¼ 0 and y00 ð0Þ ¼ 0. The solution is then yðtÞ ¼ . . . . . . . . . . . .

42

o  uðt  1Þ n uðtÞ  4t þ e2t  e2t  4t þ 3e2ðt1Þ þ e2ðt1Þ 16 16 Because Taking the Laplace transform of the left-hand side tells us that  1 1 1 1  ¼ HðsÞ ¼ 2 s 4 4 s2 sþ2 Taking the Laplace transform of the right hand side we get XðsÞ ¼

1 es es   s2 s2 s

Therefore     1 es es 1 1 1  YðsÞ ¼ 2  2  s 4 s2 sþ2 s s Breaking into partial fractions:  1 1 2 1 1 þ ¼  s2 ðs þ 2Þ 4 s2 s s þ 2  1 1 1 1 ¼  sðs þ 2Þ 2 s s þ 2  1 1 2 1 1   þ ¼ s2 ðs  2Þ 4 s2 s s  2  1 1 1 1 ¼  sðs  2Þ 2 s  2 s

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223

Therefore     1 es es 1 1 1  YðsÞ ¼ 2  2  s s s 4 s2 sþ2   ð1  es Þ 4 1 1 es 2 1 1  2þ  þ  þ  ¼ 16 s s2 sþ2 s s2 sþ2 8 Giving o  uðt  1Þ n uðtÞ  4t þ e2t  e2t  4ðt  1Þ þ e2ðt1Þ  e2ðt1Þ 16 16 o uðt  1Þ n 2ðt1Þ  2 þ e þ e2ðt1Þ 8 o  uðt  1Þ n uðtÞ  4t þ e2t  e2t  4t þ 3e2ðt1Þ þ e2ðt1Þ ¼ 16 16

yðtÞ ¼

This completes our work on continuous linear systems. Now we shall move on to consider discrete linear systems. Read on.

Responses of a discrete system The discrete unit impulse The value of the unit impulse ðtÞ in the study of continuous linear systems cannot be overestimated. It permits the system response to any input to be found once the system’s response to the unit impulse is known. In the case of discrete systems the equivalent is called the discrete unit impulse ½n which is defined as:  1 n¼0 where n is an integer. ½n ¼ 0 n 6¼ 0 Associated with the discrete unit impulse is the shifted discrete unit impulse:  1 n¼k ½n  k ¼ 0 n 6¼ 0 which enables us to select a particular component of an expression x½n via the equation: x½k ¼ x½n½n  k This is because the right-hand side x½n½n  k ¼ 0 unless n ¼ k. Indeed, any sequence x½n can be considered as consisting of a collection of scaled and shifted discrete unit impulses. For example, the geometric sequence: x½n ¼ 3n has values . . . , 32 , 31 , 1, 3, 32 , 33 , . . . and can be alternatively written as the sum x½n ¼ . . . þ 32 ½n  ð2Þ þ 31 ½n  ð1Þ þ 1½n  0 þ 3½n  1 þ 32 ½n  2 þ . . .

43

224

Programme 6

From this sum any term of the sequence can be selected. For instance: x½2 ¼ . . . þ 32 ½2  ð2Þ þ 31 ½2  ð1Þ þ 1½2  0 þ 3½2  1 þ 32 ½2  2 þ . . . ¼ . . . þ 32 ½4 þ 31 ½3 þ 1½2 þ 3½1 þ 32 ½0 þ . . . ¼ . . . þ 32  0 þ 31  0 þ 1  0 þ 3  0 þ 32  1 þ . . . ¼ 32 For this reason the discrete unit impulse is also referred to as the unit sample and it can be used to decompose any sequence into a sum of weighted and shifted unit samples. For example: x½n ¼ . . . þ x½2½n  ð2Þ þ x½1½n  ð1Þ þ x½0½n  0 þ x½1½n  1 þ x½2½n  2 þ . . . ¼

1 X

x½k½n  k

k¼1

Note the analogy with the continuous case: ð1 f ðsÞðt  sÞ ds f ðtÞ ¼ 1

When the input to a linear, shift-invariant system is the discrete unit impulse ½n the response is denoted by h½n and is referred to as the discrete unit impulse response. That is: h½n ¼ Lf½ng and, because it is shift-invariant, h½n  n0  ¼ Lf½n  n0 g Next frame

44

Arbitrary input Just like the continuous system a discrete, linear, shift-invariant system has the important property that its response to any input can be found from its response to the discrete unit impulse ½n. Recalling the discrete decomposition of a sequence as 1 X x½k½n  k x½n ¼ k¼1

then the response of a linear system to this input is y½n where: y½n ¼ Lfx½ng ( ) 1 X ¼L x½k½n  k k¼1

¼ ¼ ¼

1 X k¼1 1 X k¼1 1 X k¼1

Lfx½k½n  kg

because L is linear and so sums are preserved

x½kLf½n  kg

because L is linear and so scalar multiples are preserved

x½kh½n  k

because L is shift-invariant

Introduction to invariant linear systems 1 X

That is y½n ¼

x½kh½n  k which is referred to as the convolution sum of

k¼1

x½n and h½n, alternatively written as: x½n h½n

(also h½n x½n since the convolution sum is commutative)

So, by direct analogy with a continuous system, the response of a discrete linear system can be obtained from the convolution sum of the input with the system’s unit impulse response. For example, a discrete, linear, shift-invariant system has the unit impulse response:  1 n 0 h½n ¼ u½n  1 the discrete unit step function where u½n ¼ 0 n n  1

k¼0

¼ 2n  1

sum of the first n terms of a geometric series with common ratio 2

Try one yourself. A discrete linear shift-invariant system has a unit impulse response h½n ¼ u½n  4 and a response to the input x½n ¼ nu½n of : y½n ¼ . . . . . . . . . . . .

225

226

Programme 6

45

y½n ¼

1 ðn  4Þðn  3Þ 2

Because y½n ¼ x½n h½n 1 X ¼ x½kh½n  k ¼ ¼

k¼1 1 X

ku½ku½n  k  4

k¼1 1 X

ku½n  k  4

since ku½k ¼ 0 for k  0

k

since u½n  k  4 ¼ 0 for k > n  4

k¼0

¼

n4 X k¼0

¼

ðn  4Þðn  3Þ 2

sum of the first n  4 integers Move to the next frame

46

Exponential response When the input to a discrete, linear, shift-invariant system with impulse response h½n is the exponential x½n ¼ Azn (A being constant) the system response is given as: y½n ¼ h½n x½n 1 X h½kx½n  k ¼ ¼

k¼1 1 X

h½kAznk

k¼1

¼ Azn

1 X

h½kzk

k¼1

¼ Azn H½z where H½z ¼

1 X

h½kzk which you will recognize as the Z transform of h½k

k¼1

[refer to Programme 5].

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227

As in the continuous case we call H½z the system transfer function. For example, the transfer function H½z of a discrete, linear, shift-invariant system with impulse response: (  n 1 0n4 5 h½n ¼ 0 otherwise is HðzÞ ¼

1 X

h½kzk

k¼1

¼

4  X 1 k 5

zk

k¼0

 1  2  3  4 z0 þ 15 z1 þ 15 z2 þ 15 z3 þ 15 z4 1 1 1 1 ¼1þ þ þ þ 5z 25z2 125z3 625z4 ¼

1 0 5

So, the transfer function HðzÞ of a discrete, linear, shift-invariant system with impulse response:  nu½n 0  n  3 h½n ¼ 0 otherwise is . . . . . . . . . . . .

HðzÞ ¼

47

1 2 3 þ þ z z2 z3

Because HðzÞ ¼

1 X

h½kzk

k¼1

¼

3 X

ku½kzk

k¼0

¼ 0u½0z0 þ 1u½1z1 þ 2u½2z2 þ 3u½3z3 1 2 3 ¼ þ 2þ 3 z z z Move to the next frame

Transfer function We have seen in Frame 44 that the response y½n to the input x½n to a discrete, linear, shift-invariant system with impulse response h½n is given as the convolution sum: y½n ¼ h½n x½n 1 X ¼ h½kx½n  k k¼1

48

228

Programme 6

So the Z transform of the response YðzÞ is given as: 1 X

YðzÞ ¼

y½nzn

n¼1 1 X

n¼1

k¼1

¼

1 X

¼ ¼

!

1 X

k¼1 1 X

h½k

h½kx½n  k zn 1 X

! n

x½n  kz

interchanging sums

n¼1

h½kzk XðzÞ

by the first shift property of the Z transform

k¼1

¼ HðzÞXðzÞ 1 X

where the transfer function HðzÞ ¼

h½kzk is the Z transform of the discrete

k¼1

impulse response h½k. Consequently, for discrete, linear, shift-invariant systems the transfer function (as in the continuous case) completely characterizes the system and permits the response to any input to be obtained. Move to the next frame

49

Difference equations Given a linear, constant coefficient difference equation it is possible to derive its transfer function and from that the impulse response. For example, consider the difference equation: y½n ¼ 4y½n  2 þ x½n Taking the Z transform of both sides where Zfx½ng ¼ XðzÞ and Z fy½ng ¼ YðzÞ we see that: YðzÞ ¼ ð. . . . . . . . . . . .ÞXðzÞ

50

YðzÞ ¼

1 XðzÞ ð1  4z1 Þ

Because Z fy½ng ¼ Z f4y½n  1 þ x½ng ¼ 4Z fy½n  1g þ Z fx½ng That is:

 YðzÞ ¼ 4z1 YðzÞ þ XðzÞ so that YðzÞ 1  4z1 ¼ XðzÞ

and so: YðzÞ ¼

1 XðzÞ ð1  4z1 Þ

This means that the transfer function is: HðzÞ ¼ . . . . . . . . . . . .

Introduction to invariant linear systems

229

z z4

51

Because YðzÞ ¼

1 XðzÞ ð1  4z1 Þ

¼ HðzÞXðzÞ Giving: 1 ð1  4z1 Þ z ¼ z4

HðzÞ ¼

From this we can now determine the impulse response: h½n ¼ . . . . . . . . . . . . 4n u½n Because h½n ¼ Z 1 fHðzÞg n z o ¼ Z 1 z4 ¼ 4n u½n Now you try one. The transfer function and hence the impulse response of the difference equation: y½n ¼ 2y½n  1 þ 3y½n  2 þ x½n  1  2x½n  2 are given as: HðzÞ ¼ . . . . . . . . . . . . h½n ¼ . . . . . . . . . . . .

52

230

Programme 6

53

3=4 1=4 þ ðz þ 1Þ ðz  3Þ  n  o 1  3 ð1Þn1 þ 3n1 u½n  1 h½n ¼ 4 4

HðzÞ ¼

Because Taking the Z transform of both sides we see that: Z fy½ng ¼ 2Z fy½n  1g þ 3Z fy½n  2g þ Z fx½n  1g  2Zfx½n  2g that is: YðzÞ ¼ 2z1 YðzÞ þ 3z2 YðzÞ þ z1 XðzÞ  2z2 XðzÞ so that:   YðzÞ 1  2z1  3z2 ¼ XðzÞ z1  2z2 and so:  1 z  2z2 XðzÞ ¼ HðzÞXðzÞ giving: YðzÞ ¼ ð1  2z1  3z2 Þ  1 z  2z2 HðzÞ ¼ ð1  2z1  3z2 Þ ¼

ðz  2Þ ðz2  2z  3Þ

¼

ðz  2Þ ðz þ 1Þðz  3Þ

¼

3=4 1=4 þ ðz þ 1Þ ðz  3Þ

From this we can now determine the impulse response: h½n ¼ Z 1 fHðzÞg   3 z 1 z ¼ Z 1 z1 þ Z 1 z1 4 ðz þ 1Þ 4 ðz  3Þ 3 1 ¼ ð1Þn1 u½n  1 þ 3n1 u½n  1 4 4

54

Next frame

This procedure can be reversed. For example to find the linear, constant coefficient difference equation whose impulse response is   h½n ¼ 0:5 ð5Þnþ2 u½n we proceed as follows. Since the transfer function is the Z transform of the unit impulse then: HðzÞ ¼ Z fh½ng n   o ¼ Z 0:5 ð5Þnþ2 u½n o 1 n ¼ Z ð5Þnþ2 u½n 2  25  Z ð5Þn u½n ¼ 2 12:5z ¼ ðz þ 5Þ

Introduction to invariant linear systems

231

From this we can deduce the input-output relationship: YðzÞ HðzÞ ¼ XðzÞ 12:5z ¼ zþ5 so that: 12:5z XðzÞ YðzÞ ¼ zþ5 12:5 ¼ XðzÞ 1 þ 5z1 therefore: ð1 þ 5z1 ÞYðzÞ ¼ 12:5XðzÞ and so the resulting difference equation is: y½n þ 5y½n  1 ¼ 12:5x½n or y½n þ 1 þ 5y½n ¼ 12:5x½n þ 1 You try one now. The difference equation whose unit impulse response is given as:  h½n ¼ 3 2n2 u½n  2 is . . . . . . . . . . . . y½n þ 1  2y½n ¼ 3x½n  1 Because HðzÞ ¼ Z fh½ng    ¼ Z 3 2n2 u½n  2   ¼ 3Z 2n2 u½n  2 z ¼ 3  z2  ðz  2Þ ¼3

z1 ðz  2Þ

From this we can deduce the input-output relationship: HðzÞ ¼

YðzÞ XðzÞ

¼3

z1 ðz  2Þ

so that: YðzÞ ¼ 3

z1 XðzÞ ðz  2Þ

therefore: ðz  2ÞYðzÞ ¼ 3z1 XðzÞ and so the resulting difference equation is: y½n þ 1  2y½n ¼ 3x½n  1

55

232

Programme 6

And that is the end of the Programme on invariant linear systems. All that remain are the Review summary and the Can you? checklist. Read through these thoroughly and make sure you understand all the workings of this Programme. Then try the Test exercise; there is no need to hurry, take your time and work through the questions carefully. The Further problems then provide a valuable collection of additional exercises for you to try.

Review summary 6 1

Systems A system L is a process capable of accepting an input xðtÞ and processing the input to produce an output yðtÞ, also called the response of the system. This is written as yðtÞ ¼ LfxðtÞg.

2

Linear systems A linear system preserves sums and scalar products. If yðtÞ ¼ LfxðtÞg then L is linear if Lfax1 ðtÞ þ bx2 ðtÞg ¼ aLfx1 ðtÞg þ bLfx2 ðtÞg.

3

Time-invariance A continuous linear system is time-invariant if yðtÞ ¼ LfxðtÞg and yðt  t0 Þ ¼ Lfxðt  t0 Þg.

4

Shift-invariance A discrete linear system is shift-invariant if y½n ¼ Lfx½ng and y½n  n0  ¼ Lfx½n  n0 g.

5

Differential equations The general nth-order, linear, constant coefficient, inhomogeneous differential equation: dn yðtÞ dn1 yðtÞ þ an1 þ . . . þ a0 yðtÞ n dt dt n1 m m1 d xðtÞ d xðtÞ ¼ bm þ bm1 þ . . . þ b0 xðtÞ m m1 dt dt coupled with the values of the n boundary conditions   dn yðtÞ dn1 yðtÞ ,  , . . . , yðt0 Þ dt n  dt n1  an

t¼t0

t¼t0

describes the input-response relationship of a continuous linear system with input xðtÞ and response yðtÞ. Such an equation has a solution in the form yðtÞ ¼ yh ðtÞ þ yp ðtÞ where yh ðtÞ is complementary function solution to the homogeneous equation dn yðtÞ dn1 yðtÞ þ a þ . . . þ a0 yðtÞ ¼ 0 n1 dt n dt n1 and yp ðtÞ is a particular integral or particular solution to the inhomogeneous equation. an

Introduction to invariant linear systems

The procedure for solving such an equation is: (a) Find the homogeneous solution yh ðtÞ in terms of unknown integration constants (b) Find the particular solution yp ðtÞ and form the complete solution yðtÞ ¼ yh ðtÞ þ yp ðtÞ (c) Apply the boundary conditions to find the values of the unknown integration constants in yh ðtÞ. 6

Zero-input and zero-state The solution of the general nth-order, linear, constant coefficient, inhomogeneous differential equation can alternatively be written as yðtÞ ¼ yzi ðtÞ þ yzs ðtÞ where yzi ðtÞ is called the zero-input response and yzs ðtÞ is called the zero-state response. The zero-input response of the equation depends only on the initial conditions and is independent of the input. It is obtained by solving the homogeneous equation and applying the boundary conditions. The zerostate response depends only on the input and is independent of the initial conditions. It is obtained by solving the inhomogeneous equation but with all the boundary conditions equated to zero. Here the procedure is: (a) Find the homogeneous solution yh ðtÞ in terms of unknown integration constants (b) Find the particular solution yp ðtÞ and form the complete solution yðtÞ ¼ yh ðtÞ þ yp ðtÞ (c) Equate the boundary conditions to zero and then find the values of the unknown integration constants in yðtÞ. This is then the zero-state response yzs ðtÞ. (d) Apply the original boundary conditions to find the values of the unknown integration constants in yh ðtÞ. This is then the zero-input solution yzi ðtÞ.

7

Zero input and zero response For a linear time-invariant system zero input yields zero response. This is equivalent to all the boundary conditions having a zero value.

8

Arbitrary input If hðtÞ is the response of a continuous linear time-invariant system to the unit impulse ðtÞ, that is hðtÞ ¼ LfðtÞg then the response to an arbitrary input xðtÞ is the convolution of the input with the unit impulse response. That is: ð1 xð Þhðt  Þ d ¼ xðtÞ hðtÞ. LfxðtÞg ¼ 1

9

Exponential response The response of a linear, time-invariant system to an exponential input is a scaled exponential. That is LfAest g ¼ HðsÞðAest Þ. Therefore the exponential is an eigenfunction of the system and the scaling factor HðsÞ is the eigenvalue. This eigenvalue HðsÞ is referred to as the system transfer function.

233

234

Programme 6

10

Transfer function The transfer function HðsÞ of a linear, time-invariant system is the Laplace transform of the unit impulse response. That is: ð1 hðtÞest dt. HðsÞ ¼ 1

11

Convolution theorem The fact that the response of a continuous linear time-invariant system is the convolution of the input with the unit impulse response enables the use of the convolution theorem as it applies to the Laplace transform: yðtÞ ¼ xðtÞ hðtÞ and so lfyðtÞg ¼ lfxðtÞ hðtÞg ¼ lfxðtÞglfhðtÞg. That is YðsÞ ¼ XðsÞHðsÞ, the Laplace transform of the response, is equal to the product of the Laplace transform of the input and the system’s transfer function.

12

Arbitrary input to a discrete system If h½n is the response of a discrete linear shift-invariant system to the discrete unit impulse ½n, that is h½n ¼ Lf½ng then the response to an arbitrary input x½n is the convolution sum of the input with the unit impulse response. That is: Lfx½ng ¼

1 X

x½kh½n  k ¼ x½n h½n.

k¼1

13

Exponential response The response of a discrete linear, time-invariant system to an exponential input is a scaled exponential. That is LfAzn g ¼ HðzÞðAzn Þ. Therefore the exponential is an eigenfunction of the system and the scaling factor HðzÞ is the eigenvalue. This eigenvalue HðzÞ is referred to as the system transfer function.

14

Transfer function The transfer function HðzÞ of a discrete linear, shift-invariant system is the Z transform of the unit impulse response. That is: HðzÞ ¼

1 X

h½kzk .

k¼1

15

Difference equations The transfer function HðzÞ of a discrete linear system described by a difference equation can be derived by taking the Z transform of the equation. By taking the inverse Z transform of the transfer function the unit impulse response can be found. Alternatively, given the impulse response of a discrete system the corresponding difference equation can be derived.

Introduction to invariant linear systems

235

Can you? Checklist 6 Check this list before and after you try the end of Programme test On a scale of 1 to 5 how confident are you that you can:

Frames

. Recognize a system as a process whereby an input (either continuous or discrete) is converted to an output, also called the response of the system?

1

to

4

5

to

16

17

to

25

26

to

28

29

to

33

34

to

36

37

to

38

39

to

42

43

to

45

Yes

No

. Distinguish between linear and nonlinear systems and recognize time-invariant and shift-invariant systems?

Yes

No

. Determine the zero-input response and the zero-state response?

Yes

No

. Appreciate why zero valued boundary conditions give rise to a time-invariant system?

Yes

No

. Demonstrate that the response of a continuous, linear, time-invariant system to an arbitrary input is the convolution of the input with response of the system to a unit impulse?

Yes

No

. Understand the role of the exponential function with respect to a linear, time-invariant system?

Yes

No

. Use the convolution theorem to find the response of a continuous, linear, time-invariant system to an arbitrary input?

Yes

No

. Derive the system transfer function of a constant coefficient linear differential equation and use it to solve the equation?

Yes

No

. Demonstrate that the response of a discrete, linear, shift-invariant system to an arbitrary input is the convolution sum of the input with response of the system to a unit impulse?

Yes

No

236

Programme 6

. Understand the role of the exponential function with respect to a discrete linear, shift-invariant system?

Yes

46

to

48

49

to

53

54

to

55

No

. Derive the system transfer function of a constant coefficient linear difference equation and use it to solve the equation?

Yes

No

. Derive the constant coefficient difference equation from knowledge of its unit impulse response?

Yes

No

Test exercise 6 1 Which of the following are linear, nonlinear, time-invariant and shift-invariant: (a) yðtÞ ¼ LfxðtÞg ¼ 3xðtÞ (e) yðtÞ ¼ LfxðtÞg ¼ txðtÞ (b) y½n ¼ Lfx½ng ¼ 2x½n4 (c) yðtÞ ¼ LfxðtÞg ¼ e2t sin xðtÞ (d) y½n ¼ Lfx½ng ¼ 2x½n  cos x½n

(f) y½n ¼ Lfx½ng ¼ x½n½n  4 xðtÞ (g) yðtÞ ¼ LfxðtÞg ¼ 4 (h) y½n ¼ Lfx½ng ¼ Lfx½ng ¼ 4n x½n?

2 Find the zero-input response and the zero-state response for each of the following and determine which are time-invariant: (a) y0 ðtÞ  3yðtÞ ¼ t 2 uðtÞ : yð0Þ ¼ 2 (b) y00 ðtÞ  5y0 ðtÞ þ 4yðtÞ ¼ uðtÞ sin t : y0 ð0Þ ¼ 4, yð0Þ ¼ 0 (c) 5y0 ðtÞ þ 4yðtÞ ¼ et uðtÞ : yð0Þ ¼ 0 (d) y00 ðtÞ þ 2y0 ðtÞ þ yðtÞ ¼ uðtÞ : y0 ð0Þ ¼ 0, yð0Þ ¼ 0. 3 A linear, time-invariant system has the impulse response hðtÞ ¼ e3t uðtÞ find the system response to the input xðtÞ ¼ uðtÞ  uðt  3Þ. 4 A linear, time-invariant system has the impulse response hðtÞ ¼ tuðt  1Þ find the transfer function HðsÞ and use it to find the response to the input xðtÞ ¼ uðtÞ  2uðt  1Þ þ uðt  2Þ. 5 Given the differential equation y00 ðtÞ þ 3y0 ðtÞ  4yðtÞ ¼ 30e2t : y0 ð0Þ ¼ 0, yð0Þ ¼ 0 find the transfer function and solve the equation. 6 A linear, shift-invariant system has the impulse response h½n ¼ nu½n find the system response to the input x½n ¼ 4n u½n. 7 Find the impulse response of the difference equation y½n þ 1  3y½n þ 2y½n  1 ¼ x½n þ 1  x½n. 8 A linear, shift-invariant system has the impulse response h½n ¼ nu½n, find the difference equation.

Introduction to invariant linear systems

237

Further problems 1 For what values of  is the system y½n ¼ Lfx½ng ¼ x½n shift-invariant? 2 For what values of a and b is the system yðtÞ ¼ LfxðtÞg ¼ axðtÞ þ b linear? 3 Is the system yðtÞ ¼ LfxðtÞg ¼

1 X

xðtÞðt  nt0 Þ linear and time-invariant?

n¼0

4 A linear, time-invariant system has the impulse response hðtÞ ¼ e3t uðtÞ find the system response to the input xðtÞ ¼ e3t uðtÞ. 5 Is the system y½n ¼ Lfx½ng ¼ x½n  x½n linear and shift-invariant?

 6 Is the system y½n ¼ Lfx½ng ¼ x n3 linear and shift-invariant? 7 Show that the 8 2 > > >

6 > > : 0

sequence: n¼0 n¼1 n¼2 otherwise

can be represented as x½n ¼ 2½n þ 4½n  1 þ 6½n  2 or as x½n ¼ 2ðu½n þ u½n  1 þ u½n  2  3u½n  3Þ. 8 The sign function sgnðxÞ (called the signum function to avoid confusion with the sine function) is defined as: 8 8 x < 1 < 1 sgnðxÞ ¼

> :

0

x ¼ 0 the discrete form being sgn½n ¼

1

x>0

> :

0

n¼0

1

n>0

The signum function is essentially the sign of a number so that x ¼ jxjsgnðxÞ because if x < 0 then x ¼ jxj and if x > 0 then x ¼ jxj. Show that if x½n ¼ an u½n then the even part of x½n is:  1  jnj a þ ½n xe ½n ¼ 2 and the odd part is 1 xo ½n ¼ ajnj sgn½n. 2 9 Show that the convolution sum of a½n ¼ nu½n  1 and b½n ¼ n2 u½n is n2 ðn2  1Þ . 12 1  anþ1 10 Show that if x½n ¼ an u½n (a 6¼ 1) and y½n ¼ u½n then x½n y½n ¼ u½n. 1a 11 Show that if p½n 6¼ 0 only for m1  n  m2 and q½n 6¼ 0 only for M1  n  M2 then p½n q½n 6¼ 0 only for m1 þ M1  n  m2 þ M2 .

238

Programme 6

12

The cross-correlation of two sequences a½n and b½n is defined as: 1 X a½kb½n þ k a½n?b½n ¼ k¼1

Show that if x½n ¼ an u½n then ajnj 1  a2 [the cross-correlation of a sequence with itself is called the autocorrelation of the sequence].  n A linear, shift-invariant system has the impulse response h½n ¼ 14 u½n find the system response to the complex input x½n ¼ e jn!0 u½n. x½n?x½n ¼

13 14

Solve the differential equation y00 ðtÞ þ 2y0 ðtÞ þ yðtÞ ¼ et uðtÞ: yð0Þ ¼ 0, y0 ð0Þ ¼ 0.

15

Find the impulse response of the differential equation 1 1 y0 ðtÞ þ yðtÞ ¼ xðtÞ : yð0Þ ¼ 0. a a

16

Solve the differential equation 1 G y0 ðtÞ ¼  yðtÞ þ uðtÞ : yð0Þ ¼ 0. T T

17

Solve the difference equation y½n ¼ y½n  1 þ ð1  Þu½n : y½0 ¼ 0.

18

Solve the difference equation 7 y½n þ 1  y½n ¼  ðy½n  20Þ : y½0 ¼ 160. 100

19

Solve the difference equation 2y½n ¼ y½n  1 þ y½n þ 1 : y½0 ¼ 0, y½1 ¼ 8.

20

A continuous, linear, time-invariant system has output yðtÞ ¼ tuðtÞ when the input is xðtÞ ¼ uðtÞ. Find the impulse response of the system and the output when the input is xðtÞ ¼ uðt  1Þ.

21

A discrete, linear, shift-invariant system has output y½n ¼ nu½n when the input is x½n ¼ 2n u½n. Find the impulse response of the system and the output when the input is x½n ¼ 3n u½n.

Frames 1 to 37

Programme 7

Fourier series 1 Learning outcomes When you have completed this Programme you will be able to: Determine the period and amplitude of a periodic function Write down the harmonics of a periodic trigonometric function Give an analytic description of a non-sinusoidal periodic function Evaluate integrals with periodic integrands Demonstrate the orthogonality of the trigonometric functions sin nx and cos nx for n ¼ 0, 1, 2, . . . Describe a periodic function as a Fourier series subject to Dirichlet conditions Obtain the Fourier coefficients and hence the Fourier series of a periodic function Describe the effects of the harmonics in the construction of the Fourier series Find the value of the Fourier series at a point of discontinuity of the periodic function

Prerequisite: Engineering Mathematics (Eighth Edition) Programmes 15 Integration 1 and 17 Reduction formulas 239

240

Programme 7

Introduction 1

We have seen earlier that many functions can be expressed in the form of infinite series. Problems involving various forms of oscillations are common in fields of modern technology and Fourier series, with which we shall now be concerned, enable us to represent a periodic function as an infinite trigonometrical series in sine and cosine terms. One important advantage of a Fourier series is that it can represent a function containing discontinuities, whereas Maclaurin’s and Taylor’s series require the function to be continuous throughout.

Periodic functions A function f ðxÞ is said to be periodic if its function values repeat at regular intervals of the independent variable. The regular interval between repetitions is the period of the oscillations. y f(x) 0

x1

x1+p

x

Graphs of y ¼ A sin nx (a) y ¼ sin x The obvious example of a periodic function is y ¼ sin x, which goes through its complete range of values while x increases from 08 to 3608. The period is therefore 3608 or 2 radians and the amplitude, the maximum displacement from the position of rest, is 1. y

x

(b) y ¼ 5 sin 2x The amplitude is 5. The period is 1808 and there are thus 2 complete cycles in 3608.

y f(x) x

Fourier series 1

241

(c) y ¼ A sin nx Thinking along the same lines, the function y ¼ A sin nx has amplitude . . . . . . . . . . . .; period . . . . . . . . . . . .; and will have . . . . . . . . . . . . complete cycles in 3608.

amplitude ¼ A; period ¼

2

3608 2 ¼ ; n cycles in 3608 n n

Graphs of y ¼ A cos nx have the same characteristics. By way of revising earlier work, then, complete the following short exercise. Exercise In each of the following, state (a) the amplitude and (b) the period. 1

y ¼ 3 sin 5x

5

y ¼ 5 cos 4x

2

y ¼ 2 cos 3x x y ¼ sin 2

6

y ¼ 2 sin x

7

y ¼ 3 cos 6x

8

y ¼ 6 sin

3 4

y ¼ 4 sin 2x

2x 3

Deal with all eight. They will not take much time. No.

Amplitude

Period

No.

Amplitude

Period

1

3

2=5

5

5

=2

2

2

2=3

6

2

2

3

1

4

7

3

=3

4

4



8

6

3

Harmonics A function f ðxÞ is sometimes expressed as a series of a number of different sine components. The component with the largest period is the first harmonic, or fundamental of f ðxÞ. y ¼ A1 sin x

is the first harmonic or fundamental

y ¼ A2 sin 2x

is the second harmonic

y ¼ A3 sin 3x

is the third harmonic, etc.

and in general y ¼ An sin nx is the . . . . . . . . . . . . harmonic, with amplitude . . . . . . . . . . . . and period . . . . . . . . . . . .

3

242

Programme 7

4

nth harmonic; amplitude An ; period ¼

2 n

Non-sinusoidal periodic functions Although we introduced the concept of a periodic function via a sine curve, a function can be periodic without being obviously sinusoidal in appearance. Example In the following cases, the x-axis carries a scale of t in milliseconds. (a)

y

x

(b)

period ¼ 8 ms

y

x

(c)

period ¼ . . . . . . . . .

y

period ¼ . . . . . . . . .

x

5

(b) period ¼ 6 ms; (c) period ¼ 5 ms

Analytic description of a periodic function A periodic function can be defined analytically in many cases. Example 1 y

f (x) x

(a) Between x ¼ 0 and x ¼ 4; y ¼ 3, i.e. f ðxÞ ¼ 3

0 0 and n > 0: Putting ð1  xÞ ¼ u

; x¼1u

; dx ¼  du

Limits: when x ¼ 0; u ¼ 1; when x ¼ 1; u ¼ 0 ð1 ð0 m1 n1 u du ¼ ð1  uÞm1 un1 du ; Bðm, nÞ ¼  ð1  uÞ ¼

ð1

1

0

n1

u

m1

ð1  uÞ

du ¼ Bðn; mÞ

0

; Bðm, nÞ ¼ Bðn, mÞ

ð2Þ

Alternative form of the beta function We had Bðm, nÞ ¼

ð1

xm1 ð1  xÞn1 dx

0

If we put x ¼ sin2  , the result then becomes . . . . . . . . . . . .

Bðm, nÞ ¼ 2

ð =2

26

sin2m1  cos2n1  d

0

Because if x ¼ sin2 ; dx ¼ 2 sin  cos  d: When x ¼ 0,  ¼ 0; when x ¼ 1,  ¼ =2. 1  x ¼ 1  sin2  ¼ cos2  ð =2 ; Bðm, nÞ ¼ 2 sin2m2  cos2n2  sin  cos  d 0

; Bðm, nÞ ¼ 2

ð =2

sin2m1  cos2n1  d

0

Make a note of this result. We shall need to use it later.

ð3Þ

872

27

Programme 25

Reduction formulas In Programme 18 of Engineering Mathematics (Eighth Edition) we established useful reduction formulas relating to integrals of powers of sines and cosines, particularly when the integral limits are 0 and =2. ð ð =2 n  1 =2 n1 Sn2 (a) sinn x dx ¼ sinn2 x dx i.e. Sn ¼ ð4Þ n n 0 0 ð ð =2 n  1 =2 n1 Cn2 cosn x dx ¼ cosn2 x dx i.e. Cn ¼ ð5Þ (b) n n 0 0 A third reduction formula for products of powers of sines and cosines is ð ð =2 m  1 =2 sinm x cosn x dx ¼ sinm2 x cosn x dx (c) mþn 0 0 ð =2 sinm x cosn x dx by Im; n , the last result can be written If we denote 0

m1 Im2; n mþn ð =2 sinm x cosn x dx can be expressed as Alternatively, Im; n ¼

ð6Þ

0

ð n  1 =2 sinm x cosn2 x dx mþn 0 n1 Im; n2 ð7Þ i.e. Im; n ¼ mþn ð =2 Now Bðm, nÞ ¼ 2 sin2m1  cos2n1  d and if we apply (6) to the integral, 0

we have ð =2 sin2m1  cos2n1  d 0

ð =2 ð2m  1Þ  1 sin2m3  cos2n1  d ð2m  1Þ þ ð2n  1Þ 0 ð =2 m1 sin2m3  cos2n1  d ¼ mþn1 0

¼

Now, using (7) with the right-hand integral ð =2 sin2m1  cos2n1  d 0

ð =2 m1 ð2n  1Þ  1 sin2m3  cos2n3  d   m þ n  1 ð2m  3Þ þ ð2n  1Þ 0 ð =2 m1 n1   sin2m3  cos2n3  d ¼ mþn1 mþn2 0 ¼

Integral functions

873

ð =2 ðm  1Þðn  1Þ 2 ; Bðm, nÞ ¼ sin2m3  cos2n3  d ðm þ n  1Þðm þ n  2Þ 0 ðm  1Þðn  1Þ Bðm  1; n  1Þ ð8Þ i.e. Bðm, nÞ ¼ ðm þ n  1Þðm þ n  2Þ This is obviously a reduction formula for Bðm, nÞ and the process can be repeated as required. For example Bð4, 3Þ ¼ . . . . . . . . . . . .

Bð4, 3Þ ¼

ð3Þð2Þ ð2Þð1Þ Bð2, 1Þ ð6Þð5Þ ð4Þð3Þ

28

Because, applying (8) Bð4, 3Þ ¼

ð3Þð2Þ ð3Þð2Þ ð2Þð1Þ Bð3, 2Þ ¼ Bð2, 1Þ ð6Þð5Þ ð6Þð5Þ ð4Þð3Þ

Now we must evaluate Bð2, 1Þ for we can go no further in the reduction process, since, from the definition of Bðm, nÞ, m and n must be ............ >0

But Bð2, 1Þ ¼ 2

ð =2 0

29

4 =2 sin  1 sin3  cos  d ¼ 2 ¼ 4 0 2

ð3Þð2Þ ð2Þð1Þ 1 ð6Þð5Þ ð4Þð3Þ 2 ð3Þð2Þð1Þ  ð2Þð1Þ ð3!Þð2!Þ ¼ ¼ ð6Þð5Þð4Þð3Þð2Þð1Þ ð6!Þ

; Bð4, 3Þ ¼

Similarly, Bð5, 3Þ ¼ . . . . . . . . . . . .

Bð5; 3Þ ¼

ð4!Þð2!Þ ð7!Þ

Because ð4Þð2Þ ð4Þð2Þ ð3Þð1Þ Bð4; 2Þ ¼ Bð3; 1Þ ð7Þð6Þ ð7Þð6Þ ð5Þð4Þ

6 =2 ð =2 sin  1 Bð3; 1Þ ¼ 2 sin5  cos  d ¼ 2 ¼ 6 3 0 0 ð4Þð2Þ ð3Þð1Þ 1 ð2Þ ð4!Þð2!Þ ¼ ; Bð5; 3Þ ¼ ð7Þð6Þ ð5Þð4Þ 3 ð2Þ ð7!Þ Bð5; 3Þ ¼

30

874

Programme 25

ðm  1Þ!ðn  1Þ! ðm þ n  1Þ! ð =2 sin2k1  cos  d Bðk, 1Þ ¼ 2

ð9Þ

In general Bðm, nÞ ¼ Note that

0

¼2

ð =2 "

sin2k1  dðsin Þ

0

#=2 sin2k  1 ¼ ¼2 2k k 0

1 ; Bðk, 1Þ ¼ k ; Bðk, 1Þ ¼ Bð1, kÞ ¼

1 k

ð10Þ

We can also use the trigonometrical definition (3) to evaluate B  B 12 ; 12 ¼ . . . . . . . . . . . .

31

B Because Bðm, nÞ ¼ 2 ; B

1

1 2, 2



¼2 ¼2

ð =2 0 ð =2 0 ð =2

1

1 2;2



Bðm, nÞ ¼

ð1



¼

sin0  cos0  d

=2 1 d ¼ 2  ¼

xm1 ð1  xÞn1 dx

ð11Þ

0

Now let us summarize our various results so far.

Review

1 2;2

sin2m1  cos2n1  d

0

32

1

Next frame

m > 0; n > 0

0

Bðm, nÞ ¼ Bðn; mÞ ð =2 Bðm, nÞ ¼ 2 sin2m1  cos2n1  d 0

ðm  1Þðn  1Þ Bðm  1; n  1Þ ðm þ n  1Þðm þ n  2Þ ðm  1Þ!ðn  1Þ! Bðm, nÞ ¼ m and n positive integers ðm þ n  1Þ! 1 ; Bð1; 1Þ ¼ 1 Bðk, 1Þ ¼ Bð1; kÞ ¼ k Bð12 , 12Þ ¼  Bðm, nÞ ¼

Be sure that you are familiar with all these. We shall be using them all in due course.

Integral functions

875

33

Relationship between the gamma and beta functions If m and n are positive integers Bðm, nÞ ¼

ðm  1Þ!ðn  1Þ! ðm þ n  1Þ!

Also, we have previously established that, for n a positive integer, n! ¼ ðn þ 1Þ ; ðm  1Þ! ¼ ðmÞ and ðn  1Þ! ¼ ðnÞ and also ðm þ n  1Þ! ¼ ðm þ nÞ ; Bðm, nÞ ¼

ðm  1Þ!ðn  1Þ! ðmÞðnÞ ¼ ðm þ n  1Þ! ðm þ nÞ

The relation Bðm, nÞ ¼

ð12Þ

ðmÞðnÞ holds good even when m and n are not ðm þ nÞ

necessarily integers. We will prove this in the next frame, so move on Proof that Let ðmÞ ¼

Bðm, nÞ ¼ ð1

ðmÞðnÞ ðm þ nÞ

34

xm1 ex dx and ðnÞ ¼

0

; ðmÞðnÞ ¼ ¼

ð1 ð01

ð1

ð1

yn1 ey dy

0

xm1 ex dx yn1 ey dy 0 ð1 xm1 yn1 eðxþyÞ dx dy

0

y

0

Note that the integration is carried out over the first quadrant of the x–y plane. Putting x ¼ u2 and y ¼ v2 dx ¼ 2u du and dy ¼ 2v dv ð1 ð1 2 2 u2m2 v2n2 eðu þv Þ uv du dv ; ðmÞðnÞ ¼ 4 0 0 ð1 ð1 2 2 ¼4 u2m1 v2n1 eðu þv Þ du dv 0

0

O

x

876

Programme 25

If we now convert to polar coordinates, u ¼ r cos ; v ¼ r sin ; du dv ¼ rdr d u2 þ v2 ¼ r 2

00

0

ðx þ 1Þ ¼ xðxÞ (b) If x ¼ n, a positive integer ðn þ 1Þ ¼ n! ð1Þ ¼ 1 ð0Þ ¼ 1 pffiffiffi ð1  2 ex dx ¼ (c) 2 0

ðnÞ ¼  1

pffiffiffi  2 pffiffiffi 15  ð72Þ ¼ 8pffiffiffi p ffiffiffi 4  ð 12Þ ¼ 2  ð 32Þ ¼ 3 pffiffiffi  ð2nÞ  (e) Duplication formula  n þ 12 ¼ 2n1 2  ðnÞ pffiffiffi  pffiffiffi 3  ð52Þ ¼ 4

(d) ð12Þ ¼

2 Beta functions (a) Bðm, nÞ ¼

ð32Þ ¼

ð1

xm1 ð1  xÞn1 dx

m > 0; n > 0

0

Bðm, nÞ ¼ Bðn, mÞ ð =2 sin2m1  cos2n1  d Bðm, nÞ ¼ 2 0

ðm  1Þðn  1Þ Bðm  1, n  1Þ ðm þ n  1Þðm þ n  2Þ 1 Bðk, 1Þ ¼ Bð1, kÞ ¼ k Bð1, 1Þ ¼ 1; Bð12 , 12Þ ¼  ðmÞ  ðnÞ Bðm, nÞ ¼ ðm þ nÞ (c) m and n positive integers ðm  1Þ!ðn  1Þ! Bðm, nÞ ¼ ðm þ n  1Þ!

(b) Bðm, nÞ ¼

Integral functions

893

3 Error function

ð 2 x 2 (a) erf ðxÞ ¼ pffiffiffi et dt  0 pffiffiffi ð1  2 (b) ex dx ¼ 2 ð01 pffiffiffi 2 ex dx ¼ ; 1

ð1

ex

2

=2

1

dx ¼

pffiffiffiffiffiffi 2

Complementary error function ð 2 1 t 2 e dt ¼ 1  erf ðxÞ erfc ðxÞ ¼ pffiffiffi  x 4 Elliptic functions (a) Standard forms

d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 1  k2 sin2  ð  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  k2 sin2  d Eðk; Þ ¼

(1) of the first kind: Fðk, Þ ¼ (2) of the second kind:

ð

0

In each case,

0    =2;

0 < k < 1.

 (b) Complete elliptic integrals  ¼ 2   F k; ¼ K ðkÞ  2  ¼ E ðkÞ E k; 2 (c) Alternative forms of elliptic functions ðx du (1) of the first kind: Fðk; xÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ð1  u Þð1  k2 u2 Þ 0 ð x rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  k2 u2 (2) of the second kind: Eðk; xÞ ¼ du 1  u2 0 In each case

0  x  1; 0 < k < 1.

894

Programme 25

Can you? Checklist 25 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can:

Frames

. Derive the recurrence relation for the gamma function and evaluate the gamma function for certain rational arguments?

1

to

16

17

to

24

25

to

32

33

to

44

Yes

No

. Evaluate integrals that require the use of the gamma function in their solution?

Yes

No

. Identify the beta function and evaluate integrals that require the use of the beta function in their solution?

Yes

No

. Derive the relationship between the gamma function and the beta function?

Yes

No

. Use the duplication formula to evaluate the gamma function for half integer arguments?

Yes

No

. Recognize the error function and its relation to the Gaussian probability distribution?

Yes

44 and 45

46

54

to

66

66

to

75

No

. Use alternative forms of the elliptic functions?

Yes

53

No

. Evaluate integrals that require the use of elliptic functions in their solution?

Yes

52

No

. Recognize elliptic functions of the first and second kind?

Yes

to

No

Integral functions

895

Test exercise 25 ð 12Þ ð6Þ ð1:5Þ (b) (c) : 3ð4Þ ð2 5Þ ð12Þ ð1 ð1 2 x5 ex dx (e) x6 e4x dx. (d)

1 Evaluate (a)

0

0

2 Determine ð1 x5 ð2  xÞ4 dx (a)

ð =2 (b)

0

sin7  cos3  d

ð =8 (c)

0

3 Show that ða pffiffiffi 2 et dt ¼  erf ðaÞ (a)

ð1 (b)

a

ek

2 2

t

dt ¼

0

4 Evaluate (a) erfc ð1Þ

sin2 4 cos5 4 d.

0

pffiffiffi  , 2k

k > 0.

(b) erfc ð0Þ.

5 Express the following in elliptic functions. ð =4 ð pffiffi3=2 d du pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (a) (b) 2 4  5u2 þ u4 0 0 1  2 sin 

Further problems 25 ð12Þ ð5Þ ð2:5Þ ; (b) ; ; (c) 1 2ð3Þ ð3:5Þ ð 2Þ ð1 ð1 (d) x4 ex dx; (e) x8 e2x dx.

1 Evaluate (a)

0

ð1

2 Determine (a)

0 3 x

x e ð01

dx;

ð1

(b)

x4 e3x dx;

ð 01

pffiffiffi pffiffix dx. xe 0 0 ð1 n 3 If m and n are positive constants, show that xm eax dx can be expressed in 0   1 mþ1 the form  . n n  aðmþ1Þ=n (c)

x2 e2x dx;

4 Evaluate the following. ð 1=2 (a) x4 ð1  2xÞ3 dx 0 ð =2

(d) 0

pffiffiffiffiffiffiffiffiffiffiffiffiffi sin  cos5  d

2

(d)

ð 1=pffiffi2 (b) 0 ð =4

(e) 0

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 1  2x2 dx

sin3 2 cos6 2 d

ð =2 (c) 0 ð 1=3

(f) 0

sin5  cos4  d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 1  9x2 dx.

896

Programme 25

5 Show that

d 2 2 erf ðxÞ ¼ pffiffiffi ex . dx 

6 Show that the Laplace transform of the error function is given as ð1 2 s es =4 for s > 0. erfc erf ðtÞ est dt ¼ FðsÞ ¼ 2 s 0 7 The Fresnel integrals are defined as  2  2 ðx ðx t t dt and SðxÞ ¼ sin dt CðxÞ ¼ cos 2 2 0 0 Show that rffiffiffiffiffi! 1 j pffiffiffiffiffi erf x ¼ CðxÞ  jSðxÞ 2 2j 8 Express the following in elliptic functions. ð =2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð =2 ð 1 rffiffiffiffiffiffiffiffiffiffiffiffiffi2ffi d 4x 2 pffiffiffiffiffiffiffiffiffiffiffi (a) dx (b) 1 þ 4 sin  d (c) 1  x2 cos  0 0 0 ð2 ð2 dx dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (d) (e) 2 2 ð9  x Þð16  x Þ ð4  x2 Þð5  x2 Þ 0 0 ð =6 ð =3 d d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . (f) (g) 2 2 0 =4 sin  þ 2 cos2  sin  þ 2 cos2  9 Using the substitution x ¼ tan  prove that the integral ð1 dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ð1 þ x Þð1 þ 4x2 Þ 0 can be expressed in the form ð 1 =4 d qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 0 1  3 cos2  4

  , evaluate the integral in terms of elliptic functions. 2 Evaluate the following. ð 0:5 ð 1: 0 dx dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (a) (b) 2 þ x4 : 3  4x 3  4x2 þ x4 0 05 ð =2 ð =3 d d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . (c) (d) 2 0 0 25 þ 9 sin  4 þ 3 sin2  Hence, using  ¼

10

Frames 1 to 91

Programme 26

Vector analysis 1 Learning outcomes When you have completed this Programme you will be able to: Obtain the scalar and vector product of two vectors Reproduce the relationships between the scalar and vector products of the Cartesian coordinate unit vectors Obtain the scalar and vector triple products and appreciate their geometric significance Differentiate a vector field and derive a unit vector tangential to the vector field at a point Integrate a vector field Obtain the gradient of a scalar field, the directional derivative and a unit normal to a surface Obtain the divergence of a vector field and recognise a solenoidal vector field Obtain the curl of a vector field Obtain combinations of div, grad and curl acting on scalar and vector fields as appropriate

Prerequisite: Engineering Mathematics (Eighth Edition) Programme 6 Vectors 897

898

Programme 26

Introduction 1

The initial work on vectors was covered in detail in Programme 6 of Engineering Mathematics (Eighth Edition) and, if you are in any doubt, spend some time reviewing that section of the work before proceeding further. The current Programmes on vector analysis build on these early foundations, so, for quick reference, the essential results of the previous work are summarised in the following list.

Summary of prerequisites

1 A scalar quantity has magnitude only; a vector quantity has both magnitude and direction. 2 The axes of reference, OX, OY, OZ, form a right-handed set. The symbols i, j, k denote unit vectors in the directions OX, OY, OZ, respectively. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi If OP ¼ r ¼ ax i þ ay j þ az k then OP ¼ jrj ¼ a2x þ a2y þ a2z where jrj is the modulus of r. 3 The direction cosines [l, m, n] are the cosines of the angles between the vector r and the axes OX, OY, OZ, respectively. For any vector r ¼ ax i þ ay j þ az k ay ax az ; m¼ ; n¼ l¼ jrj jrj jrj and

l 2 þ m2 þ n2 ¼ 1.

4 Scalar product (‘dot product’) A  B ¼ AB cos  where  is the angle between A and B and where A and B are the moduli of A and B. If A ¼ ax i þ ay j þ az k and B ¼ bx i þ by j þ bz k then A  B ¼ ax bx þ ay by þ az bz

and

AB¼BA

5 Vector product (‘cross product’) A  B ¼ AB sin  in a direction perpendicular to A and B so that A, B, ðA  BÞ form a right-handed set. Therefore jA  Bj ¼ AB sin  i Also A  B ¼ ax bx

j ay by

k az where A  B ¼ B  A bz

6 Angle between two vectors cos  ¼ l1 l2 þ m1 m2 þ n1 n2 where l1 ; m1 ; n1 and l2 ; m2 ; n2 are the direction cosines of vectors r1 and r2 respectively. For perpendicular vectors l1 l2 þ m1 m2 þ n1 n2 ¼ 0 For parallel vectors

l1 l2 þ m1 m2 þ n1 n2 ¼ 1.

One or two examples will no doubt help to recall the main points.

Vector analysis 1

899

Example 1 Direction cosines z

If i; j; k are unit vectors in the directions OX, OY, OZ, respectively, then any position vector OP ð¼ rÞ can be represented in the form

az

OP ¼ r ¼ ax i þ ay j þ az k. Then jrj ¼ . . . . . . . . . . . .

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2x þ a2y þ a2z

2

The direction of OP is denoted by stating the direction cosines of the angles made by OP and the three coordinate axes.

m ¼ cos  ¼

y

x

jrj ¼

l ¼ cos  ¼

ay

O

ax

z

OL ax ¼ OP jrj

az

OM ay ¼ OP jrj

γ β

α

ON az ¼ OP jrj ; l, m, n ¼ cos , cos , cos  n ¼ cos  ¼

ax

ay

O

y

x

So, if P is the point ð3, 2, 6Þ, then jrj ¼ . . . . . . . . . . . . ; l ¼ ............; m ¼ ............; n ¼ ............ jrj ¼ 7; : l ¼ 0 429; m ¼ 0:286; n ¼ 0:857 Because ðj r jÞ2 ¼ 9 þ 4 þ 36 ¼ 49 l ¼ cos  ¼ 3 ¼ 0:4286 7

m ¼ cos  ¼ ¼ 0:2857 2 7

n ¼ cos  ¼ 67 ¼ 0:8571.

; j r j¼ 7

3

900

Programme 26

Example 2 Angle between two vectors If the direction cosines of A are l1 ; m1 ; n1 and those of B are l2 ; m2 ; n2 , then the angle between the vectors is given by cos  ¼ l1 l2 þ m1 m2 þ n1 n2 :

ð1Þ

If A ¼ 2i þ 3j þ 4k and B ¼ i  2j þ 3k, we can find the direction cosines of each and hence  which is . . . . . . . . . . . .

4

 ¼ 668 360 Because

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi 4 þ 9 þ 16 ¼ 29 2 3 4 ; l1 ¼ pffiffiffiffiffiffi ; m1 ¼ pffiffiffiffiffiffi ; n1 ¼ pffiffiffiffiffiffi 29 29 29 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi For B: j r2 j¼ 1 þ 4 þ 9 ¼ 14 1 2 3 ; l2 ¼ pffiffiffiffiffiffi ; m2 ¼ pffiffiffiffiffiffi ; n2 ¼ pffiffiffiffiffiffi 14 14 14 1 : Then cos  ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f2  6 þ 12g ¼ 0 3970 14  29 ;  ¼ 668 360 Let us now look at the question of scalar and vector products. For A:

j r1 j¼

On to the next frame

5

Example 3 Scalar product If A and B are two vectors, the scalar product of A and B is defined as A  B ¼ AB cos 

ð2Þ

where  is the angle between the two vectors. If A  B ¼ 0 then A ? B. If we consider the scalar products of the unit vectors i; j; k, which are mutually perpendicular, then and

i  j ¼ ð1Þð1Þ cos 908 ¼ 0 i  i ¼ ð1Þð1Þ cos 08 ¼ 1

; ij¼jk¼ki¼0 ; i  i ¼ j  j ¼ k  k ¼ 1:

and B ¼ bx i þ by j þ bz k I n g e n e r a l , i f A ¼ ax i þ ay j þ az k A  B ¼ ax bx þ ay by þ az bz which is, of course, a scalar quantity.

then

So, if A ¼ 2i  3j þ 4k and B ¼ i þ 2j þ 5k, then A  B ¼ ............

6

A  B ¼ 2  6 þ 20 ¼ 16 Also, since A  B ¼ AB cos ; we can determine the angle  between the vectors. In this case  ¼ . . . . . . . . . . . .

Vector analysis 1

901

7

 ¼ 578 90 A ¼ 2i  3j þ 4k B ¼ i þ 2j þ 5k

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi ; A ¼j A j¼ 4 þ 9 þ 16 ¼ 29 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi ; B ¼j B j¼ 1 þ 4 þ 25 ¼ 30

We have already found that A  B ¼ 16 and A  B ¼ AB cos  pffiffiffiffiffiffi pffiffiffiffiffiffi ; 16 ¼ 29 30 cos  ; cos  ¼ 0:5425 ;  ¼ 578 90 So, the scalar product of A ¼ ax i þ ay j þ az k and B ¼ bx i þ by j þ bz k is

A  B ¼ ax bx þ ay by þ az bz

and

A  B ¼ AB cos  where  is the angle between the vectors.

It can also be shown that and

(a) A  B ¼ B  A (b) A  ðB þ CÞ ¼ A  B þ A  C Make a note of these results

8

Example 4 Vector product If A ¼ ax i þ ay j þ az k and B ¼ bx i þ by j þ bz k the vector product A  B has magnitude jA  Bj ¼ AB sin  in the direction perpendicular to A and B such that A; B and (A  B) form a right-handed set. We can write this as A  B ¼ ðAB sin Þn

ð3Þ

where n is defined as a unit vector in the positive normal direction to the plane of A and B, i.e. forming a right-handed set. Also

i A  B ¼ ax bx

j ay by

k az bz

If we consider the vector products of the unit vectors, i, j, k, then i  j ¼ ð1Þð1Þ sin 908k ¼ k j  k ¼ i,

ki¼j

Note that j  i ¼ ði  jÞ ¼ k k  j ¼ i, i  k ¼ j Also i  i ¼ ð1Þð1Þ sin 08n ¼ 0 jj¼kk¼0

ð4Þ

902

Programme 26

It can also be shown that (a) A  ðB þ CÞ ¼ A  B þ A  C

ð5Þ

(b) A  B ¼ ðB  AÞ

and

Make a note of these results (3), (4) and (5). Then, if A ¼ 3i  2j þ 4k and B ¼ 2i  3j  2k A  B ¼ ............

9

A  B ¼ 16i þ 14j  5k We simply evaluate the determinant i

j

k

4 A  B ¼ 3 2 2 3 2 ¼ ið4 þ 12Þ  jð6  8Þ þ kð9 þ 4Þ ¼ 16i þ 14j  5k Move on to the next frame

10

We have seen therefore that but that

the scalar product of two vectors is a scalar the vector product of two vectors is a vector.

We know also that

jA  Bj ¼ AB sin 

Therefore, the angle between the vectors A and B given in Example 4 is  ¼ ............

11

 ¼ 798 400 Because A ¼ 3i  2j þ 4k; B ¼ 2i  3j  2k; and A  B ¼ 16i þ 14j  5k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi ; j A  B j ¼ 162 þ 142 þ 52 ¼ 477 ¼ 21:84 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi A ¼j A j¼ 32 þ 22 þ 42 ¼ 29 ¼ 5:385 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi B ¼j B j¼ 22 þ 32 þ 22 ¼ 17 ¼ 4:123 ; 21:84 ¼ ð5:385Þð4:123Þ sin  ; sin  ¼ 0:9838

;  ¼ 798 400

Vector analysis 1

903

So, to recapitulate: If A ¼ ax i þ ay j þ az k and B ¼ bx i þ by j þ bz k and  is the angle between them (a) Scalar product ¼ A  B ¼ ax bx þ ay by þ az bz ¼ AB cos    i j   (b) Vector product ¼ A  B ¼  ax ay   bx by and

 k   az   bz 

j A  B j ¼ AB sin .

Make a note of these fundamental results: we shall certainly need them. Then, in the next frame, we can set off on some new work

Triple products 12

We now deal with the various products that we form with three vectors.

Scalar triple product of three vectors If A; B; C are three vectors, the scalar formed by the product A  ðB  CÞ is called the scalar triple product. If A ¼ ax i þ ay j þ az k; B ¼ bx i þ by j þ bz k; C ¼ cx i þ cy j þ cz k; then

i B  C ¼ bx cx

j by cy

k bz cz

i ; A  ðB  CÞ ¼ ðax i þ ay j þ az kÞ  bx cx

j by cy

k bz cz

Multiplying the top row by the external bracket and remembering that ij¼jk¼ki¼0

and

ax we have A  ðB  CÞ ¼ bx cx

ii¼jj¼kk¼1

ay by

az bz

cy

cz

Example If A ¼ 2i  3j þ 4k; B ¼ i  2j  3k; C ¼ 2i þ j þ 2k; 2 3 4 then A  ðB  CÞ ¼ 1 2 3 2 1 2 ¼ ............

ð6Þ

904

Programme 26

13

A  ðB  CÞ ¼ 42 Because 2

3

4

A  ðB  CÞ ¼ 1 2 3 2 1 2 ¼ 2ð4 þ 3Þ þ 3ð2 þ 6Þ þ 4ð1 þ 4Þ ¼ 42 As simple as that.

14

Properties of scalar triple products (a)

bx B  ðC  AÞ ¼ cx ax

by cy ay

bz ax cz ¼  cx az bx

ay cy by

az cz bz

since interchanging two rows in a determinant reverses the sign. If we now interchange rows 2 and 3 and again change the sign, we have ax B  ðC  AÞ ¼ bx cx

ay by cy

az bz ¼ A  ðB  CÞ cz

; A  ðB  CÞ ¼ B  ðC  AÞ ¼ C  ðA  BÞ

ð7Þ

i.e. the scalar triple product is unchanged by a cyclic change of the vectors involved. (b)

bx

by

bz

ax

ay

az

B  ðA  CÞ ¼ ax cx

ay cy

az ¼  bx cz cx

by cy

bz cz ð8Þ

; B  ðA  CÞ ¼ A  ðB  CÞ

i.e. a change of vectors not in cyclic order, changes the sign of the scalar triple product. ax A  ðB  AÞ ¼ bx ax

(c)

ay by ay

az bz ¼ 0 since two rows are identical. az

; A  ðB  AÞ ¼ B  ðC  BÞ ¼ C  ðA  CÞ ¼ 0 Example If A ¼ i þ 2j þ 3k; B ¼ 2i  3j þ k; C ¼ 3i þ j  2k A  ðB  CÞ ¼ . . . . . . . . . . . .

C  ðB  AÞ ¼ . . . . . . . . . . . .

ð9Þ

Vector analysis 1

905

A  ðB  CÞ ¼ 52;

C  ðA  BÞ ¼ 52

Because 1 2 3 1 ¼ 1ð6  1Þ  2ð4  3Þ þ 3ð2 þ 9Þ ¼ 52 A  ðB  CÞ ¼ 2 3 3 1 2 C  ðB  AÞ is not a cyclic change from the above. Therefore C  ðB  AÞ ¼ A  ðB  CÞ ¼ 52

Coplanar vectors The magnitude of the scalar triple product j A  ðB  CÞ j is equal to the volume of the parallelepiped with three adjacent sides defined by A, B and C.

A

n

C

ϕ θ B

The scalar triple product A  ðB  CÞ ¼ AðBC sin  nÞ ¼ ABC sin  cos  where n is a unit vector perpendicular to the plane containing B and C,  is the angle between B and C and  is the angle between A and n. Therefore j A  ðB  CÞ j ¼ ABC j sin  cos  j Notice that in the figure both  and  are drawn as acute but in the general case this may not be so. Now, BC j sin  j is the area of the parallelogram defined by B and C. The altitude of the parallelepiped is A j cos  j and so ABC j sin  cos  j is the volume of the parallelepiped with three adjacent sides defined by A, B and C. Consequently if A  ðB  CÞ ¼ 0 then the volume of the parallelepiped is zero and the three vectors A, B and C are coplanar. Example 1 Show that A ¼ i þ 2j  3k;

B ¼ 2i  j þ 2k; and C ¼ 3i þ j  k are coplanar.

We just evaluate A  ðB  CÞ ¼ . . . . . . . . . . . . and apply the test.

15

906

Programme 26

16

A  ðB  CÞ ¼ 0 Because 1 2 3 2 ¼ 1ð1  2Þ  2ð2  6Þ  3ð2 þ 3Þ ¼ 0: A  ðB  CÞ ¼ 2 1 3 1 1 Therefore A, B, C are coplanar. Example 2 If A ¼ 2i  j þ 3k; B ¼ 3i þ 2j þ k; C ¼ i þ pj þ 4k are coplanar, find the value of p. The method is clear enough. We merely set up and evaluate the determinant and solve the equation A  ðB  CÞ ¼ 0. p ¼ ............

17

p ¼ 3 Because A  ðB  CÞ ¼ 0

2 1 3 ; 3 2 1 ¼0 1 p 4

; 2 ð8  pÞ þ 1ð12  1Þ þ 3ð3p  2Þ ¼ 0

; 7p ¼ 21

; p ¼ 3

One more. Example 3 Determine whether the three vectors A ¼ 3i þ 2j  k; B ¼ 2i  j þ 3k; C ¼ i  2j þ 2k are coplanar. Work through it on your own. The result shows that ............

18

A, B, C are not coplanar Because 3 in this case A  ðB  CÞ ¼ 2 1 ; A  ðB  CÞ 6¼ 0

2 1 1 3 ¼ 13 2 2

; A, B, C are not coplanar. Now on to something different

Vector analysis 1

907

19

Vector triple products of three vectors If A; B and C are three vectors, then ) A  ðB  CÞ are called the vector triple products. and ðA  BÞ  C

ð10Þ

C o n si d e r A  ðB  CÞ w he re A ¼ ax i þ ay j þ az k; B ¼ bx i þ by j þ bz k an d C ¼ cx i þ cy j þ cz k: Then ðB  CÞ is a vector perpendicular to the plane of B and C and A  ðB  CÞ is a vector perpendicular to the plane containing A and ðB  CÞ, i.e. coplanar with B and C. Note that, similarly, ðA  BÞ  C is coplanar with A and B and so in general A  ðB  CÞ 6¼ ðA  BÞ  C. Now i

j

ðB  CÞ ¼ bx cx

by cy

Then

k by bz ¼ i cy cz i

A  ðB  CÞ ¼

¼

bz bx j cz cx

ax by

bz

cy

cz



bx bz þk cx cz

j

k

ay

az

bx

bz

bx

by

cx

cz

cx

cy

i

j

k

ax

ay

az

by

bz

bz

bx

bx

by

cy

cz

cz

cx

cx

cy

by cy

In symbolic form, further expansion of the determinant becomes somewhat tedious. However a numerical example will clarify the method. So make a note of the definition (10) above and then go on to the next frame

20

Example 1 If A ¼ 2i  3j þ k; B ¼ i þ 2j  k; product A  ðB  CÞ. We start off with B  C ¼ . . . . . . . . . . . .

C ¼ 3i þ j þ 3k; determine the vector triple

908

Programme 26

21

B  C ¼ 7i  6j  5k Because i BC¼ 1 3

j 2 1

k ¼ ið6 þ 1Þ  jð3 þ 3Þ þ kð1  6Þ 1 3 ¼ 7i  6j  5k

Then A  ðB  CÞ ¼ . . . . . . . . . . . .

22

A  ðB  CÞ ¼ 21i þ 17j þ 9k Because i j k A  ðB  CÞ ¼ 2 3 1 7 6 5 ¼ ið15 þ 6Þ  jð10  7Þ þ kð12 þ 21Þ ¼ 21i þ 17j þ 9k That is fundamental enough. There is, however, an even easier way of determining a vector triple product. It can be proved that A  ðB  CÞ ¼ ðA  CÞB  ðA  BÞC and

ðA  BÞ  C ¼ ðC  AÞB  ðC  BÞA

ð11Þ

The proof of this is given in the Appendix. For the moment, make a careful note of the expressions: then we will apply the method to the example we have just completed.

23

A ¼ 2i  3j þ k; B ¼ i þ 2j  k; C ¼ 3i þ j þ 3k and we have A  ðB  CÞ ¼ ðA  CÞB  ðA  BÞC ¼ ð6  3 þ 3Þði þ 2j  kÞ  ð2  6  1Þð3i þ j þ 3kÞ ¼ 6 ði þ 2j  kÞ þ 5ð3i þ j þ 3kÞ ¼ 21i þ 17j þ 9k which is, of course, the result we achieved before. Here is another. Example 2 If A ¼ 3i þ 2j  2k; B ¼ 4i  j þ 3k; C ¼ 2i  3j þ k determine ðA  BÞ  C using the relationship ðA  BÞ  C ¼ ðC  AÞB  ðC  BÞA. ðA  BÞ  C ¼ . . . . . . . . . . . .

Vector analysis 1

909

24

50i  26j þ 22k Because ðA  BÞ  C ¼ ðC  AÞB  ðC  BÞA ¼ ð6  6  2Þð4i  j þ 3kÞ  ð8 þ 3 þ 3Þð3i þ 2j  2kÞ ¼ 2ð4i  j þ 3kÞ  14ð3i þ 2j  2kÞ ¼ 50i  26j þ 22k Now one more. Example 3 If A ¼ i þ 3j þ 2k; B ¼ 2i þ 5j  k; C ¼ i þ 2j þ 3k A  ðB  CÞ ¼ . . . . . . . . . . . . ðA  BÞ  C ¼ . . . . . . . . . . . . Finish them both.

25

A  ðB  CÞ ¼ 11i þ 35j  58k ðA  BÞ  C ¼ 17i þ 38j  31k Because A  ðB  CÞ ¼ ðA  CÞB  ðA  BÞC ¼ ð1 þ 6 þ 6Þð2i þ 5j  kÞ  ð2 þ 15  2Þði þ 2j þ 3kÞ ¼ 13ð2i þ 5j  kÞ  15ði þ 2j þ 3kÞ ¼ 11i þ 35j  58k and ðA  BÞ  C ¼ ðC  AÞB  ðC  BÞA ¼ ð1 þ 6 þ 6Þð2i þ 5j  kÞ  ð2 þ 10  3Þði þ 3j þ 2kÞ ¼ 13ð2i þ 5j  kÞ  9ði þ 3j þ 2kÞ ¼ 17i þ 38j  31k These two results clearly confirm that A  ðB  CÞ 6¼ ðA  BÞ  C

so beware!

Before we proceed, note the following concerning the unit vectors. (a)

ði  jÞ ¼ k ; i  ði  jÞ ¼ i  k ¼ j

z

; i  ði  jÞ ¼ j (b) ði  iÞ  j ¼ ð0Þ  j ¼ 0 ; ði  iÞ  j ¼ 0

x

O

y

and once again, we see that i  ði  jÞ 6¼ ði  iÞ  j

On to the next

910

26

Programme 26

Finally, by way of review: Example 4 If A ¼ 5i  2j þ 3k; B ¼ 3i þ j  2k; C ¼ i  3j þ 4k; determine (a) the scalar triple product A  ðB  CÞ (b) the vector triple products (1) A  ðB  CÞ (2) ðA  BÞ  C. Finish all these and then check with the next frame

27

(a) A  ðB  CÞ ¼ 12 (b) ð1Þ A  ðB  CÞ ¼ 62i þ 44j  74k ð2Þ ðA  BÞ  C ¼ 109i þ 7j  22k Here is the working. 5

2

3

(a) A  ðB  CÞ ¼ 3 1

1 3

2 4

¼ 5ð4  6Þ þ 2ð12 þ 2Þ þ 3ð9  1Þ ¼ 12 (b) (1) A  ðB  CÞ ¼ ðA  CÞB  ðA  BÞC ¼ ð5 þ 6 þ 12Þð3i þ j  2kÞ  ð15  2  6Þði  3j þ 4kÞ ¼ 23ð3i þ j  2kÞ  7ði  3j þ 4kÞ ¼ 62i þ 44j  74k (2) ðA  BÞ  C ¼ ðC  AÞB  ðC  BÞA ¼ 23ð3i þ j  2kÞ  ð8Þð5i  2j þ 3kÞ ¼ 109i þ 7j  22k Let us now move to the next topic

Differentiation of vectors 28

In many practical problems, we often deal with vectors that change with time, e.g. velocity, acceleration, etc. If a vector A depends on a scalar variable t, then A can be represented as AðtÞ and A is then said to be a function of t. If A ¼ ax i þ ay j þ az k then ax ; ay ; az will also be dependent on the parameter t. i.e.

AðtÞ ¼ ax ðtÞi þ ay ðtÞj þ az ðtÞk

Differentiating with respect to t gives . . . . . . . . . . . .

Vector analysis 1

911

d d d d fAðtÞg ¼ i fax ðtÞg þ j fay ðtÞg þ k faz ðtÞg dt dt dt dt In short

29

day dA dax daz ¼i þj þk . dt dt dt dt

The independent scalar variable is not, of course, restricted to t. In general, if u is the parameter, then dA ¼ ............ du day dA dax daz ¼i þj þk du du du du δA A (u + δu) A (u) O

30

If a position vector OP moves to OQ when u becomes u þ u, then as u ! 0, the direction of the chord PQ becomes that of the tangent to the curve at P, dA is along the tangent to the i.e. the direction of du locus of P.

T = dA du

O

A (u)

Example 1 If A ¼ ð3u2 þ 4Þi þ ð2u  5Þj þ 4u3 k, then dA ¼ ............ du dA ¼ 6ui þ 2j þ 12u2 k du If we differentiate this again, we get When u ¼ 2, Then

d2 A ¼ 6i þ 24uk du2

dA d2 A ¼ 12i þ 2j þ 48k and ¼ 6i þ 48k du du2

dA ¼ ............ du

and

d2 A ¼ ............ du2

31

912

Programme 26

   dA  :    du  ¼ 49 52;

32

 2  d A :    du2  ¼ 48 37

Because

and

dA ¼ f122 þ 22 þ 482 g1=2 ¼ f2452g1=2 ¼ 49:52 du d2 A ¼ f62 þ 482 g1=2 ¼ f2340g1=2 ¼ 48:37 du2

Example 2 If F ¼ i sin 2t þ je3t þ kðt 3  4tÞ, then when t ¼ 1 dF ¼ ............; dt

33

d2 F ¼ ............ dt 2

dF ¼ 2 cos 2i þ 3e3 j  k dt d2 F ¼ 4 sin 2i þ 9e3 j þ 6k dt 2 From these, we could if required find the magnitudes of dF ¼ ............; dt    dF  :    dt  ¼ 60 27;

34

dF d2 F and 2 . dt dt

d2 F ¼ ............ dt 2  2  d F :    dt 2  ¼ 180 9

Because dF ¼ fð2 cos 2Þ2 þ 9e6 þ 1g1=2 dt ¼ f0:6927 þ 3631 þ 1g1=2 ¼ 60:27 and

d2 F ¼ fð4 sin 2Þ2 þ 81e6 þ 36g1=2 dt 2 ¼ f13:23 þ 32 678 þ 36g1=2 ¼ 180:9

One more example. Example 3 If A ¼ ðu þ 3Þi  ð2 þ u2 Þj þ 2u3 k, determine (a)

dA du

(b)

d2 A du2

(c)

dA du

(d)

d2 A du2

at u ¼ 3.

Work through all sections and then check with the next frame

Vector analysis 1

Here is the working. (a)

913

A ¼ ðu þ 3Þi  ð2 þ u2 Þj þ 2u3 k

dA ¼ i  2uj þ 6u2 k du

At u ¼ 3,

35

dA ¼ i  6j þ 54k du

d2 A d2 A ¼ 2j þ 12uk At u ¼ 3, ¼ 2j þ 36k du2 du2 dA (c) ¼ f1 þ 36 þ 2916g1=2 ¼ ð2953Þ1=2 ¼ 54:34 du

(b)

(d)

d2 A ¼ f4 þ 1296g1=2 ¼ ð1300Þ1=2 ¼ 36:06 du2 The next example is of a rather different kind, so move on

36

Example 4 A particle moves in space so that at time t its position is stated as x ¼ 2t þ 3; y ¼ t 2 þ 3t; z ¼ t 3 þ 2t 2 . We are required to find the components of its velocity and acceleration in the direction of the vector 2i þ 3j þ 4k when t ¼ 1. First we can write the position as a vector r r ¼ ð2t þ 3Þi þ ðt 2 þ 3tÞj þ ðt 3 þ 2t 2 Þk Then, at t ¼ 1 dr ¼ ............; dt

d2 r ¼ ............ dt 2

dr ¼ 2i þ 5j þ 7k; dt

d2 r ¼ 2j þ 10k dt 2

Because

and

dr ¼ 2i þ ð2t þ 3Þj þ ð3t 2 þ 4tÞk dt dr ¼ 2i þ 5j þ 7k ; At t ¼ 1; dt 2 d r ¼ 2j þ ð6t þ 4Þk dt 2 d2 r ; At t ¼ 1; ¼ 2j þ 10k dt 2

Now, a unit vector parallel to 2i þ 3j þ 4k is . . . . . . . . . . . .

37

914

Programme 26

38

2i þ 3j þ 4k 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffi ð2i þ 3j þ 4kÞ 4 þ 9 þ 16 29 Denote this unit vector by I. Then dr in the direction the component of dt of I dr cos  ¼ dt dr ¼ I dt 1 ¼ pffiffiffiffiffiffi ð2i þ 5j þ 7kÞ  ð2i þ 3j þ 4kÞ 29 ¼ ............

39

8:73 Because 1 1 pffiffiffiffiffiffi ð2i þ 5j þ 7kÞ  ð2i þ 3j þ 4kÞ ¼ pffiffiffiffiffiffi ð4 þ 15 þ 28Þ 29 29 47 ¼ pffiffiffiffiffiffi 29 : ¼ 8 73 Similarly, the component of

40

d2 r in the direction of I is dt 2 ............ 8:54

Because d2 r d2 r cos  ¼ I dt 2 dt 2 1 ¼ pffiffiffiffiffiffi ð2j þ 10kÞ  ð2i þ 3j þ 4kÞ 29 1 ¼ pffiffiffiffiffiffi ð6 þ 40Þ 29 46 ¼ pffiffiffiffiffiffi 29 ¼ 8:54

dr dt

I

Vector analysis 1

915

Differentiation of sums and products of vectors If A ¼ AðuÞ and B ¼ BðuÞ, then d dA fcAg ¼ c du du d dA dB (b) fA þ Bg ¼ þ du du du d dB dA fA  Bg ¼ A  þ B (c) du du du d dB dA fA  Bg ¼ A  þ  B. (d) du du du (a)

These are very much like the normal rules of differentiation. However, if AðuÞ  AðuÞ ¼ a2x þ a2y þ a2z ¼ jAj2 ¼ A2 is a constant then d d d fAðuÞ  AðuÞg ¼ AðuÞ  fAðuÞg þ AðuÞ  fAðuÞg du du du d d fAðuÞg ¼ fA2 g ¼ 0 ¼ 2AðuÞ  du du Assuming that AðuÞ 6¼ 0, then since AðuÞ  that AðuÞ and

d d fAðuÞg ¼ fA2 g ¼ 0 it follows du du

d fAðuÞg are perpendicular vectors because du ............

41

d d fAðuÞg ¼j AðuÞ j fAðuÞg cos  ¼ 0 du du  ; cos  ¼ 0 ; ¼ 2 AðuÞ 

Now let us deal with unit tangent vectors.

Unit tangent vectors We have already established in Frame 30 of this Programme that if OP is a position vector AðuÞ in space, then the direction of the vector denoting d fAðuÞg is du ............

O

A (u)

916

Programme 26

42

parallel to the tangent to the curve at P Then the unit tangent vector T at P can be found from d fAðuÞg du T¼ d fAðuÞg du In simpler notation, this becomes: If r ¼ ax i þ ay j þ az k then the unit tangent vector T is given by T¼

dr=du jdr=duj

Example 1 Determine the unit tangent vector at the point ð2, 4, 7Þ for the curve with parametric equations x ¼ 2u; y ¼ u2 þ 3; z ¼ 2u2 þ 5. First we see that the point ð2, 4, 7Þ corresponds to u ¼ 1. The vector equation of the curve is r ¼ ax i þ ay j þ az k ¼ 2ui þ ðu2 þ 3Þj þ ð2u2 þ 5Þk ;

43

dr ¼ ............ du

dr ¼ 2i þ 2uj þ 4uk du and at u ¼ 1;

dr ¼ 2i þ 2j þ 4k du Hence

dr ¼ ............ du   pffiffiffi  dr    ¼ 2 6;  du 

44

1 T ¼ pffiffiffi fi þ j þ 2kg 6

Because pffiffiffi dr ¼ f4 þ 4 þ 16g1=2 ¼ 241=2 ¼ 2 6 du dr 2i þ 2j þ 4k 1 pffiffiffi T ¼ du ¼ ¼ pffiffiffi fi þ j þ 2kg dr 2 6 6 du Let us do another.

and T ¼ . . . . . . . . . . . .

Vector analysis 1

917

Example 2 Find the unit tangent vector at the point ð2; 0; Þ for the curve with parametric equations x ¼ 2 sin ; y ¼ 3 cos ; z ¼ 2. We see that the point ð2; 0; Þ corresponds to  ¼ =2: Writing the curve in vector form r ¼ . . . . . . . . . . . .

45

r ¼ 2 sin  i þ 3 cos  j þ 2 k

Then, at  ¼ =2,

dr ¼ ............ d dr ¼ ............ d T ¼ ............ Finish it off pffiffiffiffiffiffi dr dr ¼ 13 ¼ 3j þ 2k; d d 1 T ¼ pffiffiffiffiffiffi ð3j þ 2kÞ 13

46

And now Example 3 Determine the unit tangent vector for the curve x ¼ 3t;

y ¼ 2t 2 ;

z ¼ t2 þ t

at the point ð6, 8, 6Þ. On your own. T ¼ . . . . . . . . . . . . 1 T ¼ pffiffiffiffiffiffi ð3i þ 8j þ 5kÞ 98 The point ð6, 8, 6Þ corresponds to t ¼ 2 r ¼ 3ti þ 2t 2 j þ ðt 2 þ tÞk dr ¼ 3i þ 4tj þ ð2t þ 1Þk ; dt dr ¼ 3i þ 8j þ 5k dt pffiffiffiffiffiffi ¼ 98

At t ¼ 2, r ¼ 6i þ 8j þ 6k and

dr ¼ ð9 þ 64 þ 25Þ1=2 dt dr=dt 1 ¼ pffiffiffiffiffiffi ð3i þ 8j þ 5kÞ ; T¼ jdr=dtj 98

;

47

918

Programme 26

Partial differentiation of vectors 48

If a vector F is a function of two independent variables u and v, then the rules of differentiation follow the usual pattern. If F ¼ xi þ yj þ zk then x, y, z will also be functions of u and v. Then

@F @x @y @z ¼ iþ jþ k @u @u @u @u @F @x @y @z ¼ iþ jþ k @v @v @v @v @2F @2x @2y @2z ¼ 2iþ 2jþ 2k 2 @u @u @u @u @2F @2x @2y @2z ¼ iþ 2jþ 2k @v2 @v2 @v @v @2F @2x @2y @2z ¼ iþ jþ k @u@v @u@v @u@v @u@v

and for small finite changes du and dv in u and v, we have dF ¼

@F @F du þ dv @u @v

Example If F ¼ 2uvi þ ðu2  2vÞj þ ðu þ v2 Þk @F ¼ ............; @u @2F ¼ ............; @u2

49

@F ¼ 2vi þ 2uj þ k; @u @2F ¼ 2j; @u2

@F ¼ ............ @v @2F ¼ ............ @u@v @F ¼ 2ui  2j þ 2vk @v @2F ¼ 2i @u@v

This is straightforward enough.

Integration of vector functions The process is the reverse of that for differentiation. If a vector F ¼ xi þ yj þ zk where F; x; y; z are expressed as functions of u, then ðb ðb ðb ðb F du ¼ i x du þ j y du þ k z du. a

a

a

a

Vector analysis 1

919

Example 1 If F ¼ ð3t 2 þ 4tÞi þ ð2t  5Þj þ 4t 3 k, then ð3 ð3 ð3 ð3 F dt ¼ i ð3t 2 þ 4tÞ dt þ j ð2t  5Þ dt þ k 4t 3 dt ¼ . . . . . . . . . . . . 1

1

1

1

42i  2j þ 80k

50

Because 3

ð3 3 2 2 4 F dt ¼ iðt þ 2t Þ þ jðt  5tÞ þ kt 1

1

¼ ð45i  6j þ 81kÞ  ð3i  4j þ kÞ ¼ 42i  2j þ 80k Here is a slightly different one. Example 2 F ¼ 3ui þ u2 j þ ðu þ 2Þk

If

V ¼ 2ui  3uj þ ðu  2Þk ð2 evaluate ðF  VÞdu. and

0

First we must determine F  V in terms of u. F  V ¼ ............ F  V ¼ ðu3 þ u2 þ 6uÞi  ðu2  10uÞj  ð2u3 þ 9u2 Þk

51

Because i j k F  V ¼ 3u u2 ðu þ 2Þ 2u 3u ðu  2Þ which gives the result above. ð2 Then ðF  VÞ du ¼ . . . . . . . . . . . . 0 4 3 f14i

Because ð ðF  VÞdu ¼ ;

ð2 0

þ 13j  24kg

 4   3   4  u u3 u u þ þ 3u2 i   5u2 j  þ 3u3 k 4 3 3 2

ðF  VÞdu ¼ ð4 þ 83 þ 12Þi  ð83  20Þj  ð8 þ 24Þk ¼ 43 f14i þ 13j  24kg

52

920

Programme 26

Example 3 If F ¼ A  ðB  CÞ where A ¼ 3t 2 i þ ð2t  3Þj þ 4tk B ¼ 2i þ 4tj þ 3ð1  tÞk C ¼ 2ti  3t 2 j  2tk ð1 determine Fdt. 0

First we need to find A  ðB  CÞ. The simplest way to do this is to use the relationship A  ðB  CÞ ¼ . . . . . . . . . . . .

53

A  ðB  CÞ ¼ ðA  CÞB  ðA  BÞC So and

54

A  C ¼ ............ A  B ¼ ............ A  C ¼ 6t 3  6t 3 þ 9t 2  8t 2 ¼ t 2 A  B ¼ 6t 2 þ 8t 2  12t þ 12t  12t 2 ¼ 2t 2

Then F ¼ A  ðB  CÞ ¼ t 2 f2i þ 4tj þ 3ð1  tÞkg  2t 2 f2ti  3t 2 j  2tkg ð1 F dt ¼ . . . . . . . . . . . . ; 0

Finish off the simplification and complete the integration.

55

1 60 f20i

þ 132j þ 75kg

Because F ¼ A  ðB  CÞ ¼ ð2t 2  4t 3 Þi þ ð4t 3 þ 6t 4 Þj þ ð3t 2 þ t 3 Þk Integration with respect to t then gives the result stated above. Now let us move on to the next stage of our development

Vector analysis 1

921

Scalar and vector fields ϕ (x, y, z) P (x, y, z)

z

If every point P ðx; y; zÞ of a region R of space has associated with it a scalar quantity ðx; y; zÞ, then ðx; y; zÞ is a scalar function and a scalar field is said to exist in the region R.

z y y

O

x

56

x

Examples of scalar fields are temperature, potential, etc. Similarly, if every point P ðx; y; zÞ of a region R has associated with it a vector quantity Fðx; y; zÞ, then Fðx; y; zÞ is a vector function and a vector field is said to exist in the region R.

F (x, y, z)

z

P (x, y, z) z y x

O

y

x

Examples of vector fields are force, velocity, acceleration, etc. Fðx; y; zÞ can be defined in terms of its components parallel to the coordinate axes, OX, OY, OZ. That is, Fðx; y; zÞ ¼ Fx i þ Fy j þ Fz k. Note these important definitions: we shall be making good use of them as we proceed

57

grad (gradient of a scalar function) If a scalar function ðx; y; zÞ is continuously differentiable with respect to its variables x; y; z; throughout the region, then the gradient of , written grad , is defined as the vector grad  ¼

@ @ @ iþ jþ k @x @y @z

ð12Þ

Note that, while  is a scalar function, grad  is a vector function. For example, if  depends upon the position of P and is defined by  ¼ 2x2 yz3 , then grad  ¼ 4xyz3 i þ 2x2 z3 j þ 6x2 yz2 k

922

Programme 26

Notation The expression (12) above can be written  @ @ @ grad  ¼ i þ j þ k  @x @y @z   @ @ @ where i þ j þ k is called a vector differential operator and is denoted by @x @y @z the symbol r (pronounced ‘del’ or sometimes ‘nabla’)   @ @ @ i.e. r i þj þk @x @y @z Beware! r cannot exist alone: it is an operator and must operate on a stated scalar function ðx, y, zÞ. If F is a vector function, rF has no meaning. So we have:



 @ @ @  þj þk @x @y @z @ @ @ ¼i þj þk @x @y @z

r ¼ grad  ¼

i

ð13Þ

Make a note of this definition and then let us see how to use it

58

Example 1 If  ¼ x2 yz3 þ xy2 z2 , determine grad  at the point P ð1, 3, 2Þ. By the definition,

grad  ¼ r ¼

@ @ @ iþ jþ k. @x @y @z

All we have to do then is to find the partial derivatives at x ¼ 1, y ¼ 3, z ¼ 2 and insert their values. ; r ¼ . . . . . . . . . . . .

59

4ð21i þ 8j þ 18kÞ Because 

¼ x2 yz3 þ xy2 z2

@ @y

¼ x2 z3 þ 2xyz2

;

@ ¼ 2xyz3 þ y2 z2 @x @ ¼ 3x2 yz2 þ 2xy2 z @z

@ @ ¼ 48 þ 36 ; ¼ 84 @x @x @ @ ¼ 8 þ 24 ; ¼ 32 @y @y @ @ ¼ 36 þ 36 ; ¼ 72 @z @z ; grad  ¼ r ¼ 84i þ 32j þ 72k ¼ 4ð21i þ 8j þ 18kÞ

Then, at ð1, 3, 2Þ

Vector analysis 1

923

Example 2 If

A ¼ x2 zi þ xyj þ y2 zk

and

B ¼ yz2 i þ xzj þ x2 zk

determine an expression for grad ðA  BÞ: This we can soon do since we know that A  B is a scalar function of x, y and z. First then, A  B ¼ . . . . . . . . . . . . A  B ¼ x2 yz3 þ x2 yz þ x2 y2 z2

60

rðA  BÞ ¼ . . . . . . . . . . . .

Then

2xyzðz2 þ 1 þ yzÞi þ x2 zðz2 þ 1 þ 2yzÞj þ x2 yð3z2 þ 1 þ 2yzÞk Because if  ¼ A  B ¼ ðx2 zi þ xyj þ y2 zkÞ  ðyz2 i þ xzj þ x2 zkÞ ¼ x2 yz3 þ x2 yz þ x2 y2 z2 @ ¼ 2xyz3 þ 2xyz þ 2xy2 z2 ¼ 2xyzðz2 þ 1 þ yzÞ @x @ ¼ x2 z3 þ x2 z þ 2x2 yz2 ¼ x2 zðz2 þ 1 þ 2yzÞ @y @ ¼ 3x2 yz2 þ x2 y þ 2x2 y2 z ¼ x2 yð3z2 þ 1 þ 2yzÞ @z ; r ðA  BÞ ¼ 2xyzðz2 þ 1 þ yzÞi þ x2 zðz2 þ 1 þ 2yzÞj þ x2 yð3z2 þ 1 þ 2yzÞk Now let us obtain another useful relationship. z

dr dy

If OP is a position vector r where r ¼ xi þ yj þ zk and dr is a small displacement corresponding to changes dx, dy, dz in x, y, z respectively, then

dz dx

r z y x

O

y

dr ¼ dx i þ dy j þ dz k

x

If ðx; y; zÞ is a scalar function at P, we know that @ @ @ iþ jþ k @x @y @z grad   dr ¼ . . . . . . . . . . . . grad  ¼ r ¼

Then

61

924

Programme 26

62

grad   dr ¼ Because

@ @ @ dx þ dy þ dz @x @y @z



 @ @ @ grad   dr ¼ iþ jþ k  ðdx i þ dy j þ dz kÞ @x @y @z @ @ @ ¼ dx þ dy þ dz @x @y @z ¼ the total differential d of 

That is d ¼ dr  grad 

ð14Þ This will certainly be useful, so make a note of it

63

Directional derivatives Q (r + dr)

z

We have just established that d ¼ dr  grad 

P (r)

and ;

dz dy

If ds is the small element of arc between P ðrÞ and Q ðr þ dr) then ds ¼ jdrj dr dr ¼ ds jdrj

dr ds

O

dx

y

x

dr is thus a unit vector in the direction of dr. ds d dr ¼  grad  ds ds

If we denote the unit vector

dr ^ then the result becomes by a ds

d ^  grad  ¼a ds d ^ and is called the is thus the projection of grad  on the unit vector a ds ^ . It gives the rate of change of  directional derivative of  in the direction of a d ^ and ^  grad  will be a with distance measured in the direction of a ¼a ds ^ and grad  have the same direction, since then maximum when a ^  grad  ¼j a ^ jj grad  j cos  and  will be zero. a Thus the direction of grad  gives the direction in which the maximum rate of change of  occurs.

Vector analysis 1

925

Example 1 Find the directional derivative of the function  ¼ x2 z þ 2xy2 þ yz2 at the point (1, 2, –1) in the direction of the vector A ¼ 2i þ 3j  4k. We start off with  ¼ x2 z þ 2xy2 þ yz2 ; r ¼ . . . . . . . . . . . . r ¼ ð2xz þ 2y2 Þi þ ð4xy þ z2 Þj þ ðx2 þ 2yzÞk

64

Because @ ¼ 2xz þ 2y2 ; @x

@ ¼ 4xy þ z2 ; @y

@ ¼ x2 þ 2yz @z

Then, at ð1, 2,  1Þ r ¼ ð2 þ 8Þi þ ð8 þ 1Þj þ ð1  4Þk ¼ 6i þ 9j  3k ^ where A ¼ 2i þ 3j  4k Next we have to find the unit vector a ^ ¼ ............ a 1 ^ ¼ pffiffiffiffiffiffi ð2i þ 3j  4kÞ a 29

65

Because

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi A ¼ 2i þ 3j  4k ; j A j¼ 4 þ 9 þ 16 ¼ 29 A 1 ^¼ ¼ pffiffiffiffiffiffi ð2i þ 3j  4kÞ a jAj 29

1 ^ ¼ pffiffiffiffiffiffi ð2i þ 3j  4kÞ So we have r ¼ 6i þ 9j  3k and a 29 d ^  r ¼a ; ds ¼ ............ d 51 ¼ pffiffiffiffiffiffi ¼ 9:47 ds 29 Because d 1 ^  r ¼ pffiffiffiffiffiffi ð2i þ 3j  4kÞ  ð6i þ 9j  3kÞ ¼a ds 29 1 51 ¼ pffiffiffiffiffiffi ð12 þ 27 þ 12Þ ¼ pffiffiffiffiffiffi ¼ 9:47 29 29

66

926

Programme 26

That is all there is to it. (a) From the given scalar function , determine r. ^ in the direction of the given vector A. (b) Find the unit vector a (c) Then

d ^  r. ¼a ds

Example 2 Find the directional derivative of  ¼ x2 y þ y2 z þ z2 x at the point ð1,  1, 2Þ in the direction of the vector A ¼ 4i þ 2j  5k. Same as before. Work through it and check the result with the next frame

67

d 23 ¼ pffiffiffi ¼ 3:43 ds 3 5 Because  ¼ x2 y þ y2 z þ z2 x ; r ¼ ð2xy þ z2 Þi þ ðx2 þ 2yzÞj þ ðy2 þ 2zxÞk ; At ð1,  1, 2Þ, r ¼ 2i  3j þ 5k pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi A ¼ 4i þ 2j  5k ; j A j ¼ 16 þ 4 þ 25 ¼ 45 ¼ 3 5 1 ^ ¼ pffiffiffi ð4i þ 2j  5kÞ ; a 3 5 ;

d 1 ^  r ¼ pffiffiffi ð4i þ 2j  5kÞ  ð2i  3j þ 5kÞ ¼a ds 3 5 1 23 ¼ pffiffiffi ð8  6  25Þ ¼ pffiffiffi ¼ 3:43 3 5 3 5

Example 3 Find the direction from the point ð1, 1, 0Þ which gives the greatest rate of increase of the function  ¼ ðx þ 3yÞ2 þ ð2y  zÞ2 . This appears to be different, but it rests on the fact that the greatest rate of increase of  with respect to distance is in ............

68

the direction of r All we need then is to find the vector r, which is ............

Vector analysis 1

927

69

r ¼ 4ð2i þ 8j  kÞ Because  ¼ ðx þ 3yÞ2 þ ð2y  zÞ2 @ @ ¼ 2ðx þ 3yÞ; ¼ 6ðx þ 3yÞ þ 4ð2y  zÞ; @x @y @ @ @ ¼ 8; ¼ 32; ¼ 4 ; At ð1, 1, 0Þ, @x @y @z ;

@ ¼ 2ð2y  zÞ @z

; r ¼ 8i þ 32j  4k ¼ 4ð2i þ 8j  kÞ ; greatest rate of increase occurs in direction 2i þ 8j  k So on we go

70

Unit normal vectors The equation of ðx; y; zÞ ¼ constant represents a surface in space. For example, 3x  4y þ 2z ¼ 1 is the equation of a plane and x2 þ y2 þ z2 ¼ 4 represents a sphere centred on the origin and of radius 2. z dr

If dr is a displacement in this surface, then d ¼ 0 since  is constant over the surface.

r y x

O

z y

x

Therefore our previous relationship dr  grad  ¼ d becomes dr  grad  ¼ 0 for all such small displacements dr in the surface. But dr  grad  ¼j dr jj grad  j cos  ¼ 0:  ; grad  is perpendicular to dr, i.e. grad  is a vector perpendicular to ; ¼ 2 the surface at P, in the direction of maximum rate of change of . The magnitude of that maximum rate of change is given by j grad j.

928

Programme 26

The unit vector N in the direction of grad  is called the unit normal vector at P. ; Unit normal vector

z

N¼ dr

r j r j

ð15Þ

r y x

z

O

y

x

Example 1 Find the unit normal vector to the surface x3 y þ 4xz2 þ xy2 z þ 2 ¼ 0 at the point ð1; 3; 1Þ. Vector normal ¼ r ¼ . . . . . . . . . . . .

71

r ¼ ð3x2 y þ 4z2 þ y2 zÞi þ ðx3 þ 2xyzÞj þ ð8xz þ xy2 Þk Then, at ð1, 3,  1Þ,

r ¼ 4i  5j þ k

and the unit normal at ð1, 3,  1Þ is . . . . . . . . . . . .

72

1 pffiffiffiffiffiffi ð4i  5j þ kÞ 42 Because

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi 16 þ 25 þ 1 ¼ 42 r 1 ¼ pffiffiffiffiffiffi ð4i  5j þ kÞ N¼ j r j 42

j r j ¼ and

One more. Example 2 Determine the unit normal to the surface xyz þ x2 y  5yz  5 ¼ 0 at the point ð3, 1, 2Þ. All very straightforward. Complete it.

Vector analysis 1

929

1 Unit normal ¼ N ¼ pffiffiffiffiffiffi ð8i þ 5j  2kÞ 93

73

Because  ¼ xyz þ x2 y  5yz  5 ; r ¼ ðyz þ 2xyÞi þ ðxz þ x2  5zÞj þ ðxy  5yÞk pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi At ð3, 1, 2Þ, r ¼ 8i þ 5j  2k; j r j¼ 64 þ 25 þ 4 ¼ 93 r 1 ¼ pffiffiffiffiffiffi ð8i þ 5j  2kÞ ; Unit normal ¼ N ¼ j r j 93 Collecting our results so far, we have, for  ðx, y, zÞ a scalar function @ @ @ iþ jþ k @x @y @z @ @ @ dx þ dy þ dz (b) d ¼ dr  grad  where d ¼ @x @y @z d ^  grad  ¼a (c) directional derivative ds r (d) unit normal vector N ¼ . j r j (a) grad  ¼ r ¼

Copy out this brief summary for future reference. It will help

grad of sums and products of scalars

  @ @ @ ðA þ BÞ þ j ðA þ BÞ þ k ðA þ BÞ (a) rðA þ BÞ ¼ i @x @y @z   @A @A @A @B @B @B ¼ iþ jþ k þ iþ jþ k @x @y @z @x @y @z 

; rðA þ BÞ ¼ rA þ rB    @ @ @ (b) rðABÞ ¼ i ðABÞ þ j ðABÞ þ k ðABÞ @x @y @z    @B @A @B @A @B @A þB þB þB ¼i A þj A þk A @x @x @y @y @z @z   @B @B @B @A @A @A iþA jþA k þ B iþB jþB k ¼ A @x @y @z @x @y @z   @B @B @B @A @A @A iþ jþ k þB iþ jþ k ¼A @x @y @z @x @y @z ; rðABÞ ¼ AðrBÞ þ BðrAÞ Remember that in these results A and B are scalars.

74

930

75

Programme 26

Example If A ¼ x2 yz þ xz2 and B ¼ xy2 z  z3 , evaluate rðABÞ at the point ð2, 1, 3Þ. We know that rðABÞ ¼ AðrBÞ þ BðrAÞ At ð2, 1, 3Þ, rB ¼ . . . . . . . . . . . . ;

76

rB ¼ 3i þ 12j  25k;

rB ¼

rA ¼ . . . . . . . . . . . . rA ¼ 21i þ 12j þ 16k

@B @B @B iþ jþ k ¼ y2 zi þ 2xyzj þ ðxy2  3z2 Þk @x @y @z

¼ 3i þ 12j  25k at ð2, 1, 3Þ @A @A @A rA ¼ iþ jþ k ¼ ð2xyz þ z2 Þi þ x2 zj þ ðx2 y þ 2xzÞk @x @y @z ¼ 21i þ 12j þ 16k

at ð2, 1, 3Þ

Now rðABÞ ¼ AðrBÞ þ BðrAÞ ¼ . . . . . . . . . . . . Finish it

77

rðABÞ ¼ 3ð117i þ 36j  362kÞ Because rðABÞ ¼ AðrBÞ þ BðrAÞ A ¼ x2 yz þ xz2

; at ð2, 1, 3Þ,

A ¼ 12 þ 18 ¼ 30

; at ð2, 1, 3Þ, B ¼ 6  27 ¼ 21 B ¼ xy z  z ; rðABÞ ¼ 30ð3i þ 12j  25kÞ  21ð21i þ 12j þ 16kÞ 2

3

¼ 351i þ 108j  1086k ¼ 3ð117i þ 36j  362kÞ So add these to the list of results. rðA þ BÞ ¼ rA þ rB rðABÞ ¼ AðrBÞ þ BðrAÞ where A and B are scalars. Now on to the next frame

Vector analysis 1

931

div (divergence of a vector function)

78

The operator r  (notice the ‘dot’; it makes all the difference) can be applied to a vector function Aðx, y, zÞ to give the divergence of A, written in short as div A. If A ¼ ax i þ ay j þ az k

    @ @ @  ax i þ ay j þ az k i þj þk @x @y @z @ax @ay @az ; div A ¼ r  A ¼ þ þ @x @y @z div A ¼ r  A ¼

Note that (a) the grad operator r acts on a scalar and gives a vector (b) the div operation r acts on a vector and gives a scalar. Example 1 If A ¼ x2 yi  xyzj þ yz2 k then div A ¼ r  A ¼ . . . . . . . . . . . . div A ¼ r  A ¼ 2xy  xz þ 2yz

79

We simply take the appropriate partial derivatives of the coefficients of i, j and k. It could hardly be easier. Example 2 If A ¼ 2x2 yi  2ðxy2 þ y3 zÞj þ 3y2 z2 k, determine r  A, i.e. div A. Complete it.

r  A ¼ ............ rA¼0

Because A ¼ 2x2 yi  2ðxy2 þ y3 zÞj þ 3y2 z2 k @ax @ay @az þ þ rA¼ @x @y @z ¼ 4xy  2ð2xy þ 3y2 zÞ þ 6y2 z ¼ 4xy  4xy  6y2 z þ 6y2 z ¼ 0 Such a vector A for which r  A ¼ 0 at all points, i.e. for all values of x, y, z, is called a solenoidal vector. It is rather a special case.

80

932

Programme 26

curl (curl of a vector function) The curl operator denoted by r, acts on a vector and gives another vector as a result. If A ¼ ax i þ ay j þ az k, then curl A ¼ r  A.   @ @ @ i.e. curl A ¼ r  A ¼ i þ j þ k  ðax i þ ay j þ az kÞ @x @y @z i

j

k

@ @ @ @x @y @z ax ay az       @ay @ax @az @ay @ax @az    þj þk ; rA¼i @y @z @z @x @x @y ¼

curl A is thus a vector function. It is best remembered in its determinant form, so make a note of it. If r  A ¼ 0 then A is said to be irrotational. Then on for an example

81

Example 1 If A ¼ ðy4  x2 z2 Þi þ ðx2 þ y2 Þj  x2 yzk, determine curl A at the point ð1, 3,  2Þ.

curl A ¼ r  A ¼

i @ @x

j @ @y

k @ @z

y4  x2 z2

x2 þ y2

x2 yz

Now we expand the determinant   @ @ @ @ ðx2 yzÞ  ðx2 þ y2 Þ  j ðx2 yzÞ  ðy4  x2 z2 Þ rA¼i @y @z @x @z  @ 2 @ ðx þ y2 Þ  ðy4  x2 z2 Þ þk @x @y All that now remains is to obtain the partial derivatives and substitute the values of x, y, z. ; r  A ¼ ............

82

2i  8j  106k r  A ¼ ifx2 zg  jf2xyz þ 2x2 zg þ kf2x  4y3 g. ; At ð1, 3,  2Þ,

r  A ¼ ið2Þ  jð12  4Þ þ kð2  108Þ ¼ 2i  8j  106k

Vector analysis 1

933

Example 2 Determine curl F at the point (2, 0, 3) given that F ¼ ze2xy i þ 2xz cos yj þ ðx þ 2yÞk. In determinant form, curl F ¼ r  F ¼ . . . . . . . . . . . .   i   @    @x   ze2xy

j @ @y 2xz cos y

        x þ 2y  k @ @z

83

Now expand the determinant and substitute the values of x, y and z, finally obtaining curl F ¼ . . . . . . . . . . . . curl F ¼ r  F ¼ 2ði þ 3kÞ Because r  F ¼ if2  2x cos yg  jf1  e2xy g þ kf2z cos y  2xze2xy g ; At ð2, 0, 3Þ

r  F ¼ ið2  4Þ  jð1  1Þ þ kð6  12Þ ¼ 2i  6k ¼ 2ði þ 3kÞ

Every one is done in the same way.

Summary of grad, div and curl (a) grad operator r acts on a scalar field to give a vector field. (b) div operator r acts on a vector field to give a scalar field. (c) curl operator r acts on a vector field to give a vector field. (d) With a scalar function ðx; y; zÞ grad  ¼ r ¼

@ @ @ iþ jþ k @x @y @z

(e) With a vector function A ¼ ax i þ ay j þ az k @ax @ay @az þ þ @x @y @z i j k @ @ @ ð2Þ curl A = rA ¼ @x @y @z ax ay az ð1Þ div A ¼ r  A ¼

Check through that list, just to make sure. We shall need them all

84

934

85

Programme 26

By way of review, here is one further example. Example 3 If

 ¼ x2 y2 þ x3 yz  yz2

and

F ¼ xy2 i  2yzj þ xyzk

determine for the point P ð1,  1, 2Þ, (a) r,

(b) unit normal,

(c) r  F,

(d) r  F.

Complete all four parts and then check the results with the next frame

86

Here is the working in full.  ¼ x2 y2 þ x3 yz  yz2 (a) r ¼

@ @ @ iþ jþ k @x @y @z

¼ ð2xy2 þ 3x2 yzÞi þ ð2x2 y þ x3 z  z2 Þj þ ðx3 y  2yzÞk ; At ð1; 1; 2Þ

r ¼ 4i  4j þ 3k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi j r j¼ 16 þ 16 þ 9 ¼ 41

r j r j 1 ; N ¼ pffiffiffiffiffiffi ð4i þ 4j  3kÞ 41

(b) N ¼

(c) F ¼ xy2 i  2yzj þ xyzk

rF ¼

@ax @ay @az þ þ @x @y @z

; r  F ¼ y2  2z þ xy ; At ð1,  1, 2Þ

(d)

i @ rF ¼ @x xy2

r  F ¼ 1  4  1 ¼ 4 j @ @y

k @ @z

2yz

xyz

; r  F ¼ 4

; r  F ¼ iðxz þ 2yÞ  jðyz  0Þ þ kð0  2xyÞ ¼ ðxz þ 2yÞi  yzj  2xyk ; At ð1,  1, 2Þ

r  F ¼ 2j þ 2k

; r  F ¼ 2ðj þ kÞ

Now let us combine some of these operations.

Vector analysis 1

935

87

Multiple operations We can combine the operators grad, div and curl in multiple operations, as in the examples that follow. Example 1 If A ¼ x2 yi þ yz3 j  zx3 k   @ @ @ i þ j þ k  ðx2 yi þ yz3 j  zx3 kÞ then div A ¼ r  A ¼ @x @y @z Then

¼ 2xy þ z3 þ x3 ¼  say @ @ @ iþ jþ k grad (div AÞ ¼ rðr  AÞ ¼ @x @y @z ¼ ð2y þ 3x2 Þi þ ð2xÞj þ ð3z2 Þk Move on for the next example

88

Example 2 If  ¼ xyz  2y z þ x z , determine div grad  at the point ð2, 4, 1Þ. 2

2 2

First find grad  and then the div of the result. At ð2, 4, 1Þ, div grad  ¼ r  ðrÞ ¼ . . . . . . . . . . . . div grad  ¼ 6 Because we have  ¼ xyz  2y2 z þ x2 z2 grad  ¼ r ¼

@ @ @ iþ jþ k @x @y @z

¼ ðyz þ 2xz2 Þi þ ðxz  4yzÞj þ ðxy  2y2 þ 2x2 zÞk ; div grad  ¼ r  ðrÞ ¼ 2z2  4z þ 2x2 ; At ð2, 4, 1Þ, div grad  ¼ r  ðrÞ ¼ 2  4 þ 8 ¼ 6 Example 3 If F ¼ x2 yzi þ xyz2 j þ y2 zk determine curl curl F at the point (2, 1, 1). Determine an expression for curl F in the usual way, which will be a vector, and then the curl of the result. Finally substitute values. curl curl F ¼ . . . . . . . . . . . .

89

936

Programme 26

90

curl curl F ¼ r  ðr  FÞ ¼ i þ 2j þ 6k Because i @ curl F ¼ @x

j @ @y

x2 yz xyz2

k @ @z y2 z

¼ ð2yz  2xyzÞi þ x2 yj þ ðyz2  x2 zÞk

Then

curl curl F ¼

i @ @x

j @ @y

2yz  2xyz x2 y

k @ @z yz2  x2 z

¼ z2 i  ð2xz  2y þ 2xyÞj þ ð2xy  2z þ 2xzÞ k ; At (2, 1, 1),

91

curl curl F ¼ r  ðr  FÞ ¼ i þ 2j þ 6k

Remember that grad, div and curl are operators and that they must act on a scalar or vector as appropriate. They cannot exist alone and must be followed by a function. Some interesting general results appear. (a) curl grad  where  is a scalar @ @ @ iþ jþ k @x @y @z i j k @ @ @ ; curl grad  ¼ @x @y @z @ @ @ @x @y @z  2  2 @  @2 @  @2   ¼i j @y@z @z@y @z@x @x@z  2 @  @2 þk  @x@y @y@x grad  ¼

¼0 ; curl grad  ¼ r  ðrÞ ¼ 0

Vector analysis 1

937

(b) div curl A where A is a vector.

A ¼ ax i þ ay j þ az k

i j k @ @ @ curl A ¼ r  A ¼ @x @y @z ax ay az       @ay @ax @az @ay @az @ax    j þk ¼i @y @z @x @z @x @y   @ @ @ Then div curl A ¼ r  ðr  AÞ ¼ i þ j þ k  ðr  AÞ @x @y @z @ 2 az @ 2 ay @ 2 az @ 2 ax @ 2 ay @ 2 ax   þ þ  @x@y @z@x @x@y @y@z @z@x @y@z ¼0 ; div curl A ¼ r  ðr  AÞ ¼ 0 ¼

(c) div grad  where  is a scalar grad  ¼

@ @ @ iþ jþ k @x @y @z

Then div grad  ¼ r  ðrÞ     @ @ @ @ @ @ ¼ i þj þk iþ jþ k  @x @y @z @x @y @z ¼

@2 @2 @2 þ þ @x2 @y2 @z2

; div grad  ¼ r  ðrÞ ¼

@2 @2 @2 þ þ @x2 @y2 @z2

¼ r2 , the Laplacian of  The operator r2 is called the Laplacian. So these general results are (a) curl grad  ¼ r  ðrÞ ¼ 0 (b) div curl A ¼ r  ðr  AÞ ¼ 0 (c) div grad  ¼ r  ðrÞ ¼

@2 @2 @2 þ þ . @x2 @y2 @z2

That brings us to the end of this particular Programme. We have covered quite a lot of new material, so check carefully through the Review summary and Can you? checklist that follow: then you can deal with the Test exercise. The Further problems provide an opportunity for additional practice.

938

Programme 26

Review summary 26 If A ¼ ax i þ ay j þ az k; B ¼ bx i þ by j þ bz k; C ¼ cx i þ cy j þ cz k; then we have the following relationships. 1 Scalar product (dot product) AB¼BA

A  B ¼ AB cos 

A  ðB þ CÞ ¼ A  B þ A  C

and

If A  B ¼ 0 and A, B 6¼ 0 then A ? B. 2 Vector product (cross product)

A  B ¼ ðAB sin Þn

n = unit normal vector where A, B, n form a right-handed set. i A  B ¼ ax bx

j ay by

k az bz

A  B ¼  ðB  AÞ and A  ðB þ CÞ ¼ A  B þ A  C 3 Unit vectors (a) i  i ¼ j  j ¼ k  k ¼ 1 i  j ¼ j  k ¼ k  i ¼ 0. (b) i  i ¼ j  j ¼ k  k ¼ 0 i  j ¼ k,

j  k ¼ i, k  i ¼ j. A  ðB  CÞ

4 Scalar triple product ax A  ðB  CÞ ¼ bx cx

ay by

az bz

cy

cz

A  ðB  CÞ ¼ B  ðC  AÞ ¼ C  ðA  BÞ Unchanged by cyclic change of vectors. Sign reversed by non-cyclic change of vectors. 5 Coplanar vectors

A  ðB  CÞ ¼ 0:

6 Vector triple product

A  ðB  CÞ and ðA  BÞ  C

A  ðB  CÞ ¼ ðA  CÞB  ðA  BÞC and

ðA  BÞ  C ¼ ðC  AÞB  ðC  BÞA:

7 Differentiation of vectors If A, ax , ay , az are functions of u day dA dax daz ¼ iþ jþ k du du du du 8 Unit tangent vector T dA T ¼ du dA du

Vector analysis 1

939

9 Integration of vectors ðb ðb ðb ðb A du ¼ i ax du þ j ay du þ k az du a

a

a

a

10 grad (gradient of a scalar function ) @ @ @ iþ jþ k @x @y @z   @ @ @ ‘del’ ¼ operator r ¼ i þ j þ k @x @y @z d ^  grad  ¼ a ^  r where a ^ is a unit vector in a ¼a (a) Directional derivative ds grad  ¼ r ¼

stated direction. grad  gives the direction for maximum rate of change of . (b) Unit normal vector N to surface ðx; y; zÞ ¼ constant. N¼

r j r j

11 div (divergence of a vector function A) div A ¼ r  A ¼

@ax @ay @az þ þ @x @y @z

If r  A ¼ 0 for all points, A is a solenoidal vector. 12 curl (curl of a vector function AÞ i @ curl A ¼ r  A ¼ @x ax

j @ @y ay

k @ @z az

If r  A ¼ 0 then A is an irrotational vector. 13 Operators grad ðrÞ acts on a scalar and gives a vector div ðrÞ acts on a vector and gives a scalar curl ðrÞ acts on a vector and gives a vector. 14 Multiple operations (a) curl grad  ¼ r  ðrÞ ¼ 0 (b) div curl A ¼ r  ðr  AÞ ¼ 0 (c) div grad  ¼ r  ðrÞ ¼

@2 @2 @2 þ þ @x2 @y2 @z2

¼ r2 , the Laplacian of .

940

Programme 26

Can you? Checklist 26 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can:

Frames

. Obtain the scalar and vector product of two vectors?

1

to

4

5

to

11

12

to

27

28

to

48

49

to

55

56

to

77

78

to

80

80

to

86

87

to

91

Yes

No

. Reproduce the relationships between the scalar and vector products of the Cartesian coordinate unit vectors?

Yes

No

. Obtain the scalar and vector triple products and appreciate their geometric significance?

Yes

No

. Differentiate a vector field and derive a unit vector tangential to the vector field at a point?

Yes

No

. Integrate a vector field?

Yes

No

. Obtain the gradient of a scalar field, the directional derivative and a unit normal to a surface?

Yes

No

. Obtain the divergence of a vector field and recognise a solenoidal vector field?

Yes

No

. Obtain the curl of a vector field?

Yes

No

. Obtain combinations of div, grad and curl acting on scalar and vector fields as appropriate?

Yes

No

Vector analysis 1

941

Test exercise 26 1 Find (a) the scalar product and (b) the vector product of the vectors A ¼ 3i  2j þ 4k and B ¼ i þ 5j  2k: 2 If A ¼ 2i þ 3j  5k;

B ¼ 3i þ j þ 2k;

C ¼ i  j þ 3k; determine

(a)

the scalar triple product A  ðB  CÞ

(b)

the vector triple product A  ðB  CÞ.

3 Determine whether the three vectors A ¼ 2i þ 3j þ k; B ¼ i  2j þ 2k; C ¼ 3i þ j þ 3k are coplanar. 4 If A ¼ ðu2 þ 5Þi  ðu2 þ 3Þj þ 2u3 k, determine (a)

dA ; du

(b)

d2 A ; du2

(c)

dA ; du

all at u ¼ 2.

5 Determine the unit tangent vector at the point ð2, 4, 3Þ for the curve with parametric equations x ¼ 2u2 ;

y ¼ u þ 3;

z ¼ 4u2  u.

6 If F ¼ 2i þ 4uj þ u2 k and G ¼ u2 i  2uj þ 4k, determine ð2 ðF  GÞdu: 0

7 Find the directional derivative of the function  ¼ x2 y  2xz2 þ y2 z at the point ð1, 3, 2Þ in the direction of the vector A ¼ 3i þ 2j  k. 8 Find the unit normal to the surface  ¼ 2x3 z þ x2 y2 þ xyz  4 ¼ 0 at the point ð2, 1, 0Þ. 9 If A ¼ x2 yi þ ðxy þ yzÞj þ xz2 k; B ¼ yzi  3xzj þ 2xyk; and  ¼ 3x2 y þ xyz  4y2 z2  3; determine, at the point ð1, 2, 1Þ (a) r;

(b) r  A;

(c) r  B;

(d) grad div A; (e) curl curl A.

Further problems 26 1 If A ¼ 2i þ 3j  4k; B ¼ 3i þ 5j þ 2k; C ¼ i  2j þ 3k; determine A  ðB  CÞ. 2 If A ¼ 2i þ j  3k; B ¼ i  2j þ 2k; C ¼ 3i þ 2j  k; find A  ðB  CÞ: 3 If A ¼ i  2j þ 3k; B ¼ 2i þ j  2k; C ¼ 3i þ 2j þ k; find (a) A  ðB  CÞ;

(b) ðA  BÞ  C.

4 If F ¼ x2 i þ ð3x þ 2Þj þ sin xk, find (a)

dF d2 F ; (b) ; dx dx2

(c)

dF d ðF  FÞ at x ¼ 1. ; (d) dx dx

942

Programme 26

5 If F ¼ ui þ ð1  uÞj þ 3uk and G ¼ 2i  ð1 þ uÞj  u2 k, determine d d d ðF  GÞ; (b) ðF  GÞ; (c) ðF þ GÞ. (a) du du du 6 Find the unit normal to the surface 4x2 y2  3xz2  2y2 z þ 4 ¼ 0 at the point ð2,  1,  2Þ. 7 Find the unit normal to the surface 2xy2 þ y2 z þ x2 z  11 ¼ 0 at the point ð2, 1, 3Þ. 8 Determine the unit vector normal to the surface xz2 þ 3xy  2yz2 þ 1 ¼ 0 at the point ð1,  2,  1Þ. 9 Find the unit normal to the surface x2 y  2yz2 þ y2 z ¼ 3 at the point ð2,  3, 1Þ. 10

Determine the directional derivative of  ¼ xey þ yz2 þ xyz at the point ð2, 0, 3Þ in the direction of A ¼ 3i  2j þ k.

11

Find the directional derivative of  ¼ ðx þ 2y þ zÞ2  ðx  y  zÞ2 at the point (2, 1, 1) in the direction of A ¼ i  4j þ 2k.

12

Find the scalar triple product of (a) A ¼ i þ 2j  3k;

B ¼ 2i  j þ 4k;

C ¼ 3i þ j  2k.

(b) A ¼ 2i  3j þ k;

B ¼ 3i þ j þ 2k;

C ¼ i þ 4j  2k:

(c) A ¼ 2i þ 3j  2k; 13

(a) A ¼ 3i þ j  2k;

B ¼ 2i þ 4j þ 3k;

C ¼ i  2j þ k.

(b) A ¼ 2i  j þ 3k;

B ¼ i þ 4j  5k;

C ¼ 3i  2j þ k:

B ¼ 2i  3j þ 2k;

C ¼ 3i  3j þ k:

If F ¼ 4t i  2t j þ 4tk, determine when t ¼ 1 3

(a) 15

C ¼ 2i  5j þ k:

Find the vector triple product A  ðB  CÞ of the following.

(c) A ¼ 4i þ 2j  3k; 14

B ¼ 3i  j þ 3k;

dF ; dt

2

(b)

d2 F ; dt 2

(c)

d ðF  FÞ. dt

If  ¼ x2 sin z þ zey find, at the point ð1, 3, 2Þ, the values of (a) grad  and

(b) j grad j.

16

Given that  ¼ xy2 þ yz2  x2 , find the derivative of  with respect to distance at the point ð1, 2,  1Þ, measured parallel to the vector 2i  3j þ 4k:

17

Find unit vectors normal to the surfaces x2 þ y2  z2 þ 3 ¼ 0 and xy  yz þ zx  10 ¼ 0 at the point (3, 2, 4) and hence find the angle between the two surfaces at that point.

18

If r ¼ ðt 2 þ 3tÞi  2 sin 3tj þ 3e2t k, determine (a)

19

dr ; dt

(b)

d2 r ; dt 2

(c) the value of

d2 r at t ¼ 0. dt 2

(a) Show that curl ðyi þ xjÞ is a constant vector. (b) Show that the vector field ðyzi þ zxj þ xykÞ has zero divergence and zero curl.

Vector analysis 1

943

20 If A ¼ 2xz2 i  xzj þ ðy þ zÞk, find curl curl A. 21 Determine grad  where  ¼ x2 cosð2yz  0:5Þ and obtain its value at the point ð1, 3, 1Þ. 22 Determine the value of p such that the three vectors A, B, C are coplanar when A ¼ 2i þ j þ 4k; B ¼ 3i þ 2j þ pk; C ¼ i þ 4j þ 2k: 23 If A ¼ pi  6j  3k;

B ¼ 4i þ 3j  k;

C ¼ i  3j þ 2k

(a) find the values of p for which (1) A and B are perpendicular to each other (2) A; B and C are coplanar. (b) determine a unit vector perpendicular to both A and B when p ¼ 2.

Frames 1 to 83

Programme 27

Vector analysis 2 Learning outcomes When you have completed this Programme you will be able to: Evaluate the line integral of a scalar and a vector field in Cartesian coordinates Evaluate the volume integral of a vector field Evaluate the surface integral of a scalar and a vector field Determine whether or not a vector field is a conservative vector field Apply Gauss’ divergence theorem Apply Stokes’ theorem Determine the direction of unit normal vectors to a surface Apply Green’s theorem in the plane

944

Vector analysis 2

945

We dealt in some detail with line, surface and volume integrals in an earlier Programme, when we approached the subject analytically. In many practical problems, it is more convenient to express these integrals in vector form and the methods often lead to more concise working.

Line integrals Let a point P on the curve c joining A and B be denoted by the position vector r with respect to a fixed origin O. If Q is a neighbouring point on the curve with position vector r þ dr, then PQ ¼ dr: The curve c can be divided up into many (n) such small arcs, approximating to dr1 , dr2 , dr3 . . . drp . . . so that AB ¼

n X

dr

(a)

r + dr

r

O

(b)

drp

drp

dr3

p¼1

dr2 dr1

where drp is a vector representing the element of arc in both magnitude and direction.

Scalar field If a scalar field V exists for all points on the curve, then

n X

V drp with dr ! 0,

p¼1

defines the line integral of V along the curve c from A to B, ð i.e. line integral ¼ V dr c

We can illustrate this integral by erecting a continuous ordinate proportional to V at each point of the ð curve. V dr is then represented by

y

z

V

c

the area of the curved surface between the ends A and B of the curve c.

O

r x

To evaluate a line integral, the integrand is expressed in terms of x, y, z, with dr ¼ . . . . . . . . . . . .

1

946

Programme 27

2

dr ¼ i dx þ j dy þ k dz In practice, x, y and z are often expressed in terms of parametric equations of a fourth variable (say u), i.e. x ¼ xðuÞ; y ¼ yðuÞ; z ¼ zðuÞ. From these, dx, dy and dz can be written in terms of u and the integral evaluated in terms of this parameter u. The following examples will show the method. Example 1

ð

If V ¼ xy2 z, evaluate

V dr along the curve c having parametric equations c

x ¼ 3u; y ¼ 2u2 ; z ¼ u3 between A ð0, 0, 0Þ and B ð3, 2, 1Þ. V ¼ xy2 z ¼ ð3uÞð4u4 Þðu3 Þ ¼ 12u8 dr ¼ i dx þ j dy þ k dz ¼ . . . . . . . . . . . .

3

dr ¼ i 3 du þ j 4u du þ k 3u2 du Because x ¼ 3u,

; dx ¼ 3 du

y ¼ 2u , ; dy ¼ 4u du ; dz ¼ 3u2 du z ¼ u3 , Limits: A ð0, 0, 0Þ corresponds to u ¼ . . . . . . . . . . . . B ð3, 2, 1Þ corresponds to u ¼ . . . . . . . . . . . . 2

4

A ð0; 0; 0Þ  u ¼ 0 ð V dr ¼

;

ð1

c

B ð3; 2; 1Þ  u ¼ 1

12u8 ði 3 du þ j 4u du þ k 3u2 duÞ

0

¼ ............ Finish it off

5

4i þ

24 36 jþ k 5 11

Because ð ð1 V dr ¼ 12 ði 3u8 du þ j 4u9 du þ k 3u10 duÞ c

0

which integrates directly to give the result quoted above. Now for another example.

Vector analysis 2

947

Example 2

6

ð

If V ¼ xy þ y2 z, evaluate

V dr along the curve c defined by c

x ¼ t 2 ; y ¼ 2t; z ¼ t þ 5 between A ð0, 0, 5Þ and B ð4, 4, 7Þ. As before, expressing V and dr in terms of the parameter t we have V ¼ ............

V ¼ 6t 3 þ 20t 2 ;

dr ¼ . . . . . . . . . . . .

dr ¼ i 2t dt þ j 2 dt þ k dt

7

Because V ¼ xy þ y2 z ¼ ðt 2 Þð2tÞ þ ð4t 2 Þðt þ 5Þ ¼ 6t 3 þ 20t 2 : 9 Also x ¼ t 2 dx ¼ 2t dt > = ; dr ¼ i dx þ j dy þ k dz y ¼ 2t dy ¼ 2 dt > ; z ¼t þ5 dz ¼ dt ¼ i 2t dt þ j 2 dt þ k dt ð ð V dr ¼ ð6t 3 þ 20t 2 Þði 2t þ j 2 þ k Þ dt ; c

c

Limits: A ð0, 0, 5Þ  t ¼ . . . . . . . . . . . . B ð4, 4, 7Þ  t ¼ . . . . . . . . . . . . A ð0; 0; 5Þ  t ¼ 0; ð V dr ¼

; c

ð2

B ð4; 4; 7Þ  t ¼ 2

8

ð6t 3 þ 20t 2 Þði 2t þ j 2 þ k Þ dt

0

¼ . . . . . . . . . . . . Complete the integration. 8 ð444 i þ 290 j þ 145 kÞ 15 ð V dr ¼ 2 c

ð2

fð6t 4 þ 20t 3 Þi þ ð6t 3 þ 20t 2 Þj þ ð3t 3 þ 10t 2 Þkg dt

0

The actual integration is simple enough and gives the result shown. All line integrals in scalar fields are done in the same way.

9

948

10

Programme 27

Vector field If a vector field F exists for all points of the curve c, then for each element of arc we can form the scalar product F  dr. Summing these products for all elements of arc, we have n X F  drp

F dr

r + dr

c r

O

p¼1

ð Then, if drp ! 0, the sum becomes the integral

F  dr, c

i.e. the line integral of F from A to B along the stated curve ð . ¼ F  dr c

In this case, since F  dr is a scalar product, then the line integral is a scalar. To evaluate the line integral, F and dr are expressed in terms of x, y, z and the curve in parametric form. We have F ¼ Fx i þ Fy j þ Fz k dr ¼ i dx þ j dy þ k dz

and

F  dr ¼ ðFx i þ Fy j þ Fz k Þ  ði dx þ j dy þ k dzÞ ¼ Fx dx þ Fy dy þ Fz dz ð ð ð ð F  dr¼ Fx dx þ Fy dy þ Fz dz ;

Then

c

c

c

c

Now for an example to show it in operation. Example 1

ð

If F ¼ x2 yi þ xzj  2yzk, evaluate

F  dr between A ð0, 0, 0Þ and B ð4, 2, 1Þ c

along the curve having parametric equations x ¼ 4t; y ¼ 2t 2 ; z ¼ t 3 : Expressing everything in terms of the parameter t, we have dx ¼ . . . . . . . . . . . . ;

F ¼ ............ dy ¼ . . . . . . . . . . . . ;

dz ¼ . . . . . . . . . . . .

Vector analysis 2

949

11

F ¼ 32t 4 i þ 4t 4 j  4t 5 k dx ¼ 4 dt;

dy ¼ 4t dt;

dz ¼ 3t dt 2

Because x2 y ¼ ð16t 2 Þð2t 2 Þ ¼ 32t 4

x ¼ 4t

; dx ¼ 4 dt

xz ¼ ð4tÞðt 3 Þ ¼ 4t 4

y ¼ 2t 2

; dy ¼ 4t dt

z ¼t ; dz ¼ 3t 2 dt 2yz ¼ ð4t Þðt Þ ¼ 4t ð ð Then F  dr ¼ ð32t 4 i þ 4t 4 j  4t 5 k Þ  ði 4 dt þ j 4t dt þ k 3t 2 dtÞ ð ¼ ð128t 4 þ 16t 5  12t 7 Þ dt 2

3

5

3

A ð0, 0, 0Þ  t ¼ . . . . . . . . . . . . ;

Limits:

B ð4, 2, 1Þ  t ¼ . . . . . . . . . . . .

A  t ¼ 0; ð F  dr ¼

; c

ð1

12

Bt¼1

ð128t 4 þ 16t 5  12t 7 Þ dt ¼ . . . . . . . . . . . .

0

13

128 8 3 803 þ  ¼ ¼ 26:77 5 3 2 30 ð F  dr represents the

If the vector field F is a force field, then the line integral c

work done in moving a unit particle along the prescribed curve c from A to B. Now for another example. Example 2

ð

If F ¼ x yi þ 2yzj þ 3z xk, evaluate 2

F  dr between A ð0, 0, 0Þ and B ð1, 2, 3Þ

2

c

(a) along the straight lines c1 then c2 and c3 (b) along the straight line c4

from ð0, 0, 0Þ to ð1, 0, 0Þ from ð1, 0, 0Þ to ð1, 2, 0Þ from ð1, 2, 0Þ to ð1, 2, 3Þ joining ð0, 0, 0Þ to ð1, 2, 3Þ.

As before, we first obtain an expression for F  dr which is ............

950

Programme 27

14

F  dr ¼ x2 y dx þ 2yz dy þ 3z2 x dz Because ð ;

F  dr ¼ ðx2 y i þ 2yz j þ 3z2 x k Þ  ði dx þ j dy þ k dzÞ ð ð ð F  dr ¼ x2 y dx þ 2yz dy þ 3z2 x dz z

(a) Here the integration is made in three sections, along c1 , c2 and c3 .

(1, 2, 3) c3 1 x

c1

O

(1, 0, 0)

(1) c1 : y ¼ 0, z ¼ 0, dy ¼ 0, dz ¼ 0 ð ; F  dr ¼ 0 þ 0 þ 0 ¼ 0 c1

(2) c2 : The conditions along c2 are ............

15

c2 :

x ¼ 1,

z ¼ 0,

dx ¼ 0,

dz ¼ 0

ð ;

F  dr ¼ 0 þ 0 þ 0 ¼ 0 c2

(3) c3 : x ¼ 1, y ¼ 2, dx ¼ 0, dy ¼ 0 ð ; F  dr ¼ . . . . . . . . . . . . c3

16

27 Because ð ð3 F  dr ¼ 0 þ 0 þ 3z2 dz ¼ 27 c3

0

Summing the three partial results ð ð ð1; 2; 3Þ F  dr ¼ 0 þ 0 þ 27 ¼ 27 ; ð0; 0; 0Þ

c1 þc2 þc3

F  dr ¼ 27

c2

2 (1, 2, 0)

y

Vector analysis 2

951 z

(b) If t is taken as the parameter, the parametric equations of c are

(1, 2, 3)

x ¼ ............

c4

y ¼ ............ z ¼ ............

O x

x ¼ t;

y ¼ 2t:

1

2

y

z ¼ 3t

17

and the limits of t are . . . . . . . . . . . . t ¼ 0 and t ¼ 1

18

As in Example 1, we now express everything in terms of t and complete the integral, finally getting ð F  dr ¼ . . . . . . . . . . . . c4

ð F  dr ¼ c4

115 ¼ 28:75 4

Because F ¼ 2t 3 i þ 12t 2 j þ 27t 3 k dr ¼ i dx þ j dy þ k dz ¼ i dt þ j 2 dt þ k 3 dt ð1 ð F  dr ¼ ð2t 3 i þ 12t 2 j þ 27t 3 k Þ  ði þ 2j þ 3k Þ dt ; 0

c4

¼

ð1

ð2t 3 þ 24t 2 þ 81t 3 Þ dt ¼

0

ð1

ð83t 3 þ 24t 2 Þ dt

0

1 t4 115 ¼ 28:75 ¼ 83 þ 8t 3 ¼ 4 4 0

So the value of the line integral depends on the path taken between the two end points A and B ð (a) F  dr via c1 ; c2 and c3 ¼ 27 ð (b)

F  dr via c4

¼ 28:75

We shall refer to this topic later. One further example on your own. The working is just the same as before.

19

952

Programme 27

Example 3

ð

If F ¼ x y i þ y zj þ z k, evaluate 2 2

3

F  dr along the curve x ¼ 2u2 , y ¼ 3u, z ¼ u3

2

c

between A ð2,  3,  1Þ and B ð2, 3, 1Þ. Proceed as before. You will have no difficulty. ð F  dr ¼ . . . . . . . . . . . . c

ð

20

F  dr ¼ c

500 ¼ 23:8 21

Here is the working for you to check. x ¼ 2u2

y ¼ 3u

z ¼ u3

x2 y2 ¼ ð4u4 Þð9u2 Þ ¼ 36 u6

dx ¼ 4u du

y z ¼ ð27u Þðu Þ ¼ 27u

dy ¼ 3 du

3

3

3

6

z2 ¼ u6 dz ¼ 3u2 du Limits: A ð2,  3,  1Þ corresponds to u ¼ 1 B ð2, 3, 1Þ corresponds to u ¼ 1 ð1 F  dr ¼ ðx2 y2 i þ y3 zj þ z2 k Þ  ði dx þ j dy þ k dzÞ

ð ;

1

c

¼ ¼

ð1

1 ð1 1

ð36u6 i þ 27u6 j þ u6 k Þ  ði 4u du þ j 3 du þ k 3u2 duÞ ð144u7 þ 81u6 þ 3u8 Þ du

1 81u7 u9 500 ¼ 23:8 þ ¼ ¼ 18u8 þ 7 3 1 21 Now on to the next section

Volume integrals 21

If V is a closed region bounded by a surface S and F is a vector field at each point of ð V and on its boundary surface S, then F dV is the volume integral of F V

throughout the region. z

dV = dxdydz

ð F dV ¼ V

O y x

ð x2 ð y2 ð z2 F dz dy dx x1

y1

z1

Vector analysis 2

953

Example 1 ð F dV where V is the region bounded by the planes x ¼ 0, x ¼ 2, Evaluate V

y ¼ 0, y ¼ 3, z ¼ 0, z ¼ 4, and F ¼ xyi þ zj  x2 k. We start, as in most cases, by sketching the diagram, which is ............

22

z

dV O

y

x

Then F ¼ xy i þ z j  x2 k and dV ¼ dx dy dz ð ð4 ð3 ð2 ; F dV ¼ ðxyi þ zj  x2 kÞ dx dy dz 0

V

0

ð4 ð3

0

x ¼ 2 x2 y x3 i þ xzj  k dy dz ¼ 2 3 0 0 x¼0  ð4 ð3 8 ¼ 2yi þ 2zj  k dy dz 3 0 0 ¼ . . . . . . . . . . . . Complete the integral. ð F dV ¼ 4ð9i þ 12j  8kÞ V

Because y ¼ 3 ð4

ð 8 F dV ¼ dz y2 i þ 2yzj  yk 3 0 V y¼0 ð4 ¼ ð9i þ 6zj  8kÞ dz 0



4

¼ 9zi þ 3z2 j  8zk 0

¼ 36i þ 48j  32k ¼ 4ð9i þ 12j  8kÞ Now another.

23

954

Programme 27

Example 2 ð F dV where V is the region bounded by the planes x ¼ 0, y ¼ 0, z ¼ 0 Evaluate V

and 2x þ y þ z ¼ 2, and F ¼ 2zi þ yk . To sketch the surface 2x þ y þ z ¼ 2, note that when z ¼ 0,

2x þ y ¼ 2

i.e. y ¼ 2  2x

when y ¼ 0,

2x þ z ¼ 2

i.e. z ¼ 2  2x

when x ¼ 0,

yþz¼2

i.e. z ¼ 2  y

Inserting these in the planes x ¼ 0, y ¼ 0, z ¼ 0 will help. The diagram is therefore ............

24

z 2

dV O 2 1

x

–x y = 2(1

y

)

So 2x þ y þ z ¼ 2 cuts the axes at A ð1, 0, 0Þ; B ð0, 2, 0Þ; C ð0, 0, 2Þ. Also F ¼ 2zi þ yk; z ¼ 2  2x  y ¼ 2ð1  xÞ  y ð ð 1 ð 2ð1xÞ ð 2ð1xÞy ; F dV ¼ ð2zi þ yk Þ dz dy dx 0

V

¼ ¼

0

ð 1 ð 2ð1xÞ

0 0 ð 1 ð 2ð1xÞ 0

0

z ¼ 2ð1xÞy

z2 i þ yzk

dy dx z¼0

f½4ð1  xÞ2  4ð1  xÞy þ y2 i

0

þ ½ 2ð1  xÞy  y2  kg dy dx ð 1  y3 i 4ð1  xÞ2 y  2ð1  xÞy2 þ ¼ 3 0  2ð1xÞ y3 dx þ ð1  xÞy2  k 3 y¼0 ¼ ............ Finish the last stage

Vector analysis 2

955

ð F dV ¼ V

25

1 ð2i þ kÞ 3

Because ð ð1 8 4 ð1  xÞ3 i þ ð1  xÞ3 k dx F dV ¼ 3 V 0 3

1 2 1 1 ¼  ð1  xÞ4 i  ð1  xÞ4 k ¼ ð2i þ k Þ 3 3 3 0 And now one more, slightly different. Example 3 ð Evaluate F dV where F ¼ 2i þ 2zj þ yk and V is the region bounded by the V

planes z ¼ 0, z ¼ 4 and the surface x2 þ y2 ¼ 9. z

It will be convenient to use cylindrical polar coordinates ð, , zÞ so the relevant transformations are

dV z O ϕρ

x ¼ ............;

y ¼ ............

z ¼ ............;

dV ¼ . . . . . . . . . . . .

y

x

x ¼  cos ; z ¼ z;

y ¼  sin  dV ¼  d d dz

ððð

ð F dV ¼

Then

ð2i þ 2zj þ yk Þ dx dy dz:

V

V

Changing into cylindrical polar coordinates with appropriate change of limits this becomes ð ð 2 ð 3 ð 4 F dV ¼ ð2i þ 2zj þ  sin  k Þ dz  d d ¼0

V

¼ ¼

ð 2

¼0

ð3

¼0 ¼0 ð 2 ð 3



z¼0

4

2zi þ z2 j þ  sin  zk

z¼0

 d d

ð8i þ 16j þ 4 sin  k Þ d d

0

¼4

0

ð 2 ð 3 0

ð2i þ 4j þ 2 sin  k Þ d d

0

Completing the working, we finally get ð F dV ¼ . . . . . . . . . . . . V

26

956

Programme 27

27

72ði þ 2jÞ Because 3 ð ð 2

3 2 2 F dV ¼ 4  i þ 2 j þ sin  k d 3 0 V 0 ð 2 ¼4 ð9i þ 18j þ 9 sin  k Þ d 0

¼ 36

ð 2



ði þ 2j þ sin  kÞ d

0

2

¼ 36 i þ 2j  cos  k 0

¼ 36fð2i þ 4j  k Þ  ðk Þg ¼ 72ði þ 2jÞ You will, of course, remember that in appropriate cases, the use of cylindrical polar coordinates or spherical polar coordinates often simplifies the subsequent calculations. So keep them in mind. Now let us turn to surface integrals – in the next frame

Surface integrals 28

(A x B)

The vector product of two vectors A and B has magnitude jA  Bj ¼ AB sin  at right angles to the plane of A and B to form a right-handed set. B A

 If  ¼ , then jA  Bj ¼ AB in the direction of the 2 ^ is a unit normal then normal. Therefore, if n

AB n B A

^ ¼ AB n ^ A  B ¼ jAj jBjn

Vector analysis 2

957 If P ðx, yÞ is a point in the x–y plane, the element of area dx dy has a vector area dS ¼ ði dxÞ  ðj dyÞ:

z

dxdyk O

y dx

x

dS

P dy

i.e. dS ¼ dx dyði  jÞ ¼ dx dy k i.e. a vector of magnitude dx dy acting in the direction of k and referred to as the vector area.

dSn

For a general surface S in space, each element of surface dS has a vector area dS such that ^ dS ¼ dS n.

S dS

You will remember we established previously that for a surface S given by the ^ is given by equation ðx; y, zÞ ¼ constant, the unit normal n ^¼ n

grad  r ¼ jgrad j jrj

Let us see how we can apply these results to the following examples.

29

Scalar fields Example 1 A scalar field V ¼ xyz exists over the curved surface S defined by x2 þ y2 ¼ 4 ð between the planes z ¼ 0 and z ¼ 3 in the first octant. Evaluate V dS over this S

surface. We have V ¼ xyz S:

x2 þ y2  4 ¼ 0, z ¼ 0 to z ¼ 3

r jrj @ @ @ iþ jþ k ¼ 2xi þ 2yj and Now r ¼ @x @y @z qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi jr j ¼ 4x2 þ 4y2 ¼ 2 x2 þ y2 ¼ 2 4 ¼ 4

^ dS dS ¼ n

^¼ where n

Therefore r xi þ yj xi þ yj ^ dS ¼ ^¼ ¼ so that dS ¼ n dS n jr j 2 2 ð ð ^ dS ; V dS ¼ V n S S ð 1 xyzðxi þ yj Þ dS ¼ 2 S ð 1 ¼ ðx2 yzi þ xy2 zj Þ dS 2 S

ð1Þ

958

Programme 27

We have to evaluate this integral over the prescribed surface. Changing to cylindrical coordinates with  ¼ 2 z dS dSn z

O ϕ

y ¼ ............

z ¼ ............;

dS ¼ . . . . . . . . . . . .

y

ρ

x

x ¼ ............;

30

x ¼ 2 cos ; z ¼ z;

y ¼ 2 sin  dS ¼ 2 d dz

; x2 yz ¼ ð4 cos2 Þð2 sin ÞðzÞ ¼ 8 cos2  sin  z xy2 z ¼ ð2 cos Þð4 sin2 ÞðzÞ ¼ 8 cos  sin2  z Then result (1) above becomes ð ð ð 1 =2 3 V dS ¼ ð8 cos2  sin  zi þ 8 cos  sin2  zj Þ 2 dz d 2 0 0 S ð =2 ð 3 ¼4 ðcos2  sin  i þ cos  sin2  j Þ 2z dz d 0

¼4

ð =2

0

ðcos2  sin  i þ cos  sin2  j Þ 9 d

0

and this eventually gives

ð V dS ¼ . . . . . . . . . . . . S

31

ð V dS ¼ 12ði þ jÞ S

Because

=2 ð cos3  sin3  iþ j V dS ¼ 36  ¼ 12ði þ jÞ 3 3 S 0

Vector analysis 2

959

Example 2 A scalar field V ¼ x þ y þ z exists over the surface S defined by 2x þ 2y þ z ¼ 2 bounded by x ¼ 0; y ¼ 0; z ¼ 0 in the first octant. ð Evaluate V dS over this surface. S

z k

n

γ

S:

dS y R

2x þ 2y þ z ¼ 2

x¼0

z ¼ 2  2y

y¼0

z ¼ 2  2x

z¼0

y¼1x

dR x

r jrj @ @ @ iþ jþ k ¼ 2i þ 2j þ k and Now r ¼ @x @y @z pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi jr j ¼ 4 þ 4 þ 1 ¼ 9 ¼ 3

^ dS dS ¼ n

^¼ where n

Therefore ^¼ n

r 2i þ 2j þ k 1 ^ dS ¼ ð2i þ 2j þ kÞ dS ¼ so that dS ¼ n jr j 3 3

If we now project dS onto the x–y plane, dR ¼ dS cos  ^ k¼ cos  ¼ n

1 1 ð2i þ 2j þ kÞ  ðkÞ ¼ 3 3

1 ; dS ¼ 3dR ¼ 3 dxdy ; dR ¼ dS 3 ð ð ð ð 1 ^ ðx þ y þ zÞ ð2i þ 2j þ k Þ3 dx dy ; V dS ¼ V n dS ¼ 3 S S S But z ¼ 2  2x  2y ð 1 ð 1x ð V dS ¼ ð2  x  yÞð2i þ 2j þ k Þ dy dx ; S

x¼0

y¼0

¼ ............

960

Programme 27

32

2 ð2i þ 2j þ kÞ 3 Because 1x ð ð1

y2 V dS ¼ 2y  xy  ð2i þ 2j þ k Þ dx 2 0 0 S

1 3 x3 2 xx þ ¼ ð2i þ 2j þ kÞ 2 6 0 2 ¼ ð2i þ 2j þ kÞ 3

33

Vector fields Example 1 A vector field F ¼ yi þ 2j þ k exists over a surface S defined by x2 þ y2 þ z2 ¼ 9 ð bounded by x ¼ 0, y ¼ 0, z ¼ 0 in the first octant. Evaluate F  dS over the S

surface indicated. ^ dS dS ¼ n

^¼ where n

r where  ¼ x2 þ y2 þ z2  9 ¼ 0 jrj

@ @ @ iþ jþ k ¼ 2x i þ 2y j þ 2z k and @x @y @z qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi jr j ¼ 4x2 þ 4y2 þ 4z2 ¼ 2 x2 þ y2 þ z2 ¼ 2 9 ¼ 6

Now r ¼

z n

1 ð2xi þ 2yj þ 2zkÞ 6 1 ¼ ðxi þ yj þ zkÞ 3

^¼ ; n θ O

r y

ϕ x

ð

ð

ð 1 ^ F  dS ¼ F  n dS ¼ ðyi þ 2j þ k Þ  ðxi þ yj þ zk Þ dS 3 S S S ð 1 ¼ ðxy þ 2y þ zÞ dS 3 S

Before integrating over the surface, we convert to spherical polar coordinates. x ¼ ............;

y ¼ ............

z ¼ ............;

dS ¼ . . . . . . . . . . . .

Vector analysis 2

961

x ¼ 3 sin  cos ; z ¼ 3 cos ;

34

y ¼ 3 sin  sin  dS ¼ 9 sin  d d

  Limits of  and  are  ¼ 0 to ;  ¼ 0 to : 2 2 ð ð ð 1 =2 =2 ; F  dS ¼ ð9 sin2  sin  cos  þ 6 sin  sin  3 0 0 S þ 3 cos Þ 9 sin  d d ð =2 ð =2 ð3 sin3  sin  cos  þ 2 sin2  sin  ¼9 0

0

þ sin  cos Þ d d ¼ ............ Complete the integral ð

  3 F  dS ¼ 9 1 þ 4 S

35

Because  ð ð =2   1 d F  dS ¼ 9 2 sin  cos  þ sin  þ 2 2 0 S

     =2 3 ¼ 9 sin2   cos   ¼9 1þ 2 2 0 4 Example 2 ð Evaluate F  dS where F ¼ 2yj þ zk and S is the surface x2 þ y2 ¼ 4 in the first S

two octants bounded by the planes z ¼ 0, z ¼ 5 and y ¼ 0. : x2 þ y2  4 ¼ 0

^¼ n

r jr j

@ @ @ iþ jþ k ¼ 2xi þ 2yj @x @y @z qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; jr j ¼ 4x2 þ 4y2 ¼ 2 x2 þ y2 pffiffiffi ¼2 4¼4 r 2xi þ 2yj 1 ^¼ ¼ ¼ ðxi þ yj Þ ; n jr j 4 2 ð ð ^ dS ¼ . . . . . . . . . . . . ; F  dS ¼ F  n

z

r ¼

S

S

dS

z

n

O ϕ x

ρ

y

962

Programme 27

ð

36

y2 dS S

Because ð ð 1 ^ dS ¼ ð2yj þ zk Þ  ðxi þ yj Þ dS Fn 2 S S ð ð 1 ¼ ð2y2 Þ dS ¼ y2 dS 2 S S This is clearly a case for using cylindrical polar coordinates. x ¼ ............; y ¼ ............ z ¼ ............;

37

dS ¼ . . . . . . . . . . . .

x ¼ 2 cos ; y ¼ 2 sin  z ¼ z; dS ¼ 2 d dz ð ð

ð

ð F  dS ¼

; S

y2 dS ¼ S

ð ð 4 sin2  2 d dz ¼ 8

S

Limits:  ¼ 0 to  ¼ ;

sin2  d dz S

z ¼ 0 to z ¼ 5 ð ; F  dS ¼ . . . . . . . . . . . . S

38

20 Because ð ð5 F  dS ¼ 4 S

ð

z¼0 ¼0 ð5

ð1  cos 2Þ d dz

 sin 2   dz ¼4 2 0 0

5 ð5 ¼ 4  dz ¼ 4 z ¼ 20 0

0

Example 3 ð Evaluate F  dS where F is the field x2 i  yj þ 2zk and S is the surface S

2x þ y þ 2z ¼ 2 bounded by x ¼ 0, y ¼ 0, z ¼ 0 in the first octant. We can sketch the diagram by putting x ¼ 0; y ¼ 0; z ¼ 0 in turn in the equation for S. y When x¼0 y þ 2z ¼ 2 z¼1 2 y¼0 xþz¼1 z¼1x z¼0 2x þ y ¼ 2 y ¼ 2  2x So the diagram is . . . . . . . . . . . .

Vector analysis 2

963

z

k γ

39 n

y

R x

F ¼ x2 i  yj þ 2zk;

dR

:

2x þ y þ 2z  2 ¼ 0

@ @ @ iþ jþ k ¼ 2i þ j þ 2k @x @y @z ð ð ^ dS F  dS ¼ F  n

r ¼

S

jr j ¼ 3

S

¼ . . . . . . . . . . . . ðnext stageÞ 1 3

ð

40

ð2x2  y þ 4zÞ dS S

Because ð ð 1 ^ F  n dS ¼ ðx2 i  yj þ 2zk Þ  ð2i þ j þ 2k Þ dS 3 S S ð 1 ¼ ð2x2  y þ 4zÞ dS 3 S If we now project the element of surface dS onto the x–y plane dR ¼ dS cos 

^ k cos  ¼ n

^  k dS ; dR ¼ n

; dS ¼

1 2 3 ð2i þ j þ 2k Þ  ðk Þ ¼ ; dS ¼ dx dy 3 3 2 ð ð ^ dS Using these new relationships, F  dS ¼ F  n

^ k ¼ ; n

S

S

¼ ............

dx dy ^ k n

964

Programme 27

ð ð

41

R

1 ð2x2  y þ 4zÞ dx dy 2

Because ð ð 1 ^ dS ¼ Fn ð2x2  y þ 4zÞ dS 3 S S ð ð 1 3 ð2x2  y þ 4zÞ dx dy ¼ 3 R 2 ð ð 1 ¼ ð2x2  y þ 4zÞ dx dy 2 R Limits: y ¼ 0 to y ¼ 2  2x; x ¼ 0 to x ¼ 1 ð ð ð 1 1 22x ^ dS ¼ Fn ð2x2  y þ 4zÞ dy dx ; 2 0 0 S But 2x þ y þ 2z ¼ 2

1 ð2  2x  yÞ 2 ð ^ dS ¼ . . . . . . . . . . . . ; Fn

; z¼

S

Complete the integration

42

1 2 Here is the rest of the working. ð ð ð ð 1 1 22x ^ dS ¼ F  dS ¼ F  n ð2x2  y þ 4  4x  2yÞ dy dx 2 0 0 S S ð ð 1 1 22x ¼ ð2x2  4x þ 4  3yÞ dy dx 2 0 0 22x ð

1 1 3y2 ð2x2  4x þ 4Þy  dx ¼ 2 0 2 0 ð 1 1 ¼ ð4x2  8x þ 8  4x3 þ 8x2  8x  6 þ 12x  6x2 Þ dx 2 0 ð ð1 1 1 ð6x2  4x3  4x þ 2Þ dx ¼ ð3x2  2x3  2x þ 1Þ dx ¼ 2 0 0

1 4 x 1 ¼ x3   x2 þ x ¼ 2 2 0 While we are concerned with vector fields, let us move on to a further point of interest.

Vector analysis 2

965

Conservative vector fields ð F  dr between two stated points

In general, the value of the line integral c

A and B depends on the particular path of integration followed. If, however, the line integral between A and B is independent of the path of integration between the two end points, then the vector field F is said to be conservative. þ It follows that, for a closed path in a conservative field, F  dr ¼ 0: c

c

Because, if the field is conservative ð ð F  dr ¼ F  dr c ðABÞ c2 ðABÞ ð ð1 F  dr ¼  F  dr But

c c

c2 ðBAÞ

Hence, for the closed path ABc1 þ BAc2 þ ð ð F  dr þ F  dr F  dr ¼ c ðABÞ c ðBAÞ ð2 ð1 F  dr  F  dr ¼ c1 ðABÞ c2 ðABÞ ð ð F  dr  F  dr ¼ 0 ¼

c

c

c1 ðABÞ

þ ;

c2 ðABÞ

c1 ðABÞ

F  dr ¼ 0

Note that this result holds good only for a closed curve and when the vector field is a conservative field. Now for an example. Example

ð

If F ¼ 2xyzi þ x zj þ x yk , evaluate the line integral 2

2

F  dr between A ð0, 0, 0Þ

and B ð2, 4, 6Þ (a) along the curve c whose parametric equations are x ¼ u, y ¼ u2 , z ¼ 3u (b) along the three straight lines c1 : ð0, 0, 0Þ to ð2, 0, 0Þ; c3 : ð2, 4, 0Þ to ð2, 4, 6Þ. Hence determine whether or not F is a conservative field. First draw the diagram ............

c2 : ð2, 0, 0Þ to ð2, 4, 0Þ;

43

966

Programme 27

44

z

B (2, 4, 6)

c

A

c y

c x

c

(a) F ¼ 2xyzi þ x2 zj þ x2 yk x ¼ u;

y ¼ u2 ;

; dx ¼ du;

z ¼ 3u

dy ¼ 2u du;

dz ¼ 3 du.

F  dr ¼ ð2xyzi þ x2 zj þ x2 yk Þ  ði dx þ j dy þ k dzÞ ¼ 2xyz dx þ x2 z dy þ x2 y dz Using the transformations shown above, we can now express F  dr in terms of u. F  dr ¼ . . . . . . . . . . . .

45

15u4 du Because 2xyz dx ¼ ð2uÞðu2 Þð3uÞ du ¼ 6u4 du x2 z dy ¼ ðu2 Þð3uÞð2uÞ du ¼ 6u4 du x2 y dz ¼ ðu2 Þðu2 Þ3 du

¼ 3u4 du

; F  dr ¼ 6u4 du þ 6u4 du þ 3u4 du ¼ 15u4 du The limits of integration in u are ............

46

u ¼ 0 to u ¼ 2 ð F  dr ¼

; c

ð2 0

15u du ¼ 4

2 3u5 0 ¼

ð F  dr ¼ 96

96 c

Vector analysis 2

967

(b) The diagram for (b) is as shown. We consider each straight line section in turn. z

B (2, 4, 6)

c3

c1

y

c2

x

ð

ð F  dr ¼ ð2xyz dx þ x2 z dy þ x2 y dzÞ

c1 : ð0; 0; 0Þ to ð2; 0; 0Þ; y ¼ 0, z ¼ 0, dy ¼ 0, dz ¼ 0 ð F  dr ¼ 0 þ 0 þ 0 ¼ 0 ; c1

In the same way, we evaluate the line integral along c2 and c3 : ð ð F  dr ¼ . . . . . . . . . . . . ; F  dr ¼ . . . . . . . . . . . . c2

c3

ð

ð F  dr ¼ 0;

F  dr ¼ 96

c2

ð Because we have

c3

ð F  dr ¼ ð2xyz dx þ x2 z dy þ x2 y dzÞ

c2 :

ð2; 0; 0Þ to ð2; 4; 0Þ; x ¼ 2, z ¼ 0, ð F  dr ¼ 0 þ 0 þ 0 ¼ 0 ; c2 ð F  dr ¼ 0

dx ¼ 0,

dz ¼ 0

c3 :

ð2; 4; 0Þ to ð2; 4; 6Þ; x ¼ 2, y ¼ 4, dx ¼ 0, dy ¼ 0

6 ð6 ð F  dr ¼ 0 þ 0 þ 16 dz ¼ 16z ¼ 96 ; c 0 0 ð 3 F  dr ¼ 96

c2

c3

Collecting the three results together ð ð F  dr ¼ 0 þ 0 þ 96 ; c1 þc2 þc3

c1 þc2 þc3

F  dr ¼ 96

47

968

Programme 27

In this particular example, the value of the line integral is independent of the two paths we have used joining the same two end points and indicates that F may be a conservative field. It follows that ð ð þ F  dr  F  dr ¼ 0 i.e. F  dr ¼ 0 c1 þc2 þc3

c

þ

So, if F is a conservative field;

F  dr ¼ 0 Make a note of this for future use

48

Two tests can be applied to establish that a given vector field is conservative. If F is a conservative field (a) curl F = 0 (b) F can be expressed as grad V where V is a scalar field to be determined. For example, in the work we have just completed, we showed that F ¼ 2xyzi þ x2 zj þ x2 yk is a conservative field. (a) If we determine curl F in this case, we have curl F ¼ . . . . . . . . . . . .

49

curl F ¼ 0 Because i @ curl F ¼ @x 2xyz

j @ @y

k @ @z

x2 z x2 y

¼ ðx  x Þi  ð2xy  2xyÞj þ ð2xz  2xzÞk ¼ 0 ; curl F ¼ 0 2

2

(b) We can attempt to express F as grad V where V is a scalar in x, y, z. If V ¼ f ðx; y; zÞ grad V ¼

@V @V @V iþ jþ k @x @y @z

and we have F ¼ 2xyzi þ x2 zj þ x2 yk ;

@V ¼ 2xyz @x @V ¼ x2 z @y @V ¼ x2 y @z

; V ¼ x2 yz þ f ðy, zÞ ; V ¼ ............ ; V ¼ ............

We therefore have to find a scalar function V that satisfies the three requirements. V ¼ ............

Vector analysis 2

969

V ¼ x2 yz

50

Because @V ¼ 2xyz @x @V ¼ x2 z @y @V ¼ x2 y @z

; V ¼ x2 yz þ f ðy, zÞ ; V ¼ x2 yz þ gðx, zÞ ; V ¼ x2 yz þ hðx, yÞ

These three are satisfied if f ðy, zÞ ¼ gðz, xÞ ¼ hðx, yÞ ¼ 0 ; F ¼ grad V where V ¼ x2 yz So two tests can be applied to determine whether or not a vector field is conservative. They are (a) . . . . . . . . . . . . (b) . . . . . . . . . . . . (a) curl F ¼ 0 (b) F ¼ grad V

51

Any one of these conditions can be applied as is convenient. Now what about these? Exercise Determine which of the following vector fields are conservative. (a) F ¼ ðx þ yÞi þ ðy  zÞj þ ðx þ y þ zÞk (b) F ¼ ð2xz þ yÞi þ ðz þ xÞj þ ðx2 þ yÞk (c) F ¼ y sin z i þ x sin z j þ ðxy cos z þ 2zÞk (d) F ¼ 2xyi þ ðx2 þ 4yzÞj þ 2y2 zk (e) F ¼ y cos x cos z i þ sin x cos z j  y sin x sin z k. Complete all five and check your findings with the next frame. (a) No (b) Yes (c) Yes (d) No (e) Yes

52

970

Programme 27

Divergence theorem (Gauss’ theorem) z

For a closed surface S, enclosing a region V in a vector field F, ð ð div F dV ¼ F  dS

S V

V

S

O y x

In general, this means that the volume integral (triple integral) on the left-hand side can be expressed as a surface integral (double integral) on the right-hand side. Let us work through one or two examples. Example 1 Verify the divergence theorem for the vector field F ¼ x2 i þ zj þ yk taken over the region bounded by the planes z ¼ 0, z ¼ 2, x ¼ 0, x ¼ 1, y ¼ 0, y ¼ 3. Start off, as always, by sketching the relevant diagram, which is ............

53

z

dV ¼ dx dy dz dV

We have to show that ð ð div F dV ¼ F  dS

O y

V

S

x

ð (a) To find

div F dV



 @ @ @ iþ j k  ðx2 i þ zj þ ykÞ @x @y @z @ 2 @ @ ðx Þ þ ðzÞ þ ðyÞ ¼ 2x þ 0 þ 0 ¼ 2x ¼ @x @y @z ð ððð div F dV ¼ 2x dV ¼ 2x dz dy dx

V

div F ¼ r  F ¼

ð ; V

V

V

Inserting the limits and completing the integration ð div F dV ¼ . . . . . . . . . . . . V

Vector analysis 2

971

ð

54

div F dV ¼ 6 V

Because 2 ð ð1 ð3 ð2 ð1 ð3

2xz dy dx div F dV ¼ 2x dz dy dx ¼ 0

V

¼

ð1

0

0

0

0

0

0

ð S

ð

0

0

F  dS

Now we have to find

ð

^ dS Fn

F  dS i.e.

(b) To find

0

3 1

ð1 2 4xy dx ¼ 12x dx ¼ 6x ¼ 6

S

S

z

O y x

The enclosing surface S consists of six separate plane faces denoted as S1 , S2 , . . . , S6 as shown. We consider each face in turn. F ¼ x2 i þ zj þ yk (1) S1 (base):

z ¼ 0;

^ ¼ k (outwards and downwards) n

; F ¼ x2 i þ yk dS1 ¼ dx dy ðð ^ dS ¼ Fn ðx2 i þ yk Þ  ðk Þ dy dx

ð ; S1

S1

¼

ð1 ð3 0

ð1

ðyÞ dy dx

0

y2  ¼ 2 0 9 ¼ 2 (2) S2 (top):

z ¼ 2;

3

^ ¼k n

dx 0

dS2 ¼ dx dy

ð

^ dS ¼ . . . . . . . . . . . . Fn

; S2

972

Programme 27

55

9 2 Because ð ðð ^ dS ¼ Fn ðx2 i þ 2j þ yk Þ  ðk Þ dy dx S2

S2

¼

ð1 ð3 0

y dy dx ¼

0

9 2

So we go on. ^ ¼j n

(3) S3 (right-hand end): y ¼ 3;

dS3 ¼ dx dz

F ¼ x2 i þ zj þ yk ðð ð ^ dS ¼ Fn ðx2 i þ zj þ yk Þ  ðj Þ dz dx ; S3

S3

¼

ð1 ð2 z dz dx 0

0

ð1 ð 1 2 2 z ¼ dx ¼ 2 dx ¼ 2 0 2 0 0 (4) S4 (left-hand end): y ¼ 0;

^ ¼ j n

ð

dS4 ¼ dx dz

^ dS ¼ . . . . . . . . . . . . Fn

; S4

56

2 Because ð ðð ð1 ð2 ^ dS ¼ Fn ðx2 i þ zj þ yk Þ  ðj Þ dz dx ¼ ðzÞ dz dx S4

S4

¼

ð1

0

z2  2

2 dx ¼

ð1

0

0

ð2Þ dx ¼ 2

0

0

Now for the remaining two sides S5 and S6 . Evaluate these in the same manner, obtaining ð ^ dS ¼ . . . . . . . . . . . . Fn ð

S5

^ dS ¼ . . . . . . . . . . . . Fn S6

Vector analysis 2

973

ð

ð

^ dS ¼ 6; Fn S5

57

^ dS ¼ 0 Fn S6

Check: ^ ¼i n dS5 ¼ dy dz (5) S5 (front): x ¼ 1; ð ðð ðð ^ dS ¼ ; Fn ði þ zj þ yk Þ  ðiÞ dy dz ¼ 1 dy dz ¼ 6 S5

S5

S5

^ ¼ i n dS6 ¼ dy dz (6) S6 (back): x ¼ 0; ðð ðð ð ^ dS ¼ Fn ðzj þ yk Þ  ði Þ dy dz ¼ 0 dy dz ¼ 0 ; S6

S6

S6

Now on to the next frame where we will collect our results together For the whole surface S we therefore have ð 9 9 F  dS ¼  þ þ 2  2 þ 6 þ 0 ¼ 6 2 2 S

58 ð div F dV ¼ 6

and from our previous work in section (a) V

We have therefore verified as required that, in this example ð ð div F dV ¼ F  dS V

S

We have made rather a meal of this since we have set out the working in detail. In practice, the actual writing can often be considerably simplified. Let us move on to another example. Example 2 Verify the Gauss divergence theorem for the vector field F ¼ xi þ 2j þ z2 k taken over the region bounded by the planes z ¼ 0, z ¼ 4, x ¼ 0, y ¼ 0 and the surface x2 þ y2 ¼ 4 in the first octant. z

Divergence theorem ð ð div F dV ¼ F  dS V

O y x

x2 + y2 = 4

 (a) div F ¼ r  F ¼

S consists of five surfaces S1 , S2 , . . . , S5 as shown.

@ @ @ iþ jþ k @x @y @z

¼ ............

S

  ðxi þ 2j þ z2 k Þ

974

Programme 27

59

1 þ 2z ð

ð

;

ððð

div F dV ¼ V

r  F dV ¼

ð1 þ 2zÞ dx dy dz

V

V

Changing to cylindrical polar coordinates ð, , zÞ x ¼  cos 

y ¼  sin 

z¼z

dV ¼  d d dz

Transforming the variables and inserting the appropriate limits, we then have ð div F dV ¼ . . . . . . . . . . . . V

Finish it

60

20 Because ð =2 ð 2 ð 4 ð div F dV ¼ ð1 þ 2zÞ dz  d d V

0

0

0

0

0

4 ð =2 ð 2

ð =2 ð 2 2 ¼ 20 d d z þ z  d d ¼ 0

0

0

2 ð =2 ð =2

2 d ¼ 40 d ¼ 20 10 ¼ 0

ð1Þ

0

0

ð F  dS over the closed surface.

(b) Now we evaluate S

z

k (S2) –i (S4)

–j (S3) n (S5)

The unit normal vector for each surface is shown. F ¼ xi þ 2j þ z2 k

O y x

–k (S1)

^ ¼ k (1) S1 : z ¼ 0; n F ¼ xi þ 2j ð ð ^ dS ¼ ðxi þ 2j Þ  ðk Þ dS ¼ 0 Fn ; S1

S1

Vector analysis 2

975

^ ¼k (2) S2 : z ¼ 4; n F ¼ xi þ 2j þ 16k ð ð ð ^ dS ¼ ðxi þ 2j þ 16k Þ  ðk Þ dS ¼ ; Fn 16 dS S2 S2 S2   4 ¼ 16 ¼ 16 4 ð ^ dS ¼ . . . . . . . . . . . . In the same way for S3 : Fn S3 ð ^ dS ¼ . . . . . . . . . . . . Fn and for S4 : S4

ð

^ dS ¼ 16; Fn S3

ð

^ dS ¼ 0 Fn

61

S4

Because we have ^ ¼ j n F ¼ xi þ 2j þ z2 k (3) S3 : y ¼ 0; ð ð ^ dS ¼ ðxi þ 2j þ z2 kÞ  ðj Þ dS Fn ; S3 S ð3 ¼ ð2Þ dS ¼ 2ð8Þ ¼ 16 S3

^ ¼ i (4) S4 : x ¼ 0; n F ¼ 2j þ z2 k ð ð ^ dS ¼ ð2j þ z2 k Þ  ði Þ dS ¼ 0 Fn ; S4

S4

Finally we have ^ ¼ ............ n

(5) S5 : x2 þ y2  4 ¼ 0

^¼ n

1 ðxi þ yjÞ 2

Because rS 2xi þ 2yj xi þ yj ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2 2 jrS j 2 4x þ 4y   ð ð ð xi þ yj 1 ^ dS ¼ ðxi þ 2j þ z2 k Þ  ; Fn ðx2 þ 2yÞ dS dS ¼ 2 2 S5 S5 S5 x2 þ y2  4 ¼ 0

^¼ n

Converting to cylindrical polar coordinates, this gives ð ^ dS ¼ . . . . . . . . . . . . Fn S5

62

976

Programme 27

63

4 þ 16 Because we have ð ð 1 ^ dS ¼ Fn ðx2 þ 2yÞ dS 2 S5 S5 also x ¼ 2 cos ; y ¼ 2 sin  z ¼ z; dS ¼ 2 d dz ð ð ð 1 4 =2 ^ dS ¼ Fn ð4 cos2  þ 4 sin Þ 2 d dz ; 2 0 0 S5 ð 4 ð =2 fð1 þ cos 2Þ þ 2 sin g d dz ¼2 0

0

 =2 sin 2  2 cos  dz 2 0 0 ð4   þ 2 dz ¼ 4 þ 16 ¼2 0 2

¼2

ð 4 



Therefore, for the total surface S ð ^ dS ¼ 0 þ 16  16 þ 0 þ 4 þ 16 ¼ 20 Fn S ð ð div F dV ¼ F  dS ¼ 20 ; V

ð2Þ

S

Other examples are worked in much the same way. You will remember that, for a closed surface, the normal vectors at all points are drawn in an outward direction. Now we move on to a further important theorem.

Stokes’ theorem n

64

dS

If F is a vector field existing over an open surface S and around its boundary, closed curve c, then þ ð curl F  dS ¼ F  dr S

c

c

This means that we can express a surface integral in terms of a line integral round the boundary curve. The proof of this theorem is rather lengthy and is to be found in the Appendix. Let us demonstrate its application in the following examples.

Vector analysis 2

977

Example 1 A hemisphere S is defined by x2 þ y2 þ z2 ¼ 4 ðz 0Þ: A vector field F ¼ 2yi  xj þ xzk exists over the surface and around its boundary c. þ ð curl F  dS ¼ F  dr: Verify Stokes’ theorem, that c

S

k

z

γ

n

S: x2 þ y2 þ z2  4 ¼ 0 F ¼ 2yi  xj þ xzk

dS

c is the circle x2 þ y2 ¼ 4. O

y

x

c

þ

ð F  dr ¼

(a)

ð2yi  xj þ xzk Þ  ði dx þ j dy þ k dzÞ ðc

c

ð2y dx  x dy þ xz dzÞ

¼ c

Converting to polar coordinates x ¼ 2 cos ;

y ¼ 2 sin ;

dx ¼ 2 sin  d;

z¼0

dy ¼ 2 cos  d;

Limits  ¼ 0 to 2

Making the substitutions and completing the integral þ F  dr ¼ . . . . . . . . . . . . c

þ

65

F  dr ¼ 12 c

Because þ ð 2 F  dr ¼ ð4 sin ½2 sin  d  2 cos  2 cos  dÞ c

0

¼ 4

ð 2

ð2 sin2  þ cos2 Þ d

0

¼ 4

ð 2 0

ð1 þ sin2 Þ d ¼ 2

 sin 2 2 ¼ 2 3  ¼ 12 2 0

ð 2

ð3  cos 2Þ d

0

ð1Þ On to the next frame

978

66

Programme 27

ð curl F  dS

(b) Now we determine ð

S

ð

curl F  dS ¼

^ dS curl F  n

F ¼ 2yi  xj þ xzk

; curl F ¼ . . . . . . . . . . . .

67

curl F ¼ zj  3k Because i j @ @ curl F ¼ @x @y 2y x

k @ ¼ i ð0  0Þ  j ðz  0Þ þ k ð1  2Þ ¼ zj  3k @z xz

rS 2xi þ 2yj þ 2zk xi þ yj þ zk ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2 2 2 jrS j 2 4x þ 4y þ 4z   ð ð xi þ yj þ zk ^ dS ¼ ðzj  3k Þ  dS curl F  n Then 2 S S ð 1 ðyz  3zÞ dS ¼ 2 S

^¼ Now n

Expressing this in spherical polar coordinates and integrating, we get ð ^ dS ¼ . . . . . . . . . . . . curl F  n S

68

12 Because x ¼ 2 sin  cos ; y ¼ 2 sin  sin ; z ¼ 2 cos ; dS ¼ 4 sin  d d ð ð ð 1 ^ dS ¼ ð2 sin  sin  2 cos   6 cos Þ4 sin  d d ; curl F  n 2 S S ð 2 ð =2 ð2 sin2  cos  sin  þ 3 sin  cos Þ d d ¼ 4 0

0

=2 2 sin3  sin  3 sin2  d þ 3 2 0 0  ð 2  2 3 d ¼ 12 ¼ 4 sin  þ 3 2 0

¼ 4

ð 2

ð2Þ

So we have from our two results (1) and (2) ð þ curl F  dS ¼ F  dr S

c

Before we proceed with another example, let us clarify a point relating to the direction of unit normal vectors now that we are dealing with surfaces. So on to the next frame

Vector analysis 2

979

Direction of unit normal vectors to a surface S When we were dealing with the divergence theorem, the normal vectors were drawn in a direction outward from the enclosed region. With an open surface as we now have, there is in fact no inward or outward direction. With any general surface, a normal vector can be drawn in either of two opposite directions. To avoid confusion, a convention must therefore be agreed upon and the established rule is as follows. n c

c n

n

c

^ is drawn perpendicular to the surface S at any point in the A unit normal n direction indicated by applying a right-handed screw sense to the direction of integration round the boundary c. Having noted that point, we can now deal with the next example. Example 2 A surface consists of five sections formed by the planes x ¼ 0, x ¼ 1, y ¼ 0, y ¼ 3, z ¼ 2 in the first octant. If the vector field F ¼ yi þ z2 j þ xyk exists over the surface and around its boundary, verify Stokes’ theorem. n=k

z

n = –i n=j

n = –j n=i c

c

c

y

c

x

If we progress round the boundary along c1 , c2 , c3 , c4 in an anticlockwise manner, the normals to the surfaces will be as shown. ð þ We have to verify that curl F  dS ¼ F  dr c

S

þ F  dr

(a) We will start off by finding ð

c

F  dr ¼ . . . . . . . . . . . .

69

980

Programme 27

ð

70

ð F  dr ¼ ðy dx þ z2 dy þ xy dzÞ

y ¼ 0; z ¼ 0; dy ¼ 0; dz ¼ 0 ð ð F  dr ¼ ð0 þ 0 þ 0Þ ¼ 0 ;

(1) Along c1 :

c1

x ¼ 1; z ¼ 0; dx ¼ 0; dz ¼ 0 ð ð F  dr ¼ ð0 þ 0 þ 0Þ ¼ 0 ;

(2) Along c2 :

c2

In the same way ð

ð F  dr ¼ . . . . . . . . . . . .

F  dr ¼ . . . . . . . . . . . .

and

c3

c4

ð

71

ð F  dr ¼ 3; c3

F  dr ¼ 0 c4

Because (3) Along c3 :

y ¼ 3; z ¼ 0; dy ¼ 0; dz ¼ 0

0 ð ð0 ; F  dr ¼ ð3 dx þ 0 þ 0Þ ¼ 3x ¼ 3 1

c3

(4) Along c4 :

x ¼ 0; z ¼ 0; dx ¼ 0; dz ¼ 0 ð ð F  dr ¼ ð0 þ 0 þ 0Þ ¼ 0 ; c þ4 F  dr ¼ 0 þ 0  3 þ 0 ¼ 3 ; þc F  dr ¼ 3 c

ð curl F  dS:

(b) Now we have to find S

First we need an expression for curl F. F ¼ yi þ z2 j þ xyk ; curl F ¼ . . . . . . . . . . . .

1

ð1Þ

Vector analysis 2

981

curl F ¼ ðx  2zÞi  yj  k

72

Because i @ curl F ¼ r  F ¼ @x y

j @ @y

k @ @z

z2

xy

¼ i ðx  2zÞ  j ðy  0Þ þ k ð0  1Þ ¼ ðx  2zÞi  yj  k ð ð ^ dS Then, for each section, we obtain curl F  dS ¼ curl F  n (1) S1 (top):

^ ¼k n

ð

^ dS ¼ . . . . . . . . . . . . curl F  n

; S1

3 Because ð ð ^ dS ¼ fðx  2zÞi  yj  k g  ðk Þ dS curl F  n S1 S ð1 ¼ ð1Þ dS ¼ ðarea of S1 Þ ¼ 3 S1

Then, likewise ^ ¼j (2) S2 (right-hand end): n ð ð ^ dS ¼ fðx  2zÞi  yj  k g  ðj Þ dS curl F  n ; S2 S ð2 ¼ ðyÞ dS S2

But y ¼ 3 for this section ð ð ^ dS ¼ ð3Þ dS ¼ ð3Þð2Þ ¼ 6 curl F  n ; S2

S2

^ ¼ j (3) S3 (left-hand end): n ð ^ dS ¼ . . . . . . . . . . . . ; curl F  n S3

73

982

Programme 27

74

0 Because ð ð ^ dS ¼ fðx  2zÞi  yj  k g  ðj Þ dS curl F  n S3 S ð3 y dS ¼ S3

But y ¼ 0 over S3 ð ^ dS ¼ 0 ; curl F  n S3

Working in the same way ð ^ dS ¼ . . . . . . . . . . . . ; curl F  n

ð

S4

ð

75

^ dS ¼ . . . . . . . . . . . . curl F  n S5

^ dS ¼ 6; curl F  n S4

ð

^ dS ¼ 12 curl F  n S5

Because ^ ¼i (4) S4 (front): n ð ð ^ dS ¼ fðx  2zÞi  yj  k g  ðiÞ dS ; curl F  n S4 S ð4 ¼ ðx  2zÞ dS S4

But x ¼ 1 over S4 2 ð ð3 ð2 ð3

2 ^ dS ¼ ; curl F  n ð1  2zÞ dz dy ¼ dy zz 0

S4

¼

ð3

0



ð2Þ dy ¼

3

0

0

 2y ¼ 6

0

0

^ ¼ i with x ¼ 0 over S5 n ð Similar working to that above gives

(5) S5 (back):

^ dS ¼ 12 curl F  n S5

Finally, collecting the five results together gives ð ^ dS ¼ . . . . . . . . . . . . curl F  n S

Vector analysis 2

983

ð

^ dS ¼ 3  6 þ 0  6 þ 12 ¼ 3 curl F  n

(2)

76

S

So, referring back to our result for section (a) we see that þ ð curl F  dS ¼ F  dr c

S

Of course we can, on occasions, make use of Stokes’ theorem to lighten the working – as in the next example. Example 3 A surface S consists of that part of the cylinder x2 þ y2 ¼ 9 between z ¼ 0 and z ¼ 4 for y 0 and the two semicircles of radius 3 in the planes z ¼ 0 and ð z ¼ 4. If F ¼ zi þ xyj þ xzk , evaluate curl F  dS over the surface. S

The surface S consists of three sections n=k

z

(a) the curved surface of the cylinder (b) the top and bottom semicircles.

n

We could therefore evaluate ð curl F  dS S

O y

over each of these separately. However, we know by Stokes’ theorem that ð curl F  dS ¼ . . . . . . . . . . . .

x n = –k

S

þ F  dr where c is the boundary of S c

z

F ¼ zi þ xyj þ xzk þ þ ; F  dr ¼ ðzi þ xyj þ xzk Þ c c ði dx þ j dy þ k dzÞ þ ¼ ðz dx þ xy dy þ xz dzÞ

c c c O c

c

y

x

Now we can work through this easily enough, taking c1 , c2 , c3 , c4 in turn, and summing the results, which gives ð þ curl F  dS ¼ F  dr ¼ . . . . . . . . . . . . S

c

77

984

Programme 27

78

24 þ

þ F  dr ¼

Here is the working in detail. c

ðz dx þ xy dy þ xz dzÞ c

(1) c1 : y ¼ 0; z ¼ 0; dy ¼ 0; dz ¼ 0 ð ð F  dr ¼ ð0 þ 0 þ 0Þ ¼ 0 c1

c1

(2) c2 : x ¼ 3; y ¼ 0; dx ¼ 0; dy ¼ 0

4 ð ð 3z2 F  dr ¼ ð0 þ 0  3z dzÞ ¼ ¼ 24 2 0 c2 c2 (3) c3 : y ¼ 0; z ¼ 4; dy ¼ 0; dz ¼ 0 ð ð ð3 F  dr ¼ ð4 dx þ 0 þ 0Þ ¼ 4 dx ¼ 24 c3

3

c3

(4) c4 : x ¼ 3; y ¼ 0; dx ¼ 0; dy ¼ 0

2 0 ð ð 3z F  dr ¼ ð0 þ 0 þ 3z dzÞ ¼ ¼ 24 2 4 c4 c4 Totalling up these four results, we have þ F  dr ¼ 0  24 þ 24  24 ¼ 24 c

þ

ð

ð

curl F  dS ¼

But

F  dr c

S

;

curl F  dS ¼ 24 S

ð curl F  dS over the three

This working is a good deal easier than calculating S

separate surfaces direct. So, if you have not already done so, make a note of Stokes’ theorem: þ ð curl F  dS ¼ F  dr S

c

Then on to the next section of the work

Vector analysis 2

985

Green’s theorem Green’s theorem enables an integral over a plane area to be expressed in terms of a line integral round its boundary curve. We showed in Programme 19 that, if P and Q are two single-valued functions of x and y, continuous over a plane surface S, and c is its boundary curve, then  ðð  þ @Q @P  ðP dx þ Q dyÞ ¼ dx dy @y c S @x

79

where the line integral is taken round c in an anticlockwise manner. In vector terms, this becomes: y

S is a two-dimensional space enclosed by a simple closed curve c.

dS = dxdy

dS ¼ dx dy O

^ dS ¼ k dx dy dS ¼ n

c x

If F ¼ Pi þ Qj where P ¼ Pðx; yÞ and Q ¼ Qðx; yÞ then curl F ¼ . . . . . . . . . . . . 

@Q @P  k @x @y



80

Because i j k @ @ @ curl F ¼ @x @y @z P Q 0       @Q @P @Q @P  ¼i 0 j 0 þk @z @z @x @y

  @Q @P @Q @P ¼ ¼ 0. ; curl F ¼ k  But in the xy plane, @z @z @x @y ð ð ^ dS and in the xy plane, n ^ ¼k So curl F  dS ¼ curl F  n   ð ð  ðð  @Q @P @Q @P ;    ðk Þ dS ¼ dx dy curl F  dS ¼ k @x @y @y S S S @x  ð ðð  @Q @P  ; dx dy curl F  dS ¼ @y S S @x Now by Stokes’ theorem . . . . . . . . . . . .

ð1Þ

986

Programme 27

ð

81

þ curl F  dS ¼

þ

F  dr c

S

þ F  dr ¼

and, in this case;

ðPi þ Qj Þ  ði dx þ j dy þ k dzÞ þc

c

ðP dx þ Q dyÞ

¼ þc

þ F  dr ¼

; c

ðP dx þ Q dyÞ

ð2Þ

c

Therefore from (1) and (2) þ ð curl F  dS ¼ F  dr in two dimensions becomes Stokes’ theorem S c  ðð  þ @Q @P  Green’s theorem dx dy ¼ ðP dx þ Q dyÞ @y S @x c Example

þ fðx2 þ y2 Þ dx þ ðx þ 2yÞ dyg taken round

Verify Green’s theorem for the integral c

the boundary curve c defined by y

y¼0

0x2

2

x þy ¼4

0x2

x¼0

0  y  2.

2

c2 c

O

c

x

 þ ðð  @Q @P  Green’s theorem: dx dy ¼ ðP dx þ Q dyÞ @y S @x c In this case ðx2 þ y2 Þ dx þ ðx þ 2yÞ dy ¼ P dx þ Q dy ; P ¼ x2 þ y2

and

Q ¼ x þ 2y

We now take c1 , c2 , c3 in turn. (1) c1 : y ¼ 0; dy ¼ 0

3 2 ð ð2 x 8 2 ; ðP dx þ Q dyÞ ¼ x dx ¼ ¼ 3 3 c1 0 0 (2) c2 :

x2 þ y2 ¼ 4

; y2 ¼ 4  x2

; y ¼ ð4  x2 Þ1=2

x þ 2y ¼ x þ 2ð4  x2 Þ1=2 1 x dy ¼ ð4  x2 Þ1=2 ð2xÞ dx ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx 2 4  x2 ð ðP dx þ Q dyÞ ¼ . . . . . . . . . . . . ; c2

Make any necessary substitutions and evaluate the line integral for c2 .

Vector analysis 2

987

82

4 Because we have ð ð  pffiffiffiffiffiffiffiffiffiffiffiffiffiffi  x  dx ðP dx þ Q dyÞ ¼ 4 þ ðx þ 2 4  x2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4  x2 c2 c2 ð  x2 4  2x  pffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ 4  x2 c2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Putting x ¼ 2 sin , 4  x2 ¼ 2 cos  dx ¼ 2 cos  d  Limits: x ¼ 2;  ¼ ; x ¼ 0;  ¼ 0. 2 ð ð0  4 sin2  ; 2 cos  d ðP dx þ Q dyÞ ¼ 4  4 sin   2 cos  =2 c2

  1 sin 2 0  ¼ 4 2 sin   sin2   2 2 =2 h   i  ¼4 ¼4  21 4 Finally x ¼ 0; dx ¼ 0

0 ð ð0 ; ðP dx þ Q dyÞ ¼ 2y dy ¼ y2 ¼ 4

(3) c3 :

2

c3

2

; Collecting our three partial results þ 8 16 ðP dx þ Q dyÞ ¼ þ   4  4 ¼   3 3 c That is one part done. Now we have to evaluate P ¼ x2 þ y2

;

ð1Þ  ðð  @Q @P  dx dy @y S @x

@P ¼ 2y @y

@Q ¼1 Q ¼ x þ 2y ;  @x ðð  ðð @Q @P ;  dx dy ¼ ð1  2yÞ dy dx @y S @x S It will be more convenient to work in polar coordinates, so we make the substitutions x ¼ r cos ; y ¼ r sin ; dS ¼ dx dy ¼ r dr d  ð =2 ð 2 ðð  @Q @P dx dy ¼  ð1  2r sin Þr dr d ; @y S @x 0 0 ¼ ............ Complete it

988

Programme 27

83



16 3

Here it is:  ðð  ð =2 ð 2 @Q @P  ðr  2r 2 sin Þ dr d dx dy ¼ @y S @x 0 0 2 ð =2 2 r 2r 3  sin  d ¼ 2 3 0 0 ð =2  16 ¼ sin  d 2 3 0

=2 16 16 cos  ¼ 2 þ ¼ 3 3 0

ð2Þ

So we have established once again that  ðð  þ @Q @P  dx dy ðP dx þ Q dyÞ ¼ @y c S @x And that brings us to the end of this particular Programme. We have covered a number of important sections, so check carefully down the Review summary and the Can you? checklist, and then work through the Test exercise that follows. The Further problems provide valuable additional practice.

Review summary 27 1 Line integrals

ð V dr

(a) Scalar field V: c

The curve c is expressed in parametric form. dr ¼ i dx þ j dy þ k dz ð (b) Vector field F: F  dr c

F ¼ Fx i þ Fy j þ Fz k dr ¼ i dx þ j dy þ k dz F  dr ¼ Fx dx þ Fy dy þ Fz dz 2 Volume integrals F is a vector field; V a closed region with boundary surface S. ð x2 ð y2 ð z2 ð F dV ¼ F dz dy dx V

x1

y1

z1

Vector analysis 2

989

3 Surface integrals (surface defined by ðx, y, zÞ ¼ constant) (a) Scalar field Vðx, y, zÞ: ð ð ^ dS; V dS ¼ V n S

^¼ n

S

r grad  ¼ jr j jgrad  j

(b) Vector field F ¼ Fx i þ Fy j þ Fz k: ð ð r ^ dS; ^¼ F  dS ¼ F  n n jr j S S 4 Polar coordinates (a) Plane polar coordinates ðr, Þ y dθ

dS

r O

x ¼ r cos ; y ¼ r sin  dS ¼ r dr d

y

θ x

x

(b) Cylindrical polar coordinates ð, , zÞ z P (ρ, ϕ, z)

x ¼  cos  y ¼  sin 

z O y

ϕ

ρ

z¼z dS ¼  d dz dV ¼  d d dz

x

(c) Spherical polar coordinates ðr, , Þ z

x ¼ r sin  cos  y ¼ r sin  sin 

P (r, θ, ϕ)

O

θ

z ¼ r cos 

r

ϕ

y

x

5 Conservative vector fields A vector field F is conservative if þ F  dr ¼ 0 for all closed curves (a) c

(b) curl F ¼ 0 (c) F ¼ grad V where V is a scalar.

dS ¼ r 2 sin  d d dV ¼ r 2 sin  dr d d

990

Programme 27

6 Divergence theorem (Gauss’ theorem) z

Closed surface S enclosing a region V in a vector field F. ð ð div F dV ¼ F  dS

S V

V

S

y

O x

7 Stokes’ theorem z

n c

dS c

An open surface S bounded by a simple closed curve c, then þ ð curl F  dS ¼ F  dr c

S

y

O x

8 Green’s theorem y

The curve c is a simple closed curve enclosing a plane space S in the x–y plane. P and Q are functions of both x and y.

c O

x

 þ ðð  @Q @P  dx dy ¼ ðP dx þ Q dyÞ. Then @y S @x c

Can you? Checklist 27 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can:

Frames

. Evaluate the line integral of a scalar and a vector field in Cartesian coordinates?

1

to

20

21

to

27

28

to

42

Yes

No

. Evaluate the volume integral of a vector field?

Yes

No

. Evaluate the surface integral of a scalar and a vector field?

Yes

No

Vector analysis 2

991

. Determine whether or not a vector field is a conservative vector field?

Yes

43

to

52

52

to

63

64

to

68

69

to

78

79

to

83

No

. Apply Gauss’ divergence theorem?

Yes

No

. Apply Stokes’ theorem?

Yes

No

. Determine the direction of unit normal vectors to a surface?

Yes

No

. Apply Green’s theorem in the plane?

Yes

No

Test exercise 27

ð

1 If V ¼ x3 y þ 2xy2 þ yz, evaluate

V dr between A ð0, 0, 0Þ and B ð2, 1,  3Þ c

along the curve with parametric equations x ¼ 2t, y ¼ t 2 , z ¼ 3t 3 . ð 2 If F ¼ x2 y3 i þ yz2 j þ zx2 k , evaluate F  dr along the curve x ¼ 3u2 , y ¼ u, c

z ¼ 2u3 between A ð3,  1,  2Þ and B ð3; 1; 2Þ. ð F dV where F ¼ 3i þ zj þ 2yk and V is the region bounded by the 3 Evaluate V

planes z ¼ 0, z ¼ 3 and the surface x2 þ y2 ¼ 4. ð 4 If V is the scalar field V ¼ xyz2 , evaluate V dS over the surface S defined by S

x2 þ y2 ¼ 9 between z ¼ 0 and z ¼ 2 in the first octant. ð 5 Evaluate F  dS over the surface S defined by x2 þ y2 þ z2 ¼ 4 for z 0 and S

bounded by x ¼ 0; y ¼ 0; z ¼ 0 in the first octant where F ¼ xi þ 2zj þ yk. 6 Determine which of the following vector fields are conservative. (a) F ¼ ð2xy þ zÞi þ ðx2 þ 2yzÞj þ ðx þ y2 Þk (b) F ¼ ðyz þ 2yÞi þ ðxz þ 2xÞj þ ðxy þ 3Þk (c) F ¼ ðyz2 þ 3Þi þ ðxz2 þ 2Þj þ ð2xyz þ 4Þk.

ð F  dS where

7 By the use of the divergence theorem, determine S

F ¼ xi þ xyj þ 2k, taken over the region bounded by the planes z ¼ 0, z ¼ 4, x ¼ 0, y ¼ 0 and the surface x2 þ y2 ¼ 9 in the first octant.

992

Programme 27

8 A surface consists of parts of the planes x ¼ 0, x ¼ 2, y ¼ 0, y ¼ 2 and ð curl F  dS z ¼ 3  y in the region z 0. Apply Stokes’ theorem to evaluate S

over the surface where F ¼ 2xi þ xzj þ yzk where S lies in the z ¼ 0 plane. 9 Verify Green’s theorem in the plane for the integral þ  2   xy  2x dx þ x þ 2xy2 dy c

where c is the square with vertices at ð1, 1Þ, ð1, 1Þ, ð1,  1Þ and ð1,  1Þ.

Further problems 27 ð 1 If V ¼ x2 yz, evaluate

V dr between A ð0, 0, 0Þ and B ð6, 2, 4Þ c

(a) along the straight lines c1 : ð0, 0, 0Þ to ð6, 0, 0Þ c2 : ð6, 0, 0Þ to ð6, 2, 0Þ c3 : ð6, 2, 0Þ to ð6, 2, 4Þ (b) along the path c4 having parametric equations x ¼ 3t, y ¼ t, z ¼ 2t. ð V dr along the curve c 2 If V ¼ xy2 þ yz, evaluate to one decimal place c

having parametric equations x ¼ 2t 2 , y ¼ 4t, z ¼ 3t þ 5 between A ð0, 0, 5Þ and B ð8, 8, 11Þ. ð 3 Evaluate to one decimal place the integral ðxyz þ 4x2 yÞ dr along the curve c c

with parametric equations x ¼ 2u, y ¼ u2 , z ¼ 3u3 between A ð2, 1, 3Þ and B ð4, 4, 24Þ. ð 4 If F ¼ xyi þ yzj þ 3xyzk , evaluate F  dr between A ð0; 2; 0Þ and B ð3; 6; 1Þ c

where c has the parametric equations x ¼ 3u, y ¼ 4u þ 2, z ¼ u2 . ð 5 F ¼ x2 i  2xyj þ yzk . Evaluate F  dr between A ð2, 1, 2Þ and B ð4, 4, 5Þ c

where c is the path with parametric equations x ¼ 2u, y ¼ u2 , z ¼ 3u  1: 6 A unit particle is moved in an anticlockwise manner round a circle with centre ð0, 0, 4Þ and radius 2 in the plane z ¼ 4 in a force field defined as F ¼ ðxy þ zÞi þ ð2x þ yÞj þ ðx þ y þ zÞk . Find the work done. ð F dV where F ¼ i  yj þ k and V is the region bounded by the 7 Evaluate V

plane z ¼ 0 and the hemisphere x2 þ y2 þ z2 ¼ 4; for z 0.

Vector analysis 2

993

8 V is the region bounded by the planes x ¼ 0, y ¼ 0, z ¼ 0 and the surfaces y ¼ 4  x2 ðz 0Þ and y ¼ 4  z2 ðy 0Þ. ð F dV throughout the region. If F ¼ 2i þ y2 j  k , evaluate V

ð

9 If F ¼ 3i þ 2j  2xk , evaluate

F dV where V is the region bounded by the V

planes y ¼ 0, z ¼ 0, z ¼ 4  y ðz 0Þ and the surface x2 þ y2 ¼ 16. 10 A scalar field V ¼ x þ y exists over a surface S defined by x2 þ y2 þ z2 ¼ 9, bounded by the planes x ¼ 0; y ¼ 0; z ¼ 0 in the first octant. Evaluate ð V dS over the curved surface. S

11 A surface S is defined by y2 þ z ¼ 4 and is bounded by the planes x ¼ 0, x ¼ 3, ð y ¼ 0, z ¼ 0 in the first octant. Evaluate V dS over this curved surface where S

V denotes the scalar field V ¼ x2 yz. ð 12 Evaluate curl F  dS over the surface S defined by 2x þ 2y þ z ¼ 2 and S

bounded by x ¼ 0, y ¼ 0, z ¼ 0 in the first octant and where F ¼ y2 i þ 2yzj þ xyk. ð 13 Evaluate F  dS over the hemisphere defined by x2 þ y2 þ z2 ¼ 25 with S z 0, where F ¼ ðx þ yÞi  2zj þ yk : 14 A vector field F ¼ 2xi þ zj þ yk exists over a surface S defined by x2 þ y2 ðþ z2 ¼ 16, bounded by the planes z ¼ 0, z ¼ 3, x ¼ 0, y ¼ 0. F  dS over the stated curved surface.

Evaluate S

ð

F  dS, where F is the vector field x2 i þ 2zj  yk , over the curved

15 Evaluate S

surface S defined by x2 þ y2 ¼ 25 and bounded by z ¼ 0, z ¼ 6, y 3. 16 A region V is defined by the quartersphere x2 þ y2 þ z2 ¼ 16, z 0, y 0 and the planes z ¼ 0, y ¼ 0. A vector field F ¼ xyi þ y2 j þ k exists throughout and on the boundary of the region. Verify the Gauss divergence theorem for the region stated. 17 A surface consists of parts of the planes x ¼ 0, x ¼ 1, y ¼ 0, y ¼ 2, z ¼ 1 in the first octant. If F ¼ yi þ x2 zj þ xyk , verify Stokes’ theorem. 18 S is the surface z ¼ x2 þ y2 bounded by the planes z ¼ 0 and z ¼ 4. Verify Stokes’ theorem for a vector field F ¼ xyi þ x3 j þ xzk .

994

Programme 27

19

A vector field F ¼ xyi þ z2 j þ xyzk exists over the surfaces x2 þ y2 þ z2 ¼ a2 , x ¼ 0 and y ¼ 0 in the first octant. Verify Stokes’ theorem that ð þ curl F  dS ¼ F  dr. c

S

20

A surface is defined by z2 ¼ 4ðx2 þ y2 Þ where 0  z  6. If a vector field F ¼ zi þ xy2 j þ x2 zk exists over the surface and on the boundary circle c, ð þ curl F  dS. show that F  dr ¼ c

21

S

Verify Green’s theorem in the plane for the integral þ   ðx  yÞ dx  ðy2 þ xyÞ dy c

where c is the circle with unit radius, centred on the origin.

Frames 1 to 36

Programme 28

Vector analysis 3 Learning outcomes When you have completed this Programme you will be able to: Derive the family of curves of constant coordinates for curvilinear coordinates Derive unit base vectors and scale factors in orthogonal curvilinear coordinates Obtain the element of arc ds and the element of volume dV in orthogonal curvilinear coordinates Obtain expressions for the operators grad, div and curl in orthogonal curvilinear coordinates

995

996

1

Programme 28

This short Programme is an extension of the two previous ones and may not be required for all courses. It can well be bypassed without adversely affecting the rest of the work.

Curvilinear coordinates Let us consider two variables u and v, each of which is a function of x and y i.e.

u=a

y

u ¼ f ðx, yÞ v ¼ gðx, yÞ

v=b If u and v are each assigned a constant value a and b, the equations will, in O general, define two intersecting curves. x If u and v are each given several such values, the equations define a network of curves covering the x–y plane. y

a1 a2

u = constant a3 a4 b4

O

x

b3 b2

b1

v = constant

y

u = ar

(x, y) (u, v)

y

v = br O

x

x

A pair of curves u ¼ ar and v ¼ br pass through each point in the plane. Hence, any point in the plane can be expressed in rectangular coordinates ðx, yÞ or in curvilinear coordinates ðu, vÞ. Let us see how this works out in an example, so move on

Vector analysis 3

997

2

Example 1 Let us consider the case where u ¼ xy and v ¼ x2  y. 4 (a) With u ¼ xy, if we put u ¼ 4, then y ¼ and we can plot y against x x to obtain the relevant curve. Similarly, putting u ¼ 8; 16; 32; . . . we can build up a family of curves, all of the pattern u ¼ xy. x y

u¼ 4 u¼ 8 u ¼ 16 u ¼ 32

0.5

1 .0

2.0

3.0

4 .0

8 16 32 64

4 8 16 32

2 4 8 16

1.33 2.67 5.33 10.67

1 .0 2 4 8

If we plot these on graph paper between x ¼ 0 and x ¼ 4 with a range of y from y ¼ 0 to y ¼ 20, we obtain ............

3

y

u = 32

u = 16 u=8 u=4 O

x

Note that each graph is labelled with its individual u-value. (b) With v ¼ x2  y, we proceed in just the same way. We rewrite the equation as y ¼ x2  v; assign values such as 8, 4, 0, 4, 8, 12, 16, . . . to v; and draw the relevant curve in each case. If we do that for x ¼ 0 to x ¼ 4 and limit the yvalues to the range y ¼ 0 to y ¼ 20, we obtain the family of curves ............

998

Programme 28

4

y

v = –16

v = –8

v = –4 v=0

v=4 v=8

O

x

The table of function values is as follows. x y

v¼ 8 v¼ 4 v¼ 0 v¼ 4 v¼ 8 v ¼ 12 v ¼ 16

0

1

2

3

4

8 4 0 4 8 12 16

7 3 1 5 9 13 17

4 0 4 8 12 16 20

1 5 9 13 17 21 25

8 12 16 20 24 28 32

Note again that we label each graph with its own v-value. This again is a family of curves with the common pattern v ¼ x2  y, the members being distinguished from each other by the value assigned to v in each case. Now we draw both sets of curves on a common set of x–y axes, taking and

the range of x from x ¼ 0 to x ¼ 4 the range of y from y ¼ 0 to y ¼ 20.

It is worthwhile taking a little time over it – and good practice! When you have the complete picture, move on to the next frame

Vector analysis 3

y

999

v = –8

v = –16

5

v = –4

v=0

v=4

v=8 u = 32

u = 16 u=8 u=4 O

x

The position of any point in the plane can now be stated in two ways. For example, the point P has Cartesian rectangular coordinates x ¼ 2, y ¼ 8. It can also be stated in curvilinear coordinates u ¼ 16, v ¼ 4, for it is at the point of intersection of the two curves corresponding to u ¼ 16 and v ¼ 4. Likewise, for the point Q, the position in rectangular coordinates is x ¼ 2:65, y ¼ 5:0 and for its position in curvilinear coordinates we must estimate it within the network. Approximate values are u ¼ 13, v ¼ 2. Similarly, the curvilinear coordinates of R ðx ¼ 1:8, y ¼ 14Þ are approximately u ¼ ............; u ¼ 26;

v ¼ ............ v ¼ 11

6

Their actual values are in fact u ¼ 25:2 and v ¼ 10:76. Now let us deal with another example.

7

Example 2 If u ¼ x2 þ 2y and v ¼ y  ðx þ 1Þ2 , these can be rewritten as y ¼ 12 ðu  x2 Þ and y ¼ v þ ðx þ 1Þ2 . We can now plot the family of curves, say between x ¼ 0 and x ¼ 4, with u ¼ 5ð5Þ30 and v ¼ 20ð5Þ5, i.e. values of u from 5 to 30 at intervals of 5 units and values of v from 20 to 5 at intervals of 5 units. The resulting network is easily obtained and appears as ............

1000

Programme 28

8

y

v = 5 v = 0 v = –5

v = –10 v = –15

u = 30 v = –20 u = 25 u = 20 u = 15 O

x u=5

u = 10

For P, the rectangular coordinates are ðx ¼ 2:18, y ¼ 5:1Þ and the curvilinear coordinates are ðu ¼ 15, v ¼ 5Þ. For Q, the rectangular coordinates are . . . . . . . . . . . . and the curvilinear coordinates are . . . . . . . . . . . .

9

Q: ðx ¼ 3:5, y ¼ 3:0Þ;

ðu ¼ 18:5, v ¼ 17Þ

Orthogonal curvilinear coordinates If the coordinate curves for u and v forming the network cross at right angles, the system of coordinates is said to be orthogonal. The test for orthogonality is given by the dot product of the vectors formed from the partial derivatives. This is, if @u @v @u @v þ ¼ 0 then u and v are orthogonal. @x @x @y @y Example 3 Given the curvilinear coordinates u and v where u ¼ xy and v ¼ x2  y2 then u and v form a coordinate system that is . . . . . . . . . . . .

Vector analysis 3

1001

10

orthogonal Because @u @u @v @v ¼ y and ¼ x, v ¼ x2  y2 so ¼ 2x and ¼ 2y. Then @x @y @x @y @u @v @u @v þ ¼ 2xy  2xy ¼ 0 and so u and v form a coordinate system that is @x @x @y @y

u ¼ xy so

orthogonal. Example 4 Given the curvilinear coordinates u and v where u ¼ x2 þ 2y and v ¼ y  ðx þ 1Þ2 then u and v form a coordinate system that is . . . . . . . . . . . .

11

not orthogonal Because u ¼ x2 þ 2y s o and

@v ¼ 1. @y

@u @u @v ¼ 2x a n d ¼ 2, v ¼ y  ðx þ 1Þ2 s o ¼ 2ðx þ 1Þ @x @y @x

Then @u @v @u @v þ ¼ 4xðx þ 1Þ þ 2 6¼ 0 and so u and v form a coordinate @x @x @y @y system that is not orthogonal. Let us extend these ideas to three dimensions. Move on

Orthogonal coordinate systems in space Any vector F can be expressed in terms of its components in three mutually perpendicular directions, which have normally been the directions of the coordinate axes, i.e.

12

z Fz k F k

F ¼ Fx i þ Fy j þ Fz k

Fx i

Fy j

O i

j

x

where i, j, k are the unit vectors parallel to the x, y, z axes respectively.

y

1002

Programme 28

Situations can arise, however, where the directions of the unit vectors do not remain fixed, but vary from point to point in space according to prescribed conditions. Examples of this occur in cylindrical and spherical polar coordinates, with which we are already familiar. 1 Cylindrical polar coordinates ð, , zÞ Let P be a point with cylindrical coordinates ð; ; zÞ as shown. The position of P is a function of the three variables ; ; z

z

r

z

O ϕ ρ

y

x z

(a) If  and z remain constant and  varies, then P will move out along AP by an @r amount and the unit vector I in this @ direction will be given by  @r @r I¼ @ @

r O ϕ ρ

I z y

x

(b) If, instead,  and z remain constant and  varies, P will move ............

13

round the circle with AP as radius @r is therefore a vector along the tangent to @ the circle at P and the unit vector J at P will be given by  @r @r J¼ @ @

z

r O ϕ ρ x

z y

Vector analysis 3

1003 (c) Finally, if  and 

z

r O ϕ

ρ

remain constant and z @r will be to the z-axis increases, the vector @z and the unit vector K in this direction will be given by  @r @r K¼ @z @z

z y

x

Putting our three unit vectors on to one diagram, we have ............

14

z

r O ϕ

ρ

I z y

x

Note that I, J, K are mutually perpendicular and form a right-handed set. But note also that, unlike the unit vectors i, j, k in the Cartesian system, the unit vectors I, J, K, or base vectors as they are called, are not fixed in directions, but change as the position of P changes. So we have, for cylindrical polar coordinates  @r @r I¼ @ @  @r @r J¼ @ @  @r @r K¼ @z @z If F is a vector associated with P, then F(r) ¼ F I þ F J þ Fz K where F , F , Fz are the components of F in the directions of the unit base vectors I, J, K. Now let us attend to spherical coordinates in the same way.

1004

Programme 28

2 Spherical polar coordinates ðr, , Þ

15

z

P is a function of the three variables r, , .

θ r O ϕ

y

x z I q O f

r y

x

z

θ r O y

ϕ x

(a) If  and  remain constant and r increases, P moves outwards in the @r is thus a vector normal direction OP. @r to the surface of the sphere at P and the unit vector I in that direction is therefore  @r @r I¼ @r @r (b) If r and  remain constant and  increases, P will move along the @r is a tangent ‘meridian’ through P, i.e. @ vector to this circle at P and the unit vector J is given by  @r @r J¼ @ @

(c) If r and  remain constant and  increases, P will move ............

16

along the circle through P perpendicular to the z-axis z

θ r O ϕ

y

@r is therefore a tangent vector at P and the @ unit vector K in this direction is given by  @r @r K¼ @ @

x

So, putting the three results on one diagram, we have . . . . . . . . . . . .

Vector analysis 3

1005

z

O ϕ

17

I

θ r y

x

Once again, the three unit vectors at P (base vectors) are mutually perpendicular and form a right-handed set. Their directions in space, however, change as the position of P changes. A vector F associated with P can therefore be expressed as F ¼ Fr I þ F J þ F K where Fr , F , F are the components of F in the directions of the base vectors I, J, K. Both cylindrical and spherical polar coordinate systems are ............

18

orthogonal

Scale factors Collecting the recent results together, we have: 1 For cylindrical polar coordinates, the unit base vectors are  @r @r 1 @r @r where h ¼ ¼ I¼ @  @ h @ @ @r @r 1 @r @r where h ¼ J¼ ¼ @  @ h @ @ @r @r 1 @r @r where hz ¼ K¼ ¼ @z @z hz @z @z 2 For spherical polar coordinates,  @r @r 1 @r I¼ ¼ @r  @r hr @r @r @r 1 @r J¼ ¼ @  @ h @ @r @r 1 @r K¼ ¼ @ @ h @

the unit base vectors are @r @r @r where h ¼ @ @r where h ¼ @ where hr ¼

In each case, h is called the scale factor. Move on

1006

19

Programme 28

Scale factors for coordinate systems 1 Rectangular coordinates ðx, y, zÞ With rectangular coordinates, hx ¼ hy ¼ hz ¼ 1. 2 Cylindrical coordinates ð, , zÞ z

P (ρ, ϕ, z)

r ¼ xi þ yj þ zk

r

x ¼  cos  y ¼  sin 

z O x x

ϕ

z¼z

y

ρ

y

; r ¼  cos  i þ  sin  j þ z k  @r @r 1 @r @r h ¼ ¼ ¼ j cos  i þ sin  j j I¼ @ @ h @ @ ¼ ðcos2  þ sin2 Þ1=2 ¼ 1 J¼

@r @



; h ¼ 1 @r 1 @r ¼ @ h @

h ¼

@r ¼ j  sin  i þ  cos  j j @

¼ ð2 sin2  þ 2 cos2 Þ1=2 ¼  K¼

@r @z



; h ¼  @r 1 @r ¼ @z hz @z

@r ¼j k j¼ 1 @z ; hz ¼ 1 hz

¼

; h ¼ 1; h ¼ ; hz ¼ 1 3 Spherical coordinates ðr, , Þ z

P (r, θ, ϕ)

r ¼ xi þ yj þ zk x ¼ r sin  cos 

r O x x

θ

z y

ϕ

y ¼ r sin  sin  z ¼ r cos 

y

; r ¼ r sin  cos  i þ r sin  sin  j þ r cos  k Then working as before hr ¼ . . . . . . . . . . . . ;

h ¼ . . . . . . . . . . . . ;

h ¼ . . . . . . . . . . . .

Vector analysis 3

1007

hr ¼ 1;

h ¼ r;

h ¼ r sin 

Because r ¼ r sin  cos  i þ r sin  sin  j þ r cos  k I¼

@r @r



@r 1 @r ¼ @r hr @r

@r ¼ j sin  cos  i þ sin  sin  j þ cos  k j @r ¼ ðsin2  cos2  þ sin2  sin2  þ cos2 Þ1=2

hr ¼

¼ ðsin2  þ cos2 Þ1=2 ¼ 1 ; hr ¼ 1 J¼

@r @



h ¼

@r 1 @r ¼ @ h @

@r ¼ j r cos  cos  i þ r cos  sin  j  r sin  k j @ ¼ ðr 2 cos2  cos2  þ r 2 cos2  sin2  þ r 2 sin2 Þ1=2

¼ ðr 2 cos2  þ r 2 sin2 Þ1=2 ¼ r ; h ¼ r K¼

@r @

h ¼



@r 1 @r ¼ @ h @

@r ¼ j r sin  sin  i þ r sin  cos  j j @ ¼ ðr 2 sin2  sin2  þ r 2 sin2  cos2 Þ1=2 ¼ ðr 2 sin2 Þ1=2 ¼ r sin 

; h ¼ r sin  ; hr ¼ 1; So:

h ¼ r;

h ¼ r sin 

(a) for cylindrical coordinates I¼

@r ; @

J¼

1 @r ;  @

K¼

@r @z

K¼

1 @r r sin  @

(b) for spherical coordinates I¼

@r ; @r

J¼

1 @r ; r @

20

1008

Programme 28

General curvilinear coordinate system (u, v, w) 21

Any system of coordinates can be treated in like manner to obtain expressions for the appropriate unit vectors I, J, K.    @r @r @r @r @r @r I¼ ; J¼ ; K¼ @u @u @v @v @w @w These unit vectors are not always at right angles to each other. If they are mutually perpendicular, the coordinate system is ............

22

orthogonal Unit vectors I, J, K are orthogonal if IJ¼JK¼KI¼0 Exercise Determine the unit base vectors in the directions of the following vectors and determine whether the vectors are orthogonal. 1

i  2j þ 4k

2

i þ 2j þ 2k  10i  2j þ 7k

2i þ 3j þ k  2i þ j þ k

23

2i  3j þ 2k

3 4i þ 2j  k

4 3i þ 2j þ k

3i  5j þ 2k i þ 2j þ 6k

i  3j þ 3k 6i þ j  k

The results are as follows: 1 1 I ¼ pffiffiffiffiffiffi ði  2j þ 4kÞ; 21 1 K ¼ pffiffiffi ð2i þ j þ kÞ 6 I  J ¼ 0;

J  K ¼ 0;

1 J ¼ pffiffiffiffiffiffi ð2i þ 3j þ kÞ; 14

KI¼0

; orthogonal

1 1 2 I ¼ pffiffiffiffiffiffi ð2i  3j þ 2kÞ; J ¼ ði þ 2j þ 2kÞ; 3 17 1 K ¼ pffiffiffiffiffiffiffiffiffi ð10i þ 2j þ 7kÞ 153 I  J ¼ 0;

J  K ¼ 0;

KI¼0

; orthogonal

Vector analysis 3

1009

1 1 3 I ¼ pffiffiffiffiffiffi ð4i þ 2j  kÞ; J ¼ pffiffiffiffiffiffi ð3i  5j þ 2kÞ; 38 21 1 K ¼ pffiffiffiffiffiffi ði þ 2j þ 6kÞ 41 I  J ¼ 0;

J  K 6¼ 0

1 4 I ¼ pffiffiffiffiffiffi ð3i þ 2j þ kÞ; 14 1 K ¼ pffiffiffiffiffiffi ð6i þ j  kÞ 38 I  J ¼ 0;

J  K ¼ 0;

; not orthogonal 1 J ¼ pffiffiffiffiffiffi ði  3j þ 3kÞ; 19

K  I 6¼ 0 ; not orthogonal

Transformation equations 24

In general coordinates, the transformation equations are of the form x ¼ f ðu, v, wÞ;

y ¼ gðu, v, wÞ;

z ¼ hðu, v, wÞ

where the functions f, g, h are continuous and single-valued, and whose partial derivatives are continuous. Then r ¼ xi þ yj þ zk ¼ f ðu, v, wÞi þ gðu, v, wÞj þ hðu, v, wÞk and coordinate curves can be formed by keeping two of the three variables constant. Now r ¼ xi þ yj þ zk

; dr ¼

@r @r @r du þ dv þ dw @u @v @w

@r is a tangent vector to the u-coordinate curve at P @u @r is a tangent vector to the v-coordinate curve at P @v @r is a tangent vector to the w-coordinate curve at P @w  @r @r @r @r ¼ hu I where hu ¼ I¼ ; @u @u @u @u  @r @r @r @r ¼ hv J where hv ¼ J¼ ; @v @v @v @v  @r @r @r @r ¼ hw K where hw ¼ K¼ ; @w @w @w @w Then (1) above becomes dr ¼ hu du I þ hv dv J þ hw dw K where, as before, hu ; hv ; hw are the scale factors.

ð1Þ

1010

Programme 28

Element of arc ds and element of volume dV in orthogonal curvilinear coordinates 25

w

hw dw K

r

v

dr hv dvJ hu duI

u

O

(a) Element of arc ds Element of arc ds from P to Q is given by dr ¼ hu du I þ hv dv J þ hw dw K ; dr  dr ¼ ðhu du I þ hv dv J þ hw dw KÞ  ðhu du I þ hv dv J þ hw dw KÞ ; ds2 ¼ h2u du2 þ h2v dv2 þ h2w dw2 ; ds ¼ ðh2u du2 þ h2v dv2 þ h2w dw2 Þ1=2 (b) Element of volume dV dV ¼ ðhu du IÞ  ðhv dv J  hw dw KÞ ¼ ðhu du IÞ  ðhv dv hw dw IÞ ¼ hu du hv dv hw dw ; dV ¼ hu hv hw dudvdw Note also that   @r @r @r   du dv dw dV ¼ @u @v @w @ðx, y, zÞ du dv dw ¼ @ðu, v, wÞ where

@ðx; y; zÞ is the Jacobian of the transformation. @ðu; v; wÞ

Vector analysis 3

1011

grad, div and curl in orthogonal curvilinear coordinates (a) grad V ðrV Þ

26 w r dw w

v O

r dv v

r du u

I

u

Let a scalar field V exist in space and let dV be the change in V from P to Q. If the position vector of P is r then that of Q is r þ dr. Then dV ¼

@V @V @V du þ dv þ dw @u @v @w

Let grad V ¼ rV ¼ ðrV Þu I þ ðrV Þv J þ ðrV Þw K where ðrV Þu; v; w are the components of grad V in the u, v, w directions. @r @r @r du þ dv þ dw @u @v @w @r @r @r @r ¼ ¼ I ¼ hu I; J ¼ hv J; @u @u @v @v @r @r ¼ K ¼ hw K. @w @w

Also dr ¼ But and

; dr ¼ hu du I þ hv dv J þ hw dw K We have previously established that dV ¼ grad V  dr ; dV ¼ fðrV Þu I þ ðrV Þv J þ ðrV Þw Kg  fhu duI þ hv dvJ þ hw dwKg ¼ ðrV Þu hu du þ ðrV Þv hv dv þ ðrV Þw hw dw But dV ¼

@V @V @V du þ dv þ dw @u @v @w

1012

Programme 28

; Equating coefficients, we then have @V ¼ ðrV Þu hu @u

; ðrV Þu ¼

1 @V hu @u

@V ¼ ðrV Þv hv @v

; ðrV Þv ¼

1 @V hv @v

@V ¼ ðrV Þw hw @w

; ðrV Þw ¼

1 @V hw @w

; grad V ¼ rV ¼

1 @V 1 @V 1 @V Iþ Jþ K hu @u hv @v hw @w

i.e. grad operator r ¼

27

I @ J @ K @ þ þ hu @u hv @v hw @w

Other results we state without proof. (b) div F

ðr  FÞ

div F ¼ r  F

 1 @ @ @ ðhv hw Fu Þ þ ðhu hw Fv Þ þ ðhu hv Fw Þ ¼ hu hv hw @u @v @w

Example 1 Show that the curvilinear expression for div F agrees with the earlier definition in Cartesian coordinates. In Cartesian coordinates x, y, z we have hx ¼ hy ¼ hz ¼ . . . . . . . . . . . . so that div F ¼ . . . . . . . . . . . .

28

hx ¼ hy ¼ hz ¼ 1 so that @Fx @Fy @Fz þ þ div F ¼ @x @y @z (c) curl F ðr  FÞ hu I 1 @ curl F ¼ r  F ¼ hu hv hw @u hu Fu

hv J @ @v hv Fv

 hw K  @  @w  hw Fw 

Example 2 Show that the curvilinear expression for curl F agrees with the earlier definition in Cartesian coordinates. In Cartesian coordinates x, y, z we have hx ¼ hy ¼ hz ¼ . . . . . . . . . . . . and I, K ¼ . . . , . . . , . . . so that curl F ¼ . . . . . . . . . . . .

J,

Vector analysis 3

1013

hx ¼ hy ¼ hz ¼ 1 and I, J, K ¼ i, j, k so that       @Fy @Fx @Fz @Fy @Fx @Fz þj þk    curl F ¼ i @y @z @z @x @x @y

29

Because in Cartesians hx ¼ hy ¼ hz ¼ 1 and I, j, K ¼ i, j, k so that  hu I hv J hw K   1 @ @ @  rF ¼  hu hv hw @u @v @w  hu Fu hv Fv hw Fw   i j k  @ @ @  ¼  @x @y @z   F F F  x

y

z

      @Fy @Fx @Fz @Fy @Fx @Fz ¼i þj þk    @y @z @z @x @x @y

(d) Div grad V

ðr2 V Þ

div grad V ¼ r  ðrV Þ ¼ r2 V        1 @ hv hw @V @ hu hw @V @ hu hv @V    ¼ þ þ hu hv hw @u hu @u @v hv @v @w hw @w Example 3 Show that the curvilinear expression for r2 V agrees with the earlier definition in Cartesian coordinates. In Cartesian coordinates x, y, z we have hx ¼ hy ¼ hz ¼ . . . . . . . . . so that r2 V ¼ . . . . . . . . . . . . hx ¼ hy ¼ hz ¼ 1 so that r2 V ¼

@ V @ V @ V þ 2 þ 2 @x2 @y @z 2

2

2

Let’s try another example, this time in coordinates other than Cartesians. Example 4 If Vðu; v; wÞ ¼ u þ v2 þ w3 with scale factors hu ¼ 2, hv ¼ 1, hw ¼ 1, find r2 V at the point ð5, 3, 4Þ. There is very little to it. All we have to do is to determine the various partial derivatives and substitute in the expression above with relevant values. div grad V ¼ . . . . . . . . . . . .

30

1014

Programme 28

31

26 Because

       1 @ hv hw @V @ hu hw @V @ hu hv @V    r V¼ þ þ hu hv hw @u @u @v @v @w hw @w hu hv 2

In this case, V ¼ u þ v2 þ w3 ;

@V ¼ 1; @u

Also hu ¼ 2, hv ¼ 1, hw ¼ 1    1 @ 1 @ @ ð6w2 Þ þ ð4vÞ þ ; r2 V ¼ 2 @u 2 @v @w

@V ¼ 2v; @v

@V ¼ 3w2 @w

¼ 12 f0 þ 4 þ 12wg ; At w ¼ 4, r2 V ¼ 26 That is all there is to it. Here is another. Example 5 If V ¼ ðu2 þ v2 Þw3 with hu ¼ 3, hv ¼ 1, hw ¼ 2, find div grad V at the point ð2,  2, 1Þ. r2 V ¼ . . . . . . . . . . . .

32

14 29 Because V ¼ ðu2 þ v2 Þw3

;

@V @V @V ¼ 2uw3 ; ¼ 2vw3 ; ¼ 3ðu2 þ v2 Þw2 @u @v @w

hu ¼ 3, hv ¼ 1, hw ¼ 2        1 @ 2 @V @ @V @ 3 @V ; r2 V ¼ þ þ 6 6 @u 3 @u @v @v @w 2 @w      1 @ 4 3 @  @ 9 2 3 2 2 uw þ 12vw þ ðu þ v Þw ¼ 6 @u 3 @v @w 2

also

; at ð2,  2, 1Þ r2 V ¼ 16 fð43 w3 Þ þ ð12w3 Þ þ 9ðu2 þ v2 Þwg 256 ¼ 14 29 ¼ 16 f43 þ 12 þ 72g ¼ 18

Particular orthogonal systems We can apply the general results for div, grad and curl to special coordinate systems by inserting the appropriate scale factors – as we shall now see.

Vector analysis 3

1015

33

(a) Cartesian rectangular coordinate system If we replace u, v, w by x, y, z and insert values of hx ¼ hy ¼ hz ¼ 1, we obtain expressions for grad, div and curl in rectangular coordinates, so that grad V ¼ . . . . . . . . . . . . ; div F ¼ . . . . . . . . . . . . ; curl F ¼ . . . . . . . . . . . . @V @V @V iþ jþ k @x @y @z @Fx @Fy @Fz div F ¼ þ þ @x @y @z i j k @ @ @ curl F ¼ @x @y @z Fx Fy Fz

grad V ¼

r2 V ¼

34

@2V @2V @2V þ 2 þ 2 @x2 @y @z

all of which you will surely recognise. (b) Cylindrical polar coordinate system Here we simply replace u, v, w with , , z and insert hu ¼ h ¼ 1, hv ¼ h ¼ , hw ¼ hz ¼ 1 giving grad V ¼ . . . . . . . . . . . . ; curl F ¼ . . . . . . . . . . . .

div F ¼ . . . . . . . . . . . . ;

@V 1 @V @V Iþ Jþ K @  @ @z  1 @ @ @ ðF Þ þ ðF Þ þ ðFz Þ div F ¼  @ @ @z I J K @ @ 1 @ curl F ¼  @ @ @z F F Fz

grad V ¼

r2 V ¼

@ 2 V 1 @V 1 @ 2 V @ 2 V þ þ þ 2 @2  @ 2 @2 @z

(c) Spherical polar coordinate system Replacing u, v, w with r, ,  with hr ¼ 1, h ¼ r, h ¼ r sin , grad V ¼ . . . . . . . . . . . . ; curl F ¼ . . . . . . . . . . . .

div F ¼ . . . . . . . . . . . . ;

35

1016

36

Programme 28

@V 1 @V 1 @V Iþ Jþ K @r r @ r sin  @  1 @ 2 @ @ ðr sin  Fr Þ þ ðr sin  F Þ þ ðrF Þ div F ¼ 2 r sin  @r @ @ I rJ r sin  K @ @ @ 1 curl F ¼ 2 @ r sin  @r @ Fr rF r sin  F

grad V ¼

r2 V ¼

@ 2 V 2 @V 1 @ 2 V cot  @V 1 @2V þ þ þ þ @r 2 r @r r 2 @2 r 2 @ r 2 sin  @2

The results we have compiled are sometimes written in slightly different forms, but they are, of course, equivalent. That brings us to the end of this Programme which is designed as an introduction to the topic of curvilinear coordinates. It has considerable applications, but these are beyond the scope of this present course of study. The Review summary follows as usual. Make any further notes as necessary: then you can work through the Can you? checklist and the Test exercise without difficulty. The Programme ends with the usual Further problems.

Review summary 28 1 Curvilinear coordinates in two dimensions u ¼ f ðx; yÞ;

v ¼ gðx; yÞ

2 Orthogonal coordinate system in space (a) Cartesian rectangular coordinates ðx, y, zÞ F ¼ Fx i þ Fy j þ Fz k Scale factors hx ¼ hy ¼ hz ¼ 1 (b) Cylindrical polar coordinates ð, , zÞ r ¼  cos  i þ  sin  j þ z k Base unit vectors:  @r @r I¼ @ @  @r @r J¼ @ @  @r @r K¼ @z @z F ¼ F I þ F J þ Fz K

Scale factors: @r h ¼ ¼1 @ h ¼

@r ¼ @

hz ¼

@r ¼1 @z

Vector analysis 3

1017

(c) Spherical polar coordinates ðr, , Þ r ¼ r sin  cos  i þ r sin  sin  j þ r cos  k Base unit vectors:  @r @r I¼ @r @r  @r @r J¼ @ @  @r @r K¼ @ @

Scale factors: @r hr ¼ ¼1 @r h ¼

@r ¼r @

h ¼

@r ¼ r sin  @

F ¼ Fr I þ F J þ F K 3 General orthogonal curvilinear coordinates ðu, v, wÞ x ¼ f ðu, v, wÞ;

y ¼ gðu, v, wÞ;

w ¼ hðu, v, wÞ

r ¼ xi þ yj þ zk @r ¼ hu I where @u @r ¼ hv J where @v @r ¼ hw K where @w Element of arc:

@r @u @r hv ¼ @v @r hw ¼ @w hu ¼

ds ¼ ðh2u du2 þ h2v dv2 þ h2w dw2 Þ1=2

Element of volume:

dV ¼ hu hv hw du dv dw ¼

@ðx, y, zÞ du dv dw @ðu, v, wÞ

4 grad, div and curl in orthogonal curvilinear coordinates (a) grad V ¼ rV ¼

1 @V 1 @V 1 @V Iþ Jþ K hu @u hv @v hw @w

grad operator ¼ r ¼

I @ J @ K @ þ þ hu @u hv @v hw @w

 1 @ @ @ (b) div F ¼ ðhv hw Fu Þ þ ðhw hu Fv Þ þ ðhu hv Fw Þ hu hv hw @u @v @w

(c) curl F ¼

hu I

hv J

hw K

1 @ hu hv hw @u h u Fu

@ @v h v Fv

@ @w hw Fw

(d) div grad V ¼ r  rV ¼ r2 V        1 @ hv hw @V @ hu hw @V @ hu hv @V ¼ þ þ    hu hv hw @u hu @u @v hv @v @w hw @w

1018

Programme 28

5 grad, div and curl in cylindrical and spherical coordinates (a) Cylindrical coordinates ð, , zÞ @V 1 @V @V Iþ Jþ K @  @ @z   1 @ðF Þ 1 @F @Fz div F ¼ þ þ @   @ @z

grad V ¼

J

K

1 @  @

@ @

@ @z

F

F

Fz

I curl F ¼

r2 V ¼

@ 2 V 1 @V 1 @ 2 V @ 2 V þ þ þ 2 @2  @ 2 @2 @z

(b) Spherical coordinates ðr, , Þ @V 1 @V 1 @V Iþ Jþ K @r r @ r sin  @ 1 @ 1 @ 1 @ div F ¼ 2 ðr 2 Fr Þ þ ðsin  F Þ þ ðF Þ r @r r sin  @ r sin  @ I rJ r sin  K @ @ @ 1 curl F ¼ 2 @ r sin  @r @ Fr rF r sin  F grad V ¼

r2 V ¼

@ 2 V 2 @V 1 @ 2 V cot  @V 1 @2V þ 2 2 þ 2 þ þ 2 2 2 @r r @r r @ r @ r sin  @2

Can you? Checklist 28 Check this list before and after you try the end of Programme test On a scale of 1 to 5 how confident are you that you can:

Frames

. Derive the family of curves of constant coordinates for curvilinear coordinates?

1

to

11

12

to

24

Yes

No

. Derive unit base vectors and scale factors in orthogonal curvilinear coordinates?

Yes

No

Vector analysis 3

1019

. Obtain the element of arc ds and the element of volume dV in orthogonal curvilinear coordinates?

Yes

25

No

. Obtain expressions for the operators grad, div and curl in orthogonal curvilinear coordinates?

Yes

26

to

36

No

Test exercise 28 1 Determine the unit vectors in the directions of the following three vectors and test whether they form an orthogonal set. 3i  2j þ k i þ 2j þ k 2i  j þ 4k. 2 If r ¼ u sin 2 i þ u cos 2 j þ v2 k, determine the scale factors hu , hv , h . 3 If P is a point r ¼  cos  i þ  sin  j þ z k and a scalar field V ¼ 2 z sin 2  exists in space, using cylindrical polar coordinates ð, , zÞ determine grad V at the point at which  ¼ 1,  ¼ =4, z ¼ 2. 4 A vector field F is given in cylindrical coordinates by F ¼  cos  I þ  sin 2  J þ z K Determine

(a) div F;

(b) curl F.

5 Using spherical coordinates ðr, , Þ determine expressions for (a) an element of arc ds; (b) an element of volume dV. 6 If V is a scalar field such that V ¼ u2 vw3 and scale factors are hu ¼ 1, hv ¼ 2, hw ¼ 4, determine r2 V at the point ð2, 3,  1Þ.

Further problems 28 1 Determine whether the following sets of three vectors are orthogonal. (a) 4i  2j  k (b) 2i þ 3j  k 3i þ 5j þ 2k 4i  2j þ 2k i  11j þ 26k i þ 4j þ 2k 2 If Vðu; v; wÞ ¼ v3 w2 sin 2u with scale factors hu ¼ 3, hv ¼ 1, hw ¼ 2, determine div grad V at the point ð=4,  1, 3Þ. u2 e2w exists in space. If the relevant scale factors are hu ¼ 2, v hv ¼ 3, hw ¼ 1, determine the value of r2 V at the point ð1, 2, 0Þ.

3 A scalar field V ¼

1020

Programme 28

4 If r ¼ x i þ y j þ z k and x ¼ r sin  cos , y ¼ r sin  sin , z ¼ r cos  in spherical polar coordinates ðr, , Þ, prove that, for any vector field F where F ¼ Fx i þ Fy j þ Fz k ¼ Fr I þ F J þ F K then Fx ¼ Fr sin  cos  þ F cos  cos   F sin  Fy ¼ Fr sin  sin  þ F cos  sin  þ F cos  Fz ¼ Fr cos   F sin . 5 If V is a scalar field, determine an expression for r2 V (a) in cylindrical polar coordinates (b) in spherical polar coordinates. 6 Transformation equations from rectangular coordinates ðx, y, zÞ to parabolic cylindrical coordinates ðu, v, wÞ are u 2  v2 ; y ¼ uv; z ¼ w x¼ 2 V is a scalar field and F a vector field. (a) Prove that the ðu, v, wÞ system is orthogonal (b) Determine the scale factors (c) Find div F (d) Obtain an expression for r2 V.

Frames 1 to 66

Programme 29

Complex analysis 1 Learning outcomes When you have completed this Programme you will be able to: Recognize the transformation equation in the form w ¼ f ðzÞ ¼ uðx; yÞ þ jvðx; yÞ Illustrate the image of a point in the complex z-plane under a complex mapping onto the w-plane Map a straight line in the z-plane onto the w-plane under the transformation w ¼ f ðzÞ Identify complex mappings that form translations, magnifications, rotations and their combinations Deal with the nonlinear transformations w ¼ z2 , w ¼ 1=z, w ¼ 1=ðz  aÞ and w ¼ ðaz þ bÞ=ðcz þ dÞ

Prerequisite: Engineering Mathematics (Eighth Edition) Programmes 1 Complex numbers 1, 2 Complex numbers 2 and 3 Hyperbolic functions 1021

1022

Programme 29

1

The foundations of complex numbers and their application to hyperbolic functions were treated fully in Programmes 1, 2 and 3 of Engineering Mathematics, Eighth Edition, and these provide valuable revision should you feel it to be necessary before embarking on the new work. It will be assumed that you are already familiar with the material covered in those previous Programmes and it would be a wise move to work through the relevant Test exercises to refresh your memory on this all-important part of the course.

Functions of a complex variable For a function of a single real variable f ðxÞ we can construct the graph of the function by plotting points against two mutually perpendicular Cartesian axes, the x-axis and the f ðxÞ-axis. For a function of a single complex variable w ¼ f ðzÞ ¼ uðx, yÞ þ jvðx, yÞ we have four real variables, x, y, u and v. For example if z ¼ x þ jy and f ðzÞ ¼ z2 then f ðzÞ ¼ ðx þ jyÞ2 ¼ x2 þ 2jxy þ ðjyÞ2 ¼ x2  y2 þ 2jxy and so and

uðx, yÞ ¼ x2  y2 vðx, yÞ ¼ 2xy

We cannot plot the graph of the function f ðzÞ against a single set of axes because to do so we would be required to draw four mutually perpendicular axes which is not possible. Instead, we resort to plotting z-values against x- and y-axes in the complex z-plane and to plotting the corresponding values of w ¼ f ðzÞ against uand v-axes in the complex w-plane. Accordingly, values of z are plotted on one plane and the corresponding values of f ðzÞ are plotted on another plane. So in our example above for a particular value of z, for example, z ¼ 4 þ j3 u ¼ ............ v ¼ ............

Complex analysis 1

1023

u¼7

2

v ¼ 24

Because with z ¼ 4 þ j3, x ¼ 4 and y ¼ 3. Then u ¼ 16  9 ¼ 7 and v ¼ 24. y

v

z-plane

w-plane

P' (w = 7 + j24)

P (z = 4 + j3)

w = f (z) O

x

O

u

Therefore, z (where z ¼ x þ jy) and w (where w ¼ u þ jv) are two complex variables related by the equation w ¼ f ðzÞ. Any other point in the z-plane will similarly be transformed into a corresponding point in the w-plane, the resulting position P 0 depending on (a) the initial position of P (b) the relationship w ¼ f ðzÞ, called the transformation equation or transformation function.

Complex mapping The transformation of P in the z-plane onto P 0 in the w-plane is said to be a mapping of P onto P 0 under the transformation w ¼ f ðzÞ and P 0 is sometimes referred to as the image of P. Example 1 Determine the image of the point P, z ¼ 3 þ j2, on the w-plane under the transformation w ¼ 3z þ 2  j. w ¼ u þ jv ¼ f ðzÞ ¼ 3z þ 2  j ¼ 3ðx þ jyÞ þ 2  j so that, for this example, u ¼ ............; u ¼ 3x þ 2;

v ¼ ............ v ¼ 3y  1

Then the point P ðz ¼ 3 þ j2Þ transforms onto . . . . . . . . . . . .

3

1024

Programme 29

4

w ¼ 11 þ j5 Because z ¼ 3 þ j2 ; x ¼ 3; y ¼ 2 u ¼ 3x þ 2 ¼ 11; v ¼ 3y  1 ¼ 5;

; w ¼ 11 þ j5

We can illustrate the transformation thus: y

w-plane

v

z-plane

P' (w = 11 + j5) P (z = 3 + j2)

w = f (z) O

O

x

u

w = f (z) = 3z + 2 – j

Here is another. Example 2 Map the points A ðz ¼ 2 þ jÞ and B ðz ¼ 3 þ j4Þ onto the w-plane under the transformation w ¼ j2z þ 3 and illustrate the transformation on a diagram. This is no different from the previous example. Complete the job and check with the next frame.

5

A0 ðw ¼ 1  j4Þ;

B0 ðw ¼ 5 þ j6Þ v

z-plane

y

6

4 w-plane

w = f (z) 1 –2

O

3

x

–5 w = f (z)

O 1

u

–4

Because w ¼ f ðzÞ ¼ j2z þ 3 ¼ j2ðx þ jyÞ þ 3 ¼ ð3  2yÞ þ j2x w ¼ u þ jv ; u ¼ 3  2y; v ¼ 2x A: x ¼ 2, y ¼ 1

; A0 : u ¼ 3  2 ¼ 1; v ¼ 4

; A0 : w ¼ 1  j4

B: x ¼ 3; y ¼ 4

; B0 : u ¼ 3  8 ¼ 5; v ¼ 6

; B0 : w ¼ 5 þ j6

There now follows a short practice exercise. Work all four of the items before you check the results. There is no need to illustrate the transformation in each case. So move on

Complex analysis 1

1025

Exercise

6

Map the following points in the z-plane onto the w-plane under the transformation w ¼ f ðzÞ stated in each case. 1 z ¼ 4  j2

under w ¼ j3z þ j2

2 z ¼ 2  j under w ¼ jz þ 3 3 z ¼ 3 þ j2

under w ¼ ð1 þ jÞz  2

4 z¼2þj

under w ¼ z2 . 1 w ¼ 6 þ j14

2 w ¼ 4  j2

3 w ¼ 1 þ j5

4 w ¼ 3 þ j4

7

That was easy enough. Now let us extend the ideas.

Mapping of a straight line in the z-plane onto the w-plane under the transformation w = f ( z ) A typical example will show the method. Example 1 To map the straight line joining A ð2 þ jÞ and B ð3 þ j6Þ in the z-plane onto the w-plane when w ¼ 3 þ j2z. y

We first of all map the end points A and B onto the w-plane to obtain A0 and B0 as in the previous cases.

6

A0 : w ¼ . . . . . . . . . . . . B0 : w ¼ . . . . . . . . . . . .

1 –2

A0 : w ¼ 1  j4;

O

B0 : w ¼ 9 þ j6

Because ð1Þ A: z ¼ 2 þ j

w ¼ 3 þ j2z

0

; A : w ¼ 3 þ j2ð2 þ jÞ ¼ 3  j4  2 ¼ 1  j4 ð2Þ B: z ¼ 3 þ j6 ; B0 : w ¼ 3 þ j2ð3 þ j6Þ ¼ 3 þ j6  12 ¼ 9 þ j6 Then, if we illustrate the transformations on a diagram, as before, we get ............

3

x

8

1026

Programme 29

9

w-plane

z-plane

y

v 6

6 w = f(z) 1 –2

O

x

3

1 u

O

–9

–4

w = f (z)

As z moves along the line A to B in the z-plane, we cannot assume that its image in the w-plane travels along a straight line from A0 to B0 . As yet, we have no evidence of what the path is. We therefore have to find a general point w ¼ u þ jv in the w-plane corresponding to a general point z ¼ x þ jy in the z-plane. w ¼ u þ jv ¼ f ðzÞ ¼ 3 þ j2z ¼ ..................

10

w ¼ u þ jv ¼ ð3  2yÞ þ j2x Because w ¼ 3 þ j2ðx þ jyÞ ¼ 3 þ j2x  2y ¼ ð3  2yÞ þ j2x ; u ¼ 3  2y and v ¼ 2x Rearranging these results, we also have y ¼

3u ; 2

x¼

v : 2

Now the Cartesian equation of AB is y ¼ x þ 3 and substituting from the previous 3u v line, we have ¼ þ 3 which simplifies to . . . . . . . . . . . . 2 2

11

v ¼ u  3 which is the equation of a straight line, so, in this case, the path joining A0 and B0 is in fact a straight line. y

v 6

6 w = f (z) –2

O

3

x

–9

O

1 u

–4

Note that it is useful to attach arrow heads to show the corresponding direction of progression in the transformation. On to the next

Complex analysis 1

1027

12

Example 2 If w ¼ z2 , find the path traced out by w as z moves along the straight line joining A ð2 þ j0Þ and B ð0 þ j2Þ: y

Cartesian equation of AB is z = x + jy

O

y¼2x

x

First we transform the two end points A and B onto A0 and B0 in the w-plane. A0 : . . . . . . . . . . . . ; A0 : w ¼ 4 þ j0;

B0 : . . . . . . . . . . . .

13

B0 : w ¼ 4 þ j0

Because w ¼ z2

A: z ¼ 2

; A0 : w ¼ 22 ¼ 4

B: z ¼ j2

; B0 : w ¼ ðj2Þ2 ¼ 4

So we have v

–4

O

4

u

Now we have to find the path from A0 to B0 . The Cartesian equation of AB in the z-plane is y ¼ 2  x. Also w ¼ z2 ¼ ðx þ jyÞ2 ¼ ðx2  y2 Þ þ j2xy ; u ¼ x2  y2

and

v ¼ 2xy

Substituting y ¼ 2  x in these results we can express u and v in terms of x. u ¼ ............; v ¼ ............

1028

Programme 29

14

u ¼ 4x  4;

v ¼ 4x  2x2

uþ4 4     uþ4 uþ4 2 v ¼4 2 4 4 1 ¼ u þ 4  ðu2 þ 8u þ 16Þ 8 1 2 ¼  ðu  16Þ 8

x¼

So, from the first of these Substituting in the second

1 Therefore the path is v ¼  ðu2  16Þ which is a parabola for which at 8 u ¼ 0, v ¼ 2. y 2

v w = f (z) 2

O

2

x

–4

O

4

u

Note that a straight line in the z-plane does not always map onto a straight line in the w-plane. It depends on the particular transformation equation w ¼ f ðzÞ. If the transformation is a linear equation, w ¼ f ðzÞ ¼ az þ b, where a and b may themselves be real or complex, then a straight line in the z-plane maps onto a corresponding straight line in the w-plane. Example 3 A triangle in the z-plane has vertices at A (z ¼ 0), B (z ¼ 3) and C (z ¼ 3 þ j2). Determine the image of this triangle in the w-plane under the transformation equation w ¼ ð2 þ jÞz.

y 2

0

3

x

w ¼ u þ jv ¼ f ðzÞ ¼ ð2 þ jÞz ¼ ð2 þ jÞðx þ jyÞ ¼ ð2x  yÞ þ jð2y þ xÞ ; u ¼ 2x  y;

v ¼ 2y þ x

We now transform each vertex in turn onto the w-plane to determine A0 , B0 and C0 . These are A0 : . . . . . . . . . . . . ;

B0 : . . . . . . . . . . . . ;

C0 : . . . . . . . . . . . .

Complex analysis 1

1029

A0 : w ¼ 0; B0 : w ¼ 6 þ j3;

15

C0 : w ¼ 4 þ j7

The transformation is linear (of the form w ¼ az) so A0 B0 , B0 C0 and C0 A0 are straight lines and the transformation can be illustrated in the diagram ............

v z-plane

y

16

w-plane

8

2

w = f (z)

6 4 2

0

3

x

0

2

4

6

u

f (z) = (2 + j)z

All very straightforward. Let us now take a more detailed look at linear transformations.

Types of transformation of the form w = az + b where the constants a and b may be real or complex. 1

Translation

Let a ¼ 1 and b ¼ 2  j i.e. w ¼ z þ ð2  jÞ. If we apply this to the straight line joining A ð0 þ jÞ and B ð2 þ j3Þ in the z-plane, then

y

z-plane

w ¼ x þ jy þ 2  j ¼ ðx þ 2Þ þ jðy  1Þ so the corresponding end points A0 and B0 in the w-plane are A0 : . . . . . . . . . . . . ;

B0 : . . . . . . . . . . . .

O

x

1030

Programme 29

17

A0 : w ¼ 2; B0 : w ¼ 4 þ j2

v

w-plane

The transformed line A0 B0 is then as shown. The broken line (A)(B) indicates the position of the original line AB in the z-plane.

O

u

Note that the whole line AB has moved two units to the right and one unit downwards, while retaining its original magnitude (length) and direction. Such a transformation is called a translation and occurs whenever the transformation equation is of the form w ¼ z þ b. The degree of translation is given by the value of b – in this case ð2  jÞ, i.e. 2 units along the positive real axis and 1 unit in the direction of the negative imaginary axis. On to the next frame

18

2

Magnification

Consider now w ¼ az þ b where b ¼ 0 and a is real, e.g. w ¼ 2z. y

Applying the transformation to the same line AB as before, we have w ¼ u þ jv ¼ 2z ¼ 2ðx þ jyÞ ; u ¼ 2x and O

v ¼ 2y

x

Transforming the end points A ð0 þ jÞ and B ð2 þ j3Þ onto A0 and B0 in the w-plane, we have A0 : . . . . . . . . . . . . ;

B0 : . . . . . . . . . . . .

and the w-plane diagram becomes ............

Complex analysis 1

1031

19

A0 : w ¼ j2; B0 : w ¼ 4 þ j6 v

O

u

Note that (a) all distances in the z-plane are magnified by a factor 2, and (b) the direction of A0 B0 is that of AB unchanged. Any such transformation w ¼ az where a is real, is said to be a magnification by the factor a. y

So, if we apply the transformation w ¼ z=2 to AB shown here, we can map AB onto A0 B0 in the w-plane and obtain ............ –

Sketch the result

x

O

20

v

u

O

3

Rotation

Consider next w ¼ az þ b with b ¼ 0 and a complex, e.g. w ¼ jz.

y

w ¼ u þ jv ¼ jz ¼ jðx þ jyÞ ¼ y þ jx

O

x

Transforming the end points as usual, we can sketch the original line AB and the mapping A0 B0 , which gives . . . . . . . . . . . .

1032

Programme 29

21

y

z-plane

w-plane

v

w = f (z)

O

x

O

A0 is the point w ¼ 1 þ j0; pffiffiffi Note AB ¼ 2 2

B0 is the point w ¼ 3 þ j2 pffiffiffi A0 B0 ¼ 2 2

Slope of AB ¼ m ¼ 1

Slope of A0 B0 ¼ m1 ¼ 1

u

mm1 ¼ 1ð1Þ ¼ 1 Therefore in transformation by w ¼ jz, AB retains its original length but is rotated about the origin, in this case through 908 in a positive (anticlockwise) direction. Some degree of rotation always occurs when the transformation equation is of the form w ¼ az þ b with a complex. Move on to the next frame

22

4

Combined magnification and rotation

If w ¼ ða þ jbÞz, the effect of transformation is pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (a) magnification j a þ jb j¼ a2 þ b2 b (b) rotation anticlockwise through argða þ jbÞ, i.e. arctan . a Let us see this with an example. Example Map the straight line joining A ð0 þ j2Þ and B ð4 þ j6Þ in the z-plane onto the w-plane under the transformation w ¼ ð3 þ j2Þz. The working is just as before. Draw the z-plane and w-plane diagrams, which give ............

Complex analysis 1

y

1033

z-plane

w-plane

23

v

w = f(z)

(0 + j26)

(–4 + j6)

x

O

O

u

w ¼ ð3 þ j2Þz ; u þ jv ¼ ð3 þ j2Þðx þ jyÞ ¼ ð3x  2yÞ þ jð2x þ 3yÞ ; u ¼ 3x  2y and v ¼ 2x þ 3y A: z ¼ 0 þ j2, i.e. x ¼ 0, y ¼ 2 ; A0 : u ¼ 4, v ¼ 6 ; A0 : ð4 þ j6Þ B: z ¼ 4 þ j6, i.e. x ¼ 4, y ¼ 6 ; B0 : u ¼ 0, v ¼ 26

; B0 : ð0 þ j26Þ

By a simple application of Pythagoras, we can now calculate the lengths of AB and A0 B0 , and then determine the magnification factor ðA0 B0 Þ=ðABÞ. AB ¼ . . . . . . . . . . . . ; A0 B0 ¼ . . . . . . . . . . . . ; magnification ¼ . . . . . . . . . . . . pffiffiffi pffiffiffiffiffiffi AB ¼ 4 2; A0 B0 ¼ 4 26;

mag ¼

pffiffiffiffiffiffi 13

Because

pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi 16 þ 16 ¼ 32 ¼ 4 2 pffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi A0 B0 ¼ 16 þ 400 ¼ 416 ¼ 4 26 pffiffiffiffiffiffi 4 26 pffiffiffiffiffiffi ; magnification ¼ pffiffiffi ¼ 13 4 2 pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi Also j a þ jb j¼j 3 þ j2 j¼ 9 þ 4 ¼ 13 AB ¼

; mag ¼j a þ jb j

Now let us check the rotation. For AB 0 0

For A B

tan 1 ¼ 1 tan 2 ¼ 5

; 1 ¼ 458 ¼ 0:7854 radians ; 2 ¼ 788 410 ¼ 1:3733 radians

; rotation ¼ 2  1 ¼ 1:3733  0:7854 ¼ 0:5879 i.e. rotation = 0.5879 radians Also arg ða þ jbÞ ¼ arg ð3 þ j2Þ ¼ . . . . . . . . . . . .

24

1034

Programme 29

25

0.5879 radians Because arg ð3 þ j2Þ ¼ arctan 23 ¼ 338 410 ¼ 0:5879 radians. So, in transformation w ¼ ða þ jbÞz ¼ ð3 þ j2Þz pffiffiffiffiffiffi (a) AB is magnified by j a þ jb j , i.e. 13 (b) AB is rotated anticlockwise through arg ða þ jbÞ, i.e. arg ð3 þ j2Þ i.e. 0.5879 radians. 5

Combined magnification, rotation and translation

The work we have just done can be extended to include all three effects of transformation. In general, a transformation equation w ¼ az þ b, where a and b are each real or complex, results in magnification j a j ;

rotation arg a;

translation b

Therefore, if w ¼ ð3 þ jÞz þ 2  j magnification ¼ . . . . . . . . . . . . ; translation ¼ . . . . . . . . . . . .

26

mag ¼

rotation ¼ . . . . . . . . . . . . ;

pffiffiffiffiffiffi 10 ¼ 3:162; rotation ¼ 188 260 ¼ 0:3218 radians;

translation ¼ 2 units to right; 1 unit downwards Because

pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi 9 þ 1 ¼ 10 ¼ 3:162 1 (b) rotation ¼ argð3 þ jÞ ¼ arctan ¼ 188 260 ¼ 0:3218 radians 3 (c) translation ¼ 2  j, i.e. 2 to the right, 1 downwards. (a) magnification ¼ j3 þ jj ¼

Let us work through an example in detail. Example 1 The straight line joining A ð2  j3Þ and B ð3 þ jÞ in the z-plane is subjected to the linear transformation equation w ¼ ð1 þ j2Þz þ 3  j4 Illustrate the mapping onto the w-plane and state the resulting magnification, rotation and translation involved. The first part is just like examples we have already done. So, (a) transform the end points A and B onto A0 and B0 in the w-plane (b) join A0 and B0 with a straight line, since AB is a straight line and the transformation equation is linear. That can be done without trouble, the final diagram being . . . . . . . . . . . .

Complex analysis 1

1035

v y 2 1 –2 –1

A

–1 –2 –3

Check the working.

z-plane B O 1

2

3

x w = f (z)

4 2 O –2 –4 –5 –6 –10 –12

27

w-plane B' 2

4

6

8

u

A'

w ¼ ð1 þ j2Þz þ 3  j4

A: z ¼ x þ jy ¼ 2  j3 0

A : w ¼ u þ jv ¼ ð1 þ j2Þð2  j3Þ þ 3  j4 ¼ 2  j7 þ 6 þ 3  j4 ¼ 7  j11 B: z ¼ x þ jy ¼3þj B0 : w ¼ u þ jv ¼ ð1 þ j2Þð3 þ jÞ þ 3  j4 ¼ 3 þ j7  2 þ 3  j4 ¼ 4 þ j3 Now for the second part of the problem, we have to state the magnification, rotation and translation when w ¼ ð1 þ j2Þz þ 3  j4. We remember that the ‘tailpiece’, i.e. 3  j4, independent of z, represents the ............ translation So, for the moment, we concentrate on w ¼ ð1 þ j2Þz, which determines the magnification and rotation. This tells us that magnification ¼ . . . . . . . . . . . . rotation ¼ . . . . . . . . . . . .

28

1036

Programme 29

pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 1 þ 4 ¼ 5 ¼ 2:236  rotation ¼ arg a ¼ arctan 21 ¼ 638 260 ¼ 1:107 radians

29

mag ¼j a j¼ 1 þ j2 ¼

The translation is given by ð3  j4Þ, i.e. 3 units to the right, 4 units downwards. We can in fact see the intermediate steps if we deal first with the transformation w ¼ ð1 þ j2Þz and subsequently with the translation w ¼ 3  j4. v – B B' (B) O

u

(A) – A A'

Under w ¼ ð1 þ j2Þz, A and B map onto A and B where A is w ¼ 4  j7 and B is w ¼ 1 þ j7. Then the translation w ¼ 3  j4 moves all points 3 units to the right and 4 units downwards, so that A and B now map onto A0 and B0 where A0 is w ¼ 7  j11 and B0 is w ¼ 4 þ j3. Normally, there is no need to analyze the transformation into intermediate steps. Example 2 Map the straight line joining A ð1 þ j2Þ and B ð4 þ jÞ in the z-plane onto the w-plane using the transformation equation w ¼ ð2  j3Þz  4 þ j5 and state the magnification, rotation and translation involved. There are no snags. Complete the working and check with the next frame.

Complex analysis 1

1037

30

Here is the complete working. y

w ¼ ð2  j3Þz  4 þ j5 A: z ¼ 1 þ j2 B: z ¼ 4 þ j O

x

A: z ¼ 1 þ j2 A0 : w ¼ ð2  j3Þð1 þ j2Þ  4 þ j5 ¼ 2 þ j þ 6  4 þ j5 ¼ 4 þ j6 B: z ¼ 4 þ j B0 : w ¼ ð2  j3Þð4 þ jÞ  4 þ j5 ¼ 8  j10 þ 3  4 þ j5 ¼ 7  j5 So we have y

z-plane

v

w-plane

w = f (z)

O

x

Also we have (a) magnification ¼j 2  j3 j¼

O

u

pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi 4 þ 9 ¼ 13 ¼ 3:606

0 (b) rotation ¼ argð2  j3Þ ¼ arctanð3 2 Þ ¼ 568 19

¼ 0:9828 radians clockwise (c) translation ¼ 4 þ j5 i.e. 4 units to left, 5 units upwards All very straightforward. Before we move on, here is a short revision exercise. Exercise Calculate (a) the magnification, (b) the rotation, (c) the translation involved in each of the following transformations. 1 w ¼ ð1  j2Þz þ 2  j3

4

w ¼ ðj  4Þz þ j2  3

2 w ¼ ð4 þ j3Þz  2 þ j5

5

w ¼ j2z þ 4  j

3 w ¼ ð2  j3Þz  1  j

6

w ¼ ð5 þ j2Þz þ jðj3  4Þ.

Complete all six and then check the results with the next frame.

1038

31

Programme 29

Results: 1

w ¼ ð1  j2Þz þ 2  j3 (a) magnitude ¼j 1  j2 j¼

pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 1 þ 4 ¼ 5 ¼ 2:236

(b) rotation ¼ argð1  j2Þ ¼ arctanð2Þ ¼ 638 260 ¼ 1:107 radians clockwise (c) translation ¼ 2  j3, i.e. 2 units to right, 3 units downwards. The others are done in the same way and give the following results. No.

Magnitude

Rotation (rad)

Translation

2

5

0.6435 ac

2L, 5U

3

3.606

0.9828 c

1L, 1D

4

4.123

0.2450 c

3L, 2U

5

2

1.5708 ac

4R, 1D

6

5.385

0.3805 ac

3L, 4D

Now let us start a new section, so on to the next frame

Nonlinear transformations 32

So far, we have concentrated on linear transformations of the form w ¼ az þ b. We can now proceed to something rather more interesting. 1

Transformation w ¼ z2 (refer to Frame 12)

The general principles are those we have used before. An example will show the development. Example 1 The straight line AB in the z-plane as shown is mapped onto the w-plane by w ¼ z2 . y

As before, we start by transforming the end points onto A0 and B0 in the w-plane. A0 : w ¼ . . . . . . . . . . . . B0 : w ¼ . . . . . . . . . . . .

O

x

Complex analysis 1

1039

A0 : w ¼ 16;

33

B0 : w ¼ 16

v

u

O

We cannot however assume that AB maps onto the straight line A0 B0 , since the transformation is not linear. We therefore have to deal with a general point. w ¼ u þ jv ¼ z2 ¼ ðx þ jyÞ2 ¼ x2 þ j2xy  y2 ¼ ðx2  y2 Þ þ j2xy ; u ¼ x2  y2 and v ¼ 2xy The Cartesian equation of AB in the z-plane is y ¼ 4  x. So, substituting in the results of the previous line, we can express u and v in terms of x. u ¼ ............; u ¼ 8x  16; The first gives x ¼

v ¼ ............

34

v ¼ 8x  2x2

u þ 16 and substituting this in the expression for v 8

gives . . . . . . . . . . . .

35

1 2 v ¼  32 u þ8

Because 

   u þ 16 u þ 16 2 u2 u8 v¼8 ¼ u þ 16  2 8 8 32 ; v¼

u2 þ8 32

which is an ‘inverted’ parabola, symmetrical about the v-axis, with v ¼ 8 at u ¼ 0. The mapping is therefore y

z-plane

v

w-plane

w = f (z)

O

x

–

O

u

1040

36

Programme 29

Example 2 AB is a straight line in the z-plane joining the origin A to the point B ða þ jbÞ. Obtain the mapping of AB onto the w-plane under the transformation w ¼ z2 . y b

0

a

x

As always, first map the end points. A0 : w ¼ 0 B0 : w ¼ ða þ jbÞ2 ¼ ða2  b2 Þ þ j2ab v 2ab

0

(a2 – b2 )

u

Now to find the path joining A0 and B0 , we consider a general point z ¼ x þ jy. w ¼ u þ jv ¼ z2 ¼ ðx þ jyÞ2 ¼ ðx2  y2 Þ þ j2xy ; u ¼ x2  y2 and v ¼ 2xy b The equation of AB is y ¼ x. We can therefore find u and v in terms of x and a hence v in terms of u. u ¼ ............ v ¼ ............ v ¼ f ðuÞ ¼ . . . . . . . . . . . .

Complex analysis 1

1041

 u¼

 a2  b2 2 x ; a2

 v¼

 2b 2 x ; a

 v¼

 2ab u a2  b2

37

Because

 2  2  b a  b2 2 u ¼ x2  y2 ¼ x2  2 x2 ¼ x a a2     b 2b 2 x¼ x v ¼ 2xy ¼ 2x a a     a2 2b a2 From the expression for u, x2 ¼ u ; v ¼ u a a2  b2 a2  b2   2ab u which is of the form v ¼ ku. ; v¼ a2  b2

A0 B0 is therefore a straight line through the origin. w-plane

v

z-plane

y

w = f(z)

0

a

x

0

( a 2 – b2 )

u

Therefore, under the transformation w ¼ z2 , a straight line through the origin in the z-plane maps onto a straight line through the origin in the w-plane, whereas a straight line not passing through the origin maps onto a parabola. This is worth remembering, so make a note of it

38

Example 3 A triangle consisting of AB, BC, CA in the z-plane is mapped onto the w-plane by the transformation w ¼ z2 . The transformation is w ¼ z2 . ; w ¼ ðx þ jyÞ2 ¼ ðx2  y2 Þ þ j2xy ¼ u þ jv ; u ¼ x2  y2 and v ¼ 2xy First we can map the end points A, B, C onto A0 , B0 , C0 in the w-plane.

y

A0 : . . . . . . . . . . . . B0 : . . . . . . . . . . . . C0 : . . . . . . . . . . . . 0

x

1042

Programme 29

39

A0 : w ¼ 0; B0 : w ¼ 4; C0 : w ¼ 4 So we establish v

u

O

To find the paths joining these three transformed end points, we consider each of the sides of the triangle in turn. (a) AB:

Equation of AB is y ¼ 0 ; u ¼ x2 ; v ¼ 0 ; Each point in AB maps onto a point between A0 and B0 for which v ¼ 0, i.e. part of the u-axis. v

u

O

(b) BC:

Equation of BC is y ¼ 2  x Substitute in u ¼ x2  y2 and v ¼ 2xy and determine v as a function of u. u ¼ ............ v ¼ ............ v ¼ f ðuÞ ¼ . . . . . . . . . . . .

40

u ¼ 4x  4;

v ¼ 4x  2x2 ;

v¼2

Because u ¼ x2  y2 ¼ x2  ð2  xÞ2 ¼ 4x  4

; x¼

v ¼ 2xy ¼ 2xð2  xÞ ¼ 4x  2x2     uþ4 uþ4 2 u2 ; v¼4 ¼2 2 4 4 8 Therefore, the path joining B0 to C0 is an ............

uþ4 4

u2 8

Complex analysis 1

1043

41

inverted parabola v¼2

u2 8

; at u ¼ 0; v ¼ 2 and the w-plane diagram now becomes v

u

0

To complete the mapping, we have still to deal with CA. This transforms onto ............

42

the u-axis between C0 and A0 (c) CA: Equation of CA is x ¼ 0 ; u ¼ y2 ; v ¼ 0 ; Each point between C and A maps onto the negative part of the uaxis between C0 and A0 . So finally we have z-plane

y

v

w-plane

w = z2

x

0

u

0

Mapping of regions In this last example, the three lines AB, BC and CA form the boundary of a triangular region and we have seen how this boundary maps onto the boundary A0 B0 C0 A0 in the w-plane. What we do not know yet is whether the points internal to the triangle map to points internal to the figure in the w-plane or to points external to it. y

v

0

x

v

0

u

0

u

1044

Programme 29

In the z-plane, the region is on the left-hand side as we proceed round the figure in the direction of the arrows ABCA. The region on the left-hand side as we proceed round the figure A0 B0 C0 A0 in the w-plane determines that the transformed region in this case is, in fact, the internal region. So y

z-plane w = f (z) = z

O

w-plane

v 2

x

u

O

Therefore, every point in the region shaded in the z-plane maps onto a corresponding point in the region shaded in the w-plane.

43

2

Transformation w ¼

1 (inversion) z

Example 1 A straight line joining A ðjÞ and B ð2 þ jÞ in the z-plane is mapped onto the w-plane by the 1 transformation equation w ¼ . z

O

Proceeding as before w¼

1 z

; u þ jv ¼ ¼ ; u¼

x2

y

1 x þ jy x  jy x2 þ y2

x y ; v¼ 2 2 x þ y2 þy

First we map the end points A and B onto the w-plane. A0 : w ¼ . . . . . . . . . . . . B0 : w ¼ . . . . . . . . . . . .

x

Complex analysis 1

1045

A0 : w ¼ j;

B0 : w ¼

2 1 j 5 5

44

Because A: x ¼ 0, y ¼ 1

; A0 : u ¼ 0, v ¼ 1

; A0 is w ¼ j

B: x ¼ 2, y ¼ 1

; B0 : u ¼ 25 , v ¼  15

; B0 is w ¼ 25  j 15

So far then we have v

O

u

To determine the path A0 B0 , we can proceed as follows w¼

1 z

; x¼

; z¼ u u2 þ v2

1 w

and

i:e: y¼

x þ jy ¼ v u2 þ v2

1 u  jv ¼ u þ jv u2 þ v2

The equation of AB is y ¼ x  1 v u ¼ 1 ; 2 u þ v2 u2 þ v2 which simplifies into . . . . . . . . . . . . u2 þ v2  u  v ¼ 0 Because v u ¼  1 ;  v ¼ u  u2  v2 u2 þ v2 u2 þ v2 ; u2 þ v2  u  v ¼ 0 We can write this as ðu2  uÞ þ ðv2  vÞ ¼ 0 and completing the square in each bracket this becomes     1 2 1 2 1 u þ v ¼ 2 2 2 which we recognize as the equation of a . . . . . . . . . . . .

45

1046

Programme 29

  1 1 1 circle with centre and radius pffiffiffi 2;2 2

46

The path joining A0 and B0 is therefore an arc of this circle. But we still have problems, for it could be the minor arc or the major arc. v

,

u

O

To decide which is correct, we take a further convenient point on the original line AB and determine its image on the w-plane. y

O

x

For instance, for K, x ¼ 1; y ¼ 0 x ; For K0 , u ¼ 2 ¼1 x þ y2 y ¼0 v¼ 2 x þ y2 ; K0 is the point w ¼ 1

The path is, therefore, the major arc A0 K0 B0 developed in the direction indicated. v

w-plane

z-plane

O

,

w = f (z)

y

x

O

u

Complex analysis 1

1047

If we consider the line AB of the previous example extended to infinity in each direction, its image in the w-plane would then be the complete circle.

47

v y ,

w = f (z)

O

O

x

u

Furthermore, the line AB cuts the entire z-plane into two regions and (a) the region on the right-hand side of the line relative to the arrowed direction maps onto the region inside the circle in the w-plane (b) the region on the left-hand side of the line maps onto ............

48

the region outside the circle in the w-plane Let us now consider a general case. Example 2 Determine the image in the w-plane of a circle in the z-plane under the 1 inversion transformation w ¼ . z The general equation of a circle is

y

x2 þ y2 þ 2gx þ 2fy þ c ¼ 0 with centre ðg; f Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi and radius g 2 þ f 2  c.

r

O

It is convenient at times to write this as Aðx2 þ y2 Þ þ Dx þ Ey þ F ¼ 0 in which case centre is . . . . . . . . . . . . and radius is . . . . . . . . . . . .

x

1048

Programme 29



49

centre

 D E ;  ;  2A 2A

radius ¼

ffi 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi D2 þ E 2  4AF 2A

Because g¼

D ; 2A

f ¼

E ; 2A

c¼

As before we have w ¼ ; x þ jy ¼

F . A

1 z

; z¼

1 u  jv ¼ 2 u þ jv u þ v2

1 w

; x¼

u2

u ; þ v2

y¼

u2

v þ v2

Then Aðx2 þ y2 Þ þ Dx þ Ey þ F ¼ 0 becomes . . . . . . . . . . . . Simplify it as far as possible

50

A þ Du  Ev þ Fðu2 þ v2 Þ ¼ 0 Because we have Aðu2 þ v2 Þ ðu2

þ

v2 Þ 2

þ

Du Ev  þF ¼0 u2 þ v2 u2 þ v2

; A þ Du  Ev þ Fðu2 þ v2 Þ ¼ 0 Changing the order of terms, this can be written Fðu2 þ v2 Þ þ Du  Ev þ A ¼ 0 which is the equation of a circle with centre . . . . . . . . . . . . ; radius . . . . . . . . . . . .

51

 centre



 D E ; ; 2F 2F

radius

ffi 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi D2 þ E2  4FA 2F

1 Thus any circle in the z-plane transforms, with w ¼ , onto another circle z in the w-plane. We have already seen previously that, under inversion, a straight line also maps onto a circle. This may be regarded as a special case of the general result, if we accept a straight line as the circumference of a circle of . . . . . . . . . . . . radius.

Complex analysis 1

1049

52

infinite Because Aðx2 þ y2 Þ þ Dx þ Ey þ F ¼ 0 If A ¼ 0, this becomes Dx þ Ey þ F ¼ 0 i.e. a straight line   D E ;  and also the centre  becomes infinite 2A 2A ffi 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi and the radius D2 þ E 2  4AF becomes infinite. 2A Therefore, combining the results we have obtained, we have this conclusion: 1 Under inversion w ¼ , a circle or a straight line in the z-plane transforms onto z a circle or a straight line in the w-plane. Now for one more example. Example 3 A circle in the z-plane has its centre at z ¼ 3 and a radius of 2 units. 1 Determine its image in the w-plane when transformed by w ¼ . z y

Equation of the circle is ðx  3Þ2 þ y2 ¼ 4 O

x

x2  6x þ 9 þ y2 ¼ 4 x2 þ y2  6x þ 5 ¼ 0.

1 Using w ¼ , we can obtain x and y in terms of u and v. z x ¼ ............; y ¼ ............

1050

Programme 29

53

x¼

u v ; y¼ 2 u 2 þ v2 u þ v2

1 Because w ¼ , z ; z¼

1 w

; x þ jy ¼ ¼ ; x¼

1 u þ jv u  jv u2 þ v2

u v ; y¼ 2 u 2 þ v2 u þ v2

Substituting these in the equation of the circle, we get a relationship between u and v, which is ............

54

5ðu2 þ v2 Þ  6u þ 1 ¼ 0 Because the circle is x2 þ y2  6x þ 5 ¼ 0 ;

u2 ðu2

þ

v2 Þ 2

þ

v2 ðu2

þ

v2 Þ2



u2

6u þ5¼0 þ v2

1 6u  þ5¼0 u2 þ v2 u2 þ v2 5ðu2 þ v2 Þ  6u þ 1 ¼ 0 This is of the form Aðu2 þ v2 Þ þ Du þ Ev þ F ¼ 0 where A ¼ 5, D ¼ 6, E ¼ 0, F ¼ 1. Therefore, the centre is . . . . . . . . . . . . and the radius is . . . . . . . . . . . .

Complex analysis 1

1051

  3 ; 0 ; centre ¼ 5

radius ¼

55

2 5

     D E 6 3 ;  ; 0 i.e. ; 0 Because the centre is  ¼ 2A 2A 10 5 ffi 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 and the radius ¼ D2 þ E 2  4AF ¼ 36 þ 0  20 ¼ : 2A 10 5 

y

v

O

x

O

u

Taking three sample points A, B, C as shown, we can map these onto the x y w-plane using u ¼ 2 and v ¼ 2 . x þ y2 x þ y2 A0 : . . . . . . . . . . . . ;   1 ,0 ; A: 5 0

B0 : . . . . . . . . . . . . ;

0

B:



 3 2 ,  ; 13 13

C 0: . . . . . . . . . . . .

56

C 0 : ð1; 0Þ

So we finally have z-plane y w=1 z

O

3

Transformation w ¼

x

v

O

w-plane

u

1 za

An extension of the method we have just applied occurs with transformations 1 where a is real or complex. of the form w ¼ za

1052

Programme 29

Example A circle z ¼ 1 in the z-plane is mapped onto the w-plane by w ¼ y

w¼

1 z2

; z2¼

1 . z2

1 w

1 u þ jv u  jv ðx  2Þ þ jy ¼ 2 u þ v2 u v ; x¼ 2 þ 2; y ¼ 2 u þ v2 u þ v2 x þ jy  2 ¼

x

O

Cartesian equation of the circle is x2 þ y2 ¼ 1. We then substitute the expressions for x and y in terms of u and v and obtain the relationship between u and v, which is . . . . . . . . . . . .

57

3ðu2 þ v2 Þ þ 4u þ 1 ¼ 0  2  2 u þ 2ðu2 þ v2 Þ v Because we have þ 2 ¼1 u2 þ v2 u þ v2 fu þ 2ðu2 þ v2 Þg2 þ v2 ¼ ðu2 þ v2 Þ2 u2 þ 4uðu2 þ v2 Þ þ 4ðu2 þ v2 Þ2 þ v2 ¼ ðu2 þ v2 Þ2 1 þ 4u þ 4ðu2 þ v2 Þ ¼ u2 þ v2 3ðu2 þ v2 Þ þ 4u þ 1 ¼ 0 This can be expressed as 4 1 u2 þ u þ v2 þ ¼ 0 3 3  2  2 2 1 uþ þ v2 ¼ 3 3 which is a circle with centre



 2 1  ; 0 and radius . 3 3

y

O

v

x

O

u

Complex analysis 1

1053

To determine the direction of development relative to the arrowed direction in the z-plane, we consider the mapping of three sample points A, B, C as shown onto the w-plane, giving A0 ; B0 ; C 0 . A0 : . . . . . . . . . . . . ; A0 : w ¼ ð1, 0Þ;

B0 :

B0 : . . . . . . . . . . . . ;  w¼

 2 1  ;  ; 5 5

C0 : . . . . . . . . . . . . C0 :

 w¼

 1 ,0 3

58

Because A: z ¼ 1 B: z ¼ j C: z ¼ 1

1 ¼ 1 z2 1 jþ2 ¼ ; w¼ j2 5 1 ; w¼ 3

; w¼

; A0 ¼ ð1, 0Þ   2 1 ; B0 ¼  ;  5 5   1 ; C0 ¼  ; 0 3

Whereupon we have y

z-plane

w-plane

v

w= 1 z–2

x

O

O

u

We now have one further transformation which is important, so move on to the next frame for a fresh start 4

Bilinear transformation

w¼

Transformation of the form w ¼

az þ b cz þ d az þ b where a, b, c, d are, in general, complex. cz þ d

Note that (a) if cz þ d ¼ 1; w ¼ az þ b, i.e. the general linear transformation (b) if az þ b ¼ 1; w ¼

1 , i.e. the form of inversion just considered. cz þ d

59

1054

Programme 29

Example Determine the image in the w-plane of the circle z ¼ 2 in the z-plane under zþj and show the region in the w-plane onto which the the transformation w ¼ zj region within the circle is mapped. y

O

We begin in very much the same way as before by expressing u and v in terms of x and y.

x

u............;

60

u¼

x2 þ y2  1 ; x2 þ y2  2y þ 1

v¼

v ¼ ............

2x x2 þ y2  2y þ 1

Because w ¼ u þ jv ¼

z þ j x þ jðy þ 1Þ ¼ z  j x þ jðy  1Þ ¼ ¼ ¼

; u¼

fx þ jðy þ 1Þgfx  jðy  1Þg fx þ jðy  1Þgfx  jðy  1Þg x2 þ jxðy þ 1  y þ 1Þ þ y2  1 x2 þ ðy  1Þ2 x2 þ y2  1 þ j2x x2 þ y2  2y þ 1

x2 þ y2  1 x2 þ y2  2y þ 1

and

v¼

2x x2 þ y2  2y þ 1

But the equation of the circle is x2 þ y2 ¼ 4, so these expressions simplify to u ¼ . . . . . . . . . . . . and v ¼ . . . . . . . . . . . .

61

u¼

3 ; 5  2y

v¼

2x 5  2y

From these, we can form expressions for x and y in terms of u and v. x ¼ ............;

y ¼ ............

Complex analysis 1

1055

Because, from the first, y ¼ x¼

62

3v 5u  3 ; y¼ 2u 2u

x¼

5u  3 and substituting in the second gives 2u

3v : 2u

9v2 ð5u  3Þ2 þ ¼4 4u2 4u2 which can be simplified to . . . . . . . . . . . . But x2 þ y2 ¼ 4 ;

63

9ðu2 þ v2 Þ  30u þ 9 ¼ 0 Because 9v2 þ 25u2  30u þ 9 ¼ 16u2

; 9ðu2 þ v2 Þ  30u þ 9 ¼ 0.

Dividing through by 9, we can now rearrange this to   30 u þ v2 þ 1 ¼ 0 u2  9  2 5 25 ¼0 þ v2 þ 1  i.e. u 3 9    2 5 2 4 u þ v2 ¼ 3 3 which, you will recognize, is a circle in the w-plane with centre . . . . . . . . . . . . and radius . . . . . . . . . . . .  centre ¼

y

O

 5 ,0 ; 3

radius ¼

64

4 3

v

x

O

u

To find the direction of development, we map three sample points A, B, C onto A0 , B0 , C0 as usual. A0 : . . . . . . . . . . . . ;

B0 : . . . . . . . . . . . . ;

C0 : . . . . . . . . . . . .

1056

Programme 29

65

A0 : w ¼

3 4 þ j ; B0 : w ¼ 3; 5 5

C0 : w ¼

3 4 j 5 5

Because A: z ¼ 2 B: z ¼ j2 C: z ¼ 2

2 þ j ð2 þ jÞ2 4 þ j4  1 3 ¼ ¼ 5 þ j 45 ¼ 2j 5 5 j2 þ j j3 ; w¼ ¼ ¼ 3 ; w ¼ 3 i:e: B0 j2  j j

; w¼

; w¼

2 þ j 2  j ð2  jÞ2 3 ¼ ¼ ¼ 5  j 45 2  j 2 þ j 5

y

i:e: A0

i:e: C 0

v

O

O

x

u

So an anticlockwise progression in the z-plane becomes a clockwise progression in the w-plane with this particular example. Now we can complete the problem, for the region inside the circle in the z-plane maps onto . . . . . . . . . . . . . . . . . . . . . . . . in the w-plane.

66

the region outside the circle Because the enclosed region in the z-plane is on the left-hand side of the direction of progression. The region on the left-hand side of the direction of progression in the w-plane is thus the region outside the transformed circle. y

z-plane

v

w-plane

z+j w= z–j

O

x

O

u

Complex analysis 1

1057

And that brings us successfully to the end of this Programme. We shall pursue the topic further in the following Programme. Meanwhile, all that remains is to check down the Review summary and the Can you? checklist before working through the Test exercise. All very straightforward. The Further problems will give you valuable additional practice.

Review summary 29 1 Transformation equation z ¼ x þ jy

w ¼ u þ jv

The transformation equation is the relationship between z and w, i.e. w ¼ f ðzÞ. 2 Linear transformation w ¼ az þ b where a and b are real or complex. A straight line in the z-plane maps onto a corresponding straight line in the w-plane. 3 Types of transformation

w ¼ az þ b

(a) magnification – given by jaj (b) rotation – given by arg a (c) translation – given by b. 4 Nonlinear transformation (a) w ¼ z2 A straight line through the origin maps onto a corresponding straight line through the origin in the w-plane. A straight line not passing through the origin maps onto a parabola. 1 (inversion) z A straight line or a circle maps onto a straight line or a circle in the w-plane.

(b) w ¼

A straight line may be regarded as a circle of infinite radius. (c) w ¼

az þ b (bilinear transformation) – with a, b, c, d real or complex. cz þ d

5 Mapping of a region depends on the direction of development. Right-hand regions map onto right-hand regions: left-hand regions onto left-hand regions.

1058

Programme 29

Can you? Checklist 29 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can:

Frames

. Recognize the transformation equation in the form w ¼ f ðzÞ ¼ uðx; yÞ þ jvðx; yÞ?

1

and

2

2

to

7

7

to

16

16

to

31

32

to

66

Yes

No

. Illustrate the image of a point in the complex z-plane under a complex mapping onto the w-plane?

Yes

No

. Map a straight line in the z-plane onto the w-plane under the transformation w ¼ f ðzÞ?

Yes

No

. Identify complex mappings that form translations, magnifications, rotations and their combinations?

Yes

No

. Deal with the nonlinear transformations w ¼ z2 , w ¼ 1=z, w ¼ 1=ðz  aÞ and w ¼ ðaz þ bÞ=ðcz þ dÞ?

Yes

No

Test exercise 29 1 Map the following points in the z-plane onto the w-plane under the transformation w ¼ f ðzÞ. (a) z ¼ 3 þ j2; w ¼ 2z  j6 (c) z ¼ jð1  jÞ; w ¼ ð2 þ jÞz  1 (b) z ¼ 2 þ j;

w ¼ 4 þ jz

(d) z ¼ j  2;

w ¼ ð1  jÞðz þ 3Þ.

2 Map the straight line joining A ð2  jÞ and B ð4  j3Þ in the z-plane onto the w-plane using the transformation w ¼ ð1 þ j2Þz þ 1  j3. State the magnification, rotation and translation involved. 3 A triangle ABC in the z-plane as shown is mapped onto the w-plane under the transformation w ¼ z2 . y

0

Determine the image in the w-plane and indicate the mapping of the interior triangular region ABC.

x

Complex analysis 1

1059

4 Map the straight line joining A ðz ¼ jÞ and B ðz ¼ 3 þ j4Þ in the z-plane onto 1 the w-plane under the inversion transformation w ¼ . z Sketch the image of AB in the w-plane. 5 The unit circle z ¼ 1 in the z-plane is mapped onto the w-plane by 1 . w¼ z  j2 Determine (a) the position of the centre and (b) the radius of the circle obtained. 6 The circle z þ j2 w¼ . zþj

z ¼ 2 is mapped onto the w-plane by the transformation

Determine the centre and radius of the resulting circle in the w-plane.

Further problems 29 1 A triangle ABC in the z-plane with vertices A ð1  jÞ, B ð2 þ j2Þ, C ð1 þ j2Þ is mapped onto the w-plane under the transformation w ¼ ð1  jÞz þ ð1 þ j2Þ. Determine the image A0 B0 C 0 of ABC in the w-plane. 2 The straight line joining A ð1 þ j2Þ and B ð4  j3Þ in the z-plane is mapped onto the w-plane by the transformation equation w ¼ ð2 þ j5Þz. Determine (a) the images of A and B, (b) the magnification, rotation and translation involved. 3 Map the straight line joining A ð2 þ j3Þ and B ð1 þ j2Þ in the z-plane onto the w-plane using the transformation equation w ¼ ð3 þ jÞz þ 2 þ j4. State the magnification, rotation and translation occurring in the process. 4 y Transform the square ABCD in the z-plane onto the w-plane under the transformation w ¼ z2 .

0

x

5 Map the square ABCD in the z-plane onto the w-plane using the transformation w ¼ 2z2 þ 2.

y

O

x

1060

Programme 29

6

y

O

The triangle ABC in the z-plane is mapped onto the w-plane by the transformation w ¼ j2z2 þ 1. Determine the image of ABC in the w-plane.

x

7 A circle in the z-plane has its centre at the point ð 34  jÞ and radius 74. Show that its Cartesian equation can be expressed as 2ðx2 þ y2 Þ þ 3x þ 4y  3 ¼ 0 Determine the image of the circle in the w-plane under the inversion 1 transformation w ¼ . z 1 8 The transformation w ¼ is applied to the circle z ¼ 2 in the z-plane. z1 Determine (a) the image of the circle in the w-plane (b) the region in the w-plane onto which the region enclosed within the circle in the z-plane is mapped. 9 The circle z ¼ 4 is described in the z-plane in an anticlockwise manner. zþ1 and state Obtain its image in the w-plane under the transformation w ¼ z2 the direction of development. zj is applied to the circle z ¼ 3 in the z þ j2 z-plane. Determine the equation of the image in the w-plane and state its centre and radius.

10

The bilinear transformation w ¼

11

The unit circle z ¼ 1 in the z-plane is mapped onto the w-plane under the z1 . Determine the equation of its image and the region transformation w ¼ z3 onto which the region within the circle is mapped.

12

Obtain the image of the unit circle z ¼ 1 in the z-plane under the z þ j3 . transformation w ¼ z  j2

13

The circle z ¼ 2 is mapped onto the w-plane by the transformation zþj w¼ . Determine 2z  j (a) the image of the circle in the w-plane (b) the mapping of the region enclosed by z ¼ 2.

14

za where z ¼ x þ jy, a ¼ 1 þ j4 zb 2 and b ¼ 2 þ j3, transforms the circle ðx  3Þ þ ðy  5Þ2 ¼ 5 into a straight line through the origin in the w-plane. Show that the transformation equation w ¼

Frames 1 to 75

Programme 30

Complex analysis 2 Learning outcomes When you have completed this Programme you will be able to: Appreciate when the derivative of a function of a complex variable exists Understand the notions of regular functions and singularities and be able to obtain the derivative of a regular function from first principles Derive the Cauchy–Riemann equations and apply them to find the derivative of a regular function Understand the notion of an harmonic function and derive a conjugate function Evaluate line and contour integrals in the complex plane Derive and apply Cauchy’s theorem Apply Cauchy’s theorem to contours around regions that contain singularities Define the essential characteristics of and conditions for a conformal mapping Locate critical points of a function of a complex variable Determine the image in the w-plane of a figure in the z-plane under a conformal transformation w ¼ f ðzÞ Describe and apply the Schwarz–Christoffel transformation

1061

1062

Programme 30

1

In the previous Programme we introduced the ideas of mapping from one complex plane to another and considered some of the more common transformation functions. Now we pursue our consideration of the complex variable a little further.

Differentiation of a complex function In differentiation of a function of a single real variable, y ¼ f ðxÞ, the derivative of y ðy þ yÞ  y as x tends to with respect to x can be defined as the limiting value of x zero. y

y = f(x)

y + δy

y ¼ f ðxÞ y ¼ f ðx þ xÞ  f ðxÞ  dy f ðx þ xÞ  f ðxÞ ¼ Lim i:e: dx x!0 x

δy y O

δx x + δx

x

x

In considering the differentiation of a function of a complex variable, w ¼ f ðzÞ, the derivative of w with respect to z can similarly be defined as the limiting value of . . . . . . . . . . . . as z tends to zero.

2

ðw þ wÞ  w z

i.e.

f ðz þ zÞ  f ðzÞ z

Now, of course, we are dealing in vectors. y

z-plane

z0

O

+δ

z

v

δz

w-plane

+ w0

z0 x 0

O

δw

δw

w0 u

0

If P and Q in the z-plane map onto P and Q in the w-plane, then P0 Q 0 ¼ w ¼ ðw0 þ wÞ  w0 ¼ f ðz0 þ zÞ  f ðz0 Þ w Therefore, the derivative of w at P0 ðz ¼ z0 Þ is the limiting value of as z

   0 0 dw f ðz0 þ zÞ  f ðz0 Þ PQ z ! 0; i:e: ¼ Lim ¼ Lim dz z0 z!0 z PQ Q!P If this limiting value exists – which is not always the case as we shall see – the function f ðzÞ is said to be differentiable at P.

Complex analysis 2

1063

Also, if w ¼ f ðzÞ and f 0 ðzÞ has a limit for all points z0 within a given region for which w ¼ f ðzÞ is defined, then f ðzÞ is said to be differentiable in that region. From this, it follows that the limit exists whatever the path of approach from Q ðz ¼ z0 þ zÞ to P ðz ¼ z0 Þ.

Regular function A function w ¼ f ðzÞ is said to be regular (or analytic) at a point z ¼ z0 , if it is defined and single-valued, and has a derivative at every point at and around z0 . Points in a region where f ðzÞ ceases to be regular are called singular points, or singularities. A function of a complex variable that is regular over the entire finite complex plane is called an entire function. Examples of entire functions are polynomials, ez , sin z and cos z. We have introduced quite a few new definitions, so let us pause here while you make a note of them. We shall be meeting the various terms quite often. In those cases where a derivative exists, the usual rules of differentiation apply. For example, the derivative of w ¼ z2 can be found from first principles in the normal way.

3

; w þ w ¼ ðz þ zÞ2 ¼ z2 þ 2zz þ ðzÞ2 w ¼ 2z þ z ; w ¼ 2zz þ ðzÞ2 ; z

w ¼ z2

dw ¼ Lim ð2z þ zÞ ¼ 2z and does not depend on the path along which z dz z!0 tends to zero. That was elementary. Here is a rather different one.

;

Example To find the derivative of w ¼ zz where z ¼ x þ jy and z ¼ x  jy. We have w ¼ zz

; w þ w ¼ ðz þ zÞðz þ zÞ from which w ¼ ............ z

4

w z ¼ z þ z þ z z z Because w þ w ¼ ðz þ zÞðz þ zÞ ¼ zz þ zz þ zz þ zz w z ¼ z þ z þ z ; w ¼ zz þ zz þ zz ; z z Now since z ¼ x þ jy and z ¼ x  jy, we can express w ¼ ............ z

w in terms of x and y. z

1064

Programme 30

 w x  jy ¼ ðx  jyÞ þ ðx þ jyÞ þ x  jy z x þ jy

5 Because

z ¼ x þ jy ; z ¼ x þ jy z ¼ x  jy ; z ¼ x  jy

) ;

z x  jy ¼ z x þ jy

w z ¼ z þ z þ z gives the expression quoted above. z z

Then y

(z0 + δz) δz (z0)

δy

The next step is to reduce z to zero. But z consists of x þ jy and so reducing z to zero can be done as:

δx

O

x

(1) First let y ! 0 and afterwards let x ! 0.

y

δy δx O

x

If y ! 0, Then

w x ¼ x  jy þ ðx þ jyÞ þ x z x

dw ¼ Lim fx  jy þ x þ jy þ xg dz x!0 ¼ ............

Complex analysis 2

1065

dw ¼ 2x dz

6

On the other hand, we could have reduced z to zero in another way. y

(2) First let x ! 0 and afterwards let y ! 0.

δx δy

O

We have If x ! 0 Then

x

 w x  jy ¼ x  jy þ ðx þ jyÞ þ x  jy z x þ jy w ¼ x  jy þ ðx þ jyÞð1Þ  jy ¼ j2y  jy z dw ¼ Lim fj2j  jyg ¼ j2y dz y!0

dw dw ¼ 2x and in the second case ¼ j2y. dz dz These two results are clearly not the same for all values of x and y – with one exception, i.e. ............ So, in the first case,

when x ¼ y ¼ 0 Therefore w ¼ zz is a function that is differentiable at a single point (namely the origin) but nowhere else. As a consequence the function is not regular at that point because to be so it needs to be differentiable not only at the point but also at points in a region surrounding it. This is not an uncommon occurrence so it would be convenient, therefore, to have some form of test to see whether or not the function w ¼ f ðzÞ is or is not regular at all points of a region. This useful tool is provided by the Cauchy–Riemann equations.

Cauchy–Riemann equations The development is very much along the same lines as in the previous example. If w ¼ f ðzÞ ¼ u þ jv, we have to establish conditions for w ¼ f ðzÞ to have a derivative at a given point z ¼ z0 . w ¼ u þ jv Then

; w ¼ u þ jv; z ¼ x þ jy ; z ¼ x þ jy   dw u þ jv u þ jv regardless of how x ¼ Lim f 0 ðzÞ ¼ ¼ Lim ð1Þ x!0 x þ jy and y tend to zero dz z!0 z y!0

(a) Let x ! 0, followed by y ! 0 Then from (1) above,

f 0 ðzÞ ¼

dw ¼ ............ dz

7

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Programme 30

8

dw @v @u ¼ j dz @y @y Because

  u þ jv v u @v @u ¼ Lim ¼ j j jy y @y @y y!0 y!0 y

f 0 ðzÞ ¼ Lim

ð2Þ

We use the ‘partial’ notation since u and v are functions of both x and y. Or (b) Let y ! 0, followed by x ! 0. This gives . . . . . . . . . . . .

9

dw @u @v ¼ þj dz @x @x Because

  u þ jv u v @u @v f ðzÞ ¼ Lim þj þj ¼ Lim ¼ x x @x @x x!0 x!0 x 0

ð3Þ

If the results (2) and (3) are to have the same value for f 0 ðzÞ irrespective of the path chosen for z to tend to zero, then ............

10

@u @v @v @u þj ¼ j @x @x @y @y Equating real and imaginary parts, this gives @u @v ¼ @x @y

and

@v @u ¼ @x @y

These are the Cauchy–Riemann equations. So, to sum up: A necessary condition for w ¼ f ðzÞ ¼ u þ jv to be regular at z ¼ z0 is that u, v and their partial derivatives are continuous and that in the neighbourhood of z ¼ z0 @u @v ¼ @x @y

and

@v @u ¼ @x @y

Make a note of this important result – then move on to the next frame

Complex analysis 2

1067

We said earlier that where a function fails to be regular, a singular point, or singularity occurs, for example where w ¼ f ðzÞ is not continuous or where the Cauchy–Riemann test fails.

11

Exercise Determine where each of the following functions fails to be regular, i.e. where singularities occur. 1 ðz  2Þðz  3Þ

1 w ¼ z2  4

4 w¼

z z2 zþ5 3 w¼ zþ1

5 w ¼ zz

2 w¼

6 w¼

x þ jy x2 þ y2 Finish all six: then check with the next frame

12

Conclusions: 1 Putting z ¼ x þ jy, the Cauchy–Riemann conditions are satisfied everywhere. Therefore, no singularity in w ¼ z2  4. 2 The function becomes discontinuous at z ¼ 2. Singularity at z ¼ 2. 3 The function is discontinuous at z ¼ 1. Singularity at z ¼ 1. 4 Singularities at z ¼ 2 and z ¼ 3. 5 We have already seen that w ¼ zz has no derivative for all values of z apart from z ¼ 0. All points on w ¼ zz are singularities. 6 Singularity occurs where x2 þ y2 ¼ 0, i.e. x ¼ 0 and y ¼ 0 ; z ¼ 0. At all other points the Cauchy–Riemann equations do not hold.

Harmonic functions If a function of two real variables f ðx, yÞ satisfies Laplace’s equation @ f ðx, yÞ @ f ðx, yÞ þ ¼0 @x2 @y2 2

2

then we say that f ðx, yÞ is an harmonic function. It is relatively straightforward to demonstrate that the real and imaginary parts of an analytic function are both harmonic.

13

1068

Programme 30

Let f ðzÞ ¼ uðx, yÞ þ jvðx, yÞ be an analytic function in some region of the z-plane. Because f ðzÞ is analytic the Cauchy–Riemann equations hold true. That is @u @v @u @v ¼ and ¼ @x @y @y @x Differentiating the first with respect to x and the second with respect to y shows us that @2u @2v @2u @2v @2v @2u and ¼  ¼  ¼ ¼  @x2 @x@y @y2 @y@x @x@y @x2 and so

@2u @2u þ ¼0 @x2 @y2

By a similar reasoning @2 . . . @2 . . . þ ¼0 @x2 @y2

14

@2v @2v þ ¼0 @x2 @y2 Because 

@2v @2u @2v @2u @2u @2v and ¼ ¼ 2 ¼ ¼ 2 2 @x @x@y @y @y@x @x@y @x

and so

@2v @2v þ ¼0 @x2 @y2

The functions uðx, yÞ and vðx, yÞ are called conjugate functions. In addition, the curves u ¼ constant, v ¼ constant are orthogonal. Example 1 Show that the real and imaginary parts of the function defined by f ðzÞ ¼ z2 are harmonic. f ðzÞ ¼ z2 ¼ ðx þ jyÞ2 ¼ ðx2  y2 Þ þ 2jxy and so u ¼ x2  y2 and v ¼ 2xy and therefore @2u @2u @2v @2v þ 2 ¼ . . . . . . . . . . . . and þ ¼ ............ 2 @x @y @x2 @y2

Complex analysis 2

1069

@2u @2u @2v @2v þ ¼ 0 and þ ¼0 @x2 @y2 @x2 @y2

15

Because @u @2u @u @2u ¼ 2x so ¼ 2y so ¼ 2 and ¼ 2 @x @x2 @y @y2 therefore

@2u @2u þ ¼0 @x2 @y2

and @v @2v @v @2v ¼ 2y so ¼ 2x so ¼ 0 and ¼0 2 @x @x @y @y2 therefore

@2v @2v þ ¼0 @x2 @y2

Example 2 Show that uðx, yÞ ¼ x3 y  y3 x is an harmonic function and find the function vðx, yÞ that ensures that f ðzÞ ¼ uðx; uÞ þ jvðx, yÞ is analytic. That is, find the function vðx, yÞ that is conjugate to uðx, yÞ. @2u @2u þ ¼ ............ @x2 @y2 @2u @2u þ ¼0 @x2 @y2 Because @u @2u @u @2u ¼ 3x2 y  y3 so ¼ x3  3y2 x so ¼ 6xy and ¼ 6xy 2 @x @x @y @y2 therefore

@2u @2u þ ¼0 @x2 @y2

This means that uðx, yÞ ¼ x3 y  y3 x is harmonic. Now, if f ðzÞ ¼ uðx; uÞ þ jvðx, yÞ is analytic then uðx, yÞ and vðx, yÞ satisfy the . . . . . . . . . . . . equations.

16

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Programme 30

17

Cauchy–Riemann That is @u @v ¼ 3x2 y  y3 ¼ @x @y and @u @v ¼ x3  3y2 x ¼  @y @x Integrating

@v ¼ 3x2 y  y3 with respect to y gives @y vðx, yÞ ¼ . . . . . . . . . . . .

18

vðx, yÞ ¼

3 2 2 1 4 x y  y þ aðxÞ 2 4

Because @v ¼ 3x2 y  y3 and so x is treated as a constant and the integral of yn is @y ynþ1 =ðn þ 1Þ. Did you miss the constant term in the form of aðxÞ? Because x is treated as a constant, the integration determines v up to an expression involving x. Differentiate @v the result with respect to y and you will reclaim the original form for . @y Now, differentiating this expression with respect to x gives @v ¼ ............ @x

19

@v ¼ 3xy2 þ a0 ðxÞ @x Because 3 2 2 1 4 @v @u x y  y þ aðxÞ and so ¼ 3xy2 þ a0 ðxÞ and this is equal to  . 2 4 @x @y @u 3 2 ¼ x þ 3y x and so Now  @y

vðx, yÞ ¼

a0 ðxÞ ¼ . . . . . . . . . . . . giving aðxÞ ¼ . . . . . . . . . . . . Therefore vðx, yÞ ¼ . . . . . . . . . . . .

Complex analysis 2

1071

x4 þ C. 4 3x2 y2 y4 x4   þC Therefore vðx, yÞ ¼ 2 4 4 a0 ðxÞ ¼ x3 giving aðxÞ ¼ 

20

Because @v @u ¼ 3xy2 þ a0 ðxÞ and  ¼ x3 þ 3y2 x @x @y @v @u ¼ then it is seen that a0 ðxÞ ¼ x3 . where @x @y x4 3x2 y2 y4 x4 Therefore aðxÞ ¼  þ C giving vðx, yÞ ¼   þC 4 2 4 4

Comparing

Try one for yourself. Example 3 Given uðx, yÞ ¼ ex cos y, show that uðx, yÞ is an harmonic function and find the function vðx, yÞ that ensures that f ðzÞ ¼ uðx, yÞ þ jvðx, yÞ is analytic. That is, find the function vðx, yÞ that is conjugate to uðx, yÞ. @2 . . . @2 . . . þ ¼ ............ @x2 @y2 @2u @2u þ ¼0 @x2 @y2

21

Because @u @2u ¼ ex cos y and 2 ¼ ex cos y. @x @x @u @2u @2u @2u x x ¼ e sin y so 2 ¼ e cos y. Therefore 2 þ 2 ¼ 0, that is uðx, yÞ Also @y @y @x @y is harmonic. The conjugate function vðx, yÞ is then

u ¼ ex cos y so

vðx, yÞ ¼ . . . . . . . . . . . . v ¼ ex sin y þ C Because @u @v ¼ ¼ ex cos y. Integrating with @x @y respect to y gives v ¼ ex sin y þ aðxÞ. Differentiating this with respect to x @v gives ¼ ex sin y þ a0 ðxÞ. @x @v @u ¼ ¼ ex sin y, so that Now, by the other Cauchy–Riemann equation @x @y a0 ðxÞ ¼ 0 giving aðxÞ ¼ C. Therefore, v ¼ ex sin y þ C. By the Cauchy–Riemann equation

Now we shall look at complex integration. Move to the next frame

22

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Programme 30

Complex integration 23

At the beginning of this Programme, we defined differentiation with respect to z in the case of a complex function, since z is a function of two independent variables x and y, i.e. z ¼ x þ jy. Complex integration is approached in the same way. z ¼ x þ jy Also ;

ð

and

w ¼ f ðzÞ ¼ u þ jv

where u and v are also functions of x and y.

dz ¼ dx þ j dy and dw ¼ du þ j dv ð ð w dz ¼ f ðzÞ dz ¼ ðu þ jvÞðdx þ j dyÞ ð ¼ fðu dx  v dyÞ þ jðv dx þ u dyÞg ð ð ð ; f ðzÞ dz ¼ ðu dx  v dyÞ þ j ðv dx þ u dyÞ

That is, the integral reduces to two real-variable integrals ð ð ðu dx  v dyÞ and ðv dx þ u dyÞ ð ðP dx þ Q dyÞ

Note that each of these two integrals is of the general form

which we met before during our work on line integrals and, in the complex plane, this rather neatly leads us into contour integration. Let us make a fresh start

24

Contour integration – line integrals in the z-plane If z moves along the curve c in the z-plane and at each position z has associated with it a function of z, i.e. f ðzÞ, then summing up f ðzÞ for all such points between A and B means that we are evaluating a line integral in the z-plane between ð f ðzÞ dz where A ðz ¼ z1 Þ and B ðz ¼ z2 Þ along the curve c, i.e. we are evaluating c

c is the particular path joining A to B. The evaluation of line integrals in the complex plane is known as contour integration. Let us see how it works in practice.

Complex analysis 2

1073

Example

25

ð f ðzÞ dz where f ðzÞ ¼ ðz  jÞ2 and c is the straight line

Evaluate the integral c

joining A ðz ¼ 0Þ to B ðz ¼ 1 þ j2Þ. y

(1 + j2)

z ¼ x þ jy;

j

dz ¼ dx þ j dy

x

f ðzÞ ¼ ðz  jÞ2 ¼ fx þ jðy  1Þg2 ¼ x2  ðy  1Þ2 þ j 2xðy  1Þ ð ; I ¼ fðx2  y2 þ 2y  1Þ þ jð2xy  2xÞgfdx þ j dyg ð ¼ fðx2  y2 þ 2y  1Þ dx  ð2xy  2xÞ dyg ð þ j fð2xy  2xÞ dx þ ðx2  y2 þ 2y  1Þ dyg Now the equation of AB is y ¼ 2x: ; dy ¼ 2 dx and substituting these in the expression for I, between the limits x ¼ 0 and x ¼ 1, gives I ¼ ............

I¼

Finish it.

1 ð2 þ jÞ 3

Because ð1 I ¼ fðx2  4x2 þ 4x  1Þ dx  ð4x2  2xÞ2 dxg 0

þj ¼

ð1

ð1

fð4x2  2xÞ dx þ ð2x2  8x2 þ 8x  2Þ dxg

0

ð11x2 þ 8x  1Þ dx þ j

0

ð1

ð2x2 þ 6x  2Þ dx

0

and this, by elementary integration, gives I ¼ 13 ð2 þ jÞ. Now you will remember that, in general, the value of a line integral depends on the path of integration between the end points, but that the line integral ð ðPdx þ Q dyÞ is independent of the path of integration in a simply connected region if

@P @Q ¼ throughout the region. @y @x

26

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Programme 30

In our example ð I ¼ fðx2  y2 þ 2y  1Þ dx  ð2xy  2xÞ dyg ð þ j fð2xy  2xÞ dx þ ðx2  y2 þ 2y  1Þ dyg  I1 þ j I2 If we apply the test to I1 , we get . . . . . . . . . . . .

27

@P @Q ¼ @y @x Because

ð ð for I1 ¼ fðx2  y2 þ 2y  1Þ dx  ð2xy  2xÞ dxg  ðP dx þ Q dyÞ 9 @P > ¼ 2y þ 2 > ; P ¼ x2  y2 þ 2y  1 ; > = @y @P @Q ; ¼ > @y @x > @Q > ; ¼ 2y þ 2 Q ¼ 2xy þ 2x ; @x

Similarly

ð ð for I2 ¼ fð2xy  2xÞ dx þ ðx2  y2 þ 2y  1Þdyg  ðPdx þ QdyÞ 9 @P > ¼ 2x > ; P ¼ 2xy  2x ; > = @y @P @Q ¼ ; > @y @x > @Q > 2 2 ; ¼ 2x Q ¼ x  y þ 2y  1 ; @x

Therefore, in this example, the value of the line integral is independent of the path of integration. Just to satisfy our conscience, determine the value of the line integral between the same two end points, but along the parabola y ¼ 2x2 . y

f ðzÞ ¼ ðz  jÞ2 y = 2x 2 0

y ¼ 2x2

; dy ¼ 4x dx

x

As before we have ð I ¼ fðx2  y2 þ 2y  1Þ dx  ð2xy  2xÞ dyg ð þ j ð2xy  2xÞ dx þ ðx2  y2 þ 2y  1Þ dyg Substituting y ¼ 2x2 and dy ¼ 4x dx, the evaluation gives I ¼ ............

Complex analysis 2

1075

I¼

28

1 ð2 þ jÞ 3

We have ð1 I ¼ fðx2  4x4 þ 4x2  1Þ dx  ð4x3  2xÞ4x dxg 0

þj ¼

ð1

ð1

fð4x3  2xÞ dx þ ðx2  4x4 þ 4x2  1Þ4x dxg

0

ð20x4 þ 13x2  1Þ dx þ j

ð1

0

ð16x5 þ 24x3  6xÞ dx

0

The rest is easy enough, giving I ¼ 13 ð2 þ jÞ which is, of course, the same result as before. Note that all results in Frames 25–28 can be obtained very easily by integrating the function of z with respect to z. ð For example, the integral f ðzÞ dz where f ðzÞ ¼ ðz  jÞ2 and c is the straight c

line joining A ðz ¼ 0Þ to B ðz ¼ 1 þ j2Þ can be evaluated as ð ð 1þj2 f ðzÞ dz ¼ ðz  jÞ2 dz c

z¼0

" #1þj2 ðz  jÞ3 ¼ 3 0

¼ ¼

ð1 þ j2  jÞ3 ðjÞ3  3 3

!

1 ð2 þ jÞ 3 Now on to the next frame

29

Cauchy’s theorem We have already seen that if w ¼ f ðzÞ where, as usual, w ¼ u þ jv and z ¼ x þ jy, then dz ¼ dx þ jdy and ð ð f ðzÞ dz ¼ ðu þ jvÞðdx þ j dyÞ ð ð ¼ ðu dx  v dyÞ þ j ðv dx þ u dyÞ If c is a closed curve as the path of integration, then þ þ þ f ðzÞ dz ¼ ðu dx  v dyÞ þ j ðv dx þ u dyÞ c

c

c

1076

Programme 30

Applying Green’s theorem to each of the two integrals on the right-hand side in turn, we have  ðð  þ @v @u dx dy (a) ðu dx  v dyÞ ¼   @x @y c S where S is the region enclosed by the curve c. Also, if f ðzÞ is regular at every point within and on c, then the Cauchy– Riemann equations give @u @v @v @u ¼ and therefore   ¼0 @y @x @x @y þ ; ðu dx  v dyÞ ¼ 0

ð1Þ

c

(b) Similarly, with the second integral, we have ............ þ

30

ðv dx þ u dyÞ ¼ 0 c

Because  þ ðð  @u @v ðv dx þ u dyÞ ¼ dx dy  @x @y c S Again, if f ðzÞ is regular at every point within and on c, then the Cauchy–Riemann equations give @u @v @u @v ¼ and therefore  ¼0 @x @y @x @y þ ; ðv dx þ u dyÞ ¼ 0

ð2Þ

c

Combining the two results (1) and (2) we have the following result. If f ðzÞ is regular at every point within and on a closed curve c, then þ f ðzÞ dz ¼ 0 c

This is Cauchy’s theorem. Make a note of the result; then we can see an example

Complex analysis 2

1077

Example 1

31

þ f ðzÞ dz where f ðzÞ ¼ z2

Verify Cauchy’s theorem by evaluating the integral c

around the square formed by joining the points z ¼ 1, z ¼ 2, z ¼ 2 þ j, z ¼ 1 þ j. y

z ¼ x þ jy z2 ¼ x2  y2 þ j 2xy dz ¼ dx þ j dy

O

x

þ

þ

þ

f ðzÞ dz ¼

fx2  y2 þ j 2xygfdx þ j dyg þ þ 2 2 ¼ fðx  y Þ dx  2xy dyg þ j f2xy dx þ ðx2  y2 Þ dyg

c

z2 dz ¼

c

c

c

c

We now take each of the sides in turn. y ¼ 0 ; dy ¼ 0

3 2 ð2 ð x 8 1 7 2 f ðzÞ dz ¼ x dx ¼ ¼  ¼ ; 3 3 3 3 1 AB 1

(a) AB:

x ¼ 2 ; dx ¼ 0 ð ð1 ð1 ; f ðzÞ dz ¼ ð4ydyÞ þ j ð4  y2 Þ dy

(b) BC:

BC

0



1

0



y3 1 ¼ 2y2 þj 4y  3 0  0 1 11 ¼ 2 þ j ¼ 2 þ j 4  3 3

Continuing in the same way, the results for the remaining two sides are . . . . . . . . . . . . and . . . . . . . . . . . .

CD: 

4  j3; 3

DA: 1  j

Because y ¼ 1 ; dy ¼ 0 ð1 ð1 ð f ðzÞ dz ¼ ðx2  1Þ dx þ j 2x dx ;

(c) CD:

CD

2

2

1 1 x  x þ j x2 ¼  43  j 3 ¼ 3 2 2

3

2 3

32

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Programme 30

x ¼ 1 ; dx ¼ 0 ð0 ð0 ð f ðzÞ dz ¼ ð2ydyÞ þ j ð1  y2 Þ dy ;

(d) DA:

DA

1



0

1



0 y3 ¼ y þj y ¼ 1  j 23 3 1 1 þ So, collecting the four results, f ðzÞ dz ¼ . . . . . . . . . . . . 2

c

þ

33

f ðzÞ dz ¼ 0 c

Because       þ 7 11 4 2 þ   j3 þ 1  j ¼0 f ðzÞ dz ¼ þ 2 þ j 3 3 3 3 c Example 2 A region in the z-plane has a boundary c consisting of (a) OA joining z ¼ 0 to z ¼ 2 (b) AB a quadrant of the circle z ¼ 2 from z ¼ 2 to z ¼ j2 (c) BO joining z ¼ j2 to z = 0.

ð ðz2 þ 1Þ dz

Verify Cauchy’s theorem by evaluating the integral c

(1) along the arc from A to B (2) along BO and OA. y

f ðzÞ ¼ z2 þ 1 ¼ ðx þ jyÞ2 þ 1 ¼ ðx2  y2 þ 1Þ þ j 2xy z ¼ x þ jy 0

; dz ¼ dx þ j dy

x

ð So the general expression for

f ðzÞ dz ¼ . . . . . . . . . . . .

Complex analysis 2

1079

ð fðx2  y2 þ 1Þ þ j 2xygfdx þ j dyg ð ð ¼ fðx2  y2 þ 1Þ dx  2xy dyg þ j f2xy dx þ ðx2  y2 þ 1Þ dyg

(1) Arc AB: x2 þ y2 ¼ 4

; y2 ¼ 4  x2

1 dy ¼ ð4  x2 Þ1=2 ð2xÞ dx 2 ð f ðzÞ dz ;

; y¼

34

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4  x2

x ; dy ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx 4  x2

AB

ð0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi  x  ðx2  4 þ x2 þ 1Þ dx  2x 4  x2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ 4  x2 2   ð 0  pffiffiffiffiffiffiffiffiffiffiffiffiffiffi x þj 2x 4  x2 dx þ ðx2  4 þ x2 þ 1Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx 4  x2 2 ð0 ð0 11x  4x3 14 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼  þ j I1 ¼ ð4x2  3Þ dx þ j 2 3 4x 2 2 ð0 11x  4x3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx. Now we must attend to I1 ¼ 4  x2 2 Substituting x ¼ 2 sin  and dx ¼ 2 cos  d with appropriate limits we have ............

I1 ¼ 

35

2 3

Because  ð0  22 sin   32 sin3  2 cos  d I1 ¼ 2 cos  =2 ð =2 ð32 sin3   22 sin Þ d ¼ 0

¼ 32

ð =2

=2  sin  1  cos2  d þ 22 cos 

0

0

64 2 cos2   1 dðcos Þ  22 ¼  22 ¼  3 3 ¼0 2 2 2 f ðzÞ dz ¼ 4  j ¼  ð7 þ jÞ 3 3 3 AB

¼ 32 ð ;

ð =2



(2) Along BO and OA. Complete this section on your own in the same way. ð ð f ðzÞ dz ¼ . . . . . . . . . . . . ; f ðzÞ dz ¼ . . . . . . . . . . . . BO

OA

1080

Programme 30

ð

36

2 f ðzÞ dz ¼ j ; 3 BO

ð f ðzÞ dz ¼ 4 OA

2 3

Because we have BO: x ¼ 0 ð

; dx ¼ 0

;

f ðzÞ dz ¼ j BO

 y3 0 2 ð1  y2 Þ dy ¼ j y  ¼j 3 3 2 2

ð1

OA: y ¼ 0 ; dy ¼ 0 2

3 ð2 ð x 2 2 þx ¼4 ; f ðzÞ dz ¼ ðx þ 1Þ dx ¼ 3 3 OA 0 0 Collecting the results together, therefore ð 14 2 j f ðzÞ dz ¼  3 3 AB ð 2 2 14 2 þj f ðzÞ dz ¼ j þ 4 ¼ 3 3 3 3 BO þ OA þ ð ð ; f ðzÞ dz ¼ f ðzÞ dz þ f ðzÞ dz ¼ 0 c

BO þ OA

AB

which, once again, verifies Cauchy’s theorem. Just by way of revision, Cauchy’s theorem actually states that ............

37

If f ðzÞ is regular at every point within and on þ a closed curve c, then f ðzÞ dz ¼ 0 c

In our examples so far, f ðzÞ has been regular and no problems have arisen. Let us now consider a case where one or more singularities occur within the region enclosed by the curve c.

Deformation of contours at singularities If c is the boundary curve (or contour) of a region and f ðzÞ is regular for all points within and on the contour, then the þ evaluation of f ðzÞ dz around the contour is straightforward.

(z = a)

c

c

1 , where a is a complex constant, and point A za corresponds to z ¼ a, then at A, f ðzÞ ceases to be regular and a singularity occurs at that point. However, if f ðzÞ ¼

Complex analysis 2

1081

We can isolate A in a very small region within a contour c1 and then f ðzÞ will be regular at all points within the region c and outside c1 . But the original region is now no longer simply connected (it now has a ‘hole’ in it) and this was one of our initial conditions.

c c

However, all is not lost! We select a suitable point B on the contour c and join it to the inner contour c1 . If we now ð consider the integration f ðzÞ dz starting from a point K c

and proceeding anticlockwise, the path of integration can be taken as K B L M N B D E K.

c

Therefore ð f ðzÞ dz ¼ I ¼ IKB þ IBL þ ILMN þ INB þ IBDEK ¼ . . . . . . . . . . . .

38

0 The function f ðzÞ is now regular at all points within and on the deformed contour. Remember that the inner contour c1 can be made as small as we wish.

Note that INB ¼ IBL , being in opposite directions, and these therefore cancel out.

c c

The previous result then becomes i:e: IKB þ IBDEK þ ILMN ¼ 0 IKB þ ILMN þ IBDEK ¼ 0 þ ð But IKB þ IBDEK ¼ f ðzÞ dz and ILMN ¼ @ f ðzÞ dz c

c c

c1

þ

ð ; f ðzÞ dz þ@ f ðzÞ dz ¼ 0 c c þ þ 1 f ðzÞ dz  f ðzÞ dz ¼ 0 ; c c1 þ þ ; f ðzÞ dz ¼ f ðzÞ dz c

c1

1082

Programme 30

The process can, of course, be extended to cases with more than one such singularity.

c c1

c

c c

c

c

The corresponding result then becomes þ þ þ f ðzÞ dz ¼ f ðzÞ dz þ f ðzÞ dz . . . etc: c

c1

c2

Now let us apply these ideas to an example. Example 1

þ

Consider the integral

1 f ðzÞ dz where f ðzÞ ¼ , evaluated round a closed contour z c

in the z-plane. y

We first check the function f ðzÞ ¼

1 for singularities z

and find at once that ............

c O

x

39

1 ceases to be regular z and a singularity occurs at that point

At z ¼ 0, f ðzÞ ¼

The actual position of the closed contour is not specified in the problem, so there are two possibilities: either the contour does enclose the origin, or it does not. Let us consider them in turn. (a) The contour does not enclose the origin. y

No difficulty arises here and by Cauchy’s theorem ............ c

O

x

Complex analysis 2

1083

þ

40

f ðzÞ dz ¼ 0 c

(b) If the contour does enclose the origin, the singularity must be taken into account. Then þ þ þ 1 dz f ðzÞ dz ¼ f ðzÞ dz ¼ c c1 c1 z þ 1 and we attend to evaluating dz c1 z

y c

c O

where c1 is a small circle of radius r entirely within the region bounded by c: If we take an enlarged view of the small circle c1 , we have z ¼ x þ jy which can be expressed in polar form . . . . . . . . . . . . and in exponential form ............

x

y c r O

z θ x

41

z ¼ r ðcos  þ j sin Þ z ¼ re j þ Using z ¼ re j then dz ¼ jre j d and

1 dz ¼ . . . . . . . . . . . . c1 z Complete it

j 2 Because þ

ð 2 ð 2 1 1 j dz ¼ f jre gd ¼ j d ¼ j2 j c1 z 0 re 0 þ þ 1 1 dz ¼ dz ¼ j2 ; cz c1 z

So we have: þ 1 (a) dz ¼ 0 cz þ 1 (b) dz ¼ j2 cz

if the contour c does not enclose the origin if the contour c does enclose the origin. These two constitute an important result, so note them well

42

1084

43

Programme 30

Example 2

þ

1 ðn ¼ 2, 3, 4, . . .Þ. zn Again, a singularity clearly occurs at z ¼ 0 and again also we have two possible cases. f ðzÞ dz where f ðzÞ ¼

Consider the integral

c

(a) If the contour c does not enclose the origin, then by Cauchy’s theorem þ f ðzÞ dz ¼ 0. c

(b) If the contour c does enclose the origin, then we proceed very much as before. Using z ¼ re j , dz ¼ jre j d and zn ¼ r n e jn þ þ Then f ðzÞ dz ¼ f ðzÞ dz c

y r

c1

ð 2

1 ¼ f jre j g d n e jn r 0 ð j 2 jðn1Þ e d ¼ n1 r 0

2 1 jðn1Þ ¼ e ðn  1Þr n1 0

θ x

O c

¼ ............ Finish it off

44

0 Because þ 1 1 dz ¼ fejðn1Þ2  1g n ðn  1Þr n1 cz 1 ¼ fcosðn  1Þ2  j sinðn  1Þ2  1g ðn  1Þr n1 9 = ¼ 0 since cosðn  1Þ2 ¼ 1 n ¼ 2; 3; 4; . . . ; sinðn  1Þ2 ¼ 0 þ

1 dz ¼ 0 for all positive integer values of n other than n ¼ 1, where zn c is any closed contour.

So

c

The particular case when n ¼ 1 we have seen in Example 1. Now we can easily cope with this next example.

Complex analysis 2

1085

Example 3 þ f ðzÞ dz where f ðzÞ ¼ Consider

1 for n ¼ 1; 2; 3; . . . ðz  aÞn c This is a simple extension of the previous piece of work. Here we see that a singularity occurs at z ¼ a and yet again we have two cases to consider. (a)

z-plane

y

If the contour c does not enclose z ¼ a, then by Cauchy’s theorem þ f ðzÞ dz ¼ 0

c (z = a)

c

O

x

(b) y

c

If c encloses A ðz ¼ aÞ we consider separately the cases when

(z = a)

(1) n ¼ 1 and O

(2) n > 1.

x

þ

þ f ðzÞ dz ¼

(1) If n ¼ 1, c

c

1 dz za

þ

Putting z  a ¼ w ; dz ¼ dw ;

1 dz ¼ z  a c

þ

1 dw w c

and this we have already established has a value . . . . . . . . . . . .

45

j 2 þ

þ

(2) If n > 1,

f ðzÞ dz ¼ c

1 dz ¼ ðz  aÞn c

þ

1 dw ¼ 0 for n 6¼ 1. n w c

So collecting our results together, we have the following. þ 1 , n ¼ 1, 2, 3, . . . and c is a closed contour For f ðzÞ dz, where f ðzÞ ¼ ðz  aÞn c þ 1 dz ¼ 0 n 6¼ 1 ðz  aÞn c ¼0 n ¼ 1 and c does not enclose z ¼ a ¼ j2

n ¼ 1 and c does enclose z ¼ a.

You will notice that this is a more general result and includes the results obtained from Examples 1 and 2. Make a note of it, therefore: it is quite important. Then on to Example 4

1086

46

Programme 30

Example 4 Finally, we can go one stage further and consider the contour integral of functions zj4 . such as f ðzÞ ¼ ðz þ jÞðz  2Þ First we express f ðzÞ in partial fractions zj4 A B ¼ þ ðz þ jÞðz  2Þ z þ j z  2 One quick way of finding A and B is by the ‘cover up’ method. (a) To find A, temporarily cover up the denominator ðz þ jÞ in the partial fraction A zj4 and in the function and substitute z þ j ¼ 0, i.e. z ¼ j in ½z þ j ½z þ jðz  2Þ the remainder of the function. A¼

j  j  4 4 þ j2 ¼ ¼2 ; A¼2 j  2 2þj

(b) To find B, cover up the denominator ðz  2Þ in the partial fraction

B and ½z  2

zj4 and substitute z  2 ¼ 0, i.e. z ¼ 2 in the ðz þ jÞ½z  2 remainder of the function. in the function

B ¼ ............

47

B ¼ 1 Because 2j4 2þj 2  j ¼ 2þj ¼ 1

B¼

Therefore the function f ðzÞ becomes f ðzÞ ¼

zj4 2 1   ðz þ jÞðz  2Þ z þ j z  2

Now we can see that there are singularities at . . . . . . . . . . . .

Complex analysis 2

1087

48

z ¼ j and z ¼ 2 y

Denote the singularities by L and M.

þ ;

z-plane

O

x

2 1  dz z2 c zþj  þ  1 1 2 ¼  dz zþj z2 c

zj4 dz ¼ ðz þ jÞðz  2Þ c

þ

So we now have four cases to consider, depending on whether L, M, neither, or both, are enclosed within the contour c. (a) Neither L nor M enclosed y O

Then, once again, by Cauchy’s theorem þ f ðzÞ dz ¼ 0

x

c

c

(b)

L enclosed but not M y

Then, in this case þ f ðzÞ dz ¼ 2ðj 2Þ  0 ¼ j 4

x

O

c

c

(c) M enclosed but not L y O

Here

c

x

þ f ðzÞ dz ¼ 0  ðj 2Þ ¼ j 2 c

1088

Programme 30

(d) Both L and M enclosed y

In this case

þ f ðzÞ dz ¼ . . . . . . . . . . . .

x

O

c

c

49

j 2 Because, when both L and M are enclosed  þ  þ 1 1  dz f ðzÞ dz ¼ 2 zþj z2 c c ¼ 2ðj 2Þ  j 2 ¼ j 2 The key is provided by the results we established earlier. þ 1 ¼ . . . . . . . . . . . . if . . . . . . . . . . . . n dz c ðz  aÞ ¼ . . . . . . . . . . . . if . . . . . . . . . . . . ¼ . . . . . . . . . . . . if . . . . . . . . . . . .

50

þ c

1 dz ðz  aÞn

¼0

if n 6¼ 1

¼0

if n ¼ 1 and c does not enclose z ¼ a

¼ j 2

if n ¼ 1 and c does enclose z ¼ a.

Now for something somewhat different.

Complex analysis 2

1089

Conformal transformation (conformal mapping) A mapping from the z-plane onto the w-plane is said to be conformal if the angles between lines in the z-plane are preserved both in magnitude and in sense of rotation when transformed onto the corresponding lines in the w-plane. y

z-plane

w-plane

v

α

α

O

x

O

u

The angle between two intersecting curves in the z-plane is defined by the angle  ð0    Þ between their two tangents at the point of intersection, and this is preserved. y

z-plane

v

w-plane

α α

O

x

O

u

The essential characteristic of a conformal mapping is that ............ angles are preserved both in magnitude and in sense of rotation

Conditions for conformal transformation The conditions necessary in order that a transformation shall be conformal are as follows. 1 The transformation function w ¼ f ðzÞ must be a regular function of z. That is, it must be defined and single-valued, have a continuous derivative at every point in the region and satisfy the Cauchy–Riemann equations. dw must not be zero, i.e. f 0 ðzÞ 6¼ 0 at a point of intersection. 2 The derivative dz

51

1090

Programme 30

Critical points A point at which f 0 ðzÞ ¼ 0 is called a critical point and, at such a point, the transformation is not conformal. So, if w ¼ f ðzÞ is a regular function, then, except for points at which f 0 ðzÞ ¼ 0, the transformation function will preserve both the magnitude of the angle and its sense of rotation. Now for a short exercise by way of practice. Exercise Determine critical points (if any) which occur in the following transformations w ¼ f ðzÞ. 1 f ðzÞ ¼ ðz  1Þ2

5

f ðzÞ ¼ ð2z þ 3Þ3

2 f ðzÞ ¼ ez 1 3 f ðzÞ ¼ 2 z

6

f ðzÞ ¼ z3 þ 6z þ 9 zj f ðzÞ ¼ zþj

4 f ðzÞ ¼ z þ

7 1 z

8

f ðzÞ ¼ ðz þ 3Þðz  jÞ.

Finish the whole set before checking with the results in the next frame.

52

1 z¼1 2 none

5 z ¼  32 pffiffiffi 6 z ¼ j 2

3 none

7 none

4 z ¼ 1

8 z ¼ 12 ð j  3Þ

All that is required is to differentiate each function and to find for which values of z, f 0 ðzÞ ¼ 0. Now one or two simple examples on conformal mapping. Example 1 Linear transformation w ¼ az þ b, a 6¼ 0, a and b complex. (a) Cauchy–Riemann conditions satisfied. (b) f 0 ðzÞ ¼ a i.e. not zero

; no critical points.

Therefore, the transformation w ¼ az þ b provides conformal mapping throughout the entire z-plane. Example 2 Nonlinear transformation w ¼ z2 . First check for singularities and critical points. These, if any, occur at ............

Complex analysis 2

1091

53

no singularities; critical point at z ¼ 0 Because f 0 ðzÞ ¼ 2z

; f 0 ðzÞ ¼ 0 at z ¼ 0.

Therefore, the transformation is not conformal at the origin. If we choose to express z in exponential form z ¼ x þ jy ¼ re j , then w ¼ z2 ¼ r 2 e j2 , i.e. r is squared and the angle doubled. y

A'

v r

w = z2 r

2

A 2θ

θ

O

O

x

u

y

So ABCD, a section of an annulus of inner and outer radii r1 and r2 respectively, will be mapped onto θ2 O

θ1

............ r1

r2

x

54

v B

C D

2θ2

A 2θ1

O

r 12

r 22

u

The angles at the origin are doubled, but notice that the right angles at A, B, C, D are preserved at A0 ; B0 ; C0 ; D0 ; i.e. the transformation there is conformal. Example 3 Consider the mapping of the circle |z| = 1 under the transformation w ¼ z þ onto the w-plane. First, as always, check for singularities and critical points. We find ............

4 z

1092

Programme 30

55

singularity at z ¼ 0; critical points at z ¼  2 A singularity occurs at z ¼ 0, i.e. f 0 ðzÞ does not exist at z ¼ 0. Also f ðzÞ ¼ z þ

4 z

; f 0 ðzÞ ¼ 1 

4 z2

; f 0 ðzÞ ¼ 0 at z ¼  2.

Therefore the transformation is not conformal at z ¼ 0 and at z ¼  2. 4 In fact, if we carry out the transformation w ¼ z þ on the unit circle z jzj = 1, we get . . . . . . . . . . . . Complete it: it is good revision

56

the ellipse

u2 v2 þ ¼1 52 32

Because w ¼ u þ jv ¼ z þ

4 z

¼ x þ jy þ

4 x þ jy

¼ x þ jy þ

4ðx  jyÞ x2 þ y2

; u¼xþ

4x ; x 2 þ y2

v¼y

4y x 2 þ y2

jzj ¼ 1

; x2 þ y2 ¼ 1 ; u ¼ xð1 þ 4Þ ¼ 5x; u v and y ¼  ; x¼ 5 3 Then x2 þ y2 ¼ 1 gives

v ¼ yð1  4Þ ¼ 3y

u2 v2 þ ¼1 52 32

The image of the unit circle is therefore an ellipse with centre at the origin; semi major axis 5; semi minor axis 3. Now let us move on to a new section

Complex analysis 2

1093

57

Schwarz–Christoffel transformation Example 1 Consider a semi-infinite strip on BC as base, the arrows at A and D indicating that the ordinate boundaries extend to infinity in the positive ydirection and that progression round the boundary is to be taken in the direction indicated.

y

O

a

x

z to the shaded region. Let us apply the transformation w ¼  cos a z Then w ¼ u þ jv ¼  cos a ðx þ jyÞ ¼  cos a  x jy x jy  sin sin ¼  cos cos a a a a Now cos j ¼ cosh  and sin j ¼ j sinh . ; w ¼ u þ jv

x y x y cosh þ j sin sinh a a a a x y x y v ¼ sin sinh ; u ¼  cos cosh ; a a a a ¼  cos

So B and C map onto B0 and C0 where B0 ¼ . . . . . . . . . . . . ; C 0 ¼ . . . . . . . . . . . . B0 : u ¼ 1; v ¼ 0;

58

C0 : u ¼ 1; v ¼ 0

Because ð1Þ at B, x ¼ 0, y ¼ 0

; u ¼ ð1Þð1Þ ¼ 1; v ¼ ð0Þð0Þ ¼ 0

and ð2Þ at C, x ¼ a, y ¼ 0

; u ¼ ð1Þð1Þ ¼ 1; v ¼ ð0Þð0Þ ¼ 0

So we have

v

O

u

1094

Programme 30

Now we map AB, BC, CD onto the w-plane giving A0 B0 ; B0 C0 ; C0 D0 : y ; v¼0 a ; As y decreases from 1 to 0, u increases from 1 to 1.

(a) AB:

x ¼ 0 ; A0 B0 :

u ¼  cosh

v A

B

C u

O

x (b) BC: y ¼ 0 ; B0 C0 : u ¼  cos ; v ¼ 0 a ; As x increases from 0 to a, u increases from 1 to 1. v

u

O

(c) CD: In the same way you can map CD and C0 D0 in the w-plane and the mapping then becomes . . . . . . . . . . . .

59

v

u

O

Because CD: x ¼ a

; C0 D0 : u ¼ cosh

y ; a

v ¼ 0.

Therefore, as y increases from 0 to 1, u increases from 1 to 1: Notice the direction of the arrows. These correspond to the directed travel round the boundary shown in the z-plane. y

v w = f (z)

O

a

x

O

u

The shaded region in the z-plane is on the left-hand side of the boundary as traversed. This maps onto the left-hand side of the image on the w-plane, i.e. the entire upper half of the plane. dw  z dw Note that ¼ sin ; at B ðz ¼ 0Þ and C (z ¼ a), ¼ 0. dz a a dz Therefore, the conformal property does not hold at these points. The internal  angle at B and at C is , while at B0 and C0 it is  . 2

Complex analysis 2

1095

Example 2 Consider an infinite strip in the z-plane bounded by the real axis and z ¼ ja y

Note the arrows. The boundary comes from þ 1 (A) and continues to  1 (C); then returns from  1 (D) to þ 1 (F).

a

O

x

The strip can be considered as a closed figure with the left- and right-hand vertices at infinity. We now map the infinite strip onto the w-plane by the transformation w ¼ ez=a . ; w ¼ u þ jv ¼ ez=a , from which u ¼ ............; v ¼ ............ u ¼ ex=a cos

y ; a

v ¼ ex=a sin

60

y a

Because u þ jv ¼ ez=a ¼ eðxþjyÞ=a ¼ ex=a e jy=a  y y ¼ ex=a cos þ j sin a a y y v ¼ ex=a sin ; u ¼ ex=a cos ; a a Now we map points B and E onto B0 and E0 . (1) B: x ¼ 0; y ¼ a ; B0 : u ¼ 1; v ¼ 0 (2) E: x ¼ 0; y ¼ 0 ; E0 : u ¼ 1; v ¼ 0 i.e. v

O

u

1096

Programme 30

Now we map the lines AB, BC, DE, EF onto the w-plane. (a) AB: y ¼ a ; u ¼ ex=a , v ¼ 0 ; As x decreases from þ1 to 0, u increases from 1 to 1. v

O

u

(b) BC: y ¼ a ; u ¼ ex=a , v ¼ 0 (as for AB) ; As x decreases from 0 to 1, u increases from 1 to 0. v

u

O

(c) Now there is DE which maps onto ............

61

v

O

u

Because (c) DE: y ¼ 0 ; u ¼ ex=a , v ¼ 0 ; As x increases from 1 to 0, u increases from 0 to 1. (d) EF: y ¼ 0 ; u ¼ ex=a , v ¼ 0 (as for DE) ; As x increases from 0 to þ1, u increases from 1 to þ1. v

O

u

Notice that C and D map to the same point, namely u ¼ v ¼ 0. Finally, what about the shaded region in the z-plane? This maps onto ............

Complex analysis 2

1097

62

the upper half of the w-plane because it is on the left-hand side of the directed boundary in the z-plane. y v

w = f (z)

x

O

u

O

The previous two examples have been simple cases of the application of the Schwarz–Christoffel transformation under which any polygon in the z-plane can be made to map onto the entire upper half of the w-plane and the boundary of the polygon onto the real axis of the w-plane. y

w = f (z)

x

O

v

u1

u2

O

u4

u3

u

The process depends, of course, on the right choice of transformation function for any particular polygon, which can be defined by its vertices and the internal angle at each vertex. y

v α2

α1 α3

w = f (z) α4

O

x

u1

u2

O

u3

u4

u

The Schwarz–Christoffel transformation function is given by dz ¼ Aðw  u1 Þ1 =1 ðw  u2 Þ2 =1 ðw  u3 Þ3 =1 . . . dw ð

; z ¼ A ðw  u1 Þ1 =1 ðw  u2 Þ2 =1 . . . ðw  un Þn =1 dw þ B where A and B are complex constants, determined by the physical properties of the polygon. This is not as bad as it looks! Make a careful note of it: then we will apply it

1098

63

Programme 30

Here it is again. y

v α1

α2

w = f (z) α3

α4 u1

x

O

u2

O

u3

u4

u

dz ¼ Aðw  u1 Þ1 =1 ðw  u2 Þ2 =1 ðw  u3 Þ3 =1 . . . dw ð

; z ¼ A ðw  u1 Þ1 =1 ðw  u2 Þ2 =1 . . . ðw  un Þn =1 dw þ B where A and B are complex constants. Three other points also have to be noted. 1 Any three points u1 ; u2 ; u3 on the u-axis can be selected as required. 2 It is convenient to choose one such point, un , at infinity, in which case the relevant factor in the integral above does not occur. 3 Infinite open polygons are regarded as limiting cases of closed polygons where one (or more) vertex is taken to infinity.

Open polygons y

We have already introduced these in Examples 1 and 2 of this section. In Example 1, the semi-infinite strip is a case of a triangle with one vertex that is

O

............

x

64

taken to infinity in the y-direction y

In Example 2, the infinite strip is a case of a double triangle, or quadrilateral, with two vertices taken to infinity. O

x

An open polygon with n sides with one vertex at infinity will have ðn  1Þ internal angles. An open polygon with n sides with two vertices at infinity will have ðn  2Þ internal angles. Now for an example to see how all this works.

Complex analysis 2

1099

65

Example 3 To determine the transformation that will map the semi-infinite strip ABCD onto the w-plane so that the images of B and C occur at u ¼ 1 and u ¼ 1, respectively, and the shaded region maps onto the upper half of the w-plane. y v

w = f (z)

x

O

O

u

In this case, B0 is u1 ¼ 1 and C 0 is u2 ¼ 1. The corresponding internal angles are:   and at C ðz ¼ 0Þ, 2 ¼ . at B ðz ¼ jaÞ, 1 ¼ 2 2 So we have dz ¼ Aðw þ 1Þð=2Þ=1 ðw  1Þð=2Þ=1 dw ¼ Aðw þ 1Þ1=2 ðw  1Þ1=2

where A is a complex constant

¼ Aðw2  1Þ1=2 K ¼ Kð1  w2 Þ1=2 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  w2 ð K ; z ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dw ¼ . . . . . . . . . . . . 1  w2

66

z ¼ K arcsin w þ B

; arcsin w ¼

zB K

; w ¼ sin

zB K

Now we have to find B and K. (a) We require B ðz ¼ jaÞ to map onto B0 ðw ¼ 1Þ ja  B K ja  B  ; ¼ ; 2ja  2B ¼ K K 2 ;  1 ¼ sin

(b) We also require C ðz ¼ 0Þ to map onto C 0 ðw ¼ 1Þ ; 1 ¼ sin ; 

B  ¼ K 2

;  2B ¼ K

Then, from (1) and (2), B ¼ . . . . . . . . . . . . ; K ¼ . . . . . . . . . . . .

ð1Þ 0B K ð2Þ

1100

Programme 30

67

B¼

ja ; 2

K¼

ja 

  z  ðjaÞ=2   jz ¼ sin jz þ ¼ cos ; w ¼ sin ja= a 2 a But cos j ¼ cosh 

; w ¼ cosh

z a

To verify that this is the required transformation, let us apply it to the figure given in the z-plane. We will do that in the next frame

68

We have z ðx þ jyÞ ¼ cosh a a x jy x jy þ sinh sinh ; u þ jv ¼ cosh cosh a a a a

w ¼ u þ jv ¼ cosh

But cosh j ¼ cosh  and sinh j ¼ j sin  x y x y ; u þ jv ¼ cosh cos þ j sinh sin a a a a x y x y ; u ¼ cosh cos ; v ¼ sinh sin a a a a First map the points B and C onto B0 and C0 in the wplane.

y a

π 2 π 2

O

69

B0 : . . . . . . . . . . . . ; C0 : . . . . . . . . . . . . x

B0 : u ¼ 1; v ¼ 0;

C0 : u ¼ 1; v ¼ 0

Because B: x ¼ 0, y ¼ a

; B0 : u ¼ cos  ¼ 1, v ¼ 0 ; B0 : u ¼ 1, v ¼ 0

C: x ¼ 0; y ¼ 0

; C0 : u ¼ 1; v ¼ 0

; C0 : u ¼ 1; v ¼ 0.

v

O

Now we map AB, BC, CD in turn.

u

Complex analysis 2

1101

x , v¼0 a ; As x decreases from 1 to 0, u increases from 1 to 1.

(a) AB: y ¼ a

; u ¼  cosh

v

O

(b) BC:

u

)

(c) CD:

Complete the working and show the mapped region

which is . . . . . . . . . . . .

70

v

u

O

Because we have y ; v¼0 a ; As y decreases from a to 0, u increases from 1 to 1.

(b) BC: x ¼ 0 ; u ¼ cos

CD: y ¼ 0

; u ¼ cosh

x ; a

v¼0

; As x increases from 0 to 1, u increases from 1 to 1. In each plane, the shaded region is on the left-hand side of the boundary. We will now finish with one further example. So move on

71

Example 4 y v

π 3 O

x

O

u

Determine the transformation function w ¼ f ðzÞ that maps the infinite sector in the z-plane onto the upper half of the w-plane with points B and C mapping onto B0 and C0 as shown. The transformation function w ¼ f ðzÞ is given by . . . . . . . . . . . .

1102

Programme 30

72

dz ¼ Aðw  u1 Þ1 =1 ðw  u2 Þ2 =1 . . . ðw  un Þn =1 dw

At B, 1 ¼

 . 3

α2 = π

At C, 2 ¼ .

With that reminder, you can now work through on your own, just as we did before, finally obtaining w ¼ ............

73

w ¼ z3 Check with the working. dz ¼ Aðw  0Þð=3Þ=1 ðw  1Þ=1 dw ¼ Aw2=3 ðw  1Þ0 ¼ Aw2=3 ; z ¼ 3Aw1=3 þ B ¼ Kw1=3 þ B  3 zB ; w¼ K To find B and K

 3  z 3 B ; B¼0 ; w¼ K K  3 1 (b) At C: z ¼ 1 At C0 : w ¼ 1 ; 1 ¼ ; K ¼ 1 ; w ¼ z3 K

(a) At B: z ¼ 0 At B0 : w ¼ 0 ; 0 ¼

; the transformation function is w ¼ z3 Finally, as a check – and a little more valuable practice – apply the function w ¼ z3 to the region shaded in the z-plane. w ¼ u þ jv ¼ ðx þ jyÞ3 ¼ x3 þ 3x2 ðjyÞ þ 3xðjyÞ2 þ ðjyÞ3 ; u ¼ ............; v ¼ ............

Complex analysis 2

1103

u ¼ x3  3xy2 ;

74

v ¼ 3x2 y  y3

At B: x ¼ 0, y ¼ 0

; u ¼ 0, v ¼ 0 ; B0 : u ¼ 0, v ¼ 0

At C: x ¼ 1, y ¼ 0

; u ¼ 1, v ¼ 0 ; C0 : u ¼ 1, v ¼ 0 v

O

u

Now we map AB, BC, CD onto A0 B0 , B0 C0 , C0 D0 . pffiffiffi AB: y ¼ 3x ; u ¼ x3  9x3 ¼ 8x3 ; v ¼ 0 ; As x decreases from 1 to 0, u increases from  1 to 0. You can now deal with BC and CD in the same way and finally show the transformed region. So we get . . . . . . . . . . . .

75

Here is the remaining working. BC:

y ¼ 0 ; u ¼ x3 ; v ¼ 0

; As x increases from 0 to 1, u increases from 0 to 1. CD:

y ¼ 0 ; u ¼ x3 ; v ¼ 0

; As x increases from 1 to 1, u increases from 1 to 1. So we have v

O

u

The shaded region is to the left of the directed boundary in the z-plane. This therefore maps onto the region to the left of the directed real axis in the w-plane, i.e. the upper half of the plane. We have just touched on the fringe of the work on Schwarz– Christoffel transformation. The whole topic of mapping between planes has applications in fluid mechanics, heat conduction, electromagnetic theory, etc. and it is at times convenient to solve a problem relating to the z-plane by transforming to the upper half of the w-plane and later to transform back to the z-plane. The transformation function can be operated in either direction. And that is it. The Review summary follows and the Can you? checklist. Then on to the Test exercise and the Further problems for additional practice.

1104

Programme 30

Review summary 30 1 Differentiation of a complex function  dw f ðz0 þ zÞ  f ðz0 Þ ¼ f 0 ðzÞ ¼ Lim w ¼ f ðzÞ dz z z!0 2 Regular (or analytic) function w ¼ f ðzÞ is regular at z0 if it is defined, single-valued and has a derivative at every point at and around z ¼ z0 . 3 Singularities or singular points – points at which f ðzÞ ceases to be regular. 4 Cauchy–Riemann equations test whether w ¼ f ðzÞ has a derivative f 0 ðzÞ at z ¼ z0 . w ¼ u þ jv ¼ f ðzÞ where z ¼ x þ jy. @u @v @v @u ¼ and ¼ Then @x @y @x @y 5 If a function of two real variables f ðx, yÞ satisfies Laplace’s equation @ 2 f ðx, yÞ @ 2 f ðx, yÞ þ ¼ 0 then f ðx, yÞ is an harmonic function. The real and @x2 @y2 imaginary parts of an analytic function are both harmonic and form a conjugate pair of functions. 6 Complex integration ð ð ð ð wdz ¼ f ðzÞdz ¼ ðudx  vdyÞ þ j ðvdx þ udyÞ 7 Contour integration – evaluation of line integrals in the z-plane. 8 Cauchy’s theorem If f ðzÞ is regular at every point within and on closed þ curve c, then f ðzÞ dz ¼ 0. c

9 Deformation of contours

(z = a)

(a)

c

c c

c

(b)

c

(c)

(a) Singularity at A (b) Restored to a closed curve þ þ (c) f ðzÞ dz ¼ f ðzÞ dz: c c1 þ 1 For f ðzÞ dz where f ðzÞ ¼ n ¼ 1, 2, 3, . . . ðz  aÞn c þ 1 ¼0 if n 6¼ 1 n dz c ðz  aÞ ¼0 if n ¼ 1 and c does not enclose z ¼ a ¼ j 2 if n ¼ 1 and c does enclose z ¼ a.

Complex analysis 2

1105

10 Conformal transformation – mapping in which angles are preserved in size and sense of rotation. Conditions 1 w ¼ f ðzÞ must be a regular function of z. dw , ¼ 6 0 at the point of intersection. dz 0 If f ðzÞ ¼ 0 at z ¼ z0 , then z0 is a critical point.

2 f 0 ðzÞ, i.e.

11 Schwarz–Christoffel transformation maps any polygon in the z-plane onto the entire upper half of the w-plane and the boundary of the polygon onto the real axis of the w-plane. dz ¼ Aðw  u1 Þ1 =1 ðw  u2 Þ2 =1 . . . ðw  un Þn =1 dw (i) Any three points u1 ; u2 ; u3 can be selected on the u-axis. (ii) One such point can be chosen at infinity. (iii) Infinite open polygons are regarded as limiting cases of closed polygons.

Can you? Checklist 30 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can:

Frames

. Appreciate when the derivative of a function of a complex variable exists?

1

to

3

3

to

6

7

to

12

13

to

22

23

to

28

29

to

36

Yes

No

. Understand the notions of regular functions and singularities and be able to obtain the derivative of a regular function from first principles?

Yes

No

. Derive the Cauchy–Riemann equations and apply them to find the derivative of a regular function?

Yes

No

. Understand the notion of an harmonic function and derive a conjugate function?

Yes

No

. Evaluate line and contour integrals in the complex plane?

Yes

No

. Derive and apply Cauchy’s theorem?

Yes

No

1106

Programme 30

. Apply Cauchy’s theorem to contours around regions that contain singularities?

Yes

37

to

49

No

. Define the essential characteristics of and conditions for a conformal mapping?

Yes

50 and 51

No

. Locate critical points of a function of a complex variable?

Yes

51 and 52

No

. Determine the image in the w-plane of a figure in the z-plane under a conformal transformation w ¼ f ðzÞ?

Yes

52

to

56

57

to

75

No

. Describe and apply the Schwarz–Christoffel transformation?

Yes

No

Text exercise 30 1 Determine where each of the following functions fails to be regular. z2 (a) w ¼ z3 þ 4 (d) w ¼ ðz  4Þðz þ 1Þ z x  jy (e) w ¼ 2 (b) w ¼ . zþ5 x þ y2 (c) w ¼ e2zþ4 2 Demonstrate that each of the following is harmonic and obtain the conjugate function. (a) uðx, yÞ ¼ sinh x cos y (b) uðx, yÞ ¼ 4yð1 þ 3xÞ.

þ f ðzÞ dz where f ðzÞ ¼ z2 round the

3 Verify Cauchy’s theorem by evaluating c

rectangle formed by joining the points z ¼ 2 þ j, z ¼ 2 þ j4, z ¼ j4, z ¼ j: þ 3z  6  j round the contour 4 Evaluate the integral f ðzÞdz where f ðzÞ ¼ ðz  jÞðz  3Þ c jzj ¼ 2: 5 Determine critical points, if any, at which the following transformation functions w ¼ f ðzÞ fail to be conformal. 2 (a) w ¼ z4 (d) w ¼ z þ z 2 (b) w ¼ z3  3z (e) w ¼ eðz Þ zþj (c) w ¼ e1z (f) w ¼ z  j.

Complex analysis 2

1107

6 Determine the Schwarz–Christoffel transformation function w ¼ f ðzÞ that will map the semi-infinite strip shaded in the z-plane onto the upper half of the w-plane, so that the image of B is B0 (w ¼ 1) and that of C is C0 (w ¼ 0). Obtain the image of the point D.

y

O

x

Further problems 30 1 Verify Cauchy’s theorem for the closed path c consisting of three straight lines joining A ð1 þ jÞ, B ð3 þ j3Þ, C ð1 þ j3Þ where f ðzÞ ¼ z  1 þ j. 2 If z ¼ 2 þ jy is mapped onto the w-plane under the transformation 1 w ¼ f ðzÞ ¼ , show that the locus of w is a circle with centre w ¼ 0:25 and z radius 0.25. 3 Determine the image in the w-plane of the circle jz  2j ¼ 1 in the z-plane under the transformation w ¼ ð1  jÞz þ 3. 4 The unit circle jzj ¼ 1 in the z-plane is generated in an anticlockwise manner z from the point A ðz ¼ 1Þ and is transformed onto the w-plane by w ¼ . z2 Determine the locus of w and the direction in which it is generated. 5 Find the conjugate function of each of the following. (a) uðx, yÞ ¼ x2  2x  y2 (b) uðx, yÞ ¼ x3  3xy2  x2 þ y2 þ x (c) uðx, yÞ ¼ 2yðx  1Þ (d) uðx, yÞ ¼ ex y cos 2xy. þ 5z  2  j3 around the closed contour c for 6 Evaluate f ðzÞ dz where f ðzÞ ¼ ðz  jÞðz  1Þ c the two cases when 2

2

(a) c is the path jzj ¼ 2 (b) c is the path jz  1j ¼ 1. 5z þ j 7 If f ðzÞ ¼ , evaluate ðz  jÞðz þ j2Þ 3 (a) jz  1j ¼ 1; (b) jzj ¼ ; 2

þ f ðzÞdz along the contours c

(c) jzj ¼ 3.

j ðw þ zÞ is entirely real, then jwj ¼ jzj. 8 If z ¼ x þ jy and w ¼ f ðzÞ, show that, if wz þ 3z  j5 9 Evaluate f ðzÞ dz, where f ðzÞ ¼ , around the perimeter ðz þ 1  j2Þðz  2  jÞ c of the rectangle formed by the lines z ¼ 1, z ¼ j3, z ¼ 2, z ¼ j.

1108

Programme 30

10

þ 8z2  2 , evaluate f ðzÞ dz along the contour c where c is the zðz  1Þðz þ 1Þ c triangle joining the points z ¼ 2, z ¼ j, z ¼ 1  j.

11

1 (a) For the transformation w ¼ z þ , state (1) singularities, (2) critical z points.

If f ðzÞ ¼

(b) Apply w ¼ z þ 12

13

14

1 to map the circle jzj ¼ 2 onto the w-plane. z

Find the images in the w-plane of (a) the line y ¼ 0 and (b) the line y ¼ x that zj result from the mapping w ¼ . Show that the curves intersect at the zþj points ð1; 0Þ in the w-plane and determine the angle at which they intersect. j ð1 þ zÞ to map the unit circle jzj ¼ 1 in the 1z z-plane onto the w-plane. Determine also the image in the w-plane of the region bounded by jzj ¼ 1 and inside the circle. Use the transformation w ¼

Determine the transformation that will map the semi-infinite strip shown, onto the upper half of the w-plane, where the image of B is B0 ðw ¼ 1Þ and that of C is C0 ðw ¼ 1Þ.

y

–a

O

a

x

Frames 1 to 34

Programme 31

Complex analysis 3 Learning outcomes When you have completed this Programme you will be able to: Expand a function of a complex variable about the origin in a Maclaurin series Determine the circle and radius of convergence of a Maclaurin series expansion Recognize singular points in the form of poles of order n, removable and essential singularities Expand a function of a complex variable about a point in the complex plane in a Taylor series, transforming the coordinates with a shift of origin Expand a function of a complex variable about a singular point in a Laurent series Recognize the principal and analytic parts of the Laurent series and link the form of the principal part to the type of singularity Recognize the residue of a Laurent series and state the Residue theorem Calculate the residues at the poles of an expression without resort to deriving the Laurent series Evaluate certain types of real integrals using the Residue theorem

1109

1110

Programme 31

Maclaurin series 1

You will recall that the Maclaurin series expansion of the function of a real variable x with output f ðxÞ is given as f ðxÞ ¼ f ð0Þ þ xf 0 ð0Þ þ x2

f 00 ð0Þ f 000 ð0Þ f ðnÞ ð0Þ þ x3 þ    þ xn þ  2! 3! n!

This is an infinite series expansion of f ðxÞ about the point x ¼ 0. Because the series on the right-hand side of this equation contains an infinite number of terms, the right-hand side may only converge for a restricted set of values of x. Consequently, this expansion is only valid for that restricted set of values. For example, the expression f ðxÞ ¼ ð1  xÞ1 has the Maclaurin series expansion ............

2

f ðxÞ ¼ 1 þ x þ x2 þ x3 þ    þ xn þ    Because f ðxÞ ¼ ð1  xÞ1 and so f ð0Þ ¼ ð1  0Þ1 ¼ 1 f 0 ðxÞ ¼ ð1  xÞ2 and so f 0 ð0Þ ¼ ð1  0Þ2 ¼ 1 f 00 ðxÞ ¼ 2ð1  xÞ3 and so f 00 ð0Þ ¼ 2ð1  0Þ3 ¼ 2 f 000 ðxÞ ¼ 3!ð1  xÞ4 and so f 000 ð0Þ ¼ 3!ð1  0Þ4 ¼ 3! .. .. . . f ðnÞ ðxÞ ¼ n!ð1  xÞðnþ1Þ and so f ðnÞ ð0Þ ¼ n!ð1  0Þðnþ1Þ ¼ n! Therefore, substituting into the Maclaurin series expansion, we find f 00 ð0Þ f 000 ð0Þ f ðnÞ ð0Þ þ x3 þ    þ xn þ  2! 3! n! 2! 3! n! ¼ 1 þ x  1 þ x2  þ x3  þ    þ xn  þ    2! 3! n! ¼ 1 þ x þ x2 þ x3 þ    þ xn þ   

f ðxÞ ¼ f ð0Þ þ xf 0 ð0Þ þ x2

This same result could also be derived by using the binomial theorem or even by performing the long division of 1 by 1  x. However, performing the algorithmic procedure is one thing, but knowing that the result of the procedure is valid is another. To determine the validity of the expansion we resort to convergence tests, and in this case we use the ratio test. To refresh your memory, the ratio test for the infinite series f ðxÞ ¼ a0 ðxÞ þ a1 ðxÞ þ a2 ðxÞ þ a3 ðxÞ þ    þ an ðxÞ þ    is that given   anþ1 ðxÞ   ¼ L then if Lim   n!1 an ðxÞ

L < 1 the series converges L > 1 the series diverges L ¼ 1 the test fails and an alternative convergence test is required.

Complex analysis 3

1111

Applying the ratio test to the Maclaurin series expansion f ðxÞ ¼ 1 þ x þ x2 þ x3 þ    þ xn þ    tells us that The series converges for . . . . . . . . . . . . The series diverges for . . . . . . . . . . . . The test fails for . . . . . . . . . . . . The series converges for  1 < x < 1 The series diverges for x < 1 or x > 1 The test fails for x ¼ 1 Because

 nþ1    anþ1 ðxÞ x      Lim   ¼ Lim  xn  ¼ Lim jxj ¼ jxj, so a ðxÞ n!1 n n!1 n!1

if jxj < 1, that is 1 < x < 1, the series converges and so the expansion is valid jxj > 1, that is x < 1 or x > 1, the series diverges and so the expansion is invalid jxj ¼ 1, that is x ¼ 1, the ratio test fails to give a conclusion. By inspection, when x ¼ 1 the series clearly diverges and when x ¼ 1 the sum of terms alternates between 1 and 0 as each successive term is added. Clearly the series does not converge and so, therefore, it must diverge when x ¼ 1. Everything that has been said about the Maclaurin series expansion of an expression involving a real variable x can equally be said about an expression involving a complex variable z. That is, if f ðzÞ is a function in the complex variable z, analytic at z ¼ 0, then the Maclaurin series expansion is f ðzÞ ¼ f ð0Þ þ zf 0 ð0Þ þ z2

f 00 ð0Þ f 000 ð0Þ þ z3 þ  2! 3!

So, the Maclaurin series expansion of f ðzÞ ¼ sin z is . . . . . . . . . . . .

3

1112

Programme 31

4

f ðzÞ ¼ z 

z3 z5 ð1Þn z2nþ1 þ   þ þ  3! 5! ð2n þ 1Þ!

Because f ðzÞ ¼ sin z and so f ð0Þ ¼ sin 0 ¼ 0 f 0 ðzÞ ¼ cos z and so f 0 ð0Þ ¼ cos 0 ¼ 1 f 00 ðzÞ ¼  sin z and so f 00 ð0Þ ¼  sin 0 ¼ 0 f 000 ðzÞ ¼  cos z and so f 000 ð0Þ ¼  cos 0 ¼ 1 .. .

.. . Therefore

f 00 ð0Þ f 000 ð0Þ þ z3 þ  2! 3! 0 ð1Þ ¼ 0 þ z  1 þ z2  þ z3  þ  2! 3!

f ðzÞ ¼ f ð0Þ þ zf 0 ð0Þ þ z2

¼z

z3 z5 ð1Þn z2nþ1 þ   þ þ  3! 5! ð2n þ 1Þ!

Furthermore, applying the ratio test tells us that this series expansion is valid for ............

5

all finite values of z Because     ð1Þnþ1 z2ðnþ1Þþ1 =½2ðn þ 1Þ þ 1! anþ1 ðzÞ    ¼ Lim  Lim     ð1Þn z2nþ1 =½2n þ 1! n!1 an ðzÞ n!1      z2 ¼0 1 the series diverges and so the expansion is invalid jzj ¼ 1 the ratio test fails We shall look at the case jzj ¼ 1 a little later. Move to the next frame

1114

Programme 31

Radius of convergence 8

We have just seen that the Maclaurin expansion of lnð1 þ zÞ is valid for jzj < 1. This inequality defines the interior of a circle of radius 1 centred on the origin, namely z ¼ 1e j . y 1

–1

1

x

–1

This means that the expansion is valid for all z-values lying within this circle. The radius of the circle within which a series expansion is valid is called the radius of convergence of the series and the circle is called the circle of convergence. Example To find the infinite series expansion and radius of convergence of the expression z z f ðzÞ ¼ , we progress in stages, noting that ¼ zð1  3zÞ2 . We 2 ð1  3zÞ ð1  3zÞ2 expand ð1  3zÞ2 first. By the binomial theorem, the expansion of ð1  3zÞ2 is ð1  3zÞ2 ¼ . . . . . . . . . . . .

9

ð1  3zÞ2 ¼ 1 þ 6z þ 27z2 þ 108z3 þ 405z4 þ    Because ð1  3zÞ2 ¼

1 þ ð2Þ  ð3zÞ þ

ð2Þð3Þ  ð3zÞ2 2!

! ð2Þð3Þð4Þ  ð3zÞ3 þ ... þ 3!

 ¼ 1 þ 6z þ 3ð3zÞ2  4ð3zÞ3 þ 5ð3zÞ4 þ . . .

 þ ð1Þn ðn þ 1Þð3Þn zn þ . . .

¼ 1 þ 6z þ 27z2 þ 108z3 þ 405z4 þ . . . þ ðn þ 1Þ3n zn þ . . . and so zð1  3zÞ2 ¼ z þ 6z2 þ 27z3 þ 108z4 þ 405z5 þ . . . þ ðn þ 1Þ3n znþ1 þ . . . The radius of convergence is then . . . . . . . . . . . .

Complex analysis 3

1115

10

1=3 Because The general term of the expansion is an ðzÞ ¼ ðn þ 1Þ3n znþ1 and so the ratio test tells us that       nþ1 nþ2    anþ1 ðzÞ   ¼ Lim ðn þ 2Þ3 z  ¼ Lim 3ðn þ 2Þz ¼ j3zj Lim      n nþ1 ðn þ 1Þ  n!1 an ðzÞ n!1 ðn þ 1Þ3 z n!1 So, if j3zj < 1, that is jzj < 1=3, then the series converges and the expansion is valid. The radius of convergence is therefore 1=3. Move to the next frame

Singular points Any point at which f ðzÞ fails to be analytic, that is where the derivative does not exist, is called a singular point (also called a singularity). For example f ðzÞ ¼

1 z1

is analytic everywhere in the finite complex plane except at the point z ¼ 1 where not only is the derivative f 0 ðzÞ not defined but neither is f ðzÞ. Accordingly, the point z ¼ 1 is a singular point. There are different types of singular points, for now we shall look at just two of them.

Poles If f ðzÞ has a singular point at z0 and for some natural number n,   Lim ðz  z0 Þn f ðzÞ ¼ L 6¼ 0 then the singular point is called a pole of order n. z!z0

For example f ðzÞ ¼

2z ðz þ 4Þ2

has a singular point at z ¼ 4 and because n o Lim ðz þ 4Þ2 f ðzÞ ¼ Lim f2zg ¼ 8 6¼ 0 z!4

z!4

the singularity is a pole of order 2 (also called a double pole).

11

1116

Programme 31

Removable singularities If f ðzÞ has a singular point at z0 but Lim ff ðzÞg exists then the singular point is z!z0

called a removable singularity. For example f ðzÞ ¼

sin z z

 sin z ¼ 1 and so the singularity z z!0 at z ¼ 0 is a removable singularity. We can see this from the Maclaurin series expansion of f ðzÞ where   sin z 1 z3 z5 z2 z4 ¼ z  þ   ¼ 1  þ   f ðzÞ ¼ z z 3! 5! 3! 5! sin z , we can define f ð0Þ ¼ 1 in While we cannot substitute z ¼ 0 into f ðzÞ ¼ z complete consistency with the series expansion. In this sense the singularity at z ¼ 0 is removable by virtue of the fact that we can assign a value to f ðzÞ at the singularity which is consistent with the series expansion. has a singular point at z ¼ 0. However, Lim

Move to the next frame

Circle of convergence 12

When an expression is expanded in a Maclaurin series, the circle of convergence is always centred on the origin and the radius of convergence is determined by the location of the first singular point met as jzj increases from jzj ¼ 0. For example, the Maclaurin series expansion of f ðzÞ ¼ lnð1 þ zÞ is lnð1 þ zÞ ¼ z 

z2 z3 ð1Þnþ1 zn þ   þ þ  2 3 n

which is valid inside the circle of convergence jzj ¼ 1. The first singular point met by this function as jzj increases from zero is at z ¼ 1, for at that point lnð1 þ zÞ is not defined and the series 1 

1 1 1       2 3 n

diverges – it is the negative of the harmonic series. Hence the radius of convergence is 1. When z ¼ 1, substitution into the series expansion gives ln 2 ¼ 1 

1 1 ð1Þnþ1 þ   þ þ  2 3 n

Complex analysis 3

1117

The right-hand side is the alternating harmonic series which we know converges by the alternating sign test which states that if the magnitude of the terms decreases and the signs alternate then the series converges. Now we see that it converges to ln 2. Notice that the circle of convergence is identified by the location of the first singularity as jzj increases from jzj ¼ 0. This does not mean that the function is singular at all points on the circle of convergence. There are times when it is desirable to have a series expansion of an expression that is singular at the origin. Because the Maclaurin expansion requires the function to be analytic everywhere within the circle of convergence which is centred on the origin, we cannot use that method. Fortunately, we do have a method of expanding a function about any point in the complex plane – this is Taylor’s expansion. Move to the next frame

Taylor’s series Provided f ðzÞ is analytic inside and on a simple closed curve c, the Taylor series expansion of f ðzÞ about the point z0 which is interior to c is given as f ðzÞ ¼ f ðz0 Þ þ ðz  z0 Þf 0 ðz0 Þ þ þ

ðz  z0 Þn f n!

ðnÞ

ðz0 Þ

ðz  z0 Þ2 f 00 ðz0 Þ þ  2!

þ 

where here, the point z0 is the centre of the circle of convergence. The circle of convergence is given as jz  z0 j ¼ R. That is z  zo ¼ Re j or z ¼ z0 þ Re j where R is the radius of convergence. Notice that Maclaurin’s series is a special case of Taylor’s series where z0 ¼ 0.

y

R z0

x

Example 1 in a Taylor series about the point z ¼ 1 and find the values zþ1 of z for which the expansion is valid. The simplest way of doing this is to perform a coordinate transformation that moves the origin of the new coordinate to the point z ¼ 1 and then derive the series about the new origin. To do this we define a new complex variable u ¼ z  1 so that z ¼ u þ 1 and so 1 1 1 u1 becomes ¼ ð2 þ uÞ1 ¼ 1þ . zþ1 uþ2 2 2 Expand f ðzÞ ¼

13

1118

Programme 31

The expansion of this expression can now be derived using either Maclaurin or, as here, the binomial theorem to obtain   1 1 u ð1Þð2Þ u2 ¼ 1 þ ð1Þ þ þ uþ2 2 2 2! 2 ¼

1 u u2 u3  þ  þ  2 4 8 16

Transforming back to the original variable z gives 1 1 z  1 ðz  1Þ2 ðz  1Þ3 ¼  þ  þ  zþ1 2 4 8 16   u  z  1    ¼ 1 or jz  1j ¼ 2. The circle of convergence is given by   ¼ 1, that is  2 2  Consequently, this series expansion is valid provided z is inside the circle defined by z  1 ¼ 2e j that is z ¼ 1 þ 2e j By the same reasoning, the Taylor series expansion of f ðzÞ ¼ cos z about the point z ¼ =3 is ............

14

pffiffiffi 1 ðz  =3Þ2 pffiffiffi ðz  =3Þ3 ðz  =3Þ4 1  3ðz  =3Þ  þ 3 þ   2 2! 3! 4! Because If u ¼ z  =3 then cos z ¼ cosðu þ =3Þ ¼ cos u cos =3  sin u sin =3  pffiffiffi 1 cos u  3 sin u ¼ 2 



 pffiffiffi 1 u2 u4 u3 u5 1  þ    3 u  þ   ¼ 2 2! 4! 3! 5!   pffiffiffi pffiffiffi u3 pffiffiffi u5 1 u2 u4 ¼ 1  þ      3u þ 3  3     2 2! 4! 3! 5!   2 3 4 5 pffiffiffi pffiffiffi u pffiffiffi u 1 u u 1  3u  þ 3 þ  3     ¼ 2 2! 3! 4! 5! ¼

pffiffiffi 1 ðz  =3Þ2 pffiffiffi ðz  =3Þ3 1  3ðz  =3Þ  þ 3 2 2! 3! ðz  =3Þ4 þ   4!

! for z < 1

!

Complex analysis 3

1119

Laurent’s series Sometimes a valid series expansion of a function is required within a specific region of the complex plane that contains a singular point. In this case we cannot avoid the singular point as we did with Taylor’s series by expanding about an alternative non-singular point, because then we move away from part of the specified region. To accommodate this case we can use the Laurent series expansion which provides a series expansion valid within an annular region centred on the singular point. y c1

c

c2 z0

x

Let f ðzÞ be singular at z ¼ z0 and let c1 and c2 be two concentric circles centred on z0 . Then if f ðzÞ is analytic in the annular region between c1 and c2 and if c is any concentric circle lying within the annular region between c1 and c2 we can expand f ðzÞ as a Laurent series in the form a2 a1 f ðzÞ ¼    þ þ a0 þ a1 ðz  z0 Þ þ a2 ðz  z0 Þ2 þ    þ 2 ðz  z0 Þ ðz  z0 Þ 1 X ¼ an ðz  z0 Þn n¼1

where an ¼

1 2j

þ

f ðzÞ c

ðz  z0 Þnþ1

dz

Example e3z

in a Laurent series about the point z ¼ 2 and determine the nature ðz  2Þ4 of the singularity at z ¼ 2.

Expand

f ðzÞ ¼

e3z

and f 0 ðzÞ ¼

e3z ð3z  10Þ

so f ðzÞ is analytic everywhere except at ðz  2Þ5 ðz  2Þ z ¼ 2. The first thing we must do is to transform the coordinate system by shifting the origin to the point z ¼ 2 by defining u ¼ z  2 so that z ¼ u þ 2. Then 4

e3z 4

ðz  2Þ

¼

e3ðuþ2Þ e3u ¼ e6 4 . 4 u u

15

1120

Programme 31

Now we can expand using the Maclaurin series expansion ( ) e6 ð3uÞ2 ð3uÞ3 ð3uÞ4 ð3uÞ5 ¼ 4 1 þ 3u þ þ þ þ þ  u 2! 3! 4! 5! ( ) 1 3u ð3uÞ2 ð3uÞ3 ð3uÞ4 ð3uÞ5 6 þ þ þ þ þ þ  ¼e u4 u4 2!u4 3!u4 4!u4 5!u4  1 3 9 27 81 243u þ þ þ  ¼ e6 4 þ 3 þ 2 þ u u 2u 6u 24 120 ( ) 1 3 9 9 27 81ðz  2Þ 6 ¼e þ þ þ þ þ þ 40 ðz  2Þ4 ðz  2Þ3 2ðz  2Þ2 2ðz  2Þ 8 This series converges for all finite z except z ¼ 2 at which point there is a pole of order 4. The part of the Laurent series that contains negative powers of the variable is called the principal part of the series and the remaining terms constitute what is called the analytic part of the series. If, in the principal part the highest power of 1=z is n, then the function possesses a pole of order n; and if the principal part contains an infinite number of terms, the function possesses an essential singularity. Now you try one

16

The Laurent series expansion of z2 cos

1 about the point z ¼ 0 z

is . . . . . . . . . . . . valid for . . . . . . . . . . . . at which point there is . . . . . . . . . . . .

17

1 1 1 þ  þ    valid for all z 6¼ 0 2! 4!z2 6!z4 at which point there is an essential singularity z2 

Because 1 1 1 and f 0 ðzÞ ¼ 2z cos þ sin and so f ðzÞ is analytic everywhere z z z except at z ¼ 0. Expanding about z ¼ 0 gives ! 1 ð1=zÞ2 ð1=zÞ4 ð1=zÞ6 2 2 z cos ¼ z 1  þ  þ  z 2! 4! 6! f ðzÞ ¼ z2 cos

¼ z2 

1 1 1 þ  þ  2! 4!z2 6!z4

valid for all z 6¼ 0, at which point there is an essential singularity because there is an infinity of terms in the principal part of the series. z valid for Try another. The Laurent series expansion of ðz þ 2Þðz þ 4Þ 2 < jzj < 4 is . . . . . . . . . . . .

Complex analysis 3

 þ

1121

8 4 2 1 1 z z2 z3 þ  þ  þ   þ  z4 z3 z2 z 2 8 32 128

Because z 2 1 ¼  ðz þ 2Þðz þ 4Þ z þ 4 z þ 2 If jzj > 2 then we can write

(separating into partial fractions)

1 1 ð1 þ 2=zÞ1 ¼ ¼ z þ 2 zð1 þ 2=zÞ z

and because jzj > 2, that is, j2=zj < 1, we can now use the binomial theorem  1 1 1 2 4 8 1 2 4 8 ¼ ¼ 1  þ 2  3 þ  ¼  2 þ 3  4 þ  z þ 2 zð1 þ 2=zÞ z z z z z z z z and if jzj < 4 then

 2 1 1 z z2 z3 ¼ ¼ 1 þ  þ  z þ 4 2ð1 þ z=4Þ 2 4 16 64 ¼

1 z z2 z3  þ  þ  2 8 32 128

Note the expansion of ð1 þ z=4Þ1 which is valid for jz=4j < 1, that is jzj < 4. The first expansion for jzj > 2 is still valid for jzj < 4 since 4 > 2 and the second expansion for jzj < 4 is still valid for jzj > 2 since 2 < 4. Consequently, if 2 < jzj < 4, then, by subracting the first series from the second z 2 1 ¼  ðz þ 2Þðz þ 4Þ z þ 4 z þ 2   1 z z2 z3 1 2 4 8 ¼  þ  2 þ 3  4 þ   þ   2 8 32 128 z z z z ¼  þ

8 4 2 1 1 z z2 z3 þ  þ  þ   þ  z4 z3 z2 z 2 8 32 128

Take care here! You may be tempted to think that this displays an essential singularity at z ¼ 0. This is not the case because the expansion is only valid inside the annular region 2 < jzj < 4 centred on the origin. Consequently, the point z ¼ 0 is outside this region and the series expansion is invalid at that point. The series expansion of the same function valid for jzj < 2 is ............

18

1122

Programme 31

19

z 3z2 7z3  þ þ  8 32 128 Because If jzj < 2 then

 1 1 1 z z2 z3 ¼ ¼ 1  þ  þ  z þ 2 2ð1 þ z=2Þ 2 2 4 8

1 z z2 z3  þ  þ  2 4 8 16 We have already seen that if jzj < 4 then ¼

2 1 z z2 z3 ¼  þ  þ  z þ 4 2 8 32 128 This is still valid for jzj < 2 since 2 < 4. Conseqently, if jzj < 2, then, by subtracting the first series from the second z 2 1 ¼  ðz þ 2Þðz þ 4Þ z þ 4 z þ 2   1 z z2 z3 1 z z2 z3 ¼  þ  þ   þ   þ  2 8 32 128 2 4 8 16 z 3z2 7z3  þ   8 32 128 Notice that for different regions of convergence we obtain different series expansions. Furthermore, each series expansion is unique within its own particular radius of convergence. ¼

Try one more just to make sure that you can derive these expansions. 1  cosðz  6Þ

about the point z ¼ 6 is ðz  6Þ2 . . . . . . . . . . . . valid for . . . . . . . . . . . . at which point there is . . . . . . . . . . . .

The Laurent series of

20

1 ðz  6Þ2 ðz  6Þ4  þ     valid for all z 6¼ 6 2! 4! 6! at which point there is a removable singularity Because If we let u ¼ z  6 then 1  cosðz  6Þ 1  cos u ¼ u2 ðz  6Þ2    1 u2 u4 u6 ¼ 2 1  1  þ  þ  u 2! 4! 6! ¼

1 u2 u4  þ   2! 4! 6!

¼

1 ðz  6Þ2 ðz  6Þ4  þ   2! 4! 6!

Complex analysis 3

1123

This is valid for all finite values of z 6¼ 6 at which point there is a removable 1  cosðz  6Þ 1 singularity which can be removed by defining at z ¼ 6 as . 2 2! ðz  6Þ Notice that here the principal part has no terms, so that the Laurent series is identical to the Taylor series. Next frame

Residues In the Laurent series a2 a1 f ðzÞ ¼    þ þ a0 þ a1 ðz  z0 Þ þ a2 ðz  z0 Þ2 þ    þ 2 ðz  z0 Þ ðz  z0 Þ the coefficient a1 is referred to as the residue of f ðzÞ for reasons that will soon become apparent. Recall the integral in Frame 45 of Programme 28 which states that if the simple closed contour c has z0 as an interior point, then þ dz ¼ 2jn1 ðz  z0 Þn c  1 if n ¼ 1 where the Kronecker delta n1 ¼ . Applying this fact to the Laurent 0 if n 6¼ 1 series of f ðzÞ yields þ þ" a2 a1 þ a0 þ a1 ðz  z0 Þ f ðzÞ dz ¼  þ þ ðz  z0 Þ2 ðz  z0 Þ c c # þ a2 ðz  z0 Þ2 þ    dz þ ¼  þ

þ þ a1 dz þ þ a0 dz 2 c ðz  z0 Þ c ðz  z0 Þ c þ þ þ a1 ðz  z0 Þ dz þ a2 ðz  z0 Þ2 dz þ    a2 dz

c

c

¼    þ 0 þ 2ja1 þ 0 þ 0 þ 0 þ    ¼ 2ja1 That is, provided f ðzÞ is analytic at all points inside and on the simple closed contour c, apart from the single isolated singularity at z0 which is interior to c, then þ f ðzÞ dz ¼ 2ja1 c

Hence the name residue for a1 because it is all that remains when the Laurent series is integrated term by term. This statement is called the Residue theorem and it has many far reaching consequences – we shall see some of these later. For now, just try an example.

21

1124

Programme 31 y 3 c

–3

O

3

x

–3

If c is a circle, centred on the origin and of radius 3, then þ z dz ¼ ............ c ðz þ 2Þðz þ 4Þ þ

22

z dz ¼ 2j c ðz þ 2Þðz þ 4Þ

Because The circle jzj ¼ 3 lies within the annular region 2 < jzj < 4 and we have already found the Laurent series for the integrand valid for 2 < jzj < 4 in Frame 18, namely z 2 1 ¼  ðz þ 2Þðz þ 4Þ z þ 4 z þ 2 8 4 2 1 1 z z2 z3 þ  þ  þ   þ  z4 z3 z2 z 2 8 32 128 þ z dz Here the residue is a1 ¼ 1 and so ¼ 2jð1Þ ¼ 2j where c ðz þ 2Þðz þ 4Þ ¼  þ

c lies entirely within the region of convergence. The Residue theorem extends to the case where the contour contains a finite number of singularities. If f ðzÞ is analytic inside and on the simple closed contour c except at the finite number of points z0 , z1 , z2 , . . ., each with a Laurent ð0Þ

ð1Þ

ð2Þ

series expansion and each with corresponding residues a1 , a1 , a1 , . . . then  þ ð0Þ ð1Þ ð2Þ   f ðzÞ dz ¼ 2j a1 þ a1 þ a1 ¼ 2j sum of residues inside c c

y c z0 zn

z1

z2 x

What could be more straightforward? Next frame

Complex analysis 3

1125

Calculating residues When evaluating these integrals the major part of the exercise is to find the residues, and it would be very tedious if we had to find a Laurent series for each and every singularity. Fortunately there is a simpler method for poles. If f ðzÞ is analytic inside and on the simple closed contour c except at the interior point z0 at which there is a pole of order n, then " # 1 dn1  n ðz  z0 Þ f ðzÞ a1 ¼ Lim n1 z!z0 ðn  1Þ! dz

23

Example Find the residues at all the poles of f ðzÞ ¼

3z

. ðz þ 2Þ2 ðz2  1Þ

f ðzÞ has a pole of order 2 (a double pole) at z ¼ 2 and two poles of order 1 (simple poles) at z ¼ 1. " # 1 d21  2 At z ¼ 2 the residue is a1 ¼ Lim ðz þ 2Þ f ðzÞ 21 z!2 ð2  1Þ! dz

  d 3z ¼ Lim 2 z!2 dz z  1 "  # 3 z2  1  6z2 ¼ Lim 2 z!2 ðz2  1Þ 3ð4  1Þ  24 5 ¼ ¼ 2 3 ð4  1Þ At z ¼ 1 the residue is . . . . . . . . . . . .

24

1 6 Because

"

At z ¼ 1 the residue is a1

1 d11 ¼ Lim ððz  1Þf ðzÞÞ 11 z!1 ð1  1Þ! dz " !# d0 3z ¼ Lim 0 z!1 dz ðz þ 2Þ2 ðz þ 1Þ

#

The zeroth derivative of an expression is the expression itself " # 3z ¼ Lim 2 z!1 ðz þ 2Þ ðz þ 1Þ 3 1 ¼ ¼ ð3Þ2 ð2Þ 6 At z ¼ 1 the residue is . . . . . . . . . . . .

1126

Programme 31

25

3 2 Because

"

At z ¼ 1 the residue is a1

1 d11 ¼ Lim ððz þ 1Þf ðzÞÞ 11 z!1 ð1  1Þ! dz " !# d0 3z ¼ Lim 0 z!1 dz ðz þ 2Þ2 ðz  1Þ " # 3z ¼ Lim 2 z!1 ðz þ 2Þ ðz  1Þ 3 ¼ ð1Þ2 ð2Þ 3 ¼ 2

#

Move to the next frame

Integrals of real functions 26

The Residue theorem can be very usefully employed to evaluate integrals of real functions that cannot be evaluated using the real calculus. Even when an integral is susceptible to evaluation by the real calculus, the use of the residue calculus can often save a great amount of effort. We shall look at three types of real integral and in each case we shall proceed by example.

Integrals of the form

∫

2π

F(cos , sin ) d

0

Example ð 2 Evaluate 0

1 d. 4 cos   5

To evaluate this integral we make use of the exponential representation of a complex number of unit length, namely z ¼ e j ,and the exponential form of the trigonometric functions cos  ¼

e j þ ej z þ z1 e j  ej z  z1 ¼ and sin  ¼ ¼ , 2 2 2j 2j

and finally dz ¼ je j d ¼ jz d so that d ¼ dz=jz

Complex analysis 3

1127

Using these relations we can transform the real integral from 0 to 2 into a contour integral in the complex plane where the contour c is the unit circle centred on the origin. That is ð 2 þ 1 1 dz d ¼  1 4 cos   5 jz z þ z c 0 5 4 2 þ 1 ¼ j dz 2 c 2z  5z þ 2 þ 1 ¼ j dz ð2z  1Þðz  2Þ c 1 which is inside the 2 contour c and another at z ¼ 2 which is outside the contour c. Using the Residue theorem þ   1 j dz ¼ j  2j  residue at z ¼ 1=2 c ð2z  1Þðz  2Þ The complex integrand has two simple poles, one at z ¼

The residue at z ¼ 1=2 is   1 1 ¼ Lim Lim ðz  1=2Þ ð2z  1Þðz  2Þ z!1=2 z!1=2 2ðz  2Þ ¼ so that ð 2 0

1 d ¼ j 4 cos   5

1 3

þ

1 dz ð2z  1Þðz  2Þ   ¼ j  2j  residue at z ¼ 1=2 c

¼ 2=3 Now you try one

ð 2 0

d ¼ ............ 2 þ cos 

1128

Programme 31

27

2 pffiffiffi 3 Because ð 2

þ

dz=jz where c is the unit circle centred on the z þ z1 origin. 2þ 2 þ 2 dz ¼ j 2 þ 4z þ 1 z þc 2 dz pffiffiffi pffiffiffi  ¼ j  c zþ2 3 zþ2þ 3 pffiffiffi The integrand has two pffiffiffisimple poles, one at z ¼ 2 þ 3 which is inside c and another at z ¼ 2  3 which is outside c. Therefore þ n pffiffiffio 2 dz pffiffiffi ¼ j  2j  residue at z ¼ 2 þ 3 pffiffiffi  j  c zþ2 3 zþ2þ 3 0

d ¼ 2 þ cos 

c

The residue is (  pffiffiffi zþ2 3  Limpffiffi

) 2 pffiffiffi pffiffiffi  zþ2 3 zþ2þ 3 z!2þ 3 ( ) 2 1 pffiffiffi ¼ pffiffiffi and so ¼ Limpffiffi  zþ2þ 3 3 z!2þ 3 þ ð 2 d 2 dz 1 pffiffiffi ¼ j  2j  pffiffiffi pffiffiffi  ¼ j  3 c zþ2 3 zþ2þ 3 0 2 þ cos  1 2 ¼ 2  pffiffiffi ¼ pffiffiffi 3 3

28

Integrals of the form Example Evaluate

∫

∞

F(x) dx

–∞

ð1

1 dx. 4 1 1 þ x

To evaluate this integral we must consider the integral

þ

1 dz where c is the 1 þ z4 c

contour shown in the figure, so that þ ð ðR   1 dz dx dz ¼ þ ¼ 2j sum of residues inside c 4 4 4 c1þz s1þz R 1 þ x Notice that along the real axis between R and R, z ¼ x. Provided R > 1 we can evaluate this integral using the Residue theorem. That is þ   1 dz ¼ 2j  sum of residues inside c 4 1 þ z c

y S

–R

R x

Complex analysis 3

1129

29

 pffiffiffi 2 Because 1 possesses four simple poles at z ¼ ej=4 , e3j=4 , e5j=4 , e7j=4 of 1 þ z4 which only the first two are inside c.   1 z  ej=4  The residue at z ¼ ej=4 is Lim 1 þ z4 z!ej=4  1 ¼ Lim ˆ pital’s rule by L’Ho 3 z!ej=4 4z

The integrand

e3j=4 4   is Lim z  e3j=4 

¼ The residue at z ¼ e

3j=4

z!e3j=4

¼ Lim z!e3j=4

¼



1 4z3



1 1 þ z4



ˆ pital’s rule by L’Ho

e9j=4 ej=4 ¼ 4 4

Therefore   þ  1 1 3j=4 j=4 e dz ¼ 2j  þe 4 4 c1þz 3 3 1 j  j sin ¼  pffiffiffi  pffiffiffi and Now e3j=4 ¼ cos 4 4 2 2   1 j ej=4 ¼ cos  j sin ¼ pffiffiffi  pffiffiffi and so 4 4 2 2    þ 1 1 2j  pffiffiffi dz ¼ 2j  ¼ pffiffiffi 4 1 þ z 4 2 2 c We now look at the components of this integral in the next frame We now recognize that þ ðR ð 1 1 1 dz ¼ dx þ dz 4 4 1 þ z 1 þ x 1 þ z4 R c S because z ¼ x along the real line. Now we let R increase indefinitely and take limits, so that þ ð ð1 1 1 1  dz ¼ dx þ Lim dz ¼ pffiffiffi Lim 4 4 4 1 þ z 1 þ x 1 þ z 2 R!1 c R!1 S 1 because the value of the contour integral is independent of the value of R. We ð 1 shall now proceed to show that Lim dz ¼ 0. 1 þ z4 R!1 S

30

1130

Programme 31

Writing z ¼ Re j so that, on S, dz ¼ Re j d, the limit of the integral becomes ð Re j d ¼ 0 Lim 4 j4 R!1 S 1 þ R e Notice that the requirement that ensures that the integral along the semicircle vanishes in the limit is equivalent to the requirement that the degree of the denominator be at least two degrees higher than the numerator. Now you try one.

ð1 1

x2 dx 2

ðx2 þ 1Þ

31

¼ ............

 2 Because

þ

z2 dz

Consider the integral

where the contour c is the same semi-circular 2 ðz2 þ 1Þ contour as in the previous example. Here the integrand has two double poles at z ¼ j and z ¼ j but only the pole at z ¼ j is inside the contour. The residue at z ¼ j is ( ) ( ) d z2 2zðz þ jÞ2  z2 2ðz þ jÞ 2 ðz  jÞ ¼ Lim Lim dz z!j z!j ðz þ jÞ4 ðz  jÞ2 ðz þ jÞ2 c

¼

j 4

Therefore þ z2 dz

  j  ¼ 2j  ¼ 2 4 2 c ðz2 þ 1Þ

Taking limits ð1 þ ð z2 dz x2 dx z2 dz  ¼ þ Lim ¼ Lim 2 2 2 2 2 2 2 R!1 c ðz þ 1Þ R!1 S ðz þ 1Þ 1 ðx þ 1Þ Where, in the second integral on the right-hand side, the degree of the denominator is two higher than the degree of the numerator, and so ð ð1 z2 dz x2 dx  ¼ 0, therefore ¼ Lim 2 2 2 2 2 R!1 S ðz þ 1Þ 1 ðx þ 1Þ

Complex analysis 3

Integrals of the form

1131

∫

∞

F(x)

–∞

32

x dx { sin cos x

These integrals are often referred to as Fourier integrals because of their appearances within Fourier analysis. Example ð1 Evaluate

cos kx dx where a > 0 and k > 0. 2 þ x2 a 1

þ

e jkz dz where þ z2 c c is the semicircular contour of the previous problems and whose integrand possesses two simple poles at z ¼ aj and z ¼ aj of which only the first is inside the contour. Consequently þ   e jkz dz ¼ 2j residue at z ¼ aj ¼ . . . . . . . . . . . . 2 2 c a þz

To evaluate this integral we consider the contour integral

a2

eka a Because The residue at z ¼ aj is   jkz e jkz e e jkðajÞ jeka ¼ ¼  and so ¼ Lim ðz  ajÞ 2 Lim a þ z2 2aj 2a z!aj z!aj z þ aj  þ e jkz jeka eka ¼ dz ¼ 2j  2 2 2a a c a þz Taking limits as R ! 1 þ ð ð1 e jkz ejkz e jkz eka Lim dz ¼ dz þ Lim dz ¼ 2 2 2 2 2 2 a R!1 c a þ z R!1 S a þ z 1 a þ z In the second integral on the right-hand side, the degree of the denominator is two higher than the degree of the numerator, and so ð1 ð e jkz e jkx eka . That is dz ¼ 0, therefore dx ¼ Lim 2 2 2 2 a R!1 S a þ z 1 a þ x ð1   cos kx þ j sin kx eka ¼ 2j residue at z ¼ aj . dx ¼ 2 2 a þx a 1 Consequently ð1   cos kx eka ¼ 2 Im residue at z ¼ aj and dx ¼ 2 þ x2 a a 1 ð1   sin kx dx ¼ 0 ¼ 2 Re residue at z ¼ aj 2 2 1 a þ x

33

1132

Programme 31  Notice that e jkz is easier to use than cos kx ¼ e jkx þ ejkx =2, and it also gives the solution to the related integral with cos kx replaced with sin kx. Finally, to finish off the Programme, here is one for you to try. ð1 cos x dx ¼ . . . . . . . . . . . . 2þxþ1 x 1

34

0 Because

þ

e jz dz where c is the semicircular contour of the previous þzþ1 c pffiffiffi  problem. The integrand is singular at the simple poles z ¼ 1  j 3 =2 where pffiffiffi pffiffiffi   only z ¼ 1 þ j 3 =2 is inside the contour. The residue at z ¼ 1 þ j 3 =2 is Consider

z2

then Lim

pffiffi z!ð1þj 3Þ=2

¼ ¼

Limpffiffi

 h pffiffiffii  z  1 þ j 3 =2

z!ð1þj 3Þ=2 pffiffi jð1þj 3Þ=2

e

(

e jz

e jz z2 þ z þ 1 )



pffiffiffi z  1  j 3 =2

pffiffiffi j 3

pffiffi ej=2 e 3=2 pffiffiffi ¼ j 3 pffiffi e 3=2 ¼  pffiffiffi since ej=2 ¼ j 3

Therefore ( pffiffi ) pffiffi þ e jz e 3=2 2e 3=2 pffiffiffi pffiffiffi dz ¼ 2j ¼ j 2 3 3 cz þzþ1 that is þ

e jz dz ¼ 2 cz þzþ1

þ

pffiffi cos z þ j sin z 2e 3=2 pffiffiffi dz ¼ j z2 þ z þ 1 3 c

and so pffiffi þ þ cos z sin z 2e 3=2 pffiffiffi dz ¼ 0 and dz ¼  2 2 3 cz þzþ1 cz þzþ1 Note that, again, the contribution from the contour integral along the semicircle is zero. The Review summary now follows. Check it through in conjunction with the Can you? checklist before going on to the Test exercise. The Further problems provide additional practice.

Complex analysis 3

1133

Review summary 31 1 Maclaurin series The Maclaurin series expansion of a function of a complex variable z is f ðzÞ ¼ f ð0Þ þ zf 0 ð0Þ þ z2

f 00 ð0Þ f 000 ð0Þ þ z3 þ  2! 3!

2 Ratio test for convergence The ratio test for convergence of a series of terms of a complex variable f ðzÞ ¼ a0 ðzÞ þ a1 ðzÞ þ a2 ðzÞ þ a3 ðzÞ þ    þ an ðzÞ þ    is that given   anþ1 ðzÞ ¼L  Lim   n!1 an ðzÞ then if L < 1 the series converges and so the expansion is valid L > 1 the series diverges and so the expansion is invalid L ¼ 1 the ratio test fails to give a conclusion. 3 Radius and circle of convergence The radius of the circle within which a series expansion is valid is called the radius of convergence of the series and the circle is called the circle of convergence. The radius of convergence can be found using the ratio test for convergence. 4 Singular points Any point at which f ðzÞ fails to be analytic, that is where the derivative does not exist, is called a singular point. Poles If f ðzÞ has a singular point at z0 and for some natural number n   Lim ðz  z0 Þn f ðzÞ ¼ L 6¼ 0 z!z0

then the singular point (also called a singularity) is called a pole of order n. Removable singularity If f ðzÞ has a singular point at z0 but Lim ff ðzÞg exists then the singular z!z0

point is called a removable singularity. 5 Circle of convergence When an expression is expanded in a Maclaurin series, the circle of convergence is always centred on the origin and the radius of convergence is determined by the location of the first singular point met as z moves out from the origin.

1134

Programme 31

6 Taylor’s series Provided f ðzÞ is analytic inside and on a simple closed curve c, the Taylor series expansion of f ðzÞ about a point z0 which is interior to c is given as f ðzÞ ¼ f ðz0 Þ þ ðz  z0 Þf 0 ðz0 Þ þ þ

ðz  z0 Þn f n!

ðnÞ

ðz0 Þ

ðz  z0 Þ2 f 00 ðz0 Þ þ  2! þ 

where, here, the expansion is about the point z0 which is the centre of the circle of convergence. The circle of convergence is given as jz  z0 j ¼ R where R is the radius of convergence. Maclaurin’s series is a special case of Taylor’s series where z0 ¼ 0. 7 Laurent’s series The Laurent series expansion provides a series expansion valid within an annular region centred on the singular point. Let f ðzÞ be singular at z ¼ z0 and let c1 and c2 be two concentric circles centred on z0 . Then if f ðzÞ is analytic in the annular region between c1 and c2 and c is any concentric circle lying within the annular region between c1 and c2 we can expand f ðzÞ as a Laurent series in the form a2 a1 f ðzÞ ¼    þ þ a0 þ a1 ðz  z0 Þþ a2 ðz  z0 Þ2 þ    þ 2 ðz  z Þ 0 ðz  z0 Þ þ 1 X 1 f ðzÞ an ðz  z0 Þn where an ¼ dz ¼ nþ1 2j c ðz  z0 Þ n!1 8 Residues In the Laurent series a2 a1 f ðzÞ ¼    þ þ a0 þ a1 ðz  z0 Þþ a2 ðz  z0 Þ2 þ    þ ðz  z0 Þ2 ðz  z0 Þ the coefficient a1 is referred to as the residue of f ðzÞ. Residue theorem Provided f ðzÞ is analytic at all points inside and on the simple closed contour c, apart from the single isolated singularity at z0 which is interior to c, then þ f ðzÞ dz ¼ 2ja1 c

9 The Residue theorem extends to the case where the contour contains a finite number of singularities. If f ðzÞ is analytic inside and on the simple closed contour c except at the finite number of points z0 , z1 , z2 ,    each with a Laurent series expansion and each with corresponding residues ð0Þ

ð1Þ

ð2Þ

a1 ,a1 ,a1 ,    then  þ ð0Þ ð1Þ ð2Þ f ðzÞ dz ¼ 2j a1 þ a1 þ a1 þ    c

Complex analysis 3

1135

10 Calculating residues " # 1 dn1  n a1 ¼ Lim ðz  z0 Þ f ðzÞ n1 z!z0 ðn  1Þ! dz 11 Real integrals The Residue theorem can be very usefully employed to evaluate integrals of real functions. ð 2 Fðcos , sin Þ d Integrals of the form 0

Use z ¼ e and the exponential form of the trigonometric functions e j þ ej z þ z1 e j  ej z  z1 cos  ¼ ¼ , sin  ¼ ¼ and dz ¼ je j d ¼ jz d 2 2 2j 2j so that d ¼ dz=jz. Convert the integral into a contour integral around the unit circle centred on the origin and use the Residue theorem.  ð1 ð1 sin x Integrals of the form FðxÞ dx and FðxÞ dx cos x 1 1 þ þ Consider integrals of the form FðzÞ dz and FðzÞe jz dz respectively, where j

c

c

the contour c is a semicircle with the diameter lying along the real axis. The principle is that the integral can be evaluated by the Residue theorem and then the contour can be expanded to cover the required extent of the real axis, the integration along the semicircle giving a zero contribution.

Can you? Checklist 31 Check this list before and after you try the end of Programme test On a scale of 1 to 5 how confident are you that you can:

Frames

. Expand a function of a complex variable about the origin in a Maclaurin series?

1

to

7

8

to

10

Yes

No

. Determine the circle and radius of convergence of a Maclaurin series expansion?

Yes

No

. Recognize singular points in the form of poles of order n, removable and essential singularities?

Yes

No

11

1136

Programme 31

. Expand a function of a complex variable about a point in the complex plane in a Taylor series, transforming the coordinates with a shift of origin?

Yes

12

20

21 and 22

23

to

25

26

to

34

No

. Evaluate certain types of real integrals using the Residue theorem?

Yes

to

No

. Calculate the residues at the poles of an expression without resort to deriving the Laurent series?

Yes

16

No

. Recognize the residue of a Laurent series and state the Residue theorem?

Yes

15

No

. Recognize the principal and analytic parts of the Laurent series and link the form of the principal part to the type of singularity?

Yes

14

No

. Expand a function of a complex variable about a singular point in a Laurent series?

Yes

to

No

Test exercise 31 1 Expand each of the following in a Maclaurin series and determine the radius and the circle of convergence in each case. (a) f ðzÞ ¼ ez (b) f ðzÞ ¼ lnð1 þ 4zÞ. 2 Determine the location and nature of the singular points in each of the following. 3z (a) f ðzÞ ¼ ðz þ 1Þ5 (b) f ðzÞ ¼ z10 e1=z (c) f ðzÞ ¼ z sinð1=zÞ 1  cos z (d) f ðzÞ ¼ z2 3 Expand f ðzÞ ¼ sin z in a Taylor series about the point z ¼ =4 and determine the radius of convergence.

Complex analysis 3

1137

4 Expand each of the following in a Laurent series. In (a) and (c) determine the nature of the singularity from the principal part of the series. 1 (a) f ðzÞ ¼ ð5  zÞ cos about the point z ¼ 3 zþ3 2z (b) f ðzÞ ¼ valid for 1 < jzj < 3 ðz þ 1Þðz þ 3Þ 1 (c) f ðzÞ ¼ about the point z ¼ 2. 3 z ðz  2Þ2 5 Calculate the residues at each of the singularities of 3z  1 . f ðzÞ ¼ 2 z ðz þ 1Þ2 ðz  1Þ 6 Evaluate each of the following integrals. ð 2 d (a) 0 5 cos   13 ð1 dx (b) 2 1 x þ x þ 1 ð1 cos 3x dx (c) 4 2 1 x þ 2x þ 1

Further problems 31 1 For each of the following find the Maclaurin series expansion and determine the radius of convergence. (a) sinh z (b) tan z   1þz (c) ln 1z (d) az , where a > 0 (e)

15z2 ð5  3zÞ3

.

2 By using the appropriate Maclaurin series expansions, show that (a) ðcos zÞ0 ¼  sin z ez þ ez (b) cos z ¼ 2 (c) ðez Þ0 ¼ ez . 3 Given the series expansion for ð1 þ zÞ1 (a) show by integration that this is compatible with the series expansion for lnð1 þ zÞ 1 1 X X (b) by differentiation find ð1Þn nzn and ð1Þn n2 zn . n¼1

n¼1

1138

Programme 31

4 Use the ratio test to test each of the following for convergence. 1 1 X X ð2nÞ! n ðcos nÞzn z (d) (a) 2 2n  1 n¼0 ðn!Þ n¼0 1 1 X ð1Þn zn X zn (e) (b) 1  3n ðn þ 1Þ! n¼0 n¼1 (c)

1 X n2 zn 1  3n n¼0

5 Find the Taylor series about the point indicated of each of the following. (a) ez about the point z ¼ 2 (b) cos z about the point z ¼ =6 (c) ðz  3Þ sinðz þ 3Þ about the point z ¼ 3 (d) ð2z  5Þ1 about the point z ¼ 1=3 (e) ð2z  5Þ1 about the point z ¼ 3. 6 Find the series expansion of z ln z valid for jz  1j < 1. 7 Find the circle of convergence of each of the following when expanded in a Taylor series about the point indicated. (a) ez cosðz  2Þ about the point z ¼ 1 z3 about the point z ¼ 0 ðz2 þ 6Þ z2 about the point z ¼ 5 (c) ðz  6Þðz  4Þ

(b)

(d)

ðez

z2 about the point z ¼ 0. þ 1Þ

8 Locate and classify all of the singularities of each of the following. (a)

ðz  1Þ3 2

z2 ðz2  1Þ

(b) z2 e1=z . 9 Find the Laurent series about the point indicated of each of the following.   1 1 about the point z ¼ 0 (a) sin z z 1 (b) about the point z ¼ 3=2 2z  3 z about the point z ¼ 3. (c) ðz  2Þðz  3Þ 10

Find the Laurent series of (a) 2 < jzj < 5 (b) jzj > 5 (c) jzj < 2.

z1 that is valid for ðz þ 2Þðz þ 5Þ

Complex analysis 3

11 Evaluate each of the following integrals. ð 2 d (a) 2 þ sin  0 ð 2 d for  > jj (b)  þ  sin  0 ð 2 d (c) where 0 <  < 1 2 0 1 þ   2 cos  ð 2 sin2 d (d) 0 5  4 cos  ð 2 d (e) 5  3 cos  ð01 dx (f) 2 1 x þ 6x þ 13 ð1 x2 dx (g) 4 2 1 x þ 6x þ 13 ð1 2 x dx (h) 2 2 1 ðx þ 4Þ ð1 2 x þxþ1 dx (i) x4 þ x2 þ 1 ð1 1 dx (j) 6þ1 x 1 ð1 x2 sin x dx (k) 4 x þ 6x2 þ 13 ð1 1 sin x dx. (l) 4 þ x2 þ 1 x 1

1139

Frames 1 to 29

Programme 32

Optimization and linear programming Learning outcomes When you have completed this Programme you will be able to: Describe an optimization problem in terms of the objective function and a set of constraints Algebraically manipulate and graphically describe inequalities Solve a linear programming problem in two real variables graphically Use the Microsoft Add-on Solver to solve linear programming problems in two, three and four real variables Construct the algebraic form of the objective function and the constraints of a linear problem stated in words Demonstrate the use of Solver to obtain the solution to a nonlinear programming problem

1140

Optimization and linear programming

1141

Optimization An optimization problem is one requiring the determination of the optimal (maximum or minimum) value of a given function, called the objective function, subject to a set of stated restrictions, or constraints, placed on the variables concerned. In practice, for example, we may need to maximize an objective function representing units of output in a manufacturing situation, subject to constraints reflecting the availability of labour, machine time, stocks of raw materials, transport conditions, etc.

1

Linear programming (or linear optimization) Linear programming is a method of solving an optimization problem when the objective function is a linear function and the constraints are linear equations or linear inequalities. In this Programme, we shall restrict our considerations to problems of this type that form an important introduction to the much wider study of operational research. Let us consider a simple example, so move on to the next frame

2

A simple linear programming problem may look like this: Maximize subject to

P ¼ x þ 2y 9 y3 > > > xþy5 = x  2y  2 > > > ; x 0; y 0

(objective function)

(constraints)

The last two constraints, i.e. x 0 and y 0, apply to all linear programming (LP) problems and indicate that the problem variables, x and y, are restricted to nonnegative values: they may have zero or positive values, but NOT negative values. These two constraints are often combined and written x, y 0 – or omitted altogether since they are taken for granted in all LP problems. Before we proceed, we will take a brief look at linear inequalities in general. On, then, to Frame 3

1142

3

Programme 32

Linear inequalities In most respects, linear inequalities can be manipulated in the same manner as can equations. (a) Both sides may be increased or decreased by a common term, e.g. 2x  y þ 4

; 2x  y  4

(b) Both sides may be multiplied or divided by a positive factor, e.g. 4x þ 6y 12

; 2x þ 3y 6

But NOTE this: (c) If both sides are multiplied or divided by a negative factor, e.g. ð1Þ, then the inequality sign must be reversed, i.e. becomes  and vice versa. Here, then, is a short exercise. Exercise Simplify the following inequalities so that each right-hand side consists of a positive constant term only. (a) 3x  5  4y

(b) 2ðx þ 2yÞ  8

(c) 4x  6y  10

(d) 2x þ 3 ðy þ 4Þ

(e)  ðx  3y þ 5Þ 2x þ 4y  6 Check the results in the next frame

4

(a) 3x  4y  5

(b)  x  2y 4

(c)  2x þ 3y 5

(d)  2x  y  7

(e) 3x þ y  1

Graphical representation of linear inequalities Consider the inequality y  2x  3. We can add 2x to each side, so that y  2x þ 3. The equation y ¼ 2x þ 3 can be represented y by a straight line dividing the x–y plane into two parts. +3 y=

2x

For all points on the line, y ¼ 2x þ 3. For all points below the line, x

............

Optimization and linear programming

1143

5

y < 2x þ 3 ; y  2x þ 3 indicates all points on or below the straight line, but excludes all points above it. We can indicate this exclusion zone by shading the upper side of the line. y

y=

2x

+3

x

x Arguing in much the same way, x  2y  2 can be rewritten as y  1 and we 2 x can draw the line y ¼  1 and shade in the exclusion zone 2 ............ below the line y

i.e.

x –1 y= 2

x

Example 1 The problem we quoted earlier in Frame 2 was Maximize

P ¼ x þ 2y

subject to

y3 xþy5

9 > > > =

x  2y  2 > > > ; x 0; y 0

(objective function)

(constraints)

Now, on a common pair of x and y axes, we can represent the five constraints and shade in the exclusion zone for each. We then have the composite diagram ............

6

1144

Programme 32

7

y

y=

–x

+

5

y=3 x=0

x –1 y= 2

y=0 x

The coordinates of all points on the boundary of the polygon OABCD, or within the figure so formed, satisfy the system of constraints. The set of variables for each such point is called a feasible point or feasible solution and the figure OABCD is the feasible domain or feasible polygon. Note these definitions.

8

Our problem now is to find the particular point within this domain that makes the objective function P ¼ x þ 2y a maximum. The equation can be rewritten as x P y ¼  þ and this represents a set of parallel lines with different values of the 2 2 P intercept . 2 y

P 2

y=

–x 2 +P 2

x

If we draw one sample line of this set to cross the feasible polygon we have just obtained, we get, using P ¼ 3 ............

Optimization and linear programming

1145

9

y

x

We can increase the value of P and hence raise the objective line up the page until it passes through the extreme point C. Any further increase in the value of P would take the line outside the feasible polygon and hence fail to conform to the given set of constraints. y P 2

=4 max

P 2

B x

In this example, then, point C gives the optimal solution. From the graph it can be seen that the line with maximal P passes through the point of intersection of the two lines y ¼ 3 and y ¼ x þ 5. This means that y ¼ 3, x ¼ 2 and so Pmax ¼ x þ 2y ¼ 8. A graphical method of solution is clearly limited to linear programming problems involving two variables only. However, it is a useful introduction to other techniques, so let us deal with another example. Example 2 Maximize subject to

P ¼ x þ 4y x þ 2y  6 5x þ 4y  40 x, y 0

First of all, plot the appropriate straight line graphs to obtain the feasible polygon. This gives ............

1146

Programme 32

10

x +3 y= 2

y

y= 5x – 4 0

+1

x

x P The objective function P ¼ x þ 4y can be expressed in the form y ¼  þ and 4 4 its graph added to the feasible polygon, as before. We then obtain ............

11

y

P 4

y=– x +P 4 4

x

x P The line y ¼  þ is then raised to give the optimal solution, which is 4 4 ............

12

Pmax ¼ 24 with x ¼ 4, y ¼ 5 y P 4

max

P 4

y=– x +P 4 4

x

Optimization and linear programming

1147

From the graph it can be seen that the line with maximal P passes through the x 5x point of intersection of the two lines y ¼ þ 3 and y ¼  þ 10. That is x ¼ 4, 2 4 y ¼ 5 and so Pmax ¼ x þ 4y ¼ 24. As easy as that. Now this one. Example 3 Minimize

P ¼ 4x þ 6y

subject to

x þ 6y 24 2x  y  7 x þ 8y  80 x, y 0

It is very much as before. Complete it on your own. Pmin ¼ 6

y

with

13

x ¼ 6, y ¼ 5 P 2x + 6 = y 3

P 6

P 6

min

x

To obtain the minimum optimal value of P, the graph of the objective function is, of course, lowered to the appropriate extreme point. Optimizing the objective function when there are only two variables involved is really quite straightforward using these graphical techniques. When more than two are involved a more systematic method is necessary and in 1939 George Dantzig developed what is called the simplex method for just this purpose. One of the great advantages of the simplex method is that the algorithm it uses can be programmed into a computer. This then permits the long, tedious and error-prone series of arithmetic operations used when calculating by hand to be replaced by an error-free procedure that produces a result in seconds. Performing the simplex procedure adds nothing to the understanding of optimization so we shall leave any desire in the reader to know more of the hand-driven simplex method to other texts. Here we shall use a computer. Move to the next frame

1148

14

Programme 32

Solver There are many computer packages available that perform linear programming using the simplex method but the one we shall use is called Solver and is available as an Add-on to the Microsoft Excel spreadsheet package. It is accessed via the Tools drop-down menu but before we go there we shall return to Example 3 that we have just completed graphically in the previous frame. Minimize subject to

P ¼ 4x þ 6y

Objective function

 x þ 6y 24 2x  y  7 x þ 8y  80

Constraint equations

x, y 0 Now open your spreadsheet and complete as shown here where the values of the variables x and y that optimize the objective function will eventually be entered into cells A2 and B2 respectively.

The coefficients of the variables of the objective function are entered into cells A5 and B5. We now have to enter the objective function in cell A8 and the constraints in cells A16, A17, A18. In Cell A8 enter the objective function =$A$5*$A$2+$B$5*$B$2

that is 4x þ 6y (note the absolute address)

In cells A16 to A18 enter the constraint equations =$A$11*$A$2+$B$11*$B$2

that is x þ 6y

=$A$12*$A$2+$B$12*$B$2

that is 2x  y

=$A$13*$A$2+$B$13*$B$2

that is x þ 8y

Optimization and linear programming

1149

Enter the initial values for x and y in cells A2 and B2 as 0 and your sheet should now look as follows:

Having prepared the spreadsheet with the appropriate variables, objective function and constraints it is now ready to be used by Solver. Move to the next frame

Solver parameters Click Tools on the topmost bar of the Excel window to reveal the drop-down menu. From that menu select Solver to reveal Solver Parameters.

Working our way down this window the topmost box labelled Set Objective requires the address of the cell in which the value of the objective function is to be displayed. This is cell A8 so enter A8 into this box.

15

1150

Programme 32

Next we are looking for a minimum value of the objective function so click the Min button. In the box below labelled By Changing Variable Cells enter the range $A2:$B2 – these being the two cells that will take the values of the variables. Now we must enter the contraints in the large box so click the Add button to reveal the Add Constraint window:

In the Cell Reference box enter the cell reference of the first constraint equation in the list, namely $A$16. In the Constraint window enter the number 24 and in the relationship panel in the middle select >=. This completes the entry of the first constraint equation so press OK and the Solver preferences now look as follows:

Optimization and linear programming

Notice that all the cell references have become absolute references. Now add the other constraints:

Finally, select Simplex LP from the drop-down menu in the box below the Constraints window and click the Solve button. After a few seconds the solution is presented in the spreadsheet display as:

That is Pmin ¼ 6 with x ¼ 6, y ¼ 5 as we found in Frame 13. Save this sheet because you will be able to use it again. Now you try one.

1151

1152

Programme 32

Maximize

P ¼ x þ 4y

subject to

 x þ 2y  6

Objective function

5x þ 4y  40

Constraint equations

x, y 0 Place the value of the variable x in cell C2 and the various coefficients in cell C5 for the objective function and C11:C13 for the constraints. The result is: Pmax ¼ . . . . . . . . . . . . with x ¼ . . . . . . . . . . . . , y ¼ . . . . . . . . . . . .

16

Pmax ¼ 24 with x ¼ 4, y ¼ 5 This can straightforwardly be extended to problems involving more variables. Try this one: Maximize

P ¼ 2x þ 6y þ 4z

subject to

2x þ 5y þ 2z  38 4x þ 2y þ 3z  57 x þ 3y þ 5z  57

Objective function

Constraint equations

x, y, z 0 The result is Pmax ¼ . . . . . . . . . . . . with x ¼ . . . . . . . . . . . . , y ¼ . . . . . . . . . . . . , z ¼ . . . . . . . . . . . .

17

Pmax ¼ 60 with x ¼ 0, y ¼ 5, z ¼ 9 Again, save this sheet for use in the future. Before we finish let’s look at the four variable case: Maximize

P ¼xþyþzþw

subject to

x þ 2y þ 3z  5

Objective function

2x þ 3y þ w  5 3x þ z þ 3w  5

Constraint equations

y þ 3z þ 2w  5 x, y, z, w 0 The result is Pmax ¼ . . . . . . . . . . . . with x ¼ . . . . . . . . . . . . , y ¼ . . . . . . . . . . . . , z ¼ . . . . . . . . . . . . , w ¼ . . . . . . . . . . . .

Optimization and linear programming

1153

Pmax ¼ 3:20, x ¼ 0:60, y ¼ 1:00, z ¼ 0:80, w ¼ 0:80 Just to complete this section of the Prorgamme try these: 1

Maximize

P ¼ x þ 2y

subject to

xþy5

Objective function

x  2y  2

Constraint equations

y3 x, y 0 2

Minimize subject to

P ¼ 2x þ 8y

Objective function

3x þ 4y  80 3x þ 4y 8 x þ 4y 40

Constraint equations

x, y 0 3

Maximize

P ¼ 3x þ 4y þ 5z

subject to

2x þ 4y þ 3z  80 4x þ 2y þ z  48 x þ y þ 2z  40

Objective function

Constraint equations

x, y, z 0 4

Maximize

P ¼ 24x þ 21y þ 30z

subject to

12x þ 4y þ 8z  240 8x þ 3y þ 3z  140 6x þ 2y þ 3z 110

Objective function

Constraint equations

x, y, z 0 5

Maximize subject to

P ¼ 2x  3y þ z  2w

Objective function

4x þ 2y þ 3w 5 y þ 2z  5w  10 x  5y  z þ w 15

Constraint equations

2x  3y þ 4z þ 2w  55 x, y, z, w 0 The answers are in the next frame

18

1154

Programme 32

19

1 2

Pmax ¼ 8 with x ¼ 2, y ¼ 3 Pmin ¼ 48 with x ¼ 8, y ¼ 8

3 4

Pmax ¼ 113 with x ¼ 5, y ¼ 7, z ¼ 14 Pmax ¼ 750 with x ¼ 10, y ¼ 10, z ¼ 10

5

Pmax ¼ 30 with x ¼ 15, y ¼ 0, z ¼ 0, w ¼ 0

Applications 20

So far we have seen how to solve a typical linear programming problem by using Microsoft’s Excel Solver. Here the data are presented as a linear objective function and a number of linear constraints in the form of equations or inequalities. A practical problem, however, will be stated in words and must be first translated into algebraic form to enable a solution to be found and this we now demonstrate here. Let us consider the following example. Example 1 A firm manufactures two types of couplings, A and B, each of which requires processing time on lathes, grinders and polishers. The machine times needed for each type of coupling are given in the table. Coupling type

Time required (hours) Lathe

Grinder

Polisher

A

2

8

5

B

5

5

2

The total machine time available is 250 hours on lathes, 310 hours on grinders and 160 hours on polishers. The net profit per coupling of type A is £9 and of type B £10. Determine (a) the number of each type to be produced to maximize profit (b) the maximum profit. If we let x ¼ the number of type A units to be produced y ¼ the number of type B units to be produced the objective function to be maximized can be expressed as . . . . . . . . . . . .

Optimization and linear programming

1155

21

P ¼ 9x þ 10y Now we have to sort out the constraints from the given data. Total time available on lathes ¼ 250 hours ; 2x þ 5y  250 ðlathesÞ Similar statements for the grinders and polishers are . . . . . . . . . . . .

22

8x þ 5y  310 ðgrindersÞ 5x þ 2y  160 ðpolishersÞ The problem now can be expressed as Maximize subject to

P ¼ 9x þ 10y 2x þ 5y  250 8x þ 5y  310 5x þ 2y  160

ðx; y 0Þ

Using Solver this produces the result: Pmax ¼ . . . . . . . . . . . . with x ¼ . . . . . . . . . . . . , y ¼ . . . . . . . . . . . .

23

Pmax ¼ 550 with x ¼ 10, y ¼ 46 The maximum profit of £550 occurs with a manufacturing schedule of and

10 couplings of type A 46 couplings of type B.

Now for another, so move on.

24

Example 2 A firm produces three types of pumps, A, B, C, each of which requires the four processes of turning, drilling, assembling and testing. Pump type A B C Total available time (h/week)

Process time (hours) per pump Turning

Drilling

Assembling

2 1 2

1 1 1

3 4 2

98

60

145

Testing 4 3 2 160

Profit per pump £ 84 72 52

1156

Programme 32

From the information given in the table, determine (a) the weekly output of each type of pump to maximize profit (b) the maximum profit. So, if we let

x ¼ the number of pumps, type A y ¼ the number of pumps, type B z ¼ the number of pumps, type C

we can interpret the problem into its algebraic form, which is ............

25

Maximize

P ¼ 84x þ 72y þ 52z

subject to

2x þ y þ 2z  98 x þ y þ z  60 3x þ 4y þ 2z  145 4x þ 3y þ 2z  160

ðx, y, z 0Þ

Using Solver this produces the result: Pmax ¼ . . . . . . . . . . . . with x ¼ . . . . . . . . . . . . , y ¼ . . . . . . . . . . . . , z ¼ . . . . . . . . . . . .

26

Pmax ¼ 3652 with x ¼ 23, y ¼ 8, z ¼ 22 i.e. by producing 23 pumps, type A 8 pumps, type B 22 pumps, type C the maximum profit of £3652 is attained.

Nonlinear programming 27

Nonlinear programming deals with the optimization of an objective function subject to constraint equations not all of which are linear. The subject is very extensive and we shall not even pretend to do anything other than demonstrate that the Solver Add-on to Excel can be used to attack a nonlinear problem. We shall also demonstrate that when dealing with certain nonlinear problems optimization need not necessarily occur on the boundary of the feasible region but can occur within it. For example, find the minimum value of the quadratic objective function P ¼ x2 þ y2  4x  4y  13 subject to the linear constraints x  10 y  10

Optimization and linear programming

1157

The constraints define a feasible region that is the interior of a square of side length 10 with bottom left-hand corner at the origin. Enter the formulas: =$A$2 and =$B$2 in cells A8 and A9 respectively. Using Solver and selecting CRG Nonlinear in the Select a Solving Method box we find the Solver Parameters window and the solution to be given as:

and

That is, Pmin ¼ 21 when x ¼ 2, y ¼ 2 – the point ð2, 2Þ being inside the feasible region. You try one. The maximum and minimum values of the quadratic objective function P ¼ 2x2  3y2  6x þ y þ 1 subject to the linear constraints x  15 x þ y  25

ðEnter =$A$2 in A8 and =$A$2+$B$2 in A9) are . . . . . . . . . . . . The answer is in the next frame

1158

Programme 32

28

Pmax ¼ 361:08 with x ¼ 15, y ¼ 0:17 Pmin ¼ 3:42 with x ¼ 1:50, y ¼ 0:17 Now try another one. Find the range of values of the nonlinear objective function P ¼ cosðx þ 2yÞ subject to the nonlinear constraints 9  x2 þ y2  25

Enter =($A$2)^2+($B$2)^2 in A8) The result is . . . . . . . . . . . .

29

1  P  0:18 Because Pmin ¼ 1 with x ¼ 3:16, y ¼ 3:13 and Pmax ¼ 0:18 with x ¼ 2:24, y ¼ 4:47 so 1  P  0:18 Notice that whilst Pmin occurs inside the feasible region Pmax occurs on the boundary. That completes the Programme. Check down the Review summary that comes next, in conjunction with the Can you? checklist, before working through the Test exercise that follows thereafter. As usual, a set of Further problems provides further necessary practice in these useful techniques.

Review summary 32 1 Optimization – determination of an optimal value (maximum or minimum) of an objective function subject to a set of constraints. 2 Linear programming (linear optimization) – optimization where the objective function is a linear function and the constraints are linear equations or linear inequalities. 3 Inequalities – multiplying or dividing both sides by a negative factor ðkÞ reverses the inequality, i.e. becomes  and  becomes . 4 Problem variables (x, y, z, etc.) are always non-negative. 5 Feasible solution – a set of variables that satisfies all the given constraints. 6 Optimal solution – a feasible solution for which the objective function becomes a maximum (or minimum) within the constraints.

Optimization and linear programming

1159

7 Graphical solution (a) Constraints – graphs of constraints form the feasible polygon or feasible domain. y

P b

max

P b

y=– a x P b + b x

Feasible point or feasible solution – coordinates of all points within the feasible polygon or on its boundary (OABCD). a P (b) Objective function P ¼ ax þ by ; y ¼  x þ represented by a set b b a P of parallel lines, slope  , intercept . Line through the extreme point b b C gives Pmax , the optimal value of P. 8 Solver – Solver is the add-on to the Microsoft Excel package that enables linear programming problems to be solved with a minimum of effort. 9 Word problems – problems given in word form have to be converted to algebraic form for them to be solved. 10 Nonlinear programming – problems with nonlinear objective functions and/or constraints can also be attacked using Solver.

Can you? Checklist 32 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can:

Frames

. Describe an optimization problem in terms of the objective function and a set of constraints?

1

and

2

3

to

6

Yes

No

. Algebraically manipulate and graphically describe inequalities?

Yes

No

1160

Programme 32

. Solve a linear programming problem in two real variables graphically?

Yes

6

to

13

14

to

19

20

to

26

27

to

29

No

. Use the Microsoft Add-on Solver to solve linear programming problems in two, three and four real variables?

Yes

No

. Construct the algebraic form of the objective function and the constraints of a linear problem stated in words?

Yes

No

. Demonstrate the use of Solver to obtain the solution to a nonlinear programming problem?

Yes

No

Test exercise 32 1 Using a graphical method, maximize P ¼ x þ 2y subject to the constraints 3x þ 4y  8 x þ 4y  16 3x þ 2y  18 x, y 0. Note: Use Solver to solve Exercises 2 to 6. In each case, all variables are nonnegative. 2 Maximize

P ¼ 3x þ 4y

3 Maximize

P ¼ 8x þ 12y þ 10z

subject to

3x  2y  15

subject to

4x þ 3y þ 2z  64

x þ y  10

2x þ y þ 4z  48

x þ 4y  15

x þ 2y þ z  24.

2x þ y  2. 4 Maximize

P ¼ 44x þ 20y

5 Minimize

subject to

12x þ 6y  84

subject to

3x þ 2y 24.

P ¼ 3y  4x x þ 4y  60 2x þ y  22 x þ y 7.

P ¼ 2x  3y þ z  2w

6 Minimize subject to

4x þ 2y x

þ 3w 5 þ 2z  5w  10

x  5y  z þ w 15 2x  3y þ 4z þ 2w  55.

Optimization and linear programming

1161

7 A firm makes two types of containers, A and B, each of which requires cutting, assembly and finishing. The maximum available machine capacity in hours per week for each process is: cutting 50, assembly 84, finishing 72. The process times for one unit of each type are as follows: Time in hours A

B

Cutting

2

5

Assembly

4

8

Finishing

4

5

Process

If the profit margin is £600 per unit A and £1000 per unit B, determine (a) the optimum weekly output of containers (b) the maximum profit. 8 Find the maximum value of P ¼ x2  y2 þ 3xy subject to the constraints x þ y3 4 sinðxÞ  0:5

Further problems 32 All variables in the following problems are non-negative. 1 Maximize

P ¼ x þ 8y

2 Maximize

subject to

3x þ 4y  10

subject to

P ¼ 4x þ 8y x þ 3y  57

x þ 4y  14

7x þ 4y  110

3x þ 2y  21

x þ 5y  40.

3x þ y  18. 3 Maximize subject to

5 Maximize subject to

P ¼ 5x þ 4y

4 Maximize

x  2y  2

subject to

P ¼ 2x þ y x þ 4y  24

3x  4y  8

xþ y 9

5x þ 6y  45

x y 3

x þ 3y  18.

x  2y  2.

P ¼ 3x þ 4y

6 Maximize

P ¼ x þ 2y

3x  4y  12

subject to

2x þ y  1

5x þ 4y  36

x þ y  2

x þ 3y  8

xþ y6

3x þ y  0.

2x  3y  2.

1162

Programme 32

7 Maximize subject to

P ¼ 4y  3x

8 Maximize

x  2y  0

subject to

x y 2

P ¼ 3x þ 4y þ 5z 5x þ 4y þ 8z  40 3x þ 2y þ 12z  30

x þ 2y  14

 8.

y

x þ 2y  6 3x þ 2y  2. 9 Maximize

P ¼ 3x þ 4y þ 3z

10 Maximize

P ¼ 4x þ 3y þ 3z

subject to

2x þ 3y þ 4z  58

subject to

4x þ y þ 2z  40

11 Maximize subject to

4x þ 2y þ 3z  51

x þ 4y þ z  50

3x þ 4y þ 2z  62.

2x þ 3y þ 4z  60.

P ¼ 5:3x þ 3:6y þ 2:0z 2:1x þ 4:3y þ 1:5z  70 3:2x þ 1:4y þ 2:2z  60 1:6x þ 6:2y þ 3:1z  100.

12 Maximize

P ¼ 8x þ 5y

13 Maximize

subject to

2x þ y  80

subject to

P ¼ 12x þ 8y x þ 2y  20

x þ 3y  90

4x  y  8

x þ y 30.

x þ y 1.

14 Maximize

P ¼ 3x þ 4y

15 Minimize

P ¼ 4x þ 5y

subject to

x þ 4y  76

subject to

x þ 2y  63

5x þ 8y 40

3x þ y  70

x þ 4y 32.

2x þ y 42 x þ 4y 84.

16 Maximize subject to

P ¼ 65x  23y 5x 

y  30

17 Maximize

P ¼ 24x  8y

subject to

x þ 3y  360 2x þ

10x þ 4y 150.

y  850

5x þ 25y 320. 18 Maximize subject to

P ¼ 4x þ 2y

20 Maximize

x þ 2y  60

subject to

P ¼ 60x þ 45y þ 25z 4x þ 8y þ 2z  160

3x þ 2y  80

6x þ 3y þ 4z  168

3x þ 10y 40.

4x þ 3y þ 3z 128.

19 Maximize

P ¼ 18x þ 40y þ 24z

21 Maximize

P ¼ 12x þ 8y  10z

subject to

5x þ 2y þ 4z  63

subject to

4x þ 2y  3z  210

2x þ 4y þ 2z  42

6x þ 8y þ z  630

2x þ 3y þ z 35.

2x  y þ 4z 210 x þ y þ z  180.

Optimization and linear programming

22 Minimize subject to

24 Minimize

1163

P ¼ 4x þ 3y

23 Minimize

x þ 4y  20

subject to

P ¼ 5x þ 8y x þ 2y  40

2x þ y  12

3x þ 2y  48

x  y  3.

x þ 4y 40.

P ¼ 4x þ 8y

P ¼ 2x þ 8y

25 Minimize

subject to x þ 2y  24

subject to 5x þ 4y  32 7x þ 4y  80

7x þ 6y  132

x þ 8y 40.

x þ 2y

4

x þ 2y 12. 26 Minimize

P ¼ 4x  8y þ 5z

P ¼ 6x  5y  3z

27 Minimize

subject to 2x þ 3y þ z  70

subject to 5x þ 8y þ 4z  220

x þ 2y þ 2z  60

2x þ y þ 6z  154

3x þ 4y þ z  84

4x þ 2y þ z 77

x þ y þ z 33.

x þ y þ 2z 55.

28 A firm manufacturing two types of switching module, A and B, is under contract to produce a daily output of at least 35 modules in all. Assembly and testing times for each type of module are as follows: Module type

A B

Processing time (hours) Assembly

Testing

1 .0 2 .0

2.0 1.0

Available staff resources provide a daily maximum of 80 hours for assembly and 55 hours for testing. The profit on the sale of each A-module is £40 and of each B-module £50. Determine (a) the daily production schedule for maximum profit. (b) the maximum daily profit.

1164

Programme 32

29 Three different types of coupling units are produced by a firm. The times required for machining, polishing and assembling a unit of each type are included in the information given in the following table. Type of unit

Profit (£) per unit

Process time (hours) per unit Machining

Polishing

Assembling

A

4

1

2

110

B

2

3

1

100

C

3

2

4

120

320

250

280

Available time (h/week)

The firm is required to supply a total of at least 100 units of mixed types each week. Determine (a) the weekly output of each type to maximize profit (b) the maximum weekly profit. 30 A firm makes three types of wooden cabinets, A, B, C, with profit margins of £35, £30, £24 per unit respectively. Time in hours per cabinet A

B

C

Preparation

2

5

4

Assembly

2

3

2

Finishing

5

4

3

Process

The manufacturer has 25 men available for preparation, 20 men for assembly and 30 men for polishing, and all staff work a 40 hour week. To remain competitive, at least 300 cabinets in all must be produced each week. Determine (a) the number of each model to be manufactured each week in order to maximize the profit (b) the maximum weekly profit.

Appendix 1 Green’s theorem If P and Q are two functions in x and y, finite and continuous inside a region R and on its boundary c in the x–y plane, with continuous first partial derivatives, then Green’s theorem states that  þ ðð  @P @Q  dx dy ¼  fPdx þ Qdyg @y @x c R

Proof of Green’s theorem Let the lower boundary of the region be the curve y1 ¼ f ðxÞ and the upper boundary the curve y2 ¼ FðxÞ. y d

y

y2 = F (x)

c

y2 = F (x)

y1 = f (x) O

a

y1 = f(x) O

x

b

a

b

x

Using vertical strips, we then have ðð R

@P dx dy ¼ @y ¼

ð b ð y2 a

ðb a

y1

@P dy dx ¼ @y

ð b " # y2 ¼F ðxÞ a

dx

P y1 ¼f ðxÞ

fPðx; y2 Þ  Pðx; y1 Þg dx

ðb

ða

Pðx; y1 Þ dx  Pðx; y2 Þ dx ¼ a b ð b ða ¼ Pðx; y1 Þ dx þ Pðx; y2 Þ dx b þ a ¼  Pðx; yÞ dx

1165

ð1Þ

1166

Appendix

Similarly, using horizontal strips, we have y d

y d x1 = g(y) x1 = g(y)

c

x

O

ðð R

c

x2 = h(y)

@Q dx dy ¼ @x ¼

ð d ð x2 c

x1

ðd " c

O

x2 = h(y)

x

@Q dx dy @y # x2 ¼hð yÞ dy

Q x1 ¼gð yÞ

where x1 ¼ gðyÞ and x2 ¼ hðyÞ are the left-hand and right-hand portions of the boundary curve c. ðð ; R

@Q dx dy ¼ @x ¼

ðd

Qðx2 ; yÞ dy 

c

ðd þ

Qðx1 ; yÞ dy

c

Qðx2 ; yÞ dy þ

c

¼

ðd

ðc

Qðx1 ; yÞ dy

d

Qðx; yÞ dy c

;

 þ þ ðð  @P @Q  dx dy ¼  Pðx; yÞ dx  Qðx; yÞ dy @y @x c c R

þ ¼

fP dx  Q dyg c

ð2Þ

Appendix

1167

2 Proof that sec γ =

2

( ) ( )

1+ z + x

z y

2

Let , ,  be the angles that OP makes with the x, y and z axes respectively. z

(x, y, z) γ

β

α x

r z y

O

y

x

Then x ¼ r cos ; y ¼ r cos ; z ¼ r cos  Also

x2 þ y2 þ z2 ¼ r 2

If r ¼ 1 unit, then x2 þ y2 þ z2 ¼ 1 ; z2 ¼ 1  x2  y2 ; z ¼ ð1  x2  y2 Þ1=2 @z 1 ¼ ð1  x2  y2 Þ1=2 ð2xÞ @x 2 x ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  x2  y2 @z 1 ¼ ð1  x2  y2 Þ1=2 ð2yÞ @y 2 y ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  x2  y2 

  2 @z 2 @z x2 y2 ; 1þ þ ¼1þ þ @x @y 1  x2  y2 1  x2  y2 1  x2  y2 þ x2 þ y2 1  x2  y2 1 1 ¼ ¼ 2 2 2 1x y z ¼

1 ¼ sec2  z2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2  2 @z @z ; sec  ¼ 1 þ þ @x @y

But, with r ¼ 1, z ¼ cos 

;

1168

Appendix

3 Vector triple products (a) A  ðB  CÞ ¼ ðA  CÞ B  ðA  BÞ C (b) ðA  BÞ  C ¼ ðC  AÞ B  ðC  BÞ A Let

A ¼ ax i þ ay j þ az k;

B ¼ bx i þ by j þ bz k;

C ¼ cx i þ cy j þ cz k Then B  C ¼ ðbx i þ by j þ bz kÞ  ðcx i þ cy j þ cz kÞ i

j

k

¼ bx

by

bz

cx

cy

cz

¼i

Then

by

bz

cy

cz

A  ðB  CÞ ¼

( ¼i

ay bx cx

j

bx

bz

cx

cz

þk

bx

by

cx

cy

i

j

k

ax

ay

az

by

bz

bz

cy

cz

cz

by  az bz

bx

cy

cx

cz

bx )

cx

(

j

bx

by

cx

cy

ax bx cx

( þk

ax bz cz

by  az by

bz

cy

cy

cz

bx  ay by

bz

cy

cz

cx

)

)

¼ i fay ðbx cy  by cx Þ  az ðbz cx  bx cz Þg þ j faz ðby cz  cy bz Þ  ax ðbx cy  by cx Þg þ k fax ðbz cx  bx cz Þ  ay ðby cz  bz cy Þg ¼ i fbx ax cx þ bx ay cy þ bx az cz  cx ax bx  cx ay by  cx az bz g þ j fby ax cx þ by ay cy þ by az cz  cy ax bx  cy ay by  cy az bz g þ k fbz ax cx þ bz ay cy þ bz az cz  cz ax bx  cz ay by  cz az bz g ¼ i fbx ðax cx þ ay cy þ az cz Þ  cx ðax bx þ ay by þ az bz Þg þ j fby ðax cx þ ay cy þ az cz Þ  cy ðax bx þ ay by þ az bz Þg þ k fbz ðax cx þ ay cy þ az cz Þ  cz ðax bx þ ay by þ az bz Þg

Appendix

1169

A  C ¼ ðax i þ ay j þ az kÞ  ðcx i þ cy j þ cz kÞ ¼ ax cx þ ay cy þ az cz

Now

and similarly A  B ¼ ax bx þ ay by þ az bz ; A  ðB  CÞ ¼ i fbx ðA  CÞ  cx ðA  BÞg þ j fby ðA  CÞ  cy ðA  BÞg þ k fbz ðA  CÞ  cz ðA  BÞg: ; A  ðB  CÞ ¼ ðA  CÞ fibx þ jby þ kbz g  ðA  BÞ ficx þ jcy þ kcz g ; A  ðB  CÞ ¼ ðA  CÞ B  ðA  BÞ C In the same way, it can be established that ðA  BÞ  C ¼ ðC  AÞ B  ðC  BÞ A

4 Divergence theorem (Gauss’ theorem) ð

ð div F dV ¼

To prove that V

F  dS for the region V bounded by the surface S. S

z

dz O

y

dy

dx

x

Consider an element of volume dV ¼ dx dy dz and let the components of F in the x, y and z directions be denoted by Fx i, Fy j and Fz k respectively at any point P. ð We then determine F  dS over the element dV and finally sum the results for all such elements throughout the region. (a) S1 :

dS1 ¼ dy dz;

n¼i

–i S S

i

ðF  dSÞ1 ¼ ðFx i þ Fy j þ Fz kÞ  ðnÞ dS1 ¼ ðFx i þ Fy j þ Fz kÞ  ðiÞ dS1 ¼ Fx dS1

1170

Appendix

(b) S2 :

dS2 ¼ dy dz;

n ¼ i

; ðF  dSÞ2 ¼ ðFx i þ Fy j þ Fz kÞ  ðiÞ dS2 ¼ Fx dS2 Combining these two results, we have ðF  dSÞ1 þ ðF  dSÞ2 ¼ ðFx dSÞ1  ðFx dSÞ2 @ ðFx dSÞ dx @x   @Fx @Fx dS dx ¼ dx dy dz F  dS ¼ @x @x ¼

ð ; S1 þ S2

ð1Þ

Similarly, for S3 and S4 we have –j

j

S

ð

S S3 þ S4

  @Fy F  dS ¼ dx dy dz @y

ð2Þ

and for S5 and S6 k



ð

S

F  dS ¼ S

S5 þ S6

@Fz @z

 dx dy dz

ð3Þ

–k

These three results together cover the total surface of the element dV.   ð @Fx @Fy @Fz þ þ dx dy dz ¼ div F dV F  dS ¼ @x @y @z S1 ... S6

Finally, summing the results for all such elements throughout the region with dV ! 0 and dS ! 0, we obtain ð Xð div F dV ¼ F  dS with dS ! 0 V

Appendix

1171 ð

On the common boundaries between adjacent elements, the values of

F  dS

cancel out. On the boundary surface, however, there are no such adjacent faces þ and the integral F  dS remains. S

ð ;

ð

div F dV ¼ V

F  dS S

5 Stokes’ theorem If F is a single-valued vector field, continuous and differentiable over an open surface S and on the boundary c of the surface, then n

þ

ð

dS

curl F  dS ¼

F  dr c

S

c

Proof of Stokes’ theorem Consider the surface S divided into small rectangular elements and let ABCD be one such element. If axes of reference x and y be arranged to coincide with AB and AD respectively as shown, a third axis z will then be normal to the surface at A. z y dy

dx dy

dx

If

AB ¼ dx, then dx ¼ i dx and

if

AD ¼ dy, then dy ¼ j dy.

x

Let Fa denote the vector field at A; Fb that at B; Fc that at C; and Fd that at D. Now consider each side in turn. AB: F  dr ¼ Fa  dx ¼ fFax i þ Fay j þ Faz kg  fi dx g ¼ Fax dx BC: F  dr ¼ Fb  dy ¼ fFbx i þ Fby j þ Fbz kg  fj dy g ¼ Fby dy CD: F  dr ¼ Fc  dx ¼ fFcx i þ Fcy j þ Fcz kg  fi dx g ¼ Fcx dx DA: F  dr ¼ Fd  dy ¼ fFdx i þ Fdy j þ Fdz kg  fj dy g ¼ Fdy dy

1172

Appendix

(a) AB þ CD: ð F  dr ¼ Fax dx  Fcx dx dy

¼ ðFcx  Fax Þ dx

dx

ð ; ðAB þ CDÞ

¼  Fx dx @Fx dy dx ¼ @y @Fx dx dy F  dr ¼  @y

ð1Þ

(b) BC þ DA: ð F  dr ¼ Fby dy  Fdy dy dy

dx

ð ; ðBC þ DAÞ

¼ ðFby  Fdy Þ dy ¼ Fy dy @Fy dx dy ¼ @x @Fy dx dy F  dr ¼ @x

Adding these two results together for the complete rectangle, we have  ð @Fy @Fx  F  dr ¼ dx dy @x @y

ð2Þ

ð3Þ

ðABCDÞ

i j k @ @ @ Now curl F ¼ @x @y @z Fx Fy Fz       @Fy @Fx @Fz @Fy @Fz @Fx j þk    ¼i @y @z @x @z @x @y  @Fy @Fx ;  ¼ ðcurl FÞ  ðkÞ @x @y ð F  dr ¼ curl F  k dx dy ¼ curl F  dS From (3)

ð4Þ

ABCD

Summing for all such elements over the surface 8 9 ð = X< ð curl F  dS ¼ Lim F  dr : ; dr!0 S

ABCD

ð5Þ

Appendix

1173

c

ð

F  dr on boundary lines between adjacent rectangular elements will cancel out, except on the boundary curve c of the surface S. The integral then becomes þ F  dr. c

ð

þ

;

curl F  dS ¼ S

F  dr c

Answers Test exercise 1 (page 42)

pffiffiffi 1 x ¼ 1  j 3; x2 þ 2x þ 4 ¼ 0 2 x ¼ 4, 6, 3=2 3 x ¼ 1:6_ ¼ 5=3 4 x  1:710 5 x  0:454304 6 x  1:317672 7 (a) 39.375 (b) 103.392 (c) 481.528 8 12:8

Further problems 1 (page 43) 1 2 6 9 14 16 18

21 24 25

pffiffiffi pffiffiffi pffiffiffi pffiffiffi 1 þ j 3 1  j 4 , pffiffiffi , x þ ð1 þ 2Þx3 þ ð2 þ 2Þx2 þ ð1 þ 2Þx þ 1 ¼ 0 2 2 x ¼ 1, 6,  2 3 p ¼ 5, q ¼ 1 4 p ¼ 4, q ¼ 9 5 x ¼ 2, 3,  3 x ¼ 1,  3, 9 7 y3  5y2 þ 17y  13 ¼ 0 8 y3  13y2 þ 52y  60 ¼ 0 x ¼ 12 , 32 ,  1 10 x ¼ 2, 4, 8 11 2y3  15y2 þ 25y ¼ 0 13 0:8934 x ¼ 2:732,  0:732,  2:000 15 y3  3y þ 2 ¼ 0; x ¼ 4,  1,  1 x ¼ 1:646 17 (a) 0:6736 (b) 0:3717 (a) 2:3301, 0:2016, 2:1284 (b) 1,  0:50  j1:66 (c) 2:115, 0:254, 1:861 19 (a) 4:104,  0:9481  j0:5652 (b) 0:5,  1:5,  1:5 (c) 0:25, 1  j3 20 (a) 2:456 (b) 1:765 (c) 0:739 (d) 1:812 (e) 1:8175 (f) 0:5170 (g) 0:8449 (h) 0:8806 (a) 32:872 (b) 204.328 (c) 381.375 22 (a) 1:375 and 81.104 (b) 136:971 and 363:429 23 (a) 6:048 (b) 461:496 (a) 133 and 9:048 (b) 0.136 and 65:433 (c) 200:312 and 867 0:02768 26 1:0670 27 (a) 2:54846 (b) 2:41734 (c) 1:87134

Test exercise 2 (page 90) 1 (a) (e)

32  2s s2  16 6s ðs2 þ 9Þ2

sþ4 1 sþ2 (c) 4 f4s3  s2 þ 4s þ 6g (d) 2 s2þ 16 s s þ 4s þ 29 pffiffiffi pffiffiffi sþ2 5 2 (a) 2e3t  e4t (b) 2 cos 2t þ pffiffiffi sin 2t  et (f) ln sþ1 2 (b)

þ 9 sin 2tÞ  e3t g 3 (a) x ¼ e2t þ e3t   t3 t 1 2t 1 5 3t 5 4t (b) x ¼ 12 f13e  cos 2t  sin 2tg (c) x ¼ 6  3 e þ 2 e (d) x ¼ e 1  t þ 6 1 3t 2t 4 x ¼ 2 f9 cos t  7 sin t  e g y ¼ 3 sin t  2 cos t þ e (c) et ð3t þ 2Þ  e3t

(d)

1 t 8 fe ð17 cos 2t

Further problems 2 (page 91) 1 (a) (e)

s4 2 s  8s þ 20 2s3  6s ðs2 þ 1Þ3

(b)

4s

(c)

6 8 5 4s2  24s þ 38 (d) þ þ s4 s3 s ðs  3Þ3

ðs2 þ 4Þ2 rffiffiffiffiffiffiffiffiffiffiffi sþ2 2 (a) e2t þ e4t (f) ln s2

1174

(b) 3e4t þ 2

Answers

1175

 2 3t þ 2t þ 1 (d) et f2 cos t  5 sin tg  2e2t (e) 13 ðcos t  cos 2tÞ 2 4t=3 3  26 fcos 2t þ 23 sin 2tg (f) e2t fcos 4t  74 sin 4tg 3 x ¼ 4e4t  2 4 x ¼ 35 78 e (c) e2t

5 x ¼ et ð2t þ 1Þ þ 2t þ 4 þ cos t

6 x ¼ 32 e4t  e3t  12 e2t

7 x ¼ 45 cos 3t þ sin 3t þ 15 cos 2t 8 x ¼ 15 fe2t  et ðcos 2t  2 sin 2tÞg 9 x ¼ 18 f2t 2  4t þ 3 þ e2t ð4t 2 þ 6t þ 1Þg 10 x ¼ 25 f2ðe4t  1Þ cos 4t þ ðe4t þ 1Þ sin 4tg 11 x ¼ ð2t þ 1Þ cos 5t þ t sin 5t 1 1 f2e2t þ 3e2t  5ðcos 3t  sin 3tÞg y ¼ 13 f5ðcos 3t þ sin 3tÞ  3e2t  2e2t g 12 x ¼ 13

13 x ¼ 16 f7e6t þ 5g y ¼ 13 f7e6t þ 5g 14 x ¼ 10e4t þ 2 y ¼ 5e4t þ 3 15 x ¼ e2t  et þ 2t

y ¼ 3et þ 12 e2t þ t  72

17 x ¼ 4 cos t  2 sin t 

1 t 3 f8e

16 x ¼ 5et þ 3et

þ e g y ¼ 6 cos t þ 2 sin t  pffiffiffi pffiffiffi pffiffiffi 5 18 x ¼ 3 fcos 2t þ sin 2t  cosh 2t  2 sinh 2tg 2t

y ¼ 4et  et

4 t 3 f2e

þ e2t g

4 19 y ¼ 15 f3 sin 2t  4 cos 2t þ 43 sin 3t þ 48 7 cos 3tg  7 cos 4t rffiffiffiffiffiffi! rffiffiffiffiffiffi! pffiffiffi pffiffiffi 3 3 5 3 1 þ cosðt 6Þ y ¼ cos t  cosðt 6Þ 20 x ¼ cos t 10 4 4 10 4

Test exercise 3 (page 121) 1 (a) FðsÞ ¼

f(t) O

 3 1  e2s 2 s

t

(b) e –2t

FðsÞ ¼

f(t)

 1  1  e6 e3s sþ2

t

O

(c)

f(t) O

t

 2 2 1 2s 1 FðsÞ ¼ 3  2e þ þ s s3 s2 s 2 þ e3s s

(d) f(t) O

π 2

π

t

FðsÞ ¼

2 f1  es g s2 þ 4

1176

Answers 2 f ðtÞ ¼ 2  uðtÞ  5  uðt  1Þ þ 8  uðt  3Þ

f(t) O

t

3 f ðtÞ ¼ t  uðtÞ þ 3ðt  2Þ  uðt  2Þ  ðt  3Þ  uðt  3Þ  3ðt  5Þ  uðt  5Þ

f(t) O

t

4 f ðtÞ ¼ uðtÞ sinh t  uðt  1Þ sinhðt  1Þ 5

t sin 4t 4

Further problems 3 (page 122) 1 f ðtÞ ¼ 3  uðtÞ þ 2ðt  2Þ  uðt  2Þ  2ðt  5Þ  uðt  5Þ 2 f ðtÞ ¼ t  uðtÞ  ðt  1Þ  uðt  1Þ þ ðt  2Þ  uðt  2Þ  ðt  3Þ  uðt  3Þ 3 f(t) FðsÞ ¼ O

4 3e3s 2e3s þ þ 2 s s s

t

4 (a) f ðtÞ ¼ t 2  uðtÞ  ðt 2  5tÞ  uðt  3Þ (b) f ðtÞ ¼ cos t  uðtÞ þ ðcos 2t  cos tÞ  uðt  Þ þ ðcos 3t  cos 2tÞ  uðt  2Þ   1 3 1 4  e3s 2 þ 5 FðsÞ ¼ e2s 2 þ s s s s 6 (a) f ðtÞ ¼ t 2  uðtÞ  t 2  uðt  2Þ þ 4  uðt  2Þ  4  uðt  5Þ 2 2e2s 4e2s 4e5s (b) FðsÞ ¼ 3  3  2  s s s s   2t 3t 2t þe þ e2ðt1Þ uðt  1Þ uðtÞ  7 (a) t t 1   uðt  1Þ  2t 2t uðtÞ  1  e2ðt1Þ þ 2ðt  1Þe2ðt1Þ (b) 1  e þ 2te 4 4  t uðtÞ  t1 uðt  1Þ t (c) 3e  3e  sin 3t  3e  3eðt  1Þ  sin 3ðt  1Þ 20 20 ( pffiffiffi pffiffiffi ) uðt  1Þeðt1Þ=2 pffiffiffi 3t 3t pffiffiffi  sin (d) ðt  2Þuðt  1Þ þ 3 cos 2 2 3

Answers

1177

o pffiffiffi  1 1n 1  et ðcos t þ sin tÞ (b) cosh 2t  1 2 2 pffiffiffi  2t t 2 (c)  sinh pffiffiffi þ sinh pffiffiffi 5 3 2

8 (a)

Test exercise 4 (page 154)

 2 1  e2s  2se2s 2 (a) e6 1 FðsÞ ¼ s2 ð1  e4s Þ 3 (a) FðsÞ ¼ 4e3s (b) FðsÞ ¼ e2ð3þsÞ

(b) 0 (c) 11

4δ(t – 2)

4 f(t) 3δ(t)

FðsÞ ¼ 3 þ 4e2s  3e4s

O

t 3δ(t – 4)

  5 x ¼ e3t f4 sin t  cos t g 6 x ¼ 3e4 et  uðt  4Þ þ e2t 2  uðtÞ  3e8  uðt  4Þ 7 (a) f ðtÞ ¼ sin t, frequency 1 radian per unit of time, period 2 units of time pffiffiffiffiffiffi! pffiffiffiffiffiffi 53 53 18 t=6 (b) f ðtÞ ¼ pffiffiffiffiffiffi e sin t, frequency radian per unit of time, 6 6 53 12 period pffiffiffiffiffiffi units of time 53 pffiffiffi pffiffiffi  et  pffiffiffi 2 5t e 32 2 sin 2t  40 cos 2t , steady-state solution 8 Transient solution 19 19

Further problems 4 (page 155) 2 Lff ðtÞg ¼

1  e2ð1sÞ 1  es ðs þ 1Þ (c) FðsÞ ¼ s2 ð1  e2s Þ ðs  1Þð1  e2s Þ  2s 2s 1 2 2e 4e 4e3s (d) FðsÞ ¼    1  e3s s3 s3 s2 s   P E t x¼ sin !t 5 i ¼ cos pffiffiffiffiffiffiffi M! L LC x ¼ 2e2t f1 þ 10e8  uðt  4Þg  2e3t f1 þ 10e12  uðt  4Þg pffiffiffi t t 1 (a) f ðtÞ ¼ 4 3 sin pffiffiffi  cos pffiffiffi, frequency pffiffiffi radian per unit of time, 2 3 2 3 2 3 pffiffiffi pffiffiffi pffiffiffi 1 period 4 3 units of time (b) f ðtÞ ¼ 2 cos 2 3t  pffiffiffi sin 2 3t, 2 3 pffiffiffi pffiffiffi frequency 2 3 radian per unit of time, period  3 units of time  : 14 (a) f ðtÞ ¼ 4:48 sin 0:69t þ 1:06 cos 0:69t (b) f ðtÞ ¼ 1 sin½ð1 5Þ t ð3=2Þ4 (b) FðsÞ ¼

4 6 7

8

 ws að1 þ es Þ 1 w e 3 (a) FðsÞ ¼  2 s 2 ðs þ 1Þð1  e Þ s s 1  ews

1178

Answers

9 Transient solution e3t=8 steady-state solution

pffiffiffiffiffiffi pffiffiffiffiffiffi ! 421 1 23 23 pffiffiffiffiffiffi sin t  cos t , 9 8 8 9 23

1 t e 9

Test exercise 5 (page 191) 1 (a) f ðn þ 1Þ ¼ f ðnÞ þ 5, f ð1Þ ¼ 4 (b) f ðn þ 1Þ ¼ f ðnÞ  4, f ð0Þ ¼ 23 (c) f ðn þ 1Þ ¼ f ðn=3Þ, f ð2Þ ¼ 9 2 (a) order 3: 1, 1, 3, 1, 4, 13 (b) order 2: 0, 1, 5, 21, 79, 275 (c) order 2: 2, 5, 32, 25, 406, 103 1 z 4zðz  aÞ  2zðz  1Þ2 (b) 3 f ðnÞ ¼ 3  2nþ2 þ 3nþ3 þ 2n þ 5 4 (a) 2 zþ1 ðz  aÞðz  1Þ2 zð4  3zÞ 25z 5 f ðnÞ ¼ ð2n þ 1  2n ÞuðnÞ (c) (d) z5 ðz  1Þ2  z sin T 6 f ðnÞ ¼ 2n1 n  2nþ1 þ 3 uðnÞ 7 2 z  2z cos T þ 1

Further problems 5 (page 191)

 z 1 3 2 provided jzj > jaj 2 (a) uk  ð3Þk þ ð2Þk zþa 12 4 3   1 k 3 2 k 1 k k uk  þ ð1=3Þ ð3 Þ þ ð3Þ  2k (c) (b) 4 2 4 3 3   1 1 3 k1 k1 k ð1 þ jÞðjÞ þ ð1  jÞðjÞ 3 4 (a) uk þ kð2Þ 2 2 2  2 1 5 8 z z uk  kð2Þk þ ð2Þk (b) 2 (b) 5 (a) 2 9 6 9 z 1 z 1 1

z7 þ z5 þ z4 þ 1 z7 þ z6 þ z5 þ z þ 1 z7 þ z6 þ z5 þ z þ 1 (d) (e) z7 z7 z10  6 5 z þz þzþ1 1 k k k ðð3Þ (f) 6 (a) fx g ¼  2ð2Þ þ ð1Þ Þ for k 1 k z6 2  1 ðð3kþ1  ð2Þkþ2 þ ð1Þkþ1 Þ (c) fxk g ¼ f10ð3k Þ  7ð2k Þg (b) fxk g ¼ 2 2 z sinh T (d) fxk g ¼ f6ð2k Þ  3uk g 9 3 10  13 (a) 2 7 z  2z cosh T þ 1 zðz  cosh aTÞ zeaT ðzeaT  cosh bTÞ (c) 2 (b) 2 z  2z cosh aT þ 1 z  2zeaT cosh bT þ e2aT (c)

f(t)

O

T

2T

3T

4T

5T .. .

t

Answers

1179

1  1 1 þ 8ð2Þn  9ð3Þn uðnÞ (b) f ðnÞ ¼ 31n  2n þ 1 uðnÞ 12 4 1  1 10ð3n Þ þ 5ð3Þn  3 uðnÞ (d) f ðnÞ ¼ 3n þ 1  ð5Þn uðnÞ (c) f ðnÞ ¼ 12 9 1 n 81ðz  2Þ 16 gðnÞ ¼ 3  ð2Þn uðnÞ 18 5 z2 ðz  3Þ2 ( pffiffiffi!n pffiffiffi!n ) 1 1þ 5 1 5 19 f ðnÞ ¼ pffiffiffi  2 2 5 14 (a) f ðnÞ ¼

Test exercise 6 (page 236) 1 (a) (c) (e) (g) 2

4 5 6

linear and time-invariant (b) nonlinear, shift-invariant nonlinear, not time-variant (d) nonlinear, shift-invariant linear, not time-invariant (f) linear, not shift-invariant linear, time-invariant (h) linear, not shift-invariant  1  3t 2e  ð9t 2 þ 6t þ 2Þ , not time-invariant (a) yzi ðtÞ ¼ 2e3t , yzs ðtÞ ¼ 27  4 1  4t 2e  17et þ 3ð14 sin t þ 5 cos tÞ , (b) yzi ðtÞ ¼ et  e4t , yzs ðtÞ ¼ 3 102 not time-invariant (c) yzi ðtÞ ¼ 0, yzs ðtÞ ¼ e4t=5  et , time-invariant e3t 9 (d) yzi ðtÞ ¼ 0, yzs ðtÞ ¼ 1  ð1 þ tÞet , time-invariant 3 ðe  1Þ 3 es tðt þ 2Þ fuðt  1Þ  2uðt  2Þ þ uðt  3g HðsÞ ¼ 2 ðs þ 1Þ; yðtÞ ¼ 2 s 1 ; yðtÞ ¼ 2et  5e2t þ 3e4t HðsÞ ¼ ðs þ 4Þðs  1Þ 1  y½n ¼ 2  4nþ1 þ 3n þ 1 u½n 7 h½n ¼ 2n u½n n 94

8 y½n  2y½n  1 þ y½n  2 ¼ x½n  1

Further problems 6 (page 237) 1 All non-zero values 2 All values of a but only b ¼ 0 sinh 3t 4e jn!0 5 Yes 6 Yes 13 3 Linear but not time-invariant 4 3 4  e jn!0     t2 1 t 14 yðtÞ ¼ et 15 exp  uðtÞ 16 yðtÞ ¼ G 1  et=T a a 2   nþ1 17 y½n ¼ 1   uðnÞ 18 y½n ¼ 20 þ 140ð0:93Þn u½n 19 y½n ¼ nu½n 20 hðtÞ ¼ uðtÞ: yðtÞ ¼ ðt  1Þuðt  1Þ 21 h½n ¼ nu½n  2ðn  1Þu½n  1: y½n ¼

u½n ð2n  1 þ 3n Þ 4

1180

Answers

Test exercise 7 (page 266) 1 Amplitude

2

pffiffiffi 8 2, period 3

8 : 0x

: x3 : 3x

2x > > : 1 f ðx þ 5Þ ¼ f ðxÞ

: 4  x < 3 : 3  x < 1 : 1  x < 0 : 0x 3 3 y4 ¼ 480x þ 96

4

y6 ¼ fðx4  180x2 þ 360Þ cos x þ ð24x3  480xÞ sin xg

Answers

1185

5 y4 ¼ 4ex sin x 6 y3 ¼ 2xð13 þ 12 ln xÞ 8 y6 ¼ 1018 10 (a) y2n ¼ fx2 þ 2nð2n  1Þg sinh x þ 4nx cosh x (b) y2n ¼ fx3 þ 6nð2n  1Þxg cosh x þ f6nx2 þ 2nð2n  1Þð2n  2Þg sinh x pffiffiffi 11 y6 ¼ 25 e2x f2x3 þ 24x2 þ 81x þ 75g 12 y3 ¼ 2 2a3 eax fcosðax þ =4Þg   9x2 15x4 7x6 27x8 4x3 þ  þ þ . . . þ y1 x þ 14 y ¼ y0 1 þ 2 8 16 128 3 15 y ¼ Að1 þ x2 Þ þ Bex  32  x2 32  52  x4 32  52  72  x6 þ þ þ ... 16 y ¼ y0 1 þ 2! 4! 6!  42  x3 42  62  x5 þ y1 x þ þ þ ... 3! 5!  x4 x6 x8 17 y ¼ y1 x þ y0 1  x2     . . . 3 5 7  2 4 3 2x 2  x 2  x6 18 y ¼ y0 1  2 þ 2  þ ... 2 2  42 22  42  62  2x3 22  x5 23  x7  þ ... þ y1 x  2 þ 2 3 3  52 32  52  72 19 yðxÞ ¼ Ax þ Bx4  x3 pffiffiffiffi 11=2

20 yðxÞ ¼ Ax2=3 þ Bx3=2 þ pffiffiffiffi 11=2

2x3 3x2  99 56

4x3 9 pffiffi pffiffi x ð1=2Þþj 3=2 ð1=2Þj 3=2 þ Bx þ 22 yðxÞ ¼ Ax 3

21 yðxÞ ¼ Axð1=2Þþj

þ Bxð1=2Þj

þ

Test exercise 11 (page 376)   x x2 x3 x x2 x3 1  þ . . . þ Bx3 1  þ  þ ... 1 (a) y ¼ A 1  þ 2 20 480 4 56 1680   x4 x8 x12 x4 x8 x12 (b) y ¼ A 1  þ  . . . þ Bx x  þ  þ ... 12 672 88 704 20 1440 226 640  x x2 x3 (c) y ¼ ðA þ B ln xÞ 1 þ þ þ þ ... 3 24 360    x1 4x 25x2 157x3    . . . þA 2  x2 9 288 21 600

Further problems 11 (page 376)  x2 x3 x4 þ þ þ ... 1 (a) y ¼ A 1 þ x þ 2  4 ð2  3Þð4  7Þ ð2  3  4Þð4  7  10Þ  x x2 x3 2 þ þ þ ... þ Bx3 1 þ 1  5 ð1  2Þð5  8Þ ð1  2  3Þð5  8  11Þ   2 x x4 x3 (b) y ¼ a0 1  þ þ . . . þ a1 x  þ . . . 2! 4! 3!

1186

Answers  x3 x6 þ þ ... (c) y ¼ a0 1 þ 2  3 ð2  3Þð5  6Þ  x4 x7 þ a1 x þ þ þ ... 3  4 ð3  4Þð6  7Þ  x x2 x3 þ  þ ... (d) y ¼ A 1  1  4 ð1  2Þð4  7Þ ð1  2  3Þð4  7  10Þ  x x2 x3 1 þ  þ ... þ Bx 3 1  1  2 ð1  2Þð2  5Þ ð1  2  3Þð2  5  8Þ  x2 x4 3x6 ð3Þð5Þx8 (e) y ¼ a1 x þ a0 1     þ ... 2! 4! 6! 8!  4 x x5 þ  ... (f) y ¼ u þ v where u ¼ A 4! 3! 5! 3!   4    x x5 x x2 þ þ  ... þ 1 þ þ ... v ¼ B ln x 1  3 ð1  2Þð2  3Þ 4! 3! 5! 3!  3x 32  x2 33  x3 (g) y ¼ u þ v where u ¼ A 1 þ 2 þ 2 þ þ ... 1 1  22 12  22  32    3x 32  x2 33  x3 þ þ ... v ¼ B ln x 1 þ 2 þ 2 1 1  22 12  22  32   2  3x 3  32  x2 11  33  x3 þ þ 2 þ ...  12 12  22 1  22  33

Test exercise 12 (page 394) 1 P2 ðxÞ ¼

3x2  1 5x3  3x 1 4 , P3 ðxÞ ¼ 2 P0 ðxÞ  P2 ðxÞ 2 2 3 3

Further problems 12 (page 395) 1 eigenfunctions: yn ðxÞ ¼ An cos

pffiffiffiffiffi ð2n þ 1Þ2 2 n x; eigenvalues: n ¼ 4

2 H0 ¼ 1, H1 ¼ 2x, H2 ¼ 4x2  2, H3 ¼ 8x3  12x 3 L0 ¼ 1, L1 ¼ 1  x, L2 ¼ 2  4x þ x2 , L3 ¼ 6  18x þ 9x2  x3

Text exercise 13 (page 434) 1

x

y

2

x

y

0

1:0

1

0

0:1

1:1

1:2

0:204

0:2

1:211

1:4

0:4211

0:3

1:3352

1:6

0:6600

0:4

1:4753

1:8

0:9264

0:5

1:6343

2:0

1:2243

Answers

3

5

1187

x

y

4

x

y

0

1:0

2:0

3:0

0:1

1:2052

2:1

3:005

0:2

1:4214

2:2

3:0195

0:3

1:6499

2:3

3:0427

0:4

1:8918

2:4

3:0736

0:5

2:1487

2:5

3:1117

x

y

6

x

y

1:0

0

0:0

0:0000000

1:1

0:1052

0:1

0:1005000

1:2

0:2215

0:2

0:2030226

1:3

0:3401

0:3

0:3096820

1:4

0:4717

0:4

0:4227589

1:5

0:6180

0:5

0:5448011

0:6

0:6787373

0:7

0:8280166

0:8

0:9967810

0:9

1:1900859

1:0

1:4141835

Further problems 13 (page 435) 1

x

y

2

x

y

0

1:0

0

1:4

0:2

0:8

0:1

1:596

0:4

0:72

0:2

0:8707

0:6

0:736

0:3

2:2607

0:8

0:8288

0:4

2:8318

1:0

0:9830

0:5

3:7136

1188

Answers

3

x

9

x

y

1:0

2:0

0

0:5

1:2

2:0333

0:1

0:543

1:4

2:1143

0:2

0:5716

1:6

2:2250

0:3

0:5863

1:8

2:3556

0:4

0:5878

2:0

2:5000

0:5

0:5768

x

y

5

7

4

y

x

y

0

1:0

0:1

1:1022

0:2

1:2085

0:3

1:3179

0:4

1:4296

0:5

1:5428

x

y

6

8

1:0

1:0

1:1

1:1871

1:2

1:3531

1:3

1:5033

1:4

1:6411

1:5

1:7688

x

y

0

0

0

1:0

0:1

0:1002

0:2

0:8562

0:2

0:2015

0:4

0:8110

0:3

0:3048

0:6

0:8465

0:4

0:4110

0:8

0:9480

0:5

0:5214

1:0

1:1037

x

y

x

y

10

0

1:0

0

0:4

0:1

0:9138

0:2

0:4259

0:2

0:8512

0:4

0:4374

0:3

0:8076

0:6

0:4319

0:4

0:7798

0:8

0:4085

0:5

0:7653

1:0

0:3689

Answers

1189

11

13

15

17

x

y

12

x

y

1:0

2:0

0

1:0

1:2

2:4197

0:2

1:1997

1:4

2:8776

0:4

1:3951

1:6

3:3724

0:6

1:5778

1:8

3:9027

0:8

1:7358

2:0

4:4677

1:0

1:8540

x

y

14

x

y

0

1:0

0

1:0

0:2

1:1679

0:1

1:11

0:4

1:2902

0:2

1:2422

0:6

1:3817

0:3

1:4013

0:8

1:4497

0:4

1:5937

1:0

1:4983

0:5

1:8271

x

y

x

y

16

0

3:0

0

0

0:1

2:88

0:2

0:1987

0:2

2:5224

0:4

0:3897

0:3

1:9368

0:6

0:5665

0:4

1:1424

0:8

0:7246

0:5

0:1683

1:0

0:8624

x

y

x

y

18

0

1:0

0

2:0

0:2

1:1972

0:1

2:0845

0:4

1:3771

0:2

2:1367

0:6

1:5220

0:3

2:1554

0:8

1:6161

0:4

2:1407

1:0

1:6487

0:5

2:0943

1190

Answers

19

21

23

x

y

20

x

y

0

1:0

1:0

0

0:2

1:0367

1:2

0:1833

0:4

1:1373

1:4

0:3428

0:6

1:2958

1:6

0:4875

0:8

1:5145

1:8

0:6222

1:0

1:8029

2:0

0:7500

x

y

x

y

1:0

2:0000

0:0

1:0000

1:2

2:0333

0:2

0:8600

1:4

2:1121

0:4

0:8118

1:6

2:2219

0:6

0:8452

1:8

2:3522

0:8

0:9454

2:0

2:4965

1:0

1:1002

x

y

1:0

2:0000

1:2

2:4191

1:4

2:8769

1:6

3:3715

1:8

3:9018

2:0

4:4666

22

Test exercise 14 (page 478) 1 (a) solutions unique (b) infinite number of solutions 2 x1 ¼ 4, x2 ¼ 2, x3 ¼ 3 3 x1 ¼ 2, x2 ¼ 2, x3 ¼ 3 4 x1 ¼ 3, x2 ¼ 4, x3 ¼ 2 5 x1 ¼ 1, x2 ¼ 2, x3 ¼ 2 6 a ¼ 2, b ¼ 1, c ¼ 5, d ¼ 0, e ¼ 4

: 

 7 196 8 (b) 7 (a) 0:464 1

Further problems 14 (page 479) 1 x1 ¼ 1, x2 ¼ 4, x3 ¼ 3 2 (a) x1 ¼ 3, x2 ¼ 1, x3 ¼ 4 (b) x1 ¼ 4, x2 ¼ 2, x3 ¼ 1 3 (a) x1 ¼ 4, x2 ¼ 2, x3 ¼ 5, x4 ¼ 3 (b) x1 ¼ 5, x2 ¼ 4, x3 ¼ 1, x4 ¼ 3 (c) x1 ¼ 3, x2 ¼ 2, x3 ¼ 0, x4 ¼ 5 4 (a) x1 ¼ 3, x2 ¼ 1, x3 ¼ 3 (b) x1 ¼ 5, x2 ¼ 2, x3 ¼ 1 (c) x1 ¼ 4, x2 ¼ 3, x3 ¼ 1, x4 ¼ 2 6 I1 ¼ 2, I2 ¼ 3, I3 ¼ 2 7 x1 ¼ 2, x2 ¼ 0:5, x3 ¼ 1

Answers

1191

Test exercise 15 (page 510) 

       1 3t xðtÞ 2 1 þ 6t 2 e ¼ e4t þ 2 e4t 2 yðtÞ 2 3  6t pffiffiffi pffiffiffi 1 4 2 1 3 xðtÞ ¼ cos 5t þ pffiffiffi sin 5t þ cosh 2t þ sinh 2t, 3 3 3 3 5 pffiffiffi pffiffiffi 1 4 1 1 yðtÞ ¼  cos 5t  pffiffiffi sin 5t þ cosh 2t þ sinh 2t 3 3 6 3 5 1

xðtÞ yðtÞ





¼

Further problems 15 (page 510)   t    6t    1 5 1 1 xðtÞ 9 9 e e (b) ¼ yðtÞ e3t e4t 4 1 1 5 1 9         xðtÞ 1  t 2t xðtÞ 10 sin t (a) ¼ e (b) ¼ yðtÞ t yðtÞ 4 sin t  2 cos t 0 pffiffiffi pffiffiffi pffiffiffi 1 pffiffiffi 3 11 3 9   p ffiffiffi p ffiffiffi cosh 7 t þ cosh 2 t þ sinh 2t C sinh 7 t  B5 xðtÞ 5 5 7 5 2 C ¼B (a) pffiffiffi pffiffiffi A pffiffiffi pffiffiffi @3 11 2 6 yðtÞ cosh 7t þ pffiffiffi sinh 7t þ cosh 2t  pffiffiffi sinh 2t 5 5 5 7 5 2 1 0 p ffiffiffi p ffiffiffi 5 5 5 1   p ffiffiffi cos 2 2 t þ cosh 2t þ sinh 2t  sin 2 2 t þ C B 12 xðtÞ 12 12 12 2 C ¼B (b) p ffiffiffi p ffiffiffi A @ 1 1 5 1 yðtÞ cos 2 2t  pffiffiffi sin 2 2t þ cosh 2t þ sinh 2t 6 6 6 6 2 1 ¼ 0, 2 ¼ 7, 3 ¼ 13 0 1 0 1 0 1 0 1 3 1 0 xðtÞ 5 3 2 (a) @ yðtÞ A ¼ @ 1 Ae4t  @ 1 Ae2t þ @ 1 Aet 6 2 3 3 3 6 zðtÞ 0 1 0 1 0 1 0 1 xðtÞ 3 6 3 1 1 7t 2t (b) @ yðtÞ A ¼ @ 1 Ae þ @ 1 Ae  @ 4 Ae4t 3 3 zðtÞ 1 1 5 1 0 13 7 cosh 3t þ sinh 3t 0 1B C 144 1 7 7 B 48 5 C 1 C B sinh t  cosh t þ (a) @ 1 1 1 AB C C B 16 16 5 5 7 @ 5 pffiffiffi pffiffiffi A 1 cosh 3t  pffiffiffi sinh 3t 12 12 3 0 3 pffiffiffi pffiffiffi 1 1 sinh 7t 0 1B 80 cosh 7t þ 40pffiffiffi C 7 6 6 2 B C 3 3 B C @ A (b) 27 5 1 B C  cos t þ sin t B C 16 8 p ffiffiffi 10 2 0 @ A pffiffiffi pffiffiffi 19 2 3 cos 3t  sin 3t 20 5 0 1 0 1 0 1 1 1 1 1 ¼ 3, x1 ¼ @ 1 A, 2 ¼ 1, x2 ¼ @ 1 A, 3 ¼ 0, x3 ¼ @ 1 A 2 0 1 

2 (a) 3

4

5 6

7

8

xðtÞ yðtÞ



¼

1192

Answers

Test exercise 16 (page 536) 1 (a)

(b)

3

2

x ¼ 2 x¼0 x¼1

stable unstable semi-stable

Unstable, Nullclines pffiffiffiffiffiffiffiffiffiffi at xðtÞ ¼  3 sin t

Further problems 16 (page 536) 1 xðtÞ ¼ ðt þ CÞ3

150

2

100 50

–5

–4

–3

–2

0 –1 0 –100

1

2

3

4

5

–100 –150

3 Alternating stable and unstable equilibrium points when xðtÞ ¼ n (integer). Unstable when n is even, stable when n is odd. dxðtÞ ¼ xðtÞðxðtÞ þ 1Þ, xðtÞ ¼ 1=ðAet  1Þ 4 dt 5

From the phase line it can be seen that xðtÞ ¼ 1 is a stable equilibrium solution and xðtÞ ¼ 0 is an unstable equilibrium solution. If the initial value of t is t0 and: (a) xðt0 Þ < 0 then xðtÞ ! 1 as t ! 1 Asymptotically unstable solutions; (b) xðt0 Þ ¼ 0 then xðtÞ ¼ 0 as t ! 1 An unstable equilibrium solution (source); (c) 0 < xðt0 Þ < 1 then xðtÞ ! 1 as t ! 1 Asymptotically stable solutions (d) xðt0 Þ ¼ 1 then xðtÞ ¼ 1 as t ! 1 A stable equilibrium solution (sink)

Answers

1193

6

A parabolic nullcline with a horizontal axis of symmetry and equation x2 ðtÞ ¼ t. All possible solutions are asymptotically unstable in that they all diverge as t ! 1 but there are four distinct regions of behaviour. Region t < 0 The solutions increase from negative to positive but are attracted towards the parabola pffiffi Region t > 0 and xðtÞ > 2 t The parabola acts as a source of solutions pffiffi Region t > 0 and xðtÞp< ffiffi 2 t The pffiffiparabola acts as a sink of solutions Region t > 0 and 2 t < xðtÞ < 2 t The parabola acts as both a source and a sink of solutions 7 Stable equilibria become unstable and vice versa. Semi-stable equilibria remain as semi-stable. 8 (a) xðtÞ ¼ cos t ðA þ lnðsec tÞÞ (b)

(c)

9 Equilibrium solution xðtÞ ¼ 0 stable for t  0, unstable for t > 0, nullcline xðtÞ ¼ t 10 Nullcline is circle centred on the origin with radius 2. 11 Nullclines xðtÞ ¼ 1  t and t ¼ 0.  1=3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 24 4 2 t þC 12 (a) xðtÞ ¼  C  t 2 (b) xðtÞ ¼ (c) xðtÞ ¼ Cet 16

Test exercise 17 (page 577) 1 (a) x 0 ¼ 4xðtÞ  4yðtÞ and y 0 ¼ xðtÞ, improper node (b) x 0 ¼ yðtÞ and y 0 ¼ xðtÞ, saddle (c) x 0 ¼ xðtÞ and y 0 ¼ xðtÞ, singular coefficient matrix 2 (a) Concentric ellipses, stable (b) Spiral source, unstable (c) Nodal sink, asymptotically stable (d) Nodal source, unstable (e) Saddle, unstable (f) Improper nodal source, unstable (g) Star node sink, asymptotically stable (h) Line of critical points along yðtÞ ¼ xðtÞ source trajectories parallel to yðtÞ ¼ ð6=5ÞxðtÞ, unstable (i) Line of critical points along yðtÞ ¼ xðtÞ sink trajectories parallel to yðtÞ ¼ ð1=4ÞxðtÞ, asymptotically stable 3 Concentric ellipses centred on critical point at ð3,  4Þ, stable

1194

Answers

Further problems 17 (page 578) 1 x 0 ¼ yðtÞ and y 0 ¼ ðk=mÞxðtÞ  ðc=mÞyðtÞ. Unstable for c < 0, spiral source, negative damping; stable for c ¼ 0, no damping; asymptotically stable spiral pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi sink, positive damping for 0 < c < 2 km, p improper nodal source for c ¼ 2 km, ffiffiffiffiffiffiffi asymptotically stable nodal sink for c > 2 km.     0 kt 1 t e : k < 0, saddle with eigenlines x ¼ 0 and y ¼ 0: k ¼ 0, 4 XðtÞ ¼  e þ 1 0 vertical line of critical points at x ¼ 0, horizontal phase trajectories: 0 < k < 1, proper node symmetric about y ¼ 0: k ¼ 1, star node: k > 1, proper node, symmetric about x ¼ 0. 5 Phase portrait stays the same pattern but the direction arrows reverse direction for k < 0. Phase portrait identical for k > 0.     1 1 2t e2t 6 (a) (i) saddle, XðtÞ ¼  e þ 2 2     1 3jt 1 (ii) concentric ellipses, XðtÞ ¼  e þ e3jt 3j 3j        1 1 1 t t t (iii) improper node, XðtÞ ¼  e þ te þ e 1 1 0     1 pffiffiffi ð1þpffiffi2Þt 1 pffiffiffi ð1pffiffi2Þt e e (iv) saddle, XðtÞ ¼  þ 1 þ 2 1  2 (b) (i) x 00 ðtÞ  4xðtÞ ¼ 0 (ii) x 00 ðtÞ þ 9xðtÞ ¼ 0 (iii) x 00 ðtÞ þ 2x 0 ðtÞ þ xðtÞ ¼ 0 (iv) x 00 ðtÞ þ 2x 0 ðtÞ  xðtÞ ¼ 0 7 (a) T 2  4D < 0, T ¼ 0: Ellipse (b) T 2  4D > 0, D > 0, T < 0: Nodal sink; T 2  4D ¼ 0, T < 0: Improper sink node; T 2  4D < 0, T < 0: Spiral (c) T 2  4AD > 0, D > 0, T > 0: Nodal source; T 2  4D > 0 D < 0: Saddle; T 2  4D ¼ 0, T > 0: Improper source node; T 2  4D < 0, T > 0: Spiral pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi 8 R ¼ 0: stable centre. R < 2 L=C: Spiral sink, asymptotically stable. R ¼ 2 L=C: pffiffiffiffiffiffiffiffiffi Improper sink node, asymptotically stable. R > 2 L=C asymptotically stable pffiffiffi pffiffiffi nodal sink 9 (a) 2 q < p < 0 (b) 0 < p < 2 q, q > 0     1 2t 1 3t 2t 3t 10 (a) (i) xðtÞ ¼ Ae þ Be , (ii) XðtÞ ¼  e þ e 2 3

      1 0 1 2t e2t þ t þ (b) (i) xðtÞ ¼ ½A þ Bt e , (ii) XðtÞ ¼  2 1 2     1 jt 1 jt jt e þ ejt (c) xðtÞ ¼ Ae þ Be , (ii) XðtÞ ¼  j j 11 (a) k ¼ 0, centre (b) 0 < k < 1, spiral sink (c) k ¼ 1, improper sink node (d) k > 1, nodal sink     3 2t 2 3t 2t 3t e  e 12 (a) xðtÞ ¼ 3e  2e (ii) XðtÞ ¼ 6 6

    1 2 (b) xðtÞ ¼ ½1  2te2t , (ii) XðtÞ ¼ þt e2t 0 4   

   1 jt 1 1 1 jt jt (c) xðtÞ ¼ e þ e , (ii) XðtÞ ¼ ejt e þ j j 2 2

Answers pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffi pffiffiffiffi 13 (a) Saddle: xðtÞ ¼  t 3þ 17 þ  t 3 17 (b) Improper nodal sink: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffi pffiffi xðtÞ ¼ t 2 þ t 3 (c) Spiral sink: xðtÞ ¼  t 1þj 3 þ  t 1j 3 14 Nodal source at ð2, 0Þ 15 Nodal source at ð2,  3Þ 16 (a) Saddle at ð9=2, 6Þ (b) Centre at ð8=9,  3Þ (c) Improper sink node at ð14,  4Þ (d) Saddle at ð15, 6Þ    pffiffi     2 a b xðtÞ e j p3ffiffit þ , spiral sink at ð2,  4Þ. 17 ¼ e2t 4 c d yðtÞ ej 3t   pffiffiffi 1 18 Unstable spiral source: ¼ a  j b: j   1 19 ¼ a  b: : b < a unstable nodal source, b > a saddle, b ¼ a coefficient 1 matrix is singular

Test exercise 18 (page 600) 1 Saddle at spiral at ð2,   source  2Þ.   ð2,   2Þ, unstable   pffiffiffi pffiffiffi 2pffiffiffiffiffiffi ð3þ2 41Þt xðtÞ 2pffiffiffiffiffiffi ð32 41Þt 2 e e þB þ þA 2 Saddle: 5  41 2 5 þ 41 yðtÞ Spiral source:   pffi       pffi   2 pffiffiffi ð5j2 7Þt 2 pffiffiffi 2 xðtÞ 2 pffiffiffi ð5þj2 7Þt e þD þD þ e þC 3j 7 3j 7 2 yðtÞ 3þj 7 3 Saddle at ð0, 0Þ, centre at ð1, 1Þ

Further problems 18 (page 600) 1 Spiral sink at ð0, 0Þ, saddles at ð8,  2Þ, ð8, 2Þ 2 (a) Saddle at ð0, 0Þ and ð1, 0Þ (b) Saddle at ð0, 0Þ, stable spiral sink at ð1, 1Þ (c) Unstable nodal source at ð0, 0Þ (d) Asymptotically stable nodal sink at ð1, 1Þ, Saddle at ð1,  1Þ and ð2, 2Þ, unstable spiral source at ð2,  2Þ 3 Centre predicted via linearization but (a) Anticlockwise spiral (i) source, (ii) sink, (b) Clockwise spiral (i) sink, (ii) source 4 Centre at ð0, 0Þ predicted and exists.  Saddle at  ð1, 1Þ  1 1 1 1 5 Saddle at ð0, 0Þ, unstable stars at pffiffiffi ,  pffiffiffi and  pffiffiffi , pffiffiffi 2 2 2 2 6 Eigenvalues ¼ 1, 0: line of critical points predicted but proper nodal sink displayed at ð0, 0Þ 7 Predicted: For k > 0:25 Spiral sink at ð0, 0Þ, saddle at ðk, 0Þ; For k ¼ 0:25 Improper nodal sink at ð0, 0Þ, saddle at ðk, 0Þ; For 0 < k < 0:25 Nodal sink at ð0, 0Þ, saddle at ðk, 0Þ; For k ¼ 0 Line of critical points and parallel trajectories; For 0:25 < k < 0: Nodal sink at ðk, 0Þ, saddle at ð0, 0Þ; For k ¼ 0:25 Improper nodal sink at ðk, 0Þ, saddle at ð0, 0Þ; For k < 0:25 Spiral sink at ðk, 0Þ, saddle at ð0, 0Þ. Actual: For k ¼ 0 Improper nodal sink at ð0, 0Þ is the only difference 8 Predicted: Nodal sink at ð0:739, 0:739Þ approx. Actual: Improper sink node at the origin.   32 24 , 9 Proper nodal sink at ð0, 0Þ, saddle at 9 9 10 Spiral source at ð0, 0Þ, saddle at ð2, 1Þ      1 3 1 1 11 Spiral sink at  , , 0 , saddle at 0, Þ and ð0, 0Þ , spiral source at 11 11 4 3

1195

1196

Answers

Test exercise 19 (page 635) 1 Saddle at ð0, 0Þ, Centre at ð25, 12:5Þ 2 Saddle at ð0, 31:82Þ and ð26:25, 0Þ, asymptotically stable node at ð23:47, 13:89Þ, unstable node at origin ð0, 0Þ 3 Saddles at ð0,  1Þ and centre at ð0, 0Þ. System periodic between the saddle points 4pCentre ffiffiffiffiffiffiffiffiffi at ð0, 0Þ with closed orbits contained within the isoclines yðtÞ ¼ 2 xðtÞ 5 (a) k ¼ 1, (b) k < 0 unstable clockwise spiral, k ¼ 0 unstable star, k > 0 unstable anticlockwise spiral pffiffiffi 6 Attractor limit cycle with spiral source centred on the origin, radius k where k 0, stable spiral sink if k < 0

Further problems 19 (page 635) 1 Spiral sink at ð0, 0Þ, saddles at ð8,  2Þ, ð8, 2Þ and an unstable limit cycle with spiral sink centred on the origin. 2 (a) Saddle at ð0, 0Þ and ð1, 0Þ (b) Saddle at ð0, 0Þ, spiral sink at ð1, 1Þ (c) Unstable nodal source at ð0, 0Þ surrounded by a stable limit cycle (d) Asymptotically stable nodal sink at ð1, 1Þ, Saddle at ð1,  1Þ and ð2, 2Þ, spiral source at ð2,  2Þ 3 Centre at ð0, 0Þ. (a) Anticlockwise spiral (i) source, (ii) sink, (b) Clockwise spiral (i) sink, (ii) source 4 k ¼ 1 x

6 4 2 –3

–2

–1

0

1

2

–2

3 k

–4 –6

5 (a) k ¼ 0 and k ¼ 0:25 150

x

100 50 –0.4

–0.3

–0.2

–0.1

0 –50

–100 –150

1

2

3

4 k

Answers

1197

(b) k ¼ 0:25 x

2.5 2 1.5 1 0.5 0 –2.5

–2

–1.5

–1

–0.5

–0.5

0

0.5 k

–1 –1.5

6 (a) k ¼ 0 (b) k ¼ 0 7 (a) k >p 0:ffiffiffi Stable limit cycle centred on spiral source critical point at ð0, 0Þ and radius k; k ¼ 0: Centre critical point at ð0, 0Þ takes the form of a spiral source at distance away from critical point; k < 0: spiral sink critical point at ð0, 0Þ (b) k 0: Spiral source critical point at ð0, 0Þ 8 (a) Stable centre at ð0, 0Þ (b) Stable centre at ð0, 0Þ, saddles centre at ð1, 0Þ. Periodicity restricted to 1 < xðtÞ < 1 9 (a) Critical point at ð0, 0Þ, stable centre for k ¼ 0. For k 6¼ 0 it is surrounded by a limit cycle that is bounded by 2  yðtÞ  2, circular for 0:2 < k < 0:2 unstable for k < 0 and stable for k > 0. Inside the limit cycle: k < 2 asymptotically stable nodal sink; k ¼ 2 asymptotically stable improper sink node; 2 < k < 0 spiral sink; k ¼ 0 stable centre; 0 < k < 2 spiral source; k ¼ 2 unstable improper source node; k > 2 unstable nodal source 10 k ¼ 0 centre at ð0, 0Þ, k ¼ 1 stable spiral at ð0, 0Þ, k ¼ 3 asymptotically stable node at ð0, 0Þ 11 (a) Critical point at ð0, 0Þ, Eigenvalues imaginary imply a centre (b) PPLANE shows a centre for k ¼ 0 a spiral sink centered on ð0, 0Þ for k < 0 and an spiral source for k > 0 with the appearance of a circular limit cycle with reducing radius as jkj increases.  12 Centres at  þ 2n, n ¼ 0,  1,  2, . . . Saddles at n, n ¼ 0,  1,  2, . . . 3

13 Centre at ð1, =2Þ Saddles at ð0, 0Þ and ð0, Þ 14 Critical points at ð0, 0Þ where for k < 2: unstable proper node; k ¼ 2: unstable improper node; 2 < k < 2: spiral sink; k ¼ 2: asymptotically stable improper node; k > 2: asymptotically stable proper node. Saddles at ð1, 0Þ for all values of k. 15 Spiral sink at ð1000, 15Þ

1198

Answers

16 Saddle at ð10, 0Þ and ð3959, 3949Þ, spiral sink at ð51, 41Þ

17 Critical points at ðn, 0Þ are spiral sinks in the damped case. In the undamped case they are centres 18 Saddle at ð2, 0Þ and spiral sink at ð0, 0Þ 19 Saddle at ð0, 100Þ andð100=3, 0Þ an  asymptotically stable 3250 3750 , and an unstable node at ð0, 0Þ nodal sink at 93 31 pffiffiffi 20 Spiral sink for k  0 at ð0, 0Þ. Stable circular limit cycle radius k centred on ð0,0Þ for k > 0 21 Spiral sink at ð0, 0Þ, improper source node at ð1:106, 1:221Þ 22 Saddles at ð0, 0Þ, ð0, 100Þ and ð40, 0Þ and a spiral sink at ð3:272, 1:836Þ approximately 23 ða, bÞ ¼ ð1, 1Þ: Spiral sink at ð0, 0Þ; ða, bÞ ¼ ð1,  1Þ: Saddle at ð0, 0Þ and spiral sinks at ð1, 0Þ and ð1, 0Þ; ða, bÞ ¼ ð1,  1Þ: Saddle at ð0, 0Þ and centres at ð1, 0Þ and ð1, 0Þ - these appear as spiral sources with a centre in the middle and all surrounded by a stable limit cycle; ða, bÞ ¼ ð1,  1Þ: Spiral source at ð0, 0Þ surrounded by a stable limit cycle. 24 k ¼ 1: Spiral sink at ð1, 1Þ k ¼ 1:5: Spiral sink at ð1, 1:5Þ; k ¼ 2: Predicted centre at ð1, 2Þ and ð1, 0Þ – this appears as a spiral sink with a centre in the middle; k ¼ 3: Spiral source at ð1, 3Þ surrounded by a stable limit cycle 25 ðx, yÞ ¼pð0, ffiffiffi 0Þ: Proper nodal source; ðx, yÞ ¼ ð1, 0Þ: Improper nodal sink; ðx, yÞ ¼ ð0, 2Þ:!Proper nodal sink; pffiffiffi pffiffiffi pffiffiffi 1þ 5 3 5 , : Saddle; ðx, yÞ ¼ ð0,  2Þ: Saddle; ðx, yÞ ¼ 4 4 pffiffiffi pffiffiffi! 1 5 3þ 5 , : Saddle ðx, yÞ ¼ 4 4

Test exercise 20 (page 680) 2ðx þ yÞ 2 ; 2x þ 3y ð2x þ 3yÞ3 x y y x ; ; ; 5 2ðx2  y2 Þ 4ðx2  y2 Þ 2ðx2  y2 Þ 4ðx2  y2 Þ 6 (a) ð1, 1Þ, saddle; ð1,  43Þ, min (b) an infinity of maxima along the line y ¼ 5x=2 when z ¼ 4 7 1:10 m  1:10 m  0:825 m high 8 6 4 2 8 u¼ ,x¼ ,y¼ ,z¼ 7 7 7 7 2 145:7  3:1 mm

3 5:8 m/s

4

Further problems 20 (page 680) 1 ð8x cos x  6y sin xÞ=J; ð4x3 cos y þ 6x sin yÞ=J; J ¼ 4x cos x sin y þ 2x2 y sin x cos y 2 e3y =2ðxe3y þ e3y Þ; e3y =2ðxe3y þ e3y Þ; 1=3ðxe3y þ e3y Þ;

x=3ðxe3y þ e3y Þ

Answers

1199

5 ð2ex sinh 2x sin 3y þ 3yex cosh 2x cos 3yÞ=ð1 þ 3y2 Þ; f4yex sinh 2x sin 3y þ 3ex ð1 þ y2 Þ cosh 2x cos 3yg=2ð1 þ 3y2 Þ 7 (a) (4,  4; 11), min

(b) (1; 2; 4), saddle

(c) (10 7;

6 97 7 ; 7 ), max ( 23 ,  13), min

8 (0, 0), saddle; (2, 0), min; (2; 0), min 9 (2, 1), max; 10 (0, 0); (3, 3); (3; 3), all saddle points 11 (a) (1, 0), saddle; (1, 1), min; (2, 12 ), saddle; ( 75, 15 ), max

(b) (0, 0), max; (1, 1); (1; 1); (1; 1); (1; 1), all four saddle points 12 (a) A point of inflexion at the origin (b) An infinity of maxima along the line y ¼ x=4 when z ¼ 6 (c) The value of z ranges from 1 to 1 and has an infinity of stationary points lying on the circles x2 þ y2 ¼ n. When n is even the stationary points are maxima and when n is odd the stationary points are minima. There is also a single maximum at ð0, 0, 1Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi  1 p 3 14 l ¼ h ¼ 13 x ¼ 66:7 mm;  ¼ 602 V ; d ¼ l 5 15 l ¼ 1:00 cm; 3 5 2r 8r 3 0 d ¼ 4:48 cm;  ¼ 488 11 16 cube of side pffiffiffi; Vmax ¼ pffiffiffi 3 3 3 2 3 3 17 (a) x ¼ y ¼ z ¼  pffiffiffi (b) x ¼ y ¼  pffiffiffiffi , x ¼ y ¼  pffiffiffi 14 3 2

Test exercise 21 (page 717) 1 (a) u ¼ 2x4 ðt  2Þ þ 4xt þ e2t (b) u ¼ 2 sin 2x  ðey  1Þ þ sin x þ y2 1 16 X 1 r rx rt  cos 2 uðx; tÞ ¼ 2  sin  sin  r ¼ 1 r2 2 10 10 1 X 100 1 x rc 2 3 uðx; tÞ ¼ ð1Þrþ1  sin  e t where ¼  r¼1 r c 2 1 X 20 rx ry  cosech r  sin  sinh with r ¼ 1, 3, 5, . . . 4 uðx; yÞ ¼ r 2 2 r¼1 5 vðr, Þ ¼ 5r 3 cos 3

Further problems 21 (page 718) 1 32 X 1 rx 3rt  cos ðr oddÞ  sin 3 r ¼ 1 r 3 2 2 1 2 X 1 r rx 5rt  cos  sin  sin uðx; tÞ ¼ 252 r ¼ 1 r 2 2 4 2 1 25 X 1 r rx crt  cos uðx; tÞ ¼ 2  sin  sin 2 r ¼ 1 r 2 5 10 10 1 X 800 1 rx 4 2 t r uðx; tÞ ¼ 3 e  sin with r ¼ 1, 3, 5, . . . where ¼  r ¼ 1 r3 10 10 1 16 X 1 r rx r 2 c2 2 t=100 uðx; tÞ ¼ 2 e  sin  sin with r ¼ 1, 3, 5, . . .  r ¼ 1 r2 2 10

2 uðx; tÞ ¼ 3 4 5 6

1200

Answers 1 128 X 1  cosech 3  r ¼ 1 r3 1 200 X 1 8 uðx; yÞ ¼ 3  cosech  r ¼ 1 r3

7 uðx; yÞ ¼

r r rx  sinh ð2  yÞ  sin with r ¼ 1, 3, 5, . . . 2 4 4 2r rx r  sin  sinh ðy  2Þ with r ¼ 1, 3, 5, . . . 5 5 5

9 vðr, Þ ¼ 4r cos  þ r 2 sin 2 10 vðr, Þ ¼

3 ð1  r 2 cos 2Þ 2

Test exercise 22 (page 761) 1 f ð1=4, 1=3Þ ¼ 19=12, f ð1=2, 1=3Þ ¼ 5=6, f ð3=4, 1=3Þ ¼ 1=12, f ð1=4, 2=3Þ ¼ 1=12, f ð1=2, 2=3Þ ¼ 5=6, f ð3=4, 2=3Þ ¼ 19=12 2 f ð1=3, 1=3Þ ¼ 4, f ð2=3, 1=3Þ ¼ 17=3, f ð1, 1=3Þ ¼ 26=3, f ð1=3, 2=3Þ ¼ 2=3, f ð2=3, 2=3Þ ¼ 3, f ð1, 2=3Þ ¼ 16=3

3 (a) parabolic

(b) hyperbolic (c) parabolic (d) hyperbolic (e) elliptic 4 f ð1=3, 1=3Þ ¼ 1:61728, f ð2=3, 1=3Þ ¼ 1:18519, f ð1, 1=3Þ ¼ 0:82716, f ð1=3, 2=3Þ ¼ 1:61728, f ð2=3, 2=3Þ ¼ 1:18519, f ð1, 2=3Þ ¼ 0:82716 5

6

t\x

0 .0

0.2

0 .4

0 .6

0.8

1 .0

1.2

0.00 0.02

0.00000 0.00000

0.04000 0.08000

0.16000 0.20000

0.36000 0.40000

0.64000 0.68000

1.00000 0.76500

0.89000 0.93000

0.04 0.06

0.00000 0.00000

0.10000 0.12000

0.24000 0.27000

0.44000 0.41125

0.58250 0.62250

0.80500 0.70750

0.83250 0.87250

0.08 0.10

0.00000 0.00000

0.13500 0.13281

0.26563 0.29063

0.44625 0.41250

0.55938 0.59688

0.74750 0.68438

0.80938 0.84688

0.12 0.14

0.00000 0.00000

0.14531 0.13633

0.27266 0.29453

0.44375 0.41055

0.54844 0.58281

0.72188 0.67344

0.79844 0.83281

0.16

0.00000

0.14727

0.27344

0.43867

0.54199

0.70781

0.79199

t\x

0.00

0.20

0.40

0.60

0.80

1.00

0.000 0.040

1.000000 1.000000

0.840000 0.898182

0.760000 0.832727

0.760000 0.832727

0.840000 0.898182

1.000000 1.000000

0.080 0.120

1.000000 1.000000

0.929917 0.952517

0.886942 0.923125

0.886942 0.923125

0.929917 0.952517

1.000000 1.000000

0.160 0.200

1.000000 1.000000

0.967729 0.978081

0.94779 0.964533

0.94779 0.964533

0.967729 0.978081

1.000000 1.000000

Answers

1201

Further problems 22 (page 762) 1

2

3

4

5

6

x\y

0.00

0.33

0.67

1.00

0.00 0.25

3.0000 2.7500

2.3333 2.0833

1.6667 1.4167

1.0000 0.7500

0.50 0.75

2.5000 2.2500

1.8333 1.5833

1.1667 0.9167

0.5000 0.2500

1.00

2.0000

1.3333

0.6667

0.0000

x\y

0.00

0.33

0.67

1.00

0.00 0.33

4.0000 6.3333

7.3333 9.6667

10.6667 13.0000

14.0000 16.3333

0.67 1.00

8.6667 11.0000

12.0000 14.3333

15.3333 17.6667

18.6667 21.0000

x\y

0.00

0.33

0.67

1.00

0.00 0.33

1.0000 0.6667

1.0000 0.7500

1.0000 0.8000

1.0000 0.8333

0.67 1.00

0.3333 0.0000

0.5000 0.2500

0.6000 0.4000

0.6667 0.5000

x\y

0.00

0.33

0.67

1.00

0.00 0.25

0.0000 0.0000

0.0000 0.0069

0.0000 0.0694

0.0000 0.1875

0.50 0.75

0.0000 0.0000

0.0278 0.1042

0.0556 0.0417

0.2500 0.1875

1.00

0.0000

0.2222

0.2222

0.0000

x\y

0.00

0.33

0.67

1.00

0.00 0.33

15.0000 17.3333

16.6667 19.0000

18.3333 20.6667

20.0000 22.3333

0.67 1.00

19.6667 22.0000

21.3333 23.6667

23.0000 25.3333

24.6667 27.0000

x\y

0.00

0.33

0.67

1.00

0.00 0.33

21.0000 22.6667

20.0000 21.6667

19.0000 20.6667

18.0000 19.6667

0.67 1.00

24.3333 26.0000

23.3333 25.0000

22.3333 24.0000

21.3333 23.0000

1202

Answers

7

8

9

10

11

12

x\y

0.00

0.33

0.67

1.00

0.00 0.33

4.0000 4.2222

4.0000 4.1111

4.0000 3.7778

4.0000 3.2222

0.67 1.00

4.8889 6.0000

4.6667 5.6667

4.0000 4.6667

2.8889 3.0000

x\y

0.00

0.33

0.67

1.00

0.00 0.33

0.0000 0.0000

0.0000 0.0000

0.0000 0.0741

0.0000 0.2963

0.67 1.00

0.0000 0.0000

0.0741 0.2963

0.0000 0.3704

0.3704 0.0000

x\y

0.00

0.33

0.67

1.00

0.00 0.33

0.0000 0.3333

0.5556 0.2222

2.2222 1.8889

5.0000 4.6667

0.67 1.00

1.3333 3.0000

0.7778 2.4444

0.8889 0.7778

3.6667 2.0000

x\y

0.00

0.33

0.67

1.00

0.00 0.33

1.0000 1.0000

1.0000 0.7037

1.0000 0.3333

1.0000 0.1111

0.67 1.00

1.0000 1.0000

0.3333 0.1111

0.4815 1.4444

1.4444 3.0000

x\y

0.00

0.33

0.67

1.00

0.00 0.33

0.0000 0.1111

0.0000 0.1050

0.0000 0.0873

0.0000 0.0600

0.67 1.00

0.4444 1.0000

0.4200 0.9450

0.3493 0.7859

0.2401 0.5403

x\y

0.00

0.33

0.67

1.00

0.00 0.33

0.0000 0.0370

0.0370 0.1481

0.2963 0.5556

1.0000 1.4815

0.67 1.00

0.2963 1.0000

0.5556 1.4815

1.1852 2.4074

2.4074 4.0000

Answers

13

14

15

16

1203

x\y

0.00

0.33

0.67

1.00

0.00 0.33

0.0000 0.0000

0.0000 0.1111

0.0000 0.2222

0.0000 0.3333

0.67 1.00

0.0000 0.0000

0.2222 0.3333

0.4444 0.6667

0.6667 1.0000

x\y

0.00

0.33

0.67

1.00

0.00 0.33

0.0000 0.0000

0.0000 0.0000

0.0000 0.0741

0.0000 0.2222

0.67 1.00

0.0000 0.0000

0.0741 0.2222

0.0000 0.2222

0.2222 0.0000

t\x 0.00

0.00 0.0000

0.20 0.1600

0.40 0.2400

0.60 0.2400

0.80 0.1600

1.00 0.0000

0.02 0.04

0.0400 0.0800

0.1200 0.0800

0.2000 0.1600

0.2000 0.1600

0.1200 0.0800

0.0400 0.0800

0.06 0.08

0.1200 0.1600

0.0400 0.0000

0.1200 0.0800

0.1200 0.0800

0.0400 0.0000

0.1200 0.1600

0.10 0.12

0.2000 0.2400

0.0400 0.0800

0.0400 0.0000

0.0400 0.0000

0.0400 0.0800

0.2000 0.2400

0.14 0.16

0.2800 0.3200

0.1200 0.1600

0.0400 0.0800

0.0400 0.0800

0.1200 0.1600

0.2800 0.3200

0.18 0.20

0.3600 0.4000

0.2000 0.2400

0.1200 0.1600

0.1200 0.1600

0.2000 0.2400

0.3600 0.4000

t\x

0.00

0.20

0.40

0.60

0.80

1.00

0.00 0.02

0.0000 0.0000

0.1987 0.1983

0.3894 0.3886

0.5646 0.5635

0.7174 0.7159

0.8415 0.8398

0.04 0.06

0.0000 0.0000

0.1979 0.1975

0.3879 0.3871

0.5624 0.5613

0.7145 0.7131

0.8381 0.8364

0.08 0.10

0.0000 0.0000

0.1971 0.1967

0.3863 0.3855

0.5601 0.5590

0.7116 0.7102

0.8348 0.8331

0.12 0.14

0.0000 0.0000

0.1963 0.1959

0.3848 0.3840

0.5579 0.5568

0.7088 0.7074

0.8314 0.8298

0.16 0.18

0.0000 0.0000

0.1955 0.1951

0.3832 0.3825

0.5557 0.5546

0.7060 0.7046

0.8281 0.8265

0.20

0.0000

0.1947

0.3817

0.5535

0.7032

0.8248

1204

Answers

17

18

t\x

0.00

0.20

0.40

0.60

0.80

1.00

0.00 0.02

0.0000 0.0000

0.3830 0.3798

0.7596 0.7534

1.1239 1.1147

1.4698 1.4578

1.7916 1.7770

0.04 0.06

0.0000 0.0000

0.3767 0.3736

0.7473 0.7412

1.1056 1.0966

1.4459 1.4341

1.7624 1.7481

0.08 0.10

0.0000 0.0000

0.3706 0.3676

0.7351 0.7291

1.0876 1.0787

1.4223 1.4107

1.7338 1.7196

0.12 0.14

0.0000 0.0000

0.3646 0.3616

0.7232 0.7173

1.0699 1.0612

1.3992 1.3878

1.7056 1.6916

0.16 0.18

0.0000 0.0000

0.3586 0.3557

0.7114 0.7056

1.0525 1.0439

1.3764 1.3652

1.6778 1.6641

0.20

0.0000

0.3528

0.6998

1.0354

1.3541

1.6505

t\x

0.00

0.20

0.40

0.60

0.80

1.00

0.00 0.04

1.0000 0.9200

0.7600 0.6800

0.4400 0.3600

0.0400 0.0400

0.4400 0.5200

1.0000 1.0800

0.08 0.12

0.8400 0.7600

0.6000 0.5200

0.2800 0.2000

0.1200 0.2000

0.6000 0.6800

1.1600 1.2400

0.16 0.20

0.6800 0.6000

0.4400 0.3600

0.1200 0.0400

0.2800 0.3600

0.7600 0.8400

1.3200 1.4000

0.24 0.28

0.5200 0.4400

0.2800 0.2000

0.0400 0.1200

0.4400 0.5200

0.9200 1.0000

1.4800 1.5600

0.32 0.36

0.3600 0.2800

0.1200 0.0400

0.2000 0.2800

0.6000 0.6800

1.0800 1.1600

1.6400 1.7200

0.40 0.44

0.2000 0.1200

0.0400 0.1200

0.3600 0.4400

0.7600 0.8400

1.2400 1.3200

1.8000 1.8800

0.48 0.52

0.0400 0.0400

0.2000 0.2800

0.5200 0.6000

0.9200 1.0000

1.4000 1.4800

1.9600 2.0400

0.56 0.60

0.1200 0.2000

0.3600 0.4400

0.6800 0.7600

1.0800 1.1600

1.5600 1.6400

2.1200 2.2000

Answers

1205

19 0.00

t\x 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0.0000 0.9000 1.6000 0.4000 0.5000 1.2000 0.8000 0.1000 0.8000 1.2000 0.3000 0.4000

2.1000 2.4000 2.5000 2.4000 1.7000 2.0000 2.1000 2.0000 1.3000 1.6000 1.7000 1.6000 0.9000 1.2000 1.3000 1.2000

2.1000 1.6000 0.9000 1.7000 1.2000 0.5000 1.3000 0.8000 0.1000 0.9000 0.4000 0.3000

0.0000 0.4000

1.6000 0.7000 0.0000 2.0000 1.1000 0.4000 2.4000 1.5000 0.8000 2.8000 1.9000 1.2000

0.5000 0.8000 0.9000 0.8000 0.1000 0.4000 0.5000 0.4000 0.3000 0.0000 0.1000 0.0000 0.7000 0.4000 0.3000 0.4000

0.5000 0.0000 0.7000 0.1000 0.4000 1.1000 0.3000 0.8000 1.5000 0.7000 1.2000 1.9000

1.6000 2.0000

0.8000 1.2000

2.4000 2.8000

20 t\x

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

0.00 0.04

0.0000 0.0000

0.08 0.12

0.0000 0.0000

0.16 0.20

0.0000 0.0000

6.3705 4.2926

0.24 0.28

0.0000 0.0000

2.8925 1.9490

0.32 0.36

0.0000 0.0000

1.3133 0.8849

2.4980 1.6832

3.4382 2.3168

4.0419 2.7235

4.2499 2.8637

4.0419 2.7235

3.4382 2.3168

2.4980 1.6832

1.3133 0.8849

0.0000 0.0000

0.40 0.44

0.0000 0.0000

0.5963 0.4018

1.1342 0.7643

1.5611 1.0519

1.8352 1.2366

1.9296 1.3002

1.8352 1.2366

1.5611 1.0519

1.1342 0.7643

0.5963 0.4018

0.0000 0.0000

0.48 0.52

0.0000 0.0000

0.2707 0.1824

0.5150 0.3470

0.7088 0.4776

0.8332 0.5615

0.8761 0.5904

0.8332 0.5615

0.7088 0.4776

0.5150 0.3470

0.2707 0.1824

0.0000 0.0000

0.56 0.60

0.0000 0.0000

0.1229 0.0828

0.2338 0.1576

0.3218

0.3783 0.2549

0.3978

0.0000

0.2169

0.2338 0.1576

0.1229

0.2680

0.3783 0.2549

0.3218

0.2169

0.0828

0.0000

30.9017 58.7785 80.9017 95.1057 20.8224 39.6065 54.5136 64.0846 14.0306 26.6878 36.7327 43.1818 9.4542 17.9829 24.7514 29.0970

100.0000 95.1057 80.9017 58.7785 30.9017 67.3825 64.0846 54.5136 39.6065 20.8224 45.4041 43.1818 36.7327 26.6878 14.0306 30.5944 29.0970 24.7514 17.9829 9.4542 12.1174 16.6781 19.6063 20.6153 19.6063 16.6781 12.1174 6.3705 8.1650 11.2381 13.2112 13.8911 13.2112 11.2381 8.1650 4.2926 5.5018 7.5725 8.9021 9.3602 8.9021 7.5725 5.5018 2.8925 3.7072 5.1026 5.9984 6.3071 5.9984 5.1026 3.7072 1.9490

1.00 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

Test exercise 23 (page 815) 1 (a) dz ¼ 4x3 cos 3y dx  3x4 sin 3y dy (b) dz ¼ 2e2y f2 cos 4x dx þ sin 4x dyg (c) dz ¼ xw2 f2yw dx þ xw dy þ 3xy dwg 2 (a) z ¼ x3 y4 þ 4x2  5y3 (b) z ¼ x2 cos 4y þ 2 cos 3x þ 4y2 (c) not exact differential 3 9 square units 4 (a) 278.6 (b) =2 (c) 22.5 (d) 48 (e) 21 5 (f) 54 5 Area ¼ 12 square units 6 (a) 2 (b) 0

Further problems 23 (page 816) 1 14

2 1 .6 3

7 2  =2

8

1 8

pffiffiffi  f9  4 3g 36 9 14

1 4 9 f þ 4g 5 2 256 2 . 10 (a) 39 24 (b) 0 11 3 4

6

1  ln 2 2

1206

Answers

Test exercise 24 (page 858) pffiffiffi 1 4 2

2 að=2Þ2 3 (a) (1) ð4:47, 0:464, 3Þ (2) ð5:92, 0:564, 0:322Þ (b) (1) ð3:54, 3:54, 3Þ (2) ð0:832; 1:82; 3:46Þ 4 12 ðð vð1 þ uÞð1 þ u þ vÞ dv dv 5 a3 ð8  3aÞ=12 6 (a) I ¼ ððð ð2u þ vÞðv  4wÞ du dv dw (b) I ¼ vw

Further problems 24 (page 858)

pffiffiffiffiffiffi pffiffiffi 4 22  pffiffiffi 5 ð5 5  1Þ 6  5 7 16a2 3 24   pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 11 a4 4 2 2 2 12 13 2   8 2a ð  2Þ 9 4ða þ bÞ a  b 10 45 11 30 3 2 ð ð 3 3 pffiffiffi 3a a 4abc 2a 1 ðu2 þ v2 Þ du dv 15 17 f4 2  3g 16 18 14 x ¼ y ¼ z ¼ 8 3 4 3 3 a 7  1 pffiffiffi 21 22 2  23 ð 2  1Þ 19 u2 v du dv dw 20 z ¼  5 18 2 4 pffiffiffi 1 4 5

2

a a a ; ; 2 2 2

pffiffiffiffiffiffi 3 10 61

4

Test exercise 25 (page 895)

pffiffiffi 20 2 15  256 1 2 (b) (c) 2 (d) 120 (e) 2 (a) (b) (c) 1 (a) 3 3 2048 315 40 105 pffiffiffi!   pffiffiffi  3 1 1 1 1 , 4 (a) 0 (b) 1 5 (a) F 2, (b)  F 3 (a) pffiffiffi  K pffiffiffi 2 2 2 4 2 2 pffiffiffi! 1 1 3 , (b) F 2 2 2

Further problems 25 (page 895) 1 (a) 4 (a) 8 (a) (d) (g) 10 (a) (c)

pffiffiffiffiffiffi 1 315 8 2 . (c) 0 4 (d) 24 (e) 2 (a) 6 (b) (c) (d) 4 6 (b)  2 pffiffiffi 4 81 16 1 8 2 1  2 (b) (d) (e) (f) ¼ 0:00727 (c) 8960 315 7 63 432 64       pffiffiffi pffiffiffi 2 1 1 5  E pffiffiffi (b) 2  K pffiffiffi ¼ 2:622 (c) 2  E ; 1 ¼ 2:935 2 5 2       1 3 2 1 2 1 1  ¼ 0:193 (e) pffiffiffi  F pffiffiffi , 1 F , (f) pffiffiffi  F pffiffiffi ; 4 4 3 5 5 ( 2 2 6 pffiffiffi ! pffiffiffi !)      3  3  1 1  1  1 pffiffiffi  F pffiffiffi ; F ; ;  F pffiffiffi ; 9  F 2 2 2 2 4 2 2 3 2 4        1 1 1 1 1 1 1 pffiffiffi  F pffiffiffi ; ¼ 0:307 (b) pffiffiffi  F pffiffiffi ; 1  F pffiffiffi , 3 3 2 3 ( 3 3 2 rffiffiffi ! rffiffiffi !)   1 3 1 3  3  pffiffiffiffiffiffi  K pffiffiffiffiffiffi ¼ 0:2905 (d) pffiffiffi F ; F ; 7 2 7 6 34 34 7

Answers

1207

Text exercise 26 (page 941) 1 (a) 15 (b) 16i þ 10j þ 17k 2 (a) 9 (b) ð47i þ 17j þ 29kÞ 3 A  ðB  CÞ ¼ 0 ; vectors coplanar 4 (a) 4i  4j þ 24k 1 8 (b) 2i  2j þ 24k (c) 24.66 5 T ¼ pffiffiffiffiffiffi ð4i þ j þ 7kÞ 6 ð25i  6j  15kÞ 5 66 1 7 5.08 8 pffiffiffiffiffiffiffiffiffi ð2i þ 4j þ 9kÞ 9 (a) 14i  12j  30k (b) 8 101 (c) 5i  2j  4k (d) 7i þ 2j þ 3k (e) 3i þ 2j þ k

Further problems 26 (page 941) 1 61 2 29i  10j þ 16k 3 (a) 22i þ 14j þ 2k (b) 2i þ 14j  22k 4 (a) 2xi þ 3j þ cos x k (b) 2i  sin x k (c) ð4x2 þ 9 þ cos2 xÞ1=2 (d) 34 þ sin 2 5 (a) 2  2u  9u2 (b) ð3u2 þ 4u þ 3Þi þ ð3u2 þ 6Þj þ ð1  2uÞk (c) i  2j þ ð3  2uÞk 1 1 1 6 pffiffiffiffiffiffi ð2i  20j þ 11kÞ 7 pffiffiffiffiffiffiffiffiffi ð10i þ 2j  5kÞ 8 pffiffiffiffiffiffiffiffiffi ð5i  j þ 10kÞ 129 126 5 21 1 9 pffiffiffiffiffiffiffiffiffi ð12i þ 4j  21kÞ 10 8:285 11 9:165 12 (a) 15 (b) 33 (c) 7 601 13 (a) 6i þ 4j  7k (b) 62i þ 10j  38k (c) 18i  21j þ 10k 14 (a) 12i  4j þ 4k (b) 24i  4j (c) 144 15 (a) ð2 sin 2Þi þ 2e3 j þ ðcos 2 þ e3 Þk (b) ð4 sin2 2 þ cos2 2 þ 2e3 cos 2 þ 5e6 Þ1=2 1 1 16 5:014 17 p ¼ pffiffiffiffiffiffi ð3i þ 2j  4kÞ; q ¼ pffiffiffiffiffiffi ð6i  j þ kÞ;  ¼ 688 480 29 38 18 (a) ð2t þ 3Þi  ð6 cos 3tÞj þ 6e2t k (b) 2i þ ð18 sin 3tÞj þ 12e2t k (c) 12.17 20 4xi þ 4zk 21 ð2 cos 5:5Þi  ð6 sin 5:5Þj  ð6 sin 5:5Þk 22 p ¼ 6 1 23 (a) (1) p ¼ 15=4 (2) p ¼ 33 (b) ð3i  2j þ 6kÞ 7

Test exercise 27 (page 991) 1 3i þ

18 81 j k 2 12 7 8

6 all conservative

4 3 18 ð2i þ jÞ 4 24ði þ jÞ 5 8 þ 3   8 0 7 36 þ 1 4

Further problems 27 (page 992) 576 ð3i þ j þ 2kÞ 2 1771i þ 1107j þ 830:4k 5 16 416:1i þ 718:5j þ 5679k 4 46.9 5 4:18 6 8 7 ði þ kÞ 3   1  1 9 ð48i þ 64j  24kÞ 9 64  ð6i þ 4jÞ 10 fð þ 2Þi þ ð þ 2Þj þ 4kg 3 4 3 2 pffiffiffi 12 250 1 ð32j þ 15kÞ 12 1 13  14 ð117 þ 256  28 7Þ ¼ 91:58 5 3 6 a3 81 20 80 16 96 17 2 18 12 19  4 3

1 (a) 576k (b) 3 8 11 15

1208

Answers

Test exercise 28 (page 1019) 1 4 5 6

yes, an orthogonal set 2 hu ¼ 1, hv ¼ 2v, h ¼ 2u 3 4I þ K (a) ð2 cos  þ 2 cos 2 þ 1Þ (b) ð2 sin 2 þ sin ÞK (a) ðdsÞ2 ¼ ðdrÞ2 þ r 2 ðdÞ2 þ r 2 sin2  ðdÞ2 (b) dV ¼ r 2 sin  dr d d 10:5

Further problems 28 (page 1019) 5 18 2 2 @ V 1 @V 1 @ V @2V 52 V ¼ þ 2 2 þ 2 þ  2 @  @  @ @z     1 @2V 52 V ¼ 1  @ r 2 @V þ 1  @ sin  @V þ  2 2 2 2 r @r @r r sin  @ @ r 2 sin  @ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 hu ¼ hv ¼ u þ v ; hw ¼ 1  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   ffi @Fv  @Fw 1 @ @Fu @ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ v2  2 þ v2  div F ¼ 2 u u þ þ u þ v2 @u @v @u @v @w  2 2 2 1 @ V @ V @ V 52 V ¼ þ 2 þ u2 þ v2 @u2 @v @w2

1 (a) yes 5 (a) (b) 6 (b) (c) (d)

2 50:5

(b) no

3 2

Test exercise 29 (page 1058) 1 (a) w ¼ 6  j2 (b) w ¼ 3  j2 (c) w ¼ j3 (d) w ¼ 2 2 Magnification ¼ 2:236; rotation ¼ 638 260 ; translation ¼ 1 unit to right, 3 units downwards v

3

v¼

1 ð1  u2 Þ 2

u

O v

4

u

Minor arc of circle, centre   1 1 1  ,  , radius pffiffiffi ¼ 0:7071, 2 2 2

 5 (a) centre

 2 u ¼ 0, v ¼ 3

(b)

1 6 centre radius 3

  2 2 u ¼ ; v ¼ 0 ; radius 3 3

Answers

1209

Further problems 29 (page 1059) 1 Triangle A’B’C’ with A0 ð1 þ j2Þ, B0 ð5 þ j2Þ, C0 ð2 þ j5Þ 2 (a) A0 ð8 þ j9Þ; B0 ð23pþ ffiffiffiffiffiffij14Þ (b) Magnification ¼ 29 ¼ 5:385; rotation ¼ 688120 ; translation ¼ nil 3 Straight line joining A0 ð5  j7Þ to B0 ð3  jÞ; magnification ¼ 3:162; rotation ¼ 1618 340 anticlockwise; translation ¼ 2 to right, 4 upwards v 4 A’ B’: v ¼ 0 B’C’: u ¼ 1  C’D’: D0 A0 :

v2 1 4 v¼0 u¼

u

O

5

v2 4

v

1 ð4u  u2 Þ 4 1 B’C’ and D’A’: v ¼ ðu2  4uÞ 4 A’B’ and C’D’: v ¼

O

u

v2 ; 4   ðu  1Þ2 1 2 7  j , radius  8; C’A’: u ¼ 1 7 circle, centre B’C’: v ¼ 2 3 6 32   1 2 (a) circle, centre  j0 , radius (b) region outside the circle in (a) 3 3   3 þ j0 , radius 1; clockwise development circle, centre 2   22u 8 11 9 þ ¼ 0, centre þ j0 , radius circle, u2 þ v2  5 5 5 5   u 1 1 circle, u2 þ v2  ¼ 0, centre þ j0 , radius ; region inside this circle 2 4 4   7 5 circle, centre  þ j0 , radius 3 3   3 2 (a) circle, centre ; 0 , radius , developed clockwise 5 5 (b) region outside the circle in (a) u v¼ 3

6 A0 ð1  j2Þ; B0 ð23 þ j10Þ; C0 ð1  j8Þ A’B’: u ¼ 2 

8 9 10 11 12 13

14

1210

Answers

Test exercise 30 (page 1106) regular at all points (b) z ¼ 5 (c) regular at all points z ¼ 1 and z ¼ 4 (e) z ¼ 0, where z ¼ x þ jy vðx, yÞ ¼ cosh x sin y þ C (b) vðx, yÞ ¼ 6ðy2  x2 Þ  4x þ C 4 j4 pffiffiffi z ¼ 0 (b) z ¼ 1 (c) no critical point (d) z ¼  2 (e) z ¼ 0 z (f) no critical point 6 w ¼ cosh ; D’: w ¼ 1 4

1 (a) (d) 2 (a) 5 (a)

Further problems 30 (page 1107) pffiffiffi 3 circle, centre (5, 2), radius 2



4 circle, centre

 1  ; 0 , 3

2 radius , anticlockwise 5 (a) vðx, yÞ ¼ 2yðx  1Þ þ C 3 (b) vðx, yÞ ¼ 3x2 y  y3  2xy þ y þ C (c) vðx, yÞ ¼ x2  2x  y2 þ C (d) vðx, yÞ ¼ ex (c) j10

2

y2

9 j2

sin 2xy þ C 6 (a) j10 10 j10

(b) j6

11 (a) (1) z ¼ 0

(b) ellipse, centre (0, 0), semi major axis

5 2,

7 (a) 0

(b) j4

(2) z ¼ 1

semi minor axis

3 2

2

12 (a) u2 þ v2 ¼ 1 (b) u2 þ ðv  1Þ ¼ 2;  ¼ 458: 13 Unit circle becomes the real axis on the w-plane. Region within the circle maps onto the upper half z plane 14 w ¼ sin 2a

Test exercise 31 (page 1136) z2 z3 zn þ þ . . . þ þ . . . valid for jzj < 1 2! 3! n! ð4zÞ2 ð4zÞ3 ð1Þnþ1 ð4zÞn þ  ... þ þ . . . valid for jzj < 1=4 (b) f ðzÞ ¼ 4z  2 3 n 2 (a) pole of order 5 at z ¼ 1 (b) essential singularity at z ¼ 0 (c) essential ( singularity at z ¼ 0 (d) removable singularity at z ¼ 0 1 ðz  =4Þ2 ðz  =4Þ3 3 f ðzÞ ¼ pffiffiffi 1 þ ðz  =4Þ   2! 3! 2 ) ðz  =4Þ4 ðz  =4Þ5 þ  . . . ; valid for jzj < 1 þ 4! 5! 1 4 1 1  þ þ . . .; 4 (a) f ðzÞ ¼ ðz þ 3Þ þ 8 þ  2ðz þ 3Þ ðz þ 3Þ2 24ðz þ 3Þ3 3ðz þ 3Þ4 essential singularity 3 1 1 1 1 z z2 z3  ¼ ...  3 þ 2  þ 1  þ  þ ... (b) f ðzÞ ¼ zþ3 zþ1 z z z 3 9 27 1 3 3 5ðz  2Þ 15ðz  2Þ2 þ  þ þ . . .; (c) f ðzÞ ¼  32 128 8ðz  2Þ2 16ðz  2Þ 16 1 (a) f ðzÞ ¼ 1 þ z þ

pole of order 2 5 double pole at z ¼ 0; residue 4, double pole pat ffiffiffi z ¼ 1, residue 7=2, single pole at z ¼ 1, residue 1=2 6 (a) =6 (b) 2= 3 (c) 2e3

Answers

1211

Further problems 31 (page 1137) 1 (a) z þ

z3 z5 z2nþ1 þ þ ... þ þ . . . , jzj < 1 3! 5! ð2n þ 1Þ!

z3 2z5 17z7 þ þ þ . . . , jzj < =2 3 15 315  z3 z5 z2nþ1 þ . . . , jzj < 1 (c) 2 z þ þ þ . . . þ 3 5 2n þ 1

(b) z þ

(d) 1 þ z ln a þ (e)

z2 ðln aÞ2 z3 ðln aÞ3 zn ðln aÞn þ þ ... þ þ . . . , jzj < 1 2! 3! n!

3z2 27z3 162z4 810z5 þ þ þ þ . . . , jzj < 5=3; 25 125 625 3125

5 25 250 6250 z zðz  1Þ     . . . , jzj > 5=3 3 (b)  , 2 9z 9z2 27z3 243z4 ðz þ 1Þ ðz þ 1Þ3 4 (a) convergent for jzj < 1 (b) convergent for jzj < 1 (c) convergent for jzj < 1 (d) convergent for jzj < 1 (e) convergent for jzj < 1 ( ) 2 3 n ðz  2Þ ðz  2Þ ðz  2Þ þ þ ... þ þ ... 5 (a) e2 1 þ ðz  2Þ þ 2! 3! n! pffiffiffi pffiffiffi pffiffiffi 3 ðz  =6Þ 3ðz  =6Þ2 ðz  =6Þ3 3ðz  =6Þ4  þ þ þ ...  (b) 2 2  2! 2  3! 2  4! 2 

ðz  3Þ3 sin 6 ðz  3Þ4 cos 6  (c) ðz  3Þ sin 6 þ ðz  3Þ2 cos 6  2! 3! (  2 5 ðz  3Þ sin 6 3 3 þ . . . (d)  þ2 þ ðz  1=3Þ 4! 13 13 )  3  nþ1 3 3 2 n n ðz  1=3Þ þ . . . þ 2 ðz  1=3Þ þ . . . þ4 13 13

(e) 1  2ðz  3Þ þ 4ðz  3Þ2 þ . . . þ ð2Þn ðz  3Þn þ . . .

ðz  1Þ2 ðz  1Þ3 ðz  1Þ4 ðz  1Þ5  þ  þ . . . 7 (a) z ¼ 1 12 23 34 45 pffiffiffi (b) jzj ¼ 6 (c) jz  5j ¼ 1 (d) z ¼ 1 8 (a) poles of order 2 at z ¼ 0 and z ¼ 1, removable singularity at z ¼ 1 (b) essential singularity at 1 1 1 1 z ¼ 0 9 (a) 2  4 þ 6  8 þ . . ., jzj > 0 z z z z 3! 5! 7!   1 3 1 , j2z  3j > 0 (b) z 2 2 3  2f1  ðz  3Þ þ ðz  3Þ2  ðz  3Þ3 þ . . .g, 0 < jz  3j < 1 (c) z3 8 4 2 1 2 2z 2z2 2z3 þ  þ ... 10 (a) . . . þ 4  3 þ 2  þ  z z z z 5 25 125 625 1 8 46 242 1 17z 109z2 593z3 þ  (b)  2 þ 3  4 þ . . . (c)  þ  ... z z z z 10 100 1000 10000 pffiffiffi 2 2 11 (a) 2= 3 (b) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (c) 2 (d) =4 (e) =2 (f) =2 j  1j 2   2 6 ðz  1Þ þ

1212

Answers

(g) 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi pffiffiffiffiffiffi 13=8  3=8

pffiffiffi (h) =4 (i) 2= 3 (j) 2=3 (k) 0 (l) 0

Test exercise 32 (page 1160) 1 Pmax ¼ 10 ðx ¼ 4, y ¼ 3Þ 2 Pmax ¼ 13 ðx ¼ 1, y ¼ 4Þ 3 Pmax ¼ 188 ðx ¼ 10, y ¼ 4; z ¼ 6Þ 4 Pmax ¼ 296 ðx ¼ 4, y ¼ 6Þ ðx ¼ 5, y ¼ 12Þ 6 (a) 13 type A þ 4 type B (b) £11,800 5 Pmin ¼ 16

Further problems 32 (page 1161) 1 3 5 7 9 11 12 14 16 18 20 21 23 25 27 28 29 30

Pmax ¼ 32 ðx ¼ 4, y ¼ 9=2Þ 2 Pmax ¼ 64 ðx ¼ 0, y ¼ 8Þ Pmax ¼ 40 ðx ¼ 6, y ¼ 5=2Þ 4 Pmax ¼ 15 ðx ¼ 6, y ¼ 3Þ Pmax ¼ 9 ðx ¼ 1, y ¼ 3Þ 6 Pmax ¼ 10 ðx ¼ 2, y ¼ 4Þ Pmax ¼ 10 ðx ¼ 2, y ¼ 4Þ 8 Pmax ¼ 37 ðx ¼ 0, y ¼ 8, z ¼ 1Þ Pmax ¼ 67 ðx ¼ 4, y ¼ 10, z ¼ 5Þ 10 Pmax ¼ 65 ðx ¼ 5, y ¼ 10, z ¼ 5Þ Pmax ¼ 11:568 ðx ¼ 29=22, y ¼ 14=11, z ¼ 0Þ to 3 s:f : Pmax ¼ 340 ðx ¼ 30, y ¼ 20Þ 13 Pmax ¼ 112 ðx ¼ 4, y ¼ 8Þ Pmax ¼ 108 ðx ¼ 16, y ¼ 15Þ 15 Pmin ¼ 138 ðx ¼ 12, y ¼ 18Þ Pmax ¼ 240 ðx ¼ 9, y ¼ 15Þ 17 Pmax ¼ 4400 ðx ¼ 201, y ¼ 53Þ Pmax ¼ 100 ðx ¼ 20, y ¼ 10Þ 19 Pmax ¼ 410 ðx ¼ 9, y ¼ 5, z ¼ 2Þ Pmax ¼ 1560 ðx ¼ 11, y ¼ 10, z ¼ 18Þ Pmax ¼ 660 ðx ¼ 60, y ¼ 30, z ¼ 30Þ 22 Pmin ¼ 14 ðx ¼ 5, y ¼ 2Þ Pmin ¼ 56 ðx ¼ 8, y ¼ 12Þ 24 Pmin ¼ 16 ðx ¼ 8, y ¼ 6Þ Pmin ¼ 40 ðx ¼ 4, y ¼ 4Þ 26 Pmin ¼ 10 ðx ¼ 6, y ¼ 13, z ¼ 14Þ Pmin ¼ 75 ðx ¼ 8, y ¼ 12, z ¼ 21Þ (a) 10 type A þ 35 type B (b) £2150 (a) 22 type A þ 44 type B þ 48 type C (b) £12,580 (a) 129 type A þ 0 type B þ 185 type C; (b) £8955

Index Alternative form of a line integral 787 Alternative forms of elliptic functions 889 Analytic description of a periodic function 242 Application of gamma and beta functions 876 Applications 1154 Area enclosed by a curve 782 Autonomous differential equations 521 Bessel’s equation 378 Beta function 871 Bifurcation 624 First-order equations 624 Second-order equations 628 Bisection 10 Calculating residues 1125 Cartesian coordinates 829 Cauchy-Euler equi-dimensional equation 352 Cauchy-Riemann equations 1065 Cauchy’s theorem 1075 Central differences 31 Change of variables 647 Change of variables in multiple integrals 843 Characteristic equation 482 Circle of convergence 1116 Comparison of numerical methods 470 Competition within a single population 607 Complementary error function erfc ðxÞ 882 Complete elliptic functions 885 Complex analysis 1021, 1064, 1109 Calculating residues 1125 Circle of convergence 1116 Complex integration 1072 Complex mapping 1023 Conformal transformation 1089 Differentiation of a complex function 1062 Functions of a complex variable 1022 Harmonic functions 1067

Integrals of real functions 1126 Laurent’s series 1119 Maclaurin series 1110 Non-linear transformations 1038 Radius of convergence 1114 Residues 1123 Singular points 1115 Taylor’s series 1117 Complex eigenvalues 551 Complex exponentials 301 Complex Fourier series 301 Complex exponentials 301 Complex integration 1072 Cauchy’s theorem 1075 Contour integration 1072 Defromation of contours at singularities 1080 Line integrals in the complex plane 1072 Complex mapping 1023 Mapping of straight line in z-plane onto w-plane where w ¼ f ðzÞ 1025 Types of transformation w ¼ az þ b 1029 Complex spectra 306 Computational molecules 727 Conditions for conformal transformation 1089 Conformal transformation 1089 Conditions for conformal transformation 1089 Critical points 1090 Open polygons 1098 Schwarz-Christoffel transformation 1093 Conservative vector fields 965 Consistency of a set of linear equations 444 Uniqueness of solutions 445 Continuous spectra 308 Contour integration 1072 Convolution 113, 213, 325 Convolution theorem 118, 326 Coplanar vectors 905 ‘Cover up’ rule for partial fractions 68 Crank-Nicholson procedure 750

1213

1214

Index

Critical point behaviour 554, 556 Critical point moved to the origin 571 Critical points 1090 Cubic equations 7 Tartaglia’s solution for a real root 8 Transforming a cubic to reduced form 8 curl of a vector function 932 Curvilinear coordinates 845, 996 Orthogonal curvilinear coordinates 1000 Transformation in three dimensions 853 Cylindrical coordinates 830 Damped motion 145 Damped pendulum 620 Defromation of contours at singularities 1080 Dependence of a line integral on the path of integration 799 Derivative boundary conditions 734 Derivative of the unit step function 140 Derivative of the Z transform 177 DFIELD 518 Autonomous differential equations 521 Equilibrium solutions 521 Family of solutions 520 Phase line 524 Specific solution 519 Diagonalization of a matrix 498 Modal matrix 498 Spectral matrix 498 Difference equations 159, 228 Difference equations and the Z transform 156 Difference equations 159 Inverse transforms 178 Properties of Z transforms 172 Sampling 184 Sequences 157 Solving difference equations 161, 181 Table of Z transforms 172 Z transform 167 Differential equations 203, 513 Direction fields 516 Family of solutions 513 General nth-order equation 203 Time-invariance 211 Zero-input and zero-state response 204 Zero-input, zero-response 209

Differential equations 203, 220, 513 Differential equations involving the unit impulse 141 Differential equations involving the unit step function 109 Differentials 776 Exact differential 778 Integration of exact differentials 780 Differentiating and integrating a transform 55 Dividing the Laplace transform by t 58 First shift theorem 55 Multiplying the Laplace transform by t and t n 56 Differentiation 323 Differentiation of a complex function 1062 Cauchy-Riemann equations 1065 Regular function 1063 Differentiation of sums and products of vectors 915 Differentiation of vectors 910 Differentiation of sums and products of vectors 915 Unit tangent vectors 915 Dimensional analysis 756 Dirac delta – the unit impulse 135, 317 Graphical representation 136 Laplace transform of the Dirac delta 137 The derivative of the unit step function 140 Direction fields 512, 516 DFIELD 518 Differential equations 513 Non-autonomous equations 530 Direction of unit normal vectors to a surface S 979 Directional derivatives 924 Dirichlet conditions 253 Effect of harmonics 259 Gibbs’ phenomenon 261 Sum of Fourier series at a point of discontinuity 261 Discrete unit impulse 223 div (divergence) of a vector function 931 Divergence theorem 970 Dividing the Laplace transform by t 58 Double integrals 818 Surface integrals 823 Duplication formula for gamma functions 880

Index

Dynamical systems 603 Bifurcation 624 Competition within a single population 607 Dynamical systems 604 Interacting populations 610 Limit cycles 630 Non-interacting populations 608 Predator-prey problems 604 Second-order differential equations 616 Van der Pol equation 632 Effect of harmonics 259 Effect of the unit step 95 Eigenvalues and the phase plane 545 Eigenvalues of 2  2 matrices 482 Characteristic equation 482 Eigenvectors 484 Sum and product of eigenvalues 483 Eigenvectors 484 Element of arc ds in curvilinear coordinates 1010 Element of volume dV in curvilinear coordinates 1010 Element of volume in three coordinate systems 833 Elementary operations and equivalent matrices 440 Elliptic equations 738 Elliptic functions 884 Alternative forms of elliptic functions 889 Complete elliptic functions 885 Standard form of elliptic functions 885 Equilibrium solutions 521 Error function 881 Complementary error function erfc ðxÞ 882 Graph of erf ðxÞ 882 Euler second-order method 423 Euler-Cauchy calculations 415 Euler-Cauchy method 414 Euler’s method 399 Even functions 313 Exact differential 778 Exact differentials in three dimensions 804 Exact value and errors of Euler’s method 408 Exponential response 216, 226

1215

Family of solutions 513, 520 Final value theorem 176 First shift theorem 55, 173 First-order differential equations 399 Euler-Cauchy calculations 415 Euler-Cauchy method 414 Euler’s method 399 Exact value and errors of Euler’s method 408 Graphical interpretation of Euler’s method 412 Runge-Kutta method 420 First-order equations 624 Forced harmonic motion with damping 147 Fourier coefficients 289 Fourier cosine transformations 328 Fourier series 239, 270 Analytic description of a periodic function 242 Dirichlet conditions 253 Functions with periods other than 2 288 Graphs of y ¼ A sin nx 240 Harmonics 241 Integrals of periodic functions 246 Non-sinusoidal periodic functions 242 Odd and even functions and half-range series 271 Periodic functions 240 Fourier sine transformations 328 Fourier transform 300 Alternative forms 319 Complex Fourier series 301 Complex spectra 306 Continuous spectra 308 Convolution 325 Convolution theorem 326 Fourier cosine transformations 328 Fourier sine transformations 328 Fourier’s integral theorem 310 Heaviside unit step function 324 Properties of the Fourier transform 320 Special functions and their transforms 313 Table of Fourier transforms 330 Two domains 307 Fourier’s integral theorem 310 Frequency shifting 321 Frobenius’ method 360 Function increment 398

1216

Index

Function of two real variables 723 Computational molecules 727 Grid values 724 Summary of procedures 731 Functions of a complex variable 1022 Functions with period T 288 Functions with periods other than 2 288 Fourier coefficients 289 Functions with period T 288 Half-range series with arbitrary period 292 Functions with three independent variables 674 Fundamental theorem of algebra 2 Relations between coefficients and roots of a polynomial 4 Gamma and Bessel functions 379 Gamma and beta functions 862 Application of gamma and beta functions 876 Beta function 871 Duplication formula for gamma functions 880 Gamma function 862 Reduction formulas 872 Relation between the gamma and beta functions 875 Gamma function 862 Gauss’ theorem 970 Gaussian elimination method 457 General curvilinear coordinte system 1008 General nth-order equation 203 Gibbs’ phenomenon 261 grad (gradient) of a scalar field 921 grad of sums and products of scalars 929 grad, div and curl 933 grad, div and curl in orthogonal curvilinear coordinates 1011 Graph of erf ðxÞ 882 Graphical interpolation 25 Graphical interpretation of Euler’s method 412 Graphical representation 136 Graphical representation of linear inequalities 1142 Graphs of Bessel functions 384 Graphs of y ¼ A sin nx 240 Green’s theorem 805, 985

Gregory-Newton interpolation for backward differences 33 Gregory-Newton interpolation for forward differences 25 Grid values 724 Half-range series 281 Half-range series with arbitrary period 292 Harmonic functions 1067 Harmonic oscillators 143 Damped motion 145 Forced harmonic motion with damping 147 Resonance 150 Harmonics 241 Heat conduction equation 697 Solution of the heat conduction equation 698 Heaviside unit step function 94, 324 Higher derivatives 338 Leibnitz theorem 341 Higher derivatives 338 Hyperbolic equations 739 Identical eigenvalues 567 Imaginary eigenvalues 545 Implicit functions 646 Impulse response 212 Indicial equation 369 Inhomogeneous case 570 Initial conditions and bounday conditions 686 Initial value theorem 177 Input-response relationships 196 Integral functions 861 Elliptic functions 884 Error function 881 Gamma and beta functions 862 Integrals of periodic functions 246 Fourier series 250 Orthogonal functions 249 Integrals of real functions 1126 Using complex analysis 1126 Integration of vector functions 918 Integration of exact differentials 780 Interacting populations 610 Interpolation 24 Central differences 31 Graphical interpolation 25 Gregory-Newton interpolation for backward differences 33

Index

Gregory-Newton interpolation for forward differences 25 Lagrange interpolation 34 Linear interpolation 24 Invariant linear systems 194 Differential equations 203 Input-response relationships 196 Invariant linear systems 195 Linear systems 197 Responses of a continuous system 212 Responses of a discrete system 223 Shift-invariance of a discrete system 202 Systems 195 Time-invariance of a continuous system 200 Inverse functions 651 Inverse method 449 Inverse transformations 131 Inverse transforms 61, 178 ‘Cover up’ rule for partial fractions 68 Rules of partial fractions 62 Table of inverse Laplace transformations 69 Lagrange interpolation 34 Lagrange undetermined multipliers 671 Functions with three independent variables 674 Laplace transform of the Dirac delta 137 Laplace transform of uðt  cÞ 98 Laplace transform of uðt  cÞ:f ðt  cÞ 99 Laplace transforms 47, 93, 124 Convolution 113 Convolution theorem 118 Differential equations involving the unit impulse 141 Differential equations involving the unit step function 109 Differentiating and integrating a transform 55 Dirac delta – the unit impulse 135 Effect of the unit step 95 Harmonic oscillators 143 Heaviside unit step function 94 Inverse transforms 61 Laplace transform of uðt  cÞ 98 Laplace transform of uðt  cÞ:f ðt  cÞ 99 Laplace transforms of periodic functions 125

1217

Solution of differential equations by Laplace transforms 70 Unit impulse 135 Unit step at the origin 95 Laplace transforms of derivatives 71 Laplace transforms of periodic functions 125 Inverse transformations 131 Periodic functions 125 Laplace’s equation 703 Solution of Laplace’s equation 704 Laplace’s equation in plane polar coordinates 708 Separating the variables 709 Laurent’s series 1119 Legendre polynomials 385 Legendre’s equation 384, 390 Legendre polynomials 385 Rodrigue’s formula and generating function 385 Leibnitz theorem 341 Leibnitz-Maclaurin method 345 Limit cycles 630 Line integral with respect to arc length 797 Line integrals 786, 945, 948 Alternative form of a line integral 787 Dependence of a line integral on the path of integration 799 Exact differentials in three dimensions 804 Line integral with respect to arc length 797 Line integrals around a closed curve 793 Parametric equations 798 Properties of line integrals 790 Regions enclosed by closed curves 792 Scalar field 945 Vector field 948 Line integrals around a closed curve 793 Line integrals in the complex plane 1072 Linear inequalities 1142 Linear interpolation 24 Linear programming 1141 Linear systems 197 Linearity 172, 320 Linearization 585 Maclaurin series 1110 Mapping of regions 1043 Mapping of straight line in z-plane onto w-plane where w ¼ f ðzÞ 1025

1218

Index

Mass-spring system 539 Matrix algebra 437 Consistency of a set of linear equations 444 Elementary operations and equivalent matrices 440 Matrix transformation 471 Singular and non-singular matrices 438 Solution sets of linear equations 449 Matrix transformation 471 Rotation of axes 473 Maximum and minimum values 660 Modal matrix 498 Modified Newton-Raphson method 21 Multiple critical points 582 Linearization 585 Problems with linearization 594, 817 Multiple integration 767 Area enclosed by a curve 782 Curvilinear coordinates 845 Differentials 776 Double integrals 818 Green’s theorem 805 Line integrals 786 Three dimensional coordinate system 829 Volume integrals 834 Multiple operations 935 Multiplying the Laplace transform by t and t n 56 Newton-Raphson iterative method 15 Modified Newton-Raphson method 21 Tabular display of results 16 Non-autonomous equations 530 Non-interacting populations 608 Non-linear systems 581 Multiple critical points 582 Non-linear transformations 1038 Mapping of regions 1043 Non-sinusoidal periodic functions 242 Nonlinear programming 1156 Numerical approximation to derivatives 721 Numerical methods 10 Bisection 10 Numerical solution of equations by iteration 12 Relative addresses 14 Using a spreadsheet 12

Numerical solution of partial differential equations 720 Crank-Nicholson procedure 750 Derivative boundary conditions 734 Dimensional analysis 756 Function of two real variables 723 Numerical approximation to derivatives 721 Second partial derivatives 741 Second-order partial differential equations 738 Time-dependent equations 745 Numerical solutions of equations and interpolation 1 Cubic equations 7 Fundamental theorem of algebra 2 Interpolation 24 Newton-Raphson iterative method 15 Numerical methods 10 Numerical solution of equations by iteration 12 Numerical solutions of ordinary differential equations 396 First-order differential equations 399 Predictor-corrector methods 430 Second-order differential equations 423 Taylor’s series 397 Odd and even functions 271 Odd and even functions and half-range series 271 Half-range series 281 Odd and even functions 271 Products of odd and even functions 274 Series with only even harmonics 285 Series with only odd harmonics 285 Significance of the constant term 287 Odd functions 313 Open polygons 1098 Optimization 1141 Graphical representation of linear inequalities 1142 Linear inequalities 1142 Linear programming 1141 Solver 1148 Solver parameters 1149 Optimization and linear programming 1140 Applications 1154 Nonlinear programming 1156 Optimization 1141

Index

Orthogonal coordinate systems in space 1001 Orthogonal curvilinear coordinates 1000 Orthogonal functions 249 Orthogonality 389 Parabolic equations 739 Parametric equations 798 Partial differential equations 683 Heat conduction equation 697 Initial conditions and boundary conditions 686 Laplace’s equation 703 Laplace’s equation in plane polar coordinates 708 Separation of the variables 688 Solution by direct integration 685 Solution of the wave equation 688 Wave equation 687 Partial differentiation 640 Inverse functions 651 Lagrange undetermined multipliers 671 Small increments 641 stationary values of a function 659 Partial differentiation of vectors 918 Integration of vector functions 918 Particular orthogonal systems 1014 Particular solution 164 Periodic functions 125, 240 Phase line 524 Phase plane analysis 538, 544 Complex eigenvalues 551 Critical point behaviour 554, 556 Critical point moved to the origin 571 Eigenvalues and the phase plane 545 Identical eigenvalues 567 Imaginary eigenvalues 545 Inhomogeneous case 570 Mass-spring system 539 Phase plane analysis 544 PPLANE 542 Real and negative eigenvalues 556 Real and positive eigenvalues 560 Real eigenvalues of different signs 563 Singular coefficient matrix 568 Star node 568 Poles 1115 Polynomials as a finite series of Legendre polynomials 391

1219

Power series solutions 344 Cauchy-Euler equi-dimensional equation 352 Leibnitz-Maclaurin method 345 Power series solutions of ordinary differential equations 337, 359, 377 Bessel’s equation 378 Frobenius’ method 360 Gamma and Bessel functions 379 Graphs of Bessel functions 384 Higher derivatives 338 Indicial equation 369 Legendre’s equation 384 Power series solutions 344 Sturm-Liouville systems 387 PPLANE 542 Predator-prey problems 604 Predictor-corrector methods 430 Problems with linearization 594 Products of odd and even functions 274 Properties of line integrals 790 Properties of scalar triple products 904 Properties of the Fourier transform 320 Differentiation 323 Frequency shifting 321 Linearity 320 Symmetry 322 Time scaling 321 Time shifting 321 Properties of Z transforms 172 Derivative of the Z transform 177 Final value theorem 176 First shift theorem 173 Initial value theorem 177 Linearity 172 Scaling 175 Second shift theorem 174 Radius of convergence 1114 Rank of a matrix 439 Rates of change 645 Real and negative eigenvalues 556 Real and positive eigenvalues 560 Real eigenvalues of different signs 563 Reduction formulas 872 Regions enclosed by closed curves 792 Regular function 1063 Relation between the gamma and beta functions 875 Relations between coefficients and roots of a polynomial 4 Relative addresses 14

1220

Index

Removable singularities 1116 Repeated eigenvalues 492 Residues 1123 Resonance 150 Responses of a continuous system 212 Arbitrary input 212 Convolution 213 Differential equations 220 Exponential response 216 Impulse response 212 Transfer function HðsÞ 218 Responses of a discrete system 223 Arbitrary input 224 Difference equations 228 Discrete unit impulse 223 Exponential response 226 Transfer function 227 Rodrigue’s formula and generating function 385 Rotation of axes 473 Row transformation method 453 Rules of partial fractions 62 Runge-Kutta method 420 Runge-Kutta second-order method 425 Saddle point 666 Sampling 184 Scalar and vector fields 921 curl of a vector function 932 Directional derivatives 924 div (divergence) of a vector function 931 grad (gradient) of a scalar field 921 grad of sums and products of scalars 929 grad, div and curl 933 Multiple operations 935 Unit normal vectors 927 Scalar field 945, 957 Scalar triple product of three vectors 903 Scale factors 1005 Scale factors for coordinate systems 1006 Scaling 175 Schwarz-Christoffel transformation 1093 Second partial derivatives 741 Second shift theorem 174 Second-order differential equations 423, 616 Damped pendulum 620 Euler second-order method 423 Runge-Kutta second-order method 425

Undamped pendulum: no approximation 617 Undamped pendulum: small oscillations 616 Second-order equations 628 Second-order partial differential equations 738 Elliptic equations 738 Hyperbolic equations 739 Parabolic equations 739 Separating the variables 688, 709 Sequences 157 Series with only even harmonics 285 Series with only odd harmonics 285 Shift-invariance of a discrete system 202 Significance of the constant term 287 Simultaneous differential equations 80 Singular and non-singular matrices 438 Rank of a matrix 439 Singular coefficient matrix 568 Singular points 1115 Poles 1115 Removable singularities 1116 Small increments 641, 643 Change of variables 647 Implicit functions 646 Rates of change 645 Small increments 643 Taylor’s theorem for one independent variable 641 Taylor’s theorem for two independent variables 641 Solution by direct integration 685 Solution by inspection 161 Solution of differential equations by Laplace transforms 70 Laplace transforms of derivatives 71 Simultaneous differential equations 80 Solution of first-order differential equations 73 Solution of second-order equations 75 Solution of first-order differential equations 73 Solution of Laplace’s equation 704 Solution of second-order equations 75 Solution of sets of linear equations 449 Comparison of numerical methods 470 Gaussian elimination method 457 Inverse method 449 Row transformation method 453 Triangular decomposition method 460 Using a spreadsheet 466

Index

1221

Solution of the heat conduction equation 698 Solution of the wave equation 688 Solver 1148 Solver parameters 1149 Solving difference equations 161, 181 Particular solution 164 Solution by inspection 161 Special functions and their transforms 313 Dirac delta 317 Even functions 313 Odd functions 313 Top-hat function 315 Triangle function 319 Specific solution 519 Spectral matrix 498 Spherical coordinates 831 Standard form of elliptic functions 885 Star node 568 Stationary values of a function 659 Maximum and minimum values 660 Saddle point 666 Stokes’ theorem 976 Sturm-Liouville systems 387 Legendre’s equation 390 Orthogonality 389 Polynomials as a finite series of Legendre polynomials 391 Sum and product of eigenvalues 483 Sum of Fourier series at a point of discontinuity 261 Summary of procedures 731 Surface integrals 823, 956 Scalar field 957 Vector field 960 Symmetry 322 Systems 195 Systems of linear, first-order ordinary differential equations 487 Repeated eigenvalues 492 Systems of ordinary differential equations 481 Diagonalization of a matrix 498 Eigenvalues of 2  2 matrices 482 Systems of linear, first-order ordinary differential equations 487 Systems of linear, second-order ordinary differential equations 503 Table of Fourier transforms Table of inverse Laplace transformations 69

330

Table of Z transforms 172 Tabular display of results 16 Tartaglia’s solution for a real root 8 Taylor’s series 397, 1117 Function increment 398 Taylor’s theorem for one independent variable 641 Taylor’s theorem for two independent variables 641 Three dimensional coordinate systems 829 Cartesian coordinates 829 Cylindrical coordinates 830 Element of volume in three coordinate systems 833 Spherical coordinates 831 Time scaling 321 Time shifting 321 Time-dependent equations 745 Time-invariance 211 Time-invariance of a continuous system 200 Top-hat function 315 Transfer function 227 Transfer function HðsÞ 218 Transformation equations 1009 Transformation in three dimensions 853 Transforming a cubic to reduced form 8 Triangle function 319 Triangular decomposition method 460 Triple products 903 Coplanar vectors 905 Properties of scalar triple products 904 Scalar triple product of three vectors 903 Vector triple product of three vectors 907 Two domains 307 Types of transformation w ¼ az þ b 1029 Undamped pendulum: no approximation 617 Undamped pendulum: small oscillations 616 Uniqueness of solutions 445 Unit impulse 135 Unit normal vectors 927 Unit step at the origin 95 Unit tangent vectors 915 Using a spreadsheet 12, 466 Using complex analysis to solve real integrals 1126

1222

Index

Van der Pol equation 632 Vector analysis 897, 944, 995 Conservative vector fields 965 Curvilinear coordinates 996 Differentiation of vectors 910 Direction of unit normal vectors to a surface S 979 Divergence theorem 970 Element of arc ds in curvilinear coordinates 1010 Element of volume dV in curvilinear coordinates 1010 Gauss’ theorem 970 General curvilinear coordinte system 1008 grad, div and curl in orthogonal curvilinear coordinates 1011 Green’s theorem 985 Line integrals 945 Orthogonal coordinate systems in space 1001

Partial differentiation of vectors 918 Particular orthogonal systems 1014 Scalar and vector fields 921 Scale factors 1005 Stokes’ theorem 976 Surface integrals 956 Transformation equations 1009 Triple products 903 Volume integrals 952 Vector field 948, 960 Vector triple product of three vectors 907 Volume integrals 834, 952 Change of variables in multiple integrals 843 Wave equation

687

Z transform 167 Zero-input and zero-state response 204 Zero-input, zero-response 209