An Excursion Through Partial Differential Equations 9783031487835, 9783031487842

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Problem Books in Mathematics

Svetlin G. Georgiev

An Excursion Through Partial Differential Equations

Problem Books in Mathematics Series Editor Peter Winkler, Department of Mathematics, Dartmouth College, Hanover, NH, USA

Books in this series are devoted exclusively to problems - challenging, difficult, but accessible problems. They are intended to help at all levels - in college, in graduate school, and in the profession. Arthur Engels “Problem-Solving Strategies” is good for elementary students and Richard Guys “Unsolved Problems in Number Theory” is the classical advanced prototype. The series also features a number of successful titles that prepare students for problem-solving competitions.

Svetlin G. Georgiev

An Excursion Through Partial Differential Equations

Svetlin G. Georgiev Faculty of Mathematics and Informatics Sofia University Sofia, Bulgaria

ISSN 0941-3502 ISSN 2197-8506 (electronic) Problem Books in Mathematics ISBN 978-3-031-48783-5 ISBN 978-3-031-48784-2 (eBook) https://doi.org/10.1007/978-3-031-48784-2 Mathematics Subject Classification: 35-01, 35A09, 35A24, 35B30, 35B50, 35C05, 35C09, 35C15, 35E15, 35E20, 35F05, 35F10, 35F20, 35F35, 35J05, 35K05, 35L05 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland Paper in this product is recyclable.

Preface

Partial differential equations are ubiquitous in mathematically oriented scientific fields, such as physics and engineering. They are foundational in the modern scientific understanding of sound, heat, diffusion, electrostatics, electrodynamics, thermodynamics, fluid dynamics, elasticity, general relativity, and quantum mechanics. Many partial differential equations also arise from many purely mathematical theories, such as differential geometry and variational calculus. They are the fundamental tool in the proof of the Poincaré conjecture from geometric topology. There is a wide spectrum of different types of partial differential equations, and methods have been developed for dealing with many of the individual equations which arise. This book presents an introduction to the theory of partial differential equations (PDEs) with over 500 examples, exercises, and problems provided with detailed solutions or detailed hints or answers. The book is suitable for all types of basic courses on PDEs. The book contains seven chapters. Chapter 1 introduces some basic problems in the area of PDEs. Chapter 2 is devoted on first-order PDEs. They are considered classification of first-order PDEs, solvability of quasilinear first-order PDEs, the Cauchy problem for quasilinear first-order PDEs, the Pfaffian equation, and some special systems. In Chaps. 3 and 4 are considered the classification and canonical forms of second-order PDEs. The Laplace equation is introduced in Chap. 5. They are given the basic properties of elliptic problems, the fundamental solutions, integral representation of harmonic functions, mean-value formulas, strong principle of maximum, the Poisson equation, the Green functions, method of separation of variables, and theorems of Liouville and Harnack. Chapter 6 deals with the heat equation. They are considered the weak and strong maximum principles, the Cauchy problem, the mean value formula, the method of separation of variables, and the energy method. Chapter 7 is concerned with the wave equation. They are investigated one, two and three dimensional wave equations, method of separation of variables, and energy method.

v

vi

Preface

The aim of this book is to present a clear and well-organized treatment of the concept behind the development of mathematics and solution techniques. The text material of this book is presented in highly readable, mathematically solid format. Many practical problems are illustrated displaying a wide variety of solution techniques. Sofia, Bulgaria

Svetlin G. Georgiev

Contents

1

General Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 History and Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 2 7 10

2

First Order Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Classifications of First Order Partial Differential Equations . . . . . . . . . . 2.2 Solvability of Quasilinear First Order PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 The Cauchy Problem for Quasilinear First Order PDEs . . . . . . . . . . . . . . 2.4 The Pfaff Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Some Special Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13 13 21 27 30 37 39

3

Classifications of Second Order Partial Differential Equations . . . . . . . . 3.1 Classifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47 47 52

4

Classifications and Canonical Forms for Linear Second Order Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 4.1 Classifications and Canonical Forms for Linear Second Order Partial Differential Equations in Two Independent Variables . . 55 4.1.1 The Elliptic Case. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 4.1.2 The Parabolic Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 4.1.3 The Hyperbolic Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 4.2 Classification and Canonical Form of Second Order Linear Partial Differential Equations in n Independent Variables . . . . . . . . . . . . 108 4.3 Classification of First Order Systems with Two Independent Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 4.4 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

5

The Laplace Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 5.1 Basic Properties of Elliptic Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 vii

viii

Contents

5.2 5.3 5.4 5.5

The Fundamental Solution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Strong Maximum Principle: Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Green Function of the Dirichlet Problem . . . . . . . . . . . . . . . . . . . . . . . . . Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.1 Rectangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.2 Circular Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

144 151 153 161 162 175 181

6

The Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 The Cauchy Problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 The Method of Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 The Mean Value Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 The Weak and Strong Maximum Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 The Maximum Principle for the Cauchy Problem . . . . . . . . . . . . . . . . . . . . . 6.6 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

193 194 210 220 224 232 235

7

The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 The One Dimensional Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.1 The Cauchy Problem and the d’Alambert Formula . . . . . . . . . . . 7.1.2 The Cauchy Problem for the Nonhomogeneous Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.3 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.4 The Energy Method: Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 The Wave Equation in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.1 Radially Symmetric Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.2 The Cauchy Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 The Two Dimensional Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Advanced Practical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

241 241 241

8

Solutions, Hints and Answers to the Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

347 347 347 350 351 358 368 377

9

Solutions, Hints and Answers to the Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

387 387 387 393 394 403 411 416

247 255 298 302 303 311 323 333

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421

Chapter 1

General Introduction

1.1 Introduction A partial differential equation (PDE) describes a relation between an unknown function and its partial derivatives. PDEs appear in all areas of physics and engineering. Moreover, in recent years we have seen the use of PDEs in areas such as biology, chemistry, medicine, economics, computer sciences. The general form of a PDE for a function .u = u(x1 , . . . , xn ) is F (x1 , . . . , xn , u, ux1 , . . . , uxn , . . .) = 0,

.

where .xi , .1 ≤ i ≤ n, are independent variables, .u is the unknown variable and .uxi ∂u . The equation is, in general, supplemented by denotes the partial derivative . ∂xi additional conditions such as initial conditions or boundary conditions. Example 1.1 The equation ux1 + ux1 x2 = 0,

.

(x1 , x2 ) ∈ R2 ,

is a PDE. Example 1.2 The equation x 2 − 3x + 2 = 0,

.

x ∈ R,

is an algebraic equation. The analysis of PDEs has many facets. The classical approach is to develop methods for finding explicit solutions. It includes, the method of characteristics, the Fourier method, the Green method. A progress in PDEs was achieved with the introduction of numerical methods. The technical advances were followed by © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. G. Georgiev, An Excursion Through Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-031-48784-2_1

1

2

1 General Introduction

theoretical progress aimed at understanding the solution structure. The aim is to discover some of the solution properties before computing it, and sometimes even without a complete solution. There exist many equations that cannot be solved. All we can do in these cases is to obtain qualitative information on the solution. Furthermore, it is desired in many cases that the solution will be unique, and that it will be stable under small perturbations of the data. A theoretical understanding of the equation enables us to check whether these conditions are satisfied. As we will see in what follows, there are many ways to solve PDEs, each way applicable to a certain class of PDEs. Therefore it is important to have an analysis of the equation before or during solving it. The fundamental theoretical question is whether the problem consisting of the equation and its associated side conditions is well-posed. Definition 1.1 According to the definition given by Hadamard, a problem is called well-posed if it satisfies the following criteria. 1. Existence. The problem has a solution. 2. Uniqueness. There is no more than one solution. 3. Stability. A small change in the equation or in the side conditions gives rise to a small change in the solution.

If one or more conditions above does not hold, we say that the problem is illposed.

1.2 Classification PDEs are often classified into different types. In fact, there exist several such classifications. The first classification is according to the order of the equation. Definition 1.2 The order is defined to be the order of the highest derivative in the equation. If the highest derivative is of order .m, then the equation is said to be of order .m. Example 1.3 The equation ux1 x1 − 2ux1 + ux2 = u,

.

(x1 , x2 ) ∈ R2 ,

is a second order PDE. Example 1.4 The equation u + x1 x2 ux1 x2 − ux2 = x12 x2 ,

.

is a second order PDE.

(x1 , x2 ) ∈ R2 ,

1.2 Classification

3

Example 1.5 The equation ux1 + ux2 + ux3 + ux1 ux2 ux3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 ,

is a first order PDE. Exercise 1.1 Find the order of the following equations. 1. 2. 3. 4. 5.

ux1 x1 x1 − 2ux1 + ux2 − (ux3 )2 = x1 x2 x3 , .(x1 , x2 , x3 ) ∈ R3 . 2 2 2 .ux1 x1 x1 x1 − (ux1 x1 x1 ) + ux2 x2 x2 x2 x2 = u , .(x1 , x2 ) ∈ R . 2 4 3 .ux − 2x1 ux1 + 3x1 x2 ux3 = u , .(x1 , x2 , x3 ) ∈ R . 1 3 2 .ux1 x1 − 2ux1 x2 + ux2 x2 x2 = u , .(x1 , x2 ) ∈ R . 2 2 5 4 .ux1 − x ux2 − x ux3 + ux4 = u , .(x1 , x2 , x3 , x4 ) ∈ R . 1 2 .

Definition 1.3 A PDE is said to be linear if it is linear in the unknown and its derivatives. Otherwise, the equation is said to be nonlinear. Example 1.6 The equation ux1 + ux2 − x1 x2 u = 0,

.

(x1 , x2 ) ∈ R2 ,

is a linear PDE. Example 1.7 The equation (ux1 )2 + ux2 + ux3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 ,

is a nonlinear PDE. Exercise 1.2 Classify each of the following equations as linear or nonlinear. 1. 2. 3. 4. 5.

ux1 u + ux2 = 0, .(x1 , x2 ) ∈ R2 . 2 .ux1 x1 − ux1 x2 + ux2 = 0, .(x1 , x2 ) ∈ R . 3 .ux1 x1 + ux2 x2 + ux3 x3 = 0, .(x1 , x2 , x3 ) ∈ R . 2 2 2 .ux1 x1 ux2 x2 = x + x , .(x1 , x2 ) ∈ R . 1 2 3 .ux1 x1 − ux1 x2 + ux2 x2 + ux3 = x1 x2 x3 , .(x1 , x2 , x3 ) ∈ R . .

Other important classifications will be given in Chaps. 2, 3, and 4. Definition 1.4 A function in the set .C m that satisfies a PDE of order .m will be called a classical (strong) solution. Solutions that are not classical will be called weak solutions. Example 1.8 Consider the equation ux1 x1 + ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) /= (0, 0).

We will prove that the function u(x1 , x2 ) = log(x12 + x22 ),

.

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) /= (0, 0),

4

1 General Introduction

is its solution. We have .

ux1 (x1 , x2 ) =

2x1 , x12 + x22

ux2 (x1 , x2 ) =

2x2 , x12 + x22

ux1 x1 (x1 , x2 ) = = = ux2 x2 (x1 , x2 ) = = =

2(x12 + x22 ) − 2x1 (2x1 ) (x12 + x22 )2 2x12 + 2x22 − 4x12 (x12 + x22 )2 2(x22 − x12 ) (x12 + x22 )2

,

2(x12 + x22 ) − 2x2 (2x2 ) (x12 + x22 )2 2x12 + 2x22 − 4x22 (x12 + x22 )2 2(x12 − x22 ) (x12 + x22 )2

(x1 , x2 ) ∈ R2 ,

,

(x1 , x2 ) /= (0, 0).

Hence, ux1 x1 (x1 , x2 ) + ux2 x2 (x1 , x2 ) =

.

2(x22 − x12 ) (x12 + x22 )2

= 0,

+

2(x12 − x22 ) (x12 + x22 )2

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) /= (0, 0).

Example 1.9 Consider the equation ux1 x1 − ux2 x2 − 2ux2 = u,

.

(x1 , x2 ) ∈ R2 .

We will prove that the function u(x1 , x2 ) = e−x1 (x1 − x2 )2 ,

.

(x1 , x2 ) ∈ R2 ,

is its solution. We have ux1 (x1 , x2 ) = −e−x1 (x1 − x2 )2 + 2e−x1 (x1 − x2 )

.

= e−x1 (x1 − x2 )(x2 − x1 + 2),

1.2 Classification

5

ux1 x1 (x1 , x2 ) = −e−x1 (x1 − x2 )(x2 − x1 + 2) + e−x1 (x2 − x1 + 2) −e−x1 (x1 − x2 ) = −e−x1 (x1 − x2 )(x2 − x1 + 2) + 2e−x1 (x2 − x1 + 1), ux2 (x1 , x2 ) = −2e−x1 (x1 − x2 ), ux2 x2 (x1 , x2 ) = 2e−x1 ,

(x1 , x2 ) ∈ R2 .

Hence, ux1 x1 (x1 , x2 ) − ux2 x2 (x1 , x2 ) − 2ux2 (x1 , x2 )

.

= −e−x1 (x1 − x2 )(x2 − x1 + 2) + 2e−x1 (x1 − x2 ) = −e−x1 (x1 − x2 )(x2 − x1 + 2) + 2e−x1 (x2 − x1 + 1) − 2e−x1 + 4e−x1 (x1 − x2 ) = e−x1 (x1 − x2 )(2 − x2 + x1 − 2) = e−x1 (x1 − x2 )2 ,

(x1 , x2 ) ∈ R2 .

Example 1.10 Consider the equation ux1 x1 − 2ux1 x2 + ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

We will prove that the function u(x1 , x2 ) = ex1 +x2 (x1 + x2 ),

.

(x1 , x2 ) ∈ R2 ,

is its solution. We have ux1 (x1 , x2 ) = ex1 +x2 (x1 + x2 ) + ex1 +x2

.

= ex1 +x2 (x1 + x2 + 1), ux1 x1 (x1 , x2 ) = ex1 +x2 (x1 + x2 + 1) + ex1 +x2 = ex1 +x2 (x1 + x2 + 2), ux1 x2 (x1 , x2 ) = ex1 +x2 (x1 + x2 + 1) + ex1 +x2 = ex1 +x2 (x1 + x2 + 2), ux2 (x1 , x2 ) = ex1 +x2 (x1 + x2 ) + ex1 +x2 = ex1 +x2 (x1 + x2 + 1), ux2 x2 (x1 , x2 ) = ex1 +x2 (x1 + x2 + 1) + ex1 +x2 = ex1 +x2 (x1 + x2 + 2),

(x1 , x2 ) ∈ R2 .

6

1 General Introduction

Hence, ux1 x1 (x1 , x2 ) − 2ux1 x2 (x1 , x2 ) + ux2 x2 (x1 , x2 )

.

= ex1 +x2 (x1 + x2 + 2) − 2ex1 +x2 (x1 + x2 + 2) + ex1 +x2 (x1 + x2 + 2) = 0,

(x1 , x2 ) ∈ R2 .

Exercise 1.3 Prove that the given functions are solutions of the corresponding equations. 1. u(x1 , x2 ) = x2 φ(x12 − x22 ),

φ ∈ C 1 (R),

.

.

1 1 u ux + ux = 2 , x1 1 x2 2 x2

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) ∈ R2 .

2. u(x1 , x2 ) = sin x2 + φ(sin x1 − sin x2 ),

φ ∈ C 1 (R),

.

.

cos x2 ux1 + cos x1 ux2 = cos x1 cos x2 ,

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) ∈ R2 .

3. ( (u(x1 , x2 ))2 = x1 x2 φ

.

x2 x1

) ,

φ ∈ C 1 (R),

x1 uux1 + x2 uux2 = u2 ,

.

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) ∈ R2 .

4. ( 2 .x1

+ x22

+ (u(x1 , x2 )) = x2 φ 2

) u(x1 , x2 ) , x2

φ ∈ C 1 (R),

(x22 + u2 − x12 )ux1 − 2x1 x2 ux2 + 2x1 u = 0,

.

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) ∈ R2 .

5. 4x1 x2 u(x1 , x2 ) = −x14 − 2x12 + φ(x1 x2 ),

.

φ ∈ C 1 (R2 ),

x1 x2 ux1 − x22 ux2 + x1 (1 + x12 ) = 0,

.

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) ∈ R2 .

1.3 History and Applications

7

6. ( u(x1 , x2 ) = x1 x2 + x1 φ

.

x2 x1

) ,

φ ∈ C 1 (R),

x1 ux1 + x2 ux2 = x1 x2 + u,

.

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) ∈ R2 .

7. u(x1 , x2 ) = φ(x1 + x2 ),

.

φ ∈ C 2 (R),

ux1 x1 − 2ux1 x2 + ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) ∈ R2 .

8. u(x1 , x2 ) = e−x1 φ(x1 − x2 ),

.

φ ∈ C 2 (R),

ux1 x1 − ux2 x2 − 2ux2 = u,

.

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) ∈ R2 .

Remark 1.1 If each of the functions .u1 , .. . . , .ul satisfies a PDE and every linear combination of them satisfies that equation too, this property is called superposition principle. It allows the construction of complex solutions through combinations of simple solutions. In addition, we will use the superposition principle to obtain uniqueness of solutions of some PDEs.

1.3 History and Applications The study of partial differential equations (PDEs) arise in the eighteenth century in the work of Euler, d’Alambert, Lagrange and Laplace in the context of the development of models in the physical sciences, e.g. vibrating strings, elasticity, the Newtonian gravitational field of extended matter, electrostatics, fluid flows, and later by theories of heat conduction, electricity and magnetism. Problems in differential geometry gave rise to nonlinear PDEs such as the Monge-Ampére equation and the minimal surface equations. The Hamilton-Jacobi theory stimulated the analysis of first order PDEs. The classical calculus of variations gave rise to PDEs. The classical PDEs which serve as paradigms for the later development also appeared the eighteenth and early nineteenth century. The linear transport equation ut + cux = 0

.

8

1 General Introduction

is a first order partial differential equation through a constant speed c of quantity u in the presence of two variables that are spatial variable x and tempered variable t. The solution is a wave moving according to the value of c. In the case when .c > 0 the wave moves right. When .c < 0 the wave moves left. The Laplace equation Δu = 0

.

was first studied by Laplace in his work on gravitational potential fields around 1780. It is appeared in many applications such as fluid mechanics, heat conduction, electrostatics and gravitation. The heat equation ut = kΔu

.

was introduced by Fourier in his celebrated memoir “Théorie analytic de la chaleur” in 1810–1822. In the one dimensional heat equation the heat is expressed in a homogeneous medium by .u(x, t) as a function of x and t and .k > 0 is a constant. The wave equation utt = c2 Δu

.

expresses the propagation of the wave where c expresses the wave speed and u is a displacement as .u = u(x, t). This equation explains the transmission of guitar waves when vibrating. The displacement varies according to the medium in which it is located where in the case of the guitar .x ∈ R and in the case of the drum membrane .x ∈ R2 , and .utt is acceleration. The one dimensional wave equation is introduced and analyzed by d’Alambert in 1752. His work was extended by Euler in 1759 and later by D. Bernoulli in 1762 for two and three dimensional wave equations. Besides of these classical examples, a profusion of equations associated with major physical phenomena, appeared in the period between 1750 and 1900: • The Helmholtz equation in two dimensional form uxx + uyy + k 2 u = 0.

.

• The Poisson equation in two dimensional form uxx + uyy = f (x, y).

.

• The Schrödinger equation in two dimensional form in .(x, y, z) dimensions .

−

h2 (ψxx + ψyy + ψzz ) = Eψ, 8π 2 m

1.3 History and Applications

9

where h is the Planck constant. • The transverse vibrations equation a 2 uxxxx + utt = 0.

.

A central connection between PDEs and the mainstream of mathematical development in the nineteenth century arose from the role of PDEs in the theory of analytic functions of a complex variable. Cauchy observed in 1827 that two smooth real functions u and v of two real variables x and y are the real and imaginary parts of a single complex function of the complex variable z = x + iy

.

if they satisfy the Cauchy-Riemann system of first order equations ux = vy

.

uy = −vx . From this point of view, Riemann studied the properties of analytic functions by investigating harmonic functions in the plane. We have already mentioned some of the applications of linear partial differential equations, but there is a set of applications that does not apply linear PDEs. Here came the need to nonlinear partial differential equations. Now, we will review some applications of these equations. The inviscid Burgers equation ut + uux = 0

.

is used in many mathematical applications such as gas dynamics, fluid mechanics and nonlinear acoustics. This equation is most famous example of first order nonlinear equation. Other forms of Burgers equation for kinematic viscosity is the viscous Burgers equation ut + uux = νuxx .

.

The Fisher equation ut = Δu + u(1 − u)

.

is used as a model of spatial distribution of population dynamics. The porous medium equation ut = Δ(um ),

.

10

1 General Introduction

where .m > 0 is a constant, is one of most important fundamental equations in the theory of PDEs. It is used as a model for compacted soil and porous rock. When .m = 1 this equation will be the heat equation. The Korteweg-de Vries equation ut + uux + uxxx = 0

.

is used to describe the waves of water when there is a height of the water wave. The shallow water equations ut + (hu)x = 0

.

vt + vvx = ghx = 0 is used to describe the flow under the pressure surface of the liquid. Here g is the gravitation acceleration, h is height of a shallow layer of water, v is the velocity of a shallow layer of water, x is the horizontal spatial variable and t is variable which presents time.

1.4 Advanced Practical Problems Problem 1.1 Find the order of the following equations. 1. 2. 3. 4. 5.

ux1 x2 − ux2 x3 − x1 x2 x3 u2 = ux3 , (x1 , x2 , x3 ) ∈ R3 . ux1 − ux2 x2 − ux1 x3 x4 = u3 , (x1 , x2 , x3 , x4 ) ∈ R4 . ux1 x1 + 2ux2 x2 − x3 ux3 x3 = 0, (x1 , x2 , x3 ) ∈ R3 . ux1 ux2 ux1 x2 x3 = (x1 x2 x3 )2 , (x1 , x2 , x3 ) ∈ R3 . ux1 x1 − ux1 − ux2 − ux3 = u5 , (x1 , x2 , x3 ) ∈ R3 .

Problem 1.2 Classify each of the following equations as linear or nonlinear. 1. 2. 3. 4. 5.

ux1 x1 x1 + ux1 x2 + ux1 x3 = 0, (x1 , x2 , x3 ) ∈ R3 . ux1 x2 x3 − (ux1 )2 + ux3 = x1 , (x1 , x2 , x3 ) ∈ R3 . ux1 + ux2 + ux3 + ux4 = x1 x2 x3 x4 , (x1 , x2 , x3 , x4 ) ∈ R4 . ux1 ux2 x2 + ux3 x3 = u2 , (x1 , x2 , x3 ) ∈ R3 . ux1 x1 − ux1 − ux2 − ux3 x4 = 0, (x1 , x2 , x3 , x4 ) ∈ R4 .

Problem 1.3 Prove that the given functions are solutions of the corresponding equations. 1. u(x1 , x2 ) = φ(x2 − x1 ) − x1 φ ' (x2 − x1 ),

.

ux1 x1 − ux2 x2 = 2φ '' ,

.

φ ∈ C 3 (R), (x1 , x2 ) ∈ R2 .

(x1 , x2 ) ∈ R2 ,

1.4 Advanced Practical Problems

11

2. u(x1 , x2 ) = φ(x2 − ax1 ) + ψ(x2 + ax1 ),

.

ux1 x1 − a 2 ux2 x2 = 0,

φ ∈ C 2 (R),

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) ∈ R2 .

.

3. ( u(x1 , x2 ) = x1 φ

.

x2 x1

) φ ∈ C 2 (R),

,

x12 ux1 x1 + 2x1 x2 ux1 x2 + x22 ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 , (x1 , x2 ) ∈ R2 .

4. φ(x1 − x2 ) + ψ(x1 + x2 ) , x1

u(x1 , x2 ) =

.

ux1 x1 +

.

φ, ψ ∈ C 2 (R),

2 ux = ux2 x2 , x1 1

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) ∈ R2 .

5. / u(x1 , x2 ) =

.

x1 φ(x1 x2 ) + ψ x2

(

x2 x1

) ,

ψ ∈ C 2 (R),

x12 ux1 x1 − x22 ux2 x2 − 2x2 ux2 = 0,

.

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) ∈ R2 .

6. u(x1 , x2 ) = φ(x1 x2 ) log x2 + ψ(x1 x2 ),

.

φ, ψ ∈ C 2 (R),

x12 ux1 x1 − 2x1 x2 ux1 x2 + x22 ux2 x2 + x1 ux1 + x2 ux2 = 0,

.

(x1 , x2 ) ∈ R2 , (x1 , x2 ) ∈ R2 .

7. u(x1 , x2 ) = e−

.

x12 +x22 2

(φ(x1 ) + ψ(x2 )),

ux1 x2 + x2 ux1 + x1 ux2 + x1 x2 u = 0,

.

φ, ψ ∈ C 2 (R), (x1 , x2 ) ∈ R2 .

8. u(x1 , x2 ) = φ(x2 + x1 , x3 + x1 )

.

+ ψ(x2 − x1 , x3 − x1 ),

φ, ψ ∈ C 2 (R),

(x1 , x2 , x3 ) ∈ R3 ,

12

1 General Introduction

ux1 x1 = ux2 x2 + 2ux2 x3 + ux3 x3 ,

.

(x1 , x2 , x3 ) ∈ R3 .

9. φ(x1 − x2 ) + φ(x1 + x2 ) 1 .u(x1 , x2 ) = + 2 2

x 1 +x2

ψ(t)dt, x1 −x2

φ, ψ ∈ C 2 (R), (x1 , x2 ) ∈ R2 , ux1 x1 − ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

10. u(x1 , x2 ) =

.

φ(x2 − ax1 ) + ψ(x2 + ax1 ) , x1 ux1 x1 = a 2 ux2 x2 −

.

2 ux , x1 1

φ, ψ ∈ C 2 (R), (x1 , x2 ) ∈ R2 .

(x1 , x2 ) ∈ R2 ,

Chapter 2

First Order Partial Differential Equations

2.1 Classifications of First Order Partial Differential Equations Let .U ⊂ Rn . Definition 2.1 By a first order partial differential equation in .n independent variables .x1 , .. . . , .xn , we mean any equation of the form F (x1 , . . . , xn , u, ux1 , . . . , uxn ) = 0,

.

(x1 , . . . , xn ) ∈ U,

(2.1)

where .u : U → R is the unknown. Such equations arise in the construction of characteristic surfaces for hyperbolic partial differential equations, in the calculus of variations, in some geometrical problems, and in simple models for gas dynamics whose solution involves the method of characteristics. If a family of solutions of a single first order partial differential equation can be found, additional solutions may be obtained by forming envelopes of solutions in that family. General solutions can be obtained by integrating families of ordinary differential equations. Example 2.1 The continuity equation or transport equation ut + cux = 0,

.

t ≥ 0,

x ∈ R,

where c is a given constant, is a first order partial differential equation. The transport equation describes the transport of some quantity. Since mass, energy, momentum, electric charge and other natural quantities are conserved under their respective appropriate conditions, a variety of physical phenomena may be described using the transport equation. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. G. Georgiev, An Excursion Through Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-031-48784-2_2

13

14

2 First Order Partial Differential Equations

Example 2.2 Characteristic surfaces for the wave equation are level surfaces for solutions of the equation u2t = c2 (u2x1 + u2x2 + u2x3 ),

.

t ≥ 0,

(x1 , x2 , x3 ) ∈ R3 ,

where .c > 0 is a given constant, is a first order partial differential equation. Definition 2.2 If Eq. (2.1) can be written in the form n  .

ai (x1 , . . . , xn , u)uxi = b(x1 , . . . , xn , u),

(x1 , . . . , xn ) ∈ U,

i=1

where .ai , b : U × R → R, .i ∈ {1, . . . , n}, are given functions, then we say that the equation is quasilinear. Example 2.3 The Hopf equation ut + uux = 0,

.

t ≥ 0,

x ∈ R,

is a quasilinear first order PDE. Definition 2.3 If Eq. (2.1) can be written in the form n  .

ai (x1 , . . . , xn )uxi = b(x1 , . . . , xn , u),

(x1 , . . . , xn ) ∈ U,

i=1

where .ai : U → R, .i ∈ {1, . . . , n}, .b : U × R → R are given functions, then we say that the equation is semilinear. Example 2.4 The following equations .

x1 ux1 + x2 ux2 = u2 + x12 , (x1 + 1)2 ux1 + (x2 − 1)2 ux2 = (x1 + x2 )u2 ,

(x1 , x2 ) ∈ R2 ,

are semilinear first order PDEs. Exercise 2.1 Classify each of the following equations as quasilinear or semilinear. 1. 2. 3. 4. 5.

x14 ux1 + ux2 + uux3 = u2 − 1, .(x1 , x2 , x3 ) ∈ R3 . √ .x1 uux1 + uux2 + ux3 + 2x2 ux4 = 1 + u2 , .(x1 , x2 , x3 , x4 ) ∈ R4 . 4 2 .ux1 + ux2 + ux3 = u , .(x1 , x2 , x3 ) ∈ R . 3 3 .(x1 + 1)ux1 + (x1 x2 x3 − 1)ux2 + x uux3 = 0, .(x1 , x2 , x3 ) ∈ R . 2 2 2 .ux1 − 2x ux2 = sin u, .(x1 , x2 ) ∈ R . 2 .

2.1 Classifications of First Order Partial Differential Equations

15

Definition 2.4 If Eq. (2.1) can be written in the form n  .

ai (x1 , . . . , xn )uxi + b(x1 , . . . , xn )u = c(x1 , . . . , xn ),

(x1 , . . . , xn ) ∈ U,

i=1

(2.2) where .ai , b, c : U → R, .i ∈ {1, . . . , n}, are given functions, then we say that the equation is linear. Example 2.5 The equations .

x1 ux1 + x2 ux2 + x32 ux3 = (x1 + x2 + x3 )2 u, (x1 , x2 , x3 ) ∈ R3 , (x1 − 1)2 ux1 + x2 ux2 = 2x1 , (x1 , x2 ) ∈ R2 ,

are linear first order PDEs. Definition 2.5 A linear first order PDE is called homogeneous if .c(x1 , . . . , xn ) = 0 for any .(x1 , . . . , xn ) ∈ U and nonhomogeneous if .c(x1 , . . . , xn ) /= 0 for some .(x1 , . . . , xn ) ∈ U . Example 2.6 The equation x1 ux1 + x2 ux2 = 3x2 u,

.

(x1 , x2 ) ∈ R2 ,

is a linear homogeneous first order PDE. Example 2.7 The equation x12 ux1 − x1 x2 ux2 + x3 ux3 = x1 − 2x2 + x3 ,

.

(x1 , x2 , x3 ) ∈ R3 ,

is a linear nonhomogeneous first order PDE. Exercise 2.2 Classify each of the following equations as linear homogeneous or linear nonhomogeneous. 1. 2. 3. 4. 5.

x1 ux1 − ux2 = 0, .(x1 , x2 ) ∈ R2 . 3 .ux1 − 2ux2 + 3ux3 = x1 + x2 , .(x1 , x2 , x3 ) ∈ R . 2 2 3 .(x + x + 1)ux1 − x3 ux2 + ux3 = 1, .(x1 , x2 , x3 ) ∈ R . 1 2 4 .ux1 − ux2 + ux3 + ux4 = 0, .(x1 , x2 , x3 , x4 ) ∈ R . 3 .2ux1 − 3ux2 + x3 ux3 = x1 , .(x1 , x2 , x3 ) ∈ R . .

Definition 2.6 A first order PDE that is not linear is said to be nonlinear. Example 2.8 The equations .

u2x1 − 2uux1 + u2x2 = u2 , ux1 − u2x2 + u2 = x12 , (x1 , x2 ) ∈ R2 ,

are nonlinear first order PDEs.

16

2 First Order Partial Differential Equations

Exercise 2.3 Classify each of the following equations as linear or nonlinear. 1. .(ux1 )2 + ux2 + ux3 = 0, .(x1 , x2 , x3 ) ∈ R3 . 2. . 1+(u1 )2 + (ux2 )2 − ux3 = u2 + x2 + x3 , .(x1 , x2 , x3 ) ∈ R3 . x1

3. .ux1 + 2x1 ux2 + 2x2 ux3 = u + sin x1 , .(x1 , x2 , x3 ) ∈ R3 . 4. .−ux1 ux2 + ux3 = u, .(x1 , x2 , x3 ) ∈ R3 . 5. .sin(x1 + 1)ux2 − ux1 = cos x2 , .(x1 , x2 ) ∈ R2 . Definition 2.7 A function .u ∈ C 1 (U ) that satisfies (2.1) is said to be a solution of the PDE (2.1). Example 2.9 Consider the transport equation aux1 (x1 , x2 ) + bux2 (x1 , x2 ) = 0,

.

(x1 , x2 ) ∈ R2 ,

where .a and .b are constants. We will show that u(x1 , x2 ) = f (bx1 − ax2 ),

(x1 , x2 ) ∈ R2 ,

.

where .f ∈ C 1 (R), is its solution. Indeed, ux1 (x1 , x2 ) = bf ' (bx1 − ax2 ),

.

ux2 (x1 , x2 ) = −af ' (bx1 − ax2 ),

(x1 , x2 ) ∈ R2 .

Then aux1 (x1 , x2 ) + bux2 (x1 , x2 ) = abf ' (bx1 − ax2 ) − abf ' (bx1 − ax2 )

.

= 0,

(x1 , x2 ) ∈ R2 .

Example 2.10 Consider the equation x1 ux1 + x2 ux2 = 2,

.

(x1 , x2 ) ∈ R2 \{(0, 0)}.

We will prove that the function u(x1 , x2 ) = log(x12 + x1 x2 + x22 ),

.

(x1 , x2 ) ∈ R2 \{(0, 0)},

is its solution. We have ux1 (x1 , x2 ) =

.

ux2 (x1 , x2 ) =

x12

2x1 + x2 , + x1 x2 + x22

x12

x1 + 2x2 , + x1 x2 + x22

(x1 , x2 ) ∈ R2 \{(0, 0)}.

2.1 Classifications of First Order Partial Differential Equations

17

Hence, x1 ux1 (x1 , x2 ) + x2 ux2 (x1 , x2 ) = x1

.

= =

x12

2x1 + x2 x1 + 2x2 + x2 2 2 + x1 x2 + x2 x1 + x1 x2 + x22

2x12 + x1 x2 x12

+ x1 x2 + x22

+

x1 x2 + 2x22 x12

+ x1 x2 + x22

2(x12 + x1 x2 + x22 ) x12 + x1 x2 + x22

= 2,

(x1 , x2 ) ∈ R2 \{(0, 0)}.

Example 2.11 Consider the equation ux1 + ux2 + ux3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 .

We will prove that the function u(x1 , x2 , x3 ) = (x1 − x2 )(x2 − x3 )(x3 − x1 ),

.

(x1 , x2 , x3 ) ∈ R3 ,

is its solution. We have ux1 (x1 , x2 , x3 ) = (x2 − x3 )(x3 − x1 ) − (x1 − x2 )(x2 − x3 )

.

= x2 x3 − x1 x2 − x32 + x1 x3 − x1 x2 + x1 x3 + x22 − x2 x3 = x22 − x32 + 2x1 x3 − 2x1 x2 , ux2 (x1 , x2 , x3 ) = −(x2 − x3 )(x3 − x1 ) + (x1 − x2 )(x3 − x1 ) = −x2 x3 + x1 x2 + x32 − x1 x3 + x1 x3 − x12 − x2 x3 + x1 x2 = −x12 + x32 + 2x1 x2 − 2x2 x3 , ux3 (x1 , x2 , x3 ) = −(x1 − x2 )(x3 − x1 ) + (x1 − x2 )(x2 − x3 ) = −x1 x3 + x12 + x2 x3 − x1 x2 + x1 x2 − x1 x3 − x22 + x2 x3 = x12 − x22 − 2x1 x3 + 2x2 x3 ,

(x1 , x2 , x3 ) ∈ R3 .

Hence, ux1 (x1 , x2 , x3 ) + ux2 (x1 , x2 , x3 ) + ux3 (x1 , x2 , x3 )

.

= x22 −x32 +2x1 x3 −2x1 x2 −x12 +x32 +2x1 x2 −2x2 x3 +x12 − x22 − 2x1 x3 + 2x2 x3 = 0,

(x1 , x2 , x3 ) ∈ R3 .

18

2 First Order Partial Differential Equations

Exercise 2.4 Prove that the function u(x1 , x2 ) = x2 − x1 + f (x12 − x22 ),

.

(x1 , x2 ) ∈ R2 ,

where .f ∈ C 1 (R), is a solution to the equation x2 ux1 + x1 ux2 = x1 − x2 ,

.

(x1 , x2 ) ∈ R2 .

In some first order PDEs participate only the derivative of the unknown function with respect to one of the independent variables. In these cases, it is convenient to fix the rest of the independent variables and to consider the given first order PDE as a first order ordinary differential equation (ODE). This phenomena we will illustrate in the next examples. Example 2.12 We will find a solution .u = u(x1 , x2 ), .(x1 , x2 ) ∈ R2 , to the equation x1 ux1 − 2u = 2x14 ,

(x1 , x2 ) ∈ R2 .

.

Fix .x2 ∈ R and consider the given equation as a linear first order ODE with respect to .x1 . Then 2 u + 2x13 , x1

ux1 =

.

x1 ∈ R,

x1 /= 0,

and u(x1 , x2 ) = e

.



2 x1 dx1

   − f (x2 ) + 2 x13 e

2 x1 dx1

 dx1

   2 2 = elog(x1 ) f (x2 ) + 2 x13 e− log(x1 )dx1 dx1 



1 = x12 f (x2 ) + 2 x13 2 dx1 x1    = x12 f (x2 ) + 2 x1 dx1 = x12 (f (x2 ) + x12 ),



(x1 , x2 ) ∈ R2 ,

where .f ∈ C (R). When .x1 = 0, we get .u(0, x2 ) = 0, .x2 ∈ R. Example 2.13 We will find a function .u = u(x1 , x2 ), .(x1 , x2 ) ∈ R2 , that satisfies the equation x22 u2 ux2 + 1 = u,

.

(x1 , x2 ) ∈ R2 .

2.1 Classifications of First Order Partial Differential Equations

19

Fix .x1 ∈ R and consider the given equation as an ODE with separable variables with respect to .x2 . Then, we get x22 u2 ux2 = u − 1,

.

x2 ∈ R,

whereupon .

u2 1 ux = 2 , u−1 2 x2

u(x1 , x2 ) /= 1,

x2 ∈ R,

x2 /= 0.

Hence,  .

u2 du = u−1



1 dx2 + f (x1 ), x22

u(x1 , x2 ) /= 1,

x2 ∈ R,

x2 /= 0,

or  .

u2 − 1 du+ u−1



1 du= u−1



1 dx2 +f (x1 ), u(x1 , x2 ) /= 1, x2 ∈ R, x2 /= 0, x22

or  (u+1)du+ log |u(x1 , x2 )−1|=−

.

1 +f (x1 ), u(x1 , x2 ) /= 1, x2 ∈ R, x2 /= 0, x2

or .

(u(x1 , x2 ) + 1)2 1 + log |u(x1 , x2 )−1| = − +f (x1 ), u(x1 , x2 ) /= 1, x2 ∈ R, x2 /= 0, 2 x2

where .f ∈ C (R). Note that .u(x1 , x2 ) = 1, .(x1 , x2 ) ∈ R2 , is a solution to the given equation and .x2 = 0 is not its solution. Example 2.14 We will find a function .u = u(x1 , x2 ), .(x1 , x2 ) ∈ R2 , that satisfies the equation (x12 − 1)ux1 sin u + 2x1 cos u = 2x1 − 2x13 ,

.

(x1 , x2 ) ∈ R2 .

Fix .x2 and consider the given equation as a first order ODE with respect to .x1 . Set z(x1 , x2 ) = cos(u(x1 , x2 )),

.

(x1 , x2 ) ∈ R2 .

Then zx1 (x1 , x2 ) = −ux1 (x1 , x2 ) sin(u(x1 , x2 )),

.

(x1 , x2 ) ∈ R2 ,

20

2 First Order Partial Differential Equations

and we get the equation .

− (x12 − 1)zx1 + 2x1 z = 2x1 − 2x13 ,

(x1 , x2 ) ∈ R2 ,

or (x12 − 1)zx1 = 2x1 z + 2x13 − 2x1 ,

.

(x1 , x2 ) ∈ R2 ,

whereupon zx 1 =

.

2x1 z + 2x1 , −1

(x1 , x2 ) ∈ R2 ,

x12

x1 /= ±1.

Hence, 

z(x1 , x2 ) = e

.



=e =e

2x1 dx1 x12 −1





f1 (x2 ) + 2

⎛

d(x12 −1) x12 −1

⎝f1 (x2 ) + 2

log |x12 −1|



x1 e

 x1 e

−

 f1 (x2 ) + 2



−

x1 e





2x1 dx1 x12 −1

d(x12 −1) x12 −1

 dx1 ⎞

dx1 ⎠

− log |x12 −1|

 dx1

 x1 dx1 − 1) f1 (x2 ) + 2 x12 − 1    d(x12 − 1) 2 (x1 − 1) f1 (x2 ) + x12 − 1

 (x12 − 1) f1 (x2 ) + log |x12 − 1|

 (x12 − 1) log f (x2 )(x12 − 1) , (x1 , x2 ) ∈ R2 ,

=



(x12

= = =

x1 /= ±1,

i.e.,

 z(x1 , x2 ) = (x12 − 1) log f (x2 )(x12 − 1) ,

.

(x1 , x2 ) ∈ R2 ,

x1 /= ±1,

where .f1 ∈ C (R), .f (x2 ) = ef1 (x2 ) , .x2 ∈ R, is such that f (x2 )(x12 − 1) > 0,

.

and

(x1 , x2 ) ∈ R2 ,

x1 /= ±1,

(2.3)

2.2 Solvability of Quasilinear First Order PDEs

 (x12 − 1) log f (x2 )(x12 − 1) ∈ [−1, 1],

.

21

(x1 , x2 ) ∈ R2 ,

x1 /= ±1.

(2.4)

Therefore .

 cos(u(x1 , x2 )) = (x12 − 1) log f (x2 )(x12 − 1) ,

(x1 , x2 ) ∈ R2 ,

x1 /= ±1,

and

 u(x1 , x2 ) = arccos (x12 − 1) log f (x2 )(x12 − 1) ,

.

(x1 , x2 ) ∈ R2 ,

x1 /= ±1,

provided that (2.3), (2.4) hold. Note that .x1 = ±1 are not solutions. Exercise 2.5 Find a solution .u = u(x1 , x2 ), .(x1 , x2 ) ∈ R2 , to the equation ux1 + cos x1 u =

.

1 sin(2x1 ), 2

(x1 , x2 ) ∈ R2 .

Exercise 2.6 Derive a solution .u = u(x1 , x2 ), .(x1 , x2 ) ∈ R2 , of the equation aux1 + bux2 = u,

.

(x1 , x2 ) ∈ R2 ,

where .a and .b are constants, .b /= 0, by using the change of variables v(x1 , x2 ) = −bx1 + ax2 ,

.

w(x1 , x2 ) = x2 ,

(x1 , x2 ) ∈ R2 .

2.2 Solvability of Quasilinear First Order PDEs In this section, we discuss the solvability of the quasilinear first order PDE n  .

ai (x1 , . . . , xn , u)uxi = f (x1 , . . . , xn , u)

(2.5)

i=1

via the method of characteristics. Suppose that .ai , f : U × R → R, .ai , f ∈ C (U × R), .1 ≤ i ≤ n. To solve Eq. (2.5) we proceed as follows. Assume that we have found a solution .u = u(x1 , . . . , xn ), .(x1 , . . . , xn ) ∈ U , of (2.5). This solution may be interpreted geometrically as a surface in .(x1 , . . . , xn , xn+1 )-space, called the integral surface, where .xn+1 = u(x1 , . . . , xn ). This integral surface can be viewed as the level surface of the function F (x1 , . . . , xn , xn+1 ) = u(x1 , . . . , xn ) − xn+1 = 0.

.

22

2 First Order Partial Differential Equations

Let v = (a1 , . . . , an , f ),

.

that is the characteristic direction. Then Eq. (2.5) can be written as follows. v · ∇F = 0.

(2.6)

.

Note that .∇F is normal to the surface .F (x1 , . . . , xn , xn+1 ) = 0 and is pointing downward. Hence, .∇F is normal to .v and this implies that .v is tangent to the surface .F = 0 at .(x1 , . . . , xn , xn+1 ). Therefore to find a solution of (2.5) is equivalent to find a surface .S such that at every point on the surface the vector .v is tangent to the surface. If we find the integral curves of the vector field .v, then patch all these curves together to obtain the desired surface. We start by constructing a curve .Γ parameterized by .t such that at each point of .Γ the vector .v is tangent to .Γ. A parametrization of this curve is given by the vector function r(t) = (x1 (t), . . . , xn (t), u(t)).

.

Then the tangent vector is r ' (t) = (x1' (t), . . . , xn' (t), u' (t)).

.

Hence, the vectors .r ' (t) and .v are parallel, so these vectors are proportional and this leads to the ODE system

.

dx1 dt

a1

=

dx2 dt

a2

= ··· =

dxn dt

an

=

du dt

f (x1 , . . . , xn , u)

or in differential form .

dx1 dx2 dxn du . = = ··· = = a1 a2 an f

(2.7)

By solving the system (2.7), we are assured that the vector .v is tangent to the curve Γ which in turn lies in the solution surface .S.

.

Definition 2.8 Integral curves are called characteristic curves or simply characteristics of the PDE (2.5). Definition 2.9 We call (2.7) the characteristic equations. Definition 2.10 The projection of .Γ into .(x1 , . . . , xn )-plane is called the projected characteristic curve.

2.2 Solvability of Quasilinear First Order PDEs

23

Once we have found the characteristic curves, the surface .S is the union of these characteristic curves. Example 2.15 Consider the equation x2 ux1 − x1 ux2 = 0,

(x1 , x2 ) ∈ R2 .

.

The characteristic equation is .

dx1 dx2 =− , x2 x1

(x1 , x2 ) ∈ R2 ,

whereupon x1 dx1 + x2 dx2 = 0,

(x1 , x2 ) ∈ R2 ,

.

and then x12 + x22 = c,

.

(x1 , x2 ) ∈ R2 .

Therefore F (x12 + x22 , u) = 0,

(x1 , x2 ) ∈ R2 ,

.

or u(x1 , x2 ) = f (x12 + x22 ),

.

(x1 , x2 ) ∈ R2 ,

where .f ∈ C 1 (R) and .F ∈ C 1 (R2 ). Example 2.16 Consider the equation x1 x2 ux1 − x12 ux2 = x2 u,

.

(x1 , x2 ) ∈ R2 .

The characteristic equations are .

dx1 dx2 du , =− 2 = x1 x2 x2 u x1

(x1 , x2 ) ∈ R2 .

By the equation .

we get

dx1 dx2 =− 2 , x1 x2 x1

(x1 , x2 ) ∈ R2 ,

24

2 First Order Partial Differential Equations

x1 dx1 + x2 dx2 = 0,

(x1 , x2 ) ∈ R2 ,

.

and then x12 + x22 = c1 ,

(x1 , x2 ) ∈ R2 .

dx1 du , = x1 x2 x2 u

(x1 , x2 ) ∈ R2 ,

dx1 du , = x1 u

(x1 , x2 ) ∈ R2 ,

.

By the equation .

we obtain .

and then .

u = c2 , x1

(x1 , x2 ) ∈ R2 .

Here .c1 and .c2 are constants. Therefore   2 2 u = 0, .F x1 + x2 , x1

(x1 , x2 ) ∈ R2 ,

and u(x1 , x2 ) = x1 f (x12 + x22 ),

.

(x1 , x2 ) ∈ R2 ,

where .f ∈ C 1 (R) and .F ∈ C 1 (R2 ). Example 2.17 Consider the equation  2x24 ux1 − x1 x2 ux2 = x1 u2 + 1,

.

(x1 , x2 ) ∈ R2 .

The characteristic equations are .

dx1 dx2 du , =− = √ x1 x2 2x24 x1 u2 + 1

(x1 , x2 ) ∈ R2 .

By the equation .

dx1 dx2 =− , 4 x1 x2 2x2

(x1 , x2 ) ∈ R2 ,

2.2 Solvability of Quasilinear First Order PDEs

25

we get x1 dx1 + 2x23 dx2 = 0,

.

(x1 , x2 ) ∈ R2 ,

and then x12 + x24 = c1 ,

.

(x1 , x2 ) ∈ R2 .

By the equation −

.

dx2 du = √ , x1 x2 x1 u2 + 1

(x1 , x2 ) ∈ R2 ,

we obtain

  x2 u + u2 + 1 = c2 ,

.

(x1 , x2 ) ∈ R2 ,

where .c1 and .c2 are constants. Therefore



 2 4 .F x1 + x2 , x2 u + u2 + 1 = 0,

(x1 , x2 ) ∈ R2 ,

where .F ∈ C 1 (R2 ). Example 2.18 Consider the equation x1 (x22 − u2 )ux1 − x2 (u2 + x12 )ux2 = (x12 + x22 )u,

.

(x1 , x2 ) ∈ R2 .

The characteristic equations are dx2 du dx1 = = , 2 2 2 2 2 x1 (x2 − u ) −x2 (u + x1 ) u(x1 + x22 )

.

(x1 , x2 ) ∈ R2 .

The last equations we can write in the following form .

x12 (x22

du x1 dx1 + x2 dx2 + udu = , 2 2 2 2 2 2 2 2 − u ) − x2 (u + x1 ) + u (x1 + x2 ) u(x1 + x22 )

(x1 , x2 ) ∈ R2 ,

or .

du x1 dx1 + x2 dx2 + udu = , 2 0 u(x1 + x22 )

(x1 , x2 ) ∈ R2 .

Therefore x1 dx1 + x2 dx2 + udu = 0,

.

(x1 , x2 ) ∈ R2 .

26

2 First Order Partial Differential Equations

Hence, x12 + x22 + u2 = c1 ,

.

(x1 , x2 ) ∈ R2 .

Also,

.

dx1 x1 x22 − u2

dx2 x2 + u2 + x12

−

=

(x12

du , + x22 )u

(x1 , x2 ) ∈ R2 ,

or .

dx1 dx2 du , − = x1 x2 u

(x1 , x2 ) ∈ R2 ,

whereupon .

x2 u = c2 , x1

(x1 , x2 ) ∈ R2 .

Here .c1 and .c2 are constants. Consequently   x2 u(x1 , x2 ) = 0, F x12 + x22 + (u(x1 , x2 ))2 , x1 x1 u(x1 , x2 ) = f (x12 + x22 + (u(x1 , x2 ))2 ), x2

.

where .f ∈ C 1 (R) and .F ∈ C 1 (R2 ), and (x1 , x2 ) ∈ R2 . Exercise 2.7 Find a solution to the following equations 1. x1 ux1 + 2x2 ux2 = x12 x2 + u,

.

(x1 , x2 ) ∈ R2 .

2. x2 ux1 − x1 ux2 = 0,

.

(x1 , x2 ) ∈ R2 .

3. .

(x1 + 2x2 )ux1 − x2 ux2 = 0,

(x1 , x2 ) ∈ R2 .

x1 ux1 + x2 ux2 + x3 ux3 = 0,

(x1 , x2 ) ∈ R2 .

4. .

2.3 The Cauchy Problem for Quasilinear First Order PDEs

27

5. (x1 − x3 )ux1 + (x2 − x3 )ux2 + x3 ux3 = 0,

(x1 , x2 ) ∈ R2 .

.

2.3 The Cauchy Problem for Quasilinear First Order PDEs Let .C be a given curve in .Rn defined parametrical by the equations xi = xi0 (t),

.

t ∈ I,

i ∈ {1, . . . , n},

where .xi0 , .i ∈ {1, . . . , n}, are continuously differentiable functions on some interval I ⊂ R. Let .u0 ∈ C 1 (I ) be a given function. The Cauchy1 problem for Eq. (2.5) asks for a continuously-differentiable function .u = u(x1 , . . . , xn ) defined in a domain n .Ω ⊂ R containing the curve .C and such that .

1. .u = u(x1 , . . . , xn ) is a solution of (2.5) in .Ω. 2. On the curve .C, .u equals the given function .u0 , i.e., u(x10 (t), . . . , xn0 (t)) = u0 (t),

.

t ∈ I.

(2.8)

Definition 2.11 We call .C the initial curve, .u0 the initial data, and (2.8) the Cauchy data or initial condition of the problem. Example 2.19 Consider the Cauchy problem .

ux1 + ux2 = 1, (x1 , x2 ) ∈ R2 , u(x1 , x1 ) = x1 , x1 ∈ R.

The characteristic equations are dx1 = dx2 = du,

.

(x1 , x2 ) ∈ R2 ,

whereupon x1 − x2 = c1 ,

.

u − x1 = c2 ,

(x1 , x2 ) ∈ R2 ,

1 Augustin-Louis Cauchy (21 August 1789–23 May 1857) was a French mathematician reputed as a pioneer in analysis. His writings range widely in mathematics and mathematical physics.

28

2 First Order Partial Differential Equations

where .c1 and .c2 are constants. Then the general solution of the considered equation is u(x1 , x2 ) = x1 + f (x1 − x2 ),

.

(x1 , x2 ) ∈ R2 ,

(2.9)

where .f ∈ C 1 (R). We have u(x1 , x1 ) = x1 + f (0)

.

= x1 ,

x1 ∈ R.

Therefore f (0) = 0.

.

Hence, the solution of the considered Cauchy problem is given by (2.9), where f ∈ C 1 (R) is such that .f (0) = 0.

.

Example 2.20 Consider the problem .

x1 ux1 + x2 ux2 = u − x1 x2 , (x1 , x2 ) ∈ R2 , = x22 + 1, x2 ∈ R. u(2, x2 )

The characteristic equations are .

dx1 du dx2 = , = x2 u − x1 x2 x1

(x1 , x2 ) ∈ R2 .

By the equation .

dx1 dx2 , = x2 x1

(x1 , x2 ) ∈ R2 ,

we get .

x1 = c1 , x2

(x1 , x2 ) ∈ R2 ,

where .c1 is a constant. By the equation .

dx2 du , = u − x1 x2 x2

we obtain .

du 1 = u − x1 x2 dx2

(x1 , x2 ) ∈ R2 ,

2.3 The Cauchy Problem for Quasilinear First Order PDEs

=

1 u − c1 x2 , x2

29

x2 ∈ R.

Hence, u(x1 , x2 ) = x2 (c2 − c1 x2 )

.

= x2 (c2 − x1 ),

(x1 , x2 ) ∈ R2 ,

and .

u(x1 , x2 ) + x1 = c2 , x2

(x1 , x2 ) ∈ R2 ,

where .c2 is a constant. Consequently the general solution of the considered equation is   x1 .u(x1 , x2 ) = −x1 x2 + x2 f , (x1 , x2 ) ∈ R2 , x2 where .f ∈ C 1 (R). By the given data, we obtain

.

u(2, x2 ) = −2x2 + x2 f = x22 + 1,

 2 x2

x2 ∈ R,

whereupon  x2 f

.

2 x2

 = (x2 + 1)2 =

x22

  1 2 , 1+ x2

x2 ∈ R,

and .

1 f x2



2 x2



  1 2 , = 1+ x2

x2 ∈ R,

and for z=

.

2 , x2

x2 ∈ R,

we arrive at f (z) =

.

1 (z + 2)2 , 2z

x2 ∈ R.

30

2 First Order Partial Differential Equations

Therefore 2x1 (u(x1 , x2 ) + x1 x2 ) = (x1 + 2x2 )2 ,

.

x2 ∈ R.

Exercise 2.8 Solve the problem .

x1 ux1 − x2 ux2 = u2 (x1 − 3x2 ) x2 u(1, x2 ) + 1 = 0.

Exercise 2.9 Suppose that .x10 = x10 (t), .x20 = x20 (t) and .u0 = u0 (t) are continuously differentiable functions of .t on an interval .I, and a1 = a1 (x1 , x2 , u),

.

a2 = a2 (x1 , x2 , u), f = f (x1 , x2 , u) are continuously differentiable in a domain .D of .(x1 , x2 , u)-space containing the initial curve C : x1 = x10 (t),

.

x2 = x20 (t),

u = u0 (t),

t ∈ I.

If .(x10 (t), x20 (t), u0 (t)) is a point on .C that satisfies the condition ' ' a1 (x10 (t), x20 (t), u0 (t))x20 (t) − a2 (x10 (t), x20 (t), u0 (t))x10 (t) /= 0,

.

(2.10)

then by continuity this relation holds in a neighbourhood .U of .(x10 (t), x20 (t), u0 (t)) so that .C is nowhere characteristic in .U. Prove that there exists a unique solution .u = u(x1 , x2 ) of (2.5) such that the initial condition (2.8) holds for every point on .C contained in .U.

2.4 The Pfaff Equation Here we suppose that .G⊂ R3 , .P , .Q, .R : G → R are continuously differentiable functions. Definition 2.12 The equation P (x1 , x2 , x3 )dx1 + Q(x1 , x2 , x3 )dx2 + R(x1 , x2 , x3 )dx3 = 0,

.

(x1 , x2 , x3 ) ∈ G, (2.11)

2.4 The Pfaff Equation

31

will be called the Pfaff2 differential equation. Let e1 = (1, 0, 0), . e2 = (0, 1, 0), e3 = (0, 0, 1). In .G we consider the vector field F (x1 , x2 , x3 )=P (x1 , x2 , x3 )e1 +Q(x1 , x2 , x3 )e2 +R(x1 , x2 , x3 )e3 , (x1 , x2 , x3 ) ∈ G.

.

Definition 2.13 Every functional dependence of the variables .x1 , .x2 and .x3 for which .dx1 , .dx2 and .dx3 satisfy Eq. (2.11) will be called integral of Eq. (2.11). If this dependence is represented in the form u(x1 , x2 , x3 ) = 0,

.

(x1 , x2 , x3 ) ∈ G,

or parametric

.

x1 = x1 (t, τ ) x2 = x2 (t, τ ) x3 = x3 (t, τ ),

then it will be called two dimensional integral or integral surface of Eq. (2.11). If this functional dependence is represented in the form .

u(x1 , x2 , x3 ) = 0 v(x1 , x2 , x3 ) = 0

or parametric in the form x1 = x1 (t) . x2 = x2 (t) x3 = x3 (t),

t ∈ I,

then it will be called one dimensional integral or integral curve of Eq. (2.11). If r(x1 , x2 , x3 ) = x1 e1 + x2 e2 + x3 e3 ,

.

(x1 , x2 , x3 ) ∈ G,

2 Carl Friedrich Pfaff (22 December 1765–21 April 1825) was a German mathematician. He was described as one of Germany’s most eminent mathematicians during the nineteenth century.

32

2 First Order Partial Differential Equations

then dr = dx1 e1 + dx2 e2 + dx3 e3 ,

.

(x1 , x2 , x3 ) ∈ G.

Therefore F (x1 , x2 , x3 ) · dr(x1 , x2 , x3 ) = 0,

.

(x1 , x2 , x3 ) ∈ G.

Also, we have curlF =

.

∂R ∂x2

∂P

∂Q ∂R  ∂P  ∂Q  e1 + e2 + e3 . − − ∂x3 ∂x3 ∂x1 ∂x1 ∂x2

−

(2.12)

Exercise 2.10 Let Eq. (2.11) has two dimensional integral in G. Prove that curlF (x1 , x2 , x3 ) · F (x1 , x2 , x3 ) = 0,

.

(x1 , x2 , x3 ) ∈ G.

(2.13)

Remark 2.1 The equality (2.13) we can rewrite in the form P

.

∂R ∂x2

−

∂P

∂Q ∂Q  ∂R  ∂P  +Q +R = 0. − − ∂x3 ∂x3 ∂x1 ∂x1 ∂x2

Now, we suppose that .curlF = 0. Then .

∂R ∂Q = ∂x3 ∂x2 ∂P ∂R = ∂x1 ∂x3 ∂Q ∂P = . ∂x1 ∂x2

In this case, there exists a function .u ∈ C 1 (G) such that du(x1 , x2 , x3 ) = P (x1 , x2 , x3 )dx1 + Q(x1 , x2 , x3 )dx2 + R(x1 , x2 , x3 )dx3 . (2.14)

.

From here, for fixed .(x1 , x2 , x3 ) ∈ G, it follows that (x1 ,x2 ,x3 )

P (τ1 , τ2 , τ3 )dτ1 + Q(τ1 , τ2 , τ3 )dτ2

u(x1 , x2 , x3 ) =

.

(ξ1 ,ξ2 ,ξ3 )

 + R(τ1 , τ2 , τ3 )dτ3 ,

Then the solution of the Pfaff equation (2.11) is given by

(ξ1 , ξ2 , ξ3 ) ∈ G.

2.4 The Pfaff Equation

33

u(x1 , x2 , x3 ) = c,

.

(x1 , x2 , x3 ) ∈ G,

where .c is a constant. Example 2.21 We consider the equation (4x13 −2x2 x3 )dx1 +(4x23 −2x1 x3 )dx2 +(4x33 − 2x1 x2 )dx3 =0, (x1 , x2 , x3 ) ∈ R3 .

.

We have P (x1 , x2 , x3 ) = 4x13 − 2x2 x3 , .

Q(x1 , x2 , x3 ) = 4x23 − 2x1 x3 , R(x1 , x2 , x3 ) = 4x33 − 2x1 x2 ,

(x1 , x2 , x3 ) ∈ R3 .

Then .

∂R (x1 , x2 , x3 ) = −2x1 ∂x2 =

∂Q (x1 , x2 , x3 ), ∂x3

∂P (x1 , x2 , x3 ) = −2x2 ∂x3 =

∂R (x1 , x2 , x3 ), ∂x1

∂Q (x1 , x2 , x3 ) = −2x3 ∂x1 =

∂P (x1 , x2 , x3 ), ∂x2

(x1 , x2 , x3 ) ∈ R3 .

Therefore there exists a function .u ∈ C 1 (R3 ) such that du(x1 , x2 , x3 ) = P (x1 , x2 , x3 )dx1 + Q(x1 , x2 , x3 )dx2

.

+ R(x1 , x2 , x3 )dx3 ,

(x1 , x2 , x3 ) ∈ R3 .

From here, for fixed .(x1 , x2 , x3 ) ∈ R3 , we have

.

u(x1 , x2 , x3 ) = =

(x1 ,x 2 ,x3 ) (ξ1 ,ξ2 ,ξ3 ) x14 + x24

(4τ13 −2τ2 τ3 )dτ1 +(4τ23 −2τ1 τ3 )dτ2 +(4τ33 −2τ1 τ2 )dτ3 + x34 − 2x1 x2 x3 − ξ14 − ξ24 − ξ34 + 2ξ1 ξ2 ξ3

34

2 First Order Partial Differential Equations

for any .(ξ1 , ξ2 , ξ3 ) ∈ R3 . Consequently the general solution of the considered Pfaff equation is given by x14 + x24 + x34 − 2x1 x2 x3 − ξ14 − ξ24 − ξ34 + 2ξ1 ξ2 ξ3 = c

.

for any .(ξ1 , ξ2 , ξ3 ) ∈ R3 , where c is a constant. Exercise 2.11 Find the general solution of the following Pfaff equation x2 x3 dx1 + x1 x3 dx2 + x1 x2 dx3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 .

In the general case, if there is not a function .u ∈ C 1 (G) so that the equality (2.14) holds, then we multiply Eq. (2.11) with a function .μ = μ(x1 , x2 , x3 ) so that for the equation μ(x1 , x2 , x3 )P (x1 , x2 , x3 )dx1 + μ(x1 , x2 , x3 )Q(x1 , x2 , x3 )dx2

.

+ μ(x1 , x2 , x3 )R(x1 , x2 , x3 )dx3 = 0 to exists a function .u ∈ C 1 (G) such that .

du(x1 , x2 , x3 ) = μ(x1 , x2 , x3 )P (x1 , x2 , x3 )dx1 +μ(x1 , x2 , x3 )Q(x1 , x2 , x3 )dx2 + μ(x1 , x2 , x3 )R(x1 , x2 , x3 )dx3 .

Definition 2.14 The function .μ will be called integrating factor for the Pfaff equation. The integrating factor .μ satisfies the system .

∂(μP ) ∂(μR) = ∂x3 ∂x1 ∂(μP ) ∂(μQ) = ∂x1 ∂x2

(2.15)

∂(μQ) ∂(μR) = , ∂x3 ∂x2 and conversely, if a function .μ ∈ C 1 (G) satisfies the system (2.15), then it is an integrating factor for the Pfaff equation (2.11). Example 2.22 We consider the equation x2 x3 dx1 + x1 x3 dx2 + x1 x2 x3 dx3 = 0,

.

We multiply it with the function

(x1 , x2 , x3 ) ∈ R3 .

2.4 The Pfaff Equation

35

μ = μ(x1 , x2 , x3 ) =

.

1 , x1 x2 x3

(x1 , x2 , x3 ) ∈ R3 .

Exercise 2.12 Prove that .μ is an integrating factor for the considered equation. We obtain the equation .

1 1 dx1 + dx2 + dx3 = 0. x2 x1

Then there exists a function .u ∈ C 1 (R3 ) such that du(x1 , x2 , x3 ) =

.

1 1 dx1 + dx2 + dx3 . x1 x2

Consequently (x1 ,x2 ,x3 )

1 1 dτ1 + dτ2 + dτ3 τ1 τ2 (ξ1 ,ξ2,ξ3 ) x   2  x1  = log   + log   + x3 − ξ3 for any ξ1 ξ2

u(x1 , x2 , x3 ) = .

(ξ1 , ξ2 , ξ3 ) ∈ R3 .

Therefore the general solution is x  x   2  1 log   + log   + x3 − ξ3 = c ξ2 ξ1

.

for any

(ξ1 , ξ2 , ξ3 ) ∈ R3 ,

where c is a constant. Now, we will find an one dimensional integral for Eq. (2.11) no supposing (2.13). Let .x3 = x3 (x1 , x2 ) be an arbitrary surface .S. If an integral curve lies on .S, then dx3 =

.

∂x3 ∂x3 (x1 , x2 )dx1 + (x1 , x2 )dx2 . ∂x1 ∂x2

Equation (2.11) takes the form

.

 ∂x3 P (x1 , x2 , x3 (x1 , x2 )) + R(x1 , x2 , x3 (x1 , x2 )) (x1 , x2 ) dx1 ∂x1 

3 + Q(x1 , x2 , x3 (x1 , x2 )) + R(x1 , x2 , x3 (x1 , x2 )) ∂x (x , x ) dx2 = 0. 1 2 ∂x2 (2.16)

If .φ(x1 , x2 ) = 0 is an integral of the last equation, then .

is an integral for (2.11).

φ(x1 , x2 ) = 0 = x3 (x1 , x2 ) x3

36

2 First Order Partial Differential Equations

Example 2.23 Let us consider the equation 2x1 dx1 − x3 dx2 + x3 x1 dx3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 ,

for x3 (x1 , x2 ) = 1,

.

(x1 , x2 ) ∈ R2 .

Then x12 − x2 = c,

.

(x1 , x2 ) ∈ R2 ,

where c is a real constant. Consequently the integral curves of the considered equation are .

x12 − x2 = c x3 (x1 , x2 ) = 1.

If we search an integral surface for Eq. (2.11) through the point .(ξ, η, ζ ) in G, then we take .x3 = ζ and Eq. (2.16) takes the form P (x1 , x2 , ζ )dx1 + Q(x1 , x2 , ζ )dx2 = 0.

.

(2.17)

If .x2 = x2 (x1 ) is a solution to Eq. (2.17) for which .x2 (ξ ) = η, then we construct an integral curve of Eq. (2.11) lying in .x1 = τ and passing through the point .(τ, x2 (τ ), ζ ). Example 2.24 Now we consider the equation .

dx1 dx3 dx2 + = 0, + x1 x3 x1 x2 x2 x3

(x1 , x2 , x3 ) ∈ R3 .

We will find an integral surface through the point .(1, 1, 1). We set .x3 = 1. Then we obtain the equation .

dx1 dx2 = 0, + x1 x2

(x1 , x2 ) ∈ R2 ,

and we consider the problem .

which solution is given by

x1 dx1 + x2 dx2 = 0 = 1, x2 (1)

2.5 Some Special Systems

37

x12 + x22 = 2,

.

(x1 , x2 ) ∈ R2 .

In the plane .x1 = τ we have .

dx2 dx3 = 0, + τ x2 τ x3

(x2 , x3 ) ∈ R2 ,

whereupon x22 + x32 = 3 − τ 2 ,

.

(x2 , x3 ) ∈ R2 .

Consequently

.

x1 =τ = 3 − τ2 x22 + x32 2 2 2 x1 + x2 + x3 = 3, (x1 , x2 , x3 ) ∈ R3 .

Exercise 2.13 Find the general solution of the following equations 1. .(x1 − x2 )dx1 + x3 dx2 − x1 dx3 = 0, 2. .3x2 x3 dx1 + 2x1 x3 dx2 + x1 x2 dx3 = 0.

2.5 Some Special Systems In this section, we will consider the system ux1 = F1 (∇u, u, x) ux2 = F2 (∇u, u, x) . .. . uxn = Fn (∇u, u, x),

(2.18) (x1 , . . . , xn ) ∈ U,

where .u : U → R is unknown, .Fi : Rn × R × U → R, .i ∈ {1, . . . , n}, are given smooth functions, .U ⊂ Rn . We will give a way for finding of solutions of (2.18). For convenience, we will consider the case .n = 2. More precisely, we consider the system .

ux1 = F1 (ux1 , ux2 , u, x1 , x2 ) ux2 = F2 (ux1 , ux2 , u, x1 , x2 ),

(x1 , x2 ) ∈ U.

(2.19)

We fix the variable .x1 and consider the second equation of (2.19) as an ODE with respect to .x2 . Let

38

2 First Order Partial Differential Equations

φ(x2 ) = f (x1 )

(2.20)

.

be an integral of the second equation of (2.19). Here f is an arbitrary continuously differentiable function which plays the role of a constant of integration. To precise f , we put (2.20) in the first equation of the system (2.19). Example 2.25 Consider the system .

ux1 =

u x1

ux2 =

2u , x2

(x1 , x2 ) ∈ R2 .

We fix .x2 ∈ R and consider the equation ux1 =

.

u , x1

x1 ∈ R,

as an ODE with respect to .x1 . We get 



dx1 + f1 (x2 ), x1

x1 ∈ R,

log |u| = log |x1 | + f1 (x2 ),

x1 ∈ R,

du = u

.

i.e., .

i.e., u(x1 , x2 ) = f (x2 )x1 ,

.

x1 ∈ R.

To find the function f , we use the equation ux2 =

.

2u , x2

x2 ∈ R,

whereupon f ' (x2 )x1 = 2

.

f (x2 ) x1 , x2

x2 ∈ R,

i.e., .

f ' (x2 ) = 2

f (x2 ) , x2

x2 ∈ R,

2.6 Advanced Practical Problems

39

or f (x2 ) = cx22 ,

.

x2 ∈ R,

where c is a real constant. Therefore u(x1 , x2 ) = cx1 x22 ,

.

(x1 , x2 ) ∈ R2 ,

is a solution of the considered system. Exercise 2.14 Find a solution of the system .

ux1 = x2 − u ux2 = x1 u, (x1 , x2 ) ∈ R2 .

2.6 Advanced Practical Problems Problem 2.1 Classify each of the following equations as quasilinear or semilinear. 1. x23 ux1 + u3 ux2 − x1 x2 x3 x4 uux4 = u2 .  2. x12 + x1 x2 + x22 ux1 − ux2 − ux3 + u4 ux4 = 1 + u + u2 . 3.

1 u 1+x22 x1

+

2 3+cos x1 ux2

√

= cos u + 2.

1 4. (2 − u)ux1 + 1 + u4 ux2 − ux3 = 1+u 2. 2 2 2 5. (x1 + x2 + x3 )ux1 − ux2 − 4x1 x2 ux3 = eu + x12 + x32 .

Problem 2.2 Classify each of the following equations as linear homogeneous or linear nonhomogeneous. 1. ux1 + ux2 + 3x3 ux3 = 0. 2. −ux1 + 4 2 ux2 = x1 + x2 . 1+x2

3. ux1 − 2 sin x2 ux2 + ux3 = 0. 4. x1 + ux1 − 2ux2 + ux3 = 0. 5. ux1 + cos x1 ux2 − sin x3 ux3 = cos(x1 + x2 + x3 ). Problem 2.3 Classify each of the following equations as linear or nonlinear. 1. 2. 3. 4. 5.

ux1 + (x12 + x22 )ux2 = cos(x1 + x2 ). −ux1 + 3x1 ux2 − ux3 = x1 . ux1 − ux2 + ux3 = cos x3 u3 . ux1 ux2 ux3 = 1. ux1 + (ux2 )2 + cos x3 ux3 = 0.

40

2 First Order Partial Differential Equations

Problem 2.4 Prove that the function x2

u(x1 , x2 ) = x1 x2 + e x1 ,

.

(x1 , x2 ) ∈ R2 ,

x1 /= 0,

(x1 , x2 ) ∈ R2 ,

x1 /= 0.

is a solution to the equation x1 ux1 + x2 ux2 = 2x1 x2 ,

.

Problem 2.5 Prove that the function u(x1 , x2 ) = 

.

x1 x12

,

+ x22

(x1 , x2 ) ∈ R2 \{(0, 0)},

is a solution to the equation x1 ux1 + x2 ux2 = 0,

(x1 , x2 ) ∈ R2 \{(0, 0)}.

.

Problem 2.6 Prove that the function u(x1 , x2 , x3 ) = x1 +

.

x1 − x2 , x2 − x3

(x1 , x2 , x3 ) ∈ R3 ,

x2 /= x3 ,

is a solution to the equation ux1 + ux2 + ux3 = 1,

.

(x1 , x2 , x3 ) ∈ R3 ,

x2 /= x3 .

Problem 2.7 Prove that the function u = u(x1 , x2 ), defined by ax1 + bx2 + cu(x1 , x2 ) = f (x12 + x22 + (u(x1 , x2 ))2 ),

.

(x1 , x2 ) ∈ R2 ,

where a, b, c ∈ R, f ∈ C 1 (R), is a solution to the equation (cx2 − bu)ux1 + (au − cx1 )ux2 = bx1 − ax2 ,

.

(x1 , x2 ) ∈ R2 .

Problem 2.8 Find a solution u = u(x1 , x2 ), (x1 , x2 ) ∈ R2 , to the equation 1. x1 ux1 − 2u = 2x14 ,

.

(x1 , x2 ) ∈ R2 .

2. (2x2 + 1)ux2 = 4x2 + 2u,

.

(x1 , x2 ) ∈ R2 .

3.  .

 x1 u + ex1 dx1 − x1 du = 0,

(x1 , x2 ) ∈ R2 .

2.6 Advanced Practical Problems

41

4. x22 ux2 + x2 u + 1 = 0,

(x1 , x2 ) ∈ R2 .

.

5. u = x1 (ux1 − x1 cos x1 ),

(x1 , x2 ) ∈ R2 .

.

6. 2x1 (x12 + u)dx1 = du,

(x1 , x2 ) ∈ R2 .

.

7. (x1 ux1 − 1) log x1 = 2u,

.

(x1 , x2 ) ∈ R2 ,

x1 > 0.

8. x1 ux1 + (x1 + 1)u = 3x12 e−x1 ,

.

(x1 , x2 ) ∈ R2 .

9. (x2 + u2 )du = udx2 ,

(x1 , x2 ) ∈ R2 .

.

10. (2eu − x1 )ux1 = 1,

(x1 , x2 ) ∈ R2 .

ux1 + 2u = u2 ex1 ,

(x1 , x2 ) ∈ R2 .

.

11. .

12. (x1 + 1)(ux1 + u2 ) = −u,

.

(x1 , x2 ) ∈ R2 .

13. x1 u2 ux1 = x12 + u3 ,

.

(x1 , x2 ) ∈ R2 .

14. x1 udu = (u2 + x1 )dx1 ,

.

(x1 , x2 ) ∈ R2 .

42

2 First Order Partial Differential Equations

15. x2 ux2 + 2u + x25 u3 ex2 = 0,

.

(x1 , x2 ) ∈ R2 .

Problem 2.9 Find a solution to the following equations 1. x1 ux1 + x2 ux2 + x3 ux3 = 0.

.

2. x2 uux1 + x1 uux2 = x1 x2 .

.

3. x2 ux1 + x1 ux2 = x1 − x2 .

.

4. ex1 ux1 + x22 ux2 = x2 ex1 .

.

5. 2x1 ux1 + (x2 − x1 )ux2 − x12 = 0.

.

6. x1 x2 ux1 − x12 ux2 = x2 u.

.

7. x1 ux1 + 2x2 ux2 = x12 x2 + u.

.

8. (x12 + x22 )ux1 + 2x1 x2 ux2 + u2 = 0.

.

9.  2x24 ux1 − x1 x2 ux2 = x1 u2 + 1.

.

10. x12 uux1 + x22 uux2 = x1 + x2 .

.

2.6 Advanced Practical Problems

43

11. x2 uux1 − x1 uux2 = eu .

.

12. (u − x2 )2 ux1 + x1 uux2 = x1 x2 .

.

13. x1 x2 ux1 + (x1 − 2u)ux2 = x2 u.

.

14. x2 ux1 + uux2 =

.

x2 . x1

15. (x1 + u)ux1 + (x2 + u)ux2 = x1 + x2 .

.

16. (x1 u + x2 )ux1 + (x1 + x2 u)ux2 = 1 − u2 .

.

17. (x2 + x3 )ux1 + (x1 + x3 )ux2 + (x1 + x2 )ux3 = u.

.

18. x1 ux1 + x2 ux2 + (x3 + u)ux3 = x1 x2 .

.

19. (u − x1 )ux1 + (u − x2 )ux2 − x3 ux3 = x1 + x2 .

.

20. x1 uux1 + x2 uux2 = −x1 x2 .

.

44

2 First Order Partial Differential Equations

Problem 2.10 Solve the following problems 1. .

x1 ux1 − x2 ux2 = u − x12 − x22 u(x1 , −2) = x1 − x12 .

2. x1 ux1 − x2 ux2 = 0,

(x1 , x2 ) ∈ R2 ,

.

u(x1 , 1) = 2x1 ,

x1 ∈ R.

3.   ux1 + 2ex1 − x2 ux2 = 0,

(x1 , x2 ) ∈ R2 ,

.

u(0, x2 ) = x2 ,

x2 ∈ R.

4. √ 2 x1 ux1 − x2 ux2 = 0,

.

u(1, x2 ) =

(x1 , x2 ) ∈ R2 ,

x22 ,

x1 > 0,

x2 ∈ R.

5. ux1 + ux2 + 2ux3 = 0,

(x1 , x2 , x3 ) ∈ R3 ,

.

u(1, x2 , x3 ) = x2 x3 ,

(x2 , x3 ) ∈ R2 .

6. x1 ux1 + x2 ux2 + x1 x2 ux3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 ,

u(x1 , x2 , 0) = x12 + x22 ,

(x1 , x2 ) ∈ R2 .

7. x22 ux1 + x1 x2 ux2 = x1 ,

.

u(0, x2 ) = x22 ,

(x1 , x2 ) ∈ R2 , x2 ∈ R.

8. x1 ux1 − 2x2 ux2 = x12 + x22 ,

.

u(x1 , 1) = x12 ,

(x1 , x2 ) ∈ R2 ,

x1 ∈ R.

2.6 Advanced Practical Problems

45

9. x1 ux1 + x2 ux2 = u − x1 x2 ,

(x1 , x2 ) ∈ R2 ,

.

u(2, x2 ) = x22 + 1,

x2 ∈ R.

10. .

π , (x1 , x2 ) ∈ R2 , x1 ∈ 0, 2

π . u(x1 , x1 ) = x13 , x1 ∈ 0, 2

tan x1 ux1 + x2 ux2 = u,

Problem 2.11 Find the general solution of the Pfaff equation (x1 + x2 + x3 )(dx1 + dx2 + dx3 ) = 0.

.

Problem 2.12 Find the general solution of the following equations 1. (x3 + x1 x2 )dx1 − (x3 + x22 )dx2 + x2 dx3 = 0, 2. (2x2 x3 + 3x1 )dx1 + x1 x3 dx2 + x1 x2 dx3 = 0. Problem 2.13 Find a solution of the system .

ux1 = 2x2 u − u2 ux2 = x1 u.

Chapter 3

Classifications of Second Order Partial Differential Equations

Partial differential equations of second and higher order are encountered in a number of areas of importance to engineering and mathematical physics. As the process that are accounted for in the development of the governing differential equations increase in complexity, the differential equations themselves tend to acquire a higher order. The theory of second order partial differential equations has found extensive applications in the study of problems in fluid mechanics, flow in porous media, heat conduction in solids, diffusive transport of chemicals in porous media, wave propagation in strings and membranes, and in mechanics of solids. In this chapter, we shall restrict our attention to a classification of second order partial differential equations.

3.1 Classifications Let .U ⊂ Rn . Definition 3.1 By a second order partial differential equation in .n variables .x1 , .. . . , xn , we mean any equation of the form

.

F (x1 , . . . , xn , u, ux1 , . . . , uxn , ux1 x1 , ux1 x2 , . . . , uxn xn ) = 0,

.

(x1 , . . . , xn ) ∈ Rn . (3.1)

Example 3.1 The nonlinear Poisson equation −

n 

.

uxi xi = f (u),

(x1 , . . . , xn ) ∈ Rn ,

i=1

is a second order PDE. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. G. Georgiev, An Excursion Through Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-031-48784-2_3

47

48

3 Classifications of Second Order Partial Differential Equations

Example 3.2 The porous medium equation uxn −

.

n−1   γ u x x = 0, i i

(x1 , . . . , xn ) ∈ Rn ,

i=1

where .γ > 0 is a constant, is a second order PDE. Definition 3.2 If Eq. (3.1) can be written in the form n  .

aij (x1 , . . . , xn , u, ux1 , . . . , uxn )uxi xj +

i,j =1

n 

bi (x1 , . . . , xn , u)uxi

i=1

= f (x1 , . . . , xn , u, ux1 , . . . , uxn ),

(x1 , . . . , xn ) ∈ U,

then we say that the equation is quasilinear. Example 3.3 The equation ux2 ux1 x1 − ux1 x2 u2 = x12 u2 ,

.

(x1 , x2 ) ∈ R2 ,

is a quasilinear second order PDE. Example 3.4 The equation  .

x12 + x22 + x32 + u2x1 ux1 x2 + ux2 x2 + ux3 x3 = u2 ,

(x1 , x2 , x3 ) ∈ R3 ,

is a quasilinear second order PDE. Example 3.5 The equation  .

u2x1 + u2x2 ux1 x2 − ux2 x2 = u3 ,

(x1 , x2 ) ∈ R2 ,

is a quasilinear second order PDE. Definition 3.3 If Eq. (3.1) can be written in the form n  .

aij (x1 , . . . , xn )uxi xj +

i,j =1

n 

bi (x1 , . . . , xn )uxi =f (x1 , . . . , xn , u, ux1 , . . . , uxn ),

i=1

(x1 , . . . , xn ) ∈ U , then we say that the equation is semilinear.

.

Example 3.6 The equation ux1 x1 − ux1 x2 + ux3 x3 = x12 u4 ,

.

is a semilinear second order PDE.

(x1 , x2 , x3 ) ∈ R3

3.1 Classifications

49

Example 3.7 The equation  .

1 + x12 u2 ux1 + ux1 x2 + ux2 x2 = u5 + 1,

(x1 , x2 ) ∈ R2 ,

is a quasilinear second order PDE. Example 3.8 The equation ux1 + ux2 + (u4 + 1)ux1 x2 + u2 ux2 x2 = sin u,

.

(x1 , x2 ) ∈ R2 ,

is a quasilinear second order PDE. Exercise 3.1 Classify each of the following equations as quasilinear or semilinear. 1.  u2x1 + u6x2 + u4x3 ux1 x3 + ux2 x3 + ux3 x3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 .

2.  .

1 + x12 + x22 + x32 + x42 ux1 ux1 x1 + ux2 ux2 x3 + ux3 x4 + ux4 x4 = 0,

(x1 , x2 , x3 , x4 ) ∈ R4 .

3. (ux1 + ux2 )3 ux1 x1 − ux2 x2 =

.



1 + x12 + x22 ,

(x1 , x2 ) ∈ R2 .

4.  1 + u4 ux1 + x1 x2 x3 ux1 x1 + u2 ux1 x2 + ux3 x3 = sin u,

.

(x1 , x2 , x3 ) ∈ R3 .

5. (1 + x1 )ux1 x1 − ux1 + ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

Definition 3.4 If Eq. (3.1) can be written in the form n  .

i,j =1

aij (x1 , . . . , xn )uxi xj +

n 

bi (x1 , . . . , xn )uxi +c(x1 , . . . , xn )u=f (x1 , . . . , xn ),

i=1

(x1 , . . . , xn ) ∈ U , then we say that the equation is linear. Moreover, if f (x1 , . . . , xn ) = 0, .(x1 , . . . , xn ) ∈ U then the equation is said to be linear

. .

50

3 Classifications of Second Order Partial Differential Equations

homogeneous second order PDE. Otherwise, the equation is said to be linear nonhomogeneous second order PDE. Example 3.9 The Laplace equation n  .

uxi xi = 0,

(x1 , . . . , xn ) ∈ Rn ,

i=1

is a linear homogeneous second order PDE. Example 3.10 The heat(diffusion) equation ux1 −

n 

.

uxi xi = 0,

(x1 , . . . , xn ) ∈ Rn ,

i=2

is a linear homogeneous second order PDE. Example 3.11 The wave equation ux1 x1 −

n 

.

uxi xi = 0,

(x1 , . . . , xn ) ∈ Rn ,

i=2

is a linear homogeneous second order PDE. Example 3.12 The Kolmogorov equation ux1 −

n 

.

aij uxi xj +

i,j =2

n 

bi uxi = x12 + · · · + xn2 ,

(x1 , . . . , xn ) ∈ Rn ,

i=2

where .aij , .bi , .1 ≤ i, j ≤ n, are constants, is a linear nonhomogeneous second order PDE. Exercise 3.2 Classify each of the following equations as linear homogeneous or linear nonhomogeneous. 1. x1 ux1 x1 − ux2 = 0,

.

(x1 , x2 ) ∈ R2 .

2. ux1 x1 − x12 ux2 x2 = x1 + x2 ,

.

(x1 , x2 ) ∈ R2 .

3. ux1 x1 − sin x1 ux2 x2 + ux3 + ux3 x3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 .

3.1 Classifications

51

4. ux1 x1 − ux2 x2 + ux3 x3 − ux4 x4 = 0,

.

(x1 , x2 , x3 , x4 ) ∈ R4 .

5. ux1 x1 +

.

 x12 + x22 ux2 x2 = x1 ,

(x1 , x2 ) ∈ R2 .

Definition 3.5 A second order PDE that is not linear is said to be nonlinear. Example 3.13 The equation u2 u3x1 + u2 ux1 x2 + uux2 x2 = u3 − ux1 ,

.

(x1 , x2 ) ∈ R2 ,

is a nonlinear second order PDE. Example 3.14 The equation .

sin(ux1 x1 + ux2 x2 ) − cos(ux3 x3 ) = 0,

(x1 , x2 , x3 ) ∈ R3 ,

is a nonlinear second order PDE. Example 3.15 The equation ux1 x1 + ux2 x2 + cos(ux2 x2 ) = x1 + x2 ,

.

(x1 , x2 ) ∈ R2 ,

is a nonlinear second order PDE. Exercise 3.3 Classify each of the following equations as linear or nonlinear. 1. ux1 x1 + ux2 x2 = u2 + cos(ux1 ) − sin(ux2 ),

.

(x1 , x2 ) ∈ R2 .

2. (ux1 x1 − ux2 )2 − ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

3. x2 ux1 x1 + x1 ux2 x2 = (x1 + x2 )u,

.

(x1 , x2 ) ∈ R2 .

4. ux1 x1 − ux2 x2 + cos x2 ux2 x2 = x12 − x3 ,

.

(x1 , x2 , x3 ) ∈ R3 .

52

3 Classifications of Second Order Partial Differential Equations

5.  .

ux2 + (x1 + x2 )ux3

2

− ux1 x1 = 0,

(x1 , x2 , x3 ) ∈ R3 .

3.2 Advanced Practical Problems Problem 3.1 Classify each of the following equations as quasilinear and semilinear. 1.  .

u2x1 + u2x2 u2 ux1 x2 − x12 ux2 x2 = 0,

(x1 , x2 ) ∈ R2 .

2. ux1 ux2 ux3 ux1 x1 + ux2 x2 + ux3 x3 = 0,

(x1 , x2 , x3 ) ∈ R3 .

.

3. x1 ux1 + (x12 + 1)ux1 x1 − uux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

4. ux1 x1 − uux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

5. ux1 x1 + ux2 x3 + uux3 x3 = x1 + x2 + x3 ,

.

(x1 , x2 , x3 ) ∈ R3 .

Problem 3.2 Classify each of the following equations as linear homogeneous or linear nonhomogeneous. 1. ux1 x1 − ux2 x2 − ux3 x3 + ux3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 .

2. ux1 − ux1 x2 + ux2 x2 = x2 ,

.

(x1 , x2 ) ∈ R2 .

3.2 Advanced Practical Problems

53

3. ux1 x3 + ux2 x3 + sin x2 ux3 = 0,

(x1 , x2 , x3 ) ∈ R3 .

.

4. ux1 − ux1 x1 + ux2 x2 − sin(x1 + x2 ) = 0,

(x1 , x2 ) ∈ R2 .

.

5. ux1 x1 − ux2 + ux2 x2 = u − x1 + x2 ,

.

(x1 , x2 ) ∈ R2 .

Problem 3.3 Classify each of the following equations as linear or nonlinear. 1. ux1 x1 − ux1 x2 − ux2 x3 − ux3 x4 = u,

(x1 , x2 , x3 , x4 ) ∈ R4 .

.

2. ux1 x1 ux2 x2 − (ux3 x3 )2 = u,

.

(x1 , x2 , x3 ) ∈ R3 .

3. ux1 x1 + ux1 x2 + cos(ux2 ) = 0,

.

(x1 , x2 ) ∈ R2 .

4. ux1 x1 − ux2 x2 + (1 − ux1 x1 ux2 x2 )4 = sin x1 ,

.

(x1 , x2 ) ∈ R2 .

5. ux1 x1 +(x1 +x2 +x3 )ux2 x2 +(cos x1 − cos x3 )ux3 x3 =x1 +x2 +x3 , (x1 , x2 , x3 ) ∈ R3 .

.

Chapter 4

Classifications and Canonical Forms for Linear Second Order Partial Differential Equations

The objectives of a classification system of differential equations are basically threefold. Firstly, to identify the general character of the differential equation which may not be self evident from the representation provided. Secondly, to identify a method of solution of the partial differential equation which is specific to the category of equation under consideration. Finally, to establish a priori the particular attributes of the solution by appeal to expected behaviour derived from physical or mathematical considerations of a general nature. Partial differential equations can represent themselves in a variety of forms. In some cases the category of the partial differential equation is easily recognizable by a comparison with the standard forms. In other cases, the category is not evident. In developing either a method of solution, or assessing the well posed nature of the problem or establishing a priori the general nature of the solution it is desirable to find the canonical form of the partial differential equation. This is important when we develop numerical schemes for the solution of the partial differential equation where the method of solution will depend on the character of the partial differential equation.

4.1 Classifications and Canonical Forms for Linear Second Order Partial Differential Equations in Two Independent Variables Let .U be a domain in .R2 . Definition 4.1 A linear differential operator of second order for the function u = u(x1 , x2 ) is given by

.

L(u) = aux1 x1 + 2bux1 x2 + cux2 x2 ,

.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. G. Georgiev, An Excursion Through Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-031-48784-2_4

55

56

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

where the coefficients .a, b and .c are supposed to be continuously differentiable and not simultaneously vanishing functions of .x1 and .x2 in the domain .U. We consider the differential operator ˜ L(u) = L(u) + g(x1 , x2 , u, ux1 , ux2 ) = L(u) + · · · ,

.

(4.1)

where .g is not necessarily linear and does not contain second derivatives. ˜ Definition 4.2 The operator .L is called the principal part of the operator .L. Our aim is to transform the operator (4.1) or the corresponding equation L(u) + · · · = 0

.

into a simple form, called the canonical form, by introducing new independent variables. Let .ξ1 and .ξ2 be new independent variables which are connected with .x1 and .x2 in the following way .

ξ1 = φ1 (x1 , x2 ) ξ2 = φ2 (x1 , x2 ),

where .φ1 , φ2 ∈ C 2 (U ). We will denote with .u(ξ1 , ξ2 ) the transformed function .u(x1 , x2 ) into the variables .ξ1 and .ξ2 . We have the following relations. ux1 = uξ1 ξ1x1 + uξ2 ξ2x1 ,

.

ux2 = uξ1 ξ1x2 + uξ2 ξ2x2 , ) ( ux1 x1 = uξ1 ξ1x1 + uξ2 ξ2x1 x 1 ( ( ) ) = uξ1 ξ1x1 x + uξ2 ξ2x1 x 1 1 ( ) ( ) = uξ1 x ξ1x1 + uξ1 ξ1x1 x1 + uξ2 x ξ2x1 + uξ2 ξ2x1 x1 1 1 ) ( = uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1 ξ1x1 + uξ1 ξ1x1 x1 ) ( + uξ1 ξ2 ξ1x1 + uξ2 ξ2 ξ2x1 ξ2x1 + uξ2 ξ2x1 x1 ( )2 = uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ1x1 ξ2x1 + uξ1 ξ1x1 x1 ( )2 +uξ1 ξ2 ξ1x1 ξ2x1 + uξ2 ξ2 ξ2x1 + uξ2 ξ2x1 x1 ( )2 ( )2 = uξ1 ξ1 ξ1x1 + 2uξ1 ξ2 ξ1x1 ξ2x1 + uξ2 ξ2 ξ2x1 +uξ1 ξ1x1 x1 + uξ2 ξ2x1 x1 2 2 = uξ1 ξ1 φ1x + 2uξ1 ξ2 φ1x1 φ2x1 + uξ2 ξ2 φ2x 1 1

+uξ1 φ1x1 x1 + uξ2 φ2x1 x1 ,

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

) ( ux2 x2 = uξ1 ξ1x2 + uξ2 ξ2x2 x 2 ) ) ( ( = uξ1 ξ1x2 x + uξ2 ξ2x2 x 2 2 ( ) ( ) = uξ1 x ξ1x2 + uξ1 ξ1x2 x2 + uξ2 x ξ2x2 + uξ2 ξ2x2 x2 2 2 ) ( = uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 ξ1x2 + uξ1 ξ1x2 x2 ) ( + uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 ξ2x2 + uξ2 ξ2x2 x2 ( )2 = uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ1x2 ξ2x2 + uξ1 ξ1x2 x2 ( )2 +uξ1 ξ2 ξ1x2 ξ2x2 + uξ2 ξ2 ξ2x2 + uξ2 ξ2x2 x2 ( )2 ( )2 = uξ1 ξ1 ξ1x2 + 2uξ1 ξ2 ξ1x2 ξ2x2 + uξ2 ξ2 ξ2x2 +uξ1 ξ1x2 x2 + uξ2 ξ2x2 x2 2 2 = uξ1 ξ1 φ1x + 2uξ1 ξ2 φ1x2 φ2x2 + uξ2 ξ2 φ2x 2 2

ux1 x2 = = = =

+uξ1 φ1x2 x2 + uξ2 φ2x2 x2 , ) ( uξ1 ξ1x1 + uξ2 ξ2x1 x 2 ) ) ( ( uξ1 ξ1x1 x + uξ2 ξ2x1 x 2 2 ( ) ( ) uξ1 x ξ1x1 + uξ1 ξ1x1 x2 + uξ2 x ξ2x1 + uξ2 ξ2x1 x2 2 2 ) ( uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 ξ1x1 + uξ1 ξ1x1 x2 ) ( + uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 ξ2x1 + uξ2 ξ2x1 x2

= uξ1 ξ1 ξ1x1 ξ1x2 + uξ1 ξ2 ξ1x1 ξ2x2 + uξ1 ξ1x1 x2 +uξ1 ξ2 ξ1x2 ξ2x1 + uξ2 ξ2 ξ2x1 ξ2x2 + uξ2 ξ2x1 x2 ( ) = uξ1 ξ1 φ1x1 φ1x2 + uξ1 ξ2 φ1x1 φ2x2 + φ1x2 φ2x1 +uξ2 ξ2 φ2x1 φ2x2 + uξ1 φ1x1 x2 + uξ2 φ2x1 x2 . From here, L(u) = aux1 x1 + 2bux1 x2 + cux2 x2 ⎧ 2 2 = a uξ1 ξ1 φ1x + 2uξ1 ξ2 φ1x1 φ2x1 + uξ2 ξ2 φ2x 1 1 ⎨ +uξ1 φ1x1 x1 + uξ2 φ2x1 x1 ⎧ ( ) +2b uξ1 ξ1 φ1x1 φ1x2 + uξ1 ξ2 φ1x1 φ2x2 + φ1x2 φ2x1 ⎨ +uξ2 ξ2 φ2x1 φ2x2 + uξ1 φ1x1 x2 + uξ2 φ2x1 x2

.

57

58

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

⎧ 2 2 +c uξ1 ξ1 φ1x + 2uξ1 ξ2 φ1x2 φ2x2 + uξ2 ξ2 φ2x 2 2 ⎨ +uξ1 φ1x2 x2 + uξ2 φ2x2 x2 ⎧ ⎨ 2 2 + 2bφ1x1 φ1x2 + cφ1x = uξ1 ξ1 aφ1x 1 2 ( ) ) ( +2uξ1 ξ2 aφ1x1 φ2x1 + b φ1x2 φ2x1 + φ1x1 φ2x2 + cφ1x2 φ2x2 ⎧ ⎨ 2 2 + 2bφ2x1 φ2x2 + cφ2x +uξ2 ξ2 aφ2x 1 2 ( ) +uξ1 aφ1x1 x1 + 2bφ1x1 x2 + cφ1x2 x2 ( ) +uξ2 aφ2x1 x1 + 2bφ2x1 x2 + cφ2x2 x2 . Let 2 2 α = aφ1x + 2bφ1x1 φ1x2 + cφ1x , 1 2 ) ( β = aφ1x1 φ2x1 + b φ1x2 φ2x1 + φ1x1 φ2x2 + cφ1x2 φ2x2 ,

.

2 2 + 2bφ2x1 φ2x2 + cφ2x , γ = aφ2x 1 2

α1 = aφ1x1 x1 + 2bφ1x1 x2 + cφ1x2 x2 , γ1 = aφ2x1 x1 + 2bφ2x1 x2 + cφ2x2 x2 . Then the differential operator .L assumes the form L(u) = αuξ1 ξ1 + 2βuξ1 ξ2 + γ uξ2 ξ2 + α1 uξ1 + γ1 uξ2 ,

.

which is called the canonical form of the operator .L. We set Λ(u) = αuξ1 ξ1 + 2βuξ1 ξ2 + γ uξ2 ξ2 .

.

Exercise 4.1 Prove that .a, b, c and .α, β, γ are related as follows )2 ( αγ − β 2 = (ac − b2 ) φ1x1 φ2x2 − φ1x2 φ2x1 .

.

Exercise 4.2 Prove that al 2 + 2blm + cm2 = αλ2 + 2βλμ + γ μ2 ,

.

where l = λφ1x1 + μφ1x2 ,

.

m = λφ2x1 + μφ2x2 .

(4.2)

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

59

We will impose two conditions on the transformed coefficients .α, .β and .γ so that to obtain a simple canonical form of .Λ(u). We consider the following cases. 1. .α = γ , .β = 0. 2. .β = γ = 0. 3. .α = −γ , .β = 0 or .α = γ = 0. The transformations .φ1 and .φ2 satisfy one of the above cases. This depends on the algebraic character of the characteristic quadratic form Q(l, m) = al 2 + 2blm + cm2 = αλ2 + 2βλμ + γ μ2 .

.

Geometrically speaking, this depends on the character of the quadratic curve in the l, m-plane, i.e., for fixed .x1 and .x2 such that .Q(l, m) = 1, this curve may be an ellipse, a parabola or a hyperbola. From here and (4.2), we get to the following definition.

.

Definition 4.3 At a point .(x1 , x2 ) the operator .L(u) will be called 1. elliptic if .a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 > 0. 2. parabolic if .a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 = 0. 3. hyperbolic if .a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 < 0. Example 4.1 Let L(u) = e−x1 ux1 x1 + 4e−

.

x1 +x2 2

ux1 x2 + e−x2 ux2 x2 + ux1 + ux2 ,

(x1 , x2 ) ∈ R2 .

Here a(x1 , x2 ) = e−x1 ,

.

b(x1 , x2 ) = 2e−

x1 +x2 2

c(x1 , x2 ) = e−x2 ,

,

(x1 , x2 ) ∈ R2 .

Then a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 = e−x1 −x2 − 4e−x1 −x2

.

= −3e−x1 −x2 < 0,

(x1 , x2 ) ∈ R2 .

Therefore the operator .L is hyperbolic. Example 4.2 Let L(u) = ux1 x1 + sin x1 cos x2 ux1 x2 + 4ux2 x2 ,

.

(x1 , x2 ) ∈ R2 .

60

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

Here a(x1 , x2 ) = 1,

.

b(x1 , x2 ) =

1 sin x1 cos x2 , 2

c(x1 , x2 ) = 4,

(x1 , x2 ) ∈ R2 .

Then 1 a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 = 4 − (sin x1 )2 (cos x2 )2 4

.

> 0,

(x1 , x2 ) ∈ R2 .

Hence, the operator .L is elliptic. Example 4.3 Let L(u) = x12 ux1 x1 + 2x1 x2 ux1 x2 + x22 ux2 x2 + 4ux1 − ux2 ,

.

(x1 , x2 ) ∈ R2 .

Here a(x1 , x2 ) = x12 ,

.

b(x1 , x2 ) = x1 x2 , c(x1 , x2 ) = x22 ,

(x1 , x2 ) ∈ R2 .

Then a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 = x12 x22 − x12 x22

.

= 0, Therefore the operator .L is parabolic. Exercise 4.3 Determine the operator .L. 1. 2. 3. 4. 5. 6. 7.

L(u) = 2ux1 x1 − 3ux1 x2 + 4ux2 x2 + ux1 − 2ux2 , L(u) = ux1 x1 − 2ux1 x2 + 3ux2 x2 + u, .L(u) = ux1 x1 + 4ux1 x2 + ux2 x2 , .L(u) = −2ux1 x1 + ux1 x2 + 3ux2 x2 , .L(u) = −3ux1 x1 + 2ux1 x2 − 4ux2 x2 − ux1 , .L(u) = ux1 x1 − ux1 x2 + ux2 x2 , .L(u) = 2ux1 x1 + 8ux1 x2 + 8ux2 x2 . . .

(x1 , x2 ) ∈ R2 .

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

61

Example 4.4 Let us consider the operator L(u) = x1 ux1 x1 − 2ux1 x2 + x2 ux2 x2 + ux1 ,

(x1 , x2 ) ∈ R2 .

.

Here a(x1 , x2 ) = x1 ,

.

b(x1 , x2 ) = −1, c(x1 , x2 ) = x2 ,

(x1 , x2 ) ∈ R2 .

Then a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 = x1 x2 − 1,

.

(x1 , x2 ) ∈ R2 .

Hence, 1. the operator .L is elliptic for .x1 x2 > 1. 2. the operator .L is parabolic for .x1 x2 = 1. 3. the operator .L is hyperbolic for .x1 x2 < 1. Example 4.5 Consider the Tricomi operator L(u) = ux1 x1 + x1 ux2 x2 ,

.

(x1 , x2 ) ∈ R2 .

Here a(x1 , x2 ) = 1,

.

b(x1 , x2 ) = 0, c(x1 , x2 ) = x1 ,

(x1 , x2 ) ∈ R2 .

Then a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 = x1 ,

.

(x1 , x2 ) ∈ R2 .

Hence, 1. the operator .L is elliptic for .x1 > 0. 2. the operator .L is parabolic for .x1 = 0. 3. the operator .L is hyperbolic for .x1 < 0. The corresponding canonical forms of the differential operator .L(u) are as follows.

62

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

1. .α = γ ,

β = 0. Λ(u) + · · · = α(uξ1 ξ1 + uξ2 ξ2 ) + · · · .

.

2. .β = γ = 0. Λ(u) + · · · = αuξ1 ξ1 + · · · .

.

3. .α = −γ ,

β = 0. Λ(u) + · · · = α(uξ1 ξ1 − uξ2 ξ2 ) + · · · .

.

When .α = γ = 0, we have Λ(u) + · · · = 2βuξ1 ξ2 + · · · .

.

The corresponding canonical forms of the considered differential equations are as follows. 1. .α = γ ,

β = 0. uξ1 ξ1 + uξ2 ξ2 + · · · = 0.

.

2. .β = γ = 0. uξ1 ξ1 + · · · = 0.

.

3. .α = −γ ,

β = 0. uξ1 ξ1 − uξ2 ξ2 + · · · = 0.

.

When .α = γ = 0, we have uξ1 ξ2 + · · · = 0.

.

For fixed .x1 , x2 , such a canonical form can be obtained by the linear transformation which takes .Q into the corresponding canonical form. If we assume that the operator .L is of the same type in every point of the domain .U, we will search functions .φ1 and .φ2 which will transform .L(u) into a canonical form at every point of .U. To be found such functions, it depends on whether certain first order systems of linear partial differential equations can be solved. Without loss of generality, we suppose that .a /= 0 everywhere in the domain .U.

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

63

4.1.1 The Elliptic Case We suppose that .L(u) is elliptic in .U. In this case, we have .ac − b2 > 0, also, .α = γ and .β = 0. Then the linear independent variables ξ1 = φ1 (x1 , x2 )

.

and

ξ2 = φ2 (x1 , x2 )

satisfy the system .

2 2 2 2 aφ1x + 2bφ1x1 φ1x2 + cφ1x = aφ2x + 2bφ2x1 φ2x2 + cφ2x 1 2 2 )1 ( aφ1x1 φ2x1 + b φ1x2 φ2x1 + φ1x1 φ2x2 + cφ1x2 φ2x2 = 0

or )( ) ( ) ( a φ1x1 − φ2x1 φ1x1 + φ2x1 + 2b φ1x1 φ1x2 − φ2x1 φ2x2 ( )( ) + c φ1x2 − φ2x2 φ1x2 + φ2x2 = 0 ) ( aφ1x1 φ2x1 + b φ1x2 φ2x1 + φ1x1 φ2x2 + cφ1x2 φ2x2 = 0.

.

(4.3)

Next, we rewrite the system (4.3) in the form ⎨ ⎨ ⎧ ⎧ ( ) 2 2 2 2 + 2b φ =0 a φ1x + (iφ ) φ − φ φ + c φ + (iφ ) 2x 1x 1x 2x 2x 2x 1 1 2 1 2 2 1x2 1 ) ( 2iaφ1x1 φ2x1 + 2ib φ1x2 φ2x1 + φ1x1 φ2x2 + 2icφ1x2 φ2x2 =0.

.

We add both equations and we arrive at the equation )2 )( ) ( )2 ( ( a φ1x1 + iφ2x1 + 2b φ1x1 + iφ2x1 φ1x2 + iφ2x2 + c φ1x2 + iφ2x2 = 0. (4.4)

.

Let .

φ3 = φ1x1 + iφ2x1 , φ4 = φ1x2 + iφ2x2 .

Then, using (4.4), we have aφ32 + 2bφ3 φ4 + cφ42 = 0

.

and ⎧ a

.

φ3 φ4

⎨2 + 2b

φ3 + c = 0. φ4

64

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

Consequently ⎧ .

φ3 φ4

⎨ 1,2

√ −b ± i ac − b2 = a

or ⎧

b .φ3 = − ±i a

⎨ √ ac − b2 φ4 , a

or φ1x1 + iφ2x1

.

⎧

√

⎧

√

b = − ±i a

ac − b2 a

⎨

) ( φ1x2 + iφ2x2 .

Let φ1x1 + iφ2x1

.

b = − +i a

ac − b2 a

⎨

) ( φ1x2 + iφ2x2 .

Then φ1x1 + iφ2x1

.

b = − φ1x2 + i a

√ √ ac − b2 ac − b2 b φ1x2 − iφ2x2 − φ2x2 a a a

and from here, √ ac−b2 b φ1x1 = − φ − φ2x2 a √a 1x2 . ac−b2 φ2x1 = φ1x2 − ab φ2x2 . a From the first equation of the last system, we get aφ1x + bφ1x2 φ2x2 = − √ 1 . ac − b2

.

Then, using its second equation, we find φ2x1

.

√ ⎨ ⎧ aφ1x1 + bφ1x2 ac − b2 b φ1x2 − − √ = a a ac − b2 √ ac − b2 abφ1x1 + b2 φ1x2 = φ1x2 + √ a a ac − b2

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

65

(ac − b2 + b2 )φ1x2 + abφ1x1 √ a ac − b2 acφ1x2 + abφ1 x1 = √ a ac − b2 bφ1x + cφ1x2 = √1 , ac − b2 =

i.e., we obtain the system φ2x1 = .

bφ1x1 +cφ1x2

√

ac−b2 aφ1x1 +bφ1x2

φ2x2 = − √

ac−b2

(4.5)

.

Definition 4.4 Equations (4.5) will be called Beltrami differential equations. From these Beltrami differential equations by eliminating one of the unknowns, for instance .φ2 , we get ∂ . ∂x1

⎧

aφ1x1 + bφ1x2 √ ac − b2

⎨

∂ + ∂x2

⎧

bφ1x1 + cφ1x2 √ ac − b2

⎨ = 0.

Using (4.5), we find aφ1x + bφ1x2 bφ1x1 + cφ1x2 φ1x2 + √ 1 φ1x1 √ 2 ac − b ac − b2 ⎧ ⎨ 1 2 2 = √ + 2bφ φ + cφ aφ1x 1x 1x 1 2 1x2 . 1 ac − b2

φ2x1 φ1x2 − φ1x1 φ2x2 =

.

If we assume that 2 2 aφ1x + 2bφ1x1 φ1x2 + cφ1x = 0, 1 2

.

then from the first equation of (4.3), we get 2 2 aφ2x + 2bφ2x1 φ2x2 + cφ2x = 0. 1 2

.

Therefore .

φ2x φ1x1 and . 1 are the roots of the quadratic equation φ1x2 φ2x2 aν 2 + 2bν + c = 0.

.

From here,

(4.6)

66

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

.

φ1x1 2b φ2x1 =− , + φ2x2 a φ1x2 φ1x1 φ2x1 c = . φ1x2 φ2x2 a

From the last relations and from the second equation of (4.3), we obtain ⎧ ⎨ φ1x1 φ1x2 φ1x1 φ2x1 .0 = a +b + +c φ2x1 φ2x2 φ1x2 φ2x2 ⎧ ⎨ 2b c +c = a +b − a a = 2c −

2b2 a

ac − b2 a / 0. =

=2

Consequently 2 2 aφ1x + 2bφ1x1 φ1x2 + cφ1x /= 0 1 2

.

2 2 aφ2x + 2bφ2x1 φ2x2 + cφ2x /= 0 1 2

and φ2x1 φ1x2 − φ1x1 φ2x2 /= 0.

.

(4.7)

In other words, the transformation of the differential equation to the canonical form ) ( α uξ1 ξ1 + uξ2 ξ2 + · · · = 0

.

in a neighbourhood of a point is given by any pair of functions satisfying (4.5) and having nonvanishing Jacobian (4.7). Such functions are determined once we have a solution of (4.6) with nonvanishing gradient. If .a, b, c ∈ C 2 (U ) such a solution always exists, at least locally, and hence the system .φ1 , φ2 may be introduced in a neighbourhood of any point. Definition 4.5 The curves .ξ1 = φ1 (x1 , x2 ) = const and .ξ2 = φ2 (x1 , x2 ) = const are called the characteristic curves of the linear elliptic differential operator .L(u). Let φ = φ1 + iφ2 .

.

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

67

Then φx1 = φ1x1 + iφ2x1 ,

.

φx2 = φ1x2 + iφ2x2 . Hence and (4.4), we get aφx21 + 2bφx1 φx2 + cφx22 = 0,

.

whereupon ⎧ a

.

φx 1 φx2

⎨2 + 2b

φx 1 +c =0 φx2

and √ φx1 −b ± i ac − b2 = . φx2 a or ⎨ ⎧ √ aφx1 + b ± i ac − b2 φx2 = 0.

.

Therefore .

dx2 dx1 = √ a b ± i ac − b2

or √ dx2 b ± i ac − b2 , = . dx1 a and ⎧

⎨2 √ b ± i ac − b2 (dx1 )2 .a(dx2 ) − 2bdx1 dx2 + c(dx1 ) = a a √ b ± i ac − b2 (dx1 )2 + c(dx1 )2 −2b a = 0. 2

2

(4.8)

68

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

Definition 4.6 The equation a(dx2 )2 − 2bdx1 dx2 + c(dx1 )2 = 0

.

is called the characteristic equation of the elliptic operator .L(u). Note that the general integrals of Eq. (4.8) are given by φ5 (x1 , x2 ) ± iφ6 (x1 , x2 ) = const,

.

where .φ5 and .φ6 are real valued functions. In the practice, for convenience, we very often take ξ1 = φ5 (x1 , x2 )

.

ξ2 = φ6 (x1 , x2 ). Example 4.6 Consider the equation x22 ux1 x1 +2x1 x2 ux1 x2 +2x12 ux2 x2 +x2 ux2 = 0,

.

(x1 , x2 ) ∈ R2 ,

x1 /= 0,

x2 /= 0.

Here a(x1 , x2 ) = x22 ,

.

b(x1 , x2 ) = x1 x2 , c(x1 , x2 ) = 2x12 ,

(x1 , x2 ) ∈ R2 ,

x1 /= 0,

x2 /= 0.

Then a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 = 2x12 x22 − x12 x22

.

= x12 x22 (x1 , x2 ) ∈ R2 ,

> 0,

x1 /= 0,

x2 /= 0.

Therefore the considered equation is an elliptic equation. The characteristic equation is x22 (dx2 )2 − 2x1 x2 dx1 dx2 + 2x12 (dx1 )2 = 0,

.

(x1 , x2 ) ∈ R2 ,

x1 /= 0,

x2 /= 0,

whereupon ⎧ x22

.

dx2 dx1

⎨2 − 2x1 x2

dx2 + 2x12 = 0, dx1

(x1 , x2 ) ∈ R2 ,

x1 /= 0,

x2 /= 0.

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

69

Hence, ⎧

⎨

dx2 dx1

.

= 1,2

x1 x2 ± ix1 x2 , x22

(x1 , x2 ) ∈ R2 ,

x1 /= 0,

x2 /= 0.

Consider dx2 x1 + ix1 = , dx1 x2

.

(x1 , x2 ) ∈ R2 ,

x1 /= 0,

x2 /= 0.

Then x2 dx2 = (x1 + ix1 )dx1 ,

(x1 , x2 ) ∈ R2 ,

.

x1 /= 0,

x2 /= 0,

whereupon x22 = x12 + ix12 + c,

(x1 , x2 ) ∈ R2 ,

.

x1 /= 0,

x2 /= 0.

Here c is a constant. We set ξ1 (x1 , x2 ) = x12 − x22 ,

.

ξ2 (x1 , x2 ) = x12 ,

(x1 , x2 ) ∈ R2 ,

x1 /= 0,

x2 /= 0.

x1 /= 0,

x2 /= 0,

Then ξ1x1 (x1 , x2 ) = 2x1 ,

.

ξ1x2 (x1 , x2 ) = −2x2 , ξ2x1 (x1 , x2 ) = 2x1 , ξ2x2 (x1 , x2 ) = 0,

(x1 , x2 ) ∈ R2 ,

and ux1 = uξ1 ξ1x1 + uξ2 ξ2x1

.

ux1 x1

= 2x1 uξ1 + 2x1 uξ2 , ( ) = 2uξ1 + 2x1 uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1 ( ) +2uξ2 + 2x1 uξ1 ξ2 ξ1x1 + uξ2 ξ2 ξ2x1 ( ) = 2uξ1 + 2uξ2 + 2x1 2x1 uξ1 ξ1 + 2x1 uξ1 ξ2 ( ) +2x1 2x1 uξ1 ξ2 + 2x1 uξ2 ξ2 = 2uξ1 + 2uξ2 + 4x12 uξ1 ξ1 + 8x12 uξ1 ξ2 + 4x12 uξ2 ξ2 ,

70

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

( ) ux1 x2 = 2x1 uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 ( ) +2x1 uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 ( ) ( ) = 2x1 −2x2 uξ1 ξ1 + 2x1 −2x2 uξ1 ξ2 = −4x1 x2 uξ1 ξ1 − 4x1 x2 uξ1 ξ2 , ux2 = uξ1 ξ1x2 + uξ2 ξ2x2 = −2x2 uξ1 ,

( ) ux2 x2 = −2uξ1 − 2x2 uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 = −2uξ1 + 4x22 uξ1 ξ1 ,

(x1 , x2 ) ∈ R2 ,

x1 /= 0,

x2 /= 0.

Hence, 0 = x22 ux1 x1 + 2x1 x2 ux1 x2 + 2x12 ux2 x2 + x2 ux2 ⎧ ⎨ = x22 2uξ1 + 2uξ2 + 4x12 uξ1 ξ1 + 8x12 uξ1 ξ2 + 4x12 uξ2 ξ2 ( ) +2x1 x2 −4x1 x2 uξ1 ξ1 − 4x1 x2 uξ1 ξ2 ⎧ ⎨ ( ) +2x12 −2uξ1 + 4x22 uξ1 ξ1 + x2 −2x2 uξ1

.

= −4x12 uξ1 + 2x22 uξ2 + 4x12 x22 uξ1 ξ1 + 4x12 x22 uξ2 ξ2 , x1 /= 0,

(x1 , x2 ) ∈ R2 ,

x2 /= 0,

and uξ1 ξ1 + uξ2 ξ2 −

.

1 1 uξ1 + 2 uξ2 = 0, 2 x2 2x1

(x1 , x2 ) ∈ R2 ,

x1 /= 0,

x2 /= 0,

or uξ1 ξ1 + uξ2 ξ2 −

.

1 1 uξ1 + uξ = 0, ξ 2 − ξ1 2ξ2 2

(x1 , x2 ) ∈ R2 ,

is the canonical form of the considered equation. Example 4.7 Consider the equation ux1 x1 + x1 ux2 x2 = 0,

.

x1 > 0.

Here a(x1 , x2 ) = 1,

.

b(x1 , x2 ) = 0, c(x1 , x2 ) = x1 ,

(x1 , x2 ) ∈ R2 ,

x1 > 0.

x1 /= 0,

x2 /= 0,

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

71

Then a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 = x1

.

> 0,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

i.e., the considered equation is an elliptic equation. The characteristic equation is (dx2 )2 + x1 (dx1 )2 = 0,

.

(x1 , x2 ) ∈ R2 ,

x1 > 0,

whereupon √ dx2 = ±i x1 dx1 ,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

2 3 x2 = ± ix12 + c, 3

(x1 , x2 ) ∈ R2 ,

x1 > 0.

(x1 , x2 ) ∈ R2 ,

x1 > 0.

.

and .

Here c is a constant. We set ξ1 (x1 , x2 ) = x2 ,

.

ξ2 (x1 , x2 ) =

2 32 x , 3 1

Then ξ1x1 (x1 , x2 ) = 0,

.

ξ1x2 (x1 , x2 ) = 1, 1

ξ2x1 (x1 , x2 ) = x12 , ξ2x2 (x1 , x2 ) = 0,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

and ux1 = uξ1 ξ1x1 + uξ2 ξ2x1

.

1

= x12 uξ2 , 1 ( ) 1 1 u + x12 uξ1 ξ2 ξ1x1 + uξ2 ξ2 ξ2x1 1 ξ2 2 2 x1 ⎧ 1 ⎨ 1 1 2 2 x u u + x = 1 ξ2 1 1 ξ2 ξ2 2x12

ux1 x1 =

72

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

1

=

1 2

uξ2 + x1 uξ2 ξ2 ,

2x1

ux2 = uξ1 ξ1x2 + uξ2 ξ2x2 = uξ1 , ux2 x2 = uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 = uξ1 ξ1 ,

(x1 , x2 ) ∈ R2 ,

x1 > 0.

Hence, 0 = ux1 x1 + x1 ux2 x2

.

=

1 1 2

uξ2 + x1 uξ2 ξ2 + x1 uξ1 ξ1 ,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

2x1 and

uξ1 ξ1 + uξ2 ξ2 +

.

1

uξ2 = 0,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

1 uξ = 0, 3ξ2 2

(x1 , x2 ) ∈ R2 ,

x1 > 0,

3 2

2x1 and uξ1 ξ1 + uξ2 ξ2 +

.

is the canonical form of the considered equation. Example 4.8 Consider the equation (1 + x12 )ux1 x1 + (1 + x22 )ux2 x2 + x1 ux1 + x2 ux2 = 0,

.

(x1 , x2 ) ∈ R2 .

Here a(x1 , x2 ) = 1 + x12 ,

.

b(x1 , x2 ) = 0, c(x1 , x2 ) = 1 + x22 ,

(x1 , x2 ) ∈ R2 .

Then a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 = (1 + x12 )(1 + x22 )

.

> 0,

(x1 , x2 ) ∈ R2 ,

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

i.e., the considered equation is an elliptic equation. The characteristic equation is (1 + x12 )(dx2 )2 + (1 + x22 )(dx1 )2 = 0,

.

(x1 , x2 ) ∈ R2 ,

and / .

/ 1 + x12 dx2 = ±i 1 + x22 dx1 ,

(x1 , x2 ) ∈ R2 ,

and .

/

dx2

dx1 = ±i / , 1 + x22 1 + x12

(x1 , x2 ) ∈ R2 ,

and .

⎨ ⎧ ⎨ ⎧ / / log x2 + 1 + x22 = ±i log x1 + 1 + x12 + c,

(x1 , x2 ) ∈ R2 .

Here c is a constant. We set ⎨ ⎧ / ξ1 (x1 , x2 ) = log x2 + 1 + x22 ,

.

ξ2 (x1 , x2 ) = log(x1 +

/

1 + x12 ),

(x1 , x2 ) ∈ R2 .

Then ξ1x1 (x1 , x2 ) = 0,

.

1 , ξ1x2 (x1 , x2 ) = / 1 + x22 1 , ξ2x1 (x1 , x2 ) = / 1 + x12 ξ2x2 (x1 , x2 ) = 0, and ux1 = uξ1 ξ1x1 + uξ2 ξ2x1

.

= /

1 1 + x12

uξ2 ,

(x1 , x2 ) ∈ R2 ,

73

74

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

ux1 x1 = − =−

( ) 1 uξ2 + / uξ1 ξ2 ξ1x1 + uξ2 ξ2 ξ2x1 (1 + x12 ) 1 + x12 x1

3 2

x1 (1 + x12 )

3 2

uξ2 +

1 uξ2 ξ2 , 1 + x12

ux2 = uξ1 ξ1x2 + uξ2 ξ2x2 = /

ux2 x2 = − =−

1 1 + x22

uξ1 ,

( ) 1 uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 uξ1 + / (1 + x22 ) 1 + x22 x2

3 2

x2 3 (1 + x22 ) 2

uξ1 +

1 uξ1 ξ1 , 1 + x22

(x1 , x2 ) ∈ R2 .

Hence, 0 = (1 + x12 )ux1 x1 + (1 + x22 )ux2 x2 + x1 ux1 + x2 ux2 ⎧ ⎨ x 1 1 = (1 + x12 ) − uξ2 ξ2 u + 3 ξ2 1 + x12 (1 + x12 ) 2 ⎧ ⎨ 1 x2 2 uξ1 ξ1 u + +(1 + x2 ) − 3 ξ1 1 + x22 (1 + x 2 ) 2

.

2

+/

x1 1 + x12

uξ2 + /

x2 1 + x22

uξ1 ,

(x1 , x2 ) ∈ R2 ,

whereupon uξ1 ξ1 + uξ2 ξ2 = 0,

.

(x1 , x2 ) ∈ R2 ,

is the canonical form of the considered equation. Exercise 4.4 Find the canonical form of the equation x2 ux1 x1 + x1 ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0.

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

75

4.1.2 The Parabolic Case In this case, we have ac − b2 = 0 and

.

β = γ = 0.

Then .

) ( aφ1x1 φ2x1 + b φ1x2 φ2x1 + φ1x1 φ2x2 + cφ1x2 φ2x2 = 0 2 2 + 2bφ2x1 φ2x2 + cφ2x = 0. aφ2x 1 2

(4.9)

Let λ1 =

.

φ2x1 . φ2x2

Then from the second equation of (4.9), we get aλ21 + 2bλ1 + c = 0,

.

whereupon (λ1 )1,2 =

.

−b ±

√

b2 − ac a

b =− , a or .

φ2x1 b =− , φ2x2 a

or aφ2x1 + bφ2x2 = 0.

.

(4.10)

Hence and the first equation of (4.9), we find ⎨ ⎨ ⎨ ⎧ ⎧ ⎧ b b − φ2x2 + b φ1x2 − φ2x2 + φ1x1 φ2x2 + cφ1x2 φ2x2 = 0, .aφ1x1 a a or −bφ1x1 φ2x2 −

.

b2 φ1x2 φ2x2 + bφ1x1 φ2x2 + cφ1x2 φ2x2 = 0, a

76

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

or .

ac − b2 φ1x2 φ2x2 = 0, a

or 0 = 0.

.

Therefore the function .φ2 can be determined by Eq. (4.10) and the function .φ1 is an arbitrarily chosen .C 1 -function in .U which is independent of .φ1 and φ1x1 φ2x2 − φ1x2 φ2x1 /= 0 in

.

U.

Equation (4.10) gives a family of solutions of the ordinary differential equation .

dx2 b = , dx1 a

(4.11)

where .x2 is considered as a function of .x1 along the curves of the family. Definition 4.7 The curves .ξ1 = φ1 (x1 , x2 ) = const, ξ2 = φ2 (x1 , x2 ) = const are called the characteristic curves of the linear parabolic operator .L(u). Using (4.11), we get ⎧ a

.

dx2 dx1

⎨2 − 2b

b2 dx2 b2 −2 +c +c = dx1 a a =−

b2 +c a

= 0. Definition 4.8 The equation a(dx2 )2 − 2bdx1 dx2 + c(dx1 )2 = 0

.

is called the characteristic equation of the parabolic operator .L(u). Example 4.9 Consider the equation x12 ux1 x1 − 2x1 x2 ux1 x2 + x22 ux2 x2 + x1 ux1 + x2 ux2 = 0, (x1 , x2 ) ∈ R2 , Here

x1 > 0,

x2 > 0.

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

77

a(x1 , x2 ) = x12 ,

.

b(x1 , x2 ) = −x1 x2 , c(x1 , x2 ) = x22 ,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0.

Then a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 = x12 x22 − x12 x22

.

= 0,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0.

Therefore the considered equation is parabolic. The characteristic equation is x12 (dx2 )2 + 2x1 x2 dx1 dx2 + x22 (dx1 )2 = 0,

.

(x1 , x2 ) ∈ R2 ,

x1 > 0,

whereupon .

(x1 dx2 + x2 dx1 )2 = 0,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

or x1 dx2 = −x2 dx1 ,

.

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0.

Hence, .

dx2 dx1 =− , x2 x1

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

and x1 x2 = c,

.

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0.

We set ξ1 (x1 , x2 ) = x1 ,

.

ξ2 (x1 , x2 ) = x1 x2 ,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0.

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

Then ξ1x1 (x1 , x2 ) = 1,

.

ξ1x2 (x1 , x2 ) = 0, ξ2x1 (x1 , x2 ) = x2 , ξ2x2 (x1 , x2 ) = x1 ,

x2 > 0,

78

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

and ξ1x1 (x1 , x2 )ξ2x2 (x1 , x2 ) − ξ1x2 (x1 , x2 )ξ2x1 (x1 , x2 ) = x1

.

> 0,

(x1 , x2 ) ∈ R2 , x1 > 0,

x2 > 0,

and ux1 = uξ1 ξ1x1 + uξ2 ξ2x1

.

= uξ1 + x2 uξ2 ,

( ) ux1 x1 = uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1 + x2 uξ1 ξ2 ξ1x1 + uξ2 ξ2 ξ2x1 ( ) = uξ1 ξ1 + x2 uξ1 ξ2 + x2 uξ1 ξ2 + x2 uξ2 ξ2 = uξ1 ξ1 + 2x2 uξ1 ξ2 + x22 uξ2 ξ2 ,

( ) ux1 x2 = uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 + uξ2 + x2 uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 = x1 uξ1 ξ2 + uξ2 + x1 x2 uξ2 ξ2 , ux2 = uξ1 ξ1x2 + uξ2 ξ2x2 = x1 uξ2 , ( ) = x1 uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2

ux2 x2

= x12 uξ2 ξ2 ,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0.

Hence, 0 = x12 ux1 x1 − 2x1 x2 ux1 x2 + x22 ux2 x2 + x1 ux1 + x2 ux2 ⎧ ⎨ = x12 uξ1 ξ1 + 2x2 uξ1 ξ2 + x22 uξ2 ξ2 ( ) −2x1 x2 x1 uξ1 ξ2 + x1 x2 uξ2 ξ2 + uξ2 ( ) +x12 x22 uξ2 ξ2 + x1 uξ1 + x2 uξ2 + x1 x2 uξ2 , (x1 , x2 ) ∈ R2 ,

.

x1 > 0,

x2 > 0,

whereupon ξ1 uξ1 ξ1 + uξ1 = 0,

.

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

is the canonical form of the considered equation. Hence, .

( ) ξ1 uξ1 ξ = 0, 1

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

79

and ξ1 uξ1 = f (ξ2 ),

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

1 f (ξ2 ), ξ1

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

.

and uξ1 =

.

and u(x1 , x2 ) = log(ξ1 )f (ξ2 ) + g(ξ2 ),

.

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

and u(x1 , x2 ) = log(x1 )f (x1 x2 ) + g(x1 x2 ),

(x1 , x2 ) ∈ R2 ,

.

x1 > 0,

x2 > 0,

is the general solution of the considered equation, where .f and .g are .C 2 -functions. Example 4.10 Consider the equation x12 ux1 x1 − 2x1 ux1 x2 + ux2 x2 + 2ux2 − x1 ux1 = 0,

.

(x1 , x2 ) ∈ R2 .

Here a(x1 , x2 ) = x12 ,

.

b(x1 , x2 ) = −x1 , c(x1 , x2 ) = 1,

(x1 , x2 ) ∈ R2 .

Then a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 = x12 − x12

.

= 0,

(x1 , x2 ) ∈ R2 .

Therefore the considered equation is parabolic. The characteristic equation is x12 (dx2 )2 + 2x1 dx1 dx2 + (dx1 )2 = 0,

.

(x1 , x2 ) ∈ R2 ,

whereupon x1 dx2 + dx1 = 0,

.

and

(x1 , x2 ) ∈ R2 ,

80

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

x1 ex2 = c,

.

(x1 , x2 ) ∈ R2 .

We set ξ1 (x1 , x2 ) = x1 ex2 ,

.

ξ2 (x1 , x2 ) = x2 ,

(x1 , x2 ) ∈ R2 .

Then ξ1x1 (x1 , x2 ) = ex2 ,

.

ξ1x2 (x1 , x2 ) = x1 ex2 , ξ2x1 (x1 , x2 ) = 0, ξ2x2 (x1 , x2 ) = 1,

(x1 , x2 ) ∈ R2 ,

and ξ1x1 (x1 , x2 )ξ2x2 (x1 , x2 ) − ξ1x2 (x1 , x2 )ξ2x1 (x1 , x2 ) = ex2

.

/= 0,

(x1 , x2 ) ∈ R2 ,

= x1 ex2 uξ1 + x12 e2x2 uξ1 ξ1 + 2x1 ex2 uξ1 ξ2 + uξ2 ξ2 ,

(x1 , x2 ) ∈ R2 .

and ux1 = uξ1 ξ1x1 + uξ2 ξ2x1

.

ux1 x1

= ex2 uξ1 , ( ) = ex2 uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1 = e2x2 uξ1 ξ1 ,

( ) ux1 x2 = ex2 uξ1 + ex2 uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 ( ) = ex2 uξ1 + ex2 x1 ex2 uξ1 ξ1 + uξ1 ξ2 = ex2 uξ1 + x1 e2x2 uξ1 ξ1 + ex2 uξ1 ξ2 , ux2 = uξ1 ξ1x2 + uξ2 ξ2x2 = x1 ex2 uξ1 + uξ2 ,

( ) ux2 x2 = x1 ex2 uξ1 + x1 ex2 uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 +uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 ( ) = x1 ex2 uξ1 + x1 ex2 x1 ex2 uξ1 ξ1 + uξ1 ξ2 +x1 ex2 uξ1 ξ2 + uξ2 ξ2

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

81

Hence, 0 = x12 ux1 x1 − 2x1 ux1 x2 + ux2 x2 + 2ux2 − x1 ux1 ⎧ ⎨ = x12 e2x2 uξ1 ξ1 − 2x1 ex2 uξ1 + x1 e2x2 uξ1 ξ1 + ex2 uξ1 ξ2

.

+x1 ex2 uξ1 + x12 e2x2 uξ1 ξ1 + 2x1 ex2 uξ1 ξ2 + uξ2 ξ2 +2x1 ex2 uξ1 + 2uξ2 − x1 ex2 uξ1 = uξ2 ξ2 + 2uξ2 ,

(x1 , x2 ) ∈ R2 ,

is the canonical form of the considered equation. Let v = uξ2 .

.

Then vξ2 = −2v,

.

(x1 , x2 ) ∈ R2 ,

and v = e−2ξ2 f (ξ1 ),

.

(x1 , x2 ) ∈ R2 ,

and uξ2 = e−2ξ2 f (ξ1 ),

.

(x1 , x2 ) ∈ R2 ,

and 1 u(x1 , x2 ) = − f (ξ1 )e−2ξ2 + g(ξ1 ), 2

.

(x1 , x2 ) ∈ R2 ,

i.e., ) ( ) ( 1 u(x1 , x2 ) = − e−2x2 f x1 ex2 + g x1 ex2 , 2

.

(x1 , x2 ) ∈ R2 ,

is the general solution of the considered equation, where .f and .g are .C 2 -functions. Example 4.11 Consider the equation ux1 x1 − 2 sin x1 ux1 x2 + (sin x1 )2 ux2 x2 = 0,

.

Here a(x1 , x2 ) = 1,

.

(x1 , x2 ) ∈ R2 .

82

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

b(x1 , x2 ) = − sin x1 , c(x1 , x2 ) = (sin x1 )2 ,

(x1 , x2 ) ∈ R2 .

Then a(x1 , x2 )c(x1 , x2 ) − (b(x1 , x2 ))2 = (sin x1 )2 − (sin x1 )2

.

= 0,

(x1 , x2 ) ∈ R2 .

Therefore the considered equation is parabolic. The characteristic equation is (dx2 )2 + 2 sin x1 dx1 dx2 + (sin x1 )2 (dx1 )2 = 0,

.

(x1 , x2 ) ∈ R2 ,

whereupon .

(dx2 + sin x1 dx1 )2 = 0,

(x1 , x2 ) ∈ R2 ,

and dx2 + sin x1 dx1 = 0,

(x1 , x2 ) ∈ R2 ,

.

and x2 − cos x1 = c,

.

(x1 , x2 ) ∈ R2 .

We set ξ1 (x1 , x2 ) = x2 − cos x1 ,

.

ξ2 (x1 , x2 ) = x1 ,

(x1 , x2 ) ∈ R2 .

Then ξ1x1 (x1 , x2 ) = sin x1 ,

.

ξ1x2 (x1 , x2 ) = 1, ξ2x1 (x1 , x2 ) = 1, ξ2x2 (x1 , x2 ) = 0,

(x1 , x2 ) ∈ R2 ,

and ξ1x1 (x1 , x2 )ξ2x2 (x1 , x2 ) − ξ1x2 (x1 , x2 )ξ2x1 (x1 , x2 ) = 0 − 1

.

= −1 /= 0,

(x1 , x2 ) ∈ R2 ,

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

and ux1 = uξ1 ξ1x1 + uξ2 ξ2x1

.

= sin x1 uξ1 + uξ2 ,

( ) ux1 x1 = cos x1 uξ1 + sin x1 uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1 +uξ1 ξ2 ξ1x1 + uξ2 ξ2 ξ2x1 ( ) = cos x1 uξ1 + sin x1 sin x1 uξ1 ξ1 + uξ1 ξ2 + sin x1 uξ1 ξ2 + uξ2 ξ2

ux1 x2

= cos x1 uξ1 + (sin x1 )2 uξ1 ξ1 + 2 sin x1 uξ1 ξ2 + uξ2 ξ2 , ( ) = sin x1 uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 +uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 = sin x1 uξ1 ξ1 + uξ1 ξ2 ,

ux2 = uξ1 ξ1x2 + uξ2 ξ2x2 = uξ1 , ux2 x2 = uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 = uξ1 ξ1 ,

(x1 , x2 ) ∈ R2 .

Hence, 0 = ux1 x1 − 2 sin x1 ux1 x2 + (sin x1 )2 ux2 x2

.

= cos x1 uξ1 + (sin x1 )2 uξ1 ξ1 + 2 sin x1 uξ1 ξ2 + uξ2 ξ2 ( ) −2 sin x1 sin x1 uξ1 ξ1 + uξ1 ξ2 + (sin x1 )2 uξ1 ξ1 = cos x1 uξ1 + uξ2 ξ2 ,

(x1 , x2 ) ∈ R2 ,

and .

cos(ξ2 )uξ1 + uξ2 ξ2 = 0,

(x1 , x2 ) ∈ R2 ,

is the canonical form of the considered equation. Exercise 4.5 Find the canonical form of the equation ux1 x1 + 6ux1 x2 + 9ux2 x2 + ux1 + ux2 = 0,

.

(x1 , x2 ) ∈ R2 .

Exercise 4.6 Consider the equation 4x22 ux1 x1 + 4x2 ux1 x2 + ux2 x2 + 2ux1 = 0,

.

(x1 , x2 ) ∈ R2 .

83

84

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

1. Find the canonical form. 2. Find the general solution. 3. Find the solution .u(x1 , x2 ) for which u(x1 , 0) = 9 sin x1 ,

.

ux2 (x1 , 0) = ex1 ,

x1 ∈ R.

4.1.3 The Hyperbolic Case We suppose that .L(u) is hyperbolic in .U and suppose that .α = γ = 0. Then, using the definitions for .α and .γ , we obtain the system

.

2 + 2bφ 2 aφ1x 1x1 φ1x2 + cφ1x2 = 0 1 2 2 =0 aφ2x1 + 2bφ2x1 φ2x2 + cφ2x 2

(4.12)

or a

⎧φ

1x1

⎨2

φ1x2 . ⎧ ⎨ φ2x 2 a φ2x1 2

φ1x

+ 2b φ1x1 + c = 0 2

φ2x + 2b φ2x1 2

+ c = 0.

From the above system, we see that if there exist such functions .φ1 and .φ2 , then φ2x φ1x1 and . 1 satisfy the quadratic equation . φ2x2 φ1x2 ap2 + 2bp + c = 0,

.

(4.13)

where .p is unknown. Since .L(u) is hyperbolic in .U, we have that .ac − b2 < 0 and then Eq. (4.13) has two roots .p1 and .p2 . Thus, in the hyperbolic case we obtain the canonical form 2βuξ1 ξ2 + · · · = 0

.

by determining the functions .φ1 and .φ2 from the differential equations .

φ1x1 − p1 φ1x2 = 0 φ2x1 − p2 φ2x2 = 0.

These two first order linear homogeneous partial differential equations yield two families of curves

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

85

φ1 = const,

.

φ2 = const. These two families can be defined as the families of solutions of the ordinary differential equations .

dx2 = −p1 , dx1 dx2 = −p2 , dx1

and since .p1 and .p2 are roots of (4.13), we have ⎧ a

.

dx2 dx1

⎨2 − 2b

dx2 + c = 0. dx1

(4.14)

Here .x2 is considered as a function of .x1 along the curves of the family. We have that √ b ± b2 − ac .p1,2 = a and let us set p1 =

.

p2 =

b+ b−

√ √

b2 − ac , a b2 − ac . 2

Then ⨜ x2 +

.

⨜ x2 +

b+ b−

√ √

b2 − ac dx1 = const, a b2 − ac dx1 = const, a √ 2 b2 − ac . p 1 − p2 = a

86

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

Definition 4.9 The curves ⎧ ⨜ ξ1 = φ1 x2 + ⎧ . ⨜ ξ2 = φ2 x2 +

√

⎨

√

⎨

b+ b−

b2 −ac dx1 a b2 −ac dx1 a

, ,

are called the characteristic curves of the linear hyperbolic differential operator L(u).

.

Remark 4.1 For convenience, in the practice, we very often take ⨜ ξ1 = x2 +

.

⨜ ξ2 = x2 +

b+ b−

√ √

b2 − ac dx1 , a b2 − ac dx1 . a

Definition 4.10 The curves .ξ1 = φ1 (x1 , x2 ) = const and .ξ2 = φ2 (x1 , x2 ) = const which satisfy the system (4.12), are called the characteristic curves of the linear hyperbolic operator .L(u). In the case when .α = −γ , β = 0, the functions .φ1 and .φ2 satisfy the system .

2 + 2bφ 2 2 2 aφ1x 1x1 φ1x2 + cφ1x2 = −(aφ2x1 + 2bφ2x1 φ2x2 + cφ2x2 ) 1 aφ1x1 φ2x1 + b(φ1x2 φ2x1 + φ1x1 φ2x2 ) + cφ1x2 φ2x2 = 0

or .

2 + φ 2 ) + 2b(φ 2 2 a(φ1x 1x1 φ1x2 + φ2x1 φ2x2 ) + c(φ1x2 + φ2x2 ) = 0 2x1 1 aφ1x1 φ2x1 + b(φ1x2 φ2x1 + φ1x1 φ2x2 ) + cφ1x2 φ2x2 = 0.

(4.15)

Thus, we have the canonical form α(uξ1 ξ1 − uξ2 ξ2 ) + · · · = 0.

.

Definition 4.11 Equation (4.14) is called the characteristic equation. Now, we will simplify the system (4.15). Let us assume that .φ3 , φ4 ∈ C 1 (U ) and .

Then

φ1 = φ 3 + φ4 φ2 = φ 3 − φ4 .

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

φ1x1 φ1x2 . φ2x1 φ2x2

= = = =

φ3x1 φ3x2 φ3x1 φ3x2

87

+ φ4x1 , + φ4x2 , − φ4x1 , − φ4x2

and )2 ( )2 ( 2 2 φ1x + φ2x = φ3x1 + φ4x1 + φ3x1 − φ4x1 1 1

.

2 2 2 2 = φ3x + 2φ3x1 φ4x1 + φ4x + φ3x − 2φ3x1 φ4x1 + φ4x 1 1 1 1 ⎨ ⎧ 2 2 , = 2 φ3x + φ4x 1 1

( )2 ( )2 2 2 φ1x + φ2x = φ3x2 + φ4x2 + φ3x2 − φ4x2 2 2

φ1x1 φ1x2 + φ2x1 φ2x2

2 2 2 2 = φ3x + 2φ3x2 φ4x2 + φ4x + φ3x − 2φ3x2 φ4x2 + φ4x 2 2 2 2 ⎨ ⎧ 2 2 = 2 φ3x2 + φ4x2 , )( ) ( )( ) ( = φ3x1 + φ4x1 φ3x2 + φ4x2 + φ3x1 − φ4x1 φ3x2 − φ4x2

= φ3x1 φ3x2 + φ3x1 φ4x2 + φ4x1 φ3x2 + φ4x1 φ4x2 +φ3x1 φ3x2 − φ3x1 φ4x2 − φ4x1 φ3x2 + φ4x1 φ4x2 ) ( = 2 φ3x1 φ3x2 + φ4x1 φ4x2 , and ⎨ ⎨ ⎧ ⎧ ( ) 2 + φ2 2 + φ2 + 2b φ + c φ a φ1x φ + φ φ 1x 1x 2x 2x 1 2 1 2 2x 1x 2x 1 2⎧ 2 ⎧1 ⎨ ⎨ . ( ) 2 + φ2 2 + φ2 = 2a φ3x + 4b φ + 2c φ φ + φ φ 3x 3x 4x 4x 1 2 1 2 4x1 3x2 4x2 , 1 from where, ⎨ ⎨ ⎧ ⎧ ( ) 2 2 2 2 + 2b φ + c φ .a φ3x + φ4x φ + φ φ + φ 3x 3x 4x 4x 1 2 1 2 3x2 4x2 = 0. 1 1

(4.16)

Also, )( ) ( φ1x1 φ2x1 = φ3x1 + φ4x1 φ3x1 − φ4x1

.

φ1x2 φ2x2

φ1x2 φ2x1 + φ1x1 φ2x2

2 2 − φ4x , = φ3x 1 1 )( ) ( = φ3x2 + φ4x2 φ3x2 − φ4x2 2 2 − φ4x , = φ3x 2 2 )( ) ( )( ) ( = φ3x2 + φ4x2 φ3x1 − φ4x1 + φ3x1 + φ4x1 φ3x2 − φ4x2

88

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

= φ3x1 φ3x2 + φ3x1 φ4x2 − φ3x2 φ4x1 − φ4x1 φ4x2 +φ3x1 φ3x2 − φ3x1 φ4x2 + φ4x1 φ3x2 − φ4x1 φ4x2 ) ( = 2 φ3x1 φ3x2 − φ4x1 φ4x2 , and ) ( aφ1x⎧1 φ2x1 + b φ1x 2 φ2x1 + φ1x1 φ2x2 + cφ1x2 φ2x2 ⎧ ⎨ ⎨ ) ( . 2 − φ2 2 − φ2 + c φ = a φ3x φ − φ φ + 2b φ 3x1 3x2 4x1 4x2 4x1 3x2 4x2 . 1 Using the last identity and (4.15), (4.16), we obtain ⎧ ⎧ ⎨ ⎨ ) ( 2 + φ2 2 + φ2 a φ3x + 2b φ3x1 φ3x2 + φ4x1 φ4x2 + c φ3x =0 4x 4x 1 1 2 2 ⎧ ⎨ ⎨ . ⎧ ) ( 2 − φ2 2 2 a φ3x 4x1 + 2b φ3x1 φ3x2 − φ4x1 φ4x2 + c φ3x2 − φ4x2 = 0, 1 from where .

2 + 2bφ 2 aφ3x 3x1 φ3x2 + cφ3x2 = 0 1 2 + 2bφ 2 aφ4x 4x1 φ4x2 + cφ4x2 = 0. 1

In this way, we have reduced the case .α = −γ , .β = 0 to the case .α = γ = 0. Example 4.12 We will find the canonical form of the Tricomi equation ux1 x1 + x1 ux2 x2 = 0,

(x1 , x2 ) ∈ R2 ,

.

x1 < 0.

The characteristic equation is (dx2 )2 + x1 (dx1 )2 = 0,

.

(x1 , x2 ) ∈ R2 ,

x1 < 0,

whereupon ⎧ .

dx2 dx1

⎨2 + x1 = 0,

(x1 , x2 ) ∈ R2 ,

x1 < 0,

or .

√ dx2 = ± −x1 , dx1

(x1 , x2 ) ∈ R2 ,

x1 < 0,

and √ dx2 = ± −x1 dx1 ,

.

(x1 , x2 ) ∈ R2 ,

x1 < 0.

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

We integrate the last equation and we find ⨜

⨜

√ −x1 dx1 + c,

dx2 = ±

.

(x1 , x2 ) ∈ R2 ,

x1 < 0,

or 3

x2 = ∓

(−x1 ) 2

+ c,

(x1 , x2 ) ∈ R2 ,

x1 < 0,

3 2 x2 ± (−x1 ) 2 = c, 3

(x1 , x2 ) ∈ R2 ,

x1 < 0.

.

3 2

and .

Here c is a constant. We set 3 2 ξ1 (x1 , x2 ) = x2 + (−x1 ) 2 , 3 3 2 ξ2 (x1 , x2 ) = x2 − (−x1 ) 2 , 3

.

(x1 , x2 ) ∈ R2 ,

x1 < 0.

Then 1

ξ1x1 (x1 , x2 ) = −(−x1 ) 2 ,

.

ξ1x2 (x1 , x2 ) = 1, 1

ξ2x1 (x1 , x2 ) = (−x1 ) 2 , ξ2x2 (x1 , x2 ) = 1,

(x1 , x2 ) ∈ R2 ,

x1 < 0.

Hence, ux1 = uξ1 ξ1x1 + uξ2 ξ2x1

.

1

1

= −(−x1 ) 2 uξ1 + (−x1 ) 2 uξ2 , ux1 x1 =

1 2(−x1 ) −

=

1 2

) 1 ( uξ1 − (−x1 ) 2 uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1

1 2(−x1 ) 1

2(−x1 )

1 2

1 2

) 1 ( uξ2 + (−x1 ) 2 uξ1 ξ2 ξ1x1 + uξ2 ξ2 ξ2x1

uξ1 −

1 1

2(−x1 ) 2

uξ2

⎧ ⎨ 1 1 1 −(−x1 ) 2 −(−x1 ) 2 uξ1 ξ1 + (−x1 ) 2 uξ1 ξ2

89

90

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

⎧ ⎨ 1 1 1 +(−x1 ) 2 −(−x1 ) 2 uξ1 ξ2 + (−x1 ) 2 uξ2 ξ2 =

=

1

uξ1 −

1

uξ1 −

1

u 1 ξ2 2(−x1 ) 2(−x1 ) 2 −x1 uξ1 ξ1 + x1 uξ1 ξ2 + x1 uξ1 ξ2 − x1 uξ2 ξ2 1 2

1 2(−x1 )

1 2

1

2(−x1 ) 2

uξ2 − x1 uξ1 ξ1 + 2x1 uξ1 ξ2 − x1 uξ2 ξ2 ,

ux2 = uξ1 ξ1x2 + uξ2 ξ2x2 = ux1 + uξ2 , ux2 x2 = uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 + uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 = uξ1 ξ1 + 2uξ1 ξ2 + uξ2 ξ2 ,

(x1 , x2 ) ∈ R2 ,

x1 < 0.

From here, 1

ux1 x1 + x1 ux2 x2 =

.

2(−x1 )

1 2

uξ1 −

1 1

2(−x1 ) 2

uξ2 − x1 uξ1 ξ1

+2x1 uξ1 ξ2 − x1 uξ2 ξ2 + x1 uξ1 ξ1 + 2x1 uξ1 ξ2 + x1 uξ2 ξ2 =

1 2(−x1 )

1 2

uξ1 −

1 1

2(−x1 ) 2

(x1 , x2 ) ∈ R2 ,

uξ2 + 4x1 uξ1 ξ2 ,

x1 < 0.

Therefore 0 = ux1 x1 + x1 ux2 x2

.

=

1 2(−x1 )

1 2

uξ1 −

1 1

2(−x1 ) 2

uξ2 + 4x1 uξ1 ξ2 ,

(x1 , x2 ) ∈ R2 ,

x1 < 0,

whereupon 3

.

− 8(−x1 ) 2 uξ1 ξ2 + uξ1 − uξ2 = 0,

(x1 , x2 ) ∈ R2 ,

x1 < 0,

6(ξ2 − ξ1 )uξ1 ξ2 + uξ1 − uξ2 = 0,

(x1 , x2 ) ∈ R2 ,

x1 < 0,

and .

which is the canonical form of the considered Tricomi equation. Example 4.13 We will find the canonical form of the equation x25 ux1 x1 − x2 ux2 x2 + 2ux2 = 0,

.

(x1 , x2 ) ∈ R2 ,

x2 /= 0.

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

The characteristic equation is x25 (dx2 )2 − x2 (dx1 )2 = 0,

.

(x1 , x2 ) ∈ R2 ,

x2 /= 0,

whereupon x22 dx2 = ±dx1 ,

.

(x1 , x2 ) ∈ R2 ,

x2 /= 0.

Hence, ⨜

⨜ x22 dx2

.

=±

dx1 + c,

(x1 , x2 ) ∈ R2 ,

x2 /= 0,

and .

x23 = ±x1 + c, 3

(x1 , x2 ) ∈ R2 ,

x2 /= 0,

x23 ± 3x1 = c,

(x1 , x2 ) ∈ R2 ,

x2 /= 0.

or .

Here c is a constant. We set ξ1 (x1 , x2 ) = x23 + 3x1 ,

.

ξ2 (x1 , x2 ) = x23 − 3x1 ,

(x1 , x2 ) ∈ R2 ,

x2 /= 0.

Then ξ1x1 (x1 , x2 ) = 3,

.

ξ1x2 (x1 , x2 ) = 3x22 , ξ2x1 (x1 , x2 ) = −3, ξ2x2 (x1 , x2 ) = 3x22 ,

(x1 , x2 ) ∈ R2 ,

x2 /= 0.

From here, ux1 = uξ1 ξ1x1 + uξ2 ξ2x1

.

ux1 x1

= 3uξ1 − 3uξ2 , ) ( = 3 uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1 ( ) −3 uξ1 ξ2 ξ1x1 + uξ2 ξ2 ξ2x1 ( ) ( ) = 3 3uξ1 ξ1 − 3uξ1 ξ2 − 3 3uξ1 ξ2 − 3uξ2 ξ2

91

92

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

= 9uξ1 ξ1 − 18uξ1 ξ2 + 9uξ2 ξ2 , ux2 = uξ1 ξ1x2 + uξ2 ξ2x2

ux2 x2

= 3x22 uξ1 + 3x22 uξ2 , ( ) = 6x2 uξ1 + 3x22 uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 ( ) +6x2 uξ2 + 3x22 uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 ⎧ ⎨ = 6x2 uξ1 + 6x2 uξ2 + 3x22 3x22 uξ1 ξ1 + 3x22 uξ1 ξ2 ⎧ ⎨ +3x22 3x22 uξ1 ξ2 + 3x22 uξ2 ξ2 = 6x2 uξ1 + 6x2 uξ2 + 9x24 uξ1 ξ1 + 18x24 uξ1 ξ2 + 9x24 uξ2 ξ2 , (x1 , x2 ) ∈ R2 ,

x2 /= 0.

Therefore 0 = x25 ux1 x1 − x2 ux2 x2 + 2ux2

.

= 9x25 uξ1 ξ1 − 18x25 uξ1 ξ2 + 9x25 uξ2 ξ2 − 6x22 uξ1 − 6x22 uξ2 −9x25 uξ1 ξ1 − 18x25 uξ1 ξ2 − 9x25 uξ2 ξ2 + 6x22 uξ1 + 6x22 uξ2 = −36x25 uξ1 ξ2 ,

(x1 , x2 ) ∈ R2 ,

x2 /= 0,

or uξ1 ξ2 = 0,

.

(x1 , x2 ) ∈ R2 ,

x2 /= 0,

is the canonical form of the considered equation. Example 4.14 We will find the canonical form of the equation ux1 x1 − 2 sin x1 ux1 x2 − (cos x1 )2 ux2 x2 − cos x1 ux2 = 0,

.

(x1 , x2 ) ∈ R2 .

The characteristic equation is (dx2 )2 + 2 sin x1 dx1 dx2 − (cos x1 )2 (dx1 )2 = 0,

.

(x1 , x2 ) ∈ R2 ,

whereupon ⎧ .

Hence,

dx2 dx1

⎨2 + 2 sin x1

dx2 − (cos x1 )2 = 0, dx1

(x1 , x2 ) ∈ R2 .

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

⎧ .

dx2 dx1

⎨ =

− sin x1 ±

1,2

√ (sin x1 )2 + (cos x1 )2 1

= − sin x1 ± 1,

(x1 , x2 ) ∈ R2 ,

or dx2 = (− sin x1 ± 1)dx1 ,

(x1 , x2 ) ∈ R2 ,

.

from where ⨜

⨜ dx2 =

.

(− sin x1 ± 1)dx1 + c

= cos x1 ± x1 + c,

(x1 , x2 ) ∈ R2 ,

or x2 − cos x1 ∓ x1 = c,

.

(x1 , x2 ) ∈ R2 .

We set ξ1 (x1 , x2 ) = x2 − cos x1 − x1 ,

.

ξ2 (x1 , x2 ) = x2 − cos x1 + x1 ,

(x1 , x2 ) ∈ R2 .

Then ξ1x1 (x1 , x2 ) = sin x1 − 1,

.

ξ1x2 (x1 , x2 ) = 1, ξ2x1 (x1 , x2 ) = sin x1 + 1, ξ2x2 (x1 , x2 ) = 1,

(x1 , x2 ) ∈ R2 .

From here, ux1 = uξ1 ξ1x1 + uξ2 ξ2x1

.

= (sin x1 − 1)uξ1 + (sin x1 + 1)uξ2 ,

( ) ux1 x1 = cos x1 uξ1 + cos x1 uξ2 + (sin x1 − 1) uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1 ( ) +(sin x1 + 1) uξ1 ξ2 ξ1x1 + uξ2 ξ2 ξ2x1 = cos x1 uξ1 + cos x1 uξ2 ) ( +(sin x1 − 1) (sin x1 − 1)uξ1 ξ1 + (sin x1 + 1)uξ1 ξ2

93

94

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

) ( +(sin x1 + 1) (sin x1 − 1)uξ1 ξ2 + (sin x1 + 1)uξ2 ξ2 = cos x1 uξ1 + cos x1 uξ2 + (sin x1 − 1)2 uξ1 ξ1

ux1 x2

+2((sin x1 )2 − 1)uξ1 ξ2 + (sin x1 + 1)2 uξ2 ξ2 , ) ( = (sin x1 − 1) uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 ) ( +(sin x1 + 1) uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 ) ( = (sin x1 − 1)(uξ1 ξ1 + uξ1 ξ2 ) + (sin x1 + 1) uξ1 ξ2 + uξ2 ξ2 = (sin x1 − 1)uξ1 ξ1 + 2 sin x1 uξ1 ξ2 + (sin x1 + 1)uξ2 ξ2 ,

ux2 = uξ1 ξ1x2 + uξ2 ξ2x2 = uξ1 + uξ2 , ux2 x2 = uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1 + uξ1 ξ2 ξ1x1 + uξ2 ξ2 ξ2x1 = uξ1 ξ1 + 2uξ1 ξ2 + uξ2 ξ2 ,

(x1 , x2 ) ∈ R2 .

Therefore 0 = ux1 x1 − 2 sin x1 ux1 x2 − (cos x1 )2 ux2 x2 − cos x1 ux2

.

= cos x1 uξ1 + cos x1 uξ2 + (sin x1 − 1)2 uξ1 ξ1 + 2((sin x1 )2 − 1)uξ1 ξ2 +(sin x1 + 1)2 uξ2 ξ2 −2 sin x1 (sin x1 − 1)uξ1 ξ1 − 4(sin x1 )2 uξ1 ξ2 − 2 sin x1 (sin x1 + 1)uξ2 ξ2 −(cos x1 )2 uξ1 ξ1 −2(cos x1 )2 uξ1 ξ2 − (cos x1 )2 uξ2 ξ2 − cos x1 uξ1 − cos x1 uξ2 , and .

− 4uξ1 ξ2 = 0,

(x1 , x2 ) ∈ R2 ,

or uξ1 ξ2 = 0,

.

(x1 , x2 ) ∈ R2 ,

which is the canonical form of the considered equation. Exercise 4.7 Find the canonical form of the following equations. 1. ux1 x1 − 5ux1 x2 + 6ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

(x1 , x2 ) ∈ R2 ,

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

95

2. ux1 x1 + 5ux1 x2 + 4ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

3. x12 ux1 x1 − x22 ux2 x2 − 2x2 ux2 = 0,

.

(x1 , x2 ) ∈ R2 ,

x1 /= 0,

Example 4.15 Consider the equation ux1 x1 − ux2 x2 + 2ux1 + 2ux2 = 0,

.

(x1 , x2 ) ∈ R2 .

We will find 1. the canonical form. 2. the general solution. 3. the solution .u = u(x1 , x2 ), .(x1 , x2 ) ∈ R2 , for which u(0, x2 ) = 1,

.

ux1 (0, x2 ) = −1.

x2 ∈ R.

1. The characteristic equation is (dx2 )2 − (dx1 )2 = 0,

.

(x1 , x2 ) ∈ R2 ,

whereupon dx2 = ±dx1 ,

(x1 , x2 ) ∈ R2 ,

x2 ± x1 = c,

(x1 , x2 ) ∈ R2 .

.

and .

We set ξ1 (x1 , x2 ) = x2 + x1 ,

.

ξ2 (x1 , x2 ) = x2 − x1 . Then ξ1x1 (x1 , x2 ) = 1,

.

ξ1x2 (x1 , x2 ) = 1,

(x1 , x2 ) ∈ R2 .

x2 /= 0.

96

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

ξ2x1 (x1 , x2 ) = −1, ξ2x2 (x1 , x2 ) = 1,

(x1 , x2 ) ∈ R2 .

Hence, ux1 = uξ1 ξ1x1 + uξ2 ξ2x1

.

= uξ1 − uξ2 , ux1 x1 = uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1 − uξ1 ξ2 ξ1x1 − uξ2 ξ2 ξ2x1 = uξ1 ξ1 − uξ1 ξ2 − uξ1 ξ2 + uξ2 ξ2 = uξ1 ξ1 − 2uξ1 ξ2 + uξ2 ξ2 , ux2 = uξ1 ξ1x2 + uξ2 ξ2x2 = uξ1 + uξ2 , ux2 x2 = uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 + uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 = uξ1 ξ1 + uξ1 ξ2 + uξ1 ξ2 + uξ2 ξ2 = uξ1 ξ1 + 2uξ1 ξ2 + uξ2 ξ2 ,

(x1 , x2 ) ∈ R2 .

Therefore 0 = ux1 x1 − ux2 x2 + 2ux1 + 2ux2

.

= uξ1 ξ1 − 2uξ1 ξ2 + uξ2 ξ2 − uξ1 ξ1 − 2uξ1 ξ2 − uξ2 ξ2 + 2uξ1 −2uξ2 + 2uξ1 + 2uξ2 = −4uξ1 ξ2 + 4uξ1 ,

(x1 , x2 ) ∈ R2 .

Thus, uξ1 ξ2 − uξ1 = 0,

.

(x1 , x2 ) ∈ R2 ,

is the canonical form of the considered equation. 2. Fix .ξ1 and set v = uξ1 .

.

Equation (4.17) takes the form vξ2 − v = 0,

.

(4.17)

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

97

from where v = f1 (ξ1 )eξ2 .

.

Hence, uξ1 = f1 (ξ1 )eξ2

.

and u(x1 , x2 ) = eξ2 f (ξ1 ) + g(ξ2 ),

.

(x1 , x2 ) ∈ R2 ,

where .f1 and .g are .C 2 -functions and ⨜ f (ξ1 ) =

.

f1 (ξ1 )dξ1 .

Consequently u(x1 , x2 ) = f (x1 + x2 )ex2 −x1 + g(x2 − x1 ),

.

(x1 , x2 ) ∈ R2 ,

(4.18)

is the general solution of the given equation. 3. Using (4.18), we get u(0, x2 ) = f (x2 )ex2 + g(x2 ),

.

ux1 (x1 , x2 ) = f ' (x1 + x2 )ex2 −x1 − f (x1 + x2 )ex2 −x1 − g ' (x2 − x1 ), ux1 (0, x2 ) = f ' (x2 )ex2 − f (x2 )ex2 − g ' (x2 ),

(x1 , x2 ) ∈ R2 ,

i.e., we obtain the system .

f (x2 )ex2 + g(x2 ) =1 f ' (x2 )ex2 − f (x2 )ex2 − g ' (x2 ) = −1,

(x1 , x2 ) ∈ R2 .

(4.19)

We differentiate the first equation of the last system with respect to .x2 and we obtain .

f ' (x2 )ex2 + f (x2 )ex2 + g ' (x2 ) = 0 f ' (x2 )ex2 − f (x2 )ex2 − g ' (x2 ) = −1,

(x1 , x2 ) ∈ R2 ,

whereupon 2f ' (x2 )ex2 = −1,

.

x2 ∈ R,

98

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

and 1 f ' (x2 ) = − e−x2 , 2

.

x2 ∈ R.

So, f (x2 ) =

.

1 −x2 e + c, 2

x2 ∈ R,

where c is a real constant. From here and from the first equation of (4.19), we find g(x2 ) = 1 − f (x2 )ex2 ⎧ ⎨ 1 −x2 e + c e x2 = 1− 2

.

=

1 − cex2 , 2

x2 ∈ R.

Therefore ⎧ u(x1 , x2 ) =

.

=

⎨ 1 −x1 −x2 1 + c ex2 −x1 + − cex2 −x1 e 2 2

1 −2x1 1 e + , 2 2

(x1 , x2 ) ∈ R2 .

Example 4.16 Consider the equation x12 ux1 x1 − 2x1 x2 ux1 x2 − 3x22 ux2 x2 = 0,

(x1 , x2 ) ∈ R2 , x1 > 0, x2 > 0.

.

We will find 1. the canonical form. 2. the general solution. 3. the solution .u = u(x1 , x2 ), .(x1 , x2 ) ∈ R2 , .x1 > 0, .x2 > 0, for which u(x1 , 1) = 0,

.

ux2 (x1 , 1) =

4 5

,

x1 ∈ R.

3x14 1. The characteristic equation is x12 (dx2 )2 +2x1 x2 dx1 dx2 −3x22 (dx1 )2 =0,

.

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

99

whereupon ⎧ 2 .x1

dx2 dx1

⎨2 + 2x1 x2

dx2 − 3x22 = 0, dx1

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

and ⎧ .

⎨

dx2 dx1

−x1 x2 ±

=

/

x12 x22 + 3x12 x22 x12

1,2

−x1 x2 ± 2x1 x2 , x12

=

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

and .

dx2 x2 = , dx1 x1 dx2 x2 = −3 , dx1 x1

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0.

Hence, .

x2 = c, x1

x13 x2 = c, (x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0.

We set ξ1 (x1 , x2 ) =

.

x2 , x1

ξ2 (x1 , x2 ) = x13 x2 ,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0.

x1 > 0,

x2 > 0,

Then ξ1x1 (x1 , x2 ) = −

.

ξ1x2 (x1 , x2 ) =

x2 , x12

1 , x1

ξ2x1 (x1 , x2 ) = 3x12 x2 , ξ2x2 (x1 , x2 ) = x13 , whereupon

(x1 , x2 ) ∈ R2 ,

100

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

ux1 = uξ1 ξ1x1 + uξ2 ξ2x1 x2 = − 2 uξ1 + 3x12 x2 uξ2 , x1

.

) x2 ( uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1 2 x1 ( ) +6x1 x2 uξ2 + 3x12 x2 uξ1 ξ2 ξ1x1 + uξ2 ξ2 ξ2x1 ⎧ ⎨ x2 2x2 x2 2 = 3 uξ1 − 2 − 2 uξ1 ξ1 + 3x1 x2 uξ1 ξ2 x1 x1 x1 ⎧ ⎨ x2 2 2 +6x1 x2 uξ2 + 3x1 x2 − 2 uξ1 ξ2 + 3x1 x2 uξ2 ξ2 x1

ux1 x1 =

2x2

=

2x2

x13

x13

uξ1 −

uξ1 + 6x1 x2 uξ2 +

x22 x14

uξ1 ξ1 − 6x22 uξ1 ξ2 + 9x14 x22 uξ2 ξ2 ,

) 1 x2 ( uξ1 − 2 uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 2 x1 x1 ( ) +3x12 uξ2 + 3x12 x2 uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 ⎧ ⎨ 1 x2 1 = − 2 uξ1 + 3x12 uξ2 − 2 uξ1 ξ1 + x13 uξ1 ξ2 x1 x1 x1 ⎨ ⎧ 1 uξ1 ξ2 + x13 uξ2 ξ2 +3x12 x2 x1

ux1 x2 = −

=−

1 x2 uξ1 + 3x12 uξ2 − 3 uξ1 ξ1 + 2x1 x2 uξ1 ξ2 + 3x15 x2 uξ2 ξ2 , 2 x1 x1

ux2 = uξ1 ξ1x2 + uξ2 ξ2x2 1 uξ + x13 uξ2 , x1 1 ) ( ) 1 ( uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 + x13 uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 = x1 ⎧ ⎨ ⎧ ⎨ 1 1 1 3 3 3 uξ ξ + x1 uξ1 ξ2 + x1 uξ ξ + x1 uξ2 ξ2 = x1 x1 1 1 x1 1 2

= ux2 x2

=

1 uξ1 ξ1 + 2x12 uξ1 ξ2 + x16 uξ2 ξ2 , (x1 , x2 ) ∈ R2 , x12

Therefore 0 = x12 ux1 x1 − 2x1 x2 ux1 x2 − 3x22 ux2 x2

.

x1 > 0,

x2 > 0.

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

⎧ =

x12

2x2 x13

uξ1 + 6x1 x2 uξ2 +

⎧

101

⎨

x22

uξ1 ξ1 − 6x22 uξ1 ξ2 x14

+ 9x14 x22 uξ2 ξ2

x2 1 −2x1 x2 − 2 uξ1 + 3x12 uξ2 − 3 uξ1 ξ1 + 2x1 x2 uξ1 ξ2 + 3x15 x2 uξ2 ξ2 x1 x1 ⎨ ⎧ 1 2 6 2 −3x2 uξ1 ξ1 + 2x1 uξ1 ξ2 + x1 uξ2 ξ2 x12

⎨

x2 x2 uξ1 + 6x13 x2 uξ2 + 22 uξ1 ξ1 − 6x12 x22 uξ1 ξ2 x1 x1

=2

+9x16 x22 uξ2 ξ2 + 2

x2 x2 uξ1 − 6x13 x2 uξ2 + 2 22 uξ1 ξ1 x1 x1

−4x12 x22 uξ1 ξ2 − 6x16 x22 uξ2 ξ2 − 3 x2 uξ − 16x12 x22 uξ1 ξ2 = 0, x1 1

=4

x22 x12

uξ1 ξ1 − 6x12 x22 uξ1 ξ2 − 3x16 x22 uξ2 ξ2

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

and 4uξ1 ξ2 −

.

1

uξ1 x13 x2

= 0,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

and 1 uξ = 0, ξ2 1

4uξ1 ξ2 −

.

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

is the canonical form of the considered equation. 2. Fix .ξ1 and set v = uξ1 .

.

Equation (4.20) takes the form 4vξ2 −

.

1 v = 0, ξ2

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

or 4

.

vξ2 1 = , ξ2 v

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0.

(4.20)

102

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

Therefore v(ξ1 , ξ2 ) =

.

√ 4

ξ2 f1 (ξ1 ),

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

and uξ1 (ξ1 , ξ2 ) =

.

√ 4

(x1 , x2 ) ∈ R2 ,

ξ2 f1 (ξ1 ),

x1 > 0,

x2 > 0.

So, u(ξ1 , ξ2 ) =

.

√ 4

ξ2 f (ξ1 ) + g(ξ2 ),

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

i.e., u(x1 , x2 ) =

.

/ 4

⎧ x13 x2 f

x2 x1

⎨ + g(x13 x2 ),

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0, (4.21)

is the general solution of the considered equation, where .f1 and .g are .C 2 functions and ⨜ .f (ξ1 ) = f1 (ξ1 )dξ1 . 3. From (4.21), we get u(x1 , 1) =

.

/ 4

⎧ x13 f

/

1 x1 ⎧

⎨ + g(x13 ),

⎨ / ⎧ ⎨ x2 x x2 4 2 ' + + x13 g ' (x13 x2 ), f f 3 x x x x2 1 1 1 ⎧ ⎨ ⎧ ⎨ / 1 ' 1 1 14 3 + x13 g ' (x13 ), +√ ux2 (x1 , 1) = f x f 4 x x1 4 1 x1 1

1 ux2 (x1 , x2 ) = 4

(x1 , x2 ) ∈ R2 ,

.

x1 > 0,

4

x13

x2 > 0. In this way we get the system

/

⎧ ⎨ =0 x13 f x11 + g(x13 ) ⎧ ⎨ ⎧ ⎨ . / 4 1 1 3 ' 1 + x 3 g ' (x 3 ) = √4 √1 1 1 x1 4 x1 f x1 + 4 x f 34x 4

1

(x1 , x2 ) ∈ R2 , x1 > 0, system and we get

.

(4.22) 1 x1

,

x2 > 0. We differentiate the first equation of the last

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

⎧ ⎨ 1 ' 1 + 3x 2 g ' (x 3 ) = 0 − f 5 1 1 x1 4 x1 4 . / ⎧ ⎨ ⎧ ⎨ x1 1 1 4 3 ' 1 + x 3 g ' (x 3 ) = √4 √1 1 1 4 x1 f x1 + 4 x f x1 34x ⎧ ⎨

3 √ 4

f

1 x1

1

(x1 , x2 ) ∈ R2 ,

x1 > 0,

.

103

1 x1

,

x2 > 0, whereupon

/ ⎧ ⎨ 4 3 x1 f x11 − ⎧ ⎨ . / 1 1 4 3 x f 1 4 x1 + 3 4

⎧ ⎨

1 ' 1 + 3x 3 g ' (x 3 ) √ 4x f 1 1 1 ⎧ x1 ⎨ 1 1 3 3 ' ' √ 4x f x1 + x1 g (x1 ) 1

=0 =

√4 3 4 x1 x1

and f

.

'

⎧

1 x1

⎨ =

1 , x1

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

i.e., f ' (z) = z,

(x1 , x2 ) ∈ R2 ,

.

x1 > 0,

x2 > 0.

Therefore f (z) =

.

1 2 z + c, 2

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

where c is a real constant. Hence and the first equation of the system (4.22), we obtain ⎨ / ⎧ 1 4 3 . x + c + g(x13 ) = 0, (x1 , x2 ) ∈ R2 , x1 > 0, x2 > 0. 1 2x12 From here, √ g(z) = −c 4 z −

1

.

5

,

2z 12

(x1 , x2 ) ∈ R2 ,

x1 > 0,

Consequently u(x1 , x2 ) =

.

=

/ 4

/ 4

⎧ x13 x2 f ⎧ x13 x2

x2 x1

⎨ + g(x13 x2 )

⎨ / 1 x22 4 3 + c − c x1 x2 − 2 x12

1 5 4

5

2x1 x212

x2 > 0.

104

4 Classifications and Canonical Forms for Linear Second Order Partial. . . 9

1 x24 1 = , − 5 5 2 54 4 12 x1 2x1 x2

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0.

Example 4.17 Consider the equation ux1 x1 + 6ux1 x2 − 16ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

We will find 1. the canonical form. 2. the general solution. 3. the solution .u = u(x1 , x2 ),

(x1 , x2 ) ∈ R2 , for which

u(−x1 , 2x1 ) = x1

.

u(x1 , 0) = 2x1 ,

x1 ∈ R.

1. The characteristic equation is (dx2 )2 − 6dx1 dx2 − 16(dx1 )2 = 0,

.

(x1 , x2 ) ∈ R2 ,

whereupon ⎧ .

dx2 dx1

⎨2 −6

dx2 − 16 = 0, dx1

(x1 , x2 ) ∈ R2 ,

and √ dx2 3 ± 9 + 16 = . dx1 1 = 3 ± 5,

(x1 , x2 ) ∈ R2 ,

and dx2 = 8dx1 ,

.

dx2 = −2dx1 ,

(x1 , x2 ) ∈ R2 .

Therefore x2 − 8x1 = c,

.

x2 + 2x1 = c,

(x1 , x2 ) ∈ R2 .

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

We set ξ1 (x1 , x2 ) = x2 − 8x1 ,

.

ξ2 (x1 , x2 ) = x2 + 2x1 ,

(x1 , x2 ) ∈ R2 .

Then ξ1x1 (x1 , x2 ) = −8,

.

ξ1x2 (x1 , x2 ) = 1, ξ2x1 (x1 , x2 ) = 2, ξ2x2 (x1 , x2 ) = 1,

(x1 , x2 ) ∈ R2 ,

and ux1 = uξ1 ξ1x1 + uξ2 ξ2x1

.

ux1 x1

= −8uξ1 + 2uξ2 , ) ( ) ( = −8 uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1 + 2 uξ1 ξ2 ξ1x1 + uξ2 ξ2 ξ2x1 ( ) ( ) = −8 −8uξ1 ξ1 + 2uξ1 ξ2 + 2 −8uξ1 ξ2 + 2uξ2 ξ2

ux1 x2

= 64uξ1 ξ1 − 32uξ1 ξ2 + 4uξ2 ξ2 , ) ( ) ( = −8 uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 + 2 uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 ) ( ) ( = −8 uξ1 ξ1 + uξ1 ξ2 + 2 uξ1 ξ2 + uξ2 ξ2 = −8uξ1 ξ1 − 6uξ1 ξ2 + 2uξ2 ξ2 ,

ux2 = uξ1 ξ1x2 + uξ2 ξ2x2 = uξ1 + uξ2 , ux2 x2 = uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 + uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 = uξ1 ξ1 + uξ1 ξ2 + uξ1 ξ2 + uξ2 ξ2 = uξ1 ξ1 + 2uξ1 ξ2 + uξ2 ξ2 ,

(x1 , x2 ) ∈ R2 .

Therefore

.

0 = ux1 x1 + 6ux1 x2 − 16ux2 x2 = 64uξ1 ξ1 − 32uξ1 ξ2 + 4uξ2 ξ2 − 48uξ1 ξ1 − 36uξ1 ξ2 + 12uξ2 ξ2 −16uξ1 ξ1 − 32uξ1 ξ2 − 16uξ2 ξ2 , (x1 , x2 ) ∈ R2 ,

whereupon .

− 100uξ1 ξ2 = 0,

(x1 , x2 ) ∈ R2 ,

105

106

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

i.e., uξ1 ξ2 = 0,

.

(x1 , x2 ) ∈ R2 ,

(4.23)

is the canonical form of the considered equation. 2. Fix .ξ2 and set v = uξ2 .

.

Equation (4.23) takes the form vξ1 = 0,

.

(x1 , x2 ) ∈ R2 ,

from where v = f1 (ξ2 ),

.

(x1 , x2 ) ∈ R2 ,

and uξ2 = f1 (ξ2 ),

.

(x1 , x2 ) ∈ R2 .

Hence, u(ξ1 , ξ2 ) = f (ξ2 ) + g(ξ1 ),

.

(x1 , x2 ) ∈ R2 ,

i.e., u(x1 , x2 ) = f (x2 + 2x1 ) + g(x2 − 8x1 ),

.

(x1 , x2 ) ∈ R2 ,

(4.24)

is the general solution of the considered equation, where .f1 and .g are .C 2 functions and ⨜ .f (ξ2 ) = f1 (ξ2 )dξ2 . 3. From (4.24), we get u(−x1 , 2x1 ) = f (2x1 − 2x1 ) + g(2x1 + 8x1 )

.

= f (0) + g(10x1 ), u(x1 , 0) = f (2x1 ) + g(−8x1 ), In this way, we get the system

(x1 , x2 ) ∈ R2 .

4.1 Classifications and Canonical Forms for Linear Second Order Partial. . .

.

f (0) + g(10x1 ) = x1 f (2x1 ) + g(−8x1 ) = 2x1 ,

x1 ∈ R,

whereupon g(10x1 ) = x1 − f (0),

.

x1 ∈ R,

i.e., g(z) =

.

1 z − f (0), 10

z ∈ R.

From here and from the second equation of (4.25), we find f (2x1 ) = 2x1 − g(−8x1 )

.

= 2x1 + =

8x1 + f (0) 10

14 x1 + f (0), 5

i.e., f (z) =

.

7 z + f (0), 5

z ∈ R.

Consequently x2 − 8x1 7 − f (0) (x2 + 2x1 ) + f (0) + 10 5 3 = x2 + 2x1 , (x1 , x2 ) ∈ R2 . 2

u(x1 , x2 ) =

.

Exercise 4.8 Consider the equation ux1 x1 + 2ux1 x2 − 3ux2 x2 = 0,

.

1. Find the canonical form. 2. Find the general solution. 3. Find the solution .u = u(x1 , x2 ),

(x1 , x2 ) ∈ R2 .

(x1 , x2 ) ∈ R2 , for which

u(x1 , 0) = 3x12 ,

.

ux2 (x1 , 0) = 0,

(x1 , x2 ) ∈ R2 .

107

(4.25)

108

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

Exercise 4.9 Consider the equation x1 ux1 x1 + (x1 + x2 )ux1 x2 + x2 ux2 x2 = 0,

.

1. Find the canonical form. 2. Find the general solution. 3. Find a solution .u = u(x1 , x2 ),

(x1 , x2 ) ∈ R2 ,

x1 > 0, x2 > 0.

(x1 , x2 ) ∈ R2 x1 > 0, x2 > 0, for which

⎧

⎨ 1 .u x1 , = x13 , x1 ⎧ ⎨ 1 = 2x12 , ux1 x1 , x1

x1 ∈ R.

4.2 Classification and Canonical Form of Second Order Linear Partial Differential Equations in n Independent Variables The classifications in the previous sections focused on the classifications of second order partial differential equations in two independent variables. The classification concepts become more meaningful in terms of concepts based on analytic geometry when there are only two independent variables. When we consider a partial differential equation in a four dimensional space, three independent variables could be spatial dimensions and the fourth independent variable could be designated as time. The classification of second order partial differential equations in n independent variables is a generalization of the classification concepts developed for two independent variables. Consider a general second order linear partial differential equation in .n independent variables n ∑ .

i,j =1

aij uxi xj +

n ∑

bi uxi + b0 u + d = 0,

(4.26)

i=1

where the coefficients .aij , .bi , .1 ≤ i, j ≤ n, .b0 , .d and the unknown .u are functions of .x = (x1 , . . . , xn ). An important observation in the development of the theory of partial differential equations is that the most important properties of the solutions of linear partial differential equations depend only on the form of the highest order terms appearing in the equation. These terms are referred to as principal part of the equation. For

4.2 Classification and Canonical Form of Second Order Linear Partial. . .

109

example, in Eq. (4.26), the principal part of the equation involves only the second order derivatives with multiplying functions .aij , .i, j ∈ {1, . . . , n}. Let ⎞ a11 a12 · · · a1n ⎜ a21 a22 · · · a2n ⎟ ⎟. .A = ⎜ ⎠ ⎝ ··· an1 an2 · · · ann ⎛

Definition 4.12 The matrix A is said to be the coefficients matrix of the principal part of Eq. (4.26). Since the operator .

∂2 , ∂xi ∂xj

i, j ∈ {1, . . . , n},

is symmetric, i.e., .

∂2 ∂2 , = ∂xj ∂xi ∂xi ∂xj

i, j ∈ {1, . . . , n},

we assume that the matrix .A is symmetric. If the matrix A is not symmetric, we can ) 1( aij + aj i such that (4.26) can be rewritten always find a symmetric matrix .a ij = 2 in the form n ∑ .

i,j =1

a ij uxi xj +

n ∑

bi uxi + cu + d = 0.

i=1

If we consider the principal part of the partial differential equation in two variables, the coefficients matrix will be ⎨ ⎧ a11 a12 . .A = a12 a22 We can write n ∑ .

aij uxi xj = a11 ux1 x1 + a12 ux1 x2 + a21 ux2 x1 + a22 ux2 x2 .

(4.27)

i,j =1

If we compare (4.27) with the equation aux1 x1 + 2bux1 x2 + cux2 x2 = lower order terms,

.

(4.28)

110

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

we have a11 = a,

.

a12 = b, a21 = b, a22 = c. Then the coefficients matrix of the principal part takes the form ⎧ A=

.

ab bc

⎨ .

The eigenvalues of the coefficients matrix of the principal part satisfy the equation ⎧ det

.

a−λ b b c−λ

⎨ = 0.

Expanding the last equation, we get (a − λ)(c − λ) − b2 = 0

.

or λ2 − (a + c)λ + ac − b2 = 0.

.

(4.29)

The roots of Eq. (4.29) are λ1 =

.

λ2 =

a+c+ a+c−

√ √

(a + c)2 − 4(ac − b2 ) , 2 (a + c)2 − 4(ac − b2 ) , 2

or λ1 =

.

λ2 =

a+c+ a+c−

√ √

(a − c)2 + 4b2 , 2 (a − c)2 + 4b2 . 2

Since the matrix A is symmetric, the roots .λ1 and .λ2 are real numbers. Observe that

4.2 Classification and Canonical Form of Second Order Linear Partial. . .

⎧ λ1 λ2 =

.

a+c+

√

(a − c)2 + 4b2 2

⎨⎧

a+c−

=

(a + c)2 − (a − c)2 − 4b2 4

=

a 2 + 2ac + c2 − a 2 + 2ac − c2 − 4b2 4

=

4ac − 4b2 4

√

(a − c)2 + 4b2 2

111

⎨

= ac − b2 = − det A. Thus, the conclusions relating to the sign of .det A can be interpreted in relation to the sign of the eigenvalues .λ1 and .λ2 . Namely, we have the following. 1. .det A > 0 implies that .λ1 and .λ2 are nonzero and are of opposite sign. 2. .det A = 0 implies that one of .λ1 and .λ2 is zero. 3. .det A < 0 implies that .λ1 and .λ2 are nonzero and have the same sign. From here and from the knowledge by the previous sections, we conclude the following. 1. If the eigenvalues .λ1 and .λ2 are nonzero and have the same sign, Eq. (4.28) is elliptic. 2. If one of the eigenvalues .λ1 and .λ2 is zero, then Eq. (4.28) is parabolic. 3. If the eigenvalues .λ1 and .λ2 are nonzero and are of opposite sign, then Eq. (4.28) is hyperbolic. This classification concept based on the signs of the eigenvalues of the coefficients matrix of the principal part is another way to classify a second order partial differential equation. Now, consider the case of a partial differential equation with n independent variables. Denote by .λ1 , . . . , λn the eigenvalues of the matrix A. Since A is a symmetric matrix, we have that .λj , .j ∈ {1, . . . , n}, are real numbers. In this case, we have the following classification. Definition 4.13 Equation (4.26) is called 1. elliptic, if all eigenvalues .λi of .A are nonzero and have the same sign. 2. parabolic, if the eigenvalues .λi of .A are all positive or all negative, save one that is zero. 3. hyperbolic, if all eigenvalues .λi of .A are nonzero and have the same sign except for one of the eigenvalues. 4. ultrahyperbolic, if there is more than one positive eigenvalue and more than one negative eigenvalue, and there are non-zero eigenvalues.

112

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

Invoking a theorem in linear algebra, we have that the number of positive, zero and negative eigenvalues of the matrix A remain invariant under smooth nonsingular transformation of coordinates. Let Q be the diagonal matrix of A. Consider the transformation ξ = Qx,

.

where .ξ = (ξ1 , . . . , ξn ) and .Q = (qij ) is an .n × n matrix. We have that ξi =

n ∑

.

qij xj .

j =1

Using the chain rule, we have uxi

= .

n ∑

=

uxi xj = =

k=1 n ∑

uξk ξkxi uξk qki ,

k=1 n ∑

uξk ξl ξkxi ξlxj

k,l=1 n ∑

uξk ξl qki qlj .

k,l=1

This allows Eq. (4.26) to be expressed as n ∑ .

k,l=1

⎛ ⎝

n ∑

⎞ qki aij qlj ⎠ uξk ξl + lower order terms = 0.

i,j =1

The coefficient matrix of the terms .uξk ξl in this transformed expression is equal to ( .

) qki aij qlj = QT AQ.

Let .Λ be a diagonal matrix whose elements are the eigenvalues .λi , .i ∈ {1, . . . , n}, of the matrix .A. We have ) ( Λ = λi δij ,

.

where .δij are the Kronecker delta, and QT AQ = Λ

.

i, j = 1, . . . , n,

4.2 Classification and Canonical Form of Second Order Linear Partial. . .

⎛ ⎜ ⎜ =⎜ ⎝

⎞

λ1

⎟ ⎟ ⎟. ⎠

λ2 ..

. λn

The nonsingular transformation x = QT ξ

.

transforms Eq. (4.26) in the canonical form n ∑ .

λi uξi ξi + lower order terms = 0.

i=1

Example 4.18 Consider the equation ux1 x1 − ux1 x2 + ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

Then ⎧ A=

.

1 − 12 − 12 1

⎨ .

We will find the eigenvalues of the matrix .A. We have I I I1 − λ −1 I 2 I I . det(A − λI ) = I −1 1 − λI 2 = 0, whereupon (λ − 1)2 −

1 =0 4

λ2 − 2λ +

3 = 0, 4

.

and .

and 3 , 2 1 λ2 = . 2

λ1 =

.

113

114

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

Since .λ1 , λ2 > 0, the considered equation is an elliptic equation. Now we will find the matrix .Q. Let A1 = (a, b)

.

be an eigenvector of the matrix .A corresponding to the eigenvalue .λ1 = ⎧ .

− 12 − 12 − 12 − 12

3 . Then 2

⎨⎧ ⎨ ⎧ ⎨ a 0 = , b 0

i.e., a + b = 0.

.

We take 1 q1 = √ (1, −1). 2

.

Let A2 = (a, b)

.

be an eigenvector of the matrix .A corresponding to the eigenvalue .λ2 = ⎧ .

− 12

1 2

− 12

1 2

⎨⎧ ⎨ ⎧ ⎨ a 0 = , b 0

i.e., a = b.

.

We take 1 q2 = √ (1, 1). 2

.

Therefore ⎨ ⎧ Q = q1T , q2T

.

1 = √ 2

⎧

⎨ 1 1 . −1 1

1 . Then 2

4.2 Classification and Canonical Form of Second Order Linear Partial. . .

Let ξ = Qx ⎧

.

⎨⎧

⎨ x1 = x2 ⎧ 1 ⎨ √ (x1 + x2 ) 2 = . √1 (−x1 + x2 ) √1 √1 2 2 − √1 √1 2 2

2

Then 1 ξ1x1 = √ , 2 1 ξ1x2 = √ , 2 1 ξ2x1 = − √ , 2 1 ξ2x2 = √ , 2

.

and ux1 = uξ1 ξ1x1 + uξ2 ξ2x1

.

ux1 x1

ux1 x2

ux2

1 1 = √ uξ1 − √ uξ2 , 2 2 ( ) ) 1 ( 1 = √ uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1 − √ uξ1 ξ2 ξ1x1 + uξ2 ξ2 ξ2x1 2 2 ⎨ ⎨ ⎧ ⎧ 1 1 1 1 1 1 = √ √ uξ1 ξ1 − √ uξ1 ξ2 − √ √ uξ1 ξ2 − √ uξ2 ξ2 2 2 2 2 2 2 1 1 = uξ1 ξ1 − uξ1 ξ2 + uξ2 ξ2 , 2 2 ) ) 1 ( 1 ( = √ uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 − √ uξ1 ξ1 ξ1x2 + uξ2 ξ2 ξ2x2 2 2 ⎨ ⎨ ⎧ ⎧ 1 1 1 1 1 1 = √ √ uξ1 ξ1 + √ uξ1 ξ2 − √ √ uξ1 ξ2 + √ uξ2 ξ2 2 2 2 2 2 2 1 1 = uξ1 ξ1 − uξ2 ξ2 , 2 2 = uξ1 ξ1x2 + uξ2 ξ2x2

115

116

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

ux2 x2

1 1 = √ uξ1 + √ uξ2 , 2 2 ) ) 1 ( 1 ( = √ uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 + √ uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 2 2 ⎨ ⎨ ⎧ ⎧ 1 1 1 1 1 1 = √ √ uξ1 ξ1 + √ uξ1 ξ2 + √ √ uξ1 ξ2 + √ uξ2 ξ2 2 2 2 2 2 2 1 1 = uξ1 ξ1 + uξ1 ξ2 + uξ2 ξ2 . 2 2

Therefore 1 1 1 1 uξ ξ − uξ1 ξ2 + uξ2 ξ2 − uξ1 ξ1 + uξ2 ξ2 2 11 2 2 2 1 1 + uξ1 ξ1 + uξ1 ξ2 + uξ2 ξ2 2 2 1 3 = uξ1 ξ1 + uξ2 ξ2 . 2 2

ux1 x1 − ux1 x2 + ux2 x2 =

.

Hence, the canonical form of the considered equation is uξ1 ξ1 + 3uξ2 ξ2 = 0.

.

Example 4.19 Consider the equation ux1 x1 + 2ux1 x3 + ux2 x2 + ux3 x3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 .

Then ⎞ 101 .A = ⎝ 0 1 0 ⎠ . 101 ⎛

We will find the eigenvalues of the matrix .A. We have I I I1 − λ 0 1 II I I .det(A − λI ) = I I 0 1−λ 0 I I 1 0 1 − λI = 0, whereupon −(λ − 1)3 + λ − 1 = 0

.

4.2 Classification and Canonical Form of Second Order Linear Partial. . .

117

and ⎧ ⎨ −(λ − 1) (λ − 1)2 − 1 = 0,

.

and (λ − 1)(λ − 2)λ = 0,

.

and λ1 = 0,

.

λ2 = 1, λ3 = 2. Since .λ1 = 0 and .λ2 , λ3 > 0, the considered equation is a parabolic equation. Let A1 = (a, b, c)

.

be an eigenvector of the matrix .A corresponding to the eigenvalue .λ1 = 0. Then ⎛

⎞⎛ ⎞ ⎛ ⎞ 101 a 0 ⎝ ⎠ ⎝ ⎠ ⎝ . 010 b = 0⎠, 101 c 0 i.e., .

a+c = 0 b = 0.

We take 1 q1 = √ (1, 0, −1). 2

.

Let A2 = (a, b, c)

.

be an eigenvector of the matrix .A corresponding to the eigenvalue .λ2 = 1. Then ⎛

00 .⎝0 0 10

⎞⎛ ⎞ ⎛ ⎞ 1 a 0 0⎠⎝b⎠ = ⎝0⎠, 0 c 0

118

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

i.e., .

a=0 c = 0.

We take q2 = (0, 1, 0).

.

Let A3 = (a, b, c)

.

be an eigenvector of the matrix .A corresponding to the eigenvalue .λ3 = 2. Then ⎛

⎞⎛ ⎞ ⎛ ⎞ −1 0 1 a 0 .⎝ 0 −1 0 ⎠ ⎝ b ⎠ = ⎝ 0 ⎠ , 1 0 −1 c 0 i.e., .

a=c b = 0.

We take 1 q3 = √ (1, 0, 1). 2

.

Hence, ⎨ ⎧ Q = q1T , q2T , q3T

.

⎛

√1 2

0

√1 2

⎞

⎟ ⎜ =⎝ 0 1 0 ⎠ − √1 0 √1 2

2

and ⎞ ⎛ √1 0 √1 ⎞ ⎛ ⎞ ξ1 x1 2 ⎟ ⎜ 2 . ⎝ ξ2 ⎠ = ⎝ 0 1 0 ⎠ ⎝ x2 ⎠ x3 ξ3 − √1 0 √1 ⎛

2

2

4.2 Classification and Canonical Form of Second Order Linear Partial. . .

⎛ ⎜ =⎝

√1 x1 2

− √1

2

+

√1 x3 2

x2 x1 +

√1 x3 2

⎞ ⎟ ⎠.

Then 1 ξ1x1 = √ , 2 ξ1x2 = 0,

.

1 ξ1x3 = √ , 2 ξ2x1 = 0, ξ2x2 = 1, ξ2x3 = 0, 1 ξ3x1 = − √ , 2 ξ3x2 = 0, 1 ξ3x3 = √ 2 and ux1 = uξ1 ξ1x1 + uξ2 ξ2x1 + uξ3 ξ3x1

.

ux1 x1

ux 1 x 3

1 1 = √ uξ1 − √ uξ3 , 2 2 ) ( 1 = √ uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1 + uξ1 ξ3 ξ3x1 2 ) 1 ( − √ uξ1 ξ3 ξ1x1 + uξ2 ξ3 ξ2x1 + uξ3 ξ3 ξ3x1 2 ⎨ ⎨ ⎧ ⎧ 1 1 1 1 1 1 = √ √ uξ1 ξ1 − √ uξ1 ξ3 − √ √ uξ1 ξ3 − √ uξ3 ξ3 2 2 2 2 2 2 1 1 = uξ1 ξ1 − uξ1 ξ3 + uξ3 ξ3 , 2 2 ) 1 ( = √ uξ1 ξ1 ξ1x3 + uξ1 ξ2 ξ2x3 + uξ1 ξ3 ξ3x3 2 ) 1 ( − √ uξ1 ξ3 ξ1x3 + uξ2 ξ3 ξ2x3 + uξ3 ξ3 ξ3x3 2

119

120

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

ux2

⎨ ⎨ ⎧ ⎧ 1 1 1 1 1 1 = √ √ uξ1 ξ1 + √ uξ1 ξ3 − √ √ uξ1 ξ3 + √ uξ3 ξ3 2 2 2 2 2 2 1 1 = uξ1 ξ1 − uξ3 ξ3 , 2 2 = uξ1 ξ1x2 + uξ2 ξ2x2 + uξ3 ξ3x2 = uξ2 ,

ux2 x2 = uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2 + uξ2 ξ3 ξ3x2 = uξ2 ξ2 , ux3 = uξ1 ξ1x3 + uξ2 ξ2x3 + uξ3 ξ3x3

ux3 x3

1 1 = √ uξ1 + √ uξ3 , 2 2 ) 1 ( = √ uξ1 ξ1 ξ1x3 + uξ1 ξ2 ξ2x3 + uξ1 ξ3 ξ3x1 2 ) 1 ( + √ uξ1 ξ3 ξ1x3 + uξ2 ξ3 ξ2x3 + uξ3 ξ3 ξ3x3 2 ⎨ ⎨ ⎧ ⎧ 1 1 1 1 1 = √ √ uξ1 ξ1 + √ uξ1 ξ3 + √ uξ1 ξ3 + √ uξ3 ξ3 2 2 2 2 2 1 1 = uξ1 ξ1 + uξ1 ξ3 + uξ3 ξ3 . 2 2

From here, ux1 x1 + 2ux1 x3 + ux2 x2 + ux3 x3

.

⎧ ⎨ 1 1 1 1 = uξ1 ξ1 − uξ1 ξ3 + uξ3 ξ3 + 2 uξ ξ − uξ ξ 2 2 2 11 2 33 1 1 +uξ2 ξ2 + uξ1 ξ1 + uξ1 ξ3 + uξ3 ξ3 2 2 = 2uξ1 ξ1 + uξ2 ξ2 .

Therefore the canonical form of the considered equation is 2uξ1 ξ1 + uξ2 ξ2 = 0.

.

Exercise 4.10 Classify the following equation ( ) ux1 x1 − c2 ux2 x2 + ux3 x3 + ux4 x4 = 0,

.

(x1 , x2 , x3 , x4 ) ∈ R4 ,

where .c is a constant, .c /= 0. Exercise 4.11 Find the canonical form of the following equations.

4.3 Classification of First Order Systems with Two Independent Variables

121

1. ux1 x1 + 2ux1 x2 − 2ux1 x3 + 2ux2 x2 + 6ux3 x3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 .

2. ux1 x1 + 2ux1 x2 + 2ux2 x2 + 2ux2 x3 + 2ux2 x4 + 2ux3 x3 + 3ux4 x4 = 0,

.

(x1 , x2 , x3 , x4 ) ∈ R4 . 3. ux1 x2 − ux1 x4 + ux3 x3 − 2ux3 x4 + 2ux4 x4 = 0,

.

(x1 , x2 , x3 , x4 ) ∈ R4 .

4. ux1 x1 + 2

n ∑

.

uxk xk − 2

k=2

n−1 ∑

uxk xk+1 = 0,

(x1 , . . . , xn ) ∈ Rn .

k=1

5. ux1 x1 − 2

n ∑

.

(−1)k uxk−1 xk = 0,

(x1 , . . . , xn ) ∈ Rn .

k=2

4.3 Classification of First Order Systems with Two Independent Variables The problem of classification of differential equations of mathematical physics is very important in view of modeling new industrial process and creating advanced materials. Models based on the fundamental physical principles and allowing convenient use of the mathematical methods of investigation have a natural advantage. For many models it is more convenient they to be reduced to systems of partial differential equations. For instance, in the field of complex analysis in mathematics, the Cauchy-Riemann system(equations) consist of a system of two partial differential equations which form a necessary and sufficient condition for a complex function of a complex variable to be complex differentiable. These equations are ∂u ∂v = ∂x1 ∂x2 (4.30)

.

∂u ∂v =− , ∂x2 ∂x1

122

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

where u and v are real differentiable bivariate functions and they are the real and imaginary parts of a complex-valued function, respectively. Observe that if we differentiate the first equation of (4.30) with respect to .x1 and the second equation of (4.30) with respect to .x2 and then add the obtained equations, we get the Laplace equation in two independent variables .

∂ 2u ∂ 2u + 2 = 0. ∂x12 ∂x2

(4.31)

If we differentiate the first equation of (4.30) with respect to .x2 and the second equation of (4.30) with respect to .x1 and then add the obtained equations, we arrive at the Laplace equation in two independent variables .

∂ 2v ∂ 2v = 0. + ∂x22 ∂x12

Now, we consider the Laplace equation (4.31). Set ux1 = p,

.

ux2 = q. Then ux1 x1 = px1 ,

.

ux1 x2 = px2 , ux2 x1 = qx1 , ux2 x2 = qx2 . Hence, we get the Cauchy-Riemann equations px2 = qx1

.

px1 = −qx2 . In the above example, we saw that some partial differential equations are equivalent to systems of partial differential equations. The main aim in this section is to classify the systems Aux1 + Bux2 = F,

.

where

(4.32)

4.3 Classification of First Order Systems with Two Independent Variables

⎛

a11 a12 ⎜ a21 a22 .A = ⎜ ⎝ ... an1 an2 ⎛ b11 b12 ⎜ b21 b22 B=⎜ ⎝ ... bn1 bn2 ⎛ ⎞ f1 ⎜ f2 ⎟ ⎟ F =⎜ ⎝...⎠,

123

⎞ . . . a1n . . . a2n ⎟ ⎟, ⎠ . . . ann ⎞ . . . b1n . . . b2n ⎟ ⎟, ⎠ . . . bnn

fn ⎛

⎞ u1 ⎜ u2 ⎟ ⎟ u=⎜ ⎝...⎠, un ⎞ ⎛ u1xi ⎜ u2x ⎟ i ⎟ uxi = ⎜ ⎝ ... ⎠, unxi

i = 1, 2,

aij , bij , fi , i, j ∈ {1, 2, . . . , n}, are given functions, .u is unknown. Now we consider the transformation

.

v = P −1 u,

.

i.e., u = P v,

.

where .P is a nonsingular .n × n-matrix. Then uxi = (P v)xi

.

= Pxi v + P vxi ,

i = 1, 2,

and (4.32) takes the form ) ( ) ( A Px1 v + P vx1 + B Px2 v + P vx2 = F

.

or

124

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

( ) AP vx1 + BP vx2 = F − APx1 + BPx2 v

.

= G. Assuming that .A is nonsingular, we multiply the above equation by .(AP )−1 to obtain (AP )−1 AP vx1 + (AP )−1 BP vx2 = (AP )−1 G

.

or vx1 + P −1 A−1 BP vx2 = (AP )−1 G.

.

(4.33)

We set D = A−1 B,

.

H = (AP )−1 G. Then vx1 + P −1 DP vx2 = H.

.

If .P is taken to be the diagonal matrix of .D and .Λ is a diagonal matrix whose elements are the eigenvalues .λi of .D and the columns of .P are linearly independent eigenvectors of .D, .pi = (p1i , p2i , . . . , pni ), .|pi | = 1, .i ∈ {1, 2, . . . , n}, so, P = (pij ),

.

Λ = (λj δij ),

i, j ∈ {1, 2, . . . , n},

where .δij is the Kronecker delta, then P −1 DP = Λ.

.

Thus, we can write the system (4.33) in the form vx1 + Λvx2 = H

.

or vx1 + λi vx2 = hi ,

.

with .n characteristics given by .

i ∈ {1, 2, . . . , n},

dx2 = λi , .i ∈ {1, 2, . . . , n}. dx1

Definition 4.14 System (4.34) is said to be the canonical form of (4.32).

(4.34)

4.3 Classification of First Order Systems with Two Independent Variables

125

The classification of the system (4.32) is done basing on the nature of the eigenvalues .λi of .D. Definition 4.15 1. If all the .n eigenvalues of .D are complex, then the system (4.32) is said to be elliptic. 2. If all the .n eigenvalues of .D are real and some of them are repeated, then the system (4.32) is said to be parabolic. 3. If all the .n eigenvalues of .D are real and distinct, then the system (4.32) is said to be hyperbolic. 4. If some of the .n eigenvalues of .D are real and other complex, the system (4.32) is considered as hybrid of elliptic-hyperbolic type. Example 4.20 Consider the Cauchy-Riemann equations (4.30). Since the CauchyRiemann equations are equivalent to the Laplace equation in two independent variables, we expect that the Cauchy-Riemann system is an elliptic system. Here ⎧

⎨ 10 .A = , 01 ⎧ ⎨ 0 −1 B= . 1 0 Then .D = B. The eigenvalues of D satisfy the equation ⎧ .

det(D − λI ) = det

−λ −1 1 −λ

⎨

= λ2 + 1 = 0, whereupon .λ1,2 = ±i. Therefore the Cauchy-Riemann system is an elliptic system. Example 4.21 Consider the system .

u1x1 + αu1x2 + βu2x2 = 0 u2x1 + γ u1x2 + δu2x2 = 0,

Then ⎧

⎨ 10 , 01 ⎧ ⎨ αβ B= . γ δ A=

.

(x1 , x2 ) ∈ R2 .

126

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

The matrix .A is nonsingular and .D = B. We will find the eigenvalues of the matrix D. We have I I Iα − λ β I I I .det(D − λI ) = I γ δ − λI

.

= 0, whereupon (λ − α)(λ − δ) − βγ = 0,

.

or λ2 − (α + δ)λ + αδ − βγ = 0.

.

Therefore 1. if .(α − δ)2 + 4βγ < 0, then the considered system is elliptic. 2. if .(α − δ)2 + 4βγ = 0, then the considered system is parabolic. 3. if .(α − δ)2 + 4βγ > 0, then the considered system is hyperbolic. Example 4.22 Consider the system .

u1x1 − u2x2 = 0 u1x2 − u2x1 = 0.

Here ⎧

⎨ 1 0 , 0 −1 ⎧ ⎨ 0 −1 B= . 1 0 A=

.

We have that detA = −1 /= 0,

.

i.e., the matrix .A is nonsingular. Also, −1

A

.

⎧ =

⎨ 1 0 , 0 −1

D = A−1 B

4.3 Classification of First Order Systems with Two Independent Variables

⎧ =

⎨ 0 −1 . −1 0

We will find the eigenvalues of the matrix .D. We have I I I −λ −1 I I det(D − λI ) = II −1 −λ I

.

= 0, whereupon λ2 − 1 = 0

.

and λ1 = 1,

.

λ2 = −1. Therefore the considered system is hyperbolic. Example 4.23 Consider the system .

u1x2 − u2x1 =0 + (ρu1 )x1 (ρu2 )x2 = 0,

where ⎧ ⎨σ ρ = 1 + u21 + u22

.

and .σ is a constant. We have ⎧ ⎨σ −1 ( ) 2u1 u1xi + 2u2 u2xi , ρxi = σ 1 + u21 + u22 ) ( ρuj x = ρxi uj + ρuj xi , i,j =1,2. .

i

Therefore the considered system we can rewrite in the form .

Here

u1x2 − u2x1 =0 ρu1x1 + ρu2x2 = −ρx1 u1 − ρx2 u2 .

127

128

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

⎧

⎨ 0 −1 .A = , ρ 0 ⎧ ⎨ 10 B= . 0ρ Since detA = ρ /= 0,

.

the matrix .A is nonsingular and ⎧ −1

=

A

.

0 ρ1 −1 0

⎨ ,

D = A−1 B ⎧ ⎨ 0 1 = . −1 0 We will find the eigenvalues of the matrix .D. We have I I I −λ 1 I I I .det(D − λI ) = I −1 −λ I = 0, whereupon λ2 + 1 = 0

.

and λ1,2 = ±i.

.

Therefore the considered system is elliptic. Exercise 4.12 Prove that the system .

is a hyperbolic system.

u1x2 − u2x1 = 0 u1x1 − 9u2x2 = 0

4.4 Advanced Practical Problems

129

4.4 Advanced Practical Problems Problem 4.1 Determine the type of the operator L, where √ √ 1. L(u) = 3ux1 x1 − √ 4 3ux1 x2 + ux2 x2 + u, 2. L(u) = −ux1 x1 − 2 2ux1 x2 + ux2 x2 + ux2 , 3. L(u) = ux1 x1 + ux2 x2 + u, 4. L(u) = ux1 x1 − ux2 x2 + ux1 , 5. L(u) = 2ux1 x2 − u √x1 − ux2 , 6. L(u) = ux1 x1 + 2 3ux1 x2 + 3ux2 x2 , 7. L(u) = ux1 − ux2 x2 + ux2 . Problem 4.2 Find the canonical form of the following equations. 1. ux1 x1 − 3ux1 x2 + 2ux2 x2 = 0,

(x1 , x2 ) ∈ R2 .

ux1 x1 − ux1 x2 − 2ux2 x2 = 0,

(x1 , x2 ) ∈ R2 ,

.

2. .

3. ux2 x2 −

.

2 x1 ux1 x2 + ux = 0, x2 3x2 2

(x1 , x2 ) ∈ R2 ,

x2 /= 0.

4. ux1 x1 − 2ux1 x2 − 3ux2 x2 + ux2 = 0,

.

(x1 , x2 ) ∈ R2 .

5. ux1 x1 − 6ux1 x2 + 10ux2 x2 + ux1 − 3ux2 = 0,

.

(x1 , x2 ) ∈ R2 .

6. 4ux1 x1 + 4ux1 x2 + ux2 x2 − 2ux2 = 0,

.

(x1 , x2 ) ∈ R2 .

7. .

ux1 x1 − x1 ux2 x2 = 0,

(x1 , x2 ) ∈ R2 .

ux1 x1 − x2 ux2 x2 = 0,

(x1 , x2 ) ∈ R2 .

8. .

130

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

9. .

x1 ux1 x1 − x2 ux2 x2 = 0,

(x1 , x2 ) ∈ R2 .

.

x2 ux1 x1 − x1 ux2 x2 = 0,

(x1 , x2 ) ∈ R2 .

.

x12 ux1 x1 + x22 ux2 x2 = 0,

(x1 , x2 ) ∈ R2 .

.

x22 ux1 x1 + x12 ux2 x2 = 0,

(x1 , x2 ) ∈ R2 .

x22 ux1 x1 − x12 ux2 x2 = 0,

(x1 , x2 ) ∈ R2 .

10.

11.

12.

13. .

14. (1 + x12 )ux1 x1 + (1 + x22 )ux2 x2 + x2 ux2 = 0,

.

(x1 , x2 ) ∈ R2 .

15. ⎨ ⎧ ux1 x1 − 2 sin x1 ux1 x2 + 2 − (cos x1 )2 ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

16. .

x22 ux1 x1 + 2x2 ux1 x2 + ux2 x2 = 0,

(x1 , x2 ) ∈ R2 .

x12 ux1 x1 − 2x1 ux1 x2 + ux2 x2 = 0,

(x1 , x2 ) ∈ R2 .

17. .

Problem 4.3 Find the general solution of the following equations. 1. ux1 x2 = 0,

.

2.

(x1 , x2 ) ∈ R2 .

4.4 Advanced Practical Problems

ux1 x1 − a 2 ux2 x2 = 0,

.

131

a ∈ R,

(x1 , x2 ) ∈ R2 .

3. ux1 x1 − 2ux1 x2 − 3ux2 x2 = 0,

(x1 , x2 ) ∈ R2 .

ux1 x2 + aux1 = 0,

(x1 , x2 ) ∈ R2 .

.

4. .

a ∈ R,

5. 3ux1 x1 − 5ux1 x2 − 2ux2 x2 + 3ux1 + ux2 = 2,

(x1 , x2 ) ∈ R2 .

ux1 x2 + aux1 + bux2 + abu = 0,

(x1 , x2 ) ∈ R2 .

.

6. .

a, b ∈ R,

7. ux1 x2 − 2ux1 − 3ux2 + 6u = 2ex1 +x2 ,

(x1 , x2 ) ∈ R2 .

.

8. ux1 x1 + 2aux1 x2 + a 2 ux2 x2 + ux1 + aux2 = 0,

.

a ∈ R, ,

(x1 , x2 ) ∈ R2 .

Problem 4.4 Find the general solution of the following equations. 1. x2 ux1 x1 + (x1 − x2 )ux1 x2 − x1 ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

2. x12 ux1 x1 − x22 ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

3. x12 ux1 x1 + 2x1 x2 ux1 x2 − 3x22 ux2 x2 − 2x1 ux1 = 0,

.

(x1 , x2 ) ∈ R2 .

4. x12 ux1 x1 + 2x1 x2 ux1 x2 + x22 ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 .

132

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

5. ux1 x2 − x1 ux1 + u = 0,

(x1 , x2 ) ∈ R2 .

.

6. ux1 x2 + 2x1 x2 ux2 − 2x1 u = 0,

(x1 , x2 ) ∈ R2 .

.

7. ux1 x2 + ux1 + x2 ux2 + (x2 − 1)u = 0,

(x1 , x2 ) ∈ R2 .

ux1 x2 + x1 ux1 + 2x2 ux2 + 2x1 x2 u = 0,

(x1 , x2 ) ∈ R2 .

.

8. .

9. ux1 x2 = h(x1 , x2 ),

.

(x1 , x2 ) ∈ X,

where X = {(x1 , x2 ) ∈ R2 : |x1 − x10 | < a,

.

|x2 − x20 | < b},

a, b, x10 .x20 ∈ R, a, b > 0, h ∈ C 2 (X). 10. ux1 x2 + h(x1 , x2 )ux1 = 0,

.

(x1 , x2 ) ∈ X,

where X = {(x1 , x2 ) ∈ R2 : |x1 − x10 | < a,

.

|x2 − x20 | < b},

a, b, x10 , x20 ∈ R, a, b > 0, h ∈ C 1 (X). 11. .

ux1 x2 −

1 1 ux + ux = 0, x1 − x2 1 x1 − x2 2

(x1 , x2 ) ∈ R2 ,

x1 /= x2 .

ux1 x2 −

m k ux1 + ux = 0, x1 − x2 x1 − x2 2

(x1 , x2 ) ∈ R2 ,

x1 /= x2 ,

12. .

m, k ∈ N.

4.4 Advanced Practical Problems

133

13. ux1 x2 +

.

m k ux − ux = 0, x1 − x2 1 x1 − x2 2

(x1 , x2 ) ∈ R2 ,

x1 /= x2 ,

m, k ∈ N. Problem 4.5 Classify the three dimensional heat equation ( ) ux1 − α 2 ux2 x2 + ux3 x3 + ux4 x4 = 0,

.

where α is a constant, α /= 0. Problem 4.6 Prove that the system .

u1x2 − u2x1 = 0 u1x1 + 2u1x2 + 4u2x2 = 0

is an elliptic system. Problem 4.7 Find the canonical form of the following equations. 1. 4ux1 x1 − 4ux1 x2 − 2ux2 x3 + ux2 + ux3 = 0,

(x1 , x2 , x3 ) ∈ R3 .

.

2. ux1 x2 − ux1 x3 + ux1 + ux2 − ux3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 .

3. ux1 x1 + 2ux1 x2 − 2ux1 x3 + 2ux2 x2 + 2ux3 x3 = 0,

(x1 , x2 , x3 ) ∈ R3 .

ux1 x1 + 2ux1 x2 − 4ux1 x3 − 6ux2 x3 − ux3 x3 = 0,

(x1 , x2 , x3 ) ∈ R3 .

.

4. .

5. ux1 x2 + ux1 x3 + ux1 x4 + ux3 x4 = 0,

.

(x1 , x2 , x3 , x4 ) ∈ R4 .

6. ux1 x1 + 2ux1 x2 − 2ux1 x3 − 4ux2 x3 + 2ux2 x4 + ux3 x3 = 0,

.

(x1 , x2 , x3 , x4 ) ∈ R4 .

134

4 Classifications and Canonical Forms for Linear Second Order Partial. . .

7. ux1 x1 + 2ux1 x3 − 2ux1 x4 + ux2 x2 + 2ux2 x3 + 2ux2 x4 + 2ux3 x3 + 2ux4 x4 = 0,

.

(x1 , x2 , x3 , x4 ) ∈ R4 . 8. n ∑ .

n ∑

kuxk xk + 2

k=1

luxl uxk = 0,

(x1 , . . . , xn ) ∈ Rn .

k,l=1,l 0,

un (x1 , x2 ) =

.

Then 1 cos(nx1 ) sinh(nx2 ), n unx1 x1 (x1 , x2 ) = − sin(nx1 ) sinh(nx2 ), unx1 (x1 , x2 ) =

.

unx2 (x1 , x2 ) =

2 Jacques Salomon Hadamard(8 December 1865–17 October 1963) was a French mathematician who made major contributions in number theory, complex function theory, differential geometry and partial differential equations.

5.1 Basic Properties of Elliptic Problems

143

whereupon Δun (x1 , x2 ) = 0,

.

x1 ∈ R,

x2 > 0,

i.e., .un is a harmonic function satisfying (5.7). When .n is large enough, the initial conditions (5.7) describe an arbitrary small perturbation of the trivial solution .u = 0. On the other hand, .supx1 ∈Rn |un (x1 , x2 )| grows exponentially fast as .n → ∞ for any .x2 > 0. Therefore the Cauchy problem for the Laplace equation is not stable and hence, it is not well-posed with respect to the initial conditions (5.7). Definition 5.5 We define a harmonic polynomial of degree .m to be a harmonic function .Pm (x1 , . . . , xn ) of the form 

Pm (x1 , . . . , xn ) =

ai1 ...in x1i1 . . . xnin ,

.

(x1 , . . . , xn ) ∈ Rn ,

0≤i1 +···+in ≤m

where .ai1 ...in are constants. Example 5.7 The polynomial P1 (x1 , . . . , xn ) = x1 + · · · + xn ,

.

(x1 , . . . , xn ) ∈ Rn ,

is a harmonic polynomial of degree .1. Example 5.8 The polynomial P3 (x1 , x2 ) = x13 − 3x1 x22 − x2 ,

.

(x1 , x2 ) ∈ R2 ,

is a harmonic polynomial of degree .3. Definition 5.6 The harmonic polynomials Pmh (x1 , . . . , xn ) =



.

ai1 ...in x1i1 . . . xnin ,

(x1 , . . . , xn ) ∈ Rn ,

i1 +···+in =m

where .ai1 ...in are constants, are called homogeneous harmonic polynomials of order m.

.

Example 5.9 The polynomial in Example 5.7 is a homogeneous harmonic polynomial of degree 1. Example 5.10 The polynomial in Example 5.8 is not a homogeneous harmonic polynomial.

144

5 The Laplace Equation

5.2 The Fundamental Solution Clearly, there are a lot of functions which satisfy the Laplace equation. In particular, any constant function is harmonic. In addition, any function u of the form u(x1 , . . . , xn ) = a1 x1 + · · · + an xn ,

.

(x1 , . . . , xn ) ∈ Rn ,

where .aj ∈ R, .j ∈ {1, . . . , n}, are constants, is also a solution to the Laplace equation. Of course, we can list a number of others. Here, we are interested in finding a particular solution of the Laplace equation which will allow us to solve the Poisson equation. Given the symmetric nature of the Laplace equation, we look for a radial solution, i.e., we look for a harmonic function u on .Rn such that u(x) = v(r),

.

x = (x1 , . . . , xn ),

where 1

r = |x| = (x12 + · · · + xn2 ) 2 .

.

In addition, to being natural choice due to the symmetry of the Laplace equation, radial solutions are natural to look because they reduce a partial differential equation to an ordinary differential equation, which is generally easier to solve. Therefore we look for a radial solution. We have uxi = v ' (r)

.

xi , r

uxi xi = v '' (r) Δu =

n  i=1

=

n  i=1

xi2 r 2 − xi2 ' + v (r) , r2 r3

i = 1, . . . , n,

uxi xi 

r 2 − xi2 x2 v (r) i2 + v ' (r) r r3 ''

= v '' (r) +

n−1 ' v (r). r

Then .Δu = 0 if and only if v '' +

.

n−1 ' v = 0. r



5.2 The Fundamental Solution

145

1. Let .n = 2. Then 1 v '' + v ' = 0, r

.

whence v(r) = a log r + b,

a, b = const,

r > 0.

a + b, r n−2

a, b = const,

r > 0.

.

2. Let .n ≥ 3. Then v(r) =

.

Definition 5.7 The function

Φ(x) =

.

1 log |x| − 2π

n=2 n ≥ 3,

1 n(n−2)κ(n)|x|n−2

defined for .x ∈ Rn , .x /= 0, is called the fundamental solution of the Laplace n π2 n

. equation. Here .κ(n) denotes the volume of the unit ball in .R , .κ(n) = n Γ 2 +1 Exercise 5.2 Prove that there exist positive constants .c1 and .c2 such that c1 , |x|n−1 c2 |Φxj xl (x)| ≤ , x ∈ Rn , |x|n |Φxj (x)| ≤

.

x /= 0,

j, l ∈ {1, . . . , n}.

.

Exercise 5.3 Prove that the function .x → Φ(x − y), .x /= y, is a harmonic function at each .y ∈ Rn . Suppose that .f ∈ C 2 (Rn ) and its support .suppf is a compact set in .Rn . Define the function  .u(x) = Φ(x − y)f (y)dy, x ∈ Rn . Rn

Note that  u(x) =

Φ(y)f (x − y)dy,

.

Rn

x ∈ Rn .

146

5 The Laplace Equation

We will show that .u ∈ C 2 (Rn ). Let ej = (0, . . . , 0, 1, 0, . . . , 0),

.

j ∈ {1, . . . , n}.

Take .h /= 0 and fix .j, l ∈ {1, . . . , n} arbitrarily. Then ⎞ ⎛   u(x + hej ) − u(x) 1⎝ Φ(y)f (x + hej − y)dy − Φ(y)f (x − y)dy ⎠ . = h h 1 h

=





Rn

Rn

Φ(y)(f (x + hej − y) − f (x − y))dy Rn

=

Φ(y) Rn

f (x + hej − y) − f (x − y) dy, h

x ∈ Rn .

Since .

lim

h→0

f (x + hej − y) − f (x − y) = fxj (x − y), h

x, y ∈ Rn ,

uniformly on .Rn , we get u(x + hej ) − u(x) h→0 h  f (x + hej − y) − f (x − y) = lim Φ(y) dy h→0 h

uxj (x) = lim

.

Rn

 =

Φ(y) lim

h→0

Rn

f (x + hej − y) − f (x − y) dy h



=

Φ(y)fxj (x − y)dy,

x ∈ Rn .

Rn

Now, using that .

lim

fxj (x + hel − y) − fxj (x − y) h

h→0

= fxj xl (x − y),

uniformly on .Rn , we obtain uxj xl (x) = lim

.

h→0

uxj (x + hel ) − uxj (x) h

x, y ∈ Rn ,

5.2 The Fundamental Solution

147

 = lim

h→0 Rn

Φ(y)

 =

Φ(y) lim

fxj (x + hel − y) − fxj (x − y) h

fxj (x + hel − y) − fxj (x − y) h

h→0

Rn

dy

dy



Φ(y)fxj xl (x − y)dy,

=

x ∈ Rn .

Rn

Thus, .u ∈ C 2 (Rn ). By the above computations, we find  Δu(x) =

Φ(y)Δx f (x − y)dy,

.

x ∈ Rn .

Rn

By the definition of .Φ, it follows that it blows up at 0. Therefore we will need for subsequent calculations to isolate this singularity inside a small ball. Take .ϵ > 0 arbitrarily. With .B(0, ϵ) we will denote the ball with center the origin and radius .ϵ. Then  .Δu(x) = Φ(y)Δx f (x − y)dy Rn

 Φ(y)Δx f (x − y)dy

= B(0,ϵ)∪(Rn \B(0,ϵ))



=



Φ(y)Δx f (x − y)dy +

Φ(y)Δx f (x − y)dy,

x ∈ Rn .

Rn \B(0,ϵ)

B(0,ϵ)

Set  I1 =

Φ(y)Δx f (x − y)dy,

.

B(0,ϵ)



I2 =

Φ(y)Δx f (x − y)dy,

x ∈ Rn .

Rn \B(0,ϵ)

Then Δu(x) = I1 + I2 ,

.

x ∈ Rn .

Now, we will estimate .I1 and .I2 . Firstly, observe that

(5.8)

148

5 The Laplace Equation

 |Φ(y)|dy =

.

B(0,ϵ)

⎧ 1 ⎪ ⎨ 2π

≤

| log |y||dy  1

B(0,ϵ) 1 n(n−2)κ(n)

⎪ ⎩



ϵ2 2 | log ϵ| ϵ2 n(n−2) n

B(0,ϵ)

n=2

|y|n−2

dy

n≥3

n=2 ≥ 3.

Hence,        |I1 | =  Φ(y)Δx f (x − y)dy  B(0,ϵ)   ≤ |Φ(y)||Δx f (x − y)|dy B(0,ϵ)  . |Φ(y)|dy ≤ n max sup |fxj xj (x)| j ∈{1,...,n} x∈Rn B(0,ϵ)

2 ϵ | log ϵ| n = 2 ≤ n max sup |fxj xj (x)| 2 ϵ 2 j ∈{1,...,n} x∈Rn n(n−2) n ≥ 3, x ∈ Rn . To estimate .I2 , we observe that an integration by parts yields

.

 I2 =

Φ(y)Δx f (x − y)dy

.

Rn \B(0,ϵ)



=

Φ(y)Δy f (x − y)dy Rn \B(0,ϵ)



=−

∇Φ(y) · ∇y f (x − y)dy

Rn \B(0,ϵ)



+

Φ(y)∂νy f (x − y)dy,

x ∈ Rn .

∂B(0,ϵ)

Let  I21 = −

∇Φ(y) · ∇y f (x − y)dy,

.

Rn \B(0,ϵ)



I22 =

Φ(y)∂νy f (x − y)dy, ∂B(0,ϵ)

Therefore

x ∈ Rn .

(5.9)

5.2 The Fundamental Solution

149

I2 = I21 + I22 ,

.

x ∈ Rn .

(5.10)

To estimate .I2 we will estimate .I21 and .I22 . For this aim, we have a need of the following estimate  .

|Φ(y)|dsy =

∂B(0,ϵ)

⎧ 1 ⎪ ⎨ 2π ⎪ ⎩

≤



∂B(0,ϵ) 1 n(n−2)κ(n)

| log |y||dsy  1 ∂B(0,ϵ)

|y|n−2

n=2 dsy

(5.11)

ϵ| log ϵ| n = 2 ϵ n−2 n ≥ 3.

Next,

∇Φ(y) =

.

− 2π 1|y|2 y

1 − nκ(n)|y| ny

n=2 n ≥ 3,

y ∈ Rn , .y /= 0. The unit outward normal vector to .∂B(0, ϵ) is

.

y |y| y =− , ϵ

ν(y) = −

.

y ∈ ∂B(0, ϵ).

Hence, ∂νy Φ(y) = ν(y) · ∇Φ(y)

1 n=2 = 2π ϵ 1 n ≥ 3, n−1 nκ(n)ϵ

.

y ∈ ∂B(0, ϵ). Now, integrating by parts and using that .Φ is a harmonic function away from the origin, for .I21 we get

.

 I21 = −

∇Φ(y) · ∇y f (x − y)dy

.

Rn \B(0,ϵ)



=

ΔΦ(y)f (x − y)dy Rn \B(0,ϵ)



− ∂B(0,ϵ)

∂νy Φ(y)f (x − y)dy

150

5 The Laplace Equation

 = −

∂νy Φ(y)f (x − y)dy

∂B(0,ϵ)

→ −f (x),

as

ϵ → 0,

x ∈ Rn ,

i.e., I21 → −f (x),

.

as

ϵ → 0,

x ∈ Rn .

(5.12)

For .I22 , applying (5.11), we have          .|I22 | =  Φ(y)∂νy f (x − y)dy    ∂B(0,ϵ)   |Φ(y)||∂νy f (x − y)|dy ≤ ∂B(0,ϵ)

 sup |fxj (x)|

≤ n max

j ∈{1,...,n} x∈Rn

|Φ(y)|dsy

∂B(0,ϵ)

≤ n max

sup |fxj (x)|

j ∈{1,...,n} x∈Rn

→ 0,

as

ϵ| log ϵ| n = 2 ϵ n−2 n ≥ 3.

ϵ → 0.

Now, using (5.10) and applying the last estimate and the estimate (5.12), we find I2 → −f (x),

.

as

ϵ → 0,

x ∈ Rn

By the last estimate and (5.8), (5.9), we arrive at the equation Δu(x) = −f (x),

.

x ∈ Rn .

Remark 5.2 We sometimes write .

− ΔΦ(x) = δ0 ,

x ∈ Rn ,

where .δ0 denotes the Dirac measure on .Rn giving unit mass to the point 0. By this notation, we find  . − Δu(x) = − Δx Φ(x − y)f (y)dy Rn

5.3 Strong Maximum Principle: Uniqueness

151

 =

δx f (y)dy Rn

= f (x),

x ∈ Rn .

Exercise 5.4 Let .u be a harmonic function in .D and continuous in .D ∪ ∂D with its partial derivatives of first order. Prove that  u(x) =

 Φ(x − y)∂νy u(y)dsy −

.

∂D

u(y)∂νy Φ(x − y)dsy ,

x ∈ D.

(5.13)

∂D

Exercise 5.5 Let .u ∈ C 2 (D) be harmonic. Prove that  1 u(x) = u(y)dsy nκ(n)r n−1 .

1 = κ(n)r n



∂B(x,r)

u(y)dy,

(5.14) x∈R , n

B(x,r)

for each ball .B(x, r) ⊂ D. Definition 5.8 The formulas (5.14) are known as mean-value formulas for harmonic functions, for a sphere and for a ball, respectively. Exercise 5.6 (Converse to Mean-Value Formula) Let .u ∈ C 2 (D) satisfies (5.14) for each ball .B(x, r) ⊂ D. Prove that .u is harmonic in .D.

5.3 Strong Maximum Principle: Uniqueness The maximum principle refers to a collection of results and techniques in the study of partial differential equations. It is also a very valuable tool for most results concerning existence, uniqueness and qualitative properties for elliptic and parabolic equations. Example 5.11 Suppose .u ∈ C 2 (D) ∩ C (D) is harmonic in .D and there exists a point .x0 ∈ D such that u(x0 ) = max u.

.

D

We will prove that u = const

.

within D

152

5 The Laplace Equation

and max u = max u.

(5.15)

.

∂D

D

Really, let .M = max u. Suppose .u(x0 ) = M for some .x0 ∈ D. Let .0 < ϵ < dist(x0 , ∂D) D

be arbitrarily chosen. Assume that there exists .y ∈ B(x0 , ϵ) so that .u(y) < M. Since .u ∈ C (D), there is a .δ > 0 so that .B(y, δ) ⊂ B(x0 , ϵ) and .u(ξ ) < M for all .ξ ∈ B(y, δ). Also, using (5.14), we have M = u(x0 )

.

=



1 κ(n)ϵ n

u(y)dy < M, B(x0 ,ϵ)

which is a contradiction. Therefore .u(y) = M for any .y ∈ B(x0 , ϵ). Let .x ∈ D be arbitrarily chosen and .l be a continuous curve lying within .D and joining the points .x and .x0 . We take .0 < ϵ1 < dist(l, ∂D). Note that if .y ∈ l we have that .u(η) = M for any .η ∈ B(y, ϵ1 ). Therefore .u(x) = M. Because .x ∈ D was arbitrarily chosen, we conclude that .u(x) = M for all .x ∈ D. If there is .x0 ∈ D so that .u(x0 ) = max u(x), then .u(x) = u(x0 ) for any .x ∈ D, D

whereupon we get (5.15). Example 5.12 Consider the Dirichlet problem .

Δu = f u =φ

in D on ∂D,

(5.16)

where .f ∈ C (D), .φ ∈ C (∂D). Let .u1 , u2 ∈ C 2 (D) ∩ C (D) be two solutions of the Dirichlet problem (5.16). Then .v = u1 − u2 ∈ C 2 (D) ∩ C (D) satisfies the Dirichlet problem .

Δv = 0 in D v = 0 on ∂D.

Hence, applying Example 5.11, we conclude that .v = 0 in .D, i.e., .u1 = u2 in .D. Thus, the considered Dirichlet problem has at most one solution in .C 2 (D) ∩ C (D). Exercise 5.7 Let D = {(x1 , . . . , xn ) ∈ Rn : 2x12 +

n 

.

j =2

xj2 < 3}.

5.4 The Green Function of the Dirichlet Problem

153

Prove that the problem Δu =

n 

.

xj2

on

D

xj

on

∂D

j =1

u=

n  j =1

has at most one solution in .C 2 (D)



C (D).

Exercise 5.8 Let .f1 , f2 ∈ C (D) and .φ1 , φ2 ∈ C (∂D). Let also, .u1 be a solution to the Dirichlet problem Δu = f1

.

u = φ1

in D on

∂D,

and .u2 be a solution to the Dirichlet problem Δu = f2

.

u = φ2

in D on

∂D.

Prove that |u1 (x) − u2 (x)| ≤ max |φ1 (x) − φ2 (x)|,

.

x∈∂D

x ∈ D.

5.4 The Green Function of the Dirichlet Problem In physics and mathematics, the Green functions are auxiliary functions in the solution of linear partial differential equations. The Green function is named for the self-taught English mathematician George Green, who investigated electricity and magnetism in a thoroughly mathematical fashion. In 1828 Green published a privately printed booklet, introducing what is now called the Green function. This was ignored until William Thomson discovered it, recognized its great value and had it published nine years after the Green death. The Green functions technique is a method to solve a nonhomogeneous differential equation. The essence of the method consists in finding an integral operator which produces a solution satisfying all boundary conditions. The Green function is the kernel of the integral operator inverse to the differential operator generated by the given differential equation and the homogeneous boundary conditions. It reduces the study of the properties of the differential operator to the study of similar

154

5 The Laplace Equation

properties of the corresponding integral operator. The integral operator has a kernel called the Green function, usually denoted by .G(x, y). This is multiplied by the nonhomogeneous term and integrated by one of the variables. Definition 5.9 The Green3 function of the Dirichlet problem for the Laplace equation in a domain .D is a function .G(x, y) depending on two points .x, y ∈ D which possesses the following properties. 1. .G(x, y) has the form G(x, y) = Φ(x − y) + g(x, y),

.

where .g(x, y) is a harmonic function both with respect to .x, y ∈ D. 2. When .x ∈ ∂D or .y ∈ ∂D, the equality .G(x, y) = 0 is fulfilled.

We will show that the Green function is nonnegative in .D, i.e., we will show G(x, y) ≥ 0,

.

(x, y) ∈ D.

Let .y ∈ D be arbitrarily chosen. Let also, .δ > 0 be sufficiently small so that B(y, δ) ⊂ D.

.

Denote Dδ = D\B(y, δ).

.

Since .

lim G(x, y) = ∞,

x→y

then we must have, for sufficiently small .δ > 0, G(x, y) > 0,

x ∈ B(y, δ).

G(x, y) ≥ 0,

(x, y) ∈ ∂Dδ .

.

Therefore .

3 George Green(14 July 1793–31 May 1841) was a British mathematical physicist who introduced several important concepts, among them a theorem similar to the modern Green theorem, the idea of potential functions and the concept of what are now called the Green functions.

5.4 The Green Function of the Dirichlet Problem

155

Hence and the maximum principle, we conclude that G(x, y) ≥ 0

.

for all

x ∈ Dδ .

Consequently G(x, y) ≥ 0,

.

(x, y) ∈ D.

One of the next properties of the Green function is the symmetry property of the Green function , i.e., we have G(x, y) = G(y, x)

.

for any

x, y ∈ D.

To see this, let .x, y ∈ D, .x /= y, be arbitrarily chosen and fixed. Take .ϵ > 0 small enough so that .B(x, ϵ) ⊂ D, .B(y, ϵ) ⊂ D, and .B(x, ϵ) ∩ B(y, ϵ) = Ø. Denote Dϵ = D\ (B(x, ϵ) ∪ B(y, ϵ)) .

.

Note that .G(z, y) is harmonic in .D\B(y, ϵ) and .G(z, x) is harmonic in .D\B(x, ϵ). Applying (5.5) to the domain .Dϵ for .G(z, x) and .G(z, y), we get 

G(z, y)Gνz (z, x) − G(z, x)Gνz (z, y) dsz

0=

.

∂Dϵ





G(z, y)Gνz (z, x) − G(z, x)Gνz (z, y) dsz

= ∂D





G(z, y)Gνz (z, x) − G(z, x)Gνz (z, y) dsz

− ∂B(y,ϵ)



G(z, y)Gνz (z, x) − G(z, x)Gνz (z, y) dsz ,

− ∂B(x,ϵ)

whereupon 



G(z, y)Gνz (z, x) − G(z, x)Gνz (z, y) dsz

∂D .



= ∂B(y,ϵ) 

+ ∂B(x,ϵ)



G(z, y)Gνz (z, x) − G(z, x)Gνz (z, y) dsz

G(z, y)Gνz (z, x) − G(z, x)Gνz (z, y) dsz .

156

5 The Laplace Equation

Because G(z, y) = G(z, x) = 0 for

.

z ∈ ∂D,

we get 



G(z, y)Gνz (z, x) − G(z, x)Gνz (z, y) dsz

∂B(x,ϵ) 



−G(z, y)Gνz (z, x) + G(z, x)Gνz (z, y) dsz .

.

=

(5.17)

∂B(y,ϵ)

Note that 



G(z, y)Gνz (z, x) − G(z, x)Gνz (z, y) dsz

∂B(x,ϵ) 



=

  Φ(z − y) + g(z, y) Φνz (z − x) + gνz (z, x)

 ∂B(x,ϵ)  − Φ(z − x) + g(z, x) Φνz (z − y) + gνz (z, y) dsz 

= Φ(z − y)Φνz (z − x) − Φ(z − x)Φνz (z − y) dsz .

∂B(x,ϵ) 

+ ∂B(x,ϵ) 

+ ∂B(x,ϵ) 

+

(5.18)

g(z, y)Φνz (z − x) − gνz (z, y)Φ(z − x) dsz

Φ(z − y)gνz (z, x) − g(z, x)Φνz (z − y) dsz

g(z, y)gνz (z, x) − g(z, x)gνz (z, y) dsz .

∂B(x,ϵ)

Since .g(z, y) and .g(z, x) are harmonic with respect to .z in .B(x, ϵ), using Example 5.5, we get 



g(z, y)gνz (z, x) − g(z, x)gνz (z, y) dsz = 0.

.

(5.19)

∂B(x,ϵ)

Because .Φ(z − y) and .g(z, x) are harmonic with respect to .z in .B(x, ϵ), using Example 5.5, we obtain  .

∂B(x,ϵ)



Φ(z − y)gνz (z, x) − g(z, x)Φνz (z − y) dsz = 0.

(5.20)

5.4 The Green Function of the Dirichlet Problem

157

Since .Φ(z − y) is harmonic with respect to .z in .B(x, ϵ), using Exercise 5.4, we have 

Φ(z − y)Φνz (z − x) − Φ(z − x)Φνz (z − y) dsz = Φ(x − y). . (5.21) ∂B(x,ϵ)

Because .g(z, y) is harmonic with respect to .z in .B(x, ϵ), using Exercise 5.4, we obtain 

g(z, y)Φνz (z − x) − gνz (z, y)Φ(z − x) dsz = g(x, y). . (5.22) ∂B(x,ϵ)

From (5.18), (5.19), (5.20), (5.21) and (5.22), we obtain 

G(z, y)Gνz (z, x) − G(z, x)Gνz (z, y) dsz . ∂B(x,ϵ)

−→ Φ(x − y) + g(x, y) = G(x, y)

(5.23)

as .ϵ → 0. Similarly, 



.

G(z, x)Gνz (z, y) − G(z, y)Gνz (z, x) dsz −→ G(y, x)

(5.24)

∂B(y,ϵ)

as .ϵ → 0. From (5.17), (5.23) and (5.24), letting .ϵ → 0, we get G(x, y) = G(y, x).

.

Now, we suppose that .u is harmonic in .D and satisfies the boundary condition u=φ

.

on

∂D.

Then, by Exercise 5.4, we get  u(x) =

 Φ(x − y)∂νy u(y)dsy −

.

∂D

u(y)∂νy Φ(x − y)dsy .

∂D

Hence, using Example 5.5, Φ(x − y) = G(x, y) − g(x, y)

.

and the boundary condition (5.25), we obtain

(5.25)

158

5 The Laplace Equation





u(x) =

(G(x, y) − g(x, y)) ∂νy u(y)dsy −

.

∂D



∂D



=

G(x, y)∂νy u(y)dsy − ∂D



+



u(y) Gνy (x, y) − gνy (x, y) dsy

u(y)Gνy (x, y)dsy

∂D



u(y)gνy (x, y) − g(x, y)uνy (y) dsy

∂D



=−

u(y)Gνy (x, y)dsy

∂D



=−

φ(y)Gνy (x, y)dsy ,

∂D

i.e.,  u(x) = −

.

φ(y)Gνy (x, y)dsy ,

x ∈ D.

(5.26)

∂D

The harmonicity of the function .u expressed by the formula (5.26) follows from the fact that the Green function .G(x, y) is harmonic with respect to .x for .x /= y. The fact that this function satisfies the boundary condition (5.25) requires special proof. Let .D be the ball .B(0, 1) and let .x, y be two interior points of that ball. Example 5.13 We will prove that the function   x , .G(x, y) = Φ(x − y) − Φ |x|y − |x| is the Green function for the ball .D. Here   x , .g(x, y) = −Φ |x|y − |x| Note that if y=

.

x , |x|2

then |y| =

.

1 > 1, |x|

x, y ∈ D,

x, y ∈ D,

5.4 The Green Function of the Dirichlet Problem

159

i.e., |x|y −

.

x /= 0 |x|

for any .x, y ∈ B(0, 1). Therefore .g(x, y) is harmonic with respect to .x and .y in B(0, 1). If .|y| = 1, then

.

 1 2 |y − x| = |x|2 − 2xy + 1    y   = |y|x − |y|     x   , = |x|y − |x| 

.

  x i.e., .Φ(x − y) = Φ |x|y − and .G(x, y) = 0. |x| Similarly, if .|x| = 1, then .G(x, y) = 0. For .|y| = 1, we have ⎛

 ⎞ xi y |x|y − i i |x| ⎟ 1 ⎜ yi (yi − xi ) n ⎠ .Gνy (x, y) = − − |x|  ⎝ n  nκ(n) |y − x| x  |x|y − |x|  i=1 ⎞  ⎛ xi n |x|y y −  i i |x| 1 ⎝ yi (yi − xi ) − |x| ⎠ =− n n nκ(n) |y − x| |y − x| n 

i=1

1  yi2 − xi yi − yi2 |x|2 + xi yi nκ(n) |y − x|n n

=−

i=1

1  yi2 (1 − |x|2 ) nκ(n) |y − x|n n

=−

i=1

=− =

1 |y|2 (1 − |x|2 ) nκ(n) |y − x|n

1 |x|2 − 1 . nκ(n) |y − x|n

Then, applying (5.26), we obtain

160

5 The Laplace Equation

u(x) =

.

1 nκ(n)

 |y|=1

1 − |x|2 φ(y)dsy , |y − x|n

(5.27)

where .φ ∈ C (∂B(0, 1)). Definition 5.10 The formula (5.27) is known as the Poisson formula. The Poisson formula for the ball .B(x0 , r) is u(x) =

.

1 rnκ(n)

 |y−x0 |=r

r 2 − |x − x0 |2 φ(y)dsy , |y − x|n

where .φ ∈ C (∂B(x0 , r)). Exercise 5.9 Prove that .

1 nκ(n)



1 − |x|2 dsy = 1 for |y − x|n

∂B(0,1)

x ∈ B(0, 1).

Exercise 5.10 Let .x0 ∈ ∂B(0, 1) be arbitrarily chosen and fixed. Prove that .

u(x) = φ(x0 ), lim x → x0 x ∈ B(0, 1)

where .u is defined by (5.27). Exercise 5.11 Let D = {x ∈ Rn : xn > 0}.

.

1. Prove that G(x, y) = Φ(x − y) − Φ(x − y ' ),

.

where .x, y ∈ D, .y = (y1 , . . . , yn−1 , yn ), .y ' = (y1 , . . . , yn−1 , −yn ), is the Green function. 2. Let .φ ∈ C (∂D) ∩ L∞ (∂D). Prove that .

 2xn φ(y) ds = φ(x0 ), lim n y x → x0 nκ(n) ∂D |x − y| x∈D

x0 ∈ ∂D.

5.5 Separation of Variables

161

Exercise 5.12 Let .u be harmonic throughout the space .Rn and .u ≥ (≤)0 in .Rn . Then .u is identically equal to a constant in .Rn . Exercise 5.13 (The Liouville4 Theorem) Let .u be harmonic throughout .Rn and it is bounded above(below) in .Rn . Then .u is identically equal to a constant in .Rn . Exercise 5.14 Prove that the Dirichlet problem for the half space .xn > 0 cannot have more than one solution in the class of bounded functions. Exercise 5.15 (The Harnack5 Theorem) Let .um , .m ∈ N, be harmonic functions ∞  um (x) is uniformly in a domain .D which are continuous in .D and the series . m=1

convergent on the boundary .∂D. Then this series is uniformly convergent in .D and its sum u(x) =

∞ 

.

um (x)

m=1

is a harmonic function in .D.

5.5 Separation of Variables The method of separation of variables combined with the principle of superposition is widely used to solve initial boundary value problems and boundary value problems involving linear partial differential equations. Usually, the dependent variable .u(x1 , x2 ) is expressed in the separable form u(x1 , x2 ) = X(x1 )Y (x2 ),

.

where X and Y are functions of .x1 and .x2 , respectively. In many cases, the partial differential equation reduces to two ordinary differential equations for X and Y . A similar treatment can be applied to equations in three or more ordinary differential equations is by no means a trivial one. In spite of this question, the method is widely used in finding of solutions of large classes initial boundary value problems and boundary value problems. This method of solution is also known as the Fourier

4 Joseph Liouville(24 March 1809–8 September 1882) was a French mathematician who worked in a number of different fields in mathematics including number theory, complex analysis, differential geometry, topology, mathematical physics and astronomy. 5 Carl Gustav Axel von Harnack(7 May 1851–3 April 1888) was a German mathematician who contributed to potential theory Harnack’s inequality applied to harmonic functions. He also worked on the real algebraic geometry of plane curves, proving Harnack’s curve theorem for real plane algebraic curves.

162

5 The Laplace Equation

method(or the method of eigenfunction expansion). Thus, the procedure outlined above leads to the important ideas of eigenvalues, eigenfunctions and orthogonality, all of which are very general and powerful for dealing with linear problems. In this section, we introduce the method of separation of variables for the Dirichlet problem for the Laplace equation in the case when D is a rectangle and in the case when D is a circular domain.

5.5.1 Rectangles In this section, we will apply the method of separation of variables in the case when D is a rectangle. Let .u be the solution to the Dirichlet problem in a rectangular domain Δu = 0,

0 < x1 < a,

.

0 < x2 < b,

(5.28)

with the boundary conditions .

u(0, x2 ) = φ1 (x2 ), u(x1 , 0) = ψ1 (x1 ),

u(a, x2 ) = φ2 (x2 ), 0 ≤ x2 ≤ b, u(x1 , b) = ψ2 (x1 ), 0 ≤ x1 ≤ a,

(5.29)

where .φ1 , φ2 ∈ C ([0, b]), .ψ1 , ψ2 ∈ C ([0, a]), and .

φ1 (0) = ψ1 (0), φ2 (0) = ψ1 (a),

φ1 (b) = ψ2 (0), φ2 (b) = ψ2 (a).

We split .u into the form .u = u1 + u2 , where .u1 solves

.

Δu1 = 0, 0 < x1 < a, 0 < x2 < b, u1 (0, x2 ) = φ1 (x2 ), u1 (a, x2 ) = φ2 (x2 ), u1 (x1 , 0) = u1 (x1 , b) = 0, 0 ≤ x1 ≤ a,

0 ≤ x2 ≤ b,

(5.30)

and .u2 satisfies

.

Δu2 = 0, 0 < x1 < a, 0 < x2 < b, u2 (0, x2 ) = u2 (a, x2 ) = 0, 0 ≤ x2 ≤ b, u2 (x1 , 0) = ψ1 (x1 ), u2 (x1 , b) = ψ2 (x1 ),

(5.31) 0 ≤ x1 ≤ a,

under the compatibility condition φ1 (0) = ψ1 (0) = φ1 (b) = ψ2 (0) = φ2 (0) = ψ1 (a) = φ2 (b) = ψ2 (a) = 0.

.

We will seek a solution .u1 of the problem (5.30) in the form

5.5 Separation of Variables

163

u1 (x1 , x2 ) = X1 (x1 )Y 1 (x2 ),

(x1 , x2 ) ∈ [0, a] × [0, b].

.

We have '

u1x1 (x1 , x2 ) = X1 (x1 )Y 1 (x2 ),

.

''

u1x1 x1 (x1 , x2 ) = X1 (x1 )Y 1 (x2 ), '

u1x2 (x1 , x2 ) = X1 (x1 )Y 1 (x2 ), ''

u1x2 x2 (x1 , x2 ) = X1 (x1 )Y 1 (x2 ),

(x1 , x2 ) ∈ [0, a] × [0, b].

Substituting the second partial derivatives of .u1 with respect to .x1 and .x2 into the Laplace equation, we get ''

''

X1 (x1 )Y 1 (x2 ) + X1 (x1 )Y 1 (x2 ) = 0,

.

(x1 , x2 ) ∈ [0, a] × [0, b],

whereupon ''

.

''

Y 1 (x2 ) X1 (x1 ) = − , X1 (x1 ) Y 1 (x2 )

(x1 , x2 ) ∈ [0, a] × [0, b].

Since the left hand side of the last equality depends on .x1 , .x1 ∈ [0, a], and the right hand side of the last equality depends on .x2 , .x2 ∈ [0, b], we conclude that there exists a constant .λ such that ''

.

''

Y 1 (x2 ) X1 (x1 ) =λ=− 1 , 1 X (x1 ) Y (x2 )

(x1 , x2 ) ∈ [0, a] × [0, b].

Thus, we obtain the following equations ''

X1 (x1 ) − λX1 (x1 ) = 0,

.

''

Y 1 (x2 ) + λY 1 (x2 ) = 0,

.

0 < x1 < a,

(5.32)

0 < x2 < b.

(5.33)

By the boundary conditions u1 (x1 , 0) = u1 (x1 , b) = 0,

.

x1 ∈ [0, a],

we find X1 (x1 )Y 1 (0) = X1 (x1 )Y 1 (b) = 0,

.

x1 ∈ [0, a].

164

5 The Laplace Equation

Because we look for a nontrivial solution of the problem (5.30), we get Y 1 (0) = Y 1 (b) = 0.

.

Thus, using (5.33), we obtain the Sturm6 -Liouville problem for .Y 1 .

''

Y 1 (x2 ) + λY 1 (x2 ) = 0,

0 < x2 < b,

Y 1 (0) = Y 1 (b) = 0.

(5.34) (5.35)

.

Consider the Eq. (5.34). Its characteristic equation is r 2 + λ = 0.

.

For the roots of the characteristic equation we have the following cases. 1. Let .λ < 0. Then √ r1,2 = ± −λ

.

and the general solution of the Eq. (5.34) is given by Y 1 (x2 ) = c1 e

.

√

−λx2

√

+ c2 e−

−λx2

,

x2 ∈ [0, b].

2. Let .λ = 0. Then r1,2 = 0

.

and the general solution of the Eq. (5.34) is given by Y 1 (x2 ) = c1 x2 + c2 ,

.

x2 ∈ [0, b].

3. Let .λ > 0. Then √ r1,2 = ±i λ

.

and the general solution of the Eq. (5.34) is given by √ √ Y 1 (x2 ) = c1 cos( λx2 ) + c2 sin( λx2 ),

.

x2 ∈ [0, b].

6 Jacques Charles Francois Sturm(29 September 1803–15 December 1855) was a French mathematician who discovered the theorem which bears his name and which concerns the determination of the number and the localization of the real roots of a polynomial equation included between given limits.

5.5 Separation of Variables

165

Here .c1 and .c2 are constants which will be determined using the boundary conditions (5.35). For the constants .c1 and .c2 we have the following cases. 1. Let .λ < 0. Then, using (5.35), we find 0 = Y 1 (0)

.

= c1 + c2 , 0 = Y 1 (b) = c1 e

√

−λb

√

+ c2 e−

−λb

,

or we get the following system c1 + c2 = 0

.

√

c1 e

−λb

√

+ c2 e−

−λb

= 0.

Because .b /= 0, the solution of the last system is c1 = 0

.

c2 = 0. Then Y 1 (x2 ) = 0,

x2 ∈ [0, b],

.

and hence, u1 (x1 , x2 ) = 0,

.

(x1 , x2 ) ∈ [0, a] × [0, b].

Since we look for a nontrivial solution to the problem (5.30), this is not our case. 2. Let .λ = 0. Then, using (5.35), we find 0 = Y 1 (0)

.

= c2 , 0 = Y 1 (b) = c1 b + c2 , whereupon we get the system c2 = 0

.

c1 b + c2 = 0.

166

5 The Laplace Equation

Since .b /= 0, the solution of the last system is the trivial solution and then u1 (x1 , x2 ) = 0,

(x1 , x2 ) ∈ [0, a] × [0, b],

.

which is again not our case. 3. Let .λ > 0. Then, using (5.35), we find 0 = Y 1 (0)

.

= c1 , 0 = Y 1 (b)

√ √ = c1 cos( λb) + c2 sin( λb) √ = c2 sin( λb),

or we get the system .c1 = 0 √ c2 sin( λb) = 0.

Observe that when √ λb = nπ,

n ∈ N0 ,

n2 π 2 , b2

n ∈ N0 ,

.

or λ=

.

the last system has infinitely many nontrivial solutions .(0, c2 ), .c2 ∈ R, .c2 /= 0. The associated function .Y 1 is given by Y 1 (x2 ) = sin

 nπ

.

b

 x2 ,

x2 ∈ [0, b].

Hence, the solution of the eigenvalue problem (5.34) and (5.35) is an infinite sequence of nonnegative eigenvalues and the associated eigenfunctions are given by Yn1 (x2 ) = sin

.

λn =

 nπ b

n2 π 2 , b2

 x2 ,

x2 ∈ [0, b],

n ∈ N0 .

5.5 Separation of Variables

167

Consider the Eq. (5.32) for .λ = λn , .n ∈ N0 , i.e., consider the equation ''

X1 (x1 ) −

.

n2 π 2 1 X (x1 ) = 0, b2

x1 ∈ [0, a].

For its general solution we have the following representation X1 (x1 ) = αn e

.

nπ b x1

+ βn e−

nπ b x1

x1 ∈ [0, a],

,

n ∈ N0 ,

where .αn , βn , .n ∈ N0 , are constants. The product solution of the problem (5.30) is as follows u1 (x1 , x2 ) =

.

∞    nπ   nπ nπ x2 , αn e b x1 + βn e− b x1 sin b

(x1 , x2 ) ∈ [0, a] × [0, b]

n=0

To find the constants .αn and .βn , we will use the boundary conditions u1n (0, x2 ) = φ1 (x2 ),

.

u1n (a, x2 ) = φ2 (x2 ),

x2 ∈ [0, b].

In fact, we have φ1 (x2 ) = u1l (0, x2 )

.

=

∞  nπ   x2 , (αn + βn ) sin b

x2 ∈ [0, b],

n=1

whereupon b φ1 (x2 ) sin

.

 nπ b

 x2

b   nπ 2 x2 dx2 = (αn + βn ) sin dx2 b

0

0

b = (αn + βn )

1 − cos 2 nπ b x2 dx2 2

0

⎛ 1 = (αn + βn ) ⎝ 2 

b 0

1 dx2 − 2

b

⎞  nπ  cos 2 x2 dx2 ⎠ b

0

 nπ  x2 =b b b = (αn + βn ) − sin 2 x2  2 4nπ b x2 =0



168

5 The Laplace Equation

b = (αn + βn ) , 2

n ∈ N0 ,

b

 x2 dx2 ,

or 2 .αn + βn = b

φ1 (x2 ) sin

 nπ b

n ∈ N0 .

0

Next, φ2 (x2 ) = u1l (a, x2 )

.

=

∞  

αn e

nπ b

a

+ βn e−

nπ b

a

 sin

 nπ b

n=1

 x2 ,

x2 ∈ [0, b].

Hence, b φ2 (x2 ) sin

.

 nπ b





x2 dx2 = αn e

nπ b

a

+ βn e

− nπ b a

 b  sin

0

 nπ b

0



= αn e

nπ b

a

+ βn e−

nπ b

a



  nπ nπ = αn e b a + βn e− b a ⎛ ⎝1 2

b dx2 − 0



1 2

b

b

1 − cos

2 x2

dx2





2nπ b x2

2

0

⎞  nπ  cos 2 x2 dx2 ⎠ b

0

 a = αn e a + βn e   nπ  x2 =b b b  − sin 2 x2  2 4nπ b x2 =0 =

nπ b

− nπ b

 nπ nπ b αn e b a + βn e− b a , 2

n ∈ N,

or

αn e

.

nπ b

a

+ βn e

− nπ b a

2 = b

b φ2 (x2 ) sin 0

Set

 nπ b

 x2 dx2 ,

n ∈ N.

dx2

5.5 Separation of Variables

169

2 .An = b

b φ1 (x2 ) sin

 nπ b

 x2 dx2 ,

0

2 Bn = b

b φ2 (x2 ) sin

 nπ b

 x2 dx2 ,

n ∈ N.

0

Thus, we get the system .

αn + βn = An nπ nπ αn e b a + βn e− b a = Bn ,

n ∈ N.

Now, we multiply the first equation of (5.36) by .e− αn e−

.

αn e

nπ b

nπ b

a

+ βn e−

nπ b

a

= An e−

nπ b

a

+ βn e−

nπ b

a

= Bn ,

a

nπ b

, .n ∈ N, and we find a

n ∈ N,

whereupon   nπ nπ nπ αn e b a − e− b a = Bn − An e− b a ,

n ∈ N,

.

or 2αn sinh

.

 nπ  nπ a = Bn − An e− b a , b

n ∈ N,

or nπ

Bn − An e− b a

, .αn = 2 sinh nπ b a

n ∈ N.

From here, βn = An − αn

.

nπ

Bn − An e− b a

= An − 2 sinh nπ b a

nπ 2An sinh nπ Bn + An e− b a b a −

= 2 sinh nπ b a  nπ  nπ nπ An e b a − e− b a − Bn + An e− b a

= 2 sinh nπ b a

(5.36)

170

5 The Laplace Equation nπ

An e b a − Bn

, 2 sinh nπ b a

=

n ∈ N.

Therefore  nπ nπ ∞  nπ   Bn − An e− b a nπ x1 An e b a − Bn − nπ x1 nπ e b + nπ e b .u1 (x1 , x2 ) = x2 , sin b 2 sinh b a 2 sinh b a n=1 (x1 , x2 ) ∈ [0, a] × [0, b], which is a formal solution to the problem (5.30). Now we seek a solution .u2 of the problem (5.31) in the form

.

u2 (x1 , x2 ) = X2 (x1 )Y 2 (x2 ),

0 < x1 < a,

.

0 < x2 < b.

Substituting it into the Laplace equation, we get ''

X2 (x1 ) − λX2 (x1 ) = 0,

.

''

Y 2 (x2 ) + λY 2 (x2 ) = 0,

.

0 < x1 < a,

(5.37)

0 < x2 < b.

(5.38)

As above, the solution of the eigenvalue problem (5.37) is an infinite sequence of nonpositive eigenvalues and their associated eigenfunctions are given by Xn2 (x1 ) = sin

 nπ

.

λn = −

a

 x1 ,

n2 π 2 , a2

x1 ∈ [0, a],

n ∈ N0 . 2 2

As above, for the solution of the Eq. (5.38) for .λ = − naπ2 , .n ∈ N0 , using the boundary conditions u2 (x1 , 0) = ψ1 (x1 ),

.

u2 (x1 , b) = ψ2 (x1 ),

x1 ∈ [0, a],

we have the representation Y 2 (x2 ) = γn e

.

nπ a x2

+ δn e−

nπ a x2

where nπ

γn =

.

D n − Cn e − a b

, 2 sinh nπ a b nπ

Cn e a b − Dn

, δn = 2 sinh nπ a b

,

x2 ∈ [0, b],

n ∈ N0 ,

5.5 Separation of Variables

2 Cn = a

171

a ψ1 (x1 ) sin

 nπ a

 x1 dx1 ,

0

2 Dn = a

a ψ2 (x1 ) sin

 nπ a

 x1 dx1 ,

n ∈ N0 .

0

As above, the formal solution to the problem (5.31), we have the representation  nπ nπ ∞  nπ   Dn − Cn e− a b nπ x2 Cn e a b − Dn − nπ x2 nπ e a nπ e a + .u2 (x1 , x2 ) = x1 , sin a 2 sinh a b 2 sinh a b n=1 (x1 , x2 ) ∈ [0, a] × [0, b].

.

Example 5.14 Consider the problem .

Δu = 0 in 0 < x1 , x2 < π, u(0, x2 ) = x2 (x2 − π ), u(π, x2 ) = u(x1 , 0) = u(x1 , π ) = 0,

0 ≤ x1 , x2 ≤ π.

We seek a nontrivial formal solution .

u(x1 , x2 ) = X(x1 )Y (x2 ),

(x1 , x2 ) ∈ [0, π ] × [0, π ].

X'' (x1 ) Y '' (x2 ) = λ, =− Y (x2 ) X(x1 )

(x1 , x2 ) ∈ [0, π ] × [0, π ],

Then .

where .λ is a constant. Since .u is nontrivial and u(x1 , 0) = u(x1 , π ) = 0,

.

x1 ∈ [0, π ],

we obtain Y (0) = Y (π ) = 0.

.

Thus, we get the Sturm-Liouville problem .

Y '' (x2 ) + λY (x2 ) = 0, Y (0) = Y (π ) = 0.

0 < x2 < π,

Hence, Yn (x2 ) = sin(nx2 ),

.

x2 ∈ [0, π ],

172

5 The Laplace Equation

λ n = n2 ,

n ∈ N0 ,

and Xn'' (x1 ) − n2 Xn (x1 ) = 0,

0 < x1 < π,

.

n ∈ N0 .

Therefore Xn (x1 ) = An enx1 + Bn e−nx1 ,

.

x1 ∈ [0, π ],

n ∈ N0 ,

and u(x1 , x2 ) =

∞ 

An enx1 + Bn e−nx1 sin(nx2 ),

.

(x1 , x2 ) ∈ [0, π ] × [0, π ],

n=1

where .An and .Bn are constants which we will determine using the boundary conditions u(0, x2 ) = x2 (x2 − π ),

.

u(π, x2 ) = 0,

x2 ∈ [0, π ],

We get ∞  (An + Bn ) sin(nx2 ) = x2 (x2 − π ),

u(0, x2 ) =

.

x2 ∈ [0, π ],

n=1

whereupon multiplying by .sin(nx2 ), .n ∈ N, the last equality and integrating over [0, π ], we obtain

.

π (An + Bn )

π (sin(nx2 )) dx2 =

sin(nx2 )x2 (x2 − π )dx2 ,

2

.

0

n ∈ N,

0

or π (An + Bn )

.

0

x2 =π  1 − cos(2nx2 ) 2 , dx2 = 3 cos(nx2 ) 2 n x2 =0

n ∈ N,

or ⎛ (An + Bn ) ⎝

.

1 2

π dx2 − 0

1 2

π 0

⎞ cos(2nx2 )dx2 ⎠ =

2 ((−1)n − 1), n3

n ∈ N,

5.5 Separation of Variables

173

or 

x2 =π  π 2 1 .(An + Bn ) = 3 ((−1)n − 1), − sin(2nx2 ) 2 4n n x2 =0

n ∈ N,

or (An + Bn )

.

2 π = 3 ((−1)n − 1), 2 n

n ∈ N,

or An + Bn =

.

4 (−1)n − 1 ,

n3 π

n ∈ N.

Also, u(π, x2 ) =

.

∞ 

An enπ + Bn e−nπ sin(nx2 ) n=1

= 0,

x2 ∈ [0, π ],

whence An enπ + Bn e−nπ = 0,

.

n ∈ N.

From the last equality and from (5.39), we go to the system

.

An + Bn =

4 (−1)n − 1

n3 π −nπ

An enπ + Bn e

= 0,

n ∈ N.

For its solution we have 2e−nπ ((−1)n − 1) 3 π sinh(nπ ) n . 2enπ ((−1)n − 1), Bn = − 3 n π sinh(nπ ) An = −

n ∈ N.

Therefore An enx1 + Bn e−nx1 =

.

4 sinh(n(π − x1 )) ((−1)n − 1), n3 π sinh(nπ )

x1 ∈ [0, π ], n ∈ N, and

(5.39)

174

5 The Laplace Equation ∞  8 sinh((2n + 1)(π − x1 )) sin((2n + 1)x2 ), n3 π sinh((2n + 1)π )

u(x1 , x2 ) = −

.

n=0

(x1 , x2 ) ∈ [0, π ] × [0, π ]. Exercise 5.16 Find a formal solution to the problem

.

Δu = 0, 0 < x1 , x2 < 1, u(0, x2 ) = x2 (1 − x2 ), u(1, x2 ) = 0, 0 ≤ x2 ≤ 1, u(x1 , 0) = sin(π x1 ), u(x1 , 1) = 0, 0 ≤ x1 ≤ 1.

Exercise 5.17 Find a formal solution to the Neumann boundary value problem Δu = 0,

0 < x1 < a,

.

ux1 (0, x2 ) = 0,

0 < x2 < b,

ux1 (a, x2 ) = f (x2 ),

ux2 (x1 , 0) = ux2 (x1 , b) = 0,

x2 ∈ [0, b],

x1 ∈ [0, a],

where .f ∈ C ([0, b]) is such that b f (x2 )dx2 = 0.

.

0

Exercise 5.18 Find a formal solution to the following Neumann boundary value problem Δu = 0,

.

0 < x1 < 1,

ux2 (x1 , 0) = ux2 (x1 , 1) = 0, ux1 (0, x2 ) = 0,

0 < x2 < 1, x1 ∈ [0, 1],

x2 ∈ [0, 1],

1 ux1 (1, x2 ) = x2 − , 2

x2 ∈ [0, 1].

Exercise 5.19 Find a formal solution to the following mixed problem Δu = 0,

.

0 < x1 < a,

0 < x2 < b,

ux1 (0, x2 ) = ux1 (a, x2 ) = u(x1 , 0) = 0, u(x1 , b) = f (x1 ), where .f ∈ C ([0, a]).

x1 ∈ [0, a],

(x1 , x2 ) ∈ [0, a] × [0, b],

5.5 Separation of Variables

175

5.5.2 Circular Domains In this section, we will apply the method of separation of variables in the case when D is a circular domain. Consider the Dirichlet problem .

Δu = 0, (x1 , x2 ) ∈ B(0, a), u(x1 , x2 ) = φ(x1 , x2 ), (x1 , x2 ) ∈ ∂B(0, a).

(5.40)

Introduce polar coordinates x1 = r cos θ,

.

x2 = r sin θ,

r ≥ 0,

θ ∈ [0, 2π ].

Then B(0, a)   . φ(x1 , x2 )

= {(r, θ ) : 0 ≤ r ≤ a, ∂B(0,a)

0 ≤ θ ≤ 2π} ,

= φ(a cos θ, a sin θ ) = h(θ ),

θ ∈ [0, 2π ].

Thus, the problem (5.40) takes the form 1 1 urr + ur + 2 uθθ = 0 in B(0, a) r r . u(a, θ ) = h(θ ), lim u(r, θ ) exists and it is finite.

(5.41)

r→0

We will search a formal solution of the problem (5.41) in the form u(r, θ ) = R(r)Θ(θ ),

.

r ∈ [0, a],

θ ∈ [0, 2π ].

Substituting this function in (5.41), we find r 2 R '' (r) + rR ' (r) − λR(r) = 0, 0 < r < a, Θ '' (θ ) + λΘ(θ ) = 0, 0 < θ < 2π, Θ ' (0) = Θ ' (2π ) . Θ(0) = Θ(2π ), R(a)θ (θ ) = h(θ ), 0 ≤ θ ≤ 2π, lim u(r, θ ) exists and it is finite.

(5.42)

r→0

Since we search a solution .u of the class .C 2 , we need to impose the periodicity conditions

176

5 The Laplace Equation

Θ(0) = Θ(2π ),

.

Θ ' (0) = Θ ' (2π ).

Hence and the second equation of (5.42), we get Θn (θ ) = An cos(nθ ) + Bn sin(nθ ),

.

λn = n2 ,

θ ∈ [0, 2π ],

n ∈ N0 .

Substituting the eigenvalues .λn into the first equation of (5.42), we find r 2 R '' (r) + rR ' (r) − n2 R(r) = 0,

.

r ∈ [0, a],

whereupon .

Rn (r) = Cn r n + Dn r −n , n ∈ N, R0 (r) = C0 + D0 log r, n = 0, r ∈ [0, a]. θ ∈ [0, 2π ], to exists and to be finite, we get .Dn = 0,

Since we want . lim u(r, θ ), r→0

n ∈ N0 . Therefore we obtain a formal solution as follows

.

u(r, θ ) =

∞  n=0

.

=

Rn (r)Θn (θ ) ∞

α0  n + r (αn cos(nθ ) + βn sin(nθ )) , 2

r ∈ [0, a],

θ ∈ [0, 2π ].

n=1

(5.43) Formally differentiating this series term-by-term, we verify that (5.43) is indeed harmonic. Imposing the boundary condition .u(a, θ ) = h(θ ), .0 ≤ θ ≤ 2π, we obtain α0 =

.

1 π

2π h(θ )dθ, 0

1 αn = π an

2π h(θ ) cos(nθ )dθ, 0

1 βn = π an

2π h(θ ) sin(nθ )dθ,

n ∈ N.

0

Example 5.15 Consider the Dirichlet problem .

Δu = 0, x12 + x22 < 1, u(x1 , x2 ) = x22 on x12 + x22 = 1.

5.5 Separation of Variables

177

Introduce polar coordinates x1 = r cos θ,

.

x2 = r sin θ,

r ≥ 0,

θ ∈ [0, 2π ],

we get the problem

.

1 1 urr + ur + 2 uθθ = 0 in B(0, 1), r r u(cos θ, sin θ ) = (sin θ )2 , θ ∈ [0, 2π ].

Here h(θ ) = (sin θ )2 ,

.

θ ∈ [0, 2π ].

Then 1 .α0 = π

2π (sin θ )2 dθ 0

1 = π

2π

1 − cos(2θ ) dθ 2

0

⎛ 1 1 = ⎝ π 2 1 = π

2π 0

1 dθ − 2

2π

⎞ cos(2θ )dθ ⎠

0

θ=2π  1 π − sin(2θ ) 4 θ=0



= 1, 1 αn = π

2π (sin θ )2 cos(nθ )dθ 0

1 = 2π

2π (1 − cos(2θ )) cos(nθ )dθ 0

1 = 2π

2π 0

1 cos(nθ )dθ − 2π

2π cos(2θ ) cos(nθ )dθ 0

178

5 The Laplace Equation

θ=2π 2π  1 1  = − (cos((n + 2)θ ) + cos((n − 2)θ ))dθ sin(nθ ) 2nπ 4π θ=0 0

1 =− 4π

2π

1 cos((n + 2)θ )dθ − 4π

0

2π cos((n − 2)θ )dθ,

n ∈ N.

0

We have the following cases. 1. Let .n = 2. Then 2π

1 .α2 = − 4π

1 cos(4θ )dθ − 4π

0

θ=2π  1 1 − sin(4θ ) =− 16π 2 θ=0

2π dθ 0

1 =− . 2 2. Let .n ∈ N, .n /= 2. Then θ=2π θ=2π   1 1  − .αn = − sin((n + 2)θ ) sin((n − 2)θ ) 4π(n + 2) 4π(n − 2) θ=0 θ=0 = 0. Therefore  αn =

.

− 21 n = 2 0 n /= 2, n ∈ N.

Next, 1 .βn = π

2π (sin θ )2 sin(nθ )dθ 0

1 = 2π

2π (1 − cos(2θ )) sin(nθ )dθ 0

5.5 Separation of Variables

1 = 2π

2π

179

1 sin(nθ )dθ − 2π

0

2π cos(2θ ) sin(nθ )dθ 0

θ=2π 2π  1 1  − (sin((n − 2)θ ) + sin((n + 2)θ ))dθ cos(nθ ) =− 2π n 4π θ=0 0

1 =− 4π

2π

2π

1 sin((n − 2)θ )dθ − 4π

sin((n + 2)θ )dθ,

0

n ∈ N.

0

We have the following cases. 1. Let .n = 2. Then 1 .β2 = − 4π

2π sin(4θ )dθ 0

θ=2π  1 cos(4θ ) = 16π θ=0 = 0. 2. Let .n ∈ N, .n /= 2. Then βn = −

.

θ=2π θ=2π   1 1 + cos((n − 2)θ ) cos((n + 2)θ ) 4π(n − 2) 4π(n + 2) θ=0 θ=0

= 0. Thus, βn = 0,

.

n ∈ N.

Consequently u(x1 , x2 ) = u(r, θ )

.

=

1 r2 − cos(2θ ) 2 2

1 r 2 (cos θ )2 − r 2 (sin θ )2 − 2 2 1 = (1 − x12 + x22 ), x12 + x22 ≤ 1. 2 =

180

5 The Laplace Equation

Exercise 5.20 Find a formal solution to the problem

.

Δu = 0 in x12 + x22 < 1, 3 u(x1 , x2 ) = x13 + x2 − x23 on 2

x12 + x22 = 1.

Exercise 5.21 Find a formal solution to the Dirichlet problem .

Δu = 0 in R2 \B(0, a), u(x1 , x2 ) = φ(x1 , x2 ) on ∂B(0, a),

where .φ ∈ C (∂B(0, a)). Exercise 5.22 Find a formal solution to the Dirichlet problem

.

Δu = 0 in B(0, b)\B(0, a), u(x1 , x2 ) = φ1 (x1 , x2 ) on ∂B(0, a), u(x1 , x2 ) = φ2 (x1 , x2 ) on ∂B(0, b),

where .0 < a < b, .φ1 ∈ C (∂B(0, a)), .φ2 ∈ C (∂B(0, b)). Exercise 5.23 Find a formal solution .u(r, θ ) to the problem 1 1 urr + ur + 2 uθθ = 0 in Dγ = {(r, θ ) : 0 < r < a, 0 < θ < γ }, r r . u(a, θ ) = φ(θ ), 0 ≤ θ ≤ γ , u(r, 0) = u(r, γ ) = 0, 0 ≤ r ≤ a, where .φ ∈ C ([0, γ ]). Exercise 5.24 Find a formal solution to the problem

.

1 1 urr + ur + 2 uθθ = f (r, θ ) in B(0, a), r r u(a cos θ, a sin θ ) = φ(θ ), 0 ≤ θ ≤ 2π,

where ∞

f (r, θ ) =

.

φ0 (r)  + (φn (r) cos(nθ ) + ψn (r) sin(nθ )) , 2 n=1

∞

φ(θ ) =

α0  + (αn cos(nθ ) + βn sin(nθ )) , 2 n=1

r ∈ [0, a],

θ ∈ [0, 2π ].

5.6 Advanced Practical Problems

181

5.6 Advanced Practical Problems Problem 5.1 Find the expression of the Laplace operator in the following coordinates. 1. n = 2 and x1 = φ(ξ, η),

.

x2 = ψ(ξ, η). 2. n = 2 and x1 = r cos φ,

.

x2 = r sin φ. 3. n = 3 and x1 = r cos φ,

.

x2 = r sin φ, x3 = x3 . 4. n = 3 and x1 = r cos φ sin θ,

.

x2 = r sin φ sin θ, x3 = r cos θ. 5. n = 3 and x1 = ξ η sin φ,  x2 = (ξ 2 − 1)(1 − η2 ),

.

x3 = ξ η cos φ. Problem 5.2 Let u = u(x1 , . . . , xn ), x = (x1 , . . . , xn ) ∈ Rn , be a harmonic function. Check if the following functions are harmonic. 1. 2. 3. 4.

u(x + h), h = (h1 , . . . , hn ) ∈ Rn . u(λx), λ ∈ R. u(Cx), where C is an n × n orthogonal matrix. n = 2, ux1 ux2 .

182

5. 6. 7. 8. 9.

5 The Laplace Equation

n > 2, ux1 ux2 . n = 3, x1 ux1 + x2 ux2 + x3 ux3 . n = 2, x1 ux1 − x2 ux2 . n = 2, x2 ux1 − x1 ux2 . n = 2, .

ux1 . (ux1 )2 + (ux2 )2

10. n = 2, (ux1 )2 − (ux2 )2 . 11. n = 2, (ux1 )2 + (ux2 )2 . Problem 5.3 Find k ∈ R so that the following functions to be harmonic functions. 1. 2. 3. 4. 5.

x13 + kx1 x22 , (x1 , x2 ) ∈ R2 . x12 + x22 + kx32 , (x1 , x2 , x3 ) ∈ R3 . e2x1 cosh(kx2 ), (x1 , x2 ) ∈ R2 . sin(3x1 ) cosh(kx2 ), (x1 , x2 ) ∈ R2 .

.



1 n 

j =1

k ,

(x1 , . . . , xn ) ∈ Rn .

2

xj2

Problem 5.4 Let n = 1 and x0 ∈ R and f : R → R. Let also, f ' exists in a neighbourhood of x0 and f '' (x0 ) exists. Prove that 2 .f (x0 ) = lim h→0 h2 ''



 f (x0 + h) + f (x0 − h) − f (x0 ) . 2

Problem 5.5 Let x0 ∈ D and u ∈ C 2 (D). Prove that ⎛

⎞



1 2n ⎜ ⎝ 2 r→0 r nκ(n)r n−1

⎟ u(y)dsy − u(x0 )⎠ .

Δu(x0 ) = lim

.

∂B(x0 ,r)

Problem 5.6 Let x0 ∈ D and u ∈ C 2 (D). Prove that ⎛ Δu(x0 ) = lim

.

r→0

2(n + 2) ⎜ 1 ⎝ κ(n)r n r2

 B(x0 ,r)

⎞ ⎟ u(y)dy − u(x0 )⎠ .

5.6 Advanced Practical Problems

183

Problem 5.7 Let D = {(x1 , . . . , xn ) ∈ Rn : 2x12 +

n 

.

xj2 < 3}.

j =2

Prove that the problem Δu =

n 

.

xj2

on

D

xj

on

∂D

j =1

u=

n  j =1

has at most one solution in C 2 (D)



C (D).

Problem 5.8 Let D = {x = (x1 , . . . , xn ) ∈ Rn : 0 < xn < a},

.

where a is a positive constant. Prove that G(x, y) =

∞ 

m

Φ(x, y m ) − Φ(x, y ' ) ,

.

m=−∞

where y = (y1 , . . . , yn−1 , yn ), y m = (y1 , . . . , yn−1 , 2ma + yn ), 'm y = (y1 , . . . , yn−1 , 2ma − yn ), is the Green function. Problem 5.9 Find the Green functions for the following domains in R3 . 1. {(x1 , x2 , x3 ) ∈ R3 : x3 > 0}.

.

2. {(x1 , x2 , x3 ) ∈ R3 : x2 > 0,

.

x3 > 0}.

3. {(x1 , x2 , x3 ) ∈ R3 : x1 > 0,

.

x2 > 0,

x3 > 0}.

4. {(x1 , x2 , x3 ) ∈ R3 : |x| < 1,

.

x3 > 0}.

184

5 The Laplace Equation

5. {(x1 , x2 , x3 ) ∈ R3 : |x| < 1,

.

x2 > 0,

x3 > 0}.

6. {(x1 , x2 , x3 ) ∈ R3 : |x| < 1,

.

x1 > 0,

x2 > 0,

x3 > 0}.

Problem 5.10 Let D = {(x1 , x2 , x3 ) ∈ R3 : x3 > 0}.

.

Solve the following problems. 1. .Δu = 0 on D,   u = cos x1 cos x2 .

∂D

2. −x3 .Δu = −e sin x1 cos x2 ,   u = 0.

on

D,

∂D

3. .Δu = 0 on D,   1 . u =  ∂D 1 + x12 + x22

4. 2 Δu = −

2 2 2 x1 + x2 + (x3 + 1)2   1 u = . 2 2 1 + x ∂D 1 + x2 .

5. Δu = 0 on

.

D,

on

D,

5.6 Advanced Practical Problems

185

  u

 = ∂D

−1 x1 < 0 1 x1 > 0.

Problem 5.11 Let D = {(x1 , x2 , x3 ) ∈ R3 : x2 > 0,

x3 > 0}.

.

Solve the following problems. 1. .Δu = 0 on D,   = 0, u x2 =0   u = e−4x1 sin(5x2 ).

x3 =0

2.   u   u

Δu = 0 on

D,

.

x2 =0

x3 =0

= 0, =

x2 (1 + x12

3

+ x22 ) 2

.

Problem 5.12 Let D = {(x1 , x2 , x3 ) ∈ R3 : |x| < 1}.

.

Solve the following problems. 1. .Δu = −1   u = 0.

on

D,

∂D

2. Δu = −|x|n

.

on

D,

n ∈ N0 ,

186

5 The Laplace Equation

  u

= 1. ∂D

3. |x| .Δu = −e   u = 0.

on

D,

∂D

Problem 5.13 Find a formal solution to the following problems 1.

.

Δu = 0, 0 < x1 , x2 < 1, u(x1 , 0) = 1 + sin(π x1 ), u(x1 , 1) = 2, 0 ≤ x1 ≤ 1, u(0, x2 ) = u(1, x2 ) = 1 + x2 , 0 ≤ x2 ≤ 1.

2.

.

Δu = 0, 0 < x1 < 1, 0 < x2 < 2, u(x1 , 0) = 0, u(x1 , 2) = 1, 0 ≤ x1 ≤ 1, ux1 (0, x2 ) = 0, ux1 (1, x2 ) = sin(2π x2 ), 0 ≤ x2 ≤ 2.

3.

.

Δu = 0, 0 < x1 , x2 < π, ux1 (0, x2 ) = ux1 (π, x2 ) = 0,

0 ≤ x2 ≤ π, π ux2 (x1 , π ) = x1 − , 0 ≤ x1 ≤ π. 2

ux2 (x1 , 0) = 0, 4. .

Δu = 0 in x12 + x22 < 1, u(x1 , x2 ) = x12 on x12 + x22 = 1.

Problem 5.14 Solve the Dirichlet problem   u(x1 , x2 ) where

Δu = 0

.

x12 +x22 =R 2

in x12 + x22 < R 2 ,

= g(x1 , x2 ),

x12 + x22 = R 2 ,

5.6 Advanced Practical Problems

187

1. g(x1 , x2 ) = x1 + x1 x2 ,

(x1 , x2 ) ∈ R2 .

g(x1 , x2 ) = 2(x12 + x2 ),

(x1 , x2 ) ∈ R2 .

.

2. .

Problem 5.15 Solve the Dirichlet problem   u(x1 , x2 )

Δu = 0

.

x12 +x22 =R 2

in x12 + x22 > R 2 ,

= g(x1 , x2 ),

x12 + x22 = R 2 ,

where 1. g(x1 , x2 ) = x2 + 2x1 x2 ,

(x1 , x2 ) ∈ R2 .

.

2. g(x1 , x2 ) = ax1 + bx2 + c,

.

(x1 , x2 ) ∈ R2 ,

where a, b, c ∈ R. 3. g(x1 , x2 ) = x12 − x22 ,

(x1 , x2 ) ∈ R2 .

g(x1 , x2 ) = x12 + 1,

(x1 , x2 ) ∈ R2 .

.

4. .

5. g(x1 , x2 ) = x22 − x1 x2 ,

.

(x1 , x2 ) ∈ R2 .

6. g(x1 , x2 ) = x22 + x1 + x2 ,

(x1 , x2 ) ∈ R2 .

g(x1 , x2 ) = 2x12 − x1 + x2 ,

(x1 , x2 ) ∈ R2 .

.

7. .

188

5 The Laplace Equation

Problem 5.16 Solve the Dirichlet problem   u(x1 , x2 )

Δu = f (x1 , x2 ),

x12 + x22 < R 2 ,

= g(x1 , x2 ),

x12 + x22 = R 2 ,

.

x12 +x22 =R 2

where 1. f (x1 , x2 ) = 1,

.

g(x1 , x2 ) = 0,

(x1 , x2 ∈ R2 .

2. f (x1 , x2 ) = x1 ,

.

g(x1 , x2 ) = 0,

(x1 , x2 ) ∈ R2 .

3. f (x1 , x2 ) = −1,

.

g(x1 , x2 ) =

x22 , 2

(x1 , x2 ) ∈ R2 .

4. f (x1 , x2 ) = x2 ,

.

g(x1 , x2 ) = 1,

(x1 , x2 ) ∈ R2 .

5. f (x1 , x2 ) = 4,

.

g(x1 , x2 ) = 1,

(x1 , x2 ) ∈ R2 .

Problem 5.17 Solve the Dirichlet problem   u(x1 , x2 )

Δu = 0,

.

x12 +x22 =R 2

x12 + x22 < R 2 ,

= g(θ ),

θ ∈ [0, 2π ],

5.6 Advanced Practical Problems

189

where 1. g(θ ) = cos θ,

.

θ ∈ [0, 2π ].

2. g(θ ) = (cos θ )2 ,

.

θ ∈ [0, 2π ].

Problem 5.18 Solve the Dirichlet problem   u(x1 , x2 )   u(x1 , x2 )

Δu = 0,

.

x12 +x22 =1

x12 +x22 =4

1 < x12 + x22 < 4,

= f1 (θ ), = f2 (θ ),

θ ∈ [0, 2π ],

where 1. f1 (θ ) = (cos θ )2 ,

.

f2 (θ ) =

1 ((cos θ )2 − 1), 8

θ ∈ [0, 2π ].

2. f1 (θ ) = (cos θ )2 ,

.

4 f2 (θ ) = 4(cos θ )2 − , 3

θ ∈ [0, 2π ].

Problem 5.19 Find a formal solution to the Neumann problem Δu = 0,

.

0 < x1 < a,

ux1 (0, x2 ) = g1 (x2 ),

0 < x2 < b,

x2 ∈ [0, b],

ux1 (a, x2 ) = 0,

x2 ∈ [0, b],

ux2 (x1 , 0) = 0,

x1 ∈ [0, a],

ux2 (x1 , b) = 0,

x1 ∈ [0, a],

190

5 The Laplace Equation

where g1 ∈ C ([0, b]) and b g1 (x2 )dx2 = 0.

.

0

Problem 5.20 Find a formal solution to the Neumann problem Δu = 0,

.

ux1 (0, x2 ) = 0,

0 < x1 < a,

0 < x2 < b,

x2 ∈ [0, b],

ux1 (a, x2 ) = g2 (x2 ),

x2 ∈ [0, b],

ux2 (x1 , 0) = 0,

x1 ∈ [0, a],

ux2 (x1 , b) = 0,

x1 ∈ [0, a],

where g1 ∈ C ([0, b]) and b g2 (x2 )dx2 = 0.

.

0

Problem 5.21 Find a formal solution to the Neumann problem Δu = 0,

.

0 < x1 < a,

ux1 (0, x2 ) = 0,

x2 ∈ [0, b],

ux1 (a, x2 ) = 0,

x2 ∈ [0, b],

ux2 (x1 , 0) = f1 (x1 ), ux2 (x1 , b) = 0,

0 < x2 < b,

x1 ∈ [0, a],

x1 ∈ [0, a],

where f1 ∈ C ([0, a]) and a f1 (x1 )dx1 = 0.

.

0

Problem 5.22 Find a formal solution to the Neumann problem Δu = 0,

.

0 < x1 < a,

ux1 (0, x2 ) = 0,

x2 ∈ [0, b],

ux1 (a, x2 ) = 0,

x2 ∈ [0, b],

ux2 (x1 , 0) = 0,

x1 ∈ [0, a],

ux2 (x1 , b) = f2 (x1 ),

0 < x2 < b,

x1 ∈ [0, a],

5.6 Advanced Practical Problems

191

where f1 ∈ C ([0, a]) and a f2 (x1 )dx1 = 0.

.

0

Problem 5.23 Find a formal solution to the Neumann problem Δu = 0,

.

0 < x1 < a,

0 < x2 < b,

ux1 (0, x2 ) = g1 (x2 ),

x2 ∈ [0, b],

ux1 (a, x2 ) = g2 (x2 ),

x2 ∈ [0, b],

ux2 (x1 , 0) = f1 (x1 ),

x1 ∈ [0, a],

ux2 (x1 , b) = f2 (x1 ),

x1 ∈ [0, a],

where f1 , f2 ∈ C ([0, a]), g1 , g2 ∈ C ([0, b]) and b g1 (x2 )dx2 = 0,

.

0

b g2 (x2 )dx2 = 0, 0

a f1 (x1 )dx1 = 0, 0

a f2 (x1 )dx1 = 0 0

Chapter 6

The Heat Equation

The heat equation is a partial differential equation that describes the variation of temperature in a given region over a period of time. The theory of heat equation was first developed by Joseph Fourier in 1822 for the purpose of modeling how a quantity such as heat diffuses through a given region. The heat equation along with variants thereof, is important in many fields of science and applied mathematics. In probability theory the heat equation is connected with the study of random walks and Brownian motion via the Focker-Planck equation. The Black-Scholes equation of financial mathematics is variant of the heat equation, and the Schrödinger equation of quantum mechanics can be regarded as a heat equation in imaginary time. The heat equation is used to resolve pixelation and to identify edges in image analysis. Solutions of the heat equation are sometimes known as caloric functions and they have been given much attention in the numerical analysis. If U is a given open subset of .Rn and I is a subinterval of .R, one say that a function .u : U × I → R is a solution of the heat equation if ut = ux1 x1 + · · · + uxn xn ,

.

where .(x1 , . . . , xn , t) denotes a general point of the domain. We refer to t as “time” and .x1 , . . . , xn as “spatial variables”. The collection of spatial variables is often referred to simply as x. For any given value of t, the right hand side of the heat equation is the Laplacian of the function .u(·, t) : U → R. Then the heat equation is often written more compactly as ut = Δu.

.

In physics and engineering context, it is more common to fix a Cartesian coordinate system and then to consider the specific case of a function .u(x1 , x2 , x3 , t) of three spatial variables .(x1 , x2 , x3 ) and time variable t. Then one says that u is a solution of the heat equation if © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. G. Georgiev, An Excursion Through Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-031-48784-2_6

193

194

6 The Heat Equation

ut = k(ux1 x1 + ux2 x2 + ux3 x3 ).

.

Here k is a positive coefficient called the thermal diffusity of the medium. The “diffusity constant” k is often not present in mathematical studies of the heat equation while its value can be very important in engineering. In this chapter, we investigate the Cauchy problem for the heat equation. We define fundamental solution for the heat equation and we give a representation of the solutions of the Cauchy problem for the heat equation. It is applied the method of separation of variables for the heat equation. In the chapter, the mean value formula is derived. The maximum principle for the Cauchy problem of the heat equation, the weak and strong maximum principles are investigated.

6.1 The Cauchy Problem The Cauchy problem for the heat equation is a model of a situation where one seeks to compute the temperature, or heat flux of the surface of a body by using interior measurements. Consider the heat equation ut − Δu = 0,

.

x ∈ Rn ,

t > 0.

(6.1)

We will seek a solution .u of (6.1) that has the special structure  1  u(x, t) = t −α v t − 2 |x| ,

.

x ∈ Rn ,

t > 0,

where .α is a constant and .v is a function that must be found, .|x| =

(6.2)  n  i=1

set 1

y = t − 2 x.

.

We have  1   1  1 3 ut (x, t) = −αt −α−1 v t − 2 |x| − t −α− 2 v ' t − 2 |x| 2 1 = −αt −α−1 v(|y|) − t −α−1 |y|v ' (|y|), 2  1  1 xj uxj (x, t) = t −α− 2 v ' t − 2 |x| , |x| .

1 2

xi2

. We

6.1 The Cauchy Problem

195 n 

1

uxj xj (x, t) = t −α− 2

xl2

l=1 l /= j

 1  v ' t − 2 |x|

|x|3

+t −α−1

  '' − 21 v |x| t |x|2 xj2

n 

n 

uxj xj (x, t) = t

−α− 21

j =1

l=1 n l /= j  j =1

+t −α−1

|x|3

xl2  1  v ' t − 2 |x|

  '' − 21 v |x| t |x|2

n x2  j j =1

= t −α− 2

(n − 1)|x|2 ' v (|y|) + t −α−1 v '' (|y|) |x|3

= t −α−1

(n − 1) ' v (|y|) + t −α−1 v '' (|y|), |y|

1

x ∈ Rn , .t > 0. Then, by Eq. (6.1), we find

.

1 0 = −αt −α−1 v(|y|) − t −α−1 |y|v ' (|y|) 2 n−1 ' −t −α−1 v (|y|) − t −α−1 v '' (|y|) |y|  n−1 ' 1 = −t −α−1 αv(|y|) + |y|v ' (|y|) + v (|y|) + v '' (|y|) 2 |y|

.

x ∈ Rn , .t > 0, or

.

1 n−1 ' αv(|y|) + |y|v ' (|y|) + v (|y|) + v '' (|y|) = 0. |y| 2

.

We take .α =

n and set .r1 = |y|. Then 2 .

n n−1 ' 1 v (r1 ) + v '' (r1 ) = 0, v(r1 ) + r1 v ' (r1 ) + 2 r1 2

which we multiply by .r1n−1 and we find

196

6 The Heat Equation

.

1 n−1 1 nr v(r1 ) + r1n v ' (r1 ) + (n − 1)r1n−2 v ' (r1 ) + r1n−1 v '' (r1 ) = 0, 2 2 1

whereupon  ' 1  ' r n v = 0. r1n−1 v ' + 2 1

.

Thus, 1 r1n−1 v ' + r1n v = C1 2

.

for some constant .C1 . Assuming .

lim v(r1 ), v ' (r1 ) = 0,

r1 →∞

we get .C1 = 0 and 1 r1n−1 v ' + r1n v = 0, 2

.

whence 1 v ' = − r1 v 2

.

and r12

v(r1 ) = Ce− 4 ,

.

C = const.

From here, u(x, t) =

1

=



n v

.

t2

C t

n 2

|x|



1

t2 e−

|x|2 4t

x ∈ Rn ,

,

t > 0,

solves the heat equation (6.1). Definition 6.1 The function Φ(x, t) =

.

⎧ ⎨

2

1 − |x| 4t , n e (4π t) 2 ⎩ 0, x ∈ Rn ,

x ∈ Rn , t < 0,

t > 0,

6.1 The Cauchy Problem

197

is called the fundamental solution of the heat equation. For each .t > 0, we have  Φ(x, t)dx = 1.

.

Rn

Really, we have  Φ(x, t)dx =

.

Rn

=

=



1 (4π t) 1 π

n 2



n 2

π

= 1,

|x|2 4t

dx

Rn

e−|z| dz 2

Rn

n ∞ 1  n 2

e−

e−zi dzi 2

i=1−∞

t > 0.

Consider the Cauchy problem .

ut − Δu = 0 in Rn × (0, ∞), u = φ on Rn × {t = 0},

(6.3)

where .φ ∈ C (Rn ) and .|φ(x)| ≤ M for all .x ∈ Rn and for some .M > 0. Let .u be defined by u(x, t) =



1

.

(4π t)

n 2

e−

|x−y|2 4t

φ(y)dy,

x ∈ Rn ,

t > 0.

Rn

The defined function u has the following properties. 1. .u ∈ C ∞ (Rn × (0, ∞)). Really, let .t1 > 0 be arbitrarily chosen. Since .

1 n 2

e−

|x|2 4t

t is infinitely times differentiable with uniformly bounded derivatives of all orders on .Rn × [t1 , ∞) and .t1 > 0 was arbitrarily chosen, we conclude that ∞ .u ∈ C (Rn × (0, ∞)). 2. .ut (x, t) − Δu(x, t) = 0, x ∈ Rn , t > 0. Really, firstly note that  ut (x, t) − Δu(x, t) =

(Φt − Δx Φ) (x − y, t)φ(y)dy

.

Rn

= 0,

x ∈ Rn ,

t > 0,

198

6 The Heat Equation

because |x−y|2 n n n Φt (x − y, t) = − (4π )− 2 t − 2 −1 e− 4t 2 |x−y|2 n n 1 + (4π ) 2 t − 2 −2 |x − y|2 e− 4t , 4 |x−y|2 n n 1 Φxj (x − y, t) = − (4π )− 2 t − 2 −1 (xj − yj )e− 4t , 2 |x−y|2 n n 1 Φxj xj (x − y, t) = − (4π )− 2 t − 2 −1 e− 4t 2 |x−y|2 n n 1 + (4π )− 2 t − 2 −2 (xj − yj )2 e− 4t , 4 |x−y|2 n n n Δx Φ(x − y, t) = − (4π )− 2 t − 2 −1 e− 4t 2 |x−y|2 n n 1 + (4π )− 2 t − 2 −2 |x − y|2 e− 4t , x, y ∈ Rn , 4 .

3.

t > 0,

i.e., .Φ itself solves the heat equation. u(x, t) = φ(x 0 ) for each .x 0 ∈ Rn . lim 0 (x, t) → (x , 0) x ∈ Rn , t > 0 Really, fix .x 0 ∈ Rn and .ε > 0. Then there exists .δ = δ(ε) > 0 such that

.

|φ(y) − φ(x 0 )| < ε

if |y − x 0 | < δ,

.

Let .|x − x 0 | ≤

y ∈ Rn .

δ . Then 2

       0 0  .|u(x, t) − φ(x )| =   Φ(x − y, t) φ(y) − φ(x ) dy  n  R  ≤ |Φ(y − x, t)||φ(y) − φ(x 0 )|dy Rn

 Φ(x − y, t)|φ(y) − φ(x 0 )|dy

= B(x 0 ,δ)



+

Φ(x − y, t)|φ(y) − φ(x 0 )|dy

Rn \B(x 0 ,δ)

= I1 + I2 ,

x ∈ Rn ,

t > 0.

6.1 The Cauchy Problem

199

Note that  I1 ≤ ε

Φ(x − y, t)dy

.

B(x 0 ,δ)



≤ε

Φ(x − y, t)dy

Rn

= ε,

x ∈ Rn ,

and 

  Φ(x − y, t) |φ(y)| + |φ(x 0 )| dy

I2 ≤

.

Rn \B(x 0 ,δ)



≤ 2M

Φ(x − y, t)dy Rn \B(x 0 ,δ)

=

≤

≤



2M

e

n

(4π t) 2

C1



t

n 2

e−

n 2

∞

2

dy

 1 0 |x − y| ≥ |y − x | 2

Rn \B(x 0 ,δ)

2M (4π t)

− |x−y| 4t

|x 0 −y|2 16t

dy

Rn \B(x 0 ,δ) r2

e− 16t r n−1 dr → 0 as

t → 0+,

δ

δ where .C1 is a constant independent of .t. Therefore if .|x − x 0 | ≤ and .t > 0 is 2 small enough, we have that .I2 ≤ ε and |u(x, t) − φ(x 0 )| ≤ 2ε.

.

Example 6.1 Consider the Cauchy problem .

Then

ut − uxx = 0, u(x, 0) = x,

x ∈ R, x ∈ R.

t > 0,

200

6 The Heat Equation

1 .u(x, t) = √ 2 πt 1 = √ 2 πt

∞

e−

(x−y)2 4t

ydy

−∞

∞

z2

e− 4t (x − z)dz

−∞

⎞ ⎛ ∞ ∞  z2 z2 1 ⎝ e− 4t dz − e− 4t zdz⎠ x = √ 2 πt ⎛

−∞

−∞

√ 1 1 = √ ⎝2x π t − 2 2 πt ⎛

∞ e −∞

√ 1 = √ ⎝2x π t + 2t 2 πt 1 = √ 2 πt 1 = √ 2 πt = x,

2 − z4t

∞ e −∞

2 − z4t

⎞ dz2 ⎠ ⎞  2 z ⎠ d − 4t

  2 z→∞ √ − z4t  2x π t + 2te  

 2x π t + 0

z→−∞

√

x ∈ R,

t > 0.

Example 6.2 Consider the Cauchy problem .

‘ut − ux1 x1 − ux2 x2 = 0, (x1 , x2 ) ∈ R2 , t > 0, = x12 + x22 , (x1 , x2 ) ∈ R2 . u(x1 , x2 , 0)

We have u(x1 , x2 , t) =

.

1 4π t



e−

|x−y|2 4t

e−

z12 +z22 4t

(y12 + y22 )dy1 dy2

R2

=

1 4π t





 (x1 − z1 )2 + (x2 − z2 )2 dz1 dz2

R2

=

(x12 +x22 )

1 4π t

∞ ∞ e −∞ −∞

−

z12 +z22 4t

x1 dz1 dz2 − 2π t

∞ ∞ −∞ −∞

e−

z12 +z22 4t

z1 dz1 dz2

6.1 The Cauchy Problem

x2 − 2π t

=

201

∞ ∞ e ∞ 2π 0

x2 − 2π t 1 4π t

r2

e− 4t rdφdr −

1 2π 8π t

r2

∞ e

2 − r4t

⎞ ⎛ 2π ⎞ ⎛∞   r2 x 1 ⎝ e− 4t r 2 dr ⎠ ⎝ cos φdφ ⎠ d(r 2 ) − 2π t 0

r2

e− 4t r 2 d(r 2 )

= (x12 + x22 )

e

2 − r4t

 d

0

∞

r2 4t



⎞ ⎛∞ φ=2π   r2 x1 ⎝ − 4t 2 ⎠ e r dr sin φ  − 2π t φ=0 0

⎞

φ=2π 2  − r4t 2 ⎠ e r dr cos φ  φ=0

0



2

e

0

0

∞

∞

+

0

e− 4t r 3 dφdr

0

∞

r2

e− 4t r 2 cos φdφdr

⎞ ⎛ 2π ⎞ ∞  r2 ⎝ e− 4t r 2 dr ⎠ ⎝ sin φdφ ⎠

x2 ⎝ 2π t

(z12 +z22 )dz1 dz2

r2

0

+

z12 +z22 4t

e− 4t r 2 sin φdφdr

0

⎛

⎛

e−

−∞ −∞

∞ 2π

x1 2π t

0

0

x 2 + x22 2π = 1 8π t

+

∞ ∞

0

∞ 2π 0

x2 2π t

1 z2 dz1 dz2 + 4π t

0

∞ 2π 0

−

z12 +z22 4t

−∞ −∞

x12 + x22 4π t

+

−

− r4t 2

r d

r2 4t



0

=

−(x12

2

r + x22 )e− 4t

r→∞ r→∞ ∞ 2 2   r − r4t 2   −e r  + 2 e− 4t rdr  r=0

∞ = x12 + x22 + 4t 0

r2

e− 4t d

r=0



2

r 4t

0

202

6 The Heat Equation

=

x12

+ x22

=

x12

+ x22

− 4te

2

− r4t

r→∞    r=0

+ 4t,

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

Exercise 6.1 Solve the following Cauchy problems 1. 4ut − uxx = 0,

.

u(x, 0) = e

x ∈ R,

2x−x 2

,

t > 0,

x ∈ R.

2. 2ut − ux1 x1 − ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 ,

u(x1 , x2 , 0) = cos(x1 x2 ),

t > 0,

(x1 , x2 ) ∈ R2 .

3. ut − ux1 x1 − ux2 x2 − ux3 x3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 ,

u(x1 , x2 , x3 , 0) = cos(x1 x2 ) sin x3 ,

t > 0,

(x1 , x2 , x3 ) ∈ R3 .

4. ut −

n 

.

uxj xj = 0,

(x1 , . . . , xn ) ∈ Rn ,

t > 0,

j =1

⎛ ⎞ n  xj ⎠ , u(x1 , . . . , xn , 0) = cos ⎝

(x1 , . . . , xn ) ∈ Rn .

j =1

5. .

ut − ux1 x1 − ux2 x2 − ux3 x3 = 0, (x1 , x2 , x3 ) ∈ R3 , t > 0, u(x1 , x2 , x3 , 0) = 2x1 − x2 + 3x3 , (x1 , x2 , x3 ) ∈ R3 .

Next, we consider the Cauchy problem .

ut − Δu = f (x, t), x ∈ Rn , u(x, 0) = 0, x ∈ Rn ,

t > 0,

(6.4)

where .f ∈ C 2 (Rn , C 1 ([0, ∞))), .|ft |, .|fxi |, .|fxi xi | ≤ M, .i = 1, . . . , n, in n .R × [0, ∞) for some positive constant .M. Let also,

6.1 The Cauchy Problem

203

t  u(x, t) =

Φ(x − y, t − s)f (y, s)dyds,

.

x ∈ Rn ,

t ≥ 0.

0 Rn

The defined function has the following properties. 1. .u ∈ C 2 (Rn , C 1 ((0, ∞))). Really, we change the variables and we get t  u(x, t) =

Φ(y, s)f (x − y, t − s)dyds.

.

0

Rn

Since .f ∈ C 2 (Rn , C 1 ((0, ∞))) and .|f |, .|ft |, .|fxi |, .|fxi xi | ≤ M, .i = 1, . . . , n, we conclude that .u ∈ C 2 (Rn , C 1 ((0, ∞))). 2. .ut − Δu = f (x, t) in .Rn × (0, ∞). Really, we have t  ut (x, t) =

Φ(y, s)ft (x − y, t − s)dyds

.

Rn

0



Φ(y, t)f (x − y, 0)dy

+ Rn

t  =−

Φ(y, s)fs (x−y, t−s)dyds 0 Rn



Φ(y, t)f (x − y, 0)dy,

+ Rn

t  Δu(x, t) =

Φ(y, s)Δx f (x − y, t − s)dyds 0

Rn

t  =

Φ(y, s)Δy f (x − y, t − s)dyds, 0

x ∈ Rn ,

t ≥ 0.

Rn

Therefore t  ut (x, t) − Δu(x, t) =

.

0 Rn

 Φ(y, s) −fs (x − y, t−s)−Δy f (x−y, t−s) dyds

204

6 The Heat Equation

 +

Φ(y, t)f (x − y, 0)dy

Rn

t 

 Φ(y, s) −fs (x −y, t−s)−Δy f (x−y, t−s) dyds

= ε Rn

ε  +

 Φ(y, s) −fs (x−y, t−s)−Δy f (x−y, t−s) dyds

0 Rn

 +

Φ(y, t)f (x − y, 0)dy

Rn

= J1 + J2 + J3 ,

x ∈ Rn ,

t ≥ 0.

Note that t  J1 =

.

 Φ(y, s) −fs (x − y, t − s) − Δy f (x − y, t − s) dyds

ε Rn

t  =

 Φs (y, s) − Δy Φ(y, s) f (x − y, t − s)dyds

ε Rn



Φ(y, ε)f (x − y, t − ε)dy

+ Rn



Φ(y, t)f (x − y, 0)dy

− 

Rn

Φ(y, ε)f (x − y, t − ε)dy − J3 ,

=

x ∈ Rn ,

t ≥ 0.

Rn

Hence,  ut (x, t) − Δu(x, t) =

Φ(y, ε)f (x − y, t − ε)dy − J3

.

Rn

ε  +

Φ(y, s)(−fs (x − y, t − s) 0 Rn

−Δy f (x − y, t − s))dyds + J3

6.1 The Cauchy Problem

205

 =

Φ(y, ε)f (x − y, t − ε)dy Rn

ε  +

Φ(y, s)(−fs (x − y, t − s) 0 Rn

−Δy f (x − y, t − s))dyds,

x ∈ Rn ,

t ≥ 0,

and  ut (x, t) − Δu(x, t) = lim

.

ε→0 Rn

Φ(y, ε)f (x − y, t − ε)dy,

x ∈ Rn ,

t ≥ 0.

We observe that         ≤ Φ(y, ε)|f (x−y, t−ε)−f (x, t)|dy . Φ(y, ε)f (x−y, t−ε)dy−f (x, t)    n R Rn  Φ(y, ε)|f (x − y, t − ε) = B(x,δ)

−f (x, t)|dy  Φ(y, ε)|f (x − y, t − ε)

+

Rn \B(x,δ)

−f (x, t)|dy = J4 + J5 ,  J4 ≤ ε Φ(y, ε)dy Rn

= ε, J5 ≤

≤



2M (4π ε) C2 ε

∞

n 2

n 2

e−

|y|2 4ε

dy

Rn \B(x,δ) r2

e− 4ε r n−1 dr

δ

≤ C3 ε, where .C2 and .C3 are positive constants independent of .0 < ε < 1. Therefore

206

6 The Heat Equation

       ≤ (1 + C3 )ε. . Φ(y, ε)f (x − y, t − ε)dy − f (x, t)    n R

From here,  .

lim

ε→0 Rn

Φ(y, ε)f (x − y, t − ε)dy = f (x, t),

x ∈ Rn ,

t ≥ 0,

and ut (x, t) − Δu(x, t) = 0,

.

3.

.

x ∈ Rn ,

t ≥ 0.

u(x, t) = 0 for each .x 0 ∈ Rn . Really, from the definition of the lim 0 (x, t) → (x , 0) x ∈ Rn , t > 0 function .u, we have  t        .|u(x, t)| =  Φ(y, s)f (x − y, t − s)dyds     n 0 R

t  ≤

Φ(y, s)|f (x − y, t − s)|dyds 0 Rn

t  ≤M

Φ(y, s)dyds 0 Rn

= Mt,

x ∈ Rn ,

t ≥ 0.

Consequently .

u(x, t) = 0 for lim (x, t) → (x 0 , 0) x ∈ Rn , t > 0

each

x 0 ∈ Rn .

Example 6.3 Consider the Cauchy problem .

We have

ut − ux1 x1 − ux2 x2 = x1 + x2 , (x1 , x2 ) ∈ R2 , = 0, (x1 , x2 ) ∈ R2 . u(x1 , x2 , 0)

t > 0,

6.1 The Cauchy Problem

t

207

1 4π(t − s)

u(x1 , x2 , t) =

.

0

t

1 4π(t − s)

= 0

1 = 4π

t

1 t −s

0

=

1 4π

t

1 t −s

0

1 − 4π

t 0

x1 + x2 = 4

t

−

1 4π

0

|x−y|2

e− 4(t−s) (y1 + y2 )dy1 dy2 ds

R2

∞ ∞

∞ 2π 0

∞ 2π 0

1 t −s

r2

e− 4(t−s) (x1 + x2 − r(cos φ + sin φ))rdφdrds

0 r2

e− 4(t−s) r(x1 + x2 )dφdrds

0

∞ 2π 0

1 t −s

r2

e− 4(t−s) r 2 (cos φ + sin φ)dφdrds

0

∞

∞

r2

e− 4(t−s) d(r 2 )ds

0

φ=2π  r2 e− 4(t−s) r 2 (sin φ − cos φ) drds φ=0

0

 t ∞ = (x1 + x2 ) 0

z12 +z22

e− 4(t−s) (x1 + x2 − z1 − z2 )dz1 dz2 ds

−∞ −∞

1 t −s

0

t



r2

e− 4(t−s) d



0

t = −(x1 + x2 )

r2 ds 4(t − s)

r→∞ r2  ds e− 4(t−s)  r=0

0

t = (x1 + x2 )

ds 0

= t (x1 + x2 ),

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

Exercise 6.2 Solve the following Cauchy problems 1. ut − uxx = 2t (x 2 − t),

.

u(x, 0) = 0,

x ∈ R.

x ∈ R,

t > 0,

208

6 The Heat Equation

2. ut − uxx = cos t + x 3 − 6tx,

.

u(x, 0) = 0,

x ∈ R,

t > 0,

x ∈ R.

3. ut − ux1 x1 − ux2 x2 = x1 + x22 − 2t,

(x1 , x2 ) ∈ R2 ,

.

u(x1 , x2 , 0) = 0,

t > 0,

(x1 , x2 ) ∈ R2 .

4. ut −

n 

⎛ ⎞ n  = cos t ⎝ xj2 ⎠ − 2n sin t,

(x1 , . . . , xn ) ∈ Rn ,

t > 0,

ut − ux1 x1 − ux2 x2 − ux3 x3 = 2t (x1 − x2 + x3 ), (x1 , x2 , x3 ) ∈ R3 , = 0, (x1 , x2 , x3 ) ∈ R3 . u(x1 , x2 , x3 , 0)

t > 0,

.

uxj xj

j =1

j =1

u(x1 , . . . , xn , 0) = 0,

(x1 , . . . , xn ) ∈ Rn .

5. .

If .u1 is a solution to the problem (6.3) and .u2 is a solution to the problem (6.4), then u(x, t) = u1 (x, t) + u2 (x, t)

.

 =

t  Φ(x − y, t)φ(y)dy +

Rn

Φ(x − y, t − s)f (y, s)dyds, 0 Rn

where .f and .φ satisfy the hypotheses above, is a solution to the problem .

ut − Δu = f u =φ

in Rn × (0, ∞), on Rn × {t = 0}.

Example 6.4 Consider the Cauchy problem ut = ux1 x1 + ux2 x2 + x1 + x2 ,

.

u(x1 , x2 , 0) =

x12

+ x22 ,

(x1 , x2 ) ∈ R . 2

(x1 , x2 ) ∈ R2 ,

t > 0,

6.1 The Cauchy Problem

209

By Example 6.2, we have that the function u1 (x1 , x2 , t) = x12 + x22 + 4t,

(x1 , x2 ) ∈ R2 ,

t ≥ 0,

(x1 , x2 ) ∈ R2 ,

t > 0,

.

is the solution of the Cauchy problem ut = ux1 x1 + ux2 x2 ,

.

u(x1 , x2 , 0) =

x12

+ x22 ,

(x1 , x2 ) ∈ R . 2

By Example 6.3, we have that the function u2 (x1 , x2 , t) = t (x1 + x2 ),

(x1 , x2 ) ∈ R2 ,

.

t ≥ 0,

is the solution of the Cauchy problem ut = ux1 x1 + ux2 x2 + x1 + x2 ,

(x1 , x2 ) ∈ R2 ,

.

u(x1 , x2 , 0) = 0,

t > 0,

(x1 , x2 ) ∈ R2 .

Therefore u(x1 , x2 , t) = u1 (x1 , x2 , t) + u2 (x1 , x2 , t)

.

= x12 + x22 + 4t + t (x1 + x2 ) = x12 + x22 + t (x1 + x2 + 4),

(x1 , x2 ) ∈ R2 ,

t ≥ 0,

is the solution of the considered Cauchy problem. Exercise 6.3 Solve the following Cauchy problems 1. ut = 4uxx + t + et ,

x ∈ R,

.

u(x, 0) = 2,

t > 0,

x ∈ R.

2. ut = uxx + 3t 2 ,

.

u(x, 0) = sin x,

x ∈ R,

t > 0,

x ∈ R.

3. ut = ux1 x1 + ux2 x2 + et ,

.

u(x1 , x2 , 0) = cos x1 sin x2 ,

(x1 , x2 ) ∈ R2 ,

(x1 , x2 ) ∈ R . 2

t > 0,

210

6 The Heat Equation

4. 

ut = 2 ux1 x1 + ux2 x2 + ux3 x3 + t cos x1 ,

(x1 , x2 , x3 ) ∈ R3 ,

.

u(x1 , x2 , 0) = cos x2 cos x3 ,

t > 0,

(x1 , x2 , x3 ) ∈ R3 .

5. .

ut = ux1 x1 + ux2 x2 + 2t − 2t 3 + x1 + 3t 2 x22 , u(x1 , x2 , 0) = x2 , (x1 , x2 ) ∈ R2 .

(x1 , x2 ) ∈ R2 ,

t > 0,

6.2 The Method of Separation of Variables In this section, we will apply the method of separation of variables for some classes of initial boundary value problems for the heat equation. Consider the following initial boundary value problem ut − kuxx = 0,

.

.

0 < x < L,

u(0, t) = 0, u(L, t) = 0,

u(x, 0) = φ(x),

.

t > 0,

t ≥ 0, 0 ≤ x ≤ L,

(6.5) (6.6) (6.7)

where .φ is a given initial condition, .k > 0 is a given constant. Assume the compatibility condition φ(0) = φ(L) = 0.

.

We seek solutions of the problem (6.5)–(6.7) that have the special form u(x, t) = X(x)T (t).

.

(6.8)

We are not interested in the zero solution .u(x, t) = 0, .0 ≤ x ≤ L, .t ≥ 0. Therefore, we seek functions .X and .T that do not vanish identically. We substitute (6.8) into (6.5) and we find .

X'' (x) T ' (t) . = X(x) kT (t)

Since .x and .t are independent variables, from the last equality, it follows that there exists a constant .λ, which is called the separation constant, such that

6.2 The Method of Separation of Variables

.

211

X'' (x) T ' (t) = = −λ. kT (t) X(x)

(6.9)

Using the boundary conditions (6.6), because .u is not the trivial solution, it follows that X(0) = X(L) = 0.

.

Thus, the function .X should be a solution to the boundary problem .

X'' + λX = 0, X(0) = X(L) = 0.

0 < x < L,

For its solution, we find X(x) = sin

 nπ x 

.

L

λ=

,

 nπ 2 L

n ∈ N.

,

For convenience, we use the notation Xn (x) = sin

.

 nπ x 

L  nπ 2 λn = , L

, n ∈ N.

Hence, nπ 2

Tn (t) = Bn e−k ( L ) t ,

.

n ∈ N,

t ≥ 0.

Thus, we obtain the following sequence of separated solutions un (x, t) = Xn (x)Tn (t)

.

nπ 2

= Bn e−k ( L ) t sin

 nπ x  L

,

n ∈ N,

0 ≤ x ≤ L, .t ≥ 0, where .Bn , .n ∈ N, are constants. The superposition principle implies that

.

u(x, t) =

∞ 

.

n=1

Bn sin

 nπ x  L

nπ 2

e−k ( L ) t ,

0 ≤ x ≤ L, .t ≥ 0, is a formal solution of the problem (6.5), (6.6). We will find the constants .Bn using the initial condition (6.7). We have

.

212

6 The Heat Equation

u(x, 0) = φ(x) =

∞ 

.

Bn sin

 nπ x  L

n=1

Fix .m ∈ N and multiply by .sin by-term over .[0, L], we find ∞  .

L sin

Bn

n=1

 nπ x  L

 mπ x  L

sin

,

0 ≤ x ≤ L.

the last equality, then we integrate it term-

 mπ x  L

L dx =

0

sin

 mπ x  L

φ(x)dx,

0

whereupon 2 .Bm = L

L sin

 mπ x  L

φ(x)dx.

0

Example 6.5 Consider the initial-boundary value problem ut − uxx u(0, t) . u(π, t) u(x, 0)

= = = =

0, 0 < x < π, t > 0, 0, 0, t ≥ 0, x(x − π ), 0 ≤ x ≤ π.

Here .L = π, .φ(x) = x(x − π ), .0 ≤ x ≤ π , .k = 1. Then 2 .Bm = π

π sin(mx)x(x − π )dx 0

2 =− mπ

π x(x − π )d(cos(mx)) 0

x=π π  2 2  x(x − π ) cos(mx) =− + (2x − π ) cos(mx)dx mπ mπ x=0 0

2 = 2 m π

π (2x − π )d(sin(mx)) 0

=

x=π π  2 4  (2x − π ) sin(mx) − sin(mx)dx  m2 π m2 π x=0 0

6.2 The Method of Separation of Variables

213

x=π  4 = 3 cos(mx) m π x=0  4

m (−1) − 1 . m3 π

= Then

u(x, t) =

.

∞   4

2 (−1)n − 1 sin(nx)e−n t , 3 n π n=1

0 ≤ x ≤ π , .t ≥ 0, is a formal solution of the considered problem.

.

Exercise 6.4 Find a formal solution to the problem ut − uxx u(0, t) . u(π, t) u(x, 0)

= 0, 0 < x < π, t > 0, = 0, = 0, t ≥ 0, = sin x, 0 ≤ x ≤ π.

Next, we consider the problem ut − kuxx = f (x, t),

.

.

0 < x < L,

u(0, t) = 0, u(L, t) = 0,

u(x, 0) = φ(x),

.

t > 0,

(6.10) (6.11)

t ≥ 0, 0 ≤ x ≤ L,

(6.12)

where f (x, t) =

∞ 

.

fn (t) sin

n=1

φ(x) =

∞ 

An sin

n=1

nπ x , L

nπ x , L

0 ≤ x ≤ L,

t ≥ 0,

fn are given continuous functions on .[0, ∞) and .An are given constants, .n ∈ N. Let u be a formal solution to the problem (6.10)–(6.12) that has the form

. .

u(x, t) =

∞ 

.

n=1

Tn (t) sin

 nπ x  L

0 ≤ x ≤ L, .t ≥ 0. Substituting (6.13) into (6.10), we get

.

,

(6.13)

214

6 The Heat Equation

 ∞  ∞  nπ x   nπ x   n2 π 2 Tn' (t) + k 2 Tn (t) sin = , fn (t) sin L L L

.

n=1

n=1

0 ≤ x ≤ L, .t ≥ 0, whereupon

.

Tn' + k

.

n2 π 2 Tn = fn L2

and Tn (t) = e

.

  2 2 kn π t Bn + fn (t)e L2 dt ,

2π2 t L2

−k n

0 ≤ x ≤ L, .t ≥ 0, where .Bn , .n ∈ N, are constants which will be determined below. We set  2 2 kn π t .gn (t) = fn (t)e L2 dt, t ≥ 0.

.

Therefore u(x, t) =

∞ 

.

e

2π2 t L2

−k n

(Bn + gn (t)) sin

 nπ x 

n=1

L

,

0 ≤ x ≤ L, .t ≥ 0. We will find the constants .Bn using the initial condition (6.12). We have

.

u(x, 0) =

∞ 

.

(Bn + gn (0)) sin

n=1

=

∞ 

An sin

n=1

 nπ x  L

,

 nπ x  L 0 ≤ x ≤ L.

Hence, Bn = An − gn (0)

.

and u(x, t) =

∞ 

.

n=1

0 ≤ x ≤ L, .t ≥ 0.

.

e

2π2 t L2

−k n

(An − gn (0) + gn (t)) sin

 nπ x  L

,

6.2 The Method of Separation of Variables

215

Example 6.6 Consider the problem ut − uxx u(0, t) . u(π, t) u(x, 0)

= tx, 0 < x < π, t > 0, = 0, = 0, t ≥ 0, = x(x − π ), 0 ≤ x ≤ π.

Here f (x, t) = tx,

.

φ(x) = x(x − π ), k = 1, L = π. Note that φ(0) = φ(π ) = 0

.

and π .

1 x sin(nx)dx = − n

π xd(cos(nx)) 0

0

x=π π  1 1 = − x cos(nx) + cos(nx)dx n n x=0 0

x=π  1 π n = − (−1) + 2 sin(nx) n n x=0 π = (−1)n+1 , n π π 1 x 2 d(cos(nx)) x 2 sin(nx)dx = − n 0

0

x=π π  1 2 2 = − x cos(nx) + x cos(nx)dx n n x=0 0

=−

π2 2 (−1)n + 2 n n

π xd(sin(nx)) 0

216

6 The Heat Equation

x=π π  2 2 π2 n+1  + 2 x sin(nx) − 2 sin(nx)dx = (−1) n n n x=0 0

x=π  2 n+1 + 3 cos(nx) (−1) = n n x=0 π2

=

π2 2 (−1)n+1 + 3 ((−1)n − 1). n n

Then f (x, t) = t

.

∞  2 (−1)n+1 sin(nx), n

fn (t) =

n=1

2 (−1)n+1 t, n

∞   4 n ((−1) − 1) sin(nx), φ(x) = n3 π n=1

4 ((−1)n − 1), n3 π  2 2 gn (t) = (−1)n+1 ten t dt n   2 2 n+1 td en t = 3 (−1) n An =

2 2 2 (−1)n+1 ten t − 3 (−1)n+1 3 n n  2 1 2 n+1 = 3 (−1) t − 2 en t , n n =



2

en t dt

2 (−1)n , n5 4 2 Bn = 3 ((−1)n − 1) − 5 (−1)n , n π n

gn (0) =

0 ≤ x ≤ π, .t ≥ 0. Consequently

.

u(x, t) =

∞ 

.

n=1

+ 0 ≤ x ≤ π , .t ≥ 0.

.

e−n

2t

 4 2 ((−1)n − 1) − 5 (−1)n 3 n π n

  2 1 2 n+1 (−1) t − en t sin(nx), 3 2 n n

6.2 The Method of Separation of Variables

217

Exercise 6.5 Find a formal solution of the following problem ut = uxx + t sin x cos(3x),

0 < x < π,

.

t > 0,

u(0, t) = 0, u(π, t) = 0,

t ≥ 0,

u(x, 0) = x(x − π ),

x ∈ [0, π ].

Exercise 6.6 Find a formal solution to the problem ut = kuxx + f (x, t),

0 < x < L,

.

t > 0,

u(0, t) = g(t), u(L, t) = h(t),

t ≥ 0,

u(x, 0) = φ(x),

x ∈ [0, L],

where .f ∈ C ([0, L] × [0, ∞)), .g, h ∈ C 1 ([0, ∞)), .φ ∈ C ([0, L]), φ(0) = g(0),

.

φ(L) = h(0). Exercise 6.7 Find a formal solution of the following problem ut − kuxx = 0,

.

0 < x < L,

t > 0,

ux (0, t) = 0, ux (L, t) = 0,

t ≥ 0,

u(x, 0) = φ(x),

x ∈ [0, L],

where .φ ∈ C 1 ([0, L]) satisfies the condition φ ' (0) = φ ' (L) = 0.

.

Exercise 6.8 Find a formal solution of the following problem ut = kuxx + f (x, t),

.

0 < x < L,

ux (0, t) = 0, ux (L, t) = 0,

t ≥ 0,

u(x, 0) = φ(x),

x ∈ [0, L],

t > 0,

218

6 The Heat Equation

where .φ ∈ C 1 ([0, L]) satisfies the conditions φ ' (0) = φ ' (L) = 0.

.

Exercise 6.9 Find a formal solution of the following problem ut = kuxx + f (x, t),

0 < x < L,

.

t > 0,

ux (0, t) = g(t), ux (L, t) = h(t),

t ≥ 0,

u(x, 0) = φ(x),

x ∈ [0, L],

where .k > 0, .g, h ∈ C 1 ([0, ∞)), .φ ∈ C 1 ([0, L]) and φ ' (L) = h(0),

.

φ ' (0) = g(0). Exercise 6.10 Find a formal solution of the following problem ut − kuxx = 0,

.

0 < x < L,

t > 0,

u(0, t) = 0, ux (L, t) = 0,

t ≥ 0,

u(x, 0) = φ(x),

x ∈ [0, L],

where .φ ∈ C 1 ([0, L]) satisfies the condition φ(0) = φ ' (L) = 0.

.

Exercise 6.11 Find a formal solution of the following problem ut = kuxx + f (x, t),

.

0 < x < L,

t > 0,

u(0, t) = 0, ux (L, t) = 0,

t ≥ 0,

u(x, 0) = φ(x),

x ∈ [0, L],

where .φ ∈ C 1 ([0, L]) satisfies the conditions φ(0) = φ ' (L) = 0.

.

Exercise 6.12 Find a formal solution of the following problem

6.2 The Method of Separation of Variables

219

ut = kuxx + f (x, t),

0 < x < L,

.

t > 0,

u(0, t) = g(t), ux (L, t) = h(t),

t ≥ 0,

u(x, 0) = φ(x),

x ∈ [0, L],

where .k > 0, .g, h ∈ C 1 ([0, ∞)), .φ ∈ C 1 ([0, L]) and φ ' (L) = h(0),

.

φ(0) = g(0). Exercise 6.13 Find a formal solution of the following problem ut − kuxx = 0,

.

0 < x < L,

t > 0,

ux (0, t) = 0, u(L, t) = 0,

t ≥ 0,

u(x, 0) = φ(x),

x ∈ [0, L],

where .φ ∈ C 1 ([0, L]) satisfies the condition φ ' (0) = φ(L) = 0.

.

Exercise 6.14 Find a formal solution of the following problem ut = kuxx + f (x, t),

.

0 < x < L,

t > 0,

u(0, t) = 0, ux (L, t) = 0,

t ≥ 0,

u(x, 0) = φ(x),

x ∈ [0, L],

where .φ ∈ C 1 ([0, L]) satisfies the conditions φ ' (0) = φ(L) = 0.

.

Exercise 6.15 Find a formal solution of the following problem ut = kuxx + f (x, t),

.

0 < x < L,

u(0, t) = g(t), ux (L, t) = h(t), u(x, 0) = φ(x),

t ≥ 0, x ∈ [0, L],

t > 0,

220

6 The Heat Equation

where .k > 0, .g, h ∈ C 1 ([0, ∞)), .φ ∈ C 1 ([0, L]) and φ(L) = h(0),

.

φ ' (0) = g(0).

6.3 The Mean Value Formula In this section, we will derive the mean value formula for the heat equation. For fixed .x ∈ Rn , .t ∈ R, we define   1 n+1 .W (x, t, r) = (y, s) ∈ R : s ≤ t, Φ(x − y, t − s) ≥ n . r We set .W (r) = W (0, 0, r). Example 6.7 For .n = 2 we will compute  

|y|2 dyds. 2 W (1) s

.

We have  1 |y|2 e 4s ≥ 1, W (1) = (y, s) ∈ R3 : − 4π s

.

 s≤0 .

Hence,   W (1) = (y, s) ∈ R3 : |y| ≤ 2 s log(−4π s),

.

Therefore   .

|y|2 dyds = 2 W (1) s



0 1 − 4π |y|≤2

0 = 2π 1 − 4π

|y|2 dyds s2

√

s log(−4π s) 2

1 s2

√



s log(−4π s)

r 3 drds 0

−

 1 ≤s≤0 . 4π

6.3 The Mean Value Formula

221

 √ 1 4 r=2 s log(−4π s) ds r s 2 r=0

0

π = 2

1 − 4π

0

π = 2

1 16s 2 (log(−4π s))2 ds s2

1 − 4π

0 = 8π

(log(−4π s))2 ds 1 − 4π

s→0  = 8π s(log(−4π s))  2

1 s=− 4π

0 − 8π

2s log(−4π s) 1 − 4π

−4π ds −4π s

0 = −16π

log(−4π s)ds 1 − 4π

s→0  = −16π s log(−4π s)

1 s=− 4π

0 + 16π

s 1 − 4π

1 (−4π )ds −4π s

0 = 16π

ds 1 − 4π

= 16π ·

1 4π

= 4. Exercise 6.16 Consider the case .n = 3 and compute   .

|y|2 dyds. 2 W (1) s

Generalize the result for arbitrary .n. Assume that .Q ⊂ Rn is an open and bounded set, and fix .T > 0. Define the parabolic cylinder QT = Q × (0, T ].

.

222

6 The Heat Equation

Let .u ∈ C 2 (QT ) solves the heat

Example 6.8 (The Mean Value Formula) equation ut − Δu = 0,

.

(x, t) ∈ QT .

We will prove that u(x, t) =

.

 

1 4r n

u(y, s) W (x,t,r)

|x − y|2 dyds. (t − s)2

Without loss of generality, we will prove the formula for .x = 0, t = 0. Otherwise, we translate the space and time coordinates so that .x = 0, .t = 0. Let   φ(r) =

u(rz, r 2 τ )

.

W (1)

|z|2 dzdτ, τ2

r ≥ 0.

Then  n 

 

'

φ (r) =

.

W (1)

=

i=1

 

1 r n+1

|z|2 |z|2 uyi zi 2 + 2rus τ τ n 

W (r) i=1

uyi yi

 dzdτ

|y|2 2 dyds + n+1 s2 r

  us W (r)

= A + B. We introduce the function |y|2 n + n log r. ψ(y, r, s) = − log(−4π s) + 2 4s

.

We have Φ(y, −s) =

.

1 rn

on

∂W (r).

Hence, .ψ = 0 on .∂W (r). Also, ψyi =

.

n  i=1

yi ψyi =

yi , 2s n 1  2s

i=1

yi2

|y|2 dyds s

6.3 The Mean Value Formula

223

=

|y|2 , 2s

whereupon  |y|2 yi ψyi . =2 s n

.

i=1

Therefore B=

.

=

=

=

 

4

n 

us

r n+1

W (r)

 

4

W (r)



 

4 r n+1

 (yi ψ)yi − ψ dyds

i=1

 n  (yi ψ)yi dyds −nus ψ + us

W (r)

i=1



 

4

n 

us

r n+1

yi ψyi dyds

i=1

−nus ψ −

r n+1

W (r)

n 

(ψ = 0

on

∂W (r))



usyi yi ψ dyds.

i=1

Now, we integrate by parts with respect to .s and we get B=

.

=

−nus ψ +

r n+1

W (r)

r n+1

−

=− Consequently,

4

−nus ψ + W (r)

nus ψ +

r n+1 1

W (r)

 

r n+1 4 r n+1

 uyi yi ψs dyds

n 

 uyi yi

i=1



 

n  i=1



 

4

=−



 

4

W (r) i=1

n n  uyi yi dyds 2s

uyi yi

|y|2 s2

dyds



W (r)



i=1

n 

 

|y|2 n − − 2 2s 4s 

 n n  uyi yi dyds − A. nus ψ + 2s i=1

dyds

224

6 The Heat Equation

 n n  .φ (r) = − uyi yi dyds nus ψ + 2s r n+1 W (r) i=1     n 4 n  = − n+1 uyi yi dyds (ψ = 0 nΔuψ + 2s r W (r) i=1     n n  4 n  = − n+1 uyi ψyi + uyi yi dyds −n 2s r W (r) i=1 i=1     n n 4 n  n  = − n+1 uyi yi + uyi yi dyds − 2s 2s r W (r) 4

'



 

i=1

on

∂W (r))

i=1

= 0. Thus, .φ is a constant. Hence, φ(r) = lim φ(t)

.

t→0

 

= u(0, 0)

|y|2 dyds 2 W (1) s

= 4u(0, 0), whereupon u(0, 0) =

.

1 4r n

  u(y, s) W (r)

|y|2 dyds. s2

6.4 The Weak and Strong Maximum Principles Consider the heat equation for a function .u in a .n dimensional bounded domain .D, ut = kΔu,

.

(x1 , . . . , xn ) ∈ D,

t > 0,

(6.14)

where .k is a positive constant. Define the domain QT = {(x1 , . . . , xn , t) : (x1 , . . . , xn ) ∈ D,

.

0 < t ≤ T },

where .T > 0 is arbitrarily chosen. Define the parabolic boundary of .QT as follows ∂p QT = {D × {0}} ∪ {∂D × [0, T ]}.

.

6.4 The Weak and Strong Maximum Principles

225

Let .CQT be the class of functions that are twice continuously differentiable in .QT with respect to .(x1 , . . . , xn ) and once continuously differentiable with respect to .t in .QT , and continuous in .QT . Example 6.9 Let .u ∈ CQT and .ut − Δu < 0 in .QT . We will prove that .u has no local maximum in .QT and .u achieves its maximum in .∂p QT . Really, assume that .u has a local maximum at some point .(x, t) ∈ QT . Then .ut (x, t) = 0 and .Δu(x, t) ≤ 0, which is a contradiction. Since .u is continuous in .QT , then its maximum is achieved somewhere in .∂QT . If the maximum is achieved at a point .(x0 , T ) ∈ D × {T }, then .ut (x, T ) = 0 and .Δu(x, T ) ≤ 0, which is a contradiction. Exercise 6.17 Let .u ∈ CQT and .ut − kΔu > 0 in .QT . Prove that .u has no local minimum in .QT and .u achieves its minimum in .∂p QT . Example 6.10 Let .u ∈ CQT , .f ∈ C (QT ) and ut − Δu = f (x, t),

(x, t) ∈ QT .

.

We will prove that  .

min

(x,t)∈∂p QT



u(x, t) − T

|f (x, t)| + 1 ≤ u(x, t),

sup

(x, t) ∈ QT .

(x,t)∈QT

(6.15) Let K=

.

sup

|f (x, t)|.

(x,t)∈QT

Consider the function v(x, t) = u(x, t) − (K + 1)t,

.

(x, t) ∈ QT .

Then vt (x, t) = ut (x, t) − (K + 1),

.

Δv(x, t) = Δu(x, t),

(x, t) ∈ QT .

Hence, vt (x, t) − Δv(x, t) = ut (x, t) − K − 1 − Δu(x, t)

.

= f (x, t) − K − 1 < 0,

(x, t) ∈ QT .

226

6 The Heat Equation

Hence, applying Example 6.9, we get .

min

(x,t)∈∂p QT

v(x, t) ≤ v(x, t) = u(x, t) − (K + 1)t ≤ u(x, t),

(x, t) ∈ QT .

Now, using that .

min

(x,t)∈∂p QT

v(x, t) ≥ =

u(x, t) − (K + 1)T

min

(x,t)∈∂p QT

 u(x, t) − T

min

(x,t)∈∂p QT

 sup

|f (x, t)| + 1 ,

(x,t)∈QT

we get the inequality (6.15). Exercise 6.18 Let .u ∈ CQT , .f ∈ C (QT ) and ut − Δu = f (x, t),

.

(x, t) ∈ QT .

Prove that  u(x, t) ≤

.

max

(x,t)∈∂p QT



u(x, t) + T

sup

|f (x, t)| + 1

(x, t) ∈ QT .

(x,t)∈QT

Exercise 6.19 Let .u ∈ CQT , .f ∈ C (QT ) and ut − Δu = f (x, t),

.

(x, t) ∈ QT .

Prove that  |u(x, t)| ≤

.

max

(x,t)∈∂p QT

|u(x, t)| + T

 sup

|f (x, t)| + 1 ,

(x,t)∈QT

Exercise 6.20 Let .T = 2 and D = {(x1 , x2 ) ∈ R2 : x12 + x22 ≤ 4}.

.

Let also, .u ∈ C (Q2 ) be a solution to the equation ut − Δu =

.

Prove that

1 + x12

1 , + x22 + t 2

(x, t) ∈ Q2 .

(x, t) ∈ QT .

6.4 The Weak and Strong Maximum Principles

|u(x, t)| ≤

max

.

(x,t)∈∂p Q2

227

|u(x, t)| + 4,

(x, t) ∈ Q2 .

Example 6.11 (Weak Maximum Principle for the Heat Equation) Let .u ∈ CQT be a solution to the heat equation (6.14). We will prove that .u achieves its maximum(minimum) on .∂p QT . Let .ε > 0 be arbitrarily chosen and M = max u.

.

∂p QT

Define the function v(x, t) = u(x, t) − εt.

.

We have . max v ≤ M. Hence, ∂p QT

vt − Δv = ut − ε − Δu

.

= −ε < 0

in QT .

From here, using Example 6.9, we conclude that .v achieves its maximum in .∂p QT . Consequently v≤M

.

in

QT

or u ≤ M + εt ≤ M + εT

in QT .

.

Because .ε > 0 was arbitrarily chosen, we conclude that .u ≤ M in .QT . Exercise 6.21 Let .u ∈ CQT be a solution to the heat equation (6.14). If u(x, t) ≥ (≤)0 on .∂p QT , prove that .u(x, t) ≥ (≤)0 on .QT .

.

Exercise 6.22 Let .u ∈ CQT be a solution to the equation ut − kΔu = f (x, t),

.

(x1 , · · · , xn ) ∈ D,

0 < t ≤ T,

(6.16)

where .f ∈ C (QT ) and .|f | ≤ N on .QT . Let also, .|u(x, t)| ≤ m on .∂p QT . Prove that |u(x, t)| ≤ Nt + m

.

on

QT .

Exercise 6.23 Let .n = 3 and D = {(x1 , x2 , x3 ) ∈ R3 : x1 + x2 + x3 ≥ 1}.

.

(6.17)

228

6 The Heat Equation

Let also, u be a solution of Eq. (6.16) for .k = 1, .T = 2 and f (x1 , x2 , x3 , t) = 3+

.

sin t , 1 + (x1 + x2 + x3 )2 + t 2

(x1 , x2 , x3 ) ∈ R3 ,

0 ≤ t ≤ 2.

Prove that |u(x, t)| ≤ 4t +

sup

.

|u(x, t)|,

(x, t) ∈ Q2 .

(x,t)∈∂p Q2

Exercise 6.24 Let .u1 , u2 ∈ CQT be solutions to Eq. (6.16) with initial condition ui (x, 0) = φi (x), .x ∈ D, .i = 1, 2, and boundary condition .ui (x, t) = ψi (x, t), 2 2 .x ∈ ∂D, .0 ≤ t ≤ T , .i = 1, 2, respectively, where .φi ∈ C (D), .ψi ∈ C (QT ), .i = 1, 2. Set .

δ = max |φ1 − φ2 | + max |ψ1 − ψ2 |.

.

D

∂D×{0}

Prove that |u1 − u2 | ≤ δ,

.

on

(6.18)

QT .

Exercise 6.25 Let .n = 2, .T = 1 and D = {(x1 , x2 ) ∈ R2 : x12 + x22 ≤ 9}.

.

Let also, .u1 and .u2 be solutions to the IBVPs ut − ux1 x1 − ux2 x2 = t (x1 + x2 )2 ,

.

u(x1 , x2 , 0) = 0,

(x1 , x2 ) ∈ D,

0 < t ≤ 1,

(x1 , x2 ) ∈ D,

u(x1 , x2 , t) = 2t (x12 + x22 ), 2

(x1 , x2 ) ∈ ∂D,

t ∈ [0, 1],

and ut − ux1 x1 − ux2 x2 = t (x1 + x2 )2 ,

(x1 , x2 ) ∈ D,

u(x1 , x2 , 0) = (x12 + x22 )2 ,

(x1 , x2 ) ∈ D,

.

u(x1 , x2 , t) = 0,

(x1 , x2 ) ∈ ∂D,

0 < t ≤ 1,

t ∈ [0, 1],

respectively. Prove that |u1 − u2 | < 99

.

on Q1 .

Example 6.12 (The Strong Maximum Principle for the Heat Equation) Let u ∈ CQT be a solution to Eq. (6.14). Let also,

.

6.4 The Weak and Strong Maximum Principles

229

u(x 0 , t 0 ) = max u(x, t) = m > 0

.

QT

in some point .(x 0 , t 0 ) ∈ QT \∂p QT . We will prove that .u(x, t) = m for all .(x, t) ∈ Qt 0 for which there exists a continuous curve that connects .(x, t) and 0 0 .(x , t ) and lies in .Qt 0 . Really, suppose the contrary. Then there exists a point 1 0 1 1 1 .(x , t ) so that .x ∈ D, .0 ≤ t < t , and u(x 1 , t 1 ) < m1 < m.

.

Let .Q1 be the cylinder ⎧  n 1 2 ⎨  1 ≤ ρ, .Q = (x, t) : (xi − xi1 )2 ⎩

⎫ ⎬

t1 ≤ t ≤ t2 , ⎭

i=1

where .1 > ρ > 0 and .t 1 < t 2 < t 0 are chosen so that .Q1 ⊂ Qt 0 and .u(x, t 1 ) < m1 for all .x for which  n  .

1 2

≤ ρ.

(xi − xi1 )2

i=1

Let .α > 0 be chosen so that 4k 2 (n + 2)2 − 8kαρ 2 < 0.

.

We consider the function  w(x, t) = m−(m−m1 ) ρ 2 −

n 

.

2 (xi −xi1 )2

e−α(t−t ) −u(x, t), 1

(x, t) ∈ Q1 .

i=1

Then ⎛ .

⎝ ρ − 2

n 

2 ⎞ (xi − xi1 )2



 n    1 2 −2(xj − xj1 ) =2 ρ − (xi − xi )

⎠

i=1

2

xj

i=1



 n  1 2 = −4 ρ − (xi − xi ) (xj − xj1 ), 2

⎛ 2 ⎞ n  ⎝ ρ2 − (xi − xi1 )2 ⎠ i=1

xj xj

i=1

  = −4 −2(xj − xj1 ) (xj − xj1 )

230

6 The Heat Equation



n  −4 ρ − (xi − xi1 )2



2

i=1



 n  1 2 (xi − xi ) + 8(xj − xj1 )2 , = −4 ρ − 2

⎛ Δ⎝ ρ − 2

n 

i=1

2 ⎞



⎠ = −4n ρ −

(xi − xi1 )2

2

i=1

n 

 (xi − xi1 )2

i=1

+8

n  (xi − xi1 )2 ,

(x, t) ∈ Q1 ,

i=1

from where ⎛

n 

Δw(x, t) = −(m . − m1 )Δ ⎝ ρ 2 − 

2 ⎞ (xi − xi1 )2

⎠ e−α(t − t 1 ) − Δu(x, t)

i =1



  n n   = 4n(m−m1 ) ρ 2 − (xi −xi1 )2 − 8(m − m1 ) (xi − xi1 )2 i=1

×e 

−α(t−t 1 )

i=1

− Δu(x, t), 

  n n   1 2 1 2 −kΔw(x, t) = −4kn(m −m1 ) ρ − (xi −xi ) + 8k(m−m1 ) (xi −xi ) 2

i=1

×e

i=1

−α(t−t 1 )

+ kΔu(x, t),  2 n  1 2 1 2 (xi − xi ) e−α(t−t ) − ut (x, t), wt (x, t) = α(m − m1 ) ρ − i=1

(x, t) ∈ Q1 , and

.

.

2   n  2 1 2 wt (x, t) − kΔw(x, t) = (m − m1 ) α ρ − (xi − xi )  −4kn ρ − 2

n 

 (xi − xi1 )2

i=1



 n  1 1 2 2 (xi − xi ) + 8kρ e−α(t−t ) − 8k ρ − 2

i=1

i=1

⎛ 

n  = (m − m1 ) ⎝α ρ − (xi − xi1 )2 2

i=1

2

 − 4k(n + 2) ρ − 2

n  i=1

 (xi − xi1 )2

6.4 The Weak and Strong Maximum Principles

231

 +8kρ ≥ 0,

2

e−α(t−t

1)

(x, t) ∈ Q1 .

When  n 1 2  1 2 =ρ . (xi − xi ) i=1

and .t 1 ≤ t ≤ t 2 , we have w(x, t) = m − u(x, t) ≥ 0,

.

(x, t) ∈ Q1 .

When .t = t 1 and  n 1 2  1 2 ≤ ρ, . (xi − xi ) i=1

we have 

n  2 .w(x, t) = m − (m − m1 ) ρ − (xi − xi1 )2

2 − u(x, t)

i=1

≥ m − (m − m1 )ρ 4 − m1 > m − (m − m1 ) − m1 = 0,

(x, t) ∈ Q1 .

Hence and Exercise 6.21, we conclude that .w(x, t) ≥ 0 in .Q1 . Let .(x 2 , t 2 ) ∈ Q1 and n  (xi2 − xi1 )2 < ρ 2 .

.

i=1

Then w(x 2 , t 2 ) ≥ 0

.

and

232

6 The Heat Equation

 w(x , t ) = m − (m − m1 ) ρ −

.

2

2

2

n 

2 (xi2

− xi1 )2

e−α(t

2 −t 1 )

− u(x 2 , t 2 )

i=1

≥ 0, (x, t) ∈ Q1 , whereupon

.



n  .u(x , t ) ≤ m − (m − m1 ) ρ − (xi2 − xi1 )2 2

2

2

2

e−α(t

2 −t 1 )

< m,

i=1

(x, t) ∈ Q1 . Continuing this process we obtain the points .(x 1 , t 1 ), .(x 2 , t 2 ), .. . . , k k 0 0 .(x , t ), .(x , t ) so that .

t1 < t2 < . . . < t0

.

and from .u(x s , t s ) < m we get .u(x s+1 , t s+1 ) < m, .s = 1, . . . , k − 1. Therefore 0 0 .u(x , t ) < m, which is a contradiction.

6.5 The Maximum Principle for the Cauchy Problem Here we consider the Cauchy problem .

ut − Δu = 0 in Rn × (0, T ), u = φ on Rn × {t = 0},

(6.19)

where .T > 0 is fixed. Example 6.13 (The Maximum Principle for the Cauchy Problem) Suppose that .φ ∈ C (Rn ) and .u ∈ C 2 (Rn , C 1 ((0, T ])) ∩ C (Rn × [0, T ]) solves the Cauchy problem (6.19) and satisfies the growth estimate 2

u(x, t) ≤ Aea|x| ,

.

x ∈ Rn ,

0 ≤ t ≤ T,

for some constants .A, a > 0. We will prove that .

sup

Rn ×[0,T ]

u = sup φ. Rn

We consider the following cases. 1. Let .4aT < 1. Then there are .ε > 0 and .γ > 0 such that

6.5 The Maximum Principle for the Cauchy Problem

233

1 = a + γ. 4(T + ε)

4a(T + ε) < 1 and

.

We fix .y ∈ Rn and .μ > 0. Define the function v(x, t) = u(x, t) −

.

|x−y|2

μ (T + ε − t)

n 2

x ∈ Rn ,

e 4(T +ε−t) ,

0 < t ≤ T.

We have vt (x, t) = ut (x, t) −

.

|x−y|2 n n μ(T + ε − t)− 2 −1 e 4(T +ε−t) 2

|x−y|2 n μ − (T + ε − t)− 2 −2 |x − y|2 e 4(T +ε−t) , 4 |x−y|2 n μ vxi (x, t) = uxi (x, t) − (T + ε − t)− 2 −1 (xi − yi )e 4(T +ε−t) , 2 |x−y|2 n μ vxi xi (x, t) = uxi xi (x, t) − (T + ε − t)− 2 −1 e 4(T +ε−t) 2 |x−y|2 n μ − (T + ε − t)− 2 −2 (xi − yi )2 e 4(T +ε−t) , i = 1, . . . , n, 4 |x−y|2 n n Δv(x, t) = Δu(x, t) − μ(T + ε − t)− 2 −1 e 4(T +ε−t) 2 |x−y|2 n μ − (T + ε − t)− 2 −2 |x − y|2 e 4(T +ε−t) , 4

whence vt (x, t) − Δv(x, t) = 0

.

in Rn × (0, T ].

Let .D = B(y, r). Then, by Example 6.11, we get max v = max v.

.

QT

∂p QT

If .x ∈ Rn , then v(x, 0) = u(x, 0) −

.

≤ u(x, 0) = φ(x). If .|x − y| = r, .0 ≤ t ≤ T , then

|x−y|2

μ (T + ε)

n 2

e 4(T +ε)

234

6 The Heat Equation

(T + ε − t)

n 2

(T + ε − t)

a(|y|+r)2

n 2

(T + ε)

e 4(T +ε−t) r2

μ

2

≤ Aea(|y|+r) −

e 4(T +ε−t) r2

μ

2

≤ Aea|x| −

= Ae

r2

μ

v(x, t) = u(x, t) −

.

n 2

e 4(T +ε) n

− μ(4(a + γ )) 2 e(a+γ )r

2

≤ sup φ Rn

for .r > 0 sufficiently small. Consequently v(x, t) ≤ sup φ

.

Rn

for all .x ∈ Rn , .t ∈ [0, T ]. Let .μ → 0. Then u(x, t) ≤ sup φ

.

Rn

for all .x ∈ Rn and .0 ≤ t ≤ T . 2. If .4aT ≥ 1, then we apply the above result on .[0, T1 ], .[T1 , 2T1 ], .. . . , for some 1 . .T1 ∈ 0, 4a Exercise 6.26 Suppose that .φ ∈ C (Rn ) and .u ∈ C 2 (Rn , C 1 ((0, T ])) ∩ C (Rn × .[0, T ]) solves the Cauchy problem (6.19) and satisfies the growth estimate 2

u(x, t) ≥ −Aea|x| ,

.

x ∈ Rn ,

0 ≤ t ≤ T,

for some constants .A, a > 0. Prove that .

inf

Rn ×[0,T ]

u = infn φ. R

Exercise 6.27 Let .φ ∈ C (Rn ), .f ∈ C (Rn × [0, T ]). Prove that there exists at most one solution .u ∈ C 2 (Rn , C 1 ((0, T ])) ∩ C (Rn × [0, T ]) of the Cauchy problem .

ut − Δu = f (x, t) in Rn × (0, T ], u = φ on Rn × {t = 0}

(6.20)

satisfying the growth estimate 2

|u(x, t)| ≤ Aea|x| ,

.

x ∈ Rn ,

0 ≤ t ≤ T,

(6.21)

6.6 Advanced Practical Problems

235

for some constants .A, a > 0.

6.6 Advanced Practical Problems Problem 6.1 Solve the following the Cauchy problems 1. ut − uxx = 0,

x ∈ R,

.

u(x, 0) = xe

−x 2

t > 0,

x ∈ R.

,

2. 4ut − uxx = 0,

x ∈ R,

.

u(x, 0) = sin xe

−x 2

t > 0, x ∈ R.

,

3. ut −

n 

.

uxj xj = 0,

(x1 , . . . , xn ) ∈ Rn ,

t > 0,

j =1

u(x1 , . . . , xn , 0) = e

−

n  j =1

xj2

(x1 , . . . , xn ) ∈ Rn .

,

4. ut −

n 

.

j =1

uxj xj = 0, ⎛

u(x1 , . . . , xn , 0) = ⎝

(x1 , . . . , xn ) ∈ Rn ,

n 

⎞ xj ⎠ e

−

n  j =1

xj2

t > 0,

(x1 , . . . , xn ) ∈ Rn .

,

j =1

5. ut −

n 

.

j =1

uxj xj = 0,

(x1 , . . . , xn ) ∈ Rn , ⎛

u(x1 , . . . , xn , 0) = sin ⎝

n  j =1

⎞ xj ⎠ e

−

n  j =1

xj2

,

t > 0,

(x1 , . . . , xn ) ∈ Rn .

236

6 The Heat Equation

6. ut −

n 

.

uxj xj = 0,

(x1 , . . . , xn ) ∈ Rn ,

t > 0,

j =1

u(x1 , . . . , xn , 0) = e

−

n  j =1

xj2

,

(x1 , . . . , xn ) ∈ Rn .

7. .

ut − ux1 x1 − ux2 x2 − ux3 x3 = 0, (x1 , x2 , x3 ) ∈ R3 , t > 0, = 2x12 − 2x22 + 3x32 , (x1 , x2 , x3 ) ∈ R3 . u(x1 , x2 , x3 , 0)

Problem 6.2 Solve the following Cauchy problems 1. ut − uxx = 2t (x 2 − t),

.

u(x, 0) = 0,

x ∈ R,

t > 0,

x ∈ R.

2. ut − uxx = cos t − 6tx + x 3 ,

.

u(x, 0) = 0,

x ∈ R,

t > 0,

x ∈ R.

3. ut − ux1 x1 − ux2 x2 = x1 + x22 − 2t,

.

u(x1 , x2 , 0) = 0,

(x1 , x2 ) ∈ R2 ,

t > 0,

(x1 , x2 ) ∈ R2 .

4. ut −

n 

.

uxj xj

⎛ ⎞ n  = cos t ⎝ xj2 ⎠ − 2n sin t,

j =1

u(x1 , . . . , xn , 0) = 0,

(x1 , . . . , xn ) ∈ Rn ,

t > 0,

j =1

(x1 , . . . , xn ) ∈ Rn .

5. .

ut − ux1 x1 − ux2 x2 − ux3 x3 = 2tx1 + x2 + 6t 2 x3 , (x1 , x2 , x3 ) ∈ R3 , R3 , t > 0, = 0, (x1 , x2 , x3 ) ∈ R3 . u(x1 , x2 , x3 , 0)

6.6 Advanced Practical Problems

237

Problem 6.3 Solve the following Cauchy problems 1. ut = uxx + e−t cos x,

x ∈ R,

.

u(x, 0) = cos x,

t > 0,

x ∈ R.

2. ut = uxx + et sin x,

x ∈ R,

.

u(x, 0) = sin x,

t > 0,

x ∈ R.

3. ut = uxx + sin t,

.

u(x, 0) = e

−x 2

x ∈ R,

t > 0,

x ∈ R.

,

4. ut = ux1 x1 + ux2 x2 + sin t sin x1 sin x2 ,

(x1 , x2 ) ∈ R2 ,

.

u(x1 , x2 , 0) = 1,

t > 0,

(x1 , x2 ) ∈ R2 .

5. ut = ux1 x1 + ux2 x2 + cos t,

(x1 , x2 ) ∈ R2 ,

.

u(x1 , x2 , 0) = x1 x2 e

−x12 −x22

,

t > 0,

(x1 , x2 ) ∈ R2 .

6. 8ut = ux1 x1 + ux2 x2 + 1,

.

u(x1 , x2 , 0) = e−(x1 −x2 , )2

(x1 , x2 ) ∈ R2 ,

t > 0,

(x1 , x2 ) ∈ R2 .

7. ut = 3(ux1 x1 + ux2 x2 + ux3 x3 ) + et ,

.

u(x1 , x2 , 0) = sin(x1 − x2 − x3 ),

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

(x1 , x2 , x3 ) ∈ R3 .

8. 4ut = ux1 x1 + ux2 x2 + ux3 x3 + sin(2x3 ),

.

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

238

6 The Heat Equation

1 2 sin(2x3 ) + e−x1 cos(2x2 ), 4

u(x1 , x2 , 0) =

(x1 , x2 , x3 ) ∈ R3 .

9. ut = ux1 x1 + ux2 x2 + ux3 x3

.

+ cos(x1 − x2 + x3 ), u(x1 , x2 , 0) = e−(x1 +x2 −x3 , )2

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

(x1 , x2 , x3 ) ∈ R3 .

10. = ux1 x1 + ux2 x2 + ux3 x3 . +2t (x12 + x22 + x32 ) − 6t 2 , u(x1 , x2 , x3 , 0) = x1 , (x1 , x2 , x3 ) ∈ R3 . ut

(x1 , x2 , x3 ) ∈ R3 ,

Problem 6.4 Find a formal solution to the following problems 1. ut = uxx ,

0 < x < 1,

.

t > 0,

ux (0, t) = 0, u(1, t) = 0,

t ≥ 0,

u(x, 0) = 1 − x,

x ∈ [0, 1].

2. ut = uxx ,

.

0 < x < 1,

t > 0,

ux (0, t) = 0, ux (1, t) = 0,

t ≥ 0,

u(x, 0) = 1,

x ∈ [0, 1].

3. ut = uxx ,

.

0 < x < 1,

ux (0, t) = 0, ux (1, t) = 0, u(x, 0) = 2x,

t ≥ 0, x ∈ [0, 1].

t > 0,

t > 0,

6.6 Advanced Practical Problems

239

4. ut = uxx ,

.

0 < x < 1,

t > 0,

u(0, t) = 0, ux (1, t) = e−t ,

t ≥ 0,

u(x, 0) = 1,

x ∈ [0, 1].

5. ut = uxx ,

.

0 < x < 1,

t > 0,

ux (0, t) = t, ux (1, t) = 1,

t ≥ 0,

u(x, 0) = 0,

x ∈ [0, 1].

Problem 6.5 Consider the Cauchy problem .

ut − Δu = f (x, t) in QT u = φ on D × {t = 0},

(6.22)

where f ∈ C (QT ), φ ∈ C (D). Prove that there exists at most one solution u ∈ CQT of the problem (6.22). Problem 6.6 Let T = 1 and D = {(x1 , x2 ) ∈ R2 : x12 + x22 ≤ 1}.

.

Let also, u ∈ C (Q1 ) be a solution to the equation 1 , + x22 ) + t 2

(x, t) ∈ Q1 .

7 |u(x, t)| + , 3 (x,t)∈∂p Q1

(x, t) ∈ Q1 .

ut − Δu =

.

3 + 2(x12

Prove that |u(x, t)| ≤

.

max

Problem 6.7 Let n = 4 and D = {(x1 , x2 , x3 , x4 ) ∈ R4 : x1 + x2 + x3 = 2,

.

x4 ∈ R}.

240

6 The Heat Equation

Let also, u be a solution of Eq. (6.16) for k = 3, T = 5 and f (x1 , x2 , x3 , t) =

.

1 1 + x42 + t 2 +

1 , 1 + (x1 + x2 + x3 )2

(x1 , x2 , x3 , x4 ) ∈ R4 ,

0 ≤ t ≤ 5.

Prove that |u(x, t)| ≤

.

6 t + sup |u(x, t)|, 5 (x,t)∈∂p Q5

(x, t) ∈ Q5 .

Problem 6.8 Let n = 2, T = 2 and D = {(x1 , x2 ) ∈ R2 : x12 + x22 ≤ 4}.

.

Let also, u1 and u2 be solutions to the IBVPs ut − ux1 x1 − ux2 x2 = t + x1 + x2 ,

(x1 , x2 ) ∈ D,

.

u(x1 , x2 , 0) = t

2

(x12

+ x22 )2 ,

0 < t ≤ 2,

(x1 , x2 ) ∈ D,

u(x1 , x2 , t) = 3t (x12 + x22 ),

(x1 , x2 ) ∈ ∂D,

t ∈ [0, 2],

and ut − ux1 x1 − ux2 x2 = t + x1 + x2 ,

(x1 , x2 ) ∈ D,

.

u(x1 , x2 , 0) = 2t 2 (x12 + x22 )2 , u(x1 , x2 , t) =

t (x12

+ x22 ),

(x1 , x2 ) ∈ D,

(x1 , x2 ) ∈ ∂D,

respectively. Prove that |u1 − u2 | ≤ 80

.

on

0 < t ≤ 1,

Q2 .

t ∈ [0, 2],

Chapter 7

The Wave Equation

The wave equation is a hyperbolic second order linear partial differential equation for descriptions of waves or standing wave fields. It arises in fields like acoustic, electromagnetism, and fluid dynamics. The scalar wave equation describes waves in scalars by scalar function u = u(x1 , . . . , xn , t) of a time variable t and one or more spacial variables x1 , . . . , xn . The scalar wave equation is utt = c2 (ux1 x1 + · · · + uxn xn ),

.

where c is a fixed nonnegative coefficient. In other words, u is the factor representing a displacement from rest situation, t represents time, utt is a term for how the displacement accelerates, x represents space or position, uxi xi , i ∈ {1, . . . , n}, is a term for how the displacement is varying at the point x in one of the dimensions. The wave equation and its modifications play fundamental roles in continuum mechanics, quantum mechanics, plasma physics, general relativity, geophysics, and many other scientific and technical disciplines.

7.1 The One Dimensional Wave Equation 7.1.1 The Cauchy Problem and the d’Alambert Formula The homogeneous wave equation in one(spatial) dimension has the form utt − c2 uxx = 0,

.

−∞ ≤ a < x < b ≤ ∞,

t > 0,

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. G. Georgiev, An Excursion Through Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-031-48784-2_7

(7.1)

241

242

7 The Wave Equation

where c ∈ R is called the wave speed. Introducing the new variables ξ1 (x, t) = x + ct,

.

ξ2 (x, t) = x − ct,

−∞ ≤ a < x < b ≤ ∞, t > 0,

we get the canonical form of the Eq. (7.1) utt − c2 uxx = −4c2 uξ1 ξ2

.

= 0,

−∞ ≤ a < x < b ≤ ∞,

t > 0.

Therefore its general solution is given by u(x, t) = f (x + ct) + g(x − ct),

.

−∞ ≤ a < x < b ≤ ∞,

t > 0,

(7.2)

where f, g ∈ C 2 (R) are two arbitrary functions. Conversely, any two functions f, g ∈ C 2 (R) define a solution of the wave Eq. (7.1) via the formula (7.2). The function g(x − ct) represents a wave moving to the right with velocity c and it is called a forward wave. The function f (x + ct) is a wave travelling to the left with the same speed, and it is called a backward wave. The Eq. (7.2) shows the fact that any solution of the wave equation is the sum of two such travelling waves. Since for any two piecewise continuous functions f and g, the Eq. (7.2) defines a piecewise continuous function u that is a superposition of a backward and a forward wave travelling in opposite directions with speed c. Let {fn (s)}n∈N and {gn (s)}n∈N be sequences of smooth functions converging at any point t to f and g, respectively, which converge uniformly to these functions in any bounded and closed interval that does not contain points of discontinuity. The function un (x, t) = fn (x + ct) + gn (x − ct),

.

−∞ ≤ a < x < b ≤ ∞,

t > 0,

is a proper solution of the wave equation, but the limiting function u(x, t) = f (x + ct) + g(x − ct),

.

−∞ ≤ a < x < b ≤ ∞,

t >0

is not necessary to be twice differentiable. Therefore it might be not a solution of (7.1). Definition 7.1 We call a function u = u(x, t), −∞ ≤ a < x < b ≤ ∞, t > 0, that satisfies (7.2) with piecewise continuous functions f and g a generalized solution of the wave Eq. (7.1). The Cauchy problem for the one dimensional homogeneous wave equation is given by

7.1 The One Dimensional Wave Equation

utt − c2 uxx = 0,

.

243

−∞ < x < ∞,

t > 0,

u(x, 0) = φ(x), .

ut (x, 0) = ψ(x),

−∞ < x < ∞,

(7.3) (7.4)

where φ ∈ C 2 (R) and ψ ∈ C 1 (R). Definition 7.2 A classical(proper) solution of the Cauchy problem (7.3), (7.4) is a function u that is 1. twice continuously differentiable for all t > 0, 2. u and ut are continuous in the half space t ≥ 0 and such that (7.3), (7.4) are satisfied. Recall that the general solution of the wave equation is of the form (7.2). We will find the functions f and g using the initial conditions (7.4). Substituting t = 0 in (7.2), we obtain u(x, 0) = f (x) + g(x) (7.5)

.

= φ(x),

−∞ < x < ∞.

Differentiating (7.2) with respect to t and substituting t = 0, we get ut (x, 0) = cf ' (x) − cg ' (x)

.

= ψ(x),

−∞ < x < ∞.

Integrating the last equation over [0, x], we get 1 .f (x) − g(x) = c

x ψ(s)ds + C,

(7.6)

0

where C = f (0) − g(0). The Eqs. (7.5) and (7.6) are two linear algebraic equations for f and g. The solution of this system of equations is given by f (x) = 12 φ(x) +

1 2c

g(x) = 12 φ(x) −

1 2c

.

x 0 x 0

ψ(s)ds +

C 2

ψ(s)ds −

C 2,

−∞ < x < ∞.

Substituting these expressions for f and g into the general solution (7.2), we obtain

244

7 The Wave Equation x+ct 

1 φ(x + ct) + φ(x − ct) + .u(x, t) = 2c 2

−∞ < x < ∞,

ψ(s)ds,

t ≥ 0.

x−ct

(7.7) which is called the d’Alambert formula.1 Example 7.1 Consider the Cauchy problem utt − 9uxx = 0,

−∞ < x < ∞,

t > 0,

.

= ut (x, 0) = x 2 ,

u(x, 0)

−∞ < x < ∞.

Here c = 3 and φ(x) = ψ(x)

.

= x2,

−∞ < x < ∞.

Then, using the d’Alambert formula, we get 1 (x + 3t)2 + (x − 3t)2 + .u(x, t) = 6 2

x+3t 

s 2 ds x−3t

= x 2 + 9t 2 +

1 3 s=x+3t s  18 s=x−3t

= x 2 + 9t 2 +

 1  (x + 3t)3 − (x − 3t)3 18

= x 2 + 9t 2 + x 2 t + 3t 3 ,

−∞ < x < ∞,

t ≥ 0.

Example 7.2 Consider the Cauchy problem

1 Jean le Rond D’lambert (17 November 1717–29 October 1783) was a French mathematician, mechanician, physicist, philosopher, and music theorist. D’Alambert’s formula for obtaining solutions to the wave equation is named after him. The wave equation is sometimes referred as d’Alambert’s equation.

7.1 The One Dimensional Wave Equation

245

utt − 4uxx = 0,

−∞ < x < ∞,

t > 0,

= φ(x)

u(x, 0)

= .

ut (x, 0)

⎧ ⎨ 1 − x2 ⎩

|x| ≤ 1

0 otherwise,

= ψ(x)

=

⎧ ⎨ (x − 1)(x − 2) ⎩

1≤x≤2

0 otherwise.

We will find u(1, 1). Using the d’Alambert formula, we have

u(1, 1) =

.

φ(3) + φ(−1) 1 + 2 4

1 = 4

3 ψ(s)ds −1

2 (s − 1)(s − 2)ds 1

1 = 4

2 (s 2 − 3s + 2)ds 1

=

1 4

=−



s=2  1 3 s=2 3 2 s=2  s  − s  + 2s  s=1 3 s=1 2 s=1

1 . 24

Example 7.3 Consider the Cauchy problem utt − uxx = 0,

.

−∞ < x < ∞,

u(x, 0) = x, ut (x, 0) = 4x,

−∞ < x < ∞.

t > 0,

246

7 The Wave Equation

Here c = 1 and φ(x) = x,

.

ψ(x) = 4x,

−∞ < x < ∞.

Applying the d’Alambert formula, we find 1 x+t +x−t + .u(x, t) = 2 2

x+t 4sds x−t

x+t = x+2

sds

x−t

s=x+t  = x + s 2  s=x−t

= x + (x + t)2 − (x − t)2 = x + x 2 + 2xt + t 2 − x 2 + 2xt − t 2 = x + 4xt = x(1 + 4t),

−∞ < x < ∞,

t ≥ 0.

Exercise 7.1 Solve the Cauchy problem utt − uxx = 0, .

u(x, 0)

−∞ < x < ∞,

t > 0,

= x,

ut (x, 0) = cos x,

−∞ < x < ∞.

Exercise 7.2 Fix T > 0. Prove that the Cauchy problem (7.3), (7.4) in the domain −∞ < x < ∞, 0 ≤ t ≤ T , is well-posed for any φ ∈ C 2 (R), ψ ∈ C 1 (R). Remark 7.1 The d’Alambert formula is also valid for −∞ < x < ∞, T < t ≤ 0, and the Cauchy problem is well-posed in this domain.

7.1 The One Dimensional Wave Equation

247

Remark 7.2 The Cauchy problem is ill-posed on the domain −∞ < x < ∞, t ≥ 0. Indeed, consider the Cauchy problem utt − uxx = 0, .

un (x, 0) =

1 n2

−∞ < x < ∞,

t > 0,

sin(nx),

unt (x, 0) = 0,

−∞ < x < ∞.

We have that un (x, t) =

.

1 cosh(nt) sin(nx), n2

−∞ < x < ∞,

t ≥ 0,

is its solution. When n is large enough, the initial conditions describe an arbitrary small perturbation of the trivial solution u = 0. On the other hand, sup |un (x, t)| x∈R

grows fast as n → ∞ for any t > 0.

7.1.2 The Cauchy Problem for the Nonhomogeneous Wave Equation Consider the following Cauchy problem utt − c2 uxx = f (x, t),

.

−∞ < x < ∞,

t > 0,

(7.8)

u(x, 0) = φ(x), (7.9)

.

ut (x, 0) = ψ(x),

−∞ < x < ∞,

where f ∈ C (R × (0, ∞)), φ ∈ C 2 (R), and ψ ∈ C 1 (R) are given functions. This problem models, for example, the vibration of a very long string in the presence of an external force f. The initial conditions φ and ψ represent the shape and the vertical velocity of the string at time t = 0. Exercise 7.3 Prove that the Cauchy problem (7.8), (7.9) admits at most one solution. Let f, fx ∈ C (R × [0, ∞)), φ ∈ C 2 (R), ψ ∈ C 1 (R). Let also, (x0 , t0 ) ∈ R× (0, ∞) be arbitrarily chosen and ∆ be the triangle with edges the points (x0 , t0 ), (x0 − ct0 , 0) and (x0 + ct0 , 0).(see Fig. 7.1) Integrating both sides of the Eq. (7.8) over the triangle ∆, we get

248

7 The Wave Equation

Fig. 7.1 The triangle ∆

 



.

∆

   c2 uxx (x, t) − utt (x, t) dxdt = −

∆

f (x, t)dxdt.

Using the Green formula, we obtain   .−

Δ

.

=

ut (x, t)dx + c2 ux (x, t)dt

 (7.10)

∂∆

(x0  −ct0 ,0)





f (x, t)dxdt =

(x0  +ct0 ,0)





ut (x, t)dx + c ux (x, t)dt + 2

ut (x, t)dx + c2 ux (x, t)dt



(x0 −ct0 ,0)

(x0 ,t0 ) (x0 ,t0 )

+



 ut (x, t)dx + c2 ux (x, t)dt .

(x0 +ct0 ,0)

Note that (x0  −ct0 ,0)

  ut (x, t)dx + c2 ux (x, t)dt =

(x0 ,t0 ) .

(x0  −ct0 ,0)

(cut (x, t)dt + cux (x, t)dx) (x0 ,t0 ) (x0  −ct0 ,0)

=c

du (x0 ,t0 )

= c (u(x0 − ct0 , 0) − u(x0 , t0 )) = c (φ(x0 − ct0 ) − u(x0 , t0 )) , (7.11)

7.1 The One Dimensional Wave Equation

249

(x0  +ct0 ,0)

x0+ct0   2 ut (x, t)dx + c ux (x, t)dt = ut (x, 0)dx x0 −ct0

(x0 −ct0 ,0)

(7.12)

.

x0+ct0

=

ψ(x)dx, x0 −ct0

(x0 ,t0 )

(x0 ,t0 )

  ut (x, t)dx + c2 ux (x, t)dt =

(x0 +ct0 ,0)

(−cut (x, t)dt − cux (x, t)dx) (x0 +ct0 ,0) (x0 ,t0 )

= −c

.

du

(x0 +ct0 ,0)

= −c (u(x0 , t0 ) − u(x0 + ct0 , 0)) = −c (u(x0 , t0 ) − φ(x0 + ct0 )) . (7.13) We substitute (7.11), (7.12) and (7.13) into (7.10) and we find x0+ct0

  .

−

∆

f (x, t)dxdt = c (φ(x0 − ct0 ) − u(x0 , t0 )) +

ψ(x)dx

x0 −ct0

−c (u(x0 , t0 ) − φ(x0 + ct0 )) , or φ(x0 + ct0 ) + φ(x0 − ct0 ) 1 + .u(x0 , t0 ) = 2c 2

x0+ct0 x0 −ct0

1 ψ(x)dx+ 2c

  ∆

f (x, t)dxdt.

Because (x0 , t0 ) ∈ R × (0, ∞) was arbitrarily chosen, we finally obtain an explicit formula for the solutions of the Cauchy problem (7.8), (7.9) at an arbitrary point (x, t) given by 1 φ(x + ct) + φ(x − ct) + .u(x, t) = 2 2c

x+ct 

x−ct

1 ψ(s)ds + 2c

t

x+c(t−τ ) 

f (ξ, τ )dξ dτ, 0 x−c(t−τ )

(7.14)

250

7 The Wave Equation

−∞ < x < ∞, t ≥ 0. Now, we will prove that the function u given by the formula (7.14) is indeed a solution to the Cauchy problem (7.8), (7.9). We have u(x, 0) = φ(x),

.

ut (x, t) =

cφ ' (x + ct) − cφ ' (x − ct) ψ(x + ct) + ψ(x − ct) + 2 2 1 + 2

t (f (x + c(t − τ ), τ ) + f (x − c(t − τ ), τ )) dτ, 0

ut (x, 0) = ψ(x),

utt (x, t) = c2

φ '' (x + ct) + φ '' (x − ct) cψ ' (x + ct) − cψ ' (x − ct) + f (x, t) + 2 2

1 + 2

t (cfx (x + c(t − τ ), τ ) − cfx (x − c(t − τ ), τ )) dτ 0

ux (x, t) =

φ ' (x + ct) + φ ' (x − ct) ψ(x + ct) − ψ(x − ct) + 2 2c 1 + 2c

t (f (x + c(t − τ ), τ ) − f (x − c(t − τ ), τ )) dτ, 0

uxx (x, t) =

φ '' (x + ct) + φ '' (x − ct) ψ ' (x + ct) − ψ ' (x − ct) + 2 2c 1 + 2c

t (fx (x + c(t − τ ), τ ) − fx (x − c(t − τ ), τ )) dτ, 0

−∞ < x < ∞, t ≥ 0, whereupon utt (x, t) − c2 uxx (x, t) = f (x, t),

.

−∞ < x < ∞,

Therefore u is a solution to the Cauchy problem (7.8), (7.9).

t > 0.

7.1 The One Dimensional Wave Equation

251

Definition 7.3 The formula (7.14) is also called the d’Alambert formula. Remark 7.3 Note that for f = 0 both d’Alambert formulas (7.14) and (7.7) coincide. Example 7.4 Consider the Cauchy problem utt − 4uxx = ex + sin t, .

u(x, 0)

= 0,

ut (x, 0)

=

1 , 1+x 2

−∞ < x < ∞,

t > 0,

−∞ < x < ∞.

Here c = 2,

.

f (x, t) = ex + sin t, φ(x) = 0, ψ(x) =

1 , 1 + x2

−∞ < x < ∞,

t ≥ 0.

Then, using the d’Alambert formula (7.14), we get 1 .u(x, t) = 4

x+2t 

x−2t

1 1 ds + 4 1 + s2

t

x+2(t−τ ) 

 eξ + sin τ dξ dτ

0 x−2(t−τ )

s=x+2t 1  t  ξ =x+2(t−τ ) 1  = arctan s  eξ + ξ sin τ  dτ + ξ =x−2(t−τ ) s=x−2t 4 4 0

=

arctan(x + 2t) − arctan(x − 2t) 4 t   1 ex+2t−2τ − ex−2t+2τ + 4(t − τ ) sin τ dτ + 4 0

=

arctan(x + 2t) − arctan(x − 2t) 4

252

7 The Wave Equation

  t 1 x−2t+2τ τ =t 1 x+2t−2τ τ =t − e − e + (t − τ ) sin τ dτ   8 8 τ =0 τ =0 0

arctan(x + 2t) − arctan(x − 2t) 4   1 − ex − ex+2t 8

=

τ =t  t   1 x x−2t − cos τ dτ − (t − τ ) cos τ  − e −e 8 τ =0 0

arctan(x + 2t) − arctan(x − 2t) 4 τ =t    1 1 x+2t − ex + + ex−2t + t − sin τ  e 8 4 τ =0

=

arctan(x + 2t) − arctan(x − 2t) 4 1 x 1 x − e + e cosh(2t) + t − sin t, 4 4

=

−∞ < x < ∞,

Example 7.5 Consider the Cauchy problem utt − uxx = cos(x + t), .

−∞ < x < ∞,

t > 0,

= x,

u(x, 0)

ut (x, 0) = sin x,

−∞ < x < ∞.

Here c = 1,

.

f (x, t) = cos(x + t), φ(x) = x, ψ(x) = sin x,

−∞ < x < ∞,

Then, using the d’Alambert formula (7.14), we have

t ≥ 0.

t ≥ 0.

7.1 The One Dimensional Wave Equation

1 x+t +x−t + .u(x, t) = 2 2

x+t

253

1 sin sds + 2

x−t

t

x+(t−τ  )

cos(ξ + τ )dξ dτ 0 x−(t−τ )

ξ =x+(t−τ ) s=x+t 1  t 1   = x − cos s  + sin(ξ + τ ) dτ s=x−t ξ =x−(t−τ ) 2 2 0

t

1 1 = x − (cos(x + t)− cos(x − t)) + 2 2

(sin(x + t)− sin(x − t + 2τ )) dτ 0

1 1 = x + sin x sin t + t sin(x + t) − 2 2

t sin(x − t + 2τ )dτ 0

τ =t  1 1 = x + sin x sin t + t sin(x + t) + cos(x − t + 2τ ) 2 4 τ =0 1 1 1 = x + sin x sin t + t sin(x + t) + cos(x + t) − cos(x − t) 2 4 4 1 1 = x + sin x sin t + t sin(x + t) − sin x sin t 2 2 =x+

1 1 sin x sin t + t sin(x + t), 2 2

−∞ < x < ∞,

Example 7.6 Consider the Cauchy problem utt − 4uxx = tx,

.

−∞ < x < ∞,

u(x, 0) = x 2 , ut (x, 0) = x, Here c = 2,

.

f (x, t) = xt,

−∞ < x < ∞.

t > 0,

t ≥ 0.

254

7 The Wave Equation

φ(x) = x 2 , ψ(x) = x,

−∞ < x < ∞,

t ≥ 0.

Then, using the d’Alambert formula, we get (x + 2t)2 + (x − 2t)2 1 .u(x, t) = + 2 4

x+2t 

1 sds + 4

x−2t

x

x+2(t−τ ) 

ξ τ dξ dτ 0 x−2(t−τ )

 1 2 s=x+2t x 2 + 4xt + 4t 2 + x 2 − 4xt + 4t 2 + s  = 2 8 s=x−2t 1 + 8

t

ξ =x+2(t−τ )  τ ξ 2  dτ ξ =x−2(t−τ )

0

= x 2 + 4t 2 +

1 + 8

t

 1 (x + 2t)2 − (x − 2t)2 8

  τ (x + 2(t − τ ))2 − (x − 2(t − τ ))2 dτ

0

= x 2 + 4t 2 +

+

1 8

t

 1 2 x + 4xt + 4t 2 − x 2 + 4xt − 4t 2 8

  τ x 2 + 4x(t − τ ) + 4(t − τ )2 − x 2 + 4x(t − τ ) − 4(t − τ )2 dτ

0

1 = x + 4t + xt + 8 2

t τ 8x(t − τ )dτ

2

0

t = x + 4t + xt + x 2

(tτ − τ 2 )dτ

2

0

  τ 3 τ =t τ 2 τ =t −x  = x + 4t + xt + xt  2 τ =0 3 τ =0 2

2

7.1 The One Dimensional Wave Equation

255

1 1 = x 2 + 4t 2 + xt + xt 3 − xt 3 3 2 1 = x 2 + 4t 2 + xt + xt 3 , 6

−∞ < x < ∞,

t ≥ 0.

Exercise 7.4 Solve the following Cauchy problems 1. utt − uxx = xt, .

u(x, 0)

−∞ < x < ∞,

t > 0,

= 0,

ut (x, 0) = ex ,

−∞ < x < ∞.

utt − uxx = 6,

−∞ < x < ∞,

2. .

t > 0,

u(x, 0) = x 2 , ut (x, 0) = 4x,

−∞ < x < ∞.

Exercise 7.5 Let T > 0 be fixed and f, fx ∈ C (R × [0, ∞)), φ ∈ C 2 (R), ψ ∈ C 1 (R). Prove that the Cauchy problem (7.8), (7.9) is well-posed in the domain −∞ < x < ∞, 0 ≤ t ≤ T . Exercise 7.6 Suppose that f (·, t), φ(·), ψ(·) are even functions for all t ≥ 0.Prove that the solution u(·, t) of the Cauchy problem (7.8), (7.9), is also even function for every t ≥ 0. Exercise 7.7 Suppose that f (·, t), φ(·), ψ(·) are odd functions for all t ≥ 0. Prove that the solution u(·, t) of the Cauchy problem (7.8), (7.9), is also odd function for every t ≥ 0. Exercise 7.8 Let ω > 0. Suppose that f (·, t), φ(·), ψ(·) are ω−periodic functions for all t ≥ 0. Prove that the solution u(·, t) of the Cauchy problem (7.8), (7.9), is also ω−periodic function for every t ≥ 0.

7.1.3 Separation of Variables In this section, we will apply the method of separation of variables to the initial boundary value problems for the one dimensional wave equation.

256

7.1.3.1

7 The Wave Equation

Homogeneous IBVPs

We consider the initial value problem utt − c2 uxx = 0,

.

0 < x < L,

t > 0,

(7.15)

u(x, 0) = φ(x), (7.16)

.

ut (x, 0) = ψ(x),

0 < x < L,

where φ ∈ C 2 (R), ψ ∈ C 1 (R) are given functions. The IVP (7.15), (7.16) will be subject to one of the following boundary conditions 1. Neumann2 boundary conditions ux (0, t) = 0, (7.17)

.

ux (L, t) = 0,

t > 0.

2. Dirichlet boundary conditions u(0, t) = 0, (7.18)

.

u(L, t) = 0,

t > 0.

3. Mixed boundary conditions u(0, t) = 0, (7.19)

.

ux (L, t) = 0,

t > 0,

or ux (0, t) = 0, (7.20)

.

u(L, t) = 0,

t > 0.

For each type of boundary conditions, there is a need of compatibility conditions. The compatibility conditions are given as follows.

2 John von Neumann (28 December 1903–8 February 1957) was a Hungarian- American pure and applied mathematician, physicist, inventor, computer scientist and polymath.

7.1 The One Dimensional Wave Equation

257

1. Neumann boundary conditions φ ' (0) = φ ' (L) = ψ ' (0) = ψ ' (L) = 0.

.

2. Dirichlet boundary conditions φ(0) = φ(L) = ψ(0) = ψ(L) = 0.

.

3. Mixed boundary conditions φ(0) = φ ' (L) = ψ(0) = ψ ' (L) = 0,

.

or φ(L) = φ ' (0) = ψ(L) = ψ ' (0) = 0.

.

We will find nontrivial solutions of the Eq. (7.15) with separable variables, i.e., solutions of the form u(x, t) = X(x)T (t),

0 ≤ x ≤ L,

.

t ≥ 0,

(7.21)

that satisfy one of the boundary conditions (7.17), (7.18), (7.19) or (7.20). Substituting (7.21) into (7.15), we find X(x)T '' (t) = c2 X'' (x)T (t),

.

0 ≤ x ≤ L,

t ≥ 0.

By separating the variables, we see .

X'' (x) T '' (t) , = X(x) c2 T (t)

0 ≤ x ≤ L,

t ≥ 0.

Therefore there exists a constant λ such that .

X'' (x) T '' (t) = −λ, = X(x) c2 T (t)

0 ≤ x ≤ L,

X'' + λX = 0,

0 ≤ x ≤ L,

t ≥ 0,

whereupon

(7.22)

.

T '' We have the following cases.

= −λc2 T ,

t ≥ 0.

258

7 The Wave Equation

1. Neumann boundary conditions. By the boundary conditions (7.17), we get ux (0, t) = X' (0)T (t) = 0, .

ux (L, t) = X' (L)T (t) = 0,

t ≥ 0.

Since u is a nontrivial solution, it follows that X' (0) = X' (L)

.

= 0. Therefore X should be a solution of the eigenvalue problem X'' + λX = 0,

.

0 < x < L,

(7.23)

X' (0) = 0, (7.24)

.

X' (L)

= 0.

For the general solution of the first equation of (7.22), we have (a) √ √ X(x) = c1 cosh( −λx) + c2 sinh( −λx),

0 ≤ x ≤ L,

X(x) = c1 + c2 x,

λ = 0.

.

λ < 0.

(b) .

0 ≤ x ≤ L,

(c) √ √ X(x) = c1 cos( λx) + c2 sin( λx),

.

where c1 and c2 are arbitrary real constants. (a) When λ < 0, using (7.24), we find c1 = c2 = 0.

.

0 ≤ x ≤ L,

λ > 0,

7.1 The One Dimensional Wave Equation

259

Therefore X(x) = 0,

0 ≤ x ≤ L,

.

and the eigenvalue problem (7.23), (7.24) does not admit negative eigenvalues. (b) When λ = 0, using (7.24), we find X(x) = c1 ,

0 ≤ x ≤ L.

.

(c) When λ > 0, then c2 = 0

.

and √ √ c1 λ sin(L λ) = 0.

.

If c1 = 0, then X(x) = 0,

0 ≤ x ≤ L,

.

and the eigenvalue problem (7.23), (7.24) does not admit positive eigenvalues. Thus, √ λL = nπ,

n ∈ N,

n2 π 2 , L2

n ∈ N.

.

or λ=

.

The associated eigenfunction is X(x) = cos

 nπ x 

.

L

,

0 ≤ x ≤ L,

which is uniquely determined up to a multiplicative factor. Hence, the solution of the eigenvalue problem (7.23), (7.24) is an infinite sequence of nonnegative simple eigenvalues and their associated eigenfunctions will be denoted by Xn (x) = cos

 nπ x 

.

λn =

L

 nπ 2 L

,

,

0 ≤ x ≤ L,

n ∈ N0 .

260

7 The Wave Equation

Consider the second equation of (7.22) for λ = λn . The solutions are T0 (t) = α0 + β0 t, .

Tn (t) = αn cos

(7.25)

cπ n 

cπ n  L t + βn sin L t ,

t ≥ 0,

n ∈ N,

where αn , βn , n ∈ N0 , are real constants. Then the product solutions of the initial boundary value problem (7.15), (7.16), (7.17) are given by u0 (x, t) = X0 (x)T0 (t) = A0 + B0 t,

.

un (x, t) = cos

nπ x 

n ∈ N,

L

An cos

cπ nt  L

0 ≤ x ≤ L,

+ Bn sin

cπ nt  L

,

t ≥ 0.

Applying the superposition principle, the function u(x, t) = A0 + B0 t +

.

 



∞

 nπ x   cπ nt cπ nt An cos + Bn sin cos , L L L n=1

0 ≤ x ≤ L,

t ≥ 0,

is a generalized(formal) solution of the problem (7.15), (7.16), (7.17). To find the constants An and Bn , n ∈ N0 , we will use the initial conditions (7.16). Assume that the initial data φ and ψ can be expanded into generalized Fourier series with respect to the sequence of the eigenfunctions of the problem (7.23), (7.24) and these series are uniformly convergent, φ(x) = a0 + .

ψ(x) = b0 +

∞

n=1 an cos

∞

n=1 bn cos

We have that 1 .a0 = L

L φ(x)dx, 0

1 b0 = L

L ψ(x)dx, 0

nπ x  L

nπ x  L

, ,

0 ≤ x ≤ L.

7.1 The One Dimensional Wave Equation

2 am = L

L cos

261

 mπ x 

φ(x)dx,

L

0

2 bm = L

L cos

 mπ x 

ψ(x)dx,

L

m ∈ N.

0

Hence, ∞ 

u(x, 0) = A0 +

An cos

.

n=1

= a0 +

∞ 

an cos

 nπ x  L

 nπ x 

n=1

L

,

0 ≤ x ≤ L,

whereupon an = An , n ∈ N0 .

.

Moreover, ut (x, 0) = B0 +

∞ 

.

Bn

n=1

= b0 +

∞ 

 nπ x  cπ n cos L L

bn cos

n=1

 nπ x  L

,

0 ≤ x ≤ L.

Therefore B 0 = b0 ,

.

Bn =

Lbn , cπ n

n ∈ N.

Thus, the problem (7.15), (7.16), (7.17) is formally solved. 2. Dirichlet boundary conditions. Now, we will consider the IVP (7.15), (7.16) subject to the boundary conditions (7.18). Using the boundary conditions (7.18), we find

262

7 The Wave Equation

u(0, t) = X(0)T (t) = 0, .

u(L, t) = X(L)T (t) = 0,

t ≥ 0.

Because u is a nontrivial solution of the considered, we obtain X(0) = X(L)

.

= 0. Therefore X should be a solution of the eigenvalue problem for the Eq. (7.23) subject to the boundary conditions X(0) = 0, (7.26)

.

X(L) = 0. The general solution of the first equation of (7.22) is given by (a) √ √ X(x) = c1 cosh( −λx) + c2 sinh( −λx),

0 ≤ x ≤ L,

X(x) = c1 + c2 x,

λ = 0.

.

λ < 0.

(b) .

0 ≤ x ≤ L,

(c) √ √ X(x) = c1 cos( λx) + c2 sin( λx),

.

where c1 and c2 are arbitrary real constants. (a) When λ < 0, using (7.26), we find c1 = c2 = 0.

.

Therefore

0 ≤ x ≤ L,

λ > 0,

7.1 The One Dimensional Wave Equation

263

X(x) = 0,

0 ≤ x ≤ L,

.

and the eigenvalue problem (7.23), (7.26) does not admit negative eigenvalues. (b) When λ = 0, using (7.26), we find X(x) = 0,

0 ≤ x ≤ L.

.

(c) When λ > 0, then c1 = 0

.

and √ c2 sin(L λ) = 0.

.

If c2 = 0, then X(x) = 0,

0 ≤ x ≤ L,

.

and the eigenvalue problem (7.23), (7.26) does not admit positive eigenvalues. Therefore √ λL = nπ,

n ∈ N,

n2 π 2 , L2

n ∈ N.

.

or λ=

.

The associated eigenfunction is X(x) = sin

 nπ x 

.

L

,

0 ≤ x ≤ L,

which is uniquely determined up to a multiplicative factor. Hence, the solution of the eigenvalue problem (7.23), (7.26) is an infinite sequence of nonnegative simple eigenvalues and their associated eigenfunctions will be denoted by Xn (x) = sin

 nπ x 

.

λn =

L

 nπ 2 L

,

,

0 ≤ x ≤ L,

n ∈ N0 .

264

7 The Wave Equation

The solutions of the second equation of (7.22) are given by (7.25). Hence, the product solutions of the initial boundary value problem (7.15), (7.16), (7.18) are given by u0 (x, t) = X0 (x)T0 (t) = A0 + B0 t,

.

 

nπ x 

An cos cπLnt + Bn sin cπLnt , L

un (x, t) = sin n ∈ N,

0 ≤ x ≤ L,

t ≥ 0.

By the superposition principle, we obtain that the function  



∞

 nπ x   cπ nt cπ nt An cos + Bn sin sin , .u(x, t) = A0 + B0 t + L L L n=1

0 ≤ x ≤ L,

t ≥ 0,

is a generalized(formal) solution of the problem (7.15), (7.16), (7.18). To find the constants An and Bn , n ∈ N0 , we will use the initial conditions (7.16). Assume that the initial data φ and ψ can be expanded into generalized Fourier series with respect to the sequence of the eigenfunctions of the problem (7.23), (7.26) and these series are uniformly convergent, φ(x) = .

ψ(x) =

∞

n=1 an sin

∞

n=1 bn sin

nπ x  L

nπ x  L

, 0 ≤ x ≤ L.

,

We have that am =

2 L

L

sin

mπ x 

sin

mπ x 

0

L

φ(x)dx,

.

bm =

2 L

L 0

L

ψ(x)dx,

m ∈ N.

Hence, u(x, 0) = A0 +

∞ 

.

n=1

An sin

 nπ x  L

7.1 The One Dimensional Wave Equation ∞ 

=

265

an sin

 nπ x  L

n=1

,

0 ≤ x ≤ L,

whereupon A0 = 0 and an = An , n ∈ N,

.

and ut (x, 0) = B0 +

∞ 

Bn

.

n=1

=

∞ 

bn sin

 nπ x  cπ n sin L L

 nπ x 

n=1

L

,

0 ≤ x ≤ L.

Consequently B0 = 0,

.

Bn =

Lbn , cπ n

n ∈ N.

Thus, the problem (7.15), (7.16), (7.18) is formally solved. 3. Now, we consider the IBVP (7.15), (7.16), (7.19). By the boundary conditions (7.19), we get u(0, t) = X(0)T (t) = 0, .

ux (L, t) = X' (L)T (t) = 0,

t ≥ 0.

Since u is a nontrivial solution, it follows that X(0) = X' (L)

.

= 0. Therefore X should be a solution of the eigenvalue problem for the Eq. (7.23) subject to the boundary conditions

266

7 The Wave Equation

X(0) = 0, (7.27)

.

X' (L) = 0. For the general solution of the first equation of (7.22), we have (a) √ √ X(x) = c1 cosh( −λx) + c2 sinh( −λx),

0 ≤ x ≤ L,

X(x) = c1 + c2 x,

λ = 0.

.

λ < 0.

(b) .

0 ≤ x ≤ L,

(c) √ √ X(x) = c1 cos( λx) + c2 sin( λx),

.

0 ≤ x ≤ L,

λ > 0,

where c1 and c2 are arbitrary real constants. (a) When λ < 0, using (7.27), we find c1 = c2 = 0.

.

Therefore X(x) = 0,

.

0 ≤ x ≤ L,

and the eigenvalue problem (7.23), (7.27) does not admit negative eigenvalues. (b) When λ = 0, using (7.27), we find X(x) = 0,

.

0 ≤ x ≤ L.

(c) When λ > 0, then c1 = 0

.

and √ √ c2 λ cos(L λ) = 0.

.

If c2 = 0, then X(x) = 0,

.

0 ≤ x ≤ L,

7.1 The One Dimensional Wave Equation

267

and the eigenvalue problem (7.23), (7.27) does not admit positive eigenvalues. Thus, √ .

λL =

(2n + 1)π , 2

n ∈ N,

or λ=

.

(2n + 1)2 π 2 , 4L2

n ∈ N.

The associated eigenfunction is

X(x) = sin

.

(2n + 1)π x 2L

 ,

0 ≤ x ≤ L,

which is uniquely determined up to a multiplicative factor. Hence, the solution of the eigenvalue problem (7.23), (7.27) is an infinite sequence of nonnegative simple eigenvalues and their associated eigenfunctions will be denoted by

Xn (x) = sin

.

λn =

(2n + 1)π x 2L

(2n + 1)π 2L

 ,

0 ≤ x ≤ L,

2 ,

n ∈ N0 .

Consider the second equation of (7.22) for λ = λn . The solutions are T0 (t) = α0 + β0 t, .

Tn (t) = αn cos



cπ(2n+1) t 2L



+ βn sin



cπ(2n+1) t 2L

 ,

t ≥ 0,

n ∈ N, (7.28)

where αn , βn , n ∈ N0 , are real constants. Then the product solutions of the initial boundary value problem (7.15), (7.16), (7.19) are given by u0 (x, t) = X0 (x)T0 (t) .

= C0 + D0 t, un (x, t) = sin



(2n+1)π x 2L

     An cos cπ(2n+1)t +Bn sin cπ(2n+1)t , n ∈ N, 2L 2L

268

7 The Wave Equation

0 ≤ x ≤ L, t ≥ 0. Applying the superposition principle, the function

 ∞

 cπ(2n + 1)t .u(x, t) = C0 + D0 t + An cos 2L n=0 



(2n + 1)π x cπ(2n + 1)t sin , +Bn sin 2L 2L 0 ≤ x ≤ L, t ≥ 0, is a generalized(formal) solution of the problem (7.15), (7.16), (7.19). To find the constants C0 , D0 , An and Bn , n ∈ N0 , we will use the initial conditions (7.16). Assume that the initial data φ and ψ can be expanded into generalized Fourier series with respect to the sequence of the eigenfunctions of the problem (7.23), (7.27) and these series are uniformly convergent, φ(x) = .

ψ(x) =

∞



∞



n=0 an sin

n=0 bn sin

(2n+1)π x 2L (2n+1)π x 2L

 , 

0 ≤ x ≤ L.

,

We have that am =

2 L

L

 sin

0

.

bm =

2 L

L

 sin

0

(2m+1)π x 2L

(2m+1)π x 2L

 φ(x)dx, 

m ∈ N0 .

ψ(x)dx,

Hence, ∞ 

u(x, 0) = C0 +

.

n=0

=

∞ 



(2n + 1)π x An sin 2L

an sin

n=0

(2n + 1)π x 2L



whereupon C0 = 0 and an = An , n ∈ N0 .

.

Moreover,



,

0 ≤ x ≤ L,

7.1 The One Dimensional Wave Equation ∞ 

ut (x, 0) = D0 +

.

269

Bn

n=0

=

∞  n=0

 (2n + 1)π x cπ(2n + 1) sin 2L 2L 



(2n + 1)π x bn sin 2L

,

0 ≤ x ≤ L.

Therefore D0 = 0,

.

Bn =

2Lbn , cπ(2n + 1)

n ∈ N0 .

Thus, the problem (7.15), (7.16), (7.19) is formally solved. Now, we consider the IBVP (7.15), (7.16), (7.20). By the boundary conditions (7.20), we get ux (0, t) = X' (0)T (t)

.

= 0, u(L, t) = X(L)T (t) = 0,

t ≥ 0.

Since u is a nontrivial solution, it follows that X' (0) = X(L)

.

= 0. Therefore X should be a solution of the eigenvalue problem for the Eq. (7.23) subject to the boundary conditions X' (0) = 0, (7.29)

.

X(L) = 0. For the general solution of the first equation of (7.22), we have (a) √ √ X(x) = c1 cosh( −λx) + c2 sinh( −λx),

.

0 ≤ x ≤ L,

λ < 0.

270

7 The Wave Equation

(b) X(x) = c1 + c2 x,

.

0 ≤ x ≤ L,

λ = 0.

(c) √ √ X(x) = c1 cos( λx) + c2 sin( λx),

.

0 ≤ x ≤ L,

λ > 0,

where c1 and c2 are arbitrary real constants. (a) When λ < 0, using (7.29), we find c1 = c2 = 0.

.

Therefore X(x) = 0,

.

0 ≤ x ≤ L,

and the eigenvalue problem (7.23), (7.29) does not admit negative eigenvalues. (b) When λ = 0, using (7.29), we find X(x) = 0,

.

0 ≤ x ≤ L.

(c) When λ > 0, then c2 = 0

.

and √ √ c1 λ cos(L λ) = 0.

.

If c1 = 0, then X(x) = 0,

.

0 ≤ x ≤ L,

and the eigenvalue problem (7.23), (7.29) does not admit positive eigenvalues. Thus, √ .

or

λL =

(2n + 1)π , 2

n ∈ N,

7.1 The One Dimensional Wave Equation

λ=

.

271

(2n + 1)2 π 2 , 4L2

n ∈ N.

The associated eigenfunction is

X(x) = cos

.

(2n + 1)π x 2L

 ,

0 ≤ x ≤ L,

which is uniquely determined up to a multiplicative factor. Hence, the solution of the eigenvalue problem (7.23), (7.29) is an infinite sequence of nonnegative simple eigenvalues and their associated eigenfunctions will be denoted by

Xn (x) = cos

.

λn =

(2n + 1)π x 2L

(2n + 1)π 2L

 ,

0 ≤ x ≤ L,

2 ,

n ∈ N0 .

The solutions of the second equation of (7.22) for λ = λn are given by (7.28) Then the product solutions of the initial boundary value problem (7.15), (7.16), (7.20) are given by u0 (x, t) = X0 (x)T0 (t) .

= C0 + D0 t, un (x, t) = cos



(2n+1)π x 2L

     +Bn sin cπ(2n+1)t , n ∈ N0 , An cos cπ(2n+1)t 2L 2L

0 ≤ x ≤ L, t ≥ 0. Applying the superposition principle, the function 

∞

 cπ(2n + 1)t An cos 2L n=0 



(2n + 1)π x cπ(2n + 1)t cos , +Bn sin 2L 2L

u(x, t) = C0 + D0 t +

.

0 ≤ x ≤ L, t ≥ 0, is a generalized(formal) solution of the problem (7.15), (7.16), (7.20). To find the constants An and Bn , n ∈ N0 , we will use the initial conditions (7.16). Assume that the initial data φ and ψ can be expanded into generalized Fourier series with respect to the sequence of the eigenfunctions of the problem (7.23), (7.29) and these series are uniformly convergent,

272

7 The Wave Equation ∞ 

φ(x) =

 an cos

n=0 .

∞ 

ψ(x) =

 bn cos

n=0

(2n+1)π x 2L

(2n+1)π x 2L

 , 

0 ≤ x ≤ L.

,

We have that am =

L

2 L

 cos

0

.

bm =

L

2 L

 cos

0

(2m+1)π x 2L

(2m+1)π x 2L

 φ(x)dx, 

m ∈ N0 .

ψ(x)dx,

Hence, u(x, 0) = C0 +

∞ 

An cos

.

n=0

=

∞ 

an cos

n=0

(2n + 1)π x 2L

(2n + 1)π x , 2L

0 ≤ x ≤ L,

whereupon C0 = 0 and an = An , n ∈ N0 .

.

Moreover, ut (x, 0) = D0 +

∞ 

.

Bn

n=0

=

∞ 

bn cos

n=0

 (2n + 1)π x cπ(2n + 1) cos 2L 2L

(2n + 1)π x 2L

 ,

0 ≤ x ≤ L.

Therefore D0 = 0,

.

Bn =

2Lbn , cπ(2n + 1)

n ∈ N0 .

Thus, the problem (7.15), (7.16), (7.20) is formally solved.

7.1 The One Dimensional Wave Equation

273

Example 7.7 We will find a formal solution to the IVP utt − uxx = 0, .

0 < x < π,

t > 0,

= (cos x)2 ,

u(x, 0)

ut (x, 0) =

(sin x)2 ,

(7.30) 0 ≤ x ≤ π.

subject to the Neumann boundary conditions ux (0, t) = 0,

.

ux (π, t) = 0,

t ≥ 0.

Here c = 1,

.

L = π, φ(x) = (cos x)2 , ψ(x) = (sin x)2 ,

0 ≤ x ≤ π.

We have φ ' (x) = −2 sin x cos x

.

= − sin(2x), ψ ' (x) = 2 sin x cos x = sin(2x),

0 ≤ x ≤ π,

and φ ' (0) = 0,

.

φ ' (π ) = 0, ψ ' (0) = 0, ψ ' (π ) = 0.

274

7 The Wave Equation

Consequently the initial and boundary conditions are compatible. Next, Xn (x) = cos(nx),

.

λn = n2 ,

1 a0 = π

π (cos x)2 dx 0

1 = 2π

π (1 + cos(2x))dx 0

1 = 2π

π

1 dx + 2π

0

2π cos(2x)dx 0

x=2π  1 1 sin(2x) = + 2 4π x=0 =

1 2

and 1 .b0 = π

π (sin x)2 dx 0

1 = 2π

π (1 − cos(2x))dx 0

1 = 2π

π 0

1 dx − 2π

2π cos(2x)dx 0

x=2π  1 1 sin(2x) = − 2 4π x=0 =

1 , 2

7.1 The One Dimensional Wave Equation

275

and 2 .am = π

π cos(mx)(cos x)2 dx 0

1 = π

π cos(mx)(1 + cos(2x))dx 0

=

1 π

π cos(mx)dx +

π

1 π

cos(mx) cos(2x)dx

0

0

x=π π  1 1  sin(mx) = + (cos((m − 2)x) + cos((m + 2)x))dx mπ 2π x=0 0

1 = 2π

π

1 cos((m − 2)x)dx + 2π

0

=

cos((m + 2)x)dx 0

x=π x=π   1 1 sin((m − 2)x) sin((m + 2)x) + 2π(m − 2) 2π(m + 2) x=0 x=0

= 0,

a2 =

π

1 π

m ∈ N,

m /= 2,

π cos(2x)(1 + cos(2x))dx 0

1 = π

π 0

1 cos(2x)dx + π

π (cos(2x))2 dx 0

x=π π  1 1  sin(2x) = + (1 + cos(4x))dx 2π 2π x=0 0

276

7 The Wave Equation

1 = 2π

π

1 dx + 2π

0

π cos(4x)dx 0

=

x=π  1 1 + sin(4x) 2 8π x=0

=

1 , 2

and 2 .bm = π

π cos(mx)(sin x)2 dx 0

1 = π

π cos(mx)(1 − cos(2x))dx 0

1 = π

π

1 cos(mx)dx − π

0

π cos(mx) cos(2x)dx 0

x=π π  1 1  − (cos((m − 2)x) + cos((m + 2)x))dx sin(mx) = mπ 2π x=0 0

1 =− 2π

π

1 cos((m − 2)x)dx − 2π

0

π cos((m + 2)x)dx 0

x=π x=π   1 1  − sin((m − 2)x) sin((m + 2)x) =− 2π(m − 2) 2π(m + 2) x=0 x=0 = 0,

1 b2 = π

m ∈ N,

m /= 2,

π cos(2x)(1 − cos(2x))dx 0

7.1 The One Dimensional Wave Equation

1 = π

π

277

π

1 cos(2x)dx − π

(cos(2x))2 dx

0

0

x=π π  1 1  (1 + cos(4x))dx − sin(2x) = 2π 2π x=0 0

=−

1 2π

π dx −

1 2π

π cos(4x)dx 0

0

x=π  1 1 sin(4x) =− − 2 8π x=0 1 =− . 2 Therefore A0 =

1 , 2

B0 =

1 , 2

A2 =

1 , 2

.

1 B2 = − , 4 Am = Bm = 0,

m /= 2,

m ∈ N.

and 1 t .u(x, t) = + + 2 2



 1 1 cos(2t) − sin(2t) cos(2x), 2 4

0 ≤ x ≤ π,

t ≥ 0.

Example 7.8 Consider the IVP (7.30) subject to the Dirichlet boundary conditions u(0, t) = 0,

.

u(π, t) = 0,

t ≥ 0.

278

7 The Wave Equation

Since φ(0) = 1,

.

/= 0, φ(π ) = 1 /= 0, the initial and boundary conditions are not compatible. Example 7.9 Consider the IVP (7.30) subject to the mixed boundary conditions of the first kind u(0, t) = 0,

.

ux (π, t) = 0,

t ≥ 0.

Since φ(0) /= 0, we have that the initial and boundary conditions are not compatible. Example 7.10 Consider the IVP (7.30) subject to the mixed boundary conditions of the second kind ux (0, t) = 0,

.

u(π, t) = 0,

t ≥ 0.

Since φ(π ) /= 0, we have that the initial and boundary conditions are not compatible. Exercise 7.9 Find a formal solution to the problem utt − uxx = 0, ux (0, t)

u(x, 0)

t > 0,

= 0,

ux (π, t) = 0,

.

0 < x < π,

t ≥ 0,

= 0,

ut (x, 0) = (sin x)3 ,

0 ≤ x ≤ π.

Example 7.11 Consider the following IBVP utt − uxx = 0,

.

0 < x < 1,

u(x, 0) = x(1 − x),

t > 0,

7.1 The One Dimensional Wave Equation

279

ut (x, 0) = 0,

0 < x < 1,

u(0, t) = 0, u(1, t) = 0,

t > 0.

Here c = 1,

.

L = 1, φ(x) = x(1 − x), ψ(x) = 0,

0 ≤ x ≤ 1.

Then φ(0) = φ(1) = ψ(0) = ψ(1) = 0

.

and the initial conditions and the boundary conditions are compatible. We have 1 am = 2

sin(mπ x)x(1 − x)dx

.

0

1 =2

sin(mπ x)(x − x 2 )dx 0

x=1 1  2 2 2  =− + cos(mπ x)(1 − 2x)dx cos(mπ x)(x − x ) mπ mπ x=0 0

2 = mπ

1 0

=

4 cos(mπ x)dx − mπ

1 cos(mπ x)xdx 0

x=1 x=1 1   2 4 4   sin(mπ x) − sin(mπ x)x + sin(mπ x)dx   (mπ )2 (mπ )2 (mπ )2 x=0 x=0 0

280

7 The Wave Equation

=−

x=1  4  cos(mπ x)  3 (mπ ) x=0

=−

4 ((−1)m − 1) (mπ )3

 =

8 (mπ )3

0

if m

if m

is odd

,

is even

bm = 0, A m = am , Bm = 0,

m ∈ N.

Therefore u(x, t) =

∞ 

.

odd

k=1,k

8 cos(π kt) sin(π kx), (kπ )3

t ≥ 0,

0 ≤ x ≤ 1.

Exercise 7.10 Find a formal solution to the following IBVP utt − 4uxx = 0,

.

0 < x < 1,

t > 0,

u(x, 0) = sin(5π x) + 2 sin(7π x), ut (x, 0) = 0,

0 < x < 1,

u(0, t) = 0, u(1, t) = 0,

t > 0.

Example 7.12 Consider the IBVP utt − uxx = 0,

.

0 < x < 1,

t > 0,

u(x, 0) = 0, ut (x, 0) = sin

 π  3π x + sin x , 2 2

u(0, t) = 0, ux (1, t) = 0,

t > 0.

0 < x < 1,

7.1 The One Dimensional Wave Equation

281

Here c = 1,

.

L = 1, φ(x) = 0, ψ(x) = sin

 π  3π x + sin x , 2 2

0 < x < 1.

Then φ ' (x) = 0,

.

  π  3π 3π π x + cos x , ψ (x) = cos 2 2 2 2 '

0 < x < 1,

and φ(0) = 0,

.

φ ' (1) = 0, ψ(0) = 0, ψ ' (1) = 0. Thus, the initial and boundary conditions are compatible. We have am = 0,

m ∈ N0 ,

.

and 1 b0 = 2

.

 π  π  3π x sin x + sin x dx sin 2 2 2

0

 1    1 π  2 3π π =2 x x sin x dx sin dx + 2 sin 2 2 2 0

0

1 =

1 (1 − cos(π x))dx +

0

(cos(π x) − cos(2π x))dx 0

282

7 The Wave Equation

1 =

1 dx −

1 cos(π x)dx +

0

1 cos(π x)dx −

0

0

cos(2π x)dx 0

x=1  1 sin(2π x) = 1− 2π x=0 = 1, B0 =

2b0 π

=

2 , π

and

1 b1 = 2

sin

.

3π x 2



sin

 π  3π x + sin x dx 2 2

0

2

 1

1 π  3π 3π =2 dx + 2 sin x x sin x dx sin 2 2 2 0

0

1 =

1 (1 − cos(3π x))dx +

0

0

1 =

(cos(π x) − cos(2π x))dx

1 dx −

0

1 cos(3π x)dx +

0

1 cos(π x)dx −

0

cos(2π x)dx 0

x=1 x=1 x=1    1 1 1   − − sin(2π x) sin(3π x) = 1 + sin(π x) π 2π 3π x=0 x=0 x=0 = 1, B1 =

2b1 3π

=

2 , 3π

7.1 The One Dimensional Wave Equation

283

and 1 bm = 2

.



(2m + 1)π x sin 2

  

 π 3π sin x + sin x dx 2 2

0



1 =2

sin

 π  (2m + 1)π x sin x dx 2 2

0

1 +2





 (2m + 1)π 3π sin x sin x dx 2 2

0

1 (cos(mπ x) − cos((m + 1)π x))dx

= 0

1 (cos((m − 1)π x) − cos((m + 2)π x))dx

+ 0

x=1 x=1   1 1  = sin(mπ x) sin((m + 1)π x) − mπ (m + 1)π x=0 x=0 x=1  1 sin((m − 1)π x) + (m − 1)π x=0 x=1  1 sin((m + 2)π x) − (m + 2)π x=0 = 0, Bm = 0,

m ∈ N,

m /= 1.

Consequently 



π  π  3π 3π t sin x + B1 sin t sin x .u(x, t) = B0 sin 2 2 2 2



 π  π  2 3π 3π 2 t sin x + sin t sin x , = sin π 2 2 3π 2 2 0 ≤ x ≤ 1,

t ≥ 0.

284

7 The Wave Equation

Exercise 7.11 Solve the following IBVP utt − uxx = 0,

.

0 < x < 1,

u(x, 0) = sin ut (x, 0) = sin

t > 0,

 5π x , 2

π  x , 2

0 < x < 1,

u(0, t) = 0, ux (1, t) = 0,

t > 0.

Example 7.13 Consider the IBVP utt − uxx = 0,

.

0 < x < 1,

t > 0,

π  x , 2 



5π 3π ut (x, 0) = cos x + cos x , 2 2 u(x, 0) = cos

0 < x < 1,

ux (0, t) = 0, u(1, t) = 0,

t > 0.

Here c = 1,

.

π  x , 2



 3π 5π ψ(x) = cos x + cos x , 2 2 φ(x) = cos

0 ≤ x ≤ 1.

We have π  π sin x , 2 2



 3π 5π 5π 3π ' ψ (x) = − sin x − sin x , 2 2 2 2 φ ' (x) = −

.

φ(1) = 0,

0 ≤ x ≤ 1,

7.1 The One Dimensional Wave Equation

285

φ ' (0) = 0,

ψ(1) = cos

3π 2



+ cos

5π 2



= −1 + 1 = 0, ψ ' (0) = 0. Thus, the initial conditions and the boundary conditions are compatible. Next,

a0 = 2

1  cos

.

 π 2 dx x 2

0

1 =

(1 + cos(π x))dx 0

1

1

=

dx + 0

cos(π x)dx 0

= 1+

x=1  1 sin(π x) π x=0

= 1, 1 am = 2



 π  (2m + 1)π cos x cos x dx 2 2

0

1 =

(cos((m + 1)π x) + cos(mπ x))dx 0

1 =

1 cos((m + 1)π x)dx +

0

cos(mπ x)dx 0

286

7 The Wave Equation

=

x=1 x=1   1 1 + sin((m + 1)π x) sin(mπ x) (m + 1)π mπ x=0 x=0

= 0,

m∈N

1



and b1 = 2

cos

.

3π x 2





cos



 3π 5π x + cos x dx 2 2

0

1

=2



3π x cos 2

2 dx + 2

0

cos



 3π 5π x cos x dx 2 2

0

1 =



1

1 (1 + cos(3π x))dx +

0

(cos(π x) + cos(4π x))dx 0

1

1

=

dx + 0

1 cos(3π x)dx +

0

1 cos(π x)dx +

0

cos(4π x)dx 0

x=1 x=1 x=1    1 1 1   + sin(π x) + sin(3π x) sin(4π x) = 1+ 3π π 4π x=0 x=0 x=0 = 1,

1 b2 = 2

cos

5π x 2





cos



 3π 5π x + cos x dx 2 2

0

1 =2





 2

1

5π 3π 5π dx cos x cos x dx + 2 x cos 2 2 2 0

0

1 =

1 (cos(π x) + cos(4π x))dx +

0

(1 + cos(5π x))dx 0

7.1 The One Dimensional Wave Equation

1 =

287

1 cos(π x)dx +

0

1 cos(4π x)dx +

0

1 dx +

0

cos(5π x)dx 0

x=1 x=1 x=1    1 1 1   + +1+ sin(4π x) sin(5π x) = sin(π x) π 4π 5π x=0 x=0 x=0 = 1, and

1 bm = 2

cos

.

(2m + 1)π x 2





cos



 3π 5π x + cos x dx 2 2

0



1 =2

cos



 (2m + 1)π 3π x cos x dx 2 2

0



1 +2

cos



 (2m + 1)π 5π x cos x dx 2 2

0

1 =

(cos((m + 2)π x) + cos((m − 1)π x))dx 0

1 +

(cos((m + 3)π x) + cos((m − 2)π x))dx 0

1 =

1 cos((m + 2)π x)dx +

0

1 cos((m − 1)π x)dx +

0

cos((m + 3)π x)dx 0

1 +

cos((m − 2)π x)dx 0

=

x=1 x=1   1 1 + sin((m + 2)π x) sin((m − 1)π x) (m + 2)π (m − 1)π x=0 x=0

288

7 The Wave Equation

+ = 0,

x=1 x=1   1 1 + sin((m + 3)π x) sin((m − 2)π x) (m + 3)π (m − 2)π x=0 x=0 m ∈ N0 ,

m /= 1, 2.

Consequently A0 = a0

.

= 1, A m = am = 0, B0 =

m ∈ N,

2 b0 π

= 0, B1 = = B2 = = Bm =

2 b1 3π 2 , 3π 2 b2 5π 2 , 5π 2 bm (2m + 1)π

= 0,

m ∈ N,

m /= 1, 2.

Therefore u(x, t) = A0 cos

.



  π   π  3π  3π 3π t x + A1 t + B1 sin t cos x 2 2 2 2 2



 

 5π 5π 5π + A2 t + B2 sin t cos x 2 2 2

7.1 The One Dimensional Wave Equation

289



 π  π  2 3π 3π = cos t cos x + sin t cos x 2 2 3π 2 2



 2 5π 5π + sin t cos x , 5π 2 2

0 ≤ x ≤ 1,

t ≥ 0.

Exercise 7.12 Find a formal solution to the following IBVP utt − uxx = 0,

0 < x < 1,

.

t > 0,

u(x, 0) = x 2 (1 − x), ut (x, 0) = 0,

0 < x < 1,

ux (0, t) = 0, u(1, t) = 0,

7.1.3.2

t > 0.

Nonhomogeneous IBVP with Homogeneous Boundary Conditions

Now we consider the nonhomogeneous initial boundary value problem utt − c2 uxx = f (x, t),

0 < x < L,

.

t > 0,

(7.31)

u(x, 0) = φ(x), .

(7.32)

ut (x, 0) = ψ(x),

0 < x < L,

ux (0, t) = 0, .

(7.33)

ux (L, t) = 0,

t > 0,

where f (x, t) = C0 + D0 t +

∞  n=1

0 ≤ x ≤ L, .

φ(x) ψ(x)

= a0 + = b0 +

∞  n=1 ∞  n=1

an cos bn cos

Cn cos

cπ nt  L

+ Dn sin

cπ nt  L

t ≥ 0,

π nx  L

π nx  L

, ,

0 ≤ x ≤ L,

φ and ψ satisfy the compatibility conditions

t ≥ 0,

cos

π nx  L

,

290

7 The Wave Equation

φ ' (0) = φ ' (L) = ψ ' (0) = ψ ' (L) = 0,

.

Cn , Dn , an and bn , n ∈ N0 , are given constants. Let u = u(x, t), 0 ≤ x ≤ L, t ≥ 0, be a solution to the problem (7.31)–(7.32) in the following form u(x, t) = T0 (t) +

∞ 

.

Tn (t) cos

n=1

 π nx  . L

(7.34)

Substituting (7.34) in (7.31), we get

.

 ∞

 π nx   c2 π 2 n2 '' Tn + cos + T n L L2 n=1 



cπ nt  ∞ = C0 + D0 t + n=1 Cn cos L + Dn sin cπLnt cos πLnx ,

T0''

0 ≤ x ≤ L, t ≥ 0. Hence, T0'' (t) .

Tn'' (t) +

c 2 π 2 n2 L2

= C0 + D0 t,  

Tn (t) = Cn cos cπLnt + Dn sin cπLnt ,

t ≥ 0.

Therefore + D60 t 3 + β0 t + α0 ,

  . Tn (t) = αn cos cπLnt + βn sin cπLnt − n ∈ N, t ≥ 0. T0 (t) =

C0 2 2 t

Dn L 2cπ n t

cos

cπ nt  L

+

Cn L 2cπ n t

sin

cπ nt  L

,

We will find the constants α0 , αn and βn , n ∈ N, using the initial condition (7.32). We have u(x, 0) = α0 +

∞ 

.

αn cos

 π nx  L

n=1

= a0 +

∞ 

an cos

n=1

 π nx  L

,

whereupon αn = an ,

.

n ∈ N0 .

Also, ut (x, 0) = β0 +

.

 ∞

 π nx   Dn L cπ n βn cos − 2cπ n L L n=1

0 ≤ x ≤ L,

7.1 The One Dimensional Wave Equation

= b0 +

∞ 

291

bn cos

 π nx 

n=1

L

,

0 ≤ x ≤ L,

whereupon β0 = b0 ,

.

βn =

 L Dn L bn + , cπ n 2cπ n

n ∈ N.

Therefore

 ∞   cπ nt C0 2 D 0 3 u(x, t) = t + t + b0 t + a0 + an cos 2 6 L . n=1    Dn L 



cπ nt  Cn L

Dn L cπ nt L + cπ n bn + 2cπ n sin L − 2cπ n t cos L + 2cπ n t sin cπLnt cos πLnx , 0 ≤ x ≤ L, t ≥ 0. Example 7.14 We will find a formal solution to the problem utt − uxx = cos(2π x) cos(2π t), u(x, 0) .

0 < x < 1,

t > 0,

= (cos(π x))2 ,

ut (x, 0) = 2 cos(2π x), ux (0, t)

= 0,

ux (1, t)

= 0,

0 ≤ x ≤ 1,

t > 0.

Here c

= 1,

f (x, t) = cos(2π x) cos(2π t), .

φ(x)

=

ψ(x)

= 2 cos(2π x),

1 2

+

1 2

0 ≤ x ≤ 1,

cos(2π x),

Then C0 = D 0

.

= 0, C1 = 0,

0 ≤ x ≤ 1.

t ≥ 0,

292

7 The Wave Equation

C2 = 1, Cn = 0,

n ∈ N,

Dn = 0,

n ∈ N,

a0 =

n ≥ 3,

1 , 2

a1 = 0, a2 =

1 , 2

an = 0,

n ∈ N,

n ≥ 3,

n ∈ N,

n ≥ 3.

b0 = 0, b1 = 0, b2 = 2, bn = 0, Therefore u(x, t) =

.

1 + 2



 1 t +4 cos(2π t) + sin(2π t) cos(2π x), 2 4π

0 ≤ x ≤ 1,

Exercise 7.13 Find a formal solution to the problem utt − c2 uxx = f (x, t),

.

u(x, 0)

= φ(x),

ut (x, 0)

= ψ(x),

u(0, t)

= 0,

u(L, t)

= 0,

f (x, t) =

∞   cπ n   t Cn cos L n=0

0 < x < L,

t > 0.

where .

0 < x < L,

t > 0,

t ≥ 0.

7.1 The One Dimensional Wave Equation

+Dn sin φ(x) =

∞ 

 cπ n   nπ x  t sin , L L

an sin

n=0

ψ(x) =

∞ 

bn sin

n=0

293

 nπ x  L  nπ x  L

0 ≤ x ≤ L,

t ≥ 0,

,

0 ≤ x ≤ L,

,

t ≥ 0,

an , bn , Cn , Dn , n ∈ N0 , are given constants, φ and ψ satisfy the compatibility conditions φ(0) = φ(L) = ψ(0) = ψ(L) = 0.

.

Exercise 7.14 Find a formal solution to the problem utt − c2 uxx = f (x, t),

.

u(x, 0)

= φ(x),

ut (x, 0)

= ψ(x),

u(0, t)

= 0,

ux (L, t)

= 0,

0 < x < L,

t > 0,

0 < x < L,

t > 0.

where f (x, t) =

.

φ(x) =

 cπ(2n + 1) t 2L n=0



 cπ(2n + 1) (2n + 1)π x +Dn sin t sin , 2L 2L Cn cos

∞  n=0

ψ(x) =



∞



∞  n=0



(2n + 1)π x an sin 2L

(2n + 1)π x bn sin 2L

0 ≤ x ≤ L,

t ≥ 0,

 ,  ,

0 ≤ x ≤ L,

t ≥ 0,

an , bn , Cn , Dn , n ∈ N0 , are given constants, φ and ψ satisfy the compatibility conditions φ(0) = φ ' (L) = ψ(0) = ψ ' (L) = 0.

.

294

7 The Wave Equation

Exercise 7.15 Find a formal solution to the problem utt − c2 uxx = f (x, t),

.

u(x, 0)

= φ(x),

ut (x, 0)

= ψ(x),

ux (0, t)

= 0,

u(L, t)

= 0,

0 < x < L,

t > 0,

0 < x < L,

t > 0.

where f (x, t) =

.

φ(x) =



∞

 cπ(2n + 1) Cn cos t 2L n=0 



(2n + 1)π x cπ(2n + 1) t cos , +Dn sin 2L 2L ∞  n=0

ψ(x) =

∞  n=0



(2n + 1)π x an cos 2L

(2n + 1)π x bn cos 2L

0 ≤ x ≤ L,

t ≥ 0,

 ,  ,

0 ≤ x ≤ L,

t ≥ 0,

an , bn , Cn , Dn , n ∈ N0 , are given constants, φ and ψ satisfy the compatibility conditions φ ' (0) = φ(L) = ψ ' (0) = ψ(L) = 0.

.

7.1.3.3

Nonhomogeneous IBVPs with Nonhomogeneous Boundary Conditions

Consider the IBVP utt − c2 uxx = f (x, t),

.

u(x, 0)

= φ(x),

ut (x, 0)

= ψ(x),

ux (0, t)

= g1 (t),

ux (L, t)

= g2 (t),

0 < x < L,

0 < x < L,

t > 0,

t > 0,

(7.35)

7.1 The One Dimensional Wave Equation

295

where φ ∈ C 2 ([0, L]), ψ ∈ C 1 ([0, L]), g1 , g2 ∈ C 2 ([0, ∞]), c is a positive constant and the compatibility conditions φ ' (0) = g1 (0),

.

φ ' (L) = g2 (L), ψ ' (0) = g1' (0), ψ ' (L) = g2' (0) hold. We will reduce the problem (7.35) to a nonhomogeneous problem with homogeneous boundary conditions. Set v(x, t) = u(x, t) +

.

1 1 2 (x − L)2 g1 (t) − x g2 (t), 2L 2L

0 ≤ x ≤ L,

t ≥ 0. (7.36)

Then u(x, t) = v(x, t) −

.

1 1 2 x g2 (t), (x − L)2 g1 (t) + 2L 2L

ux (x, t) = vx (x, t) −

1 1 (x − L)g1 (t) + xg2 (t), L L

uxx (x, t) = vxx (x, t) −

1 1 g1 (t) + g2 (t), L L

ut (x, t) = vt (x, t) −

1 1 2 ' x g2 (t), (x − L)2 g1' (t) + 2L 2L

utt (x, t) = vtt (x, t) −

1 2 '' 1 (x − L)2 g1'' (t) + x g2 (t), 2L 2L

0 ≤ x ≤ L,

t ≥ 0.

Hence, f (x, t) = utt (x, t) − c2 uxx (x, t)

.

= vtt (x, t) −

1 2 '' 1 (x − L)2 g1'' (t) + x g2 (t) 2L 2L

−c2 vxx (x, t) + whereupon

c2 c2 g1 (t) − g2 (t), L L

0 ≤ x ≤ L,

t ≥ 0,

296

7 The Wave Equation

vtt − c2 vxx = f (x, t) +

.

c2 c2 1 2 '' 1 x g2 (t) − g1 (t) + g2 (t), (x − L)2 g1'' (t) − 2L L L 2L

0 ≤ x ≤ L, t ≥ 0. Next, v(x, 0) = u(x, 0) +

.

= φ(x) +

1 2 1 x g2 (0) (x − L)2 g1 (0) − 2L 2L

1 1 2 (x − L)2 g1 (0) − x g2 (0), 2L 2L

vt (x, 0) = ut (x, 0) + = ψ(x) +

1 1 2 ' x g2 (0) (x − L)2 g1' (0) − 2L 2L

1 1 2 ' (x − L)2 g1' (0) − x g2 (0), 2L 2L

vx (0, t) = ux (0, t) − g1 (t) = g1 (t) − g1 (t) = 0, vx (L, t) = ux (L, t) − g2 (t) = g2 (t) − g2 (t) = 0,

0 ≤ x ≤ L,

t ≥ 0.

Thus, we get the following nonhomogeneous IBVP with homogeneous boundary conditions vtt − c2 vxx = f (x, t) +

1 2L (x

− L)2 g1'' (t)

1 2 '' − 2L x g2 (t) −

c2 c2 L g1 (t) + L g2 (t),

0 ≤ x ≤ L,

v(x, 0)

= φ(x) +

1 2L (x

− L)2 g1 (0) −

1 2 2L x g2 (0),

vt (x, 0)

= ψ(x) +

1 2L (x

− L)2 g1' (0) −

1 2 ' 2L x g2 (0),

vx (0, t)

= 0,

vx (L, t)

= 0,

.

t ≥ 0,

0 ≤ x ≤ L,

t ≥ 0. (7.37)

To find a solution u with separable variables to the IBVP (7.35), firstly we find a solution v with separable variables of the IBVP (7.37) and then using (7.36) we find u.

7.1 The One Dimensional Wave Equation

297

Exercise 7.16 Reduce the following IBVP utt − c2 uxx = f (x, t),

0 < x < L,

.

t > 0,

u(x, 0) = φ(x), ut (x, 0) = ψ(x),

0 < x < L,

u(0, t) = g1 (t), u(L, t) = g2 (t),

t > 0,

where f ∈ C ([0, L] × [0, ∞)), φ ∈ C 2 ([0, L]), ψ ∈ C 1 ([0, L]), g1 , g2 ∈ C 2 ([0, ∞)), and φ(0) = g1 (0),

.

φ(L) = g2 (0),

ψ(0) = g1' (0),

ψ(L) = g2' (0).

to an IBVP with homogeneous boundary conditions. Exercise 7.17 Reduce the following IBVP utt − c2 uxx = f (x, t),

0 < x < L,

.

t > 0,

u(x, 0) = φ(x), ut (x, 0) = ψ(x),

0 < x < L,

u(0, t) = g1 (t), ux (L, t) = g2 (t),

t > 0,

where f ∈ C ([0, L] × [0, ∞)), φ ∈ C 2 ([0, L]), ψ ∈ C 1 ([0, L]), g1 , g2 ∈ C 2 ([0, ∞)), and φ(0) = g1 (0),

.

φ ' (L) = g2 (0),

ψ(0) = g1' (0),

ψ ' (L) = g2' (0),

to an IBVP with homogeneous boundary conditions. Exercise 7.18 Reduce the following IBVP utt − c2 uxx = f (x, t),

.

0 < x < L,

u(x, 0) = φ(x), ut (x, 0) = ψ(x),

0 < x < L,

t > 0,

298

7 The Wave Equation

ux (0, t) = g1 (t), u(L, t) = g2 (t),

t > 0,

where f ∈ C ([0, L] × [0, ∞)), φ ∈ C 2 ([0, L]), ψ ∈ C 1 ([0, L]), g1 , g2 ∈ C 2 ([0, ∞)), and φ ' (0) = g1 (0),

.

ψ ' (0) = g1' (0),

φ(L) = g2 (0),

ψ(L) = g2' (0),

to an IBVP with homogeneous boundary conditions.

7.1.4 The Energy Method: Uniqueness The energy method is a fundamental tool in the theory of PDEs for proving the uniqueness of the solutions of initial boundary value problems. Consider the problem utt − c2 uxx = f (x, t),

0 < x < L,

.

t > 0,

u(0, t) = g(t), .

u(L, t) = h(t),

(7.39)

t ≥ 0,

u(x, 0) = φ(x), .

ut (x, 0) = ψ(x),

(7.38)

(7.40)

0 ≤ x ≤ L,

where f ∈ C ([0, L] × [0, ∞)), g, h ∈ C 2 ([0, ∞)), φ ∈ C 2 (R), ψ ∈ C 1 (R), and g(0) = φ(0),

.

g ' (0) = ψ(0),

h' (0) = ψ(L),

h(0) = φ(L).

Let u1 and u2 be two solutions of the problem (7.38)–(7.40). We set v(x, t) = u1 (x, t) − u2 (x, t),

.

x ∈ [0, L],

t ≥ 0.

Then vtt − c2 vxx = 0,

.

v(0, t)

= 0,

v(L, t)

= 0,

v(x, 0)

= 0,

vt (x, 0)

= 0,

0 < x < L,

t ≥ 0,

0 ≤ x ≤ L.

t > 0,

7.1 The One Dimensional Wave Equation

299

Define the total energy 1 .E(t) = 2

L 

 (vt (x, t))2 + c2 (vx (x, t))2 dx.

0

Here 1 . 2

L (vt (x, t))2 dx,

t ≥ 0,

0

is the total kinetic energy, while c2 . 2

L (vx (x, t))2 dx,

t ≥ 0,

0

is the total potential energy. We have E ' (t) =

L 

.

 vt (x, t)vtt (x, t) + c2 vx (x, t)vxt (x, t) dx,

t ≥ 0.

(7.41)

0

Note that c2 vx (x, t)vxt (x, t) = c2 ((vx (x, t)vt (x, t))x − vxx (x, t)vt (x, t))

.

= c2 (vx (x, t)vt (x, t))x − c2 vxx (x, t)vt (x, t) = c2 (vx (x, t)vt (x, t))x −vtt (x, t)vt (x, t), Hence and by (7.41), we get '

L

E (t) = c

.

2

(vx (x, t)vt (x, t))x dx 0

x=L  = c2 vx (x, t)vt (x, t) x=0

= 0, where we have used that

t ≥ 0,

x ∈ [0, L], t ≥ 0.

300

7 The Wave Equation

vt (0, t) = vt (L, t) = 0,

t ≥ 0.

.

Therefore E ≡ const on [0, ∞). By the initial conditions we have vt (x, 0) = vx (x, 0) = 0,

.

0 ≤ x ≤ L.

Therefore E(0) = 0 and E ≡ 0 on [0, ∞). Consequently (vt (x, t))2 + c2 (vx (x, t))2 = 0,

.

x ∈ [0, L],

t ≥ 0.

Hence, vt (x, t) = vx (x, t) = 0,

x ∈ [0, L],

.

t ≥ 0.

From here, v ≡ const on [0, L] × [0, ∞). Using that v(x, 0) = 0, x ∈ [0, L], we conclude that v ≡ 0 on [0, L] × [0, ∞), from where u1 (x, t) = u2 (x, t),

.

x ∈ [0, L],

t ∈ [0, ∞).

Example 7.15 Consider the IBVP utt − c2 uxx = f (x, t),

.

ux (0, t)

= g(t),

ux (L, t)

= h(t),

u(x, 0)

= φ(x),

ut (x, 0)

= ψ(x),

0 < x < L,

t > 0,

t ≥ 0,

0 ≤ x ≤ L,

where f ∈ C ([0, L] × [0, ∞)), h, g ∈ C 1 ([0, ∞)), φ ∈ C 2 (R), ψ ∈ C 1 (R), and h(0) = φ ' (L),

.

g(0) = φ ' (0),

h' (0) = ψ ' (L),

g ' (0) = ψ ' (0).

Suppose that the considered problem has two solutions u1 and u2 . Set v(x, t) = u1 (x, t) − u2 (x, t),

.

Then v satisfies the IBVP

x ∈ [0, L],

t ≥ 0.

7.1 The One Dimensional Wave Equation

vtt − c2 vxx = 0,

.

vx (0, t)

= 0,

vx (L, t)

= 0,

v(x, 0)

= 0,

vt (x, 0)

= 0,

301

0 < x < L,

t > 0,

t ≥ 0,

0 ≤ x ≤ L,

where f ∈ C ([0, L] × [0, ∞)). h, g ∈ C 1 ([0, ∞)), Let E be as above. Then x=L  .E (t) = c vx (x, t)vt (x, t)  '

2

x=0

= 0,

t ≥ 0.

Thus, E ≡ const on [0, ∞). By the initial and boundary conditions, we have vx (x, 0) = 0,

.

vt (x, 0) = 0,

x ∈ [0, L].

Therefore E(0) = 0

.

and

E(t) = 0,

t ≥ 0.

x ∈ [0, L],

t ≥ 0.

Hence, vx (x, t) = 0,

.

vt (x, t) = 0,

Thus, v ≡ const on [0, L] × [0, ∞). By the initial conditions, we find v(x, t) = 0 or

.

u1 (x, t) = u2 (x, t),

x ∈ [0, L],

Exercise 7.19 Prove that the IBVP utt − c2 uxx = f (x, t),

.

u(0, t)

= g(t),

ux (L, t)

= h(t),

u(x, 0)

= φ(x),

ut (x, 0)

= ψ(x),

0 < x < L,

t ≥ 0,

0 ≤ x ≤ L,

t > 0,

t ≥ 0.

302

7 The Wave Equation

where f ∈ C ([0, L] × [0, ∞)), h, g ∈ C 1 ([0, ∞)), φ ∈ C 2 (R), ψ ∈ C 1 (R), and h(0) = φ ' (L),

.

h' (0) = ψ ' (L),

g(0) = φ(0),

g ' (0) = ψ(0),

has a unique solution. Exercise 7.20 Prove that the IBVP utt − c2 uxx = f (x, t),

.

ux (0, t)

= g(t),

u(L, t)

= h(t),

u(x, 0)

= φ(x),

ut (x, 0)

= ψ(x),

0 < x < L,

t > 0,

t ≥ 0,

0 ≤ x ≤ L,

where f ∈ C ([0, L] × [0, ∞)), h, g ∈ C 1 ([0, ∞)), φ ∈ C 2 (R), ψ ∈ C 1 (R), and h' (0) = ψ(L),

.

g ' (0) = ψ ' (0),

g(0) = φ ' (0),

h(0) = φ(L),

has a unique solution.

7.2 The Wave Equation in R3 We now turn to the three dimensional wave equation, which can be used to describe a variety of wavelike phenomena: sound waves, atmospheric waves, electromagnetic waves, and gravitational waves. The three dimensional wave equation for the displacement u = u(x1 , x2 , x3 , t) of the oscillator medium at the point labeled (x1 , x2 , x3 ) at time t is as follows utt = c2 (ux1 x1 + ux2 x2 + ux3 x3 ),

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

(7.42)

The compact form of the Eq. (7.42) is utt − c2 Δu = 0,

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0,

where Δu = ux1 x1 + ux2 x2 + ux3 x3 .

.

The three dimensional wave equation is a linear homogeneous partial differential equation with constant coefficients. It has one dependent variable u and four independent variables x1 , x2 , x3 and t.

7.2 The Wave Equation in R3

303

7.2.1 Radially Symmetric Solutions Solutions of the three dimensional wave Eq. (7.42) are not any harder to come than those of the one dimensional wave equation. Indeed, set r=

.

 x12 + x22 + x32 ,

(x1 , x2 , x3 ) ∈ R3 .

We will look for solutions of the Eq. (7.42) of the form u = u(r, t),

.

r > 0,

t > 0.

We have

.

uxi

= ur xri ,

uxi xi

= urr r i2 + ur

x2

r 2 −xi2 , r3

ux1 x1 + ux2 x2 + ux3 x3 = urr + 2r ur ,

i = 1, 2, 3,

(x1 , x2 , x3 ) ∈ R3 ,

r > 0,

t > 0,

i.e., the function u = u(r, t), r > 0, t > 0, satisfies the equation 

2 utt − c2 urr + ur = 0, r

r > 0,

t > 0,

rutt − c2 (rurr + 2ur ) = 0,

r > 0,

t > 0.

.

or .

Defining v(r, t) = ru(r, t),

.

r > 0,

t > 0,

we get vt (r, t) = rut (r, t),

.

vtt (r, t) = rutt (r, t), vr (r, t) = u(r, t) + rur (r, t), vrr (r, t) = ur (r, t) + ur (r, t) + rurr (r, t) = 2ur (r, t) + urr (r, t), and

r > 0,

t > 0,

(7.43)

304

7 The Wave Equation

vtt − c2 vrr = rutt − c2 (rurr + 2ur )

.

= 0,

r > 0,

t > 0.

This is exactly the one dimensional wave equation. Therefore the general solution for the Eq. (7.43) is given by u(r, t) =

.

1 (g1 (r + ct) + g2 (r − ct)) , r

r ≥ 0,

t > 0,

(7.44)

where g1 , g2 ∈ C 2 (R). Definition 7.4 The solutions of the form (7.44) of the three dimensional wave equation are said to be radially symmetric solutions of the Eq. (7.42). We can solve the Cauchy problem utt − c2 Δu = f (r, t), .

u(r, 0)

= φ(r),

ut (r, 0)

= ψ(r),

(x1 , x2 , x3 ) ∈ R3 ,

t > 0, (7.45)

r > 0.

Note that the source term and the initial conditions are only given along the ray r ≥ 0 and not for all r. If we suppose that φ, ψ ∈ C 1 ([0, ∞)), f ∈ C 1 ([0, ∞), C ([0, ∞))) and φ ' (0) = 0,

.

ψ ' (0) = 0, fr (0, t) = 0,

t ≥ 0,

then we can extend φ, ψ, f (·, t), t ≥ 0, to the whole line −∞ < r < ∞ by defining them to be even extensions of the given φ, ψ and f . Hence, the initial conditions and the source term for v are odd functions, and therefore v = v(r, t) is odd, which implies that u = u(r, t) is an even function. If we denote the even extensions of φ, , ψ  and f, respectively, we thus obtain the following IVP ψ and f by φ vtt − c2 vrr = f,

.

r ∈ R,

t > 0,

(r), v(r, 0) = r φ (r), vt (r, 0) = r ψ

r ∈ R,

7.2 The Wave Equation in R3

305

and the following radially symmetric solution for the three dimensional radial wave equation r+ct   1 1  ˜ ˜ ˜ (r + ct)φ(r + ct) + (r − ct)φ(r − ct) + .u(r, t) = s ψ(s)ds 2r 2cr r−ct

+

1 2cr

 t r+c(t−τ  )

ξ f(ξ, τ )dξ dτ,

0 r−c(t−τ )

r ∈ R, t ≥ 0. Example 7.16 Consider the IVP utt = Δu,

.

(x1 , x2 , x3 ) ∈ R3 ,

u(x1 , x2 , x3 , 0) = r 4 , ut (x1 , x2 , x3 , 0) = r 4 ,

r > 0.

This IVP can be rewritten in the form 2 utt − urr − ur = 0, r

.

r > 0,

t > 0,

u(r, 0) = r 4 , ut (r, 0) = r 4 ,

r > 0.

Here φ(r) = r 4 ,

.

ψ(r) = r 4 , f (r, t) = 0,

r ≥ 0,

t ≥ 0.

We have φ ' (r) = 4r 3 ,

.

ψ ' (r) = 4r 3 , and

r ≥ 0,

t > 0,

306

7 The Wave Equation

φ ' (0) = 0,

.

ψ ' (0) = 0. Thus, φ and ψ can be extended as even functions on R. Their extensions on R are as follows (r) = φ(r), φ

.

(r) = ψ(r), ψ

r ∈ R.

After we set v(r, t) = ru(r, t),

r ∈ R,

.

t ≥ 0,

we get v(r, 0) = ru(r, 0)

.

= r 5, vt (r, 0) = rut (r, 0) = r 5,

r ≥ 0,

and the IVP vtt − vrr = 0,

.

r ∈ R,

t > 0,

v(r, 0) = r 5 , vt (r, 0) = r 5 ,

r ∈ R.

Hence, for the solution of the considered IVP we have the representation 1 1 ((r + t)(r + t)4 + (r − t)(r − t)4 ) + .u(r, t) = 2r 2r

r+t s 5 ds r−t

=

 1 1 6 s=r+t ((r + t)5 + (r − t)5 ) + s  2r 12r s=r−t

=

  1  1  (r + t)5 + (r − t)5 + (r + t)6 − (r − t)6 2r 12r

7.2 The Wave Equation in R3

307

1 5 = r + 5r 4 t + 10r 3 t 2 + 10r 2 t 3 + 5rt 4 + t 5 2r  +r − 5r t + 10r t − 10r t + 5rt − t 5

+

4

3 2

2 3

4

5

1 r 6 + 6r 5 t + 15r 4 t 2 + 20r 3 t 3 + 15r 2 t 4 + 6rt 5 + t 6 12r 

−r 6 + 6r 5 t − 15r 4 t 2 + 20r 3 t 3 − 15r 2 t 4 + 6rt 5 − t 6 =

 1 1  5 (2r 5 + 20r 3 t 2 + 10rt 4 ) + 12r t + 40r 3 t 3 + 12rt 5 2r 12r

10 2 3 r t + t5 3 

1 = r 4 (1 + t) + 10r 2 t 2 1 + t + t 4 (t + 5) 3

= r 4 + 10r 2 t 2 + 5t 4 + r 4 t +

 1 1 + t + t 4 (5 + t), 3

=

(x12

+ x22

+ x32 )2 (1 + t) + 10(x12

+ x22

+ x32 )t 2

(x1 , x2 , x3 ) ∈ R3 , t ≥ 0. Example 7.17 Consider the IVP utt = Δu,

(x1 , x2 , x3 ) ∈ R3 ,

.

u(r, 0) = cos r, ut (r, 0) = cos r,

r > 0.

Here φ(r) = cos r,

.

ψ(r) = cos r,

r ≥ 0.

Then φ ' (r) = ψ ' (r)

.

= − sin r,

r ≥ 0,

t > 0,

308

7 The Wave Equation

and φ ' (0) = ψ ' (0)

.

= 0. Thus, φ and ψ have even extensions on R (r) = ψ (r) φ

.

= cos r,

r ≥ 0.

Hence, 1 1 ((r + t) cos(r + t) + (r − t) cos(r − t)) + .u(r, t) = 2r 2r

r+t s cos sds r−t

s=r+t  1 1 ((r + t) cos(r + t) + (r − t) cos(r − t)) + s sin s  = 2r 2r s=r−t 1 − 2r

r+t sin sds r−t

=

1 ((r + t) cos(r + t)+(r − t) cos(r − t)) 2r 1 + ((r + t) sin(r + t)−(r − t) sin(r − t)) 2r +

=

1 (cos(r + t) − cos(r − t)) 2r

1 (r(cos(r + t) + cos(r − t)) + t (cos(r + t) − cos(r − t))) 2r

1 (r(sin(r + t) − sin(r − t)) + t (sin(r + t) + sin(r − t))) 2r 1 − sin t sin r r 1 1 = (r cos t cos r − t sin r sin t) + (r sin t cos r + t sin r cos t) r r 1 − sin t sin r r +

7.2 The Wave Equation in R3

309

t 1 = (sin t + cos t) cos r + (cos t − sin t) sin r − sin t sin r r r     √ √ π t π 1 = 2 cos t − cos r + 2 sin − t sin r − sin t sin r 4 r 4 r

  π  √ = 2 cos x12 + x22 + x32 − t cos 4  π   √ t + 2 sin x12 + x22 + x32 − t sin 4 x12 + x22 + x32

  1 2 2 2 − sin t sin x1 + x2 + x3 , (x1 , x2 , x3 ) ∈ R3 , t ≥ 0. 2 2 2 x1 + x2 + x3 Example 7.18 Consider the IVP utt = Δu + r 2 et ,

.

r > 0,

t > 0,

u(r, 0) = 0, ut (r, 0) = 0,

r > 0.

Here φ(r) = 0,

.

ψ(r) = 0, f (r, t) = r 2 et ,

r ≥ 0,

t ≥ 0.

fr (r, t) = 2ret ,

r ≥ 0,

t ≥ 0,

We have .

fr (r, 0) = 0,

t ≥ 0.

Thus, f has an even extension with respect to r on R which is given by f(r, t) = r 2 et ,

.

r ∈ R,

t ≥ 0.

Hence, for the solution of the considered problem we have the following representation 1 .u(r, t) = 2r

t

r+(t−τ  )

ξ 3 eτ dξ dτ 0 r−(t−τ )

310

7 The Wave Equation

1 = 8r

t

ξ =r+(t−τ )  eτ ξ 4  dτ ξ =r−(t−τ )

0

1 = 8r

t

  eτ (r + (t − τ ))4 − (r − (t − τ ))4 dτ

0

1 = 8r



t e

r 4 + 4r 2 (t − τ )2 + (t − τ )4 + 4r 3 (t − τ )

τ

0

+2r 2 (t − τ )2 + 4r(t − τ )3 −r 4 − 4r 2 (t − τ )2 − (t − τ )4 + 4r 3 (t − τ ) − 2r 2 (t − τ )2  +4r(t − τ ) dτ 3

=

1 8r

t

  eτ 8r 3 (t − τ ) + 8r(t − τ )3 dτ

0

t =r

t e (t − τ )dτ +

2

eτ (t − τ )3 dτ

τ

0

0

τ =t  t τ =t t   τ τ 3  = r e (t − τ ) + e dτ + e (t − τ )  + 3 eτ (t − τ )2 dτ 2 τ

τ =0

τ =0

0

0

τ =t t  = −r t + e − 1 − t + 3e (t − τ )  + 6 eτ (t − τ )dτ 2

t

3

2

τ

τ =0

0

τ =t t   = −r t + e − 1 − t − 3t + 6e (t − τ ) + 6 eτ dτ 2

t

3

2

τ

τ =0

0

= −r 2 t + et − 1 − t 3 − 3t 2 − 6t + 6et − 6 = −r 2 t + 7et − t 3 − 3t 2 − 6t − 7, Hence,

r ∈ R,

t ≥ 0.

7.2 The Wave Equation in R3

311

u(x1 , x2 , x3 , t) = −(x12 + x22 + x32 )t + 7et − t 3 − 3t 2 − 6t − 7,

.

(x1 , x2 , x3 ) ∈ R3 , t ≥ 0. Exercise 7.21 Find the radially symmetric solution of the following IVP utt = Δu + r 2 + t,

r > 0,

.

t > 0,

u(r, 0) = 0, ut (r, 0) = cos r,

r > 0.

7.2.2 The Cauchy Problem In this section, we will give integral representations of the solutions of the Cauchy problem for the homogeneous and nonhomogeneous three dimensional wave equations. For this aim, we introduce an auxiliary function that satisfies the homogeneous three dimensional wave equation. Let (x, t) = (x1 , x2 , x3 , t) ∈ R3 × [0, ∞) be arbitrarily chosen. With S we will denote the sphere |y − x|2 = t 2 , where y = (y1 , y2 , y3 ), |x − y| is the distance between the points x and y. Let also, μ be an arbitrary real valued twice continuously differentiable function defined on S. We will prove that the function u(x, t) =

.

1 t

 μ(y1 , y2 , y3 )dsy

(7.46)

S

is a solution to the Eq. (7.42) in the case when c = 1. If c /= 1, then we make the 1 change t = τ. Really, consider the following change of variables c yi − xi = tξi ,

.

i = 1, 2, 3,

brings the expression (7.46) in the form  u(x, t) = t

μ(x1 + tξ1 , x2 + tξ2 , x3 + tξ3 )dσξ ,

.

(7.47)

σ

where σ is the unit sphere |ξ | = 1 and dσξ = By (7.47), we get Δu(x, t) = t

  3

.

σ

i=1

dsy is an element of the unit sphere. t2

μyi yi (x1 + tξ1 , x2 + tξ2 , x3 + tξ3 )dσξ

312

7 The Wave Equation

and  ut (x, t) =

μ(x1 + tξ1 , x2 + tξ2 , x3 + tξ3 )dσξ

.

σ

+t

  3 σ

=

ξi μyi (x1 + tξ1 , x2 + tξ2 , x3 + tξ3 )dσξ

i=1

u(x, t) +t t

  3

ξi μyi (x1 + tξ1 , x2 + tξ2 , x3 + tξ3 )dσξ .

i=1

σ

We set I=

  3

.

S

μyi (y)νi (y)dsy ,

i=1

where ν(y) = (ν1 (y), ν2 (y), ν3 (y)) is the outer normal vector to S at the point y. Then t

  3

.

ξi μyi (x + tξ )dσξ =

i=1

σ

1 I t

and ut (x, t) =

.

u(x, t) 1 + I. t t

We differentiate the last equation with respect to t and we find ut (x, t) u(x, t) 1 1 − − 2 I + It 2 t t t t

 1 1 u(x, t) 1 1 1 + I − 2 u(x, t) − 2 I + It = t t t t t t 1 = It . t

utt (x, t) = .

By the Gauss-Ostrogradsky formula, we have that  t π 2π I=

Δμρ 2 sin θ dφdθ dρ,

.

0

0

0

(7.48)

7.2 The Wave Equation in R3

313

where we have used the polar coordinates x1 = ρ cos φ sin θ,

.

x2 = ρ sin φ sin θ, x3 = ρ cos θ,

ρ ∈ [0, t],

φ ∈ [0, 2π ],

θ ∈ [0, π ].

Hence, π 2π It = t

.

2

Δμ sin θ dφdθ 0

0

 = t2

Δμdσξ σ

= tΔu. We substitute the last formula in (7.48) and we obtain that u satisfies the Eq. (7.42). Example 7.19 Consider the function u(x1 , x2 , x3 , t) =

.

1 t

 (y1 + y2 + y3 )dsy ,

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

S

Then 1 .u(x, t) = t

2π π (x1 +x2 +x3 +t cos φ sin θ +t sin φ sin θ +t cos θ ) t 2 sin θ dθ dφ 0

0

2π π =t

(x1 + x2 + x3 + t cos φ sin θ + t sin φ sin θ + t cos θ ) sin θ dθ dφ 0

0

2π π = t (x1 + x2 + x3 ) 0

⎞⎛ π ⎞ ⎛ 2π   sin θ dθ dφ + t 2 ⎝ cos φdφ ⎠ ⎝ (sin θ )2 dθ ⎠

0

0

0

⎞⎛ π ⎞ ⎞ ⎛ 2π ⎛ 2π ⎞ ⎛ π     +t 2 ⎝ sin φdφ ⎠ ⎝ (sin θ )2 dθ ⎠ + t 2 ⎝ dφ ⎠ ⎝ cos θ sin θ dθ ⎠ 0

0

0

0

314

7 The Wave Equation



θ =π   = t (x1 + x2 + x3 )2π − cos θ  θ =0

t2 + 2



⎞ ⎛ φ=2π  π π  ⎝ dθ − cos(2θ )dθ ⎠ sin φ  φ=0

0

0

⎞ ⎛  φ=2π  π π 2  t ⎝ dθ − cos(2θ )dθ ⎠ + − cos φ  2 φ=0 0

0

θ=π  +π t (sin θ )  2

2

θ=0

= 4π t (x1 + x2 + x3 ),

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

Hence, ut (x1 , x2 , x3 , t) = 4π(x1 + x2 + x3 ),

.

utt (x1 , x2 , x3 , t) = 0, uxj (x1 , x2 , x3 , t) = 4π t, uxj xj (x1 , x2 , x3 , t) = 0,

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0,

j ∈ {1, 2, 3}.

Therefore utt (x1 , x2 , x3 , t) −

3 

.

uxj xj (x1 , x2 , x3 , t) = 0,

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0,

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0,

j =1

i.e., u satisfies the Eq. (7.42). Exercise 7.22 Check that the following functions 1. 1 .u(x1 , x2 , x3 , t) = t

 (y1 − 2y2 + 3y3 )dsy , S

2. u(x1 , x2 , x3 , t) =

.

1 t

 (y1 + y2 )dsy , S

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0,

7.2 The Wave Equation in R3

315

3. 1 .u(x1 , x2 , x3 , t) = t

 (y1 + y3 )dsy ,

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0,

(y2 + y3 )dsy ,

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0,

S

4. u(x1 , x2 , x3 , t) =

.

1 t

 S

5. u(x1 , x2 , x3 , t) =

.

1 t

 (x1 , x2 , x3 ) ∈ R3 ,

yj dsy ,

t ≥ 0,

j ∈ {1, 2, 3},

S

satisfy the Eq. (7.42) for c = 1. We set 1 .M(μ) = t2

 μ(y1 , y2 , y3 )dsy . S

Then u(x, t) = tM(μ) is a solution to the Eq. (7.42) for c = 1. Now, we consider the Cauchy problem utt − Δu = 0,

.

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

u(x1 , x2 , x3 , 0) = φ(x1 , x2 , x3 ), .

ut (x1 , x2 , x3 , 0) = ψ(x1 , x2 , x3 ),

(x1 , x2 , x3 ) ∈ R3 ,

(7.49) (7.50)

where φ ∈ C 3 (R3 ), ψ ∈ C 2 (R3 ). We will prove that u(x, t) =

.

1 1 ∂ tM(ψ) + (tM(φ)) 4π 4π ∂t

(7.51)

is its solution. Definition 7.5 The equality (7.51) is called the Kirchhoff3 formula.

3 Gustav Robert Kirchhoff (12 March 1824–17 October 1887) was a German physicist who contributed to the fundamental understanding of electrical circuits, spectroscopy, and the emission of black-body radiation by heated objects.)

316

7 The Wave Equation

We have that the function (7.51) satisfies the Eq. (7.49). We will prove that it satisfies the initial conditions (7.50). Indeed,  1  M(φ) t=0 4π  1 φ(x1 , x2 , x3 )dσξ = 4π

u(x1 , x2 , x3 , 0) =

.

|ξ |=1

= φ(x1 , x2 , x3 ), ut (x1 , x2 , x3 , t) =

∂ ∂t



 1 1 ∂ tM(ψ) + (tM(φ)) 4π 4π ∂t

=

1 1 ∂ 1 ∂2 M(ψ) + t M(ψ) + (tM(φ)) 4π 4π ∂t 4π ∂t 2

=

1 1 ∂ 1 t M(ψ) + tΔM(φ), M(ψ) + 4π ∂t 4π 4π

 1  M(ψ) t=0 4π  1 ψ(x1 , x2 , x3 )dσξ = 4π

ut (x1 , x2 , x3 , 0) =

|ξ |=1

= ψ(x1 , x2 , x3 ). Example 7.20 Consider the Cauchy problem utt − ux1 x1 − ux2 x2 − ux3 x3 = 0, .

u(x1 , x2 , x3 , 0)

= x1 ,

ut (x1 , x2 , x3 , 0)

= x3 ,

(x1 , x2 , x3 ) ∈ R3 ,

(x1 , x2 , x3 ) ∈ R3 .

Here φ(x1 , x2 , x3 ) = x1 ,

.

ψ(x1 , x2 , x3 ) = x3 . Hence,

t > 0,

7.2 The Wave Equation in R3

317



 φ(y1 , y2 , y3 )dsy =

.

|x−y|=t

y1 dsy |x−y|=t

2π π =

(x1 + t cos φ sin θ ) t 2 sin θ dθ dφ 0

0

2π π = t x1

2π π sin θ dθ dφ + t

2

0

3

0

cos φ(sin θ )2 dθ dφ 0

0

θ=π  = −2π t x1 cos θ  2

θ=0

 +t

φ=2π  π  1 − cos(2θ ) sin φ  dθ 2

3

φ=0



0

= 4π t 2 x1 ,  ψ(y1 , y2 , y3 )dsy = y3 dsy

|x−y|=t

|x−y|=t

2π π =

(x3 + t cos θ ) t 2 sin θ dθ dφ 0

0

2π π = t x3

2π π sin θ dθ dφ + t

2

0

0

3

sin θ cos θ dθ dφ 0

0

θ=π θ=π   = −2π t 2 x3 cos θ  + π t 3 (sin θ )2  θ=0

= 4π t 2 x3 , Then, using Kirchhoff’s formula, we get

(x1 , x2 , x3 ) ∈ R3 ,

θ=0

t ≥ 0.

318

7 The Wave Equation

1 1 1 ∂ (4π t 2 x3 ) + .u(x1 , x2 , x3 , t) = 4π t 4π ∂t = tx3 + x1 ,



1 4π t 2 x1 t

(x1 , x2 , x3 ) ∈ R3 ,



t ≥ 0.

Exercise 7.23 Find a solution to the following IVPs 1. utt − ux1 x1 − ux2 x2 − ux3 x3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

u(x1 , x2 , x3 , 0) = x1 + x2 + x3 , ut (x1 , x2 , x3 , 0) = 1,

(x1 , x2 , x3 ) ∈ R3 .

2. utt − ux1 x1 − ux2 x2 − ux3 x3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

u(x1 , x2 , x3 , 0) = 0, ut (x1 , x2 , x3 , 0) = x1 x2 + x1 x3 + x2 x3 ,

(x1 , x2 , x3 ) ∈ R3 .

3. utt − ux1 x1 − ux2 x2 − ux3 x3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

u(x1 , x2 , x3 , 0) = x1 − x2 − x3 , ut (x1 , x2 , x3 , 0) = 1,

(x1 , x2 , x3 ) ∈ R3 .

utt − ux1 x1 − ux2 x2 − ux3 x3 = 0,

(x1 , x2 , x3 ) ∈ R3 ,

4. .

t > 0,

u(x1 , x2 , x3 , 0) = x1 x2 x3 , ut (x1 , x2 , x3 , 0) = x12 x22 x32 ,

(x1 , x2 , x3 ) ∈ R3 .

5. utt − ux1 x1 − ux2 x2 − ux3 x3 = 0, .

u(x1 , x2 , x3 , 0)

=

ut (x1 , x2 , x3 , 0)

= 0,

1 2 2 x1

(x1 , x2 , x3 ) ∈ R3 , +

1 2 2 x2

+ x32 ,

(x1 , x2 , x3 ) ∈ R3 .

t > 0,

7.2 The Wave Equation in R3

319

Next, we consider the Cauchy problem utt − Δu .

= f (x1 , x2 , x3 , t),

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

u(x1 , x2 , x3 , 0) = φ(x1 , x2 , x3 ), ut (x1 , x2 , x3 , 0) = ψ(x1 , x2 , x3 ),

(7.52) (x1 , x2 , x3 ) ∈ R3 ,

where φ ∈ C 3 (R3 ), ψ ∈ C 2 (R3 ), f ∈ C 2 (R3 × [0, ∞)). We set v(x1 , x2 , x3 , t) = u(x1 , x2 , x3 , t) − φ(x1 , x2 , x3 )

.

− tψ(x1 , x2 , x3 ),

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

Then v(x1 , x2 , x3 , 0) = u(x1 , x2 , x3 , 0) − φ(x1 , x2 , x3 )

.

= 0, vt (x1 , x2 , x3 , t) = ut (x1 , x2 , x3 , t) − ψ(x1 , x2 , x3 ) vt (x1 , x2 , x3 , 0) = ut (x1 , x2 , x3 , 0) − ψ(x1 , x2 , x3 ) = 0, vtt (x1 , x2 , x3 , t) = utt (x1 , x2 , x3 , t), vxi xi (x1 , x2 , x3 , t) = uxi xi (x1 , x2 , x3 , t) − φxi xi (x1 , x2 , x3 ) −tψxi xi (x1 , x2 , x3 ),

i = 1, 2, 3,

vtt (x1 , x2 , x3 , t) − Δv(x1 , x2 , x3 , t) = utt (x1 , x2 , x3 , t) − Δu(x1 , x2 , x3 , t) +Δφ(x1 , x2 , x3 ) + tΔψ(x1 , x2 , x3 ) = f (x1 , x2 , x3 , t) + Δφ(x1 , x2 , x3 ) +tΔψ(x1 , x2 , x3 ), ×(x1 , x2 , x3 ) ∈ R3 , Therefore v satisfies the Cauchy problem

t ≥ 0.

320

7 The Wave Equation

vtt − Δv .

= g(x1 , x2 , x3 , t),

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

v(x1 , x2 , x3 , 0) = 0, vt (x1 , x2 , x3 , 0) = 0,

(7.53) (x1 , x2 , x3 ) ∈ R3 ,

where g(x1 , x2 , x3 , t) = f (x1 , x2 , x3 , t) + Δφ(x1 , x2 , x3 ) + tΔψ(x1 , x2 , x3 ),

.

(x1 , x2 , x3 ) ∈ R3 , t ≥ 0. Let τ > 0 be arbitrarily chosen. Consider the Cauchy problem wtt − Δw .

= 0,

(x1 , x2 , x3 ) ∈ R3 ,

t > τ,

w(x1 , x2 , x3 , τ ) = 0,

(7.54)

wt (x1 , x2 , x3 , τ ) = g(x1 , x2 , x3 , τ ). Let z be the solution of the problem (7.54) which is supposed to be continued as identically zero for t ≤ τ. We set t h(x1 , x2 , x3 , t) =

z(x1 , x2 , x3 , t, τ )dτ,

.

(x1 , x2 , x3 ) ∈ R3 ,

0

Then h(x1 , x2 , x3 , 0) = 0,

.

t ht (x1 , x2 , x3 , t) = z(x1 , x2 , x3 , t, t) +

zt (x1 , x2 , x3 , t, τ )dτ 0

t =

zt (x1 , x2 , x3 , t, τ )dτ, 0

ht (x1 , x2 , x3 , 0) = 0, t htt (x1 , x2 , x3 , t) = zt (x1 , x2 , x3 , t, t) +

ztt (x1 , x2 , x3 , t, τ )dτ 0

t ≥ τ.

7.2 The Wave Equation in R3

321

t = g(x1 , x2 , x3 , t) +

ztt (x1 , x2 , x3 , t, τ )dτ 0

t = g(x1 , x2 , x3 , t) +

Δz(x1 , x2 , x3 , t, τ )dτ 0

= g(x1 , x2 , x3 , t)+Δh(x1 , x2 , x3 , t),

(x1 , x2 , x3 ) ∈ R3 , t ≥ τ,

i.e., h is a solution to the Cauchy problem (7.53). Now we apply the Kirchhoff formula and we obtain z(x1 , x2 , x3 , t, τ ) =

.

1 (t − τ )Mt−τ (g(x1 , x2 , x3 , τ )), 4π

(x1 , x2 , x3 ) ∈ R3 , t ≥ τ , whereupon 1 .v(x1 , x2 , x3 , t) = 4π

t (t − τ )Mt−τ (g(x1 , x2 , x3 , τ ))dτ, 0

(x1 , x2 , x3 ) ∈ R3 , t ≥ 0, and u(x1 , x2 , x3 , t) = φ(x1 , x2 , x3 ) + tψ(x1 , x2 , x3 ) 1 + 4π

.

t (t − τ )Mt−τ (f (x1 , x2 , x3 , τ ) 0

+Δφ(x1 , x2 , x3 ) + τ Δψ(x1 , x2 , x3 )) dτ, (x1 , x2 , x3 ) ∈ R3 , t ≥ 0, is a solution to the Cauchy problem (7.52). Example 7.21 Consider the Cauchy problem utt − Δu .

= x1 ,

(x1 , x2 , x3 ) ∈ R3 ,

u(x1 , x2 , x3 , 0) = x2 , ut (x1 , x2 , x3 , 0) = x3 ,

Here f (x1 , x2 , x3 , t) = x1 ,

.

φ(x1 , x2 , x3 ) = x2 ,

(x1 , x2 , x3 ) ∈ R3 .

t > 0,

(7.55)

322

7 The Wave Equation

ψ(x1 , x2 , x3 ) = x3 ,

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

Then, using (7.55), t

1 .u(x1 , x2 , x3 , t) = x2 + tx3 + 4π

1 t −τ

0

t

1 = x2 + tx3 + 4π

1 t −τ

0

 y1 dsy dτ |x−y|=t−τ

2π π (x1 + (t − τ ) cos φ sin θ) 0

0

×(t − τ )2 sin θ dθ dφdτ = x2 + tx3 +

1 x1 4π

t

2π π (t − τ )

0

1 + 4π

t

0

0

2π π (t − τ )

2

0

sin θ dθ dφdτ

cos φ(sin θ )2 dθ dφdτ 0

0

   θ=π 

 1 (t − τ )2 τ =t = x2 + tx3 + x1 − (2π ) − cos θ   4π 2 τ =0 θ=0 1 + 4π



⎞ ⎛   φ=2π  π  1 − cos(2θ ) (t − τ )3 τ =t ⎝ − dθ ⎠ sin φ   3 2 τ =0 φ=0 0

1 = x2 + tx3 + x1 t 2 . 2 Exercise 7.24 Solve the following IVPs 1. utt − ux1 x1 − ux2 x2 − ux3 x3 = x1 + t,

.

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

u(x1 , x2 , x3 , 0) = x1 x2 x3 , ut (x1 , x2 , x3 , 0) = x1 x2 + x3 ,

(x1 , x2 , x3 ) ∈ R3 .

7.3 The Two Dimensional Wave Equation

323

2. x1 x2 e cos x3 , (x1 , x2 , x3 ) ∈ R3 , 1 + t2 √ u(x1 , x2 , x3 , 0) = x3 sin( 2(x1 + x2 )),

utt − ux1 x1 − ux2 x2 − ux3 x3 =

.

ut (x1 , x2 , x3 , 0) = 0,

t > 0,

(x1 , x2 , x3 ) ∈ R3 .

3. utt − ux1 x1 − ux2 x2 − ux3 x3 =

.

x1 t (t 2 + 5) , (1 + t 2 )2

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

u(x1 , x2 , x3 , 0) = x1 sin x2 , ut (x1 , x2 , x3 , 0) = x2 cos x3 − 2x1 ,

(x1 , x2 , x3 ) ∈ R3 .

4. utt − ux1 x1 − ux2 x2 − ux3 x3 = sin tx1 x2 sin x3 ,

(x1 , x2 , x3 ) ∈ R3 ,

.

t > 0,

u(x1 , x2 , x3 , 0) = x3 + x1 x2 , ut (x1 , x2 , x3 , 0) = 0,

(x1 , x2 , x3 ) ∈ R3 .

5. utt − ux1 x1 − ux2 x2 − ux3 x3 = −4x1 − 6x2 , .

u(x1 , x2 , x3 , 0)

= x13 + x23 ,

ut (x1 , x2 , x3 , 0)

= 0,

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

(x1 , x2 , x3 ) ∈ R3 .

7.3 The Two Dimensional Wave Equation The solution of the two dimensional wave equation can be obtained by solving the three dimensional wave equation in the case where the initial data depends only on x1 and x2 , but not x3 . In this case the three dimensional solution consists of cylindrical waves. We consider the Cauchy problem for the wave equation with two spatial variables utt − ux1 x1 − ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 ,

t > 0,

(7.56)

324

7 The Wave Equation

u(x1 , x2 , 0) = φ(x1 , x2 ), .

ut (x1 , x2 , 0) = ψ(x1 , x2 ),

(x1 , x2 ) ∈ R2 ,

(7.57)

where φ ∈ C 3 (R2 ), ψ ∈ C 2 (R2 ). The solution u of the problem (7.56), (7.57) can be derived using the Kirchhoff formula. To obtain the solution of the original three dimensional problem, we need to convert the Kirchhoff formula over a sphere of radius ct to an integral over a disc of radius ct. This approach for obtaining the solution is called the method of descent, which is due to Hadamard. We have u(x1 , x2 , t) =

1 4π t

.



ψ(x1 + y1 , x2 + y2 )dsy

|y|=t

1 ∂ + 4π ∂t

1 t

 |y|=t

 φ(x1 + y1 , x2 + y2 )dsy ,

(x1 , x2 ) ∈ R2 ,

t ≥ 0,

(7.58) which is independent of x3 and satisfies (7.56), (7.57). Note that the projection dy1 dy2 of the element of the arc dsy of the sphere |y| = t, y = (y1 , y2 , y3 ), on the circle y12 + y22 ≤ t 2 is expressed by the formula dy1 dy2 =

.

|y3 | dsy . t

To compute the integrals on the right-hand side of the formula (7.58) we should project on the circle y12 + y22 ≤ t 2 both the upper hemisphere y3 > 0 and the lower hemisphere y3 < 0 of the sphere |y| = t. Therefore 1 u(x1 , x2 , t) = 2π

 

ψ(y1 , y2 )

dy1 dy2 − (y1 − x1 )2 − (y2 − x2 )2 B  φ(y1 , y2 ) 1 ∂  + dy1 dy2 , 2 2π ∂t t − (y1 − x1 )2 − (y2 − x2 )2

.

t2

B

×(x1 , x2 ) ∈ R2 ,

t ≥ 0,

where B is the circle (y1 − x1 )2 + (y2 − x2 )2 ≤ t 2 . Definition 7.6 The equality (7.59) is called the Poisson formula. Example 7.22 Consider the Cauchy problem utt − ux1 x1 − ux2 x2 = 0,

.

(x1 , x2 ) ∈ R2 ,

u(x1 , x2 , 0) = x1 , ut (x1 , x2 , 0) = x2 ,

(x1 , x2 ) ∈ R2 .

t > 0,

(7.59)

7.3 The Two Dimensional Wave Equation

325

Then, using the Poisson formula, we have 1 2π

u(x1 , x2 , t) =

.



y2  dy1 dy2 t 2 − (y1 − x1 )2 − (y2 − x2 )2

B

1 ∂ + 2π ∂t

 B

y1  dy1 dy2 , 2 t − (y1 − x1 )2 − (y2 − x2 )2

B : (y1 − x1 )2 + (y2 − x2 )2 ≤ t 2 . Let y1 = x1 + r cos φ,

.

y2 = x2 + r sin φ,

r ∈ [0, t],

φ ∈ [0, 2π ].

Hence, 1 .u(x1 , x2 , t) = 2π

 t 2π 0

t = x2 0

0

 t 2π 0

0

x1 + r cos φ rdφdr √ t2 − r2

⎞ ⎛ 2π ⎞ ⎛ t   2 r r 1 ⎝ dr + dr ⎠ ⎝ sin φdφ ⎠ √ √ 2 2 2 2π t −r t − r2 0

⎛ +

x2 + r sin φ 1 ∂ rdφdr + √ 2 2 2π ∂t t −r

∂ ⎝ x1 ∂t

0

⎞ ⎛ 2π ⎞⎞ ⎛ t   2 r r 1 ⎝ dr+ dr ⎠ ⎝ cos φdφ ⎠⎠ , √ √ 2π t 2 −r 2 t 2 −r 2

t 0

0

0

(x1 , x2 ) ∈ R2 , t ≥ 0. Note that 2π .

φ=2π  sin φdφ = − cos φ  φ=0

0

= −(1 − 1) = 0, 2π

φ=2π  cos φdφ = sin φ  φ=0

0

=0

326

7 The Wave Equation

and t .

0

1 dr = − √ 2 t2 − r2 r

t 0

d(t 2 − r 2 ) √ t2 − r2

r=t   2 2 = − t − r 

r=0

=



t2

= |t| = t,

t ≥ 0.

Therefore u(x1 , x2 , t) = x2 t +

.

∂ (x1 t) ∂t

= x1 + x2 t,

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

(x1 , x2 ) ∈ R2 ,

t > 0,

Exercise 7.25 Solve the Cauchy problem utt − ux1 x1 − ux2 x2 = 0,

.

u(x1 , x2 , 0) = x1 − x2 , ut (x1 , x2 , 0) = x2 ,

(x1 , x2 ) ∈ R2 .

Next, we consider the Cauchy problem utt − ux1 x1 − ux2 x2 = f (x1 , x2 , t),

.

(x1 , x2 ) ∈ R2 ,

t > 0,

u(x1 , x2 , 0) = φ(x1 , x2 ), ut (x1 , x2 , 0) = ψ(x1 , x2 ),

(x1 , x2 ) ∈ R2 ,

where φ ∈ C 3 (R2 ), ψ ∈ C 2 (R2 ) and f ∈ C 2 (R2 × [0, ∞)). Using the formula (7.55), for its solution u(x1 , x2 , t) we have the representation

7.3 The Two Dimensional Wave Equation

327

u(x1 , x2 , t) = φ(x1 , x2 ) + tψ(x1 , x2 ) t 

1 + 2π

.

0 Bt−τ

f (y1 , y2 , τ ) + Δφ(y1 , y2 ) + τ Δψ(y1 , y2 )  dy1 dy2 dτ, (t − τ )2 − (y1 − x1 )2 − (y2 − x2 )2

(x1 , x2 ) ∈ R2 , t ≥ 0, where Bt−τ : (y1 − x1 )2 + (y2 − x2 )2 ≤ (t − τ )2 and Δφ(y1 , y2 ) = φy1 y1 (y1 , y2 ) + φy2 y2 (y1 , y2 ),

.

Δψ(y1 , y2 ) = ψy1 y1 (y1 , y2 ) + ψy2 y2 (y1 , y2 ),

(y1 , y2 ) ∈ R2 .

Example 7.23 Consider the problem utt = ux1 x1 + ux2 x2 + 2,

.

(x1 , x2 ) ∈ R2 ,

u(x1 , x2 , 0) = x1 , ut (x1 , x2 , 0) = x2 ,

(x1 , x2 ) ∈ R2 .

Here f (x1 , x2 , t) = 2,

.

φ(x1 , x2 ) = x1 , ψ(x1 , x2 ) = x2 ,

(x1 , x2 ) ∈ R2 ,

We have φx1 (x1 , x2 ) = 1,

.

φx1 x1 (x1 , x2 ) = 0, φx2 (x1 , x2 ) = 0, '

φx2 x2 (x1 , x2 ) = 0, ψx1 (x1 , x2 ) = 0, ψx1 x1 (x1 , x2 ) = 0, ψx2 (x1 , x2 ) = 1,

t ≥ 0.

t > 0,

328

7 The Wave Equation

ψx2 x2 (x1 , x2 ) = 0,

(x1 , x2 ) ∈ R2 .

Then 1 . 2π

t 

f (y1 , y2 , τ ) + Δφ(y1 , y2 ) + τ Δψ(y1 , y2 )  dy1 dy2 dτ (t − τ )2 − (y1 − x1 )2 − (y2 − x2 )2

0 Bt−τ

t 

1 = 2π

2  dy1 dy2 dτ 2 (t − τ ) − (y1 − x1 )2 − (y2 − x2 )2

0 Bt−τ

t 

1 = π

 0 Bt−τ

1 (t

− τ )2

×(x1 , x2 ) ∈ R2 ,

− (y1 − x1 )2 − (y2 − x2 )2

dy1 dy2 dτ,

t ≥ 0.

Let y1 = x1 + r cos φ,

.

y2 = x2 + r sin φ,

φ ∈ [0, 2π ],

r ∈ [0, t − τ ],

τ ∈ [0, t],

t ≥ 0.

Then

.

t 

1 π

 0 Bt−τ

 t t−τ2π

1 = π

0

0

 t t−τ =2 0

0

0

0

0

(t

− (y1 − x1 )2 − (y2 − x2 )2

1  rdφdrdt (t − τ )2 − r 2

r  drdτ (t − τ )2 − r 2

 t t−τ =−

1 − τ )2

d((t − τ )2 − r 2 )  dτ (t − τ )2 − r 2

dy1 dy2 dτ

7.3 The Two Dimensional Wave Equation

329

r=t−τ t   = −2 (t − τ )2 − r 2  dτ r=0

0

=2

t 

(t − τ )2 dτ

0

t =2

|t − τ |dτ 0

t =2

(t − τ )dτ 0

τ =t  = −(t − τ )  2

τ =0

= t 2,

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

Therefore u(x1 , x2 , t) = φ(x1 , x2 ) + tψ(x1 , x2 )

.

1 + 2π

t  0 Bt−τ

f (y1 , y2 , τ ) + Δφ(y1 , y2 ) + τ Δψ(y1 , y2 )  dy1 dy2 dτ (t − τ )2 − (y1 − x1 )2 − (y2 − x2 )2

= x1 + tx2 + t 2 ,

(x1 , x2 ) ∈ R2 ,

t ≥ 0,

is the solution of the considered problem. Example 7.24 Consider the problem utt = ux1 x1 + ux2 x2 + e3x1 +4x2 ,

.

(x1 , x2 ) ∈ R2 ,

u(x1 , x2 , 0) = e3x1 +4x2 , ut (x1 , x2 , 0) = e3x1 +4x2 ,

(x1 , x2 ) ∈ R2 .

We will search a solution of the considered problem in the form u(x1 , x2 , t) = φ(t)e3x1 +4x2 ,

.

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

t > 0,

330

7 The Wave Equation

We have ux1 (x1 , x2 , t) = 3φ(t)e3x1 +4x2 ,

.

ux1 x1 (x1 , x2 , t) = 9φ(t)e3x1 +4x2 , ux2 (x1 , x2 , t) = 4φ(t)e3x1 +4x2 , ux2 x2 (x1 , x2 , t) = 16φ(t)e3x1 +4x2 , ut (x1 , x2 , t) = φ ' (t)e3x1 +4x2 , utt (x1 , x2 , t) = φ '' (t)e3x1 +4x2 ,

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

Next, u(x1 , x2 , 0) = φ(0)e3x1 +4x2

.

= e3x1 +4x2 , ut (x1 , x2 , 0) = φ ' (0)e3x1 +4x2 = e3x1 +4x2 ,

(x1 , x2 ) ∈ R2 .

Therefore φ(0) = 1,

.

φ ' (0) = 1. By the given equation, we get φ '' (t)e3x1 +4x2 = 9φ(t)e3x1 +4x2 + 16φ(t)e3x1 +4x2 + e3x1 +4x2

.

= (25φ(t) + 1) e3x1 +4x2 ,

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

Thus, we get the following IVP φ '' (t) − 25φ(t) = 1, .

φ(0)

= 1,

φ ' (0)

= 1.

t ≥ 0, (7.60)

7.3 The Two Dimensional Wave Equation

331

The characteristic equation of the corresponding homogeneous equation is r 2 = 25,

.

whereupon r = ±5. Hence, the general solution of the corresponding homogeneous equation of (7.60) is φ(t) = c1 e5t + c2 e−5t ,

.

t ≥ 0,

where c1 and c2 are constants. Note that φp (t) = −

.

1 , 25

t ≥ 0,

is a particular solution of the first equation of (7.60). Then, the general solution of the first equation of (7.60) is φ(t) = c1 e5t + c2 e−5t −

.

1 , 25

t ≥ 0.

We have φ(0) = c1 + c2 −

.

1 25

= 1, φ ' (t) = 5c1 e5t − 5c2 e−5t , φ ' (0) = 5c1 − 5c2 = 1. Therefore, we obtain the system c1 + c2 =

.

26 25

5c1 − 5c2 = 1, whereupon .

c1 =

31 , 50

c2 =

21 . 50

t ≥ 0,

332

7 The Wave Equation

Hence, φ(t) =

.

31 5t 21 −5t 1 e + e − , 50 50 25

t ≥ 0,

and

 31 5t 21 −5t 1 e + e e3x1 +4x2 − 50 50 25



 21 e−5t 1 31 e5t · + · − e3x1 +4x2 25 2 25 2 25



 1 e5t − e−5t 1 26 e5t + e−5t · + · − e3x1 +4x2 25 2 5 2 25



 1 1 26 cosh(5t) + sinh(5t) − e3x1 +4x2 , (x1 , x2 ) ∈ R2 , t ≥ 0, 25 5 25

u(x1 , x2 , t) =

.

=

= =

is the solution of the considered problem. Exercise 7.26 Solve the following Cauchy problems 1. utt = ux1 x1 + ux2 x2 + 6x1 x2 t,

.

(x1 , x2 ) ∈ R2 ,

t > 0,

u(x1 , x2 , 0) = x12 − x22 , ut (x1 , x2 , 0) = x1 x2 ,

(x1 , x2 ) ∈ R2 .

2. utt = ux1 x1 + ux2 x2 + x13 − 3x1 x22 ,

.

(x1 , x2 ) ∈ R2 ,

t > 0,

u(x1 , x2 , 0) = ex1 cos x2 , ut (x1 , x2 , 0) = ex2 sin x1 ,

(x1 , x2 ) ∈ R2 .

3. utt = ux1 x1 + ux2 x2 + t sin x2 ,

.

u(x1 , x2 , 0) = x12 , ut (x1 , x2 , 0) = sin x2 ,

(x1 , x2 ) ∈ R2 .

(x1 , x2 ) ∈ R2 ,

t > 0,

7.4 Advanced Practical Problems

333

4. utt = 2(ux1 x1 + ux2 x2 ),

.

(x1 , x2 ) ∈ R2 ,

t > 0,

u(x1 , x2 , 0) = 2x12 − x22 , ut (x1 , x2 , 0) = 2x12 + x22 ,

(x1 , x2 ) ∈ R2 .

5. utt = ux1 x1 + ux2 x2 + x22 ,

.

(x1 , x2 ) ∈ R2 ,

u(x1 , x2 , 0) = 1, ut (x1 , x2 , 0) = x1 ,

(x1 , x2 ) ∈ R2 .

7.4 Advanced Practical Problems Problem 7.1 Solve the following Cauchy problems 1. utt − 4uxx = 0,

.

−∞ < x < ∞,

t > 0,

u(x, 0) = ex , ut (x, 0) = x,

−∞ < x < ∞.

2. utt − uxx = 0,

.

−∞ < x < ∞,

t > 0,

u(x, 0) = ex , ut (x, 0) = e−x ,

−∞ < x < ∞.

3. utt − 2uxx = 0,

.

−∞ < x < ∞,

u(x, 0) = 0, ut (x, 0) = x,

−∞ < x < ∞.

t > 0,

t > 0,

334

7 The Wave Equation

4. utt − 16uxx = 0,

−∞ < x < ∞,

.

t > 0,

u(x, 0) = 1, ut (x, 0) = sin x,

−∞ < x < ∞.

5. utt − 25uxx = 0,

−∞ < x < ∞,

.

t > 0,

u(x, 0) = 1, ut (x, 0) = −1,

−∞ < x < ∞.

Problem 7.2 Solve the following Cauchy problems 1. utt − 4uxx = 6t, .

u(x, 0)

= x,

ut (x, 0)

= 0,

−∞ < x < ∞,

t > 0,

−∞ < x < ∞.

2. utt − uxx = sin x,

−∞ < x < ∞,

.

t > 0,

u(x, 0) = sin x, ut (x, 0) = 0,

−∞ < x < ∞.

3. utt − 9uxx = sin x,

.

−∞ < x < ∞,

u(x, 0) = 1, ut (x, 0) = 1,

−∞ < x < ∞.

t > 0,

7.4 Advanced Practical Problems

335

4. utt − a 2 uxx = sin(aωx),

−∞ < x < ∞,

.

a, ω ∈ R,

t > 0,

a, ω /=,

u(x, 0) = 0, ut (x, 0) = 0,

−∞ < x < ∞.

5. utt − a 2 uxx = sin(ωt),

.

−∞ < x < ∞,

t > 0,

a, ω ∈ R,

u(x, 0) = 0, ut (x, 0) = 0,

−∞ < x < ∞.

Problem 7.3 Find a formal solution to the following IBVPs 1. utt − uxx = 0, u(x, 0)

ux (0, t)

t > 0,

= (sin x)3 ,

ut (x, 0) = 0,

.

0 < x < π,

0 < x < π,

= 0,

ux (π, t) = 0,

t > 0.

utt − uxx = 0,

0 < x < 1,

2. .

t > 0,

u(x, 0) = 0, ut (x, 0) = sin(2π x),

0 < x < 1,

u(0, t) = 0, u(1, t) = 0,

t > 0.

3. utt − 4uxx = 0,

.

u(x, 0) = 0,

0 < x < 1,

t > 0,

a, ω /= 0,

336

7 The Wave Equation

ut (x, 0) = sin(2π x),

0 < x < 1,

u(0, t) = 0, u(1, t) = 0,

t > 0.

4. utt − 9uxx = 0,

.

0 < x < 1,

t > 0,

u(x, 0) = 0, ut (x, 0) = 1,

0 < x < 1,

ux (0, t) = 0, ux (1, t) = 0,

t > 0.

5. utt − uxx = 0,

.

0 < x < 1,

t > 0,

u(x, 0) = 0, ut (x, 0) = x 2 (1 − x),

0 < x < 1,

ux (0, t) = 0, u(1, t) = 0,

t > 0.

Problem 7.4 Find a formal solution of the following IBVPs. 1. utt − uxx = x,

.

0 < x < π,

u(x, 0) = sin(2x), ut (x, 0) = 0,

0 < x < π,

u(0, t) = 0, u(π, t) = 0,

t > 0.

t > 0,

7.4 Advanced Practical Problems

337

2. utt − uxx = e−t sin(π x),

.

0 < x < 1,

t > 0,

u(x, 0) = 0, ut (x, 0) = 0,

0 < x < 1,

u(0, t) = 0, u(1, t) = 0,

t > 0.

3. utt − uxx = xe−t ,

.

0 < x < 1,

t > 0,

u(x, 0) = 0, ut (x, 0) = 0,

0 < x < 1,

u(0, t) = 0, u(1, t) = 0,

t > 0.

4. utt − uxx = sin t,

.

u(x, 0) = 0,

0 < x < 1,

t > 0,

0 < x < 1,

ut (x, 0) = 0, u(0, t) = 0, ux (1, t) = 0,

t > 0.

5. utt − uxx = e−t cos

.

π  x , 2

u(x, 0) = 0, ut (x, 0) = 0,

0 < x < 1,

0 < x < 1,

t > 0,

338

7 The Wave Equation

ux (0, t) = 0, u(1, t) = 0,

t > 0.

Problem 7.5 Find a formal solution to the following IBVPs. 1. utt − uxx = f (x),

.

0 < x < 1,

t > 0,

0 < x < 1,

t > 0,

u(x, 0) = 0, ut (x, 0) = 0, u(0, t) = a, u(1, t) = b, where f ∈ C ([0, 1]), a, b ∈ R. 2. utt − uxx = f (x),

.

u(x, 0) = φ(x), ut (x, 0) = ψ(x),

0 < x < 1,

ux (0, t) = a, ux (1, t) = b,

t > 0,

where f ∈ C ([0, 1]), a, b ∈ R, φ ∈ C 2 ([0, 1]), ψ ∈ C 1 ([0, 1]), φ ' (0) = a,

.

φ ' (1) = b,

ψ ' (0) = ψ ' (1) = 0.

3. utt − uxx = 0,

.

0 < x < 1,

x u(x, 0) = sin , 2 ut (x, 0) = 1,

0 < x < π,

u(0, t) = t, ux (π, t) = 1,

t > 0.

t > 0,

7.4 Advanced Practical Problems

339

4. utt − uxx = 0,

0 < x < 1,

.

u(x, 0) =

t > 0,

cosh x , sinh 1

ut (x, 0) = −

cosh x , sinh 1

0 < x < 1,

ux (0, t) = 0, ux (1, t) = e−t ,

t > 0.

5. utt − uxx = sin(2t),

0 < x < 1,

.

t > 0,

u(x, 0) = 0, ut (x, 0) = −2 cos(2x),

0 < x < 1,

ux (0, t) = 0, ux (1, t) = 2 sin 2 sin(2t),

t > 0.

Problem 7.6 Let n = 3. Find the radially symmetric solution of the following IVPs. 1. utt = Δu,

r > 0,

.

t > 0,

u(r, 0) = sin(r 2 ), ut (r, 0) = 0,

r > 0.

2. utt = Δu + t,

.

r > 0,

u(r, 0) = cos r, ut (r, 0) = 1,

r > 0.

t > 0,

340

7 The Wave Equation

3. utt = Δu + r 2 ,

.

r > 0,

t > 0,

u(r, 0) = 0, ut (r, 0) = cos r,

r > 0.

4. utt = Δu + r 2 et ,

.

r > 0,

t > 0,

u(r, 0) = cos r, ut (r, 0) = 0,

r > 0.

5. utt = Δu + r 2 + t,

.

r > 0,

t > 0,

u(r, 0) = 0, ut (r, 0) = 0,

r > 0.

Problem 7.7 Let n = 3. Prove that the following functions 1. 

1 t

u(x1 , x2 , x3 , t) =

.

(y12 − 2y1 y2 + y3 )dsy ,

(x1 , x2 , x3 ) ∈ R3 ,

S

2. u(x1 , x2 , x3 , t) =

.

1 t

 (y12 + 2y22 − y32 )dsy ,

(x1 , x2 , x3 ) ∈ R3 , t ≥ 0,

S

3. u(x1 , x2 , x3 , t) =

.

1 t

 (y12 + y22 )dsy , S

(x1 , x2 , x3 ) ∈ R3 , t ≥ 0,

7.4 Advanced Practical Problems

341

4. 1 .u(x1 , x2 , x3 , t) = t

 (y12 + y32 )dsy ,

(x1 , x2 , x3 ) ∈ R3 , t ≥ 0,

(y22 + y32 )dsy ,

(x1 , x2 , x3 ) ∈ R3 , t ≥ 0,

S

5. u(x1 , x2 , x3 , t) =

.

1 t

 S

satisfy the Eq. (7.42) for c = 1. Problem 7.8 Find a solution to the following IVPs 1. utt − ux1 x1 − ux2 x2 − ux3 x3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

u(x1 , x2 , x3 , 0) = x12 + x22 + x32 , ut (x1 , x2 , x3 , 0) = x1 x2 ,

(x1 , x2 , x3 ) ∈ R3 .

2. utt − ux1 x1 − ux2 x2 − ux3 x3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

u(x1 , x2 , x3 , 0) = ex1 cos x2 , ut (x1 , x2 , x3 , 0) = x12 − x22 ,

(x1 , x2 , x3 ) ∈ R3 .

3. utt − ux1 x1 − ux2 x2 − ux3 x3 = 0,

.

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

u(x1 , x2 , x3 , 0) = x12 + x22 , ut (x1 , x2 , x3 , 0) = 1,

(x1 , x2 , x3 ) ∈ R3 .

utt − ux1 x1 − ux2 x2 − ux3 x3 = 0,

(x1 , x2 , x3 ) ∈ R3 ,

4. .

t > 0,

u(x1 , x2 , x3 , 0) = ex1 , ut (x1 , x2 , x3 , 0) = e−x1 ,

(x1 , x2 , x3 ) ∈ R3 .

342

7 The Wave Equation

5. utt − ux1 x1 − ux2 x2 − ux3 x3 = 0, .

(x1 , x2 , x3 ) ∈ R3 ,

u(x1 , x2 , x3 , 0)

= 2x1 − 3x2 + 4x3 ,

ut (x1 , x2 , x3 , 0)

= 3,

t > 0,

(x1 , x2 , x3 ) ∈ R3 .

Problem 7.9 Solve the following IVPs 1. utt − ux1 x1 − ux2 x2 − ux3 x3 = x1 x2 x3 e−t ,

(x1 , x2 , x3 ) ∈ R3 ,

.

t > 0,

u(x1 , x2 , x3 , 0) = 2x1 x2 , √ √ ut (x1 , x2 , x3 , 0) = x1 sin( 2x2 ) cos( 2x3 ),

(x1 , x2 , x3 ) ∈ R3 .

2. utt − ux1 x1 − ux2 x2 − ux3 x3 = x1 x2 x3 sin t,

.

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

u(x1 , x2 , x3 , 0) = x12 x2 x32 , ut (x1 , x2 , x3 , 0) = x2 sin x1 ex3 ,

(x1 , x2 , x3 ) ∈ R3 .

3. utt − ux1 x1 − ux2 x2 − ux3 x3 = x1 x2 x3 log(1 + t 2 ),

.

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

u(x1 , x2 , x3 , 0) = x2 ex1 sin x3 , ut (x1 , x2 , x3 , 0) = x1 x3 sin x2 ,

(x1 , x2 , x3 ) ∈ R3 .

4. utt − ux1 x1 − ux2 x2 − ux3 x3 =

.

x2 x3 t 3 , 1 + t2

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

u(x1 , x2 , x3 , 0) = x1 ex2 , ut (x1 , x2 , x3 , 0) = x2 ex3 ,

(x1 , x2 , x3 ) ∈ R3 .

5. utt − ux1 x1 − ux2 x2 − ux3 x3 = 2x1 x2 x3 ,

.

(x1 , x2 , x3 ) ∈ R3 ,

t > 0,

7.4 Advanced Practical Problems

343

u(x1 , x2 , x3 , 0) = x12 + x22 − 2x32 , ut (x1 , x2 , x3 , 0) = 1,

(x1 , x2 , x3 ) ∈ R3 .

6. utt − ux1 x1 − ux2 x2 − ux3 x3 = 6te

.

√

2x1

sin x2 cos x3 , (x1 , x2 , x3 ) ∈ R3 , t > 0,

√ u(x1 , x2 , x3 , 0) = ex1 +x2 cos( 2x3 ), ut (x1 , x2 , x3 , 0) = e3x2 +4x3 sin(5x1 ),

(x1 , x2 , x3 ) ∈ R3 .

utt − ux1 x1 − ux2 x2 − ux3 x3 = cos x1 sin x2 ex3 ,

(x1 , x2 , x3 ) ∈ R3 ,

7. .

t > 0,

u(x1 , x2 , x3 , 0) = x12 ex2 +x3 , ut (x1 , x2 , x3 , 0) = sin x1 ex2 +x3 ,

(x1 , x2 , x3 ) ∈ R3 .

8. utt − ux1 x1 − ux2 x2 − ux3 x3 = x1 et cos(3x2 + 4x3 ), (x1 , x2 , x3 ) ∈ R3 , t > 0,

.

u(x1 , x2 , x3 , 0) = x1 x2 cos x3 , ut (x1 , x2 , x3 , 0) = x2 x3 ex1 ,

(x1 , x2 , x3 ) ∈ R3 .

9. utt −ux1 x1 −ux2 x2 −ux3 x3 = 2(x12 +x22 +x32 )−6t 2 , (x1 , x2 , x3 ) ∈ R3 , t > 0,

.

u(x1 , x2 , x3 , 0) = 0, ut (x1 , x2 , x3 , 0) = 0,

(x1 , x2 , x3 ) ∈ R3 .

10. utt − ux1 x1 − ux2 x2 − ux3 x3 = −4, .

(x1 , x2 , x3 ) ∈ R3 ,

u(x1 , x2 , x3 , 0)

= x12 + x22 + x32 ,

ut (x1 , x2 , x3 , 0)

= 0,

(x1 , x2 , x3 ) ∈ R3 .

t > 0,

344

7 The Wave Equation

Problem 7.10 Solve the following Cauchy problems 1. utt = 3(ux1 x1 + ux2 x2 ) + x13 + x23 ,

.

(x1 , x2 ) ∈ R2 ,

t > 0,

u(x1 , x2 , 0) = x12 , ut (x1 , x2 , 0) = x22 ,

(x1 , x2 ) ∈ R2 .

2. utt = ux1 x1 + ux2 x2 ,

.

(x1 , x2 ) ∈ R2 ,

t > 0,

u(x1 , x2 , 0) = cos(x1 + x2 ), ut (x1 , x2 , 0) = sin(x1 + x2 ),

(x1 , x2 ) ∈ R2 .

3. utt = ux1 x1 + ux2 x2 ,

.

(x1 , x2 ) ∈ R2 ,

t > 0,

u(x1 , x2 , 0) = cos(x1 − x2 ), ut (x1 , x2 , 0) = sin(x1 − x2 ),

(x1 , x2 ) ∈ R2 .

4. utt = ux1 x1 + ux2 x2 ,

.

(x1 , x2 ) ∈ R2 ,

t > 0,

u(x1 , x2 , 0) = (x12 + x22 )2 , ut (x1 , x2 , 0) = (x12 + x22 )2 ,

(x1 , x2 ) ∈ R2 .

5. utt = ux1 x1 + ux2 x2 + (x12 + x22 )et ,

.

u(x1 , x2 , 0) = 0, ut (x1 , x2 , 0) = 0,

(x1 , x2 ) ∈ R2 .

(x1 , x2 ) ∈ R2 ,

t > 0,

7.4 Advanced Practical Problems

345

6. utt = ux1 x1 + ux2 x2 ,

.

(x1 , x2 ) ∈ R2 ,

t > 0,

u(x1 , x2 , 0) = 2x1 − 3x2 , ut (x1 , x2 , 0) = 3x1 + x2 ,

(x1 , x2 ) ∈ R2 .

7. utt = ux1 x1 + ux2 x2 + t,

.

u(x1 , x2 , 0) = x12 , ut (x1 , x2 , 0) = x22 ,

(x1 , x2 ) ∈ R2 .

(x1 , x2 ) ∈ R2 ,

t > 0,

Chapter 8

Solutions, Hints and Answers to the Exercises

Chapter 1 Exercise 1.1. Answer. 1. 2. 3. 4. 5.

3. 5. 1. 3. 1.

Exercise 1.2. Answer. 1. 2. 3. 4. 5.

Nonlinear. Linear. Linear. Nonlinear. Linear.

Chapter 2 Exercise 2.1. Answer. 1. 2. 3. 4. 5.

Quasilinear. Quasilinear. Semilinear. Quasilinear. Semilinear.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. G. Georgiev, An Excursion Through Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-031-48784-2_8

347

348

8 Solutions, Hints and Answers to the Exercises

Exercise 2.2. Answer. 1. 2. 3. 4. 5.

Linear homogeneous. Linear nonhomogeneous. Linear nonhomogeneous. Linear homogeneous. Linear nonhomogeneous.

Exercise 2.3. Answer. 1. 2. 3. 4. 5.

Nonlinear. Nonlinear. Linear. Nonlinear. Linear.

Exercise 2.5. Answer. u(x1 , x2 ) = f (x2 )e− sin x1 + sin x1 − 1,

.

(x1 , x2 ) ∈ R2 ,

where .f ∈ C (R). Exercise 2.6. Solution. We have ux1 = uv vx1 + uw wx1

.

= −buv , ux2 = uv vx2 + uw wx2 = auv + uw . Then aux1 + bux2 = −abuv + abuv + buw

.

= buw . In this way, we get the equation uw =

.

1 u. b

We fix v and consider the last equation as a first order linear ODE. The function w

u(v, w) = f (v)e b ,

.

where .f ∈ C 1 (R), is its solution. Therefore x2

u(x1 , x2 ) = f (−bx1 + ax2 )e b ,

.

(x1 , x2 ) ∈ R2 ,

8 Solutions, Hints and Answers to the Exercises

349

is a solution to the given equation. Exercise 2.7. Answer. 1. x12 x2 .u(x1 , x2 ) = + x1 f 3



x12 x2

 , (x1 , x2 ) ∈ R2 ,

f ∈ C 1 (R).

2. u(x1 , x2 ) = f (x12 + x22 ),

.

(x1 , x2 ) ∈ R2 ,

f ∈ C 1 (R).

3. u(x1 , x2 ) = f (x1 x2 + x22 ),

.

(x1 , x2 ) ∈ R2 ,

f ∈ C 1 (R).

4.  u(x1 , x2 , x3 ) = f

.

x2 x3 , x1 x1

 ,

(x1 , x2 , x3 ) ∈ R3 ,

f ∈ C 1 (R2 ).

5.  u(x1 , x2 , x3 ) = f

.

x1 − x2 (x1 + x2 + 2x3 )2 , x3 x3

(x1 , x2 , x3 ) ∈ R3 ,

 ,

f ∈ C 1 (R2 ).

Exercise 2.8. Answer. (2x1 x2 + 1 − x1 − 3x2 )u(x1 , x2 ) = 1,

.

(x1 , x2 ) ∈ R2 .

Exercise 2.9. Solution. The existence and uniqueness theorem for ODEs applied for (2.7) with initial data (2.8), guarantees the existence of a unique characteristic curve for each point on the initial curve. The family of characteristic curves forms a parametric representation of a surface. The condition (2.10) implies that the parametric representation provides a smooth surface. The surface thus constructed indeed satisfies (2.5) and there are no further integral surfaces. Exercise 2.10. Solution. Let u(x1 , x2 , x3 ) = 0,

.

(x1 , x2 , x3 ) ∈ G,

be a two dimensional integral for the Eq. (2.11) in .G, which determines the surface Π . Then the vector

.

350

8 Solutions, Hints and Answers to the Exercises

 ∂u ∂u ∂u  , , ∂x1 ∂x2 ∂x3

.

is a normal vector to the surface .Π . Therefore there exists .μ = μ(x1 , x2 , x3 ) such that ∂u P (x1 , x2 , x3 ) = μ(x1 , x2 , x3 ) ∂x (x1 , x2 , x3 ), 1 ∂u . Q(x1 , x2 , x3 ) = μ(x1 , x2 , x3 ) ∂x2 (x1 , x2 , x3 ), ∂u R(x1 , x2 , x3 ) = μ(x1 , x2 , x. 3) ∂x (x1 , x2 , x3 ), 3

(x1 , x2 , x3 ) ∈ R3 ,

Also, ∂ 2u ∂μ ∂u ∂ 2u  ∂u  ∂μ ∂u +μ − −μ ∂x1 ∂x2 ∂x3 ∂x2 ∂x3 ∂x3 ∂x2 ∂x2 ∂x3 ∂ 2u ∂μ ∂u ∂ 2u  ∂u  ∂μ ∂u +μ − −μ −μ ∂x2 ∂x1 ∂x3 ∂x1 ∂x3 ∂x3 ∂x1 ∂x1 ∂x3 ∂ 2u ∂μ ∂u ∂ 2u  ∂u  ∂μ ∂u +μ − −μ +μ ∂x3 ∂x1 ∂x2 ∂x1 ∂x2 ∂x2 ∂x1 ∂x1 ∂x2 = 0.

curlF · F = μ .

Exercise 2.11. Answer. u(x1 , x2 , x3 ) = x1 x2 x3 − ξ1 ξ2 ξ3 = c

.

for any .(x1 , x2 , x3 ), (ξ1 , ξ2 , ξ3 ) ∈ R3 , where .c is a constant. Exercise 2.13. Answer. 1. no solutions. 2. .x13 x22 x3 = c, where c is a real constant. Exercise 2.14. Answer. No solutions.

Chapter 3 Exercise 3.1. Answer. 1. 2. 3. 4. 5.

Quasilinear. Quasilinear. Quasilinear. Quasilinear. Semilinear.

8 Solutions, Hints and Answers to the Exercises

Exercise 3.2. Answer. 1. 2. 3. 4. 5.

Linear homogeneous. Linear nonhomogeneous. Linear homogeneous. Linear homogeneous. Linear nonhomogeneous.

Exercise 3.3. Answer. 1. 2. 3. 4. 5.

Nonlinear. Nonlinear. Linear. Linear. Nonlinear.

Chapter 4 Exercise 4.1. Solution. We have    2 2 2 2 2 aφ .αγ − β = aφ1x + 2bφ1x1 φ1x2 + cφ1x + 2bφ φ + cφ 2x1 2x2 2x1 2x2 1 2 2   − aφ1x1 φ2x1 + b φ1x2 φ2x1 + φ1x1 φ2x2 + cφ1x2 φ2x2 2 2 2 = a 2 φ1x φ 2 + 2abφ1x φ φ + acφ1x φ2 1 2x1 1 2x1 2x2 1 2x2 2 2 +2abφ1x1 φ1x2 φ2x + 4b2 φ1x1 φ1x2 φ2x1 φ2x2 + 2bcφ1x1 φ1x2 φ2x 1 2 2 2 2 +acφ1x φ 2 + 2bcφ1x φ φ + c2 φ1x φ2 2 2x1 2 2x1 2x2 2 2x2   2 2 2 2 2 2 2 φ −a 2 φ1x φ − b φ + 2φ φ φ φ + φ φ 1x 1x 2x 2x 1 2 1 2 1x2 2x1 1x1 2x2 1 2x1   2 2 2 φ 2 − 2ab φ1x1 φ1x2 φ2x + φ1x φ φ −c2 φ1x 2 2x2 1 1 2x1 2x2   2 2 φ φ + φ1x1 φ1x2 φ2x −2acφ1x1 φ1x2 φ2x1 φ2x2 − 2bc φ1x 2 2x1 2x2 2   2 2 2 2 φ + φ φ − 2φ φ φ φ = ac φ1x 1x 1x 2x 2x 1 2 1 2 2x 1x 2x 1 2 2 1   2 2 2 2 φ − 2φ φ φ φ + φ φ −b2 φ1x 1x 1x 2x 2x 1 2 1 2 1x1 2x2 2 2x1

2  2  = ac φ1x1 φ2x2 − φ1x2 φ2x1 − b2 φ1x1 φ2x2 − φ1x2 φ2x1 2  = (ac − b2 ) φ1x1 φ2x2 − φ1x2 φ2x1 , which completes the solution.

351

352

8 Solutions, Hints and Answers to the Exercises

Exercise 4.3. Answer. 1. 2. 3. 4. 5. 6. 7.

Elliptic. Elliptic. Hyperbolic. Hyperbolic. Elliptic. Elliptic. Parabolic.

Exercise 4.4. Answer. 3

ξ1 (x1 , x2 ) = x22 ,

.

3

ξ2 (x1 , x2 ) = x12 ,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

x2 > 0,

(x1 , x2 ) ∈ R2 ,

x1 > 0,

and uξ1 ξ1 + uξ2 ξ2 +

.

1 1 uξ1 + uξ = 0, 3ξ1 3ξ2 2

Exercise 4.5. Answer. ξ1 (x1 , x2 ) = x1 ,

.

ξ2 (x1 , x2 ) = x2 − 3x1 ,

(x1 , x2 ) ∈ R2 ,

uξ1 ξ1 + uξ1 − 2uξ2 = 0,

(x1 , x2 ) ∈ R2 .

and .

Exercise 4.6. Solution. 1. Here a(x1 , x2 ) = 4x22 ,

.

b(x1 , x2 ) = 2x2 , c(x1 , x2 ) = 1,

(x1 , x2 ) ∈ R2 .

Then (b(x1 , x2 ))2 − a(x1 , x2 )c(x1 , x2 ) = 4x22 − 4x22

.

= 0,

(x1 , x2 ) ∈ R2 .

Therefore the considered equation is parabolic. The n is 4x22 (dx2 )2 − 4x2 dx1 dx2 + (dx1 )2 = 0,

.

(x1 , x2 ) ∈ R2 ,

x2 > 0.

8 Solutions, Hints and Answers to the Exercises

353

whereupon (2x2 dx2 − dx1 )2 = 0,

(x1 , x2 ) ∈ R2 ,

.

and 2x2 dx2 − dx1 = 0,

(x1 , x2 ) ∈ R2 ,

.

and x22 − x1 = c,

.

(x1 , x2 ) ∈ R2 .

We set ξ1 (x1 , x2 ) = x22 − x1 ,

.

ξ2 (x1 , x2 ) = x2 ,

(x1 , x2 ) ∈ R2 .

Then ξ1x1 (x1 , x2 ) = −1,

.

ξ1x2 (x1 , x2 ) = 2x2 , ξ2x1 (x1 , x2 ) = 0, ξ2x2 (x1 , x2 ) = 1,

(x1 , x2 ) ∈ R2 ,

and ξ1x1 (x1 , x2 )ξ2x2 (x1 , x2 ) − ξ1x2 (x1 , x2 )ξ2x1 (x1 , x2 ) = −1

.

/= 0, and ux1 = uξ1 ξ1x1 + uξ2 ξ2x1

.

ux1 x1

= −uξ1 ,  = − uξ1 ξ1 ξ1x1 + uξ1 ξ2 ξ2x1

ux1 x2

= uξ1 ξ1 ,  = − uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 = −2x2 uξ1 ξ1 − uξ1 ξ2 ,

ux2 = uξ1 ξ1x2 + uξ2 ξ2x2 = 2x2 uξ1 + uξ2 ,

(x1 , x2 ) ∈ R2 ,

354

8 Solutions, Hints and Answers to the Exercises

 ux2 x2 = 2uξ1 + 2x2 uξ1 ξ1 ξ1x2 + uξ1 ξ2 ξ2x2 +uξ1 ξ2 ξ1x2 + uξ2 ξ2 ξ2x2  = 2uξ1 + 2x2 2x2 uξ1 ξ1 + uξ1 ξ2 + 2x2 uξ1 ξ2 + uξ2 ξ2 = 2uξ1 + 4x22 uξ1 ξ1 + 4x2 uξ1 ξ2 + uξ2 ξ2 ,

(x1 , x2 ) ∈ R2 .

Hence, 0 = 4x22 ux1 x1 + 4x2 ux1 x2 + ux2 x2 + 2ux1  = 4x22 uξ1 ξ1 + 4x2 −2x2 uξ1 ξ1 − uξ1 ξ2 + 2uξ1 + 4x22 uξ1 ξ1

.

+4x2 uξ1 ξ2 + uξ2 ξ2 − 2uξ1 ,

(x1 , x2 ) ∈ R2 ,

whereupon uξ2 ξ2 = 0,

.

(x1 , x2 ) ∈ R2 ,

is the canonical form of the considered equation. 2. Let v = uξ2 .

.

Then vξ2 = 0,

.

(x1 , x2 ) ∈ R2 ,

and v = f (ξ1 ),

.

(x1 , x2 ) ∈ R2 ,

and uξ2 = f (ξ1 ),

.

(x1 , x2 ) ∈ R2 ,

and u(x1 , x2 ) = ξ2 f (ξ1 ) + g(ξ1 ),

.

(x1 , x2 ) ∈ R2 ,

and u(x1 , x2 ) = x2 f (x22 − x1 ) + g(x22 − x1 ),

.

(x1 , x2 ) ∈ R2 ,

is the general solution of the considered equation, where .f and .g are C 2 −functions.

.

8 Solutions, Hints and Answers to the Exercises

355

3. We have u(x1 , 0) = g(−x1 ),

.

ux2 (x1 , x2 ) = f (x22 − x1 ) + 2x22 f ' (x22 − x1 ) + 2x2 g ' (x22 − x1 ), ux2 (x1 , 0) = f (−x1 ). In this way, we obtain the system .

f (−x1 ) = ex1 g(−x1 ) = 9 sin x1 ,

x1 ∈ R,

.

f (x1 ) = e−x1 g(x1 ) = −9 sin x1 ,

x1 ∈ R.

whereupon

Consequently   2 u(x1 , x2 ) = x2 ex1 −x2 − 9 sin x22 − x1 ,

.

(x1 , x2 ) ∈ R2 .

Exercise 4.7. Answer. 1. uξ1 ξ2 = 0,

.

ξ1 (x1 , x2 ) = x2 + 3x1 , ξ2 (x1 , x2 ) = x2 + 2x1 ,

(x1 , x2 ) ∈ R2 .

2. uξ1 ξ2 = 0,

.

ξ1 (x1 , x2 ) = x2 − 4x1 , ξ2 (x1 , x2 ) = x2 − x1 ,

(x1 , x2 ) ∈ R2 .

3. 1 uξ = 0, 2ξ1 2 x2 ξ1 (x1 , x2 ) = , x1

uξ1 ξ2 +

.

ξ2 (x1 , x2 ) = x1 x2 ,

(x1 , x2 ) ∈ R2 ,

x1 /= 0,

x2 /= 0.

356

8 Solutions, Hints and Answers to the Exercises

Exercise 4.8. Answer. 1. uξ1 ξ2 = 0,

.

ξ1 (x1 , x2 ) = x2 − 3x1 , ξ2 (x1 , x2 ) = x2 + x1 ,

(x1 , x2 ) ∈ R2 .

2. u(x1 , x2 ) = f (x2 − 3x1 ) + g(x2 + x1 ),

.

(x1 , x2 ) ∈ R2 ,

where .f and .g are .C 2 -functions. 3. u(x1 , x2 ) = 3x12 + x22 ,

.

(x1 , x2 ) ∈ R2 .

Exercise 4.9. Answer. 1. .

− ξ1 uξ1 ξ2 + uξ2 = 0, ξ1 (x1 , x2 ) = x2 − x1 , x2 ξ2 (x1 , x2 ) = , (x1 , x2 ) ∈ R2 x1 > 0, x2 > 0. x1

2.  u(x1 , x2 ) = (x2 − x1 )f

.

x2 x1

 + g(x2 − x1 ),

(x1 , x2 ) ∈ R2 x1 > 0, x2 > 0,

where .f and .g are .C 2 −functions. 3. u(x1 , x2 ) =

.

x12 , x2

(x1 , x2 ) ∈ R2 ,

Exercise 4.10. Answer. Hyperbolic. Exercise 4.11. Answer. 1. uξ1 ξ1 + uξ2 ξ2 + uξ3 ξ3 = 0,

.

x1 > 0, x2 > 0.

8 Solutions, Hints and Answers to the Exercises

357

ξ1 (x1 , x2 , x3 ) = x1 , ξ2 (x1 , x2 , x3 ) = x2 − x1 , 1 1 ξ3 (x1 , x2 , x3 ) = x1 − x2 + x3 , 2 2

(x1 , x2 , x3 ) ∈ R3 .

2. uξ1 ξ1 + uξ2 ξ2 + uξ3 ξ3 + uξ4 ξ4 = 0,

.

ξ1 (x1 , x2 , x3 , x4 ) = x1 , ξ2 (x1 , x2 , x3 , x4 ) = x2 − x1 , ξ3 (x1 , x2 , x3 , x4 ) = x1 − x2 + x3 , ξ4 (x1 , x2 , x3 , x4 ) = 2x1 − 2x2 + x3 + x4 ,

(x1 , x2 , x3 , x4 ) ∈ R4 .

3. uξ1 ξ1 − uξ2 ξ2 + uξ3 ξ3 + uξ4 ξ4 = 0,

.

ξ1 (x1 , x2 , x3 , x4 ) = x1 + x2 , ξ2 (x1 , x2 , x3 , x4 ) = x2 − x1 , ξ3 (x1 , x2 , x3 , x4 ) = x3 , ξ4 (x1 , x2 , x3 , x4 ) = x2 + x3 + x4 ,

(x1 , x2 , x3 , x4 ) ∈ R4 .

4. n

.

uξk ξk = 0,

k=1

ξk (x1 , . . . , xn ) =

k

xl ,

k ∈ {1, . . . , n},

(x1 , . . . , xn ) ∈ Rn .

l=1

5. .

n

(−1)k+1 uξk ξk = 0, k=1

ξk =

k

l=1

xl ,

k ∈ {1, . . . , n},

(x1 , . . . , xn ) ∈ Rn .

358

8 Solutions, Hints and Answers to the Exercises

Chapter 5 Exercise 5.1. Answer. 1. 2. 3. 4. 5.

Harmonic. Harmonic. Not harmonic. Harmonic. Harmonic.

Exercise 5.2. Hint. Use the definition of the function .Φ. Exercise 5.4. Solution. Let .x ∈ D be arbitrarily chosen. We take .ϵ > 0 so that .B(x, ϵ) ⊂ D. Denote by .Dϵ = D\B(x, ϵ). We have .∂Dϵ = ∂D ∪ ∂B(x, ϵ). Then, using Example 5.5, we have 

 Φ(x − y)∂νy u(y) − u(y)∂νy Φ(x − y) dsy

0=

.

∂Dϵ



 Φ(x − y)∂νy u(y) − u(y)∂νy Φ(x − y) dsy

= ∂D



 Φ(x − y)∂νy u(y) − u(y)∂νy Φ(x − y) dsy .

− ∂B(x,ϵ)

Hence,   Φ(x − y)∂νy u(y) − u(y)∂νy Φ(x − y) dsy ∂D



= .

 Φ(x − y)∂νy u(y) − u(y)∂νy Φ(x − y) dsy

∂B(x,ϵ) 

=

 Φ(x − y)∂νy u(y)dsy −

∂B(x,ϵ) 

−

(u(y) − u(x)) ∂νy Φ(x − y)dsy

∂B(x,ϵ)

u(x)∂νy Φ(x − y)dsy .

∂B(x,ϵ)

Note that Φ(x − y) =

.

∂νy Φ(x − y)

∂B(x,ϵ)

=

1 − 2π log |x − y|

1 n(n−2)κ(n)|x−y|n−2

n=2 n ≥ 3,

− 2π1 ϵ n = 2 1 − nκ(n)ϵ n ≥ 3. n−1

(8.1)

8 Solutions, Hints and Answers to the Exercises

359

Therefore  (u(y) − u(x)) ∂νy Φ(x − y)dsy ∂B(x,ϵ)⎧

= .

=



(u(y) − u(x)) dsy n = 2 ∂B(x,ϵ)  1 ⎪ (u(y) − u(x)) dsy n ≥ 3 ⎩ − nκ(n)ϵ n−1 ⎧ 1  ∂B(x,ϵ) (u(x + ϵz) − u(x)) dsz n = 2 ⎪ ⎨ − 2π ∂B(0,1)  1 ⎪ − nκ(n) (u(x + ϵz) − u(x)) dsz n ≥ 3 ⎩ ∂B(0,1) 1 ⎪ ⎨ − 2π ϵ

−→ 0,

as

ϵ → 0,

 u(x)∂νy Φ(x − y)dsy =

⎧ 1 ⎪ ⎨ − 2π ϵ u(x)

 ∂B(x,ϵ)

dsy 

n=2

1 ⎪ dsy ⎩ − nκ(n)ϵ n−1 u(x) ∂B(x,ϵ)  −u(x) n = 2 = −u(x) n ≥ 3,

. ∂B(x,ϵ)

(8.2)

n≥3

⎧ 



⎪ C| log ϵ| dsy n = 2

⎨

∂B(x,ϵ)  Φ(x − y)∂νy u(y)dsy ≤

C ⎪ ϵ n−2

⎩ dsy n ≥ 3

∂B(x,ϵ)

∂B(x,ϵ) .  C1 ϵ| log ϵ| n = 2 ≤ C1 ϵ n ≥ 3 −→ 0, as ϵ → 0.

(8.3)

(8.4)

(Here .C1 and .C are constants independent of .ϵ.) From (8.1), (8.2), (8.3) and (8.4) we obtain the desired result (5.13). Exercise 5.5. Solution. Note that

1 − 2π log r n = 2

.Φ(x − y) = 1 n ≥ 3, ∂B(x,r) n(n−2)κ(n)r n−2

− 2π1 r n = 2

∂νy Φ(x − y) = 1 − nκ(n)r n ≥ 3. ∂B(x,r) n−1 Hence and (5.13), we get ⎧ 1 ⎪ ⎨ − 2π log r

 ∂νy u(y)dsy + 2π1 r u(y)dsy n = 2 ∂B(x,r) ∂B(x,r)   .u(x) = 1 1 ⎪ ∂νy u(y)dsy + nκ(n)r u(y)dsy n−1 ⎩ n(n−2)κ(n)r n−2 

∂B(x,r)

∂B(x,r)

n ≥ 3.

360

8 Solutions, Hints and Answers to the Exercises

Hence and (5.6), we obtain (5.14). Exercise 5.6. Solution. Suppose that .Δu /≡ 0 in .D. Then there exist .x ∈ D and a ball .B(x, r) ⊂ D such that, without loss of generality, .Δu > 0 within .B(x, r). By (5.14) we have  1 u(y)dsy .u(x) = nκ(n)r n−1 =



1 nκ(n)

∂B(x,r)

u(x + rz)dsz , ∂B(0,1)

whereupon 0=

.



1 nκ(n)

∇u(x + rz) · zdsz ∂B(0,1)

1 = nκ(n)r n−1

 ∇u(y) ·

y−x dsy r

∂B(x,r)



1 = nκ(n)r n−1

Δu(y)dy B(x,r)

> 0, which is a contradiction. Exercise 5.7. Hint. Use Example 5.12. Exercise 5.8. Hint. Use Example 5.12. Exercise 5.9. Hint. Use the Poisson formula. Exercise 5.10. Solution. We have 1 .u(x) − φ(x0 ) = nκ(n)

 |y|=1

1 − |x|2 (φ(y) − φ(x0 )) dsy . |y − x|n

Let .ϵ > 0 be arbitrarily chosen. Then there exists .δ = δ(ϵ) > 0 so that |φ(y) − φ(x0 )| < ϵ

.

f or

y ∈ B(x0 , δ) ∩ ∂B(0, 1).

Set S = ∂B(0, 1)\ (B(x0 , δ) ∩ ∂B(0, 1)) .

.

We choose .δ > 0 small enough so that

8 Solutions, Hints and Answers to the Exercises

1 − |x|2 ≤

.

ϵnκ(n) 4MQ

361

x ∈ B(0, 1),

for

|x − x0 | ≤

δ , 2

where  Q=

.

S

1 dsy , |y − x0 |n

M = sup |φ|. ∂B(0,1)

Then for .|x − x0 | ≤

δ , .x ∈ B(0, 1), we have 2

|u(x) − φ(x0 )| ≤



1 nκ(n)

∂B(0,1)



1 = nκ(n)

.

+

1 − |x|2 |φ(y) − φ(x0 )|dsy |y − x|n

B(x0 ,δ)∩∂B(0,1)  1 − |x|2

1 nκ(n)

= I1 + I2 .

S

|y − x|n

1 − |x|2 |φ(y) − φ(x0 )|dsy |y − x|n

(8.5)

|φ(y) − φ(x0 )|dsy

Note that 

1 I1 ≤ ϵ nκ(n)

B(x0 ,δ)∩∂B(0,1) .

1 ≤ϵ nκ(n)



∂B(0,1)

=ϵ For .|x − x0 | ≤

for

1 − |x|2 dsy |x − y|n

1 − |x|2 dsy |x − y|n

|x − x0 | ≤

δ , 2

x ∈ B(0, 1).

δ , .x ∈ B(0, 1), and .|y − x0 | ≥ δ, .y ∈ ∂B(0, 1), we have 2 |y − x0 | ≤ |y − x| + |x − x0 |

.

δ 2 1 ≤ |y − x| + |y − x0 |, 2 ≤ |y − x| +

i.e., if .|x − x0 | ≤

δ , .x ∈ B(0, 1), and .|y − x0 | ≥ δ, .y ∈ ∂B(0, 1), we have 2

(8.6)

362

8 Solutions, Hints and Answers to the Exercises

|y − x| ≥

.

Hence, for .|x − x0 | ≤

1 |y − x0 |. 2

δ , .x ∈ B(0, 1), we have 2 2M .I2 ≤ nκ(n)

 S

≤

4M nκ(n)

 S

1 − |x|2 dsy |x − y|n 1 − |x|2 dsy |y − x0 |n

≤ ϵ. From the last estimate and (8.5), (8.6), we get |u(x) − φ(x0 )| ≤ 2ϵ

.

δ for .|x − x0 | ≤ , .x ∈ B(0, 1). 2 Exercise 5.11. Hint. Use the definition for the Green function. Exercise 5.12. Solution. Suppose that .u ≥ 0 in .Rn . Let .R > 0 be arbitrarily chosen. Then by (5.27) we have u(x) =

.

1 nκ(n)R

 |y|=R

R 2 − |x|2 φ(y)dsy , |y − x|n

|x| < R,

(8.7)

for .φ ∈ C (∂B(0, R)), and .u(x) → φ(x0 ) as .x → x0 , .x ∈ B(0, R), .x0 ∈ ∂B(0, R). Hence, u(0) =

.

=

1 nκ(n)R

 |y|=R

1 nκ(n)R n−1

R2 φ(y)dsy |y|n  φ(y)dsy .

|y|=R

Since R − |x| ≤ |y| − |x| ≤ |y − x| ≤ |y| + |x| ≤ R + |x|

.

for .|y| = R, .|x| < R, we get

8 Solutions, Hints and Answers to the Exercises

u(x) ≥

.

=

=

1 nκ(n)R

 |y|=R

363

R 2 − |x|2 φ(y)dsy (R + |x|)n

1 R n−2 (R 2 − |x|2 ) n (R + |x|) nκ(n)R n−1

 φ(y)dsy |y|=R

R n−2 (R 2 − |x|2 ) u(0), (R + |x|)n

and 1 .u(x) ≤ nκ(n)R =

=

 |y|=R

R 2 − |x|2 φ(y)dsy (R − |x|)n

− |x|2 ) 1 n (R − |x|) nκ(n)R n−1



R n−2 (R 2

R n−2 (R 2

− |x|2 )

(R − |x|)n

u(0),

φ(y)dsy |y|=R

|x| < R.

Therefore .

R n−2 (R 2 − |x|2 ) R n−2 (R 2 − |x|2 ) u(0) ≤ u(x) ≤ u(0), n (R + |x|) (R − |x|)n

|x| < R.

Making .R tend to .∞ we get .u(x) = u(0) for any .x ∈ Rn . Exercise 5.13. Solution. Let .u ≤ M in .Rn . Set v(x) = M − u(x),

x ∈ Rn .

.

Then .v is harmonic throughout .Rn and it is nonnegative in .Rn . Hence and Exercise 5.12, we conclude that M − u(x) = M − u(0),

.

x ∈ Rn ,

whereupon u(x) = u(0),

.

x ∈ Rn .

Exercise 5.14. Hint. Use Exercise 5.13. Exercise 5.15. Solution. Let .ϵ > 0 be arbitrarily chosen. Since

∞

.

m=1

uniformly convergent on .∂D, there exists an index .N = N(ϵ) such that

um (x) is

364

8 Solutions, Hints and Answers to the Exercises

p



. uN +i (y) < ϵ

i=1

holds for all .p ≥ 1 and .y ∈ ∂D. Note that

p

.

uN +i (x) is harmonic in .D and

i=1

continuous in .D. Hence and the maximum principle we conclude that

p



. uN +i (x) < ϵ

f or

all

x ∈ D.

i=1

Therefore .

∞

um (x) is uniformly convergent in .D. Let now, .x0 ∈ D be an arbitrary

m=1

point. We take .R > 0 so that .B(x0 , R) ⊂ D. Then 

1 .um (x) = nκ(n)R

∂B(x0 ,R)

Hence, using that .

∞

R 2 − |x − x0 |2 um (y)dsy , |y − x|n

m ∈ N,

|x − x0 | < R.

um (x) is uniformly convergent in .D, we get

m=1

u(x) =

∞

.

um (x)

m=1

=

1 nκ(n)R

 ∂B(x0 ,R)

=

1 nκ(n)R



∂B(x0 ,R)

∞ R 2 − |x − x0 |2

um (y)dsy |y − x|n m=1

R 2 − |x − x0 |2 u(y)dsy , |y − x|n

|x − x0 | < R.

Therefore .u is harmonic in .|x − x0 | < R. Because .x0 ∈ D was arbitrarily chosen, we conclude that .u is harmonic everywhere in .D. Exercise 5.16. Answer. u(x1 , x2 ) =

.

sinh(π(1 − x2 )) sin(π x1 ) sinh π ∞ 8 sinh((2n + 1)π(x1 − 1)) sin((2n + 1)π x2 ) , + 3 sinh((2n + 1)π ) π (2n + 1)3 n=0

(x1 , x2 ) ∈ [0, 1] × [0, 1].

8 Solutions, Hints and Answers to the Exercises

365

Exercise 5.17. Answer. u(x1 , x2 ) = A0 +

∞

.

An cosh

 nπ x  1

a

n=1

cos

 nπ x  2

b

,

(x1 , x2 ) ∈ [0, a] × [0, b],

where .A0 is a constant and 2  .An = nπ sinh nπb a

b f (y) cos

 nπy  b

n ∈ N.

dy,

0

Exercise 5.18. Answer. u(x1 , x2 ) = b0 −

.

∞ 4 cosh((2n − 1)π x1 ) cos((2n − 1)π x2 ) , π3 (2n − 1)3 sinh((2n − 1)π ) n=1

(x1 , x2 ) ∈ [0, 1] × [0, 1]. Exercise 5.19.

.

u(x1 , x2 ) = A0 x2 +

∞

.

An sinh

n=1

 nπ x  2

a

cos

 nπ x  1

a

,

(x1 , x2 ) ∈ [0, a] × [0, b], where

.

A0 =

.

1 ab

a f (x1 )dx1 , 0

An =

2  a sinh nπa b

a f (x1 ) cos

 nπ x  1

a

dx1 ,

n ∈ N.

0

Exercise 5.20. Answer. u(r, θ ) =

.

r r (3 sin θ − r 2 sin(3θ )) + (3 cos θ − r 2 cos(3θ )), 4 4

r ∈ [0, 1], θ ∈ [0, 2π ]. Exercise 5.21. Answer. α0  a n + (αn cos(nθ ) + βn sin(nθ )) , 2 r ∞

u(r, θ ) =

.

n=1

r ∈ [0, a], .θ ∈ [0, 2π ], where

.

366

8 Solutions, Hints and Answers to the Exercises

1 .α0 = π

2π h(θ )dθ, 0

1 αn = π

2π h(θ ) cos(nθ )dθ, 0

1 βn = π

2π h(θ ) sin(nθ )dθ,

n ∈ N,

0

h(θ ) = φ(a cos θ, a sin θ ),

θ ∈ [0, 2π ].

Exercise 5.22. Answer. u(r, θ ) = C0 +D0 log r +

.

    ∞ 

Dn Cn An r n + n cos(nθ )+ Bn r n + n sin(nθ ) , r r n=1

r ∈ [0, a], .θ ∈ [0, 2π ], where .C0 , D0 , Cn , Dn , An , Bn , .n ∈ N, are constants for which

.

2π

1 C0 + D0 log a = 2π C0 + D0 log b =

An a n +

Cn 1 = n a π

h(θ )dθ 0 2π

1 2π

g(θ )dθ 0

2π

cos(nθ )h(θ )dθ 0

.

An bn +

Bn a n +

Cn 1 = bn π 1 Dn = n a π

2π cos(nθ )g(θ )dθ 0 2π

sin(nθ )h(θ )dθ 0

Bn b n +

Dn 1 = bn π

2π sin(nθ )g(θ )dθ,

n ∈ N,

0

h(θ ) = φ1 (a cos θ, a sin θ ), g(θ ) = φ2 (b cos θ, b sin θ ),

θ ∈ [0, 2π ].

8 Solutions, Hints and Answers to the Exercises

367

Exercise 5.23. Answer. u(r, θ ) =

∞

.

αn r

nπ γ

nπ θ , γ

sin

n=1

αn =

2a

− nπ γ

γ φ(θ ) sin

γ

nπ θ dθ, γ

r ∈ [0, a],

θ ∈ [0, γ ].

0

Exercise 5.24. Hint. Seek a formal solution in the form ∞

u(r, θ ) =

.

f0 (r)

+ (fn (r) cos(nθ ) + gn (r) sin(nθ )) , r ∈ [0, a], θ ∈ [0, 2π ]. 2 n=1

Substituting it into the Poisson equation, find 1 n2 fn'' + fn' − 2 fn = φn (r), r r . n2 1 ' '' gn + gn − 2 gn = ψn (r), r r

n ∈ N0 , n ∈ N,

r ∈ [0, a].

The general solution of these equations can be written as follows fn (r) = An r n + f˜n (r),

gn (r) = Bn r n + g˜ n (r),

.

r ∈ [0, a],

where, in the case when .f˜n (a) = g˜ n (a) = 0, and .f˜n and .g˜ n are bounded at the origin, .f˜0 (r) =

r

r log φ0 (ρ)ρdρ + a

log

ρ φ0 (ρ)ρdρ, a

r

0

1 f˜n (r) = 2n

a

r  n  n    r ρ n a − φn (ρ)ρdρ a r a 0

1 + 2n

 n    a   ρ n r n a − φn (ρ)ρdρ, a ρ a r

g˜ n (r) =

1 2n

r

 r n a

−

 a n   ρ n r

a

ψn (ρ)ρdρ

0

1 + 2n

 n    a   ρ n r n a − ψn (ρ)ρdρ, a ρ a r

n ∈ N,

r ∈ [0, a].

368

8 Solutions, Hints and Answers to the Exercises

Answer. u(r, θ ) =

.

∞    A0 + f˜0 (r)  + An r n + f˜n (r) cos(nθ )+ Bn r n + g˜ n (r) sin(nθ ) , 2 n=1

r ∈ [0, a],

θ ∈ [0, 2π ], where

.

αn , an βn Bn = n , a

n ∈ N0 ,

An =

.

n ∈ N.

Chapter 6 Exercise 6.1. Answer. 1. 2x−x 2 +t 1 u(x, t) = √ e 1+t , 1+t

.

x ∈ R,

t ≥ 0.

2. 1



x1 x2 .u(x1 , x2 , t) = √ cos 2 1 + t2 1+t

 e

−

t (x12 +x22 ) 2(1+t 2 )

,

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

3.   t (x 2 +x 2 ) sin x3 x1 x2 −t− 1 22 1+4t e .u(x1 , x2 , x3 , t) = √ , cos 1 + 4t 2 1 + 4t 2 (x1 , x2 , x3 ) ∈ R3 , .t ≥ 0.

.

4. ⎛ ⎞ n

−nt .u(x1 , . . . , xn , t) = e cos ⎝ xj ⎠ ,

(x1 , . . . , xn ) ∈ Rn ,

t ≥ 0.

j =1

5. u(x1 , x2 , x3 , t) = 2x1 − x2 + 3x3 ,

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

8 Solutions, Hints and Answers to the Exercises

369

Exercise 6.2. Answer. 1. u(x, t) = t 2 x 2 ,

x ∈ R,

.

t ≥ 0.

2. u(x, t) = sin t + tx 3 ,

x ∈ R,

.

t ≥ 0.

3. u(x1 , x2 , t) = t (x1 + x22 ),

.

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

4. ⎛ u(x1 , . . . , xn , t) = sin t ⎝

n

⎞ xj2 ⎠ ,

(x1 , . . . , xn ) ∈ Rn ,

t ≥ 0.

u(x1 , x2 , x3 , t) = t 2 (x1 − x2 + x3 ),

(x1 , x2 , x3 ) ∈ |R 3 ,

t ≥ 0.

.

j =1

5. .

Exercise 6.3. Answer. 1. .

1 u(x, t) = 1 + et + t 2 , 2

x ∈ R,

t ≥ 0.

u(x, t) = t 3 + e−t sin x,

x ∈ R,

t ≥ 0.

2. .

3. u(x1 , x2 , t) = et − 1 + e−2t cos x1 sin x2 ,

.

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

4. u(x1 , x2 , x3 , t) =

.

(x1 , x2 , x3 ) ∈ R3 , .t ≥ 0.

.

  1 cos x1 e−2t − 1 + 2t + cos x2 cos x3 e−4t , 4

370

8 Solutions, Hints and Answers to the Exercises

5. u(x1 , x2 , t) = t 2 + tx1 + t 3 x22 + x2 ,

.

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

Exercise 6.4. Answer. u(x, t) = sin xe−t ,

.

0 ≤ x ≤ π , .t ≥ 0. Exercise 6.5. Answer.

.

.u(x, t)

∞

=

e−n

2t



n=1,n/=2,4

+

 4  n (−1) − 1 sin(nx) n3 π

  1  16t 1  4t e (4t − 1) + 1 e−4t sin(2x) + e (16t − 1) + 1 e−16t sin(4x), 32 512 x ∈ [0, π ], t ≥ 0.

Exercise 6.6. Hint. Use the function v(x, t) = u(x, t) +

.

x x−L g(t) − h(t), L L

x ∈ [0, L],

t ≥ 0.

Then, reduce the given problem to the problem vt = kvxx + f (x, t) +

.

x x−L ' g (t) − h' (t), L L

0 < x < L,

t > 0,

v(0, t) = 0, v(L, t) = 0,

t ≥ 0,

v(x, 0) = φ(x) +

x−L x g(0) − h(0), L L

x ∈ [0, L].

Answer. u(x, t) =

∞

.

e

2π2 t L2

−k n

n=1

(An − Gn (0) + Gn (t)) sin

 nπ x  L

,

x ∈ [0, L], .t ≥ 0, where

.

x x−L ' g (t) − h' (t), x ∈ [0, L], L L x x−L g(0) − h(0), x ∈ [0, L], Φ(x) = φ(x) + L L

F (x, t) = f (x, t) +

.

t ≥ 0,

8 Solutions, Hints and Answers to the Exercises

2 Fn (t) = L

L sin

 nπ x  L

371

F (x, t)dx,

0

2 An = L

L sin

L

Φ(x)dx,

0

 Gn (t) =

 nπ x 

Fn (t)e

2π2 t L2

kn

t ≥ 0.

dt,

Exercise 6.7. Answer. ∞

u(x, t) =

.

Bn cos

 nπ x  L

n=0

x ∈ [0, L],

nπ 2

e−k ( L ) t ,

t ≥ 0,

where 2 .Bn = L

L cos

 nπ x  L

φ(x)dx,

n ∈ N0 .

0

Exercise 6.8. Answer. u(x, t) =

∞

.

e

−k n

2π2 t L2

(An − gn (0) + gn (t)) cos

 nπ x 

n=0

L

x ∈ [0, L, t ≥ 0, where 2 .An = L  gn (t) = 2 fn (t) = L

L φ(x) cos

 nπ x  L

dx,

t ≥ 0,

0

fn (t)e

2π2 t L2

kn

dt,

L f (x, t) cos 0

 nπ x  L

dx,

t ≥ 0.

,

372

8 Solutions, Hints and Answers to the Exercises

Exercise 6.9. Hint. Use the function x2 (x − L)2 g(t) − h(t), 2L 2L

v(x, t) = u(x, t) +

.

x ∈ [0, L],

t ≥ 0,

to reduce the given problem to the problem vt = kvxx +

.

x2 ' k (x − L)2 ' g (t) − h (t) − g ' (t) 2L 2L L

k + h(t), L

0 < x < L,

t > 0,

vx (0, t) = 0, vx (L, t) = 0,

t ≥ 0,

v(x, 0) = φ(x) +

x2 (x − L)2 g(0) − h(0), 2L 2L

x ∈ [0, L].

Answer. u(x, t) =

∞

.

e

2π2 t L2

−k n

(An − Gn (0) + Gn (t)) cos

n=0

x ∈ [0, L,

 nπ x  L

,

t ≥ 0,

where 2 .An = L  Gn (t) = 2 Fn (t) = L

L Φ(x) cos

 nπ x  L

dx,

t ≥ 0,

0

Fn (t)e

2π2 t L2

kn

dt,

L F (x, t) cos

 nπ x  L

dx,

t ≥ 0,

0

F (x, t) = f (x, t) + Φ(x) = φ(x) + Exercise 6.10. Answer.

x2 ' k (x − L)2 ' k g (t) − h (t) − g(t) + h(t), 2L 2L L L

x2 (x − L)2 g(0) − h(0), 2L 2L

x ∈ [0, L],

t ≥ 0.

8 Solutions, Hints and Answers to the Exercises

u(x, t) =



∞

Bn sin

.

n=0

373

(2n + 1)π x 2L

x ∈ [0, L],

 e

−k



 (2n+1)π 2 t 2L

,

t ≥ 0,

where 2 .Bn = L



L sin

(2n + 1)π x 2L

 φ(x)dx,

n ∈ N0 .

0

Exercise 6.11. Answer. u(x, t) =

∞

.

e

−k n(2n+1)2

2π2

4L

t

 (An − gn (0) + gn (t)) sin

n=0

x ∈ [0, L],

(2n + 1)π x 2L

 ,

t ≥ 0,

where 2 .An = L  gn (t) = 2 fn (t) = L

L



(2n + 1)π x φ(x) sin 2L

 t ≥ 0,

dx,

0 2π2

fn (t)e

k (2n+1)2 4L



L f (x, t) sin

t

dt,

(2n + 1)π x 2L

 dx,

t ≥ 0.

0

Exercise 6.12. Hint. Use the function v(x, t) = u(x, t) −

.

(x − L)2 g(t) − xh(t), L2

x ∈ [0, L],

t ≥ 0,

to reduce the given problem to the problem vt = kvxx +

.

(x − L)2 ' 2k g(t) − g (t) − xh' (t), L2 L2

v(0, t) = 0, vx (L, t) = 0,

t ≥ 0,

v(x, 0) = φ(x) −

(x − L)2 g(0) − xh(0), L2

x ∈ [0, L].

0 < x < L,

t > 0,

374

8 Solutions, Hints and Answers to the Exercises

Answer. ∞

u(x, t) =

.

2π2

e

−k (2n+1)2 4L



t

(An − Gn (0) + Gn (t)) sin

n=0

x ∈ [0, L],

(2n + 1)π x 2L

 ,

t ≥ 0,

where 2 .An = L  Gn (t) =



L Φ(x) sin

 t ≥ 0,

dx,

0 2π2

Fn (t)e

2 Fn (t) = L

(2n + 1)π x 2L

k (2n+1)2 4L



L F (x, t) sin

t

dt,

(2n + 1)π x 2L

 t ≥ 0,

dx,

0

2k (x − L)2 ' g(t) − g (t) − xh' (t), 2 L L2

F (x, t) = f (x, t) +

(x − L)2 g(0) − xh(0), L2

Φ(x) = φ(x) −

x ∈ [0, L],

t ≥ 0.

Exercise 6.13. Answer. u(x, t) =

∞

.

n=0



(2n + 1)π x Bn cos 2L

x ∈ [0, L],

 e

−k



 (2n+1)π 2 t 2L

,

t ≥ 0,

where 2 .Bn = L

L



(2n + 1)π x cos 2L

 φ(x)dx,

n ∈ N0 .

0

Exercise 6.14. Answer. u(x, t) =

∞

.

2π2

e

−k (2n+1)2 4L

n=0

x ∈ [0, L], where

t



(2n + 1)π x (An − gn (0) + gn (t)) cos 2L

t ≥ 0,

 ,

8 Solutions, Hints and Answers to the Exercises

2 .An = L  gn (t) =

L



(2n + 1)π x φ(x) cos 2L

 t ≥ 0,

dx,

0 2π2

fn (t)e

2 fn (t) = L

375

L

k (2n+1)2 4L

t

dt,



(2n + 1)π x f (x, t) cos 2L

 t ≥ 0.

dx,

0

Exercise 6.15. Hint. Use the function v(x, t) = u(x, t) −

.

(x − L)2 g(t) − xh(t), L2

x ∈ [0, L],

t ≥ 0,

to reduce the given problem to the problem (x − L)2 ' 2k g (t) − xh' (t) + g(t), L L2

vt = kvxx −

.

0 < x < L,

t > 0,

vx (0, t) = 0, v(L, t) = 0,

t ≥ 0, (x − L)2 g(0) − xh(0), L2

v(x, 0) = φ(x) −

x ∈ [0, L].

Answer. u(x, t) =

∞

.

2π2

e

−k (2n+1)2 4L

t

 (An − Gn (0) + Gn (t)) cos

n=0

x ∈ [0, L],

(2n + 1)π x 2L

t ≥ 0,

where 2 .An = L  Gn (t) = Fn (t) =



L Φ(x) cos

 dx,

t ≥ 0,

0 2π2

Fn (t)e 2 L

(2n + 1)π x 2L

k (2n+1)2 4L



L F (x, t) cos 0

t

dt, (2n + 1)π x 2L

 dx,

t ≥ 0,

 ,

376

8 Solutions, Hints and Answers to the Exercises

F (x, t) = f (x, t) − Φ(x) = φ(x) −

2k (x − L)2 ' g (t) − xh' (t) + 2 g(t), 2 L L

(x − L)2 g(0) − xh(0), L2

x ∈ [0, L],

t ≥ 0.

Exercise 6.16. Answer. .4. Exercise 6.17. Hint. Use the solution of Example 6.9. Exercise 6.18. Hint. Use the function v(x, t) = u(x, t) + (K + 1)t,

.

where .K =

(x, t) ∈ QT ,

|f (x, t)| and the solution of Example 6.10.

sup (x,t)∈QT

Exercise 6.19. Hint. Use Example 6.10 and Exercise 6.18. Exercise 6.20. Hint. Use Exercise 6.19. Exercise 6.21. Hint. Use Example 6.11. Exercise 6.22. Solution. Let .ϵ > 0 be arbitrarily chosen and w± (x, t) = (N + ϵ)t + m ± u(x, t).

.

Note that .w± (x, t) ≥ 0 on .∂p QT . Also, .

(w± )t − kΔ (w± ) = N + ϵ ± ut ∓ kΔu = N + ϵ ± (ut − kΔu) = N +ϵ±f >0

on

QT .

Hence and Exercise 6.17, we conclude that .w± achieves its maximum on .QT . Because .w± ≥ 0 on .∂p QT , using Exercise 6.21, we obtain that .w± ≥ 0 on .QT , i.e., ±u(x, t) ≤ (N + ϵ)t + m on

.

QT .

Because .ϵ > 0 was arbitrarily chosen, we get (6.17). Exercise 6.23. Hint. Use Exercise 6.22. Exercise 6.24. Solution. Let .w = u1 − u2 . Then .w is a solution to the problem

.

wt − kΔw = 0 on D × (0, T ], w(x, 0) = φ1 (x) − φ2 (x), x ∈ D, w(x, t) = ψ1 (x, t) − ψ2 (x, t),

Hence and Example 6.11, we obtain that

x ∈ ∂D,

0 ≤ x ≤ T.

8 Solutions, Hints and Answers to the Exercises

.

377

min w(x, t) ≤ w(x, t) ≤ max w(x, t)

∂p QT

∂p QT

in

QT ,

whereupon we get (6.18). Exercise 6.25. Hint. Use Exercise 6.24. Exercise 6.26. Hint. Use the function v(x, t) = u(x, t) +

.

|x−y|2

μ (T + ϵ − t)

n 2

e 4(T +ϵ−t) ,

x ∈ Rn ,

0 < t ≤ T,

for some .μ > 0. Exercise 6.27. Solution. Assume that .u1 , u2 ∈ C 2 (Rn , C 1 ((0, T ])) ∩ C (Rn × [0, T ]) satisfy (6.20), (6.21). Then .u1 − u2 satisfies .

ut − Δu = 0 in Rn × (0, T ] u = 0 on Rn × {t = 0}

and (6.21). Hence and Example 6.13, applied for .±(u1 − u2 ), we get sup

.

Rn ×[0,T ]

|u1 − u2 | = 0,

which completes the solution.

Chapter 7 Exercise 7.1. Answer. u(x, t) = x + sin t cos x,

.

−∞ < x < ∞,

t ≥ 0.

Exercise 7.2. Solution. Note that the d’Alambert formula provides us with a solution and any solution of the Cauchy problem (7.3), (7.4) is necessarily equal to the d’Alambert solution. Since .φ ∈ C 2 (R), .ψ ∈ C 1 (R), we have that u ∈ C 2 (R × (0, ∞)) ∩ C 1 (R × [0, ∞)).

.

Therefore the d’Alambert solution is a classical solution. Now we will prove the stability of the Cauchy problem. Let .ϵ > 0 be arbitrarily chosen. We take ϵ . Let also, .u1 and .u2 be solutions of the Cauchy problem with initial .0 < δ < 1+T conditions .φ1 , ψ1 and .φ2 , ψ2 , respectively, such that |φ1 (x) − φ2 (x)| < δ,

.

378

8 Solutions, Hints and Answers to the Exercises

|ψ1 (x) − ψ2 (x)| < δ, for all .x ∈ R. Then for all .x ∈ R and .0 ≤ t ≤ T , we have

φ (x + ct) − φ (x + ct) φ (x − ct) − φ (x − ct) 1 2 2

1 |u1 (x, t) − u2 (x, t)| = + 2 2 x+ct 

1

+ (ψ1 (s) − ψ2 (s)) ds 2c

.

x−ct

≤

|φ1 (x + ct) − φ2 (x + ct)| |φ1 (x − ct) − φ2 (x − ct)| + 2 2 x+ct  1 |ψ1 (s) − ψ2 (s)|ds + 2c x−ct

δ δ + + δT 2 2 = δ(1 + T ) ≤

< ϵ, which shows the stability of the d’Alambert solution. Exercise 7.3. Solution. Suppose that .u1 and .u2 are solutions to the problem (7.8), (7.9). Then the function .u = u1 − u2 is a solution to the homogeneous problem utt − c2 uxx = 0,

.

.

−∞ < x < ∞,

u(x, 0) = 0, ut (x, 0) = 0,

t > 0,

(8.8) (8.9)

−∞ < x < ∞.

Note that .v = 0 is also a solution to the homogeneous problem (8.8), (8.9). Hence and Exercise 7.2, we conclude that .u ≡ v ≡ 0, i.e., .u1 ≡ u2 . Exercise 7.4. Answer. 1. 1 u(x, t) = ex sinh t + xt 3 , 6

.

−∞ < x < ∞,

t ≥ 0.

2. u(x, t) = (x + 2t)2 ,

.

−∞ < x < ∞,

t ≥ 0.

8 Solutions, Hints and Answers to the Exercises

379

Exercise 7.5. Solution. Let .ϵ > 0 be arbitrarily chosen and .0 < δ < Let also, .fi , .φi and .ψi , .i = 1, 2, be such that .fi , fix φi ∈ C 2 (R), .ψi ∈ C 1 (R), and

ϵ

. 2 1 + T + T2 ∈ C (R × [0, ∞)),

.

|f1 (t, x) − f2 (t, x)| < δ,

.

|φ1 (x) − φ2 (x)| < δ, |ψ1 (x) − ψ2 (x)| < δ, for all .x ∈ R and .0 ≤ t ≤ T . Let .u1 and .u2 be the solutions of the Cauchy problems utt − c2 uxx = f1 (x, t),

−∞ < x < ∞,

.

t > 0,

u(x, 0) = φ1 (x), ut (x, 0) = ψ1 (x),

−∞ < x < ∞,

and utt − c2 uxx = f2 (x, t),

−∞ < x < ∞,

.

t > 0,

u(x, 0) = φ2 (x), ut (x, 0) = ψ2 (x),

−∞ < x < ∞,

respectively. Then

φ (x + ct) − φ (x + ct) φ (x − ct) − φ (x − ct) 1 2 2

1 + |u1 (x, t) − u2 (x, t)| = 2 2 x+ct  1 + (ψ1 (s) − ψ2 (s)) ds 2c

.

x−ct

1 + 2c

t

x+c(t−τ ) 

(f1 (ξ, τ ) − f2 (ξ, τ )) dξ dτ

0 x−c(t−τ )

φ (x + ct) − φ (x + ct) φ (x − ct) − φ (x − ct) 2 2

1

1

≤

+

2 2 x+ct  1 |ψ1 (s) − ψ2 (s)| ds + 2c x−ct

380

8 Solutions, Hints and Answers to the Exercises

1 + 2c

t

x+c(t−τ ) 

|f1 (ξ, τ ) − f2 (ξ, τ )| dξ dτ 0 x−c(t−τ )

δ δ T2 + + δT + δ 2 2 2   2 T = δ 1+T + 2
0.

Then vt (x, t) = ut (−x, t),

.

vtt (x, t) = utt (−x, t), vx (x, t) = −ux (−x, t), vxx (x, t) = uxx (−x, t),

−∞ < x < ∞,

t > 0.

Hence, vtt (x, t) − c2 vxx (x, t) = utt (−x, t) − c2 uxx (−x, t)

.

= f (−x, t), = f (x, t),

−∞ < x < ∞,

t > 0,

and v(x, 0) = u(−x, 0)

.

= φ(−x) = φ(x), vt (x, 0) = ut (−x, 0) = ψ(−x) = ψ(x),

−∞ < x < ∞.

Therefore .v is a solution to the Cauchy problem (7.8), (7.9). Hence, it follows that

8 Solutions, Hints and Answers to the Exercises

u(x, t) = u(−x, t),

381

−∞ < x < ∞,

.

t ≥ 0.

Exercise 7.7. Hint. Use the idea of the solution of Exercise 7.6. Exercise 7.8. Solution. Since .φ(·), .ψ(·) and .f (·, t), .t > 0, are .ω-periodic functions, we have φ(x + ω) = φ(x),

.

ψ(x + ω) = ψ(x), f (x + ω, t) = f (x, t),

−∞ < x < ∞,

t > 0.

Now, applying the d’Alambert formula, we find φ(x + ω + ct) + φ(x + ω − ct) 1 .u(x + ω, t) = + 2 2c

x+ω+ct 

ψ(s)ds x+ω−ct

+

1 2c

)  t x+ω+c(t−τ 

f (ξ, τ )dξ dτ 0 x+ω−c(t−τ )

1 φ(x + ct) + φ(x − ct) + = 2 2c

x+ct 

ψ(s + ω)ds x−ct

+

1 2c

)  t x+c(t−τ 

f (ξ + ω, τ )dξ dτ 0 x−c(t−τ )

1 φ(x + ct) + φ(x − ct) + = 2 2c

x+ct 

ψ(s)ds x−ct

1 + 2c

t

x+c(t−τ ) 

f (ξ, τ )dξ dτ 0 x−c(t−τ )

= u(x, t),

−∞ < x < ∞,

t ≥ 0.

Exercise 7.9. Answer. ∞

u(x, t) =

.

12 4 t+ sin(2nt) cos(2nx). 2 3π π n(4n − 1)(4n2 − 9) n=1

382

8 Solutions, Hints and Answers to the Exercises

Exercise 7.10. Answer. u(x, t) = sin(5π x) cos(10π t) + 2 sin(7π x) cos(14π t),

0 ≤ x ≤ 1,

.

t ≥ 0.

Exercise 7.11. Answer.     π  π  2 5π 5π .u(x, t) = sin t sin x + cos t sin x , π 2 2 2 2

0 ≤ x ≤ 1,

t ≥ 0.

Exercise 7.12. Answer. .u(x, t)

∞

=

m=0

     (2m + 1)π 128 (2m + 1)π m − 3 cos (2m + 1)π(−1) t cos x , 2 2 π 5 (2m + 1)5

0 ≤ x ≤ 1,

t ≥ 0.

Exercise 7.13. Answer. u(x, t) = .

∞    an cos cπLnt n=0    cπ nt Dn L  Dn L − 2cπ n t cos cπLnt + cπLn bn + 2cπ n sin L  cπ nt  Cn L + 2cπ sin πLnx , 0 ≤ x ≤ L, t ≥ 0. n t sin L

Exercise 7.14. Answer. u(x, t) = .

  ∞   an cos cπ(2n+1)t 2L n=0     Dn L 2L bn + cπ(2n+1) sin cπ(2n+1)t − + cπ(2n+1)   2L   cπ(2n+1)t π(2n+1)x Cn L sin . + cπ(2n+1) t sin L 2L



Dn L cπ(2n+1) t

cos

Dn L cπ(2n+1) t

cos

cπ(2n+1)t 2L



Exercise 7.15. Answer. u(x, t) = .

  ∞   an cos cπ(2n+1)t 2L n=0     Dn L 2L sin cπ(2n+1)t − + cπ(2n+1) bn + cπ(2n+1)   2L   Cn L cos π(2n+1)x , t sin cπ(2n+1)t + cπ(2n+1) L 2L



cπ(2n+1)t 2L

0 ≤ x ≤ L,

t ≥ 0.

0 < x < L,

t > 0,

Exercise 7.16. Answer. 1 1 (L − x)g1'' (t) − xg2'' (t), L L 1 1 v(x, 0) = φ(x) − (L − x)g1 (0) − xg2 (0), L L

vtt − c2 vxx = f (x, t) −

.



8 Solutions, Hints and Answers to the Exercises

vt (x, 0) = ψ(x) −

383

1 1 (L − x)g1' (0) − xg2' (0), L L

0 < x < 1,

v(0, t) = 0, v(L, t) = 0,

t > 0.

Exercise 7.17. Answer. vtt − c2 vxx = f (x, t) −

.

1 1 (L − x)2 g1'' (t) − x(x − L)g2'' (t) 2 L L

2c2 2c2 g1 (t) − g2 (t), 0 < x < L, t > 0, 2 L L 1 1 v(x, 0) = φ(x) − 2 (L − x)2 g1 (0) − x(x − L)g2 (0), L L 1 1 vt (x, 0) = ψ(x) − 2 (L − x)2 g1' (0) − x(x − L)g2' (0), 0 < x < L, L L v(0, t) = 0, −

vx (0, t) = 0,

t > 0.

Exercise 7.18. Answer. vtt − c2 vxx = f (x, t) − (x − L)g1'' (t) −

.

0 < x < L,

2c 1 2 '' x g2 (t) + 2 g2 (t), 2 L L

t > 0,

1 2 x g2 (0), L2 1 vt (x, 0) = ψ(x) − (x − L)g1' (0) − 2 x 2 g2' (0), L vx (0, t) = 0, v(x, 0) = φ(x) − (x − L)g1 (0) −

v(L, t) = 0,

0 < x < L,

t > 0.

Exercise 7.21. Answer.    1 u(x1 , x2 , x3 , t) =  (t + x12 + x22 + x32 ) sin(t + x12 + x22 + x32 ) 2 x12 + x22 + x32   −( x12 + x22 + x32 − t) sin( x12 + x22 + x32 − t)   −2 sin t sin x12 + x22 + x32

.

384

8 Solutions, Hints and Answers to the Exercises

1 + 2



 1 t 2 2 2 + x1 + x2 + x3 t 2 + t 4 , 3 4

(x1 , x2 , x3 ) ∈ R3 , t ≥ 0.

Exercise 7.23. Answer. 1. u(x1 , x2 , x3 , t) = t + x1 + x2 + x3 ,

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

2. u(x1 , x2 , x3 , t) = t (x1 x2 + x1 x3 + x2 x3 ),

.

(x1 , x2 , x3 ) ∈ R3 , t ≥ 0.

3. u(x1 , x2 , x3 , t) = t + x1 − x2 + x3 ,

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

4.  1 2 2 x1 x2 + x12 x32 + x22 x32 t 3 3  1 7 1  2 + x1 + x22 + x32 t 5 + t , (x1 , x2 , x3 ) ∈ R3 , t ≥ 0. 15 105

u(x1 , x2 , x3 , t) = x1 x2 x3 + x12 x22 x32 t +

.

5. u(x1 , x2 , x3 , t) =

.

1 2 1 2 x + x + x32 + 2t 2 , 2 1 2 2

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

Exercise 7.24. Answer. 1. u(x1 , x2 , x3 ) = x1 x2 x3 +t (x1 x2 +x3 )+

.

x1 t 2 t 3 + , 2 6

(x1 , x2 , x3 ) ∈ R3 ,

2. √ u(x1 , x2 , x3 , t) = x3 cos(2t) sin( 2(x1 + x2 ))   1 2 + t arctan t − log(1 + t ) x1 ex2 cos x3 , 2

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

t ≥ 0.

8 Solutions, Hints and Answers to the Exercises

385

3. u(x1 , x2 , x3 , t) = x1 sin x2 cos t + x2 cos x3 sin t   t 2 log(1+t )−t− arctan t , (x1 , x2 , x3 ) ∈ R3 , t ≥ 0. +x1 2

.

4. u(x1 , x2 , x3 , t) = x3 +x1 x2 +x1 x2 (t −sin t) sin x3 ,

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

5. u(x1 , x2 , x3 , t) = x13 + x23 + t 2 x1 ,

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

Exercise 7.25. Answer. u(x1 , x2 , t) = x1 + x2 (t − 1),

.

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

Exercise 7.26. Answer. 1. u(x1 , x2 , t) = x1 x2 t (1 + t 2 ) + x12 − x22 ,

.

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

2. u(x1 , x2 , t) =

.

1 2 3 t (x1 −3x1 x22 )+ex1 cos x2 +tex2 sin x1 , 2

(x1 , x2 ) ∈ R2 , t ≥ 0.

3. u(x1 , x2 , t) = x12 + t 2 + t sin x2 ,

.

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

4. u(x1 , x2 , t) = 2x12 − x22 + (2x12 + x22 )t + 2t 2 + 2t 3 ,

.

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

5. 1 1 u(x1 , x2 , t) = 1 + tx1 + x22 t 2 + t 4 , 2 12

.

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

Chapter 9

Solutions, Hints and Answers to the Problems

Chapter 1 Problem 1.1. Answer. 1. 2. 3. 4. 5.

2. 3. 2. 3. 2.

Problem 1.2. Answer. 1. 2. 3. 4. 5.

Linear. Nonlinear. Linear. Nonlinear. Linear.

Chapter 2 Problem 2.1. Answer. 1. 2. 3. 4. 5.

Quasilinear. Quasilinear. Semilinear. Semilinear. Semilinear.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. G. Georgiev, An Excursion Through Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-031-48784-2_9

387

388

9 Solutions, Hints and Answers to the Problems

Problem 2.2. Answer. 1. 2. 3. 4. 5.

Linear homogeneous. Linear nonhomogeneous. Linear homogeneous. Linear nonhomogeneous. Linear nonhomogeneous.

Problem 2.3. Answer. 1. 2. 3. 4. 5.

Linear. Linear. Nonlinear. Nonlinear. Nonlinear.

Problem 2.8. Answer. 1. u(x1 , x2 ) = f (x2 )x12 + x14 ,

(x1 , x2 ) ∈ R2 ,

.

f ∈ C (R).

2. u(x1 , x2 ) = (2x2 + 1)(f (x1 ) + log |2x2 + 1|) + 1,

.

(x1 , x2 ) ∈ R2 ,

1 x2 /= − , 2

f ∈ C (R).

3. u(x1 , x2 ) = ex1 (log |x1 | + f (x2 )),

.

(x1 , x2 ) ∈ R2 ,

x1 = 0,

f ∈ C (R).

4. x2 u(x1 , x2 ) = f (x1 ) − log |x2 |,

.

(x1 , x2 ) ∈ R2 ,

x2 /= 0,

f ∈ C (R).

5. u(x1 , x2 ) = x1 (f (x2 ) + sin x1 ),

.

(x1 , x2 ) ∈ R2 ,

f ∈ C (R).

(x1 , x2 ) ∈ R2 ,

f ∈ C (R).

6. 2

u(x1 , x2 ) = f (x2 )ex1 − x12 − 1,

.

7. u(x1 , x2 ) = f (x2 )(log x1 )2 − log x1 ,

.

(x1 , x2 ) ∈ R2 ,

x1 > 0,

f ∈ C (R).

9 Solutions, Hints and Answers to the Problems

389

8. x1 u(x1 , x2 ) = (x13 + f (x2 ))e−x1 ,

(x1 , x2 ) ∈ R2 ,

.

f ∈ C (R).

9. x2 = (u(x1 , x2 ))2 + f (x1 )u(x1 , x2 ),

u(x1 , x2 ) = 0,

.

(x1 , x2 ) ∈ R2 ,

f ∈ C (R).

10. x1 = eu(x1 ,x2 ) + f (x2 )e−u(x1 ,x2 ) ,

.

(x1 , x2 ) ∈ R2 ,

f ∈ C (R).

11.   u(x1 , x2 ) ex1 + f (x2 )e2x1 = 1,

.

(x1 , x2 ) ∈ R2 ,

u(x1 , x2 ) = 0,

f ∈ C (R).

12. u(x1 , x2 )(x1 + 1)(log |x1 + 1| + f (x2 )) = 1,

.

x1 /= −1,

(x1 , x2 ) ∈ R2 ,

f ∈ C (R),

u(x1 , x2 ) = 0, .(x1 , x2 ) ∈ R2 .

.

13. (u(x1 , x2 ))3 = f (x2 )x13 − 3x12 ,

.

(x1 , x2 ) ∈ R2 ,

f ∈ C (R).

14. (u(x1 , x2 ))2 = f (x2 )x12 − 2x1 ,

.

(x1 , x2 ) ∈ R2 ,

f ∈ C (R),

x1 = 0.

15.   (u(x1 , x2 ))2 x24 2ex2 + f (x1 ) = 1,

.

u(x1 , x2 ) = 0, .(x1 , x2 ) ∈ R2 .

.

(x1 , x2 ) ∈ R2 ,

f ∈ C (R),

390

9 Solutions, Hints and Answers to the Problems

Problem 2.9. Answer. 1.  u(x1 , x2 , x3 ) = f

.

x2 x3 , x1 x1

 (x1 , x2 ) ∈ R2 ,

,

f ∈ C 1 (R).

2. (u(x1 , x2 ))2 = x22 + f (x12 − x22 ),

(x1 , x2 ) ∈ R2 ,

.

f ∈ C 1 (R).

3.   F x12 − x22 , x1 − x2 + u = 0,

.

(x1 , x2 ) ∈ R2 ,

F ∈ C (R2 ).

4. 

1 x1 − log |x2 | − ,u + x2 e−x1 − x12

F e−x1

.

= 0,

(x1 , x2 ) ∈ R2 ,

F ∈ C (R2 ).

5.   (x1 + x2 )2 2 = 0, .F x1 − 4u, x1

(x1 , x2 ) ∈ R2 ,

F ∈ C (R2 ).

6.   u = 0, F x12 + x22 , x1

(x1 , x2 ) ∈ R2 ,

.

F ∈ C (R2 ).

7.  F

.

x12 3u , x1 x2 − x2 x1

= 0,

(x1 , x2 ) ∈ R2 ,

F ∈ C (R2 ).

8. 

1 1 1 1 + , + x1 + x2 u x1 − x2 u

 = 0,

(x1 , x2 ) ∈ R2 ,

F ∈ C (R2 ).

  

F x12 + x24 , x2 u + u2 + 1 = 0,

(x1 , x2 ) ∈ R2 ,

F ∈ C (R2 ).

F

.

9. .

9 Solutions, Hints and Answers to the Problems

391

10.  F

.

1 u2 1 − , log |x1 − x2 | − x1 x2 2

 = 0,

(x1 , x2 ) ∈ R2 ,

F ∈ C (R2 ).

11.     x1 + (u + 1)e−u = 0, F x12 + x22 , arctan x2

.

(x1 , x2 ) ∈ R2 ,

F ∈ C (R2 ).

12.   F u2 − x22 , x12 + (x2 − u)2 = 0,

(x1 , x2 ) ∈ R2 ,

.

F ∈ C (R2 ).

13.  F

.

u , 2x1 − 4u − x22 x1

 = 0,

(x1 , x2 ) ∈ R2 ,

F ∈ C (R2 ).

14.   F u − log |x1 |, 2x1 (u − 1) − x22 = 0,

(x1 , x2 ) ∈ R2 ,

.

F ∈ C (R2 ).

15.  F

.

 x1 + x2 + u , (x1 − x2 )(x1 + x2 − 2u) = 0, (x1 − x2 )2

(x1 , x2 ) ∈ R2 ,

F ∈ C (R2 ).

16. F ((x1 − x2 )(u + 1), (x1 + x2 )(u − 1)) = 0,

.

(x1 , x2 ) ∈ R2 ,

F ∈ C (R2 ).

17.   x1 + x2 + x3 = 0, .F u(x1 − x2 ), u(x2 − x3 ), u2 (x1 , x2 , x3 ) ∈ R3 ,

F ∈ C (R3 ).

18.  F

.

x3 + u − x1 x2 x1 , x1 x2 − 2u, x2 x1

 = 0,

(x1 , x2 , x3 ) ∈ R3 ,

F ∈ C (R3 ).

392

9 Solutions, Hints and Answers to the Problems

19.  F

.

x1 − x2 u − x1 − x2 , (2u + x1 + x2 )x3 , x3 x32

(x1 , x2 , x3 ) ∈ R3 ,

= 0,

F ∈ C (R3 ).

20.  F

.

x1 2 , u + x1 x2 x2

 = 0,

(x1 , x2 ) ∈ R2 .

x22 8 + x1 + , 3 3x2

(x1 , x2 ) ∈ R2 ,

Problem 2.10. Answer. 1. u(x1 , x2 ) = −x12 +

.

x2 /= 0.

2. u(x1 , x2 ) = 2x1 x2 ,

.

(x1 , x2 ) ∈ R2 .

3. u(x1 , x2 ) = x2 ex1 − e2x1 + 1,

(x1 , x2 ) ∈ R2 .

.

4. u(x1 , x2 ) = x22 e2

.

√

x1 −2

,

(x1 , x2 ) ∈ R2 ,

x1 > 0.

5. u(x1 , x2 , x3 ) = (1 − x1 + x2 )(2 − 2x1 + x3 ),

.

(x1 , x2 , x3 ) ∈ R3 .

6. 

x2 x1 .u(x1 , x2 , x3 ) = (x1 x2 − 2x3 ) + x1 x2 x1 /= 0,

 ,

(x1 , x2 , x3 ) ∈ R3 ,

x2 /= 0.

7.  u(x1 , x2 ) = x22 − x12 − log x22 − x12 + log |x2 |,

.

(x1 , x2 ) ∈ R2 ,

x12 < x22 ..

9 Solutions, Hints and Answers to the Problems

393

8. 1 2 1 1 x (x2 + 1) − x22 + , 2 1 4 4

(x1 , x2 ) ∈ R2 .

(x1 + 2x2 )2 = 2x1 (u(x1 , x2 ) + x1 x2 ),

(x1 , x2 ) ∈ R2 .

u(x1 , x2 ) =

.

9. .

10. .

u(x1 , x2 ) x23

sin x1 = sin

u(x1 , x2 ) , x2

 π , x2 /= 0. (x1 , x2 ) ∈ R2 , x1 ∈ 0, 2

Problem 2.11. Answer. u(x1 , x2 , x3 ) =

.

(ξ1 + ξ2 + ξ3 )2 (x1 + x2 + x3 )2 − =c 2 2

for any .(ξ1 , ξ2 , ξ3 ) ∈ R3 , where .c is a constant. Problem 2.12. Answer. 1. .x3 = x22 − x1 x2 , 2. .x12 x2 x3 = c − x13 ,

x1 = 0, where c is a real constant.

Problem 2.13. Answer. .u = 0.

Chapter 3 Problem 3.1. Answer. 1. 2. 3. 4. 5.

Quasilinear. Quasilinear. Quasilinear. Quasilinear. Quasilinear.

Problem 3.2. Answer. 1. 2. 3. 4. 5.

Linear homogeneous. Linear nonhomogeneous. Linear homogeneous. Linear nonhomogeneous. Linear nonhomogeneous.

394

9 Solutions, Hints and Answers to the Problems

Problem 3.3. Answer. 1. 2. 3. 4. 5.

Linear. Nonlinear. Nonlinear. Nonlinear. Linear.

Chapter 4 Problem 4.1. Answer. 1. 2. 3. 4. 5. 6. 7.

Hyperbolic. Hyperbolic. Elliptic. Hyperbolic. Hyperbolic. Parabolic. Parabolic.

Problem 4.3. Answer. 1. uξ1 ξ2 = 0,

.

ξ1 (x1 , x2 ) = x1 + x2 , ξ2 (x1 , x2 ) = x2 + 2x1 ,

(x1 , x2 ) ∈ R2 .

2. uξ1 ξ2 = 0,

.

ξ1 (x1 , x2 ) = x2 + 2x1 , ξ2 (x1 , x2 ) = x2 − x1 ,

(x1 , x2 ) ∈ R2 .

3. uξ1 ξ2 +

.

1 uξ = 0, 3ξ2 1

ξ1 (x1 , x2 ) = x1 x2 , ξ2 (x1 , x2 ) = x1 ,

(x1 , x2 ) ∈ R2 ,

x2 /= 0.

9 Solutions, Hints and Answers to the Problems

395

4. uξ1 ξ2 −

.

1 (uξ − uξ2 ) = 0, 16 1 ξ1 (x1 , x2 ) = x1 − x2 , ξ2 (x1 , x2 ) = 3x1 + x2 ,

(x1 , x2 ) ∈ R2 .

5. uξ1 ξ1 + uξ2 ξ2 + uξ1 = 0,

.

ξ1 (x1 , x2 ) = x1 , ξ2 (x1 , x2 ) = 3x1 + x2 ,

(x1 , x2 ) ∈ R2 .

6. uξ2 ξ2 + uξ1 = 0,

.

ξ1 (x1 , x2 ) = x1 − 2x2 , ξ2 (x1 , x2 ) = x1 ,

(x1 , x2 ) ∈ R2 .

7. 1 (uξ + uξ2 ) = 0, 6(ξ1 + ξ2 ) 1

uξ1 ξ2 +

.

2 32 x + x2 , 3 1 2 3 ξ2 (x1 , x2 ) = x12 − x2 , 3 ξ1 (x1 , x2 ) =

(x1 , x2 ) ∈ R2 ,

x1 > 0,

and uξ1 ξ1 + uξ2 ξ2 +

.

1 uξ = 0, 3ξ1 1

ξ1 (x1 , x2 ) =

3 2 (−x1 ) 2 , 3

ξ2 (x1 , x2 ) = x2 ,

(x1 , x2 ) ∈ R2 ,

x1 < 0.

8. uξ1 ξ2 +

.

1 (uξ − uξ2 ) = 0, 2(ξ1 − ξ2 ) 1

√ ξ1 (x1 , x2 ) = x1 + 2 x2 , √ ξ2 (x1 , x2 ) = x1 − 2 x2 ,

(x1 , x2 ) ∈ R2 ,

x2 > 0,

396

9 Solutions, Hints and Answers to the Problems

and uξ1 ξ1 + uξ2 ξ2 −

.

1 uξ = 0, ξ2 2

ξ1 (x1 , x2 ) = x1 , √ ξ2 (x1 , x2 ) = 2 −x2 ,

(x1 , x2 ) ∈ R2 ,

x2 < 0.

9. uξ1 ξ1 − uξ2 ξ2 −

.

1 1 uξ + uξ = 0, ξ1 1 ξ2 2

ξ1 (x1 , x2 ) = |x1 |,

ξ2 (x1 , x2 ) = |x2 |,

(x1 , x2 ) ∈ |R2 ,

x1 > 0,

x2 > 0 or

x1 < 0,

x2 < 0,

and uξ1 ξ1 + uξ2 ξ2 −

.

1 1 uξ1 − uξ2 = 0, ξ1 ξ2

ξ1 (x1 , x2 ) = |x1 |,

ξ2 (x1 , x2 ) = |x2 |,

(x1 , x2 ) ∈ R2 ,

x1 > 0, x2 < 0,

or

x1 < 0, x2 > 0.

10. uξ1 ξ1 − uξ2 ξ2 +

.

1 1 uξ1 − uξ = 0, 3ξ1 3ξ2 2 3

ξ1 (x1 , x2 ) = |x1 | 2 , 3

ξ2 (x1 , x2 ) = |x2 | 2 ,

(x1 , x2 ) ∈ R2 ,

x1 > 0, x2 > 0,

or

x1 < 0, x2 < 0,

and uξ1 ξ1 + uξ2 ξ2 +

.

1 1 uξ1 + uξ = 0, 3ξ1 3ξ2 2 3

ξ1 (x1 , x2 ) = |x1 | 2 , 3

ξ2 (x1 , x2 ) = |x2 | 2 ,

(x1 , x2 ) ∈ R2 ,

x1 > 0, x2 < 0,

or

x1 < 0, x2 > 0.

9 Solutions, Hints and Answers to the Problems

397

11. uξ1 ξ1 + uξ2 ξ2 − uξ1 − uξ2 = 0,

.

ξ1 (x1 , x2 ) = log |x1 |, ξ2 (x1 , x2 ) = log |x2 |,

(x1 , x)2) ∈ R2 ,

x1 /= 0,

x2 /= 0.

12. uξ1 ξ1 + uξ2 ξ2 +

.

1 1 uξ + uξ = 0, 2ξ1 1 2ξ2 2 ξ1 (x1 , x2 ) = x22 , ξ2 (x1 , x2 ) = x12 ,

(x1 , x2 ) ∈ R2 .

13. uξ1 ξ2 +

.

2(ξ22

1 (ξ2 uξ1 − ξ1 uξ2 ) = 0, − ξ12 ) ξ1 (x1 , x2 ) = x22 − x12 , ξ2 (x1 , x2 ) = x12 + x22 ,

(x1 , x2 ) ∈ R2 .

14. uξ1 ξ1 + uξ2 ξ2 − tanh ξ1 uξ1 = 0,

.

ξ1 (x1 , x2 ) = log(x1 + ξ2 (x1 , x2 ) = log(x2 +

 

1 + x12 ), 1 + x22 ),

(x1 , x2 ) ∈ R2 .

15. uξ1 ξ1 + uξ2 ξ2 + cos ξ1 uξ1 = 0,

.

ξ1 (x1 , x2 ) = x1 , ξ2 (x1 , x2 ) = x2 − cos x1 ,

(x1 , x2 ) ∈ R2 .

16. uξ2 ξ2 − 2uξ1 = 0,

.

ξ1 (x1 , x2 ) = 2x1 − x22 , ξ2 (x1 , x2 ) = x2 ,

(x1 , x2 ) ∈ R2 .

398

9 Solutions, Hints and Answers to the Problems

17. uξ2 ξ2 − ξ1 uξ1 = 0,

.

ξ1 (x1 , x2 ) = x1 ex2 , ξ2 (x1 , x2 ) = x2 ,

(x1 , x2 ) ∈ R2 .

Problem 4.3. Answer. 1. u(x1 , x2 ) = f (x1 ) + g(x2 ),

.

(x1 , x2 ) ∈ R2 ,

f, g ∈ C (R2 ).

2. u(x1 , x2 ) = f (x2 + ax1 ) + g(x2 − ax1 ),

(x1 , x2 ) ∈ R2 ,

f, g ∈ C (R2 ).

u(x1 , x2 ) = f (x1 − x2 ) + g(3x1 + x2 ),

(x1 , x2 ) ∈ R2 ,

f, g ∈ C (R2 ).

.

3. .

4. u(x1 , x2 ) = f (x2 ) + g(x1 )e−ax2 ,

.

(x1 , x2 ) ∈ R2 ,

f, g ∈ C (R2 ).

5. u(x1 , x2 ) = x1 − x2 + f (x1 − 3x2 ) + g(2x1 + x2 )e

.

(x1 , x2 ) ∈ R2 ,

3x2 −x1 7

,

f, g ∈ C (R2 ).

6. u(x1 , x2 ) = (f (x1 ) + g(x2 ))e−bx1 −ax2 ,

.

(x1 , x2 ) ∈ R2 ,

f, g ∈ C (R2 ).

7. .

u(x1 , x2 ) = ex1 +x2 +(f (x1 )+g(x2 ))e3x1 +2x2 ,

(x1 , x2 ) ∈ R2 ,

f, g ∈ C (R2 ).

u(x1 , x2 ) = f (x2 − ax1 ) + g(x2 − ax1 )e−x1 ,

(x1 , x2 ) ∈ R2 ,

f, g ∈ C (R2 ).

8. .

9 Solutions, Hints and Answers to the Problems

399

Problem 4.4. Answer. 1. u(x1 , x2 ) = f (x1 + x2 ) + (x1 − x2 )g(x12 − x22 ),

.

x1 > −x2

or

x1 < −x2 ,

(x1 , x2 ) ∈ R2 ,

f, g ∈ C 2 (R).

2.

u(x1 , x2 )=f (x1 x2 )+ |x1 x2 |g



.

x1 x2

 ,

(x1 , x2 ) ∈ R2 , x2 /= 0, f, g ∈ C 2 (R).

,

(x1 , x2 ) ∈ R2 , x2 /= 0, f, g ∈ C 2 (R).

3.  3 4

u(x1 , x2 )=f (x1 x2 )+|x1 x2 | g

.

x13 x2



4.  u(x1 , x2 ) = x1 f

.

   x1 x1 +g , x2 x2

(x1 , x2 ) ∈ R2 ,

x2 /= 0,

f, g ∈ C 2 (R).

5.

x1 .u(x1 , x2 ) = x1 f (x2 ) − f (x2 ) + (x1 − ξ )g(ξ )eξ x2 dξ, '

0

(x1 , x2 ) ∈ R2 ,

f, g ∈ C 2 (R).

6. 1 ' g (x1 ) + .u(x1 , x2 ) = 2x2 g(x1 ) + x1 (x1 , x2 ) ∈ R , 2

x1 /= 0,

x2 2 (x2 − ξ )f (ξ )e−x1 ξ dξ, 0

f, g ∈ C 2 (R).

7. ⎞

x2 −x2 ⎝ .u(x1 , x2 ) = e x2 f (x1 ) + f ' (x1 ) + (x2 − η)g(η)e−x1 η dη⎠ , ⎛

0

(x1 , x2 ) ∈ R , 2

f, g ∈ C (R). 2

400

9 Solutions, Hints and Answers to the Problems

8. ⎛

⎞

x2 −x1 x2 ⎝ x2 f (x1 ) + f ' (x1 ) + (x2 − η)g(η)e−x1 η dη⎠ , .u(x1 , x2 ) = e 0

(x1 , x2 ) ∈ R ,

f, g ∈ C (R).

2

2

9.

x1 x2 u(x1 , x2 ) = f (x1 ) + g(x2 ) +

h(ξ1 , ξ2 )dξ2 dξ1 ,

.

x10

(x1 , x2 ) ∈ X,

x20

f, g ∈ C 2 (R).

10. x2

− h(ξ1 ,ξ2 )dξ2

x1 0 .u(x1 , x2 ) = f (x2 )+ g(ξ1 )e x2 dξ1 ,

(x1 , x2 ) ∈ R2 , f, g ∈ C 2 (R).

x10

11. u(x1 , x2 ) =

.

f (x1 ) + g(x2 ) , x1 − x2

(x1 , x2 ) ∈ R2 ,

x1 /= x2 ,

f, g ∈ C 2 (R).

12. u(x1 , x2 ) =

.

x2 /= x2 ,

∂ m+k−2



∂x1k−1 ∂x2m−1

 f (x1 ) + g(x2 ) , x1 − x2

f ∈ C k+1 (R),

(x1 , x2 ) ∈ R2 ,

g ∈ C m+1 (R).

13. u(x1 , x2 ) = (x1 − x2 )m+k−1

.

(x1 , x2 ) ∈ R2 ,

x1 /= x2 ,

∂ m+k ∂x1m ∂x2k



 f (x1 ) + g(x2 ) , x1 − x2

f ∈ C m+2 (R),

g ∈ C k+2 (R).

9 Solutions, Hints and Answers to the Problems

401

Problem 4.5. Answer. Parabolic. Problem 4.7. Answer. 1. uξ1 ξ1 − uξ2 ξ2 + uξ3 ξ3 + uξ2 = 0,

.

1 x1 , 2 1 ξ2 (x1 , x2 , x3 ) = x1 + x2 , 2 1 ξ3 (x1 , x2 , x3 ) = − x1 − x2 + x3 , 2

ξ1 (x1 , x2 , x3 ) =

(x1 , x2 , x3 ) ∈ R3 .

2. uξ1 ξ1 − uξ2 ξ2 + 2uξ1 = 0,

.

ξ1 (x1 , x2 , x3 ) = x1 + x2 , ξ2 (x1 , x2 , x3 ) = x2 − x1 , ξ3 (x1 , x2 , x3 ) = x2 + x3 ,

(x1 , x2 , x3 ) ∈ R3 .

3. uξ1 ξ1 + uξ2 ξ2 = 0,

.

ξ1 (x1 , x2 , x3 ) = x1 , ξ2 (x1 , x2 , x3 ) = x2 − x1 , ξ3 (x1 , x2 , x3 ) = 2x1 − x2 + x3 ,

(x1 , x2 , x3 ) ∈ R3 .

4. uξ1 ξ1 − uξ2 ξ2 − uξ3 ξ3 = 0,

.

ξ1 (x1 , x2 , x3 ) = x1 , ξ2 (x1 , x2 , x3 ) = x2 − x1 , ξ3 (x1 , x2 , x3 ) =

3 1 1 x1 − x2 + x3 , 2 2 2

5. uξ1 ξ1 − uξ2 ξ2 + uξ3 ξ3 − uξ4 ξ4 = 0,

.

ξ1 (x1 , x2 , x3 , x4 ) = x1 + x2 , ξ2 (x1 , x2 , x3 , x4 ) = x1 − x2 ,

(x1 , x2 , x3 ) ∈ R3 .

402

9 Solutions, Hints and Answers to the Problems

ξ3 (x1 , x2 , x3 , x4 ) = −2x2 + x3 + x4 , ξ4 (x1 , x2 , x3 , x4 ) = x3 − x4 ,

(x1 , x2 , x3 , x4 ) ∈ R4 .

6. uξ1 ξ1 − uξ2 ξ2 + uξ3 ξ3 = 0,

.

ξ1 (x1 , x2 , x3 , x4 ) = x1 , ξ2 (x1 , x2 , x3 , x4 ) = x2 − x1 , ξ3 (x1 , x2 , x3 , x4 ) = 2x1 − x2 + x3 , ξ4 (x1 , x2 , x3 , x4 ) = x1 + x3 + x4 ,

(x1 , x2 , x3 , x4 ) ∈ R4 .

7. uξ1 ξ1 + uξ2 ξ2 = 0,

.

ξ1 (x1 , x2 , x3 , x4 ) = x1 , ξ2 (x1 , x2 , x3 , x4 ) = x2 , ξ3 (x1 , x2 , x3 , x4 ) = −x1 − x2 + x3 , ξ4 (x1 , x2 , x3 , x4 ) = x1 − x2 + x4 ,

(x1 , x2 , x3 , x4 ) ∈ R4 .

8. n  .

uξk ξk = 0,

k=1

ξ1 (x1 , . . . , xn ) = x1 , ξk (x1 , . . . , xn ) = xk − xk−1 ,

k ∈ {2, . . . , n},

(x1 , . . . , xn ) ∈ Rn .

9. n  .

uξk ξk = 0,

k=2

 ξk (x1 , . . . , xn ) =

⎞ ⎛ n 2k ⎝ 1  xl ⎠ , xk − k+1 k l=1,l 2. . .

Problem 5.4. Hint. Use the Taylor formula of second order. Problem 5.5. Hint. Use the Taylor formula of second order. Problem 5.6. Hint. Use the Taylor formula of second order and

r u(y)dy =

.

⎛ ⎜ ⎝

0

B(x0 ,r)

⎞ ⎟ u(y)dsy ⎠ dρ.

B(x0 ,ρ)

Problem 5.7. Hint. Use Exercise 5.12. Problem 5.9. Answer. For .y = (y1 , y2 , y3 ) ∈ R3 , denote   ymnk = (−1)m y1 , (−1)n y2 , (−1)k y3 ,

.

1 ymnk , |y|2

∗ ymnk =

∗ |ymnk ||ymnk | = 1,

m, n, k ∈ N0 .

1.

.

1 1  (−1)k . 4π |x − y00k | k=0

2. 1 1  (−1)n+k . . 4π |x − y0nk | n,k=0

3.

.

1 4π

1  m,n,k=0

(−1)m+n+k . |x − ymnk |

406

9 Solutions, Hints and Answers to the Problems

4.   1 1 1  1 k . (−1) − ∗ | . 4π |x − y00k | |y||x − y00k k=0

5.

.

  1 1 1  1 (−1)n+k − ∗ | . 4π |x − y0nk | |y||x − y0nk n,k=0

6. 1 

1 . 4π

 (−1)

m+n+k

m,n,k=0

 1 1 − ∗ | . |x − ymnk | |y||x − ymnk

Problem 5.10. Answer. 1. √

u(x1 , x2 , x3 ) = e−

.

2x3

(x1 , x2 , x3 ) ∈ D.

cos x1 cos x2 ,

2.   √ u(x1 , x2 , x3 ) = e− 2x3 − e−x3 sin x1 cos x2 ,

.

(x1 , x2 , x3 ) ∈ D.

3. 1 u(x1 , x2 , x3 ) =  1 , 2 2 x1 + x2 + (x3 + 1)2 2

.

(x1 , x2 , x3 ) ∈ D.

4. u(x1 , x2 , x3 ) =

.

x12

+ x22

1 , + (x3 + 1)2

(x1 , x2 , x3 ) ∈ D.

5. u(x1 , x2 , x3 ) =

.

x1 2 arctan , π x3

(x1 , x2 , x3 ) ∈ D.

Problem 5.11. Answer. 1. u(x1 , x2 , x3 ) = e−4x1 −3x3 sin(5x2 ),

.

(x1 , x2 , x3 ) ∈ R3 .

9 Solutions, Hints and Answers to the Problems

407

2. u(x1 , x2 , x3 ) = 

x2

.

x12

+ x22

+ (x3

+ 1)2

 32 ,

(x1 , x2 , x3 ) ∈ R3 .

Problem 5.12. Answer. 1. u(x1 , . . . , xn ) =

.

1 (1 − |x|2 ), 6

(x1 , . . . , xn ) ∈ D.

2. u(x1 , . . . , xn ) = 1 +

.

1 − |x|n+2 , (n + 2)(n + 3)

(x1 , . . . , xn ) ∈ D.

3. u(x1 , . . . , xn ) = e − e|x| − 2(e − 1) +

.

 2  |x| e −1 , |x|

(x1 , . . . , xn ) ∈ D.

Problem 5.13. Answer. 1. u(x1 , x2 ) = sin(π x1 )

.

sinh(π(1 − x2 )) + 1 + x2 . sinh π

2. u(x1 , x2 ) =

.

cosh(2π x1 ) sin(2π x2 ) x2 + . 2 2π sinh(2π )

3. u(x1 , x2 ) = A0 −

.

∞ 4  cos((2n − 1)x1 ) cosh((2n − 1)x2 ) , π (2n − 1)3 sinh((2n − 1)π ) n=0

where .A0 is a constant. 4. u(x1 , x2 ) =

.

1 (1 + x12 − x22 ). 2

408

9 Solutions, Hints and Answers to the Problems

Problem 5.14. Answer. 1. u(x1 , x2 ) = x1 + x1 x2 ,

.

x12 + x2 ≤ R 2 .

2. u(x1 , x2 ) = x12 − x22 + 2x2 + R 2 ,

.

x12 + x22 ≤ R 2 .

Problem 5.15. Answer. 1. u(x1 , x2 ) =

.

x12

R2 R4 x2 + 2 2 x1 x2 , 2 + x2 (x1 + x22 )2

x12 + x22 ≥ R 2 .

2. u(x1 , x2 ) =

.

x12

R2 (ax1 + bx2 ) + c, + x22

x12 + x22 ≥ R 2 .

3. u(x1 , x2 ) =

.

(x12

R4 (x12 − x22 ), + x22 )2

x12 + x22 ≥ R 2 .

4. u(x1 , x2 ) =

.

R2 R4 2 2 + 1, (x − x ) + 1 2 2 2(x12 + x22 )2

x12 + x22 ≥ R 2 .

5. u(x1 , x2 ) =

.

R4 R2 − (x12 − x22 + 2x1 x2 ), 2 2(x12 + x22 )2

x12 + x22 ≥ R 2 .

6. u(x1 , x2 ) =

.

x12 + x22 ≥ R 2 .

.

R 4 (x12 − x22 ) R2 R2 − + (x1 + x2 ), 2 2(x12 + x22 )2 x12 + x22

9 Solutions, Hints and Answers to the Problems

409

7. u(x1 , x2 ) = R 2 +

.

R2 R4 2 2 (x − x ) − (x1 − x2 ), 1 2 (x12 + x22 )2 x12 + x22

x12 + x22 ≥ R 2 .

.

Problem 5.16. Answer. 1. u(x1 , x2 ) =

.

x12 + x22 − R 2 , 4

x12 + x22 ≤ R 2 .

2. u(x1 , x2 ) =

.

1 3 (x + x1 x22 − R 2 x1 ), 8 1

x12 + x22 ≤ R 2 .

3. u(x1 , x2 ) =

.

R 2 − x12 , 2

x12 + x22 ≤ R 2 .

4. u(x1 , x2 ) =

.

1 3 (x + x12 x2 − R 2 x2 + 8), 8 2

x12 + x22 ≤ R 2 .

5. u(x1 , x2 ) = x12 + x22 − R 2 + 1,

.

x12 + x22 ≤ R 2 .

Problem 5.17. Answer. 1.  u(x1 , x2 ) =

.

x12 + x22 R

cos θ,

θ ∈ [0, 2π ],

x12 + x22 ≤ R 2 .

2.  x12 + x22 x12 + x22 1 .u(x1 , x2 ) = (cos θ )2 , + 1− 3 R2 R2 x12 + x22 ≤ R 2 , .θ ∈ [0, 2π ].

.

410

9 Solutions, Hints and Answers to the Problems

Problem 5.18. Answer. 1. 1 3(cos θ )2 − 1 u(x1 , x2 ) =  + , 3 3(x12 + x22 ) 2 3 x12 + x22

.

1 ≤ x12 + x22 ≤ 4, .θ ∈ [0, 2π ].

.

2. ⎞ ⎛ 2 1⎝  − 1 + (x12 + x22 )(3(cos θ )2 − 1)⎠ , .u(x1 , x2 ) = 3 2 2 x +x 1

2

1 ≤ x12 + x22 ≤ 4, .θ ∈ [0, 2π ].

.

Problem 5.19. Answer. u(x1 , x2 ) = a0 +

  nπ x  nπ(a − x1 ) 2 cos , an cosh b b 

∞ 

.

n=1

(x1 , x2 ) ∈ [0, a] × [0, b], where

.

2   .an = − nπ sinh nπb a

b g1 (x2 ) cos

 nπ x  2

b

n ∈ N,

dx2 ,

0

and .a0 is a constant. Problem 5.20. Answer. u(x1 , x2 ) = a0 +

∞ 

.

an cos

 nπ x  1

b

n=1

cos

 nπ x  2

b

,

(x1 , x2 ) ∈ [0, a] × [0, b], where

.

2   .an = nπ sinh nπb a

b g2 (x2 ) cos

 nπ x  2

b

dx2 ,

n ∈ N,

0

and .a0 is a constant. Problem 5.21. Answer. u(x1 , x2 ) = a0 +

∞ 

.

n=1

an cos

 nπ x  1

a



 nπ(b − x2 ) cosh , a

9 Solutions, Hints and Answers to the Problems

411

(x1 , x2 ) ∈ [0, a] × [0, b], where

.

an = −

.

2 nπ sinh

a  nπ b 

f1 (x1 ) cos

a

 nπ x  1

a

dx1 ,

n ∈ N,

0

and .a0 is a constant. Problem 5.22. Answer. u(x1 , x2 ) = a0 +

∞ 

.

an cos

 nπ x  1

a

n=1

cosh

 nπ x  2

a

,

(x1 , x2 ) ∈ [0, a] × [0, b], where

.

an =

.

2 nπ sinh

a  nπ b  a

f2 (x1 ) cos

 nπ x  1

a

dx1 ,

n ∈ N,

0

and .a0 is a constant. Problem 5.23. Answer. Let .u1 be the solution in Problem 5.19, .u2 be the solution in Problem 5.20, .u3 be the solution in Problem 5.21 and .u4 be the solution in Problem 5.22. Then u(x1 , x2 ) = u1 (x1 , x2 ) + u2 (x1 , x2 ) + u3 (x1 , x2 ) + u4 (x1 , x2 ),

.

(x1 , x2 ) ∈ [0, a] × [0, b], is a formal solution of the considered problem.

Chapter 6 Problem 6.1. Answer. 1. u(x, t) =

.

x (1 + 4t)

x2

3 2

e− 1+4t ,

x ∈ R,

t ≥ 0.

2. 1



x .u(x, t) = √ sin 1 + t 1+t



4x 2 +t

e− 4(1+t) ,

x ∈ R,

t ≥ 0.

412

9 Solutions, Hints and Answers to the Problems

3. u(x1 , . . . , xn , t) =

.

1 (1 + 4t)

n e

n 

1 − 1+4t

j =1

xj2

(x1 , . . . , xn ) ∈ Rn ,

,

2

t ≥ 0.

4. 

1

u(x1 , . . . , xn , t) =

.

(1 + 4t)

(x1 , . . . , xn ) ∈ Rn ,

e

n+2 2

1 − 1+4t

n 

j =1

 xj2

n 

j =1

xj

,

t ≥ 0.

5. ⎛ u(x1 , . . . , xn , t) =

.

1

sin ⎝

n

(1 + 4t) 2 nt+

×e

−

n 

j =1 1+4t

1 1 + 4t

n 

⎞ xj ⎠

j =1

xj2

(x1 , . . . , xn ) ∈ Rn ,

,

t ≥ 0.

6. 

u(x1 , . . . , xn , t) = √

.

1 1 + 4nt

e

1 − 1+4nt

n 

j =1

2 xj

,

(x1 , . . . , xn ) ∈ Rn ,

t ≥ 0.

7. u(x1 , x2 , x3 , t) = 2x12 − 2x22 + 3x32 + 6t,

.

(x1 , x2 , x3 ) ∈ R3 ,

Problem 6.2. Answer. 1. u(x, t) = x 2 t 2 ,

.

x ∈ R,

t ≥ 0.

2. u(x, t) = sin t + tx 3 ,

.

x ∈ R,

t ≥ 0.

3. u(x1 , x2 , t) = t (x1 + x22 ),

.

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

t ≥ 0.

9 Solutions, Hints and Answers to the Problems

413

4. ⎛ u(x1 , . . . , xn , t) = sin t ⎝

n 

.

⎞ xj2 ⎠ ,

(x1 , . . . , xn ) ∈ Rn ,

t ≥ 0.

j =1

5. u(x1 , x2 , x3 , t) = t 2 x1 + tx2 + 2t 3 x3 ,

(x1 , x2 , x3 ) ∈ R3 ,

.

t ≥ 0.

Problem 6.3. Answer. 1. u(x, t) = (1 + t)e−t cos x,

x ∈ R,

.

t ≥ 0.

2. u(x, t) = cosh t sin x,

.

x ∈ R,

t ≥ 0.

3. u(x, t) = 1 − cos t + √

.

1 1 + 4t

x2

e− 1+4t ,

x ∈ R,

t ≥ 0.

4. 1 u(x1 , x2 , t) = 1+ sin x1 sin x2 (2 sin t −cos t +e−2t ), 5

.

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

5. u(x1 , x2 , t) = sin t +

.

x12 +x22 x1 x2 − 1+4t , e (1 + 4t)3

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

6. u(x1 , x2 , t) =

.

(x1 −x2 )2 t 1 e− 1+t , +√ 8 1+t

(x1 , x2 ) ∈ R2 ,

t ≥ 0.

7. u(x1 , x2 , x3 , t) = et − 1 + sin(x1 − x2 − x3 )e−9t ,

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

414

9 Solutions, Hints and Answers to the Problems

8. u(x1 , x2 , x3 , t) =

.

1 cos(2x2 ) −t− x12 1+t , e sin(2x3 ) + √ 4 1+t

(x1 , x2 , x3 ) ∈ R3 , .t ≥ 0.

.

9.   (x1 +x2 −x3 )2 1 1 cos(x1 − x2 + x3 ) 1 − e−3t + √ e− 1+12t , 3 1 + 12t

u(x1 , x2 , x3 , t) =

.

(x1 , x2 , x3 ) ∈ R3 , .t ≥ 0.

.

10. u(x1 , x2 , x3 , t) = t 2 (x12 + x22 + x32 ) + x1 ,

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

Problem 6.4. Answer. 1. u(x, t) =

.

 2   ∞ 1 (2k + 1)π 8  t − (2k+1)π 2 e cos x , 2 π2 (2k + 1)2 k=1

x ∈ [0, 1],

t ≥ 0.

2. u(x, t) = 1,

.

x ∈ [0, 1],

t ≥ 0.

3. u(x, t) = 1 −

.

∞ 1 8  2 2 e−(2k+1) π t cos((2k + 1)π x), 2 2 π (2k + 1) k=0

0 ≤ x ≤ 1, .t ≥ 0.

.

4. u(x, t) =

.

1 −t e sin x cos 1    ∞ 2π2  2 (−1)k (2k + 1)π t − (2k+1) 4 +2 + x , sin e (2k + 1)π 1− (2k+1)2 π 2 2 k=0

x ∈ [0, 1], .t ≥ 0.

.

4

9 Solutions, Hints and Answers to the Problems

415

5. 

1 2 .u(x, t) = − t − 2

 1 1 2 2 t + x2 x −x− 2 3 2

∞    2  1  1 k 2 2 −k 2 π 2 t 1 − 1 + (−1) e cos(kπ x), − + 4 k π 6 π k4 k=1

x ∈ [0, 1], .t ≥ 0.

.

Problem 6.5. Solution. Suppose that .u1 , u2 ∈ CQT are solutions of the problem (6.22). Let .v = u1 − u2 . Then .v satisfies the problem vt − Δv = 0 in QT ,

.

v = 0 on

D × {t = 0}.

Set

E(t) =

(v(x, t))2 dx,

.

0 ≤ t ≤ T.

D

Then

'

E (t) = 2

.

v(x, t)vt (x, t)dx D

=2

v(x, t)Δv(x, t)dx D

= −2

|∇v(x, t)|2 dx D

≤ 0,

0 ≤ t ≤ T.

Hence, E(t) ≤ E(0) = 0,

.

Consequently .u1 = u2 in .QT . Problem 6.6. Hint. Use Exercise 6.19. Problem 6.7. Hint. Use Exercise 6.22. Problem 6.8. Hint. Use Exercise 6.24.

0 ≤ t ≤ T.

416

9 Solutions, Hints and Answers to the Problems

Chapter 7. Problem 7.1. Answer. 1. u(x, t) = ex cosh(2t) + xt,

.

−∞ < x < ∞,

t > 0.

2. u(x, t) = ex cosh t + e−x sinh t,

.

−∞ < x < ∞,

t > 0.

3. u(x, t) = xt,

.

−∞ < x < ∞,

t > 0.

4. u(x, t) = 1 +

.

1 sin x sin(4t), 4

−∞ < x < ∞,

t > 0.

5. u(x, t) = 1 − t,

.

−∞ < x < ∞,

t > 0.

−∞ < x < ∞,

t > 0.

Problem 7.2. Answer. 1. u(x, t) = x + t 3 ,

.

2. u(x, t) = sin x,

.

−∞ < x < ∞,

t > 0.

3. 1 u(x, t) = 1 + t + (1 − cos(3t)) sin x, 9

.

−∞ < x < ∞,

t > 0.

4. u(x, t) =

.

1 a 2 ω2

(1 − cos(aωt)) sin(aωx),

−∞ < x < ∞,

t > 0.

9 Solutions, Hints and Answers to the Problems

417

5. t 1 sin(ωt), − ω ω2

u(x, t) =

.

−∞ < x < ∞,

t > 0.

Problem 7.5. Answer. 1. u(x, t) = v(x, t) + w(x),

0 ≤ x ≤ 1,

.

t ≥ 0,

where v(x, t) =

∞ 

ak cos(kπ t) sin(kπ x),

.

k=1

1 ak = −2

w(x) sin(kπ x)dx, 0

x y

1 y f (ξ )dξ dy + x

w(x) = − 0

0

f (ξ )dξ dy + (b − a)x + a, 0

0

0 ≤ x ≤ 1, .t ≥ 0, .k ∈ N.

.

2. u(x, t) =

.

b−a 2 F0 2 x + ax + Φ0 + ψ0 t + t 2 2   ∞   1 1 + F + Φ − F k k k k2π 2 k2π 2 k=1  ψk sin(kπ t) cos(kπ x), cos(kπ t) + kπ

0 ≤ x ≤ π , .t ≥ 0, where

.

1 F k = ɛk

(f (x) + b − a) cos(kπ x)dx,

.

0

1 Φk = ɛ k

(φ(x) − (b − a)x 2 − ax) cos(kπ x)dx, 0

418

9 Solutions, Hints and Answers to the Problems

1 ψk = ɛk

k ∈ N,

ψ(x) cos(kπ x)dx, 0

ɛ0 = 1, ɛk = 2,

k ∈ N.

3. u(x, t) = x + t + cos

.

x t sin 2 2

    ∞ 2k + 1 2k + 1 8  (−1)k t sin x , cos − π 2 2 (2k + 1)2 k=0

0 ≤ x ≤ 1,

t ≥ 0.

4. u(x, t) =

.

1 e−t cosh x, sinh 1

0 ≤ x ≤ 1,

t ≥ 0.

5. t − .u(x, t) = 2



 1 + cos(2x) sin(2t), 4

0 ≤ x ≤ 1,

t ≥ 0.

Problem 7.8. Answer. 1. u(x1 , x2 , x3 , t) = x12 + x22 + x32 + 3t 2 + x1 x2 t,

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

2. u(x1 , x2 , x3 , t) = ex1 cos x2 + t (x12 − x22 ),

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

3. u(x1 , x2 , x3 ) = x12 + x22 + t + 2t 2 ,

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

4. u(x1 , x2 , x3 , t) = ex1 cosh t + e−x1 sinh t,

.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

9 Solutions, Hints and Answers to the Problems

419

5. u(x1 , x2 , x3 , t) = 2x1 − 3x2 + 4x3 + 3t,

.

Problem 7.9. Hint. Use the Kirchhoff formula. Problem 7.10. Hint. Use the Poisson formula.

(x1 , x2 , x3 ) ∈ R3 ,

t ≥ 0.

Index

B Backward wave, 242 Beltrami differential equations, 65 C Canonical form of first order system, 124 Cauchy data, 27 Characteristic, 22 Characteristic curve, 22 Characteristic curves of linear hyperbolic differential operator, 86 Characteristic curves of linear parabolic operator, 76 Characteristic direction, 22 Characteristic equation of linear parabolic operator, 76 Characteristic equations, 22, 86 Characteristic surface, 14 Classical solution, 3, 243 Coefficients matrix of principal part, 109 D D’Alambert’s formula, 244, 251 Diffusion equation, 50 Dirichlet boundary conditions, 140, 256 Dirichlet boundary problem, 140 E Elliptic first order system, 125 Elliptic-hyperbolic first order system, 125 Elliptic linear differential operator, 59 Elliptic second order equation, 111

F First order partial differential equation, 13 Forward wave, 242 Fundamental solution of the heat equation, 197 Fundamental solution of the Laplace equation, 145

G Generalized solution, 242 Green function, 154

H Hadamard example, 142 Harmonic function, 137 Harmonic polynomial, 143 Harnack Theorem, 161 Heat equation, 50 Homogeneous harmonic polynomial, 143 Hopf equation, 14 Hyperbolic first order system, 125 Hyperbolic linear differential operator, 59 Hyperbolic second order equation, 111

I Ill-posed problem, 2 Initial condition, 27 Initial curve, 27 Initial data, 27 Integral of Pfaffian equation, 31 Integral surface, 21 Integrating factor of Pfaff equation, 34

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. G. Georgiev, An Excursion Through Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-031-48784-2

421

422 K Kirchhoff formula, 315 Kolmogorov equation, 50

L Laplace equation, 50 Linear differential operator, 56 Linear first order PDE, 15 Linear homogeneous first order PDE, 15 Linear homogeneous second order PDE, 50 Linear nonhomogeneous first order PDE, 15 Linear nonhomogeneous second order PDE, 50 Linear partial differential equation, 3 Linear second order PDE, 49 Liouville Theorem, 161

M Mean-value formula for harmonic functions for a ball, 151 Mean-value formula for harmonic functions for a sphere, 151

N Neumann boundary conditions, 141, 256 Neumann boundary problem, 141 Nonlinear first order PDE, 15 Nonlinear partial differential equation, 3 Nonlinear poisson equation, 47 Nonlinear second order PDE, 51

O One dimensional integral of Pfaffian equation, 31 Order of equation, 2

P Parabolic first order system, 125 Parabolic linear differential operator, 59 Parabolic second order equation, 111

Index Pfaffian equation, 31 Poisson equation, 140 Poisson formula, 160, 324 Projected characteristic curve, 22 Proper solution, 243

Q Quasilinear first order PDE, 14 Quasilinear second order PDE, 48

R Radially symmetric solutions, 304

S Second order partial differential equation, 47 Semilinear first order PDE, 14 Semilinear second order PDE, 48 Separation constant, 210 Strong solution, 3 Superposition principle, 7 Symmetry property of the Green function, 155

T The mean value formula, 222 The strong maximum principle for the heat equation, 228 Transport equation, 13 Two dimensional integral of Pfaff equation, 31

U Ultrahyperbolic second order equation, 111

W Wave equation, 50 Weak maximum principle for the heat equation, 227 Weak solution, 3 Well-posed problem, 2