Integral Equations and Integral Transforms [1 ed.] 9789819963591, 9789819963607

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Sudeshna Banerjea Birendra Nath Mandal

Integral Equations and Integral Transforms

Integral Equations and Integral Transforms

Sudeshna Banerjea · Birendra Nath Mandal

Integral Equations and Integral Transforms

Sudeshna Banerjea Department of Mathematics Jadavpur University Kolkata, West Bengal, India

Birendra Nath Mandal Physics and Applied Mathematics Unit Indian Statistical Institute Kolkata, West Bengal, India

ISBN 978-981-99-6359-1 ISBN 978-981-99-6360-7 (eBook) https://doi.org/10.1007/978-981-99-6360-7 Mathematics Subject Classification: 45A05, 45B05, 45D05, 45E10, 44A05, 44A10, 42A38, 44A15 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 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 Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore Paper in this product is recyclable.

Preface

This textbook introduces the theory of linear integral equations and integral transforms in a simple way with appropriate illustrative examples. It is comprehensible to any interested reader having basic knowledge in real analysis, complex analysis and functional analysis. The proposed book is a textbook meant for M.Sc. students of Mathematics and Physics in universities in India and B.Tech./M.Tech. students of engineering institutions in India. It has two parts. Part I deals with linear integral equations and consists of five chapters (Chaps. 1–5) while Part II deals with integral transforms and consists of seven chapters (Chaps. 6–11). Chapter 1 involves a brief history, classification and occurrence of integral equations. Chapter 2 deals with solution of Fredholm integral equations of second kind with degenerate kernel, Fedholm alternative. Second kind integral equations with more general kernels and their various methods of solution, Fredholm theorems, etc., have been discussed in Chap. 3. Chapter 4 discusses second kind integral equations with symmetric kernels and various related results, eigenvalues and eigenfunctions, Hilbert Schmidt theorem, etc. Chapter 5 considers solution of Abel integral equation by using elementary methods. Each chapter is supplemented with illustrative examples. Chapters 6–10 describe integral transforms such as Fourier, Laplace, Mellin, Hankel and Z-transform. Each of Chaps. 6–10 describes how a particular integral transform together with its inverse transform formula is obtained, evaluation of integral transforms of some simple functions, properties, evaluation of inverse transforms, application to appropriate physical problems. Also an adequate number of illustrative examples are included in each chapter. Chapter 11 describes a general method for the construction of various integral transforms and their inverses. The materials in this chapter are somewhat new and are not available in any text book on integral transforms.

v

vi

Preface

While teaching the post graduate course in Mathematics, it was observed that there is a dearth of textbooks covering both the topics on the theories of Integral Equations and Integral Transforms. This was the motivation behind the idea of writing this book so that it is useful as a source book for teaching M.Sc. courses in Mathematics, Physics and B.Tech./M.Tech. courses in Engineering. Kolkata, India December 2022

Sudeshna Banerjea Birendra Nath Mandal

Contents

Part I

Integral Equations

1

Integral Equations: An Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 What is an Integral Equation? . . . . . . . . . . . . . . . . . . . . . . 1.1.2 Classifications of Integral Equations . . . . . . . . . . . . . . . . . 1.2 Occurrence of Integral Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Occurrence of Volterra Integral Equations . . . . . . . . . . . . 1.2.2 Occurrence of Fredholm Integral Equations . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 3 3 6 8 8 10 14 16

2

Fredholm Integral Equation of the Second Kind with Degenerate Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Integral Equation with Degenerate Kernel . . . . . . . . . . . . . . . . . . . . 2.2 Homogeneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Nonhomogeneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17 17 23 29 34 36

3

Integral Equations of Second Kind with Continuous and Square Integrable Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Fredholm Integral Equations of Second Kind with Continuous Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Volterra Integral Equations of Second Kind with Continuous Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Illustrative Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Iterated Kernels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Fredholm Theory for Integral Equation with Continuous Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

37 37 47 50 53 59

vii

viii

Contents

3.6

4

5

Fredholm Integral Equations of Second Kind with Square Integrable Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.1 Some Important Properties of Square Integrable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.2 Method of Solution of Integral Equation with Square Integrable Kernel . . . . . . . . . . . . . . . . . . . . . . 3.7 Fredholm Theory for Integral Equation with Square Integrable Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

73 75 79

Integral Equations of the Second Kind with a Symmetric Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Symmetric Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Properties of Integral Equations with a Symmetric Kernel . . . . . . 4.3 Hilbert–Schmidt Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

81 81 82 86 95 96

67 68

Abel Integral Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 5.1 Solution Based on Elementary Integration . . . . . . . . . . . . . . . . . . . 97 5.2 Solution Based on Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . 101 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

Part II 6

66

Integral Transform

Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Integral Transform: An Introduction . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Fourier Integral Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Rigorous Justification of Fourier Integral Theorem . . . . . . . . . . . . 6.4 Fourier Cosine and Sine Transforms . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Fourier Transforms of Some Simple Functions . . . . . . . . . . . . . . . 6.6 Properties of Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 Convolution Theorem and Parseval Relation . . . . . . . . . . . . . . . . . 6.8 Fourier Transforms in Two or More Dimensions . . . . . . . . . . . . . . 6.9 Application of Fourier Transforms in Solving Linear Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.10 Application of Fourier Sine and Cosine Transforms in Solving Linear Ordinary Differential Equations . . . . . . . . . . . . . 6.11 Application to Partial Differential Equations . . . . . . . . . . . . . . . . . 6.12 Application of Fourier Sine and Cosine Transform to the Solution of Partial Differential Equations . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

109 109 110 112 114 115 120 125 133 134 140 144 148 153 159

Contents

7

ix

Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Derivation of Laplace Transform from Fourier Integral Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Laplace Inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Operational Properties of Laplace Transform . . . . . . . . . . . . . . . . . 7.4 Laplace Convolution Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Tauberian Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Method of Evaluation of Inverse Laplace Transform . . . . . . . . . . . 7.7 Application of Laplace Transform in Solving Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8 Application Laplace Transform in Solving Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

161

188 191 199

8

Mellin Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Formal Derivation of Mellin Transform . . . . . . . . . . . . . . . . . . . . . . 8.3 Theorem on Inversion of Mellin Transform . . . . . . . . . . . . . . . . . . 8.4 Properties of Mellin Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Mellin Transform of Some Simple Functions . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

201 201 202 203 205 209 214 217

9

Hankel Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Formal Derivation of Hankel Transform . . . . . . . . . . . . . . . . . . . . . 9.2 Properties of Hankel Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Hankel Transform of Some Known Functions . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

219 219 226 229 236 238

10 Z Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Properties of Z-Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Inverse Z-Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

239 239 241 245 251 253

11 Formal Construction of Integral Transforms and Their Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 The Method of Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Construction of Some Familiar Integral Transforms . . . . . . . . . . . 11.2.1 Fourier Integral Transform . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.2 Fourier Cosine Transform . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.3 Fourier Sine Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . .

255 255 257 257 259 260

161 166 168 173 174 176 183

x

Contents

11.2.4 Havelock Transform (Mixed Fourier Transform, Hybrid Fourier Transform) . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.5 Finite Fourier Sine Transform . . . . . . . . . . . . . . . . . . . . . . 11.2.6 Finite Fourier Cosine Transform . . . . . . . . . . . . . . . . . . . . 11.2.7 Finite Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.8 Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

261 263 264 264 264 265

About the Authors

Sudeshna Banerjea is Professor at the Department of Mathematics, Jadavpur University, Kolkata. She obtained her M.Sc. degree and Ph.D. degree in Applied Mathematics from the University of Calcutta, respectively, in 1985 and 1992. She was NBHM Post-doctoral Fellow at the Indian Statistical Institute, Kolkata, in 1993, before joining Jadavpur University as Lecturer in 1993. She is Fellow of the West Bengal Academy of Sciences since 2011. Her research interest includes integral equations, theory of water waves and associated mathematical methods. With more than 68 research papers, she has supervised more than eight Ph.D. students. Earlier, she was Principal Investigator for a number of major research projects sanctioned by NBHM, DST and CSIR. Birendra Nath Mandal is former NASI Senior Scientist Platinum Jubilee Fellow at the Indian Statistical Institute (ISI), Kolkata (from 2009 to 2014), where he has been Honorary Professor (from 2006 to 2008), and Faculty (from 1989 to 2005). He also had taught at the University of Calcutta (from 1970 to 1989). He earned his M.Sc. degree and Ph.D. degree in Applied Mathematics, respectively, in 1966 and 1973, from the University of Calcutta. He was Post-doctoral Commonwealth Fellow at Manchester University, England, from 1973 to 1975. His research work comprises several areas of applied mathematics including water waves, integral transforms, integral equations, inventory problems, wavelets, etc. He has supervised 27 Ph.D. students and published more than 300 research papers. He was also Chief Editor of the OPSEARCH journal (Springer) and is on the editorial board of a number of reputed journals. He co-authored and co-edited a number of advanced level research monographs published by reputed publishers.

xi

Part I

Integral Equations

Chapter 1

Integral Equations: An Introduction

1.1 Introduction The theory of Integral Equations constitute an important topic in Mathematics as this is one of the most useful mathematical tools in both pure and applied mathematics. Integral equations arise in a natural way in course of solving the initial and boundary value problems associated with mathematical modeling of physical phenomena. The solutions of integral equations play an important role to understand the qualitative features of the physical phenomena in the natural sciences. The theory of integral equations originated while studying a problem of mechanics by N.H. Abel in 1826. Abel reduced the problem of finding the path of descent of a particle along a smooth vertical curve under the action of gravity in an interval of time, to an integral equation. Later V. Volterra in 1896 developed the general theory of solution of a class of linear integral equations which were named after him as Volterra integral equations. In this type of integral equations one of the limits of integration is variable. In 1900, I. Fredholm developed the theory of integral equation in which the limits of integration are constants and these integral equations are named after him as Fredholm integral equations.

1.1.1 What is an Integral Equation? Before answering this question, let us consider the classical problem of mechanics considered by Abel in 1826. A smooth curve x = ψ(y), 0 < y < b, ψ(0) = 0 is held fixed in a vertical (x, y) plane as shown in Fig. 1.1. The problem is to determine the time a particle of mass m takes to slide freely under gravity down the curve from any point (X,Y) on the curve to the lower most point, the origin.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 S. Banerjea and B. N. Mandal, Integral Equations and Integral Transforms, https://doi.org/10.1007/978-981-99-6360-7_1

3

4

1 Integral Equations: An Introduction

Fig. 1.1 Illustration of Abel’s problem

Assuming that the particle started to slide from rest from the point (X,Y), we have from conservation of energy that 1 2 mv + mg y = mgY, 2

(1.1)

where v is the speed of the particle at time t, when its position is given by the coordinates (x, y) and g represents acceleration due to gravity. , where s represents the arc-length, measured Noting that the speed v as v = ds dt from the origin to the point (x, y), we observe that we can express the relation (1.1) as ds 1 = −[2g(Y − y)] 2 . (1.2) dt The minus sign is used because the particle is descending and therefore s decreases with time t. We now use the well-known formula ds 1 = [1 + (ψ  (y))2 )] 2 , dy where ψ  (y) =

dψ , dy

(1.3)

to express (1.2) in the form: 1

ds/dt [2g(Y − y)] 2 dy = =− 1 . dt ds/dy [1 + (ψ  (y))2 )] 2

(1.4)

Integrating the relation (1.4), we easily obtain 

0 Y



1 + (ψ  (y))2 2g(Y − y)

 21



T

dy = −

dt = −T,

(1.5)

0

where T is the total time of fall of the particle, from the point (X, Y ) to the origin (0,0). Now, writing

1.1 Introduction

5

 φ(y) =

1 + (ψ  (y))2 2g

 21 (1.6)

and T = f (Y ),

(1.7)

dy = f (Y ), 0 ≤ Y ≤ b.

(1.8)

the relation (1.5) can be written as 

φ(y)

Y

1

(Y − y) 2

0

It may be noted that f (0) = 0. Thus we find that the problem of determining the time of descent T of the particle can be solved completely, by using the formula (1.8), if the equation ( or shape) of the curve x = ψ(y), is known and hence the function φ(y), is known. Now let us consider the converse problem which is known as the historic Abel’s problem. Suppose the time of fall T [= f (Y )] is given and we are to determine the shape of the curve which is related to the function φ(y). It is important to note from Eq. (1.8) that the unknown function φ(y) appears under the sign of integration and we say that Eq. (1.8) is an integral equation which is known as Abel’s integral equation. We say that an integral equation is an equation where the unknown function appears under the sign of integration. Followings are some examples of integral equations satisfied by the unknown function u(x), where the functions K (x, t) and f (x) are known.  b K (x, t)u(t)dt = h(x), a ≤ x ≤ b, 1. u(x) + a



b

K (x, t)u(t)dt = h(x),

a ≤ x ≤ b,

K (x, t)u(t)dt = h(x),

a ≤ x ≤ b,

2. a



a

3. x



x

4. u(x) +

K (x, t)u(t)dt = h(x),

a ≤ x ≤ b,

b



b

5. u(x) +

K (x, t)[u(t)]3 dt = h(x), a ≤ x ≤ b,

a

 6. [u(x)] + 2

a

b

K (x, t)[u(t)]3 dt = h(x), a ≤ x ≤ b.

6

1 Integral Equations: An Introduction

1.1.2 Classifications of Integral Equations In the examples shown above, it is observed that the unknown function u(x) appears linearly in Examples 1–4 and nonlinearly in Examples 5 and 6. Based on these observations, we may classify an integral equation into linear and nonlinear integral equations. An integral equation in which the unknown function appears linearly is called a linear integral equation, otherwise it is a nonlinear integral equation. Here we shall restrict our discussion on linear integral equations only. In the above illustration of linear integral equations, we make the following observations: 1. Examples 1 and 4 show that the unknown function u(x) appears under an integral as well as freely without an integral. Also in Examples 2 and 3 the unknown function appears only under an integral. We say that integral equations in Examples 1 and 4 are linear integral equations of second kind while the integral equations in Examples 2 and 3 are linear integral equations of first kind. 2. It is observed that in Examples 1 and 2, the limits of integration are constants while in Examples 3 and 4, one of the limits of integration is variable. The integral equations in Examples 1 and 2 are known as Fredholm integral equation and that in Examples 3 and 4 are known as Volterra integral equations. We now summarize the above observations for a linear integral equation. The general one-dimensional linear integral equation for an unknown function ϕ(x) is of the form  βϕ(x) + λ

b

K (x, t)ϕ(t)dt = f (x), a ≤ x ≤ b

(1.9)

a

where the constant β may be either 0 or 1. The functions K (x, t) and f (x) are known functions which are termed as the kernel of the integral equation and forcing function respectively. The known constant λ is the parameter of the integral equation. If β = 0, then the integral equation (1.9) is called integral equation of first kind, i.e.,  b K (x, t)ϕ(t)dt = f (x), a ≤ x ≤ b. a

The integral equation (1.9) in which β = 1 is called integral equation of second kind, i.e.,  b ϕ(x) + λ K (x, t)ϕ(t)dt = f (x) a ≤ x ≤ b. a

1.1 Introduction

7

An integral equation is homogenous when the forcing term f (x) is equal to zero, otherwise it is a nonhomogeneous integral equation. The integral equation (1.9) is known as Fredholm integral equation, if the limits of integration are constants. If any one limit of integration is a known function of x, then the integral equation (1.9) is called Volterra integral equation. An integral equation can be further classified depending on the behavior of the functions K (x, t) and f (x) in the domain in which it is defined. If the kernel K (x, t) of integral equation (1.9) is such that 

b



a

b

|K (x, t)|2 d xdt

a

has a finite value, then the kernel is known as regular kernel and the corresponding integral equation is known as a regular integral equation otherwise it is a singular integral equation. The integral equation (1.9) is known as weakly singular integral equation if the kernel K (x, t) is of the form K (x, t) =

k(x, t) (x − t)α

where 0 < α < 1 and k(x, t) is differentiable in [a, b] × [a, b] with k(x, x) = 0. The kernel in this case is a weakly singular kernel. If α ≥ 1 then the kernel is strongly singular and the integral equation (1.9) is a strongly singular integral equation. The integral equation (1.9) is a Cauchy type singular integral equation, if α = 1. In this case, the integral is to be understood in the sense of Cauchy principal value as given by 

b

 K (x, t)ϕ(t)dt = lim

→0+

a

x−

 K (x, t)ϕ(t)dt +

a

b

 K (x, t)ϕ(t)dt .

x+

The integral equation (1.9) is known as hypersingular integral equation if α > 1.  b ψ(t) A hypersingular integral a (x−t) a ≤ x ≤ b is understood in the sense of 2 dt, two-sided Hadamard finite part integral of order two defined by  a

b

ψ(t) dt = lim →0+ (x − t)2

 a

x−

ψ(t) dt + (x − t)2 −



b x+

ψ(t) dt (x − t)2

 ψ(x + ) + ψ(x − ) . 

8

1 Integral Equations: An Introduction

1.2 Occurrence of Integral Equations 1.2.1 Occurrence of Volterra Integral Equations The initial value problems for ordinary differential equations (ODEs) can be reduced to Volterra type integral equations as shown below. Example 1 Let the function φ(x) satisfy the following initial value problem y  (x) = ψ(x, y(x)), 0 ≤ x ≤ 1, y(0) = y0 . We assume that the function ψ is integrable for all continuous functions y(x), in the closed interval [0, 1]. Integrating from 0 to x we obtain  x ψ(t, y(t))dt + y0 , 0 ≤ x ≤ 1, y(x) = 0

which is a Volterra integral equation of second kind, and in general nonlinear. Example 2 Consider the function y(x), which satisfies the following second-order ordinary differential equation (ODE) y  (x) + a(x)y  (x) + b(x)y(x) = f (x), 0 ≤ x ≤ 1,

(1.10)

with the two initial conditions y(0) = y0 , y  (0) = y  0 .

(1.11)

Here we assume that the function a  (x), b(x), y  (x) are continuous in [0,1] and a(x), b(x) are known functions. Integrating the given differential equation from 0 to x and using the initial conditions we obtain  x [{a  (t) − b(t)}y(t) + f (t)]dt. y  (x) = y0 − a(x)y(x) + a(0)y0 + 0

Integrating again and using the initial condition y(0) = y0 we obtain y(x) = x(y0 + a(0)y0 ) −

 x 0

a(t)y(t)dt +

 x s 0

0

[{a  (t) − b(t)}y(t) + f (t)]dtds.

This can be further simplified by interchanging the order of integration (cf Fig. 1.2) in a standard manner to

1.2 Occurrence of Integral Equations

9

Fig. 1.2 Interchange of order of integration

y(x) +

 x 0

[a(t) − (x − t){a  (t) − b(t)}]y(t)dt = x(y0 + a(0)y0 ) +

 x 0

(x − t) f (t)]dt.

(1.12)

Equation (1.12) is a Volterra integral equation of second kind. Thus an initial value problem gives rise to a Volterra integral equation of second kind. Examples Example 1 Consider the following initial value problem y  (x) = x y(x), 0 ≤ x ≤ 1 with

y(0) = 1, y  (0) = 0.

Solution: Integrating twice and using the initial conditions, it can be easily seen that this problem is equivalent to the following Volterra integral equation of second kind.  y(x) = 1 +

x

(x − t)t y(t)dt, 0 ≤ x ≤ 1.

0

Example 2 Convert the following initial value problem to an integral equation (x + 1)2 y  (x) + 2(x + 1)y  (x) = f (x) y(0) = α, y  (0) = β, f (x) being a known function.

0 ≤ x ≤ 1,

10

1 Integral Equations: An Introduction

Solution: The given differential equation is (t + 1)2 y  (t) + 2(t + 1)y  (t) = f (t). Integrating with respect to t from 0 to x and using the initial condition y  (0) = β we obtain  x (x + 1)2 y  (x) = f (t)dt + β. 0

Integrating once again and using the boundary condition y(0) = α we obtain 

x

(x + 1)2 y(x) = α + βx + 2



x

(t + 1)y(t)dt +

0

0



s

f (t)dtds.

0

Interchanging the order of integration in the last term, we finally obtain the following Volterra integral equation of second kind. 

x

(x + 1) y(x) − 2 2

 (t + 1)y(t)dt = α + βx +

0

x

(x − t) f (t)dt, 0 ≤ x ≤ 1.

0

1.2.2 Occurrence of Fredholm Integral Equations Consider the following boundary value problem for an ordinary differential equation of second order (1.13) y  (x) = χ(x, y(x)), 0 ≤ x ≤ 1, with the boundary conditions y(0) = y0 , y(1) = y1 . Here χ is a continuous function, y0 and y1 are known constants. The differential equation is given by y  (t) = χ(t, y(t)), 0 ≤ t ≤ 1. Integrating with respect to t from 0 to x and writing y  (0) = A we get y  (x) = A +



x 0

Here A is a constant to be determined.

χ(t, y(t))dt.

(1.14)

1.2 Occurrence of Integral Equations

11

Again integrating and using the boundary condition y(0) = y0 , we obtain  x s y(x) = χ(t, y(t))dtds + Ax + y0 . 0

0

Interchanging the order of integration, we obtain  x y(x) = (x − t)χ(t, φ(t))dt + Ax + y0 .

(1.15)

0

Using the second boundary condition in (1.14), we obtain the constant A as  1 (1 − t)χ(t, y(t))dt. (1.16) A = y1 − y0 − 0

Substituting this value of A in the relation (1.15), and rearranging the various terms, we obtain the following Fredholm integral equation of second kind 

1

y(x) = −

k(x, t)χ(t, y(t))dt + (y1 − y0 )x + y0 ,

(1.17)

0



where k(x, t) =

t (1 − x), t ≤ x, x(1 − t), t ≥ x.

(1.18)

We thus observe that the boundary value problems for ordinary differential equations can be cast into Fredholm integral equation of second kind. Examples Here we shall consider some boundary value problems and reduce them to Fredholm integral equation of second kind. Example 1 Consider the following ordinary differential equation φ (x) = −λφ(x), 0 ≤ x ≤ 1,

(1.19)

with the boundary conditions φ(0) = φ0 , φ(1) = φ1 .

(1.20)

Solution: From Eq. (1.13), we may observe that χ(x, φ(x)) = −λφ(x),

(1.21)

where λ is a constant. The boundary value problem given by Eqs. (1.19) and (1.20) can be cast into the following Fredholm integral equation.

12

1 Integral Equations: An Introduction



1

φ(x) = λ

k(x, t)φ(t)dt + (φ1 − φ0 )x + φ0

(1.22)

0



where k(x, t) =

t (1 − x), t ≤ x, x(1 − t), t ≥ x.

We mention here that the kernel k(x, t) for the above integral equation is a symmetric real function, satisfying the relationship k(x, t) = k(t, x)

(1.23)

and, it is for this reason that the integral equation (1.22) is regarded as a symmetric integral equation. Example 2 Reduce the following boundary value problem to a Fredholm integral equation y  (x) + y = 0, 0 ≤ x ≤ 1, y(0) = 0, y  (1) + 3y(1) = 1. Solution: On integration the above differential equation can be reduced to  x   y (x) = y (0) − y(t)dt.

(1.24)

0

This again on integration yields y(x) = x y  (0) −



x

(x − t)y(t)dt

(1.25)

0

after using the boundary condition y(0) = 0. For x = 1, we have from Eq. (1.24) y  (0) = y  (1) +



1

y(t)dt.

(1.26)

0

Substituting y  (0) into Eq. (1.25) gives 



y(x) = x[y (1) +

1



x

y(t)dt] −

0

(x − t)y(t)dt.

0

For x = 1, using the second boundary condition in Eq. (1.27) we obtain 



1

4y (1) = 1 − 3

t y(t)dt. 0

(1.27)

1.2 Occurrence of Integral Equations

13

Using y  (1) in Eq. (1.27) we obtain the Fredholm integral equation x y(x) = + 4



1

[min(x, t) −

0

3xt ] y(t)dt, 0 ≤ x ≤ 1. 4

Example 3 x y  (x) + y  (x) + y(x) sin x = f (x) 1 ≤ x ≤ 2, y(1) + y(2) = 0, y  (1) + y  (2) = 0. Solution: Given differential equation is rewritten as d {t y  (t)} + y(t) sin t = f (t) 1 ≤ t ≤ 2. dt Integrating with respect to t from 1 to x, we have x y  (x) − y  (1) +



x



1

At x = 2 2y  (2) − y  (1) +



x

y(t) sin tdt =

f (t)dt.

(1.28)

1

2



2

y(t) sin tdt =

1

f (t)dt.

1

Using the boundary condition y  (1) + y  (2) = 0 we have 1 y (1) = 3 



2

[y(t) sin t − f (t)]dt.

(1.29)

1

Again integrating equation (1.28) and simplifying by changing the order of integration, we have  x  x  y(t)dt − (x − 1)y (1) + (x − t)(y sin t − f (t))dt = 0. x y(x) − y(1) − 1

1

(1.30)

At x = 2, we have 

2

2y(2) − y(1) − 1

y(t)dt − y  (1) +



2 1

(2 − t)(y sin t − f (t))dt = 0.

14

1 Integral Equations: An Introduction

Using the boundary condition y(1) + y(2) = 0 we have y(1) =

1 3



2 1

1 5 [( − t) sin t − 1]y(t)dt + 3 3



2

1

5 (t − ) f (t)dt. 3

(1.31)

Substituting into y  (1) from Eq. (1.29) and y(1) from Eq. (1.31) into Eq. (1.30), we obtain after simplification the following Fredholm integral equation of second kind. 1 3

x y(x) −



k(x, t)y(t)dt = F(x), 1 ≤ x ≤ 2

1



where

2

k(x, t) =

B(x, t) + 3 − 3(x − t) sin t B(x, t) t > x,

t < x,

2 B(x, t) = ( + x − t) sin t − 1, 3 and F(x) = −

1 3



2 1

2 [(x − t) + ] f (t)dt + 3



x

(x − t) f (t)dt.

1

Exercises 1. Reduce the following boundary value problem to a Fredholm integral equation d2 y + y = 0, 0 ≤ x ≤ 1 y(0) = 0, y  (1) + 3y(1) = 1. dx2 Answer: y(x) =

4 − 3x 4



x

t y(t)dt +

0

x 4



1 x

t (4 − 3t)y(t)dt + . 4

2. Obtain an integral equation corresponding to initial value problem d2 y dy − 3y = 0 0 ≤ x ≤ 1, y(0) = 0, y  (0) = 0. − 2x 2 dx dx 

Answer:

x

y(x) = 0

(x + t)y(t)dt.

1.2 Occurrence of Integral Equations

15

3. Obtain an integral equation corresponding to initial value problem d2 y + (1 + x 2 )y = sin x, 0 ≤ x ≤ 1, y(0) = 1, y  (0) = 0. dx2 

Answer:

x

y(x) = (1 + x − sin x) −

(1 + t 2 )(x − t)y(t)dt.

0

4. Convert the following boundary value problem to an integral equation y  (x) + y(x) = f (x), 0 ≤ x ≤ 1 y(0) = 0, y(1) + y  (1) = 4 where f (x) is a known function. Answer:  y(x) −

1

k(x, t)y(t)dt =

0

4x − 2



1

k(x, t) f (t)dt, 0 ≤ x ≤ 1

0

⎧ x ⎪ ⎨ t (1 − ), t < x, 2 k(x, t) = t ⎪ ⎩ x(1 − ), t > x. 2 5. Obtain an integral equation for y  (x) + λy(x) = 0

0≤x ≤1

ay(0) = by(1), by  (0) = ay  (1) where a, b are constants. Answer:



1

y(x) + λ

K (x, t)y(t)dt = 0, 0 ≤ x ≤ 1

0

 k(x, t) = L(x, t) =

L(x, t), t > x, L(x, t) + (x − t), t < x. 1 [ax − bt − ab + b]. a−b

16

1 Integral Equations: An Introduction

References The content of this chapter is mainly based on the website epgp.inflibnet.ac.in/ of UGC E Pathsala and on the following literature. 1. A. Chakrabarti, Applied Integral Equations (Vijay Nicole Imprints Pvt Ltd, 2008) 2. A. Wazwaz, A First Course in Integral Equations (World Scientific Publishing Co. Pvt Ltd, 1997) 3. R.P. Kanwal, Linear Integral Equations Theory and Techniques (Academic Press Inc, 1971) 4. D. Porter, D.S.G. Stirling, Integral Equations (Cambridge University Press, 1971)

Chapter 2

Fredholm Integral Equation of the Second Kind with Degenerate Kernel

Consider the Fredholm integral equation of the second kind, given by  φ(x) − λ

b

k(x, t)φ(t)dt = f (x), a ≤ x ≤ b, λ is a constant

(2.1)

a

which can be rewritten in the operator form as (I − λK )φ = f,

(2.2)

where I is the identity operator and K is the linear operator defined by the relation  (K φ)(x) =

b

k(x, t)φ(t)dt, a ≤ x ≤ b.

(2.3)

a

In this chapter, we shall discuss about the method of solution of integral equation (2.1) where the kernel k(x, t) is degenerate.

2.1 Integral Equation with Degenerate Kernel What is meant by a degenerate kernel? The kernel k(x, t) of the integral equation (2.1) is said to be degenerate or separable if it has the form n  ai (x)bi (t), a ≤ x, t ≤ b, (2.4) k(x, t) = i=1

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 S. Banerjea and B. N. Mandal, Integral Equations and Integral Transforms, https://doi.org/10.1007/978-981-99-6360-7_2

17

18

2 Fredholm Integral Equation of the Second Kind with Degenerate Kernel

where ai (x) and bi (t), i = 1, 2, . . .n are two finite, linearly independent sets of continuous functions on [a, b]. Method of Solution In this section, we will describe the method of solution of the integral equation (2.1) with degenerate kernel k(x, t) given by Eq. (2.4) by applying basic concept of linear algebra. We replace k(x, t) in Eq. (2.1) by the degenerate kernel given in Eq. (2.4), so that the integral equation takes the form n 

φ(x) = f (x) + λ

ci ai (x), a ≤ x ≤ b,

(2.5)

i=1

where



b

ci =

bi (t)φ(t)dt.

(2.6)

a

Substituting the expression for φ from (2.5) into the relation (2.6), we obtain  ci =

b

bi (t)[ f (t) + λ

a

 =

b

n 

bi (t) f (t)dt + λ

a

n 

 cj

j=1

= βi + λ

c j a j (t)]dt

j=1

n 

b

bi (t)a j (t)dt

a

di j c j , i = 1, 2, . . .n,

(2.7)

j=1

where  b βi = a bi (t) f (t)dt, b di j = a bi (t)a j (t)dt, i, j = 1, 2, . . .n.

(2.8)

Equation (2.7) gives rise to system of linear algebraic equations in ci (i = 1, 2, . . . n) given by n  (δi j − di j )c j = βi , (i = 1, 2, . . . n) j=1

which can be alternatively written in usual matrix notation as → − → − DC = β

(2.9)

2.1 Integral Equation with Degenerate Kernel

19

where D = I − λA,

(2.10)

− → I being the n × n identity matrix, A being the square matrix (di j )n×n , C = − → (c1 , c2 , . . .cn )T and β = (β1 , β2 , . . .βn )T . We now consider the following two cases depending on whether the matrix D is singular or not. Case I The Matrix D is Nonsingular In this case D −1 = (I − λA)−1 exists and the unique solution of the system of linear Eq. (2.9) is given by − → C = D −1 β and the homogeneous system − → DC =0 corresponding to (2.9) has only trivial solution. Thus applying Cramer’s rule, the solution of Eq. (2.9) is obtained as   1 − λd11 −λd12   −λd21 −λd22   . .   . . ci =  . .   . .   . .   −λdn1 −λdn2

. . . −λd1i−1 . . . −λd2i−1 . . . . . . . . . . . . . −λdni−1

i = 1, 2, . . ., n where

β1 −λd1i+1 β2 −λd2i+1 . . . . . . . . . . βn −λdni+1

 . . . −λd1n  . . . λd2n  . .  . .  /J (λ), . .  . .  . .  . . . −λdnn 

J (λ) = det(D)

(2.11)

(2.12)

and J (λ) = 0, since the matrix D is nonsingular. The relation (2.11) can be rewritten if the following form as ci =

 n

 J ji β j /J (λ)

(2.13)

j=1

where J ji is the cofactor of ( ji)th element of the matrix D. Hence knowing ci from the relation (2.13), the solution of the integral equation (2.5) is given by

20

2 Fredholm Integral Equation of the Second Kind with Degenerate Kernel

 λ  ai (x) J ji β j . J (λ) i=1 j=1 n

φ(x) = f (x) +

n

(2.14)

This can be further simplified after replacing β j from first relation of (2.8) and this yields φ(x) = f (x) +

λ J (λ)



b

f (t)

 n

a

ai (x)

i=1

n 

 J ji b j (t) dt.

(2.15)

j=1

Equation (2.15) can be alternatively written as  φ(x) = f (x) + λ

b

f (t)R(x, t, λ)dt

(2.16)

a

where R(x, t, λ) =

 n i=1

ai (x)

n 

 J ji b j (t) /J (λ).

(2.17)

j=1

The function R(x, t, λ) is called the resolvent kernel of the integral equation (2.1). Thus φ(x) given by Eq. (2.16) is the solution of integral equation (2.1) with degenerate kernel when J (λ) = 0. In this case, the homogeneous integral equation corresponding to Eq. (2.1), i.e.,  b k(x, t)φ(t)dt = 0 (2.18) φ(x) − λ a

where k(x, t), given by Eq. (2.4), has only trivial solution. Remark 2.1 Thus we observe that if the homogeneous Fredholm integral equation of second kind (2.1) with degenerate kernel (2.4) has only trivial solution then the corresponding non homogeneous Fredholm integral equation of second kind with degenerate kernel has unique solution as given by the relation (2.16). Case II The Matrix D is Singular In this case, from linear algebra we know that the homogeneous system of linear equations − → DC =0 (2.19) corresponding to the system (2.9) has nontrivial solution and the system of Eq. (2.9) is either inconsistent or it has non unique solution. The non-unique solution of the − → system (2.9) will exist if and only if any solution ξ = (ξ1 , ξ2 , . . ., ξn )T of the adjoint homogeneous system of equation D∗ξ = 0

(2.20)

2.1 Integral Equation with Degenerate Kernel

satisfies the relation

n 

21

ξi βi = 0.

(2.21)

i=1

Here

D ∗ = (I − λA)T

(2.22)

is the adjoint operator of D and “bar ” denotes the complex conjugate. Equation (2.21) takes the following form after replacing βi from relation (2.8) 

 ξi

b

bi (t) f (t)dt = 0.

a

i

Writing

n 

ξi bi (t),

(2.23)

the above relation can be rewritten as  b f (t)ψ(t)dt = 0.

(2.24)

ψ(t) =

i=1

a

The system of Eq. (2.20) can be rewritten as [I − λA]T ξ = 0 which produces ξi − λ

n 

d ji ξ j = 0.

(2.25)

j=1

Multiplying Eq. (2.25) by bi (x) and summing over i from 1 to n we obtain n 

ξi bi (x) − λ

i=1

n  n 

bi (x)d ji ξ j = 0,

i=1 j=1

which after replacing the expression for d ji from the second equation of (2.8) becomes n  i=1

 ξi bi (x) − λ a

b

 n i=1

a i (t)bi (x)

n  j=1

 ξ j b j (t) dt = 0.

22

2 Fredholm Integral Equation of the Second Kind with Degenerate Kernel

Using the relations (2.24) and (2.4), we see that ψ(x) satisfies 

b

ψ(x) − λ

k(t, x)ψ(t)dt = 0.

(2.26)

a

Equation (2.26) is the adjoint equation of the given integral equation 

b

φ(x) − λ

k(x, t)φ(t)dt = 0.

a

Remark 2.2 We observe that if the homogeneous integral equation with degenerate kernel k(x, t) given by  φ(x) − λ

b

k(x, t)φ(t)dt = 0

a

possesses nontrivial solution then the corresponding non-homogeneous equation (2.1) is either inconsistent or possesses non-unique solution if and only if the forcing term satisfies the relation (2.24) where ψ(t) is the solution of adjoint homogeneous equation (2.26). Now we summarize the observations given by Remarks 2.1 and 2.2 in form of the following important theorem known as Fredholm Alternative. Theorem 2.1 (Fredholm Alternative) Let φ(x) satisfy the following Fredholm integral equation of second kind  φ(x) − λ

b

k(x, t)φ(t)dt = f (x), a ≤ x ≤ b

(2.27)

a

with degenerate kernel k(x, t) given by k(x, t) =

n 

ai (x)bi (t)

(2.28)

i=1

where ai (x) and bi (t) are two linearly independent continuous functions on [a, b]. Then either, the integral equation (2.27) possesses unique continuous solution φ(x) for every continuous function f (x) on [a, b] in which case the homogeneous equation  b φ(x) − λ k(x, t)φ(t)dt = 0, a ≤ x ≤ b (2.29) a

corresponding to Eq. (2.27) possesses trivial solution or, if the homogeneous equation (2.29) possesses nontrivial solution φ0 (x), then the nonhomogeneous integral

2.2 Homogeneous Equations

23

equation (2.27) is either inconsistent or possesses non-unique solution if and only if f (x) satisfies  b f (t)ψ(t)dt = 0 (2.30) a

where the continuous function ψ(t) is the solution of adjoint homogeneous equation 

b

ψ(x) − λ

k(t, x)ψ(t)dt = 0.

(2.31)

a

This theory is known as Fredholm alternative.

2.2 Homogeneous Equations Illustrative Examples Here we shall study the homogeneous Fredholm integral equation of second kind with degenerate kernel given by  φ(x) − λ

b

k(x, t)φ(t)dt = 0, a ≤ x ≤ b

a

n where k(x, t) = i=1 ai (x)bi (t), and ai (x) and bi (t) are two linearly independent finite set of continuous functions. It is obvious that φ(x) = 0 is a solution of the given integral equation. We shall now seek nontrivial solution of this integral equation. We consider the following examples. Example 2.1 Solve  φ(x) − λ

b

k(x, t)φ(t)dt = 0, a ≤ x ≤ b

(2.32)

a

where k(x, t) = g(x)h(t) and g(x), h(t) are continuous functions. Solution: Substituting k(x, t) = g(x)h(t)

(2.33)

in the given integral equation, we obtain  φ(x) − λ a

b

g(x)h(t)φ(t)dt = 0.

(2.34)

24

2 Fredholm Integral Equation of the Second Kind with Degenerate Kernel

This can be written as 

b

φ(x) − λg(x)

h(t)φ(t)dt = 0.

(2.35)

a

Writing

 c=

b

h(t)φ(t)dt.

(2.36)

a

Equation (2.35) becomes φ(x) = λcg(x).

(2.37)

Substituting φ(x) from Eq. (2.37) in Eq. (2.36), we obtain c = λcα where

 α=

b

g(x)h(x)d x

(2.38)

(2.39)

a

The relation (2.38) can be written as or c(1 − λα) = 0. For λ =

1 α

we obtain c = 0. In this case from Eq. (2.37) we have φ(x) = 0.

For λ = α1 , c = 0 is arbitrary. Thus in this case from Eq. (2.37) we obtain the nontrivial solution of given Eq. (2.32) as φ(x) =

c g(x) α

where c is an arbitrary constant. Example 2.2 Find the nontrivial solutions of φ(x) =

2λ π



π

cos(x − t)φ(t)dt, 0 ≤ x ≤ π.

0

Solution: The given integral equation can be rewritten as

  π  π 2λ cos x φ(x) = cos t φ(t)dt + sin x sin t φ(t)dt . π 0 0

(2.40)

2.2 Homogeneous Equations

25

This can be alternatively written as φ(x) =

2λ 2λ c1 cos x + c2 sin x, π π 

where

π

c1 =

(2.41)

φ(t) cos t dt,

(2.42)

φ(t) sin t dt.

(2.43)

0



and c2 =

π

0

We substitute φ(x) from Eq. (2.41) into Eqs. (2.42) and (2.43) and simplify to obtain the following homogeneous linear algebraic equations in c1 and c2 . (1 − λ)c1 = 0

(2.44)

(1 − λ)c2 = 0.

(2.45)

Equations (2.44) and (2.45) imply that for λ = 1, c1 and c2 are arbitrary constants. Here nontrivial solution of given equation is 2 (c1 cos x + c2 sin x) π

φ(x) =

where c1 , c2 are arbitrary constants. Example 2.3 Find nontrivial solutions of 

1

φ(x) = λ

 3−

0

 3x tφ(t)dt 0 ≤ x ≤ 1. 2

(2.46)

Solution: The given integral equation is 

1

φ(x) = λ



1

3tφ(t)dt − λx

0

0

3t φ(t)dt. 2

This can be alternatively written as φ(x) = λc1 − λxc2 , where



(2.47)

1

c1 =

3tφ(t)dt 0

(2.48)

26

2 Fredholm Integral Equation of the Second Kind with Degenerate Kernel

and

 c2 =

1

0

3t φ(t)dt. 2

(2.49)

Substituting φ(t) from Eq. (2.47) into Eqs. (2.48) and (2.49), we obtain after simplification the following two homogeneous linear algebraic equations in c1 and c2 given by   3 1 − λ c1 + λc2 = 0 (2.50) 2   3 λ c2 = 0. − λc1 + 1 + 4 2

(2.51)

This system of linear equation has trivial solution if   1 − 3 λ λ   2   3   − λ 1 + 1 λ  = 0.  4  2 This gives λ = 1.

(2.52)

Thus for λ = 1, c1 = c2 = 0 and hence from Eq. (2.47) we obtain φ(x) = 0. Now for λ = 1, we obtain from Eqs. (2.50) and (2.51), c1 = 2c2 = m 1 , where m 1 is an arbitrary constant. Hence from Eq. (2.47) we obtain the nontrivial solution of given integral equation as   x . φ(x) = m 1 1 − 2 Example 2.4 Solve  φ(x) − λ

π

sin(x + t)φ(t)dt = 0, 0 ≤ x ≤ π.

0

Solution: The given integral equation is 

π

φ(x) − λ sin x 0



π

φ(t) cos tdt − λ cos x 0

φ(t) sin tdt = 0

(2.53)

2.2 Homogeneous Equations

27

which can be alternately written as φ(x) − λc1 sin x − λc2 cos x = 0 

where

π

c1 =

(2.54)

φ(t) cos tdt.

(2.55)

φ(t) sin tdt.

(2.56)

0



π

c2 = 0

Substituting φ(t) from (2.54) into Eqs. (2.55) and (2.56), we obtain λπ c2 = 0 2

(2.57)

λπ c1 − c2 = 0. 2

(2.58)

   1 − λπ  2   = 0.  λπ  −1  2

(2.59)

c1 −

This system has trivial solution if

This gives λ = ± π2 . So for λ = ± π2 , we obtain from Eqs. (2.57) and (2.58) c1 = c2 = 0. Hence from Eq. (2.54) we find that the given integral equation has trivial solution given by φ(x) = 0. For λ = π2 , we have c1 = c2 = m 2 , where m 2 is arbitrary constant.

(2.60)

Hence from (2.60) we obtain nontrivial solution of the given integral equation as φ(x) =

2 m 2 (sin x + cos x). π

For λ = − π2 , we obtain c1 = −c2 = m 3 , where m 3 is a arbitrary constant and the nontrivial solution of corresponding integral equation is 2 φ(x) = − m 3 (sin x − cos x). π

28

2 Fredholm Integral Equation of the Second Kind with Degenerate Kernel

Example 2.5 Find nontrivial solutions of  φ(x) = λ

1 −1

(xt + x 2 t 2 )φ(t)dt

− 1 ≤ x ≤ 1.

Solution: The given integral equation is  φ(x) = λx

1

−1

 tφ(t)dt + λx

1

2

t 2 φ(t)dt.

−1

This can be alternatively written as φ(x) = λxc1 + λx 2 c2 , where

 c1 =

and

 c2 =

1

tφ(t)dt −1

1

t 2 φ(t)dt.

−1

Proceeding in usual manner, we obtain after simplification the following two homogeneous linear algebraic equations in c1 and c2 as   2λ 1− c1 = 0 3   2λ c2 = 0. 1− 5 This system of linear equation has nontrivial solution if λ1 =

3 5 and λ2 = . 2 2

Thus for λ1 = 23 ; c2 = 0 and c1 is arbitrary. So we obtain φ1 (x) = 3x2 c1 where c1 is an arbitrary constant. 2 Also for λ2 = 25 ; c1 = 0 and c2 is arbitrary and we obtain φ2 (x) = 5x2 c2 where c2 is an arbitrary constant.

2.3 Nonhomogeneous Equations

29

2.3 Nonhomogeneous Equations Illustrative Examples In this section, we will illustrate the theory of Fredholm Alternative by considering the following problems: Example 2.1 Solve the integral equation 

π

φ(x) − λ

sin(x + t)φ(t)dt = f (x), 0 ≤ x ≤ π,

(2.61)

0

whenever solutions exist. Solution: We rewrite the given Eq. (2.61) as 

π

φ(x) − λ sin x



π

cos tφ(t)dt − λ cos x

0

sin tφ(t)dt = f (x).

0

Defining the constants  c1 =

π



π

cos tφ(t)dt, and c2 =

0

sin tφ(t)dt,

(2.62)

0

we obtain φ(x) = λc1 sin x + λc2 cos x + f (x).

(2.63)

Substituting φ(x) from relation (2.63) into the relations (2.62) we obtain c = β1 , c1 − λπ 2 2 λπ c − c 1 2 = −β2 , 2 

with

π

β1 = 0

(2.64) 

f (t) cos tdt and β2 =

π

f (t) sin tdt.

(2.65)

0

We find that the unique solution of the system of equations (2.64) can be obtained when    1 − λπ  2   = 0.  λπ  −1  2 This gives λ = ± π2 .

30

2 Fredholm Integral Equation of the Second Kind with Degenerate Kernel

Thus for λ = ± π2 the unique solution is given by c1 =

λπβ2 /2 + β1 , 1 − λ2 π 2 /4

c2 =

λπβ1 /2 + β2 . 1 − λ2 π 2 /4

(2.66)

Substituting c1 , c2 from relation (2.66) into the relation (2.63) we obtain the unique solution of integral equation (2.61) for λ = ± π2 as   λ λπ λπ β1 (sin x + cos x) + β2 (cos x + sin x) . φ(x) = f (x) + 1 − λ2 π 2 /4 2 2 (2.67) Using the relations given by Eq. (2.65), this can be simplified as φ(x) = f (x) +



λ 1−

λ2 π 2 4

π

0

Here R(x, t; λ) =



λπ cos(x − t) dt. f (t) sin(x + t) + 2

sin(x + t) + 1−

λπ cos(x 2 λ2 π 2 4

− t)

is the resolvent kernel. We now consider the case when λ = ± π2 . It is observed from Eq. (2.64) that β1 + β2 = 0 for λ = and

2 π

2 β1 − β2 = 0 for λ = − , π

which yield after noting relation (2.65), π 0

π 0

⎫ f (t)(cos t + sin t)dt = 0 for λ = π2 , ⎬ f (t)(cos t − sin t)dt = 0 for λ = − π2 .

⎭

(2.68)

We observe that the homogeneous integral equation corresponding to the given integral equation (2.61) for λ = ± π2 is given by φ(x) ∓

2 π



π

sin(x + t)φ(t)dt = 0.

(2.69)

0

The kernel k(x, t) = sin(x + t) is a real symmetric kernel. Hence the homogeneous integral equation (2.69) is self-adjoint. We have already obtained the solution of the integral equation (2.69) in Example 2.4 of Sect. 2.2 and it is given by

2.3 Nonhomogeneous Equations

31

φ(x) = π2 (sin x + cos x), for λ =

2 π

and φ(x) = − π2 (sin x − cos x), for λ = − π2 .

⎫ ⎬ ⎭

(2.70)

It is observed from relations (2.68) that the functions φ(x) defined in Eq. (2.70) for λ = ± π2 satisfy the condition given by Eq. (2.24). Hence by Fredholm Alternative theory, the given integral equation (2.61) has non-unique solutions for λ = ± π2 . For λ = π2 , the system of linear equations (2.64) reduces to c1 − c2 = β1 . Taking c2 = l1 , we can write c1 = l1 + β1 , where l1 is arbitrary constant. Hence the solution of (2.61) is given by 2 2 φ(x) = l1 (sin x + cos x) + f (x) + sin x π π



π

f (t) cos tdt

0

and the first relation of Eq. (2.68) holds. Similarly for λ = − π2 , the non-unique solution of given integral equation can be similarly obtained as 2 2 φ(x) = l2 (sin x + cos x) + f (x) + sin x π π



π

f (t) cos tdt

0

where l2 is arbitrary and the second relation of Eq. (2.68) holds. Example 2.2 Solve the integral equation 

1

φ(x) −

(12xt + 6(t − x) − 2)φ(t)dt = f (x), 0 ≤ x ≤ 1,

0

where f (x) is a real-valued continuous function in [0, 1]. The given integral equation can be written as 

1

φ(x) − 6(2x + 1)



1

tφ(t)dt + 2(3x + 1)

0

φ(t)dt = f (x).

(2.71)

tφ(t)dt,

(2.72)

0

Defining the constants c1 and c2 , as  c1 = 0

1

 φ(t)dt and c2 =

1

0

we can write Eq. (2.71) as φ(x) = f (x) − 2c1 (3x + 1) + 6c2 (2x + 1).

(2.73)

32

2 Fredholm Integral Equation of the Second Kind with Degenerate Kernel

Substituting φ(x) from Eq. (2.73) into Eq. (2.72), we obtain 6c1 − 12c2 = f 1 , 3c1 − 6c2 = f 2 , where



1

f1 =

(2.74) 

f (t)dt and f 2 =

0

1

t f (t)dt.

(2.75)

0

It is observed that the solutions of system of linear equations (2.74) are not unique. Non-unique solutions of this system will exist if and only if f 1 ξ1 + f 2 ξ2 = 0,

(2.76)

where (ξ1 , ξ2 ) satisfy the adjoint homogeneous system as given by 6ξ1 + 3ξ2 = 0, −12ξ1 − 6ξ2 = 0.



We find from above system of linear equations that ξ2 = −2ξ1 .

(2.77)

Hence Eq. (2.76) yields for ξ1 = 0 f 1 − 2 f 2 = 0. Noting the relation (2.75), we obtain 

1

(1 − 2t) f (t)dt = 0.

(2.78)

0

We can easily verify that ψ(x) = 1 − 2x satisfies the adjoint homogeneous equation  ψ(x) −

1

(12xt + 6(x − t) − 2)ψ(t)dt = 0, 0 ≤ x ≤ 1

0

corresponding to the given integral equation. Hence by Fredholm Alternative theory in this case the non-unique solution of given integral equation exists. From relation (2.74), we can express c1 in terms of c2 as

2.3 Nonhomogeneous Equations

33

c1 = 2c2 +

f1 . 6

Taking c2 = l3 , this becomes c1 = 2l3 + f61 where l3 is an arbitrary constant. Substituting the value of c1 in the relation (2.73), we obtain the non-unique solution of the given integral equation as 

1

φ(x) = f (x) − (6x + 2)

f (t)dt + 2l3

(2.79)

0

where f (t) satisfies (2.78). ‘ Example 2.3 Solve the Fredholm integral equation 

1

φ(x) = 1 + λ

(1 − 3xt)φ(t)dt. 0 ≤ x ≤ 1.

(2.80)

0

We rewrite the integral equation as 

1

φ(x) = λ



1

φ(t)dt − 3xλ

0

tφ(t)dt + 1,

0

which can be alternatively written as φ(x) = λc1 − 3xλc2 + 1, where



1

c1 =

(2.81)

φ(t)dt

(2.82)

tφ(t)dt.

(2.83)

0



1

c2 = 0

Substituting φ(t) from relation (2.81) into the relations (2.82) and(2.83), we obtain the following system of linear equations. c = 1, (1 − λ)c1 + 3λ 2 2 − λ2 c1 + (1 + λ)c2 = 21 .



This system of linear equations has unique solution if and only if   1 − λ 3λ  2   λ  − (1 + λ) = 0 2 which gives λ = ±2. For λ = ±2, we obtain

(2.84)

34

2 Fredholm Integral Equation of the Second Kind with Degenerate Kernel

c1 =

1+ 1−

λ 2 λ4 4

and c2 =

1 2(1 −

λ2 ) 4

.

Hence from relation (2.81) we obtain the unique solution of the given integral equation as 

 3x 1 ], λ = ±2. (2.85) 1+λ 1− φ(x) = 2 2 1 − λ4 Now we consider the case where λ = ±2. For λ = 2, the system of linear equations (2.84) becomes −c1 + 3c2 = 1, −c1 + 3c2 =

1 2

which are inconsistent. Hence for λ = 2, the given integral equation is inconsistent. Similarly for λ = −2, the system of linear equations (2.84) becomes 3c1 − 3c2 = 1, c1 − c2 =

1 2

which is inconsistent. Hence for λ = −2 also the given integral equation does not possess any solution.

Exercises Find the nontrivial solution of following Fredholm integral equation of second kind 1.



1

φ(x) = λ

2tφ(t)dt,

0≤x ≤1

0

Answer: φ1 (x) = A, 2.



A is arbitrary an constant.

1

φ(x) = λ

4xφ(t)dt,

0≤x ≤1

0

Answer: φ1 (x) = 2 Ax,

A is arbitrary an constant.

Exercises

35

3.



1

φ(x) = λ

xet φ(t)dt,

0≤x ≤1

0

Answer: φ1 (x) = Ax, 4. φ(x) =

2 λ π



π

cos(x + t)φ(t)dt,

0≤x ≤π

0

Answer: φ1 (x) =

5.

A is arbitrary an constant.

2 2 A cos x, φ2 (x) = B sin x π π

where A and B are arbitrary constants.  1 φ(x) = λ (6x − 2t)φ(t)dt, 0 ≤ x ≤ π 0

Answer: φ1 (x) = φ2 (x) = 2 A(3x − 1), A is an arbitrary constant. 6. Find the resolvent kernel R(x, t; λ) of the following integral equation  φ(x) = x + λ

1

(xt 2 + x 2 t)φ(t)dt,

0≤x ≤1

0

Answer: R(x, t; λ) =

3xt 5x 2 t 2 + 3 − 2λ 5 − 2λ

7. Analyze and solve the following integral equation. 

π

φ(x) − λ

sin(x + t)φ(t)dt = x,

0≤x ≤π

0

Answer: For λ = ± π2 , unique solution is given by φ(x) = x +

2λ [(λπ 2 − 4) sin x + 2π(1 − λ) cos x]. 4 − λ2 π 2

For λ = π2 , non-unique solutions are given by φ(x) =

4 sin x 2A (sin x + cos x) + x − π π

where A is an arbitrary constant and for λ = − π2 , non-unique solutions are given by

36

2 Fredholm Integral Equation of the Second Kind with Degenerate Kernel

4 2B (cos x − sin x)x + cos x π π

φ(x) =

where B is an arbitrary constant. 8. Find the unique solution of 

1

φ(x) = x + λ

(1 + x + t)φ(t)dt,

0≤x ≤1

0

Answer: φ(x) = x +

λ [10 + (6 + λ)x] 12 − 24λ − λ2

9. Find the unique solution 

1

φ(x) = 1 +

(1 + e x+t )φ(t)dt,

0≤x ≤1

0

(e2 − 3) − 2(e − 1)e x 2(e − 1)2

Answer: φ(x) =

10. For what value of λ does the unique solution of the following integral equation exist?  1

φ(x) = f (x) + λ

xtφ(t)dt,

0≤x ≤1

0

Hence find the unique solution. Answer: For λ = 3, the unique solution of the integral equation exists. This is given by  1 3λx t f (t)dt. φ(x) = f (x) + 3−λ 0

References The content of this chapter is mainly based on the website epgp.inflibnet.ac.in/ of UGC E Pathsala and on the following literature. 1. A. Chakrabarti, Applied integral equations, (2008) Vijay Nicole Imprints Pvt Ltd 2. A. Wazwaz, A First Course in Integral Equations (World Scientific Publishing Co., Pvt Ltd, 1997) 3. R.P. Kanwal, Linear Integral Equations Theory and Techniques (Academic, 1971) 4. D. Porter, D.S.G. Stirling, Integral Equations (Cambridge University Press, 1971)

Chapter 3

Integral Equations of Second Kind with Continuous and Square Integrable Kernel

3.1 Fredholm Integral Equations of Second Kind with Continuous Kernel Let us consider the Fredholm integral equation of second kind  φ(x) = f (x) + λ

b

k(x, t)φ(t)dt, a ≤ x ≤ b,

(3.1)

a

which in operator form is given by (I − λK )φ = f, where I is the identity operator and K is the linear operator defined by the relation  (K φ)(x) =

b

k(x, t)φ(t)dt, a ≤ x ≤ b.

a

Here the forcing function f (x) and the kernel k(x, t) are continuous in the interval [a, b] and in the square [a, b] × [a, b] respectively.

Solution by the Method of Successive Approximations We shall now use the method of successive approximation to obtain the solution of integral equation (3.1). Let φ0 (x) = φ(0) (x) = f (x),

(3.2)

where φ0 is the zero th approximation of φ(x). © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 S. Banerjea and B. N. Mandal, Integral Equations and Integral Transforms, https://doi.org/10.1007/978-981-99-6360-7_3

37

38

3 Integral Equations of Second Kind with Continuous …

We now substitute φ0 (t) in place of φ(t) in the integral on the right-hand side of equation (3.1) to obtain first approximation φ1 (x) of φ(x) as 

b

φ1 (x) = f (x) + λ

k(x, t)φ0 (t)dt.

a

In view of equation (3.2) this can be rewritten as φ1 (x) = φ(0) (x) + λφ(1) (x) where φ(1) (x) =



b

(3.3)

k(x, t)φ(0) (t)dt = (K φ(0) )(x).

a

We obtain the second approximation φ2 (x) of φ(x) by substituting φ1 (t) in place of φ(t) in the integral of the right-hand side of equation (3.1) as  φ2 (x) = f (x) + λ

b

k(x, t)φ1 (t)dt.

a

Using the relation (3.2) and (3.3) we can write 

(0)

φ2 (x) = φ (x) + λ

b

k(x, t)[φ(0) (t) + λφ(1) (t)]dt

a

which gives

φ2 (x) = φ(0) (x) + λφ(1) (x) + λ2 φ(2) (x)

where φ(2) (x) =



b

k(x, t)φ(1) (t)dt = (K φ(1) )(x).

a

Defining ( j)



b

φ (x) =

k(x, t)φ( j−1) (t)dt = (K φ( j−1) )(x),

j = 1, 2, ....

(3.4)

a

we can continue in similar manner to obtain the Nth approximation φ N (x) of φ(x) as φ N (x) = φ(0) (x) + λφ(1) (x) + λ2 φ(2) (x) + · · · + λ N φ(N ) (x) ≡

N  i=0

λi φ(i) (x), for any positive integer N .

(3.5)

3.1 Fredholm Integral Equations of Second Kind with Continuous Kernel

39

We call φ N (x) the N th approximation to the solution φ(x) of the integral equation (3.1). We shall now show that the above sequence φ N (x) converges to the unique solution φ(x) of the integral equation (3.1) under certain special restrictions on the values of λ. If we assume f (x) and k(x, t) are continuous functions, then we easily see that φ( j) (x) is continuous in [a, b], for j = 0, 1, 2, .... Assuming m = sup | f (x)| x∈[a,b]

and M=

|k(x, t)|,

sup x∈[a,b],t∈[a,b]

we observe ⎫ |φ(0) (x)| = | f (x)| ≤ m , ⎪ ⎪   ⎪ b b |φ(1) (x)| = | a k(x, t)φ(0) (t)dt| ≤ a |k(x, t)||φ(0) (t)|dt ⎪ ⎪ ⎪ ⎪ ⎪ ≤ m M(b − a) , ⎪ ⎪  ⎬ b (2) (1) 2 2 |φ (x)| ≤ a |k(x, t)||φ (t)|dt ≤ m M (b − a) , ⎪ . ⎪ ⎪ ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪ . ⎪ ⎭ (n) n n |φ (x)| ≤ m M (b − a) , n = 1, 2, ....

(3.6)

Using |φ(i) (x)|, i = 0, 1, 2, ...n, from Eq. (3.6) we obtain from Eq. (3.5), |φ N (x)| ≤ m

N 

|λ|n M n (b − a)n .

(3.7)

n=0

n (n) Hence making N → ∞, the series ∞ n=0 λ φ (x) is absolutely and uniformly convergent in [a, b] if |λ|M|b − a| < 1; (3.8) i.e., if |λ|
x.

We now apply the method of successive approximations to Eq. (3.26). Let φ0 (x) = φ(0) (x) = f (x),

(3.27)

where φ0 is the zeroth approximation of φ(x). We now substitute φ0 (t) in place of φ(t) in the integral on the right-hand side of equation (3.26) to obtain first approximation φ1 (x) of φ(x) as 

x

φ1 (x) = f (x) + λ

k(x, t)φ0 (t)dt.

a

In view of equation (3.27), this can be rewritten as φ1 (x) = φ(0) (x) + λφ(1) (x) where φ(1) (x) =



x

(3.28)

k(x, t)φ(0) (t)dt = (K φ(0) )(x).

a

We obtain the second approximation φ2 (x) of φ(x) by substituting φ1 (t) in place of φ(t) in the integral on the right-hand side of equation (3.26) as  φ2 (x) = f (x) + λ

x

k(x, t)φ1 (t)dt.

a

Using the relation (3.27) and (3.28) we can write φ2 (x) = φ(0) (x) + λ



x

k(x, t)[φ(0) (t) + λφ(1) (t)]dt

a

which gives

φ2 (x) = φ(0) (x) + λφ(1) (x) + λ2 φ(2) (x)

where (2)



φ (x) = a

x

k(x, t)φ(1) (t)dt = (K φ(1) )(x).

3.2 Volterra Integral Equations of Second Kind with Continuous Kernel

49

Defining φ( j) (x) =



x

k(x, t)φ( j−1) (t)dt = (K φ( j−1) )(x),

j = 1, 2, ....

(3.29)

a

we can continue in similar manner to obtain the nth approximation φn (x) of φ(x) as φn (x) = φ(0) (x) + λφ(1) (x) + λ2 φ(2) (x) + ...... + λn φ(n) (x) ≡

n 

λi φ(i) (x), for any positive integer n.

(3.30)

i=0

Assuming m = sup | f (x)| x∈[a,b]

and M=

|k(x, t)|,

sup x∈[a,b],t∈[a,b]

we obtain the following estimates of φ( j) (x). ⎫ |φ(0) (x)| = | f (x)| ≤ m, ⎪ ⎪ ⎪ (1) ⎪ |φ (x)| ≤ m M(x − a), ⎪ ⎪ 2 2 ⎪ M (x−a) (2) ⎪ |φ (x)| ≤ m 2! , ⎬ . ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪ . ⎪ ⎪ n n ⎭ M (x−a) (n) |φ (x)| ≤ m n! , n = 0, 1, 2, ....

(3.31)

Using the above estimates, we find from Eq. (3.30) |φn (x)| ≤ m

n  |λ| j M j (x − a) j . j! j=0

j j j The series nj=0 |λ| M j!(x−a) converges as n → ∞, absolutely and uniformly for all values of λ to the function e|λ|M(x−a) . Consequently, the sequence φn (x) converges as n → ∞ to the solution φ(x) of the given Volterra integral equation for all values of λ, and hence from Eq. (3.30) we obtain φ(x), as n → ∞, φ(x) =

∞  n=0

which is Neumann series in this case.

λn φ(n) (x)

(3.32)

50

3 Integral Equations of Second Kind with Continuous …

3.3 Illustrative Examples As illustrative examples, we consider the following problems: Example 3.5 Solve the Volterra integral equation:  φ(x) = f (x) + λ

x

e x−t φ(t)dt, a ≤ x ≤ b.

a

Here,

φ(0) (x) = f (x), k(x, t) = e x−t , φ(1) (x) =



x

k(x, t)φ(0) (t)dt =



a



x

e x−t f (t)dt = e x

a

φ(2) (x) =



x

x

e−t f (t)dt,

a

k(x, s)φ(1) (s)ds = e x



a

x



a

s

 e−t f (t)dt ds.

a

Changing order of integration, we obtain 

(2)

φ (x) = e

x

x

e

−t

 f (t)

a

φ(3) (x) =



x

x





x

t



x



a



a

x



(n)

φ (x) = a

x

s

 (s − t)e−t f (t)dt ds

a

  (s − t)ds dt =

t

and so on

(x − t)e x−t f (t)dt,

a

k(x, s)φ(2) (s)ds = e x

e−t f (t)

x

ds dt =

a

= ex



a

x

1 (x − t)2 e x−t f (t)dt, 2!

(x − t)n−1 x−t e f (t)dt, n = 1, 2, ... (n − 1)!

Thus, Eq. (3.32) gives φ(x) =

∞  n=0

The series

n (n)



λ φ (x) = f (x) + λ a

∞ x n=1

∞  (λ(x − t))n−1 (n − 1)! n=1

(λ(x − t))n−1 x−t e f (t)dt (n − 1)!

(3.33)

3.3 Illustrative Examples

51

on the right-hand side converges to eλ(x−t) . Hence we obtain  φ(x) = f (x) + λ

x

e(λ+1)(x−t) f (t)dt,

(3.34)

a

as the solution of (3.33), valid for all values of λ. Example 3.6 Solve the following Volterra integral equation by using the method of successive approximations: 

x

φ(x) = 1 +

(x − t)φ(t)dt, 0 ≤ x ≤ 1.

(3.35)

0

Solution: Here k(x, t) = x − t, f (x) = 1, λ = 1. Thus φ(0) (x) = 1, 

(1)

x

φ (x) =



(0)

0

φ(2) (x) =



x

(x − t)dt =

0



k(x, t)φ(1) (x)dt =

0

φ(3) (x) =

x

k(x, t)φ (x)dt = x

(x − t)

0



x

k(x, t)φ(2) (x)dt

0



x

(x − t)

0

x2 , 2

x4 t2 dt = , 2 4!

x6 t4 dt = . 4! 6!

Continuing in this way, we obtain φ(n) (x) =

x 2n . (2n)!

Hence the solution of the given integral equation, by the method of successive approximation is given by ∞  φ(x) = λn φ(n) (x), n=0

i.e., φ(x) =

∞  x 2n . (2n)! n=0

The series on the right-hand side converges to the function cosh x. Thus φ(x) = cosh x

52

3 Integral Equations of Second Kind with Continuous …

is the solution of the given integral equation. Example 3.7 Solve by the method of successive approximations:  φ(x) = 1 +

x

2tφ(t)dt, 0 ≤ x ≤ 1.

(3.36)

0

Solution:

Here k(x, t) = 2t, f (x) = 1, λ = 1. Thus φ(0) (x) = 1. 

Then φ(1) (x) =

x

2tdt =

0



φ(2) (x) =

x

2t 3 dt =

x4 . 2!

t 5 dt =

x6 . 3!

0



φ(3) (x) =

x

0



φ(4) (x) =

x

0

x2 . 1!

x8 t7 dt = . 3 4!

Continuing in this way we obtain φ(n) (x) =

x 2n . n!

Hence the solution of the given integral equation, by the method of successive approximation is given by ∞  φ(x) = λn φ(n) (x), n=0

φ(x) =

∞  x 2n 2 = ex . n! n=0

Example 3.8 Solve by the method of successive approximations: 

x

y(x) = sin x +

z(t)dt, 0 ≤ x ≤ 1,

0

 z(x) = 1 − cos x − 0

x

y(t)dt, 0 ≤ x ≤ 1.

3.4 Iterated Kernels

53

Solution: Substituting z(t) from the second equation into the first one, it gives 

x

y(x) = sin x +



t

[1 − cos t −

0

y(u)du] dt, 0 ≤ x ≤ 1

0

which simplifies to



x

y(x) = x −

(x − u) y(u) du.

0

Taking y0 (x) = y 0 (x) = x, and applying the method of successive approximation we obtain  x x3 (x − u) u du = x − , y1 (x) = x − 3! 0 

x

y2 (x) = x −

(x − u)(u −

0

yn (x) =

n  j=0

x3 x5 u3 )du = x − + , 3! 3! 5!

(−1) j

x 2 j+1 (2 j + 1)!

so that y(x) = sin x. Hence knowing y(x) we obtain z(x) from the second equation of the given problem as z(x) = 0.

3.4 Iterated Kernels We have already illustrated the method of solution of Fredholm integral equation of second kind, given by  φ(x) = f (x) + λ

b

k(x, t)φ(t)dt, a ≤ x ≤ b

(3.37)

a

or equivalently in the operator form (I − λK )φ = f, by the method of successive approximations. The successive approximations to the solution φ(x) of the integral equation (3.37) are represented by the sequence φi (x), i = 0, 1, 2, ..., with

54

3 Integral Equations of Second Kind with Continuous …

φ0 (x) = f (x), b φi (x) = f (x) + λ a k(x, t)φi−1 (t)dt.

(3.38)

It has been proved that the sequence φi (x) tends to the solution φ(x) of the equation (3.37), as i → ∞, where φ(x) is represented in the form of a series known as Neumann series which is given by ∞ 

φ(x) =

λn φ(n) (x).

(3.39)

n=0

b Here φ(n) (x) = a k(x, t)φ(n−1) (t)dt, n = 1, 2, ... and φ(0) (x) = f (x).

(3.40)

Noting the last relation of equation (3.40), Neumann series given by equation (3.39) can be expressed as ∞  φ(x) = f (x) + λn φ(n) (x). (3.41) n=1

Using the first relation of equation (3.40), we can write φ(n) (x) as φ(n) (x) =



b

k1 (x, t)φ(n−1) (t)dt

(3.42)

a

where k1 (x, t) = k(x, t).

(3.43)

Again substituting φ(n−1) (t) in terms of φ(n−2) (t) using the first relation of equation (3.40) in equation (3.42), we obtain φ(n) (x) =



b

 k1 (x, t){

a

b

k(t, u 1 )φ(n−2) (u 1 )du 1 }dt.

a

Interchanging the order of integration, the above relation can be written as (n)



φ (x) =

b

φ

(n−2)

 (u 1 )

a

Writing

k1 (x, t)k(t, u 1 )dt du 1 .

a

 k2 (x, u 1 ) = a

we can express

b

b

k1 (x, t)k(t, u 1 )dt,

3.4 Iterated Kernels

55



(n)

b

φ (x) =

k2 (x, u 1 )φ(n−2) (u 1 )du 1 .

a

In a similar manner we can express 

φ(n) (x) =

b

k3 (x, u 2 )φ(n−3) (u 2 )du 2 ,

a

where

 k3 (x, u 2 ) =

b

k2 (x, u 1 )k(u 1 , u 2 )du 1 .

a

We can continue in this way to obtain φ(n) (x) =



b

kn (x, t) f (t)dt

(3.44)

a

where



b

kn (x, t) =

kn−1 (x, s)k(s, t)ds,

n≥2

(3.45)

a

and k1 (x, t) = k(x, t) as given by Eq. (3.43). We now substitute φ(n) (x) from Eq. (3.44) in Eq. (3.41) to obtain φ(x) = f (x) +

∞ 



b

λn

kn (x, t) f (t)dt

(3.46)

a

n=1

where kn (x, t) and k1 (x, t) are given by Eqs. (3.43) and (3.45) respectively. Thus we can express Eq. (3.41) or Eq. (3.46) as  φ(x) = f (x) + λ

b

R(x, t; λ) f (t)dt

(3.47)

a

where the resolvent kernel R(x, t, λ) is given by R(x, t; λ) =

∞ 

λn kn+1 (x, t).

(3.48)

n=0

It may be noted that all the above results are valid for |λ|M(b − a) < 1, where M=

sup x∈[a,b],t∈[a,b]

|k(x, t)|.

(3.49)

56

3 Integral Equations of Second Kind with Continuous …

The kernels kn (x, t), n = 1, 2... are known as iterated kernel. We now observe that under the condition (3.49), the Fredholm integral equation (3.37) will possess the unique solution, as given by either of the forms (3.46) or (3.47). This implies that the operator (I − λK ), associated with the integral equation (3.37), possesses a unique inverse, under the condition (3.49) and we obtain φ(x) = (I − λK )−1 f (x) = f (x) +

∞ 

λn (K n f )(x),

(3.50)

n=1

where using Eqs. (3.46) and (3.45) we find 

b

(K n f )(x) =

kn (x, t) f (t)dt.

(3.51)

a

The representation (3.50) shows that we have the following expansion formula involving the operator K : (I − λK )−1 = I +

∞ 

λn K n ,

(3.52)

n=1

provided the sufficient condition (3.49) is satisfied. We now consider the following examples.

Examples Example 3.9 Find the iterated kernels and solve the following integral equation 

1

φ(x) = 1 + λ

xtφ(t)dt, 0 < x < 1.

0

Solution Here the kernel k(x, t) is given by k(x, t) = k1 (x, t) = xt and f (x) = 1. The next iterated kernels are given by 

1

k2 (x, t) =

k1 (x, s)k(s, t)ds

0

 =

1

xs 2 tds 0

(3.53)

3.4 Iterated Kernels

57

= 

1

k3 (x, t) =

xt , 3

k2 (x, s)k(s, t)ds =

xt , 32

k3 (x, s)k(s, t)ds =

xt . 33

0



1

k4 (x, t) = 0

Continuing in this way, we find kn+1 (x, t) =

xt , n = 0, 1, 2.... 3n

Hence the resolvent kernel is given by R(x, t; λ) =

∞ 

λn kn+1 (x, t)

n=0

= xt

=

∞  λ ( )n 3 n=0

3xt , provided |λ| < 3. 3−λ

(3.54)

Hence the solution of the integral equation (3.53) is given by  1 φ(x) = 1 + λ R(x, t; λ)dt. 0

Noting Eq. (3.54) we can write after simplification φ(x) = 1 +

3λx . 2(3 − λ)

Example 3.10 Find the iterated kernel and solve the following integral equation 

1

φ(x) = f (x) + λ

e x−t φ(t)dt, 0 ≤ x ≤ 1.

0

Solution Here k(x, t) = k1 (x, t) = e x−t .

58

3 Integral Equations of Second Kind with Continuous …



1

Hence k2 (x, t) =

k1 (x, s)k(s, t)ds

0



1

=

e x−s es−t ds 0

= e x−t ,  k3 (x, t) =

1

k2 (x, s)k(s, t)ds = e x−t .

0

Continuing similarly we obtain kn (x, t) = e x−t

n = 1, 2, ....

Hence the resolvent kernel is given by R(x, t; λ) =

∞ 

λn kn+1 (x, t)

n=0

= e x−t

∞ 

λn

n=0

e x−t , provided |λ| < 1. 1−λ

=

Hence the solution of the given integral equation is λ x e 1−λ

φ(x) = f (x) +



1

e−t f (t)dt.

0

Example 3.11 Find the iterated kernels and hence solve 

1

φ(x) = f (x) + λ

xet φ(t)dt

0 ≤ x ≤ 1.

0

Solution Here k(x, t) = xet = k1 (x, t).  k2 (x, t) = 0

1

k1 (x, s)k(s, t)ds

3.5 Fredholm Theory for Integral Equation with Continuous Kernel



1

=

59

xes .set ds

0

= xet , 

1

k3 (x, t) =

k2 (x, s)k(s, t)ds

0



1

=

xes .set ds

0

= xet . Continuing in same way we obtain kn (x, t) = xet

n = 1, 2, ....

Hence, the resolvent kernel is given by R(x, t; λ) =

∞ 

λn kn+1 (x, t)

n=0

=

xet , |λ| < 1. 1−λ

The solution of given integral equation is φ(x) = f (x) +

λx 1−λ



1

et f (t)dt.

0

3.5 Fredholm Theory for Integral Equation with Continuous Kernel We shall continue to study Fredholm integral equation of second kind with continuous kernel and forcing function. We shall first see that how a degenerate kernel can be associated with a continuous kernel so that the theory of Fredholm alternative can be applied to Fredholm integral equation of second kind with continuous kernel. Then we will present this idea in the form of three important theorems, known as the Fredholm theorems. We now make the following remarks to derive three important theorems called Fredholm Theorems.

60

3 Integral Equations of Second Kind with Continuous …

Remark 3.1 It is known that if a function k(x, t) is continuous in [a, b] × [a, b], then k(x, t) possesses a Fourier expansion in terms of complete orthonormal set of ∞ in [a, b], the form continuous functions {bi (x)}i=1 ∞ 

k(x, t) =

ai (x)bi (t),

(3.55)

i=1 ∞ where the set of continuous functions {ai (x)}i=1 are the Fourier coefficients as given by  b ai (x) = k(x, t)bi (t) dt. (3.56) a

Remark 3.2 From the well-known Parseval’s relation for Fourier series (3.55), one can write  b ∞  |ai (x)|2 = |k(x, t)|2 dt. (3.57) a

i=1

Integrating both sides with respect to x, from a to b and noting that ai (x) and k(x, t) are continuous functions we obtain ∞   i=1

b





b

|ai (x)| d x = 2

a

a

b

|k(x, t)|2 dt d x < ∞.

(3.58)

a

This shows that given sufficiently small positive number , there exists sufficiently large positive integer N , such that for all n (> N ) the remainder in the series of the left of equation (3.58) ∞  b  |ai (x)|2 d x < 2 . (3.59) i=n+1 a

Remark 3.3 Given a complete orthonormal set of continuous functions {φn (x)}∞ n=1 in [a, b], we can expand any continuous function φ(x) in [a, b], into the Fourier series ∞  φ(x) = cn φn (x), (3.60) n=1

where the Fourier coefficients cn ’s are given by  cn = (φ, φn ) = a

b

φ(x)φn (x) d x.

(3.61)

3.5 Fredholm Theory for Integral Equation with Continuous Kernel

61

Also since {φn }∞ n=1 are orthonormal so we have 

b

φn (x)φm (x) d x =

a

We note that

∞ 

0 for m = n, 1 for m = n.

cn φn = φ −

n=N +1

Consider,





n=1



b

=

  ∞

a

=



=

m=1

cn φn (x)

  N  φ(x) − cm φm (x) d x

n=N +1 b

m=1

φ(x)φn (x)d x −

cn a

n=N +1 ∞ 

n=1

   N N   φ(x) − cn φn (x) φ(x) − cm φm (x) d x

b

a

∞ 

cn φn .

n=1

 2 N    φ(x) − cn φn (x) d x 

b

a

=

N 

(3.62)

∞  n=N +1

cn cn −

n=N +1

∞ 

cn

n=N +1

=

N  m=1

∞ 

 cm

b

cn

N 

 cm

b

φn (x)φm (x)d x

a

m=1

φn (x)φm (x)d x (by Eq. (3.61))

a

|cn |2 (by Eq. (3.62)).

n=N +1

Thus we obtain  a

b

 2 N ∞     φ(x) −  dx = c φ (x) |cn |2 . n n   n=1

(3.63)

n=N +1

Noting Eq. (3.63), we recall Remark 3.1 and observe that the function k(x, t) which possesses the Fourier expansion (3.55), involving the complete orthonormal set ∞ {bi (x)}i=1 , satisfy  a

b

 2 n ∞     k(x, t) −  a (x)b (t) dt = |ai (x)|2 . i i   i=1

i=n+1

(3.64)

62

3 Integral Equations of Second Kind with Continuous …

Integrating both sides with respect to x from a to b and noting the relation (3.59), we obtain 2  b b n    k(x, t) − ai (x)bi (t) dt d x < 2 , (3.65)  a

a

i=1

whenever n > N , where N is an approximately chosen large positive integer. The above result (3.65) shows that the given continuous function k(x, t) can be approximated by the function p(x, t) =

n 

ai (x)bi (t),

(3.66)

i=1

for a sufficiently large n, which produces the inequality 

b



a

b

|k(x, t) − p(x, t)|2 d x dt < 2 ,

(3.67)

a

for an arbitrarily chosen small positive number . Remark 3.4 We write k(x, t) = p(x, t) + q(x, t),

(3.68)

so that the result (3.67) can be expressed as 

b



a

b

|q(x, t)|2 d x dt < 2 ,

(3.69)

a

and this relation holds for all (x, t)  [a, b] × [a, b]. Hence if Mq = Sup(x,t) |q(x, t)| then noting relation (3.69), we assume that  a

b

 a

b

|q(x, t)|2 d x dt < Mq2 (b − a)2 < 2 .

Noting the result (3.66) we now express the integral equation (3.1) given by  φ(x) = f (x) + λ

b

k(x, t)φ(t) dt, a ≤ x ≤ b

a

as

 φ(x) = f (x) + λ a

b

( p(x, t) + q(x, t))φ(t) dt, a ≤ x ≤ b

(3.70)

3.5 Fredholm Theory for Integral Equation with Continuous Kernel

63

which yields  φ(x) − λ

b



b

q(x, t)φ(t)dt = f (x) + λ

a

p(x, t)φ(t) dt, a ≤ x ≤ b. (3.71)

a

This can be expressed in operator form as (I − λQ)φ(x) = f 1 (x, λ), a ≤ x ≤ b where



b

f 1 (x, λ) = f (x) + λ

p(x, t)φ(t) dt

(3.72)

a

and

 (Qφ)(x) =

b

q(x, t)φ(t) dt. a

We have seen that (I − λQ)−1 exists uniquely, if |λ|Mq (b − a) < 1. Noting relation (3.70), we find that (I − λQ)−1 exists whenever |λ|Mq (b − a) < |λ| < 1 i.e., |λ|
6 (c) |λ| < √16 (d) |λ| > 6 Answer (a). 30: Show that the integral equation 

1

φ(x) = f (x) + λ

max(x, t)φ(t)dt, 0 ≤ x ≤ 1

0

has uniquesolution if

(a) |λ| > 23 (b) |λ| < Answer (b).



3 2

(c) |λ|


√3 2

References The content of this chapter is mainly based on the website epgp.inflibnet.ac.in/ of UGC E Pathsala and on the following literature. 1. A. Chakrabarti, Applied Integral Equations (Vijay Nicole Imprints Pvt Ltd, 2008) 2. A. Wazwaz, A First Course in Integral Equations (World Scientific Publishing Co., Pvt Ltd, 1997) 3. R.P. Kanwal, Linear Integral Equations Theory and Techniques (Academic Inc., 1971) 4. D. Porter, D.S.G. Stirling, Integral Equations (Cambridge University Press, 1971)

Chapter 4

Integral Equations of the Second Kind with a Symmetric Kernel

The Fredholm integral equation of the second kind given by  φ(x) = f (x) + λ

b

k(x, t)φ(t) dt, a ≤ x ≤ b

(4.1)

a

may be expressed in the operator form as (I − λK )φ = f, where

 (K φ)(x) =

b

k(x, t)φ(t) dt.

(4.2)

a

In this chapter, the method of the solution of integral equation (4.1) is discussed when its kernel k(x, t) is symmetric.

4.1 Symmetric Kernel A complex-valued function k(x, t), a ≤ x, t ≤ b appearing in the integral equation (4.1) is called a symmetric or Hermitian kernel if k(x, t) = k(t, x), where the bar denotes the complex conjugate. The integral equation with a symmetric kernel is also called a “symmetric integral equation”. Moreover, if k(x, t) is a square integrable kernel in [a, b] × [a, b] then k is called a “Hilbert–Schmidt kernel”.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 S. Banerjea and B. N. Mandal, Integral Equations and Integral Transforms, https://doi.org/10.1007/978-981-99-6360-7_4

81

82

4 Integral Equations of the Second Kind with a Symmetric Kernel

The following are the examples of symmetric or Hermitian kernel in a ≤ x ≤ b: 1. k(x, t) = sin(x + t), 2. k(x, t) = i(x  − t), x(1 − t), x ≤ t, 3. k(x, t) = t (1 − x), x ≥ t.

4.2 Properties of Integral Equations with a Symmetric Kernel Some important properties of integral equations and operators involving Hermitian (or symmetric) kernels are presented in this section. Theorem 4.1 (K φ, ψ) = (φ, K ψ) holds for all Hermitian or symmetric kernels k(x, t) where (φ, ψ) denotes the inner product of two square integrable functions φ and ψ. Proof By definition of inner product, 

b

(K φ, ψ) =

K φ(t)ψ(t) dt

a

which after noting Eq. (4.2) can be rewritten as  (K φ, ψ) =

b



a

b

 k(t, s)φ(s) ds ψ(t) dt.

a

Interchange of the order of integration produces  (K φ, ψ) =

b

 φ(s)

a



b

k(t, s)ψ(t) dt ds, a

 = a

b

 φ(s)



b

k(t, s)ψ(t) dt ds. a

Since the kernel is symmetric, therefore by definition k(s, t) = k(t, s).

4.2 Properties of Integral Equations with a Symmetric Kernel

Hence



b

(K φ, ψ) =





b

φ(s)

83

k(s, t)ψ(t) dt ds,

a

a



b

=

φ(s)K ψ(s) ds,

a

= (φ, K ψ). Remark 4.1 In Theorem 3.4 of Chap. 3, it is proved that (K φ, ψ) = (φ, K ∗ ψ)

(4.3)

where (K ∗ φ)(x) is the adjoint operator corresponding to K φ defined as (K ∗ φ)(x) =



b

k(t, x)φ(t) dt. a

Comparing the result of Theorem 4.1 with Eq. (4.3), it is found that K = K ∗ . This shows that the operator involving Hermitian or symmetric kernel is a symmetric or a self-adjoint operator and, hence, every homogeneous Fredholm integral equation of the second kind also represents its adjoint equation. In view of this, one can say that Fredholm theory which holds for a continuous or square integrable kernel is valid for a symmetric kernel. Theorem 4.2 Iterated kernels of a symmetric kernel are symmetric. Proof Suppose the kernel k(x, t) is symmetric and hence the first iterated kernel k1 (x, t) = k(x, t) is symmetric so that k1 (x, t) = k1 (t, x). For the second and third iterated kernels k2 (x, t) and k3 (x, t), one can have  b k1 (x, s)k1 (s, t) ds k2 (x, t) = a

 =

b

k1 (s, x)k1 (t, s) ds

a

 = a

b

k1 (t, s)k1 (s, x) ds

84

4 Integral Equations of the Second Kind with a Symmetric Kernel

= k2 (t, x) and



b

k3 (x, t) =

k2 (x, u)k1 (u, t) du

a

 =

b

a

 =



b

 k1 (x, s)k1 (s, u) ds k1 (u, t) du

a b



b

k1 (x, s)

a

 k1 (s, u)k1 (u, t) du ds

a



b

=

k1 (x, s)k2 (s, t) du

a

 =

b

k1 (s, x)k2 (t, s) ds

a

 =

b

k2 (t, s)k1 (s, x) ds

a

= k3 (t, x). Thus km (x, t) is symmetric for m = 1, 2, 3. Suppose it is assumed that kn (x, t) is symmetric, i.e., kn (x, t) = kn (t, x). It is now required to show that the result is true for m = n + 1, i.e., kn+1 (x, t) = kn+1 (t, x). Now,

 kn+1 (x, t) =

b

kn (x, u)k1 (u, t) du.

a

Since kn (x, t) and k1 (x, t) are symmetric,  kn+1 (x, t) = a

b

 kn (u, x)k1 (t, u) du =

b

k1 (t, u)kn (u, x) du = kn+1 (t, x).

a

So by induction, the theorem can be shown to be true for all iterated kernels.

4.2 Properties of Integral Equations with a Symmetric Kernel

85

Theorem 4.3 Every eigenvalue of a symmetric kernel is real. Proof Suppose λ0 be an eigenvalue of a symmetric operator K , and φ0 (x) (= 0) is the corresponding eigenfunction. Then φ0 (x) satisfies φ0 (x) − λ0 K φ0 (x) = 0,

a ≤ x ≤ b.

(4.4)

Multiplying both sides of Eq. (4.4) by φ0 (x) and integrating with respect to x from a to b, one finds after noting the relation (4.4) 

b

 φ0 (x)φ0 (x) d x = λ0

a

b

K φ0 (x)φ0 (x) d x.

a

This gives  φ0 2 −λ0 (K φ0 , φ0 ) = 0. Hence λ0 =

 φ0 2 . (K φ0 , φ0 )

(4.5)

Again from Eq. (4.4), one can write φ0 (x) − λ0 K φ0 (x) = 0,

a ≤ x ≤ b.

(4.6)

Multiplication of Eq. (4.6) on its both sides by φ0 (x) and integration with respect to x from a to b produces  b  b φ0 (x)φ0 (x) d x = λ0 φ0 (x)K φ0 (x) d x. a

a

Equivalently, the above relation can be written as  φ0 2 −λ0 (φ0 , K φ0 ) = 0. Using Theorem 4.1, this gives  φ0 2 −λ0 (K φ0 , φ0 ) = 0. Hence λ0 =

 φ0 2 . (K φ0 , φ0 )

(4.7)

By comparing Eqs. (4.5) and (4.7), it is found that λ0 = λ0 .

(4.8)

86

4 Integral Equations of the Second Kind with a Symmetric Kernel

The result (4.8) shows that if λ0 and φ0 exist, then λ0 is a real quantity for φ0 = 0. Theorem 4.4 The eigenfunctions of a symmetric kernel corresponding to different eigenvalues are orthogonal. Proof Let λ1 , λ2 be two different eigenvalues and φ1 , φ2 be their corresponding eigenfunctions respectively associated with symmetric kernel k(x, t), then φ1 = λ1 K φ1 , a ≤ x ≤ b,

(4.9)

φ2 = λ2 K φ2 , a ≤ x ≤ b.

(4.10)

Keeping in mind that the eigenvalues are real, multiplication of both sides of equations (4.9) by λ2 φ2 (x) and (4.10) by λ1 φ1 (x) respectively and then integration with respect to x from a to b produce the following two relations. λ2 (φ1 , φ2 ) = λ1 λ2 (K φ1 , φ2 ).

(4.11)

λ1 (φ2 , φ1 ) = λ1 λ2 (K φ2 , φ1 ).

(4.12)

Using the result of Theorem 4.1, the relation (4.12) becomes λ1 (φ2 , φ1 ) = λ1 λ2 (φ2 , K φ1 ). By taking complex conjugate of both sides of the above equation, one obtains λ1 (φ1 , φ2 ) = λ1 λ2 (K φ1 , φ2 ).

(4.13)

Then, differences of Eqs. (4.11) and (4.13) yield

Since λ1 = λ2 , therefore,

(λ1 − λ2 )(φ1 , φ2 ) = 0.

(4.14)

(φ1 , φ2 ) = 0.

(4.15)

Hence the theorem is proved.

4.3 Hilbert–Schmidt Theorem An important theorem concerning symmetric kernel, known as the Hilbert–Schmidt theorem, will be stated now. Using this theorem, the method of solution of the Fredholm integral equation of the second kind with a symmetric kernel given by

4.3 Hilbert–Schmidt Theorem

87



b

φ(x) = f (x) + λ

k(x, t)φ(t) dt, a ≤ x ≤ b

(4.16)

a

will be demonstrated here. Theorem 4.5 (Hilbert–Schmidt Theorem) Let λ1 ,λ2 ,…,λn ,…be the eigenvalues of a symmetric kernel k(x, t), and(a ≤ x ≤ b, a ≤ t ≤ b) and φ1 (x), φ2 (x),…,φn (x),… be the corresponding orthonormalized eigenfunctions. Then, if k(x, t) is absolutely square integrable in [a, b] × [a, b] and if a known function h(x) is also absob lutely square integrable in [a, b], then the function g(x) = a k(x, t)h(t)dt can be expanded in series as ∞  gn φn (x), (4.17) g(x) = n=1

where the coefficients gn ’s are given by gn = (g, φn ),

n = 1, 2, . . . .

(4.18)

It may be noted here that the completeness of the set of eigenfunctions is not assumed. Remark 4.2 The coefficients gn ’s can be written as gn = (g, φn ) =

1 (h, φn ), λn

and g(x) =

n = 1, 2, . . .

∞  hn φn (x), λ n=1 n

(4.19)

(4.20)

where h n = (h, φn ), n = 1, 2, . . . .

(4.21)

This can be shown in the following way: φn (x) satisfies  b

φn (x) = λn

k(x, t)φn (t) dt, a ≤ x ≤ b.

a

By taking complex conjugate and noting that λn are real and k(x, t) is symmetric, one can obtain  b

φn (x) = λn

k(t, x)φn (t) dt,

a ≤ x ≤ b.

a

Multiplication of both sides of the above equation by h(x) and integration with respect to x from a to b produces

88

4 Integral Equations of the Second Kind with a Symmetric Kernel





b



b

h(x)φn (x) d x = λn

h(x)

a

a

b

 k(t, x)φn (t) dt d x.

a

Interchanging the order of integration and noting Eq. (4.21), one obtains 

b

h n = λn

 φn (t)

a

Again, g(x) =

b a



b

k(t, x)h(x) d x dt.

a

k(x, t)h(t)dt, so that 

b

h n = λn

g(t)φn (t) dt,

n = 1, 2, . . . .

a

Using relation (4.18), this gives h n = λn gn , and g(x) =

n = 1, 2, . . .

∞  hn φn (x). λ n=1 n

Here, an outline of the proof is given. The following result for a square integrable symmetric kernel k(x, t) is assumed. A detailed proof is given in the book of F. G. Tricomi, Integral Equations, Interscience publishers 1957. Let n (x, t) =

n i=1

φi (x)φi (t) . λi

Lemma 4.1 The function n (x, t) converges in the mean to the kernel k(x, t), i.e., k(x, t) = lim

n→∞

n  φi (x)φi (t) i=1

λi

in the sense that limn→∞ |k(x, t) − n (x, t)|2 = 0. Writing  g(x) =

b

 k(x, t)h(t)dt =

a

b

[k(x, t) − n (x, t) + n (x, t)]h(t)dt,

a

one can obtain  g(x) = a

b

 [k(x, t) − n (x, t)]h(t)dt + a

b

n (x, t)h(t)dt

4.3 Hilbert–Schmidt Theorem

89

so that 

b

g(x) =

[k(x, t) − n (x, t)]h(t)dt +

a

 n  φi (x) λi

i=1



b

=

[k(x, t) − n (x, t)]h(t)dt +

a

 =

[k(x, t) − n (x, t)]h(t)dt +

a

h(t)φi (t)dt

a

n  φi (x)h i

λi

i=1 b

b

n 

φi (x)gi .

i=1

By the Schwartz inequality, [g(x) −

n 



b

φi (x)gi ]2 ≤

 |k(x, t) − n (x, t)|2 dt

a

i=1

b

|h(t)|2 dt.

a

Making n → ∞ and noting Lemma 4.1, the proof follows. Use of the Hilbert–Schmidt theorem to solve the symmetric integral equation  φ(x) − λ

b

k(x, t)φ(t) dt = f (x)

(4.22)

a

is described below. Equation (4.22) can be expressed as φ(x) − λg(x) = f (x) where

 g(x) =

(4.23)

b

k(x, t)φ(t) dt. a

Using the Hilbert–Schmidt theorem, this can be written as g(x) =

∞  an φn (x) λ n=1 n

where an = (φ, φn ),

n = 1, 2, . . . ,

(4.24)

90

4 Integral Equations of the Second Kind with a Symmetric Kernel

and the φn ’s are the eigenfunctions corresponding to λn . Thus, Eq. (4.23) becomes φ(x) = f (x) + λ

∞  an φn (x). λ n=1 n

(4.25)

Multiplying Eq. (4.25) by φm (x) and integrating with respect to x from a to b, one finds  b a

φm (x)φ(x) d x =

 b a

φm (x) f (x) d x + λ

 ∞  an b φm (x)φn (x) d x, λn a

m = 1, 2, . . . ,

n=1

which can be alternatively written as am = f m + λ

∞  an (φn , φm ) λ n=1 n

m = 1, 2, . . .

where am = (φ, φm ), f m = ( f, φm ) and (φn , φm ) = Since φn ’s are orthonormalized,

b a

(4.26)

φm (x)φn (x) d x.

(φn , φm ) = 0, for m = n, = 1, for m = n. Using this relation, Eq. (4.26) becomes am = f m + so that am =

λ am , λm

λm f m , m = 1, 2, . . . . λm − λ

Hence, using this expression for am in Eq. (4.25), the solution φ(x) can be expressed as ∞  fn φn (x). φ(x) = f (x) + λ (4.27) λ −λ n=1 n The following two cases arise: Case 1 If λ = λn , then Eq. (4.27) gives the unique solution of integral equation (4.22). Case 2 If λ = λn , for any n, then Eq. (4.27) will make no sense if f n = 0 and in that case the integral equation (4.22) is inconsistent. Otherwise if f n = 0, i.e., ( f, φn ) = 0, then f is orthogonal to the eigenfunction φn , and in that case the nonunique solution of Eq. (4.22) will exist. If λ = λm , for some m, and ( f, φm ) = 0,

4.3 Hilbert–Schmidt Theorem

91

then Eq. (4.27) produces the non-unique solution of the integral equation (4.22) as ∞ 

φ(x) = f (x) + λ

n=1,n=m

fn φn (x) + αm φm (x) λn − λ

(4.28)

where αm is an arbitrary constant.

Examples We consider the following examples to illustrate the use of the Hilbert–Schmidt theorem to solve the integral equation of the second kind with a symmetric kernel. Example 4.1 Use the Hilbert–Schmidt theorem to solve the following integral equation:  1 φ(x) − λ k(x, t)φ(t) dt = f (x), 0 ≤ x ≤ 1, (4.29) 0



where the kernel k(x, t) =

x(1 − t), x < t, t (1 − x), x ≥ t.

(4.30)

f (x) are square integrable functions in [0, 1]. Solution: The kernel of the given integral equation (4.29) is symmetric. The homogeneous integral equation corresponding to (4.29) can be written as 

x

φ(x) − λ



1

t (1 − x)φ(t) dt − λ

0

x(1 − t)φ(t) dt = 0.

(4.31)

x

On differentiation, this gives  x  1 tφ(t) dt − λ (1 − t)φ(t) dt = 0. φ (x) + λ 0

x

Again differentiating we obtain φ (x) + λφ(x) = 0. From Eq. (4.31) we have φ(0) = 0 and φ(1) = 0. Thus, the homogeneous equation (4.31) is equivalent to an ODE: d 2φ + λφ = 0, 0 ≤ x ≤ 1, dx2

(4.32)

92

4 Integral Equations of the Second Kind with a Symmetric Kernel

with boundary conditions φ(0) = φ(1) = 0.

(4.33)

Solving this boundary value problem, we obtain φ(x) = A cos

√

λx + B sin

√

λx.

The boundary condition φ(0) = 0 gives A = 0, so that φ(x) = B sin

√

λx.

The boundary condition φ(1) = 0 gives for B = 0, sin

√ λ = 0,

i.e., λ = n2 π2 ,

n = 1, 2, . . . .

Hence, the eigenvalues and the corresponding orthonormal eigenfunctions can be taken as √ (4.34) λn = n 2 π 2 , φn (x) = 2 sin(nπx), n = 1, 2, . . . . We then find, by using Eq. (4.27), that if λ = n 2 π 2 then the unique solution of the given integral equation is given by φ(x) = f (x) +

∞  n=1

2λ 2 n π 2 − λ2

or

 φ(x) = f (x) +

1



1

f (t) sin(nπt) dt sin(nπx),

0

R(x, t; λ) f (t) dt, λ = n 2 π 2

0

where the resolvent R(x, t; λ) is given by the formula R(x, t; λ) = 2λ

∞  sin(nπt) sin(nπx) n=1

n2 π2 − λ

, λ = n 2 π2 .

(4.35)

If λ = m 2 π 2 for some positive integer m, then either Eq. (4.31) has no solution or has the non-unique solution if ( f, φm ) = 0, i.e., when 

1 0

f (x) sin(mπx) d x = 0.

4.3 Hilbert–Schmidt Theorem

93

The non-unique solution is given by 

1

φ(x) = f (x) + αm sin(mπx) +

l(x, t; λ) f (t) dt,

(4.36)

0

where

∞ 

l(x, t; λ) =

n=1,n=m

2λ sin(nπx) sin(nπt) , λ = n 2 π2 , n2 π2 − λ

(4.37)

where αm is an arbitrary constant. Example 4.2 Use the Hilbert–Schmidt theorem to solve the following integral equation:  1

φ(x) − λ

min(x, t)φ(t) dt = x, 0 ≤ x ≤ 1.

(4.38)

0

Solution: Equation (4.38) is of the form 

1

φ(x) − λ

k(x, t)φ(t) dt = x, 0 ≤ x ≤ 1,

0

where  k(x, t) =

t for t < x, x for x ≤ t.

Thus the kernel k(x, t) is symmetric and square integrable. The homogeneous integral equation corresponding to Eq. (4.38) can be written as 

x

φ(x) − λ



1

tφ(t) dt − λx

0

φ(t) dt = 0

x

which on differentiation gives



1

φ (x) − λ

φ(t) dt = 0.

x

Again differentiating, we obtain the following boundary value problem for φ(x):

with boundary conditions

φ (x) + λφ(x) = 0

(4.39)

φ(0) = 0, φ (1) = 0,

(4.40)

94

4 Integral Equations of the Second Kind with a Symmetric Kernel

and solution for Eq. (4.39) is φ(x) = A cos

√

λx + B sin

√

λx.

Now φ(0) = 0 gives A = 0, so that φ(x) = B sin The boundary condition φ (1) = 0 gives cos λn = (2n − 1)2

π2 , 4

√

√

λx.

λ = 0 for B = 0. This gives

n = 1, 2, . . . .

Hence, the eigenvalues and corresponding orthonormal eigenfunctions are given by λn = (2n − 1)2 φn (x) =

π2 , 4

√ πx 2 sin(2n − 1) , n = 1, 2, . . . . 2

Thus, when λ = (2n − 1)2 π4 , n = 1, 2, . . . then the unique solution of the given integral equation is ∞  fn φn (x) φ(x) = x + λ λ −λ n=1 n 2

where

√  fn = 2

1

x sin(2n − 1)

0

i.e., f n = (−1)

(n−1)

πx d x, 2

√ 4 2 . (2n − 1)2 π 2

If λ = λn , then the solution of the given integral equation does not exist as f n = 0. Example 4.3 Use the Hilbert–Schmidt theorem to solve the following integral equation:  1 φ(x) − λ (xt + x 2 t 2 )φ(t) dt = (x + 1)2 , − 1 ≤ x ≤ 1. −1

Solution: The kernel k(x, t) = (xt + x 2 t 2 ) corresponding to the given integral equation is symmetric in −1 ≤ x, t ≤ 1. The corresponding homogeneous integral equation is given by

Exercises

95

 φ(x) − λ

1

−1

(xt + x 2 t 2 )φ(t) dt = 0, − 1 ≤ x ≤ 1.

We have seen from Example 2.5 of Sect. 2.2, Chap. 2, that there are two eigenvalues λ1 = 23 and λ2 = 25 , and the corresponding orthonormalized eigen functions are



φ1 (x) = 23 x and φ2 (x) = 25 x 2 . Thus by the Hilbert–Schmidt theorem, when λ = λi , i = 1, 2, the solution of the given integral equation is given by φ(x) = (x + 1)2 + 4λ[

8x 2 x + ]. 3 − 2λ 15(5 − 2λ)

For λ = λi , i = 1, 2, the solution does not exist as ( f, φi ) = 0.

Exercises 1. Which of the following kernels k(x, t) are symmetric or Hermitian in a ≤ x ≤ b. (A) k(x, t) = cos(x − t), (B) k(x, t) = −i(x − t), (C) k(x, t) = e(x−t) . Answer:

(A), (B)

2. For all Hermitian or symmetric kernels, k(x, t), which of the following is correct? (A) (K φ, ψ) = (φ, K ψ), (B) K (φ, ψ) = (φ, K ψ), (C) (K ∗ φ, ψ) = (φ, K ψ), (D) None of the above. Answer:

(A)

3. All the iterated kernels of a symmetric kernel are − − − − −−. Answer:

Symmetric.

4. Every eigenvalue of a symmetric kernel is − − −−. Answer:

Real.

96

4 Integral Equations of the Second Kind with a Symmetric Kernel

5. The eigenfunctions of a symmetric kernel corresponding to different eigenvalues are − − − − −−. Answer:

Orthogonal.

6. Use the Hilbert–Schmidt theorem to solve the following integral equation:  3 1 (xt + x 2 t 2 )φ(t) dt = x 2 + 1, − 1 ≤ x ≤ 1. φ(x) − 2 −1 Answer:

φ(x) = 5x 2 + Ax + 1, where A is an arbitrary constant.

7. Use the Hilbert–Schmidt theorem to solve the following integral equation:  2π sin(x + t)φ(t) dt = x, − 0 ≤ x ≤ 2π. φ(x) − 0 3

Answer:

2λπ 2 φ(x) = x − (λπ sin x + cos x). 1 − λ2 π 2

References The content of this chapter is mainly based on the website epgp.inflibnet.ac.in/ of UGC E Pathsala and on the following literature. 1. A. Chakrabarti, Applied Integral Equations (Vijay Nicole Imprints Pvt Ltd, 2008) 2. A. Wazwaz, A First Course in Integral Equations (World Scientific Publishing Co. Pvt Ltd, 1997) 3. R.P. Kanwal, Linear Integral Equations Theory and Techniques (Academic, 1971) 4. D. Porter, D.S.G. Stirling, Integral Equations (Cambridge University Press, 1971) 5. F.G. Tricomi, Integral Equations (Interscience Publishers, 1957)

Chapter 5

Abel Integral Equations

5.1 Solution Based on Elementary Integration The simplest form of the Abel integral equation is given by 

φ(t)

x

1

(x − t) 2

0

dt = f (x), x > 0, f (0) = 0

(5.1)

which is a Volterra integral equation of the first kind. Here, we shall solve the Abel integral equation using a very simple method based on elementary integration. 1 We multiply both sides of Eq. (5.1) by (y − x)− 2 and then integrate with respect to x between 0 to y, to obtain  0

y



f (x) 1 2

(y − x) 

y

=

y

dx = 0



x 0

y

φ(t)

0





φ(t) (x − t)

1 2

dt

1

(y − x) 2 

dx 1

t

dx

[(x − t)(y − x)] 2

dt.

(5.2)

The inner integral can be simplified by using the substitution x = y sin2 θ + t cos2 θ so that



y

dx 1

t

[(x − t)(y − x)] 2

= π.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 S. Banerjea and B. N. Mandal, Integral Equations and Integral Transforms, https://doi.org/10.1007/978-981-99-6360-7_5

97

98

5 Abel Integral Equations

Hence, we have from Eq. (5.2), 

f (x)

y



(y − x) 2

0

y

1 dx = π

φ(t)dt

0

which on differentiation with respect to y yields the solution of Abel integral equation (5.1) as  y f (x) 1 d φ(y) = d x. (5.3) π dy 0 (y − x) 21 Another form of the Abel integral equation is given by 

φ(t)

b

1

x

(t − x) 2

dt = f (x), 0 < x < b, f (b) = 0.

(5.4)

Its solution can be obtained in a similar way as 1 d φ(y) = − π dy



b

f (x) 1

(x − y) 2

y

d x.

A more general form of the Abel integral equation is given by  x φ(t) dt = f (x), x > 0, 0 < α < 1, f (0) = 0. α 0 (x − t)

(5.5)

(5.6)

Its solution can be obtained in the following manner. Multiplying both sides of Eq. (5.6) by (y − x)−1+α and integrating with respect to x from 0 to y, we obtain 

y 0

f (x) dx = (y − x)1−α  =

y

0



y



0

0



y

φ(t) t

x

 φ(t) dx dt (x − t)α (y − x)1−α

 dx dt. (x − t)α (y − x)1−α

Now  t

Hence

y

dx (x −

t)α (y

−



y 0

x)1−α

= B(1 − α, α) =

f (x) π dx = (y − x)1−α sin πα

π (α)(1 − α) = . (1) sin πα  0

y

φ(t)dt.

5.1 Solution Based on Elementary Integration

99

Differentiating both sides with respect to y, we obtain the solution of Eq. (5.6) as φ(y) =

sin πα d π dy



y 0

f (x) d x. (y − x)1−α

(5.7)

The solution of another form of a more general Abel integral equation 

b x

φ(t) dt = f (x), 0 < x < b, f (b) = 0, 0 < α < 1 (t − x)α

can be obtained in a similar manner as  b sin πα d f (x) φ(y) = − d x, 0 < y < b. π dy y (y − x)1−α

(5.8)

The most general form of the Abel integral equation is 

x

a

φ(t) dt = f (x) a < x < b (h(x) − h(t))α

(5.9)

where f (a) = 0, 0 < α < 1 and h(x) is a strictly monotonically increasing and differentiable function of x in [a, b] and h  (x) = 0 in [a, b]. Its solution is given by  x  f (t)h  (t) sin πα d dt , a < x < b. (5.10) φ(x) = π d x a (h(x) − h(t))(1−α) The most general form of the other type of the Abel integral equation is given by 

b x

φ(t) dt = f (x), (h(t) − h(x))α

a < x < b,

(5.11)

where f (b) = 0 and 0 < α < 1. Its solution is given by φ(x) = −

sin πα d π dx



b x

 f (t)h  (t) dt , a < x < b. (h(t) − h(x))1−α

(5.12)

Illustrative Examples We now illustrate this method by considering the following examples. Example 5.1 Solve the following: 

x 0

φ(t) dt = π sin πx, 0 ≤ x ≤ 1. √ x −t

(5.13)

100

5 Abel Integral Equations

Solution: Multiplying both sides by (y − x)− 2 and integrating with respect to x between 0 to y, we obtain   y  x  y sin πx φ(t) dx dx = . π dt √ √ √ y−x y−x x −t 0 0 0 1

Interchanging the order of integration and simplifying, we obtain 

y

π 0

sin π(y − x) dx = x 1/2





y

φ(t)

0

y

t

Noting the result (5.2), we obtain  y  φ(t)dt = 0

y

0

dx

 dt. 1

{(x − t)(y − x)} 2

sin π(y − x) d x, x 1/2

which on differentiation with respect to y gives the solution as 

cos π(y − x) d x. x 1/2

(5.14)

dt = x 2 , 0 ≤ x ≤ 1,

(5.15)

y

φ(y) = π 0

Example 5.2 Solve 

φ(t)

x

(x 2

0

1

− t 2) 2

Solution: Substituting x 2 = u, t 2 = v

(5.16)

in Eq. (5.15), we obtain 

u 0

φ1 (v) dv = u, 0 ≤ u ≤ 1 √ u−v

where φ1 (v) =

√ φ( v) √ . 2 v

(5.17)

(5.18)

Equation (5.17) is in the form given by Eq. (5.1). Hence its solution is given by (cf. Eq. (5.3))  v u 1 d du. φ1 (v) = √ π dv 0 v−u

5.2 Solution Based on Laplace Transform

101

Using the relations given by Eq. (5.16), we obtain the solution of Abel integral equation (5.15) as φ(t) =

4 2 t , 0 ≤ t ≤ 1. π

(5.19)

Example 5.3 Solve 

1 x

φ(t) dt = g(x) 0 ≤ x ≤ 1, g(1) = 0. √ t2 − x2

Solution: Substituting x 2 = u, t 2 = v, this equation can be cast into  1 φ1 (v) dv = g1 (u), 0 ≤ u ≤ 1 √ v−u u √ √ where φ1 (v) = φ(√vv) , g1 (u) = g( u). This equation has the same form as given by Eq. (5.4) and its solution is given by (cf Eq. (5.5))  1 g1 (u) 1 d φ1 (v) = − du. √ π dv v u−v

Using the transformation mentioned above, we obtain the solution of the equation given the integral equation as 2 d φ(x) = − π dx



1 x

 tg(t) dt , 0 ≤ x ≤ 1. (t 2 − x 2 )

5.2 Solution Based on Laplace Transform We now solve the Abel integral equation by using the Laplace transform technique. The Laplace transform of the functions φ(x) and f (x) is defined by  L(φ(x), f (x)) = ((s), F(s)) =

∞

(φ(x), f (x))e−sx d x,

(5.20)

0

where s is a complex number such that (s) > δ, and δ is a positive real number. The inversion formula for (s) and F(s) is given by 1 L ((s), F(s)) = (φ(x), f (x)) = 2πi −1



γ+i∞ γ−i∞

((s), F(s))esx ds,

(5.21)

102

5 Abel Integral Equations

where γ is a constant which exceeds the real parts of all singularities of (s) and F(s). We may note here that γ may be different for (s) and F(s). We also define the convolution of two functions f (x) and g(x) as 

x

( f ∗ g)(x) =

f (t)g(x − t)dt.

(5.22)

0

Taking the Laplace transform of ( f ∗ g)(x), we obtain 

∞

L[( f ∗ g)(x)] =



0

x

f (t)g(x − t)dt e−sx d x = F(s)G(s).

(5.23)

0

Here, F(s) and G(s) are Laplace transforms of the functions f (x) and g(x), respectively. The inverse formula for the Laplace transform yields ( f ∗ g)(x) = L−1 [F(s)G(s)].

(5.24)

The Abel Integral equation of the first kind is given by 

φ(t) dt = f (x), x > 0, f (0) = 0. √ x −t

x 0

(5.25)

Taking Laplace transform of the above equation, we obtain  0

∞



x

0

 φ(t) dt e−sx d x = F(s). √ x −t

The inner integral is the convolution of the functions φ(t) and Eq. (5.23), we can write Eq. (5.26) as

Here

(5.26) √1 . x

In view of

(s)M(s) = F(s).

(5.27)

   π 1 = , Re(s) > 0, M(s) = L √ s x

(5.28)

1 L−1 [M(s)] = √ . x

(5.29)

so that

Thus, (s) can be written as  (s) = F(s)

s π

5.2 Solution Based on Laplace Transform

103

which can be alternately expressed as  s π 1 F(s)] = [M(s)F(s)]. (s) = s[ π s π

(5.30)

Let H (s) = M(s)F(s)

(5.31)

be the Laplace transform of the function h(x), Thus h(x) = L−1 [M(s)F(s)].

(5.32)

Noting the relation given by Eq. (5.24) and using the relation (5.29), we obtain  x f (t) h(x) = dt. (5.33) √ x −t 0 Hence (s) =

1 s H (s). π

(5.34)

   ∞ dh dh −d x = e L = s H (s) − h(0). dx dx 0

Now

From Eq. (5.33) we observe that h(0) = 0, so that dh = L−1 [s H (s)] = πL−1 [(s)]. dx

(5.35)

Now taking Laplace inverse of (s), we obtain φ(x) =

1 dh . π dx

Substituting h(x) from Eq. (5.33), we obtain the solution of Abel integral equation (5.25) as  x 1 d f (t) φ(x) = dt. (5.36) √ π dx 0 x −t Next we shall employ the Laplace transform method to obtain the solution of the general form of the Abel integral equation given by 

x 0

φ(t) dt = f (x), x > 0, 0 < α < 1, f (0) = 0. (x − t)α

(5.37)

104

5 Abel Integral Equations

We define the Laplace transform of the functions φ(x) and f (x) together with the inverse formula as given by Eqs. (5.20) and (5.21). Multiplying Eq. (5.37) by e−sx and integrating with respect to x from 0 to ∞, we obtain  0

∞



x 0

 φ(t) dt e−sx d x = F(s). (x − t)α

(5.38)

The left-hand side of Eq. (5.38) is the Laplace transform of the convolution of the functions φ(x) and x −α , so Eq. (5.38) can be written as (s)N (s) = F(s)

(5.39)

N (s) = L(x −α ) = s α−1 (1 − α)

(5.40)

where

and L−1 (s α−1 ) = Hence (s) = Let

so that

x −α . (1 − α)

s [s −α F(s)]. (1 − α)

(5.41)

(5.42)

(s) = s −α F(s) ψ(x) = L−1 [(s)] = L−1 [s −α F(s)].

(5.43)

Using relations (5.41) and (5.24), we obtain ψ(x) =

1 (α)



x

f (u)(x − u)α−1 du.

(5.44)

0

Now, from Eq. (5.42) we obtain (s) =

s (s). (1 − α)

(5.45)

Using the inverse formula for the Laplace transform, we obtain φ(x) = Now,

1 L−1 [s(s)]. (1 − α)

(5.46)

5.2 Solution Based on Laplace Transform

105

  ∞  dψ −sx dψ (x) = e d x = s(s) − ψ(0). L dx dx 0 From Eq. (5.44), we observe that ψ(0) = 0. Thus   dψ . s(s) = L dx Hence

dψ = L−1 [s(s)]. dx

(5.47)

Thus comparing Eqs. (5.46) and (5.47), we obtain φ(x) =

dψ 1 . (1 − α) d x

(5.48)

Now using the relation (5.44), we can write Eq. (5.48) as φ(x) =

1 d (1 − α)(α) d x



x

0

f (u) du. (x − u)1−α

Using the well-known relation (1 − α)(α) =

π , sin πα

we obtain the solution of Abel integral equation (5.37) as φ(x) =

1 d sin πα π dx

 0

x

f (u) du. (x − u)1−α

(5.49)

Remark: It may be noted that apart from the Abel integral equation, which involves a weakly singular kernel, there are integral equations whose kernels are strongly singular, which are also known as singular integral equations, such as Cauchy singular integral equations and hypersingular integral equations. However, these are outside the scope of this book. The interested readers may consult the books on singular integral equations such as Applied Singular Integral Equations, B. N. Mandal and A. Chakrabarti, Science Publishers/CRC Press, USA (2011) and others.

106

5 Abel Integral Equations

Exercises Solve the following Abel integral equations: 1.

π (x 2 2

− x) =

x

x 0

0 ≤ x ≤ 1,

Answer: φ(x) =

− π2

√

x( 4x3 − 1).  u x f (x) d



f (x), 0 ≤ x ≤ 1, Answer: uφ(u) = − du 0 √u 2 −x 2 d x .   u x f (x) 1 φ(x) 2 d 3. u √x 2 −u 2 d x = f (u), 0 ≤ u ≤ 1, Answer: φ(x) = − π d x 0 √u 2 −x 2 d x . 2.

4.

=

√φ(t) dt x−t

√uφ(u) du 0 x 2 −u 2

x 0

√φ(t) dt x−t

3

= 43 x 2 , 0 ≤ x ≤ 1,

Answer: φ(x) = x.

Solve the following Abel integral equations by using Laplace transform:

Given that

x

√φ(t) dt, x−t L[x] = s12 .

5. πx =

0

x √ 6. 4 x = 0 1 2

Given that L[x ] 7.

3π 2 x 8

=

x 0 2

Given that L[x ]

0 ≤ x ≤ 1,

√φ(t) dt, x−t √ = 2π 13 s2 √φ(t) dt, x−t = s23 .

0 ≤ x ≤ 1,

√ Answer: φ(x) = 2 x.

Answer: φ(x) = 2.

. 0 ≤ x ≤ 1,

3

Answer: u(x) = x 2 .

References The content of this chapter is mainly based on the website epgp.inflibnet.ac.in/ of UGC E Pathsala and on the following literature. 1. A. Chakrabarti, Applied Integral Equations (Vijay Nicole Imprints Pvt Ltd, 2008) 2. A. Wazwaz, A First Course in Integral Equations (World Scientific Publishing Co. Pvt Ltd, 1997) 3. R.P. Kanwal, Linear Integral Equations Theory and Techniques (Academic, 1971) 4. D. Porter, D.S.G. Stirling, Integral Equations (Cambridge University Press, 1971)

Part II

Integral Transform

Chapter 6

Fourier Transform

6.1 Integral Transform: An Introduction What is an integral transform? An integral transform of a function f (x) defined on (a, b) has the form  I ( f ) ≡ F(y) =

b

K (x, y) f (x) d x

a

provided the integral exists, where K (x, y) is a known function of x, y. The function K (x, y) is called the kernel of the integral transform. Note: (1) The kernel K (x, y) along with the interval (a, b) (which need not be finite) distinguishes a particular integral transform from another. (2) In general, the existence of the integral in I ( f ) depends on the functions f (x) and K (x, y). However, all integral transforms satisfy I ( f + g) = I ( f ) + I (g) and I (c f ) = cI ( f ) where c is a constant ( f and g being functions). Remark: Integral transforms are very much useful in dealing with differential equations (linear) with appropriate boundary conditions. Proper choice of the integral transform makes it possible to convert an ordinary differential equation, which may be intractable, to an algebraic equation that can perhaps be easily solved. The solution © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 S. Banerjea and B. N. Mandal, Integral Equations and Integral Transforms, https://doi.org/10.1007/978-981-99-6360-7_6

109

110

6 Fourier Transform

obtained is in fact the transform of the solution of the original differential equation and it is necessary to find the inverse transform to obtain the desired solution. Thus, we need to know the integral transform and the corresponding inverse transform. There exists quite a large number of integral transforms, but here only five transforms, viz. Fourier, Laplace, Mellin, Hankel and z-transform, will be discussed.

Fourier Transform Motivation One can obtain the Fourier transform from the study of the Fourier series. In the Fourier series, a periodic function is expressed in terms of the sum of sines and cosines. Thus, a complicated function which is periodic can be expressed as the sum of simple functions represented by sines and cosines. The Fourier transform can be thought of as an extension of the Fourier series in which the period of the function is made infinity.

6.2 Fourier Integral Theorem We assume that a function f is periodic with period 2 p and f (x), f  (x) are piecewise continuous on [− p, p], then it has the Fourier series representation f (x) = where an = and

1 p

 ∞   1 nπx nπx an cos a0 + +bn sin 2 p p n=1 

p −p

1 bn = p



p

−p

f (t) cos

nπt dt, n = 0, 1, 2, 3 . . . p

f (t) sin

nπt dt, n = 1, 2, 3 . . . . p

(6.1)

Substituting the expression for an and bn into the expression for f (x), we find that the representation of f (x) given by Eq. (6.1) is equivalent to 1 f (x) = 2p



p

−p

1 f (t)dt + p



p −p

∞ 

 nπ(t − x) f (t) cos dt. p n=1

(6.2)

We now make p → ∞ in this representation so that f (x) is defined on (−∞, ∞) and also assume that f is absolutely integrable on (−∞, ∞). Then for the first term in (6.2), we find

6.2 Fourier Integral Theorem

111

1 p→∞ 2 p



p

f (t)dt = 0.

lim

−p

For the second term in (6.2), we write s = πp so that as p → ∞, s → 0 and (6.2) gives ∞  π  1 s f (x) = lim f (t) cos(ns(t − x))s dt s→0 π − π s n=1 =

1 π





∞ −∞

 cos s(t − x)ds dt.

∞

f (t) 0

Interchanging the order of integration, we find 

1 f (x) = π

∞



∞

f (t) cos s(t − x)dtds.

−∞

0

(6.3)

Thus, we summarize this result in the following theorem. Theorem 6.1 (Fourier Integral Theorem) If a function f (x) is defined on (−∞, ∞) and is absolutely integrable on (−∞, ∞) then 

1 π

f (x) =

∞



∞

f (t) cos s(t − x)dtds.

−∞

0

This is known as the Fourier integral theorem. The integral expansion (6.3) can be written as 1 f (x) = 2π



∞

1 2π

∞

−∞

0

1 = 2π =







∞

−∞ ∞

f (t) ei(t−x)s + e−i(t−x)s dtds



∞

f (t)ei(t−x)s dtds

−∞

e−isx



−∞

∞

−∞

 f (t)eits dt ds.

(6.4)

This is in fact the exponential form of the Fourier integral theorem. Fourier Transform: In (6.4), if we write 1 F(s) = √ 2π then

1 f (x) = √ 2π





∞

f (t)eits dt,

(6.5)

F(s)e−isx ds.

(6.6)

−∞ ∞ −∞

112

6 Fourier Transform

F(s) is called the Fourier transform of f (x) defined on (−∞, ∞) and f (x) is called the inverse Fourier transform of F(s). Note: The above is a purely formal procedure to obtain Fourier transform and the inverse Fourier transform.

6.3 Rigorous Justification of Fourier Integral Theorem The Fourier integral theorem written in the form (6.3) can be justified rigorously from the following two results. (1) Lemma 6.1 Let f (x) be piecewise continuous and absolutely integrable in (−∞, ∞). Then  ∞ lim f (x)eiλx d x = 0. (6.7) λ→∞ −∞

This result is usually known as the Riemann–Lebesgue Lemma. Proof Its proof is straightforward if we assume f (x) to be continuous and its derivative is bounded on (−∞, ∞). Let − p < x < p, then by performing integration by parts, we obtain 

p

−p

f (x)e

iλx

eiλ p − e−iλ p 1 d x = f (x) − iλ iλ



p

−p

f  (x)eiλx d x.

The first term on the right side vanishes as λ → ∞ since f (x) and e±iλ p are bounded. Since f  (x) is bounded,  |

p −p





f (x)e

iλx

d x| ≤

p

−p

| f  (x)|d x < ∞

for all λ. Hence, the second term vanishes as λ → ∞. Thus  p lim f (x)ei xλ d x = 0. λ→∞ − p

Now making p → ∞, the result is proved. Remark: The assumption that f  (x) is bounded in this result is not required. However, this assumption makes the proof simple. (2) Lemma 6.2 If f (x) is piecewise smooth and absolutely integrable on (−∞, ∞), and x is a point of continuity of f (x), then 1 λ→∞ π



∞

lim

−∞

f (x + t)

sin λt dt = f (x). t

(6.8)

6.3 Rigorous Justification of Fourier Integral Theorem

113

Proof To prove this result, we first note that for p > 0 1 λ→∞ π



p

lim

−p

sin λt 1 dt = lim λ→∞ π t



λp

sin u 1 du = u π

−λ p



∞

−∞

sin u du = 1. u

(6.9)

To derive this result, we choose  f (x) =

1, −1 < x < 1, 0, otherwise.

In the Fourier integral expansion (6.3) putting x = 0, we find 1=

1 π



∞

  2 ∞ sin s ds. cos stdt ds = π 0 s −1



0

1

Using the result (6.9), we see that proving (6.8) is equivalent to proving 1 λ→∞ π



p

lim

−p

f (t + x) − f (x) sin λtdt = 0. t

(6.10)

f (x) The function f (x+t)− , regarded as a function of t, is piecewise continuous for all t t = 0. For t → 0, the above function is f  (x) which exists since f (x) is piecewise smooth. Thus, the conditions of Lemma 6.1 above are satisfied by this function, and taking the imaginary part, the integral (6.10) vanishes. The desired result follows if we allow p to become infinite. To prove (6.3) rigorously, we consider points of continuity of f (x). We consider the double integral    ∞ 1 λ f (t) cos s(t − x)dt ds π 0 −∞

=

1 π



∞

−∞

= =

1 π 1 π

 f (t)

λ

 cos s(t − x)ds dt

0



∞ −∞



∞ −∞

f (t)

sin λ(t − x) dt t−x

f (t + x)

sin λt dt. t

Making λ → ∞ and using (6.8), we obtain (6.3) somewhat rigorously. Thus, (6.4) is also justified in this sense. If x is a point of discontinuity, then f (x) in (6.3) or (6.4) is to be replaced by f (x+0)+ f (x−0) . 2

114

6 Fourier Transform

6.4 Fourier Cosine and Sine Transforms If f (x) is an even function of x, i.e., f (−x) = f (x), x > 0, then (6.3) gives

f (x) =

1 π



∞ 0





∞ 0

=

∞

f (t) cos s(t − x) ds +

 f (t) cos s(t + x) ds dt

0

2 π



∞



0

 =

 f (t) cos st dt cos sx ds

∞

0

2 π



∞

Fc (s) cos sx ds, x > 0

(6.11)

0

where 



2 π

Fc (s) =

∞

f (x) cos sx d x, s > 0.

(6.12a)

0

This is the Fourier cosine transform, the inverse cosine transform being  f (x) =

2 π



∞

Fc (s) cos sx ds, x > 0.

(6.12b)

0

Similarly, if f (x) is an odd function of x, i.e., f (−x) = − f (x), x > 0, then (6.3) gives    ∞ 1 ∞ f (t){cos s(t − x) − cos s(t + x)} ds dt f (x) = π 0 0 2 = π



∞



0

 =

∞

 f (t) sin st dt sin sx ds

0

2 π



∞

Fc (s) sin sx ds, x > 0

(6.13)

0

where  Fs (s) =

2 π

 0

∞

f (x) sin sx d x, s > 0.

(6.14a)

6.5 Fourier Transforms of Some Simple Functions

115

This is the Fourier sine transform, the inverse sine transform being  f (x) =

2 π



∞

Fs (s) sin sx ds, x > 0.

(6.14b)

0

6.5 Fourier Transforms of Some Simple Functions We have learnt that the Fourier transform of a function f (x), denoted by F[ f (x)] or F(s), is given by  ∞ 1 f (x) d x F[ f (x)] = √ 2π −∞ with its inverse transform as 1 f (x) = √ 2π



∞ −∞

F[ f (x)] d x.

We shall consider the following examples to obtain the Fourier transform of some simple functions. Example 6.1 Find the Fourier transform of f (x) where  f (x) =

1, −1 < x < 1, 0, otherwise.

Solution: 1 F[ f (x)] = √ 2π



1

−1

1.eisx d x =

1 eis − e−is = 2π is

Remark: If we write the inverse transform, then   1 ∞ sin s −isx 1, −1 < x < 1, e ds = 0, otherwise. π −∞ s For x = 0, one obtains the result  ∞ −∞

so that



∞ 0

sin s ds = π s π sin s ds = . s 2



2 sin s . π s

116

6 Fourier Transform

Example 6.2 Find the Fourier transform of f (x), where f (x) = e−a|x| , a > 0. Solution:

 ∞ 1 e−a|x| eisx d x F[e−a|x| ] = √ 2π −∞

 ∞   ∞ 1 e−ax+isx d x + e−ax−isx d x =√ 2π 0 0   1 1 1 + =√ a + is 2π a − is  =

a 2 , a > 0. 2 π s + a2

Example 6.3 Find the Fourier transform of f (x) where x2

f (x) = e− 2 . Solution:

 ∞ 1 x2 e− 2 eisx d x F[e ] = √ 2π −∞  1 − s2 ∞ − 1 (x−is)2 2 =√ e e 2 dx 2π −∞  ∞ 1 1 s2 2 = √ e− 2 e−u du, u = √ (x − is). π 2 −∞ 2

− x2

To derive I =

∞

−∞

2

eu du, we write  I = 2

∞

e

−u 2

  du

e

−∞

 =

∞



∞ −∞





∞ 0

−v 2

−∞

−∞

=

∞

0

e−(u

2π

2

+v 2 )

du dv

e−r r dθ dr, 2

 dv

6.5 Fourier Transforms of Some Simple Functions

117

changing to polar coordinates, 

∞

=

e

−r 2

 



2π

r dr

dθ

0

0

=

1 .2π = π 2

so that I = Hence

√

π.

x2

s2

F[e− 2 ] = e− 2 . x2

Thus, the function f (x) = e− 2 has the Fourier transform f (s) and is selfreciprocal in the sense that it is its own Fourier transform. Example 6.4 Find the Fourier transform of f (x) where  f (x) = Solution: Here

0, x < 0, e−ax , x > 0, a > 0.

1 F[ f (x)] = √ 2π



∞

e−ax eisx d x

0

1 1 =√ . 2π a − is

Example 6.5 Find the Fourier transform of f (x) where  f (x) = Solution: Here

√

sin x, 0 < x < π, 0, otherwise. 

2πF[ f (x)] =

π

sin xeisx d x 0

=

1 2i

1 = 2i



π

 ix  e − e−i x eisx d x

0



ei(s+1)x ei(s−1)x − i(s + 1) i(s − 1)

π 0

118

6 Fourier Transform

 1 ei(s+1)π − 1 ei(s−1)π − 1 − =− 2 s+1 s−1

isπ  e + 1 eisπ + 1 1 + =− − 2 s+1 s−1 1 eisπ + 1 . F[ f (x)] = √ 2π 1 − s 2 Remark: This integral can also be evaluated in a different manner: 

π

I =

sin x eisx d x 0



π = − cos x eisx 0 + is =e

isπ

+ 1 + is[sin x

eisx ]π0



π

cos x eisx d x 0



π

− is

sin x eisx d x 0

= eisπ + 1 + s 2 I so that I =

eisπ + 1 . 1 − s2

Example 6.6 Show that the Fourier transform of the function  f (x) =

1 − |x|, |x| < 1, 0, otherwise 

is F(s) = 2 and hence show that



∞



−∞

Solution: Here

sin x x

2 sin2 π s2

s 2

2 d x = π.

 1 √ 2π F(s) = (1 − |x|)eisx d x −1



1

=2 0

(1 − x) cos sx d x

6.5 Fourier Transforms of Some Simple Functions



(1 − x) sin sx =2 s

119

1



2 + s 0

sin sx d x 0

=

2 [− cos sx]10 s2

=

2(1 − cos sx) s2 =4

1

sin2 2s . s2

 sin2 s Thus F(s) = 2 π2 s 2 2 . For the second part, we note that 1 f (x) = F −1 [F(s)] = √ 2π



∞

F(s)eisx ds.

−∞

Putting x = 0, we obtain 1 f (0) = √ 2π i.e.,

∞

F(s) ds, −∞

  ∞ sin2 2s 2 1 ds 1= √ 2 π −∞ s 2 2π  2 ∞ sin u = 2 du π −∞ 4u 2 1 = π

Thus, we deduce





∞ −∞



∞



−∞



sin x x

sin u u

2 du.

2 d x = π.

Example 6.7 Find the Fourier transform of f (x) = 1, − ∞ < x < ∞.

120

6 Fourier Transform

Solution: The Fourier transform of f (x) defined on (−∞, ∞) exists if 

∞

−∞

| f (x)| d x < ∞.

Obviously, if f (x) = 1, then this is not satisfied. However, one can still define the Fourier transform of 1 by considering the notion of the generalized function δ(x) which is defined as δ(x) = 0 for x = 0, and 

∞

−∞

ϕ(x)δ(x − a) d x = ϕ(a)

where the integral has no meaning in the usual sense and is used to define the generalized function δ(x). Thus if we take a = 0, ϕ(x) = e−isx then 

∞

−∞

so that

δ(x)e−isx d x = 1

1 F −1 [δ(s)] = √ , 2π

and one can write F[1] =

√

2π δ(s).

6.6 Properties of Fourier Transform In this section, we shall study the properties of Fourier transform. Before proceeding further, we first make the following remark. Remark: Let f (x) satisfy the conditions of Fourier integral theorem, then its Fourier transform F(s) is unique. However, if f (x) and g(x) are two functions which are identical everywhere except for certain isolated points, then they have the same transform, say F(s). This means that the inverse transform of F(s) can be either f (x) or g(x). The distinction between functions that differ at isolated points is mostly of academic interest and we ignore this in our discussion. Thus for a given function F(s), its inverse f (x) is uniquely related to F(s) by the relation:  ∞ 1 F(s)e−isx ds. f (x) = √ 2π −∞

6.6 Properties of Fourier Transform

121

Linearity Property: If F(s) and G(s) are the Fourier transforms of f (x) and g(x) respectively, then the Fourier transform of c1 f (x) + c2 g(x), where c1 and c2 are constants, is c1 F(s) + c2 G(s). Proof The proof is obvious. However for completeness the proof is given below. Let F[ f (x)] denote the Fourier transform of f (x) so that F[ f (x)] = F(s) and F[g(x)] = G(s). Then  ∞ 1 {c1 f (x) + c2 g(x)}eisx d x F[c1 f (x) + c2 g(x)] = √ 2π −∞  ∞  ∞ 1 1 isx = c1 √ f (x)e d x + c2 √ g(x)eisx 2π −∞ 2π −∞ = c1 F(s) + c2 G(s). ∞ Note: If the inverse transform is denoted by F −1 [F(s)] = √12π −∞ F(s)e−i xs ds, then F −1 [c1 F(s) + c2 G(s)] = c1 F −1 [F(s)] + c2 F −1 [G(s)] so that the inverse Fourier transform has the linearity property. Scaling Property: 1 F(s), a = 0 where a is constant, either positive or negative. F[ f (ax)] = |a| Proof To prove this, we first take a > 0, then

 ∞ 1 f (ax)eisx d x F[ f (ax)] = √ 2π −∞  ∞ 1 1 s = √ f (u)ei a u du a 2π −∞ after putting ax = u. Thus   s 1 for a > 0. F[ f (ax)] = F a a   Similarly, if a < 0, then F[ f (ax)] =

− a1 F

s a

for a < 0.

Thus, the scaling property for Fourier transform is F[ f (ax)] = a = 0.

  1 F |a|

s a

for

122

6 Fourier Transform

Shifting Property: (i) F[eiax f (x)] = F(s + a) (ii) F[ f (x − a)] = eias F(s). Proof of (i)

 ∞ 1 F[eiax f (x) = √ eiax f (x)eisx d x 2π −∞  ∞ 1 =√ f (x)ei(s+x)x d x 2π −∞ = F(s + a).

Proof of (ii)

 ∞ 1 F[ f (x − a)] = √ f (x − a)eisx d x 2π −∞  ∞ 1 =√ f (u)eis(x+a) du 2π −∞  ∞ ias 1 =e √ f (x)eisu du 2π −∞ = eias F(s).

Remark: The above properties are useful to deduce the transform of many other functions once the transforms of some standard functions are derived. Fourier Transform of a Derivative: Let f (x) be continuous everywhere and f  (x) (the derivative of f (x)) be piecewise smooth, both f (x) and f  (x) being absolutely integrable on (−∞, ∞) and f (x) → 0 as |x| → ∞, then F[ f  (x)] = −is F(s). Proof  ∞ 1 F[ f  (x)] = √ f  (x)eisx d x 2π −∞  ∞

1 1 isx ∞ =√ f (x)eisx d x f (x)e −∞ − is √ 2π 2π −∞ = −is F(s).

6.6 Properties of Fourier Transform

123

This can be generalized as F[ f n (x)] = (−is)n F(s), n = 1, 2, . . . if f (x), f  (x), . . . . . . , f n−1 (x) are continuous and absolutely integrable on (−∞, ∞), f (x), f  (x), . . . . . . , f n−1 (x) → 0 as |x| → ∞ and f n (x) is piecewise smooth and absolutely integrable on (−∞, ∞). Remark: For sine and cosine Fourier transforms, the above results are somewhat different. Let Fc (s) and Fs (s) respectively denote the Fourier cosine and sine transforms of f (x) defined on [0, ∞), i.e.,   ∞ 2 Fc [ f (x)] ≡ Fc (s) = f (x) cos sx d x, π 0  Fs [ f (x)] ≡ Fs (s) =

2 π



∞

f (x) sin sx d x.

0



Then 

Fc [ f (x)] = s Fs (s) − Similarly

2 f (0). π

Fs [ f  (x)] = −s Fc (s). 

For second derivative f (x) of f (x), one can derive  

2



2

Fc [ f (x)] = −s Fc (s) −  Fs [ f (x)] = −s Fs (s) +

2  f (0), π 2 s f (0). π

Derivative of Fourier Transforms: ∞ Let F(s) = √12π −∞ f (x)eisx d x, then the formal differentiation of both sides with respect to s produces 1 F  (s) = √ 2π



∞

−∞

i x f (x)eisx d x

so that F[x f (x)] = −i F  (s) provided of course the Fourier transform of x f (x) exists.

124

6 Fourier Transform

Repeated differentiation w.r.t. s produces F so that

(n)

1 (s) = √ 2π



∞

−∞

(i x)n f (x)eisx d x

F[x n f (x)] = (−i)n F (n) (s)

provided the Fourier transform of x n f (x) exists. Fourier Transform of a Fourier Transform Let F[ f (x)] = F(s) then F[F(x)] = f (−s). Proof We have F[ f (x)] ≡ F(s) =

∞

√1 2π

1 f (x) = √ 2π

−∞



∞

f (x)eisx d x and

F(s)e−isx ds.

−∞

Interchanging x and s, and then replacing s by −s we find 1 f (−s) = √ 2π



∞

F(x)eisx d x

−∞

= F[F(x)] Behavior at Infinity lim |F(s)| = 0.

|s|→∞

Proof This follows from the Riemann–Lebesgue lemma. However, a simple direct proof is given below. We have  ∞ 1 f (x)eisx d x F(s) = √ 2π −∞  ∞ 1 = −√ f (x)eisx+iπ d x 2π −∞

6.7 Convolution Theorem and Parseval Relation

1 = −√ 2π



∞ −∞

125 π

f (x)eis(x+ s ) d x

so that

  ∞  ∞ 1 1 1 π f (x)eisx d x − √ f (x)eis(x+ s ) d x √ 2 2π −∞ 2π −∞

  ∞  ∞ 1 1 π isu 1 isx f (x)e d x − √ f (u − )e d x = √ 2 s 2π −∞ 2π −∞  ∞ π  1 f (x) − f (x − ) eisx d x. = √ s 2 2π −∞

F(s) =

Thus

so that

1 |F(s)| ≤ √ 2 2π



 π    f (x) − f (x − ) d x s −∞ ∞

1 lim F(s) ≤ √ lim |s|→∞ 2 2π |s|→∞



 π    f (x) − f (x − ) d x s −∞ ∞

= 0.

6.7 Convolution Theorem and Parseval Relation Convolution of Two Functions Defined on (−∞, ∞) Definition 6.1 The convolution of two functions f (x) and g(x), denoted by ( f ∗ g), is defined by  ∞ 1 ( f ∗ g)(x) = √ f (x − t)g(t) dt. 2π −∞ Properties of the Convolution of Two Functions: The convolution of two functions f (x) and g(x) has the following properties: (a) Commutative: ( f ∗ g)(x) = (g ∗ f )(x) (b) Associative : (( f ∗ g) ∗ h)(x) = ( f ∗ (g ∗ h))(x) (c) Distributive: ( f ∗ (g + h))(x) = ( f ∗ g)(x) + ( f ∗ h)(x). Convolution Theorem for Fourier Transform: If F[ f (x)] = F(s) and F[g(x)] = G(s), then (i) F[( f ∗ g)(x)] = F(s)G(s),

126

6 Fourier Transform

(ii) F −1 [F(s)G(s)] = ( f ∗ g)(x). Proof 1 F[( f ∗ g)(x)] = 2π 1 = 2π 1 = 2π



∞



−∞

∞

∞

g(t)

−∞

 g(t)

−∞

∞

−∞



−∞





∞

 f (x − t)g(t) dt eisx d x 

f (x − t)e

isx

d x dt 

∞

f (u)e

−∞

is(u+t)

du dt



  ∞  ∞ 1 1 = √ g(t)eist dt √ f (u)eisu du 2π −∞ 2π −∞ = F(s)G(s). Parseval Relation: Parseval’s relation is given by 

∞

−∞

 | f (x)| d x =

∞

2

−∞

|F(s)|2 ds.

Proof The result in the previous theorem is equivalent to 

∞

−∞

F(s)G(s)e−isx ds =



∞ −∞

f (x − t)g(t) dt.

Putting x = 0, this produces 

∞

−∞

 F(s)G(s) ds =  =

∞ −∞

∞

−∞

f (−t)g(t) dt

f (t)g(−t) dt.

If we put g(x) = f (−x), then  ∞  ∞ 1 1 g(t)eisx d x = √ f (−x)eisx d x G(s) = F[g(x)] = √ 2π −∞ 2π −∞  ∞  ∞ 1 1 =√ f (x)e−isx d x = √ f (x)eisx d x = F(s). 2π −∞ 2π −∞

6.7 Convolution Theorem and Parseval Relation

127

Thus, the above gives 



∞

−∞

F(s)F(s) ds =

∞

−∞

f (x) f (x) d x

which is equivalent to 



∞

| f (x)|2 d x =

−∞

∞

−∞

|F(s)|2 ds.

Remark: The general Parseval’s relation is 



∞

−∞

f (x)g(x) d x =

∞

F(s)G(s) ds. −∞

A formal proof is as follows. The expression on the right-hand side is 

∞

F(s)G(s) ds −∞

1 = 2π =

1 2π

=

1 2π





∞ −∞



∞

∞

f (x)e

−∞



−∞



 

∞

isx

 g(t)

−∞

−∞

f (x)

We know that F[1] = i.e., 1 2π 

so that

∞

−∞



∞

−∞

∞

 g(t)e−ist dt ds

−∞



∞

dt ds

−∞

f (x)eisx d x

−∞

g(t)eist

dx  

∞



∞

∞

 eis(x−t) ds

 d x dt.

−∞

√ 2πδ(s),

eisx d x = δ(s)

eis(x−t) ds = 2πδ(x − t).

Using this relation, we obtain 

∞

−∞

 F(s)G(s) ds =

∞

−∞

 g(t)

∞

−∞

 f (x)δ(x − t) d x dt.

128

6 Fourier Transform



Hence



∞

−∞

F(s)G(s) ds =

∞

−∞

g(t) f (t) dt.

Convolution Theorem for Fourier Cosine and Sine Transforms: 

Let Fc [ f (x)] ≡ Fc (s) =   2 ∞

and Fc [g(x)] ≡ G c (s) = 

∞

π

2 π



∞

f (x) cos sx d x

0

g(x) cos sx d x, then

0

1 2

Fc (s)G c (s) cos sx ds =

0



∞

f (t) [g(x + t) + g(|x − t|)] dt.

0

The left side is 

2 π

 = 1 = 2



∞



∞



0

 f (t) cos st dt cos sx ds

0



2 π

∞

G c (s) 

∞

f (t)

0

 f (t)

0

1 = 2

∞

 G c (s) cos st cos sx ds dt

0



2 π



∞

G c (s){cos s(x + t) + cos s(x − t)}ds dt

0 ∞

f (t) [g(x + t) + g(|x − t|)] dt.

0

Putting x = 0, this produces 

∞



∞

Fc (s)G c (s) ds =

0

f (t)g(t) dt.

0

Setting g(x) = f (x), we find  G c (s) =  =

2 π

 0

∞

2 π



∞

f (x) cos sx d x

0

f (x) cos sx d x = Fc (s)

6.7 Convolution Theorem and Parseval Relation



so that

∞

129



∞

|Fc (s)|2 ds =

0

| f (x)|2 d x.

0

This is the Parseval relation for Fourier cosine transform.

Examples Example 6.8 Let f (x) = e−ax (a > 0) and g(x) = e−bx (b > 0). Then it can be shown that  a 2 Fc (s) = , π a2 + s2  s 2 , Fs (s) = 2 π a + s2  b 2 , G c (s) = π b2 + s 2  s 2 . G s (s) = 2 π b + s2 Using these results, show that  ∞ π 1 ds = , a > 0, b > 0. 2 + s 2 )(b2 + s 2 ) (a 2ab(a + b) 0 Solution: We have the Parseval relation for cosine transforms  ∞  ∞ Fc (s)G c (s) ds = f (t)g(t) dt. 0

0

The right side is 

∞ 0

The left side is  ∞ 0

so that



∞ 0



∞

f (t)g(t) dt =

e−(a+b)t dt =

0

2 Fc (s)G c (s) ds = ab π

 0

∞

1 . a+b

1 ds (a 2 + s 2 )(b2 + s 2 )

1 π ds = , a > 0, b > 0. (a 2 + s 2 )(b2 + s 2 ) 2ab(a + b)

130

6 Fourier Transform

Similarly, using the Parseval relation for sine transforms 



∞

Fs (s)G s (s) ds =

0

we find



f (t)g(t) dt,

0

∞

0

∞

s2 π ds = . (a 2 + s 2 )(b2 + s 2 ) 2(a + b)

Remark: These integrals can be derived by other methods. Example 6.9 For a > 0 and x > 0, prove that the Fourier cosine and sine transforms of e−ax x n−1 are   ∞

2 e−ax x n−1 (cos sx, sin sx) d x = Fc,s e−ax x n−1 = π 0 

  cos nθ sin nθ 2 (n) , π rn rn

where a = r cos θ and s = r sin θ. Solution: We have

(Fc + iFs ) e−ax x n−1  =

2 π



∞

e−ax x n−1 (cos sx + i sin sx) d x

0

 =  =

2 π



2 d n−1 (−1)n−1 n−1 π dz  =

∞

e−(a−is)x x n−1 d x

0



∞

e−zx d x(z = a − is, Rez > 0)

0

n−1   1 2 n−1 d (−1) n−1 π dz z

 =  =

2 (n) π zn

2 (n) π (a − is)n

6.7 Convolution Theorem and Parseval Relation

 =

131

2 (n) (cos nθ + i sin nθ). π rn

Equating the real and imaginary parts from both sides, the results follow. Remark: One can derive some apparently complicated integrals from the above result. For example, if we write the inverse cosine transform from above, we find  e

−ax n−1

x

=

Noting cos θ = ar , sin θ = e−ax x n−1 = =



2 (n) π





π 2

0 π 2

∞

0

cosn θ an

s r

2 (n) π a n−1

2 π

=



2 cos nθ (n) n cos sxds. π r

1 , rn

s = a tan θ,

cos nθ cosn θ cos(ax tan θ)a sec2 θdθ an

cos nθ cosn−2 θ cos(ax tan θ)dθ.

0

Putting a = 1, this produces for n > 1 and x ≥ 0 

π 2

cos nθ cosn−2 θ cos(x tan θ)dθ =

0

π e−x x n−1 . 2(n)

Similarly, one can show that, for n > 1 and x ≥ 0, 

π 2

sin nθ cosn−2 θ sin(x tan θ)dθ =

0

π e−x x n−1 . 2(n)

Example 6.10 Find the Fourier cosine transforms of e−ax cos ax, and e−ax sin ax 1 x2 a > 0 and hence obtain Fourier cosine transforms of x 4 +k 4 and x 4 +k 4 (k > 0). Solution:

Fc [e−ax (cos ax − i sin ax)] = Fc [e−a(1+i)x ]  =  =

2 Re j π

 0

∞

2 π



∞

e−a(1+i)x cos sxd x

0

e−{a(1+i)− js}x d x( j does not interact with i)

132

6 Fourier Transform

 =  =

2 1 Re j π a(1 + i) − js

2 a(1 + i) + js Re j π [a(1 + i)]2 + s 2

 =

a(1 + i) 2 π [a(1 + i)]2 + s 2 

=  =  =

2 a(1 + i) π (2a 2 i + s 2 )

2 a(1 + i)(s 2 − 2ia 2 ) π (2a 2 i + s 2 )(s 2 − 2ia 2 )

2

a 2 s + 2a 2 + i(s 2 − 2a 2 ) . 4 4 π 4a + s

Taking the real and imaginary parts, we obtain  Fc [e

−ax

cos ax] = 

and Fc [e

−ax

sin ax] =

2 a(s 2 + 2a 2 ) π 4a 4 + s 4

2 a(2a 2 − s 2 ) . π 4a 4 + s 4

By addition and subtraction, we find 



and



2 2as 2 = Fc e−ax (cos ax − sin ax) 4 4 π 4a + s

(i)



2 4a 3 = Fc e−ax (sin ax + cos ax) . 4 4 π 4a + s

(ii)

Using Fourier cosine inversion, we find from (ii)  0

∞

4a 3 π cos sxds = e−ax (sin ax + cos ax). 4a 4 + s 4 2

6.8 Fourier Transforms in Two or More Dimensions

Putting 2a 2 = k 2 (a = 

∞ 0

√k ), 2

133

and interchanging x with s, 

cos sx π − √ks dx = √ e 2 4 4 x +k 2 2k 3

 ks ks sin √ + cos √ . 2 2

Similarly, from (i) we find 

∞

0

x 2 cos sx π − √ks dx = √ e 2 4 4 x +k 2k 2

  ks ks cos √ − sin √ . 2 2

6.8 Fourier Transforms in Two or More Dimensions The Fourier transform pair for a function f (x) of one variable x 

1 F(s) = √ 2π 1 f (x) = √ 2π

∞

f (x)eisx d x,

−∞



∞

F(s)e−isx ds

−∞

can be extended to a function f (x1 , x2 ) of two variables x1 , x2 as F(s1 , s2 ) = f (x1 , x2 ) =

1 2π



1 2π

∞



−∞



∞

∞

−∞



−∞

∞ −∞

f (x1 , x2 )ei(s1 x1 +s2 x2 ) d x1 d x2 , F(s1 , s2 )ei(s1 x1 +s2 x2 ) ds1 ds2 .

For a function f (x) of n variables x1 , x2 , . . . , xn , this can easily be extended as F(s) = √

1 (2π)n

1 f (x) = √ (2π)n



∞ −∞



∞

−∞

 ......

−∞

 . . .. . .

∞

∞

−∞

f (x)eixs d x1 . . .d xn , F(s)e−ixs ds1 . . .dsn ,

where x denotes a vector in the n-dimensional space having Cartesian components x1 , x2 , . . .xn and s is a similar vector having component s1 , s2 , . . ., sn .

134

6 Fourier Transform

6.9 Application of Fourier Transforms in Solving Linear Ordinary Differential Equations In this section, we shall apply Fourier transform to solve linear ordinary differential equations. We consider the following examples.

Examples Example 6.11 Consider the ODE d2 f df +a − b f = g(x), −∞ < x < ∞ dx2 dx with f (x) → 0, f  (x) → 0 as x → ±∞, where a, b are constants and g(x) is a given function whose Fourier transform exists. ∞ ∞ Solution: Let F(s) = √12π −∞ f (x)eisx d x, G(s) = √12π −∞ g(x)eisx d x. Now  ∞  ∞ 2

 d f isx isx ∞ e d x = f (x)e −∞ − is f  (x)eisx d x 2 −∞ d x −∞   isx ∞ = 0 − is [ f (x)e ]−∞ − is

∞

−∞

 f (x)e

isx

dx

= −s 2 F(s) and



∞ −∞

d f isx e d x = −is F(s). dx isx

Multiplying both sides of the ODE by √e 2π and integrating between −∞ and ∞, we find −(s 2 + ias + b)F(s) = G(s) so that F(s) = −

G(s) . s 2 + ias + b

The inverse produces f (x). To find the inverse, we note that 1 = s 2 + ias + b (s 2 + We first find the Fourier inverse of

1 (c s 2 +c2

1 ia 2 ) 2

+

> 0),

a2 4

+b

.

6.9 Application of Fourier Transforms in Solving Linear Ordinary …

135

Fig. 6.1 Complex s-plane

′

F −1 [

1 1 ]= √ s 2 + c2 2π



∞ −∞

e−i xs ds. s 2 + c2

 ∞ −i xs The evaluation of the integral −∞ se2 +c2 ds can be done by using the complex variable theory. We first choose x > 0. Consider the following contour as in Fig. 6.1 in the complex s-plane, R being a very large positive number and γ R being a halfcircle in the lower complex half-plane with radius R and center at the origin. Inside −i xs this contour, the function se2 +c2 has a pole at s = −ic. Using the residue theorem of complex variable theory, 

−R

  +

R

Now |



e−i xs c R s 2 +c2 ds|

γ R

e−i xs ds = 2πi Res at s = −ic. s 2 + c2

→ 0 as R → ∞ and Res at s = −ic is lim (s + ic)

x→−ic

=

e−i xs s 2 + c2

e−cx . −2ic

Thus making R → ∞, one finds for x > 0, 

∞

−∞

π e−i xs e−cx = e−cx . ds = −2πi 2 +c −2ic c

s2

 ∞ i(−x)s When x < 0, the integral −∞ es 2 +c2 ds can be evaluated by considering the contour as in Fig. 6.2 in the complex s-plane, where γ R is a half-circle with radius R in the upper complex half-plane.

136

6 Fourier Transform

Fig. 6.2 Complex s-plane

Inside this contour, 

ei(−x)s s 2 +c2 R

−R

Now |



ei(−x)s γ R s 2 +c2 ds|

has a simple pole at s = ic, and using residue calculus

  +

ei(−x)s ds = 2πi Res at s = ic. s 2 + c2

γR

→ 0 as R → ∞ and Res at s = ic is lim (s − ic)

s→ic

= =

ei(−x)s s 2 + c2

ecx 2is

e−c(−x) . 2ic

Thus making R → ∞, we find for x < 0, 

∞

−∞

e−i xs π ds = e−c(−x) . 2 2 s +c c

Hence for x > 0 or x < 0, 

∞ −∞

e−i xs π ds = e−c|x| + c2 c

s2

so that F

−1

1 [ 2 ]= s + c2



π e−c|x| . 2 c

Suppose L(s) is the Fourier transform of l(x), then L(s) = F[l(x)].

6.9 Application of Fourier Transforms in Solving Linear Ordinary …

137

By the shifting property of Fourier transform, we know that F[eiax l(x)] = L(s + a) so that

F −1 [L(s + α)] = l(x)eiαx .

We choose L(s) = shifting property,

F To find F −1 we choose α =

F

−1



so l(x) =

1 , s 2 +c2



ia , c2 2

−1



 π e−c|x| 2

c

. Thus, L(s + α) =

so by

 −c|x|  1 πe iαx =e . 2 2 (s + α) + c 2 c

  1 = F −1 s 2 + ias + b (s + =

1 , (s+α)2 +c2

a 2 +4b , 4



1 ia 2 ) 2

+

a2 4

+b

so that

√  √ e− 21 (ax + a 2 + 4b|x|) √ 1 = 2π ≡ 2πn(x), say √ s 2 + ias + b a 2 + 4b

where

√ 1 e− 2 (ax + a 2 + 4b|x|) n(x) = . √ a 2 + 4b

Thus f (x) = −F 1 = −√ 2π



∞

−∞

 =−

−1

∞ −∞



1 G(s) 2 s + ias + b



√ g(u) 2πn(x − u)du g(u)n(x − u)du

where n(x) is given above. Special Case: 

For a = 0, b = 1, g(x) =

−1, −1 < x < 1, 0, otherwise,

138

6 Fourier Transform

the problem is to solve the ODE d2 f − f = dx2



−1, −1 < x < 1, 0, otherwise

with f, f  → 0 as |x| → ∞. Thus

1 −|x| e . 2

n(x) = Hence the solution is



1 f (x) = 2 = Thus, 1 f (x) = 2 =

1 2



1

e−|x−u| du

−1



x+1

e−|t| dt.

x−1



0

x+1

e dt + t

x−1

−t



e dt 0

1 1 1 − e x−1 − e−(x+1) + 1 = 1 − cosh x for | x |< 1. 2 e

Also, f (x) =

f (x) =

1 2 1 2



x+1

e−t dt = e−x sinh 1, for x > 1

x−1



x+1

et dt = e x sinh 1, for x < −1.

x−1

Example 6.12 Use Fourier transform to solve the differential equation d2 f − f = e−|x| , −∞ < x < ∞ dx2 with f, f  → 0 as |x| → ∞. Using the Fourier transform of both sides, we find  −s F(s) − F(s) = 2

Thus

 F(s) = −

2 1 . π s2 + 1

1 2 . π (s 2 + 1)2

6.9 Application of Fourier Transforms in Solving Linear Ordinary …

Hence

139

   1 1 π −1 2 2 F f (x) = − 2 π (s 2 + 1) π (s 2 + 1)   ∞ π 1 =− e−|u| e−|x−u| du, (using convolution theorem) √ 2 2π −∞  1 ∞ −|u|−|x−u| =− e du 2 −∞ 1 =− 2 1 =− 2





0

e

u−|x−u|

−∞



∞

e

du +

e

−u−|x−u|

 du

0 −u−|x+u|



∞

du +

0

=−

∞

e

−u−|x−u|

 du

0



1 2

∞

  e−u e−|x+u| + e−|x−u| du.

0

∞ ∞ Let I1 (x) = 0 e−u−|x+u| du and I2 (x) = 0 e−u−|x−u| du. For x > 0,  ∞ 1 e−u−(x+u) du = e−x , I1 (x) = 2 0  I2 (x) =

x

e

−u−(x−u)



∞

du +

0

e−u−(u−x) du

x

= xe

−x

 +e

∞

x

e−2u du

x

=

1 −x e + xe−x 2

so that f (x) = − 21 (x + 1)e−x for x > 0. For x < 0, let x = −x  so that x  > 0. Then I1 (x) = I2 (x  ) =

1 −x  1  e + x  e−x = e x − xe x , 2 2

I2 (x) = I1 (x  ) =

1 −x  1 e = ex 2 2

140

6 Fourier Transform

so that f (x) = − 21 (1 − x)e x for x < 0. Hence 1 f (x) = − (1 − |x|)e−|x| . 2

6.10 Application of Fourier Sine and Cosine Transforms in Solving Linear Ordinary Differential Equations If a linear ordinary differential equation is defined in a semi-infinite interval (0, ∞), then to solve it one has to use either Fourier sine or cosine transform depending on the type of the boundary condition specified at x = 0. If f (0) is prescribed, then Fourier sine transform has to be used while Fourier cosine transform has to be used if f  (0) is given. The following two examples will clarify this. Example 6.1 Solve the ODE d2 f − f = e−x , 0 < x < ∞ dx2 given that f (0) = 0, f (x) → 0 and f  (x) → 0 as x → ∞. Solution: Taking Fourier sine transform, we obtain  −s Fs (s) − Fs (s) = 2



so that

Fs (s) = − 

Hence

f (x) = −

2 s π 1 + s2

s 2 . π (1 + s 2 )2

 s 2 −1 Fs π (1 + s 2 )2

   π −1 2 1 2 s F =− 2 t π 1 + s2 π 1 + s2 

∞

=− 0



2 1 π 1 + s2

 

2 s π 1 + s2

 sin sx ds

6.10 Application of Fourier Sine and Cosine Transforms in Solving …



∞

=−

141

G c (s)Hs (s) sin sxds

0



where G c (s) =



and Hs (s) = so that g(x) = e−x . Now



∞

2 1 π 1 + s2 2 s π 1 + s2

G c (s)Hs (s) sin sxds

0

 = 1 = 2



∞

 g(u)

0

1 = 2



∞ 0

2 π



∞ 0

∞



0



∞

g(u) cos sudu Hs (s) sin xsds

0

 Hs (s) sin s(u + x)ds +

1 g(u)h(x + u)du + 2 

∞

 g(u)

x

1 g(u)h(x + u)du + 2

1 = 2



∞

0

1 + 2 1 = 2



2 π



∞

e

−u −(x+u)

e



x 0

x

 g(u)

0

2 π



∞

2 π





∞

 Hs sin s(x − u)ds

0

 Hs sin s(u − x)ds

e

−u u−x

e



1 + x −x e . 2



∞

e x

g(u)h(u − x)du

x

∞

du +

 1 e−x e−x + xe−x + 2 2 2 =

Hs (s) sin s(x − u)ds

0

x

du +



0

0

=

∞

1 g(u)h(x − u)du + 2 

0

e−x . Hence f (x) = − 1+x 2



2 π

−u −u+x

e

 du

142

6 Fourier Transform

Note: In the sine inversion formula  h(x) =

2 π



∞

Hs (x) sin sxd x

0

x is always positive. Example 6.2 Solve the ODE d2 f − k2 f = dx2



−1, 0 < x < 1, 0, otherwise

where k > 0 and f  (0) = 0, f (x) → 0 and f  (x) → 0 as x → ∞. Solution: Use cosine transform to obtain  −s Fc (s) − k Fc (s) = − 2

2

so that Fc (s) = 

Hence f (x) =

Fc−1  =

2 sin s π s

1 sin s . s s2 + k2

2 sin s π s



k 2 2 π s + k2



π1 2k

π 1 −1 F [G(s)H (s)] 2k c

 where G(s) = − π2 sins s so that  g(x) = and H (s) = Thus



2 k π s 2 +k 2

so that h(x) = e−kx (k > 0). 

f (x) = 1 = k

−1, 0 < x < 1, 0, otherwise



∞ 0



π1 2k

2 π

 0



∞

2 π



∞

G(s)H (s) cos sxds

0

g(u) cos sudu H (s) cos sxds

6.10 Application of Fourier Sine and Cosine Transforms in Solving …

1 = 2k



∞

 g(u)

0

=

1 2k



2 π

∞

143

H (s) {cos s(u + x) + cos s(u − x)} ds du

0



∞

g(u) {h(u + x) + h(|u − x|)} du

0

=−

1 2k



 −k(u+x)  e + e−k|u−x| du.

1 0

For 0 < x < 1,



1 f (x) = − 2k =−

1

e

−k(u+x)

 du +

0

x

e

−k(x−u)



1

du +

0

e

−k(u−x)

 du

x

  1  x  1 1 e−kx e−ku du + e−kx eku du + ekx e−ku du 2k 0 0 x

=−

1 −kx e (1 − e−k ) + e−kx (ekx − 1) + ekx (e−kx − e−k ) 2 2k =−

1 −kx e − e−k(1+x) − e−kx − e−k(1−x) + 2 2 2k =−

1 1 − e−k cosh kx 2 k e−k cosh kx − 1 . k2

= For x > 1, f (x) = − =−

1 2k



1 0

e−k(u+x) du −



1

e−k(x−u) du

0

1 −kx e (1 − e−k ) − e−kx (ek − 1) 2 2k =−

1 −kx e [1 − cosh k] . k2



144

6 Fourier Transform

6.11 Application to Partial Differential Equations Here, we apply Fourier transform to solve initial or boundary value problems involving partial differential equations. We consider the following examples. Example 6.13 Solve the Laplace equation ∂2φ ∂2φ + = 0, −∞ < x < ∞, y ≥ 0 ∂x 2 ∂ y2 with the boundary condition φ(x, 0) = f (x), −∞ < x < ∞, , ∂φ → 0 f (x) being a prescribed function whose Fourier transform exists and φ, ∂φ ∂x ∂ y  as (x 2 + y 2 ) → ∞.   ∞ 1 isx Solution: Let (s; y) = F[φ(x, y)] = 2π −∞ φ(x, y)e d x. Noting that 

∞ −∞



 ∞ ∂ 2 φ isx ∂φ ∂φ isx ∞ e dx e d x = − is 2 ∂x ∂x −∞ ∂x −∞

  isx ∞ = 0 − is φe −∞ − is  = −s

∞

2

∞

 φe

isx

dx

−∞

φeisx d x,

−∞

the PDE reduces to the ODE d 2 − s 2  = 0, y ≥ 0 dy 2 with (s; 0) = F(s), F(s) being the Fourier transform of f (x), and , d → 0 as dy y → ∞. Hence (s; y) = F(s)e−|s|y , so that by inversion 1 φ(x, y) = √ 2π



∞

F(s)e−|s|y−isx ds

−∞

= F −1 [F(s)G(s)] where G(s) = e−|s|y . We note that

6.11 Application to Partial Differential Equations

1 F −1 [G(s)] = √ 2π 1 =√ 2π



∞

e

−s(y+i x)

145



∞

 ds +

0

1 =√ 2π  =



e−|s|y−isx ds

−∞



0

e

s(y−i x)

ds

−∞

 1 1 + ,y >0 y + ix y − ix

y 2 = g(x, y), say. 2 π x + y2

 ∞ 1 φ(x, y) = √ f (u)g(x − u, y)du 2π −∞  y ∞ f (u) = du, y > 0. π −∞ (x − u)2 + y 2

Thus

This is known as Poisson’s formula for a function which is harmonic in the half-plane y ≥ 0 and is prescribed at y = 0. Note: To satisfy φ(x, 0) = f (x), we have to use the relation lim

y→0

y 1 = δ(u − x). π (u − x)2 + y 2

Remark: The above problem is known as the Dirichlet Problem for a harmonic function in a half-plane. Example 6.14 Solve the PDE ∂2φ 1 ∂φ = 0, −∞ < x < ∞, t ≥ 0 − 2 2 ∂x a ∂t with boundary conditions φ(x, y), ∂φ → 0 as |x| → ∞ and initial condition ∂x φ(x, 0) = f (x), −∞ < x < ∞ where f (x) is prescribed and is such that its Fourier transform exists. Solution: Let  (s; t) = F [φ(x, t); x → s] =

1 2π



∞

φ(x, t)eisx d x.

−∞

Then using Fourier transform in x, the PDE reduces to the ODE

146

6 Fourier Transform

d + a 2 s 2  = 0, t ≥ 0 dt with (s; 0) = F(s), F(s) being the Fourier transform of f (x). Thus (s, t) = F(s)e−a

2 2

s t

  2 2 φ(x, t) = F −1 F(s)e−a s t

so that

= F −1 [F(s)G(s)] where G(s) = e−λs with λ = a 2 t. Let g(x) = F −1 [G(s), s → x] 2

 ∞ 1 2 e−λs −isx ds =√ 2π −∞  ∞ 1 i x 2 x2 =√ e−λ(s+ 2λ ) + 4λ ds 2π −∞  ∞ 1 x2 2 4λ =e √ e−λξ dξ 2π −∞ x2

e 4λ 1 =√ √ 2π λ x2

e 4λ =√ as 2λ





∞

e−ξ dξ 2

−∞

∞

−∞

e−ξ dξ = 2

√

π

x2

e 4a2 t = √ . a 2t Thus

 ∞ 1 f (u)g(x − u)du φ(x, t) = √ 2π −∞  ∞ (x−u)2 1 = √ f (u)e 4a2 t du. 2a πt −∞

6.11 Application to Partial Differential Equations

147

Remark: This problem is the mathematical model for the flow of heat in an infinite medium when the initial temperature distribution f (x) is known. Example 6.15 Solve the PDE ∂2φ 1 ∂2φ − 2 2 = 0, −∞ < x < ∞, t ≥ 0 2 ∂x c ∂t with boundary condition φ(x, t) → 0 as |x| → ∞ and initial conditions φ(x, 0) = f (x),

∂φ (x, 0) = g(x), −∞ < x < ∞ ∂t

where f (x) and g(x) are prescribed functions of x and are such that their Fourier transforms exists. Solutions: Using Fourier transforms in x, we obtain d 2 + c2 s 2  = 0, t ≥ 0 dt 2 with (s; 0) = F(s),

d (s; 0) = G(s), dt

where (s; t) = F [φ(x, t); x → s] , F(s) = F[ f ], G(s) = F[g]. Thus (s; t) = F(s) cos(cst) + Hence −1

φ(x, t) = F 1 = √ 2 2π 1 + √ 2 2π =





∞

G(s) F(s) cos(cst) + sin(cst) cs

∞

−∞





 e−is(x−ct) + e−is(x+ct) F(s)ds



 G(s) ds ics

−∞



G(s) sin(cst). cs

e−is(x−ct) − e−is(x+ct)

1 { f (x − ct) + f (x + ct)} + I 2

where I denotes the second term. To find it, note that

148

6 Fourier Transform

1 g(z) = √ 2π



∞

e−isz G(s)ds.

−∞

Integrating both sides w.r.t. z between x − ct to x + ct, we find 

x+ct x−ct

1 g(z)dz = √ 2π

Thus φ(x, t) =



∞

−∞

e−is(x+ct) − e−is(x−ct) G(s)ds. −is

1 1 { f (x − ct) + f (x + ct)} + 2 2c



x+ct

g(u)du.

x−ct

Remark: This is known as the famous D’Alembert solution of the one-dimensional wave equation.

6.12 Application of Fourier Sine and Cosine Transform to the Solution of Partial Differential Equations It is often useful to use Fourier sine or cosine transform to solve an initial or boundary value problem where the governing partial differential equation is defined in a semiinfinite interval. The following examples will illustrate this. Example 6.16 Solve the PDE ∂2φ ∂2φ + =0, ∂x 2 ∂ y2

0 ≤ x < ∞, y ≥ 0

for which φ(0, y) = 0, φ(x, 0) = f (x), φ(x, y) → 0 as



x 2 + y 2 → ∞.

Solution: Since φ(0, y) is prescribed, we use the Fourier sine transform in x. Thus, the PDE reduces to the ODE d 2 s − s 2  = 0, y ≥ 0 dy 2 given that s (s; 0) = Fs (s),  → 0 as y → ∞ where s (s; y) = Fs [φ(x, y); x → s]

6.12 Application of Fourier Sine and Cosine Transform to the Solution …





2 π

=

∞

Fs (s) = Thus

φ(x, y) sin sx d x

0



and

149



2 π

∞

f (x) sin sx d x.

0

s (s; y) = Fs (s)e−sy 

so that φ(x, y) =

=

1 π



= = =



2 π

=

0

y π

∞



∞

Fs (s)e−sy sin sx ds

0

f (u) sin su du}e−sy sin sx ds

0

f (u)[

0

1 π

 {



∞

1 π

∞

2 π

∞

{cos s(u − x) − cos s(u + x)}e−sy ds]du

0



∞



0



 eis|u−x| − eis|u+x| e−sy ds]du

0



∞



 y y du − y 2 + (u − x)2 y 2 + (u + x)2



 1 1 du. − y 2 + (u − x)2 y 2 + (u + x)2

f (u)

0



∞

f (u)[Re

∞

f (u)

0

Note: The solution for the special case f (x) = c where c is a constant can be deduced from above although the Fourier sine transform of c does not exist in the ordinary sense. In this case,  

1 yc ∞ 1 du φ(x, y) = − π 0 (u − x)2 + y 2 (u + x)2 + y 2 =

 c u−x u+x ∞ tan−1 − tan−1 π y y 0

=

 π x x c π + tan−1 − + tan−1 π 2 y 2 y =

2c x tan−1 . π y

150

6 Fourier Transform

Example 6.17 Solve the PDE ∂2φ 1 ∂φ =0, − 2 2 ∂x a ∂t along with

0 ≤ x < ∞, t ≥ 0

∂φ (0, t) = − f (t), φ(x, t) → 0 as x → ∞, ∂x

and φ(x, 0) = 0, 0 ≤ x < ∞. Since ∂φ is prescribed at x = 0, we can use the Fourier cosine transform here to ∂x obtain the solution. Using Fourier cosine transform in x, the PDE reduces to the ODE  2 2 d 2 2 +a s = a f (t), t ≥ 0 dt π 

where (s; t) =

2 π



∞

φ(x, t) cos sx d x,

0

together with (s; 0) = 0. 

Thus (s; t) = so that

2 φ(x, t) = a 2 π 2 = a2 π





2 2 a π

e−a



s (t−τ )

2 2

t

cos sx 0

f (τ ) dτ

e

−a 2 s 2 (t−τ )

 f (τ ) dτ ds

0 ∞

f (τ )

0

t

0

∞



t



e

−a 2 (t−τ )s 2

 cos sx ds dτ .

0

The inner integral is 

∞

e−λs cos sx ds 2

(λ = a 2 ( t − τ ) > 0 )

0

=

1 2



∞ 0

 2  e−λs eisx + e−isx ds

(x > 0)

6.12 Application of Fourier Sine and Cosine Transform to the Solution …

= =

1 2

1 − x2 e 4λ 2



∞

e−λs

2

+isx

151

ds

−∞



∞

e−λ(s

2

2

ix x − 2λ − 4λ 2)

ds

−∞



1 x2 = e− 4λ 2

∞

e−λ(s− 2λ ) ds ix 2

−∞



1 x2 = e− 4λ 2

∞

e−λξ dξ 2

−∞

 1 − x 2 ∞ −η2 4λ = √ e e dη 2 λ −∞ √ π x2 = √ e− 4λ . 2 λ Thus



a2 φ(x, t) = √ π a =√ π



t 0

f (τ ) − x 2 √ e 4λ dτ λ

t 0

2 f (τ ) − 4a2x(t−τ ) dτ . e √ t −τ

Special case: f (t) = const. = K , say. In this case, Ka φ(x, t) = √ π



t

2

e

− 4a 2x(t−τ )

0

√

dτ t −τ

  t 2 2 √ √ Ka x2 − x − x = √ −2 t − τ e 4a2 (t−τ ) |tτ =0 − 2 t − τ e 4a2 (t−τ ) × 2 dτ 4a (t − τ )2 π 0 



t − x 22 2 e 4a t − √ = K 2a π π



t

2

e

− 4a 2x(t−τ )

0

x2 × dτ 4a(t − τ )3/2



x If we put 2a √xt−τ = z, then 4a(t−τ dτ = −dz, and when τ = 0, z = )3/2 τ = t, z = ∞. Thus the second term inside the square bracket is

2x −√ π where er f c(z) =

√2 π

∞ z



∞

e x√ 2a t

−z 2

x dz = −x er f c √ 2a t

e−t dt. Thus in this case 2





. x√ , 2a t

when

152

6 Fourier Transform



  t − x 22 x e 4a t − x er f c φ(x, t) = K 2a . √ π 2a t 

Example 6.18 Solve the PDE ∂2φ ∂2φ − 2 =0, ∂x 2 ∂t given that

0 ≤ x < ∞, t ≥ 0

∂φ (0, t) = 0, φ(x, t) → 0 as x → ∞, ∂x

and

∂φ (x, 0) = 0, 0 < x < ∞. ∂t

φ(x, 0) = e−x ,

Solution: Since ∂φ is prescribed at one end point x = 0, we use the Fourier cosine ∂x transform. Thus, the PDE reduces to the ODE d 2 + s 2  = 0, t ≥ 0 dt 2   ∞ 2 (s; 0) = e−x cos sx d x π 0  =

1 2 π 1 + s2

∂ (x, 0) = 0, ∂t 

where (s; t) =



2 π

(s; t) =

2 φ(x; t) = π =

1 π

 0

∞

 0

φ(x, t) cos sx d x.

0



Thus

Hence

∞

∞

1 2 cos st. π 1 + s2

1 cos st cos sx ds 1 + s2

cos s(t − x) + cos s(t + x) ds 1 + s2

Exercises

153

=

1 2π



∞ 0

1 is|t−x| e + e−is|t−x| + eis|t+x| + e−is|t+x| ds 2 1+s 1 = 2π



∞

−∞

eis|t−x| + eis|t+x| ds 1 + s2

= i Res at s = i =

 1  −|t−x| e + e−|t+x| 2

(using residue theorem)

= e−t cosh x for x < t e−x cosh t for x > t.

Exercises 1. The Fourier transform of f (x) defined on (−∞, ∞) exists if ∞ (A) 0 f (x) d x < ∞, ∞ (C) 0 | f (x)| d x < ∞,

∞ (B) −∞ | f (x)| d x < ∞, 0 (D) −∞ | f (x)| d x < ∞. Answer:

2. The value of the integral (A) π

∞ 0

sin x x

B

d x is

(B) 1

(C) Answer:

π 2

(D) π4 .

C

3. The Fourier transform of which one of the functions defined below does not exist?   x, 0 < x < 1 x, −1 < x < 1 (A) f (x) = (B) f (x) = 0, otherwise 0, otherwise √ (D) f (x) = e x . (C) f (x) = 2 Answer:

C

4. Derive the Fourier cosine transform and its inverse from the Fourier integral expansion of a function. 5. Derive the Fourier sine transform and its inverse from the Fourier integral expansion of a function.

154

6 Fourier Transform

 6. Let f (x) =  (A)

1, −1 < x < 1 , then the Fourier transform of f (x) is 0, otherwise

2 sin s , π s

(B)

sin s , s

(C) sin s,

Answer:

(D)

√ π

sin s . s

A

7. Let f (x) = e−a|x| , − ∞ < x < ∞, a > 0. Then its Fourier transform is  (A)

2 a2 , π s 2 +a 2

(B)

π

a , 2 s 2 +a 2

Answer: 8. Let f (x) = (A)

1 , x 2 +a 2

π

e−as , 2 a

(B)

2 , π

2 a . π s 2 +a 2

D



π

e−a|s| , 2 a

9. The Fourier sine transform of f (x) = 

(D)

− ∞ < x < ∞, a > 0. Then its Fourier transform is (C)

Answer:

(A)



a , s 2 +a 2

(C)

(B)

1 x



2 e−a|s| , π a

(D)

2 e−as . π a

B

is

π

, 2 Answer:

 (C)

1 , π

(D)

√ π.

B

10. The Fourier sine transform of f (x) = e−x is  (A)

2 s , π 1+s 2

(B)

π

1 , 2 1+s 2

Answer:

 (C)

2 1 , π 1+s 2

(D)

π

s . 2 1+s 2

A

11. The Fourier cosine transform of f (x) = e−x is  (A)

2 s , π 1+s 2

 (B)

2 s , π 1+s 2

Answer:

(C)

B

π

1 , 2 1+s 2

(D)

π

s . 2 1+s 2

Exercises

155



1, −1 ≤ x < 1 is 0, otherwise

12. The Fourier sine transform of f (x) = (A)

 π 1−cos s 2

s

,

 (B)

2 1−cos s , π s

(C)

Answer:

(A)

 π sin s 2

s

,

(B)

 π cos s 2

s

2

 (D)

2 1+cos s . π s

1, −1 ≤ x < 1 is 0, otherwise 

,

s

,

B 

13. The Fourier cosine transform of f (x) =

 π 1+cos s

(C)

2 sin s , π s

 (D)

2 cos s . π s

Answer: C  1 − x 2 , |x| < 1 14. Find the Fourier transform of f (x) = 0, otherwise Answer:

√ 2 2 sin s − s cos s π s

15. Find the Fourier transform of f (x) = δ(x − a) + δ(x + a) where δ(x) is the Dirac delta function. 2 cos sa Answer: π 16. Let F(s) be the Fourier transform of f (x), −∞ < x < ∞. Then the Fourier transform of f (−x), −∞ < x < ∞, is (A) F(−s),

(B) − F(s),

Answer:

(C) − F(−s),

(D) F

1 s

.

A

17. Let F(s) be the Fourier transform of f (x), −∞ < x < ∞. Then the Fourier transform of f (x + 1), −∞ < x < ∞, is (A) eis F(s),

(B) F(s + 1),

Answer:

(C) F(s − 1),

D

(D) e−is F(s).

156

6 Fourier Transform

18. Let F(s) be the Fourier transform of f (x), −∞ < x < ∞. Then lim |F(s)| is |s|→∞

(A) 1,

(B) e,

(D) ∞.

(C) 0,

Answer:

C

19. Let F(s) be the Fourier transform of f (x), −∞ < x < ∞. Then the Fourier transform of F(x), −∞ < x < ∞, is (A) f (s),

(B) − f (s),

(C) f (−s),

Answer:

(D) − f (−s).

C

20. Let F(s) be the Fourier transform of f (x), −∞ < x < ∞. Then find the Fourier transform f n (x), −∞ < x < ∞, stating any assumptions which you require for the existence of the transform. 21. If the Fourier transform of x n f (x) exists (n being a nonnegative integer), show how to find the nth derivative of the Fourier transform F(s) of f (x) in terms of the Fourier transform of x n f (x). 22. Find the Fourier sine transform of the nth derivative of f (x), n being a nonnegative integer, in terms of the Fourier sine transform of f(x) and initial values of f k (x), k = 0, 1, . . . , (n − 1), which one is required. 23. Find the Fourier cosine transform of the nth derivative of f (x), n being a nonnegative integer, in terms of the Fourier cosine transform of f (x) and initial values of f k (x), k = 0, 1, . . . , (n − 1), which one is required. 24. Let F(s) be the Fourier transform of f (x), −∞ < x < ∞. Find which one of the following relations is true: ∞ ∞ (A) 0 | f (x)|2 d x = 0 |F(s)|2 ds, ∞ 0 (C) −∞ | f (x)|2 d x = −∞ |F(s)|2 ds,

∞ ∞ (B) −∞ | f (x)|2 d x = 0 |F(s)|2 ds, ∞ ∞ (D) −∞ | f (x)|2 d x = −∞ |F(s)|2 ds.

Answer:

D

25. Let Fs (s) be the Fourier sine transform of f (x), 0 ≤ x < ∞. Then is equal to (A)

∞ 0

f (x)d x,

(B)

∞ 0

(C)

∞

Answer:

B

| f (x)|d x,

0

| f (x)|2 d x,

(D)

∞ 0

∞ 0

|Fs (S)|2 ds

1

| f (x)| 2 d x.

Exercises

157

26. Using the Parseval relation of Fourier cosine transforms, show that  0

∞

πp x−p π . d x = sec x 2 + a2 2a p+1 2

27. Using the Parseval relation of Fourier transforms, show that 

∞

−∞

(x 2

x2 π dx = . + a 2 )4 16a 5

28. Using Fourier transform, solve the ordinary differential equation dy x2 = xe− 2 , − ∞ < x < ∞ dx given that y(x) → 0 as |x| → ∞. x2

y(x) = −e− 2

Answer:

29. Using Fourier transform, solve the ordinary differential equation dy + y = e−|x| , − ∞ < x < ∞ dx given that y(x) → 0 as |x| → ∞.  ex Answer:

y(x)

,

2 e−x 2

x 0.

30. Using Fourier transform, solve the ODE 

−1, −1 < x < 1 0, otherwise dy → 0 as |x| → ∞. given that y(x) → 0, dx ⎧ x ⎨ e sinh 1, −∞ < x < −1 x Answer: y(x) = 1 − cosh , −1 ≤ x ≤ 1 ⎩ −x e e sinh 1, 1 < x < ∞. d2 y −y = dx2

31. Use Fourier sine transform to solve the ordinary differential equation d2 y − f = e−x , 0 < x < ∞ dx2 given that f (0) = 0, f (x) → 0 as x → ∞, f  (x) → 0 as x → ∞.

158

6 Fourier Transform

32. Use Fourier cosine transform to solve the ordinary differential equation d2 y − f = e−x , 0 < x < ∞ dx2 given that f  (0) = 0, f (x) → 0 as x → ∞, f  (x) → 0 as x → ∞. 33. Use Fourier transform in x to solve the partial differential equation 1 ∂φ ∂2φ , − ∞ < x < ∞, t ≥ 0, = 2 ∂x k ∂t given that φ(x, 0) = f (x), − ∞ < x < ∞. Answer:

φ(x, t) = √

1 4πkt



∞

−∞

f (s)e− 4kt (x−s) ds. 1

2

34. Use Fourier transform in x to solve the partial differential equation 1 ∂φ ∂2φ , − ∞ < x < ∞, t ≥ 0, = 2 2 ∂x c ∂t ∂φ given that φ(x, 0) = f (x), (x, 0) = 0, − ∞ < x < ∞. ∂t Answer:

φ(x, t) =

1 { f (x − ct) + f (x + ct)}. 2

35. Use Fourier sine transform in x to solve the partial differential equation ∂2φ ∂2φ + = 0, 0 ≤ x < ∞, 0 ≤ y < ∞, ∂x 2 ∂ y2 given that φ(x, 0) = f (x), 0 < x < ∞, φ(0, a) = 0, φ(0, y) = 0, φ(x, y) → 0 as x → ∞ for 0 < y < a. Answer:

2 φ(x, y) = π



∞ 0



∞ 0

 f (t) sin st dt

sinh s(a − y) sin sx ds. sinh sa

36. Use Fourier sine transform in x to solve the partial differential equation ∂2φ ∂2φ + = 0, 0 ≤ x < ∞, 0 ≤ y < ∞, ∂x 2 ∂ y2 given that φ(0, y) = 1, φ(x, 0) = 0, ∂φ ∂φ 1 , → 0 as (x 2 + y 2 ) 2 → 0. ∂x ∂ y

References

159

Answer:

φ(x, y) =

x 2 tan−1 . π y

37. Use Fourier sine transform in x to solve the partial differential equation ∂2φ ∂φ = , 0 ≤ x < ∞, t > 0, 2 ∂x ∂t given that φ(0, t) = 0, t > 0,  1, 0 < x < 1 φ(x, 0) = 0, x ≥ 1 and φ(x, t) is bounded as x → ∞ and t → ∞. 38. Use Fourier cosine transform in x to solve the partial differential equation ∂2φ = ∂x 2 given that φ(0, t) = ∂φ (0, t) = ∂x

∂φ , 0 ≤ x < ∞, t > 0, ∂t 0, t > 0, −1, t > 0

and φ(x, t) is bounded as x → ∞ and t → ∞.  cos st 2 ∞ 2 ds. 1 − e−s t Answer: φ(x, t) = π 0 s 39. Use Fourier cosine transform in x to solve the partial differential equation ∂2φ ∂φ , 0 ≤ x < ∞, t > 0, = ∂x 2 ∂t given that φ(x, 0) = 0, x ≥ 0, ∂φ (0, t) = −1, t > 0. ∂x

References The content of this chapter is mainly based on the website epgp.inflibnet.ac.in/ of UGC E Pathsala and on the following literature. 1. 2. 3. 4.

I.N. Sneddon, Use of Integral Transform (Mc Graw Hill, 1972) I.N. Sneddon, Fourier Transform (Mc Graw Hill, 1951) L.C. Andrews, B.K. Shivamoggi, Integral Transform for Engineers (Prentice-Hall, 2007) L. Debnath, D. Bhatta, Integral Transforms and Their Applications, 3rd edn. (Chapman and Hall/CRC, USA, 2014)

Chapter 7

Laplace Transform

Laplace transform dates back to the French mathematician Laplace who made use of the transform integral in his work on probability theory in 1780. Oliver Heaviside (1850–1925), an English electrical engineer, popularized the use of Laplace transform as a mathematical tool in solving linear differential equations arising in electrical circuits. Heaviside was a self-made man without any academic training but made significant contributions to electromagnetic theory.

7.1 Derivation of Laplace Transform from Fourier Integral Theorem It is instructive to derive formally the Laplace transform and its inversion formula from the Fourier integral theorem. We know that for a function g(x) defined on (−∞, ∞) to have its Fourier transform is that the integral 

∞

−∞

|g(x)|d x

must be convergent. Let us change the notation x to t to introduce the concept of time. Thus, if  ∞ |g(t)|dt −∞

is not convergent then the Fourier transform G(s) of the function need not exist for real values of x. This situation arises in many cases. For example, if g(t) = sin σt, where σ is real, then

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 S. Banerjea and B. N. Mandal, Integral Equations and Integral Transforms, https://doi.org/10.1007/978-981-99-6360-7_7

161

162

7 Laplace Transform



∞ −∞

|g(t)|dt

is not convergent. We suppose that a function f (t) and its derivative f  (t) are both piecewise continuous functions for all t ≥ 0. We define  −ct e f (t), t ≥ 0, (7.1) g(t) = 0, otherwise ∞ where c is a positive real constant. Then obviously −∞ |g(t)|dt is convergent so that the Fourier transform of g(t) exists. Thus, by Fourier integral theorem 1 g(t) = 2π



∞

e



−ist



∞

e

−∞

isx

g(x)d x ds.

(7.2)

−∞

Using (7.1), we find ect 2π Thus,



∞

e−ist



−∞

ect f (t) = 2π



  f (t), t ≥ 0, eisx g(x)d x ds = 0, otherwise. −∞ ∞

∞

e

−ist





∞

e

−∞

isx−cx

f (x)d x ds, t ≥ 0.

0

If we now write p = c − is, then 1 f (t) = 2πi



c+i∞

F( p)e pt dp, t ≥ 0,

c−i∞



where

∞

F( p) =

f (t)e− pt dt.

0

Thus, in a purely formal manner, the transform formulae  F( p) ≡ L[ f (t)] =

∞

e− pt f (t)dt

(7.3)

0

called the Laplace transform of f (t), and f (t) ≡ L−1 [F( p)] =

1 2πi



c+i∞

F( p)e pt dt, t ≥ 0,

c−i∞

called the inverse Laplace transform, have been derived.

(7.4)

7.1 Derivation of Laplace Transform from Fourier Integral Theorem

163

Remark: 1. An integral of the form (7.3) in which the function f (t) is integrable over the interval (0, T ) for any positive T , is called a Laplace integral. 2. The Laplace integral may not exist for any function f (t). For example, if 2 f (t) = et , then the integral 

∞

e− pt et dt = 2

0



∞

p 2

e(t− 2 )

2

− p4

dt = e−

p2 4



∞

2

eu du

p 2

0

which is clearly divergent for all values of p. ∞ 3. For some functions f (t), the integral 0 e− pt f (t)dt  ∞ exists for some  ∞ values of p but not for others. For example, if f (t) = et , then 0 e− pt et dt = 0 e−( p−1)t dt which exists if  p > 1, but not if  p ≤ 1.

Existence of Laplace Transform Definition 7.1 Let a function f (t) defined for t ≥ 0 be piecewise continuous. It is said to be of exponential order c if lim e−ct f (t) = 0

t→∞

and it is written as f (t) = O(ect ) as t → ∞. 2

Note: The function et considered above is not of exponential order since lim e−ct et = ∞ 2

t→∞

for any constant c > −∞. Theorem 7.1 (Existence theorem) If f (t) defined on [0, ∞) is piecewise continuous and is of exponential order, i.e., f (t) = O(ect ), then the Laplace transform F( p) of f (t) exists in the half-plane  p > c. Proof 

∞

|F( p)| ≤

|e− pt f (t)|dt =

0

(α =  p, T is finite).

 0

T

e−αt | f (t)|dt +



∞ T

e−αt | f (t)|dt

164

7 Laplace Transform

The first integral exists since f is piecewise continuous. Since f (t) = O(ect ), there exists  for a given T , such that | f (t)|e−ct <  for t > T . Thus, the second integral is 

∞

c. Remark: The conditions for existence of the Laplace transform stated above are 1 sufficient rather than necessary. For example, the function f (t) = t − 2 has infinite discontinuity at t = 0 and hence is not piecewise continuous on [0, ∞). However, its Laplace transform exists (see Example 7.4 below).

Examples Example 7.1 Find the Laplace transform of f (t) = eat 

∞

L[e ] = at

e− pt eat dt.

0

This integral exists for  p > a so that L[eat ] = it diverges. Example 7.2 L[1] =

1 p−a

for  p > a, where for  p ≤ a,

for  p > 0.

1 p

Example 7.3 L[cos ωt] =

p , p2 +ω 2

  ∞ Example 7.4 L √1t = 0

− pt e√ t

L[sin ωt] =

dt =

√2 p

∞ 0

ω p2 +ω 2

for  p > 0.

e−x d x = 2



π p

√ ( pt = x).

Remark: In this example the function f (t) = t − 2 has infinite discontinuity at t = 0 and hence is not piecewise continuous on [0, ∞). However, its Laplace transform exists.This shows that the conditions for existence of the Laplace transform stated above are sufficient rather than necessary. 1

Example 7.5 L[t γ ](γ > −1) =



∞ 0

=

e− pt t γ dt =

1 p γ+1



∞

e−u u γ du (u = pt,  p > 0)

0

(γ + 1) (γ > −1,  p > 0). p γ+1

7.1 Derivation of Laplace Transform from Fourier Integral Theorem

Note: For γ = −1, the integral of 1t does not exist.

∞

165

e− pt dtt does not exist so that the Laplace transform

0

Example 7.6 

∞

L[cosh at] =

p2

p ,  p > a > 0. − a2

p2

a ,  p > a > 0. − a2

e− pt cosh at dt =

0



∞

L[sinh at] =

e− pt sinh at dt =

0

Example 7.7



1 2

L[t ] =

∞

e

1 t dt = 2p

− pt

1 2

0



π . p

Example 7.8 

t2

L[e− 4 ] =

∞

t2

e− pt− 4 dt = e p

2

0

= ep

2



∞

2

2



2p

as er fc( p) =

√2 π

∞ z

∞

e− 4 (t+2 p) dt 1

2

0

e− 4 u du = e p 1



∞

e−x d x = 2

√ p2 πe erfc( p)

p

e−x d x. 2

Example 7.9

   ∞ − pt 1 e L √ = dt √ t +a t +a 0 2 =e √ ap



∞

ap

√ ap

e

−x 2

dx =

π ap √ e er fc( ap), a > 0. p

Example 7.10 Find L[H (t − t0 )] where H (t − t0 ) is the Heaviside function given by H (t − t0 ) = 0, t < t0 , = 1, t > t0 . 

∞

L[H (t − t0 )] = 0

e− pt H (t − t0 )dt =

 t0

∞

e− pt dt =

e− pt0 . p

Theorem 7.2 If a function f (t) defined on [0, ∞) is piecewise continuous and is of exponential order O(ect ), then its Laplace transform F( p) tends to 0 as | p| → ∞.

166

7 Laplace Transform

Proof We have

 F( p) =

∞

e− pt f (t)dt

0

Since f (t) = O(ect ), we can find some T such that for any given small positive number , | f (t)|e−ct <  when t > T. Now   T

|F( p)| ≤

∞

e−αt | f (t)|dt +

0

e−αt | f (t)|dt (α =  p)

T

= I1 + I2 T ∞ where I1 = 0 e−αt | f (t)|dt → 0 as α → ∞ and I2 ≤  T e−(α−c)t = 0 as α → ∞. Thus, lim| p|→∞ |F( p)| = 0.

e−(α−c)T α−c

→

Remark: This theorem shows that if F( p) is any function for which lim| p|→∞ F( p)

= 0, then it cannot be the Laplace transform of a piecewise continuous function of exponential order. Thus, many functions such as polynomials, e p , sin p, cos p, etc., cannot represent Laplace transform.

7.2 Laplace Inversion Theorem 7.3 (Laplace inversion theorem) If F( p) is an analytic function of the complex variable p in the half-plane  p > c and is of order O( p −k ) where k is real and k > 1, then the integral 1 2πi



c+i∞

F( p)e pt dp c−i∞

converges to a function f (t) which is independent of c, provided, the Laplace transform of f (t) is F( p) for  p > c. Furthermore, the function f (t) is continuous for t ≥ 0 and is O(ect ) as t → ∞. Note: F( p) is of order O( p −k ) in the half-plane  p > c =⇒ there exists a positive real constant M such that | p k F( p)| < M whenever p is sufficiently large. Proof Let f (t) and f  (t) be continuous functions for 0 ≤ t < ∞ and that f (t) = O(ect ). Then its Laplace transform defined by  F( p) = 0

∞

e− pt f (t)dt

(7.5)

7.2 Laplace Inversion

167

exists and is analytic in the half-plane  p > c. We consider the integral 

c+iλ

e pt F( p)dp

(7.6)

c−iλ

We put F( p) from (7.5) so that 

c+iλ

 e F( p)dp =

e

c−iλ

 =

∞



c+iλ

pt

e

c−iλ



c+iλ

f (u)

0

= 2i

∞ −t

  e p(t−u) dp du = 2i

f (u)du dp

∞

f (u)ec(t−u)

0

f (t + v)e−cv

1 λ→∞ 2πi



− pu

0

c−iλ



∞

pt



c+iλ

sin λv dv, (substituting v = u − t). v 1 λ→∞ π



e pt F( p)dp = lim

Thus lim

sin λ(t − u) du t −u

c−iλ

∞

g(v)

−t

sin λv dv v

(7.7)

where g(v) = f (t + v)e−cv . It may be noted that f (t) = 0 for t < 0 so that g(v) = 0 for v < −t. Thus, the integral on the right-hand side of (7.7) can be written as =

1 π



∞

g(v)

0



sin λv dv + v

0

g(v)

−t

sin λv dv + v



−t

g(v)

−∞

sin λv dv v



Since g(v) = 0 for v < −t, we get 1 λ→∞ 2πi



c+iλ

lim

c−iλ

=

=

 Now, lim

λ→∞ 0

λ

1 π



1 π

∞



1 π

e pt F( p)dp =

∞

g(v)

0



∞

g(v)

−∞



λ

g(v)

−∞

sin λv dv + v



0 −∞

g(v)

sin λv dv v

sin λv dv v  cos zv dz dv.

0

 cos zv dz = 0

∞

cos zv dz =

1 2



∞

−∞

ei zv dz = πδ(v).



168

7 Laplace Transform

Thus, from Eq. (7.3), as λ → ∞, 1 π



∞

sin λv dv = g(v) v −∞

so that f (t) =

1 2πi





∞ −∞

c+i∞

g(v)δ(v)dv = g(0) = f (t)

e pt F( p)dp, t > 0.

c−i∞

This gives the Laplace inversion formula. Remark: 2 1. Consider the function F( p) = e p which is analytic everywhere in the complex −k p−plane. However, it is not O( p ) for any positive k and hence cannot represent the Laplace transform of a function f (t). 2. The conditions on the function F( p) are sufficient to ensure the validity of the inverse formula. 3. For the simple function F( p) = 1p ( p > 0), which is the Laplace transform of f (t) = 1, 0 ≤ t < ∞, the condition that F( p) = O( p −k ), k > 1, is not satisfied. In fact, the conditions on F( p) are quite severe. 4. Stating conditions on the function f (t) rather than on F( p), these conditions can be relaxed so that the Laplace inverse formula is valid for nearly all practical cases of interest.

7.3 Operational Properties of Laplace Transform 1. Linearity property: If F( p) = L[ f (t)] and G( p) = L[g(t)], then L[c1 f (t) + c2 g(t)] = c1 F( p) + c2 G( p), where c1 and c2 are constants. Proof Proof is obvious but for the sake of completeness, we give the proof below. 

∞

L[c1 f (t) + c2 g(t)] =

e− pt (c1 f (t) + c2 g(t))dt

0



∞

= c1

e− pt f (t)dt + c2

0



∞

e− pt g(t)dt = c1 F( p) + c2 G( p)

0

2. Scaling property: For a > 0, L[ f (at)] = a1 F( ap ). Proof

 L[ f (at)] = 0

∞

e− pt f (at)dt

7.3 Operational Properties of Laplace Transform

=

1 a



∞

169

p

e(− a )z f (z)dz =

0

  p 1 F a a

after substituting z = at. 3. Shifting Property: If F( p) is the Laplace transform of f (t), then L[eat f (t)] = F( p − a) Proof 

∞

L[eat f (t)] =

e− pt eat f (t)dt =

0



∞

e−t ( p−a) f (t)dt = F( p − a).

0

4. Translation Property: If F( p) is the Laplace transform of f (t), then the Laplace transform of  g(t) =

f (t − a), t > a, 0, otherwise

is e−ap F( p), a > 0. Proof

 L[g(t)] =

∞

e− pt f (t − a)dt

a

= e−ap



∞

e− pt f (t)dt = e−ap F( p).

0

5. Differentiation property: Let f (t) be a continuous function with a piecewise continuous derivative f  (t) for t ≥ 0 and let both f (t) and f  (t) be of exponential order O(ect ). Then L[ f  (t)] = p F( p) − f (0) where L[ f (t)] = F( p). Proof L[ f  (t)] =



∞

0

e− pt f  (t)dt = [e− pt f (t)]∞ 0 + p



∞

e− pt f (t)dt.

0

Since f (t) = O(ect ), it follows that e− pt f (t) → 0 as t → ∞ for  p > c. Thus

Similarly

L[ f  (t)] = p F( p) − f (0).

(7.8)

L[ f  (t)] = p 2 F( p) − p f (0) − f  (0).

(7.9)

170

7 Laplace Transform

Repeated application of the results in (7.8) and (7.9) produces L[ f n (t)] = p n F( p) − p n−1 f (0) − · · · − f n−1 (0)

(7.10)

provided of course f (t), f  (t), . . ., f n−1 are continuous for 0 ≤ t < ∞, f n (t) is piecewise continuous for 0 ≤ t < ∞ and all are of exponential order O(ect ). 6. Integration Property: 



t

L

f (u)du =

0

F( p) p

where L[ f (t)] = F( p) provide f (t) is piecewise continuous for 0 ≤ t < ∞ and is of exponential order O(ect ). t Proof Let g(t) = 0 f (u)du, then g(t) is continuous, g(0) = 0 and g  (t) = f (t).  Thus

L

t





0

 =

e− pt g(t) −p

∞

f (u)du =

e− pt g(t)dt

0

∞ + 0

1 p



∞

e− pt f (t)dt =

0

F( p) . p

7. Derivatives of the transform: If f (t) is piecewise continuous for 0 ≤ t < ∞ and is of exponential order O(ect ), and F( p) is its Laplace transform, then F (n) ( p) = (−1)n L[t n f (t)], n = 1, 2, . . .. Proof is obvious and follows from the fact that  ∞ dn e− pt f (t)dt = (−1)n L[t n f (t)]. F (n) ( p) = dp n 0 8. Integral of the transform: If f (t) is piecewise continuous for 0 ≤ t < ∞ and is of exponential order O(ect ) has a Laplace transform, then and f (t) t 

∞ p

Proof

  f (t) where F( p) = L[ f (t)]. F(u)du = L t



∞ p



∞

F(u)du = p

 0

∞

 e−ut f (t)dt du

7.3 Operational Properties of Laplace Transform



∞

=



∞

f (t)

0

171

  e−ut du dt =

p

∞

f (t) dt = L t

e− pt

0



 f (t) . t

Examples Example 7.11 Find L[te−at cos ωt]. p Solution: If we write f (t) = cos ωt, then F( p) = L[cos ωt] = p2 +w 2 . Now F( p) = ∞  ∞ − pt  − pt f (t)e dt so that F ( p) = − t f (t)e dt so that L[t f (t)] = −F  ( p) = 0  0  2 p −w 2 d − dp = ( pp2 +w 2 )2 . If we call g(t) = t cos ωt, then p2 +w 2

L[g(t)] ≡ G( p) =

p2 − w2 ( p 2 + w 2 )2 ( p + a)2 − w 2 . ( p + a)2 + w 2

L[e−at g(t)] = G( p + a) =

Now

 t Example 7.12 Find L 0

 sin x dx x

.

∞ Solution: We know that L[sin t] = 0 e− pt sin tdt = both sides with respect to p between p to ∞, we obtain 

∞

e− pt

0

 Now 0

∞

e− pt



t 0

=

1 , p2 +1

 p > 1. Integrating

∞ sin t π dt = tan−1 p p = − tan−1 p. t 2

  − pt  t ∞  e sin x sin x 1 ∞ − pt sin t d x dt = dx dt + e x −p 0 x p 0 t 0   1 1 π 1 − tan−1 p = tan−1 ,  p > 1. p 2 p p

Example 7.13 Find L[erf t]. t 2 Solution: er f t = √2π 0 e−x d x 2 L[er f t] = √ π 2 =√ π



e− pt −p



t

e 0



∞

e− pt

0 −x 2



t

 2 e−x d x dt

0

∞ dx 0

1 + p



∞

e 0

− pt −t 2

e

 dt

172

7 Laplace Transform

1 2 P2 √ e4 p π

=



∞



1 2 P2 √ e4 p π

e−(t− p/2) dt = 2

0

∞

e−x d x = 2

p/2

   where er fc(z) = Example 7.14 Find L erfc √at Solution:

2 =√ π

2 F( p) = √ π 

e− pt −p



∞

√ a/ t

e

−x 2

2a √ p π

∞



∞ dx

a 1 √ p π

∞



e− pt 1 + p



∞

∞

√ a/ t

0

0

= =





∞

e

∞ z

e− pt−

− pt

a2 t

e−x d x = 1 − erf z. 2

 2 e−x d x dt

(−e

−a 2 /t

0

0

p

√2 π

  1 P2 p e 4 er f c . p 2



a dt ) − 3/2 2t



dt t 3/2

  1 put √ = x t

e− x 2 −(ax) d x 2

0

To evaluate this integral, Let us write p = b2 , and denote 

∞

I (a, b) =

b2

e− x 2 −(ax) d x. 2

0

so that



∞

I (a, 0) =

e

−(ax)2

0

Thus, F( p) = F(b2 ) =

√ π . = 2a

2a √ I (a, b). p π

Now we write ax = u in I (a, b) to obtain I (a, b) =

1 a



∞

e−u

2

2 2

− a u 2b

du

0

Also, dI = −2b db



∞ 0

e−a

2

x − bx 2

2 2

dx = −2 x2



∞

2 2

e 0

−z 2 − a z 2b

dz (substituting

b = z). x

√ dI π −2ab −2ab = −2a I so that, I (a, b) = I (a, 0)e e Thus = . db 2a

7.4 Laplace Convolution Integral

Hence,

173

e−2a F( p) = p

√

p

( p > 0).

7.4 Laplace Convolution Integral Definition and Properties Definition 7.2 The Laplace convolution integral of two functions f (t) and g(t) defined on [0, ∞) is defined as 

t

( f ∗ g)(t) =

f (t − x)g(x)d x.

0

It is trivial to show that the Laplace convolution integral satisfies the following properties: (i) Commutative property: ( f ∗ g)(t) = (g ∗ f )(t). (ii) If c is a constant, then f ∗ (cg) = (c f ) ∗ g = c( f ∗ g). (iii) Distribution property f ∗ (g ∗ k) = ( f ∗ g) ∗ k. Convolution Theorem for Laplace Transform Theorem 7.4 (Convolution theorem) If F( p), G( p) are Laplace transforms of two functions defined on [0, ∞), then the inverse Laplace transform of the product f (t) and g(t), is the Laplace convolution integral of f (t) and g(t), i.e., L−1 [F( p)G( p)] = ( f ∗ g)(t). Proof We write  F( p)G( p) = 0

∞

f (x)e

− px



∞

dx 0

g(y)e− py dy

174

7 Laplace Transform



∞

=



0

∞

e− p(x+y) f (x)g(y)d xd y

0

If we put x + y = t, keeping y fixed, in the x-integral, then 

∞

F( p)G( p) =



0



∞

=

e− pt

∞

e− pt f (t − y)g(y)dtdy

y



0

t

 f (t − y)g(y)dy dt

0



∞

=

e− pt ( f ∗ g)(t)dt

0

= L[( f ∗ g)]. Thus L−1 [F( p)G( p)] = ( f ∗ g)(t) =



t

f (t − y)g(y)dy.

0 p Example: Find the Laplace inverse of ( p2 +a 2 )2 . p 1 We choose F( p) = p2 +a 2 and G( p) = p2 +a 2 . We find that F( p) = L[cos at] and G( p) = L[ sinaat ]. Thus,  t

L−1 [F( p)G( p)] =

cos a(t − y)

0

=

1 2a

1 = 2a





t

sin ay dy a

[sin a(2y − t) + sin at]dy

0

cos a(2y − t) − 2a =

t

 + t sin at

0

t sin at . 2a

7.5 Tauberian Theorems The behavior of a function f (t) in the neighborhood of the origin is, in some sense, reflected in the behavior of its Laplace transform as | p| → ∞. Theorem 7.5 (Initial value theorem) If f (t) is piecewise continuous on [0, ∞) and is of exponential order O(ect ), then its Laplace transform F( p) satisfies

7.5 Tauberian Theorems

175

lim p F( p) = lim f (t) = f (0).

| p|→∞

t→0+

Proof



∞

p F( p) =

pe− pt f (t)dt

0



= −e

− pt

f (t)

∞



0

+



∞

∞

e− pt f  (t)dt

0

= f (0) +

e− pt f  (t)dt.

0

Making | p| → ∞, we find 

∞

lim p F( p) = f (0) + lim

| p|→∞

| p|→∞ 0

e− pt f  (t)dt = f (0).

Remark: We have assumed that Laplace transform of f  (t) exists which obviously tends to 0 as | p| → ∞. Theorem 7.6 (Final value theorem) If f (t) is piecewise continuous and is of exponential order O(ect ), then its Laplace transform satisfies lim p F( p) = lim f (t).

| p|→0

t→∞

Proof



∞

lim p F( p) = lim p

p→0

p→0

e− pt f (t)dt

0



∞ = lim −e− pt f (t) 0 + p→0



∞

= f (0) +



∞

e− pt f  (t)dt

0

f  (t)dt

0

= f (0) + lim f (t) − f (0) t→∞

= lim f (t). t→∞

Example: Find the Laplace transform of Ci(t) defined by Ci(t) = Let f (t) = Ci(t), then f  (t) = cost t so that t f  (t) = cos t. Taking Laplace transform of both sides, we find L[t f  (t)] = L[cos t]

t ∞

cos u du u

t > 0.

176

7 Laplace Transform



so that

 te− pt f  (t)dt =

∞

0

−

d dp

This leads to − so that



∞

p2

 e− pt f  (t)dt =

0

p +1

p . p2 + 1

p d . [− f (0) + p F( p)] = 2 dp p +1 1 p F( p) = − ln(1 + p 2 ) + c, 2

where c is a constant. By the initial value theorem, lim p F( p) = lim f (t) = 0.

p→0

Thus, c = 0 and hence F( p) = −

t→∞

1 ln(1 + p 2 ). 2p

Watson’s Lemma Theorem 7.7 (Watson’s If f (t) defined on [0, ∞) is of exponential order  Lemma) an n (t) = ∞ t , |t| < R, then F( p) = L[ f (t)] has the asymptotes O(ect ), and if f n=0 n! an as | p| → ∞. series F( p) ∼ ∞ n=0 pn+1 Proof The result is obtained formally by applying the Laplace transform to the expansion of f (t) termwise.

7.6 Method of Evaluation of Inverse Laplace Transform ∞ We know that if F(s) = 0 e−st f (t)dt, where f (t), defined on [0, ∞) is of exponential order O(ect ), then  c+i∞ 1 f (t) = L−1 [F(s)] = F(s)est ds, t > 0. 2iπ c−i∞ Given F(s), we want to find f (t) directly by using the above complex integration formula. We consider the following examples.

7.6 Method of Evaluation of Inverse Laplace Transform

177

Example 7.15 Find L−1 [1/s]. Here, L−1 [1/s] =

1 2iπ



c+i∞

(1/s)est ds, t > 0.

c−i∞

We choose a contour as in Fig. 7.1, where C denotes the part of a circle of large radius R with center at the √ origin of the complex p-plane and L is the line c − i R1 to c + i R1 , where R1 = R 2 − c2 so that as R → ∞, R1 → ∞. By Cauchy’s residue theorem,  e pt dp = 2πi (Residue at p = 0). C+L p Now residue at p = 0 is lim p

p→0



e pt L p

 c+i∞

e pt p

dp as R → ∞.  pt It can be shown that C ep dp → 0 as R → ∞ for t > 0. The proof of this is given below.     pt  pt e e dp = dp. + + p p C AB BDE EF and

dp →

e pt =1 p

c−i∞

Fig. 7.1 Complex p-plane

178

7 Laplace Transform

Now

   

BDE

   3π  e pt   2 et (R cos θ+i R sin θ) iθ  dp  =  Re idθ  iθ π   p Re 2  ≤

 =

π

3π 2

et R cos θ dθ

π 2

e−t R sin θ dθ

(changing θ to

0



π 2

=2

π + θ) 2

e−t R sin θ dθ.

0

We shall now use Jordon’s inequality given below. Jordon’s inequality: For 0 ≤ θ ≤ π2 , Hence, the above integral     =

BDE

≤ sin θ ≤ θ.

  π  π 2 2 e pt  θ −t R sin θ dp  ≤ 2 e dθ ≤ 2 e−2t R π dθ p 0 0

π 2π(1 − e−t R ) = (1 − e−t R ) → 0, as R → ∞. 2Rt tR    

Now

2θ π

AB

  π   e pt   2 et R(cos θ+i sin θ) iθ dp  =  Re idθ iθ π p Re 2 −α  ≤ 

α

= 

α

et R cos θ dθ π − θ) 2

et R sin α dθ (as sin θ ≤ sin α for θ ≤ α ≤

0

= αe

π 2 −α

et R sin φ dφ, (where φ =

0


0. √

Here F( p) = e−a p has a branch point at p = 0 and p = ∞ in the complex p-plane. We take a branch cut from p = 0 to ∞ along the negative real axis, and choose a contour as in Fig. 7.3. Here ABC, G H I are arcs of√a circle of large radius R, D E F is the arc of a circle of small radius , R1 = R 2 − c2 , tan α = Rc1 . There is no singularity of √ e−a p inside this closed contour. Thus,  √ e−a p+t p dp = 0. ABC D E F G H I A

The left side is 

 + ABC





+ CD



+ DE F



+ FG

+ GHI

. IA

182

7 Laplace Transform

Fig. 7.3 Complex p-plane

The integrals along the arcs ABC and G H I tend to zero as R → ∞ and the integral along the curve D E F tends to zero as  → 0, while the integral along I A becomes  c+i∞ √ e−a p+t p dp. c−i∞

Now along the line C D, arg( p) = π so that along C D, p = xeiπ where x is real and positive, and along the line F G, arg( p) = −π so that along F G, p = xe−iπ . Thus,



e−a

√

 p+t p



e−a FG

Thus,

e−ai

√

x−xt

dx

0

CD

and

∞

dp = −

√

 p+t p

dp = 0

∞

eai

√

x−xt

d x.

7.7 Application of Laplace Transform in Solving Ordinary Differential Equations

183

 ∞  ∞ √  √  √ 1 1 L−1 e−a p = e−ai x−xt d x − eai x−xt d x 2πi 0 2πi 0 1 π

= 2 = π



∞



∞

√ e−xt sin(a x)d x

0

ue−tu sin au du, x = u 2 2

0

2 d =− π da



∞

e



where I (a) = a2



 cos au du

0

=−

= e− 4t

−u 2 t

1 d I (a) π da ∞

e−u

2

t+iau

du

−∞

∞

1 ia 2 a2 e−t (u+ 2t ) du = √ e− 4t t −∞



∞

e−z dz 2

−∞

√ π a2 = √ e− 4t t Thus,

 √  a a2 L−1 e−a p = √ e− 4t , a > 0. 2 πt 3

7.7 Application of Laplace Transform in Solving Ordinary Differential Equations Ordinary Differential Equations with Constant Coefficients Example 7.18 Solve

df dt

+ f (t) = 1, t ≥ 0, given that f (0) = 2.

Solution: Use of Laplace transform on both sides of the differential equation produces 1 p F( p) − 2 + F( p) = p so that ( p + 1)F( p) =

1 + 2p 1 +2= . p p

184

7 Laplace Transform

Hence, F( p) =

1 1 + p+1 p

whose Laplace inversion gives f (t) = 1 + e−t , t ≥ 0. Example 7.19 Solve the initial value problem df d2 f + b f = g(t), t ≥ 0 +a 2 dt dt with f (0) = f 0 , f  (0) = f 1 , a, b, being constants. Solution: Taking Laplace transform of both sides of the given equation, and noting that L[ f ] = F( p), L[ f  (t)] = p F( p) − f 0 , L[ f  (t)] = p 2 F( p) − p f 0 − f 1 , L[g(t)] = G( p), we find p 2 F( p) − f 0 p − f 1 + a( p F( p) − f 0 ) + bF( p) = G( p). This gives F( p) =

G( p) + ( p + a) f 0 + f 1 . p 2 + ap + b

Once g(t) is known, G( p) can be found, and then f (t) can be obtained as f (t) = L−1 [F( p)]. For example, let g(t) = 1, f 0 = 0, f 1 = 0, a = 3, b = 2. Then F( p) =

1 1 1 − + , 2p p + 1 2( p + 2)

so that f (t) =

1 (1 + e−2t ) − e−t . 2

Example 7.20 Solve the initial value problem d2 f df + 2 f = e−t , t ≥ 0 −3 dt 2 dt with f (0) = 0, f  (0) = 0.

7.7 Application of Laplace Transform in Solving Ordinary Differential Equations

185

Solution: Let F( p) = L[ f (x)]. Then   df = p F( p), L dt  2  d f L = p 2 F( p) dt 2 L[e−t ] =



∞

e−( p+1)t dt =

0

1 . ( p + 1)

Using the above relations and taking Laplace transform of the given differential equation, we find 1 ( p 2 − 3 p + 2)F( p) = p+1 so that F( p) = −

1 1 1 + + . 2( p − 1) 3( p − 2) 6( p + 1)

Hence, inverse Laplace transform gives f (t) = −

e−t e3t e−t + + . 2 3 6

Example 7.21 Solve d2 f df + b f (t) = m(t), t ≥ 0 +a 2 dt dt with f (0) = 0, f  (0) = 0, m(t) being a known function. Solution: Laplace transform of the given equation gives p 2 F( p) + ap F( p) + bF( p) = M( p) so that F( p) =

p2

M( p) where M( p) = L[m(t)]. + ap + b

Hence, using convolution theorem, −1



f (t) = L where g(t) = L−1



1

p2 +ap+b

 .

  t M( p) = g(t − u)m(u)du p 2 + ap + b 0

186

7 Laplace Transform

    1 . We know that Let a = −2, b = 5, then g(t) = L−1 p2 −21 p+5 = L−1 ( p−1) 2 +4     1 sin 2t 1 sin 2tet −1 −1 L = 2 , then g(t) = L = 2 . Thus, p2 +4 ( p−1)2 +4 1 f (t) = 2



t

sin 2(t − u)et−u m(u)du.

0

Suppose (i) m(t) = et , then  f (t) =

t

sin 2(t − u) du =

0

et (cos 2t − 1). 4

(ii) m(t) = 1, then 1 f (t) = 2



t

sin 2(t − u) e(t−u) du =

0

1 t 1 1 e cos 2t − et sin 2t − . 3 6 6

Example 7.22 Solve the ODE d2 f + 4 f (t) = dt 2



4t, 0 ≤ t ≤ 1, 0, t > 1

with f (0) = 1, f  (0) = 0. Solution: Multiplying both sides by e− pt and integrating with respect to t between 0 to ∞, we obtain  p F( p) − p + 4F( p) = 4 2

0

1

te− pt dt =

4(1 − e− p (1 + p)) p2

so that F( p) =

p + p2 + 4



1 1 − 2 p2 p +4



 −

   1 p 1 −p − 1 − e e− p − p p2 p2 + 4 p2 + 4

Thus, f (t) = cos 2t + t −

sin 2t − L−1 2



     p 1 1 1 − p − L−1 −p . e e − − p p2 p2 + 4 p2 + 4

7.7 Application of Laplace Transform in Solving Ordinary Differential Equations

Now L−1

−1

e− p ]= p2



0, t ≤ 1, t − 1, t > 1,



  0, t ≤ 1, e− p sin 2(t − 1) = , t > 1, p2 + 4 2  −p   0, t ≤ 1, −1 e = L 1, t > 1, p



  pe− p 0, t ≤ 1, = 2 cos 2(t − 1), t > 1. p +4

L

and −1

L



Hence, sin 2t − f (t) = t + cos 2t − 2



t ≤ 1, sin 2(t − 1) t− − cos 2(t − 1), t > 1. 2

0,

Ordinary Differential Equations with Variable Coefficients Example 7.23 Solve the ODE d2 f df − t f (t) = 0 − 2 dt dt

t with f (0) = 0.

Solution: Let L[ f (t)] = F( p) =

∞ 0

f (t)e− pt dt so that

   ∞ df d f − pt L = e dt = − f (0) + p F( p) = p F( p)(Rep > 0). dt dt 0 To find the L[t f (t)] and L[t 2 ddtf ] in terms of F( p), we proceed as follows: 

∞

f (t)e− pt dt = F( p),

0

differentiation of both sides with respect to p, produces L[t f (t)] = −F  ( p). Again, we know

187

188

7 Laplace Transform

 2   ∞ 2 d f − pt d f = L e dt = − f  (0) + p 2 F( p). 2 dt 2 dt 0 Differentiating both sides with respect to p, we find  2  d d f L t 2 = − ( p 2 F( p)) = −2 p F( p) − p 2 F  ( p). dt dp Thus, the use of Laplace transform reduces the given ordinary differential equation to −2 p F( p) − p 2 F  ( p) − p F( p) + F  ( p) = 0 which is equivalent to ( p 2 − 1)

dF + 3 p F( p) = 0. dp

Its solution is F( p) = Thus, −1

c ( p2



f (t) = cL

3

− 1) 2

.



1 3

( p 2 − 1) 2

= ct I1 (t),

where I1 (t) is the modified Bessel  function  of order 1. The direct evaluation of L−1

1 3

( p2 −1) 2

is somewhat clumsy. An indirect proof is

obtained by finding the Laplace transform of t I1 (t), where  2r +1 I1 (t) =

∞  r =0

t 2

r !(r + 1)!

.

Note that an arbitrary constant ‘c’ appears in the solution since only one initial condition is prescribed.

7.8 Application Laplace Transform in Solving Partial Differential Equations The Laplace transform is suitable for solving initial value problems involving heat conduction equation and wave equation in which the time variable t occurs. For Laplace equation in space variables (x, y) if x is such that 0 ≤ x < ∞, then

7.8 Application Laplace Transform in Solving Partial Differential Equations

189

 2  ∂ φ L (x, y) = p 2 ( p, y) − pφ(0, y) − φx (0, y). ∂x 2 so that knowledge of φ and φx at x = 0 are required. However, prescribing both φ and φx on the boundary x = 0 is usually not possible. Thus, Laplace transform appears to be not suitable to solve a boundary value problem involving Laplace equation. Because of this, we will only consider heat equation or wave equation. Example 7.24 Solve the PDE 1 ∂φ ∂2φ , 0 ≤ x < ∞, t ≥ 0 = ∂x 2 k ∂t with φ(0, t) = f (t), t > 0, φ(x, 0) = 0, x > 0, φ(x, t) → 0, as x → ∞ where f (t) is a known function of t. 

Solution: Let

∞

(x; p) =

φ(x, t)e− pt dt.

0

Multiplying both sides of the PDE by e− pt and integrating with respect to t between 0 to ∞, we obtain the ODE p d 2 = , 0 ≤ x < ∞ dx2 k with (0; p) = F( p), (x; p) → 0 as x → ∞, where F( p) = L[ f (t)]. The solution of the ODE is (x, p) = F( p)e− Hence,

√ k

p

x

, x ≥ 0.

√   p φ(x, t) = L−1 F( p)e− k x .

Let

g(t) = L−1 [e−

We know that

2

−1

L [e

−

√ k

p

x

]=

x √x e− 4kt k

√ 2 πt 3

√ k

p

x

]. x2

xe− 4kt = √ ≡ g(t). 2 πkt 3

190

7 Laplace Transform

Thus, using convolution theorem,  φ(x, t) =

t

0

x f (u)g(t − u)du = √ 2 πk



f (u)

t 0

(t − u)

x2

3 2

e− 4k(t−u) du.

Special case: If f (u) = c, then cx φ(x, t) = √ 2 πk If we put z =

√ x , 2 k(t−u)



t

x2

(t − u)− 2 e− 4k(t−u) du. 3

0

then

2c φ(x, t) = √ π



∞ √x 2 kt

  x 2 . e−z dz = c erfc √ 2 kt

Example 7.25 Solve the PDE ∂2φ 1 ∂2φ = , 0 ≤ x < ∞, t ≥ 0 ∂x 2 c2 ∂t 2 with φ(0, t) = f (t), where f (t) is a known function of t, φ(x, t) → 0 as x → ∞, φ(x, 0) = 0,

∂φ (x, 0) = 0, 0 < x < ∞. ∂t

∞ Solution: Let (x; p) = 0 φ(x, t)e− pt dt be the Laplace transform of φ(x, t). Using Laplace transform, the given PDE reduces to the following ODE. p2 d 2 −  = 0, 0 ≤ x < ∞, dx2 c2 with (0; p) = F( p), (x, p) → 0 as x → ∞. The solution of the ODE is (x; p) = F( p)e− c p . x

Thus,

  x φ(x, t) = L−1 F( p)e− c p .

Exercises

191



We know that

∞

F( p) =

e− pu f (u)du

0



Hence F( p)e

− xc p

=

∞

e

− p(u+ xc )

 f (u)du =

∞ x c

0

e− pt f (t −

x )dt. c

Thus, inverse Laplace transform yields 

L−1 F( p)e− c p

Hence,

x



⎧ x ⎪ t≤ , ⎨ 0, c   = x x ⎪ ,t> ⎩f t− c c

⎧ ⎪ ⎨ 0,

x , c φ(x, t) = ⎪ ⎩ f (t − x ), t > x . c c t≤

If f (t) = 1 then ⎧ x ⎪ ⎨ 0, t ≤ , c φ(x, t) = ⎪ ⎩ 1, t > x . c It may be noted that a linear partial differential equation in two variables is transformed to an ordinary differential equation by using an appropriate integral transform. After solving the ordinary differential equation, the task of finding the inverse remains. In many cases, the inverse may not be obtained analytically, and usually an appropriate numerical technique for finding the inversion is required. However, this is not discussed here.

Exercises 1. Let F( p) denote the Laplace transform of the function f (t) defined on [0, ∞). Then F( p) is defined as (A)

∞ 0

f (t)ei pt dt,

(B)

∞ 0

f (t)e− pt dt,

(C)

Answer:

∞ 0

B

f (t)e pt dt,

(D)

∞ 0

f (t)e−i pt dt.

192

7 Laplace Transform

2. The parameter p of the Laplace transform F( p) of the function f (t) defined on [0, ∞) is in general a (A) positive integer,

(B) negative integer,

(C) complex number,

Answer:

(D) real number.

C

3. Let F( p) denote the Laplace transform of the function f (t) defined on [0, ∞). For the existence of F( p) (A) f (t) must be a periodic function, (C) f (t) must be an exponential function,

(B) f (t) must be a quadratic function, (D) f (t) = (ect ).

Answer:

D

4. Let F( p) denote the Laplace transform of the function f (t) defined on [0, ∞). Then f (t) is related to F( p) by the integral (A) (C)

1 2πi 1 2πi

1  c+i∞ F( p)e−t p d p, (B) 2πi  ∞ c−i∞ −t p d p. (D) 0 F( p)e

 c+i∞ F( p)et p d p,  c−i∞ i∞ tp −i∞ F( p)e d p,

Answer:

A

5. Let F( p) denote the Laplace transform of the function f (t) defined on [0, ∞). If 2 f (t) = et then 2

(A) F( p) = e p ,

(B) F( p) = e− p , 2

(C) F( p) =

Answer:

1 , p2

(D) F( p) does not exist.

D

6. The Laplace transform of f (t) = t 2 is (A)

1 , p2

(B)

1 , p3

(C)

Answer:

C

2 , p3

(D)

1 . 2 p3

Exercises

193

7. Which of the following functions has Laplace transform? (A) f (t) = 1t ,

2

(C) f (t) = sinh(2t 2 ),

(B) f (t) = e3t ,

Answer:

(D) f (t) = 4.

D

8. The Laplace transform of f (t) = cos 2t is (A)

1 , p 2 +4

(B)

p , p 2 +4

(C)

Answer:

2 , p 2 +4

(D)

4 . p 2 +4

B

9. The Laplace transform of f (t) = t μ exists if (A) μ > −1,

(B) μ > −1,

(C) μ = −1,

Answer:

(D) μ < −2.

B

10. The Laplace transform of f (t) = t − 2 is 1

√

(A)

π

3 2p 2

,

(B)

√ π 3

(C) π3

,

p2

p2

Answer:

(D) 1 3 . 2p 2

A

11. Let F( p) denote the Laplace transform of the function f (t) defined on [0, ∞).  Then the Laplace transform of f at , (a > 0), is (A) F(ap),

(B) F

 p a

(C) a F(ap),

,

Answer:

C

(D)

1 aF

 p a

.

194

7 Laplace Transform

12. Let F( p) denote the Laplace transform of the function f (t) defined on [0, ∞). Then F( p − a), is the Laplace transform of (A) f (t − a),

(B) f (t + a),

Answer:

(D) e−at f (t).

(C) eat f (t),

C

13. Let F( p) denote the Laplace transform of the function f (t) defined on [0, ∞). Then lim |F( p)| is | p|→∞

(A) 0,

(B) 1,

(D) ∞.

(C) e,

Answer:

A

14. State which of the following functions represent a Laplace transform (A) e p ,

(B) 1p ,

  (D) sin 1p .

(C) sin p,

Answer:

B

15. Let F( p) denote the Laplace transform of the function f (t) defined on [0, ∞). , is Then the Laplace transform of f (t) t (A)

p 0

F(u)du,

(B)

∞ p

F(u)du,

Answer:

(C)

p 0

(D)

F(u)du,

∞ p

F(u) u du.

B

16. Let F( p) denote the Laplace transform of the function f (t) defined on [0, ∞). Then the Laplace transform of t 2n f (t) (n a positive integer) is (A)

d 2n F( p) , dp 2n

n

2n

F( p) (C) (−1)n d dp n ,

p) (B) (−1)n d dpF( 2n ,

Answer:

A

(D)

d n F( p) dp n .

Exercises

195

17. The Laplace transform of f (t) = (A)



1, p

(B)

√1 t

(t > 0) is

1 πp,

Answer:

(C)



π, p

C

18. The inverse Laplace transform of the function F( p) = (B) e−t ,

(A) t,

Answer:

(B) cos 4t,

1 p

(D) 1t .

C p p2 +16

(C) t sin 4t,

Answer:

is

(C) 1,

19. The inverse Laplace transform of the function F( p) = (A) sin 4t,

(D) √πp .

is (D) t cos 4t.

B

20. Let F( p) denote the Laplace transform of the function f (t) defined on [0, ∞). Then the Laplace transform of F(pp) is (A)

t 0

f (u)du,

(B)

t 0

u f (u)du,

Answer:

(C) t

t 0

f (u)du,

(D)

∞ t

f (u)du.

A

21. Let the convolution  t of two functions f (t) and g(t) defined on [0, ∞) be denoted by f (t) ∗ g(t) = 0 f (t − u)g(u)du. Then the convolution of f (t) and g(t) when f (t) = 1, g(t) = 1, is 2 (A) t2 ,

(B) t,

(C) 1,

Answer:

B

(D) −t.

196

7 Laplace Transform

22. Let the convolution  t of two functions f (t) and g(t) defined on [0, ∞) be denoted by f (t) ∗ g(t) = 0 f (t − u)g(u)du. Then the convolution of f (t) and g(t) when f (t) = sin t, g(t) = sin t, is (A) sin2 t,

(B) sin t−t2 cos t ,

(C) sin t,

Answer:

(D) cos t.

B

23. Let f (t) = t, g(t) = 1. Then the Laplace transform of the convolution of f (t) ∗ g(t) is (A) 1p ,

(B) 12 , p

Answer: 24. The Laplace inversion of (A) et − 1,

1 p( p−1)

(C) − 1p ,

(D) − 12 .

(C) et−1 ,

(D) et+1 .

p

B

is

(B) et + 1,

Answer:

A

25. Let F( p) denote the Laplace transform of the function f (t) defined on [0, ∞). Then lim f (t) is equal to t→∞

(A) lim p F( p), p→0

(B) lim F( p),

(C) lim F( p), p→∞

p→0

Answer: 26. Evaluate the inverse Laplace transform of Answer:

(a−b) ( p−a)( p−b)

eat − ebt

27. Evaluate the inverse Laplace transform of Answer:

A

( p) ( p−a)( p−b)

aeat − bebt a−b

(D) lim p F( p). t→∞

Exercises

197

28. Evaluate the inverse Laplace transform of

p ( p2 +a 2 )( p+ b2 )

(|a| = |b|)

cos at − cos bt (|a| = |b|) b2 − a 2 √ 29. Evaluate the inverse Laplace transform of πp e−2 p Answer:

Answer:

1 1 √ e− t t

30. Evaluate the inverse Laplace transform of

p 3

( p−a) 2

(1 + 2at)eat √ πt

Answer:

31. Evaluate the inverse Laplace transform of Answer:

√

p−a−

√

p − b.

ebt − eat √ 2 πt 3

32. Use Laplace transform to solve the ordinary differential equation df − 2 f = 4, t ≥ 0 dt along with f (0) = 0. Answer:

f (t) = 2(e2t − 1).

33. Use Laplace transform to solve the ordinary differential equation d2 f + f = sin 2t, t ≥ 0 dt 2 along with f (0) = 0, f  (0) = 0. Answer:

f (t) =

2 sin t − sin 2t . 2

34. Use Laplace transform to solve the ordinary differential equation d4 f − 16 f = 0, t ≥ 0 dt 4

198

7 Laplace Transform

along with f (0) = 0, f  (0) = 1, f  (0)0, f  (0) = 1. Answer:

f (t) =

2 sin 2t + 5 sinh 2t . 16

35. Use Laplace transform to solve the ordinary differential equation t

df d2 f + + t f = 0, t ≥ 0 2 dt dt

along with f (0) = 0, f  (0) = 0. Answer:

f (t) = J0 (t).

36. Use Laplace transform to solve the ordinary differential equation df d2 f +t + f = 0, t ≥ 0 dt 2 dt along with f (0) = 1, f  (0) = 0. Answer:

t2

f (t) = e− 2 .

37. Use Laplace transform to solve 1 ∂2φ ∂2φ − 2 2 = 0, x ≥ 0, t ≥ 0 2 ∂x c ∂t along with φ(0, t) = f (t), φ(x, t) → 0 as x → ∞, ∂φ (x, 0) = 0, 0 < x < ∞. φ(x, 0) = 0, ∂t  x  0, x  t ≤ xc Answer: φ(x, t) = f x − c , t > c. 38. Use Laplace transform to solve ∂ 2 φ 1 ∂φ = 0, x ≥ 0, t ≥ 0 − ∂x 2 k ∂t along with φ(0, t) = 1, φ(x, t) → 0 as x → ∞, t > 0, φ(x, 0) = 0, 0 < x < ∞.  Answer:

φ(x, t) = er f c

x √

2 kt

 .

References

199

References The content of this chapter is mainly based on the website epgp.inflibnet.ac.in/ of UGC E Pathsala and on the following literature. 1. 2. 3. 4.

I.N. Sneddon, Use of Integral Transform (Mc Graw Hill, 1972) I.N. Sneddon, Fourier Transform (Mc Graw Hill, 1951) L.C. Andrews, B.K. Shivamoggi, Integral Transform for Engineers (Prentice-Hall, 2007) L. Debnath, D. Bhatta, Integral Transforms and Their Applications, 3rd edn. (Chapman and Hall/CRC, USA, 2014)

Chapter 8

Mellin Transform

8.1 Introduction If one wants to solve the two-dimensional Laplace equation in an infinite wedge described by r > 0, − α ≤ θ ≤ α, where r, θ are plane polar co-ordinates, then Mellin transform has to be used. The Laplace equation in r, θ co-ordinate system is 1 ∂2φ ∂ 2 φ 1 ∂φ + + = 0 , r > 0, − α ≤ θ ≤ α, ∂r 2 r ∂r r 2 ∂θ2

(8.1)

where φ is a function of r and θ. There are various methods to solve this partial differential equation. One method is to integrate equation (8.1) with respect to r between 0 to ∞ after multiplying by r s+1 , where s is some parameter. We note that 

∞

 r s+1

0

 = 0

∞

 ∂ 2 φ 1 ∂φ dr + ∂r 2 r ∂r

rs

∂ ∂r

  ∂φ r dr ∂r

 ∞  ∞ s+1 ∂φ s 2 = r − sr φ +s φ(r, θ)r s−1 dr. ∂r 0 0 If both r s+1

∂φ ∂r

and r s φ → 0 as r → ∞ and as r → 0, then 



∞

r 0

s+1

 ∂ 2 φ 1 ∂φ dr = s 2 (s, θ), + ∂r 2 r ∂r

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 S. Banerjea and B. N. Mandal, Integral Equations and Integral Transforms, https://doi.org/10.1007/978-981-99-6360-7_8

201

202

8 Mellin Transform



where

∞

(s, θ) =

r s−1 φ(r, θ)dr.

(8.2)

0

The above partial differential equation (8.1) then reduces to the ordinary differential equation d 2 + s 2  = 0, − α ≤ θ ≤ α. (8.3) dθ2 This can be solved easily if φ(r, θ) is prescribed on the two sides θ = ±α of the wedge. For simplicity, suppose φ(r, α) = φ(r, −α) = f (r )

(8.4)

where f (r ) is a known function of r . Then (s, α) = (s, −α) = F(s) 

where

∞

F(s) =

(8.5)

r s−1 f (r )dr,

(8.6)

0

Thus (s, θ) =

cos sθ F(s). cos sα

(8.7)

In fact, F(s) in (8.6) is called the Mellin transform of the function f (r ), and (s, ·) in (8.7) is the Mellin transform of φ(r, ·). Once we know how to obtain f (r ) from F(s), i.e, the inverse Mellin transform formula, then φ(r, θ) can be found from (s, θ) given by (8.7) and this provides the solution of two-dimensional Laplace equation in an wedge with prescribed boundary conditions (8.4).

8.2 Formal Derivation of Mellin Transform The Mellin transform and its inversion formula can be derived formally from the Fourier integral theorem 1 g(u) = 2π



∞

e

−iku



−∞

∞

 g(v)e

ikv

−∞

dv dk, − ∞ < u < ∞.

If we put x = ev , y = eu and s = c + ik where c is a constant, then 1 g(ln y) = 2πi



c+i∞

e c−i∞

−(s−c)lny



∞ 0

g(ln x)e

(s−c)lnx

dx x

 ds

(8.8)

8.3 Theorem on Inversion of Mellin Transform

so that g(ln y)y −c = If we define

1 2πi



c+i∞

203

y −s



c−i∞

∞

 g(ln x)x −c x s−1 d x ds.

f (x) = g(ln x)x −c ,

then (8.9) leads to f (x) =

1 2πi



c+i∞

x −s F(s)ds

(8.10)

(8.11)

c−i∞



where

(8.9)

0

∞

F(s) =

x s−1 f (x)d x.

(8.12)

0

The relation (8.12) is the Mellin transform of f (x) and (8.11) is its inversion formula. Remarks: 1. This is purely a formal procedure to derive the Mellin transform and its inversion. 2. The transform parameter s is in general complex as in the case of Laplace transform. 3. The significance of the real constant c appearing in the inversion formula will be apparent from the theorem given below. In the following theorem, the existence of Mellin transform and its inversion formula is proved rigorously.

8.3 Theorem on Inversion of Mellin Transform Theorem 8.1 Let F(s) be a function of the complex variable s = σ + iτ and it be regular in the strip a < σ < b and for any arbitrary small positive number η let F(s) tend  ∞ to zero uniformly as |τ | → ∞ in the strip a + η ≤ σ ≤ b − η. Let the integral −∞ F(σ + iτ )dτ be absolutely convergent for each σ for a < σ < b. If we define f (x) =

1 2πi



c+i∞

x −s F(s)ds

(8.13)

c−i∞

for positive real values of x and a fixed c ∈ (a, b), then in the strip a < σ < b, 

∞

F(s) = 0

x s−1 f (x)d x.

(8.14)

204

8 Mellin Transform

Proof We first note that since F(s) → 0 uniformly as |τ | → ∞ in the strip a + η ≤ σ ≤ b − η, the path of integration in the integral in (8.13) may be displaced parallel to itself. This means that the integral in (8.13) is independent of a particular choice of the constant c. We now choose two values of c in (a, b), say c1 and c2 , such that c1 < σ < c2 , and we write 



∞

x

s−1



1

f (x)d x =

x

0

s−1

∞

f (x)d x +

0

x s−1 f (x)d x.

1

In the first integral, we choose in the integral representation (8.13) for f (x), c = c1 , and in the second integral, we choose c = c2 . Thus 

∞

x s−1 f (x)d x = I1 + I2 ,

(8.15)

0

where





1

I1 =

x s−1 0





∞

I2 =

x

s−1

1

Now |I1 | =

≤

1 2π

1 2πi 1 2πi



c1 +i∞

x −t F(t)dt d x,

(8.16)

c1 −i∞



c2 +i∞

−t

x

F(t)dt d x.

(8.17)

c2 −i∞

  1

1

c1 +i∞ s−t−1 F(t) x d x dt



2π c1 −i∞ 0 

∞

−∞

 |F(c1 + iτ1 )|dτ1

1

x σ−c1 −1 d x.

0

1 ∞ As c1 < σ, the integral 0 x σ−c1 −1 d x is convergent. Also, as the integral −∞ F(σ + ∞ iτ )dτ is absolutely convergent for a < σ < b, the integral −∞ |F(c1 + iτ1 )|dτ1 is convergent. Thus, |I1 | < ∞. Hence, we may change the order of integration in I1 so that  1  c1 +i∞ 1 s−t−1 I1 = F(t) x d x dt 2πi c1 −i∞ 0 =

1 2πi =



1 2πi

c1 +i∞ c1 −i∞

 C1

where C1 is the line as shown in Fig. 8.1.

F(t) dt s−t

F(t) dt, t −s

(8.18)

8.4 Properties of Mellin Transform

205

Fig. 8.1 Complex t (= σ1 + iτ1 )-plane

Similarly, we can show that I2 =

1 2πi

 C2

F(t) dt t −s

(8.19)

where C2 is the line as shown in Fig. 8.1. Thus, we obtain 

∞

x 0

s−1

1 f (x)d x = 2πi

 C1

F(t) dt + t −s

 C2

F(t) dt t −s

 (8.20)

Since |F(t)| → 0 uniformly as |I m(t)| → ∞ in the strip a < Re(t) < b, we see that each of the integrals in (8.18) and (8.19) tends to zero as |τ1 | → ∞. Hence, we can replace the two integrals in Eq. (8.20) round a closed contour  enclosing s. Since in the complex F(t) is regular in the strip a < Re(t) < b, the only singularity of F(t) t−s t-plane is a simple pole at t = s, and by Cauchy’s residue theorem of complex variable, the value of the integral on the right side is F(s).

8.4 Properties of Mellin Transform Let us denote the Mellin transform of a function f (x) defined for x > 0 by 

∞

M[ f (x)] ≡ F(s) =

f (x)x s−1 d x.

0

(i) Linearity property If F(s) and G(s) are the Mellin transforms of the function f (x) and g(x), then

206

8 Mellin Transform

M[c1 f (x) + c2 g(x)] = c1 F(s) + c2 G(s), where c1 and c2 are constants. The proof is obvious. (ii) Scaling Property F(s) , a > 0. as

M[ f (ax)] = Proof



∞

M[ f (ax)] =

x s−1 f (ax)d x

0

=

1 as



∞

t s−1 f (t)dt

0

=

F(s) . as

(iii) Translation property M[x a f (x)] = F(s + a) The proof is obvious.   s , a > 0. The proof is obvious. a

(iv) M[ f (x )] =    = F(1 − s) (v) M x1 f x1 a

1 F a

Proof

   ∞    1 1 1 s−1 1 f = f dx M x x x x x 0  =

∞

t −s f (t)dt = F(1 − s).

0

(vi) Mellin transform of the derivative of a function Let f (x) be continuous for x ≥ 0 and F(s) be its Mellin transform. Further, let lim x s−1 f (x) = 0

x→0

and

lim x s−1 f (x) = 0

x→∞

for σ1 < Re(s) < σ2 . Then M[ f  (x)] = −(s − 1)F(s − 1), σ1 < Re(s) < σ2 ,

8.4 Properties of Mellin Transform

207

provided F(s − 1) exists for σ1 < Re(s) < σ2 . Proof M[ f  (x)] =



∞

x s−1 f  (x)d x

0

= [x s−1 f (x)]∞ 0 − (s − 1)



∞

x s−2 f (x)d x

0

= −(s − 1)F(s − 1), σ1 < Re(s) < σ2 . A second application of this result leads to M[ f  (x)] = −(s − 1)M[ f  (x)](s − 1) = (s − 1)(s − 2)F(s − 2), provided

lim x s−2 f  (x) = 0.

x→0

One can prove by induction (s) F(s − n) (s − n)

M[ f (n) (x)] = (−1)n provided

lim x s−k−1 f (k) (x) = 0, k = 0, 1, . . . , n − 1.

x→0

(vii) The convolution theorem for Mellin transform Let F(s) and G(s) be the Mellin transforms of f (x) and g(x), defined for x > 0, respectively. Then M−1 [F(s)G(s)] =



∞

f 0

Proof We have



∞

  dt x g(t) . t t

f (x)g(x)x s−1 d x

0

 =

∞

 g(x)x s−1

0

1 = 2πi



1 2πi



c+i∞ c−i∞



c+i∞

F(z) c−i∞

F(z)x −z dz d x

0

∞

g(x)x

s−z−1

d x dz

(8.21)

208

8 Mellin Transform

1 = 2πi



c+i∞

F(z)G(s − z)dz.

c−i∞

This shows that 1 2πi

M[ f (x)g(x)](s) =



c+i∞

F(z)G(s − z)dz.

(8.22)

F(z)G(1 − z)dz.

(8.23)

c−i∞

If we put s = 1 in this result, then 

∞

1 2πi

f (x)g(x)d x =

0



c+i∞ c−i∞

Again,



1 M [F(s)G(s)](x) = 2πi −1

=



1 2πi 

= 0

∞

c+i∞

g(t) t



1 2πi

F(s)

0

(viii) M [ f (x ); s] =

g(t)t s−1 dt ds

c+i∞

c−i∞ ∞

= 1 F a

∞

0



a



F(s)x −s

F(s)G(s)x −s ds

c−i∞

c−i∞



c+i∞

 x −s t

)ds dt

  x g(t)dt. f t

  s , a>0 a

Proof





∞

M f (x ; s = a

x s−1 f (x a )d x

0



∞

=

t

s−1 a

0

=

1 a



∞

1

dt , a>0 a

t a −1 f (t)dt =

0

    (ix) M x1 f x1 ; s = F(1 − s). The proof is simple. (x) M[(ln x) f (x); s] = F  (s).

f (t)t a −1

s

1 s F . a a

(8.24)

8.5 Mellin Transform of Some Simple Functions

209

Proof  M[(ln x) f (x); s] =

∞

x s−1 ln x f (x)d x =

0

(xi) M[x f  (x); s] = assuming of course

∞ 0

d ds



∞

x s−1 f (x)d x = F  (s).

0

x s f  (x)d x = [x s f (x)]∞ 0 −s

∞ 0

x s−1 f  (x)d x = −s F(s)

lim x s f (x) = 0 and lim x s f (x) = 0.

x→∞

x→0

F(s) (xii) M[x n f (n) (x)] = (−1)n (s+n) (s) 2   (xiii) M[x s] = s 2 F(s).  x f (x) + x f (x);F(s+1) (xiv) M[ 0 f (t)dt; s] = − s . x ∞ x Proof M[ 0 f (t)dt; s] = 0 x s−1 { 0 f (t)dt}ds. Integration by parts yields ∞ x M[ 0 f (t)dt; s] = − 1s 0 x s f (x)d x = − F(s+1) . s ∞ . (xv) M[ x f (t)dt; s] = F(s+1) s The proof is similar to property (xiv).

8.5 Mellin Transform of Some Simple Functions Example 8.1 Find the Mellin transform of e−x , x > 0. 

Solution: M[e

−x

∞

]=

x s−1 e−x d x = (s), Re(s) > 0

0

as this is the definition of the gamma function. Remark: The Laplace transform of x s−1 is 

∞

L[x s−1 ; p] =

x s−1 e− px d x.

0

For p = 1, this gives  L[x

s−1

; p = 1] =

∞

x s−1 e−x d x = M[e−x ; s].

0

Example 8.2 Find the Mellin transform of cos x and sin x.

210

8 Mellin Transform

Fig. 8.2 Complex z-plane



Solution:

∞

M[cos x, sin x] =

x s−1 (cos x, sin x)d x.

0

The integrals can be derived by using the complex variable theory. Consider the function f (z) = z s−1 e−z (0 < Re(s) < 1) and integrate around the closed contour as shown in Fig. 8.2. As there is no singularity of the function z s−1 e−z inside the contour in the first quadrant of the complex z plane formed by arcs C , C R and the lines L 1 = ( , R), L 2 = (i R, i ), and then make → 0 and R → ∞, we must have 

R

x s−1 e−x d x +



z s−1 e−z − i s

CR



R

y s−1 e−i y dy +



z s−1 e−z dz = 0

C





Now the integrals C R dz → 0 as R → ∞ and C → 0 as → 0. Making ε → 0, R → ∞, we obtain  ∞  ∞ x s−1 e−x d x = i s y s−1 e−i y dy 0

Thus,



∞

0

y s−1 e−i y dy = i −s (s) = e

−isπ 2

(s)

0

Changing y to x this gives  0

∞

 (cos sx − i sin x)d x =



sπ cos 2





sπ − i sin 2

 (s).

8.5 Mellin Transform of Some Simple Functions

211

Equating the real and imaginary parts, we find 

s−1

 πs (s), cos sxd x = cos 2

s−1

 πs (s), sin sxd x = sin 2



∞

x 0



and



∞

x 0

where 0 < s < 1, Thus,



M[cos x; s] = (s) cos  M[sin x; s] = (s) sin Example 8.3 Find the Mellin transform of  M

Solution:

 πs , 0 < Re(s) < 1, 2

 πs , 0 < Re(s) < 1. 2 1 . 1+x

  ∞ −s 1 x ;s = dx 1+x 1 +x 0

For 0 < Re(s) < 1, this integral can be evaluated using the method of contour integration in complex variable theory. Consider the complex function f (z) =

z s−1 , 0 < Re(s) < 1. 1+z

This has a simple pole at z = 0 and a branch points at z = 0 and z = ∞. Choose a branch cut along the positive real axis from z = 0 to z = ∞ of the complex z−plane. In this cut plane, consider a closed contour as shown in Fig. 8.3, where C R is a circle of radius R, C is a circle of radius , the two circle being punctured along the real axis. There is a singularity of f (z) inside this closed in the form of a pole at z = −1. Thus, 

 +

CR

 +

Cε

R ε

+

 ε

f (z)dz = 2πi Residue of f (z) at z = −1;

R

where the integral from ε to R is just above the real axis and the integral from R to ε is just below the real axis. It can be shown that  f (z)dz → 0 as R → ∞ CR

212

8 Mellin Transform

Fig. 8.3 Complex z-plane



and

f (z)dz → 0 as ε → 0. Cε

Now 

R ε

 f (z)dz =

R ε

x s−1 dx → 1+x



∞ 0

x s−1 d x as R → ∞ and ε → 0, 1+x

and 

ε

 f (z)dz =

R

ε R

(xe2πi )s−1 2πi e d x → −e2πis 1 + xe2πi



∞

0

x s−1 d x as R → ∞ and ε → 0. 1+x

Residue of f (z) at z = −1 is lim (z + 1)

z→−1

z s−1 1+z

= (−1)s−1 = (eπi )s−1 = −eπis . Thus



∞

(1 − e2πis ) 0

x s−1 d x = −2πieπis 1+x

so that 

∞ 0

0 < Re(s) < 1.

x s−1 π 2πi = d x = − −πis = (s)(1 − s) πis 1+x e −e sin(πs)

8.5 Mellin Transform of Some Simple Functions

Example 8.4 Find the Mellin transform of

213

, m > 0.

1 (1+ax)m

1 

1  ∞ Solution: Let M 1+x ; s = F(s) = (s)(1 − s) then M 1+ax ;s = 0  ∞ s−1 = a1s 0 t1+t dt (a > 0) = F(s) = (s)(1−s) , 0 < Re(s) < 1. as as If we differentiate both sides with respect to a, then 

∞

− 0

x 1+(s−1) d d x = (s)(1 − s) (e−s ln a ) (1 + ax)2 da

= (s)(1 − s)e so that



∞

x s−1 dx 1+ax

−s ln a



−s a

 =−

s(s)(1 − s) a s+1

x 1+(s−1) (s + 1)(1 − s) dx = ; −1 < s < 1. 2 (1 + ax) a s+1

0

Another differentiation with respect to a gives 

∞

−2! 0

x 2+(s−1) d  −(s+1) ln a  e d x = (s + 1)(1 − s) 3 (1 + ax) da

= (s + 2)(1 − s)e so that



∞

0

−(s+1) ln a



s+1 − a



x 2+(s+1) (s + 2)(1 − s) dx = ; −2 < Re(s) < 1 (1 + ax)3 2!a s+2

Continuing this (m − 1) times, we find 

∞ 0

x m−1+(s−1) (s + m − 1) (1 − s) dx = , −(m − 1) < Re(s) < 1. m (1 + ax) (m − 1)! a s+m−1

If we put m − 1 + s = s  , then 

∞

0

so that



x s −1 (s  )(m − s  ) d x = , −(m − 1) < −(m − 1) + Re(s  ) < 1 (1 + ax)m (m)a s  

∞ 0

x s−1 (s)(m − s) dx = , 0 < Re(s) < m. (1 + ax)m (m)a s

214

8 Mellin Transform

Exercises 1. Let F(s) denote the Mellin transform of f (x) defined on [0, ∞). Then F(s) is defined by ∞ (B) F(s) = 0 x s−1 f (x) d x, ∞ (D) F(s) = 0 x −s−1 f (x) d x.

∞ (A) F(s) =  0 x s f (x) d x, ∞ (C) F(s) = 0 x −s f (x) d x,

Answer:

B

2. Let F(s) denote the Mellin transform of f (x) defined on [0, ∞). Then f (x) is related to F(s) by (A) f (x) = (C) f (x) =

1 2πi 1 2πi

 c+i∞ s x F(s) ds,  c−i∞ c+i∞ −s−1 x F(s) ds, c−i∞

(B) f (x) = (D) f (x) =

Answer:  3. Let f (x) =

(A) 1s ,

1 2πi 1 2πi

 c+i∞ s−1 x F(s) ds, c−i∞ c+i∞ −s x F(s) ds. c−i∞

D

1, 0 ≤ x ≤ 1 . Then the Mellin transform of f (x) is 0, x > 1 (C) s 2 ,

(B) s,

Answer:

(D)

1 . s2

A

4. Let F(s) denote the Mellin transform of f (x) defined on [0, ∞). Then the Mellin transform of f (2x) is (A) 2s F(s),

(B) 2−s F(s),

Answer:

(C) F(2s),

B

(D) F

s 2 .

Exercises

215

5. Let F(s) denote the Mellin transform of f (x) defined on [0, ∞). Then the Mellin transform of f (x α ) (α > 0) is (A)

1 αF

s α ,

(B) αF(sα),

Answer:

(C) αF

s α ,

(D)

1 α F(sα).

A

6. Let F(s) denote the Mellin transform of f (x) defined on [0, ∞). Then the Mellin transform of x f (x) is (A) F(s − 1),

(B) F(s + 1),

Answer:

(C) F(1 − s),

(D) F(−s).

B

7. Let F(s) denote the Mellin transform of f (x) defined on [0, ∞). Then the Mellin transform of log f (x) is (A) log F(s),

(B) F(log s),

Answer:

(C) F  (s),

(D) −F  (s).

C

8. Let F(s) denote  Mellin transform of f (x) defined on [0, ∞). Then the Mellin  the transform of x1 f x1 is (A) F(−s),

(B) F(1 − s),

Answer:

(C) F(1 + s),

(D) 1s F

1 s .

B

9. Let F(s) denote  x the Mellin transform of f (x) defined on [0, ∞). Then the Mellin transform of 0 f (t) dt is (A) 1s F(s),

(B) − 1s F(s),

(C) 1s F(s + 1),

Answer:

D

(D) − 1s F(s + 1).

216

8 Mellin Transform

10. Let F(s) denote the Mellin transform of f (x) defined on [0, ∞). Then the Mellin transform of f (x) = e−x (x ≥ 0) is (A) (s),

(B) (s + 1),

(C) (s − 1),

Answer:

(D) (−s).

A

11. The Mellin transform of f (x) = e−3x (x ≥ 0) is (A)

(s) 3 ,

(B)

(s) 3s ,

Answer: 12. The Mellin transform of f (x) = (A)

π sin πs ,

(B)

(B)

(s) 3s .

1 sin πs ,

(D)

πs sin πs .

B

s sin πs ,

(C)

13. The Mellin transform of f (x) = s 2s sin πs ,

(D)

(x ≥ 0) is

1 1+x

Answer:

(A)

(s) , s3

(C)

1 1+2x

A

(x ≥ 0) is

π2s sin πs ,

(C)

Answer:

2s sin πs ,

A

14. The Mellin transform of f (x) = sin 2x is Answer:

(s) πs sin . 2s 2

15. The Mellin transform of f (x) = cos 2x is Answer:

(s) πs cos . s 2 2

(D)

1 2s sin πs .

References

217

References The content of this chapter is mainly based on the website epgp.inflibnet.ac.in/ of UGC E Pathsala and on the following literature. 1. 2. 3. 4.

I.N. Sneddon, Use of Integral Transform (Mc Graw Hill, 1972) I.N. Sneddon, Fourier Transform (Mc Graw Hill, 1951) L.C. Andrews, B.K. Shivamoggi, Integral Transform for Engineers (Prentice-Hall, 2007) L. Debnath, D. Bhatta, Integral Transforms and Their Applications, 3rd edn. (Chapman and Hall/CRC, USA, 2014)

Chapter 9

Hankel Transform

For solving boundary value problems formulated in cylindrical coordinate system, Hankel transforms are employed. A formal derivation of Hankel transform is given below.

9.1 Formal Derivation of Hankel Transform The two-dimensional Fourier transform of a function f (x, y) defined for −∞ < x, y < ∞ is F(k, l) =

1 2π



∞ −∞



∞ −∞

f (x, y)ei(kx+ly) d xd y, −∞ < k, l < ∞

and the inverse formula is  ∞ ∞ 1 F(k, l)e−i(kx+ly) dkdl, −∞ < x, y < ∞. f (x, y) = 2π −∞ −∞

(9.1)

(9.2)

The rectangular cartesian co-ordinates (x, y) and (k, l) are changed to polar coordinates (r, θ) and (ρ, φ) respectively, defined by x = r cos θ, y = r sin θ, 0 < r < ∞, 0 < θ < 2π and k = ρ cos φ, l = ρ sin φ, 0 < ρ < ∞, 0 < φ < 2π, then kx + ly = r ρ cos(θ − φ). © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 S. Banerjea and B. N. Mandal, Integral Equations and Integral Transforms, https://doi.org/10.1007/978-981-99-6360-7_9

219

220

9 Hankel Transform

Thus, the Eqs. (9.1) and (9.2) are changed to 1 G(ρ, φ) = 2π 

1 2π

g(r, θ) =





∞ 0

∞

2π

r g(r, θ)eir ρ cos(θ−φ) dθdr

(9.3)

ρG(ρ, φ)e−ir ρ cos(θ−φ) dφdρ

(9.4)

0



0

2π

0

where F(k, l) ≡ G(ρ, φ) and f (x, y) ≡ g(r, θ).

(9.5)

(a) Hankel transform of order zero: It is known that if J0 (x) is the Bessel function of order zero, then J0 (x) has the integral representation  1 π i x cos θ e dθ. (9.6) J0 (x) = π 0 Let ψ = θ − φ and we consider the following integral: 

2π

eir ρ cos(θ−φ) dθ =



−φ

0



φ

=

2π−φ

eir ρ cos ψ dψ =

eir ρ cos ψ dψ +

0



2π



0 −φ

eir ρ cos ψ dψ +



φ

=

e

ir ρ cos ψ

 dψ +

0



2π

=

2π

e

ir ρ cos ψ



0

0





π

π

eir ρ cos ψ dψ

2π−φ

eir ρ cos(2π−ψ) dψ



φ

dψ −

eir ρ cos ψ dψ +

eir ρ cos ψ dψ +

0



eir ρ cos ψ dψ

0

0

=

2π

φ

0

eir ρ cos ψ dψ =

 −

0

0



2π

+



π



2π

eir ρ cos ψ dψ

π

eir ρ cos(2π−ψ) dψ

0

 =2

π

eir ρ cos ψ dψ = 2π J0 (r ρ).

(9.7)

0

Let g(r, θ) be independent of θ, so that g(r, θ) ≡ g(r ). Then the Eq. (9.3) can be written as  ∞  2π 1 r g(r )eir ρ cos(θ−φ) dθdr G(ρ, φ) = 2π 0 0

9.1 Formal Derivation of Hankel Transform





∞

=

221



1 r g(r ) 2π

0

2π

e

ir ρ cos(θ−φ)

 dθ dr.

0

Use of Eq. (9.7) produces 

∞

G(ρ, φ) =

r g(r )J0 (ρr )dr.

(9.8)

0

Thus, Eq. (9.8) shows that G(ρ, φ) is independent of φ if g(r, θ) is independent of θ. Writing G(ρ, φ) ≡ G(ρ), the Eq. (9.4) takes the form 

1 g(r ) = 2π

∞



2π

ρG(ρ)

e

0

−ir ρ cos(θ−φ)

 dφ dρ

0

which is equivalent to 

1 g(r ) = 2π where L(r, ρ) = β = θ − φ to get

 2π

∞

ρG(ρ)L(r, ρ)dρ

(9.9)

0

e−ir ρ cos(θ−φ) dφ. Now, to simplify L(r, ρ), one can substitute

0

 L(r, ρ) =  =

0

θ−2π



θ

2π

+

e−ir ρ cos β dβ

θ−2π

+

 θ

0

e−ir ρ cos β dβ.

2π

Substitution of ψ = β + 2π in the first integral and χ = π − β in the second integral produces  L(r, ρ) =

2π

e

−ir ρ cos ψ

θ

 dψ +



π

L(r, ρ) = 2

π

e −π

ir ρ cos χ

 dχ +

θ

e−ir ρ cos β dβ

2π

eir ρ cos χ dχ = 2π J0 (r ρ).

0

With this simplification, the function g(r ) in Eq. (9.9) becomes 

∞

g(r ) =

ρG(ρ)J0 (ρr )dρ

(9.10)

0

and Eq. (9.8) produce

 G(ρ) = 0

∞

r g(r )J0 (ρr )dr.

(9.11)

222

9 Hankel Transform

The function G(ρ) in relation (9.11) is called the Hankel Transform of order zero, and the function g(r ) in relation (9.11) is called the inverse formula for Hankel transform of order zero. (b) Hankel transform of order n: Let us assume that g(r, θ) in Eq. (9.4) can be expanded as ∞ 

g(r, θ) =

gn (r )einθ

n=−∞

where 1 gn (r ) = 2π



2π

g(r, θ)e−inθ dθ.

(9.12)

0

Then from Eq. (9.3), one gets 1 G(ρ, φ) = 2π





∞

2π

r

e

0

ir ρ cos(θ−φ)

 g(r, θ)dθ dr.

(9.13)

0

Interchanging the order of integration, the Eq. (9.13) can be written as 1 G(ρ, φ) = 2π



2π



∞

re 0

ir ρ cos(θ−φ)

 g(r, θ)dr dθ.

0

Thus, substituting g(r, θ) from Eq. (9.12), one finds 1 G(ρ, φ) = 2π



2π



∞

re 0

ir ρ cos(θ−φ)

0

∞ 

 gm (r )e

imθ



dr dθ

m=−∞

 2π  ∞  ∞  ir ρ cos(θ−φ) imθ  1  = e r gm (r ) e dθ dr. 2π m=−∞ 0 0 The inner integral can be evaluated by substituting θ − φ = α so that 

2π

e

ir ρ cos(θ−φ) imθ

e

 dθ = e

imφ

G(ρ, φ) =

∞ 



m=−∞

where

e

imφ

1 2π

eir ρ cos α eimα dα.

−φ

0

Hence

2π−φ

0

∞

 r gm (r )Im (ρ, r )dr

(9.14)

9.1 Formal Derivation of Hankel Transform

 Im (ρ, r ) =

223

2π−φ

eimα+ir ρ cos α dα.

−φ

Next, we simplify Im (ρ, r ),  Im (ρ, r ) =  = 

φ

=

e

−φ

−imψ+ir ρ cos ψ

−φ



0

0

π

+



 +

0

e



2π

eimα+ir ρ cos α dα

2π−φ

2π π

π

dψ + 2

 −

0

0



2π

+



2π

−



eimα+ir ρ cos α dα

2π−φ

ir ρ cos ψ



0



π

=2

φ

cos mψdψ −

e−imψ+ir ρ cos(2π−ψ) dψ

0

eir ρ cos ψ cos mψ dψ = 2πi m Jm (r ρ)

0

where Jm (x) =



1 πi m

π

ei x cos ψ cos mψ dψ

0

is Bessel function of first kind of order m. Thus, from Eq. (9.14), it is found that G((ρ, φ) has the expansion ∞ 

G(ρ, φ) =

i m eimφ G m (ρ)

(9.15)

r gm (r )Jm (r ρ)dr.

(9.16)

m=−∞



where

∞

G m (ρ) = 0

The Eq. (9.16) produces the Hankel Transform G m (ρ) of order m of the function gm (r ). To find the inverse Hankel transform, it is observed from Eq. (9.4) that g(r, θ) =

1 2π

 0

∞



2π

ρG(ρ, φ)eir ρ cos(θ−φ) dφdρ.

0

It is already seen from Eq. (9.12) that if g(r, θ) has the expansion g(r, θ) =

∞  n=−∞

where

gn (r )einθ ,

224

9 Hankel Transform

1 gn (r ) = 2π



2π

g(r, θ)e−inθ dθ,

0

then from Eq. (9.15), G(ρ, φ) has the expansion ∞ 

G(ρ, φ) =

i m eimφ G m (ρ)

(9.17)

m=−∞

so that the coefficient i m G m (ρ) is given by 

1 i G m (ρ) = 2π

2π

m

e−imφ G(ρ, φ)dφ.

0

From Eq. (9.3), one gets 

1 2π

g(r, θ) =

∞



0

2π

ρG(ρ, φ)e−ir ρ cos(θ−φ) dφdρ.

0

Substitution of the expansion for G(ρ, φ) from Eq. (9.17) produces 1 g(r, θ) = 2π



∞



0

2π

ρe

−ir ρ cos(θ−φ)

  ∞

0

m imφ

i e

 G m (ρ) dφdρ.

m=−∞

Rewriting this as g(r, θ) =

∞   m=−∞ 0

∞



1 ρi G m (ρ) 2π



m

2π

e

−ir ρ cos(θ−φ)+imφ

0

one obtains =

∞  m=−∞

 im

∞

ρG m (ρ)Fm (r, ρ, θ))dρ

0

where 1 Fm (r, ρ, θ) = 2π



2π

e−ir ρ cos(θ−φ)+imφ dφ.

0

To simplify Fm (r, ρ, θ), one can use β = θ − φ to obtain 1 Fm (r, ρ, θ) = 2π



θ

θ−2π

e−ir ρ cos β+im(θ−β) dβ

 dφ dρ,

9.1 Formal Derivation of Hankel Transform



1 imθ e = 2π

θ

1 imθ e P(r, ρ). 2π

Now,

 P(r, ρ) = 

θ−2π

θ

e−ir ρ cos β−imβ dβ

θ−2π



0

e−ir ρ cos β−imβ dβ

θ−2π

=

=

225

2π

+

 θ

+

0

e−ir ρ cos β−imβ dβ.

2π

Substitute ψ = β + 2π in the first integral to obtain  2π  2π  θ  P(r, ρ) = e−ir ρ cos ψ−imψ dψ + + θ

0

 =

2π

2π

e−ir ρ cos ψ−imψ dψ

0

 =

π

−π

eir ρ cos χ−im(π−χ) dχ, (χ = π − ψ) 

= (−1)m  = (−1)

π

eir ρ cos χ+imχ dχ

−π

0

m −π

 π

+

eir ρ cos χ+imχ dχ

0



π

= 2(−1)m

eir ρ cos χ cos mχdχ

0

= 2π(−1)m i m Jm (r ρ). Thus Fm (r, ρ, θ) = so that g(r, θ) =

1 imθ e P(r, ρ) = (−1)m i m eimθ Jm (r ρ) 2π ∞ 

m=−∞



∞

im 0

ρG m (ρ)Fm (r, ρ, θ)dρ

226

9 Hankel Transform

=

∞ 



∞

(−1)m i 2m eimθ

ρG m (ρ)Jm (r ρ)dρ

0

m=−∞ ∞ 

=

eimθ gm (r )

m=−∞



where

∞

gm (r ) =

ρG m (ρ)Jm (r ρ)dρ.

(9.18)

0

Thus, the Eqs. (9.16) and (9.18) produce the Hankel transform of order n given by



∞

Hn [g(r )] ≡ G(ρ) =

r g(r )Jn (r ρ)dr,

0

Hn−1 [G(ρ)] = g(r ) =



∞

ρG(ρ)Jn (r ρ)dρ.

0

Notes: 1. The general Hankel transform of order ν(ν > − 21 ) is defined by 

∞

Hν [ f (r )] ≡ F(ρ) =

r f (r )Jν (ρr )dr

(9.19)

ρF(ρ)Jν (ρr )dρ.

(9.20)

0

and the corresponding inversion formula is Hν−1 [F(ρ)] ≡ f (r ) =



∞

0

√ 2. For the existence of the Hankel transform defined by (9.19), r f (r ) is piecewise continuous and absolutely integrable on the positive real line. 3. The proof of the inversion formula (9.20) is similar to the corresponding proof of the Fourier inversion formula. But it is somewhat complicated as the kernel involves the Bessel functions which are more complicated than those of the kernels of Fourier transform. The inversion formula is not proved here. The proof can be found in the book of I. N. Sneddon: The Use of Integral Transforms.

9.2 Properties of Hankel Transform (i) Linearity property Hν [c1 f (r ) + c2 g(r )] = c1 F(ρ) + c2 G(ρ),

9.2 Properties of Hankel Transform

227

where c1 , c2 are constants and F(ρ) and G(ρ) are the Hankel transform of f (r ) and g(r ), respectively. Similarly Hν−1 [c1 F(ρ) + c2 G(ρ)] = [c1 f (r ) + c2 g(r )]. The proof is obvious. (ii) Scaling property Let Hν [ f (r )] = F(ρ), then   1 ρ , a > 0. F a2 a

Hν [ f (ar )] = Proof



∞

Hν [ f (ar )] =

r Jν (r ρ) f (ar )dr

0

=



1 a2



∞

r Jν 0

rρ a

 f (r )dr

  1 ρ , a > 0. F a2 a

= (iii) Parseval’s Theorem

Let Fν (ρ) ≡ Hν [ f (r )] and G ν (ρ) ≡ Hν [g(r )]. Then  ∞

ρFν (ρ)G ν (ρ)dρ

0



∞

=

r f (r )g(r )dr.

0

Proof



∞

ρF(ρ)G(ρ)dρ

0

 =

∞



0

 =

∞

ρF(ρ)

 r g(r )Jν (ρr )dr dρ

0 ∞

 r g(r )

0

∞

 ρJν (ρr )F(ρ)dρ dr

0



∞

= 0

r f (r )g(r )dr.

228

9 Hankel Transform

Note: 1. There is no shifting property of Hankel transform. 2. There is no convolution theorem for Hankel transform. These are due to nonexistence of a simple addition formula for the Bessel function. (iv) Division by r Let Fν (ρ) ≡ Hν [ f (r )], then

Hν

 1 ρ {Fν−1 (ρ) + Fν+1 (ρ)} . f (r ) = r 2ν

Proof This follows from the property 2ν Jν (z) = Jν−1 (z) + Jν+1 (z). z

We have Hν =

1 2ν

1 = 2ν =

ρ 2ν



∞



  ∞ 1 f (r ) f (r ) = dr r Jν (ρr ) r r 0 ∞

0



∞ 0

   Jν (t) t t 2ν f dt, (t = ρr ) ρ t ρ t (Jν−1 (t) + Jν+1 (t)) f ρ

  t dt, ρ

r {Jν−1 (ρr ) + Jν+1 (r ρ)} f (r )dr, (t = ρr )

0

ρ {Fν−1 (ρ) + Fν+1 (ρ)} . 2ν  f (r )) = −ρFν−1 (ρ). =

  (v) Hν r ν−1 drd (r 1−ν

Proof This follows from the property d ν (z Jν (z)) = z ν Jν−1 (z). dz We have

  ∞ d d r ν−1 (r 1−ν f (r ))r Jν (ρr )dr Hν r ν−1 {r 1−ν f (r )} = dr dr 0 =

[r f (r )Jν (r ρ)]∞ 0

 − 0

∞

r 1−ν f (r )

d ν [r Jν (ρr )]dr. dr

9.3 Hankel Transform of Some Known Functions

229

Assuming that f (r ) is such that the first term on left side is zero, 

∞

=−

r 1−ν f (r )r ν Jν−1 (ρr )ρdr

0



∞

= −ρ

r f (r )Jν−1 (ρr )dr = −ρFν−1 (ρ).

0

  (vi) Hν r −ν−1 drd {r 1+ν f (r )} = ρFν+1 (ρ) Proof This follows from the property d −ν (z Jν (z)) = −z −ν Jν+1 (z). dz

  ∞ d d Hν r −ν−1 {r 1+ν f (r )} = r −ν−1 {r 1+ν f (r )}r Jν (ρr )dr dr dr 0 =

[r f (r )Jν (ρr )]∞ 0



∞

− 0

r ν+1 f (r )

d −ν r Jν (ρr )dr. dr

Assuming that f (r ) is such that the first term on left side is zero, 

∞

=0−

  r ν+1 f (r ) −r −ν Jν+1 (ρr ) ρdr

0



∞

=ρ

r f (r )Jν+1 (νr )dr = ρFν+1 (ρ).

0

9.3 Hankel Transform of Some Known Functions Example 9.1 Find the Hankel transform of order zero for the function f (r ) = e−ar , a > 0. r

Solution: H0

  ∞ −ar e−ar e ;ρ = r J0 (ρr )dr r r 0  = 0

∞

J0 (ρr )e−ar dr

230

9 Hankel Transform

 2n =

∞   n=0

∞

(−1)

ρr 2

n

e−ar dr

(n!)2

0

 2n =

∞ (−1) 

∞  n=0

=



∞

e−ar r 2n dr

0

(−1)n ρ2n 22n (n!)2 a 2n+1



∞

e−t t 2n dt

0

∞  (−1)n ρ2n (2n + 1)

22n (n!)2 a 2n+1

n=0

=

ρ 2

(n!)2

n=0

=

n

  ∞ 1  (−1)n (2n)! ρ2 n . a n=0 22n (n!)2 a2

However, (1 + x)

− 21

=

∞  n=0



− n

1 n 2 x

for |x| < 1, where  1 (− 21 )(− 21 − 1)(− 21 − 2) . . . (− 21 − (n − 1)) − = 2 n! n   (−1)n =

H0

1 2

   + 1)( 21 + 2 . . . 21 + (n − 1) n!

= Thus,

1 2

(−1)n (2n)! (−1)n 1.3.5 . . . (2n − 1) = . n 2 n! 22n (n!)2

 e−ar 1 ρ2 1 1 ; ρ = (1 + 2 )− 2 =  , ρ < a. 2 r a a ρ + a2

If we make a → +0, then

  ∞  ∞ 1 1 1 H0 ; ρ = J0 (r ρ)dr = , ρ > 0. r J0 (r ρ)dr = r r ρ 0 0

9.3 Hankel Transform of Some Known Functions

231

Example 9.2 Find the zero-order Hankel transform of f (r ) = e−ar , a > 0. 2



Solution:

∞

F0 (ρ) ≡

e−ar r J0 (ρr )dr 2

0

The integral can be evaluated by using the series expansion of J0 (ρr ) given by  2n J0 (ρr ) =

∞ 

F0 (ρ) =

 ∞  (−1)n ρ2n n=0

22n (n!)2

∞

= But



1  (−1)n ρ2n 2 n=0 22n (n!)2

∞



∞

r e−ar r 2n dr 2

e−at t n dt.

0



a n+1

0

=

∞ 0

1

e−at t n dt =

.

(n!)2

n=0

Thus,

ρr 2

(−1)n

∞

e−s s n ds

0

(n + 1) n! = n+1 . n+1 a a

Thus,

∞

F0 (ρ) =

1  (−1)n ρ2n 2 n=0 22n (n!)a n+1  n

=

1 2a

∞ 

(−1)n n!

n=0

ρ2 4a

=

1 − ρ2 e 4a . 2a

Note: If we put 2a = 1, then r2

ρ2

H0 [e− 2 ; ρ] = e− 2 . r2

Thus, the function [e− 2 ; ρ] is self-reciprocal under the Hankel transform of order zero.

232

9 Hankel Transform

Example 9.3 Find the Hankel transform of order zero of the function e−ar , (a > 0). Solution: H0 [e−ar ; ρ] =



∞

e−ar r J0 (ρr )dr

0

=−

=− d =− da





d da

∞

e−ar J0 (ρr )dr

0

−ar  e d H0 ;ρ da r



1

using the result of Example 9.1 above

1

(ρ2 + a 2 ) 2

=

a (ρ2

3

+ a2) 2

.

Example 9.4 Find the Hankel transform of order ν of the function  f (r ) =

rν, r ≤ a 0, r > a 

Solution: Here Hν [ f (r ); ρ] =

a

r ν r Jν (ρr )dr

0

= =

1 ρν+2



ρa 0



1 ρν+2

ρa

x ν+1 Jν (x)d x, (r ρ = x)

0

 d  ν+1 d x Jν+1 (x) d x using the result (x ν Jν (x)) = x ν Jν−1 (x). dx dx

Thus Hν [ f (r ); ρ] =

1

(ρa)ν+1 Jν+1 (ρa) = ρν+2

a ν+1 Jν+1 (ρa). ρ

Example 9.5 Use the Parseval relation of Hankel transform to evaluate the integral  0

∞

1 Jν+1 (ar )Jν+1 (br ) dr for ν > − , 0 < a < b r 2

Solution: Let F(ρ) = Jν+1ρ(aρ) , G(ρ) = We note that, by Example 9.4

Jν+1 (bρ) . ρ

9.3 Hankel Transform of Some Known Functions

233

F(ρ) ≡ Hν [ f (r ); ρ] 

where f (r ) =

rν , a ν+1

r 0 ∂r 2 r ∂r ∂z 2 with boundary condition φ(r, 0) = f (r ), r > 0 f (r ) being a known prescribed function, and φ(r, z),

∂φ 1 −→ 0 as (r 2 + z 2 ) 2 −→ ∞, z > 0 ∂r

Solution: Use of zero-order Hankel transform defined by 

∞

(ρ; z) ≡ H0 [φ(r, z); ρ] = 0

φ(r, z)r J0 (ρr )dr

234

9 Hankel Transform

in the PDE, 

∞

0

∂2φ r 2 J0 (ρr )dr + ∂r 

∞

0



∞

0

∂φ d2 J0 (ρr )dr + 2 ∂r dz

∂2φ r 2 J0 (ρr )dr + ∂r



∞



∞

r φJ0 (ρr )dr = 0

0

∂φ d 2 =0 J0 (ρr )dr + ∂r dz 2

0

Integration by parts of the 1st integral gives

∂φ r J0 (ρr ) ∂r

∞



∞

− 0

0

∂φ d (r J0 (ρr ))dr + ∂r dr



∞

∂φ d 2 J0 (ρr )dr + =0 ∂r dz 2

0

and using the boundary conditions at infinity, we get 

∞

− 0

Performing 

∞

− 0

d dr

∂φ d (r J0 (ρr ))dr + ∂r dr

 0

∞

∂φ d 2 J0 (ρr )dr + =0 ∂r dz 2

in the first term, we obtain

∂φ J0 (ρr )dr − ∂r

so that



r 0



∞

−

r 0

We know

For ν = 0, this gives

∞

∂φ d (J0 (ρr ))dr + ∂r dr



∞

0

∂φ d 2 J0 (ρr )dr + =0 ∂r dz 2

∂φ d d 2 (J0 (ρr ))dr + =0 ∂r dr dz 2

d −ν (t Jν (t)) = −t −ν Jν+1 (t). dt d (J0 (t)) = −J1 (t) dt

so that taking, ρr = t, we can write d (J0 (ρr )) = −ρJ1 (ρr ) dr Therefore,

d 2 +ρ dz 2



∞

r 0

∂φ J1 (ρr )dr = 0 ∂r

Then integrating by parts and using the boundary conditions, we get

9.3 Hankel Transform of Some Known Functions

d 2 −ρ dz 2 Using the fact,

d ν (t Jν (t)) dt



∞

φ

0

235

d (r J1 (ρr ))dr = 0 dr

= t ν Jν−1 (t), we have, for ν = 1, d (t J1 (t)) = t J0 (t) dt

so that for, ρr = t, we can write d (r J1 (ρr )) = ρr J0 (ρr ). dr Using this relation, we obtain d 2 − ρ2 dz 2



∞

r φ(J0 (ρr ))dr = 0

0

d 2 − ρ2  = 0, z > 0. dz 2

(9.21)

Taking Hankel transform of the boundary conditions, we get (ρ, 0) = F(ρ) 

where F(ρ) =

∞

(9.22)

f (r )r J0 (ρr )dr

0

and (ρ, z) −→ 0 as z −→ ∞.

(9.23)

The solution of the ODE (9.21) is (ρ; z) = Aeρz + Be−ρz . The boundary condition (9.23) gives A = 0 and the boundary condition (9.22) gives B = F(ρ) so that (ρ; z) = F(ρ)e−ρz . Taking inverse Hankel transform, we get 

∞

φ(r, z) = 0

ρF(ρ)e−ρz J0 (ρr )dρ.

236

9 Hankel Transform

Exercises 1. The Hankel transform of order ν denoted by Hν [ f (r ); ρ] ≡ F(ρ) = (ρr ) dr of the function f (r ) defined on [0, ∞) exists for

(A) ν < − 21 ,

(B) ν > − 21 ,

Answer:

(C) ν < −1,

∞ 0

r f (r )Jν

(D) ν < − 23 .

B

2. Let Hν [ f (r ); ρ] ≡ F(ρ) be the Hankel transform of f (r ) defined on [0, ∞). Then Hν [ f (ar ); ρ] (a > 0) is equal to

(A)

1 a F(aρ),

(B)

1 a2

F(aρ),

Answer:

(C)

1 a2

F

ρ a

,

(D)

1 a2

F(a 2 ρ).

C

3. Let  H  ν [ f (r); ρ] ≡ F(ρ) be the Hankel transform of f (r ) defined on [0, ∞). Then Hν f r2 ; ρ is (A) 21 F

ρ 2

,

(B) 2F(2ρ),

Answer:

(C) 4F(2ρ),

(D) 4F(4ρ).

C

4. Let Hν [ f (r ); ρ] ≡ F(ρ) be the Hankel transform of f (r ) defined on [0, ∞). Let g(r ) = f (r + a). Then Hν [g(r ); ρ] is

(A) F(ρ + a),

(B) F(ρ − a),

(C) F(ρa),

Answer:

(D) None A, B, C.

D

5. Let Hν [ f (r ); ρ] ≡ Fν (ρ) be the Hankel transform of f (r ) defined on [0, ∞). Then H1 [ f  (r )); ρ] is

Exercises

237

(A) −ρF0 (ρ),

(C) ρ1 F0 (ρ),

(B) ρF0 (ρ),

Answer:

(D) − ρ1 F0 (ρ).

A

6. Let Hν [ f (r ); ρ] ≡ Fν (ρ) be the Hankel transform of f (r ) of order ν defined on [0, ∞). Then Hν [ r1 f (r ); ρ] is

(A) (C)

 ρ  , 2ν Fν−1 (ρ) + Fν+1 (ρ)  ρ 2 Fν−1 (ρ) + Fν+1 (ρ) ,

 ρ  ν Fν−1 (ρ) + Fν+1 (ρ) , 1 2ν Fν−1 (ρ) + Fν+1 (ρ) .

(B) (D)

Answer:

A

7. The zero-order Hankel transform F0 (ρ) of f (r ) = e−r is 2

(A) 21 e−

ρ2 2

(B) 21 e−

,

ρ2 4

(C) e−

,

Answer:

ρ2 2

(D) e−ρ . 2

,

B r2

8. The zero-order Hankel transform F0 (ρ) of f (r ) = e− 2 is

(A) e−

ρ2 2

,

(B) 21 e−

ρ2 2

(C) 2e−

,

Answer:

ρ2 2

,

(D)

√

2e−

ρ2 2

A

9. Use the zero-order Hankel transform to solve the partial differential equation ∂ 2 φ 1 ∂φ ∂ 2 φ + 2 = f (r ), 0 < r < ∞, z > 0 + ∂r 2 r ∂r ∂z given that φ(r, 0) = 0 1

φ(r, z) → 0 as (r 2 + z 2 ) 2 → ∞. 

∞

1 − e−ρz J0 (ρr ) dρ ρ 0 where F(ρ) is the zero-order Hankel transform of f (r ).

Answer: φ(r, z) =

F(ρ)

.

238

9 Hankel Transform

References The content of this chapter is mainly based on the website epgp.inflibnet.ac.in/ of UGC E Pathsala and on the following literature. 1. 2. 3. 4.

I.N. Sneddon, Use of Integral Transform (Mc Graw Hill, 1972) I.N. Sneddon, Fourier Transform (Mc Graw Hill, 1951) L.C. Andrews, B.K. Shivamoggi, Integral Transform for Engineers (Prentice-Hall, 2007) L. Debnath, D. Bhatta, Integral Transforms and Their Applications, 3rd edn. (Chapman and Hall/CRC, USA, 2014)

Chapter 10

Z Transform

10.1 Introduction The Z-transform is simply a power series with coefficients formed by a discrete sequence. It is a powerful technique for solving difference equations. The essential features of Z-transform date back to the early eighteenth century (1730) when De Moivre introduced the concept of generating function which is identical with that of the Z-transform. Applications of Z-transform are relatively new. It is a transform technique used for discrete time signals and systems. Definition The Z-transform of a sequence { f (n)}∞ −∞ is defined by Z[ f (n)] ≡ F(z) =

∞ 

f (n)z −n

(10.1)

n=−∞

for all complex number z for which the series converges. Remark: 1. The series in Eq. 10.1 converges at least in a ring of the form 0 ≤ r1 < |z| < r2 ≤ ∞, where the radii r1 , r2 depend on the behavior of f (n) at n = ±∞. In fact r1 = lim sup{| f (n)|}1/n , r2 = lim inf n→∞

n→∞

1 . {| f (n)|}1/n

2. The function F(z) is analytic in this ring, but it may be possible to continue it analytically beyond the boundaries of this ring. 3. If f (n) = 0 for n < 0, then r2 = ∞, and if f (n) = 0 for n ≥ 0, then r1 = 0. 4. Let z = r eiθ . Then the Z-transform evaluated at r = 1 is ∞ 

f (n)e−inθ .

−∞

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 S. Banerjea and B. N. Mandal, Integral Equations and Integral Transforms, https://doi.org/10.1007/978-981-99-6360-7_10

239

10 Z Transform

240

This is the Fourier series of the sequence { f (n)}∞ −∞ . Example 10.1 Let a be a complex number and let  f (n) =

Then Z[ f (n)] = z , |z| z−a

∞ 

a n z −n =

a n , for n ≥ 0; 0, otherwise.

∞  (a/z)n =

n=0

0

a 1 , | | < 1, so that here F(z) = 1 − a/z z

> |a|. Here r1 = |a| and r2 = ∞.

Example 10.2 Let a be a complex number and let  f (n) =

na n , for n ≥ 0; 0, otherwise,

then F(z) =

∞ 

n(a/z)n =

0

∞ ∞ a a n(a/z)n−1 = n(a/z)n−1 z 1 z 1   ∞ a d  n x= =x x dx 0 z     d 1 a =x x = , |x| < 1 dx 1 − x z   a x x = , |x| < 1 = (1 − x)2 z az = (a < |z|) . (z − a)2

Here also, r1 = |a| and r2 = ∞. Example 10.3 Let

 f (n) =

1, for n ≥ 0; 0, otherwise,

then Z[ f (z)] ≡ F(z) =

∞  0

Here r1 = 1 and r2 = ∞.

z −n =

z 1 = (|z| > |1|) . 1 − 1/z z−1

10.2 Properties of Z -Transform

241

Example 10.4 Let

 f (n) ≡ δ(n) =

1, for n = 0; 0, otherwise.

Then Z[δ(n)] = 1, Z[δ(n − k)] = z −k , k = 0, ±1, ±2, . . . . Here r1 = 1 and r2 = ∞. Example 10.5 Let

 f (n) =

 Then Z

1 , n!

for n ≥ 0; 0, otherwise.

  ∞ 1 z −n = = e1/z for allz. Here r1 = 0 and r2 = ∞. n! n! 0

10.2 Properties of Z-Transform (i) Linearity property: Let Z[ f (n)] = F(z) and the region of convergence of the Z-transform of the sequence { f (n)} be denoted by D f and similarly Z[g(n)] = G(z) and the region of convergence of the Z-transform of the sequence {g(n)} be denoted by Dg . Then Z[c1 f (n) + c2 g(n)] = c1 F(z) + c2 G(z), z ∈ D f ∩ Dg ,

(10.2)

where c1 and c2 are constants. The region D f ∩ Dg contains the ring r1 < |z| < r2 where r1 = max{r1 ( f ), r1 (g)} and r2 = min{r2 ( f ), r2 (g)}. Proof The proof is obvious. (ii) Shifting property: Z[ f (n − k)] = z −k F(z)

10 Z Transform

242

Proof Z[ f (n − k)] = =

∞ 

f (n − k)z −n

n=−∞ ∞ 

f (m)z −(m+k) (n − k = m)

m=−∞ ∞ 

= z −k

f (m)z −m

m=−∞

=z

−k

F(z).

  n (iii) Z[a f (n)] = F az , ar1 < |z| < ar2 , a > 0 Proof Z[a n f (n)] =

∞ 

∞ 

a n f (n)z −n =

n=−∞

f (n)

n=−∞

 z −n a

=F

z a

.

(iv) Z[e−an f (n)] = F(ea z) Proof Z[e−an f (n)] =

∞ 

e−an f (n)z −n =

n=−∞

∞ 

f (n)(ea z)−n = F(ea z).

n=−∞

(v) Z[n f (n)] = −z d F(z) dz Proof Z[n f (n)] =

∞ 

n f (n)z −n

n=−∞ ∞ 

=z

f (n)nz −n−1

n=−∞

= −z

∞ d  f (n)z −n dz n=−∞

= −z

d F(z) . dz

10.2 Properties of Z -Transform

243

(vi) Initial and final value theorems: (a) If f (n) = 0 for n < 0, then lim|z|→∞ F(z) = f (0). Proof Since f (n) = 0 for n < 0, F(z) = lim F(z) = f (0) + lim

|z|→∞

|z|→∞

∞ 

∞ 

f (n)z −n so that

n=0

f (n)z

−n

= f (0) + 0 = f (0).

n=1

(b) If F(z) is defined for r1 < |z| and for some integer m, lim z m F(z) = A, then |z|→∞

f (m) = A and f (n) = 0 for n < m. Proof We have z m F(z) =

∞ 

f (n)z m−n .

n=−∞

Making |z| → ∞, we find A = lim

|z|→∞

∞  n=−∞



m−1 

= lim

|z|→∞

f (n)z m−n f (n)z

(m−n)

+ f (m) +

n=−∞

= f (m) + lim

|z|→∞

∞ 

 f (n)z

−(n−m)

n=m+1 m−1 

f (n)z m−n .

n=−∞

−(n−m) As n > m so the series ∞ → 0 as |z| → ∞. As m > n, the n=m+1 f (n)z m−1 (m−n) is divergent as |z| → ∞ unless f (n) = 0 for n < m. Thus, series n=−∞ f (n)z f (m) = A and f (n) = 0 for n < m. (vii) Z[ f (−n)] = F

1 z

Proof F

   −n ∞  1 1 = f (n) z z n=−∞   −∞  1 m = f (−m) z m=∞

10 Z Transform

244

=

∞ 

f (−n) (z)−n

n=−∞

 =



∞ 

f (−n)z −n

n=−∞

= Z [ f (−n)].

Convolution of Two Sequences The convolution of two sequences { f 1 (n)} and { f 2 (n)} is defined as a sequence { f (n)}, where ∞ 

f (n) ≡ f 1 (n) ∗ f 2 (n) =

f 1 (k) f 2 (n − k).

k=−∞

(viii) Convolution theorem for Z transform The Z transform of the convolution of two discrete sequences is equal to the product of their Z transforms. Proof Let { f 1 (n)} and { f 2 (n)} be two sequences whose Z transforms are given by Z [ f i (n)] = Fi (z) =

∞ 

f i (n)z −n , i = 1, 2.

n=−∞

The Z transform of convolution of { f 1 (n)} and { f 2 (n)} is given by F(z) = Z [ f (n)] = Z [ f 1 (n) ∗ f 2 (n)] =

∞ 

{ f 1 (n) ∗ f 2 (n)}z −n

n=−∞

F(z) =

∞ 

{

∞ 

f 1 (k) f 2 (n − k)}z −n =

n=−∞ k=−∞

∞  k=−∞

f 1 (k)z −k {

∞ 

f 2 (n − k)z −(n−k) }

n=−∞

Substituting m = n − k we have F(z) =

∞  k=−∞

f 1 (k)z −k {

∞  m=−∞

f 2 (m)z −m } = F1 (z) ∗ F2 (z).

10.3 Inverse Z -Transform

245

10.3 Inverse Z-Transform −n To find the sequence { f (n)} from its Z-transform F(z) = ∞ n=−∞ f (n)z , one has to find the inversion formula. How to find the inversion formula? To answer this question, we consider two sequences { f (n)} and {g(n)} defined by  f (n) = 

and g(n) = Then F(z) =

∞ 

f (n)z −n =

n=−∞

1, n ≥ 0 0, n < 0

0, n ≥ 0 −1, n < 0

∞ 

z −n =

n=0

z for |z| > 1, z−1

and G(z) =

∞ 

g(n)z −n = −

n=−∞

−1 

z −n = −

n=−∞

∞  n=1

zn = −

z z = for |z| < 1. 1−z z−1

z This example shows that the inverse Z-transform of the function F(z) = z−1 is not unique. In general, the inverse Z-transform is not unique unless its region of convergence is specified. There are several methods by which the inverse Z-transform can be evaluated ∞  (to find { f (n)} from F(z)(= f (n)z −n )). However, here we use the following n=−∞

method to find inverse Z-transform. A Inversion by using power series: If F(z) is given by its series F(z) =

∞ 

an z −n , r1 < |z| < r2 ,

n=−∞

then the inverse Z-transform is unique and is given by { f (n) = an }.

10 Z Transform

246

Example 10.6 Find the inverse Z-transform of F(z) =

z+3 , |z| > 2. z−2

Here z 3 + z−2 z−2     2 −1 3 2 −1 = 1− 1− + z z z ∞  n ∞  m   2 2 3 = + , |z| > 2 z z z n=0 m=0

F(z) =

= 1+

∞ 

2n (z)−n +

n=1

= 1+ = 1+

∞ 

2 (z) n

n=1 ∞ 

5 2

= 1+5

−n

∞ 3  m+1 −(m+1) 2 z , |z| > 2 2 m=0 ∞

3  n −n + 2 z , |z| > 2 (n = m + 1) 2 n=1

2n z −n , |z| > 2

n=1 ∞ 

2n−1 z −n .

n=1

Thus f (0) = 1, n = 0, f (n) = 5 × 2n−1 , n ≥ 1, f (n) = 0, otherwise.

B Inversion by complex integration: If F(z) is regular in the region |z| > r , then its inverse Z-transform can be obtained by using  1 F(z)z n−1 dz, f (n) = 2πi  where  is a closed contour surrounding the origin in the domain of analyticity of F(z) in the anticlockwise sense.

10.3 Inverse Z -Transform

247

Example 10.7 Find the inverse Z-transform of F(z) = plex integration.

z+3 , |z| z−2

> 2 by using com-

Solution: Since r2 = ∞, we must have f (n) = 0 for n < 0. Now  z + 3 −1 1 z dz f (0) = 2πi  z − 2 = Residue at z = 0 + Residue at z = 2 3 5 = − + = 1, 2 2  1 z + 3 −1 f (1) = z dz 2πi  z − 2 = Residue at z = 2 =5 and for n > 1  1 z + 3 n−1 z dz 2πi  z − 2 = Residue at z = 2 = 2n−1 × 5.

f (n) =

C Inversion by using Fourier series: If the domain of analyticity of F(z) contains the unit circle |z| = 1 and if F(z) is single-valued on |z| = 1, then G(θ) ≡ F(eiθ ) is a periodic function with period 2π, and consequently it can be expanded in a Fourier series. The coefficients of this series are the inverse Z-transform and they are given explicitly by 1 f (n) = 2π



π

−π

F(eiθ )einθ dθ.

Example 10.8 Find the inverse Z-transform of F(z) =

1 1 , |z| < . 2z − 1 2

(10.3)

10 Z Transform

248

Solution: Noting equation (10.3), we have  π 1 1 einθ dθ 2π −π 2eiθ − 1    π 1 1 1 = 1 − einθ dθ 2π −π 2eiθ 2eiθ  π ∞ 1  1 = ei(n−m−1)θ dθ 2π m=0 2m+1 −π

f (n) =

Let n = −l(l > 0), then f (−l) = =

 π ∞ 1  1 e−i(l+m+1)θ dθ 2π m=0 2m+1 −π ∞ 1  1 e−i(l+m+1)π − ei(l+m+1)π 2π m=0 2m+1 −i(l + m + 1)

= 0. Let n = 0, then f (0) = =

 π ∞ 1  1 e−i(m+1)θ dθ 2π m=0 2m+1 −π ∞ 1  1 e−iπ(m+1) − eiπ(m+1) 2π m=0 2m+1 −i(m + 1)

= 0. Let n > 0, then 1 1 f (n) = 2π 2n = 2−n .



π

−π

dθ (for m = n − 1)

For m = n − 1, the integral vanishes. Thus  Z −1

 1 = f (n), 2z − 1 

where f (n) =

1, n ≤ 0 2−n , n > 0.

This result can also be obtained by using power series expansion.

10.3 Inverse Z -Transform

249

D Inversion by using partial fraction expansion: Let F(z) be a rational function of the form N 

F(z) =

P(z) k=0 = M Q(z) 

ak z k . bk z

k

k=0

To find the inverse transform, we consider the following two cases: Case (i) Let N < M. The denominator Q(z) can be factored. Let Q(z) = c(z − z 1 )k1 (z − z 2 )k2 . . . (z − z m )km , where c is a constant and k1 , k2 , . . . , km are positive m  integers satisfying k j = M. Hence, F(z) can be decomposed to partial fraction j=1

as

F(z) =

ki m   i=1 j=1

where Ai, j =

Ai, j (z − z i ) j

 d ki − j  1 lim k − j (z − z j )ki F(z) . (k j − j)! z→zi dz i

(10.4)

The inverse Z-transform of the fractional decomposition (10.4) in the region that is exterior to the smallest circle containing all the zeros of Q(z) can now be obtained by appropriate power series expansion of each term of the form (z − a)−k . Case (ii) Let N ≥ M. Then F(z) can be written as F(z) = H (z) +

R(z) , Q(z)

where R(z) has degree ≤ M − 1, and H (z) is a polynomial of degree at most N − M. The inverse Z-transform of H (z) (a polynomial of degree at most N − M) can be obtained from  1, n = −k Z −1 [z k ] = δ(n + k) = 2−n , otherwise where k = 0, ±1, ±2, . . ., and that of

P(z) Q(z)

can be obtained as in case (i).

10 Z Transform

250

Example 10.9 Find the inverse Z-transform of F(z) =

z4 + 5 , |z| > 2. (z − 1)2 (z − 2)

Solution: We can write 1 16 5 + − (z − 1)2 z−1 z−2       1 −2 5 1 −1 16 2 −1 1 1− 1− − + z+4− 2 1− z z z z z z  ∞  ∞ ∞  n    2 1 16 −2 −n 5 z − z+4− 2 z −n + , |z| > 2 z n=0 n z n=0 z n=0 z  ∞  ∞ ∞   16  n −n−1 −2 −n−2 z+4− −5 z −n−1 + 2 z z n z n=0 n=0 n=0  ∞  ∞ ∞    −2 z+4− z −n + 16 2n−1 z −n z −n − 5 n−2

F(z) = z + 4 − = = = =

n=2

n=0

n=1

Thus f (−1) = 1, f (0) = 4, f (1) = −5 + 16 = 11,   −2 f (n) = − − 5 + (16)2n−1 , n ≥ 2. n−2 E Inversion by using Convolution Theorem: Example 10.10 Using Convolution theorem, find the inverse Z-transform of F(z) =

z2 , |a| < |b| < |z|. (z − a)(z − b)

Solution: We know that for a complex number a, the Z transform of  f 1 (n) =

a n , for n ≥ 0; 0, otherwise.

Exercises

251

is F1 (z) =

z , z−a

|z| > |a|. Also, for a complex number b, the Z transform of  f 2 (n) =

is F2 (z) =

z , z−b

bn , for n ≥ 0; 0, otherwise.

|z| > |b|. Now z2 = F1 (z)F2 (z) = Z [ f 1 (n) ∗ f 2 (n)] (z − a)(z − b)

So by Convolution theorem, f 1 (n) ∗ f 2 (n) = Z −1 [F1 (z)F2 (z)] This gives Z

−1

[F1 (z)F2 (z)] =

∞ 

f 1 (k) f 2 (n − k) =

k=−∞

= bn

n 

a k b(−k) = bn

k=0

n 

a k b(n−k) , n ≥ k

k=0

n  k  a k=0

b

=

bn+1 − a n+1 , |a| < |b|. (b − a)

Exercises 1. Let F(z) denote the Z -transform of  the sequence { f (n)}∞ n=−∞ and be defined by cn ∞ e ,n≥0 F(z) = n=−∞ f (n)z −n . Let f (n) = , n being an integer. Then F(z) 0, n < 0 is z , |z| > c, (A) z−c

z , |z| > ec , (B) z−e c

Answer:

1 , |z| > ec , (C) z−e c

z , |z| < c. (D) z−c

B

∞ 2. Let F(z) denote the Z -transform of the  2sequence { f (n)}n=−∞ and be defined by n ,n≥0 −n F(z) = ∞ , n being an integer. Then F(z) n=−∞ f (n)z . Let f (n) = 0, n < 0 is

(A) z(z+1)3 , |z| > 1, (z−1)

(B) (z+1)3 , |z| > 1, (z−1)

Answer:

(C) z(z+1)3 , |z| < 1, (z−1)

A

(D) (z+1)3 , |z| < 1. (z−1)

10 Z Transform

252

∞ 3. Let F(z) denote the Z -transform of the  1 sequence { f (n)}n=−∞ and be defined by ,n≥0 −n n! , n being an integer. Then F(z) F(z) = ∞ n=−∞ f (n)z . Let f (n) = 0, n < 0 is

1

1

1

1

(B) e z for |z| > 0,

(A) e z for all z,

(D) e z 2 for |z| > 0.

(C) e z 2 for all z,

Answer:

A

4. Let F(z) denote the Z -transform of the sequence { f (n)}∞ n=−∞ and be defined −n by F(z) = ∞ n=−∞ f (n)z . Let f (n) = 0 for n < 0, n being an integer. Then lim F(z) is

|z|→∞

(A) f (0),

(C) f  (0),

(B) f (1),

Answer :

(D) f  (1).

A

5. Let F(z) Z -transform of the sequence { f (n)}∞ n=−∞ and be defined by ∞denote the −n F(z) = n=−∞ f (n)z . Then the Z -transform of the sequence {n f (n)}∞ n=−∞ is (A) z F  (z),

(B) F  (z),

(C) −z F  (z),

(D) −F  (z).

Answer : C 6. Let F(z) Z -transform of the sequence { f (n)}∞ n=−∞ and be defined by ∞denote the −n F(z) = n=−∞ f (n)z . Then the Z -transform of the sequence {2−n f (n)}∞ n=−∞ is

(A) F(2z),

(B) F

z 2 ,

(C) F(z 2 ),

Answer : 7. The inverse Z -transform of F(z) =

z z−1

(D) F

 2 z 2 .

A

for |z| > 1 is 



0, n ≥ 0 ,  −1, n < 0 −1, n ≥ 0 (D) f (n) = . 1, n < 0

0, n ≥ 0 ,  1, n < 0 1, n ≥ 0 (C) f (n) = , 0, n < 0

(B) f (n) =

(A) f (n) =

Answer : C

References

253

8. The inverse Z -transform of F(z) =

z z−1

for |z| < 1 is





0, n ≥ 0 ,  1, n < 0 1, n ≥ 0 (C) f (n) = , 0, n < 0

0, n ≥ 0 ,  −1, n < 0 −1, n ≥ 0 (D) f (n) = . 1, n < 0

(A) f (n) =

(B) f (n) =

Answer: 9. Find the inverse Z -transform of F(z) =

B

z z−1



Answer:

f (n) =

10. Find the inverse Z -transform of F(z) = 11. Find the inverse Z -transform of F(z) = 12. Find the inverse Z -transform of F(z) = 13. Find the inverse Z -transform of F(z) =

for |z|
0 0, n ≤ 0

z for |z| > 1. 1+z 2 z 2 +1 for |z| < 1. z 2 −1 z 2 −1 for |z| > 2. z 2 −2z z 2 −1 for |z| < 2. z 2 −2z

References The content of this chapter is mainly based on the website epgp.inflibnet.ac.in/ of UGC E Pathsala and on the following literature. 1. I.N. Sneddon, Use of Integral Transform (Mc Graw Hill, 1972) 2. L.C. Andrews, B.K. Shivamoggi, Integral Transform for Engineers (Prentice-Hall, 2007) 3. A.I. Zayed, Handbook of Functions and Generalized Function Transformations (CRC Press, 1996) 4. L. Debnath, D. Bhatta, Integral Transforms and Their Applications, 3rd edn. (Chapman and Hall/CRC, USA, 2014)

Chapter 11

Formal Construction of Integral Transforms and Their Inverses

11.1 The Method of Construction A formal method of construction of the integral transform together with its appropriate inversion formula is described below. Let a second-order linear differential equation of the Sturm–Liouville type be   du d p(x) + {q(x) − λr (x)}u = 0, a < x < b, Lu ≡ − dx dx

(11.1)

where p(x), q(x), r (x) are piecewise continuous in (a, b), p(x) = 0 in (a, b), λ is a parameter (in general complex) and u(x) satisfies some prescribed boundary conditions at the end points x = a, x = b. Let Green’s function G(x, ξ; λ) satisfy LG = δ(x − ξ), a < x < b

(11.2)

where δ(x) denotes the Dirac delta function and G(x, .; ..) satisfies the same boundary conditions as u(x) at x = a, x = b. Let u 1 (x, λ) and u 2 (x, λ) be the two independent solutions of the ordinary differential equation (11.1) satisfying the prescribed end conditions at x = a and at x = b, respectively. Then  G(x, ξ; λ) =

u 1 (x,λ)u 2 (ξ,λ) , A(λ) u 1 (ξ,λ)u 2 (x,λ) , A(λ)

a < x < ξ < b, a 0? Choose a branch cut along the positive real axis of the complex λ-plane as shown in Fig. 11.1.

Fig. 11.1 Complex λ-plane

258

11 Formal Construction of Integral Transforms and Their Inverses

Fig. 11.2 Complex λ-plane

√ √ θ cut complex plane,√let λ =√r eiθ , 0 < θ < 2π so that λ =√ r ei 2 = √ In this r (cos 2θ + isin 2θ ). Thus Im( λ) = r sin 2θ , 0 < 2θ < π, so that Im( λ) > 0. √ Now W (u 2 , u 1 ) = u 1 u 2 − u 2 u 1 = √ 2i λ. Thus A(λ) = pW (u 2 , u 1 ) = 2i λ. Hence G(x, ξ; λ) =

⎧ −i √λx i √λξ ⎨ e √e , −∞ < x < ξ < ∞, 2i λ ⎩ e−i

√

√ λξ i λx

√e 2i λ

, −∞ < ξ < x < ∞.

(11.9)

Thus δ(x − ξ) = δ(ξ − x) has the representation 1 δ(x − ξ) = − 2πi



ei

√



√ λξ −i λx

e √ 2i λ

dλ

(11.10)

where  is a contour in the complex λ-plane enclosing all the singularities of G(., .; λ) in the anticlockwise sense. The path  encloses all the singularities of G(., .; λ) in the complex λ-plane. The singularities of G(λ) are the branch points at λ = 0 and λ = ∞.  is chosen to be a circle of large radius with center at the origin and a slit at λ = ∞ in the complex cut λ-plane. It can be deformed into a contour C as shown in Fig. 11.2, which encloses the positive real axis in the complex cut λ-plane anticlockwise, i.e., it starts from the point ∞ei0 + i0 and ends at ∞ei2π − i0, the origin being on the left as one traverses along C starting at ∞ei0 + i0. Thus 1 f (x) = − 2πi

 C

√

e−i λx √ 2i λ



∞ −∞

f (ξ)e

√ i λξ

 dξ dλ.

√ Now put λ = μ, then the path of integration in the complex μ-plane may be taken to run parallel to, but just above, the real μ-axis so that

11.2 Construction of Some Familiar Integral Transforms

1 f (x) = − 2πi



√

∞

e−i μx 2iμ

−∞

which is equivalent to f (x) =

1 2π



F(μ) =

∞

−∞

∞

 f (ξ)e

iμξ

dξ 2μdμ

F(μ)e−iμx dμ

(11.11)

f (ξ)eiμξ dξ.

(11.12)

−∞



where



259

∞

−∞

The relation (11.12) is the usual Fourier integral transform and (5) is its inverse. Note: A rigorous proof for the integral expansion (11.11) and the class of functions for which (11.12) and (11.11) exist can be found in any standard book dealing with Fourier transform (e.g., Tichmarsh, E. C. Theory of Fourier integrals OUP 1937).

11.2.2 Fourier Cosine Transform Consider the ODE Lu = −

d 2u − λu = 0, 0 < x < ∞ dx2

with boundary conditions u  (0) = 0 and u(x) is bounded as x → ∞. Then u 1 (x) = cos

√

λx, u 2 (x) = ei

√

λx

(11.13)

√ where Im λ > 0. The branch cut in the complex λ-plane is to be chosen along the positive real axis of the complex λ-plane as shown in Fig. 11.1. √ Here A(λ) = pW (u 2 , u 1 ) = i λ so that G(x, ξ; λ) =

⎧ √ i √λξ λxe ⎨ cos √ , 0 < x < ξ < ∞, i λ ⎩ cos

√ √ λξei λx √ i λ

, −∞ < ξ < x < ∞.

Hence we can obtain 1 δ(x − ξ) = − 2πi where C is the same contour as in (i).

 C

cos

√ √ λxei λξ dλ √ i λ

260

11 Formal Construction of Integral Transforms and Their Inverses

Thus here, formally,  f (x) =

∞

0

If we put

1 δ(x − ξ) f (ξ)dξ = − 2πi

 C

√  ∞ √ cos λx ei λξ f (ξ)dξ dλ. √ i λ 0

√ λ = μ, then this produces

f (x) =

1 π



∞ −∞



 2 ∞ eiμξ f (ξ)dξ dμ = F(μ) cos μdμ π 0

∞

cos μx 0



where

∞

F(μ) =

f (ξ) cos μξ dξ.

(11.14)

(11.15)

0

The relation (11.15) is the usual Fourier cosine transform and (11.14) is its inverse.

11.2.3 Fourier Sine Transform Consider the ODE Lu = −

d 2u − λu = 0, 0 < x < ∞ dx2

with boundary conditions u(0) = 0, and u(x) bounded as x → ∞. Then here u 1 (x) = sin

√ √ λx, u 2 (x) = ei λx

(11.16)

√ where Im λ > 0. The branch cut in the complex λ-plane is to be chosen along the positive real axis √ of the complex λ-plane as shown in Fig. 11.1. Here A(λ) = pW (u 2 , u 1 ) = − λ. We obtain here 1 f (x) = − 2πi =

1 iπ





√  ∞ √ λx i λx e f (ξ)dξ dλ √ − λ 0  ∞  iμξ sin μx e dξ dμ,

sin

C ∞ −∞

so that f (x) =

0

2 π



F(μ) sin μxdμ

(11.17)

f (ξ) sin μξdξ.

(11.18)

0



where

∞

∞

F(μ) = 0

Equation (11.18) is the usual Fourier sine transform while (11.17) is its inverse.

11.2 Construction of Some Familiar Integral Transforms

261

11.2.4 Havelock Transform (Mixed Fourier Transform, Hybrid Fourier Transform) Consider the ODE Lu = −

d 2u − λu = 0, 0 < x < ∞ dx2

with boundary conditions as K u(0) + u  (0) = 0, and u(x) is bounded as x → ∞, where K is a positive real number. Note that here the boundary condition at x = 0 is mixed, and, because of this, integral transform that will be generated is called the mixed Fourier transform. Here u 1 (x) =

√ √ √ λ cos λx − K sin λx, u 2 (x) = ei

√

λx

(11.19)

√ where Im λ > 0. The branch cut in the complex λ-plane is chosen along the positive real axis of the complex λ-plane as shown in Fig. 11.1. In this case, √ √ A(λ) = − λ(K + i λ) so that G(x, ξ; λ) =

⎧√ √ √ √ i λξ ⎪ ⎨ ( λ cos √λx−K sin√ λx)e , 0 < x < ξ < ∞, − λ(K +i λ) √

⎪ ⎩(

√ √ √ λ cos √ λξ−K sin√ λξ)ei λξ , − λ(K +i λ)

0 < ξ < x < ∞.

√ Thus, apart from branch points, G(., .; λ) has a simple pole at λ = −K 2 ( λ = i K ). δ(x − ξ) has the representation 1 δ(x − ξ) = − 2πi

 

√ √ √ √ ( λ cos λx − K sin λx)ei λξ dλ. √ √ − λ(K + i λ)

Since the integrand has a pole at λ = −K 2 a contribution from the pole has to be taken into account when  is √ deformed into C (Fig. 11.3). The residue at λ = −K 2 ( λ = i K ) is −2K e−K (x+ξ) . Thus δ(x − ξ) = 2K e

−K (x+ξ)

1 − 2πi

 C

√ √ √ √ ( λ cos λx − K sin λx)ei λξ) dλ. √ √ (− λ(K + i λ)

262

11 Formal Construction of Integral Transforms and Their Inverses

Fig. 11.3 Complex λ-plane

Hence here f (x) = 2K e−K x



∞

1 2πi

e−K ξ f (ξ)dξ −

0

 C

√ √ √ √   ∞ √ ( λ cos λx − K sin λx)ei λξ) ei λξ f (ξ)dξ dλ. √ √ (− λ(K + i λ) 0

Writing λ = μ2 , we can finally obtain, for 0 < x < ∞ f (x) = A0 e

−K x

 +

∞

A(μ)(μ cos μx − K sin μx)dμ

(11.20)

0



where A0 = 2K

∞

e−K ξ f (ξ)dξ,

0

1 2 A(μ) = π μ2 + K 2



∞

(μ cos μξ − K sin μξ) f (ξ)dξ.

(11.21)

0

Equation (11.20) is known as Havelock’s integral expansion of the function f (x) and (11.21) is the Havelock transform. Note: 1. K = 0 will reduce it to the usual Fourier cosine transform. 2. This transform has immense usefulness in the mathematical study of water wave problems assuming linear theory. This transform was developed by Sir T. Havelock in (1929) Phil Mag. 3. A different type of proof of the Havelock integral transform is given in the paper: A note on Havelock’s generalization of the Fourier cosine expansion, Mandal, B. N., Int. J. Math. Educ. Sci. Technol., 20 (1989) 179–181.

11.2 Construction of Some Familiar Integral Transforms

263

11.2.5 Finite Fourier Sine Transform Consider the ODE Lu = −

d 2u − λu = 0, 0 < x < 1 dx2

with boundary conditions u(0) u(1) = 0. Then u 1 (x) = sin √ = 0,√ √ λ(1 − x). Here A(λ) = − λ sin λ so that  sin √λx sin √λ(1−x) G(x, ξ; λ) =

sin

√ , √ − λ sin √λ λξ√sin λ(1−x) , − λ sin λ

√ λx, u 2 (x) = sin

0 < x < ξ < 1, 0