The Hypergeometric Approach to Integral Transforms and Convolutions
An Introduction to Integral Transforms
Transcript of An Introduction to Integral Transforms
An Introduction to Integral Transforms
An Introduction to Integral Transforms
Baidyanath Patra, Ph. D.Ex-Professor, Bengal Engineering Science University
(Recently named as IIEST, Shibpur)
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Dedication
I dedicate this book to my beloved wife, Mrs. Minu Patra, for standing beside me throughout my career and during the writing of this book. She has been my inspi-ration and motivation to expand my knowledge and without her selfless support and encouragement this book would not have been possible.
Preface
Integral transform is one of the powerful tools in Applied Mathematics. This book provides an introduction to this subject for solving boundary value and initial value problems in mathematical physics, Engineering Science and related areas.
During my fifty years teaching and research experiences at the Indian Institute of Engineering Science and Technology, Shibpur, formerly known as Bengal Engineering College and Bengal Engineering & Science University, I have been introduced to the subject of Integral Transforms.
Here an attempt has been made to cover the basic theorems of commonly used various integral transform techniques with their applications.
The transforms that have been covered in this book in detail are Fourier Transfoms, Laplace Transforms, Hilbert and Stieltjes Transforms, Mellin Transforms, Hankel Transforms, Kontorovich-Lebedev Transforms, Legendre Transforms, Mehler-Fock Transforms, Jacobi-Gegenbauer-Laguerre- Hermite Transforms, and the Z Transform.
This textbook targets the graduate and some advanced undergraduate students. My endeavor will be considered fruitful if it serves the need of at least one reader.
Date : January, 2016 B. Patra
Acknowledgement
I would like to take this opportunity to express my gratitude to my respected teacher, Professor S C Dasgupta, D. Sc.. He introduced me to this subject and developed my career in this field of research with special emphasis to application of Integral Transforms.
I am indebted to my many research students who worked under my supervision on this subject. I also acknowledge the authors of all the books that I read and results of which have been used in preparation of this book.
I would also like to thank my sons and their families for their constant encouragement during the writing of this book. Special thanks goes to my grandsons for their support for this work despite all the time it took me away from them.
During the process of writing, I have been greatly inspired by the textbooks of I. N. Sneddon and L. K. Debnath on this subject. Many other books and research papers, all of which are not possible to be mentioned individually, have also been consulted during this process.
I am thankful to the publishers, Levant Books for their unfailing co-operation in bringing this book to light within a reasonable amount of time.
Contents
1 FOURIER TRANSFORM 1
1.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Classes of functions . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Fourier Series and Fourier Integral Formula . . . . . . . . 2
1.4 Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . 6
1.4.1 Fourier sine and cosine Transforms. . . . . . . . . . 7
1.5 Linearity property of Fourier Transforms. . . . . . . . . . 8
1.6 Change of Scale property. . . . . . . . . . . . . . . . . . . 9
1.7 The Modulation theorem. . . . . . . . . . . . . . . . . . . 10
1.8 Evaluation of integrals by means of inversion theorems. . 11
1.9 Fourier Transform of some particular functions. . . . . . . 13
1.10 Convolution or Faltung of two integrable functions. . . . . 20
1.11 Convolution or Falting or Faltung Theorem for FT. . . . . 21
1.12 Parseval’s relations for Fourier Transforms. . . . . . . . . 23
1.13 Fourier Transform of the derivative of a function. . . . . . 26
1.14 Fourier Transform of some more useful functions. . . . . . 30
1.15 Fourier Transforms of Rational Functions. . . . . . . . . . 36
1.16 Other important examples concerning derivative of FT. . 37
1.17 The solution of Integral Equations of Convolution Type. . 47
1.18 Fourier Transform of Functions of several variables. . . . . 53
1.19 Application of Fourier Transform to Boundary Value Prob-lems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
2 FINITE FOURIER TRANSFORM 79
2.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 79
2.2 Finite Fourier cosine and sine Transforms. . . . . . . . . . 79
2.3 Relation between finite Fourier Transform of the deriva-tives of a function. . . . . . . . . . . . . . . . . . . . . . . 81
2.4 Faltung or convolution theorems for finite Fourier Trans-form. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
2.5 Multiple Finite Fourier Transform. . . . . . . . . . . . . . 85
2.6 Double Transforms of partial derivatives of functions. . . . 86
2.7 Application of finite Fourier Transforms to boundary valueproblems. . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
3 THE LAPLACE TRANSFORM 102
3.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 102
3.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
3.3 Sufficient conditions for existence of Laplace Transform. . 103
3.4 Linearity property of Laplace Transform. . . . . . . . . . 104
3.5 Laplace transforms of some elementary functions . . . . . 105
3.6 First shift theorem. . . . . . . . . . . . . . . . . . . . . . . 107
3.7 Second shift theorem. . . . . . . . . . . . . . . . . . . . . 107
3.8 The change of scale property. . . . . . . . . . . . . . . . . 107
3.9 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
3.10 Laplace Transform of derivatives of a function. . . . . . . 110
3.11 Laplace Transform of Integral of a function . . . . . . . . 112
3.12 Laplace Transform of tnf(t) . . . . . . . . . . . . . . . . . 113
3.13 Laplace Transform of f(t)/t . . . . . . . . . . . . . . . . 114
3.14 Laplace Transform of a periodic function. . . . . . . . . . 115
3.15 The initial-value theorem and the final-value theorem ofLaplace Transform. . . . . . . . . . . . . . . . . . . . . . . 116
3.16 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
3.17 Laplace Transform of some special functions. . . . . . . . 121
3.18 The Convolution of two functions. . . . . . . . . . . . . . 131
3.19 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . 132
4 THE INVERSE LAPLACE TRANSFORM ANDAPPLICATION 141
4.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 141
4.2 Calculation of Laplace inversion of some elementary func-tions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
4.3 Method of expansion into partial fractions of the ratio oftwo polynomials . . . . . . . . . . . . . . . . . . . . . . . 145
4.4 The general evaluation technique of inverse Laplace trans-form. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
4.5 Inversion Formula from a different stand point : The Tri-comi’s method. . . . . . . . . . . . . . . . . . . . . . . . . 158
4.6 The Double Laplace Transform . . . . . . . . . . . . . . . 161
4.7 The iterative Laplace transform. . . . . . . . . . . . . . . 166
4.8 The Bilateral Laplace Transform. . . . . . . . . . . . . . . 166
4.9 Application of Laplace Transforms. . . . . . . . . . . . . . 168
5 Hilbert and Stieltjes Transforms 220
5.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 220
5.2 Definition of Hilbert Transform . . . . . . . . . . . . . . . 220
5.3 Some Important properties of Hilbert Transforms. . . . . 221
5.4 Relation between Hilbert Transform and Fourier Transform.225
5.5 Finite Hilbert Transform. . . . . . . . . . . . . . . . . . . 226
5.6 One-sided Hilbert Transform. . . . . . . . . . . . . . . . . 227
5.7 Asymptotic Expansions of one-sided Hilbert Transform. . 228
5.8 The Stieltjes Transform. . . . . . . . . . . . . . . . . . . . 230
5.9 Some Deductions. . . . . . . . . . . . . . . . . . . . . . . . 231
5.10 The Inverse Stieltjes Transform. . . . . . . . . . . . . . . . 232
5.11 Relation between Hilbert Transform and Stieltjes Trans-form. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
6 Hankel Transforms 238
6.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 238
6.2 The Hankel Transform. . . . . . . . . . . . . . . . . . . . 238
6.3 Elementary properties . . . . . . . . . . . . . . . . . . . . 238
6.4 Inversion formula for Hankel Transform. . . . . . . . . . . 242
6.5 The Parseval Relation for Hankel Transforms. . . . . . . . 244
6.6 Illustrative Examples: . . . . . . . . . . . . . . . . . . . . 245
7 Finite Hankel Transforms 260
7.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 260
7.2 Expansion of some functions in series involving cylinderfunctions : Fourier-Bessel Series. . . . . . . . . . . . . . . 260
7.3 The Finite Hankel Transform. . . . . . . . . . . . . . . . . 262
7.4 Illustrative Examples. . . . . . . . . . . . . . . . . . . . . 263
7.5 Finite Hankel Transform of order n in 0 � x � 1 of thederivatrive of a function. . . . . . . . . . . . . . . . . . . . 265
7.6 Finite Hankel Transform over 0 � x � 1 of order n ofd2fdx2 + 1
xdfdx , when p is the root of Jn(p) = 0. . . . . . . 266
7.7 Finite Hankel Transform of f ′′(x) + 1x f ′(x) − n2
x2 f(x), wherep is the root of Jn(p) = 0 in 0 � x � 1 . . . . . . . . . . 266
7.8 Other forms of finite Hankel Transforms. . . . . . . . . . . 267
7.9 Illustrations. . . . . . . . . . . . . . . . . . . . . . . . . . 268
7.10 Application of finite Hankel Transforms. . . . . . . . . . . 269
8 The Mellin Transform 277
8.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 277
8.2 Definition of Mellin Transform. . . . . . . . . . . . . . . . 278
8.3 Mellin Transform of derivative of a function. . . . . . . . . 281
8.4 Mellin Transform of Integral of a function. . . . . . . . . . 283
8.5 Mellin Inversion theorem. . . . . . . . . . . . . . . . . . . 285
8.6 Convolution theorem of Mellin Transform. . . . . . . . . . 286
8.7 Illustrative solved Examples. . . . . . . . . . . . . . . . . 287
8.8 Solution of Integral equations. . . . . . . . . . . . . . . . . 292
8.9 Application to Summation of Series. . . . . . . . . . . . . 293
8.10 The Generalised Mellin Transform. . . . . . . . . . . . . . 295
8.11 Convolution of generalised Mellin Transform. . . . . . . . 297
8.12 Finite Mellin Transform. . . . . . . . . . . . . . . . . . . . 297
9 Finite Laplace Transforms 302
9.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 302
9.2 Definition of Finite Laplace Transform. . . . . . . . . . . . 302
9.3 Finite Laplace Transform of elementary functions. . . . . 304
9.4 Operational Properties. . . . . . . . . . . . . . . . . . . . 307
9.5 The Initial Value and the Final Value Theorem . . . . . . 311
9.6 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . 312
10 Legendre Transforms 317
10.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 317
10.2 Definition of Legendre Transform. . . . . . . . . . . . . . 317
10.3 Elementary properties of Legendre Transforms. . . . . . . 318
10.4 Operational Properties of Legendre Transforms . . . . . . 323
10.5 Application to Boundary Value Problems. . . . . . . . . . 325
11 The Kontorovich-Lebedev Transform 328
11.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 328
11.2 Definition of Kontorovich - Lebedev Transform. . . . . . . 328
11.3 Parseval Relation for Kontorovich-Lebedev Transforms. . 329
11.4 Illustrative Examples. . . . . . . . . . . . . . . . . . . . . 330
11.5 Boundary Value Problem in a wedge of finite thickness. . 332
12 The Mehler-Fock Transform 335
12.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 335
12.2 Fock’s Theorem (with weaker restriction). . . . . . . . . . 335
12.3 Mehler-Fock Transform of zero order and its properties. . 337
12.4 Parseval type relation. . . . . . . . . . . . . . . . . . . . . 339
12.5 Mehler-Fock Transform of order m . . . . . . . . . . . . . 341
12.6 Application to Boundary Value Problems. . . . . . . . . . 342
12.6.1 First Example . . . . . . . . . . . . . . . . . . . . 342
12.6.2 Second Example . . . . . . . . . . . . . . . . . . . 344
12.6.3 Third Example . . . . . . . . . . . . . . . . . . . . 345
12.6.4 Fourth Example . . . . . . . . . . . . . . . . . . . 347
12.7 Application of Mehler-Fock Transform for solving dualintegral equation. . . . . . . . . . . . . . . . . . . . . . . . 348
13 Jacobi, Gegenbauer, Laguerre and Hermite Transforms351
13.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 351
13.2 Definition of Jacobi Transform. . . . . . . . . . . . . . . . 351
13.3 The Gegenbauer Transform. . . . . . . . . . . . . . . . . . 355
13.4 Convolution Theorem . . . . . . . . . . . . . . . . . . . . 356
13.5 Application of the Transforms . . . . . . . . . . . . . . . . 357
13.6 The Laguerre Transform . . . . . . . . . . . . . . . . . . . 359
13.7 Operational properties . . . . . . . . . . . . . . . . . . . . 361
13.8 Hermite Transform. . . . . . . . . . . . . . . . . . . . . . 364
13.9 Operational Properties. . . . . . . . . . . . . . . . . . . . 366
13.10Hermite Transform of derivative of a function. . . . . . . . 367
14 The Z-Transform 372
14.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . 372
14.2 Z - Transform : Definition. . . . . . . . . . . . . . . . . . 372
14.3 Some Operational Properties of Z-Transform. . . . . . . . 376
14.4 Application of Z-Transforms. . . . . . . . . . . . . . . . . 383
Appendix 390Bibliography 405Index 407
Chapter 1
FOURIER TRANSFORM
1.1 Introduction.
The method of integral transforms is one of the most easy and effectivemethods for solving problems arising in Mathematical Physics, AppliedMathematics and Engineering Science which are defined by differentialequations, difference equations and integral equations. The main ideain the application of the method is to transform the unknown function,say, f(t) of some variable t to a different function, say, F (p) of a complexvariable p. Then the associated differential equation can be directly re-duced to either a differential equation of lower dimension or an algebraicequation in the variable p. There are several forms of integral transformsand one form may be obtained from the other by a transformation of thecoordinates and the functions. The choice of integral transform dependson the structure of the equation and on the geometry of the domainunder consideration. This method of integral transform simplifies thecomputational techniques considerably.
Suppose that there exists a known function K(p, t) of two variablesp and t and that ∫ b
aK(p, t) f(t) dt = F (p) (1.1)
is convergent. Then F (p) is called the integral transform of the functionf(t) by the function K(p, t), which is called the kernel of the transform.Here, the variable p is a parameter, which may be real or complex. Dif-ferent forms of kernels will generate different form of transforms. Belowwe discuss a variety of integral transforms with their applications afterusing different forms of kernels in variety of domains. To begin with, weconsider Fourier Transform.
2 An Introduction to Integral Transforms
1.2 Classes of functions
A singlevalued function f(x) of the independent variable x which iscontinuous in an interval [a,b], is said to belong to a class denoted byf ∈ C [a, b].
A function f(x) is said to be piecewise continuous in an interval (a, b)if the interval can be partitioned into finite number of non-intersectingsubintervals (a, a1), (a1, a2), ..., (an−1, b), in each of which the function iscontinuous and has finite limits as x approaches the end points of eachof the sub-intervals. Such a function is said to belong to a class denotedby f ∈ P (a, b).
A piecewise continuous function f in (a, b), whose first order deriv-ative is also a piecewise continuous function in (a, b) and belongs to aclass denoted by f ∈ P 1(a, b).
The set of functions f(x) is said to be absolutely integrable over Ω,if∫Ω |f(x)|dx is finite. Then we say that this function belongs to a class
denoted by f(x) ∈ A1(Ω). Similarly, the statement f ∈ Am(Ω) implies∫Ω |f(x)|mdx is finite.
Finally, we introduce a class of functions f(x), which satisfies thefollowing conditions
(i) f(x) is defined in c < x < c + 2 l
(ii) f(x) is periodic function of period 2 l and(iii) f(x) and f ′(x) are piecewise continuous in c < x < c+2l denoting
by f(x) ∈ P 1(c, c + 2l).
This class of function f(x) is said to satisfy Dirichlet’s conditions.
A function f(x) is said to be of exponential order σ as x → ∞, ifconstants σ, m(> 0) can be so found that |e−σxf(x)| < m implying|f(x)| < m emσ for x > x0. Equivalently, we also write f(x) = 0 (emσ)as x → ∞.
1.3 Fourier Series and Fourier Integral Formula
Suppose that a function f(x) satisfies Dirichlet’s conditions over theinterval (−l, l) and it belongs to the class P 1(R) and also to the class
Fourier Transform 3
A1(R). Let the principal period of f(x) be 2 l. Then f(x) admits of theFourier series
f(x) = a0 +∞∑
n=1
[an cos
nπx
l+ bn sin
nπx
l
]
where a0 =12l
∫ l
−lf(x)dx
(an, bn) =1l
∫ l
−lf(x)
[cos
nπx
l, sin
nπx
l
]dx
so that an − i bn =1l
∫ l
−lf(x) e
−inπxl dx
Let us now set a0 = c0
an = cn + c−n
i bn = cn − c−n
Then, f(x) = c0 +∞∑
n=1
[cne
−inπxl + c−ne
inπxl
]
=+∞∑−∞
cne−inπx
l (1.2)
where, cn =12l
∫ l
−lf(x) e
inπxl dx =
12l
∫ l
−lf(t) e
inπtl dt
(1.3)
This form of Fourier series is called the complex form of Fourier seriesin (−l, l). Thus, from (1.2) and (1.3), one gets
f(x) =+∞∑−∞
[12l
∫ l
−lf(t) e
inπtl
dt
]e−
inπxl (1.4)
Let us now put πl = δξ. We note that δξ → 0 as l → ∞ such that
l.δξ = π = a finite number. Then the series (1.4), before taking thelimit as δξ → 0, becomes
f(x) =12π
+∞∑−∞
δξ
[∫ πδξ
−πδξ
f(t) eint δξ dt
]e−inx δξ
=12π
∫ πδξ
−πδξ
f(t)
[+∞∑−∞
δξ.ein(t−x)δξ
]dt
4 An Introduction to Integral Transforms
after interchanging formally the summation and integration signs. Since,by assumption f(x) ∈ P 1(R) and also f(x) ∈ A1(R), making δξ → 0and using the definition of Riemann definite integral as limit of a sumwe get
f(x) =12π
∫ +∞
−∞f(t)
[∫ +∞
−∞ei(t−x)ydy
]dt (1.5)
⇒ f(x) =1√2π
∫ +∞
−∞e−iyx[F (y)] dy, (1.6)
where F (y) =1√2π
∫ +∞
−∞f(t)eiyt dt (1.7)
Again, from eqn. (1.5) one gets
f(x) =1π
∫ ∞
0dy
∫ +∞
−∞f(t) cos y(−t + x) dt (1.8)
Eqn. (1.8) is known as the Fourier Integral formula. At a point offinite discontinuity of f(x), the left hand side of (1.8) is replaced by12 [f(x + 0) + f(x − 0)] in the sense of limiting values.
A detailed proof of this statement is included in the following corol-laries.
Coro 1.1 Fourier’s Integral Theorem.
If f(t) ∈ P 1(R) and also f(t) ∈ A1(R), then for all x ∈ R,
1π
∫ ∞
0dy
∫ +∞
−∞f(t) cos[y(x − t)]dt =
12[f(x + 0) + f(x − 0)]
Proof. If k > 0,∫ ∞
0f(t)dt
∫ λ
0cos[y(t − x)]dy −
∫ λ
0dy
∫ ∞
0f(t) cos[y(t − x)]dt
=∫ k
0f(t)dt
∫ λ
0cos[y(t − x)]dy +
∫ ∞
kf(t)dt
∫ λ
0cos[y(t − x)]dy
−∫ λ
0dy
∫ k
0f(t) cos[y(t − x)]dt −
∫ λ
0dy
∫ ∞
kf(t) cos[y(t − x)]dt
=∫ ∞
kf(t)dt
∫ λ
0cos[y(t − x)]dy −
∫ λ
0dy
∫ ∞
kf(t) cos[y(t − x)]dt
Since f(t) ∈ A1(R), for every arbitrary positive number ∈, we canfind a number K1 such that∫ ∞
k|f(t)| dt <
∈2λ
, k > K1
Fourier Transform 5
Thus, ∣∣∣∣∫ λ
0dy
∫ ∞
kf(t) cos[y(t − x)]dt
∣∣∣∣ <∫ λ
0dy
∫ ∞
k|f(t)|dt <
∈2
.
Also, since∫ λ
0cos[y(t − x)]dy =
sin[λ(t − x)]t − x
it follows that∣∣∣∣∫ ∞
kf(t)dt
∫ ∞
0cos[y(t − x)]dy
∣∣∣∣ =∣∣∣∣∫ ∞
k−xf(t + x)
sinλ t
tdt
∣∣∣∣Thus, there exists a number K2 such that if k > K2 + x,∣∣∣∣∫ ∞
k−xf(t + x)[sin λt/t]dt
∣∣∣∣ < ∈2
.∣∣∣∣∫ ∞
k−xf(t)dt
∫ λ
0cos[y(t − x)]dy −
∫ λ
0dy
∫ ∞
kf(t) cos[y(t − x)]dt
∣∣∣∣ <∈,
for k > K = max(K1,K2) even for large values of λ
In other words,
limλ→∞
∫ ∞
0f(t)dt
∫ λ
0cos[y(t − x)]dy
= limλ→∞
∫ λ
0dy
∫ ∞
0f(t) cos[y(t − x)]dt (1.9)
In an exactly similar way we can show that
limλ→∞
∫ 0
−∞f(t)dt
∫ λ
0cos[y(t − x)]dy
= limλ→∞
∫ λ
0dy
∫ 0
−∞f(t) cos[y(t − x)]dt (1.10)
Adding (1.9) and (1.10) we get
limλ→∞
∫ ∞
−∞f(t)dt
∫ λ
0cos[y(t − x)]dy
= limλ→∞
∫ λ
0dy
∫ +∞
−∞f(t) cos[y(t − x)]dt
Again since,∫ 0
−∞f(x + t)
sin λ t
tdt =
∫ ∞
0f(x − t)
sin λ t
tdt
and for all x ∈ R,
∫ +∞
−∞f(x + t)
sin λ t
tdt → π
2[f(x + 0) + f(x − 0)].
6 An Introduction to Integral Transforms
We have,
12[f(x + 0) + f(x − 0)] = lim
λ→∞1π
∫ +∞
−∞f(x + u)
sin λ u
udu
= limλ→∞
1π
∫ +∞
−∞f(t)
sin[λ(t − x)]t − x
dt
= limλ→∞
1π
∫ +∞
−∞f(t)dt
∫ λ
0cos[y(t − x)] dy
= limλ→∞
1π
∫ λ
0dy
∫ +∞
−∞f(t) cos[y(t − x)] dt
Thus, the result of the Fourier Integral Theorem follows immediately.
In addition, if f(t) is continuous at the point t = x, we have
12π
∫ +∞
−∞e−iyxdy
∫ +∞
−∞f(t) eiytdt = f(x) (1.11)
From eqn. (1.11) the result in eqn. (1.8) of article 1.3 will follow.
1.4 Fourier Transforms
We are now ready to define Fourier transform of a piecewise continuousfunction f(x) ∈ P 1(R) and f(x) ∈ A1(R). It is defined by
F [f(x);x → ξ] = F [f(x)] = F (ξ) = f(ξ) =1√2π
∫ +∞
−∞f(x)eiξxdx
(1.12)
and f(x) = F−1[F (ξ)] =1√2π
∫ +∞
−∞F (ξ) e−iξx dξ (1.13)
at a point of continuity of f(x).
F (ξ) in (1.12) is called Fourier transform of f(x) and f(x) in (1.13) iscalled the inverse Fourier transform of F (ξ).
Some authors define Fourier transform (1.12) of f(x) and inversionformula in (1.13) by
F [f(x) ; x → ξ] = F (ξ) =∫ +∞
−∞f(x) eiξxdx (1.14)
and f(x) =12π
∫ +∞
−∞F (ξ) e−iξxdξ (1.15)
respectively.
Fourier Transform 7
At a point of discontinuity of x ∈ R, the left hand sides of eqns.(1.13) and (1.15) take the form
12
[f(x + 0) + f(x − 0)]
instead of f(x).
From the definitions of Fourier transform and Inverse Fourier trans-form we can, therefore, write from (1.12)
F (ξ) = F [f(x)] = F [F−1(F (ξ))] ⇒ F F−1[F (ξ)] ≡ I[F (ξ)]
⇒ F F−1 = I, an identity operator, F and F−1 being Fourier andits inverse operators respectively and from (1.13)
f(x) = F−1[F (ξ)] = F−1[F (f(x))] = F−1F [f(x)] = I[f(x)]
⇒ F−1F = I (1.16)
Thus in operator notation FF−1 = F−1 F = I . It shows that theseoperators F and F−1 are commutative.
1.4.1 Fourier sine and cosine Transforms.
Let in addition belonging to the classes P 1(R) and A1(R), the functionf(x) be an odd function of x ∈ R. Then, clearly f(−x) = −f(x), Now,by (1.12)
F [f(x);x → ξ] = F (ξ) =1√2π
∫ ∞
0f(x)
[eiξx − e−iξx
]dx
= i
√2π
∫ ∞
0f(x) sin ξx dx
≡ i Fs(ξ), say (1.17)
where, Fs(ξ) =
√2π
∫ ∞
0f(x) sin ξx dx (1.18)
Fs(ξ) in (1.18) is defined to be Fourier sine transform of the functionf(x). Its connection with the Fourier transform is given by (1.17) pro-vided f(x) in an odd function of x ∈ R. The inversion formula of (1.18)is then defined by
f(x) =
√2π
∫ ∞
0Fs(ξ) sin ξx dξ (1.19)
8 An Introduction to Integral Transforms
Let f(x) be an even function of x ∈ R in addition belonging to classesP 1(R) and A1(R). Then, f(−x) = f(x) and therefore, by (1.12) above
F [f(x);x → ξ] = F (ξ) =1√2π
∫ ∞
0f(x)
[eiξx + e−iξx
]dx
=
√2π
∫ ∞
0f(x) cos ξx dx
≡ Fc(ξ), say, (1.20)
where Fc(ξ) =
√2π
∫ ∞
0f(x) cos ξx dx (1.21)
Fc(ξ) in (1.21) is defined to be Fourier cosine transform of the func-tion f(x). Its connection with the Fourier transform is given by eqn.(1.20), provided f(x) is an even function of x ∈ R. The inversion for-mula of Fourier cosine transform defined by (1.21) is then defined as
f(x) =
√2π
∫ ∞
0Fc(ξ) cos ξx dξ (1.22)
1.5 Linearity property of Fourier Transforms.
If c1 and c2 are constants, then
(i) F [c1 f1(x) + c2 f2(x)] = c1 F [f1(x)] + c2 F [f2(x)]
(ii) Fs[c1 f1(x) + c2 f2(x)] = c1 Fs[f1(x)] + c2 Fs[f2(x)]
(iii) Fc[c1 f1(x) + c2 f2(x)] = c1 Fc[f1(x)] + c2 Fc[f2(x)]
Proof.
(i) By definition of Fourier Transform in eqn.(1.12) of article 1.4, wehave
F [c1 f1(x) + c2 f2(x);x → ξ]
=1√2π
∫ +∞
−∞[c1 f1(x) + c2 f2(x)] · eiξx dx
= c1 · 1√2π
∫ +∞
−∞f1(x)eiξx dx + c2 · 1√
2π
∫ +∞
−∞f2(x)eiξx dx
= c1 F [f1(x);x → ξ] + c2 F [f2(x);x → ξ]
Fourier Transform 9
(ii) By definition of Fourier sine transform in eqn. (1.18) of article1.4.1 we get
Fs[c1 f1(x) + c2 f2(x);x → ξ] =
√2π
∫ ∞
0c1 f1(x) sin ξx dx
+
√2π
∫ ∞
0c2 f2(x) sin ξx dx
= c1 Fs[f1(x);x → ξ] + c2 Fs[f2(x);x → ξ].
(iii) By definition of Fourier cosine transform in eqn.(1.21) of article1.4.1 proceeding similarly as above, we have
Fc[c1 f1(x) + c2 f2(x)] = c1 Fc[f1(x)] + c2 Fc[f2(x)] .
1.6 Change of Scale property.
(i) If F [f(x);x → ξ] = F (ξ), then F [f(ax) ; x → ξ] = 1a F
(ξa
)
Proof. Since F [f(x);x → ξ] =1√2π
∫ +∞
−∞f(x)eiξx dx = F (ξ),
we have F [f(ax);x → ξ] =1√2π
∫ +∞
−∞f(ax)eiξx dx
=1√2π
∫ +∞
−∞f(t)ei ξ
at 1a
dt
=1a· 1√
2π
∫ +∞
−∞f(t)ei( ξ
a)t dt =1aF
(ξ
a
).
(ii) If Fs[f(x)] = Fs(ξ), then Fs[f(ax)] = 1a Fs
(ξa
)Proof. By definition
Fs[f(x);x → ξ] =
√2π
∫ ∞
0f(x) sin ξx dx = Fs(ξ).
Therefore,
Fs[f(ax);x → ξ] =
√2π
∫ ∞
0f(ax) sin ξx dx
=1a
√2π
∫ ∞
0f(t) sin
(ξ
at
)dt
=1aFs
(ξ
a
)
10 An Introduction to Integral Transforms
(iii) Similarly, it is easy to prove that if
Fc[f(x);x → ξ] = Fc(ξ), then Fc[f(ax);x → ξ] = 1a Fc
(ξa
).
1.6 The shifting property of Fourier Transform.
If F (ξ) is Fourier transform of f(x) in R, then Fourier transform off(x − a) is eiξaF (ξ).
Proof. By definition, we have
F (ξ) =1√2π
∫ +∞
−∞f(x) eiξx dx = F [f(x);x → ξ]
Therefore,
F [f(x − a);x → ξ] =1√2π
∫ +∞
−∞f(x − a) eiξx dx
=1√2π
∫ +∞
−∞f(t) eiξ(t+a) dt
= eiξa · 1√2π
∫ +∞
−∞f(t) eiξt dt
= eiξaF (ξ).
1.7 The Modulation theorem.
Theorem : If F [f(x);x → ξ] = F (ξ), then F [f(x) cos ax ;x → ξ]= 1
2 [F (ξ + a) + F (ξ − a)].
Proof. Since it is given that
F [f(x);x → ξ] = F (ξ) =1√2π
∫ +∞
−∞f(x)eiξx dx
we have
F [f(x) cos ax ;x → ξ] =1√2π
∫ +∞
−∞f(x) · eiax + e−iax
2· eiξx dx
=12
[1√2π
∫ +∞
−∞f(x) ei(ξ+a)x dx
+1√2π
∫ +∞
−∞f(x) ei(ξ−a)x dx
]= [F (ξ + a) + F (ξ − a)].
Other three parts of the theorem are
(a) if Fs[f(x);x → ξ] = Fs(ξ) then
Fourier Transform 11
Fs[f(x) cos ax ;x → ξ] =12[Fs(ξ + a) + Fs(ξ − a)]
(b) if Fs[f(x);x → ξ] = Fs(ξ) then
Fc[f(x) sin ax ;x → ξ] =12[Fs(ξ + a) − Fs(ξ − a)]
and (c) if Fc[f(x);x → ξ] = Fc(ξ) then
Fs[f(x) sin ax ;x → ξ] =12[Fc(ξ − a) − Fc(ξ + a)]
These results can similarly be proved as was done in the first case onreplacing cos ax and sin ax by their exponential forms in the definitionsof the corresponding transforms of the left hand side.
1.8 Evaluation of integrals by means of inversion theorems.
By definitions of Fourier transform and its inversion formula in eqns.(1.12), (1.13), (1.17), (1.19), (1.21) and (1.22) of article 1.4, we mayemploy these results to evaluate certain integrals involving trigonometricfunctions. For example, first of all we consider
I1 =∫ ∞
0e−bx cos ax dx , I2 =
∫ ∞
0e−bx sin ax dx
Integrating by parts we can now evaluate I1 and I2 as
I1 =b
a2 + b2, I2 =
a
a2 + b2
These results means that if we set f(x) = e−bx, then its cosine andsine transforms are given by
Fc
[e−bx;x → ξ
]=
√2π
b
ξ2 + b2
Fs
[e−bx;x → ξ
]=
√2π
ξ
ξ2 + b2
Substituting these expressions in eqns. (1.19) and (1.22) of article1.4.1, we get ∫ ∞
0
cos ξx
ξ2 + b2dξ =
π
2be−bx
and∫ ∞
0
ξ sin ξx
ξ2 + b2dξ =
π
2e−bx
12 An Introduction to Integral Transforms
As a second example, if we take f(x) =
{1 , 0 < x < a
0 , x > a,
then
Fc(ξ) =
√2π
∫ a
01 cos ξx dx =
√2π
sin ξa
ξ
and so it gives
2π
∫ ∞
0
sin ξa
ξcos ξx dξ =
{1, if 0 < x < a
0, if x � a
Let us consider a third example when f(x) =
⎧⎪⎨⎪⎩
0 , 0 < x < a
x , a ≤ x ≤ b
0 , x > b
Then,
Fs(ξ) =
√2π
∫ b
ax sin ξx dx
=
√2π
[a cos ξ a − b cos ξ b
ξ+
sin ξ b − sin ξ a
ξ2
]
and this result gives
2π
∫ ∞
0sin ξ x [Fs(ξ)] dξ =
⎧⎪⎨⎪⎩
0 , 0 < x < a
x , a ≤ x ≤ b
0 , x > b
In the fourth case if we choose f(x) =
{1, 0 < x < a
0, x > a
Then
Fs(ξ) =1 − cos ξ a
ξ
and it gives
2π
∫ ∞
0sin ξ x
[1 − cos ξ a
ξ
]dξ = f(x) =
{1 , 0 < x < a
0 , x > a
Similarly taking f(x) =
{1, a < x < b
0, otherwise
Fourier Transform 13
we get
2π
∫ ∞
0sin ξ x
[cos a ξ − cos b ξ
ξ
]dξ =
⎧⎪⎨⎪⎩
0 , 0 < x < a
1 , a < x < b
0 , x > b
Combining these above results, we get
2π
∫ ∞
0sin ξ x
[a − b
ξ+
cos a ξ − cos b ξ
ξ2
]dξ =
⎧⎪⎨⎪⎩
a − b , 0 < x < a
x − b , a < x < b
0 , x > b
1.9 Fourier Transform of some particular functions.
We consider the following step function f(x) defined by
f(x) =
{0 , x < 01 , x > 0
This function is called Heaviside unit step function and it is denotedby the symbol H(x). Thus, we have
H(x) =
{0 , x < 0 (or sometimes for x ≤ 0)1 , x > 0
Almost all step functions can be expressed as a combination of Heavisideunit step functions. For example,
f(x) =
{1 , |x| < a
0 , otherwise
is equivalent to f(x) = H(x + a) − H(x − a) ≡ H(a − |x|)
and the step function f(x) =
{k , x > a
0 , otherwise
is equivalent to f(x) = k H(x − a), for all x
Example 1.1. Evaluate Fourier transform of H(x + a) − H(x − a)
Solution. By definition,
F [H(x + a) − H(x − a);x → ξ] =1√2π
∫ ∞
0eiξx[H(x + a) − H(x − a)]dx
14 An Introduction to Integral Transforms
=1√2π
∫ a
−aeiξx dx
=1√2π
[eiξx
iξ
]a
−a
=1√2πiξ
[eiξa − e−iξa
]
=
√2π
sin a ξ
ξ
Therefore, using the formula for inverse F T we get
1√2π
∫ +∞
−∞
√2π
sin a ξ
ξe−iξx dξ = H(x + a) − H(x − a)
This result implies,
1π
∫ +∞
−∞
sin a ξ
ξ· [cos ξ x − i sin ξ x] dξ = H(x + a) − H(x − a)
i.e∫ +∞
−∞
sin a ξ
ξcos ξ x dξ = π[H(x + a) − H(x − a)]
=
{π , for |x| < a
0 , for |x| > a
Putting x = 0 and a = 1, in the above we get∫ +∞
−∞
sin ξ
ξdξ = π
implying,
∫ ∞
0
sin ξ
ξdξ =
π
2
and∫ +∞
−∞
sin a ξ sin ξ x
ξdξ = 0
Example 1.2. Evalute FT of
f(x) =
{x , |x| < a
0 , |x| > a
Solution. By definition,
F [f(x) ; x → ξ] =1√2π
∫ a
−ax eiξx dx
=1√2π
{[x
eiξx
iξ
]a
−a
− 1iξ
∫ a
−aeiξx dx
}
Fourier Transform 15
=1√2π
[a eiξa + a e−iξa
iξ+
1ξ2
{eiaξ − e−iaξ
}]
=1√2π
[2 a cos a ξ
iξ+
2 i sin a ξ
ξ2
]
=
√2π
i
[sin a ξ − a ξ cos a ξ
ξ2
].
Example 1.3. If F T of f(x) is F (ξ), calculate F T of(i) eiλxf(x) and (ii) f(−x) in terms of F (ξ).
Solution. We know that,
F [f(x) ; x → ξ] =1√2π
∫ +∞
−∞f(x) eiξx dx = F (ξ).
Therefore,
(i) F[eiλx f(x) ; x → ξ
]=
1√2π
∫ +∞
−∞f(x) · ei(ξ+λ)x dx = F (ξ + λ).
(ii) F [f(−x) ; x → ξ] =1√2π
∫ +∞
−∞f(−x) eiξx dx
=1√2π
∫ +∞
−∞f(y) e−iξy dy = F (−ξ) = F−1(ξ)
Example 1.4. Find the function whose cosine transform is
√2π
sin a ξ
ξ.
Solution. Let f(x) be the required function such that
Fc[f(x) ; x → ξ] =
√2π
sin a ξ
ξ
f(x) =2π
∫ ∞
0
sin a ξ
ξ· cos ξ x d ξ
=1π
∫ ∞
0
1ξ[sin ξ(a + x) + sin ξ(a − x)] dξ
=1π
[∫ ∞
0
sin ξ(a + x)ξ
dξ +∫ ∞
0
sin ξ(a − x)ξ
dξ
]
=1π
[π2
+π
2
], when x < a and
1π
[π2− π
2
], when x > a
So, f(x) =
{1 , when x < a
0 , when x > a
16 An Introduction to Integral Transforms
Example 1.5. Find the F T of f(x) =
{1 − x2 , |x| ≤ 10 , |x| > 1
and hence evaluate∫ ∞
0
x cos x − sin x
x3cos
x
2dx
Solution. From definition, we have
F (ξ) = F [f(x) ; x → ξ] =1√2π
∫ 1
−1(1 − x2) eiξx dx
=[− 2
ξ2(eiξ + e−iξ) +
2iξ3
(eiξ − e−iξ)]· 1√
2π
Therefore,
F (ξ) =−4√2π
[(ξ cos ξ − sin ξ) /ξ3]
Now, the inversion formula gives
1√2π
∫ +∞
−∞
−4√2π
[ξ cos ξ − sin ξ
ξ3
]· e−iξx dx = f(x)
Hence,∫ +∞
−∞
[−4(ξ cos ξ − sin ξ)ξ3
][cos ξ x − i sin ξ x] dξ = 2πf(x)
Equating real parts, we have∫ ∞
−∞
sin ξ − ξ cos ξ
ξ3cos ξ x dξ =
π
2f(x)
Or, 2∫ ∞
0
sin ξ − ξ cos ξ
ξ3cos ξ x dξ =
{π2 (1 − x2) , |x| ≤ 10 , |x| > 1
Now, putting x = 12 in above we get∫ ∞
0
sin ξ − ξ cos ξ
ξ3cos
ξ
2dξ =
12· π
2· 34
=3π16
This gives,∫ ∞
0
x cos x − sin x
x3cos
x
2dx = −3π
16
Example 1.6. Find FT of f(x) = [1 − |x|]H[1 − |x|]Solution.
Here f(x) =
{1 − |x| , when − 1 < x < 10 , otherwise
Fourier Transform 17
Here, f(x) is an even function of x and so
F [f(x) ; x → ξ] = Fc[f(x) ; x → ξ]
=
√2π
∫ 1
0(1 − x) cos ξ x dx
=
√2π
1 − cos ξ
ξ2
So, F−1c [Fc{f(x) ; x → ξ} ; ξ → x] = [1 − |x|] H [1 − |x|]
Or,2π
∫ ∞
0
1 − cos ξ
ξ2cos ξ x dξ =
{1 − x , 0 < x < 10 , x > 1
(i)
Some Deductions :
(a) Putting x = 0 in the final result (i) of the example 1.6 we get∫ ∞
0
1 − cos ξ
ξ2dξ =
π
2
Or,∫ ∞
0
sin2 x
x2dx =
π
2
(b) Integrating the result (i) in the example 1.6 with respect to x from0 to X, we get∫ ∞
0
1 − cos ξ
ξ3
sin ξ X
ξdξ =
π
2
[X − X2
2
](ii)
Now, putting X = 12 , one gets∫ ∞
0
2 sin3 ξ2
8(
ξ2
)3 · 2d(
ξ
2
)=
π
2
[38
]
⇒∫ ∞
0
sin3 x
x3dx =
3π8
(c) We have seen in Example 1.1 that∫ ∞
0
sin x
xdx =
π
2
Again∫ ∞
0
sin2 x
x2dx =
π
2, from (a)
Also in (b),∫ ∞
0
sin3 x
x3dx =
3π8
<π
2
18 An Introduction to Integral Transforms
These results show that
sin x
x�[sin x
x
]2�[sin x
x
]3
(d) Again integrating the result (ii) in (b) above with respect to X
from X = 0 to X = 1, we get
∫ ∞
0
1 − cos ξ
ξ2
[− cos ξ X
ξ2
]10
dξ =π
2
[X2
2− X3
6
]10
Or,∫ ∞
0
sin4 x
x4dx =
π
3
Example 1.7. Find the F T of (i) f(x) = e−a|x|, a > 0 and(ii) f(x) = |x|e−a|x|, a > 0.
Solutions (i) F [f(x) ; x → ξ] =1√2π
∫ +∞
−∞e−a|x| · eiξx dx
=1√2π
[∫ 0
−∞eax · eiξx dx +
∫ ∞
0e−ax · eiξx dx
]
=1√2π
[1
a + iξ+
1a − iξ
]
=
√2π
a
a2 + ξ2
(ii) F [ |x|e−a|x| ; x → ξ ]
=1√2π
∫ +∞
−∞|x| e−a|x| · eiξx dx
=1√2π
[− d
da
∫ +∞
−∞e−a|x| eiξx dx
]
= − d
da
[√2π
a
a2 + ξ2
]
=
√2π
[a2 − ξ2
(a2 + ξ2)2
]
Example 1.8. (a) Find the F T of f(x) =
{e−ax, x > 0, a > 0−eax, x < 0, a > 0
Solution. Clearly, the given function is an odd function of x in R.
Fourier Transform 19
Therefore
F [f(x) ; x → ξ] = i
√2π
∫ ∞
0e−ax sin ξ x dx
= i
√2π· ξ
ξ2 + a2
(b) Find F T of f(x) = x e−a|x| , a > 0.
Solution. The given function f(x) is an odd function of x ∈ R. Hence,
F [f(x) ; x → ξ] = i
√2π
∫ ∞
0x e−ax · sin ξ x dx
= i
√2π
(− d
da
)∫ ∞
0e−ax sin ξ x dx = −i
√2π
d
da
[ξ
a2 + ξ2
]
= i
√2π
2 a ξ
(ξ2 + a2)2
Example 1.9. Find the F T of f(x) = e−αx2, α > 0.
Solution. We know from definition that
F [f(x) ; x → ξ] =1√2π
∫ +∞
−∞e−αx2
eiξx dx
=1√2π
∫ ∞
−∞e−�√
αx− iξ2√
α
�2− ξ2
4α dx
=e−
ξ2
4α√2π
∫ +∞
−∞e−z2 · dz√
α
=2 · e− ξ2
4α√2πα
∫ ∞
0e−z2
dz
F [e−αx2] =
2 · e− ξ2
4α√2πα
·√
π
2=
1√2α
e−ξ2
4α (i)
Some important deductions.
(i) Let α = 1 . Then F [e−x2; x → ξ] = 1√
2e
−ξ2
4
(ii) Let α = 12 . Then F [e
−x2
2 ; x → ξ] = e−ξ2
2
This result shows that F−1 [e−ξ2
2 ; ξ → x] = e−x2
2
20 An Introduction to Integral Transforms
Such a function f(x) = e−x2
2 having the property that
F [f(x) ; x → ξ] = f(ξ)
is called self reciprocal.(iii) Differentiating (i) with respect to ξ we get
d
dξ
[√2π
∫ ∞
0e−αx2
cos ξx dx
]=
d
dξ
1√2α
· e−ξ2
4α
⇒ Fs[xe−αx2; x → ξ] =
1√8α3
ξ e−ξ2
4α
Putting , α =12
we get Fs
[x e
−x2
2 ; x → ξ
]= ξ e
−ξ2
2
Thus f(x) = x e−x2
2 is self reciprocal with regard to Fourier sine trans-form.
Putting, α =i
2we get Fc
[e
ix2
2 ; x → ξ
]=
1 − i√2
eiξ2
2
Therefore, Fc
[cos
x2
2; x → ξ
]=
1√2
Re
[(1 − i)e
iξ2
2
]
= cos(
ξ2
2− π
4
)(ii)
Also, Fc
[sin
x2
2; x → ξ
]=
1√2
[cos
ξ2
2− sin
ξ2
2
]
= − sin(
ξ2
2− π
4
)(iii)
after equating real and imaginary parts of both sides. Combining resultsin (ii) and (iii) above one gets
Fc
[cos(
x2
2− π
8
); x → ξ
]= cos
[ξ2
2− π
8
]
1.10 Convolution or Faltung of two integrable functions.
The convolution or Falting or Faltung of two integrable functions f(x)and g(x), where −∞ < x < ∞ is denoted and defined as
f ∗ g =1√2π
∫ +∞
−∞f(x − u) g(u) du (1.23)
Fourier Transform 21
It possesses many formal properties such as
f ∗ (λg) = (λf) ∗ g = λ(f ∗ g), λ = constant
f ∗ g = g ∗ f , (the commutative property)
f ∗ (g + h) = f ∗ g + f ∗ h ,
which can be verified easily directly from definition (1.23). Further, ifboth f(x) and g(x) belong to C1(R) and A1(R) classes, then so doestheir convolution h(x) = f ∗ g, since
√2π∫ ∞
−∞|h(x)|dx =
∫ +∞
−∞dx
∣∣∣∣∫ +∞
−∞f(u)g(x − u)du
∣∣∣∣<
∫ +∞
−∞dx
∫ +∞
−∞|f(u)g(x − u)| du
=∫ +∞
−∞|f(u)|du
∫ +∞
−∞|g(x − u)| dx
⇒∫ +∞
−∞|h(x)|dx <
1√2π
∫ +∞
−∞|f(u)|du
∫ +∞
−∞|g(υ)|dυ
and the result follows from the fact that both f(x) and g(x) belong tothe class A1(R).
Again, (f ∗g)∗h = f ∗(g∗h) is true by direct verification. Therefore,the convolution property is also associative.
We shall now discuss Fourier transform of the convolution of a pairof functions as detailed below, by the name Convolution or Falting orFaltung theorem for F T .
1.11 Convolution or Falting or Faltung Theorem for FT.
Theorem.
The FT of the convolution of f(x) and g(x), both belonging to theclasses C1(R) and A1(R), is the product of the F T of f(x) and g(x).That means that,
F [f ∗ g;x → ξ] = F (ξ) G(ξ)
where F (ξ) =1√2π
∫ +∞
−∞f(x) eiξx dx and
G(ξ) =1√2π
∫ +∞
−∞g(x) eiξxdx
22 An Introduction to Integral Transforms
Proof. Since f(x) and g(x) both belong to the classes C1(R) andA1(R), f ∗g also belongs to the classes C1(R) and A1(R) and hence F T
of f ∗ g also exists. Therefore,
F [f ∗ g ; x → ξ] =1√2π
∫ +∞
−∞eiξx dx√
2π
∫ +∞
−∞f(x − u)g(u)du
=12π
∫ +∞
−∞
∫ +∞
−∞f(x − u)g(u)eiξx dx du
=12π
∫ +∞
−∞g(u)
[∫ +∞
−∞eiξxf(x − u)dx
]du
=1√2π
∫ ∞
−∞g(u)eiξu
[1√2π
∫ +∞
−∞eiξυf(υ) dυ
]du
=1√2π
∫ +∞
−∞g(u)eiξu · F (ξ) du
F [f ∗ g ; x → ξ] = F (ξ) · G(ξ) (1.24)
This proves the theorem.
Coro 1.2 Another interpretation of convolution theorem from eqn.(1.24) is that
F−1 [F (ξ)G(ξ) ; ξ → x] =1√2π
∫ +∞
−∞f(x − u)g(u)du
Or,1√2π
∫ +∞
−∞F (ξ)G(ξ)e−iξxdξ =
1√2π
∫ +∞
−∞f(υ) g(x − υ)dυ
= f ∗ g (1.25)
Coro 1.3 From the result (1.25) above, we have
1√2π
∫ +∞
−∞F (ξ) G(ξ) e−iξxdξ = f ∗ g =
1√2π
∫ +∞
−∞f(υ)g(x − υ)dυ
Putting x = 0 in the above equation, we get∫ +∞
−∞F (ξ) G(ξ) dξ =
∫ +∞
−∞f(υ) g(−υ) dυ (1.26)
Coro 1.4 Let f(x) ∈ R. Then F (ξ) = F [f(x) ; x → ξ] is a complexvalued function and its conjugate, denoted by F (ξ) is then given by
F (ξ) =1√2π
∫ +∞
−∞f(x)e−iξxdx = F−1[f(x);x → ξ]
Fourier Transform 23
i.e, F{f(x);x → ξ} = F−1[f(x) ; x → ξ]
Therefore, F ≡ F−1 (1.27)
Coro 1.5 Similar results for sine and cosine transforms of the func-tions f(x) and g(x) ∈ R may also be established in regard to theirconvolution. For example, if f(x), g(x) are even functions and
Fc [f(x) ; x → ξ] =
√2π
∫ ∞
0f(x) cos ξ x dx
Fc [g(x) ; x → ξ] =
√2π
∫ ∞
0g(x) cos ξ x dξ ,
then Fc [f ∗ g ; x → ξ] = Fc(ξ) Gc(ξ) (1.28)
Also, if f(x), g(x) are odd functions of x in R and if
Fs[f(x) ; x → ξ] = Fs(ξ) and Fs[g(x) ; x → ξ] = Gs(ξ)
then Fs [f ∗ g ; x → ξ] = Fs(ξ) Gs(ξ) (1.29)
It may be noted that the convolution had already been defined inarticle 1.10 and keeping this together with evenness and oddness char-acter of the associated functions one can very easily deduce the resultsin equations (1.28) and (1.29) respectively.
1.12 Parseval’s relations for Fourier Transforms.
Theorem.
If F (ξ) and G(ξ) are complex F.T of f(x) and g(x) respectively, then
(i)1√2π
∫ +∞
−∞F (ξ)G(ξ) dξ =
1√2π
∫ +∞
−∞f(x)g(x) dx (1.30)
(ii)1√2π
∫ +∞
−∞|F (ξ)|2dξ =
1√2π
∫ +∞
−∞|f(x)|2dx (1.31)
where bar sign over function signifies complex conjugate of the complexfunctions or absolute value for real functions
Proof (i) By Fourier Inverse transform formula, we have
g(x) =1√2π
∫ +∞
−∞G(ξ)e−iξxdξ
24 An Introduction to Integral Transforms
Taking complex conjugates on both sides, the above equation gives
g(x) =1√2π
∫ +∞
−∞G(ξ) eiξx dξ
Therefore,1√2π
∫ +∞
−∞f(x) g (x)dx
=1√2π
∫ +∞
−∞f(x)
[1√2π
∫ +∞
−∞G(ξ) eiξxdξ
]dx
=1√2π
∫ +∞
−∞G(ξ)
[1√2π
∫ +∞
−∞f(x)eiξxdx
]dξ
=1√2π
∫ +∞
−∞G(ξ)F (ξ) dξ , which proves the result (i).
(ii) In the result (i) if we take g(x) = f(x), we get
1√2π
∫ +∞
−∞f(x) f(x) dx =
1√2π
∫ +∞
−∞F (ξ) F (ξ) dξ
i.e,1√2π
∫ +∞
−∞|f(x)|2dx =
1√2π
∫ +∞
−∞|F (ξ)|2 dξ
This proves the part (ii) of the Parseval’s relations.
Note : In the above parseval’s relations or identities one may drop thefactors 1√
2πfrom either sides of (1.30) and (1.31)
There exists other four Parseval’s identities given by
(iii)
√2π
∫ ∞
0Fc(ξ)Gc(ξ)dξ =
√2π
∫ ∞
0f(x)g(x)dx (1.32)
(iv)
√2π
∫ ∞
0Fs(ξ)Gs(ξ)dξ =
√2π
∫ ∞
0f(x)g(x)dx (1.33)
(v)
√2π
∫ ∞
0|Fc(ξ)|2dξ =
√2π
∫ ∞
0|f(x)|2dx (1.34)
and (vi)
√2π
∫ ∞
0|Fs(ξ)2dξ =
√2π
∫ ∞
0|f(x)|2dx (1.35)
connecting to Fourier cosine and sine transforms. Where again the con-
stant terms√
2π may be dropped from either sides in the above results.
Their proofs can similarly be deduced as were done in cases (i) or (ii)above.
Fourier Transform 25
Example 1.10. Let f(x) = e−bx , g(x) = e−ax. Use parseval’s relationof Fourier cosine transform to evaluate the integral to prove∫ ∞
0
dx
(a2 + x2)(b2 + x2)=
π
2ab(a + b)
Solution.
We have Fc [f(x) ; x → ξ] =
√2π
∫ ∞
0e−bx cos ξ x dx
=
√2π
b
b2 + ξ2≡ Fc(ξ)
Similarly, Gc(ξ) =
√2π
a
a2 + ξ2
Therefore, by the Parseval’s relation (iii) above, we get√2π
∫ ∞
0
2π
ab
(ξ2 + a2)(ξ2 + b2)(ξ2 + b2)dξ =
√2π
∫ ∞
0e−(a+b)xdx
i.e,∫ ∞
0
dξ
(ξ2 + a2)(ξ2 + b2)=
π
2ab(a + b).
Example 1.11. Let f(x) =
{1 , 0 < x < a
0 , x > a
Use Parseval′s identity to evaluate∫ ∞
0
sin2 ax
x2dx .
Solution. We have from example 4 of article 1.9 that
Fc[f(x) ; x → ∞] =
√2π
sin a ξ
ξ.
Then by the Parseval’s identity (1.34) of the corollary above√2π
∫ ∞
0
2π
∣∣∣∣sin a ξ
ξ
∣∣∣∣2
dξ =
√2π
∫ ∞
0|f(x)|2dx
i.e,∫ ∞
0
sin2 a ξ
ξ2dξ =
π
2
∫ a
0dx =
πa
2
Example 1.12. Taking g(x) = e−ax and f(x) =
{1 , 0 < x < b
0 , x > b
26 An Introduction to Integral Transforms
and using proper Parseval’s relation prove that
∫ ∞
0
sin at
t(a2 + t2)dt =
π
21 − e−a2
a2
Solution. By Fourier cosine transform
Gc(ξ) = Fc[g(x);x → ξ] =a
ξ2 + a2and Fc[f(x);x → ξ] =
sin aξ
ξ= Fc(ξ)
Then by the Parseval’s relation connecting Fourier cosine transform weget
∫ ∞
0
a sin a ξ
ξ(ξ2 + a2)dξ =
∫ a
0e−ax · 1 dx =
1 − e−a2
a
⇒∫ ∞
0
sin at
t(t2 + a2)dt =
1 − e−a2
a2
1.13 Fourier Transform of the derivative of a function.
In many applications of the theory of Fourier transform to boundaryvalue problems of Mathematical Physics it is necessary to express FT
of the derivative of a function f(x) in terms of FT of the function f(x).To this direction we now denote FT of dnf(x)
dxn by F (n)(ξ), say. Then, weget
F (n)(ξ) =−1√2π
∫ +∞
−∞(iξ)
dn−1f(x)d xn−1
eiξxdx +[
1√2π
dn−1f(x)d xn−1
· eiξx
]+∞
−∞,
which is obtained after integrating by parts the right hand side of thedefinition of
F (n)(ξ) =1√2π
∫ +∞
−∞
dnf(x)dxn
eiξx dx (1.36)
If we assume that dn−1f(x)dxn−1 tends to zero as |x| → ∞, the above result
takes the formF (n) = −iξF (n−1)
By repeated application of this rule and by the assumption
lim|x|→∞
[drf(x)
dxr
]= 0 , r = 1, 2, · · · n − 1,
Fourier Transform 27
we have finallyF (n) = (−iξ)n F
This implies that FT of the nth derivative of a function f(x) is (−iξ)n
times FT of the function, provided that the first (n − 1) derivative ofthe function vanish as |x| → ∞.
The corresponding results for Fourier cosine and Fourier sine trans-form of a function are not so simple. For discussion towards this direc-tion, we define F
(n)s and F
(n)c by the equations
F (n)s =
√2π
∫ ∞
0
dnf(x)dxn
sin ξx dx ,
F (n)c =
√2π
∫ ∞
0
dnf(x)dxn
cos ξ x dx (1.37)
Then, integrating by parts the right hand side of the second of the aboveequations in (1.37) we get
F (n)c =
√2π
[dn−1f(x)
dxn−1cos ξ x
]∞0
+ ξ
√2π
∫ ∞
0
dn−1f(x)dxn−1
sin ξ x dx
= −an−1 + ξ F (n−1)s , (1.38)
on the assumption that limx→∞
dn−1f(x)dxn−1
= 0,
limx→0
√2π
dn−1f(x)dxn−1
= an−1
Similarly, from the first of the equation in (1.37) we derive
F (n)s = −ξ F (n−1)
c (1.39)
Using the result (1.39) in (1.38) we have
F (n)c = −an−1 − ξ2 F (n−2)
c (1.40)
Thus, by repeated application of equation (1.40) one can reduce F(n)c to
a sum of ξ’s and either F(1)c of Fc according as n is odd or even integer
respectively. Therefore, we obtain the following formulae :
F (2m)c = −
m−1∑s=0
(−1)sa2m−2s−1 ξ2s + (−1)m ξ2m Fc (1.41)
F (2m+1)c = −
m∑s=0
(−1)sa2m−2s ξ2s + (−1)m ξ2m+1 Fs (1.42)
28 An Introduction to Integral Transforms
Similar results hold for sine transforms too. From (1.39) and (1.40) weobtain,
F (n)s = ξ an−2 − ξ2 F (n−2)
s
From this formula we can derive the formulas
F (2n)s = −
n∑k=1
(−1)kξ2k−1a2n−2k + (−1)n+1ξ2n Fs (1.43)
F (2n+1)s = −
n∑k=1
(−1)k ξ2k−1a2n−2k+1 + (−1)n+1 ξ2n+1Fc (1.44)
Certain special cases of the above formulae arise frequently. For example,if df
dx = d3fdx3 = 0, when x = 0, then
√2π
∫ ∞
0
d2f
dx2cos ξ x dx = −ξ2 Fc
and
√2π
∫ ∞
0
d4f
dx4cos ξ x dx = ξ4 Fc
On the other hand, if f(x) = d2f(x)dx2 = 0, when x = 0, then
√2π
∫ ∞
0
d2f
dx2sin ξ x dx = −ξ2 Fs
and
√2π
∫ ∞
0
d4f
dx4sin ξ x dx = ξ4 Fs
Corollary 1.6. If a function f(x) has a finite discontinuity at a singlepoint x = a ∈ R for simplicity, the above formula for F T of itsderivative has to be modified in the following manner :
We write
F [f ′(x) ; x → ξ] =1√2π
∫ a−0
−∞f ′(x) eiξxdx +
1√2π
∫ ∞
a+0f ′(x) eiξxdx
As before, integrating the right hand side integrals by parts, one gets
F [f ′(x) ; x → ξ] =1√2π
[f(x) eiξx
]a−0
−∞+
1√2π
[f(x)eiξx
]∞a+0
− iξ√2π
∫ a−0
−∞f(x)eiξxdx − iξ√
2π
∫ ∞
a+0f(x)eiξxdx,
Fourier Transform 29
which can be written in the form
F [f ′(x) ; x → ξ] = −iξ F [f(x) ; x → ξ] − 1√2π
eiξa[f(x)]a (1.45)
where [f(x)]a = f(a + 0) − f(a − 0), called the jump of f(x) at x = a .
As a generalisation if we now suppose that f(x) has n points of finitediscontinuities at points x = ai, i = 1, 2, · · · n, the modified form ofeqn.(1.45) is
F [f ′(x);x → ξ] = −iξ F [f(x);x → ξ] − 1√2π
n∑i=1
[f(x)]ai eiξai (1.46)
Corollary 1.7.
We know that f(x) ∈ C1(R), then regarding the function f(x) eiξx
as a function of x and ξ, we see that f(x) eiξx ∈ C1(R×R) and thereforeF (ξ) = 1√
2π
∫ +∞−∞ f(x)eiξxdx is convergent and the integral
i√2π
∫ +∞
−∞xf(x) eiξxdx =
1√2π
∫ +∞
−∞
∂
∂ξ
[f(x)eiξx
]dx
is uniformly convergent. Then
F ′(ξ) = iF [xf(x) ; x → ξ] (1.47)
We can continue the above process any finite number of times to obtain
dr
dξrF (ξ) = irF [xrf(x) ; x → ξ], r = 0, 1, · · · (1.48)
Corollary 1.8.
If a > 0, Fc [f(ax) ; x → ξ] =
√2π
∫ ∞
0f(ax) cos ξ x dx
= a−1
√2π
∫ ∞
0f(x) cos
(ξx
a
)dx
Therefore, Fc[f(ax) ; x → ξ] =1aFc[f(x) ; x → ξ
a], a > 0, (1.49)
Then, Fc[f(x) cos(ωx) ; x → ξ] =12[Fc(ξ + ω) + Fc(ξ − ω)]
(1.50)
and Fc[f(x) sin ω x ; x → ξ] =12[Fs(ω + ξ) + Fs(ω − ξ)]
(1.51)
30 An Introduction to Integral Transforms
Corollary 1.9.
As in corollary 1.8 above, we can easily deduce that
Fs[f(ax) ; x → ξ] = a−1Fs
[f(x) ; x → ξ
a
], a > 0 (1.52)
Fs[f(x) cos ωx ; x → ξ] =12[Fs(ξ + ω) + Fs(ξ − ω)]
and Fs[f(x) sin ωx ; x → ξ] =12[Fc(ξ − ω) − Fc(ξ + ω)]
Corollary 1.10.
In actual calculation of Fourier sine transform the following resultmay be found useful. Let us denote
Fs [f(x) ; x → ξ] = ϕ(ξ) , for ξ > 0
Then, for ξ < 0 we have
−ϕ(η) ≡ −ϕ(−ξ) = Fs [f(x) ; x → ξ] , ξ = −η , η > 0
Therefore, we have in general
Fs[f(x) ; x → ξ] = ϕ(|ξ|) · sgn ξ
where sgn ξ =
{+1 , for ξ > 0−1 , for ξ < 0
1.14 Fourier Transform of some more useful functions.
(a) We require to find Fc[e−axxn−1;x → ξ] and Fs[e−ax xn−1; x → ξ]where a > 0 and n > 0. For this purpose let us assume√
2π
∫ ∞
0e−axxn−1 cos ξ x dx = C
and
√2π
∫ ∞
0e−axxn−1 sin ξ x dx = S
Then, C − iS =
√2π
∫ ∞
0xn−1 · e−∝xdx, where ∝= a + iξ = reiθ, say
=
√2π
∫ ∞
0
1∝n
e−zzn−1dz
Fourier Transform 31
=
√2π
1∝n
Γ(n) ,
=
√2π
Γ(n)rn
· e−inθ
where r =√
a2 + ξ2 and θ = tan−1 ξ
a.
Thus, Fc [xn−1e−ax ; x → ξ] =
√2π
cos n θ · Γ(n)(a2 + ξ2)
n2
and Fs [xn−1e−ax ; x → ξ] =
√2π
Γ(n) · sinnθ
(a2 + ξ2)n2
Deductions :
Since Fc F−1c = I and Fs F−1
s = I, we get
xn−1 e−ax = Fc
[√2π
Γ(n)r−n cos n θ ; ξ → x
]
=2π
Γ(n)∫ ∞
0r−n cos nθ cos ξx dξ
Putting a = 1 in the above equation we get
π
2· xn−1 e−x
Γ(n)=∫ ∞
0(1 + ξ2)−
n2 cos(n tan−1 ξ) cos ξx dξ
As tan θ = ξ ⇒ sec2 θ dθ = dξ , we have
π
2xn−1 e−x
Γ(n)=∫ π
2
0cosn−2 θ cos nθ cos(x tan θ) dθ (1.53)
Similarly,
xn−1 e−x = Fs
[√2π
sin nθ Γ(n)(1 + ξ2)
n2
; ξ → x
]
=2π
∫ ∞
0
sin nθ · Γ(n)(1 + ξ2)
n2
sin ξx dξ
This result gives,
π
2· xn−1e−x
Γ(n)=∫ π/2
0cosn−2 θ · sin nθ · sin(x tan θ) dθ (1.54)
32 An Introduction to Integral Transforms
If we multiply (1.53) by e−x xm−1 and integrate the result withrespect to x from 0 to ∞, we get
π
2Γ(n)
∫ ∞
0e−2xxm+n−2dx =
∫ π/2
0cosn−2 θ · cos nθ · A dθ
where A =∫ ∞
0e−xxm−1 · cos(x tan θ) dx (1.55)
Similarly, multiplying (1.54) by e−x xm−1 and integrating theresult with respect to x from 0 to ∞, we get
π
2Γ(n)
∫ ∞
0e−2xxm+n−2 dx =
∫ π/2
0cosn−2 θ · sin nθ · B dθ
where B =∫ ∞
0e−xxm−1 sin(x tan θ) dx (1.56)
Therefore, from (1.55) and (1.56)
A − iB =∫ ∞
0e−x xm−1 e−ix tan θ dx
=∫ ∞
0e−∝x xm−1 dx, where ∝= 1 + i tan θ =
eiθ
cos θ
=1
∝mΓ(m) = Γ(m) cosm θ e−imθ
This result implies, A = Γ(m) cosm θ cos mθ,B = Γ(m) cosm θ sinmθ.
Again,π
2Γ(n)
∫ ∞
0e−2xxm+n−2dx =
π
2m+n
Γ(m + n − 1)Γ(n)
Therefore,π Γ(m + n − 1)
2m+nΓ(n)=∫ π/2
0cosn−2 θ cos nθ
[Γ(m) cosm θ cos mθ]dθ
⇒ π Γ(m + n − 1)2m+nΓ(m)Γ(n)
=∫ π/2
0cosm+n−2 θ cos mθ cos nθ dθ
Also,π Γ(m + n − 1)2m+nΓ(m)Γ(n)
=∫ π/2
0cosm+n−2 θ sin nθ sinmθ dθ
Adding these two results, we get
π Γ(m + n − 1)2m+n−1 Γ(m)Γ(n)
=∫ π/2
0cosm+n−2 θ cos(m − n)θ dθ
Fourier Transform 33
If we now set m + n = 2, then we get
π
2Γ(n)(1 − n)Γ(1 − n)=
sin nπ
2(1 − n)
Therefore, we get Γ(n)Γ(1 − n) = πsinnπ , the duplication formula
for the Gamma function.(b) We now evaluate Fc
[(a2 − x2)ν−
12 H(a − x) ; x → ξ
]Solution. Following the definition, we have
Fc[(a2 − x2)ν−12 H(a − x) ; x → ξ] =
√2π
∫ a
0(a2 − x2)ν−
12 cos ξxdx
=
√2π· I, say
Now, from the series expansion of cos ξx, we can write
I =∫ a
0(a2 − x2)ν−
12
∞∑r=0
(−1)r(ξx)2r
(2r)!dx
=∞∑
r=0
(−1)rξ2r
(2r)!
∫ a
0(a2 − x2)ν−
12 x2r dx
=∞∑
r=0
(−1)rξ2r
(2r)!· a2ν+2r
2
∫ 1
0zr− 1
2 (1 − z)ν−12 dz,where x2 = a2z
=∞∑
r=0
(−1)rξ2ra2ν+2r
2(2r)!Γ(ν + 1
2 )Γ(r + 12)
Γ(ν + r + 1)
=∞∑
r=0
(−1)rΓ(ν + 12 )
Γ(ν + r + 1)·(
ξ
2
)2r
· a2ν+2r
r!
√π
2
Thus,
Fc
[(a2 − x2)ν−
12 H(a − x) ; x → ξ
]
=
√2π
√π
2
[ ∞∑r=0
(−1)r
Γ(r + 1)Γ(ν + r + 1)
](aξ
2
)2r+ν (ξ
2
)−ν
aνΓ(
ν +12
)
=1√2
Jν (aξ) aν Γ(
ν +12
)(ξ
2
)−ν
=(
a
ξ
)ν
2ν− 12 Γ(
ν +12
)Jν(aξ)
34 An Introduction to Integral Transforms
⇒ Fc
[(a
ξ
)ν
2ν− 12 Jν(aξ) ; ξ → x
]= (a2 − x2)ν−
12 H(a − x)
=
{(a2 − x2)ν−
12 , 0 < x < a
0 , otherwise
Deduction. Putting ν = 0, a = 1 in the above, we get
Fc
[√π
2J0 (ξ); ξ → x
]=
{1√
1−x2, 0 < x < 1
0 , otherwise .
(c) Find Fourier cosine transforms of e−ax cos ax and e−ax sin ax andhence evaluate that of 1
x4+k4
Solution.
Let C = Fc[e−ax cos ax ; x → ξ]
and S = Fc[e−ax sin ax ; x → ξ]
Then, C − iS =
√2π
∫ ∞
0e−αx cos ξ x dx, where α = a(1 + i)
=
√2π
I, say.
So, I =α
ξ2− α
ξ2I
⇒ I =α
α2 + ξ2
Therefore, C − iS =
√2π
a(1 + i)ξ2 + a2(1 + i)2
=
√2π· a(1 + i)(ξ2 − 2a2i)(ξ2 + 2a2i)(ξ2 − 2a2i)
=
√2π
a[(ξ2 + 2a2) + i(ξ2 − 2a2)]ξ4 + 4a4
So, Fc[e−ax cos ax;x → ξ] =
√2π
a(ξ2 + 2a2)ξ4 + 4a4
and Fc[e−ax sin ax;x → ξ] =
√2π
a(−ξ2 + 2a2)ξ4 + 4a4
.
Then, Fc[e−ax(λ cos ax + μ sin ax) ; x → ξ]
=
√2π
a(λ − μ)ξ2 + 2a2(λ + μ)
ξ4 + 4a4
Fourier Transform 35
Now choosing, 1 = λ = μ �= 0, we get
Fc[e−ax(cos ax + sin ax);x → ξ] =
√2π
4a3
ξ4 + 4a4
∴ Fc
[√2π
4a3
ξ4 + 4a4; ξ → x
]= e−ax(cos ax + sin ax)
Thus Fc
[1
ξ4 + 4a4; ξ → x
]=√
π
2e−ax
4a3(cos ax + sin ax)
∴ Fc
[1
ξ4 + k4; ξ → x
]=√
π
2· e
−kx√2√
2k3
(cos
kx√2
+ sinkx√
2
)
Or, Fc
[1
x4 + k4; x → ξ
]=√
π
2· e
−kξ√2√
2k3
(cos
kξ√2
+ sinkξ√
2
).
(d) From Pn(x) =1
2n n!dn
dxn(x2 − 1)n evaluate
F [Pn(x) H(1 − |x|) ; x → ξ]
Solution.
F [Pn(x)H(1 − |x|);x → ξ]
=1√2π
· 12n n!
∫ 1
−1
dn
dxn(x2 − 1)neiξx dx
=(−iξ)√2π2n n!
∫ 1
−1
dn−1
dxn−1(x2 − 1)n · eiξx dx , on integration by parts
=(−iξ)2√2π2n n!
∫ 1
−1
dn−2
dxn−2(x2 − 1)n · eiξx dx
Similarly proceeding after n-th step we get
=1√2π
(−iξ)n
2n n!
∫ 1
−1(x2 − 1)neiξxdx
=inξn
√2π2n n!
∫ 1
−1(1 − x2)neiξxdx
=inξn
√2π2nn!
· 2 ·[∫ 1
0(1 − x2)n cos ξ x dx
]
But from the result of (b) above, after putting a = 1 and ν − 12 = n, we
get
F [Pn(x) H(1 − |x|) ; x → ξ]
36 An Introduction to Integral Transforms
=
√2π
inξn
2n n!·√
π
2ξ−n− 1
2 2nJn+ 12
(ξ) · Γ(n + 1)
=in√ξ
Jn+ 12
(ξ)
1.15 Fourier Transforms of Rational Functions.
Using calculus of residues FT of rational functions can be calculated.Let us assume that the function of a complex variable z admits of thefollowing properties :
(i) f(z) has finite number of singularities at z = a1, a2 · · · an in thehalf plane Im z > 0
(ii) f(z) is analytic everywhere on real z axis except at the pointsb1, b2, · · · bm, which are simple poles and
(iii)∣∣zf(z) eiξz
∣∣→ 0 as |z| → ∞, Im z � 0.
Let us now integrate eiξzf(z), ξ > 0 round the contour shown in theadjoining figure and making use of Cauchy’s residue theorem to obtain
Fig. 1.1
12πi
∫ +∞
−∞f(x)eiξxdx =
12
m∑s=1
res [f(bs)] · eibsξ +n∑
s=1
res [f(as)] · eiξas
This formula yeilds for ξ > 0
F [f(x);x → ξ] =√
2πi
[n∑
s=1
res {f(as)}eiξas +12
m∑s=1
res {f(bs)}eiξbs
]
Fourier Transform 37
To provide an application of the above result we first consider thesimplest example when f(z) = 1
z .
This function has a simple pole at z = 0 and it has no other singu-larities inside and on the contour shown above. Then, for ξ > 0
F [1x
; x → ξ] =√
2πi · 12
res [f(0)] · eiξ.0
= i
√π
2
Again f(x) = 1x is an odd function of x in R. This gives
Fs
[1x
; x → ξ
]=√
π
2, ξ > 0 .
To provide another example, consider f(z) = 1z(z2+a2)
where a is areal number. This function f(z) has simple poles inside the contour asdescribed above at z = 0 and at z = ia having residues a−2 and −1
2a−2
respectively, when ξ > 0. Then, we get
F
[1
x(x2 + a2); x → ξ
]=√
π
21a2
i(1 − e−ξ|a|
)
and from it we have
Fs
[1
x(x2 + a2); x → ξ
]=√
π
21a2
(1 − e−ξ|a|
)sgn(ξ) .
1.16 Other important examples concerning derivative of
FT .
The following example will depict different methods for finding F T ofgiven function using derivative of the transformed function whenevernecessary.
Example 1.13. Find Fourier sine and cosine transform of f(x) = x.
Solution.
Let Fc(ξ) =
√2π
∫ ∞
0x cos ξ x dx
and Fs(ξ) =
√2π
∫ ∞
0x sin ξ x dx
38 An Introduction to Integral Transforms
Thus, Fc(ξ) − i Fs(ξ) =
√2π
∫ ∞
0x e−iξxdx
=
√2π
∫ ∞
0
t e−t
iξ
dt
iξ, where iξx = t
= −Γ(2)ξ2
= − 1ξ2
Equating real and imaginary parts, we get the required result.
Example 1.14. Find f(x), if Fs[f(x);x → ξ] = e−aξ
ξ . Hence, evaluate
F−1s
[1ξ ; ξ → x
].
Solution. we know that
f(x) = F−1s [Fs(ξ)] =
2π
∫ ∞
0
e−aξ
ξsin ξ x dξ (1.57)
Therefore,df (x)dx
=2π
∫ ∞
0e−aξ cos ξ x dξ =
2π
a
a2 + x2
Integrating, we get
f(x) =2aπ
1a
tan−1 x
a+ A , A is an arbitrary constant
Putting x = 0 in the above, we get A = f(0) = 0, from (1.57).
So, f(x) =2π
tan−1 x
a= F−1
s [Fs(ξ)] = F−1s
[1ξ
e−aξ
]
Putting a = 0 in the above, we get
2π
tan−1(x
0
)=
2π· π
2= 1 = F−1
s
(1ξ· 1)
⇒ F−1s
(1ξ
)= 1
Example 1.15. Find Fourier sine transform of f(x) = 1x(x2+a2)
[An alternative method].
Solution. Let I(ξ) = Fs
[1
x(x2 + a2); x → ξ
]
=
√2π
∫ ∞
0
sin ξ x
x(x2 + a2)dx ,
dI
dξ=
√2π
∫ ∞
0
cos ξ x
x2 + a2dx
Fourier Transform 39
Therefore,d2I
dξ2= −
√2π
∫ ∞
0
[1 − a2
x2 + a2
]sin ξ x
xdx
= −√
2π
∫ ∞
0
sin ξ x
xdx + a2 I(ξ)
= −√
π
2+ a2 I(ξ)
Solving this differential equation, we get
I(ξ) = A eaξ + B e−aξ +√
π
21a2
⇒ dI(ξ)dξ
= Aa eaξ − Ba e−aξ
Taking ξ = 0, Aa − Ba =
√2π
π
2· 1a
=√
π
2· 1a
⇒ A − B =√
π
2· 1a2
Also, A + B = −√
π
2· 1a2
Solving we get, A = 0 and B = −√
π
2· 1a2
∴ Fs
[1
x(x2 + a2)
]= −
√π
21a2
e−aξ +√
π
2· 1a2
=√
π
21a2
(1 − e−aξ
)
i.e
√2π
∫ ∞
0
sin ξ x
x(x2 + a2)dx =
√π
2· 1a2
(1 − e−aξ
)
i.e∫ ∞
0
sin ξ x
x(x2 + a2)dx =
π
2a2
(1 − e−aξ
).
Example 1.16. Find Fourier cosine inverse of 11+ξ2 .
Solution.
Solution. Let f(x) = F−1c
[1
1 + ξ2; ξ → x
]
=
√2π
∫ ∞
0
cos ξ x
1 + ξ2dξ
∴ df (x)dx
= −√
2π· π
2+
√2π
∫ ∞
0
sin ξ x dξ
ξ(1 + ξ2)
40 An Introduction to Integral Transforms
= −√
π
2e−x .
Thus, f(x) =√
π2 e−x, after evaluating the constant of integration by
putting x = 0 there.
Example 1.17. Evaluate Fc[f(x) ; x → ξ] and Fc[g(x) ; x → ξ] where
f(x) = x−s, 0 < s < 1, and g(x) =(1 − x2
)ν− 12 H(1 − x),
(ν > −1
2
)
Solution. We find Fc[f(x) ; x → ξ] by evaluating the contour integral∫Γ f(z) eiξz dz, where Γ is a positively described quarter circle |z| = R
with identation at z = 0 and as depicted in the adjoining figure.
Fig. 1.2
It may be noted here that
z · f(z) eiξz −→ 0 as ∈→ 0 ⇒ z → 0
and z · f(z) eiξz −→ 0 as R → ∞, 0 < θ <π
2,where z = Reiθ on C2
Therefore,∫
c1
f(z) eiξz dz −→ 0 as |z| → 0
and∫
c2
f(z) eiξz dz −→ 0 as |z| → ∞
Then by Cauchy’s residue theorem we have∫ ∞
0x−s eiξx dx = e
−πsi2
∫ ∞
0y−se−ξy i dy
since the integrand has no singularity inside the contour depicted aboveand is regular there.
Fourier Transform 41
This gives,∫ ∞
0x−s eiξx dx = i
(cos
πs
2− i sin
πs
2
)ξs−1Γ(1 − s), which implies
Fc[f(x) ; x → ξ] =√
2π
∫∞0 x−s cos ξ x dx
=√
2π sin sπ
2 Γ(1 − s)ξs−1 and
Fs[f(x) ; x → ξ] =√
2π
∫∞0 x−s sin ξ x dx
=√
2π cos sπ
2 · Γ(1 − s) ξs−1
⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭(1.58)
Also, from the result in (b) of article 1.14 we get
Fc [g(x) ; x → ξ]
= Gc(ξ)
= 2ν− 12 Γ(
ν +12
)ξ−ν Jν(ξ)
Then by the result (iii) of article 1.12 we have∫ ∞
0Fc(ξ) Gc(ξ) dξ =
∫ ∞
0f(x) g(x) dx
implying that
2νπ− 12 sin
sπ
2Γ(1 − s) Γ
(ν +
12
)∫ ∞
0ξs−ν−1 Jν(ξ) dξ
=∫ 1
0x−s(1 − x2)ν−
12 dx , ν > −1
2
=12
∫ 1
0y−
s2− 1
2 (1 − y)ν−12 dy
= Γ(
ν +12
)Γ(
12− s
2
)/[2Γ(ν − s
2+ 1)]
Since, π cosec(sπ
2
)= Γ
(s
2
)Γ(1 − s
2
)and Γ
(12
)Γ(1 − s) = 2−s Γ
(12− s
2
)Γ(1 − s
2
)
we have∫ ∞
0ξs−ν−1 Jν(ξ) dξ =
2s−ν−1 Γ(
s2
)Γ(ν − s
2 + 1) , 0 < s < 1, ν > −1
2.
42 An Introduction to Integral Transforms
Similarly another pair of formulae∫ ∞
0Fs(ξ)Gc(ξ) sin(ξx) dξ
=12
∫ ∞
0f(u)[g(|u − x|) − g(u + x)] du
=12
∫ ∞
0g(u)[f(x + u) − f(u − x)] du
can also be used for the derivation of∫ ∞
0Fs(ξ)Gc(ξ) dξ =
∫ ∞
0f(x)g(x) dx
in this case.
Example 1.18. Find Fourier sine and cosine transforms of the function
f(x) =(eax + e−ax
)/(eπx − e−πx
)=
ch ax
sh πx.
Solution. Fs [f(x) ; x → ξ] =
√2π
∫ ∞
0
ch ax
sh πxsin ξ x dx
=
√2π
∫ ∞
0
ch ax
sh πx· eiξx − e−iξx
2idx
=
√2π· 12i
∫ ∞
0
[{e(a+iξ)x − e−(a+iξ)x
}eπx − e−πx
−{e(a−iξ)x − e−(a−iξ)x
}eπx − e−πx
]dx
=
√2π· 12i
[12
tana + iξ
2− 1
2tan
a − iξ
2
],by use of tables of integrals
=
√12π
sh ξ
cos a + ch ξ, on simplification.
Similarly, Fc [f(x) ; x → ξ] =
√2π
∫ ∞
0
ch ax
sh πxcos ξ x dx
=
√2π
12
∫ ∞
0
[{e(a+iξ)x + e−(a+iξ)x
}eπx − e−πx
+
{e(a−iξ)x + e−(a−iξ)x
}eπx − e−πx
]dx
=1√2π
[sec
a + iξ
2+ sec
a − iξ
2
], from tables of integrals
=1√2π
cos a2 ch ξ
2
cos a + ch ξ.
Fourier Transform 43
Example 1.19. Show that∫ ∞
0
cos px
1 + p2dp =
π
2e−x , x � 0 ,
Solution. We know that∫ ∞
0e−x cos px dx =
11 + p2
This means that
Fc[e−x ; x → ξ] =
√2π
11 + ξ2
Therefore, by inverse Fourier cosine transform formula we get
F−1c
[√2π
11 + ξ2
; ξ → x
]= e−x
⇒√
2π
∫ ∞
0
√2π
11 + ξ2
cos ξ x dξ = e−x
Or,∫ ∞
0
cos ξ x
1 + ξ2dξ =
π
2· e−x .
Example 1.20. Find Fourier sine and Fourier cosine transforms of thefunction
f(x) =
{sin x , 0 < x < a
0 , x > a
Solution. Fs [ f(x) ; x → ξ ]
=
√2π
∫ a
0sin x sin ξx dx
=1√2π
∫ a
0[cos(1 − ξ)x − cos(1 + ξ)x]dx
=1√2π
[sin(1 − ξ)a
1 − ξ− sin(1 + ξ)a
1 + ξ
]
Fc [f(x) ; x → ξ]
=
√2π
∫ a
0sin x cos ξ x dx
=1√2π
∫ a
0[sin(1 + ξ)x + sin(1 − ξ)x]dx
44 An Introduction to Integral Transforms
=1√2π
[−cos(1 + ξ)x
1 + ξ− cos(1 − ξ)x
1 − ξ
]a
0
=1√2π
[− cos(1 + ξ)a + 11 + ξ
+− cos(1 − ξ)a + 1
1 − ξ
]
=
√2π
[sin2(1 + ξ)a
2
1 + ξ+
sin2(1 − ξ)a2
1 − ξ
].
Example 1.21. Find f(x), if its Fourier sine transform is√
2π
ξ1+ξ2
Solution. By question, f(x) = F−1s
[√2π
ξ1+ξ2 ; ξ → x
]
Therefore, f(x) =2π
∫ ∞
0
ξ
1 + ξ2sin ξ x dξ
=2π
∫ ∞
0
[sin ξx
ξ− sin ξx
ξ(1 + ξ2)
]dξ
= 1 − 2π
∫ ∞
0
sin ξx
ξ(1 + ξ2)dξ (i)
∴ df
dx= − 2
π
∫ ∞
0
cos ξx
1 + ξ2dξ (ii)
⇒ d2f
dx2=
2π
∫ ∞
0
ξ sin ξx
1 + ξ2dξ = f(x)
⇒ f(x) = A ex + B e−x (iii)
But f(0) = 1 = A + B , by (i) and (ii) above.
Also since[
df
dx
]x=0
= −1 = A − B , by (ii) and (iii)
∴ A = 0 and B = 1
Hence, f(x) = e−x .
Example 1.22. Using Fourier transform evaluate the following integralsto prove that
(a)∫ +∞
−∞
dx
(x2 + a2)(x2 + b2)=
π
ab(a + b), a > 0 , b > 0
(b)∫ ∞
0
x−pdx
a2 + x2=
π
2a−(p+1) sec
(πp
2
)
(c)∫ ∞
0
x2 dx
(x2 + a2)(x2 + b2)=
π
2(a + b), a > 0 , b > 0
(d)∫ ∞
0
x2
(x2 + a2)4dx =
π
(2a)5, a > 0
Fourier Transform 45
Solution.
(a) Let us take f(x) = e−a|x| and g(x) = e−b|x|
Then F (ξ) =
√2π
a
ξ2 + a2, G(ξ) =
√2π
b
ξ2 + b2
Therefore,∫ +∞
−∞F (ξ) G(ξ) eiξxdξ =
∫ +∞
−∞f(t) g(x − t) dt
Hence, by putting x = 0 on both sides we get∫ +∞
−∞F (ξ) G(ξ) dξ =
∫ +∞
−∞f(t) g(−t) dt
⇒∫ +∞
−∞
dξ
(ξ2 + a2)(ξ2 + b2)=
π
2ab
∫ +∞
−∞e−|t|(a+b) dt
=π
ab
∫ ∞
0e−(a+b)tdt
=π
ab(a + b)
(b) Taking f(x) = e−ax, g(x) = xp−1 we have
Fc(ξ) = Fc
[e−ax ; x → ξ
]=
√2π
a
a2 + ξ2
and Fc
[xp−1 ; x → ξ
]=
√2π
ξ−p Γ(p) cosπp
2= Gc(ξ)
By using the Parseval’s relation for Fourier cosine transform weget ∫ ∞
0Fc(ξ) Gc(ξ) dξ =
∫ ∞
0f(x) g(x) dx
Hence,2aπ
cosπp
2Γ(p)
∫ ∞
0
ξ−p dξ
ξ2 + a2
=∫ ∞
0xp−1 · e−ax dx
=Γ(p)ap
, after putting ax = t
Therefore,∫ ∞
0
ξ−p
a2 + ξ2dξ =
π
2ap+1sec
πp
2.
46 An Introduction to Integral Transforms
(c) We know that
Fs
[e−ax ; x → ξ
]=
√2π
ξ
ξ2 + a2= Fs(ξ), say
Fs
[e−bx ; x → ξ
]=
√2π
ξ
ξ2 + b2= Gs(ξ), say,
if we choose f(x) = e−ax and g(x) = e−bx.
Then by convolution theorem we have∫ ∞
0Fs(ξ) Gs(ξ) cos ξ x dξ =
12
∫ ∞
0g(ξ) [f(ξ + x) + f(ξ − x)] dξ
Putting x = 0, we get
2π
∫ ∞
0
ξ2 dξ
(ξ2 + a2)(ξ2 + b2)
=∫ ∞
0e−(a+b)ξ dξ
=1
a + b
⇒∫ ∞
0
ξ2 dξ
(ξ2 + a2)(ξ2 + b2)=
π
2(a + b).
(d) By convolution theorem we have∫ +∞
−∞f(x) f(x) dx =
∫ +∞
−∞F (ξ) F (ξ) dξ as the Parseval′s relation,
where F (ξ) is the Fourier transform of f(x).
Let us now write f(x) =1
2(x2 + a2)⇒ f ′(x) = − x
(x2 + a2)4
and F [f(x) ; x → ξ] = F (ξ) =√
π
212a
exp (−a|ξ|) .
Hence,∫ +∞
−∞|f ′(x)|2 dx =
∫ +∞
−∞|F{f ′(x) ; x → ξ}|2 dξ
=∫ +∞
−∞|(iξ) · 1
2a
√π
2exp (−a|ξ|)|2 dξ
⇒∫ +∞
−∞
x2 dx
(x2 + a2)4=
π
22
4a2
∫ ∞
0ξ2 exp (−2aξ) dξ
=2π
(2a)5.
Fourier Transform 47
1.17 The solution of Integral Equations of Convolution
Type.
An integral equation is an equation in which the unknown function ap-pears under the integral sign. The most general type of integral equationcan, in general, be broadly classified into two types - Fredholm integralequation where limits of integral are constants and Volterra integralequation in which one of the limits of integration is variable.
The simplest integral equation that can be tackled by means ofFourier transform technique is the integral equation of convolution type
1√2π
∫ +∞
−∞f(t) k(x − t) dt = g(x) , −∞ < x < ∞ (1.59)
Here k(x) , g(x) are prescribed functions for all real values of x and areassumed to possess Fourier transforms K(ξ) and G(ξ) respectively.
Taking F T of each side of (1.59) and making use of the convolutionproperty we obtain formally the relation
F (ξ) K(ξ) = G(ξ) ⇒ F (ξ) = G(ξ) L(ξ), (1.60)
where L(ξ) =1
K(ξ)·
Now, if F−1[L(ξ) ; ξ → x] = l(x)
exists, then applying the convolution theorem we would obtain the so-lution of (1.60) and hence from (1.59) that
f(x) =1√2π
∫ +∞
−∞g(t) l(x − t) dt (1.61)
The main difficulty in the above procedure will arise if l(x), as definedabove, does not exist. However if it so happens that for some positiveinteger n, the inverse transform
F−1
[1
(−iξ)n K(ξ); ξ → x
]= m(x) (1.62)
exists, then (1.60) can be expressed as
F (ξ) = (−iξ)n G(ξ) · M(ξ) ,where M(ξ) = F [m(x) ; x → ξ]
48 An Introduction to Integral Transforms
which finally yeilds
f(x) =1√2π
∫ +∞
−∞m(x − t) g(n)(t) dt , (1.63)
where, g(n)(t) =dn
dtng(t) .
As an example, we consider the integral equation
1√2π
∫ +∞
−∞|x − t|− 1
2 f(t) dt = g(x) (1.64)
Here, k(x) = |x|− 12 and hence K(ξ) = |ξ|− 1
2 , from (1.58) of Example1.17 of article 1.16. Now, since
F−1
[1
(−iξ) K(ξ); ξ → x
]
= F−1[i sgn(ξ) |ξ|− 1
2 ; ξ → x]
= |x|− 12 sgn(x)
the solution of the integral equation in (1.63) is given by
f(x) =1√2π
∫ x
−∞
g′(t) dt√x − t
− 1√2π
∫ ∞
x
g′(t) dt√t − x
.
In cases in which the inverse transform (1.62) does not exist it is some-times possible to have a solution of eqn. (1.59), provided that the func-tion g(x) has derivatives of all orders. The corresponding method dueto Eddington consists essentially in assuming the solution of eqn. (1.59)in the form
f(x) =∞∑
n=1
Cn g(n)(x) (1.65)
where Cn (n = 0, 1, 2, · · ·) are constants to be determined. For such atype of solution we have
F [f(x) ; x → ξ] = F (ξ) =∞∑
n=0
Cn(−iξ)n G(ξ),
provided F T of all the derivatives of g(x) exist. Substituting this formof F (ξ) in eqn. (1.60) we get
[K(ξ)]−1 =∞∑
n=0
Cn(−iξ)n ,
Fourier Transform 49
so that if 1/K(ξ) can be expanded as a Maclaurin series in the neighbour-hood of ξ = 0, then (−i)nCn is the coefficient of ξn in that expansion.
If there is an Eddington solution of the form (1.65), then Ci’s areuniquely determined from k(x) ; but there exists still some kernels k(x)which exclude, for any infinitely differentiable function g(x), all possi-bility that an Eddington solution exists.
For the case in which Eddington solution exists, we consider theintegral equation
1σ√
2π
∫ +∞
−∞f(t)e−
12(x−t)2/σ2
dt = g(x),
where gn(x) exists for all n ∈ I. Here
k(x) = σ−1e−12x2/σ2 ⇒ K(ξ) = e−
12ξ2σ2
.
Assuming the solution due to Eddington we have∞∑
n=0
Cn(−iξ)n = e12ξ2σ2
.
This equation uniquely determines Cn. In fact, here
C2m−1 = 0
and C2m = (−1)m12σ2m
m!.
Then the solution of the integral equation is
f(x) =∞∑
m=0
(−1)m[
12 σ2m
m!
]· g2m (x) .
Example 1.23. Solve the integral equation∫ ∞
0f(x) cos ξ x dx =
{(1 − ξ) , 0 ≤ ξ ≤ 10 , ξ > 1
Solution. By definition, we can have
Fc[f(x) ; x → ξ] = Fc(ξ)
=
√2π
∫ ∞
0f(x) cos ξ x dx
=
{ √2π (1 − ξ) , 0 ≤ ξ ≤ 1
0 , ξ > 1
50 An Introduction to Integral Transforms
Therefore, taking inverse Fourier cosine transform we have√2π
∫ ∞
0Fc [f(x) ; x → ξ] cos ξ x dξ = f(x)
⇒ f(x) =
√2π·√
2π
∫ 1
0(1 − ξ) cos ξ x dξ
=2π
∫ 1
0(1 − ξ) cos ξ x dξ
=2π
[(1 − ξ)
sin ξ x
x
∣∣∣∣1
0
+∫ 1
0
sin ξx
xdξ
]
=2π
[−cos ξ x
x2
∣∣∣∣1
0
]
=2(1 − cos x)
πx2.
Example 1.24. Solve the integral equations
(i)∫ ∞
0f(x) cos λx dx = e−λ
and (ii)∫ ∞
0f(x) sin λx dx = e−λ
Solution. (i) By definition of F.C. transform, we have
Fc[f(x) ; x → ξ] = Fc(ξ) =
√2π
∫ ∞
0f(x) cos ξx dx
=
√2π
e−ξ , by the problem (i) .
Thus by Inverse transform, we get
f(x) =2π
∫ ∞
0e−ξ cos ξ x dξ
=2π
[e−ξ
1 + x2(− cos ξ x + x sin ξ x)
∣∣∣∣∞
0
]
=2π
[1
1 + x2
]
(ii) Fs[f(x) ; x → ξ] = Fs(ξ) =
√2π
∫ ∞
0f(x) sin ξ x dx
=
√2π
e−ξ , by the given question
Fourier Transform 51
Then on taking the inverse Fourier sine transform, we get
f(x) =2π
∫ ∞
0e−ξ sin ξ x dξ
=2π
[x
1 + x2
], by evaluating the integral.
Example 1.25. Solve the integral equation
∫ ∞
0f(x) sin ξ x dx = F (ξ) =
⎧⎪⎨⎪⎩
1 , 0 � ξ < 12 , 1 � ξ < 20 , ξ � 2 .
Solution. As in example (1.23),we can have
f(x) =2π
∫ ∞
0F (ξ) sin ξ x dξ
=2π
[∫ 1
0sin ξ x dξ +
∫ 2
12 sin ξ x dξ
]
=2π
[− cos ξ x
x
∣∣∣∣1
0
− 2 cos ξ x
x
∣∣∣∣2
1
]
=2π
[1 − cos x
x+
2x
(cos x − cos 2x)]
=2
πx[1 + cos x − 2 cos 2x]
Example 1.26. Solve the following Fredholm integral equation withconvolution type kernel∫ +∞
−∞f(t) k(x − t) dt + λ f(x) = u(x)
for the particular cases
(i) k(x) =1x
(ii) k(x) =12
x
|x| , λ = 1
(iii) k(x) = f(x) , λ = 0 and u(x) =1
x2 + a2
(iv) k(x) =1
x2 + a2, λ = 0 and u(x) =
1x2 + b2
(v) k(x) = 4 e−a(x) , λ = 1 and u(x) = e−|x|
52 An Introduction to Integral Transforms
Solution. Fourier transform of the integral equation gives√
2π F (ξ) K(ξ) + λ F (ξ) = U(ξ)
Or, F (ξ) = U(ξ)/[√
2π K(ξ) + λ]
Inverting the Fourier transformed function we get
f(x) =1√2π
∫ +∞
−∞
U(ξ) eiξx
[√
2πK(ξ) + λ]dξ
(i) Since k(x) = 1x , we have K(ξ) = i
√π2 sgn ξ .
Therefore, the solution of the particular integral equation is
f(x) =1√2π
∫ +∞
−∞
U(ξ) eiξx
λ + i π sgn (ξ)dξ
(ii) If k(x) = 12
x|x| , K(ξ) = 1√
2π1iξ , then the solution of the particular
integral equation is given by
f(x) =1√2π
∫ +∞
−∞iξ
U(ξ) eiξx
(1 + iξ)dξ
=1√2π
∫ +∞
−∞F [u′(x);x → ξ] F [
√2π e−x;x → ξ] eiξx dξ
= u′(x) ∗√
2π e−x =∫ +∞
−∞u′(ξ) exp (ξ − x) dξ
(iii) In this particular case we have
F (ξ) =1√2a
exp{−1
2a |ξ|
}
Therefore, f(x) =1√2π
1√2a
∫ +∞
−∞exp
{iξx − 1
2a |ξ|
}dξ
=1
2√
πa
[∫ ∞
0exp{−ξ(a
2+ ix
)}dξ
+∫ ∞
0exp{−ξ(a
2− ix
)}dξ
]
=√
a
π
24x2 + a2
Fourier Transform 53
(iv) For this particular case Fourier transform of the integral equationis
√2π F (ξ) F
[1
x2 + a2; x → ξ
]=√
π
2e−b|ξ|
b
Or√
2π F (ξ)√
π
2e−a|ξ|
a=√
π
2e−b|ξ|
b
Thus, F (ξ) =1√2π
a
bexp {−|ξ|(b − a)}
Inverting the Fourier transformed function we get
f(x) =a
2πb
[∫ ∞
0exp[−ξ{(b − a) + ix}] dξ
+∫ ∞
0exp[−ξ{(b − a) − ix}]dξ
]
=a
2πb
[1
(b − a) + ix+
1(b − a) − ix
]
=a
πb
[b − a
(b − a)2 + x2
]
(v) In this case Fourier transform of the integral equation is
F (ξ) =a2 + ξ2
a2 + ξ2 + 8aU(ξ)
Inverting, f(x) =1√2π
∫ +∞
−∞(a2 + ξ2)U(ξ)a2 + ξ2 + 8a
eiξx dξ
If u(x) = e−|x|, a = 1, U(ξ) =
√2π
11 + ξ2
Therefore, f(x) =1π
∫ +∞
−∞
eiξx
ξ2 + 32dξ
=13
exp {−3|x|} ,
after separately considering the cases x > 0 and x < 0 and thencombining them.
1.18 Fourier Transform of Functions of several variables.
Similar results like those in article 1.13, FT of the functions of severalvariables do hold. For simplicity, we state them for a function f(x, y)
54 An Introduction to Integral Transforms
of two independent variables. The extension to a higher number ofvariables is obvious. These extensions are sometimes called multipleFourier transforms.
Let us define
F [f(x, y) ; x → ξ] = F [ξ, y] =1√2π
∫ +∞
−∞f(x, y) eiξx dx
Fc[f(x, y) ; x → ξ] = Fc[ξ, y] =
√2π
∫ ∞
0f(x, y) cos ξ x dx
and Fs[f(x, y) ; x → ξ] = Fs[ξ, y] =
√2π
∫ ∞
0f(x, y) sin ξ x dx
The corresponding inversion formulae are then given by
F−1[F (ξ, y) ; ξ → x] = F−1(ξ, y) =1√2π
∫ +∞
−∞F (ξ, y) · e−iξx dξ,
F−1[Fc(ξ, y) ; ξ → x] = F−1c (ξ, y) =
√2π
∫ ∞
0Fc(ξ, y) cos ξ x dξ
and F−1[Fs(ξ, y) ; ξ → x] = F−1s (ξ, y) =
√2π
∫ ∞
0Fs(ξ, y) sin ξ x dξ
respectively.
The analogues of the results in 1.13 for continuous functions are
F
[∂f
∂x; x → ξ
]= −iξ F (ξ, y)
Fc
[∂f
∂x; x → ξ
]= −f(0, y) + ξ Fs(ξ, y)
Fs
[∂f
∂x; x → ξ
]= −ξ Fc(ξ, y)
Applying these results in succession we find that
F
[∂2f
∂x2; x → ξ
]= −ξ2 F (ξ, y)
Fc
[∂2f
∂x2; x → ξ
]= −ξ2 Fc(ξ, y) − ∂f
∂x(0, y)
Fs
[∂2f
∂x2; x → ξ
]= −ξ2 Fs(ξ, y) + ξ f(0, y)
Similarly, FT of other higher order partial derivatives with respect tocorresponding variables can be deduced. These results will help us in
Fourier Transform 55
reducing partial differential equation to a lower dimensional equation ofvariables. Thus, FT can be used in solving the boundary value problemsin two or higher dimensions.
The above ideas can be extended to several variables. Let f(x, y)be a function of two independent variables x, y. Temporarily treatingf(x, y) as a function of x, we have
F [f(x, y) ; x → ξ] = f(ξ, y) =1√2π
∫ +∞
−∞f(x, y) eiξx dx
Then treating f(ξ, y) as a function of the independent variable y, its FT
is given by
f=
(ξ, η) =1√2π
∫ +∞
−∞f(ξ, y) eiηy dy = F [f(ξ, y) ; y → η]
Thus, finally we have
f=
(ξ, η) =12π
∫ +∞
−∞
∫ +∞
−∞f(x, y) ei( ξx+ηy ) dx dy
and the corresponding inversion formula is
f(x, y) =12π
∫ +∞
−∞
∫ +∞
−∞f=
(ξ, η) e−i(ξx+ηy) dξ dη .
Such extensions to Fourier sine and Fourier cosine transforms in func-tions of several variables do exist and the corresponding results can bededuced easily.
Using the above definitions of FT to functions of several variables,we can apply FT to mixed partial derivatives too arising in partial dif-ferential equations in solving boundary value problems.
1.19 Application of Fourier Transform to Boundary Value
Problems
We first discuss the use of sine and cosine transform and finally discussthe use of complex Fourier transform arising in boundary value problems.
The sine and cosine transforms can be applied when the range ofvariable selected for exclusion is from 0 to ∞. The choice of sine and
56 An Introduction to Integral Transforms
cosine transform is decided by the form of the boundary conditions atthe lower limit of the variable selected for exclusion.
For example,
Fs
[∂2u(x, y)
∂x2;x → ξ
]=
√2π
∫ ∞
0
∂2u
∂x2sin ξ x dx
=
√2π
ξ
∫ ∞
0
∂u
∂xcos ξ x dx
=
√2π
ξ u(0, y) −√
2π
ξ2 us(ξ, y), (1.66)
provided u(x, y) is known at x = 0 and∂u
∂x→ 0 as x → ∞
Similarly,
Fc
[∂2u(x, y)
∂x2; x → ξ
]=
√2π
∫ ∞
0
∂2u
∂x2cos ξ x dx
= −√
2π
[∂u(x, y)
∂x
]x=0
−√
2π
ξ2
∫ ∞
0u(x, y) cos ξ x dx, (1.67)
provided∂u(0, y)
∂xis known and u,
∂u
∂x→ 0 as x → ∞.
Noting carefully the results in (1.66) and (1.67) it may be seen thatremoving a term ∂2u(x,y)
∂x2 from a partial differential equation requires theknowledge of u(0, y) for use of a sine transform while the use of a cosinetransform for the same purpose requires the knowledge of ux(0, y) to beknown.
It may be noted that a term ∂u∂x or any odd order partial derivative
can not be removed with the help of sine or cosine transforms.
Again, complex Fourier transform will be found useful for the samepurpose as above, if the range of variable is from −∞ to +∞ of thepartial differential equation.
Example 1.27. The temperature u(x, t) of a semi-infinite rod is deter-mined by the partial differential equation ∂u
∂t = ∂2u∂x2 , x > 0 , t > 0
subject to the initial condition u(x, 0) = 1 , 0 < x < 1
= 0 , x > 1
and the boundary condition u(0, t) = 0. Find the temperature at anytime t at any distance x from x = 0.
Fourier Transform 57
Solution. Since the variable x varies from 0 to ∞ and since thevalue of u(x, t) at x = 0 is prescribed, Fourier sine transform of bothsides of the partial differential equation is to be taken on the variablex which is to be excluded in the transformed equation. Then the givenequation becomes
d
dtus(ξ, t) =
√2π
∫ ∞
0
∂2u
∂x2sin ξx dx
=
√2π
[ξ u(0, t)] − ξ2 us(ξ, t)
= −ξ2 us(ξ, t)
∴ us(ξ, t) = c e−ξ2t, where c is an arbitrary constant . (i)
It is given that initially
u(x, 0) = 1 , 0 < x < 1
= 0 , x > 1
∴ us(ξ, 0) =
√2π
∫ ∞
0u(x, 0) · sin ξ x dx
=
√2π
∫ 1
0sin ξ x dx =
√2π
[− cos ξ x
ξ
]10
=
√2π
[1 − cos ξ
ξ
](ii)
Therefore, from (i) and (ii) we get
c =
√2π
1 − cos ξ
ξ
Using this value of c eqn(i) gives
u(x, t) =2π
∫ ∞
0
1 − cos ξ
ξe−ξ2t sin ξ x dξ.
This is the required temperature function u(x, t).
Example 1.28. Solve the diffusion equation ∂u∂t = ∂2u
∂x2 , x > 0, t > 0subject to the boundary condition ux(0, t) = 0 and u(x, t) is bounded.The initial condition is being
u(x, 0) =
{x , 0 ≤ x ≤ 10 , x > 1
58 An Introduction to Integral Transforms
Solution. Since the range of x is from 0 to ∞ and the value ofux(0, t) is prescribed, it is useful to apply Fourier cosine transform toremove the variable x from the PDE.
Thus we get,
d
dtuc(ξ, t) =
√2π
∫ ∞
0
∂2u
∂x2cos ξ x dx
=
√2π
[cos ξ x
∂u
∂x
∣∣∣∣∞
0
+ ξ
∫ ∞
0
∂u
∂xsin ξ x dx
]
= −√
2π
ξ2
∫ ∞
0u(x, t) cos ξ x dx = −ξ2 uc(ξ, t)
∴ uc(ξ, t) = A e−ξ2t ⇒ uc(ξ, 0) = A, arbitrary constant.
Now, uc(ξ, 0) =
√2π
∫ ∞
0u(x, 0) cos ξ x dx =
√2π
∫ 1
0x cos ξ x dx,
using given initial condition
i.e uc(ξ, 0) =
√2π
[sin ξ
ξ+
cos ξ − 1ξ2
]= A
This result gives
uc(ξ, t) =
√2π
[sin ξ
ξ− 1 − cos ξ
ξ2
]. e−ξ2t
Taking inverse Fourier cosine transform the required solution is
u(x, t) =2π
∫ ∞
0
ξ sin ξ − 1 + cos ξ
ξ2e−ξ2t cos ξ x dξ .
Example 1.29. Use Fourier transform to determine the displacementy(x, t) of an infinite vibrating string, given that the string is initially atrest and that the initial displacement is f(x),−∞ < x < ∞. Show thatthe required solution can also be put in the form
y(x, t) =12[f(x + ct) + f(x − ct)]
when the one dimensional wave equation of the vibrating string is givenby
∂2y
∂t2= c2 ∂2y
∂x2, −∞ < x < ∞ , t > 0 .
Fourier Transform 59
Solution. Since the variable x varies from −∞ to ∞, to remove thevariable from the PDE, we apply complex FT to the given equation toobtain
d2y(ξ, t)dt2
=c2
√2π
∫ +∞
−∞
∂2y(x, t)∂x2
eiξxdx
= −c2ξ2y(ξ, t)
Therefore, y(ξ, t) = A cos(cξt) + B sin(cξt) (i)
and yt(ξ, t) = A c ξ sin(cξt) + B c ξ cos(cξt) (ii)
where A and B are two arbitrary constants. Now, the given initialconditions are
y(x, 0) = initial displacement = f(x) (iii)
and∂y
∂t(x, 0) = 0, since initial velocity of vibrating string is zero.
(iv)
From (ii) and (iii) we get
y(ξ, 0) =1√2π
∫ +∞
−∞f(x) eiξx dx = f(ξ) , say (v)
and∂y
∂t(ξ, 0) = yt(ξ, 0) = 0 . (vi)
Then from eqn. (i) we have
y(ξ, 0) = A = f(ξ) , by eqn. (v)
Also, from eqn. (ii) we have
yt(ξ, 0) = B c ξ = 0, by eqn. (vi)
Thus, B = 0 .
Theny(ξ, t) = f(ξ) cos (c ξ t)
Taking inverse FT to the above equation we get
y(x, t) =1√2π
∫ +∞
−∞f(ξ)
[eicξt + e−icξt
2
]e−iξx dξ
60 An Introduction to Integral Transforms
=12
1√2π
∫ +∞
−∞f(ξ) ·
[e−iξ(x−ct) + e−iξ(x+ct)
]dξ
=12
[1√2π
∫ +∞
−∞f(ξ) e−ξ(x−ct)dξ +
1√2π
∫ +∞
−∞f(ξ) · e−iξ(x+ct)dξ
]
⇒ y(x, t) =12
[f(x − ct) + f(x + ct)] , by inverse F T formula.
Thus, the required result is proved.
Example 1.30. The steady state temperature distribution in a semi-infinite solid y > 0 is represented by the two-dimensional Laplace equa-tion
∂2u
∂x2+
∂2u
∂y2= 0, 0 < y < ∞ , −∞ < x < ∞
subject to the boundary conditions
u(x, 0) = 1 , |x| < a
and u(x, 0) = 0 , for |x| > a
Thus the temperature on the insulated surface y = 0 is kept at unityover the strip |x| < a and at zero outside the strip. Then using cosinetransform show that
u(x, y) =1π
[tan−1 a + x
y+ tan−1 a − x
y
],
after assuming the result that∫ ∞
0e−sx sin rx
xdx = tan−1 r
s, (r, s > 0).
Solution. Since the surface y = 0 is kept at a fixed temperature,there is no flow of heat normal to it and hence ∂u
∂x → 0 as x → 0 and asx → ∞. Therefore, taking Fourier cosine transform of the given partialdifferential equation in variable x we get√
2π
∫ ∞
0
∂2u
∂x2cos ξ x dx +
√2π
∫ ∞
0
∂2u
∂y2cos ξx dx = 0
Or, − ξ2 uc(ξ, y) +d2
dy2uc(ξ, y) = 0
Its general solution is
uc(ξ, y) = A eξy + B e−ξy
Fourier Transform 61
But u(x, y) is finite as y → ∞, uc(ξ, y) is also finite and hence we musthave A = 0. Thus
uc(ξ, y) = B e−ξy (i)
Again,
√2π
∫ ∞
0u(x, 0) cos ξ x dx =
√2π
∫ 1
0cos ξ x dx
=sin ξa
ξ·√
2π
Thus uc(ξ, 0) =
√2π
sin ξ a
ξ= B , from eqn. (i)
Hence, we have
uc(ξ, y) =
√2π
sin ξ a
ξe−ξy
Therefore, u(x, y) =2π
∫ ∞
0
sin ξ a
ξ· e−ξy cos ξ x dξ
=1π
∫ ∞
0
e−ξy
ξ[sin(a + x)ξ + sin(a − x)ξ] dξ
=1π
[tan−1 a + x
y+ tan−1 a − x
y
].
Example 1.31. Solve the one dimensional heat conduction problemwhere the temperature of a semi-infinite bar satisfies the PDE
∂u
∂t= k
∂2u
∂x2, t > 0, 0 < x < ∞,
the boundary and initial prescribed conditions are being
u(0, t) = f(t), u(x, 0) = 0 and u(x, t) → 0 as x → ∞.
Solution. Applying Fourier sine transform over the variable x thePDE transforms to
d
dtus(ξ, t) = k ξ
√2π
u(0, t) − k ξ2 us(ξ, t)
Or,d us
dt(ξ, t) = −k ξ2 us(ξ, t) +
√2π
k ξ f(t)
This linear ODE in us(ξ, t) has the solution given by
us(ξ, t) · ekξ2t =
√2π
kξ
∫f(t) ekξ2t dt + A
=
√2π
kξ
∫ t
f(ς) ekξ2ς dς + A (i)
62 An Introduction to Integral Transforms
Again, u(x, 0) = 0 , by given condition and therefore us(ξ, 0) = 0 .
Putting t = 0 in eqn. (i) we get
us(ξ, 0) · 1 =
√2π
kξ
∫ ς=0
f(ς) ekξ2ςdς + A (ii)
Therefore, ekξ2tus(ξ, t) =
√2π
kξ
∫ t
0f(ς) ekξ2ςdς
Or, us(ξ, t) =
√2π
kξ
∫ t
0f(ς) e−kξ2(t−ς)dς
So, u(x, t) = F−1s
[√2π
kξ
∫ t
0f(ς) e−kξ2(t−ς)dς ; ξ → x
]
=
√2π
k
∫ t
0f(ς)F−1
s [ξ e−lξ2; ξ → x] ,where l = k(t − ς)
From the deduction (iii) of example 1.9 we can have the result
F−1s
[ξe−lξ2
; ξ → x]
=1√8l3
x e−x2
4l = g(x, ς), say
Therefore, u(x, t) =
√2π
k
∫ t
0f(ς) g(x, ς) dς,
where g(x, ς) =x(t − ς)−3
√8k3
· e−x2
[4k(t−ς)]
gives the required solution of the problem.
Example 1.32. Find the solution of the Laplace’s equation
∂2u
∂x2+
∂2u
∂y2= 0
in the half-plane y � 0, subject to the boundary conditionu(x, 0) = f(x),−∞ < x < ∞ and the limiting condition u(x, y) → 0 asρ =
√x2 + y2 → ∞.
Solution. Introducing Fourier transform to remove the variable x,we get
d2 u (ξ, y)dy2
− ξ2 u(ξ, y) = 0 (i)
From the given boundary condition we get
u(ξ, 0) =1√2π
∫ +∞
−∞f(x) eiξxdx = F (ξ), say (ii)
Fourier Transform 63
The limiting condition gives u(ξ, y) → 0 , y → ∞ . (iii)
Using (ii) and (iii), the solution of (i) is given by
u(ξ, y) = F (ξ) e−|ξ|y with F (ξ) = F [f(x) ; x → ξ]
Then using the convolution theoem we can invert the above equation toobtain
u(x, y) =1√2π
∫ +∞
−∞f(t) g(x − t) dt (iv)
where g(x) = F−1[e−|ξ|y ; ξ → x
]
=
√2π
y
x2 + y2
Substituting this result in (iv) we get
u(x, y) =y
π
∫ +∞
−∞
f(t)dt
(x − t)2 + y2, y > 0 .
Example 1.33. Derive the solution of Laplace equation in the infinitestrip −∞ < x < ∞, 0 � y � a subject to the boundary conditionsu(x.0) = f(x), u(x, a) = g(x) and limiting condition u(x, y) → 0 asx → ±∞ .
Solution. To remove the variable x, we introduce Fourier transformover the variable x. Then the Laplace equation transforms to
d2
dy2u(ξ, y) − ξ2 u(ξ, y) = 0 (i)
Also, the boundary conditions under F T transform to
u(ξ, 0) = F (ξ) = F [f(x) ; x → ξ] , u(ξ, a) = G(ξ) = F [g(x) ; x → ξ] .
Therefore, the solution of eqn. (i) under transformed boundary conditionis
u(ξ, y) = F (ξ)sh ξ(a − y)
sh ξa+ G(ξ)
sh ξy
sh ξa, 0 < y < a .
By convolution theorem, the required solution of the problem is givenby
u(x, y) =12a
∫ +∞
−∞f(t)K1(x − t , a − y) dt
64 An Introduction to Integral Transforms
+12a
∫ +∞
−∞g(t)K1(x − t , y) dt (ii)
where K1(x, y) =
√2π
a F−1
[sh ξ y
sh ξ a; ξ → x
]
=
√2π
a Fc
[sh ξ y
sh ξ a; ξ → x
],
By calculus of residues, above Fourier cosine transform is evaluated as
K1 (x, y) =sin(πy
a
)ch(
πxa
)+ cos
(πya
)Substituting this result in eqn. (ii), we get
u(x, y) =12a
sin(πy
a
) ∫ +∞
−∞
f(t)dt
ch π (x−t)a − cos πy
a
+12a
sin(πy
a
)∫ +∞
−∞
g(t) dt
ch π (x−t)a + cos πy
a
Example 1.34. Consider the problem as in 1.33 subject to the bound-ary conditions u(x, 0) = f(x) and uy(x, a) = 0. Find the temperaturedistribution u(x, y) of the problem defined by the Laplace equation
∂2u
∂x2+
∂2u
∂y2= 0 , −∞ < x < ∞ , 0 � y � a
Solution. We apply F T over variable x to obtain
u(ξ, y) = F (ξ)chξ(a − y)
ch ξa, 0 � y � a (i)
where F (ξ) =1√2π
∫ +∞
−∞f(x) eiξx dx (ii)
after using the transformed value of the boundary conditionsu(ξ, 0) = F (ξ) and uy(ξ, 0) = 0. Then by convolution theorem we get
u(x, y) =∫ +∞
−∞f(t) K2(x − t , a − y) dt
where K2(x, y) =1√2π
Fc
[ch (ξy)ch (ξa)
; ξ → x
]
Fourier Transform 65
This K2(x, y) may be evaluated by method of contour integral. But thisevaluation is being quite involved, we stop here in persuing it. We onlymention the result for K2(x, y) as
K2 (x, y) =ch(
πx2a
)cos(πy
2a
)ch(
πxa
)+ cos
(πya
)which can also be found from the tables of Integral transform. Then thesolution of the problem is given by
u(x, y) = sin(πy
2a
) ∫ +∞
−∞
f(t) chπ(x−t)2a dt
chπ(x−t)a − cos πy
a
.
Example 1.35. Illustrate the method of solution of the boundary valueproblem defined by the Laplace’s equation
∂2u
∂x2+
∂2u
∂y2= 0 , x � 0 , 0 � y � a
in the semi-infinite strip subjected to the boundary conditions
(i) u(0, y) = 0, 0 � y � a ; u(x, 0) = f(x) , x � 0 ;
u(x, a) = g(x) , x � 0
and (ii) u(x, 0) = f(x), x � 0;ux(0, y) = 0, 0 � y � a;uy(x, a) = 0, x � 0.
Solution. (i) Physical interpretation of this boundary value prob-lem is the distribution of the temperature u(x, y) in the semi-infinitestrip where the faces are maintained at prescribed temperatures e.g.f(x) on y = 0 and g(x) on y = a and zero on x = 0 , 0 � y � a.These conditions tells us that Fourier sine transform over the variablex is found useful in solving the problem as below. Applying the saidtransform to the PDE we get
[d2
dy2− ξ2
]us(ξ, y) = 0 , where us(ξ, y) =
√2π
∫ ∞
0u(x, y) sin ξ x dx
This equation satisfies the boundary conditions on y = 0 and y = a.
Therefore, in 0 ≤ y ≤ a we have
us(ξ, y) = Fs(ξ)sh ξ(a − y)
sh ξa+ Gs(ξ)
sh ξy
sh ξa
66 An Introduction to Integral Transforms
where Fs(ξ) and Gs(ξ) are Fourier sine transforms of f(x) and g(x)respectively. Then proceeding as in example 1.34 we get
us(ξ, y) =1a
√π
2[Fs(ξ)Fc{K1(x, a−y) ; x → ξ}+Gs(ξ)Fc{K1(x, y) ; x → ξ}]
Now, using the convolution theorem the solution of the problem is givenby
u(x, y) =12a
∫ ∞
0f(t){K1(|t − x| , a − y) − K1(t + x , a − y)} dt
+12a
∫ ∞
0g(t){K1(|t − x| , y) − K1(t + x , y)} dt
Solution. (ii) In this case, the given boundary conditions suggestthat use of Fourier cosine transform is found useful over variable x.Hence, the PDE transforms to[
d2
dy2− ξ2
]uc(ξ, y) = 0 ,
where uc(ξ, y) = Fc[u(x, y) ; x → ξ] (i)
The transformed form of the boundary conditions are then
uc(ξ, 0) = Fc(ξ) = Fc[f(x) ; x → ξ]
and ddy uc(ξ, a) = 0
⎫⎪⎬⎪⎭ (ii)
The appropriate solution of eqn.(i) under eqns. (ii) is given by
uc(ξ, y) = Fc(ξ)ch ξ(a − y)
ch ξa=
√2π Fc (ξ) Fc [K2(x , a − y) ; x → ξ]
where K2(x, y) is same as defined in the solution of the example in 1.34.Now making use of convolution theorem we get
u(x, y) =∫ ∞
0f(t)[K2(x + t , a − y) + K2(|x − t| , a − y)] dt
A more explicit expression as in examples 1.34 and 1.35 can similarlybe obtained. But we do not persue it due to lack of space.
Example 1.36. Solve the PDE
∂u
∂t= 2
∂2u
∂x2
Fourier Transform 67
subject to the conditions u(0, t) = 0, u(x, 0) = e−x , x > 0 and u(x, t)is bounded when x > 0 , t > 0 .
Solution. Taking Fourier sine transform on x, we get the PDE as
d
dtus(ξ, t) = 2 ·
√2π
∫ ∞
0
∂2u
∂x2sin ξ x dx
= −2ξ2us(ξ, t)
Integrating, us(ξ, t) = A e−2ξ2t (i)
Now, from the condition u(x, 0) = e−x, we have
us(ξ, 0) =
√2π
ξ
(1 + ξ2)(ii)
Using (ii) in (i) we have A = ξ1+ξ2
√2π and therefore,
us(ξ, t) =
√2π
ξ
1 + ξ2e−2ξ2t
Now, taking inverse Fourier sine transform of the above result the solu-tion of the PDE under given conditions is
u(x, t) =2π
∫ ∞
0
ξ
1 + ξ2e−2ξ2t sin ξ x dξ
Example 1.37. Prove that the steady state temperature distributionu(x, y, ) inside a semi-infinite strip x > 0 , 0 < y < b under the boundaryconditions u(x, 0) = f(x) , 0 < x < ∞ ; u(x, b) = 0 , 0 < x < ∞ andu(0, y) = 0 , 0 < y < b is given by
u(x, y) =2π
∫ ∞
0f(u)du
∫ ∞
0
sh(b − y)ξsh bξ
sin ξ x sin ξ u dξ
Solution. Since u(x, y) is the steady-state temperature, it must satisfythe PDE
∂2u
∂x2+
∂2u
∂y2= 0 (i)
Taking Fourier sine transform over the variable x, eqn.(i) becomes
d2us
dy2− ξ2 us(ξ, y) = 0 , after using the value u(0, y) = 0 .
68 An Introduction to Integral Transforms
Its solution is given by, us(ξ, y) = A ch ξy + B sh(b − y)ξ, (ii)where A, B are arbitrary constants.
Taking Fourier sine transforms of u(x, 0) = f(x) and u(x, b) = 0 weget
us(ξ, 0) =
√2π
∫ ∞
0f(u) sin ξ u dx and us(ξ, b) = 0 .
Using these results in (ii) we obtain, A = 0 and
B =1
sh bξ[us(ξ, 0)] =
√2π
∫ ∞
0
f(u) sin ξ u du
sh bξ
Therefore,
us(ξ, y) =
√2π
sh(b − y)ξsh bξ
∫ ∞
0f(u) sin ξ u du
Then taking inverse Fourier sine transform we get
u(x, y) =2π
∫ ∞
0sin ξ x
[sinh(b − y)ξ
sh bξ
∫ ∞
0f(u) sin ξ u du
]dξ
=2π
∫ ∞
0f(u)du
[∫ ∞
0
sh(b − y)ξsh bξ
sin ξ x sin ξ u dξ
]
Example 1.38. The temperature u(x, t) of a semi-infinite rod0 � x < ∞ satisfies the PDE ∂u
∂t = κ∂2u∂x2 , subject to conditions
u(x, 0) = 0 , x � 0 and ∂u∂x = −λ, a constant, when x = 0 , t > 0.
Solution. Using Fourier cosine transform over the variable x, thePDE transforms to
d uc(ξ, t)dt
=
√2π
κ λ − κ ξ2 uc(ξ, t)
⇒ d uc
dt+ κ ξ2 uc(t) = κ λ
√2π
(i)
using given conditions on the boundary.
Now solving the eqn.(i) we have
uc(ξ, t) =λ
ξ2
√2π
+ A e−κξ2t (ii)
Again, u(x, 0) = 0 ⇒ us(ξ, 0) = 0 (iii)
Fourier Transform 69
Using (iii) in (ii), we get
uc(ξ, t) =
√2π
λ
ξ2
[1 − e−κξ2t
]
Therefore, by inverse cosine transform we get the required temperature
u(x, t) =2λπ
∫ ∞
0
cos ξx
ξ2(1 − e−κξ2t) dξ .
Example 1.39. Solve the PDE ∂2u∂x2 − ∂u
∂y = 0 in −∞ < x < ∞ andy > 0 under the conditions u(x, 0) = f(x) if u(x, y) is bounded when|x| → ∞ .
Solution. Taking F T over the variable x, the given PDE gives
1√2π
∫ +∞
−∞
∂2u
∂x2eiξx dx =
1√2π
∫ +∞
−∞
∂u(x, y)∂y
· eiξx dx
Thus, − ξ2 u(ξ, y) =d
dyu(ξ, y)
Solving this simple ordinary differential equation we get
u(ξ, y) = A e−ξ2y (i)
The condition u(x, 0) = f(x) gives rise to
u(ξ, 0) =1√2π
∫ +∞
−∞f(x) eiξx dx (ii)
Also, from (i), putting y = 0 we get
u(ξ, 0) = A (iii)
Comparing (ii) and (iii) and using the result in (i) we finally get
u(ξ, y) = e−ξ2y 1√2π
∫ +∞
−∞f(u) eiξu du
Taking inverse Fourier transform, we get
u(x, y) =12π
∫ +∞
−∞e−iξx
[e−ξ2y
∫ +∞
−∞f(u) · eiξudu
]dξ
=12π
∫ +∞
−∞f(u)
[∫ +∞
−∞e−ξ2y · e−iξ(x−u)dξ
]du
70 An Introduction to Integral Transforms
=12π
∫ +∞
−∞f(u)
[∫ +∞
−∞e−y{ξ+ i(x−u)
2y}2− (x−u)2
4y dξ
]du
=12π
∫ +∞
−∞f(u) · e
−(x−u)2
4y
[∫ +∞
−∞e−y{ξ+ i(x−u)
2y}2
dξ
]du
Now, putting[ξ +
i(x − u)2y
]=
η√y
, we get
dξ =dη√
y
so that the inner integral becomes∫ +∞
−∞e−η2 · dη =
√π , a standard result.
Thus we get,
u(x, y) =12π
∫ +∞
−∞
f(u)√y
e−(x−u)2
4y√
π du
=1
2√
πy
∫ +∞
−∞f(u) exp
[−(x − u)2
4y
]du .
Example 1.40. Solve the boundary value problem for the Laplaceequation uxx + uyy = 0 in the quarter plane 0 < x < ∞ , 0 < y < ∞with the boundary conditions
u(0, y) = a , u(x, 0) = 0
∇u → 0 as r =√
x2 + y2 → ∞
where a is a constant. Using Fourier transform solve it.
Solution. Applying Fourier sine transform over x the Laplace equa-tion takes the form
d2Us
dy2− ξ2 Us +
√2π
ξa = 0,
where Us =
√2π
∫ ∞
0u(x, y) sin ξx dx
Its solution is
Us(ξ, y) = A e−ξy +
√2π
a
ξ,
Fourier Transform 71
where A is a constant. Now, sine transform of u(x, 0) = 0 is
Us(ξ, 0) = 0 .
Hence, Us(ξ, y) =a
ξ
√2ξ
[1 − e−ξy
].
Inverting, the solution of the boundary value problem is
u(x, y) =2aπ
∫ ∞
0
1ξ
(1 − e−ξy) sin ξ x dξ
= a − 2aπ
[π2− tan−1 y
x
]=
2aπ
tan−1
(x
y
).
Example 1.41. Solve the Laplace equation uxx + uyy = 0 in a semi-infinite strip in the xy-plane defined by 0 < x < ∞ , 0 < y < b underthe boundary conditions
u(0, y) = 0 , u(x, y) → 0 as x → ∞ for 0 < y < b
u(x, b) = 0 , u(x, 0) = f(x) , for 0 < x < ∞ .
Solution. Sine transform of the Laplace equation with respect to x
under boundary conditions u(0, y) = 0 and u(x, y) → 0 as x → ∞ givesrise to
d2Us
dy2− ξ2Us = 0
Also sine transforms of u(x, b) = 0 , u(x, 0) = f(x) become
Us(ξ, b) = 0
Us(ξ, 0) = Fs(ξ)
Hence, the solution of the transformed differential equation under trans-formed boundary conditions is
Us(ξ, y) = Fs(ξ)sinh [ξ(b − y)]
sinh ξb
Now, inverting this equation we get
u(x, y) =
√2π
∫ ∞
0Fs(ξ)
sinh[ξ(b − y)]sinh ξb
sin ξ x dξ
=2π
∫ ∞
0
[∫ ∞
0f(t) sin ξt dt
]sinh ξ(b − y)
sinh ξbsin ξ x dξ
72 An Introduction to Integral Transforms
When ξb → ∞ ,[
sh ξ(b−y)sh ξb
]∼ exp (−ξb) and so the solution becomes
u(x, y) =1π
∫ ∞
0f(t) dt
∫ ∞
0[cos ξ(x − t) − cos ξ(x + t)] e−ξy dξ
Or, u(x, y) =1π
∫ ∞
0f(t)
[y
(x − t)2 + y2− y
(x + t)2 + y2
]dt
as the final solution of the corresponding quarter plane problem0 < x < ∞ , 0 < y < ∞ of the Laplace equation under the givenboundary conditions when ξb → ∞.
Exercises
(1) Find Fourier Transform of the function f(x), if
f(x) =sin ax
x, a > 0 .
[Ans. π , when |ξ| < a
0 , when |ξ| > a
]
(2) Find Fourier Transform of
f(x) =
{eiωx , 0 < x < a
0 , otherwise
to prove that F [f(x) ; x → ξ] =i√2π
[ei(ξ+ω)a − ei(ξ+ω)b
ξ + ω
]
(3) Prove that F [H(a − |x|) ; x → ξ] =√
2π
sin aξξ , where H(x) is the
Heaviside’s unit step function.
(4) Prove that Fc F−1c ≡ I and Fc = F−1
c , where Fc and F−1c are co-
sine transform and inverse cosine transform operators respectively.
(5) Find Fourier transform of each of the following functions :
(a) f(x) = 11+x2
(b) f(x) = ex exp (−ex)
(c) f(x) = x exp(−ax2
2
), a > 0
[Ans. (a)√
π2 e−|ξ| (b) Γ(1−iξ)/
√2π (c) Take f(x) = − 1
addx
(e
−ax2
2
)etc].
Fourier Transform 73
(6) (a) Prove that F [δ(x − ct) + δ(x + ct) ; x → ξ] =√
π2 cos ξ ct
(b) Prove that F [H(ct − |x|) ; x → ξ] =√
π2
sin ξ ctξ
(c) Prove that F [xnf(x) ; x → ξ] = in dn
dξn F (ξ)
(d) If f(x) has a finite discontinuity at x = a, show thatF [f ′(x) ; x → ξ] = iξ F (ξ)− 1√
2πexp (−iξa)[f(a+0)−f(a−0)] .
(7) (a) Prove that√
2πδ(x) ∗ f(x) = f(x)
(b) Prove that ddx [f(x) ∗ g(x)] = f ′(x) ∗ g(x) = f(x) ∗ g′(x).
(8) (a) Prove that Fs
[1x ; x → ξ
]=√
π2
(b) Prove that Fs [e−ax ; x → ξ] =√
2π
ξa2+ξ2
(c) Fc[e−ax ; x → ξ] =√
2π
aa2+ξ2 , a > 0
(9) If Fc[f(x) ; x → ξ] = Fc(ξ) , prove that
Fc[f(x) cos ω x ; x → ξ] = 12 [Fc(ξ + ω) + Fc(ξ − ω)]
(10) Prove the following :
(a) F [e−x2; x → ξ] = 1√
2e
−ξ2
2
(b) Fs
[e−ax
x ; x → ξ]
=√
2π tan−1 ξ
a ,
(11) Prove that F−1s
[e−πξ ; ξ → x
]=√
2π
xx2+π2
(12) Use shift theorem to prove that
F[e−|x−t| ; t→ξ
]= eiξxF
[e−|t| ; t → ξ
]
=
√2π
eiξx
1 + ξ2
Hence, show that eiξx =12(1 + ξ2)
∫ +∞
−∞e−|x−t|+iξtdt .
74 An Introduction to Integral Transforms
(13) Use Parseval’s theorem to prove that
∫ ∞
0
sin at
t(a2 + t2)dt =
π
21 − e−a2
a2
by choosing f(x) = e−ax and g(x) =
{1 , 0 < x < a
0 , x > a
(14) Prove that∫ ∞
0
[a − b
ξ+
cos aξ − cos bξ
ξ2
]sin ξ x dξ
=
⎧⎪⎨⎪⎩
π2 (a − b) , 0 < x < aπ2 (x − b) , a < x < b
0 , x > b
(15) From the identity
f(x) = − d
dx
∫ ∞
xf(u)du
prove that Fs[f(x);x → ξ] = ξ Fc
[∫ ∞
xf(u)du;x → ξ
]by using the result Fs [f ′(x) ; x → ξ] = −ξ Fc [f(t) ; t → ξ] .
(16) If δn(x) = n√π
e−n2x2, prove that
F [δn(x) ; x → ξ] =1√2π
e−ξ2
4n2
Hence, prove that∫ +∞
−∞δn(x) dx = 1 .
(17) Show that
F
[cosh ax
cosh πx; x → ξ
]=
1√2π
2 cos a2 · ch ξ
2
cos a + ch ξ
by using contour integration method round a rectangular contourdescribed by x = ± R , y = 0 , y = 2π, after suitably indented atpoints if needed.
Fourier Transform 75
(18) Putting f(x) = x− 12 in the sine and cosine forms of Fourier’s inte-
gral theorem show that∫ ∞
0
sin x√x
dx =∫ ∞
0
cos x√x
dx =√
π
2
and deduce that Fs
[t−
12 ; t → ξ
]= Fc
[t−
12 ; t → ξ
]= ξ−
12
(19) From Fourier cosine transforms of e−ax cos ax and of e−ax sin ax
deduce F. C. transforms of (x4+k4)−1 and x2(x4+k4)−1 , (k > 0).
(20) From Fourier sine transforms of e−ax cos ax and of e−ax sin ax
deduce F.S. transforms of (a) x(x4 +k4)−1 (b) x3(x4 +k4)−1
(c) x(x4 + k4)−2 (d) x3(x4 + k4)−2
(21) Show that if b and ξ are real,
Fs
[x2 − b2
x(x2 + b2); x → ξ
]=
√2π[e−|bξ| − 1
2
]sgn (ξ)
(22) Show that if ξ > 0∫ ∞
0
sin x cos ξx
xdx =
π
2H(1 − ξ)) .
(23) If f(x) = (1−|x|)H(1−|x|), find F [f(x) ; x → ξ] and deduce that
(a)∫ ∞
0
(sin x
x
)2
dx =π
2
(b)∫ ∞
0
(sin x
x
)3
dx =3π8
(c)∫ ∞
0
(sin x
x
)4
dx =π
3
(24) Show that F T of 34a3 (a2 − x2)H
[1 − |x|
a
], a > 0 is 1√
2πϕ(ξa),
where ϕ(t) = 3 sin tt3
− cos tt2
(25) If a > 0, b > 0 show that
(a) Fc
[e−bx−e−ax
x ; x → ξ]
= 1√2π
log a2+ξ2
b2+ξ2 , (a > b)
(b) Fs
[e−bx−e−ax
x ; x → ξ]
=√
2π tan−1 (a−b)ξ
ab+ξ2 , (a > b)
76 An Introduction to Integral Transforms
(26) If F (ξ) = Fc
[e−
x2
2 ; x → ξ
], show that dF
dξ = −ξF , F (0) = 1 and
hence duduce that F (ξ) = e−ξ2
2 .
(27) Using Parseval’s theorem show that if a and b are positive con-stants ∫ ∞
0sin ax sin bx
dx
x2=
π
2min (a, b).
(28) Making use of Fourier cosine transform and the Parseval’s relationprove that ∫ ∞
0
x−sdx
1 + x2=
π
2sec(π
2s)
, 0 < s < 1
(29) Use Fourier transform to solve the ODE given below :
(a) y′′(x)−y(x)+2 f(x) = 0 , where f(x) = 0 when x < −a andwhen x > a and y(x) and its derivatives vanish at x → ±∞(b) 2y′′(x) + xy′(x) + y(x) = 0
[Ans. (a) y(x) = e−x∫ x−a etf(t)dt + ex
∫ ax e−tf(t)dt
(b) y(x) = A exp(−x2
4
), A is a constant]
(30) Solve the following integral equation for f(x) :
(a)∫ +∞
−∞e−at2f(x − t)dt = e−bx2
, a, b > 0
(b) f(x) +∫ +∞
−∞f(x − t)e−atdt =
1x2 + b2
(c)1π
∫ +∞
−∞
f(t)x − t
dt = ϕ(x) ,
the integral is being in Cauchy principle value sense
[Ans. (a) f(x) =a√
π(a − b)exp
(−abx2
a − b
)
(b) f(x) =
√2π
{b
a(a − b)
}1
(a − b)2 + x2
(c) f(x) = − 1π
∫ +∞
−∞
ϕ(t)x − t
dt when F (ξ) =1π· iπΦ(ξ)
sgn ξ]
Fourier Transform 77
(31) Solve the PDE utt + uxxx = 0 , −∞ < x < ∞ , t > 0 under theconditions u(x, 0) = f(x) , ut(x, 0) = 0 for −∞ < x < ∞.
[Use F−1[cos(ξ2t) ; ξ → x] =1√2t
cos(
x2
4t− π
4
)for solution.]
(32) A function y(x, t) satisfies the diffusion equation k ∂2y∂x2 = ∂y
∂t
on the half line x � 0, for t > 0, the initial condition
y(x, 0) = Cx e−x2
4a2 , C and a are being constants, and boundaryconditions y(0, t) = 0 and y(x, t) → 0 as x → ∞. Prove thaty(x, t) = Cx(1 + kt/a2)
−32 exp
[− x2
4(a2+kt)
].
(33) A harmonic function u(x, y) in x � 0 , y � 0 satisfies the boundaryconditions ux(0, y) = f(y) , u(x, 0) = 0. Show that
uy(x, 0) = − 2π
∫ ∞
0
tf(t)dt
x2 + t2
Find uy(x, 0) if f(y) = q H(b − y) , where q, b > 0 .
(34) Show that θ(x, t) satisfying ∂2θ∂x2 = ∂θ
θt , x � 0, t � 0, θ(x, 0) = f(x)and ∂θ
∂x − hθ = 0 for x = 0 is given by
θ(x, t) =1√2π
∫ ∞
0ω(u, t)[f(|x − u|) + f(x + u) + h
∫ x+u
|x−u|f(υ)dυ]du,
where ω(x, t) =1√2t
e−x2
4t −√
π
2hehx+h2t Erfc
(12xt−
12 + ht−
12
).
(35) Prove that Fourier cosine transform of f(x) = 1x does not exist but
Fourier sine transform of it exists.
(36) Find Fourier sine transform of
(a) f(x) = x e−ax , a > 0
(b) f(x) = e−ax
x , a > 0
[Ans. (a) , (b) : Differentiate with respect to a and integrate w.r.ta both sides the integral
∫∞0 e−ax sin ξ x dx = ξ
ξ2+a2 to obtain theresults respectively.]
(37) If Fc(ξ) = Fc[e−ax2; x → ξ] , prove that Fc(ξ) satisfies the ODE
dFcdξ + ξ
2a Fc = 0 , when Fc(0) = 1
78 An Introduction to Integral Transforms
(38) Apply Fourier cosine transform to solve uxx+uyy = 0, 0< x, y< ∞subject to the conditions u(x, 0) = H(a − x) ,
a > x ; ux(0, y) = 0 ; 0 < x, y < ∞.
[Ans. u(x, y) = − 2π
∫ ∞
0
sin aξ
ξe−ξy cos ξ x dξ]
(39) Use Fourier sine or cosine transform properly to solve the followingintegral equations :
(a)∫ ∞
0f(x) cos ξ x dx =
√π
2ξ
(b)∫ ∞
0f(x) sin ξ x dx =
a
a2 + ξ2
(c)2π
∫ ∞
0f(x) sin ξ x dx = J0(aξ)
(d)∫ ∞
0f(x) cos ξ x dx =
sin aξ
ξ[Ans. (a) f(x) =
1√x
(b) f(x) = e−ax
(c) f(x) =H(x − a)√
x2 − a2(d) f(x) = H(a − x)
]
(40) Use Parseval formula to evaluate the following integral whena, b > 0 :
(a)∫ +∞
−∞
dx
(x2 + a2)2(b)
∫ +∞
−∞
sin ax dx
x(x2 + b2)(c)∫ +∞
−∞
sin2 ax
x2dx[
Ans. (a)πa3
2(b)
π
b2
(1 − e−ab
)(c) πa
]
(41) Solve the following free vibration problem of a semi-inifinite stringdefined by utt = c2 uxx , 0 < x < ∞ , t > 0 subject to conditionu(0, t) = 0 , u(x, 0) = f(x) , ut(x, 0) = g(x).
[ For x > ct , u(x, t) =12[f(x − ct) + f(x + ct)] +
12c
∫ x+ct
x−ctg(ξ)dξ.
Similarly result for x < ct can also be obtained ] .
Chapter 2
FINITE FOURIER TRANSFORM
2.1 Introduction.
Integral transform applied to solve the boundary value and initial valueproblems in mathematical physics in which the range of values of one ofthe independent variables say x, is finite for reducing the order of theassociated partial differential is of the type∫ b
af(x, · · ·)K(x, ξ)dx,
where a, b are finite numbers. Such a transform is termed as a finiteIntegral transform. Further, if the kernel K(x, ξ) is a Fourier kernellike sin xξ or cos xξ then the corresponding transforms are called finiteFourier transforms.
2.2 Finite Fourier cosine and sine transforms.
It is well-known result in the theory of Fourier series that, if f(x) satisfiesDirichlet’s conditions in 0 ≤ x ≤ a, a being a finite real number andhence f(x) ε ℘1 (0, a), finite Fourier cosine transform fc(n) of f(x) isdefined by
fc[f(x) ; n] = fc(n) =∫ a
0f(x) cos
nπx
adx (2.1)
and the corresponding inverse finite cosine transform is given by
f−1c [fc (n); x] = f(x) =
1a
fc(0) +2a
∞∑n=1
fc(n) cosnπx
adx (2.2)
at a point of continuity of f(x) in 0 ≤ x ≤ a. But at a pointing ofdiscontinuity x = c, 0 < c < a, the right hand side series in (2.2)
80 An Introduction to Integral Transforms
takes the value 12 [f(c + 0) + f (c − 0)] and the series takes the value
12 [f(0) + f(a)] both at x = 0 or at x = a. In the sequel, (2.2) will bereferred to as the inversion formula for finite Fourier cosine transform.
Similar results hold for finite sine transform. For a function f(x)ε ℘1(0, a),and thereby satisfying the Dirichlet’s conditions, finite Fourier sine trans-form fs (n) of f(x) in (0, a) is defined by
fs[f(x);n] = fs (n) =∫ a
0f(x) sin
nπx
adx (2.3)
and its corresponding inversion formula is given by
f−1s [fs (n); x] = f(x) =
2a
∞∑n=1
fs (n) sinnπx
a(2.4)
at a point of continuity x of f(x) in (0, a) and at a point of discontinuityx = c, 0 < c < a of f(x), the right hand side series in eqn (2.4) takes thevalue 1
2 [f(c + 0) + f(c− 0)] and the series takes the value 12 [f(0) + f(a)]
both at x = 0 or x = a .
Corollary 2.1. In particular, if a = π and that if f(x) ∈ ℘1(0, π),finite Fourier cosine transform of f(x) and the corresponding inversetransform are given by
fc(n) =∫ π
0f(x) cos nx dx (2.5)
and f−1c [fc(n)] = f−1
c [fc(n);x] = f(x) =1π
fc(0) +2π
∞∑n=1
fc(n) cos nx
(2.6)
Similarly, finite Fourier sine transform of f(x) and corresponding Inversetransform are given by
fs(n) =∫ π
0f(x) sin nx dx (2.7)
and f−1s [fs(n) ; x] = f(x) =
2π
∞∑n=1
fs(n) sin nx (2.8)
The right hand series in eqns (2.6) and (2.8) converge to 12 [f(x + 0) +
f(x−0)] at a point of discontinuity x and to f(x) at a point of continuityx and to 1
2 [f(0) + f(π)] at points x = 0 or x = π.
Finite Fourier Transform 81
2.3 Relation between finite Fourier transforms of the deriv-
atives of a function.
In applications finite Fourier transforms of the derivatives of a functionare found useful to the solution of boundary value and initial valueproblems. Accordingly, the following theorems are important.
Theorem 2.1. Let f(x) be continuous and f ′(x) be sectionally contin-uous on the interval 0 ≤ x ≤ a, then
(i) fc[f ′(x) ; x → n] = (−1)nf(a) − f(0) + nπa fs(n), n ∈ Z∗
(ii) fs[f ′(x) ; x → n] = −nπa fc(n), n ∈ N
Proof.(i) From the definition of finite Fourier cosine transform we have
fc[f ′(x) ; n] =∫ a
0
df
dx· cos nπx
adx
=[f(x) · cos nπx
a
]a0
+nπ
a
∫ a
0f(x) · sin nπx
adx
= (−1)nf(a) − f(0) +nπ
afs(n), n = 0, 1, 2, ..... = Z∗
(2.9)
(ii) Again
fs[f ′(x) ; n] =∫ a
0
df
dxsin
nπx
adx
= −nπ
a
∫ a
0f(x) cos
nπx
adx
= −nπ
afc(n) , n = N (2.10)
Theorem 2.2. Let f(x) and f ′(x) be continuous and f ′′(x) be section-ally continuous in 0 ≤ x ≤ a . Then
(i) fc[f ′′(x) ; n] = −f ′(0) + (−1)n f ′(a) − n2π2
a2fc (n)
(ii) fs[f ′′(x) ; n] =nπ
af(0) − (−1)n
nπ
af(a) − n2π2
a2fs (n)
Proof.(i) From the definition we have
fc [f ′′(x);n] =∫ a
0
d2f
dx2cos
nπx
adx =
[df
dx· cos
nπx
a
]a
0
+nπ
afs[f ′(x);n]
= (−1)n f ′(a) − f ′(0) − n2π2
a2fc(n), by (2.10) of Th. 2.1. (2.11)
82 An Introduction to Integral Transforms
(ii) Again, fs[f ′′(x) ; n] =∫ a
0
d2f
dx2sin
nπx
adx
= −nπ
a
∫ a
0
df
dx· cos
nπx
adx
= −nπ
a(−1)n f(a) +
nπ
af(0) +
n2π2
a2fc(n).
(2.12)
For finite Fourier transform for the derivatives of order greater than 2of a function may be obtained by induction. In the analysis of bound-ary value problems we often need to know the results for fourth-orderderivatives. It follows that if d2f
dx2 is zero at x = 0 and at x = a, then
∫ a
0
d4f(x)dx4
sinnπx
adx = −n2π2
a2
∫ a
0
d2f
dx2sin
nπx
adx
=n4π4
a4fs (n) (2.13)
Similarly, if d3fdx3 = df
dx = 0 when x = 0 and when x = a, it is seen that
∫ a
0
d4f(x)dx4
cosnπx
adx =
n4π4
a4fc (n) (2.14)
2.4 Faltung or convolution theorems for finite Fourier
transform.
Like ordinary Fourier transform, the convolution theorems for finiteFourier transform can not be put into such simple forms. Accordingly,we introduce two new functions f1(x) and f2(x) instead of f(x), whichare defined as odd and even periodic extensions respectively of f(x) withperiod 2π in the sense
f1(x) =
{f(x) , 0 � x � π
−f(−x) , −π < x < 0
f1(x + 2 r π) = f1(x) , r = ± 1 , ± 2, · · ·
and f2(x) =
{f(x) , 0 ≤ x ≤ π
f(−x) , −π < x < 0
f2(x + 2 r π) = f2(x) , r = ± 1 , ± 2, · · ·
Finite Fourier Transform 83
Thus with the above understanding if f1(x) and g1(x) are the oddperiodic extensions of f(x) and g(x) respectively, then∫ π
−πf1(x − u)g1(u)du =
∫ π
0f(x − u)g(u)du −
∫ π
0f(u − x)g(u)du
(2.15)
Again since f1(x) is periodic∫ π
−πf1(x − u)g1(u)du =
∫ π
0f(−x − u)g(u)du −
∫ π
0f(x + u)g(u)du
(2.16)
We call ∫ π
−πf1(x − u) g1(u)du ≡ f1(x) ∗ g1(x),
as the convolution of the functions f1(x) and g1(x), introducing thenotation introduced by Churchill.
If finite Fourier sine transforms of f(x) and g(x) in (0, π) be fs(n)and gs(n) respectively, then
2π
m∑n=1
fs(n)gs(n) cos nx =2π
∫ π
0dξ
∫ π
0f(ξ)g(η)dη
m∑n=1
cos nx sin nξ sinnη
Making use of
4 sin nξ sin nη cos nx = cos(ξ − η + x)n + cos(ξ − η − x)n
− cos(ξ + η + x)n − cos(ξ + η − x)n
andn∑
r=1
cos ry = −12
+sin(m − 1
2
)y
2 sin(
12y)
we get2π
m∑r=1
fs(n)gs(n) cos nx =14π
∫ π
0f(ξ)dξ
∫ π
0g(η)dη
[sin (m − 1
2)(ξ − η + x)sin 1
2(ξ − η + x)+
sin (m − 12)(ξ − η − x)
sin 12(ξ − η − x)
−sin (m − 12)(ξ + η + x)
sin 12(ξ + η + x)
− sin (m − 12)(ξ + η − x)
sin 12(ξ + η − x)
]
Introducing the transformations
λ = ξ − η + x, μ = η
84 An Introduction to Integral Transforms
in the above four integrals, we get the first integral
14π
∫ π
0f(ξ)dξ
∫ π
0g(η)
sin(m − 1
2
)(ξ − η + x)
sin 12(ξ − η + x)
dη
=12π
∫ π
0g(μ)dμ
∫ x+π−μ
x−μf(λ + μ − x)
sin(m − 1
2
)λ
λ
12λ
sin 12λ
dλ
→ 14
∫ π
0g(μ)f(μ − x)dμ as m → ∞ .
Similarly, for m → ∞ the other three integrals can be evaluated to give
2π
∞∑n=1
fs(n)gs(n) cos nx = −14
∫ π
0g(μ)[f(x − μ) + f(−x − μ)
−f(μ − x) − f(μ + x)]dμ
Thus, by (2.15) and (2.16) we get from the above equation that
2π
∞∑n=1
fs(n) gs(n) cos nx = −12
∫ π
−πf1(x − μ)g1(μ)dμ
= −12f1(x) ∗ g1(x)
Then by inverse finite cosine transform we express the above relationas
fc[f1(x) ∗ g1(x)] = −2 fs(n) gs(n) (2.17)
Similarly, by considering the limit as m → ∞ of the sum
2π
m∑n=1
fs(n) gc(n) sin nx
we can have the result
fs[f1(x) ∗ g1(x)] = 2 fs(n) gc(n) (2.18)
We can also show that
fc[f2(x) ∗ g2(x)] = 2 fc(n) gc(n) (2.19)
Finite Fourier Transform 85
2.5 Multiple Finite Fourier Transform.
Let f(x, y) be a function of two independent variables x and y definedin the rectangle 0 � x � a , 0 � y � b. For the time being, regard-ing f(x, y) as a function of x satisfying certain necessary conditions, itpossess a finite Fourier sine transform. We denote this transform byfs(m, y) and it is defined by
fs(m, y) = fs[f(x, y);x → m] =∫ a
0f(x, y) sin
mπx
adx (2.20)
This function will itself have a finite Fourier sine transform over variabley which we may now be denote d by fss(m,n) and defined by
fss(m,n) = f=
ss[f(x, y) ; (x, y) → (m,n)]
=∫ b
0
∫ a
0f(x, y) sin
mπx
asin
nπy
bdx dy (2.21)
We regard this equation as a double finite sine transform of the functionf(x, y) defined over the rectangular domain 0 � x � a , 0 � y � b.
The corresponding inversion formula giving f(x, y) in terms of fss(m,n),are easily given by
fs(m, y) =2b
∞∑n=1
fss(m,n) sinnπy
b(2.22)
and f(x, y) =2a
∞∑m=1
fs(m, y) sinmπx
a(2.23)
and therefore, by (2.22) and (2.23) we finally have
f(x, y) =4ab
∞∑m=1
∞∑n=1
fss(m,n) sinmπx
asin
nπy
b(2.24)
This is the inversion formula for the double finite Fourier transformfss(m,n) of f(x, y).
Proceeding as above, the double finite Fourier cosine transform fcc(m,n)of f(x, y) in the region 0 � x � a , 0 � y � b can be defined as
fcc(m,n) =∫ a
0
∫ b
0f(x, y) cos
mπx
acos
nπy
bdx dy (2.25)
86 An Introduction to Integral Transforms
In a similar way we can define double finite Fourier transforms of thetypes
fcs(m,n) = fc[fs{f(x, y); y → n};x → m]
=∫ a
0
∫ b
0f(x, y) cos
mπx
asin
nπy
bdx dy
fsc(m,n) =∫ a
0
∫ b
0f(x, y) sin
mπx
acos
nπy
bdx dy
Inversion formulas for double transforms of these types can easilybe deduced by successive applications of the inversion formulas for thefinite sine and cosine transforms. For example, in the case, in whichf(x, y) is defined on the region 0 � x � a , 0 � y � b , we get
f−1cs [fcs(m,n) ; (m,n) → (x, y)] ≡ f(x, y)
=2ab
∞∑n=1
fcs(0, n) sinnyπ
b+
4ab
∞∑m=1
∞∑n=1
fcs(m,n) cosmπx
asin
nπy
b
2.6 Double Transforms of partial derivatives of functions.
The double finite Fourier sine transforms of the partial derivatives offunction f(x, y) ∈ C2(D), where
D = {(x, y) : 0 < x < a, 0 < y < b}is given by
fs
[∂2f
∂x2; x → m]
]=∫ a
0
∂2f
∂x2sin
mπx
adx
=mπ
a
[f(0, y) + (−1)m+1 f(a, y)
]− m2π2
a2fs [f(x, y) ;x → m]
Therefore, fss
[∂2f
∂x2; (x, y) → (m,n)
]= Θ1(m,n) − m2π2
a2fss(m,n)
(2.26)
where Θ1(m,n) =mπ
a
[fs{f(0, y) ; y → n}
+ (−1)m+1fs{f(a, y); y → n}] (2.27)
Thus, fss
[∂2f
∂x2; (x, y) → (m,n)
]= −m2π2
a2fss(m,n),
if f(0, y) = f(a, y) = 0, 0 < y < b, in particular .
Finite Fourier Transform 87
A similar result holds for double finite sine transform of ∂2f∂y2 as
fss
[∂2f
∂y2; (x, y) → (m,n)
]=
nπ
b
[fs{f(x, 0) ; x → m}]
+nπ
b
[(−1)n+1 fs{f(x, b) ; x → m}]− n2π2
b2fss(m,n)
= Θ2(m,n) − n2π2
b2fss(m,n), say. (2.28)
In particular, if f(x, 0) = f(x, b) = 0, 0 < x < a, then
fss
[∂2f
∂y2; (x, y) → (m,n)
]= −n2π2
b2fss (m,n) (2.29)
From (2.26) and (2.28) we immediately get
fss
[∂2f
∂x2+
∂2f
∂y2; (x, y) → (m,n)
]
= Θ1(m,n) + Θ2(m,n) − π2
[m2
a2+
n2
b2
]fss(m,n) (2.30)
If, in particular, f(x, y) vanishes along the boundary of D, then
fss
[∇2f(x, y) ; (x, y) → (m,n)]
= −π2
[m2
a2+
n2
b2
]fss(m,n) (2.31)
The above deductions may be extended for functions of more than twovariables similarly in its corresponding region of definitions.
2.7 Application of finite Fourier transforms to boundary
value problems.
When the range of one of the variables is finite in the associated PDEof a given boundary value problem, finite Fourier transform over thatvariable may be applied to remove it. Further, the choice of sine andcosine transform is decided by the given form of the boundary / initialconditions of the problem. We shall now briefly indicate below someapplications of finite Fourier transforms to the solution of boundaryvalue and initial value problems.
Example 2.1. The temperature u(x, t) at any point x at any time t ina solid bounded by planes x = 0 and x = 4 satisfies the heat conduction
88 An Introduction to Integral Transforms
equation ∂2u∂x2 = ∂u
∂t , when the end faces x = 0 and x = 4 are kept at zerotemperature. Initially the temperature at x is 2x. It is required to findu(x, t) for all x and t.
Solution. The initial and the boundary value problem is defined by
∂2u
∂x2=
∂u
∂t, 0 ≤ x ≤ 4 , t � 0 (i)
subject to the initial condition u(x, 0) = 2x and boundary conditionsu(0, t) = u(4, t) = 0.
Since u(0, t) and u(4, t) are prescribed, we introduce finite Fouriertransform over x and then eqn. (i) becomes
∫ 4
0
∂u(x, t)∂t
sinnπx
4dx =
∫ 4
0
∂2u(x, t)∂x2
sinnπx
4dx
i.e,d us(n, t)
dt= −nπ
4[u(4, t) cos nπ − u(0, t)] − n2π2
16us(n, t)
= −n2π2
16us(n, t)
Its solution is us(n, t) = c e−n2π2
16t (ii)
Putting t = 0 , us(n, 0) = c
Now, us(n, 0) =∫ 4
0u(x, 0) sin
nπx
4dx
=32(−1)n+1
nπ(iii)
Therefore, from (ii) and (iii), we get
us(n, t) =32(−1)n+1
nπe−
n2π2
16t (iv)
Taking inverse finite Fourier transform of eqn (iv), we get
u(x, t) =24
∞∑n=1
us(n, t) sinnπx
4
=16π
∞∑n=1
(−1)n+1
ne
−n2π2
16t sin
nπx
4(v)
as the expression of the temperature of the solid at any point at anytime.
Finite Fourier Transform 89
Example 2.2. Solve the wave equation
∂2u
∂x2=
1c2
∂2u
∂t2, 0 ≤ x ≤ a , t > 0 (i)
satisfying the boundary conditions u(0, t) = u(a, t) = 0 , t > 0 and theinitial conditions u(x, 0) = 4b
a2 x(a − x), ∂u(x,0)∂t = 0 , 0 � x � a to
determine the displacement u(x, t).
Solution. Since the boundary condition u(0, t) and u(a.t) are prescribedwe may apply finite Fourier sine transform over the variable x. Thenthe given PDE (i) becomes
[d2
dt2+
c2n2π2
a2
]us(n, t) = 0, where us(n, t) = fs(u(x, t) ; x → n)
Its solution is given by
us(n, t) = A cosnπc
at + B sin
nπc
at (ii)
Further, taking finite Fourier sine transform over x on initial conditionswe get
us(n, 0) =8baπ3
1n3
[1 + (−1)n+1
], (iii)
d
dtus(n, 0) = 0. (iv)
Using (iii), (iv) in eqn. (ii), we get
us(n, t) =8ab
π3
[1 + (−1)n+1
n3
]cos
nπct
a(v)
Making use of inversion formula for finite Fourier transform, we get therequired displacement as
u(x, t) =∞∑
r=0
16b(2r + 1)−3
π3cos
(2r + 1)πct
asin
(2r + 1)πx
a.
Example 2.3. Consider the diffusion equation
∂2u
∂x2=
1k
∂u
∂t, 0 < x < a , t > 0
90 An Introduction to Integral Transforms
for its solution satisfying the boundary conditions
∂u(0, t)∂x
=∂u(a, t)
∂x= 0 , t > 0
and the initial condition u(x, 0) = f(x) , 0 � x � a , using proper finiteFourier transform.
Solution. Since ∂u∂x on the boundaries x = 0 , a are prescribed the
proper finite Fourier transform to be applied here is finite cosine trans-form over x. Hence, the diffusion equation and the initial conditiontransform to
duc
dt+
kn2π2
a2uc(n, t) = 0 (i)
and uc(n, 0) = fc(n) = fc[f(x) ; x → n] (ii)
Solving eqn.(i) under condition (ii), one gets
uc(n, t) = fc(n) e−k n2 π2
a2 t
Therefore, by inversion theorem for finite Fourier cosine transform, weget
u(x, t) =1afc(0) +
2a
∞∑n=1
fc(n) e−kn2π2
a2 t cosnπx
a.
Example 2.4. The displacement u(x, t) describing the vibration of abeam of length a with freely hinged end points satisfies the equation
∂4u
∂x4+
1c2
∂2u
∂t2=
p (x, t)EI
, 0 ≤ x ≤ a , t > 0
The displacement u(x, t) satisfies the initial conditionsu(x, 0) = ∂u
∂t (x, 0) = 0 , 0 ≤ x ≤ a and the boundary conditionsu(0, t) = ∂2u
∂x2 (0, t) = 0 and u(a, t) = ∂2u∂x2 (a, t) = 0 , t > 0
It is necessary to determine the displacement of the beam anywhere andat any time using finite Fourier transform.
Solution. Taking Fourier sine transform over x, the PDE transformsto
n4π4
a4us(n, t) +
d2
dt2us(n, t) =
c2
EIps(n, t), (i)
where ps(n, t) = fs[ p (x, t) ; x → n ] ,
Finite Fourier Transform 91
after using the four boundary conditions. Again, the transformed formof the two given initial conditions become
us(n, 0) =d us(n, 0)
dt= 0 (ii)
Therefore, the solution of the new initial value problem defined byeqn. (i) and under the initial conditions in eqn. (ii) is given by
us(n, t) =a2
n2π2c
∫ t
0
c2
EIps(n, t) sin
n2π2c(t − ς)a2
dς
Inverting the above equation we get
u(x, t) =2ac
π2EI
∞∑n=1
1n2
sinnπx
a
∫ t
0ps(n, t) sin
n2π2c
a2(t − ς) dς
Example 2.5. Find the solution of the one dimensional heat conductionequation in a bar with the temperature distribution satisfying ∂u
∂t = k ∂2u∂x2
under the boundary conditions ∂u∂x(x, t) = 0 at x = 0 and at x = π and
the given initial condition u(x, 0) = f(x) , 0 � x � π.
Solution. Taking Fourier cosine transform over x, the given PDE underthe boundary conditions becomes
d uc
dt= −kn2 uc(n, t)
Its solution is, clearly , uc(n, t) = uc(n, 0) e−kn2t
Now, since u(x, 0) = f(x), we have uc(n, 0) =∫ π0 f(x) cos nx dx
Therefore, uc(n, t) = e−kn2t∫ π0 f(x) cos nx dx
Now, taking inverse finite Fourier transform, the above eqn. gives
u(x, t) =1π
∫ π
0f(y)dy +
2π
∞∑n=1
[e−kn2t cos nx
∫ π
0f(y) cos ny dy
].
Example 2.6. The end points of a solid bounded by x = 0 and x = π
are maintained at temperatures u(0, t) = 1, u(π, t) = 3, where u(x, t)represents its temperature at any point of it at any time t. Initially,the solid was held at 1 unit temperature with its surfaces were insu-lated. Find the temperature distribution u(x, t) of the solid, given thatuxx(x, t) = ut(x, t).
92 An Introduction to Integral Transforms
Solution. The heat conduction equation of the solid is
uxx = ut (i)
with boundary conditions u(0, t) = 1, u(π, t) = 3 and the initial conditionu(x, 0) = 1.
Taking finite Fourier transform over x, the eqn.(i) becomes
dus
dt= −n [u(π, t) cos nπ − u(0, t) + nus(n, t)]
i.e,dus
dt+ n2us(n, t) = n [1 − 3 cos nπ] .
Solving this linear ODE we get
us(n, t) = c e−n2t + (1 − 3 cos nπ)/n (ii)
Putting t = 0, (ii) gives
us(n, 0) = c + (1 − 3 cos nπ)/n
Also, since us(n, 0) =∫ π
0u(x, 0) sin nx dx =
1 − cos nπ
n
we get c =2 cos nπ
nand hence eqn. (ii) takes the form
us(n, t) =2 cos nπ
ne−n2t +
1 − 3 cos nπ
n
Therefore, after inversion
u(x, t) =2π
∞∑n=1
[2 cos nπ
ne−n2t +
1 − 3 cos nπ
n
]sin nx
=4π
∞∑n=1
cos nπ
ne−n2t sin nx +
2π
∞∑n=1
1 − 3 cos nπ
nsin nx
(iii)
But fs [1 ; x → n] =∫ π
0sin nx dx =
1 − cos nπ
nand therefore
1 =2π
∞∑n=1
1 − cos nπ
nsin px (iv)
Also, fs[x ; x → n] =∫ π
0x sin nx dx = −π cos nπ
nimplying
x =2π
∞∑n=1
[−π cos nπ
n
]sin nx (v)
Finite Fourier Transform 93
Thus from (iv) and (v) one gets
1 +2xπ
=2π
∞∑n=1
1 − cos nπ
nsin nx (vi)
Using (vi) in (iii), it gives
u(x, t) = 1 +2xπ
+4π
∞∑n=1
cos nπ
ne−n2t sin nx
Example 2.7. A uniform string of length π is stretched between itsends and initially one end, say x = 0, is given a small oscillation a sin ωt
while the other end is kept fixed. Using finite Fourier transform provethat the displacement of the point x of the string at time t is given by
a sin wt sinw(π − x)
ccosec
wπ
c+
2acw
π
∞∑n=1
1w2 − n2c2
sin nx sin nct,
if the displacement u(x, t) of the string satisfies the PDE
∂2u
∂t2= c2 ∂2u
∂x2(i)
Solution. The boundary and the initial conditions of the problem aregiven by
u(0, t) = a sin wt (ii)
u(π, t) = 0 (iii)
u(x, 0) = 0 (iv)
and∂u
∂t(x, 0) = 0 (v)
The form of the boundary conditions in (ii) and (iii) imply that finiteFourier sine transform need to be applied over the variable x to reduceeqn (i) to
d2 us(n, t)(dt2)
= −c2n [u(π, t) cos nπ − u(0, t) + nus(n, t)]
Using (ii) and (iii), this equation becomes[d2
dt2+ c2n2
]us(n, t) = c2 a n sin wt
94 An Introduction to Integral Transforms
The general solution of this equation is given by
us(n, t) = A cos nct + B sin nct +c2a n sinwt
c2n2 − w2(vi)
The transformed form of (iv) and (v) becomes
us(n, 0) = 0 (vii)
and[d us(n, t)
dt
]t=0
= 0 (viii)
Using (vii) and (viii), the solution becomes
us(n, t) =acw
w2 − c2n2sin cnt − c2an sinwt
w2 − c2n2
Inverting the above transformed solution, we get
u(x, t) =2π
∞∑n=1
[acw
w2 − c2n2sin cnt − c2an sinwt
w2 − c2n2
]sin nt (ix)
To write the above solution in the required form we note that
fs
[sin
w(π − x)c
; x → n
]=∫ π
0sin
(π − x)wc
sinnx dx
=12
∫ π
0
[cos{
w(π − x)c
− nx
}− cos
{w(π − x)
c+ nx
}]dx
= sin(wπ
c
)· n
n2 − w2
c2
=nc2
n2c2 − w2sin
wπ
c.
This result gives
sinw(π − x)
c=
2π
∞∑n=1
nc2
n2c2 − w2sin
wπ
csin nx
Thus,2ac2 sin wt
π
∞∑n=1
n sin nx
n2c2 − w2= a sin wt sin
w(π − x)c
cosecwπ
c
(x)
The solution (ix), after use of eqn. (x) becomes
u(x, t) = a sin wt sinw(π − x)
ccosec
wπ
c
+2acw
π
∞∑n=1
1w2 − c2n2
sin nx sin nct
Finite Fourier Transform 95
Example 2.8. Solve the three dimensional Laplace equation
∂2V
∂x2+
∂2V
∂y2+
∂2V
∂z2= 0, 0 � x � π, 0 � y � π, 0 � z � π
with the boundary conditions
V = V0 , when y = π ; V = 0 , when y = 0 and
V = 0 , when x = 0, π and V = 0 , when z = 0, π .
Solution. The boundary conditions can be expressed as
V (0, y, z) = V (π, y, z) = 0 (i)
V (x, 0, z) = 0 (ii)
V (x, π, z) = V0 (iii)
and V (x, y, 0) = V (x, y, π) = 0 (iv)
The form of the boundary conditions in (i) and (iv) show that finiteFourier sine transforms over variables x and z successively reduce thegiven Laplace equation in the following forms :
Firstly, taking finite Fourier sine transform over the variable x weget
−m2Vs(m, y, z) +[
∂2
∂y2+
∂2
∂z2
]Vs(m, y, z) = 0, by (i). (v)
Again, taking finite Fourier transform over the variable z, the eqn.(v) gives
−m2V s(m, y, n) − n2V s(m, y, n) +d2
dy2V s(m, y, n) = 0, · · · (vi)
since V s(m, y, 0) = V s(m, y, π) = 0
Again, since∫ π
0V (x, π, z) sinmx dx =
∫ π
0V0 sin mx dx
= V0(1 − cos mπ)/m
V s(m,π, z) =
{0 , m = even2V0m , m = odd
· · · (vii)
and V s(m, 0, z) = 0
The solution of the eqn (vi) is
V s(m, y, n) = c1 ch μy + c2 sh μ y
96 An Introduction to Integral Transforms
where, μ =√
m2 + n2
Using the transformed boundary conditions (vii), the above generalsolution of eqn(vi) becomes,
V s(m, y, n) = c2 sh μy,
where c2 is given by
c2 =V s(m,π, n)
sh μ π
But since,∫ π
0V s(m,π, z) sin nz dz = 0,when m = even
and∫ π
0V s(m,π, z) sin nz dz
=∫ π
0
2 V0
msin nz dz =
2 V0
mn[1 − cos nπ]
=
{0, if m = odd, n = even4V0mn , if m = odd, n = odd.
V s(m,π, n) =4V0
mn, when m,n are both odd.
Hence, V s(m, y, n) =4V0
mn
sin μy
sin μπ, after substituting the value of c2 .
Thus, V s(m, y, n) =4V0
(2m + 1)(2n + 1)sh μ y
sh μ π,
where, μ =[√
m2 + n2]m,n=odd
=√
(2m + 1)2 + (2n + 1)2;m,n = 0, 1, 2, · · ·
Inverting firstly over z, we get
V s(m, y, z) =2π
∞∑n=0
V s(m, y, n) sin(2n + 1)z
Again, inverting over the variable x.
V (x, y, z) =16π2
∞∑m=0
∞∑n=0
sh μy
sh μπ· sin(2m + 1)x
2m + 1· sin (2n + 1)z
(2n + 1),
where μ =√ [
(2m + 1)2 + (2n + 1)2].
Example 2.9. The deflection y(x) of a uniform beam of finite lengthl satisfies the well known ODE d4y
dx4 = w(x)EI = W (x) , 0 � x � l. Find
Finite Fourier Transform 97
y(x) if the beam is freely hinged at x = 0 and at x = l, implying
y(x) = y′′(x) = 0 at x = 0, l.
Solution. Application of finite Fourier transform the given equationunder the given boundary condition gives y(n) =
(l
nπ
)4Ws(n). Inverting
the result we get
y(x) =2l3
π4
∞∑n=1
1n4
sinnπx
lW s(n)
Or y(x) =2l3
π4
∞∑n=1
1n4
sinnπx
l
∫ l
0W (ξ) sin
nπξ
ldξ.
In particular, if W (x) = W0 δ(x − α), where W0 is a constant, then
y(x) =2l3W0
π4
∞∑n=1
1n4
sinnπx
lsin
nπα
l, where 0 < α < l.
Example 2.10. The transverse displacement of an elastic membraneu(x, y, t) satisfies the PDE ∂2u
∂x2 + ∂2u∂y2 = 1
c2∂2u∂t2
, under the boundaryconditions u = 0 on the boundary, u = f(x, y), ut = g(x, y) at t = 0.Find the displacement u(x, y, t) after utilising finite Fourier transform.
Solution. Applying double finite Fourier sine transform we get
us(m,n) =∫ a
0
∫ b
0u(x, y) sin
mπx
asin
nπy
bdx dy .
Then the PDE givesd2us
dt2+ c2π2
[m2
a2+
n2
b2
]us = 0
with transformed boundary conditions
us(m,n, 0) = fs(m,n),[dus
dt
]t=0
= gs(m,n).
The solution of this transformed problem is
us(m,n, t) = fs(m,n) cos (cπwmn t) + gs(m,n) sin (cπwmn t)
where wmn =[m2
a2+
n2
b2
]1/2
98 An Introduction to Integral Transforms
Inverting this equation we get the solution of the problem as
u(x, y, t) =4ab
∞∑m=1
∞∑n=1
sinmπx
a· sin
nπx
b
[fs(m,n)
cos (cπwmn t) + gs(m,n) sin (cπwmn t)]
where[fs(m,n), gs (m,n)
]=∫ a
o
∫ b
0sin
mπξ
asin
nπη
b[f(ξ, η), g(ξ, η)] dξ dη
Exercises.
(1) Find finite sine and cosine transforms of f(x) = x2, 0 < x < π.
[Ans. fs(n) =2n3
(cos nπ − 1) − π2
ncos nπ, n = 1, 2, · · ·
=π2
3, n = 0 ;
fc(n) = 2π(cos nπ − 1)/n2]
(2) Find f(x) if (i) fs(n) = (1 − cos nπ)/n2π2 and 0 < x < 1[Ans.
2π3
∞∑n=1
1 − cos nπ
n2sin nx
]
(3) Calculate fs(n) , if f(x) =
{x , 0 � x � π
2
π − x , π2 � x � π[
Ans.2n2
sinnπ
2
]
(4) Find fc(n), if f(x) = ch c(π−x)sh πc .
[Ans. c/(c2 + n2)
]
(5) Find fs(n), if f(x) = 2π tan−1 2b sin x
1−b2
[Ans. bn [1 − (−1)n] /n, |b| ≤ 1]
(6) Prove that, if f(x) is defined in (0, a)
(a) fs [1 ; n] =a
nπ
[1 + (−1)n+1
](b) fs [x(a − x) ; n] =
2a3
n3π3
[1 + (−1)n+1
](c) f−1
s
[1n
; x
]=
π
a
[1 − x
a
]
Finite Fourier Transform 99
(7) By finding finite Fourier sine and cosine transform of
f(x) = ekx, 0 � x � a, deduce that
f−1s
[n
k2 + n2 π2
a2
; n → x
]=
a
π
sh {k(a − x)}sh{ka} , 0 � x � a
(8) Show that the solution u(x, t) of the diffusion equation∂2u∂x2 = 1
k∂u∂t , 0 � x � a , t > 0 satisfying boundary conditions
u(0, t) = u(a, t) = 0, t > 0
and the initial condition
u(x, 0) = f(x), 0 � x � a
can be expressed as
u(x, t) =2π
∞∑n=1
fs(n) e−k n2π2t
a2 sin(nπx
a
).
(9) The function u(x, y) satisfying ∂2u∂x2 + ∂2u
dy2 = 0 in the semi-infinite
strip { (x,y) : 0 ≤ x < ∞ , 0 ≤ y ≤ b}, and the boundary
conditions
u(x, 0) = 0 , u(x, b) = 0 , x � 0
u(0, y) = f(y) , 0 ≤ y ≤ b
u(x, y) → 0 as x → ∞ , 0 ≤ y ≤ b
can be expressed as
u(x, y) =2b
∞∑n=1
fs(n) · e−nπx/b cosnπy
b.
(10) u(x, y) satisfies the PDE
∂2u
∂x2+
∂2u
∂y2= 0
in the rectangle R = {(x, y) : 0 ≤ x ≤ a , 0 ≤ y ≤ b}and the boundary conditions
∂u(0, y)∂x
=∂u(a, y)
∂x= 0, 0 ≤ y ≤ b
u(x, 0) = f(x) , u(x, b) = 0 , 0 ≤ x ≤ b
100 An Introduction to Integral Transforms
Then prove that taking Fourier cosine transform over the variable x
given by fc [u(x, y) ; x → m] = fc(m), the solution of the boundaryvalue problem is given by
u(x, y) = f−1c
[fc(m)
sinh{mπ(b − y)/a}sinh(mπb/a)
; m → x
]
(11) Find the steady state temperature in a long square bar of side π
when one face of it is kept at constant temperature u0 and the otherface at zero temperature. Assume that the temperature function u(x, y)is bounded and satisfies the PDE uxx + uyy = 0.
[Hint : Take the boundary conditions as u(0, y) = u(π, y) = 0 , u(x, 0) = 0and u(x, π) = u0. Also take finite Fourier transform over x] .[
Ans. u(x, y) =4u0
π
∞∑n=0
sh(2n + 1)y sin(2n + 1)x(2n + 1)sh(2n + 1)π
]
(12) Prove that
(a) Fc [x2 ; n] =
{a3
3 , if n = 02(
anπ
)2a(−1)n, if n = 1, 2, 3, · · ·
(b) Fs
[x2;n
]= a3
nπ (−1)n+1 − 2(
anπ
)3 [1 + (−1)n+1]
when 0 < x < a
(13) Prove that the initial-boundary value problem defined by
ut = k uxx, 0 ≤ x ≤ a, t > 0 under the conditions u(x, 0) = 0,
u(0, t) = f(t), u(a, t) = 0 is given by
u(x, t) =2πk
a
∞∑n=1
n sinnπx
a
∫ t
0f(ς)exp
[−k(t − ζ)
(nπ
a
)2]
dς
(14) Prove the following results defined over (0, π)
(i) Fs
[x2 (π − x) ; x → n
]= 1
n3
[1 + (−1)n+1
](ii) Fs
[sh a(π−x)
sh πa ; x → n]
= nn2+a2 , a �= 0
(iii) Fc
[(π − x)2 ; x → n
]= 2π
n2 for n �= 0 and π3
3 for n = 0
Finite Fourier Transform 101
(iv) Fc [ch a(π − x) ; x → n] = a sh (aπ)n2+a2 , for a �= 0.
(v) Fs
[ddx{f(x) ∗ g(x)} ; x → n
]= 2nfs(n) gs(n)
(vi) Fs
[∫ x0 f(t) ∗ g(t)dt ; x → n
]= 2
n fs(n) gs(n)
(15) If fc(p) =∫ π0 f(x) cos px dx , fs(p) =
∫ π0 f(x) sin px dx, where
p is not necessarily an integer, show for any constant α that
(i) Fc [{2f(x) cos αx} ; x → n] = fc(n − α) + fc(n + α)
(ii) Fc [{2f(x) sin αx} ; x → n] = fs(n + α) − fs(n − α)
(iii) Fs [{2f(x) cos αx} ; x → n] = fs(n + α) + fs(n − α)
(iv) Fs [{2f(x) sin αx} ; x → n] = fc(n − α) − fc(n + α) .
Chapter 3
THE LAPLACE TRANSFORM
3.1 Introduction.
In article 1.3 of chapter 1 it is noted that if∫ +∞−∞ |f(t)|dt is not conver-
gent, Fourier transform F (ξ) of the function f(t) need not exist for allreal ξ. For example, when f(t) = sin ωt, ω = real, F (ξ) does not ex-ist. But such situations do arise occasionally in practice. To handle thissituation, we consider a new function f1(t) connected to f(t) defined by
f1(t) = e−γtf(t)H(t),
where γ is an arbitrary real positive constant and H(t) is the Heavisideunit step function. Clearly, here f1(t) ∈ P 1(R) and f1(t) ∈ A1(R) forall t ∈ R and therefore, Fourier transform of f1(t) exists, since∫ +∞
−∞f1(t)dt =
∫ ∞
0e−γtf(t) dt
is convergent. In fact, in this case, by Fourier Integral theorem
f1(t) =12π
∫ +∞
−∞e−iξt dξ
∫ +∞
−∞f1(u) eiξu du
implying
f(t) =12π
∫ +∞
−∞e−iξt dξ · eγt
∫ ∞
0f(u) · e−(γ−iξ)u du
If we write p = γ − iξ, the above relation can be expressed as
f(t) =1
2πi
∫ γ+i∞
γ−i∞ept
[∫ ∞
0f(u) e−pu du
]dp
=1
2πi
∫ γ+i∞
γ−i∞ept f (p) dp (3.1)
where f(p) =∫ ∞
0f(u)e−pu du, Re p = γ > 0 (3.2)
Eqns. (3.1) and (3.2) constitute a transform with K(p, u) = e−pu as thekernel of it.
The Laplace Transform 103
3.2 Definitions
We now define Laplace transform of a piecewise continuous function f(t)of the real variable t defined on the semi-axis t � 0 by
L [f(t) ; t → p] = L [f(t)] = F (p) = f(p) =∫ ∞
0f(t)e−ptdt (3.3)
and inverse Laplace transform by
f(t) = L−1 [f(p), p → t] = L−1 [F (p)] =1
2πi
∫ γ+i∞
γ−i∞eptf(p) dp (3.4)
where γ = Re p > 0.
Note 1. Some authors use variable s in place of p.
Note 2. The function f(t) must be of exponential order for the existenceof its Laplace transform.
Note 3. In operator notation eqn.(3.3) is expressed as
L [f(t) ; t → p ] = f(p) (3.5)
and relation in eqn.(3.4) is expressed as
L−1 [f(p) ; p → t] = f(t) (3.6)
Thus, L−1[L(f(t))] = f(t) = I [f(t)]
⇒ L−1 L ≡ I
Also, L[L−1(f(p))] = f(p) = I[f(p)]
⇒ LL−1 ≡ I
This means that L−1L ≡ LL−1 ≡ I, showing that these operators L andL−1 are commutative.
3.3 Sufficient conditions for existence of Laplace Trans-
form.
Theorem 3.1. If f(t) is of some exponential order for large t and ispiecewise continuous over o � t � ∞, then Laplace transform of f(t)exists.
104 An Introduction to Integral Transforms
Proof. Let f(t) be of exponential order σ such that
|f(t)| < Meσt , for t � t0
Then we have
L [f(t); t → p] =∫ ∞
0e−ptf(t) dt
=∫ t0
0e−ptf(t) dt +
∫ ∞
t0
e−ptf(t) dt
= I1 + I2 , say
Since f(t) is a piecewise continuous function on every finite interval0 � t � to , I1 exists and it is convergent Also, we have
| I2 | = |∫ ∞
t0
e−ptf(t) dt |
�∫ ∞
t0
e−pt|f(t)| dt
< M
∫ ∞
t0
e−(p−σ)tdt =Me−(p−σ)t0
(p − σ), if p > σ
Therefore, |I2| is finite for all t0 > 0 and p > σ and hence, I2 is conver-gent. Thus, L[f(t)] exists for all p > σ.
Note 3.1 Though the conditions stated in the theorem above are suf-ficient for the existence of Laplace transform, but these are not . Thismeans that even if the above conditions are not satisfied by a function,Laplace transform of that function may or may not exist. This can beshown by considering the example f(t) = 1√
t.
Here, f(t) → ∞ as t → 0.Hence, f(t) is not a piecewise continuousfunction on every finite interval for t � 0.
Now, L[ 1√t] =
∫∞0 e−pt 1√
tdt = 2√
p
∫∞0 e−x2
dx =√
πp , p > 0. This
proves the existence of Laplace transform of f(t).
3.4 Linearity property of Laplace Transform.
Theorem 3.2. If L[f1(t) ; t → p ] and L[f2(t) ; t → p ] both exist andc1, c2 are constants, then
L[c1f1(t) + c2f2(t) ; t → p ] = c1L[f1(t) ; t → p ] + c2[f2(t) ; t → p ]
The Laplace Transform 105
Proof. By definition,
L[c1f1(t) + c2f2(t) ; t → p ] =∫ ∞
0e−pt[c1f1(t) + c2f2(t)]dt
= c1
∫ ∞
0e−ptf1(t)dt + c2
∫ ∞
0e−ptf2(t)dt
= c1 L[f1(t); t → p] + c2 L [f2(t); t → p ]
3.5 Laplace Transforms of some elementary functions
We calculate below Laplace transforms of some elementary functionsfrom definition.
Example 3.1. L[H(t) ; t → p]
=∫ ∞
0e−ptH(t)dt =
∫ ∞
0e−ptdt =
1p.
Therefore, L[H(t − a) ; t → p] =∫ ∞
0H(t − a)e−ptdt
=∫ ∞
ae−ptdt =
e−ap
p
Example 3.2. L[tν ; t → p ]
=∫ ∞
0tνe−ptdt = p−ν−1
∫ ∞
0uνe−udu,when u = p t
= p−ν−1 Γ(ν + 1), which exists when Re ν > −1.
Thus in particular,
L[tn ; t → p ] =n!
pn+1, n = 1, 2, 3, · · · , p > 0
so that L[t ; t → p] =1p2
, L[t2 ; t → p] =2p3
, · · · etc., p > 0
Example 3.3. L[eat ; t → p ]
=∫ ∞
0e−pteat dt =
1p − a
, p > a.
Example 3.4.
L [sin at ; t → p ] =∫ ∞
0e−pt sin at dt
106 An Introduction to Integral Transforms
=∫ ∞
0e−pt
[eait − e−ait
2i
]dt
=
[∫ ∞
0
e−(p−ai)t − e−(p+ai)t
2i
]dt
=12i
[1
p − ai− 1
p + ai
], by linearity property
=a
p2 + a2, p > 0,
Example 3.5. L [cos at ; t → p ]
=∫ ∞
0e−pt eait + e−ait
2dt
=12
[∫ ∞
0e−(p−ai)t dt
]+
12
[∫ ∞
0e−(p+ai)t dt
], by linearity property
=12
[1
p − ai+
1p + ai
]
=p
p2 + a2, p > 0
Example 3.6. L [cosh at ; t → p ]
=∫ ∞
0e−pt
[eat + e−at
2
]dt
=12
∫ ∞
0e−(p−a)t dt +
12
∫ ∞
0e−(p+a)t dt, by linearity property
=12
[1
p − a+
1p + a
]
=p
p2 − a2, p > a > 0
Example 3.7. L [sinh at ; t → p ]
=∫ ∞
0e−pt
[eat − e−at
2
]dt
=12
∫ ∞
0e−(p−a)t dt − 1
2
∫ ∞
0e−(p+a)t dt, by linearity property
=12
[1
p − a− 1
p + a
]
=a
p2 − a2, p > a > 0
The Laplace Transform 107
3.6 First shift theorem.
Theorem 3.3. If Laplace transform of f(t) is f(p), then Laplacetransform of eatf(t) is f(p − a).
Proof. Given that
L [ f(t) ; t → p ] =∫ ∞
0f(t) e−pt dt
= f(p)
Therefore, L[
eatf(t)]
=∫ ∞
0eatf(t) e−pt dt
=∫ ∞
0f(t) · e−(p−a)t dt
= f(p − a). (3.7)
3.7 Second shift theorem.
Theorem 3.4. If Laplace transform of f(t) is f(p), then Laplace trans-form of f(t − a)H(t − a) is e−apf(p).
Proof. Given that
L [f(t) ; t → p] =∫ ∞
0f(t) e−pt dt
= f(p)
Therefore , L [ f(t − a)H(t − a) ; t → p ]
=∫ ∞
0f(t − a)H(t − a)e−ptdt
=∫ ∞
af(t − a) e−pt dt
=∫ ∞
0f(x) e−px · e−pa dx
= e−pa · f(p). (3.8)
3.8 The change of scale property.
Theorem 3.5. If L [f(t); t → p] = f(p), then L [f(at); t → p] = 1a f( p
a
)
108 An Introduction to Integral Transforms
Proof. Given that,∫ ∞
0f(t) e−pt dt = f(p)
Therefore, L [f(at) ; t → p ] =∫ ∞
0f(at) e−pt dt
=1a
∫ ∞
0f(x) · e−P
ax dx
=1a
∫ ∞
0f(x) · e−( p
a)x dx
=1a
f(p
a
). (3.9)
3.9 Examples
Example 3.8. Evaluate
L[f(t)] , where f(t) =
{t/a , 0 < t < a
1 , t > a
Solution.
L [f(t)] =∫ ∞
0e−ptf(t) dt
=∫ a
0e−pt · t
adt +
∫ ∞
ae−pt dt
=1 − e−ap
ap2
Example 3.9. Evaluate L[
1√πt
]
Solution. L
[1√πt
]=∫ ∞
0e−pt 1√
πtdt
=√
p√π
∫ ∞
0e−x x
12−1 dx
p
=1√πp
Γ(
12
)
=√
π√πp
=1√p
.
The Laplace Transform 109
Example 3.10. Evaluate L[
tνe−at ; t → p], if Re ν + 1 > 0
Solution. It is known that L [ tν ; t → p ] = Γ(ν+1)pν+1 .
Therefore, by shift theorem, we get
L[
tνe−at; t → p]
=Γ(ν + 1)
(p + a)ν+1
Example 3.11. Evaluate L [sin(t − a)H(t − a) ; t → p] .Hence, evaluate L
[e(t−a)k sin(t − a) H(t − a) ; t → p
].
Solution. We know that L [sin t ; t → p] = 1p2+1
Hence, by the second shift theorem, we get
L [sin(t − a) H(t − a)] = e−pa/(1 + p2)
Thus, L[ek(t−a) sin(t − a)H(t − a) ; t → p
]= e−(p−k)/
[1 + (p − k)2
], after using the first shift theorem.
Example 3.12. Evaluate L [ sinh bt ; t → p ]. Hence evaluateL[
eat sinh bt ; t → p]
Solution. Since, L [ sinh t ; t → p ] = 1p2−1
, we have by the change ofscale property that
L [ sinh bt ; t → p ] =b
p2 − b2
Hence, L[eat sh bt ; t → p
]=
b
(p − a)2 + b2
Example 3.13. Evaluate L[
eat cos bt ; t → p]. Hence, evaluate
L[ea(t−c) cos b(t − c)H(t − c) ; t → p
]Solution. We know that L [ cos bt ; t → p ] = p
p2+b2
So, L[
eat cos bt ; t → p]
=p − a
(p − a)2 + b2,by shift theorem.
Hence , L[
ea(t−c) cos b (t − c)H (t − c) ; t → p]
= e−pc (p − a)(p − a)2 + b2
Example 3.14. Evaluate L[(πt)−
12 e
t2
].
110 An Introduction to Integral Transforms
Solution. We know that L[
1√πt
]= 1√
p
Hence, L[
1√πt
et2 ; t → p
]= 1�
p− 12
.
Example 3.15. Evaluate(i) L [t H (t − 1) ; t → p ]and (ii) L [ H(t − 1) cos at ; t → p ]
Solution. From definition L [ f(t)H(t − a) ; t → p ]
=∫ ∞
ae−pt f(t) dt
=∫ ∞
0e−p(u+a) f(u + a) du
= e−pa
∫ ∞
0e−pu f(u + a) du
= e−paL [f(t + a) ; t → p]
(i) L [t H (t − 1) ; t → p]
=∫ ∞
1e−pt t dt , a = 1
= e−p
∫ ∞
0e−pu (t + 1) dt
= e−p
[1p2
+1p
]
=p + 1p2
e−p
(ii) L [ H(t − 1) cos at ]
= e−p L [ cos a(t + 1)]
= e−p L [ cos at. cos a − sin at. sin a]
=e−p
p2 + 1[ p cos a − a sin a ] .
3.10 Laplace Transform of derivatives of a function.
Theorem 3.9. Let f(t) be a continuous function of t � 0 and it isof exponential order for large t and if f ′(t) is a piecewise continuousfunction for t � 0, then Laplace transform of the derivative f ′(t) existswhen p > σ and is given by
L[
f ′(t)]
= p L [ f(t) ] − f(0)
The Laplace Transform 111
Proof. By definition L [ f ′(t) ]
=∫ ∞
0e−pt f ′(t) dt
=[
e−pt. f(t)]∞0
−∫ ∞
0(−p) e−pt f(t) dt
= limt→∞ e−ptf(t) − f(o) + p L [ f(t) ] (3.10)
Since, f(t) is of exponential order σ for t → ∞, we have
|f(t)| � M eσt , t � 0
Hence, |f(t) e−pt| = e−pt |f(t)| � M e−pt · eσt = M e−(p−σ)t
So, limt→∞ e−pt f(t) = 0, as p − σ > 0
Thus from (3.10), we get
L[
f ′(t)]
= p L [ f(t) ] − f(0). (3.11)
Corollary 3.1. If f ′′(t) exists for t � 0 and is a piecewise continuousfunction, then proceeding as above we can extend the result of the abovetheorem as
L[
f ′′(t)]
= p L[
f ′(t)]− f ′(0).
= p [ p{L(f(t))} − f(0)] − f ′(0)
= p2 L [(f(t)] ] − pf(0) − f ′(0) (3.12)
Corollary 3.2. In general, if fn(t) exists for t � 0 and is a piecewisecontinuous function of t, then
L [fn(t)] = pn L [f(t)] − pn−1 f (0) − pn−2f ′(0) − · · · − f (n−1)(0)
(3.13)
Theorem 3.10. Suppose f(t) is not continuous at t = a, 0 < a < ∞and f(a − 0), f(a + 0) exist and f(t) is of requisite order for existenceof L [f(t)]. Further if f ′(t) exists, for t �= a > 0 then
L [f ′(t) ] = p [ f(p) ] − f(0) − e−ap [f(a + 0) − f(a − 0)]
Proof. By definition
L[
f ′(t)]
=∫ ∞
0e−pt. f ′(t) dt
112 An Introduction to Integral Transforms
=∫ a−0
0e−pt. f ′(t) dt +
∫ ∞
a+0e−pt f ′(t) dt
=[e−pt f(t)
]a−0
0+ p
∫ a−0
0e−ptf(t) dt
+[e−pt f(t)
]∞a+0
+ p
∫ ∞
a+0e−ptf(t) dt
=[e−p(a−0) f(a − 0) − f(0)
]+[0 − e−p(a+0) f(a + 0)
]+ p
[∫ a−0
0e−ptf(t) dt +
∫ ∞
a+0e−ptf(t) dt
]= p L [f(t)] − f(0) − e−ap [f(a + 0) − f(a − 0)] (3.14)
Corollary 3.3. If f(t) is discontinuous at t = a1, a2, · · · ak, theabove theorem 3.10 can easily be extended as
L[
f ′(t)]
= p L [ f(t) ] − f(0) −k∑
i=1
e−ai p{f(ai + 0) − f(ai − 0)}
(3.15)
Remark. 3.1 In the light of the above Theorem 3.9 and Theorem3.10, one can easily get the extension of the theorems for the Laplacetransforms of f ′′(t), f ′′′(t). · · · etc, provided that the functions satisfy therequisite properties for the existence of their Laplace transforms.
3.11 Laplace Transform of Integral of a function
. Theorem 3.11. If Laplace transform of a function f(t) is f(p). thenLaplace transform of
∫ t0 f(τ)dτ is f(p)/p.
Proof. Given that
L [f(t) ; t → p] =∫ ∞
0e−pt f(t) dt = f(p)
Now, L
[∫ t
0f(τ)dτ
]
=∫ ∞
0e−p t
[∫ t
0f(τ) d τ
]dt
=∫ ∞
0f(τ)
[∫ ∞
τe−pt dt
]dτ,
The Laplace Transform 113
by changing order of integration
=∫ ∞
0f (τ).
1p
e−p τ dτ
=1p
f(p). (3.16)
An alternative proof.
Let∫ t
0f(τ)dτ = g(t). Then, g′(t) = f(t) and g(0) = 0.
Now, L
[ ∫ t
0f(τ) dτ
]. = L [g(t)] .
But, by the theorem 3.9 of article 3.10, we know that
L[g′ (t)
]= p L [g(t)] − g(0) = p L [g(t)] .
Therefore, L [g (t) ] =1p
L[
g′ (t)].
=1p
L [ f (t) ] =1p
f(p)
Hence, L
[ ∫ t
0f (τ) dτ
]=
1p
f (p).
3.12 Laplace Transform of tnf(t)
Theorem 3.12. If Laplace transform of a piecewise continuous functionf(t) is f(p), then Laplace transform of tnf(t) is (−1)n dnf(p)
dpn , when n isa positive integer.
Proof. Before proving the above result, we may restate it as thederivative of the Laplace transform of a given function satisfying requisiteconditions for existence of the necessary results.
Given that, f(p) =∫ ∞
0e−pt f(t) dt
Differentiating both sides of the above equation with respect to p, weget
114 An Introduction to Integral Transforms
d f(p)dp
=d
dp
∫ ∞
0e−pt f(t)dt
= −∫ ∞
0t e−pt f(t)dt ,
valid for assumed nature of the function f(t)
= −L [ t f(t) ]
Thus, L [tf(t)] = (−1)1d
dpf(p) , (3.17)
which is the case of the theorem, when n = 1.
To prove the general case for any positive integral value of n, let usassume that for n = k
L [ tn f(t)] = (−1)ndn
dpnf(p) (3.18)
Hence, we have
L[
tk f(t)]
= (−1)kdk
dpkf(p)
implying,
∫ ∞
0e−pt{tkf(t)} dt = (−1)k
dk
dpkf(p) (3.19)
Differentiating both sides of (3.19) with respect to p once, we get∫ ∞
0e−pt {tk+1f(t)} dt = (−1)k+1 dk+1
dpk+1f(p)
implying, L[tk+1f(t)
]= (−1)k+1 dk+1
dpk+1f(p) (3.20)
Equation (3.20) shows that the result of the theorem is true for n = k+1,if it is true for n = k. But it is seen in equation (3.17) that it is true forn = 1. Hence, it is true for n = 1 + 1 = 2, n = 2 + 1 = 3, · · ·. Therefore,by the method of mathematical induction the required general result ofthe theorem is true for all positive integral value of n.
This result is sometimes known as the derivative of the transformedfunction.
3.13 Laplace Transform of f(t)t
Theorem 3.13. If Laplace transform of a piecewise continuous functionf(t) be f(p), then Laplace transform of f(t)/t is
∫∞p f(p)dp.
The Laplace Transform 115
Proof. Given that ∫ ∞
0f(t) e−pt dt = f(p)
Integrating both sides of the above equation with respect to p from p to∞, we get∫ ∞
pf(p1)dp1 =
∫ ∞
p
[ ∫ ∞
0e−p1tf(t) dt
]dp1,
=∫ ∞
0f(t)
{∫ ∞
pe−p1t dp1
}dt, by changing order of integration.
=∫ ∞
0
f(t)t
e−pt dt
= L [ f(t)/t ] . (3.21)
Thus, the result of the theorem is proved to be true. This result issometimes known as the integral of Laplace transformed function.
3.14 Laplace Transform of a periodic function.
Theorem 3.14. Let f(t) be a periodic function of period τ , so thatf(t + nτ) = f(t), for n = 1, 2, 3 · · · . If f(t) is a piecewise continuousfunction for t > 0, then
L [ f(t) ] =1
1 − e−pτ
∫ τ
0e−pt f(t) dt
Proof.
L [ f(t) ] =∫ ∞
0e−pt f(t) dt
=∫ τ
0e−pt f(t)dt +
∫ 2τ
τe−pt f(t) +
∫ 3τ
2τe−pt f(t) dt + · · ·
=∞∑
n=0
∫ (n+1)τ
nτe−pt f(t) dt
=∞∑
n=0
∫ τ
0e−p(x+nτ) f(x + nτ)dx, putting t = x + nτ
=∞∑
n=0
e−pnτ
∫ τ
0e−px f(x) dx
116 An Introduction to Integral Transforms
=[ ∫ τ
0e−px f(x)dx
] ∞∑n=0
e−pnτ
=∫ τ
0e−px f(x)dx .
11 − e−pτ
, since p > 0, e−pτ < 1.
=1
1 − e−pτ
∫ τ
0e−pt f(t)dt, changing x to t. (3.22)
3.15 The initial-value theorem and the final-value theo-
rem of Laplace Transform.
Theorem 3.15.
Let f(t) be a continuous function for all t � 0 and is of exponentialorder for large t and f ′(t) be a piecewise continuous function of t .Then,
limt→0
f(t) = limp→∞ p L [f(t); t → p]
This result is called the initial-value theorem.
Proof. We know by theorem 3.9 of article 3.10 that
L[f ′(t)
]=∫ ∞
0e−ptf ′(t) dt = p L [f(t)] − f(0)
Taking p → ∞, the above equation gives
limp→∞
∫ ∞
0e−ptf ′(t)dt =
∫ ∞
0lim
p→∞[e−ptf ′(t)
]dt = lim
p→∞ [p L (f(t) − f(0)]
Since f ′(t) is sectionally continuous and of exponential order, wehave from above that
limp→∞ [pL{f(t)} − f(0)] = 0
which implies f(0) = limp→∞ p [L{f(t)}]
Thus, limt→0
f(t) = limp→∞ p [L{ f(t)}] (3.23)
Theorem 3.16. Let f(t) be a continuous function for all t � 0 and isof exponential order and f ′(t) be a piecewise continuous function of t.Then the result of the final-value theorem states that
limt→∞ f(t) = lim
p→0p L [ f(t) ]
The Laplace Transform 117
Proof. As in theorem 3.15 above we have
limp→0
∫ ∞
0e−ptf ′ (t) dt = lim
p→0[ pL{f(t)} − f(0) ]
This result gives∫ ∞
0
[limp→0
e−pt
]f ′(t) dt = lim
p→0p L [f(t)] − f(0)
Or∫ ∞
0f ′(t) dt = [ f(t) ]∞0 = lim
p→0p L [f(t)] − f(0)
Or limt→∞ f(t) − f(0) = lim
p→0p L [ f(t) ] − f(0)
Or limt→∞ f(t) = lim
p→0p L [ f(t) ] (3.24)
Thus the result of the final value theorem of Laplace transform isproved.
3.16 Examples
Example 3.16 Find the Laplace transform of f(t) = sin at − at cos at
Solution. By linearity property of Laplace transform, we have
L [sin at − at cos at] = L [sin at] − L [ at cos at ]
=a
p2 + a2+ a
d
dpL [ cos at ] , by article 2.12
=a
p2 + a2+ a
d
dp
[p
p2 + a2
]
=a
p2 + a2+ a
{1
p2 + a2− 2p2
(p2 + a2)2
}
=2a
(p2 + a2)2[
a2]
=2a3
(p2 + a2)2
Example 3.17. Evaluate L[
sh tt
].
Solution. From article 3.13 that L[
f(t)t
]=∫∞p f(p) dp, where
f(p) = L [ f(t)] . Now choosing f(t) = sh t , we have
L
[sh t
t
]=∫ ∞
pL [sht ; t → p] dp
=∫ ∞
p
1p2 − 1
dp =12
logp − 1p + 1
118 An Introduction to Integral Transforms
Example 3.18. Evaluate L [sin at/t]. Does L [ cos at/t ] exist ?
Solution. As in the above example 2, we have
L [ sin at/t ] =∫ ∞
pL [ sin at ; t → p ] dp
=∫ ∞
p
a
p2 + a2dp =
[tan−1 p
a
]∞p
= cot−1 p
a
Thus,∫ ∞
0e−pt sin at
t= cot−1 (p/a).
Putting p = 0 and a = 1 we have,∫ ∞
0
sin t
tdt = lim
p→0
( π
2− tan−1 p
)=
π
2, an important result
Again, L [ cos at/t ] =∫ ∞
p
p
p2 + a2dp
=12[
log(
p2 + a2) ]∞
p
But, limp→∞ log (p2 + a2) is unbounded and therefore L
[cos at
t
]does not
exist.
Example 3.19. Evaluate L[
t1n
]and then evaluate L
[sin
√t]
and
finally evaluate L[
cos√
t / i.e cos√
t√t
√t].
Solution. From definition, we have
L[
t1n
]=∫ ∞
0e−pt. t
1n dt =
Γ(
1n + 1
)p
1n
+1
Again, since sin√
t = t12 − t
32
3!+
t52
5!− t
72
7!· · · · · · ,
we have
L[
sin√
t]
=Γ(3
2)
p32
− 13!
Γ(52 )
p5/2+
15!
Γ(72 )
p7/2· · · · · ·
=√
π
2p32
[1 − 1
4p+
12!
(14p
)2
− 13!
(14p
)3
+ · · · · · ·]
=√
π
2p32
e−14p .
Now considering f(t) = sin√
t , we get
f ′(t) =cos
√t
2√
t
The Laplace Transform 119
and also L[f ′(t)
]= p L [ f(t) ] − f(0) implying
L
[cos
√t
2√
t
]= p
√π
2p32
e− 1
4p
Thus, L
[cos
√t√
t
]=√
π
pe−
14p
Example 3.20. Let f(t) be a periodic function of period 4, where
f(t) =
{3t , 0 < t < 26 , 2 < t < 4
Evaluate. L[f(t)]
Solution. By the result of article 3.14 we have here
L[f(t) ; t → p] =1
1 − e−4p
∫ 4
0f(t) · e−ptdt
=1
1 − e−4p
[∫ 2
03t e−ptdt +
∫ 4
26 e−ptdt
]
=1
1 − e−4p
[3(1 − e−2p − 2p e−4p2
)p2
], after simplification.
Example 3.21. Evaluate L[cos at ; t → p] and hence deduce L[sin at].
Solution. From the definition
L[cos at] =∫ ∞
0e−pt cos at dt =
p
p2 + a2, by actual integration.
Again,
∫ t
0cos at dt =
sin at
a·
Hence, L
[sin at
a
]= L
[∫ t
0cos at dt
]
=1p
L[cos at] =1p· p
p2 + a2
This result implies,L[sin at] =
a
p2 + a2.
Example 3.22. Verify the result of (a) initial-value theorem and(b) final-value theorem for the function f(t) = e−at cos bt, where a > 0.
120 An Introduction to Integral Transforms
Solution. (a) It is seen that f(t) satisfy the necessary condition forapplication of these theorems. We have now
limt→0
f(t) = limt→0
e−at cos bt = 1
Also, L[f(t)] =(p + a)
(p + a)2 + b2· So, pL[f(t)] =
p(p + a)(p + a)2 + b2
Therefore, limp→∞ pL [ f(t) ; t → p ] = lim
p→∞p(p + a)
(p + a)2 + b2= 1
Thus, the initial value theorem is verified.
(b) Again, limt→∞ f(t) = lim
t→∞ e−at cos bt = 0
Also, limp→0
pL [ f(t) ; t → p ]
= limp→0
p(p + a)(p + a)2 + b2
= 0
Therefore, the final-value theorem of the above function is verified.
Example 3.23. Is it possible to use the initial-value or the final-valuetheorem to evaluate lim
t→0f(t) or to evaluate lim
t→∞ f(t), when f(t) =et sin t ? Give reason for your answer. Also comment on your answer.
Solution.
L[f(t)] = L[et sin t] =1
(p − 1)2 + 1Now, lim
t→0f(t) = lim
t→0et sin t = 0
limp→∞ pL[f(t)] = lim
p→∞p
(p − 1)2 + 1= 0
For this function, IVT is applicable.
Again, limt→∞ f(t) = lim
t→∞ et sin t = does not exist
Also, limp→0
pL [f(t)] = limp→0
p
(p − 1)2 + 1= 0
This shows that FVT is not applicable here.
The reason for it is that f(t) does not satisfy the condition of expo-nential order of f(t)for large t.
The above two results show that the conditions in the theorems arethough sufficient but not necessary.
The Laplace Transform 121
Example 3.24. Verify the result L[f ′′(t)] = p2f(p) − pf(0) − f ′(0) forthe case f(t) = et sin t .
Solution. We have f ′(t) = et(sin t + cos t) , f ′′(t) = 2et cos t
Now, L[f ′′(t)] = L[2et cos t] =2(p − 1)
(p − 1)2 + 1
Also, f(p) = L[f(t)] =1
(p − 1)2 + 1, f(0) = 0 , f ′(0) = 1
Therefore, p2f(p) − p f(0) − f ′(0)
=p2
(p − 1)2 + 1− p × 0 − 1 =
p2 − (p − 1)2 − 1(p − 1)2 + 1
=2(p − 1)
(p − 1)2 + 1. Thus, the result is verified.
Example 3.25. If y(t) = 12k [ekt
∫ t0 e−kudu− e−kt
∫ t0 ekudu] , and noting
y′′(t) − k2y(t) = 1 , verify that y(p) = 1p(p2−k2)
when k �= 0, after usingLaplace transform in article 2.10.
Solution. We have,
L[y′′(t) − k2y(t)] = L[y′′(t)] − k2L[y(t)]
= p2y(p) − py(0) − y′(0) − k2y(p)
= (p2 − k2) y(p) ≡ L[1] =1p
⇒ y(p) =1
p(p2 − k2)Also, y(p) = L [y(t)]
=1
2k2
[−2
p+
2pp2 − k2
]
=1
p (p2 − k2)
Thus, value of y(p) is verified.
3.17 Laplace Transform of some special functions.
I. The Heaviside functionWe have already introduced the unit step function H(t) defined by
H(t) =
{0 , t < 01 , t > 0,
122 An Introduction to Integral Transforms
This is usually known as “Heaviside unit function”. It may further benoted that any step function f(t) can always be expressed as a combi-nation of different Heaviside functions. For example, let a step functionbe defined by the relation
f(t) =
⎧⎪⎨⎪⎩
g(t) , 0 < t < a
k , a < t < b
0 , t > b
Then we can express it as
f(t) = g(t)H(a − t) + k[H(t − a) − H(t − b)]
Further, L[H(t)] =∫ ∞
0H(t)e−ptdt
=∫ ∞
01 · e−pt =
1p
implying, L−1
[1p
]= H(t)
II. The Dirac delta function.
A practical problem that arises in application in finding the Laplaceinversion of unity or H(t). Also, similar situation arises in handling unitimpulse function. Dirac introduced it as, δ(t) which acts for a very shorttime duration τ with a finite impulse 1 satisfying the relations
δ(t) = 0 , for t �= 0 (3.25)
and∫ ∞
−∞δ(t)dt =
∫ τ
0δ(t)dt = 1 (3.26)
In mathematical analysis, such function is unlike ordinary one and intro-duction of such a function sometimes leads to inconsistency in analysis.For this reason, Dirac called it as an “improper function” . It may beused only when no inconsistency do follow afterwards in analysis. Thus,entirely formally such function may be used in classical analysis withoutsearching for rigour in treatment. The equations (3.25) and (3.26) abovedo not depict any clear picture for δ(t). Since, the precise variation ofδ(t) with t in the interval is not an important issue but its effect is thepoint of study, we can use it properly with assumption that it does nothave any unnecessary violent oscillations in the neighbourhood of t = 0.
The Laplace Transform 123
For example, if we define a step function δ∈(t) by
δ∈(t) =
⎧⎪⎨⎪⎩
0 , t < 01∈ , 0 < t <∈0 , t >∈
(3.27)
then since∫ +∞
−∞δ∈ (t) dt =
∫ ∈
0
1∈ dt = 1 , (3.28)
it follows that lim∈→0
δ∈ (t) = δ(t) (3.29)
as defined in (3.25) and (3.26)
If f(t) be a continuous function in the neighbourhood of t = 0, then∫ +∞
−∞δ∈(t)f(t)dt =
∫ ∈
0δ∈(t)f(t)dt =
1∈∫ ∈
0f(t)dt
=f(θ ∈)
∈∫ ∈
0dt = f(θ ∈) , 0 < θ < 1 , by mean value theorem.
Thus,∫ +∞
−∞f(t)δ∈(t) dt = f(θ ∈) , 0 < θ < 1
Now, letting ∈→ 0 we get∫ +∞
−∞f(t) δ(t) dt = f(0) (3.30)
In general, putting∫ +∞
−∞f(ξ) δ(ξ − a) dξ
=∫ +∞
−∞f(t + a) δ(t) dt , putting ξ − a = t
= f(0 + a) = f(a) (3.31)
This is an important result which can be used formally in analysis when-ever no confusion do arise.
The Laplace transform of δ(t) is then given by definition that
L[δ(t) ; t → p] =∫ ∞
0e−ptδ(t)dt
= e−p.0 = 1 = H(t) , t > 0 (3.32)
Thus, L[δ(t)] = H(t)
and in turn L−1[H(t)] = δ(t) (3.33)
124 An Introduction to Integral Transforms
Hence, (3.32) and (3.33) are the required result obtained as proposed inthe beginning of this analysis.
III. Laplace Transform of the sine and the cosine integrals.The sine integral Si(t) is denoted and defined by
Si(t) =∫ t
0
sin u
udu
Now, L [ Si(t) ; t → p ] =∫ ∞
0e−pt
[∫ t
0
sinu
udu
]dt
=1p
∫ ∞
0
sin t
te−ptdt
=1p
[∫ ∞
pL [ sin t ] dp
], by 3.13
=1p
[∫ ∞
p
1p2 + 1
dp
]=
1p
[π2− tan−1 p
]
=1p
cot−1 p. (3.34)
Again, the cosine integral Ci(t) is denoted by
Ci(t) = −∫ ∞
t
cos u
udu
Therefore, L [ Ci(t) ; t → p ]
= −∫ ∞
0e−pt
[∫ ∞
t
cos u
udu
]dt
= −∫ ∞
0
cos u
u
[∫ u
0e−ptdt
]du ,
by changing the order of integration
= −1p
∫ ∞
0
cos u
u(1 − e−pu)du (3.35)
Again, f(p1) =∫ ∞
0f(u) e−p1udp1
Integrating this result with respect to p1 between the limits 0 to p, weget ∫ p
0f(p1) dp1 =
∫ p
0
[∫ ∞
0f(u) e−p1u du
]dp1
The Laplace Transform 125
Therefore,∫ p
0f(p1)dp1 =
∫ ∞
0f(u)
[∫ p
0e−p1udp1
]du ,
by changing the order of integration.
=∫ ∞
0
f(u)u
(1 − e−pu)du
Assuming, f(u) = cos u , we have f(p) = L[f(u) ; u → p] =p
p2 + 1
and hence, ∫ p
0
p1
p21 + 1
dp1 =∫ ∞
0
cos u
u(1 − e−pu) du (3.36)
Using (3.36) in (3.35) above, we get
L[Ci (t) ; t → p] = −1p
∫ p
0
p1 dp1
p21 + 1
= − 12p
log (1 + p2)
An alternative method.
Since, Ci (t) = −∫ ∞
t
cos u
udu
t Ci′ (t) = cos t
So, L[t Ci′ (t)] =p
p2 + 1⇒ − d
dp[L{C ′
i (t)}] =p
p2 + 1
or,d
dp[p L{Ci (t)} − Ci (0)] = − p
p2 + 1
or,d
dp[p L{Ci (t)}] = − p
p2 + 1
Integrating both sides with respect to p,we get
p L[Ci (t)] = −12
log (p2 + 1) + c where c is a constant
Now, by the final-value theorem
limp→0
p L[Ci (t)] = limt→∞Ci(t)
we get, limp→0
[−1
2log (p2 + 1) + c
]= − lim
t→∞
∫ ∞
t
cos u
udu = 0
Thus c = 0 ⇒ p L[Ci (t)] = −12
log (p2 + 1)
⇒ L[Ci (t)] = − 12p
(p2 + 1) .
126 An Introduction to Integral Transforms
IV. Laplace Transform of the Exponential integral.
The exponential integral is defined by
Ei (t) =∫ ∞
t
e−u
udu
Then, Ei′ (t) = −e−t
t⇒ t E′
i (t) = −e−t
Thus as before,
L[t Ei′ (t)] = − 1p + 1
⇒ − d
dp[ L{Ei′ (t)} ] = − 1
p + 1
Or,d
dp[ p L{Ei (t)} − Ei (0) ] =
1p + 1
Or,d
dp[ p L{Ei (t)} ] =
1p + 1
Integrating both sides with respect to p , we get
p L[Ei (t)] = log (p + 1) + c , a constant
Now, by final-value theorem, we get
limp→0
[ log (p + 1) + c ] = limt→∞Ei(t) = 0 ⇒ c = 0
Thus, L[Ei (t)] =1p
log (p + 1) (3.37)
V. Laplace Transform of the Error function.
The error function Erf(t) is defined by
Erf(t) =2√π
∫ t
0e−x2
dx
and the error complementary function Erfc(t) is also defined by
Erfc(t) =2√π
∫ ∞
te−x2
dx = 1 − 2√π
∫ t
0e−x2
dx (3.38)
= 1 − Erf(t),
since,2√π
∫ ∞
0e−x2
dx = 1 , by use of Gamma function.
Thus, by evaluating Laplace transform of Erf(t), the same for Erfc(t)can easily be evaluated.
The Laplace Transform 127
To make the evaluation steps relatively simpler, we evaluate Laplacetransform of Erf(
√t) instead of Erf(t) in the following steps. We know
that
Erf(√
t) =2√π
∫ √t
0e−x2
dx
=2√π
∫ √t
0
[1 − x2 +
x4
2!− x6
3!+ · · ·
]dx,
=2√π
[t1/2 − t3/2
3+
t5/2
5.2!− t7/2
7.3!+ · · ·
]
Thus,
L[Erf(√
t)] =2√π
[Γ(3/2)p3/2
− Γ(5/2)3 · p5/2
+Γ(7/2)
5.2! · p7/2− Γ(9/2)
7.3!p9/2+ · · ·
]
=1
p3/2
[1 − 1
2· 1p
+1.32.4
1p2
− 1.3.52.4.6
1p3
+ · · ·]
=1
p3/2
(1 +
1p
)− 12
=1
p(p + 1)1/2(3.39)
Therefore,
L[Erfc(√
t)] = L[1 − Erf(√
t)]
=1p− 1
p(p + 1)1/2=
(p + 1)1/2 − 1p(p + 1)1/2
(3.40)
VI. Laplace Transforms of Bessel functions.
The Bessel function of first kind of order ν, Jν(t) is defined by
Jν(t) =∞∑
r=0
(−1)r(
12 t)ν+2r
r! Γ(ν + r + 1).
Its Laplace transform is given by the equation
L [tνJν (t) ; t → p] =∞∑
r=0
(−1)ν2−ν−2r
r! Γ(r + ν + 1)L[t2ν+2r ; t → p
](3.41)
The duplication formula for the Gamma function is
22z−1 Γ(z) Γ(
z +12
)=
√π Γ(2z) (3.42)
Using (3.42) we have the relation
1Γ(r + ν + 1)
L[t2ν+2r; t → p
]=
Γ(2ν + 2r + 1)Γ(ν + r + 1)
p−2ν−2r−1
128 An Introduction to Integral Transforms
= π− 12 22ν+2r Γ
(ν + r +
12
)p−2ν−2r−1,(
Re p > 0 , ν > −12
)(3.43)
Therefore, results in eqns.(3.41) and (3.43) give
L[tν Jν(t) ; t → p] = 2ν π− 12 p−2ν−1
∞∑r=0
Γ(ν + 1
2 + r)
r!(−p−2
)rIf |p| > 1, the above series is convergent and it gives
L[tν Jν(t) ; t → p] =2ν Γ
(ν + 1
2
)Γ(
12
) (p2 + 1)−ν− 12 , Re p > 1 (3.44)
after using the Binomial theorem
(1 − x)−α =∞∑
r=0
Γ(α + r)xr
r! Γ(α), |x| < 1.
In particular, putting ν = 0 in (3.44), we get
L[J0(t) ; t → p] = (p2 + 1)−12 , Re p > 1
Letting p → 0 , it follows that∫∞0 J0(t)dt = 1.
Again by the result in eqn. (3.44), we have
L [ tν Jν(at) ; t → p ]
=(2a)ν Γ
(ν + 1
2
)√
π(p2 + a2)ν+ 12
, ν > −12
, Re p > a > 0
and L[J0(at) ; t → p] = (p2 + a2)−12 , Re p > a > 0
Also, L[e−atJ0(at) ; t → p] =1√
p2 + 2ap + 2a2
Deduction 3.1.
If Re p > 1 , like eqn. (3.41) one can write
L [ tν+1 Jν(at) ; t → p ] =∞∑
r=0
(−1)r 2−ν−2r Γ(2ν + 2r + 2)r! Γ(ν + r + 1) p2ν+2r+2
The Laplace Transform 129
At this stage, using the duplication formula of the Gamma function , itis found that
2−ν−2r Γ(2ν + 2r + 2)Γ(ν + r + 1)
= 2ν+1 1√π
Γ(
ν + r +32
),
and so,
L[tν+1Jν(t) ; t → p
]= 2ν+1 · 1√
π· 1p2ν+2
∞∑r=0
(− 1
p2
)ν Γ(ν + r + 3
2
)r!
= 2ν+1 1√π
p−2ν−2(1 + p−2
)−ν− 32 Γ
(ν +
32
),
after using Binomial theorem. Thus,
L[tν+1Jν(t) ; t → p
]=
2ν+1p Γ(ν + 3
2
)√
π(p2 + 1)ν+ 32
, Re p > 1
Therefore, L[tν+1Jν(at) ; t → p
]=
2ν+1 aνp Γ(ν + 3
2
)√
π(p2 + a2)ν+ 32
, Re p > a > 0
(3.45)
In particular, if ν = 0,
L [tJ0(at)] =p
(p2 + a2)32
.
Deductions 3.2. By principle of analytic continuation we can extendthe result of the definition of Bessel function of first kind for the per-missible range of argument of the variable as
tν2 Jν(2
√t) =
∞∑r=0
(−1)rtν+r
r! Γ(ν + r + 1)
and hence
L[t
ν2 Jν(2
√t) ; t → p
]=
∞∑r=0
(−1)r
r! Γ(r + ν + 1)L[tν+r ; t → p
]
= p−ν−1∞∑
r=0
(−1
p
)r
r!, if Re p > 0
= p−ν−1 e− 1
p , Re p > 0
Also, if a > 0 , Re p > 0. (3.46)
L[t
ν2 Jν(2
√at) ; t → p
]= a
ν2 p−ν−1e
− ap (3.47)
and L[J0(2
√at) ; t → p
]=
1pe− a
p .
130 An Introduction to Integral Transforms
Deduction 3.3. Modified Bessel function of the first kind Iν(t) ofthe variable t is given by
Iν(t) =∞∑
r=0
(12 t)ν+2r
r! Γ(ν + r + 1)
Therefore, by the same procedure as above, we get
L [tν Iν(t) ; t → p] = π− 12 2νp−2ν−1
∞∑r=0
Γ(ν + r + 1
2
)r!
p−2r
This result implies that if Re p > 1 , ν > −12 ,
L [tν Iν(t) ; t → p] =1√π
2ν Γ(
ν +12
)(p2 − 1)−ν− 1
2
Also, L [tν Iν(at) ; t → p] =(2a)ν Γ
(ν + 1
2
)√
π(p2 − a2)ν+ 12
, for Re p > a > 0
(3.48)
In a similar way one can prove that if ν > −12
L[tν+1Iν(at) ; t → p
]=
2ν+1aνp Γ(ν + 3
2
)√
π(p2 − a2)ν+ 32
, for Re p > a > 0 (3.49)
Again,
L[t
ν2 Iν(2
√at) ; t → p
]= a
ν2 p−ν−1e
ap , for Re p > 0 , a > 0 , ν > −1
2(3.50)
by similar arguments made above.
The special cases for ν = 0 are given by
L [I0(at) ; t → p] = (p2 − a2)−12 , Re p > a > 0 (3.51)
L [t I0(at) ; t → p] = p(p2 − a2)−12 , Re p > a > 0 (3.52)
and L[I0(2
√at) ; t → p
]= p−1 e
ap , Re p > 0 , a > 0 (3.53)
An important result due to Tricomi [cf Tricomi (1935)] can be statedas
L
[t
ν2
∫ ∞
0x− ν
2 Jν
(2√
xt)
f(x)dx ; t → p
]= p−ν−1 f
(1p
)(3.54)
We do not persue the derivation of this last result due to lack ofscope of this book.
The Laplace Transform 131
3.18 The Convolution of two functions.
Let f(t) and g(t) be two piecewise continuous functions and are of someexponential order for large t and for every t � 0. Then the convolutionof these functions is denoted by f ∗ g(t) and is defined by
f ∗ g(t) =∫ t
0f(u)g(t − u)du (3.55)
The above relation in eqn. (3.55) is also known as faltung or resultantof f(t) and g(t). By definition,
f ∗ g(t) =∫ t
0f(u)g(t − u) du
=∫ t
0f(t − τ) g(τ) dτ
= g ∗ f(t) (3.56)
Therefore, convolution of two functions satisfies the commutative law .It also satisfies the distributive law
f ∗ [g + h](t) = f ∗ g(t) + f ∗ h(t) (3.57)
and also the associative law
[f ∗ (g ∗ h)](t) = [(f ∗ g) ∗ h](t) , (3.58)
The two results in eqns. (3.57) and (3.58) can be directly verified fromthe definition (3.55). For example,
[(f ∗ g) ∗ h](t) =∫ t
0h(t − τ) dτ
[∫ τ
0f(ξ) g(τ − ξ) dξ
]
=∫ t
0f(ξ)dξ
∫ t
ξg(τ − ξ)h(t − τ)dτ, Changing the order of integration
=∫ t
0f(ξ) dξ
∫ t−ξ
0g(η) h(t − ξ − η) dη
= f ∗ (g ∗ h) (t)
We now turn our attention to evaluate Laplace transform of the convo-lution of two functions f(t) and g(t).
L[(f ∗ g)(t) ; t → p] = L
[∫ t
0f(τ) g(t − τ) dτ ; t → p
]
132 An Introduction to Integral Transforms
=∫ ∞
0e−pt
[∫ t
0f(τ) g(t − τ) dτ
]dt
=∫ ∞
0f(τ)
[∫ ∞
τe−pt g(t − τ) dt
]dτ
=∫ ∞
0f(τ) e−pτ
[∫ ∞
0e−pη g(η) dη
]dτ
= f(p) g(p) , (3.59)
where f(p) and g(p) are Laplace transforms of f(t) and g(t) respectively.
In particular, L[f ∗ f(t) ; t → p] =[f(p)
]2As an illustration, if f(t) = ta−1, a > 0, g(t) = tb−1, b > 0
L[ta−1; t → p
]=
Γ(a)pa
, L[tb−1 ; t → p
]=
Γ(b)pb
then L[(ta−1 ∗ tb−1) ; t → p
]=
Γ(a) Γ(b)pa+b
(3.60)
Also, it is known that
L[ta+b−1 ; t → p
]=
Γ(a + b)pa+b
(3.61)
Dividing the results of (3.60) by that of (3.61) one gets
β(a, b) =Γ(a) Γ(b)Γ(a + b)
=L[(
ta−1 ∗ tb−1)
; t → p]
L [ta+b−1 ; t → p]
3.19 Applications
Example 3.26. By considering Si(t), find its Laplace transform andhence prove that ∫ ∞
0
sin u
udu =
π
2
Solution. We have, from III of article 3.17 that
L[Si(t) ; t → p] =∫ ∞
0e−pt
[∫ t
0
sin u
udu
]dt
=1p
∫ ∞
0
sin t
te−pt dt
=1p
[∫ ∞
pL{sin t ; t → p}dp
]
The Laplace Transform 133
=1p
∫ ∞
p
1p2 + 1
dp =1p
[π2− tan−1 p
]
Thus,∫ ∞
0
sin t
te−pt dt =
π
2− tan−1 p (3.62)
In eqn.(3.62) making p → 0, we get∫ ∞
0
sin t
tdt =
π
2.
Example 3.27. Prove that L
[e−
t2
4 ; t → p
]=
√π ep2
Erfc (p) .
Solution. We know that
L [ Erfc{g(t)} ; t → p ] = L [ 1 − Erf{g(t)} ; t → p ]
=1p− L [Erf{g(t)} ; t → p]
Now, L
[e−
t2
4 ; t → p
]=∫ ∞
0e−pt e−
t2
4 dt
=∫ ∞
0e−�( t
2+p)2−p2
�dt = ep2
∫ ∞
0e−( t
2+p)2
dt
= ep2
∫ ∞
pe−u2 · 2 du , putting u = p +
t
2
= 2 ep2 × Erfc(p) ×√
π
2, by eqn. (3.38) of article 3.17 (iv)
=√
π ep2Erfc(p) .
Example 3.28. Find L [ Erf (t) ; t → p ] to show that its value is
expressed as 1p e
p2
4 Erfc (12 p).
Solution. We have from definition
L [ Erf(t) ; t → p ] =∫ ∞
0e−pt
[2√π
∫ t
0e−u2
du
]dt
=2√π
∫ ∞
0e−u2
[ ∫ ∞
ue−pt dt
]du
=2
p√
π
∫ ∞
0e−u2−pu du
=2
p√
π
∫ ∞
0e−( u+ p
2 )2
· e p2
4 du
=2ep2/4
p√
π
∫ ∞
p/2e−v2
dv
134 An Introduction to Integral Transforms
=2ep2/4
p√
π· Erfc
( p
2
)·√
π
2
=1p
ep2/4 Erfc
(12p
)
Deduction.
L [ Erf (bt) ; t → p ] =1b
L[
Erf (t); t → p
b
]=
b
p· 1
be
p2
4b2 · Erfc( p
2b
), by the above result.
Example 3.29. If f(t) = | sin t|, prove that L [ | sin t| ; t → p ] is equalto coth (pπ
2 )/(1 + p2).
Solution. Clearly, f(t) = | sin t| is a periodic function of principalperiod τ = π. Hence
L [ f(t) ; t → p ] = L [ | sin t | ; t → p ]
=1
1 − e−pπ
∫ π
0e−pt | sin t| dt
=1
1 − e−pπ
∫ π
0e−pt sin t dt =
11 − e−pn
· 1 + e−pπ
1 + p2
=1
1 + p2coth
(pπ
2
)
Example 3.30. Find the Laplace transform of f(t) =∫∞0 cos tx2 dx
Solution. We have, by definition
L
[ ∫ ∞
0cos tx2dx
]=∫ ∞
0e−pt
[ ∫ ∞
0cos tx2 dx
]dt
=∫ ∞
0
[ ∫ ∞
0cos t x2 .e−pt dt
]dx ,
after changing order of integration.
=∫ ∞
0
p
p2 + x4dx
=1
2√
p
∫ π2
0
√cot θ dθ, where x2 = p tan θ
=π
2√
2p
The Laplace Transform 135
Example 3.31. Find (i) L [ f(t) ] and (ii) L [ f ′(t) ] where f(t) isdefined by
f(t) =
{2t , 0 � t < 1t , t > 1
Solution.
(i) L[f(t) ; t → p]
=∫ 1
02t · e−ptdt +
∫ ∞
1t e−ptdt
=2p2
−(
1p
+1p2
)e−p , after evaluation of integrals
(ii) Since f(t) =
{2t , 0 � t < 1t , t > 1
, we have f ′(t) =
{2, 0 � t < 11, t > 1
Therefore, L[f ′(t) ; t → p] =∫ 1
02e−pt dt +
∫ ∞
1e−pt dt
=2 − e−p
p.
Example 3.32. The Laguerre polynomial Ln(t) is defined by
Ln(t) =et
n!dn
dtn[
e−t tn]
, n = 0, 1, 2, . . .
Prove that
L [Ln(t) ; t → p ] =(p − 1)n
pn+1, n = 0, 1, 2, . . .
Solution. By definition,
L [ Ln(t) ;→ p ] =∫ ∞
0
e−t(p−1)
n!dn
dtn(e−t tn) dt
=1n!
(p − 1)∫ ∞
0e−t(p−1) dn−1
dtn−1
(e−t · tn
)dt
= · · · · · ·=
(p − 1)n
n!
∫ ∞
0e−pt · tn dt
=(p − 1)n
n!Γ(n + 1)
pn+1
=(p − 1)n
pn+1.
136 An Introduction to Integral Transforms
Example 3.33. If f(t) = | sin t| , prove that f(t) is a periodic functionof t of principal period τ = π. Find the Laplace transform of f(t).
Solution. Here
f(t) = | sin(t)|, f(t + τ) = f(t + π) = | sin(π + t)|= | − sin t| = | sin t| = f(t)
f(t + nτ) = | sin t| = f(t), n = 0, 1, 2, · · ·Hence, f(t) is a periodic function of period τ = π. Therefore,
L [ f(t) ; t → p ]
= L [ | sin t| ; t → p ]
=1
1 − e−pπ
∫ π
0f(t)e−pt dt
=1
1 − e−pπ
∫ π
0| sin t| e−pt dt
=1
1 − e−pπ
∫ π
0sin t · e−pt dt
=e−pπ + 11 − e−pπ
· 1p2 + 1
, Re p > 0
= cothp π
2/(p2 + 1) , Re p > 0
Example 3.34.
Let f(t) =
{1 , 0 � t < τ/2
−1 , τ/2 < t < τ
and f(t) is a periodic function of principal period τ . Prove that
L [ f(t) ; t → p] =1p
tanhpτ
2·
Solution. We can express f(t) = H(t) − 2 H(
t − τ2
), 0 < t < τ ,
and f(t + rτ) = f(t), for all t > 0.
L [ f(t) ; t → p ] =1
1 − e−pτ
∫ τ
0f(t)e−pt dt
=1
1 − e−pτ
[∫ τ2
0e−pt dt −
∫ τ
τ/2e−pt dt
]
=1
p ( 1 − e−pτ )
[1 − e−pτ/2
]2=
1p
tanhpτ
2, Re p > 0
The Laplace Transform 137
In particular, if τ = π we get
L [ f(t) ; t → p ]τ=π =1p
tanhpπ
2, Re p > 0
Example 3.35. By considering the Fourier series of a periodic functionf(t) of principal period 2π, find Laplace transform of f(t) to deduce
a0
2p+
∞∑n=1
pan + n bn
p2 + n2=
11 − e−2pπ
∫ 2π
0f(u) e−pu du
Hence, deduce the Mittag-Leffler identity
12
+2pπ
∞∑r=1
[p2 + (2r − 1)2
]−1 =(1 + e−pπ
)−1
for f(t) =
{1, 0 < t < π
0, π < t < 2π
Solution. The Fourier series corresponding to f(t) is given by
f(t) � a0
2+
∞∑n=1
[ an cos nt + bn sin nt ] (1)
where an = 1π
∫ 2π0 f(u) cos nu du
bn = 1π
∫ 2π0 f(u) sin nu du
⎫⎪⎬⎪⎭ (2)
Taking Laplace transform of both sides of eqn.(1) one gets
11 − e−2pπ
∫ 2π
0f(u)e−pudu =
a0
2p+
∞∑n=1
pan + nbn
p2 + n2(3)
where an and bn are given in eqns.(2).
As an example let f(t) =
{1 , 0 < t < π
0 , π < t < 2π
be a periodic function of period 2π. Then,
a0 = 1, an = 0 for n = 1, 2, · · ·
and b2n = 0, b2n−1 =(
n − 12
)−1
π, n = 1, 2, 3, · · ·
Also,∫ 2π
0f(u)e−pudu =
∫ π
0e−pudu =
1p
(1 − e−pπ
),
138 An Introduction to Integral Transforms
Substituting these results in eqn. (3), we get
12
+2pπ
∞∑n=1
1p2 + (2n − 1)2
=1
1 + e−pπ,
which is the required result.
Example 3.36. Using the definition of convolution of two functions,find
L
[∫ t
0J0(τ) J0(t − τ) dτ ; t → p
]
in its simplest form and hence evaluate∫ t0 J0(τ)J0(t − τ)dτ .
Solution. Let f(t) = J0(t) and g(t) = J0(t). Then
f(t) ∗ g(t) =∫ t
0f(τ) g(t − τ) dτ
=∫ t
0J0(τ) J0(t − τ) dτ
∴ L
[∫ t
0J0(τ) J0(t − τ)dτ
]= L [f ∗ g(t)]
= f(p) g(p)
=[f(p)
]2, since f(t) = g(t) ⇒ f(p) = g(p)
∴ L
[∫ t
0J0(τ) J0(t − τ)dτ ; t → p
]=
1p2 + 1
This result leads to
L−1
[L
{∫ t
0J0(τ) J0(t − τ) dτ
}]= L−1
[1
p2 + 1
]= sin t
Or∫ t
0J0(τ) J0(t − τ) dτ = sin t
Excercises
(1) Find the Laplace transform of
f(t) =
{0 , 0 < t < 1
(t − 1)2 , t > 1[Ans. 2e−p
p3
]
The Laplace Transform 139
(2) Evaluate
L [ f(t) ; t → p ] where
(i) f(t) =
{sin t , 0 < t < π
0 , t > π
(ii) f(t) = sin√
t[Ans. (i)(1 + e−pπ) / (p2 + 1) (ii)
√π e−1/4p / (2p
32 )]
(3) Find Laplace transform of f(t), where
(i) f(t) = sh at cos at
(ii) f(t) = t e−t(3 sh 2t − 5 ch 2t)
(iii) f(t) = [ g(t) cos wt ] if L[g(t)] = g(p)[Ans. (i) a(p2 − 2a2)/(p4 + 4a4) (ii) − d
dp
[1−5p
p2+2p−3
]
(iii) 12 [ g(p − iw) + g(p + iw) ]
](4) Evaluate L [(sin at − at cos at) ; t → p] and hence find
limt→0
L [ sin at − at cos at ; t → p ].
[Ans. 2a3/(p2 + a2)2 , p > 0 ; 0]
(5) Using Laplace transform evaluate I =∫∞0 t e−2tdt to prove
that I = 325 .
(6) Prove that∫∞0 (e−at − e−bt)/t dt = log(b/a) .Hence, find the value
of∫∞0 (e−t − e−3t)dt
t
[Ans. log 3]
(7) Prove that L [ J1(t) ; t → p ] = 1 − p√p2+1
and L [ t J1(t) ; t → p ] = 1(p2+1)3/2
(8) Prove that(i) L[t erf(2
√t) ; t → p] = (3p+8)
p2(p2+4)3/2
140 An Introduction to Integral Transforms
(ii) L[e3terf√
t ; t → p] = 1(p−3)
√p−2
(9) Prove that∫∞0
f(t)t dt =
∫∞0 f(p)dp , provided that the integrals
converge.
(10) If L[f(t) ; t → p] = f(p) , show that L[∫∞
0f(u)
u du ; t → p]
= 1p
∫∞p f(x)dx . Hence, show that L
[∫ t0
sinuu du ; t → p
]= 1
p cot−1 p
(11) Prove that L[Erf (t) ; t → p] = 2p√
π
∫∞0 e−u2−pu du
(12) Prove that L[tn eiat ; t → p] = n!(p+ia)n+1(p2 +a2)−n−1 , wheren is a positive integer.
Chapter 4
THE INVERSE LAPLACE TRANSFORM
AND
APPLICATION OF LAPLACE TRANSFORM
4.1 Introduction.
If f(t) belongs to a class A (meaning that it is a piecewise continuousfunction over 0 ≤ t < ∞ and is of some exponential order), its Laplacetransform f(p) exists. This is denoted symbolically by
L [f(t); t → p] =∫ ∞
0e−pt f(t) dt ≡ f(p) (4.1)
Then, in turn f(t), the object function is the inverse Laplace transformof the image function f(p) and is given symbolically by
L−1[f(p); p → t
] ≡ f(t) (4.2)
Thus by eqn. (4.1), one can evaluate f(p) for a given f(t), since f(p)exists. We shall now consider the inverse problem - deriving informationfrom the prescribed f(p) that enable us to derive the original functionf(t) through some formula, called the Laplace inversion formula.
Before answering this basic problem raised above, we shall at firstpresent heuristic approach of determining the inverse Laplace transformof some given f(p). Also as per need we define a new function, calledthe null function in this connection together with a connected theorem,known as Lerch’s theorem.
Definition.Null function. A function N(t) satisfying the condition
∫ t0 N (τ) dτ = 0,
for all t > 0 is called a null function.
Theorem 4.1.Lerch’s Theorem. If L [ f1(t) ; t → p ] , Re p > c1 and
142 An Introduction to Integral Transforms
L [ f2(t) ; t → p ] , Re p > c2 both exists and if
L [ f1(t) ; t → p ] = L [ f2(t) ; t → p ]
for Rep > c = max(c1, c2), then f2(t) − f1(t) = N(t)
In addition, if f1(t) and f2(t) are continuous on the whole real line, thenf1(t) = f2(t) , t > 0.
Proof. We shall not persue here the proof of this theorem. But thecomment that the inverse Laplace transform of a given f(p) is uniqueexcept for a null function, should be kept in mind.
In the proposed heuristic approach, we reduce the given function f(p)as a combination of some functions which are the Laplace transformsof known functions. Then applying the idea of the Lerch’s theorem,the inverse Laplace transform of the given f(p) can be obtained byusing the inverse results already discussed in the previous article of thischapter. This approach will be just like evaluation of the integrationas an inverse process of differentiation of a given function, called theintegrand, after reducing it to sum of functions which are the derivativesof known standard functions. Sometimes for reduction of f(p) into sumof standard expressions for manipulation of Laplace inversion, partialfraction rules are found useful. Here, again the linearity property ofLaplace inversion holds.
Theorem 4.2. If f1(p) = L [f1(t); t → p] and f2(p) = L [f2(t); t → p]and c1, c2 be two arbitrary constants, then
L−1[
c1 f1(p) + c2 f2(p) ; p → t]
= c1 L−1[f1(p) ; p → t
]+ c2 L−1
[f2(p) ; p → t
]= c1 f1(t) + c2 f2(t).
Proof. Since, L−1[f1(p) ; p → t
]= f1(t) and
L−1[f2(p) ; p → t
]= f2(t)
under the given condition, we have by the linearity property of theLaplace transform
L [c1 f1(t) + c2 f2(t); t → p] = c1L [f1(t); t → p] + c2L [f2(t) ; t → p]
= c1f1(p) + c2f2(p).
The Inverse Laplace Transform 143
Hence, L−1 L [c1 f1(t) + c2 f2(t) ; t → p]
= L−1[c1f1(p) +c2 f2(p) ; p → t
]
implying, L−1[c1 f1(p) + c2 f2(p) ; p → t
]= c1 f1 (t) + c2 f2 (t)
= c1L−1[
f1(p) ; p → t]+ c2L
−1[
f2(p) ; p → t]
Thus, the theorem is proved.
4.2 Calculation of Laplace inversion of some elementary
functions.
Before finding Laplace inversion of elementary functions which are a bitcomplex in nature, we state below some results obtained directly fromthe previous articles in 3.5 and 3.9. These results may be considered asformulae for further applications.
(i) L [ H(t ) ; t → p ] = 1p implies L−1
[1p ; p → t
]= H(t)
(ii) L [tν ; t → p] = Γ(ν+1)pν+1 implies L−1
[Γ(ν+1)pν+1 ; p → t
]= tν , Re ν > −1
(iii) L[eat; t → p
]= 1
p−a implies L−1[
1p−a ; p → t
]= eat, Re p > a
(iv) L [sinat; t → p] = ap2+a2 implies L−1
[a
p2+a2 ; p → t]
= sinata , p > 0
(v) L [cos at; t → p] = pp2+a2 implies L−1
[p
p2+a2 ; p → t]
= cos at, p > 0
(vi) L [sh at ; t → p] = ap2−a2 implies L−1
[1
p2−a2 ; p → t]
= sh ata ,
p > a > 0
(vii) L [ch at ; t → p] = pp2−a2 ; implies L−1
[p
p2−a2 ; p → t]
= ch at,p > a > 0
(viii) L[tνe−at; t → p
]= Γ(ν+1)
(p+a)ν+1 implies
L−1[
Γ(ν+1)
(p+a)ν+1 ; p → t]
= tνe−at, Re ν + 1 > 0
144 An Introduction to Integral Transforms
(ix) L [ sin(t − a)H(t − a); t → p] = e−pa
1+p2 implies
L−1[
e−pa
1+p2 ; p → t]
= sin(t − a) H(t − a)
(x) L [cos(t − a)H(t − a); t → p] = pe−pa
1+p2 implies
L−1[
pe−pa
1+p2 ; p → t]
= cos(t − a) H(t − a)
(xi) L[sh(t − a)H(t − a); t → p] = e−pa
p2−1implies
L−1[ e−pa
p2−1; p → t] = sh(t − a)H(t − a)
(xii) L [ch(t − a)H(t − a); t → p] = pe−pa
p2−1implies
L−1[
pe−pa
p2−1 ; p → t]
= ch(t − a) H(t − a)
We now turn our attention in discussing some rules of manipulationof Laplace inversion of some combinations of elementary functions of p
through the following examples.
Example 4.1. Evaluate L−1[
1(p+1)(p2+1)
; p → t]
Solution. Resolving into partial fractions under usual method we have
1(p2 + 1)(p + 1)
≡ 12
1p + 1
− 12
p
p2 + 1+
12
1p2 + 1
Therefore,
L−1
[1
(p + 1)(p2 + 1)
]=
12L−1
[1
p + 1
]− 1
2L−1
[p
p2 + 1
]+
12L−1
[1
p2 + 1
]
=12[e−t − cos t + sin t
]
Example 4.2. Evaluate L−1[
6p2+22p+18p3+6p2+11p+6
]Solution.
L−1
[6p2 + 22p + 18
p3 + 6p2 + 11p + 6
]
= L−1
[6p2 + 22p + 18
(p + 1)(p + 2)(p + 3)
]
= L−1
[1
p + 1+
2p + 2
+3
p + 3
]= e−t + 2e−2t + 3e−3t
The Inverse Laplace Transform 145
Example 4.3. Evaluate L−1
[ (√p−1p
)2]
Solution.
L−1
[ (√p − 1p
)2]
= L−1
[p − 2
√p + 1
p2
]
= L−1
[1p− 2
p3/2+
1p2
]
= H(t) − 4
√t
π+ t
4.3 Method of expansion into partial fractions of the ratio
of two polynomials
Let f(p) and g(p) be two polynomials in p such that the degree of f(p)is less than that of g(p). Then f(p)/g(p) can be expressed as
f(p)g(p)
=n∑
r=1
Ar
p − ar, where p − ar are the factors of g(p)
with Ar = limp→ar
(p − ar) f(p)g(p)
=f(ar)g′(ar)
, for r = 1, 2, · · · n.
To prove the above result, we write
f(p)g(p)
≡ A1
p − a1+
A2
p − a2+ · · · An
p − an(4.3)
Then, limp→ar
(p − ar) f(p))g(p)
= Ar, the other terms being zero in the limit.
Therefore
Ar = limp→ar
f(p)g(p)/p − ar
=f(ar)g′(ar)
, provided g′(ar) �= 0
meaning that p − ar is a non-repeated factor of g(p). Thus, the resultstated above is proved.
Example 4.4. Evaluate L−1[
p+5(p+1)(p2+1)
]
146 An Introduction to Integral Transforms
Solution. We express by partial fraction that[p + 5
(p + 1)(p2 + 1)
]≡ A1
p + 1+
A2
p + i+
A3
p − i
=2
p − 1+
i−52(1+i)
p + i+
i+52(i−1)
p − i
=2
p + 1+
−2pp2 + 1
+3
p2 + 1,
Therefore, L−1
[p + 5
(p + 1)(p2 + 1); p → t
]= 2e−t − 2 cos t + 3 sin t
Example 4.5. Evaluate L−1[
p2−6p3+4p2+3p
]Solution. We express, by partial fraction that
p2 − 6p3 + 4p2 + 3p
≡ A1
p+
A2
p + 1+
A3
p + 3=
f(p)g(p)
, say
where f(p) = p2 − 6, g(p) = p3 + 4p2 + 3p, g′(p) = 3p2 + 8p + 3
Then we have
A1 =f(0)g′(0)
= −2, A2 =f(−1)g′(−1)
=52, A3 =
f(−3)g′(−3)
=12
Hence,
L−1
[p2 − 6
p3 + 4p2 + 3p
]=
52
e−t +12
e−3t − 2H(t)
Example 4.6. Evaluate L−1[
4p+5(p−1)2(p+2)
]Solution. By method of partial fraction we express
4p + 5(p − 1)2(p + 2)
≡ A
p − 1+
B
(p − 1)2+
C
p + 2Then, 4p + 5 ≡ A(p − 1)(p + 2) + B(p + 2) + C(p − 1)2
From this identity we get
9 = 3B, ( by putting p = 1) =⇒ B = 3
−3 = 9C, ( by putting p = −2) =⇒ C = −13
and 5=−2A + 2B + C, (by equating term independent of p) =⇒ A =13
The Inverse Laplace Transform 147
Therefore , L−1
[4p + 5
(p − 1)2(p + 2)
]=
13et + 3L−1
[1
(p − 1)2
]− 1
3e−2t
=13et − 1
3e−2t + 3 t et, by article 4.2
Example 4.7. Evaluate L−1[
1√2p+3
], after proving
L−1[f (ap + b)
]=(
1a
)e−
bta f(
ta
),where L−1
[f(p)
]= f(t)
Solution. Given that
f(p) =∫ ∞
0e−ptf(t) dt , by definition.
So, f(ap + b) =∫ ∞
0e−pat . e−bt f(t) dt
=1a
∫ ∞
0e−px e−
bxa f
(x
a
)dx , putting at = x
= L
[1a
e−bta f
(t
a
) ]
⇒ L−1[
f(ap + b)]
=1a
e−bta f
(t
a
).
Now putting a = 2, b = 3 we get.
f(p) =1√p
=√
π√π√
p=
1√π
Γ(
12
)p
12
so that f(t) =1√πt
. Therefore, f
(t
a
)= f
(t
2
)=
√2√
π t
These results give,
L−1
[1√
2p + 3
]=
12
e−32t
√2√
π t=
1√2πt
e−3t2
Example. 4.8. Evaluate
L−1
[(p + 1) e−πp
p2 + p + 1
]
Solution. We have
L−1
[p + 1
p2 + p + 1
]
148 An Introduction to Integral Transforms
= L−1
⎡⎢⎣ p + 1
2 + 12
(p + 12)2 +
( √3
2
)2
⎤⎥⎦
= L−1
⎡⎢⎣ p + 1
2
(p + 12)2 +
( √3
2
)2
⎤⎥⎦+
1√3
L−1
⎡⎢⎣
√3
2
(p + 12)2 +
( √3
2
)2
⎤⎥⎦
= e−t2 L−1
⎡⎢⎣ p
p2 +( √
32
)2
⎤⎥⎦+
e−t2√3
L−1
⎡⎢⎣
√3
2
p2 +( √
32
)2
⎤⎥⎦
= e−t2 cos
( √3
2t
)+ e−
t2
1√3
sin
( √3
2t
)
∴ L−1
[(p + 1)e−πp
p2 + p + 1
]
= e−(t−π)
2
[cos
{√3
2(t − π)
}+
1√3
sin
{√3
2(t − π)
}]· H(t − π),
by the second shift theorem of Laplace transform
Example. 4.9. Evaluate L−1[
3(1+e−pπ)p2+9
]Solution. We have
L−1
[3(1 + e−pπ)
p2 + 9
]
= L−1
[3
p2 + 32+ e−pπ 3
p2 + 32
]
= L−1
[3
p2 + 32
]+ L−1
[e−pπ 3
p2 + 32
]= sin 3t + H(t − π) sin 3 (t − π)
= sin 3t − sin 3 t. H(t − π)
Example 4.10. Using the result of inverse Laplace transform evaluateL−1
[p+1
(p2+2p+1)2
].
Solution. We know that L [tn f(t)] = (−1)n dn
dpn f(p), for n = 1, 2, 3, · · · .This result implies
L−1
[dn
dpnf(p)
]= (−1)tnf(t) , where L[f(t)] = f(p)
The Inverse Laplace Transform 149
Now, L−1
[p + 1
{ (p + 1)2 + 1 }2
]
= e−t L−1
[p
(p2 + 1)2
]
= −e−t L−1
[12
d
dp.
(1
(p2 + 1)
) ]
= −e−t . (−1)1t
2sin t
= e−t t
2sin t
Example 4.11. Evaluate L−1[
log p+3p+2
]Solution. Let f(p) =
[log p+3
p+2
]. We wish to calculate f(t).
Now,d
dpf(p) =
1p + 3
− 1p + 2
L−1
[d
dpf(p)
]= e−3t − e−2t
Therefore, −L−1
[d
dpf(p)
]= tf(t) = e−2t − e−3t
implying, f(t) =e−2t − e−3t
t
Example 4.12. Evaluate L−1[
log(
1 − 1p2
) ]Solution. Let
f(p) = log(
1 − 1p2
)⇒ − d
dp
[f(p)
]=
2p− 2p
p2 − 1
∴ −L−1
[d
dpf(p)
]= L−1
[2p− 2p
p2 − 1
]⇒ tf(t) = 2H(t) − 2 cosh t
Or, f(t) = 2 [ H(t) − cosh t ] /t.
Example 4.13. Assuming that limt→∞
f(t)t do exist, prove that
L−1[
f(p)p
]=∫ t0 f(x) dx. Hence, evaluate L−1
[1
p3(p2+1)
]Solution. First part is discussed in article 2.10.
By the above result, L−1[
1p
(1
p2+1
) ]=∫ t0 sin x dx = 1 − cos t
L−1
[1p
{1
p(p2 + 1)
} ]=∫ t
0(1 − cos x) dx = t − sin t.
150 An Introduction to Integral Transforms
Hence, L−1
[1p
{1
p2(1 + p2)
}]=∫ t
0(x − sin x) dx =
t2
2+ cos t − 1.
Example 4.14. Evaluate
L−1
[1
p(p + 1)3
]
Solution. We have[1
p(p + 1)3
]= L−1
[1
{(p + 1) − 1 }(p + 1)3
]
= e−tL−1
[1
(p − 1)p3
]
Now, L−1
[1
p − 1
]= et
∴ L−1
[1
p3(p − 1)
]=∫ t
0
[ ∫ t
0
{ ∫ t
0etdt
}dt
]dt
=∫ t
0
[ ∫ t
0
(et − 1
)dt
]dt =
∫ t
0
(et − 1 − t
)dt
= et − 1 − t − t2
2
∴ L−1
[1
p(p + 1)3
]= e−t
[et −
(1 + t +
t2
2
)]
= 1 − e−t
(1 + t +
t2
2
)
Example 4.15. If L−1[
p(p2+1)2
]= 1
2 t sin t, evaluate L−1[
1(p2+1)2
]Solution. We have
L−1
[1
(p2 + 1)2
]= L−1
[1p
p
(p2 + 1)2
]
=∫ t
0
(12
x sin x
)dx
=12
[sin t − t cos t]
Example 4.16. By using the convolution theorem evaluate L−1[
1p2(p+1)2
]to show that its value is t e−t + 2 e−t + t − 2.
Solution. We have by convolution theorem
L−1[f(p) · g(p)] = f ∗ g =∫ t
0f(τ)g(t − τ) dτ
The Inverse Laplace Transform 151
Now, L−1
[1p2
]= t and L−1
[1
(p + 1)2
]= (t)e−t.
Therefore, L−1
[1p2
.1
(p + 1)2
]=∫ t
0e−u . u(t − u)du
= te−t + 2e−t + t − 2 , on simplification.
Example 4.17. Evaluate L−1[
p(p2+4)3
]by using covolution theorem.
Solution. We know that L−1[
1(p2+4)
]= sin 2t
2
and L−1
[p
(p2 + 4)2
]= L−1
[−1
2d
dp
1(p2 + 4)
]
= −12.t(−1)1 L−1
[1
p2 + 4
]
=12t
sin 2t2
. =14
t sin 2t
Therefore, L−1
[p
(p2 + 4)2.
1p2 + 4
]=∫ t
0
u sin 2u4
.12
sin 2(t − u)du
=18
∫ t
0u sin 2u( sin 2t cos 2u − cos 2t sin 2u) du
=sin 2t
8
∫ t
0u sin 2u cos 2udu − cos 2t
8
∫ t
0u sin2 2u du
=164
[ sin 2t − 2t cos 2t ] ,
after evalution of the integrals and on simplification.
Example 4.18. Prove that L−1[
1p√
p+4
]= 1
2 erf (2√
t).
Solution. We know that
L−1
[1p
]= H(t)
and L−1
[1√
p + 4
]= e−4t L−1
[1√p
]=
e−4t
√πt
Therefore, L−1
[1p
1√p + 4
]=∫ t
0
e−4τ
√πτ
H(t − τ)dτ
=∫ t
0
e−4τ
√4τ
dτ =1√π
∫ 2√
t
0e−u2
du, on putting 4τ = u2
=12
erf (2√
t) , by definition of erf(x) =2√π
∫ x
0e−α2
dα
152 An Introduction to Integral Transforms
Example 4.19. Evaluate L−1[
1√p(p−a)
]by the convolution theorem.
Hence deduce the value of L−1[
1p√
p+a
]Solution. We know that L−1
[1√p
]= 1√
πtand L−1
[1
(p−a)
]= eat
Therefore, by the convolution theorem
L−1
[1√p
· 1(p − a)
]=∫ t
0
1√πu
ea(t−u)du
=eat
√π
∫ t
0
1√u
e−au du , putting au = ς2
=eat
√a
· 2√π
∫ √at
0e−ς2dς
=eat
√a
erf(√
at).
We have now
L−1
[1
p√
p + a
]= L−1
[1
(p + a) − a· 1√
p + a
]
e−at L−1
[1
(p − a)· 1√
p
]
= e−at eat
√a
erf(√
at) =1√a
erf(√
at).
Example 4.20. A periodic function f(t) of period 2π having a finitediscontinuity at t = π is given by
f(t) =
{sin t , 0 � t < π
cos t , π < t � 2π
Evaluate its Laplace transform
Solution The Laplace transform of f(t) is given by
f(p) =1
1 − e−2πp
[∫ 2π
0f(t)e−pt dt
]
=1
1 − e−2πp
[∫ π−0
0e−pt sin t dt +
∫ 2π
π+0e−pt sin t dt
]
=1
1 − e−2pπ
[1 + e−pπ
1 + p2− pe−pπ(1 + e−pπ)
1 + p2
]
=1
1 − e−pπ
[1 − pe−pπ
1 + p2
]
The Inverse Laplace Transform 153
4.4 The general evaluation technique of inverse Laplace
Transform.
The inverse Laplace transform has already been discussed in articles 3.3and 3.4 and is given by
f(t) = L−1[f(p) ; p → t] =1
2πiP.V.
∫ c+i∞
c−i∞eptf(p)dp , t > 0 (4.4)
where Re p = c > 0 and an arbitrary constant, provided f(t) = 0(eγt),Re p > γ with
f(p) = L [ f(t); t → p ] =∫ ∞
0f(t)e−pt dt (4.5)
where f(p) is being the function of a complex variable p in the half-planeRe p > γ. The integral in (4.4) is called a Bromwich integral.
The working rule for determining the inverse Laplace transform f(t)of f(p) with the help of the contour integral of a complex variable isshown below:
Since, f(t) =1
2πiP.V.
∫ c+i∞
c−i∞eptf(p) dp, t > 0,
to evaluate f(t) by (4.4), we evaluate the contour integral
12πi
∫C
eptf(p) dp, (4.6)
where C is the contour ABCDEA described in the counter-clockwisesense and consists of the line Re p = c > γ and the sectorial part of thecircle of large radius R with centre at the origin of the complex p-plane.All the singularities of f(p) are lying to the left of the line segment fromc − iR to c + iR and therefore they are all enclosed inside the contourC. This contour C is known as Bromwich contour in turn and is shownin FIG 1 below.
154 An Introduction to Integral Transforms
For actual evaluation of the integral in (4.6) let us assume thatpn, (n = 1, 2, · · · , N) denote the singularities of f(p) inside C. Let R0
denotes the largest of the moduli of all these singular points. The para-metric representation of the semicircular part, say, CR is given by
p = c + Reiθ ,π
2� θ � 3π
2(4.7)
where R > R0 + c. Note that for each pn,
|pn − c| � |pn| + c � R0 + c < R.
Then by Cauchy’s residue theorem, theorem we get
∫Γ
eptf(p) dp = 2πiN∑
n=1
Resp=pn
[eptf(p)
]− ∫CR
eptf(p) dp (4.8)
It may be noted at this point that if f(p) has any branch point singu-larity, corresponding branch-cut must be given suitably for making f(p)analytic except for the singular points inside the modified contour C.
Suppose now that, for all point p on CR, there is a positive constantMR such that |f(p)| � MR, where MR → 0 as R → ∞. We may use theparametric representation (4.7) for CR to write
∫CR
ept f(p) dp =∫ 3π
2
π2
exp(ct + Rteiθ
)f(c + Reiθ
)· Ri eiθdθ
Then since,∣∣∣ exp (ct + Rteiθ)
∣∣∣ = ect · eRt cos θ
The Inverse Laplace Transform 155
and∣∣∣ f(c + Reiθ)
∣∣∣ � MR, we find that∣∣∣∣∫
CR
eptf)(p)dp
∣∣∣∣ � ect MR R
∫ 3π/2
π/2eRt cos θ dθ (4.9)
But the substitution φ = θ− π2 together with Jordan’s inequality reveals
that ∫ 3π/2
π/2eRt cos θ dθ =
∫ π
0e−Rt sin φ dφ <
π
Rt
Thus inequality (4.9) becomes∣∣∣∣∫
CR
eptf(p) dp
∣∣∣∣ � ect MR π
t(4.10)
and hence limR→∞
∫CR
eptf(p) dp = 0 (4.11)
Therefore, letting R → ∞ in eqn. (4.4), we see that f(t), defined bythe equation
f(t) =1
2πilim
R→∞
∫ c+i∞
c−i∞eptf(p) dp , t > 0 (4.12)
can be written as
f(t) =N∑
n=1
Resp=pn
[eptf(p)
], t > 0. (4.13)
Further it is to be noted in case that when f(p) has countable infi-nite number of singular points, the evaluation of f(t) in (4.13) may bereplaced by
f(t) =∞∑
n=1
Resp=pn
[eptf(p)
], t > 0. (4.14)
whenever it is possible to show that
limR,N→∞
∫CR
eptf(p) dp = 0 (4.15)
In many applications of Laplace transforms, such as the solution of ODE
and PDE arising in initial value problem or boundary-value problem,the form of f(p) oblained there may not be that much simple to put
156 An Introduction to Integral Transforms
it in a combination of the forms discussed in the heuristic approachwith elementary functions. Thus, in those cases the necessity of thisgeneral evaluation technique of the inverse Laplace transform may befound useful.
It may also be noted that both heuristic method and the general con-tour integration method inversion of Laplace transform lead to identicalresults.
Example 4.21. Using heuristic as well as complex Inverse formula eval-uate the inverse Laplace transform of the image function f(p) = p
(p2+1)2.
Solution. By heuristic approach, we get
L−1[
f(p)]
= L−1
[p
(p2 + 1)2
]= L−1
[−1
2d
dp
{1
(p2 + 1)
}]
=12
t L−1
[1
p2 + 1
]
=12
t sin t = f(t)
Also, by complex inversion rule we have
f(t) =1
2πi.
∫ c+i∞
c−i∞eptf(p) dp
with f(p) = p(p2+1)2
, which has singularities only at points ±i , both ofthem being double poles. Therefore, we choose the arbitrary constantc > 0 and the corresponding Bromwhich contour C as shown below.
In this case, f(t) = Sum of the residues of ept p(p2+1)2 at p = ±i
The Inverse Laplace Transform 157
= limp→i
d
dp
[(p − i)2
eptp
(p + i)2(p − i)2
]+ lim
p→−i
d
dp
[(p + i)2
eptp
(p + i)2(p − i)2
]
= − it
4eit +
14it e−it
=t
2sin t , after simplification
Example 4.22. Use complex inversion formula to evaluate inverseLaplace transform of f(p) = 2p2−4
(p+1)(p−2)(p−3)
Solution. If inverse transform of f(p) be f(t), then by complex Inversionformula for Laplace transform it is given by
f(t) =1
2πi.
∫ c+i∞
c−i∞f(p)ept dp
=1
2πi
∫ c+i∞
c−i∞
(2p2 − 4) ept dp
(p + 1)(p − 2)(p − 3)=
12πi
∫C
(2p2 − 4) ept dp
(p + 1)(p − 2)(p − 3)
where C is a positively oriented closed contour with the Bramwich linefrom c − i∞ to c + i∞ as a part and enclosing all the poles of f(p).
Since, f(p) has only three simple poles at points p = −1, 2, 3 we choosehere the arbitrary constant c > 3 for the Bromwich integral. Thus byarticle 2.21
f(t) = sum of the residues at the poles p = −1, p = 2 and p = 3 ofthe integrand
[f(p)ept
].
Now, residue of f(p)ept at p = −1
= limp→−1
[(p + 1)/ (2p2 − 4)ept
(p + 1)/ (p + 2)(p − 3)
]=
16
e−t
Residue of f(p) ept at p = 2 = limp→2
[(p − 2)(2p2 − 4)ept
(p + 1)(p − 2)(p − 3)
]
= −43e2t
and residue of f(p)ept at p = 3 = limp→3
[(p − 3)(2p2 − 4)ept
(p + 1)(p − 2)(p − 3)
]
=72e3t
Therefore, f(t) = −16
e−t − 43
e2t +72
e3t
158 An Introduction to Integral Transforms
4.5 Inversion Formula from a different stand point : The
Tricomi’s method.
The Laguerre polynomial Ln(t) is defined by
Ln(t) =n∑
r=o
cn
r
(−t)r
r!(4.16)
Therefore, L [Ln(t); t → p ] =n∑
r=0
(−1)r cn
r
1pr+1
=1p
(1 − 1
p
)2
(4.17)
Making use of change of scale and the first shift theorem of Laplacetransform, we get
L[
e−bt Ln(2bt); t → p]
=1
p + b
(p − b
p + b
)n
(4.18)
Therefore, if f(t) =∞∑
n=0
an e−bt Ln(2bt) (4.19)
then L [ f(t); t → p ] = f(p)
=∞∑
n=0
an1
p + b
(p − b
p + b
)n
(4.20)
by Lerch’s theorem.
Thus the equation (4.19) defines the Tricomi’s method of determiningLaplace inversion L−1
[f(p); p → t
], when f(p) is expressed in the
form (4.20). For this purpose we substitute p = b(
1+σ1−σ
)and then
express the given f(p) in the form
f(p) =1 − σ
2b
∞∑n=0
anσn
Thus, f(t) = L−1[
f(p); p → t]
is interpreted as
f(t) =∞∑
n=0
an Bn(bt) , (4.21)
where Bn(bt) = e−btLn(2bt) (4.22)
To verify the accuracy of the method we discuss the following specialcase as discussed by Ward [(1954) Proc. camb. Phil. Soc.,50,49 ] , where
f(p) =1
(p + 1)(p − 2)and f(t) = e−t − e−2t
The Inverse Laplace Transform 159
To evaluate f(t) by the Tricomi’s method, we substitutep = (1 + σ)/(1 − σ) , with b = 1. Then we get
f(p) =12(1 − σ) .
13(1 − σ
3)−1(1 − σ)
=1 − σ
2
[13− 2
3
∞∑n=1
( σ
3
)n]
So, f(t) =13
B0(t) − 23
∞∑n=1
3−nBn(t), (4.23)
where Bn(t) is defined in (4.22) above.
Estimated accuracy can be verified by keeping seven terms in (4.23),for fixed values of t.
A similar result can be derived involving Legendre Polynomials too.
Example 4.23. If f(t) = L−1[f(s); s → t
], then prove that
L−1
[1sf(s) ; s → t
]=∫ t
0f(x) dx
Solution. Consider g(t) = 1 ⇒ g(s) = 1s . Therefore, by convolution
theorem L−1[
1s f(s); s → t
]=∫ t0 f(t − τ) dτ if L [ f(t); t → s ] = f(s)
This equation implies,
L−1
[1sf(s); s → t
]=∫ t
0f(x)dx , where t − τ = x
Example 4.24 Evaluate
L−1
[ch x
√sa
s ch(√
sa l) ; s → t
]
Solution. By complex inversion formula
f(t) =1
2πi
∫ c+i∞
c−i∞est ch(αx)
ch(αl)ds
s, where α =
√s
a
Clearly, the integrand has poles at s = 0, s = sn = −(2n + 1)2 aπ2
4l2,
n = 0 or integer.
Residue at s = 0 is 1.
160 An Introduction to Integral Transforms
Residue at s = sn isexp(−snt)ch
{i(2n + 1) πx
2l
}[
s dds
{ch l
√sa
} ]s=sn
=4(−1)n+1
(2n + 1) πexp
[−{
(2n + 1)π2l
}2
at
]ch{
(2n + 1)πx
2l
}
Thus, f(t) = Sum of the residues at the poles of the integrand
= 1 +4π
∞∑n=0
(−1)n+1
(2n + 1)exp
[−(2n + 1)2
π2at
4l2
]ch{
(2n + 1)πx
2l
}.
Example 4.25. Using power series expansion in s due to Heavisideevaluate the Laplace inversion of f(s) = 1
ssh x
√s
sh√
s, 0 < x < 1, s > 0.
Solution. We have
f(s) =1s
ex√
s − e−x√
s
e√
s − e−√
s
=1s
e−(1−x)√
s − e−(1+x)√
s
1 − e−2√
s
=1s
[e−(1−x)
√s − e−(1+x)
√s].
∞∑n=0
exp(−2n√
s)
Or f(s) =1s
∞∑n=0
[exp{−(1 − x + 2n)
√s}− exp
{−(1 + x + 2n)√
s}]
Therefore, f(t) =∞∑
n=0
[erfc
(1 − x + 2n
2√
t
)− erfc
(1 + x + 2n
2√
t
)]
Example 4.26. Evaluate L−1[
1(s2+a2)2
; s → t]
Solution. We know that L−1[
1(s2+a2) ; s → t
]= sin at
a
Therefore, L−1
[1
(s2 + a2)2; s → t
]=
sin at
a∗ sin at
a
=1a2
∫ t
0sin a τ sin a(t − τ) dτ, by convolution theorem
=1
2a3(sin at − at cos at).
Example 4.27. Prove that L−1[
1(√
s+a); s → t
]= 1√
π t−aea2t erfc (a
√t)
The Inverse Laplace Transform 161
Solution.1
(√
s + a)=
1√s− a√
s(√
s + a)
∴ L−1
[1
(√
s + a); s → t
]
= L−1
[1√s
; s → t
]− a L−1
[(√
s − a)(s − a2)
√s
; s → t
]
= L−1
[1√s
; s → t
]− a L−1
[1
s − a2; s → t
]
+a2L−1
[1√
s(s − a2); s → t
]
=1√πt
− a exp (a2t) + a exp (a2t) erf (a√
t)
=1√πt
− a exp (a2t) erfc (a√
t).
4.6 The Double Laplace Transform
The double Laplace transform of a function of two variables f(x, y) inthe positive quadrant of the xy- plane is defined by the equation
f(p, q) = L2 [f(x, y) ; (x, y) → (p, q)]
= L [L {f(x, y) ; x → p} ; y → q] (4.24)
and hence by the double integral
f(p, q) =∫ ∞
0
∫ ∞
0f(x, y) e−(px+qy) dx dy (4.25)
whenever the integral exists .
It therefore, follows that
L2 [f(ax, by); (x, y) → (p, q)] =1ab
f(p
a,q
b
), a > 0, b > 0
(4.26)
and L2 [f(ax) g(by); (x, y) → (p, q) ] =1ab
f(p
a
)g(q
b
), a > 0, b > 0
(4.27)
In particular, we have therefore
L2 [ f(x); (x, y) → (p, q)] =1qf(p) (4.28)
162 An Introduction to Integral Transforms
The formulae (4.26)- (4.28) follow from the definitions (4.24) and(4.25) above.
To evaluate double Laplace transform of a function f(x + y), wechange the independent variables (x, y) to (ξ, η) where we definex = 1
2(ξ − η), y = 12 (ξ + η) implying dxdy = 1
2 dξdη
Thus,
L2 [f(x + y); (x, y) → (p, q)] =12
∫ ∞
0f(ξ)dξ
[∫ ξ
−ξe−
12(p+q)ξ− 1
2(p−q)ηdη
]
=12
∫ ∞
0f(ξ)
[2
p − q(e−qξ − e−pξ)
]dξ
=1
p − q
[f(q) − f(p)
](4.29)
after introducing the new region of double integration in the ξη - planeand evaluating the double integral.
To evaluate the double Laplace transform of f(x−y) no such simpleformula as above exists. For the case when f(t) is an even function oft, we have from definition that
L2 [f(x − y) ; (x, y) → (p, q)] =∫ ∫
Q1
f(x − y) e−px−qy dx dy
+∫ ∫
Q2
f(y − x) e−px−qy dx dy
(4.30)
while if f(t) is an odd function of t
L2 [f(x − y) ; (x, y) → (p, q)] =∫ ∫
Q1
f(x − y) e−px−qy dx dy
−∫ ∫
Q2
f(y − x) e−px−qy dx dy (4.31)
where Q1 and Q2 are infinite regions shown below in Fig. 1.4.4
The Inverse Laplace Transform 163
After making the transformation x = 12(ξ + η), y = 1
2(η − ξ) we findthat∫ ∫
Q1
f(x − y)e−px−qydxdy =12
∫ ∞
0f(ξ)dξ
[∫ ∞
ξe−
12(p−q)ξ− 1
2(p+q)ηdη
]= f(p)/(p + q) (4.32)
On the other hand making the transformation
x =12(η − ς), y =
12(η + ς)
we get∫ ∫
Q2
f(y − x) e−px−qy dx dy
=12
∫ ∞
0f(ς)dς
[∫ ∞
ςe−
12(q−p) ς− 1
2(p+q)η dη
]
=f(q)p + q
(4.33)
Therefore, from above when f(t) is an even function of t
L2 [ f(x − y); (x, y) → (p, q)] =1
p + q
[f(p) + f(q)
]. (4.34)
164 An Introduction to Integral Transforms
But, if f(t) is an odd function of t, we have
L2 [ f(x − y); (x, y) → (p, q)] =1
p + q
[f(p) − f(q)
]. (4.35)
Also, L2 [ f(x) H (x − y); (x, y) → (p, q) ]
=∫ ∞
0f(x)e−px
( ∫ x
0e−qydy
)dx
=1q
[f(p) − f(p + q)
], (4.36)
L2 [ f(x) H (y − x); (x, y) → (p, q) ]
=∫ ∞
0f(x)e−px
[ ∫ ∞
xe−qydy
]dx
=1q
[f(p + q)
], (4.37)
and L2 [ f(x) H (x + y) ; (x, y) → (p, q) ] =1qf(p) (4.38)
As a special case if f(x) = 1, f(p) = 1p so that
L2 [H (x − y); (x, y) → (p, q) ] =1
p(p + q)(4.39)
The formulae for double Laplace transform of the partial derivativesof a function of two variables can be obtained from the definition asgiven below.
L2
[∂u(x, y)
∂x; (x, y) → (p, q)
]=∫ ∞
0e−qydy
[∫ ∞
0
∂u
∂xe−pxdx
]
= −∫ ∞
0e−qydy
[u(0, y) − p
∫ ∞
0e−pxu(x, y)dx
]= p u (p, q) − u1(q),
where u1(q) = L [u(0, y); y → q]
and similarly,
L2
[∂u(x, y)
∂y; (x, y) → (p, q)
]= q u (p, q) − u2(p), (4.40)
where u2(p) = L [u(x, 0);x → p]
The Inverse Laplace Transform 165
Laplace transform of the higher order partial derivatives of a functionu(x, y). of the variables x, y can similarly be derived as
L2
[∂2u
∂x ∂y; (x, y) → (p, q)
]= pqu(p, q) − qu1(q) − pu2(p) + u(0, 0)
(4.41)
L2
[∂2u
∂x2; (x, y) → (p, q)
]= p2u(p, q) − pu1(q) − u3(q) (4.42)
L2
[∂2u
∂y2; (x, y) → (p, q)
]= q2u(p, q) − qu2(p) − u4(p) (4.43)
where, u3(q) = L
[∂u
∂x(0, y); y → q
]and u4(p) = L
[∂u
∂y(x, 0);x → p
]
The convolution theorem in this case is generalised as below.
The convolution of two double variable functions f(x, y) and g(x, y)is defined by
f ∗ ∗ g =∫ x
0ds
∫ y
0f(s, t) g (x − s, y − t)dt (4.44)
and its double Laplace transform is given by
L2 [ f ∗ ∗ g ; (x, y) → (p, q) ] = f(p, q) g(p, q) (4.45)
The proof of (4.45) follows directly from the definition of double Laplacetransform.
From (4.45) it therefore follows that
L−12
[f(p, q) g (p, q) ; (p, q) → (x, y)
]= f ∗ ∗g.
Example 4.28 Solve the PDE ∂u∂x = ∂u
∂y satisfying the boundaryconditions u(x, 0) = a(x), u(0, y) = b(y), for x > 0, y > 0.
Solution. Taking double Laplace transform, the PDE gives
L2
[∂u
∂x; (x, y) → (p, q)
]= L2
[∂u
∂y; (x, y) → (p, q)
]implying, pu(p, q) − b(q) = qu(p, q) − a(p).
Thus, u(p, q) =b(q) − a(p)
p − q
Or, u(x, y) = L−12
[b(q) − a(p)
p − q; (p, q) → (x, y)
]
166 An Introduction to Integral Transforms
4.7 The iterative Laplace Transform.
In case of double Laplace transform of a function of two variables andthose for its partial derivatives if the transformed variables p and q areidential, then such a double Laplace transform, in particular, is calledan iterated Laplace transform which is denoted and defined as
I [ f(x, y) ; p ] = L2 [ f(x, y); (x, y) → (p, p) ]
=∫ ∞
0
∫ ∞
0f(x, y)e−p(x+y) dx dy (4.46)
Thus, by substituting q = p in the last section 4.4 all the propertiesof iterated Laplace transform can be derived.
In this case the generalised convolution of a function f(x, y) is de-noted and defined by
f (2)(x) =∫ x
0f(x − y, y) dy (4.47)
and hence
L[f (2)(x); x → p
]=∫ ∞
0e−px dx
[∫ x
0(x − y, y) dy
](4.48)
Using the variable transformation x = u + v , y = v ⇒ ∂(x,y)∂(u,v) = 1
L[f (2)(x); x → p
]=∫ ∞
0
∫ ∞
0f(u, v)e−p(u+v)du dv , from (4.48)
= I [ f(x, y) ; p ] , [ by (4.46) ] (4.49)
4.8 The Bilateral Laplace Transform.
Here, we shall now define a transform on functions which are definedfor all real values of the independent variables with the kernel e−px asdenoted and defined by the equation
f+(p) = L+ [ f(x) ; x → p ] =∫ +∞
−∞f(x)e−pxdx (4.50)
provided that f(x) is such that integral on the right hand side of eqn.(4.50) is convergent for some values of p. f+(p) is sometimes called thetwo-sided Laplace transform of f(x) by some authors.
The Inverse Laplace Transform 167
Writing (4.50) in the form
L+ [ f(x) ; x → p ] =∫ ∞
0f(x)e−pxdx +
∫ ∞
0f(−x)epxdx
we have
L+[f(x) ; x → p] = L [ f(x) ; x → p ] + L [ f(−x) ; x → −p ]
(4.51)
Let the first integral on the r.h.s converges for Re p > σ1 . and thesecond for Re(−p) > −σ2 ⇒ Re p < σ2 . Therefore, for the bilateralLaplace transform of f(x) is to exist these two half-planes must overlapimplying that σ2 > σ1. Hence, the condition for existance of the bilateralLaplace transform defined above in eqn. (4.50) is when Re p satisfies
σ1 < Rep < σ2 (4.52)
If σ1 = σ2, the strip contracts to a line Re p = σ1 and the region ofconvergence is the line Re p = σ1. But, if σ2 < σ1, the bilateral Laplacetransform for f(x) as defined above does not exist.
For example, if f(x) = e−x2, the bilateral Laplace transform is
defined for all Re p. While if f(x) = e−a|x|, the bilateral transformexists for −a < Re p < +a. But the function f(x) = e−ax does not havebilateral Laplace transform.
Thus , if f+(p) =∫ +∞
−∞f(x)e−pxdx, σ1 < Re p < σ2,
its inversion formula is given by
f(x) =1
2πi
∫ c+i∞
c−i∞f+(p)epxdp, σ1 < c < σ2 (4.53)
Here of course the region of convergence must always be specifiedsince two distinct object functions may have the same image function.For example, L+ [ H(t); t → p ] = 1, Re p > 0 and L+ [ −H(−t) ; t → p ]= 1, Re p < 0
168 An Introduction to Integral Transforms
4.9 Application of Laplace Transforms.
A. Solution of ODE with constant coeffiecients.
Suppose that we wish to solve the nth order linear ODE
dny
dtn+ c1
dn−1y
dtn−1+ · · · + cn−1
dy
dt+ cn y = f(t) (4.54)
where ci, i = 1, 2, · · · n are given constants, subject to the initial condi-tions y(0) = k1, y
′(0) = k2, · · · yn=1(0) = kn. Taking Laplace transformof both sides of the ODE using the results of article 3.9, and the initialconditions the ODE gives,[
y(p)pn − pn−1k1 − · · · − p kn−1 − kn
]+ c1
[y(p)pn−1 − pn−2k1 − · · · kn−1
]+ · · · + cn−1 [py(p) − k1] + cn.y(p) = f(p).
Thus,
y(p)[pn + c1p
n−1 + · · · + cn
]= f(p) + k1
[pn−1 + c1p
n−2 · · · + cn−1 p]
+k2
[pn−2 + c1p
n−3 + . . .]+ · · · + kn−1 [p + c1] + kn
=⇒ y(p) =[f(p) + g(p)
]/[pn + c1p
n−1 + · · · + cn
]y(p) =
f(p)h(p)
+g(p)h(p)
, where f(p) = L [f(t)] , (4.55)
g(p) is a polynomial of degree n − 1 in p and h(p) is also a polynomialof degree n in p.
Thus, the solution of the ODE under the given initial conditions canbe obtained after taking inverse Laplace transform of the last equationin(4.55). Since, the right side is a known function, we have
y(t) = L−1
[f(p)h(p)
]+ L−1
[g(p)h(p)
](4.56)
If eqn.(4.54) is a homogeneous one, then f(t) ≡ 0 and so, the requiredsolution of the homogeneous ODE under given initial conditions is
y(t) = L−1 [ g(p)/h(p) ] . (4.57)
Example 4.29 Solve by using Laplace transform the initial -valueproblem y′′′(t) + 2y′′(t) − y′(t) − 2y(t) = 0 with initial conditionsy(0) = y′(0) = 0 and y′′(0) = 6.
The Inverse Laplace Transform 169
Solution. Taking Laplace transform of both sides of the given ODEunder the initial conditions, we get[
p3y(p) − p2y(0) − py′(0) − y′′(0)]+ 2[p2y(p) − py(0) − y′(0)
]− [py(p) − y(0)] − 2y(p) = 0
=⇒ y(p) = 6/[p3 − 2p2 − p − 2
]=⇒ y(p) =
6(p − 1)(p + 1)(p + 2)
≡ 1p − 1
− 3p + 1
+2
p + 2
Thus inverting both sides , we get
y(t) = et − 3e−t + 2e−2t .
Example 4.30. Solve the initial-value problem defined by[d2
dt2+ n2
]x(t) = a sin (nt + α) ; x(0) = x′(o) = 0 .
Solution. Taking Laplace transform of both sides of the given ODEunder the initial conditions, we get(
p2 + n2)x(p) = a cos α
n
p2 + n2+ a sin α
p
p2 + n2
⇒ x(p) = an cos α1
(p2 + n2)2+ a sin α
p
(p2 + n2)2
On inversion, the above eqn. gives
x(t) = an cos α1
2n3[ sinnt − nt cos nt ] + a sin α
t
2nsin nt
= a [ sinnt cos α − nt cos(nt + α) ] /2n2
Example 4.31. Solve the initial-value problem
dy
dt+ 2y +
∫ t
0ydt = sin t , y(0) = 1.
Solution. Taking Laplace transform of the ODE we get under the giveninitial condition that
py(p) − 1 + 2y(p) +1py(p) =
1p2 + 1
⇒ y(p) =p(2 + p2)
(p2 + 1)(p + 1)2
170 An Introduction to Integral Transforms
By partial fraction method we can express
y(p) =1
(p + 1)− 3
21
(p + 1)2+
12.
1p2 + 1
On inverting, we get
y(t) = e−t − 32t e−t +
12
sin t
as the required solution of the given problem.
Example 4.32. Solve the BVP d2xdt2
+ 9x = cos 2t, under the boundaryconditions x(0) = 1, x
(π2
)= −1
Solution. Taking Laplace transform, the given ODE leads to
(p2 + 9) x(p) =p
p2 + 4+ p + A ; where A = x′(0), say
Therefore,
x(p) =p
(p2 + 32)(p2 + 22)+
p
p2 + 32+
3A3(p2 + 32)
=15
[p
p2 + 22− p
p2 + 32
]+
p
p2 + 32+
A
3· 3p2 + 32
Inverting,
x(t) =15
[ cos 2t − cos 3t ] + cos 3t +A
3· sin 3t
=15
cos 2t +45
cos 3t +A
3sin 3t
Since, x(
π2
)= −1 is being given, the above equation gives A = 4
5
Thus the solution of the BVP is
x(t) =15
[ cos 2t + 4cos 3t + 4 sin 3t ]
Example 4.33. A voltage Ee−at is applied at t = 0 to a circuit ofinductance L and resistance R connected in series where a,E,L and R
are constants. Using Laplace transform show that the current i(t) atany time t is given by
i(t) =E
R − aL
[e−at − e
−RtL
]Solution. In an electrical network with given voltage E(t), resistanceR and inductance L, the current i(t) builds at the rate given by
Ldi
dt+ Ri(t) = E(t)
The Inverse Laplace Transform 171
Here it is given that E(t) = Ee−at and i(0) = 0. Therefore, takingLaplace transform of the above equation under given conditions, we get
L [ p i(p) − 0 ] + R i(p) = E1
p + a
⇒ i(p) =E
(Lp + R)(p + a)≡ E
L(
a − RL
)[
1p + R
L
− 1p + a
]
=E
R − aL
[1
p + a− 1
p + RL
]
Inverting, i(t) =E
R − aL
[e−at − e−
RtL
].
B. Solution of simultaneous ODE with constant co-efficeents
The ODE involving more than one dependent variables but witha single independent variable give rise to simultaneous equations. Theprocedure for solving such simultaneous equations is almost same as thatdiscussed in section A. Here also, we have to take Laplace transform ofthe simultaneous equations to reduce them to corresponding number ofalgebraic equations which can then be solved for the Laplace transformeddependent variables. Finally inverting these relations we can recoverthe dependent variables forming the required solutions. The method isillustrated through the following examples.
Example 4.34. Solve the following initial value problem defined by thesimultaneous ODE
dx
dt− y = et dy
dt+ x = sin t, given that x(0) = 1, y(0) = 0
Solution. Taking Laplace transform the two given ODE under assignedinitial conditions, we get
px(p) − 1 − y(p) =1
p − 1
py(p) + x(p) =1
p2 + 1
Solving these two equations for x(p) and y(p), we get
x(p) =p
p2 + 1+
p
(p − 1)(p2 + 1)+
1(p2 + 1)2
172 An Introduction to Integral Transforms
=⇒ x(t) = 12
[cos t + 2 sin t + et − t cos t
], on inversion of Laplace
transform.
Also, y(p) = −1 − 1p−1 + px(p)
This expression after Laplace inversion and on simplification givesrise to
y(t) =12[
t sin t − et + cos t − sin t]
Example 4.35. Solve the IVP defined by
x′(t)+y′(t) = t, x′′(t)−y(t) = e−t if x(0) = 3, x′(0) = −2 and y(0) = 0.
Solution. Taking Laplace transform of the ODE under given initialconditions we get,
px(p) − 3 + py(p) =1p2
and p2x(p) − 3p + 2 − y(p) =1
p + 1
Solving these equations for x(p) and y(p), we get
x(p) =3p2 + 1
p3(p2 + 1)+
3p2 + p − 1(p + 1)(p2 + 1)
and y(p) =3p2 + 1
p2(p2 + 1)− 3p2 + p − 1
(p + 1)(p2 + 1)
Expressing into partial fractions, we get
x(p) =2p
+1p3
− 32(p2 + 1)
+p
2(p2 + 1)+
12(p + 1)
and y(p) =1p
+1
2(p + 1)+
32(p2 + 1)
− p
2(p2 + 1)
Taking inverse Laplace transform, it is found that
x(t) = 2 +12t2 +
12et − 3
2sin t +
12
cos t
y(t) = 1 − 12e−t − 1
2cos t +
32
sin t,
as solutions of the given problem.
The Inverse Laplace Transform 173
C. Solution of ODE with variable co-efficients.
In this case let the differenial equation be a combination of terms ofthe form tnyn(t) and the Laplace transform of which are given by theformulae
L [tmyn(t)] = (−1)mdm
dpm
[L{
y(n)(t)} ]
,m = 1, 2, . . .
Then the usual method of solution, as in the case A, may be employedfor its solution.
Example 4.36. Solve the ODE with variable coefficients
ty′′(t) + y′(t) + ty(t) = 0
under the initial condition y(0) = 1
Solution. Taking Laplace transform, the differential equation gives
− d
dp
[p2 y(p) − p.1 − y′(0)
]+ [ p y(p) − 1 ] − d
dpy(p) = 0
under the given initial condition. On simplification, it gives
− d
dp
[(p2 + 1) y (p)
]+ py(p) = −1 + 1 = 0
⇒ −(p2 + 1)d y(p)
dp− py(p) = 0
or,dy(p)y(p)
= − p dp
p2 + 1
Integrating, log y(p) = −12
log(p2 + 1) + log c
or, y(p) =c√
1 + p2
or, y(t) = c J0(t)
But, by given condition y(0) = 1 ⇒ 1 = c.1 ⇒ c = 1. Therefore, therequired particular solution is given by
y(t) = J0(t).
Example 4.37. Solve the initial value problem defined by the ODEwith variable coefficients given by
ty′′(t) + 2y′(t) + ty(t) = 0, y(0) = 1
174 An Introduction to Integral Transforms
Solution. Taking Laplace transform under the given initial condition,the ODE leads to
− d
dp
[p2y(p) − py(0) − y′(0)
]+ 2 [ py(p) − y(0) ] − d
dpy(p) = 0
⇒ (p2 + 1)dy(p)dp
= −1
Or,∫
dy(p) = −∫
dp
1 + p2
Or, y(p) = cot−1 p +(c − π
2
), c is being an arbitrary constant.
After inverting, the above relation gives
y(t) = L−1(
−π
2+ c
)+
sin t
tAs t → 0, y(0) = 1 and therefore we get
1 = L−1[
c − π
2
]+ 1 ⇒ L−1
[c − π
2
]= 0
Hence, y(t) = sin tt is the required solution of the IVP.
[Note here that this last result can also be obtained by considering thelimit of y(p) as t → 0 or as p → ∞ leading to c = π
2 ].
Example 4.38. Solve the ODE
ty′′(t) + (t − 1)y′(t) − y(t) = 0 under the conditions y(0) = 5 andy′(0) = −5 , if y(∞) is finite.
Solution. Taking Laplace transform, the ODE under given initial con-ditions leads to
− d
dp
[(p2 + p)y(p) − 5p − y′(0) − 5
]= py(p) + y(p) − 5
or,dy(p)dp
+3p + 2
p(p + 1)y(p) =
10p(p + 1)
The I.F of this linear ODE in y(p) is e� (3p+2)dp
p(p+1) = p2(p + 1). Hence itssolution is given by
y(p) · p2(p + 1) =∫ [
I.F × 10p(p + 1)
]dp + A, A = arbitrary constant
⇒ (p3 + p2)y (p) = 5p2 + A
The Inverse Laplace Transform 175
Now by the final value theorem of Laplace transform as t → ∞implying p → 0 we get
limp→0
(p + p2) · limp→0
p y(p) = A
⇒ 0 · y(∞) = A
∴ A = 0 and hence y(p) =5
p + 1=⇒ y(t) = 5e−t, as the required
solution.
D. Ordinary differential equations in application to Electricalnetworks.
We consider an electrical network consisting of (i) a power source orsupplier of electromotive force E(volts) (ii) a resistor having rasistanceR(ohms) (iii) an inductor having inductance L(henrys) and (iv) a ca-pacitor having capacitance C(farads). When the circuit is switched onat time t = 0 by Kirchhoff law the circuit usually known as an L-C-Rcircuit satisfies the ODE
Ld2Q
dt2+ R
dQ
dt+
Q(t)C
= E
with the current i(t) at time t is given by i(t) = dQdt , where Q is the
charge measured in coulombs. To discuss such a problem we considerthe following examples.
Example 4.39. At time t = 0, a constant voltage E is applied to aL-C-R circuit. The initial current i and the charge Q are zero. Find thecurrent at any time t > 0, when R2 < or = or > 4L
C .
Solution. By Kirchhoff law, the ODE in Q at time t for the circuit isgiven by Ld2Q
dt2+ R dQ
dt + QC = E , i(t) = dQ
dt under i(0) = 0, Q(0) = 0.The above pair of simultaneous ODE can also expressed as
Ldi
dt+ Ri +
Q
C= E ,
dQ
dt= i
Taking Laplace transform, these equations under given initial conditionsbecome
L [ p i (p) − i(0) ] + R i (p) +1C
Q (p) =E
p
and p Q (p) − Q(0) = i(p)
176 An Introduction to Integral Transforms
Since i(0) = 0, Q(0) = 0, these equations become
i(p) [ Lp + R ] +1C
Q (p) =E
p, i (p) = p Q(p) · · · (i)
Solving for i(p), we get[Lp + R +
1Cp
]i (p) =
E
p⇒ i (p) =
E
L[
(p + R2L)2 + n2
] · · · (ii)
where we assume1
CL− R2
4L2= n2
We shall now discuss three different cases individually.
Case I. If R2 < 4LC , R2
4L2 < 1CL ⇒ n2 > 0.
Then inverting, the equation (ii) gives
i(t) = L−1
[E
L{
(p + R2L)2 + n2
}]
=E
Le−
Rt2L L−1
{1
p2 + n2
}=
E
Lne−
Rt2L sin nt
Case II. If R2 = 4LC , R2
4L2 = 1CL ⇒ n = 0
Then after inversion of Laplace transform, the eqn.(ii) gives
i(t) =E
LL−1
[1(
p + R2L
)2]
=E
Le−
Rt2L L−1
(1p2
)=
Et
Le−
Rt2L .
Case III. If R2 > 4LC , n2 < 0 ⇒ n2 = −m2, say
Then after inversion of Laplace transform, the equation(ii) gives
i(t) =E
LL−1
[1(
p + R2L
)2 − m2
]
=E
Le−
Rt2L L−1
(1
p2 − m2
)
=E
Lme−
Rt2L sinh mt
Example 4.40. In an electrical network with e.m.f E(t), rasistance R
and inductance L in series, the current i builds up at the rate given byL di
dt + Ri = E(t). If the switch of the circuit is connected at t = 0 anddisconnected at t = a, find the current i at any time.
The Inverse Laplace Transform 177
Solution. In this case, it is given that i = 0 at t = 0 and
E(t) =
{E , for 0 < t < a
0 , for t > a .
Taking Laplace transform, the given ODE in i(t) gives
(Lp + R) i(p) =∫ ∞
0e−ptE(t) dt =
∫ a
0e−pt E dt =
E
p
(1 − e−ap
)This gives i(p) =
E
p (Lp + R)− Ee−ap
p (Lp + R)
To find i(t) we have to invert the above equation.
Now, L−1
[E
p (Lp + R)
]= L−1
[E
R
{1p− 1
p + RL
}]
=E
R
(1 − e−
RtL
)
and L−1
[E e−ap
p(Lp + R)
]= L−1
[E e−ap
R
{1p− 1
p + RL
}]
=E
R
(1 − e−
R(t−a)L
)H(t − a), by 2nd shift theorem.
Therfore, we get
i(t) =E
R
(1 − e−
RtL
)− E
R
(1 − e
−R(t−a)L
)H(t − a).
where H(t − a) =
{0 , 0 < t < a
1 , t > a
E. Application to Mechanics.
In the vibration problem the displacement of a particle measuredfrom given fixed point satisfies some ODE. In case of loaded beams thetransverse deflection of the beam also satisfies some ODE. Under initialand boundary conditions for these respective cases these displacementswill be evaluated in the following examples. It may be noted that in caseof beam problems we assume the boundary conditions as (i) y = y′ = 0for clamped or built in fixed end (ii) y = y′′ = 0 for hinged or simplysupported ends and (iii) y′′ = y′′′ = 0 for free ends, y being the deflectionof the beam.
178 An Introduction to Integral Transforms
Example 4.41 A mass m moves along the x-axis under the influenceof a force which varies to its instantaneous speed and acts in a directionopposite to the direction of motion of the mass. Assuming that themass is located at x = a initially and is moving to the right side oralong positive x-axis with the speed v0 , find the position where the masscomes to rest if the position of the mass at any time is x and it satisfiesthe ODE m d2x
dt2= −μ dx
dt .
Solution. The initial conditions of the problem are x(0) = a, dxdt (0) = v0 .
Now taking Laplace transform of the ODE under the given initial con-ditions, we get
m[p2 x(p) − pa − v0
]= −μ [ px(p) − a]
This implies , x(p)[
mp2 + μp]
= mpa + mv0 + μa
or x(p) =mv0 + μa
μp− mv0
μ(p + μm)
Inverting Laplace transform, the above equation gives the positionx(t) of the mass as
x(t) =mv0 + μa
μ− mv0
μe−
μm t
Thus the velocity of the mass at any time is given by
dx
dt= v0 e−
μtm
When the mass comes to rest, dxdt = 0 ⇒ e−
μtm = 0
Thus for such a time t when e−μtm = 0, we get
x =mv0 + μa
μ.
Hence, the mass comes to rest when it is at a distance a + mv0μ from
the origin.
Example 4.42. Obtain the deflection y at a distance x from one end ofa weightless beam of length l and freely supported at both ends, whena concentrated load W acts at x = a. The differential equation fordeflection y is being
EId4y
dx4= Wδ(x − a),
The Inverse Laplace Transform 179
where E is the Young’s modulus of elasticity of the beam and I is themoment of inertia of a cross-section of the beam about its axis. Itmay be noted that for freely supported ends, the following conditionsy(0) = y′′(0) = 0 and y(l) = y′′(l) = 0 are satisfied
Solution. Taking Laplace transform in variable x the given ODE afterusing the initial conditions y(0) = y′′(0) = 0 and assuming y′(0) = c1
, y′′′′(0) = c2, we get
p4y(p) = c1p2 + c2 +
W
EIe−ap
Or, y(p) =c1
p2+
c2
p4+
W
EI
e−ap
p4
Taking inverse Laplace transform the last equation gives
y(x) = c1x + c2x3
3!+
W
EI
(x − a)3
3!H(x − a)
Therefore, y′(x) = c1 +12c2x
2 +W
2EI(x − a)2 H(x − a)
y′′(x) = c2x +W
EI(x − a) H(x − a)
Using the conditions y(l) = 0, y′′(l) = 0 we get
c2 =−W
EI
l − a
l
c1 =[
Wa(l − a)6EIl
][ 2l − a ]
Therefore, the required deflection of the beam at a distance x from theend x = 0 is
y(x) =W
6EI
[a(l − a)(2l − a)
lx − l − a
lx3 + (x − a)3 H (x − a)
]
In particular, if 0 < x < a , H(x − a) = 0
∴ y(x) =W
6(EI)
[a(l − a)(2l − a)
lx − (l − a)
lx3
]Also , if a < x < l , H(x − a) = 1
∴ y(x) =W
6EI
[a(l − a)(2l − a)
lx − l − a
lx3 + (x − a)3
]
180 An Introduction to Integral Transforms
From these above two results we have at x = a
y(a) =13
W
EI
a2(l − a)2
l.
F. Application to partial differential equation : Boundaryvalue Problem.
It is known that in case of ordinary differential equation the Laplacetransform reduces it to an algebraic equation of the transformed func-tion. But it may now be noted that by the use of Laplace transform oneof the independent variable of the equation will be removed by the trans-formed parameter p or s. Thus the PDE in two independent variableswill be reduced to an ODE and for PDE in more than two indepen-dent variables, one of the variable will be removed by the transformedparameter p , after taking account of the boundary conditions on thatvariable.
To start with the matter under discussion, let u(x, t) be a functionof two independent variables defined for a � x � b and t � 0. Alsolet u(x, t), regarded as a function t , be of some exponential order forlarge t and is a piecewise continuous function over t � 0. Then Laplacetransform over the variable t of u(x, t) is defined as
L [ u(x, t); t → p ] =∫ ∞
0e−ptu(x, t) dt = u(x, p) (4.58)
Therefore , L
[∂u
∂t
]= pu(x, p) − u(x, 0)
L
[∂2u
∂t2
]= p2u(x, p) − pu(x, 0) − ∂u
∂t(x, 0) , etc.
The proof of the above results will follow from the definition in (4.58).Similarly, for functions of more than two independent variables of whichone of them is t, say u(x, y, t), or u(x, y, z, t) etc, let it be defined fort � 0 and is of some exponential order for large t and is piecewisecontinuous function in t � 0. Then Laplace transform of such functionand of their partial derivatives can be defined as an extension of theabove case of two or more variables.
Example 4.43. Find the solution of the PDE ∂2y∂x2 − ∂2y
∂t2= xt subject
to y(x, 0) = ∂y∂t (x, 0) = 0 and y(0, t) = 0. It is given that the required
solution is bounded both for x � 0 and t � 0.
The Inverse Laplace Transform 181
Solution. Taking Laplace transform on the variable t the given PDEunder first two initial conditions gives
d2y(x, p)dx2
− [ p2 y(x, p) − p.0 − 0]
=x
p2
Or,d2y(x, p)
dx2− p2 y(x, p) =
x
p2
The general solution of the last equation is
y (x, p) = Aepx + Be−px − x
p4(i)
Since y (x, t) is bounded for all x and t, so also y (x, p) for all x andp. Hence for large x , y (x, p) is bounded and therefore, the arbitraryconstant A = 0 in the above form (i) of y (x, p). So,
y (x, p) = B e−px +−x
p4(ii)
But, by the given condition y(0, t) = 0 we get
y(0, p) = 0 (iii)
Therefore (ii) implies under (iii) that B = 0. Then eqn. (ii) implies
y (x, p) = − x
p4
Taking inverse Laplace transform the above solution reduces to
y (x, t) = −x t3
6
as the required solution of the PDE under given conditions.
Example 4.44. A bar of length a is at zero temperature. At t = 0, theend x = a is suddenly raised to a temperature u0 and the end x = 0 isinsulated. Find the temperature at any point x of the bar at any timet > 0 assuming that the surface of bar is insulated.
Solution. Let u(x, t) be the temperature at any point x and at anytime t of the bar so that it satisfies the heat conduction equation
∂u
∂t= c2 ∂2u
∂x2, (0 � x � a , t � 0) (i)
182 An Introduction to Integral Transforms
subject to the conditions
u(x, 0) = 0 (ii)
ux(0, t) = 0 (insulated end means heat flow from that end is nil) (iii)
and u(a, t) = u0 (iv)
Taking Laplace transform of eqn. (i) above over the variable t weget
pu(x, p) − u(x, 0) = c2 d2
dx2u(x, p) (v)
where u (x, p) = L [ u(x, t) ; t → p ] =∫ ∞
0e−ptu(x, t) dt
Using condition given in eqn. (ii) , the eqn (v) becomes
d2u(x, p)dx2
− p
c2u(x, p) = 0 (vi)
Also, Laplace transform of eqns. (iii) and (iv) become
ux(0, p) = 0 (vii)
and u(a, p) =u0
p(viii)
Solving eqn. (vi), we have
u(x, p) = A e√
px
c + B e−√
px
c
Under the condition (vii) we get A = B and therefore,
u(x, p) = 2A cosh√
px
c
Now, using condition in eqn. (viii) we have finally
u(x, p) =uo cosh
(√pxc
)p cosh
(√pac
) (ix)
To obtain the temperature u(x, t) we have to invert Laplace trans-form of u(x, p) given in eqn. (ix). But the inversion can not be obtainedby heuristic method. Therefore, we take recourse to general inversion
The Inverse Laplace Transform 183
formula of Laplace transform as discussed in article 4.4. By such aformula we have here,
u(x, t) =1
2πi. P.V
∫ λ+i∞
λ−i∞eptu(x, p) dp , (x)
where the Bromwich line contour Γ from λ− i∞ to λ+ i∞ , (λ is a realconstant) is such that all the singularities of u(x, p) in p-plane lie to theleft of Re p = λ. Then if C as shown in the figure below is a positivelyoriented closed contour consisting of Γ and a large (infinite) circle CR
enclosing all the singularities of u(x, p) after giving branch-cut in thep-plane required to make the integrand single valued eqn. (x) can bewritten equivalently as
u(x, t) =1
2πi
∫C
eptu(x, p) dp (xi)
=N∑
n=1
Resp=pn
[eptu(x, p)
], (xii)
where N= number of poles of u(x, p) inside C. Here also
limR→∞
∫CR
eptu(x, p) dp = 0
.
184 An Introduction to Integral Transforms
Also,∫
L1+L2
ept u (x, p) dp = 0 ,
limr→0
∫Cr
ept u (x, p) dp = 0
Thus eqn. (xii) follows from eqns. (x) and (xi). Now the poles ofthe integrand in (xi) inside C are given by
p = pn = −(2n − 1)2
4a2c2π2 , n = 0, 1, 2, · · ·
Residue at p = pn
= limp→pn
⎡⎣uo(p − pn) ept cosh
(√pxc
)p cosh
( √p a
c
)⎤⎦
= u0 limp→pn
[ {p − pn
ch√
p ac
} ]. limp→pn
[eptch(
√p x
c )p
]
= u0 limp→pn
{1
sh(√
p ac )
.2√
pc
a
}. limp→pn
[eptch(
√p x
c )p
]
=4u0(−1)n
(2n − 1)π. e
−(2n−1)2π2 c2 t
4a2 . cos(2n − 1)πx
2a
Thus, we get
u(x, t) =4u0
π
∞∑n=1
(−1)n
(2n − 1)e
−(2n−1)2π2 c2 t
4a2 cos(2n − 1)πx
2a,
as the required solution of the given heat conduction problem.
Example 4.45. A semi-infinite solid x > 0 has its initial temperaturezero. A constant heat flux H is applied at the face x = 0 and hence−kux(0, t) = A, where k is the thermal diffusivity of the material of thesolid and u(x, t) is the temperature of the solid at any point of it at timet. If the heat conduction equation of the solid is given by
∂u
∂t= k
∂2u
∂x2, 0 < x < ∞, t > 0 (i)
with the boundary condition ∂u(0,t)∂x = −H
τ and the initial conditionu(x, 0) = 0, find the temperature of the solid at any point x > 0 at anyinstant, given that
The Inverse Laplace Transform 185
L−1
[e−x
√p
p32
; p → t
]=√
tπ e
−x2
4t2 − x2 erfc
{x
(2√
t )
}Solution. Taking Laplace transform over the variable t the eqn. (i)under the initial condition takes the form
pu(x, p) = kd2
dx2u (x, p)
Its general solution is given by
u(x, p) = Aex√
p/k + Be−x√
p/k (ii)
Since the temperature u(x, t) is bounded for large x, so is u(x, p) andtherefore eqn. (ii) gives A = 0. Thus eqn.(ii) becomes
u(x, p) = Be−x√
pk (iii)
Also, Laplace transform of the boundary condition becomes
∂u
∂x(0, p) = −H
τp, (τ = thermal conductivity ) (iv)
Using (iv) in (iii), we get B = H
τ√
kp32
and hence eqn. (iii) implies
u(x, p) =√
k H e−x√
pk
τ p32
Now, inverting Laplace transform the required solution of theproblem is given by
u(x, t) = L−1
[H√
k
τ
e−x√
pk
p32
; p → t
]
Therefore, by given condition
u(x, t) =H√
k
τ
[√t
πe−
x2
4t2k − x
2√
kerfc
{x
2√
kt
}]
Example 4.46. The temperature u(x, t) of a slab of material mediumsatisfies the heat conduction equation ∂u
∂t = k ∂2u∂x2 , 0 < x < l, t > 0. If
the initial temperature at the bounding planes x = 0 and x = l of the
186 An Introduction to Integral Transforms
slab be u0, find the temperature in this solid after the face at x = 0 isinsulated and the temperature of the face at x = l is reduced to zero.
Solution. The temperature distribution u(x, t) of the slab satisfies thePDE
∂u
∂t= k
∂2u
dx2(i)
subject to the boundary conditions du(o,t)dx = 0 and u(l, t) = 0 and the
initial condition u(x, 0) = u0, 0 < x < l. Taking Laplace transformequation (i) under the initial condition gives
pu(x, p) − u0 = kd2
dx2u(x, p)
Solving the above equation we get
u(x, p) = A cosh(√
p
kx
)+ B sinh
(√p
kx
)+
u0
p(ii)
The Laplace transform of the given boundary conditions become
d
dxu(0, p) = 0 and u(l, p) = 0 (iii)
Using (iii) in (ii) we get B = 0 and A = − u0
p cosh(√
pk l)
Therefore, eqn. (ii) becomes
u(x, p) =u0
p− u0
p
cosh(√
pk x)
cosh(√
pk l)
Inverting Laplace transform the above relation gives
u(x, t) = u0 − u0L−1
⎡⎣ cosh
(√pk x)
p cosh(√
pk l)⎤⎦ (iv)
The second term in the above is almost equivalent to eqn. (ix) ofexample-4.44. Hence, following the same technique (iv) can be evaluatedto get the solution u(x, t) of the problem in terms of x and t.
The Inverse Laplace Transform 187
Note: As an alternative procedure for evaluation of Laplace inversionthe expression 1
p . ch(√
pk x)
/ch(√
pk l)
may be expanded in an infi-nite series and then term by term inversion may be obtained for solution,noting that L−1
[1p e−c
√p]
= erfc(
c2√
t
).
Example 4.47. Consider the diffusion equation in a semi-infinite line0 < x < ∞ defined by ∂2θ
∂x2 = 1k
∂θ∂t , t � 0 and subject to the boundary
condition θ(0, t) = f(t) and the initial condition θ(x, 0) = 0 togetherwith the limiting condition θ (x, t) → 0 as x → ∞. Obtain the temper-ature θ (x, t) of the line under this general case and of the particularcases (i) θ (0, t) = θ0 (T/t )
12 , θ0 and T are positive constants
(ii) θ(0, t) = θ0, a constant and (iii) ∂∂x θ(0, t) = f(t) instead of the
condition θ(0, t) = f(t) by using Laplace transform.
Solution. Taking Laplace transform the PDE under the given initialcondition reduces to
d2θ (x, p)dx2
=1k
[p θ (x, p)
]We solve the above second order ODE to obtain
θ (x, p) = Ae−√
pkx + Be
√pkx (i)
Now, taking Laplace transform to the given boundary condition
θ (0, t) = f(t) we getθ (0, p) = f(p)
Under Laplace transformed the limiting condition becomes θ(x, p) → 0as x → ∞. Using these two results, we get
θ (x, p) = f(p) e−√
pkx , since B = 0 (ii)
Therefore, inverting this equation we have by the convolution theoremthat
θ (x, t) =∫ t
0f (τ) g (x, t − τ) dτ (iii)
where g (x, t) = L−1[
e−√
pkx ; p → t
](iv)
But since , L[
t−12 e−c/t ; t → p
]=√
π
pe−2
√cp, differentiating both
188 An Introduction to Integral Transforms
sides with respect to c we obtain
L[
t−32 e− c/t ; t → p
]=√
π
ce−2
√cp (v)
Therefore, g (x, t) =1√
4kπt3e−x2/4kt
and hence the solution of this general problem is given by
θ (x, t) =1√4πk
∫ t
0f(τ)exp
[x2
4k (t − τ)
]dτ
(t − τ)32
We shall now take up the particular cases shown below.
Case-I
θ(0, t) = θ0
(T
t
) 12
So, f(p) = θ0 T12 L
[t−
12 ; t → p
]= θ0
(πT
p
)12
and hence
θ(x, p) = θ0 (πT )12 p−
12 e−x
√pk
Inverting , θ(x, t) = θ0
(T
t
)12
e−x2
4kt
Case-II
θ(0, t) = θ0 , a constant .
In this case , f(p) =θ0
pand so θ(x, p) =
θ0
pe−x
√pk
After Laplace inversion we get
θ (x, t) = θ0 L−1
[1p
e−x√
pk ; p → t
]
= θ0 Erfc
(12
x√kt
).
Case - III
In this case∂
∂x
[θ(0, p)
]= f(p) , θ(x, p) → 0 as x → ∞
Therefore, θ(x, p) = −√
kf(p)1√p
e−x√
pk
The Inverse Laplace Transform 189
Hence, θ (x, t) =∫ t
0f (τ) g (x, t − τ) dτ
where g(x, t) = −k12 L−1
[1√p
e−x
√pk ; p → t
]
= −(
k
πt
) 12
e−x2
4kt
Therefore, the required solution for this general case is given by
u(x, t) = −√
k
π
∫ t
0f(τ) exp
[− x2
4k(t − τ)
]d τ√t − τ
.
Example 4.48. An infinitely long string of negligible mass having oneend at x = 0, is initially at rest along the x-axis. The end x = 0 isgiven a transverse displacement f(t), t > 0. Find the displacement ofany point of the string at any time by using Laplace transform.
Solution. Let y(x, t) be the transverse displacement of any point x
of the string at any time t. Then, y(x, t) satisfies the wave equation∂2y∂t2 = c2 ∂2y
∂x2 , x > 0, t > 0 ; subject to the initial conditions y(x, 0) = 0,∂y∂t (x, 0) = 0 and the boundary condition y(0, t) = f(t) and y(x, t) isbounded. Laplace transform of the PDE over the variable t gives
p2y (x, p) − p y (x, 0) − ∂y (x, 0)∂t
= c2 d2y(x, p)dx2
Using the initial conditions we rewrite the above equation as
d2y
dx2=( p
c
)2y (x, p) (i)
Again, Laplace transform of the given boundary conditions are
y(0, p) = f(p) and y(x, p) is bounded.
Solving (i), we get
y (x, p) = A epxc + B e−
pxc
Since y(x, t) is bounded so is y(x, p) for all x. Then we have
y (x, p) = Be−pxc , because A = 0.
190 An Introduction to Integral Transforms
By use of Laplace transformed boundary condition y(0, p) = f(p) wehave
y (x, p) = f(p) . e−pxc
Therefore, by complex inversion formula of Laplace transform we get
y(x, t) =1
2πi
∫ λ+i∞
λ−i∞e(t−x
c)p f(p) dp
= f(t − x
c
).
Example 4.49 A light flexible string of length a is stretched betweentwo fixed points at x = 0 and at x = a. If the string is displaced intothe curve y = b sin π x
a and released from rest in that position initially,find its displacement at any time t and at any point x.
Solution. The transverse vibration y(x, t) of the string satisfies thePDE ∂2y
∂t2= c2 ∂2y
∂x2 subject to boundary conditions y(0, t) = 0, y(a, t) =0 and the initial conditions y(x, 0) = b sin πx
a , ∂y∂t (x, 0) = 0. Taking
Laplace transform of the PDE over the variable t under the initial con-ditions we get
d2y(x, p)dx2
− p2
c2y (x, p) = −bp
c2sin
πx
a
The general solution of this second order ODE is
y(x, p) = A epxc + B e−
pxc +
bp sin(
πxa
)p2 +
(πca
)2 (i)
Now, Laplace transform of the given boundary conditions are
u(0, p) = 0 and u(a, p) = 0
Using these results in (i) we get
A + B = 0 and A epac + B e−
pac = 0.
Solving these equations for A and B, we get A = B = 0, then
y(x, p) = b sinπx
a.
p
p2 +(
πca
)2
The Inverse Laplace Transform 191
Therefore, inverting this equation we get the required solution of thegiven problem as
y(x, t) = b sinπx
aL−1
[p
p2 +(
πca
)2 ; p → t
]
= b sinπx
acos
πct
a
Example 4.50. A tightly stretched flexible string has its end points atx = 0 and at x = l. At time t = 0 the string is given the shape definedby y = μ
∑∞n=1 sin nπ
l x, where μ is a constant, and then released fromrest. Find the displacement y = (x, t) of the string.
Solution. The partial differential equation satisfying the transversedisplacement y(x, t) of the string and satisfying the initial conditionsy(x, 0) = μ
∑∞n=1 sin nπx
l and ∂y (x,0)∂t = 0 and boundary conditions
y(0, t) = y(l, t) = 0 is given by
∂2y
∂t2= c2 ∂2y
∂x2,
c being a material constant of the string.
Taking Laplace transform of the PDE under the initial conditions,we have
p2y(x, p) − py(x, 0) − ∂y(x, 0)∂t
= c2 d2
dx2y(x, p)
Or[
d2
dx2− p2
c2
]y (x, p) = −pμ
c2
∞∑n=1
sinnπx
l
This equation has the solution
y(x, p) = A chp
c(l − x) + B sh
p
c(l − x) + μ
∞∑n=1
p sin nπxl
p2 +(
nπcl
)2Laplace transform of the given boundary conditions are
y(0, p) = y(l, p) = 0
Using these results, the above form of y(x, p) gives
A chpl
c+ B sh
pl
c= 0
and A = 0. Thus , B = 0 and hence
y(x, p) = μ
∞∑n=1
p sin nπxl
p2 +(
nπcl
)2
192 An Introduction to Integral Transforms
Inverting Laplace transform we get the solution of the problem as
y(x, t) = μ
∞∑n=1
cosnπct
lsin
nπx
l.
Example 4.51. An infinite string lying along the x-axis is vibratingby giving a displacement f(t) to the end x = 0. The other end of thestring at large distance from the origin remains fixed. If initially thedisplacement and the velocity of every point of the string were zero,find the displacement of every point of the string at any time, when thedisplacement satisfies the PDE
∂2y
∂t2= c2 ∂2y
∂x2(i)
Solution. As per statement of the problem the initial and the boundaryconditions are
y(x, 0) = 0,∂y (x, 0)
dt= 0 and y(0, t) = f(t), lim
x→∞ y(x, t) = 0
respectively. Taking Laplace transform over t, eqn. (i) becomes
p2y(x, p) − py(x, 0) − ∂y
∂t(x, 0) = c2 d2
dx2y(x, p)
Using the initial conditions, we have then
d2y
dx2(x, p) − p2
c2y(x, p) = 0 , p > 0.
The general solution of this ODE is
y(x, p) = Aepc
x + Be−pc
x (ii)
Since, limx→∞ y(x, 0) = 0, we must have from the above solution that A = 0.
Therefore, the solution (ii) becomes
y(x, p) = Be−pc
x (iii)
But, again from the boundary condition y(0, t) = f(t) under Laplacetransform, we get y(0, p) = f(p) = L [ f(t) ]. Hence, by (iii) we get
y(0, p) = B = f(p)
The Inverse Laplace Transform 193
Therefore, y(x, p) = f(p) e−pcx. Inverting the above transformed solu-
tion y(x, p), we have then finally
L−1 [ y(x, p) ; p → t ] = L−1[
f(p) . e−pcx ; p → t
]Or y(x, t) = f
(t − x
c
)H(
t − x
c
).
by the second shift theorem of Laplace transform. Thus the requireddisplacement is
y(x, t) = 0, t <x
c
= f(t − x
c
), t >
x
c
G. Application to solution of Integral equation and Integro-differntial equation.
An integral equation is an equation in which the unknown functionappears under integral sign. For example,
f(s) =∫ b
ak (s, t) g(t) dt ,
g(s) = f(s) +∫ b
ak (s, t) g (t) dt
and g(s) =∫ b
ak(s, t) [g (t)]2 dt ,
where b, the upper limit of the integral may be either a variable or aconstant. The function g(s) is the unknown function while all otherfunctions are known. These functions may be complex valued functionsof the real values of s and t. If the limits a and b are both constants,the integral equation is known as Fredholm integral equation. If theupper limit b is a variable and a is a constant, the corresponding in-tegral equation is called Volterra equation. The kernel k(s, t) of theintegral equation is a known function. If k(s, t) = K(s − t), a functionof the difference s− t only, the corresponding integral equation is calleda convolution type integral equation. Many interesting problems of me-chanics and physics lead to integral equation in which the kernel k(s, t)is a function of the difference (s− t) only. Thus, when k(s, t) = K(s− t),where K is a function of one variable, though k(s, t) is a function of twovariables, Laplace transform is one of the necessary tools towards thesolution of such convolution type Volterra integral equation
194 An Introduction to Integral Transforms
g(s) = f(s) +∫ s
0K(s − t)g(t) dt (4.59)
as shown below.
On applying Laplace transform to both sides of the eqn.(4.59) and usingthe convolution formula, we have
g(p) = f(p) + K(p) . g (p)
Or g(p) = f(p)/[1 − K(p)
], (4.60)
and then the inversion of eqn. (4.60) yields the solution of (4.59)
A special type convolution integral equation∫ s
0
f(u)(s − u)α
du = g(s), 0 < α < 1 (4.61)
is called an Abel’s integral equation. Here f(u) is an unknown and g(s)is a known function.
Laplace transform method may also be effectively used to solve suchan equation in (4.61) too.
The following examples may be considered for much more details inapplications of the method.
Example 4.52. Solve the integral equation
s =∫ s
0es−t g(t) dt
Solution. Taking Laplace transform of both sides, we obtain
1p2
= K(p) g (p) =1
p − 1g(p)
Or , g (p) =1p− 1
p2
Inverting , g (s) = 1 − s , as the solution of the integral equation
Example 4.53. Solve the integral equation
f(s) = s2 +∫ s
0f(u) sin (s − u) du
The Inverse Laplace Transform 195
Solution. Taking Laplace transform on both sides, we get
f(p) =2p3
+ L [ f(s) ∗ sin s ] =2p3
+ f(p)/(p2 + 1)
⇒ f(p) =2(p2 + 1)
p5=
2p3
+2p5
.
Inverting , f(s) = t2 +t4
12.
Example 4.54. Solve the integral equation
g(t) = 1 −∫ t
o( t − τ) g (τ) dτ
Solution. Taking Laplace transform, the given integral equationbecomes
g(p) =1p− L
[ ∫ t
o( t − τ) g (τ) dτ ; t → p
]
=1p− 1
p2. g (p) , by convolution theorem .
Therefore , g(p)[
1 + p2
p2
]=
1p
⇒ g(p) =p
p2 + 1
Inverting the last equation, the required solution of the integralequation is
g(t) = L−1
[p
p2 + 1; p → t
]= cos t
Example 4.55. Solve the integral equation
sin t =∫ t
0J0 (t − τ ) g (τ) dτ
Solution. Taking Laplace transform, the given integral equationbecomes
1p2 + 1
= L [ Jo (t) ; t → p] . g (p) =1√
p2 + 1g (p)
Thus , g (p) =1√
p2 + 1.
196 An Introduction to Integral Transforms
Inverting, we get the solution of the problem as
g(t) = L−1
[1√
p2 + 1
]= J0 (t) .
Example 4.56. Solve the integral equation
cos t =∫ t
0J0 (t − τ ) g ( τ ) dτ
Solution. Taking Laplace transform the integral equation gives
p
p2 + 1=
1√p2 + 1
. g(p) ⇒ g (p) =p√
p2 + 1
Inverting, g(t) = L−1
[p.
1√p2 + 1
]=∫ t
0H (τ)J0 (t − τ)dτ
=∫ t
0J0 (t − τ) dτ , t > τ > 0
since it is known that , L
[ ∫ t
0f (τ) g (t − τ)dτ
]= f(p) . g (p) .
Example 4.57. Solve the Abel’s integral equation∫ t
0
f (τ)dτ
(t − τ)13
= t (1 + t)
Solution. It is known that f(t) ∗ t−13 = t + t2. Taking the Laplace
transform of both sides we get
L [ f(t) ; t → p ] . L[t−
13
]= L [t] + L
[t2]
Or , f(p) .Γ (2
3 )
p23
=1p2
+2p3
Or , f(p) =1
Γ (23)
[1
p43
+2
p73
]
Taking inverse Laplace transform, the above equation gives the solutionof the Abel’s integral equation as
f(t) =1
Γ(
23
)[
t43−1
Γ (43 )
+ 2t
73−1
Γ(73 )
]
The Inverse Laplace Transform 197
=1
Γ (23 )
[t
13
13 Γ (1
3 )+
2 t1+13
43 . 1
3 Γ (13)
]
=t
13
Γ (13 )Γ (1 − 1
3)
[3 +
92t
]
=3t
13 sin π
3
π[1 +
32
t]
=3√
3 t13
4π(2 + 3t)
Example 4.58. Solve the integro-differential equation
f ′(t) = t +∫ t
0f (t − τ) cos τ dτ , if f(0) = 4.
Solution. Taking Laplace transform of both sides the given equationbecomes
pf(p) − 4 =1p2
+ f(p) · p
p2 + 1
⇒ pf(p)[1 − 1
1 + p2
]=
1p2
+ 4
Or,p3
1 + p2f(p) =
1p2
+ 4 =1 + 4p2
p2
Or, f(p) =1p5
+5p3
+4p
.
Inverting Laplace transform, the solution of the equation is
f(t) =t4
24+
5t2
2+ 4.
G. Application to solution of Difference and Differential-differenceeqations.
Difference equation arise in a natural way in physics and engineering.Let a, a + 1, a + 2, . . . be consequitive finite values of the argument x ofa function denoted by u(x) or ux at discrete points. Here the constant1 is called the interval of difference. The nth order finite difference ofur is defined by
Δnur ≡ Δn−1 [Δur] = Δn−1 [ur+1 − ur] =n∑
k=0
(−1)k
(n
k
)ur+n−k
(4.62)
198 An Introduction to Integral Transforms
Any equation containing the differences of different orders of the functionur is called a difference equation. The highest order finite difference ap-pearing in a difference equation is called its order. Again any differenceequation containing the derivatives of the unknown function is called adifferential-difference equation. Thus a differential - difference equationpossesses two orders - one due to the largest order difference and theother associated with the highest order derivatives. For example,
Δ2un − Δun + un = 0 (4.63)
and u′(t) + u (t − 1) = f(t) (4.64)
are second order difference equation and first order differential-differenceequation respectively. To solve difference equation it facilitates to intro-duce a step function Sn(t) defined by
Sn(t) = H (t − n) − H (t − n − 1) , n � t � n + 1 (4.65)
where H(t) is the Heaviside unit step function
H(t) =
{1 , t > 00 , otherwise
(4.66)
Then Laplace transform of Sn(t) is given by
L [Sn(t) ; t → p] =∫ ∞
0e−pt [H (t − n) − H(t − n − 1)] dt (4.67)
=∫ n+1
ne−pt dt =
1p
(1 − e−p) e−np
= S0 (p) e−np (4.68)
where S0(p) =1p
(1 − e−p)
Let us now define a function u(t) by a series
u(t) =∞∑
n=0
un Sn(t) (4.69)
where {un} is a given sequence of finite numbers. Then it follows from(4.69) that
The Inverse Laplace Transform 199
u(t) = un for n � t < n + 1
and u(t + 1) =∞∑
n=0
un Sn (t + 1)
=∞∑
n=0
un [H(t + 1 − n) − H(t − n)]
=∞∑
n=1
unSn−1(t)
=∞∑
n=0
un+1Sn(t) (4.70)
Similarly , u (t + 2) =∞∑
n=0
un+2 Sn(t) (4.71)
and more generally , u (t + j) =∞∑
n=0
un+j Sn(t) (4.72)
when j = 1, 2, 3, · · · · · · · · ·
Then from (4.69) Laplace transform of u(t) is given by
L [u(t) ; t → p ] = u (p) =∫ ∞
0e−ptu (t) dt
=∫ ∞
0e−pt [
∞∑n=0
unSn (t) ] dt
=∞∑
n=0
un
∫ ∞
0e−ptSn (t) dt
=∞∑
n=0
un Sn (p) (4.73)
Now this eqn. (4.73) gives
u (p) =∞∑
n=0
S0 (p) e−np
=1p
(1 − e−p)∞∑
n=0
un e−np
= S (p) ζ (p) , (4.74)
200 An Introduction to Integral Transforms
where ζ (p) =∞∑
n=0
un e−np (4.75)
Then , u(t) = L−1 [ S0 (p) ζ (p) ; p → t ] (4.76)
In particular, if un = an then in this case
ζ (p) =∞∑
n=0
an e−np =∞∑
n=0
(a e−p)n
=1
1 − ae−p=
ep
ep − a(4.77)
L [an] ≡ L [an (t) ] = S0 (p) ζ (p) implying , n � t < n + 1
L−1 [S0 (p)ep
ep − a] = an, by eqn. (4.77) (4.78)
Also, if un = (n + 1) an , then correspondingly
ζ (p) =∞∑
n=0
(n + 1) an e−np
=∞∑
n=0
(n + 1) (ae−p)n = (1 − ae−p)−2 (4.79)
Hence,
L [ (n + 1) an] = L [(n + 1) an (t)] = S0 (p)ζ(p), n � t < n + 1
=1
(1 − ae−p)2S0 (p)
=e2p
(ep − a)2S0 (p) (4.80)
This equation therefore implies
L−1
[e2p S0 (p)(ep − a)2
; p → t
]= (n + 1) an (4.81)
Again,
∞∑n=0
nan e−np =aep
(1 − ae−p)2
and so
L[nan] ≡ L[nan (t)] = S0 (p)ζ (p) , n � t < n + 1
=S0 (p) aep
(ep − a)2
Therefore, L−1
[aS0 (p) ep
(ep − a)2
]= n an (4.82)
The Inverse Laplace Transform 201
Theorem 4.3.
If u (p) = L[u (t) ; t → p ] , then
L [u (t + 1) ; t → p ] = ep [u (p) − u0 S0 (p) ] (4.83)
where u0 = u(0).Proof. By definition
L[u (t + 1) ; t → p ] =∫ ∞
0e−pt u (t + 1) dt
=∫ ∞
1e−pτ ep u(τ) dτ
= ep
[ ∫ ∞
0e−pτu (τ) dτ −
∫ 1
0e−pτ u(τ) dτ
]
= ep[u(p) −∫ 1
0e−pτu(0) dτ ] , since u (τ) = u0 in (0, 1)
= ep u(p) − u0 ep S0 (p)
= ep[L{u(t) ; t → p} − u0S0(p)
]Also in view of the above theorem, we get
L[u (t + 2) ; t → p ] = ep [ L{ u (t + 1) ; t → p } − u(1)S0 (p) ]
= e2p [u (p) − u0 S0(p) ] − ep u(1)S0 (p)
= e2p [ u (p) − { u0 + u1 e−p } S0 (p) ] (4.84)
More generally,
L [ u (t + j) ; t → p ] = ejp [u(p) − S0 (p)j−1∑i=0
ui e−ip ] (4.85)
Example 4.59. Solve the difference equation
yk+2 − 5y k+1 + 6yk = 0 when y0 = 0, y1 = 1.
Solution. Taking Laplace transform the given equation results in
e2p [y (p)−{ y0 + y1e−p } S0 (p)]− 5ep [ y (p)− y0 S0 (p) ] + 6y (p) = 0
We use the given initial condition to obtain[e2p − 5ep + 6
]y (p) = ep S0 (p)
Thus , y(p) =ep
ep − 3S0 (p) − ep
ep − 2S0 (p)
⇒ yk = 3k − 2k
202 An Introduction to Integral Transforms
Example 4.60. Solve the difference equation
yn+2 − 2λ yn+1 + λ2yn = 0 , when y0 = 0 and y1 = 1
Solution. Taking Laplace transform the given difference equation underthe initial condition becomes
(ep − λ)2 y (p) = ep S0 (p)]
⇒ y (p) =ep
(ep − λ )2S0(p)
=1λ
λ ep
(ep − λ)2S0(p)
Inverting this Laplace transformed equation we get
yn =1λ
n λn , by (4.83)
= n λn−1
Example 4.61. Solve the differential-difference equation y′(t) = y(t−1)under the initial condition y(0) = 1
Solution. Laplace transform of the given equation we get
p y (p) − 1 = e−p [ y (p) − S0 (p)]
⇒ y(p) [ p − e−p ] = 1 +e−p − 1
p. e−p
Therefore,y (p) =
1p − e−p
− e−p
p (p − e−p)+
e−2p
p(p − e−p)
=1p
+e−2p
p2 (1 − e−p
p )
=1p
+(
e−p
p
)2 [1 − e−p
p
]−1
=1p
+e−2p
p2+
e−3p
p3+
e−4p
p4+ · · · · · ·
On inversion of Laplace transform, the above result gives
y(t) = 1 +t − 21!
+(t − 3)2
2!+ · · · · · · + (t − n)n−1
(n − 1)!, t > n.
Example 4.62. Solve the non-homogeneous difference equationyk+1 − 3yk = 1 under the condition y0 = 1
2
The Inverse Laplace Transform 203
Solution. Laplace transform of the given difference equation results in
ep [ y (p) − y0 S (p) ] − 3y (p) =1p
⇒ (ep − 3) y (p) =12
ep S0 (p) +1p
⇒ y (p) =12
ep
ep − 3S0 (p) +
1p(ep − 3)
=12
ep
ep − 3S0 (p) +
1p
1 − e−p
(1 − e−p)(ep − 3)
=12
ep
ep − 3S0 (p) + S0 (p)
ep
e2p − 4ep + 3
=[
12
ep
ep − 3+
ep
(ep − 3) (ep − 1)
]S0 (p)
=[
12
ep
ep − 3+
12
ep
ep − 3− 1
2ep
ep − 1
]S0 (p)
=ep
ep − 3S0 (p) − 1
2ep
ep − 1S0 (p)
Inverting Laplace transform, we get
yk = 3k − 12
. 1k
Or, yk = 3k − 12.
Example 4.63. Show that the solution of the difference equation
un+2 + 4 un+1 + un = 0 with u0 = 0 and u1 = 1 is given by
un =1
2√
3
[(√
3 − 2)n + (−1)n+1 (2 +√
3)n]
Solution. Application of Laplace transform yields
e2p[u (p) − (e−pu1 + u0) S0(p) ] + 4ep[u (p) − u0 S0(p)] + u (p) = 0
Or [e2p + 4 ep + 1 ]u (p) = ep S0(p)
Or, u(p) =ep
(ep − α)(ep − β)S0(p),where α =
√3 − 2 and β = −(
√3 + 2)
=1
−β + α
[ −ep
ep − β+
ep
ep − α
]S0 (p)
=1
2√
3
[ep
ep − αS0 (p) − ep
ep − βS0 (p)
]
204 An Introduction to Integral Transforms
Inverting Laplace transform we get
un =1
2√
3
[(√
3 − 2)n − { −(√
3 + 2)}n]
=1
2√
3
[(√
3 − 2)n + (−1)n+1(√
3 + 2)n]
as the solution of the given difference equation.
Exercises
(1) Find inverse Laplace transforms of :
(a)p
(p + 1)2(p2 + 1)(b)
p2 + 6(p2 + 1)(p2 + 4)
(c)p + 2
(p2 + 4p + 5)2(d)
p
p4 + p2 + 1
(e)p
p4 + 4a4(f)
1p(p + 1)3
[Ans. (a)
12
(sin t − t e−t) (b)13
(5 sin t − sin 2 t)
(c)12
t e−2t sin t (d)2√3
sint
2· sin
√3
2t
(e)1
2a2sin at sh at (f) 1 − e−t
(1 + t +
t2
2
)]
(2) Prove the following :
(a) L [ (1 − 2t) e−2t ; t → s ] = s (s + 2)−2
(b) L [t H (t − a) ; t → s ] = (1 + sa) s−2 exp (−as)
(c) L [(1 + 2at) t−12 exp (at) ; t → s] = s
√π (s − a)−
32
(d) L
[1t
{e−at − e−bt
}; t → s
]= log
(s + b
s + a
)
(3) Prove that
(i) L−1
[1p
sin1p
]= t − t3
(3!)2+
t5
(5!)2− t7
(7!)2+ · · · and
(ii) L−1
[1p
cos1p
]= 1 − t2
(2!)2+
t4
(4!)2− t6
(6!)2+ · · ·
The Inverse Laplace Transform 205
(4) Find inverse Laplace transform of :
(a) logp + 1p − 1
(b) logp − 1
p(c)
3p + 1(p + 1)4
(d)12
logp2 + b2
p2 + a2(e) cos−1
(p
2
)(f) log
(1 +
1p2
)
[Ans. (a)
2 sh t
t(b)
1 − et
t(c) e−t
(32
t2 − 13
t3)
(d)cos at − cos bt
t(e)
sin (2t)t
(f)2(1 − cos t)
t
]
(5) Prove that
L−1
[1pf(p)
]=∫ t
0f(x)dx and hence obtain
(i) L−1
[1
p(p + 1)
]and (ii) L−1
[1
p2(p + a)
][Ans. (i)
1a
(1 − e−at) (ii)1a2
(at + e−at − 1)]
(6) Prove that
(a) L−1
[e−as
s2; s → t
]= (t − a)H(t − a)
(b) L−1
[w
s2 + w2f(s) ; s → t
]=∫ t
0f(t − τ) sin wτ dτ
(c) L−1
[1s
e−a√
s ; s → t
]= erfc
(a
2√
t
)
(d) L−1[tan−1
(a
s
); s → t
]=
1t
sh(at)
(e) L−1
[f(s)s2
; s → t
]=∫ t
0(t − τ) f(τ) dτ
(7) Show that
(i)∫ ∞
0cos x2dx =
√π
2√
2=∫ ∞
0sin x2dx and
(ii)∫ ∞
0e−x2
dx =√
π
2,
by using Laplace transform and its inversion and considering func-tions f(t) =
∫∞0 cos(t x2)dx and f(t) =
∫∞0 e−tx2
dx in cases (i)and (ii) respectively.
206 An Introduction to Integral Transforms
(8) Evaluate inverse Laplace transform of :
(i)3(1 + e−pπ)
(p2 + 9)(ii)
32p(16 p2 + 1)2
(iii)1p
logp + 2p + 2
(iv) log(1 − 1/p2)[Ans. (i) − sin 3t H(t − π) + sin 3t (ii)
t
4sin
t
4
(iii)∫ t
0
e−x − e−2x
xdx (iv)
2(1 − ch t)t
]
(9) Prove that
(i) L−1
[1p
log(
1 +1p2
)]=∫ x
∂
2x
(1 − cos x)dx and
(ii) L−1
[1
p2(1 + p2)
]= t e−t + 2 e−t + t − 2
(10) Apply convolution theorem to evaluate
(a) L−1
[p2
(p2 + a2)(p2 + b2)
](b) L−1
[1
p(p2 + 4)
]
(c) L−1
[p
(p2 + 4)3
](d) L−1
[8
(p2 + 1)3
]
(e) L−1
[1
(p + 2)2(p − 2)
]
[ Ans. (a)(a sin at − b sin bt)
(a2 − b2)(b)
14(1 − cos 2t)
(c)t
64[sin 2t − 2t cos 2t] (d) (3 − t2) sin t − 3t cos t
(e)116[e2t − (4t + 1)e−2t
](11) Using Laplace transform properly prove that∫ ∞
0
e−t − e−3t
tdt = log 3
(12) Find inverse Laplace transform of
(a)p e−p/2 + π e−p
p2 + 4(b)
p e−ap
p2 − w2, a > 0
(c)w
(1 − e−πp
w )(p2 + w2)(d)
e−p
(p − 1)(p − 2)
The Inverse Laplace Transform 207
[Ans. (a) sin πt
[H
(t − 1
2
)− H(t − 1)
](b) ch w(t − a) H(t − a)
(c) f(t),where f(t) =
{sin wt , 0 < t < π
w
0 , πw < t < 2π
w
and f(t) is a periodic function of period2πw
(d) [e2(t−1) − e(t−1)H(t − 1)]]
(13) Using L[J0(t)] prove that L[J1(t)] = 1− p√p2+1
. Also prove that
L[t J1(t)] = 1
(1+p2)32
and hence show that
L−1
[1√
p2 + 2ap + 2a2
]= e−atJ0(at)
(14) By expanding tan−1(
1p
)in ascending powers of p, and taking term
by term inverse Laplace transform show that
L−1
[tan−1
(1p
); p → t
]=
sin t
t.
(15) Evaluate L[et erf√
t] and hence evaluate L−1
[3p+2
2p2(p+1)32
][Ans.
1(p − 1)
√p
, t erf√
t
]
(16) Evaluate L−1[
1p3(1+p2)
]by using
(a) partial fractions,
(b) the convolution theorem.
[Ans. −1 + t2
2 − cos t](17) Apply convolution theorem to prove that
(a) L−1
[1
p2(p2 − a2)
]=
(sh at − at)a3
(b) L−1
[1
(p − 1)√
p
]= et erf(
√t)
(c) L−1
[1p
p
(p2 + 1)2
]=
12(sin t − t cos t),
using L−1
[p
(p2 + 1)2
]=
12
t sin t
208 An Introduction to Integral Transforms
(18) Prove that
L−1[e−
√p]
=e−
14t
2√
π t32
(19) Use convolution theorem to prove that∫ t
0J0 (τ) J0 (t − τ) dτ = sin t
(20) (a) If f(t) = H(t − π
2
)sin t, prove that L[f(t) ; t → s] = s
s2+1e−
πs2
(b) Prove that L[ | sin at| ; t → s] =a
s2 + a2coth
πs
2a, s > 0
(c) Prove that L
[d
dt(f ∗ g) ; t → s
]= g(0)f (s) + L[f ∗ g′(t) ; t → s] = sf(s) g(s)
(d) Show that f(t) = sin(a√
t) satisfies the ODE
4t f ′′(t) + 2f ′(t) + a2f(t) = 0
(e) Establish that L
[∫ ∞
t
f(x)x
dx ; t → s
]=
1s
∫ s
0f(x) dx
(f) Prove that L
[cos at − cos bt
t; t → s
]=
12
log(
s2 + a2
s2 + b2
)
(21) Find Laplace transform of the triangular wave function defined in(0, 2a) by
f(t) =
{t , 0 < t < a
2a − t , a < t < 2a[Ans. s−2 th
(as
2
)]
(22) Apply convolution theorem to show that
(a)∫ t
0sin u cos(t − u)du =
12
t sin t
(b) L−1
[1√
p(p − a)
]=
eat
√a
erf√
at and deduce that
L−1
[1p
1√p + a
]=
erf√
at√a
The Inverse Laplace Transform 209
(23) Evaluate, by the method of residues that
(a) L−1
[p
(p + 1)3(p − 1)2
]
(b) L−1
[1
(p2 + 1)2
]
(c) L−1
[2p2 − 4
(p + 1)(p − 2)(p − 3)
]
(d) L−1[e−
√p]
(e) L−1
[ch x
√p
p cosh√
p
], 0 < x < 1
(f) L−1
[1p2
thπp
2
]
[Ans. (a)e−t
16(1 − 2t2) +
et
16(2t − 1)
(b)12t cos t +
12
sin t
(c)72
e3t − 43
e2t − 16
e−t
(d) t−32 e−
14t /(2
√π)
(e)4π
∞∑n=1
(−1)n
2n − 1e−(2n−1)2 π2t
4 cos(
n − 12
)π x + 1
(f) A periodic function f(t) =
{t , 0 < t < π
2π − t , π < t < 2πof period 2π]
(24) Solve the following equations by using Laplace transform :
(a) x′′(t) − 2x′(t) + x(t) = et , when x(0) = 2 , x′(0) = −1
(b) y′′(t) + 4y′(t) + 3y(t) = e−t , when y(0) = y′(0) = 1
(c) y′′(t) − 3y′(t) + 2y(t) = 4t + e3t , when y(0) = 1 , y′(0) = −1
(d) y′′′(t) + 2y′′(t) − y′(t) − 2y(t) = 0 , when y(0) = 1, y′(0) = 2,
y′′(0) = 2
(e) y′′(t) + 2y′(t) + 5y(t) = et sin t , when y(0) = 0, y′(0) = 1
(f) x′′(t) + 9x(t) = cos 2t , when x(0) = 1, x(π
2
)= −1
(g) (D3 − 3D2 + 3D − 1)y(t) = t2e2t,where y(0) = 1, y′(0) = 0,
y′′(0) = −2
210 An Introduction to Integral Transforms
(h) (D4 + 2D2 + 1)y(t) = 0,where y(0) = 0, y′(0) = 1, y′′(0) = 2,
y′′′(0) = −3
(i) (D3 − 2D2 + 5D)x(t) = 0, where x(0) = 0, x′(0) = 1, x(π
8
)= 1
(j) (D2 − 1) x (t) = a ch nt , where x(0) = x′(0) = 0,
(k) (D3 + 1) x (t) = 1 , x(0) = x′(0) = x′′(0) = 0
(l) x′′(t) − k2x(t) = f(t) , where x(0) = x′(0) = 0, k �= 0
(m) (D2 + 1)x(t) = f(t) , x(0) = x′(0) = 0
[ Ans. (a) x = 2et − 3t et +12
t2 et (b) y =74
e−t − 34
e−3t − 12t e−t
(c) y = 2t + 3 +13(e3t − et
)− 2e2t (d) y =13(5et + e−2t) − e−t
(e) y =113
e−t(sin t + sin 2t) (f) x =15(4 cos 3t + 4 sin 3t + cos 2t)
(g) y = (t2 − 6t + 12) e2t −(
32t2 + 7t + 11
)et
(h) y = t(sin t + cos t) (i) x = 1 − et(cos at − sin 2t)
(j) x =a(ch nt − ch t)
(n2 − 1)(k) x = 1 − 1
3e−t − 2
3e
t2 cos
√3t
2
(l) x =12k
[ekt
∫ t
0e−kuf(u) du − e−kt
∫ t
0ekuf(u) du
]
(m) x =∫ t
0sin(t − u)f(u) du ]
(25) Solve the following simultaneous ODE by using Laplace transform:
(a) x′(t) − 2x(t) + 3y(t) = 0, y′(t) + 2x(t) − y(t) = 0,
when x(0) = 8 , y(0) = 3
(b) (D2 − D) x (t) + y(t) = 0,(
D − 12
)x (t) + D y (t) = 0,
when x(0) = 0, y(0) = 1, x′(0) = 0
(c) D2 x (t) + y(t) = −5 cos 2t, D2 y (t) + x(t) = 5 cos 2t,
when x(0) = x′(0) = y′(0) = y′(0) = 1 , y(0) = −1
(d) x′(t) − y′(t) − 2x(t) + 2y(t) = 1 − 2t, x′′ (t) + 2y′(t)
+x(t) = 0, x(0) = x′(0) = y(0) = 0
(e) x′(t) + 5x(t) − 2y(t) = t, y′(t) + 2x(t) + y(t) = 0,
The Inverse Laplace Transform 211
when x(0) = y(0) = 0
(f) (D2 − 1)x(t) + 5Dy(t) = t, −2Dx(t) + (D2 − 4) y (t) = 2,
if x(0) = y(0) = x′(0) = y′(0) = 0
(g) Dx(t) + Dy(t) = t, D2x(t) − y(t) = e−t,
if x(0) = 3, x′(0) = −2, y(0) = 0
(h) (D2 + 2) x (t) − Dy(t) = 1, Dx(t) + (D2 + 2) y (t) = 0,
if x(0) = x′(0) = y(0) = y′(0) = 0
[ Ans. (a) x(t) = 5e−t + 3e4t , y(t) = 5e−t − 2e4t
(b) x(t) =12(sh t − tet
), y(t) = ch t
(c) x(t) = sin t + cos 2t, y(t) = sin t − cos 2t
(d) x(t) = 2(1 − e−t − te−t
), y(t) = −t(1 + 2e−t)
+2(1 − e−t)
(e) x(t) = − 127
(1 + 6t)e−3t +127
(1 + 3t) ,
y(t) = −19(2 + 3t)e−3t +
127
(2 − 3t)
(f) x(t) = 5 sin t − 2 sin 2t − t, y(t) = 1 − 2 cos t + cos 2t
(g) x(t) = 2 +t2
2+
et
2− 3 sin t
2+
cos t
2,
y(t) = 1 − 12e−t − cos t
2+
3 sin t
2
(h) x(t) =16[3 − 2 cos t − cos 2t], y(t) =
16[−2 sin t + sin 2t] ]
(26) Solve (D2 + tD − 1)x(t) = 0, when x(0) = 0 , x′(0) = 1 using
Laplace transform to prove that x(t) = t + cL−1
[1p2 e
p2
2
]. Also
discuss for the particular solution when x(0) = 0.
[ Ans. x = t ](27) Using Laplace transform prove that x(t) = 1 + 2t is the particular
solution of the ODE x′′(t) − tx′(t) + x(t) = 1, when x(0) = 1,x′(0) = 2. Assume here that L−1(pn) = 0, for n = 0, 1, 2, 3, · · ·
(28) Using Laplace transform prove that x(t) = sin t/t is the solutionof the ODE with variable coefficients tx′′(t) + 2x′(t) + tx(t) = 0,if x(0) = 0.
212 An Introduction to Integral Transforms
(29) Prove, by using Laplace transform that y = (1 − t) e−t + 12 sin t is
the solution of the integro-differential equation given by
dy
dt+ 2y(t) +
∫ t
0y(u)du = sin t , y(0) = 1
(30) Solve the following integral/integro-differential equation :
(a) f(t) = e−t − 2∫ t
0cos(t − u)f(u)du
(b) 16 sin 4t =∫ t
0f(u)f(t − u)du
(c) f ′(t) = sin t +∫ t
0f(t − u) cos u du , if f(0) = 0
(d) f ′(t) = t +∫ t
0f(t − u) cos u du , if f(0) = 4
(e)∫ t
0f ′(t)f(t − u)du = 24t3 , if f(0) = 0
(f)∫ t
0f(u) cos(t − u)du = f ′(t), if f(0) = 1
[Ans. (a) f(t) = e−t(1 − t)2 (b) f(t) = ±8 J0(4t)
(c) f(t) =t2
2(d) f(t) = 4 +
52
t2 +t4
24
(e) f(t) = ±16 t32 /√
π (f) f(t) = 1 +t2
2]
(31) A mechanical system, with two degrees of freedom, satisfies theequations 2x′′(t) + 3y′(t) = 4 , 2y′′(t) − 3x′(t) = 0. Use Laplacetransform to determine x(t) and y(t), given that x(t), y(t), x′(t), y′(t)all vanish at t = 0.[Ans. x =
89
(1 − cos
32
t
), y =
89
(32
t − sin32
t
)]
(32) The co-ordinates (x, y) of a particle moving along a plane curve atany time t are given by
y′(t) + 2x(t) = sin 2t , x′(t) − 2y(t) = cos 2t , (t > 0)
If x(t = 0) = 1, y(t = 0) = 0, using Laplace transform show thatthe particle moves on the curve 4x2 + 4xy + 5y2 = 4
The Inverse Laplace Transform 213
(33) Solve the differential system
dx
dt= Ax , x(0) =
[01
]where x =
[x1
x2
], A =
[0 1
−2 3
][Ans. x(t) =
[e2t − et
2e2t − et
] ]
(34) Solve the system
d2x1
dt2− 3x1 − 4x2 = 0 ,
d2x2
dt2+ x1 + x2 = 0 , t > 0 with
x1 = x2 = 0 ,dx1
dt= 2 ,
dx2
dt= 0 at t = 0
[Ans. x1 = 2t ch t, x2 = sh t − t ch t]
(35) Obtain the solution of tx′′(t) + x′(t) + a2x(t) = 0 , x(0) = 1in the form
x(t) = AL−1
[1√
s2 + a2; s → t
]= A J0(at)
(36) Solve the following differential equations under the given initialconditions :
(a)dX
dt= AX,X(0) =
[10
],where X =
[x
y
], A =
[1 −2
−2 1
]
(b)dx1
dt= x1 + 2x2 + t,
dx2
dt= x2 + 2x1 + t, x1(0) = 2, x2(0) = 4
(c)dx
dt= 2x − 3y,
dy
dt= y − 2x ; x(0) = 2, y(0) = 1
[Ans. (a) x(t) =12(e3t + e−t
), y(t) =
12(e3t − e−t
)(b) x1 =
289
e3t − e−t − t
3− 1
9, x2 =
289
e3t + e−t − t
3− 1
9
(c) x =15(7e−t + 3e4t
), y =
15(7e−t − 2e4t
)]
(37) The zero-order chemical reaction satisfies the initial value problem
dC(t)dt
+ k0 = 0 , t > 0 with C(0) = c0 at t = 0
where k0 is a positive constant and C(t) is the concentration of areacting substance at time t. Show that
C(t) = c0 − k0 t
214 An Introduction to Integral Transforms
(38) Using Laplace transform investigate the motion of a particlegoverned by the equations of motion
d2x
dt2−w
dy
dt= 0 ,
d2y
dt2+w
dx
dt= w2a under initial conditions
x(0) = y(0) =d
dt(x(0)) =
d
dty(0) = 0
[Ans. x(t) = a(wt − sin wt) , y(t) = a(1 − cos wt)]
(39) A weightless beam of length l has its ends at x = 0 and x = l.The beam is hinged at these two points. If a concentrated load W
acts at point x = l3 , show that the deflection of the beam at any
point x of it is given by
EI y(x) =W
81x (5l2 − 9x2) +
W
6
(x − l
3
)3
H
(x − l
3
),
on the assumption that the deflection y(x) satisfies the ODE
EI y(iv)(x) = Wδ
(x − l
3
), E, I being the elastic constants of the
beam.(40) The deflection of a beam of length l, clamped horizontally at both
ends and loaded at x = l4 by a weight W is given by
EI y(iv)(x) = Wδ
(x − l
4
). Find the deflection curve of the beam,
given that y(x) = y′(x) = 0, when x = 0, l.
[Ans. EI y(x) =W
6
(x − l
4
)3
H
(x − l
4
)+
9128
WIx2− 964
W 2x3. ]
(41) Find Laplace transform of(i) a square-wave periodic function of period a given by
f(t) =
{1, 0 < t < a
2
−1, a2 < t < a
(ii) a rectified semi-wave periodic function of period 2πw given by
f(t) =
{sin wt, 0 < t < π
w
0, πw < t < 2π
w⎡⎣Ans. (i)
1p
th(ap
4
)(ii)
w[(1 − e
−πpw
)(p2 + w2)
]⎤⎦
The Inverse Laplace Transform 215
(42) Using Laplace transform solve the initial value problem defined by
(D2 + a2)2 x (t) = cos at, if x(0), x′(0), x′′(0), x′′′(0) all vanish.[Ans. x(t) =
18a3
[t sin at − at2 cos at
]]
(43) Solve the following ODE :
(a) ty′′(t) + (2t + 3)y′(t) + (t + 3)y(t) = a e−t , y(0) = 0 y′(0) =a
3(b) y′′(t) + at y′(t) − 2ay(t) = 1, y(0) = y′(0) = 0 , a > 0
[Ans. (a) y(t) =at e−t
3(b) y(t) =
t2
2]
(44) Solve the following initial value problems defined by
(a) x′′(t) + y′(t) + 3x(t) = 15e−t, y′′(t) − 4x′(t) + 3y(t) = 15 sin 2t,
x(0) = 35, x′(0) = −48, y(0) = 27 and y′(0) = −55
(b) x′(t) + 2y′′(t) = e−t , x′(t) + 2x(t) − y(t) = 1, x(0) = y(0) = 0
[Ans. (a) x(t) = 30 cos t − 15 sin 3t − 3e−t + 2cos 2t
y(t) = 30 cos 3t − 60 sin t − 3e−t + sin 2t
(b) x(t) = 1 + e−t − e−at − e−bt, y(t) = 1 + e−t − be−at − ae−bt
where a = 1 − 1√2
and b = 1 +1√2
]
(45) An alternating e.m.f. E sin wt is applied to an electrial networkconsisting of a constant inductance L and capacitance C connectedin series. If Q(t) be the charge at any time t and i(t) be the currentat that instant and they satisfy the ODE
LdQdt + Q
C = E sin wt , i(t) = dQdt with initial charge Q(0) = 0 and
initial current i(0) = 0, show that the current i(t) is given by
i(t) =Ew(cos wt − cos nt)
[ (n2 − w2) L ]where LC =
1n2
(46) (a) A beam which is clamped at the ends x = 0 and x = l, carriesa uniform load w0 per unit length. Show that the deflection y(x)at any point is given by
y(x) =[w0 x2(l − x)2]
(24 E I)
216 An Introduction to Integral Transforms
(b) If the beam in (a) be clamped at x = 0 but is free at x = l andcarries a uniform load w0 per unit length, show that its deflectionat any point x is given by
y(x) =(
w0 x2
24 E I
)(x2 − 4lx + 6l2)
(47) An impulsive voltage Eδ(t) is applied to a circuit consisting ofL,R,C in series with zero initial conditions. If I(t) be the currentat any subsequent time t, find the limit lim
t→0I(t) and justify your
answer. [ Ans. EL ]
(48) A semi-infinite transmission line of negligible inductance and leak-age per unit length satisfy the PDE ∂v
∂t = −Ri, ∂i∂x = −C ∂v
∂t wherev is the voltage and i is the current, R is the constant resistanceand C is the constant capacitance. If v0, the constant voltage, isapplied at the sending end x = 0 at t = 0, prove that the voltage
and current at any point are given by v(x, t) = v0 erfc
(x2
√RCt
)and i(x, t) = v0
√x
2
√CR t−3/2 e−(RCx2/4t).
(49) Using Laplace transform prove that the bounded solution of thePDE
(a) ∂y∂t = ∂2y
∂x2 − 4y under conditions y(0, t) = y(π, t) = 0 andy(x, 0) = 6 sin x−4 sin 2x is y(x, t) = 6e−5t sinx−4e−8t sin 2x
(b) ∂y∂x = y + 2∂y
∂t under condition y(x, 0) = 6e−3x is given byy(x, t) = 6 e−(3x+2t)
(50) A semi-infinite solid x > 0 is initially at temperature zero. At timet = 0, a constant temperature u0 > 0 is applied and maintained atthe face x = 0. Prove that at any point of the solid at a later timeis given by u0 erfc[x/2
√kt] provided the heat conduction equation
of the solid is given by ∂u∂t = k ∂2u
∂x2 , 0 < x < ∞ , t > 0 , k is thediffusivity of the material.
(51) The faces x = 0 and x = l of a slab of material for which k = 1 arekept at temperature zero and until the temperature distributionbecomes u(x, 0) = x. Prove that the temperature u(x, t) at asubsequent time is given by
u(x, t) =2π
∞∑n=1
(−1)n
ne−n2π2t sin nπx .
The Inverse Laplace Transform 217
[ The following result may be used for inversion of Laplace trans-formation :
L−1
[sh x
√p
p sh√
p; p → t
]= x +
2π
∞∑n=1
(−1)n
ne−n2π2t sin(nπx) ] .
(52) An infinite string having one end at x = 0 is initially at reston the x-axis. The end x = 0 undergoes a periodic transversedisplacement A0 sin nt. Find the displacement of any point on thestring at t > 0. Assume that the displacement u(x, t) satisfies thewave equation ∂2u
∂t2 = c2 ∂2u∂x2 .
[Ans. u(x, t) = A0 sin n(t − x
c
)H(t − x
c
). ]
(53) Solve the BVP ∂2u∂t2
= a2 ∂2u∂x2 − g , x > 0 , t > 0 under conditions
u(x, 0) = ut(x, 0) = 0 , u(0, t) = 0 , limx→∞
ux(x, t) = 0 , t � 0 to
show that u(x, t) = 12 g(t − x
a
)2H(t − x
a
)− 12 g t2 .
(54) Prove that, the temperature u(x, t) in the semi-infinite mediumx > 0, when the end x = 0 is maintained at zero temperature andthe initial distribution of temperature is f(x), is given by
u(x, t) =2π
∫ ∞
0f(p) e−c2p2t sin xp dp .
where f(p) is Laplace transform of f(x).
(55) If the initial temperature of an infinite bar is given by
θ(x, 0) =
{θ0 , |x| < a
0 , |x| > a, determine the temperature at
any point x and at any instant t to prove that
θ(x, t) =θ0
2
[erf
(a + x
2c√
t
)+ erf
(a − x
2c√
t
)]
(56) Prove that y(x, t) = f(t − xc ) is the displacement of an infinitely
long string having one end at x = 0 and is initially at rest alongx-axis, when the end x = 0 is given a transverse displacementf(t) , t > 0 .
218 An Introduction to Integral Transforms
(57) Show that u(x, y) = L−12
[f(p,q)p+q ; (p, q) → (x, y)
]is the solution of
the BVP defined by
∂u
∂x+
∂u
∂y= f(x, y) in x � 0 , y � 0 and which vanishes on the
co-ordinate axes.
(58) Use the method of double Laplace transform to find the solutionin the positive quadrant (x � 0 , y � 0) of the PDE
∂2u(x, y)∂x∂y
+ u(x, y) = 0
if u(0, y) = u(x, 0) = 1.[Ans. u(x, y) = L−1
2
[2
p + q; (p, q) → (x, y)
]]
(59) Use eqn. (4.26) of article 4.4 to evaluate the double Laplace trans-form of (i) sin(ax + by) and of (ii) cos(ax + by).
(60) Solve the following difference equations with prescribed initial con-ditions :
(a) un+1 − 2 un = 0 , u0 = 1 [Ans. un = 2n]
(b) yk+2−5 yk+1+6 yk = 2, y0 = 0, y1 = 2 [Ans. yk = 1−2k+3k]
(c) 2 yk+1−yk−4 = 0 , y0 = 3[Ans. yk = 4 − (1
2
)k](d) u′(t) − α u(t − 1) = β, u(0) = 0[
Ans. u(t) = β
{t +
α(t − 1)2
Γ(3)+ α2 (t − 2)3
Γ(4)+ · · · + αn(t − n)n+1
Γ(n + 2)
}]
(e) Δ2un + 3un = 0 , u0 = 0 , u1 = 1 [Ans. un = n 2n . ]
(61) Prove that the general solution of the difference equation
un+2 − 4 un+1 + 3 un = 0
is un = A . 3n + B . 2n , where A = u1 − 2u0 and B = 3u0 − u1.
If u0 = 2 and u1 = −2 , find the general solution un.
The Inverse Laplace Transform 219
(62) Solve the following integral equations :
(a) f(t) = sin 2t +∫ t
0f(t − τ) sin τ dτ
(b) f(t) =t
2sin t +
∫ t
0f ′′(τ) sin(t − τ)dτ , f (0) = f ′(0) = 0
(c)∫ t
0f(τ) J0{a(t − τ)} dτ = sin at
(d) f(t) = sin t +∫ t
0f(τ) sin(2t − 2τ)dτ
(e) f(t) = t2 +∫ t
0f ′(t − τ)e−aτdτ , f (0) = 0
[Ans. (a)12
(t +
32
sin 2t)
(b) (t − cos t) (c) a J0(at)
(d) 3 sin t −√
2 sin(√
2t) (e) t2 +2ta
]
(63) Using Laplace transform evaluate the following :
(a)∫ ∞
0
sin tx
x(x2 + a2)dx, (a, t > 0) (b)
∫ +∞
−∞
cos tx dx
x2 + a2, (a, t > 0)
(c)∫ +∞
−∞
x sin xt
x2 + a2dx , (a, t > 0) (d)
∫ ∞
0exp (−tx2)dx , (t > 0)
[Ans. (a)
π
2a2(1 − e−at) (b)
π
ae−at (c) 2π e−at (d)
√π
4t
]
(64) Solve the following difference equations using Laplace transform :
(a) Δun − 2un = 0 , u0 = 1
(b) Δ2un − 2un+1 + 3un = 0 , u0 = 0 , u1 = 1
(c) un+2 − 4un+1 + 4un = 0 , u0 = 1 , u1 = 4
(d) un+2 − 5un+1 + 6un = 0 , u0 = 1 , u1 = 4
(e) Δ2un + 3un = 0 , u0 = 0 , u1 = 0
[Ans. (a) un = 3n (b) un = n 2n−1 (c) un = (n + 1)2n
(d)un = 2(3n − 2n−1) (e) un = n2n]
Chapter 5
Hilbert and Stieltjes Transforms
5.1 Introduction.
Both Hilbert and Stieltjes transforms appear in many problems ofApplied Mathematics. Specially, Hilbert transform plays important rolein solving problems of fluid mechanics, electronics and signal processingetc. Also while solving problems of fracture mechanics through integralequation technique use of Stieltjes transform may be found useful dueto its simple inversion formula. At first this chapter deals with Hilberttransform.
5.2 Definition of Hilbert Transform
Let f(t) be defined in the real line −∞ < t < ∞. The Hilbert transformof f(t) denoted by fH(x), is defined by
fH(x) = H[f(t) ; t → x] =1π
∫ +∞
−∞
f(t)dt
t − x(5.1)
The integral on the right hand side of eqn. (5.1) is defined in the Cauchyprincipal value sense.
The inverse Hilbert transform is derived formally by use of Fouriertransform in the following steps.
Let us rewrite the formula in eqn. (5.1) as an integral equation ofthe convolution type for the determination of the function f(t) given by
fH(x) =1π
∫ +∞
−∞
f(t)dt
t − x≡ 1√
2π
∫ +∞
−∞f(t) g(x − t) dt (5.2)
where g(x) =√
2π
(− 1x
)and on the assumption that fH(x) is a known
function of x ∈ (−∞ , ∞).
Hilbert and Stieltjes Transforms 221
Since g(x) = −√
2π
x−1
its Fourier transform g(ξ) is given by
g (ξ) = − i sgn (ξ)
Now, taking Fourier transform of eqn.(5.2), it is found that
f=
H (ξ) = − i sgn (ξ) · f (ξ) , (5.3)
where f=
H(ξ) and f(ξ) are the Fourier transforms of fH(x) and f(x)respectively. Now, from eqn. (5.3) one gets
f(ξ) =−1
i sgn(ξ)f=
H(ξ) = i sgn(ξ) f=
H(ξ)
Taking Inverse Fourier transform we finally get
H−1[fH(x)
] ≡ f(x) = − 1π
∫ +∞
−∞
fH(t)dt
t − x, (5.4)
as the required inversion formula for Hilbert transform after reversingall the arguments used above.
For a rigorous proof of this formula one may be referred to the textbooks of Titchmarsh (Oxford Univ. Press , 1948) or Integral equationsby Tricomi (Q.J.Maths, Oxford, 2, 199, 1951).
5.3 Some Important properties of Hilbert Transforms.
From the definition of Hilbert transform in eqn. (5.1) and its inversionformula in eqn. (5.4) it can be derived in operator forms that
H−1 = −H (5.5)
Also, H[ f(t + a) ; t → x ] =1π
∫ +∞
−∞
f(t + a)dt
t − x
=1π
∫ +∞
−∞
f(y)dy
y − a − x=
1π
∫ ∞
−∞
f(y)dy
y − (a + x)
= H[ f(t) ; t → x + a ] = fH (x + a) (5.6)
222 An Introduction to Integral Transforms
H[ f(at) ; t → x ] =1π
∫ +∞
−∞
f(at)dt
t − x
=1π
∫ +∞
−∞
f(y)dy
a[ya − x]=
1π
∫ +∞
−∞
f(y)dy
y − (ax), (a > 0)
= H[ f(t) ; t → ax ] ≡ fH(ax) , (a > 0) (5.7)
Also, H[ f(−at) ; t → x ] =1π
∫ +∞
−∞
f(−at)dt
t − x
=1π
∫ +∞
−∞
f(y)dy
a[−y
a − x] , (a > 0)
=−1π
∫ +∞
−∞
f(y)dy
y − (−ax)= −H[ f(t) ; t → −ax ] , (a > 0)
≡ −fH(−ax) (5.8)
H[ f ′(t) ; t → x ] =1π
∫ +∞
−∞
f ′(t)dt
t − x
=1π
[ {1
t − xf(t)}t→+∞
t→−∞+∫ +∞
−∞
f(t)dt
(t − x)2
], integrating by parts
=1π
∫ +∞
−∞
f(t)dt
(t − x)2≡ d
dxfH(x), (5.9)
sinced
dx[ fH(x) ] =
d
dx
{1π
∫ +∞
−∞
f(t)dt
t − x
}
=1π
∫ +∞
−∞
f(t)(t − x)2
dt
and assuming the existence of all the integrals involved here in theCauchy principal value sense.
Finally, H[tf(t); t → x] =1π
∫ +∞
−∞
tf(t)t − x
dt =1π
∫ +∞
−∞
t − x + x
t − xf(t)dt
=1π
∫ +∞
−∞f(t)dt +
x
π
∫ +∞
−∞
f(t)dt
t − x= x fH(x) + (π)−1
∫ +∞
−∞f(t)dt
(5.10)
Hilbert and Stieltjes Transforms 223
Theorem 5.1. If f(t) is an even function of t, then
fH (x) =x
π
∫ +∞
−∞
f(t) − f(x)t2 − x2
dt (5.11)
Proof. Since by Cauchy principal value sense∫ +∞
−∞
dt
t − x= 0,
we have fH (x) =1π
∫ +∞
−∞f(t)dt
t − x
=1π
∫ +∞
−∞
f(t) − f(x)t − x
dt
=1π
∫ +∞
−∞
(t + x)[ f(t) − f(x) ]t2 − x2
dt
=1π
∫ +∞
−∞
x[ f(t) − f(x) ]t2 − x2
dt +1π
∫ +∞
−∞
t[ f(t) − f(x) ]t2 − x2
dt
So, fH (x) =x
π
∫ +∞
−∞
f(t) − f(x)t2 − x2
dt
as because the second integral of the right-hand side is zero, theintegrand is being an odd function of t.
Example 5.1.
Find the Hilbert transform of
f(t) =
{1 , for |t| < a
0 , for |t| > a
Solution. We have by definition
fH (x) =1π
∫ +a
−a
dt
t − x
If |t| < a, the integrand has a singularity at t = x and hence
fH (x) =1π
lim∈→0
[∫ x−∈
−a
dt
t − x+∫ a
x+∈
dt
t − x
]
224 An Introduction to Integral Transforms
=1π
lim∈→0
[ log | ∈ | − log |(a + x)| + log |(a − x)| − log | ∈ | ], for |x| < a
=1π
log∣∣∣∣a − x
a + x
∣∣∣∣ , for |x| < a
If |t| > a, the integrand has no singularity in −a < t < a andtherefore,
fH (x) =1π
∫ a
−a
dt
t − x=
1π
[ log |t − x| ]a−a
=1π
log∣∣∣∣a − x
a + x
∣∣∣∣ , for |x| > a
Thus for all x, in this case
fH (x) =1π
log∣∣∣∣a − x
a + x
∣∣∣∣Example 5.2. Find the Hilbert transform of
f(t) =t
t2 + a2, a > 0
Solution. We have by definition
fH (x) =1π
∫ +∞
−∞
t dt
(t2 + a2)(t − x)
=1π
∫ +∞
−∞
1a2 + x2
[a2
t2 + a2+
x
t − x− xt
t2 + a2
]dt
=1
π(a2 + x2)
[a2
∫ +∞
−∞
dt
t2 + a2+ x
∫ +∞
−∞
dt
t − x− x
∫ +∞
−∞
t dt
t2 + a2
]
=1
π(a2 + x2)· (aπ) + 0 + 0 , by Cauchy principal value of the
second and the third integrals
=a
x2 + a2·
Example 5.3. Find the Hilbert transforms of
(i) f(t) = cos wt and
(ii) f(t) = sinwt.
Hilbert and Stieltjes Transforms 225
Solution.
(i) From the definition of Hilbert transform, we get
fH (x) =1π
∫ +∞
−∞
cos wt
t − xdt
=1π
∫ +∞
−∞
cos{w(t − x) + wx}t − x
dt
=cos wx
π
∫ +∞
−∞
cos w(t − x)t − x
dt − sin wx
π
∫ +∞
−∞
sin w(t − x)t − x
dt
=cos wx
π
∫ +∞
−∞
cos wα
αdα − sin wx
π· (π)
= 0 − sin wx = − sinwx
(ii) It can similarly be shown that
fH (x) = cos wx.
5.4 Relation between Hilbert Transform and Fourier Trans-
form.
If we write
a(t) =1π
∫ +∞
−∞f(α) cos (αt) dα
b(t) =1π
∫ +∞
−∞f(α) sin (αt) dα (5.12)
Fourier integral can be expressed as
f(x) =12
∫ +∞
−∞[ a(t) + i b(t) ] e−ixt dt
=∫ ∞
0[ a(t) cos (xt) + b(t) sin (xt) ] dt
Let us now define a function of the complex variable z as
φ (z) =∫ ∞
0[ a(t) − i b(t) ] eizt dt , where z = x + iy
≡ U(x, y) + i V (x, y) , say,
226 An Introduction to Integral Transforms
We have then
limy→0
U(x, y) = f(x)
and also − limy→0
V (x, y) =∫ ∞
0[ b(t) cos (xt) − a(t) sin (xt) ] dt
Substituting the values of a(t) and b(t) from eqn. (5.12), we find that
− limy→0
V (x, y) =1π
∫ ∞
0dt
[∫ +∞
−∞f(α) sin(α − x)t dα
]
= limλ→∞
1π
∫ +∞
−∞
1 − cos λ(α − x)α − x
f(α) dα
= limλ→∞
1π
∫ ∞
0
1 − cos λ t
t{ f(x + t) − f(x − t)} dt
If f(t) is sufficiently smooth, we have
limλ→∞
∫ ∞
0cos λ t
f(x + t) − f(x − t)t
dt = 0
by the Riemann-Lebesgue lemma. Therefore,
− limy→0
V (x, y) =1π
∫ ∞
0
f(x + t) − f(x − t)t
dt (5.13)
≡ H [ f(t) ; t → x ] = fH (x) ·
This last step is obtained by changing the variable of integration in thesecond integral in eqn. (5.13) and combining it with the first integral.
5.5 Finite Hilbert Transform.
The finite Hilbert transform is defined by Tricomi (1951) as
H[ f(t) , a, b ] = fH (x, a, b) =1π
∫ b
a
f(t)t − x
dt (5.12)
when x ∈ (a < x < b). In studying the airfoil theory such transformdo arise in Aerodynamics. Also this transform is used in finding thesolution of singular integral equation of the type
1π
∫ 1
−1
f(t)t − x
dt = fH (x) , (5.13)
Hilbert and Stieltjes Transforms 227
where fH (x) and f(t) are respectively known and unknown functionssatisfying Holder conditions on (−1, 1) of the variables. Eqn. (5.13) arisein boundary value problems in solid and in fracture mechanics. In thisrespect works of the authors like Muskhelishvili (1963), Gakhov (1966),Peters (1972), Chakraborty & Williams (1980), Comninou (1977), Das,Patra and Debnath (2004) and others may be referred to.
Several authors including Okikiolu (1965) and Kober (1967)introduced the modified Hilbert transform defined by
Hα [f(t)] = fHα (x) =cosec
(πα2
)2 Γ(α)
∫ +∞
−∞(t − x)α−1 f(t) dt ,
(5.14)
where 0 < α < 1 and x ∈ (−∞ , ∞) ·
The Parseval’s relation of this modified Hilbert transform is given inthe following theorem.
Theorem 5.2. If Hα [f(t)] = fHα (x) , then∫ +∞
−∞Hα [ f(t), x ] h(x) dx = −
∫ +∞
−∞Hα [ h(t), x ] f(x) dx (5.15)
Proof.
We do not persue with the proof of this theorem due to lack of spaceand scope of this treatise.
5.6 One-sided Hilbert Transform.
The two-sided Hilbert transform of f(t) in −∞ < t < x, defined ineqn (5.1), can be written as
fH (x) =1π
∫ +∞
−∞
f(t)dt
t − x
=1π
∫ ∞
0
f(t)t − x
dt − 1π
∫ ∞
0
f(−t)t + x
dt (5.16)
when x > 0 , where the second integral is actually the Stieltjes transformof [−f(−t)] to be defined in this chapter. A similar expression for fH (x)can also be written for x < 0. We now define one-sided Hilbert transform
228 An Introduction to Integral Transforms
of f(t) as
H+ [f(t)] = f+H (x) =
1π
∫ ∞
0
f(t)t − x
dt (5.17)
Further, Mellin transform of H+{f(t)}, to be defined in the next chapter,is given by
M [ H+{f(t)} ] =∫ ∞
0xp−1
[1π
∫ ∞
0
f(t)dt
t − x
]dx
=1π
∫ ∞
0f(t)
[∫ ∞
0
xp−1 dx
t − x
]dt
= cot (πp)∫ ∞
0xp−1 f(t) dt
= cot (πp) M{f(t)}
Taking inverse Mellin transform, we obtain
H+ [ f(t) ] = f+ (x) =1
2πi
∫ c+i∞
c−i∞x−p cot(πp) f(p) dp (5.18)
where f(x) is defined on 0 < x < ∞ and p may be a complex number.
5.7 Asymptotic Expansions of one-sided Hilbert Trans-
form.
The one-sided Hilbert transform of a function f(t) that is analytic for0 < t < ∞ and having the asymptotic expansion with known an, An, Bn
and w > 0 in the form
f(t) =∞∑
n=1
an
tn+ cos wt
∞∑n=1
An
tn+ sinwt
∞∑n=1
Bn
tn, as t → ∞, (5.19)
is defined as
H+ [ f(t) ; t → x ] = f+H (x) =
∫ ∞
0
f(t)t − x
dt (5.20)
Therefore, to find the asymptotic expansion of f+H (x) for such f given
in (5.19), we need to find the expansions of
P
∫ ∞
0
cos wt
t − xdt and P
∫ ∞
0
sinwt
t − xdt
Hilbert and Stieltjes Transforms 229
Now, since P
∫ ∞
0
eiwt
t − xdt = πi (exp iwx) +
∫ ∞
0
eiwt
t − xdt ,
(5.21)
where the contour in the right hand side integral is being indented aboveat t = x. Deforming this contour into the positive imaginary axis bysetting t = iu , u > 0 we get
∫ ∞
0
eiwt
t − xdt = − i
∫ ∞
0
e−wu
x − iudu
= −∞∑
n=0
in+1
xn+1
∫ ∞
0un e−wu du , by Watson′s lemma
= −∞∑
n=0
n!(wx)n+1
exp{
(n + 1) iπ
2
}(5.22)
From (5.17) and (5.18), separating the real and imaginary parts we get
P
∫ ∞
0
cos wt
t − xdt ∼ −π sin wx −
∞∑n=0
n!(wx)n+1
cos{
(n + 1)π
2
}as x → ∞
and
P
∫ ∞
0
sinwt
t − xdt ∼ −π cos wx −
∞∑n=0
n!(wx)n+1
sin{
(n + 1)π
2
}as x → ∞
(5.23)
Then the asymptotic expansion of one-sided Hilbert transform of f(t)satisfying (5.19) is given by
f+H (x) =
∫ ∞
0
f(t)dt
t − x∼ −
∞∑n=1
dn
xn− log x
∞∑n=1
an
xn
−(π sin wx)∞∑
n=1
An
xn+ (π cos wx)
∞∑n=1
Bn
xnas x → ∞
(5.24)
where dn = M{f(t), n} =∫ ∞
0tn−1f(t) dt (5.25)
The detailed of the derivation of (5.24) and (5.25) is not taken up here.For interested readers, the paper entilled “Integrals with a large para-meter : Hilbert transforms (1983), Math Proc. Camb, Phil. Soc., 93,141-149” may be referred to.
230 An Introduction to Integral Transforms
5.8 The Stieltjes Transform.
From the definition of Laplace Transform we have
f(p) = L[ f(t) ; t → p ] =∫ ∞
0e−ptf(t) dt (5.26)
and so Laplace Transform f(p) in variable p to y is given by
L[ f(p) ; p → y ] =∫ ∞
0e−py f(p) dp
=∫ ∞
0e−py dp
∫ ∞
0f(t) e−pt dt , (5.27)
provided that the right hand side integral exists.
Now since,∫ ∞
0e−pt−py dp =
1t + y
we see that
L [ L{f(t) ; t → p} ; p → y ] =∫ ∞
0
f(t)t + y
dt (5.28)
If the right hand side integral of eqn. (5.28) is convergent for complexvalues of y belonging to some region Ω of y-plane then we representeqn. (5.28) as
f2 (y) =∫ ∞
0
f(t)t + y
dt , for y ∈ Ω
≡ S[ f(t) ; t → y ] (5.29)
This is called Stieltjes transform of f(t) for y ∈ Ω. In particular, f2 (y)is analytic in the complex y-plane with a branch cut along the negativey-axis.
Hilbert and Stieltjes Transforms 231
5.9 Some Deductions.
Since,t
t + y= 1 − y
t + y, we have formally
S [ t f(t) ; t → y ] =∫ ∞
0f(t) dt − y f2 (y) (5.30)
Also since,1
(t + a)(t + y)=
1a − y
[1
t + y− 1
t + a
],we have
S
[f(t)t + a
; t → y
]=
1a − y
[ f2 (y) − f2 (a) ] (5.31)
Again, from the change of scale we can easily deduce that
S [ f(ax) ; x → y ] = f2 (y) , for a > 0 (5.32)
Also, putting x = 1u we can show that∫ ∞
0
1x f(
1x
)x + y
dx =∫ ∞
0
f(u)du
1 + uy
This result implies that
S[
x−1 f(x−1)
; x → y]
= y−1 f2
(y−1)
(5.33)
and using (5.32) in (5.33) it is obtained that
S[
x−1 f(a
x
); x → y
]= y f2
(a
y
)(5.34)
Substituting t = u2 it is deduced that
S [f(t
12
); t → y] = 2
∫ ∞
0
uf (u)u2 + y
du
Again since,2u
u2 + y=
1u + i
√y
+1
u − i√
y, we obtain
S [f(t
12
); t → y] = f2(i
√y) + f2(−i
√y) (5.35)
Stieltjes transform of the derivative of f(t) is given by
S [f ′(t) ; t → y] =∫ ∞
0
f ′(t)dt
t + y
=[
f(t)t + y
]∞0
+∫ ∞
0
f(t) dt
(t + y)2
= −1y
f(0) − d
dyf2(y) (5.36)
232 An Introduction to Integral Transforms
If |Arg a| < π and |Arg y| < π, we have from definition
S
[1
t + a; t → y
]=
1a − y
log∣∣∣∣ay∣∣∣∣ (5.37)
If S [ f(t) ; t → y ] =∫ ∞
0
f(t) dt
t + y, then
dn
dyn[ S{f(t) ; t → y} ] = (−1)n n!
∫ ∞
0
f(t) dt
(t + y)n+1, n = 1, 2, 3, · · ·
As a matter of fact Stieltjes transform can be looked upon as a repeatedLaplace transform and the calculation of Stieltjes transform of functionscan be found easily. For example, we know that
L
[1√t
e−at ; t → p
]=
√π (p + a)−
12
L[√
π (p + a)−12 ; p → y
]=
π√y
eay Erfc (√
ay)
These results imply that
S
[1√t
e−at ; t → y
]=
π√y
eay Erfc (√
ay) (5.38)
5.10 The Inverse Stieltjes Transform.
Hilbert and Stieltjes Transforms 233
We describe below an informal proof of inverse Stieltjes transform.A rigorous proof of it may be found in the book of Widder (The LaplaceTransform, Princeton University Press, 1941).
Let us consider the contour integral of g(z) epz over an anticlockwisecontour C in the z-plane, consisting of a straight line γ from c − iR
to c + iR, the circular arc AB of large radius R, a branch cut at z = 0along the negative z-axis from B to B1, with a small circle γ1 : |z| = δ
and then from C1 to C and the circular arc CD of large radius R asshown in the above figure 1 where the function g(z) is assumed to beanalytic except for its branch singularities on the negative real z-axis.The function is also assumed to be such that
lim|z|→∞
g(z)epz = 0 , Re z � c , ( c is arbitrary ) (5.39)
lim|z|→0
z g(z) = 0 (5.40)
Then clearly, ∫C
g(z) epz dz = 0
This result implies that∫ c+iR
c−iRg(z) epz dz +
∫arcAB
g(z) epz dz +∫ δ
Rg(x eiπ) e−px(−dx)
+∮
γ1
g(z) epz dz +∫ R
δg(x e−iπ) e−px(−dx) +
∫arc CD
g(z) epz dz = 0
(5.41)
Making R → ∞ and δ → 0 when on γ1 : z = δ eiθ, we get from abovethat
limδ→0
∫γ1
g(z) epz dz = 0 , by eqn. (5.40) above
limR→∞
∫arcAB
g(z) epz dz = 0 and limR→∞
∫arc CD
g(z) epz dz = 0, by (5.39)
above
Therefore, eqn. (5.41) reduces to∫ c+i∞
c−i∞g(z) epz dz = −
∫ ∞
0g(x eiπ
)e−pxdx +
∫ ∞
0g(x e−iπ
)e−pxdx
=∫ ∞
0
[g(x e−iπ
)− g(x eiπ
) ]e−px dx
234 An Introduction to Integral Transforms
The last equation can be rewritten as
2πi L−1 [g(z) ; z → p]
= L[{
g(x e−iπ
)− g(x eiπ
)}; x → p
]Therefore,
g(z) = L[[
L (2πi)−1 {g (x e−iπ)− g
(x eiπ
)}; x → p
]; p → z
](5.42)
Again, it is known from double Laplace transform that
L [ L { f(x) ; x → p } ; p → y ] = S [ f(x) ; x → y ] (5.43)
Therefore, comparing eqns. (5.42) and (5.43), we can write
S−1 [ g(y) ; y → x ]
=1
2πi
{g(x e−πi
)− g(x eiπ
)}(5.44)
as the formula for inverse stieltjes transform
5.11 Relation between Hilbert Transform and Stieltjes
Transform.
Since Hilbert transform and Stieltjes transform are closely related weshall find a relation between them as below.
We know that
fH (x) = H [ f(t) ; t → x ] =1π
∫ +∞
−∞
f(t) dt
t − x,
where x is real and the integral is in the sense of Cauchy principal value.
If x > 0, we may write
fH (x) =1π
∫ ∞
0
f(t) dt
t − x+
1π
∫ 0
−∞
f(t) dt
t − x
=12π
∫ ∞
0
f(t) dt
t + x eiπ+
12π
∫ ∞
0
f(t) dt
t + x e−iπ− 1
π
∫ ∞
0
f(−t)(t + x)
dt
Hilbert and Stieltjes Transforms 235
∴ fH (x) = (2π)−1 S[
f(t) ; t → x eiπ]
+ (2π)−1 S[
f(t) ; t → x e−iπ]− 1
πS [ f(−t) ; t → x ]
for x > 0 (5.45)
Similarly, when x < 0, we have x = −|x| and hence we have
fH (x) =1π
∫ ∞
0
f(t) dt
t + x+
1π
∫ 0
−∞
f(t) dt
t − x
= (π)−1 S [ f(t) ; t → |x| ] − 1π
∫ ∞
0
f(−t)t + x
dt
= (π)−1 S [ f(t) ; t → −x ] − 12π
∫ ∞
0
f(−t) dt
t + (−x eiπ)
− 12π
∫ ∞
0
f(−t) dt
t + (−x e−iπ)for x < 0 (5.46)
Excercises
(1) Find Hilbert transform of the following functions :
(a) f(t) =1
(a2 + t2), Re a > 0
(b) f(t) =tα
(t + a), | Re α | < 1
(c) f(t) = exp (−at)
(d) f(t) = t−α exp (−at) , Re a > 0 , Re α < 1
(e) f(t) =1t
sin (a√
t) , a > 0
(f) f(t) =sin t
t
[ Ans. (a) (a2 + x2)−1[πx
2a− log
(x
a
)](b) (a − x)−1(aα − xα) π cosec πα
(c) − exp (ax) Ei (−ax)
(d) Γ(1 − α) x−a exp (ax) Γ (α , ax)
(e)π
x[1 − exp(−a
√x)]
(f)(cos x − 1)
x]
236 An Introduction to Integral Transforms
(2) Find Stieltjes transform of the following functions.
(a) f(t) =tα−1
(t + a)
(b) f(t) =1
(t2 + a2)
(c) f(t) =t
(t2 + a2)
[ Ans. (a)(a − x)−1(xα−1 − aα−1
)π cosec (απ)
(b) (a2 + x2)−1[πx
2a− log
(x
a
)]
(c) (a2 + x2)−1[πa
2+ x log
(x
a
)]]
(3) Show that
(a) S
[1√t
cos a√
t ; t → y
]=
π√y
exp (−a√
x) , x > 0
(b) S[sin (k
√t) ; t → y
]= π exp (−k
√y) , k > 0
(4) Prove that
lim∈ → 0+
1π
∫ ∞
∈
[f(x + u) − f(x − u)
u
]du =
1π
P
∫ ∞
−∞
f(t)t − x
dt
(5) Prove that if f(t) = H(t), then
(a) [H{f(t − a) − f(t − b)} ; t → x] =1π
log∣∣∣∣ b − x
a − x
∣∣∣∣ , b > a > 0
(b) H
[f(t − a)
t; t → x
]=
1πx
loga
|a − x| , a > 0
(c) H
[f(t)
at + b; t → x
]=
1π(ax + b)
logb
a|x| ,
(a, b > 0 , x �= b
a
)
(d) H
[αt + βa
t2 + a2; t → x
]=
αa − βx
x2 + a2, a > 0
Hilbert and Stieltjes Transforms 237
(6) Prove that
(a) S
[1
(a + x)(b + x); x → y
]=
⎧⎪⎪⎨⎪⎪⎩
(a−b) log(ay)−(a−y) log(a
b)
(a−y)(b−y)(a−b) , b �= a
a log( ya)−(a−y)
(a−y)2a , b = a
(b) S[xν−1 ; x → y
]= π
yν−1
sin ν π, 0 < ν < 1
(c) S[(a + x)−1 ; x → y
]= (a − y)−1 log
(a
y
)
(7) Show that f(t) = A
[Γ′( 1
2)√πt
− 2 log t√t
]is the solution of the integral
equation f(s) = − 1√π
∫∞0 e−st f(t) dt .
Chapter 6
Hankel Transforms
6.1 Introduction.
Hankel transform arises in the discussion of axisymmetric problems inthe cylindrical coordinates involving Bessel functions. This chapter dealswith the definitions, elementary properties, the inversion theorem andparseval relations of this transform and its relation with other trans-forms. Applications of this transform to various axisymmetric problemsin applied Mathematics are also discussed in this chapter.
6.2 The Hankel Transform.
The (infinite) Hankel transform of order ν of f(r) is formally definedand denoted by ∫∞
0 r f(r) Jν(ξr) dr ≡ Hν [ f(r); ξ ] ,
= fν(ξ),
⎫⎪⎬⎪⎭ (6.1)
if f(r) is integrable over (0,∞)
Theoretically, ν may be zero, integer or a non-integral number. Butin physical application ν is usually either zero or a positive integer.Accordingly, in discussions of elementary properties of Hankel transformν is taken as ν = n , n = 0 or a positive integer.
6.3 Elementary properties
Theorem 6.1. If f1(r) and f2(r) be two integrable functions in (0,∞)and c1, c2 are constants, then
Hn [ c1 f1(r) + c2 f2(r) ] = c1 Hn [ f1(r) ] + c2 Hn [ f2(r) ] (6.2)
Hankel Transforms 239
This property is called the linearity property of Hankel transform.
Proof. We have by definition,
Hn [ c1 f1(r) + c2 f2(r) ] =∫ ∞
0r [ c1 f1(r) + c2 f2(r) ] Jn (ξr) dr
= c1
∫ ∞
0f1(r) Jn (ξr) dr + c2
∫ ∞
0f2(r) Jn (ξr) dr
= c1 Hn [ f1(r) ; ξ ] + c2 Hn [ f2(r) ; ξ ] (6.2)
Theorem 6.2. If a is a constant and H [ f(r) ; ξ ] is Hilbert transformof f(r) of order ν, then
Hν [ f(ar) ; ξ ] =1a2
Hν
[f(r) ;
ξ
a
]=
1a2
fν
(ξ
a
)
Proof. By definition of Hankel transform in (6.1), we have
Hν [ f(ar) ; ξ ] =∫ ∞
0f(ar) · r Jν(ξr) dr
=∫ ∞
0f(x)
x
aJν
(ξ
ax
)1a
dx , taking ar = x
=1a2
∫ ∞
0x f(x) Jν
(ξ
ax
)dx
=1a2
Hν
[f(x) ;
ξ
a
]=
1a2
fν
(ξ
a
)(6.3)
This property of Hankel transform is known as the change of scaleproperty of the transform.
Theorem 6.3. If fn(ξ) be Hankel transform of f(r) of order n, thenHankel transform of f ′(r) of order n is given by
Hn
[f ′(r) ; ξ
]=
ξ
2n[
(n − 1)fn+1(ξ) − (n + 1)fn−1(ξ)], n � 1
(6.4)
H1
[f ′(r) ; ξ
]= −ξ f0 (ξ) (6.5)
provided [rf(r)] vanishes as r → ∞ and as r → 0.
Proof. We have
Hn
[f ′(r) ; ξ
]=∫ ∞
0rJn(ξr)f ′(r)dr
= [ rJn(ξr)f(r) ]∞0 −∫ ∞
0
d
dr{ rJn(ξr) } f(r) dr
240 An Introduction to Integral Transforms
Using the properties of Bessel function we have
d
dr[ rJn(ξr) ] = (1 − n)Jn(ξr) + ξrJn−1(ξr)
Therefore, if { rJn(ξr)f(r) } tends to zero as r → ∞ and also as r → 0,then
Hn
[f ′(r) ; ξ
]= −
∫ ∞
0{ (1 − n)Jn(ξr) + ξrJn−1(ξr) } f(r)dr
= (n − 1)∫ ∞
0f(r)Jn(ξr)dr − ξ
∫ ∞
0rf(r)Jn−1(ξr)dr
Again, using the recurrence relation
Jn(ξr) =ξr
2n[Jn−1(ξr) + Jn+1(ξr)]
of Bessel function, we get
Hn
[f ′(r) ; ξ
]= −ξfn−1(ξ) +
ξ(n − 1)2n
[∫ ∞
0rf(r)Jn−1(ξr) dr
+∫ ∞
0rf(r)Jn+1(ξr)dr
]
= −ξfn−1(ξ) + ξ(n − 1)
2nfn−1(ξ) + ξ
(n − 1)2n
fn+1(ξ)
= −(n + 1)2n
ξfn−1 (ξ) +(n − 1)
2nξ fn+1 (ξ)
Hn
[f ′(r) ; ξ
]=
ξ
2n[
(n − 1)fn+1(ξ) − (n + 1)fn−1(ξ)]
When n = 1 , it follows that
H1
[f ′(r) ; ξ
]= −ξ f0 (ξ)
Thus, the eqns. (6.4) and (6.5) are proved.
Repeated application of (6.4) gives
Hn
[f ′′(r) ; ξ
]=
ξ
2n[
(n − 1)f ′n+1(ξ) − (n + 1)f ′
n−1(ξ)]
=ξ2
4
[n + 1n − 1
fn−2(ξ) − 2n2 − 3n2 − 1
fn (ξ) +n − 1n + 1
fn+2 (ξ)]
(6.6)
Hankel Transforms 241
Theorem 6.4. If fn(ξ) be Hankel transform of order n of f(r), thenHankel transform of order n of f ′′(r) + 1
r f ′(r) − n2
r2 f(r) is −ξ2fn(ξ),provided that rf ′(r) → 0 and rf(r) → 0 when r → 0 and when r → ∞Proof. From the definition of Hankel transform we have
Hn
[f ′′(r) +
1rf ′(r) − n2
r2f(r) ; ξ
]
=∫ ∞
0
{f ′′(r) +
1rf ′(r) − n2
r2f(r)
}r Jn (ξr) dr
Now, integrating by parts the first integral gives∫ ∞
0r
d2
dr2f(r)Jn (ξr) dr = −
∫ ∞
0
df (r)dr
Jn (ξr) dr
−∫ ∞
0rdf (r)dr
J ′n (ξr) dr
and therefore,∫ ∞
0
{d2f
dr2+
1r
df
dr
}r Jn (ξr) dr = −
∫ ∞
0
df
drr J ′
n (ξr) dr
=∫ ∞
0f(r)
d
dr
{r J ′
n(ξr)}
dr, on integrating by parts
Since, Jn(r) is Bessel function of order n, we have
d
dr
[r
d
drJn(r)
]+(
1 − n2
r2
)rJn(r) = 0
Here replacing r by ξr we get,
d
dr
[rJ ′
n(ξr)]
= −(
ξ2 − n2
r2
)rJn(ξr)
Using this result we get∫ ∞
0
{f ′′(r) +
1rf ′(r) − n2
r2f(r)
}rJn(ξr) dr = −ξ2
∫ ∞
0rf(r)Jn(ξr)dr
Or∫ ∞
0
{f ′′(r) +
1rf ′(r) − n2
r2f(r)
}rJn(ξr) dr = −ξ2fn(ξ) (6.7)
The last equation proves the required result under the given condition.
242 An Introduction to Integral Transforms
6.4 Inversion formula for Hankel Transform.
The valid proof for Hankel inversion theorem for general order ν is com-plicated. In application of Hankel transform to the solution of boundaryvalue problems in mathematical physics it is usually only the transformsof order zero and unity are involved. So we present here a proof of Hankelinversion theorem of non-negative order n.
Theorem 6.5. If fn(ξ) is Hankel transform of the function f(r)
implying fn(ξ) = H [ f(r) ; n ] =∫ ∞
0r f(r)Jn(ξr) dr
then f(r) =∫ ∞
0ξfn(ξ)Jn(ξr) dξ (6.8)
is called Hankel inversion formula of fn(ξ) and we express it as
f(r) = H−1[
fn(ξ) ; r]
(6.9)
Proof. Extending the results of general Fourier transform and its in-version formula of a function of one variable for function of two variables,we get
f(s, t) =∫ +∞
−∞
∫ +∞
−∞f(x, y) ei(sx+ty)dx dy (6.10)
f(x, y) =1
4π2
∫ +∞
−∞
∫ +∞
−∞f(s, t) e−i(sx+ty) ds dt (6.11)
Taking the following change of variables
x = r cos θ, y = r sin θ ; s = ξ cos α, t = ξ sin α
in eqns. (6.10) and (6.11) respectively, we get the transform pair as
f(ξ, α) =∫ ∞
0rdr
∫ 2π
0f(r, θ) eiξr cos(θ−α) dθ (6.12)
and f(r, θ) =1
4π2
∫ ∞
0ξdξ
∫ 2π
0f(ξ, α) e−iξr cos(θ−α) dα (6.13)
Let us now choose
f(r, θ) = f(r)e−inθ (6.14)
Then f(ξ, α) =∫ ∞
0f(r)r dr
∫ 2π
0ei{−nθ+ξr cos(θ−α) }dθ (6.15)
Hankel Transforms 243
Now putting φ = α − θ − π
2, we get∫ 2π
0ei{−nθ+ξr cos(θ−α)} dθ = ein(π
2−α)∫ 2π
0ei(nφ−ξr sinφ) dφ
= ein(π2−α) · 2π Jn(ξr), since Jn(ξr)
=12π
∫ 2π
0ei(nφ−ξr sinφ) dφ
Therefore, eqn. (6.15) reduces to
f(ξ, α) =∫ ∞
0r f(r) ein(π
2−α) · 2π Jn(ξr) dr (6.16)
= 2π ein(π2−α) ·
∫ ∞
0r f(r) Jn(ξr) dr
= 2π ein(π2−α) · f(ξ) (6.17)
Thus using eqns, (6.14) and (6.17) in eqn. (6.13), we get
f(r) e−inθ =12π
∫ ∞
0
[ξ f(ξ) e−inθ · 2π Jn(ξr)
]dξ
= e−inθ
∫ ∞
0ξf(ξ)Jn(ξr) dξ
Or f(r) =∫ ∞
0ξ f(ξ) Jn(ξr) dr (6.18)
This result is the required inversion formula for the nth order Hankeltransformed function fn(ξ).
Note 6.1. For the case of general order ν > −12 if
√rf(r) is piecewise
continuous and absolutely integrable on the positive real line, Hankelinversion theorem is given by∫ ∞
0ξ fν(ξ) Jν(ξr)dξ =
12[f(r + 0) + f(r − 0)], (6.19)
where f ν(ξ) = Hν [f(r) ; ξ] .
244 An Introduction to Integral Transforms
6.5 The Parseval Relation for Hankel Transforms.
Theorem 6.6. The Parseval theorem.
Let f(ξ) and g(ξ) be Hankel transforms of the functions f(r) andg(r) respectively. Then∫ ∞
0r f(r) g(r) dr =
∫ ∞
0ξ f(ξ) g(ξ) dξ
Proof. It is known from eqn. (6.19) that
f(r) =∫ ∞
0f(ξ) ξ Jn(ξr) dξ
and also that g(ξ) =∫∞0 r g(r) Jn(ξr) dr from the definition of Hankel
transform.
Therefore,∫ ∞
0ξ f(ξ) g(ξ)dξ
=∫ ∞
0ξf(ξ)
[ ∫ ∞
0rg(r)Jn(ξr)dr
]dξ
=∫ ∞
0rg(r)
[ ∫ ∞
0ξf(ξ)Jn(ξr)dξ
]dr,
by interchanging the order of integration.
=∫ ∞
0rg(r)f(r)dr (6.20)
Thus the Parseval relation is formally established for the order n ofHankel transform.
Note 6.2. For the case of general order ν of Hankel transform, thecorresponding parseval relation is given by∫ ∞
0ξ f(ξ) g(ξ)dξ =
∫ ∞
0rf(r)g(r)dr (6.21)
where f(ξ) =∫ ∞
0rf(r)Jν(ξr)dr
and g(ξ) =∫ ∞
0rg(r)Jν(ξr)dr
Hankel Transforms 245
6.6 Illustrative Examples.
Example 6.1.
Find Hankel transform of order zero of f(x) =
{1 , 0 < x < a
0 , x > a
Solution. We know from definition of Hankel transform that
H0 [f(x)] =∫ ∞
0xf(x)J0 (ξx) dx
=∫ a
0xJ0 (ξx) dx
=[
x
ξJ1(ξx)
]a
0
by the recurrence relationd
dx[ xnJn(x)] = xJn−1(x)
=a
ξJ1(ξa)
Example 6.2. Evaluate (a)H1
[e−ax
x ; ξ]
(b)H0
[1x
; ξ
]
Solution.
(a) We have, H1
[e−ax
x; ξ
]=∫ ∞
ox.
e−ax
xJ1 (ξx) dx
=∫ ∞
oe−axJ1 (ξx) dx
=(a2 + ξ2)
12 − a
ξ(a2 + ξ2)12
(b) H0
[1x
; ξ
]=∫ ∞
oJ0 (ξx) dx =
∫ ∞
oe−0xJ0 (ξx) dx
= (ξ2)−12 =
1ξ, since
∫ ∞
oe−axJ0(bx)dx =
1√a2 + ξ2
Example 6.3. Find Hankel transform of order zero of e−iax
x . and hence
evaluate H0
[cos ax
x , sin (ax)x
]
246 An Introduction to Integral Transforms
Solution. We have from definition
H0
[e−iax
x; ξ
]=∫ ∞
0(cos ax − i sin ax)J0(ξx) dx
=∫ ∞
0cos axJ0(ξx)dx − i
∫ ∞
0sin ax J0(ξx) dx
= Re(−a2 + ξ2)−12 − Im(ξ2 − a2)−
12
∴ H0
[ cos ax
x; ξ]
=
{(ξ2 − a2)−
12 , ξ > a
0 , 0 < ξ < a
and H0
[sin ax
x; ξ
]=
{0 , ξ > a
(a2 − ξ2)−12 , ξ < a
Example 6.4. If
f(x) =
{xn, 0 < x < a,
0, x > a
find Hankel transform of order n of f(x).
Solution. We have, by definition
Hn [ f(x) ] =∫ ∞
0xJn(ξx)f(x) dx
=∫ a
0xn. x Jn(ξx) dx
By recurrence relation of Bessel function we have
d
dx
[xn+1Jn+1(x)
]= xn+1Jn(x)
⇒ 1ξ
d
dx
[xn+1Jn+1(ξx)
]= xn+1Jn(ξx)
Therefore,
Hn [f(x)] =∫ a
0
1ξ
d
dx
[xn+1Jn+1(ξx)
]dx
=an+1
ξJn+1(ξa).
Hankel Transforms 247
Example. 6.5. Find Hankel transform of order zero of[d2
dx2+
1x
d
dx
](e−ax
x
).
Solution. We know from (6.7) for n = 0 that∫ ∞
0
[f ′′(r) +
1rf ′(r)
]rJ0(ξr)dr = −ξ2f0(ξ)
Therefore,∫ ∞
0
[d2
dx2+
1x
d
dx
] (e−ax
x
). xJ0(ξx) dx
= −ξ2
∫ ∞
0e−axJ0 ( ξx ) dx
= − ξ2
(a2 + ξ2)12
.
Example. 6.6. Deduce that
Hν [xν(a2 − x2)μ−ν−1H(a − x) ; ξ] = 2μ−ν−1 Γ(μ − ν)aμξν−μJμ(ξa)
and hence prove that
Hν
[x ν−μ Jμ(ax); ξ
]=
ξν(a2 − ξ2) μ−ν−1
2 μ−ν−1 Γ( μ − ν) aμH(a − ξ).
Solution. From definition of Hankel transform we have
Hν
[xν(
a2 − x2)μ−ν−1
H(a − x) ; ξ]
=∫ a
0xν+1(a2 − x2)μ−ν−1Jν(xξ) dx
=∞∑
r=0
(−1)r( ξ2 )ν+2r
r! Γ(r + ν + 1)
∫ a
0x2ν+2r+1(a2 − x2)μ−ν−1 dx,
for μ > ν � 0 , a > 0 . (6.22)
The integral on the right hand side of this last equation has the value
12
a2ν+2r .Γ(r + ν + 1) Γ(μ − ν)
Γ (r + μ + 1)
and therefore,
Hν
[xν(a2 − x2)μ−ν−1 H(a − x) ; ξ
]
248 An Introduction to Integral Transforms
= 2μ−ν−1Γ(μ − ν)aμξν−μ∞∑
r=0
(−1)r(12ξa)μ+2r
r! Γ (r + μ + 1)
= 2μ−ν−1Γ(μ − ν)aμξν−μJμ(ξa)
(6.23)
Now, using Hankel inversion theorem we have if μ > ν � 0
2μ−ν−1 Γ(μ − ν)aμ Hν
[ξν−μJμ(ξa)
]= xν(a2 − x2)μ−ν−1 H(a − x)
This result in an alternative form can be expressed as
Hν [xν−μJμ(ax) ; ξ] = ξν(a2−ξ2)μ−ν−1
2μ−ν−1 Γ (μ−ν) aμ H(a − ξ)
⇒ ∫∞0 x1−μ+νJμ(ax)Jν (ξx) dx = ξν(a2−ξ2)μ−ν−1
2μ−ν−1 Γ (μ−ν) aμ H(a − ξ)
⎫⎪⎪⎬⎪⎪⎭ (6.24)
Example 6.7. Using eqns.(6.23) and (6.24) in example 6.6, deduce thecorresponding results for the following particular cases
(a) μ = ν + 1 , (b) μ = ν +12
, (c) ν = 0 , (d) ν =12
Solution. (a) Puttintg μ = ν + 1 in eqns. (6.23) and (6.24) we obtain
Hν [ xνH (a − x) ; ξ ] =aν+1
ξJν+1(ξa) , a > 0
Hν
[x−1Jν+1(ax) ; ξ
]=
ξν
aν+1H (a − ξ) , a > 0
(b) Putting μ = ν + 12 in the pair of eqns. (6.23) and (6.24) we get,
Hν
[xν(a2 − x2)−
12 H (a − x) ; ξ
]=√
π
2ξaν+ 1
2 Jν+ 12
(ξa)
Hν
[x− 1
2 Jν+ 12
(ax) ; ξ]
=
√2π
ξν H (a − ξ)
aν+ 12
√a2 − ξ2
, a > 0, ν � 0
(6.25)
(c) Taking ν = 0, the above two equations become
H0
[J1(ax)
x; ξ
]=
H (a − ξ)a
, a > 0 (6.26)
H0
[sin(ax)
x; ξ
]=
H (a − ξ)√a2 − ξ2
, a > 0, ξ < a.
Hankel Transforms 249
(d) Similarly taking ν = 12 , the above pair of equations give∫ ∞
0
√x J1 (ax) J 1
2(xξ) dx =
√2π
√ξ
a2 − ξ2
H(a − ξ)a
Interchanging a and ξ, we have
H1
[sin(ax)
x; ξ
]=
aH (ξ − a)
ξ√
ξ2 − a2
Example 6.8. Prove that H0
[1 − J0(ax)
x2; ξ
]= log
a
ξ. H (a − ξ)
Solution. From eqn. (6.26) we can write∫ ∞
0J1 (ax) J0 (ξx) dx =
H (a − ξ)a
(6.27)
Here, using the results∫ a
0J1 ( ax ) da =
1 − J0 (ax)x
and∫ a
0
H (a − ξ)a
da = H (a − ξ) log(
a
ξ
),
we find on integrating (6.27) with respect to a from 0 to a that∫ ∞
0
1 − J1 (ax )x
J0 (ξx)dx = H(a − ξ) log(
a
ξ
)
⇒ H0
[x−2 { 1 − J0 (ax) } ; ξ
]= H(a − ξ) log
(a
ξ
).
Example 6.9. Using the result
L [ xνJν (xa) ; x → p ] =2 νaν Γ (ν + 1
2)√
π(p2 + a2)ν+ 12
,
prove that Hν
[xν−1 e−px ; ξ
]=
2 νξν Γ (ν + 12)
√π(p2 + ξ2)ν+ 1
2
Solution. From the definition of Hankel transform we know that
Hν
[xν−1 e−px ; ξ
]=∫ ∞
0x. xν−1 e−px Jν (ξx) dx
=∫ ∞
0{ xν . Jν (ξx)} e−px dx
= L [ xνJν(ξx) ; x → p ]
=2νξν Γ ( ν + 1
2)√
π (p2 + ξ2)ν+ 12
, by using the given condition.
250 An Introduction to Integral Transforms
Example 6.10. Evaluate Hν
[x−2Jν(ax) ; ξ
]when ν > −1
2
Solution. Consider the following two functions
f(x) = xνH(a − x) , g(x) = Hxν H(b − x) , ν = −12
Then they give, fν(ξ) = aν+1
ξ Jν+1(ξa), gν(ξ) = bν+1
ξ Jν+1 (ξb),
by using the result of Example 6.7(a). Therefore, using Parseval relationof f(x) and g(x) we have
(ab)ν+1
∫ ∞
0ξ−1 Jν+1(ξa)Jν+1(ξb) dξ =
∫ min (a,b)
0x2ν+1 dx
Now, suppose 0 < a < b. Then the last equation becomes∫ ∞
0ξ−1 Jν+1 (ξa) Jν+1 (ξb) d ξ =
12(ν + 1)
( a
b
)ν+1, ν > −1
2
⇒ Hν
[x−2 Jν(ax) ; ξ
]=
{12ν ( ξ
a)ν , 0 < ξ < a12ν (a
ξ )ν , ξ > a , ν > −12
Example 6.11. Find Hankel inversion of order 1 for the followingfunctions
(a)1ξ
e−aξ (b)1ξ2
e−aξ
Solution.
(a) H−1
[1ξ
e−aξ ; ξ → x
]=∫ ∞
0
e−aξ
ξ. ξ J1 (ξx) dξ
=∫ ∞
0e−aξ J1 (ξx) dξ
=1x− a
x (a2 + x2)12
, by table of integrals.
(b) H−1
[1ξ2
e−aξ ; ξ → x
]=∫ ∞
0
e−aξ
ξJ1 (aξ) dξ
=(a2 + x2)
12 − a
x, by example (a) above
Example 6.12. Find Hankel transform of order zero of 1x . Then apply
the inversion formula to get the original functions.
Hankel Transforms 251
Solution.
H0
[1x
; ξ
]=∫ ∞
0
1x
. x J0 (ξx) dx
=∫ ∞
0J0 (ξx) dx =
1ξ, by table of integrals
Thus, H−10
[1ξ
; x
]=∫ ∞
0
1ξ
. ξ J0 (ξx) dξ
=∫ ∞
0J0 (ξx) dx =
1x
.
Example 6.13. Obtain zero-order Hankel transforms of
(a) r−1e−ar (b)δ(r)r
(c) H(a − r)
Solution.
(a) H0
[r−1e−ar ; ξ
]=∫ ∞
0e−ar J0 (ξr) dr =
1√ξ2 + a2
(b) H0
[δ(r)r
; r → ξ
]=∫ ∞
0δ(r) J0 (ξr) dr = 1
(c) H0 [H(a − r) ; ξ ] =∫ ∞
0rH(a − r)J0(ξr) dr =
∫ a
0rJ0 (ξr) dr
=1ξ2
∫ ar
0p J0 (p) dp =
1ξ2
[p J1(p)]aξ0 =
a
ξJ1(aξ).
Example 6.14. Find the nth order Hankel transforms for the functions
(a) rn H(a − r) (b) rne−ar2
Solution.
(a) Hn [ rn H(a − r) ; ξ ] =∫ ∞
0rn+1 H(a − r) Jn (ξr) dr
=∫ a
0rn+1 Jn(ξr) dr =
an+1
ξJn+1 (aξ)
(b) Hn
[rn e−ar2
; ξ]
=∫ ∞
0rn+1 e−ar2
Jn(ξr) dr
=ξn
(2a)n+1exp(− ξ2
2a
), from the table of integrals
252 An Introduction to Integral Transforms
Example 6.15. Solve the differential equation
∂2u
∂r2+
1r
∂u
∂r+
∂2u
∂z2= 0 , r � 0 , z � 0
satisfying the conditions (i) u → 0 as z → ∞ and as r → ∞,
(ii) u = f(r), on z = 0, r � 0.
Solution. Let u0(ξ, z) be Hankel transform of zero order of u(r, z).Then taking the zeroth order Hankel transform to
∂2u
∂r2+
1r
∂u
∂r+
∂2u
∂z2= 0
we get it as
−ξ2 u0 (ξ, z) +d2 u0
∂z2(ξ, z) = 0
Solving the last ODE we have
u0 (ξ, z) = A eξz + B e−ξz (6.28)
where A, B are constants to be calculated. Now Hankel transform oforder zero of given condition (i) becomes u0 (ξ , z) = 0 , as z → ∞ andhence (6.28) gives A = 0 implying
u0 (ξ , z) = B e−ξz (6.29)
Again, taking Hankel transform of order zero to the given condition(ii), it becomes
u0 (ξ , 0) = f0 (ξ) ≡ H0 [ f(r) ; r → ξ ] (6.30)
Then, eqn. (6.29.) by virtue of eqn. (6.30.) gives
B = f0 (ξ)
Thus, u (ξ , z) = f0 (ξ) · e−ξz
Applying Hankel inversion formula we have, therefore,
u(r , z) =∫ ∞
0f0 (ξ) e−ξz ξ J0 (ξr) dξ
as the required solution of the B V P.
Hankel Transforms 253
Example 6.16. The vibration u(r, t) of a large membrane is given by
∂2u
∂r2+
1r
∂u
∂r=
1c2
∂2u
∂t2, r � 0 , t � 0
with the conditions
u(r, 0) = f(r) ,∂u(r, 0)
∂t= g(r).
Find u(r, t) for t > 0 .
Solution. Taking zeroth order Hankel transform, the given PDEgoverning the vibration becomes
− ξ2 u0 (ξ, t) =1c2
d2 u0
dt2(ξ, t)
Its general solution is
u0 (ξ, t) = A cos (c ξ t) + B sin (c ξ t)
Let f0 (ξ) and g0 (ξ) be Hankel transform of order zero of the givenfunctions f(r) and g(r) respectively.
Then from the given conditions
u0 (ξ, 0) = f0 (ξ) =∫ ∞
0f(r) r J0 (ξr) dr (6.31)
and∂
∂ tu0 (ξ, 0) = g0 (ξ) =
∫ ∞
0g (r) rJ0 (ξr) dr (6.32)
From the general solution and (6.31) we get
f0(ξ) = A
and hence the general solution and (6.32) finally give
g0(ξ) = Bcξ
Therefore, the solution u0(ξ, t) ultimately becomes
u0 (ξ, t) = f0 (ξ) cos (c ξt) +g0 (ξ)c ξ
sin (cξt)
Now, applying the inversion formula of Hankel transform of order zerowe get
u (r, t) =∫ ∞
0
[f0 (ξ) cos (c ξt) +
g0 (ξ)c ξ
sin (c ξ t)]
ξ J0 ( ξ r) dr
254 An Introduction to Integral Transforms
Example 6.17. Consider the axisymmetric Dirichlet problem for a
thick plate |z| � b determined by the function u(ρ, z) which is harmonicand satisfies the boundary conditions.
u(ρ, b) = f(ρ) , u(ρ,−b) = g(ρ) (6.33)
Find the solution u(ρ, z). If f(ρ) = g(ρ), evaluate u(ρ, 0) and∂
∂zu(ρ, 0)
Solution. Since u(ρ, z) satisfies the PDE[∂2
∂ρ2+
1ρ
∂
∂ρ+
∂2
∂z2
][u(ρ, z) = 0 (6.34)
applying zeroth order Hankel transform to (6.34), we get
d2
dz2uo (ξ, z) − ξ2 u0 (ξ, z) = 0
The appropriate solution of this equation is
u0 (ξ, z) = A (ξ) sinh {ξ (b + z)} + B (ξ) sinh { ξ (b − z)} (6.35)
Now, Hankel transformed boundary conditions in eqn. (6.33) are
u0(ξ, b) = f(ξ) and u0(ξ,−b) = g0 (ξ)
Using these conditions, the solution results in
A (ξ) sinh (2 ξ b) = f0 (ξ)
and B (ξ) sinh (2ξb) = g0 (ξ) .
Thus the solution in (6.35) takes the form
u0 (ξ, z) = cosech (2 ξ b)[f0 (ξ) sinh {ξ(z + b)} + g0 (ξ) sinh { ξ (b − z)}]
Applying Hankel inversion theorem we obtain the solution as
u(ρ , z) = H0
[f0 (ξ)
sinh ξ(b + z)sinh (2 ξ b)
+ g0 (ξ)sinh ξ(b − z)sinh (2 ξ b)
; ξ → ρ
](6.36)
If, in particular, f(ρ) = g(ρ) ⇒ f0(ξ) = g0(ξ) , then
u(ξ , z) = f0(ξ)cosh (ξz)cosh (ξb)
Hankel Transforms 255
and the solution is
u(ρ , z) = H0
[f0(ξ)
cosh(ξz)cosh(ξb)
; ξ → ρ
](6.37)
From (6.36) we can derive the general formulae
u(ρ, 0) = H0
[12{f0 (ξ) + g0 (ξ)} sech (ξb) ; ξ → ρ
]∂u
∂z(ρ, 0) = H0
[12ξ{f0 (ξ) − g0 (ξ)}cosech (ξb) ; ξ → ρ
]
Also, from (6.37) we can derive the particular solution
u(ρ, 0) = H0 [f0 (ξ) sech (ξb) ; ξ → ρ]∂u(ρ, 0)
∂z= 0
Example 6.18. Consider the transverse displacement z(ρ, t) of a largemembrane which satisfies the non-homogeneous wave equation
∂2z
∂ρ2+
1ρ
∂z
∂ρ=
1c2
∂2z
∂t2− p (ρ, t)
T(6.38)
where T is the tension in the membrane, p (ρ, t) is the symmetricalpressure applied normally to the membrane and c2 = T/σ, where σ isthe density per unit area of the membrane.
If the initial conditions are
z (ρ , 0) = f(ρ) ,∂z(ρ, 0)
∂t= g(ρ), (6.39)
find the solution of the problem in the Hankel transformed domain.
Discuss the particular case of the problem if there is no appliedpressure and if the initial displacement is ∈ (1 + ρ2/a2)−1/2 to themembrane.
Solution. Introducing Hankel transform of order zero, the equation(6.38) becomes
d2z (ξ, t)dt2
+ c2 ξ2 z (ξ, t) =p0(ξ, t)
σ(6.40)
where, p0 (ξ, t) = H0 [ p (ρ, t) ; ρ → ξ ]
256 An Introduction to Integral Transforms
Let Hankel transform of order zero of the initial conditions in (6.39)be given by
z (ξ, 0) = f0 (ξ) ,d
dtz(ξ, 0) = g0 (ξ) (6.41)
Then, the solution of (6.40) under the initial conditions (6.41) is givenby
z (ξ, t) = f0 (ξ) cos (c t ξ) + (c ξ)−1 g0 (ξ) sin c t ξ
+(c σ ξ)−1
∫ t
0p0 (ξ, τ) sin {c ξ (t − τ)} dτ (6.42)
Particular case.
In this case p0 (ξ, t) = 0 and g0 (ξ) = 0. Also f0 (ξ) =∈ a ξ−1 e−ξa.Then the solution in eqn. (6.42) becomes
z (ξ, t) = ∈ a ξ−1 e−ξa cos (c t ξ)
Now, using the inversion theorem of Hankel transform of order zero tothe above transformed solution, we get
z (ρ, t) = ∈ a H0
[ξ−1 e−ξa cos (c t ξ) ; ξ → ρ
]= ∈ a Re
[∫ ∞
0e−ξ(a+ict) J0 (ξρ) dξ
]
= ∈ a Re[ρ2 + (a + ict)2
]− 12
= ∈ a R−1 cos ϕ
where R and ϕ are given by
R4 = (a2 + ρ2 − c2t2)2 + 4a2c2t2
and tan 2ϕ =(2 act)
(a2 + ρ2 − c2t2).
Example 6.19. Find the potential V (r, z) of a field due to a flatcircular disc of unit radius with its center at the origin, axis along thez-axis and satisfying the differential equation
∂2V
∂r2+
1r
∂V
∂r+
∂2V
∂z2= 0 , 0 � r < ∞ , z � 0 (6.43)
and the boundary conditions
V (r, z) = V0, when z = 0, 0 � r < 1 (6.44)
and∂V (r, z)
∂z= 0, when z = 0, r > 1 (6.45)
Hankel Transforms 257
Solution. Introducing Hankel transform of order zero, eqn. (6.43)becomes
d2V 0
dz2− ξ2 V 0 = 0
Its general solution is
V 0 (ξ, z) = A(ξ) eξz + B(ξ) e−ξz
where A(ξ) , B(ξ) are arbitrary functions of ξ .
On physical grounds, V 0 (ξ, z) → 0 as ξ → ∞, since V (r, z) → 0as r → ∞ . Therefore, A(ξ) = 0 implying that
V 0 (ξ, z) = B(ξ) e−ξz
Applying Hankel inversion formula, we get
V (r, z) =∫ ∞
0B(ξ) ξ e−ξz J0(ξr) dξ (6.46)
and∂V (r, z)
∂z= −
∫ ∞
0B(ξ) ξ2 e−ξz J0(ξr) dξ (6.47)
Now, the boundary conditions in eqns. (6.44) and (6.45) can beexpressed as∫∞
0 ξ B(ξ) J0 (ξr) dξ = V0 , 0 � r < 1
and∫∞0 ξ2 B(ξ) J0 (ξr) dξ = 0 , r > 1
⎫⎪⎬⎪⎭ (6.48)
after utilising eqns. (6.46) and (6.47).
But it is known that∫ ∞
0J0 (ξr)
sin ξ
ξd ξ =
π
2, 0 � r < 1
and∫ ∞
0J0 (ξr) sin ξ d ξ = 0 , r > 1
Therefore, comparing the last two equations with the eqns. in (6.48),we get
B(ξ) =2V0 sin ξ
πξ2
So, V (r, z) =2V0
π
∫ ∞
0e−ξz sin ξ
ξJ0(ξr) dξ.
258 An Introduction to Integral Transforms
Exercises.
(1) Find Hankel transform of order 1 of x−2 e−x.[Ans.
�(1+ξ2)
12 −1
�
ξ
]
(2) If f(x) =
{a2 − x2 , 0 � x < a,
0 , x > a
calculate Hankel transform of f(x) of order zero.
[Ans. 4a
ξ3 J1(aξ) − 2a2
ξ2 J0(ξa)]
.
(3) (a) Prove that H0
[dfdx ; x → ξ
]= −ξ
√a2 + ξ2, if f(x) = 1
x e−ax.
(b) Prove that H0
[δ(r)r ; r → ξ
]= 1 .
(c) Prove that H1 [e−ar ; r → ξ] = ξ(a2+ξ2)
.
(d) Prove that Hn
[rn e−ar2
; r → ξ]
= ξn
(2a)n+1 e−r2
4a .
(4) Show that the solution of the boundary value problem defined by
∂2u
∂r2+
1r
∂u
∂r+
∂2u
∂z2= 0 , 0 � r < ∞ , z > 0 ,
u(r, 0) = u0 for 0 � r < a , where u0 is a constant andu(r, z) → 0 as z → ∞ is
u(r, z) = a u0
∫ ∞
0J1 (aξ) J0 (ξr) e−zξ dξ
(5) Solve the Neumann problem for the Laplace equation
urr +1r
ur + uzz = 0 , 0 < r < ∞ , 0 < z < ∞ when
uz(r, 0) = − 1πa2
H(a − r) , 0 < r < ∞ , u(r, z) → 0 as z → ∞
to prove that u(r, z) =1πa
∫ ∞
0
1ξ
J1(aξ) J0(rξ) e−ξz dξ
(6) Heat is supplied at a constant rate Q per unit area per unit timeover a circular area of radius a in the plane z = 0 to an infinite solid
Hankel Transforms 259
of thermal conductivity k, the rest of the plane is being kept atzero temperature. Solve the steady state heat conduction problemto find the temperature field u(r, z) satisfying the equation
urr + 1rur + uzz = 0 , 0 < r < ∞ , −∞ < z < ∞
with the boundary conditions u → 0 as r → ∞, u → 0 as|z| → ∞ and −k uz = 1
2 Q H(a − r) when z = 0. Finally, provethat
u(r, z) = Qa2k
∫∞0 ξ−1 exp (−|z|ξ) J1(aξ) J0(rξ) dξ .
Chapter 7
Finite Hankel Transforms
7.1 Introduction.
Finite Hankel transform arises in discussing the solution of certain specialtype of boundary value problems. Sneddon (Phil. Mag., 37, 1946) wasthe first author who introduced this transform. Later application of thistransform was found in the works of several other authors in discussingsolutions of axisymmetric physical problems in long circular cylindersand membranes.
7.2 Expansion of some functions in series involving cylin-
der functions : Fourier-Bessel Series.
Suppose f(r) is a real piecewise continuous function in (0, a) and is ofbounded variation in every subinterval [r1, r2] where 0 < r1 < r2 < a.
Then, if the integral∫ a0
√r |f(r)|dr is finite, a Fourier series like series
expansion to f(r) represented by
f(r) =∞∑
m=1
cm Jν
(xνm
r
a
), ν � −1
2(7.1)
where xνm is given by Jν (xνm ) = 0,m = 1, 2, 3, · · · do exist at everypoint of continuity of f(r) and to
12
[ f(r + 0) + f(r − 0) ] (7.2)
at every point of discontinuity of f(r).
The co-efficient cm of the expansion of f(r) in (7.1) can be deter-mined by using the orthogonality property of the system of functions
Jν
(xνm
r
a,)
m = 1, 2, . . . . (7.3)
Finite Hankel Transforms 261
To prove this assertion, let there exist two distinct non zero real numbersα, β such that Jν (α r) , Jν (β r ) are the Bessel functions of first kind.
Then we would have[d 2
dr2+
1r
d
dr+(
α2 − ν2
r2
) ]Jν ( α r) = 0[
d 2
dr2+
1r
d
dr+(
β2 − ν2
r2
) ]Jν ( β r) = 0
Now, subtracting the second equation multiplied by r Jν (α r) fromthe first equation multiplied by r Jν (β r) , and integrating the resultwith respect to r from r = 0 to r = a, we get∫ a
0r Jν (αr) Jν (βr) dr
=[aβ Jν (aα) J ′
ν ( aβ) − aα Jν (βa) Jν′ (αa)]
(α2 − β2)(7.4)
Setting, aα = xνm and aβ = xνn we get from equation (7.4) that∫ a
0r Jν (αr) Jν (βr) dr = 0, if m �= n.
When m = n, taking the limit of eqn. (7.4) as β → α we get afterelimination of J ′′
ν that∫ a
0r J2
ν (αr) dr =a2
2
[J ′ 2
ν (αa) +(
1 − ν2
a2α2
)J2
ν (αa)]
(7.5)
Or,∫ a
0r J2
ν (xmr
a) dr =
a2
2J
′2ν (xνn) (7.6)
=a2
2J2
ν+1 (xνn) (7.7)
Thus assuming the possibility of an expansion of f(r) of the form (7.1),multiplying it by r Jν(xνn
ra) and integrating term by term from r = 0
to r = a , we obtain
cm =2
a2 J2ν+1 (xνm)
∫ a
0rf(r) Jν
(xνm
r
a
)dr, m = 1, 2, · · · (7.8)
The series (7.1) with co-efficients calculated from eqn.(7.8), is called theFourier-Bessel series of f(r).
We are now ready to define finite Hankel transform in the followingsection.
262 An Introduction to Integral Transforms
7.3 The Finite Hankel Transform.
Form I If f(r) satisfies the conditions laid down in article 7.1 overthe finite interval 0 � r � a, then finite Hankel transform of order n isdenoted by Hn[f(r)] and is defined by
Hn [ f(r) ] =∫ a
0r f(r) Jn (rpi) dr ≡ fn (pi) (7.9)
where pi is a positive root of the equation
Jn (api) = 0 (7.10)
The function f(r) can be considered as inverse finite Hankel transformof fn(pi) and it is denoted by
f(r) = H−1n
[fn (pi)
]and is defined by
f(r) =2a2
∞∑i=1
fn (pi)Jn (rpi)
J2n+1 (api)
(7.11)
where pi (0 < p1 < p2 < · · ·) are the roots of the equation Jn(api) = 0.This means that
J ′n (api) = Jn−1 (api) = −Jn+1 (api)
due to standard recurrence relation of Besel function. Here, the abbre-viation api is used instead of xνi as was used in article 7.2.
As a particular case, zeroth order finite Hankel transform of f(r)and its inversion are then defined by
H0 [ f(r) ] ≡ f0 (pi) =∫ a
0r f(r) J0 (rpi) dr (7.12)
H−10 [ f0 (pi) ] ≡ f(r) =
2a2
∞∑i=1
f0 (pi)J0 (rpi)J2
1 (api)(7.13)
where the summation is taken over all the positive roots of J0(ap) = 0.Similarly, finite Hankel transforms of other different orders of f(r) canbe defined as shown above.
Finite Hankel Transforms 263
7.4 Illustrative Examples.
Example 7.1. Find finite Hankel transform of order n of f(r) = rn,for 0 � r � a.
Solution. We know that
d
dx[xn Jn (x)] = xn Jn−1 (x)
Therefore,d
dr
[rn+1 Jn+1 (ki r)
]= ki rn+1 Jn (ki r)
and hence Hn[rn] =∫ a
0rn · r Jn (ki r) dr
=1ki
∫ a
0
d
dr
[rn+1 Jn+1 (ki r)
]dr
=an+1
kiJn+1 (ki a)
In particular if n = 0, then
Hn[ 1 ] =a
kiJ1 (aki) .
Example 7.2. Find finite Hankel transform of (a2 − r2) of order zerofor 0 � r � a.
Solution. We have
H0
[(a2 − r2)
]=∫ a
0r (a2 − r2) J0 (r ki) dr
=∫ a
0a2r J0 (rki) dr −
∫ a
0r3 J0 (rki) dr
= a2
∫ a
0
1ki
.d
dr{ rJ1 (kir) } dr
−∫ a
0r2 1
ki.d
dr{ rJ1 (kir) } dr
=a2
ki[ rJ1 ( kir ) ]a0 −
[r2
kir J1 (kir)
]a
0
+∫ a
0
2rki
rJ1 (ki r) dr
=a3
kiJ1 (ki a) − a3
kiJ1 (ki a) +
2k2
i
[r2 J 2 (ki r)
]a0
=2a2
k2i
J2(ki a)
264 An Introduction to Integral Transforms
=2a2
k2i
[2
ki aJ1(ki a) − J0 (ki a)
],by recurrence relation
=4ak3
i
J1 (ki a) − 2a2
k2i
J0 (ki a) .
Example 7.3. Prove that∫ 10
Jn (α x)Jn (α) . x Jn (ki x) dx = ki
α2−k2i
J ′n (ki),
where ki is the positive root of Jn (ki) = 0.
Solution. The Bessel function Jn(x) satisfies the ODE[x2 d2
dx2+ x
d
dx+ (x2 − n2)
]Jn(x) = 0 (i)
Replacing x by ki x and α x successively in (i), we get
x2 d2
dx2Jn (ki x) + x
d
dxJn (ki x) + (k2
i x2 − n2) Jn (ki x) = 0
and x2 d2
dx2Jn (α x) + x
d
dxJn (α x) + (α2x2 − n2) Jn (α x) = 0
Multiplying the first equation by Jn(α x) and the second by Jn (ki x)and then subtracting, we get
xd
dx
[x
{Jn (α x)
d
dxJn (ki x) − Jn (kix)
d
dxJn (α x)
}]−x2 (α2 − k2
i ) Jn (ki x) Jn (αx) = 0.
Or xJn (kix) Jn (α x)
=1
α2 − k2i
d
dx[x
{Jn (α x)
d
dxJn (ki x) − Jn (ki x)
d
dxJn (αx)
} ]
∴∫ 1
0
Jn (α x)Jn (α)
xJn (ki x) dx
=1
α2 − k2i
1Jn (α)[
ki Jn (α)J ′n (ki) − α Jn (ki) J ′
n (α)]
=ki
α2 − k2i
J ′n (ki) , since Jn (ki) = 0
Example. 7.4. Prove that finite Hankel transform of order n of
f(x) =21+n−m
Γ (m − n).xn (1 − x2)m−n−1 , 0 < x < 1,
is pn−m Jn (p).
Finite Hankel Transforms 265
Solution. By definition
Hn [f (x)] =∫ 1
0
21+n−m
Γ (m − n)xn (1 − x2)m−n−1 xJn (px) dx
=∞∑
r=0
∫ 1
0
21+n−m
Γ (m − n)xn+1 (1 − x2)m−n−1 (−1)r
r!Γ (n + r + 1)(p x)n+2r
2n+2rdx
∴ Hn [f(x)] =1
Γ(m − n)
∞∑r=0
(−1)r pn+2r
Γ(n + r + 1)2m+2r
∫ 1
0y(n+r+1)−1
(1 − y)m−n−1 dy
=1
Γ(m − n)
∞∑r=0
(−1)r pn+2r
Γ(n + r + 1)2m+2r.Γ (n + r + 1) Γ (m − n)Γ (n + r + 1 + m − n)
= pn−m∞∑
r=0
(−1)r
Γ(m + r + 1)
(p
2
)m+2r
= pn−m Jm (p)
Example. 7.5. Find H−1[
cp J1 (ap)
], where p is a positive root of
J0 (ap) = 0.Solution. If f(r) be the required function such thatH [f(r)] = fH (p) = c
p J1 (ap), then from equation (7.13) we have
[2a2
∑p
{c
pJ1 (ap)
}J0 (p r)J2
1 (p a)
]= f(r)
Or2ca2
∑p
J0 (pr)J1(p a)
= f(r)
7.5 Finite Hankel Transform of order n in 0 � x � 1 of
the derivatrive of a function.
For 0 � x � 1. Hankel transform of first order derivative df(x)dx of order
n is given by
Hn
[df
dx
]=∫ 1
0
df
dx.x Jn (px) dx , where p is a root ofJn(p) = 0
= [ f(x) . xJn (px) ]10 −∫ 1
0f(x)
d
dx{ x Jn (px) } dx
266 An Introduction to Integral Transforms
= −∫ 1
0f(x).
[px
2n{(n + 1) Jn−1(px) − (n − 1)Jn+1(px)}
]dx
=p
2n
[(n − 1)
∫ 1
0f(x) x Jn+1(px) dx
−(n + 1)∫ 1
0f(x) x Jn−1(px) dx
]
=p
2n[ (n − 1) Hn+1{f(x)} − (n + 1) Hn−1{f (x)} ] (7.14)
Corollary. 7.1. Putting n = 1 in eqn. (7.14), we get
H1
[df
dx
]= − p Ho [f(x)] (7.15)
7.6 Finite Hankel Transform over 0 � x � 1 of order n ofd2fdx2 + 1
xdfdx
, when p is the root of Jn(p) = 0.
By definition of Hankel transform of order n in 0 � x � 1, we have
Hn
[d2f
dx2+
1r
df
dx
]=∫ 1
0
[d2f
dx2+
1x
df
dx
]x Jn (px) dx
=∫ 1
0x f ′′(x) Jn (px) dx +
∫ 1
0f ′(x) Jn (px) dx
=[x Jn(px) f ′(x)
]10−∫ 1
0
d
dx{x Jn(px)}f ′(x) dx +
∫ 1
0f ′(x) J ′
n(px)dx
= −∫ 1
0px J ′
n(px) f ′(x) dx
= −p
2
∫ 1
0x f ′(x)[Jn−1(px) − Jn+1(px)] dx ,
byd
dxJn(px) =
p
2[Jn−1(px) − Jn+1(px)]
=p
2Hn+1 [f ′(x)] − p
2Hn−1 [f ′(x)] . (7.16)
7.7 Finite Hankel Transform of f ′′(x) + 1xf ′(x) − n2
x2 f(x),
where p is the root of Jn(p) = 0 in 0 � x � 1 .
We have, by the result of article 7.4 that
Hn
[f ′′(x) +
1x
f ′(x) − n2
x2f(x)
]
Finite Hankel Transforms 267
=∫ 1
0
[f ′′(x) +
1x
f ′(x)]
x Jn (px) dx −∫ 1
0
n2
x2f(x). xJn (px) dx
= −p
∫ 1
0x Jn
′ (px)f ′ (x) dx − n2
∫ 1
0
1x
f(x) Jn (px) dx
= −p
[ {xJ ′
n(px) f(x)}1
0−∫ 1
0
{J ′
n(px) + px J ′′n(px)
}f(x) dx
]
−n2
∫ 1
0
1x
f(x)Jn(px)dx
= −p J ′n(p) f(1) + p
[∫ 1
0
{J ′
n(px) + px
(n2
p2x2− 1)
Jn(px) − J ′n(px)
}
f(x) dx]− n2
∫ 1
0
1x
f(x) Jn (px) dx,
by px J ′′n (px) + J ′
n(px) = px
(n2
p2x2− 1
)Jn (px).
Therefore,
Hn
[f ′′(x) +
1x
f ′(x) − n2
x2f(x)
]
= −p Jn′(p)f(1) − p2
∫ 1
0x Jn(x) f(x) dx, after simplification.
(7.17)
Corollary 7.2 Putting n = 0 and using the result J ′0(p) = −J1(p) we
get from the above result that
H0
[f ′′(x) +
1x
f ′(x)]
= p J1 (p) f(1) − p2 H0 [f(x)] (7.18)
7.8 Other forms of finite Hankel Transforms.
For utility purpose in applications other two different forms of finiteHankel transforms of order n are introduced below.
Form II If f(r) is a piecewise continuous function in 0 � r � 1. thenits finite Hankel transform of order n is denoted by f(p) and is definedby
Hn [ f(r) ] = f(p) =∫ 1
0f(r). r Jn (pr) dr, (7.19)
where p is a positive root of
p Jn′ (p) + h Jn (p) = 0 (7.20)
268 An Introduction to Integral Transforms
Correspondingly, inverse finite Hankel transform of f(p) is given by
H−1n
[f(p)
] ≡ f(r) = 2∑
p
p2f(p)h2 + p2 − n2
.Jn (pr){Jn (p)}2 (7.21)
where the summation is being taken over all the positive roots of theequation (7.20).
Form III If the field of variation of the variable r ranges finitely as0 < a � r � b, a third form of finite Hankel transform of order n of afunction f(r), which is piecewise continuous in the range, is defined by
Hn [f(r) ] = f(p) =∫ b
arf(r) Bn (pr) dr, (7.22)
where Bn (p r) = Jn(p r) Yn(p a) − Yn(pr) Jn(p a) (7.23)
and Yn(p r) is the Bessel function of second kind of order n. Here p is apositive root of the equation
Bn (p b) = 0 ⇒ Jn(p b) Yn(p a) = Yn(p b) Jn(p a) (7.24)
Then inverse of f(p) in this case is defined and denoted by
H−1n
[f (p)
] ≡ f(r) =π2
2
∑p
p2 { Jn(pb)}2 f (p){ Jn(pa) }2 − { Jn(pb) }2 Bn (pr),
(7.25)
where the summation is being taken over all the positive roots of theequation (7.25).
7.9 Illustrations.
(i) As an illustration we evaluate finite Hankel transform of order zeroin (0, 1) of f ′′(x) + 1
xf ′(x), where, p is a positive root ofp J ′
0(p) + h J0(p) = 0 below.
By definition, we have
H0
[f ′′(x) +
1x
f ′(x)]
=∫ 1
0
[f ′′(x) +
1x
f ′(x)]
xJ0(px)dx
Finite Hankel Transforms 269
= [ f ′(x) · x J0(px) ]10 −∫ 1
0f ′(x) [ x p J ′
0(px) + J0(px) ] dx
+∫ 1
0f ′(x)J0(px)dx
= [f ′(1)J0(p)] − p
∫ 1
0f ′(x) · x J ′
0(px)dx
= f ′(1)J0(p) − p[{f(x) · x J ′0(px)}1
0 −∫ 1
0f(x)
d
dx{x J ′
0(px)}dx]
= f ′(1)J0(p) − p f(1)J ′0(p) + p
∫ 1
0f(x){J ′
0(px) + pxJ ′′0 (px)}dx
= f ′(1)J0(p) − p f(1)J ′0(p) + p
∫ 1
0(−px) J0(px) f(x)dx
= f ′(1)J0(p) − p f(1) J ′0(p) − p2 H0[f(x)]
But since p is a root of p J ′0(p) + h J0(p) = 0, we have
J ′0(p) = −h
J0(p)p
and therefore,
H0
[f ′′(x) +
1x
f ′(x)]
= [f ′(1) + h f(1)]J0(p) − p2 H0[f(x)] (7.26)
(ii) For illustration of the form-III of finite Hankel transform of ordern in 0 < a � r � b, of f ′′(r) + 1
r f ′(r) − n2f(r)r2 , where p is a root of
Jn(pb) Yn(pa) = Yn(pb) Jn(pa), is stated without proof as
Hn
[f ′′(r) +
1rf ′(r) − n2
r2f(r)
]
=2π
[f(b)
Jn(pa)Jn(pb)
− f(a)]− p2 Hn [f(r)] (7.27)
7.10 Application of finite Hankel Transforms.
We discuss below the application of finite Hankel transform to solvesome initial-boundary value problems.
Example 7.6 Solve the following initial-boundary value problemdefined by
∂2υ
∂r2+
1r
∂υ
∂r=
1κ
∂υ
dt, 0 � r < 1 , t � 0
270 An Introduction to Integral Transforms
under the conditions υ(1, t) = υ0, a costant, when t > 0 and
υ(r, 0) = 0 , 0 � r < 1, by using finite Hankel transform in 0 � r < 1.
Solution. Let υ0 (p, t) be Hankel transform of υ(r, t) over 0 � r < 1of order zero. Then, taking Hankel transform of order zero, the givenPDE under the given boundary condition υ(1.t) = υ0 becomes
p υ(1, t) J1(p) − p2 H0[υ] =1κ
∫ 1
0
∂υ
∂tr J0(pr) dr
Hence, p υ0 J1(p) − p2 υ0 =1κ
d
dtυ0
Thus, υ0(p, t) is a solution of the linear ODE
d υ0
dt+ κ p2 υ0 = κ p υ0 J1(p)
Its solutions is given by
υ0(p, t) =(
υ0
p
)J1(p) + c e−κp2t, where c is a constant.
Now, taking zero order Hankel transform of υ(r, 0) = 0 we getυ0(p, 0) = 0 and therefore, the above solution gives c = −
(υ0p
)J1(p).
Hence,υ0(p, t) =
υ0
pJ1(p)
(1 − e−κp2t
)Applying inversion formula of Hankel transform we finally get the solu-tion as
υ(r, t) = 2∑
p
υ0 (p, t)J0(pr)[J ′
0(p)]2
= 2 υ0
∑p
(1 − e−κp2t
){ J0(pr)p J1(p)
}, since J ′
0(p) = −J1(p),
where summation is taken over all p satisfying J0(p) = 0.
Example 7.7. Find the temperature distribution in a long circularcylinder defined by the axisymmetric heat conduction equation
1κ
∂u
∂t=
∂2u
∂r2+
1r
∂u
∂r, 0 � r � a , t > 0
Finite Hankel Transforms 271
under the boundary and initial conditions
u(r, t) = f(t) , on r = a , t > 0
u(r, 0) = 0 , 0 � r � a
Solution. Under finite Hankel transform of order zero, the given PDEbecomes
1κ
ut(p, t) = −p2u (p, t) + ap J1(ap) f(t) , by (7.18)
Also, u(r, 0) = 0 , 0 � r � a under the transformed domain results in
u (p , 0) = 0
The solution of the above first order ODE gives
u (p, t) = κ ap J1 (ap)∫ t
0f(τ) exp
{−κp2(t − τ)}
dτ
Thus, the formal solution of the boundary value problem is given by
u(r, t) =2κa
∑p
p J0(pr)J1(pa)
∫ t
0f(τ) exp
{−κp2(t − τ)}
dτ,
where the summation is taken over all positive root p satisfying J0(ap) = 0.In particular, if f(t) = constant = T0 , say, then
u(r, t) =2κ T0
a
∑p
p J0(pr)J1(pa)
∫ t
0exp
{−κp2(t − τ)}
dτ
=2T0
a
∑p
J0(pr)p J1(pa)
[1 − exp (−κp2t)
]
Thus, the temperature distribution consists of a transient term whichdecays to zero as t → ∞ together with a steady-state term.
Example 7.8. The axisymmetric unsteady motion of a viscous fluid inan infinitely long circular cylinder of radius a is governed by the PDE
∂u
∂t= ν
[∂2u
∂r2+
1r
∂u
∂r− u(r, t)
r2
], 0 � r � a , t > 0 ,
where u = u(r, t) is the tangential fluid velocity and ν is the constantkinematic viscosity of the fluid. The cylinder is initially at rest at t = 0,
272 An Introduction to Integral Transforms
and it is then allowed to rotate with constant angular velocity Ω. Thusthe respective boundary and initial conditions are
u(r, t) = a Ω , t > 0
and u(r, t) = 0 , t = 0
Solution. Taking Laplace transform in t and finite Hankel transformof order one over 0 < r < a jointly defined by
¯u (p, s) =∫ ∞
0e−st dt
[∫ a
0r J1(pr) u(r, t) dr
],
where p is the positive root of the equation J1(ap) = 0 and using (7.17)the given PDE under the boundary and initial conditions becomes
s¯u (p, s) = −ν[p2 ¯u(p, s)
]− ν a2 p Ωs
J ′1(ap)
Or ¯u (p, s) = −ν a2 Ω p J ′1(ap)
s(s + νp2)
The inverse Laplace transform gives
u(p, t) = −a2 Ωp
J ′1(ap)
[1 − exp (−νtp2)
]Finally, using inverse Hankel transform with J ′
1(ap) = −J2(ap) the aboveequation gives
u(r, t) = 2Ω∑
p
J1(rp)p J2(ap)
[1 − exp(−νtp2)]
Since, Hn [rn] =an+1
pJn+1(ap) , for n = 0, 1, 2, · · ·
we have r = H−11
[a2
pJ2(ap)
]= 2∑
p
J1(rp)p J2(ap)
Using this result we get the final solution of the initial-boundary valueproblem as
u(r, t) = rΩ − 2Ω∑
p
J1(pr)p J2(pa)
exp (−νtp2)
This solution consists of a steady part rΩ together with a transient part−2Ω
∑p
J1(pr)p J2(pa) exp (−νp2t) which tends to zero as t → ∞.
Finite Hankel Transforms 273
Example 7.9. A viscous fluid of kinematic viscosity ν is containedbetween two infinitely long concentric circular cylinders of radi a andb. The inner cylinder is kept at rest and the outer cylinder suddenlystarted rotating with uniform angular velocity ω. Find the tangentialvelocity u(r, t) of the fluid if the equation of motion is
∂2u
∂r2+
1r
∂u
∂r− u
r2=
1ν
du
dt, a < r < b , t > 0,
given that u = b ω when r = b , u = 0 when r = a and also u = 0 whent = 0 .
Solution. Given the boundary values and the initial condition of theinitial boundary value problem as
u(b, t) = b ω , u(a, t) = 0 (i)
and u(r, 0) = 0 (ii)
Now, taking finite Hankel transform of order 1 over a � r � b on r, letu(p, t) be given by
u(p, t) =∫ b
ar B1(pr) u(r, t) dr (iii)
where p is a positive root of the equation
J1(pb) Y1(pa) = Y1(pb) J1(pa) (iv)
Then, the given constitutive PDE becomes
H1
[∂2u
∂r2+
1r
∂u
∂r− u2
r2
]=
1ν
d
dtu (p , t)
Simplifying, we get
2π
[u(b, t)
J1(pa)J1(pb)
− u(a, t)]− p2u(p, t) =
1ν
du
dt, by (7.27)
Ordu
dt+ p2ν u(p, t) =
2bνω
π
J1(pa)J1(pb)
This linear ODE has the solution given by
u(p, t) =2bωπp2
J1(pa)J1(pb)
+ C e−νp2t (v)
274 An Introduction to Integral Transforms
But taking Hankel transform of order 1 on r, the initial condition in (ii)gives
u(p , 0) = 0 (vi)
Employing (vi) in (v), we get
C = −2bωπp2
J1(pa)J1(pb)
and thus (v) becomes
u(p, t) =2bωπp2
J1(pa)J1(pb)
[1 − e−νp2t
](vii)
Now, inverting Hankel transform of order 1, the above equation leads to
u(r, t) = π ω b∑
p
(1 − e−νp2t) J1(pa) J1(pb)[J1(pa)]2 − [J1(pb)]2
· B1(pr)
as the final solution of the initial-boundary value problem.
Example 7.10. The free symmetric vibration of a thin circular mem-brane of radius a is governed by the wave equation
∂2u
∂r2+
1r
∂u
∂r=
1c2
∂2u
∂t2, 0 < r < a , t > 0
with boundary and initial condition
u(a, t) = 0 , t � 0
u(r, 0) = f(r) ,∂u(r, 0)
∂t= g(r) , 0 < r < a
Determine the solution of the problem
Solution. Applying zeroth order finite Hankel transform over r in0 < r < a the given constitutive equation becomes
d2u
dt2+ c2 p2 u = 0 (i)
and the given boundary conditions become
u(p, 0) = f(p) andd
dt[ u (p , 0) ] = g(p) (ii)
Finite Hankel Transforms 275
The solution of this system of second order ODE in (i) under (ii) is then
u(p, t) = f(p) cos (p c t) +g(p)pc
sin (p c t) (iii)
Finally the inverse Hankel transform yeilds the solution as
u(r, t) =2a2
∑p
f(p) cos (p c t)· J0(rp)J2
1 (ap)+
2ca2
∑p
g(p) sin (p c t)J0(rp)
p J21 (ap)
,
where the summation is taken over all positive roots p of J0(pa) = 0 .
Exercises.
(1) Find the zeroth order Hankel transform of(a) f(r) = r2 (b) f(r) = J0(α r)
[Ans. (a)
a2
p2
(ap − 4
ap
)J1(ap) (b)
ap
(α2 − p2)J0(aα) J1(ap)
]
(2) Prove that H0
[1r f ′(r)
]= p H1
{1r f(r)
}− f(a) , where p is apositive root of J0(ap) = 0
(3) Find the solution of the forced symmetric vibrations of a thinelastic membrane that satisfy the initial boundary value problem
∂2u
∂r2+
1r
∂u
∂r− 1
c2
∂2u
∂t2= −p (r, t)
T0,
where p(r, t) is the applied pressure for producing the vibrationand the membrane is stretched by a constant tension T0. Themembrane is set into motion from rest from its equilibrium positionso that
u(r, 0) = 0 =[∂u
∂t
]t=0
(4) Find finite Hankel transform of (r2−b2)r with the kernel
J1(pr) Y1(pb) − J1(pb) Y1(pr).
276 An Introduction to Integral Transforms
[Hint: Start with evalvating∫ ba (r2−b2){J1(pr)Y1(pb)−Y1(pr)J1(pb)}dr]
[Ans. H1
{(r2 − b2)
r
}=
b2 − a2
ap2
J1(pb)J1(pa)
, where p is the positive
root of Y1(pb) J1(ap) − J1(pb) Y1(pa) = 0
and Y1(pb) J ′1(pa) − J1(pb) Y ′
1(pa) =1pa
Jn (pb)Jn (pa)
] .
Chapter 8
The Mellin Transform
8.1 Introduction.
This chapter deals with the origin, theory and applications of Mellintransforms. We derive the transform from its natural way in connectionwith the solution of a problem of potential theory in an infinite wedge.Riemann was the first author to discuss this transform in 1876 and manyothers too have shown its applications later.
Let us consider a problem of potential theory in an infinite wedgedefined by ρ > 0, |ϕ| < α where ρ, ϕ are plane polar co-ordinates.We wish to find the solutions u(ρ, ϕ) of the two-dimensional Laplaceequation
∂2u
∂ρ2+
1ρ
∂u
∂ρ+
1ρ2
∂2u
∂ϕ2= 0 (8.1)
under the conditions that
u (ρ , ± α) = f(ρ) , (a given function) (8.2)
and ρs u (ρ , ϕ) , ρs+1 ∂u
∂ρ(8.3)
tend to zero as ρ → 0 and also as ρ → ∞ for some specified complexnumber s. From (8.1), we can write
∂
∂ρ
[ρ
∂u
∂ρ
]+
1ρ
∂2u
∂ϕ2= 0
Now multiplying the above equation by ρs and integrating the resultwith respect to ρ from ρ = 0 to ρ → ∞ we get∫ ∞
0ρs
[∂
∂ρ
{ρ
∂u
∂ρ
}]dρ +
∫ ∞
0ρs−1 ∂2
∂ϕ2u(ρ, ϕ) dρ = 0
278 An Introduction to Integral Transforms
If we use the formula for integration by parts in the first term we getthe above equation as[
ρs+1 ∂u
∂ρ− s ρs u(ρ, ϕ)
]∞0
+ s2
∫ ∞
0ρs−1 u(ρ, ϕ) dρ
+d2
dφ2
∫ ∞
0ρs−1 u(ρ, ϕ) dρ = 0
Now, using the conditions (8.3) we get[d2
dϕ2+ s2
] ∫ ∞
0ρs−1 u(ρ, ϕ) dρ = 0 (8.4)
Defining Mellin transform of the function u(ρ, ϕ) over 0 < ρ < ∞ as
u∗(s, ϕ) ≡ M [ u(ρ, ϕ) ; ρ → s ] =∫ ∞
0ρs−1 u(ρ, ϕ) dρ
the equation (8.4) can be expressed as
d2u∗
dϕ2(s, ϕ) + s2 u∗(s, ϕ) = 0 (8.5)
which is an ordinary differential equation.
If the conditions on the faces of wedge are as specified in (8.2) beused, we have
u∗(s ,± α) = f∗(s)
after taking Mellin transforms. Then the solution of the boundary valueproblem under prescribed conditions is
u∗(s, ϕ) = f∗(s) · cos sϕ
cos sα(8.6)
Therefore, the solution u∗(s, ϕ) of the boundary-value problem satis-fying the prescribed condition is a function whose Mellin transform isgiven by
f∗(s)cos sϕ
cos sα.
8.2 Definition of Mellin Transform.
Let f(x) be defined for 0 < x < ∞. Mellin transform of this function isa new function of the complex variable s defined by
f∗(s) = M [ f(x) ; x → s ] =∫ ∞
0f(x) xs−1 dx (8.7)
The Mellin Transform 279
for those values of s for which the right hand side improper integral ineqn. (8.7) converges.
We shall now discuss some of the basic important properties of Mellintransform.
Property I. Linearity property.
Let c1, c2 be constants and let f1(x) and f2(x) be two given functions.Then the linearity property of Mellin transform states that
M [c1f1(x)+c2f2(x) ;x → s] = c1M [f1(x) ; x → s]+c2 M [f2(x) ; x → s]
Proof.
The proof of this result is very simple and can be seen in the followingsteps.
M [c1f1(x) + c2f2(x) ; x → s]
=∫ ∞
0xs−1 [c1f1(x) + c2f2(x)] dx
= c1
∫ ∞
0f1(x) xs−1 dx + c2
∫ ∞
0f2(x) xs−1 dx
= c1f∗1 (s) + c2f
∗2 (s) , (8.8)
for those s for which both the improper integrals do exist.
Property II. Change of scale property.
If M [f(x) ; x → s] = f∗(s) , then
M [f(ax) ; x → s] = a−s f∗(s)
Proof.
The proof of this result is deduced by a transformation of variablein the integral form of the transform.
M [f(ax) ; x → s] =∫ ∞
0f(ax)xs−1dx
=∫ ∞
0f(t) · ts−1
asdt , putting ax = t
= a−s f∗(s) (8.9)
Property III.
If M [f(x) ; x → s] = f∗(s) , then
M [xaf(x) ; x → s] = f∗(s + a)
280 An Introduction to Integral Transforms
Proof.
To prove the above result, we get
M [xaf(x) ; x → s] =∫ ∞
0xaf(x) · xs−1 dx =
∫ ∞
0f(x) · x(s+a)−1 dx
= f∗(s + a) , by use of (8.7) (8.10)
Property IV.
If M [f(x) ; x → s] = f∗(s) , then
M
[1x
f
(1x
); x → s
]= f∗(1 − s) .
Proof.
To prove the above result, we have by definition of Mellin transform
M
[1x
f
(1x
); x → s
]
=∫ ∞
0
1x
f
(1x
)xs−1 dx
=∫ ∞
0t−sf(t) dt , putting x =
1t
=∫ ∞
0t(1−s)−1 f(t) dt
= f∗(1 − s) , by (8.7) (8.11)
Property V.
If M [f(x) ; x → s] = f∗(s) , then
M [f(xa) ; x → s] =1a
f∗(s
a
), a > 0
Proof. To prove the above result, we have
M [f(xa) ; x → s] =∫ ∞
0xs−1 f(xa) dx
=1a
∫ ∞
0t(
sa−1) f(t) dt , where xa = t
=1a
f∗(s
a
), by (8.7) (8.12)
The Mellin Transform 281
Property VI. Derivative of Mellin Transform.
If M [f(x) ; x → s] = f∗(s), then M [log x.f(x) ; x → s] is equalto d
ds f∗(s).
Proof. We have, by definition of Mellin transform
M [log x.f(x) ; x → s] =∫ ∞
0xs−1 log x.f(x) dx
But, from the given conditiond
ds[f∗(s)] =
∫ ∞
0
d
ds
[xs−1 f(x)
]dx
=∫ ∞
0f(x) .
1x
. xs log x dx =∫ ∞
0xs−1 . f(x) log x.dx
Thus, M [ log x.f(x) ; x → s ] =d
dsf∗(s) (8.13)
8.3 Mellin Transform of derivative of a function.
The following two sets of results are found to be true in these cases.
Results I (a) M [f ′(x) ; x → s] = −(s − 1)f∗(s − 1)
(b) M [f ′′(x) ; x → s] = (s − 1)(s − 2)f∗(s − 2)
(c) M [fn(x);x → s] = (−1)nΓ(s)
Γ(s − n)f∗(s − n), n = 1, 2, 3, · · ·
where M [f(x) ; x → s] = f∗(s) and fn(x) =dn
dxnf(x)
Proof.
(a) By definition of Mellin transform
M [f ′(x) ; x → s] =∫ ∞
0xs−1f ′(x) dx
=[xs−1 f(x)
]∞0
− (s − 1)∫ ∞
0xs−2 f(x) dx
= −(s − 1)∫ ∞
0x(s−1)−1 f(x) dx ,
for the existence of real numbers λ1 and λ2 such that
lims→0
xs−1 f(x) = 0 , lims→∞ xs−1 f(x) = 0
for λ1 < Re s < λ2 and f∗(s − 1) exists.
Thus, M [ f ′(x) ; x → s ] = −(s − 1) f∗(s − 1)
(8.14)
282 An Introduction to Integral Transforms
(b) Again application of the result in (8.14) will give
M [f ′′(x) ; x → s] = M [ϕ′(x) ; x → s] , where ϕ(x) = f ′(x)
= −(s − 1) ϕ∗(s − 1) = −(s − 1) M [f ′(x) ; x → s]
= −{(s − 1)}{−(s − 2)} M [f(x) ; x → s]
= (s − 1)(s − 2) f∗(x) (8.15)
(c) For the last case (c), if n is a positive integer, by using theMathematical induction n times, we get
M [f (n)(x) ; x → s] = (−1)n (s − 1)(s − 2) · · · (s − n + 1)f∗(s − n)
= (−1)nΓ(s)
Γ(s − n)f∗(s − n) (8.16)
Results II If M [f(x) ; x → s] = f∗(s) , then
(a) M
[(x
d
dx
)f(x) ; x → s
]= −s f∗(s)
(b) M
[(x
d
dx
)2
f(x) ; x → s
]= s2 f∗(s)
(c) M
[(x
d
dx
)n
f(x) ; x → s
]= (−1)n sn f∗(s) , n = 2, 3, · · ·
Proof.
(a) By (8.10) we know that
M [ x φ (x) ; x → s ] = ϕ∗ (s + 1)
Replacing ϕ(x) ≡ f ′(x) , we get
M [ x f ′(x) ; x → s ] = M [ f ′(x) ; x → s + 1 ] (8.17)
Also, since M [ f ′(x) ; x → s ] = −(s− 1) f∗(s− 1) , replacing s
by s + 1 here we get
M [ f ′(x) ; x → s + 1 ] = −s f∗(s) (8.18)
Therefore, from eqn. (8.17) and (8.18) we get
M [ x f ′(x) ; x → s ] = −s f∗(s) , which is the result in (a).(8.19)
The Mellin Transform 283
(b) Let us assume here that(x
d
dx
)f(x) = g(x)
Then,
M [x g′(x) ; x → s] = M [g′(x) ; x → s + 1] , by (8.17)
= −s g∗(s)
= −s
{M
[x
d
dxf(x)
]}= −s {−s f∗(s)} , by (8.18)
= s2 f∗(s) (8.20)
(c) To prove this result, we use the method of Mathematical inductionto obtain
M
[(x
d
dx
)n
f(x) ; x → s
]= (−1)n sn f∗(s) (8.21)
after the nth stage.
8.4 Mellin Transform of Integral of a function.
Result I. If M [f(x) ; x → s] = f∗(s), then
(a) M
[∫ x
0f(t)dt ; x → s
]= −1
sf∗(s + 1)
(b) M
[∫ x
0dy
{∫ y
0f(t)dt
}; x → s
]=
1s(s + 1)
f∗(s + 2)
(c) M [In f(x) ; x → s] = (−1)nΓ(s)
Γ(s + n)f∗(s + n) , where by
In f(x) we mean In f(x) =∫ x
0In−1 f(t) dt
Proof. The proof of these results are given below :
(a) Let∫ x0 f(t) dt = G(x) ⇒ G′(x) = f(x)
Now, since M [G′(x) ; x → s] = −(s − 1)G∗(s − 1)
= −(s − 1)M [G(x) ; x → s − 1]
= −(s − 1)M[∫ x
0f(t)dt ; x → s − 1
]
284 An Introduction to Integral Transforms
we get M [f(x) ; x → s + 1] = −s M
[∫ x
0f(t)dt ; x → s
]
∴ M
[∫ x
0f(t)dt ; x → s
]= −1
sM [f(x) ; x → s + 1]
= −1s
f∗(s + 1) (8.22)
(b) Repeating the result in (a) we get
M
[∫ x
0dy
{∫ y
0f(t)dt
};x → s
]= −1
sM
[∫ x
0f(t)dt;x → s + 1
]
= −1s
{− 1
s + 1f∗(s + 2)
}
=1
s(s + 1)f∗(s + 2) (8.23)
(c) For proving the result in this part the method of Mathematicalinduction will be applied on the index n.
Result II. If M [f(x) ; x → s] = f∗(s) , then
(a) M
[∫ ∞
xf(t)dt ; x → s
]=
1s
f∗(s + 1)
(b) M
[∫ ∞
xdy
{∫ ∞
yf(t)dt
};x → s
]=
1s(s + 1)
f∗(s + 2)
(c) M [I∞n f(x) ; x → s] =Γ(s)
Γ(s + n)f∗(s + n) ,
where I∞n f(x) =∫ ∞
xI∞n−1 f(t)dt .
Proof. As in result I, the result of these parts is shown below.
(a) Let∫ ∞
xf(t)dt = g(x) ⇒ −f(x) = g′(x)
Therefore, M [−f(x) ; x → s] = M [g′(x) ; x → s]
= −(s − 1)M[∫ ∞
xf(t)dt ; x → s − 1
]
⇒ M [f(x) ; x → s] = (s − 1) M
[∫ ∞
xf(t)dt ; x → s − 1
]
⇒ M [f(x) ; x → s + 1] = s M
[∫ ∞
xf(t)dt ; x → s
],
(8.24)
The Mellin Transform 285
which is the proof of the result (a).
(b) Part (b) can be proved by using the result of part (a) onceagain as in Result I.
(c) To prove this part once again the method of Mathematicalinduction on index n will be used.
8.5 Mellin Inversion theorem.
Fourier Transform of g(t) , −∞ < t < ∞ is defined as
G(ξ) =1√2π
∫ +∞
−∞g(t)eiξt dt
Substituting x = et , s = c + iξ in the above we get
G (ic − is) =1√2π
∫ ∞
0x−c g(log x) . xs−1 dx (8.25)
Again in the corresponding inversion formula of Fourier transform
g(t) =1√2π
∫ +∞
−∞G(ξ) e−iξt dξ ,
if the above transform be applied we get
g(log x) =1√2π
i
∫ c+i∞
c−i∞G(ic − is) xc−s ds (8.26)
Now, writing
f(x) = (2π)−12 x−c g(log x) , f∗(s) = G(ic − is)
we have from eqns (8.25) and (8.26)
f∗(s) =∫ ∞
0xs−1 f(x) dx , (8.27)
f(x) =1
2πi
∫ c+i∞
c−i∞f∗(s) x−s ds (8.28)
as the definition of Mellin transform of f(x) and its correspondinginversion formula respectively.
It may be noted here that the left hand side of eqn. (8.28) can alsobe expressed as
f(x) = M−1 [f∗(s) ; s → x] =1
2πi
∫ c+i∞
c−i∞f∗(s) x−s ds .
286 An Introduction to Integral Transforms
8.6 Convolution theorem of Mellin Transform.
Theorem 8.1. If f∗(s) and g∗(s) be Mellin Transforms of f(x) andg(x) respectively, then
M [f(x) g(x) ; x → s] =1
2πi
∫ c+i∞
c−i∞f∗(z) · g∗(s − z)dz .
Proof.
M [f(x) g(x) ; x → s]
=∫ ∞
0f(x) g(x) · xs−1 dx
=∫ ∞
0f(x) xs−1
[1
2πi
∫ c+i∞
c−i∞g∗(z) x−z dz
]dx
=1
2πi
∫ c+i∞
c−i∞g∗(z)
[∫ ∞
0f(x) xs−z−1 dx
]dz
=1
2πi
∫ c+i∞
c−i∞g∗(z) f∗(s − z) dz
Similarly, we would get
M [f(x) g(x) ; x → s] =∫ ∞
0f(x) g(x) xs−1 dx
=1
2πi
∫ c+i∞
c−i∞f∗(z) g∗(s − z) dz
Thus we get
M [f(x) g(x) ; x → s]
= 12πi
∫ c+i∞c−i∞ f∗(z) g∗(s − z) dz
= 12πi
∫ c+i∞c−i∞ g∗(z) f∗(s − z) dz
⎫⎪⎬⎪⎭ (8.29)
as the result of the convolution theorem.
Corollary 8.1. Another form of convolution theorem of Mellin trans-form is given below :
The Mellin Transform 287
Theorem 8.2. If f∗(s) and g∗(s) are Mellin transforms of the functionsf(x) and g(x) respectively, then
M−1 [f∗(s) g∗(s) ; s → x] =∫ ∞
0f(x
t
)g(t)
dt
t
Proof. We know that
M−1 [ f∗(s) g∗(s) ; s → x ]
=1
2πi
∫ c+i∞
c−i∞x−s f∗(s) g∗(s) ds , by (8.28)
=1
2πi
∫ c+i∞
c−i∞x−s f∗(s)
{∫ ∞
0ts−1 · g(t)dt
}ds
=∫ ∞
0
g(t)t
{1
2πi
∫ c+i∞
c−i∞f∗(s)
(x
t
)−sds
}dt
=∫ ∞
0
g(t)t
f(x
t
)dt , (8.30)
since f(x
t
)=
12πi
∫ c+i∞
c−i∞
(x
t
)−sf∗(s) ds (8.31)
8.7 Illustrative solved Examples.
Example 8.1. Evaluate Mellin transform of
(a) f(x) = e−nx
(b) f(x) = (1 + x)−a
(c) f(x) = (1 + xa)−b, where 0 < Re s < Re (ab)
(d) f(x) = cos kx and f(x) = sin kx
(e) f(x) = 2(e2x−1)
Solutions. By definition of Mellin transform, we have
(a) M[e−nx ; x → s
]=∫ ∞
0xs−1 e−nx dx
=1ns
∫ ∞
0ts−1 e−t dt, where nx = t
=Γ(p)ns
288 An Introduction to Integral Transforms
(b) M[(1 + x)−a ; x → s
]=∫ ∞
0
xs−1
(1 + x)s+(a−s)dx
= B (s , a − s) , for Re s > 0 , Re (a − s) > 0
=Γ(s) Γ(a − s)
Γ(a), for Re a > Re s > 0
(c) M
[1
(1 + xa)b; x → s
]=∫ ∞
0
xs−1
(1 + xa)bdx
=∫ ∞
0
(y)sa−1dy
(1 + y)b, where xa = y,
a
xdx =
dy
y
=1a
∫ ∞
0
y( sa−1)
(1 + y)(b− sa)+ s
a
dy
=1a
B(b − s
a,s
a
)
=Γ(b − s
a
)Γ(
sa
)a Γ(b)
, where 0 < Re s < Re (ab)
(d) M [cos kx ; x → s] =∫ ∞
0xs−1 · eikx + e−ikx
2dx
=12
∫ ∞
0e−ikx · xs−1 dx +
12
∫ ∞
0eikx xs−1 dx
We know from (a) above that∫ ∞
0e−nx · xs−1 dx =
Γ(s)ns
Hence,∫ ∞
0e−ikx · xs−1 dx =
Γ(s)(ik)s
and∫ ∞
0eikx · xs−1 dx =
Γ(s)(−ik)s
Therefore,∫ ∞
0cos kx · xs−1 dx =
Γ(s)ks
cosπs
2
and similarly∫ ∞
0xs−1 sin kx dx =
Γ(s)ks
sinπs
2
The Mellin Transform 289
(e) we have M
[2
e2x − 1; x → s
]
= 2∫ ∞
0xs−1 dx
e2x − 1dx
= 2∞∑
n=1
∫ ∞
0xs−1 · e−2nx dx
= 2∞∑
n=1
Γ(s)(2n)s
= 21−s Γ(s)∞∑
n=1
1ns
= 21−s Γ(s) ζ(s) ,
where ζ(s) =∞∑
n=1
1ns
, Re s > 1 is Riemann zeta function.
Example 8.2. If f∗(s) and g∗(s) are Mellin transforms of f(x) andg(x) respectively, prove that∫ ∞
0f(x) g(x) dx =
12πi
∫ c+i∞
c−i∞f∗(z) g∗(1 − z) dz .
Solution. We know, by convolution theorem of Mellin transform that∫ ∞
0xs−1 f(x) g(x) dx =
12πi
∫ c+i∞
c−i∞f∗(z) g∗(s − z) dz
Now, putting s = 1 in this relation we get∫ ∞
0f(x) g(x) dx =
12πi
∫ c+i∞
c−i∞f∗(z) g∗(1 − z) dz .
Example 8.3. Find Mellin inversion of Γ(s) .
Solution. Let f(x) be the required Mellin inversion of Γ(s) . Then
f(x) =1
2πi
∫ c+i∞
c−i∞x−s Γ(s) ds
=1
2πi
∫ c+i∞
c−i∞x−s · π ds
Γ(1 − s) sin sπ
=12i
∫ c+i∞
c−i∞x−s ds
Γ(1 − s) sin sπ
=12i
∫C
x−s ds
Γ(1 − s) sin sπ,
290 An Introduction to Integral Transforms
where C is an anticlockwise closed large semi-circular contourwith centre at s = 0 lying to the left of the imaginary s-axis. Then theintegrand has singularities at its poles at s = 0,−1,−2, · · · . Thus, thevalue of the integral is 2πi times sum of the residues of the integrandat its poles. Hence,
f(x) =12i
2πi∞∑
n=0
xn
Γ(1 + n)[
dds (sin sπ)
]s=−n
=∞∑
n=0
(−1)nxn
n!
= e−x .
Example 8.4. Solve the boundary value problem defined by
x2 uxx + x ux + uyy = 0 , 0 � x < ∞ , 0 < y < 1
subject to boundary conditions u(x, 0) = 0 , u(x, 1) =
{A, 0 � x � 10, x > 1
where A is a constant, by applying Mellin transform over x .
Solution. Applying Mellin transform over x we get the PDE as
u∗yy + s2 u∗ = 0 , 0 < y < 1
and the boundary conditions transform to
u∗(s, 0) = 0 , u∗(s, 1) = A
∫ 1
0xs−1 dx =
A
s
Then the solution of the transformed ODE is
u∗(s, y) =A
s
sin sy
sin s, 0 < Re s < 1 .
Taking inverse Mellin transform we get
u(x, y) =A
2πi
∫ c+i∞
c−i∞
x−s
s
sin sy
sin sds .
The function u∗(s, y) is analytic in 0 < Re s < π and thus 0 < c < π.The above integrand has simple poles at s = nπ , n = 1, 2, 3, · · · and all
The Mellin Transform 291
those lie inside a large semi-circular contour in the right-half of s-plane.Therefore,
u(x, y) =A
π
∞∑n=1
1n
(−1)n x−nπ sin (nπy) .
Example 8.5. Consider the problem of determining the potentialϕ(r, θ) that satisfies the Laplace equation
r2 ϕrr + r ϕr + ϕθθ = 0
in an infinite wedge 0 < r < ∞ , −α < θ < α and the boundaryconditions
ϕ(r, α) = f(r) , ϕ(r,−α) = g(r) , 0 � r < ∞ϕ(r, θ) → 0 as r → ∞ for all θ in − α < θ < α .
Solution. Applying Mellin transform over r, the Laplace equationbecomes
s2 ϕ∗(s, θ) +d2 ϕ∗(s, θ)
dθ2= 0
and the boundary conditions transform to
ϕ∗(s, α) = f∗(s) , ϕ∗(s,−α) = g∗(s) and ϕ∗(s, θ) → 0 as s → ∞
The general solution of the transformed ODE is
ϕ∗(s, θ) = A cos sθ + B sin sθ ,
where A and B are arbitrary constants to be determined under thetransformed boundary conditions. These give
A cos s α + B sin s α = f∗(s)
A cos s α − B sin s α = g∗(s)
Solving for A and B , we get
A =f∗(s) + g∗(s)
2 cos s α, B =
f∗(s) − g∗(s)2 sin s α
Thus the solution becomes
ϕ∗(s, θ) = f∗(s)sin s(α + θ)
sin (2sα)+ g∗(s)
sin s(α − θ)sin (2sα)
292 An Introduction to Integral Transforms
Inverting the Mellin transformed equation, the solution of the problemcan be expressed as
ϕ(r, θ) = M−1 [f∗(s) h∗(s, α + θ)] + M−1 [g∗(s) h∗(s, α − θ)]
where h∗(s, θ) =sin (sθ)sin (2sα)
Now, h(r, θ) = M−1 [h∗(s, θ)] = M−1
[sin sθ
sin 2sα
]
=12α
rn sin (nθ)1 + 2rn cos nθ + r2n
,
after simplification where n =π
2αTherefore, the above form of solution of the problem is represented as
ϕ(r, θ) =n rn cos nθ
π
[∫ ∞
0
un−1 f(u) du
u2n − 2rnun sin nθ + r2n
]
+nrn cos nθ
π
[∫ ∞
0
un−1 g(u) du
u2n + 2rnun sin nθ + r2n
],− π
2n< θ <
π
2n
8.8 Solution of Integral equations.
We begin with the solution of the following integral equation of generaltype ∫ ∞
0f(x) K(xt) dx = g(t) , t > 0 (8.32)
Taking Mellin transform of both sides of eqn. (8.32) we get
f∗(1 − s) K∗(s) = g∗(s)
by using eqn. (8.30).
Now replacing s by 1 − s we find
f∗(s) = g∗(1 − s) L∗(s) , where L∗(s) =1
K∗(1 − s)
This solution can be written in the form
f(x) =∫ ∞
0g(t) L(xt) dt, (8.33)
The Mellin Transform 293
provided that the inverse Mellin transform in
L(x) = M−1
[1
K∗(1 − s); s → x
]exists.
In particular, the equation (8.32) will have the solution
f(x) =∫ ∞
0g(t) K(xt) dt , x > 0
if K∗(s) K∗(1 − s) = 1
Example 8.6.
Solve the integral equation∫ ∞
0f(ξ) g
(x
ξ
)dξ
ξ= h(x) , x > 0
Solution. Applying Mellin transform the given integral equation re-duces to
f∗(s) = h∗(s) K∗(s) , where K∗(s) =1
g∗(s)
Then inverting the Mellin transform we get the required solution of thegiven integral equation as
f(x) = M−1 [h∗(s) K∗(s)]
=∫ ∞
0h(ξ) K
(x
ξ
)dξ
ξ, by use of (8.30)
8.9 Application to Summation of Series.
A useful method of summing slowly convergent series has been given byMacfarlane (Phil Mag., 40,188,(1950)).
Theorem 8.3. If f∗(s) is Mellin transform of f(x), then∞∑
n=0
f(n + a) =1
2πi
∫ c+i∞
c−i∞f∗(s) ζ(s, a) ds
where ζ(s, a) is the generalised zeta function of Riemann, defined whenRe s > 1, by the equation
ζ(s, a) =∞∑
n=0
(s + a)−n , 0 < a � 1 , Re s > 1
294 An Introduction to Integral Transforms
Proof. It follows from inverse Mellin transform that
f(n + a) =1
2πi
∫ c+i∞
c−i∞f∗(s) (n + a)−s ds (8.34)
Summing this result over all n from n = 0 to n → ∞, it gives
∞∑n=0
f(n + a) =1
2πi
∫ c+i∞
c−i∞f∗(s) ζ(s, a) ds (8.35)
Similarly proceeding, we can have
f(nx) = M−1[n−s f∗(s)
]=
12πi
∫ c+i∞
c−i∞x−s n−s f∗(s) ds
Thus,∞∑
n=0
f(nx) =1
2πi
∫ c+i∞
c−i∞x−s f∗(s) ζ(s) ds = M−1 [f∗(s) ζ(s)]
(8.36)
where ζ(s) =∞∑
n=1
n−s
When x = 1 , eqn. (8.36) reduces to∞∑
n=1
f(n) =1
2πi
∫ c+i∞
c−i∞f∗(s) ζ(s) ds (8.37)
Illustrative Examples.
Example 8.7. Show that
(a)∞∑
n=1
(−1)n−1 n−s = (1 − 21−s) ζ(s)
(b)∞∑
n=1
sin an
n=
12
(π − a) .
Solutions.
(a) We know that
∞∑n=1
(−1)n−1 n−s · tn =∞∑
n=1
(−1)n−1 tn · 1Γ(s)
∫ ∞
0xs−1 e−nx dx
=1
Γ(s)
∫ ∞
0xs−1 dx
∞∑n=1
(−1)n−1 tn e−nx
The Mellin Transform 295
=1
Γ(s)
∫ ∞
0xs−1 t e−x
1 + t e−xdx
=1
Γ(s)
∫ ∞
0xs−1 t
ex + tdx
Thus in the limit as t → 1, we finally get
∞∑n=1
(−1)n−1 n−s =1
Γ(s)M
[1
1 + ex; x → s
]
= (1 − 21−s) ζ(s)
(b) Mellin transform of f(x) = sin axx is given by
M
[sin ax
x; x → s
]=∫ ∞
0xs−2 sin ax dx
= −Γ(s − 1)as−1
cosπs
2, using Fourier sine transform
Thus,∞∑
n=1
sin an
n=
12πi
∫ c+i∞
c−i∞
Γ(s − 1)as−1
ζ(s) cosπs
2ds .
After using the functional equation for the zeta function
(2π)s ζ(1 − s) = 2 Γ(s) ζ(s) cos(πs
2
)the above equation gives
∞∑n=1
sin an
n= −a
2· 12πi
∫ c+i∞
c−i∞
(2πa
)s ζ(1 − s)(s − 1)
ds
=12
(π − a) ,
since the integrand has two simple poles at s = 0 and at s = 1inside the large semicircular contour closed on the left side of thes-plane.
8.10 The Generalised Mellin Transform.
In order to extend the applicability of Mellin transform D. Naylor(J. Math. Mech. 12, 265 - 274, 1963) has introduced the generalised
296 An Introduction to Integral Transforms
Mellin transform in a � r < ∞ when the unknown function in theconstitutive equation is prescribed at r = a. It is defined as
M− [f(r) ; r → s] =∫ ∞
a
[rs−1 − a2s
rs+1
]f(r) dr ≡ f∗
−(s) (8.38)
together with its inverse transform
M−1− [f∗(s) ; s → r] = f(r) ≡ 1
2πi
∫ c+i∞
c−i∞r−s f∗
−(s) ds , r � a, (8.39)
where f∗−(s) is analytic in |Re s| < γ with c < γ .
Again, on the other hand if the derivative of the unknown function inthe constitutive equation is prescribed, then the corresponding transfrompair is given by
M+ [f(r) ; r → s] = f∗+(s) ≡
∫ ∞
a
[rs−1 +
a2s
rs+1
]f(r) dr, (8.40)
and M−1+ [f∗
+(s) ; s → r] = f(r) ≡ 12πi
∫ c+i∞
c−i∞r−s f∗
+(s) ds , r � a
(8.41)
where c is defined as in eqn. (8.39)
Under (8.38), the differential operator expression over f(r) becomes
M−[r2 ∂2f
∂r2+ r
∂f
∂r
]= s2 f∗
−(s) + 2sas f(a) , (8.42)
provided f(r) is appropriately behaved at infinity. More precisely
limr→∞
[(rs − a2s r−s) r f∗
−(s) − s (rs + a2s r−s) f(r)]
= 0 .
Also under the eqn. (8.40) the same expression over f(r) becomes
M+
[r2 ∂2f
∂r2+ r
∂f
∂r
]= s2 f∗
+(s) − 2as+1 f ′(a) , (8.43)
where f ′(r) exists at r = a and f(r) is appropriately behaved at infinity.
These types of generalised Mellin transforms have been applied insolving physical problems on truncated conical region by B. Patra [ Bull.De L’Academic Pol. des. Sci., 24, 10, 1976] and many others.
The Mellin Transform 297
8.11 Convolution of generalised Mellin Transform.
Theorem 8.4. If M+[f(r) ; r → s] = f∗+(s)
and M+[g(r) ; r → s] = g∗+(s), then
M+ [ g(r) f(r) ; r → s ] =1
2πi
∫L
f∗+(ξ) g∗+(s − ξ) dξ .
Proof. Assuming that f∗+(s) and g∗+(s) are analytic in |Re s| < γ , we
have
M+ [f(r) g(r) ; r → s] =∫ ∞
a
[rs−1 +
a2s
rs+1
]f(r) g(r) dr
=1
2πi
∫L
f∗+(ξ) dξ
(∫ ∞
ars−ξ−1 g(r) dr
)
+1
2πi
∫ ∞
a
a2s
rs+1g(r) dr
∫L
r−ξ f∗+(ξ) dξ
Replacing ξ by −ξ and using f∗+ (ξ) = a2ξ f∗
+ (−ξ) we obtain∫L
r−ξ f∗+ (ξ) dξ =
∫L
rξ a−2ξ f∗+ (ξ) dξ
The path of integration L : Re ξ = c becomes Re ξ = −c, but thesepaths can be reconciled if f∗
+(ξ) → 0 as [Im ξ] >> 1. Then∫ ∞
a
a2s
rs+1f(r) g(r) dr =
12πi
∫L
f∗+(ξ) dξ
∫ ∞
a
a2s−2ξ
rs−ξ+1g(r) dr
Therefore, using these results we get
M+ [f(r) g(r) ; r → s] =1
2πi
∫L
f∗+ (ξ) dξ
∫ ∞
ars−ξ−1 g(r) dr
+1
2πi
∫L
f∗+(ξ) dξ
∫ ∞
a
a2s−2ξ
rs−ξ+1g(r) dr (8.44)
=1
2πi
∫L
f∗+(ξ) g∗+(s − ξ) dξ .
Thus the proof is complete.
8.12 Finite Mellin Transform.
We conclude this chapter with an account of a finite Mellin transformdue to D. Naylor (1963) as defined below :
298 An Introduction to Integral Transforms
The finite Mellin transform of first kind over the range 0 � r � a isdefined as
M1 [ f(r); r → s ] =∫ a
0
[a2s
rs+1− rs−1
]f(r) dr ≡ f∗
1 (s) (8.45)
and its corresponding inversion formula is defined as
M−11 [ f∗
1 (s) ; s → r ] =1
2πi
∫L
r−sf∗+(s) ds ≡ f(r), (8.46)
where the contour L : (c − i∞ , c + i∞) in the s-plane in some strip|Re s| � γ and 0 < c < γ.
Also for the finite range of r in 0 � r � a , Mellin transform ofsecond kind is defined as
M2 [ f(r) ; r → s ] =∫ a
0
[a2s
rs+1+ rs−1
]f(r) dr ≡ f∗
2 (s) (8.47)
and its corresponding inversion formula is
M−12 [ f∗
2 (s) ; s → r ] =1
2πi
∫L
r−sf∗2 (s) ds ≡ f(r) (8.48)
The transform pairs shown above in eqns. (8.45) - (8.48) are useful insolving boundary value problems in finite conical regions.
Other finite Mellin transform pair, the third kind, suitable foranalysis of boundary value problems in spherical polar co ordinates aresimilarly defined.
For example, in 0 � r � a, the third kind Mellin transform and itsinversion are defined as
M3 [ f(r) ; r → s] =∫ a
0
[a2s+1
rs+1− rs
]f(r) dr ≡ f∗
3 (s) (8.49)
M−13 [ f∗
3 (s) ; s → r] =1
2πi
∫L
r−s−1 f∗3 (s) ds ≡ f(r) (8.50)
The finite Mellin transform of fourth kind and its inversion formula aredefined as
M4[f(r) ; r → s] =∫ a
0
[a2s+2
rs+1+ rs+1
]f(r) dr ≡ f∗
4 (s) (8.51)
M−14 [f∗
4 (s) ; s → r] =1
2πi
∫L
r−s−1 f∗4 (s) ds ≡ f(r) (8.52)
The Mellin Transform 299
Illustrations.
(a) It is easy to verify by application of “integration by parts” that
M1
[r2frr + r fr ; r → s
]=∫ a
0
[a2s
rs+1− rs−1
] [r2 frr + r fr
]dr
= 2s as f(a) − s2 f∗1 (s) (8.53)
Clearly, this transform is useful if f(a) is prescribed as one of theboundary conditions in the corresponding boundary value problems.
(b) Similarly it may be verified that
M2
[r2 frr + r fr ; r → s
]=∫ a
0
[a2s
rs+1+ rs−1
] [r2 frr + r fr
]dr
= 2as−1 f ′(a) + s2 f∗2 (s) . (8.54)
This transform is useful when f ′(a) is prescribed as one of theboundary conditions in the corresponding boundary value problems.
(c) When there is an axisymmetric differential operator
L =[r2 d2
dr2+ 2r
d
dr
]
in the constitutive Laplace equation of the boundary value problemwith 0 � r � a, we can similarly, as in (a) and (b), verify that
M3 [Lf(r) ; r → s] = s(s + 1) f∗3 (s) + 2as+1 f(a) (8.55)
and M4 [Lf(r) ; r → s] = s(s + 1)f∗4 (s) + 2as+3 f ′(a) (8.56)
These formulae in eqns. (8.55) or (8.56) are useful in transformingthe portion Lf(r) of the associated axisymmetric Laplace equationof the physical problem when either f(a) or f ′(a) are prescribedas one of the respective boundary conditions in the correspondingboundary value problems.
300 An Introduction to Integral Transforms
Exercises.
(1) Show that
(a) M [ cos x ; x → s ] = Γ(s) cos πs2 , 0 < Re s < 1
(b) M [ sin x ; x → s ] = Γ(s) sin πs2 , 0 < Re s < 1
(c) M[(1 + x2)−1 ; x → s
]= π
2 cosec(
πs2
), 0 < Re s < 2
(d) M[12 sech x ; x → s
]= Γ(s) L[s] , where L(s) =
∑∞n=1
1(2n−1)s
(2) Show that
M[e−x cos ϕ cos(x sin ϕ) ; x → s
]= Γ(s) cos s ϕ, Re s > 0 , |ϕ| <
π
2M[e−x cos ϕ sin(x sin ϕ) ; x → s
]= Γ(s) sin s ϕ, Re s > −1 , |ϕ| <
π
2
Deduce that if |ϕ| < π2 ,
M−1[ cos s ϕ
sin s π; s → x
]=
1π
· 1 + x cos ϕ
1 + 2x cos ϕ + x2
M−1
[sin s ϕ
sin s π; s → x
]=
1π
· x sin ϕ
1 + 2x cos ϕ + x2
and hence that if |ϕ| < π2n , n > 0
M−1
[cos s ϕ
sin sπn
; s → x
]=
n
π· 1 + xn cos (nϕ)
1 + 2xn cos (nϕ) + x2n
M−1
[sin s ϕ
sin sπn
; s → x
]=
n
π· xn sin (nϕ)
1 + 2xn cos (nϕ) + x2n
(3) Prove that
(a) M
[2√π
∫ x
0
f(t) dt√x2 − t2
; x → s
]=
Γ(−s+1
2
)Γ(1 − s
2
) f∗(s)
(b) M
[2√π
∫ ∞
x
f(t) dt√t2 − x2
; x → s
]=
Γ(
s2
)Γ(
s+12
) f∗(s)
(4) Show that the integral equation
f(x) = h(x) +∫ ∞
0g(xξ) f(ξ) dξ
The Mellin Transform 301
has the formal solution
f(x) =1
2πi
∫ c+i∞
c−i∞
[h∗(s) + g∗(s) h∗(1 − s)
1 − g∗(s) g∗(1 − s)
]x−s ds
(5) Show that∞∑
n=1
cos kn
n2=
k2
4− πk
2+
π2
6
Hence deduce that ∞∑n=1
1n2
=π2
6
(6) If f(x) =∑∞n=1
an e−nx , show that
M [f(x) ; x → s] = f∗(s) = Γ(s) g∗(s)
where g∗(s) =∞∑
n=1
an n−s .
If an = 1, for all n , derive
f∗(s) = Γ(s) ζ(s)
Show also that M
[exp (−ax)1 − e−x
; x → s
]= Γ(s) ζ(s, a) .
(7) Show that
∞∑n=1
cos an
n3=
112[a3 − 3πa2 + 2π2a
]
(8) Prove that
(a) M
[∫ ∞
0ξn f(xξ) g(ξ) dξ ; x → s
]= f∗(s) g∗(1 + n − s)
(b) M
[∫ ∞
0ξn f
(x
ξ
)g(ξ) dξ ; x → s
]= f∗(s) g∗(s + n + 1)
(9) Show that the solution of the boundary value problem defined by
r2 ϕrr + r ϕr + ϕθθ
= 0 , 0 < r < ∞ , 0 < θ < π
ϕ(r, 0) = ϕ(r, π) = f(r)
is ϕ(r, θ) =1
2πi
∫ c+i∞
c−i∞r−s f∗(s) cos
{s(θ − π
2
)}ds
cos(
πs2
)
Chapter 9
Finite Laplace Transforms
9.1 Introduction.
Laplace transform of a piecewise continuous function f(t) for 0 � t < ∞,where |f(t)| < Meσ t,M, σ being real constants defined in section 3.3by
F (s) =∫ ∞
0f(t) e−st dt
exists when Re s > σ. But from the above definition it can not beconcluded that Laplace transform of any general form of f(t) do exist.For example, if f(t) = eat2 , a > 0, its Laplace transform does notexist. Again from the physical point of view there are situations wheredetermining the response of a linear system with disturbance f(t) of anygeneral form in 0 � t � T, T is finite is also a pertinent question fordiscussion.
To study all such situations raised above the power and elegance ofFinite Laplace transform defined below may be found very useful. In thisdirection the papers of H.S. Dunn (Proc. Camb. Phil. Soc. 63, 1967)and L. Debnath and J. Thomas (ZAMM, 56, 1976) may be consulted asbasic and pioneering works.
9.2 Definition of Finite Laplace Transform.
Finite Laplace transform of a continuous function or of a piecewise con-tinuous function f(t) in 0 � t � T satisfying the Lipschitz condition isdenoted by
LT [ f (t) ; t → s ] = f(s, T )
and is defined by
LT [f(t) ; t → s] = f(s, T ) =∫ T
0f(t) e−st dt, (9.1)
Finite Laplace Transforms 303
where s is any real or complex number and T is a finite real number.
The inverse finite Laplace transform is defined by the complexintegral
f(t) = L−1T [ f (s, T ) ; s → t ] =
12πi
∫ c+i∞
c−i∞f (s, T ) est ds, (9.2)
where the contour of integration denoted by Γ is a straight line joiningpoints c − iR and c + iR as R → ∞ and when the function f(t) iscontinuous in 0 � t � T . If f(t) is a piecewise continuous functionwith a finite number of finite discontinuities in 0 � t � T , then thecorresponding inverse finite transform of f (s, T ) at t is given by
12πi
∫Γ
f (s, T ) e−st ds =12[ f (t − 0) + f (t + 0)] (9.3)
as R → ∞. This formula (9.3) is obtained due to the fact that f (s, T )is an entire function of the complex variable s.
Note. 9.1. Let∫
f (t) e−stdt = −F (s, t) e−st.
Then f (s, t) =∫ T
0f(t) e−st dt = F (s, 0) − F (s, T ) e−st, (9.4)
where F (s, 0) = f (s) =∫ ∞
0f (t). e−st dt (9.5)
Therefore, from (9.2) and (9.4) we can write
f(t) =1
2πi
∫Γ
F (s, 0) est ds − 12πi
∫Γ
F (s, T ) es(t−T ) dt, (9.6)
where Γ is a line contour from c − iR to c + iR as R → ∞.
The first integral in (9.6) may be closed on the left half plane. Fort < T, the contour of the second integral is closed in the right half plane.So, selectintg Γ such that all the poles of F (s, 0) lie to the left of Γ andhence the first integral is the solution of the initial value problem, thesecond integral being zero in this case. When t > T , the second integralis closed in the left half of the complex s-plane so that f(t) = 0. Thus,there is no need to consider the second integral and this is identical withusual Laplace transform.
Note. 9.2. Following the definitions of usual Laplace transform andFinite Laplace transform it may be noted that if the usual Laplace trans-form of a function f(t) exists, then its Finite Laplace transform must
304 An Introduction to Integral Transforms
exist. This is so because
f(s) = L [f(t); t → s] =∫ ∞
0f(t)e−stdt =
∫ T
0e−stf(t)dt +
∫ ∞
Te−stf(t)dt
and the existence of the left hand side integral implies the existence ofthe second integral of the right hand side of the above equation. Butthe converse of this statement is not true. Since, for example
L [eat2 ; t → s ] =∫∞0 eat2e−st dt , for a > 0 does not exist though∫ T
0 eat2 e−stdt does exist for a > 0.
9.3 Finite Laplace Transform of elementary functions.
Determine finite Laplace transforms of f(t) in the following examplessupplied in 9.1-9.10
Example 9.1. f(t) = 1
Solution.
LT [1; t → s ] =∫ T
01e−st dt =
1s
[1 − e−sT ] (9.7)
Example 9.2. f(t) = eat
Solution.
LT [eat ; t → s] =∫ T
0e−(s−a)t dt =
1 − e−(s−a)T
s − a(9.8)
Example 9.3. f(t) = sin at
Solution.
LT [sin at ; t → s ] =∫ T
0sin at. e−st dt
=a
s2 + a2− e−sT
s2 + a2(s sin aT + a cos aT ) (9.9)
Example 9.4. f(t) = cos at
Solution.
LT [cos at ; t → s ] =∫ T
0cos at. e−st dt
=s
s2 + a2+
e−sT
s2 + a2[a sin aT − s cos aT ] (9.10)
Finite Laplace Transforms 305
Example 9.5. f(t) = t
Solution.
LT [ t; t → s ] =∫ T
0t e−st dt
=1s2
− e−sT
s[1s
+ T ] (9.11)
Example 9.6. f(t) = tn , n = positive integer.
Solution.
LT [ tn; t → s ] =∫ T
0tn e−st dt
=n!
sn+1− e−sT
s[ T n +
n
sT n−1 +
n(n − 1)s2
T n−2
+ · · · + n! T
sn−1+
n!sn
] , (9.12)
after using integration by parts successively.
Example 9.7. f(t) = ta , a > −1
Solution.
LT [ ta; t → s ] =∫ T
0ta. e−st dt , putting u = st
= s−(a+1)
∫ sT
0ua e−u du
= s−(a+1)γ (a + 1 , sT ) , (9.13)
by use of incomplete gamma function defined by
γ (α, x) =∫ x
0e−u uα−1 du
Example 9.8. f(t) = a periodic function of t of period w.
Solution. We have f(t + nw) = f(t), for all integer value of n.
Now,
LT [f(t) ; t → s] =∫ T
0f(t) e−st dt
=∫ nw
0f(t) e−st dt
306 An Introduction to Integral Transforms
=∫ w
0f(t)e−stdt +
∫ 2w
wf(t) e−stdt + · · · +
∫ nw
(n−1) wf(t) e−st dt
=∫ w
0f(u) e−su du + e−sw
∫ w
0f(u) e−su du + · · ·
+e−s(n−1)w
∫ w
0f(u)e−su du
=1 − e−nsw
1 − e−sw
∫ w
0e−su f(u) du
=1 − e−nsw
1 − e−swf (s,w) (9.14)
When n → ∞, the above result reduces to
f(s) =
∫ w0 f(u)e−sudu
(1 − e−sw),
which is a known result when T → ∞(i, e n → ∞) , for usual LaplaceTransform.
Example 9.9. f(t) = erf(at)
Solution. LT [ erf(at) ; t → s ] =∫ T0 erf (at) e−st dt
Defining the Error function by erf(x) = 2√π
∫ x0 e−α2
dα, we have
∫ T
0erf (at).e−st dt
= −[
e−st
serf (at)
]T
0
+1s
2√π
∫ T
0exp[−{ st + a2t2 }] dt
= −e−st
serf(aT ) +
1s
exp
(s2
4a2
)2√π
∫ aT+ 12a
12a
e−u2du
= −e−sT
serf(aT ) +
1sexp (
s2
4a2)[erf(aT +
s
2a
)− erf
( s
2a
)].
(9.15)
Example 9.10. f(t) = H(t − a)
Solution.
LT [ H(t − a) ; t → s ] =∫ T
0H(t − a) e−st dt
=∫ T
ae−st dt =
1s
(e−as − e−Ta) . (9.16)
Finite Laplace Transforms 307
Example 9.11. Find finite Laplace Transform of
(i) f(t) = cosh at, a > 0 (ii) f(t) = t4 e−at, a > 0
Solutions.
(i) LT [ cosh at ; t → s ] =∫ T
0
12
(eat + e−at) . e−st dt
=12
1 − e−(s−a) T
s − a+
12
1 − e−(s+a)T
s + a
=s
s2 − a2− e−st
[eaT
s − a+
e−aT
s + a
]
(ii) LT [ t4 e−at ; t → s ] =∫ T
0t4 e−at e−st dt
=∫ T
0t4 e−(s+a)t dt
=4!
(s + a)5− e−(s+a)T
s + a
[T 4 +
4 T 3
s + a+
12 T 2
(s + a)2+
24 T
(s + a)3+
24(s + a)4
]
Example 9.12. If f(x) has a discontinuity at t = a where 0 < a < T ,prove that
LT [ f ′(t) ; t → s ] = sf (s, T )+e−sT f(T )−f(0)−e−sa [f(a + 0) − f(a − 0)]
Solution.
LT [ f ′(t) ; t → s]
=∫ a−0
0f ′(t)e−st dt +
∫ T
a+0f ′(t) e−st dt
=[e−stf(t)
]a−0
0+ s
∫ a−0
0f(t)e−stdt +
[e−stf(t)
]Ta+0
+ s
∫ T
a+0e−stf(t)dt
= sf(s, T ) + e−sT f(T ) − f(0) − e−sa [f(a + 0) − f(a − 0)] .
When f(t) has a finite number k of finite discontinuities at t = a1, a2, · · · ak
this result can be generalised after application of the above method ina similar manner.
9.4 Operational Properties.
Theorem 9.1. If LT [ f(t) ; t → s ] = f(s, T ), then
LT { e−atf(t) ; t → s } = f(s + a , T )
308 An Introduction to Integral Transforms
This property of finite Laplace transform is called its shifting property.
Proof. It is given that f(s, T ) =∫ T0 f(t)e−st dt.
Therefore,
LT
[e−at f(t) ; t → s
]=∫ T
0f(t) . e−at.e−st dt
=∫ T
0f(t) . e−(s+a)t. dt
= f (s + a, T ) (9.17)
Theorem 9.2. If LT [ f(t) ; t → s ] = f(s, T ),then LT [ f(at ; t → s) ] = 1
a f ( sa , aT )
This property of finite Laplace transform is called its scaling property.
Proof. Since, LT [ f(t) ; t → s ] = f(s, T ), we have
LT [f(at) ; t → s ] =∫ T
0f(at) . e−st dt
=1a
∫ aT
0f(x) . e−
sax. dx, where at = x
=1af(
s
a, aT ). (9.18)
Theorem 9.3. (Finite Laplace transform of Derivatives)
If LT [f(t) ; t → s ] = f (s, T ) ,
then LT [f ′(t) ; t → s ] = sf (s, T ) − f(0) + e−sT f(T )
LT [f ′′(t) ; t → s ] = s2f (s, T ) − sf(0) − f ′(0) + sf(T )e−sT + f ′(T )e−sT
and generally, LT [ f (n)(t) ; t → s ] = snf(s, T ) −n∑
k=1
sn−k f (k−1)(0)
+e−sTn∑
k=1
sn−k f (k−1) (T ) .
Proof. Applying integration by parts, we have
LT [f ′ (t) ; t → s ] =∫ T
0f ′ (t) e−st dt
= sf (s, T ) − f(0) + e−sT f(T ) (9.19)
Let f ′(t) ≡ ϕ(t) . Then f ′′(t) = ϕ′ (t) and hence
Finite Laplace Transforms 309
LT [f ′′(t) ; t → s] = LT [ϕ′(t) ; t → s]
= s LT [ϕ(t) ; t → s] − ϕ(0) + e−sT ϕ(T )
= s[sf(s, T ) − f(0) + e−sT f(T )] − f ′(0) + e−sT f ′(T )
= s2 f(s, T ) − s f(0) − f ′(0) + e−sT [s f(T ) + f ′(T )] (9.20)
Repeating the above process (n−1) times, the last formula can be provedeasily.
Theorem 9.4. (Finite Laplace transform of Integrals).If LT [f(t) ; t → s] = f(s, T ), then
LT
[∫ t
0f(u) du
]=
1s
[f(s, T ) − e−sT
∫ T
0f(u) du
]
Proof. Let the integral∫ t
0f(u) du = ϕ(t), so that ϕ′(t) = f(t)
Now, LT [ϕ′(t) ; t → s] = s LT [ϕ(t) ; t → s] − ϕ(0) + e−sT ϕ(T )
⇒ LT [f(t) ; t → s] = s LT
[∫ t
0f(u) du ; t → s
]+ e−sT ϕ(T )
⇒ LT
[∫ t
0f(u) du ; t → s
]=
1s
[f(s, T ) − e−sT
∫ T
0f(u) du
](9.21)
Theorem 9.5. (Derivative of finite Laplace transform).
If LT [f(t) ; t → s] = f(s, T ), thend
ds[f(s, T )] = LT [(−t)f(t) ; t → s]
d2
ds2[f(s, T )] = LT [(−t)2f(t) ; t → s]
and generallydn
dsn[f(s, T )] = LT [(−t)n f(t) ; t → s] .
Proof.
Since, f(s, T ) =∫ T
0e−st f(t) dt
we haved
dsf(s, T ) =
∫ T
0
∂
∂se−st · f(t) dt
310 An Introduction to Integral Transforms
=∫ T
0e−st (−t) f(t) dt
= LT [(−t) f(t) ; t → s] (9.22)d2
ds2f(s, t) =
∫ T
0
∂2
∂s2e−st · f(t) dt
=∫ T
0e−st · (−t)2f(t)dt
= LT [(−t)2f(t) ; t → s] . (9.23)
In general, similarly we get
dn
dsnf(s, T ) =
∫ T
0
∂n
∂sne−st · f(t) dt
=∫ T
0e−st (−t)n f(t) dt
= LT [(−t)nf(t) ; t → s] . (9.24)
Theorem 9.6. Finite Laplace transform of the product of tn and m thderivatives of f(t) is
(−1)ndn
dsn
[smf(s, t) −
m∑k=1
sm−kfk−1(0) + e−sTm∑
k=1
sm−kf (k−1)(T )
].
Proof.
We have LT [ tnf ′(t) ; t → s ] = (−1)ndn
dsn[ LT{f ′(t) ; t → s} ]
= (−1)ndn
dsn[s f(s, T ) − f(0) + e−sT f(T )]
= (−1)n[
dn
dsn{s f(s, T )}
]+ (T )n e−sT f(T ) (9.25)
Similarly, LT [tnf (m)(t) ; t → s]
= (−1)ndn
dsn
[smf(s, T ) −
m∑k=1
sm−k f (k−1) (0) + e−sTm∑
k=1
sm−kf (k−1)(T )
]
(9.26)
Theorem 9.7. Integral of finite Laplace transform of a function is givenby ∫ T
sf(s, T )ds =
∫ T
0
f(t)t
e−stdt −∫ T
0
f(t)t
e−tT dt.
Finite Laplace Transforms 311
Proof.
Let f(s, T ) = LT [f(t) ; t → s] =∫ T
0f(t) · e−st dt
Then∫ T
sf(s, T ) ds =
∫ T
sds
∫ T
0f(t) e−st dt
=∫ T
0f(t) dt
∫ T
se−st ds
=∫ T
0
f(t)t
e−st dt −∫ T
0
f(t)t
e−tT dt , (9.27)
provided that both the integrals on the right hand side of eqn. (9.27)exist.
9.5 The Initial Value and the Final Value Theorem .
These theorems give the behavior of the object functions in terms of thebehavior of the transformed functions
Theorem 9.8. (The Initial Value Theorem).
If f(t) be at most a piecewise continuous function, for 0 ≤ t ≤ T andif LT [f(t) ; t → s] = f(s, T ) then,
lims→∞[sf (s, T ) ] = lim
t→0f(t) (9.28)
Proof. The proof of the above theorem can be developed like that inthe theorem of section 3.15.
Theorem 9.9. (The Final value Theorem)
If f(t) be at most a piecewise continuous function for 0 � t � T andif LT [f(t) ; t → s ] = f(s, T ) then,
limt→∞ f(t) = lim
s→0sf (s, T ) (9.29)
Proof. The proof of this theorem can also be developed as that in theTheorem of section 3.16
Theorem 9.10. The solution of an initial value problem is identicalwith that of the final value problem.
Proof. Suppose fin(t) is the solution of the initial value problem, andit is given by
fin(t) =1
2πi
∫Γ
F (s, 0) est ds (9.30)
312 An Introduction to Integral Transforms
where Γ is the Bromwich contour extending from c − iR to c + iR asR → ∞.
Again suppose ffi(t) is the solution of the finial value problem andis defined by
ffi(t) =1
2πi
∫Γ
F (s, T ) es(t−T )ds (9.31)
where Γ lies to the left of the singularities of F (s, t) or of F (s, 0). Thenfrom eqns. (9.27) and (9.28) we get
fin(t) − ffi (t) =1
2πi
∫C
[ F (s, 0) − F (s, T ) e−sT ] est ds
=1
2πi
∫C
f(s, T ) est ds (9.32)
where C is a closed contour which contains all the singularities of F (s, 0)or of F (s, T ) and F (s, t) is defined by
−F (s, t)e−st =∫
f(t) e−st dt
The integrand in eqn. (9.32) is an entire function of s and hence byCauchy’s theorem, the integral must vanish.
Hence.fin (t) = ffi (t) = f(t) (9.33)
9.6 Applications
Example 9.13. Verify Theorems 9.8 and 9.9 when f(t) = exp (−t).
Solution.
Let LT [f(t) ; t → s] = f(s, T ). Then for f(t) = exp (−t),
we have f(s, T ) =1 − e−(1+s)T
(1 + s). Therefore,
lims→∞ f(s, T ) = 0, by L′Hospital′s rule. Again,
lims→∞ s f(s, T ) = lim
s→∞(1 + s − 1)(1 − e−(s+1)T )
1 + s
= lims→∞[1 − e−(s+1)T ] − lim
s→∞1 − e−(s+1)T
1 + s
Finite Laplace Transforms 313
= 1 − 0 = 1
Also, limt→0
f(t) = limt→0
e−t = 1
Thus lims→∞ s f(s, T ) = lim
t→0f(t) is verified.
Also, lims→0
f(s, T ) = lims→0
1 − e−(1+s)T
1 + s= 1 − e−T
and∫ T
0f(t) dt = 1 − e−T
∴ lims→0
f(s, T ) =∫ T
0f(t) dt is verified
Further, lims→0
s f(s, T ) = lims→0
s
[1 − e−(1+s)T
]s + 1
= 0
Thus both the theorems 9.8 and 9.9 are verified.
Example 9.14. Use finite Laplace transform to solve the initial valueproblem
di
dt= A , 0 ≤ t ≤ T ; i(0) = 0 .
Solution. Finite Laplace transform of the ODE under the initial con-dition x(0) = 0 gives
s i (s, T ) + i(T ) e−sT =A
s
[1 − e−sT
]or i (s, T ) +
e−sT
si(T ) = A
[1 − e−sT
s2
]
Here i (s, T ) is not an entire function unless we choose i(T ) = AT .Then we have
i (s, T ) =−AT
s[ e−sT ] +
A
s2(1 − e−sT )
= A
[1s2
− e−sT
s
(1s
+ T
)]
Inverting finite Laplace transform the above equation gives
i(t) = At.
Example. 9.15. The equation for vertical displacement of horizontaltaut string between two fixed points x = 0 and x = L caused by a
314 An Introduction to Integral Transforms
concentrated or distributed load W (x) normalised with the tension ofthe string is
d2y
dx2= W (x) , 0 ≤ x ≤ L
Find y(x).
Solution. The above boundary value problem under the above state-ment is subjected to the boundary conditions
y(0) = y(L) = 0
Let us assume here that the given load is a concentrated load of unitmagnitude at x = a and therefore.
W (x) = δ(x − a) , 0 < a < L
Taking finite Laplace transform over x ∈ (0, L) the above differentialequation under the loading W (x) = δ(x − a) gives
y (s, L) =1s2
[e−sa + y′(0) − e−sL y′(L)
]This function y(s, L) is not an entire function of s unless we choosey′(0) = y′(L) − 1. Using this condition the above equation can beexpressed as
y(s, L) =y′(0)s2
[1 − e−sL − sL e−sL
]+
e−sa
s2
[1 − e−s(L−a)
+ y′(0) sL e−s(L−a)]
To complete the inversion of finite Laplace transform in the above equa-tion we assume Ly′(0) = a − L so that we get
y(x) = xy′(0) + (x − a) H (x − a)
Example 9.16. In an L − R electrical circuit the current I(t) satisfiedthe ODE
LdI
dt+ RI = E0 cos wt , 0 � t � T,
under the emf E0 cos wt and I(t) = 0 at t = 0 ; L,R,E0 being constants.Find I(t) using finite Laplace transform.
Finite Laplace Transforms 315
Solution.
Applying finite Laplace transform the given ODE becomes
sI(s, T ) + e−sT I(T ) +R
LI(s, T ) =
E0
L
[s
s2 + w2
+e−sT
s2 + w2(w sin wT − s cos wT )
]
I(s, T ) = −e−sT I(T )s + R
L
+E0
L(s + RL )
[s
s2 + w2
+e−sT
s2 + w2(w sin wT − s cos wT )
]
Since I(s, T ) is not an entire function of s, we make it entire by choosing
I(T ) =E0
L
w
w2 + R2
L2
[R
wLcos wT + sin wT − R e−
RTL
wL
]
Using this result we get
I(s, T ) =w E0
L(w2 + R2
L2
) [ R
wL
{s
s2 + w2+
e−sT
s2 + w2(w sin wT − s cos wT )
}]
+w E0
L(w2 + R2
L2
) [{ w
s2 + w2− e−sT
s2 + w2(s sin wT + w cos wT )
}]
− w E0
L(w2 + R2
L2
)[
R
wL(s + R
L
) {1 − e−(s+ RL )T}]
Inverting finite Laplace transformed equation, the above relation gives
I(t) =w E0
L(w2 + R2
L2
) { R
wLcos wt + sin wt − R
wLe−
RtL
}
It may be noted that the first two terms on the right hand side of theabove equation represent the steady-state current and the third termthere represents the transient part of the current field I(t) atany time t.
316 An Introduction to Integral Transforms
Exercises.
(1) Find finite Laplace transform of
f(t) = sinh at[Ans. a
s2−a2 + e−sT
2
(eaT
a−s + e−aT
a+s
)]
(2) Use finite Laplace transform to solve the initial value problem
dx
dt+ α x = At , 0 � t � T ; x(0) = a
to prove that
x(t) =(
a +A
α2
)e−αt +
At
α− A
α2
(3) Solve the simple harmonic oscillator governed by
d2x
dt2+ w2x = F , x(0) = a , x(0) = u
where F, a and u are constants, to prove that
x(t) =(
a − F
w2
)cos wt +
u
w2sin wt +
F
w2.
Chapter 10
Legendre Transforms
10.1 Introduction.
As an aid to solve some special type of boundary value problems thischapter deals with the Legendre transform whose kernel is a Legendrepolynomial. Initially, definition and basic operational properties of thetransform are discussed based on the papers of Churchill (J.Math. andPhysics, 33. 165, 1954) and Churchill and Dolph (Proc. Amer. Math,Sec, 5, 93, 1954).
10.2 Definition of Legendre Transform.
Legendre transform of a function f(x) over the interval −1 � x � 1 isdefined by
fl (n) ≡ l [ f(x) ; x → n ] =∫ 1
−1f(x) Pn(x) dx , (10.1)
where Pn(x) is the Legendre polynomial of integral degree n(� 0) pro-vided that the integral in (10.1) exists.
The orthogonal property of Legendre polynomial is well-known as∫ 1
−1Pm(x) Pn(x) dx =
22n + 1
δmn (10.2)
and this can also be expressed as
l [ Pm(x) ; x → n ] =2
2n + 1δmn (10.3)
From eqns. (10.2) and (10.3) Fourier-Legendre expansion theorem ofa piecewise continuous function f(x) for −1 � x � 1 is given by
318 An Introduction to Integral Transforms
∞∑n=0
(n +
12
)Pn(x)fl(n) = f(x) ,
if f(x) ∈ C (−1, 1) =12
[f(x + 0) + f(x − 0)] , if f(x) ∈ P 1(−1, 1)
(10.4)
We, therefore, have the inversion formula for Legendre transform is
l−1[
fl (n) ; n → x]
=∞∑
n=0
(n +
12
)fl(n) Pn(x) (10.5)
10.3 Elementary properties of Legendre Transforms.
(1) Since P0(x) = 1 and P1(x) = x we get
fl(0) =∫ 1
−1f(x) dx (10.6)
and fl(1) =∫ 1
−1xf(x) dx (10.7)
Again, l[f ′(x) ; x → 0] =∫ 1
−1f ′(x)dx = f(1) − f(−1)
(10.8)
l[f ′(x) ; x → 1] =∫ 1
−1xf ′(x) dx = f(1) + f(−1)
−∫ 1
−1f(x) dx (10.9)
Also, l[f ′′(x) ; x → 1] = f ′(1) + f ′(−1) −∫ 1
−1f ′(x)dx
= f ′(1) + f ′(−1) − f(1) + f(−1) (10.10)
and l[C ; x → 0] =∫ 1
−1C dx = 2C (10.11)
l[C ; x → n] =∫ 1
−1C Pn(x)dx
= C
∫ 1
−1Pn(x)dx = C
∫ 1
−1P0(x) Pn(x) dx
⇒ l[C ; x → n] = 0 , when n = 1, 2, 3, · · ·(by putting m = 0 in eqn. (10.3)). (10.12)
Legendre Transforms 319
(2) Again since, Pn(−x) = (−1)n Pn(x) we have
l[f(−x) ; x → n] =∫ 1
−1f(−x) Pn(x) dx
=∫ 1
−1f(x) Pn(−x) dx
= (−1)n∫ 1
−1f(x) Pn(x) dx
= (−1)n l[f(x) ; x → n]
From this result we can write
l−1[(−1)n fl(n) ; n → x
]= f(−x) (10.13)
(3) Legendre transform of eiαx is given by
ln [eiαx ; x → n] =∫ 1
−1eiαx Pn(x) dx =
√2πα
in Jn+ 12
(α),
(10.14)from table of integrals (Copson, 1935). Similarly,
ln [eαx ; x → n] =
√2πα
In+ 12
(α) (10.15)
(4) Also
l[(1 − x2)−
12 ; x → n
]=∫ 1
−1
Pn(x)dx√1 − x2
= π P 2n(0) (10.16)
and l
[1
2(t − x); x → n
]=
12
∫ 1
−1
Pn(x) dx
t − x= Qn(t) , |t| > 1
(10.17)
These results are also obtained from “Tables of integral Trans-forms” by Harry Batemann (1954).
(5) From the generating relation of Legendre polynomial we have
(1 − 2rx + r2)−12 =
∞∑n=0
rn Pn(x) , |r| < 1
320 An Introduction to Integral Transforms
Multiplying both sides by Pn(x) and integrating the result withrespect to x from −1 to 1, we get∫ 1
−1(1 − 2rx + r2)−
12 Pn(x) dx = rn
∫ 1
−1P 2
n(x) dx
=2rn
(2n + 1)(10.18)
When r = 1, we get∫ 1
−1[2(1 − x)]−
12 Pn(x) dx =
22n + 1
⇒∫ 1
−1(1 − x)−
12 Pn(x) dx =
2√
22n + 1
Or l [(1 − x)−12 ; x → n ] =
2√
22n + 1
(10.19)
Defferentiating eqn. (10.18) with respect to r and after multiplyingthe result by r, we get
12
∫ 1
−1(1 − 2rx + r2)−
32 (2rx − 2r2) Pn(x) dx =
2nrn
2n + 1
⇒ 12
∫ 1
−1(1 − 2rx + r2)−
32[−{1 − 2rx + r2 + r2 − 1}]Pn(x)dx
=2nrn
2n + 1
⇒ −12
∫ 1
−1(1 − 2rx + r2)−
12 Pn (x)
+(1 − r2)
2
∫ 1
−1(1 − 2rx + r2)−
32 Pn (x) dx
=2nrn
2n + 1
⇒ −l[
(1 − 2rx + r2)−12 ; x → n
]+(1 − r2) l
[(1 − 2rx + r2)−
32 ; x → n
]=
4nrn
2n + 1
Now using eqn. (10.18), the above result reduces to
l[(1 − 2rx + r2)−32 ; x → n]
=[
4nrn
2n + 1+
2rn
2n + 1
]1
1 − r2
Legendre Transforms 321
=2rn
1 − r2(10.20)
(6) Legendre transform of H(x) is given by
l [ H(x) ; x → n ] =∫ 1
−1H (x) Pn(x) dx
=∫ 1
0H(x) Pn(x) dx
=∫ 1
0Pn(x) dx
If n = 0, Pn(x) = P0(x) = 1 and hence l[ H(x) ; x → n ] = 1.
If n > 1, we know from the recurrence relation of Legendre poly-nomial that
(2n + 1) Pn(x) = P′
n+1 (x) − P′
n−1 (x) (10.21)
Therefore, in this case∫ 1
0Pn (x) dx
=1
2n + 1
∫ 1
0[ P
′n+1(x) − P
′n−1 (x) ] dx
=1
2n + 1[Pn−1 (0) − Pn+1 (0)]
Thus, l[ H(x) ; x → n ] =
{1 , for n = 0
12n+1 [Pn−1 (0) − Pn+1 (0)] , for n � 1
(10.22)
(7) From eqn. (10.21) we have∫ 1
−1g(x) Pn(x) dx =
12n + 1
[ ∫ 1
−1g(x)P
′n+1 (x) dx
−∫ 1
−1g(x) P
′n−1 (x) dx
]
=1
2n + 1
[g(x) Pn+1 (x)|1−1 −
∫ 1
−1g′(x) Pn+1 (x) dx
−g(x) P′
n−1 (x)|1−1 +∫ 1
−1g′(x) Pn−1(x) dx
]
322 An Introduction to Integral Transforms
=1
2n + 1[ g(x) { Pn+1(x) − Pn−1 (x)}1
−1
−∫ 1
−1g′(x) {Pn+1(x) − Pn−1 (x)} dx ]
Thus if g(x) =∫ x−1 f(t)dt is a continuous function in (−1, 1)
l [ g(x) ; x → n ] =1
2n + 1[
l{ g′(x) ; x → (n − 1) } ]− 1
2n + 1[l{ g′ (x) ; x → n + 1 } ] (10.23)
provided the expression [· · ·]1−1 vanishes. Thus, in particular, wehave the special case
l
[ ∫ x
−1f(t) dt ; x → n
]=
12n + 1
[fl(n − 1) − fl(n + 1)
](10.24)
Therefore, by repeated application of (10.8) and (10.9) and by(10.23) we get
l[
f ′ (x) ; x → 2m]
= f(1) − f(−1) −m−1∑r=0
(4m − 4r − 1) fl (2m − 2r − 1)
(10.25)
l[
f ′ (x) ; x → 2m + 1]
= f(1) − f(−1) −m−1∑r=0
(4m − 4r + 1) fl (2m − 2r)
(10.26)
Also from the recurrence relation
(n + 1) Pn+1 (x) − (2n + 1) x Pn (x) + n Pn−1 (x) = 0
we can easily deduce that
l [ xf (x) ; x → n ] =1
2n + 1[
(n + 1) fl (n + 1) + n fl (n)]
(10.27)
Legendre Transforms 323
10.4 Operational Properties of Legendre Transforms
Theorem 10.1 If f ′(x) is continuous and f ′′(x) is bounded and inte-grable in each subinterval of −1 � x � 1 and if Legendre transform off(x) exists and
lim|x|→1
(1 − x2)[
f(x) , f ′(x)]
= 0
then l
[d
dx{ (1 − x2)
d
dxf(x) } ; x → n
]= −(n + 1) n fl (n).
Proof. We have
l
[d
dx
{(1 − x2)
d
dxf(x)
}; x → n
]
=∫ 1
−1
d
dx
{(1 − x2)
d
dxf(x)
}Pn (x) dx.
= −∫ 1
−1(1 − x2)
d
dxf(x) P ′
n(x) dx ,
on integration by parts and using the given condition.
Again on integration by part the above result becomes
= −[(1 − x2) P ′
n (x) f(x)|1−1 −∫ 1
−1
d
dx
{(1 − x2) P ′
n (x)}
f(x) dx
]
=∫ 1
−1
d
dx
{(1 − x2)
d
dxPn (x)
}f(x) dx , by using given condition.
But it is known that
d
dx
[(1 − x2)
d
dxPn (x)
]= −n(n + 1) Pn(x)
and therefore,
l
[d
dx
{(1 − x2)
df (x)dx
}; x → n
]= −n (n + 1)
∫ 1
−1f(x) Pn(x) dx
= −n(n + 1) fl(n) (10.28)
Note 10.1. If we denote the operator R by
Rf(x) ≡ d
dx
[(1 − x2)
d
dxf (x)
]
324 An Introduction to Integral Transforms
we can extend the result of theorem 10.1 as
l[
R2 f (x) ; x → n]
= (−1)2 n2 (n + 1)2fl (n)
l[
R3 f (x) ; x → n]
= (−1)3 n3 (n + 1)3fl (n)
· · · · · ·l[
Rk f(x) ; x → n]
= (−1)k nk (n + 1)k fl (n) (10.29)
Corollary 10.1. It is seen that n(n + 1) =(n + 1
2
)2 − 14 . Therefore
we have
l [R f(x) ; x → n] = −n(n + 1) fl(n)
= −[(
n +12
)2
− 14
]fl(n)
= −(
n +12
)2
fl(n) +14
fl(n)
Thus, l
[14f(x) − R(f(x)) ; x → n
]=(
n +12
)2
fl(n) (10.30)
Extending this result, we get
(−1)k l [Rk(f(x))−4k ; x → n] =k−1∑r=0
(−1)r kcr
[4−r
(n +
12
)2k−2r]
fl(n)
Theorem 10.2. (Convolution Theorem).
If l [f(x) ; x → n] = fl(n) and l[g(x) ; x → n] = gl(n)
then l−1 [fl(n)gl(n) ; n → x] = f(x) ∗ g(x) (10.31)
where f(x)∗g(x) is called the convolution of f(x) and g(x) and is definedby
f(x) ∗ g(x) =1π
∫ π
0f(cos λ) sin λ dλ
∫ π
0g (cos (η)) dβ (10.32)
with x = cos μ and cos η = cos λ cos μ + sinλ sin μ cos β
Proof. We have
fl(n) gl(n) =∫ π
0f(cos λ)Pn(cos λ)(sin λ)dλ
∫ π
0g(cos η)Pn(cos η) sin η dη
=∫ π
0f(cos λ) sin λ
[∫ π
0g(cos η)Pn(cos η)Pn(cos λ) sin η
]dλ
where f(x) = f(cos λ) and g(x) = g (cos η)
Legendre Transforms 325
Further, since
Pn(cos η) Pn(cos λ) =1π
∫ π
0Pn(cos μ) dα
where cos μ = cos η cos λ + sin η sin λ cos α , we can rewrite
fl(n)gl(n) =1π
∫ π
0f(cos λ) sin λ
[∫ π
0
∫ π
0g(cos λ)Pn(cos λ) sin η dα dη
]dλ
Now the double integral on the right hand of above equation is given by∫ π
0
∫ π
0g(cos λ cos μ + sinλ sin μ cos β) Pn(cos μ) sin μ dμ
Using this result and changing the order of integration, we get
fl(n)gl(n) =1π
∫ π
0Pn(cos μ) sin μ
[∫ π
0
∫ π
0f(cos λ) sin λ g(cos η)dλ dβ
]dμ
=∫ π
0h(cos μ) Pn(cos μ) sinμ dμ,
where cos η = cos λ cos μ + sin λ sin μ cos β
and h(cos μ) =1π
∫ π
0f(cos λ) sinλ dλ
∫ π
0g(cos η) dβ
Thus the theorem is proved.
In particular, if μ = 0 we get from (10.32) that
h(1) =∫ 1
−1f(t) g(t) dt (10.33)
and when μ = π, h(−1) =∫ 1
−1f(t) g(−t) dt (10.34)
10.5 Application to Boundary Value Problems.
As an illustration we consider application of finite Legendre transform inthe solution of interior Dirichlet problem for the potential u(r, θ) insidea unit sphere (r = 1) satisfying the partial differential equation
∂
∂r
[r2 ∂u
∂r
]+
∂
∂x
[(1 − x2)
∂u
∂x
]= 0 , 0 < r � 1 (10.35)
with boundary condition
u(1, x) = f(x) , −1 < x < 1 (10.36)
326 An Introduction to Integral Transforms
where x = cos θ .
Applying finite Legendre transform over the variable x, the PDE(10.35) and the boundary condition (10.36) reduce to
r2 d2ul(r, n)dr2
+ 2rdul(r, n)
dr− n(n + 1) ul(r, n) = 0 (10.37)
and ul(1, n) = fl(n) (10.38)
respectively, where ul(r, n) is a continuous function of r in 0 < r � 1
The bounded solution of (10.37) and (10.38) is
ul(r, n) = fl(n) rn , 0 < r � 1 , where n is a non − negative integer.
On inversion the solution of the problem is given by
u(r, x) =∞∑
n=0
(n +
12
)fl(n) rn Pn(x) , 0 < r � 1 , |x| < 1 (10.39)
Another representation of this solution can also be obtained throughapplication of convolution theorem and also the formula in equation(10.20) as
u(r, cos θ) = l−1 [ fl(n) rn ; n → x = cos θ ]
=12π
∫ π
0f(cos λ) sin λ dλ
∫ π
0
(1 − r2)dη
(1 − 2r cos μ + r2)32
(10.40)
where cos μ = cos λ cos θ + sinλ sin θ cos η
Integral in (10.40) is called the Poisson integral formula of the potentialinside the unit sphere.
The solution of exterior Dirichlet problem under boundary conditionw(1, x) = f(x) , x = cos θ of the unit sphere r = 1 is then given by
wl (r, n) =1r
fl(n) r−n , n = 0, 1, 2, · · ·On inversion it gives
w (r, cos θ) =1r
w
(1r
, cos θ
), r > 1
=12π
∫ π
0f(cos λ) sin λ dλ
∫ π
0
(r2 − 1) dη
(1 − 2r cos μ + r2)32
(10.41)
Legendre Transforms 327
where cos μ = cos λ cos θ + sin λ sin θ cos η and w(r , cos θ) is thesolution of the exterior problem.
Exercises.
(1) (a) Show that
l [ xn ; x → n ] =2n+1 (n!)2
(2n + 1)!
(b) Using the result in eqn. (10.27) find l [ x2f(x) ; x → n ]
(2) Using the definition of odd Legendre transform as
f2n+1 ≡ l0 [ f(x) ; x → n ] =∫ 1
0f(x) P2n+1 (x) dx
and the even Legendre transform as
f2n ≡ le [ f(x) ; x → n ] =∫ 1
0f(x) P2n (x) dx
and their respective inversion formulae
l0 [ f2n+1 ; n → x ] ≡ f(x) =∞∑
n=0
(4n + 3) f2n+1 P2n+1(x), 0 � x � 1,
l−1e [ f2n ; n → x ] ≡ f(x) =
∞∑n=0
(4n + 1) f2n P2n(x) , 0 � x � 1
and denoting the operator L = ∂∂x
[(1 − x2) ∂
∂x
]prove that
l0 [ L f(x) ; x → n ] = (2n + 1) P2n(0) f(0)
−(2n + 1)(2n + 2) f2n+1
and le [ L f(x) ; x → n ] = −P2n(0) f ′(0) − 2n(2n + 1) f2n
(3) Prove that the Legendre transformed solution of the Dirichletboundary value problem for u(r, θ) defined by
∂2u
∂r2+
1r
∂u
∂r+ (1 − x2)
∂2u
∂x2− 2x
∂u
∂x= 0 , 0 � r � a , 0 � θ � π
u (a, θ) = f(x) , 0 � θ � π
where x = cos θ, is given by
ul (r, n) =(r
a
)nfl(n) .
Chapter 11
The Kontorovich-Lebedev Transform
11.1 Introduction.
In discussing the variable separable solution of Boundary value Problemsin cylindrical co-ordinates for wage type region Kontorovich-Lebedevtransform has been found application. To understand the applicabilityof the transform some elementary properties of the modified Besselfunction of the second kind, also called Macdonald function, is needed.We present below only operational proof of the results at various stagesof required theorems for easy understanding of the readers in engineeringand physics.
11.2 Definition of Kontorovich - Lebedev Transform.
Kontorovich-Lebedev transform of a function f(x) over the positive realline is denoted by f(τ) and is defined by
K[f(x) ; x → τ ] ≡ f(τ) =∫ ∞
0x−1 f(x)Kiτ (x) dx, (x � 0) (11.1)
The corresponding inversion theorem of Kontorovich-Lebedev transformis given by
K−1[f (τ) ; τ → x
] ≡ f(x) =2π2
∫ ∞
0Kiτ (x) τ sinh (πτ)f (τ) dτ,
(11.2)if f(x) is a function defined on the positive real line such that 1
xf(x) iscontinuously differentiable and xf(x) and x d
dx [ 1x f(x)] are absolutely
integrable over the positive real line.
If f(x) is a piecewise continuous function over (0 � x < ∞] havinga finite discontinuity at x, then (11.2) becomes
K−1 [f (τ) ; τ → x] =2π2
∫ ∞
0Kiτ (x)τ sinh (πτ)f (τ) dτ
=12
[f(x + 0) + f(x − 0)] (11.3)
The Kontorovich-Lebedev Transform 329
We present below an operational proof of the above inversion theorem,since the rigorous proof is a bit complicated.
We know that the cosine transform of Kiτ (x) is given by
Fc[ Kiτ (x) ; τ → t ] =√
π
2. e−x cosh t (11.4)
Therefore, we have
Fc [K {f(x) ; x → τ} ; τ → t]
=√
π
2
∫ ∞
0e−xch t f(x)
xdx (11.5)
This result can also be expressed as Laplace transform of 1x f(x) as
L
[f (x)
x; x → p
]=
2π
∫ ∞
0f (τ) cos (τ cosh−1 p ) dτ (11.6)
But since, L [ f(x) ; x → p ] =−∂
∂pL
[f(x)
x; x → p
]
we have from eqn. (11.6) that
L[f(x) ; x → p ] =2π
∫ ∞
0τ f(τ)
sin (τ cosh−1 p)√p2 − 1
dτ (11.7)
Taking Laplace inversion of both sides of the above equation we obtain
K−1[f(τ) ; τ → x ] ≡ f(x) =2π2
∫ ∞
0τ sinh (πτ) Kiτ (x) f (τ) dτ
(11.8)as the inversion formula for Kontorovich-Lebedev transform.
11.3 Parseval Relation for Kontorovich-Lebedev Trans-
forms.
Theorem 11.1. If f(τ) and g(τ) are respectively Kontorovich-Labedevtransforms of f(x) and g(x) respectively, x � 0 then
2π2
∫ ∞
0τ sinh (πτ) f(τ)g(τ) dτ =
∫ ∞
0x−1f(x) g(x) dx (11.9)
330 An Introduction to Integral Transforms
Proof. We have
2π2
∫ ∞
0τ sinh (πτ) f (τ)g (τ) dτ
=2π2
∫ ∞
0τ sinh (πτ) f (τ) dτ
∫ ∞
0x−1 g(x) Kiτ (x) dx
=∫ ∞
0x−1 g(x) dx.
2π2
∫ ∞
0τ sinh (πτ) f (τ) Kiτ (x) dτ
=∫ ∞
0x−1g(x) f(x) dx , by (11.8)
This proves the Parseval relation of Kontorovich-Lebedev transform.
11.4 Illustrative Examples.
Example 11.1. Prove that
K[te−t cos x ; t → τ
]=
π sinh(xτ)sinh(πτ) sin x
and hence deduce the values of K[t ; t → τ ] and K[te−t ; t → τ ].
Solution. By definition in (11.1) we have
K[
t e−t cos x ; t → τ]
=∫ ∞
0
1t
. t e−t cos x Kiτ (t) dt
=∫ ∞
0Kiτ (t) e−pt dt , where p = cos x
= L [ Kiτ (t) ; t → p ] =π sinh (xτ)
sinh (πτ) sin x,
using table of integral transforms.
Thus, in particular, if x = π2 and x = 0 we get respectively
K[t ; t → τ ] = π2 sech πτ
2 and K[t e−t ; t → τ ] = πτ cosech (πτ)
Example 11.2. Prove that
K[x sin α e−x cos α ; x → τ
]=
πsh(ατ)sh(πτ)
and K[e−xcht ; t → τ
]=
π cos (τt)τ sh (πτ)
Solution. Proceeding as the solution of Example 11.1, the above resultscan also be proved easily.
The Kontorovich-Lebedev Transform 331
Example 11.3. Using the results of Example 11.2 deduce that
(a)∫ ∞
0e−x f(x) dx =
2π
∫ ∞
0τ2 f(τ) dτ
(b)∫ ∞
0e−x(1+cht) dx = 2
∫ ∞
0
cos (τt) dτ
sh (πτ)
and hence deduce that
Fc [ τ cosech (πτ) ; τ → t ] =1
2√
2πsech2
(t
2
)
Solutions.
(a) Taking g(x) = x sin α e−x cos α , we have from the table ofintegral transform of Copson that
g(τ) = K[ g(x) ; x → τ ] =π sh (ατ)sh (πτ)
Therefore, Parseval relation for Kontorovich-Lebedev transforms∫ ∞
0x−1 f(x) g(x) dx =
2π2
∫ ∞
0τ sh(πτ) f(τ) g(τ) dτ ,
for this g(x), gives∫ ∞
0x−1f(x) · x sin α e−x cos α dx =
2π2
∫ ∞
0τ sinh(πτ) f(τ) · πsh(ατ)
sh(πτ)dτ
⇒∫ ∞
0e−x cos α f(x) dx =
2π
∫ ∞
0τsh(ατ)sinα
f(τ) dτ
Now letting α → 0 , we get∫ ∞
0e−x f(x) dx =
2π
∫ ∞
0τ2f(τ) dτ
(b) Now taking f(x) = e−xcht we get
f(τ) =π
τ
cos(τt)sh (πτ)
Then using these relations in the final result of (a), we get∫ ∞
0e−x(1+cht)dx =
2π
∫ ∞
0τ2 π
τ
cos(τt)sh(πτ)
dτ
332 An Introduction to Integral Transforms
⇒ 12
sech2 t
2= 2∫ ∞
0
τ cos (τt)sh (πτ)
dτ
=√
2π Fc [ τ cosech (πτ) ; τ → t ]
⇒ Fc [ τ cosech (πτ) ; τ → t ] =1
2√
2πsech2
(t
2
).
11.5 Boundary Value Problem in a wedge of finite thickness.
Consider the boundary value problem of determining a harmonic func-tion u(ρ, ϕ, z) in a finite wedge defined by
ρ � 0 , 0 � ϕ � α , 0 � z � a (11.10)
when the function u satisfies the boundary conditions
u (ρ , α , z) = f (ρ , z) ,
u (ρ , ϕ , 0) = u (ρ , ϕ , a) = 0and u (ρ , 0 , z) = 0 .
⎫⎪⎬⎪⎭ (11.11)
Since u (ρ , ϕ , z) is a harmonic function, it satisfies the PDE
∂2u
∂ρ2+
1ρ
∂u
∂ρ+
1ρ2
∂2u
∂ϕ2+
∂2u
∂z2= 0 (11.12)
We seek a variable separable non-zero solution of (11.12) in the form
u (ρ , ϕ , z) = R(ρ) Φ(ϕ) Z(z) �= 0
so that after substituting the above value of u, differential equation(11.12) becomes
1R
d2R
dρ2+
1ρR
dR
dρ+
1Φ ρ2
d2Φdϕ2
= − 1Z
d2Z
dz2= σ2, say (11.13)
Therefore, Z(z) = A(σ) cos σz + B(σ) sin σz (11.14)
andρ2
R
d2R
dρ2+
ρ
R
dR
dρ− σ2ρ2 =
−1Φ
d2Φdφ2
= − τ2, say (11.15)
Therefore, Φ (φ) = C(τ) ch (τφ) + D(τ) sh(τφ) (11.16)
Also from (11.15) we have
ρ2 d2R
dρ2+ ρ
dR
dρ− (σ2ρ2 − τ2)R = 0 (11.17)
The Kontorovich-Lebedev Transform 333
Putting τ = −iλ ⇒ λ = iτ eqn. (11.17) becomes
ρ2 d2R
dρ2+ ρ
dR
dρ− (λ2 + σ2ρ2)R = 0
One solution of this equation is Kλ (ρσ) = Kiτ (ρσ)
Hence,
u(ρ, φ, z) = L (τ) Kiτ (ρσ) [A(σ) cos σz + B(σ) sin σz]
[C (τ) ch τφ + D (τ) sh τϕ ].
From (11.11) since u (ρ, φ, 0) = u (φ, φ, a) = 0, we have
A(σ) = 0 and sin σa = 0 = sinnπ, n = 1, 2, · · ·
Also, since u(ρ, 0, z) = 0 we have C (τ) = 0. Therefore, by method oflinear superposition the solution under the last three boundary condi-tions in (11.11) of the problem can be expressed as
u(ρ, φ, z) =∞∑
n=1
L(τ) B(n) D(τ) sinnπz
ash τφ.Kiτ
(nπρ
a
)
for any variable value of the positive constant τ . After little adjustmentof the constants L(τ)B(n)D(τ), on linear superposition of solutions overτ the above solution u(ρ, φ, z) can also be expressed as
u(ρ, φ, z) =2π2
∞∑n=1
sinnπz
a
∫ ∞
0τFn(τ)
sh (τφ)sh (τ a)
Kiτ (nπρ
a) sh πτ dτ
(11.18)
This final form of u (ρ, φ, z) in (11.18) will provide a solution of theboundary value problem satisfying all the boundary conditions in (11.11),provided that we can find Fn(τ) such that
f(ρ, z) =∞∑
n=1
sinnπz
a
[K−1
{Fn(τ) ; τ → nπρ
a
}]
Inverting as the finite Fourier transform we get
K−1[
Fn (τ) ; τ → nπρ
a
]=
2a
∫ a
0f (ρ, y) sin
nπy
ady
334 An Introduction to Integral Transforms
Now replacing ρ by arnπ , we see that the above equation is equivalent to
K−1 [ Fn (τ) ; τ → r ] =2a
∫ a
0f( ar
nπ, y)
sinnπy
ady
and hence
Fn(τ) =2a
∫ a
0sin
nπy
aK[
f( ar
nπ, y)
; r → τ]
dy
=2a
∫ a
0sin
nπy
ady
∫ ∞
0r−1 f
( ar
nπ, y)
Kiτ (r) dr
Changing the variable of integration in last integration we can expressit as
Fn(τ) =2a
∫ a
0sin
nπy
ady
{∫ ∞
0x−1f(x, y) Kiτ
(nπx
a
)dx
}(11.19)
Substituting this value of Fn(τ) in eqn. (11.18), the solution of theboundary value is obtained.
Exercises.
(1) Prove that
K [ 1 ; τ → x ] =π
21
x sinh(
π2 x)
(2) Prove that
(a)∫ ∞
0
{τ2 Kit (x) dt
}=
π
2x e−x
(b)∫ ∞
0Kit (x) dt =
π
2e−x
(3) Prove that
K[
e−x cos t ; x → τ]
=π ch (tτ)τ sh (πτ)
Hence deduce that
(a) K [ e−x ; x → τ ] =π
τ
1sh(πτ)
(b) K [ ex ; x → τ ] =π
τcoth (πτ)
(c) K [ e−xcht ; x → τ ] =π cos (tτ)τ sh (πτ)
Chapter 12
The Mehler-Fock Transform
12.1 Introduction.
In solving some of the boundary value problems associated with themathematical theory of elasticity application of the Mehler-Fock trans-form has been found. Specially in problems concerning analysis of stressfield in the vicinity of external crack in elastic region these transformshave found significant applications . In this chapter our main objectiveis to develop some basic concepts of this transform in an elegant mannertogether with some useful basic properties of the associated Legendrefunctions of first kind.
12.2 Fock’s Theorem (with weaker restriction).
Theorem 12.1 If a function g(x) defined on 1 � x < ∞ is such thatthe integral
∫ ∞
1
|g(x)| dx√(x + 1)
exists, then at every point x in whose neighbourhood g(x) has a boundedvariation∫ ∞
0P− 1
2+iτ (x)g∗0(τ) dτ =
12
[ g(x + 0) + g(x − 0) ] (12.1)
where g∗0(τ) = τ tanh (πτ)∫ ∞
1P− 1
2+iτ (x) g(x) dx (12.2)
However, if g(x) is continuous at x ∈ [1,∞) we have (12.1) is equivalentto ∫ ∞
0P− 1
2+iτ (x)g∗0(τ) dτ = g(x) (12.3)
336 An Introduction to Integral Transforms
Proof. To supply a formal proof of this form of Fock’s theorem wereplace g (cosh α) = f(α) and proceed as below.
Let us introduce a pair of operators Φ0 and Φ−10 such that
Φ0 [ f(α) ; α → τ ] = f∗0 (τ)
=∫ ∞
0f(α)P− 1
2+iτ (cosh α) sinhα dα (12.4)
and Φ−10 [ f∗
0 (τ) ; τ → α ] = f(α)
=∫ ∞
0τ tanh (π τ) P− 1
2+iτ (cosh α) f∗
0 (τ) dτ (12.5)
We begin by finding the forms of Φ−10 and Φ0. Determining these
operators is equivalent to finding a solution
f∗0 (τ) = Φ0 [ f(α) ; α → τ ]
of the integral equation (12.4). Now introducing the formula
P− 12+iτ (cosh α) =
√2 cth (πτ)
π
∫ ∞
α
sin (τt)√cht − chα
dt (12.6)
in (12.5) we find
f(α) =√
2π
∫ ∞
0τf∗
0 (τ) dτ
∫ ∞
α
sin (τt)√(cht − chα)
dt
=1√π
∫ ∞
α
dt√ch t − ch α
√2π
∫ ∞
0τ f∗
0 (τ) sin (τ t) dτ
=1√π
∫ ∞
αFs [ τf∗
0 (τ) ; τ → t ]dt√
(ch t − ch α)(12.7)
where Fs [· · · ] is the Fourier sine transform of τf∗0 (τ). Treating (12.7)
as an Abel integral equation, its solution is given by
Fs[ τf∗0 (τ) ; τ → t ] = − 1√
π
d
dt
∫ ∞
t
f(α) dα√ch α − cht
(12.8)
Again inverting (12.8) by means of Fourier sine inversion theorem weget
τ f∗0 (τ) = − 1√
πFs
[d
dt
∫ ∞
t
f(α) sh α dα√ch α − ch t
; t → τ
]
The Mehler-Fock Transform 337
=τ√π
Fc
[ ∫ ∞
t
f (α) sh α dα√ch α − ch t
; t → τ
]Therefore,
f∗0 (τ) =
√2
π
∫ ∞
0cos(τt) dτ
∫ ∞
t
f(α)sh α dα√(ch α − ch t)
=∫ ∞
0f(α) sh α dα
√2
π
∫ α
0
cos(τ t)√(ch α − ch t)
dt
⇒ f∗0 (τ) = Φ0[f(α);α → τ ] =
∫ ∞
0f(α)P− 1
2+iτ(chα)shα dα
(12.9)
since, P− 12 +iτ
(chα) =√
2π
∫ α
0
cos(τ t)√(ch α − ch t)
dt
Thus (12.4) is obtained after starting from (12.5).
If we replace f(α) by g(chα) and if
g(x) =∫ ∞
0P− 1
2 +iτ(x) g0 (τ) dτ , x ∈ [1,∞] (12.10)
then eqn. (12.9) implies
g∗0(τ) = τ th(πτ)
∫ ∞
1P− 1
2+iτ(x) g(x) dx (12.11)
This result proves the theorem for the case when g(x) is a continuousfunction.
There are other two forms of the Fock’s theorem which we need notpersue here.
12.3 Mehler-Fock Transform of zero order and its prop-
erties.
The Mehler-Fock integral transform of order zero of a function f(α),α ∈ [0,∞) is denoted by f∗
0(τ) and is defined by
f∗0(τ) = Φ0 [f(α);α → τ ] =
∫ ∞
0f(α) P− 1
2 +iτ(chα)shα dα (12.12)
The inverse of this transform is given by
f(α) = Φ−1
0[f∗
0(τ) ; τ → α] =
∫ ∞
0τ f∗
0(τ) th(πτ) P− 1
2+iτ (chα)dτ
(12.13)
338 An Introduction to Integral Transforms
Properties.
Considering the special case ν = −12 + iτ , θ = iα and m = 0 in the
F.G. Mehler’s formula [Math.Ann. 18,161,1881] we find that
P− 12+iτ (chα) =
√2
π
∫ α
0
cos(τt) dt√(ch α − ch t)
(12.14)
=1√π
√2π
∫ ∞
0
cos(τt)H(α − t)√(ch α − ch t)
dt (12.15)
and in the Lagrange formula [Mem. des. Sci. Math. Fasic 47, 1939]that
P− 12+iτ (chα) =
√2
πcth (πτ)
∫ ∞
α
sin(τt) dt√(ch t − ch α)
(12.16)
=1√π
√2π
cth (πτ)∫ ∞
0
sin(τt)H(t − α)√(ch t − ch α)
dt
(12.17)
Treating eqn. (12.14) as an Abel integral equation, we get after itsinversion that
d
dt
[sin τt
τ
]= cos(τt) =
1√2
d
dt
∫ t
0P− 1
2+iτ (chα)
sh α · d α√(ch t − ch α)
⇒ sin(τt)τ
=1√2
∫ t
0P− 1
2+iτ (chα)
sinhα dα√(ch t − ch α)
(12.18)
Also treating (12.16) as another Abel integral equation we get after itsinversion that√
2π
sin(τt) cth(πτ) = − 1π
d
dt
∫ ∞
t
P− 12+iτ (chα)shα dα√{(ch α − ch t)}
⇒ − d
dt
[cos τt
τ
]= sin τt = − 1√
2d
dt
∫ ∞
tP− 1
2+iτ (chα) · th(πτ)
shα dα√(ch α − ch t)
⇒ cos τt
τ=
1√2
th πτ
∫ ∞
t
P− 12+iτ (chα) · shα dα√
(ch α − ch t)(12.19)
Thus from (12.18) we can have
1√2
∫ ∞
0H(t − α) P− 1
2+iτ (chα)
shα dα√(ch t − ch α)
=sin(τt)
τ
The Mehler-Fock Transform 339
The last equation can be expressed as
Φ0
[H(t − α)√
(ch t − ch α); α → τ
]=
√2 sin τt
τ(12.20)
Again (12.19) gives∫ ∞
0H(α − t)P− 1
2+iτ (chα)
shα dα√(ch α − ch t)
=√
2 cos(τt)τ
cth (πτ)
This equation can also be expressed as
Φ0
[H(α − t)√
(ch α − ch t); α → τ
]=
√2
τcth (πτ) cos(τt) (12.21)
12.4 Parseval type relation.
From Fourier transform the corresponding parseval type relation of Mehler-Fock transform of order zero can be deduced easily [N.N. Lebedev, Dokl.Akad. Nauk. SSSR, 68,445, 1949] if we define alternatively that
φ0 [f(x) ; x → τ ] = f0(τ) =∫ ∞
1f(x)P− 1
2+iτ (x) dx (12.22)
and f(x) =∫ ∞
0τ th (πτ) f0(τ) dτ ≡ φ−1
0[f0(τ) ; τ → x]
(12.23)
For convenience if we define
f0(τ) = φ0 [f(x) ; x → τ ]
g0(τ) = φ0[g(x) ; x → τ ]
where f0(τ) =∫ ∞
1f(x) P− 1
2+iτ (x) dx, (12.24)
and g0(τ) =∫ ∞
1g(x) P− 1
2+iτ (x) dx (12.25)
then∫ ∞
0τ th(πτ) f0(τ) g0(τ) dτ
=∫ ∞
0τ th (πτ) g0 (τ) dτ
∫ ∞
1f(x)P− 1
2+iτ (x) dx
=∫ ∞
1f(x)dx
∫ ∞
0τ th (πτ) g0(τ)P− 1
2+iτ (x) dτ
=∫ ∞
1f(x) g(x) dx, (12.26)
340 An Introduction to Integral Transforms
by using inversion theorem in the inner integral.
This is the required Parseval type relation.
For example, if
f(x) =1
x + s, g(x) =
1x + t
we have f0(τ) =∫ ∞
1
1x + s
P− 12+iτ (x) dx
= πsech (πτ) P− 12+iτ (s) , |s| < 1
and g0(τ) = πsech (πτ) P− 12+iτ (t) , |t < 1
If s �= t ,
∫ ∞
1
dx
(x + s)(x + t)=
1t − s
log1 + t
1 + s
while∫ ∞
1
dx
(x + s)2=
1s + 1
Thus from (12.26) we have
∫ ∞
0
τ sh πτ
ch3 πτP− 1
2+iτ (s)P− 1
2+iτ (t) dτ =
{1
π2(t−s) log 1+t1+s , t �= s
1π2(1+s)
, t = s
where |s| < 1 , |t| < 1. (12.27)
As another example, let
f(x) = e−ax , g(x) = e−bx
Then f0(τ) =
√2πa
Kiτ (a) and g0(τ) =
√2πb
Kiτ (b)
Also,∫ ∞
1e−(a+b)x dx =
e−(a+b)
(a + b).
Therefore, by (12.26) we get
∫ ∞
0τ th (πτ) Kiτ (a)Kiτ (b) dτ =
π√
ab
2(a + b)e−(a+b) (12.28)
As a third example, let
f(x) = x− 32 , g(x) = x− 5
2
and hence f0(τ) =√
2 sech(πτ
2
)and
The Mehler-Fock Transform 341
g0(τ) =2√
23
τ cosechπτ
2
Again, since∫ ∞
1f(x) g(x) dx =
∫ ∞
1x−4 dx =
13
,
we have from (12.26) that∫ ∞
0τ2 sech π τ d τ =
18
(12.29)
12.5 Mehler-Fock Transform of order m
The Mehler-Fock transform of order m is denoted and defined by
f∗m(τ) = Φm [f(α) ; α → τ ] =
∫ ∞
0f(α)P
m
− 12+iτ (chα) dα (12.30)
The corresponding inversion formula for the transform is
f(α) = (−1)m∫ ∞
0τ th(πτ)P
m
− 12+iτ (chα)f∗
m(τ) dτ
≡ Φ−1
m [f∗m
(τ) ; τ → α] (12.31)
where Pmν (x) = (x2 − 1)
m2
dm
dxmPν(x) , m > 0
Let chα = x. Then eqn. (12.31) can be expressed as
f(ch−1x) = (−1)m(x2 − 1)m2
dm
dxm
∫ ∞
0τ th (πτ)P− 1
2+iτ (x) f∗
m (τ) dτ
(12.32)Also, if we define the Mehler-Fock transform of order m by the equation
Φm [f(α) ; α → τ ] = f∗m (τ) =
∫ ∞
0f(α) P
−m
− 12+iτ (chα) shα dα (12.33)
then the corresponding inversion formula will be
Φ−1
m[f∗
m(τ) ; τ → α] = f(α) = (−1)m∫ ∞
0τ th (πτ) f∗
m(τ)Pm
− 12+iτ (chα) dτ,
(12.34)since the associated Legendre equation does not change after replacingm by −m and therefore the corresponding Legendre functions.
342 An Introduction to Integral Transforms
Also, for some class of functions f for convenience, if we definealtarnatively
φm [f(x) ; x → τ ] ≡ fm(τ) =∫ ∞
1f(x) P
m
− 12+iτ (x) dx (12.35)
the corresponding inversion formula will be
φ−1
m[fm(τ) ; τ → x] ≡ f(x) = (−1)m
∫ ∞
0τ th (πτ) f∗
m(τ) P
m
− 12+iτ (x) dx
(12.36)
12.6 Application to Boundary Value Problems.
12.6.1 First Example
At first we consider to determine a harmonic function, say, ψ in cylin-drical co-ordinates (ρ, φ, z) under certain given boundary conditions like
ψ = f(ρ) , 0 � ρ � a (12.37)
and∂ψ
∂z= 0 , ρ > a on the boundary plane z = 0 (12.38)
If we introduce a toroidal co-ordinate system in the same region as(α, β, φ) the equations of transformations are
ρ =a shα
chα + cos β, z =
a sin β
chα + cos β, φ = φ (12.39)
implying ρ + iz = a th12(α + iβ) ≡ f(α + iβ), say , (12.40)
where f is a holomorphic function of (α + iβ). It may be noticed from(12.39) that
ρ2 + (z + a cot β)2 = a2 cosec2 β (12.41)
and (ρ − a th α)2 + z2 = a2 cosech2 α (12.42)
representing surfaces β = β1, say and α = α1, say, as toruses respectivelyabout z-axis .
In (12.41), if 0 < β1 < π , β = β1 is the part of the sphere withcentre at (0, 0,−α cot β1) and radius a cosec β1 lying above the xy-planei.e, z > 0. Again if −π < β1 < 0, the surface is that part of the above
The Mehler-Fock Transform 343
sphere which lies below the xy-plane i.e, where z < 0. In cylindricalsystem the Laplace equation in ψ
∂2ψ
∂ρ2+
1ρ
∂ψ
∂ρ+
1ρ2
∂2ψ
∂φ2+
∂2ψ
∂z2= 0 (12.43)
then takes the form
∂
∂α
[ρ∂ψ
∂α
]+
1∂β
[ρ∂ψ
∂β
]+
|f ′|2ρ
∂2ψ
∂φ2= 0 (12.44)
Again from (12.40) we have
|f ′|2 = a2(chα + cos β)−2
and so the above PDE in (12.44) can be expressed as
∂
∂α
[shα
chα + cos β
∂ψ
∂α
]+
∂
∂β
[shα
chα + cos β
∂ψ
∂β
]
+1
shα(chα + cos β)∂2ψ
∂φ2= 0 (12.45)
Let us seek a solution of (12.45) in toroidal co-ordinates in the form
ψ(α, β, φ) =√
(chα + cos β) · υ(α, β, φ) (12.46)
Then (12.45) leads to a PDE in υ(α, β, φ) as
shα∂2υ
∂α2+ chα
∂υ
∂α+ shα
∂2υ
∂β2+
14
υ shα +1
shα
∂2υ
∂φ2= 0 (12.47)
Eqn. (12.47) has the separable solutions of the form
υ(α, β, φ) = A(α) · e±τβ · e±imφ ,
if A(α) satisfy the equation
(1 − μ2)d2A
dμ2− 2μ
dA
dμ−[
m2
1 − μ2+(
14
+ τ2
)]A = 0 (12.48)
where μ = chα .
Now representing ν = −12 + iτ , we have ν(ν + 1) = − (1
4 + τ2)
andhence eqn, (12.48) can be expressed as
(1 − μ2)d2A
dμ2− 2μ
dA
dμ+[ν(ν + 1) − m2
1 − μ2
]A = 0 (12.49)
344 An Introduction to Integral Transforms
which is an associated Legendre differential equation. This equation(12.49) has a solution of the form
A(μ) = c P−mν (μ) = c P
−m
− 12+iτ (chα)
implying ν(α, β, φ) = c P−m
− 12+iτ (chα) e±τβ · e±imφ
Therefore, ψ(α, β, φ) =√
(chα + cos β) · P−m
− 12+iτ (chα) e±τβ · e±imφ
(12.50)
are the forms of the general solution of eqn. (12.44)
If we are interested to find the separable solutions of the Laplaceequation in the axisymmetric case that is in the φ-independent case, wehave
ψ(α, β) =√
(chα + cos β) · P− 12+iτ (chα) e±τβ (12.51)
Thus by the principle of superposition the general solution of the Laplaceequation in axisymmetric case is expressed as
ψ(α, β) =√
(chα + cos β) · Φ−1
0[A(τ) ch β τ + B(τ) sh β τ ; τ → α]
(12.52)
where Φ−1
0[χ(τ) ; τ → α] =
∫ ∞
0τ th (πτ) P− 1
2+iτ (chα) χ (τ) dτ
(12.53)
For the general case this solution similarly can be expressed as
ψ(α, β, φ) =√
(chα + cos β)∞∑
m=−∞eimφ Φ
−1
m[Am(τ)ch(βτ)
+Bm(τ) sh(βτ) ; τ → α] , (12.54)
where Φ−1
m[χm(τ); τ → α] = (−1)m
∫ ∞
0τ th(πτ) P
−m
− 12+iτ (chα)χm(τ)dτ
(12.55)
12.6.2 Second Example
Secondly for example, let us consider the particular problem of deter-mination of the solutions ψ of the Laplace equation in a half-spacez � 0 under the prescribed boundary conditions on ψ(ρ, z, φ) on a disk0 < ρ < a , z = 0 and ∂ψ
∂z = 0 in the region of the boundary outside
The Mehler-Fock Transform 345
that disk in cylindrical co-ordinate system. In toroidal co-ordinate sys-tem (α, β, φ), as discussed above in 12.6.1, the given boundary conditionsare transformed to
ψ (α, 0, φ) = f(α, φ) , 0 < φ < 2π , α > 0 (12.56)
and∂ψ
∂β(α, π, φ) = 0 . (12.57)
Following the result in eqn. (12.50), we take a separable solution of theLaplace equation as
ψ(α, β, φ) =√
(chα + cos β)·∞∑
m=−∞Φ
−1
m
[Am(τ)
ch(πτ − βτ)ch(πτ)
; τ → α
]eimφ
(12.58)Here we may note that this form of solution identically satisfied thecondition ∂ψ
∂β (α, π, φ) = 0 . Now, to satisfy the boundary condition ineqn. (12.56) we have to find the functions Am(τ) for m ranging fromm → −∞ to m → +∞ through integral values such that
√2 sh
(α
2
)m=+∞∑m=−∞
Φ−1
m[Am(τ) ; τ → α]eimφ = f(α, φ) (12.59)
implying that
Φ−1
m[Am(τ) ; τ → α] =
2− 3
2
πcosech
(α
2
)∫ 2π
0f(α, φ)e−imφ dφ
Therefore, Am(τ) = 2− 3
2 π−1
∫ 2π
0e−imφΦm
[cosech
(α
2
)f(α, φ);α → τ
]dφ
(12.60)
12.6.3 Third Example
Thirdly another important problem of interest is to find the solutionψ(ρ, z) of Laplace equation in a half space z � 0 for the axisymmetriccase under the boundary conditions
ψ(ρ, 0) = F (ρ) for 0 � ρ � a (12.61)
and[∂ψ
∂z
]= 0, for ρ > a (12.62)
346 An Introduction to Integral Transforms
Then in the toroidal co-ordinate system the φ independent solution isobtained from eqn. (12.58) with a little variation after putting φ = 0and m = 0 in the form
ψ(α, β) =√
(chα + cos β) Φ−1
0
[A0(τ)
ch(πτ − βτ)ch(πτ)
; τ → α
](12.63)
for identically satisfying the condition in eqn. (12.62). Also the condi-tion in eqn. (12.61) will be satisfied if we have found out A0(τ) suchthat
A0(τ) = Φ0
[1√2
sech(α
2
)F(a th
α
2
); α → τ
](12.64)
= Mehler − Fock transform of1√2
sech(α
2
)F(a tan
α
2
)
Thus eqns. (12.63) together with (12.64) will constitute the requiredsolution of the given boundary value problem in the axi-symmetric case.
This boundary value problem corresponds to the axisymmetric con-tact problem for a half-space in the theory of elasticity.
As a particular case of the above problem, let us suppose that F (α) = k,
a constant
implying1√2
sech(α
2
)F(a th
α
2
)=
k√2
sech(α
2
),
Now, A0(τ) = Φ0
[k√2
sechα
2; α → τ
]=
k√2· 2 τ−1 cosech (πτ)
=√
2 k τ−1 cosech (πτ)
∴ ψ(α, β) = k√
[2(chα + cos β)] Φ−1
0
[τ−1 2 ch(πτ − βτ)
sh(2πτ); τ → α
]
=2kπ
tan−1
[chα + cos β
1 − cos β
] 12
, (12.65)
from table of integrals.
A quantity of physical interest is the number
P = −2π∫ a
0ρ
(∂ψ
∂z
)z=0
dρ
to be evaluated now.
The Mehler-Fock Transform 347
We have, a∂ψ
∂z= −sh α sinβ
∂ψ
∂α+ (1 + chα cos β)
∂ψ
∂β
∴[∂ψ
∂z
]z=0
=1 + chα
a
(∂ψ
∂β
)β=0
Now,∂ψ
∂β=
√2 k
πcos(
β
2
)(chα + cos β)−
12
So,[∂ψ
∂z
]z=0
= − 2kπa
ch(α
2
)
∴ P = 2 k a
∫ ∞
0th
α
2sech
α
2dα , since ρ = a th
α
2= 4 a k (12.66)
12.6.4 Fourth Example
Finally we consider the problem of determining the axisymmetric stressesin an elastic body weakened by a penny-shaped crack . This problem isreduced to determining a harmonic function ψ in the half-space z � 0satisfying the conditions[
∂ψ
∂z
]z=0
= p (ρ) , 0 � ρ � a (12.67)
[Ψ]z=0 = 0 , ρ > a (12.68)
As before, introducing toroidal co-ordinates (α, β, φ) under transforma-tion
ρ =a sh α
chα + cos β, z = a
sin β
chα + cos β, φ = φ
the boundary conditions in eqns. (12.67) and (12.68) reduce to
∂ψ (α, 0)∂β
= −2a sech2(α
2
)p(a th
α
2
)(12.69)
and ψ(α, π) = 0 respectively. (12.70)
So, we choose ψ(α, β) satisfying (12.70) identically as
ψ(α, β) =√
(chα + cos β) Φ−1
0
[τ−1 f∗(τ)
sh(πτ − βτ)ch(πτ)
; τ → α
]
To satisfy (12.69) we take
f∗(τ) = Φ0 [f(α) ; α → τ ]
where f(α) =√
2 a p(a tanh
α
2
)sech3 α
2(12.71)
348 An Introduction to Integral Transforms
For a particular case, if p(ρ) = k, a constant, then
f(α) =√
2 k a sech3(α
2
)and hence f∗(τ) = 8
√2 k a τ cosech (πτ) , from table of integrals.
Therefore, ψ(α, β) = 16
√(ch2
α
2− sin2 β
2
)k a
Φ−1
0
[2 sh(πτ − βτ)
sh(2πτ); τ → α
]
12.7 Application of Mehler-Fock Transform for solving
dual integral equation.
Consider the dual integral equations for determination of f(τ) as∫ ∞
0f(τ) cos x τ dτ = g1(x) , 0 � x � a (12.72)∫ ∞
0f(τ) cth (πτ) sin x τ dτ = h2(x) , x > a , (12.73)
where g1(x) and h2(x) are two prescribed functions in their given ranges.
Multiplying both sides of the eqn. (12.72) by√
2π√
(chα − chx)and
integrating the result with respect to x from 0 to α and finally usingthe result in eqn. (12.14) we get∫ α
0
√2
π√
(chα − ch x)
[∫ ∞
0f(τ) cos xτ dτ
]dx =
∫ α
0
√2
π
g1(x)dx√(chα − ch x)
Thus∫ ∞
0f(τ) P− 1
2+iτ (chα) dτ = Ω1(α), say
=√
2π
∫ α
0
g1(x)dx√(chα − ch x)
, 0 < α < a
(12.74)
Similarly, multiplying eqn. (12.73) by√
2π (chx− chα)−
12 and integrating
the result with respect to x from α to ∞ and finally using the result ineqn (12.16) we get∫ ∞
0f(τ) P− 1
2+iτ (chα) dτ =
√2
π
∫ ∞
α
h2(x)dx√(chx − chα)
≡ Ω2(α), say, α > a
(12.75)
The Mehler-Fock Transform 349
Let us now define
Ω (α) =
{Ω1 (α) , 0 � α � a
Ω2 (α) , α > a
so that the eqns in (12.74) and (12.75) are equivalent to∫ ∞
0f(τ) P− 1
2+iτ (chα) dτ = Ω (x)
⇒ Φ−1
0
[τ−1 cth (πτ) f(τ) ; τ → α
]= Ω (α)
After Mehler-Fock inversion the above equation yields
f(τ) = τ th (πτ) Φ0 [ Ω(α) ; α → τ ]
= τ th (πτ)∫ a
0Ω1(α) P− 1
2+iτ (chα) sh α dα
+ τ th (πτ)∫ ∞
aΩ2(x) P− 1
2+iτ (chα) sh α dα (12.76)
Thus the dual integral equation is solved in closed form.
Exercises
(1) Considering the alternative notation of Mehler-Fock transform as
f0(τ) = ϕ0 [f(x) ; x → τ ] =∫ ∞
1f(x) P− 1
2+iτ (x) dx
and the inverse
f(x) = ϕ−1
0[f0(τ) ; τ → x] =
∫ ∞
0τ th (πτ)f0(τ) P− 1
2+iτ (x) dτ
prove that
(a) ϕ0 [e−ax ; x → τ ] =
√2πa
Kiτ (a) , |arg a| <π
2
(b) ϕ0 [x− 32 ; x → τ ] =
√2 sech
(πτ
2
)(c) ϕ0 [(chα + cos β)−
32 ; α → τ ] = 2
√2 cosec β
sh(βτ)sh(πτ)
,−π < β < π
(d) ϕ0
[sech3
(α
2
); α → τ
]= 8 τ cosech (πτ)
350 An Introduction to Integral Transforms
(2) Following the dual integral equations discussed in section 12.7prove that∫ ∞
0f(τ) cos(x τ) dτ =
1√2
d
dx
∫ a
0
Ω1(α)sh α dα√(ch x − chα)
+1√2
d
dx
∫ x
a
Ω2(α) sh α dα√(ch x − chα)
, for x > a
Ω1(α) , Ω2(α) are functions defined in the section.
(3) Reduce the following dual integral equations∫ ∞
0τ−1 f(τ) sin xτ dτ = j1(x) , 0 � x � a∫ ∞
0cth (πτ) f(τ) sin xτ dτ = h2(x) , x > a
as a pair of dual integral equations discussed in section 12.7.
(4) Prove that
(a) Φ0
[sech
(α
2
); α → τ
]= 2τ−1 cosech (πτ)
(b) Φ0 [√
(sech (α)) ; α → τ ] =1√2 τ
cosech(πτ
2
)
Chapter 13
Jacobi, Gegenbauer, Laguerre and Hermite
Transforms
13.1 Introduction.
In this section we present below a brief account of some important in-tegral transforms like Jacobi, Gegenbauer, Laguerre and Hermite trans-forms. These transforms are applicable to a limited class of problems ofphysical science and therefore these are sparingly discussed in literature.We only persue here their definitions, basic operational properties anda few important applications.
13.2 Definition of Jacobi Transform.
In view of the orthogonal relation of Jacobi polynomials of ordersα(> −1) and β(> −1) and degrees m and n given by
∫ 1
−1(1 − x)α (1 + x)β P (α,β)
n (x) P (α,β)m (x) dx = δn δmn (13.1)
where δnm is the Kronecker delta symbol and
δn =2α+β+1 Γ(n + α + 1) Γ(n + β + 1)
n! (α + β + 2n + 1) Γ(n + α + β + 1)(13.2)
there exist under certain restrictions a series expansion of f(x) of theform
f(x) =∞∑
n=1
an P (α,β)n (x) ,−1 < x < 1 (13.3)
with an =1δn
∫ 1
−1f(x) P (α,β)
n (x) dx =1δn
f (α,β)(n) (13.4)
352 An Introduction to Integral Transforms
Accordingly, the Jacobi transform of degree n of a function f(x) in−1 < x < 1 is defined by Debnath (Bull Cal Math Soe., 55, 1963) as
J [f(x);x → n] ≡ f (α,β)(n) =∫ 1
−1(1 − x)α (1 + x)β P (α,β)
n (x) f(x) dx
(13.5)and the inverse Jocobi transform is given by
J−1[f (α,β) (n); n → x
]≡ f(x) =
∞∑n=1
(δn)−1 f (α,β) (n) P (α,β)n (x)
(13.6)
Example 13.1. Prove that the Jacobi Transform of degree n of
(a) f(x), a polynomial of degree m < n is zero.
(b) P(α,β)m (x) is δmn
Solution.
(a) Since f(x) is a polynomial of degree m < n, it admits of a seriesexpansion in Jacobi polynomials as
f(x) =m<n∑r=1
ar P (α,β)r (x) , −1 < x < 1
and therefore,
J [f(x) ; x → n] =∫ 1
−1(1 − x)α(1 + x)βP (α,β)
n (x)
{m<n∑r=1
ar P (α,β)r (x)
}dx
=m∑
r=1
ar
∫ 1
−1(1 − x)α(a + x)βP (α,β)
n (x) P (α,β)r (x) dx
= δn δnr = 0 , since n �= r(= 1, 2, · · ·m < n)
with δn is as defined in (13.2).
(b) J[P (α,β)
m (x) ; x → n]
=∫ 1
−1(1 − x)α (1 + x)β P (α,β)
n (x) P (α,β)m (x) dx
= δnm.
[Tables of Integral Transforms, Erdelyi et al,Mc Graw Hill, NY]
Jacobi, Gegenbauer, Laguerre and Hermite Transforms 353
Theorem 13.1
Defining a differential operator R over a function f(x) by
R [f(x)] = (1 − x)−α (1 + x)−β d
dx
[(1 − x)α+1 (1 + x)β+1 d
dxf(x)
]
and assuming
lim|x|→1
[(1 − x)α+1 (1 + x)β+1 f(x)] = 0 = lim|x|→1
[(1 − x)α+1 (1 + x)β+1f ′(x)]
(13.7)if J [f(x) ; x → n] = f (α,β)(n) , then
J {R[f(x)] ; x → n} = −n(n + α + β + 1) f (α,β)(n), n = 0, 1, 2, · · ·(13.8)
Proof. By definition
J{R[f(x)]} =∫ 1
−1(1 − x)α(1 + x)β(1 − x)−α(1 + x)−β
d
dx
[(1 − x)1+α(1 + x)1+β df (x)
dx
]P (α,β)
n (x) dx
=∫ 1
−1
d
dx
[(1 − x)1+α (1 + x)1+β df (x)
dx
]P (α,β)
n (x) dx
=[P (α,β)
n (x)(1 − x)1+α(1 + x)1+βf ′(x)]1−1
−∫ 1
−1(1 − x)1+α(1 + x)1+βf ′(x)
d
dxP (α,β)
n (x) dx
= −∫ 1
−1(1 − x)1+α(1 + x)1+βf ′(x)
d
dxP (α,β)
n (x)dx, after integrating by parts
Again integrating by parts and using the orthogonal relation (13.1), theabove relation further reduces to
J{R[f(x)]} = −n(n + α + β + 1)∫ 1
−1(1 − x)α(1 + x)βP (α,β)
n (x)f(x)dx
= −n(n + α + β + 1) f (α,β) (n)
This completes the proof.
Corollary 13.1 If f(x) and R[f(x)] satisfy the conditions in (13.7) thenJ{R{R[f(x)]}} ≡ J{R2[f(x)]} exists and is given by
J{R2 [f(x)]} = (−1)2 n2(n + α + β + 1)2 f (α,β) (n)
354 An Introduction to Integral Transforms
More generally, if f(x) and Rj[f(x)] satisfy the conditions in(13.7) forj = 1, 2, · · · then J{Rk[f(x)]} = (−1)k nk(n + α + β + 1)k f (α,β) (n).
Example 13.2
Find the Jacobi Transform of the following functions :
(a) f(x) = xn
(b) f(x) = (1 + x)p−β
(c) f(x) = (1 − x)q−α
Solution.
(a) Since P (α,β)n (x) =
(1 + α + β)2n xn
2n n!(1 + α + β)n+ a polynomial of degree(n − 1),
therefore
xn =2nn!(1 + α + β)n
(1 + α + β)2nP (α,β)
n (x) + a polynomial of(n − 1)thdegree
So,
J [xn; x → n] =∫ 1
−1(1 − x)α (1 + x)βPα,β
n (x) xn dx
=∫ 1
−1(1 − x)α (1 + x)β .
[Pα,β
n (x)]2
.2nn!(1 + α + β)n
(1 + α + β)2ndx
+∫ 1
−1(1 − x)α (1 + x)βPα,β
n .(x). [a polynomial of (n − 1)th degree] dx
= 2n+α+β+1 Γ(n + α + 1) Γ(n + β + 1)Γ(n + α + β + 1)
, as the second integral vanishes
Here, (α)n is the factorial function defined by
(α)n = α(α + 1)(α + 2) · · · (α + n − 1), n � 1
=Γ(α + n)
Γ(α).
(b) J[(1 + x)p−β ; x → n
]=∫ 1
−1(1 − x)α(1 + x)p P (α,β)
n (x) dx
=
(n + α
n
)2n+p+1 Γ(p + 1) Γ(α + 1) Γ(p − β + 1)
Γ(α + p + n + 2) Γ(p − β + n + 1),
Jacobi, Gegenbauer, Laguerre and Hermite Transforms 355
follows from the Tables of Integrals [ Erdelyi et al, McGraw Hill, NY.]
In particular, if α = β = 0, then
J[(1 + x)p−β ; x → n
]=∫ 1
−1P (0,0)
n (x)(1 + x)p dx
=∫ 1
−1(1 + x)p Pn(x)dx
=2p+1{Γ(1 + p)}2
Γ(p + n + 2) Γ(p + n + 1).
This result again follows from the Tables of Integrals [Erdelyi et al,Mc Graw Hill, NY]
(c) J[(1 − x)q−α; x → n
]=∫ 1
−1(1 − x)q (1 + x)βPα,β
n (x)dx
=2 q+β+1
n!Γ(α − q)Γ(q + 1) Γ(n + β + 1)Γ(n + α − q)
Γ(n + β + q + 2)
This result also follows from the tables of integrals [ Erdelyi et al,Mc Graw Hill, NY.]
13.3 The Gegenbauer Transform.
The Gegenbauer Transform and its inversion is denoted and defined by
G [f(x);x → n] ≡ f (ν)(n) =∫ 1
−1(1 − x2)(ν−
12)Cν
n(x)f(x) dx
and G−1[f (ν)(n);n → x
]= f(x) =
∞∑n=0
δ−1n Cν
n (x)f (ν)(n)
with δn =21−2ν πΓ(n + 2ν)n!(ν + n) [Γ(ν)]2
(13.9)
When α = β = ν − 12 , the Jacobi Polynomial P
(α,β)n (x) becomes the
Gegenbauer polynomial Cνn(x). The corresponding orthogonal relation
of Gegenbauer polynomial is∫ 1
−1(1 − x2)ν−
12 Cν
m(x)Cνn(x)dx = δn δnm (13.10)
356 An Introduction to Integral Transforms
where δn =21−2νπΓ(n + 2ν)n!(n + ν)[Γ(ν)]2
Also the differential form R defined by
R[f(x)] =(1 − x2
) d2f(x)dx2
− (2ν + 1) xdf (x)dx
can be expressed as
R[f(x)] = (1 − x2)12−ν d
dx
[(1 − x2
)ν+ 12
df
dx
](13.11)
Then under the Gegenbauer transform it reduces to
G[R{f(x)};x → n] = −n(n + 2ν) f (ν)(n) (13.12)
Similarly,
G[Rk{f(x)}; x → n] = (−1)k nk(n + 2ν)k f (ν)(n),
where k = 1, 2, 3, · · · (13.13)
13.4 Convolution Theorem
If G[ f(x) ; x → n] = f (ν)(n) and G [ g(x) ; x → n] = g(ν)(n), then
f ν(n)g(ν)(n) = G [ h(x) ; x → n ] = h(ν)(n),
where h(x) is given by
h(cos ψ) = A(sin ψ)1−2ν
∫ π
0
∫ π
0f(cos θ) g (cos φ) (sin θ)2ν
(sin φ)2ν−1 (sin λ)2ν−1 dθ dα
and α is defined by
cos φ = cos θ cos ψ + sin θ sin ψ cos α
Proof. From definition
f (ν)(n) g(ν)(n)
=∫ 1
−1f(x) (1 − x2)ν−
12 Cν
n (x) dx
∫ 1
−1g(x) (1 − x2)ν−
12 Cν
n (x) dx
=∫ π
0f(cos θ)(sin θ)2ν Cν
n(cos θ)dθ.
∫ π
0g(cos φ)(sin φ)2ν Cν
n(cos φ) dφ
=∫ π
0f(cos θ)(sin θ)2ν
[∫ π
0g(cos φ) Cν
n(cos θ) Cνn(cos φ)(sin φ)2νdφ
]dθ
(13.14)
Jacobi, Gegenbauer, Laguerre and Hermite Transforms 357
By the additional formula of Gegenbauer polynomial, we have
Cνn(cos θ) Cν
n(cos φ) = A
∫ π
0Cν
n(cos ψ)(sin λ)2ν−1 dλ, (13.15)
where A ={Γ(n + 2ν)/ n! 22ν−1 Γ2(ν)
}(13.16)
and cos ψ = cos θ cos φ + sin θ sin φ cos λ (13.17)
Accordingly we get
f (ν)(n) g(ν)(n) =∫ π
0f(cos θ) (sin θ)2ν
[ ∫ π
0
∫ π
0g (cos φ) Cν
n (cos ψ)
(sin φ)2ν (sin λ)2ν−1 . dλ dφ]dθ (13.18)
If we make the variable transformation
cos φ = cos θ cos ψ + sin θ sin ψ cos α (13.19)
then from eqns (13.17) and (13.19) we get
dλ dφ = (sin ψ/ sin φ) dψ dα
and so the double integral in eqn. (13.18) becomes∫ π
0
∫ π
0(cos φ) Cν
n (cos ψ) (sin φ)2ν−1(sin λ)2ν−1 sin ψ dψ dα (13.20)
Then we get from (13.18) as
f (ν)(n) g(ν)(n) =∫ π
0(sin ψ)2ν Cν
n(cos ψ) h(cos ψ) dψ
= G[h (cos ψ)],
where h [cos ψ] = A(sin ψ)1−2ν
∫ π
0
∫ π
0f (cos θ) g (cos θ) (sin θ)2ν .
(sin φ)2ν−1 (sin λ)2ν−1 dθ dα (13.21)
Thus the convolution theorem is proved.
13.5 Application of the Transforms
Case I We consider below the one-dimensional generalised heat con-duction equation defined by
∂
∂x
[(1 − x2)
∂u
∂x
]+ [(1 − x) β − (1 + x)α]
∂u
∂x= ρc
∂u
∂t(13.22)
358 An Introduction to Integral Transforms
under the initial condition
u(x, 0) = g(x) , for − 1 < x < 1 (13.23)
This equation can be expressed as
(1 − x)−α (1 + x)−β
[∂
∂x
{(1 − x)α+1 (1 + x)β+1 ∂u
∂x
}]= ρα
∂u
∂t
(13.24)
Or R[u(x, t)] =1c
∂u
∂t,
where c =α
ρdis a positive constant depending on the density, specific
heat and thermal conductivity of the material.
We solve this boundary value problem by using the Jacobi transform.Taking Jacobi transform, the eqns. (13.24) and (13.23) give rise to
d
dtu(α,β)(n, t) = −cn(n + α + β + 1) u(α,β)(n, t) (13.25)
and u(α,β)(n, 0) = g(α,β)(n) (13.26)
respectively. Solving these equations we get
u(α,β)(n, t) = g(α,β) (n) exp [−n(n + α + β + 1) ct]
Therefore, the formal solution of the problem after inversion of Jacobitransform then becomes
u(x, t) =∞∑
n=0
δ−1n g(α,β) (n) P (α,β)
n (x) exp [−n(n + α + β + 1) ct ]
(13.27)
Case II If the heat conduction equation for a homogeneous solid beambe
∂
∂x
[(1 − x2)
∂u
∂x
]− (2ν + 1) x
∂u
∂x=
1c
∂u
∂t(13.28)
under the initial condition u (x, 0) = g(x), −1 < x < 1 (13.29)
we can apply Gegenbauer transform to solve it. After application of theGegenbauer transform the PDE (13.28) becomes
d
dtu(ν) (n, t) = −dn(n + 2ν) u(ν) (n, t) (13.30)
Jacobi, Gegenbauer, Laguerre and Hermite Transforms 359
and the initial condition (13.29) becomes
u(ν)(n, 0) = g(ν)(n) (13.31)
The solution of equation (13.30) under (13.31) is then given by
u(ν)(n, t) = g(ν)(n) exp [−n(n + 2ν) ct ] (13.32)
On inversion of (13.32) we get the formal solution of the given boundaryvalue problem as
u(x, t) =∞∑
n=0
δ−1n Cν
n (x) exp [−n(n + 2ν) ct] (13.33)
13.6 The Laguerre Transform
Debnath and McCully introduced this transform through their pioneer-ing works. Laguerre transform of a function f(x) over 0 � x < ∞ isdenoted and defined by the integral
L[f(x); x → n] ≡ fa(n) =∫ ∞
0e−xxα Lα
n (x) f(x) dx (13.34)
where Lαn(x) is the Laguerre polynomial of degree n(� 0) and order
α (> −1) which satisfy the ODE
d
dx
[e−x xα+1 d
dxLα
n (x)]
+ ne−x xα Lαn(x) = 0 (13.35)
These Laguerre polynomials satisfy the orthogonal property∫ ∞
0e−x xα Lα
n (x) Lαm (x) dx =
(n + α
n
)Γ (n + 1) δn = δn δmn
(13.36)
where δmn is the Kronecker delta function and δn is given by
δn =
(n + α
n
)Γ (n + 1) (13.37)
Accordingly, the inverse Laguerre transform is defined by
f(x) ≡ L−1[fα(n) ; n → x
]=
∞∑n=0
(δn)−1 fα (n) Lαn (x) (13.38)
360 An Introduction to Integral Transforms
If α = 0, according to McCully, Laguerre transform pair takes the form
L[f(x); x → n] ≡ f0(n) =∫ ∞
0e−xLn(x) f(x) dx (13.39)
L−1[f0(n); n → x] ≡ f(x) =∞∑
n=0
f0 (n) Ln (x) (13.40)
where Ln(x) is the Laguerre polynomial of degree n and order zero.
Example 13.3. If f(x) = Lαm(x) then prove that L[f(x)] = δnδmn
Solution. We have from definition
L [f(x); x → n ] = L[Lαm(x); x → n ]
=∫ ∞
0e−xxαLα
n (x) Lαm (x) dx
=
(n + α
n
)Γ (α + 1) δmn ,
by the orthogonal property of the Laguerre polynomial
= δn δmn
Example 13.4 Find the Laguerre transform of f(x) = e−αx, α > −1
Solution. We have L[e−αx ; x → n ]
=∫ ∞
0e−(α+1)x xα Lα
n (x) dx
=Γ(n + α + 1) αn
n!(α − 1)n+α+1,
from the tables of integrals [Erdelyi et al (1954).]
Example 13.5. Prove that∫ ∞
0e−x xk Ln(x) dx =
{0, k �= n
(−1)n n!, k = n
where Ln(x) is a Laguerre polynomial due to McCully.
Solution. We can express xk as linear combination of polynomialsL0(x), L1 (x), · · ·Lk(x). Let it be
xk =k∑
r=0
CrLr(x)
Jacobi, Gegenbauer, Laguerre and Hermite Transforms 361
Then∫ ∞
0e−xxkLn (x) dx =
k∑r=0
Cr
∫ ∞
0e−x Lr(x) Ln(x) dx
=k∑
r=0
Cr × 0 ,
by the orthogonality relation since k = r �= n.
But, if k = n , then xk = xn =n∑
r=0
Cr Lr (x)
=n−1∑r=0
Cr Lr (x) + Cn Ln (x)
Therefore,∫ ∞
0e−x xnLn(x)dx
=∫ ∞
0e−xL2
n (x).Cn dx
= Cn.1
Now, in the expansion of xn
xn = C0 L0(x) + C1L1(x) + · · · + Cn Ln(x)
= [C0L0(x) + C1 L1(x) + · · · + Cn−1 Ln−1]
+Cn
[(−1)nxn
n!+ a polynomial of degree (n − 1)
]
as xn occurs only on the single term Cn(−1)n
n! xn on the right hand sideof the above expansion we have by comparing the coefficients of xn onboth sides that
1 = Cn (−1)n/n!
This implies that Cn = (−1)nn! and therefore∫ ∞
0e−x xn Ln (x) dx = (−1)n n!
13.7 Operational properties
(a) Laguerre transform of the derivative of a function.
Let L[f(x) ; x → n] = fα(n), then
L[f ′(x) ; x → n] =∫ ∞
0e−x xαLα
n(x)f ′(x) dx
362 An Introduction to Integral Transforms
=[e−x xαLα
n (x) f(x)]∞0
+∫ ∞
0e−xxαLα
n (x)f(x) dx
−α
∫ ∞
0e−x xα−1 Lα
n(x) f(x) dx −∫ ∞
0e−xxα
{d
dxLα
n(x)}
f(x) dx
= fα(n) − α
n∑k=0
fn−1(k) +n∑
k=0
fα(k) , (13.41)
from Higher Transandental function, [Erdelyi, 1953.]
(b) If L [f(x) ; x → n] = fa(n), then
L [R {f(x)} ;x → n] = −nfa(n) (13.42)
where R is the differential operator defined by
R[f(x)] = ex x−α d
dx
[e−x xα+1 df (x)
dx
].
Solution. We have by definition
L [R{f(x)} ; x → n] =∫ ∞
0Lα
n (x)d
dx
[e−x xα+1 df (x)
dx
]dx
=[Lα
n(x){
e−xxα+1 df (x)dx
}]∞0
−∫ ∞
0e−xxα+1 dLα
n(x)dx
.df (x)dx
dx
= −[e−xxα+1 d
dxLα
n (x) f(x)]∞
0
+∫ ∞
0f(x)
d
dx
{e−xxα+1 d
dxLα
n(x)}
dx
=∫ ∞
0f(x).
{−ne−x xα Lαn (x)
}dx ,by eqn. (13.35)
= −n
∫ ∞
0e−x xα Lα
n(x) f(x)dx
= −nfα(n) (13.43)
On extendingthis proof, we can have
L[R2{f(x) ; x → n}] = (−1)2 n2fα(n)
More generally,
L[Rk {f(x)} ; x → n
]= (−1)k nkfα(n), k = 1, 2, 3, · · · (13.44)
(c) The convolution theorem of Laguerre transform of order zero(α = 0) is given by
L−1[f0(n) g0(n)] = h(x) (13.45)
Jacobi, Gegenbauer, Laguerre and Hermite Transforms 363
where h(x) is given by
h(x) =1π
∫ ∞
0e−t f(t) dt
∫ π
0exp (
√xt cos θ) cos (
√xt sin θ)
g(x + t − 2
√xt cos θ
)dθ, (13.46)
if L[f(x) ; x → n] = f0(n) and L[g(x) ; x → n] = g0(n).
Proof. We have by definition
f0(n) g0(n) =∫ ∞
0e−x Ln (x) f(x)dx
∫ ∞
0e−yLy(y)g(y)dy
=∫ ∞
0e−x f(x) dx
[∫ ∞
0e−y Ln (x) Ly(y) g(y) dy
]
Now from a formula of the product of two Laguerre functions wehave
Ln(x)Ly(y) =1π
∫ π
0e√
xy cos θ cos(√
xy sin θ) Ln(x+y−2√
xy cos θ) dθ
[Bateman, Partial differential equation, 1944]. Therefore we canexpress
πf0(n)g0(n) =∫ ∞
0e−x f(x) dx
[∫ ∞
0e−y g(y){∫ π
0exp(
√xy cos θ) cos(
√xy sin θ). Ln(x + y − 2
√xy cos θ) dθ
}dy
]
The integral within the square bracket is
∫ ∞
0e−tLn (t) dt
∫ π
0exp (
√xt cos φ) cos(
√xt sinφ)
g (x + t − 2√
xt cos φ) dφ
So,we can write, f0(n)g0(n) = L[h(t); t → n]
=∫ ∞
0e−tLn(t)h(t)dt
where, h(x) =1π
∫ ∞
0e−tf(t)dt
[∫ π
0exp (
√xt cos θ) cos(
√xt sin θ)
g(x + t − 2√
xt cos θ)dθ]
This is the convolution theorem of the Laguere transform and theproof of it is complete.
364 An Introduction to Integral Transforms
13.8 Hermite Transform.
In 1964 Debnath introduced Hermite transform through his paperentitled “on Hermite transform”, Mathematicki Vesnik, 1,(16)[].He denoted and defined it by
H[f(x); x → n] ≡ fH(n) =∫ +∞
−∞exp (−x2) Hn(x) f(x) dx,
(13.47)where Hn(x) is the Hermite polynomial of degree n defined throughthe Rodrigues formula
Hn(x) = (−1)n exp (x2)dn
dxn
{exp (−x2)
}(13.48)
The inverse Hermite transform denoted by H−1[fH(n);n → x] andis defined by
H−1[fH(n) ; n → x] ≡ f(x) =∞∑
n=0
δ−1n fH(n) Hn(x) (13.49)
where δn is given byδn =
√π n! 2n, (13.50)
Since any function f(x) can be expanded in a series of Hermitepolynomials like
f(x) =∞∑
n=0
an Hn(x)
where the coefficients an of the (n + 1)th term is given by
an = δ−1n fH(n)
after using the orthogonality relation of Hermite polynomials∫ ∞
−∞exp (−x2) Hn(x) Hm(x) dx = δnmδn (13.51)
Example 13.6. Find Hermite transform of each of the followingfunctions :
(a) f(x) = a polynomial of degree m < n
(b) f(x) = Hm(x)
(c) f(x) = exp (2xt − t2)
(d) f(x) = exp (ax)
Jacobi, Gegenbauer, Laguerre and Hermite Transforms 365
Solutions.
(a) Let f(x) = Pr(x), r < n
Then
H[f(x) ; x → n] =∫ +∞
−∞exp (−x2)
r∑k=0
ak Hk(x).
Hn(x) dx, r < n
=∞∑
k=0
ak {0} = 0 ,
from the orthogonality relation of Hermite polynomials.
(b) H [Hm (x) ; x → n]
=∫ +∞
−∞exp(−x2)Hm (x)Hn(x) dx
= δn δmn
(c) To find H[exp(2xt − t2) ; x → n
]we consider the generating
function of Hermite polynomial as
exp (2xt − t2) =∞∑
n=0
tn
n!Hn(x)
Then H[exp(2xt − t2) ; x → n
]=∫ +∞
−∞exp(−x2) Hn(x).
∞∑n=0
tn
n!Hn (x) dx
=∞∑
n=0
tn
n!
∫ +∞
−∞exp(−x2) H2
n (x) dx
=∞∑
n=0
tn
n!δn =
√π
∞∑n=0
(2t)n , |t| <12.
(d) We have from definition
H[exp(ax) ; x → n]
=∫ +∞
−∞exp(−x2) exp (ax) Hn(x) dx
Now since ,∫ +∞
−∞exp(−x2 + 2bx) Hn (x) dx
366 An Introduction to Integral Transforms
=√
π
∞∑n=0
(2b)n exp (b2),
the above relation gives
H [exp (ax) ; x → n] =√
π
∞∑n=0
an exp
(a2
4
)
13.9 Operational Properties.
(a) Theorem 13.2. If f ′(x) is continious and f ′′(x) is integrable in−∞ < x < ∞ and if H[f(x); x → n] = fH(n), then
H[R{f(x)} ; x → n] = −2nfH(n),
where R[f(x)] is defined as
R[f(x)] = exp(x2)d
dx
[exp(−x2)
df
dx
].
Proof. We have by definition,
H [R{f(x)} ; x → n] =∫ +∞
−∞Hn (x).
d
dx
[exp (−x2)
df
dx
]dx
= −∫ +∞
−∞df
dx(x) e−x2
H ′n(x) dx
=∫ +∞
−∞f(x)
[e−x2
Hn”(x) − 2xe−x2H ′
n(x)]dx
=∫ +∞
−∞e−x2
f(x) [−2n Hn (x)] dx
= −2n fH (n) (13.52)
If R [f(x)] also satisfies conditions of the theorem, then
H[R2{f(x)} ; x → n
]= (−1)2 (2n)2fH(n).
In general, when Rk−1satisfies the conditions of the theorem, then
H[Rk{f(x)} ; x → n
]= (−1)k (2n)kfH(n),
where k = 1, 2, · · ·
Theorem 13.3. If f(x) is bounded and integrable and fH(0) = 0, thenfor each constant C
H−1
[−f
H(n)
2n
]= R−1[f(x)] =
∫ x
exp (s2)∫ s
−∞exp (−t2) f(t) dt + C,
Jacobi, Gegenbauer, Laguerre and Hermite Transforms 367
where R−1 is the inverse of the differential operator R and n is a positiveinteger.
Proof.Let R−1[f(x)] = y(x) ⇒ R [y(x)] = f(x).
Thus y(x) is the solution of a second order ODE. Since,
fH (0) = 0 and H0(x) = 1, we have∫ +∞
−∞e−x2
. R [y (x)] dx =∫ +∞
−∞e−x2
R [y (x)] Ho(x) dx
=∫ +∞
−∞e−x2
f(x) H0(x) dx = fH
(0) = 0 (13.53)
Therefore, the first integral of R [y(x)] = f(x) is
exp(−x2)dy
dx=∫ x
−∞exp(−t2) f (t) dt
As |x| → ∞, the right hand side integral in above tends to zero byvirtue of eqn. (13.53). Then the second integral is
y(x) =∫ x
0exp(s2){∫ s
−∞
[exp (−t2)f(t)
]dt
}ds + C (13.54)
Therefore ,
lim|x|→∞
exp (−x2) y(x) = 0 and H [y(x)] exists.
So, H [R{y (x)} ; x → n] = −2n H [y(x)] implying
H[f (x) ; x → n] = −2n H [y (x)]
⇒ fH
(n) = −2n H[R−1{f(x)}]
⇒ H[R−1{f (x)}] = −f
H(n)
2n(13.55)
13.10 Hermite Transform of derivative of a function.
Theorem 13.4. If H[f (x) ; x → n ] = fH
(n) and m is a positiveinteger, then H[f (m) (x) ; x → n ] = fH (n + m)
Proof. we have by definition
H[f ′ (x) ; x → n ] =∫ +∞
−∞exp (−x2) Hn (x) f ′ (x) dx
368 An Introduction to Integral Transforms
=[exp (−x2) f(x) Hn (x)
]+∞−∞ −
∫ +∞
−∞f(x)
d
dx
{e−x2
Hn(x)}
dx
= 2∫ +∞
−∞xe−x2
Hn (x) f(x) dx −∫ +∞
−∞)f(x) e−x2
H ′n (x) dx
=∫ +∞
−∞e−x2
[Hn+1 (x) + 2n Hn−1 x] f(x) dx
−2n∫ +∞
−∞e−x2
Hn−1 (x) f(x) dx,
since Hn+1(x) = 2x Hn (x) − 2n Hn−1 (x) (13.56)
and H ′n (x) = 2n Hn−1 (x) (13.57)
=∫ +∞
−∞exp (−x2) Hn+1 (x) f(x) dx
= fH
(n + 1). (13.58)
Proceeding in the similar manner we can arrive at
H [f (m)(x) ; x → n] = fH
(n + m)
Thoerem 13.5. If Harmite transforms of f(x) and of x fm−1(x) exists,then
H [x f (m)(x) ; x → n] = n fH
(m + n − 1) +12fH
(m + n + 1)
Proof. We have by definition
H[x f (m)(x) ; x → n] =∫ +∞
−∞exp (−x2)Hn(x)
{x f (m)(x)
}dx
=[x exp (−x2) Hn(x) f (m−1)(x)
]+∞−∞
−∫ +∞
−∞d
dx
{x exp(−x2) Hn(x)
}fm−1(x) dx
=∫ +∞
−∞2x2 exp(−x2) Hn(x)fm−1(x)dx
−∫ +∞
−∞exp(−x2)Hn(x)f (m−1)(x)dx
−n
∫ +∞
−∞2x exp (−x2)Hn−1(x)f (m−1)(x) dx
Using eqns. (13.56) and (13.57), we therefore have
H[xfm(x);x → n] =∫ +∞
−∞x e−x2
[Hn+1(x) + 2n Hn−1(x)]f (m−1)(x)dx
Jacobi, Gegenbauer, Laguerre and Hermite Transforms 369
−n
∫ +∞
−∞e−x2
[Hn(x) + 2(n − 1)Hn−2(x)]f (m−1)(x)dx − fH
(n + m + 1)
=12
∫ +∞
−∞e−x2
[Hn+2(x) + 2(n + 1)Hn(x)] f (m−1)(x) dx
+n
∫ +∞
−∞e−x2
[Hn(x) + 2(n − 1)Hn−2(x)] f (m−1)(x) dx
−n fH
(n + m − 1) − 2n(n − 1)fH
(n + m − 3) − fH
(n + m + 1) ,
since H[f (m)(x)] = fH
(n + m)
=12
fH
(n + m + 1) + (n + 1)fH
(n + m − 1) + n[fH
(n + m − 1)
+2(n − 1)fH
(n + m − 3)] − nfH
(n + m − 1) − 2n(n − 1)fH
(n + m − 3)
−fH
(n + m + 1)
= n fH
(n + m − 1) +12
fH
(n + m + 1)
Accordingly, for m = 1 and m = 2, we get
H [x f ′(x) ; x → n] = n fH
(n) +12
fH
(n + 2)
H [x f ′′(x) ; x → n] = n fH
(n + 1) +12
fH
(n + 3)
Exercises
(1) Find Jacobi Transform of the following functions to prove that
(a) J(1 ; n) = 0
(b) J [xm ; x → n] =
{0 , if m �= n
1 , if m = n
(c) J [(1 − x)−1P (α,β)n (x) ; x → n]
=2α+β Γ(α + n + 1)Γ(β + n + 1)
n! αΓ(α + β + n + 1), Re α > 0 , Re β > −1
(d) J [(1 − x)α P (α,β)n (x) ; x → n]
=24α+β+1 Γ
(α + 1
2
) {Γ(α + n + 1)}2 Γ(β + 2n + 1)√π(n!)2 Γ(α + 1) Γ(2α + β + 2n + 2)
, Re α > −12
Re β > −1
after using the Tables of integrals, if necessary.
370 An Introduction to Integral Transforms
(2) Express∂
∂x
[(1 − x2)
∂u
∂x
]+ [β(1 − x) − α(1 + x)]
∂u
∂x
as R [u(x, t)], if R is the differential operator defined by
(1 − x)−α(1 + x)−β
[∂
∂x
{(1 − x)α+1 (1 + x)β+1 ∂
∂x
}]
(3) Prove that∫ 1
−1
[(1 − x) P (α+1,β)
n (x) + (1 + x) P (α,β+1)n (x)
](1 − x)−α(1 + x)−β P (α,β)
n (x) dx
= 2∫ 1
−1(1 − x)−α (1 + x)−β
[P (α,β)
n (x)]2
dx
= 2.
(4) Expressing xβ , β > 0 as a series using the result from Erdelyi inthe form
xβ = Γ(α + β + 1)∞∑
n=0
(−β)nΓ(n + α + 1)
Lα
n(x) , x > 0, α > −1.
prove that L[xβ ; x → n] = Γ(α + β + 1)∞∑
n=0
(−β)nΓ(n + α + 1)
δn
where δn =
(n + α
n
)Γ(α + 1)
(5) Prove that
fα+1 (n) = (n + α + 1) fα(n) − (n + 1) fα(n + 1)
by using the recurrence relation of Laguerre polynomial
x Lα+1
n(x) = (n + α + 1)L
α
n(x) − (n + 1) L
α
n+1 (x)
(6) Prove that the zero order Laguerre transfrom of
(a) f(x) = H(x − a), a � 0 is exp (−a) if n = 0 and is
e−a[Ln(a) − Ln−1(a)], if n � 1
(b) f(x) = e−ax , a > −1 is an(1 + a)n+1
(c) f(x) = xm is 0 if n > m and is (−1)n(
m
n
), if m � n
(d) f(x) = Ln(x) is 1, if n is a non zero positive integer.
Jacobi, Gegenbauer, Laguerre and Hermite Transforms 371
(7) If L [f(x) ; x → n] = f0(n) =∫ ∞
0e−x Ln(x) f(x) dx,
prove that L
[ex d
dx
{x e−x f ′(x)
}; x → n
]= −n f0 (n)
(8) Prove that H[xm ; x → n] = 0 , for m = 0, 1, 2, · · · (n − 1)
(9) Show that H[xn ; x → n] =√
π(n!) Pn(1) ,where Pn(x)
is a Legendre Polynomial.
(10) Prove that H[eax(1 + ax) ; x → n] = H[x eax ; x → n].
Chapter 14
The Z-Transform
14.1 Introduction.
Analysing the behavior of functions whose values are known on a finiteor infinite set of discrete points in a given domain cannot be achievedby using the existing Fourier or Laplace Transform techniques. Accord-ingly, some efficient procedure for the numerical evaluation of the systemwill be aimed at. For example, such necessity do arise for the determi-nation of the output relation when the discrete input function in knownin a signal processing system. The Z-transform technique is one suchmethods to answer the problem proposed above.
14.2 Z - Transform : Definition.
We have learnt that, under certain restrictions, the Laplace transformof a function f(t) is defined as
F (s) =∫ ∞
0f(t)e−st dt (14.1)
After writing the integral on the right hand side of eqn. (14.1) as aRiemann sum we have
F (s) = T
∞∑k=0
f(kT )exp(−kTs), (14.2)
where T is the length of the subinterval chosen to get a proper repre-sentation of f(t) with its sampled values at the discrete set of pointstk = kT, given by f(kT ), k = 0, 1, 2, . . . and the value of the integralunder (14.1) with desired accuracy.
The Z-Transform 373
Then putting z = exp(sT ), we have F (s) in (14.2) is expressed as
F (s) = T∞∑
k=0
f(kT )z−k (14.3)
Here, it shows that F (s) is expanded in a series in power of z. Thisrelation in (14.3) can further be generalized by considering a sequence{xk} for k = 0, 1, 2, · · · and write a series in powers of z correspondingto the above as
∞∑k=0
xkz−k = x(z) (14.4)
after conveniently setting T = 1 as the interval between samples. Thissetting and representations have no effect on the properties of this trans-form.
This relation in (14.4) defines the Z - transform of the sequence {xk}.Symbolically, we write
Z [{xk}] = X(z) =+∞∑k=0
xkz−k (14.5)
as the Z-transform of {xk} and
Z−1[X(z)] = {xk} (14.6)
as the inverse Z-transform of X(z). It means that with a given sequencewe can associate a series in a domain of complex z-plane and vice versa.
To obtain the inversion of Z-transform in terms of a complex integralin the z-plane, we consider eqn. (14.5) as
X(z) =∞∑
k=0
xk z−k ≡∞∑
k=0
x(k) z−k , say (14.7)
Or X(z) = x(0) + x(1)z−1 + x(2)z−2 + · · · + x(k)z−k + · · · (14.8)
Multiplying both sides by zk−1 and integrating with respect to z
along the closed positive contour C, which encloses all singularities ofX(z), we get
374 An Introduction to Integral Transforms
12πi
∮C
X(z) zk−1 dz =1
2πi
[∮C
x(0) zk−1 dz +∮
Cx(1)zk−2dz + · · ·+
+∮
Cx(k)z−1 dz +
∫C
x(k + 1)z−2 dz + · · · · · ·]
By Canchy’s integral formula, the above relation gives
12πi
∮C
X(z) zk−1 dz =1
2πi
∮C
x(k)z
dz
= x(k) ,
since all other integrals on the right hand side vanish.
Thus we get,
12πi
∮C
X(z) zk−1 dz = x(k) (14.9)
implying that
z−1[X(z)] = x(k) = {xk} =1
2πi
∮C
X(z) zk−1 dz (14.10)
Example 14.1.
If x(n) = H(n) =
{1, n � 00, n < 0
,find the Z − transform of x(n).
Solution.
Z[x(n)] =∞∑
n=0
[x(0) + x(1)z−1 + · · · + x(n)z−n + · · ·]
=α∑
n=0
(1z
)n
=1
1 − 1z
=z
z − 1, |z| > 1.
Example 14.2. If x(n) = an, then find its Z-transform.
Solution.
X(z) = Z[x(n)] =∞∑
n=0
(a
z
)n=
z
z − a, |z| > a.
The Z-Transform 375
Example 14.3. Find the Z-transform of the following functions
(a) x(n) = n
(b) x(n) = n2
(c) x(n) = 1n!
(d) x(n) = einx
(e) x(n) = cosh nx
(f) x(n) = H(n), where H(n) =
{0, if n < 01, if n > 0
and n is an integer.
Solution.
(a) Z[n] =∞∑
n=0
n z−n = z∞∑
n=0
n z−n−1 = −zd
dz
[ ∞∑n=0
z−n
]
= z/(z − 1)2, |z| > 1
(b) Z[n2] = Z[n.n] =∞∑
n=0
n.nz−n
= z∞∑
n=0
n.n z−(n+1)
= z
∞∑n=0
n.
[− d
dzz−n
]
= −zd
dz
∞∑n=0
n z−n
= −zd
dz
[z
(z − 1)2
], |z| > 1,by using result in (a)
=z(z + 1)(z − 1)3
, |z| > 1
(c) Z
[1n!
]=
∞∑n=0
1n!
z−n = exp(
1z
), for all z
(d) Z[einx]
=∞∑
n=0
einx
zn=
∞∑n=0
(eix)n
z−n =∞∑
n=0
(eix
z
)n
=1
1 − eix
z
=z
z − eix
Similarly Z[e−inx
]=
z
z − e−ix
376 An Introduction to Integral Transforms
Therefore, Z [cos n x] =z(z − cos x)
z2 − 2z cos x + 1
and Z [sin nx] =z sin x
z2 − 2z cos x + 1
(e) Z [cosh nx] =12Z[enx + e−nx
]=
12
[z
z − ex+
z
z − e−x
]=
z(z − chx)z2 − 2z chx + 1
(f) [H(n)] =∞∑
n=0
H(n) z−n =∞∑
n=0
z−n =z
z − 1, |z| > 1
Example 14.4. If f(n) is a periodic sequence of integral period N ,then prove that
F (z) = Z[f(n)] =zN
zN − 1F1(z)
where F1(z) =N−1∑k=0
f(k)z−k
Solution. By definition,
F (z) = Z[f(n)] =∞∑
n=0
f(n)z−n
=[f(0) + f(1)z−1 + f(2)z−2 + · · · + f(N − 1)z−N+1
]+[f(0)z−Nz0 + f(1)z−N .z−1 + · · · + f(N − 1) z−N z−N+1
]+ · · ·
=N−1∑k=0
f(k)z−k.1 + z−NN−1∑k=0
f(k)z−k + z−2NN−1∑k=0
f(k)z−k + · · ·
=N−1∑k=0
f(k)z−k
[1 +
1zN
+1
z2N+ · · ·
]
=1
1 − 1zN
F1(z) =zN
zN − 1F1(Z)
14.3 Some Operational Properties of Z-Transform.
Theorem 14.1 : (Translation). If Z[f(n)] = F (z) and m � 0, then
Z [f(n − m)] = z−m
[F (z) +
−1∑k=−m
f(k)z−k
], (14.11)
The Z-Transform 377
Z [f(n + m)] = zm
[F (z) −
m−1∑k=0
f(k)z−k
]. (14.12)
In particular, if m = 1, 2, 3, · · · ., then from the first relation
Z[f(n − 1)] = z−1 F (z)
Z[f(n − 2)] = z−2 F (z) +∑−1
k=−2
f(k)z−k
⎫⎬⎭ (14.13)
Similarly, it follows from the second relation (14.12) that
Z[f(n + 1)] = z[F (z) − f(0)]Z[f(n + 2)] = z2[F (z) − f(0)] − zf(1)Z[f(n + 3)] = z3[F (z) − f(0)] − z2f(1) − zf(2)
⎫⎪⎬⎪⎭ (14.14)
Result in (14.12) will be useful in solving difference and differential equa-tion for initial value problems.
Proof. By definition
Z[f(n − m)] =∞∑
m=0
f(n − m) z−n
=∞∑
k=−m
f(k)z−m−k
= z−m∞∑
k=−m
f(k) z−k
= z−m
[ −1∑k=−m
f(k)z−k +∞∑
k=0
f(k) z−k
]
= z−m
[ −1∑k=−m
f(k)z−k + F (z)
]
If m = 1, then Z [f(n − 1)] = z−1 F (z),when f(k) = 0, k < 0
Also, Z [f(n + m)] =∞∑
m=0
f(n + m)z−n
= zm∞∑
k=m
f(k)z−k
378 An Introduction to Integral Transforms
= zm
[ ∞∑k=0
f(k)z−k −m−1∑k=0
f(k) z−k
]
= zm
[F (z) −
m−1∑k=0
f(k) z−k
]
For m = 1, 2, 3, · · · the results in (14.14) will follow immediately.
Theorem 14.2 : (Multiplication)
If Z [f(n)] = F (z), then Z [anf(n)] = F(z
a
), |z| > |a|,
Z[e−bnf(n)
]= F (zeb) and Z [nf(n)] = −z
d
dzF (z).
Proof. From definition Z [anf(n)] =∞∑
n=0
an f(n) z−n
=∞∑
n=0
f(n)(z
a
)−n= F
(z
a
),|z||a| > 1 (14.15)
Again, Z[e−bn f(n)
]=
∞∑n=0
f(n)z−ne−bn
=∞∑
n=0
f(n)(zeb)−n
= F(zeb)
(14.16)
Also, Z [nf(n)] =∞∑
n=0
nf(n) z−n
= z∞∑
n=0
f(n){− d
dzz−n
}
= −zd
dz
[ ∞∑n=0
f(n)z−n
]
= −zd
dz[F (z)] (14.17)
Thus, all the results of the theorem is proved.
Theorem 14.3 : (Division)
If Z [f(n)] = F (z), then Z
[f(n)
n + m
]= −zm
∫ z
0
F (ξ)ξm+1
dξ
The Z-Transform 379
Proof. By definition, we have
Z
[f(n)
n + m
]=
∞∑n=0
f(n)n + m
z−n, m � 0
=∞∑
n=0
f(n)[−∫ z
0ξ−(m+n+1)dξ
]
= −zm
∫ z
0ξ−(m+1)
[ ∞∑n=0
f(n)ξ−n
]dξ
= −zm
∫ z
0ξ−(m+1)F (ξ) dξ (14.18)
Theorem 14.4 : (The convolution Theorem).
If Z [f(n)] = F (z) and Z [g(n)] = G(z) , then Z-transform of theconvolution f(n) ∗ g(n) defined by
f(n) ∗ g(n) =∞∑
m=0
f(n − m)g(m) (14.19)
is given byZ [f(n) ∗ g(n)]
= F (z) G(z)
= Z [f(n)]Z [g(n)]
Equivalently, Z−1 [F (z) G(z)] = f(n) ∗ g(n) =∞∑
m=0
f(n − m) g(m)
Proof. From definition
Z [f(n) ∗ g(n)] =∞∑
n=0
z−n
[ ∞∑m=0
f(n − m) g(m)
]
=∞∑
m=0
g(m)∞∑
n=0
f(n − m)z−n
=∞∑
m=0
g(m)z−m∞∑
k=−m
f(k)z−k
=∞∑
m=0
g(m)z−m∞∑
k=0
f(k)z−k , on the assumption that f(k) = 0, k < 0
= Z [f(n)] Z [g(n)] (14.20)
Thus the theorem is proved.
380 An Introduction to Integral Transforms
Theorem 14.5 : (The initial value Theorem).
If Z [f(n)] = F (z), then f(0) = limz→∞F (z).
Also if f(0) = 0, then f(1) = limz→∞ zF (z).
Proof. we have by definition
F (z) =∞∑
n=0
f(n)z−n = f(0) +f(1)
z+
f(2)z2
+ · · ·
Then letting z → ∞, we get from this relation that
limz→∞F (z) = f(0) (14.21)
If f(0) = 0, then we get
F (z) =f(1)
z+
f(2)z2
+ · · ·Therefore, lim
z→∞ z F (z) = f(1) (14.22)
Theorem 14.6 : (The Final Value Theorem)
If Z [f(n)] = F (z), then limn→∞ f(n) = lim
z→1[(z − 1)F (z)] ,
provided the limits exist.
Proof. We have from (14.14)
Z [f(n + 1) − f(n)] = z[F (z) − f(0)] − F (z)
This implies that
∞∑n=0
[f(n + 1) − f(n)] z−n = (z − 1)F (z) − zf(0)
In the limit as z → 1, we obtain
limz→1
∞∑n=0
[f(n + 1) − f(n)] z−n = limz→1
(z − 1) F (z) − f(0)
Or f(∞) − f(0) = limz→1
(z − 1) F (z) − f(0)
Thus , limz→∞f(n) = lim
z→1(z − 1) F (z), (14.23)
The Z-Transform 381
provided the limits exist. Hence the result of this theorem is proved.
Theorem 14.7 : The Z-transform of the partial derivatives of afunction.
The Z-transform of ∂∂a [f(n, a)] is given by
Z
[∂
∂af(n, a)
]=
∂
∂a[Z{f(n, a)}]
Proof. We have by definition
Z
[∂
∂af (n, a)
]=
∞∑n=0
[∂
∂af(n, a)
]z−n
=∂
∂a
[ ∞∑n=0
f(n, a)z−n
]
=∂
∂aZ[f(n, a)] (14.24)
For example , if f(n, a) = nena then
Z [nena] = Z
[∂
∂aena
]
=∂
∂a[Z{ena}]
=∂
∂a
[z
z − ea
]=
zea
(z − ea)2
Example 14.5 If F (z) = z(z−a)(z−b) , find f(0) and f(1) by using the
initial value theorem.
Solution. By initial value theorem we get from eqns. (14.21) and(14.22) that
f(0) = limz→∞F (z) = lim
z→∞z
(z − a)(z − b)= 0
Again,
f(1) = limz→∞ zF (z)
= limz→∞
z2
(z − a)(z − b)= 1
Example. 14.6
Find the inverses of the following Z-transformed functions:
(a) F (z) =z
z − 2
382 An Introduction to Integral Transforms
(b) F (z) = exp (1z)
(c) F (z) =3z2 − z
(z − 1)(z − 2)2
(d) F (z) =z(z + 1)(z − 1)3
Solution.
(a) Here F (z) = zz−2
=1
1 − 2z
=(
1 − 2z
)−1
= 1 +2z
+22
z2+
23
z3+ · · ·
Therefore, Z−1[F (z)] = Z−1
[1 +
2z
+22
z2+ · · ·
]Here f(0) = 1, f(1) = 2, f(2) = 22, f(3) = 23, · · ·
Therefore, f(n) = 2n
So, Z−1[F (z)] = f(n) = 2n
(b) We have in this case
F (z) = 1 +1z
+12!
1z2
+ · · ·
Therefore, f(0) = 1, f(1) = 1, f(2) =12!
, · · ·
∴ f(n) =1n!
∴ Z−1[F (z)] = f(n) =1n!
Thus Z−1[e
1z
]=
1n!
(c) Here, by partial fraction method
3z2 − z
(z − 1)(z − 2)2≡ 2
z
z − 1− 2
z
z − 2+
52
2z(z − 2)2
Then, Z−1
[3z2 − z
(z − 1)(z − 2)2
]
= Z−1
[2z
z − 1
]− Z−1
[2z
z − 2
]+
52
Z−1
[2z
(z − 2)2
]
The Z-Transform 383
= 2.1 − (2)n+1 +52
. n 2n, by theorem (14.7)
= 2 − 2n+1 + 5n 2n−1
(d) We can express
z(z + 1)(z − 1)3
≡ z
(z − 1)2
[z
z − 1+
1z − 1
]
Let F (z) =z
(z − 1)2and G(z) =
z
z − 1+
1z − 1
Then by inversion of Z-transform we have
f(n) = n and g(n) = H(n) + H(n − 1)
where H(n) is the unit function .
Therefore, by convolution theorem
Z−1
[z(z + 1)(z − 1)3
]= f(n) ∗ g(n)
=n∑
m=0
m[H(n − m) + H(n − m − 1)]
= n2.
14.4 Application of Z-Transforms.
(a) Solution of difference equations.
Example 14.7
Solve the following initial value problems :
(i) f(n + 1) + 3f(n) = n, f(0) = 1
(ii) f(n + 2) − f(n + 1) − 6f(n) = 0, f(0) = 0, f(1) = 3
(iii) un+1 = un + un−1, u1 = u(0) = 1
(iv) u(n + 2) − u(n + 1) + u(n) = 0, u(0) = 1, u(1) = 2
384 An Introduction to Integral Transforms
Solutions.
(i) f(n + 1) + 3 f(n) = n , f(0) = 1The Z-transform of this difference equation yields
(z + 3) F (z) = z +z
(z − 1)2
Or F (z) =z
z + 3
(1 +
116
)+
14
z
(z − 1)2− 1
16z
z − 1
Upon inverting this transformed equation by Z-transform weget
f(n) =1716
(−3)n − 116
(1)n +14n.
(ii) f(n + 2) − f(n + 1) − 6 f(n) = 0 , f(0) = 0 , f(1) = 3.The Z-transform of the difference equation under the initialconditions yeilds
F (z)[z2 − z − 6] = 3z
Or F (z) =3z5
[1
z − 3− 1
z + 2
]
=35
z
z − 3− 3
5z
z + 2Upon inversion the Z-transformed equation yeilds
f(n) =353n − 3
5(−2)n.
(iii) un+1 = un + un−1 , u1 = u(0) = 1This equation defines a sequence which is known as theFibonacci Sequence.The Z-transform of the above difference equation yeilds
U(z) =z2
z2 − z − 1Upon inversion of the above transformed equation we get
un = Z−1
[z2
z2 − z − 1
]
≡ Z−1
[z2
(z − a)(z − b)
], say
where (a, b) =
(1 +
√5
2,
1 −√5
2
)
The Z-Transform 385
Therefore,
un = Z−1
[az
z − a
1a − b
]− Z−1
[bz
z − b
1a − b
]
=a
a − b(a)n − b
a − b(b)n
=an+1 − bn+1
a − b
(iv) u(n + 2) − u(n + 1) + u(n) = 0, u(0) = 1, u(1) = 2The Z-transform of this difference equation under the initialconditions yeilds
U(z)[z2 − z + 1
]= (z2 + z)
Or U(z) =z2 − z
2
z2 − z + 1+
√3(√
32 z)
z2 − z + 1
Upon inversion of the Z-transformed equation we get
u(n) = cosnπ
3+
√3 sin
(nπ
3
)as the solution of the given initial value problem.
(b) Summation of Infinite Series
Theorem 14.8 If Z[f(n)] = F (z), then
n∑k=1
f(k) = Z−1
[z
z − 1F (z)
]
and∞∑
k=1
f(k) = limz→1
F (z) = F (1).
Proof. Let h(n) =∑nk=1
f(k). Then clearly,
h(n) − h(n − 1) = f(n)
Applying Z-transform this relation gives
H(z) − z−1 H(z) = F (z) , by (14.13).
Therefore,H(z) =
z
z − 1F (z)
386 An Introduction to Integral Transforms
Inverting Z-transform we get
h(n) =n∑
k=1
f(k)
= Z−1
[z
z − 1F (z)
]
Also, if z → 1 we have from the Final Value Theorem that
limn→∞h(n) = lim
n→∞
n∑k=1
f(k)
= limz→1
[(z − 1).z
z − 1F (z)
]
Or∞∑
k=1
f(k) = limz→1
[zF (z)] = F (1).
Thus the proof is complete.
Example 14.8 Find the sum of the following series using Z-transform:
(a)∞∑
n=0
aneinx
(b)∞∑
n=0
e−(2n+1)x
(c)∞∑
n=0
(−1)nxn+1
n + 1
(d)∞∑
n=0
an cos nx
Solutions.
(a) Let f(n) = aneinx
We know that Z[einx]
=z
z − eix
Therefore , Z[f(n)] = Z[aneinx
]=
z
z − aeix
So,∞∑
n=0
aneinx = limz→1
z
z − aeix
=(1 − aeix
)−1
The Z-Transform 387
(b)∞∑
n=0
e−(2n+1)x = e−x∞∑
n=0
e−2nx
So, Z[e−(2n+1)x
]= Z
[e−xe−2nx
]=
e−x
1 − ze−2x
Therefore ,
∞∑n=0
e−(2n+1)x
= limz→1
e−x
1 − ze−2x
=1
ex − e−x= (2 sinh x)−1
(c) We know that Z[xn+1
]=
zx
z − x
Hence , Z
[xn+1
n + 1
]= z
∫ ∞
z
zx
z − x
dz
z2
= xz
∫ ∞
z
dz
z(z − x)
= xz
[1x
logz − x
z
]∞z
= −z log[z − x
z
]Now replacing x by − x, this above result gives
Z
[(−1)n
xn+1
n + 1
]= z log
[z + x
z
]Therefore ,
∞∑n=0
(−1)nxn+1
n + 1= lim
z→1z log
[z + x
z
]
= log(1 + x)
(d) We know that
Z[cos nx] =z(z − cos x)
z2 − 2z cos x + 1
∴ Z [an cos nx] =za
(za − cos x
)z2
a2 − 2za cos x + 1
=z(z − a cos x)
z2 − 2az cos x + a2
388 An Introduction to Integral Transforms
So ,
∞∑n=0
an cos nx = limz→1
[z(z − a cos x)
z2 − 2az cos x + a2
]
=1 − a cos x
1 − 2a cos x + a2
Exercises.
(1) Prove the following :
(a) Z[n3]
= (z3 + 4z2 + z)/(z − 1)4
(b) Z [H(n) − H(n − 2)] = 1 +1z
(c) Z[n2an
]=
z(z + a)(z − a)2
(d) Z [sinhna] =z sinh a
z2 − 2z cosh a + 1
(e) Z [nanf(n)] = −zd
dz
[F(z
a
)](f) Z
[f(n)
n + m
]= zm
∫ ∞
z
F (ξ)dξ
ξm+1,
Hence duduce from the result of (f) that
Z
[1
n + 1
]= z log
(z
z − 1
)
(2) Find the inverse Z transform of the following functions :
(a)1
(z − a)2(b)
z3
(z2 − 1)(z − 2)(c)
z2
(z − 1)(z − 12 )
(d)z
(z − a)m+1(e) z log
z + 1z
(f)z
z − 2[Ans. (a) (n − 1) an−2 H(n − 1) (b)
16[(−1)n + 2n+3 − 3(1)n
](c) (2 − 2−n) (d) n(n − 1) · · · (n − m + 1) an−m /m!
(e)(−1)n
(n + 1)(f) 2n
]
(3) Solve the following difference equations :
(a) f(n + 1) + 2f(n) = n, f(0) = 1
The Z-Transform 389
(b) f(n + 2) − 3f(n + 1) + 2f(n) = 0, f(0) = 1, f(1) = 2
(c) f(n + 2) − 5f(n + 1) + 6f(n) = 2n, f(0) = 1, f(1) = 0
(d) f(n + 2) − 2xf(n + 1) + f(n) = 0 , |x| ≤ 1 ,
f(0) = f0, f(1) = 0, n � 2
(e) f(n + 3) − 3f(n + 2) + 3f(n + 1) − f(n) = 0,
f(0) = 1, f(1) = 0 ; f(2) = 1
Ans. (a)19
[10(−2)n + 3n − 1] (b) 2n (c) 2n+1 − 3n − n 2n−1
(d)f0 z[(z − x) − 2x(1 − x2)−
12 ]
(z2 − 2xz + 1)(e) (n − 1)2
(4) Show that∞∑
n=0
(−1)n e−n
n + 1= e log (1 + e−1)
(5) (a) If F (z) = zz−α , prove by using initial value theorem that
f(0) = 1
(b) Use final value theorem to evaluate
limn→∞ f(n) if F (z) =
(3z2 − z)[(z − 1)(z − 2)2]
[Ans.2](6) Solve the difference equation
yk+2 − 3yk+1 + 2yk = 0
to prove that Y (z) =1
1 − 2z− 1
1 − z
and hence deduce that yk
= 2k − 1 , if y(0) = 0 , y(1) = 1.
Appendix : Tables of Integral Transforms
We collect here some important formulae derived in this text or elsewherewhich are of common use. For an exhaustive list of transforms the readershould also consult G.A. Campbell and R.M. Foster (Bell Telephone-system, Technical Publication, 1931), W. Magnus and F. Oberhettinger(Springer, Berlin, 1948), R.V. Churchill (Modern Operational Mathe-matics in Engineering, Mc Graw-Hill, NY, 1944), Erdelyi et al. (Tablesof Integral Transforms, Mc Graw-Hill, NY, 1954), Oberhettinger andHiggens (Boeing Sc. Res. Lab. Maths Note 246, Seattle, 1961), Ian N.Sneddon (Mc Graw-Hill Book Company, 1951), etc. Unless otherwisestated here a, t are assumed to be positive constants.
A1 : Fourier Transforms.f(t) F [f(t) ; t → ξ]
1√
2π δ(ξ)
tkH(t)√
2π ik[
12δ(k)(ξ) + (−1)kk!
2πiξk+1
]e−a|t|
√2π
aa2+ξ2
te−α|t|√
2π2aiξ (a2 + ξ2)−2
|t|e−a|t|√
2π (a2 − ξ2) (a2 + ξ2)−2
e−a2t2 1√2a
e−14
ξ2
a2
δ(t) 1√2π
1t −i
√π2 sgn ξ
Appendix : Tables of Integral Transforms 391
A1 : Fourier Transforms. (Continued)
f(t) F [f(t) ; t → ξ]
tk sgn t√
2k (−iξ)−k−1k!
sin att
{ √π2 , if |ξ| < a
0 , if |ξ| > a
chatchπt ,−π < a < π
√2π cos a
2 ch ξ2/[chξ + cos a]
shatshπt ,−π < a < π 1√
2πsin a
chξ+cos a
{1√
a2−t2, |t| < a
0 , |t| > aπ2 J0(aξ)
|t|−1 1|ξ|
H(a − |t|)√
2π
sin aξξ
t−1 i√
π2 sgn ξ
(1 − t2)(1 + t2)−2√
π2 ξ e−ξ
A2 : Fourier Cosine Transforms.
f(t) fc [f(t) ; t → ξ]
{1 , 0 < t < a
0 , t > a
√2π
sin aξξ
a(t2 + a2)−1, a > 0√
π2 e−aξ
e−at√
2π
aa2+ξ2
tp−1, 0 < p < 1√
2π Γ(p)ξ−p cos pπ
2
392 An Introduction to Integral Transform
A2 : Fourier Cosine Transforms. (Continued)
f(t) fc [f(t) ; t → ξ]
e−a2t2 1√2|a| e−
14
ξ2
a2
cos t2
21√2
[cos ξ2
2 + sin ξ2
2 ]
sin t2
21√2
[cos ξ2
2 − sin ξ2
2 ]
sh(αt)sh(βt) , 0 < α < β
√π2
1β
sin( παβ
)
ch(πξβ
)+cos�
παβ
�
ch(αt)ch(βt) , 0 < α < β
√2π 1
β
cos�
πα2β
�ch�
πξ2β
�
ch�
πξβ
�+cos
�παβ
�
t−νJν(at), ν > −12
(a2−ξ2)ν− 12 H(a−ξ)
2ν− 12 aν Γ(ν+ 1
2)
K0(at)√
π2
(a2 + ξ2
) 12
2e−t sin tt
√2π tan−1
(2ξ2
)(1−t2)(1+t2)2
√π2 ξ e−ξ
A3 : Fourier Sine Transforms.
f(t) Fs [f(t) ; t → ξ]
1t
√π2 sgn ξ
e−at√
2π
ξξ2+a2
t e−at√
2π
2aξ(ξ2+a2)2
Appendix : Tables of Integral Transforms 393
A3: Fourier Sine Transforms. (Continued)
f(t) Fs [f(t) ; t → ξ]
t−1 e−at√
2π tan−1
(ξa
)
tp−1 , 0 < p < 1√
2π ξ−p Γ(p) sin
(πp2
)
t e−a2t2 2−32 a−3 ξ e
�− 1
4ξ2
a2
�
{0 , 0 < t < a
(t2 − a2)−12 , t > a
J0(aξ)
1t(t2+a2)
√π2
1a2 (1 − e−|ξa|) sgn ξ
H(a − t) , a > 0√
2π
1−cos aξξ
cosech t√
π2 tanh(1
2πξ)
cosech(αt)sinh(βt) , 0 < α < β
√π2
1β
sinh�
πξβ
�
cosh�
πξβ
�+ cos
�παβ
�
sinh(αt)cosh(βt) , 0 < α < β
√2πβ
sin�
πα2β
�sinh
�πξ2β
�
cos�
παβ
�+ cosh
�πξβ
�
J0(at)t
{ √π2 , a < ξ < ∞√2π sin−1
(ξa
), 0 < ξ < a
t J0(at)k2+t2
√π2 e−kξ I0 (ak) , ξ > a
394 An Introduction to Integral Transform
A4 : Laplace Transforms.
f(t) L [f(t) ; t → p]
1 1p , p > 0
tν , ν > −1 p−ν−1 Γ (ν + 1)
eat 1p−a , Re p > a
cosh (at) p(p2−a2)
, Re p > a
sin (at) a(p2−a2)
, Re p > a
t cosh (at) (p2 + a2)(p2 − a2)−2 , Re p > a
t sinh (at) 2ap((p2 − a2)−2 , Re p > a
cos (at) p (p2 + a2)−1
sin (at) a (p2 + a2)−1
t cos (at) (p2 − a2) (p2 + a2)−2
t sin (at) 2ap (p2 + a2)−2
t−1 sin (at) cot−1 ( pa)
e−14b2t2 √
π b−1ep2
b2 Erfc (pb ) , b > 0
Erf(√
t) p−1(p + 1)−12
Erf( a√t) p−1e−2a
√p
tνJν(at) (2a)ν Γ (ν+ 12)
√π(p2+a2)ν+1
2, ν > −1
2
tν+1Jν(at) (2a)ν+1aν p Γ (ν+ 32)
√π(p2+a2)ν+3
2, R ep > a > 0
tνI0(at) p (p2 − a2)−32 , R ep > a > 0
eat−1a
1p(p−a)
Appendix : Tables of Integral Transforms 395
A4 : Laplace Transforms. (Continued)
f(t) L [f(t) ; t → p]
sin(a√
t) 12 a√
πp3 e
−a2
4p , Re p > a > 0
sin(a√
t)t Erf(1
2a√p) , Re p > a > 0
{0 , 0 < t < b
tn , t > be−bp
∑nm=0
n! bm
m!pn−m+1 , Re p > 0
{0 , 0 < t < b
t/√
t2 − b2 , t > bbK1(bp) , Re p > 0.
e−14
at
√a p−
12 K1(
√pa) , Re a � 0.
log t −p−1 log(γp)
2t sinh(at) log p+a
p−a , Re p > |Re a|
t cos at (p2−a2)(p2+a2)2
t cosh at (p2 + a2)/(p2 − a2)−2
sin(at)t tan−1 (a
p )
J0 (at) (p2 + a2)−12
I0(at) (p2 − a2)−12
δ(t − a) exp (−ap) , a > 0
| sin at| , a > 0 a(p2+a2)
coth (πp2a )
396 An Introduction to Integral Transform
A5 : Hilbert Transforms.f(t) H [f(t) ; t → x]
tt2+a2
ax2+a2
H(t − a) − H(t − b) 1π log | b−x
a−x | , b > a > 0
H(t−a)t
1πx log a
|a−x| , a > 0
(at + b)−1 H(t) 1π(ax+b) log b
a|x| , a, b > 0, x �= ba
|t|ν−1 , 0 < Re ν < 1 −ctn (νπ2 ) |x|ν−1 sgn x
eiat i eiax
sin(at)/t {cos(ax)−1}x
sin atJ1(at) cos(ax) J1 (a x)
cos atJ1(at) − sin(ax) J1 (a x)
A6 : Stieltjes Transforms.
f(x) S [ f(x) ; x → y ] =∫∞
0
f(x)x+y dx
1x+a
1a−y log |ay |
e−ax√x
π√y eay Erfc (
√ay).
1a2+x2
1a2+y2
[ny2a − log y
a
]xν , −1 < Re ν < 0 −πyν cosec (πν)
xν
a+x , −1 < Re ν < 1 π(aν−yν)(a−y) sin ν π
x−νe−ax , Re α > 0, Re ν < 1 Γ(1 − ν) y−ν eay Γ (ν, ay)
Appendix : Tables of Integral Transforms 397
A6 : Stieltjes Transforms. (Continued)
f(x) S [ f(x) ; x → y ] =∫∞
0
f(x)x+y dx
log x(a+x) , |arg a| < π 1
2 (y − a)−1[(log y)2 − (log a)2
]sin(a
√x) πexp(−a
√y)
x−1 sin(a√
x) πy−1[1 − exp (−a
√y)]
1√x
cos(a√
x) π√y exp (−a
√y )
A7 : Hankel Transforms.
f(x) Hν [ f(x) ; x → ξ ]
{xν , 0 < x < a
0 , x > a(ν > −1) aν+1
ξ Jν+1 (ξa)
(a2 − x2) H (a − x) 4aξ3 J1(ξa) − 2a2
ξ2 J0(ξa)
e−px , (ν = 0) p (ξ2 + p2)−12
e−px
x , (ν = 0) (ξ2 + p2)−12
e−px , (ν = 1) ξ (ξ2 + p2)−12
e−px
x , (ν = 1) 1ξ − p
ξ√
ξ2+p2
a√a2+x2
, (ν = 0) e−aξ
sin(ax)x , (ν = 0) (a2 − ξ2)−
12 H(a − ξ)
sin(ax)x , (ν = 1) a
ξ√
ξ2−a2H (ξ − a)
398 An Introduction to Integral Transform
A7: Hankel Transforms. (Continued)
f(x) Hν [ f(x) ; x → ξ ]
xνe−x2
a2(
12a2)ν+1
ξν e−ξ2a2
4
x e−ax , (ν = 1) ξ(a2 + ξ2
)− 32
1x , (ν = 0) 1
ξ
√x (a2 + x2)−
12 , (ν = 0) 1√
ξe−aξ
√x (a2 − x2)−
12 H(a − x) , (ν = 0) 1√
ξsin (−aξ)
√x (x2 − a2)−
12 H (x − a) , (ν = 0) 1√
ξcos (−aξ)
1√x
log x , (ν = 0) − 1√ξ
log (2ξ)
1√x
sin (ax) , (ν = 0) ξ−12
(a2 − ξ2
)− 12 H (a − ξ)
1√x
cos (ax) , (ν = 0) ξ12
√(ξ2 − a2) H (ξ − a)
√x cos
(a2x2
2
), (ν = 0)
√ξ
2a2
H(a − x) , (ν = 0) aξ J1(aξ)
1x2 (1 − cos ax) , (ν = 0) cosh−1 (a
ξ ) H (a − ξ)
(x2 + a2)−12 , (ν = 0) 1
ξ e−aξ{√
ξ2 + a2 − a}2
e−ax
x , (ν = 1) 1ξ
[1 − a
(ξ2+a2)12
]
Appendix : Tables of Integral Transforms 399
A8 : Finite Hankel Transforms.
f(r)∫ a0 f(r) r Jn (rξ)dr = Hν [ f(r) ; r → ξ ]
c, (n=0) caξ J1(ξa)
(a2 − r2) , (n = 0) 4aξ3 J1(ξa) − 2a2
ξ2 J0(ξa)
rn , (n > −1) an+1
ξ Jn+1 (ξa)
J0(ar)J0 (a) − 1 , (n = 0) −J1(ξ)
ξ�1− ξ2
a2
�
1r , (n = 1) 1
ξ{1 − J0(aξ)}
(a2 − r2)−12 , (n = 0) 1
ξ sin aξ
1r (a2 − r2)−
12 , (n = 0) (1−cos aξ)
(aξ)
A9 : Mellin Transforms.
f(x) M [f(x) ; x → s] =∫∞
0f(x) xs−1dx
e−px p−s Γ (s) , Re s > 0
e−x2 12 Γ
(s2
)(1 + x)−a Γ (s) Γ(a − s)/Γ(a) , 0 < Re s < Re a
(1 + xa)−1 πa cosec πs
a , 0 < Re s < Re a
sin x Γ (s) sin(
sπ2
), 0 < Re s < 1
cos x Γ (s) cos(
sπ2
), 0 < Re s < 1
H(a − x) as
s
400 An Introduction to Integral Transform
A9 : Mellin Transforms. (Continued)
f(x) M [f(x) ; x → s] =∫∞
0f(x) xs−1dx
(1 − x)p−1 H (1 − x) Γ (s)Γ(p)Γ(s+p) , Re p > 0
log (1 + x) πs cosec sπ
(1 − x)−1 π cot (sπ) , 0 < Re s < 1
xν H (1 − x) (s + ν)−1 , Re s > −Re ν
log x . H (a − x) 1sas
(log a − 1
s
), Re s > 0
{log x}(a+x) , |arg a| < π πas−1 cosec πs [log a − π cot πs] ,
0 < Re s < 1
xν log x H (1 − x) −(s + ν)−2, Re s > −Re ν
e−x(log x)n dn
dsn [Γ (s)] , Re s > 0
log |1 − x| πs−1 cot (πs), −1 < Re s < 0
log |1+x1−x | π
s tan πs2 , −1 < Re s < 1
Jν (ax) , (a > 0) 2s−1 Γ(
s2 + ν
2
)as Γ
(ν2 − s
2 + 1)
,
Re ν < Re s < 32
log axH(a − x) as
s2
x−1 log (1 + x) π (1 − s)−1 cosec πs
Appendix : Tables of Integral Transforms 401
A9 : Mellin Transforms. (Continued)
f(x) M [f(x) ; x → s] =∫∞
0f(x) xs−1dx
xa (1 + x)−b Γ(a + s) Γ (b − a − s)Γ(b)
(1 + ax)−n Γ(s) Γ(n−s)asΓ (x) , 0 < Re s < n
A10 : Kontorovich - Lebedev Transforms.
f(x) K [f(x) ; τ ] ≡ f(τ) =∫∞
0x−1 f(x) Kiτ (x) dx
1 π2τ cosech
(πτ2
)x π
2 sech(
πτ2
)e−x π
τ cosech (τπ)
ex πτ coth (τπ)
xe−x πτ cosech (τπ)
e−x cosh t π cos(τt)τ sinh(πτ)
x e−x cos t π sinh(τt)cosec tsinh(πτ)
x e−x cosh t π sin(τt)sinh(πτ) sinh t
402 An Introduction to Integral Transform
A11 : Mehler - Fock Transforms (of zero order)f(α) f∗
0 (τ) = Φ[f(α) ; α → τ ]
H(t − α) (ch t − ch α)−12
√2 τ−1 sin tτ
H(α − t) (ch α − ch t)−12
√2 τ−1 cth (πτ) cos(τ t)
(ch α + cos β)−1 π sech (τπ) P− 12+iτ (cos β), −π < β < π
(ch α + cos β)−12
√2
τch (βτ)sh (πτ) , π < β < π
sech (α2 ) 2 τ−1 cosech (πτ)
√sech (α) 1√
2τcosech
(π τ2
)
A11 : Another form of the Mehler-Fock Transforms.
f(x) f0(τ) = Φ0[f(x) ; x → τ ] =∫∞1 f(x) P− 1
2+iτ(x) dx
e−ax√
2πa Kiτ (a) , |arg a| < π
2
x− 12
√2 sech (π τ
2 )
x− 32
2√
23 τ cosech(πτ
2 )
1x+t πsech (πτ)P− 1
2+iτ (t), |t| < 1
1√x+t
√2 ch(τ cos−1 t)
τ sh (πτ)
Appendix : Tables of Integral Transforms 403
A12 : The Z - Transformsyn
∑ynzn
{1 , n = k0 , n �= k zk
c c1−z
n z(1−z)2
n2 z(1+z)(1−z)3
n3 z(z2+4z+1)(1−z)4
cn 1(1−cz)
ncn cz(1−cz)
(n + k
k
)cn 1
(1−cz)k+1
(b
n
)cn ab−n (a + cz)b
cn
n , (n = 1, 2, 3, · · ·) −ln(1 − cz)
cn
n , (n = 1, 3, 5, · · ·) 12 ln
(1+cz1−cz
)= tanh−1 (cz)
cn
n , (n = 2, 4, 6, · · ·) −12 ln
(1 − c2z2
)cn/n! ecz
cn/n! , (n = 1, 3, 5, · · ·) sinh (cz)
cn/n! , (n = 0, 2, 4, · · ·) cosh (cz)
(lnc)n/n! cn
404 An Introduction to Integral Transform
A12 : The Z - Transforms (Continued)
yn∑
ynzn
sin (cn) z sin cz2+1−2z cos c
cos(cn) 1−z cos cz2+1−2z cos c
b−an sin(cn) z sin cba+b−az2−2z cos c
b−an cos (cn) ba−z cos cb−az2+ba−2z cos c
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Company, NY.(21) Tricomi, F.G. (1951). On the finite Hilbert Transformation, Q.J.
Maths., Oxford, 2, 199-211(22) Tricomi, F.G. (1955). Integral Equations, Dover Publications, Inc,
NY.(23) Titchmarsh, E.C. (1959). An Introduction to the Theory of Fourier
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Index
AAbel’s integral equation, 196absolutely integrable, 2analytic, 154analytic continuation, 129Application to Mechanics, 177associated Legendre equation, 341associative law, 131asymptotic expansion, 228axisymmetric contact problem, 346
BBessel function, 127Bilateral Laplace Transform, 166Binomial theorem, 128Boundary value Problem, 55, 180Boundary Value Problem in a
wedge, 332branch point singularity, 154branch-cut, 154Bromwich integral, 153BVP, 170
CCanchy’s integral formula, 374Cauchy principal value sense, 220,
222Cauchy’s residue theorem, 36, 154Cauchy’s theorem, 312Change of Scale, 9, 107, 239, 279circular disc, 256Classes of functions, 2commutative law, 131complex form of Fourier series, 3complex Fourier transform, 56
complex Inverse formula, 156, 159concentrated load, 178contour, 37contour integral, 153Convolution, 20, 131, 324Convolution of generalised Mellin
transform, 297convolution theorem, 22, 150, 286,
324, 356convolution theorems for finite Fourier
transform, 82cosine integral, 124cosine transforms, 7
DDerivative of finite Laplace trans-
form, 309derivative of FT, 37Derivative of Mellin Transform, 281Difference, 197difference equations, 1, 198differential - difference equation, 197,
198differential equations, 1diffusion equation, 57, 89, 187Dirac delta function, 122Dirichlet’s conditions, 2distributive law, 131double finite Fourier cosine trans-
form, 85double finite Fourier transforms, 86double finite sine transform, 85Double Laplace Transform, 161Double Transforms of partial
derivatives, 86
408 book name
duplication formula, 127, 129
EEddington solution, 49elastic membrane, 97electrical network, 170, 175entire function, 314error complementary function, 126Error function, 126Estimated accuracy, 159even function, 8, 163existence, 103exponential, 104Exponential integral, 126exponential order, 2exterior Dirichlet problem, 326external crack, 335
FFalting, 20, 21Faltung, 20, 82faltung or resultant, 131Faltung Theorem, 21Fibonacci Sequence, 384final-value theorem, 116, 311finite Hankel transform, 268finite difference, 197finite Fourier cosine transform, 79,
80finite Fourier sine transform, 80finite Fourier transforms, 79Finite Hankel transform, 260, 262,
269finite Hankel transform of order n,
267finite Hilbert transform, 226Finite Laplace transform, 302Finite Laplace transform of Deriv-
atives, 308
Finite Laplace transform of Inte-grals, 309
Finite Mellin transform, 297finite sine transform, 80First shift theorem, 107flexible string, 190Fourier Integral Formula, 2, 4Fourier kernel, 79Fourier Series, 2Fourier sine transform, 7Fourier transform, 6, 7Fourier’s integral theorem, 75Fourier-Bessel Series, 260, 261Fredholm integral equation, 47, 51,
193freely hinged, 97freely supported ends, 179Functions of several variables, 53
GGamma function, 33, 127, 129Gegenbauer polynomial, 355Gegenbauer Transform, 355general evaluation technique of in-
verse Laplace transform, 153,156
Generalised Mellin transform, 295
HHankel inversion formula, 242Hankel transform, 238, 243harmonic, 254heat conduction equation, 88 91,
92, 181, 184heat conduction problem, 61Heaviside function, 121Heaviside unit step function, 13, 121,
198Hermite transform, 364
Index 409
Hermite transform of derivative ofa function, 367
heuristic approach, 141Hilbert transform, 220hinged end, 90
Iidentity operator, 7image function, 141improper function, 122infinite strip, 63infinite wedge, 277Initial Value, 311initial-boundary value
problems, 269initial-value problem, 169initial-value theorem, 116insulated surface, 60integral equation, 1, 47, 193integral equation of convolution
type, 47Integral of a function, 112Integral of finite Laplace
transform, 310integral of Laplace transformed
function, 115integral transforms, 1Integro-differntial equation, 193, 197interior Dirichlet problem, 325inverse finite cosine transform, 79inverse finite Hankel
transform, 262, 268inverse Fourier transform, 6, 7inverse Hilbert transform, 220, 221inverse Laplace transform, 103, 141Inverse Stieltjes
Transform, 232, 233, 234
inversion formula for, 243inversion formula for Legendre
transform, 318inversion of Z-transform, 373iterative Laplace transform, 166
JJacobi polynomials, 351Jacobi Transform, 351
Kkernel, 1, 193kinematic viscosity, 271Kontorovich-Lebedev transform, 328
LL’ Hospital’s rule, 312L-C-R circuit, 175Laguerre polynomial, 135, 158Laguerre polynomial due to McCully
Mittag-Leffler, 137, 360Laguerre transform, 359Laguerre transform of order zero,
362Laguerre transform of the deriva-
tive, 361Laplace equation, 291Laplace equation in a half space,
345Laplace inversion formula, 141Laplace transform, 103Legendre Transform, 317Lerch’s theorem, 141, 158linear superposition, 333Linearity property, 8, 104, 142, 279linearity property of Hankel trans-
form, 239Lipschitz condition, 302
410 book name
MMaclaurin series, 49mathematical induction, 114, 284mean value theorem, 123Mehler-Fock inversion, 349Mehler-Fock transform of
order m, 341Mehler-Fock transform of zero or-
der, 337Mellin Inversion theorem, 285Mellin transform, 278Mellin transform of derivative, 281Mellin transform of Integral, 283Modified Bessel function, 130modified Hilbert transform, 227Modulation theorem, 10moment of inertia, 179Multiple Finite Fourier
Transform, 85multiple Fourier transforms, 54
NNull function, 141
Oodd and even periodic extensions,
82odd function, 7, 164ODE with variable co-efficients, 173one-sided Hilbert transform, 228Operational Properties of
Z-transform, 376Operational Properties, 307, 323operators, 7, 103original function, 141orthogonality property, 260
PParseval relation, 23, 244, 329
Parseval type relation, 339Parseval’s identities, 24partial differential equation, 180partial fractions, 145penny-shaped crack, 347periodic extensions, 83periodic function, 115, 136, 152piecewise continuous, 2Poisson integral formula, 326power series expansion, 160principal period, 3problem of potential theory, 277properties of Legendre
transforms, 318
Qquarter plane, 70
RRational Functions, 36Riemann sum, 372Rodrigues formula, 364
Sscaling property, 308Second shift theorem, 107semi-infinite strip, 99shift theorem, 158shifting property, 10, 308simple poles, 3, 36simultaneous ODE, 171sine integral Si(t), 124singlevalued function, 2singular integral equation, 226singular points, 154singularities, 36, 37slowly convergent series, 293Solution of difference equations, 383Solution of Integral equations, 292
Index 411
Solution of ODE, 168special functions, 121steady state, 60steady state temperature, 67, 100step function, 13Stieltjes Transform, 230strip, 60Sufficient conditions, 103, 104Summation of Infinite Series, 385Summation of Series, 293
TThe convolution Theorem, 82, 379The Final value Theorem, 311, 380The Initial Value Theorem, 311, 380The inverse finite Laplace
transform, 303The Z-transform of the partial
derivatives of a function, 381thermal diffusivity, 184three dimensional Laplace
equation, 95to Eddington, 48toroidal co-ordinate system, 342transverse displacement, 189, 255Tricomi’s method, 158two-sided Hilbert transform, 227two-sided Laplace transform, 166
Uunit step function, 102
Vvibrating string, 58vibration, 90viscous fluid, 271Volterra integral equation, 47, 193
Wwave equation, 89, 255, 274
YYoung’s modulus, 179
ZZ - Transform , 372zeta function, 293