The Laplace Transformation

THE LAPLACE TRANSFORMATION

THE LAPLACETRANSFORMATION

by FRANCIS D. MURNAGHANConsultant, Applied Mathematics LaboratoryThe David Taylor Model Basin

SPARTAN BOOKS WASHINGTON, D. C. 1962

Library of Congress Catalog Card No. 62-19096

Copyright Q 1962 by Fsnncas D. MURNAGHAN

Printed in the United States of AmericaAll rights reserved

This book or parts thereof, may not be reproduced in any formwithout permission of the publishers

LECTURES ON APPLIED MATHEMATICS

VOLUME I

Volume I: The Laplace Transformation

Volume II: The Calculus of Variations

Volume III: Unitary and Rotation Groups

Preface

This book is based on lectures, given at the Applied Mathematics Labora-tory of the David Taylor Model Basin. It is devoted to the Laplace Trans-formation and its application to linear ordinary differential equations withvariable coefficients, to linear partial differential equations, with two inde-pendent variables and constant coefficients, and to the determination ofasymptotic series. The treatment of the Laplace Transformation is basedon the Fourier Integral Theorem and the ordinary differential equationsselected for detailed treatment are those of Laguerre and Bessel. Thepartial differential equation governing the motion of a tightly stretchedvibrating string and a generalization of this equation are fully treated.Asymptotic series for the integral f exp (-z') dz and for the modifiedBessel function I,.(p), } arg p I < (r/2), are obtained by means of theLaplace Transformation and, finally, asymptotic series useful in the calcu-lation of the ordinary Bessel functions J (t) are treated.

Care has been taken to make the treatment self-contained, and detailsof the proofs of the basic mathematical theorems are given..

Washington, D. C. FwAxcis D. MUENAGRANApril, 1962

Contents

Preface ................................................... vii1. Absolutely Integrable Piecewise Continuous Functions.......... 1

2. The Fourier Transform of an Absolutely Integrable Piecewise Con-tinuous Function .......................................... 5

3. The Fourier Integral Theorem ............................... 94. Completion of The Proof of The Fourier Integral Theorem. The

Laplace Version of The Fourier Integral Theorem .............. 155. The Laplace Transform of a Right-sided Function .............. 226. The Laplace Transform of exp (- t2) .......................... 287. The Laplace Transform of the Product of a Right-sided Function

by t and of the Integral of a Right-sided Function over the Inter-val [0,1] .................................................. 35

8. Functions of Exponential Type .............................. 409. The Characterization of Functions of Exponential Type......... 45

10. The Polynomials of Laguerre ................................ 5111. Bessel's Differential Equation ................................ 5712. The Recurrence and other Relations Connecting Bessel Functions. 6513. The Problem of the Vibrating String ......................... 7514. The Solution of the Problem of the Vibrating String............ 8115. The Generalized Vibrating String Problem .................... 8716. The Solution of the Generalized Vibrating String Problem...... 9417. The Asymptotic Series for j; exp (-0) dz .................... 9918. The Asymptotic Series for (27p)i exp (- p)I (p), arg p 1 < it/2,

The Hankel Functions ...................................... 106

19. The Asymptotic Series for and 114Bibliography .............................................. 124

Index .................................................... 125

1

Absolutely Integrable

Piecewise Continuous Functions

Let f (t) be a complex-valued function of the unrestricted real variablet, - oo < t < eo, it being understood that real-valued functions of t areincluded in the class of complex-valued functions of t, a real-valued func-tion being a complex-valued function whose imaginary part is identicallyzero. The class of continuous functions is too restricted for our purposeand we shall merely suppose that the number of points of discontinuityof f(t), if any such exist, in any finite interval is finite. This will be thecase if the number of points of discontinuity of f(t) is finite, but this suffi-cient condition is not necessary; for example, f(t) may be discontinuousfor all integral values of t, or it may be a periodic function, of period T,with a finite number of points of discontinuity in the interval 0 S t < T.When f(t) possesses not more than a finite number of points of discon-tinuity in any finite interval we shall say that it possesses Property 1 andwe shall term any function f(t) which possesses Property 1 a piecewisecontinuous function.

In enlarging the class of functions which we propose to consider fromcontinuous to pieoewise continuous functions, we lose some of the mostconvenient properties of the class of continuous functions. Although everycontinuous function is bounded over every finite closed interval, the sameis not true for pieoewise continuous functions. For example, the functionf (t) which is equal to t-' if t p 0 and which is assigned any value at t - 0(the particular value assigned to it at t - 0 being immaterial) is pieoewisecontinuous, since it has only one point of dsoontinuity, but it is not boundedover any interval which contains the point t - 0. Furthermore, every

1

$ Lectures on Applied Mathematics

continuous function is integrable, in the sense of Riemann, over any closedinterval a < t < b, but this is not necessarily true for a piecewise con-tinuous function if the interval contains a point of discontinuity of thefunction at which the function is unbounded. If c is such a point of dis-continuity of f(t) in the interval a < t < b, and if f(t) is continuous atall other points of this interval, we say that f (t) is improperly integrable,in the sense of Riemann, over the interval a < t < b if the function F(61, 86)of the two non-negative variables 81 and 8, which is furnished by the sumof the integrals of f(t) over the intervals a < t < c - 81 and c + dh < t < b,where 81 < c - a and ds < b - c, possesses a limit as 81 and 8: tend, in-dependently of each other, to zero. We term this limit the improper integralof f(t) over the interval a < t < b. (Note. We assume that the non-negativevariables 81 and 4 are actually positive unless c = a, in which case 61 = 0and 8s is positive, or c = b, in which case 8, = 0 and 81 is positive. Thereis no lack of generality in assuming that f (t) has only one point of dis-continuity, at which it is unbounded, in the interval a < t < b, since thenumber of its points of discontinuity in this interval is, by hypothesis,finite; if it has more than one point of discontinuity at which it is un-bounded in the interval a < t < b, we can break this interval down intoa number of subintervals each containing only one point of discontinuityof f(t), at which it is unbounded, and define the improper integral. of f(t)over the interval a < t < b as the sum of the improper integrals of f(t)over the subintervals, assuming that each of these improper integralsexists. If a single one of these improper integrals fails to exist, the im-proper integral of f(t) over the interval a < t < b fails to exist.)

The function f(t) of the unrestricted real variable t which is 0 if t < 0and is 0 if t > 0, the value assigned to f (t) at t = 0 being immaterial, isimproperly integrable over any interval a < t < b which contains thepoint t = 0. For example, the improper integral of f (t) over the interval0 < t < b is 20. On the other hand, the function f(t) which is 0 if t < 0and is t' if t > 0, the value assigned to f(t) at t - 0 being, again, im-material, is not improperly integrable over any interval which containsthe point t = 0. In general the point t = 0 is a point of discontinuity off(t) = t°, t > 0, a real, f(t) = 0, t < 0, at which f(t) is unbounded, ifa < 0. If a > -1, this piecewise continuous function is improperly in-tegrable over any interval a < t < b which contains the point t = 0 and,if a < -1, f (t) is not improperly integrable over any such interval.

If f(t) is a piecewise continuous function, so also is I f(t) 1, and it iseasy to see that if I AO I is improperly integrable over an an intervala < t < b which contains a single point c of discontinuity of f(t) at whichf(t) is unbounded, then f(t) is also improperly integrable over the intervala < t < b. Indeed to prove this we must show that each of the two in-tegrals fcc=a;' f(t) dt and fa+e;. f(t) dt where 0 < Si' < 81 and 0 < 8,' <

The Laplace Transformation 3

52, may be made arbitrarily small by making 61 and &, respectively,sufficiently small. The moduli of these integrals are dominated by, i.e.,

are not greater than, I At) I dt and f +i; I At) I dt, respectively, andour hypothesis that f(t) is absolutely integrable (i.e., that I f(t) I is in-tegrable) over a < t < b assures us that each of these two dominatingnumbers may be made arbitrarily small by merely making Sl and &,respectively, sufficiently small. We shall assume that our piecewise con-tinuous functions f(t) are such that I f(t) I is improperly integrable overany finite interval a s t < b which implies, as we have just seen, thatf (t) is improperly integrable over any. finite interval a < t < b. (Note.If the interval a < t < b does not contain a point of discontinuity of f(t)at which f (t) is unbounded, both f (t) and I f (t) I are properly integrableover the interval a < t < b, since they are bounded over this interval andcontinuous save possibly for a finite number of points.)

We now make a final assumption concerning the class of complex-valuedfunctions of the unrestricted real variable t which we propose to consider.We assume that not only is I f (t) I integrable, properly or improperly, overevery finite interval a < t < b but that the function F(a, b) = fa I f(t) I diof the two real variables a and b possesses a finite limit as a and b tend,independently, to - ao and + ao , respectively. When this is the case wesay that f(t) is absolutely integrable over - co < t < co and we term thelimit of F(a, b), as a -+ - co and b --), co, the integral of I f(t) I from - coto cc, this integral being denoted by the symbol f-'Qo I f (t) I dt. When acomplex-valued function f (t) of the unrestricted real variable t is such thatit is absolutely integrable over - co < t < oo we say that it possessesProperty 2. The functions f (t) which we propose to consider are thosethat possess both Property 1 and Property 2; in other words, they arepiecewise continuous functions which are absolutely integrable over-c <t< Co.

It is clear that if f (t) possesses Properties 1 and 2, then the functionfb f(t) dt of the two real variables a and b possesses a finite limit as a andb tend, independently, to - ao and + co, respectively. Indeed, in order toprove this we have to show that f e f (t) dt and f', f (t) dt, where b' > band a' < a, may be made arbitrarily small by making b and -a sufficientlylarge. However I f b f (t) dt I and I f;. f (t) dt I are dominated byf bbl

I

f (t) I dt and f:, I f (t) I dt, respectively, and each of these dominatingnumbers may be made arbitrarily small by making b and -a, respectively,sufficiently large (since f (t) is, by hypothesis, absolutely integrable over- so < t < ao) . Thus f% f (t) dt exists. The converse of this result isnot true; f'° f (t) dt may well exist without f I f (t) I dt existing. Anexample is furnished by the everywhere continuous function f(t) =(sin t) It, t 0 0, f (O) = 1. f (t) is an even function and so it suffices toconsider fu I(sin t)/tIdt. If nr < b < (n + 1)r, n = 0, 1, 2, - , we have

4 Lectures on Applied Mathematics

f'sintfsintt t

+ f2w sin tdtT t

+J

wY sin t dt + f ° on tdtt wf t

=I1-_Il+... t'(-1}w-lTw+ 81n1 dll fir t

where Il = fo ( (sin t)/t} dt > 0, Is =-f!' ((sin t)/t dt) > 0 and so on.On writing t = u + v in the formula for 12 we have

Is = f o { (sin u) /(u + r) } du,

so that Is < I1. Similarly Is < Is , 14 < Is and, generally, Iw+1 < I.,n - 1, 2, 3, . In order to appraise the integral f' . ( (sin t)/t} dt we usethe second theorem of the mean of integral calculus which tells us that,since 1/t is monotone decreasing over nx < t < b and continuous at t =nr, ft. {(sin t)/t} dt = (1/nr) f;,, sintdt = (cosnr - cosc)/nr, nr < c < b.Thus J ft. . { (sin t)/t} dt J < 2/nv and this may be made arbitrarily small,no matter what is the value of b > nr, by merely making n sufficientlylarge. In particular may be made arbitrarily small by making n suffi-ciently large so that the alternating infinite series I1 - Is + Is - isconvergent, its sum being the infinite integral fo ( (sin t)/t} dt. It followsthat this infinite integral, whose existence we have just proven, lies betweenIl - I1 and I1 . A simple application of Simpson's Rule shows that I, _1.86, 12 = 0.44, approximately, so that f o' { (sin t) It } dt lies between 1.42(approximately) and 1.86 (approximately) and this implies that theintegral of (sin t) /t over - co < t < co exists and lies between 2.84 (ap-proximately) and 3.72 (approximately)- We shall shortly see thatf°m { (sin t) It) dt = r. That (sin t) It is not absolutely integrable over - Co <t < w is clear since f 'o { J sin t J/t } dt = I1 + Is + + I. + f,. { J sin t I It } dtand Il > 1/r Jo sin t dt = 2/r, Is > (/2r) 5o' sin u du = 2/2r and, gen-erally, I. > 2/nr so that f o { J sin t J/t} dt > (2/r) (1 + + -}- (1/n)),n = 1, 2, . Thus, since the partial sums of the infinite series 1 + j +} + are unbounded, f o { ( sin t J/t } dt may be made arbitrarily large bymaking b sufficiently large and the infinite integral fo I (sin t J/t} dt doesnot exist. This implies that I (sin t) J/ t is not integrable over - co < t < eo.

EXMCLSE

Show that if co is any real number the infinite integral f"m { (sin wt)/t}dt exists, its value being C, 0, - C according as w > 0, to = 0, w < 0, re-spectively, where C is the value of the infinite integral ft. { (sin t)/t}dt.

2

The Fourier Transform

of an Absolutely Integrable

Piecewise Continuous Function

If f(t) is a complex-valued function of the unrestricted real variable twhich possesses Properties 1 and 2, so that it is piecewise continuous andabsolutely integrable over - co < t < co, it does not lose these propertieson multiplication by exp(-iwt), w any real number. Indeed exp(-iwt) iseverywhere continuous so that f (t) exp (-iwt) is piecewise continuous andI exp(-iwt) I = 1 so that I f(t) exp(-iwt) I = I f(t) 1. since f(t) exp(--iwt)is absolutely integrable over - c < i < oo, the infinite integral f-f(t)exp(-iwt) de exists for each value of co. Introducing, for our later con-venience, the numerical factor (2r)-4 we set

g(w) = (2r)-i f(t) exp (-iwt) dt

and we term p(w) the Fourier Transform of f(t). Since

f` f(t) exp (-iwt) dt <- f (f(g) exp (-iwt)1 dt

= f f(t)Idi :5 f(t)Idtwe have I f_'» f(t) exp(-iwt) dt 1 < f"-. I f(t) I dt, so that

(Q(w) ( <- (2r)-4 I f(t) I dte

5


Thus g(w) is bounded over - co < w < co . It is easy to see thatf° f(t) exp( -iwt) dt is an everywhere continuous function of w, no matterwhat is the interval a < t < b. To prove this we first consider the casewhere (f(t) ( is bounded, say <M, over a < t < b (so that the integralfa f(t) exp(-iwt) dt is a proper Riemann integral). If w and w + Aw areany two real numbers we have

e

skff()exP(_ttfb

f (t)exp(-iwt)exp(-iw.t) -1dt11

< M f b (1 exp (--itlw.t) - 1 } (dt0

Since the function exp z of the complex variable z is continuous at z = 0,where it has the value 1, we can make ( exp (-iAw.t) - 1 I arbitrarilysmall, say <e/M(b - a), by making -iAw.t ( sufficiently small, say<6, a being an arbitrarily assigned positive number. Denoting, for amoment, by a the greater of the two numbers (a ( and ( b (, ( -iAw.t ( < aover the interval a < t < 6 if ( -iAw.a i < a, i.e., if (Aw I < (a/a); hence,

lAfbf(t) exp (-iwt) dt < e if I Aw < (8/a)

so that f .b f(t) exp (- iwt) dt is an everywhere continuous function of w.If f(t) is not bounded over a < t < b it suffices to consider the case wheref(t) has a single point of dscontinuity c, at which it is unbounded, in theinterval a < t < b. We write f;':f(t) exp (-iwt) dt in the form

f f(t)exp(-iwt)dt+f_ f(t)exp(-iwt)dta

+ fd

f(t) exp (-iwl) dt,e+t:

where al and a2 are any two positive numbers which are less than c - aand b - c, respectively, save when c = a, in which case al = 0 and a2 isany positive number <b - a, or when c = b, in which case a2 = 0, and6, is any positive number <b - a. It suffices to consider the first case,where a < c < b, the argument in the other two cases being precisely thesame. The integrals fa' f(t) exp (-iwt) dt and f.+#, f(t) exp (-iwt) dtare everywhere continuous functions of co, since f(t) is continuous, byhypothesis, over the intervals a < t < c - a1 and c + a2 < t < b, andso we have merely to consider the integral f,' f(t) exp (-iwt) dt. Themodulus of this integral, being dominated by f a±a; (f(t) (dt, may be madearbitrarily small, say < e, since f (t) possesses Property 2, by making a, anda2 sufficiently small, the choice of a1 and a: being independent of w. Once


this choice of 62 and 62 has been made, it follows that the modulusof A f0+a; f(t) exp (-iwt) dt is less than 2E, no matter what are the valuesassigned tow and Aw. Hence I A f; f(t) exp (-iwt) dt ! may be madearbitrarily small, no matter what is the value of w, by making ! Aw I suffi-ciently small so that f a f (t) exp (-iwt) dt is an everywhere continuousfunction of w. This implies, in view of the fact that f(t) is, by hypothesis,absolutely integrable over - oo < t < -, that the infinite integralf% f(t) exp (-iwt) dt is an everywhere continuous function of w; indeed,the modulus of the difference between f f(t) exp (-iwt) dt andf a f (t) exp (- iwt) dt is dominated by the sum of the two infinite integralsf ! f (t) I dt, f b j f(t) ! dt and this sum may be made arbitrarily small,say < E, by making a negative and b positive and choosing -a and bsufficiently large, the choice of a and b being independent of w. Once thischoice of a and b has been made, it follows that the modulus ofA(f f (t) exp (- iwt) di + f' f (t) exp (-iwt) dt) is less that 2e nomatter what are the values assigned to w and to + Aw. Hence! A f f (t) exp (-iwt) dt ! may be made arbitrarily small, no matterwhat is the value of w, by making ! Aw I sufficiently small, so thatf"° f(t) exp (-iwt) dt is an everywhere continuous function of co. This,combined with the boundedness of g(w) over - co < w < 00, is our firstresult which may be stated as follows:

The Fourier Transform g(w) = (2w) - f°° f(t) exp (-iwt) dt of anypiecewise continuous complex-valued function f(t) of the unrestricted realvariable t which is absolutely integrable over - oo < t < co is an every-where continuous function of the unrestricted real variable w; moreover,g(w) is bounded over - -o < w < oo.

EXAMPLE I

Let f(t) = 0 if t < -b and if t > b, where b is any positive real number,and let f (t) = 1 if - b < t < b, the values assigned to f (t) when t = - band when t = b being immaterial. Then

g(w) _ (A) sin b``' , if w 0, while g(0) = (2)} br(Note. The above Example shows that, while the Fourier Transform

operation is a smoothing or strengthening one as far as Property 1 is con-cerned, (g(w) being everywhere continuous and bounded over - oo <w < oo while f (t) may be only piecewise continuous and may not bebounded over - oo < I < -o), it is a roughening, or weakening, one asfar as Property 2 is concerned: g(w) may not be absolutely integrable over- oo < w < . In the present example g(w) is integrable, in the Riemannsense, over - ao < w < co, but the following Example shows that g(w)may not be integrable over - oo < w < oo. )


Ex tupii 2Let f(t) - 0 if t < 0 and = exp (izt), where z = x + iy is any complex

number whose imaginary part y is positive, if t > 0, the value assigned tof(t) at t - 0 being immaterial. The function f(t) possesses Property 1,since its only point of discontinuity is t - 0, and it possesses Property 2,i.e., it is absolutely integrable over - ao < t < ao, since I f(t) I - 0 ift < 0 and I f(t) I = exp (-yt) if t > 0 and t' exp (-yt) is arbitrarilysmall, say < 1, if t is sufficiently large since y is, by hypothesis, positive.The Fourier Transform of f(t) is

g(w) _ (2r)1 £ exp (-i(w - z)t] dt = 1(2r)ii(w - z)and, since Iw - zI IwI + IzI < 2Iwl,if IwI > Iz1,wehave

g(w) > (2r)i2 w j' if IwI > I z I

so that g(w) I is not integrable over - co < w < co. Also, since log (w - z)is unbounded at ca - ao and at w - co, g(w) is not integrable over- oo < w < ao .

(Note. We shall see shortly that if f(t), in addition to possessing Prop-erties 1 and 2, is such that its real and imaginary parts are monotone oversufficiently small intervals to the right and to the left of t = 0, or if f(t)possesses a right-hand and a left-hand derivative at t - 0, then the Cauchyprincipal value, Um.-. f%, g(w) dw of the integral of g(w) over - co <co < ao exists, despite the fact that the integral of g(w) over - co < w < comay not exist. We shall denote this Cauchy principal value of the integralof g(w) over - ao < w < co by the symbol f g(w) dw. Thus in thepresent Example we have

I`p3 g(w) dw = 1 lim dw a 1 lim log (w - z)Iii(2r)

ii « w - z (2r) 'since the argument of (a - z) / (- a - z) tends to r as a - co while themodulus of (a - z)/(-a - z) tends to 1 as a -- co.)

3

The Fourier Integral Theorem

Let f(t) s fl(t) + 'f,(') be a complex-valued function, whose real andimaginary parts are fi(t) and f:(t), respectively, of the unrestricted realvariable t, and let f(t) possess Properties 1 and 2. At any point t at whichf(t) is continuous the two limits

f(t+0) - limf(t+a), a> 0"-.o

f(t-0)=limf(t-a), a>08.0

exist and are equal, their common value being f(t) (this being the definitionof the concept of continuity). At a point t where f(t) fails to be continuousthe limits f(t + 0), f(t -- 0) need not exist, and, if they do exist, theyneed not be f(t). However, these limits will exist, by hypothesis, if f(t)possesses a right-hand derivative and a left-hand derivative at t,the definition of the right-hand derivative, for example, beinglim,..o V(9 + $) - .f (t + 0))/6, a > 0; they will exist also if the real andimaginary parts, fl(g) and fi(t), respectively, of f(t) are monotone andbounded over sufficiently small intervals to the right and to the left oft, f(t + 0), for example, being fi(t + 0) + of2(t + 0) where fi(t + 0),for example, is the greatest lower bound, or least upper bound, of fi(t + 6),a > 0 and sufficiently small, according as fl(t) is monotone nondecreasing,or monotone nonincreasing, over a sufficiently small interval to the rightof t. The values assigned to f (t) at its points of discontinuity are immaterial,as far as the definition of the Fourier Transform g(w) of f(t) is concerned,but for the purposes of the Fourier Integral Theorem, which we now pro-pose to study, it is convenient to assign to f(t) at any of its points of dis-

8


continuity at which both the limits f(t + 0) and f(t - 0) exist the meanof these two limits; i.e., we set

.f(t) = k(f(t + 0) -f- At - 0)1For example, if f(t) = 0, t < 0, and f(t) = exp (izt), t > 0, where theimaginary part y of z = x + yi is positive, we have f(0 - 0) = 0,f(0 + 0) = 1, and so we set f(0) = J. We have seen that, for this par-ticular function, f g(w) dw = (r/2)*, so that (2,r)' f .., g(w) dro4 = f(0). This result is not an accident, peculiar to this particular function;if f(t) is any complex-valued function of the unrestricted real variable t,which possesses Properties 1 and 2, and which, in addition, is such thatits real and imaginary parts are monotone and bounded over sufficientlysmall intervals to the right and to the left of t = 0, or such thatit possesses a right-hand and a left-hand derivative at t = 0, then(2r)-' f g(w) dw exists, its value being f(0), on the understandingthat f (O) is defined as the mean of the two limits f (O + 0), f (O --- 0). TheFourier Integral Theorem is merely the extension of this result fromt = 0 to an arbitrary value t = r of the unrestricted real variable I. Tomake this extension we must multiply g(w) by exp (iwr) before takingthe Cauchy principal value of the integral over - co < w < -. Thusthe Fourier Integral Theorem (in the form in which we propose to proveit and which is satisfactory for our purposes) may be stated as follows:

Let f(t) be any complex-valued piecewise continuous function of theunrestricted real variable t which is absolutely integrable over - co <t < co, and let r be any value of t at which f (t) possesses either of thetwo properties:

(a) The real and imaginary parts of f(t) are monotone and boundedover sufficiently small intervals to the right and to the left of r.

(b) f (t) possesses a right-hand and a left-hand derivative at r.Then

(2r)-J

g(w) exp (iwr) dwc-m)

exists with the value f(r) (on the understanding that f(r) is defined asthe mean of the two limits f(r + 0), f(r - 0)).

(Note. Pay attention to the fact that g(w) is multiplied by exp (iwr)and not by exp ( -iwr) while, in the definition of g(w), f(t) was multipliedby exp (- iwt) and not by exp (iwt). Furthermore, note that mere con-tinuity of f(t) at t = r does not suffice for the validity of the FourierIntegral Theorem. When f (t) is continuous at t = r both the limits f (,r + 0)and f(r - 0) exist and are equal, their common value being f(r), but ourproof of the Fourier Integral Theorem requires either Property (a) or


Property (b) above and these are not guaranteed by mere continuity off(t) at t = T. If f(t), in addition to being continuous at t = r, is differ-entiable at t = r it possesses Property (b), with the added equality of theright-hand and left-hand derivatives, and the Fourier Integral Theorem isvalid at t = r.

We begin the proof of the Fourier Integral Theorem by writingb

(2r)-4 f f(t) exp (-iwt) dt = gab(w) so that b

aI 9(w) - 9 (w)

may be made arbitrarily small, say < e, by choosing the positive numbers-a and b to be sufficiently large, the choice of a and b being independentof w. If r is any real number and a any positive real number we denote(2r)-4 f_`a g(w) exp (irw) dw, which exists since g(w) is everywhere con-tinuous, by Fa(r) so that

I Fa(r) - (2r)-1f a

ga (w) exp (irw) dwa

_ (2r)If [g(w) -ga(w)]exp(irw)dwla

(2w)'I

g(w) - ga (w) (dw < (2a)- 2a<a

on the understanding that -a and b have been chosen sufficiently largeto ensure that I g(w) - gab(w) 1 < e, w arbitrary. Thus the differencebetween Fa( r) and (2r) -4 f fa gab (w) exp (irw) dw may be made arbi-trarily small, once the positive number a is given, by making -a andb sufficiently large. The integral (2r)-4 f"a gab( w) exp (irw) dw is the re-peated integral (2r) -' f [ P. f(t) exp (-iwt) dt[ exp (irw) dw and weconsider the associated double integral of (2r)-'f(t) exp [-iw(t - r)jover the rectangle a < t < b, -a <- w < a. If f(t) is continuous over theinterval a < t < b, the integrand of this double integral is a continuousfunction of the variables (t, w) over the rectangle of integration and theorder of integration in the repeated integral may be changed. This changeof order is also valid when f(t) is not continuous over a <- t < b by virtueof the fact that f(t) is, by hypothesis, piecewise continuous and absolutelyintegrable over a < t < b. To see this it is sufficient to consider the casewhere f (t) has a single point c of discontinuity in the interval a < t < b,c being an interior point of this interval. Writing f; f(t) exp (-iwi) dt inthe form

[f(t) exp (-iwt) dt + f f(t) exp (-iwt) di

+ f f(t) exp (-iwt)dt,e+a=


where a, and a, are any positive numbers which are less than c - a andb - c, respectively, we have to consider three repeated integrals whoseassociated double integrals are extended over the rectangles a < t < c -ai,-a<to <a;c-a,<t<c+32,-a<w<a;c+a:<t<b,-a < w < a, respectively, and, since (2r)-f(t) exp [-iw(t - r)] is acontinuous function of the two variables (t, w) over the first and third ofthese rectangles, the order of integration in the first and third of theserepeated integrals may be changed. The modulus of the second of ourrepeated integrals may, since A0 is absolutely integrable over a < t < b,be made arbitrarily small, say <a, by taking a, and a, sufficiently small.Thus the difference between (2ar)-1 f` exp (irw){ J f(t) exp (--iwt) 0 dwand the sum of the two repeated integrals

(tar)-, f(t) exp [--iw(t - r)] dw dt

and

(2r)-1 [1(t) {1. exp [-iw(t - r)] dwj dt

may be made arbitrarily small by choosing 6, and a: to be sufficiently small.Furthermore, since I exp [-iw(t - r)] I - 1, so that

aexp[-iw(t - r)]dwI :52a,

the product of f-. exp [-iw(t - r)J dw by f( t) is integrable over the in-tervals a < t < c and c < t <( b and the differences between

(tar)-1 f f(t){fexp[-iw(t - r)]dwl dtand

and between

and

(21r)-, f ~'` f(t) exp r)] dw j dt

(2w)-1 [f(t {L: exp [--tw(t - r)] dw} dt

(2r)-' J,iy f(t) {. atop -iw(t - r) dw dt

may be made arbitrarily small by choosing a, and a, , respectively, to besufficiently small. Thus the difference between

(tar)-'1.

exp (iI.) Vot) exp (-it) dtj dw

The Lapiace Transformation i3

and the sum of the two repeated integrals

and

(24r)-' f e f(t) {L exp [-iw(t - r)] dw) dto a 1

()-1 f(t)1ta exp -iw(t - r) dwr dt

may be made arbitrarily small by choosing dl and as to be sufficiently small.Since this difference is independent of dj and 8s it must be zero and so

(2fr)-'La

exp (izw) f(t) exp (-iwt) dt} dw

(24r)-1J a f(t) exp [-iw(t -- r)] dw dt

+ (21r)-1 [ f(t) {f: exp [-iw(t - T)] dw1 di

(2r)-' f1t{fa

exp [--iw(t -- r)] dwj CU

which proves the legitimacy of the interchange of the order of integrationin the repeated integral

(,.)-' La exp (irw) { f f(t) exp (-iwt) dt} dw

even when f (f) is not continuous over the interval a < t < b. We have,then, proved that the difference between Fa(r) and

}(2ir)-' f f(t) { to exp [-uu(t - r)] dwj dt

may be made arbitrarily small, once the positive real number a is given,by choosing the numbers -a and b to be positive and sufficiently large.In other words, the infinite integral

(2x)-' `tl f(f) { L exp -iw(t -- r) dw} dt

exists, no matter what is the positive number a, its value being Fa(r).Since

Q exp [-iw(t - r)] dw = 2 sin a(t - r)a t - T

this result may be stated as follows:


T h e infinite integral (1/i) f ., f(t) ([sin a(t - r)]/[t - r)) dt exists,no matter what is the positive number a, its value being

F. (r) = (2w)' J

a

9(w) exp (irw) dw

This is the first, and most crucial, step in the proof of the Fourier IntegralTheorem. In the next chapter we shall complete the proof of this theoremby showing that, if f (t) is either (a) such that its real and imaginary partsare monotone and bounded over sufficiently small intervals to the rightand to the left of r or (b) such that it possesses a right-hand and a left-hand derivative at r, then lima-. Fa(r) exists with the value f(r), itbeing understood that, if r is a point of discontinuity of f(t), the valueassigned to f(t) at t = r is the mean of the two limits f (r + 0) and f(7 -- 0).

4

Completion of The Proof of

The Fourier Integral Theorem.

The Laplace Version of

The Fourier Integral Theorem.

We now examine the behavior of the infinite integral

1/ir f f(t){[sin a(t - r)]/[t - r]] dt

as a -- oo . The function f (t) is a complex-valued function fl (t) + tf=(t )of the unrestricted real variable t and so this infinite integral is the sumof the two infinite integrals

1 f nft (4)sina(t-r)dtx _ t - r

and

i f°1 (t) sina(t - r)dtr . t-r

It suffices to treat the first of these two infinite integrals, the treatment of thesecond being precisely the same. We write (1/.Y) f°_° fl(t) [ [sin a(t - r)

16


It - 71) dt as the sum of the two infinite integrals

li 1 f-' A(t) sin a(t - T) dt., t - r

s o fl(r u)82n au dti;t1 = T - tf U

fl(t)sin a(s

_T

dtI= = I

TOV t - r4f fi(r+v)sin aydv;v=t-r

and it again suffices to treat the first of these two infinite integrals, thetreatment of the second being precisely the same. We write Il as the sumof the three integrals

s

J1 = -f fi(r - u)suaudu;

J:=s Jbfi(r-u)n°`udu;U

J,= i fi(r - u)sinaudu

where a and b are any two positive numbers which are such that a < b.It is clear that I Js 1 < (1/ib) Je }fi(r -- u) J du - (1/orb) F--.b J fi(t) I dt <(1/,rb) f°.o fi(t) I dt, so that I Js I may be made arbitrarily small, say< e, by choosing b sufficiently large, the choice of b being independent ofa. If A(t) is monotone and bounded over a sufficiently small interval tothe left of t - r, fi(r - 0) exists and

Jl - h(r- 0) ]" sin au du j{fi(r - u) - fi(r - 0)} sin au du

If a is sufficiently small, the function fi(r - u) - fi(r - 0) of u is mono-tone over the interval 0 < u < a, being either positive and nondecreasing(when fl(t) is monotone nonincreasing to the left of t = r) or negativeand nonincreasing (when fl(t) is monotone nondecreasing to the left of( = r). Hence we may apply the second Theorem of the Mean of integralcalculus to obtain

J1 - Mr- 0) ('° sin au du = - jfi(r - a + 0) - fi(r - 0) f s uau du

where a' is some positive number` <a. The integral f . ( (sin au)/u} du is


the difference of the two integrals r a { (sin au)/u} du = fee { (sin t)/t} dt, t =au, and f o { (sun au) /u) du - f e"' { (sin t) It) dt; each of these integralsis dominated, no matter what are the values of the positive numbers a',a and a by the number f o { (sin t)/t }dt. HenceJtand this assures us, in view of the existence of the limit fi(r - 0), that

lJl - fi(r - 0) f"'duu

may be made arbitrarily small, say < e, by choosing a sufficiently small,the choice of a being independent of a. If fi(t) possesses a left-hand deriva-tive, d, say, at t = r, the function

IA(T -- u) -A(r - 0) - dl-u {

is arbitrarily small, say < 1, over the interval 0 _< u < a if a is sufficientlysmall, and so

I'- u) - fI(r 0)i (d) +, 1 over 0 < u < au

if a is sufficiently small, the choice of a being independent of a. Hence

f {fi(r - u) - Mr - 0) } sin °ru dul < (tdl + 1) a,

if a is sufficiently small, so that

-A(r - 0) f sin au dul

= To u

may be made arbitrarily small, say <*, by choosing a sufficiently small,the choice of a being, again, independent of a. Supposing, then, that aand 6 are so chosen that, for all values of the positive real number a,

(1) lJ,l < e

(2) l Jl _ fi(r7

0) j'dtsf < e

we have, for every a > 0,

lJ + -fi(rr 0) f` sin au dul < 2e


The integral

sin au du sin vdv' v = au

u V

and as a -i oo, a remaining fixed, this tends to fo f (sin v)/v} do whosevalue we denote, for the moment, by C. Thus

I fi(r 0) A° sin au du _ fi(r _. 0)C1

may be made arbitrarily small, say < e, by choosing a sufficiently largeso that

IJ,+J:-f'(r-0)CI<3eif a is sufficiently large.

It remains to investigate the behaviorof J2 = (1/r) f; fi(r - u) f ((sin au)/u]) du as a --> eo. If fi(r - u) is bounded over a < u < b, so also isfl(r - u)/u which we denote, for a moment, by h(u), so that h(u) isbounded and, hence, since it is piecewise continuous, properly integrableover the interval a < u < b. If, then, e' is an arbitrarily given positivenumber we may construct a net of points a = uo < ui < . < ua - bon the interval a < u < b with the following property: Let h*(u) be thefunction defined by setting, over any open cell u, < u < ut+4 , j - 0, ,

n - 1, of the net, h'°(i) equal to the greatest lower bound m; of h(u) overthe corresponding closed cell u; < u < u;44 and setting, at the pointsa, ul , , u,, , b of the net, h'(u) equal to the greatest lower bound ofh(u) over a < u < b. Then h(u) - h'(u) > 0 over a < u < b and thenet a, u2 , , u.-t , b can be so chosen that 0 < f; f h(u) k*(u)) du <e'. Since I sin au I < 1, it follows that I f; f h(u) _ h*(u)} sin (au) du I < e'and it is easy to see that I f: h*(u) sin (au) du I may be made arbitrarilysmall, say <e', by choosing the positive number a sufficiently large. In-deed fa h'(u) sin (au) du is the sum of n terms of the form

"f+1 cos (au;) - cos (auj.,,)m;1 sin(au)du=m;k, a

so that

IJ e(u) sin (au) du 2a a j_o

Hence

f. h(u) sin (au) du 1 :5 2c'

if a is sufficiently large. This result remains valid even when f'(r - u)

The Laplaee Transformation 19

fails to be bounded over a < u < b; to show this, it suffices to considerthe case where fi(r - u) is unbounded at a single interior point c of theinterval a < u < b. Writing

f b h(u) sin (au) du = f ' h(u) sin (au) dus s

+ f+2h(u) sin (au) du -I- fb

h(u) sin (au) du--sl 0+11

the first and third of the integrals on the right may be made arbitrarilysmall by choosing a sufficiently large, since h(u) is properly integrable overthe intervals a :5u <- c - b, , c + th < u < b. The modulus of the secondintegral on the right is dominated by

Ei,

i4s

J h(u) J du

and this is again dominated by

1 f J u) I duc - 1K e--1,

which may be made arbitrarily small by choosing b1 and 62 sufficiently small,the choice of 3, and 8h being independent of a. Thus I J2 J may be madearbitrarily small, say < e, by choosing a sufficiently large and, since1, = J, + J : + Js, it follows that J 11 - [fi(r - 0)/rJC J < 4e if a issufficiently large. Similarly, J 12 - V &r + 0)/,r]C J < 4e if a is sufficientlylarge and, since

r fW A(t)sin

f rr) dt = 11 -}- 12

this implies that

1I fl(t) Bin a(t - r) dt - 2f1(T) C

a 7 -

< 8e<

if a is sufficiently large (it being understood that at any point r of discon-tinuity of fl(t) at which the limits f,(r + 0) and fl(r - 0) exist, fl(r) isdefined as the mean of these two limits). Thus

lim 1 1' f,(t) sin a(t - r) dt _ 20fl (T)

a-,= r ro t - r rand similarly for f2(t), so that

lim1( f(t)sina(t-r)dt=2Cf(r)a-= r .00 t - r


which implies that (2x)`4f((') g(w) exp (irw) dw = (2C/r)f(r). The con-stant C is independent of the function f(t) and, to determine it, we choosethe function f(t) which is 0 if t < 0 and = exp (izt), where the imaginarypart y of z is positive, if t > 0. Setting r = 0 and using the already provedfact that (21r)-4f g(w) dw = f(0) we see that C = it/2. Thus

(2rr)- f g(w) exp (ir(.,) dw = f(r)(-:)

which completes the proof of the Fourier Integral Theorem.The Fourier Integral Theorem is one of the most useful theorems of ap-

plied mathematics, but in the form here stated, it suffers from a serious dis-advantage. The class of complex-valued functions of the unrestricted realvariable t which possess Property 2 is too restricted. For example, theHeaviside unit-function u(t) which is defined as follows:

u(t) = 0 if t < 0; u(t) = 1 if t > 0; u(0) _ +}

while possessing Property 1 (since it is continuous save at t = 0) does notpossess Property 2. To remove this disadvantage we introduce a complexvariable p = c + iw whose real part c is not, necessarily, zero. Then w =i(c - p) and the Fourier Integral Theorem may be written in the form

f(r) = f g[i(c - p)] exp [r(p - c)] dpi (c+ao)

the integration in the complex p-plane being along the line p = c which isparallel to the w-axis and the Cauchy principal value of the integral beingtaken. On multiplying by (27r) exp (cr) and setting (2r)-4 exp (cr)f(r)= h(r), g[i(c - p)] = k(p) we obtain()

h(t) = 1. f k(p) exp (rp) dp2xi <r-soo)

where

k(p) = 9(w) = (2x)-41:1(t) (--iwt) 1h(t) exp (-pt) dt

O

Here k(p) is termed the Iaplace Transform of the complex-valued functionh(t) of the unrestricted real variable t and the relation

h(r)2a-i Jtk(p) exp (rp) dp

is the Laplace version of the Fourier Integral Theorem. The grit advan-tage of this version is that h(t) is not required, as is f(t), to be absolutelyintegrable over - co < t < oo . It suffices that there exist a real number c

The Iaplace Transformation 21

such that exp (-ct)h(t) is absolutely integrable over - oo < t < W. Thus,for example, if h(t) is zero when t < 0, h(t) may be furnished, if c > 0,for positive values of t by any polynomial function of t. In particular, theHeaviside unit function u(t) possesses, if c > 0, the Laplace Transformf "o exp (-pt) dt = 1/p and the Laplace version of the Fourier IntegralTheorem tells us that

1 [(4+iO) exp ('rp)_ dp2xi ce-+b) P

wherec> O,islif r > 0,+4if 7 = 0,and0if r <0.

5

The Laplace Transform of

a Right-sided Function

The function h(t) which appears in the Laplace version of the FourierIntegral Theorem is connected with the function f(t) which appeared inthe original version by the relation

h(t) = (27r)_'exp(ct)f(t)

Since exp(ct) is everywhere continuous, h(t) possesses, likef(t), Property 1,i.e., it is piecewise continuous. Since f(t) possesses, by hypothesis, Prop-erty 2, h(t), which need not possess this property, must be such that thereexists a real number c such that exp(-ct)h(t) possesses Property 2, i.e., isabsolutely integrable over - - < t < ao. For example, h(t) may be theHeaviside unit-function u(t) which is defined as follows:

u(1) = 0, t < 0; u(t) = 1, t > 0; u(0) =

since, if c is any positive real number, exp(-ct)u(t) is absolutely integra-ble over - ao < t < oo, the value of the infinite integral

J : exp (-ct)u(t) I dt = - exp (-ct) dtm

being 1/c. The product of any complex-valued function of the unrestrictedreal variable t by u(t) is zero if t < 0 and we term any piecewise continuouscomplex-valued function of t which is zero if t < 0 a right-sided function.Similarly, we term any piecewise continuous complex-valued function of twhich is zero if t > 0 a left-sided function; for example, the product of any

The Laplace TranFformation 33

piecewise continuous complex-valued function of t by u(-t) is a left-sidedfunction. The Laplace Transform, Lh, of a right-sided function h(t) isdefined by the formula

Lh = fl&(t)exp(-pt)dt

where p is any complex number for which the infinite integral on the rightexists (it being not required that this infinite integral converge absolutely,i.e., that the infinite integral f o I h(t) I exp( -ct) dt, where c is the real partof p, exist). Let us now suppose that Lh exists at some point Cl of the realaxis of the complex p-plane. We propose to prove that this implies theexistence of Lh at any point p of the complex p-plane whose real part c is> cl ; not only this, but also that Lh In an analytic function of the complexvariable p over the half-plane c > cl .

Since h(t) exp(-ct) is integrable, by hypothesis, over - ao < t < oo itis integrable over the interval 0 < t < T, where T is any positive real num-ber, and we denote by H, (T) the integral for h(t) exp(-clt) dt so thatHf, (T) is everywhere continuous and, at every point of continuity of h(t),differentiable with the derivative h(T) exp(-c1T). In view of the conti-nuity of HC, (T ), H,,, (T) is bounded over any interval 0 < 7' < b, whereb is any positive number, and this implies, since the infinite integraifo h(t) exp(-c1t) dt exists, by hypothesis, that is bounded over0 < T < m ; in other words, there exists a positive number M whichdominates I H., (T) I , T any non-negative real number. On writing Lh =f 'o h(t) exp(-pt) dt in the form fo h(t) exp(-clt) exp [- (p - cl)t] dtwe obtain, on integration by parts,

Lh = (t) exp [-(p - c)t] to + (p - cl) fH.,(t) exp [-(p - cl)t] dt

(p - Cl) J Hel(t) exp [-(p - Ci)t] dt

provided that the real part c of p > cl . Since

H.,(t) exp [-(p - cl)tJ ! < M exp [-(c - cl)tJ,

the infinite integral f o H,, (t) exp [ - (p - cl )tl dt exists over the half-plane c > c1 , its convergence being absolute. Thus, although the conver-gence of the infinite integral r o h(t) exp(-pt) di which defines Lh neednot be absolute at p = c1 , nor at points of the half-plane c > c1, Lh existsover this half-plane and may be expressed, over this half-plane, as theproduct of p - c1 by an infinite integral f o' H, (t) exp [ - (p - cl) t] dlwhich converges absolutely over the half-plane. Let, now, T be any positive


real number and let us consider the integral

dir(p) = f HH,(t) exp [-(p - c,)t] dt.0

Since H, (t) is bounded over the interval 0 < t < T, 41. (p) is a differen-tiable function, i.e., an analytic function, of the complex variable p, itsderivative being - f o' tH,,, (t) exp (- (p - cl) t] dt no matter what is thevalue of p. Assigning to T, in turn, the values 1, 2, 3, , we obtain asequence of functions 4,(p), 02(p), , of the complex variable p whichare analytic over the entire finite complex p-plane. At any point of thehalf-plane c > cl this sequence converges to

f H.,(t) exp (-(p cl)tl dt =

say, and it is easy to see that the convergence is uniform over the half-planec, + S < c, where S is any positive number. Indeed, the modulus of

4(p) - 4,.(p) = Ie H.,(t) exp (-(p - c,) t] dt - 4,.(p)

fQ*

is dominated, over the half-plane cl ± S < c, by M f exp (c - c,) t] diwhich is, in turn, dominated by Mf. R exp(-bt)dt = (M/S) exp( -nd), whichis arbitrarily small if n is sufficiently large, the choice of n being independentof c. Since 4,.(p) is analytic over the half-plane c > c, the integralf c4n(p) dp, where C is any simple closed curve of finite length which iscovered by this half-plane, is zero. Since the points of C constitute a closedset their distances from the line c = c, possess a positive lower bound sothat C is covered by a half-plane c, + a < c, if S is sufficiently small, andso fc4(p) dp, which is the same as fc {4(p) - 4 (p)J dp is dominated by(M/S) exp( -nS)l, where t is the length of C. Since f c4(p) dp is independentof n it follows that f c 4(p) dp = 0 and this implies that 4(p) is an analyticfunction of the complex variable p over the half-plane c > c, . HenceLh = (p - c,)4(p) is an analytic function of the complex variable p overthe half-plane c > c, .

EXAMPLE 1

h(t) = u(t)Here

c>0P


(Note. Lh does not exist at p = 0, but is exists at c = cl where ci is anypositive number and, if c > 0, there exists a positive number c, , Jc forexample, such that c > cl .

ExAMPLE 2

h(t) = exp (at)u(t), a an arbitrary complex number

Here

Lh=exp[-(p-a)t]dt= p - aand a > real part a, of a.

(Note. Similarly, if, for any right-sided function h(t), Lh = .(p), c > cl,and hl(t) = exp(at)h(t), then Lh' _ b(p - a), c > cl + a, . This usefulproperty of the Laplace Transform of a right-sided function is known asthe Translation Theorem.)

EXAMPLE 3

h(t) = tau(t), a a complex number a, -i- ia;.The complex power, ta, of a positive real number i is defined by the rela-

tion t° = exp (a log t) so that I t° = exp(a,. log t) = ta', where a, is thereal part of a. In order that Lh = f o to exp( - pt) dt exist at the pointp = cl of the real axis in the complex p-plane we must have a, > -1(to take care of the small values of t) and c, > 0 (to take care of the largevalues of t). Thus the Laplace Transform of tau(t), where the real parta, of a is > -1, exists, and is an analytic function of the complex variablep, over the half-plane c > 0, c being the reai part of p. On setting pt = sin the infinite integral f o to exp( -pt) dt which furnishes this LaplaceTransform this infinite integral appears as (1/pa+l) fo sa exp(-a) de, theintegration being along the ray from 0 to co in the complex p-plane whichpasses through the point p. If R and 0 are the modulus and argument, re-spectively, of any point s in the complex p-plane, sa = exp(a log a) =exp { (a, log R - atB) + i(a,B + a; log R) I so that

sa ( = exp(a, log R - a.B) = R"' exp(-a;B)

and, since I exp(-s) I = exp(-R cos 0), we have 1 sa exp(-a) I =Ra, exp(-a.B)exp(-R cos 0). Denoting, for a moment, by S the argumentof p, so that -(ir/2) < 6 < (ir/2), it follows that along the are of thecircle s = R exp(Bi) in the complex p-plane from 0 = 0 to B = f8, this arelying in the first quadrant if 8 > 0 and in the fourth quadrant if 6 < 0,sa exp(-a) 1 < Rap exp(I a4 I)exp(-R cos 0) and this implies, since

Ra'+1 exp(-R cos $) tends to zero as R --> -, that the integral of


as exp(-8) along this are of the circle 8 = R exp(8i) tends to zero asR -- co, or, equivalently, that the integral of s° exp(-s) along the ray ofargument 6 from 0 to ao in the complex p-plane is the same as the integralof s° exp(-s) along the ray of argument zero from 0 to co in the complexp-plane. This integral, f o' t° exp( -t) dt, is the Gamma Function, r(a + 1),of argument a + 1, and hence it follows that the Laplace Transform oftau(t), where the real part of a is > -1, is r(a + 1)/pa+i, over thehalf-plane c > 0.

A simple integration by parts shows that if the real part of a is not only>-1 but also >0, then r(a + 1) - ar(a) and, since r(1) =f o' exp( -t)dt = 1, it follows that, if a is a positive integer, r(a + 1) = a!.The Laplace version of the Fourier Integral Theorem tells us that

p (pt) dp.ta

4--i'*)u(t) = r(a + 1) cf..0> ex

2xi pa+1

real part of a > -1; c > 0 and in particular, on setting t = 1, that

r(a + 1) exp (p)1 = f 1 dp; real part of a> - 1;c> 0

2a-i (c-: o)

Thus r(a + 1) is never zero over the half-plane a, > -1, where a, is thereal part of a, for which the Laplace Transform of tau(t) is defined. If ais real and > -1, r (a + 1) is real, and since it is continuous and neverzero, it is one-signed. Since r (1) = 1 is positive, it follows that r (a + 1)is positive for every real value of a > -1.

EXERCISE 1

Show that the Laplace Transform operator L h = f h(t)exp(-pt) dtis linear, i.e., L(h, + h2) = Lh1 + Lh:, L(ah) = aLh, a any complexnumber, and use this property to determine the Laplace Transforms of theright-sided functions sin(flt)u(t), cos($t)u(t), where ,6 is any complex num-ber, indicating in each case the half-planes over which the Laplace Trans-forms are analytic functions of the complex variable p.

EXERCISE 2

Show that if the Laplace Transform, Lh, of a right-sided function h(t)exists at a point p, = c, + iwz of the complex p-plane then Lh exists, andis an analytic function of the complex variable p, over the half-planec > c, . (Hint: The Laplace Transform of h(t) at p, is the same as theLaplace Transform of h(t)exp(-iw,t) at cl and the Laplace Transform ofh(t)exp(-iw,t) at p - iw2 is the same as the Laplace Transform of h(t)at p.)


An important consequence of the Laplace version of the Fourier IntegralTheorem is the following uniqueness theorem: If two pieeewise continuousright-sided functions h,(t), hz(t), possess Laplace Transforms at a point c,of the real axis in the complex p-plane and if their Laplace Transformscoincide over the half-plane c > c, , then h2(t) coincides with hi(t) at allpoints t which are not discontinuity points of either hl (t) or h! (t) . To provethis, we observe that the relation

Lh = (p - ci)fo H.,(t)exp [-(p - ci)t] dt,c > c, , tells us that the Laplace Transform, over the half-plane c > c, , ofH.,(t)exp(c,t) is (Lh)/(p - c,). Since the convergence of the infiniteintegral f o* Hal(t)exp [-(c - c,)tj dt, c > c, , is absolute we may applythe Laplace version of the Fourier Integral Theorem to obtain

1 f(`+"0) LhH, (t) exp (c,t) = - exp (pt) dp271 J c-:.,) p - c,this equality being valid at any continuity point of h(t), sinceH,,(t)exp(c,t) is differentiable at any such continuity point. Thus HH,(t)is unambiguously determined, at any continuity point of h(t), by the valuesof Lh at the points of the complex p-plane whose real parts have anycommon value c > c, or, equivalently, by the values of Lh over the half-plane c > c, . Since the derivative of H,, (t), at any continuity point ofh(t), is h(t)exp(-c,t) it follows that h(t) is unambiguously determined atany point where it is continuous by the values of Lh over the half-planec > c, . In particular, if Lh = 0 over the half-plane c > c, , then h(t) = 0at all its continuity points.

(Note. It is not necessary, for the validity of this uniqueness theorem,that the Laplace Transforms, at the point p = c, , of hi(t) and h2(t) beabsolutely convergent.)

6

The Laplace Transform

of exp (-t')

We have seen that if the Laplace Transform, Lh, of a piecewise continuousright-sided function h(t) exists at a point cl of the real axis of the complexp-plane then Lh exists, and is an analytic function of the complex variablep, over the half-plane c > c, , so that Lh possesses a derivative with respectto p over this half-plane. If h(t) is left-sided, instead of right-sided,Lh h(t)exp(-pt)dt = fo h(-t')exp(pt') dt', t' = -t,

= fo h(-t')eXp(-p't') dt', p' = -p, = fo h(-t)exp(-p't) dtand h(-t) is a right-sided function of t so that, if Lh exists at p' = c,' _-c, it exists, and is an analytic function of the complex variable p', overthe half-plane c' > c,'. This is the same as saying that if Lh exists at p = c,it exists, and is an analytic function of the complex variable p, over thehalf-plane, c < c, . If h(t) is neither right-sided nor left-sided we may writeit, since u(t) + u(-t) is the constant function 1, as the sum of a right-sided and a left-sided function as follows: h(t) = h(t)u(t) + h(t)u(-t).If, then, the Laplace Transform of the piecewise continuous right-sidedfunction h(t)u(t) exists at a point p = c, of the real axis in the complexp-plane, and if the Laplace Transform of the piecewise continuous left-sided function h(t)u(-t) exists at a point p = c2 of this real axis, and ifc2 > c, , Lh exists, and is an analytic function of the complex variable p,over the strip c, < c < C2 parallel to the imaginary axis in the complexp-plane. For example, exp( -t2) is the sum of the right-sided functionexp(-t2)u(t) and the left-sided function exp(-t2)u(-t) and each of these

.E8


functions possesses a Laplace Transform at any point ex of the real axis inthe complex p-plane. Indeed, exp(-t2)exp(-cxt) = exp (c12/4) exp(-v2),v = t + ex/2, and so both of the infinite integrals

f exp (- t2) exp (-c1 t) dt = exp (L4) exp (-v2) dvo (c1)/2

0f exp (-t2) exp (-cx t) t* = exp/( 4 f exp (-v2) dv

exist, exp(-v2) being dominated by (1 + v2)-1 no matter what the valueof the real variable v. Thus the Laplace Transform of exp(-t2) exists, andis an analytic function of the complex variable p, over any strip c1 < c < c2parallel to the imaginary axis in the p-plane; in other words, the LaplaceTransform of exp(-t2) is an analytic function of p over the entire finitecomplex p-plane. At any point c1 of the real axis in the complex p-planethis Laplace Transform has the value A exp(c12/4) where

r(e1)/R eo

A = L exp (-v2) dv + exp (-v2) dv = exp (-v2) dvw

f(c,)/2 EW

In order to evaluate this infinite integral we observe that, if (r, 0) are planepolar coordinates, the double integral of exp(-r2) over the circle of radiusR with center at the origin is f exp(-r2)rdrde = jr(1 - exp(-R2)) andthat the double integral of exp( -r2) over the square of side 2b with centerat the origin and with sides parallel to the coordinate axes is, since r2 =x2 + y2, the square of the integral f_o exp(-t2) dt. Since this square of side2b is covered by the circle of radius R, if R is large enough, we know thatthe square of j°_e exp( -t2)dt < Rr(1 - exp( -R2)}, if R is large enough,and so the square of jb6 exp(-t2) dt is less than ,r, no matter what is thevalue of b. Hence the infinite integral f°° exp(-9) dt exists with a value< T. On the other hand, the square of side 2b covers the circle of radiusR, if b is large enough, and this leads to the opposite inequality

f exp(-t2)dt > r}.

Thus f! exp(-t2)dt = ,rl so that the Laplace Transform of exp( -t2)assumes the value ir} exp(cx2/4) at any point c1 of the real axis in the com-plex p-plane. Hence it coincides with the analytic function a exp(p2/4) onthe real axis in the complex p-plane and this implies, since it is analyticover the entire finite complex p-plane, that it is ,r} exp (p2/4) over the entirefinite complex p-plane:

exp (-t2) exp (-pt) dt -- r*exp (f',), p arbitrary.


On setting t = kit', p - kip', where k is any positive real number, we ob-tain f "., exp (-kt")exp(-p't')dt' = (r/k) exp(p''/4k) so that theLaplace Transform of exp(-kt'), k > 0, is (r/k)ii exp(p2/4k). In particu-lar, on setting k = }, the Laplace Transform of exp(-t'/2) is

(2r)3 exp(p /2)

On evaluating the Laplace Transform of exp(-kt'), k > 0, at any pointp - io of the imaginary axis of the complex p-plane, we obtain

f`-.e exp(-kt=)exp(-iot)dt = (r/k)l exp(-w2/4k)

and it follows, on multiplication by (2r)3, that the Fourier Transform ofexp(-kt') is (2k)-4exp(-w'/4k). In particular, on setting k - 1, theFourier Transform of exp(-4t') is exp(-W) ; this result is expressed bythe statement that exp(-4t') is its own Fourier Transform.

The Laplace Transform of the one-sided functions exp(-kt')u(t),exp(-kt2)u(-t), k > 0, are not as simple as the Laplace Transform oftheir sum exp (- kt') . For example, the Laplace Transform ofexp(-kt')u(t), k > 0, is

jexp(_kt2)exp(_pt)dt = exp(4k)f exp[-k(t+k)]dt

= 0 exp p f r exp (--z') dz,4k fki

Z = k} (t + 2)the integration in the complex z-plane being along the ray of argument zerofrom p/20 to ao. Similarly, the Laplace Transform of

= 9

exp (-kt')u(-t) is k7* exp k()frexp (-z') dz,

=the integration being along the ray of argument zero from - COexp(ir) to p/20. In particular, when k - }, the Laplace Transform of

exp (- 4 ) u(t)

is

2 exp (p') f exp (-zs) dzV

CO

and the Laplace Transform of/

exp (- 4, u(-t)


is

2exp(p') f exp(-z')dz

If we are certain that it is permissible to differentiate with respect to k,under the sign of integration, the infinite integral f., exp(-kt')exp(-pt)dtwhich furnishes the Laplace Transform, (T/k)} exp(p2/4k), of exp(-W),k > 0, we may obtain the relation

t' exp (-kt') exp (-pt) dt = (kJ{ exp (4k\ [2k + 21which furnishes us with the Laplace Transform of t' exp(-k9) and, con-tinuing this process, we may obtain the Laplace Transform of the productof exp( -kt') by any even power of t (always provided that the differentia-tion of the infinite integral involved with respect to k, under the integralsign, is legitimate). Similarly, if we are certain that it is legitimate to dif-ferentiate, under the integral sign, the infinite integral f,°., exp(-k9)exp(- pt) dt with respect to p or, equivalently, with respect to the realpart c of p, we may obtain the relation

; 2

t exp (-kt') exp (-pt) dt = (k>2k

expf kwhich furnishes us with the Laplace Transform of t exp( -kt'), k > 0, andcontinuing this process we may obtain, always under the same proviso, theLaplace Transform of the product of exp(-kt') by any positive integralpower, odd or even, of I. In order to formulate, in as convenient a manneras possible, conditions which guarantee the validity of this differentiationof an infinite integral under the integral sign we shall consider the casewhere we propose to differentiate the infinite integral f_'., h(t)exp(-pt) dt,which furnishes the Laplace Transform of h(t), with respect to the realpart c of the complex variable p = c + io. The integrand, h(t)exp(-pt),of this infinite integral is a function F(t, c) of the two real variables (t, c),the imaginary part w of p being supposed held constant, and the derivative,FF(t, c), of F(t, c) with respect to c, being -th(t)exp(-pt), is since h(t) is,by hypothesis, a piecewise continuous function of the unrestricted realvariable t, either a continuous function of the two variables (t, c) over anystrip - oo < t < co, c, < c S ce parallel to the taxis in the (t, c)-plane,or else its points of discontinuity in any rectangle a < t < b, cl < c < c, ,where a and b are any two real Numbers which are such that a < b andci, c, are any two real numbers which are such that c, < o2, He on a finitenumber of lines parallel to the c-axis. We make now the following two addi-tional hypotheses concerning the function F,(t, c) of the two real vari-ables (t, c) :

3o Lectures on Applied Mathematics

(1) F. (t, c) is absolutely integrable with respect to t, for every value ofc in a given closed interval c, < c < c, , over - oo < t < cc.

(2) The convergence of the infinite integral f," F.(t, c)dt is uniform withrespect to c over the interval c, < c < c,,and a single additional hypothesis concerning F(t, c) :

(3) The infinite integral f% F(t, c,) dt exists.

We shall show in the following paragraph that these three hypotheses aresufficient to guarantee the following three facts:

(1') The infinite integral f, F(t, c) dt exists for each value of c in theinterval c, < c < 4 .

(2') f°° F(t, c) dt is a differentiable function of c over the intervalCl < c < ez.

(3') The derivative of f," F(t, c) dt with respect to c, where c, < c < c, ,is furnished by the formula

(f°F(t,c)dl) f F.(1,e)dl

In other words, differentiation of the infinite integral f°°,. F(t, c) dt, withrespect to c, under the integral sign is legitimate.

Writing f°_°., Fe(t, c) dt in the form fa F.(t, c) dt + R.b(c), a < b, weknow from (2) that I Rab(c) I may be made arbitrarily small, say <e, forevery c in the interval c, < c < c, by making -a and b positive and suffi-ciently large, the choice of a and b being independent of c. If, then, a and bare so chosen that ( Rb(c) I < e, c, < c < rt , and c and c + Ac are anytwo values of c in the interval c, < c < co we have

Af '0F.(t,c)dt1<1A fbF.(t,c)dtj+2e

where Af.., F.(t, c) dt denotes

and similarly for Af.b F.(t, c) dt. If F.(t, c) is a continuous function of thetwo variables (t, c) over the rectangle a < t < b, c, <- c < c, , it is a uni-formly continuous function of the two variables (t, c) over this rectangleand so I AF.(t, c) I = I F.(t, c + Ac) - F.(t, c) I may be made arbitrarilysmall, say <e/(b - a), by making I Ac I sufficiently small, the choice ofAc being independent of either t or c. Supposing Ac so chosen, we have

b b b

Itl../Fc(t,c)dt=1fAJPc(t,c)dtif1tJIFc(t,c)Idt<(

The Lapiaee Transformation. 88

which implies that I nf'. F,(t, c) dt I < 3e. Thus the two integralsf; F,(t, c) dt, f°. F,(t, c) dt and, hence, their difference R,b(c), are con-tinuous functions of c over the interval c, < c < cs . This conclusion remainstrue, by virtue of (1), when F,(t, c) fails to be continuous over the rectanglea < t <- b, c, < c < c: , since, by hypothesis, its points of discontinuity inthis rectangle lie on a finite number of lines parallel to the c-axis in the(t, c)-plane. To prove this it suffices to treat the case where the points ofdiscontinuity lie on a single line t = d, where a < d < b. Wewrite fa F,(t, c) dl in the form

as

F. (t, c) dt ++a,

F. (t, c) dt + fb

F, (t, c) dt, a+a,raa--a

where a, and 62 are positive numbers which are less than d - a and b - d,respectively. The first and third of these three integrals are continuousfunctions of c over the interval c, < c < ci , since F,(t, c) is a continuousfunction of the two variables (t, c) over the rectangles a < t < d - a,,c, < c < c: and d + 82 < t < b, c, < c < C2 , and we direct our attentionto the second. The modulus of this second integral may be made arbitrarilysmall, say <e, by choosing 8, and 8: sufficiently small, the choice of a, and8: being, by virtue of (2), independent of c. Supposing 8, and 6, so chosen,we have I Af al-i; F,(t, c) dl < 2e and it follows that, if Lc is chosen sosmall that

a, aIF,(t,c)dtI<e andI.

f+a!F,(t,c)dtI <ea

thatb1,6 f F. (t, c) dtl < 4e

a

which implies that ; c) dt < 6e proving the continuity of thetwo integrals fa F,(t, c) dl, f-', F,(t, c) dt and, hence, of their differenceR,b(c), over the interval c, < c < c2. Hence the three functionsf; F,(t, c) dt, f_b, F,(t, c) dt and R,b(c) of c are integrable over the intervalc, < c < and, if c' is any point of this interval, we have the relation

c{f 00

F.(t,c)ddo=fe' {jb

F,(t,c)4do+ f R,b(c)do111J

If F,(t, c) is a continuous function of the two variables (t, c) over `,herectangle a < t < b, c, < c < ca, the order of integration may be changedin the repeated integral on the right and the same argument as before showsthat this remains true, by virtue of (1) and (2), when F,(t, c) fails to becontinuous over the rectangle a < t < b, c, < c <- c:. Hence

e' .o a

f {J_ F, (t, c) dt} de = f{f F.: (t, c) dc} dl + R,° (c) do

$4 Lectures on Applied Mathematics

so thatr

{L:F. (t,c) di} dc - fi F.(t,c)dc}dtl

< ( c ' - cl) S e(cs - Cl)

proving the existence of the infinite integral f'°.o f% F.(t, c)dc) dt with thevalue f,, {f-.% F°(t, c)dt) dc. Since

F. (t, c) do = F(t, c') - F(t, cl)

it follows, by virtue of (3), that the infinite integral f°.. F(t, c') dt exists,for every point c' of the interval cl < c' < cs , its value being

CI

f L_.F.(t,c)dt}dc+ fF(t,ct)dl

The fret of these two terms is a differentiable function of c', sincef!.. F.(t, c) di is a continuous function of c, and the second is a constantfunction of c'. Hence the infinite integral fr.. F(t, c') di is a differentiablefunction of c' over the interval c1 < c' < c, , its derivative beingf_.. F.(t, c') dt . This completes the proof of the legitimacy of differentiatingthe infinite integral f!!. F(t, c) dt with respect to c under the sign of integra-tion, when F.(t, c) satisfies conditions (1) and (2) and F(t, c) satisfiescondition (3).

7

The Laplace Transform of the

Product of a Right-sided

Function by t and of the Integral

of a Right-sided Function over

the Interval [0, t]

We have seen that if a piecewise continuous right-sided function h(t)possesses a Laplace Transform Lh, which need not be absolutely conver-gent, at a point cl of the real axis in the complex p-plane then Lh may bewritten, over the half-plane c > cl in the form

Lh = (p-c,)fH.1(t) e xp[-(p-c!)t]dt

where

Ha,(t) = f h(s) exp (-ci s) ds`0

the convergence of the infinite integral which multiplies p - c, being abso-lute over this half-plane. The integrand, H., (t )exp (- (p - c,) t], of thisinfinite integral is a function F(t, c) of the two real variables (t, c), wherec is the real part of p, it being understood that the imaginary part w of p

$5


is held constant, and the derivative of this function with respect to c, beingthe same as its derivative with respect to p, exists at every point p of thefinite complex p-plane, with the value -tHH,(t)exp [-(p - c,)t]. Since0<texp[-(c2 -c)t] <[1/(c2-c)],if c1>cand t>0,tHq (t)exp [ - (ci - ct)t]

= tH.,(t)exp [-(c2 - c)tj exp [-(c -- c,)t], cl < c < c j,

is absolutely integrable over 0 < t < co (since H.,(t)exp [-c - c,)ij,c, < c < c2, is absolutely integrable over 0 < t < co ). Thus the infiniteintegral f o F,(t, c) dt converges absolutely over the half-plane c > c, .Moreover, the convergence of this infinite integral is uniform over thehalf-plane c > cl + 6, where b is any positive number, since, over thishalf-plane, I exp [-(p - c,)t] I = exp [-(c -- cl)t] < exp(-at), and so itis permissible to differentiate with respect to c or, equivalently, with respectto p, under the sign of integration, the infinite integral

fo H.,(t)exp [-(p - c,) t] dt,p being any point of the half-plane c > c, ; indeed, if c > c, , we may setS = (1/2) (c - c,) and ensure that c > c, + a. Thus

(Lh), = foo H,, (t) exp [- (p - c,)11 dt -

(p - c,) Jo tH.,(t) exp [-(p - c,) t] dt,

fHci(t)ttexp[-(p-c,)t]itdt, c>ci

and the right-hand side of this equation reduces, on integration by parts,since H,,(co ) exists and since t exp -- (c - c,)t tends to zero as t --> oo, to- f th (t) exp ( - pt) dt. Thus we have the following useful result :

If the piecewise continuous right-sided function h(t) possesses, at a pointc, of the real axis in the complex p-plane, a Laplace Transform, whose con-vergence need not be absolute, then the product, th(t) of h(t) by t pos-sesses, over the half-plane c > c, , the Laplace Transform - (Lh ), . Weexpress this result by the statement that multiplication of a piecewise con-tinuous right-sided function by t is reflected, in the domain of LaplaceTransforms, by differentiation with respect to p followed by a changeof sign.

EXAMPLE

exp(at)u(t) possesses, over the half-plane c > a,, where a, is the realpart of a, the Laplace Transform 1/(p - a). Hence t exp(at)u(t) possesses,over the half-plane c > a,, the Laplace Transform 1/(p - a)=. Continuing


this process we see that, if n is any positive integer, t" exp(at)u(t) possesses,over the half-plane c > a the Laplace Transform n!/(p - a)"+1

Note. This is a special case of the result that to exp(at) u (t) , where 0 isany complex number whose real part is > -1, possesses, over the half-plane c > a, , the Laplace Transform r(P + 1)/(p - a) o+', which resultis an immediate consequence of an application of the Translation Theoremto the result that tou(t) possesses, over the half-plane c > 0, the LaplaceTransform r(ft + 1)/po+

Let us now consider a piecewise continuous right-sided function h(t)which is such that the integral f o' h(s)da = Ho(t), which we shall denotesimply by H(t), exists over 0 < t < oo. We do not assume the existence ofH( ao ), i.e., that h(t) possesses at p = 0 a Laplace Transform, but we doassume the existence of a positive real number cl such that h(t) possesses atp = c, a Laplace Transform which need not be absolutely convergent.H(t) is an everywhere continuous right-sided function, which is, in addi-tion, differentiable, with derivative h(t), at the points of continuity of h(t).The function 40(t) = fo H(v)exp(-c1v) dv exists, since H(t) is everywherecontinuous, over 0 < t < oo and is an everywhere differentiable function,its derivative being H(t)exp( --c,t). Since H(O) = 0, we obtain, on integra-tion by parts,

H(t) exp (-c, t) `fi(t) +Cl f h(v) exp (-c,v) &

Cl

H(t) exp (c, t) `fi(t) exp (c, t)Cl

+Cl

f h(v) exp (-c,v) du

Hence, at the points of continuity of h(t),

{fi(t) exp (c, t) ), = -hit) + exp (c, t) f g h(v) exp (-c,v) dv

+ hct) = exp (c, t) fh(v) exp (-c, v) dv

and this implies, since the derivative of 0(t) is continuous over 0 < t < -o,that

{#(t) exp (c, t) {, = exp (c, t) f 'h(v) exp(-cjv) dv over 0 < t < CO

0

Since h(t) possesses, by hypothesis, a Laplace Transform at p = C1, itfollows that the quotient of {-0(t)exp(c1t)) tby exp(c1t) = {(1/c1) exp(c1t){:has, at t - ao, the limit (Lh),, , and we shall show in a later paragraphthat this implies that the quotient of 0(t)exp(c1t) by (1/c1) exp(c1t) has, att = co, the same limit, Assuming this, for the moment, it follows

M Lectures on Applied Mathematics

that clo(t) = -H(t)exp( -cat) + fo h(v)exp(-clv) dv has, at t = ao, thelimit (Lh),., and, since the second term on the right has, at t = m, thelimit (Lh),..,,1 , this implies that H(t)exp(-clt) has at t = oo, the limitzero. The existence of the limit, at t = ao, of q6(t) = f `o H (v) exp (- civ) duassures us that H(t) possesses, at p = cl, a Laplace Transform and thisimplies that the Laplace Transform of H(t) exists, and is an analytic func-tion of the complex variable p, over the half-plane c > cl . Since the limit,at t = -, of

H(t) exp - (c1 t) 1 `¢(t} h(v) exp (-civ) dvCl cl

is

h(v)exp(-clv)dvCl

we see that

(LH) q = c (Lh)p-.,

The number ci may be replaced throughout the entire preceding argumentby any real number > ci , and so the value of (LH) at any point p = c ofthe real axis in the complex p-plane which lies to the right of the pointp = c, is the quotient of the value of Lh at p - c by c. Since both (Lh)/pand LH are analytic functions of the complex variable p over the half-planec > cl , it follows that LH = (1 /p) (Lh) over this half-plane. We expressthis result, which is the central one in the theory of Laplace Transforms,as follows:

Integration with respect to t, over the interval [0, t], of a right-sided functionit reflected, in the domain of Laplace Transforms, by division by p.

To complete the proof of this fundamental theorem let us denote0(t)exp(clt) by f(t) and exp(cit) by g(t). Since g(t) is monotone increasingand unbounded at t = ao, we may associate with any positive real numbert a real number T >- t such that, if 9 > T,

f(t) g(t)g(t') II< E' I g(t') - 031

< E

where e is an arbitrary positive number. Applying the Theorem of theMean of differential calculus to the function

f(s) - f(t) - f (t') - f (t) (g(s) - g(t) ]g(t) - g(t)of the real variable e, which function vanishes when s = t and when s = 9,


we see that

f,(t") = f(t') - f(t) _ 1f(i) _ f(t) g(i)g,(t") g(t') - g(t) lgg(t') g(t') g(t') - g(t)

where t" is some real number between t and t'. As t -+ oo so also do t' andt" and so f e (t") /g a (r) is of the form l + 'j , where I P, I is arbitrarily small,say <e, if t is sufficiently large, l being the limit, at t of f e (t) /gt (t) .

Alsog(t') = 1 +

'g(t)+ va

g(t') - g(t) g(t) - g(t)where I r= I < e for every t, and f(t)/g(t') = iv=, where I Ya I < e forevery t; thus

f(t') - vs l+P= i+PI - i_R

g(t') 1+P2 1+P2so that I [ f (t')/g(t' )j - i I is arbitrarily small if t' > T is sufficiently large.Hence f(t)/g(t) has, at t = oo, the limit 1.

The theorem which states that integration of piecewise continuous right-aided functions over the interval [0, tj is reflected, in the domain of LaplaceTransforms, by division by p may be presented in a different form which isuseful in the application of the Laplace Transformation to differential equa-tions. Let us suppose that the right-sided function h(t) is continuous over0 < t < co, without being, necessarily, continuous at t = 0, so that h(+0)may be different from sero, and that h(t) possesses over 0 < t < co a piece-wise continuous derivative hi(t). Writing fo h,(s)ds = h(t) - h(+0),h(t) - h(+0) plays the role of H(t), and so the mere assumption thatL(he) exists at a point c, of the positive part of the real axis in the complexp-plane guarantees that L(h(t) - h(+O)} exists, and is an analytic func-tion of the complex variable p, over the half-plane c > c, and, furthermore,that L{h(t) - h(+0)} - L(ht)/p over this half-plane. Since L{h(+0)} =h(+0)/p over the half-plane c > 0, it follows that L(he) = pLh - h(+0) ;c > cl > 0. We express this result by the statement that differentiation of aright-sided function h(t) is reflected, in the domain of Laplace Transforms,by multiplication by p followed by subtraction of h(+0).

Similarly, if h, is continuous over 0 < t < co and h,, is piecewise con-tinuous and possesses a Laplace Transform at p = ex > 0, then

L(h,e) = pL(h,) - h,(+0) - p2Lh -- ph(+0) - h,(+0)and so on.

8

Functions of Exponential Type

The right-sided function exp(at)u(t), where a is any complex number,possesses, over the half-plane c > a,, where a, is the real part of a, theLaplace Transform 1/(p - a). This Laplace Transform is not only ananalytic function of the complex variable p over the half-plane c > a,., butit possesses, in addition, the following two properties:

(a) It is zero at p = co .(b) It is analytic over the neighborhood I p I > I a ( of p = ao. It follows

that any finite linear combination, C, exp(alt)u(t) + C2 exp(a t)u(t) ++ C. exp(a.t)u(t), where C1, , C. are any complex constants, of

right-sided functions of the form exp(at)u(t) possesses the two additionalproperties (a) and (b).

We term any piecewise continuous right-sided function h(t) a function ofexponential type if it shares with any such finite linear combination ofright-sided functions of the form exp(at)u(t) the following three prop-erties:

(1) Lh exists at some point c, of the real axis in the complex p-plane.(2) The analytic function, f(p), of the complex variable p which is fur-

nished, over the half-plane c > c, , by Lh is zero at p = ao .

(3) f(p) is analytic over some neighborhood I p I > R > 0 of p = .

We proceed to investigate what properties of a given piecewise continuousright-sided function h(t) are sufficient to ensure that h(t) is of exponentialtype. On denoting by (ao/p) + (a,/p=) + + (a./p"+') + ... thepower series development of f(p) over the neighborhood I p ( > R ofP = w this it .mite series converges absolutely at p = R + S where S is

40


any positive number, and it may also converge (not, necessarily, absolutely)at p = R. For example, if

..AP) _ (1 + e)-* = p ( + P!/ p \12p1 1.3

= + 24p4

a,, is zero when n is odd while ao = 1, a2 = -4, a# = 1.3/2.4, . Thusr/f =r2

a... _ (-1)'" f I coo' 8d9 and 2 I cost"` a dB

is a monotone decreasing function of m which has the limit zero at m = co(since it is dominated by (2/v) [a + (cos 6)2-((v/2) - 6)] where b is anypositive number < (ir/2)) so that the infinite series 1 - 4 + (1.3/2.4) -

.. , being alternating, is convergent. Thus the infinite series (1/p) -(1/2p') + (1.3/2.4p') - .. which converges, with the sum (1 + pt)-*,over the neighborhood I p I > 1 of p = eo, also converges at p = 1. Onthe other hand, if f(p) _ (1 - the corresponding infinite series(1/p) + (1/2p') + (13/2.4p') + , which also converges, but withthe sum (1 - pt)- 1, over the neighborhood I p I > 1 of p = oo, fails toconverge at p = 1; indeed

so that the infinite series 1 + 4 + (1.3/2.4) + does not converge.We use the symbol r to denote R, if the infinite series (ao/p) + (a,/p2) +

converges at p = R, and to denote R + 8, where S is an arbitrarilysmall positive number, otherwise. Thus the infinite series (ao/r) +(a,/r') + . converges and this implies that I a,.

I/r"+1 is arbitrarilysmall, say < E, if n is sufficiently large, say > N. If, then, M is any positivenumber which dominates each of the N + 1 numbers I ao I, I a, I /r, -,I aN-, I/rx-1, er, we have I a,. 1/r" < M, n = 0, 1, 2, , and this implies thatthe two infinite power series in the complex variable z:

ao + a, z + 2! zt + ... + (na"-')I z,.--1

+...;Iaol+jai Iz+Iaslzt+...+ Ia,.-,I +...2! (n - 1)1

converge over the entire finite complex z-plane, their sums at any pointz of this plane being each dominated by M exp (r I z I). Assigning realvalues to z we obtain two functions k(t), k*(t) of the unrestricted realvariable t, where

(na.-I) t"-1+...

k*(t) iaoI+Ia1it+ + (na"-1)!t"_'+ ...


and we know that both f k(t) I and I k*(t) f < M exp (r f t 1). For example,when f(p) = (1 + p2)-#, k(t) = 1 (t2/22) + (t`/22.42) _ -.. andk* (t) = 1 + (t2/22) + (i4/22.42) + . In this case k(t) is known asthe Bessel Function, Jo(t), of the first kind, of index zero, and k*(t) isknown as the modified Bessel Function, to(t) = Jo(it), of the first kind,of index zero, and since r = 1 and we can take M = 1, we know that bothJo(t) and Io(t) are dominated, over - co < t < oo, by exp f t I. We shallstudy the functions Jo(t) and I0(t) in detail and shall see that the in-equality f Jo(t) f < exp f t f

is very crude for large values off t f, it beingpossible to replace this inequality by the inequality f Jo(t) f < 1, but wecannot thus improve on the inequality f Io(t) f < exp f t 1. The inequalityf Jo(t) f < 1 assures us that the right-sided function ho(t) = Jo(t)u(t)possesses, over the half-plane c > 0, a Laplace Transform and we shallsee that this Laplace Transform is (1 + p2) 4. Similarly the right-sidedfunction ho*(t) = Io(t)u(t) possesses, over the half-plane c > 1, the La-place Transform (1 - p2)-;. The fact that ho(t) = Jo(t)u(t) possesses aLaplace Transform over the half-plane c > 0, as contrasted with ho*(t) _Io(t)u(t), which does not possess a Laplace Transform over the half-plane c > c, if cl < 1, is because (1 + p2)-; does not have a singular pointin the half-plane c > 0, whereas p = 1 is a singular point of (1 - p2)4.

Since the two functions

k(t) = ao + a1 t + ... + (na^-i)1 o-1 + .. .

k*(t) = f ao f + f al f t + ... (man-1, ! to-1 + .. .

of the unrestricted real variable t are dominated, over - co < t < 00, byM exp (r f t 1), the two right-sided functions

h'(t) = k(t)u(t); h*(t) = k*(t)u(t)

the latter of which is a non-negative real-valued function of t, possessLaplace Transforms which are analytic functions of the complex variablep over the half-plane c > r or, equivalently, since S, if it is not zero, isarbitrarily small, over the half-plane c > R. If we denote by h1'(t) andhi(t) the real and imaginary parts, respectively, of h'(t) both h1'(t) andh2'(t), being dominated by f h'(t) f, also possess Laplace Transforms whichare analytic functions of the complex variable p over the half-plane c > Rand this implies that each of the following four non-negative real-valuedright-sided functions of the unrestricted real variable t, h*(t) f h1'(t),h*(t) f h2'(t), whose non-negativeness follows from the inequalityf h'(t) f < h*(t), possesses a Laplace Transform which is an analyticfunction of the complex variable p over the half-plane c > R. Denoting


by Ao + All + . [AA_i/(n - 1)!]t"`' + the everywhere convergentinfinite series whose sum is k*(t) + ki(t), for example, where k1(t) is thereal part of k(t), the coefficients Ao , A1i - of this series are all non-negative real numbers, since I I + the real part of n = 1, 2,is a non-negative real number, and we propose to show that the LaplaceTransform, over the half-plane c > R, of h*(t) + hl'(t) is furnished by thesum of the infinite series, (Ao/p) + (Al/p2) + , the non-negativenessof the real numbers Ao , A1i playing an essential role in our proof.A similar result holds for the other three non-negative real-valued right-sided functions h*(t) - hi (t), h*(t) + h2'(t), h*(t) - h,'(t) and, sinceh'(t) is the linear combination

4 {h*(t) + h1'(t)I - 4 {h*(t) - h1'(t)l + {h*(t) + h2'(t))

-2{h*(t)-h2'(t)}

of the four non-negative real-valued right-sided functions h*(t) t hi (t),h*(t) f h2'(t), it follows that the Laplace Transform, over the half-planec > R, of h'(t) is the sum of the infinite power series (ao/p) + (a,/p2) +

. In other words, the two right-sided functions h(t) and h'(t) possess,over the half-plane c > R, coincident Laplace Transforms, and this impliesthat h(t) coincides with h' (t) at any point t which is a continuity point ofboth h(t) and h'(t). h'(t) is everywhere continuous, save, possibly, att = 0 and so h(t) coincides with h'(t) at every continuity point save,possibly, t = 0 of h(t). Hence h(t + 0) and h(t - 0) exist, with the com-mon value h'(t), at any discontinuity point, if one exists, of h(t) and,since we have agreed to set h(t) = 4{h(t + 0) + h(t - 0)} at any dis-continuity point of h(t) where both h(t + 0) and h(t - 0) exist, it followsthat h(t) is coincident with h'(t) = k(t)u(t).

Once, then, we shall have proved that the Laplace Transform, over thehalf-plane c > R, of h*(t) + hl'(t) is (Ao/p) + (A1/p2) + we shallknow that h(t) is of the form k(t)u(t) where k(t) is the sum of an every-where convergent power series in t, this power series being such that both

k(z) I and I k*(z) 1, where z is an arbitrary complex number and k*(z)is the sum of the power series obtained from the power series ao + alz +(a2/2!)z2 + whose sum is k(z) by replacing each of its coefficients byits absolute value, are dominated, over the entire finite complex z-plane,by a constant times exp r J z', r being =R if the power series (ao/R) -i-(a1/R2) + converges and being =R + S, where b is an arbitrary posi-tive number, otherwise. For example, the function of exponential typewhose Laplace Transform, over the half-plane c > 1, is (1 + p2)-, L

4.4 Lectures on Applied Mathematics

J,(t)u(t) and the function of exponential type whose Iaplace Transform,over the half-plane c > 1, is (1 - p1) is Jo*(t)u(t) - lo(t)u(t) andJo(z) is dominated, over the entire finite complex z-plane, by a constant(actually 1) times exp I z ( while lo(z) is dominated, over the entire finitecomplex z-plane, by a constant times exp (1 + d) ( z (, d any positivenumber. Actually, since lo(z) = Jo(iz), lo(s) is dominated, over theentire finite complex z-plane, by exp I z I. We shall prove in the followingchapter that the Laplace Transform, over the half-plane c > R, of h*(t) +hi'(t) is (Ao/p) + (A,/p') + and, furthermore, that if h(t) is theproduct of u(t) by the sum, k(t), of an everywhere convergent infiniteseries ao + alt + (a=/2!)9 + - - which is such that k(z) = ao + a1z +(%/2!)z2 + - is dominated, over the entire finite complex s-plane, bya constant times exp (r ( z I), where r is a positive real number, then h(t)is of exponential type. In other words, this property of right-sided functionsis characteristic of functions of exponential type; every function of ex-ponential type possesses it and every right-sided function which possessesit is of exponential type.

We conclude with the observation that the convergence of the LaplaceTransform, over the half-plane c > R, of a function h(t) of exponentialtype is absolute, since I h(t) I < h*(t) and h*(t) possesses, over the half-plane c > R, the Laplace Transform I ao I/p -f- I ai I/p' +

9

The Characterization of

Functions of Exponential Type

Our first task is to show that, if the product of u(t) by the sum of an every-where convergent infinite series Ao + A,t + + [A.-,/(n -- 1) l]t-'

. with non-negative coefficients, Ao , A 1, .. , possesses, over a hsif -plane c > R >_ 0, a Laplace Transform, this Laplace Transform is thesum of the infinite series (Ao/p) + (A,/?) + + (A,.-i/p") +To do this we observe that the infinite integral

{Ao+Ait+....+ "-1 C- +...}exp(-ct)dt

which furnishes, at the point p - c > R of the positive real axis in thecomplex p-plane, the Laplace Transform in question is the sum by col-umns E:- u,." of the double series of non-negative terms which isdefined by the formula

u,."= A"t"exp(_ct)dt, m=0,1,2,---0,1,2,,. n!

(the sum of the infinite series which is furnished by the (m + 1)st columnof the cc X co matrix which has u,." as the element in its (m + 1)stcolumn and (n + 1)st row being

f ' Ao+Ai9+ ... + (A"-i t"-1+ }exP(_c)d)

Denoting by s this sum by columns of the non-negative double series we4,6


write o, = F-'...o L ..o um", j = 0, 1, 2, , so that the sequence oo ,01 , a2 . is monotone nondecreasing and o, is dominated by s, no matterwhat is the value of j. Thus the monotone sequence oo , of , has alimit o, its least upper bound, and o < s. From the definition of s we knowthat the non-negative number s - ER's-0 E LO u,,," may be made arbi-trarily small, say <e, by choosing j sufficiently large and, once j has beenso chosen, each of the j + I non-negative numbers F..o u," rk,,..o um",m = 0, 1, , j, may be made arbitrarily small, say <e/(j + 1), bychoosing k sufficiently large. Hence the non-negative number s -

,.4 E n..o u,,,' is less than 2e if j and k are large enough so that, inparticular, on denoting by p the larger of the two numbers j and k, thenon-negative number s - o, is less than 2e if p is sufficiently large, provingthat a = s. Turning now to the rows of our - X oo matrix we observethat any partial sum EM-4 um" of the infinite series furnished by theelements in the (n + 1)st row of this matrix is dominated by o, wherep is the greater of the two integers n and j and this implies that E" um"is dominated by a = s no matter what are the integers n and j so that theinfinite series -a um" is convergent, with sum <s, no matter what isthe value of n = 0, 1, 2, - . The argument given above tells us that the

k ,e " knon-negative number "-0 m..0U. - "-o EL o u,,," is arbitrarily

small if j is large enough and, since the two finite sums F':,-4 Ek,.o U.0and Fn _a Em are equal, it follows that the difference between sand ?,n_o .o u,,," is arbitrarily small if k is sufficiently large. In otherwords, the sum by rows F1R-o E:-0 u.,", of the non-negative doubleseries (um"1 exists with the same value, s, as its sum by columnsEm-o E..o u,,,". Since

= f A! t" exp (- ct) dt,

the sum of the infinite series E. um", being the infinite integral

t" exp (-ct) dtf W

is A"/c"+' (the Laplace Transform of 1"u(t) being nt/p"') and so wehave proved that the Laplace Transform of

Ao+Ait-f- ... + A*-1)1t.

I+.}u(t)

is furnished, over the part of the positive real axis in the complex p-planewhich is covered by the half-plane c > R > 0, by the sum of the con-vergent infinite series (Ao/c) + (Al/c2) + + (A"_,/c") + .


Since both the Laplace Transform of

t+ .....{,.{A +A ...}u(t)o l

and the sum of the infinite series

Ao + Al + + .. .p p2 p"

are analytic functions of the complex variable p over the half-plane c > Rit follows that the Laplace Transform of

Ao + Ai t + ... (n .-,)' "-1 + ...l u(i)

is furnished, over the half-plane c > R, by the sum of the convergentinfinite series (Ao/p) + (A1/p2) + - - + (A"-1/p") + - - . For example,once we are assured that Io(t) = 1 + (t2/22) + (14/22.42) + - - possessesa Laplace Transform over the half-plane c > 1, the result we have justproved tells as that this Laplace Transform is (1/p) + (1/2p8) +(1.3/2.4p6) + ... _ (1 - p2)-'

We have now completed the proof of the theorem stated in our lastlecture, namely, that, if a piecewise continuous right-sided function h(t)is of exponential type i.e., if it possesses a Laplace Transform which iszero at p = cc and is an analytic function f(p) _ (ao/p) + (a1/p2) + - -

of the complex variable p over a neighborhood J p ( > R of p = cc, thenh(t) is the product of u(t) by the sum of the everywhere convergent in-finite series ao + alt + + [a"-1/(n - 1)1Jt"-1 + . We now proposeto show, conversely, that if the sum k (z) = ao + a1z + - - +[a._1/(n - 1)!]z" ' + - of an everywhere convergent power series in acomplex variable z is such that I k(z) 1 < M exp (R I z 1), where M andR are positive real constants, then the right-sided function h(t) = k(t)u(t),- co < t < ao, is of exponential type. Since k(z) is analytic over the entirefinite complex z-plane the integral of k(z)/z"+' around the circle I z I = bin the positive sense, b being any positive number, is 2xi (a,./n !) and,since I k(z) 1 < M exp (Rb) at all points of the circle I zI = b, we havea" 1 < [n! M exp (Rb)J/b" no matter what is the positive number b.

Setting b = n/R we obtain

nl M exp (n)R"

n

or, equivalently, log I a" I < log (n1) + log M + n + n log R - n log n.To appraise the expression on the right-hand side of this inequality weconsider the curve y = log x, x > 0. Since f; dr/x = log b - log a, 0 <


a < b, and since (1/x) < (l/a) over the interval a < x < b, we have(log b - log a)/(b - a) < (1/a) so that the secant of the curve y =log x, which passes through the two points P. : (a, log a), Pb : (b, log b),is less steep than the tangent at P. to the curve and, similarly, this secantis steeper than the tangent at Pb to the curve. It follows that the secantin question does not intersect the curve in more than two points; for, ifP , Pb , P. , where a < b < c, were three collinear points of the curvey = log x the secant PQPa would be at once steeper than, and less steepthan, the tangent to the curve at Pb . Over the interval a < x :5 b -theordinate of the curve y = log x is > the ordinate of the secant PaPb ,the equality holding only at the end points of the interval, and sof b (log x) dx > 1(log a + log b) (b - a) or, equivalently, b log b - b -a log a + a > (1 /a) (log a + log b) (b - a). Setting b = a + 1 and thenassigning to a, in turn, the values 1, - , n we obtain the following ninequalities, n being any positive integer,

2log2 - 1 > 1log231og 3 - 2 log 2 - I > 1 log 2 + } log 3

(n+ 1) log (n+ 1) - nloga -I> 1logn-+ -2log(n+ 1)and these yield, on addition, the inequality (n + 1) log (n + 1) -- n >log (n!) + 1 log (n + 1) or, equivalently, (n + 1) log (n + 1) -- n >log (n!). Hence log (a j < (n + 1) log (n + 1) + log M + n log R --nlogn so that log(j a jI") < (1/2n) log (n + 1) + log (1 + (1/n)) +log R + (1/n) log M. If, then, n is sufficiently large, j log ({ a" jig") -log R j is arbitrarily small, say <log (1 + e), where a is an arbitrarypositive number, so that j an { < {R(1 + e) { " if n is sufficiently large.Thus there exists a positive number M' such that the quotient of j an {by (R(1 + e) { " is less than M' for all non-negative integral values 0, 1,2, of n, and this implies that the sum k*(z) of the everywhere con-vergent infinite series j as j + { a2 j z + (j az 1/2 !) z2 + is dominatedover the entire finite complex z-plane by M' exp [R(1 + e) ( z p. Hence,by the argument of the preceding paragraph, h(t) = k(t)u(t) possesses,over the half-plane c > R, the Laplace Transform (ao/p) + (al/p') + . .

which is zero at p = oo and also, by virtue of the inequality j a,, j <M'{R(1 + e)}', an analytic function of the complex variable p over theneighborhood j p j > R of p = co. In other words, h(t) is of exponentialtype.

If the coefficient ae of 1/p in the expansion, over the neighborhoodp j > R of p = ao , of f (p) as a power series in 1/p is zero we may in-

tegrate f(p) from any point p for which j p j > R to p = co, obtaining

The Laplace Transformation

the new function

ff(q)dq+ `a + aaa+...P 2p2 3p

49

which is the Laplace Transform, over the half-plane c > R, of the productof k(t)/t = a1 + (a2/21)t + (as/3!)e + - by u(t) where k(t)u(t) =(alt + (a2/2 !) t2 + - - - ) u (t) is the right-sided function whose LaplaceTransform, over the half-plane c > R, is f(p) = (al/p2) + (a2/p') + - - - .

Thus, when a function of exponential type is zero at t = 0 its quotient byt is also of exponential type and division by t is reflected, in the domain ofLaplace Transforms, by integration from the point p whose real part c is>R to p = co. For instance, (sin 1)u(t) is a function of exponential typewhich is zero at t = 0, its Laplace Transform, over the half-plane c > 0,being (1 + pz)-1 which is an analytic function of the complex variable pover the neighborhood I p ( > 1 of p = oo. Hence the Laplace Transformof [(sin t)/t]u(t), over the half-plane c > 1, is

fo dq _ f" dep 1+q2Jo 1+s''8-q

= arc tanP

We have already seen that the Laplace Transform of [(sin t)/t]u(t)exists at p = 0 with the value r/2 and so the Laplace Transformof [(sin t)/t]u(t) exists, and is an analytic function of the complex variablep, over the half-plane c > 0. Since are tan 1/p is also an analytic functionof p over the half-plane c > 0 it follows that the Laplace Transform of[(sin t)/t]u(t), over the half-plane c > 0, is are tan 1/p.

EXERCISE

Show that, if the real part a, of a is positive, the Laplace Transform of[(exp (at) - 1)/t]u(t), over the half-plane c > a, , is - log (1 - (a/p)).Show, also, that the Laplace Transform of [(exp (at) - 1 - at)/t2]u(t),over the half-plane c > a,, is a - (a - p) log (1 - (a/p) ).

Denoting by f(p) the function (ao/p) + (al/p2) + which is analyticover the neighborhood I p I > R of p = co and which is furnished by theLaplace Transform of a function (ao + alt + (a:/2!)t= + - -)u(t) =k(t)u(t) of exponential type let us consider the integral fo f(p) exp (pt) dp,

where C is any simple closed curve, of finite length 1, all of whose pointslie in the region I p I > R of the complex p-plane and which encircles thecircle I p I = R and, hence, all the singular points of f (p) . The infinite


series (ao/p) + (a,/p2) + converges uniformly along C and so itmay be integrated along C, after multiplication by exp (pt), term by

term. Since f [(exp (pt))/p"+'] dp = 2rri(t"/n1), it follows thatc

2afof(p)exp(pt)dp=ao-+ -a,t+2 t2+... =k(t)

We already know, from the Laplace version of the Fourier Integral Theo-rem, that

I (e+ioo)f(p) exp (pt) dp = k(t) u(t),c > R

2rri (e-iao)

the convergence of the infinite integral which furnishes, over the half-plane c > R, the Laplace Transform of k(t)u(t) being absolute, and soI/2wi times the integral of f (p) exp (pt) around the circle I p I = r > Rin the positive sense from c + i(r2 c2) } to c - i (r2 - c2);, where R <c < r, has the limit k(t) - k(t)u(t) = k(t)u(-t) as r --> oo. Indeed, theintegral of f(p) exp (pt) around the circle I p I = r in the positive sensefrom c - i(r2 - c2); to c + i(r2 - c2)1 has the limit

.

(o++ac )

f(p) exp (pt) dt = 2iri k(t)u(t)(c-ioo)

as r -> oo, this integral being the same as the integral of f (p) exp (pt)along the segment of the straight-line, real part of p = c, fromc - i(r2 - c2); to c + i(r2 - c2)4.

10

The Polynomials of Laguerre

We are now ready to study the application of Laplace Transforms toordinary linear differential equations with variable coefficients and webegin with the equation

txtt + (1 - t)xt + ax = 0; h = a real constant

which occurs in the wave-mechanical treatment of the hydrogen atom.We assume that there exists a right-sided function h(t) which satisfiesthis differential equation save, possibly, at t = 0, where h(t) may not bedifferentiable. Furthermore, we assume that L(htt) exists at a point p =C, of the positive real axis of the complex p-plane. Then all three of theright-sided functions htt, At and h possess Laplace Transforms which areanalytic functions of the complex variable p over the half-plane c > c,and L(ht) = pLh - h(+0), L(ht,) = p2Lh - ph(+O) - ht(+0). Wedenote Lh simply by f and observe that L(th,) = -{L(ht){, = -f -pf L(th,) = -2pf - p2f, + h(+0), over the half-plane c > c,. Since,by hypothesis, thtt + (1 - t)ht + Ah = 0, save, possibly, at t = 0, itfollows that f satisfies, over the half-plane c > c, , the first-order lineardifferential equation

p(p-1)fv+(p-1-X)f=0Writing this equation in the form

fp a _1 __ a +1f p(p-1} p p-1 p

we see that f is a constant times (p - 1)1/p''+i = (1/p) (1 - (1/p))" sothat f is zero at p = oo and is an analytic function of the complex variable

51


p over the neighborhood I p ( > I of p = o. Thus h(t) is a function ofexponential type which is dominated over 0 < t < co by a constant timesexp [(1 + S)t], where a is an arbitrary positive number. The power seriesdevelopment of (1/p) (1 - (1/p))", over the neighborhood I p { > I ofp= 0, is

1 a X(X - 1) X(X _ 1)(X - 2)

P pa+ 21pe 31 p'-+

and h(t) is thus a constant times the product of u(t) by the sum ofthe everywhere convergent infinite series

e-a(a--1)(i2)e+...(2!)z (3!)2

It is now easy to justify our hypotheses concerning the existence of h(t)by verifying that k,,(t)u(t) satisfies these hypotheses with cl any number> 1. Since each term of ka(t) is dominated by a constant times the cor-responding term of exp [(1 + S)t], and since a power series may be differ-entiated term-by-term the second derivative of k,,(t) is dominated, over0 < t < -, by a constant times exp [(1 + S)t] (the constant being theproduct of the previous constant by (1 + 6)2) so that [k,,(t)u(t)J,t pos-sesses a Laplace Transform at p = I + S', S' > S. It remains only to verifythat x = k,,(t)u(t) satisfies, for every value of t 0 0, the differential equa-tion Ix,, + (1 - t)x, + Ax = 0. To do this we avail ourselves of the rela-tions L(tx,,) _ -2pf - p=fp + x(+0), L(x,) = pf - x(+0), L(txe) =-f - pfp which are valid over the half-plane c > 1, since Lx = f overthis half-plane. It follows, since p(p - 1)fp + (p - 1 - X)f = 0, thatthe Laplace Transform of tx + (1 - t)x, + Ax, over the half-planec > 1, is zero and this implies that tx,t + (1 - t)z, + Ax = 0, - oo <t < co. In particular, t( kA) $, + (1 - t) (ka ), + Xk = 0, 0 < t < -, and,since the left-hand side of this equation is the sum of an everywhere con-vergent power series in t, it follows that this equation is valid over theextended range - co < i < -. Indeed, if we replace t in the power serieswhose sum is k.%(t) by an arbitrary complex number z, we obtain a functionka(z) = 1 - Az + [X(X - 1)/(21)']? - of the complex variable zwhich is analytic over the entire finite complex z-plane, and we know thatz(ka(z)J,. + (1 - z)(k),(z)Js + ak),(z) is zero over the positive part of thereal axis in the complex z-plane. However, this implies that it is zero overthe entire finite complex z-plane and, in particular, over the part - 00 <t < 0 of the real axis in the complex z-plane. Thus x = ka(t) is a solution,over - co < t < oo, of the differential equation txu + (1 - t)x, + ax = 0and we know that x is dominated, over - co < t < a o, by a constant timesexp [(1 + 6) 1 t (]. We now raise the following question: For what values


of A, if any, is x dominated, over -- o < t < CO, by a constant timesexp (a I t 1) where a is any positive number less than 1? For this to bethe case k(t)u(t) must possess, over the half-plane c > a, and not merelyover the half-plane c > 1, a Laplace Transform and so (1/p) (1 - (1/p))*can have no singularities in the half-plane c > a. Hence A must be a non-negative integer for, if not, p = 1 would be a singular point of(1/p) (1 - (1/p))". For example, if X = -1, p = I is a pole of (1/p) (1 -(1/p))" while, if A = 1, p = 1 is a branch point of (1/p) (1 - (1/p)the function (1/p) (1 - (1/p) )t of the complex variable p not being uni-form over any neighborhood of p = 1. When A is a non-negative integern = 0, 1, 2, , the power series whose sum is k), (t) is a constant timesthe polynomial function of t, of degree n, 1 - nt + [n (n - 1) / (2 !) 2]t2 -

. + Choosing the as yet undetermined multiplicativeconstant to be n!, so that the coefficient of t" in this polynomial functionbecomes I)", we obtain the following sequence of polynomial functionsof t:

L.(t) = (-I)" { 1" - netn-1 + n2(n2 1)2 t"-2 - n2(n- 13)'(n - 2)2 t"-a

+ ... .+(-1)"nl}, n=0,1,2,...

For example,

Lo(t) = 1, L1(t) _ -(t - 1), Le(t) = t' - 4t + 2

Le(t) = - (t' - 9t2 +18t - 6)

L,(t) = t' - 16t' + 72t2 - 961 +24

and so on. These polynomials are known as the polynomials of Laguerreand the differential equation tx tt + (1 - t) x t + nx = 0 is known as La-guerre's differential equation of index n. The Laplace Transform, over thehalf-plane c > 0, of L.(t)u(t) is (n!/p) (1 - (1/p))". The restriction ofA to the non-negative integers 0, 1, 2, . , which is imposed by the re-quirement that e- "kx(t), 0 < a < 1, be bounded over - cc < t < -,furnishes the quantisation of the radial coordinate in the wave-mechanicaltheory of the hydrogen atom.

It is easy to show, on availing ourselves of the fact that the LaplaceTransform of L"(t)u(t), over the half-plane c > 0, is (n!/p) (1 - (1/p))",that the function x" = of the unrestricted real variable t satisfies thelinear second-order difference equation

x.+1 + (t - 2n - 1)x" + n x"_1 = 0, n = 1, 2, 3,

54 Lectures an Applied Mathematics

Indeed the Laplace Transform of tL.(t)u(t), over the half-plane c > 0, is

n! 1 - 1N' _ n(n.f) 1 -On-1

P ( _P) _P

and this appears, on writing

1 I=I 1-2 (1 p+\1pp- p

in the form

t

-(n p1)!+ 1 - 1) 1 + (2n +p1)(n!)

\1 - 1p P)

n'(n p1)1

(1 _ pYwhich is the Laplace Transform, over the half-plane c > 0, of

(2n + n0La_1(t)u(t). Hence, by virtue ofthe uniqueness theorem,

tL.(t)u(t) = (2n + 1)L"(t)u(t) - n'L,,_1(t)u(t)

or, equivalently, tL,,(t) = (2n + 1)L,.(t) - n2L,_1(t), 0 <t < co. Since this equation connecting polynomials is valid for more thana finite number of values of t it must be an identity so that, on collectingterms,

L.+1(t) + (i - 2n - 1)L"(t) + n'L,,.-1(t) = 0, - oc < t < ocSimilarly, we can show that L.(t) is the product of the nth derivative,D"[t" exp (-t)], of t" exp (-t) by exp t. Indeed the Laplace Transform,over the half-plane c > -1, of exp (-t)u(t) is 1/(p + 1) and thus theLaplace Transform, over the half-plane c > -1, of t" exp (-t)u(t) isn !/ (p + 1) "+1. Since t" exp (- t) is zero, together with its derivatives upto the order n -- 1, inclusive, at t = 0, if follows that the Laplace Trans-form, over the half-plane c > -1, of D"[t" exp (-t)]u(t) is nlp"/(p + 1)n+'so that, by virtue of the Translation Theorem, the Laplace Transform,over the half-plane c > 0, of exp (t) D"[t" exp (-t)]u(t) is

n! (p-1). -n! 1-p p

"

Hence exp (t) D"[t" exp (-t)] = L" (t ), 0 < t < co, which implies, sinceboth sides of the equation are polynomial functions of t, that

exp (t) D"[t" exp (--t)], - oo < t < aoLet us denote by L.*(t) the polynomial function of t, of degree n, ob-


tamed by replacing each coefficient of L"(t) by its absolute value. Forexample,

Lo*(t) = 1, Ll*(t) = t + 1, Lz*(t) = e+ 4t + 2

W (t) = t'+9t'+ 18t + 6and so on, L"* (t) being L. (- t), n = 0, 1, 2, . Then the Laplace Trans-form of L"*(t)u(t), over the half-plane c > 0, being obtained from theLaplace Transform of L"(t)u(t) by replacing each coefficient in the de-velopment of this latter as a power series in 1/p by its absolute value, is(n!/p)(1 + (1/p))". Since L"*(t) = L"(-t), x"* = L"*(t) satisfies thedifferential equation

t (x"*) t t + (1 + t) (x"*) t - nx"* = 0

which we term the modified Laguerre differential equation of index n.Similarly x"* satisfies the linear second order difference equation

x"+, - (t + 2n + 1)x"* + n'x*.-, = 0

and

L.* (t) = exp ( - t) D"(t" exp t), n = 0, 1, 2,

Let x be any non-negative real number less than 1 and let us considerthe non-negative double series (u/k) where

u,(j) fix'

Then the - X co matrix which has u/k as the element in its (k + 1)strow and (j + 1) st column is triangular with zeros above the diagonal, the

binomial coefficient being zero if k < j. The sum of the elements in

the (k + 1)st row of t his co X co triangular matrix is (xk/k!)Lk*(t) whilethe sum of the elements in the (j + 1)st column is

1 + (j + 1)x + (.7 + 22(.7 + 1) xa + ...} =j!

(lOx'x),

Hence the sum by columns of the non-negative double series (u/k) existswith the value 11/(1 - x) J exp tx/(1 - x) and this implies that the sumby rows, namely,

0o kx

Lk* t


exists with the same value. Thus

kkLk'(t)=1 xexp1- x 0 <x<1,0<-t<

If we multiply u/k by (-1)' we obtain a new double series (v/k), where

k (klv;

A0,1,2,...k=0,1,2,---

This double series is no longer non-negative over 0 < t < o' but easbof the two double series (u/k ± v/k) is non-negative and the sum by columnsof each of these two non-negative double series exists. Since v/1 =I(u/k + v/k) - +)(u/k - v t) it follows that, despite the fact that thedouble series (v/k) is not non-negative, its sum by rows exists and has thesame value as its sum by columns. Thus

xkLk(t) =

1 eRp -tx,-o k! 1 x 1 x

which is the same thing as saying that the previous relation

xkLk

_ 1 txk-o k!

(t) 1-xexp 1-xis valid over the entire t-axis, - - < t < c o. The same argument showsthat we may change the sign of x and so

totXk! Lk'(t) = 1 1

xe x p 1

x, -1 < x < -1, - ao < t < 00

or, equivalently,

E xkLk(t) 1 exp -1 < z < -1 00 < t < ao.

k_ok! 1-x 1-x' '

We conclude with the remark that Laguerre's differential equation txet +(1 - t )z j + nx = 0 possesses a second solution, linearly independent ofL.(t), but this solution does not have as much importance as L.(t) inapplications to physical problems since it possesses a logarithmic singularityat t = 0. For example, when n = 0, this second solution may be taken tobe the indefinite integral, f' ((exp a)/s} ds, of (exp t)/t. If this second solu-tion is required it may be obtained by writing x = yL.(t) in Laguerre'sdifferential equation; on doing this we find that y, is a constant times thequotient of exp t by IL.'(t).

11

Bessel's Differential Equation

As a second application of the Laplace Transformation to ordinary lineardifferential equations with variable coefficients we consider the second-order linear differential equation

Oxee+txe+(t=-n')x=0which is known as Bessel's differential equation, of index n, n = 0, 1, 2, ,being a non-negative integer. It is not difficult to see that x,.(t)(1/ir) f o oos (t sin 8 - na) d8 is a solution of this differential equation.Indeed,

{x,(t) {e = -- * sin (t sin 0 - ne)(sin 8) d8

{u,.(t) In = -- * cos (t sin 8 - ns)(sin' 8) d8

so that

{x.(t) }ee + 4(t) _ Cog (1 sin 8 - na)(cos' B) de

A simple integration by parts yields

IX;.(t);e = sin (t sin a - no) cos a 10'

--I cos(tsina-n8)(cos 8)(tcos8-n)d8

= -t I* coo (t sin a -- no) (cos' 0) dO

+ cos (t sin 8 - no) (coo 0) do

67


so that

t{xn(t)},I + tx,(t) 1* coo (t sin e - no) (cos e) do

Now

{cos(tsin0-n8)}(tcose-n)do= sin(tsinO-n0)l=0Iand so

t fcos(tsin0-nO)}(cos0)doa

= n k* {coo (t sin 9 - n6) } d6 = naxn(t),

so thatt2{xn}« + t{x (t)}, + t'x(t) = n2x,(t)

which proves that x (t) is a solution of Bessel's equation of index n. Wedenote this solution by Jn(t) and observe that

I Jn(t) I< 1, 11 J. (0) e i< 1,1 {J'.MI a (< 1

so that the right-sided function An(t) = is a solution, save,possibly, at t = 0, where h.(t) may not be differentiable, of Bessel's differ-ential equation of index n, this solution being such that the three piecewisecontinuous right-sided functions hn(t), {h,(t)} _ , t, all possess LaplaceTransforms which are analytic functions of the complex variable p overthe half-plane c > 0. On denoting the Laplace Transform of h,(t), overthe half-plane c > 0, by f we have I f 1 < f 'o exp (-ct) dt = 1/c, c > 0,so that I f I tends to zero as c -p x' . Furthermore

L[{hn(t)}=} = pf - h.(+0),

L[{h (t)} al = p2f - phn(+0) - {h,.}t(+0),

L(thn) _ -ft, , L(t2h,.) = frn ,

L[t{hn(t)):) = -f - pf, -p2fv - 2pf + hn(+0),L[t2{hn(t)) ) = p2ff, + 4pff + 2f,

all these equations being valid over the half-plane c > 0. Sincet2{hn(t)} + t{h (t)}, + (t2 - n2)h,.(t) = 0, t 5 0,

it follows that, over the half-plane c > 0,

(1 + p2)f, + 3pf, + (1 - n')f = 0


This homogeneous second-order linear differential equation is readilysolved by setting p = sinh z so that f, _ (cosh z)f, , f = (cosh2z)f , +(sinh z)f, . Thus f, = (sech z)f, , (1 + p2)f,,, = f.. - (tanh z)f, , so thatf + 2(tanh z)f, + (1 - n2)f = 0. Writing, finally, (cosh z)f = f', wehave (cosh z)f, + (sinh z)f = f,', (cosh z)f,, + 2(sinh z)f, + (cosh z)f =f:, so that f;, = n2(cosh z)f = n2f . If, then, n = 1, 2, 3, , f' is a linearcombination, with constant coefficients, of exp nz and exp (-nz). Ifx and y are the real and imaginary parts, respectively, of z, c = sinh x cos yand so, if - (r/2) < y < (v/2), c -+ co as x - co. Furthermore, thequotient of exp (nx) by cosh x -i 2 as x --* co , if n = 1, and -- oo, asx --> ao , if n = 2, 3, - , while the quotient of exp (- nx) by cosh x --- 0as x --, oo, n = 1, 2, 3, . . Since f = f'/cosh z tends to zero as c -+ o0it follows that the coefficient of exp (nz) in the linear combination ofexp (nz) and exp (-nz) which furnishes f must be zero. Thus, if n =1, 2, 3, , f(p) is a constant times exp (-nz) divided by cosh z, i.e.,a constant times (cosh z - sink z)" = ( (1 + p2); - p) " divided by (1 +p')}. On the other hand, if n = 0, f is a linear function of z so that f isof the form (a + bz)/cosh z, where a and b are constants. Hence at anypoint c = sinh x of the positive real axis in the complex p-plane, cf =c(a + bx) /cosh x = tanh x(a + bx) and this is not bounded at x = Gounless b = 0 (since tanh x -+ 1 as x --' co ). Thus, for all non-negativeintegral values, 0, 1, 2, . , of n, f is a constant times exp (-nz) dividedby cosh z, i.e., a constant, C, times { (1 + p2)4 - p] n/(1 + p2)4. Sincethis function of the complex variable p is zero at p = 0 and analytic overthe neighborhood I p j > 1 of p = oo, its development as a power seriesin 1/p starting out with p2); - p being p(1 + p -Y -p = (1/2p) + - ), h,,(t) is a function of exponential type, namely, theproduct of u(t) by an everywhere convergent power series in t whichstarts out with the term Thus vanishes, together withits derivatives up to the (n - 1)st, inclusive, at t = 0, while the nthderivative of J"(t) does not vanish at t - 0. The nth derivative of

J"(t) = 1 k coo (t sin 0 - no) de, at t = 0ir

is

sin (no) sin" 0 do011/f0'ra

if n = 2k + 1 is odd, and(_1)k ft

a ocos (n9) sin" 8 dA


if n - 2k is even. In the first case, sin" o = (2i)-"{exp (io) - exp (-io))"is a linear combination of sines of odd integral multiples of B, the coefficientof sin no in this linear combination being (2i)-"+' = (-1)k/2"-` while,in the second case, sin" 0 is a linear combination of cosines of even integralmultiples of 0, the coefficient of cos no in this linear combination being2(2i)-% = (-1)`/2"-'. Now (1/r) fo (sin no) (sin mode = 0, if the oddnumbers m and n are unequal, and (1/r) f o' (cos no) (cos mo)do = 0, if theeven numbers in and n are unequal, while both (1/r) f o' {sin no}' do, n =1, 3, 5, and (1/r) f o' { cos no}' do, n = 2, 4, . have the common value 1.Hence, if n is any positiv.' integer, the nth derivative of J,(t) has, at t = 0,the value 1/2" and this is also true when n = 0, the value of Jo(t) _ (1/7)fo cos(t sin 0) do at t = 0 being 1. Thus the multiplicative constant Cisland

L{J,(t)u(t) } ej'P "z) p = sinh z, n = 0, 1, 2, , c > 0

{(1 + e)i - pin c > 0(1 +P');In particular,

L{Jo(t)u(t) } = (1 + p')+' L{Ja(t)u(t) } 1 - 1 -1', c >0

Since J,(0) = 0, it follows that

L{tJ,(t)u(t) } = p 1

In general, the Iaplace Transform, over the half-plane c > 0, of t"J,(t)u(t)is simpler, when n > 0, than that of J,(t)u(t). To obtain this LaplaceTransform we first set up the differential equation which is satisfied byy = t"x, where x = J,(t). Thusz = 97'Y, x, = t "y, - ni "-'1/'

x = t "yet - 2nt `-1y. + n(n + 1)t-'2yand, since t'x + ft, + (t' - n')x = 0, it follows that ty - (2n - 1)y, +ty = 0. Denoting L{yu(t)} by g and noting that y(0) = 0, since n > 0,we have

L[{yu(t)} e] = pg, L[{yu(t)} ] = e q - y'(0)so that

L[1 { yu(t)1 u] '° -p=g, - 2Pg


and, since L{tyu(t)} = -g,, we obtain

(1+p2)ga+(2n+1)pg=0so that g is a constant times (1 + p2)-"-4 and, since the power series de-velopment of y starts out with the term t="/2"(n1), the multiplying con-stant is

(2n)1 - (2n - 1)(2n - 3) 3.1. Hence L{t"J"(t)u(t)}25(n!)

_(2n-1)(2n-3) .3.1 c>0,n=0,1,2 ,(1 + p2)"+(1/2) f

(it being understood that, when n = 0, (2n - 1) , 3.1 is replaced by 1) .Since the development of [(2n - 1) ... 3.1]/(1 + p2) '+' as a power seriesin 1/p is

1 n+,} (n+})(n+1) _(2n - 1) ... 3.1 p:"+1 - +.+a + 21 ps-+a

the development of t"J"(t) as a power series in t is

(2n- 1) ... 3.1{ t'" _ (n+!})e`" (n+4)(n+$)t"+` _l(2n)1 (2n + 2) ! + (2n + 4)121

and so the development of J (t) as a power series in t is

1 (t1

W1 (2)" (n + 1)1(2}"+s + (n +12) ! 2! It

+a

(n + 3)13 ! (2)± .. .

For example

Jo(t) = 1 - 22+2242 -22.4 62+

Je(t) = t - t' t°2224+22426- ...

so that J;(t) is the negative of the derivative of Jo(t) with respect to t;this result could have been predicted, without computation, from the factthat L{J,(t)u(t)}, namely 1 - [p/(1 + is the negative of

pL{Jo(t)u(t)) - Jo(0).

EXERCISE

Show that the product of u(t) by J"(t)/t" is a function of exponential


type whose Laplace Transform, over the half-plane c > 0, is2"-1(cosh z} f (sech v)2n dv, p = sinh z

2"-1(n - 1)! .

and, in particular, that the Laplace Transform, over the half-plane c > 0,of (J,(t)/t]u(t) is cosh z - sinh z = (1 + p2)1 - p. (Hint: The differentialequation satisfied by to = C"x, where x = J"(t), is twe, + (2n + 1)w$ +tw = 0 and it follows, since w(O) = that the differential equa-tion satisfied by L(wu(t)) = g' is (1 + p2)gn - (2n - 1) pg' --11/(2"-l (n - 1)1)]. The solution of this differential equation which iszero at p = oo is g' = (1 + p2)"mss, where

1: "+}) 89 = -(1 + p1)!

and s is zero at p = co. Thus

"1 f00 t4 1 2= (sech v) dvs 1)! f, (1 + q2)"+ ` 1fo

where q = sinh v, p = sinh z.)If we replace each coefficient of the everywhere convergent power series

in t whose sum is Jn(t) by its absolute value we obtain a new everywhere-convergent power series in t whose sum, J"*(t), is the product of by(-i)". Jn*(t) is termed the modified Bessel function, of the first kind, ofindex n and is usually denoted by I"(t) but we shall use, for the present,the notation J"* (t) to recall the manner in which the modified Besselfunction, J"*(t) or I" (t) , is derived from J" (t) . The expansion, over theneighborhood I p I > 1 of p = -, of the Laplace Transform of J"*(t)u(t)as a power series in 1/p is obtained from the corresponding expansion ofthe Laplace Transform of by replacing each coefficient of thelatter expansion by its absolute value. Now the expansion of the LaplaceTransform of Jn(t)u(t) as a power series in 1/p is of the form

f" (p) = a" + a1+2 + 1+4 + .. .P+1n pn+a pn+a

where an = 1/2" and the coefficients an , a"+2 , are alternately positiveand negative. Hence

f"(ip) _ (-011+1 an - an+2 + a"+,p"+1 i;;i p"+s

(-i)"+11I a" I + I an+2 I .+ ...}p"+1 p"+s


so that f"*(p), the sum, over I p I > 1, of the power series (I a" I/p"+`) +(I a"+2 I /p"+') + ... , is

i+'f (i) = i+1 ((1-p2)I_ip}"" p (1 - p2)1

Writing (1 - p2); = i(p2 - 1); this takes the form

Ip - (p2 - 1)iI"(p2-1)a

which is exp (-nz*)/sinh z* where cosh z* = p. Thus, side by side withthe result

L(J"(t)u(t) } = exp (-nz)cosh z

we have the result

L(J"*(t)u(t) } = exp (-nz*)sinh z*

p = sinh z, c > 0

p = cosh z*,c>0

EXERCISE 1

Show that x"* = J"*(t) = I"(t) satisfies the modified Bessel equation,of index n, t2(x"*),, + t(x"*): - t2 + n2)x"* = 0 (Hint: Denoting J"(t)by x"(t), x"(z), where z is a complex variable, satisfies the differential equa-tion z2(x")ss + z(x")= + (z2 - n2)x" = 0. Thus along the imaginary axisz = iy in the complex z-plane, along which (x"), = i(x")t, (x"),,, =i2(x"):s , we have y2(x")vv + (y2 + n2)x" = 0 and, since x"(iy)is a constant times x"*(y) it follows that y2(x"*),,,, + y(x"*)v(y2 + n2)xn* = 0, - ao < y < cc .)

EXERCISE 2

Show that if n = 1, 2, both I J"(z) I and I J"*(z) are dominatedover the entire complex z-plane by a constant times

IZI-"eXp[(1 + 6) Izllwhere a is an arbitrary positive number. (Hint:

L{t"J"(t)u(t) } _ (2n - 1) ... 3.1(1 + p2)"+(1I2)

and the development of (1 + p2)-"_t as a power series in 1/p convergesatp= 1+5.)

Note. If we substitute p = 1 in this series we obtain a series whose termssteadily increase in numerical value and which, therefore, fails to converge.

64 Lectures on A pplied Mathematics

EXExcisJ 3

Show that both 1 Jo(z) I and ( Jo*(z) f = 1 Io(z) I are dominated, overthe the entire complex z-plane by exp I z j . (Hint: The developmentof (1 + p')`} as a power series in 1/p reduces, when p = 1, to 1 - I +(1.3/2.4) - which converges; furthermore each of its terms isnumerically < 1.

12

The Recurrence and

other Relations Connecting

Bessel Functions

The Laplace Transforms, over the half-plane c > 0, of the two right-sidedfunctions t "!'J"(2ti)u(t), each of which is of exponential type, are particu-larly simple. To obtain them we set 20 = a, t > 0, and we denote J"(a) byx(a) and write x(a) y(t). Then y, = x,t4 so that x, = ttye - Jays andx.. _ jys + *eyes = Ie + tyee. Since a=x.. + ax. + (a' - ns)x = 0 itfollows that dyes + tye + It - (n'/4) }y - 0. We next set y = £-*"v so thaty,

= t-"nve - (n/2) tc-"rn-ev, yes = f"twee - ntc-"nE-'ve + (n/2) [(n/2) +1]v and find that v satisfies the differential equation tvee + (1 - n) v: + v =0. All three of the right-sided functions v(t)u(t), v,(t)u(t), vt,(t)u(t) pos-sess Laplace Transforms over the half-plane c > 0 and, on denoting by gthe first of these three Laplace Transforms, we have L[ve(t)u(t)] = pg --

- - - ,v(0),L[vee(t)u(t)1 - p'g - pv(0) - ve(0) wherev(0) = O if n = 1,2,3,while v(0) = 1 if n - 0. Since L[tv,,(t)u(t)1 = - 2pg - peg, + v(0), gsatisfies the differential equation p=g, + { (n + I) p - 1}g - nv(0) = 0 or,equivalently, since rw(0) = 0, n = 0, 1, 2, , p'g, + { (n + 1) p - 1 }g =0. Thus g is a constant times exp (-1 /p) /p"+' and, since the developmentof t"12J" (2t}) as a power series in t starts out with the term (1 /n 1) t", themultiplying constant is 1. Thus

lexpL[t"12J"(2t)u(t)1

p"+4 c > 0, n = 0, 1, 2, ...

66

66 Lectures on Applied Mathwmatic8

From this we obtain the Laplace Transform, over the half-place c > 0,of L[t ""J" (2t})u(t)] by integrating n times, with respect to p, from p toco (t-""Jn(2t}) being the quotient of to"J"(2t}) by t"). Writing

1)exp (_p = 1 - 1 _ 1 - ...p"+, p*+, p*+2 21 p*+*

we obtain

1 1 in1p- (n+1)!p2+(n+2)!p'

which is the part involving negative powers of p in the Laurent develop-ment of (-1) 'p"-' exp (-1 /p) over 0 < I z I < ao, i.e., the finite complexz-plane punctured at the origin:

L[t-"r:JA(2t})u(t))= nlp

(n-+11

)1 p; + (n +2)!p' -...

c>0,n=0,1,2,Thus, multiplication of the Laplace Transform of t""J"(20)u(t), n =1, 2, 3, , by p is equivalent to replacing n by n - 1 while multiplicationof the Laplace Transform of t ""J"(2t!)u(t), n = 0, 1, 2, , by p, fol-lowed by subtraction of 1/n!, the value of t"J%(2t;) at t = 0, is equivalentto replacing n by n + 1 and changing the sign of the Laplace Transform.This implies, by virtue of the uniqueness theorem, that

(a) The derivative of tn'V.(2t}), n > 0, with respect to t ist("-,)r2J -,(2t})

and

(b) The derivative of F"'%(20), n > 0, with respect to t is-tt("+,)i:t J"+1(20)

Since differentiation with respect to t is the same as differentiation withrespect to s = 20 followed by multiplication by t } = 2/s, these resultscan be expressed as follows:

(a') The derivative of (8/2)"J"(s), n = 1, 2, 3, , with respect to 8,is (a/2)"J"_,(s)

(b') The derivative of (s/2)-"J"(s), n = 0, 1, 2, , with respect to s,is -(8/2)-"J"+,(s)

The above results have been proven only for non-negative values of sbut they remain valid if 8 is replaced by an arbitrary complex number z,


since each side of each of the resulting equations is an analytic function ofthe complex variable z over the entire finite complex z-plane. In this waywe obtain the following two sequences of relations:

z{J"(z)}, + nJ,.(z) = zJ.-,(z), n = 1,2,3,z arbitrary

and these yield, on addition and subtraction, the sequences of relations:

2{J"(z)), = J"_1(z) - n = 1, 2, 3, ;

2nJ,.(z) = z{J,._,(z) + J"+I(z)}, n = 1, 2,3Y . ;z arbitrary

The second of these two sequences of relations furnishes what are knownas the recurrence relations connecting the sequence Jo(z), Jj(z), J2(z),of Bessel functions of the first kind. These recurrence relations express thefact that J,.(z) is a solution of the linear second-order difference equation

J"+1(z) - 2n J,.(z) + J,.-,(z) = 0; n = 1, 2, 3, ; z arbitrary.z

EXERCISE 1Show that the sequence of modified Bessel functions I"(z) = J"*(z),

satisfies the two sequences of relations

2{I"(z)}, = I"-,(z) + I"+,(z), 2nI%(z) = z{I"-a(z) - I"+i(z)},

n=1,2,3,EXRacISE 2

Show that the Laplace Transforms, over the half-plane c > 1, oft"r:If(2t})u(t) and t-""I,.(20)u(t) are exp (1/p)/p"+' and [1/(n1p)] +[1/((n + 1) !p')) + [1/((n + 2) !p')) + ... , respectively.

We have seen that the Laplace Transform, over the half-plane c > 1,of J""(t)u(t) = I"(t)u(t) is exp(-nz*)/sinh z*, where p = cosh z*. Ontaking the real and imaginary parts of the equation p = cosh z*, we obtainc = (cosh x*) cos y*, w - (sink x*) sin y* where c, to are the real and imag-inary parts, respectively, of p and x*, y* are the real and imaginary parts,respectively, of z*. Thus the relation p = cosh z* maps the strip - 7/2 <y* < a/2 in the complex z*-plane onto the half-plane c > 0, the points ofthis strip in the complex z*-plane for which x* has any given value otherthan zero mapping into the points of the ellipse [c'/ ((cosh x*) 2) ] + [ws/((sinh x*)') ] - 1 in the complex p-plane, and the points of the strip forwhich x* = 0 mapping into the line segment 0 < c < 1, w = 0. Thus therelation p = cosh z* furnishes a one-to-one mapping of the positive half,x* > 0, of the strip - r/2 < y* < r/2 onto the half-plane c > 0 with theline-segment 0 < c < 1, w = 0, removed. Over the positive half, x* > 0,


of the strip - r/2 < y < ,r/2, ( exp(-z*) I = exp(-x*) is < 1 and so, 0being any real number, I exp[- (z* -- ie) ] I < I so that the infinite series1 + 2 exp[-2(z* - ie)] + 2 exp[-4(z* - ie)] + 2 exp[-6(z* - 19)] +converges, its sum being 1 + 12 exp[-2(z* - ie)]/1 - exp[_2(z* -ie)]) = cosh(z* - ie)/sinh(z* - ie). If z* = x* is real and positive itfollows, on taking the real parts of the terms of the infinite series, that

1 + 2 (eos 20) exp (-2x*) + 2 (cos 40) exp (- 4x*) + .. .

1 cosh (x* - io) cosh Cr + ie)2 sinh (x* - io) + sinh (x* + ie)

_= sinh x* cosh x*sinh 2x*2 sinh' x* cos= 0 + cosh2 x* sin= e] cosh' x* - cos' 0

When z = x* is real and positive, p = c is real and > 1 and exp (- nx*) /sink x* is the value at p = c of the Laplace Transform, f.* (p), of J.* (t) u(t)and so we find, on division by sinh x*, that

fo*(c) + 2 (cos 2s)f:*(c) + 2 (coa 40f.*(c) + ... - c= - cos' 0

In particular, when 0 = 0,

fo c' C 1

The coefficients of the development of as a power series in 1/c arenon-negative real numbers and we construct the non-negative doubleseries (u,k) where u! , j = 0, 1, 2, , k = 0, 1, 2, is the term involving/ (), g , while Eo = 1. The oo X co1 c"*1in8 st c St bein 2ifk = 1 2

matrix which has uit as the element in its (j + 1) at column and (k + 1) strow is triangular, with zeros below the diagonal, since fu* (c) starts out withthe 1/e+1 term. The sum of the elements in the (k + 1)st row of thematrix is 8.f(c) and so we know that the sum by rows of the non-negativedouble series (ujt) exists with the value c/(c' - 1) and this implies thatthe sum by columns exists with the same value. The sum by columns is apower series in 1/c whose sum, over the part c > 1 of the real axis in thecomplex p-plane, is (1/c) + (I/c') + and so the sum of the elementsin the Cl + 1) st column is 1/c''+1. Let us now consider the non-negativedouble series (v;4), where vft = c"+'[tsf/((2j)1)]uI, t any real number; thesum of the elements in the (j + 1) at column of the co X oo matrix whichhas v!t as the element in its (j + 1) at column and (k + 1) at row, j =0, 1, 2, , k = 0, 1, 2, , is ?'/((2j)1) so that the sum by columns ofthe non-negative double series (v;t) exists with the value cosh t and thisimplies that the sum by rows of the non-negative double series (v;") exists


with the value cosh t. The sum of the elements in the (k + 1) at row isdkJa2h(t) and so we have the relation

Jo*(t) + 2J2*(t) + 2J4*(t) + = cosh t, - o < t <which implies, in particular, that Jo*(t) < cosh t over - oo < t < -. Wemay now apply the argument just given, which depended only on the non-negativeness of Sk , k = 0, 1, 2, , to the two series

fo*(c) + {1 + (cos 20))f2* (c) + {1 + (cos 48) If,* (c) +{1 -- (cos2B)}f2*(c) + {1 - (cos48)}f,,*(c) +

whose sums are

andC

c

-l+cs-cos= 042 7-1 c _ c

c2-1 c2-cos2Brespectively. We find that the two series

Jo*(t) + { 1 + (cos 26) J2*(t) + { 1 + (cos 48) } J4*(t) +

{1 - (cos2B)}J2*(t) + {1 - (cos 46)}Jik*(9) +

converge over - < t < -, their sums being J{cosh t + cosh (t cos B)}and cosh t - cosh (t cos B) }, respectively, the Laplace Transformof cosh (t cos 8)u(t), at p = c > 1, being c/(c2 - cos2B). Hence, on sub-traction,

Jo*(t) + 2(cos 20) J2*(t) + 2(cos 40) Jj*(t) + = cosh(t cos 0),

-00 <t<-,--<0<00EXERCISE 3

Show that (cos 0)Jl*(t) + (cos 30) Js*(t) + _ +} sinh (t cos 0),- - < t < -, - - < 0 < co .(Hint. The infinite series, exp[-(x* -i8] + exp[-3(x* - i8) ] + , is convergent, with the sum 1/[2 sinh(x* -10) ], if x* > 0.)

The facts that J*k(t), k = 0, 1, 2, , is a non-negative continuousfunction of the unrestricted real variable t and that the sum, cosh t, of theeverywhere convergent infinite series Jo*(t) + 2J2*(t) + is every-where continuous assure us that the convergence of this infinite series isuniform over the closed interval 0 < t < T, where T is an arbitrary positivenumber. Indeed, the remainder, R.(t), after n terms of this infinite seriespossesses the following two properties:

(1) It is continuous over the interval 0 < t < T.(2) 0 < R..(t) < R.(t), - ao < t < oo, n' > n.

70 Leeturee on Applied Mathematics

By virtue of (1) R (t) assumes its maximum value, over the interval0 < t < T, at some point, t say, of this interval and the infinite sequenceof numbers tl , t2, - possesses at least one accumulation point, l say,in the interval 0 < t < T. Since R,.(t), n = 1, 2, - , is continuous at lwe know that I .(t) - R,.(l) I is arbitrarily small, say < e, if I t - I Iis sufficiently small, say < S . Since the infinite series Jo*(t) + 2J,*(t) +

is convergent at 1, 0 < R,.(l) < e if n is sufficiently large and we denoteby N any such sufficiently large value of n so that 0 < Rr,(l) < e. Then0 < RN(t) < 2e if t - I I < 5,, and this implies that 0 < R,.(t) < 2eif n > N and I t - l < 8,, . There exists, since I is an accumulation pointof the sequence of numbers tl , t2, , an n, say n', > N suchthat 14, -- 1 I < 8, and so 0 < R,.' 2e which implies, by virtue ofthe definition of t,. , that 0 < RA'(t) < 2e, 0 < t -< T, and, hence, that0 < R. (t) < 2e over 0 < t < T if n >- n', the choice of n' being independentof t. In other words, the convergence, over the interval 0 < t < T, of theinfinite series Jo*(t) + 2J,*(t) + 2J4*(t) + - - is uniform. If z is anycomplex number, Jl*(z), k = 0, 1, 2, , is dominated by J k(I z I) and,so the infinite series Jo*(z) + 2(cos 8) J2* (z) + 2(cos 48) J4* (z) + ,

being dominated by the infinite series Jo* (I z I) + 2 J,* (I z I) + con-verges uniformly over the disc, 0 < I z I < T, with center at the origin,in the. complex z-plane. Each term of this infinite series is an analy-tic. function of the complex variable z over the disc and so the sumof the infinite series is an analytic function of z over the disc as is alsocosh(z cos 0). Since these two analytic functions of z coincide overthe diameter -- T < t < T of the disc they coincide over the entiredisc and this implies, since the positive number T is arbitrary, that theycoincide over the entire finite complex z-plane. Thus Jo*(z) +2(cos 20)J,*(z) + 2(cos 46)J4*(z) + = cosh(z cos 0), z arbitrary,- - < 8 < -. Assigning purely imaginary values it to z we obtain

Jo(t) - 2(cos 28)J,(t) + 2(cos 48)/4(1) - = cost cos 8),

-00<t<-, -00<0<00and, in particular, on setting a - = 0,

Jo(t) - 2J2(t) + 2J4(t) - = cos t, - oo < t < 00.On setting 8 = ir/3 in the relation Jo*(t) + 2(cos 28) J,* (t) + _

cosh(t cos 8) we obtain Jo*(t) - J,*(t) - J4*(t) + 2Je*(t) -cosh (t/2) and, on' combining this relation with the relation Jo*(t) +2J,*(t) + = cosh t, we obtainJ,*(t) + 2J,*(t) + 2Jli(t) + _ 1{cosh t + 2 cosh(t/2)},

-- oo < t < Go


Thus if cosh t + 2 cosh(t/2)} is an upper bound, over - oo < t < co,for Jo*(t), the excess of this upper bound over J6*(t) being twice the sumof the infinite series J6* (t) + J i (t) + . If j t f < 5 and n > 6,

1 t()J,.*t, (t)"2(n + 1)! 2 + (n -+2)! 2R 2-+ .. .

is less than

1)2

()4±...}nt 0)" + n 1 t2)2 + (n +1

For example, when t = 1, J6* (1) < (2.3)10-b, J111(1) < 6.10-'a, andso on, so that J{cosh 1 + 2 cosh 1} is greater than 42(1), the excess beingless than (4.7) 10-6. Actually, }{cosh I + 2 cosh ,)} = 1.26611, Jo*(1) _1.26607, both to 5 decimals.

On replacing 0 by 0 - (r/2) in the relation

Jo*(t) + 2(cos 20)Js*(t) + . . = cosh(t cos 0),

we obtain

Jo*(t) - 2(cos 28) J2*(t) + 2(cos 40) J4*(t) - = cosh(t sin 8)

and on setting, in turn, 0 = 0 and 0 = r/3 in this relation we obtain thetwo relations

Jo*(t) - 2Jz*(t) + 2J4*(t) - = 1

Jo*(t) + J:*(t) - J4*(t) - 2J6*(t) - = cosh (2t)

On combining these two relations we obtain

iJ0*(t) - 2J6*(t) + 2J 12* (1) - ... = 3 1 + 2 cosh

2t)Now it folows from the recurrence relation t{ J;_x(t) - JA-, (t) },n = 1, 2, 3, , that J*_,(t) > J:+,(t), the equality holding only whenn > 1 and t = G. Thus J6*(t) > J6*(t) > Jo(t) > Jiz(t) so thatthe sum of the infinite series J6*(t) - Jli(t) + is non-negative.Hence 4;1 + 2 cosh(3tt/2) } is a lower bound, over - oo < t < c, forJo. (t), the numerical value of the difference between Jo*(t) and thislower bound being twice the sum of the infinite series j6*(".J1 (t) + . We have, then, obtained an upper and a lower bound, oats


- oo < t < ao, for Jo `(t) and we know that the mean of these two bounds,namely,

11 + cosh t + cosh C2) + cosh ()}

is an upper bound, over - ao < t < co, for Jo*(t), the excess of this upperbound over Jo*(t) being twice the sum of the infinite series J (t) +

Je(t)+...

On combining the two relations Jo*(t) + 2(cos 20) Ja*(t) + acosh(t cos B), Jo*(t) - 2(cos 20)J,*(t) + ... = cosh(t sin B), we obtainthe relation

Jo*(t) + 2(cos 40) J*(t) + _ i{cosh(t cos e) + cosh(t sin B)}and from this we deduce that i{cosh(t cos(r/12)) + cosh(t (sin r/12)) +cosh(2it) I is a lower bound, over - ao < t < oo, for Jo*(t), the differencebetween Jo* (t) and this lower bound being twice the sum of the infinite seriesJl,(t) - Ji(t) + - . Hence i{[(1 + cosh t)/2] + cosh(t cos it/12) +cosh(t sin(r/12)) + cosh(t/2) + cosh(3}t/2) + cosh(2-4t)} is an upperbound, over - co < t < ao, for Jo* (t) the excess of this upper bound overJo*(t) being twice the sum of the infinite series J ;(t) + J*a(t) + .

If (t 1 < 9 this excess is less than 10-'. Continuing this process one stepfurther we see that

\12 {Oosh 2 + 1 + cosh Ct cos m) + cosh Ct sun )

+ cosh ` t we i2) + cosh (t sin i2} + cosh [ t oos g1

+ cosh (t sin

8)+ cosh (2 + cosh () + cosh (I we )

+ cosh (t sin } + cosh (27*t) }

is an upper bound, over - oo < t < ao, for Jo*(t) the excess of this upperbound over Jo*(t) being twice the sum of the infinite series J*a(t) +J ;(t) + . If It ( < 10 this excess is less than 1.2 X 10-".

The same argument may be applied to Jo(t), the hyperbolic cosinesbeing replaced by ordinary cosines, the only difference being that we can-not say, since J,a(t), k = 1, 2, , may assume negative values, thatour approximations are upper, or lower, bounds as the can may be. Forexample, }{cos t + 2 cost/2)) is an approximation, over - ao < t < a*,to Jo(t), the difference J{cos t + 2 cost/2)) - Je(t) being the productof the sum of the infinite series Je(t) - J1,(t) + by -2. Since

The Laplace Transformation 7$

I J (t) 1 S 1 J.*(t) 1, n = 0, 1, 2, , the approximations to Jo(t) whichwe obtain in this way are at least as good as the approximations we haveobtained to Jo*(t). When t = it/2 the approximation to Jo(t) which isfurnished by *(cos t + 2 cos(t/2)1 is }21 = 0.4714, to 4 decimals, whileJo(r/2) - 0.4720, to 4 decimals. When t = r/4, the approximation toJ,(,r/4) is 0.85162, to 5 decimals, while Jo(-Y/4) = 0.85163, to 5 decimals.

Ex cisn 4Show that

J,*(t) + (cos48)[J,*(t) + Js*(t)J + (cos 88)[J,*(t) + J,*(t)1 + .._ j{cos B sinh(t cos B) + sin B sinh(t sin B)}

and deduce that

J1*(t) + [Ji(t) + Ji(t)1 + [Js,(t) + Jia(t)) +

3

[si2 t+ 21 s (inh + 1 sinh2 2)

Ji*(t) - vu(t) + Ji`i\(t)l + [Js (t) + Ji(t)J -\

=3 C(cos

12) sinh (t cos 12) + (sin 121 sink l t sin i2

+ 2--* sinh (2it)]

ExMciSa 5Show that

1[(sinh t/2) + (cos it/12) sinh(t cos(ir/12)) + (sin r/12) sinh(t sin(i/12))

+ (31/2) sinh(31t/2) + I sinh(t/2) + 2-* sinh(2-4t))

is an upper bound, over 0 < t < oo, for Jj*(t), the excess of this upperbound over Jl*(t) being the sum of the infinite series [4g(t) + J 5(t)1 +[J: (t) + J +(t)1 + .. .ExnacIBB 6

Show that

e[(cos xr/24) sinh(t cos(rr/24)) + (sin it/24) sinh(t ein(r/24))

+ (cos r/8) sinh(t cos(ar/8)) + (sin r/8) sinh(t sin(v/8))

+ (cos 5,x/24) siinh(t cos(5x/24)) + (sin 5w/24) sinh(t sin(5T/24))1

is a lower bound, over 0 < t < co, for Jl*(t), the numerical value of the


difference between J1*(t) and this lower bound being the sum of the infiniteseries [J a(t) + J 6(t)] - [Ju(t) + J9(t)] +ExaRcIsE 7

Show that the mean of the upper and lower bounds for J,' (t) , given inExercises 5 and 6, respectively, is an upper bound, over 0 < t < ao , forJl" (t) the excess of this upper bound over Jl*(t) being the sum of theinfinite series [J ; (t) + J; (t) ] + [J 5 (t) + Jo (t) ] +

EXERCISE 8Write down approximations to JI(t) analogous to the approximations

to J,*(t) which are furnished by Exercises 5, 6, and 7.

13

The Problem of the Vibrating

String

We shall discuss in this chapter the application of the Laplace Transforma-tion to the problem of a vibrating string, of length 1, with fixed end-points.This is one of the simplest instances of what is known as a boundary-valueand initial-condition problem. In the first place, the transverse displace-ment d = d (x, t), 0 < x <_ 1, 0 <_ t < oo, must satisfy the linear second-order partial differential equation with constant coefficients:

0<x<l, 05 t<co, a>0In the second place the boundary values of d, i.e., the values of d whenz = 0 and when z = 1, are specified as follows:

B: d(0, t) = 0; d(l, t) = 0, 0 < t < 00Finally, d must satisfy the following initial conditions:

I: d(z, 0) = i(x) ; d,(x, 0) = v(z), 0 <- x <- l

¢(x) and v(x) being given continuous functions of x over the interval0 < x :< 1. We assume that this boundary-value and initial-conditionproblem possesses a solution d(x, t) with the following properties:

(1) dtt is piecewise continuous, for all values of x in the interval0 < z < 1, over 0 < t < - and the Laplace Transform of dt,u(t) existsat a point p = cl of the positive real axis in the complex p-plane.

(2) The infinite integral fo d(x, t) exp ( -- pt) dt which furnishes, overthe half-plane c > cl , the Laplace Transform f = f(x, p) of d(x, t) u(j) is

75


twice differentiable with respect to x under the integral sign, so thatd. (x, t) u(t) possesses, over the half-plane c > c, , the Laplace Transformf.. Since L[dt(x, t)u(t)] = pf -- 0, L[d,,(x, t)u(t) ] = p'f - pu" -- v, c > cl,and since a d.. - d = 0, 0 < t < cc, f must satisfy the nonhomogeneoussecond-order ordinary linear differential equation

D':afx-pf=-po--v; 0<x<lp playing the role of a constant parameter. Furthermore, the boundaryvalues f (O, p) and f (l, p) are zero for all points p in the half-plane c > c, :

B': f(0, p) = 0; Al' p) = 0; c > c,

Thus the boundary-value and initial-condition problem D, B, I, is re-placed by the boundary-value problem D', B'. We shall solve this simplerproblem and shall then determine a function d(x, t) which is such that theLaplace Transform of d(x, t)u(t), over some half-plane c > c, > 0, is thesolution, f = f (x, p), of the boundary-value problem D', B' which wehave obtained. All that remains, then, is to verify that d(x, t) is a solutionof the boundary-value and initial-condition problem D, B, I, and to showthat this problem does not possess any other solution.

Our first step in solving the boundary-value problem D', B', is to con-sider the associated homogeneous boundary-value problem D", B', whereD" is the homogeneous second-order linear differential equation, withconstant coefficients, a'k1, - p'k = 0. D" does not have, no matter whatis the value of c, > 0, a nontrivial solution, i.e., a solution which does notvanish identically, which satisfies the boundary conditions B'. Indeedsinh(gx) and sinh[q(l - x)], q - p/a, are two linearly independent solu-tions of D" so that the general solution of D" is A sinh(qx) +A' sinh[q(l -- x)] where A and A' are undetermined constants; for thisto be zero at x = 0 we must have A' - 0, since sinh(gl) 0 0, and A sinh(gi)is not zero if A is not zero. In order to avoid this dilemma of the non-existence of a nontrivial solution of the homogeneous boundary valueproblem D", B' we lighten our requirements on the function k(x, p) asfollows: we do not insist that k satisfy D" at all points of the interval0 < x < 1; we require merely that, in addition to satisying the boundaryconditions B', it satisfy D" at all the points of this interval save one, s say,at which it does not possess a second derivative, s being an interior pointof the interval, so that 0 < s < 1. We. do require that k be continuous atx - a and that it possess both a right-hand derivative, ks(a + 0) and aleft-hand derivative, kz(s -- 0). In order to completely determine thisfunction k of x, which depends on the parameter s and which we denote by

r (8), we prescribe the difference ks(s - 0) - ks(s + 0) = I',8_0)_


r, (a -[- 0) to be 1/a', the reciprocal of the value at s of the coefficient ofa J

k,= in D" (this coefficient being actually, in the particular problem weare discussing, independent of s). \

Over the interval 0 < x < a, r (8) is a linear combination, A, sinh(gx) +

Ai sinh[q(l - z) ], with constant coefficients, of the two linearly independentsolutions sinh(qx), sinh[q(l - x)] of D", and we denote this linear com-

bination by r, Similarly, over the interval a < x < 1, r (8) is of the

form r, (8) = A, sinh(qx) + - A,' sinh[q(l - x) ]. Since r, (0) = 0,A,' = 0,

and, since r, (8) = 0, A, = 0 and, finally, since r, (r28) = (e),

A, sinh(q8) = A,'sinh[q(l - a)]. Thus r11 3) = A sinh[q(1 - s)] sinh qx,

r, (8) = A sinh(qs) sinh[q(l - x) ], where A is an undetermined function

of 8. Then r, (a3

0) =[{r1 (x)}].... = Aq sinh[q(l - s) ] cosh(ga) and,

similarly, r, ( 8 0) = -Aq sinh(gs) cosh[q(l - s)] so that

Aq{sinh[q(l - a)] cosh(qs) + sinh(qs) cosh[q(l - s)]},

i.e., Aq sinhgl, = 1/a'. Thus A = (a2q sinhgl) ' and

x sink q(l - s) sink (qx) sinh (qs) sinh q(l - x)(s) a2q sinh (ql) r' (s a'q sinh ql

0 < x < 1; 0 < s < 1. We define r `8) when a = 0 and s = l by there-

quirement that r { 8) be a continuous function of s at s = 0 and s = 1;

thus r ` ) = r, (4) = 0 and, similarly, r 01) = r1(i) = 0. We observe

thatr2

(8) may be obtained from r, (M)8by merely interchanging x and

a and this implies that r l s) = r (8), i.e., that r (8) is a symmetric

function of the two/real variables (x, a) over the square 0 < x < 1, 0 < a < 1.

Indeed, if x < a, r (x) = r, (x), while r (8) = r, (8); similarly, if x > a,

r (x) = r, (x) and r (8) = r, ($). The function r (8) of the two real


variables (x, s) is the second of two functions known as the Green's func-tions of the boundary-value problem D", B'; the first of these two Green'sfunctions, with which we shall not be concerned since p 0 0, is the func-

tion G ($ to which r (8) reduces when p = 0:

(x)=(1G(8)=G l )x 0<x<s,a

G(8)G,(8)s(la=1z), s<x<1We now turn to the non-homogeneous differential equation

D': a'f==-p2f= -p/ - v, 0 <x<1On combining this with the homogeneous equation

D": a2ru-p'r=0; xs 8in such a way as to eliminate the undifferentiated functions f and r weobtain

( \l (x)a2 [r (Z)j= -- f (x) {r (8)}= _ - pr .o(x) - r(z)v(z), x 0

Now r (8)fu - f(x) 5 r (3)T= _ r(8 ) f= - f(x) {r (:)}z]x'and so the integral of r (x) 8f= - f(x) { r ()}

x =over the interval 0 <

z<lis l

[r ($ f= - f(x) lr f + [r (7)f= -- f(x) {r

it being necessary to break the interval of integration into the two parts

0 < x < s, s < x < 1, since { r (s)=1 is discontinuous at x = s. Regarding

r (x) as an integral operator we denote f o r t 8) h(s) ds, where h(x) is

any function which is integrable over 0 </ x\ < 1, simply by rh. Then

for (a) O(x) dx, being the same as f' r (x) -0 (x) dx, is the value of

r. at x = s and, similarly, far (8) v(x) dx is the value of rv at x = 8.

Since both r and f are zero at x = 0 and at x = 1 the expression

M. - fr=)lo + (rf= - fr.)1.


reduces to

and so-fr=I:+o = -f(s)/as

f(s) = p(ry) (s) + (rv) (8).It was to obtain this simple expression that the particular value 1/a2 ofthe discontinuity r=I;+o in the first derivative of r at x = 8 was prescribed.Since s is any point of the open interval 0 < s < 1 we may write the resultjust obtained in the form f(x) = pr¢ + rv, 0 < x < 1, and, since f(x),r4, and rv are all continuous at x = 0 and at x = 1, we have

f= prc6+rv, 0<x<1What we have proved so far is a uniqueness theorem; granting the

existence of a solution of the boundary-value problem D', B', this solutionis unambiguously determined by the formula f = pro + rv. We mustnow remove the existence hypothesis by verifying that pro + rv is actuallya solution of the boundary-value problem D', B'. To do this we first ob-serve that the continuity of O(z) and of v(x) over the interval 0 < x < 1assures us that f = pro + rv is twice differentiable over this interval.

Indeed, on writing r.0, for example, as Jo r, ()it(a) ds + f= r: (x) 4(s) da

we see that rr is differentiable over the interval 0 < x < 1, its derivative,(r') a , being furnished by

{riJ6 lr=

lm(8) ds + fx)}

(s) ds +\r2 (x) - rl (x/} (x)

which reduces to

f{r1 (')}.()+

f1{r1\x/1 (s) ds

(z),s

since r, \x) = ri by virtue of the continuity of r (x) at x = 8.Hence (r#), is differentiable over 0 5 z < 1, its derivative, be-ing furnished by

as s\x/} 0(s) do + f rlx 0(8)

do, 8 :tf{r'

+ [{r(:)}= - rl 1 \/111

z

--a4+(x)=a r-a ¢

80 Lectures on Applied Mathematica

Applying this result to the function f = pro + rv, we see that f is twicedifferentiable over the interval 0 < x < 1, its second derivative, f,, ,being furnished by the formula

ro-a=0+a rv - a v

=a'f d.0 -a v, 0<x<1Thus a=f,. - p=f = - pvi - v, so that f satisfies the differential equationD'. That f = pr,+ rv satisfies the boundary conditions B' is evident,

since r (0) = 0, r (.1)1) = 0, 0 < s < i, so that, if h(x) is any function which

is integrable over 0 < x < 1, rh is zero at x = 0 and at z = 1. Thus wehave the following definitive result:

The unambiguously determinate solution, f = f (x, p), of the boundary-value problem

D':a'f,,-p'f=-pO V

B': f(0, p) = 0; Al' p) - 0is

wheref=pry+rv

x sinh q(l - a) sinh qx pr a}= a2gsinhgl qa,0<x<asinhgasinhq(1-x) a<x<1

a'q sinh ql '

14

The Solution of the Problem

of the Vibrating String

We have seen that the unambiguously determinate solution of theboundary value problem:

D':a'f1-p'f=-pu-v; 05x51B': f(O, p) - 0; Al' p) = 0

is f - pry + rv, where the integral operator r is furnished by the formulas

rM_rlrxl sinkq(l-a)ainh(gz) q=p'0<x5aa `a / a'q sinh ql) a

r t z ) = r i (xl = sink ( a) sink q(l - z) < <a a J &q sin (ql)

and our first task now is the determination of a function d(z, 1) which issuch that the product of d(x, t) by u(t) has, over some half-plane c > cl ,

the Laplace Transform pr# + rv. If we regard x as fixed, r(8z) is a func-

tion of a and p which is analytic save at the points (nra/l)i, n .. 0, f1,f2, , on the imaginary axis in the complex p-plane, at which sinh qlsinh pl/a is sero, and so c, 0. If d(z, t) is, for each value of z in theinterval 0 < z < 1, bounded over 0 < t < co, d(z, t) u(t) possesses a LaplaceTransform over the half-plane c > 0 and we set e3 = 0. We first examinein detail the first term

81

89 Lectures on Applied Matlismatica

pro=p f rs(s}¢(s)d8+p f r1(3)o(e)ds

i f sinh [q(l - x)] sinh (gs)0(s) dsa sink ql L.

+f:sinh

(qx) sinh [q(l - s)]o(s) de

off - pro + rv. On writingsinh q(l - x) = I exp (ql){exp (-qx) - exp [- q(21 -- x)]},

sinh (qs) _ {exp (qs) - exp (-q8)

and so on, we obtain[ILI

pro= exi(4l) {exp[-q(z -s)] exp[-q(x+ s)]4a sinh (ql)

- exp [-q(21 - x - s)] + exp [-q(21 - x + s)]}o(s) ds

+ f (exp (-q(s - z)) - exp [-q(s + x)] - exp [-q(21 - s - x)]

+ exp [-q(21 - s + x)]}0(8) do]

exp (ql) Chx {exp [-q(x - s)] + exp [-q(21 - x + s)]}4i(s) ds4a sinh (ql)

-I' {exp[-q(x+ 8)]+exp[--q(21 -x -s)]}¢(s)de

+ f a {exp [-q(s -' x)] + exp [-q(2l - e + x)1}4,(8) ds]

This complicated expression takes a simpler appearance if we extend therange of definition of O(x) from the interval 0 < x < 1 to theentire x-axis, - - < x < -, by saying that o(x) is an odd periodic functionof period 21, it being permissible to do this since o(0) = 0 and -0(l) = 0.The oddness of ¢(z) furnishes the values of o(x) over the interval - l <x < 0, since its values over the interval 0 < x < 1 are known, and theperiodicity, with period 21, of o(x) furnishes the values of o(x) over- cc < x < co since its values over the interval - l < x < t are known.In the various integrals which appear in the expression furnishing pro wemake changes, of the type s = a + Rot, of the variable of integration froms to t in such a way that the exponential factor in each of the transformedintegrals is exp(- pt) = exp(-aqt). For example, in the integral j o exp


[.- q(x - s) ]O(s) ds we write s = x - at so that it appears as a fo °exp (- pt)o(x - at)dt; in the integral fo exp[-q(21 - x + s)10(s)ds we writee = x - 21 + at so that it appears as

4 1 exp (-pt)¢(x - 21 + at) dt = f(221fa exp (-pt)0(x + at) dt-x)/. r-:via

whereas in the integral

I1 exp [-q(x + 8)]4,(s) ds

we writes = - x + at so that it appears as

exp (-pt)¢(at - x) dt = -1/a exp (-pt)0(x - at) dt

and so on. Continuing in this way we obtain

_ exp (ql):u.

pry 4 sinh (ql){O(x - at) + O(x + at) } exp (-pt) dt

Now

exp (ql) _ 11 - exp (-2ql)}-'2 sinh (ql)

and, since the real part of q = p/ a is positive, this may be written as thesum of the convergent infinite series 1 + exp (- 2ql) + exp (- 4q1) + - .

The product of{/s

{o(x - at) + o(x + at)) exp (-pt) dt by exp (-2kql),k = 1, 2, ...

appears, on writing t = s - 2k (l/a) , as

fuf(k+I (i/.)

{o(x - as + 2k1) + o(x + as + 2k1) } exp (-ps) ds[l/a)

and this is the same, since O (x) is periodic with period 21, as

/.)

(k+1)(!/a){¢(x -- at) + O(x + at) ) exp (-pt) dt

fU

!

a

Since (4(x - at) + O(x + at))exp(-pt), being continuous, is boundedover the interval 0 < t < 21/a the infinite series obtained by multiplyingeach term of the infinite series 1 + exp(-2q1) + exp(-4q1) + - by{o(x - at) + ¢(x + at)) exp(- pt) is uniformly convergent over the

84 Lector es on Applied Mathematics

interval 0 < t 5 21/a and so term-by-term integration over the intervalis legitimate. Hence

.CIA)pro -

2 E {#(x - at) + #(x + at)) exp (-pt) at

2+ {#(z - at) + O(x + at)} exp (-pt) at

so that the Laplace Transform, over the half-plane c > 0, of J10(x -- at) +(x + at) } u(t) is pry. Similarly the Laplace Transform, over the half-planec > 0, of J{v(x - at) + v(x + at)}u(t) is ply, it being understood thatv(x) is an odd periodic function, with period 21, of the unrestricted realvariable x, and this implies that the Laplace Transform, over the half-planec > 0, of {jfo{v(x - as) + v(x + aa)}ds]u(t) is rv. Hence the LaplaceTransform, over the half-plane c > 0, of the product of u(t) by

da]2

[,O(X - at) + O(x + at) + I' {v(x - as) + v(x + as))

is pr* + rv. We now proceed to show that the function

d(x, t) - 2 at) + #(x + at) + L' {v(x - as) + v(x + as)) do]

0 < x < 1, - ao < t < ao, is a solution of the boundary-value and initial-condition problem D, B, I and that this problem possesses no other solu-tion. On introducing the variables t - x - at, r - z + at, D takes theform de, = 0 and d(x, t) becomes }j¢() + #(r) + (1/a) f v(s)do] so thatdE is a function of t alone; hence d(x, t) is a solution of the differentialequation 1). That d(0, t) is 0 over 0 S t < .o and, indeed, over - co < t< co, is an immediate consequence of the fact thatO(t) and v(t) are, aftertheir range of definition has been extended, odd functions of the unre-stricted real variable t. Similarly, since *(t) and v(t) are not only odd butalso periodic functions, with period 21, d(l, t) - 0 over - co < t < -;indeed, 0(1 - at) - 0(--i - at) - --0(l + at) and r(I as) - -v(l +as). Thus d(x, t) satisfies the boundary conditions B. Finally, d(x, 0)o(x) and d,(x, 0) - }{-a#,(x) + a¢,(x) + 2v(x)] - v(x) so that d(x, t)satisfies the initial conditions I. Thus d(x, t) is a solution of the boundary-value and initial-condition problem D, B, I. If this problem possessed twodifferent solutions their difference d(z, t) would be a solution of the as-sociated homogeneous problem D, B, I' where

I': d(x, 0) - 0; d,(x, 0)

Being a solution of the differential equation D, which appears, when written


in terms of the variables E, r, as dE, 0, A (z, t) is of the form F(f) + G(r)and the initial conditions I' yield F(x) + G(x) = 0, F=(x) - G.(x) = 0.On differentiating the first of these two relations and combining the resultwith the second, we see that F and G are constant functions of their argu-ments so that o(x, t) is a constant function of the two variables (z, t).Being zero when t - 0, it is identically zero. Thus we have the followingresult:

The unambiguously determinate solution of the problem of the vibratingstring is

d(z,t) Zt4,(x -at)+O(z+at)} +2L (v(x-as)+v(x+(s)}d8

- 1 ((x - at) + #(x + at) } + 1 v(Q) do2 2a zs

If we denote fo v(s) ds by V(x), it is clear that V(x) is an even periodicfunction, with period 21, of the unrestricted real variable x. Indeed V(x) -V(-z) = J ,v(#) de is sera by virtue of the oddness of v(x) and

-"iV (z + 24) - V W - F.= v(s) de

= f 1 v(s) ds + f' v(s) d8a I

- f { f c-v(s)d+v(o)d,,, s-2l,I

- f v(s) ds = 0

The unambiguously determinate solution, d(x, t), of the boundary-valueand initial-condition problem may be written as the sum of the followingtwo functions of x - at = and x + at = r, respectively,

dl(x, t) - 2 4(x - at) - 2a V(x - at)

d2(x, t) - 2 o(x + at) + 2a V (x + at)

and we observe that d2(-x, t) - - di(x, t), d2(l - x, t) = - d,(t + x, t)-d, (x - at) is constant so long as x - at remains constant and we say thatit represents a wave travelling, in the direction of the positive x-axis, withvelocity a. When x attains the value I and begins to assume values > Iwe must replace x, t) by - d2(l - z, t) which represents a v.ave


travelling, in the direction of the negative x-axis, with velocity a. When x,in the expression d,(z, t), attains the value 21 and begins to assume values> 21,1 - x, in the expression d:(1 - x, t), attains the value 0 and begins toassume negative values and we must replace x, t) by dl(z - 1, t)and so on. We express this result by the statement that the solution of theproblem of the vibrating string is the sum of two waves, one travelling withvelocity a in the direction of the positive x-axis and the other with velocitya in the direction of the negative x-axis, these waves being subjected tocontinual reflections at the ends of the string.

The level curves of the functions Z = x - at, r = z + at play a dominantrole in the theory of the partial differential equation

and they are known as the characteristics of this differential equation. Letus suppose that the values of the two first-order derivatives, d, and d, ,of d(x, t) are assigned, as continuously differentiable functions of a param-eter a, along some smooth curve x = z(a), t = t(a) in the (x, t) -plane.Then (di) , = dm. + d,sta , (d,) a = d,.XQ + de,t& along this curve andthese relations, together with the relations a=du - d« = 0, dr. = ds, enableus to unambiguously determine the three second-order derivatives, d.,,.,

d (x, t) along the curve at any point of the curve at which the3 X 3 matrix

is nonsingular. Since the determinant of this matrix is a=(ta)2 - (x.)2 _(ata - xa) (ata + xa) we see that d,s , d.,, d are unambiguously deter-minate at any point of the curve x = x(a), t = t(a), at which this curve isnot tangent to any characteristic of the differential equation D, i.e., to anymember of either of the two families of straight lines x - at const, x + at =constant.

15

The Generalized

Vibrating String Problem

The simplest generalization of the partial differential equation a2d= - d,, =0, which occurs in the theory of the vibrating string, is the partial dif-ferential equation a2du - d,, + pd. + qd, + rd = 0, where a > 0 and p, q,r are given constants. Setting d = d'exp(ax + at), .where a and,8 are un-determined constants, we have

d,=(d1'+ad')exp(ax +at); dc= (d,'+ ')exp(ax +$e)d. = (dam + 2adx + a2d') exp(ax + $t) ;

d,, = (d;, + 2.6d,' + ed') exp(ax + ftt)

so that d' satisfies the partial differential equation

a2d;= - dee + (2a2a + p)dz' + (q - 20) d.' +

(aTa2-$2+pct +qi+r)d'=0Setting a = - p/2a2, 0 = q/2 the terms involving the first-order derivativesof d' disappear so that d' = d[exp[(px/2a2) - (qt/2))] is a solution of thepartial differential equation

2 za2du-di,+(r`-4a2+9}d' = 0

If 'r = }[(p2/a2) - q21 this is the partial differential equation which we havealready met in the theory of the vibrating string. Assuming that r -

87


(p'/4a') + (q'/4) is not zero we denote its absolute value by k`', k > 0,and we write x - kx', t - kt', so that d;, - k,4, d:,., - k'4 and, similarly,de,*, - k'd . Thus a'4..' - do, f d' - 0, the upper, or lower, sign beingused according as r - (p'/4a') + (q'/4) is positive, or negative, respec-tively. Dropping the primes attached to x, t and d we are confronted by oneor other of the two partial differential equations a d,. - ds, f d - 0 andwe first consider the equation

D:a'd..-dis+d-0We take the boundary and initial conditions to be the same as in the prob-lem of the vibrating string, namely,

B: d(0, t) = 0; d(l, t) = 0; 0:59< 0oI: d(x, 0) = -0 (z) ; d, (z, 0) - v(x) ; 0 < x < i

and we term the boundary-value and initial condition problem D, B, I, thegeneralized vibrating string problem.

Proceeding as in the case of the vibrating string problem we encounterthe second-order ordinary differential equation a'f.. - (p' - I) f =- pp - v, rather than a'f - p'f - - pqb - v and we set q = (1/a)(p' --- 1)4, rather than q = p/a. Thus the boundary-value problem D", B'has the same formal appearance as in the case of the vibrating string prob-lem, the difference between the two problems lying entirely in the definition

of q as a function of p. The integral operator r = r C8) which we encounter

is, then, the same function of q as before but this implies that it is a dif-ferent function of p. Its singularities, instead of lying on the imaginary axisin the complex p-plane, are the points p for which p' - 1 = - n'r a'/l',n - 0, 1, 2, ... , so that p - 1, corresponding to n - 0, is a singular pointof r. Thus the half-plane over which d(x, 1)u(t) possesses, we assume, aLaplace Transform f cannot be, as it was in the problem of the vibratingstring, the half-plane c > 0; since r does not possess any singularities in thehalf-plane c > 1 we assume that d(x, t)u(t) possesses, over the half-planec > 1, a Laplace Transform f. The same argument as in the problem of thevibrating string shows that agr4 - if o' (4,(x - at) + O(x + at) } exp(-aqt) dt provided that the real part of q is > 0 and that the range ofdefinition of ¢(x) has been extended from the interval 0 < x < 1 to theentire x-axis by the statement that 4,(x) is an odd periodic function, withperiod 21, of x. Since aq is no longer p, the integral f e (0(x - at) + o(x +at) j exp(-aqt) dt is no longer the Laplace Transform of {#(x - at) +O(x + at)) u(t) and we proceed as follows, writing

r* = 2.a {*(x - at) + O(x + at) } exp (g a¢) dt


2a .b {+(x - a) + #(x + a) }exp (Q aq) a = at

We try to determine an integral operator, K - K(;), which is such thatK(;)u(t), a any non-negative constant, possesses, over the half-plane c > 1,the Laplace Transform exp(-aq) /q, q = (p= - 1);/a. Once we have deter-mined K we may write rF in the form

rms , f"([#(x-8)+.O(x+8)4fQOK()expPt)dt}>dsF,d

and, provided that the order of integration in this repeated infinite integralmay be changed, it follows that

r# - Lifof**

K () [O(x - a) + O(x + a)) & eRp (-pt) di

We shall see, when we determine K, that K(;) is zero if s > at so that rris the Laplace Transform, over the half-plane c > 1, of

1ia2 - f K(8)[O(x-a)+O(x+a)deand this implies that the Laplace Transform, over the half-plane c > 1, of

2a K (att

p) [4(x - at) + 4(x + at) }

f° K,(')[t(x-a)+O(x+a)]ds

is p r4,.We turn, now, to the determination of the integral operator K. Setting,

in the relation q = (1/a) (p= - 1) 4, p - cosh z*, we have

q = a [p - exp (-z*)]

and thus

exp(-sq) = exp(_4 p)exp[aexp(-z') I

exp (.- ap) 1

+a exp (- z* )

+ 21a! exp (- 2z*) +

Now exp(-nz*)/sinh z* = ,..(p) is the Laplace Transform, over the


half-plane c > 1, of JA*(t)u(t) = In(t)u(t) and, since the coefficients of the

infinite series fo*(p) + (s/a)f,*(p) +(s2/2!a2)f,*(p) + are all non-

negative, s being, by hypothesis, non-negative, it follows that the LaplaceTransform, over the half-plane c > 1, of the product of the sum of theeverywhere convergent infinite series lo(t) + (s/a)1,(t) + (82/2!a2)I2(t) +

by u(t), is expfs/a exp(- z*) ]/sinh z*. We shall show in the next para-graphthat the sum of this infinite seriesis [Io(1 + (2s/at))*t]; admitting this,for the moment, exp(-aq)/q is, over the half-plane c > 1, the product ofthe Laplace Transform of alo[(1 + (2s/at))it]u(t) by exp(- (8/a) p) :

\\exp 84) /4 = a exp (-- p J f Io [(1

+ 2s tj exp (- pt) dt

=a [t r2_.a jexp(-Pr)dr, r=t+8+!a ` / J a

In other words,

K(1) =aloRe -!a-)Iu\t-a)so that, in particular, `K(;) = 0 if 8 > at. Since the derivative of Io(t) withrespect to f is 11(t),

Kt (18) =at

82\1} Il R42/ u (t - a)\t a2/

and so, since KG, .` o) = a, the Laplace Transform, over the half-planec > 1, of

[.O(z-at)+#(z+at)]

2+ 2a Jo f z _t a2 hI [(t2 - ai}} s) - O(x + 8)) ds

(t a2)

is pro. Thus the solution of the boundary-value and initial condition prob-lem D, B, I, which is suggested by the application of the Laplace trans-formation, is

d(x, t) _ j [¢(z - at) + ¢(x + at)]

+ T+t

) -} O(x -f 8)1 daI [(t_)] Wx --_2a ( t s -

a4)4

+ a.f*'Io[(t2-x-2);{v(x-s)+v(x+s))ds


We must, before verifying that the function d(x, t) furnished by thisformula is a solution of the boundary-value and initial condition problemD, B, I, justify the statement, made in the preceding paragraph, that thesum of the everywhere-convergent infinite series Io(t) + (s/a)I1(t) +(s'/2!a2)I:(t) + . is Io{(1 + (28/at));t]. To do this we recall that theLaplace Transform, over the half-plane c > 0, of t"12Xn(2t')u(t), n =0, 1, 2, , is exp (1/p) /p"+'. On replacing t by (1 + a) t' and p by p'/(1 + a), where a is any positive real number, in the relationGD f

t"I2I (2ti) exex (- t) dt > 0pp " p ,+iP-0c

and then dropping the primes, we obtain(1 a\ = ('° tn/:Iar/2

ex [2(1 + ex t) dt(- > 0pp

+ p p , c

so that the Laplace Transform, over the half-plane c > 0, of

(1 , a)-niztn':In[2(1 + a)ttiJu(t) is +I exp 1

pa

Pft

>z

(p!Since the coefficients of the infinite series on the right-hand side are allnon-negative, a being, by hypothesis, non-negative, the infinite series

stni:I,(21) + at(n+ur,l"+1(2t;) +2a

1

tc°+z)1zI.+:(2t;) + ...

is everywhere convergent and the product of its sum by u(t) has, over thehalf-plane c > 0, the Laplace Transform (1/p"+') exp (1 + a)/p. Hence,by the uniqueness theorem,

Stn/2II(2t1) + atcn+l)1:In+l(2t{) + a t(ni3)l"Iz+2(2t) +2!

(1 + a)4ti] 0 < t < CO, 0 < a < ao

On dividing through by to/2 and writing 2t} = t', and then dropping theprime, we obtain the equation

I.(t) + 2 In+I(t) + 2' (!!!)'! 2

In+2(t) + ... = (1 + a)-,.nln[(1 + a)}t],

0 < 6 < oo,0 < OD

For any given value of t > 0 the left-hand side of this equation is a powerseries in a which converges for every positive value of a and so, if we regarda as a complex, rather than a real, variable the stun of this p>>w. r :ies hi


an analytic function of a over the entire finite complex a-plane. The right-hand side, (1 + a) (1 + a) 1tl, of our equation is also, being the sumof an everywhere-convergent power series in (I + a), an analytic functionof a over the entire finite complex a-plane and, since these two analyticfunctions of a coincide over the non-negative real axis in the complex a-plane, they coincide over the entire finite complex a-plane. In the sameway we see that, a being any given complex number, I. (z) + (az/2) (z)+ (1/21) (az/2)'I,,+2(z) + is an analytic function of the complex vari-able z which coincides over the entire finite complex z-plane with (1 + a)1.((l + a)z). Thus

1.(z) + 2 1.+I(z) '+1ill (?0' i.+:(z) + (1 + ar"I.I (1 + a )}z)

where a and z are arbitrary complex numbers. On setting z = zz' and thendropping the prime we see that this relation is equivalent to the relation

J.(z) - 2 J.+i(z) + 21(2) J.-12(Z) - ... = (1 + a)_J2J.[(I + a)}zl

On setting n = 0, z = t, a = 2s/at, we obtainr/ \

lo(t) + Ii(t) + a I:(t) -}- ... = to (1 } ) t]Q L

`/

which is the relation we wished to prove.

EXERCISE IShow that

z 1(zl'+21 2I.+s(z)-... na091,2,...

with z an arbitrary complex number. (Hint. Set a = - 1.)EXERCISE 2

Show thatz'J.(z) = IA(z) - zl.+I(z) + 2 I.4-2(Z) - ... , n - 0, 1, 2, ...

with z an arbitrary complex number, and deduce that 1.(z) - J.(z) +(z'/21)J,.+i(z) + . . , n = 0, 1, 2, .. , with z an arbitrary

complex number. (Hint. Set a 2.)

Ex acxsa 3Show that

Jo(31s) - Je(z) - 5J1(z) + 2! J11(c) - jME) + ... ,z being an arbitrary complex number.

The Laplace Transformation 9$

Eaaxciss 4Show that

Jo(2}z) - Jo(z) - 2 J1(z) + 2I (z) J. (Z)

JI(2;$) - 2i {JA (z) - 2 J2(z) + 2f (z JS(z)

Z 2!

(z)2 J(z) - ...(J,(2;s) - 2 { J2(z) - Js(z) +

and so on, z being an `arbitrary complex number.

In the next chapter we shall verify that the function d(z, t) which wehave obtained in this lecture is a solution of the boundary value problemD, B, I and shall show that this boundary value problem does not possessany other solution.

16

The Solution of the Generalized

Vibrating String Problem

The solution, d(x, t), of the generalized vibrating string problem which hasbeen suggested by applying the Laplace transformation may be written inthe form d,(x, t) + d2(x, t) where

d, (x, t) _ {0(x -- at) + ¢(x + at) }

+ 2a

lae

Il«a) {¢(x - s) + 4, (x + s) } d.9

d2(x, t) = Za

foeIo(a) {v(x - s) + v(x + s)) ds; a = (t2 - a2);

In order to verify that d(x, t) is a solution of the partial differential equationa2d - d + d = 0 it is sufficient to verify that d2(x, t) is a solution ofthis differential equation. Indeed, d,(x, t) is the derivative with respect to tof (1/2a) f o=Io(a) {0(x - s) + ¢(x + s) l ds andif d2(x, t) is a solution of thedifferential equation a2du - d + d = 0, so also is (1/2a) f o Io(a) {0(x - s)+ O(x + s) } ds, which implies that the derivative of this expression withrespect to t, namely d,(x, t), is a solution of the differential equationa2dzx - d + d = 0. On making the substitution s = x - s' in the integralf olo(a) t; (x -- s) ds, and the substitutions = s' - x in the integral foIo(a)v(x + s) ds, and then dropping the prime, d2(x, t) appears in the form

d2(x, t) =1 z+atIo($)v(s)

ds;- j2 - (x __ 8)21;

J2a r,, a2

a4


Upon introducing as new independent variables the two functions t _x - at, r = x + at of x and t whose level curves are the characteristics ofthe differential equation D: a2d=z - d + d = 0, D appears in the form

D*: 4asd,, + d = 0

and we have to verify that

1G(, r) = f Io($)v(s) ds; R = (r - s) (s - )}/a

is a solution of the partial differential equation D*. Since ,6 = 0 when 8 = r,

2af Io (S) (r - 8)-}(s - t);v(8) ds + v(r)

where the prime attached to Io denotes differentiation with respect to itsargument,o. On differentiating 4,T with respect to t we obtain

I,,E _ -2 f {Io"(0) + aIo ($)(r - s)-'(s - t)-4]v(s) A

-4a2 jT{i,,(S ) + ' 1o'(0) j v(s) ds

1 = Io($), that y'rE = - (1/4a2) ',and it follows, since Io"(0) + (1/f)Io'(0)which proves that 4,Q, r) is a solution of the partial differential equationD*. This completes the proof of the fact that d(x, t) = di(x, t) + d2(x, t)is a solution of the partial differential equation D. That d(x, t) satisfies theboundary condition4 B and the initial conditions is proved in the same wayas in the problem of the vibrating string. Thus d,(0, t) = 0, since O(x) isan odd function of the unrestricted real variable x and d2(0, t) = 0 sincev(x) is also an odd function of the unrestricted real variable x; similarlyd,(l, t) = 0, d2(l, t) = 0 since ¢(x) and v(x) are not only odd functions ofx but are also periodic with period 21. d,(x, 0) is evidently 4(x) and d2(x, 0)is evidently 0 so that d(x, 0) = 0(x) . Finally, [d, (x, t) ], is zero when t = 0and [ds(x, t)], is v(x) when t = 0 so that d,(x, 0) = v(x). This completesthe proof of the fact that d(x, t) is a solution of the generalized vibratingstring boundary-value and initial condition problem.

The proof that the generalized vibrating string boundary-value andinitial-condition problem does not possess a solution differing from d(x, t)is not as simple as the proof of the corresponding uniqueness theorem forthe vibrating string boundary-value and initial-condition problem. Wefirst observe that, if r,) is any point in the (t, r)-plane, the functionw = Io(8), where 5 = (E - )(r - r,)'/a, of the two variables (t, r)


satisfies the differential equation 4a'wf, + w 0. Indeed, on denotingdifferentiation with respect to d by a prime,

we = 2a I0, ( - 6)3(r - r,)}

" i1 (r r,)Io +4aIo(f - t,)

' 4ao { lo" +a

tol = 4ds I. = -L W

On combining the two differential equations 4a'4 + d = 0, 4a'we, -l- w = 0in such a way as to eliminate the undifferentiated functions d and w weobtain wd5 - dw1, - 0, or, equivalently, (wdf), _ (dw,) f . If, then, C isany piecewise smooth closed curve in the (, r) -plane the integral of wdtwith respect to t around C is the negative of the integral of dw, with respectto r around C, both integrals being taken in the positive sense. Now w = 1when S = 0, i.e., when f = 6 or r = re, and w, = (1 /2a) Io' (k - &)I(r - r,) -4 is zero when E _ and so, if C consists partly of segments of thelines f = f, and r - r, and if we denote the remainder of C by r, we have

fpj

PZ

P]

PS

d(P,)-d(Ei,ri)+wdfdt+J dw,dr=0

where P, and Ps are the points where the lines r = r, and = , , respec-tively, intersect the curve r, and both the integrals from P, to P: are takenalong the curve F. Thus d( a , r,) is unambiguously determined by thevalues of d and of df along the curve r. If the curve r is a segment of theline E - r = 0, which corresponds tot = 0, d and df - }(d, - (1/a)d,)1(0, (1 /a) v) are given along r, and we we that d(& , r,) is unambigu-osly determinate. If the curve r is not a segment of the line Z- r = 0,we have to deal with the phenomenon of reflection at the ends x = 0 andx = 1 and it is not hard to see that df is ]mown along the lines f + r = 0and t + r = 21 which correspond to x = 0 and x - 1, respectively. It willsuffice to consider the first reflection at the end z = 0 for which r consistsof a segment, lying in the second quadrant and ending at the origin, of theline t + r = 0 and a segment, lying in the first quadrant and beginning atthe origin, of the line E - r = 0. We take , to be in the interval 0 < t < 1and move the point (, , r,) towards the point P2 along the line t _ , .

Observing that df =. - d, along the line t + r = 0 we obtain

d(PI) + d,(Pl) - dF(P1) - d,(P:) + w(Pi)df(Pi) - d(Pl)w,(P,) = 0In our boundary-value and initial-condition problem d(x, t) is identicallyzero when x = 0 and so both d and d, = (1/a) (d, - df) are zero along the

The Laplace Tranaformation 97

line t + i - 0. Hence w(Pl)d1(Pl) = d,(P:), and since w is zero only at(& , T,) which is distinct from Pl , the E-coordinate of P, being negative, itfollows that dd(P1) is the quotient of

d,(P2) - 2{+a

de(P:)} = 2'S ,(x) + I v(x by w(PI)

Thus dE is known along t and the solution of our boundary-value and initial-condition problem which is furnished by an application of the Laplacetransformation is the only one which exists.

The modifications necessary to deal with the differential equation a2d -de, - d - 0, rather than a'd., - d,, + d = 0, the boundary and initial con-ditions being the same as before, are minor. Setting q = (1/a) (1 + p') },

instead of (1/a) (1 - p') } as before, the boundary-value problem D', B,which we encounter has the same formal appearance as before, the dif-ference between the two problems receding entirely in the different defini-tions of q as a function of p. All the singularities of sinh ql, regarded as afunction of p, being furnished by the formula, p' = - 1 - (n'r'a'/1'),n - 0, 1, 2, , lie on the imaginary axis in the complex p-plane, and sothe half-plane over which duu(t) is supposed to possess a Laplace Trans-form is the half-plane c > 0, rather than the half-plane c > 1 as before. Inorder to find the integral operator K - K(:), a > 0, which is such that theLaplace Transform, over the half-plane c > 0, of K(!)u(t) is exp(-qs)/qwe set p - sinh z so that aq = cosh z - p + exp (- z) and exp (- qs) = exp(-ap/a) { 1 - (a/a) exp(-z) + (a'/21a') exp(-2z) - 1. Sinceexp(-nz)/cosh z, n = 0, 1, 2, , is the Laplace Transform, over thehalf-plane c > 0, of J.(t)n(t) it follows that exp(-qs)/q is the product ofthe Laplace Transform, over the half-plane c > 0, of

{J0t) - a Jilt) +2!82

a' J:(t) - ... u(t)

by a exp (- (s/a) p). The sum of the infinite series Jo(t) - (a/a) J, (t) +(8'/214')J:(t) - is Jo((1 + (2a/at))}t] and it follows that

K(81 =a .(e-a=)}u(t-a), 8>0and, since the derivative of

Jo(t)\\with

respect to t is -J,(t), this impliesthat the Laplace Transform, over the half-plane c > 0, of

I [O(z - at) + IO(x + at)]rse

Re}

-ij2a }Ja ] 8) + o(x + s)1 de


is pIo. Thus the solution of our boundary-value and initial-conditionproblem which is suggested by the application of the Laplace transforma-tion is

d(x, t) = [4,(x - at) + #(x + at))

- Jai

Is=2a 6

a2)

+ fat Jo L(2 - a (v(x - s) + v(x + 8)) ds

it being understood that the rangeofdefiniti/oooon

of 4(x) and v(x) is extendedby the statement that both O(x) and v(x) are odd periodic functions, withperiod 21, of the unrestricted real variable z.

The verification that this function of x and t actually is a solution of our .

boundary-value and initial-condition problem and the proof of the fact thatthis problem does not possess more than one solution are the same as before(the function Jo(8), d = ( - &i)}(r - 7-1) */a, playing, in the proof of theuniqueness theorem, the role previously played by lo(o)).

17

The Asymptotic Series.

*0 exp (-z2) dzforfp

Let h(t) be a piecewise continuous right-sided function which is zero overthe interval 0 < t < b, where a is any positive number, and let h(t) possessa Laplace Transform at a point p = c1 > 0 of the positive real axis in thecomplex p-plane. We suppose, further, that H(t) = f h(s) ds is definedover 0 < t < c but we do not assume the existence of the infinite integralf h(s) ds. We have seen that H(t) exp(-c1t) has the limit 0 at t = CO andthat implies, since H(t) exp(-clt) is everywhere continuous, that H(t)exp(-c1t) is bounded over 0 < t < -, i.e., that there exists a positiveconstant M such that I H(t) I exp(-clt) < M for every non-negativevalue of t. Thus H(t) possesses, over the half-plane c > cl, an absolutelyconvergent Laplace Transform and Lh = p(LH), c > c1 . Since H(t) is,like h(t), zero over the interval 0 < t < S, we have

LH - f H(t) exp (-pt) dt = f H(t) exp (-clt) exp [-(p - cl)tJ dt

so that

ILHI <M f"exp[-(c-c1)t) dt M exp[-(c-c1)8]8 C-C1

and the right-hand side of this inequality <2M/c exp[- (c - c1)6J ifc >- 2c1 . On denoting arg p by e, so that I p I = c(sec 6), it follows thatL(h) < 2M (see B) exp[-(c - c1) a] or, equivalently, that

Lh I exp(cd) < 2M(sec 0) exp(c1S)99


so that I Lh I exp(cb) is bounded as p -+ oo along the ray 0--1, p. If p liesin the sector - (,r/2) + P < 0 < ,r/2 - {4, 0 < 6 < rr/2, sec 9 < we f3and I Lh I exp(ca) is bounded over the part of the half-plane c > c, whichis covered by the sector - Or/2) + 0 < 8 < (,r/2) - 0. Now the productof any positive power of p by exp(-ca) tends to zero as p -+ co along anycurve which lies in the sector - (x/2) + 0 < 6 < (r/2) - P, the con-vergence to zero being uniform over the sector, and so we have the followingresult:

The product of Lh by any positive power of p tends to zero as p -> coalong any curve which lies in the sector - (,r/2) + R S a < (,r/2)the convergence to zero being uniform over this sector.

We next consider a piecewise-continuous right-sided function h(t) whichpossesses a Laplace Transform at p = c, and which, while not zero over anyinterval 0 <- t < 8, can be written, if S is sufficiently small, in the form0 °{A + e(t)}, where

(1) a is a constant whose real part a,. is > - 1 and A is any constant;(2) e(t) is continuous over 0 < t < a and arbitrarily small, say I e(t) I < E.

if b is sufficiently small, say 8 < b, .

The right-sided function hi(t) which = t°(A + e(t)) over the interval0 < t < a, and = 0 if t > a, possesses, over the half-plane c > 0, the La-place Transform jot°{ A + e(t)) exp(- pt) dl and we may write this inthe form

A O° t° exp (-pt) dt - A f° t° exp (-pt) dt

+J

t°e(t) exp (-pt) dt = pQ+ + I: + Io.0

12 is the Laplace Transform of a piecewise continuous right-sided functionwhich is zero over the interval 0 < t < a, so that the product of I I2 I byany positive power of p tends to zero as p -+ oo along any curve which liesin the sector - (a/2) + 0 < 8 < (w/2) - X

r(a, 1)I13I< J` ta'exp(-ct)dt<eJ t°'exp(-ct)dl +=e

and, since pa+' = exp[(a + 1) log p] so that

pa" exp { (a. + 1) log I p ac8) < (c sec #) +1 eRp\I a, 2/

it follows that pa+'I3 is arbitrarily small if 5, is sufficiently small. Since bothh(t) and h,(t) possess Laplace Transforms at p = c, , so also does their


difference h2(t) = h(t) - h1(t) and, since h2(t) is zero over the interval0 < t < S1 , the product of Lh2 by any positive power of p tends to zero asp - co along any curve which lies in the sector - (1r/2) + < 0 < (,r/2)- fl. Since Lh = Lh1 + Lhz , it follows that p°`+1(Lh) - A r (a + 1) isarbitrarily small if (1) S1 is sufficiently small, and (2) c is sufficiently large.Since p`+(Lh) - Ar(a + 1) is independent of b1 the proviso (1) may beomitted and so we have the following important result:

p"+1(Lh) - Ar(a + 1) tends to zero as p - co along any curve whichlies in the sector - (ir/2) + # < 0 < j3, the convergence to zerobeing uniform over this sector.

The right-sided function h(t) = exp(-t3/4)u(t) possesses, at any pointp of the complex p-plane, the Laplace Transform 2 exp (p2) f' exp (- z2) da,the integral being extended along the ray, from p to co in the complex z-plane, whose angle is zero. We denote by an(t) the sum of the first it termsof the power series development of exp(-t2/4) near t = 0:

snit) = 1 - (t)2 + 21()4 + (-.1)n-' (n1 1) t

(t)2n-2

72

and observe that a.(t)u(t) possesses, over the half-plane c > 0, the LaplaceTransform

3)p - 2pa +1.3 - ... + (-1)n-1 1.3 2A-(p2n-I

The right-sided function h.(t) = {exp(-t2/4) - an(t)IU(t) is of the formten{[(-1)"/n22n] + e(t)ju(t), where e(t) is continuous over any interval0 < t < S and, furthermore, I e(t) I is arbitrarily small if S is sufficientlysmall, by virtue of the continuity, at t = 0, of exp(-t2/4) - andthe fact that the value, at t = 0, of exp(-t2/4) - an(t) is 0. Hence p2n-'-1L

(-1) "[(2n) i/nl22?D tends to zero as p ---' ac along any curvewhich lies in the sector - (x/2) + # < 8 < (ir/2) - /3, 0 < # < jr/2.The Laplace Transform, over the half-plane c > 0, of hn(t) is

2exp(p2) f, exp(-z2)dz-(p-21 +2Q

+ 1.3...(2n - 3)2''p-'

and so the product of

1.32(exp p2) fD exp (-z2) dz __ { - 2I +

+ (-1),, 1.3 ...(2n - 1)2,p2-+1


by p°+1 tends to zero as p ---> oo along any curve which lies in the sector- (x/2) + S < 0 < (,r/2) - p. We express this result by the statementthat the infinite series (l/p) - (1/2p3) + (1.3/2'p6) - (1.3.5/2'p') + ... ,which fails to converge at any point p of the finite complex p-plane, is anasymptotic series, over the sector -a/2 < arg p < v/2, for the function2 exp(p2) f; exp(-z2) dz of the complex variable p and we write

2 exp (p2) J exp (-z2) dz1

-l-1=3 - ... , - r < arg p < 2

p 2p° _ 2 pb 2 2

The sector over which this asymptotic series is valid may be enlarged to-3,r/4 < arg p < 31r/4. To see this, let a be any number in the interval0 < a < r/4 and set t = v exp (-- ia) in the infinite integral f e' exp (-- t2/4 )exp(-pt) dt which defines L(exp(-t2/4)u(t)); then L(exp(-t'/4)u(t))appears in the form

exp (-ia) J exp [-- 4 exp (-2ia)] exp [-p exp(-ia)vj dv

the integral being along the ray from 0 to oo in the complex v-plane whoseangle is a. The modulus of the integrand of this integral, at any point vwhose modulus and argument are R and 0, respectively, is exp - [R'/4 cos2(a - 0) + f p ( R cos(a - 0 - 6) j, where a is the argument of p, and,if 0 < 0 < a, so that cos 2(a - 4') is positive, the product of this modulusby R tends to zero as R -a -, the convergence being uniform with respectto 0. Hence the integral of exp [-v2/4 exp(-2ia)] exp[-p exp(-ia)vjalong the are of the circle [ v t = R from v = R to v = R exp(ia) tends tozero as R -- co, and this implies that the integral of exp[-v'/4 exp(-2ia) jexp[-p exp(-ia)v) along the ray from 0 to -, in the complex v-plane,whose angle is a is the same as the integral of the same integrand along thepositive real axis in the complex v-plane. Thus 2 exp(p') f r exp(-z') de,which is the Laplace Transform of exp(-t2/4)u(t), may be written in theform

exp (-ia)J

exp L- 4 exp (-2ia) I exp [-p exp (-ia)t] dt0

where 0 < a < r/4 and the same argument shows that it may be writtenin this same form if - r/4 < a < 0 so that

2 exp (p') exp (-z2) dz

exp (-ia) f exp 1-e4exp (=2ia)I exp [--pexp (-ia)tjdi;

A x-4< a <

4


On denoting by a»'(t) the sum of the first n terms of the power series de-velopment of exp[- (t2/4) exp (-2ia)) near t = 0:

a»'(t) = 1 - exp (-2ia)/r

2J

\: + exp (2! 4ia) 4

(21)

+ e x p - (2n - 2) is t :0-2

(n - I) ! 2

and denoting p exp (- ia) by p' we have

exp(-ia) fp a»'(t) exp (-p't) dt = exp sa) _ 2 exp(PI),

3ia +

+ 1.3 . (2n - 3) exp - (2n - 1) ia I I

+. +(_I)»-z I.33I ,._1

provided that - w/2 < arg p' < w/2 or, equivalently, that - (w/2) +a < arg p < (w/2) + a. Since a may be assigned any value in the interval- 7/4 < a < w/4 it follows, by a repetition of the argument already givenin the case a = 0, that the asymptotic series (1/p) - (1/2P') + (1.3/22p6) - for 2 exp(p') f v exp(-z2) dz is valid over the sector -3w/4 <arg p < 3x/4.

In order to deal with the sector 3x/4 < arg p < 5w/4 of the complexp-plane we set z = - z' in the infinite integral f ;exp(-z2) dz and thendrop the prime:fbexp(-z')dz=fexp (_2) dz = exp (-i)dz

w ao

--- L exp (-z') dz

Now j'°., exp(-z') dz = exp(y2) f°° exp(-x2) exp (-2iyx) dx, z =x + iy, and f°° exp(-x2) exp(-2iyx) dx, being the Laplace Transformat p = 2iy of exp(-t2), is w} exp(-y2), so that f°°., exp(-z2) dz = ,r.When 3x/4 < arg p < 5w/4, -w/4 < arg(-p) < w/4, and so we mayuse the asymptotic series we have already obtained for 2 exp(-p)'f°pexp(-z2) dz to obtain the result

2x}exp(e)+I

P (P2) -P_ 2( P)= + ...1

V_.+2'ip6-...; 3w<algP<4w

We may use this trick if w/2 < arg p < 37/4 or if - 57/4 < arg p <


- r/2 and so we obtain two different asymptotic expressions for 2 exp(p')j; exp (- z') dz over these sectors, the difference being that one of theasymptotic expressions contains the additive term 2r; exp(p'). When p liesin either of these two sectors the real part of p' is negative, so that theproduct of exp(p') by any positive power of p tends to zero as p _, CO alongany curve which lies in the sector. Thus we see that we may use, for thefunction 2 exp(p') f, exp(-z') dz, the asymptotic series (1/p) --- (1/2p')+ (1.3/22p') - , with or without the additive term 2r exp(p'), overthe sectors r/4 < arg p < 3r/4, - 37x/4 < arg p < - it/4 while over thesector 3v/4 < arg p < 5r/4 the additive term must be used and over thesector --T/4 < arg p < r/4 it must not be used. The fact that 2 exp(p')f; exp(-z') dz has different asymptotic expressions over different sectorsof the complex p-plane is an instance of what is known as Stokes' phenome-non.

In order to appraise, when - T/2 < arg p < Tr/2, the differencebetween 2 exp(p') f; exp( -z2) dz and the sum of the first n + 1 terms ofthe asymptotic series (1/p) - (1/2p') + (1.3/2'p6) - we observethat, if r is any real number,

2 w r*+1expr=1+r+

r2l+... +exp(Or)(n+

1)1

where 0 < 0 < 1, 0 varying with r. Setting r = - t'/4 we obtain

exp (- 4 J = 1 - (2)2 + ! (2)`+ (_ 1)^ n' (2)

exp +2

(n + 1) ! (2}so that A.+,, which is the Laplace Transform, over the half-plane c > 0, of

[_ 1 - G1 ...(-1)n n! (2)'J ] u(t)

may be written as

21"+' 1)1 1 exp(_E4) t n+* exp (-pt) dt

Since 0 < 0 < 1, 0 < exp(-0t'/4) < 1, no matter what is the value of t.and so

1 "t:n+2 exp(-ct)dt = 1.3 ... (2n+ 1)a*+1 <_21 2(n -}- 1) ! 2n+lcs,.+s

Thus is dominated by the first term omitted of the asymptotic series.


Furthermore, if p = c > 0 is real, has the sign of this first term omittedsince fo

exp (- et'/4) e"' exp (- ct) t* is positive. For example, 2 exp (c)f; exp(-t') dt, c > 0, lies between 1/c and (1/c) - (1/2c'), so that if weuse exp (-c') /2c as an approximation to f ? exp (- t') dt, c > 0, this ap-proximation is in excess by less than 100/ (2c' - 1) per cent; if we useexp(-c') j (1/2c) - (1/4ca) J as an approximation to f o' exp(-t') di,c > 0, this approximation is too small by less than 150/(c2(2c' - 1)) percent and so on. If c 1, the term of the asymptotic series (1/c) - (1/2c')+ (1.3/2'c') - , whose numerical value is least is the second term andthe asymptotic series cannot guarantee a better approximation than thatgiven by its first term; if c - 2, the term whose numerical value is least isthe fifth and the asymptotic series cannot guarantee a better approximationthan that given by the sum of its first 4 terms. In general, if c is an integer,the asymptotic series cannot guarantee a better approximation than thatgiven by the sum of its first c' terms.

18

The Asymptotic Series for

(2irp)l exp (-p)In(p), I arg p I < 2;

The Hankel Functions

If p is any complex number, the infinite series Io(p) + 2(cos 8)I2(p) +2(cos 48)14(p) + converges, with the sum cosh(p cos 8), and, sincethis infinite series is dominated by the infinite series Io(I p I) + 212(1 p I) +. which converges, with the sum cosh I p I , the convergence is uniformover any closed interval. Hence the infinite series may be integrated term-by-term, after multiplication by cos 2m8, m = 0, 1, 2, , over the interval0 < 8 < ,r so that

I2.(p) = 1 f * cosh (p cos 0) (cos 2m9) do, m = 0, 1, 2,

If, then, f (O) is any linear combination, co + c, cos 20 + + " cos 2n8,of the functions cos 2m8, m = 0, 1, , n, we have

1 f cosh (p cos 8)f(0) do = colo(p) + c,I2(p) + +Tf

Now(-1)n22n-1 sin2n8 = i{exp(i8) - exp(-i8)}2n, n = 1, 2, . .

is such a linear combination, the coefficients c , c, being thefirst n coefficients of the binomial expansion (1 - x) 2n = 1 - 2nx +

108


()'2 z- .. and co = (-1) "}(2n) being one-half the (n + 1) at term of

thisbinomial expansion. For example, when n = 1, -2 sin'O = cos 20 - 1;when n - 2, 2' sin'B = cos 40 - 4 cos; 20 + 3, and so on. Hence

cosh (p cos 0) sin'"B d9 = c" I, (p)

+ I,,,_,(P) + + colo(P)

where c" , , cl are the first n coefficients of the binomial expansion of(1 -- x)'" and co is one-half the (n + 1) at coefficient of this expansion.Setting n = 1, we obtain

-2 I r cosh (p cos 0) (sin' 0) do = I=(P) - lo(p) = -2 Ii(P)o

so that

1 * cosh (p cos 8) (sin' 0) dO = Ii(p)rSetting n - 2, we obtain

s

2 L cosh (p cos 0) (sin' B) dO = I4(p) - 4I,(p) + 31o(p)W

Now

I4(P) _" 1,(P) 6I3(p) -'2(P) - {I) - I1(P)

= I2 (P) + 3{I2 (P) - Io(P)) + %,(P)

so that I4(p) - 4IL008hp0086)(8mfl40)d0,(p) + 3Io(p) = (24/p')I,(p). Hence

1 - I,(p).

These two results are special cases, corresponding to n - I and n = 2, ofthe general formula

(pcos0)(sin'"0)do = I* (p)

where A. = 1.3.5 (2n -- 1), n = 1, 2, . getting Ao - 1, this formulacovers the case n = 0 so that

1 f'-O

cosh (p cos 0) (sin' 0) d9 = `9A* I"(p), n 0 1 2x


This general formula may be easily verified by multiplying the infiniteseries Is(p) + 2(cos 28)I,(p) + - , whose sum is cosh(p cos 0), bysing"8 and integrating term-by-term. We prefer, however, to derive it byan extension of the method by which we proved it in the can n - 2 sincewe obtain in this way a useful generalization of the recurrence relation

I.(p) = I,.--:(p) 2(n p 1) n = 2, 3, .. .

Replacing n by n - 1 in this relation we obtain

3,4,(p) = I" (p) - 2(np

2) I.-2(P), n =

I.(p) - I.-,(P) - 2(n 1) (p) + (n - 2)p P,

and since

2(n - 3) I,w(p) = I.-4(P)p

n= 3,4,...

n=4,5,the right-hand side of the relation just derived may be replaced, whenn= 4, 5, , by

n - 3 {I"-4(p) -.1,2(p), + 1)(n - 2) I.-s(p)

Thus

I.(p)-2n-3I.-s(P)+n-3l.-.a(p)=2:(n- 1)(n-2)I,.-s(P),

Ps

This is our first extension of the recurrence relation

I.(p) -- I.-s(p) _ 2(np

1) In-gy(p),

and it yields, when n = 4, the relation

I4(p) - 4Is(p) + 3Io(p) _ 2.37 II(p)

On setting

n=4,5,

n - 2,3, --

I.-3(P) = I.-+(p) - 2(n p 3) I,.-gy(p)

The Laplace Tranaformotion 109

on the right-hand side of this extension of the recurrence relation and usingthe fact that

I..-2(p) - 2 n1 --4- 5 I,.-.(p) + n _52=(n -(n - 4) I.-4(P), n = 6, 7, .. .

we obtain

I.(P) _2n-3I,.-2(P)+n-3I.-4(P)

0 (n-1)(n-2)fl..-:(p)-2n-4I,..-4(p)+n-3l,.-e(P)}(n - 3)(n - 4) ` n - 5 n - b

2'(n - Mn - 2)(n - 3)P;

or, equivalently,

1.(P)3(n - 2) I,.-2(p) + 3(n - 1)

I.-4(P)n-4 n-5_(n-1)(n-2)

(n - 4)(n - 5) I"-e(P) = -P, (n --- 1)(n - 2)(n -- 3)I,.-a(p),.. .n 6,7,8,

This is our second extension of the recurrence relation 1.(p) -- I.-j(p)- 2[(n - 1) /pjI.-i (p) ; it yields, when n = 6, the relation

6I4(p) + 15I2(P) - 10I4(P) = ----.5 r (p)

On setting n = 3 in the relation

(-1)" 2b`-1 cosh (p oos e) (sin'" 9) d9 = c,.It.(p) + ... + celo(p)

we obtain

1 * cosh (p cos 9) (sine 9) d9 - JIs() ° A-po' I.(p)

Proceeding in this way, we obtain the following generalized recurrencerelation whose validity may` be verified by an induction proof :

I.(p) - \2) a.'I..-.4(p) + (-1)"«,.°I..-:.(P)/1),.2(m - 1)(m - 2) ... (m - n) Im(p)

p"


w h e r e n = 1, 2, - , m = 2n, 2n + 1, .. The coefficient, a.k, of (--1) kon the left-hand side of this relation is a fraction whose de-

n: ninator is (m - n - 1) (»c - n - k) and whose numerator is inde-pendent of n, this numerator being determined by the fact that

k (m-1)...(m-k+1)ak =(m k - 1) ... (m - 2k -}- 1)

while al' = 1. Thus

1 m-2°G" m-n 1

and

ifk=2,3,- -,

k 2) - (m-k+1)(m-2k)a, (m-n-k)k = 2, 3, - , n. When m = 2n, this generalised recurrence formula yieldsthe relation

I2.(P) - 2nI:-:(p) + ( 2 2 ) I:".-.(P) - + (-1)" 2 ( ) Io(p)

(-1)"2'1A"1"(p}

and this implies that (1/r) f o' cash (p cos 8) (s&"#) do = (A"/p") I. (p),n = 0, 1, 2, - . On replacing cosh (p cos 0) by if exp (p cos 6) + exp (- pcos 9) j and observing that

exp (p cos 0) (sins" 0) do - lo* exp (--p cos 9') (sin'" 9') d9',fe

"

we obtain

rA" p-"I"(p) (sin' 0) exp(--pcos0)dot11

(1 - v')""«' exp (-pv) dv, v - coo*1

(exp p) ft"_(R) (2 - t)"-« ) exp (-pt) dt, t + 1

0

so that TA"p" exp(_p)

I"(p) is the Laplace Transform, over the entirefinite complex p-plane, of the right-sided function which = t"i(2 - t) "iover the interval 0 < t < 2 and which = 0, if t > 2. If n - 0, this right-sided function is unbounded at t = 0 and t - 2. Over the interval 0 < t < 2,1"(2 -- t) "-4 may be written in the form


2"-}t"`4 1 -t + (n 2) to 2)(

\n5>

2! (2) - .. .

/ 2k-3(k - 1)!

2\2/

ri-k+(}) 1 21c - 12 (n -21 ... n-2

where e(t) is continuous over 0 < t < 2 and is arbitrarily small over0 < t < b, if S is sufficiently small. Hence, if the real part c of p is positive,the product of

I'[n+2)- n2)r(n+23

}VA. p exp (-p)I,. (p) - 2n }

L2p-4

.. .

2J... (n2k21

+. +) I'{n+k-f-

2k . k ! \ p,.+k+}

by p" +k+} tends to zero as p -+ oo along the ray 0 -i p, the convergencebeing uniform over the sector - (ir/2) + $ < arg p < (7r/2) - d, where 6is any positive number less than a/2. Since r(n + 1) = (n --- ,}) r(n - 1)

_ (n - 1) (n - $) jir} = it follows, on multiplicationby that the product of

Ok+i = (2wp)} exp p) 1. (p) 1 - 4W1 + (4n2 -- 1) (4n2 - 3')

2!(8p)2

- -}- (-1)k (4n2 - 1) (4n2 - 32) ... (4n2 - (2k - 1)2}

k!(8p)k

by pk tends to zero as p - along any curve which is covered by thesector -(ir/2) + $ < arg p < (ir/2) - $ so that the infinite series I -[(4n2 - 1) /8p) + [(4n' - 1) (4n2 -- 32)/(2!(8p)2)) - ... , which fails toconverge at any point of the finite complex p-plane, is an asymptotic series,over the sector -- (w/2) + 0 < arg p < (vr/2) --- d, for the function (2,irp)exp (- p) I. (p) of the complex variable p.

The asymptotic series which we have just obtained for (2irp)-} exp(-p)I. (p) is not valid when p = it is a pure imaginary and so it fails to providean asymptotic series for J. (t) . To obtain an asymptotic series useful in thecalculation of we proceed as follows: Let v be a complex variable and

i1 g Lectures on Applied Mathematics

let the complex v-plane be cut along the segment 0 < v < 2 of its real axisso as to make v"(2 - v) "-} exp(-pv) uniform over the two-sheetedRiemann surface so obtained, the value of this function at any point of thelower sheet of this Riemann surface being the negative of its value at thecorresponding point of the upper sheet. Let us consider the closed curve Con this two-sheeted Riemann surface which consists of the following fourparts:

(1) The line segment from the point 2 - 8 in the lower sheet to the point6 in the lower sheet, 0 < 8 < 1.

(2) The circumference I v ) = 8 from the point 8 in the lower sheet to thepoint 8 in the upper sheet.

(3) The line segment from the point 8 in the upper sheet to the point2 - 6 in the upper sheet.

(4) The circumference I v - 2 8 from the point 2 - 8 in the uppersheet to the point 2 - 8 in the lower sheet.The integral of v"-; (2 - v) "4 exp (- pv) along C is independent of 8 and as6 -i 0 the contributions from the parts (2) and (4) of C tend to 0 while thecontributions from the parts (1) and (3) of C tend to fot"-4(2 - t)`4exp( - pt) dt, so that the integral of v"-*(2 - v)" 4 exp( -pv) along C hasthe value 2TAp " exp(-p)I"(p).

We next consider the closed curve C' on our two-sheeted Riemann surfacewhich consists of the following five parts:

(1') The line segment from the point R exp(ia) in the lower sheet to thepoint 6 in the lower sheet where R is any positive number, a is any numberwhich is such that I a + arg p < r/2 and 8 is any positive number leesthan 1.

(2') The circumference ( v = 8 from the point 8 in the lower sheet tothe point 8 in the upper sheet.

(3') The line segment from the point 8 in the upper sheet to the point2 - 8 in the upper sheet.

(4') The circumference I v - 2 8 from the point 2 - 8 in the uppersheet to the point 2 - 6 in the lower sheet.

(5') The line segment from the point 2 - 8 in the lower sheet to thepoint R exp(ia) in the lower sheet.

The integral of v"-*(2 - v) "`4 exp(-v) along C' is the same as the integralof the same integrand along C and is independent of R, a and 8. As 8 -- 0,the contributions to this integral from the parts (2') and (4') of C' tendto 0 while the contribution from the part (3') of C' tends to fU-'(2 -t) "i exp(-pt) dt - rA "p7" exp(-p) I"(p; . As R --> -o, the contributionfrom the part (I') of C' tends to f;v"-4 (2 - u; "_i

exp( -pv) dv, the inte-gral being extended along the ray of angle a from 0 to co (this infinite

The Laplace Tranoformation 113

integral existing since I a + arg p I < v/2) and the contribution from thepart (5') of Ctends to -f2-v"(2 - v) "-4exp (-- pv) dv, the integral beingextended along the ray of angle a from 2 to cc . Thus, for example,

.oa9hrTA. pr" exp (-p)I" (p) = f v"-i(2 - v)'" f exp (-pv) dv

0

fo , iav"(2-v)"exp(-pv)dv

If r/2 < arg p < r, we set a = - r/2 and, if - r < arg p < - r/2 weset a = ,r/2. Treating the first case, we set v = i&in Il , and v - 2 - it inI, so that, in each of the two integrals, 0 < t < ao. Il appears as 2"-}(-i)" (it/2))"i exp (-p't) dt, where p' ip so that thereal part of p' is positive, and 12 `appears as

f t"' (/1 - 2)"-} exp (-p't) dt X exp (-2p)

On multiplying through by exp p we obtain

rA. (iP)-"I"(iP) exp (ip') t' (i+)f mo

exp (-P t) dt + 2"-4i'. exp (-ip) 10r-' (1 - 2 )"' exp (-p t) dt

or, equivalently, since i "I.(ip') = J.(p'), we have

2J.(p') = 2"44w 'A;'(P )" exp [ (0, - (n + )](1+ )"exp(-pt)di+ 2'. 'A*'(P)"

expL-i\p - \n+2/ )]ft"-(1- 22 exp(-pt)dt

-= H.<1'(p) + H.(2)(g )where H.(')(p') is the first member, and H.("(p') is the second member, onthe right-hand side and 0 < arg p' < r/2. H.c')(p') and H.(2) (P') are knownas Hankel Functions. A similar argument shows that the relation 2J.(p')H."%(p') + H.("(p') remains valid when -r/2 < arg p' < O,inwhich case--r < arg p < - r/2 and we set a = r/2. We shall derive in the nextchapter, from this representation of J.(p') as the mean of the two Hankelfunctions, H.(')(p') and H.("(p'), a formula furnishing J.(t), when t is realand positive, as the sum of two asymptotic series, each of which has theconvenient property that the error made in stopping at any term has thesame sign as the next term and is dominated by this term.

19

The Asymptotic Series

for P.(c) and Q.(c)

If p is any complex number whose real part c is positive the Hankel Func-tion H.(')(p) is defined by the formula

H,,(') (r) = 2"*ix 'A, 'p" exp [iCp - (n + 2} 2}]

/ *

2 exp(-pt)dt

In particular, when p = c is real and positive,we have/, on making thesubstitution t = t'/c and then dropping the prime,

H"(') (c) =

Similarly,

2"44x 'A"'c } exp [ Cc - Cn + 2/]

+ exp(-t)dt

H"(" (c) = 2n4iW 'A 'c-4 exp I -i !c -(n + 2) 2)]

J exp(-t)dt114

The Lapfaoe Transformation 116

so that

(it2g.a)(c) ) 2) P"(c) cos Ic - + 2/ 2lJ

wherel

P. (C) _ A"(1 exp (-t) dt\ JJJ n=0,1,2,...

)" - [ 1 -)*-}I exp (-t) dtQ"(c) m_t" 1 +w-TA-. f til 51

Ifn= 1,2, ,

J.-I(C) (!); { - P, (c) sin [c -\n + 2) 21

-Q,r,(c)cosic-(n+

and, if n = 0, 1, 2,

J"+1(c) = (c); {P*+i(c) sin c - n + 2) J- Q.+1(c) oos C -

\n +'I

ZJ}

and it follows, since J.-I(c) - J.+, (c) = (2n/c) J.(c) , n = 1, 2, , that

P.+1(c) - P.-I(c)2n

Q.(c); Q.+1(c) - Q.-I(c) =cn P.(c);

n = 1, 2, .. .

We shall obtain asymptotic series for Po(c), QW(c), Pl(c), Qi(c) and shalldeduce from these, by means of the recurrence relations just derived,asymptotic series for and Q"(c), n = 2, 3, .

In order to obtain asymptotic series for Po(c) and Qo(c) we observe that,if v is any nonreal complex number,

2,It

(1 -v) =710 1 - vsinto

11 6 Lectures on Applied Mathematics

Indeed,n do ,It 2d4, do

J6 1 -vsing02-v+vvcoe2¢-Je 2--v+vcos8'8-20,

= 1f' d8

2 ,2-v+vcos 8Setting exp iO = z, so that d8 = dz/iz,

1 f' do 1!

dz2 v+vcos8 iv c.4 +2[2-1+1

\v

where C is the circumference (z 1. The two zeros, z1 and Z2, of the quad-ratic polynomial z' + 2((2/v) - 1)z + 1 are such that 2122 = 1 and, sincetheir sum, 2(1 - (2/v)), is not real, neither can be a complex number ofunit modulus for, if f z1 I = 1, for example, the reciprocal z2 of z1 would be itsconjugate, 11 , and z1 + z2 would be real. Writing z1 = 1 - (2/v) + (2/v)(1 - v)4 we see that, when I v I < 1, z1 = - Jv + - tends to zero withv. Hence I z1 I < 1 if I v I is sufficiently small and this implies, since I z1 I isnever 1, that { z1 ( < 1 no matter what is the nonreal complex number v;hence ( z, I > 1 no matter what is the nonreal complex number v so that z1is the only zero of z2 + 2((2/v) - 1)z + 1 which lies inside C. The coef-ficient of 1/(z - z1) in the development of 1/[(z - z,)(z - z2)] as an in-finite series of the type [c/(z - z1)] + co + c1(z - z1) + is 1/(z1 -z:) _ (v/4) (1 - v) and thus we have

dz (1-v)'-1 z+l' 2

tv

proving that

2 doTI,/2

1 - v sin' 4

The usefulness of this result lies in the fact that it provides us with a con-venient expression for the remainder in the binomial expansion of (1 - v)Writing

(1 - vsin'4')-1 = l +vsin'4+

+ v:..-1 .. ' 4 +Vm sin`" ' ' m = 1, 2 .. .

1 -vsin=0 '


we obtain

(1-v)-* l+IV -{-1_3v2+... +1.3... (4m--3)v2m-1

2.4 2.4 (4m - 2)

+ 2 2M [1/2 (Sin" 0)d4)7v 1 - vsin2-0

On changing the sign of v we obtain, by addition and subtraction, the rela-tions

++v)-;} +2.3U2+...

1.3 ... (4m - 5) ,,,,_2 2 2,"+2.4... (4m-4)v +v

o

1{(1-v)-* +v)-411+1.3.5v,+...

2 2 2.4.6

sin4m

1 - v2 sin4d

+ 1.3 ... (4m - 3) v2".-, + 2 v2,"+, f/2 sin4,,:+2 o d4,

2.4 (4m - 2) ir 1 - v2sin44,

Setting v =it/2c, 0 < t < oo, these yield the relations

21 + (1 += 1 - 23

)2+ .. .

+ (-1)M_1 1.3 (4m - 5) t2'"-2

(2m - 2)1 (4c)2"-2

," 2 t s.. =/2 sin4,"

ir XL

1 -}-4c2

sin4

2i {(1 2c)-} (1 + 4ct

- 1315 (4c)'+ .. .

+ (-1)..-1 1.3 ... (4m - 3) t2".-1

(2m - 1) ! (4c)2,t-1

2 / t \ law+l v12 4w44 4,

s 2c ° 1 +4 sin4 4

Since 1 + (t2/4c2) sin44 > 1, the integral

2 f"2 sin'" 4,f0

do

1 + sin4 ¢

118 Leetvrea on Applied Mathematics

which is positive, is dominated by

2 f,1'sin" 1.3 ... 4m - 1)

2.4 ... (4m

and, similarly, the integral

2 j,/2 Sin"+2 # d4o

'r 1 +jsw4which is also positive, is dominated by [1.3 - .. (4m + 1)1/[2.4 . . . (4m +2) j. On denoting by $(t) any positive function of t which is dominated by 1we have the following two equations which we shall refer to as the equationsAo:

Ao:

2 ( 2c) 2c) 21 (4c)'' - I - !--3

+ 1.3 - - (4m - 5)(2m - 2) ! (4c)'

+ (-1)m 1.3 ... (4m - 1) t2'9(t)

(2m) ! (4c)='"

2i{(1 2c} - (1 + c)- 4c 3! (4c)' ' ...+, (-001-11.3 ... (4m - 3) en-1

(2m - 1) ! (4c)2 -1

+ (-1)m 1.3 ... (4m+ 1) t2"+1

(2m + 1)! (4c)21^+19(t)

where the positive function B(t) which appears on the right-hand side ofthese two equations is not the same in the two equations. Upon integratingeach of the equations Ao over the interval [0, t] and using the fact that theintegral of t"O(t) over this interval is positive and is dominated by e,+1/(2m + 1) we obtain the following two equations which we shall refer to asthe equations At :

_t 1.3 t' +...2i 2c 2c )

-4c - 3 ! (4c)'


Al

+ 1.3 ... (4m - 1) is-+'e(t)(2m + 1) ! (4c)'u1+'

21W)a1415(`k)`+...

+ 1.3 .. (44n - 3) t''"

(2m) !, (4c)$-

+ (-1)" 1.3 ... (4m + 1) t21e+2B(t)(2m + 2)1 (4c)'"'+'

where the positive function B(t) which appears on the right-hand side ofeach of the equations A,1 is not the same in each of the two equations northe same as the positive function 9(t) which appeared on the right-handside of the equations Ao.

Upon multiplying the equations Ao by the non-negative function t4exp (- t) and integrating over the positive real axis we obtain, on denotingby 0 any positive number which is dominated by 1,

Po(c) z 1

r(21) _ 1.3 r(1) + .. .:i 21 (4c)'

+1.3 (4m- 5)r(2m-4)(2m - 2) ! (4c)2" -2

+ 1.3 ... (47n -- 1) r(2m + 4)B = 1 - 12.31 + ...

(2m) ! (4c) 2m 21(8c)2

+ 12.32 ... (4m - 5)1 + (-1)" 11.32 ... (4m - 1)'(2m - 2) !(8c)2"'-' (2m) 1(8c)2-

Qo(c)_ 1 rr(4) _ 1.3.5 r(i) + .. .P 4c 31 (4c)'L

+1.3... (4m-3)r(2m-4)(2m - 1) ! (4c)2-1

+3... (4m+1)r(2m+4)B = -1 +12.32.52-(2m + 1) 1 (4c)2"'+1 8c 3I(8c)'

...+ (-1)" 12.32 ... (4m - 3)2 + (-1)'"+1 12.32 ... (4m + 1) B(2m - 1) (2m + 1) !(8c)2-+1

where the positive number 0 which appears on the right-hand side of eachof these two equations is not the same in the two equations. Neither of thetwo infinite series

1 '1.3 ' ' ' 21 .3.5.7 ' : '1 1 .3.5 2 ' ' ' '1 .3 .b .7 .91 -

2!(8c)2-+ 4'(8c)l

8c + 3!(8c)3 5!(8c)5+


converges for any finite value of c but the first of these two infinite series is anasymptotic series for NO and the second is an asymptotic series forQo(c). Each of these two asymptotic series possesses the property that theerror made in stopping at any term has the sign of the next term and isdominated by this term.

Upon multiplying the equations Al by the non-negative function itexp(--t) and integrating over the interval [0, t] we obtain, similarly,

QI(c) - 3 3'.5.7 + ... + (_ ... (4m - 5)1(4yn - 3) (4m - 1)Sc 3!(8c)' (2m - 1)1(8c)2 '

+ ...(4m - 1)'(4m + 1) (4m 3) e(2m + 1) !(8c)2"+1

Pl(c) + 3.5 3'.5'.7 + ... +21(8c)= 41(8c)d

(_ ... (4m - 3)_(4m _ 1) (4m + 1)(2m) !(8c)21

+ (2m + 2) !(8c)2"

where, again, the positive number a which appears on the right-hand sideof these equations is not the same in each of the two equations. Thus thetwo infinite series

3.5 _ 32.52.7.9 3'.52.72.92.11.13

1 + 2!(8c)' 41(8c)' + 61(8c)'

3 3'.5.7 3'.5'.7'.9.118c 3!(8c)' + 51(8c)'

each of which fails to converge for any finite value of c, are asymptoticseries for PI(c) and Q,(c), respectively. The asymptotic series for Q,(c) isalternating while the asymptotic series for Pl(c) is alternating if we removeits first term. The asymptotic series for Q,(c) possesses the same propertyas the asymptotic series for NO and Qo(c) : The error made in stopping atany term has the sign of the next term and is dominated by this next term.On the other hand, the asymptotic series for P, (c) does not, necessarily, pos-sess this property if we stop at the first term; for this asymptotic series allwe can claim is that the error made in stopping at any term, after the first, hasthe sign of the next term and is dominated by this next term.

The asymptotic series which we have obtained for Po(c) and PI(c) arespecial cases, corresponding to n = 0 and n = 1, respectively, of the series

1(4n'-11)(4n'-3') + (4n=-12)(4n2 -3')(4'-5')(4'n'-7')

2I(8c)= 4!(8c)4


and the asymptotic series which we have obtained for Qo(c) and Q1(c)are special cases, corresponding to n - 0 and n - 1, respectively, of theseries

(4n' - 12) (4n' - 1') (4n' - 3') (4n' + ..8c 31(8c)i

.

We proceed to show that these series are asymptotic series for PA(c) andQ (c), respectively, and, furthermore, that, if n = 2k is even, theseasymptotic series possess the property that the error made in stoppingat any term after the kth has the sign of the next term and is dominated bythis next term; if n = 2k + 1 is odd the error made in stopping at anyterm after the (k + 1)st of the asymptotic series for and at anyterm after the kth of the asymptotic series for has the sign of thenext term and is dominated by this term. We have shown that these state-ments are true when n = 0 and when n = 1 and, assuming that they aretrue for any two consecutive values, j - 1 and j, of n we proceed to showthat this assumption implies that they are true for the next consecutivevalue, j + 1, of n. Thus we assume that

P%(c) = 1 - (4n' - 1') (4n' - 3') -- ... -4-2l(8c)A

8(2m 1(8c)3

if n - j - 1 and m is sufficiently large, and that

Wc)(4t'-1)_ ...+(-1),*(4n'-1')...{4n'-(4m,...7)2

8c (2m - 351(8c)i°-_71_

+1 1(8c)21-1

if n = j and m is sufficiently large, both the positive numbers 8 and 8'being dominated by 1. The coefficient of (-1)'{ (2r)1(8c)2'}-', r = 0, 1, ,m - 1, in the asymptotic series for PI-I(c) is the product (2j - 4r - 1) .

(2j + 4r - 3) of all the odd integers, not necessarily positive, beginningwith 2j - 4r - 1 and ending with 2j + 4r - 3 and the coefficient of(-1) '{ (2r) l (8c) "'} -, r = 1, , m - 1, in the product of the asymptoticseries for Q,(c) by - (2j/c) is 32jr times the product (2j - 4r + 3)(2j + 4r -- 3) of all the odd integers beginning with 2j - 4r + 3 andending with 2j + 4r - 3. Since (2j - 4r --- 1) (2j - 4r + 1) + 32jr =

1Y*_I(4n'..._ 12)... (4n2 - (4m - 5)2}

(2m - 2)1(8c)b"+(-1)'"(41e _ 1')...{4m'_ (4m- 1)2}

itE Lectures on Applied Mathematua

(2j + 4r - 1)(2j + 4r + 1) the coefficient of (-1)'{(2r)!(&}''}-',r - 1, , m - 1, in the result of subtracting 2j/c times the asymptoticseries for Q,(c) from the asymptotic series for P;....,(c) is the product,(2j - 4r + 3) ... (2j + 4r + 1), of all the odd integers beginning with2j - 4r + 3 and ending with (2j + 4r + 1) and this product is the value,when n = j + 1, of (4n' - 12) {4n' - (4r - 1)1. Since P,+, (c)P,-,(c) - (2j/c)Qj(c) it follows that, when n j + 1, P. (c) is

1- (4n'-1')(4n'-3)+...+2 t(8c)'

(-1)W-1 (4n' - 1')... {4n' - (4m - 5)'}(2m - 2)1(8c)2--2

plus a remainder term, this remainder term being (-1) "{ (2m) !(8c)''"}times the product of (2j - 4m - 1) (2j - 4m + 1) 8 + 32jm8' by theproduct, (2j - 4m + 3) (2j + 4m - 3), of all the odd integers beginningwith 2j - 4m + 3 and ending with 2j + 4m - 3. Since (2j - 4m - 1)(2j-4m+1)+32jm= (2j + 4m - 1) (2j + 4m + 1), (2j - 4m - 1) -(2i- 4m+1)8+32jm8' is of theform (2j+4m- 1)(2j+4m+1)8"where 8" is positive and dominated by 1 provided that m is large enough tomake 2j - 4m + 1 negative (so that (2j - 4m - 1) (2j - 4m + 1)is positive). Thus the remainder term is the value, when n = j + 1, of

(-1) (4n' - 12) {4n' - (4m - 1)'} B" which roves the validity,(2m)1(8c)2"' P

when n = j + 1, of the statement made concerning the asymptotic seriesfor P,.(c). In the same way we prove the validity, when n - j + 1, ofthe statement made concerning the asymptotic series for Q,,(c). Thiscompletes the proof, by mathematical induction, of the validity, for allnon-negative integral values of n, of the statements made concerning theasymptotic series for P,.(c) and Q,.(c).

Now the product of any term, say the rth, of the asymptotic series forQj(c) by -2j/c becomes, by virtue of the relation P;+,(c) =P,-,(c) - (2j/c)Qj(c), part of the (r + 1)st term of the asymptoticseries for P,+,(c) while the product of the rth term of the asymptoticseries for P,(c) becomes, by virtue of the relation Q,+,(c) = Q,-,(c) +(2j/c)P,(c) part of the rth term of the asymptotic series for Q,+, (c). Thus,in order to be assured that the error made in stopping at any term of theasymptotic series in question has the sign of the next term and is dominatedby this next term we must take

(1) more than 1 term of the asymptotic series for P,(c), Q2(c) andQS (C)


(2) more than 2 terms of the asymptotic series for Pl(c), P4(c), Q4(c)and Qi(c), and so on.

In general, if n - 2k is even, we must take more than k termsof the asymptotic series for and while, if n = 2k + 1 is odd,we must take more than k + 1 terms of the asymptotic series forand more than k terms of the asymptotic series for QA(c).

Bibliography

1. Doetsch, G., Einftlhrung in Theorie Land Anwendtasg der Laplaws-Tranaformatton, Leipzig, Birkhauser, 1958.

2. Doetach, G., Handbuch der Laplace Transformation, Vols. 1-3, Leipzig,Birkhauaer, 1950-56.

3. Pol, van der B., Operational Calculus Based on the Two-Sided LaplaceIntegral, London, Cambridge University Press, 1955.

4. Erd6lyi, A., Operational Calculus, California Institute Technology, 1956.5. Carslaw, H. S., and J. C. Jaeger, Operational Methods in Applied Mathe-

matics, London, Oxford University Press, 1941.6. Jaeger, J. C., Introduction to the Laplace Transformation with Engineer-

ing Applications, Melbourne, 1946.7. Churchill, R. V., Modern 0operational Mathematfca in Engineering,

New York, McGraw-Hill Book Co., Inc., 1944.8. McLachlan, N. W., Modern Operational Calculus, New York, The

Macmillan company, 1948.

194

INDEX

Absolutely integrable function, 3Asymptotic series, 102, 111, 119, 121

Bessel function, 42Bessel'e differential equation, 67Boundary value problem, 76Boundary value and Initial condition

problem, 75

Cauchy principal value of an integral, 8Characteristics of a partial differential

equation, 86Complex-valued function, 1

Exponential type, 40

Fourier Integral Theorem, 10Fourier Transform, 5

Gamma function, 26Green's functions, 78

Hankel functions, 113Heaviside unit function, 20

Improper integral, 2Improperly integrable, 2Integral operator, 78, 81, 89

Laguerre polynomials, 53Laguerre's differential equation, 53Laplace Transform, 20, 28Laplace version of the Fourier Integral

Theorem, 20Left-sided function, 22

Modified Bessel equation, 63Modified Bessel function, 42, 62Modified Laguerre differential equation,55

Piecewise continuous function, 1

Real-valued function, 1Recurrence relations for Bessel func-

tions, 67Reflection of waves, 86Right-aided function, 22

Stokes' phenomenon, 104

Translation theorem, 25, 37

Uniqueness theorem, 27

Waves, 85, 86

125

The Laplace Transformation

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