Chapter 3 Integral Transforms - School of Mathematics

1 Chapter 3 Integral TransformsThis part of the course introduces two extremely powerful methods to solvingdifferential equations: the Fourier and the laplace Transforms . Beside itspractical use, the Fourier transform is also of fundamental importance inquantum mechanics, providing the correspondence between the position andmomentum representations of the Heisenberg commutation Integral transform is useful if it allows one to turn a complicatedproblem into a simpler one. The Transforms we will be studying in this partof the course are mostly useful to solve differential and, to a lesser extent, Integral equations. The idea behind a transform is very simple. To be definitesuppose that we want to solve a differential equation, with unknown functionf. One first applies the transform to the differential equation to turn it intoan equation one can solve easily: often an algebraic equation for the transformFoff.

2 One then solves this equation forFand finally applies the inversetransform to findf. This circle (or square!) of ideas can be representeddiagrammatically as follows:differential equation forfalgebraic equation forFsolution:Fsolution:ftransforminverse transform6?--We would like to follow the dashed line, but this is often very we follow the solid line instead: it may seem a longer path, but ithas the advantage of being straightforward. After all, what is the purpose ofdeveloping formalism if not to reduce the solution of complicated problemsto a set of simple rules which even a machine could follow?We will start by reviewing Fourier series in the context of one particularexample: the vibrating string. This will have the added benefit of introduc-ing the method of separation of variables in order to solve partial differentialequations. In the limit as the vibrating string becomes infinitely long, theFourier series naturally gives rise to the Fourier Integral transform , which wewill apply to find steady-state solutions to differential equations.

3 In partic-ular we will apply this to the one-dimensional wave equation. In order todeal with transient solutions of differential equations, we will introduce theLaplace transform . This will then be applied, among other problems, to thesolution of initial value Fourier seriesIn this section we will discuss the Fourier expansion of periodic functions ofa real variable. As a practical application, we start with the study of thevibrating string, where the Fourier series makes a natural The vibrating stringConsider a string of lengthLwhich is clamped at both ends. Letxdenotethe position along the string: such that the two ends of the string are atx= 0 andx=L, respectively. The string has tensionTand a uniformmass density , and it is allowed to vibrate. If we think of the string asbeing composed of an infinite number of infinitesimal masses, we model thevibrations by a function (x, t) which describes the vertical displacement attimetof the mass at positionx.

4 It can be shown that for small verticaldisplacements, (x, t) obeys the following equation:T 2 (x, t) x2= 2 (x, t) t2,which can be recognised as theone-dimensional wave equation 2 x2 (x, t) =1c2 2 t2 (x, t),( )wherec= T/ is the wave velocity. This is a partial differential equationwhich needs for its solution to be supplemented by boundary conditions for186xand initial conditions fort. Because the string is clamped at both ends,the boundary conditions are (0, t) = (L, t) = 0,for allt.( )As initial conditions we specify that att= 0, (x, t) t t=0= 0 and (x,0) =f(x),for allx,( )wherefis a continuous function which, for consistency with the boundaryconditions ( ), must satisfyf(0) =f(L) = 0. In other words, the string isreleased from rest from an initial shape given by the functionf. This is not the only type of initial conditions that could be imposed. For example, in thecase of, say, a piano string, it would be much more sensible to consider an initial conditionin which the string is horizontal so that (x,0) = 0, but such that it is given a blow att= 0, which means that (x,t) t|t=0=g(x) for some functiong.

5 More generally still, wecould consider mixed initial conditions in which (x,0) =f(x) and (x,t) t|t=0=g(x).These different initial conditions can be analysed in roughly the same will solve the wave equation by the method ofseparation of consists of choosing as an Ansatz for (x, t) the product of two functions,one depending only onxand the other only ont: (x, t) =u(x)v(t). Wedo not actually expect the solution to be of this form; but because, as wewill review below, the equation is linear and one can use the principle ofsuperposition to construct the desired solution out of decomposable solutionsof this type. At any rate, inserting this Ansatz into ( ), we haveu (x)v(t) =1c2u(x)v (t),where we are using primes to denote derivatives with respect to the variableon which the function depends:u (x) =du/dxandv (t) =dv/dt. We nowdivide both sides of the equation byu(x)v(t), and obtainu (x)u(x)=1c2v (t)v(t).

6 Now comes the reason that this method works, so pay close that the right-hand side does not depend onx, and that the left-handside does not depend ont. Since they are equal, both sides have to be equalto a constant which, with some foresight, we choose to call 2, as it will bea negative number in the case of interest. The equation therefore breaks upinto two ordinary differential equations:u (x) = 2u(x) andv (t) = 2c2v(t).187 The boundary conditions say thatu(0) =u(L) = us consider the first equation. It has three types of solutions depend-ing on whether is nonzero real, nonzero imaginary or zero. (Notice that 2has to be real, so that these are the only possibilities.) If = 0, thenthe solution isu(x) =a+b x. The boundary conditionu(0) = 0 means thata= 0, but the boundary conditionu(L) = 0 then means thatb= 0, whenceu(x) = 0 for allx. Clearly this is a very uninteresting solution.

7 Let usconsider imaginary. Then the solution is nowaexp(| |x) +bexp( | |x).Again the boundary conditions forcea=b= 0. Therefore we are left withthe possibility of real. Then the solution isu(x) =acos x+bsin x .The boundary conditionu(0) = 0 forcesa= 0. Finally the boundary condi-tionu(L) = 0 implies thatsin L= 0 = =n Lfornan 0 is an uninteresting solution, and because of the fact that thesine is an odd function, negative values ofngive rise to the same solution(up to a sign) as positive values ofn. In other words, all nontrivial distinctsolution are given (up to a constant multiple) byun(x) sin nx ,with n=n Land wheren= 1,2,3, . ( )Let us now solve forv(t). Its equation isv (t) = 2c2v(t),whencev(t) =acos ct+bsin ct .The first of the two initial conditions ( ) says thatv (0) = 0 whenceb= for any positive integern, the function n(x, t) = sin nxcos nct ,with n=n L,satisfies the wave equation ( ) subject to the boundary conditions ( )and to the first of the initial conditions ( ).

8 Now notice something important: the wave equation ( ) islinear; thatis, if (x, t) and (x, t) are solutions of the wave equation, so is any linearcombination (x, t) + (x, t) where and are then, any linear combination of the n(x, t) will also be a so-lution. In other words, the most general solution subject to the boundaryconditions ( ) and the first of the initial conditions in ( ) is given by alinear combination (x, t) = n=1bnsin nxcos nct .Of course, this expression is formal as it stands: it is an infinite sum whichdoes not necessarily make sense, unless we chose the coefficients{bn}in sucha way that the series converges, and that the convergence is such that we candifferentiate the series termwise at least can now finally impose the second of the initial conditions ( ): (x,0) = n=1bnsin nx=f(x).( )At first sight this seems hopeless: can any functionf(x) be represented asa series of this form?

9 The Bernoullis, who were the first to get this far,thought that this was not the case and that in some sense the solution wasonly valid for special kinds of functions for which such a series expansion ispossible. It took Euler to realise that, in a certain sense, all functionsf(x)withf(0) =f(L) = 0, can be expanded in this way. He did this by showinghow the coefficients{bn}are determined by the functionf(x).To do so let us argue as follows. Letnandmbe positive integers andconsider the functionsun(x) andum(x) defined in ( ). These functionssatisfy the differential equations:u n(x) = 2nun(x) andu m(x) = 2mum(x).Let us multiply the first equation byum(x) and the second equation byun(x)and subtract one from the other to obtainu n(x)um(x) un(x)u m(x) = ( 2m 2n)un(x)um(x).We notice that the left-hand side of the equation is a total derivativeu n(x)um(x) un(x)u m(x) = (u n(x)um(x) un(x)u m(x)) ,whence integrating both sides of the equation fromx= 0 tox=L, we obtain( 2m 2n) L0um(x)un(x)dx= (u n(x)um(x) un(x)u m(x)) L0= 0,189sinceun(0) =un(L) = 0 and the same forum.

10 Therefore we see that unless 2n= 2m, which is equivalent ton=m(sincen,mare positive integers), theintegral L0um(x)un(x)dxvanishes. On the other hand, ifm=n, we havethat L0un(x)2dx= L0(sinn xL)2dx= L0(12 12cos2n xL)dx= , in summary, we have theorthogonality propertyof the functionsum(x): L0um(x)un(x)dx={L2,ifn=m, and0,otherwise.( )Let us now go back to the solution of the remaining initial condition ( ).This condition can be rewritten asf(x) = n=1bnun(x) = n=1bnsinn xL.( )Let us multiply both sides byum(x) and integrate fromx= 0 tox=L: L0f(x)um(x)dx= n=1bn L0un(x)um(x)dx ,where we have interchanged the order of integration and summation the orthogonality relation ( ) we see that of all the termsin the right-hand side, only the term withn=mcontributes to the sum,whence L0f(x)um(x)dx=bmL2,or in other words,bm=2L L0f(x)um(x)dx ,( )a formula due to Euler. Finally, the solution of the wave equation ( ) withboundary conditions ( ) and initial conditions ( ) is (x, t) = n=1bnsinn xLcosn ctL,( )1 This would have to be justified, but in this part of the course we will be much morecavalier about these things.}