[Ablowitz, Fokas] Complex Variables - UCM FacultyWeb

1 applications of Transforms to Differential Equations28519. Establish the following result by formally inverting the Laplace transform F(s)=1ssinh(sy)sinh(s ) >0f(x)=y + n=12( 1)nn sin(n y )cos(n x )See the remark at the end of Problem 18, which explains how to showhow the inverse Laplace transform can be proven to be valid in a situationsuch as this where there are an infinite number of poles. applications of Transforms to Differential EquationsA particularly valuable technique available to solve differential equations ininfinite and semiinfinite domains is the use of Fourier and Laplace this section we describe some typical examples. The discussion is not in-tended to be complete. The aim of this section is to elucidate the transformtechnique, not to detail theoretical aspects regarding differential equations. Thereader only needs basic training in the calculus of several Variables to be ableto follow the analysis.

2 We shall use various classical partial differential equa-tions (PDEs) as vehicles to illustrate methodology. Herein we will considerwell-posed problems that will yield unique solutions. More general PDEs andthe notion of well-posedness are investigated in considerable detail in courseson state heat flow in a semiinfinite domain obeys Laplace sequation. Solve for the bounded solution of Laplace s equation 2 (x,y) x2+ 2 (x,y) y2=0( )in the region <x< ,y>0, where ony=0 we are given (x,0)=h(x)withh(x) L1 L2( , |h(x)|dx< and |h(x)|2dx< ).This example will allow us to solve Laplace s Equation ( ) by Fouriertransforms. Denoting the Fourier transform inxof (x,y)as (k,y): (k,y)= e ikx (x,y)dxtaking the Fourier transform of Eq. ( ), and using the result from Section the Fourier transform of derivatives, Eqs. ( ,b) (assuming the validity2864 Residue Calculus and applications of Contour Integrationof interchangingy-derivatives and integrating overk, which can be verifieda posteriori), we have 2 y2 k2 =0( )Hence (k,y)=A(k)eky+B(k)e kywhereA(k)andB(k)arearbitraryfunctionso fk, tobespecifiedbytheboundaryconditions.

3 We require that (k,y)be bounded for ally>0. In order that (k,y)yield a bounded function (x,y), we need (k,y)=C(k)e |k|y( )Denoting the Fourier transform of (x,0)=h(x)by H(k)fixesC(k)= H(k),so that (k,y)= H(k)e |k|y( )From Eq. ( ) by direct integration (contour integration is not necessary) wefind that F(k,y)=e |k|yis the Fourier transform off(x,y)=1 yx2+y2, thusfrom the convolution formula Eq. ( ) the solution to Eq. ( ) is given by (x,y)=1 yh(x )(x x )2+y2dx ( )Ifh(x)were taken to be a Dirac delta function concentrated atx= ,h(x)=hs(x )= (x ), then H(k)=e ik , and from Eq. ( ) directly (orEq. ( )) a special solution to Eq. ( ), s(x,y)is s(x,y)=G(x ,y)=1 (y(x )2+y2)( )FunctionG(x ,y)is called aGreen s function; it is a fundamental solutionto Laplace s equation in this region. Green s functions have the property ofsolving a given equation with delta function inhomogeneity.

4 From the boundaryvalueshs(x ,0)= (x )we may construct arbitrary initial values (x,0)= h( ) (x )d =h(x)( ) applications of Transforms to Differential Equations287and, because Laplace s equation is linear, we find by superposition that thegeneral solution satisfies (x,y)= h( )G(x ,y)d ( )which is Eq. ( ), noting that orx are dummy integration Variables . Inmany applications it is sufficient to obtain the Green s function of the underlyingdifferential formula ( ) is sometimes referred to as thePoisson formula for ahalf plane. Although we derived it via transform methods, it is worth notingthat a pair of such formulae can be derived from Cauchy s integral formula. Wedescribe this alternative method now. Letf(z)be analytic on the real axis andin the upper half plane and assumef( ) 0 uniformly as . Using alarge closed semicircular contour such as that depicted in Figure we havef(z)=12 i Cf( ) zd 0=12 i Cf( ) zd where Imz>0 (in the second formula there is no singularity because thecontour closes in the upper half plane and = zin the lower half plane).

5 Adding and subtracting yieldsf(z)=12 i Cf( )(1 z 1 z)d The semicircular portion of the contourCRvanishes asR , implyingthe following on Im =0 for the plus and minus parts of the above integral,respectively: callingz=x+iyand =x +iy ,f(x,y)=1 i f(x ,y =0)(x x(x x )2+y2)dx f(x,y)=1 f(x ,y =0)(y(x x )2+y2)dx Callingf(z)=f(x,y)=u(x,y)+iv(x,y),Ref(x, y=0)=u(x,0)=h(x)2884 Residue Calculus and applications of Contour Integrationand taking the imaginary part of the first and the real part of the second of theabove formulae, yields the conjugate Poisson formulae for a half plane:v(x,y)= 1 h(x )(x x(x x )2+y2)dx u(x,y)=1 h(x )(y(x x )2+y2)dx Identifyingu(x,y)as (x,y), we see that the harmonic functionu(x,y)(be-causef(z)is analytic its real and imaginary parts satisfy Laplace s equation) isgiven by the same formula as Eq. ( ). Moreover, we note that the imaginarypart off(z),v(x,y), is determined by the real part off(z)on the see that we cannot arbitrarily prescribe both the real and imaginary partsoff(z)on the boundary.

6 These formulae are valid for a half plane. Similarformulae can be obtained by this method for a circle (see also Example 10,Section ).Laplace s equation, ( ), is typical of a steady state situation, for example,as mentioned earlier, steady state heat flow in a uniform metal plate. If we havetime-dependent heat flow, the diffusion equation t=k 2 ( )is a relevant equation withkthe diffusion coefficient. In Eq. ( ), 2is theLaplacian operator, which in two dimensions is given by 2= 2 x2+ 2 dimension, takingk=1 for convenience, we have the following initialvalue problem: (x,t) t= 2 (x,t) x2( )The Green s function for the problem on the line <x< is obtainedby solving Eq. ( ) subject to (x,0)= (x ) for the Green s function of Eq. ( ). Define (k,t)= e ikx (x,t) applications of Transforms to Differential Equations289whereupon the Fourier transform of Eq. ( ) satisfies (k,t) t= k2 (k,t)( )hence (k,t)= (k,0)e k2t=e ik k2t( )where (k,0)=e ik is the Fourier transform of (x,0)= (x ).

7 Thus,by the inverse Fourier transform, and callingG(x ,t)the inverse transformof ( ),G(x ,t)=12 eik(x ) k2tdk=e (x )2/4t 12 e (k ix 2t)2tdk=e (x )24t2 t( )where we use e u2du= . Arbitrary initial values are included byagain observing that (x,0)=h(x)= h( ) (x )d which implies (x,t)= G(x ,t)h( )d =12 t h( )e (x )24td ( )The above solution to Eq. ( ) could also be obtained by using Laplacetransforms. It is instructive to show how the method proceeds in this case. Webegin by introducing the Laplace transform of (x,t)with respect tot: (x,s)= 0e st (x,t)dt( )Taking the Laplace transform intof Eq. ( ), with (x,0)= (x ),yields 2 x2(x,s) s (x,s)= (x )( )2904 Residue Calculus and applications of Contour IntegrationHence the Laplace transform of the Green s function to Eq. ( ) satisfiesEq. ( ). We remark that generally speaking, any functionG(x )satis-fyingLG(x )= (x )whereLis a linear differential operator, is referred to as a Green s function.

8 Thegeneral solution corresponding to (x,0)=h(x)is obtained by superposition: (x,t)= G(x )h( )d . Equation ( ) is solved by first findingthe bounded homogeneous solutions on <x< , for(x )>0 and(x )<0: +(x ,s)=A(s)e s1/2(x )forx >0 (x ,s)=B(s)es1/2(x )forx <0( )where we takes1/2to have a branch cut on the negative real axis; that is,s=rei , < . This will allow us to readily invert the Laplace trans-form (Res>0).The coefficientsA(s)andB(s)in Eq. ( ) are found by (a) requiringcontinuity of (x ,s)atx= and by (b) integrating Eq. ( ) fromx= ,tox= + , and taking the limit as 0+. This yields a jumpcondition on ( x(x ,s):[ x(x ,s)]x =0+x =0 = 1( )Continuity yieldsA(s)=B(s), and Eq. ( ) gives s1/2A(s) s1/2B(s)= 1( )henceA(s)=B(s)=12s1/2( )Using Eq. ( ), (x ,s)is written in the compact form: (x ,s)=e s1/2|x |2s1/2( )The solution (x,t)is found from the inverse Laplace transform: (x,t)=12 i c+i c i e s1/2|x |est2s1/2ds( ) applications of Transforms to Differential Equations291forc>0.)

9 To evaluate Eq. ( ), we employ the same keyhole contour asin Example in Section (see Figure ). There are no singularitiesenclosed, and the contoursCRandC at infinity and at the origin vanish in thelimitR , 0, respectively. We only obtain contributions along thetop and bottom of the branch cut to find (x,t)= 12 i 0 e ir1/2|x |e rt2r1/2ei /2ei dr+ 12 i 0eir1/2|x |e rt2r1/2e i /2e i dr( )In the second integral we putr1/2=u; in the first we putr1/2=uand then takeu u, whereupon we find the same answer as before (see Eq. ( )): (x,t)=12 e u2t+iu|x |du=12 e (u i|x |2t)2te (x )24tdu=e (x )2/4t2 t( )The Laplace transform method can also be applied to problems in which thespatial variable is on the semiinfinite domain. However, rather than use Laplacetransforms, for variety and illustration, we show below how the sine transformcan be used on Eq. ( ) with the following boundary conditions: (x,0)=0, (x=0,t)=h(t),limx x(x,t)=0,( )Define, following Section (x,t)=2 0 s(k,t)sinkx dk( ) s(k,t)= 0 (x,t)sinkx dx( )2924 Residue Calculus and applications of Contour IntegrationWe now operate on Eq.

10 ( ) with the integral 0dxsinkx, and via integra-tion by parts, find 0 2 x2sinkx dx=[ x(x,t)sinkx] x=0 k 0 xcoskx dx=k (0,t) k2 s(k,t)( )whereupon the transformed version of Eq. ( ) is s t(k,t)+k2 s(k,t)=kh(t)( )The solution of Eq. ( ) with (x,0)=0 is given by s(k,t)= t0h(t )ke k2(t t )dt ( )If (x,0)were nonzero, then Eq. ( ) would have another term. For sim-plicity we only consider the case (x,0)=0. Therefore (x,t)=2 0dksinkx{ t0h(t )e k2(t t )kdt }( )By integration we can show that (use sinkx=(eikx e ikx)/2 and integrateby parts to obtain integrals such as those in ( ))J(x,t t )= 0ke k2(t t )sinkx dk= xe x2/4(t t )4(t t )3/2( )hence by interchanging integrals in Eq. ( ), we have (x,t)=2 t0h(t )J(x,t t )dt Whenh(t)=1, if we call =x2(t t )1/2, thend =x4(t t )3/2dt , and we have (x,t)=2 x2 te 2d erfc(x2 t)( ) applications of Transforms to Differential Equations293We note that erfc(x)is a well-known function, called thecomplementaryerror function: erfc(x) 1 erf(x), where erf(x) 2 x0e should be mentioned that the Fourier sine transform applies to problemssuch as Eq.