Transcription of 8 Some Additional Examples Laplace Transform
1 Additional ExamplesIn addition to the Fourier Transform and eigenfunction expansions, it is sometimesconvenient to have the use of the Laplace Transform for solving certain problems in partialdifferential equations. We will quickly develop a few properties of the Laplace Transform anduse them in solving some example TransformDefinition of the TransformStarting with a given function of t,f t ,we can define a new functionf s of the variable new function will have several properties which will turn out to be convenient forpurposes of solving linear constant coefficient ODE s and PDE s. The definition off s is asfollows:DefinitionLetf t be defined fort 0and let the Laplace Transform off t be defined by,L f t 0 e stf t dt f s For example:f t 1, t 0,L 1 0 e stdt e st s|t 0t 1s f s for s 0f t ebt, t 0,L ebt 0 e b s tdt e b s t s b |t 0t 1s b f s ,for s Laplace Transform is defined for all functions ofexponential type.
2 That is, any functionf t which is(a) piecewise continuous has at most finitely many finite jump discontinuities on anyinterval of finite length(b) has exponential growth: for some positive constants M and k|f t | Mektfor all t 0,.Properties of the Laplace TransformThe Laplace Transform has the followinggeneral C1f t C2g t C1f s C2 s f at 1af safora of the DerivativeL f t sf s f 0 L f t s2f s sf 0 f 0 of the TransformL tf t f s L t2f t 1 2f s etcSome Special TransformsThere are some Transform pairs that are useful in solving problems involving the heatequation. The derivations are given in an appendix.
3 F t k4 t3e k2/4t,t 0L f t e ks,k 0 f t 1 te k2/4t,t 0L f t 1se ks,k 0 f t erfck2t,t 0L f t 1se ks,k 0 Hereerfc z 1 erf z anderf z 2 0ze Properties of the TransformLetf t be a function of exponential type and suppose that for someb 0,h t 0if0 t bf t b ift bThenh t is just the functionf t , delayed by the amount b. ThenL h t 0 h t e stdt b f t b e stdtLetz t bso thatL h t 0 f z e s z b dz e bs 0 f z e szdz e bsf s .If we defineH t b 0if0 t b1if t bthenh t H t b f t b and we of a DelayL H t b f t b e bsf s ,for b related results is the followingL ebtf t 0 ebtf t e stdt 0 f t e s b tdt f s b.
4 , of a TransformL ebtf t f s b Results 5 and 6 assert that a delay in the function induces an exponential multiplier in thetransform and, conversely, a delay in the Transform is associated with an exponentialmultiplier for the final property of the Laplace Transform asserts of a ProductL f g t f s s where f g t : 0tf t g d The product, f g t , is called theconvolution productof f and g. Life would be simplerif the inverse Laplace Transform off s s was the pointwise productf t g t ,but it isn t, itis the convolution product. The convolution product has some of the same properties as thepointwise product, namely f g t g f t and h f g t h f g t.
5 We will not give the proof of the result 7 but will make use of it to PDE s1. Consider the IBVP tu x,t k xxu x,t x 0,t 0,u x,0 0,x 0,u 0,t f t ,f 0 0,t x,s L u x,t Laplace Transform in x,s 0 kU x,s ,x 0,U x,s d2dx2 x,s 0,s f s .Solving this ODE in x, we find x,s Ae xs/k Bexs/kx want x,s to remain bounded for all positive x, which requires thatB theboundary condition at x 0 leads to3 x,s f s e xs/k,x together with property 7 of the Laplace Transform , givesu x,t f K x, 0tx4 k t 3e x2/4k t f d .as the unique solution of the IBVP. Suppose now that we wish to compute the flux throughx 0,Flux at 0 k xu 0,t.
6 Differentiating the integral expression for u does not seem like a pleasant , note thatL k xu x,t k x x,s ksf s e xs/k,andFlux at 0 k xu 0,t L 1ksf s .In order to use our inversion formulas, we writeksf s kssf s and, recalling thatf 0 0,we havesf s L f t .In addition, with k 0, givesL 11s 1 t,and so,Flux at 0 L 1kssf s k 0tf t d .Note that iff t At,thenFlux at 0 k 0tA t d 2 Akt .Problem 1 Show that ifv x,t solves tv x,t k xxv x,t x 0,t 0,v x,0 0,x 0, k xv 0,t g t ,t 0,then4v 0,t 0tg k t d ,t 0,and ifg t B,thenv 0,t 2Bt Consider the IBVP ttu x,t a2 xxu x,t x 0,t 0,u x,0 tu x,0 0,x 0,u 0,t f t ,t x,s L u x,t Laplace Transform in x,s 0 kU x,s ,x 0, 0,s f s.
7 Solving this ODE in x, we find a general solution of the form, x,s Aexas Be xas,and both x,s f s exas x,s f s e xassolve the transformed equation and the boundary condition at x 0. To see how to choosethe correct solution, recall that forx 0,property 5 implies thatf s e xasis the Transform off t delayed by the amountxa 0. On the other hand,f s exasis the Transform off t advanced in time by the amountxa way to say this is to sayu x,t L 1f s e xas ft xaHt xa,represents the wave formf propagating from L to R into the region x 0 whileu x,t L 1f s exas ft xaHt xarepresents the wave formf propagating from R to L out of the region x 0.
8 Then thesolution that is relevant for our problem is the wave that travels from L to R into the region x 0 .3. Consider the IBVP tu x,t K xxu x,t 0 x 1,t 0,u x,0 0,0 x 1,u 0,t f t xu 1,t 0t , we may use the Laplace Transform , or if we prefer, we can use eigenfunctionexpansion after a suitable modification of the problem. We will solve the problem first bythis means. Since the boundary conditions are not homogeneous, the method does notapply directly but if we letv x,t u x,t f t then tv x,t tu x,t f t K xxu x,t f t K xxv x,t f t ,v x,0 u x,0 f 0 0,v 0,t u 0,t f t 0, xv 1,t xu 1,t the method of eigenfunction expansion applies directly to the problem forv x,t sincethe boundary conditions are homogeneous.
9 Sincev 0,t xv 1,t 0,it follows that theeigenfunctions are the eigenfunctions of example , namely n n 122 2 n2, n x sinn 12 x sin nx,n 1,2,..We writev x,t n 1 vn t n x andf t f t 1 f t n 1 Cn n x ,whereCn 1, n n, n 01sinn 12 xdx 01sin2n 12 xdx 4 2n 1 2 these expansions into the IBVP forv x,t , we conclude thatvn t K n2vn t f t Cn,vn 0 0, t Cn 0te K n2 t s f s example, iff t mt,thenvn t mCn 0te K n2 t s ds. mCnK n2e K n2t 1andu x,t f t n 1 mCnK n2e K n2t 1 n x , mt 2mK n 1 1 n3e K n2t 1 n x .Alternatively, we may use the Laplace Transform to solve this same problem. Let x,s denote the Laplace Transform ofu x,t.
10 Then,6s x,s 0 K x,s , 0,s f s , 1,s general solution of the equation may be written as x,s Aexp xs/K Bexpxs/K,and then the boundary conditions lead toA exps/Kexp s/K exps/Kf s ,B exp s/Kexp s/K exps/Kf s .Then x,s exp x 1 s/Kexp s/K exps/Kf s exp 1 x s/Kexp s/K exps/Kf s , exp xs/Kexp 2s/K 1f s exp 2 x s/Kexp 2s/K 1f s ,and since the formula for the sum of a geometric series implies that,1exp 2s/K 1 n 0 1 nexp 2ns/K,this becomes x,s f s n 0 1 nexp x 2n s/K f s n 0 1 nexp 2n 2 x follows from ( ) thatL 1exp xKs x4 Kt3exp x24Kt: G x,Kt ,t 0,Then, using property 7, we findu x,t n 0 1 n 0tG x 2n,K t f d n 0 1 n 0tG 2n 2 x,K t f d 0tG x,K t f d 0tG 2 x,K t f d 0tG x 2,K t f d Notice that this representation for the solution looks nothing at all like the eigenfunction7expansion obtained previously.
