1 1D heat and wave equations on a finite interval


 Ada Harmon
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1 1 1D het nd wve equtions on finite intervl In this section we consider generl method of seprtion of vribles nd its pplictions to solving het eqution nd wve eqution on finite intervl ( 1, 2. Since by trnsltion we cn lwys shift the problem to the intervl (, we will be studying the problem on this intervl. 1.1 Boundry conditions The most common types of boundry conditions re Dirichlet: u(, t = h(t, Neumnn: u x (, t = h(t, u(, t = g(t. u x (, t = g(t. Mixed: u x (, t = h(t, u(, t = g(t or u(, t = h(t, u x (, t = g(t. Periodic: It is more convenient to consider the problem with periodic boundry conditions on the symmetric intervl (,. Therefore boundry conditions in this cse re u(, t = u(, t, u x (, t = u x (, t. There is generliztion of mixed boundry condition sometimes clled obin boundry condition αu(, t + u x (, t = h(t, βu(, t + u x (, t = g(t. We will not be considering it here but the methods used below work for it s well. 1.2 Het eqution Our gol is to solve the following problem nd u stisfies one of the bove boundry conditions. u t = Du xx + f (x, t, x (,, (1 u(x, = φ(x, (2 In order to chieve this gol we first consider problem when f (x, t =, h(t =, g(t = nd use the method of seprtion of vribles to obtin solution. To illustrte the method we solve the het eqution with Dirichlet nd Neumnn boundry conditions. Mixed nd Periodic boundry conditions re treted in the similr wy nd we will use them in the section for wve eqution. 1
2 1.2.1 Homogeneous het eqution with Dirichlet boundry conditions Here we re solving the following problem u t = Du xx, for x (,, t > (3 u(x, = φ(x, (4 u(, t =, u(, t =, for t >. (5 The ide behind the method of seprtion of vribles is to look for solution in the following form u(x, t = X(xT(t, where X(x nd T(t will be determined. Let s ssume tht we hve solution u(x, t in the bove form. Plugging it into the eqution we obtin X(xT (t = DX (xt(t, Dividing both prts by DT(tX(x we rrive to the following equlity T (t DT(t = X (x X(x. Since the left side is function of t only nd the right side is the function of x only the bove equlity is possible if nd only if T (t DT(t = X (x X(x = k for some constnt k. Therefore if u(x, t = X(xT(t is solution of the het eqution then X (x = kx(x, nd T (t = kdt(t. Moreover, from the boundry conditions (9 we know tht X(T(t = nd X(T(t = for ll t >. This implies X( = nd X( =. We cn esily solve the problem for T(t to obtin T(t = Ce kdt, where we don t know constnts C nd k. The problem for X(x is clled n eigenvlue problem X (x = kx(x on (,, X( =, X( =. (6 In order to solve it we hve to find ll ks nd ll nontrivil solutions X(x, corresponding to these ks. It is not difficult to see tht k <. Indeed, if we multiply both prts of the eqution (34 by X(x nd integrte by prts we obtin X (x 2 + k X(x 2 dx =. Obviously, if k then X(x = on (,. Since we re interested in nontrivil solutions we hve to consider only k = λ 2 <. In tht cse the generl solution of (34 is X(x = A cos(λx + B sin(λx 2
3 nd constnts A, B, λ will be found from the boundry conditions X( = X( =. Let s consider X( = : obviously X( = A nd therefore A =. Tking this into ccount nd using X( = we obtin B sin(λ =. It s cler tht B = s then the solution is zero everywhere nd we re interested only in nonzer solutions. Therefore λ = πn for ll n Z \ {}. The solution of the eigenvlue problem is k = π2 n 2 ( 2, n N + πn = {1, 2, 3,...}, X(x = B sin x. Combining our informtion bout k, X(x nd T(t we rrive to the conclusion tht we hve infinitely mny solutions of the het eqution (7, stisfying the boundry conditions (9. Indeed, for ll n N + function u n (x, t = e π2 n 2 ( 2 Dt πn sin x solves (7 nd (9. Therefore, ny liner combintion of u n s lso solves (7 nd (9 nd we represent u(x, t = n e π2 n 2 ( 2 Dt πn sin x, where n re rbitrry constnts. We will find n s using initil dt. We cn represent φ(x in terms of sine Fourier series where φ(x = b n = 2 Since we know tht u(x, = φ(x we obtin Therefore n b n sin x, n φ(x sin x dx. n n n sin x = φ(x = b n sin x. n ( n b n sin x =. Multiplying both prts of the bove equlity by sin m x, m N + nd integrting over (, we hve m = b m for ll m N +. Therefore we obtin tht n = 2 n φ(x sin x dx. Finlly, the solution to the problem (7, (36, (9 is where u(x, t = n = 2 n e π2 n 2 ( 2 Dt πn sin x, n φ(x sin x dx. 3
4 1.2.2 Homogeneous het eqution with Neumnn boundry conditions Here we re solving the following problem u t = Du xx, for x (,, t > (7 u(x, = φ(x, (8 u x (, t =, u x (, t =, for t >. (9 We use the sme method os seprtion of vribles nd obtin the following eigenvlue problem for X(x X (x = kx(x, on (,, X ( =, X ( =. (1 In this cse we cn show by the sme methods s before tht k = λ 2 nd therefore the generl solution is X(x = A cos(λx + B sin(λx. Using X ( = we obtin B = nd using X ( = we obtin A sin(λ =. Therefore λ = πn for ll n Z nd the solution of eigenvlue problem (1 is k = π2 n 2 n 2, n N = {, 1, 2, 3,...}, X(x = A cos x. Note tht here we llow for n = in which cse X(x = A is solution. As before we obtin the solution u(x, t = n e π2 n 2 ( 2 Dt πn cos x, n= where we need to find n s from initil dt. We cn represent φ(x in terms of cosine Fourier series where b n = 2 φ(x = n b n cos n= x, n φ(x cos x b = 1 φ(x dx. dx for n >, As before using initil dt we know tht n = b n nd therefore the solution is u(x, t = n= n e π2 n 2 ( 2 Dt πn cos x, where n = 2 n φ(x cos x = 1 φ(x dx. dx for n >, 4
5 1.2.3 Het eqution with f (x, t = nd zero Dirichlet boundry conditions Now let us explin how we cn solve the following problem u t = Du xx + f (x, t, for x (,, t > (11 u(x, = φ(x, (12 u(, t =, u(, t =, for t >. (13 We know tht Fourier series with bsis functions sin n x, n N + is consistent with the boundry conditions. We lso know tht for fixed t ny function u(x, t nd f (x, t cn be expnded in Fourier series s follows n f (x, t = f n (t sin x, where f n (t = 2 n f (x, t sin x dx for n N +, nd n u(x, t = u n (t sin x, where u n (t will be determined lter using eqution. Plugging these representtions for u(x, t nd f (x, t in the (11 we obtin (u n(t + D π2 n 2 n u n (t f n (t sin x =. 2 Multiplying both prts of equlity by sin m x nd integrting over (, we hve Using initil dt we clerly hve u m(t + D π2 m 2 u m (t = f m (t. u(x, = 2 n u n ( sin x = φ(x, nd therefore u m ( = 2 m φ(x sin x dx for m N +. Now for ech m N + we hve the following ODE u m(t + D π2 m 2 2 u m (t = f m (t, u m ( = 2 m φ(x sin x dx. It is cler tht we cn solve it to obtin u m (t = u m (e D π2 m 2 2 t + e D π2 m 2 2 t t e D π2 m 2 2 s fm (s ds. Since we know u m ( nd f m (t we obtin u m (t nd therefore we hve solution of the problem (11(13 s n u(x, t = u n (t sin x. We cn use the sme method to tret ll other boundry conditions. I omit the detils here. 5
6 1.2.4 Inhomogeneous boundry conditions In this subsection we wnt to understnd how to solve the following problem We cn construct n uxiliry function u t = Du xx, for x (,, t > (14 u(x, =, (15 u(, t = h(t, u(, t = g(t, for t >. (16 v(x, t = g(t h(t x + h(t. It is cler tht w(x, t = u(x, t v(x, t stisfies the following problem w t = Dw xx v t (x, t, for x (,, t > (17 w(x, = v(x,, (18 w(, t =, w(, t =, for t >. (19 Using results from the previous subsection we cn find w(x, t nd therefore we cn find tht u(x, t = w(x, t + v(x, t, where we know w nd v. Note tht the sme ide works for Neumnn nd Mixed boundry conditions. We just need to construct n pproprite uxiliry function v(x, t stisfying boundry conditions Generl problem Now we re redy to solve the following generl problem u t = Du xx + f (x, t, for x (,, t > (2 It is strightforwrd to check tht if v(x, t stisfies nd w(x, t stisfies u(x, = φ(x, (21 u(, t = h(t, u(, t = g(t, for t >. (22 v t = Dv xx + f (x, t, for x (,, t > (23 v(x, = φ(x, (24 v(, t =, v(, t =, for t >, (25 w t = Dw xx, for x (,, t > (26 w(x, =, (27 w(, t = h(t, w(, t = g(t, for t >, (28 then u(x, t = v(x, t + w(x, t solves (2, (21, (22. Since we lredy know from the previous subsections how to solve the problems for v(x, t nd w(x, t we re done. The sme method works for ll boundry conditions, not just Dirichlet. 6
7 1.3 Wve eqution Our gol is to solve the following problem nd u stisfies one of the boundry conditions. u tt = c 2 u xx + f (x, t, x (,, (29 u(x, = φ(x, u t (x, = ψ(x (3 In order to chieve this gol we proceed s with het eqution, first consider problem when f (x, t =, h(t =, g(t = nd use the method of seprtion of vribles to obtin solution. To illustrte the method we solve the wve eqution with Mixed nd Periodic boundry conditions. Dirichlet nd Neumnn re treted in the similr wy nd hve been considered before for het eqution Homogeneous wve eqution with mixed boundry conditions Here we re solving the following problem We look for solution in the following form u tt = c 2 u xx, for x (,, t > (31 u(x, = φ(x, u t (x, = ψ(x (32 u x (, t =, u(, t =, for t >. (33 u(x, t = X(xT(t, where X(x nd T(t will be determined. Let s ssume tht we hve solution u(x, t in the bove form. Plugging it into the eqution we obtin X(xT (t = c 2 X (xt(t, Dividing both prts by c 2 T(tX(x we rrive to the following equlity T (t c 2 T(t = X (x X(x. Since the left side is function of t only nd the right side is the function of x only the bove equlity is possible if nd only if T (t c 2 T(t = X (x X(x = k for some constnt k. Therefore if u(x, t = X(xT(t is solution of the wve eqution then X (x = kx(x, nd T (t = kc 2 T(t. Moreover, from the boundry conditions (9 we know tht X (T(t = nd X(T(t = for ll t >. This implies X ( = nd X( =. 7
8 The problem for X(x is clled n eigenvlue problem X (x = kx(x on (,, X ( =, X( =. (34 In order to solve it we hve to find ll ks nd ll nontrivil solutions X(x, corresponding to these ks. It is not difficult to see tht k <. Indeed, if we multiply both prts of the eqution (34 by X(x nd integrte by prts we obtin X (x 2 + k X(x 2 dx =. Obviously, if k then X(x = on (,. Since we re interested in nontrivil solutions we hve to consider only k = λ 2 <. In tht cse the generl solution of (34 is X(x = A cos(λx + B sin(λx nd constnts A, B, λ will be found from the boundry conditions X ( = X( =. Let s consider X ( = : obviously X ( = Bλ nd therefore B =. Tking this into ccount nd using X( = we obtin A cos(λ =. It s cler tht A = s then the solution is zero everywhere nd we re interested only in nonzero solutions. Therefore λ = πn for ll n Z. The solution of the eigenvlue problem is k = 1 ( πn, n N = {, 1, 2, 3,...}, X(x = A cos 2 + πn x. Solving eqution for T(t we obtin ( c ( c T(t = C cos 2 + πn t + D sin 2 + πn t. Combining our informtion bout X(x nd T(t we rrive to the conclusion tht we hve infinitely mny solutions of the wve eqution (35, stisfying the boundry conditions (33. Indeed, for ll n N function u n (x, t = ( ( c ( c n cos 2 + πn t + b n sin 2 + πn t cos ( πn x solves (35 nd (33. Therefore, ny liner combintion of u n s lso solves (35 nd (33 nd we represent u(x, t = ( ( c ( c ( 1 n cos n= 2 + πn t + b n sin 2 + πn t cos 2 + πn x, where n, b n re rbitrry constnts. We will find n, b n s before using initil dt Homogeneous wve eqution with periodic boundry conditions Here we re solving the following problem u tt = c 2 u xx, for x (,, t > (35 8
9 u(x, = φ(x, u t (x, = ψ(x (36 u(, t = u(, t, u x (, t = u x (, t, for t >. (37 We look for solution in the following form u(x, t = X(xT(t, where X(x nd T(t will be determined. Let s ssume tht we hve solution u(x, t in the bove form. Plugging it into the eqution we obtin X(xT (t = c 2 X (xt(t, Dividing both prts by c 2 T(tX(x we rrive to the following equlity T (t c 2 T(t = X (x X(x. Since the left side is function of t only nd the right side is the function of x only the bove equlity is possible if nd only if T (t c 2 T(t = X (x X(x = k for some constnt k. Therefore if u(x, t = X(xT(t is solution of the wve eqution then X (x = kx(x, nd T (t = kc 2 T(t. Moreover, from the boundry conditions (9 we know tht X( T(t = X(T(t nd X ( T(t = X (T(r for ll t >. This implies X( = X( nd X ( = X (. Solving eqution for X(x we obtin X(x = A cos( kx + B sin( kx. Using boundry conditions we hve A cos( k B sin( (k = A cos( k + B sin( k; k(a sin( k + B cos( k = k( A sin( k + B cos( k. Solving these equtions we obtin Therefore k = πn sin( k =. for ll n N = {, 1, 2,...} nd X n (x = A n cos Solving eqution for T(t we obtin T n (t = C n cos n x + B n sin Now we cn find the generl solution u(x, t to be u(x, t = + n x. nc nc t + D n sin t. ( nc nc ( n n C n cos t + D n sin t A n cos x + B n sin x. Using initil dt we cn recover, A n, B n, C n, D n s before. 9
10 1.3.3 Generl problem In order to solve the generl problem (29, (3 with inhomogeneous boundry conditions we follow exctly the sme ides s in sections nd The detils re omitted. 2 1D het nd wve equtions on In this section we consider two generl methods of solving het nd wve equtions on, nmely, method of fundmentl solutions (it is lso sometimes clled method of Green s functions. We lso cn construct solution on the semiinfinite domin + bsed on the knowledge of solution on. 2.1 Boundry conditions When we re on the whole we cnnot prescribe boundry conditions. Usully we ssume some informtion bout behviour of the solution t infinity, for instnce, tht u x (x, t or/nd u(x, t vnish s x ±. We will spell out these ssumptions for prticulr problems. For semiinfinite domin + we prescribe Dirichlet or Neumnn boundry conditions only t one end x = nd tret infinity s described bove. 2.2 Het eqution on. Method of fundmentl solution. Our gol is to solve the following problem u t = Du xx + f (x, t, x, t >, (38 u(x, = φ(x, x. (39 We lso wnt to mke resonble ssumptions bout behviour of f (x, t nd φ(x t infinity: φ(x dx <, f (x, t dx < for ll t >. Firstly, we will try to solve homogeneous eqution without ny initil dt u t = Du xx, x, t > using similrity solutions. It is cler tht rescling of by constnt will lwys give the sme domin. Therefore if we chnge spce vrible z = x nd chnge time vrible s = 2 t (it s clled diltion, we will obtin exctly the sme domins for x nd t. Moreover we see tht if u(x, t is solution of the het eqution then v(z, s = Au(x, 2 t 1
11 is lso solution of the bove het eqution for generl prmeters, A. We choose A = α, where we will determine α lter. Now we hve two different solutions v(z, s nd u(x, t nd we notice tht z 2 s = x2 t, v(z, s u(x, t s α/2 = t α/2. This suggests to us tht there re solutions such tht these quntities re invrint with respect to diltion nd therefore we cn define ξ = x t, f (ξ = u(x, t t α/2. Therefore we look for similrity solution in the form ( x u(x, t = f t α/2. (4 t Plugging it into the het eqution we obtin the following eqution for f α 2 f (ξ 1 2 ξ f (ξ D f (ξ =. (41 Now we hve choice of α tht will give use solution to the (41 nd hence solution to het eqution. We would like to look for solution tht is integrble nd decys t infinity together with its xderivtives, i.e. u(x, t dx <, u x (x, t for x ±. If we impose these ssumptions then integrting het eqution over we obtin Therefore d u(x, t dx = D u xx (x, t dx = D(u x (, t u x (, t =. dt u(x, t dx = const for ll t >. Without loss of generlity (by rescling u we cn choose the bove constnt to be 1. Plugging in our nztz (4 for u(x, t we obtin ( x t α/2 f dx = 1, t or t (α+1/2 Therefore we hve to tke α = 1. The eqution (41 becomes nd we cn solve it s follows f (ξ dξ = 1 for ll t >. 1 2 f (ξ 1 2 ξ f (ξ D f (ξ = 1 2 (ξ f (ξ + D f (ξ = ( D f (ξ ξ f (ξ =. 11
12 Using behvior t infinity we obtin D f (ξ ξ f (ξ = nd Now reclling tht f (ξ = Ae ξ2 4D. f (ξ dξ = 1 we hve A = 1 4πD nd u(x, t = The solution in (42 is clled the fundmentl solution of het eqution. 1 4πDt e x2 4Dt. (42 In wht follows below we will use fundmentl solution to construct solutions to het eqution. We will denote the fundmentl solution s Φ(x, t = nd note tht it hs the following properties 1 4πDt e x2 4Dt 1. Φ t = DΦ xx for ll x, t > ; 2. Φ(x, t = Φ( x, t >, for ll x, t > ; 3. Φ(x, t is smooth (C function of (x, t for x, t > ; 4. Φ(x, t dx = 1 for ll t >. 5. As t +, Φ(x, t for ll x = nd Φ(, t. Looking t the properties 4 nd 5 we note tht s t +, Φ(x, t δ(x, where δ(x is deltfunction. Note tht we don t specify the convergence here s δfunction is not function in usul sense. In fct the convergence will be in sense of mesures but we don t discuss it here Homogeneous het eqution In this section we solve the following problem u t = Du xx, x, t >, (43 u(x, = φ(x, x. (44 For simplicity, we ssume tht φ(x is uniformly continuous nd bounded function. The result is true for ny integrble function φ(x. We note tht if Φ(x, t is the fundmentl solution of het eqution then for ny fixed y the function Φ(x y, t lso solves het eqution (59. Moreover we cn tke convolution of Φ nd φ nd obtin tht v(x, t = Φ(x y, tφ(y dy 12
13 stisfies (59. eclling tht s t, Φ(x y, t δ(x y we expect tht v(x, t δ(x yφ(y dy s t. eclling properties of δfunction we hve δ(x yφ(y dy = φ(x nd therefore v(x, = φ(x. It seems tht if we tke u(x, t = Φ(x y, tφ(y dy, then it will be solution of the problem (59, (6. Let s verify it. It is cler tht u(x, t = Φ(x y, tφ(y dy stisfies (59 s u t Du x x = [Φ t (x y, t DΦ xx (x y, t] φ(y dy =. We just need to check tht Using property 4 of Φ it is cler tht u(x, t φ(x = u(x, t φ(x s t +. Φ(x y, t(φ(y φ(x dy. As φ(x is uniformly continuous function we know tht for ny ɛ > there exists δ > such tht φ(x φ(y < ɛ for ll x y < δ. We now split our integrl s follows Φ(x y, t(φ(y φ(x dy = Φ(x y, t(φ(y φ(x dy + It is cler tht y x <δ y x <δ Φ(x y, t(φ(y φ(x dy y x <δ y x δ ɛ Φ(x y, t dy ɛ y x <δ Φ(x y, t(φ(y φ(x dy. (45 Φ(x y, t φ(y φ(x dy Φ(x y, t dy = ɛ (46 For the second integrl we notice tht s φ(x is bounded, i.e. φ(x C for ll x then Φ(x y, t(φ(y φ(x dy 2C Φ(x y, t dy = 4C Φ(z, t dz. y x δ y x δ δ Using explicit expression for Φ we cn esily compute δ Φ(z, t dz = 1 4πDt δ e 4Dt z2 dz = 1 e s2 ds 1 π δ 4Dt 2 π e Combining everything we obtin tht for ny ɛ > there exists δ > such tht Φ(x y, t(φ(y φ(x dy ɛ + 2C e π 13 δ 4Dt. δ 4Dt.
14 Tking limit s t we hve lim t + Φ(x y, t(φ(y φ(x dy ɛ. Since ɛ > ws rbitrry, we tke ɛ + to obtin Φ(x y, tφ(y dy = φ(x. lim t + We just showed tht u(x, t = Φ(x y, tφ(y dy solves (59, ( Het eqution with f (x, t = nd zero initil dt In this section we solve the following problem u t = Du xx + f (x, t, x, t >, (47 u(x, =, x. (48 We will use Duhmel s principle to find the solution of the bove problem. Let us strt with the following uxiliry problem. Fix ny s > nd consider v t = Dv xx, x, t > s, (49 v(x, s = f (x, s, x. (5 Due to trnsltion invrince of the het eqution we cn esily check tht for ech fixed s v(x, t; s = Φ(x y, t s f (y, s dy solves the problem (64, (65 (note tht v(x, s; s = f (x, s. Now we define the following function w(x, t = t nd clim tht it solves (47, (48. Let s verify it. v(x, t; s ds It is cler tht w(x, = nd therefore (48 is stisfied. Now nd Therefore for ll t > We just showed tht solves (47, (48. w t (x, t = v(x, t; t + w t Dw xx = f (x, t + u(x, t = t w xx (x, t = t t v t (x, t; s ds = f (x, t + t v xx (x, t; s ds. t v t (x, t; s ds (v t (x, t; s Dv xx (x, t; s ds = f (x, t. Φ(x y, t s f (y, s dy ds 14
15 2.2.3 Generl problem Now we re redy to solve the following generl problem It is strightforwrd to check tht if v(x, t stisfies nd w(x, t stisfies u t = Du xx + f (x, t, for x, t > (51 u(x, = φ(x, for x. (52 v t = Dv xx, for x, t > (53 v(x, = φ(x, (54 w t = Dw xx + f (x, t, for x, t > (55 w(x, =, (56 then u(x, t = v(x, t + w(x, t solves (51, (52. Since we lredy know from the previous subsections how to solve the problems for v(x, t nd w(x, t we obtin u(x, t = Φ(x yφ(y dy + t Φ(x y, t s f (y, s dy ds. 2.3 Wve eqution on. Method of fundmentl solution. Our gol is to solve the following problem u tt = c 2 u xx + f (x, t, x, t > (57 u(x, = φ(x, u t (x, = ψ(x, x. (58 We strt by solving homogeneous problem with f (x, t Homogeneous wve eqution. D Alembert s solution. In this section we solve the following problem u tt = c 2 u xx, x, t >, (59 u(x, = φ(x, u t (x, = ψ(x, x. (6 Using the following chnge of vribles ξ = x ct, η = x + ct we obtin u t = cu ξ + cu η, u x = u ξ + u η, u tt = c 2 u ξξ 2c 2 u ξη + c 2 u ηη, u xx = u ξξ + 2u ξη + u ηη. Therefore u tt c 2 u xx = 2c 2 u ξη =. It follows tht u = f (ξ + g(η nd therefore returning to originl vribles u(x, t = f (x ct + g(x + ct. 15
16 Using initil dt we obtin f (x + g(x = φ(x, c f (x + cg (x = ψ(x, for ll x. After strightforwrd clcultion it follows tht for ll x nd some fixed f (x = 1 2 φ(x 1 2c Now we see tht x u(x, t = f (x ct + g(x + ct = 1 2 φ(x ct 1 2c Finlly we obtin D Alembert s solution ψ(s ds, g(x = 1 2 φ(x + 1 2c x ct u(x, t = 1 2 φ(x ct φ(x + ct + 1 2c x ψ(s ds. ψ(s ds φ(x + ct + 1 2c x+ct x ct x+ct ψ(s ds. ψ(s ds. ( Wve eqution with f (x, t = nd zero initil dt In this section we solve the following problem u tt = c 2 u xx + f (x, t, x, t >, (62 u(x, =, u t (x, = x. (63 We gin use Duhmel s principle to find the solution of the bove problem. Let us strt with the following uxiliry problem. Fix ny s > nd consider v tt = c 2 v xx, x, t > s, (64 v(x, s =, v t (x, s = f (x, s, x. (65 Due to trnsltion invrince of the wve eqution we cn esily check tht for ech fixed s v(x, t; s = 1 2c x+c(t s x c(t s f (r, s dr. solves the problem (64, (65 (note tht v(x, s; s =, v t (x, s; s = f (x, s. Now we define the following function w(x, t = t v(x, t; s ds nd clim tht it solves (62, (63. Let s verify it. It is cler tht w(x, = nd therefore (63 is stisfied. Now w t (x, t = v(x, t; t + t v t (x, t; s ds = t v t (x, t; s ds, nd w tt (x, t = v t (x, t; t + t w xx (x, t = v tt (x, t; s ds = f (x, t + t v xx (x, t; s ds. t v tt (x, t; s ds. 16
17 Therefore for ll t > w tt c 2 w xx = f (x, t + t (v tt (x, t; s c 2 v xx (x, t; s ds = f (x, t. We just showed tht solves (62, (63. u(x, t = 1 2c t x+c(t s x c(t s f (r, s dr ds Generl problem Now we re redy to solve the following generl problem u tt = c 2 u xx + f (x, t, for x, t > (66 u(x, = φ(x, u t (x, = ψ(x for x. (67 It is strightforwrd to check tht if v(x, t stisfies nd w(x, t stisfies v t = Dv xx, for x, t > (68 v(x, = φ(x, v t (x, = ψ(x for x. (69 w t = Dw xx + f (x, t, for x, t > (7 w(x, =, w t (x, = for x (71 then u(x, t = v(x, t + w(x, t solves (66, (67. Since we lredy know from the previous subsections how to solve the problems for v(x, t nd w(x, t we obtin u(x, t = 1 2 φ(x ct φ(x + ct + 1 2c x+ct x ct ψ(r dr + 1 2c t x+c(t s x c(t s f (r, s dr ds. 17
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