Wave Equation Dirichlet Boundary Conditions

Transcription

1 Wave Equation Dirichet Boundary Conditions u tt x, t = c u xx x, t, < x <, t > 1 u, t =, u, t = ux, = fx u t x, = gx Look for simpe soutions in the form ux, t = XxT t Substituting into 13 and dividing both sides by XxT t gives T t kt t = X x Xx Since the eft side is independent of x and the right side is independent of t, it foows that the expression must be a constant: T t kt t = X x Xx = λ Here T means the derivative of T with respect to t and X means means the derivative of X with respect to x We seek to find a possibe constants λ and the corresponding nonzero functions X and T We obtain Furthermore, the boundary conditions give X λx =, T kλt = XT t =, XT t = for a t Since T t is not identicay zero we obtain the desired eigenvaue probem X x λxx =, X =, X = We have soved this probem many times and we have λ = µ so that Xx = a cosµx + b µx Appying the boundary conditions we have = X = a a = = X = b µ From this we concude µ = which impies µ = nπ 1

2 and therefore nπ λ n = µ n =, Xn x = µ n x, n = 1,, 3 The soution of T c λ n T = is then T t = a cos + b where a and b are arbitrary constants Next we ook for u as an infinite sum ux, t = a n cos + b n 4 5 The ony probem remaining is to somehow pick the constants a n and b n so that the initia conditions ux, = fx and u t x, = gx are satisfied Setting t = in 5, we seek to obtain a n } satisfying fx = ux, = a n This nothing more than a Sine expansion of the function fx on the interva, a n = fx dx 6 Next we differentiate formay 5 with respect to t to obtain nπc u t x, t = a n + b n cos Again setting t = in 5, we seek to obtain b n } satisfying gx = u t x, = nπc b n This is amost a Sine expansion of the function gx on the interva, Namey we obtain After simpifying we have nπc 1 b n = b n = nπc gx dx gx dx 7

3 At this point we note an important difference between the heat and wave equation soutions For the heat equation the soutions were of the form c n e λnt ϕ n x and, at east for t >, there is no question about the convergence of this series due to exponentia decay of the terms e λnt But for the wave equation the series does not incude such terms Indeed, the individua terms ook ike a n cos + b n and these do not decay rapidy Therefore to justify that we have a soution to 13 we must take another approach First notice that cos 8 = 1 [ ] nπx + ct nπx ct + and = 1 [ ] nπx ct nπx + ct cos cos = nπ x+ct x ct 9 nπξ dξ 1 Ug 8, 9, 6 and 7 we can rewrite 5 as ] nπx + ct nπx ct ux, t = a n [ x+ct nπc nπξ b n dξ c x ct 11 This expression can be reduced further as foows: Let F x and G x denote the odd -periodic extensions of f and g, ie, et F x = fx < x < f x < x <, G x = gx < x < g x < x < Then we extend F and G to be periodic functions on R, which we denote by F x and G x, respectivey Then the Fourier Sine series of F x is where f n = 1 F x = f n F x dx = 3 fx dx = a n

4 Simiary, Then the Fourier Sine series of G x is G x = g n where g n = 1 G x dx = In other words the series 11 can be written as nπc gx dx = b n ux, t = 1 [ F x + ct + F x ct ] + 1 x+ct G ξ dξ 1 c Now we notice that, for exampe, if fx is C and gx is C 1 on,, then 1 gives the soution to 13 Let us use 13, assuming f and g are sufficienty smooth, to check the conditions First we note that 1 formay satisfies the wave equation from our work on the D Aembert form of the soution Next we note that ux, = 1 [ F x + F x ] = F x = fx, < x < Exampe 1 u t x, = 1 x ct [ G x + G x ] = G x = gx, < x < u tt x, t = c u xx x, t, < x <, t > 13 u, t =, u, t = x, x / ux, = fx = x, / x u t x, = gx = For this exampe 5 becomes ux, t = a n cos In this case we have a n = / x dx + / ] x dx n =, 4, = 4 / x dx = 4 nπ n π n = 1, 3, 4

5 Here we have used the foowing fact Ug the change of variabes s = x we can write nπ s x dx = s ds So / = / / s nπ nπs ds / nπs = 1 n+1 s ds / = 1 n+1 x dx / ] x dx + x dx / = / ] / x dx + 1 n+1 x dx n =, 4, = 4 / x dx n = 1, 3, Finay, keeping in mind that n is odd, we empoy integration by parts to evauate the integra / / x dx = x nπ cos dx = x / nπ cos / nπ cos dx = nπ / cos dx / = nπ = nπ nπ So setting n = k 1 for k = 1,,, we arrive at We arrive at the soution ux, t = 4 π k=1 1 k k 1 cos k 1πct 5 k 1πx 14