General Solution of PDE using Fourier's method

• Oct 17th 2009, 09:39 PM
Maccaman
General Solution of PDE using Fourier's method
Hello,
Can someone please have a look at this problem and tell me if I have answered it correctly (I do get the solution so I'm more concerned about the process being correct).

Show that using the Fourier's method for $0 < x < L$ and $t > 0$ that the PDE

$u_t(x,t) = c^2u_{xx}(x,t)$

with boundary conditions

$u(0,t) = u_x (L,t) = 0$

has the general solution

$u(x,t) = \sum_{n=1}^\infty A_ne^{-(\frac{(2n+1)\pi c}{2L})^2t} \sin (\frac{(2n+1)\pi x}{2L})$
• Oct 17th 2009, 09:40 PM
Maccaman

$u_t(x,t) = c^2u_{xx}(x,t)$

$u(0,t) = 0$, $u_x(L,t) = 0$

$u(x,t) = F(x) G(t)$

$u_x(x,t) = F'(x) G(t)$

$u_{xx}(x,t) = F''(x) G(t)$

$u_t(x,t) = F(x) \dot{G}(t)$

Substitute back into PDE

$F\dot{G} = c^2 F''G$

$\Rightarrow \ \frac{\dot{G}}{c^2G} = \frac{F''}{F}$

Both must equal so constant $k = -p^2$

This gives 2 ODE's:

$F'' + p^2 F = 0$

$\dot{G} + c^2 p^2 G = 0$

$F(x) = A \cos (px) + B \sin (px)$

$u(0,t) = F(0) G(t) = 0$

$\Rightarrow \ F(0) = 0$

$F(0) = A \ \Rightarrow \ A=0$

$F(x) = B \sin (px)$

$F'(x) = Bp \cos (px)$

$u_x(L,t) = F'(L) G(t) = 0$

$\Rightarrow \ F'(L) = 0$

$F'(L) = Bp \cos(pL)$

$\Rightarrow \ p = \frac{z \pi}{2L}$ for some $z = 2n+1$ for $n = 1,2,3,....$

$F_n(x) = \sin (\frac{(2n+1)\pi x}{2L})$ , (Let $B = 1$)

Let $\lambda_n^2 = c^2 p^2$

$= \frac{c^2 (2n+1)^2 \pi}{4L^2}$

$\Rightarrow \ \lambda_n = \frac{c(2n+1) \pi}{2L}$

$\dot{G} + \lambda_n^2 G = 0$

$\Rightarrow G_n(t) = A_n e^{-\lambda_n^2 t}$ where $A_n$ is some co-efficient

Then $u_n(x,t) = F_n(x) G_n(t)$

$= A_n e^{-\lambda_n^2 t} \sin (\frac{(2n+1)\pi x}{2L})$

$= A_n e^{-(\frac{(2n+1)c \pi}{2L})^2 t} \ \sin (\frac{(2n+1)\pi x}{2L})$

Therefore

$u(x,t) = \sum_{n=1}^\infty u_n(x,t) = \sum_{n=1}^\infty A_n e^{-(\frac{(2n+1)c \pi}{2L})^2 t} \ \sin (\frac{(2n+1)\pi x}{2L})$