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Dirichlet Problem

Posted December 2nd, 2007 by vic777
in

What is the solution for this equation?
$ \nabla^2u=0, 0<x<a, 0<y<b, 0<z<c,<br /> u(0,y,z)=sin(\frac{\pi*y}{b})sin(\frac{\pi*z}{c}), u(a,y,z)=0,<br /> u(x,0,z)=0, u(x,b,z)=0,<br /> u(x,y,0)=0, u(x,y,c)=0 $

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Comments

Short hint

On December 3rd, 2007 Structure says:

Short hint
Separating first variable z you get fot $ u(x,y,z)=V(x,y)Z(z) $ $ Z"(z)+\lambda^2Z(z)=0 $ and $ Z(0)=Z(c)=0 $ with solution $ \lambda_n=\frac{n\pi}{c} $ and $ Z_n(z)=\sin \frac{n\pi}{c}z $
Separating the rest gives
$ V(x,y)=X(x)Y(y) $

$$\frac{X"(x)y(y)+X(x)Y"(y)}{X(x)Y(y)}=\lambda_n^2$$
$$\frac{X"(x)}{X(x)}-\lambda_n^2=-\frac{Y"(y)}{Y(y)}=\mu^2$$

$ Y"(y)+\mu^2 Y(y)=0 $ with $ Y(0)=Y(b)=0 $, $ \mu_m=\frac{m\pi}{b} $ and
$ Y_m(y)=\sin \frac{m\pi}{b}y $.
For the rest you have for
$ X"(x)-(\lambda_n^2+\mu_m^2)X(x)=0 $
Let $ \tau_{m,n}^2=\lambda_n^2+\mu_m^2 $
You have
$ X_{m,n}(x)=A_{m,n}\cosh\tau_{m,n}x+B_{m,n}\sinh\tau_{m,n}x $
Then look for

$$u(x,y,z)=\sum_{n,m=1}^{+\infty}(A_{m,n}\cosh\tau_{m,n}x+B_{m,n}\sinh\tau_{m,n}x)\sin \frac{m\pi}{b}y\sin \frac{n\pi}{c}z$$

Using initial condition you get

$$u(x,y,z)=\frac{1}{\sinh\tau_{1,1}a}\sinh\tau_{1,1}(a-x)\sin \frac{\pi}{b}y\sin \frac{\pi}{c}z$$

where of course

$$\tau_{ \: 1,1}=\pi\sqrt {\frac{1}{b^2}+\frac{1}{c^2}}$$

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