Solve the boundary value problem $$u_{xx}+u_{yy}=0, \ 0<x,\ y<1$$ $$u_x(0,y)=0, \ u(1,y)=0, \ 0<y<1$$ $$u(x,0)=1,u_y(x,1)=0, \ 0<x<1.$$

I have, $$\frac{\phi''(x)}{\phi(x)}=-\frac{h''(y)}{h(y)}=-\lambda$$ $$\phi''(x)+\lambda\phi(x)=0$$ $$h''(u)-\lambda h(y)=0$$

Solving the ODE for $\phi$ and applying the boundary conditions I got $$\phi_n(x)=\cos\sqrt{\lambda_n}x, \ \text{where} \ \lambda_n=\left[\frac{(2n+1)\pi}{2}\right].$$

Therefore, $$u(x,y)=\sum_n\cos\sqrt{\lambda_n}x\left[a_n\cos h\sqrt{\lambda_n}y+b_n\sin h\sqrt{\lambda_n}y\right]$$ $$u(x,0)=\sum_n\left[a_n\right]\cos\sqrt{\lambda_n}x=1$$ $$a_n=\frac{1}{\pi}\int^{\pi}_0f(x)\cos\sqrt{\lambda_n}xdx=\frac{1}{\pi}\int^{\pi}_0\cos\sqrt{\lambda_n}xdx$$ and using the other initial condition I got that $b_n=0$.

Did I do this right?


You have successfully found that $$ \psi(y)\sim\{\cosh(\lambda_ny),\sinh(\lambda_ny)\} $$ The BC are $u(x,0) = 1$ and $u_y(x,1)=0$. It would be better to have the zero BC zero out a hyperbolic term. That is, we need to make the change of variables $y\mapsto 1- y$ so $\psi(y)\sim\{\cosh(\lambda_n(1-y)),\sinh(\lambda_n(1-y))\}$ $$ \psi_y(1) = \lambda_nB =0 $$ Thus, $\psi\sim\cosh(\lambda_n(1 - y))$. The general form of the PDE is then $$ u(x,y) = \sum_{n=0}^{\infty}A_n\cos\Bigl[\frac{\pi}{2}(2n + 1)x\Bigr]\cosh\Bigl[\frac{\pi}{2}(2n+1)(1-y)\Bigr] $$ Now we can apply the final BC $u(x,0) = 1$. $$ u(x,0) = \sum_{n=0}^{\infty}A_n\cos\Bigl[\frac{\pi}{2}(2n + 1)x\Bigr]\cosh\Bigl[\frac{\pi}{2}(2n+1)\Bigr] = 1 $$ Now solve for $A_n$ and note that $\cosh$ is treated as a constant.

  • $\begingroup$ Thank you! A couple of questions can you elaborate a bit more why we must do a change of variables? Also, to find $A_n$ would this be correct $$A_n=\frac{1}{\pi}\int^{\pi}_0f(x)\cos\sqrt{\lambda_n}xdx=\frac{1}{\pi}\int^{\pi}_0\cos\sqrt{\lambda_n}xdx$$ since $\cos h$ vanished because of the initial condition $u(x,0)$. $\endgroup$ – Robben Nov 22 '14 at 4:36
  • $\begingroup$ $\cosh$ doesnt vanish. $$A_n = \frac{2}{\cosh\bigl[\frac{\pi}{2}(2n+1)\bigr]}\int_0^11\cos\Bigl[\frac{\pi}{2}(2n+1)x\Bigr] dx$$ Note that your domain is $0<x,y<1$ not to $\pi$ as well. $\endgroup$ – dustin Nov 22 '14 at 4:38
  • $\begingroup$ Oh yea, because you did the change of variables. Also, can you elaborate a bit more why the change of variables is necessary for this problem? $\endgroup$ – Robben Nov 22 '14 at 4:40
  • $\begingroup$ With the Laplace equation, you want the nonzero boundary conditions to be on the ends; that is, at $x=1$ and/or $y=1$. When one or both are at $x,y=0$, you can should make the change of variables $\text{end point } - x,y$ what ever the variable may be. $\endgroup$ – dustin Nov 22 '14 at 4:41
  • $\begingroup$ Hm, I never knew that. Thank you very much for clarifying! $\endgroup$ – Robben Nov 22 '14 at 4:43

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