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So, my problem is as follows:

$$ U_{xx} +U_{yy}=0; U(x,0)=0, U_x (0,y)=0 ,U_x (L,y)=0 $$

I need to use Separation of Variables, which I did, and got:

$$ U(x,y) = (Ae^{\sqrt{\lambda}x}+Be^{-\sqrt{\lambda}x})(C\cos{\sqrt{\lambda}y}+D\sin{\sqrt{\lambda}y}) $$

for ${\lambda}>0$. My question is how to plug in the boundary conditions, and why this first solution does/doesn't satisfy the conditions.

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You repeat $U_x(0,y)=0$ twice. –  user17090 Nov 14 '11 at 6:18
    
fixed, sorry... –  Chris Nov 14 '11 at 6:19

1 Answer 1

You use first $U(x,0)=0$ to get $C=0$ and then use $U_x(0,y)=0$ to get \begin{equation*} (A {\sqrt{\lambda}}e^{\sqrt{\lambda}x}-B \sqrt{\lambda} e^{-\sqrt{\lambda}x})D\cos \sqrt{\lambda}y|_{x=0}=0 \end{equation*} and then $A-B=0$ or $A=B$. Similarily use $U_x(L,y)=0$ to get $A\sqrt{\lambda}[2 sh\sqrt{\lambda} L]=0$ which implies $e^{2\sqrt{\lambda} L}=1$ which gives $\lambda=0$ which is not what you want. So you made an error in choosing the solution. The solution should be: $$U(x,y)=(A\cos \sqrt{\lambda}x +B \sin \sqrt{\lambda}x)(Ce^{\sqrt{\lambda}y} +De^{-\sqrt{\lambda}y}).$$ Now we repeat the above steps to get the values of $A,B,C$ and $D$ (up to a multiplicative factor).

More precisely, plug $U(x,0)=0$ to get $C+D=0$. Then use $U_x(0,y)=0$ to get $\sqrt{\lambda}B=0$ so $B=0$. Also, $U_x(L,y)=0$ implies $\sqrt{\lambda}A\sin \sqrt{\lambda}L =0$ hence $\lambda= \frac{n^2\pi^2}{L^2}$. Finally we put: \begin{equation*} U_n(x,y)= A_n \cos {\frac{n\pi}{L}x} \sinh {\frac{n\pi}{L}y} \end{equation*} which is the required solution and where we absorbed $C$ and $D$ into $A$. The full solution to the problem is of the form \begin{equation*} U(x,y)=\sum_n A_n\cos {\frac{n\pi}{L}x} \sinh {\frac{n\pi}{L}y}. \end{equation*}

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Thanks Willie:) –  user17090 Nov 14 '11 at 15:25

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