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Let $\Omega := [0, 1] \times [0,\pi]$. We are searching for a function $u$ on $\Omega$ s.t. $$ \Delta u =0 $$ $$ u(x,0) = f_0(x), \quad u(x,1) = f_1(x), \quad u(0,y) = u(\pi,y) = 0 $$ with $$ f_0(x) = \sum_{k=1}^\infty A_k \sin kx \quad, f_1(x) = \sum_{k=1}^\infty B_k \sin kx $$ If I use seperation of variables, say $u(x,y) = f(x)g(y)$ I get $$ f''(x)+\lambda f(x) = 0 , \quad g''(y)-\lambda g(y) = 0 $$ with $f(0) = f(\pi) = 0$ where I use that $f,g \neq 0$. $\lambda$ is some constant. How can I proceed ?

Thanks in advance.

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up vote 2 down vote accepted

Hint: which is the most general solution of

$$f''(x)=-\lambda f(x)$$


$$g''(y)=\lambda g(y)?$$

You need to consider linear combinations of exponentials. Such exponentials have real or complex exponents depending on the sign of $\lambda$, i.e. $\lambda >0$, $\lambda<0$ (not necessarily in this order!). Try quickly to see what happens if $\lambda=0$, instead.

To determine which choice of sign for $\lambda$ is the correct one for your problem, you need to apply the boundary conditions you wrote for $f$ and $g$ at $0$ and $\pi$. Once you are there apply superposition and the boundary conditions with the Fourier series. You are done.

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please tell me if you need more details. – Avitus Jun 24 '13 at 18:51
I guess $f(x) = e^{i \sqrt \lambda x}$ and $g(x) = \sinh( \sqrt \lambda x)$. Since $f(0)=f(\pi)=0$ we have that $\lambda = k^2$ for $k \in \mathbb Z$ I guess. – Andre Jun 24 '13 at 18:54
so you chose $\lambda>0$. In general all $f(x)=A\exp(i\sqrt{\lambda}x)+B\exp(-i\sqrt{\lambda}x)$ are solutions of your ODE. Similarly for $g$. Quickly check it and try the computations again applying the boundary conditions. Ok? What do you get for $\lambda$? – Avitus Jun 24 '13 at 19:00
I am not sure what your intention is. If I use $u(0,y) = u(\pi,y) = 0$ I get $f(0)=f(\pi)=0$ s.t. $A+B = 0$ and $B(\exp(-i \pi \sqrt \lambda) - \exp(i \pi \sqrt \lambda )) =0$ But $B \neq 0$ since $f \neq 0$ s.t. $0 = \sin \pi \sqrt \lambda \iff \lambda = k^2$ for some $ k\in \mathbb Z $ ? – Andre Jun 24 '13 at 19:18
Exactly. You need to use that $f(x)$ is the linear combinations of exponentials with exponents $-i...$ and $+i...$. The condition on $\lambda$ is correct: you need $\sqrt{\lambda}\pi=\pi k$, with $k\in\mathbb Z$, which implies $\lambda=k^2$. – Avitus Jun 24 '13 at 19:27

After some time, work and help by Avitus I finally got it :

Assume $u(x,y) = A(x)B(y)$ with $A,B \neq 0$ on $\Omega$. This yields to $$ A''(x)+\lambda A(x) = 0 , \quad B''(y)-\lambda B(y) =0 $$ General solutions are $$ A(x) = \gamma_1 e^{\sqrt \lambda i x} + \gamma_2 e^{-\sqrt \lambda ix} $$ which gives with the conditions $u(0,y)=u(0,\pi) = 0$ that $A_k(x) =\gamma_k \sin |k| x$ for $k \in \mathbb Z$. For $B(y)$ we find then $$ B_k(y) = c_k \sinh |k|x + d_k \cosh |k|y $$ Then we have that $$ u(x,y) = \sum_{n \in \mathbb Z} A_k(x)B_k(y) = \sum_{n=1}^\infty \sin kx \left ( \lambda_k \sinh kx + \mu_k \cosh ky \right ) $$ where $\lambda_k = \gamma_kc_k +\gamma_{-k}c_{-k}$ and similar for $\mu_k$. By using $u(x,0)= f_0(x)$ and $u(x,1)=f_1(x)$ we find by comparing coefficients that $$ \mu_k = A_k, \quad \lambda_k = \frac{B_k -A_k\cosh k}{\sinh k} $$ Writing out $\lambda_k \sinh ky + \mu_k \cosh ky$ and using that $\sinh(a-b) = \sinh a\cosh b - \sinh b \cosh a$ we find that $$ u(x,y) = \sum_{k=1}^\infty \sin kx \left ( A_k \frac{\sinh k(1-y)}{\sinh k} + B_k \frac{ \sinh ky }{\sinh k} \right ) $$

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