Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Hint: which is the most general solution of

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

and

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

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.

share|improve this answer
    
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. –  André 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 $ ? –  André 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 $$ u(x,y) = \sum_{k=1}^\infty \sin kx \left ( A_k \frac{\sinh k(1-y)}{\sinh k} + B_k \frac{ \sinh ky }{\sinh k} \right ) $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.