Separation of variables - Laplace's Equation

1. May 21, 2012

hermish

1. The problem statement, all variables and given/known data

Use separation of variables to find the solution to Laplaces equation satisfying the boundary conditions

u(x,0)=0 (0<x<2)
u(x,1)=0 (0<x<2)
u(0,y)=0 (0<y<1)
u(2,y)= asin2πy(0<y<1)

3. The attempt at a solution

I am able to perform the separation of variables technique on the wave equation. The heat equation is a little harder, I struggle a bit, but eventually I get there. Laplace's equation is pretty much impossible. From my understanding, the method is very similar in all three cases, but I think there are some differences which I don't see, which is why I can't do this question.

So I managed to separate the variables, deriving two ODE's, one in terms of x and one in terms of y, with the separation constant λ.

F''(x) - λF(x) = 0
G''(x) - λG(x) = 0

For the case where λ=0, there are no solutions because nothing can satisfy the last boundary condition listed.

For the case where λ<0, I think there are no solutions, I could sort of tell by having a look at the final answer given. I don't understand how to show this?

For the case where λ>0
I get F(x) = A*cosh(sigma*x) + B*sinh(sigma*y)
G(y) = (Ccos(sigma*y) + Dsin(sigma*y))

where sigma is the roots of the ODE's.

so now u(x,t) = F(x)*G(y)
I have this function with 5 unknowns, A,B,C,D, and sigma
When I apply all the boundary conditions, I don't really get anywhere. No helpful information appears.

What am I doing wrong? Or, what am I not doing?

2. May 21, 2012

vela

Staff Emeritus
You have a sign error in one of your ODEs. Also, one of them should have ys in it. It looks like what you meant to write was:
\begin{align*}
F''(x) - \lambda F(x) &= 0 \\
G''(y) + \lambda G(y) &= 0
\end{align*} Depending on the sign of $\lambda$, you get regular trig functions for one solution and hyperbolic functions for the other.

To see why you can't have $\lambda<0$, try to satisfy the boundary conditions when you have $G(y) = C\sinh \sigma y + D\cosh \sigma y$. To satisfy the first two boundary conditions, you need G(0)=0 and G(1)=0. You should find the only way you can do this is if C=D=0, which isn't a useful solution.

If $\lambda>0$, you have $G(y) = C\sin \sigma y + D\cos \sigma y$. What do you get from the first two boundary conditions in this case?