How Does Separation of Variables Solve the DFQ Dirichlet Problem?

[jake2]

Homework Statement

∇^{2}u=0 on 0<x<∏, 0<y<2∏
subject to u(0,y)=u(∏,y)=0
and u(x,0)=0, u(x,2∏)=1

Homework Equations

--

The Attempt at a Solution

I've solved the SLP, and now I am trying to solve the Y-equation that results from separation of variables:

Y''-λY=0, Y(0)=0
Y_{n}(y)=Acosh(ny)+Bsinh(ny)
Y(0)=(0)=Acosh(n*0)+Bsinh(n*0)\RightarrowA=0

Doesn't this effectively "kill" the problem? Or is this the solution:
Y_{n}(y)=Bsinh(ny)

 

[LCKurtz]

Homework Statement

∇^{2}u=0 on 0<x<∏, 0<y<2∏
subject to u(0,y)=u(∏,y)=0
and u(x,0)=0, u(x,2∏)=1

Homework Equations

--

The Attempt at a Solution

I've solved the SLP, and now I am trying to solve the Y-equation that results from separation of variables:

Y''-λY=0, Y(0)=0
Y_{n}(y)=Acosh(ny)+Bsinh(ny)
Y(0)=(0)=Acosh(n*0)+Bsinh(n*0)\RightarrowA=0

Doesn't this effectively "kill" the problem? Or is this the solution:
Y_{n}(y)=Bsinh(ny)

Well, so far you have $X_n(x) = \sin(nx)$ and $Y_n(y) = \sinh ny$ so your potential solution is $$
u(x,y)=\sum_{n=1}^\infty b_n\sin(nx)\sinh(ny)$$ which satisfies 3 of the four boundary conditions. You still have all the $b_n$ to use. So you need$$
u(x,2\pi)=\sum_{n=1}^\infty b_n\sin(nx)\sinh(2n\pi)=1$$I'm guessing you know how to do that.