# Partial DFQ Dirichlet Problem

## 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

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## 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)$\Rightarrow$A=0

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

Last edited:

LCKurtz
Homework Helper
Gold Member

## 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

--

## 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)$\Rightarrow$A=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.

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