Separation of variables and generalised fourier series

[catcherintherye]

Homework Statement

if \nabla^2u = 0 in 0 \leq x \leq \pi, 0\leq y \leq \pi,

boundary conditions u(0,y)=0, u(\pi,y)=cos^2y, u_y(x,0) = u_y(x,\pi)=0

Homework Equations

I am required to show that u(x,y) = \frac{x}{2\pi} + \frac{cos2ysinh2x}{2sinh2\pi}

The Attempt at a Solution

I have done similar questions before for example consider then following problem:

\nabla^2u = 0 in 0 \leq x \leq a, 0\leq y \leq b,

but with boundary conditions:

u(0,y)=0, u(a,y)=0, u(x,0)=0, u(x,b)=f(x)


I derived the general solution to be u(x,y) = \sum_{n=0}^\infty D_nsin\frac{n\pix}{a}sinh\frac{n\piy}{a}

with D_n =\frac{2}{asinh\frac{n\pib}{a}}\int_{0}^{a}f(x)sin\frac{n\pix}{a}dx n=1,2,3...



...This example I understand but in the first example I am confused so far I have done the following,

let u(x,y) =X(x)Y(y)

X(x) =Acospx + Bsinpx
Y(y) =Ccoshpy + Dsinhpy
then Y'(y) =Esinhpy + Fcoshpy

B.C's ->

X(0) =A=0

X(\pi)= Bsinp\pi=cos^2y


Y&#039;(0) = F=0, Y&#039;(\pi)=0<br /> <br /> so Esinhp\pi=0 this seems to tell me nothing about E I have tried also type three solutions viz.

X(x)=Acoshpx + Bsinhpx

Y(y)= Ccospy + D sinpy
Y'(y)=Ecospy-Fsinpy

X(0)=A=0
X(\pi)= Bsinhp\pi =cos^2(y)

Y'(0)=E=0

Y&#039;(pi)= -Fsinppi=0

when p = 1,2,3

so u(x,y) = \sum_{p=0}^\infty F_nsinpxsinhpy

which is markedly different to the form of solution required, what am I doing wrong??

 

[cristo]

If you only need show that the expression you give is the solution, can you not just show it satisfies the equation and that it satisfies the boundary conditions?