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## Homework Statement

if [tex]\nabla^2u = 0[/tex] in [tex] 0 \leq x \leq \pi, 0\leq y \leq \pi, [/tex]

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

## Homework Equations

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

## The Attempt at a Solution

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

[tex]\nabla^2u = 0[/tex] in [tex] 0 \leq x \leq a, 0\leq y \leq b, [/tex]

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 [tex] u(x,y) = \sum_{n=0}^\infty D_nsin\frac{n\pix}{a}sinh\frac{n\piy}{a} [/tex]

with [tex] D_n =\frac{2}{asinh\frac{n\pib}{a}}\int_{0}^{a}f(x)sin\frac{n\pix}{a}dx [/tex] 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

[tex] X(\pi)= Bsinp\pi=cos^2y [/tex]

[tex]Y'(0) = F=0, Y'(\pi)=0

so Esinhp\pi=0 [/tex] 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

[tex] X(\pi)= Bsinhp\pi =cos^2(y) [/tex]

Y'(0)=E=0

[tex] Y'(pi)= -Fsinppi=0 [/tex]

when p = 1,2,3

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

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

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