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## Homework Statement

Find the harmonic function in the square {0<x<1, 0<y<1} with the boundary conditions

u(x,0)=x

u(x,1)=0

ux(0,y)=0

ux(1,y)=y²t

## Homework Equations

## The Attempt at a Solution

__Part1:__[/B]

We first solve the problem in the square with BC

u(x,0)=0 ; u(x,1)=0 ; ux(0,y)=0 ; ux(1,y)=y²

The solution is

u(x,y)= ∑ An cosh(nπx) sin(nπy) for n=1,2,....

**Part2:**Then solving the problem in the square with BC

u1(x,0)=x ; u(x,1)=0 ; ux(0,y)=0 ; ux(1,y)=0

By separation of variable we solve

X''=-λX with X'(0)=X'(1)=0

Here the solution says that λ=(nπ)² n≥0 with eigenfunctions Xo=constant and Xn=cos(nπx) for n>0

and Y''=λY

For λ=0, we get Y''=0, solving and applying BC Yo(y)=A(y-1)

And for n=1,2,... Yn(y)=A(cosh(nπy)-cothan(nπy)*sinh(nπy))

u2(x,y)=Ao(y-1)+ΣAn (cosh(nπy)-cothan(nπy)*sinh(nπy))*cos(nπx)

What I don't understant, why in part 1, n starts at 1 and not 0 ?

And why in part 2 it starts at n=0, I understand that the sum is not defined for n=0 (coth(0)=oo) but how can I know the values that take n when i solve for the functions X and Y ?

Thanks