Why Does Solving Laplace's Equation in a Square Yield an Infinite Sum?

[cgmeytanperos]

Homework Statement

i need to solve the laplace equation in square with length side 1 i tried to solve by superposition and i got infinite sum enen thouth i know that the answer should be finite

Homework Equations

1.ψ(x=0,0≤y≤1)=0
2.ψ(y=0,0≤x≤1)=0
3.ψ(x=1,0≤y≤1)=10sin(∏*y)+3x
4.ψ(y=1,0≤x≤1)=9sin(2∏*y)+3x

The Attempt at a Solution

from 1 and to 2 have (by superposition):
ψ n=Asin(n∏x)sinh(n∏y) or ψ n=Asin(n∏y)sinh(n∏x)
and after multiply by sin(n∏x)
A n=2∫((9sin(2∏x)+3x)*sin(n∏x))/sinh(n∏) (the integral from 0 to 1)
the problem is that i heve the infinite sum from ∫3x*sin(n∏x)
thank you very much!

 

[BvU]

Hello cg, and welcome to PI.
Are you sure your relevant equations are relevant equations ? They look like boundary conditions to me.
In that case, filling in x=1 changes 10sin(∏*y)+3x to 10sin(∏*y)+3 and 9sin(2∏*y)+3x changes into 3x.
And then the relevant eqations still have to be found out ...

 

[cgmeytanperos]

hi BvU and thanks.
you were right this is boundry conditions and they are also incorrect (i am sorry)
3.ψ(x=1,0≤y≤1)=10sin(∏*y)+3y
4.ψ(y=1,0≤x≤1)=9sin(2∏*x)+3x

 

[BvU]

And then the relevant eqations still have to be found out ... What do you have available ?