# Laplace's eqn solution

1. Dec 6, 2005

### sachi

We have laplace's eqn in 2-d polar co-ordinates.

We find separable solutions where V(theta) = V(theta + 2Pi)
This gives us the general solution
V(r,theta) = A + Blnr + sum from 1 to infinity of {(Cn*sin(n*theta) + Dn * cos(n*theta))*(En*r^-n + Fn *r^n) }

where A,B,C,D,E,F are all constants

we are given the boundary conditions V tends to zero as r tends to zero, therefore we conclude that A=B=Dn=0 therefore we end up with
V = sum of {(Cn*sin(n*theta) + Dn*cos(n*theta))*r^-n}

We are given further B.C's that at r= r0

V= 2*Vo*theta/Pi for -Pi/2<theta<=Pi/2
V=2*Vo*(1-theta/Pi) for -Pi/2<theta<=Pi/2

N.b Vo is a constant

at this point we clearly need to take an F.S but I'm a bit confused as the function has so symmetry about theta=0 and this messes up the algebra a lot.
I think the solution might be to translate the theta function by Pi/2 in the -ve theta direction to make it even, and then take the Fourier cosine series, then using the substitution theta = theta + Pi/2 to turn in back into the original function. Is this is appropriate method, or is there a different way? Thanks very much for your help.

Sachi