# Possion equation in circular annulus-II

1. Jan 4, 2012

### smoger

I previosuly asked about possion equation in circular annulus.
I found the general soultion of the problem myself. but i meet a another problem and still can not find the solution.

The problem is :

2A/∂r2 + 1/r ∂A/∂r + 1/r22A/∂θ2= f(r,θ) in the circular annulus (a<r<b)

where f(r,θ)=1/r + Xn/r cos(nθ)

The boundary condition is :

1. ∂A/∂θ=0 (at r=a)
2. ∂A/∂r =0 (at θ=±β)

I solved the problem as ,

Seperating variables as :
A(r,θ)=G(r)+R(r)T(θ) , assuming that
G(r) should satisfy 1/r , and R(r)T(θ) should satisfy Xn/r cos(nθ) term in f(r,θ)
Then the problem for G(r) is :

2G(r)/∂r2 + 1/r ∂G(r)/∂r = 1/r (1)

and Considering the boundary coundition 2,

∂G(r)/∂r =0 at θ=±β (2)

But substituding (2) in (1) becomes 0.
How can i solve the problem? thx in advance