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Possion equation in circular annulus-II

  1. Jan 4, 2012 #1
    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
     
  2. jcsd
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