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Laplace's eqn

  1. Dec 20, 2005 #1
    I need to solve laplace's equation in 2-d polar co-ordinates, and I just get the standard V(r,theta) = A + Blnr + sum to infinity of [An*sin(n*theta) + Bn * cos(n*theta)]*[Cn*(r^-n) + Dn*(r^n)]
    by using separation of variables and considering all values of the separation constant which give sinusoids in theta, and also letting the separation constant equal zero. N.b A,B,An,Bn,Cn,Dn are all constant.

    I need to impose boundary conditions

    1) V Tends to zero as r tends to infinity (i.e A=Dn=0)

    2) for r=a V=2V1*theta/pi for -pi/2<theta<pi/2
    V=2V1*(1-theta/pi) for pi/2<theta<3pi/2
    where v1 is a constant

    I know that I need to use a fourier series but because V(a) does not have its centre at the origin, the algerba is messy and doesn't really go anywhere. The only way I can think of solving this problem is to translate V(a) so that it is symmetric about the origin, take the fourier cosine series, and then translate it back using the substitution theta = theta + pi/2. Is this the correct way of doing it?
    Thanks very much
     
  2. jcsd
  3. Dec 20, 2005 #2

    HallsofIvy

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    I'm not sure what you mean by "take the fourier cosine series". Typically, the solution to Laplace's equation can be written as a summation of Bessel functions with coefficients depending on the angle [itex]\theta[/itex]. When you "separate variables" the r equation will be Bessel's equation. You appear to be doing the "outer" problem: find a solution, defined outside a circle of radius a that has a given value on the circle r= a and goes to 0 as r goes to infinity. If I remember correctly, it is the Bessel function of the first kind that gives the "inner solution" (defined at r= 0) and the Bessel function of the second kind that gives the "outer solution".
     
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