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Laplaces equation in polar coordinates

  1. Mar 20, 2007 #1
    The function [tex] u(r,\theta) [/tex]

    satisfies Laplace's equation in the wedge [tex] 0 \leq r \leq a, 0 \leq \theta \leq \beta [/tex]

    with boundary conditions [tex] u(r,0) = u(r,\beta) =0, u_r(a,\theta)=h(\theta) [/tex]. Show that

    [tex] u(r,\theta) = \sum_{n=0}^\infty A_nr^{n\pi/\beta}sin(\frac{n\pi\theta}{\beta}) [/tex]

    [tex]A_n=a^{1-\frac{n\pi}{\beta}\frac{2}{n\pi}\int_{0}^{\beta}h(\theta)sin\frac{n\pi\theta}{\beta}d\theta [/tex]
    Last edited: Mar 21, 2007
  2. jcsd
  3. Mar 21, 2007 #2


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    You have posted here before so surely you know the basic rules!

    No one is going to do your homework for you and it wouldn't help you if they did! Show us what you have tried so we can see where you went wrong or got stuck.
  4. Mar 21, 2007 #3
    I saw the question, realised I was incapable and thought I'd put it up for some hints... I do have a similar problem viz laplace in the unit circle.....

    [tex] \nabla^2U=0 [/tex]

    Boundary conditions are 1) U=0 at r=0

    2) [tex] U(1,\theta)=2cos\theta [/tex]

    now I have quoted from the notes that the general solution is

    [tex] U(r,\theta) =C_0lnr + D_0 + \sum_{0}^\infty(C_0r+\frac{D_0}{r^n}).(A_ncosn\theta + B_nsinn\theta) [/tex]

    now I am told that B.C 1 implies [tex] D_0 = 0=C_n [/tex] for n=0,1,2,3,4....

    I am immediately confused why it is necessary to have these two coeffiecient set to zero, surely we could have some situation whereby the three terms could cancel to zero without insisting their coefficients are zero??
    Last edited: Mar 21, 2007
  5. Mar 21, 2007 #4


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    The sines, cosines and the constant function are linearly independent on the unit circle.
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