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Laplace's eqn of wedge

  1. Sep 23, 2007 #1
    I have a quick question about solving Laplace's equation for a wedge with radius a and angle 60º. I need to make the periodicity conditions correctly, so that I can have a reasonable problem to solve. For a circular ring you would simply say that the equation should not differ whether you come from the top or bottom, so that

    [tex]u (r, \pi) = u(r, -\pi)[/tex]

    for which the rates would have to be the same too

    [tex]\frac{\partial u}{\partial \theta} (r, \pi) = \frac{\partial u}{\partial \theta} (r, -\pi)[/tex]

    So, this is good because these periodicity conditions give the eigenfunctions
    [tex]sin n\theta[/tex] and [tex] cos n \theta[/tex]

    My question is whether this still holds for the wedge. Obviously anywhere on the circle it all still applies, but what about the radial parts of the wedge? Should those be treated with their own periodicity conditions? If so, would it just be that

    u(r, 0) = u(r, pi/3)

    But that worries me if it is so because that won't give nice eigenvalues and eigenfunctions for the radial parts.
     
    Last edited: Sep 23, 2007
  2. jcsd
  3. Sep 23, 2007 #2
    Are you aware of the general polar solutions to Laplace's equation?
     
  4. Sep 23, 2007 #3
    Maybe I should have been more explicit, but that is exactly what I am wondering, if the general polar solution, which comes from periodicity, can still be applied such that

    [tex]u(r, \theta) = a_0 + b_0 lnr + \sum_{n=1}^\infty [(a_n r^n + b_n r^{-n})(\alpha_n cosn\pi + \beta_n sinn\pi)][/tex]
     
  5. Sep 23, 2007 #4
    The answer is indeed that some modifications have to be made. Do you know how that solution is derived?
     
  6. Sep 23, 2007 #5
    Okay, that's what I thought made the most sense, but wanted to double check. I do know how it is derived, and I can easily do it, but no way am I going to post a page of LaTeX eqns. :p
     
  7. Sep 23, 2007 #6
    :wink: Have fun
     
  8. Sep 23, 2007 #7
    Ah crap, problem with the fourier series, (I'm solving the boundary value problem of u(a,theta) = f(theta)) just make 2L = pi/3 such that -L = -pi/6 and L = pi/6?
     
    Last edited: Sep 23, 2007
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