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Laplace's equation on a wedge

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the solution of Laplace's equation for [itex]\phi(r,\theta)[/itex] in the circular sector [itex]0 < r < 1; 0 < \theta < \alpha[/itex] with the boundary conditions [itex]\phi(r,0) = f(r), \phi(r,\alpha) = 0, \phi(1,\theta) = 0.[/itex] (also, implicitly, the solution is bounded at r = 0). Use two different spectral representations. (note: this just means do the problem twice, expanding in a different variable each time). Below is a crude MS Paint drawing of the boundary conditions, just to summarize:

    http://tjackson3.webs.com/laplace.png [Broken]

    2. Relevant equations

    In polar coordinates, Laplace's equation is

    [tex]\frac{\partial^2 \phi}{\partial r^2} + \frac{1}{r}\frac{\partial \phi}{\partial r} + \frac{1}{r}^2\frac{\partial^2\phi}{\partial \theta^2}[/tex]

    3. The attempt at a solution

    As I said, the goal of this problem is to use an eigenfunction expansion. We have to solve the problem twice - once using the eigenfunction corresponding to r, once using the eigenfunction corresponding to θ. Despite the fact that the θ variable is the inhomogeneous one, I was able to complete the eigenfunction expansion in that one. The problem kicked in for the r variable.

    If you do the usual separation of variables ([itex]\phi(r,\theta) = u(r)v(\theta)[/itex], and collect the r terms, you find that the r eigenfunction has to satisfy

    [tex]r^2 \frac{d^2u}{dr^2} + r\frac{du}{dr} - \lambda u = 0[/tex]

    with the boundary conditions that [itex]u(1) = 0, |u(0)| < \infty[/itex]. Unlike the normal case when you deal with Laplace's equation on a circle, we can't stipulate that [itex]\lambda[/itex] is an integer. There are two cases to consider, and they both seem unlikely. Prior experience leads me to assume [itex]\lambda > 0[/itex], so for simplicity, let [itex]\lambda = \mu^2[/itex]. Then the solution to the above ODE is

    [tex]u(r) = c_1 r^{\mu} + c_2 r^{-\mu} = c_1 e^{\mu\ln r} + c_2 e^{-\mu\ln r}[/tex]

    Since [itex]u(0)[/itex] has to be bounded, we know that [itex]c_2 = 0[/itex]. But now the problem is that there's no way to make [itex]u(1) = 0[/itex] without setting [itex]c_1 = 0[/itex], which we obviously can't do. Does anyone have any experience with how to get around this problem?

    Thanks!

    edit: If you assume lambda is negative, then you would get oscillatory functions: [itex]u(r) = c_1\sin(\mu\ln r) + c_2\cos(\mu\ln r)[/itex], which could solve the problem at r = 1, but obviously has trouble as r goes to zero.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 10, 2012 #2
    I have the feeling that your radial component eigenfunction is the Bessel function of some non-integer order, try to compare Bessel equation to your radial ODE and see if there is a match
     
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