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Homework Help: Laplace Equation - Wedge

  1. Apr 17, 2010 #1
    1. The problem statement, all variables and given/known data I need to solve the Laplace equation, u_rr + (1/r)u_r + (1/r^2)u_{theta}{theta} = 0,
    on a circular wedge with radius R, angle {alpha}, where u(r,0) = 0, u(R,{theta}) = 0, and
    u(r,{alpha}) = 50.

    2. Relevant equations



    3. The attempt at a solution Separate variables - u(r,{theta}) = P(r)Q({theta}), and we have the equations

    Q'' + (k^2)Q = 0, (r^2)P'' + rP' - (k^2)P = 0, where k is a constant. Q_n = Asin(n{theta})+Bcos(n{theta}), P_n = C(r^n).

    Since u(r,0) = 0, B = 0. Now, since u(R,{theta}) = 0, this must mean R^n = 0, hence
    R = 0, so u = 0 for all r, {theta}. Is this reasoning correct?
     
  2. jcsd
  3. Apr 17, 2010 #2

    LCKurtz

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    No, it isn't. You don't get just rn but you also get r-n. And I would suggest using something other than n since it isn't restricted to integer values with only one theta boundary condition. Perhaps μ would be better. So you have sin(μθ) and you have {rμ,r}, or equivalently {eμln(r),e-μln(r)} or, even better, {cosh(μln(r)),sinh(μln(r)}. This would give a two parameter family Acosh(μln(r)) + Bsinh(μln(r)), which looks like an addition formula. Since you have a zero boundary condition when r = R, this suggests even another choice for a two parameter family:

    P(r) = A sinh(μln(r) + B)

    Then P(R) = A sinh(μln(R) + B) and this can be made to be 0 by picking B = -μln(R).

    This gives P(r) = A sinh(μln(r)-μln(R)) = A sinh(μln(r/R))

    So at this point you have:

    Qμ = sin(μθ) and Pμ=sinh(μln(r/R)).

    These work in your DE and solve the homogeneous BC's. Hopefully, you can take it from there because it's been too long since I have looked at this stuff.
     
  4. Apr 17, 2010 #3
    I thought the r^(-n) part was removed because then the solution would go to infinity at r = 0.
     
  5. Apr 17, 2010 #4

    LCKurtz

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    You didn't give careful statements of your boundary conditions but I would expect they are something like this:

    u(r,0) = 0, 0 < r < R
    u(R,θ) = 0, 0 < θ < α
    u(r,α) = 50, 0 < r < R

    In particular you aren't giving u(0,0) and the first and third conditions would disagree if you tried. So I don't see where keeping the r is ruled out. And you need it to get your second boundary condition as I have shown. Your separated equation for P(r) has a singular point at r =0.

    You can verify that u(r,θ) = Asin(μθ)sinh(μln(r/R)) satisfies the DE and the homogeneous BC's. But, like I said, you're on your own from here.
     
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