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Laplace electric potential

  1. Oct 8, 2012 #1
    Solve Laplace's equation on a circular disk of radius [itex]a[/itex] subject to the piecewise boundary condition
    $$
    u(a,\theta) = \begin{cases}
    1, & \frac{\pi}{2} - \epsilon < \theta < \frac{\pi}{2} + \epsilon\\
    0, & \text{otherwise}
    \end{cases}
    $$
    where [itex]\epsilon \ll 1[/itex].
    Physically, this would reflect the electric potential distribution on a conducting disk whose edge is almost completely grounded except a small portion of angular extent [itex]\Delta\theta = 2\epsilon[/itex] around the location [itex]\theta = \frac{\pi}{2}[/itex].
    Obtain the solution to this problem and plot the solution for the case of [itex]a = 1[/itex] and [itex]\epsilon = 0.05[/itex].

    By separation of variables, we have
    $$
    \begin{cases}
    \Theta(\theta) = A\cos\lambda\theta + B\sin\lambda\theta\\
    R(r) = r^{\pm\lambda}
    \end{cases}
    $$
    So how do I use the conditions now?
     
  2. jcsd
  3. Oct 8, 2012 #2

    LCKurtz

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    The cosines and sines should have square roots over the lambdas, no? And can't you eliminate the ##r^{-\lambda}## by assuming the solution is bounded at the origin? And I'm guessing you have some unstated condition that will give you the eigenvalues ##\lambda##. Once you have that you should be able to finish the problem with a Fourier Series to fit the boundary condition at ##r=a##.
     
  4. Oct 8, 2012 #3
    That is everything. I have [itex]\lambda[/itex] because I set the ODEs to [itex]-\lambda^2[/itex]
    I am still not sure on how to proceed with the conditions.
    $$
    \sum_{n=1}^{\infty}r^{\lambda_n}(A\cos\lambda_n \theta + B\sin\lambda_n\theta)
    $$
     
  5. Oct 8, 2012 #4

    LCKurtz

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    Then ##R(r) = r^{\lambda_n}##. If you had the ##\lambda_n## you would just set$$
    u(a,\theta)=\sum_{n=1}^{\infty}a^{\lambda_n}(A_n \cos \lambda_n \theta + B_n\sin\lambda_n\theta)$$and solve for the ##A_n## and ##B_n## with Fourier series. But first you need to figure out the distinct eigenvalues ##\lambda_n=\mu_n^2## from some explicit or implicit condition in the problem, which I don't know.
     
    Last edited: Oct 8, 2012
  6. Oct 8, 2012 #5
    All the information is there. Typo I didn't mean -\lambda^2
    $$
    \frac{r^2R''+rR'}{R} = \lambda^2\iff r^2R''+rR'-\lambda^2R
    $$
    The problem is the Cauchy-Euler type.
    $$
    r^n[n(n-1)+n-\lambda^2] = 0\Rightarrow n^2 = \lambda^2\Rightarrow n = \lambda
    $$
     
  7. Oct 8, 2012 #6

    LCKurtz

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    OK, I agree about the non-squared powers on the ##\lambda_n## and edited my post accordingly. But I'm afraid I can't help you with regard to determining the distinct values of the ##\lambda's##. There is something we are missing which is likely implicit in the polar coordinate function, but I don't know what it is.
     
  8. Oct 8, 2012 #7
    We have to have ##u(r,\pi) = u(r,-\pi)## as well as the derivatives equal. I am glad you said that. I felt I missing some BC.
    $$
    u(r,\theta) = \alpha_0 + \beta_0\ln r + \sum_{n=1}^{\infty}r^{\lambda_n}(A_n\cos\lambda_n \theta + B_n\sin\lambda_n\theta)
    $$
    So now how do I use the piecewise condition?
     
    Last edited: Oct 8, 2012
  9. Oct 8, 2012 #8

    LCKurtz

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    Where did that ## \alpha_0 + \beta_0\ln r ## suddenly come from? I know the answer to that, but do you have something you need to show me?

    So use your new boundary conditions on ##\Theta(\theta)## to figure out the eigenvalues ##\lambda_n##. What do you get? You need to do that before you are ready to write the form of the solution for ##u(r,\theta)## and address the boundary condition.
     
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