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Laplaces Equation with Fourier Series

  1. Apr 12, 2006 #1
    Ok, I'm going to bed. But I have to ask ANOTHER question about my homework... so I can get up early and work on it.

    Q: Around the unit circle suppose [itex] u [/itex] is a square wave:

    [tex] u_0 = \left\{\begin{array}{c} +1 \,\,\,\, on\,the\,upper\,semicircle \,\,\,\, 0<\theta < \pi \\ -1 \,\,\,\, on\,the\,lower\,semicircle \,\,\,\, -\pi < \theta < 0 \end{array} [/tex]

    From the Fourier series for the square wave write down the Fourer series for u (the solution (21) to Laplace's equation). What is the value of [itex] u [/itex] at the orgin?

    I'm seriously having trouble with Fourier series. Not clicking yet. So ANY help on this problem would be awesome. Stratagies on how to go about solving something like this. Insight of any sort. Anything !

    I think this is what I have to do... but I'm not sure :(

    What I have so far:
    [tex] \hat u(\theta) = \frac{4}{\pi} \left(\frac{\sin \theta}{1}+\frac{\sin 3\theta}{3}+\frac{\sin 5\theta}{5}+\ldots \left) [/tex]

    So this is the fourier series for the boundary condition.
    I now need a function [itex] u(r,\theta) [/itex] that satisfies Laplaces equation and the boundary condition.

    The Fourier series function [itex] \hat u(r,\theta) [/itex] of the form:
    [tex] u = a_0 + a_1 r \cos \theta + b_1 r \sin \theta +a_2 r^2 \cos 2\theta + b_2 r^2 \sin 2\theta [/itex] satisfies Laplaces equation.

    So now I have to find a way to make [itex] \hat u [/itex] satisfy the boundary condition such that:
    [tex] \hat u = 1 [/tex] when [tex] 0 < \theta \pi [/tex]
    [tex] \hat u = -1 [/tex] when [tex] -pi < \theta < 0 [/tex]

    I'm not really sure if my thought process is even right. So a little guidance to get on track would be awesome. Have a good night everyone.
    Last edited: Apr 12, 2006
  2. jcsd
  3. Apr 14, 2006 #2
    I figured this out. Sorry, if you read all that.
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