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An interior Dirichlet problem for a circle

  1. Dec 1, 2016 #1
    1. The problem statement, all variables and given/known data
    $$
    \bigtriangledown^2=0 for : 0<r<1 \\
    BC : u(1,\Theta)= sin(\Theta), 0<\Theta<\pi \\ u(1,\Theta)= 0, pi<\Theta<2\pi \\

    $$
    Basically its an interior dirichlet problem for a circle.



    2. Relevant equations


    3. The attempt at a solution

    The answer is supposed to be $$U(r,\Theta) = \Sigma r^n[a_n cos(n\Theta) + b_n sin(n\Theta)$$
    and the a_n, b_n is basically a fourier expansion of the boundary conditions.
    The books answer is :

    $$ \frac{r}{2}sin\Theta + \frac{2}{\pi}(\frac{1}{2} - \frac{r^2}{3}cos(2\Theta)-\frac{r^4}{15}cos(4\Theta)-\frac{r^6}{35}cos(6\Theta)...)$$

    Now, I cant seem to get hte fourier expansion right I guess, because I dont get this answer.

    Heres my attempt ....

    $$a_n = \frac{2}{\pi}\int_{0}^{\pi}sin(n\Theta)cos(n\Theta)d\Theta = 0

    \\
    b_n = \frac{2}{\pi}\int_{0}^{\pi}sin^2(n\Theta) = \frac{1}{\pi}
    \\
    a_0 = \frac{1}{\pi}\int_{0}^{\pi}sin\Theta d\Theta = 0

    $$

    Cant think of how to make this fourier expansion into the books answer...
     
  2. jcsd
  3. Dec 2, 2016 #2

    Orodruin

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    Why are you computing the integral with ##\sin(n\Theta)##? The boundary condition is ##\sin(\Theta)## ...

    Also, the integral of sine over half a period is not zero.
     
  4. Dec 2, 2016 #3
    Thanks!
     
  5. Dec 12, 2016 #4
    'okay I am getting 2/pi + summation [4/pi * [0.5 - r^n/(1-n^2)]cos(nx)]
    The Bn term goes to 0 (per wolfram). A0 = 2/pi, An = 2/pi * cos(n*pi +1)/(1-n^2)

    Is the books answer wrong? I dont know where their sin term comes from because the Bn integral is 0...
     
  6. Dec 13, 2016 #5
    NVM i figured it out cant delete last post for some reason.
     
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