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Math Help - PDE non seperables

  1. #1
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    PDE non seperables

    Hey I have been given the below question but am really struggling to get it still I understand how to solve problems using seperation of variables but this has got me stumped

    We want to find the value of the electrostatic potential \phi(r,\theta) in the quadrant.

    0 \leq r \leq b , 0 \leq \theta \leq \frac{\pi}{2}

    where r and \theta are the standard polar co-ordinates and b is a given positive real number. Assuming electrostatic potential is the solution of Laplaces equation in polar co-ordinates

    \frac{\delta^2 \phi}{\delta r^2} +\frac{1}{r}\frac{\delta\phi}{\delta r} + \frac{1}{r^2} \frac{\delta^2\phi}{\delta\theta^2} = 0

    The boundary conditions on \phi(r,\theta) are

    \phi(b,\theta) = \frac{4}{\pi^2} \theta(\pi - \theta),

    \phi(r,0) = 0,     \phi(r,\frac{\pi}{2}) = 1

    (a) show that the function \phi(\theta) = \frac{2\theta}{\pi} is a solution of the initial equation that satisfies the first set of boundary conditions

    Any help would be great

    Thanks
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  2. #2
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    Quote Originally Posted by jezzyjez View Post
    Hey I have been given the below question but am really struggling to get it still I understand how to solve problems using seperation of variables but this has got me stumped

    We want to find the value of the electrostatic potential \phi(r,\theta) in the quadrant.

    0 \leq r \leq b , 0 \leq \theta \leq \frac{\pi}{2}

    where r and \theta are the standard polar co-ordinates and b is a given positive real number. Assuming electrostatic potential is the solution of Laplaces equation in polar co-ordinates

    \frac{\delta^2 \phi}{\delta r^2} +\frac{1}{r}\frac{\delta\phi}{\delta r} + \frac{1}{r^2} \frac{\delta^2\phi}{\delta\theta^2} = 0

    The boundary conditions on \phi(r,\theta) are

    \phi(b,\theta) = \frac{4}{\pi^2} \theta(\pi - \theta),

    \phi(r,0) = 0, \phi(r,\frac{\pi}{2}) = 1

    (a) show that the function \phi(\theta) = \frac{2\theta}{\pi} is a solution of the initial equation that satisfies the first set of boundary conditions

    Any help would be great

    Thanks
    Actually, it's the second BC.

    if \phi(r,\theta) = \frac{2 \theta}{\pi} then

     <br />
\phi(r,0) = \frac{2 0}{\pi} = 0<br />

     <br />
\phi(r,\frac{\pi}{2}) = \frac{2}{\pi} \cdot \frac{\pi}{2} = 1.<br />
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  3. #3
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    Ok yeh I understand this next part askes to write a PDE and BC's for the function

    \psi(r,\theta) = \phi(r,\theta) - \phi_0(\theta)

    Then using seperation of variables solve \psi (r,\theta) and hence write the solution \phi (r,\theta) of the original eqn/BC.

    Any help would be great.
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  4. #4
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    Quote Originally Posted by jezzyjez View Post
    Ok yeh I understand this next part askes to write a PDE and BC's for the function

    \psi(r,\theta) = \phi(r,\theta) - \phi_0(\theta)

    Then using seperation of variables solve \psi (r,\theta) and hence write the solution \phi (r,\theta) of the original eqn/BC.

    Any help would be great.
    THe idea here is to transform the two BC to have them both zero so a separation of variables will work. IF we let

     <br />
\phi = \psi + \frac{2}{\pi} \theta<br />

    the PDE becomes

    <br />
\frac{\partial^2 \psi}{\partial r^2} +\frac{1}{r}\frac{\partial\psi}{\partial r} + \frac{1}{r^2} \frac{\partial^2\psi}{\partial\theta^2} = 0<br />

    and the BCs

     <br />
\psi(r,0) = 0,\;\; \psi\left(r ,\frac{\pi}{2}\right) = 0<br />

    and the IC

    <br />
\psi(b,\theta) = \frac{2 \theta}{\pi} \left(1 - 2 \frac{\theta}{\pi}\right).

    Now separation of variables will work.
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