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Thread: 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 $\displaystyle \phi(r,\theta)$ in the quadrant.

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

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

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

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

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

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

    (a) show that the function $\displaystyle \phi(\theta) = \frac{2\theta}{\pi}$ 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 $\displaystyle \phi(r,\theta)$ in the quadrant.

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

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

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

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

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

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

    (a) show that the function $\displaystyle \phi(\theta) = \frac{2\theta}{\pi}$ 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 $\displaystyle \phi(r,\theta) = \frac{2 \theta}{\pi}$ then

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

    $\displaystyle
    \phi(r,\frac{\pi}{2}) = \frac{2}{\pi} \cdot \frac{\pi}{2} = 1.
    $
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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

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

    Then using seperation of variables solve $\displaystyle \psi (r,\theta)$ and hence write the solution $\displaystyle \phi (r,\theta)$ 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

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

    Then using seperation of variables solve $\displaystyle \psi (r,\theta)$ and hence write the solution $\displaystyle \phi (r,\theta)$ 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

    $\displaystyle
    \phi = \psi + \frac{2}{\pi} \theta
    $

    the PDE becomes

    $\displaystyle
    \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
    $

    and the BCs

    $\displaystyle
    \psi(r,0) = 0,\;\; \psi\left(r ,\frac{\pi}{2}\right) = 0
    $

    and the IC

    $\displaystyle
    \psi(b,\theta) = \frac{2 \theta}{\pi} \left(1 - 2 \frac{\theta}{\pi}\right).$

    Now separation of variables will work.
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