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Thread: PDE Separation Problem

  1. #1
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    PDE Separation Problem

    Hey I have been given the below question but am really struggling to get even the first part:

    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) State why the differential equation with the given boundary conditions can't be solved by the method of separating variables.


    The problem I am having is that I can separate the equation into to parts so I am assuming it is to do with the boundary conditions why it cannot be solved, but I don't know what aspect of the conditions effects this?
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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 even the first part:

    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) State why the differential equation with the given boundary conditions can't be solved by the method of separating variables.


    The problem I am having is that I can separate the equation into to parts so I am assuming it is to do with the boundary conditions why it cannot be solved, but I don't know what aspect of the conditions effects this?
    Let's suppose it can. So the solution would be of the form

    $\displaystyle \phi = R(r) T(\theta) $.

    The boundary conditions become

    $\displaystyle \phi(r,0) = R(r) T(0) = 0\;\;\; \Rightarrow\;\;\; T(0) = 0 $

    and

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

    but this can't happen since $\displaystyle R(r)$ varies.
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  3. #3
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    Quote Originally Posted by Danny View Post
    Let's suppose it can. So the solution would be of the form

    $\displaystyle \phi = R(r) T(\theta) $.

    The boundary conditions become

    $\displaystyle \phi(r,0) = R(r) T(0) = 0\;\;\; \Rightarrow\;\;\; T(0) = 0 $

    and

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

    but this can't happen since $\displaystyle R(r)$ varies.
    I still dont understand the boundary conditions you have mentioned, dont they just mean if $\displaystyle \theta = 0$ the potential is 0 and if $\displaystyle \theta = \frac{\pi}{2}$ potential is 1. surely the other boundary cond is the one I need to look at???

    **************EDIT 15:24*********************
    Last edited by jezzyjez; Apr 13th 2010 at 06:19 AM.
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  4. #4
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    Quote Originally Posted by jezzyjez View Post
    I still dont understand the boundary conditions you have mentioned, dont they just mean if $\displaystyle \theta = 0$ the potential is 0 and if $\displaystyle \theta = \frac{\pi}{2}$ potential is 0. surely the other boundary cond is the one I need to look at???
    But you said that $\displaystyle \phi = 1$ at $\displaystyle \theta = \frac{\pi}{2}$ not $\displaystyle \phi = 0.$
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  5. #5
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    Sorry thats what i meant but i still dont fully understand
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  6. #6
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    $\displaystyle
    \phi(r,\frac{\pi}{2}) = R(r) T(\frac{\pi}{2}) = 1
    $
    $\displaystyle
    R(r) = \frac{1}{T(\frac{\pi}{2})} = const
    $
    and
    $\displaystyle R(r) = const$ this is not possible
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