Results 1 to 6 of 6

Math Help - PDE Separation Problem

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    30

    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 \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) 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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    56
    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 \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) 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

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

    The boundary conditions become

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

    and

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

    but this can't happen since R(r) varies.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2009
    Posts
    30
    Quote Originally Posted by Danny View Post
    Let's suppose it can. So the solution would be of the form

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

    The boundary conditions become

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

    and

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

    but this can't happen since R(r) varies.
    I still dont understand the boundary conditions you have mentioned, dont they just mean if \theta = 0 the potential is 0 and if \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; April 13th 2010 at 07:19 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    56
    Quote Originally Posted by jezzyjez View Post
    I still dont understand the boundary conditions you have mentioned, dont they just mean if \theta = 0 the potential is 0 and if \theta = \frac{\pi}{2} potential is 0. surely the other boundary cond is the one I need to look at???
    But you said that \phi = 1 at \theta = \frac{\pi}{2} not \phi = 0.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Dec 2009
    Posts
    30
    Sorry thats what i meant but i still dont fully understand
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    <br />
\phi(r,\frac{\pi}{2}) = R(r) T(\frac{\pi}{2}) = 1<br />
    <br />
R(r) = \frac{1}{T(\frac{\pi}{2})} = const<br />
    and
    R(r) = const this is not possible
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Axiom of Separation Problem
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: August 13th 2011, 08:41 AM
  2. simple integral separation problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 19th 2011, 12:59 AM
  3. Having problem with this separation
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: February 5th 2011, 11:54 AM
  4. Simple Separation of Variable Problem
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: January 22nd 2011, 01:19 PM
  5. Separation of Variables Problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 26th 2010, 06:06 AM

Search Tags


/mathhelpforum @mathhelpforum