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Solving Poisson's equation

  1. Feb 19, 2007 #1

    I have a problem seeing how a solution is reached??

    The expression is poisson's equation for a positive charge which is debye shielded.

    [tex]\nabla . \psi = (2 / \lambda^2) \psi [/tex]

    The solution of which is..

    \psi = (Q / 4\pi\epsilon r )* exp(-2 \sqrt{r}/\lambda)

    I was hoping I could express this as a Legendre polynomial but alas, this is only for Laplace's equation it seems.

    How can I arrive at this solution?


    Last edited: Feb 19, 2007
  2. jcsd
  3. Feb 19, 2007 #2


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    Does [itex] \psi [/itex] depend upon [itex] \theta [/itex] and [itex]\varphi [/itex] ?
  4. Feb 20, 2007 #3
    No [tex] \theta [/tex] or [tex] \varphi [/tex] dependence.
  5. Feb 20, 2007 #4


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    Then the Poisson's eq becomes a second order ODE. Which can be solved.
  6. Feb 20, 2007 #5
    Oh :blushing: , yeah, I see that now...


  7. Feb 26, 2007 #6
    What am I missing here...

    The first equation expands out to...

    V'' + (2/r)V' - (2/c)V = 0

    Where I have replaced psi with V.

    My standard repository of all mathematical knowledge (Mary L Boas. Mathematical Methods in the Physical Sciences) only covers ODEs which have constant coefficients.

    I'm sure there must be a standard technique out there...
  8. Feb 26, 2007 #7
    Well, first thing's first, you have the wrong equation. You have "divergence of scalar equals scalar".

    As for your resulting ODE, expand [tex]\phi = \sum_j a_j r^j[/tex]. That should get you a nice recursion relation for the coefficients, and then you're golden.
    Last edited: Feb 26, 2007
  9. Feb 26, 2007 #8
    As StatMechGuy said, the equation given isn't Poisson (Laplacian of a function equals something that can only depend on position and time)... and it's weird since a scalar function doesn't have a divergence (did you mean gradient?).
  10. Feb 27, 2007 #9


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    First of all

    [tex] \nabla^2 \psi(r)=\frac{2}{\lambda^2}\psi(r) [/tex]

    is not Poisson equation, but a Helmholtz equation.

    It can be written

    [tex] \frac{1}{r}\frac{d^2}{dr^2}\left[r\psi(r)\right]=\frac{2}{\lambda^2}\psi(r) [/tex]

    [tex] \frac{d^2}{dr^2}\left[r\psi(r)\right]=\frac{2}{\lambda^2}\left[r\psi(r)\right] [/tex]

    Can you solve it now ?
  11. Mar 6, 2007 #10
    Ok I got it (said I.... after a little delay).

    Thanks Dextercioby, StatMechGuy and Ahmes, I think thats the first time I've used a series solution in anger.

    Oh and yes... I initially wrote the equation down incorrectly.. well done to those who spotted it (sorry).
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