Solving the Poisson equation (second order ODE)

Posted on 2009-12-17
Last Modified: 2012-05-08
I need to solve Poisson's equation in spherical coordinates (which is modelling something related to plasmas). The equation is:

   1/r^2 d/dr(r^2 * dp(r)/dr) = 2p(r)/k^2

where k is a known constant. So the first thing seems to be to take the r^2 to the other side, and apply the product rule on the left, to get the second order ODE:

   2r p' + r^2 p'' = 2r^2 p/k^2


   p'' + (2/r)p' - (2/k^2)p = 0.                                                 (*)

How do I solve this? According to Wolfram, this has the general solution:

   p(r) = c1*exp(-sqrt(2)*r/k)/r + c2*exp(sqrt(2)*r/k)/(2*sqrt(2)*r/k)                           *EDITED*

which indeed leads to the correct solution when I apply some known conditions, but how can I derive this solution?

Thanks very much
Question by:InteractiveMind
    LVL 26

    Expert Comment

    "how can I derive this solution?"
    Differentiate it?
    LVL 25

    Author Comment

    Differentiate (*)? What then?
    LVL 26

    Expert Comment

    You said you thought you had the general solution
    " p(r) = c1*exp(-sqrt(2)*r/k)/r + c2*exp(sqrt(2)*r/k)/(2*sqrt(2)*r/k)"
    so if you differentiate it and get
    "  p'' + (2/r)p' - (2/k^2)p = 0.       "
    you thought correctly
    LVL 25

    Author Comment

    Sorry, no, I wanted to know if the general solution can be derived. I always hate it when we have to pull a solution out of the hat and justify it on the basis that it works (Ansatz).
    LVL 26

    Accepted Solution

    "I always hate it when we have to pull a solution out of the hat and justify it on the basis that it works (Ansatz)."
    I do too but it is often the easiest way. Integration is the inverse of differentiation and inverse operations are always harder. Consider long division, the inverse of multiplication. It is just an organized trial and error procedure. The same with integration. Poisson's equation can just be looked at as the result of an organized trial and error.
    In general physics SHM equation is usually just plopped down (and checked by differentiation).
    The whole practice of integration is the result of trail and errors. 1st order, 2nd order, homogeneous, non-homogeneous, separable, etc)
    You may be able to get somebody to help you further (but not me)'

    Note which might lead to something
    If the variables cannot be separated, the method of integral transforms might help.

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