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Whittaker's solution and separable variables

  1. Jan 11, 2016 #1
    So It is well known that the 2D solution to the Laplace equation can be obtained by changing to complex coordinates ##u=x+iy## and ##v=x-iy##. This can be extended to n dimensions as long as the complex coordinates chosen also solve the Laplace equation. For example in 3D

    ##w=i\sqrt{2}z+xcos\vartheta +iysin(\vartheta)##

    along with the complex conjugate of w. This allows the 3D laplace equation to be solved in the same way as the 2D case. I.e the solution of
    ##\nabla^{2}f=0## can be found through

    ##f=g(w)+h(\tilde{w})##

    A general solution to the Laplace equation is given by

    ##f=\int f(w,\vartheta)d\vartheta##

    this is called Whittaker's solution and from this we can obtain

    ##f=\frac{2\pi x}{(x^{2}+y^{2}+z^{2})^{3/2}}##

    which is a solution to the Laplace equation...


    Ok so my question is I am unsure how to show that they are equivalent?

    Whittaker's solution must be able to be expressed as

    ##f=\int f(w,\vartheta)d\vartheta=g(w)+h(\tilde{w})=\frac{2\pi x}{(x^{2}+y^{2}+z^{2})^{3/2}}##

    I am sure there is a simple explanation? probably involving a taylor expansion or something? but I am not sure how to get there...I believe these all have some context and relation to twistor theory and the like but before i delve cohmology etc I am hoping for an easier solution =]
    Matt
     
  2. jcsd
  3. Jan 16, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Jan 17, 2016 #3
    As far as I could figure out by reading some old old document Whittaker shewed that solutions to the 3d laplace equation can be put on the form ## \int_{0}^{2\pi} f(x + iy\cos(v) + iz\sin(v)) dv ##. By pure guessing I found that

    ## \int_{0}^{2\pi} f(x,y,z,v) dv = \frac{2\pi x }{\left(x^2 + y^2 + z^2\right)^{3/2}}## with ## f(x,y,z,v) = \left(x + iz\cos(v) + iy\sin(v)\right)^{-2}##. Now you just have to find g(w) and h(w), right?
     
  5. Jan 17, 2016 #4
    Pretty much, but where the solution to the Laplace equation is the addition of some function ##g(w)## and ##h(\tilde{w})## where ##\tilde{w}## is the complex conjugate of ##w##...
    The more I'm reading about it, it seems as if I need to look at twistor theory, global and local solutions etc...I wonder if take a laurent expansion would help, but I havent really been able to get very far
     
  6. Feb 24, 2016 #5
    Would you show me how you solved the integral? I appear to be having some trouble
     
  7. Feb 28, 2016 #6
    Maple syntax.

    f:=(x,y,z,v) -> (x + I*z*cos(v) + I*y*sin(v))^-2;
    int(f(x,y,z,v),v=0..2*Pi) assuming(x,real,y,real,z,real,v,real)
     
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