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Fourier transform of Green's function

  1. Sep 1, 2009 #1
    By taking the Fourier transform of the fundamental Helmholtz equation


    one finds that




    However, I can't figure out how to directly confirm that this Fourier transform pair is correct. I tried directly transforming [itex]e^{ikr}/r[/itex] as if it were a regular function, but I ended up with something which doesn't converge:

    [tex]\frac{4\pi}{\xi}\int\limits_0^\infty e^{ikr}sin(\xi r)dr[/tex]

    I didn't really expect that to work, since distributions are involved. So I tried doing it with distributions:

    &\equiv\int \frac{e^{ikr}}{r}\mathcal{F}\left\{u(\vec{\xi})\right\}d\vec{x}\\
    &=\int \frac{e^{ikr}}{r}\int u(\vec{\xi})e^{-i\vec{\xi}\cdot\vec{x}}d\vec{\xi}d\vec{x}

    At this point I'm thinking about using a Gaussian convergence factor, but I'm not sure exactly how to do it. Can anyone help out?
  2. jcsd
  3. Sep 3, 2009 #2


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    It turns out that you can invert the Fourier transform of your Green's function without extra convergence factors or the theory of distributions. This is a pretty standard problem you will find in some textbooks - I don't have any of them handy right now as I am on travel, but the trick is that you can argue that you can perform a coordinate rotation int he Fourier domain to make inverting the transforms easier. I'm pretty sure that "Functions of a complex variable" by Carrier et al. has this, either in the text or as a problem (most likely as a problem, since almost everything in that book is in a problem :).

    Good luck.

  4. Sep 3, 2009 #3
    Thanks for your reply JasonRF. I am actually specifically trying to forward transform the Green's function, as opposed to inverting the transform, which I am relatively comfortable with. Does your method still apply? I will try to get a hold of that text.
  5. Sep 6, 2009 #4
    Is it possible that this Fourier transform is defined purely through the Helmholtz equation? I.e. no definition of a Fourier transform can be directly applied to the function itself?
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