# Fourier transform of Green's function

## Main Question or Discussion Point

By taking the Fourier transform of the fundamental Helmholtz equation

$$(\nabla^2+k^2)G(\vec{x})=-\delta(\vec{x})$$,

one finds that

$$G(\vec{x})=\frac{e^{ikr}}{r}$$

and

$$\tilde{G}(\vec{\xi})=\frac{1}{k^2-\xi^2}$$.

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

$$\frac{4\pi}{\xi}\int\limits_0^\infty e^{ikr}sin(\xi r)dr$$

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

\begin{align*} &=\int\mathcal{F}\left\{\frac{e^{ikr}}{r}\right\}u(\vec{\xi})d\vec{\xi}\\ &\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} \end{align*}

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?

jasonRF
Gold Member
bdforbes,

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.

Jason

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.

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?