# Fourier transform of Green's function

1. Sep 1, 2009

### bdforbes

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?

2. Sep 3, 2009

### jasonRF

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

3. Sep 3, 2009

### bdforbes

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.

4. Sep 6, 2009

### bdforbes

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?