- #1
bdforbes
- 149
- 0
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:
<br /> \begin{align*}<br /> &=\int\mathcal{F}\left\{\frac{e^{ikr}}{r}\right\}u(\vec{\xi})d\vec{\xi}\\<br /> &\equiv\int \frac{e^{ikr}}{r}\mathcal{F}\left\{u(\vec{\xi})\right\}d\vec{x}\\<br /> &=\int \frac{e^{ikr}}{r}\int u(\vec{\xi})e^{-i\vec{\xi}\cdot\vec{x}}d\vec{\xi}d\vec{x}<br /> \end{align*}<br />
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?
(\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:
<br /> \begin{align*}<br /> &=\int\mathcal{F}\left\{\frac{e^{ikr}}{r}\right\}u(\vec{\xi})d\vec{\xi}\\<br /> &\equiv\int \frac{e^{ikr}}{r}\mathcal{F}\left\{u(\vec{\xi})\right\}d\vec{x}\\<br /> &=\int \frac{e^{ikr}}{r}\int u(\vec{\xi})e^{-i\vec{\xi}\cdot\vec{x}}d\vec{\xi}d\vec{x}<br /> \end{align*}<br />
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?