[Click here for a PDF version of this post, and the previous posts in this series.]
We used a complicated limiting argument to show that the \( \mathrm{sgn}(x – x’) \) factor in the contour integral derivation of the Helmholtz operator Green’s function was wrong.
Having discovered, even if slightly by accident, what the correct form of that Green’s function is, we can check it more directly. This time, we use the Heaviside theta technique that we used to verify the 1D Laplacian Green’s function.
The goal is to show that
\begin{equation}\label{eqn:helmholtzGreens:2060}
\lr{ \spacegrad^2 + k^2 } G(x, x’) = \delta(x – x’),
\end{equation}
where
\begin{equation}\label{eqn:helmholtzGreens:2080}
G(x, x’) = \frac{e^{j k \Abs{x – x’}}}{2 j k}.
\end{equation}
Let’s start with an \( r = x – x’ \) change of variables, for which
\begin{equation}\label{eqn:helmholtzGreens:2100}
\frac{d}{dx} = \frac{dr}{dx} \frac{d}{dr} = \frac{d}{dr}.
\end{equation}
This means that
\begin{equation}\label{eqn:helmholtzGreens:2120}
\spacegrad^2 e^{j k \Abs{x – x’}} = \frac{d^2}{dr^2} e{j k \Abs{r}}
\end{equation}
Starting with the first derivative we have
\begin{equation}\label{eqn:helmholtzGreens:2140}
\begin{aligned}
\frac{d}{dr} e^{j k \Abs{r} }
&=
j k e^{j k \Abs{r} } \frac{d\Abs{r}}{dr} \\
&=
j k e^{j k \Abs{r} } \frac{d}{dr} \lr{ r \Theta(r) – r \Theta(-r) } \\
&=
j k e^{j k \Abs{r} } \lr{ \Theta(r) – \Theta(-r) + 2 r \delta(r) } \\
&=
j k e^{j k \Abs{r} } \lr{ \Theta(r) – \Theta(-r) } \\
&=
j k e^{j k \Abs{r} } \mathrm{sgn}(r).
\end{aligned}
\end{equation}
Using that Heaviside representation of the sign function we have \( sgn(r)’ = 2 \delta(r) \), so
\begin{equation}\label{eqn:helmholtzGreens:2160}
\frac{d^2}{dr^2} e{j k \Abs{r}}
=
\lr{ j k \mathrm{sgn}(r) }^2 e^{j k \Abs{r} } + 2 j k e^{j k \Abs{r} } \delta(r).
\end{equation}
We can identify \( e^{j k \Abs{r} } \delta(r) = \delta(r) \), just as we identified \( r \delta(r) = 0 \), by application to a test function. That is
\begin{equation}\label{eqn:helmholtzGreens:2180}
\begin{aligned}
\int e^{j k \Abs{r} } \delta(r) f(r) dr
&=
\evalbar{e^{j k \Abs{r} } f(r)}{r = 0} \\
&=
f(0) \\
&=
\int \delta(r) f(r) dr.
\end{aligned}
\end{equation}
With that identification
\begin{equation}\label{eqn:helmholtzGreens:2200}
\spacegrad^2 e^{j k \Abs{r} } = -k^2 e^{j k \Abs{r} } + 2 j k \delta(r),
\end{equation}
or
\begin{equation}\label{eqn:helmholtzGreens:2220}
\boxed{
\lr{ \spacegrad^2 + k^2 } \frac{e^{j k \Abs{x – x’} }}{2 j k} = \delta(x – x’).
}
\end{equation}