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While it was difficult to attempt to verify the 2D Green’s function, it actually turns out to be fairly easy to derive it, provided we pick an alternate pole displacement from the 1D evaluation to make our lives easier.
With \( \Br = \Bx – \Bx’ \), and \( \tau = t – t’ \), and \( \epsilon > 0 \), we can form
\begin{equation}\label{eqn:waveEquationGreens:1340}
G_\epsilon(\Br, \tau) = \frac{c^2}{\lr{2 \pi}^3} \int d^2 \Bk d\omega \frac{ e^{j \Bk \cdot \Br + j \omega \tau}}{\lr{\omega -j \epsilon}^2 – \Bk^2 c^2 }
\end{equation}
This pole displacement has the nice property that both poles live in the upper half plane, so for \( \tau > 0 \), we have
\begin{equation}\label{eqn:waveEquationGreens:1360}
\begin{aligned}
G_\epsilon(\Br, \tau)
&= \frac{c^2}{\lr{2 \pi}^3} \int d^2 \Bk d\omega \frac{ e^{j \Bk \cdot \Br + j \omega \tau}}{
\lr{\omega -\lr{ \Abs{\Bk} c – j \epsilon}}
\lr{\omega -\lr{ -\Abs{\Bk} c – j \epsilon}}
} \\
&=
\Theta(\tau) \frac{c^2 j}{\lr{2 \pi}^2} \int d^2 \Bk e^{j \Bk \cdot \Br}
\lr{
\evalbar{
\frac{e^{ j \omega \tau}}{ \lr{\omega -\lr{ -\Abs{\Bk} c – j \epsilon}} }
}
{\omega = \Abs{\Bk} c – j \epsilon}
+
\evalbar{
\frac{e^{ j \omega \tau}}{ \lr{\omega -\lr{ \Abs{\Bk} c – j \epsilon}} }
}
{\omega = -\Abs{\Bk} c – j \epsilon}
}
\\
&=
\Theta(\tau) \frac{c^2 j}{\lr{2 \pi}^2} \int d^2 \Bk e^{j \Bk \cdot \Br} e^{ -j \epsilon \tau }
\lr{
\frac{e^{ j \Abs{\Bk} c \tau}}{ 2 \Abs{\Bk} c }
+
\frac{e^{ -j \Abs{\Bk} c \tau}}{ -2 \Abs{\Bk} c }
} \\
&=
\Theta(\tau) \frac{j^2 c}{\lr{2 \pi}^2} \int_{k=0}^\infty k dk \int_{\phi=0}^{2 \pi} d\phi e^{j k\Abs{\Br} \cos\phi } e^{ -j \epsilon \tau } \frac{\sin\lr{ k c \tau }}{k}.
\end{aligned}
\end{equation}
We’ve now successfully removed the singularity, and can evaluate the \(\epsilon \rightarrow 0 \) limit. We may also evaluate the \( \phi \) integral, remembering that
\begin{equation}\label{eqn:waveEquationGreens:1380}
\int_0^{2 \pi} e^{j \Abs{a} \cos\phi} d\phi = 2 \pi J_0(\Abs{a}),
\end{equation}
to find
\begin{equation}\label{eqn:waveEquationGreens:1400}
G(\Br, \tau) = -\Theta(\tau) \frac{c}{2 \pi} \int_{k=0}^\infty dk J_0(k\Abs{\Br}) \sin\lr{ k c \tau }.
\end{equation}
This integral yields easily to Mathematica, and we find
\begin{equation}\label{eqn:waveEquationGreens:1420}
G(\Br, \tau) = -\Theta(\tau) \frac{c}{2 \pi} \frac{\Theta(c \tau – \Abs{\Br})}{\sqrt{(c\tau)^2 – \Br^2}}.
\end{equation}
However, since \( \Theta(c \tau – \Abs{\Br}) = 1 \) only for \( \tau > \Abs{\Br}/c \), the \( \Theta(\tau) \) factor is redundant, and we find
\begin{equation}\label{eqn:waveEquationGreens:1440}
\boxed{
G(\Br, \tau) = – \frac{1}{2 \pi} \frac{\Theta(c \tau – \Abs{\Br})}{\sqrt{\tau^2 – \Br^2/c^2}},
}
\end{equation}
which matches the retarded Green’s function claimed by Grok.
Repeating this analysis for \( \tau < 0, \epsilon < 0 \), we find
\begin{equation}\label{eqn:waveEquationGreens:1460}
G(\Br, \tau) = -\Theta(-\tau) \frac{c}{2 \pi} \frac{\Theta(-c \tau – \Abs{\Br})}{\sqrt{(c\tau)^2 – \Br^2}},
\end{equation}
which we also see matches the Grok result for the advanced Green’s function. Both of these computations can be trivially performed in Mathematica following the same steps (taking all the fun from the story.) The advanced integral evaluation is shown in fig. 1 as an example.
fig. 1. Advanced 2D Green’s function for wave equation operator.
