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I was questioning the correctness of the 1D and 2D Helmholtz Green’s functions derived above, since they are both seemingly malformed for \( k \rightarrow 0 \).
Let’s try to carefully expand the 2D Green’s function in the neighbourhood of \( k = 0 \) to validate that result, and as a side effect, obtain the Green’s function for the 2D Laplacian.
In [1], section 9.1.7, 9.1.8, we have for \( z \rightarrow 0 \)
\begin{equation}\label{eqn:helmholtzGreens:980}
\begin{aligned}
J_\nu(z) &\sim \lr{\frac{z}{2}}^\nu/\Gamma(\nu+1) \\
Y_0(z) &\sim \frac{2}{\pi} \ln z,
\end{aligned}
\end{equation}
so for \( k \ll r \)
\begin{equation}\label{eqn:helmholtzGreens:1000}
H_0^{(1)}(k r) \sim 1 + \frac{2j}{\pi} \ln\lr{k r},
\end{equation}
and
\begin{equation}\label{eqn:helmholtzGreens:1020}
\begin{aligned}
G(\Br)
&\sim -\frac{j}{4} \lr{ 1 + \frac{j}{2 \pi} \ln\lr{k r} } \\
&= -\frac{j}{4} + \frac{1}{2 \pi} \ln k + \frac{1}{2 \pi} \ln r.
\end{aligned}
\end{equation}
Here is where we have to get sneaky. Since we seek a Green’s function for the Laplacian operator, we are free to add any solution \( f(x,y) \) that satisfies \( \spacegrad^2 f = 0 \). Constants are clearly in that homogeneous solution space, so we may adjust this expansion of the Green’s function, throwing away the leading constant imaginary term, and treating \( k \) as a small constant, the \( \ln k \) term. That leaves us with
\begin{equation}\label{eqn:helmholtzGreens:1040}\boxed{
G(\Bx, \Bx’) = \frac{1}{2 \pi} \ln \Abs{\Bx – \Bx’}.
}
\end{equation}
Verifying the Laplacian Green’s function.
Let’s try to verify that this Green’s function is correct, since we’ve had lots of opportunities to screw up signs. We want to evaluate the Laplacian of the convolution and if all goes well, it should be \( V(\Bx) \). That is
\begin{equation}\label{eqn:helmholtzGreens:1060}
\begin{aligned}
\spacegrad^2 \int G(\Bx, \Bx’) V(\Bx’) d^2 \Bx’
&= \inv{2 \pi} \int \spacegrad^2 \ln \Abs{\Bx – \Bx’} V(\Bx’) d^2 \Bx’ \\
&= \inv{2 \pi}
\int V(\Bx’) \lr{\spacegrad’}^2 \ln \Abs{\Bx – \Bx’} d^2 \Bx’ \\
\end{aligned}
\end{equation}
We can verify that \( \lr{\spacegrad’}^2 \ln \Abs{\Bx – \Bx’} \) is zero whenever \( \Abs{\Bx – \Bx’} \ne 0 \). A nice way of doing that is in polar coordinates. Write
\begin{equation}\label{eqn:helmholtzGreens:1160}
\begin{aligned}
\Br &= \Bx’ – \Bx \\
r &= \Abs{r} \\
\end{aligned}
\end{equation}
and recall that
\begin{equation}\label{eqn:helmholtzGreens:1180}
\spacegrad^2 f = \inv{r} \PD{r}{} \lr{ r \PD{r}{f} } + \frac{\partial^2 f}{\partial \theta^2},
\end{equation}
but \( r \PD{r}{\ln r} = 1 \), and \( \ln r \) has no angular dependence. That means that
\begin{equation}\label{eqn:helmholtzGreens:1200}
\spacegrad^2 \int G(\Bx, \Bx’) V(\Bx’) d^2 \Bx’
= \inv{2 \pi} \int_{\Abs{\Bx – \Bx’} < \epsilon} V(\Bx’) \lr{\spacegrad’}^2 \ln \Abs{\Bx – \Bx’} d^2 \Bx’,
\end{equation}
where we let \( \epsilon \rightarrow 0 \). Such a region is illustrated in fig. 7.
fig. 7. Neighborhood around x
We can now apply Green’s theorem, which for 2D is
\begin{equation}\label{eqn:helmholtzGreens:1080}
\int_A \lr{ u \spacegrad^2 v – v \spacegrad^2 u } dA = \int_{\partial A} \lr{ u \spacegrad v – v \spacegrad u } \cdot \mathbf{\hat{n}} \, dS,
\end{equation}
or
\begin{equation}\label{eqn:helmholtzGreens:1090}
\int_A v \spacegrad^2 u \, dA = \int_A u \spacegrad^2 v \, dA -\int_{\partial A} \lr{ u \spacegrad v – v \spacegrad u } \cdot \mathbf{\hat{n}} \, dS,
\end{equation}
With
\begin{equation}\label{eqn:helmholtzGreens:1161}
\begin{aligned}
\mathbf{\hat{r}} &= \Br/r = \mathbf{\hat{n}} \\
u &= \ln r \\
v &= V(\Bx’) \\
dA &= r dr d\theta \\
dS &= r d\theta,
\end{aligned}
\end{equation}
we have
\begin{equation}\label{eqn:helmholtzGreens:1100}
\begin{aligned}
\int V(\Bx’) \lr{ \lr{\spacegrad’}^2 \ln \Abs{\Bx – \Bx’} } d^2 \Bx’
&=
\int_{r=0}^\epsilon r dr d\theta \ln r \lr{ \spacegrad’}^2 V(\Bx’) \\
&\quad-
\int_{\theta = 0}^{2 \pi}
\epsilon d\theta
\evalbar{
\lr{
\ln \epsilon \spacegrad’ V(\Bx’) – V(\Bx’) \spacegrad’ \ln r
}
\cdot \mathbf{\hat{r}}
}
{\,r = \epsilon}
\end{aligned}
\end{equation}
We have \( r \ln r \), or \( \epsilon \ln \epsilon \) dependence in two of the integrand terms, and with \( r < \epsilon \), and \( \epsilon \ln \epsilon \rightarrow 0 \), in the limit, we are left with
\begin{equation}\label{eqn:helmholtzGreens:1120}
\begin{aligned}
\int V(\Bx’) \lr{\spacegrad’}^2 \ln \Abs{\Bx – \Bx’} d^2 \Bx’
&=
\int_{\theta = 0}^{2 \pi} \epsilon \evalbar{ d\theta V(\Bx’) \lr{ \spacegrad’ \ln r } \cdot \mathbf{\hat{r}} }{\,r = \epsilon} \\
&=
\int_{\theta = 0}^{2 \pi}
\epsilon
\evalbar{
d\theta V(\Bx’) \lr{ \lr{ \mathbf{\hat{r}} \partial_r + \frac{\thetacap}{r^2} \partial_\theta } \ln r } \cdot \mathbf{\hat{r}}
}{\,r = \epsilon}
\\
&=
\int_{\theta = 0}^{2 \pi}
\epsilon
\evalbar{
d\theta V(\Bx’) \frac{\mathbf{\hat{r}}}{r} \cdot \mathbf{\hat{r}}
}{\,r = \epsilon}
\\
&=
\int_{\theta = 0}^{2 \pi} d\theta V(\Bx + \epsilon \mathbf{\hat{r}}).
\end{aligned}
\end{equation}
In the limit we are left with just \( 2 \pi V(\Bx) \), so
\begin{equation}\label{eqn:helmholtzGreens:1140}
\spacegrad^2 \int G(\Bx, \Bx’) V(\Bx’) d^2 \Bx’ = V(\Bx),
\end{equation}
as desired.
References
[1] M. Abramowitz and I.A. Stegun. Handbook of mathematical functions with formulas, graphs, and mathematical tables, volume 55. Dover publications, 1964.
