Month: September 2025

Correcting the errors: Green’s functions for the 1D Helmholtz and Laplacian operators.

September 28, 2025 math and physics play , , , , , , , , , , , , , , , ,

[Click here for a PDF version of this post, and others in this series]

The following recent posts explored 1D Green’s functions for the Helmholtz and Laplacian operators.  There was a sign error (wrong residue sign for a negatively oriented contour) that I made near the beginning that caused a lot of trouble.  Having found the error, I’ve now reworked all that exploratory content into a more coherent form.  That reworked content can be found in today’s blog post below (or in the PDF above, which includes all of this, plus the 2D and 3D derivations.)

Part 1. Green’s functions for the Helmholtz (wave equation) operator in various dimensions.

 

A trilogy in five+ parts: Confirming an error in the derived 1D Helmholtz Green’s function.

 

A trilogy in six+ parts: 1D Laplacian Green’s function

A trilogy in 7+ parts: A better check of the 1D Helmholtz Green’s function.

 

 

Helmholtz Green’s function in 1D.

Evaluating the integral.

For the one dimensional case, we want to evaluate
\begin{equation}\label{eqn:helmholtzGreensV2:200}
G(r) = -\inv{2 \pi} \int \inv{p^2 – k^2} e^{j p r} dp,
\end{equation}
an integral which is unfortunately non-convergent. Since we are dealing with delta functions, it is not surprising that we have convergence problems. The technique used in the book is to displace the pole slightly by a small imaginary amount, and then take the limit.

That is
\begin{equation}\label{eqn:helmholtzGreensV2:220}
G(r) = \lim_{\epsilon \rightarrow 0} G_\epsilon(r),
\end{equation}
where
\begin{equation}\label{eqn:helmholtzGreensV2:240}
\begin{aligned}
G_\epsilon(r)
&= -\inv{2 \pi} \int_{-\infty}^\infty \inv{p^2 – \lr{k + j \epsilon}^2} e^{j p r} dp \\
&= -\inv{2 \pi} \int_{-\infty}^\infty \frac{e^{j p r}}{\lr{ p – k -j \epsilon}\lr{p + k + j \epsilon}} dp.
\end{aligned}
\end{equation}
Our contours, for \( \epsilon > 0 \), are illustrated in fig 1.

fig 1. Contours for 3D Green’s function evaluation

For \( r > 0 \) we can use an upper half plane infinite semicircular contour integral, with \( R \rightarrow \infty \).

The residue calculation for this contour gives
\begin{equation}\label{eqn:helmholtzGreensV2:260}
\begin{aligned}
G_\epsilon(r)
&= -\frac{(+2 \pi j)}{2 \pi} \evalbar{\frac{e^{j p r}}{p + k + j \epsilon} }{p = k + j \epsilon} \\
&= -j \frac{e^{j \lr{k + j \epsilon} r}}{2\lr{k + j \epsilon}} \\
&= -j \frac{e^{j k r} e^{-\epsilon r}}{2\lr{k + j \epsilon}} \\
&\rightarrow -\frac{j}{2k} e^{j k r}.
\end{aligned}
\end{equation}

For \( r < 0 \) we use the lower half plane infinite semicircular contour For this contour, we find \begin{equation}\label{eqn:helmholtzGreensV2:280} \begin{aligned} G_\epsilon(r) &= -\frac{2 \pi j}{(-2 \pi)} \evalbar{\frac{e^{j p r}}{p – k – j \epsilon} }{p = -k – j \epsilon} \\ &= -j \frac{e^{-j \lr{k + j \epsilon} r}}{2\lr{k + j \epsilon}} \\ &= -j \frac{e^{-j k r} e^{\epsilon r}}{2\lr{k + j \epsilon}} \\ &\rightarrow -\frac{j}{2k} e^{-j k r}. \end{aligned} \end{equation} Combining both results, our Green’s function, after a positive pole displacement \( \epsilon > 0 \), is
\begin{equation}\label{eqn:helmholtzGreensV2:300}
G(r) = \frac{1}{2 j k} e^{j k \Abs{r}}.
\end{equation}

Similarly, should we pick \( \epsilon < 0 \), the same sort of calculation yields an incoming wave solution \begin{equation}\label{eqn:helmholtzGreensV2:2240} G(r) = -\frac{1}{2 j k} e^{-j k \Abs{r}}. \end{equation} Allowing for either, we have Green’s functions for both the incoming and outgoing wave cases \begin{equation}\label{eqn:helmholtzGreensV2:2260} \boxed{ G_{\pm}(x – x’) = \pm \frac{1}{2 j k} e^{ \pm j k \Abs{x – x’}}. } \end{equation} With two Green’s functions, we can also make a linear combination. Specifically \begin{equation}\label{eqn:helmholtzGreensV2:2280} \begin{aligned} G(r) &= \inv{2}\lr{ G_{+}(r) + G_{-}(r) } \\ &= \inv{4 j k}\lr{ e^{ j k \Abs{r}} – e^{ – j k \Abs{r}} } \\ &= \inv{2 k} \sin\lr{ k \Abs{r} } \end{aligned} \end{equation} This real valued Green’s function is plotted in fig. 2.

fig. 2. Green’s function for 1D Helmholtz operator.

The convolution is now fully specified, providing a specific solution to the non-homogeneous equation \begin{equation}\label{eqn:helmholtzGreensV2:400} U(x) = \inv{2k} \int_{-\infty}^\infty \sin( k\Abs{r} ) V(x + r) dr. \end{equation}

The general solution may also include any solutions to the homogeneous Helmholtz equation \begin{equation}\label{eqn:helmholtzGreensV2:2300} U(x) = A e^{j k x} + B e^{-j k x} + \inv{2k} \int_{-\infty}^\infty \sin( k\Abs{r} ) V(x + r) dr. \end{equation}

A strictly causal solution.

We can split the convolution kernel a “causal” part, where only the spatially-“past” values of \( V \) contribute, and an “acausal” part \begin{equation}\label{eqn:helmholtzGreensV2:2320} U(x) = \inv{2k} \int_{-\infty}^0 \sin( k\Abs{r} ) V(x + r) dr + \inv{2k} \int_{0}^\infty \sin( k\Abs{r} ) V(x + r) dr. \end{equation} In a sense, we are averaging causal and acausal portions of the convolution. Suppose that we form a convolution with a built in cut-off, so that values of \( V(x’), x’ > x \) do not contribute to \( U(x) \). That is
\begin{equation}\label{eqn:helmholtzGreensV2:440}
f(x) = \int_{-\infty}^x \frac{\sin\lr{k \lr{x – x’}}}{k} V(x’) dx’.
\end{equation}
Here the one-half factor has been dropped, since we are no longer performing a QFT like average of causal and acausal terms.

Intuition suggests this should be a solution to the Helmholtz equation, but let’s test that guess. We start with the identity
\begin{equation}\label{eqn:helmholtzGreensV2:460}
\frac{d}{dx} \int_a^x g(x, x’) dx’
=
\evalbar{g(x, x’) }{x’ = x} + \int_a^x \frac{\partial g(x,x’)}{dx} dx’.
\end{equation}
Taking the first derivative of \( f(x) \), we find
\begin{equation}\label{eqn:helmholtzGreensV2:480}
\begin{aligned}
\frac{df}{dx}
&= \evalbar{ \frac{\sin\lr{k \lr{x – x’}}}{k} V(x’) }{x’ = x} + k \int_{-\infty}^x \frac{\cos\lr{k \lr{x – x’}}}{k} V(x’) dx’ \\
&= k \int_{-\infty}^x \frac{\cos\lr{k \lr{x – x’}}}{k} V(x’) dx’,
\end{aligned}
\end{equation}
where we have, somewhat lazily, treated the infinite limit as a constant. Effectively, this requires that the forcing function \( V(x) \) is zero at \( -\infty \). Taking the second derivative, we have
\begin{equation}\label{eqn:helmholtzGreensV2:500}
\begin{aligned}
\frac{d^2f}{dx^2}
&=
\evalbar{ k \frac{\cos\lr{k \lr{x – x’}}}{k} V(x’) }{x’ = x} – k^2 \int_{-\infty}^x \frac{\sin\lr{k \lr{x – x’}}}{k} V(x’) dx’ \\
&= V(x) – k^2 f(x),
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:helmholtzGreensV2:520}
\frac{d^2}{dx^2} f(x) + k^2 f(x) = V(x).
\end{equation}
This verifies that \ref{eqn:helmholtzGreensV2:440} is also a specific solution to the wave equation, as expected and desired.

It appears that the general solution is likely of the following form
\begin{equation}\label{eqn:helmholtzGreensV2:380b}
U(x) =
A’ \cos\lr{ k x } +
B’ \sin\lr{ k x }
+\alpha \int_{-\infty}^x \frac{\sin\lr{k \lr{x – x’}}}{k} V(x’) dx’
+(1-\alpha)\int_x^\infty \frac{\sin\lr{k \lr{x – x’}}}{k} V(x’) dx’,
\end{equation}
with \( \alpha \in [0,1] \).

It’s pretty cool that we can completely solve the 1D forced wave equation, for any forcing function, from first principles. Yes, I took liberties that would make a mathematician cringe, but we are telling a story, and leaving the footnotes to somebody else.

Verification of the 1D Helmholtz Green’s function.

Let’s show that the outgoing Green’s function has the desired delta function semantics. That is
\begin{equation}\label{eqn:helmholtzGreensV2:2060}
\lr{ \spacegrad^2 + k^2 } G(x, x’) = \delta(x – x’),
\end{equation}
where
\begin{equation}\label{eqn:helmholtzGreensV2:2080}
G(x, x’) = \frac{e^{j k \Abs{x – x’}}}{2 j k}.
\end{equation}
Making a \( r = x – x’ \) change of variables gives
\begin{equation}\label{eqn:helmholtzGreensV2:2120}
\spacegrad^2 e^{j k \Abs{x – x’}} = \frac{d^2}{dr^2} e^{j k \Abs{r}}
\end{equation}

The function \( \Abs{r} \) formally has no derivative at the origin, but we may use the physics trick, rewriting the absolute in terms of the Heaviside theta function
\begin{equation}\label{eqn:helmholtzGreensV2:2340}
\Abs{r} = r \Theta(r) – r \Theta(-r).
\end{equation}
We then use the delta function identification for the derivative
\begin{equation}\label{eqn:helmholtzGreensV2:2360}
\Theta'(r) = \delta(r).
\end{equation}
In particular
\begin{equation}\label{eqn:helmholtzGreensV2:2380}
\begin{aligned}
\frac{d}{dr} \Abs{r}
&= \Theta(r) – \Theta(-r) + r \delta(r) – (-1)\delta(-r) \\
&= \Theta(r) – \Theta(-r) + 2 r \delta(r),
\end{aligned}
\end{equation}
using the symmetric property of the delta function \( \delta(-r) = \delta(r) \). The delta function contribution to this derivative is actually zero, as seen when we operate with \( r \delta(r) \) against a test function
\begin{equation}\label{eqn:helmholtzGreensV2:2400}
\begin{aligned}
\int r \delta(r) f(r) dr
&=
\evalbar{r f(r)}{r = 0} \\
&= 0.
\end{aligned}
\end{equation}
We’ve now found that
\begin{equation}\label{eqn:helmholtzGreensV2:2420}
\frac{d}{dr} \Abs{r} = \Theta(r) – \Theta(-r) = \mathrm{sgn}(r),
\end{equation}
the sign function. The derivative of the sign function is
\begin{equation}\label{eqn:helmholtzGreensV2:2440}
\begin{aligned}
\mathrm{sgn}'(r)
&= \lr{ \Theta(r) – \Theta(-r) }’ \\
&= \delta(r) -(-1)\delta(-r) \\
&= 2 \delta(r).
\end{aligned}
\end{equation}

The derivatives are
\begin{equation}\label{eqn:helmholtzGreensV2:2140}
\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} } \mathrm{sgn}(r).
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:helmholtzGreensV2:2160}
\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:helmholtzGreensV2:2180}
\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:helmholtzGreensV2:2200}
\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:helmholtzGreensV2:2220}
\boxed{
\lr{ \spacegrad^2 + k^2 } \frac{e^{j k \Abs{x – x’} }}{2 j k} = \delta(x – x’).
}
\end{equation}

Verifying the Green’s function with convolution.

Avoiding the physics tricks, we may use a limiting argument to validate our Green’s function.

We first want to show that at points \( x’ \ne x \) the Helmholtz operator applied to the Green’s function is zero:
\begin{equation}\label{eqn:helmholtzGreensV2:1220}
\lr{ \spacegrad^2 + k^2} G(x,x’) = 0,
\end{equation}
Since we are avoiding the origin
\begin{equation}\label{eqn:helmholtzGreensV2:1240}
\lr{ k^2 + \frac{d^2}{dx^2} } e^{j k \Abs{x – x’}}.
\end{equation}
We expect that this will be zero. Making a change of variables \( r = x’ – x \), we want to evaluate
\begin{equation}\label{eqn:helmholtzGreensV2:1260}
\lr{ k^2 + \frac{d^2}{dr^2} } e^{j k \Abs{r}},
\end{equation}
assuming that we are omitting a neighbourhood around \( r = 0 \) where the absolute value causes trouble. For \( r > 0 \)
\begin{equation}\label{eqn:helmholtzGreensV2:1280}
\begin{aligned}
\frac{d}{dr} e^{j k \Abs{r}}
&= \frac{d}{dr} e^{j k r} \\
&= j k e^{j k r},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:helmholtzGreensV2:1300}
\begin{aligned}
\frac{d^2}{dr^2} e^{j k \Abs{r}}
&= \frac{d}{dr} j k e^{j k r} \\
&= (j k)^2 e^{j k r}.
\end{aligned}
\end{equation}
Similarly, for \( r < 0 \), we have
\begin{equation}\label{eqn:helmholtzGreensV2:1320}
\frac{d^2}{dr^2} e^{j k \Abs{r}} = (-jk)^2 e^{j k \Abs{r}}.
\end{equation}
In both cases, provided we are in a neighbourhood that omits \( r \ne 0 \), we have
\begin{equation}\label{eqn:helmholtzGreensV2:1340}
\lr{ k^2 + \frac{d^2}{dr^2} } e^{j k \Abs{r}} = 0,
\end{equation}
as desired.

The takeaway is that we have
\begin{equation}\label{eqn:helmholtzGreensV2:1360}
\begin{aligned}
\lr{ \spacegrad^2 + k^2 }\int_{-\infty}^\infty G(x, x’) V(x’) dx’
&=
\int_{\Abs{x’ – x} \le \epsilon} V(x’) \lr{ \frac{d^2}{d{x’}^2} + k^2 } G(x, x’) dx’ \\
&=
\int_{-\epsilon}^\epsilon V(x + r) \lr{ \frac{d^2}{dr^2} + k^2 } G(r) dr \\
\end{aligned}
\end{equation}
for some arbitrarily small value of \( \epsilon \). Observe that after bringing the operator into the integral, we also made a change of variables, first to \( x’ \) for the Laplacian, and then to \( r = x’ – x \).

We’d like the 1D equivalent of Green’s theorem to reduce this, so let’s work that out first.
\begin{equation}\label{eqn:helmholtzGreensV2:1380}
\begin{aligned}
\int dx\, v \frac{d^2 u}{dx^2} – \int dx\, u \frac{d^2 v}{dx^2}
&=
\int dx\,
\lr{
\frac{d}{dx} \lr{
v \frac{du}{dx}
}
– \frac{dv}{dx} \frac{du}{dx}
}
–
\int dx\,
\lr{
\frac{d}{dx} \lr{
u \frac{dv}{dx}
}
–
\frac{du}{dx} \frac{dv}{dx}
}
\\
&=
\int dx\,
\frac{d}{dx}
\lr{
v \frac{du}{dx}
}
–
\int dx\,
\frac{d}{dx} \lr{
u \frac{dv}{dx}
}
\\
&=
v \frac{du}{dx}
–
u \frac{dv}{dx},
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:helmholtzGreensV2:1400}
\boxed{
\int_a^b dx\, v \frac{d^2 u}{dx^2}
=
\int_a^b dx\, u \frac{d^2 v}{dx^2}
+
\evalrange{v \frac{du}{dx}}{a}{b}
–
\evalrange{u \frac{dv}{dx}}{a}{b}.
}
\end{equation}

This gives us
\begin{equation}\label{eqn:helmholtzGreensV2:1420}
\begin{aligned}
\lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} e^{j k \Abs{x – x’}} V(x’) dx’
&=
\int_{-\epsilon}^\epsilon e^{j k \Abs{r}} \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) dr \\
&\quad +
\evalrange{ V(x + r) \frac{d}{dr} e^{j k \Abs{r}} }{-\epsilon}{\epsilon}
–
\evalrange{ e^{j k \Abs{r}} \frac{d}{dr} V(x+r) }{-\epsilon}{\epsilon}.
\end{aligned}
\end{equation}
If we can assume that \( V \) and it’s first and second derivatives are all continuous over this small interval, then the first integral is approximately
\begin{equation}\label{eqn:helmholtzGreensV2:1440}
\begin{aligned}
\int_{-\epsilon}^\epsilon e^{j k \Abs{r}} \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) dr
&\sim
\lr{ \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) }
\int_{-\epsilon}^\epsilon e^{j k \Abs{r}} dr \\
&=
\frac{2 j}{k} \lr{ 1 – e^{ j k \epsilon} }
\lr{ \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) } \\
&\rightarrow 0.
\end{aligned}
\end{equation}
Similarly, with \( dV/dr \) continuity condition, that last term is also zero. We are left, for \( \epsilon \) sufficiently small, we are left with
\begin{equation}\label{eqn:helmholtzGreensV2:1460}
\lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} e^{j k \Abs{x – x’}} V(x’) dx’
=
V(x)
\evalrange{ \frac{d}{dr} e^{j k \Abs{r}} }{-\epsilon}{\epsilon}.
\end{equation}
Because we are evaluating this derivative only at points \( r = \pm \epsilon \ne 0 \), that derivative is
\begin{equation}\label{eqn:helmholtzGreensV2:2460}
\frac{d}{dr} e^{j k \Abs{r}}
=
j k e^{j k \Abs{r}} \mathrm{sgn}(r),
\end{equation}
leaving us with
\begin{equation}\label{eqn:helmholtzGreensV2:2480}
\begin{aligned}
\evalrange{ \frac{d}{dr} e^{j k \Abs{r}} }{-\epsilon}{\epsilon}
&=
j k e^{j k \Abs{\epsilon}} \mathrm{sgn}(\epsilon) – j k e^{j k \Abs{-\epsilon}} \mathrm{sgn}(-\epsilon) \\
&=
j k e^{j k \Abs{\epsilon}} \lr{ 1 – (-1) } \\
2 j k e^{j k \Abs{\epsilon}}.
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:helmholtzGreensV2:2500}
\lr{ \spacegrad^2 + k^2 }\int_{-\infty}^\infty e^{j k \Abs{x – x’} } V(x’) dx’
=
2 j k e^{j k \Abs{\epsilon}} V(x).
\end{equation}
Taking limits and dividing through by \( 2 j k \) proves the result.

1D Laplacian Green’s function.

Having blundered our way to what appears to be the correct Green’s function for the 1D Helmholtz operator, let’s further validate that by deriving the Green’s function for the 1D Laplacian. We should also be able to verify that it has the correct delta function semantics.

Expansion in series and taking the limit.

Expanding the Helmholtz Green’s function in series around \( k \Abs{r} \) we have
\begin{equation}\label{eqn:helmholtzGreensV2:1780}
\begin{aligned}
G(r)
&= -\frac{j}{2k} \lr{ 1 + j k \Abs{r} + O((k \Abs{r})^2) } \\
&= -\frac{j}{2} \lr{ \inv{k} + j \Abs{r} + \inv{k} O((k \Abs{r})^2) } \\
\end{aligned}
\end{equation}
This means that to first order in \( k \), we have
\begin{equation}\label{eqn:helmholtzGreensV2:1800}
G(r) + \frac{j}{2k} = \frac{\Abs{r}}{2}.
\end{equation}
As before, we are free to add constant terms to the Green’s function for the Laplacian, and we conclude that the 1D Green’s function for the Laplacian is
\begin{equation}\label{eqn:helmholtzGreensV2:1820}
\boxed{
G(r) = \frac{\Abs{r}}{2}.
}
\end{equation}

Alternatively, we may use the real sine form of the Green’s function, which has a nice expansion around \( k = 0 \), and arrive at the same result.

Observe that
\begin{equation}\label{eqn:helmholtzGreensV2:2520}
\frac{d^2}{dr^2} \frac{\Abs{r}}{2} =
\frac{d}{dr} \frac{\mathrm{sgn}(r)}{2} =
\delta(r),
\end{equation}
which verifies that this is a valid Green’s function for the 1D Laplacian.

Verifying the Green’s function with convolution.

We can also operate on the convolution with the Laplacian, to verify correctness. We are interested in evaluating
\begin{equation}\label{eqn:helmholtzGreensV2:1840}
\spacegrad^2 \int \frac{\Abs{x – x’}}{2} V(x’) dx’ = \int V(x + r) \frac{d^2}{dr^2} \frac{\Abs{r}}{2} dr.
\end{equation}
If all goes well, this should evaluate to \( V(x) \), indicating that \( \spacegrad^2 \Abs{x – x’}/2 = \delta(x – x’) \). As a first step, we expect \( \spacegrad^2 G = 0 \), for \( x \ne x’ \). Consider first \( r > 0 \), where
\begin{equation}\label{eqn:helmholtzGreensV2:1860}
\frac{d}{dr} \Abs{r}
=
\frac{d}{dr} r
= 1,
\end{equation}
and for \( r < 0 \) where
\begin{equation}\label{eqn:helmholtzGreensV2:1880}
\frac{d}{dr} \Abs{r}
=
\frac{d}{dr} (-r)
= -1.
\end{equation}
This means that, away from the origin \( d\Abs{r}/dr = \mathrm{sgn}(r) \), and \( d^2 \Abs{r}/dr^2 = 0\). We can conclude that, for some non-zero positive epsilon that we will eventually let approach zero, we have
\begin{equation}\label{eqn:helmholtzGreensV2:1900}
\begin{aligned}
\spacegrad^2 \int \frac{\Abs{x – x’}}{2} V(x’) dx’
&= \int_{-\epsilon}^\epsilon V(x + r) \frac{d^2}{dr^2} \frac{\Abs{r}}{2} dr \\
&= \int_{-\epsilon}^\epsilon \lr{
\frac{d}{dr} \lr{ V(x + r) \frac{d}{dr} \frac{\Abs{r}}{2} }
– \frac{dV(x + r)}{dr} \frac{d}{dr} \frac{\Abs{r}}{2}
}
dr \\
&=
\evalrange{ V(x + r) \frac{d}{dr} \frac{\Abs{r}}{2} }{-\epsilon}{\epsilon}
– \int_{-\epsilon}^\epsilon
\lr{
\frac{d}{dr} \lr{ \frac{dV(x + r)}{dr} \frac{\Abs{r}}{2} }
–
\frac{d^2V(x + r)}{dr^2} \frac{\Abs{r}}{2}
} dr \\
&=
\evalrange{ V(x + r) \frac{d}{dr} \frac{\Abs{r}}{2} }{-\epsilon}{\epsilon}
-\evalrange{ \frac{dV(x + r)}{dr} \frac{\Abs{r}}{2} }{-\epsilon}{\epsilon}
+
\int_{-\epsilon}^\epsilon \frac{d^2V(x + r)}{dr^2} \frac{\Abs{r}}{2} dr \\
&=
\inv{2} \lr{ V(x + \epsilon) + V(x – \epsilon) } \\
&-\quad \frac{\epsilon}{2}\lr{
\frac{dV(x + \epsilon)}{dr}
–
\frac{dV(x – \epsilon)}{dr}
} \\
&+
\frac{\epsilon^2}{2} \lr{
\frac{d^2V(x + \epsilon)}{dr^2}
+
\frac{d^2V(x + \epsilon)}{dr^2}
}.
\end{aligned}
\end{equation}
In the limit we have
\begin{equation}\label{eqn:helmholtzGreensV2:1920}
\boxed{
\spacegrad^2 \int G(x, x’) V(x’) dx’ = \inv{2} \lr{ V(x^+) + V(x^-) }.
}
\end{equation}

If the test (or driving) function is continuous at \( x’ = x \), then this is exactly the delta-function semantics that we expect of a Green’s function. It’s interesting that this check provides us with precise semantics for the Green’s function for discontinuous functions too.

A trilogy in 7+ parts: A better check of the 1D Helmholtz Green’s function.

September 26, 2025 math and physics play , , , , ,

[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}

A trilogy in six+ parts: 1D Laplacian Green’s function

September 25, 2025 math and physics play , , , , , , , , ,

[Click here for a PDF version of this post, and the others in this series.]

Having blundered our way to what appears to be the correct Green’s function for the 1D Helmholtz operator, let’s further validate that by deriving the Green’s function for the 1D Laplacian. We should also be able to verify that it has the correct delta function semantics.

The 1D Laplacian Green’s function.

Expanding the Helmholtz Green’s function in series around \( k \Abs{r} \) we have
\begin{equation}\label{eqn:helmholtzGreens:1780}
\begin{aligned}
G(r)
&= -\frac{j}{2k} \lr{ 1 + j k \Abs{r} + O((k \Abs{r})^2) } \\
&= -\frac{j}{2} \lr{ \inv{k} + j \Abs{r} + \inv{k} O((k \Abs{r})^2) } \\
\end{aligned}
\end{equation}
This means that to first order in \( k \), we have
\begin{equation}\label{eqn:helmholtzGreens:1800}
G(r) + \frac{j}{2k} = \frac{\Abs{r}}{2}.
\end{equation}
As before, we are free to add constant terms to the Green’s function for the Laplacian, and we conclude that the 1D Green’s function for the Laplacian is
\begin{equation}\label{eqn:helmholtzGreens:1820}
\boxed{
G(r) = \frac{\Abs{r}}{2}.
}
\end{equation}

Observing the delta-function semantics of our Laplacian Green’s function through convolution.

We can now attempt to validate that this has the desired delta function semantics, operating on the convolution with the Laplacian. We are interested in evaluating
\begin{equation}\label{eqn:helmholtzGreens:1840}
\spacegrad^2 \int \frac{\Abs{x – x’}}{2} V(x’) dx’ = \int V(x + r) \frac{d^2}{dr^2} \frac{\Abs{r}}{2} dr.
\end{equation}
If all goes well, this should evaluate to \( V(x) \), indicating that \( \spacegrad^2 \Abs{x – x’}/2 = \delta(x – x’) \). As a first step, we expect \( \spacegrad^2 G = 0 \), for \( x \ne x’ \). Consider first \( r > 0 \), where
\begin{equation}\label{eqn:helmholtzGreens:1860}
\frac{d}{dr} \Abs{r}
=
\frac{d}{dr} r
= 1,
\end{equation}
and for \( r < 0 \) where
\begin{equation}\label{eqn:helmholtzGreens:1880}
\frac{d}{dr} \Abs{r}
=
\frac{d}{dr} (-r)
= -1.
\end{equation}
This means that, away from the origin \( d\Abs{r}/dr = \mathrm{sgn}(r) \), and \( d^2 \Abs{r}/dr^2 = 0\). We can conclude that, for some non-zero positive epsilon that we will eventually let approach zero, we have
\begin{equation}\label{eqn:helmholtzGreens:1900}
\begin{aligned}
\spacegrad^2 \int \frac{\Abs{x – x’}}{2} V(x’) dx’
&= \int_{-\epsilon}^\epsilon V(x + r) \frac{d^2}{dr^2} \frac{\Abs{r}}{2} dr \\
&= \int_{-\epsilon}^\epsilon \lr{
\frac{d}{dr} \lr{ V(x + r) \frac{d}{dr} \frac{\Abs{r}}{2} }
– \frac{dV(x + r)}{dr} \frac{d}{dr} \frac{\Abs{r}}{2}
}
dr \\
&=
\evalrange{ V(x + r) \frac{d}{dr} \frac{\Abs{r}}{2} }{-\epsilon}{\epsilon}
– \int_{-\epsilon}^\epsilon
\lr{
\frac{d}{dr} \lr{ \frac{dV(x + r)}{dr} \frac{\Abs{r}}{2} }
–
\frac{d^2V(x + r)}{dr^2} \frac{\Abs{r}}{2}
} dr \\
&=
\evalrange{ V(x + r) \frac{d}{dr} \frac{\Abs{r}}{2} }{-\epsilon}{\epsilon}
-\evalrange{ \frac{dV(x + r)}{dr} \frac{\Abs{r}}{2} }{-\epsilon}{\epsilon}
+
\int_{-\epsilon}^\epsilon \frac{d^2V(x + r)}{dr^2} \frac{\Abs{r}}{2} dr \\
&=
\inv{2} \lr{ V(x + \epsilon) + V(x – \epsilon) } \\
&-\quad \frac{\epsilon}{2}\lr{
\frac{dV(x + \epsilon)}{dr}
–
\frac{dV(x – \epsilon)}{dr}
} \\
&+
\frac{\epsilon^2}{2} \lr{
\frac{d^2V(x + \epsilon)}{dr^2}
+
\frac{d^2V(x + \epsilon)}{dr^2}
}.
\end{aligned}
\end{equation}
In the limit we have
\begin{equation}\label{eqn:helmholtzGreens:1920}
\boxed{
\spacegrad^2 \int G(x, x’) V(x’) dx’ = \inv{2} \lr{ V(x^+) + V(x^-) }.
}
\end{equation}

If the test (or driving) function is continuous at \( x’ = x \), then this is exactly the delta-function semantics that we expect of a Green’s function. It’s interesting that this check provides us with precise semantics for the Green’s function for discontinuous functions too.

Extracting the delta-function semantics of the Laplacian Green’s function directly.

There’s a more direct, but less satisfying way to do this same computation. We can compute \( d^2 G(r)/dr^2 \). We need the trick
\begin{equation}\label{eqn:helmholtzGreens:1940}
\Abs{r} = r \Theta(r) – r \Theta(-r),
\end{equation}
and the identification \(\Theta'(r) = \delta(r) \). We find
\begin{equation}\label{eqn:helmholtzGreens:1960}
\begin{aligned}
\Abs{r}’
&= \Theta(r) – \Theta(-r) + r \delta(r) – r (-1) \delta(-r) \\
&= \Theta(r) – \Theta(-r) + 2 r \delta(r).
\end{aligned}
\end{equation}
To give \( r \delta(r) \) meaning, we can apply it to a test function
\begin{equation}\label{eqn:helmholtzGreens:1980}
\int r \delta(r) f(r) dr = \evalbar{ r f(r) }{r = 0} = 0,
\end{equation}
so
\begin{equation}\label{eqn:helmholtzGreens:2000}
\Abs{r}’ = \Theta(r) – \Theta(-r).
\end{equation}
Now we can take second derivatives
\begin{equation}\label{eqn:helmholtzGreens:2020}
\Abs{r}” = \delta(r) + \delta(-r) = 2 \delta(r).
\end{equation}
This means that
\begin{equation}\label{eqn:helmholtzGreens:2040}
\boxed{
\frac{d^2}{dr^2} \frac{\Abs{r}}{2} = \delta(r).
}
\end{equation}

A trilogy in five+ parts: Confirming an error in the derived 1D Helmholtz Green’s function.

September 24, 2025 math and physics play , , , , , , , ,

[Click here for a PDF version of this, and previous, posts in this series].

The discontinuity in the derived 1D Helmholtz Green’s function is somewhat surprising. Let’s try to verify that this works or find what does. The first thing to check is that
\begin{equation}\label{eqn:helmholtzGreens:1220}
\lr{ \spacegrad^2 + k^2} G(x,x’) = 0,
\end{equation}
at locations where \( x \ne x’ \). Since we are avoiding the origin (where the annoying sign function kicks in), means that we want to evaluate:
\begin{equation}\label{eqn:helmholtzGreens:1240}
\lr{ k^2 + \frac{d^2}{dx^2} } e^{j k \Abs{x – x’}},
\end{equation}
and expect that this will be zero. Let’s make a change of variables \( r = x’ – x \), and evaluate
\begin{equation}\label{eqn:helmholtzGreens:1260}
\lr{ k^2 + \frac{d^2}{dr^2} } e^{j k \Abs{r}},
\end{equation}
assuming that we are omitting a neighbourhood around \( r = 0 \) where the absolute value causes trouble. For \( r > 0 \)
\begin{equation}\label{eqn:helmholtzGreens:1280}
\begin{aligned}
\frac{d}{dr} e^{j k \Abs{r}}
&= \frac{d}{dr} e^{j k r} \\
&= j k e^{j k r},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:helmholtzGreens:1300}
\begin{aligned}
\frac{d^2}{dr^2} e^{j k \Abs{r}}
&= \frac{d}{dr} j k e^{j k r} \\
&= (j k)^2 e^{j k r}.
\end{aligned}
\end{equation}
Similarly, for \( r < 0 \), we have \begin{equation}\label{eqn:helmholtzGreens:1320} \frac{d^2}{dr^2} e^{j k \Abs{r}} = (-jk)^2 e^{j k \Abs{r}}. \end{equation} In both cases, provided we are in a neighbourhood that omits \( r \ne 0 \), we have \begin{equation}\label{eqn:helmholtzGreens:1340} \lr{ k^2 + \frac{d^2}{dr^2} } e^{j k \Abs{r}} = 0, \end{equation} as desired. The takeaway is that we have \begin{equation}\label{eqn:helmholtzGreens:1360} \begin{aligned} \lr{ \spacegrad^2 + k^2 }\int_{-\infty}^\infty G(x, x’) V(x’) dx’ &= \int_{\Abs{x’ – x} \le \epsilon} V(x’) \lr{ \frac{d^2}{d{x’}^2} + k^2 } G(x, x’) dx’ \\ &= \int_{-\epsilon}^\epsilon V(x + r) \lr{ \frac{d^2}{dr^2} + k^2 } G(r = x’ – x) dr \\ \end{aligned} \end{equation} for some arbitrarily small value of \( \epsilon \). Observe that after bringing the operator into the integral, we also made a change of variables, first to \( x’ \) for the Laplacian, and then to \( r = x’ – x \). We’d like the 1D equivalent of Green’s theorem to reduce this, so let’s work that out first. \begin{equation}\label{eqn:helmholtzGreens:1380} \begin{aligned} \int dx\, v \frac{d^2 u}{dx^2} – \int dx\, u \frac{d^2 v}{dx^2} &= \int dx\, \lr{ \frac{d}{dx} \lr{ v \frac{du}{dx} } – \frac{dv}{dx} \frac{du}{dx} } – \int dx\, \lr{ \frac{d}{dx} \lr{ u \frac{dv}{dx} } – \frac{du}{dx} \frac{dv}{dx} } \\ &= \int dx\, \frac{d}{dx} \lr{ v \frac{du}{dx} } – \int dx\, \frac{d}{dx} \lr{ u \frac{dv}{dx} } \\ &= v \frac{du}{dx} – u \frac{dv}{dx}, \end{aligned} \end{equation} so \begin{equation}\label{eqn:helmholtzGreens:1400} \boxed{ \int_a^b dx\, v \frac{d^2 u}{dx^2} = \int_a^b dx\, u \frac{d^2 v}{dx^2} + \evalrange{v \frac{du}{dx}}{a}{b} – \evalrange{u \frac{dv}{dx}}{a}{b}. } \end{equation} Let’s try applying that to the function \( G(r) = e^{j k \Abs{r} } \), and see what happens. That is \begin{equation}\label{eqn:helmholtzGreens:1420} \begin{aligned} \lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} e^{j k \Abs{x – x’}} V(x’) dx’ &= \int_{-\epsilon}^\epsilon e^{j k \Abs{r}} \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) dr \\ &\quad + \evalrange{ V(x + r) \frac{d}{dr} e^{j k \Abs{r}} }{-\epsilon}{\epsilon} – \evalrange{ e^{j k \Abs{r}} \frac{d}{dr} V(x+r) }{-\epsilon}{\epsilon}. \end{aligned} \end{equation} If we can assume that \( V \) and it’s first and second derivatives are all continuous over this small interval, then the first integral is approximately \begin{equation}\label{eqn:helmholtzGreens:1440} \begin{aligned} \int_{-\epsilon}^\epsilon e^{j k \Abs{r}} \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) dr &\sim \lr{ \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) } \int_{-\epsilon}^\epsilon e^{j k \Abs{r}} dr \\ &= \frac{2 j}{k} \lr{ 1 – e^{ j k \epsilon} } \lr{ \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) } \\ &\rightarrow 0. \end{aligned} \end{equation} Similarly, with \( dV/dr \) continuity condition, that last term is also zero. We are left, for \( \epsilon \) sufficiently small, we are left with \begin{equation}\label{eqn:helmholtzGreens:1460} \lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} e^{j k \Abs{x – x’}} V(x’) dx’ = V(x) \evalrange{ \frac{d}{dr} e^{j k \Abs{r}} }{-\epsilon}{\epsilon}. \end{equation} but this is an extremely problematic derivative around the origin. The core problem is evaluating \begin{equation}\label{eqn:helmholtzGreens:1480} \frac{d}{dr} e^{j k \Abs{r}} = j k e^{j k \Abs{r} } \frac{d\Abs{r}}{dr}. \end{equation} In conventional mathematics, we’d have to say that this is undefined at the origin. In physics, on the other hand, where we play fast and loose with the mathematics, we express the absolute value in terms of Heavyside theta functions \begin{equation}\label{eqn:helmholtzGreens:1500} \Abs{r} = r \Theta(r) – r \Theta(-r). \end{equation} We may now take derivatives \begin{equation}\label{eqn:helmholtzGreens:1520} \begin{aligned} \Abs{r}’ &= \Theta(r) – \Theta(-r) + r \delta(r) + r \delta(-r) \\ &= \mathrm{sgn}(r) + 2 r \delta(r). \end{aligned} \end{equation} Evaluating \( e^{j k \Abs{r} } \Abs{r}’ \) over the \( [-\epsilon, \epsilon] \) range, we have \begin{equation}\label{eqn:helmholtzGreens:1540} \begin{aligned} \evalrange{ e^{j k \Abs{r} } \Abs{r}’ }{-\epsilon}{\epsilon} &= e^{j k \epsilon} \lr{ \evalrange{ \mathrm{sgn}(r) + 2 r \delta(r) }{-\epsilon}{\epsilon} } \\ &= e^{j k \epsilon} \lr{ 2 + 2 \epsilon \delta(\epsilon) }. \end{aligned} \end{equation} Again, playing fast and loose, we evaluate this range before taking the limit, where \( \delta(\epsilon) = 0 \) for \( \epsilon > 0 \). We are left with
\begin{equation}\label{eqn:helmholtzGreens:1560}
\lim_{\epsilon \rightarrow 0} \lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} e^{j k \Abs{x – x’}} V(x’) dx’
=
2 j k V(x),
\end{equation}
provided \( V \) and its first and second derivatives are continuous.

Under those constraints, the implication is that one valid Green’s function for the 1D Helmholtz operator is
\begin{equation}\label{eqn:helmholtzGreens:1580}
G(r) = -\frac{j}{2k} e^{j k \Abs{r} }.
\end{equation}
The \( \mathrm{sgn}(r) \) scale factor that was part of the Green’s function that we derived using contour integration does not appear to be required.

What happens if we retain the sign function factor? Doing so, we have
\begin{equation}\label{eqn:helmholtzGreens:1600}
\begin{aligned}
\lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} \mathrm{sgn}(x – x’) e^{j k \Abs{x – x’}} V(x’) dx’
&=
-\int_{-\epsilon}^\epsilon \mathrm{sgn}(r) e^{j k \Abs{r}} \lr{ k^2 + \frac{d^2}{dr^2} } V(x + r) dr \\
&\quad
–
\evalrange{ V(x + r) \frac{d}{dr} \mathrm{sgn}(r) e^{j k \Abs{r}} }{-\epsilon}{\epsilon}
+
\evalrange{ \mathrm{sgn}(r) e^{j k \Abs{r}} \frac{d}{dr} V(x+r) }{-\epsilon}{\epsilon}.
\end{aligned}
\end{equation}
This time, we note that
\begin{equation}\label{eqn:helmholtzGreens:1620}
\int_{-\epsilon}^\epsilon \mathrm{sgn}(r) e^{j k \Abs{r}} dr = 0, \quad \forall \epsilon \ne 0,
\end{equation}
even without evaluating the limit. However, we have problems with the other two terms. The last term doesn’t zero out as desired, instead
\begin{equation}\label{eqn:helmholtzGreens:1640}
\evalrange{ \mathrm{sgn}(r) e^{j k \Abs{r}} \frac{d}{dr} V(x+r) }{-\epsilon}{\epsilon} \rightarrow 2 V'(x).
\end{equation}
To evaluate the \( V(x) \) factor, we write
\begin{equation}\label{eqn:helmholtzGreens:1660}
\mathrm{sgn}(r) = \Theta(r) – \Theta(-r),
\end{equation}
so
\begin{equation}\label{eqn:helmholtzGreens:1680}
\begin{aligned}
\mathrm{sgn}(r)’
&= \delta(r) + \delta(-r) \\
&= 2 \delta(r).
\end{aligned}
\end{equation}
That means that
\begin{equation}\label{eqn:helmholtzGreens:1700}
\begin{aligned}
\frac{d}{dr} \lr{ \mathrm{sgn}(r) e^{j k \Abs{r}} }
&=
2 \delta(r) e^{j k \Abs{r}} + j k \mathrm{sgn}(r) e^{j k \Abs{r}} \lr{ \mathrm{sgn}(r) + 2 r \delta(r) } \\
&=
e^{j k \Abs{r}} \lr{ 2 \delta(r) + j k \lr{ 1 + 2 r \mathrm{sgn}(r) \delta(r)} } \\
&=
j k e^{j k \Abs{r}},
\end{aligned}
\end{equation}
for \( r \ne 0 \), so
\begin{equation}\label{eqn:helmholtzGreens:1760}
–
\evalrange{ V(x + r) \frac{d}{dr} \mathrm{sgn}(r) e^{j k \Abs{r}} }{-\epsilon}{\epsilon}
\rightarrow V(x) j k \lr{ e^{j k \epsilon} – e^{j k \epsilon} } = 0.
\end{equation}

All in, we are left with
\begin{equation}\label{eqn:helmholtzGreens:1720}
\lim_{\epsilon \rightarrow 0} \lr{ \spacegrad^2 + k^2 }\int_{\Abs{x’- x} \le \epsilon} \frac{-j}{2k} \mathrm{sgn}(x – x’) e^{j k \Abs{x – x’}} V(x’) dx’
= -\frac{j}{k} V'(x),
\end{equation}
but for a Green’s function, we expected just \( V(x) \).

It seems that the sign factor in the contour integration result is definitively wrong. That result was
\begin{equation}\label{eqn:helmholtzGreens:300b}
G(u) = -\frac{j \mathrm{sgn}(u)}{2k} e^{j k \Abs{u}},
\end{equation}
but what we really want is
\begin{equation}\label{eqn:helmholtzGreens:1740}
\boxed{
G(u) = -\frac{j}{2k} e^{j k \Abs{u}}.
}
\end{equation}

Unfortunately, I don’t see any errors in the original contour integration, so I’m at a loss where things went wrong.

A trilogy in four+ parts: The 2D Laplacian Green’s function.

September 22, 2025 math and physics play , , , , , , ,

[Click here for a PDF version of this post]

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.