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.

Part 3/3. 2D Green’s functions for the Helmholtz (wave equation) operator.

September 20, 2025 math and physics play , , , , , , , , ,

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Having found the 1D and 3D Green’s function for the wave equation (Helmholtz) operator, we are now ready to attempt the harder 2D case again.

2D Green’s function.

Our starting place is
\begin{equation}\label{eqn:helmholtzGreens:680}
G(\Br) = -\inv{(2 \pi)^2} \int \frac{e^{j \Bp \cdot \Br}}{\Bp^2 – k^2} d^2 p.
\end{equation}
With a change of variables to polar coordinates, letting
\begin{equation}\label{eqn:helmholtzGreens:700}
\begin{aligned}
\Bp &= p \lr{ \cos\phi, \sin\phi } \\
\Br &= \Abs{\Br} \Be_2,
\end{aligned}
\end{equation}
we can make the integral explicit
\begin{equation}\label{eqn:helmholtzGreens:720}
G(\Br) = -\inv{(2 \pi)^2} \int_0^\infty \frac{p dp}{p^2 – k^2} \int_0^{2 \pi} d\phi e^{j p \Abs{\Br} \sin\phi}.
\end{equation}
Unlike the 3D case, where the angular dependence could be trivially evaluated, we are no longer so lucky. What on earth can we do with the \( \phi \) integral? Just like Hilter’s lament about “undoable integrals in Jackson”, we are faced with the same enemy. As it turns out, due to the cylindrical symmetry of the problem, we are also staring down the gun of Bessel functions. Both Mathematica and Grok point out that we can evaluate integrals of this form, like so:
\begin{equation}\label{eqn:helmholtzGreens:740}
\int_0^{2 \pi} d\phi e^{j a \sin\phi} = 2 \pi J_0(a).
\end{equation}

From [2] we have two representations of \( J_n \), a series representation and integral representation
\begin{equation}\label{eqn:helmholtzGreens:760}
J_n(z) = \sum_{m=0}^\infty \frac{(-1)^m (z/2)^{2m + n}}{(n + m)!m!} = \inv{\pi} \int_0^\pi \cos(n \theta – z \sin\theta) d\theta.
\end{equation}

In particular, this means that
\begin{equation}\label{eqn:helmholtzGreens:800}
J_0(z) = \sum_{m=0}^\infty \frac{(-1)^m (z/2)^{2m}}{(m!)^2} = J_0(z) = \inv{\pi} \int_0^\pi \cos(z \sin\theta) d\theta.
\end{equation}
This is a damped sine like function, as illustrated in fig. 4.

fig. 4. Bessel function of zeroth order.

 

In section 6.9, both of these are derived from a generating function representation of the Bessel functions, and one of the intermediate steps in that construction has
\begin{equation}\label{eqn:helmholtzGreens:840}
J_n(z) = \inv{2\pi} \int_{-\pi}^\pi e^{-j(n\theta – z \sin\theta)} d\theta,
\end{equation}
where the \( [-\pi, \pi] \) range was the result of a contour integration using a unit circle parameterization, which could have also used \( [0, 2 \pi] \). That means, sure enough, that we have
\begin{equation}\label{eqn:helmholtzGreens:860}
J_0(z) = \inv{2\pi} \int_{0}^{2\pi} e^{j z \sin\theta} d\theta,
\end{equation}
as claimed by both Grok and Mathematica.

This means that the evaluation of the Green’s function is now reduced to the limit of one final integral
\begin{equation}\label{eqn:helmholtzGreens:880}
G(\Br) = -\inv{2 \pi} \int_0^\infty \frac{p J_0(p \Abs{\Br} ) dp}{p^2 – \lr{k + j \epsilon}^2},
\end{equation}
where we’ve also displaced the problematic pole by a small imaginary amount as before. Grok incorrectly claimed that this was an even integral, and then argued that the end result is a Hankel function (that may be the case, but it’s reasoning to get there was clearly wrong.) Mathematica, on the other hand, can evaluate this integral
\begin{equation}\label{eqn:helmholtzGreens:900}
G(\Br) = -\inv{2 \pi} K_0\lr{\frac{\Abs{\Br}}{\sqrt{\frac{1}{(\epsilon – j k)^2}}}}, \epsilon \neq 0.
\end{equation}
It’s not clear to me why Mathematica writes the argument as 1 over a reciprocal root. Perhaps that has something to do with the branch cut that Mathematica uses for it’s square root function? If I plug in representitive numeric values, it simplifies in the expected way, as illustrated in fig. 5.

fig. 5. Mathematica weird Bessel argument.

The take away appears to be that the limiting form of the 2D Green’s function, for \( k > 0 \), is
\begin{equation}\label{eqn:helmholtzGreens:920}
G(\Bx, \Bx’) = -\inv{2 \pi} K_0\lr{-j k \Abs{\Bx – \Bx’} }.
\end{equation}
A peek at [1] shows that \( K_0 \) can be expressed in terms of a Hankel function of the first kind (order 0) \( H_0^{(1)}(z) = J_0(z) + j Y_0(z) \), plotted in fig. 6.

fig. 6. Hankel function of the first kind (order 0).

For real positive \( \alpha \), we have
\begin{equation}\label{eqn:helmholtzGreens:940}
K_0(-j \alpha) = \frac{j\pi}{2} H_0^{(1)}(\alpha),
\end{equation}
so
\begin{equation}\label{eqn:helmholtzGreens:960}
\boxed{
G(\Bx, \Bx’) = -\frac{j}{4} H_0^{(1)}(k \Abs{\Bx – \Bx’}).
}
\end{equation}

References

[1] M. Abramowitz and I.A. Stegun. Handbook of mathematical functions with formulas, graphs, and mathematical tables, volume 55. Dover publications, 1964.

[2] F.W. Byron and R.W. Fuller. Mathematics of Classical and Quantum Physics. Dover Publications, 1992.