Subsections


D.2 Some Green’s functions


D.2.1 The Poisson equation

The so-called Poisson equation is

\begin{displaymath}
- \nabla^2 u = f \qquad
\nabla \equiv {\hat\imath}\frac{...
...c{\partial}{\partial y} + {\hat k}\frac{\partial}{\partial z}
\end{displaymath}

Here $f$ is supposed to be a given function and $u$ an unknown function that is to be found.

The solution $u$ to the Poisson equation in infinite space may be found in terms of its so-called Green’s function $G({\skew0\vec r})$. In particular:

\begin{displaymath}
u({\skew0\vec r}) = \int_{{\rm all\ }{\underline{\skew0\ve...
...c{1}{4\pi\vert{\skew0\vec r}-{\underline{\skew0\vec r}}\vert}
\end{displaymath}

Loosely speaking, the above integral solution chops function $f$ up into spikes $f({\underline{\skew0\vec r}}){\rm d}^3{\underline{\skew0\vec r}}$. A spike at a position ${\underline{\skew0\vec r}}$ then makes a contribution $G({\skew0\vec r}-{\underline{\skew0\vec r}})f({\underline{\skew0\vec r}}){\rm d}^3{\underline{\skew0\vec r}}$ to $u$ at ${\skew0\vec r}$. Integration over all such spikes gives the complete $u$.

Note that often, the Poisson equation is written without a minus sign. Then there will be a minus sign in $G$.

The objective is now to derive the above Green’s function. To do so, first an intuitive derivation will be given and then a more rigorous one. (See also chapter 13.3.4 for a more physical derivation in terms of electrostatics.)

The intuitive derivation defines $G({\skew0\vec r})$ as the solution due to a unit spike, i.e. a delta function, located at the origin. That means that $G$ $\vphantom0\raisebox{1.5pt}{$=$}$ $G({\skew0\vec r})$ is the solution to

\begin{displaymath}
\nabla^2 G = \delta^3
\qquad \mbox{with}\quad G({\skew0\vec r})\to0 \quad\mbox{when}\quad {\skew0\vec r}\to\infty
\end{displaymath}

Here $\delta^3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\delta^3({\skew0\vec r})$ is the three-di­men­sion­al delta function, defined as an infinite spike at the origin that integrates to 1.

By itself the above definition is of course meaningless: infinity is not a valid number. To give it meaning, it is necessary to define an approximate delta function, one that is merely a large spike rather than an infinite one. This approximate delta function $\delta^3_\varepsilon$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\delta^3_\varepsilon({\skew0\vec r})$ must still integrate to 1 and will be required to be zero beyond some small distance $\varepsilon$ from the origin:

\begin{displaymath}
\int\delta^3_\varepsilon({\skew0\vec r}){\,\rm d}^3{\skew0...
...vec r}\vert \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon
\end{displaymath}

In the above integral the region of integration should at least include the small region of radius $\varepsilon$ around the origin. The approximate delta function will further be assumed to be nonnegative. It must have large values in the small vicinity around the origin where it is nonzero; otherwise the integral over the small vicinity would be small instead of 1. But the key is that the values are not infinite, just large. So normal mathematics can be used.

The corresponding approximate Green’s function $G_\varepsilon$ $\vphantom0\raisebox{1.5pt}{$=$}$ $G_\varepsilon({\skew0\vec r})$ of the Poisson equation satisfies

\begin{displaymath}
-\nabla^2 G_\varepsilon = \delta^3_\varepsilon
\qquad \m...
...\varepsilon\to0 \quad\mbox{when}\quad {\skew0\vec r}\to\infty
\end{displaymath}

In the limit $\varepsilon\to0$, $\delta^3_\varepsilon({\skew0\vec r})$ becomes the Dirac delta function $\delta^3({\skew0\vec r})$ and $G_\varepsilon({\skew0\vec r})$ becomes the exact Green’s function $G({\skew0\vec r})$.

To find the approximate Green’s function, it will be assumed that $\delta^3_\varepsilon({\skew0\vec r})$ only depends on the distance $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert{\skew0\vec r}\vert$ from the origin. In other words, it is assumed to be spherically symmetric. Then so is $G_\varepsilon$. (Note that this assumption is not strictly necessary. That can be seen from the general solution for the Poisson equation given earlier. But it should at least be assumed that $\delta^3_\varepsilon({\skew0\vec r})$ is nonnegative. If it could have arbitrarily large negative values, then $G_\varepsilon$ could be anything.)

Now integrate both sides of the Poisson equation over a sphere of a chosen radius $r$:

\begin{displaymath}
- \int_{\vert{\underline{\skew0\vec r}}\vert\,\mathrel{\ra...
...r} \delta^3_\varepsilon {\,\rm d}^3{\underline{\skew0\vec r}}
\end{displaymath}

As noted, the delta function integrates to 1 as long as the vicinity of the origin is included. That means that the right hand side is 1 as long as $r$ $\raisebox{-.5pt}{$\geqslant$}$ $\varepsilon$. This will now be assumed. The left hand side can be written out. That gives

\begin{displaymath}
- \int_{\vert{\underline{\skew0\vec r}}\vert\,\mathrel{\ra...
...{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon
\end{displaymath}

According to the [divergence] [Gauss] [Ostrogradsky] theorem, the left hand side can be written as a surface integral to give

\begin{displaymath}
- \int_{\vert{\underline{\skew0\vec r}}\vert=r} {\vec n}\c...
...{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon
\end{displaymath}

Here $S$ stands for the surface of the sphere of radius $r$. The total surface is $4{\pi}r^2$. Also ${\vec n}$ is the unit vector orthogonal to the surface, in the outward direction. That is the radial direction. The total differential of calculus then implies that ${\vec n}\cdot\nabla{G}_\varepsilon$ is the radial derivative $\partial{G}_\varepsilon$$\raisebox{.5pt}{$/$}$$\partial{r}$. So,

\begin{displaymath}
- \frac{\partial G_\varepsilon}{\partial r} 4\pi r^2 = 1
...
...{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon
\end{displaymath}

Because $G_\varepsilon$ is required to vanish at large distances, this integrates to

\begin{displaymath}
G_\varepsilon = \frac{1}{4\pi r} \quad\mbox{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon
\end{displaymath}

The exact Green’s function $G$ has $\varepsilon$ equal to zero, so

\begin{displaymath}
G = \frac{1}{4\pi r} \quad\mbox{if}\quad r \ne 0
\end{displaymath}

Finally the rigorous derivation without using poorly defined things like delta functions. In the supposed general solution of the Poisson equation given earlier, change integration variable to $\vec\rho$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline{\skew0\vec r}}-{\skew0\vec r}$

\begin{displaymath}
u({\skew0\vec r}) = \int_{{\rm all\ }{\skew0\vec r}} G({\s...
...c\rho
\qquad G(\vec\rho) = \frac{1}{4\pi\vert\vec\rho\vert}
\end{displaymath}

It is to be shown that the function $u$ defined this way satisfies the Poisson equation $\nabla^2u({\skew0\vec r})$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f({\skew0\vec r})$. To do so, apply $\nabla$ twice:

\begin{displaymath}
\nabla^2 u({\skew0\vec r})
= \int_{{\rm all\ }\vec\rho} ...
... \nabla_\rho^2 f({\skew0\vec r}+\vec\rho) {\,\rm d}^3\vec\rho
\end{displaymath}

Here $\nabla_\rho$ means differentiation with respect to the components of $\vec\rho$ instead of the components of ${\skew0\vec r}$. Because $f$ depends only on ${\skew0\vec r}+\vec\rho$, you get the same answer whichever way you differentiate.

It will be assumed that the function $f$ is well behaved, at least continuous, and becomes zero reasonably quickly at infinity. In that case, you can get a valid approximation to the integral above if you exclude very small and very large values of ${\skew0\vec r}$:

\begin{displaymath}
\nabla^2 u({\skew0\vec r}) \approx \int_{\varepsilon<\vert...
... \nabla_\rho^2 f({\skew0\vec r}+\vec\rho) {\,\rm d}^3\vec\rho
\end{displaymath}

In particular, this approximation becomes exact in the limits where the constants $\varepsilon\to0$ and $R\to\infty$. The integral can now be rewritten as

\begin{displaymath}
\nabla^2 u({\skew0\vec r}) \approx \int_{\varepsilon<\vert...
... r}+\vec\rho) \nabla_\rho^2 G(\vec\rho)
{\,\rm d}^3\vec\rho
\end{displaymath}

as can be verified by explicitly differentiating out the three terms of the integrand. Next note that the third term is zero, because as seen above $G$ satisfies the homogeneous Poisson equation away from the origin. And the other two terms can be written out using the [divergence] [Gauss] [Ostrogradsky] theorem much like before. This produces integrals over both the bounding sphere of radius $R$, as well as over the bounding sphere of radius $\varepsilon$. But the integrals over the sphere of radius $R$ will be vanishingly small if $f$ becomes zero sufficiently quickly at infinity. Similarly, the integral of the first term over the small sphere is vanishingly small, because $G$ is 1$\raisebox{.5pt}{$/$}$$4\pi\varepsilon$ on the small sphere but the surface of the small sphere is $4\pi\varepsilon^2$. However, in the second term, the derivative of $G$ in the negative radial direction is 1/$4\pi\varepsilon^2$, which multiplies to 1 against the surface of the small sphere. Therefore the second term produces the average of $f({\skew0\vec r}+\vec\rho)$ over the small sphere, and that becomes $f({\skew0\vec r})$ in the limit $\vert\vec\rho\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\varepsilon\to0$. So the Poisson equation applies.


D.2.2 The screened Poisson equation

The so-called screened Poisson equation is

\begin{displaymath}
- \nabla^2 u + c^2 u = f \qquad
\nabla \equiv {\hat\imat...
...c{\partial}{\partial y} + {\hat k}\frac{\partial}{\partial z}
\end{displaymath}

Here $f$ is supposed to be a given function and $u$ an unknown function that is to be found. Further $c$ is a given constant. If $c$ is zero, this is the Poisson equation. However, nonzero $c$ corresponds to the inhomogeneous steady Klein-Gordon equation for a particle with nonzero mass.

The analysis of the screened Poisson equation is almost the same as for the Poisson equation given in the previous subsection. Therefore only the differences will be noted here. The approximate Green’s function must satisfy, away from the origin,

\begin{displaymath}
- \nabla^2 G_\varepsilon + c^2 G_\varepsilon = 0
\quad\mbox{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon
\end{displaymath}

The solution to this that vanishes at infinity is of the form

\begin{displaymath}
G_\varepsilon = C \frac{e^{-cr}}{r}
\quad\mbox{if}\quad r \mathrel{\raisebox{-1pt}{$\geqslant$}}\varepsilon
\end{displaymath}

where $C$ is some constant. To check this, plug it in, using the expression (N.5) for $\nabla^2$ in spherical coordinates. To identify the constant $C$, integrate the full equation

\begin{displaymath}
- \nabla^2 G_\varepsilon + c^2 G_\varepsilon = \delta^3_\varepsilon
\end{displaymath}

over a sphere of radius $\varepsilon$ around the origin and apply the divergence theorem as in the previous subsection. Taking the limit $\varepsilon\to0$ then gives $C$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1/$4\pi$, which gives the exact Green’s function as

\begin{displaymath}
G({\skew0\vec r}) = \frac{e^{-cr}}{4\pi r} \quad\mbox{if}\quad r = \vert{\skew0\vec r}\vert \ne 0
\end{displaymath}

The rigorous derivation is the same as before save for an additional $c^2Gf$ term in the integrand, which drops out against the $-f\nabla_\rho^2G$ one.