D.2 Some Green’s functions

D.2.1 The Poisson equation

The so-called Poisson equation

is

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

The solution to the Poisson equation in infinite space may be
found in terms of its so-called Green’s function

. In particular:

Loosely speaking, the above integral solution chops function up into spikes . A spike at a position then makes a contribution to at . Integration over all such spikes gives the complete .

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

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 as the solution due to a
unit spike, i.e. a delta function,

located at the
origin. That means that is the solution to

Here 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
must still integrate
to 1 and will be required to be zero beyond some small distance
from the origin:

In the above integral the region of integration should at least include the small region of radius 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
of the Poisson equation satisfies

In the limit , becomes the Dirac delta function and becomes the exact Green’s function .

To find the approximate Green’s function, it will be assumed that only depends on the distance from the origin. In other words, it is assumed to be spherically symmetric. Then so is . (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 is nonnegative. If it could have arbitrarily large negative values, then could be anything.)

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

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 . This will now be assumed. The left hand side can be written out. That gives

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

Here stands for the surface of the sphere of radius . The total surface is . Also 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 is the radial derivative . So,

Because is required to vanish at large distances, this integrates to

The exact Green’s function has equal to zero, so

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

It is to be shown that the function defined this way satisfies the Poisson equation . To do so, apply twice:

Here means differentiation with respect to the components of instead of the components of . Because depends only on , you get the same answer whichever way you differentiate.

It will be assumed that the function 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 :

In particular, this approximation becomes exact in the limits where the constants and . The integral can now be rewritten as

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 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 , as well as over the bounding sphere of radius . But the integrals over the sphere of radius will be vanishingly small if becomes zero sufficiently quickly at infinity. Similarly, the integral of the first term over the small sphere is vanishingly small, because is 1 on the small sphere but the surface of the small sphere is . However, in the second term, the derivative of in the negative radial direction is 1/, which multiplies to 1 against the surface of the small sphere. Therefore the second term produces the average of over the small sphere, and that becomes in the limit . So the Poisson equation applies.

D.2.2 The screened Poisson equation

The so-called screened Poisson equation

is

Here is supposed to be a given function and an unknown function that is to be found. Further is a given constant. If is zero, this is the Poisson equation. However, nonzero 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,

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

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

over a sphere of radius around the origin and apply the divergence theorem as in the previous subsection. Taking the limit then gives 1/, which gives the exact Green’s function as

The rigorous derivation is the same as before save for an additional term in the integrand, which drops out against the one.