D.2 Some Green’s functions
D.2.1 The Poisson equation
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
. 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
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,
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
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