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The Green's function for the Poisson (inhomogeneous Laplace) equation:

(11.36) 
is the solution to:

(11.37) 
Thus
satisfies the homogeneous Laplace PDE
everywhere but at the single point
. The solution to the
Laplace equation that has the right degree of singularity is the
``potential of a unit point charge'':

(11.38) 
located at
. Hence:

(11.39) 
which is just exactly correct.
Note well that the inhomogeneous term
solves the
homogeneous Laplace equation and has various interpretations. It
can be viewed as a ``boundary term'' (surface integral on
, the surface bounding the volume (Green's Theorem) or, as we
shall see, as the potential of all the charges in the volume exterior to
, or as a gauge transformation of the potential. All are true, but
the ``best'' way to view it is as the potential of exterior charges as
that is what it is in nature even when it is expressed, via
integration by parts, as a surface integral, for a very sensible choice
of asymptotic behavior of the potential.
Note equally well that the Green's function itself has precisely
the same gauge freedom, and can be written in its most general form as:

(11.40) 
where
is any bilinear
(symmetric in both coordinates) solution to the Laplace equation!
However, we will not proceed this way in this part of the course
as it is in a sense unphysical to express the PDEs this way even
though it does upon occasion facilitate the solution algebraically.
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Robert G. Brown
20140819