The Green's function for the Poisson (inhomogeneous Laplace) equation: &phis#phi;= -&rho#rho;&epsi#epsilon;_0 is the solution to: G(,_0) = &delta#delta;(-_0) 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'': G(,_0) = -14&pi#pi;|- _0| located at . Hence: &phis#phi;() = &chi#chi;_0() + 14&pi#pi;&epsi#epsilon;_0 &int#int;_V &rho#rho;(_0)|- _0| d^3x_0 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:
G(,_0) = F(,_0) + -14&pi#pi;|- _0|
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.