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Poisson Equation

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

\begin{displaymath}
\nabla^2\phi = -\frac{\rho}{\epsilon_0}
\end{displaymath} (11.36)

is the solution to:
\begin{displaymath}
\nabla^2G(\mbox{\boldmath$x$},\mbox{\boldmath$x$}_0) = \delta(\mbox{\boldmath$x$}- \mbox{\boldmath$x$}_0)
\end{displaymath} (11.37)

Thus $G(\mbox{\boldmath$x$},\mbox{\boldmath$x$}_0)$ satisfies the homogeneous Laplace PDE everywhere but at the single point $\mbox{\boldmath$x$}_0$. The solution to the Laplace equation that has the right degree of singularity is the ``potential of a unit point charge'':
\begin{displaymath}
G(\mbox{\boldmath$x$},\mbox{\boldmath$x$}_0) = \frac{-1}{4\pi\vert\mbox{\boldmath$x$}- \mbox{\boldmath$x$}_0\vert}
\end{displaymath} (11.38)

located at $\mbox{\boldmath$x$}_0$. Hence:
\begin{displaymath}
\phi(\mbox{\boldmath$x$}) = \chi_0(\mbox{\boldmath$x$}) + \...
...{\vert\mbox{\boldmath$x$}- \mbox{\boldmath$x$}_0\vert} d^3x_0
\end{displaymath} (11.39)

which is just exactly correct.

Note well that the inhomogeneous term $\chi_0(\mbox{\boldmath$x$})$ solves the homogeneous Laplace equation and has various interpretations. It can be viewed as a ``boundary term'' (surface integral on $S = \partial
V$, the surface $S$ bounding the volume $V$ (Green's Theorem) or, as we shall see, as the potential of all the charges in the volume exterior to $V$, 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:

\begin{displaymath}
G(\mbox{\boldmath$x$},\mbox{\boldmath$x$}_0) = F(\mbox{\bold...
...-1}{4\pi\vert\mbox{\boldmath$x$}- \mbox{\boldmath$x$}_0\vert}
\end{displaymath} (11.40)

where $\nabla^2F(\mbox{\boldmath$x$},\mbox{\boldmath$x$}_0) = \nabla^2_0 F(\mbox{\boldmath$x$},\mbox{\boldmath$x$}_0) = 0$ 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 2013-01-04