old App A lemma 0 REVEIWED

This original first section of Appendix A got amped up into a longer section now called A.0. The new section deals with all charges including those on pieces of metal. Here I just keep a copy of the old. 0. Lemma: The Poisson equation 2φ = ρ has a unique solution. Needs a rewrite, Stak has lots to say. Proof: By Poisson equation we mean 2φ = ρ, the way it usually appears in electrostatics, ignoring constants. The usual method of solving this equation is to first solve 2g(x,x') = δ(x-x') for the Green's Function g(x,x'). The answer is g(x,x') = -1/(4π|x - x'|) in the case that there are no boundary conditions. -- this solution is known as the free-space propagator, or the fundamental solution. From this fact, one concludes that the Poisson solution is given by, φ(x) = - To show this satisfies 2φ = ρ, apply 2 to both sides. Inside the integral, 2 hits 1/|x - x'| = - 4πg(x,x') to give -4πδ(x-x') which then picks out -4πρ(x), and the -4π's cancel. Since we have an explicit formula for the solution of the Poisson equation, it must have a unique solution [ why?] . In the presence of boundary conditions, the uniqueness proof requires a bit more work. For example the Poisson equation is valid inside a closed region of space on whose bounding surface φ is specified (interior Dirichlet boundary value problem), the solution is also unique, see Stakgold Vol 2 p 102.