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\begin{document}

The green's function for the poisson equation is defined by \( \nabla ^{2}G(r^{i},r'^{j})=-4\pi \delta (r^{i}-r'^{i}) \). Using
greens theorem, you get: \( \phi (r^{i})=\int _{V}d^{3}r'\frac{\rho (r')}{|r^{i}-r'^{i}|}+\frac{1}{4\pi }\int _{S}\frac{1}{|r^{i}-r'^{i}|}\frac{\partial \phi }{\partial n'}da'-\frac{1}{4\pi }\int _{S}\phi (r'^{i})\frac{\partial }{\partial n'}(\frac{1}{|r^{i}-r'^{i}|})da' \).

This means that if S goes to infinity, the solution will return to
the equation for the scalar potential without boundary conditions.


For the dirichlet case, want to get rid of 2nd integral. So want \( G=0 \)
on the surface. Then \( \phi (r^{i})=\int _{V}d^{3}r'G_{D}(r^{i},r'^{j})\rho (r'^{i})-\frac{1}{4\pi }\int _{S}\phi (r'^{i})\frac{\partial }{\partial n'}G_{D}(r^{i},r'^{j})da' \). If S is a perfect conductor, then last term
becomes \( \phi _{S} \). \( \phi (r^{i})=\int _{V}d^{3}r'G_{D}(r^{i},r'^{j})\rho (r'^{i})+\phi _{S} \).

For the neumann problem, choose, \( \frac{\partial }{\partial n'}G_{n}(r^{i},r'^{j})=-4\pi /A_{s} \) on the boundary. Then \( \frac{1}{4\pi }\int _{S}\phi (r'^{i})\frac{\partial G_{N}}{\partial n'}da'=<\phi >_{S} \). so \( \phi (r^{i})=\int _{V}d^{3}r'\frac{G_{N}\rho (r')}{|r^{i}-r'^{i}|}+\frac{1}{4\pi }\int _{S}G_{N}\frac{\partial \phi }{\partial n'}da'+<\phi >_{S} \).

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