# Uniqueness of Poisson's equation in presence of unknown bound charge

I am familiar with the usual tidy uniqueness proofs for solutions of Poisson's equation, $$\nabla^2 \phi = -\rho/\epsilon$$ which show that if we have two solutions $$\phi'$$, $$\phi''$$, then $$\int_{\Gamma} dV \nabla(\phi' - \phi'') \nabla(\phi' - \phi'') = \int_{\partial \Gamma}(\phi' - \phi'') \nabla(\phi' - \phi'')\cdot \vec{dA} - \int_{\Gamma} dV (\phi' - \phi'') \nabla^2(\phi' - \phi'')$$ I'll be assuming throughout that I am considering a nice, simple volume $$\Gamma$$, and that I've specified boundary conditions on the boundary $$\partial \Gamma$$. The magic is that for appropriate boundary conditions, like von Neumman and Dirichlet or a mixture of the two on different parts of $$\partial \Gamma$$, the surface integral on the right hand side vanishes. Furthermore, $$\nabla^2(\phi' - \phi'')$$ vanishes everywhere, so the volume integral on the right vanishes, leaving

$$\int_{\Gamma} dV \nabla(\phi' - \phi'') \nabla(\phi' - \phi'') = 0$$ which implies that $$\nabla(\phi' - \phi'')$$ is zero in the volume, so $$\phi' - \phi''$$ is constant. All is well.

My trouble is that $$\rho$$ is often not fully specified! For example, if we have a boundary between two dielectric media within $$\Gamma$$, I expect that the boundary will have bound charge from the material response. That is, $$\rho = \rho_{\rm free,\, specified} + \rho_{\rm bound,\, unspecified}$$. For simple materials, the problem gains additional boundary conditions in place of specifying $$\rho_{\rm bound,\, unspecified}$$ at the onset.

That is, one expects $$\nabla^2 \phi = -\rho_{\rm free, specified}/\epsilon$$ away from the boundary between the simple materials, along with the additional perpendicular-to-the-boundary boundary condition $$\epsilon_{a} \frac{\partial \psi}{ \partial n}\big{|}_{a} = \epsilon_{b} \frac{\partial \psi}{ \partial n}\large|_{b}$$ at the boundary between the simple materials. This latter boundary is not a part of $$\partial \Gamma$$, and this boundary condition is of a different flavor than Dirichlet or von Neumann, as it is merely a consistency equation and not a specification.

How do I prove uniqueness (given the usual types of Dirichlet, von Neumann b.c. on $$\partial \Gamma$$) for such a problem, where additional boundary conditions within $$\Gamma$$ supplant fully specifying $$\rho$$ in $$\Gamma$$? My struggle is that I do not see that $$\phi' - \phi''$$ would satisfy Laplace's equation, since a priori the bound charge could be different for the two solutions.

Here is a simple variant on the proof in the OP that occurred to me after posting the bounty. I am once again considering two solutions $$\phi'$$ and $$\phi''$$, with $$\vec{D} = \epsilon \vec{E} = -\epsilon \nabla \phi$$.
The divergence theorem and a little rearranging nets us $$\int_{\Gamma} dV \nabla(\phi' - \phi'') (\vec{D}'-\vec{D}'') = \int_{\partial \Gamma}(\phi' - \phi'') (\vec{D}'-\vec{D}'')\cdot \vec{dA} - \int_{\Gamma} dV (\phi' - \phi'') \nabla \cdot (\vec{D}'-\vec{D}'')$$
Then, once again, for Dirichlet boundary conditions, $$\phi' - \phi''$$ is zero on the boundary, while for von Neumann boundary conditions, $$\vec{D}' - \vec{D}''$$ is zero on the boundary (since it's proportional to $$0$$). Thus, for these boundary conditions or an appropriate mixture, the boundary term vanishes.
Furthermore, $$\nabla \cdot \vec{D}' =\nabla \cdot \vec{D}'' = \rho_{\rm free,\, specified}$$, so the second term on the right hand side also vanishes. Thus, $$\int_{\Gamma} dV \nabla(\phi' - \phi'') (\vec{D}'-\vec{D}'') = 0$$ which means $$\int_{\Gamma} dV \epsilon |\nabla(\phi' - \phi'')|^2 = 0$$
If $$\epsilon$$ does not change sign, then $$\nabla(\phi' - \phi'')=0$$, proving uniqueness of the electric field.