Uniqueness of Poisson equation solution with dielectrics

Question

Let's suppose we have an isotropic homogeneous dielectric $D$ (with given relative constant $\epsilon_r$) surrounded by the void.

Inside and outside $D$ we can obviously write the Poisson equations:

$\nabla^2 V=\frac{\rho}{\epsilon_0\epsilon_r}$ (inside $D$), $\quad$ $\nabla^2 V=\frac{\rho}{\epsilon_0}$ (outside $D$).

Here $\rho$ simply represents a limited free charge distribution in the space (inside and/or outside $D$, but not on the boundary $\partial D$ of $D$).

On the boundary $\partial D$ (which separates the void and the dielectric) obviously we cannot write the Poisson equation. However on $\partial D$ we obviously have the conditions:

$\hat{n} \cdot (\vec{E}_1-\epsilon_r\vec{E}_2)=0, \quad \hat{n} \times (\vec{E}_1-\vec{E}_2)=0$,

where $\hat{n}$ is the versor normal to the surface $\partial D$, while $\vec{E}_1$ and $\vec{E}_2$ are respectively the electric fields outside and inside $D$.

We moreover impose the Dirichlet boundary condition at infinity such that $V=0$.

---

My question

Are those conditions enough to say that there exists a unique solution $V$?

i.e. we solve $\nabla^2 V=\frac{\rho}{\epsilon_0}$ outside $D$ (with boundary condition $V=0$ at infinity) obtaining the (non unique) solution $V_1$ ($V_1$ is not unique because we have not given the value of $V$ over the boundary $\partial D$, so the uniqueness theorem does not apply);

then we solve $\nabla^2 V=\frac{\rho}{\epsilon_0\epsilon_r}$ inside $D$ obtaining the (non unique) solution $V_2$;

then we connect the two (non unique) solutions through the conditions $\hat{n} \cdot (\vec{E}_1-\epsilon_r\vec{E}_2)=0$ and $\hat{n} \times (\vec{E}_1-\vec{E}_2)=0$ on the boundary $\partial D$, obtaining the total solution $V$.

Is this total solution $V$ unique?

An example of the procedure I used can be found here (page 12, example 2): https://unlcms.unl.edu/cas/physics/tsymbal/teaching/EM-913/section4-Electrostatics.pdf

Answer 1

If the permittivity is smooth (i.e., $\epsilon(\mathbf{r})$ goes from $\epsilon_0$ outside $D$ to some other value inside $D$ continuously, rather than having a discontinuous jump), then the standard proof of the uniqueness to solutions of Poisson's equation applies. Let $W$ be the difference between two solutions $V_A$ and $V_B$ satisfying $\nabla^2 V = \rho_f/\epsilon$, where $\epsilon$ is a smooth function of $\mathbf{r}$. It is not hard to see that $W = V_A - V_B$ satisfies $\nabla^2 W = 0$ and goes to zero at infinity, and so $W = 0$ everywhere. Thus, $V_A = V_B$ everywhere.

If you're willing to accept that the "discontinuous permittivity" in your original problem should be viewed as the limit of a series of smooth permittivity functions, then uniqueness should apply to the discontinuous case as well. However, it would be nice to see a rigorous proof where the discontinuity is taken into account from the start. I will update this answer if I manage to figure out how to do so.