# Uniqueness of Poisson equation solution with dielectrics

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

• The only thing that comes to my mind is that once you fixed the potential on one region, the other is uniquely determined by the Neumann boundary condition $\epsilon_r \frac{\partial}{\partial \hat n} V_2 = \frac{\partial}{\partial \hat n} V_1$. It seems to me that we need another constraint and I think that we may get it from the tangential boundary condition $\underline{E}_1 \cdot \hat t = \underline{E}_2 \cdot \hat t$ which is indeed derived from the fact that the electric field must be conservative (hence it should be a new independent constraint). However I do not know how to prove it. Apr 7 at 9:56

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.
• Thank you for your answer. I am supposing that the permittivity $\epsilon_r>1$ is constant inside $D$ (the dielectric is homogeneous) and is discontinuous at $\partial D$ (the permittivity outside $D$ in the void is $\epsilon_r=1$). Jan 8 at 15:26
• @Leonardo: Right, that's what I was trying to address in the second paragraph. If we view the function $$\epsilon(\mathbf{r}) = \begin{cases}\epsilon_r\epsilon_0&\mathbf{r}\in D\\ \epsilon_0&\mathbf{r}\not \in D \end{cases}$$as the limit of a series of smooth permittivity functions $\epsilon(\mathbf{r})$, and $V$ is unique for each of the smooth permittivity functions in the series, then perhaps we can wave our hands and believe that uniqueness continues to hold in the limit. But I freely admit this is far from rigorous. Jan 8 at 15:30
• Ok, it's clear now. Just one question: if $\epsilon(\vec{r})$ is continuous, we can use uniqueness theorem because on the boundary $\partial D$ the electric field is now continuous and so we don't have to solve the Poisson equation in two separate regions (inside and outside $D$)? Jan 8 at 15:35