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

  • $\begingroup$ 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. $\endgroup$ Commented Apr 7, 2023 at 9:56

1 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.

  • $\begingroup$ 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 $\epsilon_r=1$). $\endgroup$
    – Leonardo
    Commented Jan 8, 2023 at 15:26
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Jan 8, 2023 at 15:30
  • $\begingroup$ 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 $D$)? $\endgroup$
    – Leonardo
    Commented Jan 8, 2023 at 15:35
  • $\begingroup$ @Leonardo: Yes, that's correct. $\endgroup$ Commented Jan 8, 2023 at 15:53
  • $\begingroup$ @MichaelSeifert I have yet to see Laplace Equation rigorously (I have only seen it in physics), however my intuition tells me that such a limiting procedure does not yield a unique solution. Think for example of two different sequences converging to the same limit. Consider also the fact that as the border of the volume of integration of Laplace Equation tends to infinity (say by a sequence of balls of ever growing radius) we need to impose "normal conditions at infinity" to guarantee uniqueness, despite the fact that for each ball the solution is unique. $\endgroup$ Commented Apr 7, 2023 at 9:45

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