Let us consider a sphere made of a linear, homogeneous and isotrope dielectric with constant $\epsilon_r$ which is inserted in a vacuum where a uniform electric field $\underline{E}_0$ was applied (now it has obviously been distorted by the field $\underline{E}_r$ exerted by the polarized sphere). Say we want to find the total field $\underline{E} = \underline{E}_0 + \underline{E}_0$ in the whole space. In order to do so one may develop some intuition by considering a sphere made of conductor (which is approximately a sphere made of dielectric with $\epsilon_r \to +\infty$) thus concluding that the field inside of the sphere might be uniform. After some direct computation (or after having exploited the analogy with the sphere made of a conductor some more) one thus reach an expression for $\underline{E}$.

Now, here is the problem: the solution was found starting from the hypothesis that $\underline{E}_r$ be uniform inside of the sphere, therefore we need to check that the solution is actually correct by applying the uniqueness theorem. However I do not see why is this theorem true when there is a dielectric: indeed, one may suggest applying the one we know in the absence of dielectrics to the region inside the sphere and to the region outside the sphere but then we do not have proper boundary conditions. We only know the continuity of the tangential component and the discontinuity of the normal component of the electric field which however seems to me may be satisfied by different pairs of solutions in the two regions.

I would really appreciate if anyone knows how to prove the uniqueness theorem in this and in more general situations where dielectrics are present. Also, I know that similar questions where asked (see e.g. Uniqueness of Poisson equation solution with dielectrics) but they did not receive complete answers, especially concerning the case I have considered above in which $\epsilon_r$ is discontinuous.


1 Answer 1


First, I suppose that the discontinuous $\epsilon_r$ is not a significant problem, because the discontinuous material properties can be expressed in terms of the Heaviside step function, and the Heaviside step function has its own derivative; this does not seem to be a difficulty. For example, see here, where the magnetic permeability is discontinuous.

In any case, I will be writing about the electrostatic problem where there are dielectric discontinuities. The target equation is given as $\nabla\cdot\vec{D}=\rho$. As usual, we include the scalar electric potential $\phi$ as the fundamental variable. \begin{equation} -\nabla\cdot\left[\epsilon(x,y,z)\nabla\phi(x,y,z)\right]-\rho(x,y,z)=0,\tag{1} \end{equation} where $\epsilon(x,y,z)$ is the coordinate dependent electric permittivity (positive everywhere) and the charge densitiy distribution $\rho(x,y,z)$ is assumed to be given.

Let $\Omega$ be the target domain, and let Dirichlet and Neuman boundary conditions be specified at some outer boundary of $\Omega$. The outer boundary surface of the domain is represented as $\partial\Omega$. \begin{equation} \begin{split} \phi&=\text{prescribed value on }\partial\Omega_D \\ \frac{\partial \phi}{\partial n}&=0\text{ on }\partial\Omega_N\\ \partial\Omega&=\partial\Omega_D\cup\partial\Omega_N,\;\;\partial\Omega_D\cap\partial\Omega_N=\text{null} \end{split}\tag{2} \end{equation} And we add the condition that at the outer boundary surface, $\epsilon$ is a constant and its value is in vacuum (all dielectric materials are inside the domain). \begin{equation} \epsilon=\epsilon_0 \text{ on }\partial\Omega \tag{3} \end{equation}

Let's start with the proof of uniqueness. Suppose there are 2 different solutions $\phi_1$ and $\phi_2$ to (1) and (2) and let us define the difference as \begin{equation} \phi_D:=\phi_2-\phi_1\tag{4} \end{equation} Since $\phi_1$ and $\phi_2$ satisfy equation (1), the difference $\phi_D$ satisfies \begin{equation} \nabla\cdot\left[\epsilon(x,y,z)\nabla\phi_D(x,y,z)\right]=0.\tag{5} \end{equation} Consider solving this equation. To introduce the weak convergence argument, we have an arbitrary (i.e. all) weight function $\delta\phi$. Think of $\delta\phi$ here as a one-digit quantity, not the product of two. Note that $\delta\phi$ must vanish at the Dirichlet boundary: \begin{equation} \delta\phi=0\text{ on }\partial\Omega_D\tag{6} \end{equation} Multiply the weight function by equation (5) and integrate by volume in the domain. \begin{equation} \int_{\Omega}\delta\phi\nabla\cdot\left[\epsilon(x,y,z)\nabla\phi_D(x,y,z)\right]dV=0.\tag{7} \end{equation} Apply the partial integration to equation (7) as we always do. \begin{equation} -\int_{\Omega}\nabla\delta\phi\cdot\left[\epsilon(x,y,z)\nabla\phi_D\right]dV +\int_{\partial\Omega}\epsilon_0\delta\phi\frac{\partial\phi_D}{\partial n}\cdot\vec{n}dS =0.\tag{8} \end{equation} The second term of (8), which is the surface integration, is zero. This is because of equation (6) and the Neumann boundary condition. Since $\delta\phi$ is arbitrary, we can choose $\delta\phi=\phi_D$. Substituting this into (8) we get, \begin{equation} \begin{split} \int_{\Omega}\epsilon(x,y,z)(\nabla\phi_D)^2dV=0. \\ \therefore \phi_D=\phi_1-\phi_2=0 \end{split} \tag{9} \end{equation} This proves the uniqueness of $\phi$ in the case that dielectric materials are included.

Returning to your concern, that is, the problem of a dielectric sphere in a uniform electric field. It is when the dielectric constant is given as \begin{equation} \epsilon(x,y,z)=\epsilon_0\left[\epsilon_r1_{\text{IN}}(x,y,z)+1_{\text{OUT}}(x,y,z)\right]\tag{10}, \end{equation} Here $1_{\text{IN}}(x,y,z)$ and $1_{\text{OUT}}(x,y,z)$ are simple extensions of Heaviside's step function, which is a function of 1 inside the sphere and 0 outside, or 0 inside the sphere and 1 outside. This case is included in the above described uniqueness theorem.

  • $\begingroup$ Thank you very much for the detailed answer. I think I get your proof. Just two questions: 1. the key idea to solve my doubt is thus to use a regular representation of the Heaviside’s step function so that the main equation still holds? 2. I am not sure if you ever used the boundary conditions on the electric field between different dielectrics, is it ok anyway? $\endgroup$ Apr 12, 2023 at 7:38
  • $\begingroup$ The answers to the questions are 'yes' and 'ok anyway'. Maxwell's equations are valid for the whole domain. If it holds for the whole domain, it holds for any domain, even if it is divided into subdomains. To obtain a particular analytic solution, the solutions for each subdomain are combined by the conditions between the subdomains. Note that the imposed condition, for example that the tangential component of the electric field is continuous, is derived from the fact that Maxwell's equations can be applied in the whole domain. $\endgroup$
    – HEMMI
    Apr 13, 2023 at 3:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.