How to define ##\nabla \cdot D ## at the interface between dielectrics

[coquelicot]

TL;DR Summary

Defining $\nabla \cdot D$ or $\nabla \cdot E$ at the interface between dielectrics

It is believed that Maxwell equations (together with other relations depending on the materials) are sufficient to account for any electromagnetic macroscopic effect.
The problem is that, for a Maxwell equation to hold, it must at least be defined.
Consider for example the case of two dielectrics of distinct permittivities, say $\epsilon_1$ and $\epsilon_2$. We assume for the sake of simplicity that $D_{conductor 1} = \epsilon_1 E_{conductor 1}$ and $D_{conductor 2} = \epsilon_2 E_{conductor 2},$ which already covers a lot of dielectrics.
We also assume that the dielectrics carry no free charge, so Maxwell equation reads $\nabla \cdot D = 0$.
Let finally restrict ourselves to the electrostatic case.

If the two dielectrics interface at a surface S, then the normal component of E is discontinuous at S, its tangential component is continuous, while the normal component of D is continuous and the tangential components of D are, in general, discontinuous.
For the field E, we could define $\nabla \cdot E$ using the Dirac distribution.
I guess something like that is possible for defining $\nabla \cdot D$, despite I don't see exactly how (the difference between E and D is that E has only one discontinuous component).

Now, even if a Dirac distribution artifice is possible whenever the interface between two dielectrics is a surface, I see no way to define $\nabla \cdot E$ or $\nabla \cdot D$ whenever three or more dielectrics interface at an edge, or whenever several dielectric interface at a vertex point.

Any idea/knowledge to save Maxwell equations?

 

[vanhees71]

At boundaries of two media you have to apply the integral definition of the differential operators, possibly generalized to their forms describing discontinuities. If you have no free charge, i.e., $\vec{\nabla} \cdot \vec{D}$ using a Gaussian pill box with two of its surfaces parallel to the boundary tells you that the normal component of $\vec{D}$ is continuous, while the normal component of $\vec{E}$ is not. That's because there's a non-zero net-surface charge along the boundary, because of the different polarizability (permittivities) in the different media. from $\vec{\nabla} \times \vec{E}$ using a surface with two sides parallel to the boundary and using Stokes's integral theorem tells you that the components tangential to the boundary are continuous.

 

[coquelicot]

At boundaries of two media you have to apply the integral definition of the differential operators, possibly generalized to their forms describing discontinuities. If you have no free charge, i.e., $\vec{\nabla} \cdot \vec{D}$ using a Gaussian pill box with two of its surfaces parallel to the boundary tells you that the normal component of $\vec{D}$ is continuous, while the normal component of $\vec{E}$ is not. That's because there's a non-zero net-surface charge along the boundary, because of the different polarizability (permittivities) in the different media. from $\vec{\nabla} \times \vec{E}$ using a surface with two sides parallel to the boundary and using Stokes's integral theorem tells you that the components tangential to the boundary are continuous.

Thanks for your answer.
Yes, that's classic and I wrote that in the question too. Not of any help to define $\nabla \cdot D$ at the interfaces between several dielectrics unfortunately.

 

[vanhees71]

If you have a singularity at the surface, of course, you can't define a quantity on the surface. You have to describe what happens at the surface by describing the corresponding singularities, i.e., in this case by the surface divergence and surface curl.

 

[coquelicot]

If you have a singularity at the surface, of course, you can't define a quantity on the surface. You have to describe what happens at the surface by describing the corresponding singularities, i.e., in this case by the surface divergence and surface curl.

Actually the problem is not surfaces but edges and vertices. The essential point in my question is contained in the last 3 lines.