Physical meaning of terms in Poisson equation representation solution

Question

Given a bounded normal domain V, and a scalar potential $u(r_0)$, $r_0$ being a $3D$ position vector, the representation theorem for solutions to Poisson's equation states: $$u(r_0) = \frac{1}{4 \pi} \int_{\partial V} (\frac{1}{|r - r_0|} \frac{\partial u(r)}{\partial \nu} - u(r)\frac{\partial(\frac{1}{|r - r_0|})}{\partial \nu}) dS - \frac{1}{4\pi} \int_{V} \frac{\nabla^2 u(r)}{|r-r_0|} dV$$

Here, r is another $3D$ vector, the variable of integration. $\nu$ is the unit outward normal vector on the surface. This is a representation form for the solution of the Poisson equation as can be found in e.g, Zachmanoglou pg 191 theorem 5.1, or here pg 2, equation 4.3.

Now, let's say that $u(r_0)$ is an electrostatic potential. I want to understand the physical significance of the terms in this equation.

The volume integral term is the potential due to the volume charge density, which makes sense.

We see that if u is the electrostatic potential, $\frac{\partial u(r)}{\partial \nu} = \nabla u \cdot \nu$ is the normal part of the electric field at the surface of the region. By Gauss's law with $\epsilon_0 = 1$, the normal part of the electric field on the surface is equal to the surface charge density (Griffiths Electrodynamics) so that the term in $\frac{\partial u}{\partial \nu}$ is the potential due to the surface charges.

But then what is the $u(r)\frac{\partial(\frac{1}{|r - r_0|})}{\partial \nu}$ term supposed to be? It seems like the other two terms have already accounted for the necessary physical effects.

Answer 1

When you say that the surface charge density is given by the normal component of the electric field, you are assuming the field is zero outside the surface. Similarly if you take the potential itself to be zero outside the surface, then the other term can be interpreted as the electric dipole layer necessary to give the discontinuity in potential required by your boundary conditions.

Added to address comment

You can understand the dipole layer by looking at a plane at $z=d/2$ with surface charge given by charge per area $\sigma$, above a second plane at $z=-d/2$ with charge per area $-\sigma$. The electric field will $-\sigma\hat z$ for $|z|<d/2$. The potential difference between $z = d/2$ and $z=-d/2$ will be $\sigma d$. Taking the limit that $d\rightarrow 0$, $\sigma\rightarrow \infty$ with $\sigma d$ fixed will give an elementary dipole layer with dipole moment per unit area of $\sigma d$, and this same discontinuity in the potential. Getting infinitesimally close to the surface, a local surface dipole moment density looks like this plane of surface dipole moment.