Double layer potential

Remarks

The integral is a surface integral with an integrand that is smooth in $x$ off $\partial\Omega$, and $D\varphi$ is harmonic on $\R^n \setminus \partial\Omega$. For $n = 3$, $\Phi(x) = 1/(4\pi\abs{x})$ and the kernel is $\dfrac{\nu(y) \cdot (x - y)}{4\pi \abs{x - y}^3}$.

Example (Gauss's integral): let $\varphi \equiv 1$. By the divergence theorem, $D1(x) = -1$ for $x \in \Omega$ and $D1(x) = 0$ for $x \notin \overline{\Omega}$. For the unit ball in $\R^3$ and $x = 0$ this is a direct check: $\nu(y) = y$, so the kernel is $y \cdot (-y)/(4\pi) = -1/(4\pi)$, and the sphere has area $4\pi$.

Non-example: replacing the normal derivative of $\Phi$ by $\Phi$ itself gives the single layer potential instead. For the unit ball in $\R^3$ with $\varphi \equiv 1$, that potential equals $1$ inside the ball, while the double layer potential equals $-1$ there.