Jump relations for layer potentials

Remarks

Proof idea: for $\varphi \equiv 1$, the divergence theorem gives Gauss's integral, $D1 = -1$ in $\Omega$, $D1 = 0$ outside $\overline{\Omega}$ and $K1 = -\tfrac12$ on $\partial\Omega$. For general $\varphi$, write $D\varphi(x) = D\big(\varphi - \varphi(x_0)\big)(x) + \varphi(x_0)\, D1(x)$ for $x$ near $x_0 \in \partial\Omega$. The first term is continuous across $\partial\Omega$ at $x_0$, because the $C^2$ boundary makes the kernel weakly singular and the density vanishes at $x_0$. So the whole jump comes from $\varphi(x_0)\, D1$, and the normal derivative of $S\varphi$ is handled in the same way.

The jumps can be read off: $D_+\varphi - D_-\varphi = \varphi$ and $\partial_\nu S_-\varphi - \partial_\nu S_+\varphi = \varphi$. Example: for the unit ball in $\R^3$ and $\varphi \equiv 1$, $S1 = 1$ inside and $S1 = 1/\abs{x}$ outside. So $\partial_\nu S_-1 = 0$ and $\partial_\nu S_+1 = -1$ on the sphere, and both equal $K'1 \mp \tfrac12$ with $K'1 = -\tfrac12$.

Going deeper

For a student

These formulas are the bridge between the Laplace equation in a domain and integral equations on its boundary. If one looks for the solution of the Dirichlet problem as $D\varphi$, part 2 turns the boundary condition into the equation $K\varphi - \tfrac12\varphi = f$; this is the boundary integral method. For the Neumann problem one uses $S\varphi$ and part 3 in the same way. Because $K$ and $K'$ are compact operators on $C(\partial\Omega)$ and adjoint to each other for the pairing of $L^2(\partial\Omega)$, the Fredholm alternative then governs solvability.