Boundary integral equation for the Neumann problem

Remarks

Part 2 follows from the jump relations for layer potentials: the normal derivative of the single layer potential from inside is $K'\psi + \frac12 \psi$. For part 1, $K'$ is compact on $C(\partial D)$ and is the adjoint of the double layer operator $K$ with respect to the pairing $\int_{\partial D} \varphi \psi \, ds$. The null space of $I + 2K$ consists of the constants, since $K1 = -\frac12$ by Gauss's formula for the double layer potential of a constant density. By the Fredholm alternative, $\psi + 2K'\psi = 2g$ is solvable if and only if $2g$ is orthogonal to the constants.

Example: on the unit disc in $\R^2$, $\frac{\partial \Phi(x - y)}{\partial \nu(x)} = -\frac{1}{4\pi}$ for $x \ne y$ on the unit circle, so $K'\psi = -\frac{1}{4\pi} \int \psi \, ds$. For $g(\theta) = \cos\theta$ the equation $\psi + 2K'\psi = 2g$ has the solution $\psi = 2\cos\theta$, since $\int_0^{2\pi} \cos\theta \, d\theta = 0$; the resulting single layer potential is $u(x) = x_1$ in the disc, whose normal derivative on the circle is $\cos\theta$. For $g = 1$ the condition $\int g \, ds = 0$ fails, and indeed no harmonic $u$ in $D$ has $\partial_\nu u = 1$, because $\int_{\partial D} \partial_\nu u \, ds = \int_D \Delta u \, dx = 0$.

Going deeper

For a student

This is the companion of the boundary integral equation for the Dirichlet problem, which uses the double layer potential. The two boundary operators $K$ and $K'$ are adjoint to each other, and the solvability condition comes from the Fredholm alternative applied to this pair. The same equations, with the signs changed, treat the exterior problems. Numerically, both equations are discretized by boundary element or Nyström methods, which reduce the dimension of the problem by one.