Boundary integral method for the Dirichlet problem

Remarks

By the jump relations, the interior boundary values of $D\varphi$ are $K\varphi - \tfrac12\varphi$, so step 2 is exactly the condition that $u$ is continuous on $\overline{\Omega}$ with $u = f$ on $\partial\Omega$. Since $\partial\Omega$ is $C^2$, the kernel of $K$ is weakly singular and $K$ is a compact operator on $C(\partial\Omega)$. So the equation is a Fredholm equation of the second kind. By the Fredholm alternative, it is solvable for every $f$ once the homogeneous equation has only the solution $\varphi = 0$. This holds when $\R^n \setminus \overline{\Omega}$ is connected.

Example: let $\Omega$ be the unit ball in $\R^3$ and $f \equiv 1$. Gauss's integral gives $K1 = -\tfrac12$ on the sphere, so $\varphi \equiv -1$ solves $-\tfrac12\varphi + K\varphi = \tfrac12 + \tfrac12 = 1$. Then $u = D(-1) = 1$ in the ball, the correct solution.

Non-example: let $\Omega$ be the shell $\set{x \in \R^3 : 1 < \abs{x} < 2}$, whose complement has the bounded component $\set{\abs{x} \le 1}$. Let $\varphi = 1$ on the inner sphere and $\varphi = 0$ on the outer sphere. The outward normal of $\Omega$ on the inner sphere points toward the origin, so $D\varphi = 1$ in the inner ball and $D\varphi = 0$ outside it, in particular on $\Omega$. Thus $\varphi \ne 0$ solves the homogeneous equation, and the equation of step 2 is not uniquely solvable.