Symm's integral equation

Example

Let $\Gamma \subset \R^2$ be a closed $C^2$ curve, and let $f$ be a continuous function on $\Gamma$. Symm's integral equation asks for a density $\psi$ on $\Gamma$ with $$-\frac{1}{2\pi} \int_\Gamma \ln \abs{x - y} \, \psi(y) \, ds(y) = f(x), \quad x \in \Gamma.$$ The left side is the single layer potential of $\psi$ in the plane, a logarithmic potential, restricted to $\Gamma$. The unknown appears only under the integral, so it is an integral equation of the first kind, with the weakly singular kernel $-\frac{1}{2\pi}\ln\abs{x - y}$. Solving it gives a single layer potential that solves the interior Dirichlet problem for the Laplace equation with boundary values $f$.

Check on the circle $\Gamma = \set{x : \abs{x} = a}$, with $y = a e^{i\theta}$ and $x = a e^{i\phi}$: by the mean value property, $\frac{1}{2\pi} \int_0^{2\pi} \ln \abs{x - a e^{i\theta}} \, d\theta = \ln a$ for $\abs{x} = a$, and $\ln \abs{e^{i\phi} - e^{i\theta}} = -\sum_{n \ge 1} \frac{\cos n(\phi - \theta)}{n}$. Hence the operator maps $\psi = 1$ to the constant $-a \ln a$, and $\psi(\theta) = e^{in\theta}$ with $n \ne 0$ to $\frac{a}{2\abs{n}} e^{in\phi}$. So on a circle of radius $a \ne 1$ the equation is uniquely solvable for every trigonometric polynomial $f$. For $a = 1$ the constant density $\psi = 1$ solves the homogeneous equation.