Method of images

Remarks

Example: for the half-space $\Omega = \set{y : y_n > 0}$ and $x = (x', x_n)$, the single point $x_1 = (x', -x_n)$ with $c_1 = 1$ is an image system, since $\abs{y - x} = \abs{y - x_1}$ when $y_n = 0$. This gives $G(x, y) = \Phi(y - x) - \Phi(y - x_1)$. For the unit ball and $x \ne 0$, the inverse point $x^* = x/\abs{x}^2$ with $c = \abs{x}^{2-n}$ is an image system. Non-example: for the half-space, the point $(x', -x_n)$ with $c = -1$ fails clause 2, since then $h = -\Phi(y - x) \ne \Phi(y - x)$ on the boundary. In the plane the same idea works with $\Phi(z) = \frac{1}{2\pi} \log \frac{1}{\abs{z}}$, where $h$ may also contain an additive constant; for the unit disc this gives the Green's function $\frac{1}{2\pi} \log \abs{\frac{1 - \bar x y}{y - x}}$ in complex notation.