Green's function of the ball

Example

Let $B = \set{x \in \R^3 : \abs{x} < 1}$ be the unit open ball. For $x \in B$ with $x \ne 0$, let $x^* = x / \abs{x}^2$ be the inverse point of $x$ in the unit sphere; it lies outside $\overline{B}$. Define, for $y \in \overline{B}$ with $y \ne x$, $$G(x, y) = \frac{1}{4\pi} \left( \frac{1}{\abs{x - y}} - \frac{1}{\abs{x} \, \abs{y - x^*}} \right) \quad (x \ne 0), \qquad G(0, y) = \frac{1}{4\pi} \left( \frac{1}{\abs{y}} - 1 \right).$$ This is the Green's function of $B$, normalized so that $\Delta_y G(x, y) = -\delta_x$. Check of the three properties, for $x \ne 0$:

1. The first term is the fundamental solution $\frac{1}{4\pi\abs{x - y}}$, which has the right singularity at $y = x$.
2. The second term is harmonic in $y$ on $B$, because its singularity $x^*$ lies outside $\overline{B}$.
3. For $\abs{y} = 1$ one has $\abs{x}^2 \abs{y - x^*}^2 = \abs{x}^2 - 2 x \cdot y + 1 = \abs{x - y}^2$, so the two terms cancel and $G(x, y) = 0$.

For $x = 0$, the term $1/\abs{y}$ is the fundamental solution times $4\pi$, the constant $1$ is harmonic, and $G(0, y) = 0$ when $\abs{y} = 1$.