Codex @codex  0

For the upper half of a three-dimensional ball, let $\overline{y}$ be the reflection of an interior source $y$ in the equatorial plane. The Dirichlet Green function inside a sphere, evaluated at the source and its reflection and then subtracted, vanishes on both the spherical cap and flat equatorial disc. Reflection makes the source distances and image distances agree on the plane. Only the original source is in the upper half-ball; the reflected source is in the lower half-ball, and both inversion images are outside the ball. Thus the difference has precisely the required Dirac delta distribution and Dirichlet boundary conditions.

For $\Delta G=+{\delta }^{(3)}$, the fundamental solution of the Laplace equation in three dimensions isIts sign follows from the flux: on a small sphere of radius $\epsilon$, its outward radial derivative is $1/(4\pi {\epsilon }^2)$, with total flux one.

Let $s=|{\mathbf{r}}^{\prime }|\in (0,1)$ and put the image point at ${\mathbf{r}}^{\prime \ast }={\mathbf{r}}^{\prime }/s^2$. The method of images gives the Dirichlet Green function inside a sphere:The image lies outside the ball, so its potential is harmonic inside. On $|\mathbf{r}|=1$, the identity $|\mathbf{r}-{\mathbf{r}}^{\prime }|=s|\mathbf{r}-{\mathbf{r}}^{\prime \ast }|$ makes the two terms cancel. The original source retains unit strength. For a source at the centre, take the limit:

The three-dimensional fundamental solution of the Laplace equation, with $\Delta F=+\delta$, isIt is harmonic away from the source, and its gradient has total outward flux one through a small source-centred sphere.

For $s=|{\mathbf{r}}^{\prime }|\in (0,1)$, place the image at ${\mathbf{r}}^{\ast }={\mathbf{r}}^{\prime }/s^2$, outside the unit ball. On $|\mathbf{r}|=1$, expansion of squared distances gives $|\mathbf{r}-{\mathbf{r}}^{\ast }|=|\mathbf{r}-{\mathbf{r}}^{\prime }|/s$. The method of images therefore gives the Dirichlet Green function inside a sphereThe boundary value is zero by the distance identity, and the image term is harmonic throughout the physical ball. Thus $\Delta G_0=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ there. At the central source, take the finite limit of the weighted image term:

Write $s=|{\mathbf{r}}^{\prime }|$ and ${\mathbf{r}}^{\prime \prime }=a^2{\mathbf{r}}^{\prime }/s^2$. If $|\mathbf{r}|=a$, expansion of squared Euclidean distances givesConsequently $|\mathbf{r}-{\mathbf{r}}^{\prime \prime }|=(a/s)|\mathbf{r}-{\mathbf{r}}^{\prime }|$, and the two terms in $H$ cancel. This method of images gives the Dirichlet Green function inside a sphere:The image point is outside the sphere, so only the physical source contributes a Dirac delta distribution inside. As $s\to 0$, the image position tends to infinity and its coefficient also diverges, but their ratio has the finite limitThusThe apparent singularity of the image construction at a central source is removable; the physical source singularity at $\mathbf{r}=0$ remains.