Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / 3

Use the Dirichlet Green function with the sign conventionThe normal derivative is taken with respect to the integration variable and the outward normal of $V$. Green's second identity givesThe Dirac delta distribution evaluates the first volume term at ${\mathbf{r}}^{\prime }$. Substituting $\Delta u=\rho$, $u=f$ on $S$ and $G=0$ there yieldsHenceOne can justify applying the identity to the singular Green function by removing a small sphere around the source and then taking its radius to zero.

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:

Reflect the physical source across $z=0$, writing $\overline{{\mathbf{r}}^{\prime }}=(x^{\prime },y^{\prime },-z^{\prime })$. Its radius is still $s=|{\mathbf{r}}^{\prime }|$, and its spherical image is $\overline{{\mathbf{r}}^{\ast }}$. The Dirichlet Green function inside a hemisphere isExplicitly this four-source method of images isBoth $G_0$ terms vanish on the spherical cap. On the flat equatorial disc, reflection equates the two physical-source distances and also the two image-source distances, so both pairs cancel. The reflected physical source lies in the lower half-ball and the two inverted images lie outside the sphere. Consequently the only source in the upper half-ball is ${\mathbf{r}}^{\prime }$, with the correct coefficient:The Dirichlet uniqueness for the Poisson equation then identifies this as the required Green function.