Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 3

Apply the product rule to the vector field $\Phi \nabla \Psi -\Psi \nabla \Phi$:The gradient cross terms cancel. Integrating and applying the divergence theorem with outward unit normal $\mathbf{n}$ proves Green's second identity:Choose the Dirichlet Green function $G_D(\mathbf{r},{\mathbf{r}}^{\prime })$ with ${\Delta }_{\mathbf{r}}{G}_D=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ and $G_D=0$ on $S$. Taking $\Psi =G_D$ and using the Dirac delta distribution, the left side becomes $\Phi ({\mathbf{r}}^{\prime })$ while the second boundary term vanishes. HenceThe plus sign is tied to the convention $\Delta G_D=+\delta$, whose free-space fundamental solution is negative in three dimensions. The singularity may equivalently be handled by excising a small sphere about ${\mathbf{r}}^{\prime }$ before applying the smooth identity.

Put $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|$. Away from the source, the radial Laplacian in three dimensions givesThus $\boxed{\Delta G_0=0\text{ for }\mathbf{r}\ne {\mathbf{r}}^{\prime }}$. At the source it is a distribution, not a smooth function; its outward gradient flux through a small sphere is $4\pi s^2/(4\pi s^2)=1$, giving $\Delta G_0=\delta$.

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.

Let $\overline{{\mathbf{r}}^{\prime }}=(x^{\prime },y^{\prime },-z^{\prime })$ be the reflection of the source in the equatorial plane. For a source in the upper half-ball, the Dirichlet Green function inside a hemisphere isBoth terms vanish on the spherical boundary. On the flat boundary $z=0$, the two source distances are equal, and the distances to their spherical image points are also equal, so the difference vanishes there too. The reflected physical source is in the lower half-ball and both inverted image points are outside the sphere. Therefore the only source singularity in the desired region is the original one, with coefficient $+1$ in $\Delta G$. This combines reflection with the spherical method of images.

Reflecting repeatedly in the two parallel planes creates an infinite method of images. For ${\mathbf{r}}^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$, introduceThe Dirichlet Green function between parallel planes isEach bracket is $O(n^{-2})$, so the paired series converges absolutely away from the sources. At $x=-1$, reflection pairs the positive image indexed by $n$ with the negative image indexed by $-n$. At $x=1$ it pairs that positive image with the negative image indexed by $1-n$. The distances agree in each pairing, making both boundary values zero. This can be checked with symmetric finite partial sums first; the unmatched end terms tend to zero, so no subtraction of separately divergent series is needed. Only ${\mathbf{r}}_0^{+}={\mathbf{r}}^{\prime }$ lies within the slab, so ${\Delta }_{\mathbf{r}}{G}_{\mathrm{slab}}=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$. This is the usual Green function decaying at large transverse separation; such a condition selects the physical solution in the unbounded region.