Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 3 / 10A (Vector Calculus)

Write $p={\sum }_{j=0}^n{a}_j{x}^{n-j}{y}^j$. Equating the coefficient of $x^{n-j-2}{y}^j$ in ${\nabla }^{2}p$ givesThus all even coefficients are determined by $a_0$ and all odd coefficients by $a_1$, with no further constraints. These two choices give two independent harmonic homogeneous polynomials.

For radial functions, ${\nabla }^{2}h=h^{\prime \prime }+2h^{\prime }/r$. Since ${\nabla }^2{e}^{-r}=e^{-r}-2e^{-r}/r$ and ${\nabla }^2{r}^{-4}=12r^{-6}$, decay and the boundary value give

The angular Laplacian sends $\sin \theta$ to $\cos (2\theta )/\sin \theta$, while $1/r$ is radially harmonic. Therefore

The $\varphi$ part contributes $-\sin \varphi /(r^2{\sin }^2\theta )$, and again $1/r$ is radially harmonic. Hence

Set $A=x(1-x)$, $B=y(1-y)$ and $C=z(1-z)$. The forcing is $AB+AC+BC$, while ${\nabla }^2(ABC)=-2(AB+AC+BC)$. Thuswhich vanishes on every face.

Each sine product is a Dirichlet eigenfunction. The squared wave-number sums are $24{\pi }^2$ and $30{\pi }^2$, respectively, so