Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3 / 9B (Vector Calculus)

Suppose ${\varphi }_1,{\varphi }_2$ solve the Dirichlet problem and put $u={\varphi }_1-{\varphi }_2$. Then ${\nabla }^{2}u=0$ in $V$ and $u=0$ on $S$. The divergence theorem applied to $u\nabla u$ gives Green's identityThus $\nabla u=0$, so $u$ is constant; its zero boundary value makes it zero. The Dirichlet solution is therefore unique.

For homogeneous Neumann data the same calculation again shows that $u$ is constant, but the boundary condition does not determine that constant. Hence, whenever a Neumann solution exists, adding any constant produces another solution. This is the standard uniqueness of Poisson equation distinction.

For any rotation $R$ about the centre, ${\varphi }_R(x)=\varphi (Rx)$ satisfies the same equation and the same Dirichlet data because $V,f,g$ are rotationally invariant. Uniqueness from part (a) gives ${\varphi }_R=\varphi$ for every $R$. Thus $\varphi$ is constant on spheres and is a function of $r$ alone.

For a radial function in three dimensions, the radial Laplacian givesHere regularity at zero selectsA further integration giveswhose apparent singularity is removable, with limit $3+C$ at zero. Since the nonconstant term has value $4e-8$ at $r=1$, the boundary condition $\varphi (1)=3$ gives $C=11-4e$. Therefore

The radial equation isIts solutions satisfying decay at infinity have the formThe boundary value $\varphi (1)=1$ gives $C=0$, soIt tends to zero as $r\to \infty$ and directly satisfies ${\nabla }^2\varphi =-r^{-3}$.

Regularity at zero givesIntegrating once more,where $\arctan r/r$ has removable limit one at zero. At $r=1$, the nonconstant terms sum to $\frac{1}{2}\log 2+\frac{\pi }{4}$. Hence