Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 3 / 12C / Solution

For a continuously differentiable vector field on a volume and its boundary, the divergence theorem equates ${\int }_V\nabla \cdot FdV$ to the outward flux ${\int }_{\partial V}F\cdot ndS$. For the given field, $\nabla \cdot F=5(x^2+y^2+z^2)=5r^2$, soOn the sphere, $n=(x,y,z)/R$, andThus the surface flux is also $4\pi R^5$, verifying the theorem by explicit calculation.

Apply the product rule to $\nabla \cdot (\varphi \nabla \psi )=\varphi \Delta \psi +\nabla \varphi \cdot \nabla \psi$, then apply the divergence theorem. This proves Green's first identity:Swap $\varphi$ and $\psi$ and subtract. The gradient products cancel, giving Green's second identity:For uniqueness, set $w={\varphi }_1-{\varphi }_2$. It is a harmonic function with zero Dirichlet boundary conditions. Under the stated smoothness assumptions, Green's first identity with $\varphi =\psi =w$ gives ${\int }_V|\nabla w{|}^{2}dV=0$. Thus $w$ is constant on each connected component, and its boundary value makes every constant zero. The two solutions agree throughout $V$. This is Dirichlet uniqueness for the Poisson equation. For classical solutions merely continuous up to the boundary and twice continuously differentiable in the interior, the maximum principle for harmonic functions gives the same conclusion without requiring boundary derivatives.