Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics V / 2

For a bounded region $R\subseteq {\mathbb{R}}^3$ with piecewise smooth boundary and a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of its closure, the divergence theorem sayswhere $\mathbf{n}$ is the outward unit normal vector. The left side integrates divergence over the volume; the right side is the outward flux.

Apply the divergence theorem to $\mathbf{F}=u\nabla v-v\nabla u$. The product rule givesbecause the two gradient inner products cancel. Also $\mathbf{F}\cdot \mathbf{n}=u{\partial }_{n}v-v{\partial }_{n}u$. ThusThis is Green's second identity, under the usual regularity assumptions permitting these derivatives and boundary integrals.

Take $v=|\mathbf{x}{|}^2$ in Green's second identity. Since ${\nabla }^{2}u=0$, ${\nabla }^{2}v=6$, $v=1$ on the unit sphere, and ${\partial }_{n}v=2$ there,The final term vanishes by the divergence theorem, because it equals ${\iiint }_B{\nabla }^{2}udV=0$. ThereforeWe assume sufficient boundary regularity for the stated boundary values and integrals, or use the corresponding trace/limit formulation. This is the equality of volume and surface averages for a harmonic function, since the unit ball and sphere have volume $4\pi /3$ and area $4\pi$.

Parametrize the surface of revolution byAn oriented vector surface area element isUse the vector field $\mathbf{F}=(x/2,y/2,0)$, whose divergence is one. If this parametrization has the outward orientation on an enclosing surface, the divergence theorem givesThe second equality is integration by parts, using the endpoint values. For example this outward convention applies when the positive-$t$ branch lies above the negative-$t$ branch, $f(t)>f(-t)$ for $0<t<\pi /2$.

The stated $C^1$ regularity and endpoint conditions alone do not guarantee an embedded enclosing surface: branches can cross or coincide. Without an appropriate orientation assumption the flux formula is an algebraic signed volume. If “inside” is defined as the region between the two graph branches for each radius, its nonnegative geometric volume is insteadThis explicitly handles a reversal or change in the order of the two branches.

For $f(t)=\sin 2t$, the positive-$t$ branch is above the negative one, so the orientation in part i is outward. ConsequentlyThe integral follows from ${\sin }^{2}2t=(1-\cos 4t)/2$.

If two solutions exist, their difference $w$ satisfies ${\nabla }^{2}w-w=0$ inside $B$ and $w=0$ on the boundary. Suppose $w$ has a positive value. Continuity on the closed ball gives a positive maximum at an interior point, where the Hessian matrix is negative semidefinite and hence ${\nabla }^{2}w\le 0$. But the equation gives ${\nabla }^{2}w=w>0$, a contradiction. A negative minimum similarly contradicts ${\nabla }^{2}w=w<0$.

Thus $\boxed{w=0}$, proving uniqueness by the maximum principle for the modified Helmholtz equation. Here a classical solution is understood to be $C^2$ inside and continuous up to the boundary; mere first differentiability would not define the displayed Laplacian classically. With enough boundary regularity, the alternative Green's first identity proof is $0={\int }_{B}w({\nabla }^{2}w-w)=-{\int }_B(|\nabla w{|}^2+w^2)$.