Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 107 / 1

For a harmonic function $u$, letThe divergence theorem givesSince $M(r)\to u(x)$ as $r\downarrow 0$, $M(r)=u(x)$. Integrating the spherical averages in the radial variable gives the corresponding ball average, proving the mean value property for harmonic functions. If $u$ attains its maximum at an interior point, the average of the nonnegative function $\max u-u$ on every sufficiently small centred sphere is zero. Continuity makes $u$ constant on those spheres, and connectedness propagates that value through the domain. Thus the weak maximum principle for elliptic operators gives

For the derivative estimate, choose $r>0$ smaller than half the distance from ${\Omega }^{\prime }$ to $\partial \Omega$, and let ${\rho }_r$ be a smooth radial mollifier supported in $B_r(0)$. Writing its convolution in polar coordinates and using the spherical mean value property shows that $u\ast {\rho }_r=u$ on ${\Omega }^{\prime }$. Hence, for every multi-index $\alpha$,so Holder inequality gives

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Suppose $u_1$ and $u_2$ are weak solutions with the same trace, and put $w=u_1-u_2\in H_0^1(\Omega )$. The weak formulation permits $w$ itself as a test function, givingThus $w$ is almost everywhere constant, and its zero trace makes that constant zero. This proves uniqueness.

The weak identity also says that $\Delta u=0$ in the sense of distributions. The Weyl lemma therefore gives $u\in C^{\infty }(\Omega )$ and $\Delta u=0$ pointwise. The assumed continuity on $\overline{\Omega }$ retains the prescribed boundary values, so the weak solution is the unique classical solution in $C^{\infty }(\Omega )\cap C^0(\overline{\Omega })$.

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Fix a closed ball $\overline{B_R(a)}\subset \Omega$, and let $v$ solve the Dirichlet problem for the Laplace equation in this ball with boundary data $u$. Put $w=u-v$. The function $w$ is continuous, vanishes on the boundary, and inherits the restricted spherical mean identity because $v$ has the full mean value property for harmonic functions.

Suppose $M={\max }_{\overline{B_R(a)}}w>0$. Its maximum set $E$ is a nonempty compact subset of the open ball. Choose $x\in E$ maximizing $|x-a|$. For every sufficiently small radius in the sequence attached to $x$,Equality of the average with the maximum and continuity imply that the whole sphere belongs to $E$. Its point in the direction from $a$ through $x$ lies farther from $a$ than $x$ does; if $x=a$, any point on the sphere does. Both cases contradict the choice of $x$. Hence $w\le 0$, and applying the same argument to $-w$ gives $w=0$.

Thus $u=v$ on every relatively compact ball. It is consequently harmonic and smooth locally, proving the local converse to the mean value property.

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