Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 3 / 11A (Vector Calculus)

Apply the divergence theorem to $\mathbf{F}=\psi \nabla \varphi$, for functions smooth enough on the closure of the bounded piecewise smooth volume. The product rule givesThus Green's first theorem, also called Green's first identity, isThe normal is outward. This identifies the boundary term and its sign directly from the divergence theorem.

For the usual Dirichlet problem on a bounded regular volume, suppose ${\varphi }_1,{\varphi }_2$ both satisfy the stated data. Their difference $u={\varphi }_1-{\varphi }_2$ is a harmonic function, with ${\nabla }^{2}u=0$ in $V$ and $u=0$ on $\partial V$. In Green's first identity, choose both functions equal to $u$:The nonnegative continuous integrand forces $\nabla u=0$. Thus $u$ is constant on each connected component, and its zero boundary values force each constant to vanish. This proves uniqueness for the Dirichlet problem.

If the boundary value is a constant $c$, the constant function $c$ is itself a harmonic function with those data. Uniqueness therefore yieldsBoundedness, or an appropriate condition at infinity replacing it, is essential. On the upper half-space both $u=0$ and $u=z$ are harmonic functions with zero data on its boundary plane. They show why unrestricted uniqueness must not be asserted for an unbounded volume.

Put $\mathbf{x}=r\mathbf{\omega }$, with $|\mathbf{\omega }|=1$. The Jacobian determinant in spherical coordinates gives $dS=r^{2}d\omega =r^2\sin \theta d\theta d\phi$, soDifferentiation under this fixed-domain integral yieldsThe last step is the divergence theorem and the Laplace equation. Hence $f$ is constant on radii for which the closed ball lies inside $V$. Continuity at the origin gives ${\lim }_{r\to 0}f(r)=\varphi (0)$, proving the mean value property for harmonic functions:

The continuous harmonic function has $m\le \varphi (\mathbf{x})\le M$ on the compact sphere. Integrate these inequalities over its area $4\pi r^2$ and use the mean value property for harmonic functions:The centre value is an average with positive weights, so it lies between the smallest and largest boundary-sphere values.