Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 2 / 30A / Solution

A fundamental solution of a constant-coefficient operator $P(D)$ is a distribution $E$ satisfying $P(D)E={\delta }_0$. The proposed $N$ is locally integrable in three dimensions and harmonic away from zero. For a smooth compactly supported test function $\phi$, apply Green's identity outside the ball of radius $\eta$. On the inner boundary the outward normal is $-\hat{r}$, so ${\partial }_{\nu }N=1/(4\pi {\eta }^2)$. The outer boundary contributes nothing. ThusThe first term tends to $\phi (0)$ and the second is $O(\eta )$. Local integrability permits taking the limit on the left. Therefore $-\Delta N={\delta }_0$ in distributions.

For a harmonic function on a ball, define its spherical mean $M(r)=(4\pi {)}^{-1}{\int }_{S^2}u(x_0+r\omega )d\omega$. Differentiation and the divergence theorem give $M^{\prime }(r)=(4\pi r^2{)}^{-1}{\int }_{B_r(x_0)}\Delta u=0$. Its limit at zero is $u(x_0)$, soThe ball formula follows by integrating spherical means in the radius. If two solutions of the prescribed Poisson equation tend to zero at infinity, their difference is globally harmonic and also tends to zero. Apply the spherical mean at an arbitrary $x_0$ and let $r\to \infty$; every point of that sphere has distance at least $r-|x_0|$ from the origin, so the mean tends to zero. Hence the difference is zero everywhere, proving uniqueness.