Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 30C (Partial Differential Equations)

A fundamental solution of a linear differential operator $P(D)$ is a distribution $E$ satisfying $P(D)E={\delta }_0$. For the three-dimensional negative Laplacian,Away from zero the function is harmonic. For a test function $\varphi$, integrate $E(-\Delta \varphi )$ outside a ball of radius $\epsilon$ and use Green's identity. The small inner sphere contributes $\varphi (0)+o(1)$ because the normal derivative of $E$ has total outward-from-the-ball flux $-1$; the other boundary term is $O(\epsilon )$. Letting $\epsilon \to 0$ verifies the distributional equation and its normalization.

The mean value property for harmonic functions states that if $u$ is harmonic on a neighborhood of a closed ball, then its average on the ball or its boundary sphere equals its value at the center. To prove it in three dimensions, define $M(r)=(4\pi {)}^{-1}{\int }_{S^2}u(x+r\omega )d\omega$. Differentiation and the divergence theorem giveSince $M(r)\to u(x)$ as $r\to 0$, the spherical average equals $u(x)$. Integrating $4\pi r^{2}M(r)$ from zero to the ball radius gives the volume version,For balls merely contained in the open harmonicity domain, apply the result first to smaller radii and pass to the radius of interest by continuity.

Apply the volume mean value property to the two balls and use nonnegativity together with inclusion:For any $w,z\in B(x,r)$, the triangle inequality gives $B(z,r)\subset B(w,3r)\subset B(x,4r)\subset \Omega$. Taking $R=3r$ in the previous bound therefore gives $u(w)\ge u(z)/27$. Take the infimum over $w$ and supremum over $z$ to obtainNo positivity stronger than nonnegativity is needed; if a value vanishes, this argument forces every value in the smaller ball to vanish too.