Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 3

Use the sign convention in which a fundamental solution of the Laplace equation satisfiesin the sense of distributions. It is a harmonic function away from the source. Writing $\rho =|\mathbf{r}-{\mathbf{r}}^{\prime }|$, the radial equation is $(\rho G^{\prime }{)}^{\prime }=0$, so $G=A\log \rho +B$. The divergence theorem on a small circle about the source fixes its total flux:ConsequentlyThe arbitrary constant is a harmonic function. For the alternative convention $-\Delta G={\delta }^{(2)}$, the logarithm has the opposite sign; subsequent Green's function formulas must use that convention consistently.

A Dirichlet Green function satisfies ${\Delta }_{\mathbf{x}}{G}_D(\mathbf{x},\mathbf{y})={\delta }^{(2)}(\mathbf{x}-\mathbf{y})$ in the domain and $G_D=0$ when $\mathbf{x}$ lies on its boundary. For a source $\mathbf{y}\ne 0$ in the unit disc, put its opposite image at ${\mathbf{y}}^{\ast }=\mathbf{y}/|\mathbf{y}{|}^2$. The method of images givesThe image lies outside the unit disc, so its logarithm is a harmonic function inside. The original logarithm has precisely the unit source strength found in part (a). On $|\mathbf{x}|=1$,so the boundary value is zero. Equivalently, in complex coordinates, the Dirichlet Green function inside the unit disk isThis also covers a source at the origin, where $G_D(z,0)=(2\pi {)}^{-1}\log |z|$.

Fix the observation point $\mathbf{x}$ and apply Green's second identity in the source variable $\mathbf{y}$. Since $\Delta u=0$, ${\Delta }_{\mathbf{y}}{G}_D={\delta }_{\mathbf{x}}$ and $G_D=0$ on the boundary, it givesThe identity can be justified by deleting a small circle about $\mathbf{x}$ and letting its radius tend to zero; the logarithmic flux contributes $u(\mathbf{x})$.

Set $\mathbf{x}=(r,\varphi )$ and $\mathbf{y}=(\rho ,{\varphi }^{\prime })$, and write $c=\cos (\varphi -{\varphi }^{\prime })$. The symmetric expression for the Green's function isAt the boundary, the outward normal derivative is ${\partial }_{\rho }$ at $\rho =1$, and henceSince $ds=d{\varphi }^{\prime }$ on the unit circle, the Poisson integral on the unit disk follows, with Poisson kernel

At the origin, the Poisson kernel is one, so the mean value property for harmonic functions givesFor $0<r<1$, the denominator in the Poisson kernel satisfies $1-2r\cos \vartheta +r^2\ge (1-r{)}^2$. ThusThe nonnegativity of $f$ allows this upper bound to be inserted under the Poisson integral on the unit disk:This is the upper bound in the Harnack inequality on the unit disk. The same argument with the minimum of the Poisson kernel gives the corresponding lower bound $(1-r)u(0)/(1+r)$.