Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 3

A fundamental solution of the Laplace equation satisfies ${\Delta }_{\mathbf{r}}{G}_0=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ in the sense of distributions, without a boundary condition. Choose it rotationally symmetric about the source. Away from the source, with $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|$, the radial Laplace equation is $(s{G}_0^{\prime }{)}^{\prime }=0$, giving $G_0=A\log s+B$.

Integrating over a small disc around the source and using the divergence theorem fixes the normalization:ThereforeThe additive constant is arbitrary; adding other entire harmonic functions is also possible if rotational symmetry is not imposed. The positive sign here follows from the convention $\Delta G_0=+\delta$.

The Dirichlet Green function is the solution, with the source point held fixed, ofIts singularity is the logarithmic one found in (a); subtracting that fundamental solution leaves a harmonic function. An unbounded region also requires a condition at infinity, here boundedness.

Green's second identity, with the outward normal derivative, isSet $u=\Phi$ and $v=G(\cdot ;{\mathbf{r}}^{\prime })$, interpreting the logarithmic singularity by a small excised disc or distribution theory. The volume integral is $\Phi ({\mathbf{r}}^{\prime })$ because $\Delta \Phi =0$ and $\Delta G=\delta$. On the boundary $G=0$ and $\Phi =f$, leavingFor the exterior problem below, apply the identity first in a large truncated region. For bounded harmonic functions the extra circle at infinity contributes zero: $G$ tends to a constant, its radial derivative is $O(r^{-2})$, and a bounded exterior harmonic function also has radial derivative $O(r^{-2})$. This is the condition under which the same boundary representation applies there.

Let $s=|{\mathbf{r}}^{\prime }|>0$ and $\mathbf{p}={\mathbf{r}}^{\prime }/s^2$. For $|\mathbf{r}|=1$, expanding both squared Euclidean distances givesTaking nonnegative square roots provesThis is the distance identity behind inversion in a circle and the method of images for a circular boundary.

For a source in the exterior, its inverse point $\mathbf{p}$ lies strictly inside the excluded disc. The method of images construction of the bounded exterior disk Green function therefore allows an opposite logarithmic source there without adding a singularity to the domain. The candidate Dirichlet Green function isIts Laplacian in the exterior is exactly $\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$, since the image singularity is outside that domain. The identity in (d) makes $G=0$ on the unit circle. Moreoverso it obeys the required boundedness condition. These checks establish the stated logarithm-of-a-ratio expression with $A=1/(2\pi )$. Bounded exterior Dirichlet uniqueness for the Poisson equation follows by inversion in a circle and removal of the bounded singularity at the origin.

The outward normal of the exterior region points into the excluded disc, so it is $-\mathbf{r}$ on the unit circle. Differentiating the Dirichlet Green function from (e), and writing $s=|{\mathbf{r}}^{\prime }|$ and $d=|\mathbf{r}-{\mathbf{r}}^{\prime }|$, givesHere $|\mathbf{r}-\mathbf{p}{|}^2=d^2/s^2$ was used. The concentrated boundary data have unit integral, so the boundary representation evaluates this exterior Poisson kernel at $\mathbf{r}=(-1,0)$. Thus the bounded solution on either part of the exterior $x$-axis isIt is positive on both intervals, diverges as $x^{\prime }\to -1^{-}$ at the boundary source, and tends to $1/(2\pi )$ as $|x^{\prime }|\to \infty$. Boundedness at infinity selects this solution; otherwise one could add an arbitrary multiple of $\log |{\mathbf{r}}^{\prime }|$ without changing the circle's boundary data.