Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 3

Write $\rho =|\mathbf{x}-\mathbf{y}|$. A rotationally symmetric fundamental solution of the Laplace equation is harmonic for $\rho >0$. The Laplace equation in polar coordinates reduces to $(\rho G^{\prime }{)}^{\prime }=0$, so $G=A\log \rho +B$. Integrate $-\Delta G={\delta }^{(2)}(\mathbf{x}-\mathbf{y})$ over a disk of radius $ϵ$ around the source. The divergence theorem givesTaking $B=0$ therefore givesChanging the reference length inside the logarithm merely adds a constant. Notice the sign convention: the potential equation in this question is $\Delta V=q{\delta }^{(2)}$, whereas this Green function satisfies $-\Delta G={\delta }^{(2)}$. Thus the point-source potential used below is $-qG=+(q/2\pi )\log \rho$.

For the two insulating walls, successive Euclidean reflections in $x=b$ and $x=-b$ generate sources of strength $q$ at $(2nb,0)$ for every integer $n$. Only $n=0$ lies inside the physical strip. This is the Neumann image lattice for a point source in a strip. SetThe raw sum of logarithmic potentials diverges by additive constants. A well-defined potential is obtained by pairing opposite images and subtracting their constant reference terms:Each paired summand is $O(n^{-2})$ on compact subsets away from the source. The series therefore has local uniform convergence, as do the required derivative series. It givesIn the corresponding $x$-derivative, the images pair as $n$ and $1-n$ at $x=b$, and as $n$ and $-1-n$ at $x=-b$. Their opposite horizontal separations make the derivatives cancel. Thus both homogeneous Neumann boundary conditions hold. The constructed solution is even in $y$; without a condition at infinity the side conditions alone would allow further homogeneous harmonic functions.

For fixed $y>0$, uniform convergence on $[-b,b]$ permits integration term by term. The $n$th term has integralHere $\arctan$ is the ordinary real inverse trigonometric function, taking values in $(-\pi /2,\pi /2)$. Summing from $-N$ to $N$ makes a telescoping series:The source flux splits equally between the upper and lower parts of the strip. The next two parts give the single-wall constructions that underlie the image lattice.

For the grounded wall, use an image charge $-q$ at $(2b,0)$. Define the Euclidean distances $r=\sqrt{x^2+y^2}$ and $r_{\ast }=\sqrt{(x-2b{)}^2+y^2}$. The method of images givesThe real source has the required Dirac delta function coefficient, and the image is outside the domain. On $x=b$, $r=r_{\ast }$, so $V_D=0$ exactly. This verifies the Dirichlet boundary condition rather than only its derivative.

For the insulating wall, the image instead has the same strength $q$. With the distances defined above,It has the required source in the physical domain. Its normal derivative isAt $x=b$ the terms are $b/(b^2+y^2)$ and $-b/(b^2+y^2)$, so they cancel. The Neumann boundary condition leaves the additive constant $C$ free.