Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 2

For separation of variables, let $\Phi =S(\sigma )T(\tau )$. The Laplace equation givesA nonzero $2\pi$-periodic angular solution requires $\lambda =n^2$, $n=0,1,\dots$. For $n\ge 1$, $S$ is a linear combination of $\cos (n\sigma )$ and $\sin (n\sigma )$, while $T^{\prime \prime }-n^{2}T=0$ gives hyperbolic sine and cosine. For $n=0$, periodicity makes $S$ constant, and $T=A_0+B_0\tau$. Negative separation constants would give nonperiodic real exponentials in $\sigma$.

Superposition yieldsCompleteness follows by expanding a harmonic function in a Fourier series in the periodic coordinate: its Fourier coefficients must satisfy exactly these radial equations. As usual, the coefficients are restricted by the convergence required of the series on compact subregions. The bipolar coordinates domain imposes no boundary condition at $\tau =0$ or $\tau =\infty$, so both hyperbolic branches remain available.

The only nonzero angular Fourier mode is $\cos (2\sigma )$. Its radial equation is $T^{\prime \prime }-4T=0$. A solution vanishing at $\tau =a$ is proportional to $\sinh [2(\tau -a)]$; making its value one at $b$ fixes the constant. The Dirichlet problem between bipolar coordinate circles therefore givesIts second derivatives in $\sigma$ and $\tau$ cancel, and the endpoint values are respectively zero and $\cos (2\sigma )$. The difference of any two solutions would be a harmonic function with zero Dirichlet boundary conditions, so the maximum principle proves uniqueness.

Geometric wording. In the printed bipolar coordinates, the circle radius $1/\sinh \tau$ decreases with $\tau$. For $0<a<b$, the $b$ circle is contained inside the $a$ circle. Thus the source’s “inner” and “outer” labels are interchanged geometrically. The solution above follows the explicit coordinate boundary values, which unambiguously specify the problem.