Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 2

Use separation of variables with $\Phi =M(\mu )N(\nu )$. A nonzero product givesThe angular factor must be a periodic function of period $2\pi$. If $\lambda <0$, its exponential solutions cannot be periodic unless they vanish. If $\lambda =0$, $N=c+d\nu$ forces $d=0$, while $M=A_0+B_0\mu$. If $\lambda >0$, periodicity requires $\sqrt{\lambda }=n\in \mathbb{N}$, giving angular factors $\cos (n\nu ),\sin (n\nu )$ and radial factors $\cosh (n\mu ),\sinh (n\mu )$.

Superposing these separated harmonic functions yieldsTo see completeness, expand an arbitrary smooth harmonic function in a Fourier series in $\nu$ on each ellipse. Substitution into the equation forces every Fourier coefficient to obey the corresponding radial equation above. The series is interpreted with the convergence needed for a harmonic function, for example locally on compact subannuli. No boundedness condition at infinity or extension across $\mu =0$ has been imposed, so there is no reason to discard either radial branch in these elliptic coordinates.

Let $d=b-a$. In the mixed Dirichlet-Neumann problem in an elliptic annulus, the angular Fourier modes separate. The outer data excite the $\cos \nu$ mode, while the inner derivative excites $\cos 2\nu$.

For the first mode, a radial factor with zero derivative at $\mu =a$ is $\cosh (\mu -a)$. Normalizing its outer value gives $F_1(\mu )=\cosh (\mu -a)/\cosh d$. For the second mode, a radial factor vanishing at $\mu =b$ is $\sinh [2(b-\mu )]$. Its derivative at $a$ must be $-1$, fixing the coefficient to $1/[2\cosh (2d)]$. ThereforeEach term satisfies Laplace's equation; at $\mu =b$ the second term vanishes and the first is $\cos \nu$. At $\mu =a$, differentiating gives $0-\cos 2\nu$, as required. All other modes, including the constant mode, vanish by the homogeneous mixed conditions. Alternatively, integration by parts applied to the difference of two solutions gives a zero integral of $|\nabla \Phi {|}^2$, proving uniqueness.

Boundary convention. The calculation follows the explicitly printed coordinate derivative ${\partial }_{\mu }\Phi$. In elliptic coordinates, the scale factor is $h=\sqrt{{\sinh }^2\mu +{\sin }^2\nu }$ and the outward unit normal at the inner boundary points toward decreasing $\mu$. A physical normal derivative there would be $-h^{-1}{\partial }_{\mu }\Phi$. Thus the phrase “normal derivative” in the source must be understood in its displayed coordinate sense; imposing the same numerical data on a unit-normal derivative would be a different problem.