Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 345 / 1

Let the velocity perturbation be $(u,w)$, pressure perturbation be $p$, and density perturbation be ${\rho }^{\prime }$ about the hydrostatic background $\hat{\rho }(z)$. The linearized inviscid Boussinesq approximation givesStable stratification means ${\hat{\rho }}_z<0$, and the buoyancy frequency isDifferentiate the momentum equations to eliminate $p$, use incompressibility, and then use the density equation to eliminate ${\rho }^{\prime }$. This yieldsFor $w\propto e^{i(kx+mz-\omega t)}$, the internal gravity wave dispersion relation is ${\omega }^2=N^2{k}^2/(k^2+m^2)$.

A localized monochromatic cylinder emits four narrow beams, one in each quadrant, forming a St Andrew's cross. If $\alpha$ is the beam angle to the horizontal,equivalently, its angle $\theta =\pi /2-\alpha$ to the vertical obeys $\cos \theta =\omega /N$. As $\omega$ rises from zero to $N$, the beams rotate from horizontal toward vertical. For $\omega >N$ no freely propagating internal wave exists and the response is evanescent.

The wavevector and phase velocity are parallel. The dispersion relation is homogeneous of degree zero in $(k,m)$, so Euler's theorem gives $\mathbf{k}\cdot {\mathbf{c}}_g=0$: the group velocity is perpendicular to both the wavevector and phase velocity. Energy travels along the beams in the group-velocity direction.

Write $\mu =\tan \theta =m/k$. The incident down-right ray from $(0,s)$ has slope $-\cot \theta$ and meets the parabolic bottom $H(x)=2x^2$ at $x=x_0\in (0,1/2)$ providedAt the impact point the bottom slope is $S=H^{\prime }(x_0)=4x_0$. A stationary reflection preserves frequency and tangential wavenumber $k+Sm$. For the incident wavevector $(k,m)=a(1,\mu )$ and a reflected up-left ray $(k_r,m_r)=a_r(-1,\mu )$, tangential matching givesA positive reflected magnitude $a_r$ therefore exists exactly when $S\mu >1$. This is the supercritical-slope condition for reflection of an internal-wave ray and produces a group velocity in the second quadrant.

Since $x_0<1/2$, such a point exists whenFor any such $\theta$, chooseThe initial ray then hits the parabola at a supercritical point and its reflected ray travels up and left, as required.