Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 71 / 2

Take $U_0>0$, $k=2\pi /L>0$, and represent the stationary topographic internal gravity wave by phase $kx+mz$. Its laboratory angular frequency is zero, and its intrinsic frequency is $\hat{\omega }=-k{U}_0$. The internal gravity wave dispersion relation givesThe radiation condition selects $m>0$, since this makes the vertical group velocity upward. The lower kinematic boundary condition is $w=U_0{h}_x$, so a corresponding linear wave field iswhere $\zeta$ is the vertical fluid displacement. The constant-phase lines of an internal gravity wave slope down to the right, with $dz/dx=-k/m$.

Let $K^2=k^2+m^2$. In the frame fixed to the hills the phase pattern is stationary: its phase velocity is zero, while the group velocity isEnergy travels up and downwind, with ray slope $dz/dx=m/k$. In coordinates $x^{\prime }=x-U_{0}t$ moving with the wind, the phase is $k{x}^{\prime }+mz+k{U}_{0}t$. ThusThe intrinsic phase velocity is upstream and downward; the intrinsic group velocity is upstream and upward, perpendicular to the wave vector. Adding $(U_0,0)$ to the group velocity recovers the laboratory energy ray.

Vertically propagating waves require $k{U}_0<N_0$, equivalently $\boxed{L>2\pi U_0/N_0}$. At equality the vertical group velocity vanishes; above the mountain-wave cutoff the bounded response is an evanescent wave. The linearization requires small terrain slope $k{h}_0\ll 1$ and small wave-induced displacement gradients $|m|h_0\ll 1$, often summarized by $N_0{h}_0/U_0\ll 1$. The model also neglects rotation, viscosity and compressibility; its scales must permit those approximations.

For a stationary component, the stationary internal-wave WKB solution has local vertical wavenumber and vertical-velocity amplitudewith the upward branch $m>0$. The WKB approximation requires $|m^{\prime }|\ll m^2$ and sufficient smoothness, with $|m^{\prime \prime }|\ll |m{|}^3$ a useful additional criterion. The background changes little over a vertical wavelength; this excludes a turning level of a stationary internal wave.

For increasing $N=N_0(1+z/H)$, the vertical wavenumber increases. The crest slope $-k/m$ tends toward zero, and successive crests get closer vertically. The laboratory energy ray obeys $dx/dz=k/m$, so it becomes more nearly vertical as it rises. There is no upper turning level in this ideal increasing-stratification profile.

For decreasing $N=N_0(1-z/H)$, the vertical wavenumber decreases to zero atThis assumes propagation at the bottom, $k{U}_0<N_0$. The crests become more nearly vertical and the incident energy ray becomes horizontal at $z_t$. Above it the wave is evanescent, and the incident wave is reflected downward. The divergent WKB amplitude at $m=0$ is an approximation failure, resolved by an Airy function transition, rather than an infinite physical wave amplitude. The decreasing formula for $N$ is only a positive-frequency background below $z=H$; the relevant turning height lies below that level.

Let $\pi$ be pressure divided by the reference density and $b$ the buoyancy perturbation. The linear, inviscid Boussinesq equations about $U(z)$ areFor a stationary Fourier component $e^{ikx}$, incompressibility gives $u=i{w}^{\prime }/k$. The horizontal momentum equation then gives $\pi =i(U^{\prime }w-U{w}^{\prime })/k$, while buoyancy conservation gives $b=i{N}^{2}w/(kU)$. Substitute these into the vertical momentum equation to obtain the stationary Taylor–Goldstein equation:Hence its local WKB approximation isFor $U^{\prime \prime }$ negligible, an outgoing upward wave has $mU/k>0$. In particular $m>0$ for $k,U>0$. In an upper evanescent region choose the branch with $\mathrm{Im}m>0$ in $e^{i\int mdz}$ so the solution decays upward. These choices express the radiation condition; a reflected wave has the opposite real branch and downward group velocity.

There are two different limiting levels. A critical layer in a shear flow occurs at $U=0$, where the intrinsic frequency vanishes, $|m|\to \infty$, and upward propagation slows while the vertical wavelength contracts. Dissipation or nonlinear effects can absorb the wave and transfer its momentum to the mean flow. For a stable, slowly varying critical layer with Richardson number $\mathrm{Ri}=N_0^2/(U^{\prime }{)}^2>1/4$, the usual causal inviscid continuation suppresses transmission; it is not the same as reflection at $m=0$. When the Richardson number is small, a general absorption-only conclusion needs extra stability assumptions.

Reflection instead occurs at the turning level of a stationary internal wave $k|U|=N_0$, beyond which $m^2<0$. It is sometimes called a critical reflection height, but the wind there is nonzero. Assuming $0<k{U}_0<N_0$, the specified profiles giveThe decreasing-wind profile never encounters $m=0$ before its zero-wind layer. Both profiles have $U^{\prime \prime }=0$; a well-separated WKB description requires $N_{0}H/U_0\gg 1$, and a turning-layer transition is still required for the increasing profile.

The internal-wave ray tracing equations preserve the horizontal wavenumber $k$ and laboratory frequency. For the increasing wind, the upward branch satisfiesThe group-velocity ratio therefore givesIntegrating from $(0,0)$ yieldsThis is the ascending branch of the circular mountain-wave ray in linear shear:It reaches $z_t=H(q_0^{-1}-1)$ at $x=x_t$, where the energy direction is horizontal. After reflection, $m<0$ and the ray continues downwind and downward along the other branchreturning to $z=0$ at $x=2x_t$. The last panel of the figure in part (b) shows this path. These are WKB energy rays, with the short neighborhood of the turning height supplied by its transition solution.