Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 43 / 2

Use complex amplitudes $(U,W,\Sigma ,P)(z)e^{i(kx-\omega t)}$, with $k\ne 0$ and $\omega \ne 0$, and take physical real parts. The two-dimensional linearized Boussinesq equations giveThe first and third equations imply $U=i{W}^{\prime }/k$ and $P=i{\rho }_0\omega W^{\prime }/k^2$; the buoyancy equation gives $\Sigma =-i{N}^{2}W/\omega$. Insert these into the vertical momentum equation. The amplitude of the internal gravity wave satisfiesOscillatory vertical propagation requires $N^2>{\omega }^2$; in a well-mixed region $N=0$, the bounded solution is evanescent rather than a vertically propagating internal wave.

The vertical parcel displacement has ${\partial }_t\eta =w$, so its complex amplitude is $\mathcal{E}=iW/\omega$. The horizontal momentum equation and incompressibility giveAt a material density interface the displacement is continuous, hence $[W]=0$. Continuity of pressure must be imposed on the displaced interface. Since the mean hydrostatic gradient is ${\bar{p}}_z=-\rho g$, continuity of Lagrangian pressure gives $[P]+\mathcal{E}[{\bar{p}}_z]=0$. A stable upward jump $[\bar{b}]=\Delta b$ gives $[{\bar{p}}_z]={\rho }_0\Delta b$, and thereforeThe same derivative jump follows by integrating the amplitude equation through the delta-function contribution $N^2={\bar{b}}_z$ at the interface. At a continuous-buoyancy stratification change, $\Delta b=0$ and both $W$ and $W^{\prime }$ are continuous. Assuming Eulerian pressure itself continuous across a density jump would miss the restoring interfacial force.

The lower-layer dispersion relation givesThe incident phase is $e^{-i{m}_{0}z}$; its vertical group velocity is positive, even though its vertical phase velocity is negative. Write the lower field as $W=w_0{e}^{-i{m}_{0}z}+r{e}^{i{m}_{0}z}$ and the upper bounded field as $W=T{e}^{-kz}$. Interface continuity gives $T=w_0+r$ andThus the internal-wave response of a sharp buoyancy interface isAt fixed incident $k,\omega ,N_0$ this is largest at $S=1$, where $|T|=2|w_0|$. Consequently ${\omega }^2=\Delta bk=2{\omega }_{\mathrm{free}}^2$, or $\omega =\sqrt{2}{\omega }_{\mathrm{free}}$. The maximum does not coincide with the free frequency for two homogeneous layers: the stratified lower radiating layer has a different dynamic pressure response. The upper field is evanescent, so in this ideal lossless problem it carries no mean outgoing vertical energy flux; the reflected wave has $|r|=|w_0|$.

For the upward incident internal gravity wave, $P={\rho }_0\omega m_{0}W/k^2$. Using real physical fields, its time-averaged vertical energy flux per unit area isThe assumed mixing power is $\mathcal{I}/4$. A deepening law also requires the gravitational potential energy of the mixed layer. The wave problem specifies a well-mixed half-space above the interface, but supplies neither a finite upper mixed-layer depth nor a replacement buoyancy/energy boundary condition. There is therefore no unique finite-layer $d(t)$ from the printed information alone.

The missing parameter can be displayed explicitly. Let the initially mixed upper region have finite depth $a$, with a fixed upper boundary at $z=a$, and conserve its integrated buoyancy while homogenizing down to $z=-d$. Its new buoyancy isIntegrate $-{\rho }_{0}zb$ over the initial and final profiles in $-d<z<a$. The potential energy of mixed-layer deepening with an initial buoyancy jump isWith $d(0)=0$ and a maintained constant incident flux, the resulting law isThe right-hand side is positive and determines a unique depth once $a$ is supplied. Initially $\dot{d}=\omega m_0|w_0{|}^2/(4k^2\Delta ba)$ for $a>0$, already showing why the upper depth matters. If the added assumptions instead specify a zero initial mixed depth, the cubic term gives $d=[3\omega m_0|w_0{|}^{2}t/(2k^2{N}_0^2){]}^{1/3}$; the initial jump then has no finite volume above it to mix with. Neither assumption is contained in the original half-space wave statement. Literal homogenization of the entire upper half-space is the $a\to \infty$ limit and costs infinite energy for any fixed positive $d$. This limitation should not be concealed by a guessed entrainment depth or by keeping the upper buoyancy fixed while claiming conservation.