Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 70 (Fluid Dynamics of the Environment)

Let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation and $p$ the pressure perturbation divided by ${\rho }_0$. For stable stratification, $N>0$. The nonrotating Linearized Boussinesq equations areEliminating $u,p,b$ gives $({\partial }_x^2+{\partial }_z^2)w_{tt}+N^2{w}_{xx}=0$. Since $w={\zeta }_t$, a nonzero-frequency plane wave obeys the same equation for its displacement. Substituting its phase yields the dispersion relation for a plane internal gravity wave:Thus the frequency depends on the wavevector direction rather than its magnitude.

Advection of the background density gives ${\rho }^{\prime }=-\zeta {\bar{\rho }}_z$ to first order. With constant ${\bar{\rho }}_z<0$, the instantaneous density gradient is ${\rho }_z={\bar{\rho }}_z(1-{\zeta }_z)$. A region has unstable density stratification when this becomes positive, namely when ${\zeta }_z>1$. The maximum of ${\zeta }_z$ is $|m{A}_{\zeta }|$, so the monochromatic internal-wave overturning criterion isEquality gives a locally vanishing gradient. This is the prediction of the displacement field extrapolated to overturning; the small-amplitude approximation itself ceases to be reliable there.

For the rising packet, distinguish its conserved absolute frequency $\omega$ from its actual intrinsic frequency $\sigma =\omega -kU(z)$. The printed terminology calls $\omega$ intrinsic while also assigning it to a stationary observer; the stationary-observer interpretation is the one consistent with the displayed Doppler shift. On the positive-frequency branch, the ray Hamiltonian isThe Hamiltonian ray-tracing equations giveThe last identity follows also by differentiating the Hamiltonian along its canonical trajectory: the spatial and wavevector terms cancel in pairs. Thus absolute-frequency conservation in steady shear gives constant $k$, constant $\omega$, and constant stationary-observer horizontal phase speed $c_x=\omega /k$. In contrast, $\sigma =\omega -ksz$ decreases as the packet rises. At its initial height,This is the critical level of an internal gravity wave. In fact $m(t)=m_0-kst$ and $z(t)=[\omega -\sigma (k,m(t))]/(ks)$, so the inviscid ray approaches $z_c$ as $t\to \infty$, rather than reaching it at a finite time.

Write $\theta =|\Theta |=\arctan (|m|/k)$, so $\sigma =N\cos \theta$ with $0<\theta <\pi /2$. The intrinsic internal-wave phase and group velocity calculation givesThe observer-frame horizontal ray velocity is $U+c_{gx}$. Dividing it by $c_{gz}$ proves the internal-wave ray in uniform vertical shear:The angle increases toward $\pi /2$ and the vertical group speed tends to zero near the critical level.

The wave-action conservation law fixes the prescribed upward flux. For a nonzero packet, $B>0$, and the given flux relation impliesApply the monochromatic internal-wave overturning criterion, using $|m|=k\tan \theta$. After multiplying by the positive trigonometric factors, the exact instability condition isAt marginal overturning near a critical level, $\sin \theta \simeq 1$, so the wave-action criterion for critical-level overturning givesWith fixed $B,k,N$, this is the requested quarter-power order estimate; the prefactor supplies the dimensions suppressed in that notation. It is an onset balance, not a replacement for $\cos \theta =\sigma /N$. Combining the two relations instead gives ${\cos }^3\theta =(B{k}^3/N^2)\sin \theta$ at onset. Since $m^2{A}_{\zeta }^2$ diverges as ${\sigma }^{-3}$ toward $z_c$, any nonzero packet flux eventually violates the linear overturning criterion before reaching that level, within this nondissipative ray model.

Use $F=F_s>0$ and constant entrainment parameter $\alpha >0$. The printed integrals are normalized by $\pi$, so a top-hat plume model has $Q=b^{2}w$, $M=b^2{w}^2$ and $F=b^{2}w{g}^{\prime }$; the physical area is still $\pi b^2$. Keeping this normalization avoids spurious factors of $\pi$ in the flux equations.

Eliminate height between the two flux balances. The plume flux-balance invariant follows fromsoFor positive fluxes, both $M$ and $Q$ increase. In particular $Q^{\prime }\ge 2\alpha \sqrt{M_s}>0$, so $Q\to \infty$ as $z\to \infty$. Therefore $C/(\lambda Q^2)\to 0$, whatever the finite source imbalance, and $M\sim {\lambda }^{2/5}{Q}^{4/5}$. Integrating $Q^{\prime }\sim 2\alpha {\lambda }^{1/5}{Q}^{2/5}$ gives the attracting similarity solutionThis proves attraction to pure plume similarity for physical positive source fluxes. Sources with a vanishing initial volume or momentum flux enter the positive regime immediately and can be obtained by the corresponding limiting solutions; a point buoyancy source has the same similarity form with zero virtual-origin offset. The statement assumes positive buoyancy flux: a jet with $F=0$ is a different asymptotic problem.

Define the plume balance parameterPure plume balance means $\Gamma =1$ at each height, or equivalently $C=0$: momentum, buoyancy and entrainment have their similarity relation, even when the source has finite radius. A forced plume has $\Gamma <1$, meaning excess momentum relative to this balance; a lazy plume has $\Gamma >1$, meaning insufficient momentum. The invariant ensures that the sign of the imbalance is preserved while $\Gamma \to 1$. In particular, a strictly positive source remains pure at every height precisely whenFor this source the exact solution is the same similarity form with $z$ replaced by $Z=z+z_0$. Writing $a=(9\alpha F_s/10{)}^{1/3}$,Thus the radius growth of an axisymmetric pure plume is $b^{\prime }=6\alpha /5$ and the plume virtual origin lies at $-z_0$, with $z_0=5b_s/(6\alpha )$. Equivalently, in terms of top-hat source properties, pure plume balance is $g_s^{\prime }{b}_s/w_s^2=8\alpha /5$.

For the stated source-to-source geometry, the relevant horizontal distance is $L_s$. Consequently the edge criterion is $b_m=L_s$, rather than the radius at which two expanding edges first touch. Since $b(z)=b_s+6\alpha z/5$,Here $b_s=Q_s/\sqrt{M_s}={\lambda }^{-1/5}{Q}_s^{3/5}$. A positive merging height requires $L_s>b_s$; equality puts it at the source, while an already wider source satisfies the geometric condition from the start. The no-interaction premise extrapolates independent plumes even though their edges would first touch earlier, when each radius is $L_s/2$.

Use $M^{5/2}=\lambda Q^2$ with $Q=b^{2}w$ and $M=b^2{w}^2$ to get $F_s=(8\alpha /5)b{w}^3$. The individual plume properties at the specified height are thereforeThe source volume flux affects the height through the virtual origin, but it does not affect these properties at a prescribed radius in an exactly pure plume.

For the stipulated merger of two equal pure plumes, the new radius is $b_c=\sqrt{2}{b}_m$. This factor is outside $b_m$, as required by the area. Since speed and reduced gravity are held fixed, the merged fluxes are $Q_c=2Q_m$, $M_c=2M_m$ and $F_c=2F_s$. Their balance parameter isThe immediate combined plume is lazy. The same result follows from $\Gamma =5g^{\prime }b/(8\alpha w^2)$: enlarging the radius at fixed $g^{\prime },w$ increases this parameter.

At fixed $w_m,g_m^{\prime }$, the radius required for pure plume balance isThat is half the stipulated combined area $2\pi L_s^2$. This fixed-speed pure-plume merger and flux conservation comparison exposes the limitation of demanding immediate pure balance with unchanged speed and reduced gravity: reducing the area to this value would also halve all three merged fluxes. A conservative merger must instead adjust its properties and begin out of pure balance. Under the given constant-entrainment equations, the resulting lazy plume subsequently approaches the attracting similarity solution with total buoyancy flux $2F_s$.

Use dimensionless variables throughout this calculation and write $c$ for $\tilde{c}$. Take $k>0$ without loss of generality, so $\alpha =kh/2>0$; for the opposite Fourier orientation the decay rate is $|k|h/2$. The density profile printed alongside the velocity profile is $\tilde{\rho }=-1$ above, $0$ in the middle, and $+1$ below. It is absent from the TeX transcription but essential to the interface calculation.

The nondimensional jump conditions for stratified inviscid shear flow also follow directly from the stated scales. Divide the dimensional stress bracket by $\Delta U/h$ times the streamfunction scale. The density contrast supplies $J\tilde{\rho }\tilde{\psi }/(\tilde{U}-c)$. The term from the constant reference density is proportional to $\tilde{\psi }/(\tilde{U}-c)$ and drops out of its jump because the first condition makes this ratio continuous. This explains why only the density contrast remains.

For a vertically localized disturbance, take decaying exterior solutions and set $P=\tilde{\psi }(1)$, $Q=\tilde{\psi }(-1)$. The velocity is continuous at the interfaces, so the first jump condition gives continuity of $\tilde{\psi }$. The outer solutions are $P{e}^{-\alpha (\tilde{z}-1)}$ above and $Q{e}^{\alpha (\tilde{z}+1)}$ below. In the middle the endpoint derivative map for an evanescent wave layer givesThese expressions follow by fitting a linear combination of $\cosh (\alpha \tilde{z})$ and $\sinh (\alpha \tilde{z})$ to its two endpoint values.

At the upper interface the shear drops from one to zero and the density drops from zero to minus one. At the lower interface they change from zero to one and from one to zero. Substitute those jumps, including the exterior derivatives $-\alpha P$ and $+\alpha Q$. DefineMultiplication by the respective nonzero $U-c$ gives the homogeneous systemFor growing modes $\mathrm{Im}c>0$ the denominators cannot vanish. Neutral limiting values are interpreted by continuation of the matching calculation.

A nonzero disturbance requires the determinant to vanish. Put $T=1+J$, $\epsilon =e^{-2\alpha }$ and $d=a^2-b^2=2\alpha a$, with $a=2\alpha /(1-{\epsilon }^2)$ and $b=a\epsilon$. The determinant isDividing by $d$ and collecting powers proves the quartic dispersion relation for a three-layer stratified shear flow:whereHere $P$ as a polynomial coefficient is distinct from the endpoint amplitude used above. The result is the desired dispersion equation, derived from both shear and density jumps rather than a density-only matching rule.

Treat this as a quadratic in $X=c^2$. Whenever $Q_0<0$, its two roots are real with opposite signs: the discriminant $P^2-4Q_0$ is strictly positive and their product is negative. The negative root gives $c=+i\sqrt{-X}$, which grows under the convention $e^{ik(x-ct)}$. Factor the constant term:The unstable band of a three-layer stratified shear flow follows immediately:or equivalently $\alpha e^{\alpha }/\cosh \alpha <1+J<\alpha e^{\alpha }/\sinh \alpha$. The endpoints are neutral $c=0$ limits of this growing branch.

For large $\alpha$, the overlap of the two interface disturbances is exponentially small: $b\simeq 2\alpha e^{-2\alpha }$. Dropping this coupling leaves the two independent gravity-vorticity interface waves. Their relevant laboratory phase speeds areThe upper wave propagates backward relative to its current and the lower wave forward relative to its current. They share a stationary laboratory phase when $J=2\alpha -1$. Restoring their small evanescent coupling produces counterpropagating wave resonance in a three-layer shear flow, with the unstable intervalNear this resonance, the waves lock in phase and extract energy from the mean shear. At large wavenumber the resonance band is exponentially narrow; large wavenumber by itself does not make an arbitrary fixed-$J$ profile unstable.

Let $a=({\rho }_b-{\rho }_t)/H={\rho }_0{N}^2/g$. The initial density is $\rho (z)={\rho }_t-az$. Conservation of mass in the material homogenized over $-h<z<0$ fixes its mean density to ${\rho }_u={\rho }_t+ah/2$, while the untouched lower layer has interfacial density ${\rho }_i={\rho }_t+ah$. Hence the reduced gravity difference across the interface isThe reference contribution involving ${\rho }_t$ cancels from the difference.

Only the upper layer changes its potential energy. The potential-energy cost of homogenizing a linear stratification issoThe positive sign reflects work against stable stratification.

The rotating boundary supplies energy on scales set by $\Omega$ and $R$. Turbulent drag and dissipation remove energy, and entrainment adds initially stationary fluid that must be accelerated. If upper-layer energy grows, the increasing turbulent loss provides a restoring tendency; after a short adjustment, input and losses can approximately balance while the interface moves on a slower time scale. This gives a physical rationale for a fixed-energy turbulent mixed layer, not a consequence of mass conservation alone. Its characteristic kinetic energy is $K\sim {\rho }_0\pi R^{2}h{u}_u^2/2$. With the stipulated $N$-independent energy closure,so the characteristic speed decreases as $h^{-1/2}$ rather than remaining constant.

The stress does work at a rate proportional to $\pi R^2\tau u_u=c_D{\rho }_0\pi R^2{u}_u^3$. Combine this with the energy increase and absorb fixed drag/geometrical factors into $C>0$:Here the exponents $\alpha ,\beta$ are unrelated to the plume entrainment coefficient in Question 2. The interfacial Richardson number isSet $q=\Omega /N$, $d=\delta /R$ and $r=h/R$. Then $u_u^2/(N^2{h}^2)=d/(2r{\mathrm{Ri}}_i)$ and $r=q(2A{\mathrm{Ri}}_i/d{)}^{1/2}$. Substitution givesThe assumption that $E$ depends on the local Richardson number alone excludes separate dependences on $q,d$. The entrainment exponent from a local interfacial Richardson closure is consequentlyThis comparison treats $q$ and $d$ as independently variable external controls; fitting only a single apparatus would not by itself determine the exponents.

Finally, use the fixed-energy closure again:Integrating from a positive initial mixed-layer depth $h_0$ at time $t_0$ givesWhen the growth term dominates the initial offset, the rotating-disc mixed-layer depth law isTaking the effective time origin to be zero gives the stated scaling. Its validity is limited to the regime of the closures and $h<H$; the singular velocity predicted at $h=0$ is not a model for the initial boundary-layer formation.