Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 331 / 2

Use the Boussinesq approximation with a constant inertial reference mass density ${\rho }_0$, keeping the background mass density gradient in buoyancy. Write $V(z)=\overline{U}(z)-c$ and $P=\hat{p}/{\rho }_0$. For $k\ne 0$, the linearized normal mode equations areThe reference mass density is chosen close to the background values, whose relative variation must be small. Hydrostatic background pressure satisfies ${\overline{p}}^{\prime }=-g\overline{\rho }$.

From incompressibility, $\hat{u}=i{\hat{w}}^{\prime }/k$. The horizontal momentum equation and the density equation then giveSubstitute these into the vertical momentum equation and multiply by $ik$:For $V\ne 0$, rearrangement gives the Taylor–Goldstein equationThe buoyancy frequency $N$ is real for stable stratification, where ${\overline{\rho }}^{\prime }<0$. Points where $\overline{U}=c$ are critical levels of an internal gravity wave; the derivation there is understood through a limiting or piecewise formulation. The printed density gradient is ${\overline{\rho }}^{\prime }$, not the pressure gradient accidentally substituted in the local TeX.

Let $\hat{\eta }$ be the displacement of a material interface and use $[f{]}_{-}^{+}=f_{+}-f_{-}$. On each side, the kinematic boundary condition isThere is one displaced material interface, so its displacement is continuous. Thus the first jump condition isIn particular, vertical velocity need not be continuous when the background velocity jumps.

With no surface tension, pressure continuity applies on the displaced interface, rather than at its undisplaced height. Linearizing gives $[\hat{p}+\hat{\eta }{\overline{p}}^{\prime }{]}_{-}^{+}=0$. Since ${\overline{p}}^{\prime }=-g\overline{\rho }$, this becomes $[\hat{p}-g\overline{\rho }\hat{\eta }{]}_{-}^{+}=0$. The horizontal momentum calculation in part (a) givesSubstitute this and $\hat{\eta }=\hat{w}/[ik(\overline{U}-c)]$ to obtain the second jump condition:These jump conditions for stratified inviscid shear flow use one-sided values of ${\overline{U}}^{\prime }$ and apply to jumps of mass density, vorticity, or velocity. If $\overline{U}$ is continuous, the first condition reduces to continuity of $\hat{w}$; a jump of vorticity generally still prevents continuity of ${\hat{w}}^{\prime }$. This derivation avoids multiplying singular distributional derivatives of a discontinuous velocity in the Taylor–Goldstein equation.

Take $k>0$, without loss for this symmetric real base state, and write $s=1+J$, $e=e^{-2\alpha }$, and $y={\tilde{c}}^2$. The given quartic becomes a real quadratic $y^2+Ay+B=0$. Its constant term factors asTherefore $B<0$ exactly whenA negative product means the two quadratic roots are real and of opposite sign: the discriminant is $A^2-4B>0$. The negative root gives a pair $\tilde{c}=\pm i\sqrt{-y_{-}}$. One sign has positive temporal growth rate $k{c}_i$ and is unstable. Using the definitions of the hyperbolic functions gives the required unstable band:This proves the unstable band of a three-layer stratified shear flow directly, without needing to solve all four quartic roots.

For the wave interpretation, consider either interface in isolation with decaying normal modes $\hat{w}=w_0{e}^{-k|z-z_i|}$. Put $S=\Delta U/h$, $U_i=\overline{U}(z_i)$, and $v=c-U_i$, the intrinsic phase velocity. At either interface $[\overline{\rho }]=-\Delta \rho /2$, while $[{\overline{U}}^{\prime }]=-S$ at the upper interface and $+S$ at the lower one. The jump conditions for stratified inviscid shear flow give the gravity-vorticity interface wave relationFor stable density jumps, $J\ge 0$, both intrinsic branches are real. In units $C=2c/\Delta U$, the isolated wave speeds areThe upper wave travelling against its positive background current has $C_u=1-[1+\sqrt{1+8\alpha J}]/(4\alpha )$. The lower wave travelling against its negative background current has the opposite speed, $C_l=-C_u$. Their counterpropagating wave resonance in a three-layer shear flow occurs at $C_u=C_l=0$, which givesAt large wavenumber, coupling across the separation $h$ is exponentially weak, of order $e^{-kh}=e^{-2\alpha }$. The unstable band becomeswith full width $4\alpha e^{-2\alpha }+O(\alpha e^{-6\alpha })$. It is a counterpropagating wave instability: the two waves propagate oppositely relative to their local currents but have almost equal laboratory phase velocities, allowing weak coupling to lock their phases and extract mean-flow energy. The unstable band of a three-layer stratified shear flow thus becomes narrowly concentrated around this resonance. For fixed $J$, arbitrarily large $k$ lies outside that band; large-$k$ resonance requires $J$ to grow with $\alpha$.