Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 331 / 1

Separate the uniform stream and an arbitrary time-dependent gauge from each velocity potential: ${\varphi }_j=U_{j}x+f_j(t)+{\varphi }_j^{\prime }$. Only gradients of the velocity potential determine the velocity field. The remote fluid must remain in its unperturbed uniform stream, soAfter fixing the harmless gauge, the nonzero-wavenumber perturbation velocity potentials themselves vanish at the corresponding infinities. Incompressible flow and irrotational flow give the Laplace equation for each perturbation. For a normal mode with $k>0$, the vertical amplitude solves ${\Phi }_j^{\prime \prime }-k^2{\Phi }_j=0$, henceThe opposite exponentials are excluded because they make the perturbation grow at infinity. It is the perturbation, rather than the total potential $U_{j}x+f_j(t)$, that decays.

Let $F(x,z,t)=z-\eta (x,t)$. The interface is a material surface for both fluids: its material derivative vanishes on $F=0$. With $u_j={\partial }_x{\varphi }_j$ and $w_j={\partial }_z{\varphi }_j$, this givesThus the kinematic boundary condition says that particles stay on the moving interface. The tangential velocities can differ; each side has its own material derivative. Replacing both derivatives by one background speed would be incorrect.

The dynamic boundary condition for an inviscid interface is pressure continuity, because the fluids have no viscous normal stress and there is no surface tension. The unsteady Bernoulli equation in each layer of irrotational flow readsEquating these pressures at $z=\eta$ gives the required dynamic boundary condition for an inviscid interface. Choose the potential gauges so that ${\rho }_1{C}_1={\rho }_2{C}_2$, or set both constants to zero relative to a common reference pressure. For example, ${\varphi }_j=U_{j}x-\frac{1}{2}{U}_j^{2}t+{\varphi }_j^{\prime }$ removes the distinct uniform-stream Bernoulli constants without changing either velocity. Material-interface kinematics and pressure continuity supply the two matching conditions, with the Bernoulli gauge understood.

Use ${\varphi }_j=U_{j}x-\frac{1}{2}{U}_j^{2}t+{\varphi }_j^{\prime }$ and retain terms linear in the displacement and perturbation potentials. Evaluation at $z=\eta$ may then be replaced by evaluation at $z=0$: the correction to a perturbation is second order. The kinematic boundary conditions becomeThe linearized pressure continuity isThe decaying normal modes give $-k{A}_1=ik(U_1-c)B$ and $k{A}_2=ik(U_2-c)B$. Thus $A_1=-i(U_1-c)B$, $A_2=i(U_2-c)B$. Substitution into the dynamic condition yields the Kelvin-Helmholtz dispersion relation with gravity,Put $\rho ={\rho }_1+{\rho }_2$ and $\overline{U}=({\rho }_1{U}_1+{\rho }_2{U}_2)/\rho$. Completing the square givesFor $k>0$, the temporal growth rate is $k{c}_i$. A growing complex conjugate branch therefore exists precisely when the radicand of the square root is negative:At equality the two phase velocities coincide and there is no exponential growth. Below it the roots describe real internal gravity waves. The denser lower fluid gives stable density stratification, but the tangential shear can overcome it. The assumption $U_1\ne U_2$ is needed for the displayed threshold; equal streams have no such shear instability.

Write ${\varphi }^{\prime }=\Phi (z)e^{ik(x-ct)}$ and let $B_{\pm }$ be the displacement amplitudes at $z=\pm L$. Use subscripts $I$ for the jet interior and $O$ for the adjoining exterior. Each interface is a material vortex sheet, so its linearized kinematic boundary conditions areThe equal mass densities make the hydrostatic displacement terms cancel. The dynamic boundary condition for an inviscid interface therefore givesThe division form assumes $c\ne 0,V$; the preceding undivided equations remain the proper conditions in those exceptional cases. Potentials themselves need not be continuous, and their vertical derivatives need not be equal: the same material displacement is advected by different tangential base velocities. At infinity the exterior potentials decay. These displacement and pressure conditions, rather than continuity of $\Phi$ and ${\Phi }^{\prime }$, determine the equal-density top-hat planar-jet dispersion relation.

For an odd velocity potential, chooseAt $z=L$ the two kinematic conditions give $A=ic{B}_{+}$ and $C\cosh (kL)=i(V-c)B_{+}$. The dynamic condition is $-cA=(V-c)C\sinh (kL)$. Eliminating $A,C,B_{+}$ gives, with $s=\tanh (kL)$,The lower interface gives the same equation by odd parity of the velocity potential. The interior vertical velocity is even, so $B_{-}=B_{+}$: the displacement is a sinuous mode of a planar jet even though the potential is odd.

For $L>0$, $V\ne 0$ and $k>0$, $0<s<1$ and one root has $c_i>0$. Thus every nonzero wavenumber is unstable in this ideal top-hat planar jet. The phrase “all choices” is understood with these assumptions: $k=0$ has no exponential wave growth, and $V=0$ has no shear.

For an even velocity potential, take ${\Phi }_I=C\cosh (kz)$ and the decaying exterior amplitudes equal at the upper and lower interfaces. At $z=L$, kinematics gives $A=ic{B}_{+}$ and $C\sinh (kL)=i(V-c)B_{+}$. The dynamic condition becomes $-cA=(V-c)C\cosh (kL)$. ThereforeWriting $s=\tanh (kL)$, its two roots areThe interior vertical velocity is odd, so $B_{-}=-B_{+}$; this is a varicose mode of a planar jet. As with the odd potential, $k>0$, finite $L>0$ and $V\ne 0$ give a growing root. The parity of the potential and the parity of the interface displacements are opposite because the vertical velocity is its derivative.

The odd-potential branch uses $s=\tanh (kL)$, whereas the even-potential branch uses $1/s$. The imaginary parts are identical becauseConsequently the equality of top-hat jet parity growth rates givesHere $\gamma =k{c}_i$ is the positive temporal growth rate. The real phase velocities differ: $c_{r,\mathrm{odd}}=\frac{V}{2}(1-e^{-2kL})$ and $c_{r,\mathrm{even}}=\frac{V}{2}(1+e^{-2kL})$. Thus equal amplification does not imply equal propagation. As $kL\to \infty$, both approach the isolated-vortex sheet phase velocity $V/2$ and growth rate $k|V|/2$.

For a single equal-density vortex sheet with stream speeds $0$ and $V$, the gravitational contribution vanishes. The Kelvin-Helmholtz dispersion relation with gravity reduces to $c^2+(V-c{)}^2=0$, with roots $c=V/2\pm i|V|/2$. Its positive temporal growth rate is ${\gamma }_{\mathrm{sheet}}=k|V|/2$.

Both jet parity branches have the common growth rate derived above, henceEquivalently, $2\sqrt{s}\le 1+s$ follows from $(\sqrt{s}-1{)}^2\ge 0$. Equality is approached when the two interfaces are separated by many disturbance decay lengths, $kL\to \infty$. The zero-shear case has both growth rates zero. For small $kL$, the growth rate ratio is approximately $2\sqrt{kL}$: interaction between the sheets suppresses amplification relative to an isolated sheet.