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

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.