Past exam of the fluids and waves course of the University of Oxford / 2014 / 3

The total velocity fields on the two sides areEach base flow is divergence-free. Incompressibility therefore makes both disturbance velocity potentials satisfy the Laplace equation:At the fixed horizontal walls, impermeability requires zero vertical velocity:The discontinuity in tangential base velocity at the interface is a vortex sheet. The assumption of irrotational flow is understood separately in each open fluid region; it cannot hold distributionally across that sheet when $U\ne 0$.

The interface is material. Applying the kinematic boundary condition to $y-\eta (x,t)=0$ gives, exactly,The disturbance velocities and $\eta$ are small. Products of two disturbance quantities are second order, and a Taylor expansion moves their evaluation to $y=0$ at linear order. HenceThe extra upper-fluid term is advection of the displaced interface by the base uniform flow.

Use the Unsteady Bernoulli equation separately on each side, with constant mass density $\rho$. There are no restoring terms specified at this internal interface, and its pressure is continuous. Choose the time-dependent potential gauges so that the undisturbed pressure is $p_0$ on both sides. Subtracting the base values from Bernoulli's principle gives, at linear order,The quadratic disturbance-velocity terms have been discarded, while the cross term with the base velocity is first order and must remain. Equating the perturbation pressures yields the third condition:

Use complex normal modes and take their real parts at the end. Write $\eta =A{e}^{i(kx-\omega t)}$, with $k>0$. The Laplace equation and the wall Neumann boundary conditions give the vertical formsAt $y=0$, the two kinematic boundary conditions becomeThusThe dynamic condition isSubstituting the two amplitudes and cancelling the common nonzero depth factor gives the dispersion relation for a vortex sheet in a symmetric rigid channel:Equivalently,For the specified $U>0$, the plus sign gives positive temporal growth rate $\gamma =kU/2$ in the convention $e^{-i\omega t}$. A real growing interface displacement is thereforeThe other branch decays at the same rate. Consequently the interface is linearly unstable, exhibiting Kelvin-Helmholtz instability. There is no nonzero-amplitude mode with a purely real temporal frequency for $k>0$ and $U\ne 0$; the paper's sinusoidal notation must allow a complex frequency, with the physical displacement obtained by a real part.

The wave pattern is advected at speed $U/2$, the mean of the two base velocities. The growth rate is independent of $h$ here because both layers have the same mass density and the same depth, making their factors $\coth (kh)$ cancel. In an asymmetric channel those inertia factors would differ. In the zero-shear limit $U=0$, the two roots merge at zero and there is no exponential growth.