Past exam of the options II course of the University of Oxford / 2012 / K / K4

A material interface obeys $D(y-\eta (x,t))/Dt=0$. In each fluid this gives ${\varphi }_{jy}={\eta }_t+(U_j+{\varphi }_{jx}){\eta }_x$ at $y=\eta$, where $U_1=U$ and $U_2=V$. Discarding products of perturbations and evaluating at the undisturbed interface givesFor an inviscid irrotational fluid, unsteady Bernoulli is ${\Phi }_t+\frac{1}{2}|\nabla \Phi {|}^2+p/\rho +gy=C(t)$, with total potential $\Phi =U_{j}x+{\varphi }_j$. Subtract the undisturbed relation and linearize at the moving interface. The pressure perturbation there is $-{\rho }_j({\varphi }_{jt}+U_j{\varphi }_{jx}+g\eta )$. With no surface tension the two pressures agree, givingUse complex amplitudes and take real parts at the end. For $k>0$, solutions of Laplace's equation decaying away from the interface have the formPut ${\Omega }_1=\omega -Uk$ and ${\Omega }_2=\omega -Vk$. The kinematic conditions imply $B_1=-i{\Omega }_{1}A/k$ and $B_2=i{\Omega }_{2}A/k$. Substituting into the pressure condition givesA nonzero amplitude therefore requiresTo determine stability, let $R={\rho }_1+{\rho }_2$ and $\overline{U}=({\rho }_{1}U+{\rho }_{2}V)/R$. Completing the square yieldsFor $U\ne V$, the right-hand side is negative exactly whenThen the two frequencies have opposite nonzero imaginary parts. The root with positive imaginary part gives exponential growth through $e^{-i\omega t}$. In terms of wavelength $\lambda =2\pi /k$, the unstable range isThis is the short-wavelength Kelvin-Helmholtz instability for a density-stable interface without surface tension. Equality is the neutral threshold; if $U=V$, the gravitational term is positive and this shear instability is absent.