Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 1 / 17B (Fluid Dynamics)

Distinguish the full velocity potentials ${\Phi }_i=U_{i}x+{\phi }_i$ from their perturbations ${\phi }_i$. The linear boundary conditions use the perturbations. This distinction reconciles the notation in the printed conditions with the full potentials written in the mode ansatz. All equations below neglect products of perturbations. Assume positive densities and no surface tension.

The interface is a material surface, so its defining function $F=z-\zeta (x,y,t)$ obeys $DF/Dt=0$ on each side. Since ${\mathbf{u}}_i=\nabla {\Phi }_i$, the exact kinematic boundary condition isSubstitute ${\Phi }_i=U_{i}x+{\phi }_i$. Products such as ${\phi }_{ix}{\zeta }_x$ are second order, as is the correction from evaluating ${\phi }_{iz}$ at $z=\zeta$ instead of zero. Linearization therefore gives

Without surface tension, the exact dynamic boundary condition for an inviscid interface is equality of pressure, $p_1=p_2$, at $z=\zeta$. The Unsteady Bernoulli equation in each irrotational fluid isChoose the potential gauge so that $C_i=p_0/{\rho }_i+U_i^2/2$, its unperturbed value. Expanding the kinetic term as $U_i^2/2+U_i{\partial }_x{\phi }_i$ and evaluating gravity at the displaced interface givesto first order. Equality of these pressures gives

Incompressibility makes each perturbation velocity potential a harmonic function. A horizontal mode $e^{i(kx+\ell y)+st}$ has $q=\sqrt{k^2+{\ell }^2}>0$. Decay away from the interface selects ${\phi }_1=A_1{e}^{qz}{e}^{i(kx+\ell y)+st}$ below and ${\phi }_2=A_2{e}^{-qz}{e}^{i(kx+\ell y)+st}$ above. The kinematic boundary condition givesInsert these into the dynamic boundary condition for an inviscid interface and cancel a nonzero mode amplitude to obtainWriting ${\rho }_T={\rho }_1+{\rho }_2$ and $\bar{U}=({\rho }_1{U}_1+{\rho }_2{U}_2)/{\rho }_T$, completion of the square givesConsequentlyThe imaginary advection term does not change growth rates. A mode grows exponentially precisely when the radicand is positive, equivalentlyIf the upper fluid is heavier, every nonzero wavevector is unstable even without shear: this is Rayleigh-Taylor instability. If the lower fluid is heavier, gravity stabilizes sufficiently long waves, but any nonzero velocity difference destabilizes sufficiently short waves with $\ell =0$: this is Kelvin-Helmholtz instability. For equal densities, shear destabilizes modes with $k\ne 0$. A negative radicand gives purely imaginary frequencies and oscillatory modes; equality is the marginal threshold for exponential growth.