 Solution

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.