Past exam of the options II course of the University of Oxford / 2013 / K4

The factor ${\eta }_{xx}/(1+{\eta }_x^2{)}^{3/2}$ is the signed curvature of the graph, so the pressure jump is surface tension times curvature, the planar Young–Laplace equation.

Take total velocity potentials ${\Phi }_{-}={\varphi }_{-}$ and ${\Phi }_{+}=Ux+{\varphi }_{+}$. Each disturbance potential satisfies the Laplace equation. The interface moves with the fluid: for $F=y-\eta (x,t)$, the material equation $F_t+v\cdot \nabla F=0$ gives ${\eta }_t+u{\eta }_x=v_y$. Keeping only first-order terms and evaluating at $y=0$ yieldsThe unsteady Bernoulli equation is ${\Phi }_t+|\nabla \Phi {|}^2/2+p/\rho =C(t)$, with any common gravitational contribution canceling across this equal-density interface. Subtracting the basic flow and linearizing gives $p_{-}^{\prime }=-\rho {\varphi }_{-,t}$ and $p_{+}^{\prime }=-\rho ({\varphi }_{+,t}+U{\varphi }_{+,x})$. The linearized pressure jump is $T{\eta }_{xx}$. HenceUse complex modes with real parts understood:The decaying vertical factors solve the Laplace equation in their respective half-planes. The kinematic conditions give $B_{-}=-i\omega A/k$ and $B_{+}=i(\omega -kU)A/k$. Substitution in the dynamic condition givesWhen the discriminant is nonnegative, the phase speeds are $U/2\pm \sqrt{Tk/(2\rho )-U^2/4}$. They are two dispersive capillary-wave branches advected by the mean velocity $U/2$.

When $\boxed{U^2>2Tk/\rho }$, the frequencies are complex conjugates and one mode grows exponentially, with growth rate $k\sqrt{U^2/4-Tk/(2\rho )}$. This is Kelvin-Helmholtz instability. Surface tension stabilizes sufficiently short waves, while for any nonzero shear, sufficiently long waves are unstable in this gravity-free equal-density model. A real cosine with real frequency describes the stable branches; the unstable case is represented by the real part of the corresponding growing complex mode.