 Solution

The free surface is material, so applying the material derivative to $y-\eta (x,t)$ gives the exact kinematic boundary conditionFor an incompressible fluid with constant density, the Euler equations areSince the flow is irrotational, $\mathbf{u}=\nabla \varphi$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\nabla \varphi {|}^2/2)$. Integrating the resulting gradient equation gives a function of time on the right of Bernoulli equation. Adding a time-dependent constant to $\varphi$ removes that function, without changing velocity, and choosing the atmospheric-pressure reference givesThe pressure immediately above the surface is atmospheric; the liquid-side pressure includes the stated surface tension jump.

At first order in wave amplitude, the products ${\varphi }_x{\eta }_x$ and $|\nabla \varphi {|}^2$, the slope-squared correction to curvature, and displacement corrections in evaluating $\varphi$ at $y=\eta$ are all higher order. Hence, on $y=0$,For $\eta =A\cos (kx-\omega t)$, a harmonic velocity potential decaying as $y\to -\infty$ and satisfying the first condition isIndeed $\Delta \varphi =0$ and ${\varphi }_y{|}_0=A\omega \sin (kx-\omega t)={\eta }_t$. Substituting in the second condition yields the dispersion relation for a gravity-capillary wave:Taking the positive-frequency branch, the phase velocity and group velocity areSubtracting givesSince $\omega >0$,The threshold is also the wavenumber at the minimum phase speed of a gravity-capillary wave.