 Solution

The exact free-surface kinematic condition is ${\eta }_t+{\varphi }_x{\eta }_x={\varphi }_y$ evaluated at $y=\eta$. Linearizing about a flat surface at rest drops products of small quantities and evaluates at $y=0$, giving ${\varphi }_y={\eta }_t$. Bernoulli's equation with constant atmospheric pressure gives ${\varphi }_t+g\eta =0$ to the same order. The rigid bed has no normal flow, hence ${\varphi }_y=0$ at $y=-h$.

In the fluid, incompressibility and the velocity-potential representation give $\Delta \varphi =0$. For $k>0$, takeIt is harmonic, satisfies the bed condition and the linear dynamic condition. At the surface its vertical derivative is $Agk\tanh (kh)\sin (kx-\omega t)/\omega$, while ${\eta }_t=A\omega \sin (kx-\omega t)$. Therefore the kinematic condition is equivalent toFinally $\tanh q=q-q^3/3+O(q^5)$. Taking the square root givesThe first term is nondispersive shallow-water speed; the next term is the leading finite-depth dispersion correction.