 Solution

The velocity potential gives horizontal and vertical velocity components $({\varphi }_x,{\varphi }_z)$. The kinematic boundary condition at each fixed impermeable wall is zero normal fluid velocity, henceLet $\eta (x,t)$ be the surface elevation. To first order in its amplitude, the kinematic boundary condition for a free-surface graph gives ${\eta }_t={\varphi }_z$ at $z=0$. The dynamic boundary condition for an inviscid interface fixes the surface pressure to atmospheric pressure. The unsteady Bernoulli equation, after choosing the time-dependent gauge of $\varphi$, then gives ${\varphi }_t+g\eta =0$ at $z=0$: the quadratic velocity term is second order in amplitude. Differentiate this last equation in time and eliminate ${\eta }_t$ to obtain the linearized free-surface boundary conditions in combined form,