= Solution
Linearize about fluid at rest, with mean <free surface> $y=0$ and mean bottom $y=-d$. Regard $a$, $\eta$ and the perturbation <velocity potential> as first-order quantities. A <Taylor expansion> from $y=\eta$ to $y=0$ produces products of first-order quantities, as do $\phi_x\eta_x$ and $|\nabla\phi|^2$. Discarding these second-order terms gives the <linearized free-surface boundary conditions>
$$
\boxed{\phi_y=\eta_t,\qquad\phi_t+g\eta=0\quad(y=0).}
$$
Equivalently, differentiate the second condition in time and substitute the first to obtain $\phi_{tt}+g\phi_y=0$ at $y=0$.
The bottom is also impermeable and moving. Its exact <kinematic boundary condition> follows from $D(y-h)/Dt=0$:
$$
\phi_y=h_t+\phi_xh_x\quad(y=h).
$$
Here $h_t=-akc\cos(k(x-ct))$, while $\phi_xh_x$ and the change from evaluating $\phi_y$ at $h$ to evaluating it at $-d$ are second order. Hence
$$
\boxed{\phi_y(x,-d,t)=-akc\cos(k(x-ct)).}
$$
The interior <Laplace equation> remains $\phi_{xx}+\phi_{yy}=0$ on $-d<y<0$. This <linearization> requires small slopes and small resulting excursions as well as $a\ll d$; close to <resonance> a small prescribed bottom displacement can produce a response too large for it.
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