= Solution
{c}
For an <irrotational flow>, write $\mathbf u=\nabla\phi$. The <incompressible flow> condition then gives the <Laplace equation>
$$
\boxed{\phi_{xx}+\phi_{yy}=0,\qquad -h<y<\eta(x,t)}.
$$
At leading order in a small disturbance this equation holds on $-h<y<0$. Including <gravity>, the <Unsteady Bernoulli equation> is
$$
\phi_t+\frac12|\nabla\phi|^2+\frac p\rho+gy=B(t).
$$
Choose the time-dependent additive freedom in the <velocity potential> so that $B(t)=p_0/\rho$. Neglecting <surface tension>, <pressure continuity> at the <free surface> gives the exact dynamic condition
$$
\phi_t(x,\eta,t)+\frac12|\nabla\phi(x,\eta,t)|^2
+\frac{p_1}{\rho}\cos(x-ct)+g\eta=0.
$$
The <free surface> is a material boundary: applying the <material derivative> to $y-\eta(x,t)=0$ yields the <kinematic boundary condition for a free-surface graph>
$$
\eta_t+\phi_x(x,\eta,t)\eta_x=\phi_y(x,\eta,t).
$$
For a small disturbance, the <Taylor expansion> from $y=\eta$ to $y=0$ changes the first-order <velocity potential> and its <partial derivatives> only at second order. Both $|\nabla\phi|^2$ and $\phi_x\eta_x$ are also second-order terms. The <linearized free-surface boundary conditions> are therefore
$$
\boxed{\phi_t+\frac{p_1}{\rho}\cos(x-ct)+g\eta=0,
\qquad \phi_y=\eta_t\quad (y=0)}.
$$
The fixed bed has the <no-penetration boundary condition>
$$
\boxed{\phi_y=0\quad(y=-h)}.
$$
The required approximation is small <surface elevation>, small <wave steepness>, and small <velocity> perturbations. The stated $p_1\ll p_0$ alone does not ensure this approximation near <resonance>.
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