= Solution
The velocity is $\mathbf u=\nabla\phi$ because the flow is irrotational. <Incompressible flow> requires $\nabla\cdot\mathbf u=0$, hence the <velocity potential> satisfies <Laplace's equation>:
$$
\phi_{xx}+\phi_{yy}=0,\qquad y<\eta(x,t).
$$
The exact <kinematic boundary condition> at the free surface is $\eta_t+\phi_x\eta_x=\phi_y$. With constant atmospheric <pressure>, the dynamic condition from <Bernoulli's principle> is $\phi_t+\tfrac12|\nabla\phi|^2+g\eta=0$, after choosing the time-dependent potential gauge. For small amplitudes, the products of perturbations are second order. Evaluating the remaining first-order terms at the undisturbed level $y=0$ gives
$$
\boxed{\phi_y=\eta_t,\qquad\phi_t+g\eta=0\quad(y=0)}.
$$
Substitution of $\phi=f(y)\sin(kx-\omega t)$ into <Laplace's equation> gives $f''-k^2f=0$. For $k\ne0$, decay as $y\to-\infty$ selects $f(y)=Be^{|k|y}$. The two boundary conditions for $\eta=A\cos(kx-\omega t)$ yield
$$
|k|B=A\omega,\qquad \omega B=gA.
$$
For a nonzero wave amplitude these are compatible precisely when
$$
\boxed{\omega^2=g|k|,\qquad \phi=\frac{gA}{\omega}e^{|k|y}\sin(kx-\omega t)}.
$$
With the conventional $k>0$ and positive frequency, the <dispersion relation> is $\omega(k)=\sqrt{gk}$. The exponential decay is the infinite-depth boundary condition; retaining a growing exponential would violate it.
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