= Solution
Along $x=Vt+x_0$, the phase is $t\Phi(k)+kx_0$, with $\Phi(k)=kV-\omega(k)$. Graphically, a line of slope $V$ intersects the <dispersion relation>[dispersion curve] transversely except where it is tangent. On a transverse region, neighboring <wavenumbers> acquire order-one phase differences over a width $\Delta k\sim t^{-1}$, so cancellation leaves an $O(t^{-1})$ contribution. More formally, <integration by parts> uses
$$
e^{it\Phi}=\frac1{it\Phi'}\frac d{dk}e^{it\Phi}
$$
and gives that order when the amplitude and its derivative are integrable and $\Phi'$ is bounded away from zero on the region.
At a nondegenerate tangent point $k_*$, $\Phi'(k_*)=0$, equivalently $\boxed{V=\omega'(k_*)}$, and $\Phi(k)=\Phi(k_*)-\omega''(k_*)(k-k_*)^2/2+\cdots$. Coherent <wavenumbers> now occupy width $O(t^{-1/2})$, yielding the stationary-phase contribution
$$
A(k_*)e^{ik_*x_0+it\Phi(k_*)-i\operatorname{sgn}(\omega''(k_*))\pi/4}\sqrt{\frac{2\pi}{t|\omega''(k_*)|}}.
$$
These are the packets whose <group velocity> follows the observer. The displayed decay rates require the usual regularity and nondegeneracy hypotheses; $A\to0$ alone is not a complete analytic hypothesis, and a vanishing amplitude or degenerate stationary point changes the leading order.
\Image[/past-exam-of-the-theoretical-mechanics-course-of-the-university-of-oxford/2008-stationary-phase.png]
{title=A tangent of slope V selects the gravity-wave packet that remains near the moving observer}
For deep water take a <velocity> potential satisfying <Laplace's equation> below the undisturbed surface and decaying with depth. A Fourier mode has vertical factor $e^{|k|z}$. The linearized kinematic and dynamic surface conditions are $\eta_t=\phi_z$ and $\phi_t+g\eta=0$ at $z=0$. Thus
$$
\partial_t^2\overline\eta(k,t)+g|k|\overline\eta(k,t)=0.
$$
Release from rest gives $\overline\eta_t(k,0)=0$, so inversion of the <Fourier transform> gives
$$
\boxed{\eta(x,t)=\frac1{2\pi}\int_{-\infty}^\infty\overline\eta_0(k)\cos(t\sqrt{g|k|})e^{-ikx}\,dk.}
$$
Evenness of the initial profile makes its transform even and the reconstructed surface even; the same formula also <works> without that symmetry when the usual reality condition holds.
Split the cosine into its two time phases. Along the moving observer, their stationary conditions are $\pm\omega'(k)=V$, where $\omega(k)=\sqrt{g|k|}$. Since $|\omega'(k)|=\sqrt g/(2\sqrt{|k|})$, for $V\ne0$,
$$
\boxed{|k_*|=\frac g{4V^2},\qquad k_*=\pm\frac g{4V^2}.}
$$
These are the possible dominant stationary-phase packets, provided the initial Fourier amplitude at the selected <wavenumber> is nonzero. For $V=0$ there is no finite nonzero stationary <wavenumber>; the formula must not be applied by division by zero.
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