Past exam of the theoretical mechanics course of the University of Oxford / 2011 / B / 5

Take the nondegenerate stationary point in the interior of the integration interval. Choose a smooth cutoff equal to one near $k_{\star }$. Away from its support, ${\psi }^{\prime }$ is bounded away from zero. Integration by parts using $e^{it\psi }=(it{\psi }^{\prime }{)}^{-1}{\partial }_k{e}^{it\psi }$ makes that nonstationary contribution $O(t^{-1})$, under the usual smooth-phase assumptions.

Near the stationary point use the signed coordinateSince ${\psi }^{\prime \prime }(k_{\star })>0$, this is locally invertible, with $ds/dk=\sqrt{{\psi }^{\prime \prime }(k_{\star })}$ at the point. The local integral becomesScale $s=\sqrt{2/t}r$. The supplied Fresnel integral gives ${\int }_{-\infty }^{\infty }{e}^{i{r}^2}dr=e^{i\pi /4}\sqrt{\pi }$. Approximating the smooth amplitude by its value at zero, with a cutoff and a nonstationary estimate for the tails, therefore givesIf $f(k_{\star })\ne 0$, this is the requested asymptotic equivalence. The phase shift $\pi /4$ comes from the positive quadratic curvature. A stationary point at an integration endpoint instead needs a one-sided formula.

The printed auxiliary estimate asserting an $O({\gamma }^{-1})$ Fourier integral for every continuous amplitude is too strong without extra regularity. The argument here uses integration by parts only for the actual smooth, cutoff nonstationary amplitudes, where that estimate is justified; continuity alone generally guarantees convergence to zero, not this rate.

For a negative second derivative, the corresponding local normal form is $\psi =\psi (k_{\star })-s^2/2$. The oscillatory Gaussian integral is the complex conjugate of the positive-curvature one, so the phase shift changes sign. ThusWhen the leading amplitude is nonzero, the equality may again be written as an asymptotic equivalence. Both signs are summarized by the factor $e^{i\pi \mathrm{sgn}({\psi }^{\prime \prime })/4}$.

Irrotationality permits a velocity potential with $\mathbf{u}=({\varphi }_x,{\varphi }_z)$. Incompressibility gives Laplace's equation in the moving domain $z<\eta (x,t)$. Neglecting surface tension and taking the free-surface pressure as constant atmospheric pressure, the full surface conditions areThe kinematic condition makes the free surface material; the dynamic condition is Bernoulli's equation on that surface. A time-dependent additive gauge in $\varphi$ has absorbed the spatially uniform Bernoulli constant. Deep below the surface, impose $\nabla \varphi \to 0$ as $z\to -\infty$, together with suitable horizontal decay. Initially $\eta (x,0)={\eta }_0(x)$ and the fluid is at rest, so $\nabla \varphi (x,z,0)=0$; its initial constant may be chosen as zero.

Linearize about a flat quiescent surface. The product ${\varphi }_x{\eta }_x$ and the kinetic-energy term are quadratic, and the boundary conditions can be evaluated at $z=0$ to first order. Thus ${\eta }_t={\varphi }_z$ and ${\varphi }_t+g\eta =0$ there. Fourier transformation in $x$ changes ${\partial }_x^2$ into multiplication by $-k^2$, givingThe transform uses $e^{-ikx}$ in its forward definition; the inverse transform therefore uses $e^{ikx}/(2\pi )$. The zero-wavenumber potential has an irrelevant spatially constant gauge freedom.

For $k\ne 0$, the solution decaying at depth is $\hat{\varphi }(k,z,t)=C(k,t)e^{|k|z}$. The surface conditions give ${\hat{\eta }}_t=|k|C$ and $C_t=-g\hat{\eta }$, henceInitial rest implies ${\hat{\eta }}_t(k,0)=0$. Define the transform of the initial elevation by ${\hat{\eta }}_0(k)={\int }_{\mathbb{R}}{\eta }_0(x)e^{-ikx}dx$. Solving this oscillator equation and inverting the transform givesThe formula also includes the continuous zero-frequency limit. For real even initial data, the transform is real and even, ensuring a real even surface profile at every time.

For wavenumber selection by a moving observer of deep-water waves, an observer with nonzero velocity $V$ follows $x=Vt$ up to a constant initial displacement. Splitting the cosine into exponentials gives phase functions ${\psi }_{\pm }(k)=kV\pm \sqrt{g|k|}$. Their stationary points satisfyso the selected wavenumbers areThe two signs arise in the conjugate branches needed for a real wave. Equivalently the observer selects wave packets whose group velocity equals $V$. Stationary-phase contributions decay as $t^{-1/2}$, whereas smooth nonstationary contributions decay more rapidly, so these are the dominant wavenumbers when the initial spectrum is nonzero there.

For real even data, one may make this explicit. Set $k_{\ast }=g/(4V^2)$. The stationary curvature has magnitude $2|V{|}^3/g$, yieldingwhen its leading coefficient is nonzero. If that spectral value vanishes, the leading stationary term vanishes too. At $V=0$ there is no finite nonzero stationary wavenumber selected by this condition.