Past exam of the waves and compressible flow course of the University of Oxford / 2023 / 2

Write the density and pressure as ${\rho }_0(z)+{\rho }^{\prime }(x,z,t)$ and $p_0(z)+p^{\prime }(x,z,t)$, respectively. The background satisfies hydrostatic balance, $p_0^{\prime }=-g{\rho }_0$, and ${\rho }_0^{\prime }/{\rho }_0=-\alpha$. Linearizing the Euler equations for an inviscid fluid, incompressibility and advection of density givesDifferentiate the horizontal equation in $x$ and use incompressibility to obtain $p_{xx}^{\prime }={\rho }_0{w}_{zt}$. Differentiate the vertical equation in time and twice in $x$:Dividing by ${\rho }_0>0$ gives the equation for internal waves in an exponentially stratified incompressible fluid,The first-derivative term retains the variation of inertial density with height; no Boussinesq approximation has been made.

For $w=e^{\alpha z/2}W$, direct differentiation givesThus the transformed partial differential equation and its impermeable-wall boundary conditions areThe positive exponential factor does not change where the vertical velocity vanishes.

Substitute $W=f(z)e^{i(kx-\omega t)}$ into the transformed equation. It yieldsThe nonzero eigenfunctions of this Dirichlet boundary value problem are $f=A\sin (n\pi z/h)$, $n\ge 1$. Hence, with $N=\sqrt{\alpha g}$ and $b_n^2=(n\pi /h{)}^2+{\alpha }^2/4$,The physical vertical velocity includes the factor $e^{\alpha z/2}\sin (n\pi z/h)$. The zero horizontal wavenumber is the zero-frequency limiting case, not a propagating oscillation.

Choose the rightgoing signed dispersion relation $\omega (k)=Nk/\sqrt{k^2+b_n^2}$. The horizontal phase velocity and group velocity areThe leftgoing branch has their negatives. For a convention using positive frequency at both signs of $k$, multiply each displayed speed by $\mathrm{sgn}k$. For every nonzero $k$, $0<c_g<c_p$ on the rightgoing branch; the horizontal group-speed bound for a confined internal wave is $c_g\le N/b_n$, approached as $k\to 0$.

Use the Fourier transform convention $\hat{W}(k,z,t)={\int }_{\mathbb{R}}W(x,z,t)e^{-ikx}dx$, with inverse factor $1/(2\pi )$. The initial profile occupies only the vertical mode $\sin (\pi z/h)$. Its transform isThe transformed evolution is a harmonic oscillator for each $k$, withTherefore inversion givesUsing the nonnegative even frequency instead gives the same cosine. At $t=0$, ${\int }_{\mathbb{R}}{e}^{-a|k|}{e}^{ikx}dk=2a/(a^2+x^2)$ with $a=ϵh$, recovering the specified initial profile and its zero time derivative.

Write the cosine in i as two complex exponentials. On a ray $x=vt$, the oscillatory phases are $kv\pm \omega (k)$. The stationary phase method requires $v=\mp {\omega }^{\prime }(k)$. By ii, the largest absolute horizontal group velocity for the occupied $n=1$ mode isInside this region there are stationary wavenumbers supporting appreciable wave packet contributions. Outside it no stationary wavenumber exists; oscillatory cancellation suppresses the large-time response. This explains "mainly confined": the initial Lorentzian profile already has nonzero algebraic tails, so the displayed region is not the exact support of the solution. Near its boundary the stationary points meet $k=0$ and the usual nondegenerate stationary-phase formula needs a transition approximation.