 Solution

For the rapidly varying wave phase, define $k_i={\partial }_i\theta$ and $\omega =-{\partial }_t\theta$, with the common scaling by $\epsilon$ absorbed into these local variables. A packet moves with group velocity $c_{g,i}=\partial \Omega /\partial k_i$. Along that trajectory, $d/dt={\partial }_t+c_g\cdot {\nabla }_x$. Compatibility gives ${\partial }_t{k}_i=-{\partial }_i\omega$ and ${\partial }_i{k}_j={\partial }_j{k}_i$. Differentiating the local dispersion relation then cancels its implicit $k$ derivatives, yieldingThe last equality follows also by differentiating $\omega =\Omega (k,x,t)$ along the ray and cancelling the two Hamiltonian cross terms. The partial derivatives of $\Omega$ hold its other arguments fixed.

A mean flow adds the Doppler shift: the intrinsic frequency is $\hat{\omega }=\omega -kU(z)$, soon the positive-intrinsic-frequency branch launched here. The medium is stationary and independent of $x$, so $\omega$ and $k$ are constant. Initially $m=0,U=0$, hence $\omega =N$. Moreover $\dot{m}=-k{U}^{\prime }<0$, and then $\dot{z}=-Nkm/(k^2+m^2{)}^{3/2}>0$. The initial vertical group velocity is zero, but $m$ immediately becomes negative and the ray accelerates upwards.

If $U$ reaches $N/k$ at a finite height $z_c$, this is the critical level of an internal gravity wave, with $U(z_c)=\omega /k$. Below it,Assuming $0<U^{\prime }(z_c)<\infty$, put $\delta =z_c-z$. Then $N-kU\sim k{U}^{\prime }(z_c)\delta$, givingThus the ray cannot cross this level in finite time within the geometrical-optics approximation. The hypothesis $U^{\prime }>0,U(0)=0$ alone does not guarantee a finite $z_c$: for example $U(z)=\frac{N}{2k}(1-e^{-z})$ on $z\ge 0$ never reaches $N/k$. Existence of a finite critical level is an additional assumption. Very close to one, diverging wavenumber can invalidate the inviscid, linear, slowly varying approximation; the result is the formal ray prediction.