 Solution

Take $U_0>0$, $k=2\pi /L>0$, and represent the stationary topographic internal gravity wave by phase $kx+mz$. Its laboratory angular frequency is zero, and its intrinsic frequency is $\hat{\omega }=-k{U}_0$. The internal gravity wave dispersion relation givesThe radiation condition selects $m>0$, since this makes the vertical group velocity upward. The lower kinematic boundary condition is $w=U_0{h}_x$, so a corresponding linear wave field iswhere $\zeta$ is the vertical fluid displacement. The constant-phase lines of an internal gravity wave slope down to the right, with $dz/dx=-k/m$.

Let $K^2=k^2+m^2$. In the frame fixed to the hills the phase pattern is stationary: its phase velocity is zero, while the group velocity isEnergy travels up and downwind, with ray slope $dz/dx=m/k$. In coordinates $x^{\prime }=x-U_{0}t$ moving with the wind, the phase is $k{x}^{\prime }+mz+k{U}_{0}t$. ThusThe intrinsic phase velocity is upstream and downward; the intrinsic group velocity is upstream and upward, perpendicular to the wave vector. Adding $(U_0,0)$ to the group velocity recovers the laboratory energy ray.

Vertically propagating waves require $k{U}_0<N_0$, equivalently $\boxed{L>2\pi U_0/N_0}$. At equality the vertical group velocity vanishes; above the mountain-wave cutoff the bounded response is an evanescent wave. The linearization requires small terrain slope $k{h}_0\ll 1$ and small wave-induced displacement gradients $|m|h_0\ll 1$, often summarized by $N_0{h}_0/U_0\ll 1$. The model also neglects rotation, viscosity and compressibility; its scales must permit those approximations.