 Solution

Let $\pi$ be pressure divided by the reference density and $b$ the buoyancy perturbation. The linear, inviscid Boussinesq equations about $U(z)$ areFor a stationary Fourier component $e^{ikx}$, incompressibility gives $u=i{w}^{\prime }/k$. The horizontal momentum equation then gives $\pi =i(U^{\prime }w-U{w}^{\prime })/k$, while buoyancy conservation gives $b=i{N}^{2}w/(kU)$. Substitute these into the vertical momentum equation to obtain the stationary Taylor–Goldstein equation:Hence its local WKB approximation isFor $U^{\prime \prime }$ negligible, an outgoing upward wave has $mU/k>0$. In particular $m>0$ for $k,U>0$. In an upper evanescent region choose the branch with $\mathrm{Im}m>0$ in $e^{i\int mdz}$ so the solution decays upward. These choices express the radiation condition; a reflected wave has the opposite real branch and downward group velocity.

There are two different limiting levels. A critical layer in a shear flow occurs at $U=0$, where the intrinsic frequency vanishes, $|m|\to \infty$, and upward propagation slows while the vertical wavelength contracts. Dissipation or nonlinear effects can absorb the wave and transfer its momentum to the mean flow. For a stable, slowly varying critical layer with Richardson number $\mathrm{Ri}=N_0^2/(U^{\prime }{)}^2>1/4$, the usual causal inviscid continuation suppresses transmission; it is not the same as reflection at $m=0$. When the Richardson number is small, a general absorption-only conclusion needs extra stability assumptions.

Reflection instead occurs at the turning level of a stationary internal wave $k|U|=N_0$, beyond which $m^2<0$. It is sometimes called a critical reflection height, but the wind there is nonzero. Assuming $0<k{U}_0<N_0$, the specified profiles giveThe decreasing-wind profile never encounters $m=0$ before its zero-wind layer. Both profiles have $U^{\prime \prime }=0$; a well-separated WKB description requires $N_{0}H/U_0\gg 1$, and a turning-layer transition is still required for the increasing profile.