= Solution
Let $\pi$ be pressure divided by the reference density and $b$ the buoyancy perturbation. The linear, inviscid <Boussinesq equations> about $U(z)$ are
$$
u_t+Uu_x+U'w=-\pi_x,\quad
w_t+Uw_x=-\pi_z+b,\quad
b_t+Ub_x+N^2w=0,\quad
u_x+w_z=0.
$$
For a stationary Fourier component $e^{ikx}$, <incompressibility> gives $u=iw'/k$. The horizontal momentum equation then gives $\pi=i(U'w-Uw')/k$, while buoyancy conservation gives $b=iN^2w/(kU)$. Substitute these into the vertical momentum equation to obtain the stationary <Taylor–Goldstein equation>:
$$
w''+\left(\frac{N^2}{U^2}-\frac{U''}{U}-k^2\right)w=0.
$$
Hence its local <WKB approximation> is
$$
\boxed{m^2=\left(\frac{N^2}{k^2U^2}-\frac{U''}{k^2U}-1\right)k^2.}
$$
For $U''$ 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 $\operatorname{Im}m>0$ in $e^{i\int m\,dz}$ 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> $\operatorname{Ri}=N_0^2/(U')^2>1/4$, 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 $0<kU_0<N_0$, the specified profiles give
$$
\boxed{\begin{array}{ll}
U=U_0(1+z/H):&\text{reflection at }\displaystyle\zeta=H\left(\frac{N_0L}{2\pi U_0}-1\right),\\
U=U_0(1-z/H):&\text{critical-layer absorption at }\zeta=H.
\end{array}}
$$
The decreasing-wind profile never encounters $m=0$ before its zero-wind layer. Both profiles have $U''=0$; a well-separated WKB description requires $N_0H/U_0\gg1$, and a turning-layer transition is still required for the increasing profile.
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