= Solution
For a stationary component, the <stationary internal-wave WKB solution> has local vertical <wavenumber> and vertical-velocity amplitude
$$
m^2(z)=\frac{N^2(z)}{U_0^2}-k^2,\qquad
w\sim C\,m^{-1/2}(z)\exp\left(i kx+i\int_0^z m(s)\,ds\right),
$$
with the upward branch $m>0$. The <WKB approximation> requires $|m'|\ll m^2$ and sufficient smoothness, with $|m''|\ll |m|^3$ a useful additional criterion. The background changes little over a vertical wavelength; this excludes a <turning level of a stationary internal wave>.
For increasing $N=N_0(1+z/H)$, the vertical <wavenumber> increases. The crest slope $-k/m$ tends toward zero, and successive crests get closer vertically. The laboratory energy ray obeys $dx/dz=k/m$, so it becomes more nearly vertical as it rises. There is no upper turning level in this ideal increasing-stratification profile.
For decreasing $N=N_0(1-z/H)$, the vertical <wavenumber> decreases to zero at
$$
\boxed{z_t=H\left(1-\frac{kU_0}{N_0}\right).}
$$
This assumes propagation at the bottom, $kU_0<N_0$. The crests become more nearly vertical and the incident energy ray becomes horizontal at $z_t$. Above it the wave is <evanescent>, and the incident wave is reflected downward. The divergent WKB amplitude at $m=0$ is an approximation failure, resolved by an <Airy function> transition, rather than an infinite physical wave amplitude. The decreasing formula for $N$ is only a positive-frequency background below $z=H$; the relevant turning height lies below that level.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-71-mountain-wkb.png]
{title=Mountain-wave crests, energy rays, and turning levels for slowly varying stratification and wind}
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