Neutral mode of the equal-width Hazel model ($J=k(1-k)$)

The zero-phase velocity ansatz $w=(\mathrm{sech}z{)}^k(\tanh z{)}^{1-k}$ solves the Taylor–Goldstein equation away from its critical level of an internal gravity wave exactly when $J=k(1-k)$. Let $y=\tanh z$ and $a=1-k$. Then $w^{\prime }/w=a/y-y$ and $w^{\prime \prime }/w=a(a-1)/y^2-a-1+2y^2$, so the differential-equation residual divided by $w$ is $[J-k(1-k)](y^{-2}-1)$. The curve has maximum $1/4$ at $k=1/2$. The endpoint $k=0$ does not decay at infinity; fractional powers at the critical level require a branch and regularity qualification.

For $0<k<1$, a neutral mode of the equal-width Hazel model behaves as $w\sim z^{1-k}$ near $z=0$. Its derivative diverges as $z^{-k}$, so it is not a classical continuously differentiable mode across the critical level of an internal gravity wave. The horizontal velocity is proportional to $w^{\prime }/k$; its local kinetic energy is finite only for $k<1/2$, because ${\int }_0^{ϵ}{z}^{-2k}dz$ converges precisely then. One possible neutral-mode convention is the boundary value from $U-c$ with $c_i\downarrow 0$, which fixes the phase of the power on the negative-$z$ side. The Miles–Howard theorem concerns growing modes with nonreal $c$ and does not rule out such singular neutral limits.