= Solution
For the equal-width <Hazel model> with stable <density stratification> $J\geq0$,
$$
\mathrm{Ri}(z)=\frac{N^2}{(\overline U')^2}
=J\cosh^2z,\qquad \min_z\mathrm{Ri}(z)=J.
$$
The <Miles–Howard theorem> therefore excludes exponentially growing regular <normal modes> for $J\geq1/4$. The candidate <neutral modes of the equal-width Hazel model> instead lie on
$$
\boxed{J=k(1-k)\leq\frac14,\qquad \max J=\frac14\text{ at }k=\frac12.}
$$
The parabola touches the sufficient-stability threshold at its maximum. There is no contradiction: neutrality has $c_i=0$, whereas the proof in part (a) assumes $c_i>0$. Moreover, the $k=1/2$ profile has a singular critical-layer <derivative> and logarithmically divergent horizontal <kinetic energy>; it is not a regular growing mode satisfying the proof's hypotheses.
For $J<1/4$, the local <gradient Richardson number> condition permits instability but does not establish it. Substitution of a neutral ansatz alone also does not determine on which side of the curve unstable eigenvalues lie. It identifies the formal neutral curve; concluding a full stability boundary requires additional continuation analysis of the <eigenvalues>. The nondecaying $k=0$ endpoint and the smooth $k=1,J=0$ endpoint have the distinct qualifications described above.
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