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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.

For a stratified parallel shear flow, the gradient Richardson number compares the squared buoyancy frequency with squared vertical shear. The Miles–Howard theorem excludes exponentially growing inviscid normal modes when this ratio is at least $1/4$ everywhere. Where $U^{\prime }=0$, the equivalent criterion is written directly as $N^2\ge (U^{\prime }{)}^2/4$.

Let $V=U-c$ and suppose a growing normal mode has $\mathrm{Im}c=c_i>0$, taking $\alpha >0$ without loss of generality. Since $V$ never vanishes on the real channel, choose a continuous square-root branch and write $\varphi =V^{1/2}H$. Direct differentiation givesSubstitution into the Taylor–Goldstein equation and division by $V^{1/2}$ givesThus the required transformation isThe impermeable walls give $\varphi (\pm 1)=0$, hence $H(\pm 1)=0$. Multiply by $\overline{H}$, integrate across the channel, and use integration by parts. With the boundary term zero, the resulting power-transformed Taylor–Goldstein energy identity isBoth $U^{\prime \prime }$ and $J$ are real. Since $\mathrm{Im}V=-c_i$ and $\mathrm{Im}(1/V)=c_i/|V{|}^2$, its imaginary part, divided by $-c_i$, isIf $J\ge U^{\prime 2}/4$ everywhere, the integrand is nonnegative and the first two terms have positive integral for any nonzero mode. This contradicts the identity. Therefore instability requiresThis is the necessary instability condition of the Miles–Howard theorem. Where $U^{\prime }\ne 0$, it says that the gradient Richardson number $J/U^{\prime 2}$ must be below $1/4$ somewhere. It is not a sufficient condition for instability, and the conclusion concerns exponential normal-mode growth, not exclusion of the algebraic lift-up effect in part (a). The algebra uses the given nondimensional $J$; in dimensional notation its counterpart is the squared buoyancy frequency.

Divide the equilibrium equation by $q>0$ to get $b{q}^2-q+a=0$. Hence both positive branches, when $0<4ab<1$, areAt equilibrium, ${\mathrm{Ri}}_f=a/q=1-bq$. Therefore the larger-speed branch $q_{+}$ gives the printed flux Richardson numberwhile the smaller-speed branch gives the same formula with a plus sign before the square root. Solving the quadratic alone does not discard that second branch.

The larger-speed branch is selected by local kinetic-energy stability. If the same closures are used in the local energy evolution, $\frac{1}{2}d(q^2)/dt=\mathcal{P}-B-ϵ$. For $q>0$, divide by $q$ to obtainIts linearization at $q_{\pm }$ has derivative $C_{u}S-2q_{\pm }/L_u=\mp C_{u}S\sqrt{1-4ab}$. The upper branch is therefore stable and the lower unstable in this local closure model.

The discriminant gives the critical gradient Richardson numberAt criticality the branches meet and ${\mathrm{Ri}}_f=1/2$. At zero stratification the stable positive solution is $q=C_{u}S{L}_u$ with ${\mathrm{Ri}}_f=0$; the lower algebraic root is $q=0$. This closure-dependent critical value is not automatically the inviscid $1/4$ criterion of the Miles–Howard theorem.

Write $S=U_z$ and $N^2=d\bar{b}/dz$. The gradient Richardson number and Flux Richardson number areThe first compares the restoring buoyancy gradient with mean shear; the second is the fraction of shear production consumed by stable buoyancy work. With this convention, both are positive for stable mixing and negative for unstable stratification. In local steady equilibrium, $ϵ=P+B_0=P(1-{\mathrm{Ri}}_f)$, so a positive dissipation requires ${\mathrm{Ri}}_f<1$. The inviscid Miles–Howard theorem concerning ${\mathrm{Ri}}_g\ge 1/4$ is a linear stability criterion, not a universal equality for turbulent fluxes.

Use the gradient-diffusion closures $-\overline{u^{\prime }{w}^{\prime }}=K_{m}S$ and $B_0=-K_b{N}^2$. The turbulent Prandtl number is ${\mathrm{Pr}}_t=K_m/K_b$, givingUnder the ideal Reynolds analogy $K_m=K_b$, the two Richardson numbers coincide. It is an additional closure, not an automatic consequence of the definitions. If the buoyancy-gradient similarity function is defined by $N^2=-B_0{\varphi }_h/(\kappa z{u}_{\ast })$, then

Let $D=\overline{U}-c$ and $F=\hat{w}/D^{1/2}$. For $c_i>0$, $D$ never vanishes, so an analytic branch of a square root exists along the real flow domain. Take real smooth velocity and buoyancy frequency coefficients, nonzero $k$ (chosen positive for the growing-wave convention), a nontrivial regular normal mode, and boundary decay strong enough to remove the integration-by-parts term.

Substitution into the Taylor–Goldstein equation gives the half-power case of the power-transformed Taylor–Goldstein energy identity:For completeness, differentiating $\hat{w}=D^{1/2}F$ yields ${\hat{w}}^{\prime \prime }=D^{1/2}{F}^{\prime \prime }+{\overline{U}}^{\prime }{D}^{-1/2}{F}^{\prime }+[\frac{1}{2}{\overline{U}}^{\prime \prime }{D}^{-1/2}-\frac{1}{4}({\overline{U}}^{\prime }{)}^2{D}^{-3/2}]F$, which explains the coefficient $1/4$.

Multiply by $F^{\ast }$ and integrate over $z$. The boundary term $[F^{\ast }D{F}^{\prime }]$ is zero for the given homogeneous endpoint conditions, or for sufficiently decaying finite-energy modes at infinity. ThereforeSince $\mathrm{Im}D=-c_i$, $\mathrm{Im}(1/D)=c_i/|D{|}^2$, and $S,{\overline{U}}^{\prime \prime }$ are real, taking the imaginary part and dividing by $-c_i$ givesThe first two terms have strictly positive integral for a nonzero mode. Thus $S$ cannot be nonnegative everywhere:This proves the Miles–Howard theorem in contrapositive form. Where ${\overline{U}}^{\prime }\ne 0$, the corresponding gradient Richardson number must fall below $1/4$ somewhere. Failure of this sufficient-stability criterion does not prove instability. The derivation uses $c_i>0$; it cannot be applied unchanged to a neutral singular critical layer.

For the equal-width Hazel model with stable density stratification $J\ge 0$,The Miles–Howard theorem therefore excludes exponentially growing regular normal modes for $J\ge 1/4$. The candidate neutral modes of the equal-width Hazel model instead lie onThe 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.

Choose $a=1/2$ in the power-transformed Taylor–Goldstein energy identity and let $Q=N^2-(U^{\prime }{)}^2/4$. The result isBecause $\mathrm{Im}V=-c_i$ and $\mathrm{Im}(1/V)=c_i/|V{|}^2$, its imaginary part givesIf $4N^2>(U^{\prime }{)}^2$ everywhere, the integral is strictly positive for every nonzero eigenfunction, so a mode with $c_i\ne 0$ is impossible. Thus the flow is neutrally stable to these inviscid normal modes. This is the Miles–Howard theorem, expressed as a lower bound $\mathrm{Ri}>1/4$ on the gradient Richardson number. It excludes exponential modal growth; it does not by itself exclude transient growth.

Put $D={\partial }_t+U{\partial }_x$, write $p$ for perturbation pressure divided by ${\rho }_0$, and set $b=-g{\rho }^{\prime }/{\rho }_0$. The linearized Boussinesq approximation, incompressibility and material derivative giveTaking the curl eliminates pressure; differentiating the resulting vorticity equation in $x$ and using incompressibility gives $D{\nabla }^{2}w-U^{\prime \prime }{w}_x=b_{xx}$. A further application of $D$ therefore yieldsFor a stationary nonzero horizontal Fourier mode, $D=U{\partial }_x$, so cancellation of ${\partial }_x^2$ gives the stationary Taylor–Goldstein equationThis division requires $U\ne 0$ on the interval considered: a zero of $U$ is a critical level of an internal gravity wave. The mean profiles must be sufficiently smooth for the displayed derivatives, with background hydrostatic pressure and stable density stratification, $N^2>0$, for propagating internal gravity waves. The horizontally uniform component is excluded from the cancellation. For a horizontal wavenumber $k$, local vertical propagation additionally requires $N^2/U^2-U^{\prime \prime }/U>k^2$; this is distinct from the smoothness restrictions. If stability of the background against other disturbances is needed, the Miles–Howard theorem supplies the sufficient condition $N^2\ge (U^{\prime }{)}^2/4$, rather than a necessary condition for deriving the equation.

For a mode with nonreal phase velocity $c$, let $V=U-c$, choose a continuous branch of $V^a$, and put $\hat{w}=V^{a}q$. Under impermeable boundary conditions, multiplying the transformed Taylor–Goldstein equation by $V^{2a}\overline{q}$ and applying integration by parts givesChoosing $a=1/2$ gives the Miles–Howard theorem; choosing $a=1$ makes the real phase velocity of an unstable mode a weighted mean of $U$.