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A bulk Richardson number compares a characteristic buoyancy difference across a layer with the squared characteristic flow speed. It uses large-scale differences rather than the pointwise gradient Richardson number. Its normalization depends on the chosen velocity, density and length scales.

For $S<0$ and constant positive buoyancy frequency, the critical height is $U_0/|S|$. A stationary upward ray approaches it with vertical group velocity proportional to $U^2$, requiring infinite ideal ray travel time. Large gradient Richardson number can preserve the short-wavelength WKB approximation, but parcel-displacement amplitude grows and eventually invalidates linearity.

The ratio of stable buoyancy destruction to shear production of turbulent kinetic energy. With down-gradient transport and turbulent Prandtl number ${\mathrm{Pr}}_t$, it obeys ${\mathrm{Ri}}_f={\mathrm{Ri}}_g/{\mathrm{Pr}}_t$, rather than automatically equaling the gradient Richardson number. Unstable buoyancy production gives a negative value in this convention.

The Hazel model is a smooth stratified shear flow whose base velocity has a hyperbolic-tangent profile and whose squared buoyancy frequency has a hyperbolic-secant-squared profile. The equal-width nondimensional example is $U=\tanh z$ and $N^2=J{\mathrm{sech}}^{2}z$. For $J\ge 0$, its gradient Richardson number is $J{\cosh }^{2}z$, with minimum $J$.

A smooth inviscid stratified parallel flow with gradient Richardson number at least $1/4$ everywhere has no exponentially growing two-dimensional normal modes. Put $a=1/2$ in the power-transformed Taylor–Goldstein energy identity and take its imaginary part:The integral is positive for a nonzero mode when the numerator is nonnegative, forcing $c_i=0$. This is a modal stability theorem, not a prohibition on transient growth. Maslowe's review discusses the theorem and the role of critical layers.

If the similarity functions of a turbulent boundary layer are differentiable at zero and have finite positive neutral values, expanding $f=a+dz/\ell +O((z/\ell {)}^2)$ gives a shear proportional to $1/z-d/(a\ell )$. Integration yields a logarithmic term plus a linear correction. Dimensional reasoning alone does not fix the coefficients or guarantee this regular neutral limit. The gradient Richardson number is of order $z/\ell$, so $|z/\ell |\ll 1$ is a shear-dominated, nearly neutral regime on either side of stability.

Choose the length convention ${\ell }_m=-U_{\ast }^3/B_0$; a dimensionless multiplier can be absorbed into the similarity functions. Dimensional analysis under the stated local dependence givesTo obtain a log-linear expansion one also needs a regular neutral limit: take $f_M=a_M+d_M\zeta +O({\zeta }^2)$ and $f_B=a_B+d_B\zeta +O({\zeta }^2)$, with positive $a_M,a_B$. Dimensions alone do not specify these constants or guarantee differentiability. Let $s_u=-J_M/U_{\ast }^2$ denote the shear sign. In the high-number regime of part (a), integrate the reciprocal expansions from a reference height $z_r>0$ to obtain the near-neutral log-linear surface-layer profilesTheir leading terms are logarithmic; the first buoyancy-dependent corrections are linear in height. The additive constants require boundary or reference data and cannot be obtained from the prescribed fluxes alone.

The gradient Richardson number isThus $|z/{\ell }_m|\ll 1$ is nearly neutral, shear-dominated turbulence: the buoyancy contribution is small relative to shear production. Stable flux has $B_0<0$, ${\ell }_m>0$ and positive Richardson number; heating has $B_0>0$, ${\ell }_m<0$ and negative Richardson number. The smallness condition is on the magnitude, not an automatically satisfied signed inequality for every negative $z/{\ell }_m$.

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.

Set $S={\partial }_z\overline{u}>0$ and $N^2=-(g/{\rho }_0){\partial }_z\overline{\rho }>0$, using the constant reference mass density ${\rho }_0$ of the Boussinesq approximation. The mean-shear production is $\mathcal{P}=-\overline{u^{\prime }{w}^{\prime }}S>0$. The stable buoyancy destruction is $B=(g/{\rho }_0)\overline{{\rho }^{\prime }{w}^{\prime }}>0$. ThereforeThe Flux Richardson number compares stable buoyancy destruction with shear production; the gradient Richardson number compares restoring stratification with mean shear rate. They are generally different. The stationary energy budget gives $ϵ=\mathcal{P}(1-{\mathrm{Ri}}_f)$, so positive dissipation in this balance requires ${\mathrm{Ri}}_f<1$.

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

The WKB approximation treats the vertical wavenumber as locally constant over one vertical wavelength. Write $w=\Re \{A(z)e^{ikx+i{\int }^{z}m(s)ds}\}$, with $|m^{\prime }|\ll m^2$ and $|A^{\prime }|\ll |mA|$. For the stationary internal-wave WKB solution, $A\propto m^{-1/2}$ away from singular or turning levels. A small terrain slope and small perturbations are also needed for Linearized Boussinesq equations.

Let $p$ be the perturbation fluid pressure divided by reference mass density, and $b$ the buoyancy perturbation. The steady Linearized Boussinesq equations and incompressibility condition areFor an $e^{ikx}$ component, elimination gives $u=-w^{\prime }/(ik)$, $b=-N^{2}w/(ikU)$ and $p=(U{w}^{\prime }-U^{\prime }w)/(ik)$. Substitution into vertical momentum gives the exact stationary Taylor–Goldstein equationThe leading WKB approximation therefore identifiesThe kinematic boundary condition fixes the horizontal wavenumber to $k=k_T$ for the sinusoidal forcing.

For explicit heights assume a constant positive buoyancy frequency $N$, $0<k{U}_0<N$, and $|S|/N\ll 1$. The latter is a large gradient Richardson number. The linear shear flow has $U^{\prime \prime }=0$, so the propagation condition is $|U|<N/k$.

For $S<0$, the first obstruction is the critical level of an internal gravity wave atAs $z\uparrow z_c$, $m\sim N/U\to \infty$. The critical-height approach of a stationary wave in linear shear has upward group velocity $c_{gz}\sim k{U}^2/N$, so the travel time to $z_c$ diverges: the height is a supremum, approached in the ideal ray model. Here $|m^{\prime }|/m^2\to |S|/N$, which can remain small; it would be incorrect to say that the short-wavelength WKB approximation necessarily fails solely because $m$ diverges. Nevertheless $w\propto U^{1/2}$ and displacement $w/(ikU)\propto U^{-1/2}$, so a finite disturbance eventually violates linearity. Dissipation and overturning also matter near this singular level.

For $S>0$, the turning height of a stationary wave in linear shear isAt $z_t$, $m=0$ and the WKB approximation fails. An Airy function matching region describes reflection and an upper evanescent wave. Thus $z_t$ is the highest height of vertical propagation, rather than a sharp cut-off of every disturbance.

The problem initially allows $N(z)$ to vary. Without the extra constant-$N$ assumption, replace the explicit turning height by the first root of $N(z)=k|U_0+Sz|$ encountered on a propagating branch. A declining $N(z)$ can cause a turning level before the negative-shear critical level. The two closed-form heights require the stated constant-stratification assumption.

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.

The ratio of turbulent momentum diffusivity to buoyancy or heat diffusivity. Their ratio determines how the gradient Richardson number is related to the energy-budget Flux Richardson number. It is generally a closure-dependent function of the flow, not necessarily one.