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The dispersion relation gives the zero-group velocity saddleThus the absolute growth rate is $\sigma (0)=\mu -U^2/[4(1+c_d^2)]$. More generally the saddle growth rate along a ray isso a frame travelling at $V=U$ sees the maximal rate $\mu$.

For $U>0$, the three open regions areThe line $\mu =0$ is the temporal marginal boundary; the parabola $\mu =U^2/[4(1+c_d^2)]$ is the convective-to-absolute boundary. The Green function of the linear complex Ginzburg-Landau equation has the prefactor $t^{-1/2}$, so zero exponential rate on the latter boundary still allows algebraic decay of the impulse response. This is the stability diagram of the linear complex Ginzburg-Landau equation.

Take $\beta >0$, a uniform eastward $U>0$, and the nondivergent Barotropic Rossby wave model. Its dispersion relation is $\omega =Uk-\beta k/(k^2+l^2)$. A stationary wave with $k\ne 0$ hasThe directional component wavelengths are $2\pi /|k|$ and $2\pi /|l|$ when nonzero. Differentiating the full dispersion relation before imposing stationarity gives its ground-frame group velocity:Writing $k=K\cos \theta$, $l=K\sin \theta$ shows ${\mathbf{c}}_g=U(1+\cos 2\theta ,\sin 2\theta )$. Therefore the stationary Rossby-wave ray envelope at time $\tau$, for wave packets emitted from a localized mountain throughout $0<t<\tau$, is the diskThe circle is the front of the earliest emissions; later emissions fill ray segments back to the source. The sketch represents the geometrical wave packet envelope, not compact support of the full balanced near field. All propagating wave packet energy lies downstream in this model.

If the finite-depth free-surface model of the preceding part is retained consistently, its basic surface slope adds $U/R_D^2$ to the potential-vorticity gradient. Stationarity still gives $K^2=\beta /U$ and the same wavelength, but the group velocity is reduced by $\gamma =K^2/(K^2+R_D^{-2})$; the disk center and radius become $\gamma U\tau$. This tends to the plotted nondivergent result as $R_D\to \infty$.

A westward wind has no nontrivial stationary propagating planetary Rossby-wave train in this model: $U<0$ would require $K^2=\beta /U<0$. Its advective phase motion and intrinsic westward Rossby wave phase motion cannot cancel. A local forced, evanescent response and time-dependent disturbances can still occur. The degenerate $k=0$ limit is not a propagating wave train from an isolated mountain.

Use the geometrical optics conventionBecause ${\partial }_t=\epsilon {\partial }_T$ and ${\partial }_x=\epsilon {\partial }_X$, differentiating $e^{i\theta /\epsilon }$ gives an order-one physical angular frequency and order-one physical wavenumbers. Absorb a constant leading amplitude phase into $\theta$ and take $A_0$ real; for a complex amplitude the corresponding quadratic density is $|A_0{|}^2$.

Substitution into the wave equation gives the eikonal equation for a variable-speed wave at leading order:Choose the positive-frequency branch $\omega =\Omega =cK$, $K=\sqrt{k^2+{\ell }^2}$; the negative branch can be treated in exactly the same way. The imaginary first-order terms giveMultiplying by $A_0$ puts this wave-action transport for a variable-speed wave in conservation form:Its density is $\omega A_0^2$ and its flux is that density times the group velocity. Independently, equality of mixed wave phase derivatives givesThis phase compatibility in geometrical optics is essential when converting the differentiated dispersion relation into equations along rays.

For completeness let $\omega =\Omega (\mathbf{k};\mathbf{X},T)$ be a general local dispersion relation. Define ray tracing by $d\mathbf{X}/dT={\mathbf{c}}_g={\nabla }_{\mathbf{k}}\Omega$. Using the chain rule in ${\nabla }_{\mathbf{X}}\omega$ and the symmetry ${\partial }_{X_i}{k}_j={\partial }_{X_j}{k}_i$ givesLikewise ${\omega }_T={\Omega }_T+{\mathbf{c}}_g\cdot {\mathbf{k}}_T$ cancels its advective term because ${\mathbf{k}}_T=-\nabla \omega$. Hence the Hamiltonian ray equations for a local dispersion relation areThe derivatives on the right hold the wavevector fixed. For this wave equation they reduce to

When $c=c(Y)$, the ray invariants are $k$ and $\omega$, while $d\ell /dT=-K{c}_Y<0$. For a monochromatic component with these fixed invariants write $\theta =kX-\omega T+S(Y)$, soUsing the stated independence of $A_0$ from $X,T$, the wave-action transport for a variable-speed wave becomes $(c^2\ell A_0^2{)}_Y=0$. ThereforeThis is the stationary wave amplitude in a stratified wavespeed; $\mu$ is the constant for that component or ray family.

At the proposed simple turning point of a variable-speed wave, take $c_s=c(Y_s)=\omega /k$ and $c_s^{\prime }=c_Y(Y_s)>0$. On the chosen positive branch this entails $k>0$. DefineAs the ray approaches from below, $\ell \sim [q(Y_s-Y){]}^{1/2}$ andThe amplitude diverges and the vertical group velocity $c^2\ell /\omega$ vanishes. The WKB approximation also fails its slow-variation condition $\epsilon |{\ell }_Y|/{\ell }^2\ll 1$ when $Y_s-Y=O({\epsilon }^{2/3})$. The turning-point enhancement of wave amplitude shows that this divergence is a failure of the outer ray approximation, rather than an actual infinite solution.

To derive the inner wavefield directly, separate the conserved horizontal wave phase and frequency in the original differential equation:Dividing by $c^2$ gives ${\epsilon }^2{\Psi }_{YY}+2{\epsilon }^2(c_Y/c){\Psi }_Y+Q(Y)\Psi =0$. Set $Y-Y_s=\delta \eta$. The second derivative is of size ${\epsilon }^2/{\delta }^2$, whereas $Q\sim -q\delta \eta$, so the Airy scaling at a variable-speed wave turning point isThe first-derivative term is smaller by $O(\delta )$, and higher coefficients in the expansion of $Q$ give the same relative order. Thus the general leading inner profile is a linear combination of the two Airy functions $\mathrm{Ai}(\eta )$ and $\mathrm{Bi}(\eta )$. For decaying Airy continuation and ray reflection, if the physical continuation is bounded and decays into $Y>Y_s$, its growing $\mathrm{Bi}$ coefficient is zero. In that usual evanescent continuation the wavefield has the formWithout a condition in the evanescent region the corresponding $\mathrm{Bi}$ term remains admissible locally; the given existence of a wavefield below the turning point alone does not select its coefficient.

Finally, the outer amplitude at distance $\delta$ is $O({\delta }^{-1/4})=O({\epsilon }^{-1/6})$ for $\mu =O(1)$ and fixed nonzero $q$. The oscillatory Airy turning-point connection formula has exactly the same $(-\eta {)}^{-1/4}$ envelope on the allowed side. Therefore the inner wavefield has magnitude $O({\epsilon }^{-1/6})$. The decaying Airy continuation matches an incident and reflected pair of rays, so a single upward $\ell >0$ branch must be supplemented by its reflected branch near the turning point; no formal coefficient matching is needed to identify this magnitude.

Write $\zeta =(\nabla \times \mathbf{u})\cdot {\mathbf{e}}_z$ and $q=\zeta -f\eta /h_0$, so the PDF's vector potential vorticity is $\mathbf{Q}=q{\mathbf{e}}_z$. The local TeX loses some of this boldface. Taking the vertical curl of the momentum equation gives ${\zeta }_t=-f\nabla \cdot \mathbf{u}$, while mass conservation gives ${\eta }_t=-h_0\nabla \cdot \mathbf{u}$. ThusThis is the linear perturbation of shallow-water potential vorticity, scaled by $h_0$; it should not be confused with the full nonlinear conserved ratio $(f+\zeta )/h$.

Since $\nabla \cdot ({\mathbf{e}}_z\times \mathbf{u})=-\zeta$, taking the divergence of momentum gives $(\nabla \cdot \mathbf{u}{)}_t=f\zeta -g{\nabla }^2\eta$. Differentiate mass conservation and use $\zeta =q+f\eta /h_0$ to obtainFor $\mathbf{Q}=0$, substitution of the stated plane wave gives the linear rotating shallow-water dispersion relation, with $c=\sqrt{g{h}_0}$:The two even branches approach $\pm c|k|$ at large $|k|$ and have values $\pm |f|$ at $k=0$. The original dispersion diagram below also compares the speed magnitudes.

For $k\ne 0$, the phase velocity and group velocity magnitudes areThe printed convention $e^{i(kx+\omega t)}$ means the signed phase velocity is $-\omega /k$ and the signed group velocity is $-d\omega /dk$. The speed magnitudes above are independent of that sign convention.

For $f\ne 0$, longer wavelengths have greater phase velocity but smaller group velocity: individual crests and wave packets therefore answer “faster or slower” differently. At long wavelength, $|k|\ll |f|/c$, the frequency is nearly the inertial frequency $|f|$, $|c_p|\sim |f|/|k|$, and $|c_g|\sim c^2|k|/|f|$. Rotation has its largest relative effect here, and the waves are strongly dispersive. At short wavelength, $|k|\gg |f|/c$, both speeds tend to $c$ and rotation gives only a small, weakly dispersive correction. If $f=0$, the gravity waves are nondispersive, with speed $c$; at $k=0$ the phase velocity formula is undefined.

At the absolute wavenumber, the group velocity $d\omega /dk$ is zero. The corresponding accessible saddle therefore governs the disturbance seen at a fixed position. Its exponential growth rate isSaddle accessibility can be checked directly here. With $D=1+i{c}_d$, whose real part is positive, Fourier inversion gives the Green function of the linear complex Ginzburg-Landau equation,At fixed $x$ its exponential rate is $\mathrm{Im}{\omega }_0$, while along the packet centre $x=Ut$ it is $\mu$. For real wavenumbers of Fourier modes, the temporal growth rate is $\mathrm{Im}\omega (k)=\mu -k^2$, so the flow is temporally unstable exactly when $\mu >0$.

Consequently the classifications for a localized disturbance in this laboratory frame areFor convective hydrodynamic instability, a travelling wave packet amplifies but the response at each fixed position decays. For absolute hydrodynamic instability, that fixed-position response amplifies. The equality is the marginal absolute threshold: its exponential rate is zero and the impulse response has a $t^{-1/2}$ prefactor. For $\mu <0$ there is no temporal growth; $\mu =0$ is temporally marginal. When $U=0$ the convective window is empty. These conclusions use the infinite-line impulse problem; a general dispersion relation requires its own spatial-branch or saddle selection.

The normalized Bloch states $|k⟩=N^{-1/2}{\sum }_{n=1}^N{e}^{ikna}|n⟩$ satisfy periodicity when $k=2\pi j/(Na)$. Acting with the hopping terms shifts the coefficients by one site, givingThe first Brillouin zone is a fundamental cell of reciprocal space, for example $[-\pi /a,\pi /a)$; wavevectors differing by $2\pi /a$ label the same state. It contains exactly $N$ distinct allowed wavevectors and hence $N$ orbital states. The written Hamiltonian has no spin index; if electron spin degeneracy is included there are $2N$ states.

For a narrow wave packet, group velocity and band-curvature effective mass areFor the usual $J>0$, the mass is negative in the outer half of the zone, $\pi /(2a)<|k|\le \pi /a$, taking one endpoint representative. At $|k|=\pi /(2a)$ the inverse mass vanishes and the mass diverges. For arbitrary nonzero $J$, the sign criterion is $J\cos ka<0$; $J=0$ is a flat band.

For electron charge $-e$ with $e>0$ in a constant electric field $\mathcal{E}$, the single-band semiclassical equations are $\hslash \dot{k}=-e\mathcal{E}$, $\dot{x}=v(k)$. Thus $k(t)=k_0-e\mathcal{E}t/\hslash$, and for $\mathcal{E}\ne 0$,These are Bloch oscillations of period $2\pi \hslash /(|e\mathcal{E}|a)$. The zero-field limit is $x-x_0=v(k_0)t$. This assumes a narrow packet and neglects scattering and interband transitions.

For many electrons obeying the Pauli exclusion principle, a partially filled band has available nearby states and can carry current under a displaced occupation distribution: it describes a metal within band theory. A completely filled band has cancelling velocities over the zone, so its current is zero; if separated from empty higher bands by a nonzero gap, it describes an insulator at sufficiently low temperature and weak electric field. With spin degeneracy this isolated band is filled at two electrons per site. The single-electron Hamiltonian alone does not specify the filling or a gap to a next band, so those extra physical assumptions are required for classifying a material.

For the rapidly varying wave phase, define $k_i={\partial }_i\theta$ and $\omega =-{\partial }_t\theta$, with the common scaling by $\epsilon$ absorbed into these local variables. A packet moves with group velocity $c_{g,i}=\partial \Omega /\partial k_i$. Along that trajectory, $d/dt={\partial }_t+c_g\cdot {\nabla }_x$. Compatibility gives ${\partial }_t{k}_i=-{\partial }_i\omega$ and ${\partial }_i{k}_j={\partial }_j{k}_i$. Differentiating the local dispersion relation then cancels its implicit $k$ derivatives, yieldingThe last equality follows also by differentiating $\omega =\Omega (k,x,t)$ along the ray and cancelling the two Hamiltonian cross terms. The partial derivatives of $\Omega$ hold its other arguments fixed.

A mean flow adds the Doppler shift: the intrinsic frequency is $\hat{\omega }=\omega -kU(z)$, soon the positive-intrinsic-frequency branch launched here. The medium is stationary and independent of $x$, so $\omega$ and $k$ are constant. Initially $m=0,U=0$, hence $\omega =N$. Moreover $\dot{m}=-k{U}^{\prime }<0$, and then $\dot{z}=-Nkm/(k^2+m^2{)}^{3/2}>0$. The initial vertical group velocity is zero, but $m$ immediately becomes negative and the ray accelerates upwards.

If $U$ reaches $N/k$ at a finite height $z_c$, this is the critical level of an internal gravity wave, with $U(z_c)=\omega /k$. Below it,Assuming $0<U^{\prime }(z_c)<\infty$, put $\delta =z_c-z$. Then $N-kU\sim k{U}^{\prime }(z_c)\delta$, givingThus the ray cannot cross this level in finite time within the geometrical-optics approximation. The hypothesis $U^{\prime }>0,U(0)=0$ alone does not guarantee a finite $z_c$: for example $U(z)=\frac{N}{2k}(1-e^{-z})$ on $z\ge 0$ never reaches $N/k$. Existence of a finite critical level is an additional assumption. Very close to one, diverging wavenumber can invalidate the inviscid, linear, slowly varying approximation; the result is the formal ray prediction.

For $-\beta /{\ell }^2<V\le 0$, there is one positive stationary wavenumber $k_0\in (0,\ell ]$ satisfying $\beta (k_0^2-{\ell }^2)/(k_0^2+{\ell }^2{)}^2=V$, and its negative. Since ${\omega }^{\prime \prime }(k_0)>0$, their paired stationary-phase contribution isAt $V=0$ the finite stationary points are $\pm \ell$ and are nondegenerate, so the same formula applies, assuming sufficient decay to control the large-wavenumber tail.

For $V>\beta /(8{\ell }^2)$ or $V<-\beta /{\ell }^2$, there are no real stationary points: no group-velocity ray reaches those slopes. Integration by parts gives much smaller contributions; for a Schwartz amplitude they decay faster than every inverse power of $t$ at a fixed such $V$. For less regular data the rate depends on the amplitude, and the printed even function assumption alone does not specify it. At the excluded endpoint values of the group velocity stationary points coalesce and an ordinary $t^{-1/2}$ formula fails, typically requiring an Airy-type transition rather than the expressions above.

Substitution of $e^{i(kx-\omega t)}$ givesThe phase velocity is negative for all nonzero wavenumbers, so crests travel left. The group velocity is negative for $|k|<\ell$, zero at $|k|=\ell$, and positive for $|k|>\ell$. Its minimum is $-\beta /{\ell }^2$ at zero and maximum $\beta /(8{\ell }^2)$ at $|k|=\sqrt{3}\ell$; it tends to zero from above at infinity. The odd dispersion relation has extrema at $k=\pm \ell$, with $\omega (\ell )=-\beta /(2\ell )$.

The Fourier solution is $\phi (x,t)={\int }_{\mathbb{R}}A(k)e^{i[kx-\omega (k)t]}dk$. The stationary-phase statements require enough amplitude regularity and decay, for example $A$ smooth and rapidly decreasing; realness and evenness alone do not guarantee a convergent integral or a stationary-phase expansion. For $x=Vt$, set ${\Phi }_V(k)=kV-\omega (k)$ and solve ${\omega }^{\prime }(k)=V$ implicitly.

Write $H_{\pm }=H_0\pm d$, with the upper sign referring to $y>0$, and use an f-plane with Coriolis parameter $f_0>0$ for the stated directions. The formulas retain the signed $f_0$; a positive Rossby deformation radius means $R_{\pm }=\sqrt{g{H}_{\pm }}/|f_0|$. Let $\rho$ be the constant fluid density.

Far from the depth step the problem is independent of $y$. The linearized shallow water equations imply conservation of $v_x-f_0\eta /H_{\pm }$. Initially this is $-f_0{\eta }_0\mathrm{sgn}(x)/H_{\pm }$. In geostrophic balance, $u_s=0$ and $v_s=g{\eta }_{s,x}/f_0$, soBoundedness and matching of height and its derivative at $x=0$ give geostrophic adjustment of a surface-height jump:The potential energy release per unit length in $y$ is the finite difference between two individually infinite energies:For comparison, the final kinetic energy is $\rho g{\eta }_0^2{R}_{\pm }/2$, so the energy radiated in inertia-gravity waves is $\rho g{\eta }_0^2{R}_{\pm }$. Integrating the depth-weighted velocity over $x$ gives the volume transportsFor $f_0,{\eta }_0>0$, $Q_{-}$ is towards the step and $Q_{+}$ is away from it. Their difference is $4g{\eta }_{0}d/f_0$; a global steady state cannot simply join the two far-field currents without additional transport along the step.

To obtain the rotating shallow-water height equation with variable depth, set $\zeta =v_x-u_y$ and $J_H={\partial }_x(Hv)-{\partial }_y(Hu)$. Direct differentiation of the original momentum and continuity equations givesConsequentlyAt the step, $H_y=2d\delta (y)$ is a Dirac delta distribution; the original matching conditions are used rather than assuming a smooth depth there. Velocity elimination raises the time order: initially ${\eta }_t=0$ and ${\eta }_{tt}=g{\nabla }_h\cdot (H{\nabla }_h\eta )$, as well as the prescribed initial height.

For a decaying harmonic disturbance, the height equation on each constant-depth side givesSolving the two momentum equations yieldsContinuity of $H\hat{v}$ at the step gives the implicit step-trapped topographic Rossby wave dispersion relationThis derivation assumes ${\omega }^2\ne f_0^2$ and positive decay rates; it describes the low-frequency trapped branch, rather than the radiating inertia-gravity waves.

In the quasi-geostrophic approximation, replace $f_0^2-{\omega }^2$ by $f_0^2$. The resulting explicit frequency, and then its leading small-$d/H_0$ form, areFor $d>0$, the phase velocity has the sign of $f_0$. Thus propagation is eastward in the Northern Hemisphere, with the shallower side on the right. Differentiating the leading dispersion relation givesIn the long-wave limit the group velocity and phase velocity coincide:

To find the slow amplitude, integrate the low-frequency height equation across the step. This gives ${\partial }_t[H{\eta }_y]=f_0[H]{\eta }_x$, where brackets mean the upper-side minus lower-side value. For the given exponentially trapped outer profile, $[H{\eta }_y]=2H_{0}A/R_D$ and ${\eta }_x(x,0)=2{\eta }_0\delta (x)-A_x$. Hence long-wave transport along a depth step obeysThe method of characteristics, interpreted for the Dirac delta distribution source, givesFor $c_T>0$ this is $2{\eta }_0$ in $0<x<c_{T}t$ and zero outside: a surface depression is left in the wake of the wave. The sharp $x$-jump is a long-wave outer approximation. The fast adjustment region of width $O(R_D)$ near the origin, and the detailed wavefront, are unresolved. Keeping a smoothed background ${\eta }_s(x)$ instead would give $A_t+c_T{A}_x=c_T{\eta }_s^{\prime }(x)$ and $A={\eta }_s(x)-{\eta }_s(x-c_{T}t)$ at this order.

The along-step geostrophic flow is $u=-g{\eta }_y/f_0$. Integrating separately over the two sides givesIn the wake, $T_{-}=Q_{-}$ and $T_{+}=-Q_{+}$. The shallow-side transport is eastward and the deep-side transport westward, with magnitudes equal to the respective far-field currents. The net along-step transport is $T_{-}+T_{+}=-4g{\eta }_{0}d/f_0$, balancing their mismatch. These comparisons use the stated leading long-wave and small-depth-contrast approximations.

In an adiabatic quasi-geostrophic approximation, hydrostatic approximation and geostrophic balance give $b^{\prime }=f_0{\psi }_z$. The leading buoyancy equation is $D_g{b}^{\prime }+N^{2}w=0$; hence $w=-D_g(f_0{\psi }_z/N^2)$. At the rigid surface $z=\alpha y$, the no-normal-flow condition is $D_g(z-\alpha y)=0$, so $w=\alpha v_g=\alpha {\psi }_x$. The small boundary slope allows the disturbance condition to be evaluated at the flattened boundary $z=0$ to the retained order.

For thermal wind $U=\Lambda z$, choose the basic quasi-geostrophic streamfunction $\overline{\psi }=-\Lambda zy$. Its buoyancy anomaly is $\overline{b}=-f_0\Lambda y$, and its interior quasi-geostrophic potential vorticity is constant. Therefore the linear disturbance equations areThe term $-\Lambda {\psi }_x^{\prime }$ is the advection of the basic meridional buoyancy gradient. Since $q^{\prime }$ is simply advected at each fixed height, $q^{\prime }=0$ initially implies $q^{\prime }=0$ forever. At the lower boundary define $a=\alpha N^2/f_0$; thenThe PDF uses $\alpha$ for the lower-boundary slope throughout; the TeX's isolated $z=cy$ is a transcription error.

For nonzero zonal wavenumber let $\mu =N|k|/|f_0|>0$. The zero-interior-potential vorticity condition gives ${\hat{\psi }}_{zz}-{\mu }^2\hat{\psi }=0$. In the semi-infinite domain choose decay as $z\to \infty$. Writing $B_0(t)={\hat{\psi }}_z(0,t)$, potential-vorticity inversion then givesThis sloping-boundary Eady edge wave has all its time dependence in the boundary buoyancy equation; the interior responds instantaneously through elliptic potential-vorticity inversion. A single boundary wave oscillates without exponential growth. Vertical shear and topography contribute oppositely to its propagation. The contributions cancel when $\alpha =f_0\Lambda /N^2$, the slope of a background isopycnal. For $f_0>0$, positive $\Lambda$ without slope gives eastward propagation, while a positive slope without shear gives westward propagation. The phase velocity varies as $1/|k|$, and the trapping depth is $1/\mu$. Its frequency is independent of $|k|$ on each signed branch, so its zonal group velocity is zero in this ideal semi-infinite model.

With a rigid horizontal upper boundary, put $B_D(t)={\hat{\psi }}_z(D,t)$ and $M=\mu D$. Solving the same second-order inversion problem with both derivative data givesDifferentiation at $0$ and $D$ checks the prescribed values. At the upper boundary $w^{\prime }=0$, so $({\partial }_t+ik\Lambda D)B_D=ik\Lambda \hat{\psi }(D)$. Combining this with the lower-boundary condition givesThus two Boundary Rossby waves interact across the layer. For normal modes proportional to $e^{-ikct}$ their dimensional wave-speed matrix isFor $\Lambda \ne 0$, define $C=c/(\Lambda D)$ and $\tilde{\alpha }=a/\Lambda$. Its characteristic polynomial is the Eady model with a sloping lower boundary dispersion relationThe source writes $\tilde{\mu }=NkD/f_0$; for $k,f_0>0$ this is $M$, and the dispersion relation is unchanged by replacing that signed quantity by its absolute value. Its two roots areIf $\tilde{\alpha }>1$, the second term is positive, so $\Delta >0$ for every $M>0$: all modes are spectrally stable. If $\tilde{\alpha }<1$, the increasing function $M\tanh M$ runs from zero to infinity, so there is a unique $M$ withThere the squared term is zero and $\Delta <0$, giving an exponentially growing member of the complex conjugate pair. By continuity it lies in an unstable band. This proves both the requested instability for $0\le \tilde{\alpha }<1$ and instability for negative slopes in the sloping-boundary Eady model. It does not assert that every wavelength is unstable. Strongly negative slopes move the resonant band to large $M$, where the coupling is exponentially weak.

At $\tilde{\alpha }=1$ the speeds are real, but their crossing at $M=\coth M$ can give a defective neutral mode and algebraic growth; absence of exponential instability is weaker than boundedness of every initial disturbance. The nondimensional slope is undefined when $\Lambda =0$; the dimensional matrix remains valid and then has two real speeds. The instability for $\tilde{\alpha }<1$ is the counterpropagating wave instability mechanism: lower and upper boundary waves can match their laboratory phase speeds and exchange energy with the vertical shear.

Use the corrected PDF continuity equation ${\overline{v}}_{a,y}+{\overline{w}}_{a,z}=0$. Set $S=d{\rho }_s/dz=-{\rho }_0{N}^2/g<0$, $M=\overline{u^{\prime }{v}^{\prime }}$, and $B=\overline{{\rho }^{\prime }{v}^{\prime }}/S$. The transformed Eulerian mean replaces the Eulerian mean meridional circulation byThese substitutions preserve nondivergence, while density evolution becomes ${\overline{\rho }}_t+S{\overline{w}}_a^{\ast }=0$. Substituting ${\overline{v}}_a={\overline{v}}_a^{\ast }+B_z$ into the momentum equation givesThe geostrophic balance and hydrostatic approximation retain their original form. The vector $\mathbf{F}$ is the Eliassen–Palm flux. It gathers both eddy stresses into one mean-flow forcing rather than treating the eddy density flux as direct heating of the mean density.