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In a homogeneous layer under the hydrostatic approximation, the horizontal pressure gradient is independent of depth. The depth derivatives of velocity obey $(u_z{)}_t-f{v}_z=0$, $(v_z{)}_t+f{u}_z=0$. If the motion starts without shear, these derivatives remain zero. Hydrostatic pressure alone does not eliminate a pre-existing freely oscillating shear.

The hydrostatic approximation neglects vertical acceleration relative to gravity and vertical pressure gradients. It reduces vertical momentum balance to $p_z=-\rho g$.

For a positive-frequency non-rotating internal gravity wave with $\omega =N|k|/|m|$, differentiation gives group velocity $(\omega /k,-\omega /m)$. Its wavevector $(k,m)$ is orthogonal to this vector, so energy travels along constant-phase lines and perpendicular to normal phase velocity. The horizontal crest-intersection speed $\omega /k$ is distinct from the horizontal component $\omega k/(k^2+m^2)$ of that normal phase velocity. The hydrostatic approximation requires $|k/m|\ll 1$.

For a weak horizontal forcing proportional to $e^{imz}$ in an ideal uniformly stratified nonrotating fluid, the low-frequency hydrostatic approximation gives the displayed one-dimensional wave equation for the streamfunction amplitude. Two oppositely propagating characteristic integrals determine the response. A source confined to $|x|\le a$ and switched on at time zero produces no response beyond $|x|\le ct+a$ in this reduced equation. The bound on mixed forcing derivative $|f_{xt}|\ll |mN{f}_{\max }|$ controls the neglected vertical acceleration in characteristic magnitude.

For a thin buoyant layer beneath an inclined impermeable roof, the hydrostatic approximation and Darcy's law give $u=kg\Delta \rho \sin \theta /\mu$ and the volume flux per unit width $F=uh-u\cot \theta h{h}_x$. The first term transports fluid upslope; the second spreads the layer through its hydrostatic pressure gradient. The porosity multiplies the stored fluid volume, so the advection speed is $u/\varphi$, not the Darcy velocity $u$.

Use a plane wave with phase $kx+mz-\omega t$, and take $N>0$, $k,m\ne 0$. The positive-frequency branch of the given hydrostatic dispersion relation is $\omega =N|k|/|m|$. The wavevector is $\mathbf{K}=(k,m)$. Differentiating with respect to its components gives the group velocity, which carries the wave packet and its energy:The normal phase velocity, which follows a crest in its normal direction, isConsequently,Thus the group velocity lies along the constant-phase lines $z=-(k/m)x+\mathrm{constant}$, perpendicular to the phase velocity. For $k,m>0$, crests move upward and rightward, whereas energy moves downward and rightward. Choosing $m<0$ gives upward and rightward energy propagation and downward and rightward normal phase propagation. Conjugate or oppositely directed components supply the other outgoing directions from a localized source.

The speeds areThe hydrostatic limit is $|k/m|\ll 1$. In that limit energy propagates almost horizontally, with $|c_{g,x}|=N/|m|$ and $|c_{g,z}|=N|k|/m^2$, while normal phase propagation is almost vertical and has speed approximately $N|k|/m^2$. The horizontal speed at which a crest intersects a fixed height is $\omega /k=c_{g,x}$; this is not the horizontal component of the normal phase velocity vector. For example, $N=10^{-2}{\mathrm{s}}^{-1}$, vertical wavelength $2\pi /|m|=10\mathrm{km}$ and horizontal wavelength $2\pi /|k|=1000\mathrm{km}$ give $|c_{g,x}|\simeq 15.9\mathrm{m}{\mathrm{s}}^{-1}$ and $|c_{g,z}|\simeq 0.159\mathrm{m}{\mathrm{s}}^{-1}$.

These are local speeds in a uniform medium, without a background current. The non-rotating approximation requires $|f|\ll \omega$, while the hydrostatic approximation requires $\omega \ll N$. At lower frequencies, rotation changes the dispersion relation to that of inertia-gravity waves.

Take $x$ eastward, $y$ northward and $z$ upward, and let $u$ be the eastward geostrophic wind. With Coriolis parameter $f$ independent of $z$, geostrophic balance, hydrostatic balance, and the ideal gas law giveHere $R_d$ is the specific gas constant for dry air. Taking a vertical derivative of the first equation and interchanging mixed partial derivatives givesIt matters that $\rho$ is not constant in the pressure gradient force. Substitutingcancels the two terms proportional to $gfu/(R_{d}T)$ and yields the height-coordinate thermal wind relationIf $|u{T}_z|\ll |T{u}_z|$, this reduces toThe discarded term has relative scale $H|T_z|/T$ when $u_z\sim u/H$. Taking $H=10\mathrm{km}$, $T=250\mathrm{K}$ and a atmospheric lapse rate of $6.5\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ gives about $0.26$. Thus this height-coordinate formula is a useful estimate, with potentially tens-of-percent compressibility-related correction through the vertical temperature gradient; it is not an exact consequence of the two balances. If the temperature gradient vanishes vertically, the displayed approximation is exact within those balances.

The hydrostatic approximation is very good on the broad, shallow atmospheric scales considered. The geostrophic balance is normally justified by a small Rossby number and fails as a general approximation near the equator. At $30^{\circ }\mathrm{N}$, $f=7.292\times 10^{-5}{\mathrm{s}}^{-1}$. With the ideal $134\mathrm{m}{\mathrm{s}}^{-1}$ wind and a $1000$--$2000\mathrm{km}$ horizontal scale, $\mathrm{Ro}=U/(fL)\simeq 0.9$--$1.8$: quantitative accuracy is limited for this especially strong ideal jet. Small vertical aspect ratio alone does not ensure geostrophic balance, although a straight, steady zonal flow can satisfy it even when this scale estimate of the Rossby number is not small.

For the requested estimate, suppose the upper jet stream is $H=10\mathrm{km}$ above a weak surface wind, has temperature $T\simeq 250\mathrm{K}$, and spans a meridional width $L$. Integrating the approximate thermal wind relation in scale form givesThe northward temperature gradient is negative because the eastward wind strengthens upward. A $1000$--$2000\mathrm{km}$ jet width gives roughly $25$--$50\mathrm{K}$. No unique numerical difference follows without specifying the jet width, height and representative temperature; these are explicitly estimated quantities.

For a stationary solid skeleton, the Darcy velocity $\mathbf{u}$ is liquid volume transport per unit total cross-sectional area, while the pore velocity $\mathbf{v}$ is the intrinsic mean fluid velocity within the connected pore space. ThusFor an incompressible liquid in a saturated region, conservation of mass is ${\varphi }_t+\nabla \cdot \mathbf{u}=0$. With time-independent but spatially varying porosity, it becomesnot generally $\nabla \cdot \mathbf{v}=0$. Neglecting dispersion, a fluid interface is advected by the pore velocity; its mean normal speed is $V_n=\mathbf{v}\cdot \mathbf{n}=\mathbf{u}\cdot \mathbf{n}/\varphi$.

For the thin layer on the inclined plane, the gravitational acceleration components are $g\sin \theta$ downslope and $-g\cos \theta$ normal to the plane. The hydrostatic approximation and constant pressure $p_a$ above the current giveWith permeability of a porous medium $k(z)$, Darcy's law therefore givesHere $u_x,u_y$ denote the components of the Darcy velocity. Integrate over depth, using the prescribed power laws:Both integrals are finite for the positive exponents in the model. Conservation of the stored fluid gives ${\mathcal{V}}_t+{\nabla }_{x,y}\cdot \mathbf{Q}=0$, orFor $0<\theta <\pi /2$, set $X=x\tan \theta$, $Y=y\tan \theta$ andSince $\cos \theta {\tan }^2\theta =\sin \theta \tan \theta$, the depth-dependent inclined porous-current equation isThe longitudinal flux carries fluid downslope, while the thickness-gradient terms spread it. The degenerate horizontal or vertical slope limits need their own coordinate scaling.

Let $S(z)=d{\rho }_s/dz$, $N^2=-gS/{\rho }_0>0$, and define the quasi-geostrophic streamfunction by $\psi =\tilde{p}/({\rho }_0{f}_0)$. Use horizontal scale $L$, vertical scale $D$, velocity scale $U$ and advective time $L/U$. The required quasi-geostrophic approximation has small Rossby number $\epsilon =U/(|f_0|L)$, $\beta L/|f_0|=O(\epsilon )$, and Burger number $N^2{D}^2/(f_0^2{L}^2)=O(1)$. Hydrostatic, thin-layer scaling has $D/L\ll 1$; density variations are small compared with ${\rho }_0$, as required by the Boussinesq approximation. The leading horizontal flow is nondivergent, with vertical velocity $w=O(\epsilon UD/L)$. Retain forcing of the same slow order as density advection, $r=O(U\tilde{\rho }/L)$; stronger forcing need not preserve this balanced scaling.

Leading geostrophic balance and the hydrostatic approximation giveWrite $D_g/Dt={\partial }_t-{\psi }_y{\partial }_x+{\psi }_x{\partial }_y$. The density equation becomesTaking the curl of the horizontal momentum equations at the next order and using continuity gives the stretching equationSubstitute the expression for $w$. Vertical differentiation commutes with the geostrophic material derivative acting on ${\psi }_z$ here: the extra Jacobian is $J({\psi }_z,{\psi }_z)=0$. Since $N$ depends only on height, this yields the diabatically forced quasi-geostrophic potential vorticity equationThe minus sign is appropriate to the specified density forcing; a positive upward buoyancy forcing is $R=-gr/{\rho }_0$ and gives $D_{g}q/Dt=f_0{\partial }_z(R/N^2)$.

Now take constant $N$ and linearize about rest. Put $s=f_0^2/N^2$ and $S_0=g{f}_0\mu r_0/({\rho }_0{N}^2)$. The forced equation isFor the harmonic heating response of quasi-geostrophic flow, represent the two traveling Fourier components with frequencies ${\omega }_{+}={\omega }_0$ and ${\omega }_{-}=-{\omega }_0$:Each component of the forcing has amplitude $S_0/2$, soEquivalently its vertical structure obeysThese are the time-periodic forced responses; unspecified initial data may add free Rossby waves. Selecting no incoming waves describes forcing switched on causally and then its persistent harmonic part.

For $\beta >0$, the eastward component always has ${\lambda }_{+}^2>0$. Let ${\lambda }_{+}>0$. The growing exponential would introduce an unbounded velocity and pressure disturbance at infinity, so discard it. The bottom condition ${\varphi }_{+}(0)=0$ then givesIf $\mu ={\lambda }_{+}$, this expression has the regular limit $-C_{+}z{e}^{-\mu z}/(2\mu )$, rather than a physical singularity.

For ${\omega }_0>\beta /k$, the westward component also has positive ${\lambda }_{-}^2$. Choose ${\lambda }_{-}>0$ and use the same decaying expression with minus subscripts, including the same coincident-exponent limit. Thus both components are vertically evanescent above the heating.

For $0<{\omega }_0<\beta /k$, putBoth homogeneous solutions $e^{\pm imz}$ are bounded, so boundedness is insufficient. A free Rossby wave with phase $kx+mz-\omega t$ hasThe radiation condition demands energy propagation away from the heating, upward into $z>0$. For $\beta ,k>0$ that selects positive $m$, hence $e^{imz}$ in the westward component. The resulting solution isIt satisfies ${\varphi }_{-}(0)=0$ and, away from the decaying forcing, ${\varphi }_{-}^{\prime }\sim im{\varphi }_{-}$. Its far-field amplitude does not decay in this nondissipative model: continuous forcing supplies an outgoing wave train. Thus the low-frequency westward component propagates vertically while the eastward component remains evanescent.

At the threshold ${\omega }_0=\beta /k$, the westward vertical wavenumber is zero. Excluding the linearly growing homogeneous solution and retaining the bottom condition gives ${\varphi }_{-}=C_{-}(e^{-\mu z}-1)/{\mu }^2$. This bounded limiting response tends to a constant and has zero vertical group velocity; demanding both bottom zero and decay to zero at infinity would overdetermine this threshold problem. For $\beta \le 0$, the branch labels and propagation inequality must be reconsidered; the comparison stated here assumes the usual positive beta-plane gradient.

Here write $\sigma$ for the buoyancy perturbation about the resting constant-$N$ stratification; the total buoyancy also includes its background gradient $N^2$. Use kinematic perturbation pressure $p$. The weak-forcing Linearized Boussinesq equations areWeakness means that perturbation advection is small compared with the retained linear terms. With the specified streamfunction convention, the transverse vorticity isDifferentiate the horizontal momentum equation in $z$ and subtract the $x$ derivative of vertical momentum. The pressure derivatives cancel, givingThe second equation uses ${\sigma }_t=N^2{\psi }_x$ and is the forced internal-wave equation.

For free waves at frequency $|\omega |\ll N$, the internal gravity wave dispersion relation gives $|k|/|m|\simeq |\omega |/N\ll 1$. Thus the horizontal response scale is much larger than the vertical scale. Vertical acceleration is small in the vertical momentum equation, leaving the hydrostatic approximation $p_z=\sigma$. Correspondingly the transverse vorticity is predominantly $u_z$: neglect $-w_x={\psi }_{xx}$ relative to ${\psi }_{zz}$ in the inertial vorticity term. The restoring term $N^2{\psi }_{xx}$ must still be retained, since it balances the small time derivatives.

For $m\ne 0$ put $\psi =e^{imz}P(x,t)$, $\sigma =e^{imz}B(x,t)$ and interpret physical fields as real parts. The reduced equation isOne can solve it while obtaining the buoyancy simultaneously. Define $U=imP$ and $H=iB/(cm)$. The first-order hydrostatic equations becomeHence $R_{+}=U+H$ and $R_{-}=U-H$ obeyApply the method of characteristics with zero fields in the past. Withthe solution is $R_{\pm }=I_{\pm }$. Consequently the hydrostatic response to a localized horizontal force isThese are exactly the required two characteristic integrals, derived with their relative sign fixed by the buoyancy equation. Quiet-past initial conditions exclude additional freely propagating solutions.

For the consistency check, set $\tau =s-t$, so both integrals run over the fixed range $(-\infty ,0)$. Differentiation at fixed $x,z$ now givesThis change of variable is important: it includes the time dependence of both the sampled forcing and the moving integration geometry without omitting a term. Each straight path intersects $|x|\le a$ for a time interval of length at most $2a/c$. If $L=\sup |f_{xt}|$, thenThe characteristic size of the hydrostatic buoyancy and pressure-gradient balance is $|m|a{f}_{\max }$, as follows from the analogous bound on $B$ in its integral formula. Thus the omitted acceleration has relative scale at most a constant timesTogether with the low-frequency assumption, this verifies the requested hydrostatic consistency. It is a comparison of characteristic magnitudes, rather than division by a buoyancy value at a node where both terms may vanish. The capital $N$ here is the buoyancy frequency, as printed in the PDF.

For $t\ge 0$, only $0\le s\le t$ contributes. If $|x|>ct+a$, every point $x\mp c(s-t)$ lies outside the source interval, so both integrals vanish:This finite propagation cone belongs to the reduced hydrostatic wave equation.

The hypotheses do not require an even forcing, so no single symmetric graph follows for every permitted $f$. For an illustrative smooth even nonnegative profile, $U$ is even and $H$ is odd. At late time and fixed $x$, if $A_f={\int }_{-a}^{a}f(r,\infty )dr$, thenThus the streamfunction develops a broad nearly level interior response with outgoing fronts, while the buoyancy changes sign across the source and approaches opposite interior plateaus to its left and right. Both vanish beyond the fronts. For $m>0$, a vertical phase with $\sin (mz)=1$ gives real fields $\psi =U/m$ and $\sigma =cmH$. The figure plots those profiles up to positive normalization factors, for an even compact forcing increasing slowly from zero.

Define the quasi-geostrophic streamfunction by $p^{\prime }/{\rho }_0=f_0\psi$. Leading horizontal geostrophic balance, $f_0\hat{\mathbf{z}}\times {\mathbf{u}}_g=-{\nabla }_{H}P$, gives $(u_g,v_g)=(-{\psi }_y,{\psi }_x)$. The hydrostatic approximation gives $P_z=\sigma$, henceThus $D_g$ is the material derivative along the leading geostrophic flow, and the vertical derivative of pressure determines the buoyancy perturbation. This also gives thermal wind: $u_{gz}=-{\sigma }_y/f_0$ and $v_{gz}={\sigma }_x/f_0$.

The balanced regime is favored by small Rossby number, a shallow aspect ratio, and a Burger number $N^2{H}^2/(f_0^2{L}^2)$ of order one. Equivalently the stratified Froude number $U/(NH)=\mathrm{Ro}/\sqrt{\mathrm{Bu}}$ is small. On a beta plane, require $\beta L/|f_0|=O(\mathrm{Ro})$, and take slow, initially balanced motion away from the equator. Under the traditional approximation, the horizontal Coriolis force generated by vertical velocity is small because vertical velocity is small, while the vertical rotational force is small compared with the leading hydrostatic pressure and gravity terms. Strong stratification and small aspect ratio justify retaining only the vertical component of Earth's rotation in this regime; this argument is not uniform near the equator.

Put $A(z)=f_0^2/N^2(z)$. The buoyancy equation gives $w=-f_0{D}_g{\psi }_z/N^2$. Substitution into the vertical vorticity equation yieldsThere is an important commutator cancellation. With $J_{xy}(a,b)=a_x{b}_y-a_y{b}_x$,Also $D_{g}f=\beta {\psi }_x$ for $f=f_0+\beta y$. Adding terms therefore proves potential-vorticity conservation:This is the three-dimensional quasi-geostrophic potential vorticity.

At an impermeable stationary lower boundary $z=b(x,y)$, the kinematic condition is $w={\mathbf{u}}_H\cdot {\nabla }_{H}b=D_{g}b$. For small elevation and slope, evaluate this at the reference level $z=0$ to leading order. Combining it with the buoyancy equation gives the rigid-boundary buoyancy condition for quasi-geostrophic wavesThe quantity in parentheses is the boundary displacement plus its buoyancy-equivalent displacement.

Now take constant $N,f$ and the mean quasi-geostrophic streamfunction $\bar{\psi }=-(U_0+\Lambda z)y$. Its mean buoyancy perturbation is $-f\Lambda y$, consistent with thermal wind, and its interior potential vorticity is constant. In the boundary condition, the perturbation velocity $v^{\prime }={\psi }_x^{\prime }$ advects the mean boundary buoyancy gradient. Keeping this contribution givesIn particular the $-\Lambda {\psi }_x^{\prime }$ term must not be omitted.

The linear interior potential-vorticity equation is $({\partial }_t+U(z){\partial }_x)Q^{\prime }=0$. Starting with zero interior perturbation potential vorticity, and with forcing only through the boundary, it remains zero. For a steady solution this is also the regular zero-interior-PV choice, continued through any isolated zero of $U$. The inversion equation for a horizontal mode isThe printed convention $K>0$ takes $f>0$; for either hemisphere the decay rate is defined using $|f|$. Boundedness aloft selects the decaying mode. Writewith real part understood. The linear boundary condition becomesFor $K{U}_0+\Lambda \ne 0$, a steady quasi-geostrophic mountain wave is thereforeOn $y=0$, its real meridional velocity isFor the positive-parameter convention $k,ϵ,f,U_0>0$ and $K{U}_0+\Lambda >0$, the velocity minimum lies one quarter wavelength downstream of each topographic maximum; its maximum lies one quarter wavelength downstream of each topographic minimum. If the velocity prefactor changes sign, maxima and minima interchange. The condition $K{U}_0+\Lambda >0$ alone does not fix that prefactor if $U_0$ is allowed to be negative. The upper panel shows the positive-current convention.

At $K{U}_0+\Lambda =0$, the free boundary Eady edge wave has zero frequency in the ground frame. If $kϵU_0\ne 0$, a steady amplitude cannot satisfy the forced amplitude equation. With $A(0)=0$, integration givesConsequentlyThe symbol multiplying $i$ here is the horizontal wavenumber $k$, as in the PDF. Taking the real part and differentiating in $x$ givesFor $U_0=-\Lambda /K>0$, positive $ϵ,f$, and $t>0$, the meridional-velocity minima coincide with topographic maxima, and its maxima with topographic minima. This is the lower panel. The change from quadrature to opposition is the phase of a resonantly forced topographic edge wave.

The stationary relief continuously excites a boundary Rossby wave whose intrinsic propagation exactly cancels mean advection. The forcing therefore remains coherent with that wave; with no damping or radiation loss for this trapped mode, its linear amplitude grows secularly. This is a resonant topographic quasi-geostrophic wave. Unlimited growth is a conclusion of the linear idealization: eventually the small-disturbance approximation fails.

The assertion that resonance must be unsteady needs nonzero forcing. If $k=0$, then $b_x=0$ and the boundary current does not cross the relief; if $U_0=0$, the forcing also vanishes. In either case the zero disturbance is a steady solution even if $K{U}_0+\Lambda =0$. For example, choose $k=0$, $\ell \ne 0$, and $\Lambda =-K{U}_0$; this is a concrete exception to an unconditional resonance claim. When both horizontal wavenumbers vanish, $K=0$ and the stated decaying-wave ansatz itself does not apply.

The geostrophic material derivative isIt advects horizontally with the geostrophic flow. The exact material derivative additionally includes horizontal ageostrophic flow and vertical advection $w{\partial }_z$. The leading balances are geostrophic balance and the hydrostatic approximation:The conventional quasi-geostrophic streamfunction is $p=f_0\psi$, after fixing the horizontally uniform reference pressure consistently with the basic buoyancy profile. ThereforeHere and below pressure is divided by reference density, and $\sigma$ is the anomaly about the prescribed background buoyancy.

For the zonal average assume periodicity or vanishing boundary terms, so that averages of $x$ derivatives vanish and averaging commutes with $y,z$ derivatives. Put $S(z)=f_0^2/N^2(z)$. Since $\overline{{\psi }_x^{\prime }}=0$, the disturbance quasi-geostrophic potential vorticity is $Q^{\prime }={\psi }_{xx}^{\prime }+{\psi }_{yy}^{\prime }+{\partial }_z(S{\psi }_z^{\prime })$. The zonal part of $\overline{{\psi }_x^{\prime }{Q}^{\prime }}$ vanishes as $\frac{1}{2}\overline{{\partial }_x({\psi }_x^{\prime }{)}^2}$. For the other parts, integration by parts givesAs $u^{\prime }=-{\psi }_y^{\prime }$, $v^{\prime }={\psi }_x^{\prime }$, and ${\sigma }^{\prime }=f_0{\psi }_z^{\prime }$, the Eliassen–Palm flux isWe have therefore proved the exact Taylor identity for quasi-geostrophic flux,without imposing small amplitude. This identity is algebraic within the quasi-geostrophic approximation; it does not assert validity of that approximation for arbitrary physical flows.

Now linearize about a steady zonal basic flow $\bar{u}=U(y,z)$, with ${\bar{Q}}_y\ne 0$ and no mean meridional velocity. The linear quasi-geostrophic potential-vorticity equation is $Q_t^{\prime }+U{Q}_x^{\prime }+v^{\prime }{\bar{Q}}_y=0$. Multiply by $Q^{\prime }/{\bar{Q}}_y$ and average. The advection term is a zonal derivative and vanishes, givingThis is quasi-geostrophic wave-activity conservation law. The basic-state denominator is time-independent at the retained order; its change caused by the waves is higher order. The wave activity is signed if ${\bar{Q}}_y<0$, and the formula cannot be used where that gradient vanishes. The original PDF includes the zonal average over the squared disturbance in this definition.

For the resonant solution it is helpful to avoid squaring an imaginary coefficient as though it were real. Writewhere $B_0$ is real. Assume $f_0\ne 0$, $\beta \ne 0$, and $k\ne 0$. Define $c(z)=\cos (mz)$ and $r(z)=(z-H)\sin (mz)$. The complex coefficient of the physical streamfunction disturbance isThe original PDF contains $Bt\cos (mz)$: the converted TeX's missing $t$ would incorrectly remove the resonance and fails the differential equations.

For the vertical potential-vorticity inversion operator, $c^{\prime \prime }=-m^{2}c$ and $r^{\prime \prime }=-m^{2}r+2mc$. ConsequentlySubstitution in the linear potential-vorticity equation givessince ${\kappa }^2{B}_0=2k{f}_{0}U$. Thus this is a genuine resonant topographic quasi-geostrophic wave, with linear amplitude growth rather than an assumed stationary wave.

At the flat upper boundary $h_z(H,t)=0$, so $w^{\prime }(H)=0$. At the lower boundary, linearization of the impermeability condition gives $w^{\prime }(0)=U{\partial }_x{h}_b$, where $h_b=\mathrm{Re}\{\epsilon H{e}^{ikx}\sin (\ell y)\}$ is the bottom corrugation. Here $h_z(0,t)=-CHm$ is time-independent, so the specified vertical velocity givesusing $f_{0}Cm/N^2=1$. Both boundary conditions are verified. The linear solution is valid while its growing displacement and velocity remain small; it is not a large-time nonlinear solution.

To evaluate all fluxes, use $\overline{\mathrm{Re}(a{e}^{ikx})\mathrm{Re}(b{e}^{ikx})}=\frac{1}{2}\mathrm{Re}(a{b}^{\ast })$. The coefficients of $u^{\prime },v^{\prime },{\sigma }^{\prime }$ are respectively $-\epsilon \ell \cos (\ell y)h$, $i\epsilon k\sin (\ell y)h$, and $\epsilon f_0\sin (\ell y)h_z$. HenceThe products of the $B_{0}t$ terms alone and the $C$ terms alone are imaginary in the vertical flux and contribute nothing. Only the cross terms remain. Sincewe obtainDifferentiating $R$ gives $R^{\prime }=m(1+\cos (2mz))=2m{\cos }^2(mz)$. As $SmC=f_0$,For an independent evaluation of the right side of the Taylor identity for quasi-geostrophic flux, write $q=(2f_{0}c-{\kappa }^{2}Cr)-i{\kappa }^2{B}_{0}tc$. Multiplication of $ikh$ by $q^{\ast }$ gives real part $-2k{f}_0{B}_{0}t{c}^2$, because the remaining real cross terms cancel. Thereforewhich verifies the flux identity pointwise in $y,z$.