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Put $b=\delta B_{\parallel }/B_0$ and ${\mathbf{b}}_{\perp }=\delta {\mathbf{B}}_{\perp }/B_0=\hat{\mathbf{z}}\times {\nabla }_{\perp }\Psi /v_A$. Define the two-dimensional Poisson bracket byThen, to the retained anisotropic order, the derivative along the perturbed magnetic field isThe omitted $b{\partial }_z$ term is smaller by $ϵ$ than either retained term. The nonlinear field-line-bending term is of the same order as ${\partial }_z$, which is why it must remain.

Use ${\mathbf{u}}_e=-\alpha \nabla \times \mathbf{B}$ with $\alpha B_0=v_A{d}_i$. The leading parallel and perpendicular Electron velocities areIn detail, $\nabla \times (B_{0}b\hat{\mathbf{z}})=-B_0\hat{\mathbf{z}}\times {\nabla }_{\perp }b$, while the parallel curl of $B_0\hat{\mathbf{z}}\times {\nabla }_{\perp }\Psi /v_A$ is $B_0{\nabla }_{\perp }^2\Psi /v_A$. The perpendicular curl involving ${\partial }_z\delta {\mathbf{B}}_{\perp }$ is order-smaller. Exact Electron incompressibility follows from the divergence of a curl; its small velocity correction supplies any divergence omitted by this leading representation.

The electron magnetohydrodynamics equation is frozen-in induction,because both $\mathbf{B}$ and ${\mathbf{u}}_e$ are solenoidal. Keeping its leading order-${ϵ}^2$ perpendicular terms givesLet $X_f=\hat{\mathbf{z}}\times {\nabla }_{\perp }f=(-f_y,f_x)$. Expanding the two components gives $(X_f\cdot \nabla )X_g-(X_g\cdot \nabla )X_f=X_{\{f,g\}}$. Substitution therefore reduces the perpendicular equation toThe perpendicular-constant freedom in $\Psi$ has no magnetic effect and can be fixed so that

The parallel induction component at the same order isThe advection term vanishes since ${\mathbf{u}}_{e\perp }\cdot {\nabla }_{\perp }b=v_A{d}_i\{b,b\}=0$. ThusTogether these are both required reduced electron magnetohydrodynamics equations. Terms such as $u_{e\parallel }{\partial }_{z}b$ and $b{\partial }_z{\mathbf{u}}_e$ are order-${ϵ}^3$ and consistently absent.

Use the Goldreich–Sridhar turbulence model in its balanced form. Assume a uniform strong guide field; anisotropic fluctuations $k_{\parallel }\ll k_{\perp }$; comparable counterpropagating Alfvén wave amplitudes; a statistically steady inertial range with negligible forcing and dissipation within it; transfer mainly between comparable perpendicular scales; and a scale-independent energy flux per unit mass $\mathcal{E}$. The estimate also assumes no extra scale-dependent alignment factor in the nonlinear interaction and neglects intermittency corrections.

Let ${\ell }_{\perp }$ and ${\ell }_{\parallel }$ be an eddy's perpendicular and parallel scales, and $\delta u_{\ell }\sim \delta B_{\ell }/\sqrt{4\pi {\rho }_0}$ its Alfvénic velocity amplitude. The nonlinear interaction is supplied by the oppositely propagating Elsässer variable, so comparable populations giveThe Alfvén propagation time is ${\tau }_A\sim {\ell }_{\parallel }/v_A$. For strong turbulence, critical balance makes these times comparable; the decorrelation and cascade time is therefore of order ${\tau }_{\mathrm{nl}}$, not a much longer time obtained by adding many weak encounters.

A constant perpendicular energy flux givesDefine the one-dimensional perpendicular turbulent energy spectrum so that $\int E(k_{\perp })d{k}_{\perp }$ measures fluctuation energy per unit mass. Locality of the spectrum then givesHencewhere $C$ is a dimensionless normalization constant not determined by the scaling argument. The magnetic and kinetic spectra have the same leading scaling under the assumed Alfvénic amplitude relation.

Critical balance also determines the scale-dependent anisotropy:Thus the eddies become relatively more elongated along the guide field at smaller perpendicular scales. The $-5/3$ exponent follows under the listed cascade assumptions; it is not established for every imbalanced or aligned MHD state solely by invoking critical balance.

In the ideal, leading anisotropic system, there are five separately conserved quadratic cascade quantities: the two oppositely propagating Alfvén populations, the two slow-wave populations, and the entropy fluctuation. With the amplitudes of part (b), one convenient normalization iswhere ${\mathbf{z}}_{\perp }^{\pm }={\mathbf{u}}_{\perp }\pm \delta {\mathbf{B}}_{\perp }/\sqrt{4\pi {\rho }_0}$. Constant background factors convert the velocity variances into modal energies and the entropy variance into an entropy-related free-energy contribution. These are the five quadratic cascade invariants of anisotropic magnetohydrodynamic turbulence.

To verify conservation rather than just count polarizations, the perpendicular Elsässer variables obeyDot each equation with $2{\mathbf{z}}_{\perp }^{\pm }$ and integrate. The propagation and nonlinear terms become surface fluxes of $|{\mathbf{z}}_{\perp }^{\pm }{|}^2$, and the pressure term becomes a surface flux because the corresponding field is solenoidal. For periodic boundaries, or boundaries with vanishing invariant flux, all these terms vanish and $d{I}_A^{\pm }/dt=0$. Their sum is proportional to perpendicular kinetic plus magnetic energy, while their difference is proportional to perpendicular cross-helicity; the two Elsässer energy invariants contain both pieces of information.

Likewise the slow-mode equations giveHere ${\mathbf{u}}_{\perp }$ is divergence-free and ${\nabla }_{\parallel }$ differentiates along the divergence-free vector $\hat{\mathbf{z}}+\delta {\mathbf{B}}_{\perp }/B_0$, so both integrals are boundary fluxes. Finally $Ds=0$ gives $d{I}_e/dt=-\int {\mathbf{u}}_{\perp }\cdot {\nabla }_{\perp }{s}^{2}dV=0$. This proves all five quadratic conservation laws under the same boundary assumptions.

The two Alfvén cascades interact dynamically: each population strains the other and is needed for its nonlinear transfer between scales. Their integrated invariants are nevertheless conserved separately, so this interaction does not transfer one population's invariant into the other. The slow and entropy cascades are driven by the Alfvénic mixing and field-line motion, with no feedback on the perpendicular cascade at the retained order. Their injection rates set independent fluxes; those fluxes need not be equal, even when the resulting perpendicular spectra share the $-5/3$ scaling.

The count is a count of independent quadratic modal cascade channels. Ideal passive advection also preserves higher moments, such as $\int F(s)dV$ for arbitrary smooth $F$, and analogous integrals for the slow characteristic scalars. These are not additional quadratic modal energies. Forcing and dissipative terms make the corresponding invariants source-and-sink balances; a stationary inertial range conserves their flux through scales rather than their total content in a continually forced domain.

For $\omega >N$, the dispersion relation gives $k^2=-m^2{\omega }^2/({\omega }^2-N^2)$. Put $\kappa =m\omega /\sqrt{{\omega }^2-N^2}>0$. Boundedness as $x\to +\infty$ selects $k=i\kappa$, rather than the exponentially growing root. The internal-wave polarization relations now yieldAgain take real parts and set $v=0$. This is an evanescent wave, decaying over distance ${\kappa }^{-1}$; there is no propagating far-field internal gravity wave at this frequency. The pressure amplitude $P=-i\omega u/\kappa$ is in quadrature with the horizontal velocity, so the time-averaged horizontal pressure-work flux vanishes. The equality $\omega =N$ is a singular limiting case and is outside the two frequency ranges considered here.

In the Boussinesq approximation, density departures from a constant reference density ${\rho }_0$ are small enough to neglect in inertia and mass continuity, but their gravitational effect is retained. Write the total buoyancy acceleration as $b=-g(\rho -{\rho }_0)/{\rho }_0$, choose a hydrostatic background $\bar{b}(z)$, and set $\sigma =b-\bar{b}$. Thus $\sigma$ is the buoyancy perturbation, positive for a parcel lighter than the background at its present height. The squared buoyancy frequency isStable stratification has $N^2>0$. A small vertical displacement $\xi$ of a parcel conserving its buoyancy gives $\sigma =-N^2\xi$, explaining the restoring acceleration and the name of $N$.

Let $P=p^{\prime }/{\rho }_0$ be pressure with the background hydrostatic pressure removed, $\mathbf{u}=(u,v,w)$, and $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ the material derivative. The ideal, nonrotating Boussinesq equations areThe last equation is equivalent to $Db/Dt=0$. There is neither viscosity nor buoyancy diffusion. The reference density occurs in the pressure normalization, while the small density anomaly occurs only through buoyancy.

For the two-dimensional disturbance, linearization about rest givesA plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$ therefore satisfiesThese internal-wave polarization relations and the dispersion relation determine the individual responses in parts i and ii. A radiating response has no incoming internal gravity wave from $x=+\infty$; an evanescent response is bounded there. The harmonic solutions describe the response after excluding arbitrary initial transients.

For the localized oscillatory boundary, define $\alpha =\omega /\sqrt{N^2-{\omega }^2}$ and $I(z)=ϵ{\int }_0^{\infty }{\hat{u}}_0(m)e^{imz}dm=f(z)+ig(z)$. Both signs of vertical wavenumber must radiate into $x>0$. Their horizontal wavenumber is $k=\alpha |m|$, so the positive-$m$ contribution involves $Z_1=z+\alpha x$ and the negative-$m$ contribution involves $Z_2=z-\alpha x$. Consequently the complex horizontal velocity isTaking its real part provesThe positive-$m$ internal-wave polarization has $w=-\alpha u$ and the negative-$m$ one has $w=+\alpha u$. HenceIn particular, $2f(z)=ϵu_0(z)$, so the boundary condition is recovered exactly. This is two-beam radiation from a localized oscillating boundary.

Take a smooth localized profile, for example $u_0(z)=e^{-z^2/H^2}$. Its effective support is a neighbourhood of width a few $H$; increasing that neighbourhood makes the tails arbitrarily small. The boundary displacement is $ϵu_0(z)\sin \omega t/\omega$ to first order, so at $t=0$ it is flat and moving at maximum speed. At that instant,There are two narrow oblique beams centered on $z=-\alpha x$ and $z=+\alpha x$. Away from their overlap, the velocity is parallel to the relevant beam, in the directions $(1,-\alpha )$ and $(1,+\alpha )$. The figure uses $\alpha =3/4$ and measures velocities in units of $ϵ$.

At other phases the $g$ terms contribute. For the positive-frequency convention above, $g$ is the Hilbert transform of $f$; it generally has longer tails than $f$. Thus even when the boundary is nearly stationary, the interior need not be stationary or confined to the bright cores shown here. Half a period later all velocities reverse. The beam directions remain the same, but the instantaneous velocity profile and its quadrature component change.

A short pulse has a broad frequency spectrum. Frequencies below $N$ produce outgoing internal gravity wave packets with frequency-dependent ray slopes $\pm \omega /\sqrt{N^2-{\omega }^2}$; frequencies above $N$ contribute an evanescent near-boundary response. Moreover the group velocity depends on wavelength as well as direction. The radiated packet therefore spreads into a dispersive fan or wake, with different frequencies traveling at different angles and different scales separating along those directions. After the forcing ends the propagating packet travels away, rather than maintaining the two permanently driven monochromatic beams.

Use the streamfunction convention $u={\psi }_z$, $w=-{\psi }_x$, and take $v=0$. With $J(A,B)=A_x{B}_z-A_z{B}_x$, the material derivative is $D={\partial }_t-J(\psi ,\cdot )$. This convention differs from the horizontal quasi-geostrophic streamfunction convention used below. The nonlinear Boussinesq equations areThe $y$ component of vorticity is $\zeta =u_z-w_x={\nabla }^2\psi$. Differentiate the first momentum equation in $z$ and subtract the $x$ derivative of the second. The pressure terms cancel; the derivatives of the advecting velocity cancel by $u_x+w_z=0$. Thus $D\zeta =-{\sigma }_x$, orConservation of total buoyancy, with background gradient $N^2(z)$, givesBoth equations retain the nonlinear advection terms.

Write $\psi =\bar{\psi }(z)+\varphi$, with ${\bar{\psi }}_z=U(z)=\bar{u}(z)$, and take the background buoyancy anomaly to be zero. The linear equations for $(\varphi ,\sigma )$ areFor a normal mode $\varphi =\hat{\psi }(z)e^{ik(x-ct)}$, the second equation gives $\hat{\sigma }=N^2\hat{\psi }/(U-c)$. Substitute this into the first to obtain the Taylor–Goldstein equationHere $k\ne 0$ and the calculation is initially away from critical levels $U=c$. The first term is the restoring effect of stable buoyancy on vertically displaced parcels. The second comes from advection of the background vorticity gradient $U^{\prime \prime }$: a displacement creates a vorticity anomaly, whose induced velocity acts back on the displacement. This is the shear-flow counterpart of a Rossby wave restoring mechanism; its sign can also permit instability, so it is not invariably stabilizing.

Rigid boundaries imply $w=-{\varphi }_x=0$, hence $\hat{\psi }(0)=\hat{\psi }(\pi H)=0$. For a real, regular normal mode, multiply the Taylor–Goldstein equation by $\hat{\psi }$, integrate, and integrate its second derivative by parts. The endpoint term vanishes, leavingFor a smooth real dispersion branch, differentiate this identity with respect to $k$. The stated stationarity of $I$ removes the term involving $d\hat{\psi }/dk$, so $2k=I_{c}dc/dk$. Direct differentiation at fixed $\hat{\psi }$ givesSince the frequency is $kc$, the group velocity is $d(kc)/dk$, and thereforeThis is variational group velocity of a Taylor-Goldstein mode. Singular critical levels or a vanishing denominator require separate limiting analysis; no stationarity proof is needed here.

For the sinusoidal profiles, choose $c=0$. In the interior,The profile ratios extend to the endpoints in this reduced equation. Thus $n=1$ always gives a nonzero real $k$, with $k_1^2=N_0^2/U_0^2$. For this normal mode,The denominator in the group velocity expression is $2H(2N_0^2/U_0^3+1/(U_0{H}^2))$. It follows thatAlthough $U$ vanishes at the endpoints, this weighted integral is finite; the vanishing of ${\hat{\psi }}_1$ supplies the required cancellation. The formula is the limiting derivative approached from the regular real branch with $c<0$, before any interior critical level is encountered.

If instead $N=N_0$ and $U=U_0$ are constant, still take $c=0$. The same vertical eigenfunctions satisfy the Taylor–Goldstein equation, now withThere is no value of $n\ge 1$ guaranteed to give real $k$ for arbitrary positive parameters: if $N_{0}H<U_0$, all these squared wavenumbers are negative. A nonzero real $k$ exists exactly when $N_{0}H>U_0$; equality gives only the degenerate $n=1$, $k=0$ limit. For a real branch the same group velocity calculation gives $c_g=k_n^2{U}_0^3/N_0^2$.

To test exact nonlinear validity, take a real normal mode $\varphi =A\sin (nz/H)\cos (kx)$ and its corresponding buoyancy perturbation $\sigma =a(z)\varphi$, where $a=N^2/(U-c)$. Becausethe quadratic vorticity self-advection is zero: $J(\varphi ,{\nabla }^2\varphi )=0$. The quadratic buoyancy self-advection, however, isThis nonlinear exactness test for a stratified streamfunction mode distinguishes the cases. For constant $N,U$, $a=N_0^2/U_0$ is constant, so both quadratic terms vanish. The background-mode terms already satisfy the linear equations, hence the constant-profile modes are exact finite-amplitude solutions of the ideal Boussinesq equations. Pressure can be recovered from the momentum equations: the vorticity equation supplies their compatibility condition in this simply connected strip.

For the sinusoidal profiles, $a=(N_0^2/U_0)\sin (z/H)$ has a nonzero derivative in the interior. For every nonzero $A$ and nonzero real $k$, $a^{\prime }\varphi {\varphi }_x$ is not identically zero; it produces a second horizontal harmonic that is absent from the proposed solution. Thus none of these nontrivial sinusoidal-profile modes is an exact finite-amplitude solution. A $k=0$ field is merely an added horizontal shear with no vertical motion and is a degenerate exception to the wave calculation, not a propagating member of it.

For the shallow-water approximation, let the vertical scale be much smaller than the horizontal scale, so vertical acceleration is negligible and pressure is hydrostatic. Assume constant density, a flat impermeable bottom, a free surface under constant atmospheric pressure, and a horizontally moving column with no leading-order vertical shear. Neglect friction. Taking the bottom at $z=0$, hydrostatic pressure is $p=p_a+\rho g(h-z)$, so its horizontal gradient is independent of $z$. The horizontal momentum equations are thereforeIntegrating incompressibility from the bottom to the free surface, using zero bottom vertical velocity and the surface kinematic condition $w=h_t+u{h}_x+v{h}_y$, givesTogether these are the nonrotating shallow water equations.

Rotation with angular velocity $f\hat{\mathbf{z}}/2$ adds Coriolis force terms $-fv$ and $+fu$ to the left sides of the $u$ and $v$ equations, respectively. The centrifugal potential is absorbed into the background pressure and effective gravity. ThusWrite $\zeta =v_x-u_y$ and $d=u_x+v_y$. Taking the horizontal curl, including the derivatives of the advecting velocity, yieldsFor example the curl of the Coriolis force contribution is $fd$, and the curl of nonlinear advection supplies $\zeta d$. Since $q_a=f+\zeta$ and $D_{H}h=-hd$, division givesThis is exact shallow-water potential vorticity conservation within the rotating shallow water equations.

For the quasi-geostrophic approximation, take a horizontal scale $L$, speed $U$, mean depth $H$, and slow time $L/U$. Require small Rossby number $\mathrm{Ro}=U/(|f|L)\ll 1$ and small surface departure $\eta =h-H$. Introduce the Rossby deformation radius $L_R=\sqrt{gH}/|f|$ and Burger number $\mathrm{Bu}=L_R^2/L^2=O(1)$. Geostrophic balance gives $g\eta /L\sim |f|U$, hence $\eta /H\sim \mathrm{Ro}/\mathrm{Bu}\ll 1$. The Froude number is $U/\sqrt{gH}=\mathrm{Ro}/\sqrt{\mathrm{Bu}}\ll 1$. These assumptions select slow balanced motion, with fast gravity-wave transients absent at leading order.

Define the quasi-geostrophic streamfunction by $\psi =g\eta /f$. The leading geostrophic flow is $(u_g,v_g)=(-{\psi }_y,{\psi }_x)$, which is horizontally nondivergent. Expanding the exact shallow-water potential vorticity givesThe correction is smaller than $f/H$ by $O(\mathrm{Ro})$. Advecting it by the leading geostrophic flow, while the constant leading term has zero gradient, givesHere potential-vorticity conservation means that $Q$ is carried unchanged by the leading horizontal trajectories. Potential-vorticity inversion means solving the elliptic equation $({\nabla }_H^2-L_R^{-2})\psi =Q-f$, with appropriate boundary or decay conditions, to recover $\psi$ and thus the velocity. Conservation evolves the scalar field; inversion obtains its velocity field.

For a mean current $U(y)=-{\bar{\psi }}_y$, substitute $Q=\bar{Q}+Q^{\prime }$ and $\psi =\bar{\psi }+{\psi }^{\prime }$ into potential-vorticity conservation. Retaining first-order terms givesLet $\Delta Q=\bar{Q}(0^{+})-\bar{Q}(0^{-})>0$. Differentiating the mean inversion relation gives ${\bar{Q}}_y=-U^{\prime \prime }+L_R^{-2}U$. Away from $y=0$ this is zero, so decay at both infinities gives $U=U_0{e}^{-|y|/L_R}$. A velocity jump would introduce a derivative of a delta function, absent from ${\bar{Q}}_y$, so $U$ is continuous. Integrating through zero now gives $-[U^{\prime }]=\Delta Q$, and therefore

For the regular interfacial normal mode ${\psi }^{\prime }=\varphi (y)e^{ik(x-ct)}$, the linear equation readsIn each region, choose zero perturbation potential vorticity, giving ${\varphi }^{\prime \prime }-(k^2+L_R^{-2})\varphi =0$. This supplies a regular solution even if $U=c$ at isolated locations away from the interface. Continuity and decay require $\varphi =A{e}^{-\gamma |y|}$, where $\gamma =\sqrt{k^2+L_R^{-2}}$. Integrating the equation across zero givesSince $[{\varphi }^{\prime }]=-2\gamma A$, the nonzero normal mode hasThis potential-vorticity step edge wave is an interfacial Rossby wave. Displacing the potential vorticity jump creates alternating anomalies; their inverted velocity displaces the interface further with an intrinsic propagation opposite to the positive mean current. The mechanism requires a nonzero jump, or more generally a gradient, of the mean potential vorticity. It does not require planetary variation of $f$. Reversing the jump reverses the intrinsic propagation; setting the jump to zero removes this interfacial restoring mechanism.

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.