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The basic quasi-geostrophic streamfunction $\bar{\psi }=-\Lambda yz$ gives $\bar{u}=\Lambda z$, $\bar{v}=0$ and constant quasi-geostrophic potential vorticity $\bar{Q}=f$. DefineBecause the background has no potential-vorticity gradient, the linear potential-vorticity equation is simplyIts characteristics have constant $y,z,x-\Lambda zt$, so the general solution is $q^{\prime }=\mathcal{F}(y,z,x-\Lambda zt)$. In particular, $q^{\prime }=F(y,z)G(x-\Lambda zt)$ is the requested arbitrary separated family; one product is not the most general initial field.

Linearizing the vertical velocity requires retaining advection of ${\bar{\psi }}_z=-\Lambda y$ by $v^{\prime }={\psi }_x^{\prime }$. ThereforeThe impermeable boundary at $z=0$ imposesThis is conservation of boundary buoyancy, not merely a condition that ${\psi }_z^{\prime }$ vanish.

For zero interior potential vorticity, write ${\psi }^{\prime }=\mathrm{Re}\{\varphi (z,t)e^{i(kx+ly)}\}$. With $K_h=\sqrt{k^2+l^2}>0$, stratified quasi-geostrophic inversion givesThe decaying solution isNo time dependence has been assumed to obtain this vertical structure. Applying the boundary condition gives $-\mu \dot{A}-i\Lambda kA=0$, henceThus the Eady edge wave travels eastward with horizontal phase velocity $\omega /k=\Lambda /\mu$ for $k\ne 0$.

The propagation mechanism can be made explicit using boundary parcels. Thermal wind gives the background buoyancy gradient ${\bar{b}}_y=f{\bar{\psi }}_{zy}=-f\Lambda$. A small meridional parcel displacement $d$ produces boundary buoyancy perturbation $b^{\prime }=-d{\bar{b}}_y=f\Lambda d$. Meanwhile the decaying potential-vorticity inversion gives $b^{\prime }=f{\psi }_z^{\prime }=-f\mu {\psi }^{\prime }$ at the boundary. Consequently ${\psi }^{\prime }=-\Lambda d/\mu$ there, and the parcel equation $d_t=v^{\prime }={\psi }_x^{\prime }$ becomesA displaced buoyancy contour therefore induces a velocity field that advances that displacement eastward. No interior potential-vorticity gradient is needed because the boundary buoyancy supplies the invertibility data.

Vortex stretching is what sets the induced flow's vertical reach. In the interior, the zero-potential vorticity condition requiresBoundary isopycnal displacement decreases with height as $e^{-\mu z}$, so fluid columns are stretched or compressed over depth ${\mu }^{-1}=f/(N{K}_h)$. Conservation of potential vorticity makes this stretching balance the relative vorticity. The resulting inversion factor $b^{\prime }=-f\mu {\psi }^{\prime }$ gives the propagation speed: stronger density stratification reduces penetration and speed, while larger $f$ increases them. Zero interior potential vorticity therefore does not imply zero interior flow.

For imposed boundary vertical velocity, retain the same decaying structure and writeThe boundary relation becomes the forced Eady edge wave equationThe integrating factor $e^{i\omega t}$ gives, for ${\omega }_0\ne \omega$,The arbitrary homogeneous coefficient $B$ sets the initial disturbance. With $A(0)=0$, this is $A=C(e^{-i{\omega }_{0}t}-e^{-i\omega t})/[i(\omega -{\omega }_0)]$.

At resonance, ${\omega }_0=\omega$, integration instead yieldsAn initially undisturbed fluid has ${\psi }^{\prime }=Ct{e}^{-\mu z}\cos (kx+ly-\omega t)$. The imposed pumping reinforces the same boundary normal mode every cycle, with no dissipative loss, so the linear amplitude grows secularly. The divergence describes resonance of the undamped linear model; sufficiently large wave amplitude invalidates the assumed small-disturbance approximation.

Let $W$ be the complex vertical velocity amplitude in $w=\mathrm{Re}\{W(z)e^{ikx}\}$. It satisfies $W^{\prime \prime }+m^{2}W=0$, with $W(0)=ikU{\eta }_0$ and $W(H)=0$. Away from resonance, solve these two boundary conditions:The resulting field is a standing vertical superposition of equal-amplitude upward and downward internal gravity waves. Write $W=A_{+}{e}^{imz}+A_{-}{e}^{-imz}$. The rigid lid gives $A_{-}=-A_{+}{e}^{2imH}$, and the bed givesThe upward vertical-velocity amplitude is $|kU{\eta }_0|/[2|\sin (mH)|]$. For real positive ${\eta }_0$, its phase relative to the bed-height coefficient is $-mH+\pi$ if $\sin (mH)>0$ and $-mH$ if $\sin (mH)<0$, modulo $2\pi$. Relative to the imposed bed-velocity coefficient $ikU{\eta }_0$, the phase is instead $-mH+\pi /2$ in the positive-sine case, with an additional $\pi$ when the sine changes sign. The complex coefficient gives the answer without convention ambiguity. The upward displacement coefficient, if desired, is $A_{+}/(ikU)$, since $w=U{\zeta }_x$ for the stationary wave.

A reflecting lid reverses the vertical energy direction while retaining horizontal wavenumber and laboratory frequency. In the terrain frame the incident and reflected energy rays therefore travel downstream while zigzagging between the boundaries. The velocity field has nodes at the lid and a vertically standing pattern; the sketch marks this ray geometry, rather than falsely drawing the reflected ray as a specular optical ray.

The intrinsic group velocity makes an angle $\theta$ with vertical satisfying $\tan \theta =|c_{gx}^{\mathrm{fluid}}|/c_{gz}=m/k$. Consequently the denominator vanishes whenThese are the rigid-lid resonance of stationary topographic internal waves. At such a nonzero vertical mode the homogeneous rigid-lid eigenfunction is also zero at the bed, so it cannot satisfy the nonzero prescribed bed velocity; a bounded stationary inviscid forced solution does not exist. Time-dependent forcing produces secular modal growth until dissipation or nonlinearity intervenes. The formal $n=0$, $m=0$ limit is not this resonance: $\sin [m(H-z)]/\sin (mH)\to (H-z)/H$ is finite. Propagating resonance also requires $k<N/U$; an evanescent field has no real standing-wave resonances of this kind.

The Wronskian satisfies Abel identity. Differentiating and substituting the homogeneous equations givesso $W(x)=-\exp [-{\int }_0^{x}p(r)dr]\ne 0$. The two homogeneous solutions are therefore independent.

Define the variation of parameters kernelFor fixed $s$, it satisfies the homogeneous differential equation in $x$. Direct evaluation gives $K(s,s)=0$ and $K_x(s,s)=1$. For $y(x)={\int }_0^{x}K(x,s)f(s)ds$, differentiation under the integral therefore yieldsCombining these identities proves $y^{\prime \prime }+p{y}^{\prime }+qy=f$; both initial values vanish. Continuity of the coefficients gives uniqueness of this initial value problem, establishing the proposed formula rather than merely recognizing it as a standard method.

For the oscillator, $y_1=\sin x$, $y_2=\cos x$ and $W=-1$, so the kernel is $\sin (x-s)$. HenceUsing ${\int }_0^x\cos s\sin sds={\sin }^{2}x/2$ and ${\int }_0^x{\sin }^{2}sds=x/2-\sin (2x)/4$ givesThe factor $x\cos x$ reflects resonance with the homogeneous oscillator frequency.

For unit incident plane-wave amplitude, define the scattering amplitude by $\psi \sim e^{ikz}+f(\theta )e^{ikr}/r$ at large $r$. Comparing outgoing radial flux with incident flux gives $d\sigma /d\Omega =|f(\theta ){|}^2$. Outside the potential, a real regular radial solution is proportional to $j_{\ell }(kr)\cos {\delta }_{\ell }-n_{\ell }(kr)\sin {\delta }_{\ell }$, with asymptotic phase $\sin (kr-\ell \pi /2+{\delta }_{\ell })/(kr)$. This defines the scattering phase shift modulo $\pi$.

The incident plane-wave expansion supplies coefficient $(2\ell +1)i^{\ell }$ for $j_{\ell }{P}_{\ell }$. To preserve its incoming part, replace $j_{\ell }$ by $e^{i{\delta }_{\ell }}(j_{\ell }\cos {\delta }_{\ell }-n_{\ell }\sin {\delta }_{\ell })$. Writing the asymptotic sine in exponentials shows that its incoming coefficient is unchanged and its outgoing coefficient is multiplied by $e^{2i{\delta }_{\ell }}$. The outgoing difference from the plane wave therefore gives

Put $\rho =ka$, $t=\tan {\delta }_{\ell }$, and abbreviate $j=j_{\ell }(\rho )$, $n=n_{\ell }(\rho )$. Logarithmic-derivative matching gives $a{R}_{\ell }^{\prime }(a)/R_{\ell }(a)=\rho (j^{\prime }-t{n}^{\prime })/(j-tn)$. The supplied asymptotics determine the constant Wronskian: ${\rho }^2(j{n}^{\prime }-j^{\prime }n)=1$. ConsequentlySolving for $t$ givesAt zeros of denominators it is interpreted by limits; the logarithmic-derivative definition itself presumes its stated denominators nonzero.

Near a dominant resonant partial wave, ${\sin }^2{\delta }_{\ell }\approx {\gamma }^2/[(k-k_0{)}^2+{\gamma }^2]$. Neglecting the small other partial waves yieldsThis is a resonant peak of width $2|\gamma |$ in $k$, with the partial-wave angular pattern and maximum allowed ${\sin }^2{\delta }_{\ell }=1$ at resonance. Integrating gives $\sigma \approx 4\pi (2\ell +1){\gamma }^2/[k^2((k-k_0{)}^2+{\gamma }^2)]$. A small background may matter at angular zeros of the dominant wave.

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 $y,z$.

The basic potential-vorticity gradient is ${\bar{Q}}_y=\beta$, so the wave activity and its derivative areHere ${\kappa }^4{B}_0^2/(2\beta )=k{f}_0{B}_0$ follows from $U=\beta /{\kappa }^2$. Thus ${\mathcal{A}}_t+F_y+G_z=0$. The growth term involves $B_0^2=|B{|}^2=-B^2$; the distinction matters because the printed complex coefficient $B$ is imaginary. These formulas evaluate both flux components, their divergence, the buoyancy and PV covariances, and wave activity. In particular $G(H)=0$ and $G(0)={\epsilon }^{2}k{f}_0{B}_{0}Ht{\sin }^2(\ell y)/2$, so for $\beta >0$ the growing activity is supplied by the positive flux from the lower boundary.

Finally put $W=\pi /\ell$ and write the height-independent mean velocity change as $\delta U(y)$. Since $\delta {\bar{\psi }}_y=-\delta U$ and $\delta \bar{\psi }$ can be chosen independent of height, the mean potential vorticity becomesTo be uniform in $y$ this requires $\delta U_y=\beta y-K$ for some constant $K$. Integrating and applying both endpoint conditions gives $\delta U(0)=0$, $\delta U(W)=0$, and hence $K=\beta W/2$. ThereforeThe constant is the initial channel-mean potential vorticity, as also follows from ${\int }_0^W\delta Qdy=-[\delta U{]}_0^W=0$. This is the momentum loss from uniform potential-vorticity mixing. Averaging the velocity change across the channel gives the momentum change per unit mass,For positive $\beta$ the mean current is reduced everywhere in the interior. The imposed bottom can exchange momentum with the fluid, so this change is consistent with the complete conservation of momentum budget.

The linear acoustic momentum and continuity equations are $\rho u_t=-p_x$ and $p_t=-\rho c^2{u}_x$. For a right-moving plane wave depending on $x-ct$, integration gives $\boxed{p=\rho cu}$, with zero perturbation constants. For a left-moving wave the sign is reversed. The factor $Z=\rho c$ is the acoustic impedance. Proportionality applies to a traveling wave; a superposition of opposite directions generally has no single such constant.

Use time dependence $e^{-i\omega t}$ and set $Z_j={\rho }_j{c}_j$. In the outer regions write pressures $I{e}^{i{k}_{0}x}+R{e}^{-i{k}_{0}x}$ and $TI{e}^{i{k}_{0}x}$, taking $I=1$ for convenience. The corresponding velocities are $(e^{i{k}_{0}x}-R{e}^{-i{k}_{0}x})/Z_0$ and $T{e}^{i{k}_{0}x}/Z_0$. Pressure and normal velocity are continuous at both layer boundaries. Within the cooled layer a decomposition into right and left waves gives, with $\delta =k_{1}L$,This matrix follows by writing $p=A{e}^{i{k}_{1}x}+B{e}^{-i{k}_{1}x}$ and $Z_{1}u=A{e}^{i{k}_{1}x}-B{e}^{-i{k}_{1}x}$. Applying its inverse and using $p(0)+Z_{0}u(0)=2$ eliminates $R$, givingConsequentlyFor $\lambda \gg 1$, transmission is weak except near $k_{1}L=n\pi$, where $|T|=1$. The cooled layer acts as a resonant filter: an integer number of half-wavelengths fits within it at each resonance. The half-power width in $\delta$ is asymptotically $2/\lambda$ on either side. Since the two outer impedances agree, the transmitted power fraction is $|T{|}^2$.

The applied pressure changes the dynamic condition to ${\varphi }_t+g\eta =-p/\rho$. For its specified spatial mode, the two amplitude equations becomeWhen $\Omega \ne \omega$, the corresponding particular oscillation isFree oscillations with amplitudes $C\cos \omega t+D\sin \omega t$ may be added. If the mode starts from rest at $t=0$, its total amplitude instead isNear resonance, the response grows too large for the linearization. At exact resonance the rest-start limit is $A=(\epsilon h/2)(\omega t\cos \omega t-\sin \omega t)$, with secular growth. Dissipation or nonlinear effects would be required to limit it.

Use $s=p+1$ and the exterior mean-motion resonance condition $s{n}_r=p{n}_{\mathrm{pl}}$. Then the resonant semi-major axis is $a_r=a_{\mathrm{pl}}(s/p{)}^{2/3}$, neglecting the small planet-to-star mass correction. For the geometric construction, take $a_{\mathrm{pl}}=n_{\mathrm{pl}}=1$, $p=4$, $e=0.2$, $\varpi =\pi /4$ and start the particle at pericentre. Solve Kepler's equation $M=E-e\sin E$ with $M=(4/5)t$. Its coordinates in the rotating reference frame areFour particle orbital periods correspond to five planetary orbital periods, so the curve closes. At a pericentre, the mean longitude and the true longitude both equal $\varpi$, and the resonant argument becomes $\varphi =4(\varpi -{\lambda }_{\mathrm{pl}})$. Thus the four pericentre directions are $\varphi /4$ modulo $\pi /2$. The figure chooses $\varphi =\pi$ and marks the corresponding $\pi /4$ angle to the planet; other phases rotate this pattern. Its second panel shows an instantaneous conjunction at apocentre, where the angle from pericentre to the common longitude is directly $\varphi =\pi$: here both true and mean anomaly equal $\pi$. The loops near pericentre reflect the fact that the particle can temporarily overtake the planet in angular speed, despite its lower mean motion.

For the evolution equations, $\lambda$ must be the mean longitude. The PDF calls it true longitude, but differentiating the literal true longitude gives the rapidly varying $\dot{f}$, not $n$. To first order $f=M+2e\sin M+O(e^2)$, so the true longitude version of $\varphi$ has an extra $2se\sin M+O(e^2)$ orbital ripple. The stated Lagrange planetary equations and slow complex eccentricity equation are interpreted consistently as orbit-averaged equations in mean longitude. The pericentre geometry of an exterior resonance above is unchanged by this distinction.

Let $F_r,F_{\theta }$ be perturbing forces, and let $m$ be the particle's mass. The angular momentum about the star satisfiesThe central gravitational force and the radial perturbation have zero torque. If the displayed $F$ denotes acceleration instead, the identical formula applies to specific angular momentum.

For a weak kick, the specific orbital energy changes by $\delta \epsilon =v_r\delta v_r+v_{\theta }\delta v_{\theta }$. A planet leading the particle pulls it forwards, while a trailing planet pulls it backwards. For a nearly circular exterior Kepler orbit, a forward tangential kick raises angular momentum and semi-major axis, lowers mean motion, and makes the next conjunction occur sooner in the synodic cycle. Its longitude retreats relative to the unperturbed conjunction sequence. A backward kick produces the opposite shift. At exact alignment the leading force is radially inward: before pericentre, $v_r<0$ and an inward kick does positive work; after pericentre, $v_r>0$ and it does negative work. These likewise raise or lower semi-major axis. At an apsis, a purely radial infinitesimal kick has no first-order energy change but changes the eccentricity vector, so angular momentum alone does not determine every timing change.

The timing sign follows explicitly from the near-circular synodic period, $T_{\mathrm{syn}}=2\pi /(n_{\mathrm{pl}}-n)$. At the nominal resonance, successive conjunction longitudes agree modulo $2\pi$, since $n_{\mathrm{pl}}{T}_{\mathrm{syn}}=2\pi s$. A persistent small change givesThis is the accumulated mean timing shift; an actual impulsive encounter can also change the instantaneous orbital phase and longitude of periapsis.

Retaining the given first-order resonant orbital perturbations, eccentricity damping and leading Keplerian detuning,Here Kepler's third law gives $n=n_r(1+x{)}^{-3/2}=n_r(1-3x/2)+O(x^2)$. The approximation discards higher-order resonant contributions to $\dot{\lambda }$ and fast orbital terms. Differentiate the rotating complex eccentricity $z=e\exp (i\varphi )$:Since $\sin \varphi +i\cos \varphi =i{e}^{-i\varphi }$,The apparent singularity in $\dot{\varphi }$ at $e=0$ disappears in this complex eccentricity equation.

Take $A>0$, and the conventional positive resonant coefficient $C>0$. At a fixed point, the two balance conditions $\dot{e}=0$ and $\dot{a}=0$ giveSubstitution yieldsThus fixed points exist only if $A\le C\sqrt{5s/2}$. With strict inequality, put $S=\sqrt{(5/2)s{C}^2-A^2}$ and ${\psi }_f=\arctan (A/S)\in (0,\pi /2)$. Thenrespectively. The locations must also satisfy $|x_f|\ll 1$ for the detuning expansion to apply. At equality the phases merge at $3\pi /2$ and $x_f=0$. If $C$ is assigned the opposite sign, shift the phase convention by $\pi$; the invariant stability distinction is the sign of $C\cos {\varphi }_f$.

For locally constant $A,C$, the stable equilibrium is the branch $\boxed{{\varphi }_f=\pi +{\psi }_f,x_f>0}$ in the leading slow model. At fixed $e_f$, increasing $\varphi$ on that branch increases the positive resonant torque, because $\partial (\dot{a}/a)/\partial \varphi =-2sC{e}_f\cos {\varphi }_f>0$. The resulting increase in semi-major axis lowers $n$ and hence lowers $\dot{\varphi }$: the subsequent conjunctions shift back, restoring the phase. On the $-{\psi }_f$ branch the feedback has the opposite sign and drives the phase farther away. Eccentricity damping removes the oscillatory disturbance on the restoring branch.

One can verify that this feedback stabilizes the coupled variables rather than just the phase. Treat $A,C$ as locally constant, and write $z=u+iv$. Then $\dot{u}=-Au+Bxv$, $\dot{v}=-C-Av-Bxu$ and $\dot{x}\simeq -2sCv-4A/5$. At a fixed point $Bxv=Au$. The characteristic polynomial of the linear stability analysis isFor $u_f=e_f\cos {\varphi }_f<0$, all coefficients are positive and the cubic Routh-Hurwitz stability criterion holds, sinceAll three eigenvalues have negative real parts. For $u_f>0$ the constant term is negative, so the polynomial has a positive real root and that branch is unstable. The merged case has a zero eigenvalue. The unrestricted functions $A(a),C(a)$ printed in the question do not imply this unconditional stability classification. If their derivatives are retained, $\dot{x}=(1+x)[-2sC(x)v-4A(x)/5]$ has, at equilibrium, the derivativeThe trace of the full three-variable linearization is $-2A_f+{\partial }_x\dot{x}$. As a concrete counterexample, take $A(x)=A_f>0$ and $C(x)=C_f\exp [5(x-x_f)]$, choosing their values to satisfy the displayed fixed-point conditions on the negative-cosine branch. Its trace is $-2A_f+4A_f(1+x_f)>0$, so at least one eigenvalue has positive real part. Both coefficients are smooth. Thus the restoring conjunction feedback identifies the stable branch of the usual locally constant model, but additional assumptions on coefficient variation are necessary to assert stability for every allowed function.

The separation of the two bodies is $a$, as is fixed by the supplied Kepler third law ${\omega }_o^2=G(M_1+M_2)/a^3$. The phrase “orbit of radius $a$” cannot denote the first body's distance from the center of mass in that formula.

Expand the companion's Newtonian potential of a point mass about the first body's center. The constant term produces no force; the linear term produces the uniform acceleration of the center and disappears in the translating frame. The leading differential acceleration therefore comes from the quadrupolar tidal forcingwhere $\mathbf{n}(t)$ points toward the companion. It is a degree-two solid spherical harmonic. For a circular nonrotating-frame orbit, its time-dependent components have $m=\pm 2$ and frequency $2{\omega }_o$, because the squared direction cosines repeat twice per orbit. There is also a static $m=0$ component. The real part of one complex harmonic represents the pair of time-dependent components; its normalization and orientation are absorbed into the order-one coefficient $\psi$. This gives $\Psi =\psi (G{M}_2/a^3)r^2{Y}_2^m$ and $\omega =2{\omega }_o$. The expansion is the leading term for $R_1/a$ small; spin is being neglected.