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Consider the permanent stationary topographic internal gravity wave component of a localized source, with constant buoyancy frequency in each layer and positive $U_1,U_2$. A sufficiently broad obstacle spectrum is assumed. The dispersion relation on the upward branch in a uniform layer isFor a stationary wave, $K=(k^2+m^2{)}^{1/2}=N/U$. Differentiating before imposing stationarity gives the laboratory group velocityThe vertical component is at most $U/2$, attained at $\theta =\pi /4$. The first arrival of a stationary internal wavefront at a layer boundary is thusIf the obstacle spectrum lacks that wavenumber, the first arrival of its actual permanent wave packets is later. This statement concerns the stationary part; arbitrary long-wave startup transients are not bounded by this particular envelope.

In the lower layer an elapsed propagation time $t$ gives $x=U_{1}t{\cos }^2{\theta }_1$, $z=U_{1}t\sin {\theta }_1\cos {\theta }_1$. Elimination gives the causal envelope of stationary internal wavesLater-emitted packets fill the interior. At $t=T$ this semicircle just touches $H$ at $x=H$. At $2T$ its unobstructed continuation would reach $2H$, but rays that hit the interface must instead be refracted or reflected.

For a lower-layer ray labelled by ${\theta }_1$, its interface arrival time and point areThe refraction of a stationary internal-wave causal envelope conserves horizontal wavenumber, so $\cos {\theta }_2=(U_2{N}_1)/(U_1{N}_2)\cos {\theta }_1$. If ${\tau }_1\le t$ and $0<\cos {\theta }_2<1$, its transmitted front is parametrized byThe reflected front instead has $x=U_{1}t{\cos }^2{\theta }_1$ and $z=2H-U_{1}t\sin {\theta }_1\cos {\theta }_1$, until it reaches the ground. At $2T$ that first return just touches the ground. If the $\pi /4$ incident ray is evanescent upstairs, upper-layer propagating arrival is delayed until an allowed ray reaches $H$; this does not change the first lower-layer arrival time $T$.

The stationary phase of topographic internal waves also gives the phase sketch. In a uniform lower layer the Fourier transform integrand has phase $kx+m(k)z$. The stationary phase method requires $x=kz/m$, giving phase $\Phi =(N_1/U_1)(x^2+z^2{)}^{1/2}$. Thus the dominant permanent phase lines are downstream circular arcs centred on the obstacle. Reflected phase arcs have the image source $(0,2H)$. Upstairs the ray arriving at $(x_H,H)$ has phasewhich is continuous at the interface and determines refracted constant-phase curves. Individual monochromatic components have straight phase lines of slope $-k/m_j$ instead; the arcs describe their localized-source superposition.

The schematic uses $H=U_1=N=1$ and $U_2=U_1/2$. It shows the direct front below $H$, the transmitted front above it and the reflected front below it; dotted arcs are permanent phase curves, not material trajectories.

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.

For a zonal basic geostrophic flow $U(y)$, take ${\psi }_0^{\prime }(y)=-U(y)$ and write $\psi ={\psi }_0+\varphi$. The flat-bottom shallow-water quasi-geostrophic potential vorticity gradient of the basic state isLinearizing its material conservation gives the Rossby-wave equation for a sheared zonal current,For a general nonuniform jet, a global two-dimensional plane wave is not an exact normal mode: the coefficients depend on $y$. The exact zonal normal mode problem, $\varphi =\hat{\varphi }(y)e^{i(kx-\omega t)}$, iswith appropriate transverse boundary or radiation conditions. If the jet varies slowly compared with a wavelength, a local plane wave freezes these coefficients at $y_0$. Its local dispersion relation isIn the nondivergent Barotropic Rossby wave model under a rigid-lid approximation, this becomes $\omega =kU-k(\beta -U^{\prime \prime })/(k^2+l^2)$. It is exact for uniform $U$ and otherwise a local relation. Retaining the free-surface term also requires retaining the basic surface slope in $Q_0^{\prime }$; simply adding $R_D^{-2}$ to the denominator while dropping $R_D^{-2}U$ from the numerator would describe a different prescribed-background model.

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.

In a source-free vacuum, Maxwell equations areTaking a curl of the last two equations and using $\nabla \times (\nabla \times \mathbf{A})=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$ gives the electromagnetic wave equationsSubstitution of the complex plane wave amplitudes into Maxwell equations, then taking real parts, givesFor a nonzero propagating wave these imply the dispersion relation $\boxed{{\omega }^2=c^2|\mathbf{k}{|}^2}$. Conversely choose any nonzero transverse complex $\mathbf{e}$ and set $\mathbf{b}=\mathbf{k}\times \mathbf{e}/\omega$ to obtain such a wave. The static uniform-field sector at $\omega =|\mathbf{k}|=0$ is separate from propagation.

At a perfect conductor, the oscillatory boundary conditions are $\mathbf{n}\times \mathbf{E}=0$ and $\mathbf{n}\cdot \mathbf{B}=0$. Surface charge and surface current allow nonzero normal electric and tangential magnetic fields. The magnetic condition here concerns the time-dependent wave field; a perfect conductor can retain a pre-existing static magnetic field.

Let $R=I-2\mathbf{n}{\mathbf{n}}^T$ be the orthogonal reflection across the boundary plane, so $\det R=-1$. The proposed reflected amplitudes are ${\mathbf{k}}^{\prime }=R\mathbf{k}$, ${\mathbf{e}}^{\prime }=-R\mathbf{e}$, and ${\mathbf{b}}^{\prime }=R\mathbf{b}$. For an orthogonal reflection, $(R\mathbf{a})\times (R\mathbf{d})=-R(\mathbf{a}\times \mathbf{d})$. HenceAlso $|{\mathbf{k}}^{\prime }|=|\mathbf{k}|$, so the reflected wave obeys the same dispersion relation. At $\mathbf{n}\cdot \mathbf{x}=0$, the incident and reflected phases coincide because ${\mathbf{k}}^{\prime }-\mathbf{k}$ is parallel to $\mathbf{n}$. Their summed amplitudes satisfyThe first has zero tangential part and the second zero normal part, proving both boundary conditions. Finally ${\mathbf{k}}^{\prime }\cdot \mathbf{n}=-\mathbf{k}\cdot \mathbf{n}>0$, as required for reflection into the vacuum.

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.

Take $z$ upwards and write $\zeta =z+Wt$, so that the unperturbed interface is $\zeta =0$, with carbon dioxide above and water below. Use the sharp-interface, incompressible flow version of Darcy's law, with constant porosity and permeability of a porous medium, no capillary pressure, and effectively unbounded layers. The distinction between Darcy flux and interface speed matters: the base flux is $q_z=-\varphi W$, since the printed $W$ is the speed of the interface. Thus the base pressure gradients satisfyThe mass density ordering is stabilizing, whereas injection of the less viscous fluid into the more viscous fluid produces the Saffman–Taylor instability.

For linear stability analysis, displace the interface by $\eta =\hat{\eta }{e}^{\sigma t+i\alpha x}$. Taking the divergence of Darcy's law shows that the perturbation pressure satisfies the Laplace equation. Decay away from the interface gives, for $a_{\alpha }=|\alpha |>0$,The kinematic boundary condition and continuity of normal Darcy flux giveMeanwhile pressure continuity, evaluated on the displaced interface, givesEliminating $A,B$ yields the buoyancy-modified Darcy fingering dispersion relation:Every nonzero wavenumber decays when $W<W_c$, grows when $W>W_c$, and is neutral at equality, whereThe zero wavenumber is a uniform displacement and is neutral. Without a short-scale regularization such as surface tension, this model has no fastest-growing finite wavenumber: on the unstable side its growth rate increases without bound as $|\alpha |$ increases. Finite layer depths would change the dispersion relation; none are specified here.

Taking $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and a year of $365.25$ days, the data giveThe numerical paragraph supplies no porosity, so it cannot determine a unique numerical interface speed. If $W$ were instead intended to mean Darcy velocity, the numerical critical value would be $103.2\mathrm{m}\mathrm{y}{\mathrm{r}}^{-1}$, and the interface would move at $W/\varphi$. These two velocity conventions must not be conflated.

Linearized mass conservation, momentum balance and the isentropic pressure-density relation are ${\rho }_t^{\prime }+{\rho }_0\nabla \cdot \mathbf{u}=0$, ${\rho }_0{\mathbf{u}}_t=-\nabla \tilde{p}$, and $\tilde{p}=c_0^2{\rho }^{\prime }$. Differentiate mass conservation in time and take the divergence of momentum balance to eliminate velocity. This gives the acoustic pressure wave equationFor the convention $\tilde{p}=\mathrm{Re}{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, momentum balance gives the complex velocity amplitude $\mathbf{k}/({\rho }_0\omega )$, with $\omega =c_0|\mathbf{k}|$. Thus the incident velocity field is $\mathrm{Re}[{\mathbf{k}}_I{e}^{i({\mathbf{k}}_I\cdot \mathbf{r}-\omega t)}/({\rho }_1\omega )]$ in the lower gas.

Let $a=k_I\sin \theta$, $l_1=k_I\cos \theta >0$, $\omega =c_1{k}_I$, and $l_2=\sqrt{(\omega /c_2{)}^2-a^2}$ with nonnegative real or imaginary part. This chooses an outgoing propagating wave or an evanescent wave decaying as $y\to +\infty$. Conservation of tangential wavevector is the acoustic Snell law for elastic and acoustic waves. Write the complex pressures asInviscid kinematic matching requires equal normal velocities at the plate, not equal tangential velocities:Define the normal acoustic impedance ratio $\delta ={\rho }_2{l}_1/({\rho }_1{l}_2)$, so $T=\delta (1-R)$. For a propagating transmitted wave this is ${\rho }_2{c}_2\cos \theta /({\rho }_1{c}_1\cos {\theta }_2)$. The dynamic condition giveswhich is exactly the printed dimensionless parameter. Solving yields the pressure reflection coefficient and transmission coefficientAt $l_2=0$ use the preceding velocity equations, or take their limit, rather than substituting an infinite $\delta$. For real $l_2>0$, the reflected energy fraction is $|R{|}^2$, the transmitted fraction is $|T{|}^2/\delta$, and these sum to one. An evanescent transmitted wave carries no mean normal energy flux.

When the gases match, $\delta =1$, so $R=-i\beta /(2-i\beta )$ and $T=2/(2-i\beta )$. For a nongrazing incident wave of nonzero frequency, perfect transmission means $R=0$, equivalent toThe plate's inertial and bending terms cancel: the incident frequency and tangential wavevector coincide with its free flexural wave dispersion relation. Then $T=1$, including its phase. With positive $m,B$, this condition is possible only at angles and frequencies satisfying that relation; for normal incidence at nonzero frequency it cannot occur unless the mass is zero.

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.

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.

For an axisymmetric vertical mode of a shearing sheet in this Boussinesq approximation, the perturbations have no horizontal gradients, so background-shear advection vanishes. The linear Coriolis acceleration and perturbation advection of the background shear givewhile thermal conduction and incompressibility giveHere $\beta$ is the thermal diffusion rate of a Fourier mode, not the cooling exponent in Question 2. Since $k\ne 0$, incompressibility forces $u_z^{\prime }=0$, and the vertical momentum equation then forces $P^{\prime }=0$ for this nonzero-wavenumber amplitude.

The three remaining amplitudes satisfyA nontrivial solution requires the determinant to vanish, giving the convective overstability dispersion relationNo division by $s$ or $s+\beta$ was needed, so the thermal energy mode of a shearing sheet is retained. The notation $N^2$ denotes a signed buoyancy frequency squared and can be negative for an adverse stratification; it must not be restricted to the square of a real positive frequency.

Treat $n^2=N^2/{\Omega }^2$ as a signed small parameter, with $|n^2|\ll 1$, as printed in the original PDF. At $n^2=0$, the dispersion relation factors asThus the two epicyclic motion modes have $s_0=\pm i\Omega$, while the thermal energy mode of a shearing sheet hasIt decays on the thermal diffusion time ${\beta }^{-1}$. For completeness, its next correction is $s_E=-\beta +\beta N^2/({\beta }^2+{\Omega }^2)+O(\Omega n^4)$ at fixed $\beta /\Omega$.

For either epicyclic motion root, the implicit function theorem applied to the dispersion relation givesHenceSince $\beta >0$, the oscillations undergo overstability precisely whenThe convective overstability growth rate tends to zero both for very slow diffusion and for very fast diffusion. Differentiating $\beta /({\beta }^2+{\Omega }^2)$ shows that its maximum occurs at $\beta =\Omega$, corresponding to $k^2=\Omega /\xi$. ThereforeThis is the growth rate of the perturbation amplitude; a quadratic perturbation energy grows at twice that rate. The convective overstability arises when an adverse radial specific entropy gradient couples to epicyclic motion with a finite thermal conduction lag.