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For the local phase $S$, define $k_i={\partial }_{i}S$ and $\omega =-{\partial }_{t}S$. The dispersion relation is the Hamilton-Jacobi equation $S_t+\Omega (\nabla S,\mathbf{x},t)=0$. Differentiate it in space and follow a ray with group velocity ${\dot{x}}_i={\Omega }_{k_i}$; symmetry of the phase Hessian cancels the terms involving ${\partial }_i{k}_j$. Thus the ray-tracing equations areHere $d/dt={\partial }_t+\dot{\mathbf{x}}\cdot \nabla$ means differentiation along the moving ray; subscripts on $\Omega$ mean partial derivatives holding its other independent arguments fixed.

The depth is stationary and independent of $y$, so $\omega ={\omega }_{\infty }$ and $l$ are constant. Choose the incident direction with positive $l$; the other sign reflects the picture. In the far deep water,Since $\Omega$ depends isotropically on the horizontal wavevector, the group-velocity ratio is $l/k$. Therefore the ray equation isNear shore, $\kappa h\to 0$, so the shallow-water dispersion relation gives $\kappa \sim {\omega }_{\infty }/(\sqrt{g\alpha }{x}^{p/2})$ and $k\sim -\kappa$. Integrating the ray slope yieldsFor $p<2$, the near-shore phase is $S\sim ly-{\omega }_{\infty }t-{\omega }_{\infty }{x}^{1-p/2}/[\sqrt{g\alpha }(1-p/2)]$. A constant-phase wave crest thus hasIts tangent becomes parallel to the shoreline, while rays approach perpendicular to it. The exponent is between zero and one, giving the curved crests sketched below for $p=1$.

These are the formal geometrical optics asymptotics. For fixed $\alpha$ and $p<2$, $\kappa x\to 0$ arbitrarily close to the shore, so the slowly varying wave approximation eventually fails there; the sketch describes the ray-theory regime before that final inner region.

For the self-gravitating incompressible slab, let the fluid occupy $-H<z<H$, with mass density ${\rho }_0=\Sigma /(2H)$ and a vacuum exterior. Poisson equation and hydrostatic equilibrium giveA constant ambient pressure could be added without changing the result. Choose perturbations proportional to $e^{ikx-i\omega t}$, using rotational symmetry in the horizontal plane to put the wavevector along $x$.

For the fluid displacement, the linearized momentum equation inside the uniform slab is $-{\omega }^2\mathbf{\xi }=-\nabla (p^{\prime }/{\rho }_0+{\varphi }^{\prime })$. For $\omega \ne 0$ it follows that $\mathbf{\xi }=\nabla \chi$, with $p^{\prime }/{\rho }_0+{\varphi }^{\prime }={\omega }^2\chi$. Incompressibility gives $\nabla \cdot \mathbf{\xi }=0$, henceNeutral limits of these surface modes follow by continuity. Stationary vorticity perturbations form a separate zero-frequency sector; the nonzero-frequency surface modes are irrotational by the momentum equation itself, without an extra irrotational-flow assumption.

The Eulerian mass density perturbation vanishes in the bulk because the equilibrium mass density is uniform and the displacement is divergence free. If the two vertical surface displacements are ${\eta }_{+}={\xi }_z(H)$ and ${\eta }_{-}={\xi }_z(-H)$, the surface density perturbation of a displaced uniform interface givesThus Poisson equation reduces to Laplace equation for ${\varphi }^{\prime }$ inside and outside. The potential is continuous, while integrating through each surface givesThe perturbing potential must decay at vertical infinity. The kinematic conditions are ${\eta }_{\pm }={\partial }_z\chi (\pm H)$. The free surface pressure condition is the vanishing Lagrangian pressure perturbation, $p^{\prime }+{\eta }_{\pm }{p}_0^{\prime }=0$; at the top this gives $p^{\prime }(H)/{\rho }_0=g_s{\eta }_{+}$.

Put $q=kH>0$. For even scalar-potential symmetry, chooseThen ${\eta }_{-}=-{\eta }_{+}$: this is a thickness-changing mode, not a vertical bending mode. Write $\eta ={\eta }_{+}$; the top derivative jump gives $-kC(\cosh q+\sinh q)=4\pi G{\rho }_0\eta$. Since $\cosh q+\sinh q=e^q$,Substitute these values into ${\omega }^2\chi (H)=g_s\eta +{\varphi }^{\prime }(H)$. Using $4\pi G{\rho }_0=g_s/H$ gives the even-mode dispersion relation for the surface modes of a self-gravitating incompressible slab:The positive term comes from the restoring hydrostatic free-surface pressure in the background gravitational field. The negative term is the attraction due to the perturbed gravitational potential of the displaced surfaces, which tends to reinforce the disturbance. Both effects ultimately involve the slab's gravity, but enter the boundary condition through different terms.

For odd scalar-potential symmetry, instead takeNow ${\eta }_{-}={\eta }_{+}$, so the surfaces bend in the same direction. The jump again gives $C=-4\pi G{\rho }_0\eta /(k{e}^q)$, but $\chi (H)=\eta \tanh q/k$ and ${\varphi }^{\prime }(H)=-4\pi G{\rho }_0\eta /[k(1+\coth q)]$. The same pressure condition yieldsThe bottom kinematic, jump and pressure conditions follow with the indicated parity and give the same dispersion relation, so no boundary condition has been discarded.

To determine the gravitational instability of an incompressible slab, factor the two dimensionless expressions asFor the even mode the bracket is strictly increasing, with derivative $1+e^{-2q}>0$, is negative at zero and tends to positive infinity. It therefore has a unique positive zeroThe even mode is unstable for $0<kH<q_c$, neutral at $q_c$, and oscillatory for $kH>q_c$. Negative ${\omega }_e^2$ gives a growing mode with $\omega =i\sqrt{-{\omega }_e^2}$ under the stated time convention. For the odd mode, $1-e^{-2q}<2q$ for every $q>0$, so the bracket is always positive: all odd modes with positive wavenumber are stable.

At long wavelengths $F_e=-q+2q^2+O(q^3)$ and $F_o=q-2q^2/3+O(q^3)$: a thickness or column-density disturbance has a gravitational instability, whereas a bending disturbance is restored. At short wavelengths both approach $q-1/2$, so both surface branches are stable. The slab equilibrium is therefore linearly unstable overall, because it admits the long-wavelength even modes despite the stability of its odd sector.

Substitute the axisymmetric Fourier mode into the linearized equations. With hats denoting amplitudes,For a nondegenerate nonzero frequency, eliminate the azimuthal velocity and useContinuity then givesEquivalently, taking the determinant of the original four amplitude equations avoids division by zero and gives the full dispersion relationThe two squared frequencies areThe discriminant is $({\Omega }^2-c^2{k}^2{)}^2+4{\Omega }^2{c}^2{k}_x^2\ge 0$. For $c^2>0$, both squared frequencies are nonnegative. The high-frequency branch is predominantly acoustic and the low-frequency branch is an inertial wave. Zero-frequency limiting modes, including $k_z=0$, are retained by the determinant calculation.

Take $ck\gg \Omega$, retaining a finite low-frequency branch and a nonzero wavevector. In the dispersion relation, neglect ${\omega }^4$ against $c^2{k}^2{\omega }^2$ and neglect ${\Omega }^2$ against $c^2{k}^2$ in their coefficient. The inertial wave limit isChoose the signed branches $\omega =ϵ\Omega k_z/k$, $ϵ=\pm 1$. The phase velocity normal to the phase planes and the group velocity areTherefore $\mathbf{k}\cdot {\mathbf{c}}_g=ϵ\Omega (k_z/k-k_z{k}^2/k^3)=0$, soThe geometric reason is that this limiting dispersion relation is homogeneous of degree zero in $\mathbf{k}$: its frequency depends on direction, not magnitude. Euler homogeneous function theorem gives $\mathbf{k}\cdot {\nabla }_{\mathbf{k}}\omega =0$. Exact orthogonality holds for this inertial wave limit; finite compressibility introduces corrections. At a zero-frequency degeneracy the usual propagating-wave interpretation needs a limiting treatment.

Put $A=(Gm/h^3)F(hk)$. The two amplitude equations have coefficient matrixIts determinant gives the dispersion relationWriting $q=A/{\Omega }^2\ge 0$, its squared growth rates areand the four growth rates are both square roots of each displayed value. DefineFor $0<q<q_{-}$, the discriminant is positive, the sum of the squared rates is $q-1<0$, and their product is $2q(3-q)>0$; both squared rates are negative. These modes have purely imaginary $s$. For $q_{-}<q<q_{+}$, the squared rates form a nonreal complex conjugate pair; their square roots include roots with positive real part. This is overstability, meaning growing oscillations. For $q\ge q_{+}$, at least the plus squared rate is positive because $q>1$, so a real growing root exists. Consequently every $q>q_{-}$ is exponentially unstable.

Take the maximizing mode $hk=\pi$, so $q=(Gm/(h^3{\Omega }^2))F(\pi )$. It is unstable wheneverThis is also the threshold for the onset of exponential growth among all lattice modes, since $0\le F\le F(\pi )$. At equality the maximizing mode has repeated imaginary roots and can have algebraic growth; it is a marginal collision of oscillatory modes, not a strict-decay assertion. The $k=0$ mode has $F=0$, an epicyclic motion sector and a zero-frequency epicyclic guiding center sector with possible secular shear drift. These neutral or algebraic behaviors are distinct from the strict exponential instability established by the requested inequality.

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.