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Let $p$ denote the fluid pressure perturbation divided by the reference mass density, and let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation. The buoyancy frequency satisfies $N^2=-g{\hat{\rho }}_z/{\rho }_0$. Linearization requires small boundary slope $k|{\eta }_0|\ll 1$, small displacements compared with the vertical scales of the disturbance and background, and advection small compared with the oscillatory acceleration. For a propagating internal gravity wave this includes $|m{\eta }_0|\ll 1$; boundary slope alone is insufficient when $|m|\gg k$. The Boussinesq approximation also requires small relative mass density differences over the region of interest. These conditions must hold for the resulting disturbance, including any amplification by resonance.

The inviscid Linearized Boussinesq equations and linear kinematic boundary condition areWrite the velocity field as $(u,w)=\mathrm{Re}\{(U(z),W(z))e^{i(kx-\omega t)}\}$. Incompressibility and horizontal momentum balance giveThe vertical momentum balance therefore yieldsFor a real vertical wavenumber $m$, this gives the dispersion relationWhen $0<\omega <N$, define $\ell =k\sqrt{N^2/{\omega }^2-1}$. The radiation condition selects $m=-\ell$, because energy must leave the boundary upward. The boundary-forced internal gravity wave isIts wavevector is $\mathbf{K}=(k,m)$, its phase velocity is ${\mathbf{c}}_p=\omega \mathbf{K}/(k^2+m^2)$, and its group velocity isThus the phase velocity points down and right, while the group velocity points up and right. They are perpendicular. Internal-wave polarization makes the particle motion an oscillation along the group velocity direction, with zero first-order mean transport. The constant-phase lines of an internal gravity wave are also parallel to the group velocity. Their inclination $\vartheta$ above the horizontal satisfies $\sin \vartheta =\omega /N$.

When $\omega >N$, set $\kappa =k\sqrt{1-N^2/{\omega }^2}$. Boundedness at infinity selects the evanescent waveThe horizontal and vertical velocity components are in quadrature: fluid particles describe small ellipses, and the response decays over ${\kappa }^{-1}$. There is no upward time-averaged energy flux, because $P$ and $W$ are in quadrature. A real vertical group velocity is not defined for this evanescent wave. The pattern travels horizontally with phase velocity $\omega /k$.

For $N=0$ the response is the evanescent wave with $\kappa =k$; the ratios $\omega /N$ should not be used. At the cutoff $\omega =N>0$, the bounded harmonic solution has $W=-i\omega {\eta }_0$, $U=0$: it neither decays nor has nonzero upward group velocity. It is the limiting cutoff response, rather than a localized radiating disturbance. A uniform nonzero buoyancy frequency in an infinitely deep Boussinesq approximation is itself a local idealization of the background mass density.

The arrows distinguish the phase velocity, group velocity, and oscillatory particle motion of the internal gravity wave. The right panel shows the decay envelope and particle ellipses of the evanescent wave.

Choose $a=1$. The power-transformed Taylor–Goldstein energy identity reduces toThe right side is real. For an unstable mode, $c_i>0$, so taking the imaginary part givesConsequently the phase velocity is a weighted average:The weights are nonnegative and not identically zero. For a nonconstant shear profile, equality at an extremum would force $q$ to vanish on an interval where $U$ differs from that extremum; uniqueness for the regular ordinary differential equation would then give the zero mode. This is the phase-speed bound for unstable stratified shear modes.

For large $\alpha$, the coupling $e^{-kh}=e^{-2\alpha }$ is exponentially weak. If each interface were isolated, its interfacial gravity waves would have dimensional phase velocities $U(z_{\pm })\pm \sqrt{d}$, orThe upper-interface wave travelling against its positive local current has speed $1-\sqrt{J/(2\alpha )}$. The lower-interface wave travelling against its negative current has speed $-1+\sqrt{J/(2\alpha )}$. Their laboratory speeds match at zero when $J=2\alpha$. Weak interaction then produces counterpropagating wave instability rather than two independent travelling waves.

The band has the large-wavenumber form $J=2\alpha \pm 2\alpha e^{-2\alpha }+O(\alpha e^{-4\alpha })$. Exactly at resonance, $\mathrm{Im}\tilde{c}\sim e^{-2\alpha }/2$, soThus the instability window and growth rate both become exponentially small as the interfaces become weakly coupled.

Take $f_0>0$ for definiteness and define $K=(k^2+l^2{)}^{1/2}>0$. Thermal-wind balance gives $U_z=-b_y/f_0=\Lambda$, and the lower boundary condition fixes $U=\Lambda z$. The basic interior quasi-geostrophic potential vorticity has no gradient, so a boundary-supported normal mode ${\psi }^{\prime }=\mathrm{Re}[\varphi (z)e^{i(kx+ly-\omega t)}]$ has zero interior potential-vorticity perturbation. Its vertical equation isDecay upward selects $\varphi =A{e}^{-{\mu }_{0}z}$.

The buoyancy perturbation is $b^{\prime }=f_0{\psi }_z^{\prime }$. At a rigid boundary $w^{\prime }=0$, linearized material conservation of buoyancy givesAt $z=0$, this is $c{\varphi }^{\prime }(0)+\Lambda \varphi (0)=0$, where $c=\omega /k$ for $k\ne 0$. The Eady edge wave therefore hasIt is an exponentially trapped boundary-buoyancy wave, not a vertically propagating interior mode. The formula for $\omega$ extends to $k=0$, where the perturbation is stationary.

For the two-region problem, absence of a density jump at the common material interface requires continuity of buoyancy:Thus its basic slope is $a=h_y=f_0({\Lambda }_1-{\Lambda }_2)/(N_1^2-N_2^2)$. For a nondegenerate stratification contrast, its variation over horizontal scale $L$ is $aL/D=O(\mathrm{Ro}/\mathrm{Bu})$, which is small in the usual quasi-geostrophic approximation with Burgers number of order one. The matching conditions can therefore be applied at the constant reference level $z=0$ to leading order. An exceptionally small stratification contrast would require checking this flat-interface ordering separately.

Pressure and normal velocity must be continuous because there is no membrane or singular interfacial force. Since base buoyancy is continuous, expanding pressure continuity at the displaced interface gives simply ${\varphi }_1(0)={\varphi }_2(0)$. The base horizontal velocity is also continuous there; call it $U_I$. Decay away from the interface then givesLinearized buoyancy conservation in each region givesMatching normal velocity gives $w_1^{\prime }=w_2^{\prime }$: the common slope correction to normal velocity cancels because pressure continuity also makes the horizontal velocity perturbation continuous. HenceThe quasi-geostrophic wave at a stratification interface has intrinsic phase velocityAs $N_1/N_2\to \infty$ with $N_2$ and the shears bounded, the lower region becomes effectively rigid andrecovering the upper-fluid Eady edge wave. The matching calculation describes regular interface-supported modes; arbitrary interior potential-vorticity initial disturbances can additionally be advected by the shear.

For the long filament, take the time-periodic solution of the dimensionless bending equation, after transients have decayed. Write $y=\mathrm{Re}[Y(x)e^{it}]$. Then $Y^{(4)}=-iY$, so the spatial exponents satisfy ${\lambda }^4=-i$. Define $C=\cos (\pi /8)$ and $S=\sin (\pi /8)$. The two roots with negative real part are ${\lambda }_1=-C+iS$ and ${\lambda }_2=-S-iC$; the other roots violate decay at infinity.

Hence $Y=A{e}^{{\lambda }_{1}x}+B{e}^{{\lambda }_{2}x}$. The driven displacement gives $A+B=y_0$, while the zero bending moment gives ${\lambda }_1^{2}A+{\lambda }_2^{2}B=0$. Since ${\lambda }_2^2=-{\lambda }_1^2$, $A=B=y_0/2$. The oscillatory bending of a moment-free semi-infinite filament isHolding each wave phase constant gives its dimensionless phase velocity:The first travels toward the actuator and attenuates over length $1/C$; the second travels away and attenuates over the longer length $1/S$. These are spatially damped phase patterns in an overdamped bending equation, not two undamped inertial beam waves. Their combination satisfies both displacement and zero-moment conditions at the driven end.

The linear dispersion relation for a mode $e^{i(kx-\omega t)}$ is $5{\omega }^2=k^4+4$. The proposed modes share phase velocity $c$, so $5c^2{k}_j^2=k_j^4+4$. A quadratic wave interaction generates the second harmonic and the difference harmonic. With only two positive wavenumbers, phase matching for a quadratic wave interaction requires $k_2=2k_1$. Equating their phase velocities givesand thereforeThis is a two-to-one resonance of dispersive waves. Its amplitude changes accumulate on $t=O({ϵ}^{-1})$. Choose the slow time $T=ϵt/5$; this harmless constant rescaling makes the later amplitude formulas simple. With $\xi =x-t$, the leading real field is $A_0{e}^{i\xi }+B_0{e}^{2i\xi }+{\overline{A}}_0{e}^{-i\xi }+{\overline{B}}_0{e}^{-2i\xi }$.

For the fundamental, the order-$ϵ$ cross derivative in $5{\psi }_{tt}$ is $-2i(A_0{)}_T{e}^{i\xi }$, while the fundamental coefficient in ${\psi }_0{\psi }_{0x}$ is $i{\overline{A}}_0{B}_0$. For the second harmonic the corresponding terms are $-4i(B_0{)}_T{e}^{2i\xi }$ and $i{A}_0^2$. The solvability condition in the method of multiple scales removes the resonant forcing and givesThe nonresonant third and fourth harmonics enter the correction. They do not change these leading amplitude equations.

Write $A_0=R{e}^{i\theta }$, $B_0=P{e}^{i\varphi }$ and $\delta =\varphi -2\theta$. Taking real and imaginary parts gives the explosive two-to-one amplitude system in polar form:The polar phases are used where their corresponding amplitudes are nonzero. These equations implyFor the second identity, differentiate using ${\delta }_T=(R^2/(4P)+P)\sin \delta$: the terms proportional to $\sin \delta \cos \delta$ cancel. Thereforeare first integrals, and in particular

For constant phases with $\varphi =2\theta +\pi$, these real equations reduce to $R_T=RP/2$ and $P_T=R^2/4$. The initial data give $R^2-2P^2=8$, so the Riccati equation for $P$ is $2P_T=P^2+4$. Integrating and using $P(0)=2$ givesThe tangent and secant function have a pole at $T=\pi /4$, corresponding to $t=5\pi /(4ϵ)$. Both modes grow through their locked resonant interaction. This is formal finite-time blowup of the reduced amplitude equations; the weakly nonlinear expansion loses validity as the amplitudes become large, so it cannot establish a singularity of the full partial differential equation.

The cutoff occurs at $k=0$:For $k>0$, the phase velocity and group velocity areTheir product is $c_{p,n}{c}_{g,n}=c_S^2$. On a plot against positive $k$, $c_{p,n}$ decreases from $+\infty$ to $c_S$, while $c_{g,n}$ increases from zero to $c_S$; larger mode number shifts both transitions to larger wavenumber.

The unstable branch has temporal growth rate and phase velocityFor short waves, $|k|h\gg 1$, soFor long waves, $|k|h\ll 1$, soThus the growth-rate curve starts at the origin with zero slope, rises like $|k{|}^{3/2}$, and approaches the straight-line asymptote $|k|U/2$. The inviscid zero-thickness model consequently has no short-wave cutoff.

Without self-gravity and for $k>0$,The phase velocity and group velocity areso

With self-gravity restored,The product changes sign atFor $k>k_{\mathrm{crit}}$, crests and a localized wave packet travel in the same radial direction. For $k<k_{\mathrm{crit}}$, the group velocity and phase velocity have opposite signs: the envelope and its wave energy propagate opposite to the individual crests. At $k_{\mathrm{crit}}$ the group velocity vanishes and neighboring wave components cause the packet to spread.

At $O(ϵ\gamma )$, the slowly varying wave envelope obeyswhere the group velocity obtained from $\omega =Nk/(k^2+m^2{)}^{1/2}$ isIt satisfies $\mathbf{k}\cdot {\mathbf{c}}_g=0$. If initially $\mathbf{k}\cdot \hat{\nabla }\tilde{\psi }=0$, the envelope structure varies across the wavevector and translates unchanged in the direction ${\mathbf{c}}_g$ at speedThus internal-wave crests propagate with phase velocity parallel to $\mathbf{k}$, while the packet and its energy propagate perpendicular to $\mathbf{k}$ with the group velocity.

For ${\mathbf{u}}_0=-3\Omega x{\mathbf{e}}_y/2$, advection vanishes and the Coriolis acceleration exactly cancels $3{\Omega }^{2}x{\mathbf{e}}_x$, so constant $P_0$ and ${\theta }_0=0$ complete the equilibrium.

Write perturbations $(\mathbf{v},p,\vartheta )e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$ with $\mathbf{k}=(k_x,0,k_z)$. Incompressibility gives $\mathbf{k}\cdot \mathbf{v}=0$. It also makes the quadratic terms $\mathbf{v}\cdot \nabla \mathbf{v}=i(\mathbf{k}\cdot \mathbf{v})\mathbf{v}$ and $\mathbf{v}\cdot \nabla \vartheta$ vanish exactly; the background-advection terms vanish because $k_y=0$. The amplitude equations areEliminating the amplitudes yields the inertia-gravity wave dispersion relation

For $N^2=0$, $\omega =\pm \Omega k_z/k$, so the group velocity isIt is perpendicular to $\mathbf{k}$ and hence to the phase velocity. An inertial wave packet transports energy along beams lying in its phase surfaces.

If $N^2>0$ and $k_x/k_z\to 0$, incompressibility suppresses vertical motion and $\omega \to \Omega$: horizontal epicyclic motion and rotation dominate. If $k_z/k_x\to 0$, radial motion is suppressed and $\omega \to N$: vertical buoyancy oscillations dominate. Intermediate ratios give hybrid inertia-gravity waves.

If $N^2<0$, exponential growth occurs exactly whenOnly modes with sufficiently large radial wavenumber permit enough vertical displacement for unstable buoyancy to overcome rotational restoration; the other orientations remain stabilized by the Coriolis force.

On the forcing $e^{-i\omega t}$,The other three factors have roots with real decay rate $-1/\tau$, so a real-frequency gas wave can resonate only with the undamped factor $\mathcal{D}$. The resonance condition is thereforeAt resonance, $\mathcal{D}$ annihilates the forcing, so a particular solution acquires one power of time:This is a streaming instability resonance: the gas wave's phase velocity along $\mathbf{k}$ matches the dust drift projected along the same wavevector.

Write the base-state gradients asThe base heat flux and Stefan condition relations are

Under the quasi-stationary approximation, and with advective heat transport neglected, each temperature perturbation satisfies the Laplace equation. The decaying normal-mode solutions areKeeping the displaced ice interface at $T_m$ omits the curvature correction described by the Gibbs--Thomson relation.
The fixed melting temperature at the displaced ice interface givesExpanding temperature and conductive heat flux continuity at the displaced outer interface gives

Put $r=k_a/k_w$. Since $G_a=G_w/r$, the temperature condition in the limits $r\ll 1$ and $\alpha h\ll 1$ givesAlthough $r$ is small, the product $rC$ is therefore order one and cannot be discarded. The flux condition then givesThis is why the stated asymptotic warning matters.

The perturbed Stefan condition at $x=0$ isUsing $B\sim {\eta }_2{G}_w$ and $-k_w{G}_w={\rho }_i{L}_{f}V$ yieldsSubstitution of the interface-amplitude ratio from part i produces the thin-film icicle-ripple instability dispersion relation

Set $b=\Gamma h{\alpha }^3$. Rationalizing this complex number givesso its growth rate and imaginary part areDifferentiating the growth rate with respect to the wavenumber shows that its positive stationary point satisfiesIt is the unique global maximum, and henceAt this wavenumber $b=1/\sqrt{5}$. With the convention $e^{i\alpha z+\sigma t}$, a constant phase travels with phase velocityThereforeBecause $z$ increases downward, the negative sign means that the icicle ripples migrate upward with speed $\sqrt{5}V/6$.

A localized monochromatic cylinder emits four narrow beams, one in each quadrant, forming a St Andrew's cross. If $\alpha$ is the beam angle to the horizontal,equivalently, its angle $\theta =\pi /2-\alpha$ to the vertical obeys $\cos \theta =\omega /N$. As $\omega$ rises from zero to $N$, the beams rotate from horizontal toward vertical. For $\omega >N$ no freely propagating internal wave exists and the response is evanescent.

The wavevector and phase velocity are parallel. The dispersion relation is homogeneous of degree zero in $(k,m)$, so Euler's theorem gives $\mathbf{k}\cdot {\mathbf{c}}_g=0$: the group velocity is perpendicular to both the wavevector and phase velocity. Energy travels along the beams in the group-velocity direction.

If a differentiable dispersion relation $\omega (\mathbf{k})$ is homogeneous of degree zero at nonzero $\mathbf{k}$, the Euler homogeneous function theorem gives $\mathbf{k}\cdot {\nabla }_{\mathbf{k}}\omega =0$. Therefore phase velocity $\omega \mathbf{k}/k^2$ is orthogonal to group velocity ${\nabla }_{\mathbf{k}}\omega$. The result applies exactly to limiting inertial waves; finite compressibility can alter the dispersion.

A resonant interaction requires matching both wavenumber and angular frequency: $k_3=k_1+k_2$ and ${\omega }_3={\omega }_1+{\omega }_2$ for a sum interaction, with corresponding sign changes for differences. A common phase velocity alone does not guarantee that a generated harmonic lies on the dispersion relation.

The magnitude of the phase velocity of a real-frequency plane Fourier mode is $|\omega |/|\mathbf{k}|$. In an anisotropic medium it depends on the direction of the wave vector and need not equal the group velocity magnitude.