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The long step-trapped topographic Rossby wave has equal phase velocity and group velocity, $c_T=fd{R}_D/H_0$. For an outer surface profile $\eta ={\eta }_0\mathrm{sgn}(x)-A(x,t)e^{-|y|/R_D}$, cross-step matching gives $A_t+c_T{A}_x=2c_T{\eta }_0\delta (x)$. With $A(x,0)=0$, its weak solution is $A={\eta }_0[\mathrm{sgn}(x)-\mathrm{sgn}(x-c_{T}t)]$. This describes the slowly varying outer response; the fast adjustment region at the origin is unresolved.

A nondispersive wave has constant phase velocity on a propagation branch. For $\omega =ck$, its group velocity is also $c$. A frequency relation with a nonzero offset, such as $\omega =ck+b$, has constant group velocity but a phase velocity $c+b/k$ depending on wavenumber; constant group velocity alone is therefore not sufficient for nondispersion.

With no forcing, the linear dispersion relation has positive phase velocity $\sigma k^2/(\beta {\alpha }_1)$ and group velocity three times this. A decaying permanent wave with $\xi =x+ct$ instead obeys $\sigma (A^{\prime }{)}^2=A^2(\beta {\alpha }_{1}c+{\alpha }_{2}A/3)$. For positive coefficients a nonzero homoclinic wave requires $c>0$ and is negative, travelling upstream. Its inverse width is $k_s=\frac{1}{2}\sqrt{\beta {\alpha }_{1}c/\sigma }$. The opposite directions reflect the nonlinear amplitude-dependent propagation speed.

The lower-layer dispersion relation givesThe incident phase is $e^{-i{m}_{0}z}$; its vertical group velocity is positive, even though its vertical phase velocity is negative. Write the lower field as $W=w_0{e}^{-i{m}_{0}z}+r{e}^{i{m}_{0}z}$ and the upper bounded field as $W=T{e}^{-kz}$. Interface continuity gives $T=w_0+r$ andThus the internal-wave response of a sharp buoyancy interface isAt fixed incident $k,\omega ,N_0$ this is largest at $S=1$, where $|T|=2|w_0|$. Consequently ${\omega }^2=\Delta bk=2{\omega }_{\mathrm{free}}^2$, or $\omega =\sqrt{2}{\omega }_{\mathrm{free}}$. The maximum does not coincide with the free frequency for two homogeneous layers: the stratified lower radiating layer has a different dynamic pressure response. The upper field is evanescent, so in this ideal lossless problem it carries no mean outgoing vertical energy flux; the reflected wave has $|r|=|w_0|$.

Use the usual minor and major symmetries of the elastic stiffness tensor, positive density, and strong ellipticity in elasticity. For the plane elastic wave $u=a{e}^{ik(n\cdot x-vt)}$, substitution into momentum balance gives the acoustic tensor problemThe major symmetry makes $\Gamma$ a real symmetric matrix. Moreover $a^{\mathsf{T}}\Gamma a=c_{ijpq}{a}_i{n}_j{a}_p{n}_q>0$ for nonzero $a$ under strong ellipticity in elasticity. Hence its three eigenvalues are positive and the three positive wave speeds are real. Symmetry alone proves only reality of $v^2$: an unstable isotropic tensor with negative shear modulus has negative transverse $v^2$. The requested reality therefore uses stability as part of the physical elastic-solid assumption.

The elastic slowness surface consists of three sheets, with $s=n/v_{\alpha }(n)$, satisfyingIt is centrosymmetric. Sheets can meet at degenerate polarizations; they need not be spherical or convex in an anisotropic solid. The group velocity is normal to the corresponding smooth sheet: for a sheet tangent $ds$, differentiation of $a^{\mathsf{T}}[c_{ijpq}{s}_j{s}_q-\rho I]a=0$ gives $a_i{c}_{ijpq}{a}_p{s}_{q}d{s}_j=0$. The coefficient vector is proportional to the group velocity, so its scalar product with every tangent vanishes.

Fix horizontal slowness $p=(n_1/c,n_2/c)$ and put $q=s_3$. DefineThe symbol $p$ on a tensor index is an index, whereas $p_{\alpha }$ is a horizontal slowness component. The vertical slowness equation is the vertical slowness sextic of an anisotropic solidThe leading coefficient is $\det T>0$, so it is a degree-six polynomial with real coefficients. Nonreal roots occur in conjugate pairs, leaving between zero and six real roots counted with multiplicity. Generically the counts are $0,2,4,6$; a grazing repeated root can give an odd count of distinct real roots. Geometrically these are the intersections of a vertical line of fixed horizontal slowness with the three sheets.

All four generic counts are actually possible in a stable material. For example, take $\rho =1$ and orthotropic Voigt stiffnessesThe stiffness matrix is positive definite. For horizontal slowness $(p,0)$, the three diagonal acoustic equations areThe choices $p=0.1,0.3,0.8,1.2$ give respectively six, four, two and zero real vertical roots.

For the general depth-dependent amplitude, substitution of the horizontal harmonic factor yieldsFor six simple roots $q_{\ell }$, choose nonzero null vectors $a_{\ell }$ of the corresponding matrix. The complete solution isComplex vertical roots give evanescent waves. At defective repeated roots the usual polynomial-in-depth factors multiplying exponentials must be included, or the solution can be defined by a limiting procedure.

To classify the modes, take $x_3$ positive downwards and $\omega >0$. The reduced traction for a mode isMechanical energy crosses a plane at rate $-{\sigma }_{i3}{u}_{i,t}$. Averaging the real harmonic fields uses $⟨\mathrm{Re}(f{e}^{-i\omega t})\mathrm{Re}(g{e}^{-i\omega t})⟩=\mathrm{Re}(f{g}^{\ast })/2$. With velocity amplitude $-i\omega a_{\ell }$ and traction amplitude $i\omega b_{\ell }$, the time-averaged vertical acoustic energy flux isFor a real propagating mode, write its physical wavevector as $k=\omega (p_1,p_2,q)$. Differentiate $a^{\mathsf{T}}\Gamma (k)a=\rho \omega (k{)}^2{a}^{\mathsf{T}}a$ in $k_3$. The derivatives of $a$ cancel by the eigenvalue equation, while $a^{\mathsf{T}}({\partial }_{k_3}\Gamma )a=2\omega a^{\mathsf{T}}(Tq+R)a$. Thus its vertical group velocity and flux areThus the elastic mode classification by vertical energy flux is upgoing if $J_3<0$, downgoing if $J_3>0$. In anisotropy, the sign of vertical phase slowness alone need not classify energy propagation. For evanescent modes, select $\mathrm{Im}q>0$ for decay into a downward half-space and $\mathrm{Im}q<0$ for decay into an upward half-space. Away from grazing and degeneracy, continuation from zero horizontal slowness, with a small radiation damping if needed, separates three outgoing modes on each side. Superpose only the appropriate three to construct the most general upgoing or downgoing field; limiting modes handle exceptional cases.

For a free surface at $x_3=0$ with solid occupying $x_3>0$, let $(q_0,a_0)$ be a unit-amplitude incident upgoing mode. The reflected field consists of the three downgoing propagating or depth-decaying modes:Frequency and horizontal slowness are common to all modes, by translation invariance of the surface. Zero surface traction imposes the three equationsHence the free-surface anisotropic reflection amplitudes arewhenever $B$ is invertible. This specifies mode conversion as well as reflection. If $B$ is singular, a nonzero outgoing homogeneous surface mode exists; these are surface-wave poles, including Rayleigh waves in the isotropic case. At such parameters one must use the appropriate limiting radiation solution and check solvability instead of assuming a unique finite inverse.

Let $H=h_{00}+\zeta$ and write the horizontal shallow water equations on the constant f-plane asKeeping only terms linear in the disturbance about rest gives the linearized shallow water equationsDefine the horizontal divergence $\delta =u_x+v_y$ and relative vorticity $q=v_x-u_y$. Taking respectively the divergence and vertical curl of the momentum equation yieldsTogether with ${\zeta }_t=-h_{00}\delta$, this provesThe nonlinear shallow-water potential vorticity is $Q=(f+q)/H$. The nonlinear vertical vorticity equation gives $D(f+q)/Dt=-(f+q)\delta$, while mass conservation gives $DH/Dt=-H\delta$; their ratio is therefore materially conserved. Expanding about rest,Advection of its constant background contributes nothing at first order, explaining why the displayed linear anomaly is conserved at each fixed position.

A further time derivative eliminates $q$ and $\zeta$ from the divergence equation:For a plane wave with horizontal wavevector $(k,l)$, write $K=\sqrt{k^2+l^2}$. More directly, substituting the plane wave into the three linear equations gives a coefficient matrix with determinant proportional to $\omega [{\omega }^2-f^2-c_0^2{K}^2]$. Thus the full linear rotating shallow-water dispersion relation has one balanced branch and two wave branches:The latter are inertia-gravity waves. The positive branch starts at $f$ with zero slope and tends to the straight line $c_{0}K$; the negative branch is its reflection. The requested sketch, including the stationary branch, is

For the positive-frequency branch, the magnitudes of the phase velocity and group velocity areThe group velocity vector is $c_0^2(k,l)/\omega$; the negative branch reverses its direction. Near $\omega =f$, namely $K{R}_D\ll 1$ with the Rossby deformation radius $R_D=c_0/f$, $c_p\sim f/K$ is large while $c_g\sim c_0^{2}K/f$ is small. The $K=0$ limit is an inertial oscillation, for which a spatial phase speed is not defined. Far above $f$, both speeds approach $c_0$, the shallow water gravity-wave speed, with $c_p>c_0$ and $c_g<c_0$ at finite $K$.

On the zero-frequency branch, geostrophic balance givesFor a nonconstant plane wave, $\hat{q}=-(g{K}^2/f)\hat{\zeta }$: relative vorticity and elevation have opposite signs. They can both be nonzero and spatially structured, whereas the horizontal divergence vanishes identically. A constant elevation is also a zero-frequency solution, with zero velocity and zero relative vorticity.

For the initially resting ridge, set ${\zeta }_0(x)=ϵh_{00}{e}^{-x^2/L^2}$ and assume an unbounded domain, uniform in $y$. Initial potential vorticity fixesIn the balanced remainder, insert $q_s=(g/f){\zeta }_s^{\prime \prime }$ to obtainThe decaying Green's function of this modified Helmholtz equation gives the Geostrophic adjustment of a Gaussian height ridge explicitly:The initial outward pressure force creates a cross-ridge current; the Coriolis acceleration turns that current and builds the along-ridge geostrophic flow. The remainder is radiated inertia-gravity waves, not a dissipative disappearance of energy.

For precision, use the Fourier transform $\hat{\zeta }(k)=\int \zeta (x)e^{-ikx}dx$. Initial rest implies ${\zeta }_t(x,0)=0$, and the conserved anomaly givesConsequentlyThese formulas satisfy both initial conditions. They also show that the waves carry no linear potential vorticity anomaly: $q=(f/h_{00})(\zeta -{\zeta }_0)$ and $\delta =-{\zeta }_t/h_{00}$ reconstruct the remaining fields, with $u$ and $v$ obtained from their derivatives. On the infinite line, dispersive wave packets propagate away and decay locally, leaving the balanced ridge and its oppositely directed flanking currents. Conservation of energy still holds globally. In a reflecting finite basin, standing waves may persist, so a permanent pointwise relaxation is not guaranteed without dissipation.

If $L\gg R_D$, most of the initial elevation remains balanced: ${\zeta }_s\simeq {\zeta }_0$ with small smoothing corrections. If $L\ll R_D$, the balanced ridge spreads to width $R_D$, and away from the original narrow coreThus the final central height is much smaller than the initial one and the wave component is substantial. The integral of the balanced elevation equals the initial integral, since the Green's function has unit integral; the oscillatory component has zero net added volume.

Let $p$ be kinematic pressure, with the resting hydrostatic part subtracted, and let $\sigma$ be the buoyancy perturbation about background buoyancy $N^{2}z$. The nonrotating inviscid Boussinesq equations, with no buoyancy diffusion, areHere the third equation is mass conservation for incompressible flow, and the last expresses buoyancy conservation following a parcel. Linearization about rest removes the products of disturbances. For a plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, the resulting Linearized Boussinesq equations becomeWith $K=(k^2+m^2{)}^{1/2}$ and $k\ne 0$, the first and third equations give $\hat{p}=-m\omega \hat{w}/k^2$. Substitution in vertical momentum and buoyancy gives the dispersion relationSince the velocity is the time derivative of the displacement field, $\hat{\mathbf{u}}=-i\omega \hat{\mathbf{\xi }}$. Thus the complete buoyancy polarization of a plane internal gravity wave isBoth velocity and displacement are perpendicular to the wavevector; pressure and velocity are in quadrature with buoyancy and displacement.

For upward/rightward phase velocity, choose $k,m>0$ and $\omega =Nk/K$. Differentiating this dispersion relation gives the group velocity, while the vector phase velocity points along the wavevector:Their dot product is zero and ${\mathbf{c}}_p+{\mathbf{c}}_g=(N/K,0)$. Draw ${\mathbf{c}}_g$ from the tip of ${\mathbf{c}}_p$: the right-triangle geometry of internal-wave velocities has horizontal hypotenuse $N/K$. This hypotenuse is distinct from the horizontal phase velocity $\omega /k$.

At an instant with real complex amplitude $\hat{\sigma }=A$, write $\varphi =kx+mz-\omega t$. The spatial fields areThe constant-phase lines of an internal gravity wave slope down to the right, with $dz/dx=-k/m$. Buoyancy maxima lie a quarter wavelength from pressure maxima; the oscillating displacement and velocity run along the phase lines. The group velocity points down and right along them, despite upward/rightward motion of the wave phase.

Let $P=\overline{{\xi }_{x}u+{\zeta }_{x}w}$, using horizontal averaging over a wavelength. This is the displacement pseudomomentum of an internal gravity wave, whose sign convention is opposite to $k\overline{E}/\omega$. For complex coefficients, $\overline{ab}=\frac{1}{2}\mathrm{Re}(\hat{a}{\hat{b}}^{\ast })$. The above wave polarization consequently givesSince $c_{gz}=-\omega m/K^2$, these obey $\overline{{\zeta }_{x}p}=c_{gz}P$. With weak kinematic viscosity, each leading monochromatic velocity component satisfies ${\nabla }^{2}u=-K^{2}u$, ${\nabla }^{2}w=-K^{2}w$. The dissipative term is therefore $-\nu K^{2}P$, so the supplied averaged balance reduces toFor constant $k,m,N$ and a steady wave envelope, $c_{gz}{P}_z=-\nu K^{2}P$. As $P$ is proportional to squared wave amplitude, viscous attenuation of an internal gravity wave givesHere $c_{gz}<0$, so amplitude decreases below a source at $z_s$: the source is above the receiving level. The same conclusion follows directly because downward group velocity carries energy from the source downward. Upward phase velocity does not locate the source.

For a slowly varying mean shear flow $U(z)$, the local intrinsic frequency is $\hat{\omega }=\omega -kU(z)$. A local WKB approximation uses ${\hat{\omega }}^2=N^2{k}^2/(k^2+m(z{)}^2)$, with varying vertical wavenumber and group velocity. The appropriate transported quantity is wave action, or horizontal wave pseudomomentum $kE/\hat{\omega }$; its vertical flux has viscous loss, and the changing polarization factors must be retained. In particular, $P/|\hat{\sigma }{|}^2$ is no longer constant.

Near a critical level of an internal gravity wave, $\hat{\omega }\to 0$, so $|m|\sim N|k|/|\hat{\omega }|$ and $|c_{gz}|\sim {\hat{\omega }}^2/(N|k|)$. The damping per vertical distance, of order $\nu K^2/|c_{gz}|$, grows rapidly. The waves are strongly attenuated over a narrow altitude range and transfer momentum to the mean flow there or before reaching the critical level. The inviscid WKB approximation eventually fails; viscosity and possibly internal-wave breaking regulate the small scales.

For a homogeneous lossless elastic solid, insert the plane wave $u_i=\mathrm{Re}\{U{a}_i{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}\}$ into the equations of linear elasticity. Put $\mathbf{k}=\kappa \mathbf{n}$, $|\mathbf{n}|=1$, and $v=\omega /\kappa$. The resulting eigenvalue problem isThe minor and major symmetries make this acoustic tensor a real symmetric matrix. For a stable material it is positive-definite, so its three positive eigenvalues give three real positive phase velocities and therefore three slownesses $s_a=1/v_a$. Coincident eigenvalues are possible, so there need not be three distinct speeds. For distinct eigenvalues the displacement polarization are orthogonal; in an anisotropic solid they are generally quasi-longitudinal and quasi-transverse rather than exactly longitudinal and transverse. Stability, or at least strong ellipticity in elasticity, is required for the assertion about positive real speeds.

The elastic slowness surface is the locus of the three vectors $\mathbf{p}=\mathbf{n}/v_a(\mathbf{n})=\mathbf{k}/\omega$. Its equation isFor each wavefront normal there are three radial sheets, which may meet at degeneracies. To obtain the energy flux directly, multiply the elastic equation by ${\dot{u}}_i$. Forthe symmetry of stress gives ${\partial }_{t}W={\partial }_j({\sigma }_{ij}{\dot{u}}_i)$. Thus the instantaneous elastic-wave energy flux is $F_j=-{\sigma }_{ij}{\dot{u}}_i$. With a real branch polarization $\mathbf{a}$, harmonic averaging givesNow differentiate $c_{ijkl}{p}_j{p}_l{a}_k=\rho a_i$ along a regular slowness surface sheet and contract with $a_i$. The terms containing $d\mathbf{a}$ cancel by symmetry and the original eigenvalue equation, leavingEvery tangent $d\mathbf{p}$ is consequently perpendicular to $⟨\mathbf{F}⟩$: the energy flux is normal to the slowness surface. At a degenerate point this statement is made on the limiting branches, rather than assuming a unique normal.

For completeness, equality of the mean kinetic and strain energies gives $⟨W⟩=\rho {\omega }^2|U{|}^2|\mathbf{a}{|}^2/2$. Differentiating the acoustic tensor equation in the wavevector variables similarly givesThus the group velocity is the elastic energy velocity. The homogeneous elastic dispersion relation is degree one in $\mathbf{k}$, so differentiating $\omega (\lambda \mathbf{k})=\lambda \omega (\mathbf{k})$ at $\lambda =1$ gives $\mathbf{p}\cdot \mathbf{g}=1$.

The elastic wave surface is the unit-time locus of these group velocities. From an impulsive point source, the propagating wavefronts at time $t$ are its sheets scaled to $\mathbf{x}=t\mathbf{g}$. A particular source may fail to excite a mode in some directions, and its amplitudes require the source polarization and geometrical spreading; the wave surface determines the arrival geometry. It does not specify all near-field motion between the radiative fronts. Geometrically it is the envelope of the planes $\mathbf{p}\cdot \mathbf{x}=1$: tangent variation gives $d\mathbf{p}\cdot \mathbf{x}=0$, so $\mathbf{x}$ is the appropriately normalized normal to the slowness surface.

In a symmetry-plane cross-section write $\mathbf{p}=r(\theta )\mathbf{n}$, $\mathbf{n}=(\cos \theta ,\sin \theta )$ and $\mathbf{t}=(-\sin \theta ,\cos \theta )$. The two envelope conditions giveThe numerator is the signed curvature numerator of the slowness surface cross-section. At a simple inflection point, $h=v+v^{\prime \prime }=0$ and $h^{\prime }\ne 0$. Then ${\mathbf{g}}^{\prime \prime }=h^{\prime }\mathbf{t}$ and ${\mathbf{g}}^{\prime \prime \prime }=h^{\prime \prime }\mathbf{t}+2h^{\prime }{\mathbf{t}}^{\prime }$ are linearly independent. These are precisely the local conditions for an ordinary cusp. The slowness curvature and wave-surface cusps relation explains why a smooth nonconvex slowness surface can generate a cusped wavefront and multiple ray arrivals.

The example below uses a positive orthotropic stiffness in normalized units: $\rho =1$, $C_{11}=C_{22}=C_{33}=4$, $C_{44}=C_{55}=1.2$, $C_{66}=1$, and all normal off-diagonal stiffnesses zero. In the symmetry plane $n_3=0$, its two in-plane acoustic tensor eigenvalues are $v_{\pm }^2=(5\pm \sqrt{5+4\cos 4\theta })/2$, and its out-of-plane shear speed is $\sqrt{1.2}$. The in-plane shear sheet is nonconvex and produces the marked cusps; the construction uses an actual stable elastic tensor.

Near a spatially extended Hopf bifurcation, a critical oscillatory normal mode has a slowly varying complex amplitude. For a selected travelling-wave branch, a method of multiple scales expansion takes the formHere the distance from onset is of order ${\epsilon }^2$, $v_0$ is the critical eigenvector, and $v_g$ is the group velocity. The comoving coordinate removes the first spatial derivative of the envelope. Projection at the first resonant order gives $A_T=\rho A+D_0{A}_{XX}-G_0|A{|}^{2}A$. Spatial translation and the temporal phase of the Hopf oscillation give a constant-phase symmetry $A\mapsto e^{i\theta }A$, which permits the cubic term $|A{|}^{2}A$ but forbids a generic $A^2$ term. The coefficients are complex because the envelope has both growth and frequency detuning. For a supercritical branch with positive real diffusion, real rescalings and a uniform phase rotation give the complex Ginzburg–Landau equationA subcritical Hopf branch requires higher saturation terms instead. If spatial reflection makes opposite travelling waves critical together, two coupled amplitude equations are required initially; the scalar equation applies to a selected single-wave branch when its competing amplitude is stable.

Distant boundaries can still determine which travelling pattern is observed. They can select its phase and wavenumber, reflect a travelling disturbance, inject a competing wave, or let a growing wave packet leave the domain. In particular, amplification in a comoving frame can be convective wave-packet instability rather than absolute wave-packet instability at a fixed laboratory point. A finite-domain outcome then need not reflect intrinsic bulk modulation dynamics. Periodic boundary conditions are the appropriate idealization when studying bulk instabilities without end selection or reflected-wave forcing: they eliminate physical end layers and make the allowed envelope Fourier modes discrete. They do not assert that real distant boundaries are always negligible. Choose a periodic length compatible with the carrier, fix the phase winding, and compare perturbations within that same periodic problem. On an open physical domain the group velocity and actual boundary conditions must be restored.

The optimum carrier wavenumber corresponds to the spatially uniform envelope $A_f=e^{-icT}$, often called a flat state. Its phase is arbitrary. Write $A=e^{-icT}(1+r)e^{i\varphi }$ with real small amplitude and phase disturbances. To first order,A Fourier mode $e^{\lambda T+ikX}$ therefore has matrix and dispersion relationThe trace is negative. The amplitude mode at $k=0$ has eigenvalue $-2$, while the phase mode is neutral because constant phase shifts preserve the solution. Put $D=1+bc$. For every nonzero real $k$, the determinant is positive if $D\ge 0$, whereas $D<0$ makes it negative for $0<k^2<-2D/(1+b^2)$. ThusThis is the Benjamin-Feir stability condition. At $D=0$, nonzero modes still decay, but the leading long-wave decay is fourth order rather than diffusive. On a periodic envelope interval of length $L$, the smallest nonzero wavenumber is $2\pi /L$, so strict stability of all allowed nonzero modes requiresA sufficiently short periodic domain can therefore exclude the unstable band even when the infinite-domain Benjamin-Feir stability condition fails.

The amplitude is damped while the phase is slow. For a slowly varying phase, the leading slaved amplitude is $r=-\frac{1}{2}\{b{\varphi }_{XX}+({\varphi }_X{)}^2\}+\cdots$. Substitution in the phase equation gives the diffusion term $D{\varphi }_{XX}$ and the nonlinear frequency shift $(c-b)({\varphi }_X{)}^2$. The latter also follows directly from a constant phase gradient: an envelope with gradient $q$ has amplitude squared $1-q^2$ and frequency $c+(b-c)q^2$. Constant phase symmetry rules out undifferentiated $\varphi$; translation invariance gives constant coefficients. The leading comoving Ginzburg-Landau equation also has spatial parity, so odd linear derivatives are absent at this order. Its neutral phase eigenvalue expands asThis coefficient follows by inserting a power series for $\lambda$ in the quadratic dispersion relation. Close to $D=0$, it becomes $E=(1+b^2)/2>0$ at leading order. The long-wave phase reduction of the complex Ginzburg-Landau equation is consequentlyThe fourth derivative regularizes negative phase diffusion. Terms omitted here have higher order in the long-wave expansion. First-order advection has been removed by the group-velocity frame; higher odd derivatives in a more general travelling-wave problem are beyond this leading amplitude approximation.

Take $D=-d$ with $d>0$ small. At the instability boundary $c=-1/b$, so $g=-(1+b^2)/b$ is nonzero. Set $\xi =\sqrt{d/E}X$, $\tau =d^{2}T/E$, and $\varphi =(d/g)\Phi$. At leading order the phase dynamics become the potential Kuramoto-Sivashinsky equation,With $u=-2{\Phi }_{\xi }$ this isThe variable $u$ is a phase-gradient disturbance, not the original physical field. Periodicity of the phase fixes ${\int }_0^{P}ud\xi =0$ for the zero-winding branch; the Kuramoto-Sivashinsky equation conserves this mean. The scaled period is $P=L\sqrt{d/E}$.

At fixed period $P$, the flat phase has growth rates ${\lambda }_n=k_n^2-k_n^4$, $k_n=2\pi n/P$. Its first loss of stability occurs at $P=2\pi$. To find what replaces it, put $k=2\pi /P$ and consider $P$ just above $2\pi$. After choosing an origin by translation, expand a small steady modulation as $u=a\sin (k\xi )+b_2\sin (2k\xi )+\cdots$. Projection of $-u{u}_{\xi }$ onto the first two harmonics yieldsSince ${\lambda }_2=-12$ at onset, this harmonic is slaved to $b_2=k{a}^2/(2{\lambda }_2)+\cdots$. The fundamental amplitude obeysThe primary periodic bifurcation of the Kuramoto-Sivashinsky equation is therefore supercritical. For $0<P-2\pi \ll 1$,The radial eigenvalue of this nonzero branch is $-2{\lambda }_1+O({\lambda }_1^2)<0$; all higher harmonics remain damped and the modulation's spatial phase is neutral. Thus the small finite-amplitude modulation is stable in the sense of orbital stability, at fixed mean and fixed period, despite the flat state's instability. The two signs of $a$ represent translations of the same modulation. In the potential formulation the spatial mean of $\Phi$ advances at $⟨({\Phi }_{\xi }{)}^2⟩$: the original wave acquires a nonlinear frequency correction rather than requiring a time-independent absolute phase.

For larger $P$, further linear modes become active and the small-amplitude argument no longer determines stability. A periodic modulation $\bar{u}$ must then be tested with the operator $v_{\tau }=-v_{\xi \xi }-v_{\xi \xi \xi \xi }-{\partial }_{\xi }(\bar{u}v)$ on mean-zero periodic disturbances. Its translation mode is always neutral, and secondary oscillatory or more complicated modulation can occur. Moreover, stability against perturbations of the same period does not prove stability against disturbances of a larger period; those require a separate sideband or Floquet theory calculation. The controlled conclusion near the first threshold is the existence of a stable small modulation family, not universal stability of all periodic Kuramoto-Sivashinsky solutions.

Use $p$ for disturbance pressure divided by the constant inertial density. The ideal, nonrotating Linearized Boussinesq equations areHere $N^2>0$ is a prescribed time-independent buoyancy frequency squared. Taking the scalar product of momentum with velocity gives ${\partial }_t(|\mathbf{u}{|}^2/2)=-\nabla \cdot (p\mathbf{u})+\sigma w$, using incompressibility. Multiplying the buoyancy equation by $\sigma /N^2(z)$ gives ${\partial }_t[{\sigma }^2/(2N^2)]=-\sigma w$. No spatial derivative of $N$ is needed in this cancellation. ThusThis is positive disturbance energy per unit inertial density; its flux is pressure work.

For constant $N$, write a plane internal gravity wave with wave vector $\mathbf{K}=({\mathbf{k}}_h,m)$ and phase $\theta ={\mathbf{k}}_h\cdot {\mathbf{x}}_h+mz-\omega t$. Let $k_h=|{\mathbf{k}}_h|>0$ and $K^2=k_h^2+m^2$. In complex amplitudes, incompressibility and horizontal momentum giveConsequently $\hat{p}=-\omega m\hat{w}/k_h^2$. Substitution into vertical momentum, $-i\omega \hat{w}=-im\hat{p}+\hat{\sigma }$, yieldsThe same dispersion relation follows directly from the displacement equation: its two terms become $K^2{\omega }^2\hat{\zeta }$ and $-N^2{k}_h^2\hat{\zeta }$. The positive-frequency branch is $\omega =N{k}_h/K$; the other branch is its sign reversal.