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First find the positive spatially homogeneous equilibrium. Setting the reactions to zero gives $v=u^2$ and $u/v=b$, and hence the unique positive equilibrium isThe zero pair is not an admissible positive state and is singular in the $u^2/v$ term. For homogeneous deviations, the reaction Jacobian matrix at this equilibrium isThus $\mathrm{tr}J=b-1$, $\det J=b$, and its eigenvalues areFor $0<b<1$, both real parts are negative: a real pair has positive product and negative sum, while a complex pair has real part $(b-1)/2$. The reaction equilibrium is therefore asymptotically stable, with no growing oscillatory mode. This does not exclude damped oscillations. Indeed, for $3-2\sqrt{2}<b<1$ the eigenvalues are complex; for example, $b=1/2$ gives $-1/4\pm i\sqrt{7}/4$, a stable focus with oscillatory transients. Thus a literal exclusion of every oscillatory solution is false; the intended exclusion of sustained or growing oscillations is the meaningful statement.

One can strengthen this to exclude nonlinear periodic oscillations in the positive quadrant without relying only on the local linear stability analysis. For the reactions $f=u^2/v-bu$, $g=u^2-v$, use the Dulac function $H=u^{-2}$. ThenWere there a periodic orbit wholly in the positive quadrant, the flux of $H(f,g)$ through it would vanish, since this vector is tangent to the orbit; Green's theorem would equate that flux to a strictly negative interior integral. This contradiction proves there is no sustained positive periodic oscillation when $0<b<1$, the positive-quadrant Dulac multiplier for quadratic activation argument. It leaves the damped transients entirely consistent with the conclusion.

Now introduce a perturbation $(\delta u,\delta v)=\mathbf{a}{e}^{\sigma t+ikx}$ in the quadratic activator-inhibitor model. The Fourier mode of squared wavenumber $q=k^2$ evolves underIts matrix trace and determinant areA Turing instability must start with reaction stability, requiring $0<b<1$. Then $\tau (q)<0$ for every $q\ge 0$, so a mode is unstable exactly when $D(q)<0$: its two real eigenvalues have opposite signs. This also proves the diffusion-driven onset is stationary rather than oscillatory.

The upward-opening quadratic $D(q)$ has its minimum at $q_m=(db-1)/(2d)$. To have a negative value at a physically allowed positive squared wavenumber, one needsWriting $r=db$, the second inequality is $r^2-6r+1>0$, with roots $3\pm 2\sqrt{2}$. The first inequality discards the lower branch, givingThese are the Turing threshold of the quadratic activator-inhibitor model conditions. Equivalently, in the $(b,d)$ plane the instability region lies strictly above $d=(3+2\sqrt{2})/b$ and strictly to the left of $b=1$, with $b>0$. The boundary diverges as $b\to 0^{+}$ and approaches $3+2\sqrt{2}$ as $b\to 1^{-}$.

At equality, the determinant has a double root and one growth rate is zero, while the other remains negative. Substituting $db=3+2\sqrt{2}$ into its minimum location givesBeyond onset, the unstable band isThe stated parameter region assumes continuously available wavenumbers, as on an infinite domain. With specified finite-domain boundary conditions, an allowed nonzero spatial eigenmode must additionally lie in this band; homogeneous stability alone does not ensure that a finite box admits the critical wavelength.

Substitute $\varphi =e^{i(kx-\omega t)}$ into the mixed-derivative equation. Its left side is $k\omega \varphi$, so for $k\ne 0$ the dispersion relation, phase velocity and group velocity areThe graph of a function for each velocity is symmetric about $k=0$, singular at zero, and approaches the horizontal axis as $|k|$ grows. Energy packets travel right while their crests travel left.

At a distant fixed observer $x>0$, stationary components obey $x/t=c_g(k)$, so $|k|=\sqrt{\alpha t/x}$. As time increases their wavelength $2\pi /|k|$ decreases and their frequency magnitude $|\omega |=\sqrt{\alpha x/t}$ decreases; the crests pass toward decreasing $x$.

For the given initial spectrum the solution isThe value of a regular spectrum at the single point $k=0$ is irrelevant to this integral; a nonzero spatially constant mode is incompatible with the PDE. For asymptotic calculation assume the usual smooth localized spectrum, with sufficient integrability for the following stationary-phase steps.

At $x=Vt$, $V>0$, the phase $\Phi (k)=Vk+\alpha /k$ has stationary points $k=\pm k_0$, $k_0=\sqrt{\alpha /V}$. Their values are $\pm 2\sqrt{\alpha V}$ and their second derivatives are $\pm 2\alpha /k_0^3$. Expanding quadratically and using the oscillatory Gaussian integral gives the stationary phase leading termFor real data this is twice the real part of the positive-frequency term. If a stationary coefficient vanishes, the formula instead asserts that its contribution has smaller order. At $x=-Vt$, the phase derivative is $-V-\alpha /k^2<0$, so there is no real stationary point and no corresponding $t^{-1/2}$ wave packet. For smooth compactly supported spectra away from zero, repeated integration by parts gives decay faster than any power of $t$. More general spectra require endpoint/tail control; absence of stationary points alone does not justify that stronger rate for arbitrary $A$.

For $\bar{\nu }=C{r}^a{\Sigma }^b$, the response is $q=b+1$. A nonaccreting background has ${\bar{\nu }}_0{\Sigma }_0\propto r^{-1/2}$, while ${\Sigma }_0\propto r^{-p}$, soConsequently ${\bar{\nu }}_0\propto r^{p-1/2}$. On this steady background, introduce the square-root-radius diffusion transformSince $q$ is constant and ${\partial }_r=(2x{)}^{-1}{\partial }_x$, the linear equation becomesThus $A(x)\propto x^{2p-3}$, and the required choice isFor a Fourier mode $g\propto e^{\lambda t+i{k}_{x}x}$, $\lambda =-A{k}_x^2$. Positive kinematic viscosity makes $A$ have the sign of $q$, so $q<0$ gives growing modes whose rate increases with $k_x^2$. The formal equation is a backward heat equation and predicts arbitrarily rapid small-scale amplification. Physically the thin disc diffusion closure applies only to wavelengths sufficiently larger than the thickness and stress-relaxation scales; this formal limit identifies the need for a cutoff, rather than a finite fastest wavelength absent from the model. The boundary case $q=0$ has vanishing linear transport response.

Assume $\kappa ,c_s,{\Sigma }_0>0$, and use the positive Toomre parameter. The dimensionless dispersion relation isAt fixed dimensionless gravitational softening $\delta$, maximum growth corresponds to maximizing $F(s)=2s{e}^{-\delta s}-s^2$. Differentiating gives the most unstable softened disk wavenumberFor $\delta >0$, the right side is strictly decreasing on $0<s<1/\delta$, while the left side increases; there is exactly one positive solution. The equation implies $\delta s_{\ast }<1$ and $s_{\ast }<1-\delta s_{\ast }$, soInstability occurs precisely when $Q^2<F(s_{\ast })$. Thus the critical Toomre parameter with exponential softening isThe printed description of a minimum $Q$ for instability reverses the threshold: $Q_c$ is the upper boundary of unstable values and the lower boundary of stable ones. Because $s_{\ast }<1/(1+\delta )\le 1$ and $2s-s^2$ increases on $[0,1]$,Both strict comparisons become equality in the unsoftened limit $\delta =0$: then $s_{\ast }=1$ and $Q_c=1$. Gravitational softening suppresses short-wave gravity, shifts the most dangerous mode to a longer wavelength and requires stronger self-gravity, or smaller $Q$, for instability.

The plot holds $Q$ fixed and changes gravitational softening. If $Q$ is varied physically at fixed $\kappa ϵ/c_s$, remember $\delta =(\kappa ϵ/c_s)/Q$: the marginal curve must be evaluated at its corresponding $\delta$, rather than treating these two dimensionless parameters as independently fixed along that physical variation.

Use the homogeneous equilibrium point condition $k({\rho }_0){\rho }_0=a_{0}f({\rho }_0)$. Evaluate $D_1,D_2$ at $(a_0,{\rho }_0)$ and abbreviateThe quantity $c$ is the chemical relaxation rate with the cell density held fixed. Derivatives of $D_1,D_2$ do not enter this linear stability analysis: they multiply spatial derivatives of the homogeneous background or products of perturbations. For a Fourier mode with time dependence $e^{st}$, put $z=|\mathbf{q}{|}^2$ and writeConsequentlyAt $z=0$, the eigenvalues are $0$ and $-c$. The neutral cell-density mode reflects mass conservation, not decay of every homogeneous perturbation.

Write $a=D_2>0$, $b=D_{\rho }>0$, $C=D_1{f}_0$, and interpret production physically as $f_0\ge 0$. If $c\ge 0$, the trace-determinant stability criterion shows that an unstable nonzero wavenumber exists precisely when $C>ac$. The unstable band isIf $c<0$, the chemical field is already unstable at zero wavenumber; arbitrarily small positive wavenumbers are unstable too. Since $C\ge 0$, this also implies $C>ac$. Thus, on the infinite plane, the undivided Keller--Segel aggregation threshold isFor $\kappa >0$, dividing by $D_2\kappa$ gives the requested formAt equality there is no strictly growing mode when $c\ge 0$; nonzero spatial modes decay. On a finite domain, a permitted nonzero wavenumber must actually lie in the unstable band. If one allows a signed production function, the displayed matrix and trace-determinant stability criterion remain valid, but the simplification using $C\ge 0$ must be revisited.

The printed hypotheses do not ensure $\kappa >0$. Positivity of the degradation rate $k$ alone is insufficient. For a concrete counterexample, take $k(\rho )=e^{-2\rho }$, $f(\rho )=e^{-2}$, $a_0={\rho }_0=1$, and $D_1=D_2=D_{\rho }=1$. The homogeneous equilibrium point condition holds, all rates and transport coefficients are positive, but $\kappa =c=-e^{-2}$. The printed left-hand side equals $-1$, although $\det M=z(z-2e^{-2})<0$ for $0<z<2e^{-2}$. These spatial modes grow. If $\kappa =0$, the printed expression is undefined. The undivided criterion and the matrix above resolve both cases.

To find the fastest-growing Keller--Segel mode, use the larger eigenvalueFor $C>0$ the square root is real. Put $p=a+b$ and $d=b-a$. Differentiation givesA positive maximizing wavenumber exists in either of two growing cases: $c\ge 0$, $C>ac$, or $c<0$, $C>-bc$. Equivalently, $C>\max (ac,-bc)$. In this range $C+dc>0$, andThus the stationary point is the unique maximum. The derivative condition yieldsSolving it, with the root that satisfies the unsquared derivative equation, givesFor $c=0$, the unequal-diffusivity expression simplifies to $z_{\ast }=C/[\sqrt{ab}(\sqrt{a}+\sqrt{b}{)}^2]$, also agreeing with the equal-diffusivity limit. The formula maximizes the full two-field growth rate; it makes no instantaneous-chemical approximation. Every direction of $\mathbf{q}$ with this length is equivalent by rotational symmetry.

If $c<0$ but $0\le C\le -bc$, the maximum instead occurs at $z=0$: the fastest mode is homogeneous, with $s_{\ast }=-c$ and infinite wavelength. For $C>0$, its initial derivative is $-b+C/(-c)\le 0$. If $C+dc\ge 0$ the derivative thereafter decreases, while if $C+dc<0$ it increases towards the still-negative large-$z$ limit; either way no positive-$z$ maximum is missed. At $C=0$, the two eigenvalues are simply $-az$ and $-c-bz$, giving the same conclusion. In a finite box, maximize $s_{+}$ over the allowed Fourier modes; excluding the homogeneous mode can change the selected length. In a stable parameter range there is no fastest-growing mode.

The first ratio compares the positive feedback loop “more cells produce more attractant, which draws in more cells” with spreading by cell diffusion and removal of chemical perturbations. Its numerator $D_1{f}_0$ measures chemotaxis together with attractant production; its denominator $D_2\kappa$ measures dispersal together with incremental degradation. This competition between directed chemotaxis and cell diffusion produces spatial aggregation. The second ratio compares the concentration dependence of chemical production, $a_0{f}_0^{\prime }$, with incremental degradation $\kappa$. Positive $f_0^{\prime }$ amplifies chemical fluctuations, whereas negative $f_0^{\prime }$ suppresses them. It changes the chemical relaxation available to the chemotaxis feedback loop and can itself destabilize the homogeneous chemical field. Chemical diffusion $D_{\rho }$ suppresses short scales and sets the selected wavelength, but does not change the infinite-plane long-wave threshold. These interpretations as ratios of stabilizing and destabilizing processes assume $\kappa >0$.

Write $w(x,t)$ for the vertical displacement of the sea ice, and take the undisturbed water surface as $z=0$, with water occupying $z<0$. The elastic plate has areal mass $m={\rho }_{i}h$ and bending stiffnessHere $E$ is Young's modulus and $\eta$ is Poisson's ratio. We neglect in-plane prestress, viscosity and plate shear deformation, and linearize about hydrostatic equilibrium. These are important assumptions: perfect elasticity alone does not specify every term in a floating-plate model.

For a plane wave $w=\hat{w}{e}^{i(kx-\omega t)}$, $k>0$, potential flow in deep water has velocity potential $\varphi =\hat{\varphi }{e}^{kz}{e}^{i(kx-\omega t)}$. This solves Laplace's equation and decays downwards. The kinematic boundary condition gives $k\hat{\varphi }=-i\omega \hat{w}$. Linearizing the water pressure at the displaced interface gives an upward excess loadThe elastic plate equation is $m{w}_{tt}+D{w}_{xxxx}=p$. Substitution and multiplication by $k$ therefore give the flexural-gravity wave dispersion relationEquivalently, with $\alpha ={\rho }_{i}h/{\rho }_w$ and $\beta =D/{\rho }_w$,This determines the positive wavenumber implicitly for prescribed positive angular frequency. It is unambiguous: the derivative of ${\omega }^2$ is $(g+5\beta k^4+4\alpha \beta k^5)/(1+\alpha k{)}^2>0$, while ${\omega }^2$ runs from zero to infinity.

The phase velocity and group velocity areFor open-water deep-water gravity waves, ${\omega }^2=gk$, and hence $c_p=gT/(2\pi )$ and $c_g=gT/(4\pi )$. At large period the ice-covered curves approach these straight lines. At shorter period, plate bending raises the speeds, so both ice-covered curves turn upward as period decreases. In the bending regime with negligible plate inertia, $\omega \propto k^{5/2}$ and $c_g\simeq (5/2)c_p$; in the formal plate-inertia-dominated limit, $\omega \propto k^2$ and $c_g\simeq 2c_p$. The latter extrapolation eventually leaves thin-plate validity and should not be read as a prediction at arbitrarily small wavelength.

The plotted parameters are illustrative rather than measured at the observation site. They give a group-velocity minimum of a flexural-gravity wave near $15.1\mathrm{m}{\mathrm{s}}^{-1}$ at period $17.6\mathrm{s}$. There is no arbitrarily slow wave-energy branch under a continuous elastic sheet. Energy put into a localized disturbance travels away at at least this minimum group velocity; a slowly moving wind system cannot retain a wave packet indefinitely beneath itself. This reduces the opportunity for sustained local growth compared with slow, short open-water waves, and the continuous cover also prevents direct wind forcing of an exposed water surface. Incoming long swell can still propagate.

A group-velocity minimum of a flexural-gravity wave is not by itself a universal minimum wind speed for wave generation. A steadily translating forcing pattern requires a phase velocity matching its translation speed, so the minimum of $c_p$, not $c_g$, supplies the corresponding resonance threshold. Random wind forcing, dissipation and aerodynamic coupling must be specified before making an absolute generation claim.

It is important to distinguish loss of forward-going surface-gravity-wave energy from conversion of mechanical energy into heat. Scattering attenuation by ice floes redirects energy; it can attenuate a coherent transmitted wave without dissipating the total energy.

For fixed floe geometry, increasing frequency usually increases attenuation over the relevant swell range: shorter wavelengths respond more strongly to the contrast between water and the elastic plate, and to repeated floe edges. Long surface gravity waves have weak curvature and often penetrate much farther. This is a trend over a specified frequency range, not a theorem excluding resonances.

The diameter dependence is governed by $d/\lambda$. An ice floe much smaller than the wavelength moves nearly with the water and scatters weakly. Scattering becomes appreciable when floe size is comparable with the wavelength, and interference between its two edges can give maxima and minima. At fixed ice concentration, larger ice floes also mean fewer edges per unit propagation distance, roughly proportional to $p/d$. Consequently the attenuation coefficient need not increase monotonically with diameter: the single-floe reflection and the number of encounters must both be considered. Thickness increases areal inertia as ${\rho }_{i}h$ and bending stiffness as $h^3$, generally increasing wave mismatch and reflection, although detailed frequency-dependent resonances again prevent a universal monotonic law.

When $d\ll \lambda$, particularly for frazil ice and pancake ice, weak individual scattering leaves other processes dominant. Relative crystal and water motion causes viscous dissipation; an aggregate layer can behave as a viscous or viscoelastic material, and pancake ice collisions, rubbing and overwash remove energy. Their importance depends on concentration and wave amplitude.

For a uniform continuous sheet with horizontal dimensions much greater than the wavelength, there are no repeated floe edges in its interior. A perfectly elastic sheet over inviscid water supports undamped flexural-gravity waves, so internal scattering is not an explanation of decay there. Real attenuation can instead arise from internal ice anelasticity or viscoelasticity, a dissipative sub-ice viscous boundary layer, turbulence, cracks and brine-related processes. Small-floe mixtures and continuous sheets require dissipation models beyond the isolated-floe scattering picture.

Use a local, homogeneous razor-thin disk approximation in a frame rotating with constant $\Omega$. The unperturbed planar velocity is zero in this frame, and the large-scale gravitational and centrifugal forces balance. Neglect viscosity, magnetic fields, thickness and background gradients across a wavelength. Take small planar disturbances, with wavelength short compared with the galaxy's background scale but long enough for a fluid description. The barotropic closure of a razor-thin disk is $P=K{\Sigma }^{\gamma }$, with $K$ fixed and a positive derivativeSolid-body rotation has no shear and has radial epicyclic frequency $\kappa =2|\Omega |$. By rotational symmetry of the local model choose a wavevector along $x$, and write perturbations proportional to $e^{i(kx-\omega t)}$.

Let ${\Sigma }_1,u,v,{\Phi }_1$ be the surface density, $x$-velocity, $y$-velocity and Newtonian gravitational potential amplitudes. The linearized continuity equation and Euler equations with the Coriolis force areThe perturbing Newtonian gravitational potential solves the Poisson equation for Newtonian gravityIts decaying solution is proportional to $e^{-|k||z|}$. The jump in its derivative is $-2|k|{\Phi }_1(0)=4\pi G{\Sigma }_1$, giving the razor-thin disk Poisson kernelEliminating $v$ from the two momentum equations, and using ${\Sigma }_1=k{\Sigma }_{0}u/\omega$, gives the density-wave branchThe complete linear system also has a zero-frequency balanced mode; it is not the growing density-wave branch, and division by $\omega$ in this elimination excludes it. The result is the uniformly rotating gas-sheet dispersion relation: pressure opposes compression at large wavenumber, self-gravity promotes compression, and rotation provides epicyclic support.

A mode is exponentially unstable when ${\omega }^2<0$. With $\Omega =0$ and $c>0$, this occurs atThe uniform $k=0$ perturbation is marginal rather than growing in this local calculation. With $c=0$, rotation fails to stabilize sufficiently short waves:These limits show why both pressure and rotation are needed for stability at all wavelengths.

For nonzero pressure and rotation, put $q=|k|$. Complete the square:Its minimum lies at $q_{\ast }=\pi G{\Sigma }_0/c^2$. Thus a growing mode exists precisely whenThe printed inequality has the opposite physical meaning: it is the condition for no exponentially growing density wave. Equality is marginal. In the usual gas Toomre stability criterion, $Q=\kappa c/(\pi G{\Sigma }_0)=2|\Omega |c/(\pi G{\Sigma }_0)$, so instability is $Q<1$ and stability is $Q\ge 1$. The unstable wavenumber band of a rotating gas sheet iswhen $Q<1$.

If ${\Sigma }_0$ and $\Omega \ne 0$ remain fixed while $c^2$ decreases slowly, first reach marginality atThe first wavelength to become unstable just below this threshold is the marginal fragmentation wavelength of a rotating sheet:Density maxima are separated by approximately this wavelength. The expected fragment size is of this order; an overdense half-wave has width about ${\ell }_{\mathrm{crit}}/2$. Linear theory fixes a preferred wavelength, not an exact nonlinear clump radius or shape. Further cooling shifts the fastest-growing wavelength to $2c^2/(G{\Sigma }_0)$. A rough fragment mass is consequently of order ${\Sigma }_0{\ell }_{\mathrm{crit}}^2$, with a geometrical factor depending on the nonlinear fragmentation pattern.

Use the Minkowski metric convention $+---$. A particle's four-momentum is $P=(E/c,\mathbf{p})$, with $P\cdot P=m^2{c}^2$ for rest mass $m$. For a photon, its energy and wavelength obey $E=hc/\lambda$, and its four-momentum is null. Here $h$ is the Planck constant.

The initially stationary electron has $P=(mc,0,0,0)$. Choose the scattering plane as the $xy$ plane. The incoming and outgoing photon four-momenta areBy four-momentum conservation, the final electron has $P^{\prime }=P+K_1-K_2$. Its rest mass is unchanged, so the relativistic energy-momentum relation gives $P^{\prime 2}=P^2=m^2{c}^2$. Since $K_1^2=K_2^2=0$, expansion yieldsorMultiplying through gives the Compton scattering shift:It is nonnegative and vanishes for forward scattering, consistent with the recoil energy being supplied by the photon.