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For $A_t=A+(1+ib)A_{xx}-(1+ic)|A{|}^{2}A$, the uniform oscillation $A=e^{-ict}$ has phase diffusion coefficient $D=1+bc$. Its amplitude-phase dispersion relation is ${\lambda }^2+(2+2k^2)\lambda +2D{k}^2+(1+b^2)k^4=0$. Thus $D<0$ gives a long-wavelength modulational instability, whereas $D>0$ gives linear stability of all nonzero Fourier modes. On a fixed periodic domain, instability requires an allowed wavenumber with $k^2<-2D/(1+b^2)$.

For a spatially periodic nonlinear pattern, a linear stability disturbance is written $e^{\lambda t+iq\cdot x}v(x)$ with $v$ periodic on the pattern cell. The conjugated operator $L(q)$ acts on that fixed cell and determines the sideband eigenvalues. Translation invariance supplies zero eigenfunctions at $q=0$, namely spatial derivatives of the base pattern. Their small-$q$ eigenvalues determine long-wavelength phase stability; other parts of the spectrum can give independent instabilities.

For cell diffusion, self-produced chemotactic agent and logistic proliferation, Neumann cosine modes reduce linear stability of the uniform state to a two-by-two matrix indexed by $s={\pi }^2(n^2/L_1^2+m^2/L_2^2)$. Its trace is negative and its determinant is $D(s)=\alpha s^2+(1+\alpha \kappa -\beta \kappa \eta )s+\kappa$. Chemotaxis can destabilize a mode precisely when $D(s)<0$, despite stability of the spatially uniform reaction system.

For $b_t=d_b{b}_{xx}-{\chi }_b{\partial }_x[b(1+n{)}^{-2}{n}_x]+b(1-b)$ and $n_t=d_n{n}_{xx}+\lambda b^2/(b_0^2+b^2)-\mu bn$, the positive homogeneous equilibrium is $(1,n_{\ast })$, where $n_{\ast }=\lambda /[\mu (1+b_0^2)]$. Set $\kappa ={\chi }_b/(1+n_{\ast }{)}^2$ and $\eta =\lambda (b_0^2-1)/(1+b_0^2{)}^2$. The linear stability matrix at squared wavenumber $q$ is $\left(\begin{array}{cc}-1-d_{b}q& \kappa q\\ \eta & -\mu -d_{n}q\end{array}\right)$. Its trace is negative and its determinant is $d_b{d}_n{q}^2+(d_n+\mu d_b-\kappa \eta )q+\mu$. Minimizing this quadratic polynomial over $q>0$ proves the displayed continuous-wavenumber growth criterion. It requires $b_0>1$. On a bounded domain an actual permitted nonzero Fourier mode must lie strictly inside the negative-determinant interval. This is a criterion for growth of infinitesimal perturbations, not a general exclusion of finite-amplitude patterns.

For a small, rapid horizontal displacement $a\cos \omega t$, the leading fast angular excursion is $\xi =-(a/l)\cos \Theta \cos \omega t$. Correlating that excursion with the driving torque gives the averaged force $\ddot{\Theta }=-(g/l)\sin \Theta +(a^2{\omega }^2/2l^2)\sin \Theta \cos \Theta$. Its effective potential is $-mgl\cos \Theta +m{a}^2{\omega }^2{\cos }^2\Theta /4$. The downward state loses linear stability above $a\omega =\sqrt{2gl}$ and two tilted minima appear. Small $a/l$ and slow dynamics compared with $\omega$ are required; horizontal and vertical support vibration produce different effective potentials.

Ice-ocean albedo feedback is the ocean-surface case of ice-albedo feedback: loss of bright sea ice, often covered by snow, exposes darker water, lowers surface albedo and increases absorbed sunlight. The extra stored ocean heat content can promote further melting and delay refreezing. Reversing the sequence can amplify cooling. This is a positive climate feedback, meaning that it reinforces the initial change; the word positive does not mean beneficial, and it does not guarantee runaway warming.

For unchanged downward solar irradiance $S_{\downarrow }$ on a newly exposed opaque patch, the extra absorbed heat flux isFor illustrative local surface albedos $0.60$ and $0.10$, and $S_{\downarrow }=200\mathrm{W}{\mathrm{m}}^{-2}$, the patch absorbs an additional $100\mathrm{W}{\mathrm{m}}^{-2}$. This is a local sunlight calculation, not a whole-planet radiative forcing: newly exposed area, illumination, clouds and season determine its global contribution. With no incident sunlight this particular heating pathway vanishes. NASA's notes describe the contrast between reflective ice and snow and darker water. ^[ref]

To isolate how a positive climate feedback competes with Planck feedback, retain the assumptions of the zero-dimensional planetary energy balance and prescribe the smooth broadband Bond albedoThe exponential function makes a smooth transition from a bright cold state near $0.6$ to a darker warm state near $0.3$. The midpoint $240\mathrm{K}$ and width $3\mathrm{K}$ are chosen to illustrate the dynamics. They are not a seawater freezing law, an observed global relation, or a calibration of modern Earth. No evolving ice thickness or latent heat is represented.

An equilibrium satisfies $F(T_{\ast })=0$. For a small perturbation $\delta T$, linear stability follows fromSince $C>0$, $F^{\prime }<0$ gives an asymptotically stable equilibrium, while $F^{\prime }>0$ gives an unstable equilibrium. With $x=(T-240\mathrm{K})/(3\mathrm{K})$, the derivative isThus the optical term $-S{\alpha }^{\prime }/4$ is destabilizing, while the thermal radiation term $-4\sigma T^3$ is stabilizing. Their sum, not the optical term alone, decides the linear stability.

For the same $S=1360\mathrm{W}{\mathrm{m}}^{-2}$ and Stefan-Boltzmann constant used above, the bisection method on the intervals $(220,223)$, $(238,241)$ and $(253,255)$, with temperature in kelvins, gives:

Unrounded numerical zeros of a function have $|F(T_{\ast })|<10^{-10}\mathrm{W}{\mathrm{m}}^{-2}$; the table rounds them for readability. In the positive-temperature domain the sign of $F$ alternates between heating, cooling, heating and cooling across these three equilibria. The middle equilibrium separates the basins of attraction of the two stable equilibria. For example, a starting temperature of $235\mathrm{K}$ cools toward the cold state, whereas $245\mathrm{K}$ warms toward the warm state: this is bistability at identical external illumination.

The same model illustrates hysteresis. Its equilibrium curve can be written as $S(T)=4\sigma T^4/[1-\alpha (T)]$. The two turning points, where $4[1-\alpha (T)]+T{\alpha }^{\prime }(T)=0$, are saddle-node bifurcations: approximately $(T,S)=(232.637\mathrm{K},1567.801\mathrm{W}{\mathrm{m}}^{-2})$ and $(245.893\mathrm{K},1250.481\mathrm{W}{\mathrm{m}}^{-2})$. For illumination strictly between these thresholds there are two stable equilibria and one unstable equilibrium. Under sufficiently slow increase of $S$, the cold branch ends at the upper threshold; under sufficiently slow decrease, the warm branch ends at the lower threshold. The ensuing warming and cooling transitions differ, so the state depends on its history. Finite sweep rates can delay the transitions; these thresholds describe this model's static equilibrium curve, not predicted tipping points for modern Earth.

The striking feature is that a smooth albedo law can produce distinct long-time states and abrupt transitions, although no discontinuity was inserted into the optical law. It is useful for understanding why positive climate feedback can amplify a response while stabilizing thermal radiation still limits it. Here $0.3<\alpha (T)<0.6$ bounds the incoming heating, whereas outgoing thermal radiation grows as $T^4$; high enough temperatures therefore cool rather than run away indefinitely.

Historically, M. I. Budyko's 1969 paper coupled changing albedo from expanding ice cover to a latitude-dependent temperature and heat transfer calculation, showing how that coupling amplified the effect of changed sunlight. ^[ref] William D. Sellers's 1969 paper constructed an annual energy-budget model in 10-degree latitude belts, including planetary albedo, atmospheric infrared transmission and atmospheric and oceanic exchange. ^[ref] These primary papers provide historical examples of ice-sensitive energy-budget modeling; the smooth single-temperature example above is a separate illustration, not a reproduction of either paper's equations or predictions.

For $0<I_1<I_2<I_3$, the transverse linear stability rates about steady rotation with signed component $\Omega \ne 0$ areAxes one and three have transverse eigenvalues $\pm i{\nu }_1$ and $\pm i{\nu }_3$, producing bounded oscillations. Axis two has transverse eigenvalues $\pm {\lambda }_2$, giving growing and decaying exponential functions. A neutral longitudinal perturbation corresponds to a small change of $\Omega$. Generic intermediate-axis disturbances contain a growing component; a precisely chosen decaying direction is exceptional. For the normalized intermediate-axis model with $|\Omega |=1$, the values are ${\nu }_1=1/\sqrt{3}$, ${\nu }_3=1$ and ${\lambda }_2=1/\sqrt{3}$.

Proof outline: the linearization of the Euler equations for a torque-free rigid body about the third principal axis gives transverse perturbations obeyingand henceCyclic permutation gives the other rates. For nonlinear Lyapunov stability of the extreme rotations, use the torque-free rigid-body invariants: the conserved quadratic forms $L^2-2I_{1}E$ and $2I_{3}E-L^2$ have strictly positive transverse coefficients about axes one and three respectively. They bound transverse angular velocity, while $E$ bounds its longitudinal change near nonzero rotation. Imaginary transverse eigenvalues alone would not establish this nonlinear conclusion.^[ref]

The practical lesson is to choose a stable principal axis when using rotation to maintain an instrument's pointing direction. Both extreme principal moments of inertia are stable in the ideal torque-free model; if internal damping lowers rotational kinetic energy while approximately preserving angular momentum, the minimum energy $E=L^2/(2I_3)$ instead favors the largest principal moment of inertia. A control system must account for torque, changing mass distributions and flexibility before applying this ideal result.

Define the dimensionless right-hand side of part (d) by $F(\kappa )$. Then ${\omega }^2=-4\pi G\rho F(\kappa )$. Because $I_0^{\prime }(\kappa )=I_1(\kappa )>0$ and $I_0(\kappa )>0$ for $\kappa >0$, its sign is the sign of $I_0(\kappa )K_0(\kappa )-1/2$. The original plot is positive for sufficiently long wavelengths and negative at short wavelengths; the nonzero crossing is approximately ${\kappa }_c\simeq 1.07$. Numerically solving the displayed equation gives ${\kappa }_c\simeq 1.0667$.

For $0<\kappa <{\kappa }_c$, ${\omega }^2<0$. Write $\omega =i\sigma$ with $\sigma =\sqrt{4\pi G\rho F}>0$. The time factor $e^{-i\omega t}$ then contains an exponentially growing solution. For $\kappa >{\kappa }_c$, ${\omega }^2>0$ and the ideal inviscid modes oscillate stably, without damping. At the crossing the frequency is zero and the mode is marginal. In terms of the axial wavelength ${\lambda }_z=2\pi /k$,The growth rate tends to zero as $kR\to 0^{+}$, so the very longest wavelengths are not the fastest growing. The source plot has its maximum near $kR\simeq 0.58$, corresponding to ${\lambda }_z\simeq 11R$ and $\sigma \simeq 0.25\sqrt{4\pi G\rho }$. Thus a characteristic fragmentation spacing is several cylinder diameters; a quantitative nonlinear fragment mass is beyond this linear stability calculation.

For unstable modes the physical displacement must be interpreted as $\mathrm{Re}[ϵe^{i(kz-\omega t)}]$ with complex amplitudes handled consistently throughout. The paper's real-part notation is harmless for the squared dispersion relation, but it should not be read as imposing real $\omega$ in a gravitational instability calculation. The exactly uniform mode $k=0$ is excluded from the finite-$k$ derivation; the limit of the curve alone does not establish a separate volume-preserving uniform-radius mode.

At an equilibrium $(v_{\ast },k{v}_{\ast })$, set $p=f^{\prime }(v_{\ast })$. The Jacobian matrix, its trace and its determinant areThe eigenvalues and their discriminant areThus the complete local linear stability test isIn the last two cases, $\mathcal{D}>0$ gives a node, while $\mathcal{D}<0$ gives a focus; the sign of $T$ distinguishes inward and outward trajectories. At $\mathcal{D}=0$, a nonzero repeated eigenvalue gives a degenerate node. At $p=\gamma <k$ the eigenvalues are purely imaginary, so linear stability analysis alone does not decide asymptotic stability; with the required crossing and nondegeneracy this is a Hopf bifurcation. At $p=k$ there is a zero eigenvalue, requiring a nonlinear test and generically marking the saddle-node bifurcation in the preceding answer.

For a unique simple equilibrium, $p<k$, so it can be a sink equilibrium or a source equilibrium. The number of equilibria alone does not decide stability. If $k\le \gamma$, the unique simple equilibrium is always a sink equilibrium. If $\gamma <k$, it can become a source equilibrium where $\gamma <p<k$. The same conclusion follows particularly simply if $\gamma >M$: the trace is negative at every nonsaddle equilibrium.

For three simple equilibria ordered by their $v$ coordinates, the middle one has $p>k$ and is a saddle equilibrium; the two outer ones have $p<k$. When both outer equilibria are sink equilibria, each has an attracting neighborhood, giving bistability. In the usual two-basin portrait, the stable manifold of the saddle equilibrium separates their basins. Each branch of its unstable manifold initially leaves towards a different side; its eventual destination requires a global argument. If $k\le \gamma$, both outer equilibria must be sink equilibria. If $\gamma <k$, either or both outer equilibria can instead be source equilibria; the same test $p\lessgtr \gamma$ applies separately to each. The three lower panels illustrate all three possibilities up to exchanging left and right.

The arrows in the phase plane follow directly from the nullclines: below $w=f(v)+I$ the horizontal component points right, and above it left; below $w=kv$ the vertical component points up, and above it down. Near a stable focus trajectories spiral inward; near an unstable focus they spiral outward. Near a saddle equilibrium they approach along its stable manifold and leave along its unstable manifold. These directions, together with the local tests, determine the sketches.

There is also a useful global conclusion when the unique equilibrium is a source equilibrium. Withthe leading term in $v$ is $-\beta v^4$, so sufficiently large ellipses are trapping regions. A small repelling neighborhood of the hyperbolic source equilibrium can be removed to form an invariant annulus. The Poincaré-Bendixson theorem then guarantees at least one periodic orbit in that annulus. The numerical sketch illustrates the resulting sustained oscillation; this argument does not assert uniqueness of a limit cycle. A stable equilibrium can instead represent a resting membrane potential, with perturbations producing damped relaxation; near a threshold, a sufficiently large perturbation can produce a large excursion before recovery. Global basin boundaries and coexistence of periodic orbits need more information than the number and local stability of equilibria.

For a spatially uniform steady state with nonzero bacteria abundance, $b(1-b)=0$ forces $b_{\ast }=1$. The nutrient balance then givesTo test spatially uniform linear stability, writeAt the positive equilibrium, the Jacobian matrix isIts eigenvalues are $\boxed{-1,-\mu }$, both negative, so the positive equilibrium is asymptotically stable against spatially uniform perturbations. This remains true at $\mu =1$, where a repeated eigenvalue may produce a decaying $t{e}^{-t}$ contribution. The printed stability claim must be understood in this spatially uniform sense: sufficiently strong chemotaxis can destabilize nonuniform perturbations, as the next answer shows.

The voltage nullcline is $g(v,n)=I^{\ast }$ and the recovery nullcline is $n=n_{\infty }(v)$. In the relevant region $g_n>0$, so below the voltage curve $\dot{v}>0$ and above it $\dot{v}<0$. Below the recovery curve $\dot{n}>0$, while above it $\dot{n}<0$.

For small $ϵ$, the outer voltage branches with $g_v>0$ attract fast voltage motion; the middle branch with $g_v<0$ repels it. Arrange the intersections so the equilibrium is on that middle branch, with recovery increasing along the high-voltage attracting branch and decreasing along the low-voltage attracting branch. The trajectory then drifts towards a fold, jumps rapidly to the other attracting branch at nearly constant $n$, and repeats. This is a relaxation oscillation.

The local linear stability calculation supports the construction. At an equilibrium,A sufficiently steep recovery curve gives positive determinant, while $g_v<-ϵ$ gives positive trace. Hence the equilibrium is a repeller. The bounded physical rectangle can be chosen forward invariant, and a small neighborhood of the repeller removed to give a trapping annulus. The Poincaré-Bendixson theorem then supplies a periodic orbit. These conditions can be realized because the recovery function and the applied current control the intersection and its slope; oscillation is possible, not forced for every choice.

For the illustration, ${\gamma }_K=0.1$, $H=0.8$, $v_K=0.2$, $ϵ=0.01$, $I^{\ast }=0$, $M(v)=(1+e^{-25(v-0.2)}{)}^{-1}$ and $n_{\infty }(v)=0.8/(1+e^{-20v})$. The rectangle $[-0.2,1]\times [0,0.8]$ has inward-pointing flow, and the displayed numerical trajectory realizes the folded-branch mechanism.

Take $\gamma >{\gamma }_c$ and let $\epsilon \to 0^{+}$ with the other parameters fixed. The nullclines are the cubic polynomial graph $w=f(v)$ and the line $w=\gamma v$. In the phase plane, $v$ increases below the cubic polynomial graph and decreases above it; $w$ increases below the line and decreases above it. The fast ordinary differential equation, in time $s=t/\epsilon$, freezes $w$ and gives $dv/ds=f(v)-w$.

At $w=0$, the fast roots are $0,a,1$. Their linear stability follows from $f^{\prime }(0)=-a<0$, $f^{\prime }(a)=a(1-a)>0$ and $f^{\prime }(1)=a-1<0$. Thus a fixed $0<v^{\ast }<a$ falls towards the resting branch, whereas a fixed $a<v^{\ast }<1$ rises towards $v=1$. If $v^{\ast }>1$, the first fast segment instead settles down towards $1$, followed by the same recovery excursion. The threshold $a$ is the leading singular perturbation threshold; finite-$\epsilon$ corrections matter for initial values approaching $a$ with $\epsilon$.

LetThese are the roots of $f^{\prime }(v)=0$. The outer branches $v<v_L$ and $v>v_R$ of the critical manifold are attracting under fast motion; the middle branch is repelling. The full excursion has four stages.

For this cubic polynomial, the landing point is explicit. Since $f(v)-f(v_R)=-(v-v_R{)}^2(v-v_J)$, comparison of the quadratic coefficients givesThus a graph of $v(t)$ has a rapid rise, a slowly falling elevated plateau, a rapid fall through zero to a negative undershoot, and a slow approach back to zero. The figure shows both the phase plane and a computed time trace. The smaller initial perturbation stays near the resting branch and produces no elevated plateau.

The four-stage description concerns the singular limit. Close to a simple fold point of a critical manifold, balancing the local quadratic activation term with the slow drift gives $\Delta v=O({\epsilon }^{1/3})$ and $\Delta t=O({\epsilon }^{2/3})$ for fixed parameters. These layers shrink compared with the slow phases and do not alter the leading time integrals below.

The decreasing production term represents negative feedback: low red-cell abundance stimulates production, whereas high abundance suppresses it. The linear loss term models removal at constant per-cell rate. SetThen $u^{\prime }=f(u)-\delta u$, with $f(u)=1/(1+u^n)$. On the physical half-line $u\ge 0$, $f$ decreases from one to zero while $\delta u$ increases from zero to infinity. Their unique intersection $u_{\ast }>0$ is the steady state. Its linear derivative is $f^{\prime }(u_{\ast })-\delta <0$, so it is asymptotically stable.

For small $\delta$, the intersection lies at large $u$, where $f\sim u^{-n}$; for large $\delta$ it lies at small $u$, where $f\sim 1$. HenceImplicit differentiation of $f(u_{\ast })=\delta u_{\ast }$ gives $d{u}_{\ast }/d\delta =u_{\ast }/[f^{\prime }(u_{\ast })-\delta ]<0$. The plot joins these two decreasing asymptotic regimes.

The delay $t_0$ represents the time between feedback sensing and the appearance of newly produced mature cells. In dimensionless time it is $d=F_0{t}_0/\theta$, so the equation becomesIt requires a nonnegative initial history on $[-d,0]$, not just one initial value. Its constant equilibrium is unchanged. Put $a=-f^{\prime }(u_{\ast })=n{u}_{\ast }^{n-1}/(1+u_{\ast }^n{)}^2>0$. A small perturbation obeys $v^{\prime }=-\delta v-av(\tau -d)$, whose characteristic equation isThere is no real root with $\sigma \ge 0$, because all three terms would have positive sum; zero is never a root either. Starting from the stable zero-delay case, loss of linear stability can therefore occur only when a complex conjugate pair crosses the imaginary axis at a nonzero frequency, the linear signature of a Hopf bifurcation.

For completeness, substituting $\sigma =i\omega$ gives $\delta +a\cos (\omega d)=0$ and $\omega =a\sin (\omega d)$. Such a crossing requires $a>\delta$, withThe first crossing is transversal: $\mathrm{Re}(d\sigma /dd)={\omega }^2/[(1+\delta d{)}^2+{\omega }^2{d}^2]>0$. If $a\le \delta$, the magnitude relation rules out roots in the closed right half-plane for every delay. A nondegenerate nonlinear Hopf conclusion additionally uses the usual bifurcation hypotheses. This is delayed negative feedback in red-cell production.

For $\mu >0$, the positive equilibrium isSet $S=S_{\ast }+\epsilon S_1$ and $N=N_{\ast }+\epsilon N_1$ and retain first-order terms. The linear system isDifferentiating its first row and using the second givesIn particular the trace is positive and determinant is $4{\mu }^{7/6}>0$. The steady drainage state is linearly unstable for every positive supply, whether its instability is oscillatory or nonoscillatory.

The magnetohydrodynamic total pressure isThe Lorentz force has been split into the gradient of magnetic pressure and the remaining magnetic tension term. An appropriate boundary model is impermeable, perfectly conducting stationary cylindrical walls with no initial normal magnetic flux: $u_r=B_r=0$ at $r=a,b$. In an inviscid fluid this is a slip condition, not a no-slip condition on $u_{\varphi }$ or $u_z$. It also makes the tangential ideal electric field vanish at the walls. Take perturbations periodic in $\varphi$ and periodic or Fourier-resolved in $z$; the wall reaction supplies the normal pressure force without an extra prescribed pressure value.

Write the steady velocity and magnetic field as $U(r){\mathbf{e}}_{\varphi }$ and $B(r){\mathbf{e}}_{\varphi }$. They satisfy both divergence constraints. Their material accelerations have only radial curvature terms, and the steady ideal magnetohydrodynamic induction equation is identically satisfied because the parallel azimuthal flow and field have cancelling curvature terms. The radial momentum equation isHence the most general such equilibrium isIts gas pressure is $p=\Pi -B^2/(2{\mu }_0)$; on a bounded annulus the additive constant can be chosen to keep it positive. Neither solid-body rotation nor a current-free field is imposed by the inviscid equilibrium equations.

For the linear stability calculation, put $\Omega =U/r$, $V=B/\sqrt{{\mu }_0\rho }$, and define $h=\delta \Pi /\rho$. Use axisymmetric normal modes proportional to $e^{ikz-i\omega t}$. Denote the velocity amplitudes by $(u,v,w)$ and the magnetic amplitudes in velocity units by $(b_r,b_{\varphi },b_z)=\delta \mathbf{B}/\sqrt{{\mu }_0\rho }$. Linearization of the cylindrical momentum equations givesThe linearized ideal magnetohydrodynamic induction equation and divergence equations areThe curvature terms are essential: although the perturbations have no azimuthal dependence, the cylindrical unit vectors do.

For $\omega \ne 0$, $b_r=b_z=0$. Introduce the radial Lagrangian fluid displacement by $u=-i\omega \xi$. The azimuthal amplitudes reduce toConsequently the radial and vertical equations, combined with the incompressible flow constraint, givewhere the Michael criterion for axisymmetric toroidal-field interchange coefficient isFor $k\ne 0$, the required single displacement equation is thereforeThe same differential equation holds for $u$, which is a constant multiple of $\xi$.

Set $y=r\xi$. Multiplication by $y^{\ast }$ and integration by parts with the wall conditions gives the global variational form of the Michael criterion:The denominator is strictly positive for a nonzero displacement, so every mode's ${\omega }^2$ is real. If $\mathcal{D}\ge 0$ everywhere, no exponentially growing mode exists. Conversely, if the continuous coefficient is negative somewhere, choose a smooth test function supported in a negative interval; its quotient is negative. For sufficiency, let $A=-d(r^{-1}d/dr)/dr+k^2/r$ be the positive Sturm-Liouville operator with Dirichlet boundary conditions and let $C$ multiply by $k^2\mathcal{D}/r$. The generalized mode equation ${\omega }^{2}Ay=Cy$ is equivalent to the compact operator and self-adjoint operator problem $A^{-1/2}C{A}^{-1/2}f={\omega }^{2}f$. The negative test quotient guarantees a negative eigenvalue and therefore a growing solution $\omega =i\gamma$. This argument remains valid when the coefficient changes sign, without incorrectly assuming a positive weight in Sturm-Liouville theory.

Thus, for any fixed nonzero vertical wavenumber,Equality is marginal. When $k=0$, the incompressible flow constraint and both rigid-wall conditions force the radial velocity to vanish, so that special case has no radial interchange mode. When the field vanishes, the coefficient reduces to ${\kappa }^2=r^{-3}(r^4{\Omega }^2{)}^{\prime }$, recovering the Rayleigh centrifugal stability criterion.

For a Keplerian disk, ${\kappa }^2={\Omega }^2$. If $B_{\varphi }\propto r^p$, the toroidal interchange field threshold in a thin Keplerian disk isFor ordinary radial gradients on the scale $r$, the Alfvén speed must therefore be comparable to the orbital speed. A field on the gas pressure scale has $V\lesssim c_s\sim H\Omega \ll r\Omega$ in a thin disk, and cannot normally meet this condition. Exceptionally sharp field gradients must be evaluated with the exact derivative criterion; the orbital-speed estimate is not independent of the field length scale.

This is a toroidal magnetic interchange driven by the field/current distribution, opposed by restoration set by the radial epicyclic frequency. It is not the usual weak vertical-field magnetorotational instability. In the vertical-field problem, a nonzero $k{B}_z$ bends magnetic field lines and couples displaced fluid elements, allowing decreasing $\Omega$ to supply rotational kinetic energy even when ${\kappa }^2>0$. Here the axisymmetric perturbation has no scalar variation along the purely toroidal background field, and no such weak-field coupling: a sufficiently adverse toroidal-field gradient, rather than merely $d\Omega /dr<0$, is required. Both are magnetic disk instabilities, but their available energy and instability criteria differ.

The spatial mean is conserved: averaging every term in the partial differential equation over the periodic cell gives ${\partial }_t⟨T⟩=0$. Adding a constant to $T$ also preserves the equation. We therefore fix the mean, conveniently at zero. This qualification is essential: without it the constant Fourier mode contributes an additional neutral direction, so the centre subspace would have five real dimensions rather than four.

On a Fourier mode $e^{i(mx+ny)}$, the linearized linear operator has eigenvalueOn the fixed-mean subspace, the first eigenvalues to reach zero are those with $m^2+n^2=1$. At $\mu =1$ these are the modes $e^{\pm ix},e^{\pm iy}$; every other nonconstant Fourier mode is strictly damped. The reality condition leaves two complex amplitudes, hence a four-real-dimensional centre manifold on the fixed-mean subspace. The trivial state loses linear stability as $\mu$ increases through one.

Use slow time $\tau ={\epsilon }^{2}t$, and write $T=\epsilon W+{\epsilon }^3{T}_3+\cdots$, whereThere is no quadratic nonlinearity, so no forced second-order correction is needed. At order ${\epsilon }^3$, with $\mu =1+{\epsilon }^2\nu$, projection onto the critical Fourier modes givesThe correction $T_3$ does not contribute to the critical projection because the linear operator vanishes on those modes. To find the nonlinear coefficient of $e^{ix}$, split the divergence term intoThe first term contributes $-3|A{|}^{2}A$. In the second, only the zero-$y$ Fourier coefficient of $W_y^2$ contributes; it is $2|B{|}^2$, and differentiating $W_x$ once more gives $-2|B{|}^{2}A$. The third term has zero $y$-average, and the fourth has no $x$-dependent critical component. Interchanging $x,y$ gives the other amplitude equation. Thus the conserved-mean convection amplitude equations areTo establish the selected planform, put $A=r{e}^{i\theta }$, $B=s{e}^{i\psi }$. The phases are constant and the magnitudes obeyFor $\nu >0$, the square state has $r=s=\sqrt{\nu /5}$. Its magnitude Jacobian matrix is $-(2\nu /5)\left(\begin{array}{cc}3& 2\\ 2& 3\end{array}\right)$, with eigenvalues $-2\nu$ and $-2\nu /5$. Its two neutral phases correspond to translation symmetry. The pure roll has $r^2=\nu /3,s=0$; an infinitesimal $B$ disturbance grows at rate $\nu -2\nu /3=\nu /3>0$. The other pure roll is equivalent, while the trivial state is unstable for $\nu >0$. Consequently squares, with $|A{|}^2=|B{|}^2=\nu /5$, are the stable small-amplitude planform, modulo translations. For $\nu <0$ the trivial state is stable.

A Hopf crossing of the transverse stability matrix requires zero trace and strictly positive determinant. Since ${\beta }_R<0$, the trace condition gives the positive thresholdThe determinant there isThus the required coefficient condition and the rotating-frame Hopf frequency areThe trace derivative $-2{\beta }_R$ is strictly positive, so with $d,\beta$ fixed the real parts cross zero with nonzero speed. Under this condition the determinant stays positive from $r=0$ to $r_H$; the pure branch first loses linear stability through this Hopf bifurcation. When $Q<Q_{\ast }$ its first instability is instead the steady one of part (a), and $D_H<0$ prevents a Hopf pair at the trace-zero point. Equality gives the already noted double-zero degeneracy.

Generically the new branch is periodic in the symmetry-reduced rotating frame, with a small oscillating odd-mode amplitude and a modulation of the even-mode amplitude. Reconstructing the common temporal phase combines the modulation frequency with the carrier frequency. The original amplitudes therefore describe modulated periodic oscillations, generically a two-frequency quasiperiodic state. If the two frequencies are rationally related, the reconstructed motion can be periodic instead. The spatial pattern no longer has a fixed mixture of the even and odd modes: their relative amplitudes and phase oscillate. As with the steady case, nonlinear nondegeneracy is needed for an ordinary Hopf branch, and its criticality or stability cannot be inferred from the trace and determinant alone.

A periodic pattern breaks continuous translation symmetry: translating it changes its phase without changing its shape or its energy. Consequently a uniform phase change is neutral. A slowly varying phase change alters the local spacing or orientation and can grow even while ordinary uniform amplitude perturbations decay. These are long-wavelength sideband instabilities. The relevant wavelengths are long compared with the carrier pattern, so their leading dynamics is often a phase modulation equation. Positive phase diffusion smooths spacing errors; negative phase diffusion amplifies them. Near onset, an amplitude equation computes this diffusion explicitly. Away from onset, one must calculate it from the full nonlinear periodic pattern.

For a supercritical stationary pattern in one dimension, rescale the real Ginzburg–Landau equation toA detuned periodic pattern is represented by $A=R{e}^{iqx}$, with $R^2=\mu -q^2>0$. Perturb its magnitude and phase by $A=(R+u)e^{i(qx+\varphi )}$. Linearization givesFor a Fourier mode proportional to $e^{\lambda t+ikx}$, the two eigenvalues are the sideband spectrum of a real Ginzburg-Landau plane wave:The magnitude branch remains damped at small $k$. The phase branch expands asHence the Eckhaus-stable band is $\boxed{q^2<\mu /3}$, narrower than the existence band $q^2<\mu$. This Eckhaus instability is longitudinal: it compresses and dilates the pattern along its wavevector. The coupling to the relaxing magnitude is crucial; examining phase diffusion while holding the magnitude fixed would miss the destabilizing $2q^2/R^2$ correction. At the stability boundary the $k^2$ coefficient vanishes, and the next term is negative, $-2q^4{k}^4/R^6$. Beyond the boundary, a phase distortion can create amplitude zeros and phase slips, changing the number of periods. These nonlinear events are possible outcomes, not conclusions of the linear calculation.

A fixed periodic domain allows only discrete sideband wavenumbers. Indeed, the displayed spectrum is unstable only if $0<k^2<6q^2-2\mu$. A domain too short to support such a Fourier mode can suppress the nominal long-wave instability. Thus an infinite-domain stability band and a finite-domain threshold must be distinguished.

The zigzag instability concerns transverse bending of rolls in an isotropic two-dimensional system. Let $k_c$ be the preferred carrier wavenumber, use an $x$-directed carrier, and denote its slow longitudinal detuning by $q$. Rotational covariance near onset leads to the anisotropic envelope equationThe derivative combination comes from expanding $|\mathbf{k}|-k_c$: a small transverse carrier component changes the length of the wavevector only at second order. A straight-roll envelope $A=R{e}^{iqX}$ has $g{R}^2=\mu -\xi q^2$. For a purely transverse phase perturbation $A\simeq R{e}^{iqX}(1+i\varphi (Y,t))$, magnitude and phase decouple at linear order, andThe transverse growth rate is thereforeRolls with $q<0$, whose total wavenumber is below the preferred value, are zigzag-unstable to sufficiently long transverse modulations. Bending increases their local wavevector length and can relieve this mismatch. Rolls with $q>0$ are transversely stable in this near-onset model; at $q=0$ the restoring force first appears at fourth spatial order. Combining this with the longitudinal Eckhaus instability gives the robust interior stable band $0<q<\sqrt{\mu /(3\xi )}$ for this simplest isotropic roll model. Anisotropy, a pinned orientation, additional mean fields, or other pattern symmetries can modify this conclusion.

For oscillatory patterns the coefficients of the amplitude equation are generally complex. Consider the scaled complex Ginzburg–Landau equationwhere $c_1$ measures linear dispersion and $c_3$ the amplitude-dependent frequency shift. The uniform oscillation is $A=e^{-i{c}_{3}t}$. Put $A=e^{-i{c}_{3}t}(1+u+iv)$, with real infinitesimal $u,v$. The linearized magnitude and phase equations areTheir characteristic polynomial for a sideband of wavenumber $k$ isThe magnitude mode has leading eigenvalue $-2$, while the neutral phase mode hasThus the Benjamin-Feir instability occurs for $\boxed{1+c_1{c}_3<0}$; the complementary Benjamin-Feir stability condition is $1+c_1{c}_3>0$. Both dispersive couplings matter: the phase perturbs the magnitude through $c_1$, and the magnitude feeds back into phase through $c_3$. In the unstable regime, higher spatial derivatives regularize sufficiently short disturbances. On a finite domain, an unstable mode must satisfy $0<k^2<-2(1+c_1{c}_3)/(1+c_1^2)$. At equality of the phase-diffusion coefficient to zero, the leading phase damping is fourth order, not exponential growth at order $k^2$.