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A Floquet multiplier is an eigenvalue of a one-period linear evolution map. For a homogeneous linear differential equation with continuous coefficients of period $T>0$, this map is the monodromy matrix of a periodic linear system $N$. If $Nv=mv$ for a nonzero eigenvector $v$, the associated solution of a differential equation with $Y(0)=v$ satisfies $Y(t+T)=mY(t)$. Its exponential rate is $T^{-1}\log |m|$, the corresponding Floquet growth rate. A Floquet multiplier equal to one gives a $T$-periodic solution in that eigenvector direction; a nontrivial Jordan block at a unit-modulus Floquet multiplier can still produce polynomial growth.

For a nonlinear periodic orbit, the Floquet multipliers are eigenvalues of the one-period map of its variational equation. A nonconstant periodic orbit of a smooth autonomous differential equation has a Floquet multiplier equal to one in its tangent direction: differentiating the solution of a differential equation with respect to its starting phase gives a periodic solution of the variational equation. The derivative of a transverse Poincaré map removes this phase direction and retains the transverse Floquet multipliers. Transverse Floquet multipliers strictly inside the unit circle give local attraction, while one outside gives instability; a transverse Floquet multiplier on the unit circle requires further analysis. For a planar periodic orbit, the transverse Floquet multiplier is given by the divergence test for a planar periodic orbit.

Use a Poincaré map sampled once per forcing period. A hyperbolic periodic trajectory has stable manifolds containing parcels that approach it under forward iteration and unstable manifolds containing parcels that approach it under backward iteration. In the integrable limit the relevant branches coincide along a heteroclinic orbit and form a barrier between neighboring cells.

A transverse crossing in the perturbed map means that this barrier has split: a parcel on the intersection belongs to both asymptotic manifolds, and adjoining branches enclose fluid lobes transported across the former boundary. Iteration stretches and folds those lobes. In the connected heteroclinic network this creates complicated intercell transport and chaotic advection. The crossing is between invariant curves at the same forcing phase; it does not mean that different trajectories cross at one time in violation of uniqueness.

The incompressible flow gives an area-preserving map, so lobe exchange preserves parcel area. Without diffusion, an advected passive scalar also keeps its value on each trajectory. Therefore manifold intersections can generate fine filaments and efficient advective exchange but do not by themselves produce molecular homogenization, nor prove that every region of the flow becomes chaotic.

The homogeneous fixed point must satisfy $v=u^2$. The singular point $u=v=0$ is outside the domain of the reaction term, so substitution in the other equation gives the unique regular spatially homogeneous equilibriumThis is the basally forced quadratic activator-inhibitor model. The linearization of its reaction terms gives the Jacobian matrixPut $a=(r-1)/(r+1)$, so $-1<a<1$. Its trace and determinant are $a-q$ and $q$, and its eigenvalues areSince $q>0$, both eigenvalues have negative real part exactly when $q>a$. HenceFor $r\le 1$ the equilibrium is always strictly linearly stable. For $r>1$, equality $q=a$ is an oscillatory threshold with ${\lambda }_{\pm }=\pm i\sqrt{a}$; the nonlinear qualification at this threshold is proved below.

Now insert a perturbation proportional to $e^{\lambda t+ikx}$. The diffusion terms subtract $k^2\mathrm{diag}(1,p)$ from $J$. With $s=k^2$ the mode trace and determinant areFor a strictly stable homogeneous state, $q>a$ makes $\tau (s)<0$ for all $s\ge 0$. A nonzero Fourier mode can therefore grow only if $\Delta (s)<0$, in which case its two real eigenvalues have opposite signs. The determinant parabola has its minimum at $s_{\ast }=(pa-q)/(2p)$, and that minimum is $q-(pa-q{)}^2/(4p)$. The two-species Turing criterion is consequently derived here asThese force $a>0$, hence $r>1$. Equivalently, solving $pa-q>2\sqrt{pq}$ for $\sqrt{p}$ givesThus the inhibitor in a reaction-diffusion system must diffuse sufficiently fast. Inside this region, the growing wavenumbers obeyAt onset, the minimum touches zero and the two determinant roots coalesce. ThereforeHere $p$ is evaluated on the onset surface. These are the Turing threshold of a basally forced quadratic activator-inhibitor model and its continuum critical wavelength. On a finite or domain with periodic boundary conditions an admissible nonzero spatial mode must additionally fall in the stated band. Equality in the diffusion criterion is marginal onset, while $q=a$ is the distinct homogeneous oscillatory threshold.

For completeness, the weak focus at the Hopf threshold of a basal activator-inhibitor model resolves the nonlinear homogeneous stability when $q=a>0$. Let $h=(1+a)/2$, $\beta =(1-a)/2$, $\omega =\sqrt{a}$, and set $u=(1+r)(1+x)$, $v=(1+r{)}^2(1+y)$, $X=x$, $Y=(hy-ax)/\omega$. Expanding the rational reaction term to cubic order giveswhereWrite $X=\varrho \cos \theta$, $Y=\varrho \sin \theta$ and evaluate these homogeneous polynomials on the unit circle. If $A=\cos \theta F_2+\sin \theta G_2$, $B=\cos \theta F_3+\sin \theta G_3$, and $C=\cos \theta G_2-\sin \theta F_2$, division of the radial and angular equations givesThe angular integral of $A$ is zero because it changes sign under $\theta \mapsto \theta +\pi$. Iterating its quadratic term once over a full cycle contributes the square of this zero integral, so the first nonzero return map term is cubic. Using $⟨{\cos }^4\theta ⟩=3/8$, $⟨{\cos }^2\theta {\sin }^2\theta ⟩=1/8$ and the corresponding sixth moments gives $⟨B⟩=a/4$ and $⟨AC⟩/\omega =(a+1)/8$. The Poincaré map isBecause $0<a<1$, sufficiently small positive radii decrease on every return. The equilibrium at $q=a$ is therefore nonlinearly asymptotically stable, with weak rather than exponential attraction, despite its neutral linear eigenvalues. The strict inequalities above describe the ordinary diffusion-driven instability from a strictly linearly stable homogeneous state.

At $q=a=(r-1)/(r+1)>0$, the reaction eigenvalues are $\pm i\sqrt{a}$. This linear marginal case is a weakly attracting focus, not a nonlinear center. To check it directly, put $u=(1+r)(1+x)$, $v=(1+r{)}^2(1+y)$, $h=(1+a)/2$, $\beta =(1-a)/2$, $\omega =\sqrt{a}$, and $X=x$, $Y=(hy-ax)/\omega$. The equations become $\dot{X}=-\omega Y+F_2+F_3+O(4)$, $\dot{Y}=\omega X+G_2+G_3+O(4)$, withOn the unit circle write $A=\cos \theta F_2+\sin \theta G_2$, $B=\cos \theta F_3+\sin \theta G_3$, $C=\cos \theta G_2-\sin \theta F_2$. In polar coordinates, $d\varrho /d\theta =(A/\omega ){\varrho }^2+(B/\omega -AC/{\omega }^2){\varrho }^3+O({\varrho }^4)$. The quadratic angular average vanishes. Integrating the quadratic correction over a whole cycle contributes its final squared integral, also zero, leaving the cubic average. Using the even trigonometric moments gives $⟨B⟩=a/4$ and $⟨AC⟩/\omega =(a+1)/8$. The Poincaré map is therefore $\varrho \mapsto \varrho +[2\pi /\omega ](a-1){\varrho }^3/8+O({\varrho }^4)$. Since $0<a<1$, it contracts sufficiently small positive radii and proves nonlinear asymptotic stability at the threshold.