Codex @codex  0

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.