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A quadratic activator can have both basal production and activation inhibited by a second species. A normalized example has reaction rates $f(u,v)=1+r{u}^2/v-u$ and $g(u,v)=q(u^2-v)$, with $r,q>0$ and positive concentrations. At its spatially homogeneous equilibrium, $v=u^2$, so $u_{\ast }=1+r$ and $v_{\ast }=(1+r{)}^2$. Writing $a=(r-1)/(r+1)$, its reaction Jacobian matrix is $J=\left(\begin{array}{cc}a& -r/(1+r{)}^2\\ 2q(1+r)& -q\end{array}\right)$. Thus $\mathrm{tr}J=a-q$ and $\det J=q>0$, giving linear asymptotic stability when $q>a$ and instability when $q<a$. Basal production makes $a$ change sign at $r=1$; a positive activation feedback needed for a conventional Turing instability requires $r>1$.

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.

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.

A positive spatially homogeneous equilibrium satisfies $v=c+u$ and $au=bv$, givingIt exists iff $a>b$. The reaction Jacobian matrix isThus linear stability analysis gives strict homogeneous stability precisely whenThe boundaries are not included: at $a=1$ the linearization has purely imaginary exponents, and at $a=b$ the positive equilibrium ceases to exist.

For a mode $\cos (kx)$, put $q=k^2$. The perturbation matrix is $J-\mathrm{diag}(q,dq)$. In the homogeneously stable region its trace is still negative, so a growing mode exists iff its determinant is negative:This upward quadratic is negative for some $q>0$ precisely whenSuch a Turing instability region is nonempty iff $d>1$. The unstable band isFor a finite domain only the admissible boundary-condition wavenumbers in this band grow. The region shown below describes a continuous wavenumber spectrum.

At onset the minimum of $D$ is zero and $q_c=u_{\ast }(d-a)/(2d)$. On the threshold $b=4d{a}^2/(d+a{)}^2$, substitution gives $u_{\ast }=4dac/(d-a{)}^2$. HenceThis last expression is the threshold wavenumber; away from onset the fastest-growing mode need not coincide with the minimum of the determinant.

The $u^2/v$ term makes $u$ locally autocatalytic, while increasing $v$ suppresses this production. Thus $u$ is the activator and $v$ is the inhibitor.

At a positive spatially homogeneous equilibrium,Hence $u=2bv$ and $v=u^2$, givingThe reaction Jacobian matrix there isBecause $b>0$, the trace-determinant stability criterion shows that the spatially uniform reaction kinetics are stable exactly when

For a perturbation proportional to $e^{\lambda t+ikx}$, put $q=k^2$. Its growth rates are the eigenvalues ofThe trace is $2b-1-(d+1)q<0$ in the reaction-stable regime, while the determinant isA Turing instability occurs exactly when this upward-opening quadratic is negative for some $q>0$. By the two-species diffusion-driven instability criterion, this requiresTogether these reduce toThus the unstable region in the $(b,d)$-plane lies below the straight line $d=(6-4\sqrt{2})b$ and to the left of $b=1/2$.

Inside this region, the unstable squared-wavenumber band isAt the bifurcation boundary the two roots coincide, soSubstituting $d=(6-4\sqrt{2})b$ yields

At a positive spatially homogeneous equilibrium,soThe Jacobian matrix of the reaction terms there isThe linear stability analysis of a two-dimensional system requires negative trace and positive determinant. Since $a>0$, the homogeneous equilibrium is stable exactly when

For a spatial Fourier mode proportional to $\cos (kx)$, put $q=k^2$. The linearized matrix isIts trace is still negative when $0<a<1$, whileA Turing instability occurs for some $q>0$ exactly when this quadratic has a negative minimum. This requiresWriting $d=\sqrt{D}$, the second inequality reduces to $d<a(\sqrt{2}-1)$. Hence, for continuously available wave numbers, the condition isOn a bounded spatial domain, one additionally requires an allowed wave number $k$ for which $\det J_{k^2}<0$.