Basally forced quadratic activator-inhibitor model ($(u_{\ast },v_{\ast })=(1+r,(1+r{)}^2)$)

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 $r>1$.

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.

Add diffusivities $1,p$ to the basally forced quadratic activator-inhibitor model. A Fourier mode with $s=k^2$ has matrix $J-s\mathrm{diag}(1,p)$ and determinant $D(s)=p{s}^2+(q-pa)s+q$. When $q>a$, its trace is negative for every $s\ge 0$. A positive-growth mode therefore occurs precisely when $D(s)<0$ for some positive $s$. The parabola has a positive minimizer only if $pa>q$, and its minimum is negative exactly when $(pa-q{)}^2>4pq$. Thus the ordinary Turing instability requires $a>0$, $q>a$ and $p>q/(\sqrt{1+a}-1{)}^2$. At onset, $k_c^2=(pa-q)/(2p)=\sqrt{q/p}$ and the critical wavelength is $2\pi (p/q{)}^{1/4}$. A bounded spatial domain further requires an allowed mode inside the unstable band.