Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 13A / Solution

The species $u$ is the activator, since increasing $u$ increases its own production $u^2/v$; $v$ is the inhibitor, since increasing $v$ reduces that production. The unique positive uniform steady state solves $v=u^2$ and $u^2/v=bu$, giving $\boxed{u_{\ast }=1/b,v_{\ast }=1/b^2}$.

The reaction Jacobian matrix at this state isIts eigenvalues solve ${\lambda }^2+(1-b)\lambda +b=0$. Since $b>0$, both have negative real part precisely for $0<b<1$. This proves stability of the spatially uniform reaction kinetics.

A perturbation proportional to $e^{\lambda t+ikx}$ replaces $J$ by $J-k^2\mathrm{diag}(1,d)$. Put $q=k^2$. Its trace is $b-1-(1+d)q<0$ in the kinetically stable range, and its determinant isA growing spatial mode exists exactly when this determinant is negative for some $q>0$. Its minimum occurs at $q=(bd-1)/(2d)$, which is positive only if $bd>1$, and the minimum is negative precisely if $(bd-1{)}^2>4bd$. Together these conditions reduce to the Turing threshold of the quadratic activator-inhibitor model,The unstable band isAt the Turing instability threshold the band shrinks to its double root, yielding $\boxed{k_c^2=(1+\sqrt{2})/d}$. On a bounded spatial interval, an allowed wavenumber must additionally lie in the unstable band; the displayed parameter condition assumes the continuous spatial spectrum.