Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 14C / Solution

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