Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 14A (Mathematical Biology)

For a spatially homogeneous equilibrium of the Brusselator, the reaction terms obeyPositivity gives $uv=\beta$, and substitution into the first equation gives

The reaction Jacobian there issoThe linear stability analysis therefore makes the equilibrium asymptotically stable. It is a stable node whenand a stable focus when the reverse strict inequality holds. Since the upper-right entry of $J$ is positive and the lower-left entry is negative, focus trajectories rotate clockwise. Thus the local phase portrait consists of trajectories approaching the fixed point, either directly as a node or while spiralling clockwise as a focus.

If $D=1$, both species have the same diffusion coefficient. For a perturbation proportional to $e^{\sigma t+ikx}$, diffusion replaces $J$ by $J-k^{2}I$, shifting both reaction eigenvalues to the left by $k^2$. Since the homogeneous equilibrium is already stable, every spatial mode remains stable.

For general $D$, put $q=k^2$. The mode matrix isIts trace iswhile its determinant isWith negative trace, instability occurs exactly when $\Delta (q)<0$ for some $q>0$. The two-species diffusion-driven instability criterion says that this upward-opening quadratic has a negative minimum precisely whenEquivalently, the condition isThis is the Turing instability region.

At the first loss of stability, the minimum of $\Delta (q)$ just touches zero. The threshold condition from part (b) issoThe minimizing squared wavenumber isAt threshold, $\beta -1-D_c{\alpha }^2=2\alpha \sqrt{D_c}$, and thereforeThus the Turing threshold of the Brusselator selects