Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 74 3 b Solution Created 2026-10-03 Updated 2026-10-14
For linear stability analysis of a general stable two-species reaction–diffusion system, take Fourier modes proportional to . Their matrix is . Put , , and . ThenThe full system is stable precisely when and for every allowed mode. If the homogeneous state is stable, the trace already decreases with ; only the determinant can change sign. If , its minimum for is at zero. If , it is at . Thus the two-species Turing criterion for instability on a continuum of wavenumbers isStrict stability has together with the homogeneous conditions. Equality at positive marks a stationary neutral mode. At that first onset,Beyond threshold, the growing band lies between the positive roots , as long as the homogeneous trace stays negative.
For the Brusselator, . Write . The stationary determinant threshold gives the Turing threshold of the Brusselator in dimensional diffusivities:For this to be a genuine diffusion-driven loss of stability, it must occur before the homogeneous Hopf bifurcation threshold. The Brusselator Turing-before-Hopf diffusivity condition isIndeed is equivalent to . With this strict inequality, contains a spatially unstable but homogeneously stable interval. If the ratio condition fails, the first linear instability as increases is homogeneous at , rather than a Turing instability; equality gives simultaneous neutral homogeneous and finite-wavenumber modes. Equal diffusivities cannot produce a Turing instability.
These continuum formulas assume the onset wavenumber is available. On a finite domain with a specified discrete mode set, the discrete-mode Turing threshold for the Brusselator isprovided this minimum is below . The question supplies no boundary geometry, so the continuous-mode onset above is the usual answer; this last expression states how a finite-size restriction changes it.