Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 74 / 3

Adding the two homogeneous reaction rates gives $a-u$, so the unique positive Brusselator equilibrium isThe reaction Jacobian matrix at that state isA real two-dimensional linear system has strict asymptotic stability exactly when $\tau <0$ and $\Delta >0$: the eigenvalues solve ${\lambda }^2-\tau \lambda +\Delta =0$. Real eigenvalues then have positive product and negative sum; a complex conjugate pair has real part $\tau /2$. Conversely, stable eigenvalues require those signs.

Since $a>0$, the Brusselator is homogeneously stable precisely forAt $b_H=1+a^2$, the linear eigenvalues are $\pm ia$, so strict decay is lost through a homogeneous Hopf bifurcation threshold. For larger $b$ the equilibrium is unstable. The source's matrix label $A$ is here written $J$ to distinguish it from the filament modulus in the preceding question.

For linear stability analysis of a general stable two-species reaction–diffusion system, take Fourier modes proportional to $e^{\lambda t+ikx}$. Their matrix is $J_k=J-k^2\mathrm{diag}(D_u,D_v)$. Put $s=k^2$, $\tau =\mathrm{tr}J$, $\Delta =\det J$ and $B=D_v{a}_{11}+D_u{a}_{22}$. ThenThe full system is stable precisely when ${\tau }_k<0$ and ${\Delta }_k>0$ for every allowed mode. If the homogeneous state is stable, the trace already decreases with $s$; only the determinant can change sign. If $B\le 0$, its minimum for $s\ge 0$ is at zero. If $B>0$, it is at $s_{\ast }=B/(2D_u{D}_v)$. Thus the two-species Turing criterion for instability on a continuum of wavenumbers isStrict stability has $B<2\sqrt{D_u{D}_v\Delta }$ together with the homogeneous conditions. Equality at positive $B$ marks a stationary neutral mode. At that first onset,Beyond threshold, the growing band lies between the positive roots $s_{\pm }=[B\pm \sqrt{B^2-4D_u{D}_v\Delta }]/(2D_u{D}_v)$, as long as the homogeneous trace stays negative.

For the Brusselator, $B=D_v(b-1)-D_u{a}^2$. Write $r=D_u/D_v$. 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 $a^{2}r+2a\sqrt{r}<a^2$ is equivalent to $\sqrt{r}<(\sqrt{1+a^2}-1)/a$. With this strict inequality, $b_T<b<b_H$ contains a spatially unstable but homogeneously stable interval. If the ratio condition fails, the first linear instability as $b$ increases is homogeneous at $b_H$, 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 $b_H$. 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.