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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.

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