Turing threshold of the Brusselator

When $u$ and $v$ have diffusivities $D$ and $1$, respectively, and the homogeneous Brusselator equilibrium is stable, a Turing instability occurs precisely whenAt onset,

On a specified finite domain, only its boundary-compatible wavenumbers can grow. Solve the Brusselator mode determinant for its neutral value of $b$ and minimize over those nonzero modes. Minimizing over all continuous $k$ instead gives the usual threshold and $k_c^2=a/\sqrt{D_u{D}_v}$. The finite-mode minimum must still precede the homogeneous threshold to describe a Turing instability.

For the Brusselator, the stationary spatial threshold is $b_T=(1+a\sqrt{D_u/D_v}{)}^2$ and the homogeneous oscillatory threshold is $b_H=1+a^2$. A genuine Turing instability appears first only if $b_T<b_H$, which rearranges to the displayed diffusivity-ratio bound. Equality gives simultaneous stationary spatial and homogeneous oscillatory neutral modes. Otherwise homogeneous loss of stability occurs first.