Diffusion-driven stationary instability

For reaction Jacobian $J=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with $\mathrm{tr}J<0$ and $\det J>0$, a Fourier mode evolves under $J-q^2\mathrm{diag}(D_u,D_v)$. Its determinant is $D_u{D}_v{q}^4-(D_{v}a+D_{u}d)q^2+\det J$. A finite unstable band requires $D_{v}a+D_{u}d>0$ and $(D_{v}a+D_{u}d{)}^2>4D_u{D}_v\det J$, with an allowed domain mode in that band.

In a bounded domain, boundary conditions select a discrete set of spatial eigenvalues. If $-\Delta {\varphi }_j=q_j^2{\varphi }_j$, the linearized two-species reaction-diffusion equation evolves this mode with matrix $J-q_j^{2}D$, where $J$ is the reaction Jacobian matrix and $D$ the diagonal diffusion matrix. An unstable band of continuous wavenumbers produces an actual growing perturbation only if it contains an allowed $q_j$.

For a one-dimensional interval of length $L$ with Neumann boundary conditions, $q_j=j\pi /L$. An instability band $(q_{-},q_{+})$ is therefore active precisely when some nonnegative integer $j$ has $q_{-}<j\pi /L<q_{+}$. A sufficiently small domain can exclude the entire band even though the corresponding infinite-domain calculation predicts instability.

If both diffusion coefficients equal $D$, each spatial mode has matrix $J-D{q}^{2}I$. Its eigenvalues are those of $J$ shifted by the negative real number $D{q}^2$. Thus a reaction-stable homogeneous state remains stable under this equal-diffusion linear mechanism.