Turing pattern

Remarks

The perturbation $u = u^* + e^{\lambda t} e^{i q \cdot x} w$ with a wave vector $q \in \R^n$, $\abs{q} = k$, solves the linearised system exactly when $\lambda$ is an eigenvalue of $J - k^2 D$ with eigenvector $w$, which explains clause 2. Example: $m = 2$, $J = \begin{pmatrix} 1 & -1 \\ 3 & -2 \end{pmatrix}$, $D = \operatorname{diag}(1, 10)$. Then $\tr J = -1 < 0$ and $\det J = 1 > 0$, so clause 1 holds. For $k^2 = 0.3$, $\det(J - k^2 D) = (0.7)(-5) + 3 = -0.5 < 0$, so $J - k^2 D$ has a positive real eigenvalue and clause 2 holds. Non-example: if $D$ is a multiple $d I$ of the identity, the eigenvalues of $J - k^2 D$ are those of $J$ shifted by $-d k^2$, so clause 2 fails whenever clause 1 holds. For $m = 2$, a necessary condition is that the two diffusion coefficients differ.