Turing pattern

A Turing pattern is a spatially structured state produced by interactions between local reactions and diffusion. A Turing instability explains linear pattern onset when a homogeneous equilibrium is stable without diffusion but unstable to a nonzero spatial mode.

A Turing, or diffusion-driven, instability occurs when a spatially homogeneous equilibrium is stable to homogeneous perturbations but unstable to a nonzero spatial Fourier mode after unequal diffusion is introduced.

If a one-dimensional reaction-diffusion model has a strict unstable squared-wavenumber band $(q_{-},q_{+})$, its nonzero Neumann modes on $[0,L]$ have $q_n=(n\pi /L{)}^2$. Increasing $L$ from zero first reaches the band with $n=1$ at the displayed threshold. Equality is neutral; the first mode is strictly unstable for $\pi /\sqrt{q_{+}}<L<\pi /\sqrt{q_{-}}$. Thus positive growth has an infimum in domain length rather than an attained smallest length. The zero mode must be checked separately through reaction stability.

For a reaction-stable two-species equilibrium with both Jacobian matrix diagonal entries negative and positive diagonal diffusivities, each spatial mode has more negative trace after diffusion is added. Its determinant is the positive reaction determinant plus positive coefficients of $k^2$ and $k^4$. Hence every mode stays stable. Unequal diffusion alone is insufficient for a Turing instability; the reaction feedback must also permit a destabilizing weighted diagonal sum.

Suppose a two-species reaction-diffusion instability has an unstable squared-wave-number band $(q_{-},q_{+})$. On a rectangle with Neumann boundary conditions, $q=(n_x\pi /L_x{)}^2+(n_y\pi /L_y{)}^2$, with both indices nonnegative. The displayed thin-width condition excludes every $n_y\ge 1$ mode but can leave $n_y=0$ modes in the band for sufficiently large $L_x$. Omitting zero indices would incorrectly remove the longitudinal pattern. Linear onset selects allowed directions; it does not alone establish the final nonlinear pattern.

For $u_t=D{u}_{xx}-u(u-r)(u-1)-\rho (v-u)$ and $0=v_{xx}-v+u$, a Fourier mode perturbation about zero has $v=u/(1+k^2)$ and growth rate $\sigma (k)=-r-D{k}^2+\rho k^2/(1+k^2)$. If $\rho >D$, its maximum is $-r+(\sqrt{\rho }-\sqrt{D}{)}^2$ at $k^2=\sqrt{\rho /D}-1$; if $\rho \le D$, its maximum is $-r$ at zero. Replacing $(u,v,r)$ by $(1-u,1-v,1-r)$ gives the corresponding criterion at the other homogeneous equilibrium.

Let a homogeneous two-species reaction equilibrium have Jacobian matrixand let $D=\mathrm{diag}(D_u,D_v)$ with $D_u,D_v>0$. A spatial mode with Laplacian eigenvalue $-q$, $q\ge 0$, evolves by $J-qD$. Its trace remains negative, while its determinant isFor continuous wavenumbers, a Turing instability exists exactly whenThese conditions put the minimum of $P$ at positive $q$ and below zero. The unstable band isOn a finite domain, an actual Laplacian mode satisfying the boundary conditions must lie strictly in this band. For example, Neumann boundary conditions on $[0,L]$ allow $q_n=(n\pi /L{)}^2$, rather than every real $q$. The inequalities alone may therefore predict an instability band containing no available mode. At the band endpoints the mode is neutral. This separates the continuous-wavenumber kinetic criterion from finite-domain mode selection.^[ref]

When $D_u=D_v=D$, every spatial Fourier mode replaces the reaction Jacobian $J$ by $J-D{k}^{2}I$. Each eigenvalue is shifted left by $D{k}^2$, so diffusion cannot destabilize a stable homogeneous equilibrium.

Let $0<\delta <1$ and consider the reaction Jacobian matrixIt is reaction-stable because its trace is $-\delta$ and its determinant is $2\delta >0$. Choose the diffusion matrix $\mathrm{diag}(d,1)$, so $d$ is the first species' diffusivity divided by the second's. For squared wavenumber $q$, the mode determinant isIts minimum is negative at positive $q$ exactly when $d(1-\delta )-1>2\sqrt{2\delta d}$. Solving this inequality for $\sqrt{d}$ gives the thresholdThus the required diffusivity ratio can approach one as the reaction equilibrium approaches marginal stability. There is no universal gap from one independent of the kinetics. The threshold uses continuous wavenumbers; a finite domain must also have an allowed mode in the unstable band. Swapping species changes which physical ratio $d$ denotes.