Past exam of the topics in applied mathematics course of the University of Oxford / 2010 / A / 3

A uniform steady state satisfies $f(u^{\ast },v^{\ast })=g(u^{\ast },v^{\ast })=0$. Set $u=u^{\ast }+\tilde{u}$, $v=v^{\ast }+\tilde{v}$ and keep only first-order terms in the perturbations. Since the base state is constant, its Laplacian vanishes. Taylor expansion givesThe zero-flux boundary conditions linearize to homogeneous Neumann conditions, ${\partial }_n\tilde{u}={\partial }_n\tilde{v}=0$. The off-diagonal entries of the Jacobian describe the cross-effects of the two chemicals, while the diagonal entries describe local self-response.

A diffusion-driven instability means that the homogeneous reaction equilibrium is asymptotically stable against spatially uniform perturbations, but adding diffusion makes at least one admissible spatially nonuniform mode grow. Thus the eigenvalues of $J$ must have negative real parts, while $J-k^2\mathcal{D}$ has an eigenvalue with positive real part for some positive Neumann Laplacian eigenvalue $k^2$ on the actual domain.

The instability is driven by different diffusion responses acting together with reaction feedback. It is not merely an equilibrium that was already unstable without diffusion. On a bounded domain the permitted wave numbers are discrete, so an algebraically unstable continuous band must also contain an actual spatial eigenmode.

Write $a=f_u$, $b=f_v$, $c=g_u$, $d=g_v$, $\Delta =ad-bc$. Stability without diffusion requires $a+d<0$ and $\Delta >0$. For a mode with $q=k^2$, the characteristic equation isFor positive diffusivities the trace stays negative for every $q\ge 0$, so growth can occur only if $h(q)<0$. Put $S=D_{v}a+D_{u}d$. This upward-opening quadratic has a negative interval at positive $q$ exactly when $S>0$ and $S^2>4D_u{D}_v\Delta$, orIf $D_u=D_v=D_0$, the right side is $D_0(a+d)<0$, impossible. Equivalently diffusion would merely subtract $D_{0}q$ from both reaction eigenvalues. Thus unequal diffusivities are necessary.

The potentially unstable band has endpointsThese are necessary continuous-wave-number conditions. Actual instability on the specified bounded domain additionally requires an allowed $q$ strictly between them. The strict inequalities exclude neutral onset and presume both diffusivities are positive, as in an ordinary two-chemical diffusion model.

After relabelling the chemicals if necessary, take $f_u>0$ and $g_v<0$. At least one diagonal entry must be positive because $D_v{f}_u+D_u{g}_v>0$, whereas their sum is negative. Since their product is negative and $\det J>0$, the off-diagonal product must be negative and sufficiently large in magnitude. The two possibilities areIn the first, $u$ activates its own production and stimulates $v$, while $v$ suppresses $u$: an activator-inhibitor mechanism. In the substrate-depletion pattern mechanism, $v$ promotes the autocatalytic species $u$, which consumes or suppresses $v$: a substrate-depletion mechanism. These are different biological interaction signs, not merely two drawings of the same feedback. Suitable relative magnitudes and sufficiently different diffusivities are still required; the sign pattern alone does not guarantee a Turing instability.

For Neumann boundaries the correct eigenfunctions include zero indices:Their squared wave number is $q=(n_x\pi /L_x{)}^2+(n_y\pi /L_y{)}^2$. The constant pair $(0,0)$ is reaction-stable. Instability requires $q_{-}<q<q_{+}$, using the band derived above. If both side lengths are large, the wave-number spacings are small enough that admissible modes can enter the band in either direction, including mixed modes. This does not mean the single $(1,1)$ mode necessarily stays unstable as lengths tend to infinity: its wave number eventually tends to zero below the band. Rather, increasingly fine wave-number spacing permits some suitable higher integers.

Thin-domain Turing mode selection applies if $L_y<\pi /\sqrt{q_{+}}$, every mode with $n_y\ge 1$ has $q\ge (\pi /L_y{)}^2>q_{+}$ and is stable. A sufficiently large $L_x$ can nevertheless admit an integer $n_x\ge 1$ with $q_{-}<(n_x\pi /L_x{)}^2<q_{+}$ and $n_y=0$. HenceThe printed range starting both indices at one excludes these purely longitudinal modes and cannot imply the requested conclusion. It also omits the constant mode. Further, each allowed spatial eigenfunction generally has two temporal eigenvalue branches of $J-q\mathcal{D}$, with coefficients along their eigenvectors; a single unrestricted scalar $\lambda (q)$ is shorthand for those branches rather than a complete general expansion.

Domain geometry can select biological pattern dimension even when the same reaction parameters and diffusion coefficients are used. A sufficiently wide developing tissue can admit instabilities with variation in two spatial directions; nonlinear development may produce spots, labyrinths or other two-dimensional patterns. A thin strip suppresses transverse unstable modes, allowing primarily bands whose concentration varies along its long axis and is nearly uniform across its width.

This is a mechanism for shape-dependent pattern selection during growth: increasing tissue width can admit additional transverse modes without changing the local chemical interactions. Linear instability predicts initial wavelengths and permitted directions, not the eventual stable spot or stripe pattern; nonlinear saturation and boundary geometry determine that later outcome. Zero-flux boundaries also mean these modes arise through internal redistribution and reaction rather than externally prescribed chemical concentrations.