Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 14C (Mathematical Biology)

Letbe the reaction Jacobian at the homogeneous equilibrium. Stability without diffusion givesFor a spatial Fourier mode with $-{\nabla }^2$ eigenvalue $k^2$, the linearized matrix isIts trace is even more negative than $\mathrm{tr}M$, so instability occurs exactly when its determinant becomes negative. Put $z=D_v{k}^2$ and $d=D_u/D_v$. ThenThis upward-opening quadratic is negative for some $z>0$ exactly when its minimum occurs at positive $z$ and lies below zero:The second strict inequality implies the first when written with a positive square root, so the two-species diffusion-driven instability criterion becomes

If $d=1$, the condition would requireBut homogeneous stability requires $f_u+g_v<0$. These inequalities are incompatible, so equal diffusivities cannot produce a Turing instability.

When $D_u=D_v=D$, the linearized operator isFor a Fourier mode of wavenumber $k$, its two growth rates areBoth homogeneous growth rates have negative real part, and subtracting the nonnegative scalar $D{k}^2$ moves them farther into the left half-plane. Thus no spatial mode can become unstable. This is the impossibility of a two-species Turing instability at equal diffusivities.

Forwe haveso the homogeneous equilibrium is stable for $\delta >0$. The Turing condition isWriting $y=\sqrt{d}$, equality at threshold isIts positive root isFor $d>d_c$ sufficiently close to the threshold the strict instability condition holds, whileThus unequal diffusivities are necessary, but their ratio can be arbitrarily close to one when the stable reaction matrix is itself arbitrarily close to marginal stability. This is the near-unity diffusivity ratio for a Turing instability.