Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 2 / 13E (Mathematical Biology)

Choose the nondimensionalizationThe chain rule gives ${\partial }_{\tau }={\beta }_u{\partial }_t$ and ${\partial }_x^2=({\beta }_u/D_u){\partial }_s^2$. Dividing the first equation by ${\beta }_u$ and the second by ${\beta }_v$ gives the required scaled reaction–diffusion system. Positive degradation and diffusion constants are assumed.

A nonzero spatially constant steady state satisfies $v=u$ and $v=u^2$, hence $(u,v)=(1,1)$. The first derivatives of $u^2/v-u$ there are $1,-1$, while those of $Q(u^2-v)$ are $2Q,-Q$. ThusSince $Q>0$, both eigenvalues have strictly negative real part exactly when $Q>1$. For $0<Q<1$ the equilibrium is unstable. At $Q=1$ its linear eigenvalues are $\pm i$; this is the marginal boundary and does not give asymptotic stability from linearization. The positive-trace criterion in the hint tests positive real parts; stability requires applying it to $-J$.

Set $u=1+\hat{u}(t)\cos (s/L)$ and $v=1+\hat{v}(t)\cos (s/L)$ and retain linear terms. With $k=1/L$, the Fourier mode has second derivative $-k^2\cos (ks)$. HenceThe trace and determinant of this linearization are

Assume $Q>1$, so $T(k)<0$ for every wavenumber. A Turing instability therefore occurs exactly when $D(k)<0$ for some $k$. Put $z=k^2\ge 0$. The upward quadratic $1+(1-P)z+P{z}^2$ has a negative minimum on this half-line precisely whenCombining these givesThe unstable band is $z_{-}<k^2<z_{+}$, whereThe degradation-limited diffusion distances are proportional to ${\ell }_u=\sqrt{D_u/{\beta }_u}$ and ${\ell }_v=\sqrt{D_v/{\beta }_v}$. Thus $P=({\ell }_v/{\ell }_u{)}^2$ and the criterion says the inhibitor must travel more than $(1+\sqrt{2})$ times as far as the activator before degradation. Short-range activation and longer-range inhibition permit a local peak to reinforce itself. On a finite domain the allowed wavenumbers must also intersect the unstable band; the displayed condition assumes continuously available modes.

At the threshold $P_{\ast }=3+2\sqrt{2}$, the determinant quadratic has a double zero atConsequently the scaled disturbance length in $\cos (s/L)$ isThe physical length is ${\ell }_u{L}_{\mathrm{crit}}=\sqrt{{\ell }_u{\ell }_v}$ at threshold; its full spatial wavelength is $2\pi$ times this length, not this length itself. For much shorter lengths, the large negative diffusion terms smooth both components. For much longer lengths, diffusion contributes little and the stable homogeneous reaction feedback, especially the inhibitor response, suppresses disturbances. The selected length lies between those two stabilization mechanisms.