Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 13B (Mathematical Biology)

A homogeneous equilibrium has $v_0=u_0^2$. Besides $u_0=0$, a nonzero equilibrium satisfies $r=u_0^2-u_0$, giving $u_{\pm }=(1\pm \sqrt{1+4r})/2$. Concentrations are nonnegative, so only nonnegative roots are physically admissible. At a positive equilibrium, the reaction Jacobian matrix isNegative trace and positive determinant are equivalent toThus a positive homogeneous stable branch exists precisely when $s>1/2$ and $-1/4<r<s^2-s$, on the upper root $u_{+}$. The lower positive branch has negative determinant and is unstable. At $r=-1/4$ the branches meet and have a zero eigenvalue; at $u_0=s$ the strict linear stability condition also fails.

The trivial equilibrium $(0,0)$ has eigenvalues $r,-s$, so it is asymptotically stable for $r<0$. It exists for all $r$, independently of the nonzero branches. If negative $u_0$ were allowed mathematically, the negative branch for $r>0$ would also be homogeneously stable, but it does not represent a chemical concentration.

A Fourier mode with $q=k^2\ge 0$ evolves under $J-q\mathrm{diag}({\kappa }_1,{\kappa }_2)$. Its trace remains negative for a homogeneously stable positive branch. Its determinant isIf $u_0{\kappa }_2\le s{\kappa }_1$, this is positive for every $q\ge 0$. Otherwise its minimum is attained at $q_{\ast }=(u_0{\kappa }_2-s{\kappa }_1)/(2{\kappa }_1{\kappa }_2)$. Stability requires $D(q_{\ast })>0$, orIn terms of $d=\sqrt{{\kappa }_2/{\kappa }_1}$, the boundary is the positive root of $u_0{d}^2-s=2d\sqrt{s(2u_0^2-u_0)}$:Within $1/2<u_0<s$, all modes decay for $d<d_c$; a finite band grows for $d>d_c$, giving a Turing instability. Equality is neutral at one nonzero wavenumber.

For $s=2$, the positive lower branch $0<u_0<1/2$ is unstable for every $d$. The upper branch is homogeneously stable for $1/2<u_0<2$: it is stable to all modes to the left of $d_c=2+\sqrt{4-2/u_0}$ and unstable to spatial modes to the right. For $u_0>2$ it is already homogeneously unstable. The zero branch is stable to every spatial mode when $r<0$, because its mode eigenvalues are $r-{\kappa }_1{k}^2$ and $-s-{\kappa }_2{k}^2$.

At onset the determinant has a double root, soSquaring the neutral-boundary equation and canceling the positive factors gives $u_0=s/[d(2\sqrt{2s}-d)]$. Substitution yieldsFor a positive branch with $u_0>1/2$, the actual neutral curve lies strictly between $\sqrt{2s}$ and $2\sqrt{2s}$. If $d<\sqrt{2s}$, then $d<d_c(u_0)$ for every such branch: a homogeneously stable state cannot have this Turing instability. If $d>2\sqrt{2s}$, then $d>d_c(u_0)$ for every finite positive $u_0$, so every homogeneously stable positive state is spatially unstable. The negative value yielded by the displayed formula outside its boundary domain is not a physical critical wavenumber and does not imply renewed stability. One must also keep $u_0<s$, which can restrict the attainable neutral curve further.