Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 13B / c

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.