Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 6A / Solution

The positive steady state satisfies $1+b(N^{\ast }{)}^2=r$, soWith $N_t=N^{\ast }+{\eta }_t$, the linearization isIts characteristic polynomial is ${\lambda }^2-\lambda +d$, where $d=2(r-1)/r$. The roots are inside the unit disk precisely when $0<d<1$: when they are real they are positive with sum one, and when complex conjugate their common modulus is $\sqrt{d}$. Therefore the fixed point is linearly asymptotically stable for $1<r<2$ and unstable for $r>2$. At $r=2$,This proves period six for the neutral linearized oscillations. The stronger printed claim of a nonlinear period-six branch is not valid for the stated recurrence. It is important to distinguish these two assertions.

Here is a local obstruction, which also checks that the qualification is not merely absence of a proof. Normalize by the moving fixed point, let $x=N_{t-1}/N^{\ast }(r)-1$, $y=N_t/N^{\ast }(r)-1$, and definePut $v=(x,y{)}^T$, $\delta =r-2$, and $Q=x^2-xy+y^2>0$ for $v\ne 0$. The quadratic and cubic terms of $F_2$'s second component are $x^2/2-xy+x^{2}y/2$. Composing six times givesIn particular the quadratic terms cancel, while the cubic components are $-(x+2y)Q/2$ and $(2x-3y)Q/2$. If a small nonzero $v$ were a period-six point, invertibility of $B$ and the norm of this equation would imply $|\delta |=O(|v{|}^2)$. Taking its determinant with $Bv$ then givesSince $Q\ge |v{|}^2/2$, this is impossible for sufficiently small nonzero $v$. Thus no nontrivial nonlinear six-cycle bifurcates locally at $r=2$. At the threshold the same expansion gives $Q(F_2^6(v))=Q(v)-2Q(v{)}^2+O(|v{|}^5)$, so the fixed point is actually locally asymptotically stable nonlinearly, with slow decay. The valid intended linear-stability conclusion is the period-six neutral oscillation above; a literal nonlinear assertion requires a changed model or wording. This is an example of the distinction between linear resonance and nonlinear periodic bifurcation.