Past exam of the short options course of the University of Oxford / 2021 / 9 / b / iii / Solution

For $r>s$, $a>0$. When $0<a\le 1/4$, the roots of ${\mu }^2-\mu +a$ are real and lie strictly between zero and one. When $a>1/4$, they are a complex-conjugate pair of modulus $\sqrt{a}$. Thus the positive steady state is linearly asymptotically stable exactly when $0<a<1$, and unstable when $a>1$. Since $a=2s-2s^2/r$ increases strictly with $r$, the crossing occurs atAt the crossing,Every nonzero real linearized perturbation has period six. This establishes the intended linear stability threshold and linearized period six.

The nonlinear period-six claim in the printed paper does not follow, and is false as a local bifurcating-branch claim for this map. One can check this directly rather than infer a nonlinear period from the eigenvalues. At $r=r_c$ normalize $N_t=N^{\ast }(1+u)$, $N_{t-1}=N^{\ast }(1+v)$. The exact planar map isIts cubic Taylor expansion isHere is a concrete check of the sixfold return. Put $z=(u,v{)}^T$, $Q=u^2-uv+v^2$, andComposing the displayed cubic expansion six times cancels all quadratic terms and givesIn particularThe quartic coefficient is strictly negative. Hence, sufficiently near the equilibrium, every nonzero point has smaller $Q$ after six iterations: the critical nonlinear map has no nearby nontrivial period-six orbit, and the equilibrium itself is locally attracting despite its neutral linearization.

One can also rule out a branch of small period-six orbits on either side of the crossing. Normalize about the positive equilibrium at each nearby $r$, and put $\eta =a-1$. The derivative of the sixth power of $\left(\begin{array}{cc}1& -a\\ 1& 0\end{array}\right)$ at $a=1$ is $3I+K$. Smooth dependence on $r$ therefore givesThe leading matrix has determinant $(3\eta +\alpha Q{)}^2+3(\eta +\beta Q{)}^2$. Both terms could vanish with $Q>0$ only if $\alpha =3\beta$, butThus its norm on nonzero $z$ is bounded below by a positive constant times $(|\eta |+\Vert z{\Vert }^2)\Vert z\Vert$, and the smaller remainder cannot cancel it. There is no nearby nonzero solution of $F_{\eta }^6(z)=z$. The eigenvalues satisfy the nonresonance conditions through order four, their moduli cross outward with increasing $r$, and the cubic radial coefficient is negative. The standard Neimark–Sacker bifurcation theorem therefore gives a supercritical invariant closed curve. Its rotation is near the linear resonance $1/6$; it is not a local branch of exact period-six orbits.

Finally, the discrete population map does not preserve nonnegativity for arbitrary large lagged populations: $r/(1+b{N}_{t-1}^2)-\delta$ can be negative. The positive-equilibrium stability conclusions above are local and do not assume a global positivity property that the formula lacks.