Solution
= Solution
A fraction $1-k$ of adults survives each time step; $k$ is the adult mortality fraction. Adults produce $\mu a_n$ larvae, which mature one step later with recruitment reduced by adult-dependent competition through $1/(1+a_n)$. Eliminating larvae gives the <second-order difference equation>
$$
a_{n+1}=(1-k)a_n+\frac{\mu a_{n-1}}{1+a_n}.
$$
At an <equilibrium point>, either $(a,b)=(0,0)$ or
$$
\boxed{a_*=\frac\mu k-1,\qquad b_*=\mu\left(\frac\mu k-1\right),\qquad \mu>k.}
$$
Only nonnegative populations are admissible. At extinction the <Jacobian matrix> is $\begin{pmatrix}1-k&1\\\mu&0\end{pmatrix}$ and its <characteristic polynomial> is $\lambda^2-(1-k)\lambda-\mu$. Its positive root exceeds $1$ exactly when $\mu>k$. For $0<\mu<k$ both roots have modulus less than $1$; at $\mu=k$ the roots are $1,-k$, so linearization is marginal. On the nonnegative <state space> extinction is still attracting at this equality: if two consecutive adult values are bounded by $M>0$, the next is at most $(1-k)M+kM/(1+a_n)\leq M$, with strict decrease away from zero; the maximum of two consecutive adult values strictly decreases after two steps whenever it is positive; continuity on its bounded state region then excludes a positive limiting maximum. Thus the extinction <equilibrium point> is unstable exactly when the positive <equilibrium point> exists, with equality understood through nonlinear rather than strict linear stability.
At the positive <equilibrium point>, the trace and <determinant> of the <Jacobian matrix> are $T=1-2k+k^2/\mu$ and $-k$. The <Jury stability criterion> for $\lambda^2-T\lambda-k$ requires $1-T-k>0$, $1+T-k>0$ and $1-k>0$. The strict <linear stability analysis> conditions are
$$
\boxed{\mu>k,\qquad (3k-2)\mu<k^2.}
$$
For $k\leq2/3$ there is no upper bound on $\mu$; for $k>2/3$ the stable region is $k<\mu<k^2/(3k-2)$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-3-stability.png]
{title=Stability regions and equilibrium branches}
At the lower boundary $\mu=k$, the positive <equilibrium point> merges with extinction, with a multiplier $+1$ and slow recovery. At the upper boundary a multiplier reaches $-1$ (the other is $k$), giving alternating adult/larval fluctuations. The nonlinear map confirms a supercritical <period-doubling bifurcation>, as follows. For $k>2/3$, put $h=1-k$ and $\mu_c=k^2/(3k-2)$. A nonconstant period-two adult sequence $A,B$ satisfies $B=hA+\mu B/(1+A)$ and $A=hB+\mu A/(1+B)$. Subtracting and adding these equations gives
$$
S=A+B=\frac{\mu-1-h}{h},\qquad P=AB=\frac{hS(S+1)}{1+h},\qquad
(A-B)^2=\frac{S[(1-3h)S-4h]}{1+h}.
$$
Thus positive unequal values emerge for $\mu>\mu_c$, with amplitude proportional to $\sqrt{\mu-\mu_c}$. The two larval phases are $\mu B,\mu A$. If $T_2,\Delta_2$ are the trace and determinant of the two-step <Jacobian matrix>, direct multiplication gives
$$
\Delta_2=\frac{(1+h)\mu}{S+1},\qquad
1-T_2+\Delta_2=\frac{Sh(1+h)[(1-3h)S-4h]}{(S+1)\mu}>0.
$$
At onset $\Delta_2=k^2<1$ and $T_2=1+k^2$, so the other two strict <Jury stability criterion> inequalities hold for sufficiently small positive $\mu-\mu_c$. Hence \b[just beyond the upper boundary, a stable period-two population cycle replaces the stable equilibrium]. This is a local conclusion, not stability for arbitrarily large reproduction rates.
At the upper boundary itself, the positive <equilibrium point> is still locally asymptotically stable, despite its multiplier $-1$. To decide this equality case, apply the <centre manifold theorem for a discrete dynamical system>. Put $u=a-a_*$ and write the centre graph as $b-b_*=H(u)$ with $H'(0)=-\mu_c$. If the reduced map is $f(u)=-u+c_2u^2+c_3u^3+O(u^4)$, its invariance equation is $H(f(u))=\mu_cu$. Expanding the original recruitment map and this equation through cubic order gives
$$
c_2=\frac{(2-k)(3k-2)}{k(1-k)},\qquad
c_3+c_2^2=\frac{(2-k)(3k-2)^2}{k^2(k+1)}>0.
$$
Consequently $f^2(u)=u-2(c_3+c_2^2)u^3+O(u^4)$. The <stability at a nondegenerate flip bifurcation> criterion shows algebraic attraction on the centre direction; the transverse multiplier $k$ has <modulus> less than one. Including this boundary, the complete local asymptotic-stability condition for a positive population is therefore
$$
\boxed{\mu>k,\qquad (3k-2)\mu\leq k^2.}
$$
The shaded figure shows the strict linear-stability region; its upper boundary adds this nonhyperbolic attracting case.
<Demographic stochasticity> near the lower boundary can lead to absorption at extinction; near the upper boundary it excites alternating fluctuations and can blur the deterministic <period-doubling bifurcation>. These qualitative predictions depend on the chosen stochastic recruitment and mortality rules.