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For the Hénon map $(x,y)\mapsto (y,\mu -bx-y^2)$ with $b>-1$, fixed points exist at $y_{\pm }=-(1+b)/2\pm \sqrt{\mu +(1+b{)}^2/4}$. Their multipliers satisfy ${\lambda }^2+2y\lambda +b=0$. The Jury stability criterion gives asymptotic stability only for the upper branch, with $-1<b<1$ and $-(1+b{)}^2/4<\mu <3(1+b{)}^2/4$. The two-cycle has alternating values $y,z$ satisfying $y+z=1+b$ and $(y-z{)}^2/4=\mu -3(1+b{)}^2/4$. Its second-iterate trace is $4[(1+b{)}^2-\mu ]-2b$ and determinant $b^2$, so it is attracting for $-1<b<1$ until $\mu =(5b^2+6b+5)/4$. The fixed-point fold and first flip are nondegenerate away from endpoints and persist under small smooth perturbations.

Suppose $r>1$ and set $q=\log r/(r-1)$. Substitution of the positive equilibrium into the Jacobian matrix givesPart (b) showed $f(x)<0$ whenever $x\ne 1$. In particular,gives $\log r<r-1$, whereasgives $\det J_{\ast }>1$. The latter already violates the Jury stability criterion condition $1-\det J_{\ast }>0$. More directly, the product of the two eigenvalues is greater than one, so at least one has modulus greater than one. Therefore the positive equilibrium is unstable. This is the instability of the Nicholson-Bailey equilibrium; parasitoid feedback in this simple model does not stabilize coexistence.

At a fixed point $x=y=p$, the quadratic equation is $p^2+(1+b)p-\mu =0$. ConsequentlyThe Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -b& -2p\end{array}\right)$, and its Floquet multipliers solve ${\lambda }^2+2p\lambda +b=0$. The Jury stability criterion requires $|b|<1$ and $1+2p+b>0$, $1-2p+b>0$. With $b>-1$, this becomes $-1<b<1$, $-(1+b)/2<p<(1+b)/2$. The lower branch never satisfies it; the upper branch is attracting precisely whenAt the lower boundary the two fixed points merge with Floquet multipliers $1,b$, a saddle-node bifurcation for $b\ne 1$. At the upper boundary the upper point has Floquet multipliers $-1,-b$, giving a generic period-doubling bifurcation for $-1<b<1$.

Its direction can be established explicitly. A distinct two-cycle alternates values $y,z$ satisfying $(1+b)z=\mu -y^2$, $(1+b)y=\mu -z^2$. Subtracting gives $y+z=1+b$, and substitution givesIt exists beyond the fixed-point period-doubling bifurcation. The second-iterate Jacobian matrix has determinant $b^2$ and trace $4yz-2b=4[(1+b{)}^2-\mu ]-2b$. Applying the same stability inequalities shows attraction for $-1<b<1$ andThus the first period-doubling bifurcation is supercritical; the next upper boundary has a Floquet multiplier $-1$ for the two-cycle. These are the fixed-point and two-cycle thresholds of the Hénon map.

At $b=1$ the map preserves phase-space area, so attraction is impossible. The fixed-point fold is at $\mu =-1$, with double Floquet multiplier $+1$, and is a conservative saddle-centre bifurcation. The upper branch $p_{+}=-1+\sqrt{1+\mu }$ is linearly elliptic for $-1<\mu <3$; the lower branch is a saddle fixed point of a map. At $\mu =3$, the upper branch has double Floquet multiplier $-1$ and becomes a saddle fixed point of a map for larger $\mu$. The two-cycle $y,z=1\pm \sqrt{\mu -3}$ is elliptic for $3<\mu <4$, then reaches double Floquet multiplier $-1$ for the second iterate at $\mu =4$. The partial sketch distinguishes elliptic, neutrally stable Floquet multipliers from attracting states. Resonances may require additional nonlinear analysis; unit Floquet multiplier moduli alone do not prove nonlinear Lyapunov stability.

The $b=1$ segment is a conservative Hénon stability boundary, not an ordinary nondegenerate dissipative Neimark–Sacker bifurcation. Indeed for $b>0$ constant phase-space area scaling is $\mathrm{area}(F(D))=b\mathrm{area}(D)$. A simple invariant closed curve would enclose a bounded invariant set $D$ of positive phase-space area, forcing $b=1$. Thus no such closed invariant curve can bifurcate into $b\ne 1$ within this constant-determinant family.

Generic folds and flips away from these degeneracies persist under sufficiently small smooth perturbations: their simple unit Floquet multiplier, transverse parameter crossing and nonzero leading nonlinear coefficient persist, while the other Floquet multiplier stays away from the unit circle. The implicit function theorem then continues the bifurcation curves and their criticality. The unqualified perturbation claim needs this restriction. Arbitrary perturbations need not preserve the conservative $b=1$ behavior. For example, in local complex coordinates at a nonresonant elliptic fixed point, the area-preserving map property gives a cubic normal form $z^{\prime }=e^{i\theta }z(1+i\tau |z{|}^2)+\cdots$ with zero cubic radial damping. Composing with the arbitrarily small radial perturbation $z\mapsto (1-\eta |z{|}^2)z$ introduces radial coefficient $-\eta$. Varying the modulus through one can now give a genuine Neimark–Sacker bifurcation. Conservative degeneracies and resonances are therefore not covered by a blanket structural-stability assertion.

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 equationAt an equilibrium point, either $(a,b)=(0,0)$ orOnly nonnegative populations are admissible. At extinction the Jacobian matrix is $\left(\begin{array}{cc}1-k& 1\\ \mu & 0\end{array}\right)$ 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 $(1-k)M+kM/(1+a_n)\le M$, 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 areFor $k\le 2/3$ there is no upper bound on $\mu$; for $k>2/3$ the stable region is $k<\mu <k^2/(3k-2)$.

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 givesThus 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 givesAt 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 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 $u=a-a_{\ast }$ and write the centre graph as $b-b_{\ast }=H(u)$ with $H^{\prime }(0)=-{\mu }_c$. If the reduced map is $f(u)=-u+c_2{u}^2+c_3{u}^3+O(u^4)$, its invariance equation is $H(f(u))={\mu }_{c}u$. Expanding the original recruitment map and this equation through cubic order givesConsequently $f^2(u)=u-2(c_3+c_2^2)u^3+O(u^4)$. 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 thereforeThe 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.

The Jury stability criterion for ${\lambda }^2-\lambda +r$ gives both roots inside the unit disc exactly whenThe decay is monotone when both roots are real and positive. Since the discriminant is $1-4r$, this occurs for

The Jacobian matrix at the origin and its characteristic polynomial areThe fixed-point multipliers areBy the Jury stability criterion, both lie strictly inside the unit circle exactly whenThe last condition always holds, since ${\mu }^2+\mu +2=(\mu +1/2{)}^2+7/4$. Therefore the origin is asymptotically stable exactly for $-1<\mu <0$ among the hyperbolic parameter values.

For completeness, its linear stability throughout the parameter line is as follows. When $\mu <-1$, the complex-conjugate Floquet multipliers have modulus $|\mu |>1$, so the origin repels. When $-1<\mu <-1/3$, their modulus is $|\mu |<1$, so it attracts with oscillation. At $\mu =-1/3$ the repeated Floquet multiplier is $1/3$, still attracting. For $-1/3<\mu <0$, both Floquet multipliers are real and in $(0,1)$. For $0<\mu <1$, $p(1)<0$ and $p(0)>0$, so one Floquet multiplier lies in $(0,1)$ and the other exceeds one: the origin is a saddle fixed point of a map. For $\mu >1$, the complex-conjugate Floquet multipliers have modulus $\mu >1$, so it repels.

Possible changes at the unit circle are exhausted by $p(1)=0$, $p(-1)=0$, or a nonreal pair with product one. They giveAt $\mu =0$ the Floquet multipliers are $0,1$; the nonzero fixed point branches in part (a) meet the origin, and part (c) identifies a subcritical pitchfork bifurcation. Indeed at $\mu =0$ the exact restricted map is $y\mapsto y+y^3$, which makes the origin unstable despite its neutral linear direction.

At $\mu =-1$ the Floquet multipliers are $i,-i$: this is a 1:4 strong resonance of a planar map, coinciding with the change from repelling to attracting as $\mu$ increases. A generic nonresonant Neimark–Sacker bifurcation conclusion would be unjustified. At $\mu =1$ there is a double Floquet multiplier $+1$; the matrix has a nontrivial Jordan block, and the two further nonzero fixed point branches meet the origin. Linear stability alone is inconclusive at these unit-modulus boundary values. The repeated Floquet multiplier at $\mu =-1/3$ is strictly inside the unit circle and gives no bifurcation.

A fixed point $z_{\ast }$ of a map $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ satisfies $F(z_{\ast })=z_{\ast }$. Its linear perturbations evolve by the Jacobian matrix $DF(z_{\ast })$: both eigenvalues must lie strictly inside the unit disk for linear asymptotic stability. One inside and one outside gives a saddle; both outside give a repeller. Unit-modulus eigenvalues require nonlinear analysis, unlike the continuous-time test using signs of real parts.

Here a fixed point has $x=y=p$ and $2p=a(p-p^3/3)$. ThereforeAt zero the characteristic equation is ${\lambda }^2+\lambda -a=0$. Its negative root is less than $-1$ for every $a>0$, while its positive root crosses $+1$ at $a=2$. Thus zero is a saddle for $0<a<2$ and a real repeller for $a>2$, with a pitchfork of fixed-point branches at $a=2$.

At either nonzero point the characteristic equation is ${\lambda }^2+\lambda +2a-6=0$, whose discriminant is $25-8a$. For $2<a<3$ the fixed points are saddles. At $a=3$ their periodic-orbit multipliers are $0,-1$, giving a flip threshold. For $3<a<25/8$ they are attracting real nodes; at $25/8$ repeated-eigenvalue degenerate nodes; and for $25/8<a<7/2$ attracting foci. The Jury stability criterion conditions reduce to $a>3$ and $a<7/2$, confirming the full stable interval.

At $a=7/2$ the periodic-orbit multipliers are $e^{\pm 2\pi i/3}$. For larger $a$ they form a repelling focus equilibrium. This is a unit-circle crossing at a strong resonance of a planar map with ratio $1:3$; linearization alone does not justify asserting a generic nonresonant Neimark-Sacker bifurcation. The period-two branch at the flip threshold is derived next, including an additional family absent from the source's proposed quadratic.

A genuine two-cycle must have the form $(x,y)\mapsto (y,x)$ with $x\ne y$. Hence $a(x-x^3/3)=x+y=a(y-y^3/3)$. Subtracting givesLet $S=x+y$ and $P=xy$. Using $x^3=3x-PS$ in either original equation gives $S(aP/3-1)=0$. The cubic swap-map two-cycle factorization exhibits two cases, so cancellation of $S$ would lose an actual branch.

If $S\ne 0$, then $P=3/a$ and $S^2=3+3/a$. Thus the source's quadratic holds for this family, withThere are two reflected genuine cycles for $a>3$, coalescing with the fixed points $p=\pm 1$ at $a=3$. For their two-step Jacobian matrix, $\mathrm{tr}D{F}^2=4-a$ and $\det D{F}^2=-2a^2+3a+9$. The unit-disk necessary inequality $1-\mathrm{tr}D{F}^2+\det D{F}^2>0$ fails, since it is $-2(a-3)(a+1)<0$. Therefore these quadratic-family cycles are unstable for every $a>3$. For these unstable asymmetric cycles of a cubic swap map, their periodic-orbit multipliers are real: one exceeds $1$, while the other crosses $-1$ at $a=(1+\sqrt{29})/2$. They are saddles before that secondary threshold and repelling beyond it. Their presence on the stable side of the fixed-point flip identifies the unstable branch of a subcritical flip.

If $S=0$, then $y=-x$ and the preceding quadratic identity gives $x^2=3$. Hence there is also the exact cycleThis antipodal cycle of a cubic swap map does not satisfy the printed $P=3/a$ condition: its product is $-3$. Along it the one-step Jacobian matrix is the same matrix $B=\left(\begin{array}{cc}0& 1\\ -2a& -1\end{array}\right)$ at both points, so the cycle periodic-orbit multipliers are the squares of the roots of ${\lambda }^2+\lambda +2a=0$. The Jury stability criterion conditions giveAt $a=1/2$ its periodic-orbit multipliers are on the unit circle and linearization is inconclusive. Its real-to-complex transition occurs at $a=1/8$. Thus the printed assertion about all period-two components is incomplete: it describes only $S\ne 0$, whereas the omitted antipodal cycle supplies the actual stable period-two interval.