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For the adult-survival annual pulse-breeding predator-prey model, a positive annual fixed point exists when $r>1$ and $0<s<1$:At this point the Jacobian matrix isIf $0<(1-s)\log r<4$, the two eigenvalues are a complex conjugate pair of moduli one. The annual samples show linearly neutral seasonal oscillations. If $(1-s)\log r>4$, they form a reciprocal negative real pair and the fixed point is a saddle fixed point of a map. Equality at four is a degenerate boundary needing nonlinear analysis. For replacement rather than surviving adults the same calculation gives trace $2-\log r$ and determinant one.

The absence of an attracting annual cycle is stronger than a linear observation. Put $u=\log X$, $v=\log Y$. The update becomesIt is the composition of two nonlinear shear maps, each of determinant one: an area-preserving seasonal predator-prey map. No isolated positive periodic orbit has an open attracting basin. Neutral linear oscillations alone also do not prove nonlinear stability at every resonance. In particular, this density-independent model does not select an asymptotically stable oscillation amplitude.

The Nicholson-Bailey model instead counts parasitoid offspring from attacked hosts:Its positive equilibrium has $Y_{\ast }=\log r/\kappa$ and $X_{\ast }=Y_{\ast }/[c(1-1/r)]$. Its Jacobian matrix hasSince the determinant exceeds one while the positive trace is below two, the eigenvalues are complex with moduli greater than one: instability of the Nicholson-Bailey equilibrium produces outward oscillations near it. The two maps share an exponential prey-survival factor but have different reproduction laws and stability properties.

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.

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.

Adjoin ${\mu }_{n+1}={\mu }_n$. At the origin the stable direction is the $x$ axis, with Floquet multiplier zero, while the $y$ and parameter directions have Floquet multiplier one. By the centre manifold theorem for a discrete dynamical system, write the extended centre manifold asOn it the scalar map is $Y=(1+\mu )y-\mu h(y,\mu )+y^3$. Its discrete centre-manifold invariance equation isPut $h=h_2+O({\rho }^3)$ with $\rho =|y|+|\mu |$. Since $Y=y+O({\rho }^2)$, the quadratic equation is simply $h_2(y,\mu )=\mu y$. ThusIn particular, setting the stable variable identically to zero would lose the leading parameter correction. The reduced map through total degree three isIn the natural pitchfork bifurcation scaling $\mu =O(y^2)$ this has leading form $y_{n+1}=y_n+\mu y_n+y_n^3$. Its nonzero fixed points satisfy $y^2\sim -\mu$, and their reduced fixed-point multipliers satisfy $1-2\mu +O({\mu }^2)>1$ for small $\mu <0$. The transverse Floquet multiplier remains inside the unit circle. Therefore the bifurcation at $\mu =0$ is a subcritical pitchfork bifurcation: the origin attracts for small negative $\mu$, each nonzero branch is a saddle fixed point of a map, and the origin is a saddle fixed point of a map for small positive $\mu$.

An exact check uses the full Jacobian matrix on either nonzero branch. Its trace is $T_{\ast }=1-2\mu +3{\mu }^2$ and its determinant is $D_{\ast }={\mu }^2$. Thuswhenever a nonzero branch exists. Also $p_{\ast }(0)={\mu }^2>0$, so one real Floquet multiplier lies in $(0,1)$ and the other in $(1,\infty )$. This proves that each nonzero branch is a saddle fixed point of a map throughout $\mu <0$ and $\mu >1$, in particular near the requested bifurcation.

Use $y$ vertically and $\mu$ horizontally in the bifurcation diagram; the other coordinate is recovered from $x=\mu y$. The origin is the horizontal branch $y=0$. Draw it solid on $-1<\mu <0$, where it is asymptotically stable, and dashed on the remaining hyperbolic intervals, where it is unstable. Its three nonhyperbolic fixed points in parameter space are marked at $\mu =-1,0,1$; open markers distinguish these boundary cases from the strict linear stability classifications on the intervening intervals.

The nonzero fixed point branches are $y=\pm \sqrt{\mu (\mu -1)}$ for $\mu <0$ and $\mu >1$. Each is dashed because it is a saddle fixed point of a map, as proved in part (c). Near zero their shape is $y\sim \pm \sqrt{-\mu }$: the unstable branches lie on the same parameter side as the attracting origin, displaying the subcritical pitchfork bifurcation. The branches issuing from $\mu =1$ have $y\sim \pm \sqrt{\mu -1}$. At $\mu =-1$ no additional fixed point branch appears; the Floquet multipliers cross the unit circle at $\pm i$.

A hyperbolic saddle fixed point of a map has one eigenvalue of modulus less than one and one greater than one. The corresponding stable and unstable directions persist locally as invariant sets or manifolds under the usual smoothness and invertibility assumptions. For a noninvertible map a zero stable fixed-point multiplier is possible, and the word saddle fixed point of a map describes forward contraction in one direction and expansion in the other; backward dynamics may not be unique.

A fixed point $X_{\ast }$ satisfies $F(X_{\ast })=X_{\ast }$. The Jacobian matrix is $DF=\left(\begin{array}{cc}{f}_x& f_y\\ g_x& g_y\end{array}\right)$; its eigenvalues are the linear fixed-point multipliers used in (i)–(iii).

For the Lozi map, $y=bx$ and $x=1+bx-a|x|$. At a positive point this gives $(1+a-b)x=1$, and at a negative point $(1-a-b)x=1$. ThereforeThere is no fixed point at $x=0$, because then $y=0$ and the first component would be one. On an equality boundary the corresponding scalar equation is $0=1$, so that branch is absent, not a line of fixed points.

For $x\ne 0$, $DF=\left(\begin{array}{cc}-a\mathrm{sgn}x& 1\\ b& 0\end{array}\right)$. If $a=0$, there is exactly one fixed point $(1,b)/(1-b)$ unless $b=1$, when there is none. Its eigenvalues satisfy ${\lambda }^2=b$. It is a sink fixed point of a map for $|b|<1$ and a source fixed point of a map for $|b|>1$. At $b=-1$, the affine map about its fixed point is a quarter turn, $D{F}^2=-I$ and $F^4=I$, so it is neutral, not attracting. At $b=1$ no fixed point exists. The positive/negative branch labels follow the sign of $1-b$.

If $b=0$, $X_{+}=(1/(1+a),0)$ exists for $a>-1$ with fixed-point multipliers $-a,0$. It is a sink fixed point of a map for $-1<a<1$ and a forward saddle fixed point of a map for $a>1$. At $a=1$ the fixed-point multiplier $-1$ is neutral; on the invariant axis the interval $0\le x\le 1$ is mapped by $x\mapsto 1-x$, giving two-cycles and the fixed midpoint. The negative point $X_{-}=(1/(1-a),0)$ exists only for $a>1$ and has fixed-point multipliers $a,0$, again a forward saddle fixed point of a map. For $a\le -1$ there is no fixed point. At $a=b=0$, both fixed-point multipliers vanish and two iterates send all initial points to the fixed point.

For a genuine two-cycle let the first coordinates be $x_1,x_2$ and put $h=1-b$. The two equations are $h{x}_2=1-a|x_1|$, $h{x}_1=1-a|x_2|$. If both first coordinates had the same sign $s$, subtracting would give $(h-as)(x_2-x_1)=0$. Since distinct cycle points must have different first coordinates, this contradicts $a^2\ne h^2$. If one coordinate were zero, the remaining equations would instead force $a=\pm h$, also excluded. Hence $\boxed{x_1{x}_2<0}$.

Choose $x_1>0>x_2$. Solving $a{x}_1+h{x}_2=1$, $h{x}_1-a{x}_2=1$ givesThe required signs hold exactly when $\boxed{a>|1-b|}$. This is the unique two-cycle up to interchanging its two points, under the stipulated nondegeneracy. Borderline parameter lines can carry different degenerate families and were explicitly excluded. These are Lozi fixed points and the alternating-sign two-cycle.

For integers $k,l$,Thus reducing coordinates modulo one gives a well-defined function on the torus ${\mathbb{R}}^2/{\mathbb{Z}}^2$. Its determinant is one and its inverse matrix is the integer matrix $\left(\begin{array}{cc}1& -1\\ -1& 2\end{array}\right)$, so it induces a bijection of the torus. The Jacobian determinant one makes this bijection area-preserving.

Its eigenvalues solve ${\lambda }^2-3\lambda +1=0$, givingThere are therefore expanding and contracting eigendirections and no eigenvalue on the unit circle: the function is a hyperbolic toral automorphism. The Lyapunov exponents per iterate areThey sum to zero, as required by area preservation.

The four vertices of the unit square map to $(0,0),(2,1),(3,2),(1,1)$, the vertices of an area-one parallelogram. To reduce it modulo one, cut it by the integer grid. For each piece in the cell $[j,j+1]\times [k,k+1]$, translate by $(-j,-k)$. Up to boundary identifications these translated pieces partition the unit square, because the inverse integer matrix gives exactly one torus preimage to every point. The colored pieces in the original diagram show the same pieces before and after these translations.

For the Fibonacci identity, at $n=1$ the claimed matrix is $\left(\begin{array}{cc}{F}_2& F_1\\ F_1& F_0\end{array}\right)=A$. If the identity holds at $n$, multiplication on the right by $A$ givesEach equality uses $F_{j+1}=F_j+F_{j-1}$, with the specified indexing $F_0=F_1=1$. This proves the formula for every positive integer $n$.

A fixed point satisfies $(A-I)v\in {\mathbb{Z}}^2$. Since $A-I=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ has an inverse integer matrix, this forces $v\in {\mathbb{Z}}^2$, so the origin is the only fixed point modulo one.

For period dividing two,The second congruence gives $y=-3x(\mathrm{mod}1)$; inserting it into the first gives $5x=0(\mathrm{mod}1)$. Thus the complete list isThe first is the fixed origin. Direct application of $A$ exchanges the other four in exactly two pairs:This both verifies the periodic orbits and excludes additional ones.

Finally, if $v$ is fixed by $F^p$, then $(A^p-I)v=m$ for an integer vector $m$. The eigenvalues of $A^p-I$ are ${\lambda }_{+}^p-1$ and ${\lambda }_{-}^p-1$, neither zero. Hence its inverse exists and has rational entries by the adjugate matrix formula for an integer matrix. Therefore $v=(A^p-I{)}^{-1}m$ has rational coordinates. The derivative of the return map is $A^p$, with eigenvalues ${\lambda }_{+}^p>1$ and $0<{\lambda }_{-}^p<1$. Every periodic point is a hyperbolic saddle fixed point of a map and is unstable, although perturbations in its stable eigendirection contract.