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The displayed map has a symmetric periodic orbit $\{\pm \sqrt{\mu +1}\}$ for $\mu >-1$, with multiplier $(-2\mu -3{)}^2>1$. For $\mu >2$, put $s_{\pm }=(\mu \pm \sqrt{{\mu }^2-4})/2$. The two additional period-two orbits are $\{\sqrt{s_{+}},\sqrt{s_{-}}\}$ and their negatives. Their multiplier is $9-2{\mu }^2$, so they are stable for $2<\mu <\sqrt{5}$ and undergo a period-doubling bifurcation at $\sqrt{5}$. At $\mu =2$ these points coincide with fixed points, not genuine two-cycles.

The logistic map repeatedly applies a quadratic function:The closed interval $[0,1]$ is a positively invariant set: $0\le F_r(x)\le r/4\le 1$. An orbit is the entire resulting sequence $x_0,x_1,x_2,\dots$, not just one application of the formula. For example, with $r=2$ and $x_0=0.2$, the first states are $0.2,0.32,0.4352,0.49160192,\dots$, approaching the fixed point $1/2$.

As a simple model of population dynamics, $x_n$ is a scaled abundance in generation $n$. When abundance is small, $x_{n+1}/x_n=r(1-x_n)$ is close to $r$; increasing abundance reduces this per-capita multiplier. The falling side of the quadratic function represents sufficiently severe crowding reducing the next generation. The scale $x=1$ is not generally the carrying capacity: the positive fixed point is $1-1/r$. The logistic map is a discrete-generation model; the continuous-time logistic growth equation has a different evolution rule and does not produce its period-doubling cascade.

The practical question is how abundance behaves after the initial adjustment: does it settle, alternate, or continue varying? The same formula permits all these possibilities. The following figure shows three orbits, their cobweb plots, and a bifurcation diagram. In a cobweb plot, moving vertically to $y=F_r(x)$ calculates the next state, and moving horizontally to $y=x$ makes that value the input for the next step.

The fixed points solve $F_r(x)=x$:The second fixed point is in $[0,1]$ for $r\ge 1$ and coincides with zero at $r=1$. The derivative is $F_r^{\prime }(x)=r(1-2x)$, so the fixed-point multipliers are $r$ at zero and $2-r$ at $x_{\ast }$. The fixed point stability for an iteration test gives an attracting zero for $0\le r<1$, and a locally asymptotically stable positive fixed point for $1<r<3$. A multiplier of modulus one needs a nonlinear argument; the strict derivative test alone settles neither $r=1$ nor $r=3$.

This derivative test has a direct interpretation. A small error $e_n=x_n-x_{\ast }$ obeysAt $r=2.8$, the leading multiplier is $-0.8$: the error changes sign, explaining the alternating approach, while its magnitude eventually shrinks by approximately a factor $0.8$ per step. At $r=3.2$, the multiplier is $-1.2$, and the positive fixed point repels nearby states. However, a different outcome has asymptotic stability: the logistic map two-cycle attracts nearby orbits.

For $3<r<1+\sqrt{6}$, the attracting period-two orbit has the two valuesAt $r=3.2$ they are approximately $0.513045$ and $0.799455$, with $F_r(x_{-})=x_{+}$ and $F_r(x_{+})=x_{-}$. Their periodic-orbit multiplier is $4+2r-r^2=0.16$, so a small displacement shrinks over two steps even though the fixed point is unstable. The brief algebraic justification is to factor $F_r^2(x)-x$, remove the fixed point factors, and multiply the two derivatives using the chain rule.

At $r=3$, the positive fixed-point multiplier passes through $-1$ and a period-doubling bifurcation creates the period-two orbit. At $r=1+\sqrt{6}\approx 3.44949$, its periodic-orbit multiplier passes through $-1$ and an attracting period-four periodic orbit emerges. At $r=3.5$, the period-four values are approximately $0.382820,0.826941,0.500884,0.874997$ in that order. Further period-doubling bifurcations accumulate near $r\approx 3.56995$. Beyond that value there are also parameter windows with attracting periodic orbits, including a period-three window near $r=3.83$. Neither a crowded numerical plot nor a parameter beyond the accumulation value proves sensitive dependence on initial conditions on the attracting set.

The endpoint $r=4$ supplies an analytically transparent example. The homeomorphism $h(u)={\sin }^2(\pi u/2)$ satisfieswhere $T$ is the full tent map. Thus these maps have topological conjugacy; stretching and folding in the tent map explain sensitive dependence on initial conditions and Devaney chaos for $F_4$ on $[0,1]$. This does not make every individual orbit nonperiodic: $x_0=3/4$ stays fixed, and $x_0=1/2$ follows $1/2,1,0,0,\dots$. Sensitivity means that these exceptional orbits also have arbitrarily close initial states that eventually depart substantially.

Robert M. May's "Simple mathematical models with very complicated dynamics" (1976), Nature 261, pp. 459–467, reviewed how simple deterministic recurrence relations can pass from stable abundances through period-doubling cascades to irregular behaviour. May presented the logistic map as an especially accessible example and discussed the relevance of such models to population dynamics. The paper explicitly describes itself as a review of results found by several researchers, bringing their findings together for a broad scientific audience.^[ref]

For the logistic map $F(u)=\lambda u(1-u)$, factor $F(F(u))-u$ and remove its two fixed points. The remaining quadratic equation is ${\lambda }^2{u}^2-\lambda (\lambda +1)u+\lambda +1=0$. Its distinct roots form a periodic orbit of period two when $\lambda >3$. The periodic-orbit multiplier is $F^{\prime }(u_{-})F^{\prime }(u_{+})=4+2\lambda -{\lambda }^2$, so this orbit is asymptotically stable for $3<\lambda <1+\sqrt{6}$. It is born at a period-doubling bifurcation at $\lambda =3$.

For the quadratic unimodal map, an attracting fixed point lies on the positive branchIts multiplier has modulus less than one for $0\le \mu <3/4$ and reaches $-1$ at the first period-doubling bifurcation $b_1=3/4$. The resulting two-cycle has pointsIts periodic-orbit multiplier is $4{\mu }^2{x}_{+}{x}_{-}=4(1-\mu )$, so it is attracting for $3/4<\mu <5/4$ and doubles at $b_2=5/4$. Successive attracting periods are $4,8,16,\dots$, with their parameter intervals shrinking toward $s_{\infty }\simeq 1.401155189$.

A superstable periodic orbit contains the critical point zero, so its derivative product vanishes. Along the first period-doubling cascade, $s_n$ is the first-cascade parameter with a critical orbit of exact period $2^n$, satisfying $f_{s_n}^{2^n}(0)=0$. For example, $s_1=1$, $s_2\simeq 1.310702641$, and $s_3\simeq 1.381547484$. These are interior points of stability intervals, distinct from the bifurcation parameters $b_n$.

The following bifurcation diagram shows attracting orbit values, including the first bifurcations and superstable points. At the accumulation parameter the limiting critical orbit is not a finite attracting cycle; its closure is the period-doubling limit set.

The Feigenbaum constants use the positive spatial-scaling convention. Their parameter definition isThe same limit is obtained from successive bifurcation parameters. To define spatial scaling without ambiguity about signs, letThis is the signed central return separation in the superstable $2^n$ orbit; its sign alternates between successive levels of the primary cascade. ThenEquivalently, the ratio of the corresponding unsigned separations tends to $\alpha$. Some conventions call the signed ratio itself $\alpha$; here the spatial scale of the Feigenbaum renormalization fixed point is $a=-1/\alpha <0$, matching the orientation reversal in the doubling operator.

The quadratic discrete predator-prey map equals the identity plus the vector field of part (a), so its admissible fixed points are the same $O,E,P$. Their Jacobian matrices are the identity plus the corresponding continuous-time Jacobian matrices, but map stability requires eigenvalues inside the unit circle rather than the left half-plane.

At $O$ the multipliers are $1/a,0$: it is a sink for $a>1$ and a saddle for $a<1$. At $E$ they are $2-1/a$ and $(1-a)/b$. ThereforeFor $a<1/3$, the prey multiplier is below $-1$: $E$ is a saddle when $b>1-a$ and a repeller when $b<1-a$. For $1/3<a<1$, $b<1-a$ makes it a saddle. The $+1$ crossings at $a=1$ and $a+b=1$ are the two transcritical bifurcations already encountered. The crossing at $a=1/3$ is a prey period-doubling bifurcation.

At positive coexistence,For clarity, derive the relevant Schur stability criterion directly: setting $z=(1+s)/(1-s)$ in $z^2-Tz+D$ givesThe open unit disk maps to $\mathrm{Re}s<0$. A real quadratic has both roots there precisely when all three displayed coefficients have the same positive sign. Since $1-T+D=c>0$, the coexistence fixed point is a sink precisely whenThis region exists only for $a>1/9$. In the coexistence triangle $a+b<1$, put $K=1+T+D=(3a+1-3b)/a$. If $K<0$, one root lies below $-1$ and the other in $(-1,1)$, so $P$ is a saddle. If $K>0$ but $D>1$, it is a repeller; if $K>0,D<1$, it is a sink. These signs give the complete local fixed-point partition, including real and complex multipliers.

The coexistence period-doubling bifurcation lies at $b=a+1/3$; its segment adjoining the sink region has $1/9<a<1/3$. The other multiplier there is $2-1/(3a)$ and lies strictly inside the unit circle on this segment. This coexistence flip is subcritical on that segment. To calculate its direction, normalize the $-1$ eigenvector by $q_f=(1,(2-b/a)/b{)}^T$ and the left eigenvector by $p_f{q}_f=1$. The symmetric quadratic derivative $\mathcal{B}$ is given explicitly in part (d). Eliminating the quadratic term from the one-dimensional centre manifold map gives the effective cubic flip coefficientFor a scalar reduced map $u^{\prime }=(-1+\delta )u+c_f{u}^3+\cdots$, its second iterate satisfies $u^{\prime \prime }-u=-2\delta u-2c_f{u}^3+\cdots$. Thus $u^2=-\delta /c_f>0$ on the stable-fixed-point side $\delta >0$, and the two-cycle multiplier is $1+4\delta +\cdots >1$. An unstable coexistence two-cycle meets the stable fixed point as $b$ increases to the flip curve. The other multiplier remains inside the unit circle. The complex unit-modulus crossing lies at $a+2b=1$, with $1/9<a<1$. The two curves meet at $(a,b)=(1/9,4/9)$ with a double multiplier $-1$. The prey flip and predator invasion curves meet at $(1/3,2/3)$ with multipliers $-1,+1$.

The sketches mark the sink regions and all these local thresholds. They are a local atlas, not a claim that the map has only these global attractors. Indeed the full nonnegative quadrant is not invariant: $x^{\prime }=x(1-x-ay)/a$ becomes negative when $x+ay>1$. On the prey axis the logistic map already supplies new dynamics. Its nontrivial two-cycle, obtained by factoring $F^2(x)-x$ after removing the fixed-point factors, isIts longitudinal multiplier is $4+2/a-1/a^2$, so it is stable on the axis for $1/(1+\sqrt{6})<a<1/3$. The transverse multiplier is $x_{+}{x}_{-}/b^2=a(1+a)/b^2$, making the cycle stable in the plane when additionally $b>\sqrt{a(1+a)}$. Further flips and more complicated logistic dynamics show why the differential equation's three global regions cannot be transferred unchanged to its unit-step map.

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 origin's Jacobian matrix is triangular, $J=\left(\begin{array}{cc}a+1& 1\\ 0& b-1\end{array}\right)$, so its Floquet multipliers are $a+1$, $b-1$. It is attracting for $-2<a<0$, $0<b<2$. The candidate codimension-one boundaries are $a=0$, $a=-2$, $b=2$, $b=0$: respectively a $+1$ or $-1$ Floquet multiplier. No complex-pair crossing is possible at the origin.

The map is odd. To locate the nearby steady branches, write $S=x^2>0$. The first fixed-point equation gives $y=x(S-a)$; substituting into the second givesFor a symmetric two-cycle solve $F(x,y)=-(x,y)$. Then $y=x(S-a-2)$ andThese exact orbit equations for a triangular cubic map determine the amplitude and parameter side without confusing the original cubic coefficients with their centre manifold values.

On $a=0$, with $b\ne 2$, steady states have $x^2=a(2-b)+o(a)$ and $y=-ax+x^3$. The pitchfork bifurcation is supercritical for $b<2$ and subcritical for $b>2$. Its centre manifold cubic coefficient is $-1/(2-b)$. In the stable-origin boundary segment $0<b<2$, an attracting pair appears on $a>0$.

On $a=-2$, with $b\ne 0$, the two-cycle has $x^2=b(a+2)+o(a+2)$. The centre manifold map has cubic coefficient $-1/b$ and Floquet multiplier $-1+(a+2)$. Since its unstable side is $a+2<0$, the period-doubling bifurcation is supercritical for $b<0$ and subcritical for $b>0$. In particular the boundary segment $0<b<2$ is subcritical: the nearby two-cycle lies on $a>-2$, where the origin is still attracting, and is unstable in the critical direction.

On $b=2$, with $a\ne 0$, the steady branch hasIts scalar centre manifold map is $y^{\prime }=(1+b-2)y+(1+a^{-3})y^3+\cdots$. It is supercritical for $-1<a<0$, and subcritical for $a<-1$ or $a>0$. At $a=-1$ the cubic vanishes: the exact equation gives $b-2=-3S^2+O(S^3)$, so this is a degenerate pitchfork bifurcation with destabilizing quintic coefficient and subcritical quartic amplitude scaling. It is not a generic point of the codimension-one curve.

On $b=0$, with $a\ne -2$, put $A=a+2$. The two-cycle satisfiesThe centre manifold map's cubic coefficient is $1-A^{-3}$. Thus the period-doubling bifurcation is subcritical for $-2<a<-1$, and supercritical for $a<-2$ or $a>-1$. In the stable-origin edge $-1<a<0$, a stable small two-cycle appears for $b<0$. At $a=-1$ the cubic vanishes, requiring the calculation below. All criticality labels concern the centre manifold direction; a bifurcating orbit is fully attracting only if the remaining Floquet multiplier is also stable. The sketch marks the small branches on the appropriate sides of each boundary of the stable-origin rectangle.

For clarity, these stability labels follow directly from one-dimensional derivatives. At a pitchfork bifurcation $z^{\prime }=(1+\eta )z+c{z}^3$, the nonzero state has $z^2=-\eta /c$ and Floquet multiplier $1-2\eta$ to first order. At a period-doubling bifurcation $z^{\prime }=-(1+\eta )z+c{z}^3$, the symmetric cycle has $z^2=\eta /c$ and one-step derivative $-1+2\eta$; its two-step Floquet multiplier is $1-4\eta +o(\eta )$. A supercritical branch lies on $\eta >0$ and is attracting in the centre manifold direction; a subcritical branch lies on $\eta <0$ and is repelling there.

At $(a,b)=(-1,0)$ the Floquet multipliers are $0,-1$, so the degeneracy is not two simultaneous unit Floquet multipliers. On its centre manifold, $x=-y-y^3+O(y^7)$ and direct substitution gives $y^{\prime }=-y-3y^5+O(y^7)$. With $\alpha =a+1$ small, the unfolding to leading weighted order isThis is a generalized flip bifurcation: one parameter crosses $-1$, and the other changes the cubic sign. Writing $S\simeq y^2$, its two-cycles obey $b+3\alpha S-3S^2=0$. For $\alpha >0$ a stable smaller two-cycle and an unstable larger two-cycle collide atThe intervening range $-3{\alpha }^2/4<b<0$ contains both. For $\alpha <0$ the period-doubling bifurcation is subcritical. At $\alpha =0$, the exact orbit equation is $3S^2-3S^3+S^4=b$, so the unstable two-cycle exists for $b>0$ with $|y|\sim (b/3{)}^{1/4}$. This proves both the codimension and the required quintic effect.

The nonzero fixed point of the logistic map is $x_{\ast }=(\mu -1)/\mu$, with multiplier $F^{\prime }(x_{\ast })=2-\mu$. It is stable for $1<\mu <3$, and its multiplier crosses $-1$ at $\mu =3$. To identify the emerging orbit explicitly, factorThe remaining two roots, interchanged by $F$, areThey coincide with $x_{\ast }$ at three and are distinct for $\mu >3$. Their cycle multiplier isIt lies between $-1$ and $1$ for $3<\mu <1+\sqrt{6}$. This establishes the first period-doubling bifurcation and the stable bifurcating two-cycle.

At ${\mu }_2=1+\sqrt{6}$, ${\Lambda }_2=-1$ and $d{\Lambda }_2/d\mu =-2\sqrt{6}\ne 0$. The second flip is nondegenerate: for $H=F^2$ at either cycle point, the cubic flip coefficient is $c=(H^{\prime \prime \prime }/6)+(H^{\prime \prime }/2{)}^2=-\mathcal{S}H/6$ when $H^{\prime }=-1$. Here $\mathcal{S}$ denotes the Schwarzian derivative. Since $\mathcal{S}F=-6/(1-2x{)}^2<0$ and $\mathcal{S}(F^2)=(\mathcal{S}F\circ F)(F^{\prime }{)}^2+\mathcal{S}F$, this coefficient is positive. Locally, after following the fixed point of $H$, its second iterate has the form $H^2(y)-y=2a(\mu -{\mu }_2)y-2c{y}^3+\cdots$, with $a=2\sqrt{6}>0$. Nonzero roots therefore appear with $y^2\sim a(\mu -{\mu }_2)/c$ on the increasing-parameter side. These are a four-cycle of $F$, proving the second period-doubling bifurcation occurs at $\mu =1+\sqrt{6}$.

The fixed-point equation is $x[a{x}^2+bx+\mu -1]=0$. Thuswhen the discriminant is nonnegative. At zero the multiplier is $F^{\prime }(0)=\mu$, so it is linearly stable for $-1<\mu <1$ and unstable for $|\mu |>1$. At a nonzero fixed point, eliminate $\mu$ to obtain the branch and its multiplier:The branch is stable precisely when $-2<x(b+2ax)<0$. These formulas specify stability throughout the diagram, including any extra flip points outside the region requested.

If $b=0$ and $a>0$, the two nonzero branches $x=\pm \sqrt{(1-\mu )/a}$ exist for $\mu <1$ and have multiplier $3-2\mu >1$. Thus $\mu =1$ is a subcritical pitchfork bifurcation. If $b\ne 0$, the branch through zero has $x\sim (1-\mu )/b$ and $M\sim 2-\mu$; its stability exchanges with zero at $\mu =1$, a transcritical bifurcation. The two nonzero roots meet atwhere $M=1$. The graph of $\mu (x)$ has nonzero second derivative $-2a$, proving a saddle-node bifurcation. At zero and $\mu =-1$, the multiplier is $-1$, indicating a period-doubling bifurcation; generically its nondegeneracy coefficient is $a+b^2$. If $a+b^2=0$, necessarily $b\ne 0$ here, composition at $\mu =-1$ instead gives $F^2(x)-x=4b^4{x}^5+O(x^6)$. For $\delta =\mu +1>0$ small, the leading return equation is $-2\delta x+4b^4{x}^5=0$, producing a two-cycle with leading amplitudes $x\sim \pm [\delta /(2b^4){]}^{1/4}$. Its return multiplier is $1+8\delta +o(\delta )>1$. Thus this special case still has a flip bifurcation, but it is degenerate and subcritical rather than the generic cubic normal form.

For $a,b>0$, the fold lies above $\mu =1$. Between it and the transcritical point, the branch $-b/(2a)<x<0$ is stable provided its multiplier remains above $-1$, while the other branch is unstable. For $a,b<0$, the fold lies below $\mu =1$: immediately above it the branch $x<-b/(2a)$ is stable and the branch between $-b/(2a)$ and zero is unstable; the branch crossing zero becomes stable for $\mu >1$. All these statements are subject to the displayed exact multiplier criterion if the diagram is extended. The sketch uses representative parameter values and stops before additional flips, as stipulated.

Let $\theta (z)=\Omega \log z+\Phi$. A fixed point of the Shilnikov return map is a periodic orbit with one marked global excursion, and satisfiesIts local passage takes logarithmically long. If the global travel time tends to $T_g$, then its physical period isEliminating $z$ and absorbing section sizes and $T_g$ into constants $C_1,C_2>0$ and a phase ${\varphi }_0$ gives the leading period–parameter relationFor the idealized leading map with a constant global travel time, the displayed two terms follow exactly; for the original flow, local and global approximation errors qualify their accuracy.

If $\delta >1$, the oscillatory term is $o(z)$ and $f_S^{\prime }(z)\to 0$. In a sufficiently small interval the map is a contraction. For $\mu <0$ there is a unique small attracting fixed point $z=-\mu +O(|\mu {|}^{\delta })$; it is an attracting periodic orbit also in the transverse direction of the full return. For $\mu \ge 0$, $f_S(z)-z=-\mu -z+O(z^{\delta })<0$ for sufficiently small positive $z$, so no periodic return orbit can remain wholly in this neighborhood. As $\mu \uparrow 0$ from below, the attracting cycle becomes the homoclinic connection andContraction also excludes higher-period cycles wholly inside that small return neighborhood. This does not exclude remote periodic orbits elsewhere in the vector field.

If $1/2<\delta <1$, the oscillatory term dominates $z$. At $\mu =0$ the fixed-point equation isThere are infinitely many positive solutions approaching zero, with phases approaching successive zeros of cosine. Their leading ratio is $z_{j+1}/z_j\to e^{-\pi /\Omega }$, so their periods satisfy $T_{j+1}-T_j\to \pi /\omega$. At these roots the sine tends to $\pm 1$, and the fixed-point multiplier has magnitude asymptotic to $A\Omega z^{\delta -1}\to \infty$. Thus these are longitudinally unstable cycles, with the thin transverse direction contracting: infinitely many saddle periodic orbits accumulate on the homoclinic connection. Expanding oscillatory lobes produce interval-covering walks and horseshoe dynamics in suitable iterates; this is the positive-saddle-value case of the Shilnikov bifurcation.

Away from the connection, the one-passage branches wiggle in the period–parameter plane. To leading order,There is no single monotone logarithmic law: the phase matters, and the exponentially smaller $-C_2{e}^{-{\lambda }_{+}T}$ term matters near cosine zeros. Folds occur when $f_S^{\prime }(z)=1$, asymptotically at $\tan \theta =\delta /\Omega$. Their parameter values alternate in sign and accumulate geometrically toward zero. On a fixed-point branch, stability is $|f_S^{\prime }(z)|<1$; small stable windows occur near stationary points of the oscillatory map. Crossing $f_S^{\prime }=-1$ gives period-doubling bifurcations, while $f_S^{\prime }=1$ gives saddle-node bifurcations of periodic orbits. Subsequent doublings and chaotic invariant sets can occur; neither all periodic orbits being stable nor a particular universal sequence for every coefficient choice is implied.

For completeness, an $m$-cycle of the return map, corresponding to a periodic orbit with $m$ global excursions, obeysand its total period isIts longitudinal multiplier is ${\prod }_j{f}_S^{\prime }(z_j)$. This is the relevant period relation for multipassage cycles; substituting their total $T$ into the one-passage cosine law would generally be incorrect.

The Ricker population map has equilibria $N=0$ and $N=K$. Since $f^{\prime }(N)=e^{r(1-N/K)}(1-rN/K)$, their linearization multipliers areThus extinction is unstable and the positive equilibrium is locally asymptotically stable for $0<r<2$, becoming unstable for $r>2$.

To verify an actual period-doubling bifurcation, rather than only a multiplier crossing, put $N/K=1+x$ and $r=2+\epsilon$. The centered map isIts second iterate minus $x$ is $2\epsilon x-\frac{4}{3}{x}^3$ to leading order. After rescaling $x=\sqrt{\epsilon }y$, the limiting nonzero roots are simple, so the implicit function theorem continues them. Besides $x=0$, two nearby points therefore appear for $\epsilon >0$, with $x_{\pm }=\pm \sqrt{3\epsilon /2}+O(\epsilon )$. The map exchanges these points, and the derivative of its second iterate there is $1-4\epsilon +O({\epsilon }^{3/2})$, so the emerging two-cycle is attracting. The bifurcation at $r=2$ is supercritical period doubling.

For nonnegative populations, $f$ increases up to $N=K/r$ and decreases thereafter to zero. Hence every update after the initial time satisfiesThe bound is attained by choosing $N_0=K/r$; it is not a bound on an arbitrary initial population.

The map normal forms at multiplier $+1$ are given in the numbered subparts below. For the multiplier $-1$ problem, the second iterate instead exposes the emerging period-two cycle.

For the multiplier $-1$ problem, write $x=X+\alpha /2$. To the required weighted order, with $X=O({\mu }^{1/2})$, the shifted map is $X_{n+1}=-X_n+b{X}_n+\gamma X_n^2+\delta X_n^3$, where $b=\beta +\alpha \gamma$. Composing it with itself cancels the quadratic term. The cubic contribution is $-2\delta X^3-2{\gamma }^2{X}^3$, while the linear correction is $-2bX$. Consequently the quadratic-cubic flip criticality calculation givesThus the second iterate has normal form $X_{n+2}=X_n+\hat{\mu }{X}_n-A{X}_n^3$, up to terms of weighted order at least four. This is the asymptotic interpretation of the stated remainder under $X=O({\mu }^{1/2})$; a uniformly valid local expansion about the exact fixed point also retains mixed parameter remainders.

Assuming $\hat{\mu }$ crosses zero transversely and $A\ne 0$, the small nonzero fixed points of the second iterate satisfy $X^2\sim \hat{\mu }/A$. They form a period-two orbit. For $A>0$ they exist on the side $\hat{\mu }>0$, where the original fixed point has become unstable, and the second-iterate multiplier is $1-2\hat{\mu }+o(\hat{\mu })$, so the cycle is stable. The period-doubling bifurcation is therefore supercritical precisely when $\delta +{\gamma }^2>0$. Equality is a degenerate case requiring higher-order terms.

The fixed points satisfy $x[rx(1-x)-1]=0$. There is always $x=0$, and the other branches areThey coincide at $(r,x)=(4,1/2)$. For $0<x<1$, $F(x)>0$ and its maximum occurs at $x=2/3$, with value $4r/27$. Consequently $\boxed{F((0,1))\subset (0,1)⟺0<r<27/4}$; equality at the upper endpoint fails because the maximum is attained inside the interval.

The fixed-point multiplier at zero is zero, so zero is attracting for every $r>0$. On either nonzero branch,The lower branch has multiplier greater than one for $r>4$ and is repelling. The upper branch is attracting for $4<r<16/3$, reaches multiplier $-1$ at $x=3/4$, and is repelling for $r>16/3$. At $r=4$ the multiplier is $+1$, and the two branches are created in a saddle-node bifurcation. The fixed point there is attracting from one side and repelling from the other: $F(x)-x=-4x(x-1/2{)}^2$.

The second change is a supercritical period-doubling bifurcation at $\boxed{(r,x)=(16/3,3/4)}$. To check its type, expand at that fixed point. The local map has quadratic coefficient $a=-20/3$ and cubic coefficient $b=-16/3$ when the multiplier is $-1$. Its second iterate has cubic coefficient $-2(a^2+b)=-704/9<0$. The multiplier decreases through $-1$ as $r$ increases, so a small attracting two-cycle is born on the $r>16/3$ side. These exhaust the fixed-point bifurcations, since their multipliers have no other crossings of $\pm 1$.

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.

For $b=0,a>0$, the nonzero branches $x=\pm \sqrt{(1-\mu )/a}$ exist for $\mu <1$ and have multiplier $3-2\mu >1$, so both repel. Zero changes from attracting below one to repelling above one: the bifurcation at one is a subcritical pitchfork bifurcation. At minus one its multiplier crosses $-1$, producing a supercritical period-doubling bifurcation; the newly stable small two-cycle lies on the side $\mu <-1$.

For $F(x)=x(\mu -x^2)$, the fixed points are zero for all $\mu$ and $\pm \sqrt{\mu -1}$ for $\mu >1$. Their multipliers are respectively $\mu$ and $3-2\mu$, giving attraction for $-1<\mu <1$ and $1<\mu <2$ respectively.

Remove these fixed points from the roots of $F^2(x)=x$. The factor $\mu +1-x^2$ gives the genuine two-cycleIts multiplier is $(2\mu +3{)}^2>1$, so it is always unstable. It collapses to zero at $\mu =-1$, the subcritical period-doubling bifurcation of zero.

The remaining equation is $x^4-\mu x^2+1=0$, whose positive roots in $x^2$ exist only for $\mu \ge 2$. At $\mu =2$ they give the fixed points $\pm 1$, not genuine two-cycles. For $\mu >2$, put $r=\sqrt{(\mu +\sqrt{{\mu }^2-4})/2}>1$. Since $\mu =r^2+r^{-2}$, $F(r)=r^{-1}$ and $F(r^{-1})=r$. There are exactly two further cycles,Their common multiplier is $(\mu -3r^2)(\mu -3r^{-2})=9-2{\mu }^2$. They are attracting for $2<\mu <\sqrt{5}$ and repelling for $\mu >\sqrt{5}$, with marginal multiplier $-1$ at $\sqrt{5}$. Thus there are no cycles for $\mu \le -1$, one for $-1<\mu \le 2$, and three for $\mu >2$.