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The fixed points of this cubic iteration of a map satisfy $u=0$ or $u^2=1-1/\lambda$. The nonzero real pair exists for $\lambda <0$ or $\lambda >1$. Its derivative multipliers are $\lambda$ at zero and $3-2\lambda$ at either nonzero point. The fixed point stability for an iteration criterion therefore proves local asymptotic stability of zero for $-1<\lambda <1$ and of the nonzero pair for $1<\lambda <2$. A multiplier of modulus one requires a nonlinear argument rather than this strict derivative test.

Entrainment of circadian rhythm synchronizes a circadian clock to a repeating cue, a zeitgeber. One-to-one entrainment of circadian rhythm gives one cycle per environmental cycle with a stable relative circadian phase.^[ref] The stable relation need not put an activity onset exactly at lights-off: a constant offset also expresses phase locking.

For an original local model, let $x_n$ be the activity-onset offset from lights-off on environmental cycle $n$, with a negative value meaning an earlier onset. Choose an intrinsic period $\tau =23.5\mathrm{h}$ and environmental period $T=24\mathrm{h}$. Let $D(x)$ be a stipulated daily delay, measured in hours; positive $D$ moves the next onset later. The recurrence relation isA fixed phase relationship $x_{\ast }$ requires $D(x_{\ast })=T-\tau$: here a half-hour delay balances the half-hour intrinsic advance each cycle. This equality alone describes balance; restoration after a small disturbance also requires fixed point stability for an iteration.

Choose the illustrative local law $D(x)=0.5\mathrm{h}-0.4x$. Thenfor the chosen starting offset $x_0=-3\mathrm{h}$. The error shrinks by a factor of $0.6$ each cycle, proving fixed point stability for an iteration in this model. After five cycles $x_5=-0.23328\mathrm{h}$, about fourteen minutes early. At the fixed point, the imposed delay is $0.5\mathrm{h}$ and the corrected onset interval is $\tau +D(x_{\ast })=24\mathrm{h}$, although the uncoupled period remains $23.5\mathrm{h}$.

The uncorrected comparison has $D=0$ and $x_n=-3\mathrm{h}+n(\tau -T)$, so after five cycles it is $5.5$ hours early. The two traces separate ongoing timekeeping from adjustment to a zeitgeber. The chosen linear delay law is only a local teaching construction: it is not a measured response to light, does not account for wrapping of circadian phase, and does not predict an organism's entrainment range. Nor does it include masking of circadian rhythm. In a real study, tracking several cycles after removal of the cue or after a timed perturbation supplies information that a single activity value under illumination cannot provide.

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]

Differentiate the squared-denominator population map:At the positive fixed point, $b+\bar{x}=\sqrt{r}$, so its fixed-point multiplier isSince $0<b/\sqrt{r}<1$, we have $-1<F^{\prime }(\bar{x})<1$. A small perturbation ${\epsilon }_t$ obeys ${\epsilon }_{t+1}=F^{\prime }(\bar{x}){\epsilon }_t+O({\epsilon }_t^2)$, and hence shrinks locally. Thus the positive fixed point is locally asymptotically stable by fixed point stability for an iteration.

For the cubic population iteration, put $g(u)=\lambda u(1-u^2)$. Its fixed points solveThus $u_{\ast }=0$ always exists. For $\lambda \ne 0$ there are alsowhenever the expression under the square root is positive, namely $\lambda <0$ or $\lambda >1$. At $\lambda =1$ these coincide with zero; at $\lambda =0$ only zero exists.

The criterion for fixed point stability for an iteration is $|g^{\prime }(u_{\ast })|<1$. To justify it, continuity of $g^{\prime }$ gives a small interval around the fixed point where $|g^{\prime }|\le q<1$. The mean value theorem then gives $|g(u)-u_{\ast }|\le q|u-u_{\ast }|$, so this interval is invariant and the error decreases geometrically. This proves local asymptotic stability rather than merely linear boundedness.

Here $g^{\prime }(u)=\lambda (1-3u^2)$. At zero its value is $\lambda$, so zero is locally asymptotically stable for $-1<\lambda <1$. At a nonzero fixed point it isThe strict inequality $|3-2\lambda |<1$ is equivalent to $1<\lambda <2$. Both nonzero real fixed points are therefore locally asymptotically stable on that interval. This proves the two requested existence ranges. The strict derivative criterion alone makes no conclusion at the endpoint multipliers of modulus one.

For $\lambda >1$, the multiplier at the origin exceeds one, so it is unstable. The other fixed points areUsing fixed point stability for an iteration, $|3-2\lambda |<1$ exactly when $1<\lambda <2$, and $|3-2\lambda |>1$ when $\lambda >2$. HenceThe endpoint $\lambda =2$ has multiplier $-1$ and requires nonlinear analysis, which is not part of the asserted open intervals.

For the iteration of a map $f(x)=\lambda x(1-x^2)$, a fixed point satisfies $x[1-\lambda +\lambda x^2]=0$. If $0<\lambda <1$, the bracket is strictly positive, so the sole fixed point is $0$. Its multiplier of a periodic orbit of an iteration is $f^{\prime }(0)=\lambda$, whose modulus is below one. Thus $\boxed{x=0\text{ is locally asymptotically stable}}$ by fixed point stability for an iteration.

A constant orbit is a fixed point of $F(u)=u(1+u)/2$. Solving $F(u)=u$ gives $u(u-1)=0$, henceFor fixed point stability for an iteration, perturb a fixed point $u_{\ast }$ by $v_n$. Its local evolution is $v_{n+1}=F^{\prime }(u_{\ast })v_n+O(v_n^2)$, where $F^{\prime }(u)=u+1/2$. At zero, $F^{\prime }(0)=1/2$, so perturbations contract: zero is locally asymptotically stable. For example, if $|u|\le r<1$, then $|F(u)|\le (1+r)|u|/2$; this gives an invariant neighborhood and geometric convergence to zero.

At one, $F^{\prime }(1)=3/2$, so perturbations expand: one is unstable. More explicitly, putting $u_n=1+v_n$ gives $v_{n+1}=3v_n/2+v_n^2/2$, whose magnitude increases while a nonzero perturbation remains sufficiently small. The derivative test concerns local stability; it does not assert attraction from every real initial value.

LetAt low density, $F(N)/N\to r$, so $r$ is the density-independent per-generation multiplication factor. The parameter $b$ has units of inverse population and sets the density scale at which crowding suppresses recruitment.

The fixed points satisfy $N=F(N)$. Besides extinction $N=0$, for $r>1$ there is the positive fixed pointSinceits multiplier isFor every $r>1$, $-1<F^{\prime }(N_{\ast })<1$, so fixed point stability for an iteration proves that $N_{\ast }$ is locally asymptotically stable.

Scale the population by $x_{\tau }=b{N}_{\tau }$. ThenThe map increases on $0<x<1$, decreases on $x>1$, and has global maximum $f(1)=r/4$. Starting from $x_1=1$ therefore givesFor $r>4$, writing $s=\sqrt{r}>2$ givesThus $1<L<N_{\ast }b=s-1<U$. Since $f$ decreases on $[L,U]$, $f(U)=L$ and $f(L)>f(s-1)=s-1>L$, while the global maximum gives $f(L)\le U$. Hence $f([L,U])\subseteq [L,U]$. The cobweb plot consequently remains in this invariant interval after the first iterate, and both endpoints are attained at $\tau =2,3$:The printed bound cannot include $\tau =1$, because $N_1=1/b<L/b$ when $r>4$.

A fixed point satisfiesso either $x=0$ or $x^2=1-1/\lambda$. When $0<\lambda <1$, the latter equation has no real solution, and the origin is the only fixed point. Sincethe criterion for fixed point stability for an iteration shows that the origin is stable for $\lambda <1$.

Differentiate the map:At the positive equilibrium, $k{e}^{-\lambda x_{\ast }}=1-r$, soThe fixed point stability for an iteration requires $|g^{\prime }(x_{\ast })|<1$. Existence already makes the logarithm positive, so the upper inequality is automatic. The lower inequality givesCombining it with existence yields the stability interval of the survival-augmented Ricker equilibrium:

For the real logistic map, $f(x)=x$ factors asThere is always the fixed point $x_0=0$. If $\mu \ne 0$ there is also $x_{\ast }=1-1/\mu$, distinct from zero unless $\mu =1$. At $\mu =0$ the map is constant and zero is its only fixed point. Negative $\mu$ must be retained because the paper specifies $x\in \mathbb{R}$.

Since $f^{\prime }(x)=\mu (1-2x)$, the fixed-point multipliers areThe fixed point stability for an iteration criterion gives $x_0$ attracting for $-1<\mu <1$, and $x_{\ast }$ attracting for $1<\mu <3$. The respective fixed points are repelling when $|\mu |>1$ and $|2-\mu |>1$. At the unit-modulus boundaries the linear criterion needs a nonlinear supplement.

At $\mu =1$, the two fixed point branches meet and exchange stability: this is a transcritical bifurcation. Explicitly, with $\delta =\mu -1$,which has the required nonzero parameter and quadratic terms. At the critical value itself, $f(x)=x-x^2$ attracts small positive points and pushes negative points away, so the merged fixed point is not stable from both sides.

At $\mu =-1$ the origin’s fixed-point multiplier crosses $-1$, while at $\mu =3$ the nonzero fixed-point multiplier crosses $-1$. Both are period-doubling bifurcations. To check existence and stability of the new periodic orbit, factor the second iteration of a map:The two non-fixed roots areThey form a real period-two orbit for $\mu <-1$ or $\mu >3$. Its periodic-orbit multiplier isIt is less than one in absolute value just below $-1$ and just above $3$, respectively. Thus these are supercritical bifurcations in the direction where the parent fixed point loses stability. At the critical values the fixed points themselves are weakly attracting: at $\mu =-1$, $f^2(x)=x-2x^3+x^4$; at $\mu =3$, setting $u=x-2/3$ gives $u\mapsto u-18u^3-27u^4$ after two steps. Both shrink sufficiently small nonzero absolute values. The bifurcation values for finite fixed points are thereforeThe disappearance of the formula for $x_{\ast }$ at $\mu =0$ corresponds to that branch going to infinity, not a local bifurcation at a finite fixed point.