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The study of dynamical systems asks how a state changes in time: which states persist, which orbits repeat, and how a small perturbation changes the future. A dynamical system combines a state space with an evolution rule. For a seasonal population, the state might be its abundance and one step one generation; for a harmonic oscillator, a state vector includes both position and velocity. Choosing a sufficient state is essential: position alone does not determine where an oscillator moves next.

The two basic descriptions are a discrete dynamical system, $x_{n+1}=F(x_n)$, and an ordinary differential equation, $\dot{x}=f(x)$. Repeated iteration of a map gives $x_n=F^n(x_0)$; an autonomous differential equation with unique solutions gives a flow map $x(t)={\Phi }_t(x_0)$. In either case, evolving for two successive durations agrees with evolving for their sum. The logistic map provides a particularly simple discrete dynamical system; the Lorenz system supplies a continuous-time example with a very different geometry.

Deterministic evolution says that the exact rule and exact initial condition specify one future. Predicting that future from measurements is a separate question. An instrument supplies a range of possible initial conditions, and a computation adds small errors. Near an asymptotically stable fixed point, nearby orbits converge and the long-term state can become easier to predict. With sensitive dependence on initial conditions, arbitrarily close starting states can eventually lead to a visible disagreement, even though each follows the same rule.

Much of the usefulness of dynamical systems comes from describing outcomes without writing every solution explicitly. Fixed points identify possible resting states; periodic orbits identify repeating behaviour; stability theory asks whether those behaviours survive small disturbances. A bifurcation diagram shows how changing a bifurcation parameter changes the possible outcomes. This is useful in population dynamics, where a stable abundance and an alternating abundance have different consequences, and in thermal convection, where a steady flow and a changing flow transport heat differently.

The beauty of the subject is that a small formula can organize a large variety of behaviour. The logistic map moves from an attracting fixed point to attracting periodic orbits and then to parameter values supporting sensitive motion. Its period-doubling cascade has a recognizable structure across many unimodal maps. Nevertheless, Devaney chaos, Glendinning chaos, and a positive Lyapunov exponent express different properties; a claim of chaos should specify its definition, its state space, and the parameter values under consideration.

Use the interval exactness of a tent-map core from the preceding part. Choose two separated nondegenerate closed intervals $J_0,J_1$ inside $J=(1-s,1)$. There are integers $n_i$ with $T_s^{n_i}(J_i)=A$. Taking $N=\max (n_0,n_1)$ works for both, because $T_s(A)=A$:Within each $J_i$, the continuous function $F$ attains both endpoint levels $1-s$ and $1$. Choose a segment between those levels, reversing its orientation if necessary. Starting from a point at level $1-s$, take the first point at level one, and then the last point at level $1-s$ before it. Between these two points, continuity and the first-passage choice give $1-s<F(x)<1$, with all intermediate levels attained. The interior $K_i$ of that segment therefore satisfies $F(K_i)=J$. The intervals $K_0,K_1$ are disjoint and lie inside $J$.

Thus interval exactness produces a horseshoe:This is a horseshoe for an interval map for an iterate of $T_s$, proving Glendinning chaos for every $\sqrt{2}<s\le 2$. The proof does not require the same horseshoe iterate $N$ to work for every parameter.

A horseshoe for an interval map consists of an open interval $J$ containing two disjoint open subintervals each mapped onto $J$. Glendinning chaos means that some positive iterate has a horseshoe.

Solving on the appropriate branches of the tent map givesHere $F(x_{-2})=x_{-1}$ and $F(x_{-1})=x_0$. On $J=[x_{-1},x_0]$ the second iterate isBoth endpoints map to $x_0$, and the minimum is $v=\mu -{\mu }^2/2$. Both monotone branches cover $J$ when $v\le x_{-1}$, equivalently $(\mu -1)({\mu }^2-2)\ge 0$. Restricting them to preimages of the interior proves a horseshoe for $F^2$ when $\mu \ge \sqrt{2}$.

For $1<\mu \le \sqrt{2}$, $G(J)\subseteq J$. Set $d=(\mu -1)/(\mu +1)$ and $h(x)=(x_0-x)/d$. This maps $J$ onto $[0,1]$; substitution in each linear piece gives $hG{h}^{-1}=F_{{\mu }^2}$. The renormalization of the tent map near its fixed point therefore makes $F^4$ conjugate on $J$ to $F_{{\mu }^2}^2$. Applying the preceding threshold proves its horseshoe when $2^{1/4}\le \mu <\sqrt{2}$.

For $\mu \ge \sqrt{2}$, the inverse branch of $G$ fixing $x_0$ contracts distance to $x_0$ by ${\mu }^{-2}$. Compose $m$ copies of it with the other inverse branch. This contraction has a fixed point with a periodic itinerary, distinct from $x_0$ and at distance $O({\mu }^{-2m})$. Hence nontrivial periodic points accumulate at $x_0$.

For $\mu <\sqrt{2}$, the conjugate tent parameter $\nu ={\mu }^2<2$ has every nonzero periodic point in $[\nu -{\nu }^2/2,\nu /2]$. This interval is forward invariant; below it, a nonzero point increases until entering it. The positive lower bound separates those periodic points from zero, which corresponds to $x_0$. Thus no nontrivial periodic points are sufficiently close to $x_0$ from the left. A point sufficiently close on the right maps to such a left-hand point; periodicity would pass to its image and is impossible. Thus $x_0$ is isolated among periodic points, even though a fourth-iterate horseshoe exists elsewhere.

In Glendinning chaos, a continuous interval map is chaotic when some positive iterate has a horseshoe for an interval map. In the interval version, one may take an open interval $J$ containing two disjoint open intervals $K_0,K_1$ with $F^m(K_0)=F^m(K_1)=J$.

Let a three-cycle be ordered $a<b<c$. There are only two possible cyclic orderings; reflection of the coordinate interchanges them, so assume $F(a)=b$, $F(b)=c$, $F(c)=a$. ThenAlso $F(b)=c$ and $F(c)=a$, so the intermediate value theorem gives $d\in (b,c)$ with $F(d)=b$, whence $F^2(d)=c$. Both $[a,b]$ and $[b,d]$ therefore map across $[a,c]$ under $F^2$.

To obtain exact open-interval images, use the following elementary first-passage observation. A continuous function on $[r,s]$ with endpoint values $a,c$ has a subinterval on which its endpoint values are $a,c$ and all interior values lie strictly between them: take the first hit of the second level, then the last hit of the first level before it. The intermediate value theorem makes its interior image exactly $(a,c)$. Apply this to the two intervals, reversing endpoint order on the first. Their selected interiors $K_0\subset (a,b)$ and $K_1\subset (b,d)$ are disjoint and lie in $J=(a,c)$, withThis proves the three-cycle forces a two-iterate interval horseshoe result and establishes the required horseshoe for $F^2$.

Use binary expansions ending in zeros at dyadic rationals. The doubling map shifts the expansion left:In ordinary coordinates it is $2x$ on $[0,1/2)$ and $2x-1$ on $[1/2,1)$. The branches are continuous; the left limit at $1/2$ is $1$ while its value and right limit are $0$. This is its only discontinuity in the half-open interval.

For every $N\ge 2$, the repeating binary block $0^{N-1}1$ givesIts least period is $N$: a shorter repeating block would make more than one digit equal to one in a length-$N$ period. The orbit avoids the discontinuity, and its multiplier is $(F^N{)}^{\prime }=2^N>1$, so these periodic orbits are repelling and unstable. The same multiplier argument applies to every periodic orbit, with the endpoint fixed point treated one-sidedly.

In the two-branch horseshoe formulation of Glendinning chaos, disjoint subintervals are each mapped onto the same interval by an iterate. Here $J=(0,1)$, $K_0=(0,1/2)$ and $K_1=(1/2,1)$ already satisfy $F(K_0)=F(K_1)=J$. Their inverse branches $x\mapsto x/2$ and $x\mapsto (x+1)/2$ realize every binary itinerary. This is a horseshoe for a piecewise continuous map; on the circle the endpoint jump disappears. The symbolic explanation also shows dense periodic points, by repeating any prescribed finite prefix, and topological transitivity, because an iterate maps any sufficiently small dyadic cylinder onto $(0,1)$. Finally every neighborhood contains points sharing an arbitrarily long prefix but with different subsequent digits; an appropriate iterate separates their images by a fixed amount, for instance more than $1/4$. This proves sensitive dependence on initial conditions as well. The binary shift has full two-symbol chaotic dynamics.

A horseshoe for an interval map consists of two branches covering a common interval: there are an open interval $J$ and disjoint open subintervals $K_0,K_1\subseteq J$ with $f(K_0)=f(K_1)=J$. Equivalently one can work with two closed intervals with disjoint interiors whose images cover their union, and then restrict the branches. Glendinning chaos means that some positive iterate $f^m$ has such a horseshoe, rather than requiring one for $f$ itself.

For the three-cycle, use the intermediate value theorem on endpoint images. The interval covering relations are $I_a\to I_b$ and $I_b\to I_a,I_b$. Consequently both $f^2(I_a)$ and $f^2(I_b)$ cover $I_a\cup I_b$. This proves $f^2$ has a horseshoe. The directed covering graph of an interval map has exactly these forced arrows; a closed walk $a\to b\to b\to a$ has length three and a primitive itinerary. Pulling back the intervals successively and using the fixed-point property produces a point fixed by $f^3$ with this itinerary. It cannot be fixed by $f$, and a shared endpoint belongs to the already given three-cycle, so a genuine period-three orbit results.

For the increasing spatial order of the five-cycle, let $I_j=[x_{j-1},x_j]$, $1\le j\le 4$. The forced graph isThe primitive loops $I_4$, $I_3{I}_4$, $I_2{I}_3{I}_4$, and $I_1{I}_2{I}_3{I}_4$ force exact periods $\boxed{1,2,3,4}$. For periods below five the known five-cycle endpoints cannot be responsible, so the interval itineraries give distinct interior orbits. Both $g^2(I_3)$ and $g^2(I_4)$ cover $I_3\cup I_4$, so $g$ is guaranteed to be Glendinning-chaotic.

A horseshoe for $g$ itself is not guaranteed. As a counterexample put $x_j=j$ on $[0,4]$ and interpolate by $g(x)=x+1$ for $0\le x\le 3$, $g(x)=16-4x$ for $3\le x\le 4$. This has the required five-cycle. The unique fixed point is $16/5$, with $g(x)>x$ to its left and $g(x)<x$ to its right. Any nondegenerate closed interval whose image covers itself must contain this fixed point in its interior: an interval entirely on either side, or with the fixed point as an endpoint, fails to cover its extreme endpoint. Two intervals with disjoint interiors therefore cannot both cover themselves, as a horseshoe would require. Hence this example has no one-iterate horseshoe.

For the alternative spatial order, set $y_0=x_4,y_1=x_2,y_2=x_1,y_3=x_3,y_4=x_0$ and $J_j=[y_{j-1},y_j]$. Endpoint images giveThe self-loop at $J_2$, the two-cycle $J_1{J}_4$, and the four-cycle $J_2{J}_3{J}_1{J}_4$ force $\boxed{1,2,4}$; period three is not guaranteed. To see this last claim, take the connect-the-dots map at $y_j=j$ with values $(4,3,1,0,2)$. Its exact interval transitions are those displayed. Any length-three closed itinerary is the repeated self-loop in $J_2$; there the map is the affine decreasing function $5-2x$, whose third iterate has only its original fixed point. Endpoints belong to the five-cycle. Thus no exact three-cycle exists in this example.

Chaos is still guaranteed: from $J_2$ there are two different length-four closed walks, four self-loops and $J_2{J}_3{J}_1{J}_4{J}_2$. Pullbacks give disjoint branches of $g^4$ over a common interval in $J_2$, hence a horseshoe for an interval map for an iterate. A horseshoe for $g$ itself is again not guaranteed: the same connect-the-dots example is strictly decreasing on $[0,3]$ and increasing on $[3,4]$ with image $[0,2]$. Its unique fixed point is $5/3$, again with $g(x)>x$ to the left and $g(x)<x$ to the right. The same fixed-point argument rules out two self-covering intervals with disjoint interiors. These counterexamples distinguish the three separate questions about periods, chaos and a one-iterate horseshoe.

The map is chaotic in Glendinning's sense when some positive iterate $F^n$ has a horseshoe.

An interval map $F:I\to I$ has a horseshoe if there are two closed subintervals $J_0,J_1\subseteq I$ with disjoint interiors such thatIt is chaotic in Glendinning's sense if some positive iterate $F^n$ has a horseshoe.

Suppose $a<b<c$ are the points of a 3-cycle and put $J_0=[a,b]$, $J_1=[b,c]$. There are two possible cyclic orders. Ifthen the intermediate value theorem givesConsequently both $F^2(J_0)$ and $F^2(J_1)$ contain $J_0\cup J_1$. If insteadthenand again both second images contain $J_0\cup J_1$. Thus in either cyclic order $F^2$ has a horseshoe.

TakeNowsoThe recurrent pieces of this covering graph are the two-cycle $I_0\leftrightarrow I_2$ and the loop at $I_1$; neither contains two competing closed routes. Alsoso this ordering forces no 3-cycle.

The bound is attained by the piecewise linear functionIndeed,is a 4-cycle with the required spatial order. On $I_0\cup I_2$, the fourth iterate is the identity function. Within $I_1$, the fixed point is $5/9$ andso every other point eventually leaves $I_1$ and enters $I_0\cup I_2$. The map therefore has no horseshoe in any iterate and is not chaotic. The minimum number of distinct 3-cycles is consequently $\boxed{0}$.

SetTheir covering graph iswith adjacency matrixThe closed walks $I_1\to I_1\to I_1$ and $I_1\to I_0\to I_1$ give a horseshoe from two closed covering walks for $F^2$, so $F$ must be chaotic. Sincecounting cycles in an interval covering graph givesdistinct 3-cycles in the minimum case, attained by the corresponding connect-the-dots interval map.

LetThe covering relations areThus the covering graph has adjacency matrixThe two closed length-three walksgive a horseshoe from two closed covering walks for $F^3$, so $F$ must be chaotic. Moreover,By counting cycles in an interval covering graph, the number of primitive length-three cyclic itineraries isHence $F$ must have at least, and the connect-the-dots realization shows that it need have only, $\boxed{2}$ distinct 3-cycles.

For a continuous interval map $F$ with a three-cycle $a<b<c$, after reflecting the coordinate if needed its ordering is $F(a)=b$, $F(b)=c$, $F(c)=a$. Then $F^2(a)=c$, $F^2(b)=a$. Also the intermediate value theorem gives $d\in (b,c)$ with $F(d)=b$, hence $F^2(d)=c$. Each of $[a,b]$ and $[b,d]$ therefore maps across $[a,c]$ under $F^2$. Choose first-passage subintervals between endpoint levels to avoid overshoots; their interiors are disjoint and both map exactly onto $(a,c)$. This is a horseshoe for an interval map, and establishes Glendinning chaos.