Alligood K., Sauer T., Yorke J.A. Chaos: An Introduction to Dynamical Systems

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C HAOS

Corollary 3.30 The logistic map g(x) ⫽ ax(1 ⫺ x), where 0 ⱕ a ⱕ 4,

has at most one periodic sink.

Proof: The map g has negative Schwarzian, so Theorem 3.29 applies. All

orbits that begin outside [0, 1] tend toward ⫺

⬁

, so no points in [0, 1] have an

inﬁnite basin. Since the only critical point of g is x ⫽ 1 2, there can be at most

one attracting periodic orbit.

We found earlier in this chapter that for a ⫽ 4, all periodic orbits are sources.

Here is another way to see that fact, since when a ⫽ 4, the orbit with initial value

x ⫽ 1 2 maps in two iterates to the ﬁxed source 0.

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C HALLENGE 3

☞ C HALLENGE 3

Sharkovskii’s Theorem

CONTINUOUS MAP of the unit interval [0, 1] may have one ﬁxed point and

no other periodic orbits (for example, f(x) ⫽ x 2). There may be ﬁxed points,

period-two orbits, and no other periodic orbits (for example, f(x) ⫽ 1 ⫺ x).

(Recall that f has a point of period p if f

(x) ⫽ x and f

(x) ⫽ x for 1 ⱕ k ⬍ p.)

As we saw in Exercise T3.10, however, the existence of a periodic orbit of

period three, in addition to implying sensitive dependence on initial conditions

(Challenge 1, Chapter 1), implied the existence of orbits of all periods. We found

that this fact was a consequence of our symbolic description of itineraries using

transition graphs.

If we follow the logic used in the period-three case a little further, we can

prove a more general theorem about the existence of periodic points for a map

on a one-dimensional interval. For example, although the existence of a period-5

orbit may not imply the existence of a period-3 orbit, it does imply orbits of all

other periods.

Sharkovskii’s Theorem gives a scheme for ordering the natural numbers in

an unusual way so that for each natural number n, the existence of a period-n

point implies the existence of periodic orbits of all the periods higher in the

ordering than n. Here is Sharkovskii’s ordering:

3 Ɱ 5 Ɱ 7 Ɱ 9 Ɱ ...Ɱ 2 ⭈ 3 Ɱ 2 ⭈ 5 Ɱ ...Ɱ 2

⭈ 3 Ɱ 2

⭈ 5 Ɱ ...

...Ɱ 2

⭈ 3 Ɱ 2

⭈ 5 Ɱ ...Ɱ 2

⭈ 3 Ɱ 2

⭈ 5 Ɱ ... Ɱ 2

Ɱ 2

Ɱ 2 Ɱ 1.

Theorem 3.31 Assume that f is a continuous map on an interval and has a

period p orbit. If p Ɱ q, then f has a period-q orbit.

Thus, the existence of a period-eight orbit implies the existence of at least

one period-four orbit, at least one period-two orbit, and at least one ﬁxed point.

The existence of a periodic orbit whose period is not a power of two implies the

existence of orbits of all periods that are powers of two. Since three is the “smallest”

natural number in the Sharkovskii ordering, the existence of a period-three orbit

implies the existence of all orbits of all other periods.

The simplest fact expressed by this ordering is that if f has a period-two

orbit, then f has a period-one orbit. We will run through the reason for this fact, as

it will be the prototype for the arguments needed to prove Sharkovskii’s Theorem.

135

C HAOS

Let x

and x

⫽ f(x

) be the two points of the period-two orbit. Since f(x

) ⫽ x

and f(x

) ⫽ x

, the continuity of f implies that the set f([x

]) contains [x

(Sketch a rough graph of f to conﬁrm this.) By Theorem 3.17, the map f has a

ﬁxed point in [x

The proof of Sharkovskii’s theorem follows in outline form. We adopt the

general line of reasoning of (Block et al.; 1979). In each part, you are expected to

ﬁll in an explanation. Your goal is to prove as many of the propositions as possible.

Assume f has a period p orbit for p ⱖ 3. This means that there is an x

such that f

) ⫽ x

holds for n ⫽ p but not for any other n smaller than p.

Let x

⬍ ⭈⭈⭈ ⬍ x

be the periodic orbit points. Then f(x

)isoneofthex

but we do not know which one. We only know that the map f permutes the

.Inturn,thex

divide the interval [a, b] ⫽ [x

]intop ⫺ 1 subintervals

], [x

],...,[x

p⫺1

]. Note that the image of each of these subintervals

contains others of the subintervals. We can form a transition graph with these

p ⫺ 1 subintervals, and form itineraries using p ⫺ 1 symbols.

Let A

be the rightmost subinterval whose left endpoint maps to the right

of itself. Then f(A

) contains A

(see Figure 3.14 for an illustration of the p ⫽ 9

case).

Step 1 Recall that the image of an interval under a continuous map is an

interval. Use the fact that A 債 B ⇒ f(A) 債 f(B) to show that

債 f(A

) 債 f

) 債 ....

(We will say that the subintervals A

,f(A

),f

),... form an increasing

“chain” of subintervals.)

Step 2 Show that the number of orbit points x

lying in f

)isstictly

increasing with j until all p points are contained in f

) for a certain k. Explain

why f

) contains [x

]. Use the important facts that the endpoints of each

Figure 3.14 Deﬁnition of

is chosen to be the rightmost subinterval whose left-hand endpoint maps to the

right under f.

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C HALLENGE 3

subinterval are obliged to map to the subinterval endpoints from the partition,

and that each endpoint must traverse the entire period-p orbit under f.

As a consequence of Step 2, the endpoints of A

cannot simply map among

themselves under f—there must be a new orbit point included in f(A

), since

p ⫽ 2. So at least one endpoint maps away from the boundary of A

, implying

that f(A

) contains not only A

but another subinterval, which we could call A

Step 3 Prove that either (1) there is another subinterval (besides A

)

whose image contains A

,or(2)p is even and f has a period-two orbit. [Hints:

By deﬁnition of period, there are no periodic orbits of period less than p among

the x

. Therefore, if p is odd, the x

on the “odd” side of A

cannot map entirely

amongst themselves, and cannot simply exchange points with the “even” side of

for arithmetic reasons. So for some subinterval other than A

, one endpoint

must be mapped to the odd side of A

and the other to the even side. If p is even,

the same argument shows that either there is another subinterval whose image

contains A

,orelsethex

on the left of A

map entirely to the x

on the right of

, and vice versa. In this case the interval consisting of all points to the left of

maps to itself under f

Step 4 Prove that either (1) f has a periodic orbit of period p ⫺ 2in

], (2) p is even and f has a period-two orbit, or (3) k ⫽ p ⫺ 2. Alternative

(3) means that f(A

) contains A

and one other interval from the partition called

, f

) contains those two and precisely one more interval called A

,andso

on. [Hint: If k ⱕ p ⫺ 3, use Step 3 and the Fixed-Point Theorem (Theorem 3.17)

to show that there is a length p ⫺ 2 orbit beginning in A

Now assume that p is the smallest odd period greater than one for which f

has a periodic orbit. Steps 5 and 6 treat the concrete example case p ⫽ 9.

Step 5 Show that the endpoints of subintervals A

,...,A

map as in

Figure 3.15, or as its mirror image. Conclude that A

債 f(A

), and that the

transition graph is as shown in Figure 3.16 for the itineraries of f. In particular,

maps over A

for all odd i.

Step 6 Using symbol sequences constructed from Figure 3.16, prove the

existence of periodic points of the following periods:

(a) Even numbers less than 9;

(b) All numbers greater than 9;

This proves Sharkovskii’s Theorem for maps where 9 is the smallest odd period.

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C HAOS

Figure 3.15 A map with a period-nine orbit and no period-three, -ﬁve, or

-seven orbits.

It must map subintervals as shown, or as the mirror image of this picture.

Step 7 In Steps 5 and 6, we assumed that the smallest odd period was 9.

Explain how to generalize the proof from 9 to any odd number greater than 1.

Note that Step 4 is not required for the p ⫽ 3 case.

Step 8 Prove that if f has a periodic orbit of even period, then f has a

periodic orbit of period-two. [Hint: Let p be the smallest even period of f.Either

Step 3 gives a period-two orbit immediately, or Step 4 applies, in which case Steps

5 and 6 can be redone with p even to get a period-two orbit.]

Figure 3.16 Transition graph for period-nine map.

The existence of orbits of periods 1, 2, 4, 6, 8, and all numbers greater than 9 is

implied by this graph.

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C HALLENGE 3

Step 9 Let k be a positive integer and let f be any map such that f

has

a period r point x. Prove that x is a period 2

r point of f for some 0 ⱕ j ⱕ k.

Moreover, prove that if r is an even number, then j ⫽ k. (As always, by period we

mean minimum period.)

Step 10 Prove that if f has a periodic orbit of period 2

,thenf has periodic

orbits of periods 2

k⫺1

,...,4, 2, 1. (Since f

k⫺2

has a period-four point, it has a

period-two point, by Step 8. Ascertain the period of this orbit as an orbit of f.)

Step 11 Assume p ⫽ 2

q is the leftmost number on the list for which f

has a period-p point, where q is an odd number greater than 1. The integers to

the right of p in Sharkovskii’s ordering are either powers of 2 or are of form 2

where r is either an even number or a number greater than q.Sincef

has a

period-q orbit, Step 7 implies that f

has orbits of this form. Use Step 9 to check

that these orbits are orbits of f of period 2

r. Use these periodic orbits and Step

10 to complete the proof of Sharkovskii’s Theorem.

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C HAOS

E XERCISES

3.1. Let f

(x) ⫽ a ⫺ x

,wherea is a constant.

(a) Find a value a

of the parameter a for which f

has exactly one ﬁxed point.

(b) Describe the limit of all orbits of f

for a ⬍ a

has an attracting ﬁxed point for a in the open interval (a

Find a

(d) The map f

has an attracting period-two point for a in the open interval

). Find a

(e) Describe the dynamics of f

for a ⫽ 2.

3.2. Decide on a partition for the map f(x) ⫽ 2x (mod 1) on [0, 1], and draw its transition

graph and schematic itineraries as in Figure 3.2(a)(b). How do they differ from those

for the tent map?

3.3. (a) Find a conjugacy C between G(x) ⫽ 4x(1 ⫺ x)andg(x) ⫽ 2 ⫺ x

(b) Show that g(x) has chaotic orbits.

3.4. Show that g(x) ⫽ 2.5x(1 ⫺ x) has no chaotic orbits.

3.5. (a) Sketch a graph of the map f(x) ⫽⫺2x

⫹ 8x ⫺ 5.

(b) Find a set of two subintervals that form a partition.

orbits?

3.6. Repeat the previous exercise with f(x) ⫽ 2x(1 ⫺ x).

3.7. Let T be the tent map. Prove that the periodic points of T are dense in I.

3.8. Assume that 兵x

,...,x

其 is a periodic orbit for a continuous map f on the real

line, with x

⬍ ⭈⭈⭈ ⬍ x

. Assume f(x

) ⫽ x

i⫹1

if i ⬍ 9andf(x

) ⫽ x

(a) What periods does Sharkovskii’s Theorem guarantee on the basis of the

period-nine orbit?

(b) Draw the transition graph for this map. Which periods are certain to exist?

3.9. Assume that x

⬍ x

are points on the real line, and that f is a continuous

map satisfying f(x

) ⫽ x

,f(x

) ⫽ x

,f(x

) ⫽ x

and f(x

) ⫽ x

. For simplicity,

assume that f is monotonic (increasing or decreasing) except possibly at the four

points mentioned.

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E XERCISES

(a) Sketch a graph of f.

(b) Draw the transition graph for f.

3.10. Assume that x

⬍ ⭈⭈⭈ ⬍ x

are points on the real line, and that f is a continuous

map satisfying f(x

) ⫽ x

,f(x

) ⫽ x

,f(x

) ⫽ x

,f(x

) ⫽ x

and f(x

) ⫽ x

. Assume

that f is monotonic (increasing or decreasing) except possibly at the ﬁve points

mentioned.

(a) Sketch a graph of f.

(b) Draw the transition graph for f.

3.11. Let f be a continuous map from the unit interval onto itself (that is, such that

f([0, 1]) ⫽ [0, 1]).

(a) Prove that f must have at least one ﬁxed point.

(b) Prove that f

must have at least two ﬁxed points. (Hint: Explain why either

f or f

must have points 0 ⱕ x

⬍ x

ⱕ 1suchthatx

mapsto0andx

maps

to 1.)

must have at

least 3 ﬁxed points.

3.12. Let f(x) ⫽ rx(1 ⫺ x), r ⬎ 2 ⫹



5. Show that the Lyapunov exponent of any orbit

that remains in [0, 1] is greater than zero, if it exists.

3.13. Let n be a positive integer, and f(x) ⫽ nx (mod 1) on [0, 1]. Which points are

periodic, eventually periodic, and asymptotically periodic? Which orbits of f are

chaotic orbits?

3.14. (From Seth Patinkin) Let I be an interval in the real line and f a continuous

function on I. The goal of this exercise is to prove the following theorem:

Assume there is an initial value whose orbit is dense in I and assume f

has two distinct ﬁxed points. Then there must be a point of period 3.

The following two facts are to be used as help in proving the theorem.

(i) Suppose there is a point a in I such that a ⬍ f(a)andthereisab ⬍ a in I

such that f(b) ⱕ b and there is a k ⬎ 1suchthatf

(a) ⱕ b. Then there is a point of

period 3. (Of course it is possible to reverse the inequalities, f(a) ⬍ a ⬍ b ⱕ f(b),

and b ⱕ f

(a), to again get a true statement.)

(ii) Assume there is no period 3 point and that there is some orbit dense in I.

Then there is a unique ﬁxed point c and x ⫺ c and f(x) ⫺ x always have opposite

signs for every x ⫽ c in I.LetA ⫽ 兵x : f(x) ⬎ x其 and B ⫽ 兵x : f(x) ⬍ x其.Use(i)to

show that A and B must each be an interval.

141

C HAOS

Discussion: In Chapters 1 and 3 we discussed that implications of having a

period 3 orbit. It implies the existence of periodic orbits of all other periods and

it implies sensitivity to initial data. This exercise provides a partial converse. The

assumption that some orbit is dense in an interval is quite reasonable. It can be

shown that for any piecewise expanding map F (see Chapter 6), the orbit of almost

every initial point is dense in the union of a ﬁnite number of intervals, and for some

k the piecewise expanding map F

has the property that the orbit of almost every

point in dense in an interval. The proof of these results is beyond the scope of this

book. We could choose such an interval to be I and f ⫽ F

to be our function. It

might or might not have two ﬁxed points.

3.15. (Party trick.) (a) A perfect shufﬂe is performed by dividing a 52-card deck in half,

and interleaving the halves, so that the cards from the top half alternate with the

cards from the bottom half. The top card stays on top, and so it and the bottom

card are ﬁxed by this operation. Show that 8 perfect shufﬂes return the deck to its

original order. [Hint: Number the original card order from 0 to 51. Then a perfect

shufﬂe can be expressed as the map

f(n) ⫽



2n if 0 ⱕ n ⱕ 25

2n ⫺ 51 if 26 ⱕ n ⱕ 51

The goal is to show that all integers are ﬁxed points under f

. First show that

(n) ⫽ 2

n ⫺ 51k for some integer k,wherek may be different for different n.]

Caution: when demonstrating at actual parties, be sure to remove the jokers ﬁrst!

If the deck consists of 54 cards, then 52 perfect shufﬂes are required.

(b) If the bottom card 51 is ignored (it is ﬁxed by the map anyway), the

above map is f(x) ⫽ 2x (mod 51), where we now consider x to be a real number.

Nonperiodic orbits have Lyapunov number equal to 2, yet every integer point is

periodic with period a divisor of 8. Sharkovskii’s Theorem shows that it is typical for

chaotic maps to contain many periodic orbits. Find all possible periods for periodic

orbits for this map on the interval [0, 51].

3.16. Deﬁne the map

f(x) ⫽



1 ⫺ 3x if 0 ⱕ x ⱕ 1 3

⫺1 3 ⫹ x if 1 3 ⱕ x ⱕ 1

Note that x ⫽ 0 is a period 4 point. Let I ⫽ [0, 1 3], J ⫽ [1 3, 2 3] and K ⫽

[2 3, 1].

(a) Draw the transition graph.

(b) For each periodic orbit of period p ⬍ 6, label the points of the orbit as

⬍ a

⬍ ... ⬍ a

and list the itinerary of a

and show where each point of

the orbit maps. For example, there is a period two point IJ which is the itinerary

including a

⬍ a

,wheref(a

) ⫽ a

,andf(a

) ⫽ a

. (In this case, a

⫽ 1 6and

⫽ 1 2, but you need not list values for the a

in general.)

142

L AB V ISIT 3

☞ L AB V ISIT 3

Periodicity and Chaos in a Chemical Reaction

N ELEMENTARY chemistry classes, a great deal of emphasis is placed on

ﬁnding the equilibrium state of a reaction. It turns out that equilibria present only

one facet of possible behavior in a chemical reaction. Periodic oscillations and

even more erratic behaviors are routinely observed in particular systems.

Studies on oscillating reactions originally focused on the Belousov-

Zhabotinskii reaction, in which bromate ions are reduced by malonic acid

in the presence of a catalyst. The mechanism of the reaction is complicated.

More than 20 species can be identiﬁed at various stages of the reaction. Experi-

ments on this reaction by a group of researchers at the University of Texas were

conducted in a continuous-ﬂow stirred tank reactor (CSTR), shown schemati-

cally in Figure 3.17. The solution is stirred at 1800 rpm by the impeller. There is

a constant ﬂow in and out of the tank, perfectly balanced so that the total ﬂuid

volume does not change as the reaction proceeds. The feed chemicals are ﬁxed

concentrations of malonic acid, potassium bromate, cerous sulfate, and sulfuric

acid. The ﬂow rate is maintained as a constant throughout the reaction, and the

bromide concentration is measured with electrodes immersed in the reactor. The

output of the experiment is monitored solely through the bromide measurement.

The constant ﬂow rate can be treated as a system parameter, which can be

changed from time to time to look for qualitative changes in system dynamics,

or bifurcations. Figure 3.18 shows several different periodic behaviors of the

bromide concentration, for different settings of the ﬂow rate. The progression

of ﬂow rate values shown here is decreasing in the direction of the arrows and

results in periodic behavior of periods 6, 5, 3, 5, 4, 6, and 5. Each oscillation (an

up and down movement of the bromide concentration) takes around 2 minutes.

Approximately one hour was allowed to pass between changes of the ﬂow rate to

allow the system to settle into its asymptotic behavior.

Roux, J.-C., Simoyi, R.H., Swinney, H.L., “Observation of a strange attractor”. Physica

D 8, 257-266 (1983).

Coffman, K.G., McCormick, W.D., Noszticzius, Z., Simoyi, R.H., Swinney, H.L., “Uni-

versality, multiplicity, and the effect of iron impurities in the Belousov-Zhabotinskii

reaction.” J. Chemical Physics 86, 119-129 (1987).

143