A Chaos Lemma
Author(s): Judy Kennedy, Sahin Koçak and James A. Yorke
Source: The American Mathematical Monthly, Vol. 108, No. 5 (May, 2001), pp. 411-423
Published by: Mathematical Association of America
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A Chaos Lemma
Judy Kennedy,Sahin Kosak, and James A. Yorke
1. INTRODUCTION. Manyphysicalsystemsthatareeasy to describeexhibitcomplicatedbehavior.A reasonablegoal is to describethe differentkinds of behaviora
systemcan have.
In a studyattemptingto determineif the solarsystemis stable,Poincaredescribed
the first "chaotic"system,knownas the "restrictedthreebody problem".Here three
bodies move in a plane undergravitationalforces as describedby Newton. One of
the bodies is of negligible mass, and does not influencethe other two. Those two
move in circles aboutthe centerof mass. Poincare'sideas were appliedby NASA to
send a spacecraftwith minimalfuel throughthe tail of a comet. In this application
the threebodies were the Earth,the Moon, anda spacecraft.The spacecrafthadbeen
launchedfor other purposesand was parkedabout 1 million miles from the Earth
in the directionof the Sun, collectinginformation.Near the end of the spacecraft's
mission, the Jacobi-Zinnercomet was discoveredto be headingtowardthe Sun in a
trajectorythatwouldtakeit nearthe earth'sorbit(passingabout60 millionmiles from
the earth).Even thoughthe spacecrafthad enoughfuel remainingfor only smallorbit
corrections,Dr. RobertFarquharrealizedafterextendedcomputersimulationsthata
tiny pushcouldputthe spacecrafton a complicatedtrajectorythatwouldinterceptthe
comet's tail; see Figure 1. Poincarehad made it clear that in such chaotic systems,
therewere infinitelymany trajectories,each with long termbehaviorthat was quite
differentfromthe others.
S6=B12/22
S3=A9/27
6/10/82
\D
10121
~~~~~~S,=B
S2=B4/23
Figure1.
THE FARQUHAR TRAJECTORY.The initial change in velocity was on 6/10/82. The other dates, all
in 1983, denote near passes of the Moon.
Farquhar'strajectorywas startedwith a firingof the rocketsthatchangedthe velocity of the craftby only 10 miles perhour.On this trajectory,the spacecrafthadfive
nearpassesof the Moon,finallypassingabout60 miles fromthe surfaceof the Moon.
This last pass was sufficientto kick the spacecraftout of the Earth-Moonneighborhood andtowardthe rendezvouswith the comet. The spacecraftmadethe firstdirect
measurementsof the compositionof a comet'stail.
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We denotePoincare'stwo large masses by E and M. The thirdbody of negligible mass is denotedby S (for spacecraft).Our approachis slightly differentfrom
Poincare's.First,with the origin as the centerof mass, we representthe position of
S relativeto E and M using polarcoordinatesr, 0, with 0 = 0 denotingthe direction
coordinates,E
of M andr = 1 the distanceof M fromthe origin.In these "rotating"
andM remainfixed;see Figure2.
+'~~~~'
Figure 2. A trajectoryis shown intersecting set B with the distance from 0 decreasing.
Mathematiciansand scientistsoften use objectscalled "symbolsequences"to describethe differentkinds of trajectoriesa chaotic system can have. For an example
of this in the E-M-S system, we let SA denote the set of points with r = 1 and
0 < 0 < 0.1, andlet SB denotethe set of pointswith r = 1 and27 -0.1 < 0 < 27r.
If p, denotesthe nth elementof the associatedsymbolsequence,for the Earth-MoonSpacecraftsystem (whichis not quiteour ideal system)Pnis in SAif the Moon's nth
inwardcrossingis aheadof the Moon;it is in SB if the crossingis behindthe Moon.
The Farquhartrajectorywould be describedby the symbol sequenceABABB, after
which the craft left the Earth-Moonenvironment.Most trajectories(Pn) that don't
collide with the Earthor the Moon would leave the Earth-Moonsystem aftera finite
numberof symbols.Nonethelesssome trajectorieswould yield an infinitelylong sequence(Pn)of the symbolsA andB, andthereareuncountablymanysuchsequences
thatdo not repeatin anyperiodicpattern.
We presenta simple methodof studyingchaos that considersall the possible sequencesthatdifferenttrajectoriescan have when passingthroughsome collectionof
sets.In ourexampleof sets, SAandSB, representingpartsof the Moon'sorbit,we have
two symbols,butothersituationsrequiremoresets.
The spacecraftproblempresentsan opportunityto explainwhy discretetime maps
can be interesting.We investigatemaps, not differentialequations.Poincareargued
thatthe motioncouldbe best understoodas a mapon a two-dimensionalspace.Each
time the body S (with position written as (0(t), r(t))), crosses the circle r = 1 with
derivative r' < 0, we record two numbers: 0(t) and its derivative 0'(t). The trajec-
tory of S has a generalizedtotal energythatis constant,which we assumeis known.
Knowingthis energyand 0, 0', andhavingr = 1, we can solve for the derivativer'.
Hence,we can solve the differentialequationuntilthe next crossingif thereis one. If
Pn := (0, 0'))nE S1 x R denotes the nth such crossing, we can write
Pn+i = f (Pn4)
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To evaluatef, we mustsolve the differentialequationnumerically.Using this kindof
map, Poincareshowedthatthe behaviorof some trajectoriesis very complicatedfor
some ratioof the massesof M andE.
We do not analyzethe E-M-S systemfurther,butratherexplorea frameworkthat
is applicableto a wide varietyof problems.Chaos is a fairly well understoodphenomenonin dynamics:consequencesof its presenceoften include the "curses"of
and practicalincomputabilityof the long-termbehaviorin a system
unpredictability
governedby deterministicdynamics[1]. This propertyis usuallyformulatedin terms
of "sensitivedependenceon initialconditions",whichwe definein Section4.
Next we analyzethe celebratedSmaleHorseshoe[9].
To con2. THE SMALE HORSESHOE. SMALEHORSESHOECONSTRUCTION.
structa simplifiedSmale Horseshoe,begin with a unit squareQ := ABCD in the
plane; as picturedin Figure 3. Write Q = I(x, y) E R2: 0 < x < 1, 0 < y < 1}.
Next let Si and S2 denotetwo disjointhorizontalstripsin Q. and let Qo denotethe
unionS1U S2of the two stripsas shownin Figure3. The horseshoemap f sendseach
horizontalstripaffinelyontoa verticalstrip.The mapcanbe thoughtof as a mapfrom
the planeto itself, or a mapfrom Q into the plane,or a mapfrom Qo into Q; herewe
care only aboutthose pointsin Qo whose trajectoriesremainin Qo, so the behavior
of f outside Q does not concernus. To view f as a map from Q into the plane,we
canthinkof f as a mapthatinvolveslinearhorizontalcontractionof Q. linearvertical
D
C
Q
D
C
A
B
s
s
S2
S2
1/.///./SI
SI
A
B
D
C
f(Q)
D
C
f(S1)
2)
~~~~~~~~~~f(S
S2
Si
B
A
A
B
A
B
0
D
Figure3. The horseshoe map f is obtained by horizontally squeezing the rectangle in (a) to obtain (b); then
vertically stretchingit to obtain (c); then folding it to obtain (d). The images under f of the points A, B, C,
D are denoted by A*, B*, C*, D*, respectively. The set f (Q) := {x E Q: f (x) E Q} is the union of two
horizontal strips, Si and S2. They are mapped onto the two vertical strips of Q n f (Q).
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stretchingof Q, andfinallyfoldingthe stretchedandcontractedQ once beforeplacing
it acrossitself as indicatedin Figure3a. For a moredetailed,but elementary,discussion of this map, see [1]. For a samplingof the large literatureon such horseshoes,
see [8].
Definition. A sequence S :(Sio, Si, ..., Sin, ...) of symbol sets is called aforward
itinerary(in {Si, S21).We say a pointx follows S if ffn (x) E Sinfor all n = 0, 1, 2, ....
Proposition 1. (The Horseshoe.) Let Si and S2 be as described above, and let f be
the horseshoemap.Thenallforwarditinerariesin {Si, S21 are followed.
The proofillustratesthe techniquewe use throughoutthe paper:We introducethe
terms"expander"
and"familyof expanders"in the followingdefinition;the familyof
verticalcurvesdiscussedin the proofis a familyof expandersfor ISi, S21.
By a verticalcurvewe meanthe graphin the plane of a continuousfunctionx =
v(y). We definethe familyof sets
: {E: E is a verticalcurvein Q connectingthe upperandlower sides of Q}.
Note the followingtwo properties:
(F1) S is a nonemptyfamilyof nonemptysubsetsof Q, and
(F2) foreach E E 5, andeach Si thereis a compactset Pi C E n Sisuchthatf (Pi) E
5, that is, f (Pi) "expands"Pi to a memberof S.
(Actually,in this case we can choose Pi := E n Si.)
Definition. WheneverE is a set of sets satisfying(F1) and (F), we call E a family
of expandersfor tSj, S21.The verticalcurves E E E are called expanders,and any
compactsubsetP of an expanderis calleda pre-expanderif f (P) E E.
The proof of Proposition1 uses the CoveringPrinciple,which is given afterthat
proof.
Definition. We say a set P covers E for f if f (P) D E. This term is motivated by
askingwhenthe equationf (p) = e has a solutionp E P for everye E E. The answer,
of course,is thatP mustcover E. If, for example,A is an n x n matrix,we can solve
Ap = e for each e if A covers RW.
Proof of Proposition 1. Now, let S = (Sio9Si 9... 9Sin, .. .) be any forward itinerary.
Let Eo be any expander;property(F2)ensuresthat Eo containsa pre-expanderPo C
Sio (namely EoSn io). Thenf (PO)is anexpander,whichwe denoteby E1. Now we re-
peat the procedure:The expander E1 contains a pre-expander P1 C Si,. Then f (P1) =
E2 is an expander. Proceeding in this way we obtain the sequence of nonempty, compactsets Pn C EnBwhereEn+1 := f (Pn) for all positiveintegersn. Since Pn C S, it
follows fromthe followingCoveringPrinciplethatS is followedby some xo.
Proposition 2. (The Covering Principle.) Let Q be a metric space. Assume that f:
Qo -? Q is continuous, where Qo C Q is compact. Let
Po1
PM
C EMT P2 C E2A ...,A Pn C EnA
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I
Mt
1
be nonempty,compact subsets of Qo such that f (Pn) D En+1for each positive integer
n. Then there exists a point xo E Po that is mapped indefinitely, i.e., Xn: fn(Xo) is
definedfor all n, and xn E Pn-
Proof. WritePi' for Pi for all i. We constructfamiliesof compactsets pi+l c P' for
j = 1, 2, . . . by settingPii:
{x E P/ f(x) E P/.Hencef(Pii+)
= Pi,.+ Let
Pi' be the nonemptycompactset Rm1
jP/. It follows that f (Pi?) = Pj+1, so for any
x0 E PO, fn(x0) is definedandis in Pn.
This proofis quitedifferentfrom otherproofsof topologicalhorseshoetheorems,
andgeneralizesargumentsin [6]. Othermathematicians
who haveinvestigatedhorseshoeswithouthyperbolicityincludeA. Szymczak[10], K. Bums andH. Weiss[2], and
K. Mischaikowandvariouscoauthors(see [4], [3], and [7]). Theirproofsarebasedon
some form of topologicaldegreeargumentsuch as the Conley index. Our approach
is an extensionof the constructionof Cantorsets and is an extensionof arguments
oftenused in demonstrating
chaosfor one-dimensionalmaps,e.g., in [5]. Approaches
via topologicaldegreetheoryrequireassumingthe existenceof an isolatingneighborhood;oursdoes not.
A minorvariantof the proof given of Proposition1 illustratesthatit is more importantthatS satisfy(F1)and(F2)thanit is to worryaboutwhatsets arein S. Define
con:= {E C Q: E is compactand has a connectedcomponentthatintersectsboth
the bottomAB andthe top CD of Q}.
Another proof of Proposition 1. We use Sconin the above proof. Here again Pn can be
chosento be En n Sin
f(Q)
f(E)
E
E
The map f shown here is non-invertible.Again Sl U S2 = f (Q), but now,
Figure4. BULGINGHORSESHOE.
because of the bulge, S2 has two connected components (b). In (c) a possible E E E is shown along with its
image. Let Ei = f (E n si) for i = 1, 2. While E n s2 is not connected, its image E2 is.
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c
D
L:=f(S,).
s~:-fS2
Figure5.
n
EXML.HeeS
Fgr5.PGSTAL
PIG' S TAIL EXAMPLE.
Here
r
4 piees
hsen- wih3_n
Sl and S2 are chosen
with 3 and 4 pieces,
respectv\ely, sotatte
respectively,
so that they
map onto left and right parts of the cross-hatched part of f (Q). For any sequence (Si,) each member of which
is Si or S2, there is a trajectory through some x that follows it. Then f (x) is the initial point of a trajectory that
follows (f (Sin)),a sequence of sets L and R on the left or right.
The ExpanderLemmais applicablein many othersituationsfor an appropriately
chosenfamilyof expanders.Consider,for example,the "BulgingHorseshoe"(Figure
4), the "Pig'sTail"(Figure5), andthe "MutatedHorseshoe"(Figure6). We can take
the same symbol sets S1 and S2 for these examples,and applythe argumentwe used
for the SmaleHorseshoemap.We encapsulatethe argumentin Lemma3.
Smalerecognizedthatif f (Q) crosses Q fromtop to bottomK times, theremust
be a trajectoryfollowingeach sequenceof K symbolsets; see Figure7. Forthis case
D
__C
_
Q
(S2)
S)
D
f(S1)
*~~~C
B A*
A
B
Scrn is an appropriate family of expanders for this map. The image of Q under f is
SI U S2, and this is the set that maps to the cross-hatched pieces. In (b), we show possible shapes for Sl and
S2. Of the families of expanders discussed in Sections 2 and 3, only Eccnis a family of expanders here.
Figure 6. Here again
D* 0*
A*B*
Figure 7. A horseshoe map with K = 3.
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also, all itinerariesarefollowedbutwiththreesymbolsets. Againthe horizontalstrips
thatmapinto Q serveas symbolsets, while verticalcurvesserveas expanders.
3. FAMILIESOF EXPANDERS. In the examplesof the previoussection,the role
playedby the families S and Scon are fundamentalin provingthe existenceof chaos.
and
They act as sensorsof expansionin the dynamics(hencethe name "expanders")
theyhave specializedsubsensorslocatedin all symbolsets (thatis, subsetslying in the
symbolsets) that,when mapped,becomeexpanders(whichis why we call them"preexpanders").This soundslike a vicious circle, but it isn't! On the contrary,it works
variesfromproblemto problem;what
surprisinglywell. The definitionof "expander"
"families".
each
other
in
how
relate
to
they
is constantis
The basic settingfor the remainingresultsof this paperis the following:
HypothesisH. Let Q be a metricspace,let Qo C Q a compactsubset,let S1, S2,....
Sk be nonempty,compactsubsetsof Qo, with K > 2, and assume f: Qo -> Q is a
continuousmap.
In the next sectionwe addthe hypothesisthatthe Si arepairwisedisjoint,butthere
areinterestingcases wherethey arenot.
Definition. AssumeHypothesisH. Let S be a collectionof subsetsof Q suchthat
(F1) S is a nonemptyfamilyof nonemptysubsetsof Q, and
(F2) for each E E S and each Si, there are compact sets Pi C E n si such that
f (Pi) E S.
Thenwe say thatE is a family of expandersandeach memberE E E is an expander
(for f and {Si, S2,
...,
Ski).
Now we arereadyfor the ExpanderLemma.
Lemma 3. (The Expander Lemma.) Assume Hypothesis H and let E be an associatedfamily of expanders. For each sequence S := (Sin)n>oof symbol sets there exists
a point xo E Sio thatfollows S.
Proof. The proofis the same as the proof of the HorseshoePropositionfromthe bea
ginningof the secondparagraphon.
4. SENSITIVEDEPENDENCEON INITIAL DATA. If we assumeonly Hypothesis H, then we could have a trivialsituationin which all Si are equal. If nsi contains a fixed point x, i.e., f (x) = x, then x triviallyfollows every sequenceS since
f n (X) E Sin .
Under the following pairwise disjointedness Hypothesis PD, the situation
is quiteinteresting.
We write dist(x, y) for the distancebetweenx and y in Q. Often we requirethe
followingadditionalassumption:
HypothesisPD. (PairwiseDisjoint) AssumeHypothesisH, andassumeadditionally
thatS1, ... ., Sk arepairwisedisjointsets. Let a denotethe minimumdistancebetween
these sets, i.e., a = mindist(x, y), wherethe minimumis takenover all x and y that
arein differentsymbolsets.
We investigatesome cases whereeveryforwarditineraryis followed;thatis, every
forwarditineraryhas a pointthatfollows it. In most of our examplesf is definedon
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a largersets, butwe areconcernedonly withtrajectorieslying entirelyin Qo.Loosely
speaking,the map f is "chaotic"if it possesses the following sensitivityproperty:
Thereis a compactset Qchaos C Qo thatis invariant,i.e., f ( Qchaos) = Qchaos, andfor
everyitineraryS thereis a "sensitive"pointx E Qchaos thatfollows S. See Lemma4
for moredetail.
This definitionsuggeststhe gameof pinball,wherebumpersor reflectorsforcefully
repel a ball that hits them. We might label these bumpers S,, . . ., Sk. For some pinball
games, once the ball is launched,it could conceivablybounce foreverbetween the
bumpersif perfectlypositioned,and the ball could collide with the bumpersin any
sequence.It could even collide with only a single bumper(bouncingoff the wall and
backto the bumper).It is muchmorelikely thataftera few bounces,the ball falls into
a hole thatendsplay.Thisrichpatternof possibilitiesmakesthe gamefun andjustifies
callingits dynamics"chaotic".
Definition. We say a compactset Q* C Qois invariantif f is definedat all pointsof
Q*, and f (Q*) = Q*.Let Q* C Qo,bea nonemptycompactinvariantset. If x E Q*,
f n(X) is definedfor all positiveintegersn. Let x E Q* and xi E Q* for i a positive
integer.We say the sequence{xiI separatesfromx (or moreprecisely,the trajectories
of xi separate from the trajectory of x) if xi -- x as i -- oo and there is a 8 > 0 such
that for all i there is a positive integer m
=
m(i) such that dist(fm (xi),
ftm(x)) > 8 for
all i. An x havingsuch a sequencewith all xi in Q* is called sensitiveto initialdata
(in Q*).We say a set Q* is chaoticif it is nonemptyandinvariant,andeveryx in Q*
is called sensitive to initial data in Q*.
In practicewe assumePD andhavexi -+ x as i -> oo wherex andeachxi follow
we assumeS is not equalto anyof the
itinerariesS andSi, respectively.Furthermore,
Si. Hence, for each i, thereis a time m such that fm (x) and fm (xi) are in different
symbol sets. Hence, dist(fm (xi), ftm(x)) > ac.
Lemma 4. (The Chaos Lemma.) Assume Hypothesis PD. Then there is a chaotic set
Q* C Qo such that each forward itinerary is followed by some point in Q*.
Outlineof Proof Lemma4 impliesthatthereis a chaoticinvariantset Q*.Ratherthan
give a detailedproof,we outlinethe proof andchallengethe readerto completeeach
step.
1. (Invariance)If x E Q andif ftm(x) is definedandin Q for all positiveintegers
n, thendefine
o(x) :- {z : z is alimitpointof {fm(x)J1.
Thenco(x) is a compactinvariantset.
2. (Following)Let S* be an itinerarywith the propertythatevery finite sequence
(Sin)m i occurs in S* in order;that is, for some j, the j + 1 through j + m terms
in S* areequalto S' throughS/m,respectively.Letx[l] be a pointthatfollows S*.
Thenfor eachitinerarythereis a pointX[2]E co(x[1])thatfollows it. In particular,
some X[2] E co(x[l]) follows S*. Notice that W(x[2])C W(x[l]).
3. (Minimality)Thereis a pointx* in Qo whichfollows S* andalso satisfiesx* E
a)(x*).WriteQ* := co(x*). Hint:Considerthe set L of all compactsets L equal
to co(x)for a pointx thatfollows S*. Showthatthereis a smallestL, thatis, an
L thatcontainsno otherset in L.
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4. (Sensitivity)Write x* for fi(x*) for positive integer i. Then for i :] j, the
itinerariesof xi*andx* aredifferent.Hence,if some subsequencex.* converges
to y, then y is sensitiveto initialdatain Q*. But every y in Q* has such a seU
quence.
5. ONE-SIDEDTRAJECTORIESIMPLYDOUBLE-SIDEDTRAJECTORIES.
We have so farconsideredonly one-sidedtrajectories,butin the settingof all the lemmas, double-sidedtrajectoriesexist. Actually,this is a directconsequenceof existence
of one-sidedtrajectories.We now formulateandproveit. Let Z denotethe set of integers.
Proposition 5. Assume Hypothesis H. Let Q* c Q be a compact invariant set. Assume thatfor every one-sided sequence (Sin)n%=O
of symbol sets, there is a point xo E Q*
that follows S. Then for every double-sided sequence D := (Sin)nEz of symbol sets,
there exists a sequence (xn)nEzin Q* such that xn E Sin and f (Xn) = Xn+l, for all
n E Z.
Proof. Givenany double-sidedsymbolsequenceD := (Sin)nEzwe considerthe truncated, one-sidedsymbol sequenceS_n -(Sin)jt*n. By assumptionthereis a point
Y-n E Q* that follows S_n (Of course, Y-n E Si-n.) Next define Zn f= fn (Y-n) E SioThe sequence(Zn)nEz
has a subsequence(znj)thatconvergesto a pointxo Esio n Q
Since each pointZnfollows Do, the pointxo also follows Doby continuity.Set xm
fm (xo) for m a positive integer.
Ourgoal is to choosea subsequence(nk) suchthatthetrajectorieswithinitialpoints
converge to a trajectory defined for all time. Define Zn,m:= fn+m(Y(n) E Sim
for each m with n + m > 0. For fixed m E Z thereis a sequenceni -> oo such that
the sequence Zm,njconverges to a point xm E simn Q*, and its trajectory follows Sm.
Using a diagonalizationargument,the sequence(ni) canbe chosenindependentof m.
Y-nk
By continuity xm+l = f (xm) and Xm E sim n Q*.
m
6. SUBSHIFTS OF FINITE TYPE. The Chaos Lemma can be extendedto the
"subshift of finite type" case where not all of the subsets Si, i = 1, 2, ..., K, are
interrelated
considera transition
by the mapf . To definea restrictedinterrelationship,
matrix M
=
(mij)[Kc=1;
that is, each entry mij of M is either 0 or 1.
Definition. Let S := (Sin)be an itinerary(eitherone-sidedor two-sided).We say that
S is M-admissible if mij = 1 whenever Si and Sj are consecutive terms in S with Si
precedingSj.
So farM couldbe incrediblydull, say if mij = 0 for all i andj, or if mij = 1 if and
only if i = j. The following ensures interesting dynamics.
Hypothesis M. Suppose that M
=
(mij),f'l=
is a transition matrix. There is an M-
admissiblesequenceS := (Sin)in whicheach of the sets Sj occursfor infinitelymany
n andfor at least one i, thereare j andk with j 0 k suchthatmik = 1 = mijDefinition. Assume Hypothesis H and let M
=
(mij)lKj=I
be a transition matrix. Let
S be a nonemptycollectionof nonemptycompactsubsetsof Q. We say thatthe map
f has a set S of expandersfor {S,, S2, . S.., SK and M if the following two conditions
hold:
1. Foreveryi = 1, 2, . . ., K thereexists an E E S with Efn si # 0.
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2. If Efn si 0 for an E E ?, and mij = 1, then thereis a compactset PiJ C
EnlSi suchthatf(P1j)nSsj #0.
With the more generaldefinitionof a set of expanders,we can now formulatea
moregeneralversionof the ChaosLemma.
Lemma 6. (Chaos Lemma for Subshifts of Finite Type.) AssumeHypothesisM. If
the map f possesses an expander set 8 for {S,, S2, . . ., SK} and M, then there is a
chaotic set Q* C Qo such that for every M-admissible sequence S, there is a point
x E Q* such thatx follows S.
Proof Combineandadaptthe argumentsof Lemmas3 and4.
U
As a simpleandnice example,considera quadraticmap f : [0, 1] -* [0, 1] with a
period-threeorbitx1, X2, X3, whereX2 = f (xI), andX3 = f (X2), andx1 < X2 < X3. If
we choosetwo symbolsets, S, = [xI, x2] andS2 = [X2, X3], thenf (S,) S2, f (S2) B
S,, and f (S2) D S2. By settingS := IS1,S21 andM=( l), it follows fromLemma
6 thatfor any sequenceS = (Sin)of symbolsets in which no pairof consecutiveSin
are both S2, thereis a point xo E Si, that follows S. In this case S is finite and the
symbolsets arethe expanders.
7. WHEN DOES Q CONTAINAN INVARIANTSET? Mathematicscoursesfor
mathematicsmajorsor graduatestudentspresentpearlsof wisdom.Exercisesstressthe
abilityto proveresultsratherthanhow to findformulations,a frequentlydifficultart.
Forthe mathematician,
a proofis onlyhalf thebattleof creatinga new theorem.Before
theproofmustcome thediscoveryof a new statementorconjecture.Studentsarerarely
askedto come up with theoremsor conjecturesthatarenew to them.In this section,
we do not know what the final conjectureshouldbe, but we have some examples
and ideas. While mathematiciansdo not have a widely acceptedterm for "pseudoconjecture",i.e., a statementthat gives only the generalform of the conjecture,it is
mustmaster.
preciselythis processof creatingthatproto-mathematicians
Until now we have avoidedthe case K = 1 because it seems a bit trivial:there
would be only one possible itinerary,the constantitineraryS,. But saying that the
itineraryis followedimpliesthatthereis an invariantset in Q (andmore specifically
in SI), so the case K = 1 has a non-trivialconclusion.We havegeneralizedthe notion
of horseshoemaps in this paper,but furthergeneralizationscould be possible if the
case K = 1 werebetterunderstood.Figure8 showsit is necessaryto be careful.
Q
f(Q)
Figure8. Theset f (Q) is obtainedby rotatingQ by 90?,withthatrotationabouta pointnotin Q. Inthiscase
thereis no fixedpointin Q. Furthermore,
if x E Q, thenf2 (X) , Q, so Q containsno nonemptyinvariant
set.
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When does f have an invariantset Q? We define S1 to be {x E Q: f (x) E QI.
The simplestcase is f (Q) C Q, so the family of expanderscan be chosen to be all
nonemptycompactsets. The ExpanderLemmasays that all sequencesare followed
(extendedto the K = I case), so thereis a compactinvariantset, following the argumentsof the ChaosLemma(even thoughthereis no chaos guaranteed).It might
containjust one point,a fixedpoint,andit mightnot containa fixed point.Now suppose SI is a propersubsetof Q. In the two-dimensionalexampleour sets E in some
sense connect the top to the bottom of Q, that is, CD to AB. Figure9 is such an
example.
DD
D*C
A~~~~~~~
Figure 9.
Pseudo-Conjecture. If for each set E that"stretchesacross"Q the image f (E) n Q
also stretchesacrossQ, thenthereis a compact,nonemptyinvariantset in Q.
Wedo not knowhow "stretchesacross"shouldbest be defined,butwe believeFigures 10 and 11 areexamplesthatcouldbe covered.We conjecturethattheseexamples
have invariantsets. In these f(Q) n Q,the top of Q is connectedto its bottomindirectlyvia some kindof linkingof componentsof f (Q) n Q. We hope some reader(s)
can showthatFigures10 and 11 containinvariantsets. If so, it is probablya smallstep
to showthatFigure12 has chaoswith K = 2.
Figure 10. We conjecture that there is a nonempty invariantset in Q.
REFERENCES
1. K. T. Alligood, T. D. Sauer,and J. A. Yorke,An Introductionto Chaos, Springer,NewYork, 1997.
2. K. Bums and H. Weiss, A geometric criterionfor positive topological entropy,Commun.Math. Phys. 172
(1995) 95-118.
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All use subject to JSTOR Terms and Conditions
S
B
C
Figure 11. The three pieces of f(Q) n Q are parts of the three Borromeanrings. The circle, denoted C, is
not the boundaryof a disk Q\f (Q),but is the boundaryof a manifold (and so is homologically trivial, but not
homotopically trivial). The boundaryof Q contains top T, bottom B, and side S. We conjecture that if f (Q)
does not intersect S but C is not homotopically trivial in Q\f (Q),then there is an invariantset in Q and there
is a family of expanderswith K = 1.
f(Q)
Figure 12. We conjecturethat this example has a family of expandersfor K = 2. We would choose St and S2
as indicatedby their images.
3. M. Carbinatto,J. Kwapisz, and K. Mischaikow, Horseshoes and the Conley index, preprint.
4. M. Carbinatto,J. Kwapisz, and K. Mischaikow, Horseshoes and the Conley index spectrumII-the theorem is sharp,preprint.
5. T. Y. Li and J. A. Yorke,Period three implies chaos, Amer.Math. Monthly 82 (1975) 985-992.
6. J. Kennedy and J. A. Yorke, Topological horseshoes, Trans.Amer.Math. Soc., to appear.
7. K. Mischaikow and M. Mrozek, Isolating neighborhoods and chaos, Japan J. IndustrialAppl. Math. 12
(1995) 205-236.
8. C. Robinson, Dynamical Systems, Stability, Symbolic Dynamics and Chaos, CRC Press, Boca Raton,
1995.
9. S. Smale, Differentiable dynamical systems, Bull. Amer.Math. Soc. 73 (1967) 747-817.
10. A. Szymczak, The Conley index and symbolic dynamics, Topology35 (1996) 287-299.
JUDY KENNEDY received her B.S. in mathematicsin 1969 and her Ph.D. in 1975, both from AuburnUniversity. She fell in love with mathematicsat Auburn,and still is. Trainedas a topologist, she retainsher interest
in topology, but nows studies the topology of dynamical systems and considers herself an applied topologist.
She has been on the faculty of the University of Delaware Departmentof MathematicalSciences since 1986.
Departmentof Mathematical Sciences, Universityof Delaware, Newark DE 19716
[email protected]
SAHIN KOCAK studied mathematicsat Heidelberg University,from which he obtainedhis B.A. in 1973 and
his Ph.D. in 1979. Since then he has been on the faculty at Anadolu University in Turkey.He is interested in
chaos, gauge theory,history of mathematics,and wanderingin and wondering about nature.
Departmentof Mathematics,Anadolu University,Anadolu, Turkey
[email protected]
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[Monthly108
JAMES A. YORKE received his A.B. from Columbiain 1963 and his Ph.D. from the University of Maryland
in 1966. He is still at the University of Marylandas a Distinguished University Professor and Director of The
Institutefor Physical Science and Technology, a departmentthat emphasizes interdisciplinaryresearch. He is
on the faculty of the Mathematicsand Physics Departments.He would like to think he finds the pure mathematical ideas the scientist needs. He is working on improving weather prediction using ideas from dynamics,
writing a book on the populationdynamics of HIV/AIDS transmission,and having fun.
Institutefor Physical Science and Technology,Universityof Maryland, College ParkMD 20742
[email protected]
Mathematical
Constance
by Arthur Benjamin (with apologies to Joyce Kilmer)
I think that I shall never see
A constant lovelier than e.
Whose digits are too great to state;
They're 2.71828 ...
And e has such amazing features.
It's loved by all (but mostly teachers).
And Calculus, we'd not do well in
Without such terms as exp and In.
For e has such nice properties,
Most initegralsare done with ... ease.
Theorems are proved by fools like me,
But only Euler could make an e.
I guess, though, if I had to try
To choose another constant, I
Might offer i or bor ,
But none of those would satisfy.
Of all the constanitsI know well,
There's only one that rinigsthe Bell.
Not 7, not i, nor even e,
In fact, mnyConstance is a she.
It's Constance Reid. I wouldn't fool ya'.
With books like Hilbert, Coura,it, and Julia,
Of all the Constance you will need,
There's only one that you should Reid.
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