An Introduction to Dynamical Systems - 2nd Edition

12.4. Hyperbolic Toral Automorphisms 579

Now, assume that p is periodic, so F‘; (p) = p. Then, if 1r(:1:, y) = p,

~ (I) = <$> - oU 1/ 77'
-1<$>=<m>i( ) y n
Since the eigenvalues of A" —I are Af — 1 gé 0 and A‘; — 1 96 0, A'° —I is invertible
with integer entries, so the inverse has rational entries with denominator equal to
the determinant of A" — I. This shows that the vector

(Z) =<A"-"1 <2‘)

in R2 has rational coordinates, and p = 1r(:z:, y) is the projection of a vector with

rational coordinates. This is what we wanted to prove. El

Example 12.28. For (1/3,2/3)-r and the matrix A given in equation (12.1),

_al /&liaL2 modl

A § = 1% = modl)’
(\1/\ \ ) (_)(DOhi!-I

A 3/_\ = /_
\5) \§
A (2\ /§ _ (21:31). as
\“) \§ _
2:31)/2\ (4 _
3
\‘/ \ _

COJQIOUOINIJ-I

gives an orbit of period four.

Stable manifolds

lf two points 5 and F1 in IR’ differ by a scalar multiple of the stable eigenvector,
OI‘

— —_ s
q _ p '_ tv I
then the distance between their iterates goes to zero under forward iteration:

NA" F1 — A" ll = lil /\§ |lv”|l-

Therefore, in R2, F1 = Ii + tv‘ is on the stable manifold of I5:

W’(1A) = {F+tv’ = 16 IR}-

In R2, this stable manifold is a line. Because the stable eigenvector has irrational
slope, the stable manifold in T2 is dense; that is,

Wa(p| FA) = 7rW0(T)-1 A):

where 1r]? = p.

580 12. Higher Dimensional Maps

In the same way, W“(p,A) = {p-l-tv“ ;t e IR}
and W“(p.FA) = it W“(F. A)
is dense in 'll‘2.

'

Exercises 12.4

_1_. Find the eigenvectors for

21
1 1'

Use these to draw the local stable and unstable manifolds of the origin in the
plane.

2. The slopes of the eigenvectors of the matrix

21
11

are in"ationa.l. Explain why this forces the stable and unstable manifolds of
the origin to be dense in the torus. Hint: Consider the intersection of these
manifolds with a: = 0 (mod 1), which are multiples of an irrational number.

12.5. Applications

12.5.1. Markov Chains. In discussing expanding maps with Markov partitions
in Section 11.4.1, we encountered matrices that have row sum equal to one. These
matrices lead to Markov chains, which we discuss in more detail in this section.

Assume that some material is spread out among the n sites, with rcsol Z 0 the
amount of material at the i‘h site at time 0. Assume that mi; Z 0 is the probability

of going from the i"h site to the jtl‘ site, so that zgolmij is the amount of material
from the i°l‘ site that is returned to the j“‘ site at time 1. The total amount at the
j"" site at time 1 is the sum of the material from all the sites, or

$.21) = .’B£o)7"I'lij.
1

Let M = (rn.,-_,-) be the corresponding n X n matrix (where i gives the row and j

gives the column of the entry), and let

x("l = (:1,-S"), . . . ,:c$,"))

be the row vector of the amount of material at time q at all the sites. Using this

matrix notation,
x(1) = x(0) M,

and, more generally,

x(q) = x(q—1) M = x(0) Ma,

for the transition from the distribution at time q — 1 to time q.

12.5. Applications 581

The sum of the probabilities of going from the ii“ site to some other site is one,
so 27- mij = 1 and each of the rows sums of M is one. We assume that is possible
to make a transition to each of the j"h sites from some other site, so each column
of M has some nonzero entry. Finally, we assume that from at least one site there
are more than one possible following site, so some row has more than one positive
entry.

Notice that the total amount of material at time q is the same as at time q — 1
and so the same as at time 0:

2, = 2, (>:,¢sq-1>m..)

= Zr mijl ‘Tlqm

= Zi I£q—1)

(we use the fact that the row sums are 1). Call the total amount X = 21- 2:;-0) =
23- 1:;-q). Then,

ptq) : Eli
‘X

is the proportion of the material at the i‘h site at time q. Letting

P“) = (plow - - ~ :phq)) = %'x(q)

be the row vector of these proportions, we see that

(q—1)M = L(11M-1) = _1 ( (q—1)M) = _1_ (<1) = (11)
P xX xx P ‘

also transforms through multiplication by the matrix M.

We summarize the assumptions on the matrix M in the next de nition.

De nition 12.29. An n x n matrix M with real entries m,-j is called a stochastic
matrix or probability transition matrix provided that the following conditions are
satis ed:

(i) All the entries m,-_,- satisfy 0 5 mi, 3 1.
(ii) Each row sums to one, E]. mi, = 1 for each i.
(iii) Each column has some nonzero entry.
(iv) Some row has more than one nonzero entry.

De nition 12.30. A stochastic matrix M is called aperiodic (or eventually posi-
tive) provided that there is a qo > 0 such that M"° has all positive entries (i.e., for
this iterate, it is possible to make a transition from any site to any other site). It
then follows that M“ has all positive entries for q Z qo. An aperiodic stochastic ma-
trix automatically satis es conditions (iii) and (iv) in the de nition of a stochastic
matrix.

De nition 12.31. As in the case for a transition matrix, a stochastic matrix is
called irreducible provided that it is possible to get from each site i1 to each other
site i2 by making a nite number of transitions. In other words, for any pair of

582 12. Higher Dimensional Maps

sites (i1,i2), there are indices jk, with 1 g jk g n for k = 0, . . . ,q (where q can
depend on (i1,i2)), such that jg = ii, jq = 2'2, and mj,,_,,~,, > 0 for ls =1,...,q
(i.e., the (i;,i2)-entry of M" is nonzero).

One way to insure that a stochastic matrix is aperiodic is for it to be irreducible
and also to have one positive entry on the diagonal. We concentrate on aperiodic
matrices, which are therefore irreducible.

For a stochastic matrix M, 1 is always an eigenvalue with COlUII111 eigenvector
(1,...,1T)..

1 E]-'l’TL1j 1

.1 Z,-mm" '_ 1

Since MT and M have the sa.me eigenvalues, MT always has 1 as an eigenvalue,
with eigenvector (pi, . . . ,pj,)T. This is equivalent to saying that the row vector
(p§,. . . ,p;‘,) satis es

(Pi,---,P.'.) = (Pi----.PZ)M-

Theorem 12.7 says that this vector can be chosen with p; > 0 and 21- p; = 1.
When M is aperiodic, Theorem 12.7 further states that (i) all other eigenvalues
satisfy |/\;,~| < 1 and if (p1,.. .,p,,) is any initial probability distribution, then
(p1, . . . ,p,,)M" converges to (pf, . . . ,p,f,) as q goes to in nity.

Before stating the general result, we give some examples.

Example 12.32. Let

0.5 0.3 0.2
M= 0.2 0.8 0 .

0.3 0.3 0.4

This matrix has eigenvalues 1, 0.5, and 0.2. (We do not give the characteristic
polynomial, but do derive an eigenvector v = (v1,v2, v3) for each of these values.)

For)\=1,

L 53 01 .0no -0.4 - . -3
-0.03 0.48 ~ -6l—l@ .
MT - 1 = 0.3 -0.2 0.3 ~

0.2 0 -0.6 0.08 -0.4 OQr- 0 0

Thus, v1 = 3v3 and U2 = 6113. Since we want 1 = v1+v2+v3 = (3+6+1)v3 = 10113,
v3 = 0.1, and p‘ = v1 = (0.3,0.6, 0.1).

For /\2 = 0.5, 0.2 0.3 -
0.3 0.3 ~
MT - 0.51 =

l\)OO@ 0 _ c> F- |—'@Gl\)

( _1c>c>ro_c>_o »—roQ E" 010:» Q\0/»-= ' i-ca»-

—©©@ @lQ@ mc>cQo»—»-zo LM‘2

Thus, 2v1 = v3, 2v; = -3113, and v2 = (1,—3,2). Notice that v1 + ‘U2 + vs =
1 — 3 + 2 = 0. This is always the case for the eigenvectors of the other eigenvalues.

12.5. Applications 583

F01‘ /\3 = 0.2,

0.3 0.2 0.3
MT-o.2I= 0.3 0.6 0.3 ~
0.2 0 0.2
C0095‘ O0 -Jv—\£_.Z

ZI\4

O<D>- RQUDO OQI-I L) ©@P-‘ Qr—lQ/,.§ OOI-IL-/100:0

Thus, v1 = -113, 112 = 0, and v3 = (l,0,—1). Again, v1 +11; +113 = 1+0 -1 = O.

If the original distribution is given by

p(°> = (0.4s,0.4s,0.1) = (0.3,o.6,0.1) + 21~o(1, -3,2) + 51 0,0, -1),

then

pan Mq= (0.s,o.s,0.1)+i20(21)” (1, _3,2)+Lm(5E)“ (1,0, _ 1),

which converges to the distribution v1 = (0.3,0.6, 0.1) as q goes to in nity. This
convergence of the iterates holds for any initial distribution p(°). See Theorem 12.7.

Example 12.33 (Complex Eigenvalues). The following stochastic matrix illus-
trates the fact that an aperiodic stochastic matrix can have complex eigenvalues.

Let
0.6 0.1 0.3

M= 0.3 0.6 0.1 .
0.1 0.3 0.6

The eigenvalues are A = 1 and 0.4 zt i0.1 Notice that |0.4 :t i0.1\/§| =

\/O.l6+0.03= \/O.19< 1. '

Example 12.34 (Not Aperiodic). An example of a stochastic matrix that is not
aperiodic (nor irreducible) is given by

0.8 0.2 0 0
M: 0.3 0.7 O O

0 O 0.6 0.4 ’
0 0 0.3 0.7

which has eigenvalues A = 1, 1, 0.5, and 0.3. Notice that it is possible to go between
sites 1 and 2, and it is possible to go between and sites 3 and 4, but it is not possible
to go from the sites 1 and 2 to the sites 3 and 4.

An example of a stochastic matrix that is irreducible, but not aperiodic, is

given by

0.8 0.2

M= 0.3 0.7

>-IOO O Q

0 HOOO 0 O

which has eigenvalues /\ = 1, -1, and :i;\/i . Here, it is possible to get from any
site to any other site, but starting at site one, the odd iterates are always at either
sites 3 or 4 and the even iterates are always at either sites 1 or 2. Thus, there is
no one power for which all the transition probabilities are positive. Therefore, M

584 12. Higher Dimensional Maps

is not aperiodic. Also, this matrix has another eigenvalue -1 with absolute value
equal to one.

Theorem 12.7 (Perron—P\‘obenius). Let M be an aperiodic stochastic matrirc.
a. The matria: M has 1 as an eigenvalue of multiplicity one (i.e., 1 is a

simple root of the characteristic equation). A row eigenvector p‘ for eigenvalue 1
can be chosen with all positive entries and Zj p; = 1.

b. All the other eigenvalues Aj have |A,-| < 1. If vk is a row eigenvector for
Ah, then 27- '0? = O.

c. If p is any probability distribution with all p_,- > 0 and 21- pj = 1, then
11.

P = P‘ + Z y:=v_"

j=2
for some yz, ..., y,,. Also, pM‘1 converges to p‘ as q -goes to in nity.

Sketch of the proof. We give a sketch of the proof based on the proof using ideas
from dynamical systems given in [Rob99].

We asswne in what follows that all the mi, > 0, which can be done by taking
a power of M if necessary.

(a) As noted previously, M always has 1 as an eigenvalues, so it is always an
eigenvalue of MT (i.e., M has a row eigenvector for the eigenvalue 1). To discuss the
multiplicity, we assume that there is another column eigenvector v, with Mv = v
and not all the v_,- are equal. Assume that It is the index for which |v;,| is the largest
component. By scalar-"multiplication by -1, if necessary, we can ta.ke vk positive.
Thus, vi, = |vk| Z Iv,-l for all j and vk > |v¢| for some E. Then,

vk = 2_1TL|¢jUj < E _TT74¢jU;, = Uk.
.7 .7

The strict inequality uses the fact that all the m,-j > 0 (i.e., that M is aperiodic).
Since this shows vi, > vk, the contradiction implies that there are no such other
vectors, and so that, there can be only one eigenvector for the eigenvalue 1.

To complete the proof, we would have to consider the case in which 1 is a
multiple eigenvalue with only one eigenvector. We leave this detail to the reference.

(b) Case (i): Assume that A 75 1 is a real eigenvalue. Again, assume that
Mv = Av. Let k be such that vk = |v;,| Z Iv,-| for all j and vi, > lvgl for some 8.
Then,

Auk = Zj nik,-v_, < Xi my,-vk = vk.

This shows that Avk < vk, so A < 1.
We now show that A > -1. Notice that Uj Z —v;, for all j and v; > -vk £01"

some E. Therefore,

AU); = 2-’_ my,-vj > Zj TTt|¢j(—’Uk) = —’Uk.

This shows that Auk > —v;,, so A > -1. Combining, for a real eigenvalue that is
not equal to 1, -1 < A < 1.

12.5. Applications 585

Case (ii): Assume that A = 're?"“"' is a complex eigenvalue with complex
eigenvector v. Here, 'r = |A| and e2”“" is complex. Assume that the vj are chosen
with vk real and uh 2 R,e(vJ-) for all j. Since Mqv = Aqv,

Tq Re(e2"°"”)vk = Re(A")v;, = Re((M"v),,) = (M°Re(v))k

__ Xi (m(g) R.e(v_, ,)) < Zj (mg( ,l Uk) -_ vk.

Therefore, r"Re(e2"”"”) < 1 for all q. Since we can nd a. q, for which Re(e2’""“")
is very close to 1, we have r‘" < 1 so r = |A| < 1.

(c) Let p be a probability distribution with Z5 pj = 1. The eigenvectors are a
basis, so there exist y1,...,y,, such that

Tl

P = Zvivh

i=1

Here, all the vectors are row vectors, and v1 = p‘. Then,

n fl

1=2_Pj=y12v}+Zy¢Zv§=y1+Zyr-0=y1-

J J i=2 _1 i=2

Thus, n

P = V1 + vii
i=2

as claimed.

Writing the iteration as if all the eigenvalues are real, we have

71 71

pM" = v1M‘7 + Z1/,vlM‘1 = v1+ Zy; A:-’v",

‘i=2 i=2

which tends to v1 because all the IA? I < 1 for i 2 2. Cl

12.5.2. Newton Map in R". We discussed the Newton map of one scalar vari-
able in Section 9.3.1. A similar method applies to nding zeroes of a function from
IR“ to R".

Let F be a map from lR" to R". Assume that xj approximates a zero of F.
The linear approximation of F at xj is given by

y = F(x,-) + DF(,,,)(x — xj).

Setting y = 0 and solving for the variable x de nes x,-+1:

0 = F(xj) + DF(,,J)(x:,-.,.; — xj)

-F(X1) = DF(==,)(Xj+1 — X1")

— (DF("j))_1 F(’<1'l= X1"+1 - X1"

X1+1 = X1" — (DF(><,)l_l F99)-
This equation de nes the associated Newton map by

N,~(><) = X - (1JF,,,,)"‘ F(x).

Note that NF(X) = x if and only if F(x) = 0.

586 12. Higher Dimensional Maps

Just as in the case in one variable, the derivative of NF at a zero of F is zero.
Assume that F(p) = 0. Let I be the identity matrix, which is the diagonal matrix
with ones down the diagonal. Then,

D (NFlp = I " (DF<p>)_1 (DF<p>) - D[(DF<->)_1](p)F(P)
= I - I - D [(0F(.,)“] (P) 0
= o.

The mysterious term D [(DF(.)) _1]( ) is the derivative of the derivative (second
P

derivative), where the dot represents the variable with which it is differentiated.

Because this term acts on the zero vector, we, do not need to know what it is.

Using the Taylor expansion to degree two, -

llNF(x) — P|| 5 6' IIX — Pll2

for x near p, where C > 0 is some constant depending on the second partial
derivatives. Thus, starting near the zero leads to orbits that rapidly converge to p
(i.e., p is supemttracting).

12.5.3. Beetle Population Model. In Example 8.5, we introduced a. model
of the our beetle with three stages of population: larva, pupa, and adult. The
function for the model given there is

Ln“ = bAne—cLAAn e-Cttbn,

Pn+1=(1 _ /1'L)Lrn

A,,,, = P,, e_C‘“"A" + (1 - ,i,,)A,,,

whereb>0,0§u[_$1,0§;.iA51,CL,4 20,C;,[_20,andCp,1Z0. The
quantities exp(—C;,AA,,) and exp(—CLLL,,) are the probabilities that an egg is not
eaten by the adult population An and the larvae population Ln, respectively; the
quantity exp(—Cp,4A,,) is the survival probability of the pupae into adulthood; b is
the birth rate of larva in terms of the adult population, at, is the death rate of larva
(that do not transform into pupa), and /1,4 is the death rate of adults. The analysis
of the xed points for this system is given in the book |Bra01] by F. Brauer and
C. Castillo—Chavez. We follow their treatment for the simpli ed case presented.

To analyze this discrete system, we ma.ke the simplifying assumption that
Cu, = 0 (i.e., we neglect the cannibalism of the eggs by the larva). With this
assumption, the function becomes

L1-‘+1 Ln bAn€_CLAA"

P,,.,., =F P,-, = (1 —u;,) L,, .

An.+1 An Pn e_CPAAn +(1_ FLA) An

The xed points of F satisfy and
L= bAe_C‘-‘A,
P= (1 —u;,)L,

P = A/J.A BCPAA.

12. 5. Applications 587

Eliminating P gives

L = bAe_c'"‘A and

(1 - /11.) L = All/1 e°"""-

Certainly, L = A = P = 0 is one solution, called the extinction xed point. If these
variables a.re not zero, then we can take the ratio of these two equations, which
yields

(1 _ ML) = if e(C'r.,\+Cp,,)_4-’

e(cLA+CPA)A. = , or

A. 1P1-A1(0)

CLA + CPA ’

where

0 = b _ I'LL)

HA
Once we have solved for A‘, it follows that

L‘ = bA'e'C'-‘A’ and

P‘ =(1- /.tL)L' = (1 - ;rL)bA¢-CM”.

When 0 > 1, this gives a second xed point (L‘,P", A‘), with all the populations
positive, which is called a sunrival ased point.

The matrix of partial derivatives of F is

0 0 bB—c""A(1 —- CLAA)

(1- I11.) 0 0-
0 e_c"*"* 1 — /4,4 — PCpA8_CPAA

At the extinction xed point (0, 0, 0), it reduces to

U 0b
(1—/LL) 0 O ,

0 1 1-/1,4

which has characteristic equation

0 = A3 +a1/\2 + a2,\ + a3 with
411 = —(1—#A),
(12 =0, and

as = *b(1- FL)-

The next lemma, which appeared in the paper [Sarn41], by P. Samuelson, gives a
criterion for stability of a xed point for a function of three variables. (Also see
[Bra01].)

Lemma 12.8. Assume that p(/\) = A3 + a1/\2 + 0-2/\ + a3.

a. If either 1 + a1 +02 +a3 < 0 or 1 — a1 + 0.; — 0.3 < O, then the absolute value
of at least one of the roots -is greater than one.

588 V 12. Higher Dimensional Maps

b. If

1+a1+ag+a3>0, 1—a1+a2—a;;>0,
3+a1—az—3a3>0, 1+a1a3-0-2—a§>0,

then the absolute value of each of the roots (real 01" compleaz) is less than one.

Note that p(1) = 1+a1+ag+a3 and the coe icient of /\3 is positive, so if p(1) <
0, then there is a real root greater than one. Similarly, p(—1) = -1 + a1 — a2 + a3,
so if p(—1) > 0, then there is a real root less than minus one that has absolute
value greater than one.

-- For the characteristic equation for the extinction xed point, if 0 > 1, then

1+0.1+t1g+0.3=1—(1—p.);)—b(1—/$1,)

= P-A - b(1- ML)
< 0,

and the xed point is unstable. If 0 < 1, then the rst quantity is greater than
zero. The second and third quantities are positive as follows:

1—0»1-l-02-G3=1+(1—[J.A)+b(1—/J,L)>1>0 and
3+a1—oz—3a3 =3—(1-p.,4)+3b(1—u1,)>3—1 >0.

For the last quantity,

1 + a.1a3 —._ a2 — ag = 1 + (1 — ;.lA)b(1— #1,) -— b2(1 — pL)2

' >1—b’(1—/1L)’

> 1- pf;
> O,

since 0 < 1. Thus, the extinction xed point (0, 0, 0) is attracting for 0 < 1 and is
unstable for 0 > 1.

VVhen 0 > 1, the extinction xed point is unstable and the survival xed point
has all positive populations. Using the equations of the xed point, we can write
the matrix of partial derivatives at the survival xed point as

' - amt’

(1 - Hz.) 9 _ 9

0 6'0"“ 1- I-M — M/aC'PAA'

This matrix has characteristic equation

0: /\3+(1|/\2+0.2/\+0.3 Wit-ll
01 = -1+ IIA + /IACPA/1', 3-Ild
I12 = O,

as = —(1 - la) — 6'1.AL') ¢‘C’°""-

12.5. Applications 589

So the quantities from Lemma 12.8 for the characteristic polynomial of the survival
xed point are

l.+U-1+0.g+G3
L‘ -

= /1.4 + #AC'PAA' - (1 - /1L) (F - C'1.AL")6_c""A ,

1 — a1 + G2 — a3

L‘ _ -
= 2 - /-1-A r I1ACPA1‘1' + (1 - FL) (F - CLAY) 6 CPAA 3

3 + 0.1- 112 " 3413

=2+ll~A+lL_4Cp_4A - ++3(1—/JL) FU4-3CL_4L- 6 -cPA 4' ,

1+a1a3—a2—a§

L‘ -

= 1 '1‘ (1 - #4 - /1ACPA-4'))(1- /-‘Ll (Q + C1.AL'> 6_c""A

L‘ 2 -

- (1 - 111.): <2: + CL/113') 6_2c"‘A -

The values of these four quantities for b = 4.88, /11, = 0.2, p.,4 = 0.01, OLA = 0.01,
and CPA = 0.005 are all positive, 0.5967. . . , 1.9403..., 1.9405 . . ., and 0.9702...,
so the xed point is attracting. If two of the parameters are changed to /M = 0.96
and CPA = 0.5 and the other parameter values are left the same, then 1 + a1a3 —
(Z2 — a§ = —1.0669..., and the survival xed point becomes unstable. Figure 13
shows the plot of a single orbit for the last set of parameter values; notice that this
orbit winds around in a complicated manner (i.e., it appears chaotic).

12.5.4. A Discrete Epidemic Model. Some populations have xed time inter-
vals between generations, so a map is a better model for the population than a
differential equation. In this subsection, we consider a single population that can
become infected, and then recover to become susceptible again. In this setting, it
would be a matter of distinct periods when the infection takes place. This is based
on Section 2.9 of [Bra01] and on the work of C. Castillo—Ché.vez and A.A. Yakubu
referenced there. Compare with Section 4.7.2, where we considered an epidemic
model based on a system of differential equations.

At the nu‘ stage, .S',, is the size of the susceptible population and L, is the size
of the infected population. The model takes the form

S,,_H = A + Sn e_"e"°"" + 1,, e_" [1 — e“’],

I,,+; = Sn e"‘[1 — e_""'] + In e"‘e_°,

where A, [J., a, and a are all positive parameters. The total population Tn = S,,+I,,
satis es the iteration T,,+1 = A + e‘*‘ T,, E g(T,,), which has a unique xed point
at

A
T‘ = —i,

1 — e"‘
a.nd this xed point is attracting.

590 12. Higher Dimensional Maps

35 A 35 A

N L P
no 0 90
0 if "WT
(a)
(bl

9P

35 P

\, -- ~--,,.. ~ 90

0

L _\

ii“ L
O no I10

(C) (<1)

Figure 13. Plot of an orbit for the beetle population for b = 4.88, [J]; = 0.2,

p.,q = 0.96, CLA = 0.01, and CPA = 0.5. (a) A as a function of L, (b) A as a
function of P, (c) P as a function of L, and (d) three-dimensional plot of L,
P, and A

One xed point of the two-dimensional system is S = T‘ and I = 0. The
matrix of partial derivatives of the two-dimensional map at this xed point is

I e‘“ e_“[1 — e"’]
0 e_"‘_° + aT'e_" ’

which has eigenvalues e'" and e_““’ + cxT‘e"". The rst eigenvalue is always less
than one, 0 < e‘" < 1, and corresponds to the fact that T‘ is an attracting xed
point for g(T,-,)_. The second eigenvalue is always positive and is less than one for
some parameter values:

1 § e-'*-° + aT‘e*",

1 ~ e_"_" ; aT‘e““,

1 >? Za-T,‘Te'ta‘ = R0-

This last quantity R0 is called the reproductive number. Thus, for R0 < 1, this
xed point is attracting. For R0 > 1, this xed point is a saddle and is unstable.

If the total population is at equilibrium with Tn = T‘, then Sn = T‘ - In. and
we can get a function of a single variable,

I,,_,_1 = (T" — I,,)e_"|1— e_°""] + In e_t‘_" E f(I,,).

The derivative at I = 0 is f’(0) = e_““’ +aT‘e‘". For R0 < 1, 0 < f'(0) <1 and
I = 0 is an attracting xed point for f. In fact for 0 < I 3 T‘, 0 < f(I) < I and
I = 0 is a globally attracting xed point for f, attracting all points on (0, T"]-

12.5. Applications 591

For R0 > 1, 1 < f’(0) and I = 0 is unstable. Since f(T‘) = T‘ e'““’ < T‘,
there is a second xed point I‘ with 0 < I" < T‘. We do not discuss the stability
of this xed point.

12.5.5. One-Locus Genetic Model. A one-locus genetic model from mathe-
matical biology provides a relatively simple situation in which a nonlinear system
has a globally attracting xed point. Our treatment is based on that in J. Hofbauer
and K. Sigmund, [Hof88].

The characteristics of an organism are determined by a pair of strands of chro-
mosomes. In humans, each strand contains 46 chromosomes. A single trait can be
determined by several pairs of genes occurring at one or several sites of chromo-
somes, called the locus of the trait We consider the simplest case in which the locus
is a single site or single locus. Usually, there can be several types of genes that may
occupy a locus. These genes are called the alleles. We label them A1, . . . , A,,. We
assume that the allele A, occurs with frequency Pi Z 0, with p1 + - -- + pn = 1, so
the pair A,-A, occurs with frequency pip, in the original generation. The frequency
vector is given by

p = (p1,...,p,,)T.

By random mating, we assume that the gene pair (A,,A,-) also occurs with
probability p,-p, in the next generation. If N is the total number of alleles in this
next generation, then there are p,-p,-N of the gene pair (A,-, A,-). Let w,-j 2 0 be
the probability that the gene pair (A,,A,-) survives to adulthood. The order in
which the genes appear on the two strands is not assumed to affect their viability,
so w,-5 = w,-i. The selection matrix

W = ("H1")
is therefore symmetric. Let T(p) = p’ be the frequency of the alleles in the next
generation at adulthood, and p j be the frequency of the gene pair (A,, A,-). Then,

p.__ = w1";"P4P;'N = wijpipj _ = wa;"P="P;'
‘J EH wk€PkPtN 2k,l wkzlmpz PT WP

and

' _ /__ = . EwJupj : . i(WtP)-

pr _ $171] pl 2,“: wkfpkpt P1 p'|'Wpv

where v,- denotes the 17"‘ component of the vector v.

The state space of the possible p is the simplex

S={p=(p1,...,p,,)T :Zp;=1, 0$p,$1fori= 1,...,n}.
i

We are not going to check stability by means of the linearization, but by means
of a real-valued function. This approach is similar to the use of Lyapunov functions,
which we present for systems of ordinary differential equations in Section 5.3.

Theorem 12.9 (Fundamental theorem of natural selection). Let T be the map
from S to itself given by

(T(p)). = P.

592 12. Higher Dimensional Maps

Let (p) = pTWp be the average tness of the population. Then, with each itera-
tion, the average tness is nondecreasing, or

w(T(P)) 2 WP),

and there is equality if and only if T(p) = p is a xed point.
Therefore, if there is a xed point p" with all pf > 0, then any p with p,~ > 0

-is in the basin of attraction of p‘ and the w-limit ofp is p‘, w(p; T) = p‘.

Proof. We need to use some facts about inequalities. First, the geometric mean is

no larger than the arithmetic-mean, or '.

\/Z1_-5<ia2+b 1

for a,b > 0.

The function f(:r:) = a:°‘, with Oz > 1, has f”(:c) > 0 and is convex: so, for pl,
pg>0,p1+p2=1,a.nd0<a:1<a:2,

(PNI1 + P2112)“ = f(P11=1 +P2=12)< P1f(11)+p2f(12)= P1(-'61)“ +Pz(=r2)°'-

Similarly, for n points ':c,- > 0 and pi > 0, with p1 + - - - +p,, = 1,

(12-2) O

(gpia-ii) = f (ZPHH) S ZPif(1vi) = ZPi(-'Ii)a-

Equality occurs if and only if all the 2:; are equal.
Now we can start the calculation of the change in the average tness:

w(T(p)) : 213,- Pi(WP)iwijPj(WP)j'

WP)’

We ca.n multiply across by the denominator and get

"-T1(Pl2T17(T(P)l = Z PiwikPk'wijPj (WP)j-

i,_7',k

Interchanging the roles of j and Ic, we get

W(p)’W(T(1>)) = 2piwijpjw'il:pk(Wp)k-

i,j,I:

12.5. Applications 593

Ta.king the average and then applying the inequality for the arithmetic and geo-
metric means yields

W(P)2w(T(P)) = Zpipipkwijwik [ ]

i,j,k
L

2 2 PiPjPkwiiwiklwPl; (WP) I'\-ll-'

t',j,lc

= 2% wijPj(WP) 2 wakPk(WPlE jl

1" j e

2

= 21% wijPj(WP),%] -
i J'

Applying inequality (12.2) for a = 2, we get 2
-

l

w(P)2@(T(P)) 2 Zn Ewe-P1 (WP)? ]

i 1'
-2

= ZP1(WP) Zr.-wt]

-j i
T Z

= Z:Pj(WPl (W1-1%]

:' 2

= Zjpj<wp>§l -

J'

Applying inequality (12.2) again for a = 3/2. we obtain

$4

(i>)’W(T(P)) 2 [Z P,-(WP),-l = WP)“-

1

Dividing by ’l._ll(]))2, we get the desired result.

In the last inequality, there is equality if and only if all the (Wp)j are equal.
However, if (Wp),- = c, then 1T(p) = Zip,-c = c and

(T(P))¢ = Pig = Pa Cl
for all 2', so p is a xed point.

12. Higher Dimensional Maps

Exercises 12.5

Consider the stochastic matrix

0.8 0.1 0.1
M = 0.2 0.7 0.1 .

0.1 0.3 0.6

Find the steady-state probability distribution p‘.
A simpler SIS epidemic model, which has been proposed by L.J.S. Allen for
certain situations, [A1194], is

s,,+-1 = s, (1- % 1.,)'+ 11,,

1",, =1, (1-v+%s,,),

where N = I0 + S0, and the parameters are a > 0, 'y > 0. and a < 1+ 'y.
a. Show that Sn + 1,, = N for all n 2 0.
b. Show that making the substitution S,-, = N — In and setting p = 1 — 'y+ cr
results in the following function of a single variable:

1,,,, = f(1,,) = p1,, <1- % 1,).

c. Show that the xed point I = 0 is attracting if and only if R = °‘/1 < 1.
d. For R > 1, show that there is a second xed point at I‘ = N(P_1)/Q.

Show that this xed point is attracting if a < 2 + 7.

A model for in ation and unemployment, proposed by Ahemed, El-Misiery,
and Agiza [Al1m99], is given by the function

<Un+l) = < Un_b(In ‘ml )
I
In+1 In ‘l’ (C _ 1)f(Un) 'l' f (Un _ bun _

where f(:r) = —a + e“, where Oz, [3, m, and b are all positive parameters, and
0 < c < 1.

61> = <'"a. Show that the unique xed point is given by
b. Let A be the determinant of the matrix of partial derivatives at the xed
point. For what restriction on the parameter values is O < A < 1?

c. Show that the eigenvalues of the xed point are

,\=|, =1-i2ib-\/221 ba=~4abc.

d. For what restriction on the parameter values are the eigenvalues complex
(i.e., not real)? For these parameter values, is the xed point attracting
or repelling? Hint: Remember that the product of the eigenvalues is equal
to the determinant.

12. 6. Theory and Proofs 595

e. Assume that the parameters are chosen so that the eigenvalues at the xed
point are real. (i) Show that, if Orb < 2, then 0 < /\+ < 1. (ii) Show that,
if ab < 2 and A > 0, then A- > 0. In this last case, is the xed point
attracting or repelling?

12.6. Theory and Proofs

Linearization near periodic points

To a large extent the linearized equations at a xed point determine the features
of the phase portrait near the xed point. In particular, two maps are conjugate
provided that there is a. change of coordinates that takes the orbits of one system
into the orbits of the other. In this section, we make this idea more precise; in
particular, we need to be more speci c about how many derivatives the change of
coordinates has. The results in this section have proofs beyond the scope of this
text. We merely state the theorems and provide references for the proofs.

The statements of the theorems do not involve a conjugacy on the whole space
IR", but from neighborhoods of the xed point to a neighborhood of the origin.
Therefore, we repeat the de nition of conjugacy to introduce this idea of taking a
subset.

De nition 12.35. We say that F on U is topologically conjugate to G on V
provided that there is a homeomorphism h from U to V such that

h o F(x) = G o h(x)

for all x in U. Such a map h is a continuous change of coordinates.

For an integer r 2 1, these maps are called C’ conjugate on open sets U and
V provided that there is a C’ diffeomorphism h from U to V such that

h o F(x) = G 0 h(x);

thus, both h and h“ are C’. Such a map h is a differentiable change of coordinates.

In the theorems involving the linearization of a map F, we assume that the
derivative is invertible at the xed point x", det(DF(x.)) 76 0. By the inverse
function theorem, for such maps, there are two neighborhoods U1 and U2 about
x"' such that F is a diffeomorphism from U1 to U2. We call F a local d1l 'eomorphism
at x".

As we state below, there is a general theorem called the Grobman-Hartman
theorem which states that a nonlinear map with a hyperbolic xed point is topo-
logically conjugate to the linear map induced by the derivative at the xed point.
Unfortunately, this topological conjugacy does not tell us much about the features
of the phase portrait. Any two linear maps in the same dimension for which the
origin is attracting are topologically conjugate on all of IR". See [R0b99]. In partic-
ular, a linear system for which the origin is a stable focus is topologically conjugate
to one for which the origin is a stable node. Thus, a continuous change of coor-
dinates does not preserve the property that trajectories spiral in toward a xed
point or do not spiral. A di erentiable change of coord.inates does preserves such
features. Unfortunately, the theorems that imply the existence of a differentiable
change of coordinates require more assumptions.

596 12. Higher Dimensional Maps

We now state these results. They also apply to periodic points, but we state
them only for xed points, where the statements are simpler. In the theorems,

A = DF(,,.)

is the matrix of partial derivatives, and

DF(x-) y

is the associated linear map.

Theorem 12.10 (Grobma.n—Hartman). Let F be a C1 map with a hyperbolic xed
point x‘ such that det(Dh(,<-)) 96 0.

Then, there are open sets U containing x‘ and V containing the origin, such
that F(x) on U is topologically conjugate to the associated linear map DF(,-) y on
V. '

See [Rob99] for a. proof. _

To get a. di erentiable change of coordinates we need to treat special cases or
add more assumptions. The result with more derivatives is due to Sternberg, but
requires a “nonresonance” requirement on the eigenvalues.

Theorem 12.11 (Sternberg). Let F be a C'°'° map with a hyperbolic red point x‘
such that det(Dh(x-)) 96 0.

Let )\1,...,,\,, be the eigenvalues of DF(,‘-), and assume that each /\;, has
algebraic multiplicity one. Moreover, assume that

/\k 96 /\1"‘/\§"’ AZ“,

for each It and any nonnegative integers m_,- with 21- mj 2 2. {This condition is
called multiplicative‘ nonresonance of the eigenvalues.)

Then there are open sets U containing x" and V containing the origin such
that F on U is C°° conjugate to the linear map DF(,,-)y on V.

The proof of this theorem is much harder than Grobma.n—Ha.rtman theorem
proof. It a.lso,requires the nonresonance of the eigenvalues. The book [Har82]
contains a proof. The next two theorems state results about a C1 conjugacy.
Ha.rtman’s theorem has no assumption of the nonresonance of the eigenvalues.

Theorem 12.12 (Hartman). Let F be a C1 map with a hyperbolic attracting xed
point x‘ such that det(Dh(,‘-)) 96 0. Then, there exist open sets U containing x‘
and V containing the origin such that F on U is C1 conjugate to the linear map

DF(x») y O71.

Theorem 12.13 (Belickii [Bel72]). Let F be a C2 map with a hyperbolic attracting
red point x‘ such that det(Dh(,¢-)) 96 0. Assume that the eigenvalues of DF(,-)
satisfy a multiplicative nonresonance assumption that says that /\k 96 /\i /\_,- for any
three eigenvalues (including Ak 96 Ag).

Then, there exist open sets U containing x‘ and V containing the origin
such that the map F on U is C1 conjugate to the linear map DF(,,-) y on V.

i
Chapter 13

Invariant Sets for Higher
Dimensional Maps

In this chapter, we consider invariant sets for maps with two or more variables.
By a treatment analogous to the one used for the tent map and the logistic map
G in one dimension, we use symbolic dynamics to show that a nonlinear map has
sensitive dependence on initial conditions and a. dense orbit in an invariant set.
We start with the geometric horseshoes of Smale, which has an invariant set that
is a two-dimensional analog of the Cantor sets arising from tent maps considered
in Chapter 10. We follow by de ning symbolic dynamics for a nonlinear map in
higher dimensions, using correctly aligned boxes. Just as the geometric horseshoe
has certain rectangles whose images cross each other, a map is said to have a
Markov partition if there are nonlinear boxes whose images are correctly aligned
with the set of boxes. The toral automorphisms were among the rst maps for which
Markov partitions were constructed; we illustrate ideas by giving a Markov partition
for some simple examples of toral automorphisms in two dimensions. We also
make the connection between correctly aligned boxes and shadowing (i.e., nding
a. true orbit near an orbit with errors). Going back to ideas of Poincaré, we show
how an intersection of stable and unstable manifolds leads to a. horseshoe and, so,
complicated dynamics. These topics all relate to carrying over symbolic dynamics
to maps in two or more dimensions.

We discuss chaotic attractors for maps in multidimensions: the de nitions are
essentially the same as those in one dimension, but the examples give new insights
into the concepts and the manner in which they arise. We introduce Lyaprmov
exponents for maps in more dimensions, which is more complicated than in one
dimension. Just as for one-dimensional maps, Lyapunov exponents give a more
computable quantity than the de nition of sensitive dependence on initial condi-
tions. We relate Lyapunov exponents and computed orbits to a test for a chaotic
attractor. This treatment carries over the concepts presented in Chapter 11 in one

W

597

598 13. Invariant Sets

dimension to maps in two or more dimensions. This discussion broadens the arena
where these ideas can be applied.

13.1. Geometric Horseshoe

To start our consideration of multidimensional maps with invariant sets more com-
plicated than a periodic orbit, we give an example that is piecewise a ine (linear
plus a constant) in the regions that contain the invariant set. In this regard, this ex-
ample is like the tent map with slope 3, which produced the standard middle-third
Cantor set. S. Smale introduced this example as a model for the type of invariant
sets that arise from homoclinic intersections of stable and unstable manifolds as
discussed in Section 13.3. The map is speci ed as ai ne on two horizontal strips, is
given by a geometric description on the gap between the strips, and is extended to
the whole plane so other points eventually enter the region where the description
is given. Since the map is not given by a formula in all of 1R2, but its properties
are described geometrically, the invariant set is called a geometric horseshoe, or
sometimes, the Smale horseshoe.

Description of the Horseshoe Map

The horseshoe di eomorphisrn F is rst prescribed on a stadium shaped region

U = S’ U E0 U E1, where

is a square region E0 and E1 are semicircular regions added at the top and
bottom of S’ as indicated in Figure 1(a). The map is described geometrically on U
as the composition of two maps, F(a:,y) = FQ(L(a:,y)): (i) The linear map L from
U to R2 contracts the rst coordinate by 1/3 and expands the second coordinate by
4. (ii) The second map F0 bends the region L(U) into a horseshoe shaped object
and places the image inside U. See Figure 1(b). The name of the map derives
from the fact that F(U) has the shape of a horseshoe. By advanced mathematical
reasons (methods from differential topology), the map can be extended to the whole
plane in a manner so that other points eventually enter the region U.

To give a more precise description, we divide the region S’ into three subregions,

Hf, = {(a:,y) eS’ : "1/1651131/4},

G’={(¢,z/) E$'=1/451/53/4}. and
Hi = {(11.11) E 5'=3/4S2/S 17/16}-

The horseshoe diffeomorphism F is given by

(§:|:,4y) for (:i:, y) G H3,
F(==.z/)= _l 4_4 f H,

(1 31:» 1/) or ($1y)€ ‘I:

and F bends in a nonlinear fashion in G’. The region E0 is mapped into itself by a
contraction, so we can specify the map to have a single attracting xed point po in
this region whose basin of attraction includes all of E0. The region E1 is mapped
into E0 so it is also in the basin of attraction of po. In the analysis, it is very

13.1. Geometric Horseshoe 599

if a;,,:;IWIBLLISIJ . Fl.I.I'l ~Y}I

ietasse;.

53$;

E0 F(E0) F(E1)

('1) (b)

Figure 1. Region U = S’ U E9 U E1 and the image of F(U) for the Geometric Horseshoe

important that inside Hf, U H’1, vertical directions are taken to vertical directions
and horizontal directions are taken to horizontal directions.

Because points (:z:,y) with '1/16 5 y < 0 or 1 < y 3 17/16 leave S’ under

forward iteration, and points (:z:,y) with -1/16 3 :1: < 0 or 1 < a: _<_ 17/16 leave

under backward iteration, the only points which stay in S’ for all iterates also stay

in the unit square '

S =l0a1lxl0\1ll

that is,

Q F"(S’) = Q F1" (s).

J'=-0° J'=-0°

Therefore, we switch over to using the unit square.

There are two horizontal strips in S whose images also lie in S:

H0=Hf, S={(:r,y):05:r5land0$3/51/4} and

H;=H§ S={(:r:,y):05a:§1and3/43;/$1}.

The images by F of these two horizontal strips are the two vertical strips in S given

by

V0:{(:z,y):0§:r51/3and0§y§1} and
V1={(a:,y):2/3§:c§1and0§y§1}.

The map F takes H0 to V0 by means of a contraction of 1/3 in the rst coordinate
and an expansion of 4 in the second coordinate. It takes H1 to V1 by means of
a rotation by 180 degrees and a contraction of 1/3 in the rst coordinate and an
expansion of 4 in the second coordinate. See Figure 2.

600 13. Invariant Sets

aI s-1a-nu-m4~ i--ii%'i--g_A~‘$¢-'.|i'_-._1:‘Axi '“ .*9:3v.9=,I’-
~§4\*v~~¢ I, =1 ll}?-9(~'_K.

ysq W‘...ieqtsgxf
\f _1.~n.~.
__o,.H.' 4.-*-2..'»7~w-~unK~'1e \v!<--P‘
~-r'\- I-.,\ kg

»\;~\.4|~>n»zlt‘l2r~.¢b ‘I’-!1_£\'g|0i\1
i '1-1r.?-_'>n'uwqn.|.~ir \ Mtwnvi a '.m'/'=-=a'_t'»»f\)-
Prq-4~ v~~~ buy. '3'!-Z" _ Q10}-

<-.,-.-,r.,-~a-,->{n5’|v--€__r‘¥§' .-.\- <e*¢_w§en¢\»7e,;.w.-‘Iv1u.'-l_11.P-Q.-vI a».sg,-:.o_‘,._'-1.,Y"_xI¥<=v~-E*.i‘0'n,<.-r‘_I‘n>;,a_4'911¢--4_*.'-,.1‘-:_9--\~._< 1;-.:x--'__ \| -f-J)‘=-,-‘\ H4g1xn P~.v2.t I-' -~1i'<nb~»'s1_b'-\Ih.‘45'-.~c|_g1,-n.u -OI'-P>_TIVJe-.,<$1?'-z.vi-ezsvih:-v\-v'i-'4»a-'~r.~\ l 7'
-1._‘-_aQ‘”.o/ae'_wdj,:.a1n¢41.-n_aI-$»un"-m A5. A‘1naa.s)gh-_tsaI‘-!='Y~-;4an~-‘_'.\-1'}‘!,-&f4'ln.~

(*1) (b)

Figure 2. Horizontal and Vertical Strips in S

The inverse map is given by
F_,(I'y) = (3a:, g) 1 for (a:,y) E V0,
(3 — 311:, 1 — Zy) for (a:,y) E V1,

and F“ maps V0 onto H0 and V1 onto H1.
Invariant Set for the Horseshoe Map

Before discussing the symbolic dynamics, we indicate which points stay in S
for all iterates. We start with notation for the nite intersections of iterates by

1|.

S;= n Fj(S)={x:F_j(x)GSform§j§n}.
J='"

The next lemma describes the sets S,'.‘,, for various choices of m and n.

Lemma 13.1. a. For forward iteration with n Z 0, S3‘ is the union of 2" vertical
strips, each of width (1/3)", and S8° is a Cantor set of vertical line segments,
C1 X [0,1].

b. For backward iteration with —n 5 0, S2,, is the union of 2" horizontal
strips, each of height (1/4)", and SQ“, is a Cantor set of horizontal line segments,
[0,1] X C2.

c. The points that stay in S for all iterates are in the invariant set

A=S2.°°°=C1XCg,
which is the Cartesian product of two Cantor sets.

Proof. As can be seen from the de nitions,
S}, = SnF(S) =V0uV,.

13.1. Geometric Horseshoe 601

This set is the union of 2 vertical strips each of width 1/3.
Putting the set F(S) in the intersection twice, we have

S5 = [S n F(S)] n [F(S) n F’(s)]
= [S n F(S)] n F (S n F(S))

= SQ F1 F(S(1,).

Using S5 = V0 U V1 and F(H,-) = V,-, we get

s3 = [vo nF(s,§)] u [V1nF(S§)]
=F(H0nS,§) uF(H,nS$).

On H0, the map is a contraction by 1/3 in the horizontal direction and an expansion
by 4 in the vertical direction; therefore, the set F (Ho Pl S5) has 2 vertical strips in
V0, each of width 1/3 of the width of the strips in S6, so of width (1/3) 2 . These two
strips each reach from y = 0 to y = 1. Similarly, on H1, the map is a contraction
by 1/3 in the horizontal direction and an expansion by 4 in the vertical direction,
followed by a rotation of 180 degrees; therefore, the set F (H1 0 S6) has 2 vertical
strips in V1, each of width (1/3)2. Combining, we see that S3 has 2+2 = 22 vertical
strips each of width (1/3)2, going all the way from y = 0 to y = 1. See Figm"e 3.

,D_ -

(D

—_I ON!

1I.~:"?+;Pz:'r¢+"?-.:‘ex»-1": G‘i-es?‘ -i-anass-" 11-'?=~'-;:“~'.*o~.-T

Figure s. The set s5 = 1-‘1(s) n F(S) n s

Continuing by induction, at the next stage, S5‘ is formed by intersecting the
previous set of vertical strips S{,'_1 with H0 and taking their image together with
intersecting S3“ with H1 and taking their image. Each of these sets gives 2"“
vertical strips for a total of 2"“ + 2"“ = 2 - 2"'1 = 2". The width of the strips
in S34 is (1/3)n_1, by induction, so their images are of width 1/3 (1/3)n_l = (1/3)n-
The strips Hj Pl S8“ go all the way from the top to the bottom of H,-, so their

602 13. Invariant Sets

images go all the way from y = O to y = 1. The more formal derivation of the
images is as follows:

S3‘ = [S O F(S)] O [F(S) F1 - - - O F"(S)]

= sf, n F(s3"1)

= [V0 "F($3_1)l U [V1 "F($6‘_‘)l
= F (Ho 0S3“) uF(H1n sg-1).

Taking the limit as n goes to in nity, we see that S8° is a Cantor set of vertical
line segments,

" s3° = 0, >< [0,1].

Turning to the negative iterates, since F"1 maps V0 to Ho and V1 to H1,

s9, = SnF"(S)
= F'1(F(S)) nF-‘(s)

= 1-"1 (St)

= F_1 (V0 U V1)
= Ho U H1.

This set is the union of two horizontal strips, each of height 1/4. By induction, we
see that

s‘l,, =l'[s-"(s) n - - - n F-1(s)] n [F-1(s)ns]

= F_1 (Scln+1)n S91

= [F" (s9,,+,) n H0] u [F-1 (s‘1,,+,) n H1]
= F-1 (s‘1,,+, n vo) u F"(S‘1,,+1 n v,)

is the union of’2~2"'1 = 2" horizontal strips. Since F"‘ is a vertical contraction by
1/4 and a hori.zontal expansi.on by 3, each stri.p 1.11 SQ“ has a he.ight of 1/4 (1/4)"-1 =
(1/4)", and extends all the way from rs = O to zr = 1. Taking the limit as n goes to
in nity, SQ“ is a Cantor set of horizontal line segments:

sic, = [0, 1] >< 0,.

Next, we take the intersections over both forward and backward iterates:

A= S‘f‘°° =S§° S‘l°,, X C2

=C1 X
= C1 X Cg.

The set A is the Cartesian product of two Cantor sets and is invariant by both F

and F'1. It is also the set of all points which stay in S for all iterates. U

13.1. Geometric Horseshoe 603

Itinerary map

We want to describe the dynamics of F on its invariant set A by means of
symbolic dynamics. This description is possible because F is an expansion in the
y-coordinate and a contraction in the :2-coordinate. For a point that stays in S
for all iterates, its y-coordinate is determined by the rough location of its forward
iterates, just as in the case for the tent map with slope 3. The rough location of an
iterate is speci ed by a 0 or a 1, which speci es whether it is in H0 or H1. The map
is a contraction in the :0-coordinate, but its inverse is an expansion in this direction.
Therefore, for a point that stays in S for all iterates, its :i:-coordinate is determined
by the rough location of its backward iterates. The location of a backward iterate
is given by specifying a 0 or a 1 on the negative indices. This bi-in nite sequence
of 0's and 1's speci es the point by means of symbolic dynamics. These symbol
sequences are used to show the existence of periodic points and a point with a dense
orbit in A.

A point in S \ [Ho U H1] leaves S under the rst iteration. Therefore, A C
H0 U H1. Similarly, for a point x to be in A, F-"(x) must be in H0 U H1 for all
j. For each positive or negative j, let s_,- be either 0 or 1, depending on which
horizontal strip contains F3 (x); that is,

s_ 2 0 ifF1'(x) e H0, or

9 1 ifFJ(x)EH1,

Fj(x) E H_,, for all j.

We put the sequences of 0's and 1’s into a single bi-in nite sequence of symbols:

s = - - - s_gs_1.s0s1s2 .

The “decimal point" separates the sj with zero or positive j from those with neg-
ative j . It would be better to put a box around so to indicate that it gives the
ciurent location, but we stick with the notation that is easier to write.

The itinemry map

h(x) = s = -"S_2$_1.80$]82 - --

takes a point x in the invariant set A and assigns the bi-in nite sequence of 0’s and
1's by the rule

F-"(x) G H8, for all j.

The set of all such bi-in nite sequences is called the two-sided full shift on two
symbols and is denoted by E2, where the 2 indicates that there are two symbols.
(Remember that E; is used to denote one-sided sequences of two symbols.) Z2 is
also called the symbol space for the geometric horseshoe. The shift map 0 on Z22 is
de ned by

l5 = 0(8) = 0'(- - - S_28_1.Sg81-92 - ' = - - - 8_2S__18().S18g ' ' -
or

tj = s_,-+1 for all j.

604 13. Invariant Sets

The distance on E2 is de ned as in Section 10.3 for E3",
d(s, t) _- $ 5(%3]".»tj) .

Theorem 13.2. The shift map 0 is continuous on E; and has the following prop-
erties:

a. There is a point s‘ in Z2 such that both the forward orbit Oj'(x‘) and the
backward orbit O; (x") are dense in E2.

b. The shift map a has sensitive dependence on initial conditions.

P-roof. (a) Let s‘ be the bi-in nite sequence with slj = s; and let the positive

symbols be the same as those used in Theorem 10.10 for the one-sided shift space.

We leave the details of the proof that the forward and backward orbits are dense

to the reader. _

(b) The proof is essentially the same as for Theorem 10.15. E1

Orbits from symbol sequences
Starting with a sequence s in E2, we form the intersection

{x:F-"'(x)€H,_, for —oo<j<oo}= n F‘-"(H,,.).
j— oo

The two preceding sets are equal because Fj(x) is in H,, if and only if x is
in F“-l(H,,). For any symbol sequence, we want to see that this intersection is
nonempty and that it is a single point. For any choice of so, s1, .. . s,,_1, let

n—1 .
B-3031"‘-9n—1 = nj=o F J(H81)

={x:F-l(x)EH_,, foi"05j5n-1}.

Notice that so just to the right of the decimal point is the symbol corresponding
to the present location. From the de nition of B__.,,,, it follows that

B_Q=Hg and B_1=H1.

Since F‘1 maps V9 to B_0 = H0 by a contraction in the second coordinate by 1/4,
we have

13.00 = Ho Fl F'1(Ho) = F_1(Vo C Ho) and
B.o1 = Ho Q F-1(H1) = F_1(vo C Hi)-
Each horizontal strip H, is intersected with V0 and its image is a strip of height
equal to 1/4 of the height of H,-), (1/4)’. Similarly,

B_1() = H1 Pl F_1(Hg) = F‘1(V1r'iH0) and

B_11= H1 1'1 F_1(H1) = F_1(V1 Pl

Each of these strips is also of height (1/4)2. Altogether,

592 = 13.00 U 3.01 U 3.10 U 3.11-

13.1. Geometric Horseshoe

where each horizontal strip is labeled with the two symbols that give its present
location and the location of the rst iterate. See Figure 4(b).

B.1o rie£¥='*+i.'-Vat“ __:(:.L.1 a
.l",i.'*;’f1£:§("'_"! ¢' '‘4.>- ~_.,
..
.,.,. 8;.1‘1‘P 4.
;?: -=5"‘l*'.v~ “$6,._. i-Q!x_o,.__._~¢v ‘H18-‘I "'h' bin \i._'.
w-in 1‘; ‘xQ5Qw_.;‘e.-
ll-‘ 1'"! 7
*2.-.1-it. *“'*.* *i’»’.3 ".-=:'13.11
-.‘< -m-4.<-e': -'*v"..i..--11-4" *4“1u 311 u ;~‘i_s. § "\i1*J--h; i
Y
M "';V‘-.-4a-;.e,; -_.

ll" &:_'?*4~5-4'.‘v_.e'.'.‘5r'.-:
- 'M
,.
v- *.w»-.2,‘1¢,;7»~‘/*€l19N;‘-§1i,'1-a-eJ'm-z.*'"-_-i:t 1' - h. 25¢» L‘;.H,,.i-s.1- '>-L
.;isei%:;€Z§':*¥3fi:;“":w1~in.-»',.r-'-' 1"*1s.v0,--u.-‘._s_-".'l“‘yf'"w-t.- ,_i_i'4;:1.\'1_v-_y.§‘5_;‘¢'i:_. B.Ol ‘-_as“litt- ‘.=\.'Z \.L_ F'r.bk :1-|:VQ-QI3-'i.~ _ ‘,,, _- -.
§§%h {-1. 95%;.‘
er;
ti;=>.,g’*.;-'a'+3‘1;.,+-1’“.§"'~*'-.:‘_r 1‘I-_.

_._., ’ -*»;+¥;:z.*g~2'*.“-ir~e-:~,".~"+:"‘£il1~-r-'r 13.00 $5?!.“:@~*1.~'.-|n"‘-1'Ce- la‘-\g-. .~.-:.,.,,, '0-‘L .'»\'.3>."¢‘-X4my 1?‘
4-4- Pf.‘ st»K;.-.., “-4-"51 an ___.‘ Q-‘ IAn.
Irra

Boo. B10. B11. B01.

(a) (b)

Figure 4. (a) Boxes B_,o,, and (b) boxes B,_,,_,

Continuing by induction, we see that

B.,,,,...,,_, = L1,, n F-1(B__,,...,__,) = F-'(v,, n B__.,,...,__,)

is a strip of height

§(lieight of B,.,..._.,__,) = g (5) ‘11-— = (g) 11

The union of all these strips gives all of S0_,,,

Slln = U B.s0...s,,_1 1

sg,s1,...,s,,_|=O,1

and eachhorizontal strip is labeled with the n symbols that give its present location
and the location of the n — 1 iterates.

We now repeat the process for negative‘ indexes. For any choice of s_1, s_2
..., s_,,, let

1n

B_,_,_...,_,_= Q F'j(H,,.)= [']FJ'(H,_,)

j=—n j=1

={x:Fj(x)GH,, for —n§j§—1}.

Notice that

Ba-i- = F(H8-1) = V5-1

606 i g 13. Invariant Sets

is a vertical strip of width 1/3. Continuing by induction, since F maps H; to V, by
a contraction in the rst coordinate by 1/3, it follows that

= F (H,_,) nF=* (H,,_,) n - - - nF" (H,_,,)
= F (H,_,) n F (F (H,_,) n - - - n F"‘1(H,_,,))
= F (H,_,) n F (B,_,_...,_,_)

= F (H,_, Fl B_,_n..._.,_,_)

is a strip of width

-- %(width of B_.,__...,_,_) =§ (§)""‘ = (§)".

Again, the union of all these vertical strips gives all of S5‘; that is,

s9, = U B8_n...8_|.!

s_h““s_"=0J

and each horizontal strip is labeled with the n symbols that give its present location
and the location of the rst n — 1 backward iterates. See Figure 4(a).

Combining the forward and backward iterates shows that

n—l

B3—n3—n+l"'5—l-30' 37:-l = n F-j(Ha,-) = B3-n3l"'5—1-OB-503l"3n-l
j=—n

is a "box" of size (1/3)” by (1/4)”. This box includes its boundary, so it is closed in
the usual mathematical terminology. See Figure 5.

mij‘ -&s$ 300.10 1310.10 1311.10
1300.11 3e?a-
t..B*‘ ‘-' 4111 H B01 10
‘in ge ~L-v '9 301.11
~-noio. w 1310.11
ri- Q7<\0‘l#\Q\¥ l31L11
,,_,_B1.~*‘! .9u21,.1-'1-_- 4.'».',. 'ecu?‘a"_x_noQ‘'7s--Pe4ww1'-
’-'n~v ~-)¢:4->y-"9 . : . <.41-I-.'
-'2 '" =~".v/ 3!‘5. *7.‘ I'l‘~\:-A at-F .-41$

~BE-hith5.‘-gq=F at-.‘N1I .POv'- B1O.0l B1l.01

\"r—‘ 2-L,‘-é gvn 1%‘-'-_|;--is ‘ena- '0u» -_._ k Pvt saris B0001 rat. 1301.01
B0000 sh! P15;.1?
'1-'?.P_t0i-i‘i-f-Hi-»C=-'5eeoEi
'\5I 0:? uqi 301.00
_=‘.} .“ ;_.-;._
" ~ '-isL7' -1 '-'. -i‘ l.
191‘47 B1000
B1100
(a)

Figure 5. Boxes B_,_,_,° and Boxes B,_,,_, 80,,

Taking the limit as n goes to in nity yields

II X

n F—J(H8;) = n BS_n---$_18° 8y|_]

j=—co 3 n-I

13.1. Geometric Horseshoe 607

is nonempty because it is the nested intersection of closed boimded sets (compact
sets). The sets are nested because

B,___,..._.,_,__,,,...,,_ C B,_,_...,_,_,,,...,,,_,.

The size goes to zero in both directions as n goes to in nity, so this is a single point.
We denote this point by k(s):

{k(s)}= 6 F-i(H,,.)= B,_,_...,_,,,° ,,_,

J'=-e=» =3 ..-

This map It goes from the symbol space E2 into the invariant set A in the plane;
it is the inverse of the itinerary map h de ned earlier. If x is a point in A and
s = h(x), then the point x is in J3";_°° F‘J(H,,), so, by uniqueness, x = k;(s), or

koh(x)=x and
hoIc(s) = s.

This shows that both maps are one to one and onto.
The next theorem summarizes the properties of the maps h and It.

Theorem 13.3. Let F be the map for the geometric horseshoe with invariant set
A described previously. Then the following properties are true:

a. For any bi-in nite sequence s in 22, the intersection ;_°°F‘5(H_.,,.) is
a unique point that we denote by Ic(s). With the distance on E2 de ned de ned in
this section, It is a continuous map from the symbol space E; into the invariant
set A in the plane. Also, the itinerary h is continuous from A to Z12.

b. The maps h and It are inverses'of each other, It o h = id and h 0 k = id.
Moreover, (i) both k and h are one to one, (ii) the map h is onto Z2 and k is onto
A, and (iii) the maps h and It are both continuous, so they are homeomorphisms.

c. The map k: is a conjugacy, Fok = koa. So also, hoF=0'0h.

d. A symbol s is periodic if and only if k(s) is periodic for F.

e. There is a. point x‘ in A such that both the forward orbit O§(x‘) and the
backward orbit O; (x') are dense in A.

f. The map F has sensitive dependence on initial conditions when restricted to
A.

Proof. Parts (a) — (d) and (f) follow by the construction of the symbolic dynamics

and the treatment in one dimension. (Compare with Theorems 10.11 and 10.18.)

Parts (e) and (f) follow from Theorem 13.2. U

Stable and Unstable Manifolds

Before leaving this example, we want to discuss the stable and unstable mani-
folds of points. For points in A C S, the derivative is of the form

ii 0

DF(£,y) = ( 03 '

608 13. Invariant Sets

Since these matrices all have the same form and are diagonal, the derivative of an
iterate is also of a simple form:

DFlc¢w¥) : < 0

'

Therefore, all points in a horizontal line segment through a point p = (111,112) in A
are contracted under forward iterates, a.nd

W’ (P,F)3 [-1i6, 5,6] ><{P2} .

This line segment is a piece of the stable manifold even though it is not of a speci ed
radius; so, we just call it a local stable manifold and use the notation

wnm=}§§xmi

Using backwa.rd iterates, the vertical line segments are contracted, so a local unstable
manifold through a point p in A is a vertical line segment-,

w@m=mw}§Q-

We denote the unstable manifold of the invariant set by the union of the unstable
manifolds of points in A as

W"(A, F) = U W"(x,F).

xEA

If we use the trapping region U = S’ U E0 U E1, then

W@=Wamwmt

j=0

where pg is the attracting xed point in E0. (See Sections 11.2 and 13.4 for a
discussion of trapping regions.) Clearly, the local unstable manifolds of points in A
are in the neighborhood U. Since this trapping region is positively invariant, all of
the unstable manifolds are contained in all forward iterates of U and hence are also
contained in the intersection. Other points in U are in the stable manifold of either
a point in A or of po. This set is an attracting set, but is not indecomposable,
since both A and pq are proper invariant subsets with isolating neighborhoods of
the type speci ed in the de nition of indecomposable given in Section 13.4.

13.1.1. Basin Boundarim. Section 12.3.2 discusses a simple example in which
the boundary of the basin of an attracting xed point is the stable manifold of
a saddle xed point. In this section, we give an example in which the bound -FY
between the basin of two attracting xed points is made up of the stable manifold
of a horseshoe, so the basin boundary is much more complicated.

The map considered is like the geometric horseshoe in which a box is maPP°d
across itself three times. Let F be a map that takes the box S across itself three
times as given in Figure 6. We extend the map to regions E1 below S and E2 b°"°
S so that F contracts ea-ch E_,- into itself. The map ca.n be speci ed so that each E1‘

13. 1. Geometric Horseshoe

1/2%.

-2‘

»e-.~:»\.»-_-.,.-"‘W"*i"7-'i"n”"<"9-x<" 1‘'".8?.;~'-.*¢i.7i‘.,*_"“'*~:»-<='L.“;?c.‘-*'!"(-1i»='.‘;,.*“"‘.\*~l-.;"‘

1*?"4"3J‘—-?."~-»\'¢..1‘.5-". k.‘5l"",_.*1Q-.5"9~,

Flgure 6. Horseshoe with three strips

contains a xed point sink pj and W’(p,-; F) 1) E,-. The set of points in S whose
rst iterates are also in S consists of three horizontal substrips, so that

SnF"1(S) = H1 UH; uH3

These horizontal strips are separated from each other by two gaps G1 and G2 which
are labeled so that F(G,-) C E,-. See Figure 7.

‘I.
~a¢-,,,

“ 1; ,;,2i

t Q"- M?’-.',1._:.-__. .1,,_:_; ''-.-,._ie_’--l"'->r1-._-._,;‘-"-4.r-_s.-,,"“"'.1'I'.,“1.--I?4 '?si.n-<a»~§_,~'.»
>-4..,_N>_"-I‘-a.-tt9.-‘Q_V2_ *.'T»~,-A»''*i+v~*\-in‘-\~;'-->\v0»i-.11a.*"In-r,-~4.,, "i-H~-* __'1o,--1: {-W4»+y.e-.t

‘\-.<-:,-._,t_,,-‘

Fig-ure 7. Horizontal strips and gaps

The set of points in S whose rst two iterates are in S is made of 32 horizontal
strips separated by

G1 u G2 u (H, nF-1(c;,)) u (H1 nr-1(G,)) (H, nF-1(c:,))

U (H3 Pl F_1(G2)) U (H3 fl F_1(G1)) CC (H3 F1 F_1(G2))

The intersection of all the backward iterates gives the set of points, all of whose
forward iterates stay in S; that is,

F1'(s) = [0, 1] >< C2,

610 13. Invariant Sets

where C2 is a Cantor set created using three subintervals at each stage. The forward
iterates of S give vertical strips of points whose backward iterates stay in S, or

I

Q Fits) = C1><l0.1l.

j=O

where C1 is a Cantor set created using three subintervals at each stage. Then,

X

A= Q Fj(S)=C1xC2

.1'=-0° If
is the set of points that stay in S for all iterates.

__ The local stable manifolds of points in A are horizontal intervals [0, I] X
we denote the stable manifold of the invariant set by

W’(A,F) = U W"(x, F),

XEA

then 0

W‘(A) nS = [0,1] X c, = Q r1(s).

J'=—<>=>

The basin of pi contains all the points in E,-. Since G,» maps into Ei, it follows
that

E; U G5 C Ws(p5).

This basin also includes all the points that map into E, U G,-, so it includes points
m k_l .

F4‘ (E, u G.) n Q F-1 (H,,.),
.1'=0

where each sj is a 1, 2, or 3. Since the sets

k—l

F"‘ (E1 u G1) n Q F-1'(H_.,,) and

i=0

k—1 _

F-'= (E2 u G2) n Q F-1(H,,)

.1'=0

are on either side of k1

F-" (H2) n Q F-1'(H.,).
J' 1 O

each of the basins W‘(p,-) has W’(A) Q S in its boundary:

bd(W’(P1)lfl bd(W’(P2)) 3 W"(/\) Fl 5 = [0.1] >< C2-

Thus, the stable manifold of A, which contains a Cantor set of CLIIVBS, forms
the boundary between the basins of the two sinks and separates points that tend
toward one sink from points that tend toward the other sink. Therefore, it may be
dii cult to actually land on the horseshoe invariant set or its stable manifold, but
it still is observable as the boundary between the basins of one sink and another.

Exercises 13.1 611

Exercises 13.1

1 Consider the geometric horseshoe map given in the text.
a. Find the coordinates in the plane of the two xed points. Hint: Consider
the two different de nitions of F in H0 and H1:

F,-,(a:,y) = :c,4y) for (a:,y) G H0,

F1(:v,y)= (1—§a:,4—4y) for (a:,y)EH1.

So, one xed point satis es F@(:r,y) = (a:, y), and the other xed point
satis es F1(:1:,y) = (a:,y).
b. Find the points on the period-2 orbit. Hint: One point q on the orbit
satis es F1 o F0(a:,y) = (:i:, y), and the other point is Fq(q).

Consider the geometric horseshoe map given in Section 13.1. Describe the part
of the stable and unstable manifolds W‘(0, F) S’ and W"(0, F) |"|S’, that are
in S’. Note that this is not just the local stable and unstable manifold, but the
whole manifold.

Consider the map G from 1R2 to itself given by the formulas

G(m,y)= (§I,::y) £01-y< %,

(5+5a:,3y—2) foryz 5.

Notice that G is discontinuous along y = 1/2.
a. Let S = [0, 1] x [0,1] and S2, = ;'=mGj(S). Describe the sets S6,

sg, sg, sgs = n;';oo1(s), s‘1,', s92, s<1,,, sic, = n§?=_m <;1'(s), and

$2.. = n;“;_.. (ms)-

b. How does this example differ from the geometric horseshoe?

Let S = [0,1] X [0, 1] be the unit square. Consider the map G from S into IR2
given by the formulas

(%:r,5y) for0$y<%,

G(a:,y) = + %$,5y—2) for 7% gy < §,

(5; +-_.§:r:,5y—4) for § gyg 1.

Notice that G is discontinuous along y = 1/3 and 2/3.

a. Describe the sets sg, = nlzo c:1'(s), sg = n§=,, oJ(s), sg = n;.‘=,, G1'(S),

Sr = n;:.,G1<s>. S9. = n?=-.G1<s>. s2. = n3=-.@1(s>. S9. =

n§=_,, Gi(S), S2,, = n§=_,,° o1'(s), and sag, = n§;_°° Gi(S).

b. How does this example differ from the geometric horseshoe?

Explain more fully why the geometric horseshoe has a point x‘ with both
(x") and O;(x') being dense in A.

Consider the map F for the geometric horseshoe. Let L’ = I/I/lf,c(0) and L“ =
W1';c(0) be the local stable and unstable manifolds of the xed point at the
origin. Draw the images of L“ by F3 and L’ by F—3.

612 13. Invariant Sets

13.2. Symbolic Dynamics

The goal of this section is to identify the essential properties of the manner in which
the rectangles H; are mapped across each other for the geometric horseshoe, so that
we can get symbolic dynamics for any nonlinear maps satisfying them. A collection
of rectangles with these essential properties is called a Ma.rkov partition.

A Markov partition in one dimension is a collection of closed intervals {Ij };-‘=1

such that, if f(int(L)) F1 int(I,-) = ill, then _f(I,-) D I,-. A Markov partition in two

dimensions is a collection of closed sets {Rj };-‘:1 such that, if f (int(R.,-)) int(R,-) =

ll, then there is an interval in one direction in R; that maps across the interval in

the corresponding direction in R,-, and there is an interval in a second direction

in R4 that maps inside the interval in the second direction in R,-. Often these

conditions are given in terms of the derivative of the nonlinear map and a condition

called hyperbolicity that generalizes a saddle xed point. We give the condition in

terms of just the featmes of the function itself. .

When there is one rectangle, we want to nd at least one point that stays in
R for all iterates, and we want to know that there is a xed point in R. For a
sequence of rectangles, we want to nd at least one point that goes through the
sequence of rectangles R,-; if the sequence of rectangles R, is periodic, we want to
nd at least one periodic point that goes through the sequence of rectangles.

In the rst subsection, we use model rectangles for which the rst coordinate is
the in owing direction and the second coordinate is the over owing coordinate. In
the second subsection, we consider images of these model rectangles by homeomor-
phisms. This latter context is the one that is needed when applying the concepts to
nonlinear maps. Also, "the emphasis of the rst subsection is on nding xed points,
while in the second-subsection we use the methods to induce symbolic dynamics
for an invariant set.

13.2.1. Correctly Aligned Rectangle. We start by giving the de nitions in
two dimensions and then consider higher dimensions brie y.
Two dimensions

We use model rectangles for which the rst coordinate is the in owing direction
and the second coordinate is the over owing coordinate. This rst de nition is long
and involves the compatibility of the map and properties of the rectangle chosen.
We write the conditions in the case in which the rst coordinate is in owing and
the second coordinate is over owing. The role of these coordinates can be reversed
as follows from the treatment of Markov partitions.

De nition 13.1. Assume that F is a continuous map from R2 onto itself, with
a continuous inverse (i.e., F is a homeomorphism of 1R2). The image by F Of 8
rectangle

R =la'l1b1l X la'2)b2l

is said to be correctly aligned with itself provided that there is a choice of over owing
and in owing directions such that the four conditions that follow a.re satis ed. See
Figure 8. To indicate that choices of over owing and in owing directions have been
made, we call R an M-rectangle.

13.2. Symbolic Dynamics 613

We separate the boundary bd(R) of R into the sides that are assumed to map
“inside” R,

<9i"(R) = ({@1} X P121521) U ({b1} >< [02-52])
and the top and bottom edges that are assumed to map “outside” R,

¢9°“‘(R)=([@1»b1l>< {‘12})U([¢1,b1l >< {b2})-
The four conditions are as follows:

a. The intersection

F (int (R)) O int (R)

is one connected piece.

b. The image by F of the interior of R does not intersect incoming sides 6i“(R),

F (mt (R)) n 8"'(R) = 0.

c. The image F (z9°"‘ (R)) of the outgoing top and bottom does not intersect the
interior of R,
F (8°“°(R)) O int (R) = 0.

d. For any a: in [a1,b1], the image of each vertical line segment {a:} x [a2,b2]
stretches Erom the bottom to the top of R, separating the two sides; in par-
ticular, there is a subinterval [a',, bg] C [a2,b2] depending on x such that

F({-1'} X l'1'=»bQl)C R. and
F (:c,a;) E <9°“‘(R),

F (I, 1);) E 8°"‘(R),

where the two image points F (:1:,a/I) and F (cc, bg) are in opposite ends of R.
Thus, we assume that F takes the over owing direction of R across R in the
over owing direction.

Fm) \
\ aout

. W

F (<9‘“(R))

g F(9°“‘(R))

ain(R) _~

F({I} X lazlbzll

‘. 'MR’

Fig-ure 8. Aligned image of an M-rectangle R with itself

614 13. Invariant Sets

Remark 13.2. In the de nition, we use the interiors of the rectangles, just as we
used the open intervals in the de nition in one dimension. A more straightforward
assumption to replace (a) would be the stronger assumption that

a1. the intersection F (R) F1 R is one comiected piece.

However, in our consideration of hyperbolic toral automorphisms and the solenoid,
we can use fewer M-rectangles if we use the wea.ker assumption given in the de ni-
tion.

Remark 13.3. If the image of R by F is correctly aligned with itself, it follows
that the image of R by F“1 is correctly aligned with itself.

"a’ . The intersection

F“1 (mt(R)) o mt(R)

is one connected piece. _

b’. The interior of the inverse image F‘1(R) does not intersect the outgoing top

or bottom of R,

int (F-1 (R)) n 8°“‘(R) = 0.

c’. The inverse image F‘1(0i“(R)) of the incoming sides does not intersect the
interior of the previous M-rectangle R,

F“ (6"‘(R)) H int (R) = 0.

d’. For any y in [a2, b2], the image by F'1 of the horizontal line segment [a.,,b1] x
{y} stretches from the right side to the left side of R, separating the top from
the bottom; in particular, there is a subinterval [a;,,b;] of [a1,b1] depending
on y such that 1

F-1([a;,,b',] >< {y}) c R,
F" (a;,,y) e a*"(R), and
F“ (b;,,y) e a‘"(R),

where the two inverse image points are in opposite sides of R.

Now, we assume that the image of an M-rectangle R is correctly aligned with
itself. In this discussion, we make assumption (a1) rather than (a), so we can avoid
using interiors and closures. In the next subsection on Markov partitions, we return
to the more general assumption. The second iterate of R is a vertical strip that
still separates the two sides:

2

Q Fj(R) = Rn F(R) n F2(R) = Rn F (rm F(R)).

j=0

Continuing by induction shows that

n n—l

F1'(R) = Rr'1F(n F-1(a))

j=0 j=0

13.2. Symbolic Dynamics 615

is a vertical strip that separates the two sides, and so does

I X71

Q rim) = Q Q Fj(R).

_1'=0 n=0j=0

This in nite intersection is often just a curve connecting the top of R to the bottom

F<RnF<R>> F F201)

ix 1 F0‘)
lg!
2. -7-
Pl -°~,_,\ nJ=° F (R)

; 1 5:4
1;’-i?'.7'

Figure 9. The intersection ?=° F-'l(R)

F-1(RnF-1(a))

- ihtll” 1,5’; '5 Yr -‘".,J57' _‘.Z _ Z -2

.’§:'9;-‘a_n‘ ‘II: F (R)

‘ F‘1(R)

n2--. F1‘(R)

Flgure 10. The intersection ?=_2 F7 (R)

Taking backward iterates, R O F“ (R) 1' S a-hori'zontal strip that separates the
top from the bottom. The second backward iterate is a horizontal strip that still
separates the top from the bottom, or

Q F5(R) = Rn F-1(a) r'1F‘2(R) = R F“'(Rn F-1(R)).

j=—2

616 13. Invariant Sets

Continuing by induction, we see that

00

Q F1(R)=RnF“ Q F1(R)

j=—n j=-n+1

is a horizontal strip that separates the top from the bottom. The in nite intersec-
tion,

0 co 0

Q F"(R>= O F] F"(R>,

j=—00 3 = Q \-'=—n

also separates the top of R from the bottom and, in the simplest cases, is a curve

connecting the two sides. See Figure 11. .

n’ame) ‘ n?--..Fj<R>

\_I\

“grit!
III- \k\.\

\\\I_=ii§i><
I1II:!IIIT n;-";-..F1(R>

Figure 11. The intersection [']§';_w Fla)

Because ?=_m F7(R) separates the top from the bottom and Q;-'10 F7(R)
separates the two sides, they must intersect and f];_°o F1 (R) is nonempty. See
Figure 11. We summarize this result in the next theorem.

Theorem 13.4. Assume that F is a homeomorphism of R2 and the image of an
M-rectangle R is correctly aligned with itself. Then, the intersection

Fl Fj(R)

j=—oo

is nonempty. Thus, there is at least one point xo such that F7 (xo) is in R for an

J.

13.2. Symbolic Dynamics 617

Fixed points in two dimensions

We would like to know that there is a xed point in the nonempty intersection
guaranteed by Theorem 13.4. There are some general results in topology which
insure that this is true. (These sets satisfy the Lefschetz property.) There are also
some arguments using conditions on the derivatives of F, such as those given in
the proof of the stable manifold theorem, which prove that the intersection is a
single point, and therefore a xed point. These assumptions involve the fact that
the map F is a contraction in the rst coordinate and an expansion in the second
coordinate. See [Rob99] for a detailed presentation of these ideas. We give a
topological argument that we only describe in two dimensions. We consider the
manner in which the displacement vector x - F(x) winds a.ro1md as the point x
varies around the boundary of an M-rectangle.

De nition 13.4. Assume that F is a map on R2 and that R is a region in the
plane such that the following properties hold: (i) The boundary of F, bd(R), is
made up of a single closed curve, and (ii) the map F has no xed points on the
boundary of R, F(x) aé x for x in bd(R). De ne an induced map G from the
boundary bd (R) to the unit circle by

G(x) = x — F(x)

|lX — F(X)|| '

As the point x goes around the boundary in a counterclockwise direction, measure

the increase in the angle of G(x). It has to return to the original angle plus some

integer i multiple of 21r. Call this integer i the index of F for the region R. This

integer can be positive, zero, or negative, depending on whether the displacement

vector winds around in the same direction as the boundary is traversed, has no net

rotation, or winds around in the opposite direction as the boundary is traversed.

' III \\ @<==~>

X2 X1

G(x3) G (X4)

X2.5 G(X4.s)
K45 G(X2.5)

G(X1l

X3 I z X4

G(X1.s)
(*1) (bl

Figure 12. Index of an M-rectangle correctly aligned with itself. (a) The
vectors x — F(x). (b) The vectors G(x).

Consider the image of an IvI-rectangle that is correctly aligned with itself as
given in Figure 12(a). In the gure, the corners of the rectangle are labeled x1,
X2, x3, and x4. A point on the edge between x1 a.nd X2 is labeled with a number

618 13. Invariant Sets

, s Gee)

I \ \ *_ G(X1.a)

X2 X1-8 ‘ l X1‘. \\ \‘. G(X1)
G(x4 )
‘\ \\ I '\
I\
I\ 1| /I 1'
\ 1 /1
| I /1
I’ I’ 1
‘\
\ , I’

I\ ,,, /I
I'

I. \ \

.. X3 X _ waars X4 G(x1_6)
3 1‘
(b)
(*1)

Figure 13. Index of an M-rectangle not-correctly aligned with itself. (a) The
vectors x — F(x). (b) The vectors G(x).

between 1 and 2 _(e.g., x1_5). Part (b) of the gure shows the images of these
points by the index map G. Since the boundary of the rectangle is transversed
in a coimterclockwise direction and the image of the boundary is transversed once
in a clockwise direction, the index is —1. The index is still nonzero even if there
is a flip in some of the coordinates. For a map with a source having eigenvalues
A1 > A2 > 1, lZl'18 index is 1.

It is possible for the image of a rectangle to have the same shape as one that
is correctly aligned, but have index zero and not be correctly aligned. For the map
indicated in Figure /13, the index is zero, since the image of G does not go all the
way around the circle. The map need not have a xed point in R. The map for
Figure 13(a) is taken so that the image of R F(R) is the shaded area. Since the
image of R O F(R) is disjoint from itself, there can be no xed points. Also,

R n F(R) n F2(R) = 0,

so no orbits stay in the rectangle for all iterates. Such a map is

1

F(I:y) = (‘£31,837 + r

with M-rectangle [-1, 1] x [-1, 1]. Its only xed point is (*3/5,6/5), which is outside

[-1,1]><[-1,1].

The next result holds for any M-rectangle whose image is correctly aligned with
itself.

Theorem 13.5. Let F by a. homeomorphism of R2 that has an M-rectangle R
whose image is correctly aligned with itself. Then, either (i) F has a xed point 0"
the boundary of R {and the indea: is not de ned), or (ii) the index of R for F is
nonzero.

The next theorem relates the index and a xed point in R.

Theorem 13.6. Assume that F is a map of IR2 with a nonzero indea: for a 1'¢9‘i°"'
R whose boundary is a single closed curve. Then, F has a xed point in R-

13.2. Symbolic Dynamics 619

In particular, i_f the image of an M-rectangle R is correctly aligned with itself
by F, then F has a xed point in R..

Proof. We assume that there is no xed point in R and get a contradiction.

For an M-rectangle, we can shrink the boundary of R to a point. Assume

that 'Ys(t) is a one-pa.ra.meter family of closed curves in R such that 'y0(t) goes

around the boundary of R in a counterclockwise direction as t varies from 0 to 21r,

'y,(0) = 'y,(21r) for all 0 § s 5 1, and 71 (t) is a single point in R, independent of t.

If there are no xed points in R, then for each value of s with 0 3 s $ 1, we can

de ne the map

and Gd‘) u'1m8 (i)) —- FF((%v.((ti))))ll

0, (t) = angle(G,(t)).

By the de nition of the index, 0Q(21r) — 00(0) = k21r, where lc is the index. As s
varies from 0 to 1, 0%2 must remain an integer, so it must remain equal to k =,é O.

However, 'y1 (t) does not move, so G1(t) is a constant and 01(21r) — 01(0) = 0. This

contradiction proves that F must have a xed point in R. Cl

Higher dimensions

We now turn to stating the de nition of a correctly aligned M-rectangle in
higher dimensions. Given a homeomorphism, the choice of an M-rectangle depends
on the number of over owing and in owing directions. We also use round balls
about the origin in the over owing and in owing subspaces, but these could be
replaced by rectangular regions. (Also, in their use in Markov partitions, we use
homeomorphic images of this standard M-rectangle.) For an integer lc and r > 0,
let

B'°(0,r)={xElR":||xl|5r} and

bd (B'=(o,¢)) = {x e 112*; |lx||= T}

be the closed ball and its boundary in IR".

De nition 13.5. We assume that F is a_homeomorphism map from IR" onto itself.

The image by F of an M-rectangle, R = B"1(0, 1'1) x B"? (0, 1'2), where n1 +11; = n,

is correctly aligned with itself, provided that the four conditions listed momentarily

are satis ed. We separate the boimdary bd(R) of R into the part that maps

“inside” R, _ _

3"‘(R) = (BM (0,T1)) X B": (0,Tg),

and the part that maps “outside” R,

8°“‘(R) = BM (0,T]) X (BM (0,1‘2)) .

The four conditions are the following:

a. The intersection F (int(R)) O int(R)
is one connected piece.

620 13. Invariant Sets

b. The image by F of the interior of R does not intersect the incoming part of
the boundary of R,

F (int (R)) O 6i“(R) = 9).

c. The image of the outgoing part of the boundary, F (6°“"(R)), does not inter-
sect the interior of the next M-rectangle R,
F (6°“‘(R)) n int (R) = 0.

d. For any x in Rm (0,1'1), the image of the vertical disk {x} XBM (0, 1'2) stretches
across R in all the over owing directions.

See Figure 8 for the two-dimensional gure.

There is a variety of ways to make more precise the condition that the image of
a vertical disk is stretched across the next M-rectangle in the over owing directions.
One way is in terms of assumptions on the derivatives. Another way is in terms of
“homology theory” from algebraic topology. The homology generalizes the idea of
index that we used in two dimensions. A third way was recently developed by M.
Gidea and P. Zgliczynski. See [Gid02] or [Gid03]. They assume that the nonlinear
map can be deformed or changed into one in a standard form.

d1. There is a continuous map G from [0, 1] X R into IR” that satis es the following
three conditions:
(i) G(0,z) = F(z).
(ii) For every 0 3 s 3 1, the image of R by G(s, -) satis es the preceding
conditions (a) to (c), namely,

"G (s, int(R)) F1 int(R) is connected,

mt (G(s, R)) n a‘"(R) = 0, and

G(s,8°“"(R)) O int (R) = (D.

(iii) The map for s = 1 is an af ne map in the over owing direction; that is,

G(1»(X1.X2)) = 3 + (0, A32)»
withA an n2 >< n2 matrix and sign (det(A)) gé 0.

Because the deformation does not pull the image o ' the R or through the sides,
the map G(0, -) crosses the M-rectangle in the same way that G(1, -) does.

From these assumptions, it follows that the inverse F“ satis es the conditions,
with the roles of the over owing and in owing directions reversed.

a’. The intersection F-1 (int(R)) n int(R)

is one connected piece.

b’. The interior of the image F_1(R) does not intersect the outgoing part of the
boundary of R,

mt (F"‘ (R)) n 6°"‘(R) = w.

c’. The image of the incoming part of the boundary by F’1 does not interS9¢l'-
the interior of the M-rectangle R,

F-1 (a‘"(R)) n int (R) = 0.

13.2. Symbolic Dynamics 621

d’. For any y in B” (0, T2), the image by F" of a horizontal disk I_3"1(0,1-1) X {y}
stretches across R in all the in owing directions.

With these changes in the de nitions, Theorems 13.4 and 13.6 are still true.
To prove these results in higher dimensions, we need to either use more advanced
ideas from algebraic topology, or assume the map is a contraction and expansion
in the different directions. (cf. [Rob99].)

13.2.2. Markov Partition. For a nonlinear map, the boxes that are correctly
aligned are usually not be as simple as those given in the preceding section but are
images of actual rectangles.

De nition 13.6. Assume that F is a homeomorphism of 1R". A nite collection
of closed sets { Rj }J'-'=, in IR" is called a Markov partition for F, or is said to have
the Markov property, provided that the following conditions are satis ed:

(1) Assume that n1 and n2 _are positive integers with n1 + Ttg = n, 1?" (0, 1) is a
closed ball in lR"‘ , and B"? (0, 1) is a closed ball in lR"’. The M-rectangle

B = B"1(0, 1) >< B"=(0,1),

is a model for the Markov rectangles. In two dimensions with n1 = n; = 1,
B is a square, B = [-1, 1] X [—1,1].

(2) For each R,-, there is a homeomorphism ¢,- from the standard box B onto
R,-). The images by ¢j of the different parts of the boundary of B are given
similar labels for R,-:

5‘"(Rj) = ¢j(<9i“(B)) and 3°“'(Rj) = ¢1(3°‘"(B))-

(3) The interiors of the R, are disjoint,‘ int(R,-) |’1int(R,-) = ll] for 11 75 j.

(4) If the image by of the interior of R; intersects the interior of R,-, F(int(R;)) O
int(R,-) 96 0, then the image of R4 by F is correctly aligned with Rj in the
sense that the image of B by 4);‘ o F o 41¢ is correctly aligned with B, and the
following four conditions are satis ed:
a. The intersection F (int (R,)) F‘! int (R,-) is one connected piece.
b. The image of the interior of R; does not intersect the incoming part of
the boundary 8"‘(R_,-),

F (am; (R.,)) n a'"(R,-) = 0.

c. The image F (6 °“'(R;)) of the outgoing boimdary does not intersect the
interior of the next M-rectangle R,-,

F (6°"‘(R,)) n int (R,-) = 0.

d. For any x in E"1(0,1), the image of ¢,~({ x><1§"° (0, 1) by F stretches
across R5 in all the over owing directions.

The sets R; in the Markov partition are called Markov rectangles.

Remark 13.7. We allow F(int(R;)) F1 int(R)) to be empty for some -i and j. We
allow the image of R; by F to cross R_,- at most once, because in the de nition of
being correctly aligned we assume that F(int(R;)) F1 int(R)-) is connected.

622 13. Invariant Sets

Remark 13.8. Usually, each R, is a diifeomorphic image and frequently an affine
image of B. In the latter case, there is a matrix Aj and a constant point pj such

that Rj = Pj + Aj

Remark 13.9. The de nition of the Markov property is compatible with the one
given in one dimension. However, a one-dimensional map has only an over owing
direction and no in owing direction.

Remark 13.10. In choosing a Markov partition for a map, the rectangles must
be chosen with their over owing and in owing directions so that the map takes
the over owing direction in one rectangle across the over owing direction of any
rectangle it intersects.

Remark 13.11. The Markov property has been used in a probabilistic sense for
many years. Ya. Sinai and R. Bowen carried these ideas over to dynamical systems
in the late 1960s and early 1970s.

De nition 13.12. For a collection of boxes with the Markov property for F, we

de ne an associated transition graph £4 by letting the vertices be the labels of the

boxes in the partition, { 1, . . . ,J }, and putting a directed edge from i to j if a.nd

only if F(in‘t(R.;))f'linl3(R-j)1'é 0

(i.e., the image of R; by F has a nonempty crossing with R_,- that is correctly
aligned).

We also form the transition ma.tri.1: T = (t,-J-) by

t_.__ _ 0 if F(int(R;)) O int(R,-) = ll),

‘J ._ 1 if F(int(Ri)) n int(RJ-) ¢ 0.

The nonzero entries give the allowable transitions between symbols.

De nition 13.13. Fix a transition graph if with J vertices or a J >< J transition
matrix T. The full two-sided shift space on J symbols is the set SJ of all bi-in nite
sequences of the symbols 1, . . . , J. A symbol sequence s in E; is allowable for if
or T provided that there is an edge in g from s,- to s,-+1 for every i (i.e., t,,_,,+, = 1
in the transition matrix). Let Eg be the set of all allowable symbol sequences.
(It is also written as ET if it is speci ed by the transition matrix T.) The space
Eg is called a subshift of nite type because there is a nite set of rules given by
the transition graph that indicate which symbols are allowed to follow which other
symbols.

Notice, that for the horseshoe, all the entries of the transition matrix are 1 and
there are edges connecting all the vertices of the transition graph.

Theorem 13.7. Assume that $4 is an irreducible transition graph and some vertea:
has more than one edge going out. Let E9 be the associated two-sided subshift of
nite type. Then the shift map is topologically transitive on E9 and has sensitive
dependence on initial conditions when restricted to Eg.

The proof is essentially the same as that of Theorem 10.21 for the one-sided
subshift.

13.2. Symbolic Dynamics 623

Theorem 13.8. Assume that F is a homeomorphism on R". Assume that { R4 };/=1
is a Markov partition for F with transition graph £4 and associated subshift of nite
type E9.

a. Write S° for U;-L1 int (R4), and

co It

A: Qc1( Q F-1'(s°)).

I=—0 j_—k

Then, there is an itinerary function h from A to Eg (i.e., if x is in A, then the
itinerary h(x) is an allowable sequence). Note that, if the point F1 (x) is on the
boundary of two or more rectangles R,-, then it is necessary to make a choice for
Sj.

b. Assume that s in Eg is an allowable bi-in nite symbol sequence. Then,

oo I:

Q cl Q F-1' (int (R,,)) ¢ 0

k=0 j=—k

and there exists a point x, such that h(xs) = s (i.e., F-7(xs) is in R,’ for all j).
Thus, the itinerary function h is onto all of Eg.

c. Assume that s in E9 is an allowable bi-in nite symbol sequence that has
period-p (i.e., s,-.41, = s_.,- for all j and there is no shorter period than p). Then,
there exists a point x, such that h(x,) = s and F7’(xs) = x,. Consequently, the
period of xs divides p. If the point X, is not on the boundary of R”, then the period
is exactly p.

-d. Assume that the transition graph is irreducible. Further, assume that

Q cl ( Q F~1' (int(R,,.)))

k=0 j=—k

is a single point for each allowable sequence s. Then, the map F is topologically
transitive on A, and F restricted to A has sensitive dependence on initial conditions.

Remark 13.14. The intersections in the theorem can be simpli ed if we add the
following additional assumption to the list in De nition 13.6:

5. If
F(int(R;)) n int(R_,~) ;é 0,

then
F (R.,) n R, = cl (F (int (R1-)) n int (R,-)).

This condition says that there are no extra intersections on the ends that are not
related to the images crossing. With assumption (5) added, we can replace the
intersections in the theorem as follows: we replace

co I: 00

Q cl ( Q F-1'(s°)) with Q F” (s)

1==o j=—k j=—oo

B24 13. Invariant Sets

and

Qcl F-1(im(R,,))) with Q F-1' (11.1).

k: '=— ‘ '=—

where S = U;-1:1 R4.

Remark 13.15. In part (d) of the theorem, we need to assume that all the inter-
sections

Q cl ( Q F-1‘ (am (R,,)))

are points in order to get sensitive dependence, because the points of this set stay

close together for all iterates. ~

We apply the preceding theorem to the Hénon map with large value of a.

F<:>;<~:~*>,Example 13.16 (Hénon map). We apply these ideas to the Hénon map

for b = 0.3 and a = 5. We consider the square
S = [—-3,3] >< [—3,3].

The images oftheFFco<r<neafr:s=>a<r:e<5“=:::e~99>>==<<=e:s:>>. and

The image of a vertical line segment in S is a horizontal line segment, and the

F(¢) F(d) b . .3

__ __ _-r F(a:,y0) F(0’y)

1 _we
I

F(b) F(a) a <1

Figure 14. Image of the square S by the Hénon map

13.2. Symbolic Dynamics 625

image of a horizontal line segment in S is a parabola that cuts across S twice:

F (1:0) : (5 — 0.3y —
1'! fl-'0 '

F (rt) = (5—0.3y0—:v)_
110 I

See Figure 14.

The intersection S H F‘1(S) is the union of two strips R0 and R1 that reach
£rom the bottom to the top of S. The image of R0 U R1 equals S Q F(S), which is
the union of two strips that reach between the two sides of S. These later strips
play the roles of V0 and V1 for the horseshoe, while R4; and R1 play the roles of
H0 and H1. These strips R0 and R1 are the M-rectangles in the Markov partition.
These M-rectangles can be determined by the inverse map

<@> = (—— 5ii
y 0.3

The sides of the strips that are mapped to as = 3:3 are the parabolas

1/
F_1= (2—y2) and

0.3

< > = 8» 2 1_3 U
,, __v_
0.3

These sides are the “outgoing” part of the botmdary. The part of the boundary
on y = :i:3 is the “incoming” part of the boundary, which gets mapped to the
horizontal part of the boundaries of F(S) Q S.

The pair {R0, R1 } forms a Markov partition. The transition matrix has all
1’s. For each bi-in nite string of 0's and 1’s, s, the intersection

@

n F—J (R31)
j=—oo

is nonempty. It is not obvious that it is a single point, but this can be shown by
further detailed analysis, which is beyond the scope of this book. See [Rob99].
However, even without knowing that the intersection is a point, the preceding
results show that periodic symbol sequences correspond to periodic points. We can
also determine nonperiodic points by using symbol sequences that are not periodic.

13.2.3. Markov Partitions for Toral Automorphisms. Toral automorphisms
were among the rst maps for which Markov partitions were constructed. The
Markov partition connects the dynamics of the hyperbolic toral automorphism FA
on 'lI‘2 with a subshift of nite type for a transition matrix T. R. Adler and B.
Weiss (1970) showed that, if the original matrix A inducing the hyperbolic toral
automorphism has all positive entries, then it is possible to nd a Markov partition
with two M-rectangles having a transition matrix T that is the same as A. (See
Remark 13.20 for comments about transition matrices with entries bigger than one.)

626 13. Invariant Sets

Several other people, including M. Snavely [Sna91] and E. Rykken [Ryk98], have
given further details on constructing such a Markov partition. In this section, we
consider only the case, in which the construction is easiest to give.

The Markov partition can be constructed in a more speci c manner than using
the general de nitions we have given. Some of the properties are closer to Bowen's
original de nition, but the use of the covering map from the plane to the torus is
particular to toral automorphisms.

For the toral automorphisms, we take a Markov partition Q = },-"Q, to

cover the whole torus:

771

-- 1r’=UR,.

i=1 '

Each M-rectangle can be given as R4 = rr(R;), where is a_parallelogram in R2
and the projection rr is a homeomorphism on the interior of R,-. We represent the
projection of the interior of R, by Rf and call it the interior; that is,

R? = r (1=lt(Ri)) -

We require that

R;-’ Rf=0 ifjgéi.

We call

5 (Ra) = R4 \ R?

the boundary of the rectangle in T2 even though it might be larger than the bound-
ary in T2 in the usual sense of the term. In the Example 13.17, Rf is larger than
the interior in the torus int(R1) since it touches itself along 6 (R1); also, 6 (R1) is
larger than the usual topological boundary of R1 in the torus for the same reason.

The boundary of this parallelogram is made up of pieces of stable and unstable
manifolds of the induced map on IR2. In fact, for our examples, the boundaries can
be made up of line segments L such that rr(L) is contained in W’ (0) U W“(0), and

3"‘ (Ra) C W"(1f(0)) and
3°“ (Rt) C W’(1r(°))-

We allow two points on the boundary of R, to be equivalent modulo 1 (i.e., the map
1r can take two points on the boundary to the same point in the torus). The map
rr can be many to one on the boundary of As a consequence, an M-rectangle
R, can touch itself on the boundary.

For the covering property, we merely require that F (Rf) QR; is one connected
piece if this intersection is not empty. (Thus, we allow other contacts on the image
of the boundary.)

In the plane, the stable and unstable manifolds are lines that do not return
to a rectangle, so the intersection W°(i) Q Rj is a line segment for a = u or s.
Therefore, for 2 in R,-, z = rr(2) in R,-, and 0' = u or s, we de ne

W°(z, Rj) = rr Q R,-).

13.2. Symbolic Dynamics 627

These stable and unstable manifolds in the rectangles then satisfy the following
property: If z G Rf and F(z) 6 R;-’, then

D Wu(F(z)IRj) and

F(W°(z.Ri)) C W‘(F(Z).R,~)-

(Bowen had essentially this property as part of his de nition of an M-rectangle.)

Examples

Example 13.17. We nd a Markov partition for the toral automorphism F = FA

induced by the matrix A = . The eigenvalues are Au = % (1+ \/5) > 1 with

eigenvector v“ = (2, -1 + \/§)T and —1 < /\,, = é (1 — \/'5) < 0 with eigenvector
vs = (2, -1 — \/5)-r. The eigenvector v“ has positive slope and v‘ has negative
slope.

The lattice points in JR2 are all the points with both coordinates equal an integer;
these are the points p with 1r(p) = -rr(0).

To form the rectangles R, in R2, ta.ke the part of the unstable manifold that
goes above and to the right of the lattice point as shown in Figure 15. Take the part
of the stable manifold from the lattice point downward to the point 5, where it hits
the unstable line segment drawn above. Let [0, 5], be the line segment in the stable
manifold from 0 to 5, and in general, let [:Tc, 5'], be the line segment in the stable
manifold from 1': to 5'. Also, extend the stable manifold upward from a lattice point
to the point B, where it hits the unstable line segment drawn above. See Figure
15. These can be chosen so that F(a) = b. Let E = F(b) and E’ = 6 + (1,1)
so 1r(é) = 1r(6'). Finally, extend the unstable manifold to the point 6’, where it
hits the line segment [§,b]_, in the stable manifold. These line segments, [5, B], in
W’(0) and [0, E’],, in W“(0) (and their translates in R2), de ne two rectangles R1
and R2 in R2, and hence, R1 and R; in T2. See Figure 15.

To determine the images of the rectangles by F, we rst consider the images of
the points 5, B, and E: we have F(a) = b, F(b) = E, and e = F(f:) E [0,b]$. See
Figure 15. We have labeled multiple points that project down on same point in the
torus with the same letter, distinguishing them with primes. With these images, it
follows that

F(R1) crosses R1 and R2 in T2 and

F(R2) crosses R1 in T2.

See Figure 16. The incoming part of the boundary of each R,-, 6i"(R,-), is made
up of pieces of unstable manifolds, and the outgoing part of the boundary of each
R,-, 6°"‘(Rj), is made up of pieces of stable manifolds. The pair of rectangles
{R1,R¢} have the properties of a Markov partition for F: (1) each rectangle is a
parallelogram, (2) the collection of rectangles covers T2 and the interiors of R1 and
R2 are disjoint, and (3) if F(R§) OR; 95 ll, then the following hold. (a) F(R§-’) OR;
is one connected piece. (b) F(R.§’) does not intersect 8"‘ (Ri). (c) F(R,-) reaches all
the way across R, in the imstable direction so F (6°“" (R,-)) does not intersect Rf.
(d) For any z in Ri, F(W"(z,R¢) stretches across R, in the unstable direction.

13. Invariant Sets

QO
El

_ =1’
é = F("=) t e" = F(c ‘
*0
c-_- F' (b_ ) ~ Em

5
Figure 15. Rectangles for Example 13.17

" 7'
, // '.

r . .
//I

,1

F(R1) I."

F R2 \., \\1‘_ ~'~_>’ i
\._\-.\~.‘-..\’ \
\ ‘_\-/1x-\'\-\\- \\-~.'
..‘\a\-\ \

Figure 16. Images of rectangles for Example 13.17

The transition matrix is

T