Absolutely Continuous Invariant Measures of Piecewise Linear Lorenz Maps

0.26 MB·English

by  Yi Ming Ding

ABSOLUTELY CONTINUOUS INVARIANT MEASURES OF
PIECEWISE LINEAR LORENZ MAPS
YIMINGDING,AIHUAFANANDJINGHUYU
0
1
Abstract. Consider piecewise linear Lorenz maps on [0, 1] of the following
0
form
2 ax+1−ac x∈[0,c)
n fa,b,c(x)=(cid:26) b(x−c) x∈(c,1].
a We prove that fa,b,c admits an absolutely continuous invariant probability
J measure(acim)µwithrespecttotheLebesguemeasureifandonlyiffa,b,c(0)≤
8 fa,b,c(1), i.e. ac+(1−c)b≥1. Theacim isunique and ergodicunless fa,b,c
1 isconjugatetoarationalrotation. Theequivalencebetweentheacimandthe
Lebesguemeasureisalsofullyinvestigated viatherenormalizationtheory.

]
S
D
.
h 1. Introduction
t
a
m
Lorenz maps are one-dimensional maps with a single singularity, which arise
[ as Poincar´e return maps for flows on branched manifolds that model the strange

attractors of Lorenz systems. A Lorenz map on the interval I := [0,1] is a map
1
v f :I I such that for some critical point c (0,1) we have
→ ∈
4 (i) f is continuous and strictly increasing on [0,c) and on (c,1];
1 (ii) lim f(x)=1, lim f(x)=0.
x↑c x↓c
0
A Lorenz map f is said to be piecewise linear if it is linear on both intervals [0,c)
3
and (c,1]. Such a map is of the form
.
1
0 ax+1 ac x [0,c)
0 (1. 1) fa,b,c(x)=(cid:26) b(x c)− x∈(c,1],
1 − ∈
: where a>0,b>0, 0<c<1, ac 1 and b(1 c) 1.
v ≤ − ≤
i Let β >1. The map Tβ(x)=βx ( mod 1) is the well known β-shift related to
X
β-expansion ([25]). Assume 0 α<1. The transformation T defined by
β,α
r ≤
a T (x)=βx+α ( mod 1)
β,α
is a natural generalization of β-shift. There are many works done on T (see
β,α
[11, 13, 16, 21, 22, 23, 24, 26]). When 1<β 2, T is a piecewise linear Lorenz
β,α
≤
map. In fact, T = f with c = (1 α)/β. Recently, Dajani et al [5] studied
β,α β,β,c
−
another variation S of β-shift. For 0<α<1 and 1<β <2,
β,α
βx x [0, 1/β)
(1. 2) Sβ,α(x)=(cid:26) α(x 1/β) x∈(1/β, 1],
− ∈
which is the piecewise linear Lorenz map f .
β,α,1/β
Lorenz maps arise as return maps to a cross-section of a semi-flow on a two
dimensional branched manifold (cf. [1], [15], [27]). The flow lines starting from c
never return to I. So usually the map is considerednot defined at c (cf. [12]). But
it is also convenient to regard c as two points c+ and c , the right and left of c,
−
1
2 AcimsofpiecewiselinearLorenzmaps
so that the Lorenz map is a continuous map defined on the disconnected compact
space [0, c ] [c+, 1]. Differentdynamicalaspects of Lorenzmaps are studied in
−
theliteraturesSsuchasrotationinterval,asymptoticperiodicity,topologicalentropy
and renormalization etc (see [2], [7], [8], [12]). In this paper we shall study the
absolutely continuous invariant probability measures (acim for short) of piecewise
linearLorenzmaps. Theexistenceofacimandtheequivalenceofacimwithrespect
to the Lebesgue measure are studied.
The Lebesque measure is clearly quasi invariant under f . Let P be the
a,b,c a,b,c
associated Perron-Frobenius operator and let
n−1
1
A (h)= Pi h, h L1([0,1]).
n n a,b,c ∈
Xi=0
Our results are stated in the following two theorems.
Theorem A. The piecewise linear Lorenz map f admits an absolutely contin-
a,b,c
uous invariant probability measure µ with respect to the Lebesgue measure if and
only if f (0) f (1), i.e. ac+b(1 c) 1. More precisely,
a,b,c a,b,c
≤ − ≥
(1) If ac+b(1 c) = 1 and loga/logb is rational, then there exists positive
−
integer n such that fn (x) = x for all x [0,1]. Consequently, for each
a,b,c ∈
density g on [0, 1], A (g) is the density of an invariant measure of f .
n a,b,c
(2) If ac+b(1 c) = 1 and loga/logb is irrational, then the acim is unique
−
and its density is bounded from below and from above by the two constants
a 4 a −4
and .
b b
(cid:16) (cid:17) (cid:16) (cid:17)
(3) If ac+b(1 c) > 1, then the acim is unique and its density is of bounded
−
variation.
(4) If ac+b(1 c)<1, then f admits no acim.
a,b,c
−
It is well known to Lasota and Yorke ([19]) that a strongly expanding interval
map f (i.e. f′(x) > λ > 1 except finite points) admits an acim with respect to
| |
the Lebesgue measure. It is also known that a piecewise linear Lorenz map with a
fixed point also admits an acim with respect to the Lebesgue measure (cf. [5, 6]).
These results don’t apply to the Lorenz maps defined by (1. 1) which, in general,
are not strongly expanding and admit no fixed point.
Suppose thatf admits a unique acimwith respectto the Lebesgue measure.
a,b,c
Iff isahomeomorphism(i.e.,ac+b(1 c)=1)withirrationalrotationnumber,
a,b,c
−
we shall see from the proof of Theorem A that the acim of f is equivalent to
a,b,c
the Lebesgue measure. If ac+b(1 c) > 1, the acim is not necessarily equivalent
−
to the Lebesgue measure, even if f is strongly expanding (i.e.a>1 and b>1).
a,b,c
For example,Parry[24]provedthat the acimof symmetric piecewise linear Lorenz
map f is not equivalent to the Lebesgue measure if and only if 1 < a < √2.
a,a,1/2
We point out that the support of the acim of T was studied in [11, 16].
β,α
Assume that ac+ b(1 c) > 1. As we shall see in Lemma 4.1, the acim of
−
f is equivalent to the Lebesgue measure if and only if f is transitive, i.e.
a,b,c a,b,c
fn (U) is dense in I for each non-empty open set U I. In general, the
n≥0 a,b,c ⊂
tSransitivityofaLorenzmapisnoteasytocheck. Palmer[21]studiedthetransitivity
of T by using so-called primary cycle (see also [11]). Alves et al introduced a
β,α
topologicalinvariant to study the transitivity of T ([3]). The conditions of both
β,α
primary cycle and the topologicalinvariantof Alves et al are difficult to check too.
We will provide a rather simple criterionof the transitivity for the piecewise linear
Lorenz maps f with ac+b(1 c)>1.
a,b,c
−
Yi-MingDingAi-HuaFanandJing-HuYu 3
Let us describe our criterion. Assume ac+b(1 c) > 1. Then f admits
a,b,c
−
periodic points, because it admits positive topological entropy ([2]). Let κ be the
minimal period of the periodic points of f . Assume 2 κ < . Then f
a,b,c a,b,c
≤ ∞
admits a unique κ periodic orbit. Let P and P be adjacent κ periodic points
L R
− −
such that c [P , P ]. It can be proved that fκ is linear on [P , c) and on
∈ L R a,b,c L
(c, P ] (Lemma 2.6). Write
R
A:=fκ (c+), B :=fκ (c ).
a,b,c a,b,c −
That κ = 2 means f has no fixed point, f (0) = A < B = f (1) and
a,b,c a,b,c a,b,c
c [A, B]. When κ=2, let
∈
c A B c
M :=min − , − .
(cid:26)B c c A(cid:27)
− −
Theorem B. Suppose ac+b(1 c) > 1. If κ = 1, then the acim of f is
a,b,c
−
equivalent to the Lebesgue measure. If 2 κ < , then the acim of f is
a,b,c
≤ ∞
equivalent to the Lebesgue measure if and only if
[A, B] [P , P ]= or [A, B]=[P , P ].
L R L R
\ 6 ∅
In particular, when κ=2, the acim of f is equivalent to the Lebesgue measure
a,b,c
if and only if
ab>1+M if M <1
(cid:26) ab 2 if M =1.
≥
See Figure 1 for a piecewise linear map whose acim is not equivalent to the
Lebesgue measure.
Parry [24] proved that the acim of f (1 < a 2) is not equivalent to the
a,a,1/2
≤
Lebesgue measure if and only if 1 < a < √2. This may be obtained as a special
case of Theorem B.
We shall collect some basic useful facts in 2, including rotation number, Lya-
§
punov exponent, Frobenius-Perron operator and renormalization. Theorem A is
provedin 3 andTheoremB in 4. Densities ofsome piecewise linear Lorenzmaps
§ §
will be presented in 5.
§
2. Preliminaries
In this section, we present some facts concerning the rotation number, Lya-
punov exponent, Frobenius-Perron operator and the renormalization of Lorenz
maps, which will be useful later.
2.1. Rotation number and Lyapunov exponent.
We denote by e : R S1 = z C : z = 1 the natural covering map
→ { ∈ | | }
e(x) = exp(2πix). Let f be a Lorenz map, not necessarily linear. There exists a
map F : R R such that e F = f e and F(x+1) = F(x)+1. F is called
→ ◦ ◦
a degree one lifting of f. Furthermore, if F(0) = f(0), then there exists a unique
such lifting (cf. [2]).
The rotation number of f at x is defined by
Fn(x) x
ρ(x)=limsup − .
n→∞ n
4 AcimsofpiecewiselinearLorenzmaps
f(1)
f(0)
0 PL C PR 1
Figure 1. A piecewise linear Lorenz map with κ = 2, and
[f2(c+), f2(c )] = [f(0), f(1)] (P , P ), whose acim is not
L R
− ⊆
equivalent to the Lebesgue measure.
Itisknownthatthesetofallrotationnumbersρ(x)off isanintervalandthatthis
interval is reduced to a singleton when f(0)=f(1) ([17]). The rotation number is
tightly relate to the number of returns of x into the interval (c, 1], defined by
(2. 1) m (x)=# 0 i<n:fi(x) (c, 1]
n
≤ ∈
(cid:8) (cid:9)
Lemma 2.1. Let f be a Lorenz map (not necessarily piecewise linear). Let F be
the unique degree one lifting map of f such that F(0)=f(0). Then for any n 1
≥
and any x [0, 1] we have
∈
Fn(x)=m (x)+fn(x).
n
Proof. The proof of this Lemma would be found in the literatures, we give a proof
for completeness. We prove it by induction. First note that
F(x) x [0,c)
f(x)= ∈
(cid:26) F(x) 1 x (c,1]
− ∈
which implies the desired equality for n=1:
F(x)=1 (x)+f(x).
(c, 1]
Supposenowthattheequalityistrueforanarbitraryn. SinceF(x+k)=F(x)+k
for all positive integers k, by the hypothesis of induction we have
n−1
Fn+1(x)=F(Fn(x))= 1 (fi(x))+F(fn(x)).
(c, 1]
Xi=0
Yi-MingDingAi-HuaFanandJing-HuYu 5
According to what we have proved for n = 1, we get F(fn(x)) = 1 (fn(x))+
[c,1)
fn+1(x) so that
n
Fn+1(x)= 1 (fi(x))+fn+1(x)=m (x)+fn+1(x).
(c, 1] n+1
Xi=0
(cid:3)
Write
C = f−n(c).
f
n[≥0
When f =f we write C :=C . The Lyapunov exponent of f at x / C
a,b,c a,b,c fa,b,c ∈ f
is defined by
1
λ(f,x)=limsup log(fn)′(x).
n→∞ n
For piecewise linear Lorenz map f ,
a,b,c
dfn (x)
(2. 2) a,b,c =an−mn(x)bmn(x), x C .
a,b,c
dx ∀ ∈
For any linear Lorenz map such that f (0) = f (1), its rotation number
a,b,c a,b,c
and its Lyapunov exponent are determined in the following way.
Lemma 2.2. Let f be a piecewise linear Lorenz map such that f (0) =
a,b,c a,b,c
f (1). Let ρ be the rotation number of f . Then we have
a,b,c a,b,c
Fn(x) x
(2. 3) lim sup − ρ =0.
n→∞x∈[0,1](cid:12)(cid:12) n − (cid:12)(cid:12)
(cid:12) (cid:12)
The rotational number ρ is the soluti(cid:12)on of the equati(cid:12)on
a1−ρbρ =1.
The number ρ is rational if and only if loga/logb is rational. Furthermore, we
have
λ(f ,x)=0 ( x / C ).
a,b,c a,b,c
∀ ∈
Proof. The uniform convergence follows from the observation
Fn(0)<Fn(x)<Fn+1(0) ( x [0, 1])
∀ ∈
Fn(0)
and the fact that ρ= lim .
n→∞ n
According to Lemma 2.1, the rotation number ρ of f is nothing but the
a,b,c
frequency of visits to (c, 1] of any given point of [0, 1]. Since f is piecewise
a,b,c
linear, for x / C we have
a,b,c
∈
n−1
(2. 4) (fn )′(x)= f′ (fi (x))=an−mn(x)bmn(x)
a,b,c a,b,c a,b,c
iY=0
where m (x) is defined by (2. 1). It follows that
n
1
λ(f ,x)= lim logan−mn(x)bmn(x) =loga1−ρbρ.
a,b,c
n→∞n
Let λ = loga1−ρbρ. We will prove a1−ρbρ = 1 by showing λ = 0, which implies
that
logb
ρ=1+ .
loga logb
−
6 AcimsofpiecewiselinearLorenzmaps
So ρ is rational if and only if loga/logb is rational. Suppose λ > 0. According to
Lemma 2.1, (2. 3) and (2. 4), there exists a positive integer N such that
(fN )′(x)>1 ( x / C ).
a,b,c ∀ ∈ a,b,c
This and the piecewise linearity of f imply that fN is piecewise expanding.
a,b,c a,b,c
However,it is not possible because fN is a homeomorphism. Thus λ 0. In the
a,b,c ≤
same way, one proves λ 0 by considering f−1 . Hence we get λ=0. (cid:3)
≥ a,b,c
Lemma 2.3. Let f be a piecewise linear Lorenz map such that f (0) =
a,b,c a,b,c
f (1). Let ρ be the rotation number of f . If ρ is irrational, then
a,b,c a,b,c
(2. 5) m (x) nρ 4 ( x [0, 1], n 1).
n
| − |≤ ∀ ∈ ∀ ≥
Proof. Since ρ is irrational, f is topologically conjugate to the rigid irrational
a,b,c
rotation R , i.e. R (x) = ρ+x (cf. [20], p. 38-39). In other words, there exists a
ρ ρ
continuous strictly increasing function π on [0, 1] onto [0, 1] such that
(2. 6) π f π−1 =R .
a,b,c ρ
◦ ◦
Let F, G and G−1 be the degree one lifting map of f , π and π−1 respectively.
a,b,c
The lifted form of (2. 6) is
G F G−1(x)=x+ρ.
◦ ◦
By induction, we have
(2. 7) G Fn G−1(x)=x+nρ.
◦ ◦
Write
m (x) nρ m (x) Fn(x) + Fn(x) G Fn(x)
n n
| − | ≤ | − | | − ◦ |
(2. 8) + G Fn(x) G Fn G−1(x)
| ◦ − ◦ ◦ |
+ G Fn G−1(x) nρ.
| ◦ ◦ − |
Now we estimate the four terms on the righthand side. Notice first that F, G and
G−1 areincreasingfunctions onRandthatF(x) x, G(x) xandG−1(x) x are
− − −
1-periodic functions on R taking with values in [0,1]. So when x y 1 we have
| − |≤
(2. 9) F(x) F(y) 1, G(x) G(y) 1, G−1(x) G−1(y) 1.
| − |≤ | − |≤ | − |≤
According to Lemma 2.1, we have the following estimate for the first term:
m (x) Fn(x) 1.
n
| − |≤
The fact 0 G(x) x 1 implies immediately an estimate for the second term:
≤ − ≤
Fn(x) G Fn(x) 1.
| − ◦ |≤
Repeatingthefirstinequalityin(2. 9)weget Fn(x) Fn(y) 1when x y 1.
| − |≤ | − |≤
This and the fact G−1(x) x 1 imply an estimate for the third term:
| − |≤
G Fn(x) G Fn G−1(x) 1.
| ◦ − ◦ ◦ |≤
A direct consequence of (2. 7) is the following estimate for the fourth term:
G Fn G−1(x) nρ 1.
| ◦ ◦ − |≤
The estimation (2. 5) is thus proved. (cid:3)
Yi-MingDingAi-HuaFanandJing-HuYu 7
a
a
a
0
b
b
0
b
0 c 1 0 c 1
Figure 2. The comparisons between piecewise linear Lorenz maps.
2.2. Comparison of rotation numbers in different maps. Let f be a
a,b,c
piecewise linear Lorenz map. If a > 1 and b > 1, it is well known that f is
a,b,c
expanding and then admits a unique acim. If a<1 and b<1, f is contracting
a,b,c
and then admits no acim. So we may assume that a > 1 b or b > 1 a. In
≥ ≥
thesecases,wewillcomparef withhomeomorphicpiecewiselinearLorenzmaps
a,b,c
f andf ,wherea = 1−b(1−c) andb = 1−ac. SeeFigure2forthepictures
a0,b,c a,b0,c 0 c 0 1−c
of f and f . More precisely, in the case a>1 b, we compare f with
a0,b,c a,b0,c ≥ a,b,c
a x+1 ac x [0, c)
(2. 10) fa0,b,c(x)=(cid:26) b(0x c)− x∈(c, 1]
− ∈
if f (0)<f (1), and compare f with
a,b,c a,b,c a,b,c
ax+1 ac x [0, c)
(2. 11) fa,b0,c(x)=(cid:26) b0(x −c) x∈(c, 1]
− ∈
if f (0)>f (1).
a,b,c a,b,c
In the case b > 1 a, we compare f with f when f (0) > f (1),
≥ a,b,c a,b0,c a,b,c a,b,c
and compare f with f when f (0)<f (1).
a,b,c a0,b,c a,b,c a,b,c
Lemma 2.4. Let f and f be defined as above. We have the following
a0,b,c a,b0,c
conclusions:
(1) If f (0)<f (1), then for all x [0,1],
a,b,c a,b,c
∈
ρ(f ,x) ρ(f ) if a>1 b
a,b,c ≤ a0,b,c ≥
(cid:26) ρ(fa,b,c,x)≥ρ(fa,b0,c) if a≤1<b.
(2) If f (0)>f (1), then for all x [0,1],
a,b,c a,b,c
∈
ρ(f ,x) ρ(f ) if a 1>b
a,b,c ≥ a0,b,c ≥
(cid:26) ρ(fa,b,c,x)≤ρ(fa,b0,c) if a<1≤b.
Proof. Let F be the degree one lifting of f and F be the degree one lifting of
a,b,c 0
f . For the case f (0) < f (1) and a > 1 b, since f (x) f (x)
a0,b,c a,b,c a,b,c ≥ a,b,c ≤ a0,b,c
forallx∈[0, 1],wehaveF(y)≤F0(y)forally ∈R. ItfollowsthatFn(y)≤F0n(y)
ofotrheyr∈inRequaanlditifeosrceavnerbyepsoimsitiliavrelyinptergoevredn.. Therefore ρ(fa,b,c,x) ≤ ρ(fa,b0,c). Th(cid:3)e
8 AcimsofpiecewiselinearLorenzmaps
2.3. Frobenius-Perronoperatorandinvariantdensity. TheFrobenius-Perron
operator associated with f is defined as follows: for any h L1(I),
a,b,c
∈
1[1−ac, 1](x) x (1 ac) 1[0, b(1−c)](x) x+bc
P h(x)= h − − + h .
a,b,c
a (cid:18) a (cid:19) b (cid:18) b (cid:19)
The invariant density h of the Frobenius-Perronoperator corresponds to an acim
∗
µ of f , µ(A) = h dm, where A is a Borel set and m is the Lebesgue
a,b,c A ∗ ∈ B
measure on I. R
Lemma 2.5. Let P be the Frobenius-Perron operator associated with f . Then
a,b,c
we have the following statements:
(1) If there exists positive integer n 1 such that fn (x) = x for all x I,
≥ a,b,c ∈
then for each density g on [0, 1], A (g) is an invariant density of P .
n a,b,c
(2) If there exists a constant r >0 such that for all positive integer n,
rm(A) m(f−n (A)) m(A)/r, A ,
≤ a,b,c ≤ ∀ ∈B
then f admits a unique acim whose density is bounded by r and 1/r.
a,b,c
(3) Ifthereexists apositive integer nsuchthat fn is stronglyexpanding, i.e.,
a,b,c
(fn )′(x) >λ >1 for all x I except finite points, then f admits an
a,b,c ∈ a,b,c
acim whose density is of bounded variation.
(4) If there exists a positive integer n such that (fn )′(x) < λ < 1 for all
a,b,c
x I except finite points, then f admits no acim.
a,b,c
∈
Proof. The assertions (1) and (4) are obvious. The assertion (3) is a direct con-
sequence of Lasota and Yorke’s Theorem ([18, 19]). Now we prove (2). The as-
sumption in this case means that r Pn 1 1/r. So A (1) is weakly
≤ a,b,c ≤ { n }n≥0
precompactin L1(I). From the weaklycompactness of A (1) we canextract
n n≥0
{ }
a subsequence A (1) that converges weakly to g and P g =g. By the abstract
nk a,b,c
ergodic Theorem of Kakutani and Yosida ([18]), A (1) converges strongly to g.
n
This implies that g is an invariant density of P and r g(x) 1/r. (cid:3)
a,b,c
≤ ≤
In the third case, f is said to be eventually piecewise expanding [14].
a,b,c
2.4. Renormalization of Lorenz map. Let f be a piecewise linear Lorenz
a,b,c
map satisfying ac+b(1 c) > 1. The equivalence between the acim of f and
a,b,c
−
the Lebesgue measure is a question of transitivity of f (see Lemma 4.1). One
a,b,c
can describe the transitivity of f by using the device of renormalization.
a,b,c
A Lorenz map f : I I is said to be renormalizable if there is a proper subin-
→
terval [u, v] which contains the critical point c, and integers ℓ,r>1 such that the
map g :[u, v] [u, v] defined by
→
fℓ(x) x [u, c),
(2. 12) g(x)= ∈
(cid:26) fr(x) x (c, v],
∈
is itself a Lorenz map on [u, v].
A Lorenz map f is said to be expanding if the preimages of the critical point is
denseinI. TherenormalizationtheoryofexpandingLorenzmapiswellunderstood
(see [7, 12]). The transitivity of an expanding Lorenz map can be characterizedby
its renormalization. For example, f is transitive if it is not renormalizable ([7]).
Letf be anexpandingLorenzmap. The renormalizabilityoff is closelyrelated
to the periodic orbit with minimal period. Denote κ the smallest period of the
periodic points of f. If κ=1 (i.e., f admits a fixed point), we must have f(0)=0
or f(1)=1 because f is expanding. It follows that f is transitive ([7]). If κ = ,
∞
Yi-MingDingAi-HuaFanandJing-HuYu 9
i.e. f admits no periodic point, then f is topologically conjugates to an irrational
rotation on the circle because f is expanding ([12]). For the case 1 < κ < , we
∞
have the following Lemma.
Lemma 2.6. ([7]) Let f : [0, 1] [0, 1] be an expanding Lorenz map with 1 <
→
κ< .
∞
(1) The minimal period of f is equal to κ=m+2, where
m=min i 0:f−i(c) [f(0), f(1)] .
{ ≥ ∈ }
(2) f admits a unique κ-periodic orbit O.
(3) Let P and P be adjacent points in O such that c [P , P ]. Then fκ is
L R L R
∈
continuous on [P , c) and on (c, P ]. Moreover, we have
L R
κ−1
(2. 13) fi([P , P ])=I.
L R
i[=0
For general expanding Lorenz map f, it is difficult to check wether f is renor-
malizable or not. However, for piecewise linear Lorenz map f satisfying ac+
a,b,c
b(1 c)>1, one can check the renormalizability easily. According to the proof of
−
Theorem A, f is expanding. Denote O as the κ-periodic orbit, and
a,b,c
D := f−n (O).
a,b,c
n[≥0
Lemma 2.7. ([4]) If D =O, then f is not renormalizable.
a,b,c
6
3. Existence of absolutely continuous invariant measure
Now we prove Theorem A by distinguishing four cases: f (0)=f (1) and
a,b,c a,b,c
loga/logb is rational, f (0) = f (1) and loga/logb is irrational, f (0) <
a,b,c a,b,c a,b,c
f (1) andf (0)>f (1). In the firstcase,we will show that some powerof
a,b,c a,b,c a,b,c
f is identity, i.e., there exists n>0 such that fn (x)=x for all x I. In the
a,b,c a,b,c ∈
second case we will prove
rm(A) m(f−n (A)) m(A)/r, A ,
≤ a,b,c ≤ ∀ ∈B
for some constant r and n 0. In the third case we will show that some power
≥
of f is expanding. In the forth case, we will compare f with a suitable
a,b,c a,b,c
homeomorphic piecewise linear Lorenz map and prove that some power fn is
a,b,c
contracting.
3.1. Proof of Theorem A when f (0) = f (1) and loga/logb is ratio-
a,b,c a,b,c
nal.
According to Lemma 2.5, it suffice to prove the following proposition.
Proposition 3.1. Suppose f (0) = f (1) and loga/logb is rational. Then
a,b,c a,b,c
there exists positive integer n such that fn (x)=x for all x I.
a,b,c ∈
Proof. In this case, f can be regarded as a homeomorphism on the unit circle.
a,b,c
Since loga/logb is rational, the rotation number of f is also rational (Lemma
a,b,c
m
2.2). Write ρ(f )= with (m, n) = 1. We shall prove that fn (x) = x for
a,b,c n a,b,c
all x [0, 1].
∈
10 AcimsofpiecewiselinearLorenzmaps
m
Sinceρ(f )= , f admitsann periodicorbit([2]). Letp <p < <
a,b,c a,b,c 1 2
n − ···
p be an n-periodic orbit. The orbit forms a partition of I:
n
[p , p ), ,[p , p ), [p ,1] [0,p ),
1 2 n−1 n n 1
··· ∪
and f maps one subinterval onto the next one in the partition. Each subin-
a,b,c
terval in the partition contains only one point in C . Since ρ(f ) = m,
a,b,c a,b,c n
c [p ,p ).
n−m n−m+1
∈
If c doesn’t belong to the periodic orbit, c (p , p ). Consider the
n−m n−m+1
∈
interval[p ,c),itfollowsthatfn iscontinuousandlinearon[p ,c)because
n−m a,b,c n−m
fn has only one discontinuity c in [p ,p ]. Notice that fn (p ) =
a,b,c n−m n−m+1 a,b,c n−m
p and (fn )′(p ) = 1, we obtain that fn (x) = x on [p ,c), which
n−m a,b,c n−m a,b,c n−m
implies that c is an n-periodic point. So 1 is also an n-periodic point.
−
We denote the n-periodic orbit of 1 as 0 < q < q < < q < q =
1 2 n−m−1 n−m
···
c < q < < q = 1. Since fn is linear on [q ,q ), it follows that
n−m+1 ··· n a,b,c i i+1
fn (x)=x on [q ,q ), i=1,2,...,n. So fn (x)=x on I. (cid:3)
a,b,c i i+1 a,b,c
3.2. Proof of Theorem A when f (0) = f (1) and loga/logb is irra-
a,b,c a,b,c
tional.
Inthiscase,accordingtoLemma2.5,wehaveonlytoprovethefollowingpropo-
sition.
Proposition 3.2. Suppose f (0) = f (1) and loga/logb is irrational, there
a,b,c a,b,c
exists a constant r >0 such that
rm(A) m(f−n (A)) m(A)/r, A .
≤ a,b,c ≤ ∀ ∈B
1 b
Proof. The condition f (0) = f (1) means c= − . Consider f :=
a,b,c a,b,c a b a1,b1,c1
f−1 , the inverse map of f . It is also a piecewise lin−ear Lorenz map such that
a,b,c a,b,c
f (0)=f (1). In fact, we have
a1,b1,c1 a1,b1,c1
1 1 (a 1)b
(3. 1) a = , b = , c = − .
1 1 1
b a a b
−
Let ρ:=ρ(f ) be its rotation number. Write
a1,b1,c1
m∗(x)=# 0 i<n:fi (x) (c , 1] .
n ≤ a1,b1,c1 ∈ 1
By Lemma 2.2, we have a1−ρb(cid:8)ρ =1. Thus for all x / C ,(cid:9)we have
1 1 ∈ a1,b1,c1
(fn )′(x) = an−m∗n(x)bm∗n(x)
a1,b1,c1 1 1
m∗(x)−nρ m∗(x)−nρ
b n b n
= an(1−ρ)bnρ 1 = 1 .
1 1 ·(cid:18)a (cid:19) (cid:18)a (cid:19)
1 1
Accordingto Lemma 2.3, m∗(x) nρ 4. Itfollowsthat for all x / C and
| n − |≤ ∈ a1,b1,c1
all n 0 we have
≥ r (fn )′(x) 1/r,
≤ a1,b1,c1 ≤
where r = min b4a−4, b−4a4 = min b4a−4, b−4a4 . Consequently, by making a
{ 1 1 1 1} { }
change of variables we get
rm(A) m(f−n (A)) = 1 (fn (x))dx
≤ a,b,c Z A a,b,c
= 1 (y) (fn )′(y)dy m(A)/r.
Z A · a1,b1,c1 ≤
(cid:3)