´
ON THE ESTIMATION OF POINCARE MAPS OF THREE-DIMENSIONAL
VECTOR FIELDS NEAR A HYPERBOLIC CRITICAL POINT
By
Yuting Zou

A DISSERTATION
Submitted to
Michigan State University
in partial fulﬁllment of the requirements
for the degree of
DOCTOR OF PHILOSOPHY
Mathematics
2011

ABSTRACT
´
ON THE ESTIMATION OF POINCARE MAPS OF
THREE-DIMENSIONAL VECTOR FIELDS NEAR A HYPERBOLIC
CRITICAL POINT
By
Yuting Zou
We study the estimation of Poincar´ maps of three-dimensional vector ﬁelds
e
near a hyperbolic critical point, which involves linearization problems. Standard
linearization theorems have several defects in applications. They usually require
complicated non-resonance conditions on the eigenvalues of the vector ﬁeld at the
critical point. Even when one has these non-resonance conditions, as one gets close to
a resonance, the size of the neighborhood where the C 1 linearization exists typically
gets too small for practical uses. We seek for a linearization theorem that overcomes
these shortcomings and may have broad practical applications.
We have proved a partial linearization theorem that gives a C 1 linearization
h near a hyperbolic critical point p on a two-dimensional invariant surface Σ of a
three-dimensional vector ﬁeld X. Let the eigenvalues of DX(p) be a, b and c, where
a > 0 > b > c. Essentially our theorem only requires that 2b > c to obtain h in some
neighborhood U of p in Σ. In addition, the explicit size of U is found, which depends
on the C 2 information of X, as well as the C 0 and C 1 sizes of h. Based on our
partial linearization theorem, we obtain desired estimation of Poincar´ maps from
e
some transversal curve to the stable manifold of p to another transversal surface to
the unstable manifold of p.
Our estimation of such Poincar´ maps will have many applications, including an
e

in-depth study of the famous Lorenz equations. For example, it seems likely that
we will be able to substantially improve results of Tucker on the existence of the
Lorenz strange attractor, and obtain rigorous results on the existence of chaos near
the ﬁrst homoclinic bifurcation as numerically investigated in the well-known book
of Colin Sparrow.

ACKNOWLEDGMENT

I would like to express my deepest and sincerest gratitude to my advisor, Dr. Sheldon
Newhouse, for his exceptional support and unfailing patience. Without his guidance,
this thesis would not have been possible. It is a valuable opportunity for me to
learn from such a great mathematician and a wonderful person. He has a very broad
perspective, sharp and quick mathematical mind, good sense of humor, and a variety
of interests besides mathematics. More than I could express my appreciation, he is a
strikingly philosophical, inspiring and open-minded person. Not only have I learned
abundantly from him, but also his unique personality has a life-long inﬂuence on
me. I am very thankful for his help that has pulled me through the most challenging
times of my doctoral program. It has been an enormously enjoyable experience to
work with him.
My gratitude goes to my defense committee members, Dr. Martin Berz, Dr.
Kyoko Makino, Dr. Yang Wang and Dr. T.Y. Li, for their expertise and precious
time. Special thanks to Dr. Martin Berz, Dr. Kyoko Makino and Dr. Alexander
Wittig for their intense interests in our work, which gives me a strong encouragement
to carry on my work to its ﬁnest details. Their tool, COSY VI-integrator, that made
veriﬁed numerical integration possible, is an important source of inspiration for this
thesis.
Finally I would like to thank Dr. Zhangfang Zhou for his unceasing care and
support from the beginning of my doctoral program to the end. He is one of the
kindest and most generous people I have ever met. Whenever I have diﬃculties, he
iv

whole-heartedly oﬀers his helps. It is hard to imagine completing my Ph.D. track
without his presence.

v

TABLE OF CONTENTS

List of Figures

. . . . . . . . . . . . . . . . . . . . . . . . . .

vii

1 Introduction

1

2 Linearization theorems

6

3 A quantitative partial linearization theorem in R3
3.1 A sketch of the proof . . . . . . . . . . . . . . . . . . . . . . .
3.2 Linear estimates in dimension two . . . . . . . . . . . . . . . .
3.3 Conjugacy restricted to the stable manifold . . . . . . . . . . .
3.4 Local C 1 linearization for two dimensional diﬀeomorphisms .
3.5 Three dimensional vector ﬁelds . . . . . . . . . . . . . . . . .
3.6 Straightening Invariant manifolds of a three dimensional map .

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Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . .

vi

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12
15
17
18
25
52
89
97

LIST OF FIGURES

3.1

Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.2

D0 . For interpretation of the references to color in this and all other
ﬁgures, the reader is referred to the electronic version of this dissertation 31

3.3

F 1 (y) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.4

F 2 (y) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

vii

Chapter 1
Introduction
Let M be a smooth (C ∞ ) three dimensional Riemannian manifold with associated
topological metric d, and let X be a C 2 vector ﬁeld on M with a hyperbolic critical
point p0 ∈ M with real distinct eigenvalues a, b, c satisfying

a > 0 > b > c.

(1.1)

Assume that X is forward complete in the sense that, if φ(t, x) denotes the local
ﬂow of X, then, φ(t, x) is deﬁned for all t ≥ 0 and all x ∈ M.
Let W s (p0 ), W ss (p0 ), W u (p0 ) denote, respectively, the stable, strong stable,
and unstable manifolds of p0 .
These are deﬁned as follows.

W s (p0 ) = {x ∈ M : φ(t, x) exists for all t ≥ 0 and lim d(φ(t, x), p0 ) = 0}
t→∞
1

W ss (p0 ) = {x ∈ M : φ(t, x) exists for all t ≥ 0
1
and lim sup log d(φ(t, x), p0 ) < b}
t→∞ t

W u (p0 ) = {x ∈ M : φ(t, x) exists for all t ≤ 0 and

lim d(φ(t, x), p0 ) = 0}.
t→−∞

It is well-known that W s (p0 ), W ss (p0 ) and W u (p0 ) are C 2 injectively immersed
copies of the plane and line, respectively, in M.
Let us use the notation Comp(E, x) for the connected component of the set E
containing the point x. For a point p ∈ W s (p0 ) and a neighborhood U of p in M, let
W s (p, U) = Comp(W s (p0 )

U, p). One can choose arbitrarily small neighborhoods

U so that W s (p, U) is a topological 2-disk and U \ W s (p, U) consists of a disjoint
pair of open 3-balls.
Let p ∈ W s (p0 ) \ {p0 }, q ∈ W u (p0 ) \ {p0 } and let N be a smoothly embedded
2-disk in M, such that N is transversal to W u (p0 ) at q. An application of the
Grobman-Hartman theorem (C 0 local linearization theorem) shows that there is a
small neighborhood U of p in M and a connected component U0 of U \ W s (p, U)
such that there is a well-deﬁned Poincar´ map P from U0 into N. This map simply
e
takes a point in U0 to the ﬁrst point on its positive orbit which hits N.
In many applications (e.g. Section 2.2 in [7], [9], and the books [6], [1]) one
encounters the following problem.
Let p ∈ W s (p0 ) \ W ss (p0 ) and let γ be a C 2 embedded arc in U meeting

2

W s (p0 ) transversely at p and assume that U is small enough so that for each y ∈ γ,
d(y, p) is boundedly related to the arc length of the subarc of γ joining p to y. Let
′
U0 , and, for a point y ∈ γ0 , let γ0 (y) denote the unit tangent vector to
′
γ0 at y. One seeks an estimate of the derivative P ′ (y) = DPy (γ0 (y)).
γ0 = γ

If the vector ﬁeld X is C 1 linearizable near p0 , then it is well-known and easy
to prove that one can ﬁnd constants 0 < C1 < C2 such that

b
C1 |P ′ (y)| ≤ d(y, p)| a |−1 ≤ C2 |P ′ (y)|.

(1.2)

In most applications using estimates of the type (1.2), one merely assumes that
X is, in fact, at least C 1 linearizable near p0 . In view of the Sternberg linearization
theorem, a generic C ∞ vector ﬁeld is C ∞ linearizable near a hyperbolic critical
point, so it does not seem to be a big assumption to assume linearity. However, it
should be noted that there are several defects to the application of the Sternberg
theorem even to get C 1 linearizations. Among these defects are the following.

• The theorem requires so-called non-resonance conditions on the eigenvalues
a, b, c;
• Even in the case that X is C ∞ , as one gets close to a resonance, the size of the
neighborhood on which the C 1 linearization exists typically gets very small;
• The amount of smoothness required on X depends on the eigenvalues a, b, c.
c
Our main result shows that, with the mild additional assumption that b > 2 ,
we obtain the estimate (1.2) above for general C 2 vector ﬁelds in dimension three.
3

Theorem 1.0.1. Let X, p0 , p, γ0 and P be as above. Also, as above, for y ∈ γ0 , let

′
P ′(y) = DPy (γ0 (y))
′
be the derivative of P on the unit tangent vector γ0 (y) to γ0 at y.
Assume that the eigenvalues a, b, c of X at the critical point p0 satisfy

a > 0 > 2b > c.

(1.3)

Then, there are constants 0 < C1 < C2 such that

b
C1 |P ′ (y)| ≤ d(y, p)| a |−1 ≤ C2 |P ′ (y)|.

(1.4)

The main tools we use are modiﬁcations of techniques in so-called normal hyperbolicity theory to get a C 2 invariant surface Σ which contains both γ0 and p0
and is tangent at p0 to the sum of the eigenspaces of a, b. Since, by a theorem of
P. Hartman [2], two dimensional C 2 vector ﬁelds with a hyperbolic saddle point p0
are C 1 linearizable near p0 , one can get the required estimate.
Since we obtain a linearization of X restricted to an invariant two-dimensional
surface through p0 and not in a full neighborhood of p0 , we will refer to our result
as a partial linearization theorem.
Remarks.

• It should be noted that the general tools of normal hyperbolicity in the literature give some two dimensional surfaces tangent at p0 to the sum of the
4

eigenspaces of a, b. However, these surfaces are far from unique, and are usually constructed by means of globalization techniques. Since we need such
surfaces which contain an a priori given curve γ0 , the globalization technique
is not applicable as far as we know. The existence of C 2 invariant surfaces
through p0 which contain an arbitrary given transverse curve γ0 seems to be
a new result, and our proof avoids the use of globalizations.
• For various applications, it is important to know the sizes of the constants
C1 , C2 in (1.2). We take care to give explicit estimates for these in our work
below.
Using appropriate local coordinates near p0 and replacing p and q by points in
their orbits, it suﬃces to prove our results for vector ﬁelds in R3 with a hyperbolic
critical point at the origin. The precise statement needed is in Theorem 3.0.6.
We ﬁrst review some of the history of linearization theorems.

5

Chapter 2
Linearization theorems
The ﬁrst general linearization theorem was given by Poincar´ in his thesis. He
e
obtained a theorem that can be rephrased as follows:

Theorem 2.0.2. Let

dxi
= λi xi + fi (x1 , . . . , xn ), 1 ≤ i ≤ n.
dt

(2.1)

be a system of diﬀerential equations such that λi is a complex number, and each fi is
a complex analytic function in a neighborhood of the origin which vanishes together
with its ﬁrst order partial derivatives at the origin.
Suppose all the λi lie in the same open half-plane about the origin and
n
λi =

mj λj

(2.2)

j=1
for any non-negative integers mj with 2 ≤
6

k m .
j=1 j

Then, there exists a complex analytic change of coordinates yi = ψi (x1 , x2 , · · · , xn )
transforming (2.1) into the linear system

dyi/dt = λi yi , 1 ≤ i ≤ n
The conditions (2.2) are called non-resonance conditions and, without them,
there are examples even of polynomials fi where the theorem fails.
Let us recall the deﬁnitions of local linearizations in both the ﬂow and diﬀeomorphism contexts.
Let f : Rn → Rn be a C r diﬀeomorphism, r ≥ 1, from a neighborhood U of 0
in Rn into Rn such that

f (0) = 0, Df (0) = L.
Then, L is, of course, a linear automorphism of Rn .
A local C k linearization of f near 0 is a C k homeomorphism h from a neighborhood V of 0 in U to another neighborhood V ′ of 0 such that

Lh(x) = hf (x)
for x ∈ V

f −1 (V ) .

Next, consider a C r local ﬂow near 0 in Rn with a ﬁxed point (or critical point)
at 0. This is a C r map φ(t, x) from a neighborhood (−ǫ, ǫ) × U into Rn where ǫ is
a positive real number such that

7

1. φ(0, x) = x for all x ∈ U,
2. φ(t, 0) = 0 for all t ∈ (−ǫ, ǫ), and
3. φ(t + s, x) = φ(t, φ(s, x)) for t, s ∈ (−ǫ, ǫ) such that t + s ∈ (−ǫ, ǫ).

We often use φ or φt to denote the local ﬂow where φt (x) = φ(t, x). Sometimes
we reverse the t and x and write φ(x, t).
A C r vector ﬁeld X near 0 in Rn is a C r map X : U0 → Rn from a neighborhood
U0 of 0.
If X is a C r vector ﬁeld near 0 such that X(0) = 0 (with r ≤ 1), then, for every
T > 0 there is a neighborhood U of 0 in Rn and a local ﬂow φ(t, x) = φX (t, x)
deﬁned on (−T, T ) × U such that

∂φ(t, x)
= X(φ(t, x))
∂t
for all t ∈ (−T, T ) and x ∈ U.
The local ﬂow φ(t, x) is called the ﬂow of X near 0.
The local ﬂow φ(t, x) is called linear if, in addition to the usual ﬂow properties,
we have x → φ(t, x) is a linear automorphism of Rn for each t.
Let φt and ψt , |t| < ǫ be two local ﬂows near 0 with ﬁxed points at 0. Let k ≥ 0
be a positive integer.
A local C k conjugacy near 0 between φt and ψt is a C k homeomorphism h from
a neighborhood V of 0 to another neighborhood V ′ of 0 such that, there is a positive
real number ǫ1 such that
8

1. ψt is deﬁned on V ′ for all |t| ≤ ǫ1 ,
2. h ◦ φt is deﬁned for all |t| ≤ ǫ1 , and
3. ψt ◦ h = h ◦ φt for all |t| ≤ ǫ1 .
If ψt is linear, then a local C k conjugacy between φt and ψt near 0 is called a
local linearization of φt near 0.
If X is a C 1 local vector ﬁeld near 0 with associated local ﬂow φt and derivative
L = DX(0), then a local C k linearization of X near 0 is a local C k conjugacy
between φt and the linear ﬂow given by L. In this case, we may assume that the
local ﬂow φt is deﬁned for |t| ≤ T where T is arbitrary. For large T , we simply have
to shrink the neighborhood on which the linearization is deﬁned.
We are now in a position to state the well-known Grobman-Hartman Theorem
[6], [3].
Theorem 2.0.3. (Grobman-Hartman).
1. Let f be a C 1 local diﬀeomorphism of Rn with a hyperbolic ﬁxed point at 0
(i.e., the eigenvalues of Df (0) have norm diﬀerent from 1). Then, f has a
local C 0 linearization near 0.
2. Let X be a C 1 local vector ﬁeld near 0 with a hyperbolic critical point at 0
(i.e. X(0) = 0 and the eigenvalues of DX(0) have non-zero real parts). Let
L = DX(0) and let ψt be the ﬂow of L. Then, for any T > 0, there is a local
C 0 conjugacy h between φt and ψt for all |t| ≤ T .
9

This is a beautiful and powerful theorem. One could ask for smooth local linearizations for smoother f and/or X.
In this direction there is another well-known theorem due to Sternberg. See the
appendix starting on page 256 in [3] for a proof of this theorem as well as related
results, and further references.
Theorem 2.0.4. (Sternberg)
1. Let f be a C ∞ local diﬀeomorphism of Rn with a hyperbolic ﬁxed point at 0
(i.e., the eigenvalues of Df (0) have norm diﬀerent from 1). Let λ1 , λ2 , . . . , λn
be the eigenvalues of Df (0).
Suppose that, for every positive 1 ≤ i ≤ n,and any n-tuple of non-negative
n m , we have
j=1 j

integers (m1 , m2 , . . . , mn ) with 2 ≤

n
λi =
j=1

mj
λj .

(2.3)

Then, f has a local C ∞ linearization near 0.
2. Let X be a C ∞ local vector ﬁeld near 0 with a hyperbolic critical point at 0
(i.e. X(0) = 0 and the eigenvalues of DX(0) have non-zero real parts). Let
L = DX(0) and let ψt be the ﬂow of L.
Suppose that, for every positive 1 ≤ i ≤ n,and any n-tuple of non-negative
n m , we have
j=1 j

integers (m1 , m2 , . . . , mn ) with 2 ≤
n
λi =

mj λj .
j=1

10

(2.4)

Then, for any T > 0, there is a local C ∞ conjugacy h between φt and ψt for
all |t| ≤ T .
Again the conditions (2.3) and (2.4) are called non-resonance conditions. Without them, even a real-analytic system may not be C 1 linearizable.
Sternberg’s theorem is very important, and has many applications. However, as
we mentioned in our introduction, it does have some shortcomings.
Finally, we mention the two dimensional C 1 linearization theorems of Hartman
[2] which is of major relevance to this thesis. Here, fortunately, there are no nonresonance conditions to worry about.
Theorem 2.0.5.

1. Let f be a C 2 local diﬀeomorphism of R2 with a hyperbolic

saddle ﬁxed point at 0 (i.e., a pair of real eigenvalues λ1 , λ2 of Df (0) satisfy
0 < |λ2 | < 1 < |λ1 |). Then, f has a local C 1 linearization near 0.
2. Let X be a C 2 local vector ﬁeld near 0 in R2 with a hyperbolic saddle critical
point at 0 (i.e. X(0) = 0 and the eigenvalues λ1 , λ2 of L = DX(0) satisfy
λ2 < 0 < λ1 ). Let ψt be the ﬂow of L. Then, for any T > 0, there is a local
C 1 conjugacy h between φt and ψt for all |t| ≤ T .
As previously mentioned, we will give a new proof of Theorem 2.0.5 with explicit
estimates for the C 0 and C 1 sizes of the linearization.

11

Chapter 3
A quantitative partial linearization
theorem in R3
Let x = (x1 , x2 , x3 ) ∈ R3 , and X(x) be a forward complete C 2 vector ﬁeld which
vanishes at the origin and can be written as

¯
¯
¯
X(x) = (X1 (x), X2 (x), X3 (x)) = (ax1 + X1 (x), bx2 + X2 (x), cx3 + X3 (x))

¯
where c < 2b < 0 < a, and Xi,x (0, 0, 0) = 0 for i, j = 1, 2, 3.
j
For small positive ε0 , let
B1 = B1ε = {x ∈ R3 ||x2 | ≤ ε0 , |x1 | ≤ ε0 , |x3 | ≤ ε0 }
0

and let | · | = supx∈B | · | denote the maximum norm of various quantities in
1
¯
B1 . Then, there is a ﬁnite constant M > 0, such that |Xi,x ,x (x)| ≤ M for
j k

12

i, j, k = 1, 2, 3, and, using the mean value theorem, we have some small constants
¯
¯
ǫ1 > 0 and ǫ2 > 0, such that |Xi,x (x)| ≤ M · ε0 = ǫ2 , and |Xi (x)| ≤ ǫ2 · ε0 = ǫ1
j
for i, j = 1, 2, 3.
Let

B1top = {x ∈ B1 |x2 = ε0 },

B1bot = {x ∈ B1 |x2 = −ε0 },

+
B1 = {x ∈ B1 |x1 = ε0 },
and

−
B1 = {x ∈ B1 |x1 = −ε0 }.
and let ϕ(x, t) = ϕt (x) be the local ﬂow of X, so we may write

ϕ(x, t) = (ϕ1 , ϕ2 , ϕ3 ) = (eat x1 + ϕ1 , ebt x2 + ϕ2 , ect x3 + ϕ3 )
¯
¯
¯

where ϕi for i = 1, 2, 3 are the higher order terms. Set ψt (x) to be the ﬂow of the
¯
corresponding linear vector ﬁeld of X, i.e,

ψt (x) = (eat x1 , ebt x2 , ect x3 )

13

Let us denote the origin (0, 0, 0) by 0, and let W s (0) denote its stable manifold.
s
s
Let Wloc = Wloc (0) = Comp(W s (0)
of the W s (0)

B1 , 0) denote the connected component

B1 containing 0.

In the following we choose B1 small enough so that the indicated statements
are valid.
s
Let Γ = Wloc

B1top . Then Γ divides B1top into two connected open sets

each of whose closures is a topological 2-disk. By the Grobman-Hartman theorem,
−
+
if x ∈ B1top \ Γ, then there exists τφ (x) > 0, such that ϕ(x, τφ (x)) ∈ B1 ∪ B1 ,
−
+
and for 0 ≤ s < τ (x), ϕ(x, s) is not in B1 ∪ B1 .
Let
+
+
B1top = {x ∈ B1top |ϕ(x, τφ (x)) ∈ B1 },
and
−
−
B1top = {x ∈ B1top |ϕ(x, τφ (x)) ∈ B1 }.
+
−
Then, the map P (x) = ϕ(x, τφ (x)) is called a Poincar´ map from B1top ∪ B1top \ Γ
e
−
+
to the sides B1 ∪ B1 .

Recall that we use the maximum norm

|(v1 , v2 , v3 )| = max(|v1 |, |v2 |, |v3 |)
on vectors v = (v1 , v2 , v3 ) ∈ R3 and its induced norm on various other quantities.
For instance, the arclength of curves, lengths of tangent vectors, and norms of linear

14

maps are all deﬁned relative to this maximum norm.
Consider a C 2 curve γ in B1top which is transverse to Γ at the point p ∈ Γ.
±
Let γ ± = γ ∩ B1top . Assume that γ is parametrized by arclength and, for y ∈ γ,
let ℓy denote the arclength of the subarc of γ from p to y. Assume that there are
constants A2 > A1 > 0 such that

A1 ℓy ≤ |y − p| ≤ A2 ℓy .
Let vy be the unit tangent vector to γ ± at the point y ∈ γ ± , and let P ′ (y) =
DPy (vy ).
The main result needed to prove Theorem 1.0.1 is then the following

Theorem 3.0.6. There are constants C2 > C1 > 0 such that

b
C1 |P ′ (y)| ≤ |y − p|| a |−1 ≤ C2 |P ′(y)|.

3.1

A sketch of the proof

The proof of Theorem 3.0.6, which we call the quantitative partial linearization
theorem, is divided into several parts, shown step by step in the subsequent sections.
s
u
First, we straighten Wloc (0) and Wloc (0) and then obtain a C 2 surface Σ in B1
which meets B1top in the curve γ and is invariant under the ﬂow φ = φ(x, t) on orbit
pieces which stay in B1 . This allows us to transform the problem to an analogous
problem for a ﬂow η = η(x, t) in R2 with a hyperbolic critical point at (0, 0) with a
15

transverse curve γ through a point p in the stable manifold of η. Now, a well-known
˜
˜
theorem of Hartman [2] gives a C 1 linearization of η in some neighborhood of (0, 0).
Using this linearization, we can complete the proof of the Theorem 3.0.6.
General comment. We emphasize that, while many of the general ideas in
our construction have been available in the dynamics literature, we believe that our
results are new in at least two ways.
• the construction of the invariant C 2 surface Σ containing an arbitrary transverse curve γ.
• the detailed estimates of the C 1 sizes in our linearization.

Comment on the two dimensional local linearization. Following Sternberg [8] to linearize a ﬂow, one ﬁrst linearizes the time T map for some T > 0 and
uses the so-called integral technique. Thus the problem reduces to ﬁnding a local C 1
linearization near a hyperbolic ﬁxed point for a C 2 planar diﬀeomorphism f . This
theorem is also proved in Hartman [3].
A new proof for the diﬀeomorphism case was given by Palis and Takens in [5].
Their method consists of ﬁrst C 1 linearizing along the stable and unstable manifolds, and then contructing two transverse C 1 foliations containing these manifolds
which are used to deﬁne the coordinates of the required C 1 linearization. Their
proof relies on a clever use of the stable manifold theorem for the derivative map
Df on the tangent bundle, and, hence involves a four dimensional local diﬀeomorphism. While their proof can be modiﬁed to work in our case, it seems somewhat
complicated to obtain estimates of the size of the domain of the linearization and
16

its C 1 size from that approach. Since these types of estimates are important for
applications, we proceed to construct our own proof of the Hartman theorem. The
only thing that we borrow from Palis and Takens is the idea of transverse foliations.
Our construction of the foliations and the ensuing estimates are new.

3.2

Linear estimates in dimension two

Here let us derive some properties of a linear ﬂow in R2 :

ψt (x) = (ψ1 , ψ2 ) = (eat x1 , ebt x2 )

where a > 0 > b. Here in R2 , we only consider B1 = {x = (x1 , x2 ) ∈ R2 |0 ≤
x1 ≤ ε0 , 0 ≤ x2 ≤ ε0 }, with B1top = {(x1 , ε0 ) : 0 ≤ x1 ≤ ζε0 , 0 < ζ < 1},
B1r = {x ∈ B1 |x1 = ε0 }. Since other quadrants can be done similarly, we do
not show them here. Let the point p = (0, ε0 ), as is stated above, for any initial condition x = (x1 , ε0 ) ∈ B1top \ p, there is a ﬁnite time τ (x) > 0, such that
ψ(x, τ (x)) = (ε0 , y2 ) ∈ B1r . Let us name such map Pl (x1 , ε0 ) = (ε0 , y2 ), where
x1 = 0. We want to ﬁnd the size of DPl , so we proceed as follows.

We have Pl (x1 , ε0 ) = (ε0 , y2 ) = (eat x1 , ebt ε0 ), where x1 = 0. By comparing
the ﬁrst component, we have
ε 1
et = ( 0 ) a .
x1
17

Plugging to the second component, we have
ε b
y2 = ε0 ( 0 ) a .
x1
So we can write
ε b
Pl (x1 , ε0 ) = (ε0 , ε0 ( 0 ) a ).
x1
Since
−ε0
ε b
b ε b
b ε b
∂(ε0 ( 0 ) a )/∂x1 = ε0 · · ( 0 ) a −1
= − ( 0 ) a +1
x1
a x1
a x1
(x1 )2
we have









0 
0
0
 0
=
DPl = 
.
 ∂y

b
b ( ε0 ) a +1 0
2 0
−a x
∂x1
1
So

3.3

  

0
1 

DPl ·   = 
.
b +1 
b ( ε0 ) a
0
−a x
1

Conjugacy restricted to the stable manifold

In this section, we will adapt the proof of Sternberg’s linearization theorem [1956]
into the following quantitative version, with explicit conditions on the neighborhood
and sizes of the linearization worked out.
′
Theorem 3.3.1. Let f : R → R be a C 2 function such that f (0) = 0 and f (0) = a
′′
where |a| ∈ (0, 1). If in Bε0 (0) = {y ∈ R : |y| ≤ ε0 }, we have |f | ≤ M and
β = |a| + Mε0 /2 < 1, then there is a unique C 1 map u : Bε0 → R, such that
18

u′ (0) = 0, and if h(y) = (id + u)(y) then

h(f (y)) = ah(y).

Moreover,
|h| ≤ ε0 exp(

C
)
1−β

and

′
2C
|h | ≤ exp(
)
1−β

where C = Mε0 /(2|a|).

Proof: Using Taylor’s theorem, in a small neighborhood of the origin Bε , there
is a real number θy ∈ (0, y) such that
′′
f (θy ) 2
y .
f (y) = ay +
2

For k ≥ 0, let
fk (y) = f ◦ f ◦ · · · ◦ f (y)
k
so that
k
fk (y) = y

fi (y)/fi−1 (y)
i=1

and
k
fk (y)
=y
fi (y)/(afi−1 (y)).
ak
i=1
Since
′′
f (θi )
fi (y) = f (fi−1 (y)) = afi−1 (y) +
(fi−1 (y))2
2
19

(3.1)

we have

′′
f (θi )
f
(y),
fi (y)/(afi−1 (y)) = 1 +
2a i−1

and, hence,
fk (y)
=y
ak

k

k
fi (y)/(afi−1 (y)) = y

i=1

i=1

′′
f (θi )
[1 +
f
(y)].
2a i−1

Let us deﬁne a function h(y) by

f (y)
h(y) = lim k .
k→∞ ak
Assuming this, notice that it provides a linearization since

f
(y)
f
(y)
f (f (y))
h(f (y)) = lim k
= lim k+1
= lim a k+1
= ah(y).
k→∞
k→∞ ak+1
k→∞
ak
ak

Let us proceed to prove that h(y) is well deﬁned; i.e., that the limit exists.
We have

f (y)
h(y) = lim k
= lim y
k→∞ ak
k→∞

k
i=1

′′
f (θi )
[1 +
f
(y)]
2a i−1

(3.2)

so, it suﬃces to show that

∞
i=1

′′
f (θi )
f
(y) |< ∞.
|
2a i−1

20

(3.3)

Let

|·| =

sup | · |.
y∈Bε0

Now,
′′
′′
f (θ)
f (θ) 2
y = (a +
y)y
f (y) = ay +
2
2
′′
and, since |f | ≤ M and |a| ∈ (0, 1), we have

|f (y)| ≤ (|a| +

M
ε )|y|.
2 0

By induction, for i ≥ 1 we then get

M
M
|fi−1 (y)| ≤ (|a| + ε0 )i−1 |y| ≤ (|a| + ε0 )i−1 ε0 .
2
2

Hence
′′
f (θi )
M
M
fi−1 (y)| ≤
(|a| + ε0 )i−1 ε0 = Cβ i−1
|
2a
2|a|
2

(3.4)

ε M
where the constants C = 0
and β = |a| + M ε0 ∈ (0, 1) are those given in the
2
2|a|
theorem. The above inequality implies that the inﬁnite product in (3.2) converges
uniformly, and thus, the linearization h(y) exists. As the uniform limit of continuous
functions is continuous, we conclude that h(y) is continuous.
Using log(1 + x) ≤ x for x > 0, we have, for any positive real numbers pn
21

∞

∞
(1 + pn ) = exp(log(

n=1

∞

∞
(1 + pn ))) = exp(

n=1

≤ exp(

log(1 + pn ))
n=1

pn ).
n=1

Using that
f (y)
h(y) = lim k
k→∞ ak
where
fk (y)
=y
ak

k

k
fi (y)/(afi−1 (y)) = y

i=1

i=1

′′
f (θi )
[1 +
f
(y)]
2a i−1

we get


′′

∞



f (θi )
| h(y) | = | y 
[1 +
f
(y)] |
2a i−1
i=1
∞
C
≤ ε0 exp(
Cβ i−1 ) ≤ ε0 exp(
).
1−β
i=1
Further, for each y ∈ (0, ε0 ) there is a τi ∈ (0, ε0 ) such that

f ′ (y) = a + f ′′ (τi )y.
Since, for each k > 0 we have (with a possibly diﬀerent τi )

22

′
fk (y)
=
ak

k

k a + f ′′ (τ )f
f ′ (fi−1 (y))
i i−1 (y) )
(
)=
(
a
a
i=1
i=1
k
f ′′ (τi )
=
(1 +
fi−1 (y))
a
i=1

(3.5)

and, as above,

|

f ′′ (τi )
fi−1 (y)) |≤ 2C β i−1
a

we get h′ (y) exists and

f ′ (y)
h′ (y) = lim k .
k ak
In addition, with C as above, we get

2C
| h′ (y) |≤ exp(
).
1−β
Notice that, from the last expression in (3.5), we have that

′
fk (y)
=1
ak
for y = 0. Thus, h(y) = y + u(u) where u is a C 1 function such that u′ (0) = 0.
By now we have shown the existence of a C 1 linearization h described in the
theorem. It remains to show such h is unique. Let L(y) = ay. Suppose there is

23

another function g : Bε0 → R of the same properties as h; i.e., g −1 f g = L, g(0) = 0,
and g ′ (0) = 1. Then f = gLg −1 and so h−1 gLg −1 h(y) = ay. Let ρ = h−1 g, we
have ρLρ−1 = L, ρ(0) = 0, and ρ′ (0) = 1. So we can write ρ(y) = y + v(y)
with v(y) ∈ C 1 , v(0) = 0 and v ′ (0) = 0. Then let us show for such ρ, we have
ρ(y) = y for all y ∈ Bε0 , i.e., v(y) = 0 for all y ∈ Bε0 . By ρLρ−1 (y) = ay, we have
ρL(y) = aρ(y), which is
ay + v(ay) = a(y + v(y))
so v(ay) = av(y), for all y ∈ Bε0 . Taking derivatives, we get av ′ (y) = av ′ (ay), i.e.,
v ′ (y) = v ′ (ay) for all y ∈ Bε0 . We claim that v ′ (y) = 0 for all y ∈ Bε0 . Suppose
there is z ∈ Bε0 such that v ′ (z) = 0, then v ′ (z) = v ′ (az) = 0. Since |a| < 1,
ai z ∈ Bε0 for integer i ≥ 0. We have
v ′ (an z) = · · · = v ′ (az) = v ′ (z) = 0.

But limn→∞ an z = 0, by the continuity of v ′ we have v ′ (z) = v ′ (0) = 0 which is
a contradiction. So the claim is proved, which says v ′ (y) = 0 for all y ∈ Bε0 , i.e,
v(y) is a constant function. Since v(0) = 0, we have v(y) = 0, giving ρ(y) = y or
thus h = g. The uniqueness is proved, and so the proof of Theorem 3.3.1 is complete.

Now consider a local C 2 ﬂow φt , |t| ≤ 1, on R, with a hyperbolic critical point at
the origin, i.e., if f = φ1 , then f satisﬁes the assumptions of Theorem 3.3.1. Thus,
˜
˜
f (x) = ax + f (x), where 0 < a < 1, x ∈ R, and f (x) stands for the higher order
terms. Using the uniqueness of h, which is derived from a standard technique as in

24

[ref Newhouse Notes] show that h also linearizes {φt }|t|≤1 . That is, if ψt (x) = eat x,
then hφt (x) = ψt h(x) for x ∈ Bε0 .

3.4

Local C 1 linearization for two dimensional diffeomorphisms

Remark: It is well-known that linearizations for ﬂows follow from consideration of
associated time−1 maps. Hence we restrict here to the case of diﬀeomorphisms (see
p. 817 in [8]).

In this section, we consider the C 2 diﬀeomorphism f : R2 → R2 with a hyperbolic ﬁxed point at (0, 0). We assume the local unstable and stable manifolds of f ,
denoted by W u (0), W s (0), are contained in the coordinate lines {(x1 , x2 ) : x2 =
0}, {(x1 , x2 ) : x1 = 0}, respectively.
Let ε0 > 0 and consider the square

Bε0 = {(x1 , x2 ) : |x1 | ≤ ε0 , |x2 | ≤ ε0 }.
For x = (x1 , x2 ) ∈ Bε0 , we assume that

˜
˜
f (x1 , x2 ) = (λ1 x1 + f1 (x1 , x2 ), λ2 x2 + f2 (x1 , x2 )) = (f1 , f2 )
˜
where fi , i = 1, 2, are the nonlinear terms, and λ1 > 1, 0 < λ2 < 1.
25

(3.6)

Let us write the Jacobian matrix of Dfx as


Ax Bx 
Dfx = 
.
C x Dx
In addition, we assume that there are a small ǫ > 0 and a constant K > 0 such
that

˜
˜
fi (0, 0) = 0 and the ﬁrst partial derivatives fix (0, 0) = 0, for i, j = 1, 2
j

(3.7)

|B| ≤ ǫ, |C| ≤ ǫ, 0 < λ2 − ǫ ≤ D ≤ λ2 + ǫ < 1, 1 < λ1 − ǫ ≤ A ≤ λ1 + ǫ, (3.8)

and,

|dA−1 | ≤ K, |dB| ≤ K, |dC| ≤ K, |dD| ≤ K.

(3.9)

Using the previous section, suppose we have found C 1 linearizations hs : U → R
′
of f |W s (0) and hu : U → R of f |W u (0) . Then in this section, we will proceed to
linearize f : R2 → R2 in some neighborhood B1 of W s (0)

W u (0) in Bε0 . This

will be done via a modiﬁcation of ideas in [[5]]. We will construct the neighborhood
B1 and C 1 submersions πu : B1 → W s (0) and πs : B1 → W u (0) which commute
with f , and, together with the linearizations hs and hu restricted to the local stable

26

and unstable manifolds, will give a linearization h in B1 as
h(q) = (h1 , h2 ) = (hu πs (q), hs πu (q)) = q
¯
−1
−1
q
q
h−1 (¯) = πu (hs )−1 (¯y ) ∩ πs (hu )−1 (¯x ).
q
The inverse images of the submersions πs , πu will deﬁne two transverse folations
which are invariant by f . These foliations, after adjustments using hu , hs , become
the local coordinate curves of the linearization h.
We actually only deal with the construction of πu , leaving the analogous construction of πs to the reader.
Thus, we prove the following theorem.

Theorem 3.4.1. Let f : R2 → R2 be a C 2 diﬀeomorphism satisfying (3.6)-(3.9)
above. Deﬁne

α=

λ + 2ǫ
M(λ2 + 2ǫ) + ǫ
1
, M= 1
, µ=
.
(λ1 − ǫ)(λ2 − ǫ)
λ1 − ǫ
ǫ2 )
α(1 − α

(3.10)

Assume that ǫ is small enough so that

λ2 + 3ǫ
< 1 and µ < 1.
λ1 − ǫ
Then, there is a neighborhood B1 of W s (0)
sion πu : B1 → W s (0) such that
27

(3.11)

W u (0) in Bε0 and a C 1 submer-

πu f = f πu on B1

f −1 (B1 ).

(3.12)

Moreover, one has the estimates

|πu (x1 , x2 ) − x2 | ≤ ε0 ,

(3.13)

and

ε F
∂πu
F
| ≤ ε0 (1 +
) · exp( 0 )
∂x1
1−µ
1−µ
ε ·F
ε ·F
∂πu
− 1| ≤ ( 0
) exp( 0
)
|
∂x2
1−µ
1−µ
|

where
F = M · K(

3
+ λ2 + 3ǫ).
λ1 − ǫ

Let us start proving the theorem. We write f (x1 , x2 ) = (f1 (x1 , x2 ), f2 (x1 , x2 )).
−
+
Consider the sets Bε and Bε deﬁned by
0
0

+
Bε = {(x1 , x2 ) ∈ Bε0 |x2 ≥ 0},
0

(3.14)

−
Bε = {(x1 , x2 ) ∈ Bε0 |x2 ≤ 0}
0

(3.15)

28

+
−
We will construct πu on Bε and leave the analogous construction on Bε to
0
0
the reader.
We choose 0 < y0 < ε0 , and a small δ > 0, such that
+
f −1 (0, y0 + δ) ∈ Bε .
0
If we write p = (0, y0 ), and p± = (0, y0 ± δ), then the above condition becomes
+
f −1 (p+ ) ∈ Bε .
0
We write I1 to be the interval between p− and p+ on the y-axis, and similarly write
I2 to be the interval between f (p− ) and f (p+ ), and I0 to be the interval between
f (p− ) and p+ (see ﬁgure 3.1). In addition, we use I−1 to denote the set of the
second coordinates of f −1 (0 × I0 ) (see ﬁgure 3.1). Then we use l0 to represent the
+
horizontal line segment that passes through p inside Bε . And we take D0 as the
0
+
region bounded above and below by l0 and f (l0 ), which stays inside Bε , including
0
its boundaries (see ﬁgure 3.2).
∞ (f n (D ) ∩ B + ) that
ε0
0
n=0
contains p. Now let us proceed to deﬁne a C 2 horizontal foliation F u over the base
We deﬁne B1 to be the connected component of

I0 , which can then be extended to the whole B1 by the iteration of f .

First, we may choose a smooth bump function ρ(x2 ) over I0 such that, ρ(x2 ) = 1
over I1 ; ρ(x2 ) = 0 over I2 ; and over I0 \ (I1 ∪ I2 ), ρ(x2 ) is monotonically increasing
w.r.t. x2 with 0 < ρ(x2 ) < 1.
29

Figure 3.1: Intervals

Second, we deﬁne the ﬁrst horizontal foliation F 1 over the base I0 ∪ I−1 (see
ﬁgure 3.3). For every given x2 ∈ I0 ∪ I−1 , deﬁne the leaf through (0, x2 ) as the set
F 1 (x2 ) = {(x1 , x2 )| |x1 | ≤ ε0 }

which can also be rewritten as

F 1 (x2 ) = {(x1 , η1 (x1 )| |x1 | ≤ ε0 }

where η1 (x1 ) = x2 for every given x2 ∈ I0 ∪ I−1 , i.e., η1 is a constant function.

Next, let us deﬁne our second horizontal foliation F 2 (see ﬁgure 3.4) over the
30

Figure 3.2: D0 . For interpretation of the references to color in this and all other
ﬁgures, the reader is referred to the electronic version of this dissertation

+
base I0 . We write f −1 (0, x2 ) = (0, x2 ) ∈ Bε . And for every given x2 ∈ I0 , deﬁne
˜
0
the leaf through (0, x2 ) to be
+
F 2 (x2 ) = f (F 1 (˜2 )) ∩ Bε .
x
0

We may also write it as

F 2 (x2 ) = {(x1 , η2 (x1 )| |x1 | ≤ ε0 }

where η2 (0) = x2 .
Finally, our desired horizontal foliation F u over I0 is a combination of F 1 and
F 2 via the bump function ρ(x2 ). As for every given x2 ∈ I0 , we deﬁne the leaf of
31

Figure 3.3: F 1 (y)

F u through (0, x2 ) to be:
F u (x2 ) = {(x1 , ρ(x2 )η1 (x1 ) + (1 − ρ(x2 ))η2 (x1 )) : |x1 | ≤ ε0 }.

Observe that, by our deﬁnition of ρ(x2 ), F u (x2 ) agrees with F 1 (x2 ) for x2 ∈ I1 ,
and agrees with F 2 (x2 ) for x2 ∈ I2 . For every given x2 , the tangent vectors along
the leaf F 1 (x2 ) equals 0; and the tangent vectors along the leaf F 2 (x2 ) can be
written as



  

1 
f1x1 f1x2  1 f1x1  rescaled 

=
−− − − f
− − −→ 

 

2x1 
f2x f2x
0
f2x
f1x
1
2
1
1
where we know f1x = 0. So the tangent vectors to the leaf F u (x2 ) can be repre1
32

Figure 3.4: F 2 (y)

sented as
 


 
1

 1  
1
=
ρ(x2 )   + (1 − ρ(x2 ))  f
f2x  .
 2x1  
1
(1 − ρ(x2 ))
0
f1x
f1x
1
1

(3.16)

The computation above also implies that our leafs of F u (x2 ) for diﬀerent x2 ’s do
not intersect, i.e, the tangent vector at every point to a given leaf is uniquely deﬁned. Moreover, in the next paragraph we show the slope of each tangent vector is
bounded by 1, which implies our foliation F u (x2 ) is well deﬁned.

Let



 
1

Xu,1  
Xu = 
=
 
f2x  .
1
(1 − ρ(x2 ))
Xu,2
f1x
1

We will need the following estimate, which says the absolute value of the slope of
33

each tangent vector above is less than 1. Recall our assumption in (3.8), we know
f1x ≥ λ1 − ǫ > 1 and |f2x | ≤ ǫ, so we have
1
1
|Xu,2 |

= |(1 − ρ(x2 ))

|Xu,1 |

f2x
1|
f1x
1

f2x
1|
f1x
1
ǫ
≤
< 1.
λ1 − ǫ
≤ |

We now extend Xu to B1 . For each x ∈ B1 \ B1bot , there is a least integer
n(x), such that f −n(x) (x) ∈ D0 . Let
n(x)
Xu (x) = Df
· Xu (f −n(x) (x)).
−n(x) (x)
f
A λ-lemma (inclination lemma) argument shows that the absolute value of the slope
|Xu,2 (x)/Xu,1 (x)| of Xu (x) remains bounded by 1 for all x ∈ B1 and converges to
0 as x approaches B1bot . Let
Px =

Xu,2 (x)
Xu,1 (x)

be this slope. For each x we deﬁne




1
Yu =  
Px
for x ∈ B1 \ B1bot , and

 
1
Yu =  
0
34

(3.17)

for x ∈ B1bot . This Px is a continuous function from B1 to R, such that |Px | ≤ 1
for all x ∈ B1 .

Our main result of this section is the following.
Theorem 3.4.2. The vector ﬁeld Yu constructed above is C 1 in B1 . Equivalently,
the function x → Px is C 1 in B1 .
As we indicated above the integral curves of Yu will give our desired foliation
F u . The proof that Yu is C 1 , or equivalently, the function x → Px is C 1 , will
require several steps. The proof will be obtained by ﬁnding a sequence g1 , g2 , · · · ,
of C 1 functions, such that

1. gn converges to P uniformly on B1 ,
2. Dgn converges uniformly on B1 .
Following this, it is standard that P is C 1 and D P = limn→∞ Dgn .

We will use the technique of the Fiber Contraction Theorem familiar from the
Invariant Manifold Theory as in [4]. This involves deﬁnitions of certain function
spaces and associated mappings. Let us proceed to deﬁne our ﬁrst function space
G.

Consider the space C 0 (B1 , R) of bounded continuous functions from the box
B1 to R with supremum norm supx∈B | · | that makes it a Banach space. Let
1
35

0
C1 (B1 , R) be the closed subset of C 0 (B1 , R) with each of its element bounded by
0
1. Then C1 (B1 , R) is a complete metric space with the metric induced by this norm.
Let
0
˜
G = {P |P ∈ C1 (B1 , R), P (x) = P (x) f or x ∈ D0 ∪ B1bot }.
G is also a complete metric space equipped with the same norm supx∈B | · |.
1

The ﬁrst step is to show that Px is the unique ﬁxed point of a contraction map
Γ on G, which we now deﬁne. Let f = φX,T , by construction we have








1  c 
Dfy ·   = 

Py
c · Px

(3.18)

where y = f −1 (x) and c is a scalar. Using the notation at the beginning of this
section, we write



 







 Ay By   1   c 

· =
.
C y Dy
Py
c · Px
By solving the above equations for P , we have

Ay + By Py = c
C y + Dy P y = c · P x .

So
Px =

C y + Dy P y
Ay + By Py

36

(3.19)

which is equivalent to

Ay Px + Py By Px = Cy + Py Dy .

So we have
Px = A−1 Cy + A−1 Dy Py − A−1 By Py Px .
y
y
y
Given P ∈ G, we deﬁne Γ(P ) as follows. For x ∈ D0 ∪ B1bot , set

Γ(P )x = Px .

For x ∈ B1 \ (D0 ∪ B1bot ), let y = f −1 (x), and
Γ(P )x = A−1 Cy + A−1 Dy Py − A−1 By Px Py .
y
y
y

(3.20)

Clearly, Γ(P )x is continuous in x. To show that Γ(P ) ∈ G, it suﬃces to show
that
sup |Γ(P )x | ≤ 1.
x∈B1
For each x ∈ B1 and y = f −1 (x), we have
|Γ(P )x| ≤ |A−1 Cy | + |A−1 Dy Py | + |A−1 By Px Py |
y
y
y
≤

λ +ǫ
ǫ
ǫ
+ 2
+
.
λ1 − ǫ λ1 − ǫ λ1 − ǫ

37

If the size of B1 is properly chosen, we will have
λ2 + 3ǫ
< 1.
λ1 − ǫ
So
|Γ(P )| < 1
which means Γ maps G into itself.

Moreover, the Lipschitz constant λ of Γ can be computed as follows

|Γ(P1 )x − Γ(P2 )x | ≤ |A−1 Cy + A−1 Dy P1 (y) − A−1 By P1 (x)P1 (y)
y
y
y
−A−1 Cy − A−1 Dy P2 (y) + A−1 By P2 (x)P2 (y)|
y
y
y
≤ |A−1 Dy ||P1 (y) − P2 (y)|
y
+|A−1 By ||P1 (x)P1 (y) − P2 (x)P2 (y)|
y

where

|P1 (x)P1 (y) − P2 (x)P2 (y)| ≤ |P1 (x)P1 (y) − P1 (x)P2 (y)
+P1 (x)P2 (y) − P2 (x)P2 (y)|
≤ |P1 (x)||P1 (y) − P2 (y)| + |P2 (y)||P1(x) − P2 (x)|.

Since |P1 | ≤ 1 and |P2 | ≤ 1,

|P1 (x)P1 (y) − P2 (x)P2 (y)| ≤ 2|P1 − P2 |
38

we have
|Γ(P1 ) − Γ(P2 )| ≤ (|A−1 D| + 2|A−1 B|)|P1 − P2 |
Let
λ + 3ǫ
λ +ǫ
ǫ
+2
= 2
λ = |A−1 D| + 2|A−1 B| ≤ 2
λ1 − ǫ
λ1 − ǫ
λ1 − ǫ
If in B1 , we have λ < 1, then Γ is a contraction mapping with contraction constant
λ. That means, for any initial choice of P ∈ G, Γn (P ) converges uniformly to a
unique ﬁxed point. Since Γ(P ) = P , we know P is the unique attracting ﬁxed point.
So we may choose our converging sequence as follows.

To get our initial P0 , we may use a smooth bump function ρ : R2 → R such
that ρ(x) = 1 on a neighborhood of D0 and ρ(x) = 0 on a neighborhood of B1bot .
Then we deﬁne a C 1 function g0 , such that g0 |B ∈ G:
1
˜
˜
g0 = ρ(x)P + (1 − ρ(x)) · 0 = ρ(x)P

˜
for all points in a neighborhood U0 of B1 . Let P0 = g0 |B , we see that P0 |D = P ,
1
0
P0 | B
= 0 and is C 1 on B1 . Now for y = f −1 (x), let
1bot
g1 (x) = Γ(g0 ) = A−1 Cy + A−1 Dy g0 (y) − A−1 By g0 (x)g0 (y)
y
y
y

which is C 1 on a neighborhood of B1 . Inductively, set
gn (x) = Γ(gn−1 ) = A−1 Cy + A−1 Dy gn−1 (y) − A−1 By gn−1 (x)gn−1 (y)
y
y
y

39

This gives us a sequence of C 1 functions gn with the following properties:
1. gn is deﬁned and is C 1 on a neighborhood Un of B1 ,
2. gn |B ∈ G,
1
3. Γ(gn ) converges uniformly to P on B1 .
Let Pn = gn |B for n ≥ 0, we have Γn (P0 ) = Pn , and we can ﬁnd the C 0 size
1
of P , that is
|P | ≤

λn
|P − P0 | + |Pn |.
1−λ 1

(3.21)

Also it can be shown that

|P | ≤

λ
|P − P0 |.
1−λ 1

We wish to show that DPn converges uniformly on B1 . For this purpose we will
use the Fiber Contraction Theorem on a suitable space G × H which we proceed to
deﬁne. First of all, let us brieﬂy review some deﬁnitions and the Fiber Contraction
Theorem.
Deﬁnition 3.4.3. Let (G, dG ) and (H, dH ) be metric spaces. The map Λ : G ×H →
G × H of the form Λ(P, Q) = (Γ(P ), Ψ(P, Q)) is called a bundle map on G × H over
the base Γ : G → G with principle part Ψ : G × H → H.
Deﬁnition 3.4.4. The bundle map is called a ﬁber contraction on G × H if there
is a k ∈ [0, 1), such that, for every P ∈ G, the map Q → Ψ(P, Q) is a contraction
mapping with contraction constant k.
40

Theorem 3.4.5. [Fiber contraction theorem ] Let (G, dG ) and (H, dH ) be metric spaces and Λ a continuous ﬁber contraction on G × H over the base Γ : G → G
with principle part Ψ : G × H → H. If P and Q are unique attracting ﬁxed points
of Γ and Q → Ψ(P , Q) respectively, then (P , Q) is a unique attracting ﬁxed point of
Λ.

Now let us deﬁne the set of candidates of DP where P is C 1 , and then construct an operator which has a unique ﬁxed point as DP in this set. Let H =
C 0 (B1 , L(R2 , R)) be the space of bounded continuous functions from B1 to the linear maps from R2 to R. Let Q be any element in H, H is a Banach space equipped
with the norm
|Q| =

sup
|Qx · v|.
2 ,|v|=1
x∈B1 ,v∈R

We now deﬁne an operator Q → Ψ(P, Q) for Q ∈ H, P ∈ G. For a C 1 function P ,
we have

−1
dΓ(P )x = d(A−1
−1 (x) Cf −1 (x) ) + d(Af −1 (x) Df −1 (x) Pf −1 (x) )
f
−d(A−1
B
Px P −1 ).
f (x)
f −1 (x) f −1 (x)
Let y = f −1 (x) as before,

dΓ(P )x = {dA−1 df −1 (x)Cy + A−1 dCy df −1 (x) + d(A−1 Dy )df −1 (x)Py
y
y
y
+A−1 Dy dPy df −1 (x) − d(A−1 By )df −1 (x)(Px Py )}
y
y
−(A−1 By )(dPx Py + Px dPy df −1 (x)).
y
41

In this form, we deﬁne

Ψ(P, Q)(x) = {dA−1 df −1 (x)Cy + A−1 dCy df −1 (x)
y
y

(3.22)

+d(A−1 Dy )df −1 (x)Py + A−1 Dy Qy df −1 (x)
y
y
−d(A−1 By )df −1 (x)(Px Py )}
y
−(A−1 By )(Qx Py + Px Qy df −1 (x)).
y

Consider the bundle map Λ(P, Q) : G × H → G × H deﬁned by Λ(P, Q) =
(Γ(P ), Ψ(P, Q)). Observe that, since Γn (P0 ) = Pn , we have

Λ(P0 , dP0 ) = (Γ(P0 ), Ψ(P0 , dP0 )) = (Γ(P0 ), dΓ(P0 )) = (P1 , dP1 ).

Suppose
Λn−1 (P0 , dP0 ) = (Pn−1 , dPn−1 )
Inductively,

Λn (P0 , dP0 ) = (Γ(Pn−1 ), Ψ(Pn−1 , dPn−1 )) = (Γ(Pn−1 ), dΓ(Pn−1 ))
= (Pn , dPn ).

which converges uniformly if Λ is a ﬁber contraction, by the Fiber Contraction
Theorem. And thus dPn converges uniformly. Now let us show that Λ is a ﬁber
contraction.

42

First, we prove that for any given P ∈ G, the map Q → Ψ(P, Q) maps H to
itself. It is easy to see that Ψ(P, Q), for a given P , is continuous with respect to
x ∈ B1 , since f is C 2 . We will need to show Ψ(P, Q) is bounded. We proceed as
follows

|Ψ(P, Q)| ≤ {|dA−1 ||C| + |A−1 ||dC| + |d(A−1 D)||P | − |d(A−1 B)|P |2 }|df −1 |
+|A−1 D · Q||df −1 | + |A−1 B||Q · P + P · Q · df −1 |
≤ {|dA−1 ||C| + |A−1 ||dC| + |d(A−1 D)| + |d(A−1 B)|}|df −1 |
+|A−1 D||Q||df −1| + |A−1 B|(|Q| + |Q||df −1 |)
≤ {|dA−1 ||C| + |A−1 ||dC| + |d(A−1 D)| + |d(A−1 B)|}|df −1 | (3.23)
+{|A−1 D||df −1| + |A−1 B|(1 + |df −1 |)}|Q|

where we have used |P | ≤ 1. Now, we need to compute |df −1 | to proceed. For




 A B 
df = 

C D
we know
df −1 =





1
 D −B 

.
AD − BC
−C A

Recall the deﬁnitions of α and M from (3.10), we have

α=

1
(λ1 − ǫ)(λ2 − ǫ)

and
43

(3.24)

M=

λ1 + 2ǫ
λ1
1
≈
=
,
λ1 λ2
λ2
ǫ2 )
α(1 − α

(3.25)

We claim that

|df −1 | ≤ M.

(3.26)

2
Indeed, since |AD| > α and | BC | < ǫ , we have
α
AD

|df −1 | ≤ |





1
 D −B 
|| 
|
AD − BC
−C A

1
|(|A| + |C|)
AD − BC
1
≤
(|A| + |C|)
|AD||1 − BC |
AD
1
≤
(λ1 + ǫ + ǫ)
|AD||1 − BC |
AD
1
(λ1 + 2ǫ) = M.
≤
2
α(1 − ǫ )
α

≤ |

In addition, from (3.9) we have

{|dA−1 ||C| + |A−1 ||dC| + |d(A−1 D)| + |d(A−1 B)|}|df −1 |
K
K
K
+ K(λ2 + ǫ) +
+ Kǫ +
)
λ1 − ǫ
λ1 − ǫ
λ1 − ǫ
3
= M · K(
+ λ2 + 3ǫ).
λ1 − ǫ

≤ M(ǫK +

44

To simplify the expression, let us write

F = M · K(

3
+ λ2 + 3ǫ).
λ1 − ǫ

(3.27)

Now, going back to (3.23), we have

λ +ǫ
ǫ
|Ψ(P, Q)| ≤ F + (M 2
+
(1 + M))|Q|.
λ1 − ǫ λ1 − ǫ
Since Q ∈ H is also bounded, we have Ψ(P, Q) is bounded. So we have proved
the map Q → Ψ(P, Q) maps H to itself.
Next, we will show that the map Q → Ψ(P, Q) is a contraction mapping. For
any given P ∈ G and any Q, Q ∈ H, we know

Ψ(P, Q)x − Ψ(P, Q)x = A−1 Dy Qy df −1 (x) − A−1 Dy Qy df −1 (x)
y
y
−(A−1 By ) · (Qx · Py + Px · Qy · df −1 (x))
y
+(A−1 By ) · (Qx · Py + Px · Qy · df −1 (x))
y
= A−1 Dy (Qy − Qy )df −1 (x) − A−1 By (Qx − Qy )Py
y
y
−A−1 By Px (Qy − Qy )df −1 (x).
y

45

Since |P | ≤ 1, we have

λ +ǫ
ǫ
|Ψ(P, Q) − Ψ(P, Q)| ≤ M 2
|Q − Q| +
|Q − Q|
λ1 − ǫ
λ1 − ǫ
ǫ
+M
|Q − Q|
λ1 − ǫ
M(λ2 + 2ǫ) + ǫ
|Q − Q|.
=
λ1 − ǫ
µ
Plugging in M ≈ 1 from (3.25), we have
λ2

0<µ≈

1
< 1.
λ1

(3.28)

Assuming the contraction µ, we have proved that, for any given P ∈ G, the map
Q → Ψ(P, Q) is a contraction mapping. This implies that, for P ∈ G, the map Q →
Ψ(P , Q) has a unique ﬁxed point Q ∈ H. If we choose Λ(P, Q) = (Γ(P ), Ψ(P, Q))
as the bundle map on G × H. Then, by deﬁnition, Λ(P, Q) is a ﬁber contraction
on G × H. Since it has been proved that P is the unique attracting ﬁxed point of
Γ, then by the Fiber Contraction Theorem, (P , Q) is the unique globally attracting
ﬁxed point of Λ. It remains to show that Q = dP , so that P ∈ C 1 .
We may start with (P0 , dP0 ), where P0 ∈ C 1 . Let Γn (P0 ) = Pn , we have

Λ(P0 , dP0 ) = (Γ(P0 ), Ψ(P0 , dP0 )) = (Γ(P0 ), dΓ(P0 )) = (P1 , dP1 ).

Suppose
Λn−1 (P0 , dP0 ) = (Pn−1 , dPn−1 ).

46

Then

Λn (P0 , dP0 ) = (Γ(Pn−1 ), Ψ(Pn−1 , dPn−1 )) = (Γ(Pn−1 ), dΓ(Pn−1 ))
= (Pn , dPn ).

We know Pn converges to P uniformly, and dPn converges to Q uniformly. It is
standard that Q = dP , and thus, P is C 1 . Now we proceed to ﬁnd the size of dP .
Since

dΓ(P )x = {dA−1 df −1 (x)Cy + A−1 dCy df −1 (x) + d(A−1 Dy )df −1 (x)Py
y
y
y
−1
−d(A−1 By )df −1 (x)Px Py } + Ay Dy dPy df −1 (x)
y
−(A−1 By ) · (dPx · Py + Px · dPy · df −1 (x)).
y

In addition, we have |P | ≤ 1 and |df −1 | ≤ M, so we can write the form of the size
of |dΓ(P )| for any P ∈ G as follows
F
|dΓ(P )| ≤ {|dA−1 ||C| + |A−1 ||dC| + |d(A−1 D)||P | + |d(A−1 B)||P |2 }M
+M|A−1 D||dP | + |A−1 B|(|dP | + M|dP |)
≤ F + (M|A−1 D| + (1 + M)|A−1 B|)|dP |
≤ F + (τ1 τ2 M + (1 + M)τ1ε )|dP |
0
= F + µ|dP |

47

where
3
+ λ2 + 3ǫ) ≤ M · K(4 + 3ǫ).
λ1 − ǫ

(3.29)

=

|dΓ(Pm−1 )|

(3.30)

≤

F + µ|dPm−1 |

≤

F + µ(F + µ|dPm−2 |)

=

F + µF + µ2 |dPm−2 |

F = M · K(
For m > n we have

|dPm|

···
m−n−1
≤

F

µi + µm−n |dPn|.

i=0
So
|dP | =

F
F
+ lim µm−n |dPn | =
.
lim |dPm | ≤
m→∞
1 − µ m→∞
1−µ

(3.31)

Therefore, not only have we proved that P ∈ C 1 , but also found the explicit
formula for the C 0 and C 1 sizes of P which can be implemented in a computer.

Our next task is to ﬁnd the C 0 and C 1 sizes of πu , which projects B1 to the
x2 -axis along the horizontal foliation F u . Recall that F u consists of the integral
curves of the vector ﬁeld deﬁned in (3.17), i.e.,








 1  Yu,1 
Yu =   = 
.
Px
Yu,2

48

Let x = (x1 , x2 ), and φ(x, t) be the solution of the vector ﬁeld Yu with initial
value x ∈ B1 , and let τ (x) be the real number such that the ﬁrst coordinate of
φ(x, τ (x)) is zero. Of course, if x1 > 0, then τ (x) < 0 and if x1 < 0, then τ (x) > 0.
For simplicity, let us assume that x1 < 0. Since the ﬁrst coordinate of Yu is 1 by
deﬁnition, the maximum time τ (x) for any x ∈ B1 is ε0 .
So we can write πu as

(0, πu (x)) = φ(x, τ (x)) = (φ1 (x), φ2 (x))

with


x1 +
φ(x, t) = 
x2 +

 
 

t Y (φ (x))ds
x1 + t
0 u,1 s
 
 φ1 (x, t)
=
=

t Y (φ (x))ds
t P ds
x2 + 0 x
φ2 (x, t)
0 u,2 s

giving












0
 0   x1 + τ (x)  


=
=
.
−x1
τ (x)
πu (x)
x2 + 0
Px dt
x2 + 0
Px dt.
We have

|πu (x) − x2 | = |

−x1
0

Px dt| ≤ |x1 | · |P | ≤ ε0 ,

49

(3.32)

∂φ1
∂φ1
∂φ2
∂φ1
∂πu
= 1,
= 0,
= 0,
=
for j = 1, 2,
∂x1
∂x2
∂x2
∂xj
∂xj

τ (x) = −x1 , ∂τ (x)/∂x1 = −1, and, ∂τ (x)/∂x2 = 0.
Hence, by (3.31) and the Gronwall inequality, we have

|

−x1
∂φ ∂φ
∂πu
(x)| = |
D P (φ(x, τ (x))) · ( 1 , 2 )ds
∂x1
∂x1 ∂x1
0
∂τ
(x)P (φ(x, τ (x)))|
+
∂x1
−x1 ∂ P
∂ P ∂πu
= |
(
+
)ds − P (φ(x, τ (x)))|
∂x1 ∂x2 ∂x1
0
|x1 |
ε F
F
∂πu
|
|ds
≤ |P | + 0 +
1−µ
1 − µ ∂x1
0
ε F
F
≤ ε0 (1 +
)exp( 0 ),
1−µ
1−µ

(3.33)

when |P | ≤ ε0 in small B1 , which is possible due to the Inclination Lemma, as we
have |P | = 0 along our initial B1top . And we have

50

∂πu
(x) − 1|
∂x2
τ (x)
∂φ ∂φ
∂τ
|
D P (φ(x, τ (x))) · ( 1 , 2 )ds +
(x)P (φ(x, τ (x)))|
∂x2 ∂x2
∂x2
0
τ (x) ∂ P ∂πu
τ (x) ∂ P
(
− 1)ds +
ds|
(3.34)
|
∂x2 ∂x2
∂x2
0
0
|x1 | F
ε0 F
∂πu
+
(
)|
(x) − 1|ds
(3.35)
1−µ
1 − µ ∂x2
0
ε F
ε0 F
exp( 0 ).
(3.36)
1−µ
1−µ
|

=
=
≤
≤

This completes the proof of Theorem 3.4.1.

51

3.5

Three dimensional vector ﬁelds

Let us recall our assumptions on the vector ﬁeld introduced at the beginning. We
write x = (x1 , x2 , x3 ) ∈ R3 , and X(x) is a forward complete C 2 vector ﬁeld :

¯
¯
¯
X(x) = (X1 (x), X2 (x), X3 (x)) = (ax1 + X1 (x), bx2 + X2 (x), cx3 + X3 (x))

where

c < 2b < 0 < a,

(3.37)

¯
Xi,x (0, 0, 0) = 0 for i, j = 1, 2, 3.
j

(3.38)

and

Let
B1 = {x ∈ R3 |0 ≤ x2 ≤ ε0 , |x1 | ≤ ε0 , |x3 | ≤ ε0 }
and let | · | = supx∈B | · |. Then we have a ﬁnite constant M > 0 such that
1
¯
|Xi,x ,x (x)| ≤ M for i, j, k = 1, 2, 3. By the mean value theorem, we have some
j k
¯
small constants ǫ1 > 0 and ǫ2 > 0, such that, |Xi,x (x)| ≤ M · ε0 = ǫ2 , and
j
¯
|Xi (x)| ≤ ǫ2 · ε0 = ǫ1 for i, j = 1, 2, 3.
Let
B1top = {x ∈ B1 |x2 = ε0 },
52

+
B1 = {x ∈ B1 |x1 = ε0 },
−
B1 = {x ∈ B1 |x1 = −ε0 }.
Letting ϕt (x) be the local ﬂow of X, so we may write

ϕ(x, t) = (ϕ1 , ϕ2 , ϕ3 ) = (eat x1 + ϕ1 , ebt x2 + ϕ2 , ect x3 + ϕ3 )
¯
¯
¯

where ϕi for i = 1, 2, 3 are the higher order terms. Set ψt (x) to be the ﬂow of the
¯
corresponding linear vector ﬁeld of X, i.e,

ψt (x) = (eat x1 , ebt x2 , ect x3 )

Let f be the time-T map ϕT , for some 0 < T ≤ 1. So f : R3 → R3 is C 2
with a hyperbolic ﬁxed point at 0. Let L = Df (0), and R3 = E u ⊕ E s ⊕ E ss
be the splitting given by the eigenspaces of L. For convenience, we also write
L = (Lu , Ls , Lss ) = (Lc , Lss ) with respect to the splitting E c ⊕ E ss , where E c =
E u ⊕ E s and E u , E s and E ss correspond to the x1 , x2 and x3 axes respectively.
In addition, we assume that the local unstable and stable manifolds of f have been
u
s
straightened, i.e., Wloc (0) is in the x1 -axis and Wloc (0) is in the x2 x3 -plane inside
B1 .
Let
B1ε = {x ∈ B1 : x3 = 0}
0
B2ε = {x ∈ B1 : x1 = 0, x2 = 0}.
0

53

For the most part of this section, unless we specify the index i = 1, 2, 3, we will
always use this representation x = (x1 , x2 ) ∈ B1ε , and y = x3 ∈ B2ε , so that,
0
0

˜
˜
f (x, y) = (Lc x + f1 (x, y), Lss y + f2 (x, y)) = (f1 (x, y), f2(x, y))

(3.39)

˜
˜
where f1 (x, y) and f2 (x, y) are the nonlinear terms. In B1 , we assume f = (f1 , f2 )
satisﬁes these conditions for λ1 > 1 and 0 < λ2 , λ3 < 1:



∂f1
λ1 0 
(0) = 

∂x
0 λ2
∂f
0 < 2 (0) = λ3 < (λ2 )2 < 1
∂y

(3.40)

for i = 1, 2

˜
˜
˜
fi (0) = 0 and the ﬁrst partial derivatives fix (0) = fiy (0) = 0,

(3.41)

and

˜
˜
˜
˜
˜
|fix | ≤ ǫ, |fiy | ≤ ǫ, |fixx | ≤ M1 , |fixy | ≤ M1 , |fiyy | ≤ M1 .

(3.42)

In this section, we consider a short initial line segment in the cone of angle less
than π about the x1 -axis on the plane x2 = ε0 .
4

54

Choose small positive numbers w, ζ satisfying

ε
0 < w < ζ < 0,
2

(3.43)

and, consider the line segment

ℓ : l(s) = (l1 , l2 , l3 ) = l(0) + sv ⊂ B1
where l(0) = (0, ε0 , w),

−ζ < s < ζ,

(3.44)

 
1
 
 
v = 0 ,
 
 
θ

(3.45)

and |θ| < 1.
Let D0 =

0≤t≤T ϕt (l) ⊂ int(B1 ) which is possible when ℓ is short (i.e., ζ

is small). And let D1 be the connected component of f (D0 ) ∩ B1 . Inductively,
let Dn be the connected component of f (Dn−1 ) ∩ B1 . Deﬁne Σ =

∞ D , and
j=0 j

let Π(x1 , x2 , x3 ) = (x1 , x2 ) be the projection onto the x1 x2 −plane. Notice that Σ
invariant almost by deﬁnition: f (Σ) ∩ B1 ⊂ Σ. The next result shows that Σ is the
graph of a C 2 function g ⋆ from Π(Σ) to B2ε .
0

Theorem 3.5.1. Suppose that f, ℓ, and Σ satisfy the conditions (3.39)–(3.45) and
55

that the following inequalities hold,

λ2 − ǫ > 0 , (λ3 + ǫ)(λ2 − ǫ)−1 < 1
(λ3 + ǫ)(1 +

ǫ
)<1
λ2 − ǫ

λ3 + 2ǫ
λ3 + ǫ
λ + 2ǫ
<1
<1 ,
+ǫ 3
λ2 − 2ǫ
λ2 − 2ǫ
(λ2 − 2ǫ)2
µ=

λ3 + ǫ

(λ2 − 2ǫ)2

+

ǫ(λ3 + 2ǫ)
< 1.
(λ2 − 2ǫ)3

(3.46)
(3.47)
(3.48)

(3.49)

Then, there is a C 2 function g ⋆ : Π(Σ) → B2ε , such that g ⋆ (0, 0) = 0 and
0
Σ = {(x, g ⋆ (x))|x ∈ Π(Σ) ⊂ B1ε }.
0
Moreover, we have

|g ⋆ | ≤ ε0
|Dg ⋆| ≤
|D 2 g ⋆ | ≤

ǫ
λ2 − λ3 − 3ǫ

4M1 (λ2 + λ3 )
.
(λ2 − 2ǫ)3 (1 − µ)

(3.50)
(3.51)

(3.52)

Once we have obtained g ⋆ as in the above theorem, we consider the graph map:
h(x1 , x2 ) = (x1 , x2 , g ⋆ (x1 , x2 )). This is a C 2 diﬀeomorphism, and we can conjugate
f to f : R2 → R2 deﬁned by
56

h−1 f h(x1 , x2 ) = f (x1 , x2 ).
This means, of course, that we can replace f by the two dimensional diﬀeomorphism f .
We can also replace the ﬂow φt restricted to Σ to a two-dimensional ﬂow φt =
h−1 φt h.
The ﬁrst step in the proof of Theorem 3.5.1 is to show that the upper part D0
of Σ is a C 2 graph. That is,

Lemma 3.5.2. The set D0 is the graph

{(x1 , x2 , x3 ) : x3 = g0 (x1 , x2 ))}
of a C 2 function g0 deﬁned on Π(D0 ).
As we will show, this holds because X is C 1 close to its linear part, and the
analogous statement is true for the linear part.
Indeed, let us consider the linear vector ﬁeld L(x1 , x2 , x3 ) = (ax1 , bx2 , cx3 ).
Let ψ(t, x) = (ψ1 (t, x), ψ2 (t, x), ψ3 (t, x)) denote the ﬂow of L, and let T > 0 be
such that if x0 ∈ ℓ and |t| < T , then

bε0
.
|ψ1 (t, x0 )| <
c−a
Let
57

(3.53)

D0 = {ψ(t, x) : |t| < T, x ∈ ℓ}.
Proposition 3.5.3. Under the conditions above, the set D0 is the graph of the
(real-analytic) function

x c−a
x3 = θx1 ( 2 ) b
ε0
whose ﬁrst order partial derivatives are bounded by 1.

Proof: First, let us write down the solution of the linear ODE with initial
condition (x10 , x20 , x30 ) ∈ B1 as follows:

x1 = eat x10

(3.54)

x2 = ebt x20

(3.55)

x3 = ect x30 .

(3.56)

We want to represent x3 in terms of x1 and x2 , i.e., x3 = g0 (x1 , x2 ), the graph
of which passes through the initial line segment l. By (3.55), we have

x 1
et = ( 2 ) b .
ε0
58

(3.57)

Plugging it into (3.54), we have
x −a
x10 = x1 ( 2 ) b .
ε0
Since x30 = θx10 , together with the above equation and (3.57), solution (3.56)
becomes
x c
x −a
x c−a
x3 = ( 2 ) b · θx1 ( 2 ) b = θx1 ( 2 ) b .
ε0
ε0
ε0
Let β = c−a , which satisﬁes β > 2, because of (3.37).
b
It is straightforward to compute that

x
∂x3
= θ( 2 )β
∂x1
ε0
and
∂x3
θβx1 x2 β−1
=
( )
.
∂x2
ε0 ε0
ε
Now, we have 0 ≤ x2 ≤ ε0 , and |θ| ≤ 1. Further, (3.53) gives |x1 | < 0 , so we
β
∂x3
∂x3
get that |
| and |
| are bounded by 1, and the proposition is proved.
∂x1
∂x2
Using Lemma 3.5.5, let us prove the following lemma.

Lemma 3.5.4. Given such D0 that contains l in B1 , D0 can be represented as the
graph of a C 2 function g0 (x1 , x2 ) for (x1 , x2 ) ∈ Π(D0 ), where Π is the projection
into the x1 x2 -plane. Moreover, we have

Lip(g0 ) =

sup |Dg0 | ≤ 1
x∈Π(D0 )
59

and D 2 g0 (x) is uniformly bounded in Π(D0 ).

Proof:
Let the solution of the associated linear vector ﬁeld of X be:

ψ1 (t, l(0) + sv) = eat s
ψ2 (t, l(0) + sv) = ebt ε0
ψ3 (t, l(0) + sv) = ect (w + θs).

We have the solution of the vector ﬁeld X:

x1 (t, s) = ϕ1 (t, l(0) + sv) = ψ1 + ξ1
x2 (t, s) = ϕ2 (t, l(0) + sv) = ψ2 + ξ2
x3 (t, s) = ϕ3 (t, l(0) + sv) = ψ3 + ξ3

where ψi ’s and ξi ’s are evaluated at (t, l(0) + sv). In order to ﬁnd g0 (x1 , x2 ), we
express (t, s) as functions of (x1 , x2 ), so that we can write

g0 (x1 , x2 ) = ϕ3 (t(x1 , x2 ), l(0) + s(x1 , x2 )v).
60

This requires that the determinant of the Jacobian does not equal to zero:


∂(x1 , x2 )
x1,t x1,s 
| = det 
DET = det |

∂(t, s)
x2,t x2,s


ψ1,t + ξ1,t ψ1,s + ξ1,s 
= det 

ψ2,t + ξ2,t ψ2,s + ξ2,s


at s + ξ
at + ξ
1,t e
1,s 
 ae
= det 

bt ε + ξ
be 0
2,t 0 + ξ2,s

= (aeat s + ξ1,t )ξ2,s − (eat + ξ1,s )(bebt ε0 + ξ2,t )

We have
DET ≈ bε0 e(a+b)t = 0

(3.58)

when ξi ’s are small enough.

∂x
∂x
Here we use the notation xi,t = i and xi,s = i for i = 1, 2.
∂t
∂s

In addition, since








tx2 
∂(t, s)
1  x2,s −x1,s 
 tx
= 1
=


∂(x1 , x2 )
DET
sx1 sx2
−x2,t x1,t
we have








at − ξ
ξ2,s
−e
1 
1,s 
 tx1 tx2 

=

.
DET
bt ε − ξ
at s + ξ
sx1 sx2
−be 0
2,t ae
1,t
61

So we may compute the ﬁrst derivatives of g0 :

|

∂g0 (x1 , x2 )
∂ϕ
∂ϕ3 ∂s
∂t
| = | 3·
+
·
|
∂x1
∂t ∂x1
∂s ∂x1
(ψ3,t + ξ3,t ) · ξ2,s (ψ3,s + ξ3,s )(−bebt ε0 − ξ2,t )
= |
+
|
DET
DET
(cect (w + θs) + ξ3,t ) · ξ2,s (ect θ + ξ3,s )(−bebt ε0 − ξ2,t )
= |
+
|
DET
DET
−bθε0 e(b+c)t
| , by (3.58)
≈ |
bε0 e(a+b)t
= |θe(c−a)t | < 1

where ξi,j for i = 2, 3, j = t, s are small enough.

Similarly, we have

|

∂g0 (x1 , x2 )
∂ϕ
∂ϕ3 ∂s
∂t
| = | 3·
+
·
|
∂x2
∂t ∂x2
∂s ∂x2
(ψ3,t + ξ3,t ) · (−eat − ξ1,s ) (ψ3,s + ξ3,s )(aeat s + ξ1,t )
= |
+
|
DET
DET
(cect (w + θs) + ξ3,t ) · (−eat − ξ1,s )
= |
DET
ct θ + ξ )(aeat s + ξ )
(e
3,s
1,t
+
|
DET
−ce(a+c)t (w + θs) + a · sθe(a+c)t
≈ |
| , by (3.58)
bε0 e(a+b)t
θ(a − c)s · e(a+c)t
≈ |
| , where w is close to 0
bε0 e(a+b)t
(a − c)s (c−b)t
≤ |θ| · |
|e
<1
bε0

62

(a−c)s
where ξi,j for i = 1, 3, j = t, s are small and |
| < 1.
bε0

Since g0 is C 2 , all its second derivatives are uniformly bounded over Π(D0 ) ⊂
B1ε , the lemma is proved.
0

Lemma 3.5.5. Let φt (x) be the solution of the vector ﬁeld

¯
¯
¯
¯
X(x) = (ax1 + X1 (x), bx2 + X2 (x), cx3 + X3 (x)) = (L + X)(x),

where a > 0 > b > c and −c > a, with

¯
¯
¯
|Xi | ≤ ǫ1 , |Xi,x | ≤ ǫ2 , |D 2 X| ≤ M1 , for i, j = 1, 2, 3.
j

Let ψt (x) be the solution of the corresponding linear vector ﬁeld, then in B1 , there
exists small constants δ0 , δ and a constant M2 , such that,

|φt (x) − ψt (x)| ≤ δ0 ,

|Dφt(x) − Dψt (x)| ≤ δ,
and
|D 2 φt (x) − D 2 ψt (x)| = |D 2 φt (x)| ≤ M2 .
63

Proof: By deﬁnition, we have
t

φt (x) − ψt (x) = x +
=

t
0

0

¯
(L + X)(φs (x))ds − x −

¯
X(φs (x))ds +

t
0

t
0

Lψs (x)ds

L(φs − ψs )(x)ds.

So by the Gronwall’s inequality, we have

|φt (x) − ψt (x)| ≤

t
0

¯
|X(φs (x))|ds +

t
0

|L||φs (x) − ψs (x)|ds

≤ ǫ1 te|c|t ≤ ǫ1 T e|c|T = δ0

Next, let us compare Dφt (x) with Dψt (x), which come from the ﬁrst variational
equations of φt (x) and ψt (x). We have

˙
¯
D φt(x) = DXφ (x) · Dφt (x) = (DL + D Xφ (x) )Dφt (x).
t
t

so
˙
|D φt(x)| ≤ |Id| +

t
s=0

¯
|DL + D Xφ (x) ||Dφs (x)|ds.
s

¯
By our assumption in this lemma, we have each entry Xi,x (φs (x)), i, j = 1, 2, 3,
j
¯
of the 3 × 3 matrix D Xφ (x) is bounded by ǫ2 , thus
s
¯
¯
¯
¯
|D Xφ (x) | = max {|Xi,x | + |Xi,x | + |Xi,x |} ≤ 3ǫ2 .
s
1
2
3
i=1,2,3

64

Then by the Gronwall’s inequality we have

¯
|Dφt (x)| ≤ exp(|DL + D X|t) ≤ exp((|c| + 3ǫ2 )t).

(3.59)

Now

Dφt(x) − Dψt (x)
t
¯
(DL + D Xφ (x) )Dφs (x)ds − Id −
DL · Dψs (x)ds
s
0
0
t
t
¯
D Xφ (x) Dφs (x)ds +
DL(Dφs (x) − Dψs (x))ds.
s
0
0

= Id +
=

t

By the Gronwall’s inequality again and the inequality (3.59), we have

¯
|Dφt(x) − Dψt (x)| ≤ 3ǫ2 t exp(|DL + D X|t) exp(|DL|t)
¯
≤ 3ǫ2 t exp((2|DL| + |D X|)T )
≤ 3ǫ2 T exp((2|c| + 3ǫ2 )T )
= δ.

Finally, let us ﬁnd the size of |D 2 φt (x) − D 2 ψt (x)| = |D 2 φt (x) − 0|. We have
t
∂
∂φ(s, x)
∂2φ
(t, x) =
(ei +
DXφ (x)
ds)
s
∂xi ∂xj
∂xj
∂xi
0
t
∂φ ∂φ
∂ 2 φ(s, x)
=
D 2 Xφ (x) · (
)(
) + DXφ (x) ·
ds.
s
s
∂xi ∂xj
∂xi ∂xj
0
¯
By using (3.59) and the Gronwall’s inequality again, and since D 2 Xφ (x) = D 2 Xφ (x)
s
s

65

we have

|

∂2φ
(t, x)| ≤
∂xi ∂xj
≤

t
0

M1 exp(2s(|c| + 3ǫ2 ))ds +

t
0

(|c| + 3ǫ2 ) · |

∂ 2 φ(s, x)
|ds
∂xi ∂xj

M1
[exp(2T (|c| + 3ǫ2 )) − 1] · exp(T (|c| + 3ǫ2 ))
2(|c| + 3ǫ2 )

= M2 .

Thus the lemma is proved.

Let us be brieﬂy reminded of the deﬁnition of our surface Σ. We have D0 =
0≤t≤T ϕt (l) ⊂ int(B1 ) for our initial line segment l. By now we have shown that
D0 can be represented as a graph of a C 2 function g0 . Now let D1 be the connected
component of f (D0 ) ∩ B1 . And inductively, let Dn be the connected component of
f (Dn−1 ) ∩ B1 which is connected to Dn . We deﬁne Σ =

∞ D .
j=0 j

Theorem 3.5.6. Σ can be represented as the graph of a C 2 function g ⋆ .

We will use the same Fiber Contraction method as in the previous section. The
proof will require several steps. It involves obtaining a sequence g1 , g2 , · · · , of C 2
functions, such that

gn converges to g ⋆ uniformly on B1 ,

(3.60)

Dgn converges uniformly on B1 ,

(3.61)

66

D 2 gn converges uniformly on B1 .

(3.62)

The proof involves deﬁnitions of certain function spaces and associated mappings. Let us proceed to deﬁne our ﬁrst function space G. Here we recall that B1ε
0
is B1 ∩ x1 x2 -plane, and B2ε is B1 ∩ x3 -axis. Let
0
0
B1 = {g|g ∈ C 0 (Π(Σ), B2ε ), Lip(g) = sup
0
ξ1 =ξ2

|g(ξ1) − g(ξ2 )|
≤ 1}.
|ξ1 − ξ2 |

0
B1 is a complete metric space with respect to the supremum norm (deﬁned
above). The subspace

0
G = {g| g ∈ B1 , g(x1 , 0) = 0 f or |x1 | ≤ ε0 , g|Π(D ) = g0 }
0
is also a complete metric space equipped with the metric

d(g1 , g2 ) =

sup |g1 (x) − g2 (x)|.
x∈B1ε
0

For x ∈ Π(Σ) ∩ Π(f (Σ)), we have the following graph transform formula

Γf (g) = f2 ◦ (1, g) ◦ [f1 ◦ (1, g)]−1 .

We need to show that Γf : G → G is a well deﬁned contraction map. This will give
us a continuous function g as the unique attractive ﬁxed point of Γ.
Lemma 3.5.7. Given f satisfying the conditions at the beginning of this section, if
67

λ2 − ǫ > 0 and (λ3 + ǫ)(λ2 − ǫ)−1 ≤ 1 in B1 , then Γf (g) is well-deﬁned and maps
G to itself. If, in addition,

η = (λ3 + ǫ)(1 +

ǫ
)<1
λ2 − ǫ

then Γf (g) is a contraction mapping with contraction µ. The unique ﬁxed point of
Γf (g) is a C 0 function g ⋆ with
|g ⋆ | ≤

η
|g − g0 |.
1−η 1

Proof: To show that Γf is well-deﬁned, it is suﬃcient to show that f1 ◦ (1, g)
is invertible. Let E denote the x1 x2 −plane. For u = (u1 , u2 ) ∈ E, (L | E)(u) =
(λ1 u1 , λ2 u2 ).
Let

min(L | E) = inf |(L | E)u| > λ2 > 0.
|u|=1
For two points u = v ∈ B1ε , we have
0
68

˜
˜
|f1 ◦ (1, g)(u) − f1 ◦ (1, g)(v)| = |(L | E)(u − v) + f1 (u, gu) − f1 (v, gv)|
˜
˜
≥ min(L | E)|u − v| − |f1 (u, gu) − f1 (v, gv)|
˜
≥ λ2 |u − v| − Lip(f1 )Lip(1, g)|u − v|.

˜
Since Lip(f1 ) ≤ ǫ and Lip(1, g) ≤ 1, we have

|f1 ◦ (1, g)(u) − f1 ◦ (1, g)(v)| ≥ (λ2 − ǫ)|u − v| > 0.

(3.63)

Hence, on B1ε , f1 ◦ (1, g) is injective, and thus, invertible. That also implies
0
Lip([f1 ◦ (1, g)]−1 ) ≤ (λ2 − ǫ)−1 .

In order to have Γf (g) map G to itself, we ﬁrst prove that for x1 ∈ [0, ε0 ] we
have
Γf (g)(x1 , 0) = 0.
u
Let us be reminded that, as a part of our assumptions on f , we have Wloc (0) of f is
in the x1 -axis inside B1 . In addition, we have had Γf (g) = f2 ◦ (1, g) ◦ [f1 ◦ (1, g)]−1
being well deﬁned, i.e., f1 ◦ (1, g) is a one-to-one invertible map. So we know there
is a unique (¯1 , 0) ∈ B1ε with |¯1 | ≤ |x1 | such that
x
x
0

f1 ◦ (1, g)(¯1 , 0) = (x1 , 0)
x

69

where |¯1 | = |x1 | is achieved when x1 = 0. So
x

Γf (g)(x1 , 0) = f2 ◦ (1, g)(¯1 , 0) = f2 (¯1 , 0, g(¯1 , 0)).
x
x
x

For g ∈ G, we have g(¯1 , 0) = 0, so
x

Γf (g)(x1 , 0) = f2 (¯1 , 0, 0) = 0.
x
Next, let us prove Lip(Γf (g)) ≤ 1 as follows.
Again, for u = (u1 , u2 ), v = (v1 , v2 ) ∈ B1ε , we have
0

˜
˜
|f2 ◦ (1, g)(u) − f2 ◦ (1, g)(v)| = |λ3 g(u) − λ3 g(v) + f2 (u, gu) − f2 (u, gv)|. (3.64)

˜
Since Lip(f2 ) ≤ ǫ, Lip(1, g) ≤ 1, we have

|f2 ◦ (1, g)(u) − f2 ◦ (1, g)(v)| ≤ (λ3 + ǫ)|u − v|.

(3.65)

As we know the Lipschitz constant of the composite function of any f and g
over domain U can be computed as

Lip(f ◦ g(x)) =

sup
x,y∈U

|f (g(x)) − f (g(y))|
|x − y|

≤

sup
x,y∈U

Lip(f )|g(x) − g(y)|
= Lip(f )Lip(g)
|x − y|

70

we have
Lip(Γf (g)) ≤ (λ3 + ǫ)(λ2 − ǫ)−1 .
Under the condition that (λ3 + ǫ)(λ2 − ǫ)−1 ≤ 1, we have Lip(Γf (g)) ≤ 1, and
thus, Γf (g) maps G to itself.

Now let us show Γ is a contraction mapping in G. Let ξ1 = [f1 ◦ (1, g1 )]−1 ,
ξ2 = [f1 ◦ (1, g2 )]−1 , we have

|Γf (g1 ) − Γf (g2 )|

(3.66)

= |f2 ◦ (1, g1 ) ◦ [f1 ◦ (1, g1 )]−1 − f2 ◦ (1, g2 ) ◦ [f1 ◦ (1, g2 )]−1 |
= |f2 ◦ (1, g1 )ξ1 − f2 ◦ (1, g2 )ξ2 )|
≤ |f2 ◦ (1, g1 )ξ1 − f2 ◦ (1, g2 )ξ1 | + |f2 ◦ (1, g2 )ξ1 − f2 ◦ (1, g2 )ξ2 )|
≤ Lip(f2 )|(ξ1 , g1 ξ1 ) − (ξ1 , g2 ξ1 )| + Lip(f2 )Lip((1, g2 ))|ξ1 − ξ2 |
≤ Lip(f2 )|g1 ξ1 − g2 ξ1 | + Lip(f2 )|ξ1 − ξ2 |
= Lip(f2 )|g1 − g2 | + Lip(f2 )|ξ1 − ξ2 |.

So we need to ﬁnd |ξ1 − ξ2 |. Claim:

|ξ1 − ξ2 | ≤

ǫ
|g − g2 |.
λ2 − ǫ 1

(3.67)

Let us prove this claim. In our notations, ξ1 = [f1 ◦ (1, g1 )]−1 , ξ2 = [f1 ◦ (1, g2 )]−1 ,
−1
−1
let τ1 = ξ1 x and τ2 = ξ2 x, then we have x = ξ1 τ1 = f1 ◦ (1, g1 )τ1 , x = ξ2 τ2 =

71

f1 ◦ (1, g2 )τ2 . So
f1 ◦ (1, g1 )τ1 = f1 ◦ (1, g2 )τ2 .
As before, we write

˜
f1 (τ1 , g1 τ1 ) = Lc τ1 + f1 (τ1 , g1 τ1 ),

and
˜
f1 (τ2 , g2 τ2 ) = Lc τ2 + f1 (τ2 , g2 τ2 ).
So we have

˜
˜
Lc τ1 + f1 (τ1 , g1 τ1 ) = Lc τ2 + f1 (τ2 , g2 τ2 )
˜
˜
Lc (τ2 − τ1 ) = f1 (τ1 , g1 τ1 ) − f1 (τ2 , g2 τ2 )

then,

˜
˜
˜
˜
Lc (τ2 − τ1 ) = f1 (τ1 , g1 τ1 ) − f1 (τ2 , g1 τ2 ) + f1 (τ2 , g1 τ2 ) − f1 (τ2 , g2 τ2 )

which implies

˜
˜
˜
˜
Lc (τ2 − τ1 ) − f1 (τ1 , g1 τ1 ) + f1 (τ2 , g1 τ2 ) = f1 (τ2 , g1 τ2 ) − f1 (τ2 , g2 τ2 )
˜
˜
f1 ◦ (1, g1 )τ2 − f1 ◦ (1, g1 )τ1 = f1 (τ2 , g1 τ2 ) − f1 (τ2 , g2 τ2 ).

72

By (3.63), i.e.,

|f1 ◦ (1, g1 )τ2 − f1 ◦ (1, g1 )τ1 | ≥ (λ2 − ǫ)|τ1 − τ2 |

we have

˜
˜
(λ2 − ǫ)|τ1 − τ2 | ≤ |f1 (τ2 , g1 τ2 ) − f1 (τ2 , g2 τ2 )|
˜
≤ Lip(f1 )|(1, g1)τ2 − (1, g2 )τ2 |
≤ ǫ|g1 τ2 − g2 τ2 |
≤ ǫ|g1 − g2 |.

By our notations τ1 = ξ1 x and τ2 = ξ2 x, so

|ξ1 − ξ2 | ≤

ǫ
|g − g2 |
λ2 − ǫ 1

and the claim (3.67) is proved. Plugging (3.67) to (3.66), we have

|Γf (g1 ) − Γf (g2 )| ≤ Lip(f2 )|g1 − g2 | + Lip(f2 )

ǫ
|g − g2 |.
λ2 − ǫ 1

˜
˜
As we write f2 (x1 , x2 ) = Lss x2 + f2 (x1 , x2 ) with Lip(f2 ) ≤ ǫ, we have

Lip(f2 ) ≤ λ3 + ǫ.

Hence,
|Γf (g1 ) − Γf (g2 )| ≤ (λ3 + ǫ)(1 +
73

ǫ
)|g − g2 |.
λ2 − ǫ 1

We may observe that (λ3 + ǫ)(1 + ǫ ) is approximately equal to λ3 , which
λ2 −ǫ
is much less than 1. If the neighborhood is properly chosen, we will have

η = (λ3 + ǫ)(1 +

ǫ
) < 1.
λ2 − ǫ

So the Lemma is proved.

Thus, there is a unique globally attracting ﬁxed point g ⋆ ∈ G for the operator
Γ.
To show that g ⋆ is, in fact, C 2 , and to estimate its C 2 size, we modify the ﬁber
contraction procedure of Hirsch and Pugh familiar from invariant manifold theory
[4].
We construct two product bundles G × H and G × H × K and maps

Φ : G × H → G × H,

Ψ :G ×H×K →G ×H×K

satisfying the following properties:

1. Φ is a ﬁber contraction over Γ,
2. Ψ is a ﬁber contraction over Φ,
3. if g is any map in G which happens to be C 2 and (g, Dg, D 2g) is in G × H × K,
74

then

(a) Φ(g, Dg) = (Γ(g), DΓ(g)), and
(b) Ψ(g, Dg, D 2g) = (Γ(g), DΓ(g), D 2Γ(g)).

It will follow that, if g ∈ G is C 2 , and we set gn = Γn (¯) for each n > 0, then
¯
g
the convergence conditions (3.60), (3.61), and (3.62) will hold, and the proof that
g ⋆ is C 2 will be complete.
In the process, we will also obtain bounds for the ﬁrst and second derivatives of
g ⋆.
We have several tasks ahead.
First, we ﬁnd a C 2 initial function g in G.
¯
Recall that our original function g0 on Π(D0 ) was C 2 .
Now, choose a smooth bump function ρ : R2 → R such that ρ(x) = 1 on a
neighborhood of Π(D0 ) and ρ(x) = 0 on a neighborhood of B1bot = {(x1 , 0, 0)|0 ≤
x1 ≤ ε0 }. Then, letting ν(x) be the identically 0 function on R2 , and deﬁne g by
¯

g = ρ(x)g0 + (1 − ρ(x))ν(x).
¯
The function g is clearly C 2 and is evidently in G.
¯
Next, we proceed to deﬁne the function space H and the map Φ.
0
Let H = C1 (Π(Σ), L(R2 , R)) be the set of continuous functions H from Π(Σ)
to the linear maps from R2 to R, and |H| ≤ 1. H becomes a complete metric space
75

equipped with the metric d(H1 , H2 ) = |H1 − H2 | induced by the norm

|H| =

sup
|Hx (y)|.
2 ,|y|=1
x∈B1ε ,y∈R
0

We can see that H is the set of candidates of Dg. To get the ﬁber contraction
operator over the functional space G × H, we proceed as follows. Let u(x) = [f1 ◦
(1, g)]−1 (x). By the chain rule, we take the derivative of Γf (g) = f2 ◦ (1, g) ◦ [f1 ◦
(1, g)]−1 with respect to x, we have

Dx Γf (g) = (f2x + f2y Du(x) g)(f1x + f1y Du(x) g)−1

where the partial derivatives of f are evaluated at the point (u(x), g(u(x))). In this
form, we deﬁne

R(g, H) = (f2x + f2y Hu(x) )(f1x + f1y Hu(x) )−1

where the partial derivatives of f are evaluated at the point (u(x), g(u(x))). And
let Φ : G × H → G × H be

Φ(g, H) = (Γf (g), R(g, H)).
By construction, if g is C 1 , then

Φ(g, Dg) = (Γf (g), D(Γf (g))).

76

We shall show that Φ(g, H) = (Γf (g), R(g, H)) is a ﬁber contraction on G × H
with the maximum metric in this product space.

Lemma 3.5.8. Given f satisfying the conditions at the beginning of this section, if
the following conditions are satisﬁed in B1

λ3 + ǫ
λ + 2ǫ
λ3 + 2ǫ
< 1 and
+ǫ 3
< 1,
λ2 − 2ǫ
λ2 − 2ǫ
(λ2 − 2ǫ)2

(3.68)

then the function g ⋆ in Lemma 3.5.7 is C 1 with

|Dg ⋆ | ≤

ǫ
.
λ2 − λ3 − 3ǫ

(3.69)

Proof: Let us prove this lemma by ﬁrst showing that Φ(g, H) maps G × H to
itself. Since Γf (g) : G → G, we only need to show R(g, H) : G × H → H. As we can
write
−1
f1x + f1y Hu(x) = f1x (I + f1x f1y Hu(x) ),
77

for |H| ≤ 1, we have

−1
|(f1x + f1y Hu(x) )−1 | = |{f1x (I + f1x f1y Hu(x) )}−1 |
−1
−1
= |(I + f1x f1y Hu(x) )−1 f1x |
−1
−1
≤ |(I + f1x f1y Hu(x) )−1 ||f1x |
1
−1
|f1x |
−1 f H
min(I + f1x 1y u(x) )
1
−1
|f1x |
≤
−1 f H
1 − |f1x 1y u(x) |
−1
|f1x |
≤
−1
1 − |f1x ||f1y |
1
=
.
−1
|f1x |−1 − |f1y |
=

−1
Since |f1y | ≤ ǫ and |f1x |−1 = min(f1x ) ≥ λ2 − ǫ, we have
1
|(f1x + f1y Hu(x) )−1 | ≤
.
λ2 − 2ǫ
Moreover,

|R(g, H)| ≤ |f2x + f2y Hu(x) ||(f1x + f1y Hu(x) )−1 |
≤ (ǫ + (λ3 + ǫ)|Hu(x) |)
≤

(3.70)

1
λ2 − 2ǫ

λ3 + 2ǫ
.
λ2 − 2ǫ

λ3 +2ǫ
< 1 in some neighborhood of 0, we have proved that Φ(g, H)
λ2 −2ǫ
maps G × H to itself.
If we have

78

Now we shall show Φ(g, H) is a ﬁber contraction map. Let R2 (g, H)x(y) =
(f2x + f2y Hu(x) )(y), R1 (g, H)x(y) = ((f1x + f1y Hu(x) )−1 )(y), both of which are
linear operators in y. So we may write R(g, H)x = R2 (g, H) ◦ R1(g, H)x, where the
partial derivatives of f are evaluated at the point (u(x), g(u(x))). By what we have
just computed, we know

1
|R1 (g, H)| = |(f1x + f1y Hu(x) )−1 | ≤
λ2 − 2ǫ

(3.71)

|R2 (g, H)| = |f2x + f2y Hu(x) | ≤ λ3 + 2ǫ.

(3.72)

and

And by the linearity of R1 and R2 , we have

|R(g, H1)x − R(g, H2 )x |
≤ |R2 (g, H1) ◦ R1 (g, H1)x − R2 (g, H2) ◦ R1 (g, H1)x |
+|R2 (g, H2) ◦ R1 (g, H1 )x − R2 (g, H2) ◦ R1 (g, H2 )x |
≤ |R2 (g, H1) − R2 (g, H2 )||R1 (g, H1 )x |
+|R2 (g, H2) ◦ (R1 (g, H1 )x − R1 (g, H2 )x )|
1
≤ |f2y H2 − f2y H1 |
λ2 − 2ǫ
+|R2 (g, H2)||R1 (g, H1)x − R1 (g, H2)x |
1
≤ (λ3 + ǫ)|H2 − H1 |
λ2 − 2ǫ
+(λ3 + 2ǫ)|R1 (g, H1)x − R1 (g, H2)x |.

79

Here |R1 (g, H1 )x − R1 (g, H2 )x | = |(f1x + f1y H1 )−1 − (f1x + f1y H2 )−1 | which
involves inverse functions. So let us consider this trick:

−1
−1
−1
−1
F1 − F2 = F1 (F2 − F1 )F2

which implies
−1
−1
−1 −1
|F1 − F2 | ≤ |F1 ||F2 ||F1 − F2 |
By the above inequality, we have

|R1 (g, H1 )x − R1 (g, H2 )x |
= |(f1x + f1y H1 )−1 − (f1x + f1y H2 )−1 |
≤ |(f1x + f1y H1 )−1 ||(f1x + f1y H2 )−1 ||(f1x + f1y H1 ) − (f1x + f1y H2 )|
1
)2 |f1y H1 − f1y H2 |
λ2 − 2ǫ
1
≤ (
)2 ǫ|H1 − H2 |.
λ2 − 2ǫ
≤ (

Therefore,

λ +ǫ
λ + 2ǫ
|R(g, H1 ) − R(g, H2)| ≤ ( 3
+ǫ 3
)|H1 − H2 |.
λ2 − 2ǫ
(λ2 − 2ǫ)2
By (3.68), we have

µ=

λ3 + ǫ
λ + 2ǫ
+ǫ 3
< 1.
λ2 − 2ǫ
(λ2 − 2ǫ)2

Hence, Φ is indeed a ﬁber contraction, and the lemma is proved.
80

Moreover, we can ﬁnd the size of Dg ⋆ . By a similar argument in the previous
section, we start from (3.70), which says

|Dgn| = |R(gn−1 , Dgn−1 )| ≤ (

λ +ǫ
ǫ
+ 3
|Dgn−1 |)
λ2 − 2ǫ λ2 − 2ǫ

Let
J=

ǫ
,
λ2 − 2ǫ

and N =

λ3 + ǫ
,
λ2 − 2ǫ

we have
|Dgn| ≤ J + N|Dgn−1 |.
Similar to the estimate in the inequality (3.30), we have

|Dg ⋆ | ≤

ǫ
J
=
.
1−N
λ2 − λ3 − 3ǫ

In order to have g ⋆ in C 2 , we need another ﬁber contraction to compute
D 2 g ⋆ . We still use the same notation and deﬁnition of G and H. Recall that
0
H = C1 (Π(Σ), L(R2 , R)), now we deﬁne another functional space K to be the set
of candidates of D 2 g:
K = C 0 (Π(Σ) × (R2 )2 , R)
which is the set of bounded continuous functions K from Π(Σ) to the bilinear maps
from R2 to R. K is a complete metric space equipped with the metric d(K1 , K2 ) =

81

|K1 − K2 | induced by the norm

|K| =

sup
|Kx (v1 , v2 )|.
2
x∈B1ε ,|v1 |=1,|v2 |=1,v1 ,v2 ∈R
0

Similar to the previous ﬁber contraction constructions, we ﬁrst use the chain
2
rule to ﬁnd Dx Γf (g) which gives the form of our bundle map. As we know
Γf (g) = f2 ◦ (1, g) ◦ [f1 ◦ (1, g)]−1 .
Let u(x) = [f1 ◦ (1, g)]−1 , we have

Dx Γf (g) = f2x (u(x), g(u(x)))Du(x) + f2y (u(x), g(u(x)))Dg(u(x))Du(x).

Now we take the derivative of the above with respect to x one more time to obtain
the following formula, all the derivatives of f are evaluated at (u(x), g(u(x))):

2
Dx Γf (g) = f2xx (Du(x))2 + f2xy Dg(u(x))(Du(x))2 + f2x D 2 u(x) (3.73)
+{f2yx Du(x) + f2yy Dg(u(x))Du(x)}Dg(u(x))Du(x)
+f2y {D 2 g(u(x))(Du(x))2 + Dg(u(x))D 2 u(x)}.

Now let us explicitly compute Du(x) and D 2 u(x). We have

Du(x) = [f1x (u(x), g(u(x))) + f1y (u(x), g(u(x)))Dg(u(x))]−1

82

and with all the derivatives of f evaluated at (u(x), g(u(x))):

D 2 u(x) = −Du(x){f1xx (Du(x))2 + 2f1xy Dg(u(x))(Du(x))2
+f1yy (Dg(u(x)))2(Du(x))2 + f1y (Du(x))2 D 2 g(u(x))}.

In this form, we deﬁne

D1 = D1 (u, H)x = [f1x + f1y Hu(x) ]−1

and

D2 = D2 (u, H, K)x = −D1 {f1xx (D1 )2 + 2f1xy Hu(x) (D1 )2
+f1yy (Hu(x) )2 (D1 )2 + f1y (D1 )2 Ku(x) }.

So we deﬁne the following functional derived from (3.73)

¯
R(g, H, K) = f2xx (D1 )2 + f2xy Hu(x) (D1 )2 + f2yxD1 Hu(x) D1
+f2yy Hu(x) D1 Hu(x) D1 + f2x D2
+f2y Ku(x) (D1 )2 + f2y Hu(x) D2

where the partial derivatives of f are evaluated at the point (u(x), g(u(x))), and
only the last three terms contain K. To get the ﬁber contraction operator in the
functional space G × H × K, we let Ψ : G × H × K → G × H × K satisfy this form

Ψ(g, Dg, D 2g) = (Γf (g), DΓf (g), D 2 Γf (g)).
83

So we write
¯
Ψ(g, H, K) = (Γf (g), R(g, H), R(g, H, K))
and we shall show that Ψ(g, H, K) is a ﬁber contraction on G × H × K with the
maximum metric in this product space.

Lemma 3.5.9. Given f satisfying the conditions at the beginning of this section.
In addition, each one of the second derivatives of f is bounded by some constant
M1 > 0. If in B1 the following condition is satisﬁed

µ=

λ3 + ǫ

(λ2 − 2ǫ)2

+

ǫ(λ3 + 2ǫ)
< 1,
(λ2 − 2ǫ)3

(3.74)

then g ⋆ is C 2 with

|D 2 g ⋆ | ≤

4M1 (λ2 + λ3 )
.
(λ2 − 2ǫ)3 (1 − µ)

(3.75)

Proof: Let us ﬁrst prove that Ψ maps G × H × K to itself, and it is clear that
¯
we only need to show, for given g and H, R(g, H, K) is still in K. By our deﬁnition

D1 = [f1x + f1y Hu(x) ]−1

and what we have obtained in inequalities (3.71), we have

|D1 | = |(f1x + f1y Hu(x) )−1 | ≤
84

1
.
λ2 − 2ǫ

Also we have

|D2 | ≤ |D1 |{|f1xx ||D1 |2 + 2|f1xy ||D1 |2
+|f1yy ||D1 |2 + |f1y ||D1 |2 |K|}
= |D1 |3 (|f1xx | + 2|f1xy | + |f1yy | + |f1y ||K|)
≤

4M1 + ǫ|K|
.
(λ2 − 2ǫ)3

In addition, since |H| ≤ 1 and each one of the second derivatives of f is bounded
by some constant M1 , we have
¯
|R(g, H, K)| ≤ |f2xx ||D1 |2 + 2|f2xy ||D1 |2 + |f2yy ||D1 |2

(3.76)

+(|f2x | + |f2y |)|D2 | + |f2y ||K||D1 |2
4M1
4M1 + ǫ|K| (λ3 + ǫ)|K|
+ (λ3 + 2ǫ)
+
2
(λ2 − 2ǫ)
(λ2 − 2ǫ)3
(λ2 − 2ǫ)2
4M1
4M1 (λ3 + 2ǫ)
=
+
(λ2 − 2ǫ)2
(λ2 − 2ǫ)3
(λ3 + ǫ)
ǫ(λ3 + 2ǫ)
+{
+
}|K|.
(λ2 − 2ǫ)3 (λ2 − 2ǫ)2

≤

¯
Since K is also bounded, we know R(g, H, K) is bounded. Hence, we have proved Ψ
maps G × H × K to itself. Now let us show the bundle map Ψ is a ﬁber contraction
¯
on G × H × K. By deﬁnition, what we need to show is the map K → R(g, H, K) is

85

a contraction mapping. Again, from

¯
R(g, H, K) = f2xx (D1 )2 + f2xy Hu(x) (D1 )2 + f2yxD1 Hu(x) D1
+f2yy Hu(x) D1 Hu(x) D1 + f2y Ku(x) (D1 )2
+(f2x + f2y Hu(x) )D2

where

D2 = −D1 {f1xx (D1 )2 + 2f1xy Hu(x) (D1 )2 + f1yy (Hu(x) )2 (D1 )2 }
−D1 f1y (D1 )2 Ku(x)

¯
we notice that only the last two terms of R(g, H, K) involve K. We may have the
terms that do not contain K canceled if we perform the following substraction, i.e.,

¯
¯
R(g, H, K1 ) − R(g, H, K2)
= f2y K1 (D1 )2 − f2y K2 (D1 )2
+(f2x + f2y H){−D1 f1y (D1 )2 K1 + D1 f1y (D1 )2 K2 }.

86

By using |H| ≤ 1, we have

¯
¯
|R(g, H, K1) − R(g, H, K2 )|
≤ |f2y K1 (D1 )2 − f2y K2 (D1 )2 |
+|f2x + f2y H||D1 f1y (D1 )2 K1 − D1 f1y (D1 )2 K2 |
≤ (λ3 + ǫ)|K1 (D1 )2 − K2 (D1 )2 |
(λ3 + 2ǫ)|D1 f1y (D1 )2 ||K1 − K2 |
≤ (λ3 + ǫ)|K1 − K2 ||D1 |2
(λ3 + 2ǫ)ǫ|D1 |3 |K1 − K2 |
≤ (

λ3 + ǫ

(λ2 − 2ǫ)2

+

ǫ(λ3 + 2ǫ)
) |K1 − K2 |.
(λ2 − 2ǫ)3

µ
We may observe that

µ=

λ3 + ǫ

(λ2 − 2ǫ)2

+

ǫ(λ3 + 2ǫ)
λ3
≈
.
3
(λ2 − 2ǫ)
(λ2 )2

If close enough to the saddle ﬁxed point and the strong contraction is much smaller
than the weak contraction, we may well have µ < 1. So under the condition that

λ3 + ǫ

(λ2 − 2ǫ)2

+

ǫ(λ3 + 2ǫ)
<1
(λ2 − 2ǫ)3

¯
we have proved that the map K → R(g, H, K) is a contraction mapping. Moreover,
we can ﬁnd the size of D 2 g ⋆ by essentially the same argument used in the previous
section about the iteration (3.70). So we do not repeat the full argument here, the

87

complete details of which has been shown in the previous section. Here we start
with the following estimate, derived from (3.76), which says

¯
|R(g, H, K)| ≤

4M1 (λ3 + 2ǫ)
(λ3 + ǫ)
4M1
ǫ(λ3 + 2ǫ)
+
+
} |K|.
+{
2
3
3 (λ − 2ǫ)2
(λ2 − 2ǫ)
(λ2 − 2ǫ)
(λ2 − 2ǫ)
2
J
N

Let
J=

4M1
4M1 (λ3 + 2ǫ)
4M1 (λ2 + λ3 )
+
=
2
3
(λ2 − 2ǫ)
(λ2 − 2ǫ)
(λ2 − 2ǫ)3

and from the above
µ=

λ3 + ǫ
ǫ(λ3 + 2ǫ)
+
< 1.
(λ2 − 2ǫ)3 (λ2 − 2ǫ)2

Since K’s are candidates of D 2 g ⋆ , the above inequality gives us a recursive
relation as

¯
|D 2 gn | = |R(gn−1 , Dgn−1 , D 2 gn−1 )| ≤ J + µ|D 2 gn−1 |.

Then we have the estimate similar to the one in the inequality (3.30), i.e.,

|D 2 g ⋆ | ≤

4M1 (λ2 + λ3 )
J
=
1−µ
(λ2 − 2ǫ)3 (1 − µ)

.

88

3.6

Straightening Invariant manifolds of a three
dimensional map

In this section, we ﬂatten out the local stable and unstable manifolds of a C 2 map
f : R3 → R3 near its saddle ﬁxed point at 0. We will make a C 2 transformation of
the neighborhood so that the local invariant manifolds of f are ﬂattened.

First of all, let us state the Hadamard-Perron Theorem (Stable manifold Theorem).
Theorem 3.6.1. Let f : M → M be a C r diﬀeomorphism of a smooth manifold M,
u
s
and let p be a hyperbolic ﬁxed point of f with associated splitting Tp M = Ep ⊕ Ep .
s
then W s (p) is a C r injectively immersed copy of Ep and W u (p) is a C r injectively
u
s
immersed copy of Ep . Moreover, W s (p) is tangent at p to Ep and W u (p) is tangent
u
at p to Ep .
Now let us start by assuming f : R3 → R3 is C 2 and satisﬁes the following
conditions. Let L = Df (0) = (Lu , Ls ), and R2 = E u ⊕ E s be the splitting
given by the hyperbolicity of L, where E u and E s correspond to the x1 −axis and
x2 x3 −plane respectively. So we write
˜
˜
f (x, y) = (Lu x + f1 (x, y), Ls y + f2 (x, y)) = (f1 (x, y), f2 (x, y))

˜
˜
where f1 and f2 are the higher-order nonlinear terms of f . And let Bε0 be the box
around the origin of radius ε0 , B1ε = Bε0 ∩x1 -axis, and B2ε = Bε0 ∩x2 x3 -plane.
0
0
89

And we assume f = (f1 , f2 , f3 ) satisﬁes the following conditions in Bε0 :
1. Lu = λ1 > 1, Ls = (λ2 , λ3 ), 0 < λ3 < λ2 < 1;
˜
˜
2. fi (0) = 0 and the ﬁrst partial derivatives fix (0) = 0, for i, j = 1, 2, 3;
j
˜
˜
3. |fix | ≤ ǫ and |fix x | ≤ M2 , for i, j, k = 1, 2, 3.
j
j k
u
We wish to ﬁnd a C 2 function g : B1ε → B2ε whose graph is Wloc (0) and is
0
0
s
invariant by f . In order to ﬁnd another function whose graph is Wloc (0), one may
set g : B2ε → B1ε . Since the work are essentially the same, we only state the
0
0
u
results of g that corresponds to Wloc (0). By using the same graph transform method
appeared in the previous section, it can be shown that there is such a function g
and it is C 2 . As before, we set

Γf (g) = f2 ◦ (1, g) ◦ [f1 ◦ (1, g)]−1

and
G = {g : B1ε → B2ε | g(0) = 0, Lip(g) ≤ 1}
0
0
which is a complete metric space with the metric

d(g1 , g2 ) =

sup |g1 (x) − g2 (x)|.
x∈B1ε
0

We start with g0 ≡ 0, then g1 = Γf (g0 ), and inductively, gn = Γf (gn−1 ), the
sequence of which converges to some function g ⋆ ∈ G. Since the proof is essentially
the same as is shown in the previous section, we do not repeat the work here. We
90

will adapt a few constants, and state the lemmas and theorem accordingly.

As in the previous section,

˜
˜
f (x, y) = (Lc x + f1 (x, y), Lss y + f2 (x, y)) = (f1 (x, y), f2 (x, y))

with m(Lc ) be its weak contraction eigenvalue, and |Lss | be its strong contraction
eigenvalue.

In this section, we have

˜
˜
f (x, y) = (Lu x + f1 (x, y), Ls y + f2 (x, y)) = (f1 (x, y), f2 (x, y))

where
m(Lu ) = Lu = λ1 ,

|Ls | = λ2 .

Hence, we may adapt Lemma 3.5.7 to have

Lemma 3.6.2. Given f satisfying the conditions at the beginning of this section, if
λ1 − ǫ > 0 and (λ2 + ǫ)(λ1 − ǫ)−1 ≤ 1 in Bε0 , then Γf (g) is well-deﬁned and maps
G to itself. If, in addition,

η = (λ2 + ǫ)(1 +

ǫ
)<1
λ1 − ǫ

then Γf (g) is a contraction mapping with contraction µ. The unique ﬁxed point of
91

Γf (g) is a C 0 function g ⋆ with
|g ⋆ | ≤

η
|g − g0 |.
1−η 1

We get the following Lemma based on Lemma 3.5.8.
Lemma 3.6.3. Given f satisfying the conditions at the beginning of this section, if
the following conditions are satisﬁed in Bε0
λ2 + 2ǫ
λ2 + ǫ
λ + 2ǫ
<1
< 1 and
+ǫ 2
λ1 − 2ǫ
λ1 − 2ǫ
(λ1 − 2ǫ)2
then the function g ⋆ in Lemma 3.5.7 is C 1 with

|Dg ⋆ | ≤

ǫ
.
λ1 − λ2 − 3ǫ

Subsequently, we adapt Lemma 3.5.9 into
Lemma 3.6.4. Given f satisfying the conditions at the beginning of this section,
in addition, each one of the second derivatives of f is bounded by some constant
M2 > 0. If in Bε0 the following condition is satisﬁed

µ=

λ2 + ǫ

(λ1 − 2ǫ)2

+

ǫ(λ2 + 2ǫ)
<1
(λ1 − 2ǫ)3

then we have g ⋆ is C 2 with

|D 2 g ⋆ | ≤

4M2 (λ1 + λ2 )
(λ1 − 2ǫ)3 (1 − µ)
92

.
In summary, analogous to Theorem 3.5.1 we conclude the theorem as follows.
Theorem 3.6.5. Given that f satisﬁes the above conditions, if in Bε0 we have the
following conditions satisﬁed

λ1 − ǫ > 0 , (λ2 + ǫ)(λ1 − ǫ)−1 ≤ 1

η = (λ2 + ǫ)(1 +

ǫ
)<1
λ1 − ǫ

λ2 + 2ǫ
λ2 + ǫ
λ + 2ǫ
<1 ,
+ǫ 2
<1
λ1 − 2ǫ
λ1 − 2ǫ
(λ1 − 2ǫ)2
µ=

λ2 + ǫ

(λ1 − 2ǫ)2

+

ǫ(λ2 + 2ǫ)
<1
(λ1 − 2ǫ)3

u
then we have a unique C 2 function g ⋆ : B1ε → B2ε , whose graph is Wloc (0) of
0
0
f . Moreover, we have

|g ⋆ | ≤ ε0 |Dg ⋆|
|Dg ⋆| ≤
|D 2 g ⋆ | ≤

ǫ
λ1 − λ2 − 3ǫ

4M2 (λ1 + λ2 )
(λ1 − 2ǫ)3 (1 − µ)

.
After we have obtained C 2 parameterizations of both local stable and unstable
manifolds of f at 0, which are now denoted by gs , and gu respectively. Then, in
s
u
general, Wloc (0) (respectively Wloc (0)) is not in E s (respectively E u ). However,
93

we will show there is a C 2 diﬀeomorphism f which conjugates to f and has both
local stable and unstable manifolds straightened. Let us consider the map

ρ : Bε0 → V ⊂ R3

with
ρ(x, y) = (x − gs (y), y − gu (x)).

(3.77)

By this deﬁnition, ρ is C 2 , ρ(0) = 0 as gs (0, 0) = gu (0) = 0, and Dρ(0) is the
identity



 

′
−gs (0) 1 0
 1
Dρ(0) = 
=

′
−gu (0)
1
0 1

′
′
where we used the properties of the local manifolds that gs (0) = gu (0) = 0. Thus
ρ is a diﬀeomorphism in some neighborhood of 0 ∈ R3 by the inverse function
′
theorem. In Bε0 ∩ f −1 (Bε0 ), such that ρ : Bε0 ∩ f −1 (Bε0 ) → V , let
f = ρf ρ−1 .

(3.78)

′
Then f : V → V is a C 2 diﬀeomorphism, with f (0) = 0 and D f (0) = L. Moreover, in Bε0 the local stable (respectively, unstable) manifold of f is in E s , i.e.,
x2 x3 −plane (respectively E u , i.e., x1 -axis). We can verify that property as follows:

ρ(gs (y), y) = (gs (y) − gs (y), y − gu (gs (y))) = (0, y − gu (gs (y)))

94

and
ρ(x, gu (x)) = (x − gs (gu (x)), gu (x) − gu (x)) = (x − gs (gu (x)), 0)
Thus, up to a smooth conjugacy, the local manifolds of f is straightened.

95

BIBLIOGRAPHY

96

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