MATH 8430
Fundamental Theory
of
Ordinary Differential Equations
Lecture Notes
Julien Arino
Department of Mathematics
University of Manitoba
Fall 2006
Contents
1 General theory of ODEs
1.1 ODEs, IVPs, solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 Ordinary differential equation, initial value problem . . . . . . . . . .
1.1.2 Solutions to an ODE . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.3 Geometric interpretation . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Existence and uniqueness theorems . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 Successive approximations . . . . . . . . . . . . . . . . . . . . . . . .
1.2.2 Local existence and uniqueness – Proof by fixed point . . . . . . . . .
1.2.3 Local existence and uniqueness – Proof by successive approximations
1.2.4 Local existence (non Lipschitz case) . . . . . . . . . . . . . . . . . . .
1.2.5 Some examples of existence and uniqueness . . . . . . . . . . . . . .
1.3 Continuation of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.1 Maximal interval of existence . . . . . . . . . . . . . . . . . . . . . .
1.3.2 Maximal and global solutions . . . . . . . . . . . . . . . . . . . . . .
1.4 Continuous dependence on initial data, on parameters . . . . . . . . . . . . .
1.5 Generality of first order systems . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Generality of autonomous systems . . . . . . . . . . . . . . . . . . . . . . . .
1.7 Suggested reading, Further problems . . . . . . . . . . . . . . . . . . . . . .
3
3
3
4
8
9
9
10
13
16
21
24
27
28
29
32
34
34
2 Linear systems
2.1 Existence and uniqueness of solutions . . . . . .
2.2 Linear systems . . . . . . . . . . . . . . . . . .
2.2.1 The vector space of solutions . . . . . .
2.2.2 Fundamental matrix solution . . . . . .
2.2.3 Resolvent matrix . . . . . . . . . . . . .
2.2.4 Wronskian . . . . . . . . . . . . . . . . .
2.2.5 Autonomous linear systems . . . . . . .
2.3 Affine systems . . . . . . . . . . . . . . . . . . .
2.3.1 The space of solutions . . . . . . . . . .
2.3.2 Construction of solutions . . . . . . . . .
2.3.3 Affine systems with constant coefficients
2.4 Systems with periodic coefficients . . . . . . . .
2.4.1 Linear systems: Floquet theory . . . . .
35
35
36
37
38
41
43
43
46
46
46
47
48
48
i
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Fund. Theory ODE Lecture Notes – J. Arino
ii
CONTENTS
. . . .
. . . .
linear
. . . .
54
3 Stability of linear systems
3.1 Stability at fixed points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Affine systems with small coefficients . . . . . . . . . . . . . . . . . . . . . .
57
57
57
4 Linearization
4.1 Some linear stability theory . . .
4.2 The stable manifold theorem . . .
4.3 The Hartman-Grobman theorem .
4.4 Example of application . . . . . .
4.4.1 A chemostat model . . . .
4.4.2 A second example . . . . .
.
.
.
.
.
.
63
63
65
69
74
74
75
.
.
.
.
.
79
79
81
82
84
84
2.5
2.4.2 Affine systems: the Fredholm alternative . . . .
Further developments, bibliographical notes . . . . . .
2.5.1 A variation of constants formula for a nonlinear
component . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
5 Exponential dichotomy
5.1 Exponential dichotomy . . . . . . . .
5.2 Existence of exponential dichotomy .
5.3 First approximate theory . . . . . . .
5.4 Stability of exponential dichotomy . .
5.5 Generality of exponential dichotomy
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. . . .
. . . .
system
. . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. . .
. . .
with
. . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
a
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
51
54
References
85
A Definitions and results
A.1 Vector spaces, norms . . . . . . . . . . . . .
A.1.1 Norm . . . . . . . . . . . . . . . . .
A.1.2 Matrix norms . . . . . . . . . . . . .
A.1.3 Supremum (or operator) norm . . . .
A.2 An inequality involving norms and integrals
A.3 Types of convergences . . . . . . . . . . . .
A.4 Asymptotic Notations . . . . . . . . . . . .
A.5 Types of continuities . . . . . . . . . . . . .
A.6 Lipschitz function . . . . . . . . . . . . . . .
A.7 Gronwall’s lemma . . . . . . . . . . . . . . .
A.8 Fixed point theorems . . . . . . . . . . . . .
A.9 Jordan normal form . . . . . . . . . . . . . .
A.10 Matrix exponentials . . . . . . . . . . . . . .
A.11 Matrix logarithms . . . . . . . . . . . . . . .
A.12 Spectral theorems . . . . . . . . . . . . . . .
87
87
87
87
87
88
88
89
89
90
91
93
94
96
97
99
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
B Problem sheets
Homework sheet 1 – 2003
Homework sheet 2 – 2003
Homework sheet 3 – 2003
Homework sheet 4 – 2003
Final examination – 2003
Homework sheet 1 – 2006
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
iii
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
101
103
113
117
125
137
145
Introduction
This course deals with the elementary theory of ordinary differential equations. The word
elementary should not be understood as simple. The underlying assumption here is that, to
understand the more advanced topics in the analysis of nonlinear systems, it is important
to have a good understanding of how solutions to differential equations are constructed.
If you are taking this course, you most likely know how to analyze systems of nonlinear
ordinary differential equations. You know, for example, that in order for solutions to a
system to exist and be unique, the system must have a C 1 vector field. What you do not
necessarily know is why that is. This is the object of Chapter 1, where we consider the
general theory of existence and uniqueness of solutions. We also consider the continuation
of solutions as well as continuous dependence on initial data and on parameters.
In Chapter 2, we explore linear systems. We first consider homogeneous linear systems,
then linear systems in full generality. Homogeneous linear systems are linked to the theory
for nonlinear systems by means of linearization, which we study in Chapter 4, in which
we show that the behavior of nonlinear systems can be approximated, in the vicinity of
a hyperbolic equilibrium point, by a homogeneous linear system. As for autonomous systems, nonautonomous nonlinear systems are linked to a linearized form, this time through
exponential dichotomy, which is explained in Chapter 5.
1
Chapter 1
General theory of ODEs
We begin with the general theory of ordinary differential equations (ODEs). First, we define
ODEs, initial value problems (IVPs) and solutions to ODEs and IVPs in Section 1.1. In
Section 1.2, we discuss existence and uniqueness of solutions to IVPs.
1.1
1.1.1
ODEs, IVPs, solutions
Ordinary differential equation, initial value problem
Definition 1.1.1 (ODE). An nth order ordinary differential equation (ODE) is a functional
relationship taking the form
d2
dn
d
F t, x(t), x(t), 2 x(t), . . . , n x(t) = 0,
dt
dt
dt
that involves an independent variable t ∈ I ⊂ R, an unknown function x(t) ∈ D ⊂ Rn of
the independent variable, its derivative and derivatives of order up to n. For simplicity, the
time dependence of x is often omitted, and we in general write equations as
F t, x, x0 , x00 , . . . , x(n) = 0,
(1.1)
where x(n) denotes the nth order derivative of x. An equation such as (1.1) is said to be in
general (or implicit) form.
An equation is said to be in normal (or explicit) form when it is written as
x(n) = f t, x, x0 , x00 , . . . , x(n−1) .
Note that it is not always possible to write a differential equation in normal form, as it can
be impossible to solve F (t, x, . . . , x(n) ) = 0 in terms of x(n) .
Definition 1.1.2 (First-order ODE). In the following, we consider for simplicity the more
restrictive case of a first-order ordinary differential equation in normal form
x0 = f (t, x).
3
(1.2)
4
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
Note that the theory developed here holds usually for nth order equations; see Section 1.5.
The function f is assumed continuous and real valued on a set U ⊂ R × Rn .
Definition 1.1.3 (Initial value problem). An initial value problem (IVP) for equation (1.2)
is given by
x0 = f (t, x)
(1.3)
x(t0 ) = x0 ,
where f is continuous and real valued on a set U ⊂ R × Rn , with (t0 , x0 ) ∈ U.
Remark – The assumption that f be continuous can be relaxed, piecewise continuity only is
needed. However, this leads in general to much more complicated problems and is beyond the
scope of this course. Hence, unless otherwise stated, we assume that f is at least continuous. The
function f could also be complex valued, but this too is beyond the scope of this course.
◦
Remark – An IVP for an nth order differential equation takes the form
x(n) = f (t, x, x0 , . . . , x(n−1) )
(n−1)
x(t0 ) = x0 , x0 (t0 ) = x00 , . . . , x(n−1) (t0 ) = x0
,
i.e., initial conditions have to be given for derivatives up to order n − 1.
◦
We have already seen that the order of an ODE is the order of the highest derivative
involved in the equation. An equation is then classified as a function of its linearity. A linear
equation is one in which the vector field f takes the form
f (t, x) = a(t)x(t) + b(t).
If b(t) = 0 for all t, the equation is linear homogeneous; otherwise it is linear nonhomogeneous.
If the vector field f depends only on x, i.e., f (t, x) = f (x) for all t, then the equation is
autonomous; otherwise, it is nonautonomous. Thus, a linear equation is autonomous if
a(t) = a and b(t) = b for all t. Nonlinear equations are those that are not linear. They too,
can be autonomous or nonautonomous.
Other types of classifications exist for ODEs, which we shall not deal with here, the
previous ones being the only one we will need.
1.1.2
Solutions to an ODE
Definition 1.1.4 (Solution). A function φ(t) (or φ, for short) is a solution to the ODE
(1.2) if it satisfies this equation, that is, if
φ0 (t) = f (t, φ(t)),
for all t ∈ I ⊂ R, an open interval such that (t, φ(t)) ∈ U for all t ∈ I.
The notations φ and x are used indifferently for the solution. However, in this chapter,
to emphasize the difference between the equation and its solution, we will try as much as
possible to use the notation x for the unknown and φ for the solution.
Fund. Theory ODE Lecture Notes – J. Arino
1.1. ODEs, IVPs, solutions
Definition 1.1.5 (Integral form of the solution). The function
Z t
φ(t) = x0 +
f (s, φ(s))ds
5
(1.4)
t0
is called the integral form of the solution to the IVP (1.3).
Let R = R((t0 , x0 ), a, b) be the domain defined, for a > 0 and b > 0, by
R = {(t, x) : |t − t0 | ≤ a, kx − x0 k ≤ b} ,
where k k is any appropriate norm of Rn . This domain is illustrated in Figures 1.1 and
1.2; it is sometimes called a security system, i.e., the union of a security interval (for the
independent variable) and a security domain (for the dependent variables) [19]. Suppose
x
x0+b
(t ,x )
0 0
x0
x0−b
t0−a
t
t0+a
t0
Figure 1.1: The domain R for D ⊂ R.
y
y
0
t0
t
x0
x
Figure 1.2: The domain R for D ⊂ R2 : “security tube”.
6
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
that f is continuous on R, and let M = maxR kf (t, x)k, which exists since f is continuous
on the compact set R.
In the following, existence of solutions will be obtained generally in relation to the domain
R by considering a subset of the time interval |t − t0 | ≤ a defined by |t − t0 | ≤ α, with
a
if M = 0
α=
b
min(a, M ) if M > 0.
This choice of α = min(a, b/M ) is natural. We endow f with specific properties (continuity,
Lipschitz, etc.) on the domain R. Thus, in order to be able to use the definition of φ(t) as
the solution of x0 = f (t, x), we must be working in R. So we require that |t − t0 | ≤ a and
kx − x0 k ≤ b. In order to satisfy the first of these conditions, choosing α ≤ a and working on
|t − t0 | ≤ α implies of course that |t − t0 | ≤ a. The requirement that α ≤ b/M comes from
the following argument. If we assume that φ(t) is a solution of (1.3) defined on [t0 , t0 + α],
then we have, for t ∈ [t0 , t0 + α],
Z t
kφ(t) − x0 k = f (s, φ(s))ds
t
Z t0
≤
kf (s, φ(s))k ds
t0
Z t
≤M
ds
t0
= M (t − t0 ),
where the first inequality is a consequence of the definition of the integrals by Riemann
sums (Lemma A.2.1 in Appendix A.2). Similarly, we have kφ(t) − x0 k ≤ −M (t − t0 ) for all
t ∈ [t0 − α, t0 ]. Thus, for |t − t0 | ≤ α, kφ(t) − x0 k ≤ M |t − t0 |. Suppose now that α ≤ b/M .
It follows that kφ − x0 k ≤ M |t − t0 | ≤ M b/M = b. Taking α = min(a, b/M ) then ensures
that both |t − t0 | ≤ a and kφ − x0 k ≤ b hold simultaneously.
The following two theorems deal with the localization of the solutions to an IVP. They
make more precise the previous discussion. Note that for the moment, the existence of
a solution is only assumed. First, we establish that the security system described above
performs properly, in the sense that a solution on a smaller time interval stays within the
security domain.
Theorem 1.1.6. If φ(t) is a solution of the IVP (1.3) in an interval |t − t0 | < α̃ ≤ α, then
kφ(t) − x0 k < b in |t − t0 | < α̃, i.e., (t, φ(t)) ∈ R((t0 , x0 ), α̃, b) for |t − t0 | < α̃.
Proof. Assume that φ is a solution with (t, φ(t)) 6∈ R((t0 , x0 ), α̃, b). Since φ is continuous, it
follows that there exists 0 < β < α̃ such that
kφ(t) − x0 k < b for |t − t0 | < β and kφ(t0 + β) − x0 k = b or kφ(t0 − β) − x0 k = b , (1.5)
i.e., the solution escapes the security domain at t = t0 ± β. Since α̃ ≤ α ≤ a, β < a. Thus
(t, φ(t)) ∈ R for |t − t0 | ≤ β.
Fund. Theory ODE Lecture Notes – J. Arino
1.1. ODEs, IVPs, solutions
7
Thus kf (t, φ(t))k ≤ M for |t − t0 | ≤ β. Since φ is a solution, we have that φ0 (t) = f (t, φ(t))
and φ(t0 ) = x0 . Thus
Z t
φ(t) = x0 +
f (s, φ(s))ds for |t − t0 | ≤ β.
t0
Hence
Z t
kφ(t) − x0 k = f (s, φ(s))ds for |t − t0 | ≤ β
t0
≤ M |t − t0 | for |t − t0 | ≤ β.
As a consequence,
kφ(t) − x0 k ≤ M β < M α̃ ≤ M α ≤ M
b
= b for |t − t0 | ≤ β.
M
In particular, kφ(t0 ± β) − x0 k < b. Hence contradiction with (1.5).
The following theorem is proved using the same sort of technique as in the proof of
Theorem 1.1.6. It links the variation of the solution to the nature of the vector field.
Theorem 1.1.7. If φ(t) is a solution of the IVP (1.3) in an interval |t − t0 | < α̃ ≤ α, then
kφ(t1 ) − φ(t2 )k ≤ M |t1 − t2 | whenever t1 , t2 are in the interval |t − t0 | < α̃.
Proof. Let us begin by considering t ≥ t0 . On t0 ≤ t ≤ t0 + α̃,
Z t1
Z t2
φ(t1 ) − φ(t2 ) = x0 +
f (s, φ(s))ds − x0 −
f (s, φ(s))ds
t0
t0
Z t2
=−
f (s, φ(s))ds if t2 > t1
t1
Z t2
f (s, φ(s))ds if t1 > t2
t1
Now we can see formally what is needed for a solution.
Theorem 1.1.8. Suppose f is continuous on an open set U ⊂ R × Rn . Let (t0 , x0 ) ∈ U,
and φ be a function defined on an open set I of R such that t0 ∈ I. Then φ is a solution of
the IVP (1.3) if, and only if,
i) ∀t ∈ I, (t, φ(t)) ∈ U.
ii) φ is continuous on I.
Rt
iii) ∀t ∈ I, φ(t) = x0 + t0 f (s, φ(s))ds.
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
8
Proof. (⇒) Let us suppose that φ0 = f (t, φ) for all t ∈ I and that φ(t0 ) = x0 . Then for all
t ∈ I, (t, φ(t)) ∈ U (i ). Also, φ is differentiable and thus continuous on I (ii ). Finally,
φ0 (s) = f (s, φ(s))
so integrating both sides from t0 to t,
Z
t
φ(t) − φ(t0 ) =
f (s, φ(s))ds
t0
and thus
Z
t
φ(t) = x0 +
f (s, φ(s))ds
t0
hence (iii ).
(⇐) Assume i ), ii ) and Riii ). Then φ is differentiable on I and φ0 (t) = f (t, φ(t)) for all t ∈ I.
t
From (3), φ(t0 ) = x0 + t00 f (s, φ(s))ds = x0 .
Note that Theorem 1.1.8 states that φ should be continuous, whereas the solution should
of course be C 1 , for its derivative needs to be continuous. However, this is implied by point
iii ). In fact, more generally, the following result holds about the regularity of solutions.
Theorem 1.1.9 (Regularity). Let f : U → Rn , with U an open set of R × Rn . Suppose that
f ∈ C k . Then all solutions of (1.2) are of class C k+1 .
Proof. The proof is obvious, since a solution φ is such that φ0 ∈ C k .
1.1.3
Geometric interpretation
The function f is the vector field of the equation. At every point in (t, x) space, a solution
φ is tangent to the value of the vector field at that point. A particular consequence of this
fact is the following theorem.
Theorem 1.1.10. Let x0 = f (x) be a scalar autonomous differential equation. Then the
solutions of this equation are monotone.
Proof. The direction field of an autonomous scalar differential equation consists of vectors
that are parallel for all t (since f (t, x) = f (x) for all t). Suppose that a solution φ of
x0 = f (x) is non monotone. Then this means that, given an initial point (t0 , x0 ), one the
following two occurs, as illustrated in Figure 1.3.
i) f (x0 ) 6= 0 and there exists t1 such that φ(t1 ) = x0 .
ii) f (x0 ) = 0 and there exists t1 such that φ(t1 ) 6= x0 .
Fund. Theory ODE Lecture Notes – J. Arino
1.2. Existence and uniqueness theorems
t0
t2
t1
t0
t2
9
t1
Figure 1.3: Situations that would lead to a scalar autonomous differential equation having
nonmonotone solutions.
Suppose we are in case i), and assume we are in the case f (x0 ) > 0. Thus, the solution curve
φ is increasing at (t0 , x0 ), i.e., φ0 (t0 ) > 0. As φ is continuous, i) implies that there exists
t2 ∈ (t0 , t1 ) such that φ(t2 ) is a maximum, with φ increasing for t ∈ [t0 , t2 ) and φ decreasing
for t ∈ (t2 , t1 ]. It follows that φ0 (t1 ) < 0, which is a contradiction with φ0 (t0 ) > 0.
Now assume that we are in case ii). Then there exists t2 ∈ (t0 , t1 ) with φ(t2 ) = x0 but
such that φ0 (t2 ) < 0. This is a contradiction.
Remark – If we have uniqueness of solutions, it follows from this theorem that if φ1 and φ2 are
two solutions of the scalar autonomous differential equation x0 = f (x), then φ1 (t0 ) < φ2 (t0 ) implies
that φ1 (t) < φ2 (t) for all t.
◦
Remark – Be careful: Theorem 1.1.10 is only true for scalar equations.
1.2
◦
Existence and uniqueness theorems
Several approaches can be used to show existence and/or uniqueness of solutions. In Sections 1.2.2 and 1.2.3, we take a direct path: using either a fixed point method (Section 1.2.2)
or an iterative approach (Section 1.2.3), we obtain existence and uniqueness of solutions
under the assumption that the vector field is Lipschitz. In Section 1.2.4, the Lipschitz
assumption is dropped and therefore a different approach must be used, namely that of
approximate solutions, with which only existence can be established.
1.2.1
Successive approximations
Picard’s successive approximation method consists in using the integral form (1.4) of the
solution to the IVP (1.3) to construct a sequence of approximation of the solution, that
converges to the solution. The steps followed in constructing this approximating sequence
are the following.
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
10
Step 1. Start with an initial estimate of the solution, say, the constant function φ0 (t) =
φ0 = x0 , for |t − t0 | ≤ h. Evidently, this function satisfies the IVP.
Step 2. Use φ0 in (1.4) to define the second element in the sequence:
Z t
φ1 (t) = x0 +
f (s, φ0 (s))ds.
t0
Step 3. Use φ1 in (1.4) to define the third element in the sequence:
Z t
φ2 (t) = x0 +
f (s, φ1 (s))ds.
t0
...
Step n. Use φn−1 in (1.4) to define the nth element in the sequence:
Z t
φn (t) = x0 +
f (s, φn−1 (s))ds.
t0
At this stage, there are two major ways to tackle the problem, which use the same idea:
if we can prove that the sequence {φn } converges, and that the limit happens to satisfy
the differential equation, then we have the solution to the IVP (1.3). The first method
(Section 1.2.2) uses a fixed point approach. The second method (Section 1.2.3) studies
explicitly the limit.
1.2.2
Local existence and uniqueness – Proof by fixed point
Here are two slightly different formulations of the same theorem, which establishes that if the
vector field is continuous and Lipschitz, then the solutions exist and are unique. We prove
the result in the second case. For the definition of a Lipschitz function, see Section A.6 in
the Appendix.
Theorem 1.2.1 (Picard local existence and uniqueness). Assume f : U ⊂ R × Rn → D ⊂
Rn is continuous, and that f (t, x) satisfies a Lipschitz condition in U with respect to x.
Then, given any point (t0 , x0 ) ∈ U, there exists a unique solution of (1.3) on some interval
containing t0 in its interior.
Theorem 1.2.2 (Picard local existence and uniqueness). Consider the IVP (1.3), and assume f is (piecewise) continuous in t and satisfies the Lipschitz condition
kf (t, x1 ) − f (t, x2 )k ≤ Lkx1 − x2 k
for all x1 , x2 ∈ D = {x : kx − x0 k ≤ b} and all t such that |t − t0 | ≤ a. Then there exists
0 < δ ≤ α = min a, Mb such that (1.3) has a unique solution in |t − t0 | ≤ δ.
To set up the proof, we proceed as follows. Define the operator F by
Z t
F : x 7→ x0 +
f (s, x(s))ds.
t0
Fund. Theory ODE Lecture Notes – J. Arino
1.2. Existence and uniqueness theorems
11
Note that the function (F φ)(t) is a continuous function of t. Then Picard’s successives
approximations take the form φ1 = F φ0 , φ2 = F φ1 = F 2 φ0 , where F 2 represents F ◦ F .
Iterating, the general term is given for k = 0, . . . by
φk = F k φ0 .
Therefore, finding the limit limk→∞ φk is equivalent to finding the function φ, solution of the
fixed point problem
x = F x,
with x a continuously differentiable function. Thus, a solution of (1.3) is a fixed point of F ,
and we aim to use the contraction mapping principle to verify the existence (and uniqueness)
of such a fixed point. We follow the proof of [14, p. 56-58].
Proof. We show the result on the interval t − t0 ≤ δ. The proof for the interval t0 − t ≤ δ
is similar. Let X be the space of continuous functions defined on the interval [t0 , t0 + δ],
X = C([t0 , t0 + δ]), that we endow with the sup norm, i.e., for x ∈ X,
kxkc =
max kx(t)k
t∈[t0 ,t0 +δ]
Recall that this norm is the norm of uniform convergence. Let then
S = {x ∈ X : kx − x0 kc ≤ b}
Of course, S ⊂ X. Furthermore, S is closed, and X with the sup norm is a complete metric
space. Note that we have transformed the problem into a problem involving the space of
continuous functions; hence we are now in an infinite dimensional case. The proof proceeds
in 3 steps.
Step 1. We begin by showing that F : S → S. From (1.4),
Z t
(F φ)(t) − x0 =
f (s, φ(s))ds
t0
Z t
=
f (s, φ(s)) − f (s, x0 ) + f (s, x0 )ds
t0
Therefore, by the triangle inequality,
Z t
kF φ − x0 k ≤
kf (s, φ(s)) − f (t, x0 )k + kf (t, x0 )kds
t0
As f is (piecewise) continuous, it is bounded on [t0 , t1 ] and there exists M = maxt∈[t0 ,t1 ] kf (t, x0 )k.
Thus
Z t
kF φ − x0 k ≤
kf (s, φ(s)) − f (t, x0 )k + M ds
t
Z 0t
≤
Lkφ(s) − x0 k + M ds,
t0
12
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
since f is Lipschitz. Since φ ∈ S for all kφ − x0 k ≤ b, we have that for all φ ∈ S,
Z t
kF φ − x0 k ≤
Lb + M ds
t0
≤ (t − t0 )(Lb + M )
As t ∈ [t0 , t0 + δ], (t − t0 ) ≤ δ, and thus
kF φ − x0 kc = max kF φ − x0 k ≤ (Lb + M )δ
[t0 ,t0 +δ]
Choose then δ such that δ ≤ b/(Lb + M ), i.e., t sufficiently close to t0 . Then we have
kF φ − x0 kc ≤ b
This implies that for φ ∈ S, F φ ∈ S, i.e., F : S → S.
Step 2. We now show that F is a contraction. Let φ1 , φ2 ∈ S,
Z t
k(F φ1 )(t) − (F φ2 )(t)k = f
(s,
φ
(s))
−
f
(s,
φ
(s))ds
1
2
t0
Z t
≤
kf (s, φ1 (s)) − f (s, φ2 (s))kds
t0
Z t
≤
Lkφ1 (s) − φ2 (s)kds
t0
Z t
≤ Lkφ1 − φ2 kc
ds
t0
and thus
ρ
L
Thus, choosing ρ < 1 and δ ≤ ρ/L, F is a contraction. Since, by Step 1, F : S → S, the
contraction mapping principle (Theorem A.11) implies that F has a unique fixed point in
S, and (1.3) has a unique solution in S.
Step 3. It remains to be shown that any solution in X is in fact in S (since it is on X that
we want to show the result). Considering a solution starting at x0 at time t0 , the solution
leaves S if there exists a t > t0 such that kφ(t) − x0 k = b, i.e., the solution crosses the border
of D. Let τ > t0 be the first of such t’s. For all t0 ≤ t ≤ τ ,
Z t
kφ(t) − x0 k ≤
kf (s, φ(s)) − f (s, x0 )k + kf (s, x0 )kds
t0
Z t
≤
Lkφ(s) − x0 k + M ds
t0
Z t
≤
Lb + M ds
kF φ1 − F φ2 kc ≤ Lδkφ1 − φ2 kc ≤ ρkφ1 − φ2 kc for δ ≤
t0
Fund. Theory ODE Lecture Notes – J. Arino
1.2. Existence and uniqueness theorems
13
As a consequence,
b = kφ(τ ) − x0 k ≤ (Lb + M )(τ − t0 )
As τ = t0 + µ, for some µ > 0, it follows that if
µ>
b
Lb + M
then the solution φ is confined to D.
Note that the condition x1 , x2 ∈ D = {x : kx − x0 k ≤ b} in the statement of the theorem
refers to a local Lipschitz condition. If the function f is Lipschitz, then the following theorem
holds.
Theorem 1.2.3 (Global existence). Suppose that f is piecewise continuous in t and is
Lipschitz on U = I × D. Then (1.3) admits a unique solution on I.
1.2.3
Local existence and uniqueness – Proof by successive approximations
Using the method of successive approximations, we can prove the following theorem.
Theorem 1.2.4. Suppose that f is continuous on a domain R of the (t, x)-plane defined,
for a, b > 0, by R = {(t, x) : |t − t0 | ≤ a, kx − x0 k ≤ b}, and that f is locally Lipschitz in x
on R. Let then, as previously defined,
M = sup kf (t, x)k < ∞
(t,x)∈R
and
α = min(a,
b
)
M
Then the sequence defined by
|t − t0 | ≤ α
Z t
φi (t) = x0 +
f (s, φi−1 (s))ds,
φ0 = x0 ,
i ≥ 1,
|t − t0 | ≤ α
t0
converges uniformly on the interval |t − t0 | ≤ α to φ, unique solution of (1.3).
Proof. We follow [20, p. 3-6].
Existence. Suppose that |t − t0 | ≤ α. Then
Z t
kφ1 − φ0 k = f
(s,
φ
(s))ds
0
t0
≤ M |t − t0 |
≤ αM ≤ b
14
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
Rt
from the definitions of M and α, and thus kφ1 − φ0 k ≤ b. So t0 f (s, φ1 (s))ds is defined for
|t − t0 | ≤ α, and, for |t − t0 | ≤ α,
Z t
Z t
kf (s, φ1 (s))kds ≤ αM ≤ b.
kφ2 (t) − φ0 k = f (s, φ1 (s))ds ≤ k
t0
t0
All subsequent terms in the sequence can be similarly defined, and, by induction, for |t−t0 | ≤
α,
kφk (t) − φ0 k ≤ αM ≤ b, k = 1, . . . , n.
Now, for |t − t0 | ≤ α,
Z t
Z t
kφk+1 (t) − φk (t)k = x0 +
f (s, φk (s))ds − x0 −
f (s, φk−1 (s))ds
t0
Z t t0
=
f
(s,
φ
(s))
−
f
(s,
φ
(s))
ds
k
k−1
t0
Z t
≤L
kφk (s) − φk−1 (s)kds,
t0
where the inequality results of the fact that f is locally Lipschitz in x on R.
We now prove that, for all k,
kφk+1 − φk k ≤ b
(L|t − t0 |)k
for |t − t0 | ≤ α
k!
(1.6)
Indeed, (1.6) holds for k = 1, as previously established. Assume that (1.6) holds for k = n.
Then
Z t
kφn+2 − φn+1 k = f (s, φn+1 (s)) − f (s, φn (s))ds
t0
Z t
≤
Lkφn+1 (s) − φn (s)kds
t0
Z t
(L|s − t0 |)n
ds for |t − t0 | ≤ α
≤
Lb
n!
t0
s=t
Ln+1 |t − t0 |n+1 ≤b
n!
n + 1 s=t0
(L|t − t0 |)n+1
≤b
(n + 1)!
and thus (1.6) holds for k = 1, . . ..
Thus, for N > n we have
N
−1
X
N
−1
X
N
−1
X
(Lα)k
(L|t − t0 |)k
kφN (t) − φn (t)k ≤
kφk+1 (t) − φk (t)k ≤
b
≤b
k!
k!
k=n
k=n
k=n
Fund. Theory ODE Lecture Notes – J. Arino
1.2. Existence and uniqueness theorems
15
The rightmost term in this expression tends to zero as n → ∞. Therefore, {φk (t)} converges
uniformly to a function φ(t) on the interval |t − t0 | ≤ α. As the convergence isPuniform, the
limit function is continuous.
Moreover φ(t0 ) = x0 . Indeed, φN (t) = φ0 (t) + N
k=1 (φk (t) −
P∞
φk−1 (t)), so φ(t) = φ0 (t) + k=1 (φk (t) − φk−1 (t)).
The fact that φ is a solution of (1.3) follows from the following result. If a sequence
of functions {φk (t)} converges uniformly and that the φk (t) are continuous on the interval
|t − t0 | ≤ α, then
Z
Z
t
lim
n→∞
t
φn (s)ds =
t0
lim φn (s)ds
t0 n→∞
Hence,
φ(t) = lim φn (t)
n→∞
Z t
= x0 + lim
f (s, φn−1 (s))ds
n→∞ t
0
Z t
= x0 +
lim f (s, φn−1 (s))ds
t0 n→∞
Z t
= x0 +
f (s, φ(s))ds,
t0
which is to say that
Z
t
f (s, φ(s))ds for |t − t0 | ≤ α.
φ(t) = x0 +
t0
As the integrand f (t, φ) is a continuous function, φ is differentiable (with respect to t), and
φ0 (t) = f (t, φ(t)), so φ is a solution to the IVP (1.3).
Uniqueness. Let φ and ψ be two solutions of (1.3), i.e., for |t − t0 | ≤ α,
Z t
φ(t) = x0 +
f (s, φ(s))ds
t0
Z t
ψ(t) = x0 +
f (s, ψ(s))ds.
t0
Then, for |t − t0 | ≤ α,
Z t
kφ(t) − ψ(t)k = f
(s,
φ(s))
−
f
(s,
ψ(s))ds
t0
Z t
≤L
kφ(s) − ψ(s)kds.
(1.7)
t0
We now apply Gronwall’s Lemma A.7) to this inequality, using K = 0 and g(t) = kφ(t) −
ψ(t)k. First, applying the lemma for t0 ≤ t ≤ t0 + α, we get 0 ≤ kφ(t) − ψ(t)k ≤ 0, that is,
kφ(t) − ψ(t)k = 0,
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
16
and thus φ(t) = ψ(t) for t0 ≤ t ≤ t0 + α. Similarly, for t0 − α ≤ t ≤ t0 , kφ(t) − ψ(t)k = 0.
Therefore, φ(t) = ψ(t) on |t − t0 | ≤ α.
Example – Let us consider the IVP x0 = −x, x(0) = x0 = c, c ∈ R. For initial solution, we choose
φ0 (t) = c. Then
Z t
φ1 (t) = x0 +
f (s, φ0 (s))ds
0
Z t
=c+
−φ0 (s)ds
0
Z t
=c−c
ds
0
= c − ct.
To find φ2 , we use φ1 in (1.4).
Z
t
φ2 (t) = x0 +
f (s, φ1 (s))ds
0
Z t
=c−
(c − cs)ds
0
= c − ct + c
t2
.
2
Continuing this method, we find a general term of the form
φn (t) =
n
X
(−1)i ti
.
c
i!
i=0
This is the power series expansion of ce−t , so φn → φ = ce−t (and the approximation is valid on
R), which is the solution of the initial value problem.
Note that the method of successive approximations is a very general method that can be
used in a much more general context; see [8, p. 264-269].
1.2.4
Local existence (non Lipschitz case)
The following theorem is often called Peano’s existence theorem. Because the vector field is
not assumed to be Lipschitz, something is lost, namely, uniqueness.
Theorem 1.2.5 (Peano). Suppose that f is continuous on some region
R = {(t, x) : |t − t0 | ≤ a, kx − x0 k ≤ b},
with a, b > 0, and let M = maxR kf (t, x)k. Then there exists a continuous function φ(t),
differentiable on R, such that
i) φ(t0 ) = x0 ,
Fund. Theory ODE Lecture Notes – J. Arino
1.2. Existence and uniqueness theorems
17
ii) φ0 (t) = f (t, φ) on |t − t0 | ≤ α, where
a
if M = 0
α=
b
min a, M if M > 0.
Before we can prove this result, we need a certain number of preliminary notations and
results. The definition of equicontinuity and a statement of the Ascoli lemma are given in
Section A.5. To construct a solution without the Lipschitz condition, we approximate the
differential equation by another one that does satisfy the Lipschitz condition. The unique
solution of such an approximate problem is an ε-approximate solution. It is formally defined
as follows [8, p. 285].
Definition 1.2.6 (ε-approximate solution). A differentiable mapping u of an open ball J ∈ I
into U is an approximate solution of x0 = f (t, x) with approximation ε (or an ε-approximate
solution) if we have
ku0 (t) − f (t, u(t))k ≤ ε,
for any t ∈ J.
Lemma 1.2.7. Suppose that f (t, x) is continuous on a region
R = {(t, x) : |t − t0 | ≤ a, kx − x0 k ≤ b}.
Then, for every positive number ε, there exists a function Fε (t, x) such that
i) Fε is continuous for |t − t0 | ≤ a and all x,
ii) Fε has continuous partial derivatives of all orders with respect to x1 , . . . , xn for |t−t0 | ≤
a and all x,
iii) kFε (t, x)k ≤ maxR kf (t, x)k = M for |t − t0 | ≤ a and all x,
iv) kFε (t, x) − f (t, x)k ≤ ε on R.
See a proof in [12, p. 10-12]; note that in this proof, the property that f defines a
differential equation is not used. Hence Lemma 1.2.7 can be used in a more general context
than that of differential equations. We now prove Theorem 1.2.5.
Proof of Theorem 1.2.5. The proof takes four steps.
1. We construct, for every positive number ε, a function Fε (t, x) that satisfies the requirements given in Lemma 1.2.7. Using an existence-uniqueness result in the Lipschitz case
(such as Theorem 1.2.2), we construct a function φε (t) such that
(P1) φε (t0 ) = x0 ,
(P2) φ0ε (t) = Fε (t, φε (t)) on |t − t0 | < α.
(P3) (t, φε (t)) ∈ R on |t − t0 | ≤ α.
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
18
2. The set F = {φε : ε > 0} is bounded and equicontinuous on |t − t0 | ≤ α. Indeed,
property (P3) of φε implies that F is bounded on |t − t0 | ≤ α and that kFε (t, φε (t))k ≤ M
on |t − t0 | ≤ α. Hence property (P2) of φε implies that
kφε (t1 ) − φε (t2 )k ≤ M |t1 − t2 |,
if |t1 − t0 | ≤ α and |t2 − t0 | ≤ α (this is a consequence of Theorem 1.1.7). Therefore, for a
given positive number µ, we have kφε (t1 ) − φε (t2 )k ≤ µ whenever |t1 − t0 | ≤ α, |t2 − t0 | ≤ α,
and |t1 − t2 | ≤ µ/M .
3. Using Lemma A.5, choose a sequence {εk : k = 1, 2, . . .} of positive numbers such that
limk→∞ εk = 0 and that the sequence {φεk : k = 1, 2, . . .} converges uniformly on |t − t0 | ≤ α
as k → ∞. Then set
φ(t) = lim φεk (t) on |t − t0 | ≤ α.
k→∞
4. Observe that
Z
t
φε (t) = x0 +
Fε (s, φε (s))ds
t
Z 0t
= x0 +
Z
t
Fε (s, φε (s)) − f (s, φε (s))ds,
f (s, φε (s))ds +
t0
t0
and that it follows from iv) in Lemma 1.2.7 that
Z t
Fε (s, φε (s)) − f (s, φε (s))ds ≤ ε|t − t0 | on |t − t0 | ≤ α.
t0
This is true for all ε ≥ 0, so letting ε → 0, we obtain
Z t
f (s, φ(s))ds,
φ(t) = x0 +
t0
which completes the proof.
See [12, p. 13-14] for the outline of two other proofs of this result. A proof, by Hartman
[11, p. 10-11], now follows.
Proof. Let δ > 0 and φ` (t) be a C 1 n-dimensional vector-valued function on [t0 − δ, t0 ]
satisfying φ` (t0 ) = x0 , φ0` (t0 ) = f (t0 , x0 ) and kφ` (t) − x0 k ≤ b, kφ0` (t)k ≤ M . For 0 < ε ≤ δ,
define a function φε (t) on [t0 − δ, t0 + α] by putting φε (t) = φ` (t) on [t0 − δ, t0 ] and
Z t
f (s, φε (s − ε))ds on [t0 , t0 + α].
(1.8)
φε (t) = x0 +
t0
The function φε can indeed be thus defined on [t0 − δ, t0 + α]. To see this, remark first that
this formula is meaningful and defines φε (t) for t0 ≤ t ≤ t0 + α1 , α1 = min(α, ε), so that
φε (t) is C 1 on [t0 − δ, t0 + α1 ] and, on this interval,
kφε (t) − x0 k ≤ b,
kφε (t) − φε (s)k ≤ M |t − s|.
(1.9)
Fund. Theory ODE Lecture Notes – J. Arino
1.2. Existence and uniqueness theorems
19
It then follows that (1.8) can be used to extend φε (t) as a C 1 function over [t0 − δ, t0 + α2 ],
where α2 = min(α, 2ε), satisfying relation (1.9). Continuing in this fashion, (1.8) serves to
define φε (t) over [t0 , t0 + α] so that φε (t) is a C 0 function on [t0 − δ, t0 + α], satisfying relation
(1.9).
Since kφ0ε (t)k ≤ M , M can be used as a Lipschitz constant for φε , giving uniform continuity of φε . It follows that the family of functions, φε (t), 0 < ε ≤ δ, is equicontinuous.
Thus, using Ascoli’s Lemma (Lemma A.5), there exists a sequence ε(1) > ε(2) > . . ., such
that ε(n) → 0 as n → ∞ and
φ(t) = lim φε(n) (t) exists uniformly
n→∞
on [t0 − δ, t0 + α]. The continuity of f implies that f (t, φε(n) (t − ε(n)) tends uniformly to
f (t, φ(t)) as n → ∞; thus term-by-term integration of (1.8) where ε = ε(n) gives
Z t
φ(t) = x0 +
f (s, φ(s))ds
t0
and thus φ(t) is a solution of (1.3).
An important corollary follows.
Corollary 1.2.8. Let f (t, x) be continuous on an open set E and satisfy kf (t, x)k ≤ M .
Let E0 be a compact subset of E. Then there exists an α = α(E, E0 , M ) > 0 with the
property that if (t0 , x0 ) ∈ E0 , then the IVP (1.3) has a solution and every solution exists on
|t − t0 | ≤ α.
In fact, hypotheses can be relaxed a little. Coddington and Levinson [5] define an εapproximate solution as
Definition 1.2.9. An ε-approximate solution of the differential equation (1.2), where f is
continuous, on a t interval I is a function φ ∈ C on I such that
i) (t, φ(t)) ∈ U for t ∈ I;
ii) φ ∈ C 1 on I, except possibly for a finite set of points S on I, where φ0 may have simple
discontinuities (g has finite discontinuities at c if the left and right limits of g at c exist
but are not equal);
iii) kφ0 (t) − f (t, φ(t))k ≤ ε for t ∈ I \ S.
Hence it is assumed that φ has a piecewise continuous derivative on I, which is denoted
by φ ∈ Cp1 (I).
Theorem 1.2.10. Let f ∈ C on the rectangle
R = {(t, x) : |t − t0 | ≤ a, kx − x0 k ≤ b}.
Given any ε > 0, there exists an ε-approximate solution φ of (1.3) on |t − t0 | ≤ α such that
φ(t0 ) = x0 .
20
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
Proof. Let ε > 0 be given. We construct an ε-approximate solution on the interval [t0 , t0 +ε];
the construction works in a similar way for [t0 − α, t0 ]. The ε-approximate solution that we
construct is a polygonal path starting at (t0 , x0 ).
Since f ∈ C on R, it is uniformly continuous on R, and therefore for the given value of
ε, there exists δε > 0 such that
kf (t, φ) − f (t̃, φ̃)k ≤ ε
(1.10)
if
(t, φ) ∈ R, (t̃, φ̃) ∈ R and |t − t̃| ≤ δε
kφ − φ̃k ≤ δε .
Now divide the interval [t0 , t0 + α] into n parts t0 < t1 < · · · < tn = t0 + α, in such a way
that
δε
.
(1.11)
max |tk − tk−1 | ≤ min δε ,
M
From (t0 , x0 ), construct a line segment with slope f (t0 , x0 ) intercepting the line t = t1 at
(t1 , x1 ). From the definition of α and M , it is clear that this line segment lies inside the
triangular region T bounded by the lines segments with slopes ±M from (t0 , x0 ) to their
intercept at t = t0 + α, and the line t = t0 + α. In particular, (t1 , x1 ) ∈ T .
At the point (t1 , x1 ), construct a line segment with slope f (t1 , x1 ) until the line t = t2 ,
obtaining the point (t2 , x2 ). Continuing similarly, a polygonal path φ is constructed that
meets the line t = t0 + α in a finite number of steps, and lies entirely in T .
The function φ, which can be expressed as
φ(t0 ) = x0
φ(t) = φ(tk−1 ) + f (tk−1 , φ(tk−1 ))(t − tk−1 ),
t ∈ [tk−1 , tk ], k = 1, . . . , n,
(1.12)
is the ε-approximate solution that we seek. Clearly, φ ∈ Cp1 ([t0 , t0 + α]) and
kφ(t) − φ(t̃)k ≤ M |t − t̃| for t, t̃ ∈ [t0 , t0 + α].
(1.13)
If t ∈ [tk−1 , tk ], then (1.13) together with (1.11) imply that kφ(t) − φ(tk−1 )k ≤ δε . But from
(1.12) and (1.10),
kφ0 (t) − f (t, φ(t))k = kf (tk−1 , φ(tk−1 )) − f (t, φ(t))k ≤ ε.
Therefore, φ is an ε-approximation.
We can now turn to their proof of Theorem 1.2.5.
Proof. Let {εn } be a monotone decreasing sequence of positive real numbers with εn → 0
as n → ∞. By Theorem 1.2.10, for each εn , there exists an εn -approximate solution φn of
(1.3) on |t − t0 | ≤ α such that φn (t0 ) = x0 . Choose one such solution φn for each εn . From
(1.13), it follows that
kφn (t) − φn (t̃)k ≤ M |t − t̃|.
(1.14)
Fund. Theory ODE Lecture Notes – J. Arino
1.2. Existence and uniqueness theorems
21
Applying (1.14) to t̃ = t0 , it is clear that the sequence {φn } is uniformly bounded by
kx0 k + b, since |t − t0 | ≤ b/M . Moreover, (1.14) implies that {φn } is an equicontinuous
set. By Ascoli’s lemma (Lemma A.5), there exists a subsequence {φnk }, k = 1, . . ., of {φn },
converging uniformly on [t0 − α, t0 + α] to a limit function φ, which must be continuous since
each φn is continuous.
This limit function φ is a solution to (1.3) which meets the required specifications. To
see this, write
Z
t
φn (t) = x0 +
f (s, φn (s)) + ∆n (s)ds,
(1.15)
t0
where ∆n (t) = φ0 (t) − f (t, φn (t)) at those points where φ0n exists, and ∆n (t) = 0 otherwise.
Because φn is an εn -approximate solution, k∆n (t)k ≤ εn . Since f is uniformly continuous on
R, and φnk → φ uniformly on [t0 − α, t0 + α] as k → ∞, it follows that f (t, φnk ) → f (t, φ(t))
uniformly on [t0 − α, t0 + α] as k → ∞.
Replacing n by nk in (1.15) and letting k → ∞ gives
Z t
φ(t) = x0 +
f (s, φ(s))ds.
(1.16)
t0
Clearly, φ(t0 ) = 0, when evaluated using (1.16), and also φ0 (t) = f (t, φ(t)) since f is continuous. Thus φ as defined by (1.16) is a solution to (1.3) on |t − t0 | ≤ α.
1.2.5
Some examples of existence and uniqueness
Example – Consider the IVP
2
x0 = 3|x| 3
x(t0 ) = x0
(1.17)
Here, Theorem 1.2.5 applies, since f (t, x) = 3x2/3 is continuous. However, Theorem 1.2.2 does not
apply, since f (t, x) is not locally Lipschitz in x = 0 (or, f is not Lipschitz on any interval containing
0). This means that we have existence of solutions to this IVP, but not uniqueness of the solution.
The fact that f is not Lipschitz on any interval containing 0 is established using the following
argument. Suppose that f is Lipschitz on an interval I = (−ε, ε), with ε > 0. Then, there exists
L > 0 such that for all x1 , x2 ∈ I,
kf (t, x1 ) − f (t, x2 )k ≤ L|x1 − x2 |
that is,
2
2
3 |x1 | 3 − |x2 | 3 ≤ L|x1 − x2 |
Since this has to hold true for all x1 , x2 ∈ I, it must hold true in particular for x2 = 0. Thus
2
3|x1 | 3 ≤ L|x1 |
Given an ε > 0, it is possible to find Nε > 0 such that n1 < ε for all n ≥ Nε . Let x1 = n1 . Then for
n ≥ Nε , if f is Lipschitz there must hold
2
1 3
L
3
≤
n
n
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
22
So, for all n ≥ Nε ,
L
3
= ∞, and so f is not Lipschitz on I.
1
n3 ≤
This is a contradiction, since limn→∞ n1/3
Let us consider the set
E = {t ∈ R : x(t) = 0}
The set E can be have several forms, depending on the situation.
1)
E = ∅,
2)
E = [a, b], (closed since x is continuous and thus reaches its bounds),
3)
E = (−∞, b),
4)
E = (a, +∞),
5)
E = R.
Note that case 2) includes the case of a single intersection point, when a = b, giving E = {a}. Let us
now consider the nature of x in these different situations. Recall that from Theorem 1.1.10, since
(1.17) is defined by a scalar autonomous equation, its solutions are monotone. For simplicity, we
consider here the case of monotone increasing solutions. The case of monotone decreasing solutions
can be treated in a similar fashion.
1)
2)
Here, there is no intersection with the x = 0 axis. Thus if follows that
> 0, if x0 > 0
x(t) is
< 0, if x0 < 0
In this case,

 < 0, if t < a
= 0, if t ∈ [a, b]
x(t) is

> 0, if t > b
3)
Here,
= 0, if t < b
> 0, if t > b
< 0, if t < a
= 0, if t > a
x(t) is
4)
In this case,
x(t) is
5)
In this last case, x(t) = 0 for all t ∈ R.
Now, depending on the sign of x, we can integrate the equation. First, if x > 0, then |x| = x and
so
x0 = 3x2/3
1
⇔ x−2/3 x0 = 1
3
⇔ x1/3 = t + k1
⇔ x(t) = (t + k1 )3
Fund. Theory ODE Lecture Notes – J. Arino
1.2. Existence and uniqueness theorems
23
for k1 ∈ R. Then, if x < 0, then |x| = −x, and
x0 = 3 (−x)2/3
1
⇔ (−x)−2/3 (−x0 ) = −1
3
⇔ (−x)1/3 = −t + k2
⇔ x(t) = −(−t + k2 )3
for k2 ∈ R. We can now use these computations with the different cases that were discussed earlier,
depending on the value of t0 and x0 . We begin with the case of t0 > 0 and x0 > 0.
1)
2)
The case E = ∅ is impossible, for all initial conditions (t0 , x0 ). Indeed, as x0 > 0, we have
x(t) = (t + k1 )3 . Using the initial condition, we find that x(t0 ) = x0 = (t0 + k1 )3 , i.e.,
1/3
1/3
k1 = x0 − t0 , and x(t) = (t + x0 − t0 )3 .
If E = [a, b], then

 −(−t + k2 )3 if t < a
0 if t ∈ [a, b]
x(t) =

(t + k1 )3 if t > b
Since x0 > 0, we have to be in the t > b region, so t0 > b, and (t0 + k1 )3 = x0 , which implies
1/3
that k1 = x0 − t0 . Thus

3
 −(−t + k2 ) if t < a
0 if t ∈ [a, b]
x(t) =

1/3
(t + x0 − t0 )3 if t > b
Since x is continuous,
1/3
lim (t + x0
t→b,t>b
− t0 )3 = 0
and
lim −(−t + k2 )3 = 0
t→a,t<a
1/3
This implies that b = t0 − x0
and k2 = a. So finally,

−(−t + a)3 if t < a


1
1
0 if t ∈ [a, t0 − x03 ]
(a ≤ t0 − x03 )
x(t) =

1

1/3
(t + x0 − t0 )3 if t > t0 − x03
1/3
Thus, choosing a ≤ t0 − x0 , we have solutions of the form shown in Figure 1.4. Indeed, any
ai satisfying this property yields a solution.
3)
4)
The case [a, +∞) is impossible. Indeed, there does not exist a solution through (t0 , x0 ) such
that x(t) = 0 for all t ∈ [a, +∞); since we are in the case of monotone increasing functions,
if x0 > 0 then x(t) ≥ x0 for all t ≥ t0 .
E = R is also impossible, for the same reason.
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
24
Figure 1.4: Case t0 , x0 > 0, subcase 2, in the resolution of (1.17).
5)
For the case E = (−∞, b], we have
x(t) =
0 if t ∈ (−∞, b]
(t + k1 )3 if t > b
1/3
Since x(t0 ) = x0 , k1 = x0
1/3
− t0 , and since x is continuous, b = −k1 = t0 − x0 . So,
(
1/3
0 if t ∈ (−∞, t0 − x0 ]
x(t) =
1/3
1/3
(t + x0 − t0 )3 if t > t0 − x0
The other cases are left as an exercise.
Example – Consider the IVP
x0 = 2tx2
x(0) = 0
(1.18)
Here, we have existence and uniqueness of the solutions to (1.18). Indeed, f (t, x) = 2tx2 is continuous and locally Lipschitz on R.
1.3
Continuation of solutions
The results we have seen so far deal with the local existence (and uniqueness) of solutions to
an IVP, in the sense that solutions are shown to exist in a neighborhood of the initial data.
The continuation of solutions consists in studying criteria which allow to define solutions on
possibly larger intervals.
Consider the IVP
x0 = f (t, x)
(1.19)
x(t0 ) = x0 ,
Fund. Theory ODE Lecture Notes – J. Arino
1.3. Continuation of solutions
25
with f continuous on a domain U of the (t, x) space, and the initial point (t0 , x0 ) ∈ U.
Lemma 1.3.1. Let the function f (t, x) be continuous in an open set U in (t, x)-space, and
assume that a function φ(t) satisfies the condition φ0 (t) = f (t, φ(t)) and (t, φ(t)) ∈ U, in an
open interval I = {t1 < t < t2 }. Under this assumption, if limj→∞ (τj , φ(τj )) = (t1 , η) ∈ U
for some sequence {τj : j = 1, 2, . . .} of points in the interval I, then limτ →t1 (τ, φ(τ )) =
(t1 , η). Similarly, if limj→∞ (τj , φ(τj )) = (t2 , η) ∈ U for some sequence {τj : j = 1, 2, . . .} of
points in the interval I, then limτ →t2 (τ, φ(τ )) = (t2 , η).
Proof. Let W be an open neighborhood of (t1 , η). Then (t, φ(t)) ∈ W in an interval τ1 <
t < τ (W) for some τ (W) determined by W. Indeed, assume that the closure of W, W̄ ⊂ U,
and that |f (t, x)| ≤ M in W for some positive number M . For every positive integer j and
every positive number ε, consider a rectangular region
Rj (ε) = {(t, x) : |t − tj | ≤ ε, kx − φ(tj )k ≤ M ε}
ε
Then there exists an ε > 0 and a j such that (τ1 , η) ∈ Rj (ε) ⊂ W, with ε = min ε, M
and
M
tj − ε ≤ τ1 .
From Theorem 1.1.6 applied to the solution φ of the IVP x0 = f (t, x), x(τj ) = φ(τj ), we
obtain that (τ, φ(τ )) ∈ Rj (ε) ∈ U on the interval t1 < τ ≤ τj . Since U is an arbitrary open
neighborhood of (t1 , η), we conclude that limj→∞ (τj , φ(τj )) = (t1 , η) ∈ R.
From the previous result, we can derive a result concerning the maximal interval over
which a solution can be extended. To emphasize the fact that the solution φ of an ODE
exists in some interval I, we denote (φ, I). We need the notion of extension of a solution.
It is defined in the classical manner (see Figure 1.5).
Definition 1.3.2 (Extension). Let (φ, I) and (φ̃, Ĩ) be two solutions of the same ODE. We
say that (φ̃, Ĩ) is an extension of (φ, I) if, and only if,
I ⊂ Ĩ,
where
|I
φ̃|I = φ
denotes the restriction to I.
Theorem 1.3.3. Let f (t, x) be continuous in an open set U in (t, x)-space, and the function
φ(t) be a function satisfying the condition φ0 (t) = f (t, φ(t)) and (t, φ(t)) ∈ U, in an open
interval I = {t1 < t < t2 }. If the following two conditions are satisfied:
i) φ(t) cannot be extended to the left of t1 (or, respectively, to the right of t2 ),
ii) limj→∞ (τj , φ(τj )) = (t1 , η) (or, respectively, (t2 , η)) exists for some sequence {τj : j =
1, 2, . . .} of points in the interval I,
then the limit point (t1 , η) (or, respectively, (t2 , η)) must be on the boundary of U.
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
26
φ
~
φ
~
φ
I
~
I
Figure 1.5: The extension φ̃ on the interval Ĩ of the solution φ (defined on the interval
I).
Proof. Suppose that the hypotheses of the theorem are satisfied, and that (t1 , η) ∈ U (respectively, (t2 , η) ∈ U). Then, from Lemma 1.3.1, it follows that
lim (τ, φ(τ )) = (t1 , η)
τ →t1
(or, respectively, limτ →t2 (τ, φ(τ )) = (t2 , η)). Thus we can apply Theorem 1.2.5 (Peano’s
Theorem) to the IVP
x0 = f (t, x)
x(t1 ) = η,
(or, respectively, x0 = f (t, x), x(t2 ) = η). This implies that the solution φ can be extended
to the left of t1 (respectively, to the right of t2 ), since Theorem 1.2.5 implies existence in a
neighborhood of t1 . This is a contradiction.
A particularly important consequence of the previous theorem is the following corollary.
Corollary 1.3.4. Assume that f (t, x) is continuous for t1 < t < t2 and all x ∈ Rn . Also,
assume that there exists a function φ(t) satisfying the following conditions:
a) φ and φ0 are continuous in a subinterval I of the interval t1 < t < t2 ,
b) φ0 (t) = f (t, φ(t)) in I.
Then, either
i) φ(t) can be extended to the entire interval t1 < t < t2 as a solution of the differential
equation x0 = f (t, x), or
ii) limt→τ kφ(t)k = ∞ for some τ in the interval t1 < t < t2 .
Fund. Theory ODE Lecture Notes – J. Arino
1.3. Continuation of solutions
1.3.1
27
Maximal interval of existence
Another way of formulating these results is with the notion of maximal intervals of existence.
Consider the differential equation
x0 = f (t, x)
(1.20)
Let x = x(t) be a solution of (1.20) on an interval I.
Definition 1.3.5 (Right maximal interval of existence). The interval I is a right maximal
interval of existence for x if there does not exist an extension of x(t) over an interval I1
so that x remains a solution of (1.20), with I ⊂ I1 (and I and I1 having different right
endpoints). A left maximal interval of existence is defined in a similar way.
Definition 1.3.6 (Maximal interval of existence). An interval which is both a left and a
right maximal interval of existence is called a maximal interval of existence.
Theorem 1.3.7. Let f (t, x) be continuous on an open set U and φ(t) be a solution of (1.20)
on some interval. Then φ(t) can be extended (as a solution) over a maximal interval of
existence (ω− , ω+ ). Also, if (ω− , ω+ ) is a maximal interval of existence, then φ(t) tends to
the boundary ∂U of U as t → ω− and t → ω+ .
Remark – The extension need not be unique, and ω± depends on the extension. Also, to say, for
example, that φ → ∂U as t → ω+ is interpreted to mean that either ω+ = ∞ or that ω+ < ∞ and
if U 0 is any compact subset of U, then (t, φ(t)) 6∈ U 0 when t is near ω+ .
◦
Two interesting corollaries, from [11].
Corollary 1.3.8. Let f (t, x) be continuous on a strip t0 ≤ t ≤ t0 + a (< ∞), x ∈ Rn
arbitrary. Let φ be a solution of (1.3) on a right maximal interval J. Then either
i) J = [t0 , t0 + a],
ii) or J = [t0 , δ), δ ≤ t0 + a, and kφ(t)k → ∞ as t → δ.
Corollary 1.3.9. Let f (t, x) be continuous on the closure Ū of an open (t, x)-set U, and let
(1.3) possess a solution φ on a maximal right interval J. Then either
i) J = [t0 , ∞),
ii) or J = [t0 , δ), with δ < ∞ and (δ, φ(δ)) ∈ ∂U,
iii) or J = [t0 , δ) with δ < ∞ and kφ(t)k → ∞ as t → δ.
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
28
1.3.2
Maximal and global solutions
Linked to the notion of maximal intervals of existence of solutions is the notion of maximal
and global solutions.
Definition 1.3.10 (Maximal solution). Let I1 ⊂ R and I2 ⊂ R be two intervals such that
I1 ⊂ I2 . A solution (φ, I1 ) is maximal in I2 if φ has no extension (φ̃, Ĩ) solution of the
ODE such that I1 ( Ĩ ⊂ I2 .
Definition 1.3.11 (Global solution). A solution (φ, I1 ) is global on I2 if φ admits a extension φ̃ defined on the whole interval I2 .
U
φ1
φ2
I
Figure 1.6: φ1 is a global and maximal solution on I; φ2 is a maximal solution on I, but
it is not global on I.
Every global solution on a given interval I is maximal on that same interval. The converse
is false.
Example – Consider the equation x0 = −2tx2 on R. If x 6= 0, x0 x−2 = −2t, which implies that
x(t) = 1/(t2 − c), with c ∈ R. Depending on c, there are several cases.
• if c < 0, then x(t) = 1/(t2 − c) is a global solution on R,
√ √
√
√
• if c > 0, the solutions are defined on (−∞, − c), (− c, c) and ( c, ∞). The solutions are
maximal solutions on R, but are not global solutions.
• if c = 0, then the maximal non global solutions on R are defined on (−∞, 0) and (0, ∞).
Another solution is x ≡ 0, which is a global solution on R.
Lemma 1.3.12. Every solution φ of the differential equation x0 = f (t, x) is contained in a
maximal solution φ̃.
Fund. Theory ODE Lecture Notes – J. Arino
1.4. Continuous dependence on initial data, on parameters
29
The following theorem extends the uniqueness property to an interval of existence of the
solution.
Theorem 1.3.13. Let φ1 , φ2 : I → Rn be two solutions of the equation x0 = f (t, x), with f
locally Lipschitz in x on U. If φ1 and φ2 coincide at a point t0 ∈ I, then φ1 = φ2 on I.
Proof. Under the assumptions of the theorem, φ1 (t0 ) = φ2 (t0 ). Suppose that there exists a
t1 , t1 6= t0 , such that φ1 (t1 ) 6= φ2 (t1 ). For simplicity, let us assume that t1 > t0 .
By the local uniqueness of the solution, it follows from φ1 (t0 ) = φ2 (t0 ) that there exists
a neighborhood N of t0 such that φ1 (t) = φ2 (t) for all t ∈ N . Let
E = {t ∈ [t0 , t1 ] : φ1 (t) 6= φ2 (t)}
Since t1 ∈ E, E 6= ∅. Let α = inf(E), we have α ∈ (t0 , t1 ], and for all t ∈ [t0 , α), φ1 (t) = φ2 (t).
By continuity of φ1 and φ2 , we thus have φ1 (α) = φ2 (α). This implies that there exists
a neighborhood W of α on which φ1 = φ2 . This is a contradiction, since φ1 (t) 6= φ2 (t) for
t > α, hence there exists no such t1 , and φ1 = φ2 on I.
Corollary 1.3.14 (Global uniqueness). Let f (t, x) be locally Lipschitz in x on U. Then by
any point (t0 , x0 ) ∈ U, there passes a unique maximal solution φ : I → Rn . If there exists a
global solution on I, then it is unique.
1.4
Continuous dependence on initial data, on parameters
Let φ be a solution of (1.3). To emphasize the fact the this solution depends on the initial
condition (t0 , x0 ), we denote it φt0 ,x0 . Let η be a parameter of (1.3). When we study the
dependence of φt0 ,x0 on η, we denote the solution as φt0 ,x0 ,η .
We suppose that kf (t, x)k ≤ M and |∂f (t, x)/∂xi | ≤ K for i = 1, . . . , n for (t, x) ∈ U,
with U ∈ R × Rn . Note that these conditions are automatically satisfied on a closed bounded
region of the form R = {(t, x) : |t − t0 | ≤ a, kx − x0 k ≤ b}, where a, b > 0.
Our objective here is to characterize the nature of the dependence of the solution on the
initial time t0 and the initial data x0 .
Theorem 1.4.1. Suppose that f and ∂f /∂x are continuous and bounded in a given region
U. Let φt0 ,x0 be a solution of (1.3) passing through (t0 , x0 ) and ψt̂0 ,x̂0 be a solution of (1.3)
passing through (t̂0 , x̂0 ). Suppose that φ and ψ exist on some interval I.
Then, for each ε > 0, there exists δ > 0 such that if |t − t̂| < δ and kx0 − x̂0 k < δ, then
kφ(t) − ψ(t̂)k < ε, for t, t̂ ∈ I.
Proof. The prooof is from [2, p. 135-136]. Since φ is the solution of (1.3) through the point
(t0 , x0 ), we have, for all t ∈ I,
Z t
f (s, φ(s))ds
(1.21)
φ(t) = x0 +
t0
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
30
As ψ is the solution of (1.3) through the point (t̃0 , x̃0 ), we have, for all t ∈ I,
Z
t
f (s, ψ(s))ds
ψ(t) = x̃0 +
(1.22)
t̃0
Since
t
Z
Z
t̃0
f (s, φ(s))ds =
Z
t
f (s, φ(s))ds +
t0
t0
f (s, φ(s))ds,
t̃0
substracting (1.22) from (1.21) gives
Z
t̃0
φ(t) − ψ(t) = x0 − x̃0 +
Z
t
f (s, φ(s)) − f (s, ψ(s))ds
f (s, φ(s))ds +
t0
t̃0
and therefore
Z
Z
t
t̃0
kφ(t) − ψ(t)k ≤ kx0 − x̃0 k + f (s, φ(s))ds + f (s, φ(s)) − f (s, ψ(s))ds
t0
t̃0
Using the boundedness assumptions on f and ∂f /∂x to evaluate the right hand side of the
latter inequation, we obtain
Z t
kφ(t) − ψ(t)k ≤ kx0 − x̃0 k + M |t̃0 − t0 | + K φ(s) − ψ(s)ds
t̃0
If |t0 − t̃0 | < δ, kx0 − x̃0 k < δ, then we have
Z t
kφ(t) − ψ(t)k ≤ δ + M δ + K φ(s) − ψ(s)ds
(1.23)
t̃0
Applying Gronwall’s inequality (Appendix A.7) to (1.23) gives
kφ(t) − ψ(t)k ≤ δ(1 + M )eK|t−t̃0 | ≤ δ(1 + M )eK(τ2 −τ1 )
using the fact that |t − t̃0 | < τ2 − τ1 , if we denote I = (τ1 , τ2 ). Since
Z t
kψ(t) − ψ(t̃)k < f (s, ψ(s))ds ≤ M |t − t̃| ≤ M δ
t̃
if |t − t̃| < δ, we have
kφ(t) − ψ(t̃)k ≤ kφ(t) − ψ(t)k + kψ(t) − ψ(t̃)k ≤ δ(1 + M )eK(τ2 −τ1 ) + δM
Now, given ε > 0, we need only choose δ < ε/[M + (1 + M )K(τ2 −τ1 ) ] to obtain the desired
inequality, completing the proof.
Fund. Theory ODE Lecture Notes – J. Arino
1.4. Continuous dependence on initial data, on parameters
31
What we have shown is that the solution passing through the point (t0 , x0 ) is a continuous
function of the triple (t, t0 , x0 ). We now consider the case where the parameters also vary,
comparing solutions to two different but “close” equations.
Theorem 1.4.2. Let f, g be defined in a domain U and satisfy the hypotheses of Theorem 1.4.1. Let φ and ψ be solutions of x0 = f (t, x) and x0 = g(t, x), respectively, such
that φ(t0 ) = x0 , ψ(t0 ) = x̂0 , existing on a common interval α < t < β. Suppose that
kf (t, x) − g(t, x)k ≤ ε for (t, x) ∈ U. Then the solutions φ and ψ satisfy
kφ(t) − ψ(t)k ≤ kx0 − x̂0 keK|t−t0 | + ε(β − α)eK|t−t0 |
for all t, α < t < β.
The following theorem [6, p. 58] is less restrictive in its hypotheses than the previous
one, requiring only uniqueness of the solution of the IVP.
Theorem 1.4.3. Let U be a domain of (t, x) space, Iµ the domain |µ − µ0 | < c, with c > 0,
and Uµ the set of all (t, x, µ) satisfying (t, x) ∈ U, µ ∈ Iµ . Suppose f is a continuous function
on Uµ , bounded by a constant M there. For µ = µ0 , let
x0 = f (t, x, µ)
x(t0 ) = x0
(1.24)
have a unique solution φ0 on the interval [a, b], where t0 ∈ [a, b]. Then there exists a δ > 0
such that, for any fixed µ such that |µ − µ0 | < δ, every solution φµ of (1.24) exists over [a, b]
and as µ → µ0
φµ → φ0
uniformly over [a, b].
Proof. We begin by considering t0 ∈ (a, b). First, choose an α > 0 small enough that the
region R = {|t − t0 | ≤ α, kx − x0 k ≤ M α} is in U; note that R is a slight modification
of the usual security domain. All solutions of (1.24) with µ ∈ Iµ exist over [t0 − α, t0 + α]
and remain in R. Let φµ denote a solution. Then the set of functions {φµ }, µ ∈ Iµ is an
uniformly bounded and equicontinuous set in |t − t0 | ≤ α. This follows from the integral
equation
Z
t
f (s, φµ (s), µ)ds (|t − t0 | ≤ α)
φµ (t) = x0 +
(1.25)
t0
and the inequality kf k ≤ M .
Suppose that for some t̃ ∈ [t0 − α, t0 + α], φµ (t̃) does not tend to φ0 (t̃). Then there
exists a sequence {µk }, k = 1, 2, . . ., for which µk → µ0 , and corresponding solutions φµk
such that φµk converges uniformly over [t0 − α, t0 + α] as k → ∞ to a limit function ψ, with
ψ(t̃) 6= φ0 (t̃). From the fact that f ∈ C on Uµ , that ψ ∈ C on [t0 − α, t0 + α], and that φµk
converges uniformly to ψ, (1.25) for the solutions φµk yields
Z t
ψ(t) = x0 +
f (s, ψ(s), µ0 )ds (|t − t0 | ≤ α)
t0
32
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
Thus ψ is a solution of (1.24) with µ = µ0 . By the uniqueness hypothesis, it follows that
ψ(t) = φ0 (t) on |t − t0 | ≤ α. Thus ψ(t̃) = φ0 (t̃). Thus all solutions φµ on |t − t0 | ≤ α tend
to φ0 as µ → µ0 . Because of the equicontinuity, the convergence is uniform.
Let us now prove that the result holds over [a, b]. For this, let us consider the interval
[t0 , b]. Let τ ∈ [t0 , b), and suppose that the result is valid for every small h > 0 over [t0 , τ − h]
but not over [t0 , τ + h]. It is clear that τ ≥ t0 + α. By the above assumption, for any small
ε > 0, there exists a δε > 0 such that
kφµ (τ − ε) − φ0 (τ − ε)k < ε
(1.26)
for |µ − µ0 | < δε . Let H ⊂ U be defined as the region
H = {|t − τ | ≤ γ,
kx − φ0 (τ − γ)k ≤ γ + M |t − τ + γ|}
with γ small enough that H ⊂ U. Any solution of x0 = f (t, x, µ) starting on t = τ − γ with
initial value ξ0 , |ξ0 − φ0 (τ − γ)| ≤ γ will remain in H as t increases. Thus all solutions can
be continued to τ + γ.
By choosing ε = γ in (1.26), it follows that for |µ − µ0 | < δε , the solutions φµ can all
be continued to τ + ε. Thus over [t0 , τ + ε] these solutions are in U so that the argument
that φµ → φ0 which has been given for |t − t0 | ≤ α, also applies over [t0 , τ + ε]. Thus the
assumption about the existence of τ < b is false. The case τ = b is treated in similar fashion
on τ − γ ≤ t ≤ τ .
A similar argument applies to the left of t0 and therefore the result is valid over [a, b].
Definition 1.4.4 (Well-posedness). A problem is said to be well-posed if solutions exist, are
unique, and that there is continuous dependence on initial conditions.
1.5
Generality of first order systems
Consider an nth order differential equation in normal form
x(n) = f t, x, x0 , . . . , x(n−1)
(1.27)
This equation can be reduced to a system of n first order ordinary differential equations, by
proceeding as follows. Let y0 = x, y1 = x0 , y2 = x00 , . . . , yn−1 = x(n) . Then (1.27) is equivalent
to
y 0 = F (t, y)
(1.28)
with y = (y0 , y1 , . . . , yn−1 )T and

y1
y2
..
.







F (t, z) = 





yn−1
f (t, y0 , . . . , yn−1 )
Fund. Theory ODE Lecture Notes – J. Arino
1.5. Generality of first order systems
33
Similarly, the IVP associated to (1.27) is given by
x(n) = f t, x, x0 , . . . , x(n−1)
x(t0 ) = x0 , x0 (t0 ) = x1 , . . . , x(n−1) (t0 ) = xn−1
(1.29)
is equivalent to the IVP
y 0 = F (t, y)
y(t0 ) = y0 = (x0 , . . . , xn−1 )T
(1.30)
As a consequence, all results in this chapter are true for equations of order higher than 1.
Example – Consider the second order IVP
x00 = −2x0 + 4x − 3
x(0) = 2, x0 (0) = 1
To transform it into a system of first-order differential equations, we let y = x0 . Substituting (where
possible) y for x0 in the equation gives
y 0 = −2y + 4x − 3
The initial condition becomes x(0) = 2, y(0) = 1. So finally, the following IVP is equivalent to the
original one:
x0 = y
y 0 = 4x − 2y − 3
x(0) = 2, y(0) = 1
Note that the linearity of the initial problem is preserved.
Example – The differential equation
x(n) (t) = an−1 (t)x(n−1) (t) + · · · + a1 (t)x0 (t) + a0 (t)x(t) + b(t)
is an nth order nonhomogeneous linear differential equation. Together with the initial condition
(n−1)
x(n−1) (t0 ) = x0
(n−1)
where x0 , x00 , . . . , x0
equations by setting
, . . . , x0 (t0 ) = x00 , x(t0 ) = x0
∈ R, it forms an IVP. We can transform it to a system of linear first order
y0 = x
y1 = x0
..
.
yn−1 = x(n−1)
yn = x(n)
34
Fund. Theory ODE Lecture Notes – J. Arino
1. General theory of ODEs
The nth order linear equation is then equivalent to the following system of n first order linear
equations
y00 = y1
y10 = y2
..
.
0
yn−2
= yn−1
0
yn−1
= yn
yn0 = an−1 (t)yn (t) + an−2 (t)yn−1 (t) + · · · + a1 (t)y1 (t) + a0 (t)y0 (t) + b(t)
under the initial conditions
(n−1)
yn−1 (t0 ) = x0
, . . . , y1 (t0 ) = x00 , y0 (t0 ) = x0
1.6
Generality of autonomous systems
A nonautonomous system
x0 (t) = f (t, x(t))
can be transformed into an autonomous system of equations by setting an auxiliary variable,
say y, equal to t, giving
x0 = f (y, x)
y 0 = 1.
However, this transformation does not always make the system any easier to study.
1.7
Suggested reading, Further problems
Most of these results are treated one way or another in Coddington and Levinson [6] (first
edition published in 1955), and the current text, as many others, does little but paraphrase
them.
We have not seen here any results specific to complex valued differential equations. As
complex numbers are two-dimensional real vectors, the results carry through to the complex
case by simply assuming that if, in (1.2), we consider an n-dimensional complex vector, then
this is equivalent to a 2n-dimensional problem. Furthermore, if f (t, x) is analytic in t and
x, then analytic solutions can be constructed. See Section I-4 in [12], ..., for example.
Chapter 2
Linear systems
Let I be an interval of R, E a normed vector space over a field K (E = Kn , with K = R
or C), and L(E) the space of continuous linear maps from E to E. Let k k be a norm on
E, and ||| ||| be the induced supremum norm on L(E) (see Appendix A.1). Consider a map
A : I → L(E) and a map B : I → E. A linear system of first order equations is defined by
x0 (t) = A(t)x(t) + B(t)
(2.1)
where the unknown x is a map on I, taking values in E, defined differentiable on a subinterval of I. We restrict ourselves to the finite dimensional case (E = Kn ). Hence we
consider A ∈ Mn (K), n × n matrices over the field K, and B ∈ Kn . We suppose that A and
B have continuous entries. In most of what follows, we assume K = R.
The name linear for system (2.1) is an abuse of language. System (2.1) should be called
an affine system, with associated linear system
x0 (t) = A(t)x(t).
(2.2)
Another way to distinguish systems (2.1) and (2.2) is to refer to the former as a nonhomogeneous linear system and the latter as an homogeneous linear system. In order to lighten
the language, since there will be other qualificatives added to both (2.1) and (2.2), we use
in this chapter the names affine system for (2.1) and linear system for (2.2).
The exception to this naming convention is that we refer to (2.1) as a linear system if we
consider the generic properties of (2.1), with (2.2) as a particular case, as in this chapter’s
title or in the next section, for example.
2.1
Existence and uniqueness of solutions
Theorem 2.1.1. Let A and B be defined and continuous on I 3 t0 . Then, for all x0 ∈ E,
there exists a unique solution φt (x0 ) of (2.1) through (t0 , x0 ), defined on the interval I.
35
Fund. Theory ODE Lecture Notes – J. Arino
36
2. Linear systems
Proof. Let k(t) = |||A(t)||| = supkxk≤1 kA(t)xk. Then for all t ∈ I and all x1 , x2 ∈ K,
kf (t, x1 ) − f (t, x2 )k = kA(t)(x1 − x2 )k
≤ |||A(t)||| kx1 − x2 k
≤ k(t)kx1 − x2 k,
where the inequality
kA(t)(x1 − x2 )k ≤ |||A(t)||| kx1 − x2 k
results from the nature of the norm ||| ||| (see Appendix A.1). Furthermore, k is continuous
on I. Therefore the conditions of Theorem 1.2.2 hold, leading to existence and uniqueness
on the interval I.
With linear systems, it is possible to extend solutions easily, as is shown by the next
theorem.
Theorem 2.1.2. Suppose that the entries of A(t) and the entries of B(t) are continuous on
an open interval I. Then every solution of (2.1) which is defined on a subinterval J of the
interval I can be extended uniquely to the entire interval I as a solution of (2.1).
Proof. Suppose that I = (t1 , t2 ), and that a solution φ of (2.1) is defined on J = (τ1 , τ2 ),
with J ( I. Then
Z t
kφ(t)k ≤ kφ(t0 )k + A(s)φ(s) + B(s)ds
t0
for all t ∈ J , where t0 ∈ J . Let
K = kφ(t0 )k + (τ2 − τ1 ) maxτ1 ≤t≤τ2 kB(t)k
L = maxτ1 ≤t≤τ2 kA(t)k
Then, for t0 , t ∈ J ,
Z t
Z t
kφ(t)k ≤ K + L kφ(s)kds.
φ(s)ds ≤ K + L
t0
t0
Thus, using Gronwall’s Lemma (Lemma A.7), the following estimate holds in J ,
kφ(t)k ≤ KeL|t−t0 | ≤ KeL(τ2 −τ1 ) < ∞
This implies that case ii) in Corollary 1.3.4 is ruled out, leaving only the possibility for φ to
be extendable over I, since the vector field in (2.1) is Lipschitz.
2.2
Linear systems
We begin our study of linear systems of ordinary differential equations by considering homogeneous systems of the form (2.2) (linear systems), with x ∈ Rn and A ∈ Mn (R), the set
of square matrices over the field R, A having continuous entries on an interval I.
Fund. Theory ODE Lecture Notes – J. Arino
2.2. Linear systems
2.2.1
37
The vector space of solutions
Theorem 2.2.1 (Superposition principle). Let S 0 be the set of solutions of (2.2) that are
defined on some interval I ⊂ R. Let φ1 , φ2 ∈ S 0 , and λ1 , λ2 ∈ R. Then λ1 φ1 + λ2 φ2 ∈ S 0 .
Proof. Let φ1 , φ2 ∈ S 0 be two solutions of (2.2), λ1 , λ2 ∈ R. Then for all t ∈ I,
φ01 = A(t)φ1
φ02 = A(t)φ2 ,
from which it comes that
d
(λ1 φ1 + λ2 φ2 ) = A(t)[λ1 φ1 + λ2 φ2 ],
dt
implying that λ1 φ1 + λ2 φ2 ∈ S 0 .
Thus the linear combination of any two solutions of (2.2) is in S 0 . This is a hint that S 0
must be a vector space of dimension n on K. To show this, we need to find a basis of S 0 .
We proceed in the classical manner, with the notable difference from classical linear algebra
that the basis is here composed of time-dependent functions.
Definition 2.2.2 (Fundamental set of solutions). A set of n solutions of the linear differential equation (2.2), all defined on the same open interval I, is called a fundamental set of
solutions on I if the solutions are linearly independent functions on I.
Proposition 2.2.3. If A(t) is defined and continuous on the interval I, then the system
(2.2) has a fundamental set of solutions defined on I.
Proof. Let t0 ∈ I, and e1 , . . . , en denote the canonical basis of Kn . Then, from Theorem 2.1.1,
there exists a unique solution φ(t0 ) = (φ1 (t0 ), . . . , φn (t0 )) such that φi (t0 ) = ei , for i =
1, . . . , n. Furthermore, from Theorem 2.1.1, each function φi is defined on the interval I.
Assume that {φi }, i = 1, . . . ,P
n, is linearly dependent. Then there exists αi ∈ R, i =
n
1, . . . , n, not all zero,
αi φi (t) = 0 for all t. In particular, this is true for
Pn such that Pi=1
n
t = t0 , and thus i=1 αi φi (t0 ) = i=1 αi ei = 0, which implies that the canonical basis of
Kn is linearly dependent. Hence a contradiction, and the φi are linearly independent.
Proposition 2.2.4. If F is a fundamental set of solutions of the linear system (2.2) on the
open interval I, then every solution defined on I can be expressed as a linear combination
of the elements of F.
Let t0 ∈ I, we consider the application
Φt0 : S 0 → Kn
Y 7→ Φt0 (x) = x(t0 )
Lemma 2.2.5. Φt0 is a linear isomorphism.
Fund. Theory ODE Lecture Notes – J. Arino
38
2. Linear systems
Proof. Φt0 is bijective. Indeed, let v ∈ Kn , from Theorem 2.1.1, there exists a unique solution
passing through (t0 , v), i.e.,
∀v ∈ Kn , ∃!x ∈ S 0 , x(t0 ) = v ⇒ Φt0 (x) = v,
so Φt0 is surjective. That Φt0 is injective follows from uniqueness of solutions to an ODE.
Furthermore, Φt0 (λ1 x1 +λ2 x2 ) = λ1 Φt0 (x1 )+λ2 Φt0 (x2 ). Therefore dim S 0 = dim Kn = n.
2.2.2
Fundamental matrix solution
Definition 2.2.6. An n × n matrix function t 7→ Φ(t), defined on an open interval I, is
called a matrix solution of the homogeneous linear system (2.2) if each of its columns is a
(vector) solution. A matrix solution Φ is called a fundamental matrix solution if its columns
form a fundamental set of solutions. If in addition Φ(t0 ) = I, a fundamental matrix solution
is called the principal fundamental matrix solution.
An important property of fundamental matrix solutions is the following, known as Abel’s
formula.
Theorem 2.2.7 (Abel’s formula). Let A(t) be continuous on I and Φ ∈ Mn (K) be such
that Φ0 (t) = A(t)Φ(t) on I. Then det Φ satisfies on I the differential equation
(det Φ)0 = (trA)(det Φ),
or, in integral form, for t, τ ∈ I,
Z
t
trA(s)ds .
det Φ(t) = det Φ(τ ) exp
(2.3)
τ
Proof. Writing the differential equation Φ0 (t) = A(t)Φ(t) in terms of the elements ϕij and
aij of, respectively, Φ and A,
n
X
0
ϕij (t) =
aik (t)ϕkj (t),
(2.4)
k=1
for i, j = 1, . . . , n. Writing
ϕ11 (t) ϕ12 (t) . . . ϕ1n (t) ϕ21 (t) ϕ22 (t) . . . ϕ2n (t) ,
det Φ = ϕn1 (t) ϕn2 (t) . . . ϕnn (t)
we see that
0
ϕ11 ϕ012 . . . ϕ01n ϕ11 ϕ12 . . . ϕ1n ϕ11 ϕ12 . . . ϕ1n 0
ϕ21 ϕ22 . . . ϕ2n ϕ21 ϕ022 . . . ϕ02n ϕ21 ϕ22 . . . ϕ2n 0
+
+ ··· + .
(det Φ) = 0
0
0 ϕn1 ϕn2 . . . ϕnn ϕn1 ϕn2 . . . ϕnn ϕ
ϕ
.
.
.
ϕ
n1
n2
nn
Fund. Theory ODE Lecture Notes – J. Arino
2.2. Linear systems
39
Indeed, write det Φ(t) = Γ(r1 , r2 , . . . , rn ), where ri is the ith row in Φ(t). Γ is then a linear
function of each of its arguments, if all other rows are constant, which implies that
d
d
d
d
det Φ(t) = Γ
r1 , r2 , . . . , rn + Γ r1 , r2 , . . . , rn + · · · + Γ r1 , r2 , . . . , rn .
dt
dt
dt
dt
(To show this, use the definition of the derivative as a limit.) Using (2.4) on the first of the
n determinants in (det Φ)0 gives
P
P
P
k a1k ϕk2 . . .
k a1k ϕkn k a1k ϕk1
ϕ21
ϕ22
...
ϕ2n
.
ϕn1
ϕn2
...
ϕnn Adding −a12 times the second row, −a13 times the first row, etc., −a1n times the nth row,
to the first row, does not change the determinant, and thus
P
P
P
a11 ϕ11 a11 ϕ12 . . . a11 ϕ1n a
ϕ
.
.
.
a
ϕ
a
ϕ
1k
k2
1k
kn
1k
k1
k
k
k
ϕ21
ϕ21
ϕ22
. . . ϕ2n ϕ22
...
ϕ2n
=
= a11 det Φ.
ϕn1
ϕn2 . . . ϕnn ϕn2
...
ϕnn ϕn1
Repeating this for each of the terms in (det Φ)0 , we obtain (det Φ)0 = (a11 + a22 + · · · +
ann ) det Φ, giving finally (det Φ)0 = (trA)(det Φ). Note that this equation takes the form
u0 − α(t)u = 0, which implies that
Z t
α(s)ds = constant,
u exp
τ
which in turn implies the integral form of the formula.
Remark – Consider (2.3). Suppose that τ ∈ I is such that det Φ(τ ) 6= 0. Then, since ea 6= 0
for any a, it follows that det Φ 6= 0 for all t ∈ I. In short, linear independence of solutions for a
t ∈ I is equivalent to linear independence of solutions for all t ∈ I. As a consequence, the column
vectors of a fundamental matrix are linearly independent at every t ∈ I.
◦
Theorem 2.2.8. A solution matrix Φ of (2.2) is a fundamental solution matrix on I if,
and only if, det Φ(t) 6= 0 for all t ∈ I
Proof. Let Φ be a fundamental matrix with column vectors φi , and suppose that φ is any
nontrivial solution of (2.2). Then there exists c1 , . . . , cn , not all zero, such that
φ=
n
X
cj φj ,
j=1
or, writing this equation in terms of Φ, φ = Φc, if c = (c1 , . . . , cn )T . At any point t0 ∈ I,
this is a system of n linear equations with n unknowns c1 , . . . , cn . This system has a unique
Fund. Theory ODE Lecture Notes – J. Arino
40
2. Linear systems
solution for any choice of φ(t0 ). Thus det Φ(t0 ) 6= 0, and by the remark above, det Φ(t) 6= 0
for all t ∈ I.
Reciproqually, let Φ be a solution matrix of (2.2), and suppose that det Φ(t) 6= 0 for
t ∈ I. Then the column vectors are linearly independent at every t ∈ I.
From the remark above, the condition “det Φ(t) 6= 0 for all t ∈ I” in Theorem 2.2.8 is
equivalent to the condition “there exists t ∈ I such that det Φ(t) 6= 0”. A frequent candidate
for this role is t0 .
To conclude on fundamental solution matrices, remark that there are infinitely many
of them, for a given linear system. However, since each fundamental solution matrix can
provide a basis for the vector space of solutions, it is clear that the fundamental matrices
associated to a given problem must be linked. Indeed, we have the following result.
Theorem 2.2.9. Let Φ be a fundamental matrix solution to (2.2). Let C ∈ Mn (K) be a
constant nonsingular matrix. Then ΦC is a fundamental matrix solution to (2.2). Conversely, if Ψ is another fundamental matrix solution to (2.2), then there exists a constant
nonsingular C ∈ Mn (K) such that Ψ(t) = Φ(t)C for all t ∈ I.
Proof. Since Φ is a fundamental matrix solution to (2.2), we have
(ΦC)0 = Φ0 C = (A(t)Φ)C = A(t)(ΦC),
and thus ΦC is a matrix solution to (2.2). Since Φ is a fundamental matrix solution to
(2.2), Theorem 2.2.8 implies that det Φ 6= 0. Also, since C is nonsingular, det C 6= 0. Thus,
det ΦC = det Φ det C 6= 0, and by Theorem 2.2.8, ΦC is a fundamental matrix solution to
(2.2).
Conversely, assume that Φ and Ψ are two fundamental matrix solutions. Since ΦΦ−1 = I,
0
0
taking the derivative of this expression gives Φ0 Φ−1 + Φ (Φ−1 ) = 0, and therefore (Φ−1 ) =
−Φ−1 Φ0 Φ−1 . We now consider the product Φ−1 Ψ. There holds
0
0
Φ−1 Ψ = Φ−1 Ψ + Φ−1 Ψ0
= −Φ−1 Φ0 Φ−1 Ψ + Φ−1 A(t)Ψ
= −Φ−1 A(t)ΦΦ−1 + Φ−1 A(t) Ψ
= −Φ−1 A(t) + Φ−1 A(t) Ψ
= 0.
0
Therefore, integrating (Φ−1 Ψ) gives Φ−1 Ψ = C, with C ∈ Mn (K) is a constant. Thus,
Ψ = CΦ. Furthermore, as Φ and Ψ are fundamental matrix solutions, det Φ 6= 0 and
det Ψ 6= 0, and therefore det C 6= 0.
Remark – Note that if Φ is a fundamental matrix solution to (2.2) and C ∈ Mn (K) is a constant
nonsingular matrix, then it is not necessarily true that CΦ is a fundamental matrix solution to
(2.2). See Exercise 2.3.
◦
Fund. Theory ODE Lecture Notes – J. Arino
2.2. Linear systems
2.2.3
41
Resolvent matrix
If t 7→ Φ(t) is a matrix solution of (2.2) on the interval I, then Φ0 (t) = A(t)Φ(t) on I. Thus,
by Proposition 2.2.3, there exists a fundamental matrix solution.
Definition 2.2.10 (Resolvent matrix). Let t0 ∈ I and Φ(t) be a fundamental matrix solution
of (2.2) on I. Since the columns of Φ are linearly independent, it follows that Φ(t0 ) is
invertible. The resolvent (or state transition matrix) of (2.2) is then defined as
R(t, t0 ) = Φ(t)Φ(t0 )−1 .
It is evident that R(t, t0 ) is the principal fundamental matrix solution at t0 (since
R(t0 , t0 ) = Φ(t0 )Φ(t0 )−1 = I). Thus system (2.2) has a principal fundamental matrix
solution at each point in I.
Proposition 2.2.11. The resolvent matrix satisfies the Chapman-Kolmogorov identities
1)
R(t, t) = I,
2)
R(t, s)R(s, u) = R(t, u),
as well as the identities
3)
R(t, s)−1 = R(s, t),
∂
4) ∂s
R(t, s)
= −R(t, s)A(s),
∂
5) ∂t
R(t, s)
= A(t)R(t, s).
Proof. First, for the Chapman-Kolmogorov identities. 1) is R(t, t) = Φ(t)Φ−1 (t) = I. Also,
2) gives
R(t, s)R(s, u) = Φ(t)Φ−1 (s)Φ(s)Φ−1 (u) = Φ(t)Φ−1 (u) = R(t, u).
The other equalities are equally easy to establish. Indeed,
−1
−1
Φ(t)−1 = Φ(s)Φ−1 (t) = R(s, t),
= Φ−1 (s)
R(t, s)−1 = Φ(t)Φ−1 (s)
whence 3). Also,
∂
∂
R(t, s) =
Φ(t)Φ−1 (s)
∂s
∂s ∂ −1
= Φ(t)
Φ (s)
∂s
As Φ is a fundamental matrix solution, Φ0 exists and Φ is nonsingular, and differentiating
ΦΦ−1 = I gives
∂
∂
∂ −1
−1
−1
Φ(s)Φ (s) = 0 ⇔
Φ(s) Φ (s) + Φ(s)
Φ (s) = 0
∂s
∂s
∂s
∂ −1
∂
⇔ Φ(s)
Φ (s) = −
Φ(s) Φ−1 (s)
∂s
∂s
∂ −1
∂
−1
⇔
Φ (s) = −Φ (s)
Φ(s) Φ−1 (s).
∂s
∂s
Fund. Theory ODE Lecture Notes – J. Arino
42
2. Linear systems
Therefore,
∂
R(t, s) = −Φ(t)Φ−1 (s)
∂s
∂
∂
−1
Φ(s) Φ (s) = −R(t, s)
Φ(s) Φ−1 (s).
∂s
∂s
Now, since Φ(s) is a fundamental matrix solution, it follows that ∂Φ(s)/∂s = A(s)Φ(s), and
thus
∂
R(t, s) = −R(t, s)A(s)Φ(s)Φ−1 (s) = −R(t, s)A(s),
∂s
giving 4). Finally,
∂
∂
R(t, s) = Φ(t)Φ−1 (s)
∂t
∂t
= A(t)Φ(t)Φ−1 (s) since Φ is a fundamental matrix solution
= A(t)R(t, s),
giving 5).
The role of the resolvent matrix is the following. Recall that, from Lemma 2.2.5, Φt0
defined by
Φt0 : S → Kn
x 7→ x(t0 ),
is a K-linear isomorphism from the space S to the space Kn . Then R is an application from
Kn to Kn ,
R(t, t0 ) : Kn → Kn
v 7→ R(t, t0 )v = w
such that
R(t, t0 ) = Φt ◦ Φ−1
t0
i.e.,
(R(t, t0 )v = w) ⇔ (∃x ∈ S, w = x(t), v = x(t0 )) .
Since Φt and Φt0 are K-linear isomorphisms, R is a K-linear isomorphism on Kn . Thus
R(t, t0 ) ∈ Mn (K) and is invertible.
Proposition 2.2.12. R(t, t0 ) is the only solution in Mn (K) of the initial value problem
d
M (t) = A(t)M (t)
dt
M (t0 ) = I,
with M (t) ∈ Mn (K).
Fund. Theory ODE Lecture Notes – J. Arino
2.2. Linear systems
43
Proof. Since d(R(t, t0 )v)/dt = A(t)R(t, t0 )v,
d
R(t, t0 ) v = (A(t)R(t, t0 )) v,
dt
for all v ∈ Rn . Therefore, R(t, t0 ) is a solution to M 0 = A(t)M . But, by Theorem 2.1.1, we
know the solution to the associated IVP to be unique, hence the result.
From this, the following theorem follows immediately.
Theorem 2.2.13. The solution to the IVP consisting of the linear homogeneous nonautonomous system (2.2) with initial condition x(t0 ) = x0 is given by
φ(t) = R(t, t0 )x0 .
2.2.4
Wronskian
Definition 2.2.14. The Wronskian of a system {x1 , . . . , xn } of solutions to (2.2) is given
by
W (t) = det(x1 (t), . . . , xn (t)).
Let vi = xi (t0 ). Then we have
xi (t) = R(t, t0 )vi ,
and it follows that
W (t) = det(R(t, t0 )v1 , . . . , R(t, t0 )vn )
= det R(t, t0 ) det(v1 , . . . , vn ).
The following formulae hold
Z
t
∆(t, t0 ) := det R(t, t0 ) = exp
trA(s)ds
t
Z 0t
W (t) = exp
(2.5a)
trA(s)ds det(v1 , . . . , vn ).
(2.5b)
t0
2.2.5
Autonomous linear systems
At this point, we know that solutions to (2.2) take the form φ(t) = R(t, t0 )x0 , but this
was obtained formally. We have no indication whatsoever as to the precise form of R(t, t0 ).
Typically, finding R(t, t0 ) can be difficult, if not impossible. There are however cases where
the resolvent can be explicitly computed. One such case is for autonomous linear systems,
which take the form
x0 (t) = Ax(t),
(2.6)
that is, where A(t) ≡ A. Our objective here is to establish the following result.
Fund. Theory ODE Lecture Notes – J. Arino
44
2. Linear systems
Lemma 2.2.15. If A(t) ≡ A, then R(t, t0 ) = e(t−t0 )A for all t, t0 ∈ I.
This result is deduced easily as a corollary to another result developped below, namely
Theorem 2.2.16. Note that in Lemma 2.2.15, the notation e(t−t0 )A involves the notion of
exponential of a matrix, which is detailed in Appendix A.10.
Because the reasoning used in constructing solutions to (2.6) is fairly straightforward,
we now detail this derivation. Using the intuition from one-dimensional linear equations, we
seek a λ ∈ K such that φλ (t) = eλt v be a solution to (2.6) with v ∈ Kn \ {0}. We have
φ0λ = λeλt v,
and thus φλ is a solution if, and only if,
λeλt v = Aeλt v
= eλt Av
⇔ λv = Av
⇔ (A − λI)v = 0 (with I the identity matrix).
As v = 0 is not the only solution, this implies that A − λI must not be invertible, and so
φλ is a solution ⇔ det(A − λI) = 0,
i.e., λ is an eigenvalue of A.
In the simple case where A is diagonalizable, there exists a basis (v1 , . . . , vn ) of Kn , with
v1 , . . . , vn the eigenvectors of A corresponding to the eigenvalues λ1 , . . . , λn . We then obtain
n linearly independent solutions φλi (t) = eλi (t−t0 ) , i = 1, . . . , n. The general solution is given
by
φ(t) = eλ1 (t−t0 ) x01 , . . . , eλn (t−t0 ) x0n ,
where x0i is the ith component of x0 , i = 1, . . . , n. In the general case, we need the notion
of matrix exponentials. Defining the exponential of matrix A as
A
e =
∞
X
An
k=0
n!
(see Appendix A.10), we have the following result.
Theorem 2.2.16. The global solution φ on K of (2.6) such that φ(t0 ) = x0 is given by
φ(t) = e(t−t0 )A x0 .
Proof. Assume φ = e(t−t0 )A x0 . Then φ(t0 ) = e0A x0 = Ix0 = x0 . Also,
!
∞
X
1
(t − t0 )n An x0
φ(t) =
n!
n=0
∞
X
1
=
(t − t0 )n An x0 ,
n!
n=0
Fund. Theory ODE Lecture Notes – J. Arino
2.2. Linear systems
45
so φ is a power series with radius of convergence R = ∞. Therefore, φ is differentiable on R
and
∞
X
1
φ (t) =
n(t − t0 )n−1 An x0
n!
n=1
0
=
∞
X
n=0
1
(n + 1)(t − t0 )n An+1 x0
(n + 1)!
∞
X
1
=
(t − t0 )n An+1 x0
n!
n=0
∞
X
1
=A
(t − t0 )n An x0
n!
n=0
!
= Aφ(t)
so φ is solution of (2.6). Since (2.6) is linear, solutions are unique and global.
The problem is now to evaluate the matrix etA . We have seen that in the case where A
is diagonalizable, solutions take the form
φ(t) = eλ1 (t−t0 ) x01 , . . . , eλn (t−t0 ) x0n ,
which implies that, in this case, the matrix R(t, t0 ) takes the form


eλ1 (t−t0 )
0
0
 0
eλ2 (t−t0 )
0 


R(t, t0 ) =  ..
.
.
.
 .

.
λn (t−t0 )
0
0
e
In the general case, we need the notion of generalized eigenvectors.
Definition 2.2.17 (Generalized eigenvectors). Let λ be an eigenvalue of the n × n matrix
A, with multiplicity m ≤ n. Then, for k = 1, . . . , m, any nonzero solution v of
(A − λI)k v = 0
is called a generalized eigenvector of A.
Theorem 2.2.18. Let A be a real n × n matrix with real eigenvalues λ1 , . . . , λn repeated
according to their multiplicity. Then there exists a basis of generalized eigenvectors for Rn .
And if {v1 , . . . , vn } is any basis of generalized eigenvectors for Rn , the matrix P = [v1 · · · vn ]
is invertible,
A = D + N,
where
P −1 DP = diag(λj ),
the matrix N = A − D is nilpotent of order k ≤ n, and D and N commute.
Fund. Theory ODE Lecture Notes – J. Arino
46
2. Linear systems
2.3
Affine systems
We consider the general (affine) problem (2.1), which we restate here for convenience. Let
x ∈ Rn , A : I → L(E) and B : I → E, where I ⊂ R and E is a normed vector space, we
consider the system
x0 (t) = A(t)x(t) + B(t)
(2.1)
2.3.1
The space of solutions
The first problem that we are faced with when considering system (2.1) is that the set of
solutions does not constitute a vector space; in particular, the superposition principle does
not hold. However, we have the following result.
Proposition 2.3.1. Let x1 , x2 be two solutions of (2.1). Then x1 − x2 is a solution of the
associated homogeneous equation (2.2).
Proof. Since x1 and x2 are solutions of (2.1),
x01 = A(t)x1 + B(t)
x02 = A(t)x2 + B(t)
Therefore
d
(x1 − x2 ) = A(t)(x1 − x2 )
dt
Theorem 2.3.2. The global solutions of (2.1) that are defined on I form an n dimensional
affine subspace of the vector space of maps from I to Kn .
Theorem 2.3.3. Let V be the vector space over R of solutions to the linear system x0 =
A(t)x. If ψ is a particular solution of the affine system (2.1), then the set of all solutions of
(2.1) is precisely
{φ + ψ, φ ∈ V }.
Practical rules:
1. To obtain all solutions of (2.1), all solutions of (2.2) must be added to a particular
solution of (2.1).
2. To obtain all solutions of (2.2), it is sufficient to know a basis of S 0 . Such a basis is
called a fundamental system of solutions of (2.2).
2.3.2
Construction of solutions
We have the following variation of constants formula.
Fund. Theory ODE Lecture Notes – J. Arino
2.3. Affine systems
47
Theorem 2.3.4. Let R(t, t0 ) be the resolvent of the homogeneous equation x0 = A(t)x associated to (2.1). Then the solution x to (2.1) is given by
Z t
x(t) = R(t, t0 ) +
R(t, s)B(s)ds
(2.7)
t0
Proof. Let R(t, t0 ) be the resolvent of x0 = A(t)x. Any solution of the latter equation is
given by
x(t) = R(t, t0 )v, v ∈ Rn
Let us now seek a particular solution to (2.1) of the form x(t) = R(t, t0 )v(t), i.e., using a
variation of constants approach. Taking the derivative of this expression of x, we have
d
[R(t, t0 )]v(t) + R(t, t0 )v 0 (t)
dt
= A(t)R(t, t0 )v(t) + R(t, t0 )v 0 (t)
x0 (t) =
Thus x is a solution to (2.1) if
since R(t, s)−1
by
A(t)R(t, t0 )v(t) + R(t, t0 )v 0 (t) = A(t)R(t, t0 )v(t) + B(t)
⇔ R(t, t0 )v 0 (t) = B(t)
⇔ v 0 (t) = R(t0 , t)B(t)
Rt
= R(s, t). Therefore, v(t) = t0 R(t0 , s)B(s)ds. A particular solution is given
Z
t
x(t) = R(t, t0 )
R(t0 , s)B(s)ds
t0
Z t
=
R(t, t0 )R(t0 , s)B(s)ds
t0
Z t
=
R(t, s)B(s)ds
t0
2.3.3
Affine systems with constant coefficients
We consider the affine equation (2.1), but with the matrix A(t) ≡ A.
Theorem 2.3.5. The general solution to the IVP
x0 (t) = Ax(t) + B(t)
x(t0 ) = x0
(2.8)
is given by
(t−t0 )A
x(t) = e
Z
t
x0 +
e(t−t0 )A B(s)ds
t0
Proof. Use Lemma 2.2.15 and the variation of constants formula (2.7).
(2.9)
Fund. Theory ODE Lecture Notes – J. Arino
48
2. Linear systems
2.4
2.4.1
Systems with periodic coefficients
Linear systems: Floquet theory
We consider the linear system (2.2) in the following case,
x0 = A(t)x
A(t + ω) = A(t), ∀t,
(2.10)
with entries of A(t) continuous on R.
Definition 2.4.1 (Monodromy operator). Associated to system (2.10) is the resolvent R(t, s).
For all s ∈ R, the operator
C(s) := R(s + ω, s)
is called the monodromy operator.
Theorem 2.4.2. If X(t) is a fundamental matrix for (2.10), then there exists a nonsingular
constant matrix V such that, for all t,
X(t + ω) = X(t)V.
This matrix takes the form
V = X −1 (0)X(ω),
and is called the monodromy matrix.
Proof. Since X is a fundamental matrix solution, there holds that X 0 (t) = A(t)X(t) for all t.
Therefore X 0 (t+ω) = A(t+ω)X(t+ω), and by periodicity of A(t), X 0 (t+ω) = A(t)X(t+ω),
which implies that X(t + ω) is a fundamental matrix of (2.10). As a consequence, by
Theorem 2.2.9, there exists a matrix V such that X(t + ω) = X(t)V .
Since at t = 0, X(ω) = X(0)V , it follows that V = X −1 (0)X(ω).
Theorem 2.4.3 (Floquet’s theorem, complex case). Any fundamental matrix solution Φ of
(2.10) takes the form
Φ(t) = P (t)etB
(2.11)
where P (t) and B are n × n complex matrices such that
i) P (t) is invertible, continuous, and periodic of period ω in t,
ii) B is a constant matrix such that Φ(ω) = eωB .
Proof. Let Φ be a fundamental matrix solution. From 2.4.2, the monodromy matrix V =
Φ−1 (0)Φ(ω) is such that Φ(t + ω) = Φ(t)V . By Theorem A.11.1, there exists B ∈ Mn (C)
Fund. Theory ODE Lecture Notes – J. Arino
2.4. Systems with periodic coefficients
49
such that eBω = V . Let P (t) = Φ(t)e−Bt , so Φ(t) = P (t)eBt . It is clear that P is continuous
and nonsingular. Also,
P (t + ω) = Φ(t + ω)e−B(t+ω)
= Φ(t)V e−B(ω+t)
= Φ(t)eBω e−Bω e−Bt
= Φ(t)e−Bt
= P (t),
proving the P is ω-periodic.
Theorem 2.4.4 (Floquet’s theorem, real case). Any fundamental matrix solution Φ of (2.10)
takes the form
Φ(t) = P (t)etB
(2.12)
where P (t) and B are n × n real matrices such that
i) P (t) is invertible, continuous, and periodic of period 2ω in t,
ii) B is a constant matrix such that Φ(ω)2 = e2ωB .
Proof. The proof works similarly as in the complex case, except that here, Theorem A.11.1
implies that there exists B ∈ Mn (R) such that e2ωB = V 2 . Let P (t) = Φ(t)e−Bt , so
Φ(t) = P (t)etB . It is clear that P is continuous and nonsingular. Also,
P (t + 2ω) = Φ(t + 2ω)e−(t+2ω)B
= Φ(t + ω)V e−(2ω+t)B
= Φ(t)V 2 e−(2ω+t)B
= Φ(t)e2ωB e−2ωB e−tB
= Φ(t)e−tB
= P (t),
proving the P is ω-periodic.
See [12, p. 87-90], [4, p. 162-179].
Theorem 2.4.5 (Floquet’s theorem, [4]). If Φ(t) is a fundamental matrix solution of the
ω-periodic system (2.10), then, for all t ∈ R,
Φ(t + ω) = Φ(t)Φ−1 (0)Φ(ω).
In addition, for each possibly complex matrix B such that
eωB = Φ−1 (0)Φ(ω),
there is a possibly complex ω-periodic matrix function t 7→ P (t) such that Φ(t) = P (t)etB for
all t ∈ R. Also, there is a real matrix R and a real 2ω-periodic matrix function t → Q(t)
such that Φ(t) = Q(t)etR for all t ∈ R.
Fund. Theory ODE Lecture Notes – J. Arino
50
2. Linear systems
Definition 2.4.6 (Floquet normal form). The representation Φ(t) = P (t)etR is called a
Floquet normal form.
In the case where Φ(t) = P (t)etB , we have dP (t)/dt = A(t)P (t) − P (t)B. Therefore,
letting x = P (t)z, we obtain x0 = P (t)x0 + dP (t)/dtx = P (t)A(t)x + A(t)P (t)x − P (t)Bx
−1
−1
z = P −1 (t)x, so z 0 = dP dt (t) x + P −1 (t)x0 = dP dt (t) P (t)z + P −1 (t)A(t)P (t)z
Definition 2.4.7 (Characteristic multipliers). The eigenvalues λ1 , . . . , λn of a monodromy
matrix B are called the characteristic multipliers of equation (2.10).
Definition 2.4.8 (Characteristic exponents). Numbers µ such that eµω is a characteristic
multiplier of (2.10) are called the Floquet exponents of (2.10).
Theorem 2.4.9 (Spectral mapping theorem). Let K = R or C. If C ∈ GLn (K) is written
C = eB , then the eigenvalues of C coincide with the exponentials of the eigenvalues of B,
with same multiplicity.
Definition 2.4.10 (Characteristic exponents). The eigenvalues λ1 , . . . , λn of a monodromy
matrix B are called the characteristic exponents of equation (2.10). The exponents ρ1 =
exp(2ωλ1 ), . . . , ρn = exp(2ωλn ) of the matrix Φ(ω)2 are called the (Floquet) multipliers of
(2.10).
Proposition 2.4.11. Suppose that X, Y are fundamental matrices for (2.10) and that X(t+
ω) = X(t)V , Y (t + ω) = Y (t)U . Then the monodromy matrices U and V are similar.
Proof. Suppose that X(t + ω) = X(t)V and Y (t + ω) = Y (t)U . But, by Theorem 2.2.9,
since X and Y are fundamental matrices for (2.10), there exists an invertible matrix C such
that X(t) = Y (t)C for all t. Thus, in particular, X(t + ω) = Y (t + ω)C, and so
C −1 U CX(t + ω) = Y (t + ω)C = Y (t)U C = X(t)C −1 U C,
since Y (t) = X(t)C −1 . It follows that V = C −1 U C, so U and V are similar.
From this Proposition, it follows that monodromy matrices share the same spectrum.
Corollary 2.4.12. All solutions of (2.10) tend to 0 as t → ∞ if and only if |ρj | < 1 for all
j (or <(λj ) < 0 for all j).
Let p be an eigenvector of Φ(ω)2 associated with a multiplier ρ. Then the solution
φ(t) = Φ(t)p of (2.10) satisfies the condition φ(t + 2ω) = ρφ(t). This is the origin of the
term multiplier.
Fund. Theory ODE Lecture Notes – J. Arino
2.4. Systems with periodic coefficients
2.4.2
51
Affine systems: the Fredholm alternative
We discuss here an extension of a theorem that was proved implicitly in Exercise 4, Assignment 2. Let us start by stating the result in question. We consider here the system
x0 = A(t)x + b(t),
(2.13)
where x ∈ Rn , A ∈ Mn (R) and b ∈ Rn , with A and b continuous and ω-periodic.
Theorem 2.4.13. If the homogeneous equation
x0 = A(t)x
(2.14)
associated to (2.13) has no nonzero solution of period ω, then (2.13) has for each function
f , a unique ω-periodic solution.
The Fredholm alternative concerns the case where there exists a nonzero periodic solution
of (2.14). We give some needed results before going into details. Consider (2.14). Associated
to this system is the so-called adjoint system, which is defined by the following differential
equation,
y 0 = −AT (t)y
(2.15)
Proposition 2.4.14. The adjoint equation has the following properties.
i) Let R(t, t0 ) be the resolvent matrix of (2.14). Then, the resolvent matrix of (2.15) is
RT (t0 , t).
ii) There are as many independent periodic solutions of (2.14) as there are of (2.15).
iii) If x is a solution of (2.14) and y is a solution of (2.15), then the scalar product
hx(t), y(t)i is constant.
∂
∂
Proof. i) We know that ∂s
R(t, s) = −R(t, s)A(s). Therefore, ∂s
RT (t, s) = −AT (s)RT (t, s).
As R(s, s) = I, the first point is proved.
ii) The solution of (2.15) with initial value y0 is RT (0, t)y0 . The initial value of a periodic
solution of (2.15) is y0 such that
RT (0, ω)y0 = y0
This can also be written as
T
R (0, ω) − I y0 = 0
or, taking the transpose,
y0T [R(0, ω) − I] = 0
Now, since R(0, ω) = R−1 (ω, 0), it follows that
y0T [R(0, ω) − I] = 0 ⇔ y0T R−1 (ω, 0) − I = 0
This is equivalent to y0T [R(0, ω) − I] = 0. The latter equation has as many solutions as
[R(0, ω) − I] x0 = 0; the number of these depends on the rank of R(ω, 0) − I.
Fund. Theory ODE Lecture Notes – J. Arino
52
2. Linear systems
iii) Recall that for differentiable functions a, b,
d
d
d
ha(t), b(t)i = h a(t), b(t)i + ha(t), b(t)i
dt
dt
dt
Thus
d
hx(t), y(t)i = hA(t)x(t), y(t)i + hx(t), −AT (t)y(t)i = 0
dt
Before we carry on to the actual Fredholm alternative in the context of ordinary differential equations, let us consider the problem in a more general setting. Let H be a Hilbert
space. If A ∈ L(H, H), the adjoint operator A∗ of A is the element of L(H, H) such that
∀u, v ∈ H,
hAu, vi = hu, A∗ vi
Let Img(A) be the image of A, Ker(A∗ ) be the kernel of A∗ . Then we have H = Img(A) ⊕
Ker(A∗ ).
Theorem 2.4.15 (Fredholm alternative). For the equation Af = g to have a solution, it is
necessary and sufficient that g be orthogonal to every element of Ker(A∗ ).
We now use this very general setting to prove the following theorem, in the context of
ODEs.
Theorem 2.4.16 (Fredholm alternative for ODEs). Consider (2.13) with A and f continuous and ω-periodic. Suppose that the homogeneous equation (2.14) has p independent
solutions of period ω. Then the adjoint equation (2.15) also has p independent solutions of
period p, which we denote y1 , . . . , yp . Then
i) If
Z
ω
hyk (t), b(t)idt = 0,
k = 1, . . . , p
(2.16)
0
then there exist p independent solutions of (2.13) of period ω, and,
ii) if this condition is not fulfilled, (2.13) has no nontrivial solution of period ω.
Proof. First, remark that x0 is the initial condition of a periodic solution of (2.13) if, and
only if,
Z ω
R(0, s)b(s)ds
(2.17)
[R(0, ω) − I] x0 =
0
By Theorem 2.3.4, the solution of (2.13) through (0, x0 ) is given by
Z t
x(t) = R(t, 0)x0 +
R(t, s)b(s)ds
0
Hence, at time ω,
Z
x(ω) = R(ω, 0)x0 +
ω
R(ω, s)b(s)ds
0
Fund. Theory ODE Lecture Notes – J. Arino
2.4. Systems with periodic coefficients
53
If x0 is the initial condition of a ω-periodic solution, then x(ω) = x0 , and so
Z ω
x0 − R(ω, 0) =
R(ω, s)b(s)ds
0
On the other hand, yk (0) is the initial condition of an ω-periodic solution yk if, and only
if,
T
R (0, ω) − I yk (0) = 0
Let C = R(0, ω) − I. We have that Rn = Img(C) ⊕ Ker(C T ). We now use the Fredholm
alternative in this context. There exists x0 such that
Z ω
Cx0 =
R(0, s)b(s)ds
0
if, and only if,
Z
ω
R(0, s)b(s)ds ∈ Img(C)
0
Rω
Indeed, from the Fredholm alternative, setting f = x0 and g = 0 R(0, s)b(s)ds, we have
that Cf = g has a solution if, and only if, g is orthogonal to every element of Ker(C T ), i.e.,
since Rn = Img(C) ⊕ Ker(C T ), if, and only if, g ∈ Img(C).
Now, y1 (0), . . . , yp (0) is a basis of Ker(C T ). It follows that there exists a solution of
(2.13) if, and only if, for all k = 1, . . . , p,
Z ω
∀k = 1, . . . , p, h
R(0, s)b(s)ds, yk (0)i = 0
0
Z ω
⇔ ∀k = 1, . . . , p,
hR(0, s)b(s), yk (0)ids = 0
0
Z ω
⇔ ∀k = 1, . . . , p,
hb(s), RT (0, s)yk (0)ids = 0
0
Z ω
⇔ ∀k = 1, . . . , p,
hb(s), yk (s)ids = 0
0
If these relations are satisfied, the set of vectors v such that
Z ω
Av =
R(0, s)b(s)ds
0
is of the form v0 + Ker(C T ), where v0 is one of these vectors; hence there exist p of them
which are independent and are initial conditions of the p independent ω-periodic solutions
of (2.13).
Example – The equation
x00 = f (t)
(2.18)
Fund. Theory ODE Lecture Notes – J. Arino
54
2. Linear systems
where f is ω-periodic, has solutions of period ω if, and only if,
Z ω
f (s)ds = 0
0
Let y =
x0 .
Then, differentiating y and substituting into (2.18), we have
y 0 = f (t)
Hence the system is
0 x
0 1
x
0
=
+
y
0 0
y
f (t)
Hence,
T
A =
0 0
1 0
and the adjoint equation ξ 0 = AT ξ has the periodic solution (0, a)T .
2.5
2.5.1
Further developments, bibliographical notes
A variation of constants formula for a nonlinear system with
a linear component
The variation of constants formula given in Theorem 2.3.4 can be extended.
Theorem 2.5.1 (Variation of constants formula). Consider the IVP
x0 = A(t)x + g(t, x)
x(t0 ) = x0 ,
(2.19a)
(2.19b)
where g : R × Rn → Rn a smooth function, and let R(t, t0 ) be the resolvent associated to the
homogeneous system x0 = A(t)x, with R defined on some interval I 3 t0 . Then the solution
φ of (2.19) is given by
Z t
φ(t) = R(t, t0 )x0 +
R(t, s)g(φ(s), s)ds,
(2.20)
t0
on some subinterval of I.
Proof. We proceed using a variation of constants approach. It is known that the general
solution to the homogeneous equation x0 = A(t)x associated to (2.19) is given by
φ(t) = R(t, t0 )x0 .
We seek a solution to (2.19) by assuming that φ(t) = R(t, t0 )v(t). We have
d
0
φ (t) =
R(t, t0 ) v(t) + R(t, t0 )v 0 (t)
dt
= A(t)R(t, t0 )v(t) + R(t, t0 )v 0 (t),
Fund. Theory ODE Lecture Notes – J. Arino
2.5. Further developments, bibliographical notes
55
from Proposition 2.2.11. For φ to be solution, it must satisfy the differential equation (2.19),
and thus
φ0 (t) = A(t)φ(t) + g(t, φ(t)) ⇔ A(t)R(t, t0 )v(t) + R(t, t0 )v 0 (t) = A(t)R(t, t0 )v(t) + g(t, φ(t))
⇔ R(t, t0 )v 0 (t) = g(t, φ(t))
⇔ v 0 (t) = R(t, t0 )−1 g(t, φ(t))
⇔ v 0 (t) = R(t0 , t)g(t, φ(t))
Z t
⇔ v(t) =
R(t0 , s)g(s, φ(s))ds + C,
t0
using Proposition 2.2.11 again. Therefore,
Z t
φ(t) = R(t, t0 )
R(t0 , s)g(s, φ(s))ds + C .
t0
Evaluating this expression at t = t0 gives φ(t0 ) = C, so C = x0 . Therefore,
Z t
φ(t) = R(t, t0 )x0 + R(t, t0 )
R(t0 , s)g(s, φ(s))ds
t0
Z t
= R(t, t0 )x0 +
R(t, t0 )R(t0 , s)g(s, φ(s))ds
t0
Z t
= R(t, t0 )x0 +
R(t, s)g(s, φ(s))ds,
t0
from Proposition 2.2.11.
Chapter 3
Stability of linear systems
3.1
Stability at fixed points
We consider here the autonomous equation (not necessarily linear),
x0 = f (x).
(3.1)
To emphasize the fact that we are dealing with flows, we write x(t, x0 ) the solution to (3.1),
at time t and satisfying at time t = 0 the initial condition x(0) = x0 .
Definition 3.1.1 (Fixed point). A fixed point of (3.1) is a point x∗ such that f (x∗ ) = 0.
This is evident, as a point such that f (x∗ ) = 0 satisfies (x∗ )0 = f (x∗ ) = 0, so that the
solution is constant when x = x∗ . Note also that this implies that x(t) = x∗ is a solution
defined on R.
Definition 3.1.2 (Stable equilibrium point). The fixed point x∗ is ( positively) stable if the
following two conditions hold:
i) There exists r > 0 such that if kx0 − x∗ k < r, then the solution x(t, x0 ) is defined for
all t ≥ 0. (This is automatically satisfied for flows).
ii) For any ε > 0, there exists δ > 0 such that kx0 − x∗ k < δ implies kx(t, x0 ) − x∗ k < ε.
Definition 3.1.3 (Asymptotically stable equilibrium point). If the equilibrium x∗ is (positively) stable and that additionally, there exists γ > 0 such that kx0 − x∗ k < γ implies
limt→∞ x(t, x0 ) = x∗ , then x∗ is (positively) asymptotically stable.
3.2
Affine systems with small coefficients
We consider here a linear system of the form
x0 = Ax + b(t)x
where b is continuous and small, i.e., limt→∞ b(t) = 0, with A ∈ Mn (R) and b ∈ R.
57
(3.2)
58
Fund. Theory ODE Lecture Notes – J. Arino
3. Stability of linear systems
Theorem 3.2.1. Suppose that all eigenvalues of A have negative real parts, and that b is
continuous and such that limt→∞ b(t) = 0. Then 0 is a g.a.s. equilibrium of (3.2).
The proof comes from [2, p. 156-157].
Proof. For any given (t0 , x0 ), t0 > 0, we have, from [2, Th 2.1, p. 37] about the existence
and uniqueness of the solutions to the linear equation x0 = A(t)x + g(t), that the (unique)
solution φt (x0 ) satisfying the initial condition φt0 (x0 ) = x0 exists for all t ≥ t0 .
by the variation of constants formula, using b(t)x as the inhomogeneous term, we can
express the solution by means of the equivalent integral equation, for t0 ≤ t < ∞,
Z t
(t−t0 )A
φt (x0 ) = e
x0 +
e(t−s)A b(s)φs (x0 )ds
(3.3)
t0
by the hypothesis on A, e(t−t0 )A is such that for t0 ≤ t < ∞, kΨ(t, t0 )k ≤ Ke−σ(t−t0 ) for
K > 0, σ > 0, where R(t, t0 ) = e(t−t0 )A is the fundamental matrix of the homogeneous part
of (3.2).
Since limt→∞ b(t) = 0, given any η > 0, there exists a number T ≥ t0 such that |b(t)| < η
for t ≥ T . We now use the variation of constants formula (3.3) with the point (T, φT (x0 ))
for initial condition. We have, for T ≤ t < ∞,
Z t
(t−T )A
φt (x0 ) = e
φT (x0 ) +
e(t−s)A b(s)φs (x0 )ds
T
Thus, using kΦ(t)k ≤ Ke−σ(t−t0 ) (with t0 = T ) and |b(t)| < η for t ≥ T , we obtain, for
T ≤ t < ∞,
Z t
−σ(t−T )
kφt (x0 )k ≤ Ke
kφT (x0 )k + Kη
e−σ(t−s) kφs (x0 )kds
T
σt
Multiplying both sides of this inequality by e and using Gronwall’s inequality (Appendix A.7)
with the function kφt (x0 )keσt , we obtain, for T ≤ t < ∞,
kφt (x0 )k ≤ KkφT (x0 )ke−(σ−Kη)(t−T )
(3.4)
From this we conclude that if 0 < η < σ/K, the solution φt (x0 ) will approach zero exponentially. This does not yet prove that the zero solution of (3.2) is stable. To do this, we compute
a bound on kφT (x0 )k. Returning to (3.3) and restricting t to the interval t0 ≤ t ≤ T , we
have
Z t
−σ(t−t0 )
e−σ(t−s) kφ(s, t0 , x0 )kds
kφt (x0 )k ≤ Ke
kx0 k + K1 K
t0
σt
where K1 = maxt0 ≤t≤T |b(t)|. Multiplying by e
obtain
and applying the Gronwall inequality we
kφt (x0 )k ≤ Kkx0 ke−σ(t−t0 )+K1 K(t−t0 )
≤ Kkx0 keK1 K(t−t0 ) ,
t0 ≤ t ≤ T
(3.5)
Fund. Theory ODE Lecture Notes – J. Arino
3.2. Affine systems with small coefficients
59
Therefore,
kφT (x0 )k ≤ Kkx0 keK1 K(T −t0 ) ,
t0 ≤ T
(3.6)
Thus we can make |φT (x0 )| small by choosing |x0 | sufficiently small. This together with (3.4)
gives the stability.
Indeed, substituting (3.6) into (3.4) gives, for T ≤ t < ∞,
kφt (x0 )k ≤ K 2 kx0 keK1 K(T −t0 ) e−(σ−Kη)(t−T )
Let then K2 = max KeK1 K(T −t0 ) , K 2 eK1 K(T −t0 ) . From (3.5) and (3.7) we have
K2 kx0 k
if t0 ≤ t ≤ T
kφt (x0 )k ≤
K2 kx0 ke−(σ−Kη)(t−T ) if T ≤ t < ∞
(3.7)
(3.8)
For a given matrix A, we can compute K and σ; we next pick any 0 < η < σ/K and then
T ≥ t0 so that |b(t)| < η for t ≥ T . We then compute K1 and K2 . Now, given any ε > 0,
choose δ < ε/K2 . Then from (3.8), if kx0 k < δ, kφt (x0 )k < ε for all t ≥ t0 so that the zero
solution is stable. From (3.8), it is clear that the zero solution is globally asymptotically
stable.
Corollary 3.2.2. Let all eigenvalues of A have negative real part, so that |eAt | ≤ Ke−σt for
some constants K > 0, σ > 0 and all t ≥ 0. Let b(t) be continuous for 0 ≤ t < ∞ and
suppose that there exists T > 0 such that |b(t)| < σ/K for t ≥ T . Then the zero solution of
(3.2) is globally asymptotically stable.
Theorem 3.2.3. Let all eigenvalues
of A have negative real part, and let b(t) be continuous
R∞
for 0 ≤ t < ∞ and such that 0 |b(s)|ds < ∞. Then the zero solution of (3.2) is globally
asymptotically stable.
60
Fund. Theory ODE Lecture Notes – J. Arino
3. Stability of linear systems
We give some notions of linear stability theory, in the case of the autonomous linear
system (2.6), repeated here for convenience:
x0 (t) = Ax(t).
(2.6)
We let wj = uj + ivj be a generalized eigenvector of A corresponding to an eigenvalue
λj = aj + ibj , with vj = 0 if bj = 0, and
B = {u1 , . . . , uk , uk+1 , vk+1 , . . . , um , vm }
be a basis of Rn , with n = 2m − k.
Definition 3.2.4 (Stable, unstable and center subspaces). The stable, unstable and center
subspaces of the linear system (2.6) are given, respectively, by
E s = Span{uj , vj : aj < 0},
E u = Span{uj , vj : aj > 0}
and
E c = Span{uj , vj : aj = 0}.
Definition 3.2.5. The mapping eAt : Rn → Rn is called the flow of the linear system (2.6).
The term flow is used since eAt describes the motion of points x0 ∈ Rn along trajectories
of (2.6).
Definition 3.2.6. If all eigenvalues of A have nonzero real part, that is, if E c = ∅, then
the flow eAt of system (2.6) is called a hyperbolic flow, and the system (2.6) is a hyperbolic
linear system.
Definition 3.2.7. A subspace E ⊂ Rn is invariant with respect to the flow eAt , or invariant
under the flow of (2.6), if eAt E ⊂ E for all t ∈ R.
Theorem 3.2.8. Let E be the generalized eigenspace of A associated to the eigenvalue λ.
Then AE ⊂ E.
Theorem 3.2.9. Let A ∈ Mn (R). Then
Rn = E s ⊕ E u ⊕ E c .
Furthermore, if the matrix A is the matrix of the linear autonomous system (2.6), then E s ,
E u and E c are invariant under the flow of (2.6).
Definition 3.2.10. If all the eigenvalues of A have negative (resp. positive) real parts, then
the origin is a sink (resp. source) for the linear system (2.6).
Theorem 3.2.11. The stable, center and unstable subspaces E S , E C and E U , respectively,
are invariant with respect to eAt , i.e., let x0 ∈ E S , y0 ∈ E C and z0 ∈ E U , then eAt x0 ∈ E S ,
eAt y0 ∈ E C and eAt z0 ∈ E U .
Fund. Theory ODE Lecture Notes – J. Arino
3.2. Affine systems with small coefficients
61
Definition 3.2.12 (Homeomorphism). Let X be a metric space and let A and B be subsets
of X. A homeomorphism h : A → B of A onto B is a continuous one-to-one map of A
onto B such that h−1 : B → A is continuous. The sets A and B are called homeomorphic
or topologically equivalent if there is a homeomorphism of A onto B.
Definition 3.2.13 (Differentiable manifold). An n-dimensional differentiable manifold M
(or a manifold of class C k ) is a connected metric space with an open covering {Uα } ( i.e.,
M = ∪α Uα ) such that
i) for all α, Uα is homeomorphic to the open unit ball in Rn , B = {x ∈ Rn : |x| < 1},
i.e., for all α there exists a homeomorphism of Uα onto B, hα : Uα → B,
ii) if Uα ∩ Uβ 6= ∅ and hα : Uα → B, hβ : Uβ → B are homeomorphisms, then hα (Uα ∩ Uβ )
and hβ (Uα ∩ Uβ ) are subsets of Rn and the map
h = hα ◦ h−1
β : hβ (Uα ∩ Uβ ) → hα (Uα ∩ Uβ )
is differentiable (or of class C k ) and for all x ∈ hβ (Uα ∩ Uβ ), the determinant of the
Jacobian, det Dh(x) 6= 0.
Remark – The manifold is analytic if the maps h = hα ◦ h−1
β are analytic.
◦
Next, recall that x∗ is an equilibrium point of (4.1) if f (x∗ ) = 0. An equilibrium point
x∗ is hyperbolic if the Jacobian matrix Df of (4.1) evaluated at x∗ , denoted Df (x∗ ), has no
eigenvalues with zero real part. Also, recall that the solutions of (4.1) form a one-parameter
group that defines the flow of the nonlinear differential equation (4.1). To be more precise,
consider the IVP consisting of (4.1) and an initial condition x(t0 ) = x0 . Let I(x0 ) be the
maximal interval of existence of the solution to the IVP. Let then φ : R × Rn → Rn be
defined as follows: For x0 ∈ Rn and t ∈ I(x0 ), φ(t, x0 ) = φt (x0 ) is the solution of the IVP
defined on its maximal interval of existence I(x0 ).
Example – Consider the (linear) ordinary differential equation x0 = ax, with a, x ∈ R. The
solution is φ(t, x0 ) = eat x0 , and satisfies the group property
φ(t + s, x0 ) = ea(t+s) x0 = eat (eas x0 ) = φ(t, eas x0 ) = φ(t, φ(s, x0 ))
For simplicity and without loss of generality since both results are local results, we assume
hereforth that x∗ = 0, i.e., that a change of coordinates has been performed translating x∗
to the origin. We also assume that t0 = 0.
Chapter 4
Linearization
We consider here the autonomous nonlinear system in Rn
x0 = f (x)
(4.1)
The object of this chapter is to show two results which link the behavior of (4.1) near a
hyperbolic equilibrium point x∗ to the behavior of the linearized system
x0 = Df (x∗ )(x − x∗ )
(4.2)
about that same equilibrium.
4.1
Some linear stability theory
We now give some notions of linear stability theory, in the case of the autonomous linear
system (2.6), repeated here for convenience:
x0 (t) = Ax(t).
(2.6)
We let wj = uj + ivj be a generalized eigenvector of A corresponding to an eigenvalue
λj = aj + ibj , with vj = 0 if bj = 0, and
B = {u1 , . . . , uk , uk+1 , vk+1 , . . . , um , vm }
be a basis of Rn , with n = 2m − k.
Definition 4.1.1 (Stable, unstable and center subspaces). The stable, unstable and center
subspaces of the linear system (2.6) are given, respectively, by
E s = Span{uj , vj : aj < 0},
E u = Span{uj , vj : aj > 0}
and
E c = Span{uj , vj : aj = 0}.
63
Fund. Theory ODE Lecture Notes – J. Arino
64
4. Linearization
Definition 4.1.2. The mapping eAt : Rn → Rn is called the flow of the linear system (2.6).
The term flow is used since eAt describes the motion of points x0 ∈ Rn along trajectories
of (2.6).
Definition 4.1.3. If all eigenvalues of A have nonzero real part, that is, if E c = ∅, then
the flow eAt of system (2.6) is called a hyperbolic flow, and the system (2.6) is a hyperbolic
linear system.
Definition 4.1.4. A subspace E ⊂ Rn is invariant with respect to the flow eAt , or invariant
under the flow of (2.6), if eAt E ⊂ E for all t ∈ R.
Theorem 4.1.5. Let E be the generalized eigenspace of A associated to the eigenvalue λ.
Then AE ⊂ E.
Theorem 4.1.6. Let A ∈ Mn (R). Then
Rn = E s ⊕ E u ⊕ E c .
Furthermore, if the matrix A is the matrix of the linear autonomous system (2.6), then E s ,
E u and E c are invariant under the flow of (2.6).
Definition 4.1.7. If all the eigenvalues of A have negative (resp. positive) real parts, then
the origin is a sink (resp. source) for the linear system (2.6).
Theorem 4.1.8. The stable, center and unstable subspaces E S , E C and E U , respectively,
are invariant with respect to eAt , i.e., let x0 ∈ E S , y0 ∈ E C and z0 ∈ E U , then eAt x0 ∈ E S ,
eAt y0 ∈ E C and eAt z0 ∈ E U .
Definition 4.1.9 (Homeomorphism). Let X be a metric space and let A and B be subsets
of X. A homeomorphism h : A → B of A onto B is a continuous one-to-one map of A
onto B such that h−1 : B → A is continuous. The sets A and B are called homeomorphic
or topologically equivalent if there is a homeomorphism of A onto B.
Definition 4.1.10 (Differentiable manifold). An n-dimensional differentiable manifold M
(or a manifold of class C k ) is a connected metric space with an open covering {Uα } ( i.e.,
M = ∪α Uα ) such that
i) for all α, Uα is homeomorphic to the open unit ball in Rn , B = {x ∈ Rn : |x| < 1},
i.e., for all α there exists a homeomorphism of Uα onto B, hα : Uα → B,
ii) if Uα ∩ Uβ 6= ∅ and hα : Uα → B, hβ : Uβ → B are homeomorphisms, then hα (Uα ∩ Uβ )
and hβ (Uα ∩ Uβ ) are subsets of Rn and the map
h = hα ◦ h−1
β : hβ (Uα ∩ Uβ ) → hα (Uα ∩ Uβ )
is differentiable (or of class C k ) and for all x ∈ hβ (Uα ∩ Uβ ), the determinant of the
Jacobian, det Dh(x) 6= 0.
Fund. Theory ODE Lecture Notes – J. Arino
4.2. The stable manifold theorem
Remark – The manifold is analytic if the maps h = hα ◦ h−1
β are analytic.
65
◦
Next, recall that x∗ is an equilibrium point of (4.1) if f (x∗ ) = 0. An equilibrium point
x is hyperbolic if the Jacobian matrix Df of (4.1) evaluated at x∗ , denoted Df (x∗ ), has no
eigenvalues with zero real part. Also, recall that the solutions of (4.1) form a one-parameter
group that defines the flow of the nonlinear differential equation (4.1). To be more precise,
consider the IVP consisting of (4.1) and an initial condition x(t0 ) = x0 . Let I(x0 ) be the
maximal interval of existence of the solution to the IVP. Let then φ : R × Rn → Rn be
defined as follows: For x0 ∈ Rn and t ∈ I(x0 ), φ(t, x0 ) = φt (x0 ) is the solution of the IVP
defined on its maximal interval of existence I(x0 ).
∗
Example – Consider the (linear) ordinary differential equation x0 = ax, with a, x ∈ R. The
solution is φ(t, x0 ) = eat x0 , and satisfies the group property
φ(t + s, x0 ) = ea(t+s) x0 = eat (eas x0 ) = φ(t, eas x0 ) = φ(t, φ(s, x0 ))
For simplicity and without loss of generality since both results are local results, we assume
hereforth that x∗ = 0, i.e., that a change of coordinates has been performed translating x∗
to the origin. We also assume that t0 = 0.
4.2
The stable manifold theorem
Theorem 4.2.1 (Stable manifold theorem). Let E be an open subset of Rn containing the
origin, let f ∈ C 1 (E), and let φt be the flow of the nonlinear system (4.1). Suppose that
f (0) = 0 and that Df (0) has k eigenvalues with negative real part and n − k eigenvalues
with positive real part. Then there exists a k-dimensional differentiable manifold S tangent
to the stable subspace E s of the linear system (4.2) at 0 such that for all t ≥ 0, φt (S) ⊂ S
and for all x0 ∈ S,
lim φt (x0 ) = 0
t→∞
and there exists an (n − k)-dimensional differentiable manifold U tangent to the unstable
subspace E u of (4.2) at 0 such that for all t ≤ 0, φt (U ) ⊂ U and for all x0 ∈ U ,
lim φt (x0 ) = 0
t→−∞
There are several approaches to the proof of this result. Hale [10] gives a proof which uses
functional analysis. The proof we give here comes from [18, p. 108-111], who derives it from
[6, p. 330-335]. It consists in showing that there exists a real nonsingular constant matrix
C such that if y = C −1 x then there are n − k real continuous functions yj = ψj (y1 , . . . , yk )
defined for small |yi |, i ≤ k, such that
yj = ψj (y1 , . . . , yk ) (j = k + 1, . . . , n)
define a k-dimensional differentiable manifold S̃ in y space. The stable manifold in S space
is obtained by applying the transformation P −1 to y so that x = P −1 y defines S in terms of
k curvilinear coordinates y1 , . . . , yk .
Fund. Theory ODE Lecture Notes – J. Arino
66
4. Linearization
Proof. System (4.1) can be written as
x0 = Df (0)x + F (x)
with F (x) = f (x) − Df (0)x. Since f ∈ C 1 (E), F ∈ C 1 (E), and F (0) = f (0) = 0 since
f (0) = 0. Also, DF (x) = Df (x) − Df (0) and so DF (0) = 0. To continue, we use the
following lemma (which we will not prove).
Lemma 4.2.2. Let E be an open subset of Rn containing the origin. If F ∈ C 1 (E), then
for all x, y ∈ Nδ (0) ⊂ E, there exists a ξ ∈ Nδ (0) such that
|F (x) − F (y)| ≤ kDF (ξ)k |x − y|
From Lemma 4.2.2, it follows that for all ε > 0 there exists a δ > 0 such that |x| ≤ δ and
|y| ≤ δ imply that
|F (x) − F (y)| ≤ ε|x − y|
Let C be an invertible n × n matrix such that
B=C
−1
P 0
Df (0)C =
0 Q
where the eigenvalues λ1 , . . . , λk of the k × k matrix P have negative real part and the
eigenvalues λk+1 , . . . , λn of the (n − k) × (n − k) matrix Q have positive real part. Let α > 0
be chosen small enough that for j = 1, . . . , k, <(λj ) < −α < 0. Let y = C −1 x, we have
y 0 = C −1 x0
= C −1 Df (0)x + C −1 F (x)
= C −1 Df (0)Cy + C −1 F (Cy)
= By + G(y)
where G(y) = C −1 F (Cy). Since F ∈ C 1 (E), G ∈ C 1 (Ẽ), where Ẽ = C −1 (E). Also,
Lemma 4.2.2 applies to G.
Now consider the system
y 0 = By + G(y)
(4.3)
and let
Pt e
0
0 0
U (t) =
and V (t) =
0 0
0 eQt
Then U 0 = BU , V 0 = BV and eBt = U (t)+V (t). Choosing as previously noted α sufficiently
small, we can then choose K > 0 large enough and σ > 0 small enough that
kU (t)k ≤ Ke−(α+σ)t for all t ≥ 0
and
kV (t)k ≤ Ke−σt for all t ≤ 0
Fund. Theory ODE Lecture Notes – J. Arino
4.2. The stable manifold theorem
67
Consider now the integral equation
t
Z
Z
∞
U (t − s)G(u(s, a))ds −
u(t, a) = U (t)a +
V (t − s)G(u(s, a))ds
0
(4.4)
t
where a, u ∈ Rn and a is a constant vector. We can solve this equation using the method of
successive approximations. Indeed, let
u(0) (t, a) = 0
and
u
(j+1)
t
Z
Z
(j)
∞
V (t − s)G(u(j) (s, a))ds
U (t − s)G(u (s, a))ds −
(t, a) = U (t)a +
0
(4.5)
t
We show by induction that for all j = 1, . . . and t ≥ 0,
|u(j) (t, a) − u(j−1) (t, a)| ≤
K|a|e−αt
2j−1
(4.6)
Clearly, (4.6) holds for j = 1 since
Z
(1)
|u (t, a)| ≤ kU (t)ak +
t
Z
kU (t − s)G(u(s, a))kds +
0
∞
kV (t − s)G(u(s, a))kds
t
Now suppose that (4.6) holds for j = k. We have
Z
t
|u(k+1) (t, a) − u(k) (t, a)| = U (t − s) G(u(k+1) (s, a)) − G(u(k) (s, a)) ds
0
Z ∞
−
V (t − s) G(u(k+1) (s, a)) − G(u(k) (s, a)) ds
t
Z t
≤
kU (t − s)k G(u(k+1) (s, a)) − G(u(k) (s, a)) ds
Z 0∞
+
kV (t − s)k G(u(k+1) (s, a)) − G(u(k) (s, a)) ds
t
which, since G verifies a Lipschitz-type condition as given by Lemma 4.2.2, implies that
there exists ε > 0 such that
|u
(k+1)
Z
(k)
(t, a) − u (t, a)| ≤ ε
t
kU (t − s)k u(k+1) (s, a) − u(k) (s, a) ds
Z 0∞
+ε
t
kV (t − s)k u(k+1) (s, a) − u(k) (s, a) ds
Fund. Theory ODE Lecture Notes – J. Arino
68
4. Linearization
Using the bounds on kU k and kV k as well as the induction hypothesis (4.6), it follows that
|u
(k+1)
(k)
t
K|a|e−αs
ds
2k−1
0
Z ∞
K|a|e−αs
+ε
Keσ(t−s)
ds
2k−1
t
εK 2 |a|e−αt εK 2 |a|e−αt
≤
+
σ2k−1
σ2k−1
Z
(t, a) − u (t, a)| ≤ ε
Ke−(α+σ)(t−s)
which, if we choose ε < σ/(4K), i.e., εK/σ < 1/4, implies that
1 1 K|a|e−αt
K|a|e−αt
(k+1)
(k)
=
|u
(t, a) − u (t, a)| <
+
4 4
2k−1
2k
(4.7)
Note that for G to satisfy a Lipschitz-type condition, we must choose K|a| < δ/2, i.e.,
|a| < δ/(2K). Then, by induction, (4.6) holds for all t ≥ 0 and j = 1, . . ..
As a consequence, for t ≥ 0, n > m > N ,
|u(n) (t, a) − u(m) (t, a)| ≤
∞
X
|u(j+1) (t, a) − u(j) (t, a)|
j=N
∞
X
1
K|a|
=
≤ K|a|
2j
2N −1
j=N
As this last quantity approaches 0 as N → ∞, it follows that {u(j) (t, a)} is a Cauchy sequence
(of continuous functions).
It follows that
lim u(j) (t, a) = u(t, a)
j→∞
uniformly for all t ≥ 0 and |a| < δ/(2K). From the uniform convergence, we deduce that
u(t, a) is continuous. Now taking the limit as j → ∞ in both sides of (4.5), it follows that
u(t, a) satisfies the integral equation (4.4) and as a consequence, the differential equation
(4.3).
Since G ∈ C 1 (Ẽ), it follows from induction on (4.5) that u(j) (t, a) is a differentiable
function of a for |a| < δ/(2K) and t ≥ 0. Since u(j) (t, a) → u(t, a) uniformly, it then follows
that u(t, a) is differentiable for t ≥ 0 and |a| < δ/(2K). The estimate (4.7) implies that
|u(t, a)| ≤ 2K|a|e−αt
(4.8)
for t ≥ 0 and |a| < δ/(2K).
It is clear from (4.4) that the last n − k components of a do not enter the computation
of u(t0 , a) and may thus be taken as zero. So the components uj (t, a) of the solution u(t, a)
satisfy the initial conditions
uj (0, a) = aj for j = 1, . . . , k
Fund. Theory ODE Lecture Notes – J. Arino
4.3. The Hartman-Grobman theorem
69
and
Z
∞
uj (0, a) = −
V (−s)G(u(s, a1 , . . . , ak , 0, . . . , 0))ds
0
for j = k + 1, . . . , n
j
where ( )j denotes the jth component. For j = k + 1, . . . , n, define the functions
ψj (a1 , . . . , ak ) = uj (0, a1 , . . . , ak , 0, . . . , 0)
(4.9)
The initial values yj = uj (0, a1 , . . . , ak , 0, . . . , 0) then satisfy
yj = ψj (y1 , . . . , yk ) for j = k + 1, . . . , n
p
which defines a differentiable manifold S̃ in y space for y12 + · · · + yk2 < δ/(2K). Furthermore, if y(t) is a solution of the differential equation (4.3) with y(0) ∈ S̃, i.e., with
y(0) = u(0, a), then
y(t) = u(t, a)
It follows from the estimate (4.8) that if y(t) is a solution of (4.3) with y(0) ∈ S̃, then
y(t) → 0 as t → ∞. It can also be shown that if y(t) is a solution of (4.3) with y(0) 6∈ S̃,
then y(t) 6→ 0 as t → ∞; see [6, p. 332].
This implies, as φt satisfies the group property φs+t (x0 ) = φs (φt )(x0 ), that if y(0) ∈ S̃,
then y(t) ∈ S̃ for all t ≥ 0. And it can be shown as in [6, Th 4.2, p. 333] that
∂ψj
(0) = 0
∂yi
for i = 1, . . . , k and j = k + 1, . . . , n, i.e., that the differentiable manifold S̃ is tangent to
the stable subspace E s = {y ∈ Rn : y1 = · · · = yk = 0} of the linear system y 0 = By at 0.
The existence of the unstable manifold Ũ of (4.3) is established the same way, but considering a reversal of time, t → −t, i.e., considering the system
y 0 = −By − G(y)
The stable manifold for this system is the unstable manifold Ũ of (4.3). In order to determine
the (n − k)-dimensional manifold Ũ using the above process, the vector y has to be replaced
by the vector (yk+1 , . . . , yn , y1 , . . . , yk ).
4.3
The Hartman-Grobman theorem
Adapted from [4, p. 311].
Theorem 4.3.1 (Hartman-Grobman). Suppose that 0 is an equilibrium point of the nonlinear system (4.1). Let ϕt be the flow of (4.1), and ψt be the flow of the linearized system
x0 = Df (0)x. If 0 is a hyperbolic equilibrium, then there exists an open subset D of Rn
containing 0, and a homeomorphism G with domain in D such that G(ϕt (x)) = ψt (G(x))
whenever x ∈ D and both sides of the equation are defined.
Fund. Theory ODE Lecture Notes – J. Arino
70
4. Linearization
We follow here [18].
Theorem 4.3.2 (Hartman-Grobman). Let E be an open subset of Rn containing the origin,
let f ∈ C 1 (E), and let φt be the flow of the nonlinear system (4.1). Suppose that f (0) = 0
and that the matrix A = Df (0) has no eigenvalue with zero real part.
Then there exists a homeomorphism H of an open set U containing the origin onto an
open set V containing the origin such that for each x0 ∈ U , there is an open interval I0 ⊂ R
containing 0 such that for all x0 ∈ U and t ∈ I0 ,
H ◦ φt (x0 ) = eAt H(x0 );
i.e., H maps trajectories of (4.1) near the origin onto trajectories of x0 = Df (0)x near the
origin and preserves the parametrization by time.
Proof. Suppose that f ∈ C 1 (E), f (0) = 0 (i.e., 0 is an equilibrium) and A = Df (0) the
jacobian matrix of f at 0.
1. As we have assumed that the matrix A has no eigenvalues with zero real part (i.e., 0
is an hyperbolic equilibrium point), we can write A in the form
P 0
A=
0 Q
where P has only eigenvalues with negative real parts and Q has only eigenvalues with
positive real parts.
2. Let φt be the flow of the nonlinear system (4.1). Let us write the solution as
y(t, y0 , z0 )
x(t, x0 ) = φt (x0 ) =
z(t, y0 , z0 )
where
y
x 0 = 0 ∈ Rn
z0
has been decomposed as y0 ∈ E s , the stable subspace of A, and z0 ∈ E u , the unstable
subspace of A.
3. Let
Ỹ (y0 , z0 ) = y(1, y0 , z0 ) − eP y0
and
Z̃(y0 , z0 ) = z(1, y0 , z0 ) − eQ z0
Then Ỹ (0) = Ỹ (0, 0) = y(1, 0, 0) = 0. The same is true of Z̃(0) = 0. Also, DỸ (0) =
DZ̃(0) = 0. Since f ∈ C 1 (E), Ỹ (y0 , z0 ) and Z̃(y0 , z0 ) are continuously differentiable. Thus
kDỸ (y0 , z0 )k ≤ a
and
kDZ̃(y0 , z0 )k ≤ a
Fund. Theory ODE Lecture Notes – J. Arino
4.3. The Hartman-Grobman theorem
71
on the compact set |y0 |2 + |z0 |2 ≤ s20 . By choosing s0 sufficiently small, we can make a as
small as we like. We let Y (y0 , z0 ) and Z(y0 , z0 ) be smooth functions, defined by
Ỹ (y0 , z0 ) for |y0 |2 + |z0 |2 ≤ s20
0
for |y0 |2 + |z0 |2 ≥ s20
2
Z̃(y0 , z0 ) for |y0 |2 + |z0 |2 ≤ s20
0
for |y0 |2 + |z0 |2 ≥ s20
2
Y (y0 , z0 ) =
and
Z(y0 , z0 ) =
By the mean value theorem,
p
|Y (y0 , z0 )| ≤ a |y0 |2 + |z0 |2 ≤ a(|y0 | + |z0 |)
and
|Z(y0 , z0 )| ≤ a
p
|y0 |2 + |z0 |2 ≤ a(|y0 | + |z0 |)
for all (y0 , z0 ) ∈ Rn . Let B = eP and C = eQ . Assuming that P and Q have been normalized
in a proper way, we have
b = kBk < 1 and c = kC −1 k < 1
4. For
define the transformations
and
y
x=
∈ Rn
z
By
L(y, z) =
Cz
By + Y (y, z)
T (y, z) =
Cz + Z(y, z)
i.e., L(x) = eA x and, locally, T (x) = φ1 (x). Then the following lemma holds, which we
prove later.
Lemma 4.3.3. There exists a homeomorphism H of an open set U containing the origin
onto an open set V containing the origin such that
H ◦T =L◦H
5. We let H0 be the homeomorphism defined above and Lt and T t be the one-parameter
families of transformations defined by
Lt (x0 ) = eAt x0 and T t (x0 ) = φt (x0 )
Define
Z
H=
0
1
L−s H0 T s ds
Fund. Theory ODE Lecture Notes – J. Arino
72
4. Linearization
It follows from the above lemma that there exists a neighborhood of the origin for which
Z 1
t
LH=
Lt−s H0 T s−t dsT t
0
Z 1−t
L−s H0 T s dsT t
=
−t
Z 0
Z 1−t
−s
s
−s
s
=
L H0 T ds +
L H0 T ds T t
−t
0
Z 1
=
L−s H0 T s dsT t
0
= HT t
since by the above lemma, H0 = L−1 H0 T which implies that
Z 0
Z 0
−s
s
L H0 T ds =
L−s−1 H0 T s+1 ds
−s
−t
Z 1
=
L−s H0 T s ds
1−t
Thus H ◦ T t = Lt H or equivalently
H ◦ φt (x0 ) = eAt H(x0 )
and it can be shown that H is a homeomorphism on Rn . The outline of the proof is
complete.
We now prove Lemma 4.3.3.
Proof. We use the method of successive approximations. For x ∈ Rn , let
Φ(y, z)
H(x) =
Ψ(y, z)
Then H ◦ T = L ◦ H is equivalent to the pair of equations
BΦ(y, z) = Φ(By + Y (y, z), Cz + Z(y, z))
CΨ(y, z) = Ψ(By + Y (y, z), Cz + Z(y, z))
(4.10a)
(4.10b)
Successive approximations for (4.10b) are defined by
Ψ0 (y, z) = z
Ψk+1 (y, z) = C −1 Ψk (By + Y (y, z), Cz + Z(y, z))
(4.11)
It can be shown by induction that for k = 0, 1, . . . the functions Ψk are continuous and such
that Ψk (y, z) = z for |y| + |z| ≥ 2s0 .
Fund. Theory ODE Lecture Notes – J. Arino
4.3. The Hartman-Grobman theorem
73
Let us now prove that {Ψk } is a Cauchy sequence. For this, we show by induction that
for all j ≥ 1,
|Ψj (y, z) − Ψj−1 (y, z)| ≤ M rj (|y| + |z|)δ
(4.12)
where r = c[2 max(a, b, c)]δ with δ ∈ (0, 1) chosen sufficiently small that r < 1 (which is
possible since c < 1) and M = ac(2s0 )1−δ /r. Inequality (4.12) is satisfied for j = 1 since
|Ψ1 (y, z) − Ψ0 (y, z)| = |C −1 Ψ0 (By + Y (y, z), Cz + Z(y, z)) − z|
= |C −1 (Cz + Z(y, z)) − z|
= |C −1 Z(y, z)|
≤ kC −1 k |Z(y, z)|
≤ ca(|y| + |z|)
≤ M r(|y| + |z|)δ
since Z(y, z) = 0 for |y| + |z| ≥ 2s0 . Now assuming that (4.12) holds for j = k gives
|Ψk+1 (y, z) − Ψk (y, z)| = |C −1 Ψk (By + Y (y, z), Cz + Z(y, z)) − C −1 Ψk−1 (By + Y (y, z), Cz + Z(y, z))|
= |C −1 (Ψk − Ψk−1 )|
≤ kC −1 k |Ψk − Ψk−1 |
which, using induction hypothesis (4.12) and c = kC −1 k gives
δ
≤ cM rk |By + Y (y, z)| + |Cz + Z(y, z)|
δ
≤ cM rk b|y| + 2a(|y| + |z|) + c|z|
δ
δ
≤ cM rk 2 max(a, b, c) |y| + |z|
≤ M rk+1 |y| + |z|)δ
Using the same type of argument as in the proof of the stable manifold theorem, Ψk is
thus a Cauchy sequence of continuous functions that converges uniformly as k → ∞ to a
continuous function Ψ(y, z). Also, Ψ(y, z) = z for |y| + |z| ≥ 2s0 . Taking limits in (4.11)
and left-multiplying by C shows that Ψ(y, z) is a solution of (4.10b).
Now for (4.10a). This equation can be written
B −1 Φ(y, z) = Φ(B −1 y + Y1 (y, z), C −1 z + Z1 (y, z))
(4.13)
where Y1 and Z1 occur in the inverse of T , which exists provided that a is small enough (i.e.,
s0 is sufficiently small),
−1
B y + Y1 (y, z)
−1
T (y, z) =
C −1 z + Z1 (y, z)
Successive approximations with Φ0 (y, z) = y can then be used as above (since b = kBk < 1)
to solve (4.13).
Fund. Theory ODE Lecture Notes – J. Arino
74
4. Linearization
Therefore, we obtain the continuous map
Φ(y, z)
H(y, z) =
Ψ(y, z)
and it follows as in [11, p. 248-249] that H is a homeomorphism of Rn onto Rn .
4.4
4.4.1
Example of application
A chemostat model
To illustrate the use of the theorems in this chapter, we take an example of nonlinear
system, a system of two nonlinear differential equations modeling a biological device called
a chemostat. Without going into details, the system is the following.
dS
= D(S 0 − S) − µ(S)x
dt
dx
= (µ(S) − D)x
dt
(4.14a)
(4.14b)
The parameters S 0 and D, respectively the input concentration and the dilution rate, are
real and positive. The function µ is the growth function. It is generally assumed to satisfy
µ(0) = 0, µ0 > 0 and µ00 < 0.
To be complete, one should verify that the positive quadrant is positively invariant under
the flow of (4.14), i.e., that for S(0) ≥ 0 and x(0) ≥ 0, solutions remain nonnegative for all
positive times, and similar properties. But since we are here only interested in applications
of the stable manifold theorem, we proceed to a very crude analysis, and will not deal with
this point.
Note that in vector form, the system is noted
ξ 0 = f (ξ)
with ξ = (S, x)T and
f (ξ) =
D(S 0 − S) − µ(S)x
(µ(S) − D)x
Equilibria of the system are found by solving f (ξ) = 0. We find two, the first one situated
on one of the boundaries of the positive quadrant,
ξT∗ = (ST∗ , x∗T ) = (S 0 , 0)
the second one in the interior,
ξI∗ = (S ∗ , x∗ ) = (λ, S 0 − λ)
where λ is such that µ(λ) = D. Note that this implies that if λ ≥ S 0 , ξT∗ is the only
equilibrium of the system.
Fund. Theory ODE Lecture Notes – J. Arino
4.4. Example of application
75
At an arbitrary point ξ = (S, x), the Jacobian matrix is given by
−D − µ0 (S)x −µ(S)
Df (ξ) =
µ0 (S)x
µ(S) − D
Thus, at the trivial equilibrium ξT∗ ,
Df (ξT∗ )
=
−D
−µ(S 0 )
0 µ(S 0 ) − D
We have two eigenvalues, −D and µ(S 0 ) − D. Let us suppose that µ(S 0 ) − D < 0. Note that
this implies that ξT∗ is the only equilibrium, since, as we have seen before, ξI∗ is not feasible
if λ > S 0 .
As the system has dimensionality 2, and that the matrix Df (ξT∗ ) has two negative eigenvalues, the stable manifold theorem (Theorem 4.2.1) states that there exists a 2-dimensional
differentiable manifold M such that
• φt (M) ⊂ M
• for all ξ0 ∈ M, limt→∞ φt (ξ0 ) = ξT∗ .
• At ξT∗ , M is tangent to the stable subspace E S of the linearized system ξ 0 = Df (ξT∗ )(ξ−
ξT∗ ).
Since there are no eigenvalues with positive real part, there does not exist an unstable
manifold in this case. Let us now caracterize the nature of the stable subspace E S . It is
obtained by studying the linear system
ξ 0 = Df (ξT∗ )(ξ − ξT∗ )
−D
−µ(S 0 )
S − S0
=
x
0 µ(S 0 ) − D
0
0
−D(S − S ) − µ(S )x
=
(µ(S 0 ) − D)x
(4.15)
Of course, the Jacobian matrix associated to this system is the same as that of the nonlinear
system (at ξT∗ ). Associated to the eigenvalue −D is the eigenvector v1 = (1, 0)T , to µ(S 0 )−D
is v2 = (−1, 1)T .
The stable subspace is thus given by Span (v1 , v2 ), i.e., the whole of R2 .
In fact, the stable manifold of ξT∗ is the whole positive quadrant, since all solutions limit
to this equilibrium. But let us pretend that we do not have this information, and let us try
to find an approximation of the stable manifold.
4.4.2
A second example
This example is adapted from [18, p. 111]. Consider the nonlinear system
x0 = −x − y 2
y 0 = x2 + y
(4.16)
Fund. Theory ODE Lecture Notes – J. Arino
76
4. Linearization
From the nullclines equations, it is clear that (x, y) = (0, 0) is the only equilibrium point.
At (0, 0), the Jacobian matrix of (4.16) is given by
−1 0
J=
0 1
The linearized system at 0 is
x0 = −x
y0 = y
(4.17)
So the eigenvalues are 1 and −1, with associated eigenvectors (1, 0)T and (0, 1)T , respectively. Therefore, the stable manifold theorem (Theorem 4.2.1) implies that there exists a
1-dimensional stable (differentiable) manifold S such that φt (S) ⊂ S and limt→∞ φt (x0 ) = 0
for all x0 ∈ S, and a 1-dimensional unstable (differentiable) manifold U such that φt (U ) ⊂ U
and limt→−∞ φt (x0 ) for all x0 ∈ U . Furthermore, at 0, S is tangent to the stable subspace
E S of (4.17), and U is tangent to the unstable subspace E U of (4.17).
The stable subspace E S is given by Span (v1 ), with v1 = (0, 1)T , i.e., the y-axis. The
unstable subspace E U is Span (v2 ), with v2 = (1, 0)T , i.e., the x-axis. The behavior of this
system is illustrated in Figure 4.1.
0.05
0.04
0.03
0.02
0.01
y
0
−0.01
−0.02
−0.03
−0.04
−0.05
−0.05
−0.04
−0.03
−0.02
−0.01
0
0.01
0.02
0.03
0.04
0.05
x
Figure 4.1: Vector field of system (4.16) in the neighborhood of 0.
To be more precise about the nature of the stable manifold S, we proceed as follows.
First of all, as A is in diagonal form, we have
−1 0
A=B=
0 1
and C = I. Also, F (ξ) = G(ξ) =
−y 2
. Here, the matrices P and Q are in fact scalars,
x2
Fund. Theory ODE Lecture Notes – J. Arino
4.4. Example of application
77
P = −1 and Q = 1. Thus
U (t) =
e−t 0
,
0 0
V (t) =
0 0
0 et
Finally, a = (a1 , 0)T . So now we can use successive approximations to find an approximate
solution to the integral equation (4.4), which here takes the form
−t Z t −(t−s) 2 Z ∞
0
e a1
−e
u2 (s)
u(t, a) =
+
ds −
ds
0
0
e(t−s) u21 (s)
0
t
To construct the sequence of successive approximations, we start with u(t, a) = (0, 0)T , then
compute the successive terms using equation (4.5), which takes the form
−t Z t −(t−s) Z ∞
0
0
e a1
e
0
(j)
(j+1)
(j)
G u (s) ds −
u
(t, a) =
+
G
u
(s)
ds
0
0
0
0 e(t−s)
t
0
2 
2 


(j)
(j)
−t Z t −(t−s) Z
∞
u (s)
0
0
e a1
e
0  u2 (s) 
 2
2 ds −
2  ds
+
=
(t−s)
(j)
(j)
0
0
e
0
0
t
0
u (s)
u (s)
1
=
e−t a1
0
Z
t
+
−e−(t−s)
1
2 !
(j)
u2 (s)
Z
0
0
0
∞
ds −
2
(j)
e(t−s) u1 (s)
t
!
ds
Therefore,
(1)
u (t, a) = U (t)a =
since u(0) (t, a) =
e−a a1
0
! (0)
0
u1 (t, a)
=
.
(0)
0
u2 (t, a)
Then,
(2)
u (t, a) =
−t
Z
−
e a1
0
e−t a1
− 13 a21 e−2t
=
=
e−t a1
0
t
∞
0
2 ds
e(t−s) (e−s a1 )
0
− 1 2 −2t
ae
3 1
and continuing this process, we find
(3)
u (t, a) =
1
e−t a1 + 27
(e−4t − e−t )a41
− 13 a21 e−2t
Also, it is possible to show that u(4) (t, a) − u(3) (t, a) = O(a51 ).
The stable manifold S is 1-dimensional, so here it has the form ψ2 (a1 ) = u2 (0, a1 , 0), and
is here approximated by
1
ψ2 (a1 ) = − a21 + O(a51 )
3
Fund. Theory ODE Lecture Notes – J. Arino
78
4. Linearization
as a1 → 0. Thus S is approximated by
y=−
as x → 0.
x2
+ O(x5 )
3
Chapter 5
Exponential dichotomy
Our aim here is to show the equivalent of the Hartman-Grobman theorem for linear systems
with variable coefficients. Compared to other results we have seen so far, this is a much
more recent field. The first results were shown in the 60s by Lin. We give here only the
most elementary results. For more details, see, e.g., [13].
We consider the linear system of differential equations
dx
= A(t)x
dt
(5.1)
where the n × n matrix A(t) is continuous on the real axis.
5.1
Exponential dichotomy
Definition 5.1.1 (Exponential dichotomy). Let X(t) be a the fundamental matrix solution
of (5.1). If X(t) and X −1 (s) can be decomposed into the following forms
X(t) = X1 (t) + X2 (t)
X (s) = Z1 (s) + Z2 (s)
X(t)Z −1 (s) = X1 (t)Z1 (s) + X2 (t)Z2 (s)
−1
and satisfy the conditions that there exists α, β, positive constants such that
kX1 (t)Z1 (s)k ≤ βe−α(t−s) , t ≥ s
kX2 (t)Z2 (s)k ≤ βeα(t−s) ,
s≥t
where
X1 (t) = (X11 (t), 0), X2 (t) = (0, X12 (t)),
Z11 (s)
0
Z1 (s) =
, Z2 (s) =
,
0
Z21 (s)
79
(5.2)
80
Fund. Theory ODE Lecture Notes – J. Arino
5. Exponential dichotomy
or there is a projection P on the stable manifold such that
kX(t)P X −1 (s)k ≤ βe−α(t−s) , t ≥ s
kX(t)(I − P )X −1 (s)k ≤ βeα(t−s) ,
s≥t
(5.3)
then we say that the system (5.1) admits exponential dichotomy.
Remark – A matrix P defines a projection if it is such that P 2 = P .
◦
Another definition [1].
Definition 5.1.2. Let Φ(t, s), Φ(t, t) = I, be the principal matrix solution of (5.1). We
say that (5.1) has an exponential dichotomy on the interval I if there are projections P (t) :
Rn → Rn , t ∈ I, continuous in t, such that if Q(t) = I − P (t), then
i) Φ(t, s)P (s) = P (t)φ(t, s), for t, s ∈ I.
ii) kΦ(t, s)P (s)k ≤ Ke−α(t−s) , for t ≥ s ∈ I.
iii) kΦ(t, s)Q(s)k ≤ Keα(t−s) , for s ≥ t ∈ I.
A more general definition is the following (see, e.g., [16]).
Definition 5.1.3 ((µ1 , µ2 )-dichotomy). If µ1 , µ2 are continuous real-valued functions on
the real interval I = (ω− , ω+ ), system (5.1) is said to have a (µ1 , µ2 )-dichotomy if there
exist supplementary projections P1 , P2 on Rn such that
Z t
−1
kX(t)Pi X (s)k ≤ Ki exp( µi ), if (−1)i (s − t) ≥ 0, i = 1, 2.
s
where K1 , K2 are positive constants.
In the case that µ1 , µ2 are constants, the system (5.1) is said to have an exponential
dichotomy if µ1 < 0 < µ2 and an ordinary dichotomy if µ1 = µ2 = 0.
The following remark is from [7].
Remark – The autonomous system
x0 = Ax
has an exponential dichotomy on R+ if and only if no eigenvalue of A has zero real part. It
has ordinary dichotomy is and only if all eigenvalues of A with zero real part are semisimple
(their algebraic multiplicities are equal to their geometric multiplicities, i.e., the dimension of their
eigenspace). In each case, X(t) = etA and we can take P to be the spectral projection defined by
Z
1
P =
(zI − A)−1 dz
2πi γ
with γ any rectifiable simple closed curve in the open left half-plane which contains in its interior
all eigenvalues of A with negative real part.
◦
Fund. Theory ODE Lecture Notes – J. Arino
5.2. Existence of exponential dichotomy
5.2
81
Existence of exponential dichotomy
To check that the previous definitions hold can be a very tedious task. Some authors have
thus worked on deriving simpler conditions that imply exponential dichotomy.
Theorem 5.2.1. If the matrix A(t) in (5.1) is continuous and bounded on R, and there
exists a quadratic form V (t, x) = xT G(t)x, where the matrix G(t) is symmetric, regular,
bounded and C 1 , such that the derivative of V (t, x) with respect to (5.1) is positive definite,
then (5.1) admits exponential dichotomy.
The converse is true, without the requirement that A(t) be bounded.
A result of [7].
Theorem 5.2.2. Let A(t) be a continuous n × n matrix function defined on an interval I
such that
i) A(t) has k eigenvalues with real part ≤ −α < 0 and n − k eigenvalues with real part
≥ β > 0 for all t ∈ I,
ii) kA(t)k ≤ M for all t ∈ I.
For any positive constant ε < min(α, β), there exists a positive constant δ = δ(M, α + β, ε)
such that, if
kA(t2 ) − A(t1 )k ≤ δ for |t2 − t1 | ≤ h
where h > 0 is a fixed number not greater than the length of I, then the equation
x0 = A(t)x
has a fundamental matrix X(t) satisfying the inequalities
kX(t)P̃ X −1 (s)k ≤ Ke−(α−ε)(t−s) for t ≥ s
kX(t)(I − P̃ )X −1 (s)k ≤ Le−(β−ε)(s−t) for s ≥ t
where K, L are positive constants depending only
I
P̃ = k
0
on M, α + β, ε and
0
0
The following result, due to Muldowney [16], gives a criterion for the existence of a
(µ1 , µ2 )-dichotomy.
Proposition 5.2.3. Suppose there is a continuous real-valued function ρ on I and constants
li , 0 ≤ li < 1, i = 1, 2, such that for some m, 0 ≤ m ≤ n,
(
)
m
n
X
X
max l1 <(ajj ) + l1
|aij | +
|aij | : j = 1, . . . , m ≤ l1 ρ,
i=1,i6=j
i=m+1
Fund. Theory ODE Lecture Notes – J. Arino
5. Exponential dichotomy
82
(
min l2 <(ajj ) −
m
X
)
n
X
|aij | − l2
i=1
|aij | : j = m + 1, . . . , n
≥ l2 ρ.
i=m+1,i6=j
Then the system (5.1) has a (µ1 , µ2 )-dichotomy, where
(
)
m
n
X
X
µ1 = max l1 <(ajj ) +
|aij | + l2
|aij | : j = 1, . . . , m ,
i=m+1
i=1,i6=j
(
µ2 = min <(ajj ) − l1
m
X
n
X
|aij | −
i=1
)
|aij | : j = m + 1, . . . , n .
i=m+1,i6=j
The same sort of theorem can be proved with sums of the columns replaced by sums of
the rows.
Example – Consider

−1 0 1/2
t
t2  ,
A(t) = t/2
2
t/2 −t
t

t>0
5.3
First approximate theory
We consider the nonlinear system
dx
= A(t)x + f (t, x)
dt
(5.4)
where f : R × Rn → Rn is continuous, f (t, x) = O(kxk2 ), kxk = o(1), kf (t, x1 ) − f (t, x2 )k ≤
Lkx1 − x2 k with L small enough.
Let x(t) be a non trivial solution of (5.1); define
λ̄u (x(t)) = lim sup
t−s→∞
and
λu (x(t)) = lim inf
t−s→∞
1
kx(t)k
log
t−s
kx(s)k
1
kx(t)k
log
t−s
kx(s)k
The numbers λ̄u (x(t)) and λu (x(t)) are called the uniform upper characteristic exponent and
uniform lower characteristic exponent of x(t), respectively.
Remark – If λ̄(x) ≤ −α < 0, then lims→−∞ kx(s)k = ∞. If λ(x) ≥ α > 0, then limt→∞ kx(t)k =
∞.
◦
Then we have the following theorem.
Fund. Theory ODE Lecture Notes – J. Arino
5.3. First approximate theory
83
Theorem 5.3.1. If (5.1) admits the exponential dichotomy, then the linear system (5.1)
and the nonlinear system (5.4) are topologically equivalent, i.e.,
i) if the solution x(t) of (5.4) remains in a neighborhood of the origin for t ≥ 0, or t ≤ 0,
then limt→∞ x(t) = 0, or limt→−∞ x(t) = 0, respectively;
ii) there exists positive constants α0 and β0 such that if a solution x(t) of (5.4) is such
that limt→∞ x(0) = 0, or limt→−∞ x(t) = 0, then
kx(t)k ≤ β0 kx(s)ke−α0 (t−s) ,
t≥s
kx(t)k ≤ β0 kx(s)keα0 (t−s) ,
s≥t
or
respectively. At this time, λ̄u (x(t)) ≤ −α0 < 0, or λu (x(t)) ≥ α0 > 0;
iii) for a k-dimensional solution x of (5.1) with λ̄u (x(t)) ≤ −α < 0, or an (n − k)dimensional solution x(t) of (5.1) with λu (x(t)) ≥ α > 0, there is a unique kdimensional or (n − k)-dimensional y(t), solution of (5.4), such that λ̄u (x(t)) ≤ −α <
0, or λu (x(t)) ≥ α > 0 respectively.
A different statement of the same sort of result is given by Palmer [17].
Theorem 5.3.2. Suppose that A(t) is a continuous matrix function such that the linear
equation x0 = A(t)x has an exponential dichotomy. Suppose that f (t, x) is a continuous
function of R × Rn into Rn such that
kf (t, x)k ≤ µ,
kf (t, x1 ) − f (t, x2 )k ≤ Lkx1 − x2 k
for all t, x, x1 , x2 . Then if
4LK ≤ α
there is a unique function H(t, x) of R × Rn into Rn satisfying
i) H(t, x) − x is bounded in R × Rn ,
ii) if x(t) is any solution of the differential equation x0 = A(t)x + f (t, x), then H(t, x(t))
is a solution of x0 = A(t)x.
Moreover, H is continuous in R × Rn and
kH(t, x) − xk ≤ 4Kµα−1
for all t, x. For each fixed t, Ht (x) = H(t, x) is a homeomorphism of Rn . L(t, x) = Ht−1 (x)
is continuous in R × Rn and if y(t) is any solution of x0 = A(t)x, then L(t, y(t)) is a solution
of x0 = A(t)x + f (t, x).
Fund. Theory ODE Lecture Notes – J. Arino
5. Exponential dichotomy
84
5.4
Stability of exponential dichotomy
Theorem 5.4.1. Suppose that the linear system (5.1) admits exponential dichotomy. Then
there exists a constant η > 0 such that the linear system
dx
= (A(t) + B(t))x
dt
(5.5)
also admits exponential dichotomy, when B(t) is continuous on R and kB(t)k ≤ η.
Another version of this theorem.
Theorem 5.4.2. Let B : Mn (R+ ) be a bounded, continuous matrix function. Suppose that
(5.1) has an exponential dichotomy on R+ . If δ = sup kB(t)k < α/(4K 2 ), then the perturbed
equation
x0 = (A(t) + B(t))z
also has an exponential dichotomy on R+ with constants K̃ and α̃ determined by K, α and
δ. Moreover, if P̃ (t) is the corresponding projection, then kP (t) − P̃ (t)k = O(δ) uniformly
in t ∈ R+ . Also, |α − α̃| = O(δ).
5.5
Generality of exponential dichotomy
The exposition has been done here in the case of a system of ODEs. But it is important to
realize that exponential dichotomies exist in a much more general setting.
Bibliography
[1] A. Acosta and P. Garcı́a. Synchronization of non-identical chaotic systems: an exponential dichotomies approach. J. Phys. A: Math. Gen., 34:9143–9151, 2001.
[2] F. Brauer and J.A. Nohel. The Qualitative Theory of Ordinary Differential Equations.
Dover, 1989.
[3] H. Cartan. Cours de calcul différentiel. Hermann, Paris, 1997. Reprint of the 1977
edition.
[4] C. Chicone. Ordinary Differential Equations with Applications. Springer, 1999.
[5] E.A. Coddington and N. Levinson. Theory of Ordinary Differential Equations. McGrawHill, 1955.
[6] E.A. Coddington and N. Levinson. Theory of Ordinary Differential Equations. Krieger,
1984.
[7] W.A. Coppel. Dichotomies in Stability Theory, volume 629 of Lecture Notes in Mathematics. Springer-Verlag, 1978.
[8] J. Dieudonné. Foundations of Modern Analysis. Academic Press, 1969.
[9] N.B. Haaser and J.A. Sullivan. Real Analysis. Dover, 1991. Reprint of the 1971 edition.
[10] J.K. Hale. Ordinary Differential Equations. Krieger, 1980.
[11] P. Hartman. Ordinary Differential Equations. John Wiley & Sons, 1964.
[12] P.-F. Hsieh and Y. Sibuya. Basic Theory of Ordinary Differential Equations. Springer,
1999.
[13] Z. Lin and Y-X. Lin. Linear Systems. Exponential Dichotomy and Structure of Sets of
Hyperbolic Points. World Scientific, 2000.
[14] H.J. Marquez. Nonlinear Control Systems. Wiley, 2003.
[15] R.K Miller and A.N. Michel. Ordinary Differential Equations. Academic Press, 1982.
85
Fund. Theory ODE Lecture Notes – J. Arino
86
BIBLIOGRAPHY
[16] J.S Muldowney. Dichotomies and asymptotic behaviour for linear differential systems.
Transactions of the AMS, 283(2):465–484, 1984.
[17] K.J. Palmer. A generalization of Hartman’s linearization theorem. J. Math. Anal. Appl.,
41:753–758, 1973.
[18] L. Perko. Differential Equations and Dynamical Systems. Springer, 2001.
[19] L. Schwartz. Cours d’analyse, volume I. Hermann, Paris, 1967.
[20] K. Yosida. Lectures on Differential and Integral Equations. Dover, 1991.
Appendix A
A few useful definitions and results
Here, some results that are important for the course are given with a somewhat random
ordering.
A.1
A.1.1
Vector spaces, norms
Norm
Consider a vector space E over a field K. A norm is an application, denoted k k, from E to
R+ that satisfies the following:
1) ∀x ∈ E, kxk ≥ 0, with kxk = 0 if and only if x = 0.
2) ∀x ∈ E, ∀a ∈ K, kaxk = |a| kxk.
3) ∀x, y ∈ E, kx + yk ≤ kxk + kyk.
A vector space E equiped with a norm k k is a normed vector space.
A.1.2
Matrix norms
A.1.3
Supremum (or operator) norm
The supremum norm is defined by
∀L ∈ L(E),
|||L||| =
kL(x)k
= sup kL(x)k.
x∈E−{0} kxk
kxk≤1
sup
The inequality
kA(t)(x1 − x2 )k ≤ |||A(t)||| kx1 − x2 k
87
88
Fund. Theory ODE Lecture Notes – J. Arino
A. Definitions and results
results from the nature of the norm ||| |||. See appendix A.1. is best understood by indicating
the spaces in which the various norms are defined. We have
x
A
kAxka = kxkb kxkb a
≤ kxkb |||A|||
= kAkkxkb ,
since
x kxkb = 1.
b
A.2
An inequality involving norms and integrals
Lemma A.2.1. Suppose f : Rn → Rn is a continuous function. Then
Z b
Z b
≤
f
(x)dx
kf (x)k dx.
a
a
Proof. First, note that we have
Z b
Z b
Z b
f (x)dx = f1 (x)dx , . . . , fn (x)dx
.
a
a
a
For a given component function fi , i = 1, . . . , n, using the definition of the Riemann integral,
Z b
k
X
fi (x)dx = lim
fi (x∗j )∆xj ,
k→∞
a
j=1
where x∗j is the sample point in the interval [xj−1 , xj ] with width ∆xj . Therefore,
Z b
k
k
X
X
∗
∗
f
(x
f
(x)dx
=
lim
)∆x
=
lim
f
(x
)∆x
,
i j
i
i j
k→∞
k→∞
a
j=1
j=1
since the norm is a continuous function. The result then follows from the triangle inequality.
A.3
Types of convergences
Definition A.1 (Pointwise convergence). Let X be any set, and let Y be a topological space.
A sequence f1 , f2 , . . . of mappings from X to Y is said to be pointwise convergent (or simply
convergent) to a mapping f : X → Y , if the sequence fn (x) converges to f (x) for each x in
X. This is usually denoted by fn → f . In other words,
(fn → f ) ⇔ (∀x ∈ X, ∀ε > 0, ∃N ≥ 0, ∀n ≥ N, d(fn (x), f (x)) < ε) .
Fund. Theory ODE Lecture Notes – J. Arino
A.4. Asymptotic Notations
89
Definition A.2 (Uniform convergence). Let X be any set, and let Y be a topological space.
A sequence f1 , f2 , . . . of mappings from X to Y is said to be uniformly convergent to a
mapping f : X → Y , if given ε > 0, there exists N such that for all n ≥ N and all x ∈ X,
d(fn (x), f (x)) < ε.
u
This is usually denoted by fn → f . In other words,
u
(fn → f ) ⇔ (∀ε > 0, ∃N ≥ 0, ∀n ≥ N, ∀x ∈ X, d(fn (x), f (x)) < ε) .
An important theorem follows.
Theorem A.3.1. If the sequence of maps {fn } is uniformly convergent to the map f , then
f is continuous.
A.4
Asymptotic Notations
Let n be a integer variable which tends to ∞ and let x be a continuous variable tending
to some limit. Also, let φ(n) or φ(x) be a positive function and f (n) or f (x) any function.
Then
i) f = O(φ) means that |f | < Aφ for some constant A and all values of n and x,
ii) f = o(φ) mean that f /φ → 0,
iii) f ∼ φ means that f /φ → 1.
Note that f = o(φ) implies f = O(φ).
A.5
Types of continuities
Definition A.3 (Uniform continuity). Let (X, dX ) and (Y, dY ) be two metric spaces, E ⊆ X
and F ⊆ Y . A function f : E → F is uniformly continuous on the set E ⊂ X if for every
ε > 0, there exists δ > 0 such that
dY (f (x), f (y)) < ε whenever x, y ∈ E and dX (x, y) < δ.
In other words,
(f : E ⊆ (X, dX ) → F ⊆ (Y, dY ) uniformly continuous on E)
⇔ (∀ε > 0, ∃δ > 0, ∀x, y ∈ E, dX (x, y) < δ ⇒ dY (f (x), f (y)) < ε) .
Definition A.4 (Equicontinuous set). A set of functions F = {f } defined on a real interval
I is said to be equicontinuous on I if, given any ε > 0, there exists a δε > 0, independent
of f ∈ F and also t, t̃ ∈ I such that
kf (t) − f (t̃)k < ε whenever |t − t̃| < δε
90
Fund. Theory ODE Lecture Notes – J. Arino
A. Definitions and results
An interpretation of equicontinuity is that a sequence of functions is equicontinuous if
all the functions are continuous and they have equal variation over a given neighbourhood.
Equicontinuity of a sequence of functions has important consequences.
Theorem A.5.1. Let {fn } be an equicontinuous sequence of functions. If fn (x) → f (x) for
every x ∈ X, then the function f is continuous.
Lemma A.5 (Ascoli). On a bounded interval I, let F = {f } be an infinite, uniformly
bounded, equicontinuous set of functions. Then F contains a sequence {fn }, n = 1, 2, . . .,
which is uniformly convergent on I.
Theorem A.6. Let C(X) be the space of continuous functions on the complete metric space
X with values in Rn . If a sequence {fn } in C(X) is bounded and equicontinuous, then it has
a uniformly convergent subsequence.
A.6
Lipschitz function
Definition A.6.1 (Lipschitz function). A map f : U ⊂ R × Rn → Rn is Lipschitz in x if
there exists a real number L such that for all (t, x1 ) ∈ U and (t, x2 ) ∈ U,
kf (t, x1 ) − f (t, x2 )k ≤ Lkx1 − x2 k.
In the case of a Lipschitz function as defined above, the constant L is independent of x1
and x2 , it is given for U. A weaker version is local Lipschitz functions.
Definition A.6.2 (Locally Lipschitz function). A map f : U ⊂ R × Rn → Rn is locally
Lipschitz in x if, for all (t0 , x0 ) ∈ U, there exists a neighborhood V ⊂ U of (t0 , x0 ) and a real
number L, such that for all (t, x1 ), (t, x2 ) ∈ V ,
kf (t, x1 ) − f (t, x2 )k ≤ Lkx1 − x2 k.
In other words, f is locally Lipschitz if the restriction of f to V is Lipschitz.
Thus, a locally Lipschitz function is Lipschitz if it is locally Lipschitz on U with everywhere the same Lipschitz constant L. Another definition of a locally Lipschitz function is as
follows.
Definition A.6.3. A function f : U ⊂ Rn+1 → Rn is locally Lipschitz continuous if for
every compact set V ⊂ U, the number
L=
sup
(t,x)6=(t,y)∈V
kf (t, x) − f (t, y)k
kx − yk
is finite, with L depending on V .
Property A.6.4. Let f (t, x) be a function. The following properties hold.
Fund. Theory ODE Lecture Notes – J. Arino
A.7. Gronwall’s lemma
91
i) f Lipschitz ⇒ f uniformly continuous in x.
ii) f uniformly continuous 6⇒ f Lipschitz.
iii) f (t, x) has continuous partial derivative ∂f /∂x on a bounded closed domain D ⇒ f is
locally Lipschitz on D.
Proof. i) Suppose that f is Lipschitz, i.e., there exists L > 0 such that kf (t, x1 )−f (t, x2 )k ≤
Lkx1 − x2 k. Recall that f is uniformly continuous if for every ε > 0, there exists δ > 0 such
that for all x1 , x2 , kx1 − x2 k < δ implies that kf (t, x1 ) − f (t, x2 )k < ε. So, given an ε > 0,
choose δ < ε/L. Then kx1 − x2 k < δ implies that
kf (t, x1 ) − f (t, x2 )k < Lkx1 − x2 k < Lδ < Lε/L = ε.
Thus f is uniformly continuous (see Definition A.3).
ii ) This is left as an exercise. Consider for example the function f defined by f (x) =
1/ ln x on (0, 12 ], f (0) = 0.
iii ) Notice that ∂f /∂x continuous on D implies that k∂f /∂xk is continuous on (the
bounded closed domain) D, and thus k∂f /∂xk is bounded on D. Let
∂f (t, x) L = sup ∂x (t,x)∈D
If (t, x1 ), (t, x2 ) ∈ U, by the mean-value theorem, there exists η ∈ [x1 , x2 ] such that f (t, x2 ) −
f (t, x1 ) = (x2 − x1 ) ∂f
(t, η). As η ∈ U, it follows that k ∂f
(t, η)k ≤ L, and thus kf (t, x2 ) −
∂x
∂x
f (t, x1 )k ≤ Lkx2 − x1 k.
A.7
Gronwall’s lemma
The name of Gronwall is associated to a certain number of inequalities. We give a few of
them here. We prove the most simple one (as it is an easy proof to remember), as well as
the most general one (Lemma A.9). In [12, p. 3], the lemma is stated as follows.
Lemma A.7 (Gronwall). If
i) g(t) is continuous on t0 ≤ t ≤ t1 ,
ii) for t0 ≤ t ≤ t1 , g(t) satisfies the inequality
Z
t
0 ≤ g(t) ≤ K + L
g(s)ds.
t0
Then
0 ≤ g(t) ≤ KeL(t−t0 ) ,
for t0 ≤ t ≤ t1 .
Fund. Theory ODE Lecture Notes – J. Arino
A. Definitions and results
92
Proof. Let v(t) =
be written
Rt
t0
g(s)ds. Then v 0 (t) = g(t), which implies that the assumption on g can
0 ≤ v 0 (t) ≤ K + Lv(t).
The right inequality is a linear differential inequality, with integrating factor exp(−
Also, v(t0 ) = 0. Hence,
d −L(t−t0 )
e
v(t) ≤ Ke−L(t−t0 )
dt
and therefore,
K
e−L(t−t0 ) v(t) ≤
1 − e−L(t−t0 ) .
L
Thus Lv(t) ≤ K eL(t−t0 ) − 1 , and g(t) ≤ K + Lv(t) ≤ KeL(t−t0 ) .
Rt
t0
Lds)
In [4, p. 128-130], Gronwall’s inequality is stated as
Lemma A.8. Suppose that a < b and let g, K and L be nonnegative continuous functions
defined on the interval [a, b]. Moreover, suppose that either K is a constant function, or K
is differentiable on [a, b] with positive derivative K 0 . If, for all t ∈ [a, b]
t
Z
g(t) ≤ K(t) +
L(t)g(s)ds,
a
then
Z
g(t) ≤ K(t) exp
t
L(s)ds ,
a
for all t ∈ [a, b].
Finally, the most general formulation is in [8, p. 286].
Lemma A.9. If ϕ and ψ are two nonnegative regulated functions on the interval [0, c], then
for every nonnegative regulated function w in [0, c] satisfying the inequality
Z
w(t) ≤ ϕ(t) +
t
ψ(s)w(s)ds,
0
we have, in [0, c],
Z
w(t) ≤ ϕ(t) +
t
Z
ϕ(s)ψ(s) exp
0
t
ψ(ξ)dξ ds.
(A.1)
s
Before proving the result, let us recall that a function f from an interval I ⊂ R to a
Banach space F is regulated if it admits in each point in I a left limit and a right limit. In
particular, every continuous mapping from I ⊂ R to a Banach space is regulated, as well as
monotone maps from I to R; see, e.g., [8, Section 7.6].
Fund. Theory ODE Lecture Notes – J. Arino
A.8. Fixed point theorems
93
Rt
Rt
Proof. Let y(t) = 0 ψ(s)w(s)ds; y is continuous, and since w(t) ≤ ϕ(t) + 0 ψ(s)w(s)ds, it
follows that, except maybe at a denumerable number of points of [0, c], we have
y 0 (t) − ψ(t)y(t) ≤ ϕ(t)ψ(t)
Rt
from [8, Section 8.7]. Let z(t) = y(t) exp(− 0 ψ(s)ds). Then (A.2) is equivalent to
Z t
0
z (t) ≤ ϕ(t)ψ(t) exp −
ψ(s)ds .
(A.2)
0
Using a mean-value type theorem (see, e.g., [8, Th. 8.5.3]) and using the fact that z(0) = 0,
we get, for t ∈ [0, c],
Z s
Z t
z(t) ≤
ϕ(s)ψ(s) exp −
ψ(ξ)dξ ds,
0
0
whence by definition
Z
y(t) ≤
t
Z
ϕ(s)ψ(s) exp
0
t
ψ(ξ)dξ ds,
s
and inequality (A.1) now follows from the relation w(t) ≤ ϕ(t) + y(t).
A.8
Fixed point theorems
Definition A.10 (Contraction mapping). Let (X, d) be a metric space, and let S ⊂ X. A
mapping f : S → S is a contraction on S if there exists K < 1 such that, for all x, y ∈ S,
d(f (x), f (y)) ≤ Kd(x, y)
Every contraction is uniformly continuous on X (from Proposition A.6.4, since a contraction is Lipschitz).
Theorem A.11 (Contraction mapping principle). Consider the complete metric space (X, d).
Every contraction mapping f : X → X has one and only one x ∈ X such that f (x) = x.
Proof. Existence We use successive approximations. Let x0 ∈ X. Define x1 = f (x0 ), x2 =
f (x1 ), . . . , xn = f (xn−1 ), . . . This defines an infinite sequences of elements of X. As f is a
contraction,
d(x2 , x1 ) = d(f (x1 ), f (x0 ))
≤ Kd(x1 , x0 ).
Similarly,
d(x3 , x2 ) = d(f (x2 ), f (x1 ))
≤ Kd(x2 , x1 )
≤ K 2 d(x1 , x0 ).
Fund. Theory ODE Lecture Notes – J. Arino
A. Definitions and results
94
Iterating,
d(xn+1 , xn ) ≤ K n d(x1 , x0 ).
Therefore,
d(xn+p , xn ) ≤ d(xn+p , xn+p−1 ) + d(xn+p−1 , xn+p−2 ) + · · · + d(xn+1 , xn )
≤ (K p−1 + K p−2 + · · · + K + 1)K n d(x1 , x0 )
Kn
d(x1 , x0 ).
≤
1−K
Therefore d(xn+p , xn ) tends to 0 as n → ∞, so {xn } is a Cauchy sequence. Since X is a
complete space, it follows that {xn } admits a limit `. As limn→∞ xn = `, it follows from
continuity of f that xn+1 = f (xn ) tends to f (`). But xn+1 also converges to `, so f (`) = `,
that is, ` is a fixed point of f .
Uniqueness Suppose `1 and `2 are two fixed points. Then there must hold that
d(`1 , `2 ) ≤ Kd(`1 , `2 ) < d(`1 , `2 ),
if d(`1 , `2 ) 6= 0. Therefore d(`1 , `2 ) = 0, and `1 = `2 .
In the case that f : S ⊂ X → S, the theorem takes the form of Theorem A.12. Closedness
of S is an implicit requirement, since the set S in the complete metric space X is closed if,
and only if, S is complete.
Theorem A.12. Let S be a closed subset of the complete metric space (X, d). Every contraction mapping f : S → S has one and only one x ∈ S such that f (x) = x.
Theorem A.8.1. Consider a mapping f : X → X, where X is a complete metric space.
Suppose that f is not necessarily a contraction, but that one of the iterates f k of f , is a
contraction. Then f has a unique fixed point.
A.9
Jordan normal form
Theorem A.9.1. Every complex n × n matrix A is similar to a matrix of the form


J0 0 0 · · · 0
 0 J1 0 · · · 0 

J =
· · · ·
·
0 0 0 · · · Js
where J0 is a diagonal matrix with diagonal λ1 , . . . , λn , and,

λq+i
1 0 0 ···
0
 0 λq+i 1 0 · · ·
0


·
· · ···
·
Ji =  ·
 0
0 0 0 · · · λq+i
0
0 0 0 ···
0
for i = 1, . . . , s,

0
0 

· 

1 
λq+i
Fund. Theory ODE Lecture Notes – J. Arino
A.9. Jordan normal form
95
The λj , j = 1, . . . , q + s, are the characteristic roots of A, which need not all be distinct.
If λj is a simple root, then it occurs in J0 , and therefore, if all the roots are distinct, A is
similar to the diagonal matrix


λ1 0 0 · · · 0
 0 λ2 0 · · · 0 

J =
·
· · ·
·
0 0 0 · · · λn
An algorithm to compute the Jordan canonical form of an n × n matrix A [15].
i) Compute the eigenvalues of A. Let λ1 , . . . , λm be the distinct eigenvalues of A with
multiplicities n1 , . . . , nm , respectively.
ii) Compute n1 linearly independent generalized eigenvectors of A associated with λ1 as
follows. Compute
(A − λ1 En )i
for i = 1, 2, . . . until the rank of (A−λ1 En )k is equal to the rank of (A−λ1 En )k+1 . Find
a generalized eigenvector of rank k, say u. Define ui = (A−λ1 En )k−1 u, for i = 1, . . . , k.
If k = n1 , proceed to step 3. If k < n1 , find another linearly independent generalized
eigenvector with rank k. If this is not possible, try k − 1, and so forth, until n1 linearly
independent generalized eigenvectors are determined. Note that if ρ(A − λ1 En ) = r,
then there are totally (n − r) chains of generalized eigenvectors associated with λ1 .
iii) Repeat step 2 for λ2 , . . . , λm .
iv) Let u1 , . . . , uk , . . . be the new basis. Observe that
Au1 = λ1 u1 ,
Au2 = u1 + λ1 u2 ,
..
.
Auk = uk−1 + λ1 uk
Thus in the new basis, A has the representation

α1 1

...

1

α
1


α2 1


...

1
J =

α
2


α3 1


...

1

α3


...

















Fund. Theory ODE Lecture Notes – J. Arino
A. Definitions and results
96
where each chain of generalized eigenvectors generates a Jordan block whose order
equals the length of the chain.
v) The similarity transformation which yields J = Q−1 AQ is given by Q = [u1 , . . . , uk , . . .].
A.10
Matrix exponentials
Let A ∈ Mn (K). The exponential of A is defined by
A
e =
∞
X
An
n=0
n!
(A.3)
P 1 n
We have eA ∈ Mn (K). Also n!1 An ≤ n!1 |||A|||n , so that the series
A is absolutely
n!
A
convergent in Mn (K). Thus e is well defined.
Property A.10.1. Let A, B ∈ Mn (K). Then
i) eA ≤ e|||A||| .
ii) If A and B commute ( i.e., AB = BA), then eA+B = eA eB .
P
1 n
iii) eA = I + ∞
n=1 n! A .
iv) e0 = I.
v) eA is invertible with inverse e−A .
vi) eAt is differentiable, and
d At
e
dt
= AeAt .
vii) If P and T are linear transformations on Kn , and S = P T P −1 , then eS = P eT P −1 .
Proof. v) For any matrix A, we have A(−A) = −AA = (−A)A, so A and −A commute.
Therefore, eA e−A = eA−A = e0 = I. Therefore, for any A, eA is invertible with inverse
e−A .
There are several shortcuts to computing the exponential of a matrix A:
P
1) If A is nilpotent, that is, if there exists q ∈ N such that Aq = 0, then eA = qk=1 Ak /k!.
A nilpotent matrix has several interesting properties. A is nilpotent if and only if all
its eigenvalues are zero. A nilpotent matrix has zero determinant and trace.
2) If A is such that there exists q ∈ N such that Aq = A, then this can sometimes be
exploited to simplify the computation of eA .
3) Any matrix A can be written in the form A = D + N , where D is diagonalizable, N is
nilpotent, and D and N commute. Therefore, eA = eD+N = eD eN .
4) Other cases require the use of the Jordan normal form (explained below).
Fund. Theory ODE Lecture Notes – J. Arino
A.11. Matrix logarithms
97
Use of the Jordan form to compute the exponential of a matrix. Suppose that
J = P −1 AP is the Jordan form of the matrix A. For a block diagonal matrix

B1

0
...

B=

,
0
Bs
B1k
0
we have, for k = 0, 1, . . .,


Bk = 
...
0
Bsk
eJ0 t
0


,
Therefore, for t ∈ R,

..

eJt = 
.
0
eJs t
eλ1 t
0


,
with

..

eJ0 = 
0
.
eλk t


.
Now, since Ji = λk+i Ii + Ni , with Ni a nilpotent matrix, and since Ii and Ni commute, there
holds
eJi t = eλk+i t eNi t .
For k ≥ Ni , Nik = 0, so

etJi
A.11
1 t ···

0 1 · · ·
=

0
···
tni −1
(ni −1)!
tni −2 

(ni −2)!  .


1
Matrix logarithms
Theorem A.11.1. Suppose that M is a nonsingular n × n matrix. Then there is an n × n
matrix B (possibly complex), such that eB = M . If, additionally, M ∈ Mn (R), then there
is B ∈ Mn (R) such that eB = M 2 .
Let u ∈ C, we have, for |u| < 1,
1
= 1 − u + u2 − u3 + · · ·
1+u
98
Fund. Theory ODE Lecture Notes – J. Arino
A. Definitions and results
and, for |u| < 1,
t2 t3 t4
+ − + ···
2
3
4
∞
k
X
t
(−1)k+1
=
k
k=1
ln(1 + u) = t −
For any z ∈ C,
exp(z) = 1 + z +
=
z2 z3
+
+ ···
2!
3!
∞
X
zn
n=1
n!
Therefore, for |u| < 1, u ∈ C,
1 + u = exp(ln(1 + u))
"∞
#n
∞
k
X
1 X
u
(−1)k+1
=
n!
k
n=1
k=1
Suppose that
where
∞
X
(−1)k+1 1 k
B = (ln λ)I +
Z
k
λk
k=1


0 1 ··· 0
 .. ..


.
. 
Z=

0
0 1
0
0
is an m × m-matrix. Since Z is nilpotent (Z N = 0 for all N ≥ m), the above sum is finite.
Observe that
!
∞
X
(−1)k+1 1 k
exp(B) = exp((ln λ)I) exp
Z
k
k
λ
k=1
k !
∞
k+1
X (−1)
Z
= λ exp
k
λ
k=1
Z
=λ I+
λ
= λI + Z
=J
We say ln J = B.
Fund. Theory ODE Lecture Notes – J. Arino
A.12. Spectral theorems
A.12
99
Spectral theorems
When studying systems of differential equations, it is very important to be able to compute
the eigenvalues of a matrix, for instance in order to study the local asymptotic stability of
an equilibrium point. This can be a very difficult problem, that often becomes intractable
even for systems with low dimensionality. However, even if it is not possible to compute an
explicit solution, it is often possible to use spectrum localization theorems. We here give
two of the most famous ones: the Routh-Hurwitz criterion, and the Gershgorin theorem.
Let A be a n × n matrix, denote its elements by (aij ). The set of all eigenvalues of A is
called its spectrum, and is denoted Sp(A).
Theorem A.12.1 (Routh-Hurwitz). If n = 2, suppose that det A > 0 and trA < 0. Then
A has only eigenvalues with negative real part.
Theorem A.12.2 (Gershgorin). Let
Rj =
n
X
|aij |
i=1,i6=j
Let λ ∈ Sp(A). Then
λ∈
[
{|λ − ajj | ≤ Rj }
j
Gershgorin’s theorem is extremely
P helpful in certain cases. Suppose that A is stictly
diagonally dominant, i.e., |aii | > ni=1,i6=j |aij |. Then A has no eigenvalues with zero real
part. Also, the number of eigenvalues with negative real part is equal to the number of
negative entries on the diagonal of A, and conversely for eigenvalues with positive real parts
and the number of positive entries on the diagonal of A.
Appendix B
Problem sheets
Contents
Homework sheet 1 – 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . 103
Homework sheet 2 – 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . 113
Homework sheet 3 – 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . 117
Homework sheet 4 – 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . 125
Final examination – 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . 137
Homework sheet 1 – 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . 145
101
Fund. Theory ODE Lecture Notes – J. Arino
103
McMaster University – Math4G03/6G03
Fall 2003
Homework Sheet 1
Exercise 1.1 – We consider here the equation
x0 (t) = −αx(t) + f (x(t))
(B.1)
where α ∈ R is constant and f is continuous on R.
i) Show that x is a solution of (B.1) on R+ if, and only if,

 x is continuous on R+
and
Rt

x(t) = e−αt x(0) + e−αt 0 eαs f (x(s))ds ∀t ∈ R+
(B.2)
Suppose now that α > 0 and that f is such that
∃a, k ∈ R, a > 0, 0 < k < α :
∀u ∈ R, |u| ≤ a ⇒ |f (u)| ≤ k|u|
(B.3)
Part I. Suppose that there exists a solution x of (B.1), defined on R+ and satisfying the
inequality
|x(t)| ≤ a, t ∈ R+
(B.4)
i) Prove the inequality
|x(t)| ≤ |x(0)|e−(α−k)t ,
t ∈ R+
[Hint: Use Gronwall’s lemma with the function g(t) = eαt |x(t)|].
ii) Deduce that x admits the limit 0 as t → ∞.
Part II.
i) Show that any solution of (B.1) on R+ that satisfies |x(0)| < a, satisfies (B.4).
ii) Deduce from the preceding questions the two following properties.
a) Any solution x of (B.1) on R+ satisfying the condition |x(0)| < a, admits the
limit 0 when t → ∞.
b) The function x ≡ 0 is the unique solution of (B.1) on R+ such that x(0) = 0.
Part III. Application. Show that, for α > 1, all solutions of the equation
x0 = −αx + ln 1 + x2
tend to zero when t → ∞.
◦
Fund. Theory ODE Lecture Notes – J. Arino
104
B. Problem sheets
Exercise 1.2 – Let f : [0, +∞) → R, f ∈ C 1 , and a ∈ R. We consider the initial value
problems
and
x0 (t) + ax(t) = f (t),
x(0) = 0
t≥0
x0 (t) + ax(t) = f 0 (t),
x(0) = 0
t≥0
(B.5)
(B.6)
As these equations are linear, the initial value problems (B.5) and (B.6) admit unique solutions. We denote φ the solution to (B.5) and ψ the solution to (B.6). Find a necessary and
sufficient condition on f such that φ0 = ψ.
[Hint: Use a variation of constants formula].
◦
Exercise 1.3 – Let f : Rn → Rn be continuous. Consider the differential equation
x0 (t) = f (x(t))
(B.7)
i) Let x be a solution of (B.7) defined on a bounded interval [0, α), with α > 0. Suppose
that t 7→ f (x(t)) is bounded on [0, α).
a) Consider the sequence
zα,n = x(α −
1
),
n
n ∈ N∗
Show that (zα,n )n∈N∗ is a Cauchy sequence.
b) Deduce that there exists xα ∈ Rn such that,
kx(t) − xα k ≤ Mα |t − α|
with M = supt∈[0,α) kf (x(t))k.
c) Show that x admits a finite limit when t → α, t < α.
ii) Show that there exists an extension of x that is a solution of (B.7) on the interval
[0, α].
◦
Fund. Theory ODE Lecture Notes – J. Arino
105
McMaster University – Math4G03/6G03
Fall 2003
Homework Sheet 1 – Solutions
Solution – Exercise 1 – 1) Let us first show that x solution of (B.1) implies (B.2).
If x is a solution of (B.1) on R+ , then x is differentiable on R+ , and so x is continuous on
R+ . Furthermore,
x0 (s) = −αx(s) + f (x(s)), for all s ∈ R+
⇔ eαs x0 (s) = −αeαs + eαs f (x(s))
Z t
Z t
Z t
αs 0
αs
⇒
e x (s)ds = −α
e x(s)ds +
eαs f (x(s))ds
0
0
0
Z t
Z t
Z t
t
αs
αs
αs
[e x(s)]0 − α
e x(s)ds = −α
e x(s)ds +
eαs f (x(s))ds
0
0
0
Z t
eαt x(t) − x(0) =
eαs f (x(s))ds, for all t ∈ R+
0
Z t
−αt
−αt
x(t) = e x(0) + e
eαs f (x(s))ds, for all t ∈ R+
0
Let us now prove the converse, i.e., that if a function x satisfies (B.2), then it is a solution
to the IVP (B.1). Since x and f are continuous on R+ , t 7→ eαt f (x(t)) is continuous on R+ .
This implies that
Z
t
eαs f (x(s))ds
t 7→
0
is differentiable on R+ , and, differentiating the expression for x(t) as given in (B.2),
Z t
0
−αt
−αt
x (t) = −αe x(0) − αe
eαs f (x(s))ds + e−αt eαt f (x(t))
0
Z t
αs
0
−αt
−αt
e f (x(s))ds +f (x(t))
⇒ x (t) = −α e x(0) + e
0
|
{z
}
x(t)
And thus
x0 (t) = −αx(t) + f (x(t))
which implies that x is a solution to (B.1).
Part I. We now assume that (B.3) is also satisfied, and that there exists a solution x on R+
satisfying (B.4).
1) If x is a solution of (B.1), then
Z t
−αt
−αt
x(t) = e x(0) + e
eαs f (x(s))ds
0
Fund. Theory ODE Lecture Notes – J. Arino
106
B. Problem sheets
This implies that
−αt
|x(t)| ≤ e
−αt
t
Z
eαs |f (x(s))|ds
|x(0)| + e
0
αt
From (B.3) and multiplying both sides by e ,
t
Z
αt
keαs |x(s)|ds
e |x(t)| ≤ |x(0)| +
0
We use Gronwall’s Lemma (Lemma A.2) as follows,
Z t
αt
e |x(t)| ≤ |x(0)| +
k eαs |x(s)| ds
|{z}
| {z } | {z }
| {z }
0
g(t)
L(t)
K(t)
g(s)
Thus,
Z
αt
e |x(t)| ≤ |x(0)| exp
t
kds
0
kt
≤ |x(0)|e
and so finally, for all t ∈ R+ , we have
|x(t)| ≤ |x(0)|et(k−α) = |x(0)|e−(α−k)t
2) From (B.3), 0 < k < α, hence α − k > 0, which implies, together with the result of
the previous question, that limt→∞ x(t) = 0.
Part II.
1) Let us suppose that x is a solution of (B.1) that is such that |x(0)| < 0. Let A = {t :
|x(t)| ≤ a}. Let us show that A = [0, +∞).
First of all, notice that |x(0)| < a and x continuous on R+ implies that there exists
t0 ∈ R+ − {0} such that, for all t ∈ [0, t0 ], |x(t)| ≤ a. Indeed, suppose this were not the case.
Then, for all ε > 0, there exists tε ∈ [0, ε] such that |x(tε )| > a. This means that for all
n ∈ N − {0}, there exists un ∈ [0, n1 ] such that |x(un )| > a. As un → 0 as n → ∞ and that x
is continuous, |x(0)| ≥ a, since taking the limit implies that strict inequalities become loose.
This is a contradiction with |x(0)| < a. Thus [0, t0 ] ⊂ A.
Let us now show that for all t1 ∈ A, [0, t1 ] ⊂ A, i.e., A is an interval. First, if t1 ≤ t0 then
[0, t1 ] ⊂ [0, t0 ] ⊂ A. Secondly, in the case t1 > t0 , suppose that [0, t1 ] 6⊂ A. This means that
∃η ∈ [0, t1 ] such that η 6∈ A. More precisely, ∃η ∈ (t0 , t1 ) such that η 6∈ A, since [0, t0 ] ⊂ A
and t1 ∈ A. Let β be the smallest such η, that is, β = inf{t ∈ (t0 , t1 ); t 6∈ A}. Note that β
can also be defined as β = sup{t ∈ (t0 , t1 ); t ∈ A}.
Thus
β = inf{t ∈ (t0 , t1 ); |x(t)| > a} = sup{t ∈ (t0 , t1 ); |x(t)| < a}
Since x is continuous, this implies that |x(β)| ≥ a and |x(β)| ≤ a, hence x(β) = ±a. But,
with its sup definition, this implies that β = t1 , whereas with its inf definition, this implies
that β < t1 .
Fund. Theory ODE Lecture Notes – J. Arino
107
Hence A is an interval. We now want to show that it is an unbounded interval. Assume
it is bounded, hence let c = sup(A) < ∞. Since x is continuous, A = {t ≥ 0, |x(t)| ≤ a}
implies that c ∈ A. Thus A = [0, c]. Therefore, for all t ∈ [0, c],
|x(t)| ≤ a
so, by Part I, 1),
|x(t)| ≤ |x(0)|e−(α−k)t
⇒ |x(t)| ≤ |x(0)| < a
⇒ |x(c)| < a
Since x is continuous on R+ , there exists t > c such that |x(t)| ≤ a, and thus there exists
t > c such that t ∈ A, which is a contradiction. Therefore, A = [0, ∞), and we can conclude
that ∀t ∈ R+ , |x(t)| ≤ a.
Another proof, contributed by Guihong Fan, proceeds by contradiction, using the fact
the (B.1) is an autonomous scalar equation. Notice that equation (B.1) can be written in
the form x0 = g(x), with g(u) = −αu + f (u). Thus, since this mapping is continuous,
we can apply Theorem 1.1.8 on the monotonicity of the solutions to an autonomous scalar
differential equation. Assume that x(t) is a solution of (B.1) on R+ that satisfies |x(0)| < a,
but that (B.4) is violated.
Then, since the solution x(t) is monotone, there exists t0 ∈ R+ such that one of the
following holds.
i) x(t0 ) = a and x0 (t0 ) > 0,
ii) x(t0 ) = −a and x0 (t0 ) < 0.
Let us treat case i). From (B.3), it follows that |f (x(t0 ))| = |f (a)| ≤ ka. Therefore, using
(B.1),
x0 (t0 ) = −αx(t0 ) + f (x(t0 ))
= −αa + f (a)
≤ −αa + ka
≤ −(α − k)a
<0
since α > k. This is a contradiction with x0 (t0 ) > 0. Case ii) is treated similarly, and thus
it follows that (B.4) holds for all t ∈ R+ .
2 – a) If |x(0)| < a, then we have just proved that for all t ∈ R+ , |x(t)| ≤ a. From Part
I, 1), this implies that |x(t)| ≤ |x(0)|e−(α−k)t . Therefore, since α > k, limt→∞ x(t) = 0.
2 – b) To show that x ≡ 0 is the only solution of (B.1) such that x(0) = 0, we first show
that x ≡ 0 is a solution of (B.1). Condition (B.3) applied to 0 states that |0| < a implies
|f (0)| ≤ k|0| = 0.
Fund. Theory ODE Lecture Notes – J. Arino
108
B. Problem sheets
Uniqueness: let φ be a solution of (B.1) such that φ(0) = 0. This implies that |φ(0) < a,
and as a consequence, it follows from Part I, 1) that for all t ∈ R+ , |φ(t)| ≤ |φ(0)|e−(α−k)t ,
hence for all t ∈ R+ , φ(t) = 0.
Part III. All solutions of the nonlinear equation x0 = −αx + ln(1 + x2 ) tend to zero as
t → ∞, when α > 1. Indeed, let f (u) = ln(1 + u2 ). We have
|f 0 (u)| =
2|u|
≤1
1 + u2
since (u − 1)2 = u2 + 1 − 2u ≥ 0 (and hence 2u/(1 + u2 ) ≤ 1). Then, |f (u) − f (0)| ≤ |u − 0|
implies that |f (u)| ≤ |u|, for all u ∈ R. We thus have k = 1 < α, the hypotheses of the
exercise are satisfied and all solutions of the equation tend to zero.
◦
Solution – Exercise 2 – Using a variation of constants formula, we obtain
−at
φ(t) = e
Z
t
c1 +
e−a(t−s) f (s)ds,
t≥0
e−a(t−s) f 0 (s)ds,
t≥0
0
and
−at
ψ(t) = e
Z
c2 +
t
0
Let us show this for the solution of (B.5), the solution of (B.6) is obtained using exactly the
same method. The solution to a linear differential equation consists of the general solution
to the homogeneous equation together with a particular solution to the nonhomogeneous
equation. Here, the homogeneous equation is
x0 = −ax
and basic integration yields the general solution x(t) = c1 e−at . To obtain a particular solution
to the nonhomogeneous equatiob (B.5), we use a variation of constants formula: assume that
the constant in the solution x(t) = ce−at is a function of time, hence
x(t) = c(t)e−at
Taking the derivative of this expression, we obtain
x0 = c0 e−at − ace−at
Substituting both this expression and x = ce−at into (B.5), we get
c0 e−at − ace−at + ace−at = f (t),
and hence
c0 = eat f (t)
Integrating both sides of this expression gives
Z t
c(t) =
eas f (s)ds
0
t≥0
Fund. Theory ODE Lecture Notes – J. Arino
109
so that a particular solution to (B.5) is given by
Z t
Z t
−at
as
−at
x(t) = c(t)e
=
e f (s)ds e
=
e−a(t−s) f (s)ds
0
0
Summing the general solution to the homogeneous equation with this last expression gives
the desired result.
Using the initial conditions yields
φ(0) = c1 = 0
ψ(0) = c2 = 0
Hence the system under consideration is
Z t
φ(t) =
e−a(t−s) f (s)ds, t ≥ 0
Z0 t
ψ(t) =
e−a(t−s) f 0 (s)ds, t ≥ 0
0
Recall that if g(t) =
Rt
t0
h(s, t)ds, then for all t0 ,
Z
0
t
∂h
(s, t)ds
∂t
g (t) = h(t, t) +
t0
This implies that
Z
0
t
φ (t) = f (t) − a
e−a(t−s) f (s)ds
0
It follows that
0
Z
t
−a(t−s)
φ (t) = ψ(t) ⇔ f (t) − a
e
Z
f (s)ds =
0
Z
t
e−a(t−s) f 0 (s)ds
0
t
s=t
⇔ f (t) − a
e
f (s)ds = e−a(t−s) f (s) s=0 − a
0
−a(t−s)
s=t
⇔ f (t) = e
f (s) s=0
−a(t−s)
Z
t
e−a(t−s) f (s)ds
0
⇔ f (t) = f (t) − e−at f (0)
⇔ f (0) = 0
◦
Solution – Exercise 3 – 1 – a) For a vector-valued function, there is no mean value
theorem with an equal sign. But the following holds (see, e.g., [3, p. 44], [9, p. 209] or
Rudin, p.113).
Fund. Theory ODE Lecture Notes – J. Arino
110
B. Problem sheets
Theorem B.1.3. Let f : [a, b] → F be a continuous mapping, with F a Banach space.
Suppose that f admits a right derivative fr0 (x) for all x ∈ (a, b), and that kfr0 (x)k ≤ k, where
k ≥ 0 is a constant. Then
kf (b) − f (a)k ≤ k(b − a)
and more generally, for all x1 , x2 ∈ [a, b],
kf (x2 ) − f (x1 )k ≤ k|x2 − x1 |
Let us denote M = supt∈[0,α) kf (t, x)k, and let n, p > N , where N is sufficiently large
1
that α − n+p
∈ [0, α) and α − n1 ∈ [0, α). Using Theorem B.1.3, we obtain that
kx(α −
1
1
1
1
) − x(α − )k ≤ M | −
|
n+p
n
n n+p
kzα,n+p − zα,n k ≤ M |
1
1
−
|
n n+p
So the sequence (zα,n )n∈N∗ is a Cauchy sequence.
1 – b) For all t ∈ [0, α),
Z
t
1
kx(t) − x(α − )k ≤ kf (x(s))kds
α− 1
n
n
1
|
(B.8)
n
is a Cauchy sequence, there exists xα = limn→∞ (zα,n ). Thus
kx(t) − zα,n k ≤ Mα |t − α +
Since the sequence (zα,n )n∈N∗
taking n → ∞ in (B.8) gives
kx(t) − xα k ≤ Mα |t − α|
1 – c) According to 1.b), we have
lim kx(t) − xα k ≤ 0
t→α,t<α
Hence
lim x(t) = xα
t→α,t<α
2) Let
z(t) =
x(t)
xα
if t ∈ [0, α)
if t = α
Let us show that z is a solution of (B.7) on [0, α] if z is continuous on [0, α] and satisfies the
integral equation
Z t
z(t) = z(t0 ) +
f (z(s))ds
t0
for all t ∈ [0, α], and with an arbitrary t0 ∈ [0, α]. We know by construction that z is
continuous (since limt→∞ x(t) = xα ).
Fund. Theory ODE Lecture Notes – J. Arino
111
Let t ∈ [0, α),
t
Z
z(t) = x(t) = x(t0 ) +
f (x(s))ds
t0
because x is a solution. For t = α,
z(α) = xα
= lim x(t)
t→α,t<α
Z
=
=
t
lim x(t0 ) +
t→α,t<α
f (x(s))ds
t
Z 0t
lim z(t0 ) +
t→α,t<α
f (z(s))ds
t0
since for t < α, we have z(t) = x(t).
Furthermore, t 7→ f (z(t)) is bounded on [0, α), which implies that
Z α
Z t
f (z(s))ds = lim
f (z(s))ds
t0
t→α
So
Z
t0
α
z(α) = z(t0 ) +
f (z(s))ds
t0
So z is solution to (B.7) on [0, α].
◦
Fund. Theory ODE Lecture Notes – J. Arino
113
McMaster University – Math4G03/6G03
Fall 2003
Homework Sheet 2
Exercise 2.1 – Consider the system
0 x1
a b
x1
=
x2
c d
x2
where a, b, c, d ∈ R are constants such that ad − bc = 0. Discuss all possible behaviours of
the solutions and sketch the corresponding phase plane trajectories.
◦
Exercise 2.2 – Let A be a constant n × n matrix.
i) Show that |eA | ≤ e|A| .
ii) Show that det eA = etrA .
iii) How should α be chosen so that lim e−αt eAt = 0.
t→∞
◦
Exercise 2.3 – Let X(t) be a fundamental matrix for the system x0 = A(t)x, where A(t)
is an n × n matrix with continuous entries on R. What conditions on A(t) and C guarantee
that CX(t) is a fundamental matrix, where C is a constant matrix.
◦
Exercise 2.4 – Consider the system
x0 = A(t)x
(B.9)
where A(t) is a continuous n × n matrix on R, and A(t + ω) = A(t), ω > 0.
i) Show that P(ω), the set of ω-periodic solutions of (B.9), is a vector space.
ii) Let f be a continuous n × 1 matrix function on R with f (t + ω) = f (t). Show that,
for the system
x0 = A(t)x + f (t)
(B.10)
the following conditions are equivalent.
a) System (B.10) has a unique ω-periodic solution,
b) [X −1 (ω) − X −1 (0)] is nonsingular,
c) dim P(ω) = 0.
◦
Fund. Theory ODE Lecture Notes – J. Arino
114
B. Problem sheets
McMaster University – Math4G03/6G03
Fall 2003
Homework Sheet 2 – Solutions
Solution – Exercise 1 – The characteristic polynomial of the matrix
A=
a b
c d
is (a − λ)(d − λ) − bc = λ2 − (a + d)λ + ad − bc = λ2 − (a + d)λ since ad − bc = 0. Thus A
has the eigenvalues 0 and a + d. Hence solutions are of the form
x1 = c1
x2 = c2 e(a+d)t
with c1 , c2 ∈ R, and there are three possibilities.
• If a + d < 0, then all points on the x1 axis are equilibria, and all trajectories in
(x1 , x2 )-space go to them parallely to the x2 axis.
• If a + d = 0, then all points are equilibria.
• If a + d > 0, then the x1 axis is invariant and any solution that does not start on the
x1 axis diverges to ±∞ parallely to the x2 axis.
◦
Solution – Exercise 2 – 1) We have eA =
A
ke k = k
∞
X
P∞
k=0
Ak /k!. Taking the norm,
Ak /k!k,
k=0
P∞
whence, by the triangle inequality, keA k ≤ k=0 kAk /k!k = ekAk .
2) Let J be the Jordan form of A, i.e., there exists P nonsingular such that P −1 AP =
J, where J has the form diag(Bj )j=1,...,p , with Bj the Jordan block corresponding to the
eigenvalue λj of multiplicity mj . Then, since A and J are similar,
det eA = det eJ
= det(eB1 )m1 . . . det(eBp )mp
= eλ1 m1 · · · eλp mp
p
X
= exp(
λk m k )
k=1
= trA
Fund. Theory ODE Lecture Notes – J. Arino
115
3) We can write
lim e−αt eAt = lim e(A−αI)t
t→∞
t→∞
Let Sp (A) be the spectrum of A, i.e., the set of eigenvalues of A. Then, if λ ∈ Sp (A),
λ − α ∈ Sp (A − αI). We have limt→∞ e(A−αI)t = 0 if, and only if, <(µ) < 0 for all
µ ∈ Sp (A − αI), i.e., <(µ + α) < 0 for all µ ∈ Sp (A), i.e., <(µ) < α for all µ ∈ Sp (A).
Hence, choosing α greater than the eigenvalue of A with maximal real part ensures that
limt→∞ e(A−αI)t = 0.
◦
Solution – Exercise 3 – We have
(CX(t))0 = CX 0 (t)
= CA(t)X(t)
For CX(t) to be a fundamental matrix for x0 = A(t)x requires that (CX(t))0 = A(t)(CX(t)).
So C and A(t) must commute. Also, a fundamental matrix must be nonsingular. As X(t)
is a fundamental matrix, it is nonsingular. Thus C must be nonsingular for CX(t) to be
a fundamental matrix. So, to conclude, if X(t) is a fundamental matrix for the system
x0 = A(t)x, then CX(t) is a fundamental matrix for x0 = A(t)x if C is nonsingular and
commutes with A(t).
◦
Solution – Exercise 4 – 1) For x ∈ Rn , x ∈ P(ω) if it satisfies (B.9). Let x1 , x2 ∈ Rn
and α1 , α2 ∈ R. Then, as Rn is a vector space, α1 x1 + α2 x2 ∈ Rn . Now assume that,
moreover, x1 , x2 ∈ P(ω). Then,
(α1 x1 + α2 x2 )0 = α1 x01 + α2 x02
= α1 A(t)x1 + α2 A(t)x2
= A(t)(α1 x1 + α2 x2 )
and therefore, α1 x1 + α2 x2 ∈ P(ω), and P(ω) is a vector space.
2) There were of course several approaches to this problem. The simplest one required
the use of Theorem B.2.4, stated and proved later.
c) ⇒ b) Let V be the nonsingular matrix such that X(t + ω) = X(t)V , that we know
to exist from Theorem 2.4.2. Then X −1 (t + ω) = V −1 X −1 (t), and X −1 (t + ω) − X −1 (t) =
(V −1 − I)X −1 (t).
Suppose that dim P(ω) = 0. Then 1 is not an eigenvalue of V . This implies that 1 is
neither an eigenvalue of V −1 ; in turn, 0 is not an eigenvalue of V −1 − I. This means that
V −1 − I is nonsingular, and since X(t) is nonsingular, (V −1 − I)X −1 (t) is nonsingular. Thus
we conclude that if dim P(ω) = 0, then (X −1 (t + ω) − X −1 (t)) is nonsingular.
b) ⇒ c) The previous argument works the other way as well.
c) ⇒ a) Suppose dim P(ω) = 0 and that x1 , x2 are two solutions to (B.10). Then
0
x1 = A(t)x1 + f (t) and x02 = A(t)x2 + f (t). Therefore,
(x1 − x2 )0 = A(t)x1 + f (t) − A(t)x2 − f (t) = A(t)(x1 − x2 )
Fund. Theory ODE Lecture Notes – J. Arino
116
B. Problem sheets
which implies that x1 −x2 is a solution to (B.9), and therefore, dim P(ω) 6= 0, a contradiction.
Thus the solution to (B.10) is unique.
a) ⇒ c) Let x be the unique ω-periodic solution of (B.10), and assume that dim P(ω) 6=
0, i.e., there exists y, non trivial ω-periodic solution to (B.9). Then
(x + y)0 = A(t)x + f (t) + A(t)y
= A(t)(x + y) + f (t)
and so x + y is a solution to (B.10), which is a contradiction since x is the unique solution
to (B.10).
◦
Theorem B.2.4. Consider the system
x0 = A(t)x
where A(t) has continuous entries on R and is such that A(t + ω) = A(t) for some ω ∈ R.
Then 1 is an eigenvalue of A(t).
Proof. For some constant vector c 6= 0, we have x(t) = X(t)c. Also, because of periodicity,
x(t + ω) = X(t + ω)c. As x is periodic of period ω, x(t) = x(t + ω), so that, using the
previously obtained forms,
X(t)c = X(t + ω)c
X(t)c = X(t)V c
c=Vc
Hence c is an eigenvector of V with corresponding eigenvalue (Floquet multiplier) λ = 1.
Fund. Theory ODE Lecture Notes – J. Arino
117
McMaster University – Math4G03/6G03
Fall 2003
Homework Sheet 3
Exercise 3.1 – Compute the solution of the differential equation
x0 (t) = x(t) − y(t) − t2
y 0 (t) = x(t) + 3y(t) + 2t
(B.11)
◦
Exercise 3.2 – Consider the initial value problem
x0 (t) = A(t)x(t)
x(t0 ) = x0
(B.12)
We have seen that the solution of this initial value problem is given by
x(t) = R(t, t0 )x0
where R(t, t0 ) is the resolvent matrix of the system. Suppose that we are in the case where
the following condition holds
∀t, s ∈ I,
A(t)A(s) = A(s)A(t)
(B.13)
with I ⊂ R.
i) Show that M (t) = exp
R
t
t0
A(s)ds is a solution of the matrix initial value problem
M 0 (t) = A(t)M (t)
M (t0 ) = In
where In is the n × n identity matrix. [Hint: Use the formal definition of a derivative,
i.e., M 0 (t) = limh→0 (M (t + h) − M (t))/h]
ii) Deduce that, when (B.13) holds,
Z
t
R(t, t0 ) = exp
A(s)ds
t0
iii) Deduce the following theorem.
Theorem B.3.5. Let U, V be constant matrices that commute, and suppose that A(t) =
f (t)U + g(t)V for scalar functions f, g. Then
Z t
Z t
R(t, t0 ) = exp
f (s)dsU exp
g(s)dsV
(B.14)
t0
t0
Fund. Theory ODE Lecture Notes – J. Arino
118
B. Problem sheets
◦
Exercise 3.3 – Find the resolvent matrix associated to the matrix
A(t) =
a(t) −b(t)
b(t) a(t)
(B.15)
◦
where a, b are continuous functions on R.
Exercise 3.4 – Consider the linear system
1
x0 = x + ty
t
y0 = y
(B.16)
with initial condition x(t0 ) = x0 , y(t0 ) = y0 .
i) Solve the initial value problem (B.16).
ii) Deduce the formula for the principal solution matrix R(t, t0 ).
iii) Show that in this case,
Z
t
R(t, t0 ) 6= exp
A(s)ds
t0
with
1
A(t) =
t
0 1
t
◦
Fund. Theory ODE Lecture Notes – J. Arino
119
McMaster University – 4G03/6G03
Fall 2003
Solutions – Homework Sheet 3
Solution – Exercise 3.1 – Let ξ(t) = (x(t), y(t))T ,
A=
1 −1
1 3
and B(t) = (−t2 , 2t)T . The system (B.11) can then be written as
ξ 0 = Aξ + B(t)
This is a nonhomogeneous linear system, so we use the variation of constants,
Z t
(t−t0 )A
ξ(t) = e
ξ0 +
e(t−s)A B(s)ds
t0
where ξ0 = ξ(t0 ). Let us assume for simplicity that t0 = 0.
Eigenvalues of A are (λ − 2)2 , with associated subspace
x
−1 −1
x
0
ker(A − 2I) =
;
=
y
1
1
y
0
x
=
; x+y =0
y
Thus dim ker(A − 2I) = 1 6= 2, and A is not diagonalisable. Let us compute the Jordan
canonical form of A. There exists P nonsingular such that
2 α
−1
P AP =
0 2
where α is a constant that has to be determined.
2 1
−1
P AP =
0 2
1 −1
0 −1
−1
P =
, P =
−1 0
−1 −1
Then
eAt = P e(P
−1 AP )t
P −1
with
e(P
−1 AP )t
= eP
−1 (At)P
Fund. Theory ODE Lecture Notes – J. Arino
120
B. Problem sheets
We have P −1 AP = 2I + N , where
0 1
0 0
−1
(P AP )t
Therefore,
= e(2It+N t) = e2t eN t .
P∞ tn n e
n=0 n! N = I + N t. As a consequence,
e(P
−1 AP )t
Now, N is nilpotent (N 2 = 0), so eN t =
= e2t (I + N t)
1 0
0 1
2t
=e
+
0 1
0 0
2t
2t
e te
=
0 e2t
Thus
At
e
e2t te2t
P −1
0 e2t
=P
(1 − t)e2t
−te2t
=
te2t
(1 + t)e2t
−t
2t 1 − t
=e
t
1+t
Rt
We still have to compute 0 e−As B(s)ds. We have
2
3
2
2
s
−s
s2 − s3
−As
−2s 1 + s
−2s −s − s + 2s
−2s
e B(s) = e
=e
=e
−s 1 − s
2s
s3 + 2s − 2s2
s3 − 2s2 + 2s
Rt
Rt
Rt
Let I1 (t) = 0 e−2s sds, I2 (t) = 0 e−2s s2 ds and I3 (t) = 0 e−2s s3 ds. Then
Z t
I1 (t) − I3 (t)
−As
e B(s)ds =
I3 (t) − 2I2 (t) + 2I1 (t)
0
Evaluating the integrals, we obtain
−2t
Z t
e
−As
e B(s)ds = −2t
e
0
1 2
t
4
1 2
t
4
+ 14 t + 18 + 12 t3 − 81
− 34 t − 83 − 12 t3 + 83
As a conclusion,
ξ(t) =
=
−2t
1 − t −t
1 − t −t
e
ξ0 +
t
1+t
t
1+t
e−2t
1 2
t + 14 t + 81 + 21 t3 −
4
e2t
1 2
t − 34 t − 83 − 12 t3 +
4
−2t
e x0 − e−2t tx0 − e−2t ty0 + 21 e−2t t + 18 e−2t − 81 − 4t + 34 e−2t t2
e−2t tx0 + e−2t y0 + e−2t ty0 − 41 e−2t t2 − e−2t t + 4t − 83 e−2t + 38
is the solution of (B.11) going through (x0 , y0 ) at time 0.
1
8
3
8
◦
Fund. Theory ODE Lecture Notes – J. Arino
121
Solution – Exercise 3.2 – 1) We have
1
1
(M (t + h) − M (t)) =
h
h
Z
t+h
Z
A(s)ds − exp
exp
A(s)ds
t0
t0
Since, for all s1 , s2 ∈ I, A(s1 )A(s2 ) = A(s2 )A(s1 ), we have
Z s 1
Z s2
Z s 2
Z
A(u)du
A(u)du =
A(u)du
t0
t
t0
t0
s1
A(u)du
t0
It follows that
1
1
(M (t + h) − M (t)) =
h
h
Z
t+h
Z
t
Z
exp
A(s)ds exp
A(s)ds − exp
t
t0
Z t+h
1
A(s)ds − I
= M (t) exp
h
t
Z t+h
1
1
=
exp
A(s)ds − I M (t)
h
h
t
t
A(s)ds
t0
The second term in this equality tends to zero as h → 0, and thus
1
lim (M (t + h) − M (t)) = A(t)M (t)
h→0 h
therefore M 0 (t) = A(t)M (t), hence the result.
2) The implication is trivial.
3) If U and V commute and that A(t) = U f (t) + V g(t), then
A(s)A(t) = (U f (s) + V g(s))(U f (t) + V g(t))
= U 2 f (s)f (t) + U V f (s)g(t) + V U g(s)f (t) + V 2 g(s)g(t)
= U 2 f (t)f (s) + V U g(t)f (s) + U V f (t)g(s) + V 2 g(t)g(s)
= (U f (t) + V g(t))U f (s) + (U f (t) + V g(t))V g(s)
= (U f (t) + V g(t))(U f (s) + V g(s))
that is, A(s) and A(t) commute for all t, s. Then
Z t
Z t
R(t, t0 ) = exp
A(s)ds = exp
U f (s) + V g(s)ds
t0
t0
Z t
Z t
= exp
f (s)dsU +
g(s)dsV
t0
t0
Z t
Z t
f (s)dsU exp
g(s)dsV
= exp
t0
t0
◦
Fund. Theory ODE Lecture Notes – J. Arino
122
B. Problem sheets
Solution – Exercise 3.3 – Writing
1 0
0 −1
A(t) = a(t)
+ b(t)
= a(t)I + b(t)V,
0 1
1 0
it is obvious the Theorem B.3.5 can be used here, since I commutes with all matrices. Thus,
Z t
Z t
R(t, t0 ) = exp
a(s)dsI exp
b(s)dsV
t0
t0
Rt
Rt
Let α(t) = t0 a(s)ds and β(t) = t0 b(s)ds. Then R(t, t0 ) = eα(t)I eβ(t)V = eα(t) Ieβ(t)V . Now
notice that V 2 = I, V 3 = −V , etc., so that we can write that
(−1)p I if n = 2p,
n
V =
(−1)p V if n = 2p + 1.
This implies that
β(t)V
e
∞
X
1
β(t)n V n
=
n!
n=0
=
∞
X
(−1)p
p=0
(2p)!
!
2p
β(t)
I+
∞
X
(−1)p
β(t)2p+1
(2p + 1)!
p=0
!
V
= cos(β(t))I + sin(β(t))V
Thus
R(t, t0 ) = eα(t) (cos(β(t))I + sin(β(t))V )
α(t)
e cos(β(t)) −eα(t) sin(β(t))
=
eα(t) sin(β(t)) eα(t) cos(β(t))
◦
Solution – Exercise 3.4 – 1) Notice that the y 0 equation in (B.16) does not involve
x. Therefore, we can solve it directly, giving y(t) = Cet , with C ∈ R. Substituting this into
the equation for x0 , we have
1
x0 = x(t) + tCet
t
To solve this nonhomogeneous first-order scalar equation, we start by solving the homogeneous part, x0 = x/t. This equation is separable, giving the solution x(t) = Kt, for K ∈ R.
Now we use a variation of constants approach to find a particular solution to the nonhomogeneous problem. We use the ansatz x(t) = K(t)t, which, when differentiated and substituted
into the nonhomogeneous equation, gives K 0 (t) = Cet , and hence, K(t) = Cet is a particular
solution, giving the general solution x(t) = Kt + Cet .
Fund. Theory ODE Lecture Notes – J. Arino
123
Let t0 6= 0 (to avoid problems with 1/t), and suppose x(t0 ) = x0 , y(t0 ) = y0 . Then
x0 = Kt0 + Ct0 et0 and y0 = Cet0 . It follows that K = x0 /t0 − y0 and C = e−t0 y0 , and the
solution to the equation going through the point (x0 , y0 ) at time t0 is given by
ξ(t) =
x0
x0
t−t0
( t0 − y0 )t + e−t0 y0 tet
t
+
y
t(e
−
1)
x(t)
0
=
.
= t0
y(t)
y0 et−t0
e−t0 y0 et
2) The solution to the IVP
ξ 0 = A(t)ξ
ξ(t0 ) = ξ0
is given by ξ(t) = R(t, t0 )ξ0 . Thus, the resolvent matrix (or principal solution matrix) for
(B.16) is given by
t
−t + tet−t0
t
0
R(t, t0 ) =
0
et−t0
R t 3) First of all, notice that A(t) and A(s) do not commute. Let us compute B(t) =
A(s)ds.
t0
Rt
!
Rt
sds
Rt0t
ds
t0
1 2
2
(t
−
t
)
0
2
t − t0
1
ds
t0 s
B(t) =
0
ln tt0
0
α β
0 γ
=
which, for convenience, we denote
B(t) =
Eigenvalues of B(t) are α and γ. As R(t0 , t0 ) = B(t0 ) = I, we are concerned with finding
a t 6= t0 such that B(t) is diagonalizable. If a t exists such that α 6= γ, then B(t) is
diagonalizable, i.e., there exists P nonsingular such that
α 0
−1
P B(t)P = D =
0 β
Then
B(t)
e
=P
eα 0
P −1
0 eβ
We find
P =
1
β
,
0 γ−α
P
−1
1
=
γ−α
γ − α −β
0
1
Fund. Theory ODE Lecture Notes – J. Arino
124
Thus, after a few computations,
α
e
B(t)
e
=
0
B. Problem sheets
β
(eγ −
γ−α
γ
e
eα )
t
=
t0
∆
0 et−t0
The element ∆ in this matrix is the only one different from the elements in R(t, t0 ). We have
t
t2 − t20
t−t0
e
−
∆=
6= t(et−t0 − 1)
t0
2(t − t0 − ln tt0 )
◦
Fund. Theory ODE Lecture Notes – J. Arino
125
McMaster University – Math4G03/6G03
Fall 2003
Homework Sheet 4
Exercise 3.1 – A differential equation of the form
x0 = f (t, x(t), x(t − ω))
(B.17)
for ω > 0, is called a delay differential equation (or also a differential difference equation,
or an equation with deviating argument), and ω is called the delay. The basic initial value
problem for (B.17) takes the form
x0 = f (t, x(t), x(t − ω))
x(t) = φ0 (t), t0 − ω ≤ t ≤ t0
(B.18)
i) Use the method of steps to construct the solution to (B.18) on the interval [t0 , t0 + ω],
that is, find how to construct the solution to the non delayed problem
x0 = f (t, x(t), φ0 (t − ω))
x(t0 ) = φ0 (t0 ), t0 ≤ t ≤ t0 + ω
(B.19)
ii) Discuss existence and uniqueness of the solution on the interval [t0 , t0 + ω], depending
on the nature of φ0 and f .
iii) Suppose that φ0 ∈ C 0 ([t0 − ω, t0 ]). Discuss the regularity of the solution to (B.18) on
the interval [t0 + kω, t0 + (k + 1)ω], k ∈ N.
◦
Exercise 3.2 – Consider the delay initial value problem
x0 (t) = ax(t − ω)
x(t) = C, t ∈ [t0 − ω, t0 ]
(B.20)
with a, C ∈ R, ω ∈ R∗+ . Using the ideas of the previous exercise, find the solution to (B.20)
on the interval [t0 + kω, t0 + (k + 1)ω], k ∈ N.
◦
Exercise 3.3 – Compute Ani and etAi for the following matrices.
A1 =
0 1
,
1 0
A2 =
1 1
,
0 1


0
1
− sin(θ)
0
cos(θ)  .
A3 =  −1
− sin(θ) cos(θ)
0
◦
Fund. Theory ODE Lecture Notes – J. Arino
126
B. Problem sheets
Exercise 3.4 – Compute etA for the matrix

1
1
0
A = −1 0 −1 .
0 −1 1

◦
Exercise 3.5 – Let A ∈ Mn (R) be a matrix (independent of t), k · k be a norm on Rn
and |||·||| the associated operator norm on Mn (R).
i)
a) Show that for all t ∈ R and all k ∈ N∗ , there exists Ck (t) ≥ 0 such that,
t A
t
e k − I + A ≤ 1 Ck (t).
k 2
k
with limk→∞ Ck (t) =
t2
2
|||A2 |||.
b) Show that for all t ∈ R and all k ∈ N∗ ,
I + t A ≤ e |t|k |||A||| .
k c) Deduce that
tA
e
= lim
k→∞
t
I+ A
k
k
.
ii) Suppose now that A is symmetric and that its eigenvalues are > −α, with α > 0.
a) Show by induction that, for k ≥ 0,
−(k+1)
(αI + A)
Z
=
∞
e−t(αI+A)
0
tk
dt.
k!
b) Deduce that for all u > 0,
(I + uA)−k ≤ M if, and only if, e−tA ≤ M,
∀t > 0.
c) Show that
∀t > 0,
e−tA ≥ 0 ⇔ ∃λ0 ,
∀λ > λ0 ,
(λI + A)−1 ≥ 0
where by convention, for B = (bij ) ∈ Mn (R), writing that B ≥ 0 means that
bij ≥ 0 for all i, j = 1, . . . , n.
iii) Do the results of part 2) hold true if A is a nonsymmetric matrix?
◦
Fund. Theory ODE Lecture Notes – J. Arino
127
McMaster University – Math4G03/6G03
Fall 2003
Homework Sheet 4 – Solutions
Solution – Exercise 1 – 1) The proposed method consists in considering (B.18) as a
nondelayed IVP on the interval [t0 , t0 + ω]. Indeed, on this interval, we can consider (B.19).
That the latter is a nondelayed problem is obvious if we rewrite the differential equation as
x0 (t) = g(t, x(t))
(B.21)
with g(t, x(t)) = f (t, x(t), φ0 (t − ω)), which is well defined on the interval [t0 , t0 + ω] since
for t ∈ [t0 , t0 + ω], t − ω ∈ [−ω, 0], on which the function φ0 is defined.
We can then use the integral form to construct the solution on the interval [t0 , t0 + ω],
Z t
x(t) = x(t0 ) +
g(s, x(s))ds
t0
Z t
= φ0 (t0 ) +
f (s, x(s), φ0 (s − ω))ds
t0
2) Obviously, the discussion to make is on the nature of the function f . As problem
(B.19) is an ordinary differential equations initial value problem, existence and uniqueness
of solutions on the interval [t0 , t0 + ω] follow the usual scheme. To discuss the required
properties on f and φ0 , the best is to use (B.21). Recall that a vector field has to be
continuous both in t and in x for solutions to exist. Thus to have existence of solutions to
the equation (B.21), g must be continuous in t and x. This implies that f (t, x, φ0 (t − ω))
must be continuous in t, x. Thus φ0 has to be continuous on [t0 − ω, t0 ].
Now, for uniqueness of solutions to (B.21), we need g to be Lipschitz in x, i.e., we require
the same property from f . Note that this does not imply either φ0 or the way f depends on
φ0 .
3) Every successive integration raises the regularity of the solution: x is C 1 on [t0 , t0 + ω],
2
C on [t0 + ω, t0 + 2ω], etc. Hence, x is C n on [t0 + (n − 1)ω, t0 + nω].
◦
Solution – Exercise 2 – We proceed as previously explained. We assume for simplicity that t0 = 0. To find the solution on the interval [0, ω], we consider the nondelayed
IVP
x01 = ax0 (t)
x1 (0) = C
where x0 (t) = C for t ∈ [0, ω]. The solution to this IVP is straightforward, x1 (t) = C +aCt =
C(1 + at), defined on the interval [0, ω]. To integrate on the second interval, we consider the
IVP
x02 = a[C(1 + at)]
x2 (ω) = x1 (ω) = C + aCω
Fund. Theory ODE Lecture Notes – J. Arino
128
B. Problem sheets
Hence we find the solution to the differential equation to be, on the interval [ω, 2ω],
1
1
x2 (t) = C 1 + at + a2 t2 − a2 ω 2
2
2
Iterating this process one more time with the IVP
1 22 1 2 2
= a C 1 + at + a t − a ω
2
2
3
x3 (2ω) = x2 (2ω) = a2 Cω 2 + 2aCω + C
2
x03
we find, on the interval [2ω, 3ω], the solution
1
1
1
1
1
x3 (t) = C 1 + at + a2 t2 + a3 t3 − ta3 ω 2 − a3 ω 3 − a2 ω 2
2
6
2
3
2
We develop the intuition that the solution at step n (i.e., on the interval [(n − 1)ω, nω])
must take the form
n
X
(t − (k − 1)ω)k
(B.22)
ak
xn (t) = C
k!
k=0
This can be proved by induction (we will not do it here).
◦
Solution – Exercise 3 – For matrix A1 , we have A21 = I, A31 = A1 , etc. Hence,
2n+1
A2n
= A1 , which implies
1 = I and A1
∞
∞
X
t2n 2n X t2n+1
eAt =
A1 +
A12n+1
(2n)!
(2n
+
1)!
n=0
! n=0 ∞
!
∞
2n
X t2n+1
X
t
=
I+
A1
(2n)!
(2n
+
1)!
n=0
n=0
= cosh tI + sinh tA
cosh t sinh t
=
sinh t cosh t
For matrix A2 , remark that it can be written as A2 = I + N , where
N=
0 1
0 0
is a nilpotent matrix. Thus A22 = (I + N )2 = I + 2N + N 2 = I + 2N , A32 = (I + 2N )(I + N ) =
I + 2N + N + 2N 2 = I + 3N , and it is easily shown by induction that (I + N )n = I + nN .
Fund. Theory ODE Lecture Notes – J. Arino
129
It follows that
∞
X
tn
eA2 t =
n=0
t
!
I+
n!
=e I +t
∞
X
ntn
n=0
∞
X
n=0
tn−1
(n − 1)!
!
n!
!
N
N
= et I + tet N
t 1 t
=e
0 1
Finally, for matrix A3 , we have that


− cos2 θ − 12 sin 2θ cos θ
A23 = − 21 sin 2θ − sin2 θ sin θ 
− cos θ
− sin θ
1
and A33 = 0, i.e., A3 is nilpotent for the index 3. Therefore, eA3 t = I + tA3 +
t2 2
A.
2 3
◦
Solution – Exercise 4 – The Jordan canonical form of A is


0 0 0
J = 0 1 1
0 0 1
Now, to compute eJ t, remark that J has the form


00 0



0
A1
0
where eA1 t has been computed in Exercise 3. Hence,
 
 0t

e 0 0
1 0 0


 0 A1 t  = 0 et tet 
e
0 0 et
0
We have J = P −1 AP , where


−1 −1 2
0 −1 ,
P = 1
1
1 −1

P −1

1 1 1
= 0 −1 1
1 0 1
are the matrices of change of basis that transform A to its Jordan canonical form. Then
A = P JP −1 , and eAt = P eJt P −1 , i.e.,


(2 − t)et − 1
et − 1
(1 − t)et − 1
1
1 − et 
eAt =  1 − et
1 + (t − 1)et 1 − et 1 + tet
Fund. Theory ODE Lecture Notes – J. Arino
130
B. Problem sheets
◦
Solution – Exercise 5 – This exercise was far from trivial.
1–a) Consider the map
t 7→ eAt
R → Mn (R)
We have (eAt )0 = AeAt and (eAt )(k) = Ak eAt , where u(k) denotes the kth derivative of u.
t
t
t2
t2 t
e k A = I + A + 2 A2 + 2 ε( )
k
2k
k k
Thus
t
t2
t2 t
t
e k A − I − A = 2 A2 + 2 ε( )
k
2k
k k
Therefore, taking the norm |||·||| of this expression,
2
2 t t A
t
1
t
2
e k − (I + A) ≤
A + t ε( ) = 1 Ck (t)
k 2
k
k
2
k2
We have
t2 2 A
k→∞
2
P
j
tj
Let Sk = ∞
j=3 kj−2 j! |||A||| . This series is uniformly convergent, which implies that we can
change the order in the following limit,
j
∞
X
t
j
lim
lim Sk =
|||A||| = 0
k→∞
k→∞
k j−2 j!
j=3
1–b) We have already seen (Exercise 2, Assignment 2) that eA ≤ e|||A||| . Therefore,
t t
k A e ≤ e k |||A|||
lim Ck (t) =
But
e
|t|
|||A|||
k
∞
X
1 |t|k |
|||A|||k
=
k
k! k
k=0
=1+
which, since
|t|2
2k2
|t|
|t|2 |
|||A||| + 2 |||A|||2 + · · ·
k
2k
|||A|||2 + · · · ≥ 0, implies that
e
|t|
|||A|||
k
t |t|
≥ 1 + |||A||| = |||I||| + A
k
k
t
≥ I + A
k
Fund. Theory ODE Lecture Notes – J. Arino
131
(since |||I||| = supkvk≤1 kIV k = supkvk≤1 kvk = 1).
1–c) We skip the scalar case, and consider the case n ≥ 2. We have
t
t
t
eAt − (I + A)k = (e k A )k − (I + A)k
k
k
X
k−1
t
t
t
t
A
k
(e k A )k−1−j (I + A)j
= e − (I + A)
k
n
j=0
since
for two matrices E, F ∈ Mn (R) (or Mn (C)) that commute, E n − F n = (E −
Pk−1
F ) j=0 E k−1−j F j .
Therefore,
k−1 At
t A−(I+ t A) X
t(k−1−j) t
t
k
A
j
(I + A) e − (I + A) ≤ e k
k
k
e
(B.23)
k
k
j=0
t(k−1−j) |t|(k−1−j)
Now, we have e k A ≤ e k |||A||| . Also,
j j
t
t
|||A|||
j
(I + A) ≤ I + A ≤ e |t|k |||A||| = e j|t|
k
k
k
where the last inequality results from question 1–b). Therefore,
k−1
X
At
j|t|
t
|t| k−1−j
k
|||A||| k |||A|||
e − (I + A) ≤ 1 Ck (t)
k
e
e
k
k2
j=0
k−1
X
k−1
1
= 2 Ck (t)
e|t| k |||A|||
k
j=0
k−1
1
Ck (t)ke|t| k |||A|||
2
k
k−1
1
= Ck (t)e|t| k |||A|||
k
=
We thus have
At
1
|||A|||
e − (I + t A)k ≤ 1 Ck (t)e|t| k−1
k
= Dk (t)
k
k
k
2
k−1
2
As k → ∞, Ck (t) → t2 |||A2 ||| and e|t| k |||A||| → e|t||||A||| . Therefore, limk→∞ Dk (t) = t2 |||A2 ||| e|t||||A||| ,
which in turn implies that limk→∞ (I + kt A)k = eAt .
2–a) We now suppose that A is a symmetric matrix. Recall that any symmetric matrix
is diagonalizable, with real eigenvalues. Furthermore, there exists a matrix P such that
P −1 = P T and that P T AP = diag(λi ), with λi ∈ Sp (A).
We assume that the eigenvalues λi , i = 1, . . . , m, are such that λi > −α, for α > 0. We
want to show by induction that the following holds.
Z ∞
tk
−(k+1)
∀k ≥ 0, (αI + A)
=
e−(αI+A)t dt
(B.24)
k!
0
Fund. Theory ODE Lecture Notes – J. Arino
132
B. Problem sheets
Suppose that k = 0. Equation (B.24) reads
Z
−1
(αI + A) =
∞
e−(αI+A)t dt
0
−1
The matrix (αI + A) is nonsingular. Indeed, suppose that det(αI + A) = 0. This is
equivalent to det(−αI −A) = 0, which implies that −α is an eigenvalue of A, a contradiction
with the hypothesis on the localization of the spectrum.
Now remark that if B is a nonsingular matrix, the equality dtd eBt = BeBt implies that
B dtd eBt = eBt . Since (αI + A)−1 is nonsingular, we thus have
d −(αI+A)t
e
= −(αI + A)e−(αI+A)t
dt
and therefore
e−(αI+A)t = −(αI + A)−1
and so, integrating,
Z ∞
−(αI+A)t
e
Z
d −(αI+A)t e
dt
∞
d −(αI+A)t −(αI + A)−1
e
dt
dt
0
Z i
d −(αI+A)t −1
e
= −(αI + A)
nf ty
dt
dt
0
∞
= −(αI + A)−1 e−(αI+A)t 0
h
i
= −(αI + A)−1 lim e−(αI+A)t − I
ds =
0
t→∞
(B.25)
Now,

e−(α+λ1 )t
0
..

e−(αI+A)t = P 
0
.
e−(α+λn )t

 −1
P
Since for all i = 1, . . . , n, λi > −α, it follows that limt→∞ e−(α+λi )t = 0, which in turn implies
that limt→∞ e−(αI+A)t = 0. Using this in (B.25) gives (B.24) for k = 0.
Now assume (B.24) holds for k = j, i.e.,
Z ∞
tj−1
−j
(αI + A) =
e−(αI+A)t
dt
(j − 1)!
0
Then
Z
∞
j
−(αI+A)t t
e
0
j!
∞
d −(αI+A)t tj
−(αI + A)−1
e
dt
dt
j!
0
Z ∞
d −(αI+A)t tj
−1
= −(αI + A)
e
dt
dt
j!
0
Z ∞
j ∞
j−1
−1
−(αI+A)t t
−(αI+A)t t
= −(αI + A)
e
−
e
dt
j! 0
(j − 1)!
0
Z
dt =
Fund. Theory ODE Lecture Notes – J. Arino
133
As we did in the k = 0 case, we now use the bound on the eigenvalues to get rid of the term
 tj
tj −(αI+A)t

e
=P
j!
e−(α+λ1 )t
j!

0
...
tj
0
j!
e−(α+λn )t
 −1
 P −→0
as t → ∞
Therefore,
Z
∞
j
−(αI+A)t t
e
j!
0
Z
−1
∞
dt = −(αI + A)
e−(αI+A)t
0
tj−1
dt
(j − 1)!
−1
= −(αI + A) (αI + A)−j
= −(αI + A)−(j+1)
from which we deduce that (B.24) holds for all k ≥0.
2–b) Let us begin with the implication (∀u > 0, (I + uA)−k ≤ M ) ⇒ (∀t > 0, e−At ≤ M ).
We know from 1–c) that eAt = limk→∞ (I + kt A)k . Thus
−At
e
= lim
k→∞
t
I+ A
k
k !−1
= lim
k→∞
t
I+
k
−k
(B.26)
Let u = t/k with k ∈ N∗ and t > 0. Then
t
−k
∀t > 0, ∀k ∈ N , (I + A) ≤ M
k
t
−k
⇒ ∀t > 0, lim (I + A) ≤ M
k→∞
k
t
−k
⇒ ∀t > 0, lim (I + A) ≤ M
k→∞
k
∗
which, using (B.26), implies that
∀t > 0, e−At ≤ M
We now treat the reverse implication, (∀t > 0, e−At ≤ M ) ⇒ (∀u > 0, (I + uA)−k ≤ M ).
We have
1
1
(I + uA)−k = [u( I + A)]−k = u−k [ I + A]−k
u
u
Suppose that −α > −1/u, i.e., 0 < u < 1/α. Then, from 2–a), it follows that
−k
(I + uA)
=u
−k
Z
0
∞
1
e−( u I+A)t
tj−1
dt
(j − 1)!
Fund. Theory ODE Lecture Notes – J. Arino
134
B. Problem sheets
which, when taking the norm, gives
(I + uA)−k ≤ u−k
Z
∞
0
−k
Mu
≤
(k − 1)!
−At − t
e e u
∞
Z
tj−1
dt
(j − 1)!
t
e− u tk−1 dt
0
Let χ = t/u, then dt = udχ, and
−k Z ∞
(I + uA)−k ≤ M u
e−χ uk−1 χk−1 dχ
(k − 1)! 0
Z ∞
M
e−χ χk−1 dχ
≤
(k − 1)! 0
The latter integral is Γ(k), the Gamma Function. It is well known1 that for k ∈ N, Γ(k) =
(k − 1)!. Thus,
(I + uA)−k ≤ M
2–c) Let us begin with the forward implication (⇒). To apply 2–b) with k = 0, it
suffices that the eigenvalues of A be greater than −α. Take λ0 = α. Then λ > α = λ0 , and
so
Z ∞
−1
(λI + A) =
e−(λI+A)t dt
Z0 ∞
=
e−λt e−At dt
≥0
0
since the eigenvalues λ ∈ R and by hypothesis on e−At .
Now for the reverse implication (⇐). That there exists λ0 ∈ R such that for all k ∈ N∗ ,
(λI + A)−1 ≥ 0 implies that
∀λ > λ0 ,
∀k ∈ N∗ ,
(λI + A)−k ≥ 0
for λ sufficiently large. Take λ = k/t, the previous expression can be written
∀t > 0,
∀k ≥ k0 ,
k
( I + A)−k ≥ 0
t
where k0 is sufficiently large. This implies that
−k
k
t
∀t > 0, ∀k ≥ k0 ,
(I + A)−k ≥ 0
t
k
As (k/t)−k > 0,
∀t > 0,
1
∀k ≥ k0 ,
t
(I + A)−k ≥ 0
k
See, e.g., M. Abramowitz and I.E. Stegun, Handbook of Mathematical Functions. Dover, 1965.
Fund. Theory ODE Lecture Notes – J. Arino
135
so
∀t > 0,
t
lim (I + A)−k ≥ 0
k→∞
k
So that this finally implies that
∀t > 0,
e−At ≥ 0
3) The results of the previous part hold. However, in the case of a nonsymmetric matrix,
we need to ask for the real part of the eigenvalues to be greater than −α.
◦
Fund. Theory ODE Lecture Notes – J. Arino
137
MATH
4G03/6G03
Duration Of Examination: 72 Hours
McMaster University Final Examination
Julien Arino
December 2003
This examination paper includes 4 pages and 3 questions. You are responsible
for ensuring that your copy of the paper is complete.
Detailed Instructions
You have 72 hours, from the time you pick up and sign for this examination sheet, to
complete this examination. You are to work on this examination by yourself. Any hint of
collaborative work will be considered as evidence of academic dishonesty. You are not to
have any outside contacts concerning this subject, except myself.
You can use any document that you find useful. If using theorems from outside sources,
give the bibliographic reference, and show clearly how your problem fits in with the conditions
of application of the theorem. When citing theorems from the lecture notes, refer to them
by the number they have on the last version of the notes, as posted on the website on the
first day of the examination period.
Pay attention to the form of your answers: as this is a take-home examination, you are
expected to hand back a very legible document, in terms of the presentation of your answers.
Show your calculations, but try to be concise.
This examination consists of 1 independent question and 2 problems. In questions or
problems that have multiple parts, you are always allowed to consider as proved the results
of a previous part, even if you have not actually done that part.
Foreword to the correction. This examination was long, but established the results in
a very guided way. Exercise 1 was almost trivial. Both of the Problems dealt with Sturm
theory. This comes as an illustration of the richness of behaviors that can be observed in
differential equations: simple equations such as (B.29) can have very complex behaviors.
Concerning the difficulty of the problems, it was not excessive. Problem 2 is a shortened
and simplified version of a review problem for the CAPES, a French competition to hire
high school teachers. The original problem comprised 23 questions, and was written by
candidates in 5 hours. Problem 3 introduced the Wronskian, which we did not have time to
cover during class. It also established further results of Sturm type.
Fund. Theory ODE Lecture Notes – J. Arino
138
B. Problem sheets
Exercise 5.1 – Consider the mapping A : t 7→ A(t), continuous from R to Mn (R),
periodic of period ω and such that A(t)A(s) = A(s)A(t) for all t, s ∈ R. Consider the
equation
x0 (t) = A(t)x
(B.27)
Let R(t, t0 ) be the resolvent of (B.27).
i) Show that R(t + ω, t0 + ω) = R(t, t0 ) for all t, t0 ∈ R.
ii) Let u be an eigenvector of R(ω, 0), associated to the eigenvalue λ. Show that the
solution of (B.27) taking the value u for t0 = 0 is such that
x(t + ω) = λx(t),
∀t ∈ R.
(B.28)
iii) Conversely, show that if x is a nontrivial solution of (B.27) such that (B.28) holds,
then λ is an eigenvalue of R(ω, 0).
◦
Problem 5 2 – The aim of this problem is to study some properties of the solutions of
the differential equation
x00 + q(t)x = 0,
(B.29)
where q is a continuous function from R to R.
i) Show that for t0 , x0 , y0 ∈ R, there exists a unique solution of (B.29) such that
x(t0 ) = x0 ,
x0 (t0 ) = y0
Preliminary results : convex functions. A function f : I ⊂ R → R is convex if, for all
x, y ∈ I and all λ ∈ [0, 1],
f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y).
Before proceeding with the study of the solutions of (B.29), we establish a few useful results
on convex functions.
ii) Let f be a function defined on R, convex and nonnegative. Suppose that f has two
zeros t1 , t2 and that t1 < t2 . Show that f is zero on the interval [t1 , t2 ].
Let c ∈ R and f be a convex function that is bounded from above on the interval [c, +∞).
It can then be shown that f is decreasing on [c, +∞). Using this fact, show the following.
iii) Every convex function that is bounded from above on R is constant.
Part I.
Fund. Theory ODE Lecture Notes – J. Arino
139
iv) Let a, b ∈ R, a < b. We assume that (B.29) has a solution x, zero at a and at b and
positive on (a, b). Show that
Z b
4
|q(t)|dt >
.
b−a
a
R∞
v) We suppose that 0 |q(t)|dt converges. Let x be a bounded solution of (B.29). Determine the behaviour of x0 as t → ∞.
vi) We suppose that q ∈ C 1 and is positive and increasing on R+ . Show that all solutions
of (B.29) are bounded on R+ .
Part II. We suppose in this part that q is nonpositive and is not the zero function.
vii) Let x be a solution of (B.29). Show that x2 is a convex function.
viii) Show that if x is a solution of (B.29) that has two distinct zeros, then x ≡ 0.
ix) Show that if x is a bounded solution of (B.29), then x ≡ 0.
Part III.
x) Let x, y be two solutions of (B.29). Show that the function xy 0 − x0 y is constant.
xi) Let x1 and x2 be the solutions of (B.29) that satisfy
x1 (0) = 1,
x2 (0) = 0,
x01 (0) = 0,
x02 (0) = 1.
Show that (x1 , x2 ) is a basis of the vector space S on R of the solutions of (B.29).
What is the value of x1 x02 − x01 x2 ? Can the functions x1 and x2 have a common zero?
Justify your answer.
xii) Discuss the results of question 11) in the context of linear systems, i.e., transform
(B.29) into a system of first-order differential equations and express question 11) and
its answer in this context.
xiii) Show that if q is an even function, then the function x1 is even and the function x2 is
odd.
Problem 5 3 – The aim of this problem is to show some elementary properties of the
Wronskian of a system of solutions, and to use them to study a second-order differential
equation.
Consider the nth order ordinary differential equation
x(n) = a0 (t)x + a1 (t)x0 + · · · + an−1 (t)x(n−1) (t)
where x(k) denotes the kth derivative of x,
dk x
.
dtk
(B.30)
Fund. Theory ODE Lecture Notes – J. Arino
140
B. Problem sheets
i) Find the matrix A(t) such that this system can be written as the first-order linear
system y 0 = A(t)y.
Part I : Wronskian Let f1 , . . . , fn be n functions from R into R that are n − 1 times differentiable. We define W (f1 , . . . , fn ), the Wronskian of f1 , . . . , fn , by


f1 (t)
···
fn (t)
 f10 (t)
···
fn0 (t) 


W (f1 , . . . , fn )(t) = det 
.
..
..


.
.
(n−1)
f1
(n−1)
(t) · · · fn
(t)
If f1 , . . . , fn are linearly dependent, then W (f1 , . . . , fn ) = 0. The converse is false. Remember
that the set of solutions of (B.30) forms a vector space S of dimension n.
ii) Using the equivalent linear system y 0 = A(t)y, show that every system of n solutions
of (B.30) whose Wronskian is nonzero at a time τ constitutes a basis of S.
iii) Using the linear system x0 = A(t)x, show that for every set of n solutions,
Z t
W (t) = W (s) exp
an−1 (u)du .
s
Part II : a theorem of Sturm Let us now consider the second-order differential equation
a2 (t)x00 + a1 (t)x0 + a0 (t)x = 0
(B.31)
The objective here is to show the following theorem of Sturm.
Theorem B.5.6 (Sturm). Let f1 , f2 be two independent solutions of (B.31). Between two
consecutive zeros of f1 , there is exactly one zero of f2 .
We suppose f1 , f2 are two independent solutions of (B.31).
iv) Let u and v be two consecutive zeros of f1 . Using the Wronskian W (f1 , f2 ), show that
u and v cannot be zeros of f2 .
v) Deduce the theorem. [Hint: consider the function f1 /f2 .]
Part III : another theorem of Sturm. Let us assume that we confine ourselves to segments
of R where a0 (t) 6= 0.
vi) Let
Z
x(t) = u(t) exp
0
00
Show that (B.31) becomes u + q(t)u = 0.
We want to show the following theorem of Sturm.
t
a1 (s)
ds
a2 (s)
Fund. Theory ODE Lecture Notes – J. Arino
141
Theorem B.5.7 (Sturm). Let
x00 + q(t)x = 0,
q(t) ≤ 0
(B.32)
in an interval (t1 , t2 ). Every solution of (B.32) that is not identically zero has at most one
zero in the interval [t1 , t2 ].
vi) Let φ be a solution of (B.32) on (t1 , t2 ), and v be a zero of φ in this interval. Discuss the
properties of φ. [Hint: Use of Problem 2, Part II is possible, but not strictly necessary.]
vii) Let u < v be another zero of φ in the interval (t1 , t2 ). Discuss properties of φ. Conclude.
Fund. Theory ODE Lecture Notes – J. Arino
142
B. Problem sheets
Solution – Exercise 1 – 1) We have
Z
t+ω
R(t + ω, t0 + ω) = exp
A(s)ds
t +ω
Z 0t
= exp
t
Z 0t
= exp
A(s + ω)ds
A(s)ds
t0
= R(t, t0 )
2) Let u be an eigenvector associated to the eigenvalue λ, i.e.,
R(ω, 0)u = λu
Let x be the solution of (B.27) such that x(t0 ) = x(0) = u. As x is a solution of (B.27), we
have that, for all t,
x(t) = R(t, t0 )u = R(t, 0)u
Therefore,
x(t + ω) = R(t + ω, 0)u
= R(t + ω, ω)R(ω, 0)u
= R(t + ω, ω)λu
= R(t, 0)λu
= λR(t, 0)u
= λx(t)
and hence (B.28).
3) Let x be a nonzero solution of (B.27) such that, for all t ∈ R,
x(t + ω) = λx(t)
We assume that, for all t ∈ R, x(t + ω) = λx(t). This is true in particular for t = 0, and
hence x(ω) = λx(0). As x ≡
6 0, there exists v ∈ R − {0} such that x(0) = v. Therefore,
x(t) = λv
and as a consequence,
R(ω, 0)v = λv
and λ is an eigenvalue of R(ω, 0).
◦
Solution – Problem 2 – This problem concerns what are called Sturm theory type
results, that is, results dealing with the behavior of second order differential equations.
Fund. Theory ODE Lecture Notes – J. Arino
143
1) This is a standard application of the existence-uniqueness result of Cauchy-Lipschitz.
To see that, transform the second order equation into a system of first order equations, by
setting y = x0 . Then, differentiating y, we obtain
y 0 + qx = 0
Therefore, (B.27) is equivalent to the system of first order equations
x0 = y
y 0 = −qx
This is a linear system, hence satisfies a Lipschitz condition, and we can apply the CauchyLipschitz theorem.
2) Since f is nonnegative and convex, we have that, for all λ ∈ [0, 1],
0 ≤ f ((1 − λ)t1 + λt2 ) ≤ (1 − λ)f (t1 ) + λf (t2 )
But we have supposed that f (t1 ) = f (t2 ) = 0. Hence we have that for all λ ∈ [0, 1],
f ((1 − λ)t1 + λt2 ) = 0
and so f is zero on [t1 , t2 ].
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
◦
Solution – Problem 3 – This problem was also about Sturm results. But it also
introduced the notion of Wronskian, which is a very general tool intimately linked to the
notion of resolvent matrix.
0
1) We let y1 = x, y2 = x0 , . . . , yn = x(n−1) . As a consequence, y10 = y2 , y20 = y3 , . . . , yn−1
=
0
yn , and yn = a0 (t)y1 + a1 (t)y2 + · · · + an−1 (t)yn−1 . Written in matrix form, this is equivalent
to y 0 = A(t)y, with y = (y1 , . . . , yn )T and


0
1
0
...
0
 0
0
1 0 ...
0 
 .
.. 
 .

. 
 .
A(t) =  .
(B.33)

 ..
1
0 


 0
1 
a0 (t) a1 (t)
. . . an−1 (t)
Fund. Theory ODE Lecture Notes – J. Arino
144
B. Problem sheets
2) We know that the system is equivalent to y 0 = A(t)y, with A(t) given by (B.33). To
every basis (φ1 , . . . , φn ) of the vector space of solutions of
y 0 = A(t)y
(B.34)
there corresponds a basis (ϕ1 , . . . , ϕn ) of (B.30), where ϕi is the first coordinate of the vector
φi for every i. The converse is also true.
We know that a system (φ1 , . . . , φn ) of solutions of (B.34) is a basis if det(φ1 , . . . , φn ) 6= 0,
and it suffices for this that det(φ1 , . . . , φn ) be nonzero at one point.
Since we have det(φ1 , . . . , φn ) = W (ϕ1 , . . . , ϕn ), the result follows.
3) This is a direct application of Liouville’s theorem, which states that if R(t, s) is the
resolvent of A(t), then
Z
t
det R(t, s) = exp
trA(u)du
s
And if a system of coordinates is fixed, for every fundamental matrix Φ̂,
Z t
det Φ̂(t) = det Φ̂(s) exp
trA(u)du
s
From Liouville’s theorem,
−1
t
Z
det(Φ(t)Φ (s)) = exp
trA(u)du
s
which implies that
Z
det Φ(t) = det Φ(s) exp
t
trA(u)du
s
Now, note that det Φ(t) = W (t) and det Φ(s) = W (s). This implies the result.
Note that for a system of solutions of (B.30), W (ϕ) 6= 0 iff ϕ are linearly independent
(i.e., we have the converse implication).
4)
5)
6)
7)
◦
Fund. Theory ODE Lecture Notes – J. Arino
145
University of Manitoba – Math 8430
Fall 2006
Homework Sheet 1
Periodic solutions of differential equations
In this problem, we will study the solutions of some differential equations, and in particular,
their periodic solutions.
Let T > 0 be a real number, P be the vector space of real valued, continuous and
T -periodic functions defined on R, and let a ∈ P . Define
Z t
Z T
A=
a(t)dt,
g(t) = exp
a(u)du ,
0
0
and endow P with the norm
kxk = sup |x(t)|.
t∈R
First part
1. For what value(s) of A does the differential equation
x0 (t) = a(t)x(t)
(E1)
admit non trivial T -periodic solutions?
We now let b ∈ P , and consider the differential equation
x0 (t) = a(t)x(t) + b(t).
(E2)
2.a. Describe the set of maximal solutions to (E2) and the intervals of definition of these
solutions.
2.b. Describe the set of maximal solutions to (E2) that are T -periodic, first assuming
A 6= 0, then A = 0.
Fund. Theory ODE Lecture Notes – J. Arino
146
B. Problem sheets
Second part
In this part, we let H be a real valued C 1 function defined on R2 , and consider the
differential equation
x0 (t) = a(t)x(t) + H(x(t), t).
(E3)
3. Check that a function x is solution to (E3) if and only if it satisfies the condition
Z t
−1
x(t) = g(t) x(0) +
g(s) H(x(s), s)ds .
0
4. Suppose that H is T -periodic with respect to its second argument, and that A 6= 0.
Show that, for all functions x ∈ P , the formula
Z t+T
eA
U (H x)(t) =
g(t)
g(s)−1 H(x(s), s)ds,
A
1−e
t
defines a function UH x ∈ P , and that x est solution to (E3) if and only if UH x = x.
In the rest of the problem, we let F be a real-valued C 1 function defined on R2 , T -periodic
with respect to its second argument; for all ε > 0, define Hε = εF and Uε = UHε , so that
the differential equation (E3) is written
x0 (t) = a(t)x(t) + εF (x(t), t).
(E4)
Assume that A 6= 0. For all r > 0, we denote Br the closed ball with centre 0 and radius r
in the normed space P . We want to show the following assertion: for all r > 0, there exists
ε1 > 0 such that, for all ε ≤ ε1 , the differential equation (E4) has a unique solution x ∈ Br ,
that we will denote xε .
(v, s)|), where
We denote αr (resp. βr ) the upper bound of the set |F (v, s)| (resp. | ∂F
∂v
v ∈ [−r, r] and s ∈ [0, T ].
5.a. Find a real ε0 > 0 such that, for all ε ≤ ε0 , Uε (Br ) ⊂ Br .
5.b. Find a real ε1 ≤ ε0 such that, for all ε ≤ ε1 , the restriction of Uε to Br be a
contraction of Br .
5.c. Conclude.
6. Study the behavior of the function xε when ε → 0, the number r being fixed.
7. We now suppose that the function a is a constant k 6= 0 et that the function F takes
the form F (v, s) = f (v). Determine the solution xε of (E4).
8. We now consider T = 1, k = −1 and f (v) = v 2 , and thus (E4) takes the form
x0 (t) = −x(t) + εx(t)2 .
(E5)
8.a. Give possible values of ε0 and ε1 .
8.b. Determine the xε of (E5).
8.c. Let α ∈ R. Show that there exists a unique maximal solution ϕα of (E5) such that
ϕα (0) = α. Determine precisely this solution, and graph several of these solutions.
Fund. Theory ODE Lecture Notes – J. Arino
147
Third part
Here, we consider the differential equation
x0 (t) = kx(t) + εf (x(t)),
(E6)
where k < 0, f is C 1 and zero at zero. We let
λ = sup |f 0 (u)|,
u∈[−1,1]
and assume that ελ < −k.
We propose to show the following result: if x is a maximal solution of (E6) such that
|x(0)| < 1, then it is defined on [0, ∞) and, for all t ≥ 0,
|x(t)| ≤ |x(0)|e(k+ελ)t .
9. In this question, we suppose that the set of t such that |x(t)| > 1 is non-empty, and
we denote its lower bound by θ. Show that, for all t ∈ [0, θ],
|x(t)| ≤ |x(0)|e(k+ελ)t .
10. Conclude.
N.B. This result expresses the stability and the asymptotic stability of the trivial solution
of (E6).
Fund. Theory ODE Lecture Notes – J. Arino
149
University of Manitoba – Math 8430
Fall 2006
Homework Sheet 1 – Solutions
1. The equation (E1) is a separable equation, so we write
x0 (t)
= a(t)
x(t)
Z t
⇔ ln |x(t)| =
a(s)ds + C
0
Z t
⇔ |x(t)| = exp
a(s)ds + C
0
Z t
⇔ x(t) = K exp
a(s)ds ,
x0 (t) = a(t)x(t) ⇔
0
where it was assumed that integration starts at t0 = 0, and where the sign of |x(t)| is
absorbed into K ∈ R. Since x(0) = K, the general solution to (E1) is thus
Z t
x(t) = x(0) exp
a(s)ds .
(B.35)
0
A nontrivial solution (B.35) is T -periodic if it satisfies x(t + T ) = x(t) for all t ≥ 0. In
particular (for simplicity), there must hold that x(T ) = x(0). This leads to
Z
T
x(T ) = x(0) ⇔ x(0) exp
a(s)ds = x(0)
0
Z T
a(s)ds = 1
⇔ exp
0
Z T
⇔
a(s)ds = 0
0
⇔ A = 0.
2.a. We know that the general solution to the homogeneous equation (E1) associated to
(E2) is given by (B.35). To find the general solution to (E2), we need a particular solution
to (E2), or to use integrating factors or a variation of constants approach. We do the
latter, since we already have the solution (B.35) to (E1). Returning to the solution with
undetermined value for K, we consider the ansatz
Z t
φ(t) = K(t) exp
a(s)ds = K(t)g(t),
0
Fund. Theory ODE Lecture Notes – J. Arino
150
B. Problem sheets
where the second equality uses the definition of g(t). We have
φ0 (t) = K 0 (t)g(t) + K(t)g 0 (t)
= K 0 (t)g(t) + K(t)a(t)g(t).
The function φ is solution to (E2) if and only if it satisfies (E2); therefore, φ is solution if
and only if
φ0 (t) = a(t)φ(t) + b(t) ⇔ K 0 (t)g(t) + K(t)a(t)g(t). = a(t)K(t)g(t) + b(t)
⇔ K 0 (t)g(t) = b(t)
b(t)
, for g(t) 6= 0
⇔ K 0 (t) =
g(t)
Z t
b(s)
⇔ K(t) =
ds + C.
0 g(s)
Note that the remark that g(t) 6= 0 is made “for form”: as it is defined, g(t) > 0 for all
t ≥ 0. We conclude that the general solution to (E2) is given by
t
Z
x(t) =
0
Z t
b(s)
a(s)ds .
ds + C exp
g(s)
0
Since it will be useful to have information in terms of x(0) (as in question 1.), we note that
C = x(0). Thus, the solution to (E2) through x(0) = 0 is given by
Z
x(t) =
0
t
Z t
b(s)
a(s)ds .
ds + x(0) exp
g(s)
0
With integrating factors, we would have done as follows: write the equation (E2) as
x0 (t) − a(t)x(t) = b(t).
The integrating factor is then
Z
µ(t) = exp − a(t)dt ,
and the general solution to (E2) is given by
1
x(t) =
µ(t)
Z
t
µ(s)b(s)ds + C
0
Maximal solutions are solutions that are the restriction of no other solution.
(B.36)
Fund. Theory ODE Lecture Notes – J. Arino
151
2.b. Solutions to (E2) are T -periodic if x(T ) = x(0); therefore, a T -periodic solution
satisfies
Z T
Z T
b(s)
x(T ) = x(0) ⇔
ds + x(0) exp
a(s)ds = x(0)
0 g(s)
0
Z T
b(s)
ds + x(0) = x(0)e−A
⇔
g(s)
0
Z T
b(s)
⇔
ds = e−A − 1 x(0)
0 g(s)
Second part
3. Before proceeding, note that
d
exp
g (t) =
dt
0
t
Z
a(s)ds
0
t
Z
a(s)ds
= a(t) exp
0
= a(t)g(t).
Rt
We differentiate x(t) = g(t) x(0) + 0 g(s)−1 H(x(s), s)ds . This gives
Z
t
H(x(t), t)
x (t) = g (t) x(0) +
g(s) H(x(s), s)ds + g(t)
g(t)
0
Z t
= g 0 (t) x(0) +
g(s)−1 H(x(s), s)ds + H(x(t), t)
0
Z t
−1
= a(t)g(t) x(0) +
g(s) H(x(s), s)ds + H(x(t), t)
0
0
−1
0
= a(t)x(t) + H(x(t), t),
and thus x(t) = g(t) x(0) +
Rt
0
−1
g(s) H(x(s), s)ds is solution to (E3).
4. Let x ∈ P . Then UH x ∈ P if and only if UH x is T -periodic. We have
eA
g(t + T )
(UH x)(t + T ) =
1 − eA
Z
t+2T
t+T
g(s)−1 H(x(s), s)ds.
Fund. Theory ODE Lecture Notes – J. Arino
152
B. Problem sheets
Remark that
t+T
Z
g(t + T ) = exp
a(s)ds
0
t
Z
Z
t+T
= exp
a(s)ds +
a(s)ds
t
Z t+T
= g(t) exp
a(s)ds
0
t
A
= e g(t),
since a(t) is T -periodic. Therefore,
Z t+2T
eA A
e g(t)
g(s)−1 H(x(s), s)ds
(UH x)(t + T ) =
1 − eA
t+T
Z t+T
A
e
eA g(t)
g(s − T )−1 H(x(s − T ), s − T )ds.
=
A
1−e
t
Now
Z
s−T
g(s − T ) = exp
a(u)du
0
Z
s
Z
s−T
a(u)du +
a(u)du
s
Z s
= g(s) exp −
a(u)du
= exp
0
s−T
= e−A g(s).
So, finally,
eA 2A
e g(t)
(UH x)(t + T ) =
1 − eA
= e2A (UH x)(t),
Z
t+T
g(s)−1 H(x(s), s)ds
t
since H is T -periodic in its second argument and x ∈ P . Therefore, UH x ∈ P for x ∈ P .
Suppose that x(t) = (UH x)(t). Then,
Z t+T
H(x(s), s)
H(x(t + T ), t + T ) H(x(t), t)
eA
0
0
g (t)
ds + g(t)
−
x (t) =
1 − eA
g(s)
g(t + T )
g(t)
t
Z t+T
A
e
H(x(s), s)
H(x(t), t) H(x(t), t)
a(t)g(t)
ds + g(t)
−
=
1 − eA
g(s)
eA g(t)
g(t)
t
Z t+T
A
A
A
e
H(x(s), s)
e
(1 − e )H(x(t), t)
= a(t)
g(t)g(t)
ds +
g(t)
A
A
1−e
g(s)
1−e
eA g(t)
t
= a(t)x(t) + H(x(t), t).
Fund. Theory ODE Lecture Notes – J. Arino
153
5.a. We seek ε0 > 0 such that for all ε ≤ ε0 , kxk ≤ r ⇒ kUε xk ≤ r. Therefore, we
compute kUε xk. We have, letting H(x, s) = εF (x, s),
kUε xk = sup |(Uε x)(t)|
t∈R
A
Z t+T
e
−1
= sup g(t)
g(s) εF (x(s), s)ds
A
t∈R 1 − e
t
Z t+T
A
F
(x(s),
s)
e
sup
|g(t)|
ds
=ε
|1 − eA | t∈R
g(s)
t
Z
t+T eA
F (x(s), s) ds.
≤ε
sup
|g(t)|
A
|1 − e | t∈R
g(s) t
Note that we keep the absolute value of |1 − eA |, since A could be negative, leading to a
negative value for 1 − eA . Let kg −1 k = supt∈R |g −1 (t)|. We then have
Z t+T
eA
−1
kgkkg k sup
|F (x(s), s)| ds
kUε xk ≤ ε
|1 − eA |
t∈R t
Z t+T
eA
−1
≤ε
kgkkg k
αr ds,
|1 − eA |
t
since x(s) ∈ [−r, r], and so
kUε xk ≤ ε
eA
kgkkg −1 kαr T.
|1 − eA |
Letting
ε0 =
eA
kgkkg −1 kαr T
|1 − eA |
−1
r,
we see that if ε ≤ ε0 , then kUε xk ≤ r.
5.b. For the restriction of Uε to be a contraction, we must have the inequality obtained
above, as well as, for x, y ∈ Br , d(Uε x, Uε y) < d(x, y). In terms of the induced norm, this
means that kUε x − Uε yk < kx − yk. Therefore, letting x, y ∈ P be such that kxk ≤ r and
Fund. Theory ODE Lecture Notes – J. Arino
154
B. Problem sheets
kyk ≤ r, we compute
kUε x − Uε k = sup |(Uε x)(t) − (Uε y)(t)|
t∈R
A
Z t+T
e
εF (x(s), s) − εF (y(s), s) = sup g(t)
ds
A
g(s)
t∈R 1 − e
t
A Z t+T
e
F
(x(s),
s)
−
F
(y(s),
s)
sup |g(t)| = ε ds
1 − eA t∈R
g(s)
t
A Z t+T e
F (x(s), s) − F (y(s), s) sup |g(t)|
ds.
≤ ε A
1 − e t∈R
g(s)
t
For s ∈ [0, T ] and x(s) ∈ [−r, r], we have, picking a y(s) ∈ [−r, r],
|F (x(s), s)| = |F (x(s), s) − F (y(s), s) + F (y(s), s)|
≤ |F (x(s), s) − F (y(s), s)| + |F (y(s), s)|
≤ βr |x(s) − y(s)| + αr ,
from the mean value theorem, and thus
|F (x(s), s)| ≤ 2βr r + αr .
5.c. We used the contraction mapping principle (Theorem ??):
• for ε ≤ ε1 , Uε is a contraction of Br ,
• P is complete (it is closed in C(R, R) ∩ B(R) endowed with the supremum norm) and
Br is closed in P .
Therefore, we conclude that for a given r > 0, for all ε ≤ ε1 , there exists a unique solution
xε of (E4) in Br .
6. This is the contraction mapping theorem with a parameter:
kxε − xε0 k = kUε xε − Uε0 xε0 k ≤ kUε xε − Uε xε0 k +kUε xε0 − Uε0 xε0 k.
|
{z
}
≤Kkxε −xε0 k
Fund. Theory ODE Lecture Notes – J. Arino
155
But we have
Z t+T
eA
kUε xε − Uε0 xε0 k(t) ≤ |ε − ε |
|g(t)g(s)−1 F (x(s), s)|ds
A
|1 − e | t
eA
≤ |ε − ε0 |
eT A T αr .
|1 − eA |
|
{z
}
0
=K 0
|ε − ε0 |K 0
and therefore ε ∈ R 7→ xε ∈ P is continuous; it follows
1−K
that lim xε = x0 . But the only periodic solution of (E1) when A 6= 0 is the zero solution.
ε→0
Therefore, xε → 0 when ε → 0.
Thus, we have kxε − xε0 k ≤
7. Let x0 (t) = c0 , then g(t) = ekt and A = kT .
t+T
Z t+T
εekT
εekT kt
1 −ks
kt
−ks
Uε x0 (t) =
e f (t0 )
f (c0 )e − e
e ds =
1 − ekT
1 − ekT
k
t
t
kt
kT
kT
e
f
(c
)
εe
εe
ε
0
−kt
−kt −kT
−kT
f
(c
)
e
−
e
e
1
−
e
f (c0 )
=
=
=
−
0
1 − ekT
k
1 − ekT k
k
The constant function xε (t) = c0 is solution (where c0 is the unique solution of the equation
εf (x) + kx = 0 for ε sufficiently small).
ε
ε
Note : letting g(x) = − f (x), it follows that g 0 (x) = − f 0 (x). Thus, for r > 0 given,
k
k
there exists ε0 > 0 such that ε ≤ ε0 implies g([−r, r]) ⊂ [−r, r], and there exists ε1 ≤ ε0 such
that sup |g 0 (x)| < 1, the fixed point theorem can be applied easily.
x∈[−r,r]
8.a. Using the formula obtained in 5.a. with αr = r2 , βr = 2r, g(t) = e−t , A = −T then
ε0 =
1 − e−T
r(1 − e−T )eT e−T
=
,
r2 T
rT
ε1 =
1 1 − e−T
ε0
= .
2 2rT
4
8.b. The zero function is clearly a 1-periodic solution of (E5). By uniqueness of solutions,
xε = 0 is the only solution of (E5).
Fund. Theory ODE Lecture Notes – J. Arino
156
B. Problem sheets
8.c. The vector field −x+εx2 is C 1 , and therefore existence and uniqueness of a maximal
solution ϕα is a direct consequence of the theorem of Cauchy-Lipschitz.
We solve the equation x0 = −x + εx2 without constraint of periodicity. There are two
constant solutions , x(t) = 0 and x(t) = 1/ε. By uniqueness, any other solution never takes
the values 0 and 1/ε. We have:
x0
d x =
ln
= −1
x(1 − εx)
dt
1 − εx
x
= λe−t
1 − εx
⇒
⇒
x(t) =
λe−t
,
1 + ελe−t
λ ∈ R∗ .
α
The condition x(0) = α gives λ =
if α =
6 0 and α 6= 1/ε, and as a consequence,
1 − εα
1
letting β = − ε we obtain
α
1
ϕα (t) = t
.
βe + ε
• If β ≥ 0 then ϕα is defined on R.
• If β < 0, we let t0 = ln − βε .
ε
> 1, that is, t0 > 0, ϕα is defined on ] − ∞, t0 [.
β
ε
– If α < 0 then − < 1, that is, t0 < 0, ϕα is defined on ]t0 , +∞[.
β
– If α > 0 then −
Here are a few representative solutions obtained when setting ε = 1.
3
2
alpha>1/epsilon
y
1
0<alpha<1/epsilon
–3
–2
–1
0
1
2
x
–1
alpha<0
–2
–3
Third part
3
Fund. Theory ODE Lecture Notes – J. Arino
157
9. θ > 0 since |x(0)| < 1, and we have x(s) ∈ [−1, 1] for all s ∈ [0, θ], by definition of
θ. Since f (0) = 0 and |f 0 | is bounded by λ on [−1, 1], we have, from the inequality of finite
variations, |f (u) − f (0) | ≤ λ|u| for all u ∈ [−1, 1], that is, |f (x(s)) − f (0) | ≤ λ|x(s)|.
|{z}
|{z}
=0
=0
Let t ∈ [0, θ]. From 4.,
−kt
|e
Z
x(t)| = x(0) + ε
t
−ks
e
s=0
Z
f (x(s)) ds ≤ |x(0)| + ελ
t
e−ks |x(s)| ds,
s=0
from which |e−kt x(t)| ≤ |x(0)|eελt (using Gronwall’s lemma with ϕ(t) = e−kt |x(t)|, η = |x(0)|,
ζ = ελ), giving the inequality.
10. Since ελ < 0, it follows that |x(0)|e(k+ελ)t ≤ |x(0)| < 1. Letting E = {t > 0 | |x(t)| >
1}, which is assumed non empty, then θ = inf E > 0 (by continuity, since |x(0)| < 1 there
exists η > 0 such that |x(t)| < 1 on [0, η], and thus θ ≥ η > 0). Since lim− |x(t)| ≤ 1 and
t→θ
lim+ |x(t)| ≥ 1, it follows |x(θ)| = 1.
t→θ
On [0, θ], |x(t)| ≤ |x(0)|e(k+ελ)t and, taking the limit, |x(θ)| < 1, which is impossible.
First conclusion : E = ∅ and, if J is the interval of definition of x, then ∀t ∈ J ∩ [0, +∞[,
|x(t)| ≤ 1.
If J admits an upper bound b ∈ R, then x0 is bounded in a neighborhood of b. Thus x
admits a limit in b. The same is therefore true for x0 . We then know that x can be extended
beyond b, contradicting the maximality of J.
Final conclusion : J ∩[0, +∞[= [0, +∞[, x is defined on [0, +∞[ and the proof of question
9. holds true for all t ≥ 0, i.e.,
∀t ∈ [0, +∞[, |x(t)| ≤ |x(0)|e(k+ελ)t .
N.B. This result expresses the stability and the asymptotic stability of the trivial solution
of (E6).
This subject was the Première composition de mathématiques for the contest determining
admission to École Polytechnique in France, for MP (Math-Physics) track students, in 2004.
Students, in their second year of university, have 4 hours to write this première composition.
The original subject comprised another question, originally question 3, which was suppressed in this homework sheet. To be complete, this question is included here:
Fund. Theory ODE Lecture Notes – J. Arino
158
B. Problem sheets
2’. We suppose here that T = 2π and that the function a is a constant k.
2’.a. Assuming k 6= 0, express the Fourier coefficients x̂(n), n ∈ Z, of a solution x of
(E2) belonging to P , as a function of k and the Fourier coefficients of b. What is the mode
of convergence of the Fourier series of x?
2’.b. What happens when k = 0?
2’.a. If k 6= 0 then A = 2πk 6= 0, and from 2., there exists a unique 2π-periodic solution.
Since the mapping x 7→ x
b(n) is linear, and from the relation xb0 (n) = inb
x(n), we have
bb(n)
xb0 (n) = kb
x(n) + bb(n) ⇒ x
b(n) =
.
in − k
Since x is C 1 , we know that the
Fourier series of x is normally convergent. Since bb(n) → 0,
1
.
we ca also say that x
b(n) = o
n
2’.b. Applying here again the result of 2.b.,
• If bb(0) = 0 then all solutions are 2π-periodic. In this case, solutions satisfy x̂(n) =
b̂(n)/in for n ∈ Z non zero and x̂(0) varies with the solutions under consideration.
• If bb(0) 6= 0 then no solution is periodic.