ordinary differential equations: fundamental theory...

MATH 8430

Fundamental Theoryof

Ordinary Differential Equations

Lecture Notes

Julien ArinoDepartment of Mathematics

University of Manitoba

Fall 2006

Contents

1 General theory of ODEs 31.1 ODEs, IVPs, solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Ordinary differential equation, initial value problem . . . . . . . . . . 31.1.2 Solutions to an ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.3 Geometric interpretation . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2 Existence and uniqueness theorems . . . . . . . . . . . . . . . . . . . . . . . 91.2.1 Successive approximations . . . . . . . . . . . . . . . . . . . . . . . . 91.2.2 Local existence and uniqueness – Proof by fixed point . . . . . . . . . 101.2.3 Local existence and uniqueness – Proof by successive approximations 131.2.4 Local existence (non Lipschitz case) . . . . . . . . . . . . . . . . . . . 161.2.5 Some examples of existence and uniqueness . . . . . . . . . . . . . . 21

1.3 Continuation of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.3.1 Maximal interval of existence . . . . . . . . . . . . . . . . . . . . . . 271.3.2 Maximal and global solutions . . . . . . . . . . . . . . . . . . . . . . 28

1.4 Continuous dependence on initial data, on parameters . . . . . . . . . . . . . 291.5 Generality of first order systems . . . . . . . . . . . . . . . . . . . . . . . . . 321.6 Generality of autonomous systems . . . . . . . . . . . . . . . . . . . . . . . . 341.7 Suggested reading, Further problems . . . . . . . . . . . . . . . . . . . . . . 34

2 Linear systems 352.1 Existence and uniqueness of solutions . . . . . . . . . . . . . . . . . . . . . . 352.2 Linear systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.2.1 The vector space of solutions . . . . . . . . . . . . . . . . . . . . . . 372.2.2 Fundamental matrix solution . . . . . . . . . . . . . . . . . . . . . . 382.2.3 Resolvent matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.2.4 Wronskian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.2.5 Autonomous linear systems . . . . . . . . . . . . . . . . . . . . . . . 43

2.3 Affine systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.3.1 The space of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 462.3.2 Construction of solutions . . . . . . . . . . . . . . . . . . . . . . . . . 462.3.3 Affine systems with constant coefficients . . . . . . . . . . . . . . . . 47

2.4 Systems with periodic coefficients . . . . . . . . . . . . . . . . . . . . . . . . 482.4.1 Linear systems: Floquet theory . . . . . . . . . . . . . . . . . . . . . 48

i

iiFund. Theory ODE Lecture Notes – J. Arino

CONTENTS

2.4.2 Affine systems: the Fredholm alternative . . . . . . . . . . . . . . . . 51

2.5 Further developments, bibliographical notes . . . . . . . . . . . . . . . . . . 54

2.5.1 A variation of constants formula for a nonlinear system with a linearcomponent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3 Stability of linear systems 57

3.1 Stability at fixed points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3.2 Affine systems with small coefficients . . . . . . . . . . . . . . . . . . . . . . 57

4 Linearization 63

4.1 Some linear stability theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.2 The stable manifold theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.3 The Hartman-Grobman theorem . . . . . . . . . . . . . . . . . . . . . . . . . 69

4.4 Example of application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

4.4.1 A chemostat model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

4.4.2 A second example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

5 Exponential dichotomy 79

5.1 Exponential dichotomy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

5.2 Existence of exponential dichotomy . . . . . . . . . . . . . . . . . . . . . . . 81

5.3 First approximate theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

5.4 Stability of exponential dichotomy . . . . . . . . . . . . . . . . . . . . . . . . 84

5.5 Generality of exponential dichotomy . . . . . . . . . . . . . . . . . . . . . . 84

References 85

A Definitions and results 87

A.1 Vector spaces, norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

A.1.1 Norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

A.1.2 Matrix norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

A.1.3 Supremum (or operator) norm . . . . . . . . . . . . . . . . . . . . . . 87

A.2 An inequality involving norms and integrals . . . . . . . . . . . . . . . . . . 88

A.3 Types of convergences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

A.4 Asymptotic Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

A.5 Types of continuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

A.6 Lipschitz function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

A.7 Gronwall’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

A.8 Fixed point theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

A.9 Jordan normal form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

A.10 Matrix exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

A.11 Matrix logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

A.12 Spectral theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

B Problem sheets 101Homework sheet 1 – 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103Homework sheet 2 – 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113Homework sheet 3 – 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117Homework sheet 4 – 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125Final examination – 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137Homework sheet 1 – 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

iii

Introduction

This course deals with the elementary theory of ordinary differential equations. The wordelementary should not be understood as simple. The underlying assumption here is that, tounderstand the more advanced topics in the analysis of nonlinear systems, it is importantto 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 nonlinearordinary differential equations. You know, for example, that in order for solutions to asystem to exist and be unique, the system must have a C1 vector field. What you do notnecessarily know is why that is. This is the object of Chapter 1, where we consider thegeneral theory of existence and uniqueness of solutions. We also consider the continuationof 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 theoryfor nonlinear systems by means of linearization, which we study in Chapter 4, in whichwe show that the behavior of nonlinear systems can be approximated, in the vicinity ofa hyperbolic equilibrium point, by a homogeneous linear system. As for autonomous sys-tems, nonautonomous nonlinear systems are linked to a linearized form, this time throughexponential 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 defineODEs, initial value problems (IVPs) and solutions to ODEs and IVPs in Section 1.1. InSection 1.2, we discuss existence and uniqueness of solutions to IVPs.

1.1 ODEs, IVPs, solutions

1.1.1 Ordinary differential equation, initial value problem

Definition 1.1.1 (ODE). An nth order ordinary differential equation (ODE) is a functionalrelationship taking the form

F

(t, x(t),

d

dtx(t),

d2

dt2x(t), . . . ,

dn

dtnx(t)

)= 0,

that involves an independent variable t ∈ I ⊂ R, an unknown function x(t) ∈ D ⊂ Rn ofthe independent variable, its derivative and derivatives of order up to n. For simplicity, thetime dependence of x is often omitted, and we in general write equations as

F(t, x, x′, x′′, . . . , 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 ingeneral (or implicit) form.

An equation is said to be in normal (or explicit) form when it is written as

x(n) = f(t, x, x′, x′′, . . . , x(n−1)

).

Note that it is not always possible to write a differential equation in normal form, as it canbe 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 morerestrictive case of a first-order ordinary differential equation in normal form

x′ = f(t, x). (1.2)

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4Fund. 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

x′ = f(t, x)

x(t0) = x0,(1.3)

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 isneeded. However, this leads in general to much more complicated problems and is beyond thescope of this course. Hence, unless otherwise stated, we assume that f is at least continuous. Thefunction 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, x′, . . . , x(n−1))

x(t0) = x0, x′(t0) = x′0, . . . , x

(n−1)(t0) = x(n−1)0 ,

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 derivativeinvolved in the equation. An equation is then classified as a function of its linearity. A linearequation 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 isautonomous ; otherwise, it is nonautonomous. Thus, a linear equation is autonomous ifa(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, theprevious 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

φ′(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 aspossible to use the notation x for the unknown and φ for the solution.

1.1. ODEs, IVPs, solutionsFund. Theory ODE Lecture Notes – J. Arino

5

Definition 1.1.5 (Integral form of the solution). The function

φ(t) = x0 +

∫ t

t0

f(s, φ(s))ds (1.4)

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, ‖x− x0‖ ≤ b} ,

where ‖ ‖ is any appropriate norm of Rn. This domain is illustrated in Figures 1.1 and1.2; it is sometimes called a security system, i.e., the union of a security interval (for theindependent variable) and a security domain (for the dependent variables) [19]. Suppose

0

t −a t

x

0 t0 t +a0

x +b0

0x

x −b0

(t ,x )0

Figure 1.1: The domain R for D ⊂ R.

t

x

y

t0

x

y0

0

Figure 1.2: The domain R for D ⊂ R2: “security tube”.



that f is continuous on R, and let M = maxR ‖f(t, x)‖, which exists since f is continuouson the compact set R.

In the following, existence of solutions will be obtained generally in relation to the domainR by considering a subset of the time interval |t− t0| ≤ a defined by |t− t0| ≤ α, with

α =

{a if M = 0

min(a, bM

) 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) asthe solution of x′ = f(t, x), we must be working in R. So we require that |t − t0| ≤ a and‖x−x0‖ ≤ 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 fromthe 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],

‖φ(t)− x0‖ =

∥∥∥∥∫ t

t0

f(s, φ(s))ds

∥∥∥∥≤∫ t

t0

‖f(s, φ(s))‖ ds

≤M

∫ t

t0

ds

= M(t− t0),

where the first inequality is a consequence of the definition of the integrals by Riemannsums (Lemma A.2.1 in Appendix A.2). Similarly, we have ‖φ(t)− x0‖ ≤ −M(t− t0) for allt ∈ [t0 − α, t0]. Thus, for |t− t0| ≤ α, ‖φ(t)− x0‖ ≤M |t− t0|. Suppose now that α ≤ b/M .It follows that ‖φ − x0‖ ≤ M |t − t0| ≤ Mb/M = b. Taking α = min(a, b/M) then ensuresthat both |t− t0| ≤ a and ‖φ− x0‖ ≤ b hold simultaneously.

The following two theorems deal with the localization of the solutions to an IVP. Theymake more precise the previous discussion. Note that for the moment, the existence ofa solution is only assumed. First, we establish that the security system described aboveperforms properly, in the sense that a solution on a smaller time interval stays within thesecurity domain.

Theorem 1.1.6. If φ(t) is a solution of the IVP (1.3) in an interval |t− t0| < α ≤ α, then‖φ(t)− x0‖ < b in |t− t0| < α, i.e., (t, φ(t)) ∈ R((t0, x0), α, b) for |t− t0| < α.

Proof. Assume that φ is a solution with (t, φ(t)) 6∈ R((t0, x0), α, b). Since φ is continuous, itfollows that there exists 0 < β < α such that(‖φ(t)− x0‖ < b for |t− t0| < β

)and

(‖φ(t0 + β)− x0‖ = b or ‖φ(t0− β)− x0‖ = b

), (1.5)

i.e., the solution escapes the security domain at t = t0 ± β. Since α ≤ α ≤ a, β < a. Thus

(t, φ(t)) ∈ R for |t− t0| ≤ β.

1.1. ODEs, IVPs, solutionsFund. Theory ODE Lecture Notes – J. Arino

7

Thus ‖f(t, φ(t))‖ ≤M for |t− t0| ≤ β. Since φ is a solution, we have that φ′(t) = f(t, φ(t))and φ(t0) = x0. Thus

φ(t) = x0 +

∫ t

t0

f(s, φ(s))ds for |t− t0| ≤ β.

Hence

‖φ(t)− x0‖ =

∥∥∥∥∫ t

t0

f(s, φ(s))ds

∥∥∥∥ for |t− t0| ≤ β

≤M |t− t0| for |t− t0| ≤ β.

As a consequence,

‖φ(t)− x0‖ ≤Mβ < Mα ≤Mα ≤Mb

M= b for |t− t0| ≤ β.

In particular, ‖φ(t0 ± β)− x0‖ < b. Hence contradiction with (1.5).

The following theorem is proved using the same sort of technique as in the proof ofTheorem 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| < α ≤ α, then‖φ(t1)− φ(t2)‖ ≤M |t1 − t2| whenever t1, t2 are in the interval |t− t0| < α.

Proof. Let us begin by considering t ≥ t0. On t0 ≤ t ≤ t0 + α,

φ(t1)− φ(t2) = x0 +

∫ t1

t0

f(s, φ(s))ds− x0 −∫ t2

t0

f(s, φ(s))ds

= −∫ t2

t1

f(s, φ(s))ds if t2 > t1∫ t2

t1

f(s, φ(s))ds if t1 > t2

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 ofthe IVP (1.3) if, and only if,

i) ∀t ∈ I, (t, φ(t)) ∈ U .

ii) φ is continuous on I.

iii) ∀t ∈ I, φ(t) = x0 +∫ t

t0f(s, φ(s))ds.



Proof. (⇒) Let us suppose that φ′ = f(t, φ) for all t ∈ I and that φ(t0) = x0. Then for allt ∈ I, (t, φ(t)) ∈ U (i). Also, φ is differentiable and thus continuous on I (ii). Finally,

φ′(s) = f(s, φ(s))

so integrating both sides from t0 to t,

φ(t)− φ(t0) =

∫ t

t0

f(s, φ(s))ds

and thus

φ(t) = x0 +

∫ t

t0

f(s, φ(s))ds

hence (iii).(⇐) Assume i), ii) and iii). Then φ is differentiable on I and φ′(t) = f(t, φ(t)) for all t ∈ I.From (3), φ(t0) = x0 +

∫ t0t0f(s, φ(s))ds = x0.

Note that Theorem 1.1.8 states that φ should be continuous, whereas the solution shouldof course be C1, for its derivative needs to be continuous. However, this is implied by pointiii). 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 thatf ∈ Ck. Then all solutions of (1.2) are of class Ck+1.

Proof. The proof is obvious, since a solution φ is such that φ′ ∈ Ck.

1.1.3 Geometric interpretation

The function f is the vector field of the equation. At every point in (t, x) space, a solutionφ is tangent to the value of the vector field at that point. A particular consequence of thisfact is the following theorem.

Theorem 1.1.10. Let x′ = f(x) be a scalar autonomous differential equation. Then thesolutions of this equation are monotone.

Proof. The direction field of an autonomous scalar differential equation consists of vectorsthat are parallel for all t (since f(t, x) = f(x) for all t). Suppose that a solution φ ofx′ = f(x) is non monotone. Then this means that, given an initial point (t0, x0), one thefollowing 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.

1.2. Existence and uniqueness theoremsFund. Theory ODE Lecture Notes – J. Arino

9

1t t t t t t0 12 0 2

Figure 1.3: Situations that would lead to a scalar autonomous differential equation havingnonmonotone solutions.

Suppose we are in case i), and assume we are in the case f(x0) > 0. Thus, the solution curveφ is increasing at (t0, x0), i.e., φ′(t0) > 0. As φ is continuous, i) implies that there existst2 ∈ (t0, t1) such that φ(t2) is a maximum, with φ increasing for t ∈ [t0, t2) and φ decreasingfor t ∈ (t2, t1]. It follows that φ′(t1) < 0, which is a contradiction with φ′(t0) > 0.

Now assume that we are in case ii). Then there exists t2 ∈ (t0, t1) with φ(t2) = x0 butsuch that φ′(t2) < 0. This is a contradiction.

Remark – If we have uniqueness of solutions, it follows from this theorem that if φ1 and φ2 aretwo solutions of the scalar autonomous differential equation x′ = f(x), then φ1(t0) < φ2(t0) impliesthat 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 Sec-tions 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 solutionsunder the assumption that the vector field is Lipschitz. In Section 1.2.4, the Lipschitzassumption is dropped and therefore a different approach must be used, namely that ofapproximate 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 thesolution to the IVP (1.3) to construct a sequence of approximation of the solution, thatconverges to the solution. The steps followed in constructing this approximating sequenceare the following.



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:

φ1(t) = x0 +

∫ t

t0

f(s, φ0(s))ds.

Step 3. Use φ1 in (1.4) to define the third element in the sequence:

φ2(t) = x0 +

∫ t

t0

f(s, φ1(s))ds.

. . .Step n. Use φn−1 in (1.4) to define the nth element in the sequence:

φn(t) = x0 +

∫ t

t0

f(s, φn−1(s))ds.

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 satisfythe 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) studiesexplicitly 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 thevector field is continuous and Lipschitz, then the solutions exist and are unique. We provethe result in the second case. For the definition of a Lipschitz function, see Section A.6 inthe 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 intervalcontaining t0 in its interior.

Theorem 1.2.2 (Picard local existence and uniqueness). Consider the IVP (1.3), and as-sume f is (piecewise) continuous in t and satisfies the Lipschitz condition

‖f(t, x1)− f(t, x2)‖ ≤ L‖x1 − x2‖

for all x1, x2 ∈ D = {x : ‖x − x0‖ ≤ b} and all t such that |t − t0| ≤ a. Then there exists0 < δ ≤ α = min

(a, b

M

)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

F : x 7→ x0 +

∫ t

t0

f(s, x(s))ds.


11

Note that the function (Fφ)(t) is a continuous function of t. Then Picard’s successivesapproximations take the form φ1 = Fφ0, φ2 = Fφ1 = F 2φ0, where F 2 represents F ◦ F .Iterating, the general term is given for k = 0, . . . by

φk = F kφ0.

Therefore, finding the limit limk→∞ φk is equivalent to finding the function φ, solution of thefixed point problem

x = Fx,

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 ≤ δis 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,

‖x‖c = maxt∈[t0,t0+δ]

‖x(t)‖

Recall that this norm is the norm of uniform convergence. Let then

S = {x ∈ X : ‖x− x0‖c ≤ b}

Of course, S ⊂ X. Furthermore, S is closed, and X with the sup norm is a complete metricspace. Note that we have transformed the problem into a problem involving the space ofcontinuous functions; hence we are now in an infinite dimensional case. The proof proceedsin 3 steps.Step 1. We begin by showing that F : S → S. From (1.4),

(Fφ)(t)− x0 =

∫ t

t0

f(s, φ(s))ds

=

∫ t

t0

f(s, φ(s))− f(s, x0) + f(s, x0)ds

Therefore, by the triangle inequality,

‖Fφ− x0‖ ≤∫ t

t0

‖f(s, φ(s))− f(t, x0)‖+ ‖f(t, x0)‖ds

As f is (piecewise) continuous, it is bounded on [t0, t1] and there existsM = maxt∈[t0,t1] ‖f(t, x0)‖.Thus

‖Fφ− x0‖ ≤∫ t

t0

‖f(s, φ(s))− f(t, x0)‖+Mds

≤∫ t

t0

L‖φ(s)− x0‖+Mds,



since f is Lipschitz. Since φ ∈ S for all ‖φ− x0‖ ≤ b, we have that for all φ ∈ S,

‖Fφ− x0‖ ≤∫ t

t0

Lb+Mds

≤ (t− t0)(Lb+M)

As t ∈ [t0, t0 + δ], (t− t0) ≤ δ, and thus

‖Fφ− x0‖c = max[t0,t0+δ]

‖Fφ− x0‖ ≤ (Lb+M)δ

Choose then δ such that δ ≤ b/(Lb+M), i.e., t sufficiently close to t0. Then we have

‖Fφ− x0‖c ≤ b

This implies that for φ ∈ S, Fφ ∈ S, i.e., F : S → S.Step 2. We now show that F is a contraction. Let φ1, φ2 ∈ S,

‖(Fφ1)(t)− (Fφ2)(t)‖ =

∥∥∥∥∫ t

t0

f(s, φ1(s))− f(s, φ2(s))ds

∥∥∥∥≤∫ t

t0

‖f(s, φ1(s))− f(s, φ2(s))‖ds

≤∫ t

t0

L‖φ1(s)− φ2(s)‖ds

≤ L‖φ1 − φ2‖c

∫ t

t0

ds

and thus‖Fφ1 − Fφ2‖c ≤ Lδ‖φ1 − φ2‖c ≤ ρ‖φ1 − φ2‖c for δ ≤ ρ

L

Thus, choosing ρ < 1 and δ ≤ ρ/L, F is a contraction. Since, by Step 1, F : S → S, thecontraction mapping principle (Theorem A.11) implies that F has a unique fixed point inS, 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 thatwe want to show the result). Considering a solution starting at x0 at time t0, the solutionleaves S if there exists a t > t0 such that ‖φ(t)−x0‖ = b, i.e., the solution crosses the borderof D. Let τ > t0 be the first of such t’s. For all t0 ≤ t ≤ τ ,

‖φ(t)− x0‖ ≤∫ t

t0

‖f(s, φ(s))− f(s, x0)‖+ ‖f(s, x0)‖ds

≤∫ t

t0

L‖φ(s)− x0‖+Mds

≤∫ t

t0

Lb+Mds


13

As a consequence,b = ‖φ(τ)− x0‖ ≤ (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 : ‖x− x0‖ ≤ b} in the statement of the theoremrefers to a local Lipschitz condition. If the function f is Lipschitz, then the following theoremholds.

Theorem 1.2.3 (Global existence). Suppose that f is piecewise continuous in t and isLipschitz on U = I × D. Then (1.3) admits a unique solution on I.

1.2.3 Local existence and uniqueness – Proof by successive ap-proximations

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, ‖x− x0‖ ≤ b}, and that f is locally Lipschitz in xon R. Let then, as previously defined,

M = sup(t,x)∈R

‖f(t, x)‖ <∞

and

α = min(a,b

M)

Then the sequence defined by

φ0 = x0, |t− t0| ≤ α

φi(t) = x0 +

∫ t

t0

f(s, φi−1(s))ds, i ≥ 1, |t− t0| ≤ α

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

‖φ1 − φ0‖ =

∥∥∥∥∫ t

t0

f(s, φ0(s))ds

∥∥∥∥≤M |t− t0|≤ αM ≤ b



from the definitions of M and α, and thus ‖φ1 − φ0‖ ≤ b. So∫ t

t0f(s, φ1(s))ds is defined for

|t− t0| ≤ α, and, for |t− t0| ≤ α,

‖φ2(t)− φ0‖ =

∥∥∥∥∫ t

t0

f(s, φ1(s))ds

∥∥∥∥ ≤ ‖∫ t

t0

‖f(s, φ1(s))‖ds ≤ αM ≤ b.

All subsequent terms in the sequence can be similarly defined, and, by induction, for |t−t0| ≤α,

‖φk(t)− φ0‖ ≤ αM ≤ b, k = 1, . . . , n.

Now, for |t− t0| ≤ α,

‖φk+1(t)− φk(t)‖ =

∥∥∥∥x0 +

∫ t

t0

f(s, φk(s))ds− x0 −∫ t

t0

f(s, φk−1(s))ds

∥∥∥∥=

∥∥∥∥∫ t

t0

f(s, φk(s))− f(s, φk−1(s)) ds

∥∥∥∥≤ L

∫ t

t0

‖φk(s)− φk−1(s)‖ds,

where the inequality results of the fact that f is locally Lipschitz in x on R.We now prove that, for all k,

‖φk+1 − φk‖ ≤ b(L|t− t0|)k

k!for |t− t0| ≤ α (1.6)

Indeed, (1.6) holds for k = 1, as previously established. Assume that (1.6) holds for k = n.Then

‖φn+2 − φn+1‖ =

∥∥∥∥∫ t

t0

f(s, φn+1(s))− f(s, φn(s))ds

∥∥∥∥≤∫ t

t0

L‖φn+1(s)− φn(s)‖ds

≤∫ t

t0

Lb(L|s− t0|)n

n!ds for |t− t0| ≤ α

≤ bLn+1

n!

|t− t0|n+1

n+ 1

∣∣∣∣s=t

s=t0

≤ b(L|t− t0|)n+1

(n+ 1)!

and thus (1.6) holds for k = 1, . . ..Thus, for N > n we have

‖φN(t)− φn(t)‖ ≤N−1∑k=n

‖φk+1(t)− φk(t)‖ ≤N−1∑k=n

b(L|t− t0|)k

k!≤ b

N−1∑k=n

(Lα)k

k!


15

The rightmost term in this expression tends to zero as n→∞. Therefore, {φk(t)} convergesuniformly to a function φ(t) on the interval |t− t0| ≤ α. As the convergence is uniform, thelimit function is continuous. Moreover φ(t0) = x0. Indeed, φN(t) = φ0(t) +

∑Nk=1(φk(t) −

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

limn→∞

∫ t

t0

φn(s)ds =

∫ t

t0

limn→∞

φn(s)ds

Hence,

φ(t) = limn→∞

φn(t)

= x0 + limn→∞

∫ t

t0

f(s, φn−1(s))ds

= x0 +

∫ t

t0

limn→∞

f(s, φn−1(s))ds

= x0 +

∫ t

t0

f(s, φ(s))ds,

which is to say that

φ(t) = x0 +

∫ t

t0

f(s, φ(s))ds for |t− t0| ≤ α.

As the integrand f(t, φ) is a continuous function, φ is differentiable (with respect to t), andφ′(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,

φ(t) = x0 +

∫ t

t0

f(s, φ(s))ds

ψ(t) = x0 +

∫ t

t0

f(s, ψ(s))ds.

Then, for |t− t0| ≤ α,

‖φ(t)− ψ(t)‖ =

∥∥∥∥∫ t

t0

f(s, φ(s))− f(s, ψ(s))ds

∥∥∥∥≤ L

∫ t

t0

‖φ(s)− ψ(s)‖ds. (1.7)

We now apply Gronwall’s Lemma A.7) to this inequality, using K = 0 and g(t) = ‖φ(t) −ψ(t)‖. First, applying the lemma for t0 ≤ t ≤ t0 + α, we get 0 ≤ ‖φ(t)− ψ(t)‖ ≤ 0, that is,

‖φ(t)− ψ(t)‖ = 0,



and thus φ(t) = ψ(t) for t0 ≤ t ≤ t0 + α. Similarly, for t0 − α ≤ t ≤ t0, ‖φ(t) − ψ(t)‖ = 0.Therefore, φ(t) = ψ(t) on |t− t0| ≤ α.

Example – Let us consider the IVP x′ = −x, x(0) = x0 = c, c ∈ R. For initial solution, we chooseφ0(t) = c. Then

φ1(t) = x0 +∫ t

0f(s, φ0(s))ds

= c +∫ t

0−φ0(s)ds

= c− c

∫ t

0ds

= c− ct.

To find φ2, we use φ1 in (1.4).

φ2(t) = x0 +∫ t

0f(s, φ1(s))ds

= c−∫ t

0(c− cs)ds

= c− ct + ct2

2.

Continuing this method, we find a general term of the form

φn(t) =n∑

i=0

c(−1)iti

i!.

This is the power series expansion of ce−t, so φn → φ = ce−t (and the approximation is valid onR), which is the solution of the initial value problem. �

Note that the method of successive approximations is a very general method that can beused 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 isnot 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, ‖x− x0‖ ≤ b},

with a, b > 0, and let M = maxR ‖f(t, x)‖. Then there exists a continuous function φ(t),differentiable on R, such that

i) φ(t0) = x0,


17

ii) φ′(t) = f(t, φ) on |t− t0| ≤ α, where

α =

{a if M = 0

min(a, b

M

)if M > 0.

Before we can prove this result, we need a certain number of preliminary notations andresults. The definition of equicontinuity and a statement of the Ascoli lemma are given inSection A.5. To construct a solution without the Lipschitz condition, we approximate thedifferential equation by another one that does satisfy the Lipschitz condition. The uniquesolution of such an approximate problem is an ε-approximate solution. It is formally definedas follows [8, p. 285].

Definition 1.2.6 (ε-approximate solution). A differentiable mapping u of an open ball J ∈ Iinto U is an approximate solution of x′ = f(t, x) with approximation ε (or an ε-approximatesolution) if we have

‖u′(t)− f(t, u(t))‖ ≤ 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, ‖x− x0‖ ≤ 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) ‖Fε(t, x)‖ ≤ maxR ‖f(t, x)‖ = M for |t− t0| ≤ a and all x,

iv) ‖Fε(t, x)− f(t, x)‖ ≤ ε on R.

See a proof in [12, p. 10-12]; note that in this proof, the property that f defines adifferential equation is not used. Hence Lemma 1.2.7 can be used in a more general contextthan 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 re-

quirements 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) φ′ε(t) = Fε(t, φε(t)) on |t− t0| < α.

(P3) (t, φε(t)) ∈ R on |t− t0| ≤ α.



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 ‖Fε(t, φε(t))‖ ≤ Mon |t− t0| ≤ α. Hence property (P2) of φε implies that

‖φε(t1)− φε(t2)‖ ≤M |t1 − t2|,

if |t1 − t0| ≤ α and |t2 − t0| ≤ α (this is a consequence of Theorem 1.1.7). Therefore, for agiven positive number µ, we have ‖φε(t1)− φε(t2)‖ ≤ 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 thatlimk→∞ εk = 0 and that the sequence {φεk

: k = 1, 2, . . .} converges uniformly on |t− t0| ≤ αas k →∞. Then set

φ(t) = limk→∞

φεk(t) on |t− t0| ≤ α.

4. Observe that

φε(t) = x0 +

∫ t

t0

Fε(s, φε(s))ds

= x0 +

∫ t

t0

f(s, φε(s))ds+

∫ t

t0

Fε(s, φε(s))− f(s, φε(s))ds,

and that it follows from iv) in Lemma 1.2.7 that∥∥∥∥∫ t

t0

Fε(s, φε(s))− f(s, φε(s))ds

∥∥∥∥ ≤ ε|t− t0| on |t− t0| ≤ α.

This is true for all ε ≥ 0, so letting ε→ 0, we obtain

φ(t) = x0 +

∫ t

t0

f(s, φ(s))ds,

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 C1 n-dimensional vector-valued function on [t0 − δ, t0]satisfying φ`(t0) = x0, φ

′`(t0) = f(t0, x0) and ‖φ`(t)− x0‖ ≤ b, ‖φ′`(t)‖ ≤ 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

φε(t) = x0 +

∫ t

t0

f(s, φε(s− ε))ds on [t0, t0 + α]. (1.8)

The function φε can indeed be thus defined on [t0 − δ, t0 + α]. To see this, remark first thatthis formula is meaningful and defines φε(t) for t0 ≤ t ≤ t0 + α1, α1 = min(α, ε), so thatφε(t) is C1 on [t0 − δ, t0 + α1] and, on this interval,

‖φε(t)− x0‖ ≤ b, ‖φε(t)− φε(s)‖ ≤M |t− s|. (1.9)


19

It then follows that (1.8) can be used to extend φε(t) as a C1 function over [t0 − δ, t0 + α2],where α2 = min(α, 2ε), satisfying relation (1.9). Continuing in this fashion, (1.8) serves todefine φε(t) over [t0, t0 +α] so that φε(t) is a C0 function on [t0−δ, t0 +α], satisfying relation(1.9).

Since ‖φ′ε(t)‖ ≤ M , M can be used as a Lipschitz constant for φε, giving uniform con-tinuity 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) > . . ., suchthat ε(n) → 0 as n→∞ and

φ(t) = limn→∞

φε(n)(t) exists uniformly

on [t0 − δ, t0 + α]. The continuity of f implies that f(t, φε(n)(t − ε(n)) tends uniformly tof(t, φ(t)) as n→∞; thus term-by-term integration of (1.8) where ε = ε(n) gives

φ(t) = x0 +

∫ t

t0

f(s, φ(s))ds

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 ‖f(t, x)‖ ≤ M .Let E0 be a compact subset of E. Then there exists an α = α(E,E0,M) > 0 with theproperty 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 ε-approximate solution as

Definition 1.2.9. An ε-approximate solution of the differential equation (1.2), where f iscontinuous, on a t interval I is a function φ ∈ C on I such that

i) (t, φ(t)) ∈ U for t ∈ I;

ii) φ ∈ C1 on I, except possibly for a finite set of points S on I, where φ′ may have simplediscontinuities (g has finite discontinuities at c if the left and right limits of g at c existbut are not equal);

iii) ‖φ′(t)− f(t, φ(t))‖ ≤ ε for t ∈ I \ S.

Hence it is assumed that φ has a piecewise continuous derivative on I, which is denotedby φ ∈ C1

p(I).

Theorem 1.2.10. Let f ∈ C on the rectangle

R = {(t, x) : |t− t0| ≤ a, ‖x− x0‖ ≤ b}.

Given any ε > 0, there exists an ε-approximate solution φ of (1.3) on |t− t0| ≤ α such thatφ(t0) = x0.



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 weconstruct 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 ofε, there exists δε > 0 such that

‖f(t, φ)− f(t, φ)‖ ≤ ε (1.10)

if(t, φ) ∈ R, (t, φ) ∈ R and |t− t| ≤ δε ‖φ− φ‖ ≤ δε.

Now divide the interval [t0, t0 + α] into n parts t0 < t1 < · · · < tn = t0 + α, in such a waythat

max |tk − tk−1| ≤ min

(δε,

δεM

). (1.11)

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 thetriangular region T bounded by the lines segments with slopes ±M from (t0, x0) to theirintercept 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 thatmeets 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 ∈ C1p([t0, t0 + α]) and

‖φ(t)− φ(t)‖ ≤M |t− t| for t, t ∈ [t0, t0 + α]. (1.13)

If t ∈ [tk−1, tk], then (1.13) together with (1.11) imply that ‖φ(t)− φ(tk−1)‖ ≤ δε. But from(1.12) and (1.10),

‖φ′(t)− f(t, φ(t))‖ = ‖f(tk−1, φ(tk−1))− f(t, φ(t))‖ ≤ 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 → 0as 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

‖φn(t)− φn(t)‖ ≤M |t− t|. (1.14)


21

Applying (1.14) to t = t0, it is clear that the sequence {φn} is uniformly bounded by‖x0‖ + b, since |t − t0| ≤ b/M . Moreover, (1.14) implies that {φn} is an equicontinuousset. 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 sinceeach φn is continuous.

This limit function φ is a solution to (1.3) which meets the required specifications. Tosee this, write

φn(t) = x0 +

∫ t

t0

f(s, φn(s)) + ∆n(s)ds, (1.15)

where ∆n(t) = φ′(t)− f(t, φn(t)) at those points where φ′n exists, and ∆n(t) = 0 otherwise.Because φn is an εn-approximate solution, ‖∆n(t)‖ ≤ εn. Since f is uniformly continuous onR, 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

φ(t) = x0 +

∫ t

t0

f(s, φ(s))ds. (1.16)

Clearly, φ(t0) = 0, when evaluated using (1.16), and also φ′(t) = f(t, φ(t)) since f is contin-uous. 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 IVPx′ = 3|x|

23

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 notapply, since f(t, x) is not locally Lipschitz in x = 0 (or, f is not Lipschitz on any interval containing0). 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 followingargument. Suppose that f is Lipschitz on an interval I = (−ε, ε), with ε > 0. Then, there existsL > 0 such that for all x1, x2 ∈ I,

‖f(t, x1)− f(t, x2)‖ ≤ L|x1 − x2|

that is,3∣∣∣|x1|

23 − |x2|

23

∣∣∣ ≤ L|x1 − x2|

Since this has to hold true for all x1, x2 ∈ I, it must hold true in particular for x2 = 0. Thus

3|x1|23 ≤ L|x1|

Given an ε > 0, it is possible to find Nε > 0 such that 1n < ε for all n ≥ Nε. Let x1 = 1

n . Then forn ≥ Nε, if f is Lipschitz there must hold

3(

1n

) 23

≤ L

n



So, for all n ≥ Nε,

n13 ≤ L

3This is a contradiction, since limn→∞ n1/3 = ∞, and so f is not Lipschitz on I.

Let us consider the setE = {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 usnow 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, weconsider here the case of monotone increasing solutions. The case of monotone decreasing solutionscan be treated in a similar fashion.

1) Here, there is no intersection with the x = 0 axis. Thus if follows that

x(t) is{

> 0, if x0 > 0< 0, if x0 < 0

2) In this case,

x(t) is

< 0, if t < a= 0, if t ∈ [a, b]> 0, if t > b

3) Here,

x(t) is{

= 0, if t < b> 0, if t > b

4) In this case,

x(t) is{

< 0, if t < a= 0, if t > a

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 andso

x′ = 3x2/3

⇔ 13x−2/3x′ = 1

⇔ x1/3 = t + k1

⇔ x(t) = (t + k1)3


23

for k1 ∈ R. Then, if x < 0, then |x| = −x, and

x′ = 3 (−x)2/3

⇔ 13(−x)−2/3(−x′) = −1

⇔ (−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) The case E = ∅ is impossible, for all initial conditions (t0, x0). Indeed, as x0 > 0, we havex(t) = (t + k1)3. Using the initial condition, we find that x(t0) = x0 = (t0 + k1)3, i.e.,k1 = x

1/30 − t0, and x(t) = (t + x

1/30 − t0)3.

2) If E = [a, b], then

x(t) =

−(−t + k2)3 if t < a0 if t ∈ [a, b](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 impliesthat k1 = x

1/30 − t0. Thus

x(t) =

−(−t + k2)3 if t < a0 if t ∈ [a, b](t + x

1/30 − t0)3 if t > b

Since x is continuous,lim

t→b,t>b(t + x

1/30 − t0)3 = 0

andlim

t→a,t<a−(−t + k2)3 = 0

This implies that b = t0 − x1/30 and k2 = a. So finally,

x(t) =

−(−t + a)3 if t < a

0 if t ∈ [a, t0 − x130 ] (a ≤ t0 − x

130 )

(t + x1/30 − t0)3 if t > t0 − x

130

Thus, choosing a ≤ t0−x1/30 , we have solutions of the form shown in Figure 1.4. Indeed, any

ai satisfying this property yields a solution.

3) The case [a,+∞) is impossible. Indeed, there does not exist a solution through (t0, x0) suchthat 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.

4) E = R is also impossible, for the same reason.



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

Since x(t0) = x0, k1 = x1/30 − t0, and since x is continuous, b = −k1 = t0 − x

1/30 . So,

x(t) =

{0 if t ∈ (−∞, t0 − x

1/30 ]

(t + x1/30 − t0)3 if t > t0 − x

1/30

The other cases are left as an exercise. �

Example – Consider the IVPx′ = 2tx2

x(0) = 0(1.18)

Here, we have existence and uniqueness of the solutions to (1.18). Indeed, f(t, x) = 2tx2 is contin-uous 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 toan 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 onpossibly larger intervals.

Consider the IVPx′ = f(t, x)

x(t0) = x0,(1.19)

1.3. Continuation of solutionsFund. Theory ODE Lecture Notes – J. Arino

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, andassume that a function φ(t) satisfies the condition φ′(t) = f(t, φ(t)) and (t, φ(t)) ∈ U , in anopen interval I = {t1 < t < t2}. Under this assumption, if limj→∞(τj, φ(τj)) = (t1, η) ∈ Ufor 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, . . .} ofpoints 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 , W ⊂ U ,and that |f(t, x)| ≤ M in W for some positive number M . For every positive integer j andevery positive number ε, consider a rectangular region

Rj(ε) = {(t, x) : |t− tj| ≤ ε, ‖x− φ(tj)‖ ≤Mε}

Then there exists an ε > 0 and a j such that (τ1, η) ∈ Rj(ε) ⊂ W, with ε = min(ε, Mε

M

)and

tj − ε ≤ τ1.From Theorem 1.1.6 applied to the solution φ of the IVP x′ = 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 openneighborhood 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 overwhich a solution can be extended. To emphasize the fact that the solution φ of an ODEexists 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 (φ, I) be two solutions of the same ODE. Wesay that (φ, I) is an extension of (φ, I) if, and only if,

I ⊂ I, φ|I = φ

where |I 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 functionφ(t) be a function satisfying the condition φ′(t) = f(t, φ(t)) and (t, φ(t)) ∈ U , in an openinterval 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 .



φ~

φ~

I

I~

φ

Figure 1.5: The extension φ on the interval I of the solution φ (defined on the intervalI).

Proof. Suppose that the hypotheses of the theorem are satisfied, and that (t1, η) ∈ U (re-spectively, (t2, η) ∈ U). Then, from Lemma 1.3.1, it follows that

limτ→t1

(τ, φ(τ)) = (t1, η)

(or, respectively, limτ→t2(τ, φ(τ)) = (t2, η)). Thus we can apply Theorem 1.2.5 (Peano’sTheorem) to the IVP

x′ = f(t, x)

x(t1) = η,

(or, respectively, x′ = f(t, x), x(t2) = η). This implies that the solution φ can be extendedto the left of t1 (respectively, to the right of t2), since Theorem 1.2.5 implies existence in aneighborhood 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 φ′ are continuous in a subinterval I of the interval t1 < t < t2,

b) φ′(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 differentialequation x′ = f(t, x), or

ii) limt→τ ‖φ(t)‖ = ∞ for some τ in the interval t1 < t < t2.

1.3. Continuation of solutionsFund. Theory ODE Lecture Notes – J. Arino

27

1.3.1 Maximal interval of existence

Another way of formulating these results is with the notion of maximal intervals of existence.Consider the differential equation

x′ = 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 maximalinterval 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 rightendpoints). 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 aright 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 ofexistence (ω−, ω+). Also, if (ω−, ω+) is a maximal interval of existence, then φ(t) tends tothe 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, forexample, that φ → ∂U as t → ω+ is interpreted to mean that either ω+ = ∞ or that ω+ < ∞ andif U0 is any compact subset of U , then (t, φ(t)) 6∈ U0 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 ‖φ(t)‖ → ∞ as t→ δ.

Corollary 1.3.9. Let f(t, x) be continuous on the closure U 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 ‖φ(t)‖ → ∞ as t→ δ.



1.3.2 Maximal and global solutions

Linked to the notion of maximal intervals of existence of solutions is the notion of maximaland global solutions.

Definition 1.3.10 (Maximal solution). Let I1 ⊂ R and I2 ⊂ R be two intervals such thatI1 ⊂ I2. A solution (φ, I1) is maximal in I2 if φ has no extension (φ, I) solution of theODE such that I1 ( I ⊂ I2.

Definition 1.3.11 (Global solution). A solution (φ, I1) is global on I2 if φ admits a exten-sion φ defined on the whole interval I2.

2

I

U

φ1

φ

Figure 1.6: φ1 is a global and maximal solution on I; φ2 is a maximal solution on I, butit is not global on I.

Every global solution on a given interval I is maximal on that same interval. The converseis false.

Example – Consider the equation x′ = −2tx2 on R. If x 6= 0, x′x−2 = −2t, which implies thatx(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 aremaximal 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 x′ = f(t, x) is contained in amaximal solution φ.

1.4. Continuous dependence on initial data, on parametersFund. Theory ODE Lecture Notes – J. Arino

29

The following theorem extends the uniqueness property to an interval of existence of thesolution.

Theorem 1.3.13. Let φ1, φ2 : I → Rn be two solutions of the equation x′ = f(t, x), with flocally 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 at1, 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 existsa 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) fort > 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 byany point (t0, x0) ∈ U , there passes a unique maximal solution φ : I → Rn. If there exists aglobal solution on I, then it is unique.

1.4 Continuous dependence on initial data, on param-

eters

Let φ be a solution of (1.3). To emphasize the fact the this solution depends on the initialcondition (t0, x0), we denote it φt0,x0 . Let η be a parameter of (1.3). When we study thedependence of φt0,x0 on η, we denote the solution as φt0,x0,η.

We suppose that ‖f(t, x)‖ ≤ 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 boundedregion of the form R = {(t, x) : |t− t0| ≤ a, ‖x− x0‖ ≤ b}, where a, b > 0.

Our objective here is to characterize the nature of the dependence of the solution on theinitial time t0 and the initial data x0.

Theorem 1.4.1. Suppose that f and ∂f/∂x are continuous and bounded in a given regionU . Let φt0,x0 be a solution of (1.3) passing through (t0, x0) and ψt0,x0

be a solution of (1.3)

passing through (t0, x0). Suppose that φ and ψ exist on some interval I.Then, for each ε > 0, there exists δ > 0 such that if |t− t| < δ and ‖x0 − x0‖ < δ, then

‖φ(t)− ψ(t)‖ < ε, for t, t ∈ 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,

φ(t) = x0 +

∫ t

t0

f(s, φ(s))ds (1.21)



As ψ is the solution of (1.3) through the point (t0, x0), we have, for all t ∈ I,

ψ(t) = x0 +

∫ t

t0

f(s, ψ(s))ds (1.22)

Since ∫ t

t0

f(s, φ(s))ds =

∫ t0

t0

f(s, φ(s))ds+

∫ t

t0

f(s, φ(s))ds,

substracting (1.22) from (1.21) gives

φ(t)− ψ(t) = x0 − x0 +

∫ t0

t0

f(s, φ(s))ds+

∫ t

t0


and therefore

‖φ(t)− ψ(t)‖ ≤ ‖x0 − x0‖+

∥∥∥∥∥∫ t0

t0

f(s, φ(s))ds

∥∥∥∥∥+

∥∥∥∥∫ t

t0


∥∥∥∥Using the boundedness assumptions on f and ∂f/∂x to evaluate the right hand side of thelatter inequation, we obtain

‖φ(t)− ψ(t)‖ ≤ ‖x0 − x0‖+M |t0 − t0|+K

∥∥∥∥∫ t

t0

φ(s)− ψ(s)ds

∥∥∥∥If |t0 − t0| < δ, ‖x0 − x0‖ < δ, then we have

‖φ(t)− ψ(t)‖ ≤ δ +Mδ +K

∥∥∥∥∫ t

t0

φ(s)− ψ(s)ds

∥∥∥∥ (1.23)

Applying Gronwall’s inequality (Appendix A.7) to (1.23) gives

‖φ(t)− ψ(t)‖ ≤ δ(1 +M)eK|t−t0| ≤ δ(1 +M)eK(τ2−τ1)

using the fact that |t− t0| < τ2 − τ1, if we denote I = (τ1, τ2). Since

‖ψ(t)− ψ(t)‖ <∥∥∥∥∫ t

t

f(s, ψ(s))ds

∥∥∥∥ ≤M |t− t| ≤Mδ

if |t− t| < δ, we have

‖φ(t)− ψ(t)‖ ≤ ‖φ(t)− ψ(t)‖+ ‖ψ(t)− ψ(t)‖ ≤ 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 desiredinequality, completing the proof.

1.4. Continuous dependence on initial data, on parametersFund. Theory ODE Lecture Notes – J. Arino

31

What we have shown is that the solution passing through the point (t0, x0) is a continuousfunction 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 Theo-rem 1.4.1. Let φ and ψ be solutions of x′ = f(t, x) and x′ = g(t, x), respectively, suchthat φ(t0) = x0, ψ(t0) = x0, existing on a common interval α < t < β. Suppose that‖f(t, x)− g(t, x)‖ ≤ ε for (t, x) ∈ U . Then the solutions φ and ψ satisfy

‖φ(t)− ψ(t)‖ ≤ ‖x0 − x0‖eK|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 previousone, 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 functionon Uµ, bounded by a constant M there. For µ = µ0, let

x′ = 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 > 0such 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 theregion R = {|t − t0| ≤ α, ‖x − x0‖ ≤ Mα} is in U ; note that R is a slight modificationof 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 anuniformly bounded and equicontinuous set in |t − t0| ≤ α. This follows from the integralequation

φµ(t) = x0 +

∫ t

t0

f(s, φµ(s), µ)ds (|t− t0| ≤ α) (1.25)

and the inequality ‖f‖ ≤M .Suppose that for some t ∈ [t0 − α, t0 + α], φµ(t) does not tend to φ0(t). Then there

exists a sequence {µk}, k = 1, 2, . . ., for which µk → µ0, and corresponding solutions φµk

such that φµkconverges uniformly over [t0−α, t0 +α] as k →∞ to a limit function ψ, with

ψ(t) 6= φ0(t). 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 φµkyields

ψ(t) = x0 +

∫ t

t0

f(s, ψ(s), µ0)ds (|t− t0| ≤ α)



Thus ψ is a solution of (1.24) with µ = µ0. By the uniqueness hypothesis, it follows thatψ(t) = φ0(t) on |t − t0| ≤ α. Thus ψ(t) = φ0(t). Thus all solutions φµ on |t − t0| ≤ α tendto 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 smallε > 0, there exists a δε > 0 such that

‖φµ(τ − ε)− φ0(τ − ε)‖ < ε (1.26)

for |µ− µ0| < δε. Let H ⊂ U be defined as the region

H = {|t− τ | ≤ γ, ‖x− φ0(τ − γ)‖ ≤ γ +M |t− τ + γ|}

with γ small enough that H ⊂ U . Any solution of x′ = f(t, x, µ) starting on t = τ − γ withinitial value ξ0, |ξ0 − φ0(τ − γ)| ≤ γ will remain in H as t increases. Thus all solutions canbe continued to τ + γ.

By choosing ε = γ in (1.26), it follows that for |µ − µ0| < δε, the solutions φµ can allbe continued to τ + ε. Thus over [t0, τ + ε] these solutions are in U so that the argumentthat φµ → φ0 which has been given for |t − t0| ≤ α, also applies over [t0, τ + ε]. Thus theassumption about the existence of τ < b is false. The case τ = b is treated in similar fashionon 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, areunique, 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, x′, . . . , x(n−1)

)(1.27)

This equation can be reduced to a system of n first order ordinary differential equations, byproceeding as follows. Let y0 = x, y1 = x′, y2 = x′′, . . . , yn−1 = x(n). Then (1.27) is equivalentto

y′ = F (t, y) (1.28)

with y = (y0, y1, . . . , yn−1)T and

F (t, z) =

y1

y2...

yn−1

f(t, y0, . . . , yn−1)

1.5. Generality of first order systemsFund. Theory ODE Lecture Notes – J. Arino

33

Similarly, the IVP associated to (1.27) is given by

x(n) = f(t, x, x′, . . . , x(n−1)

)x(t0) = x0, x

′(t0) = x1, . . . , x(n−1)(t0) = xn−1

(1.29)

is equivalent to the IVPy′ = 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

x′′ = −2x′ + 4x− 3x(0) = 2, x′(0) = 1

To transform it into a system of first-order differential equations, we let y = x′. Substituting (wherepossible) y for x′ in the equation gives

y′ = −2y + 4x− 3

The initial condition becomes x(0) = 2, y(0) = 1. So finally, the following IVP is equivalent to theoriginal one:

x′ = y

y′ = 4x− 2y − 3x(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)x′(t) + a0(t)x(t) + b(t)

is an nth order nonhomogeneous linear differential equation. Together with the initial condition

x(n−1)(t0) = x(n−1)0 , . . . , x′(t0) = x′0, x(t0) = x0

where x0, x′0, . . . , x

(n−1)0 ∈ R, it forms an IVP. We can transform it to a system of linear first order

equations by setting

y0 = x

y1 = x′

...

yn−1 = x(n−1)

yn = x(n)



The nth order linear equation is then equivalent to the following system of n first order linearequations

y′0 = y1

y′1 = y2

...y′n−2 = yn−1

y′n−1 = yn

y′n = 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

yn−1(t0) = x(n−1)0 , . . . , y1(t0) = x′0, y0(t0) = x0

�

1.6 Generality of autonomous systems

A nonautonomous systemx′(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

x′ = f(y, x)

y′ = 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] (firstedition published in 1955), and the current text, as many others, does little but paraphrasethem.

We have not seen here any results specific to complex valued differential equations. Ascomplex numbers are two-dimensional real vectors, the results carry through to the complexcase by simply assuming that if, in (1.2), we consider an n-dimensional complex vector, thenthis is equivalent to a 2n-dimensional problem. Furthermore, if f(t, x) is analytic in t andx, 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 = Ror C), and L(E) the space of continuous linear maps from E to E. Let ‖ ‖ be a norm onE, and ||| ||| be the induced supremum norm on L(E) (see Appendix A.1). Consider a mapA : I → L(E) and a map B : I → E. A linear system of first order equations is defined by

x′(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 sub-interval of I. We restrict ourselves to the finite dimensional case (E = Kn). Hence weconsider A ∈Mn(K), n× n matrices over the field K, and B ∈ Kn. We suppose that A andB 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 calledan affine system, with associated linear system

x′(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 nonhomo-geneous linear system and the latter as an homogeneous linear system. In order to lightenthe language, since there will be other qualificatives added to both (2.1) and (2.2), we usein 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 weconsider the generic properties of (2.1), with (2.2) as a particular case, as in this chapter’stitle 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


2. Linear systems

Proof. Let k(t) = |||A(t)||| = sup‖x‖≤1 ‖A(t)x‖. Then for all t ∈ I and all x1, x2 ∈ K,

‖f(t, x1)− f(t, x2)‖ = ‖A(t)(x1 − x2)‖≤ |||A(t)||| ‖x1 − x2‖≤ k(t)‖x1 − x2‖,

where the inequality‖A(t)(x1 − x2)‖ ≤ |||A(t)||| ‖x1 − x2‖

results from the nature of the norm ||| ||| (see Appendix A.1). Furthermore, k is continuouson I. Therefore the conditions of Theorem 1.2.2 hold, leading to existence and uniquenesson the interval I.

With linear systems, it is possible to extend solutions easily, as is shown by the nexttheorem.

Theorem 2.1.2. Suppose that the entries of A(t) and the entries of B(t) are continuous onan open interval I. Then every solution of (2.1) which is defined on a subinterval J of theinterval 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

‖φ(t)‖ ≤ ‖φ(t0)‖+

∥∥∥∥∫ t

t0

A(s)φ(s) +B(s)ds

∥∥∥∥for all t ∈ J , where t0 ∈ J . Let{

K = ‖φ(t0)‖+ (τ2 − τ1) maxτ1≤t≤τ2 ‖B(t)‖L = maxτ1≤t≤τ2 ‖A(t)‖

Then, for t0, t ∈ J ,

‖φ(t)‖ ≤ K + L

∥∥∥∥∫ t

t0

φ(s)ds

∥∥∥∥ ≤ K + L

∫ t

t0

‖φ(s)‖ds.

Thus, using Gronwall’s Lemma (Lemma A.7), the following estimate holds in J ,

‖φ(t)‖ ≤ 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 tobe 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 ho-mogeneous systems of the form (2.2) (linear systems), with x ∈ Rn and A ∈Mn(R), the setof square matrices over the field R, A having continuous entries on an interval I.

2.2. Linear systemsFund. Theory ODE Lecture Notes – J. Arino

37

2.2.1 The vector space of solutions

Theorem 2.2.1 (Superposition principle). Let S0 be the set of solutions of (2.2) that aredefined on some interval I ⊂ R. Let φ1, φ2 ∈ S0, and λ1, λ2 ∈ R. Then λ1φ1 + λ2φ2 ∈ S0.

Proof. Let φ1, φ2 ∈ S0 be two solutions of (2.2), λ1, λ2 ∈ R. Then for all t ∈ I,

φ′1 = A(t)φ1

φ′2 = A(t)φ2,

from which it comes that

d

dt(λ1φ1 + λ2φ2) = A(t)[λ1φ1 + λ2φ2],

implying that λ1φ1 + λ2φ2 ∈ S0.

Thus the linear combination of any two solutions of (2.2) is in S0. This is a hint that S0

must be a vector space of dimension n on K. To show this, we need to find a basis of S0.We proceed in the classical manner, with the notable difference from classical linear algebrathat 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 differ-ential equation (2.2), all defined on the same open interval I, is called a fundamental set ofsolutions 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, . . . , n, is linearly dependent. Then there exists αi ∈ R, i =1, . . . , n, not all zero, such that

∑ni=1 αiφi(t) = 0 for all t. In particular, this is true for

t = t0, and thus∑n

i=1 αiφi(t0) =∑n

i=1 αiei = 0, which implies that the canonical basis ofKn 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 theopen interval I, then every solution defined on I can be expressed as a linear combinationof the elements of F .

Let t0 ∈ I, we consider the application

Φt0 : S0 → Kn

Y 7→ Φt0(x) = x(t0)

Lemma 2.2.5. Φt0 is a linear isomorphism.


2. Linear systems

Proof. Φt0 is bijective. Indeed, let v ∈ Kn, from Theorem 2.1.1, there exists a unique solutionpassing through (t0, v), i.e.,

∀v ∈ Kn, ∃!x ∈ S0, x(t0) = v ⇒ Φt0(x) = v,

so Φt0 is surjective. That Φt0 is injective follows from uniqueness of solutions to an ODE.Furthermore, Φt0(λ1x1+λ2x2) = λ1Φt0(x1)+λ2Φt0(x2). Therefore dimS0 = 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, iscalled 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 columnsform a fundamental set of solutions. If in addition Φ(t0) = I, a fundamental matrix solutionis called the principal fundamental matrix solution.

An important property of fundamental matrix solutions is the following, known as Abel’sformula.

Theorem 2.2.7 (Abel’s formula). Let A(t) be continuous on I and Φ ∈ Mn(K) be suchthat Φ′(t) = A(t)Φ(t) on I. Then det Φ satisfies on I the differential equation

(det Φ)′ = (trA)(det Φ),

or, in integral form, for t, τ ∈ I,

det Φ(t) = det Φ(τ) exp

(∫ t

τ

trA(s)ds

). (2.3)

Proof. Writing the differential equation Φ′(t) = A(t)Φ(t) in terms of the elements ϕij andaij of, respectively, Φ and A,

ϕ′ij(t) =n∑

k=1

aik(t)ϕkj(t), (2.4)

for i, j = 1, . . . , n. Writing

det Φ =

∣∣∣∣∣∣∣∣ϕ11(t) ϕ12(t) . . . ϕ1n(t)ϕ21(t) ϕ22(t) . . . ϕ2n(t)

ϕn1(t) ϕn2(t) . . . ϕnn(t)

∣∣∣∣∣∣∣∣ ,we see that

(det Φ)′ =

∣∣∣∣∣∣∣∣ϕ′11 ϕ′12 . . . ϕ′1n

ϕ21 ϕ22 . . . ϕ2n

ϕn1 ϕn2 . . . ϕnn

∣∣∣∣∣∣∣∣+∣∣∣∣∣∣∣∣ϕ11 ϕ12 . . . ϕ1n

ϕ′21 ϕ′22 . . . ϕ′2n


∣∣∣∣∣∣∣∣+ · · ·+

∣∣∣∣∣∣∣∣ϕ11 ϕ12 . . . ϕ1n

ϕ21 ϕ22 . . . ϕ2n

ϕ′n1 ϕ′n2 . . . ϕ′nn

∣∣∣∣∣∣∣∣ .


39

Indeed, write det Φ(t) = Γ(r1, r2, . . . , rn), where ri is the ith row in Φ(t). Γ is then a linearfunction of each of its arguments, if all other rows are constant, which implies that

d

dtdet Φ(t) = Γ

(d

dtr1, r2, . . . , rn

)+ Γ

(r1,

d

dtr2, . . . , rn

)+ · · ·+ Γ

(r1, r2, . . . ,

d

dtrn

).

(To show this, use the definition of the derivative as a limit.) Using (2.4) on the first of then determinants in (det Φ)′ gives∣∣∣∣∣∣∣∣

∑k a1kϕk1

∑k a1kϕk2 . . .

∑k a1kϕkn

ϕ21 ϕ22 . . . ϕ2n


∣∣∣∣∣∣∣∣ .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∣∣∣∣∣∣∣∣

∑k a1kϕk1

∑k a1kϕk2 . . .

∑k a1kϕkn

ϕ21 ϕ22 . . . ϕ2n


∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣a11ϕ11 a11ϕ12 . . . a11ϕ1n

ϕ21 ϕ22 . . . ϕ2n


∣∣∣∣∣∣∣∣ = a11 det Φ.

Repeating this for each of the terms in (det Φ)′, we obtain (det Φ)′ = (a11 + a22 + · · · +ann) det Φ, giving finally (det Φ)′ = (trA)(det Φ). Note that this equation takes the formu′ − α(t)u = 0, which implies that

u exp

(∫ t

τ

α(s)ds

)= constant,

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= 0for any a, it follows that det Φ 6= 0 for all t ∈ I. In short, linear independence of solutions for at ∈ I is equivalent to linear independence of solutions for all t ∈ I. As a consequence, the columnvectors 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 anynontrivial solution of (2.2). Then there exists c1, . . . , cn, not all zero, such that

φ =n∑

j=1

cjφj,

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


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= 0for all t ∈ I.

Reciproqually, let Φ be a solution matrix of (2.2), and suppose that det Φ(t) 6= 0 fort ∈ 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 isequivalent to the condition “there exists t ∈ I such that det Φ(t) 6= 0”. A frequent candidatefor this role is t0.

To conclude on fundamental solution matrices, remark that there are infinitely manyof them, for a given linear system. However, since each fundamental solution matrix canprovide a basis for the vector space of solutions, it is clear that the fundamental matricesassociated 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 aconstant nonsingular matrix. Then ΦC is a fundamental matrix solution to (2.2). Con-versely, if Ψ is another fundamental matrix solution to (2.2), then there exists a constantnonsingular 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)′ = Φ′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, detC 6= 0. Thus,det ΦC = det Φ detC 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,taking the derivative of this expression gives Φ′Φ−1 + Φ (Φ−1)

′= 0, and therefore (Φ−1)

′=

−Φ−1Φ′Φ−1. We now consider the product Φ−1Ψ. There holds(Φ−1Ψ

)′=(Φ−1

)′Ψ + Φ−1Ψ′

= −Φ−1Φ′Φ−1Ψ + Φ−1A(t)Ψ

=(−Φ−1A(t)ΦΦ−1 + Φ−1A(t)

)Ψ

=(−Φ−1A(t) + Φ−1A(t)

)Ψ

= 0.

Therefore, integrating (Φ−1Ψ)′

gives Φ−1Ψ = C, with C ∈ Mn(K) is a constant. Thus,Ψ = CΦ. Furthermore, as Φ and Ψ are fundamental matrix solutions, det Φ 6= 0 anddet Ψ 6= 0, and therefore detC 6= 0.

Remark – Note that if Φ is a fundamental matrix solution to (2.2) and C ∈Mn(K) is a constantnonsingular matrix, then it is not necessarily true that CΦ is a fundamental matrix solution to(2.2). See Exercise 2.3. ◦


41

2.2.3 Resolvent matrix

If t 7→ Φ(t) is a matrix solution of (2.2) on the interval I, then Φ′(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 solutionof (2.2) on I. Since the columns of Φ are linearly independent, it follows that Φ(t0) isinvertible. 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 (sinceR(t0, t0) = Φ(t0)Φ(t0)

−1 = I). Thus system (2.2) has a principal fundamental matrixsolution 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) ∂∂sR(t, s) = −R(t, s)A(s),

5) ∂∂tR(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,

R(t, s)−1 =(Φ(t)Φ−1(s)

)−1=(Φ−1(s)

)−1Φ(t)−1 = Φ(s)Φ−1(t) = R(s, t),

whence 3). Also,

∂

∂sR(t, s) =

∂

∂s

(Φ(t)Φ−1(s)

)= Φ(t)

(∂

∂sΦ−1(s)

)As Φ is a fundamental matrix solution, Φ′ exists and Φ is nonsingular, and differentiatingΦΦ−1 = I gives

∂

∂s

(Φ(s)Φ−1(s)

)= 0 ⇔

(∂

∂sΦ(s)

)Φ−1(s) + Φ(s)

(∂

∂sΦ−1(s)

)= 0

⇔ Φ(s)

(∂

∂sΦ−1(s)

)= −

(∂

∂sΦ(s)

)Φ−1(s)

⇔ ∂

∂sΦ−1(s) = −Φ−1(s)

(∂

∂sΦ(s)

)Φ−1(s).


2. Linear systems

Therefore,

∂

∂sR(t, s) = −Φ(t)Φ−1(s)

(∂

∂sΦ(s)

)Φ−1(s) = −R(t, s)

(∂

∂sΦ(s)

)Φ−1(s).

Now, since Φ(s) is a fundamental matrix solution, it follows that ∂Φ(s)/∂s = A(s)Φ(s), andthus

∂

∂sR(t, s) = −R(t, s)A(s)Φ(s)Φ−1(s) = −R(t, s)A(s),

giving 4). Finally,

∂

∂tR(t, s) =

∂

∂tΦ(t)Φ−1(s)

= 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 fromKn to Kn,

R(t, t0) : Kn → Kn

v 7→ R(t, t0)v = w

such that

R(t, t0) = Φt ◦ Φ−1t0

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. ThusR(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

dtM(t) = A(t)M(t)

M(t0) = I,

with M(t) ∈Mn(K).


43

Proof. Since d(R(t, t0)v)/dt = A(t)R(t, t0)v,(d

dtR(t, t0)

)v = (A(t)R(t, t0)) v,

for all v ∈ Rn. Therefore, R(t, t0) is a solution to M ′ = A(t)M . But, by Theorem 2.1.1, weknow 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 nonau-tonomous 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 givenby

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)

= detR(t, t0) det(v1, . . . , vn).

The following formulae hold

∆(t, t0) := detR(t, t0) = exp

(∫ t

t0

trA(s)ds

)(2.5a)

W (t) = exp

(∫ t

t0

trA(s)ds

)det(v1, . . . , vn). (2.5b)

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 thiswas 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 wherethe resolvent can be explicitly computed. One such case is for autonomous linear systems,which take the form

x′(t) = Ax(t), (2.6)

that is, where A(t) ≡ A. Our objective here is to establish the following result.


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, namelyTheorem 2.2.16. Note that in Lemma 2.2.15, the notation e(t−t0)A involves the notion ofexponential 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, weseek a λ ∈ K such that φλ(t) = eλtv be a solution to (2.6) with v ∈ Kn \ {0}. We have

φ′λ = λeλtv,

and thus φλ is a solution if, and only if,

λeλtv = Aeλtv

= eλtAv

⇔ 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 obtainn linearly independent solutions φλi

(t) = eλi(t−t0), i = 1, . . . , n. The general solution is givenby

φ(t) =(eλ1(t−t0)x01, . . . , e

λn(t−t0)x0n

),

where x0i is the ith component of x0, i = 1, . . . , n. In the general case, we need the notionof matrix exponentials. Defining the exponential of matrix A as

eA =∞∑

k=0

An

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)Ax0.

Proof. Assume φ = e(t−t0)Ax0. Then φ(t0) = e0Ax0 = Ix0 = x0. Also,

φ(t) =

(∞∑

n=0

1

n!(t− t0)

nAn

)x0

=∞∑

n=0

1

n!(t− t0)

nAnx0,


45

so φ is a power series with radius of convergence R = ∞. Therefore, φ is differentiable on Rand

φ′(t) =∞∑

n=1

1

n!n(t− t0)

n−1Anx0

=∞∑

n=0

1

(n+ 1)!(n+ 1)(t− t0)

nAn+1x0

=∞∑

n=0

1

n!(t− t0)

nAn+1x0

= A

(∞∑

n=0

1

n!(t− t0)

nAnx0

)= 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 Ais diagonalizable, solutions take the form

φ(t) =(eλ1(t−t0)x01, . . . , e

λn(t−t0)x0n

),

which implies that, in this case, the matrix R(t, t0) takes the form

R(t, t0) =

eλ1(t−t0) 0 0

0 eλ2(t−t0) 0...

. . .

0 0 eλn(t−t0)

.

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 matrixA, with multiplicity m ≤ n. Then, for k = 1, . . . ,m, any nonzero solution v of

(A− λI)kv = 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 repeatedaccording 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,

whereP−1DP = diag(λj),

the matrix N = A−D is nilpotent of order k ≤ n, and D and N commute.


2. Linear systems

2.3 Affine systems

We consider the general (affine) problem (2.1), which we restate here for convenience. Letx ∈ Rn, A : I → L(E) and B : I → E, where I ⊂ R and E is a normed vector space, weconsider the system

x′(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 ofsolutions does not constitute a vector space; in particular, the superposition principle doesnot 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 theassociated homogeneous equation (2.2).

Proof. Since x1 and x2 are solutions of (2.1),

x′1 = A(t)x1 +B(t)

x′2 = A(t)x2 +B(t)

Therefored

dt(x1 − x2) = A(t)(x1 − x2)

Theorem 2.3.2. The global solutions of (2.1) that are defined on I form an n dimensionalaffine 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 x′ =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 particularsolution of (2.1).

2. To obtain all solutions of (2.2), it is sufficient to know a basis of S0. Such a basis iscalled a fundamental system of solutions of (2.2).

2.3.2 Construction of solutions

We have the following variation of constants formula.

2.3. Affine systemsFund. Theory ODE Lecture Notes – J. Arino

47

Theorem 2.3.4. Let R(t, t0) be the resolvent of the homogeneous equation x′ = A(t)x asso-ciated to (2.1). Then the solution x to (2.1) is given by

x(t) = R(t, t0) +

∫ t

t0

R(t, s)B(s)ds (2.7)

Proof. Let R(t, t0) be the resolvent of x′ = A(t)x. Any solution of the latter equation isgiven 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 avariation of constants approach. Taking the derivative of this expression of x, we have

x′(t) =d

dt[R(t, t0)]v(t) +R(t, t0)v

′(t)

= A(t)R(t, t0)v(t) +R(t, t0)v′(t)

Thus x is a solution to (2.1) if

A(t)R(t, t0)v(t) +R(t, t0)v′(t) = A(t)R(t, t0)v(t) +B(t)

⇔ R(t, t0)v′(t) = B(t)

⇔ v′(t) = R(t0, t)B(t)

since R(t, s)−1 = R(s, t). Therefore, v(t) =∫ t

t0R(t0, s)B(s)ds. A particular solution is given

by

x(t) = R(t, t0)

∫ t

t0

R(t0, s)B(s)ds

=

∫ t

t0

R(t, t0)R(t0, s)B(s)ds

=

∫ t

t0

R(t, s)B(s)ds

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

x′(t) = Ax(t) +B(t)

x(t0) = x0

(2.8)

is given by

x(t) = e(t−t0)Ax0 +

∫ t

t0

e(t−t0)AB(s)ds (2.9)

Proof. Use Lemma 2.2.15 and the variation of constants formula (2.7).


2. Linear systems

2.4 Systems with periodic coefficients

2.4.1 Linear systems: Floquet theory

We consider the linear system (2.2) in the following case,

x′ = 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 resolventR(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 nonsingularconstant 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 ′(t) = A(t)X(t) for all t.Therefore X ′(t+ω) = A(t+ω)X(t+ω), and by periodicity of A(t), X ′(t+ω) = A(t)X(t+ω),which implies that X(t + ω) is a fundamental matrix of (2.10). As a consequence, byTheorem 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)

2.4. Systems with periodic coefficientsFund. Theory ODE Lecture Notes – J. Arino

49

such that eBω = V . Let P (t) = Φ(t)e−Bt, so Φ(t) = P (t)eBt. It is clear that P is continuousand nonsingular. Also,

P (t+ ω) = Φ(t+ ω)e−B(t+ω)

= Φ(t)V e−B(ω+t)

= Φ(t)eBωe−Bωe−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.11.1implies that there exists B ∈ Mn(R) such that e2ωB = V 2. Let P (t) = Φ(t)e−Bt, soΦ(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 2e−(2ω+t)B

= Φ(t)e2ωBe−2ωBe−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 theω-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 forall 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.


2. Linear systems

Definition 2.4.6 (Floquet normal form). The representation Φ(t) = P (t)etR is called aFloquet 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 x′ = P (t)x′ + dP (t)/dtx = P (t)A(t)x+ A(t)P (t)x− P (t)Bx

z = P−1(t)x, so z′ = dP−1(t)dt

x+ P−1(t)x′ = dP−1(t)dt

P (t)z + P−1(t)A(t)P (t)z

Definition 2.4.7 (Characteristic multipliers). The eigenvalues λ1, . . . , λn of a monodromymatrix B are called the characteristic multipliers of equation (2.10).

Definition 2.4.8 (Characteristic exponents). Numbers µ such that eµω is a characteristicmultiplier 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 writtenC = 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 monodromymatrix 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 suchthat X(t) = Y (t)C for all t. Thus, in particular, X(t+ ω) = Y (t+ ω)C, and so

C−1UCX(t+ ω) = Y (t+ ω)C = Y (t)UC = X(t)C−1UC,

since Y (t) = X(t)C−1. It follows that V = C−1UC, 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 allj (or <(λj) < 0 for all j).

Let p be an eigenvector of Φ(ω)2 associated with a multiplier ρ. Then the solutionφ(t) = Φ(t)p of (2.10) satisfies the condition φ(t + 2ω) = ρφ(t). This is the origin of theterm multiplier.


51

2.4.2 Affine systems: the Fredholm alternative

We discuss here an extension of a theorem that was proved implicitly in Exercise 4, Assign-ment 2. Let us start by stating the result in question. We consider here the system

x′ = 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

x′ = A(t)x (2.14)

associated to (2.13) has no nonzero solution of period ω, then (2.13) has for each functionf , a unique ω-periodic solution.

The Fredholm alternative concerns the case where there exists a nonzero periodic solutionof (2.14). We give some needed results before going into details. Consider (2.14). Associatedto this system is the so-called adjoint system, which is defined by the following differentialequation,

y′ = −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) isRT (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〈x(t), y(t)〉 is constant.

Proof. i) We know that ∂∂sR(t, s) = −R(t, s)A(s). Therefore, ∂

∂sRT (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 thatRT (0, ω)y0 = y0

This can also be written as [RT (0, ω)− I

]y0 = 0

or, taking the transpose,yT

0 [R(0, ω)− I] = 0

Now, since R(0, ω) = R−1(ω, 0), it follows that

yT0 [R(0, ω)− I] = 0 ⇔ yT

0

[R−1(ω, 0)− I

]= 0

This is equivalent to yT0 [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.


2. Linear systems

iii) Recall that for differentiable functions a, b,

d

dt〈a(t), b(t)〉 = 〈 d

dta(t), b(t)〉+ 〈a(t), d

dtb(t)〉

Thusd

dt〈x(t), y(t)〉 = 〈A(t)x(t), y(t)〉+ 〈x(t),−AT (t)y(t)〉 = 0

Before we carry on to the actual Fredholm alternative in the context of ordinary differ-ential equations, let us consider the problem in a more general setting. Let H be a Hilbertspace. If A ∈ L(H,H), the adjoint operator A∗ of A is the element of L(H,H) such that

∀u, v ∈ H, 〈Au, v〉 = 〈u,A∗v〉

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 isnecessary 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 ofODEs.

Theorem 2.4.16 (Fredholm alternative for ODEs). Consider (2.13) with A and f con-tinuous and ω-periodic. Suppose that the homogeneous equation (2.14) has p independentsolutions of period ω. Then the adjoint equation (2.15) also has p independent solutions ofperiod p, which we denote y1, . . . , yp. Then

i) If ∫ ω

0

〈yk(t), b(t)〉dt = 0, k = 1, . . . , p (2.16)

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, andonly if,

[R(0, ω)− I]x0 =

∫ ω

0

R(0, s)b(s)ds (2.17)

By Theorem 2.3.4, the solution of (2.13) through (0, x0) is given by

x(t) = R(t, 0)x0 +

∫ t

0

R(t, s)b(s)ds

Hence, at time ω,

x(ω) = R(ω, 0)x0 +

∫ ω

0

R(ω, s)b(s)ds


53

If x0 is the initial condition of a ω-periodic solution, then x(ω) = x0, and so

x0 −R(ω, 0) =

∫ ω

0

R(ω, s)b(s)ds

On the other hand, yk(0) is the initial condition of an ω-periodic solution yk if, and onlyif, [

RT (0, ω)− I]yk(0) = 0

Let C = R(0, ω)− I. We have that Rn = Img(C)⊕Ker(CT ). We now use the Fredholmalternative in this context. There exists x0 such that

Cx0 =

∫ ω

0

R(0, s)b(s)ds

if, and only if, ∫ ω

0

R(0, s)b(s)ds ∈ Img(C)

Indeed, from the Fredholm alternative, setting f = x0 and g =∫ ω

0R(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(CT ), i.e.,since Rn = Img(C)⊕Ker(CT ), if, and only if, g ∈ Img(C).

Now, y1(0), . . . , yp(0) is a basis of Ker(CT ). It follows that there exists a solution of(2.13) if, and only if, for all k = 1, . . . , p,

∀k = 1, . . . , p, 〈∫ ω

0

R(0, s)b(s)ds, yk(0)〉 = 0

⇔ ∀k = 1, . . . , p,

∫ ω

0

〈R(0, s)b(s), yk(0)〉ds = 0

⇔ ∀k = 1, . . . , p,

∫ ω

0

〈b(s), RT (0, s)yk(0)〉ds = 0

⇔ ∀k = 1, . . . , p,

∫ ω

0

〈b(s), yk(s)〉ds = 0

If these relations are satisfied, the set of vectors v such that

Av =

∫ ω

0

R(0, s)b(s)ds

is of the form v0 + Ker(CT ), where v0 is one of these vectors; hence there exist p of themwhich are independent and are initial conditions of the p independent ω-periodic solutionsof (2.13).

Example – The equationx′′ = f(t) (2.18)


2. Linear systems

where f is ω-periodic, has solutions of period ω if, and only if,∫ ω

0f(s)ds = 0

Let y = x′. Then, differentiating y and substituting into (2.18), we have

y′ = f(t)

Hence the system is (xy

)′=(

0 10 0

)(xy

)+(

0f(t)

)Hence,

AT =(

0 01 0

)and the adjoint equation ξ′ = AT ξ has the periodic solution (0, a)T . �

2.5 Further developments, bibliographical notes

2.5.1 A variation of constants formula for a nonlinear system witha 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

x′ = A(t)x+ g(t, x) (2.19a)

x(t0) = x0, (2.19b)

where g : R×Rn → Rn a smooth function, and let R(t, t0) be the resolvent associated to thehomogeneous system x′ = A(t)x, with R defined on some interval I 3 t0. Then the solutionφ of (2.19) is given by

φ(t) = R(t, t0)x0 +

∫ t

t0

R(t, s)g(φ(s), s)ds, (2.20)

on some subinterval of I.

Proof. We proceed using a variation of constants approach. It is known that the generalsolution to the homogeneous equation x′ = 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

φ′(t) =

(d

dtR(t, t0)

)v(t) +R(t, t0)v

′(t)

= A(t)R(t, t0)v(t) +R(t, t0)v′(t),

2.5. Further developments, bibliographical notesFund. Theory ODE Lecture Notes – J. Arino

55

from Proposition 2.2.11. For φ to be solution, it must satisfy the differential equation (2.19),and thus

φ′(t) = A(t)φ(t) + g(t, φ(t)) ⇔ A(t)R(t, t0)v(t) +R(t, t0)v′(t) = A(t)R(t, t0)v(t) + g(t, φ(t))

⇔ R(t, t0)v′(t) = g(t, φ(t))

⇔ v′(t) = R(t, t0)−1g(t, φ(t))

⇔ v′(t) = R(t0, t)g(t, φ(t))

⇔ v(t) =

∫ t

t0

R(t0, s)g(s, φ(s))ds+ C,

using Proposition 2.2.11 again. Therefore,

φ(t) = R(t, t0)

(∫ t

t0

R(t0, s)g(s, φ(s))ds+ C

).

Evaluating this expression at t = t0 gives φ(t0) = C, so C = x0. Therefore,

φ(t) = R(t, t0)x0 +R(t, t0)

∫ t

t0

R(t0, s)g(s, φ(s))ds

= R(t, t0)x0 +

∫ t

t0

R(t, t0)R(t0, s)g(s, φ(s))ds

= R(t, t0)x0 +

∫ t

t0

R(t, s)g(s, φ(s))ds,

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),

x′ = 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∗)′ = f(x∗) = 0, so that thesolution is constant when x = x∗. Note also that this implies that x(t) = x∗ is a solutiondefined on R.

Definition 3.1.2 (Stable equilibrium point). The fixed point x∗ is (positively) stable if thefollowing two conditions hold:

i) There exists r > 0 such that if ‖x0 − x∗‖ < r, then the solution x(t, x0) is defined forall t ≥ 0. (This is automatically satisfied for flows).

ii) For any ε > 0, there exists δ > 0 such that ‖x0 − x∗‖ < δ implies ‖x(t, x0)− x∗‖ < ε.

Definition 3.1.3 (Asymptotically stable equilibrium point). If the equilibrium x∗ is (pos-itively) stable and that additionally, there exists γ > 0 such that ‖x0 − x∗‖ < γ implieslimt→∞ 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

x′ = Ax+ b(t)x (3.2)

where b is continuous and small, i.e., limt→∞ b(t) = 0, with A ∈Mn(R) and b ∈ R.

57


3. Stability of linear systems

Theorem 3.2.1. Suppose that all eigenvalues of A have negative real parts, and that b iscontinuous 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 existenceand uniqueness of the solutions to the linear equation x′ = 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 canexpress the solution by means of the equivalent integral equation, for t0 ≤ t <∞,

φt(x0) = e(t−t0)Ax0 +

∫ t

t0

e(t−s)Ab(s)φs(x0)ds (3.3)

by the hypothesis on A, e(t−t0)A is such that for t0 ≤ t < ∞, ‖Ψ(t, t0)‖ ≤ Ke−σ(t−t0) forK > 0, σ > 0, where R(t, t0) = e(t−t0)A is the fundamental matrix of the homogeneous partof (3.2).

Since limt→∞ b(t) = 0, given any η > 0, there exists a number T ≥ t0 such that |b(t)| < ηfor 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 <∞,

φt(x0) = e(t−T )AφT (x0) +

∫ t

T

e(t−s)Ab(s)φs(x0)ds

Thus, using ‖Φ(t)‖ ≤ Ke−σ(t−t0) (with t0 = T ) and |b(t)| < η for t ≥ T , we obtain, forT ≤ t <∞,

‖φt(x0)‖ ≤ Ke−σ(t−T )‖φT (x0)‖+Kη

∫ t

T

e−σ(t−s)‖φs(x0)‖ds

Multiplying both sides of this inequality by eσt and using Gronwall’s inequality (Appendix A.7)with the function ‖φt(x0)‖eσt, we obtain, for T ≤ t <∞,

‖φt(x0)‖ ≤ K‖φT (x0)‖e−(σ−Kη)(t−T ) (3.4)

From this we conclude that if 0 < η < σ/K, the solution φt(x0) will approach zero exponen-tially. This does not yet prove that the zero solution of (3.2) is stable. To do this, we computea bound on ‖φT (x0)‖. Returning to (3.3) and restricting t to the interval t0 ≤ t ≤ T , wehave

‖φt(x0)‖ ≤ Ke−σ(t−t0)‖x0‖+K1K

∫ t

t0

e−σ(t−s)‖φ(s, t0, x0)‖ds

where K1 = maxt0≤t≤T |b(t)|. Multiplying by eσt and applying the Gronwall inequality weobtain

‖φt(x0)‖ ≤ K‖x0‖e−σ(t−t0)+K1K(t−t0)

≤ K‖x0‖eK1K(t−t0), t0 ≤ t ≤ T (3.5)

3.2. Affine systems with small coefficientsFund. Theory ODE Lecture Notes – J. Arino

59

Therefore,‖φT (x0)‖ ≤ K‖x0‖eK1K(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 <∞,

‖φt(x0)‖ ≤ K2‖x0‖eK1K(T−t0)e−(σ−Kη)(t−T ) (3.7)

Let then K2 = max(KeK1K(T−t0), K2eK1K(T−t0)

). From (3.5) and (3.7) we have

‖φt(x0)‖ ≤{

K2‖x0‖ if t0 ≤ t ≤ TK2‖x0‖e−(σ−Kη)(t−T ) if T ≤ t <∞ (3.8)

For a given matrix A, we can compute K and σ; we next pick any 0 < η < σ/K and thenT ≥ 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 ‖x0‖ < δ, ‖φt(x0)‖ < ε for all t ≥ t0 so that the zerosolution is stable. From (3.8), it is clear that the zero solution is globally asymptoticallystable.

Corollary 3.2.2. Let all eigenvalues of A have negative real part, so that |eAt| ≤ Ke−σt forsome constants K > 0, σ > 0 and all t ≥ 0. Let b(t) be continuous for 0 ≤ t < ∞ andsuppose 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 continuousfor 0 ≤ t < ∞ and such that

∫∞0|b(s)|ds < ∞. Then the zero solution of (3.2) is globally

asymptotically stable.


3. Stability of linear systems

We give some notions of linear stability theory, in the case of the autonomous linearsystem (2.6), repeated here for convenience:

x′(t) = Ax(t). (2.6)

We let wj = uj + ivj be a generalized eigenvector of A corresponding to an eigenvalueλj = 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 centersubspaces of the linear system (2.6) are given, respectively, by

Es = Span{uj, vj : aj < 0},

Eu = Span{uj, vj : aj > 0}and

Ec = 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 trajectoriesof (2.6).

Definition 3.2.6. If all eigenvalues of A have nonzero real part, that is, if Ec = ∅, thenthe flow eAt of system (2.6) is called a hyperbolic flow, and the system (2.6) is a hyperboliclinear system.

Definition 3.2.7. A subspace E ⊂ Rn is invariant with respect to the flow eAt, or invariantunder the flow of (2.6), if eAtE ⊂ 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 = Es ⊕ Eu ⊕ Ec.

Furthermore, if the matrix A is the matrix of the linear autonomous system (2.6), then Es,Eu and Ec are invariant under the flow of (2.6).

Definition 3.2.10. If all the eigenvalues of A have negative (resp. positive) real parts, thenthe origin is a sink (resp. source) for the linear system (2.6).

Theorem 3.2.11. The stable, center and unstable subspaces ES, EC and EU , respectively,are invariant with respect to eAt, i.e., let x0 ∈ ES, y0 ∈ EC and z0 ∈ EU , then eAtx0 ∈ ES,eAty0 ∈ EC and eAtz0 ∈ EU .

3.2. Affine systems with small coefficientsFund. Theory ODE Lecture Notes – J. Arino

61

Definition 3.2.12 (Homeomorphism). Let X be a metric space and let A and B be subsetsof X. A homeomorphism h : A → B of A onto B is a continuous one-to-one map of Aonto B such that h−1 : B → A is continuous. The sets A and B are called homeomorphicor 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 Ck) is a connected metric space with an open covering {Uα} ( i.e.,M = ∪αUα) 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−1β : hβ(Uα ∩ Uβ) → hα(Uα ∩ Uβ)

is differentiable (or of class Ck) and for all x ∈ hβ(Uα ∩ Uβ), the determinant of theJacobian, detDh(x) 6= 0.

Remark – The manifold is analytic if the maps h = hα ◦ h−1β are analytic. ◦

Next, recall that x∗ is an equilibrium point of (4.1) if f(x∗) = 0. An equilibrium pointx∗ is hyperbolic if the Jacobian matrix Df of (4.1) evaluated at x∗, denoted Df(x∗), has noeigenvalues with zero real part. Also, recall that the solutions of (4.1) form a one-parametergroup 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 themaximal interval of existence of the solution to the IVP. Let then φ : R × Rn → Rn bedefined as follows: For x0 ∈ Rn and t ∈ I(x0), φ(t, x0) = φt(x0) is the solution of the IVPdefined on its maximal interval of existence I(x0).

Example – Consider the (linear) ordinary differential equation x′ = ax, with a, x ∈ R. Thesolution is φ(t, x0) = eatx0, and satisfies the group property

φ(t + s, x0) = ea(t+s)x0 = eat(easx0) = φ(t, easx0) = φ(t, φ(s, x0))

�

For simplicity and without loss of generality since both results are local results, we assumehereforth 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

x′ = f(x) (4.1)

The object of this chapter is to show two results which link the behavior of (4.1) near ahyperbolic equilibrium point x∗ to the behavior of the linearized system

x′ = 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 linearsystem (2.6), repeated here for convenience:

x′(t) = Ax(t). (2.6)

We let wj = uj + ivj be a generalized eigenvector of A corresponding to an eigenvalueλj = 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 centersubspaces of the linear system (2.6) are given, respectively, by

Es = Span{uj, vj : aj < 0},

Eu = Span{uj, vj : aj > 0}and

Ec = Span{uj, vj : aj = 0}.

63


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 trajectoriesof (2.6).

Definition 4.1.3. If all eigenvalues of A have nonzero real part, that is, if Ec = ∅, thenthe flow eAt of system (2.6) is called a hyperbolic flow, and the system (2.6) is a hyperboliclinear system.

Definition 4.1.4. A subspace E ⊂ Rn is invariant with respect to the flow eAt, or invariantunder the flow of (2.6), if eAtE ⊂ 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 = Es ⊕ Eu ⊕ Ec.

Furthermore, if the matrix A is the matrix of the linear autonomous system (2.6), then Es,Eu and Ec are invariant under the flow of (2.6).

Definition 4.1.7. If all the eigenvalues of A have negative (resp. positive) real parts, thenthe origin is a sink (resp. source) for the linear system (2.6).

Theorem 4.1.8. The stable, center and unstable subspaces ES, EC and EU , respectively,are invariant with respect to eAt, i.e., let x0 ∈ ES, y0 ∈ EC and z0 ∈ EU , then eAtx0 ∈ ES,eAty0 ∈ EC and eAtz0 ∈ EU .

Definition 4.1.9 (Homeomorphism). Let X be a metric space and let A and B be subsetsof X. A homeomorphism h : A → B of A onto B is a continuous one-to-one map of Aonto B such that h−1 : B → A is continuous. The sets A and B are called homeomorphicor 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 Ck) is a connected metric space with an open covering {Uα} ( i.e.,M = ∪αUα) 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−1β : hβ(Uα ∩ Uβ) → hα(Uα ∩ Uβ)

is differentiable (or of class Ck) and for all x ∈ hβ(Uα ∩ Uβ), the determinant of theJacobian, detDh(x) 6= 0.

4.2. The stable manifold theoremFund. Theory ODE Lecture Notes – J. Arino

65

Remark – The manifold is analytic if the maps h = hα ◦ h−1β are analytic. ◦

Next, recall that x∗ is an equilibrium point of (4.1) if f(x∗) = 0. An equilibrium pointx∗ is hyperbolic if the Jacobian matrix Df of (4.1) evaluated at x∗, denoted Df(x∗), has noeigenvalues with zero real part. Also, recall that the solutions of (4.1) form a one-parametergroup 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 themaximal interval of existence of the solution to the IVP. Let then φ : R × Rn → Rn bedefined as follows: For x0 ∈ Rn and t ∈ I(x0), φ(t, x0) = φt(x0) is the solution of the IVPdefined on its maximal interval of existence I(x0).

Example – Consider the (linear) ordinary differential equation x′ = ax, with a, x ∈ R. Thesolution is φ(t, x0) = eatx0, and satisfies the group property

φ(t + s, x0) = ea(t+s)x0 = eat(easx0) = φ(t, easx0) = φ(t, φ(s, x0))

�

For simplicity and without loss of generality since both results are local results, we assumehereforth 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 theorigin, let f ∈ C1(E), and let φt be the flow of the nonlinear system (4.1). Suppose thatf(0) = 0 and that Df(0) has k eigenvalues with negative real part and n − k eigenvalueswith positive real part. Then there exists a k-dimensional differentiable manifold S tangentto the stable subspace Es of the linear system (4.2) at 0 such that for all t ≥ 0, φt(S) ⊂ Sand for all x0 ∈ S,

limt→∞

φt(x0) = 0

and there exists an (n − k)-dimensional differentiable manifold U tangent to the unstablesubspace Eu of (4.2) at 0 such that for all t ≤ 0, φt(U) ⊂ U and for all x0 ∈ U ,

limt→−∞

φt(x0) = 0

There are several approaches to the proof of this result. Hale [10] gives a proof which usesfunctional 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 matrixC such that if y = C−1x 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 S in y space. The stable manifold in S spaceis obtained by applying the transformation P−1 to y so that x = P−1y defines S in terms ofk curvilinear coordinates y1, . . . , yk.


4. Linearization

Proof. System (4.1) can be written as

x′ = Df(0)x+ F (x)

with F (x) = f(x) − Df(0)x. Since f ∈ C1(E), F ∈ C1(E), and F (0) = f(0) = 0 sincef(0) = 0. Also, DF (x) = Df(x) − Df(0) and so DF (0) = 0. To continue, we use thefollowing lemma (which we will not prove).

Lemma 4.2.2. Let E be an open subset of Rn containing the origin. If F ∈ C1(E), thenfor all x, y ∈ Nδ(0) ⊂ E, there exists a ξ ∈ Nδ(0) such that

|F (x)− F (y)| ≤ ‖DF (ξ)‖ |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−1Df(0)C =

[P 00 Q

]where the eigenvalues λ1, . . . , λk of the k × k matrix P have negative real part and theeigenvalues λk+1, . . . , λn of the (n− k)× (n− k) matrix Q have positive real part. Let α > 0be chosen small enough that for j = 1, . . . , k, <(λj) < −α < 0. Let y = C−1x, we have

y′ = C−1x′

= C−1Df(0)x+ C−1F (x)

= C−1Df(0)Cy + C−1F (Cy)

= By +G(y)

where G(y) = C−1F (Cy). Since F ∈ C1(E), G ∈ C1(E), where E = C−1(E). Also,Lemma 4.2.2 applies to G.

Now consider the systemy′ = By +G(y) (4.3)

and let

U(t) =

[ePt 00 0

]and V (t) =

[0 00 eQt

]Then U ′ = BU , V ′ = BV and eBt = U(t)+V (t). Choosing as previously noted α sufficientlysmall, we can then choose K > 0 large enough and σ > 0 small enough that

‖U(t)‖ ≤ Ke−(α+σ)t for all t ≥ 0

and‖V (t)‖ ≤ Ke−σt for all t ≤ 0

4.2. The stable manifold theoremFund. Theory ODE Lecture Notes – J. Arino

67

Consider now the integral equation

u(t, a) = U(t)a+

∫ t

0

U(t− s)G(u(s, a))ds−∫ ∞

t

V (t− s)G(u(s, a))ds (4.4)

where a, u ∈ Rn and a is a constant vector. We can solve this equation using the method ofsuccessive approximations. Indeed, let

u(0)(t, a) = 0

and

u(j+1)(t, a) = U(t)a+

∫ t

0

U(t− s)G(u(j)(s, a))ds−∫ ∞

t

V (t− s)G(u(j)(s, a))ds (4.5)

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

|u(1)(t, a)| ≤ ‖U(t)a‖+

∫ t

0

‖U(t− s)G(u(s, a))‖ds+

∫ ∞

t

‖V (t− s)G(u(s, a))‖ds

Now suppose that (4.6) holds for j = k. We have

|u(k+1)(t, a)− u(k)(t, a)| =

∣∣∣∣∣∫ t

0

U(t− s)[G(u(k+1)(s, a))−G(u(k)(s, a))

]ds

−∫ ∞

t

V (t− s)[G(u(k+1)(s, a))−G(u(k)(s, a))

]ds

∣∣∣∣∣≤∫ t

0

‖U(t− s)‖∣∣G(u(k+1)(s, a))−G(u(k)(s, a))

∣∣ ds+

∫ ∞

t

‖V (t− s)‖∣∣G(u(k+1)(s, a))−G(u(k)(s, a))

∣∣ dswhich, since G verifies a Lipschitz-type condition as given by Lemma 4.2.2, implies thatthere exists ε > 0 such that

|u(k+1)(t, a)− u(k)(t, a)| ≤ ε

∫ t

0

‖U(t− s)‖∣∣u(k+1)(s, a)− u(k)(s, a)

∣∣ ds+ ε

∫ ∞

t

‖V (t− s)‖∣∣u(k+1)(s, a)− u(k)(s, a)

∣∣ ds


4. Linearization

Using the bounds on ‖U‖ and ‖V ‖ as well as the induction hypothesis (4.6), it follows that

|u(k+1)(t, a)− u(k)(t, a)| ≤ ε

∫ t

0

Ke−(α+σ)(t−s)K|a|e−αs

2k−1ds

+ ε

∫ ∞

t

Keσ(t−s)K|a|e−αs

2k−1ds

≤ εK2|a|e−αt

σ2k−1+εK2|a|e−αt

σ2k−1

which, if we choose ε < σ/(4K), i.e., εK/σ < 1/4, implies that

|u(k+1)(t, a)− u(k)(t, a)| <(

1

4+

1

4

)K|a|e−αt

2k−1=K|a|e−αt

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)| ≤∞∑

j=N

|u(j+1)(t, a)− u(j)(t, a)|

≤ K|a|∞∑

j=N

1

2j=K|a|2N−1

As this last quantity approaches 0 as N →∞, it follows that {u(j)(t, a)} is a Cauchy sequence(of continuous functions).

It follows thatlimj→∞

u(j)(t, a) = u(t, a)

uniformly for all t ≥ 0 and |a| < δ/(2K). From the uniform convergence, we deduce thatu(t, a) is continuous. Now taking the limit as j → ∞ in both sides of (4.5), it follows thatu(t, a) satisfies the integral equation (4.4) and as a consequence, the differential equation(4.3).

Since G ∈ C1(E), it follows from induction on (4.5) that u(j)(t, a) is a differentiablefunction of a for |a| < δ/(2K) and t ≥ 0. Since u(j)(t, a) → u(t, a) uniformly, it then followsthat 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

4.3. The Hartman-Grobman theoremFund. Theory ODE Lecture Notes – J. Arino

69

and

uj(0, a) = −(∫ ∞

0

V (−s)G(u(s, a1, . . . , ak, 0, . . . , 0))ds

)j

for j = k + 1, . . . , n

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

which defines a differentiable manifold S in y space for√y2

1 + · · ·+ y2k < δ/(2K). Fur-

thermore, if y(t) is a solution of the differential equation (4.3) with y(0) ∈ S, i.e., withy(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) ∈ S, theny(t) → 0 as t → ∞. It can also be shown that if y(t) is a solution of (4.3) with y(0) 6∈ S,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) ∈ S,then y(t) ∈ S for all t ≥ 0. And it can be shown as in [6, Th 4.2, p. 333] that

∂ψj

∂yi

(0) = 0

for i = 1, . . . , k and j = k + 1, . . . , n, i.e., that the differentiable manifold S is tangent tothe stable subspace Es = {y ∈ Rn : y1 = · · · = yk = 0} of the linear system y′ = By at 0.

The existence of the unstable manifold U of (4.3) is established the same way, but con-sidering a reversal of time, t→ −t, i.e., considering the system

y′ = −By −G(y)

The stable manifold for this system is the unstable manifold U of (4.3). In order to determinethe (n− k)-dimensional manifold U using the above process, the vector y has to be replacedby 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 non-linear system (4.1). Let ϕt be the flow of (4.1), and ψt be the flow of the linearized systemx′ = 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.


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 ∈ C1(E), and let φt be the flow of the nonlinear system (4.1). Suppose that f(0) = 0and 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 anopen set V containing the origin such that for each x0 ∈ U , there is an open interval I0 ⊂ Rcontaining 0 such that for all x0 ∈ U and t ∈ I0,

H ◦ φt(x0) = eAtH(x0);

i.e., H maps trajectories of (4.1) near the origin onto trajectories of x′ = Df(0)x near theorigin and preserves the parametrization by time.

Proof. Suppose that f ∈ C1(E), f(0) = 0 (i.e., 0 is an equilibrium) and A = Df(0) thejacobian matrix of f at 0.

1. As we have assumed that the matrix A has no eigenvalues with zero real part (i.e., 0is an hyperbolic equilibrium point), we can write A in the form

A =

[P 00 Q

]where P has only eigenvalues with negative real parts and Q has only eigenvalues withpositive real parts.

2. Let φt be the flow of the nonlinear system (4.1). Let us write the solution as

x(t, x0) = φt(x0) =

[y(t, y0, z0)z(t, y0, z0)

]where

x0 =

[y0

z0

]∈ Rn

has been decomposed as y0 ∈ Es, the stable subspace of A, and z0 ∈ Eu, the unstablesubspace of A.

3. LetY (y0, z0) = y(1, y0, z0)− ePy0

andZ(y0, z0) = z(1, y0, z0)− eQz0

Then Y (0) = Y (0, 0) = y(1, 0, 0) = 0. The same is true of Z(0) = 0. Also, DY (0) =DZ(0) = 0. Since f ∈ C1(E), Y (y0, z0) and Z(y0, z0) are continuously differentiable. Thus

‖DY (y0, z0)‖ ≤ a

and‖DZ(y0, z0)‖ ≤ a


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

Y (y0, z0) =

{Y (y0, z0) for |y0|2 + |z0|2 ≤

(s0

2

)20 for |y0|2 + |z0|2 ≥ s2

0

and

Z(y0, z0) =

{Z(y0, z0) for |y0|2 + |z0|2 ≤

(s0

2

)20 for |y0|2 + |z0|2 ≥ s2

0

By the mean value theorem,

|Y (y0, z0)| ≤ a√|y0|2 + |z0|2 ≤ a(|y0|+ |z0|)

and|Z(y0, z0)| ≤ a

√|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 normalizedin a proper way, we have

b = ‖B‖ < 1 and c = ‖C−1‖ < 1

4. For

x =

[yz

]∈ Rn

define the transformations

L(y, z) =

[ByCz

]and

T (y, z) =

[By + Y (y, z)Cz + Z(y, z)

]i.e., L(x) = eAx and, locally, T (x) = φ1(x). Then the following lemma holds, which weprove later.

Lemma 4.3.3. There exists a homeomorphism H of an open set U containing the originonto 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-parameterfamilies of transformations defined by

Lt(x0) = eAtx0 and T t(x0) = φt(x0)

Define

H =

∫ 1

0

L−sH0Tsds


4. Linearization

It follows from the above lemma that there exists a neighborhood of the origin for which

LtH =

∫ 1

0

Lt−sH0Ts−tdsT t

=

∫ 1−t

−t

L−sH0TsdsT t

=

[∫ 0

−t

L−sH0Tsds+

∫ 1−t

0

L−sH0Tsds

]T t

=

∫ 1

0

L−sH0TsdsT t

= HT t

since by the above lemma, H0 = L−1H0T which implies that∫ 0

−s

L−sH0Tsds =

∫ 0

−t

L−s−1H0Ts+1ds

=

∫ 1

1−t

L−sH0Tsds

Thus H ◦ T t = LtH or equivalently

H ◦ φt(x0) = eAtH(x0)

and it can be shown that H is a homeomorphism on Rn. The outline of the proof iscomplete.

We now prove Lemma 4.3.3.

Proof. We use the method of successive approximations. For x ∈ Rn, let

H(x) =

[Φ(y, z)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)) (4.10a)

CΨ(y, z) = Ψ(By + Y (y, z), Cz + Z(y, z)) (4.10b)

Successive approximations for (4.10b) are defined by

Ψ0(y, z) = z

Ψk+1(y, z) = C−1Ψk(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 suchthat Ψk(y, z) = z for |y|+ |z| ≥ 2s0.


73

Let us now prove that {Ψk} is a Cauchy sequence. For this, we show by induction thatfor all j ≥ 1,

|Ψj(y, z)−Ψj−1(y, z)| ≤Mrj(|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 ispossible 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−1Ψ0(By + Y (y, z), Cz + Z(y, z))− z|= |C−1(Cz + Z(y, z))− z|= |C−1Z(y, z)|≤ ‖C−1‖ |Z(y, z)|≤ ca(|y|+ |z|)≤Mr(|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−1Ψk(By + Y (y, z), Cz + Z(y, z))− C−1Ψk−1(By + Y (y, z), Cz + Z(y, z))|= |C−1(Ψk −Ψk−1)|≤ ‖C−1‖ |Ψk −Ψk−1|

which, using induction hypothesis (4.12) and c = ‖C−1‖ gives

≤ cMrk(|By + Y (y, z)|+ |Cz + Z(y, z)|

)δ

≤ cMrk(b|y|+ 2a(|y|+ |z|) + c|z|

)δ

≤ cMrk(2 max(a, b, c)

)δ(|y|+ |z|)δ

≤Mrk+1(|y|+ |z|)δ

Using the same type of argument as in the proof of the stable manifold theorem, Ψk isthus a Cauchy sequence of continuous functions that converges uniformly as k → ∞ to acontinuous 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−1Φ(y, z) = Φ(B−1y + Y1(y, z), C−1z + 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),

T−1(y, z) =

[B−1y + Y1(y, z)C−1z + Z1(y, z)

]Successive approximations with Φ0(y, z) = y can then be used as above (since b = ‖B‖ < 1)to solve (4.13).


4. Linearization

Therefore, we obtain the continuous map

H(y, z) =

[Φ(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 Example of application

4.4.1 A chemostat model

To illustrate the use of the theorems in this chapter, we take an example of nonlinearsystem, a system of two nonlinear differential equations modeling a biological device calleda chemostat. Without going into details, the system is the following.

dS

dt= D(S0 − S)− µ(S)x (4.14a)

dx

dt= (µ(S)−D)x (4.14b)

The parameters S0 and D, respectively the input concentration and the dilution rate, arereal and positive. The function µ is the growth function. It is generally assumed to satisfyµ(0) = 0, µ′ > 0 and µ′′ < 0.

To be complete, one should verify that the positive quadrant is positively invariant underthe flow of (4.14), i.e., that for S(0) ≥ 0 and x(0) ≥ 0, solutions remain nonnegative for allpositive times, and similar properties. But since we are here only interested in applicationsof the stable manifold theorem, we proceed to a very crude analysis, and will not deal withthis point.

Note that in vector form, the system is noted

ξ′ = f(ξ)

with ξ = (S, x)T and

f(ξ) =

(D(S0 − 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 situatedon one of the boundaries of the positive quadrant,

ξ∗T = (S∗T , x∗T ) = (S0, 0)

the second one in the interior,

ξ∗I = (S∗, x∗) = (λ, S0 − λ)

where λ is such that µ(λ) = D. Note that this implies that if λ ≥ S0, ξ∗T is the onlyequilibrium of the system.

4.4. Example of applicationFund. Theory ODE Lecture Notes – J. Arino

75

At an arbitrary point ξ = (S, x), the Jacobian matrix is given by

Df(ξ) =

(−D − µ′(S)x −µ(S)

µ′(S)x µ(S)−D

)Thus, at the trivial equilibrium ξ∗T ,

Df(ξ∗T ) =

(−D −µ(S0)0 µ(S0)−D

)We have two eigenvalues, −D and µ(S0)−D. Let us suppose that µ(S0)−D < 0. Note thatthis implies that ξ∗T is the only equilibrium, since, as we have seen before, ξ∗I is not feasibleif λ > S0.

As the system has dimensionality 2, and that the matrix Df(ξ∗T ) has two negative eigen-values, the stable manifold theorem (Theorem 4.2.1) states that there exists a 2-dimensionaldifferentiable manifold M such that

• φt(M) ⊂M

• for all ξ0 ∈M, limt→∞ φt(ξ0) = ξ∗T .

• At ξ∗T , M is tangent to the stable subspace ES of the linearized system ξ′ = Df(ξ∗T )(ξ−ξ∗T ).

Since there are no eigenvalues with positive real part, there does not exist an unstablemanifold in this case. Let us now caracterize the nature of the stable subspace ES. It isobtained by studying the linear system

ξ′ = Df(ξ∗T )(ξ − ξ∗T )

=

(−D −µ(S0)0 µ(S0)−D

)(S − S0

x

)=

(−D(S − S0)− µ(S0)x

(µ(S0)−D)x

)(4.15)

Of course, the Jacobian matrix associated to this system is the same as that of the nonlinearsystem (at ξ∗T ). Associated to the eigenvalue −D is the eigenvector v1 = (1, 0)T , to µ(S0)−Dis 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 ξ∗T 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 tryto 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

x′ = −x− y2

y′ = x2 + y(4.16)


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

J =

(−1 00 1

)The linearized system at 0 is

x′ = −xy′ = y

(4.17)

So the eigenvalues are 1 and −1, with associated eigenvectors (1, 0)T and (0, 1)T , respec-tively. Therefore, the stable manifold theorem (Theorem 4.2.1) implies that there exists a1-dimensional stable (differentiable) manifold S such that φt(S) ⊂ S and limt→∞ φt(x0) = 0for all x0 ∈ S, and a 1-dimensional unstable (differentiable) manifold U such that φt(U) ⊂ Uand limt→−∞ φt(x0) for all x0 ∈ U . Furthermore, at 0, S is tangent to the stable subspaceES of (4.17), and U is tangent to the unstable subspace EU of (4.17).

The stable subspace ES is given by Span (v1), with v1 = (0, 1)T , i.e., the y-axis. Theunstable subspace EU is Span (v2), with v2 = (1, 0)T , i.e., the x-axis. The behavior of thissystem is illustrated in Figure 4.1.

−0.05 −0.04 −0.03 −0.02 −0.01 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

y

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

A = B =

(−1 00 1

)

and C = I. Also, F (ξ) = G(ξ) =

(−y2

x2

). Here, the matrices P and Q are in fact scalars,

4.4. Example of applicationFund. Theory ODE Lecture Notes – J. Arino

77

P = −1 and Q = 1. Thus

U(t) =

(e−t 00 0

), V (t) =

(0 00 et

)Finally, a = (a1, 0)T . So now we can use successive approximations to find an approximatesolution to the integral equation (4.4), which here takes the form

u(t, a) =

(e−ta1

0

)+

∫ t

0

(−e−(t−s)u2

2(s)0

)ds−

∫ ∞

t

(0

e(t−s)u21(s)

)ds

To construct the sequence of successive approximations, we start with u(t, a) = (0, 0)T , thencompute the successive terms using equation (4.5), which takes the form

u(j+1)(t, a) =

(e−ta1

0

)+

∫ t

0

(e−(t−s) 0

0 0

)G(u(j)(s)

)ds−

∫ ∞

t

(0 00 e(t−s)

)G(u(j)(s)

)ds

=

(e−ta1

0

)+

∫ t

0

(e−(t−s) 0

0 0

)(u(j)2 (s)

)2(u

(j)1 (s)

)2

ds−∫ ∞

t

(0 00 e(t−s)

)(u(j)2 (s)

)2(u

(j)1 (s)

)2

ds

=

(e−ta1

0

)+

∫ t

0

(−e−(t−s)

(u

(j)2 (s)

)2

0

)ds−

∫ ∞

t

(0

e(t−s)(u

(j)1 (s)

)2

)ds

Therefore,

u(1)(t, a) = U(t)a =

(e−aa1

0

)since u(0)(t, a) =

(u

(0)1 (t, a)

u(0)2 (t, a)

)=

(00

).

Then,

u(2)(t, a) =

(e−ta1

0

)−∫ ∞

t

(0

e(t−s) (e−sa1)2

)ds

=

(e−ta1

0

)−(

013a2

1e−2t

)=

(e−ta1

−13a2

1e−2t

)and continuing this process, we find

u(3)(t, a) =

(e−ta1 + 1

27(e−4t − e−t)a4

1

−13a2

1e−2t

)Also, it is possible to show that u(4)(t, a)− u(3)(t, a) = O(a5

1).The stable manifold S is 1-dimensional, so here it has the form ψ2(a1) = u2(0, a1, 0), and

is here approximated by

ψ2(a1) = −1

3a2

1 +O(a51)


4. Linearization

as a1 → 0. Thus S is approximated by

y = −x2

3+O(x5)

as x→ 0.

Chapter 5

Exponential dichotomy

Our aim here is to show the equivalent of the Hartman-Grobman theorem for linear systemswith variable coefficients. Compared to other results we have seen so far, this is a muchmore recent field. The first results were shown in the 60s by Lin. We give here only themost elementary results. For more details, see, e.g., [13].

We consider the linear system of differential equations

dx

dt= A(t)x (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 solutionof (5.1). If X(t) and X−1(s) can be decomposed into the following forms

X(t) = X1(t) +X2(t)

X−1(s) = Z1(s) + Z2(s)

X(t)Z−1(s) = X1(t)Z1(s) +X2(t)Z2(s)

and satisfy the conditions that there exists α, β, positive constants such that

‖X1(t)Z1(s)‖ ≤ βe−α(t−s), t ≥ s

‖X2(t)Z2(s)‖ ≤ βeα(t−s), s ≥ t(5.2)

where

X1(t) = (X11(t), 0), X2(t) = (0, X12(t)),

Z1(s) =

(Z11(s)

0

), Z2(s) =

(0

Z21(s)

),

79


5. Exponential dichotomy

or there is a projection P on the stable manifold such that

‖X(t)PX−1(s)‖ ≤ βe−α(t−s), t ≥ s

‖X(t)(I − P )X−1(s)‖ ≤ 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). Wesay 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) ‖Φ(t, s)P (s)‖ ≤ Ke−α(t−s), for t ≥ s ∈ I.

iii) ‖Φ(t, s)Q(s)‖ ≤ 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 onthe real interval I = (ω−, ω+), system (5.1) is said to have a (µ1, µ2)-dichotomy if thereexist supplementary projections P1, P2 on Rn such that

‖X(t)PiX−1(s)‖ ≤ Ki exp(

∫ t

s

µi), if (−1)i(s− t) ≥ 0, i = 1, 2.

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 systemx′ = Ax

has an exponential dichotomy on R+ if and only if no eigenvalue of A has zero real part. Ithas 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 theireigenspace). In each case, X(t) = etA and we can take P to be the spectral projection defined by

P =1

2πi

∫γ(zI −A)−1dz

with γ any rectifiable simple closed curve in the open left half-plane which contains in its interiorall eigenvalues of A with negative real part. ◦

5.2. Existence of exponential dichotomyFund. Theory ODE Lecture Notes – J. Arino

81

5.2 Existence of exponential dichotomy

To check that the previous definitions hold can be a very tedious task. Some authors havethus 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 thereexists a quadratic form V (t, x) = xTG(t)x, where the matrix G(t) is symmetric, regular,bounded and C1, 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 Isuch 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) ‖A(t)‖ ≤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

‖A(t2)− A(t1)‖ ≤ δ for |t2 − t1| ≤ h

where h > 0 is a fixed number not greater than the length of I, then the equation

x′ = A(t)x

has a fundamental matrix X(t) satisfying the inequalities

‖X(t)PX−1(s)‖ ≤ Ke−(α−ε)(t−s) for t ≥ s

‖X(t)(I − P )X−1(s)‖ ≤ Le−(β−ε)(s−t) for s ≥ t

where K,L are positive constants depending only on M,α+ β, ε and

P =

(Ik 00 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 constantsli, 0 ≤ li < 1, i = 1, 2, such that for some m, 0 ≤ m ≤ n,

max

{l1<(ajj) + l1

m∑i=1,i6=j

|aij|+n∑

i=m+1

|aij| : j = 1, . . . ,m

}≤ l1ρ,



min

{l2<(ajj)−

m∑i=1

|aij| − l2

n∑i=m+1,i6=j

|aij| : j = m+ 1, . . . , n

}≥ l2ρ.

Then the system (5.1) has a (µ1, µ2)-dichotomy, where

µ1 = max

{l1<(ajj) +

m∑i=1,i6=j

|aij|+ l2

n∑i=m+1

|aij| : j = 1, . . . ,m

},

µ2 = min

{<(ajj)− l1

m∑i=1

|aij| −n∑

i=m+1,i6=j

|aij| : j = m+ 1, . . . , n

}.

The same sort of theorem can be proved with sums of the columns replaced by sums ofthe rows.

Example – Consider

A(t) =

−1 0 1/2t/2 t t2

t/2 −t2 t

, t > 0

�

5.3 First approximate theory

We consider the nonlinear system

dx

dt= A(t)x+ f(t, x) (5.4)

where f : R×Rn → Rn is continuous, f(t, x) = O(‖x‖2), ‖x‖ = o(1), ‖f(t, x1)− f(t, x2)‖ ≤L‖x1 − x2‖ with L small enough.

Let x(t) be a non trivial solution of (5.1); define

λu(x(t)) = lim supt−s→∞

1

t− slog

‖x(t)‖‖x(s)‖

and

λu(x(t)) = lim inft−s→∞

1

t− slog

‖x(t)‖‖x(s)‖

The numbers λu(x(t)) and λu(x(t)) are called the uniform upper characteristic exponent anduniform lower characteristic exponent of x(t), respectively.

Remark – If λ(x) ≤ −α < 0, then lims→−∞ ‖x(s)‖ = ∞. If λ(x) ≥ α > 0, then limt→∞ ‖x(t)‖ =∞. ◦

Then we have the following theorem.

5.3. First approximate theoryFund. Theory ODE Lecture Notes – J. Arino

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 suchthat limt→∞ x(0) = 0, or limt→−∞ x(t) = 0, then

‖x(t)‖ ≤ β0‖x(s)‖e−α0(t−s), t ≥ s

or

‖x(t)‖ ≤ β0‖x(s)‖eα0(t−s), s ≥ t

respectively. At this time, λu(x(t)) ≤ −α0 < 0, or λu(x(t)) ≥ α0 > 0;

iii) for a k-dimensional solution x of (5.1) with λu(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 k-dimensional or (n− k)-dimensional y(t), solution of (5.4), such that λu(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 linearequation x′ = A(t)x has an exponential dichotomy. Suppose that f(t, x) is a continuousfunction of R× Rn into Rn such that

‖f(t, x)‖ ≤ µ, ‖f(t, x1)− f(t, x2)‖ ≤ L‖x1 − x2‖

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 x′ = A(t)x+ f(t, x), then H(t, x(t))is a solution of x′ = A(t)x.

Moreover, H is continuous in R× Rn and

‖H(t, x)− x‖ ≤ 4Kµα−1

for all t, x. For each fixed t, Ht(x) = H(t, x) is a homeomorphism of Rn. L(t, x) = H−1t (x)

is continuous in R×Rn and if y(t) is any solution of x′ = A(t)x, then L(t, y(t)) is a solutionof x′ = A(t)x+ f(t, x).



5.4 Stability of exponential dichotomy

Theorem 5.4.1. Suppose that the linear system (5.1) admits exponential dichotomy. Thenthere exists a constant η > 0 such that the linear system

dx

dt= (A(t) +B(t))x (5.5)

also admits exponential dichotomy, when B(t) is continuous on R and ‖B(t)‖ ≤ 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 ‖B(t)‖ < α/(4K2), then the perturbedequation

x′ = (A(t) +B(t))z

also has an exponential dichotomy on R+ with constants K and α determined by K, α andδ. Moreover, if P (t) is the corresponding projection, then ‖P (t) − P (t)‖ = O(δ) uniformlyin t ∈ R+. Also, |α− α| = 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 torealize that exponential dichotomies exist in a much more general setting.

Bibliography

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[2] F. Brauer and J.A. Nohel. The Qualitative Theory of Ordinary Differential Equations.Dover, 1989.

[3] H. Cartan. Cours de calcul differentiel. Hermann, Paris, 1997. Reprint of the 1977edition.

[4] C. Chicone. Ordinary Differential Equations with Applications. Springer, 1999.

[5] E.A. Coddington and N. Levinson. Theory of Ordinary Differential Equations. McGraw-Hill, 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 Mathe-matics. Springer-Verlag, 1978.

[8] J. Dieudonne. 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 ofHyperbolic 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


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 randomordering.

A.1 Vector spaces, norms

A.1.1 Norm

Consider a vector space E over a field K. A norm is an application, denoted ‖ ‖, from E toR+ that satisfies the following:

1) ∀x ∈ E, ‖x‖ ≥ 0, with ‖x‖ = 0 if and only if x = 0.

2) ∀x ∈ E, ∀a ∈ K, ‖ax‖ = |a| ‖x‖.

3) ∀x, y ∈ E, ‖x+ y‖ ≤ ‖x‖+ ‖y‖.

A vector space E equiped with a norm ‖ ‖ 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||| = supx∈E−{0}

‖L(x)‖‖x‖

= sup‖x‖≤1

‖L(x)‖.

The inequality

‖A(t)(x1 − x2)‖ ≤ |||A(t)||| ‖x1 − x2‖

87


A. Definitions and results

results from the nature of the norm ||| |||. See appendix A.1. is best understood by indicatingthe spaces in which the various norms are defined. We have

‖Ax‖a = ‖x‖b

∥∥∥∥A( x

‖x‖b

)∥∥∥∥a

≤ ‖x‖b |||A|||= ‖A‖‖x‖b,

since ∥∥∥∥ x

‖x‖b

∥∥∥∥b

= 1.

A.2 An inequality involving norms and integrals

Lemma A.2.1. Suppose f : Rn → Rn is a continuous function. Then∥∥∥∥∫ b

a

f(x)dx

∥∥∥∥ ≤ ∫ b

a

‖f(x)‖ dx.

Proof. First, note that we have∥∥∥∥∫ b

a

f(x)dx

∥∥∥∥ =

(∥∥∥∥∫ b

a

f1(x)dx

∥∥∥∥ , . . . ,∥∥∥∥∫ b

a

fn(x)dx

∥∥∥∥) .For a given component function fi, i = 1, . . . , n, using the definition of the Riemann integral,∫ b

a

fi(x)dx = limk→∞

k∑j=1

fi(x∗j)∆xj,

where x∗j is the sample point in the interval [xj−1, xj] with width ∆xj. Therefore,∥∥∥∥∫ b

a

fi(x)dx

∥∥∥∥ =

∥∥∥∥∥ limk→∞

k∑j=1

fi(x∗j)∆x

∥∥∥∥∥ = limk→∞

∥∥∥∥∥k∑

j=1

fi(x∗j)∆x

∥∥∥∥∥ ,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 simplyconvergent) to a mapping f : X → Y , if the sequence fn(x) converges to f(x) for each x inX. 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) .

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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 amapping 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.

This is usually denoted by fnu→ f . In other words,

(fnu→ 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 , thenf is continuous.

A.4 Asymptotic Notations

Let n be a integer variable which tends to ∞ and let x be a continuous variable tendingto 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 ⊆ Xand F ⊆ Y . A function f : E → F is uniformly continuous on the set E ⊂ X if for everyε > 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 intervalI is said to be equicontinuous on I if, given any ε > 0, there exists a δε > 0, independentof f ∈ F and also t, t ∈ I such that

‖f(t)− f(t)‖ < ε whenever |t− t| < δε



An interpretation of equicontinuity is that a sequence of functions is equicontinuous ifall 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) forevery x ∈ X, then the function f is continuous.

Lemma A.5 (Ascoli). On a bounded interval I, let F = {f} be an infinite, uniformlybounded, 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 spaceX with values in Rn. If a sequence {fn} in C(X) is bounded and equicontinuous, then it hasa uniformly convergent subsequence.

A.6 Lipschitz function

Definition A.6.1 (Lipschitz function). A map f : U ⊂ R × Rn → Rn is Lipschitz in x ifthere exists a real number L such that for all (t, x1) ∈ U and (t, x2) ∈ U ,

‖f(t, x1)− f(t, x2)‖ ≤ L‖x1 − x2‖.

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 locallyLipschitz in x if, for all (t0, x0) ∈ U , there exists a neighborhood V ⊂ U of (t0, x0) and a realnumber L, such that for all (t, x1), (t, x2) ∈ V ,

‖f(t, x1)− f(t, x2)‖ ≤ L‖x1 − x2‖.

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 every-where the same Lipschitz constant L. Another definition of a locally Lipschitz function is asfollows.

Definition A.6.3. A function f : U ⊂ Rn+1 → Rn is locally Lipschitz continuous if forevery compact set V ⊂ U , the number

L = sup(t,x) 6=(t,y)∈V

‖f(t, x)− f(t, y)‖‖x− y‖

is finite, with L depending on V .

Property A.6.4. Let f(t, x) be a function. The following properties hold.

A.7. Gronwall’s lemmaFund. Theory ODE Lecture Notes – J. Arino

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 islocally Lipschitz on D.

Proof. i) Suppose that f is Lipschitz, i.e., there exists L > 0 such that ‖f(t, x1)−f(t, x2)‖ ≤L‖x1 − x2‖. Recall that f is uniformly continuous if for every ε > 0, there exists δ > 0 suchthat for all x1, x2, ‖x1 − x2‖ < δ implies that ‖f(t, x1) − f(t, x2)‖ < ε. So, given an ε > 0,choose δ < ε/L. Then ‖x1 − x2‖ < δ implies that

‖f(t, x1)− f(t, x2)‖ < L‖x1 − x2‖ < 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/ lnx on (0, 12], f(0) = 0.

iii) Notice that ∂f/∂x continuous on D implies that ‖∂f/∂x‖ is continuous on (thebounded closed domain) D, and thus ‖∂f/∂x‖ is bounded on D. Let

L = sup(t,x)∈D

∥∥∥∥∂f(t, x)

∂x

∥∥∥∥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∂x

(t, η). As η ∈ U , it follows that ‖∂f∂x

(t, η)‖ ≤ L, and thus ‖f(t, x2) −f(t, x1)‖ ≤ L‖x2 − x1‖.

A.7 Gronwall’s lemma

The name of Gronwall is associated to a certain number of inequalities. We give a few ofthem here. We prove the most simple one (as it is an easy proof to remember), as well asthe 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

0 ≤ g(t) ≤ K + L

∫ t

t0

g(s)ds.

Then0 ≤ g(t) ≤ KeL(t−t0),

for t0 ≤ t ≤ t1.



Proof. Let v(t) =∫ t

t0g(s)ds. Then v′(t) = g(t), which implies that the assumption on g can

be written

0 ≤ v′(t) ≤ K + Lv(t).

The right inequality is a linear differential inequality, with integrating factor exp(−∫ t

t0Lds)

Also, v(t0) = 0. Hence,d

dt

(e−L(t−t0)v(t)

)≤ Ke−L(t−t0)

and therefore,

e−L(t−t0)v(t) ≤ K

L

(1− e−L(t−t0)

).

Thus Lv(t) ≤ K(eL(t−t0) − 1

), and g(t) ≤ K + Lv(t) ≤ KeL(t−t0).

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 functionsdefined on the interval [a, b]. Moreover, suppose that either K is a constant function, or Kis differentiable on [a, b] with positive derivative K ′. If, for all t ∈ [a, b]

g(t) ≤ K(t) +

∫ t

a

L(t)g(s)ds,

then

g(t) ≤ K(t) exp

(∫ t

a

L(s)ds

),

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], thenfor every nonnegative regulated function w in [0, c] satisfying the inequality

w(t) ≤ ϕ(t) +

∫ t

0

ψ(s)w(s)ds,

we have, in [0, c],

w(t) ≤ ϕ(t) +

∫ t

0

ϕ(s)ψ(s) exp

(∫ t

s

ψ(ξ)dξ

)ds. (A.1)

Before proving the result, let us recall that a function f from an interval I ⊂ R to aBanach space F is regulated if it admits in each point in I a left limit and a right limit. Inparticular, every continuous mapping from I ⊂ R to a Banach space is regulated, as well asmonotone maps from I to R; see, e.g., [8, Section 7.6].

A.8. Fixed point theoremsFund. Theory ODE Lecture Notes – J. Arino

93

Proof. Let y(t) =∫ t

0ψ(s)w(s)ds; y is continuous, and since w(t) ≤ ϕ(t) +

∫ t

0ψ(s)w(s)ds, it

follows that, except maybe at a denumerable number of points of [0, c], we have

y′(t)− ψ(t)y(t) ≤ ϕ(t)ψ(t) (A.2)

from [8, Section 8.7]. Let z(t) = y(t) exp(−∫ t

0ψ(s)ds). Then (A.2) is equivalent to

z′(t) ≤ ϕ(t)ψ(t) exp

(−∫ t

0

ψ(s)ds

).

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(t) ≤∫ t

0

ϕ(s)ψ(s) exp

(−∫ s

0

ψ(ξ)dξ

)ds,

whence by definition

y(t) ≤∫ t

0

ϕ(s)ψ(s) exp

(∫ t

s

ψ(ξ)dξ

)ds,

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. Amapping 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 con-traction 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 acontraction,

d(x2, x1) = d(f(x1), f(x0))

≤ Kd(x1, x0).

Similarly,

d(x3, x2) = d(f(x2), f(x1))

≤ Kd(x2, x1)

≤ K2d(x1, x0).



Iterating,d(xn+1, xn) ≤ Knd(x1, x0).

Therefore,

d(xn+p, xn) ≤ d(xn+p, xn+p−1) + d(xn+p−1, xn+p−2) + · · ·+ d(xn+1, xn)

≤ (Kp−1 +Kp−2 + · · ·+K + 1)Knd(x1, x0)

≤ Kn

1−Kd(x1, x0).

Therefore d(xn+p, xn) tends to 0 as n → ∞, so {xn} is a Cauchy sequence. Since X is acomplete space, it follows that {xn} admits a limit `. As limn→∞ xn = `, it follows fromcontinuity 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. Closednessof 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 con-traction 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 fk of f , is acontraction. 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

J =

J0 0 0 · · · 00 J1 0 · · · 0· · · · ·0 0 0 · · · Js

where J0 is a diagonal matrix with diagonal λ1, . . . , λn, and, for i = 1, . . . , s,

Ji =

λq+i 1 0 0 · · · 0 00 λq+i 1 0 · · · 0 0· · · · · · · · ·0 0 0 0 · · · λq+i 10 0 0 0 · · · 0 λq+i

A.9. Jordan normal formFund. Theory ODE Lecture Notes – J. Arino

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 issimilar to the diagonal matrix

J =

λ1 0 0 · · · 00 λ2 0 · · · 0· · · · ·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 withmultiplicities n1, . . . , nm, respectively.

ii) Compute n1 linearly independent generalized eigenvectors of A associated with λ1 asfollows. Compute

(A− λ1En)i

for i = 1, 2, . . . until the rank of (A−λ1En)k is equal to the rank of (A−λ1En)k+1. Finda generalized eigenvector of rank k, say u. Define ui = (A−λ1En)k−1u, for i = 1, . . . , k.If k = n1, proceed to step 3. If k < n1, find another linearly independent generalizedeigenvector with rank k. If this is not possible, try k−1, and so forth, until n1 linearlyindependent generalized eigenvectors are determined. Note that if ρ(A − λ1En) = 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 = λ1u1,

Au2 = u1 + λ1u2,

...

Auk = uk−1 + λ1uk

Thus in the new basis, A has the representation

J =

α1 1. . . 1

α1

α2 1. . . 1

α2

α3 1. . . 1

α3

. . .



where each chain of generalized eigenvectors generates a Jordan block whose orderequals the length of the chain.

v) The similarity transformation which yields J = Q−1AQ is given byQ = [u1, . . . , uk, . . .].

A.10 Matrix exponentials

Let A ∈Mn(K). The exponential of A is defined by

eA =∞∑

n=0

An

n!(A.3)

We have eA ∈ Mn(K). Also∣∣∣∣∣∣ 1

n!An∣∣∣∣∣∣ ≤ 1

n!|||A|||n, so that the series

∑1n!An is absolutely

convergent in Mn(K). Thus eA 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 = eAeB.

iii) eA = I +∑∞

n=11n!An.

iv) e0 = I.

v) eA is invertible with inverse e−A.

vi) eAt is differentiable, and ddteAt = AeAt.

vii) If P and T are linear transformations on Kn, and S = PTP−1, then eS = PeTP−1.

Proof. v) For any matrix A, we have A(−A) = −AA = (−A)A, so A and −A commute.Therefore, eAe−A = eA−A = e0 = I. Therefore, for any A, eA is invertible with inversee−A.

There are several shortcuts to computing the exponential of a matrix A:

1) If A is nilpotent, that is, if there exists q ∈ N such that Aq = 0, then eA =∑q

k=1Ak/k!.

A nilpotent matrix has several interesting properties. A is nilpotent if and only if allits 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 beexploited to simplify the computation of eA.

3) Any matrix A can be written in the form A = D+N , where D is diagonalizable, N isnilpotent, and D and N commute. Therefore, eA = eD+N = eDeN .

4) Other cases require the use of the Jordan normal form (explained below).

A.11. Matrix logarithmsFund. Theory ODE Lecture Notes – J. Arino

97

Use of the Jordan form to compute the exponential of a matrix. Suppose thatJ = P−1AP is the Jordan form of the matrix A. For a block diagonal matrix

B =

B1 0. . .

0 Bs

,

we have, for k = 0, 1, . . .,

Bk =

Bk1 0

. . .

0 Bks

,

Therefore, for t ∈ R,

eJt =

eJ0t 0

. . .

0 eJst

,

with

eJ0 =

eλ1t 0

. . .

0 eλkt

.

Now, since Ji = λk+iIi +Ni, with Ni a nilpotent matrix, and since Ii and Ni commute, thereholds

eJit = eλk+iteNit.

For k ≥ Ni, Nki = 0, so

etJi =

1 t · · · tni−1

(ni−1)!

0 1 · · · tni−2

(ni−2)!

0 · · · 1

.

A.11 Matrix logarithms

Theorem A.11.1. Suppose that M is a nonsingular n× n matrix. Then there is an n× nmatrix B (possibly complex), such that eB = M . If, additionally, M ∈ Mn(R), then thereis B ∈Mn(R) such that eB = M2.

Let u ∈ C, we have, for |u| < 1,

1

1 + u= 1− u+ u2 − u3 + · · ·



and, for |u| < 1,

ln(1 + u) = t− t2

2+t3

3− t4

4+ · · ·

=∞∑

k=1

(−1)k+1 tk

k

For any z ∈ C,

exp(z) = 1 + z +z2

2!+z3

3!+ · · ·

=∞∑

n=1

zn

n!

Therefore, for |u| < 1, u ∈ C,

1 + u = exp(ln(1 + u))

=∞∑

n=1

1

n!

[∞∑

k=1

(−1)k+1uk

k

]n

Suppose that

B = (lnλ)I +∞∑

k=1

(−1)k+1

k

1

λkZk

where

Z =

0 1 · · · 0

. . . . . .

0 0 10 0

is an m×m-matrix. Since Z is nilpotent (ZN = 0 for all N ≥ m), the above sum is finite.Observe that

exp(B) = exp((lnλ)I) exp

(∞∑

k=1

(−1)k+1

k

1

λkZk

)

= λ exp

(∞∑

k=1

(−1)k+1

k

(Z

λ

)k)

= λ

(I +

Z

λ

)= λI + Z

= J

We say ln J = B.

A.12. Spectral theoremsFund. Theory ODE Lecture Notes – J. Arino

99

A.12 Spectral theorems

When studying systems of differential equations, it is very important to be able to computethe eigenvalues of a matrix, for instance in order to study the local asymptotic stability ofan equilibrium point. This can be a very difficult problem, that often becomes intractableeven for systems with low dimensionality. However, even if it is not possible to compute anexplicit solution, it is often possible to use spectrum localization theorems. We here givetwo 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 iscalled its spectrum, and is denoted Sp(A).

Theorem A.12.1 (Routh-Hurwitz). If n = 2, suppose that detA > 0 and trA < 0. ThenA has only eigenvalues with negative real part.

Theorem A.12.2 (Gershgorin). Let

Rj =n∑

i=1,i6=j

|aij|

Let λ ∈ Sp(A). Then

λ ∈⋃j

{|λ− ajj| ≤ Rj}

Gershgorin’s theorem is extremely helpful in certain cases. Suppose that A is stictlydiagonally 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 ofnegative entries on the diagonal of A, and conversely for eigenvalues with positive real partsand the number of positive entries on the diagonal of A.

Appendix B

Problem sheets

ContentsHomework 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. Arino103

McMaster University – Math4G03/6G03

Fall 2003Homework Sheet 1

Exercise 1.1 – We consider here the equation

x′(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

x(t) = e−αtx(0) + e−αt∫ t

0eαsf(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 theinequality

|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 thelimit 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

x′ = −αx+ ln(1 + x2

)tend to zero when t→∞. ◦


B. Problem sheets

Exercise 1.2 – Let f : [0,+∞) → R, f ∈ C1, and a ∈ R. We consider the initial valueproblems

x′(t) + ax(t) = f(t), t ≥ 0

x(0) = 0(B.5)

andx′(t) + ax(t) = f ′(t), t ≥ 0

x(0) = 0(B.6)

As these equations are linear, the initial value problems (B.5) and (B.6) admit unique solu-tions. We denote φ the solution to (B.5) and ψ the solution to (B.6). Find a necessary andsufficient condition on f such that φ′ = ψ.

[Hint: Use a variation of constants formula]. ◦

Exercise 1.3 – Let f : Rn → Rn be continuous. Consider the differential equation

x′(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. Supposethat 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,

‖x(t)− xα‖ ≤Mα|t− α|

with M = supt∈[0,α) ‖f(x(t))‖.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].

◦



Fall 2003Homework 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 onR+. Furthermore,

x′(s) = −αx(s) + f(x(s)), for all s ∈ R+

⇔ eαsx′(s) = −αeαs + eαsf(x(s))

⇒∫ t

0

eαsx′(s)ds = −α∫ t

0

eαsx(s)ds+

∫ t

0

eαsf(x(s))ds

[eαsx(s)]t0 − α

∫ t

0

eαsx(s)ds = −α∫ t

0

eαsx(s)ds+

∫ t

0

eαsf(x(s))ds

eαtx(t)− x(0) =

∫ t

0

eαsf(x(s))ds, for all t ∈ R+

x(t) = e−αtx(0) + e−αt

∫ t

0

eαsf(x(s))ds, for all t ∈ R+

Let us now prove the converse, i.e., that if a function x satisfies (B.2), then it is a solutionto the IVP (B.1). Since x and f are continuous on R+, t 7→ eαtf(x(t)) is continuous on R+.This implies that

t 7→∫ t

0

eαsf(x(s))ds

is differentiable on R+, and, differentiating the expression for x(t) as given in (B.2),

x′(t) = −αe−αtx(0)− αe−αt

∫ t

0

eαsf(x(s))ds+ e−αteαtf(x(t))

⇒ x′(t) = −α[e−αtx(0) + e−αt

∫ t

0

eαsf(x(s))ds

]︸︷︷︸

x(t)

+f(x(t))

And thusx′(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

x(t) = e−αtx(0) + e−αt

∫ t

0

eαsf(x(s))ds


B. Problem sheets

This implies that

|x(t)| ≤ e−αt|x(0)|+ e−αt

∫ t

0

eαs|f(x(s))|ds

From (B.3) and multiplying both sides by eαt,

eαt|x(t)| ≤ |x(0)|+∫ t

0

keαs|x(s)|ds

We use Gronwall’s Lemma (Lemma A.2) as follows,

eαt|x(t)|︸︷︷︸g(t)

≤ |x(0)|︸︷︷︸K(t)

+

∫ t

0

k︸︷︷︸L(t)

eαs|x(s)|︸︷︷︸g(s)

ds

Thus,

eαt|x(t)| ≤ |x(0)| exp

(∫ t

0

kds

)≤ |x(0)|ekt

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 ofthe 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 existst0 ∈ 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 alln ∈ N−{0}, there exists un ∈ [0, 1

n] 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] ⊂ Aand t1 ∈ A. Let β be the smallest such η, that is, β = inf{t ∈ (t0, t1); t 6∈ A}. Note that βcan also be defined as β = sup{t ∈ (t0, t1); t ∈ A}.

Thusβ = 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 impliesthat β < t1.


Hence A is an interval. We now want to show that it is an unbounded interval. Assumeit 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 existst > c such that t ∈ A, which is a contradiction. Therefore, A = [0,∞), and we can concludethat ∀t ∈ R+, |x(t)| ≤ a.

Another proof, contributed by Guihong Fan, proceeds by contradiction, using the factthe (B.1) is an autonomous scalar equation. Notice that equation (B.1) can be written inthe form x′ = 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 scalardifferential 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 thefollowing holds.

i) x(t0) = a and x′(t0) > 0,

ii) x(t0) = −a and x′(t0) < 0.

Let us treat case i). From (B.3), it follows that |f(x(t0))| = |f(a)| ≤ ka. Therefore, using(B.1),

x′(t0) = −αx(t0) + f(x(t0))

= −αa+ f(a)

≤ −αa+ ka

≤ −(α− k)a

< 0

since α > k. This is a contradiction with x′(t0) > 0. Case ii) is treated similarly, and thusit 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 PartI, 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 showthat x ≡ 0 is a solution of (B.1). Condition (B.3) applied to 0 states that |0| < a implies|f(0)| ≤ k|0| = 0.


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 x′ = −αx + ln(1 + x2) tend to zero ast→∞, when α > 1. Indeed, let f(u) = ln(1 + u2). We have

|f ′(u)| = 2|u|1 + u2

≤ 1

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 theexercise are satisfied and all solutions of the equation tend to zero. ◦

Solution – Exercise 2 – Using a variation of constants formula, we obtain

φ(t) = e−atc1 +

∫ t

0

e−a(t−s)f(s)ds, t ≥ 0

and

ψ(t) = e−atc2 +

∫ t

0

e−a(t−s)f ′(s)ds, t ≥ 0

Let us show this for the solution of (B.5), the solution of (B.6) is obtained using exactly thesame method. The solution to a linear differential equation consists of the general solutionto the homogeneous equation together with a particular solution to the nonhomogeneousequation. Here, the homogeneous equation is

x′ = −ax

and basic integration yields the general solution x(t) = c1e−at. To obtain a particular solution

to the nonhomogeneous equatiob (B.5), we use a variation of constants formula: assume thatthe 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

x′ = c′e−at − ace−at

Substituting both this expression and x = ce−at into (B.5), we get

c′e−at − ace−at + ace−at = f(t), t ≥ 0

and hencec′ = eatf(t)

Integrating both sides of this expression gives

c(t) =

∫ t

0

easf(s)ds


so that a particular solution to (B.5) is given by

x(t) = c(t)e−at =

(∫ t

0

easf(s)ds

)e−at =

∫ t

0

e−a(t−s)f(s)ds

Summing the general solution to the homogeneous equation with this last expression givesthe desired result.

Using the initial conditions yields

φ(0) = c1 = 0

ψ(0) = c2 = 0

Hence the system under consideration is

φ(t) =

∫ t

0

e−a(t−s)f(s)ds, t ≥ 0

ψ(t) =

∫ t

0

e−a(t−s)f ′(s)ds, t ≥ 0

Recall that if g(t) =∫ t

t0h(s, t)ds, then for all t0,

g′(t) = h(t, t) +

∫ t

t0

∂h

∂t(s, t)ds

This implies that

φ′(t) = f(t)− a

∫ t

0

e−a(t−s)f(s)ds

It follows that

φ′(t) = ψ(t) ⇔ f(t)− a

∫ t

0

e−a(t−s)f(s)ds =

∫ t

0

e−a(t−s)f ′(s)ds

⇔ f(t)− a

∫ t

0

e−a(t−s)f(s)ds =[e−a(t−s)f(s)

]s=t

s=0− a

∫ t

0

e−a(t−s)f(s)ds

⇔ f(t) =[e−a(t−s)f(s)

]s=t

s=0

⇔ f(t) = f(t)− e−atf(0)

⇔ f(0) = 0

◦

Solution – Exercise 3 – 1 – a) For a vector-valued function, there is no mean valuetheorem with an equal sign. But the following holds (see, e.g., [3, p. 44], [9, p. 209] orRudin, p.113).


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 f ′r(x) for all x ∈ (a, b), and that ‖f ′r(x)‖ ≤ k, wherek ≥ 0 is a constant. Then

‖f(b)− f(a)‖ ≤ k(b− a)

and more generally, for all x1, x2 ∈ [a, b],

‖f(x2)− f(x1)‖ ≤ k|x2 − x1|

Let us denote M = supt∈[0,α) ‖f(t, x)‖, and let n, p > N , where N is sufficiently large

that α− 1n+p

∈ [0, α) and α− 1n∈ [0, α). Using Theorem B.1.3, we obtain that

‖x(α− 1

n+ p)− x(α− 1

n)‖ ≤M | 1

n− 1

n+ p|

‖zα,n+p − zα,n‖ ≤M | 1n− 1

n+ p|

So the sequence (zα,n)n∈N∗ is a Cauchy sequence.1 – b) For all t ∈ [0, α),

‖x(t)− x(α− 1

n)‖ ≤

∣∣∣∣∣∫ t

α− 1n

‖f(x(s))‖ds

∣∣∣∣∣‖x(t)− zα,n‖ ≤Mα|t− α+

1

n| (B.8)

Since the sequence (zα,n)n∈N∗ is a Cauchy sequence, there exists xα = limn→∞(zα,n). Thustaking n→∞ in (B.8) gives

‖x(t)− xα‖ ≤Mα|t− α|1 – c) According to 1.b), we have

limt→α,t<α

‖x(t)− xα‖ ≤ 0

Hencelim

t→α,t<αx(t) = xα

2) Let

z(t) =

{x(t) if t ∈ [0, α)xα 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 theintegral equation

z(t) = z(t0) +

∫ t

t0

f(z(s))ds

for all t ∈ [0, α], and with an arbitrary t0 ∈ [0, α]. We know by construction that z iscontinuous (since limt→∞ x(t) = xα).


Let t ∈ [0, α),

z(t) = x(t) = x(t0) +

∫ t

t0

f(x(s))ds

because x is a solution. For t = α,

z(α) = xα

= limt→α,t<α

x(t)

= limt→α,t<α

x(t0) +

∫ t

t0

f(x(s))ds

= limt→α,t<α

z(t0) +

∫ t

t0

f(z(s))ds

since for t < α, we have z(t) = x(t).Furthermore, t 7→ f(z(t)) is bounded on [0, α), which implies that∫ α

t0

f(z(s))ds = limt→α

∫ t

t0

f(z(s))ds

So

z(α) = z(t0) +

∫ α

t0

f(z(s))ds

So z is solution to (B.7) on [0, α]. ◦




Exercise 2.1 – Consider the system(x1

x2

)′=

(a bc d

)(x1

x2

)where a, b, c, d ∈ R are constants such that ad − bc = 0. Discuss all possible behaviours ofthe 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 limt→∞

e−αteAt = 0.

◦

Exercise 2.3 – Let X(t) be a fundamental matrix for the system x′ = A(t)x, where A(t)is an n× n matrix with continuous entries on R. What conditions on A(t) and C guaranteethat CX(t) is a fundamental matrix, where C is a constant matrix. ◦

Exercise 2.4 – Consider the system

x′ = 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

x′ = 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) dimP(ω) = 0.

◦


B. Problem sheets



Solution – Exercise 1 – The characteristic polynomial of the matrix

A =

(a bc d

)is (a− λ)(d− λ)− bc = λ2 − (a+ d)λ+ ad− bc = λ2 − (a+ d)λ since ad− bc = 0. Thus Ahas the eigenvalues 0 and a+ d. Hence solutions are of the form

x1 = c1

x2 = c2e(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 thex1 axis diverges to ±∞ parallely to the x2 axis.

◦

Solution – Exercise 2 – 1) We have eA =∑∞

k=0Ak/k!. Taking the norm,

‖eA‖ = ‖∞∑

k=0

Ak/k!‖,

whence, by the triangle inequality, ‖eA‖ ≤∑∞

k=0 ‖Ak/k!‖ = e‖A‖.2) Let J be the Jordan form of A, i.e., there exists P nonsingular such that P−1AP =

J , where J has the form diag(Bj)j=1,...,p, with Bj the Jordan block corresponding to theeigenvalue λj of multiplicity mj. Then, since A and J are similar,

det eA = det eJ

= det(eB1)m1 . . . det(eBp)mp

= eλ1m1 · · · eλpmp

= exp(

p∑k=1

λkmk)

= trA


3) We can writelimt→∞

e−αteAt = limt→∞

e(A−αI)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 allµ ∈ 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 thatlimt→∞ e(A−αI)t = 0. ◦

Solution – Exercise 3 – We have

(CX(t))′ = CX ′(t)

= CA(t)X(t)

For CX(t) to be a fundamental matrix for x′ = A(t)x requires that (CX(t))′ = 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 bea fundamental matrix. So, to conclude, if X(t) is a fundamental matrix for the systemx′ = A(t)x, then CX(t) is a fundamental matrix for x′ = A(t)x if C is nonsingular andcommutes 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, α1x1 + α2x2 ∈ Rn. Now assume that,moreover, x1, x2 ∈ P(ω). Then,

(α1x1 + α2x2)′ = α1x

′1 + α2x

′2

= α1A(t)x1 + α2A(t)x2

= A(t)(α1x1 + α2x2)

and therefore, α1x1 + α2x2 ∈ 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 −1X−1(t), and X−1(t + ω) −X−1(t) =(V −1 − I)X−1(t).

Suppose that dimP(ω) = 0. Then 1 is not an eigenvalue of V . This implies that 1 isneither an eigenvalue of V −1; in turn, 0 is not an eigenvalue of V −1 − I. This means thatV −1− I is nonsingular, and since X(t) is nonsingular, (V −1− I)X−1(t) is nonsingular. Thuswe conclude that if dimP(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 dimP(ω) = 0 and that x1, x2 are two solutions to (B.10). Then

x′1 = A(t)x1 + f(t) and x′2 = A(t)x2 + f(t). Therefore,

(x1 − x2)′ = A(t)x1 + f(t)− A(t)x2 − f(t) = A(t)(x1 − x2)


B. Problem sheets

which implies that x1−x2 is a solution to (B.9), and therefore, dimP(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 dimP(ω) 6=0, i.e., there exists y, non trivial ω-periodic solution to (B.9). Then

(x+ y)′ = 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 solutionto (B.10). ◦

Theorem B.2.4. Consider the system

x′ = 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 thepreviously obtained forms,

X(t)c = X(t+ ω)c

X(t)c = X(t)V c

c = V c

Hence c is an eigenvector of V with corresponding eigenvalue (Floquet multiplier) λ = 1.




Exercise 3.1 – Compute the solution of the differential equation

x′(t) = x(t)− y(t)− t2

y′(t) = x(t) + 3y(t) + 2t(B.11)

◦

Exercise 3.2 – Consider the initial value problem

x′(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 wherethe 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(∫ t

t0A(s)ds

)is a solution of the matrix initial value problem

M ′(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 ′(t) = limh→0(M(t+ h)−M(t))/h]

ii) Deduce that, when (B.13) holds,

R(t, t0) = exp

(∫ t

t0

A(s)ds

)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

R(t, t0) = exp

(∫ t

t0

f(s)dsU

)exp

(∫ t

t0

g(s)dsV

)(B.14)


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

x′ =1

tx+ ty

y′ = 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,

R(t, t0) 6= exp

(∫ t

t0

A(s)ds

)with

A(t) =

(1t

t0 1

)◦


McMaster University – 4G03/6G03

Fall 2003Solutions – Homework Sheet 3

Solution – Exercise 3.1 – Let ξ(t) = (x(t), y(t))T ,

A =

(1 −11 3

)and B(t) = (−t2, 2t)T . The system (B.11) can then be written as

ξ′ = Aξ +B(t)

This is a nonhomogeneous linear system, so we use the variation of constants,

ξ(t) = e(t−t0)Aξ0 +

∫ t

t0

e(t−s)AB(s)ds

where ξ0 = ξ(t0). Let us assume for simplicity that t0 = 0.Eigenvalues of A are (λ− 2)2, with associated subspace

ker(A− 2I) =

{(xy

);

(−1 −11 1

)(xy

)=

(00

)}=

{(xy

); x+ y = 0

}Thus dim ker(A − 2I) = 1 6= 2, and A is not diagonalisable. Let us compute the Jordancanonical form of A. There exists P nonsingular such that

P−1AP =

(2 α0 2

)where α is a constant that has to be determined.

P−1AP =

(2 10 2

)P =

(1 −1−1 0

), P−1 =

(0 −1−1 −1

)Then

eAt = Pe(P−1AP )tP−1

withe(P

−1AP )t = eP−1(At)P


B. Problem sheets

We have P−1AP = 2I +N , where (0 10 0

)Therefore, e(P

−1AP )t = e(2It+Nt) = e2teNt. Now, N is nilpotent (N2 = 0), so eNt =∑∞n=0

tn

n!Nn = I +Nt. As a consequence,

e(P−1AP )t = e2t(I +Nt)

= e2t

((1 00 1

)+

(0 10 0

))=

(e2t te2t

0 e2t

)Thus

eAt = P

(e2t te2t

0 e2t

)P−1

=

((1− t)e2t −te2t

te2t (1 + t)e2t

)= e2t

(1− t −tt 1 + t

)We still have to compute

∫ t

0e−AsB(s)ds. We have

e−AsB(s) = e−2s

(1 + s s−s 1− s

)(−s2

2s

)= e−2s

(−s3 − s2 + 2s2

s3 + 2s− 2s2

)= e−2s

(s2 − s3

s3 − 2s2 + 2s

)Let I1(t) =

∫ t

0e−2ssds, I2(t) =

∫ t

0e−2ss2ds and I3(t) =

∫ t

0e−2ss3ds. Then∫ t

0

e−AsB(s)ds =

(I1(t)− I3(t)

I3(t)− 2I2(t) + 2I1(t)

)Evaluating the integrals, we obtain∫ t

0

e−AsB(s)ds =

(e−2t

(14t2 + 1

4t+ 1

8+ 1

2t3)− 1

8

e−2t(

14t2 − 3

4t− 3

8− 1

2t3)

+ 38

)As a conclusion,

ξ(t) = e2t

(1− t −tt 1 + t

)ξ0 +

(1− t −tt 1 + t

)(e−2t

(14t2 + 1

4t+ 1

8+ 1

2t3)− 1

8

e−2t(

14t2 − 3

4t− 3

8− 1

2t3)

+ 38

)=

(e−2tx0 − e−2ttx0 − e−2tty0 + 1

2e−2tt+ 1

8e−2t − 1

8− t

4+ 3

4e−2tt2

e−2ttx0 + e−2ty0 + e−2tty0 − 14e−2tt2 − e−2tt+ t

4− 3

8e−2t + 3

8

)is the solution of (B.11) going through (x0, y0) at time 0. ◦


Solution – Exercise 3.2 – 1) We have

1

h(M(t+ h)−M(t)) =

1

h

(exp

(∫ t+h

t0

A(s)ds

)− exp

(∫ t

t0

A(s)ds

))Since, for all s1, s2 ∈ I, A(s1)A(s2) = A(s2)A(s1), we have(∫ s1

t0

A(u)du

)(∫ s2

t0

A(u)du

)=

(∫ s2

t0

A(u)du

)(∫ s1

t0

A(u)du

)It follows that

1

h(M(t+ h)−M(t)) =

1

h

(exp

(∫ t+h

t

A(s)ds

)exp

(∫ t

t0

A(s)ds

)− exp

(∫ t

t0

A(s)ds

))=

1

hM(t)

(exp

(∫ t+h

t

A(s)ds

)− I

)=

(1

hexp

(∫ t+h

t

A(s)ds

)− 1

hI

)M(t)

The second term in this equality tends to zero as h→ 0, and thus

limh→0

1

h(M(t+ h)−M(t)) = A(t)M(t)

therefore M ′(t) = A(t)M(t), hence the result.2) The implication is trivial.3) If U and V commute and that A(t) = Uf(t) + V g(t), then

A(s)A(t) = (Uf(s) + V g(s))(Uf(t) + V g(t))

= U2f(s)f(t) + UV f(s)g(t) + V Ug(s)f(t) + V 2g(s)g(t)

= U2f(t)f(s) + V Ug(t)f(s) + UV f(t)g(s) + V 2g(t)g(s)

= (Uf(t) + V g(t))Uf(s) + (Uf(t) + V g(t))V g(s)

= (Uf(t) + V g(t))(Uf(s) + V g(s))

that is, A(s) and A(t) commute for all t, s. Then

R(t, t0) = exp

(∫ t

t0

A(s)ds

)= exp

(∫ t

t0

Uf(s) + V g(s)ds

)= exp

(∫ t

t0

f(s)dsU +

∫ t

t0

g(s)dsV

)= exp

(∫ t

t0

f(s)dsU

)exp

(∫ t

t0

g(s)dsV

)◦


B. Problem sheets

Solution – Exercise 3.3 – Writing

A(t) = a(t)

(1 00 1

)+ b(t)

(0 −11 0

)= a(t)I + b(t)V,

it is obvious the Theorem B.3.5 can be used here, since I commutes with all matrices. Thus,

R(t, t0) = exp

(∫ t

t0

a(s)dsI

)exp

(∫ t

t0

b(s)dsV

)Let α(t) =

∫ t

t0a(s)ds and β(t) =

∫ t

t0b(s)ds. Then R(t, t0) = eα(t)Ieβ(t)V = eα(t)Ieβ(t)V . Now

notice that V 2 = I, V 3 = −V , etc., so that we can write that

V n =

{(−1)pI if n = 2p,(−1)pV if n = 2p+ 1.

This implies that

eβ(t)V =∞∑

n=0

1

n!β(t)nV n

=

(∞∑

p=0

(−1)p

(2p)!β(t)2p

)I +

(∞∑

p=0

(−1)p

(2p+ 1)!β(t)2p+1

)V

= cos(β(t))I + sin(β(t))V

Thus

R(t, t0) = eα(t)(cos(β(t))I + sin(β(t))V )

=

(eα(t) 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′ equation in (B.16) does not involvex. Therefore, we can solve it directly, giving y(t) = Cet, with C ∈ R. Substituting this intothe equation for x′, we have

x′ =1

tx(t) + tCet

To solve this nonhomogeneous first-order scalar equation, we start by solving the homoge-neous part, x′ = 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 nonhomoge-neous problem. We use the ansatz x(t) = K(t)t, which, when differentiated and substitutedinto the nonhomogeneous equation, gives K ′(t) = Cet, and hence, K(t) = Cet is a particularsolution, giving the general solution x(t) = Kt+ Cet.


Let t0 6= 0 (to avoid problems with 1/t), and suppose x(t0) = x0, y(t0) = y0. Thenx0 = Kt0 + Ct0e

t0 and y0 = Cet0 . It follows that K = x0/t0 − y0 and C = e−t0y0, and thesolution to the equation going through the point (x0, y0) at time t0 is given by

ξ(t) =

(x(t)y(t)

)=

((x0

t0− y0)t+ e−t0y0te

t

e−t0y0et

)=

(x0

t0t+ y0t(e

t−t0 − 1)

y0et−t0

).

2) The solution to the IVP

ξ′ = 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

R(t, t0) =

(tt0

−t+ tet−t0

0 et−t0

)3) First of all, notice that A(t) and A(s) do not commute. Let us compute B(t) =∫ t

t0A(s)ds.

B(t) =

(∫ t

t0

1sds

∫ t

t0sds

0∫ t

t0ds

)

=

(ln t

t012(t2 − t20)

0 t− t0

)which, for convenience, we denote

B(t) =

(α β0 γ

)Eigenvalues of B(t) are α and γ. As R(t0, t0) = B(t0) = I, we are concerned with findinga t 6= t0 such that B(t) is diagonalizable. If a t exists such that α 6= γ, then B(t) isdiagonalizable, i.e., there exists P nonsingular such that

P−1B(t)P = D =

(α 00 β

)Then

eB(t) = P

(eα 00 eβ

)P−1

We find

P =

(1 β0 γ − α

), P−1 =

1

γ − α

(γ − α −β

0 1

)


B. Problem sheets

Thus, after a few computations,

eB(t) =

(eα β

γ−α(eγ − eα)

0 eγ

)=

(tt0

∆

0 et−t0

)The element ∆ in this matrix is the only one different from the elements in R(t, t0). We have

∆ =t2 − t20

2(t− t0 − ln tt0

)

(et−t0 − t

t0

)6= t(et−t0 − 1)

◦




Exercise 3.1 – A differential equation of the form

x′ = 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 valueproblem for (B.17) takes the form

x′ = 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

x′ = 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], dependingon the nature of φ0 and f .

iii) Suppose that φ0 ∈ C0([t0 − ω, t0]). Discuss the regularity of the solution to (B.18) onthe interval [t0 + kω, t0 + (k + 1)ω], k ∈ N.

◦

Exercise 3.2 – Consider the delay initial value problem

x′(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 11 0

), A2 =

(1 10 1

), A3 =

0 1 − sin(θ)−1 0 cos(θ)

− sin(θ) cos(θ) 0

.

◦


B. Problem sheets

Exercise 3.4 – Compute etA for the matrix

A =

1 1 0−1 0 −10 −1 1

.

◦

Exercise 3.5 – Let A ∈ Mn(R) be a matrix (independent of t), ‖ · ‖ 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,∣∣∣∣∣∣∣∣∣∣∣∣e tkA −

(I +

t

kA

)∣∣∣∣∣∣∣∣∣∣∣∣ ≤ 1

k2Ck(t).

with limk→∞Ck(t) = t2

2|||A2|||.

b) Show that for all t ∈ R and all k ∈ N∗,∣∣∣∣∣∣∣∣∣∣∣∣I +t

kA

∣∣∣∣∣∣∣∣∣∣∣∣ ≤ e|t|k|||A|||.

c) Deduce that

etA = limk→∞

(I +

t

kA

)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,

(αI + A)−(k+1) =

∫ ∞

0

e−t(αI+A) tk

k!dt.

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 thatbij ≥ 0 for all i, j = 1, . . . , n.

iii) Do the results of part 2) hold true if A is a nonsymmetric matrix?

◦




Solution – Exercise 1 – 1) The proposed method consists in considering (B.18) as anondelayed 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

x′(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] sincefor 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],

x(t) = x(t0) +

∫ t

t0

g(s, x(s))ds

= φ0(t0) +

∫ t

t0

f(s, x(s), φ0(s− ω))ds

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 uniquenessof solutions on the interval [t0, t0 + ω] follow the usual scheme. To discuss the requiredproperties on f and φ0, the best is to use (B.21). Recall that a vector field has to becontinuous both in t and in x for solutions to exist. Thus to have existence of solutions tothe 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 requirethe same property from f . Note that this does not imply either φ0 or the way f depends onφ0.

3) Every successive integration raises the regularity of the solution: x is C1 on [t0, t0 +ω],C2 on [t0 + ω, t0 + 2ω], etc. Hence, x is Cn on [t0 + (n− 1)ω, t0 + nω]. ◦

Solution – Exercise 2 – We proceed as previously explained. We assume for sim-plicity that t0 = 0. To find the solution on the interval [0, ω], we consider the nondelayedIVP

x′1 = 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 theIVP

x′2 = a[C(1 + at)]

x2(ω) = x1(ω) = C + aCω


B. Problem sheets

Hence we find the solution to the differential equation to be, on the interval [ω, 2ω],

x2(t) = C

(1 + at+

1

2a2t2 − 1

2a2ω2

)Iterating this process one more time with the IVP

x′3 = a

[C

(1 + at+

1

2a2t2 − 1

2a2ω2

)]x3(2ω) = x2(2ω) =

3

2a2Cω2 + 2aCω + C

we find, on the interval [2ω, 3ω], the solution

x3(t) = C

(1 + at+

1

2a2t2 +

1

6a3t3 − 1

2ta3ω2 − 1

3a3ω3 − 1

2a2ω2

)We develop the intuition that the solution at step n (i.e., on the interval [(n − 1)ω, nω])must take the form

xn(t) = C

n∑k=0

ak (t− (k − 1)ω)k

k!(B.22)

This can be proved by induction (we will not do it here). ◦

Solution – Exercise 3 – For matrix A1, we have A21 = I, A3

1 = A1, etc. Hence,A2n

1 = I and A2n+11 = A1, which implies

eAt =∞∑

n=0

t2n

(2n)!A2n

1 +∞∑

n=0

t2n+1

(2n+ 1)!A2n+1

1

=

(∞∑

n=0

t2n

(2n)!

)I +

(∞∑

n=0

t2n+1

(2n+ 1)!

)A1

= cosh tI + sinh tA

=

(cosh t sinh tsinh t cosh t

)For matrix A2, remark that it can be written as A2 = I +N , where

N =

(0 10 0

)is a nilpotent matrix. Thus A2

2 = (I+N)2 = I+2N+N2 = I+2N , A32 = (I+2N)(I+N) =

I + 2N +N + 2N2 = I + 3N , and it is easily shown by induction that (I +N)n = I + nN .


It follows that

eA2t =

(∞∑

n=0

tn

n!

)I +

(∞∑

n=0

ntn

n!

)N

= etI + t

(∞∑

n=0

tn−1

(n− 1)!

)N

= etI + tetN

= et

(1 t0 1

)Finally, for matrix A3, we have that

A23 =

− cos2 θ −12sin 2θ cos θ

−12sin 2θ − sin2 θ sin θ

− cos θ − sin θ 1

and A3

3 = 0, i.e., A3 is nilpotent for the index 3. Therefore, eA3t = I + tA3 + t2

2A2

3. ◦

Solution – Exercise 4 – The Jordan canonical form of A is

J =

0 0 00 1 10 0 1

Now, to compute eJt, remark that J has the form 0 0 0

00A1

where eA1t has been computed in Exercise 3. Hence, e0t 0 0

00

eA1t

=

1 0 00 et tet

0 0 et

We have J = P−1AP , where

P =

−1 −1 21 0 −11 1 −1

, P−1 =

1 1 10 −1 11 0 1

are the matrices of change of basis that transform A to its Jordan canonical form. ThenA = PJP−1, and eAt = PeJtP−1, i.e.,

eAt =

(2− t)et − 1 et − 1 (1− t)et − 11− et 1 1− et

1 + (t− 1)et 1− et1 + tet


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)′ = AeAt and (eAt)(k) = AkeAt, where u(k) denotes the kth derivative of u.

etkA = I +

t

kA+

t2

2k2A2 +

t2

k2ε(t

k)

Thus

etkA − I − t

kA =

t2

2k2A2 +

t2

k2ε(t

k)

Therefore, taking the norm |||·||| of this expression,∣∣∣∣∣∣∣∣∣∣∣∣e tkA − (I +

t

kA)

∣∣∣∣∣∣∣∣∣∣∣∣ ≤ 1

k2

{t2

2

∣∣∣∣∣∣A2∣∣∣∣∣∣+ t2

∣∣∣∣∣∣∣∣∣∣∣∣ε( tk )

∣∣∣∣∣∣∣∣∣∣∣∣} =1

k2Ck(t)

We have

limk→∞

Ck(t) =t2

2

∣∣∣∣∣∣A2∣∣∣∣∣∣

Let Sk =∑∞

j=3tj

kj−2j!|||A|||j. This series is uniformly convergent, which implies that we can

change the order in the following limit,

limk→∞

Sk =∞∑

j=3

limk→∞

(tj

kj−2j!|||A|||j

)= 0

1–b) We have already seen (Exercise 2, Assignment 2) that∣∣∣∣∣∣eA

∣∣∣∣∣∣ ≤ e|||A|||. Therefore,∣∣∣∣∣∣∣∣∣e tkA∣∣∣∣∣∣∣∣∣ ≤ e

tk|||A|||

But

e|t|k|||A||| =

∞∑k=0

1

k!

|t|k|kk

|||A|||k

= 1 +|t|k|||A|||+ |t|2|

2k2|||A|||2 + · · ·

which, since |t|22k2 |||A|||2 + · · · ≥ 0, implies that

e|t|k|||A||| ≥ 1 +

|t|k|||A||| = |||I|||+

∣∣∣∣∣∣∣∣∣∣∣∣ tkA∣∣∣∣∣∣∣∣∣∣∣∣

≥∣∣∣∣∣∣∣∣∣∣∣∣I +

t

kA

∣∣∣∣∣∣∣∣∣∣∣∣


(since |||I||| = sup‖v‖≤1 ‖IV ‖ = sup‖v‖≤1 ‖v‖ = 1).1–c) We skip the scalar case, and consider the case n ≥ 2. We have

eAt − (I +t

kA)k = (e

tkA)k − (I +

t

kA)k

=

(e

tkA − (I +

t

kA)

) k−1∑j=0

(etkA)k−1−j(I +

t

nA)j

since for two matrices E,F ∈ Mn(R) (or Mn(C)) that commute, En − F n = (E −F )∑k−1

j=0 Ek−1−jF j.

Therefore,∣∣∣∣∣∣∣∣∣∣∣∣eAt − (I +t

kA)k

∣∣∣∣∣∣∣∣∣∣∣∣ ≤ ∣∣∣∣∣∣∣∣∣e tkA−(I+ t

kA)∣∣∣∣∣∣∣∣∣ k−1∑

j=0

[∣∣∣∣∣∣∣∣∣e t(k−1−j)k

A∣∣∣∣∣∣∣∣∣ ∣∣∣∣∣∣∣∣∣∣∣∣(I +

t

kA)j

∣∣∣∣∣∣∣∣∣∣∣∣] (B.23)

Now, we have∣∣∣∣∣∣∣∣∣e t(k−1−j)

kA∣∣∣∣∣∣∣∣∣ ≤ e

|t|(k−1−j)k

|||A|||. Also,∣∣∣∣∣∣∣∣∣∣∣∣(I +t

kA)j

∣∣∣∣∣∣∣∣∣∣∣∣ ≤ ∣∣∣∣∣∣∣∣∣∣∣∣I +t

kA

∣∣∣∣∣∣∣∣∣∣∣∣j ≤ (e |t|k|||A|||)j

= ej|t|k|||A|||

where the last inequality results from question 1–b). Therefore,∣∣∣∣∣∣∣∣∣∣∣∣eAt − (I +t

kA)k

∣∣∣∣∣∣∣∣∣∣∣∣ ≤ 1

k2Ck(t)

k−1∑j=0

e|t|k−1−j

k|||A|||e

j|t|k|||A|||

=1

k2Ck(t)

k−1∑j=0

e|t|k−1

k|||A|||

=1

k2Ck(t)ke

|t| k−1k|||A|||

=1

kCk(t)e

|t| k−1k|||A|||

We thus have ∣∣∣∣∣∣∣∣∣∣∣∣eAt − (I +t

kA)k

∣∣∣∣∣∣∣∣∣∣∣∣ ≤ 1

kCk(t)e

|t| k−1k|||A||| =

1

kDk(t)

As k →∞, Ck(t) → t2

2|||A2||| and e|t|

k−1k|||A||| → e|t||||A|||. Therefore, limk→∞Dk(t) = t2

2|||A2||| e|t||||A|||,

which in turn implies that limk→∞(I + tkA)k = eAt.

2–a) We now suppose that A is a symmetric matrix. Recall that any symmetric matrixis diagonalizable, with real eigenvalues. Furthermore, there exists a matrix P such thatP−1 = P T and that P TAP = diag(λi), with λi ∈ Sp (A).

We assume that the eigenvalues λi, i = 1, . . . ,m, are such that λi > −α, for α > 0. Wewant to show by induction that the following holds.

∀k ≥ 0, (αI + A)−(k+1) =

∫ ∞

0

e−(αI+A)t tk

k!dt (B.24)


B. Problem sheets

Suppose that k = 0. Equation (B.24) reads

(αI + A)−1 =

∫ ∞

0

e−(αI+A)tdt

The matrix (αI + A)−1 is nonsingular. Indeed, suppose that det(αI + A) = 0. This isequivalent to det(−αI−A) = 0, which implies that −α is an eigenvalue of A, a contradictionwith the hypothesis on the localization of the spectrum.

Now remark that if B is a nonsingular matrix, the equality ddteBt = BeBt implies that

B ddteBt = eBt. Since (αI + A)−1 is nonsingular, we thus have

d

dte−(αI+A)t = −(αI + A)e−(αI+A)t

and therefore

e−(αI+A)t = −(αI + A)−1 d

dt

(e−(αI+A)t

)and so, integrating,∫ ∞

0

e−(αI+A)tds =

∫ ∞

0

−(αI + A)−1 d

dt

(e−(αI+A)t

)dt

= −(αI + A)−1

∫ i

0

nftyd

dt

(e−(αI+A)t

)dt

= −(αI + A)−1[e−(αI+A)t

]∞0

= −(αI + A)−1[limt→∞

(e−(αI+A)t

)− I]

(B.25)

Now,

e−(αI+A)t = P

e−(α+λ1)t 0

. . .

0 e−(α+λn)t

P−1

Since for all i = 1, . . . , n, λi > −α, it follows that limt→∞ e−(α+λi)t = 0, which in turn impliesthat 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.,

(αI + A)−j =

∫ ∞

0

e−(αI+A)t tj−1

(j − 1)!dt

Then ∫ ∞

0

e−(αI+A)t tj

j!dt =

∫ ∞

0

−(αI + A)−1 d

dt

(e−(αI+A)t

) tjj!dt

= −(αI + A)−1

∫ ∞

0

d

dt

(e−(αI+A)t

) tjj!dt

= −(αI + A)−1

{[e−(αI+A)t t

j

j!

]∞0

−∫ ∞

0

e−(αI+A)t tj−1

(j − 1)!dt

}


As we did in the k = 0 case, we now use the bound on the eigenvalues to get rid of the term

tj

j!e−(αI+A)t = P

tj

j!e−(α+λ1)t 0

. . .

0 tj

j!e−(α+λn)t

P−1−→0 as t→∞

Therefore, ∫ ∞

0

e−(αI+A)t tj

j!dt = −(αI + A)−1

∫ ∞

0

e−(αI+A)t tj−1

(j − 1)!dt

= −(αI + A)−1(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 + tkA)k. Thus

e−At = limk→∞

((I +

t

kA

)k)−1

= limk→∞

(I +

t

k

)−k

(B.26)

Let u = t/k with k ∈ N∗ and t > 0. Then

∀t > 0, ∀k ∈ N∗,

∣∣∣∣∣∣∣∣∣∣∣∣(I +t

kA)−k

∣∣∣∣∣∣∣∣∣∣∣∣ ≤M

⇒ ∀t > 0, limk→∞

∣∣∣∣∣∣∣∣∣∣∣∣(I +t

kA)−k

∣∣∣∣∣∣∣∣∣∣∣∣ ≤M

⇒ ∀t > 0,

∣∣∣∣∣∣∣∣∣∣∣∣ limk→∞(I +

t

kA)−k

∣∣∣∣∣∣∣∣∣∣∣∣ ≤M

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

(I + uA)−k = [u(1

uI + A)]−k = u−k[

1

uI + A]−k

Suppose that −α > −1/u, i.e., 0 < u < 1/α. Then, from 2–a), it follows that

(I + uA)−k = u−k

∫ ∞

0

e−( 1u

I+A)t tj−1

(j − 1)!dt


B. Problem sheets

which, when taking the norm, gives∣∣∣∣∣∣(I + uA)−k∣∣∣∣∣∣ ≤ u−k

∫ ∞

0

∣∣∣∣∣∣e−At∣∣∣∣∣∣ e− t

utj−1

(j − 1)!dt

≤ Mu−k

(k − 1)!

∫ ∞

0

e−tu tk−1dt

Let χ = t/u, then dt = udχ, and∣∣∣∣∣∣(I + uA)−k∣∣∣∣∣∣ ≤ Mu−k

(k − 1)!

∫ ∞

0

e−χuk−1χk−1dχ

≤ M

(k − 1)!

∫ ∞

0

e−χχk−1dχ

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, itsuffices that the eigenvalues of A be greater than −α. Take λ0 = α. Then λ > α = λ0, andso

(λI + A)−1 =

∫ ∞

0

e−(λI+A)tdt

=

∫ ∞

0

e−λte−Atdt ≥ 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

tI + A)−k ≥ 0

where k0 is sufficiently large. This implies that

∀t > 0, ∀k ≥ k0,

(k

t

)−k

(I +t

kA)−k ≥ 0

As (k/t)−k > 0,

∀t > 0, ∀k ≥ k0, (I +t

kA)−k ≥ 0

1See, e.g., M. Abramowitz and I.E. Stegun, Handbook of Mathematical Functions. Dover, 1965.


so

∀t > 0, limk→∞

(I +t

kA)−k ≥ 0

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. ◦


MATH4G03/6G03

Julien Arino

Duration Of Examination: 72 HoursMcMaster University Final Examination December 2003

This examination paper includes 4 pages and 3 questions. You are responsiblefor ensuring that your copy of the paper is complete.

Detailed InstructionsYou 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 ofcollaborative work will be considered as evidence of academic dishonesty. You are not tohave 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 conditionsof application of the theorem. When citing theorems from the lecture notes, refer to themby the number they have on the last version of the notes, as posted on the website on thefirst day of the examination period.

Pay attention to the form of your answers: as this is a take-home examination, you areexpected 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 orproblems that have multiple parts, you are always allowed to consider as proved the resultsof a previous part, even if you have not actually done that part.

Foreword to the correction. This examination was long, but established the results ina very guided way. Exercise 1 was almost trivial. Both of the Problems dealt with Sturmtheory. This comes as an illustration of the richness of behaviors that can be observed indifferential 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 shortenedand simplified version of a review problem for the CAPES, a French competition to hirehigh school teachers. The original problem comprised 23 questions, and was written bycandidates in 5 hours. Problem 3 introduced the Wronskian, which we did not have time tocover during class. It also established further results of Sturm type.


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 theequation

x′(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 thesolution 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 ofthe differential equation

x′′ + 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, x′(t0) = y0

Preliminary results : convex functions. A function f : I ⊂ R → R is convex if, for allx, 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 resultson convex functions.

ii) Let f be a function defined on R, convex and nonnegative. Suppose that f has twozeros 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.


iv) Let a, b ∈ R, a < b. We assume that (B.29) has a solution x, zero at a and at b andpositive on (a, b). Show that ∫ b

a

|q(t)|dt > 4

b− a.

v) We suppose that∫∞

0|q(t)|dt converges. Let x be a bounded solution of (B.29). Deter-

mine the behaviour of x′ as t→∞.

vi) We suppose that q ∈ C1 and is positive and increasing on R+. Show that all solutionsof (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′ − x′y is constant.

xi) Let x1 and x2 be the solutions of (B.29) that satisfy

x1(0) = 1, x′1(0) = 0,

x2(0) = 0, x′2(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 x1x

′2 − x′1x2? 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) andits answer in this context.

xiii) Show that if q is an even function, then the function x1 is even and the function x2 isodd.

Problem 5 3 – The aim of this problem is to show some elementary properties of theWronskian of a system of solutions, and to use them to study a second-order differentialequation.

Consider the nth order ordinary differential equation

x(n) = a0(t)x+ a1(t)x′ + · · ·+ an−1(t)x

(n−1)(t) (B.30)

where x(k) denotes the kth derivative of x, dkxdtk

.


B. Problem sheets

i) Find the matrix A(t) such that this system can be written as the first-order linearsystem y′ = A(t)y.

Part I : Wronskian Let f1, . . . , fn be n functions from R into R that are n − 1 times differ-entiable. We define W (f1, . . . , fn), the Wronskian of f1, . . . , fn, by

W (f1, . . . , fn)(t) = det

f1(t) · · · fn(t)f ′1(t) · · · f ′n(t)

......

f(n−1)1 (t) · · · f

(n−1)n (t)

.

If f1, . . . , fn are linearly dependent, thenW (f1, . . . , fn) = 0. The converse is false. Rememberthat the set of solutions of (B.30) forms a vector space S of dimension n.

ii) Using the equivalent linear system y′ = A(t)y, show that every system of n solutionsof (B.30) whose Wronskian is nonzero at a time τ constitutes a basis of S.

iii) Using the linear system x′ = A(t)x, show that for every set of n solutions,

W (t) = W (s) exp

(∫ t

s

an−1(u)du

).

Part II : a theorem of Sturm Let us now consider the second-order differential equation

a2(t)x′′ + a1(t)x

′ + 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 twoconsecutive 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 thatu 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 segmentsof R where a0(t) 6= 0.

vi) Let

x(t) = u(t) exp

(∫ t

0

a1(s)

a2(s)ds

)Show that (B.31) becomes u′′ + q(t)u = 0.

We want to show the following theorem of Sturm.


Theorem B.5.7 (Sturm). Let

x′′ + 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 onezero 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 theproperties 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.


B. Problem sheets

Solution – Exercise 1 – 1) We have

R(t+ ω, t0 + ω) = exp

(∫ t+ω

t0+ω

A(s)ds

)= exp

(∫ t

t0

A(s+ ω)ds

)= exp

(∫ t

t0

A(s)ds

)= 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), wehave 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, andhence 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 typeresults, that is, results dealing with the behavior of second order differential equations.


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, bysetting y = x′. Then, differentiating y, we obtain

y′ + qx = 0

Therefore, (B.27) is equivalent to the system of first order equations

x′ = y

y′ = −qx

This is a linear system, hence satisfies a Lipschitz condition, and we can apply the Cauchy-Lipschitz 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 alsointroduced the notion of Wronskian, which is a very general tool intimately linked to thenotion of resolvent matrix.

1) We let y1 = x, y2 = x′, . . . , yn = x(n−1). As a consequence, y′1 = y2, y′2 = y3, . . . , y

′n−1 =

yn, and y′n = a0(t)y1 + a1(t)y2 + · · ·+ an−1(t)yn−1. Written in matrix form, this is equivalentto y′ = A(t)y, with y = (y1, . . . , yn)T and

A(t) =

0 1 0 . . . 00 0 1 0 . . . 0...

...... 1 00 1

a0(t) a1(t) . . . an−1(t)

(B.33)


B. Problem sheets

2) We know that the system is equivalent to y′ = A(t)y, with A(t) given by (B.33). Toevery basis (φ1, . . . , φn) of the vector space of solutions of

y′ = A(t)y (B.34)

there corresponds a basis (ϕ1, . . . , ϕn) of (B.30), where ϕi is the first coordinate of the vectorφi 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

detR(t, s) = exp

(∫ t

s

trA(u)du

)And if a system of coordinates is fixed, for every fundamental matrix Φ,

det Φ(t) = det Φ(s) exp

(∫ t

s

trA(u)du

)From Liouville’s theorem,

det(Φ(t)Φ−1(s)) = exp

(∫ t

s

trA(u)du

)which implies that

det Φ(t) = det Φ(s) exp

(∫ t

s

trA(u)du

)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) ◦


University of Manitoba – Math 8430Fall 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 andT -periodic functions defined on R, and let a ∈ P . Define

A =

∫ T

0

a(t)dt, g(t) = exp

(∫ t

0

a(u)du

),

and endow P with the norm‖x‖ = sup

t∈R|x(t)|.

First part

1. For what value(s) of A does the differential equation

x′(t) = a(t)x(t) (E1)

admit non trivial T -periodic solutions?

We now let b ∈ P , and consider the differential equation

x′(t) = a(t)x(t) + b(t). (E2)

2.a. Describe the set of maximal solutions to (E2) and the intervals of definition of thesesolutions.

2.b. Describe the set of maximal solutions to (E2) that are T -periodic, first assumingA 6= 0, then A = 0.


B. Problem sheets

Second part

In this part, we let H be a real valued C1 function defined on R2, and consider thedifferential equation

x′(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

x(t) = g(t)

(x(0) +

∫ t

0

g(s)−1H(x(s), s)ds

).

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

U(Hx)(t) =eA

1− eAg(t)

∫ t+T

t

g(s)−1H(x(s), s)ds,

defines a function UHx ∈ P , and that x est solution to (E3) if and only if UHx = x.

In the rest of the problem, we let F be a real-valued C1 function defined on R2, T -periodicwith respect to its second argument; for all ε > 0, define Hε = εF and Uε = UHε , so thatthe differential equation (E3) is written

x′(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 rin the normed space P . We want to show the following assertion: for all r > 0, there existsε1 > 0 such that, for all ε ≤ ε1, the differential equation (E4) has a unique solution x ∈ Br,that we will denote xε.

We denote αr (resp. βr) the upper bound of the set |F (v, s)| (resp. |∂F∂v

(v, s)|), wherev ∈ [−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) = v2, and thus (E4) takes the form

x′(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.


Third part

Here, we consider the differential equation

x′(t) = kx(t) + εf(x(t)), (E6)

where k < 0, f is C1 and zero at zero. We let

λ = supu∈[−1,1]

|f ′(u)|,

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, andwe 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 solutionof (E6).


University of Manitoba – Math 8430Fall 2006

Homework Sheet 1 – Solutions

1. The equation (E1) is a separable equation, so we write

x′(t) = a(t)x(t) ⇔ x′(t)

x(t)= a(t)

⇔ ln |x(t)| =∫ t

0

a(s)ds+ C

⇔ |x(t)| = exp

(∫ t

0

a(s)ds+ C

)⇔ x(t) = K exp

(∫ t

0

a(s)ds

),

where it was assumed that integration starts at t0 = 0, and where the sign of |x(t)| isabsorbed into K ∈ R. Since x(0) = K, the general solution to (E1) is thus

x(t) = x(0) exp

(∫ t

0

a(s)ds

). (B.35)

A nontrivial solution (B.35) is T -periodic if it satisfies x(t + T ) = x(t) for all t ≥ 0. Inparticular (for simplicity), there must hold that x(T ) = x(0). This leads to

x(T ) = x(0) ⇔ x(0) exp

(∫ T

0

a(s)ds

)= x(0)

⇔ exp

(∫ T

0

a(s)ds

)= 1

⇔(∫ T

0

a(s)ds

)= 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 solutionto (E2), or to use integrating factors or a variation of constants approach. We do thelatter, since we already have the solution (B.35) to (E1). Returning to the solution withundetermined value for K, we consider the ansatz

φ(t) = K(t) exp

(∫ t

0

a(s)ds

)= K(t)g(t),


B. Problem sheets

where the second equality uses the definition of g(t). We have

φ′(t) = K ′(t)g(t) +K(t)g′(t)

= K ′(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 ifand only if

φ′(t) = a(t)φ(t) + b(t) ⇔ K ′(t)g(t) +K(t)a(t)g(t). = a(t)K(t)g(t) + b(t)

⇔ K ′(t)g(t) = b(t)

⇔ K ′(t) =b(t)

g(t), for g(t) 6= 0

⇔ K(t) =

∫ t

0

b(s)

g(s)ds+ C.

Note that the remark that g(t) 6= 0 is made “for form”: as it is defined, g(t) > 0 for allt ≥ 0. We conclude that the general solution to (E2) is given by

x(t) =

(∫ t

0

b(s)

g(s)ds+ C

)exp

(∫ t

0

a(s)ds

).

Since it will be useful to have information in terms of x(0) (as in question 1.), we note thatC = x(0). Thus, the solution to (E2) through x(0) = 0 is given by

x(t) =

(∫ t

0

b(s)

g(s)ds+ x(0)

)exp

(∫ t

0

a(s)ds

). (B.36)

With integrating factors, we would have done as follows: write the equation (E2) as

x′(t)− a(t)x(t) = b(t).

The integrating factor is then

µ(t) = exp

(−∫a(t)dt

),

and the general solution to (E2) is given by

x(t) =1

µ(t)

(∫ t

0

µ(s)b(s)ds+ C

)Maximal solutions are solutions that are the restriction of no other solution.


2.b. Solutions to (E2) are T -periodic if x(T ) = x(0); therefore, a T -periodic solutionsatisfies

x(T ) = x(0) ⇔(∫ T

0

b(s)

g(s)ds+ x(0)

)exp

(∫ T

0

a(s)ds

)= x(0)

⇔(∫ T

0

b(s)

g(s)ds+ x(0)

)= x(0)e−A

⇔∫ T

0

b(s)

g(s)ds =

(e−A − 1

)x(0)

Second part

3. Before proceeding, note that

g′(t) =d

dtexp

(∫ t

0

a(s)ds

)= a(t) exp

(∫ t

0

a(s)ds

)= a(t)g(t).

We differentiate x(t) = g(t)(x(0) +

∫ t

0g(s)−1H(x(s), s)ds

). This gives

x′(t) = g′(t)

(x(0) +

∫ t

0


)+ g(t)

H(x(t), t)

g(t)

= g′(t)

(x(0) +

∫ t

0


)+H(x(t), t)

= a(t)g(t)

(x(0) +

∫ t

0


)+H(x(t), t)

= a(t)x(t) +H(x(t), t),

and thus x(t) = g(t)(x(0) +

∫ t

0g(s)−1H(x(s), s)ds

)is solution to (E3).

4. Let x ∈ P . Then UHx ∈ P if and only if UHx is T -periodic. We have

(UHx)(t+ T ) =eA

1− eAg(t+ T )

∫ t+2T

t+T

g(s)−1H(x(s), s)ds.


B. Problem sheets

Remark that

g(t+ T ) = exp

(∫ t+T

0

a(s)ds

)= exp

(∫ t

0

a(s)ds+

∫ t+T

t

a(s)ds

)= g(t) exp

(∫ t+T

t

a(s)ds

)= eAg(t),

since a(t) is T -periodic. Therefore,

(UHx)(t+ T ) =eA

1− eAeAg(t)

∫ t+2T

t+T


=eA

1− eAeAg(t)

∫ t+T

t

g(s− T )−1H(x(s− T ), s− T )ds.

Now

g(s− T ) = exp

(∫ s−T

0

a(u)du

)= exp

(∫ s

0

a(u)du+

∫ s−T

s

a(u)du

)= g(s) exp

(−∫ s

s−T

a(u)du

)= e−Ag(s).

So, finally,

(UHx)(t+ T ) =eA

1− eAe2Ag(t)

∫ t+T

t


= e2A(UHx)(t),

since H is T -periodic in its second argument and x ∈ P . Therefore, UHx ∈ P for x ∈ P .Suppose that x(t) = (UHx)(t). Then,

x′(t) =eA

1− eA

(g′(t)

∫ t+T

t

H(x(s), s)

g(s)ds+ g(t)

(H(x(t+ T ), t+ T )

g(t+ T )− H(x(t), t)

g(t)

))=

eA

1− eA

(a(t)g(t)

∫ t+T

t

H(x(s), s)

g(s)ds+ g(t)

(H(x(t), t)

eAg(t)− H(x(t), t)

g(t)

))= a(t)

eA

1− eAg(t)g(t)

∫ t+T

t

H(x(s), s)

g(s)ds+

eA

1− eAg(t)

(1− eA)H(x(t), t)

eAg(t)

= a(t)x(t) +H(x(t), t).


B. Problem sheets

‖y‖ ≤ r, we compute

‖Uεx− Uε‖ = supt∈R

|(Uεx)(t)− (Uεy)(t)|

= supt∈R

∣∣∣∣ eA

1− eAg(t)

∫ t+T

t

εF (x(s), s)− εF (y(s), s)

g(s)ds

∣∣∣∣= ε

∣∣∣∣ eA

1− eA

∣∣∣∣ supt∈R

|g(t)|∣∣∣∣∫ t+T

t

F (x(s), s)− F (y(s), s)

g(s)ds

∣∣∣∣≤ ε

∣∣∣∣ eA

1− eA

∣∣∣∣ supt∈R

|g(t)|∫ t+T

t

∣∣∣∣F (x(s), s)− F (y(s), s)

g(s)

∣∣∣∣ ds.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)| ≤ 2βrr + 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) andBr is closed in P .

Therefore, we conclude that for a given r > 0, for all ε ≤ ε1, there exists a unique solutionxε of (E4) in Br.

6. This is the contraction mapping theorem with a parameter:

‖xε − xε′‖ = ‖Uεxε − Uε′xε′‖ ≤ ‖Uεxε − Uεxε′‖︸︷︷︸≤K‖xε−xε′‖

+‖Uεxε′ − Uε′xε′‖.


But we have

‖Uεxε − Uε′xε′‖(t) ≤ |ε− ε′| eA

|1− eA|

∫ t+T

t

|g(t)g(s)−1F (x(s), s)|ds

≤ |ε− ε′| eA

|1− eA|eTATαr︸︷︷︸

=K′

.

Thus, we have ‖xε−xε′‖ ≤|ε− ε′|K ′

1−Kand therefore ε ∈ R 7→ xε ∈ P is continuous; it follows

that limε→0

xε = x0. But the only periodic solution of (E1) when A 6= 0 is the zero solution.

Therefore, xε → 0 when ε→ 0.

7. Let x0(t) = c0, then g(t) = ekt and A = kT .

Uεx0(t) =εekT

1− ekTektf(t0)

∫ t+T

t

e−ksds =εekT

1− ekTf(c0)e

kt

[−1

ke−ks

]t+T

t

=εekT

1− ekTf(c0)

ekt

k

(e−kt − e−kte−kT

)=

εekT

1− ekT

f(c0)

k

(1− e−kT

)= − ε

kf(c0)

The constant function xε(t) = c0 is solution (where c0 is the unique solution of the equationεf(x) + kx = 0 for ε sufficiently small).

Note : letting g(x) = − εkf(x), it follows that g′(x) = − ε

kf ′(x). Thus, for r > 0 given,

there exists ε0 > 0 such that ε ≤ ε0 implies g([−r, r]) ⊂ [−r, r], and there exists ε1 ≤ ε0 suchthat sup

x∈[−r,r]

|g′(x)| < 1, the fixed point theorem can be applied easily.

8.a. Using the formula obtained in 5.a. with αr = r2, βr = 2r, g(t) = e−t, A = −T then

ε0 =r(1− e−T )eT e−T

r2T=

1− e−T

rT, ε1 =

1

2

1− e−T

2rT=ε0

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).


B. Problem sheets

8.c. The vector field −x+εx2 is C1, and therefore existence and uniqueness of a maximalsolution ϕα is a direct consequence of the theorem of Cauchy-Lipschitz.

We solve the equation x′ = −x + εx2 without constraint of periodicity. There are twoconstant solutions , x(t) = 0 and x(t) = 1/ε. By uniqueness, any other solution never takesthe values 0 and 1/ε. We have:

x′

x(1− εx)=

d

dt

(ln∣∣∣ x

1− εx

∣∣∣) = −1 ⇒ x

1− εx= λe−t ⇒ x(t) =

λe−t

1 + ελe−t, λ ∈ R∗.

The condition x(0) = α gives λ =α

1− εαif α 6= 0 and α 6= 1/ε, and as a consequence,

letting β =1

α− ε we obtain

ϕα(t) =1

βet + ε.

• If β ≥ 0 then ϕα is defined on R.

• If β < 0, we let t0 = ln(− ε

β

).

– If α > 0 then − εβ> 1, that is, t0 > 0, ϕα is defined on ]−∞, t0[.

– If α < 0 then − εβ< 1, that is, t0 < 0, ϕα is defined on ]t0,+∞[.

Here are a few representative solutions obtained when setting ε = 1.

alpha<0

alpha>1/epsilon

0<alpha<1/epsilon

–3

–2

–1

0

1

2

3

y

–3 –2 –1 1 2 3

x

Third part


9. θ > 0 since |x(0)| < 1, and we have x(s) ∈ [−1, 1] for all s ∈ [0, θ], by definition ofθ. Since f(0) = 0 and |f ′| is bounded by λ on [−1, 1], we have, from the inequality of finitevariations, |f(u)− f(0)︸︷︷︸

=0

| ≤ λ|u| for all u ∈ [−1, 1], that is, |f(x(s))− f(0)︸︷︷︸=0

| ≤ λ|x(s)|.

Let t ∈ [0, θ]. From 4.,

|e−ktx(t)| =∣∣∣x(0) + ε

∫ t

s=0

e−ksf(x(s)) ds∣∣∣ ≤ |x(0)|+ ελ

∫ t

s=0

e−ks|x(s)| ds,

from which |e−ktx(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 thereexists η > 0 such that |x(t)| < 1 on [0, η], and thus θ ≥ η > 0). Since lim

t→θ−|x(t)| ≤ 1 and

limt→θ+

|x(t)| ≥ 1, it follows |x(θ)| = 1.

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 x′ is bounded in a neighborhood of b. Thus x

admits a limit in b. The same is therefore true for x′. We then know that x can be extendedbeyond b, contradicting the maximality of J .

Final conclusion : J∩[0,+∞[= [0,+∞[, x is defined on [0,+∞[ and the proof of question9. 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 solutionof (E6).

This subject was the Premiere composition de mathematiques for the contest determiningadmission to Ecole Polytechnique in France, for MP (Math-Physics) track students, in 2004.Students, in their second year of university, have 4 hours to write this premiere composition.

The original subject comprised another question, originally question 3, which was sup-pressed in this homework sheet. To be complete, this question is included here:


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 x(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 modeof 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(n) is linear, and from the relation x′(n) = inx(n), we have

x′(n) = kx(n) + b(n) ⇒ x(n) =b(n)

in− k.

Since x is C1, we know that the Fourier series of x is normally convergent. Since b(n) → 0,

we ca also say that x(n) = o

(1

n

).

2’.b. Applying here again the result of 2.b.,

• If b(0) = 0 then all solutions are 2π-periodic. In this case, solutions satisfy x(n) =b(n)/in for n ∈ Z non zero and x(0) varies with the solutions under consideration.

• If b(0) 6= 0 then no solution is periodic.

ordinary differential equations: fundamental theory...

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