Matematica 100 Note della lezione

7/29/2019 Matematica 100 Note della lezione

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MATH 8ddtR(t, t0)

v(t) + R(t, t0)v0(t)= A(t)R(t, t0)v(t) + R(t, t0)v0(t),2.5. Further developments, bibliographical notesFund. Theory ODE Lecture Notes J. Arino55from Proposition 2.2.11. For to be solution, it must satisfy the differentialequation (2.19),and thus0(t) = A(t)(t) + g(t, (t)) , A(t)R(t, t0)v(t) +ekt ektekT="ekT1 ekTf(c0)k

1 ekT430FundddtR(t, t0)


1 ekTament


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ddtR(t, t0)

v(t) + R(t, t0)v0(t)= A(t)R(t, t0)v(t) + R(t, t0)v0(t),

2.5. Further developments, bibliographical notesFund. Theory ODE Lecture Notes J. Arino55from Proposition 2.2.11. For to be solution, it must satisfy the differentialequation (2.19),and thus0(t) = A(t)(t) + g(t, (t)) , A(t)R(t, t0)v(t) +ekt ektekT="ekT1 ekTf(c0)k

1 ekTal TheoryofOrdinary

ddtR(t, t0)


1 ekT Difd


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dtR(t, t0)

v(t) + R(t, t0)v0(t)= A(t)R(t, t0)v(t) + R(t, t0)v0(t),2.5. Further developments, bibliographical notes

Fund. Theory ODE Lecture Notes J. Arino55from Proposition 2.2.11. For to be solution, it must satisfy the differentialequation (2.19),and thus0(t) = A(t)(t) + g(t, (t)) , A(t)R(t, t0)v(t) +ekt ektekT="ekT1 ekTf(c0)k

1 ekTferddtR(t, t0)

v(t) + R(t, t0)v0(t)

= A(t)R(t, t0)v(t) + R(t, t0)v0(t),2.5. Further developments, bibliographical notesFund. Theory ODE Lecture Notes J. Arino55from Proposition 2.2.11. For to be solution, it must satisfy the differentialequation (2.19),and thus0(t) = A(t)(t) + g(t, (t)) , A(t)R(t, t0)v(t) +ekt ektekT="ekT

1 ekTf(c0)k

1 ekTential EquationsLectured

dtR(t, t0)


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v(t) + R(t, t0)v0(t)= A(t)R(t, t0)v(t) + R(t, t0)v0(t),2.5. Further developments, bibliographical notesFund. Theory ODE Lecture Notes J. Arino55from Proposition 2.2.11. For to be solution, it must satisfy the differential

equation (2.19),and thus0(t) = A(t)(t) + g(t, (t)) , A(t)R(t, t0)v(t) +ekt ektekT="ekT1 ekTf(c0)k

1 ekT NotesJulien

ddtR(t, t0)

v(t) + R(t, t0)v0(t)= A(t)R(t, t0)v(t) + R(t, t0)v0(t),

2.5. Further developments, bibliographical notesFund. Theory ODE Lecture Notes J. Arino55from Proposition 2.2.11. For to be solution, it must satisfy the differentialequation (2.19),and thus0(t) = A(t)(t) + g(t, (t)) , A(t)R(t, t0)v(t) +ekt ektekT="ekT1 ekT

f(c0)k1 ekT

ddtR(t, t0)

v(t) + R(t, t0)v0(t)

= A(t)R(t, t0)v(t) + R(t, t0)v0(t),2.5. Further developments, bibliographical notesFund. Theory ODE Lecture Notes J. Arino


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55from Proposition 2.2.11. For to be solution, it must satisfy the differentialequation (2.19),and thus0(t) = A(t)(t) + g(t, (t)) , A(t)R(t, t0)v(t) +ekt ektekT

="ekT1 ekTf(c0)k

1 ekTticsUniversiddtR(t, t0)

v(t) + R(t, t0)v0(t)= A(t)R(t, t0)v(t) + R(t, t0)v0(t),2.5. Further developments, bibliographical notesFund. Theory ODE Lecture Notes J. Arino55from Proposition 2.2.11. For to be solution, it must satisfy the differential

equation (2.19),and thus0(t) = A(t)(t) + g(t, (t)) , A(t)R(t, t0)v(t) +ekt ektekT="ekT1 ekTf(c0)k

1 ekTty of ManiddtR(t, t0)

v(t) + R(t, t0)v0(t)= A(t)R(t, t0)v(t) + R(t, t0)v0(t),2.5. Further developments, bibliographical notesFund. Theory ODE Lecture Notes J. Arino

55from Proposition 2.2.11. For to be solution, it must satisfy the differentialequation (2.19),


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and thus0(t) = A(t)(t) + g(t, (t)) , A(t)R(t, t0)v(t) +ekt ektekT="ekT1 ekT

f(c0)k

1 ekTtobaFalddtR(t, t0)

v(t) + R(t, t0)v0(t)= A(t)R(t, t0)v(t) + R(t, t0)v0(t),2.5. Further developments, bibliographical notesFund. Theory ODE Lecture Notes J. Arino55from Proposition 2.2.11. For to be solution, it must satisfy the differentialequation (2.19),and thus0(t) = A(t)(t) + g(t, (t)) , A(t)R(t, t0)v(t) +

ekt ektekT="ekT1 ekTf(c0)k

1 ekT l 2ddtR(t, t0)

v(t) + R(t, t0)v0(t)= A(t)R(t, t0)v(t) + R(t, t0)v0(t),2.5. Further developments, bibliographical notesFund. Theory ODE Lecture Notes J. Arino55from Proposition 2.2.11. For to be solution, it must satisfy the differentialequation (2.19),

and thus0(t) = A(t)(t) + g(t, (t)) , A(t)R(t, t0)v(t) +ekt ektekT


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="ekT1 ekTf(c0)k

1 ekT006

Contents1 General theory of ODEs 31.1 ODEs, IVPs, solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.1 Ordinary differential equation, initial value problem . . . . . . . . . . 31.1.2 Solutions to an ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.3 Geometric interpretation . . . . . . . . . . . . . . . . . . . . . . . . . 81.2 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 . . . . . . . . . . . . . . 211.3 Continuation of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.3.1 Maximal interval of existence . . . . . . . . . . . . . . . . . . . . . . 271.3.2 Maximal and global solutions . . . . . . . . . . . . . . . . . . . . . . 281.4 Continuous dependence on initial data, on parameters . . . . . . . . . . . . . 29

1.5 Generality of first order systems . . . . . . . . . . . . . . . . . . . . . . . . . 321.6 Generality of autonomous systems . . . . . . . . . . . . . . . . . . . . . . . . 341.7 Suggested reading, Further problems . . . . . . . . . . . . . . . . . . . . . . 342 Linear systems 352.1 Existence and uniqueness of solutions . . . . . . . . . . . . . . . . . . . . . . 352.2 Linear systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.2.1 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 . . . . . . . . . . . . . . . . . . . . . . . 432.3 Affine systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.3.1 The space of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 462.3.2 Construction of solutions . . . . . . . . . . . . . . . . . . . . . . . . . 462.3.3 Affine systems with constant coefficients . . . . . . . . . . . . . . . . 472.4 Systems with periodic coefficients . . . . . . . . . . . . . . . . . . . . . . . . 482.4.1 Linear systems: Floquet theory . . . . . . . . . . . . . . . . . . . . . 48iiiFund. Theory ODE Lecture Notes J. ArinoCONTENTS

2.4.2 Affine systems: the Fredholm alternative . . . . . . . . . . . . . . . . 512.5 Further developments, bibliographical notes . . . . . . . . . . . . . . . . . . 542.5.1 A variation of constants formula for a nonlinear system with a linear


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component . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543 Stability of linear systems 573.1 Stability at fixed points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.2 Affine systems with small coefficients . . . . . . . . . . . . . . . . . . . . . . 574 Linearization 634.1 Some linear stability theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.2 The stable manifold theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.3 The Hartman-Grobman theorem . . . . . . . . . . . . . . . . . . . . . . . . . 694.4 Example of application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.4.1 A chemostat model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.4.2 A second example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755 Exponential dichotomy 795.1 Exponential dichotomy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795.2 Existence of exponential dichotomy . . . . . . . . . . . . . . . . . . . . . . . 815.3 First approximate theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 825.4 Stability of exponential dichotomy . . . . . . . . . . . . . . . . . . . . . . . . 845.5 Generality of exponential dichotomy . . . . . . . . . . . . . . . . . . . . . . 84References 85A Definitions and results 87A.1 Vector spaces, norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87A.1.1 Norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87A.1.2 Matrix norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87A.1.3 Supremum (or operator) norm . . . . . . . . . . . . . . . . . . . . . . 87A.2 An inequality involving norms and integrals . . . . . . . . . . . . . . . . . . 88A.3 Types of convergences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88A.4 Asymptotic Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89A.5 Types of continuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

A.6 Lipschitz function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90A.7 Gronwalls lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91A.8 Fixed point theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93A.9 Jordan normal form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94A.10 Matrix exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96A.11 Matrix logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97A.12 Spectral theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99B Problem sheets 101Homework sheet 1 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103Homework sheet 2 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113Homework sheet 3 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

Homework sheet 4 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125Final examination 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137Homework sheet 1 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145iii

IntroductionThis 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 isimportant


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to have a good understanding of how solutions to differential equations areconstructed.If you are taking this course, you most likely know how to analyze systems ofnonlinearordinary differential equations. You know, for example, that in order forsolutions to a

system to exist and be unique, the system must have a C1 vector field. Whatyou do notnecessarily know is why that is. This is the object of Chapter 1, where weconsider thegeneral theory of existence and uniqueness of solutions. We also consider thecontinuationof solutions as well as continuous dependence on initial data and onparameters.In Chapter 2, we explore linear systems. We first consider homogeneous linearsystems,then linear systems in full generality. Homogeneous linear systems are linkedto 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 thevicinity ofa hyperbolic equilibrium point, by a homogeneous linear system. As forautonomous systems,nonautonomous nonlinear systems are linked to a linearized form, this timethroughexponential dichotomy, which is explained in Chapter 5.

1

Chapter 1General theory of ODEsWe begin with the general theory of ordinary differential equations (ODEs).First, we defineODEs, initial value problems (IVPs) and solutions to ODEs and IVPs in Section1.1. InSection 1.2, we discuss existence and uniqueness of solutions to IVPs.

1.1 ODEs, IVPs, solutions1.1.1 Ordinary differential equation, initial value problemDefinition 1.1.1 (ODE). An nth order ordinary differential equation (ODE) is afunctionalrelationship taking the formF

t, x(t),ddtx(t),d2dt2 x(t), . . . ,


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dndtn x(t)

= 0,that involves an independent variable t 2 I R, an unknown function x(t) 2 D Rnof

the independent variable, its derivative and derivatives of order up to n. Forsimplicity, thetime dependence of x is often omitted, and we in general write equations asFt, x, x0, x00, . . . , x(n)= 0, (1.1)where x(n) denotes the nth order derivative of x. An equation such as (1.1) issaid to be ingeneral (or implicit) form.An equation is said to be in normal (or explicit) form when it is written asx(n) = ft, x, x0, x00, . . . , x(n1).Note that it is not always possible to write a differential equation in normalform, 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 simplicitythe morerestrictive case of a first-order ordinary differential equation in normal form

x0 = f(t, x). (1.2)34Fund. Theory ODE Lecture Notes J. Arino1. General theory of ODEsNote that the theory developed here holds usually for nth order equations; seeSection 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) forequation (1.2)is given by

x0 = f(t, x)x(t0) = x0,(1.3)where f is continuous and real valued on a set U R Rn, with (t0, x0) 2 U.Remark The assumption that f be continuous can be relaxed, piecewise continuityonly isneeded. However, this leads in general to much more complicated problems and isbeyond thescope of this course. Hence, unless otherwise stated, we assume that f is at leastcontinuous. 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 formx(n) = f(t, x, x0, . . . , x(n1))


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x(t0) = x0, x0(t0) = x00, . . . , x(n1)(t0) = x(n1)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 highestderivative

involved in the equation. An equation is then classified as a function of itslinearity. A linearequation is one in which the vector field f takes the formf(t, x) = a(t)x(t) + b(t).If b(t) = 0 for all t, the equation is linear homogeneous; otherwise it is linearnonhomogeneous.If the vector field f depends only on x, i.e., f(t, x) = f(x) for all t, then theequation isautonomous; otherwise, it is nonautonomous. Thus, a linear equation isautonomous 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 ODEDefinition 1.1.4 (Solution). A function (t) (or , for short) is a solution to the ODE(1.2) if it satisfies this equation, that is, if0(t) = f(t, (t)),for all t 2 I R, an open interval such that (t, (t)) 2 U for all t 2 I.

The notations and x are used indifferently for the solution. However, in thischapter,to emphasize the difference between the equation and its solution, we will tryas much aspossible to use the notation x for the unknown and for the solution.1.1. ODEs, IVPs, solutionsFund. Theory ODE Lecture Notes J. Arino5Definition 1.1.5 (Integral form of the solution). The function(t) = x0 +Z t

t0f(s, (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, byR = {(t, x) : |t t0| a, kx x0k b} ,where k k is any appropriate norm of Rn. This domain is illustrated in Figures1.1 and1.2; it is sometimes called a security system, i.e., the union of a securityinterval (for theindependent variable) and a security domain (for the dependent variables)[19]. Suppose0t a t

x0 t 0 t +a 0x +b 00 x


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x b 0(t ,x )0

Figure 1.1: The domain R for D R.t

xy

t0

x

y0

0

Figure 1.2: The domain R for D R2: security tube.6Fund. Theory ODE Lecture Notes J. Arino1. General theory of ODEsthat f is continuous on R, and let M = maxR kf(t, x)k, which exists since f iscontinuouson the compact set R.In the following, existence of solutions will be obtained generally in relation tothe domainR by considering a subset of the time interval |t t0| a defined by |t t0| ,with

=

a if M = 0min(a, bM ) if M > 0.This choice of = 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 definitionof (t) asthe solution of x0 = f(t, x), we must be working in R. So we require that |t t0|

a andkxx0k b. In order to satisfy the first of these conditions, choosing a andworking on|t t0| implies of course that |t t0| a. The requirement that b/M comesfromthe following argument. If we assume that (t) is a solution of (1.3) defined on[t0, t0 + ],then we have, for t 2 [t0, t0 + ],k(t) x0k =

Z t

t0f(s, (s))ds

Z tt0

kf(s, (s))k dsMZ tt0

ds= M(t t0),where the first inequality is a consequence of the definition of the integrals byRiemannsums (Lemma A.2.1 in Appendix A.2). Similarly, we have k(t) x0k M(t t0)


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for allt 2 [t0 , t0]. Thus, for |t t0| , k(t) x0k M|t t0|. Suppose now that b/M.It follows that k x0k M|t t0| Mb/M = b. Taking = min(a, b/M) then ensuresthat both |t t0| a and k x0k b hold simultaneously.The following two theorems deal with the localization of the solutions to an IVP.They

make more precise the previous discussion. Note that for the moment, theexistence ofa solution is only assumed. First, we establish that the security systemdescribed aboveperforms properly, in the sense that a solution on a smaller time interval stayswithin thesecurity domain.Theorem 1.1.6. If (t) is a solution of the IVP (1.3) in an interval |t t0| < ,thenk(t) x0k < b in |t t0| < , i.e., (t, (t)) 2 R((t0, x0), , b) for |t t0| < .Proof. Assume that is a solution with (t, (t)) 62 R((t0, x0), , b). Since iscontinuous, itfollows that there exists 0 < < such thatk(t)x0k < b for |tt0| t1Z t2t1

f(s, (s))ds if t1 > t2Now 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) 2U,and be a function defined on an open set I of R such that t0 2 I. Then is asolution ofthe IVP (1.3) if, and only if,i) 8t 2 I, (t, (t)) 2 U.ii) is continuous on I.iii) 8t 2 I, (t) = x0 +R tt0

f(s, (s))ds.8Fund. Theory ODE Lecture Notes J. Arino1. General theory of ODEsProof. ()) Let us suppose that 0 = f(t, ) for all t 2 I and that (t0) = x0. Then for allt 2 I, (t, (t)) 2 U (i ). Also, is differentiable and thus continuous on I (ii ). Finally,0(s) = f(s, (s))so integrating both sides from t0 to t,(t) (t0) =Z tt0

f(s, (s))dsand thus(t) = x0 +Z tt0

f(s, (s))dshence (iii ).(() Assume i ), ii ) and iii ). Then is differentiable on I and 0(t) = f(t, (t)) for all t


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2 I.From (3), (t0) = x0 +R t0t0

f(s, (s))ds = x0.Note that Theorem 1.1.8 states that should be continuous, whereas the

solution shouldof course be C1, for its derivative needs to be continuous. However, this isimplied by pointiii ). In fact, more generally, the following result holds about the regularity ofsolutions.Theorem 1.1.9 (Regularity). Let f : U ! Rn, with U an open set of RRn. Supposethatf 2 Ck. Then all solutions of (1.2) are of class Ck+1.Proof. The proof is obvious, since a solution is such that 0 2 Ck.

1.1.3 Geometric interpretationThe function f is the vector field of the equation. At every point in (t, x) space, a

solutionis tangent to the value of the vector field at that point. A particularconsequence of thisfact is the following theorem.Theorem 1.1.10. Let x0 = 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 consistsof vectorsthat are parallel for all t (since f(t, x) = f(x) for all t). Suppose that a solution of

x0 = 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 (t1) = x0.ii) f(x0) = 0 and there exists t1 such that (t1) 6= x0.1.2. Existence and uniqueness theoremsFund. Theory ODE Lecture Notes J. Arino9t t t t t t1 0 2 1 0 2

Figure 1.3: Situations that would lead to a scalar autonomous differential equationhavingnonmonotone solutions.Suppose we are in case i), and assume we are in the case f(x0) > 0. Thus, thesolution curveis increasing at (t0, x0), i.e., 0(t0) > 0. As is continuous, i) implies that thereexistst2 2 (t0, t1) such that (t2) is a maximum, with increasing for t 2 [t0, t2) anddecreasingfor t 2 (t2, t1]. It follows that 0(t1) < 0, which is a contradiction with 0(t0) > 0.Now assume that we are in case ii). Then there exists t2 2 (t0, t1) with (t2) = x0butsuch that 0(t2) < 0. This is a contradiction.

Remark If we have uniqueness of solutions, it follows from this theorem that if1 and 2aretwo solutions of the scalar autonomous differential equation x0 = f(x), then 1(t0) < 2(t0)implies


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that 1(t) < 2(t) for all t.Remark Be careful: Theorem 1.1.10 is only true for scalar equations.

1.2 Existence and uniqueness theoremsSeveral approaches can be used to show existence and/or uniqueness ofsolutions. In Sections1.2.2 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 ofsolutionsunder the assumption that the vector field is Lipschitz. In Section 1.2.4, theLipschitzassumption 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 approximationsPicards 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 thesolution, thatconverges to the solution. The steps followed in constructing thisapproximating sequenceare the following.10Fund. Theory ODE Lecture Notes J. Arino1. General theory of ODEsStep 1. Start with an initial estimate of the solution, say, the constant function0(t) =0 = x0, for |t t0| h. Evidently, this function satisfies the IVP.Step 2. Use 0 in (1.4) to define the second element in the sequence:1(t) = x0 +Z tt0

f(s, 0(s))ds.Step 3. Use 1 in (1.4) to define the third element in the sequence:2(t) = x0 +Z tt0

f(s, 1(s))ds.

. . .Step n. Use n1 in (1.4) to define the nth element in the sequence:n(t) = x0 +Z tt0

f(s, n1(s))ds.At this stage, there are two major ways to tackle the problem, which use thesame idea:if we can prove that the sequence {n} converges, and that the limit happens tosatisfythe differential equation, then we have the solution to the IVP (1.3). The first

method(Section 1.2.2) uses a fixed point approach. The second method (Section 1.2.3)studies


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explicitly the limit.

1.2.2 Local existence and uniqueness Proof by fixed pointHere are two slightly different formulations of the same theorem, whichestablishes that if thevector field is continuous and Lipschitz, then the solutions exist and are unique.We prove

the result in the second case. For the definition of a Lipschitz function, seeSection 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 respectto x.Then, given any point (t0, x0) 2 U, there exists a unique solution of (1.3) onsome intervalcontaining t0 in its interior.

Theorem 1.2.2 (Picard local existence and uniqueness). Consider the IVP (1.3),and assumef is (piecewise) continuous in t and satisfies the Lipschitz conditionkf(t, x1) f(t, x2)k Lkx1 x2kfor all x1, x2 2 D = {x : kx x0k b} and all t such that |t t0| a. Then thereexists0 < = mina, bM

such that (1.3) has a unique solution in |t t0| .To set up the proof, we proceed as follows. Define the operator F byF : x 7! x0 +Z tt0

f(s, x(s))ds.1.2. Existence and uniqueness theoremsFund. Theory ODE Lecture Notes J. Arino11Note that the function (F)(t) is a continuous function of t. Then Picardssuccessives

approximations take the form 1 = F0, 2 = F1 = F20, where F2 represents F F.Iterating, the general term is given for k = 0, . . . byk = Fk0.Therefore, finding the limit limk!1 k is equivalent to finding the function , solutionof thefixed point problemx = Fx,with x a continuously differentiable function. Thus, a solution of (1.3) is a fixedpoint 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 . The proof for the interval t0 t


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is similar. Let X be the space of continuous functions defined on the interval [t0,t0 + ],X = C([t0, t0 + ]), that we endow with the sup norm, i.e., for x 2 X,kxkc = maxt2[t0,t0+]

kx(t)k

Recall that this norm is the norm of uniform convergence. Let thenS = {x 2 X : kx x0kc b}Of course, S X. Furthermore, S is closed, and X with the sup norm is acomplete metricspace. Note that we have transformed the problem into a problem involvingthe space ofcontinuous functions; hence we are now in an infinite dimensional case. Theproof proceedsin 3 steps.Step 1. We begin by showing that F : S ! S. From (1.4),(F)(t) x0 =

Z tt0

f(s, (s))ds=Z tt0

f(s, (s)) f(s, x0) + f(s, x0)dsTherefore, by the triangle inequality,kF x0kZ tt0

kf(s, (s)) f(t, x0)k + kf(t, x0)kdsAs f is (piecewise) continuous, it is bounded on [t0, t1] and there existsM =maxt2[t0,t1] kf(t, x0)k.ThuskF x0kZ tt0

kf(s, (s)) f(t, x0)k +Mds

Z tt0

Lk(s) x0k +Mds,

12Fund. Theory ODE Lecture Notes J. Arino1. General theory of ODEssince f is Lipschitz. Since 2 S for all k x0k b, we have that for all 2 S,kF x0kZ tt0

Lb +Mds(t t0)(Lb +M)As t 2 [t0, t0 + ], (t t0) , and thuskF x0kc = max[t0,t0+]kF x0k (Lb +M)Choose then such that b/(Lb +M), i.e., t sufficiently close to t0. Then we have


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kF x0kc bThis implies that for 2 S, F 2 S, i.e., F : S ! S.Step 2. We now show that F is a contraction. Let 1, 2 2 S,k(F1)(t) (F2)(t)k =

Z t

t0f(s, 1(s)) f(s, 2(s))ds

Z tt0

kf(s, 1(s)) f(s, 2(s))kds

Z tt0

Lk1(s) 2(s)kdsLk1 2kcZ tt0

dsand thuskF1 F2kc Lk1 2kc k1 2kc for

LThus, choosing < 1 and /L, F is a contraction. Since, by Step 1, F : S ! S, thecontraction mapping principle (Theorem A.11) implies that F has a unique fixedpoint 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 onX thatwe want to show the result). Considering a solution starting at x0 at time t0, thesolutionleaves S if there exists a t > t0 such that k(t)x0k = b, i.e., the solution crossesthe borderof D. Let > t0 be the first of such ts. For all t0 t ,k(t) x0kZ tt0

kf(s, (s)) f(s, x0)k + kf(s, x0)kds

Z tt0

Lk(s) x0k +Mds

Z tt0

Lb +Mds1.2. Existence and uniqueness theoremsFund. Theory ODE Lecture Notes J. Arino13

As a consequence,b = k( ) x0k (Lb +M)( t0)As = t0 + , for some > 0, it follows that if


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>bLb +Mthen the solution is confined to D.Note that the condition x1, x2 2 D = {x : kxx0k b} in the statement of thetheorem

refers to a local Lipschitz condition. If the function f is Lipschitz, then thefollowing theoremholds.Theorem 1.2.3 (Global existence). Suppose that f is piecewise continuous in tand isLipschitz on U = I D. Then (1.3) admits a unique solution on I.

1.2.3 Loc

ddt

R(t, t0

)

v(t) + R(t, t0)v0(t)= A(t)R(t, t0)v(t) + R(t, t0)v0(t),2.5. Further developments, bibliographical notesFund. Theory ODE Lecture Notes J. Arino55from Proposition 2.2.11. For to be solution, it must satisfy the differentialequation (2.19),and thus0(t) = A(t)(t) + g(t, (t)) , A(t)R(t, t0)v(t) +

ekt ektekT="ekT1 ekTf(c0)k

1 ekT al existence and uniqueness Proof by successiveapproximationsUsing 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)-planedefined,for a, b > 0, by R = {(t, x) : |t t0| a, kx x0k b}, and that f is locallyLipschitz in xon R. Let then, as previously defined,M = sup(t,x)2R

kf(t, x)k < 1and= min(a,

bM)


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Then the sequence defined by0 = x0, |t t0|i(t) = x0 +Z tt0

f(s, i1(s))ds, i 1, |t t0|

converges uniformly on the interval |t t0| to , unique solution of (1.3).Proof. We follow [20, p. 3-6].Existence. Suppose that |t t0| . Thenk1 0k =

Z tt0

f(s, 0(s))ds

M|t t0|M b

14Fund. Theory ODE Lecture Notes J. Arino1. General theory of ODEsfrom the definitions of M and , and thus k1 0k b. SoR tt0

f(s, 1(s))ds is defined for|t t0| , and, for |t t0| ,k2(t) 0k =

Z tt0

f(s, 1(s))ds

kZ tt0

kf(s, 1(s))kds M b.All subsequent terms in the sequence can be similarly defined, and, byinduction, for |tt0|,kk(t) 0k M b, k = 1, . . . , n.Now, for |t t0| ,kk+1(t) k(t)k =

x0 +Z tt0

f(s, k(s))ds x0 Z tt0

f(s, k1(s))ds

=

Z tt0


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f(s, k(s)) f(s, k1(s)) ds

LZ tt0

kk(s) k1(s)kds,

where the inequality results of the fact that f is locally Lipschitz in x on R.We now prove that, for all k,kk+1 kk b(L|t t0|)kk!for |t t0| (1.6)Indeed, (1.6) holds for k = 1, as previously established. Assume that (1.6) holdsfor k = n.Thenkn+2 n+1k =

Z tt0

f(s, n+1(s)) f(s, n(s))ds

Z tt0

Lkn+1(s) n(s)kds

Z tt0

Lb(L|s t0|)nn!ds for |t t0|bLn+1n!|t t0|n+1n + 1

s=ts=t0

b(L|t t0|)n+1(n + 1)!and thus (1.6) holds for k = 1, . . ..Thus, for N > n we havekN(t) n(t)kNX1k=n

kk+1(t) k(t)kNX1k=n

b(L|t t0|)kk!


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bNX1k=n

(L)kk!1.2. Existence and uniqueness theorems

Fund. Theory ODE Lecture Notes J. Arino15The rightmost term in this expression tends to zero as n ! 1. Therefore, {k(t)}convergesuniformly to a function (t) on the interval |t t0| . As the convergence isuniform, thelimit function is continuous. Moreover (t0) = x0. Indeed, N(t) = 0(t) +PNk=1(k(t) k1(t)), so (t) = 0(t) +P1

k=1(k(t) k1(t)).The fact that is a solution of (1.3) follows from the following result. If asequenceof functions {k(t)} converges uniformly and that the k(t) are continuous on theinterval|t t0| , thenlimn!1

Z tt0

n(s)ds =

Z tt0limn!1

n(s)dsHence,(t) = limn!1

n(t)= x0 + limn!1

Z tt0

f(s, n1(s))ds= x0 +Z tt0

limn!1

f(s, n1(s))ds= x0 +Z tt0

f(s, (s))ds,

which is to say that(t) = x0 +Z tt0


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f(s, (s))ds for |t t0| .As the integrand f(t, ) is a continuous function, is differentiable (with respect tot), and0(t) = f(t, (t)), so is a solution to the IVP (1.3).Uniqueness. Let and be two solutions of (1.3), i.e., for |t t0| ,(t) = x0 +

Z tt0f(s, (s))ds(t) = x0 +Z tt0

f(s, (s))ds.Then, for |t t0| ,k(t) (t)k =

Z tt0

f(s, (s)) f(s, (s))ds

LZ tt0

k(s) (s)kds. (1.7)We now apply Gronwalls Lemma A.7) to this inequality, using K = 0 and g(t) =k(t) (t)k. First, applying the lemma for t0 t t0 + , we get 0 k(t) (t)k 0, that is,k(t) (t)k = 0,16Fund. Theory ODE Lecture Notes J. Arino1. General theory of ODEsand thus (t) = (t) for t0 t t0 + . Similarly, for t0 t t0, k(t) (t)k = 0.Therefore, (t) = (t) on |t t0| .Example Let us consider the IVP x0 = x, x(0) = x0 = c, c 2 R. For initial solution, wechoose0(t) = c. Then1(t) = x0 +Z t0

f(s, 0(s))ds

= c +Z t0

0(s)ds= c cZ t0

ds= c ct.

To find 2, we use 1 in (1.4).2(t) = x0 +Z t0

f(s, 1(s))ds= c Z t


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0

(c cs)ds= c ct + ct22 .Continuing this method, we find a general term of the formn(t) =

Xni=0

c(1)itii! .

This is the power series expansion of cet, so n ! = cet (and the approximation isvalid onR), which is the solution of the initial value problem.Note that the method of successive approximations is a very general methodthat 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 Peanos existence theorem. Because thevector field isnot assumed to be Lipschitz, something is lost, namely, uniqueness.Theorem 1.2.5 (Peano). Suppose that f is continuous on some regionR = {(t, x) : |t t0| a, kx x0k b},with a, b > 0, and let M = maxR kf(t, x)k. Then there exists a continuousfunction (t),differentiable on R, such thati) (t0) = x0,1.2. Existence and uniqueness theorems

Fund. Theory ODE Lecture Notes J. Arino17ii) 0(t) = f(t, ) on |t t0| , where=

a if M = 0mina, bM

if M > 0.Before we can prove this result, we need a certain number of preliminarynotations andresults. The definition of equicontinuity and a statement of the Ascoli lemmaare given inSection A.5. To construct a solution without the Lipschitz condition, weapproximate thedifferential equation by another one that does satisfy the Lipschitz condition.The uniquesolution of such an approximate problem is an "-approximate solution. It is

formally definedas follows [8, p. 285].Definition 1.2.6 ("-approximate solution). A differentiable mapping u of an open


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ball J 2 Iinto U is an approximate solution of x0 = f(t, x) with approximation " (or an "-approximatesolution) if we haveku0(t) f(t, u(t))k ",for any t 2 J.

Lemma 1.2.7. Suppose that f(t, x) is continuous on a regionR = {(t, x) : |t t0| a, kx x0k b}.Then, for every positive number ", there exists a function F"(t, x) such thati) F" is continuous for |t t0| a and all x,ii) F" has continuous partial derivatives of all orders with respect to x1, . . . , xnfor |tt0|a and all x,iii) kF"(t, x)k maxR kf(t, x)k = M for |t t0| a and all x,iv) kF"(t, x) f(t, x)k " on R.See a proof in [12, p. 10-12]; note that in this proof, the property that f definesadifferential equation is not used. Hence Lemma 1.2.7 can be used in a moregeneral 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 ", a function F"(t, x) that satisfiesthe requirementsgiven in Lemma 1.2.7. Using an existence-uniqueness result in the Lipschitzcase(such as Theorem 1.2.2), we construct a function "(t) such that(P1) "(t0) = x0,

(P2) 0"(t) = F"(t, "(t)) on |t t0| < .(P3) (t, "(t)) 2 R on |t t0| .18Fund. Theory ODE Lecture Notes J. Arino1. General theory of ODEs2. The set F = {" : " > 0} is bounded and equicontinuous on |t t0| . Indeed,property (P3) of" implies that F is bounded on |t t0| and that kF"(t, "(t))k Mon |t t0| . Hence property (P2) of" implies thatk"(t1) "(t2)k M|t1 t2|,if |t1 t0| and |t2 t0| (this is a consequence of Theorem 1.1.7). Therefore,

for agiven positive number , we have k"(t1)"(t2)k whenever |t1 t0| , |t2 t0| ,and |t1 t2| /M.3. Using Lemma A.5, choose a sequence {"k : k = 1, 2, . . .} of positive numberssuch thatlimk!1 "k = 0 and that the sequence {"k : k = 1, 2, . . .} converges uniformly on |tt0|as k ! 1. Then set(t) = limk!1

"k(t) on |t t0| .

4. Observe that"(t) = x0 +Z t


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t0

F"(s, "(s))ds= x0 +Z tt0

f(s, "(s))ds +Z tt0

F"(s, "(s)) f(s, "(s))ds,and that it follows from iv) in Lemma 1.2.7 that

Z tt0

F"(s, "(s)) f(s, "(s))ds

"|t t0| on |t t0| .This is true for all " 0, so letting " ! 0, we obtain(t) = x0 +

Z tt0

f(s, (s))ds,which completes the proof.See [12, p. 13-14] for the outline of two other proofs of this result. A proof, byHartman[11, p. 10-11], now follows.Proof. Let > 0 and `(t) be a C1 n-dimensional vector-valued function on [t0 ,t0]satisfying `(t0) = x0, 0`(t0) = f(t0, x0) and k`(t) x0k b, k0

`(t)k M. For 0 < " ,define a function "(t) on [t0 , t0 + ] by putting "(t) = `(t) on [t0 , t0] and"(t) = x0 +Z tt0

f(s, "(s "))ds on [t0, t0 + ]. (1.8)The function " can indeed be thus defined on [t0 , t0 + ]. To see this, remarkfirst thatthis formula is meaningful and defines "(t) for t0 t t0 + 1, 1 = min(, "), so that"(t) is C1 on [t0 , t0 + 1] and, on this interval,k"(t) x0k b, k"(t) "(s)k M|t s|. (1.9)

1.2. Existence and uniqueness theoremsFund. Theory ODE Lecture Notes J. Arino19It then follows that (1.8) can be used to extend "(t) as a C1 function over [t0 ,t0 + 2],where 2 = min(, 2"), satisfying relation (1.9). Continuing in this fashion, (1.8)serves todefine "(t) over [t0, t0+] so that "(t) is a C0 function on [t0, t0+], satisfyingrelation(1.9).Since k0"(t)k M, M can be used as a Lipschitz constant for ", giving uniform continuityof". It follows that the family of functions, "(t), 0 < " , is equicontinuous.Thus, using Ascolis Lemma (Lemma A.5), there exists a sequence "(1) > "(2) >


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. . ., suchthat "(n) ! 0 as n ! 1 and(t) = limn!1

"(n)(t) exists uniformlyon [t0 , t0 + ]. The continuity of f implies that f(t, "(n)(t "(n)) tends uniformly

tof(t, (t)) as n ! 1; thus term-by-term integration of (1.8) where " = "(n) gives(t) = x0 +Z tt0

f(s, (s))dsand thus (t) is a solution of (1.3).An important corollary follows.Corollary 1.2.8. Let f(t, x) be continuous on an open set E and satisfy kf(t, x)kM.Let E0 be a compact subset of E. Then there exists an = (E,E0,M) > 0 with the

property that if (t0, x0) 2 E0, then the IVP (1.3) has a solution and every solutionexists on|t t0| .In fact, hypotheses can be relaxed a little. Coddington and Levinson [5] definean "-approximate solution asDefinition 1.2.9. An "-approximate solution of the differential equation (1.2),where f iscontinuous, on a t interval I is a function 2 C on I such thati) (t, (t)) 2 U for t 2 I;ii) 2 C1 on I, except possibly for a finite set of points S on I, where 0 may have

simplediscontinuities (g has finite discontinuities at c if the left and right limits of g atc existbut are not equal);iii) k0(t) f(t, (t))k " for t 2 I \ S.Hence it is assumed that has a piecewise continuous derivative on I, which isdenotedby 2 C1p (I).Theorem 1.2.10. Let f 2 C on the rectangleR = {(t, x) : |t t0| a, kx x0k b}.Given any " > 0, there exists an "-approximate solution of (1.3) on |t t0|such that(t0) = x0.20Fund. Theory ODE Lecture Notes J. Arino1. General theory of ODEsProof. Let " > 0 be given. We construct an "-approximate solution on theinterval [t0, t0+"];the construction works in a similar way for [t0 , t0]. The "-approximatesolution that we

construct is a polygonal path starting at (t0, x0).Since f 2 C on R, it is uniformly continuous on R, and therefore for the givenvalue of


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", there exists " > 0 such thatkf(t, ) f(t, )k " (1.10)if(t, ) 2 R, (t, ) 2 R and |t t| " k k ".Now divide the interval [t0, t0 + ] into n parts t0 < t1 < < tn = t0 + , in such away

thatmax |tk tk1| min

","

M

. (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 and M, it is clear that this line segment liesinside thetriangular region T bounded by the lines segments with slopes M from (t0, x0)to theirintercept at t = t0 + , and the line t = t0 + . In particular, (t1, x1) 2 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 isconstructed thatmeets the line t = t0 + in a finite number of steps, and lies entirely in T.The function , which can be expressed as

(t0) = x0(t) = (tk1) + f(tk1, (tk1))(t tk1), t 2 [tk1, tk], k = 1, . . . , n,(1.12)is the "-approximate solution that we seek. Clearly, 2 C1p ([t0, t0 + ]) andk(t) (t)k M|t t| for t, t 2 [t0, t0 + ]. (1.13)If t 2 [tk1, tk], then (1.13) together with (1.11) imply that k(t) (tk1)k ". Butfrom(1.12) and (1.10),k0(t) f(t, (t))k = kf(tk1, (tk1)) f(t, (t))k ".Therefore, is an "-approximation.

We can now turn to their proof of Theorem 1.2.5.Proof. Let {"n} be a monotone decreasing sequence of positive real numberswith "n ! 0as n ! 1. By Theorem 1.2.10, for each "n, there exists an "n-approximatesolution n of(1.3) on |t t0| such that n(t0) = x0. Choose one such solution n for each "n.From(1.13), it follows thatkn(t) n(t)k M|t t|. (1.14)1.2. Existence and uniqueness theoremsFund. Theory ODE Lecture Notes J. Arino21Applying (1.14) to t = t0, it is clear that the sequence {n} is uniformly boundedby


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kx0k + b, since |t t0| b/M. Moreover, (1.14) implies that {n} is anequicontinuousset. By Ascolis lemma (Lemma A.5), there exists a subsequence {nk}, k =1, . . ., of {n},converging uniformly on [t0, t0+] to a limit function , which must becontinuous since

each n is continuous.This limit function is a solution to (1.3) which meets the requiredspecifications. Tosee this, writen(t) = x0 +Z tt0

f(s, n(s)) + n(s)ds, (1.15)where n(t) = 0(t) f(t, n(t)) at those points where 0n exists, and n(t) = 0 otherwise.Because n is an "n-approximate solution, kn(t)k "n. Since f is uniformly

continuous onR, and nk ! uniformly on [t0 , t0 +] as k ! 1, it follows that f(t, nk) ! f(t, (t))uniformly on [t0 , t0 + ] as k ! 1.Replacing n by nk in (1.15) and letting k ! 1 gives(t) = x0 +Z tt0

f(s, (s))ds. (1.16)Clearly, (t0) = 0, when evaluated using (1.16), and also 0(t) = f(t, (t)) since f iscontinuous.Thus as defined by (1.16) is a solution to (1.3) on |t t0| .

1.2.5 Some examples of existence and uniquenessExample Consider the IVPx0 = 3|x|23

x(t0) = x0(1.17)Here, Theorem 1.2.5 applies, since f(t, x) = 3x2/3 is continuous. However, Theorem1.2.2 does notapply, since f(t, x) is not locally Lipschitz in x = 0 (or, f is not Lipschitz on any intervalcontaining0). 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 thefollowingargument. Suppose that f is Lipschitz on an interval I = (", "), with " > 0. Then, thereexistsL > 0 such that for all x1, x2 2 I,kf(t, x1) f(t, x2)k L|x1 x2|that is,3

|x1|2

3 |x2|23


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L|x1 x2|Since this has to hold true for all x1, x2 2 I, it must hold true in particular for x2 = 0.

Thus3|x1|2

3 L|x1|Given an " > 0, it is possible to find N" > 0 such that 1

n < " for all n N". Let x1 = 1n. Then forn N", if f is Lipschitz there must hold3

1n2

3

Ln

22Fund. Theory ODE Lecture Notes J. Arino1. General theory of ODEsSo, for all n N",n1

3

L3

This is a contradiction, since limn!1 n1/3 = 1, and so f is not Lipschitz on I.Let us consider the setE = {t 2 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 = (1, b),4) E = (a,+1),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 Theorem1.1.10, since(1.17) is defined by a scalar autonomous equation, its solutions are monotone. Forsimplicity, weconsider here the case of monotone increasing solutions. The case of monotonedecreasing solutionscan be treated in a similar fashion.1) Here, there is no intersection with the x = 0 axis. Thus if follows thatx(t) is

> 0, if x0 > 0< 0, if x0 < 02) In this case,x(t) is8 0, if t > b3) Here,x(t) is

= 0, if t < b> 0, if t > b4) In this case,x(t) is

< 0, if t < a= 0, if t > a5) In this last case, x(t) = 0 for all t 2 R.Now, depending on the sign of x, we can integrate the equation. First, if x > 0, then |x|= x andsox0 = 3x2/3,1

3x2/3x0 = 1, x1/3 = t + k1, x(t) = (t + k1)31.2. Existence and uniqueness theoremsFund. Theory ODE Lecture Notes J. Arino23for k1 2 R. Then, if x < 0, then |x| = x, andx0 = 3 (x)2/3,13(x)2/3(x0) = 1, (x)1/3 = t + k2, x(t) = (t + k2)3for k2 2 R. We can now use these computations with the different cases that werediscussed 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, wehavex(t) = (t + k1)3. Using the initial condition, we find that x(t0) = x0 = (t0 + k1)3, i.e.,k1 = x1/30 t0, and x(t) = (t + x1/30 t0)3.

2) If E = [a, b], thenx(t) =8 bSince x0 > 0, we have to be in the t > b region, so t0 > b, and (t0 +k1)3 = x0, whichimpliesthat k1 = x1/30 t0. Thusx(t) =

8 bSince x is continuous,limt!b,t>b

(t + x1/3

0 t0)3 = 0andlimt!a,t:(t + a)3 if t < a0 if t 2 [a, t0 x13

0 ] (a t0 x13

0 )(t + x1/30 t0)3 if t > t0 x13

0

Thus, choosing a t0 x1/30 , we have solutions of the form shown in Figure 1.4. Indeed, anyai satisfying this property yields a solution.3)

The case [a,+1) is impossible. Indeed, there does not exist a solution through (t0, x0)suchthat x(t) = 0 for all t 2 [a,+1); since we are in the case of monotone increasingfunctions,if x0 > 0 then x(t) x0 for all t t0.4) E = R is also impossible, for the same reason.24Fund. Theory ODE Lecture Notes J. Arino1. General theory of ODEsFigure 1.4: Case t0, x0 > 0, subcase 2, in the resolution of (1.17).5) For the case E = (1, b], we havex(t) =

0 if t 2 (1, b](t + k1)3 if t > bSince x(t0) = x0, k1 = x1/30 t0, and since x is continuous, b = k1 = t0 x1/30 . So,x(t) =(0 if t 2 (1, t0 x1/30 ](t + x1/3

0 t0)3 if t > t0 x1/30The other cases are left as an exercise.Example Consider the IVP


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x0 = 2tx2x(0) = 0(1.18)Here, we have existence and uniqueness of the solutions to (1.18). Indeed, f(t, x) =2tx2 is continuousand locally Lipschitz on R.

1.3 Continuation of solutionsThe 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 theinitial data.The continuation of solutions consists in studying criteria which allow to definesolutions onpossibly larger intervals.Consider the IVPx0 = f(t, x)x(t0) = x0,

(1.19)1.3. Continuation of solutionsFund. Theory ODE Lecture Notes J. Arino25with f continuous on a domain U of the (t, x) space, and the initial point (t0, x0)2 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 (t) satisfies the condition 0(t) = f(t, (t)) and (t, (t)) 2 U, inan

open interval I = {t1 < t < t2}. Under this assumption, if limj!1(j , (j)) = (t1, ) 2 Ufor some sequence {j : j = 1, 2, . . .} of points in the interval I, then lim!t1(, ( )) =(t1, ). Similarly, if limj!1(j , (j)) = (t2, ) 2 U for some sequence {j : j = 1, 2, . . .} ofpoints in the interval I, then lim!t2(, ( )) = (t2, ).Proof. Let W be an open neighborhood of (t1, ). Then (t, (t)) 2 W in an interval 1 0, there exists > 0 such that if |t t| < and kx0 x0k < ,thenk(t) (t)k < ", for t, t 2 I.Proof. The prooof is from [2, p. 135-136]. Since is the solution of (1.3) throughthe point(t0, x0), we have, for all t 2 I,(t) = x0 +Z tt0

f(s, (s))ds (1.21)30Fund. Theory ODE Lecture Notes J. Arino1. General theory of ODEsAs is the solution of (1.3) through the point (t0, x0), we have, for all t 2 I,(t) = x0 +Z tt0

f(s, (s))ds (1.22)Since Z tt0

f(s, (s))ds =Z t0t0

f(s, (s))ds +


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Z tt0

f(s, (s))ds,substracting (1.22) from (1.21) gives(t) (t) = x0 x0 +Z t0t0

f(s, (s))ds +Z tt0

f(s, (s)) f(s, (s))dsand thereforek(t) (t)k kx0 x0k +

Z t0t0

f(s, (s))ds

+

Z tt0

f(s, (s)) f(s, (s))ds

Using the boundedness assumptions on f and @f/@x to evaluate the right handside of thelatter inequation, we obtaink(t) (t)k kx0 x0k +M|t0 t0| + K

Z tt0

(s) (s)ds

If |t0 t0| < , kx0 x0k < , then we havek(t) (t)k +M + K

Z tt0

(s) (s)ds

(1.23)Applying Gronwalls inequality (Appendix A.7) to (1.23) givesk(t) (t)k (1 +M)eK|tt0| (1 +M)eK(21)using the fact that |t t0| < 2 1, if we denote I = (1, 2). Sincek (t) (t)k 0,and U the set of all (t, x, ) satisfying (t, x) 2 U, 2 I. Suppose f is acontinuous functionon U, bounded by a constant M there. For = 0, letx0 = f(t, x, )x(t0) = x0(1.24)

have a unique solution 0 on the interval [a, b], where t0 2 [a, b]. Then thereexists a > 0such that, for any fixed such that |0| < , every solution of (1.24) existsover [a, b]and as ! 0 ! 0uniformly over [a, b].Proof. We begin by considering t0 2 (a, b). First, choose an > 0 small enoughthat theregion R = {|t t0| , kx x0k M} is in U; note that R is a slight modificationof the usual security domain. All solutions of (1.24) with 2 I exist over [t0 ,

t0 + ]and remain in R. Let denote a solution. Then the set of functions {}, 2 I isanuniformly bounded and equicontinuous set in |t t0| . This follows from theintegralequation(t) = x0 +Z tt0

f(s, (s), )ds (|t t0| ) (1.25)and the inequality kfk M.

Suppose that for some t 2 [t0 , t0 + ], (t) does not tend to 0(t). Then thereexists a sequence {k}, k = 1, 2, . . ., for which k ! 0, and correspondingsolutions k


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such that k converges uniformly over [t0 , t0 +] as k ! 1 to a limit function ,with(t) 6= 0(t). From the fact that f 2 C on U, that 2 C on [t0 , t0 + ], and that

k

converges uniformly to , (1.25) for the solutions k yields(t) = x0 +

Z tt0f(s, (s), 0)ds (|t t0| )32Fund. Theory ODE Lecture Notes J. Arino1. General theory of ODEsThus is a solution of (1.24) with = 0. By the uniqueness hypothesis, itfollows that(t) = 0(t) on |t t0| . Thus (t) = 0(t). Thus all solutions on |t t0| tendto 0 as ! 0. 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 2 [t0, b), and suppose that the result is valid for every small h > 0over [t0, h]but not over [t0, + h]. It is clear that t0 + . By the above assumption, for anysmall" > 0, there exists a " > 0 such thatk( ") 0( ")k < " (1.26)for | 0| < ". Let H U be defined as the regionH = {|t | , kx 0( )k +M|t + |}with small enough that H U. Any solution of x0 = f(t, x, ) starting on t = withinitial value 0, |0 0( )| will remain in H as t increases. Thus all solutions canbe continued to + .By choosing " = in (1.26), it follows that for | 0| < ", the solutions can allbe continued to + ". Thus over [t0, + "] these solutions are in U so that theargumentthat ! 0 which has been given for |t t0| , also applies over [t0, + "]. Thus theassumption about the existence of < b is false. The case = b is treated insimilar fashionon t .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 ifsolutions exist, areunique, and that there is continuous dependence on initial conditions.

1.5 Generality of first order systemsConsider an nth order differential equation in normal formx(n) = ft, x, x0, . . . , x(n1)(1.27)

This equation can be reduced to a system of n first order ordinary differentialequations, byproceeding as follows. Let y0 = x, y1 = x0, y2 = x00, . . . , yn1 = x(n). Then (1.27)


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is equivalenttoy0 = F(t, y) (1.28)with y = (y0, y1, . . . , yn1)T andF(t, z) =2

666664y1y2...yn1f(t, y0, . . . , yn1)37777751.5. Generality of first order systemsFund. Theory ODE Lecture Notes J. Arino33Similarly, the IVP associated to (1.27) is given byx(n) = ft, x, x0, . . . , x(n1)x(t0) = x0, x0(t0) = x1, . . . , x(n1)(t0) = xn1(1.29)is equivalent to the IVPy0 = F(t, y)y(t0) = y0 = (x0, . . . , xn1)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 IVPx00 = 2x0 + 4x 3x(0) = 2, x0(0) = 1

To transform it into a system of first-order differential equations, we let y = x0.Substituting (wherepossible) y for x0 in the equation givesy0 = 2y + 4x 3

The initial condition becomes x(0) = 2, y(0) = 1. So finally, the following IVP isequivalent to theoriginal one:x0 = y

y0 = 4x 2y 3x(0) = 2, y(0) = 1Note that the linearity of the initial problem is preserved.Example The differential equationx(n)(t) = an1(t)x(n1)(t) + + a1(t)x0(t) + a0(t)x(t) + b(t)is an nth order nonhomogeneous linear differential equation. Together with the initialconditionx(n1)(t0) = x(n1)0 , . . . , x0(t0) = x00, x(t0) = x0where x0, x00

, . . . , x(n1)

0 2 R, it forms an IVP. We can transform it to a system of linear first orderequations by settingy0 = x


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y1 = x0...yn1 = x(n1)yn = x(n)34Fund. Theory ODE Lecture Notes J. Arino

1. General theory of ODEsThe nth order linear equation is then equivalent to the following system of n first orderlinearequationsy00 = y1y01 = y2...y0n2 = yn1y0

n1 = yny0n = an1(t)yn(t) + an2(t)yn1(t) + + a1(t)y1(t) + a0(t)y0(t) + b(t)under the initial conditionsyn1(t0) = x(n1)0 , . . . , y1(t0) = x00, y0(t0) = x0

1.6 Generality of autonomous systemsA nonautonomous systemx0(t) = f(t, x(t))

can be transformed into an autonomous system of equations by setting anauxiliary variable,say y, equal to t, givingx0 = f(y, x)y0 = 1.However, this transformation does not always make the system any easier tostudy.

1.7 Suggested reading, Further problemsMost of these results are treated one way or another in Coddington andLevinson [6] (first

edition published in 1955), and the current text, as many others, does little butparaphrasethem.We have not seen here any results specific to complex valued differentialequations. Ascomplex numbers are two-dimensional real vectors, the results carry throughto the complexcase by simply assuming that if, in (1.2), we consider an n-dimensionalcomplex vector, thenthis is equivalent to a 2n-dimensional problem. Furthermore, if f(t, x) is analyticin t andx, then analytic solutions can be constructed. See Section I-4 in [12], ..., forexample.


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Chapter 2Linear systemsLet I be an interval of R, E a normed vector space over a field K (E = Kn, with K= R

or C), and L(E) the space of continuous linear maps from E to E. Let k k be anorm 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 definedbyx0(t) = A(t)x(t) + B(t) (2.1)where the unknown x is a map on I, taking values in E, defined differentiable ona subintervalof I. We restrict ourselves to the finite dimensional case (E = Kn). Hence weconsider A 2Mn(K), nn matrices over the field K, and B 2 Kn. We suppose thatA 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) shouldbe calledan affine system, with associated linear systemx0(t) = A(t)x(t). (2.2)Another way to distinguish systems (2.1) and (2.2) is to refer to the former as anonhomogeneouslinear system and the latter as an homogeneous linear system. In order tolighten

the 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 linearsystem if weconsider the generic properties of (2.1), with (2.2) as a particular case, as inthis chapterstitle or in the next section, for example.

2.1 Existence and uniqueness of solutionsTheorem 2.1.1. Let A and B be defined and continuous on I 3 t0. Then, for all x0

2 E,there exists a unique solution t(x0) of (2.1) through (t0, x0), defined on theinterval I.3536Fund. Theory ODE Lecture Notes J. Arino2. Linear systemsProof. Let k(t) = |||A(t)||| = supkxk1 kA(t)xk. Then for all t 2 I and all x1, x2 2 K,kf(t, x1) f(t, x2)k = kA(t)(x1 x2)k|||A(t)||| kx1 x2kk(t)kx1 x2k,

where the inequalitykA(t)(x1 x2)k |||A(t)||| kx1 x2kresults from the nature of the norm ||| ||| (see Appendix A.1). Furthermore, k is


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continuouson I. Therefore the conditions of Theorem 1.2.2 hold, leading to existence anduniquenesson the interval I.With linear systems, it is possible to extend solutions easily, as is shown by thenext

theorem.Theorem 2.1.2. Suppose that the entries of A(t) and the entries of B(t) arecontinuous onan open interval I. Then every solution of (2.1) which is defined on asubinterval 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 of (2.1) is defined on J = (1,2),with J ( I. Thenk(t)k k(t0)k +

Z tt0

A(s)(s) + B(s)dsfor all t2J,where t0 2J.Let

K = k(t0)k + (2 1) max1t2 kB(t)kL = max1t2 kA(t)kThen, for t0, t 2 J ,k(t)k K + L

Z tt0

(s)ds

K + LZ tt0

k(s)kds.Thus, using Gronwalls Lemma (Lemma A.7), the following estimate holds in J ,k(t)k KeL|tt0| KeL(21) < 1This implies that case ii) in Corollary 1.3.4 is ruled out, leaving only thepossibility for tobe extendable over I, since the vector field in (2.1) is Lipschitz.

2.2 Linear systemsWe begin our study of linear systems of ordinary differential equations by

considering homogeneoussystems of the form (2.2) (linear systems), with x 2 Rn and A 2Mn(R), the setof square matrices over the field R, A having continuous entries on an interval


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I.2.2. Linear systemsFund. Theory ODE Lecture Notes J. Arino37

2.2.1 The vector space of solutionsTheorem 2.2.1 (Superposition principle). Let S0 be the set of solutions of (2.2)

that aredefined on some interval I R. Let 1, 2 2 S0, and 1, 2 2 R. Then 11 + 22 2 S0.Proof. Let 1, 2 2 S0 be two solutions of (2.2), 1, 2 2 R. Then for all t 2 I,0

1 = A(t)10

2 = A(t)2,from which it comes thatddt

(11

+22

) = A(t)[11

+22

],implying that 11 + 22 2 S0.Thus the linear combination of any two solutions of (2.2) is in S0. This is a hintthat S0must be a vector space of dimension n on K. To show this, we need to find abasis of S0.We proceed in the classical manner, with the notable difference from classicallinear 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 lineardifferential

equation (2.2), all defined on the same open interval I, is called a fundamentalset 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 thesystem(2.2) has a fundamental set of solutions defined on I.Proof. Let t0 2 I, and e1, . . . , en denote the canonical basis of Kn. Then, fromTheorem 2.1.1,there exists a unique solution (t0) = (1(t0), . . . , n(t0)) such that i(t0) = ei, for i =1, . . . , n. Furthermore, from Theorem 2.1.1, each function i is defined on theinterval I.Assume that {i}, i = 1, . . . , n, is linearly dependent. Then there exists i 2 R, i =1, . . . , n, not all zero, such thatPni=1 ii(t) = 0 for all t. In particular, this is true fort = t0, and thusPni=1 ii(t0) =Pni=1 iei = 0, which implies that the canonical basis ofKn is linearly dependent. Hence a contradiction, and the i are linearly

independent.Proposition 2.2.4. If F is a fundamental set of solutions of the linear system(2.2) on the


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open interval I, then every solution defined on I can be expressed as a linearcombinationof the elements of F.Let t0 2 I, we consider the applicationt0 : S0 ! KnY 7! t0(x) = x(t0)

Lemma 2.2.5. t0 is a linear isomorphism.38Fund. Theory ODE Lecture Notes J. Arino2. Linear systemsProof. t0 is bijective. Indeed, let v 2 Kn, from Theorem 2.1.1, there exists aunique solutionpassing through (t0, v), i.e.,8v 2 Kn, 9!x 2 S0, x(t0) = v ) t0(x) = v,so t0 is surjective. That t0 is injective follows from uniqueness of solutions to anODE.Furthermore, t0(1x1+2x2) = 1t0(x1)+2t0(x2). Therefore dim S0 = dimKn = n.

2.2.2 Fundamental matrix solutionDefinition 2.2.6. An n n matrix function t 7! (t), defined on an open interval I,iscalled a matrix solution of the homogeneous linear system (2.2) if each of itscolumns is a(vector) solution. A matrix solution is called a fundamental matrix solution if itscolumnsform a fundamental set of solutions. If in addition (t0) = I, a fundamental matrixsolutionis called the principal fundamental matrix solution.

An important property of fundamental matrix solutions is the following, knownas Abelsformula.Theorem 2.2.7 (Abels formula). Let A(t) be continuous on I and 2 Mn(K) besuchthat 0(t) = A(t)(t) on I. Then det satisfies on I the differential equation(det)0 = (trA)(det ),or, in integral form, for t, 2 I,det (t) = det( ) expZ t

trA(s)ds

. (2.3)Proof. Writing the differential equation 0(t) = A(t)(t) in terms of the elements 'ijandaij of, respectively, and A,'0ij(t) =Xnk=1

aik(t)'kj(t), (2.4)

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


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'11(t) '12(t) . . . '1n(t)'21(t) '22(t) . . . '2n(t)'n1(t) 'n2(t) . . . 'nn(t)

,we see that

(det)0 =

'011 '012 . . . '01n

'21 '22 . . . '2n'n1 'n2 . . . 'nn

+

'11 '12 . . . '1n'021 '022 . . . '02n

'n1 'n2 . . . 'nn

+ +

'11 '12 . . . '1n'21 '22 . . . '2n

'0n1 '0n2 . . . '0nn

.2.2. Linear systemsFund. Theory ODE Lecture Notes J. Arino39Indeed, write det (t) = (r1, r2, . . . , rn), where ri is the ith row in (t). is then alinear

function of each of its arguments, if all other rows are constant, which impliesthatddtdet (t) =

ddtr1, r2, . . . , rn

+

r1,d


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dtr2, . . . , rn

+ +

r1, r2, . . . ,

ddtrn

.(To show this, use the definition of the derivative as a limit.) Using (2.4) on thefirst of then determinants in (det )0 gives

Pk a1k'k1Pk a1k'k2 . . .Pk a1k'kn'21 '22 . . . '2n'n1 'n2 . . . 'nn

.Adding a12 times the second row, a13 times the first row, etc., a1n times thenth row,

to the first row, does not change the determinant, and thus

Pk a1k'k1Pk a1k'k2 . . .Pk a1k'kn'21 '22 . . . '2n'n1 'n2 . . . 'nn

=

a11'11 a11'12 . . . a11'1n'21 '22 . . . '2n'n1 'n2 . . . 'nn

= a11 det.Repeating this for each of the terms in (det )0, we obtain (det )0 = (a11 + a22 + +ann) det , giving finally (det )0 = (trA)(det ). Note that this equation takes theformu0 (t)u = 0, which implies thatu expZ t


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(s)ds

= constant,which in turn implies the integral form of the formula.Remark Consider (2.3). Suppose that 2 I is such that det ( ) 6= 0. Then, since ea 6=0for any a, it follows that det 6= 0 for all t 2 I. In short, linear independence of solutionsfor at 2 I is equivalent to linear independence of solutions for all t 2 I. As a consequence,the columnvectors of a fundamental matrix are linearly independent at every t 2 I.Theorem 2.2.8. A solution matrix of (2.2) is a fundamental solution matrix on Iif,and only if, det (t) 6= 0 for all t 2 IProof. Let be a fundamental matrix with column vectors i, and suppose that isanynontrivial solution of (2.2). Then there exists c1, . . . , cn, not all zero, such that=Xnj=1

cjj ,or, writing this equation in terms of , = c, if c = (c1, . . . , cn)T . At any point t0 2I,this is a system of n linear equations with n unknowns c1, . . . , cn. This systemhas a unique40Fund. Theory ODE Lecture Notes J. Arino2. Linear systemssolution for any choice of (t0). Thus det (t0) 6= 0, and by the remark above,det(t) 6= 0for all t 2 I.Reciproqually, let be a solution matrix of (2.2), and suppose that det(t) 6= 0fort 2 I. Then the column vectors are linearly independent at every t 2 I.From the remark above, the condition det (t) 6= 0 for all t 2 I in Theorem2.2.8 isequivalent to the condition there exists t 2 I such that det (t) 6= 0. Afrequent candidate

for this role is t0.To conclude on fundamental solution matrices, remark that there are infinitelymanyof them, for a given linear system. However, since each fundamental solutionmatrix canprovide a basis for the vector space of solutions, it is clear that thefundamental matricesassociated to a given problem must be linked. Indeed, we have the followingresult.Theorem 2.2.9. Let be a fundamental matrix solution to (2.2). Let C 2 Mn(K) bea

constant nonsingular matrix. Then C is a fundamental matrix solution to (2.2).Conversely,if is another fundamental matrix solution to (2.2), then there exists a


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constantnonsingular C 2Mn(K) such that (t) = (t)C for all t 2 I.Proof. Since is a fundamental matrix solution to (2.2), we have(C)0 = 0C = (A(t))C = A(t)(C),and thus C is a matrix solution to (2.2). Since is a fundamental matrix solutionto

(2.2), Theorem 2.2.8 implies that det 6= 0. Also, since C is nonsingular, detC6= 0. Thus,detC = detdetC 6= 0, and by Theorem 2.2.8, C is a fundamental matrixsolution to(2.2).Conversely, assume that and are two fundamental matrix solutions.Since 1 = I,taking the derivative of this expression gives 01 + (1)0 = 0, and therefore (1)0=101. We now consider the product 1 . There holds1

0

=10

+ 1 0= 101 + 1A(t)=1A(t)1 + 1A(t)

=1A(t) + 1A(t)

= 0.Therefore, integrating (1 )0 gives 1 = C, with C 2 Mn(K) is a constant.Thus,

= C. Furthermore, as and are fundamental matrix solutions, det 6= 0

anddet 6= 0, and therefore detC 6= 0.Remark Note that if is a fundamental matrix solution to (2.2) and C 2Mn(K) is aconstantnonsingular matrix, then it is not necessarily true that C is a fundamental matrixsolution to(2.2). See Exercise 2.3.2.2. Linear systemsFund. Theory ODE Lecture Notes J. Arino41

2.2.3 Resolvent matrixIf t 7! (t) is a matrix solution of (2.2) on the interval I, then 0(t) = A(t)(t) on I.Thus,by Proposition 2.2.3, there exists a fundamental matrix solution.


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Definition 2.2.10 (Resolvent matrix). Let t0 2 I and (t) be a fundamental matrixsolutionof (2.2) on I. Since the columns of are linearly independent, it follows that (t0)isinvertible. The resolvent (or state transition matrix) of (2.2) is then defined asR(t, t0) = (t)(t0)1.

It is evident that R(t, t0) is the principal fundamental matrix solution at t0 (sinceR(t0, t0) = (t0)(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-Kolmogorovidentities1) R(t, t) = I,2) R(t, s)R(s, u) = R(t, u),as well as the identities3) 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) = (t)1(t) = I.Also,2) givesR(t, s)R(s, u) = (t)1(s)(s)1(u) = (t)1(u) = R(t, u).The other equalities are equally easy to establish. Indeed,R(t, s)1 =(t)1(s)1

=1(s)1

(t)1 = (s)1(t) = R(s, t),whence 3). Also,@@sR(t, s) =@

@s(t)1(s)

= (t)

@@s1(s)

As is a fundamental matrix solution, 0 exists and is nonsingular, and

differentiating1 = I gives@


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@s(s)1(s)

= 0 ,

@@s(s)

1(s) + (s)

@@s1(s)

= 0, (s)

@@s1(s)

=

@@s

(s)

1(s),@@s1(s) = 1(s)

@@s(s)

1(s).42Fund. Theory ODE Lecture Notes J. Arino2. Linear systemsTherefore,@@sR(t, s) = (t)1(s)

@@s(s)


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1(s) = R(t, s)

@@s(s)

1(s).Now, since (s) is a fundamental matrix solution, it follows that @(s)/@s = A(s)(s), andthus@@sR(t, s) = R(t, s)A(s)(s)1(s) = R(t, s)A(s),giving 4). Finally,@@tR(t, s) =@@t(t)1(s)= A(t)(t)1(s) since 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,t0

defined byt0 : S ! Kn

x 7! x(t0),is a K-linear isomorphism from the space S to the space Kn. Then R is anapplication fromKn to Kn,R(t, t0) : Kn ! Knv 7! R(t, t0)v = wsuch thatR(t, t0) = t 1t0

i.e.,(R(t, t0)v = w) , (9x 2 S, w = x(t), v = x(t0)) .

Since t and t0 are K-linear isomorphisms, R is a K-linear isomorphism on Kn. ThusR(t, t0) 2Mn(K) and is invertible.Proposition 2.2.12. R(t, t0) is the only solution in Mn(K) of the initial valueproblemddtM(t) = A(t)M(t)M(t0) = I,with M(t) 2Mn(K).2.2. Linear systemsFund. Theory ODE Lecture Notes J. Arino

43Proof. Since d(R(t, t0)v)/dt = A(t)R(t, t0)v,


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ddtR(t, t0)

v = (A(t)R(t, t0)) v,for all v 2 Rn. Therefore, R(t, t0) is a solution to M0 = 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 homogeneousnonautonomoussystem (2.2) with initial condition x(t0) = x0 is given by(t) = R(t, t0)x0.

2.2.4 WronskianDefinition 2.2.14. The Wronskian of a system {x1, . . . , xn} of solutions to (2.2)is given

byW(t) = det(x1(t), . . . , xn(t)).Let vi = xi(t0). Then we havexi(t) = R(t, t0)vi,and it follows thatW(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) = expZ tt0

trA(s)ds

(2.5a)W(t) = expZ tt0

trA(s)ds

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

2.2.5 Autonomous linear systemsAt this point, we know that solutions to (2.2) take the form (t) = R(t, t0)x0, but

thiswas obtained formally. We have no indication whatsoever as to the preciseform of R(t, t0).Typically, finding R(t, t0) can be difficult, if not impossible. There are howevercases wherethe resolvent can be explicitly computed. One such case is for autonomouslinear systems,which take the formx0(t) = Ax(t), (2.6)that is, where A(t) A. Our objective here is to establish the following result.44

Fund. Theory ODE Lecture Notes J. Arino2. Linear systemsLemma 2.2.15. If A(t) A, then R(t, t0) = e(tt0)A for all t, t0 2 I.


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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(tt0)A involves thenotion 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 linearequations, weseek a 2 K such that (t) = etv be a solution to (2.6) with v 2 Kn \ {0}. We have0

= etv,and thus is a solution if, and only if,etv = Aetv= etAv, v = Av, (A I)v = 0 (with I the identity matrix).As v = 0 is not the only solution, this implies that A I must not be invertible,and sois a solution , det(A I) = 0,i.e., is an eigenvalue of A.In the simple case where A is diagonalizable, there exists a basis (v1, . . . , vn)of Kn, withv1, . . . , vn the eigenvectors of A corresponding to the eigenvalues 1, . . . , n. Wethen obtainn linearly independent solutions i(t) = ei(tt0), i = 1, . . . , n. The general solutionis given

by(t) =e1(tt0)x01, . . . , en(tt0)x0n

,where x0i is the ith component of x0, i = 1, . . . , n. In the general case, we needthe notionof matrix exponentials. Defining the exponential of matrix A aseA =X1k=0Ann!(see Appendix A.10), we have the following result.Theorem 2.2.16. The global solution on K of (2.6) such that (t0) = x0 is given by(t) = e(tt0)Ax0.Proof. Assume = e(tt0)Ax0. Then (t0) = e0Ax0 = Ix0 = x0. Also,(t) =X1n=0

1n!(t t0)nAn


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!x0=X1n=0

1

n!(t t0)nAnx0,2.2. Linear systemsFund. Theory ODE Lecture Notes J. Arino45so is a power series with radius of convergence R = 1. Therefore, isdifferentiable on Rand0(t) =X1n=1

1n!n(t t0)n1Anx0=X1n=0

1(n + 1)!(n + 1)(t t0)nAn+1x0=X1n=0

1n!(t t0)nAn+1x0= AX1n=0

1n!(t t0)nAnx0!= A(t)so 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 casewhere Ais diagonalizable, solutions take the form(t) =e1(tt0)x01, . . . , en(tt0)x0n

,

which implies that, in this case, the matrix R(t, t0) takes the formR(t, t0) =0


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BBB@e1(tt0) 0 00 e2(tt0) 0.... . .0 0 en(tt0)

1CCCA.In the general case, we need the notion of generalized eigenvectors.Definition 2.2.17 (Generalized eigenvectors). Let be an eigenvalue of the n nmatrixA, with multiplicity m n. Then, for k = 1, . . . ,m, any nonzero solution v of(A I)kv = 0is called a generalized eigenvector of A.Theorem 2.2.18. Let A be a real n n matrix with real eigenvalues 1, . . . , nrepeatedaccording to their multiplicity. Then there exists a basis of generalizedeigenvectors for Rn.And if {v1, . . . , vn} is any basis of generalized eigenvectors for Rn, the matrix P= [v1 vn]is invertible,A = D + N,whereP1DP = diag(j),the matrix N = A D is nilpotent of order k n, and D and N commute.46

Fund. Theory ODE Lecture Notes J. Arino2. Linear systems

2.3 Affine systemsWe consider the general (affine) problem (2.1), which we restate here forconvenience. Letx 2 Rn, A : I ! L(E) and B : I ! E, where I R and E is a normed vector space, weconsider the systemx0(t) = A(t)x(t) + B(t) (2.1)

2.3.1 The space of solutionsThe 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 superpositionprinciple 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 solutionof theassociated homogeneous equation (2.2).Proof. Since x1 and x2 are solutions of (2.1),x01 = A(t)x1 + B(t)x02 = A(t)x2 + B(t)Therefored


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dt(x1 x2) = A(t)(x1 x2)Theorem 2.3.2. The global solutions of (2.1) that are defined on I form an ndimensionalaffine subspace of the vector space of maps from I to Kn.Theorem 2.3.3. Let V be the vector space over R of solutions to the linear

system x0 =A(t)x. If is a particular solution of the affine system (2.1), then the set of allsolutions of(2.1) is precisely{ + , 2 V }.Practical rules:1. To obtain all solutions of (2.1), all solutions of (2.2) must be added to aparticularsolution of (2.1).2. To obtain all solutions of (2.2), it is sufficient to know a basis of S0. Such abasis iscalled a fundamental system of solutions of (2.2).2.3.2 Construction of solutionsWe have the following variation of constants formula.2.3. Affine systemsFund. Theory ODE Lecture Notes J. Arino47Theorem 2.3.4. Let R(t, t0) be the resolvent of the homogeneous equation x0 =A(t)x associatedto (2.1). Then the solution x to (2.1) is given byx(t) = R(t, t0) +

Z tt0R(t, s)B(s)ds (2.7)Proof. Let R(t, t0) be the resolvent of x0 = A(t)x. Any solution of the latterequation isgiven byx(t) = R(t, t0)v, v 2 RnLet 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

x0(t) =ddt[R(t, t0)]v(t) + R(t, t0)v0(t)= A(t)R(t, t0)v(t) + R(t, t0)v0(t)Thus x is a solution to (2.1) ifA(t)R(t, t0)v(t) + R(t, t0)v0(t) = A(t)R(t, t0)v(t) + B(t), R(t, t0)v0(t) = B(t), v0(t) = R(t0, t)B(t)since R(t, s)1 = R(s, t). Therefore, v(t) =R tt0

R(t0, s)B(s)ds. A particular solution is givenby


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x(t) = R(t, t0)Z tt0

R(t0, s)B(s)ds=Z tt0

R(t, t0)R(t0, s)B(s)ds=Z tt0

R(t, s)B(s)ds

2.3.3 Affine systems with constant coefficientsWe consider the affine equation (2.1), but with the matrix A(t) A.Theorem 2.3.5. The general solution to the IVPx0(t) = Ax(t) + B(t)x(t0) = x0

(2.8)is given byx(t) = e(tt0)Ax0 +Z tt0

e(tt0)AB(s)ds (2.9)Proof. Use Lemma 2.2.15 and the variation of constants formula (2.7).48Fund. Theory ODE Lecture Notes J. Arino2. Linear systems

2.4 Systems with periodic coefficients

2.4.1 Linear systems: Floquet theoryWe consider the linear system (2.2) in the following case,x0 = A(t)xA(t + !) = A(t), 8t,(2.10)with entries of A(t) continuous on R.Definition 2.4.1 (Monodromy operator). Associated to system (2.10) is theresolvent R(t, s).For all s 2 R, the operatorC(s) := R(s + !, s)

is called the monodromy operator.Theorem 2.4.2. If X(t) is a fundamental matrix for (2.10), then there exists anonsingularconstant matrix V such that, for all t,X(t + !) = X(t)V.This matrix takes the formV = X1(0)X(!),and is called the monodromy matrix.Proof. Since X is a fundamental matrix solution, there holds that X0(t) = A(t)X(t)for all t.Therefore X0(t+!) = A(t+!)X(t+!), and by periodicity of A(t), X0(t+!) =

A(t)X(t+!),which implies that X(t + !) is a fundamental matrix of (2.10). As aconsequence, by


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Theorem 2.2.9, there exists a matrix V such that X(t + !) = X(t)V .Since at t = 0, X(!) = X(0)V , it follows that V = X1(0)X(!).Theorem 2.4.3 (Floquets theorem, complex case). Any fundamental matrixsolution of(2.10) takes the form(t) = P(t)etB (2.11)

where P(t) and B are n n complex matrices such thati) P(t) is invertible, continuous, and periodic of period ! in t,ii) B is a constant matrix such that (!) = e!B.Proof. Let be a fundamental matrix solution. From 2.4.2, the monodromymatrix V =1(0)(!) is such that (t + !) = (t)V . By Theorem A.11.1, there exists B 2 Mn(C)2.4. Systems with periodic coefficientsFund. Theory ODE Lecture Notes J. Arino49such that eB! = V . Let P(t) = (t)eBt, so (t) = P(t)eBt. It is clear that P iscontinuousand nonsingular. Also,P(t + !) = (t + !)eB(t+!)= (t)V eB(!+t)= (t)eB!eB!eBt= (t)eBt= P(t),proving the P is !-periodic.Theorem 2.4.4 (Floquets theorem, real case). Any fundamental matrix solutionof (2.10)takes the form

(t) = P(t)etB (2.12)where P(t) and B are n n real matrices such thati) P(t) is invertible, continuous, and periodic of period 2! in t,ii) B is a constant matrix such that (!)2 = e2!B.Proof. The proof works similarly as in the complex case, except that here,Theorem A.11.1implies that there exists B 2 Mn(R) such that e2!B = V 2. Let P(t) = (t)eBt, so(t) = P(t)etB. It is clear that P is continuous and nonsingular. Also,P(t + 2!) = (t + 2!)e(t+2!)B= (t + !)V e(2!+t)B= (t)V 2e(2!+t)B

= (t)e2!Be2!BetB= (t)etB= P(t),proving the P is !-periodic.See [12, p. 87-90], [4, p. 162-179].Theorem 2.4.5 (Floquets theorem, [4]). If (t) is a fundamental matrix solutionof the!-periodic system (2.10), then, for all t 2 R,(t + !) = (t)1(0)(!).In addition, for each possibly complex matrix B such thate!B = 1(0)(!),there is a possibly complex !-periodic matrix function t 7! P(t) such that (t) =P(t)etB forall t 2 R. Also, there is a real matrix R and a real 2!-periodic matrix function t !


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Q(t)such that (t) = Q(t)etR for all t 2 R.50Fund. Theory ODE Lecture Notes J. Arino2. Linear systemsDefinition 2.4.6 (Floquet normal form). The representation (t) = P(t)etR is called

aFloquet normal form.In the case where (t) = P(t)etB, we have dP(t)/dt = A(t)P(t) P(t)B. Therefore,letting x = P(t)z, we obtain x0 = P(t)x0 + dP(t)/dtx = P(t)A(t)x + A(t)P(t)x P(t)Bxz = P1(t)x, so z0 = dP1(t)dt x + P1(t)x0 = dP1(t)dt P(t)z + P1(t)A(t)P(t)zDefinition 2.4.7 (Characteristic multipliers). The eigenvalues 1, . . . , n of amonodromymatrix B are called the characteristic multipliers of equation (2.10).Definition 2.4.8 (Characteristic exponents). Numbers such that e! is acharacteristicmultiplier 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 2 GLn(K) iswrittenC = eB, then the eigenvalues of C coincide with the exponentials of theeigenvalues of B,with same multiplicity.Definition 2.4.10 (Characteristic exponents). The eigenvalues 1, . . . , n of amonodromy

matrix B are called the characteristic exponents of equation (2.10). Theexponents 1 =exp(2!1), . . . , n = exp(2!n) of the matrix (!)2 are called the (Floquet) multipliersof(2.10).Proposition 2.4.11. Suppose that X, Y are fundamental matrices for (2.10) andthat X(t+!) = X(t)V , Y (t + !) = Y (t)U. Then the monodromy matrices U and V aresimilar.Proof. Suppose that X(t + !) = X(t)V and Y (t + !) = Y (t)U. But, by Theorem2.2.9,

since X and Y are fundamental matrices for (2.10), there exists an invertiblematrix C suchthat X(t) = Y (t)C for all t. Thus, in particular, X(t + !) = Y (t + !)C, and soC1UCX(t + !) = Y (t + !)C = Y (t)UC = X(t)C1UC,since Y (t) = X(t)C1. It follows that V = C1UC, so U and V are similar.From this Proposition, it follows that monodromy matrices share the samespectrum.Corollary 2.4.12. All solutions of (2.10) tend to 0 as t ! 1 if and only if |j | < 1 forallj (or


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Fund. Theory ODE Lecture Notes J. Arino51

2.4.2 Affine systems: the Fredholm alternativeWe discuss here an extension of a theorem that was proved implicitly inExercise 4, Assignment2. Let us start by stating the result in question. We consider here the system

x0 = A(t)x + b(t), (2.13)where x 2 Rn, A 2Mn(R) and b 2 Rn, with A and b continuous and !-periodic.Theorem 2.4.13. If the homogeneous equationx0 = A(t)x (2.14)associated to (2.13) has no nonzero solution of period !, then (2.13) has foreach functionf, a unique !-periodic solution.The Fredholm alternative concerns the case where there exists a nonzeroperiodic 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 followingdifferentialequation,y0 = 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 scalarproducthx(t), y(t)i 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 aperiodicsolution of (2.15) is y0 such thatRT (0, !)y0 = y0This can also be written asRT (0, !) I

y0 = 0or, taking the transpose,yT0 [R(0, !) I] = 0Now, since R(0, !) = R1(!, 0), it follows thatyT0 [R(0, !) I] = 0 , yT0

R1(!, 0) I


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= 0This is equivalent to yT0 [R(0, !) I] = 0. The latter equation has as many solutions as[R(0, !) I] x0 = 0; the number of these depends on the rank of R(!, 0) I.52Fund. Theory ODE Lecture Notes J. Arino

2. Linear systemsiii) Recall that for differentiable functions a, b,ddtha(t), b(t)i = hddta(t), b(t)i + ha(t),ddtb(t)iThusddthx(t), y(t)i = hA(t)x(t), y(t)i + hx(t),AT (t)y(t)i = 0Before we carry on to the actual Fredholm alternative in the context of ordinarydifferentialequations, let us consider the problem in a more general setting. Let H be aHilbertspace. If A 2 L(H,H), the adjoint operator Aof A is the element of L(H,H) suchthat

8u, v 2 H, hAu, vi = hu,AviLet 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 asolution, 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 thecontext ofODEs.Theorem 2.4.16 (Fredholm alternative for ODEs). Consider (2.13) with A and f

continuousand !-periodic. Suppose that the homogeneous equation (2.14) has pindependentsolutions of period !. Then the adjoint equation (2.15) also has p independentsolutions ofperiod p, which we denote y1, . . . , yp. Theni) If Z !0

hyk(t), b(t)idt = 0, k = 1, . . . , p (2.16)then there exist p independent solutions of (2.13) of period !, and,ii) if this condition is not fulfilled, (2.13) has no nontrivial solution of period !.

Proof. First, remark that x0 is the initial condition of a periodic solution of (2.13)if, andonly if,


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[R(0, !) I] x0 =Z !0

R(0, s)b(s)ds (2.17)By Theorem 2.3.4, the solution of (2.13) through (0, x0) is given byx(t) = R(t, 0)x0 +

Z t0R(t, s)b(s)dsHence, at time !,x(!) = R(!, 0)x0 +Z !0

R(!, s)

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