Notes on Ordinary Differential Equations - X

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1 Notes on Ordinary Differential Equations - X Pedro Martins Rodrigues November 16, Stability of Orbits and non-autonomous linear equations The stability of an equilibrium was studied before (see Notes IX) using the linearization of the vector field at that point. The same approach may be applied, in principle, to the study of stability of other orbits: suppose that Ψ(t, x) is an orbit of the flow of ẋ = F (x); the stability or asymptotic stability of this orbit is equivalent to the same property of the constant zero solution of the non-autonomous equation u = F (Ψ(t, x + u)) F (Ψ(t, x)); Exercise 1 Verify the previous statement. Defining A(t) = DF (Ψ(t, x)), this may be written as u = A(t)u + g(t, u) g(t,u) where lim u 0 u = 0. This lead us back, as a first step, to the study of the non-autonomous linear homogeneous equation u = A(t)u, now focused on stability issues. We restate the definitions of stability and asymptotic stability in the general context of non-autonomous equations: Definition 2 A solution u(t) of ẋ = f(t, x), with u(t 0 ) = x 0 is stable if it is defined for all t t 0 and, for every ε > 0 there exists δ > 0 such that, if y 0 x 0 < δ then i) the solution v(t) with v(t 0 ) = y 0 is defined for all t t 0, and ii) v(t) u(t) < ε, t t 0. 1

2 With the same notations, the solution is asymptotically stable if it is stable and there exists δ 1 > 0 such that, if y 0 x 0 < δ 1 then lim v(t) u(t) = 0. t + Back to the linear homogeneous equation, assume from now on that A(t) is a continuous matrix function defined for all t R (or at least, all t t 0 for some t 0 ). The case of a constant matrix A has been dealt with: in that case, 1. all solutions are asymptotically stable if and only if all eigenvalues of A have negative real part; 2. all solutions are stable if and only if all eigenvalues of A have non-positive real part and the eigenvalues with zero real part have geometric multiplicity equal to the algebraic multiplicity. Exercise 3 Prove the previous statements. When A(t) is not constant, stability results are much less general: as we will see, even if the eigenvalues of A(t) have negative real part, for each t, the solutions may not be stable. We present a complementary general result on stability and an example of a condition implying asymptotic stability. Proposition 4 All solutions of ẋ = A(t)x are stable if and only if all solutions are bounded. Exercise 5 Prove the proposition: let U(t) be the fundamental matrix with U(t 0 ) = I. Fix a norm in the vector space of n n matrices, for instance, the usual operator norm defined by A = max{ Av : v = 1}, where the norm used in the righthand side is any fixed norm in R n. Recall that then Av A v. Suppose first that all solutions are bounded: for any solution u there exists a K such that u(t) K t t 0 ; what may be concluded about U(t)? Use your conclusion to bound the distance u 1 (t) u 2 (t) where u 1 (t) and u 2 (t) are any two solutions with u 1 (t 0 ) = x 1 and u 2 (t 0 ) = x 2, and justify that the bound implies that all solutions are stable. Suppose now that all solutions are stable; use the stability of the zero solution to show that U(t) is bounded by a constant, for all t t 0, and conclude that then all solutions are bounded. Show that the proposition and its proof imply the following: 2

3 i) if there exists an unbounded solution u(t) = U(t)x 0 then any other solution v(t) is unstable. ii) Every solution is stable (respectively, asymptotically stable) if and only if the zero solution is stable (respectively, asymptotically stable). Proposition 6 Suppose A(t) = A + C(t) and that i) All eigenvalues of A have negative real part; ii) + t 0 C(t) dt < M < +. Then all solutions of ẋ = A(t)x are asymptotically stable. Exercise 7 Prove the proposition along the same lines used in Notes IX, Proposition 13: If u(t) is the solution with u(t 0 ) = x 0, then (why?) t u(t) = e ta x 0 + e (t s)a C(s)u(s) ds; t 0 let t > 0 and use the bound e ta x 0 Ke at x 0 (see Notes V, proposition 18) to show that φ(t) = e ta u(t) satisfies the conditions for the application of Gronwall s Inequality. Finally, use the hypothesis on C(t) to obtain a constant bound for φ(t) and obtain the conclusion. Exercise 8 Show that the same conclusion is valid if C(t) < c, for all t and a sufficiently small c. 1.1 Floquet Theory We consider now the very important case of a T -periodic homogeneous linear equation ẋ = A(t)x, A(t + T ) = A(t) t. Let U(t) be a fundamental matrix. Defining V (t) = U(t + T ), we have V (t) = U(t + T ) = A(t + T )U(t + T ) = A(t)V (t), ie, V (t) is also a fundamental matrix. We know then (see Notes IV, Proposition 5) that U(t + T ) = V (t) = U(t)M 3

4 for some constant matrix M. Replacing t = 0 we get M = U 1 (0)U(T ). If S(t) = U(t)C is a different fundamental matrix, we have S(t + T ) = S(t)C 1 MC. Fix a initial time t 0. Given a a solution φ(t) with initial condition φ(t 0 ) = x, the monodromy operator of the equation is defined, similarly to the onedimensional case, by The formula deduced above gives L t0 (x) = φ(t 0 + T ) = U(t 0 + T )U 1 (t 0 )x. L t0 (x) = U(t 0 )U 1 (0)U(T )U 1 (t 0 ), showing that monodromy operators defined for different initial times are all conjugated: the same definition gives and so L t1 (x) = U(t 1 )U 1 (0)U(T )U 1 (t 1 ), L t1 (x) = P U 1 (0)U(T )P 1 for P = U(t 1 )U 1 (t 0 ). We notice, on the other hand, that the operator does not depend on the fundamental matrix chosen: if S(t) = U(t)C is a different fundamental matrix, S(t 0 + T )S 1 (t 0 ) = U(t 0 + T )C(U(t0)C) 1 = U(t 0 + T )U 1 (t 0 ). We may thus choose the fundamental matrix U(t) such that U(0) = I and obtain the simple expression for the monodromy operator L 0 (x) = U(T )x. Whatever expression is chosen, the significance of the monodromy operator is unchanged: it is a linear operator that maps an initial condition x to the position in its orbit after T units of time. We may, as described in the beggining of these lectures, turn the equation into an autonomous one, adding t to the state variables { ẋ = A(t)x ṫ = 1 In this case, as the so defined vector field is periodic in the variable t, the phase space is a cylinder R n T where T is a circle of lenght T, more precisely the set R/(T Z) of equivalence classes of real numbers, where two numbers t and s are equivalent if t s T Z. The flow is, using the fundamental matrix U(t) such that U(0) = I, Ψ(t, (x, t 0 )) = (U(t)x, t 0 + t mod T ). 4

5 The monodromy operator, defined to time t 0, should then be seen as a map from the set R n {t 0 } to itself: starting at (x, t 0 ), we travel around T, along the orbit of that point, until we reach Ψ(T, (x, t 0 )) = (U(T )x, t 0 + T ) = (U(T )x, t 0 ). The set R n {t 0 } is a Poincaré section to the flow (recall Notes VI, Definition 29) and x U(T )x is a Poincaré map. This concept will be considered in a more general context later. We recall now the following result (see Notes IV, proposition 46): Proposition 9 Let A M n (R) be invertible. Then there exists B M n (C) such that A = e B, and C M n (R) such that A 2 = e C. We apply this result to state Theorem 10 (Floquet Representation) Let U(t) be the fundamental matrix of the equation ẋ = A(t)x, A(t + T ) = A(t) t R, with U(0) = I. There exists B M n (C) and R M n (R) such that U(T ) = e T B, U(2T ) = U 2 (T ) = e 2T R. The matrix P (t) = U(t)e tb is T -periodic and Q(t) = U(t)e tr is 2T -periodic. Proof. The first statement is a direct consequence of the previous proposition. The statements about periodicity follow from a computation and the definitions: P (t + T ) = U(t + T )e (t+t )B = U(t)U(T )e (t+t )B = U(t)e tb = P (t), and similarly for Q(t). These representations U(t) = P (t)e tb = Q(t)e tr are important because of the information about solutions that may be obtained from them. Of course, an explicit presentation of the matrices B and P (t) (as of R and Q(t)) depends on the knowledge of the fundamental matrix U(t). However, as we ll see briefly, information about the eigenvalues of B (or of R) may determine stability or symptotic stability of solutions of the equation. Exercise 11 Show that the Floquet representations may be seen as a timedependent change of variables, transforming the initial equation into a linear equation with constant coefficients: if, for each t, we define the change of variables x = Q(t)y, and if x(t) is a solution of ẋ = A(t)x, then the corresponding function y(t) is a solution of ẏ = Ry. 5

6 Definition 12 The eigenvalues of the monodromy matrix U(T ) are the characteristic multipliers or Floquet multipliers. The complex numbers µ such that λ = e T µ, where λ is a Floquet multiplier, are the characteristic exponents or Floquet exponents. Although there are n Floquet multipliers (counted with multiplicity), there are infinitely many Floquet exponents: if λ = e T µ 2kπ T (µ+ then also λ = e T i), for any integer k. If µ is an eigenvalue of B then it is a Floquet exponent; this is a direct consequence of the definitions: if Bv = µv, then e T B v = T k B k v = T k µ k v = e T µ v, k! k! k 0 k 0 ie, λ = e T µ is an eigenvalue of e T B. On the other hand, if µ is a Floquet exponent, there is a matrix B and a T -periodic matrix P (t) such that U(t) = P (t)e tb and µ is an eigenvalue of B. Exercise 13 Prove the last statement. Proposition 14 There exists a solution of ẋ = A(t)x of the form u(t) = e µt p(t) where p(t) is a T -periodic function, if and only if µ is a Floquet exponent. This solution satisfies u(t + T ) = λu(t), where λ is the Floquet multiplier. The solutions of ẋ = A(t)x are asymptotically stable if and only if all the Floquet multipliers have norm less than 1, equivalently, all Floquet exponents have negative real part. The solutions of ẋ = A(t)x are stable if and only if all the Floquet multipliers have norm less or equal than 1 (equivalently, all Floquet exponents have non-positive real part) and the Floquet multipliers of norm 1 have geometric multiplicity equal to the algebraic multiplicity. Exercise 15 Prove the last proposition: any solution is of the form u(t) = U(t)u 0, for some vector u 0 ; use the Floquet representation. It should be noticed that the solutions u(t) = e µt p(t) mentioned in the proposition are, in general, complex functions. As A(t) is real, the real and imaginary parts of those solutions are (real) solutions as well. We may consider the different cases: Suppose that the Floquet multiplier λ is a real positive number; then we may choose the exponent µ to be real also; we obtain solutions e µt r(t), e µt s(t) 6

7 where r(t) and s(t) are, respectively, the real and imaginary part of the T - periodic function p(t). These solutions are periodic if and only if λ = 1 (equivalently if µ = 0); otherwise, either λ < 1 and these solutions have ω-limit equal to (0, 0) and are unbounded when t, or λ > 1 and the same solutions have α-limit equal to (0, 0) and are unbounded when t +. If λ < 0, then we may take µ = a + π T i, with a R, and we have solutions e at r(t)(cos( πt ) s(t) sin(πt T T )), eat r(t)(sin( πt T ) + s(t) cos(πt T )); if a = 0 (ie, λ = 1) these solutions are 2T periodic; otherwise, we have again a similar asymptotic behaviour as described above, depending on the sign of a. Finally, if λ is complex, we have µ = a + ib, and obtain the real solutions e at [cos(bt)r(t) sin(bt)s(t)] and e at [cos(bt)s(t) + sin(bt)r(t)]. The asymptotic behaviour when a 0 is again similar to the one described before. When a = 0 the nature of those solutions depends on the relation between b and T : if there exist integers m and n (that we may take to be relatively prime) such that nt = m 2π b, then the solutions are nt -periodic. If not, the solutions are bounded, not periodic, but they return arbitrarily near to the initial value, for an increasing sequence of quasi -periods. Exercise 16 Verify and justify the statements in the previous paragraphs. The Floquet multipliers or exponents may be approximated with sufficient accuracy to determine its norm, even if we don t have good approximations of the solutions. We have also, as a direct consequence of Liouville s formula Lemma 17 det(u(t )) = e T 0 tra(s) ds and tr(b) = 1 T T 0 tra(s) ds. The previous proposition and lemma could convince us that we obtain conclusions about the stability or asymptotic stability of the solutions from the eigenvalues of A(t). A famous example shows that this isn t so: Exercise 18 (The Markus-Yamabe example) Consider the π-periodic equation ẋ = A(t)x with ( ) 1+3 cos(2t) A(t) = sin(2t) sin(2t) 1+3 cos(2t) 4 7

8 i) Compute the characteristic polynomial of A(t) and confirm that its eigenvalues have negative real part. ii) A few iterations of the Picard approximation method give the following approximate values for the determinant and trace of the monodromy matrix M: det(m) = , tr(m) = Prove that, under the assumption that the above values are approximately correct, the Floquet multipliers are different and real. What may we conclude about the stability of solutions? Does there exist a solution with (0, 0) as its ω-limit? iii) Confirm the value of the determinant of the monodromy matrix by Liouville s formula. iv) Verify that e t/2 ( cos(t), sin(t)) is a solution of the equation. v) Let µ 1 and µ 2 be the Floquet exponents. Show that the solutions u 1 = e µ1t p 1 (t) and u 2 = e µ2t p 2 (t) with p i (t) π-periodic, refered to in a previous proposition, are a basis for the space of solutions. Writing e t/2 ( cos(t), sin(t)) as a linear combination, and comparing the values at t = 0 and t = π, prove that we may take e t/2 ( cos(t), sin(t)) to be one of the u i (t). vi) Find the exact values of the Floquet multipliers and check the approximate values computed in ii). Exercise 19 Consider Hill s equation ẍ + a(t)x = 0 where a : R R is T - periodic. i) Show that the Floquet multipliers of the associated first-order equation are the solutions of z 2 2bz + 1 for a certain real number b. ii) Discuss the different cases 1 < b, b < 1, 1 < b < 1, b = 1 and b = 1. determine, in each case, the stability or instability of the solutions, and the existence of periodic or bounded, non-periodic orbits. 1.2 Non-homogeneous linear periodic equations We consider now the non-homogeneous equation ẋ = A(t)x + b(t) where both A(t) and b(t) are T -periodic. 8

9 Exercise 20 Define the Monodromy operator L (at t = 0) of the equation ẋ = Ax + b(t) as in the 1-dimensional case: given x 0 R n, define L(x 0 ) = u(t ), where u(t) is the solution of the equation with initial condition u(0) = x 0. Use the fundamental matrix U(t) of the homogeneous equation satisfying U(0) = I and the formula obtained by the method of variation of constants to prove that L(x 0 ) = U(T )x 0 + c and give the formula for the vector c. Proposition 21 If 1 is not a Floquet multiplier of the homogeneous equation, then the equation above has a unique T -periodic solution. Exercise 22 Prove the proposition: first, show that if u(t) is a solution and u(t ) = u(0) then u is periodic; let u(0) = x 0. Then use the previous exercise. For the proof of uniqueness, reason by contradiction. Using again the formula obtained by the method of variation of constants, prove that this periodic orbit is asymptotically stable if and only if the zero solution is asymptotically stable for the associated homogeneous equation. Corollary 23 If A(t) = A is an hyperbolic linear vector field then the equation ẋ = Ax + b(t), where b(t) is T -periodic, has at least one T -periodic solution. 2 Stability of periodic orbits We return now to the analysis of stability of orbits, in particular of periodic orbits, of autonomous equations. We say that a T -periodic orbit O of the flow Ψ(t, x) is asymptotically stable if for any open neighborhood U of O, there exists an open set V such that O V U and, for any x V where Ψ(t, x) U t 0 and lim dist(ψ(t, x), O) = 0, t + dist(ψ(t, x), O) = min{ Ψ(t, x) p : p O}. Notice that this is implied by, but not necessarily equivalent to asymptotic stability of Ψ(t, p) for each p O. This distinction will be clarified later. A consequence of asymptotic stability is the following Proposition 24 If O is asymptotically stable, there exists a neighborhood V of O such that, for any x V lim Ψ(t + T, x) Ψ(t, x) = 0. t + We say, in this case, that x is said to have asymptotic period T. 9

10 Proof. Fix ɛ > 0 and take V to be the set mentioned in the definition of asymptotic stability given above and a x V. There exists δ < ε such that p O y y p < δ = Ψ(T, y) Ψ(T, p) = Ψ(T, y) p < ε; this is a consequence of the fact that a continuous function, in this case x Ψ(T, x), is uniformly continuous on a compact set, in this case a compact neighborhood of O. On the other hand, from the hypothesis of asymptotic stability of O, there exists t 0 such that, for any t > t 0, there exists p t O for which Ψ(t, x) p t < δ. Now Ψ(t + T, x) Ψ(t, x) Ψ(t + T, x) p t + p t Ψ(t, x) ; the first summand is Ψ(T, Ψ(t, x)) p t < ε, by the first observation; the second is less than δ < ε; so Ψ(t + T, x) Ψ(t, x) < 2ε and, as ε is arbitrary, we have the desired conclusion. To study the stability of the periodic orbit we ll use Poincaré maps. Let S be a transversal section to O at a point p (a curve if n = 2, a surface if n = 3, etc.). We assume S is contained in some open neighborhood W of p and that it is defined as a regular level set of a differentiable function, ie, there exists a C 1 function g : W R such that S = {x W : g(x) = 0} and < g(x), F (x) > 0, for every x S. This last condition assures that S is indeed a regular level set, but also that S is transversal to the flow, as it implies that F (x) does not belong to the tangent space of S at x. We now consider in the extended phase space D the subset and define the C 1 function D = {(t, x) : Ψ(t, x) W } h : D R h(t, x) = g(ψ(t, x)). The point (T, p) is a solution of the equation h(t, x) = 0; the conditions of the Implicit Function Theorem are satisfied for this equation at (T, p): we have and so, in particular, t h(t, x) = Dg(Ψ(t, x)) F (Ψ(t, x)) t h(t, p) = Dg(p)F (p) 0; in fact, Dg(p)F (p) may be interpreted as the inner product g(p), F (p), which is, as we saw before, not zero. This means that in a neighborhood of (T, p), the set {(t, x) : h(t, x) = 0} is the 10

11 graph of a C 1 function t = τ(x); more precisely, there exists an open set U W containing p and an open interval I containing T and a C 1 function τ τ : U I τ(p) = T, g(ψ(τ(x), x)) = 0, ie, Ψ(τ(x), x) S. We are interested in the restriction of τ to S, so we define S = S U and ρ : S I ρ(x) = τ(x). The function ρ is thus defined at those points of x S such that the orbit of x returns to S and ρ(x) gives precisely the time of the first return. We will need later the derivative of ρ; it is easier to compute the derivative of τ, as it is defined in a open set U; this is obtained in the usual way for implicitely defined functions: from g(ψ(τ(x), x)) = 0 we get, by the chain rule or and so Dg(Ψ(τ(x), x))[ t Ψ(τ(x), x)dτ(x) + D x Ψ(τ(x), x)] = 0 Dg(Ψ(τ(x), x))f (Ψ(τ(x), x))dτ(x) + Dg(Ψ(τ(x), x))d x Ψ(τ(x), x) = 0 Dτ(x) = Dg(Ψ(τ(x), x))d xψ(τ(x), x) Dg(Ψ(τ(x), x))f (Ψ(τ(x), x)) ; Notice that the denominator is a scalar - it is the result of applying a 1 n matrix, the derivative of g, to a vector - and it is not zero, as we saw before; the numerator is the product of a 1 n matrix with a n n matrix. In particular, we have Dτ(p) = Dg(p)D xψ(t, p). Dg(p)F (p) The derivative Dρ(x) is the restriction of Dτ(x) to the tangent space of S at the point x; if we were to parametrize S as the image of an open set in R n 1 by a map, we would obtain accordingly an expression for Dρ(x) as a (n 1) (n 1) matrix; but we will not use parametrizations and will allways use the coordinates of the phase space. Definition 25 The Poincaré map, or first return map, of the flow Ψ(t, x), associated to S, is π : S S π(x) = Ψ(ρ(x), x). The map π is a diffeomorphism: it is obviously differentiable; to see that it is invertible, and that the inverse is also differentiable, we apply the above construction to the flow Φ(t, x) = Ψ( t, x), which is the flow associated to ẋ = F (x); the orbits of Φ are the same ones of Ψ but traversed in the opposite 11

12 direction (ie, we move backwards in time along the orbits of Ψ); it is easy to see that the Poincaré map π 1, associated to S for the flow Φ(t, x) is defined for all x in the image of π and that π 1 π(x) = x, ie π 1 (or, more precisely, its restriction to the image of π) is π 1. A first reason to consider Poincaré maps in connection with the problem of stability of periodic orbits is given by the following: Proposition 26 Suppose that π n (x) is defined for all positive integers n and lim n + π n (x) = p. Then lim dist(ψ(t, x), O) = 0. t + Here, π n denotes the n-th iterate of π, ie, the composition of π with itself n times; the points π n (x) are the intersections of the orbit of x with S. Although we won t develop the subject here, we give the definition of a discrete dynamical system: Definition 27 Given a set X and an invertible map f : X X, the discrete dynamical system associated to f is the map Θ : Z X X Θ(n, x) = f n (x). So a discrete dynamical system is a discrete flow where the orbit of a point x is the sequence, f 1 (x), x, f(x), f 2 (x),, f n (x), Notice, as in the continuous flows, the validity of the formula Θ(n + m, x) = Θ(n, Θ(m, x)). The equivalent concept of equilibrium point here is a fixed point, ie f(x) = x; more generally, the orbit of x is periodic if there exists n such that f n (x) = x. The notions of stability, asymptotic stability, etc, are easily adapted to the discrete case as well. The proposition above relates properties of the flow Ψ(t, x), near the periodic orbit, with properties of the discrete dynamical system defined by the Poincaré map in the section S. This is, as hopefully we ll see later, a very fruitfull approach in the qualitative theory of differential equations. Proof. Let s denote x n = π n (x) and ρ n = ρ(x n ). The sequence ρ n is bounded above by some r, as it converges to T (recall that ρ is continuous). We claim that δ N : n > N = Ψ(s, x n ) Ψ(s, p) < δ, s [0, r]; this is a consequence of uniform continuity of continuous functions in compact sets: we consider the continuous function Ψ(s, x) restricted to [0, r] K for some 12

13 compact neighborhood of p containing the points x n. Given any t > 0, there exists some s(t) [0, r] and some n(t) N such that Ψ(t, x) = Ψ(s(t), x n(t) ) : we just have to travel along the orbit, backwards in time, from Ψ(t, x) until we reach S at a point x n(t) = Ψ(t s(t), x) for some time value s(t); then Ψ(t, x) = Ψ(s(t), x n(t) ) and obviously 0 s(t) < ρ n r. We may then estimate dist(ψ(t, x), O) Ψ(t, x) Ψ(s(t), p) = Ψ(s(t), x n(t) ) Ψ(s(t), p) < δ for sufficiently large n(t). As δ is arbitrary and n(t) tends to + as t +, we have the desired limit. So, if π n (x) p, as n +, for all x sufficiently near p in a transversal section, the orbits of those x have O as ω-limit. This strongly suggests (and it will be confirmed later) that the orbit O is asymptotically stable when p is an asymptotically stable fixed point for the Poincaré map π. We will find a sufficient condition, depending on the derivative of π, for this to happen. And we will show that this condition is in fact equivalent to a condition depending on properties of the flow that may be, in some cases, easier to verify than asymptotic stability directly. This is the general plan; before we start to develop it, we will see that the section S may be choosen in a way that considerably simplifies our work. Lemma 28 Suppose S 1 and S 2 are two Poincaré sections, transversal to the orbit O at points p 1 and p 2, respectively, and that π 1 : S 1 S 1 and π 2 : S 2 S 2 are the Poincaré maps. Then, there exists S S 1 and a diffeomorphism γ : S S 2 that conjugates (the restrictions of) π 1 and π 2 : γ π 1 (x) = π 2 γ(x). Proof. The idea for the definition of γ is obvious: for each x in S 1 such that its orbit intersect S 2 at a time s(x), we put γ(x) = Ψ(s(x), x). A detailed proof imitates the reasoning followed for the definition of the Poincaré map given above. The Lemma implies that the derivatives of the two Poincaré maps are conjugated, in particular, as γ(p 1 ) = p 2 Dγ(p 1 )Dπ 1 (p 1 ) = Dπ 2 (p 2 )Dγ(p 1 ), 13

14 and so the eigenvalues of the derivative of the Poincaré map at the fixed point p (the intersection of the orbit O with the section S) do not depend on the particular point p or on the section S. We want to relate the derivative of π at p with the space derivative of the flow along the periodic orbit, so, in order to simplify our reasoning, we will take S to be an affine set, ie S = p + E where E is a (n 1)-dimensional vector space, and will choose E to be invariant under the action of D x Ψ(T, p). The reason for this is the following: Recall that D x Ψ(t, p) is the fundamental matrix U(t) of ẋ = A(t)x, where A(t) = DF (Ψ(t, p)), that satisfies U(0) = I. According to the theory developed in the previous section, D x Ψ(T, p) is then the monodromy matrix of the equation. As we saw after the definition of the monodromy matrix in the beggining of the section on Floquet theory, the monodromy operator could be seen exactly as a Poincaré map for ẋ = A(t)x, ie, for the linearized equation along the periodic orbit O. So the idea behind the following steps is basically to show that the derivative of the Poincaré map (of the non-linear flow) is the Poincaré map of the linearized flow. We start by establishing a general property of D x Ψ. Fix some (t, x) in the extended phase space. We have F (Ψ(t, x)) = t Ψ(t, x) = lim [Ψ(t + s, x) Ψ(t, x)] = s lim s 0 s [Ψ(t, Ψ(s, x)) Ψ(t, x)] s which is the derivative of the function φ(s) = Ψ(t, Ψ(s, x)) at s = 0. But this may also be computed, by the chain rule, as D x Ψ(t, x) s Ψ(s, x) s=0 = D x Ψ(t, x)f (x). This means that the time derivative of the flow at (t, x) transforms the vector field at x into the vector field at Ψ(t, x). We apply this to (T, p) where p belongs to a T -periodic orbit, and obtain D x Ψ(T, p)f (p) = F (p), ie F (p) is an eigenvector of the monodromy operator D x Ψ(T, p), associated to the eigenvalue 1. We define E to be the (n 1)-dimensional vector space generated by all other generalized eigenvectors of D x Ψ(T, p); of course, E is invariant under the action of this matrix. Because p+e is obviously transversal to the flow at p, we may take the Poincaré section S to be an open neighborhood of p in p+e (with respect to the topology of this affine set!), and denote π the associated Poincaré map. This already 14

15 simplifies things in that the tangent space of S at p (or any other point) is E. We compute now the derivative Dπ(p): as π(x) = Ψ(ρ(x), x), we must have Dπ(x) = t Ψ(ρ(x), x)dρ(x)+d x Ψ(ρ(x), x) = F (Ψ(ρ(x), x))dρ(x)+d x Ψ(ρ(x), x) restricted to E. But ρ is the restriction of the map τ to S and we saw before that Dτ(p) = Dg(p)D xψ(t, p), Dg(p)F (p) where g is a function having S as a regular level set. Our choice allows us to put, for instance, g(x) = v, x p where v is a vector orthogonal to E. This implies Dg(x) = v where v denotes the row matrix (a linear functional) with the same entries of the vector v. We have then, for any vector u E, Dπ(p)u = F (p)dρ(p)u + D x Ψ(T, p)u = F (p) vd xψ(t, p) u + D x Ψ(T, p)u; vf (p) but the first summand is zero: D x Ψ(T, p)u belongs to E because this space is invariant under the action of D x Ψ(T, p), and vu = 0. So, with this choice of transversal section Dπ(p) coincides with the restriction of D x Ψ(T, p) to E. In particular, the eigenvalues of Dπ(p) coincide with the eigenvalues of the restriction of D x Ψ(T, p) to E. As we ve seen already, the extra eigenvalue of this matrix is 1 with eigenvector F (p) tangent to the orbit O. But, by the last Lemma, the eigenvalues of Dπ(p) do not depend on the particular Poincaré section and Poincaré map, so we may state Proposition 29 If O is a T -periodic orbit of the flow Ψ(t, x), p O and π is a Poincaré map associated to a transversal section to O at p, the eigenvalues of D x Ψ(T, p) coincide with the eigenvalues of Dπ(p) plus the eigenvalue 1 with associated eigenvector F (p). Remark 30 An immediate consequence is that the determinants of Dπ(p) and of D x Ψ(T, p) are equal and so, by Liouville s formula that can also be written as det Dπ(p) = e T 0 tra(s)ds, where A(t) = DF (Ψ(t, p)), e T 0 divf (Ψ(s,p)) ds. In the case of a 2-dimensional equation the Poincaré map is a real function of a real variable and this expression gives just the derivative π (p). 15

16 Remark 31 Notice that we could confirm directly that the characteristic multipliers (ie, the eigenvalues of DΨ(T, p)) do not depend on the point p: if q = Ψ(s, p), on the other hand D x [Ψ(T + s, x)] x=p = D x [Ψ(T, Ψ(s, x))] x=p = = D x Ψ(T, Ψ(s, x)) x=p D x Ψ(s, p) = D x Ψ(T, q)d x Ψ(s, p); D x [Ψ(T + s, x)] x=p = D x [Ψ(s, Ψ(T, x))] x=p = D x Ψ(s, p)d x Ψ(T, p) and so the two monodromy matrices, corresponding to two different points in the orbit, are conjugated. To follow with our plan, we discuss the problem of when does π n (x) p, as n +, for all x sufficiently near p, ie, when is the fixed point p asymptotically stable for the discrete dynamical system defined by π. As before, we may suppose π is defined in some open set of a vector space E. A sufficient condition is that the derivative L = Dπ(p) is a contraction. This is a general property for discrete dynamical systems: Lemma 32 Suppose π(p) = p, L = Dπ(p), and that there exists 0 < b < 1 such that Lv b v for some norm defined in E. Then p is asymptotically stable for π. Proof. By Taylor s formula where lim x p then π(x) = p + Dπ(p)(x p) + r(x) r(x) x p = 0. Given ε < 1 b, we take δ such that x p < δ = r(x) < ε x p ; π(x) p Dπ(p)(x p) + r(x) < (b + ε) x p ; this inequality shows that we may apply the same estimate to π(x) in the place of x and obtain (by induction) π m (x) p < (b + ε) m x p for all m; as b + ε < 1 we have asymptotic stability of p for the map π. Lemma 33 Let L be a linear nonsingular map in R n 1. If all eigenvalues of L have norm less than 1, then there exists a norm for which L is a contraction. 16

17 Proof. We know that there exists a M such that L = e M. If the eigenvalues of L are less than 1 in norm, then the eigenvalues of M have negative real part. From our discussion of Linear equations with constant coefficients, there exists a norm and a > 0 such that Applying this to t = 1 we get that e tm v e at v, t 0. Lv e a v, ie, L is a contraction with constant b = e a. Proposition 34 Suppose O is a periodic orbit of Ψ(t, x) and π is a Poincaré map associated with a transversal section to O at some point p. If the eigenvalues λ of Dπ(p) all satisfy λ < 1, then O is asymptotically stable. Proof. We generalize the argument used in the proof of Proposition 26. Let U be any neighborhood of O. There exists U U such that Ψ(T, x) U, x U, t [ 2T, 2T ]. We may assume the transversal section is of the form S = p + E, as before, and that the norm in E satisfies the condition of the lemma: there exists c < 1 such that π(x) p < c x p for all x sufficiently near p. More precisely, we take B r (p) S U such that Consider now the set ρ(x) < 2T and π(x) p < c x p x B r (p). V = {Ψ(t, x) : x B r (p), t 0}; V is positively invariant and V U, because Ψ(t, x) = Ψ(s, π m (x)), for some m N and 0 s < 2T ; by the contraction property π m (x) B r (p) U, for all m, and by the definition of U, Ψ(s, π m (x)) U for 0 s < 2T. Now dist(ψ(t, x), O) = dist(ψ(s, π m (x)), O) Ψ(s, π m (x)) Ψ(s, p) and this converges to zero, uniformly in the variable s, for 0 s 2T, because π m (x) p. We have in fact a stronger form of asymptotic stability. The convergence of the orbit of x V to O is in phase : Proposition 35 Under the same conditions of the previous proposition, for each x V there exists y O such that lim Ψ(t, x) Ψ(t, y) = 0. t + 17

18 Proof. We use again the transversal section S = p + E with the respective Poincaré map π, and assume that x S satisfies π m (x) p c m x p, for some positive constant c < 1. Define a sequence t m of time values by Ψ(mT, x) = Ψ(t m, π m (x)). It is easy to verify that the sequence is recursively defined by the formula t 0 = 0, t m+1 = t m + T ρ m, where ρ m = ρ(π m (x)) is the return time to S: the initial value is obvious and Ψ(ρ m, π m (x)) = π m+1 (x); Ψ((m + 1)T, x) = Ψ(T, Ψ(mT, x)) = Ψ(T, Ψ(t m, π m (x))) = = Ψ(T, Ψ(t m ρ m + ρm, π m (x))) = Ψ(T, Ψ(t m ρ m, Ψ(ρm, π m (x)))) = = Ψ(T, Ψ(t m ρ m, π m+1 (x))) = Ψ(T + t m ρ m, π m+1 (x)). We shall prove now that t m is a Cauchy sequence: the function ρ is C 1 so it satisfies a Lipschitz condition: there exists a constant K such that ρ(x 1 ) ρ(x 2 ) K x 1 x 2, for every x 1 and x 2 in an open ball around p in S. Then, for large enough m, implying that, for any k, t m+1 t m = ρ m T = ρ(π m (x)) ρ(p) t m+k t m K π m (x) p Kc m x p, k t m+i t m+i 1 i=1 k K x p c m+i which is less than K x p cm 1 c. So t m is Cauchy, thus converges to some s. Let y = Ψ(s, p). We prove that Ψ(t, x) converges to O in phase with y: given any t > 0, we write t = mt + u with 0 u < T ; of course, u depends on t but we have uniform convergence on compact sets: for any ε > 0, there exists δ such that We have i=1 x 1 x 2 < δ = Ψ(u, x 1 ) Ψ(u, x 2 ) ε u [0, T ]. Ψ(t, x) Ψ(t, y) = Ψ(u, Ψ(mT, x)) Ψ(u, y) = Ψ(u, Ψ(t m, π m (x))) Ψ(u, Ψ(s, p)). 18

19 As t +, (t m, π m (x)) (s, p); so, by continuity of Ψ, given δ, there exists t satisfying t > t = Ψ(t m, π m (x)) Ψ(s, p) < δ, and so Ψ(t, x) Ψ(t, y) < ε. We give an example to show that we may in fact have an asymptotically stable periodic orbit that doesn t satisfy the in phase condition: Exercise 36 Consider the equation defined in polar coordinates ṙ = (r 1) 3 θ = r. i) Prove that the 2π-periodic orbit O = {(1, θ)} is asymtotically stable. ii) Prove that the Floquet multiplier of the orbit O, ie, the derivative of any Poincaré map π at the point p O, is 1, and so the previous Propositions do not apply. Hint: Use the expression in Remark 30 above. It is easier to use the expression for the divergence of a vector field in polar coordinates. iii) Given an initial condition (r 0, θ 0 ), find the explicit expression for the solution u(t) = (r(t), θ(t)). iv) Confirm that u(t) has asymptotic period 2π as predicted in Proposition 24. v) Verify that, nevertheless, the convergence to O is not in phase: use the expression found in iii) for θ(t) to show that for any p = (cos(α), sin(α)) O and any T R there exists t > T such that θ(t) = α + t + π mod 2π. Justify why this proves the claim. We describe next an important example whit a different application of Poicaré maps. Example 37 (The van der Pol equation) Consider the second-order scalar equation ü + g(u) u + u = 0. We may, as usually, transform this into the first order equation u = v v = u g(u)v 19

20 In this case, however, a different, equivalent, equation is used: define G(u) = u g(z)dz, and the change of variables 0 x = u the new equation is then y = v + G(u) ẋ = y G(x) ẏ = x. These are called the Liénard coordinates and the corresponding equation is the Liénard equation. It is clear that (0, 0) is the only equilibrium point; the linearized vector field at the origin is g(0) which implies for the Liénard equation that the origin is a sink if g(0) > 0 and a source if g(0) < 0. Exercise 38 Verify the steps in the deduction of Liénard equation and the analysis of the equilibrium point. Prove that, if xg(x) > 0 for all x 0, then V (x, y) = x2 +y 2 2 is a Liapunov function. From this point on, we consider a concrete example, the van der Pol equation, given in Liénard coordinates, by ẋ = y (x 3 x) ẏ = x According to the above analysis, the origin is a source. As it is not clear how to apply Poincaré-Bendixon theorem to investigate the possible existence of a periodic orbit, we follow a different path, based on the analysis of the phase portrait. Denoting y + = {(0, y) : y > 0} y = {(0, y) : y < 0}, we show that the solution passing by a point (0, p) y + must cross the graph of G(x) on the righthand side of the plane, then cross y and the graph of G(x) on the lefthand side, coming back to y +. First, if the solution crosses the graph of G(x) at a point (x 0, y 0 ) (at time t = 0, with no loss of generality), it can not stay in the strip 0 < x < x 0 : it suffices to verify that the segment {(x, a + x) : 0 x x 0 }, for a sufficiently small a is a lower fence for the flow. 20

21 Exercise 39 Confirm the last statement, finding an appropriate value of a. Now, as F ( x, y) = F (x, y), we have that the flow is symmetric with respect to the origin: if (x(t), y(t)) is a solution, so is ( x(t), y(t)). Exercise 40 Confirm the last statement. This implies that the orbit, after crossing y must come back to y +. We may define a Poincaré map π on the transversal section y + ; a periodic orbit of the flow corresponds to a fixed point of the Poincaré map. We will in fact prove that π has a unique attracting fixed point. We ll make use of the semi -Poincaré map α : y + y where α(p) is the second coordinate of the next intersection of the orbit of (0, p) with y. In the following, we ll identify the points in y + and y with their y-coordinate. The symmetry of the flow implies the following relation between p, α(p) and π(p): 1. if α(p) > p then π(p) < α(p) < p; 2. if α(p) < p then π(p) > α(p) > p; so p is a fixed point of π if and only if α(p) = p. Exercise 41 Confirm the relation described above. We will study the function δ(p) = α2 (p) p 2 2. Obviously, p is a fixed point of π if and only if δ(p) = 0. We have δ(p) = V ((0, α(p)) V (0, p) = T 0 t V (x(t), y(t))dt, for a certain T > 0, where (x(t), y(t)) is the solution with initial condition (0, p). We have, as computed above, T 0 t V (x(t), y(t))dt = = T 0 T 0 x(t)[x 3 (t) x(t)]dt = x 2 (t)[1 x 2 (t)]dt. Let (0, p ) y + be the point whose orbit next crossing with the righthand half of the graph of g(x) is (1, 0). By the previous discussion it is clear that for all 0 < p < p we have δ(p) > 0. We will now prove that for p > p, δ(p) decreases to as p + ; this, together with the previous statement, proves the existence of a unique zero of the function δ, ie, a unique fixed point of the Poincaré map π, ie, a unique periodic 21

22 orbit of the flow. We want to show that, for p > p the integral T 0 x 2 (t)[1 x 2 (t)]dt, an integral of an unknown function over an unknown interval - decreases to. As the integral depends on t only through x(t), the idea will be to replace it by an integral on the variable x; to that end, we write it as t1 0 t2 T x 2 (t)[1 x 2 (t)]dt + x 2 (t)[1 x 2 (t)]dt + x 2 (t)[1 x 2 (t)]dt, t 1 t 2 where t 1 satisfies (x(t 1 ), y(t 1 )) = (1, y 1 ), with y 1 > 0, and t 2 satisfies (x(t 2 ), y(t 2 )) = (1, y 2 ), with y 2 < 0. As in [0, t 1 ] x(t) is an injective function of t, mapping that interval to [0, 1], we may use this function to change variables in the first integral: as x (t) = y (x 3 x), = t1 0 t1 0 x 2 (t)[1 x 2 (t)]dt = x 2 (t)[1 x 2 x (t) (t)] y (x 3 x) dt = = 1 0 x 2 [1 x 2 ] y (x 3 x) dx. Of course, in this integral y is a function of x that we don t know explicitly, but it is clear from the phase portrait that, as p increases, the value of y (x 3 x) increases as well. So the first integral decreases as p increases. Exercise 42 apply the same reasoning to the third integral and prove that it is also a decreasing function of p. In the second integral we may use y as a new variable for integration, as clearly in the interval [t 1, t 2 ] y is a monotonous function of t; so t is a monotonous function of y with derivative t (y) = 1 y (t) = 1 x, where now x is a function of y; we have then t2 y2 x 2 (t)[1 x 2 x 2 (1 x 2 ) y1 (t)]dt = dy = x(1 x 2 )dy. t 1 y 1 x This is a negative function of the initial value p, as along the piece of orbit between (1, y 1 ) and (1, y 2 ), x > 1; moreover, this integral tends to infty as y 2 22

23 p increases: one way of confirming this is to notice that, for sufficiently large p, there is an interval [y 2, y 1] [y 2, y 1 ] such that for y(t) [y 2, y 1] we have x(t) > 2, ie, the orbit with initial condition (0, p) crosses the line x = 2 at two points (2, y 1) and (2, y 2); this implies that the integral is, for p large enough, less than 6(y 1 y 2); as p increases, y 1 y 2 +, implying that the integral tends to. So, there exists a unique y 0 > 0 such that the orbit through (0, y 0 ) is periodic. If y > y 0, we have δ(y) < 0, ie, y < α(y); by symmetry, we have also then β( y) = α(y) < y, and so π(y) < y. This means that π m (y) is a decreasing sequence bounded below by y 0 ; it has to converge and its limit is a fixed point of π, so π m (y) y 0. As we saw in a general context this implies that the orbit of (0, y) converges to the periodic orbit. A similar reasoning proves the same convergence for an initial condition (0, y) with 0 < y < y 0. The arguments used in the example of the van der Pol equation may be applied to all Liénard equations ẋ = y G(x) satisfying the conditions 1. lim x ± G(x) = ; ẏ = x. 2. there exists b such that G(x) > 0 and increasing for x > b; 3. there exists a such that G(x) < 0 for 0 < x < a. Exercise 43 Prove that under these conditions the Liénard equation has a unique and asymptotically stable periodic orbit. 23