On integral separation of nonautomous differential equations

MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY DEPARTMENT OF MATHEMATICS Nguyen Thi Thu Huyen ON INTEGRAL SEPARATION OF NONAUTONOMOUS DIFFERENTIAL EQUATIONS GRADUATION THESIS Hanoi – 2019 MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY DEPARTMENT OF MATHEMATICS Nguyen Thi Thu Huyen ON INTEGRAL SEPARATION OF NONAUTONOMOUS DIFFERENTIAL EQUATIONS GRADUATION THESIS SUPERVISOR: Assoc.Prof- Dr.Sc Doan Thai Son Hanoi – 2019 i Thesis Acknowledgement I would like to express my gratitudes to the teachers of the Department of Mathematics, Hanoi Pedagogical University 2, the teachers in the Analytics group as well as the teachers involved The lecturers have imparted valuable knowledge and facilitated for me to complete the course and the thesis In particular, I would like to express my deep respect and gratitude to Assoc.Prof- Dr.Sc Doan Thai Son, who has direct guidance, help me complete this thesis Due to time, capacity and conditions are limited, so the thesis can not avoid errors Then, I look forward to receiving valuable comments from teachers and friends Graduation thesis Nguyen Thi Thu Huyen Thesis Assurance I assure that the data and the results of this thesis are true and not identical to other topics I also assure that all the help for this thesis has been acknowledged and that the results presented in the thesis has been identified clearly Ha Noi, May 5, 2019 Nguyen Thi Thu Huyen i Contents Thesis Assurance i Lyapunov exponent and basic results 1.1 Definition and main properties of characteristic exponents 1.2 The Lyapunov spectrum of a linear system 11 Continuity of the Lyapunov exponent 2.1 2.2 22 Characterization of continuity of Lyapunov spectrum 23 2.1.1 Simple Lyapunov spectrum 23 2.1.2 One point spectrum 25 Explicit examples 27 REFERENCE 32 ii Graduation thesis Nguyen Thi Thu Huyen Preface Theory exponent Lyapunov has a long history and is known to be an important tool in studying the stability of differential equations Specifically, Lyapunov’s exponent measures the separation velocity of the nearderivative solutions of the differential equation and when the exponent is negative, the solutions converge when the time is infinite In fact, there are many equations that we not know exactly the vector field and then the question is how the Lyapunov exponent will change if the vector field of the system is small One of the profound results associated with this problem is integral separation which is a necessary and sufficient condition for the continuity of the Lyapunov exponent when these exponents separate Within the framework of the thesis, student wished to present systematically the problem of continuity of the exponent Lyapunov and its relation to integral separation To this, the thesis will focus on: • Introduction sketchy to the Lyapunov exponent and basis results • Continuity of the Lyapunov exponent The thesis consits of two chapter : Chapter 1: Lyapunov exponent and basis results Chapter 2: Continuity of the Lyapunov exponent I sincerely thanks Assoc.Prof- Dr.Sc Doan Thai Son devotedly instructed the author to read the materials and practice the research I Graduation thesis Nguyen Thi Thu Huyen would also like to thank Assoc.Prof- Dr.Sc Doan Thai Son gave detailed comments on how to present some results in the thesis I sincerely thank the teachers of the Mathematics Department at Hanoi Pedagogical University 2, especially the Analysis Team, for creating favorable conditions for me in the process of studying at the University and implementing the thesis Ha Noi, 06/05/2019 Author of the thesis Nguyen Thi Thu Huyen Chapter Lyapunov exponent and basic results The aim of this chapter is to introduce the notion of Lyapunov exponent of an arbitrary function Next, we study the Lyapunov spectrum of a linear nonautonomous differential equation We refer the reader to [2, 3] for more details 1.1 Definition and main properties of characteristic exponents We start this section by giving a definition of the Lyapunov characteristic exponent of a non-zero function Roughly speaking this exponent measures the growth of this function in comparison with the exponential function Definition 1.1.1 (Lyapunov Characteristic exponent) For any nonzero function f : R → R, its Lyapunov characteristic exponent (for short Graduation thesis Nguyen Thi Thu Huyen characteristic exponent) is defined as χ[f ] = lim sup ln |f (t)| t→∞ t The following properties is straightforward from Definition 1.1.1: If |f (t)| |F (t)| for t a, then χ[f ] χ[F ] χ[f ] = χ[|f |] χ[f ] = lim sup 1t ln |f (t)| = lim sup 1t (ln |c|+ln |f |) = lim sup 1t ln |cf | = t→∞ t→∞ t→∞ χ[cf ], where c is an arbitrary non-zero constant For a non-zero constant function c, we have χ[c] = Now, we study the property of the Lyapunov characteristic exponent of a function given as a sum or a product of a finite number of functions For this purpose, we need the following technical lemma Lemma 1.1.2 Let f : R → R be a non-zero function Then, χ[f ] = α ∈ (−∞, ∞) if and only if for any ε > the following two conditions hold simultaneously: |f (t)| (α+ε)t e t→∞ (1) lim = 0, (t)| (2) lim sup e|f(α−ε)t = ∞ t→∞ Proof (Necessity) Suppose that χ[f ] = lim sup ln |f (t)| = α t→∞ t (1.1) Then, we will show that two statements (1) and (2) hold Choose and Graduation thesis Nguyen Thi Thu Huyen fix ε > By (1.1), there exists T > such that ε ln |f (t)| < α + t for all t > T Equivalently, ε |f (t)| < e(α+ )t for all t > T, which implies that |f (t)| ε ε t→∞ e(α+ )t e t = ε t→∞ e t lim lim k→∞ Hence, (1) is verified To prove (2), let a sequence tk −−−→ ∞ realizing equality (1.1), it means that ln |f (tk )| = α k→∞ tk lim Then, there exists N > such that ε ln |f (tk )| > (α − )tk for k > N Thus, we have lim sup k→∞ |f (tk )| |f (tk )| ε tk = lim sup e (α− 2ε )tk e(α−ε)tk k→∞ e ε lim e tk = ∞, k→∞ which proves (2) (Sufficiency) Suppose that two statements (1) and (2) hold We will show that χ(f ) = α By condition (1), for any ε > there exists T such that |f (t)| < e(α+ε)t for all t ≥ T Graduation thesis Nguyen Thi Thu Huyen Therefore, ξ(t) is solution of (1.4) with ξ(0) = x0 This complete the proof So far, we have proved that for any initial valued problem x(0) = x0 equation (1.4) has a unique solution In the remainning of this chapter, we examine the characteristic exponents of solutions of (1.4) Firstly, we show that the characteristic exponent of a non-zero slution of (1.4) is finite Theorem 1.2.3 Any non-trivial solution x(t) of the linear system (1.4) −M ≤ χ[x] ≤ M has a finite characteristic exponent, i.e Proof Let Rn be endowed with the Euclidean norm A driect computation yields that d x(t) dt = |< x(t), ˙ x(t) > + < x(t), x(t) ˙ >| = 2| < x, ˙ x>| = |< A(t)x(t), x(t) >| ≤ A(t)x(t) x(t) ≤ A(t) x(t) ≤ 2M x(t) Therefore, −2M x(t) d x(t) dt Thus, d x x dt −2M 2M x(t) 2M Integrating the last inequality from t0 to t yields that t − t 2M dτ t0 t0 d x dτ x dt 18 t 2M dτ t0 Graduation thesis Nguyen Thi Thu Huyen Therefore, −2M (t − t0 ) ln x(t) − ln x(t0 ) 2M (t − t0 ) Consequently, −M (1 − t ) t0 1 ln x(t) − ln x(t0 ) t t M (1 − t ) t0 Letting t → ∞, we get That means −M −M lim sup ln x(t) t→∞ t χ[x] M The proof is complete M Next, we show that the set of Lyapunov characteristic exponents of all non-zero solutions of (1.4) has no more than n elements Before proving this fact, we need the following technical lemmas Lemma 1.2.4 Vector functions x1 (t), , xm (t) defined on [0; ∞) and having different finite characteristic exponents are linearly independent Proof Let us consider the linear combination of these vectors with a m nontrivial set of coefficient is ci xi (t) According to Theorem 1.1.3, i=1 the characteristic exponent of this sum is equal to max χ[χi (t)] That is, i m ci xi (t)] = max χ[ci xi (t)] = max χ[χi (t)] = −∞ χ[ i i=1 i m If ∃ ck = such that ci xi (t) = 0, then i=1 m ck xk (t)] = χ[0] = −∞ χ[ k=1 19 Graduation thesis Nguyen Thi Thu Huyen This is contradiction Therefore, {xi }m i=1 is linearly independent Lemma 1.2.5 System (1.4) has n solutions x1 (t), , xn (t) which are linearly independent Proof Suppose the contrary, i.e., there exists n+1 solutions x1 (t), , xn+1 (t) of (1.4) which are linearly dependently Then, there exists c1 , , cn+1 such that c21 + + c2n+1 = and c1 x1 (0) + + cn+1 xn+1 (0) = Without loss of generality, we assume that c1 = Then, x1 (0) = Put ξ(t) = −c2 c1 c3 cn+1 −c2 x2 (0) − x3 (0) − − xn+1 (0) c1 c1 c1 x2 (t) − c3 c1 x3 (t) − − cn+1 c1 xn+1 (t) We have ˙ = −c2 x˙ (t) − c3 x˙ (t) − − cn+1 x˙ n+1 (t) ξ(t) c1 c1 c1 −c2 c3 cn+1 A(t)x2 (t) − A(t)x3 (t) − − A(t)xn+1 (t) = c1 c1 c1 c2 c3 cn+1 = A(t).[− x2 (t) − x3 (t) − − xn+1 (t)] c1 c1 c1 = A(t).ξ(t) Therefore, we have for all t   ξ(0) = x1 (0), ξ(t) is the solution of equation x(t) ˙ = A(t)x(t) Hence, ξ(t) ≡ x1 (t), 20 Graduation thesis Nguyen Thi Thu Huyen That is, c2 c3 cn+1 x1 (t) = − x2 (t) − x3 (t) − − xn+1 (t) c1 c1 c1 Or, c1 x1 (t) + c2 x2 (t) + + cn+1 xn+1 (t) = Therefore,{xi }n+1 i=1 is linearly independent The proof is complete Theorem 1.2.6 The number of elements of the spectrum doew not exceed n Proof By Lemma 1.2.5, system equation (1.4) has n solutions {xi (t)}ni=1 which is linearly independent That is, any solution of (1.4) can be represented by n x(t) = ci xi (t) i=1 We have n χ[x(t)] = χ[ ci xi (t)] = max χ[ci xi (t)] i=1 Thus, spectrum of (1.4) consists of finite element α1 αm , (m α2 n), where αi = χ[xi (t)] So, the number of elements of the spectrum does not exceed n 21 Chapter Continuity of the Lyapunov exponent Consider a linear planar differential equation x˙ = A(t)x, (2.1) where A ∈ C([0; +∞)) such that sup A(t) M By Theorem 1.2.3 and Theorem 1.2.6, system (2.1) has at most two Lyapunov exponents denoted by −∞ < λ1 λ2 < ∞, (2.2) Consider a pertubed system of (2.1) of the form y˙ = [A(t) + Q(t)]y, where Q ∈ C([0; +∞)), with sup Q(t) (2.3) σ The Lyapunov spectrum of (2.3) is denoted by with the spectrum: −∞ < λ1 λ2 < ∞ 22 (2.4) Graduation thesis Nguyen Thi Thu Huyen Under the influence of perturbation Q(t) , the characteristic exponents of system (2.3) vary, generally speaking, discontinuously Definition 2.0.1 The characteristic exponents of system (2.1) are said to be stable if for any ε > there exists a σ > such that the inequality sup Q(t) < σ implies the inequality: t≥0 λi − λi < ε, i = 1, (2.5) Our aim in this chapter is to give a characterization of the stability of Lyapunov spectrum of a linear planar system We distinguish the cases simple spectrum, i.e., λ1 < λ2 and one point spectrum, i.e λ1 = λ2 (Section 2.2) 2.1 Characterization of continuity of Lyapunov spectrum 2.1.1 Simple Lyapunov spectrum Firstly, we have the following result for a sufficient condition of continuity of simple Lyapunov spectrum Theorem 2.1.1 If system (2.1) has different characteristic exponent λ1 < λ2 , then their stability follows from the existence of a Lyapunov transformation x = L(t)z of this system to a diagonal system z˙ = diag[p1 (t), p2 (t)]z = P (t)z 23 (2.6) Graduation thesis Nguyen Thi Thu Huyen Where the functions p1 (t) and p2 (t) are integrally separared, i.e., t (p2 (τ ) − p1 (τ ))dτ a(t − s) − d, t s 0, a > 0, d s Proof See [2] It turns out the integral separatedness is also a necessary condition for stability of simple Lyapunov spectrum Theorem 2.1.2 If the exponents λ1 < λ2 of a two dimensional system: x˙1 = p1 (t)x1 , x˙2 = p2 (t)x2 , dx dt = diag[p1 (t), p2 (t)]x = P (t)x, are stable, then the functions p1 (t) and p2 (t) are integrally separated Here: t λ1 = lim t→∞ t p1 (τ )dτ t λ2 = lim t→∞ t p2 (τ )dτ Proof The idea of the proof is as follows Assume that the diagond is not integrally separated Fixing β ∈ (λ1 λ2 ), we show that under this assumption for any σ > the perturbation: Q(t) < σ Can be chosen such that there exists a solution y(t) of the perturbed sytem with x[y] = β, and this contradicts the stability of the exponents The method for constructing these perturbations was developed 24 Graduation thesis Nguyen Thi Thu Huyen by Millionshchikov and is called the medthod of rotations or the method of perturbations by rotations 2.1.2 One point spectrum We pass to the case of a two-dimensional diagonal system with equal characteristic exponent To formulate the characterization of continuity of the spectrum in this case, we first need to introduction the notion of central exponent Definition 2.1.3 Bounded measeureable on Rt function r(t) and R(t) are said to be lower and upper functions for system (2.1), respectively, if for any solution x(t) of this system the following estimates hold for any σ > 0: t t (r(τ ) − ε)dτ ) dr,ε exp( x(t) x(s) DR,ε exp( s s where t s (R(τ ) + ε)dτ ) (2.7) and the quantities dr,ε , DR,ε not depend on growth of the solutions from below and from above, respectively Definition 2.1.4 The number Ω defined as: t Ω = inf { lim R t→∞ t R(τ )dτ } (2.8) is called the upper central exponent of system (2.1) Here the infimum is taken over the set of all the upper functions R of system 25 Graduation thesis Nguyen Thi Thu Huyen Definition 2.1.5 The number ω defined as t ω = sup{ lim t→∞ t r r(τ )dτ } (2.9) Here r(t) is the set of all the lower functions of system (2.1) Theorem 2.1.6 If a linear two-dimensional has equal characteristic exponents λ1 = λ2 = λ, then they are stable if and only if ω=λ=Ω Proof Necessity Let the exponent of the system be stable but: max{Ω − λ, λ − ω} = γ > For the sake of definiteness, we assume that Ω − λ = γ > According to Millionshchikov’s Theorem (see [1]), there exists a perturbation Q(t) of the linear system such that Q(t) < σ, t Where σ is arbitrary small, and the perturbed system has a characteristic exponent λ Ω This contradicts the stabililty of the exponents Similar arguments can be carried out for the case λ − ω = γ > The fact that ω is attainable was established by Millionshchikov Hence Ω−λ=λ−ω =0 Or ω=λ=Ω 26 Graduation thesis Nguyen Thi Thu Huyen Sufficiency Since the greatest exponent is rigid upwards if and only if it coincides with the upper central exponent; the condition λ = Ω guarantees that the exponent is rigid upward By annalugous arguments, it can be shown that the inequality λ = ω implies that the exponent is rigid downwards Hence, the characteristic exponents of this system are stable 2.2 Explicit examples Our aim in this section is to verify some explicit example of discontinuity of Lyapunov spectrum for planar differential equations However, we start by proving the fact that if we reduce our attention to scalar systems then the Lyapunov exponent always depend continuously on the coefficients Theorem 2.2.1 The characteristic exponent of a linear scalar equation is always stable Proof The equation x˙ = a(t)x has the solution t x(t) = x(0) exp a(τ )dτ Therefore, t λ = lim t→∞ t a(τ )dτ The perturbed system: y˙ = [a(t) + q(t)]y,, where |q(t)| 27 σ for t ∈ Rt , Graduation thesis Nguyen Thi Thu Huyen has the solution t y(t) = y(0) exp [a(τ ) + q(τ )]dτ By the condition |q(t)|  σ for t ∈ R, we have:   t y(t) y(0) a(τ )dτ − σt exp  a(τ )dτ + σt exp   t and according to the properties of the exponent, we have:   t lim  t→∞ t lim [ln y(t) − ln y(0)] t→∞ t a(τ )dτ − σt and  lim [ln y(t) − ln y(0)] t→∞ t  t lim  t→∞ t a(τ )dτ + σt or λ−σ λ λ + σ ⇒ |λ − λ | Where σ t χ[y] = λ = lim t→∞ t (a(τ ) + q(τ ))dτ Now we consider the following example of planar linear system Example 2.2.2 Consider the following equation x˙1 = −ax1 x˙2 = (sin ln t + cos ln t − 2a)x2 , 28 t 0, > a > 1/2 Graduation thesis Nguyen Thi Thu Huyen λ1 = −a < λ2 = − 2a < we have For < 2a < + exp(−π), (2.10) the perturbation  Q(t) =  e −at 0   generates a solution, unbounded as t → ∞ This implies the instability of the exponents of the system, since stable exponents not change under the perturbations Q(t) → We show that the diagonal of this system is not integrally separated if (2.10) holds Indeed, in the opposite case, the inequality: t (−a + sin ln τ + cos ln t − 2a)dτ α(t − s) − d s i.e, the inequality: −a(t − s) + (t sin ln t − s sin ln s) holds for all t s α(t − s) − d with same constants α > 0, d 0, Take the sequences: sm = exp(2m − 1)π tm = exp(2m + 1)π, 29 (2.11) Graduation thesis Nguyen Thi Thu Huyen then from (2.11) for any m ∈ N we have: −a.e2mπ (eπ − e−π ) αe2mπ (eπ − e−π ) − d Therefore, taking into account (2.10), we obtain: α(eπ − e−π ) −a.(eπ − e−π ) + d.e−2mπ < − (eπ − e−π ) + d.e−2mπ m → ∞; < 0, This contradicts (2.11) Example 2.2.3 Consider the following linear system x˙ = x˙ = (1 + √ π sin π t)dτ Here λ1 = and the other Lyapunov exponent can be computed as follows t λ2 = lim t→∞ t (1 + √ π sin π t)dτ √ √ √ 1 = lim (1 + sin π t − t cos π t) = t→∞ t π At the same time, the diagonal is not integrally separated Indeed, let n ∈ N, then (2n) (1 + (2n−1) √ π sin π t)dτ ) = 2 30 Graduation thesis Nguyen Thi Thu Huyen therefore, we cannot indicate a > 0, d t (1 + √ π sin π t)dτ s would hold for all t s 31 such that : a(t − s) − d References [1] V.M.Millionshchikov,A proof of attainability of Central exponents of linear systems, Sibirsk Math.2h.10(1969) no 10,99-104, English transl in Siberian Math.J.10(1969) [2] L Ya Andrianova, Introduction to Linear Systems of Differential Equations, 1995 [3] Peter Kloeden and Martin Rasmussen, Nonautonomous Dynamiacal Systems, AMS 2011 32 ...MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY DEPARTMENT OF MATHEMATICS Nguyen Thi Thu Huyen ON INTEGRAL SEPARATION OF NONAUTONOMOUS DIFFERENTIAL EQUATIONS GRADUATION THESIS... problem of continuity of the exponent Lyapunov and its relation to integral separation To this, the thesis will focus on: • Introduction sketchy to the Lyapunov exponent and basis results • Continuity... Specifically, Lyapunov’s exponent measures the separation velocity of the nearderivative solutions of the differential equation and when the exponent is negative, the solutions converge when the time