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Stability criteria for linear Hamiltonian dynamic systems on time scales Advances in Difference Equations 2011, 2011:63 doi:10.1186/1687-1847-2011-63 Xiaofei He hexiaofei525@sina.com Xia

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Stability criteria for linear Hamiltonian dynamic systems on time scales

Advances in Difference Equations 2011, 2011:63 doi:10.1186/1687-1847-2011-63

Xiaofei He (hexiaofei525@sina.com) Xianhua Tang (tangxh@mail.csu.edu.cn) Qi-Ming Zhang (zhqm20082008@sina.com)

Article type Research

Submission date 5 August 2011

Acceptance date 20 December 2011

Publication date 20 December 2011

Article URL http://www.advancesindifferenceequations.com/content/2011/1/63

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Stability criteria for linear Hamiltonian dynamic systems on time scales

Xiaofei He1,2, Xianhua Tang ∗1 and Qi-Ming Zhang1

1School of Mathematical Sciences and Computing Technology, Central South University, Changsha 410083, Hunan, P.R China

2College of Mathematics and Computer Science,

Jishou University, Jishou 416000, Hunan, P.R.China

∗Corresponding author: tangxh@mail.csu.edu.cn

Email address:

XH: hexiaofei525@sina.com Q-MZ: zhqm20082008@sina.com

Abstract

In this article, we establish some stability criteria for the polar linearHamiltonian dynamic system on time scales

x 4 (t) = α(t)x(σ(t))+β(t)y(t), y 4 (t) = −γ(t)x(σ(t))−α(t)y(t), t ∈ T

by using Floquet theory and Lyapunov-type inequalities

2000 Mathematics Subject Classification: 39A10

Keywords: Hamiltonian dynamic system; Lyapunov-type inequality;Floquet theory; stability; time scales.

A time scale is an arbitrary nonempty closed subset of the real numbers R We

assume that T is a time scale For t ∈ T, the forward jump operator σ : T → T

is defined by σ(t) = inf{s ∈ T : s > t}, the backward jump operator ρ : T → T is defined by ρ(t) = sup{s ∈ T : s < t}, and the graininess function µ : T → [0, ∞)

is defined by µ(t) = σ(t) − t For other related basic concepts of time scales, we

refer the reader to the original studies by Hilger [1–3], and for further details,

we refer the reader to the books of Bohner and Peterson [4, 5] and Kaymakcalan

et al [6]

Definition 1.1 If there exists a positive number ω ∈ R such that t + nω ∈ T for all t ∈ T and n ∈ Z, then we call T a periodic time scale with period ω Suppose T is a ω-periodic time scale and 0 ∈ T Consider the polar linear

Hamiltonian dynamic system on time scale T

x 4 (t) = α(t)x(σ(t)) + β(t)y(t), y 4 (t) = −γ(t)x(σ(t)) − α(t)y(t), t ∈ T,

(1.1)

where α(t), β(t) and γ(t) are real-valued rd-continuous functions defined on T.

Throughout this article, we always assume that

1 − µ(t)α(t) > 0, ∀ t ∈ T (1.2)

For the second-order linear dynamic equation

[p(t)x 4 (t)] 4 + q(t)x(σ(t)) = 0, t ∈ T, (1.4)

if let y(t) = p(t)x 4 (t), then we can rewrite (1.4) as an equivalent polar linear

Hamiltonian dynamic system of type (1.1):

Recently, Agarwal et al [7], Jiang and Zhou [8], Wong et al [9] and He et

al [10] established some Lyapunov-type inequalities for dynamic equations ontime scales, which generalize the corresponding results on differential and differ-ence equations Lyapunov-type inequalities are very useful in oscillation theory,stability, disconjugacy, eigenvalue problems and numerous other applications inthe theory of differential and difference equations In particular, the stabilitycriteria for the polar continuous and discrete Hamiltonian systems can be ob-tained by Lyapunov-type inequalities and Floquet theory, see [11–16] In 2000,Atici et al [17] established the following stablity criterion for the second-orderlinear dynamic equation (1.4):

Theorem 1.2 [17] Assume p(t) > 0 for t ∈ T, and that

on T.

In this article, we will use the Floquet theory in [18, 19] and the type inequalities in [10] to establish two stability criteria for system (1.1) andequation (1.4) Our main results are the following two theorems

Lyapunov-Theorem 1.3 Suppose (1.2) and (1.3) hold and

α(t + ω) = α(t), β(t + ω) = β(t), γ(t + ω) = γ(t), ∀ t ∈ T. (1.10)

Assume that there exists a non-negative rd-continuous function θ(t) defined on

T such that

|α(t)| ≤ θ(t)β(t), ∀ t ∈ T[0, ω] = [0, ω] ∩ T, (1.11)

Then system (1.1) is stable.

Theorem 1.4 Assume that (1.6) and (1.7) hold, and that

Then equation (1.4) is stable.

Remark 1.5 Clearly, condition (1.14) improves (1.8) by removing term p0.

We dwell on the three special cases as follows:

1 If T = R, system (1.1) takes the form:

Condition (1.17) is the same as (3.10) in [12], but (1.11) and (1.16) are better

than (3.9) in [12] by taking θ(t) = |α(t)|/β(t) A better condition than (1.17)

can be found in [14, 15]

2 If T = Z, system (1.1) takes the form:

4x(n) = α(n)x(n+1)+β(n)y(n), 4y(n) = −γ(n)x(n+1)−α(n)y(n), n ∈ Z.

n=0 β(n) ω−1X

n=0

γ+(n)

#1/2

Conditions (1.19), (1.20), and (1.21) are the same as (1.17), (1.18) and (1.19) in

[16], i.e., Theorem 1.3 coincides with Theorem 3.4 in [16] However, when p(n) and q(n) are ω-periodic functions defined on Z, the stability conditions

Then system (1.1) takes the form:

x 0 (t) = α(t)x(t)+β(t)y(t), y 0 (t) = −γ(t)x(t)−α(t)y(t), t ∈ [

k∈Z [kω, kω +δ),

(1.25)and

4x(t) = α(t)x(t + 1) + β(t)y(t), 4y(t) = −γ(t)x(t + 1) − α(t)y(t),

t ∈ [k∈Z {kω + δ + n : n = 0, 1, , N − 2}. (1.26)

In this case, the conditions (1.11), (1.12), and (1.13) of Theorem 1.3 can betransformed into

Then, we can rewrite (1.1) as a standard linear Hamiltonian dynamic system

u 4 (t) = A(t)u σ (t), t ∈ T. (2.1)

Let u1(t) = (x10(t), y10(t)) > and u2(t) = (x20(t), y20(t)) > be two solutions

of system (1.1) with (u1(0), u2(0)) = I2 Denote by Φ(t) = (u1(t), u2(t)) Then Φ(t) is a fundamental matrix solution for (1.1) and satisfies Φ(0) = I2 Sup-

pose that α(t), β(t) and γ(t) are ω-periodic functions defined on T (i.e (1.10) holds), then Φ(t + ω) is also a fundamental matrix solution for (1.1) ( see [18]).

Therefore, it follows from the uniqueness of solutions of system (1.1) with initialcondition ( see [9, 18, 19]) that

It follows that det Φ(t) = det Φ(0) = 1 for all t ∈ T Let λ1and λ2 be the roots

(real or complex) of the characteristic equation of Φ(ω)

Let v1= (c11, c21)> and v2= (c12, c22)>be the characteristic vectors associated

with the characteristic roots λ1 and λ2 of Φ(ω), respectively, i.e.

This shows that v1(t) and v2(t) are two solutions of system (1.1) which satisfy

(2.7) Hence, we obtain the following lemma

Lemma 2.1 Let Φ(t) be a fundamental matrix solution for (1.1) with Φ(0) =

I2, and let λ1and λ2be the roots (real or complex) of the characteristic equation (2.4) of Φ(ω) Then system (1.1) has two solutions v1(t) and v2(t) which satisfy (2.7).

Similar to the continuous case, we have the following lemma

Lemma 2.2 System (1.1) is unstable if |H| > 2, and stable if |H| < 2.

Instead of the usual zero, we adopt the following concept of generalized zero

f (t) = cg(t) for t ∈ T[a, b].

Lemma 2.6 Let v1(t) = (x1(t), y1(t)) > and v2(t) = (x2(t), y2(t)) > be two solutions of system (1.1) which satisfy (2.7) Assume that (1.2), (1.3) and (1.10) hold, and that exists a non-negative function θ(t) such that (1.11) and (1.12) hold If H2 ≥ 4, then both x1(t) and x2(t) have generalized zeros in T[0, ω].

Proof Since |H| ≥ 2, then λ1and λ2are real numbers, and v1(t) and v2(t) are also real functions We only prove that x1(t) must have at least one generalized zero in T[0, ω] Otherwise, we assume that x1(t) > 0 for t ∈ T[0, ω] and so (2.7)

implies that x1(t) > 0 for t ∈ T Define z(t) := y1(t)/x1(t) Due to (2.7), one sees that z(t) is ω-periodic, i.e z(t + ω) = z(t), ∀ t ∈ T From (1.1), we have

Combining (2.14) with (2.15), we have

Lemma 2.7 Let v1(t) = (x1(t), y1(t)) > and v2(t) = (x2(t), y2(t)) > be two solutions of system (1.1) which satisfy (2.7) Assume that

If H2≥ 4, then both x1(t) and x2(t) have generalized zeros in T[0, ω].

Proof Except (1.12), (2.18), and (2.19) imply all assumptions in Lemma 2.6hold In view of the proof of Lemma 2.6, it is sufficient to derive an inequalitywhich contradicts (2.20) instead of (1.12) From (2.11), (2.13), and (2.18), wehave

1 + µ(t)β(t)z(t) = 1 + µ(t)β(t) y1(t)

x1(t) =

x1(σ(t))

x1(t) > 0 (2.21)

z 4 (t) = −γ(t) − β(t)z

2(t)

1 + µ(t)β(t)z(t) . (2.22)Since z(t) is ω-periodic and γ(t) 6≡ 0, it follows from (2.22) that z2(t) 6≡ 0 on T[0, ω] Integrating equation (2.22) from 0 to ω, we obtain

Lemma 2.8 [10] Suppose that (1.2) and (1.3) hold and let a, b ∈ T k with σ(a) ≤ b Assume (1.1) has a real solution (x(t), y(t)) such that x(t) has a generalized zero at end-point a and (x(b), y(b)) = (κ1x(a), κ2y(a)) with 0 <

κ2≤ κ1κ2≤ 1 and x(t) 6≡ 0 on T[a, b] Then one has the following inequality

is not identically zero on T[a, b] Then one has the following inequality

Z b

a β(t)4t

x(τ ) = (1 − ξ)µ(a)β(a)y(a) +

Z τ

σ(a) β(t)y(t)4t, σ(a) ≤ τ ≤ b, (2.26)

ϑ1µ(a)β(a)y2(a) +

Z b

σ(a) β(t)y2(t)4t =

Z b

a γ(t)x2(σ(t))4t, (2.27)and

2|x(τ )| ≤ ϑ2µ(a)β(a)|y(a)| +

Z b

σ(a) β(t)|y(t)|4t, σ(a) ≤ τ ≤ b, (2.28)

where ξ ∈ [0, 1), and

ϑ1= 1 − ξ + κ2ξ, ϑ2= 1 − ξ + |κ|ξ. (2.29)

Let |x(τ ∗ )| = max σ(a)≤τ ≤b |x(τ )| There are three possible cases:

(1) y(t) ≡ y(a) 6= 0, ∀ t ∈ T[a, b];

(2) y(t) 6≡ y(a), |y(t)| ≡ |y(a)|, ∀ t ∈ T[a, b];

(3) |y(t)| 6≡ |y(a)|, ∀ t ∈ T[a, b].

Case (1) In this case, κ = 1 It follows from (2.25) and (2.26) that

x(b) = (1 − ξ)µ(a)β(a)y(a) +

Z b

σ(a) β(t)y(t)4t

#

= x(a) + y(a)

Z b

a β(t)4t 6= x(a),

which contradicts the assumption that x(b) = κx(a) = x(a).

Case (2) In this case, we have

2|x(τ )| < ϑ2µ(a)β(a)|y(a)| +

Z b

σ(a) β(t)|y(t)|4t, σ(a) ≤ τ ≤ b (2.30)

instead of (2.28) Applying Lemma 2.5 and using (2.27) and (2.30), we have

2|x(τ ∗ )|

< ϑ2µ(a)β(a)|y(a)| +

Z b

σ(a) β(t)|y(t)|4t

# "

ϑ1µ(a)β(a)y2(a) +

Z b

σ(a) β(t)y2(t)4t

# Z b

a γ(t)x2(σ(t))4t

)1/2

≤ |x(τ ∗ )|

"Ã

ϑ2 2

ϑ1µ(a)β(a) +

Z b

σ(a) β(t)4t

# "

ϑ1µ(a)β(a)y2(a) +

Z b

σ(a) β(t)y2(t)4t

# Z b

a γ(t)x2(σ(t))4t

Dividing the latter inequality of (2.33) by |x(τ ∗ )|, we also obtain (2.32) It is

easy to verify that

ϑ2

ϑ1 = [1 − ξ + |κ|ξ]

2

1 − ξ + κ2ξ ≤ 1.

Substituting this into (2.32), we obtain (2.24) ¤

Proof of Theorem 1.3 If |H| ≥ 2, then λ1 and λ2 are real numbers and

λ1λ2 = 1, it follows that 0 < min{λ2

1, λ2

2} ≤ 1 Suppose λ2

1 ≤ 1 By Lemma 2.6, system (1.1) has a non-zero solution v1(t) = (x1(t), y1(t)) > such that (2.7)

holds and x1(t) has a generalized zero in T[0, ω], say t1 It follows from (2.7)

that (x1(t1+ ω), y1(t1+ ω)) = λ1(x1(t1), y1(t1)) Applying Lemma 2.8 to the

solution (x1(t), y1(t)) with a = t1, b = t1+ ω and κ1= κ2= λ1, we get

Competing interests

The authors declare that they have no competing interests

Authors’ contributions

XH carried out the theoretical proof and drafted the manuscript Both XT and

QZ participated in the design and coordination All authors read and approvedthe final manuscript

Acknowledgments

The authors thank the referees for valuable comments and suggestions Thisproject is supported by Scientific Research Fund of Hunan Provincial Educa-tion Department (No 11A095) and partially supported by the NNSF (No:11171351) of China

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