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Stability criteria for linear Hamiltonian dynamic systems on time scales

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Research

Abstract

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

x Δ ( t ) = α ( t ) x ( σ ( t ) ) + β ( t ) y ( t ) , y Δ ( t ) = - γ ( t ) x ( σ ( t ) ) - α ( t ) y ( t ) , t T Open image in new window

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 

1 Introduction

A time scale is an arbitrary nonempty closed subset of the real numbers ℝ. We assume that T Open image in new window is a time scale. For t T Open image in new window, the forward jump operator σ : T T Open image in new window is defined by σ ( t ) : inf { s T : s > t } Open image in new window, the backward jump operator ρ : T T Open image in new window is defined by ρ ( t ) : sup { s T : s < t } Open image in new window, and the graininess function μ : T [ 0 , } Open image in new window 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, 2, 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 + n ω T Open image in new window for all t T Open image in new window and n∈ ℤ, then we call T Open image in new window a periodic time scale with period ω.

Suppose T Open image in new window is a ω-periodic time scale and 0 T Open image in new window. Consider the polar linear Hamiltonian dynamic system on time scale T Open image in new window
x Δ ( t ) = α ( t ) x ( σ ( t ) ) + β ( t ) y ( t ) , y Δ ( t ) = - γ ( t ) x ( σ ( t ) ) - α ( t ) y ( t ) , t T , Open image in new window
(1.1)
where α(t), β(t) and γ(t) are real-valued rd-continuous functions defined on T Open image in new window. Throughout this article, we always assume that
1 - μ ( t ) α ( t ) > 0 , t T Open image in new window
(1.2)
and
β ( t ) 0 , t T . Open image in new window
(1.3)
For the second-order linear dynamic equation
[ p ( t ) x Δ ( t ) ] Δ + q ( t ) x ( σ ( t ) ) = 0 , t T , Open image in new window
(1.4)
if let y(t) = p(t)xΔ (t), then we can rewrite (1.4) as an equivalent polar linear Hamiltonian dynamic system of type (1.1):
x Δ ( t ) = 1 p ( t ) y ( t ) , y Δ ( t ) = - q ( t ) x ( σ ( t ) ) , t T , Open image in new window
(1.5)
where p(t) and q(t) are real-valued rd-continuous functions defined on T Open image in new window with p(t) > 0, and
α ( t ) = 0 , β ( t ) = 1 p ( t ) , γ ( t ) = q ( t ) . Open image in new window

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 on time scales, which generalize the corresponding results on differential and difference equations. Lyapunov-type inequalities are very useful in oscillation theory, stability, disconjugacy, eigenvalue problems and numerous other applications in the theory of differential and difference equations. In particular, the stability criteria for the polar continuous and discrete Hamiltonian systems can be obtained by Lyapunov-type inequalities and Floquet theory, see [11, 12, 13, 14, 15, 16]. In 2000, Atici et al. [17] established the following stablity criterion for the second-order linear dynamic equation (1.4):

Theorem 1.2 [17]. Assume p(t) > 0 for t T Open image in new window, and that
p ( t + ω ) = p ( t ) , q ( t + ω ) = q ( t ) , t T . Open image in new window
(1.6)
If
0 ω q ( t ) Δ t 0 , q ( t ) 0 Open image in new window
(1.7)
and
p 0 + 0 ω 1 p ( t ) Δ t 0 ω q + ( t ) Δ t 4 , Open image in new window
(1.8)
then equation (1.4) is stable, where
p 0 = max t [ 0 , ρ ( ω ) ] σ ( t ) - t p ( t ) , q + ( t ) = max { q ( t ) , 0 } , Open image in new window
(1.9)

where and in the sequel, system (1.1) or Equation (1.4) is said to be unstable if all nontrivial solutions are unbounded on T Open image in new window; conditionally stable if there exist a nontrivial solution which is bounded on T Open image in new window; and stable if all solutions are bounded on T Open image in new window.

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

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 . Open image in new window
(1.10)
Assume that there exists a non-negative rd-continuous function θ (t) defined on T Open image in new window such that
| α ( t ) | θ ( t ) β ( t ) , t T [ 0 , ω ] = [ 0 , ω ] T , Open image in new window
(1.11)
0 ω [ γ ( t ) - θ 2 ( t ) β ( t ) ] Δ t > 0 , Open image in new window
(1.12)
and
0 ω | α ( t ) | Δ t + 0 ω β ( t ) Δ t 0 ω γ + ( t ) Δ t 1 2 < 2 . Open image in new window
(1.13)

Then system (1.1) is stable.

Theorem 1.4. Assume that (1.6) and (1.7) hold, and that
0 ω 1 p ( t ) Δ t 0 ω q + ( t ) Δ t 4 . Open image in new window
(1.14)

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. 1.
    If T = Open image in new window, system (1.1) takes the form:
    x ( t ) = α ( t ) x ( t ) + β ( t ) y ( t ) , y ( t ) = - γ ( t ) x ( t ) - α ( t ) y ( t ) , t . Open image in new window
    (1.15)
    In this case, the conditions (1.12) and (1.13) of Theorem 1.3 can be transformed into
    0 ω γ ( t ) - θ 2 ( t ) β ( t ) d t > 0 , Open image in new window
    (1.16)
    and
    0 ω | α ( t ) | d t + 0 ω β ( t ) d t 0 ω γ + ( t ) d t 1 2 < 2 . Open image in new window
    (1.17)

    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. 2.
    If T = Open image in new window, system (1.1) takes the form:
    Δ x ( n ) = α ( n ) x ( n + 1 ) + β ( n ) y ( n ) , Δ y ( n ) = - γ ( n ) x ( n + 1 ) - α ( n ) y ( n ) , n . Open image in new window
    (1.18)
    In this case, the conditions (1.11), (1.12), and (1.13) of Theorem 1.3 can be transformed into
    | α ( n ) | θ ( n ) β ( n ) , n { 0 , 1 , , ω - 1 } , Open image in new window
    (1.19)
    n = 0 ω - 1 γ ( n ) - θ 2 ( n ) β ( n ) > 0 , Open image in new window
    (1.20)
    and
    n = 0 ω - 1 | α ( n ) | + n = 0 ω - 1 β ( n ) n = 0 ω - 1 γ + ( n ) 1 2 < 2 . Open image in new window
    (1.21)
    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 Open image in new window, the stability conditions
    0 n = 0 ω - 1 q ( n ) n = 0 ω - 1 q + ( n ) 4 n = 0 ω - 1 1 p ( n ) , q ( n ) 0 , n { 0 , 1 , , ω - 1 } Open image in new window
    (1.22)
    in Theorem 1.4 are better than the one
    0 < n = 0 ω - 1 q ( n ) n = 0 ω - 1 q + ( n ) < 4 n = 0 ω - 1 1 p ( n ) Open image in new window
    (1.23)

    in [16, Corollary 3.4]. More related results on stability for discrete linear Hamiltonian systems can be found in [20, 21, 22, 23, 24].

     
  3. 3.
    Let δ > 0 and N ∈ {2, 3, 4, ...}. Set ω = δ + N, define the time scale T Open image in new window as follows:
    T = k [ k ω , k ω + δ ] { k ω + δ + n : n = 1 , 2 , , N - 1 } . Open image in new window
    (1.24)
    Then system (1.1) takes the form:
    x ( t ) = α ( t ) x ( t ) + β ( t ) y ( t ) , y ( t ) = - γ ( t ) x ( t ) - α ( t ) y ( t ) , t k [ k ω , k ω + δ ) , Open image in new window
    (1.25)
    and
    Δ x ( t ) = α ( t ) x ( t + 1 ) + β ( t ) y ( t ) , Δ y ( t ) = - γ ( t ) x ( t + 1 ) - α ( t ) y ( t ) , t k { k ω + δ + n : n = 0 , 1 , , N - 2 } . Open image in new window
    (1.26)
    In this case, the conditions (1.11), (1.12), and (1.13) of Theorem 1.3 can be transformed into
    | α ( t ) | θ ( t ) β ( t ) , t [ 0 , δ ] { δ + 1 , δ + 2 , , δ + N - 1 } , Open image in new window
    (1.27)
    0 δ γ ( t ) - θ 2 ( t ) β ( t ) d t + n = 0 N - 1 γ ( δ + n ) - θ 2 ( δ + n ) β ( δ + n ) > 0 , Open image in new window
    (1.28)
    and
    0 δ | α ( t ) | d t + n = 0 N - 1 | α ( δ + n ) | + 0 δ β ( t ) d t + n = 0 N - 1 | β ( δ + n ) | 0 δ γ + ( t ) d t + n = 0 N - 1 | γ + ( δ + n ) | 1 2 < 2 . Open image in new window
    (1.29)
     

2 Proofs of theorems

Let u(t) = (x(t), y(t)), u σ (t) = (x(σ(t)), y(t)) and
A ( t ) = α ( t ) β ( t ) - γ ( t ) - α ( t ) . Open image in new window
Then, we can rewrite (1.1) as a standard linear Hamiltonian dynamic system
u Δ ( t ) = A ( t ) u σ ( t ) , t T . Open image in new window
(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. Suppose that α(t), β(t) and γ(t) are ω-periodic functions defined on T Open image in new window (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 initial condition ( see [9, 18, 19]) that
Φ ( t + ω ) = Φ ( t ) Φ ( ω ) , t T . Open image in new window
(2.2)
From (1.1), we have
( det Φ ( t ) ) Δ = x 10 Δ ( t ) x 20 Δ ( t ) y 10 ( t ) y 20 ( t ) + x 10 ( σ ( t ) ) x 20 ( σ ( t ) ) y 10 Δ ( t ) y 20 Δ ( t ) = 0 , t T . Open image in new window
(2.3)
It follows that det Φ(t) = det Φ(0) = 1 for all t T Open image in new window. Let λ1 and λ2 be the roots (real or complex) of the characteristic equation of Φ(ω)
det ( λ I 2 - Φ ( ω ) ) = 0 , Open image in new window
which is equivalent to
λ 2 - H λ + 1 = 0 , Open image in new window
(2.4)
where
H = x 10 ( ω ) + y 20 ( ω ) . Open image in new window
Hence
λ 1 + λ 2 = H , λ 1 λ 2 = 1 . Open image in new window
(2.5)
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.
Φ ( ω ) v j = λ j v j , j = 1 , 2 . Open image in new window
(2.6)
Let v j (t) = Φ(t)v j , j = 1, 2. Then it follows from (2.2) and (2.6) that
v j ( t + ω ) = λ j v j ( t ) , t T , j = 1 , 2 . Open image in new window
(2.7)
On the other hand, it follows from (2.1) that
v j Δ ( t ) = Φ Δ ( t ) v j = u 1 Δ ( t ) , u 2 Δ ( t ) v j = A ( t ) u 1 σ ( t ) , u 2 σ ( t ) v j = A ( t ) v j σ ( t ) , j = 1 , 2 . Open image in new window
(2.8)

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 λ1 and λ2 be 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 on time scales.

Definition 2.3. A function f : T Open image in new window is said to have a generalized zero at t 0 T Open image in new window provided either f(t0) = 0 or f(t0)f(σ(t0)) < 0.

Lemma 2.4. [4] Assume f , g : T Open image in new window are differential at t T k Open image in new window. If fΔ(t) exists, then f(σ(t)) = f(t) + μ(t)fΔ(t).

Lemma 2.5. [4] (Cauchy-Schwarz inequality). Let a , b T Open image in new window. For f,gC rd we have
a b f ( t ) g ( t ) Δ t a b f 2 ( t ) Δ t a b g 2 ( t ) Δ t 1 2 . Open image in new window

The above inequality can be equality only if there exists a constant c such that f(t) = cg(t) for t T [ a , b ] Open image in new window.

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 ] Open image in new window.

Proof. Since |H| ≥ 2, then λ1 and λ2 are 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 ] Open image in new window. Otherwise, we assume that x1(t) > 0 for t T [ 0 , ω ] Open image in new window and so (2.7) implies that x1(t) > 0 for t T Open image in new window. 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 Open image in new window. From (1.1), we have
z Δ ( t ) = x 1 ( t ) y 1 Δ ( t ) - x 1 Δ ( t ) y 1 ( t ) x 1 ( t ) x 1 ( σ ( t ) ) = - γ ( t ) x 1 ( t ) x 1 ( σ ( t ) ) - α ( t ) [ x 1 ( t ) + x 1 ( σ ( t ) ) ] y 1 ( t ) - β ( t ) y 1 2 ( t ) x 1 ( t ) x 1 ( σ ( t ) ) = - γ ( t ) - α ( t ) y 1 ( t ) x 1 ( t ) + y 1 ( t ) x 1 ( σ ( t ) ) - β ( t ) y 1 ( t ) x 1 ( t ) y 1 ( t ) x 1 ( σ ( t ) ) = - γ ( t ) - α ( t ) z ( t ) + y 1 ( t ) x 1 ( σ ( t ) ) - β ( t ) z ( t ) y 1 ( t ) x 1 ( σ ( t ) ) . Open image in new window
(2.9)
From the first equation of (1.1), and using Lemma 2.4, we have
[ 1 - μ ( t ) α ( t ) ] x 1 ( σ ( t ) ) = x 1 ( t ) + μ ( t ) β ( t ) y 1 ( t ) , t T . Open image in new window
(2.10)
Since x1(t) > 0 for all t T Open image in new window, it follows from (1.2) and (2.10) that
1 + μ ( t ) β ( t ) z ( t ) = 1 + μ ( t ) β ( t ) y 1 ( t ) x 1 ( t ) = [ 1 - μ ( t ) α ( t ) ] x 1 ( σ ( t ) ) x 1 ( t ) > 0 , Open image in new window
(2.11)
which yields
y 1 ( t ) x 1 ( σ ( t ) ) = [ 1 - μ ( t ) α ( t ) ] z ( t ) 1 + μ ( t ) β ( t ) z ( t ) . Open image in new window
(2.12)
Substituting (2.12) into (2.9), we obtain
z Δ ( t ) = - γ ( t ) + [ - 2 α ( t ) + μ ( t ) α 2 ( t ) ] z ( t ) - β ( t ) z 2 ( t ) 1 + μ ( t ) β ( t ) z ( t ) . Open image in new window
(2.13)
If β(t) > 0, together with (1.11), it is easy to verify that
[ - 2 α ( t ) + μ ( t ) α 2 ( t ) ] z ( t ) - β ( t ) z 2 ( t ) 1 + μ ( t ) β ( t ) z ( t ) α 2 ( t ) β ( t ) θ 2 ( t ) β ( t ) ; Open image in new window
(2.14)
If β(t) = 0, it follows from (1.11) that α(t) = 0, hence
[ - 2 α ( t ) + μ ( t ) α 2 ( t ) ] z ( t ) - β ( t ) z 2 ( t ) 1 + μ ( t ) β ( t ) z ( t ) = 0 = θ 2 ( t ) β ( t ) . Open image in new window
(2.15)
Combining (2.14) with (2.15), we have
[ - 2 α ( t ) + μ ( t ) α 2 ( t ) ] z ( t ) - β ( t ) z 2 ( t ) 1 + μ ( t ) β ( t ) z ( t ) θ 2 ( t ) β ( t ) . Open image in new window
(2.16)
Substituting (2.16) into (2.13), we obtain
z Δ ( t ) - γ ( t ) + θ 2 ( t ) β ( t ) . Open image in new window
(2.17)
Integrating equation (2.17) from 0 to ω, and noticing that z(t) is ω-periodic, we obtain
0 - 0 ω γ ( t ) - θ 2 ( t ) β ( t ) Δ t , Open image in new window

which contradicts condition (1.12). □

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
α ( t ) = 0 , β ( t ) > 0 , γ ( t ) 0 , t T , Open image in new window
(2.18)
β ( t + ω ) = β ( t ) , γ ( t + ω ) = γ ( t ) , t T , Open image in new window
(2.19)
and
0 ω γ ( t ) Δ t 0 . Open image in new window
(2.20)

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

Proof. Except (1.12), (2.18), and (2.19) imply all assumptions in Lemma 2.6 hold. In view of the proof of Lemma 2.6, it is sufficient to derive an inequality which contradicts (2.20) instead of (1.12). From (2.11), (2.13), and (2.18), we have
1 + μ ( t ) β ( t ) z ( t ) = 1 + μ ( t ) β ( t ) y 1 ( t ) x 1 ( t ) = x 1 ( σ ( t ) ) x 1 ( t ) > 0 Open image in new window
(2.21)
and
z Δ ( t ) = - γ ( t ) - β ( t ) z 2 ( t ) 1 + μ ( t ) β ( t ) z ( t ) . Open image in new window
(2.22)
Since z(t) is ω-periodic and γ(t) ≢ 0,, it follows from (2.22) that z2(t) ≢ 0 on T [ 0 , ω ] Open image in new window. Integrating equation (2.22) from 0 to ω, we obtain
0 = - 0 ω γ ( t ) + β ( t ) z 2 ( t ) 1 + μ ( t ) β ( t ) z ( t ) Δ t < - 0 ω γ ( t ) Δ t , Open image in new window

which contradicts condition (2.20). □

Lemma 2.8. [10] Suppose that (1.2) and (1.3) hold and let a , b T k Open image in new window 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 < κ 1 2 κ 1 κ 2 1 Open image in new window and x(t) ≢ 0 on T [ a , b ] Open image in new window. Then one has the following inequality
a b α ( t ) Δ t + a b β ( t ) Δ t a b γ + ( t ) Δ t 1 2 2 . Open image in new window
(2.23)
Lemma 2.9. Suppose that (2.18) holds and let a , b T k Open image in new window 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)) = (κx(a), κy(a)) with 0 < κ2 ≤ 1 and x(t) is not identically zero on T [ a , b ] Open image in new window. Then one has the following inequality
a b β ( t ) Δ t a b γ + ( t ) Δ t > 4 . Open image in new window
(2.24)
Proof. In view of the proof of [10, Theorem 3.5] (see (3.8), (3.29)-(3.34) in [10]), we have
x ( a ) = - ξ μ ( a ) β ( a ) y ( a ) , Open image in new window
(2.25)
x ( τ ) = ( 1 - ξ ) μ ( a ) β ( a ) y ( a ) + σ ( a ) τ β ( t ) y ( t ) Δ t , σ ( a ) τ b , Open image in new window
(2.26)
ϑ 1 μ ( a ) β ( a ) y 2 ( a ) + σ ( a ) b β ( t ) y 2 ( t ) Δ t = a b γ ( t ) x 2 ( σ ( t ) ) Δ t , Open image in new window
(2.27)
and
2 | x ( τ ) | ϑ 2 μ ( a ) β ( a ) | y ( a ) | + σ ( a ) b β ( t ) | y ( t ) | Δ t , σ ( a ) τ b , Open image in new window
(2.28)
where ξ ∈ [0, 1), and
ϑ 1 = 1 - ξ + κ 2 ξ , ϑ 2 = 1 - ξ + | κ | ξ . Open image in new window
(2.29)
Let |x(τ*)| = maxσ(a)≤τb|x(τ)|. There are three possible cases:
  1. (1)

    y(t) ≡ y(a) ≠ 0, t T [ a , b ] Open image in new window;

     
  2. (2)

    y(t) ≢ y(a), |y(t)| ≡ |y(a)|, t T [ a , b ] Open image in new window;

     
  3. (3)

    |y(t)| ≢ |y(a)|, t T [ a , b ] Open image in new window.

     
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 ) + σ ( a ) b β ( t ) y ( t ) Δ t = y ( a ) ( 1 - ξ ) μ ( a ) β ( a ) + σ ( a ) b β ( t ) Δ t = x ( a ) + y ( a ) a b β ( t ) Δ t x ( a ) , Open image in new window

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

Case (2). In this case, we have
2 | x ( τ ) | < ϑ 2 μ ( a ) β ( a ) | y ( a ) | + σ ( a ) b β ( t ) | y ( t ) | Δ t , σ ( a ) τ b Open image in new window
(2.30)
instead of (2.28). Applying Lemma 2.5 and using (2.27) and (2.30), we have
2 | x ( τ * ) | < ϑ 2 μ ( a ) β ( a ) | y ( a ) | + σ ( a ) b β ( t ) | y ( t ) | Δ t ϑ 2 2 ϑ 1 μ ( a ) β ( a ) + σ ( a ) b β ( t ) Δ t ϑ 1 μ ( a ) β ( a ) y 2 ( a ) + σ ( a ) b β ( t ) y 2 ( t ) Δ t 1 2 = ϑ 2 2 ϑ 1 μ ( a ) β ( a ) + σ ( a ) b β ( t ) Δ t a b γ ( t ) x 2 ( σ ( t ) ) Δ t 1 2 | x ( τ * ) | ϑ 2 2 ϑ 1 μ ( a ) β ( a ) + σ ( a ) b β ( t ) Δ t a b γ + ( t ) Δ t 1 2 . Open image in new window
(2.31)
Dividing the latter inequality of (2.31) by |x(τ*)|, we obtain
ϑ 2 2 ϑ 1 μ ( a ) β ( a ) + σ ( a ) b β ( t ) Δ t a b γ + ( t ) Δ t 1 2 > 2 . Open image in new window
(2.32)
Case (3). In this case, applying Lemma 2.5 and using (2.27) and (2.28), we have
2 | x ( τ * ) | ϑ 2 μ ( a ) β ( a ) | y ( a ) | + σ ( a ) b β ( t ) | y ( t ) | Δ t < ϑ 2 2 ϑ 1 μ ( a ) β ( a ) + σ ( a ) b β ( t ) Δ t ϑ 1 μ ( a ) β ( a ) y 2 ( a ) + σ ( a ) b β ( t ) y 2 ( t ) Δ t 1 2 = ϑ 2 2 ϑ 1 μ ( a ) β ( a ) + σ ( a ) b β ( t ) Δ t a b γ ( t ) x 2 ( σ ( t ) ) Δ t 1 2 | x ( τ * ) | ϑ 2 2 ϑ 1 μ ( a ) β ( a ) + σ ( a ) b β ( t ) Δ t a b γ + ( t ) Δ t 1 2 . Open image in new window
(2.33)
Dividing the latter inequality of (2.33) by |x(τ*)|, we also obtain (2.32). It is easy to verify that
ϑ 2 2 ϑ 1 = [ 1 - ξ + | κ | ξ ] 2 1 - ξ + κ 2 ξ 1 . Open image in new window

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 { λ 2 1 , λ 2 2 } 1 Open image in new window. Suppose λ 1 2 1 Open image in new window. 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 ] Open image in new window, 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
t 1 t 1 + ω | α ( t ) | Δ t + t 1 t 1 + ω β ( t ) Δ t t 1 t 1 + ω γ + ( t ) Δ t 1 2 2 . Open image in new window
(2.34)
Next, noticing that for any ω-periodic function f(t) on T Open image in new window, the equality
t 0 t 0 + ω f ( t ) Δ t = 0 ω f ( t ) Δ t Open image in new window
holds for all t 0 T Open image in new window. It follows from (3.1) that
0 ω | α ( t ) | Δ t + 0 ω β ( t ) Δ t 0 ω γ + ( t ) Δ t 1 2 2 . Open image in new window
(2.35)

which contradicts condition (1.13). Thus |H| < 2 and hence system (1.1) is stable. □

Proof of Theorem 1.4. By using Lemmas 2.7 and 2.9 instead of Lemmas 2.6 and 2.8, respectively, we can prove Theorem 1.4 in a similar fashion as the proof of Theorem 1.3. So, we omit the proof here. □

Notes

Acknowledgements

The authors thank the referees for valuable comments and suggestions. This project is supported by Scientific Research Fund of Hunan Provincial Education Department (No. 11A095) and partially supported by the NNSF (No: 11171351) of China.

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This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.School of Mathematical Sciences and Computing TechnologyCentral South UniversityChangshaP.R. China
  2. 2.College of Mathematics and Computer ScienceJishou UniversityJishouP.R.China

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