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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 ()
Xianhua Tang ()
Qi-Ming Zhang ()
ISSN 1687-1847
Article type Research
Submission date 5 August 2011
Acceptance date 20 December 2011
Publication date 20 December 2011
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Stability criteria for linear Hamiltonian
dynamic systems on time scales
Xiaofei He
1,2
, Xianhua Tang
∗1
and Qi-Ming Zhang
1
1
School of Mathematical Sciences and Computing Technology,

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

(t) = α(t)x(σ(t)) + β(t)y(t), y

(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)
2
and
β(t) ≥ 0, ∀ t ∈ T. (1.3)
For the second-order linear dynamic equation
[p(t)x

(t)]

+ q(t)x(σ(t)) = 0, t ∈ T, (1.4)
if let y(t) = p(t)x

(t), then we can rewrite (1.4) as an equivalent polar linear
Hamiltonian dynamic system of type (1.1):
x

(t) =


p
0
+

ω
0
1
p(t)
t


ω
0
q
+
(t)t ≤ 4, (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}, (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; conditionally stable if there exist a
nontrivial solution which is bounded on T; and stable if all solutions are bounded

ω
0
γ
+
(t)t

1/2
< 2. (1.13)
Then system (1.1) is stable.
Theorem 1.4. Assume that (1.6) and (1.7) hold, and that

ω
0
1
p(t)
t

ω
0
q
+
(t)t ≤ 4. (1.14)
Then equation (1.4) is stable.
Remark 1.5. Clearly, condition (1.14) improves (1.8) by removing term p
0
.
We dwell on the three special cases as follows:
1. If T = R, system (1.1) takes the form:
x



1/2
< 2. (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].
5
2. If T = Z, system (1.1) takes the form:
x(n) = α(n)x(n+ 1)+ β(n)y(n), y (n) = −γ(n)x(n+1)−α(n)y(n), n ∈ Z.
(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}, (1.19)
ω− 1

n=0

γ(n) − θ
2
(n)β(n)

> 0, (1.20)
and
ω− 1

n=0
|α(n)| +

ω −1


p(n)
, q(n) ≡ 0, ∀ n ∈ {0, 1, . . . ,ω − 1}
(1.22)
in Theorem 1.4 are better than the one
0 <
ω −1

n=0
q(n) ≤
ω −1

n=0
q
+
(n) <
4

ω −1
n=0
1
p(n)
(1.23)
in [16, Corollary 3.4]. More related results on stability for discrete linear Hamil-
tonian systems can be found in [20–24].
3. Let δ > 0 and N ∈ {2, 3, 4, . . .}. Set ω = δ + N, define the time scale T
as follows:
T =

k∈Z
[kω, kω + δ] ∪ {kω + δ + n : n = 1, 2, . . . , N − 1}. (1.24)


n=0

γ(δ + n) − θ
2
(δ + n)β(δ + n)

> 0, (1.28)
and


δ
0
|α(t)|dt +
N−1

n=0
|α(δ + n)|

+


δ
0
β(t)dt +
N−1

n=0
|β(δ + n)|


−γ(t) −α(t)




.
7
Then, we can rewrite (1.1) as a standard linear Hamiltonian dynamic system
u

(t) = A(t)u
σ
(t), t ∈ T. (2.1)
Let u
1
(t) = (x
10
(t), y
10
(t))

and u
2
(t) = (x
20
(t), y
20
(t))

be two solutions


x

10
(t) x

20
(t)
y
10
(t) y
20
(t)








+









− Φ(ω)) = 0,
which is equivalent to
λ
2
− Hλ + 1 = 0, (2.4)
where
H = x
10
(ω) + y
20
(ω).
8
Hence
λ
1
+ λ
2
= H, λ
1
λ
2
= 1. (2.5)
Let v
1
= (c
11
, c
21
)


v
j
(t), ∀ t ∈ T, j = 1, 2. (2.7)
On the other hand, it follows from (2.1) that
v

j
(t) = Φ

(t)v
j
=

u

1
(t), u

2
(t)

v
j
= A(t) (u
σ
1
(t), u
σ
2
(t)) v

Definition 2.3. A function f : T → R is said to have a generalized zero at
t
0
∈ T provided either f(t
0
) = 0 or f(t
0
)f(σ(t
0
)) < 0.
Lemma 2.4. [4] Assume f, g : T → R are differential at t ∈ T
k
. If f

(t)
exists, then f(σ(t)) = f(t) + µ(t)f

(t).
Lemma 2.5. [4] (Cauchy-Schwarz inequality). Let a, b ∈ T. For f, g ∈ C
rd
we
have

b
a
|f(t)g(t)| t ≤


b
a


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 H
2
≥ 4, then both x
1
(t) and x
2
(t) have generalized zeros in
T[0, ω].
Proof. Since |H| ≥ 2, then λ
1
and λ
2
are real numbers, and v
1
(t) and v
2
(t) are
also real functions. We only prove that x
1
(t) must have at least one generalized
zero in T[0, ω]. Otherwise, we assume that x
1
(t) > 0 for t ∈ T[0, ω] and so (2.7)
10
implies that x
1

(σ(t)) − α(t)[x
1
(t) + x
1
(σ(t))]y
1
(t) − β(t)y
2
1
(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))

(t)
x
1
(σ(t))

. (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. (2.10)
Since x
1
(t) > 0 for all t ∈ T, 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

1 + µ(t)β(t)z(t)
≤
α
2
(t)
β(t)
≤ θ
2
(t)β(t); (2.14)
If β(t) = 0, it follows from (1.11) that α(t) = 0, hence
[−2α(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). (2.15)
11
Combining (2.14) with (2.15), we have
[−2α(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). (2.16)

2
(t), y
2
(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, (2.18)
β(t + ω) = β(t), γ(t + ω) = γ(t), ∀ t ∈ T, (2.19)
and

ω
0
γ(t)t ≥ 0. (2.20)
If H
2
≥ 4, then both x
1
(t) and x
2
(t) have generalized zeros in T[0, ω].
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

β(t)z
2
(t)
1 + µ(t)β(t)z(t)

t < −

ω
0
γ(t)t,
which contradicts condition (2.20). 
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)) = (κ
1
x(a), κ
2
y(a)) with 0 <
κ
2
1
≤ κ
1
κ
2
≤ 1 and x(t) ≡ 0 on T[a, b]. Then one has the following inequality

b

a
γ
+
(t)t > 4. (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) = −ξµ(a)β(a)y(a), (2.25)
13
x(τ) = (1 − ξ)µ(a)β(a)y(a) +

τ
σ (a)
β(t)y(t)t, σ(a) ≤ τ ≤ b, (2.26)
ϑ
1
µ(a)β(a)y
2
(a) +

b
σ(a)
β(t)y
2
(t)t =

b
a
γ(t)x
2
(σ(t))t, (2.27)

= y(a)

(1 − ξ)µ(a)β(a) +

b
σ (a)
β(t)t

= x(a) + y(a)

b
a
β(t)t
= x(a),
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)| +

b
σ(a)
β(t)|y(t)|t, σ(a) ≤ τ ≤ b (2.30)
14
instead of (2.28). Applying Lemma 2.5 and using (2.27) and (2.30), we have
2|x(τ
∗
)|
< ϑ
2

1/2
=

ϑ
2
2
ϑ
1
µ(a)β(a) +

b
σ (a)
β(t)t


b
a
γ(t)x
2
(σ(t))t

1/2
≤ |x(τ
∗
)|

ϑ
2
2
ϑ



b
a
γ
+
(t)t

1/2
> 2. (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)| +

b
σ (a)
β(t)|y(t)|t
<

ϑ
2
2
ϑ
1
µ(a)β(a) +



b
a
γ(t)x
2
(σ(t))t

1/2
≤ |x(τ
∗
)|

ϑ
2
2
ϑ
1
µ(a)β(a) +

b
σ (a)
β(t)t


b
a
γ
+
(t)t

= 1, it follows that 0 < min{λ
2
1
, λ
2
2
} ≤ 1. Suppose λ
2
1
≤ 1. By Lemma
2.6, system (1.1) has a non-zero solution v
1
(t) = (x
1
(t), y
1
(t))

such that (2.7)
holds and x
1
(t) has a generalized zero in T[0, ω], say t
1
. It follows from (2.7)
that (x
1
(t
1
+ ω), y
1

1
+ω
t
1
|α(t)|t +


t
1
+ω
t
1
β(t)t

t
1
+ω
t
1
γ
+
(t)t

1/2
≥ 2. (2.34)
Next, noticing that for any ω-periodic function f(t) on T, the equality

t
0
+ω

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. 
Competing interests
The authors declare that they have no competing interests.
16
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 approved
the final manuscript.
Acknowledgments
The authors thank the referees for valuable comments and suggestions. This
project 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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