Volume 2009, Article ID 562329, 10 pagesdoi:10.1155/2009/562329 Research Article Existence of Nonoscillatory Solutions to Second-Order Neutral Delay Dynamic Equations on Time Scales Tong
Trang 1Volume 2009, Article ID 562329, 10 pages
doi:10.1155/2009/562329
Research Article
Existence of Nonoscillatory Solutions to
Second-Order Neutral Delay Dynamic Equations
on Time Scales
Tongxing Li,1 Zhenlai Han,1, 2 Shurong Sun,1, 3 and Dianwu Yang1
1 School of Science, University of Jinan, Jinan, Shandong 250022, China
2 School of Control Science and Engineering, Shandong University, Jinan, Shandong 250061, China
3 Department of Mathematics and Statistics, Missouri University of Science and Technology, Rolla,
MO 65409-0020, USA
Correspondence should be addressed to Zhenlai Han,hanzhenlai@163.com
Received 5 March 2009; Revised 24 June 2009; Accepted 24 August 2009
Recommended by Alberto Cabada
We employ Kranoselskii’s fixed point theorem to establish the existence of nonoscillatory solutions
to the second-order neutral delay dynamic equation xt ptxτ0tΔΔ q1txτ1t −
q2txτ2t et on a time scaleT To dwell upon the importance of our results, one interesting example is also included
Copyrightq 2009 Tongxing Li et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited
1 Introduction
The theory of time scales, which has recently received a lot of attention, was introduced by
dynamic equations on time scales For the notation used below we refer to the next section
In recent years, there has been much research activity concerning the oscillation of solutions of various equations on time scales, and we refer the reader to Erbe5, Saker 6,
Trang 2In this work, we will consider the existence of nonoscillatory solutions to the second-order neutral delay dynamic equation of the form
x t ptxτ0tΔΔ q1txτ1t − q2txτ2t et 1.1
on a time scaleT an arbitrary closed subset of the reals
positive solutions of the delay equation
x t ptxt − τ q1txt − σ1 − q2txt − σ2 et. 1.2 Recently,24 established the existence of nonoscillatory solutions to the neutral equation
Neutral equations find numerous applications in natural science and technology For instance, they are frequently used for the study of distributed networks containing lossless transmission lines So, we try to establish some sufficient conditions for the existence of
solutions for neutral delay dynamic equations on time scales
onT and satisfies 1.1 for t ≥ t1 ≥ t0 A solution of1.1 is said to be eventually positive or eventually negative if there exists c ∈ T such that xt > 0 or xt < 0 for all t ≥ c in T
A solution of1.1 is said to be nonoscillatory if it is either eventually positive or eventually negative; otherwise, it is oscillatory
2 Main Results
In this section, we establish the existence of nonoscillatory solutions to1.1 For T0, T1 ∈ T,
letT0,∞T: {t ∈ T : t ≥ T0} and T0, T1T: {t ∈ T : T0≤ t ≤ T1} Further, let CT0,∞T,R
BC T0,∞T:
x : x ∈ CT0,∞T, R, sup
t ∈T0, ∞T|xt| < ∞
Endowed on BCT0,∞Twith the normx sup t ∈T0,∞T|xt|, BCT0,∞T, · is a Banach spacesee 24 Let X ⊆ BCT0,∞T, we say that X is uniformly Cauchy if for any given ε > 0,
there exists T1∈ T0,∞Tsuch that for any x ∈ X, |xt1 − xt2| < ε, for all t1, t2∈ T1,∞T
X is said to be equicontinuous on a, bTif for any given ε > 0, there exists δ > 0 such that for any x ∈ X, and t1, t2∈ a, bTwith|t1− t2| < δ, |xt1 − xt2| < ε.
Trang 3Also, we need the following auxiliary results.
Lemma 2.1 see 24, Lemma 4 Suppose that X ⊆ BCT0,∞Tis bounded and uniformly Cauchy Further, suppose that X is equicontinuous on T0, T1T for any T1 ∈ T0,∞T Then X is relatively compact.
Lemma 2.2 see 25, Kranoselskii’s fixed point theorem Suppose that Ω is a Banach space
and X is a bounded, convex, and closed subset of Ω Suppose further that there exist two operators
U, S : X → Ω such that
i Ux Sy ∈ X for all x, y ∈ X;
ii U is a contraction mapping;
iii S is completely continuous.
Then U S has a fixed point in X.
Hτ i t ∈ C rd T, T, τ i t ≤ t, lim t→ ∞τ i t ∞, i 0, 1, 2, pt,q j t ∈ C rd T, R, q j t >
0,∞
t0 σ sq j sΔs < ∞, j 1, 2, and there exists a function Et ∈ C2
rd T, R such that EΔΔt
e t, lim t→ ∞E t e0∈ R.
Theorem 2.3 Assume that H holds and |pt| ≤ p < 1/3 Then 1.1 has an eventually positive
solution.
Proof From the assumption H, we can choose T0 ∈ T T0 ≥ 1 large enough and positive
1 < M2< 1− p − 2M1
such that
∞
T0
σ sq1sΔs ≤
1− pM2− 1
∞
T0
σ sq2sΔs ≤ 1− p1 2M2 − 2M1
∞
T0
σ sq1s q2sΔs ≤ 3
1− p
|Et − e0| ≤ 1− p
Furthermore, fromH we see that there exists T1 ∈ T with T1 > T0such that τ i t ≥
T0, i 0, 1, 2, for t ∈ T1,∞T.
Define the Banach space BCT0,∞Tas in2.1 and let
Trang 4It is easy to verify that X is a bounded, convex, and closed subset of BCT0,∞T.
4 − ptxτ0T1 ET1 − e0, t ∈ T0, T1T,
4 − ptxτ0t Et − e0, t ∈ T1,∞T,
Sxt 1− p
∞
T1
q1sxτ1s − q2sxτ2sΔs, t ∈ T0, T1T,
Sxt 1− p
∞
t
q1sxτ1s − q2sxτ2sΔs
t
T1
σ sq1sxτ1s − q2sxτ2sΔs, t ∈ T1,∞T.
2.8
x ≤ M2, M1 ≤ y ≤ M2 For any x, y ∈ X and t ∈ T1,∞T, in view of2.3, 2.4 and 2.6, we have
Uxt Sy
t ≥ 3
1− p
∞
t
q2sxτ2sΔs −
t
T1
σ sq2sxτ2sΔs
∞
T1
σ sq2sΔs ≥ M1,
Uxt Sy
t ≤ 3
1− p
∞
t
q1sxτ1sΔs
t
T1
σ sq1sxτ1sΔs
T1
σ sq1sΔs ≤ M2.
2.9
t ∈ T0, T1T Hence, Ux Sy ∈ X for any x, y ∈ X.
ii We prove that U is a contraction mapping Indeed, for x, y ∈ X, we have
0T1 − yτ0T1
t ∈T0, ∞T x t − yt 2.10
for t ∈ T0, T1Tand
0t − yτ0t
Trang 5for t ∈ T1,∞T Therefore, we have
iii We will prove that S is a completely continuous mapping First, by i we know that S maps X into X.
x ∈ X and |x n t − xt| → 0 as n → ∞ for any t ∈ T0,∞T Consequently, by2.5 we have
|Sx n t − Sxt|
≤ t
∞
t
q1s|x n τ1s − xτ1s|Δs
t
q2s|x n τ2s − xτ2s|Δs
t
T1
σ sq1s|x n τ1s − xτ1s|Δs
t
T1
σ sq2s|x n τ2s − xτ2s|Δs
≤ x n − x
t
σ sq1s q2sΔs
t
T1
σ sq1s q2sΔs
x n − x
∞
T1
σ sq1s q2sΔs ≤ 3
1− p
2.13
for t ∈ T0,∞T So, we obtain
1− p
which proves that S is continuous on X.
Finally, we prove that SX is relatively compact It is sufficient to verify that SX satisfies
all conditions inLemma 2.1 By the definition of X, we see that SX is bounded For any ε > 0, take T2∈ T1,∞Tso that
∞
T2
Trang 6For any x ∈ X and t1, t2 ∈ T2,∞T, we have
|Sxt1 − Sxt2|
≤ t1
t1
q1sxτ1s − q2sxτ2sΔs − t2
∞
t2
q1sxτ1s − q2sxτ2sΔs
t1
T1
σ sq1sxτ1s − q2sxτ2sΔs −
t2
T1
σ sq1sxτ1s − q2sxτ2sΔs
t1
σ sq1s q2sΔs M2
∞
t2
σ sq1s q2sΔs
t2
t1
σ sq1s q2sΔs < 3M2ε.
2.16
Thus, SX is uniformly Cauchy.
The remainder is to consider the equicontinuous onT0, T2T for any T2 ∈ T0,∞T.
Without loss of generality, we set T1< T2 For any x ∈ X, we have |Sxt1 − Sxt2| ≡ 0 for
t1, t2∈ T0, T1Tand
|Sxt1 − Sxt2|
≤ t1
t1
q1sxτ1s − q2sxτ2sΔs − t2
∞
t2
q1sxτ1s − q2sxτ2sΔs
t1
T1
σ sq1sxτ1s − q2sxτ2sΔs −
t2
T1
σ sq1sxτ1s − q2sxτ2sΔs
t2
t1
σ sq1s q2sΔs
t1− t2
t1
q1sxτ1s − q2sxτ2sΔs
t2
∞
t1
q1sxτ1s − q2sxτ2sΔs − t2
t2
q1sxτ1s − q2sxτ2sΔs
≤ M2t2 M2
t2
t1
σ sq1s q2sΔs
M2|t1− t2|
∞
t1
σ sq1s q2sΔs
2.17
for t1, t2∈ T1, T2T.
Trang 7Now, we see that for any ε > 0, there exists δ > 0 such that when t1, t2 ∈ T1, T2Twith
|t1− t2| < δ,
for all x ∈ X This means that SX is equicontinuous on T0, T2Tfor any T2∈ T0,∞T.
that S is a completely continuous mapping.
ByLemma 2.2, there exists x ∈ X such that U Sx x Therefore, we have
x t 3
1− p
∞
t
q1sxτ1s − q2sxτ2sΔs
t
T1
σ sq1sxτ1s − q2sxτ2sΔs Et − e0, t ∈ T1,∞T,
2.19
Theorem 2.4 Assume that H holds and 0 ≤ pt ≤ p1 < 1 Then1.1 has an eventually positive
solution.
Proof From the assumption H, we can choose T0 ∈ T T0 ≥ 1 large enough and positive
1− M4< p1< 1− 2M3
1 2M4
such that
T0
σ sq1sΔs ≤ p1 M4− 1
M4 ,
∞
T0
σ sq2sΔs ≤ 1− p11 2M4 − 2M3
,
∞
T0
σ sq1s q2sΔs ≤ 3
1− p1
|Et − e0| ≤ 1− p1
2.21
Furthermore, fromH we see that there exists T1 ∈ T with T1 > T0such that τ i t ≥
T0, i 0, 1, 2, for t ∈ T1,∞T.
Define the Banach space BCT0,∞Tas in2.1 and let
It is easy to verify that X is a bounded, convex, and closed subset of BCT0,∞T.
Trang 8Now we define two operators U and S as inTheorem 2.3with p replaced by p1 The
Theorem 2.5 Assume that H holds and −1 < −p2 ≤ pt ≤ 0 Then 1.1 has an eventually
positive solution.
Proof From the assumption H, we can choose T0 ∈ T T0 ≥ 1 large enough and positive
2M5 p2< 1 < M6, 2.23 such that
T0
σ sq1sΔs ≤
1− p2
M6 ,
∞
T0
σ sq2sΔs ≤ 1− p2− 2M5
,
∞
T0
σ sq1s q2sΔs ≤ 3
1− p2
|Et − e0| ≤ 1− p2
2.24
Furthermore, fromH we see that there exists T1 ∈ T with T1 > T0such that τ i t ≥
T0, i 0, 1, 2, for t ∈ T1,∞T.
Define the Banach space BCT0,∞Tas in2.1 and let
It is easy to verify that X is a bounded, convex, and closed subset of BCT0,∞T.
We will give the following example to illustrate our main results
Example 2.6 Consider the second-order delay dynamic equations on time scales
x t ptxτ0tΔΔ 1
t α σ t x τ1t − 1
t β σ t x τ2t
−t σt
t2σ2t , t ∈ t0,∞T,
2.26
where t0 > 0, α > 1, β > 1, τ i t ∈ C rd T, T, τ i t ≤ t, lim t→ ∞τ i t ∞, i 0, 1, 2, |pt| ≤
p < 1/3 Then q1t 1/t α σ t, q2t 1/t β σ t, et −t σt/t2σ2t Let Et
t
eventually positive solution
Trang 9The authors sincerely thank the reviewers for their valuable suggestions and useful comments that have lead to the present improved version of the original manuscript
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