/ / Article

Research Article | Open Access

Volume 2014 |Article ID 768215 | 10 pages | https://doi.org/10.1155/2014/768215

# Nonoscillatory Solutions for System of Neutral Dynamic Equations on Time Scales

Accepted02 Feb 2014
Published16 Mar 2014

#### Abstract

We will discuss nonoscillatory solutions to the -dimensional functional system of neutral type dynamic equations on time scales. We will establish some sufficient conditions for nonoscillatory solutions with the property .

#### 1. Introduction

The theory of dynamic equations on time scales was introduced by its founder Hilger in his PHD thesis [1] in 1988. The study of dynamic equations on time scales is an area of mathematics that has recently received a lot of attention. It has been created in order to unify continuous and discrete analysis. In recent years there has been much research activity concerning the oscillation and nonoscillation of solutions of dynamic equations on time scales; we refer the reader to the papers [2â€“12]. In [13, 14], authors studied nonoscillatory solutions to the -dimensional functional differential systems of neutral type and obtained some sufficient conditions for nonoscillatory solutions with the property , . Using the idea and method of [13, 14], in this paper, we will study the nonoscillatory solutions for systems of neutral dynamic equations on time scales, which have the following form: where natural number , , is a continuous, real-valued positive function defined on the time scale and through this paper we assume that (a) is a continuous and increasing function with and for some ;(b), , are continuous and increasing functions; not identically equal to zero in any neighbourhood of infinity; , , hold for any ;(c) is a continuous and increasing function with ;(d) is a continuous function, and for , where is a positive constant.

#### 2. Some Preliminary Results

A time scale is an arbitrary nonempty closed subset of the real numbers. Throughout this paper, as a matter of convenience, for any , , we denote the sets by , which is called a close interval in . Open intervals and half-open intervals and so forth are defined accordingly.

A function is defined for as

A vector function is a solution to the system (1) if there is a such that functions , , are continuously differentiable and satisfies (1) on . Let be the set of all solutions to the system (1) satisfying for any . A solution is called nonoscillatory if there exists a such that every component is different from zero for . Otherwise a solution is said to be oscillatory.

Since we restrict our attention to asymptotic properties of nonoscillatory solutions to the system (1), we suppose that the time scale under consideration is not bounded above; that is, it is a time scale interval of the form . On any time scale we define the forward operator and delta derivative as follows.

Definition 1 (see [15]). Let be a time scale. For any , we define the forward jump operator by

Definition 2 (see [15]). Let be a time scale. Assume is a function and let . The (delta) derivative of at is defined by

For some other preliminary concepts on time scales, one can refer to [15]. In the remainder of this section, we present some lemmas indispensable which will be used later.

Lemma 3. Let be a solution to (1) with on . Then is nonoscillatory and are monotone on for some .

Proof. Because on , using the conditions (b) and (d), does not change its sign and not identically equal to zero on for some . This implies that is monotone and on for some . Continue similarly as above and then conclude the desire results. The proof is complete.

Lemma 4. Assume that (a) holds and for . Let be a nonoscillatory solution to the functional inequality where is a continuous function. If for , then is bounded on . Moreover, ifâ€‰â€‰ for , then .

Proof. We firstly claim that there exists a negative integer such that for some . In fact, if not the case, then there is such that for any . We need only to consider two cases: Case 1, for each . Then there exists satisfying , which implies , a contradiction. Case 2, . Then, by the monotonicity of , , also a contradiction.
Without loss of generality, we may assume that is a positive solution of the functional inequality on . Then for all .
If for , then , for any and each . For any , from the claim above, let for some and some . Then . Therefore, is bounded on .
If for , then for any and each . For any , from the claim above, let for some and some . Then , which implies that .

Similarly as Lemma 4, we can prove the following lemma.

Lemma 5. Assume that (a) holds and for . Let be a nonoscillatory solution to the functional inequality: where is a continuous function. If for , then is bounded on . Moreover, if for , then .

Lemma 6. Assume that and are continuous functions and for all . If , then the following two statements hold.(i)The functional inequality has no eventually positive solution.(ii)The functional inequality has no eventually negative solution.

Proof. We only prove (i) for the proof of (ii) is similar. Suppose that the functional inequality (7) has an eventually positive solution . Without loss of generality, we may assume that for any . Then, from (7), for any . It follows that is nondecreasing on .
Let be the linear extension of the function , then is continuous, is delta differentiable, and on . By [15, Theorem 1.87], for any , there is such that . Because for all , for all . Therefore, By (7), we have Integrating the inequality above from to on , we get, for , By the hypothesis, there exist and a constant such that holds for . So we have for , Here, to give the last inequality, we have used the inequality for all . From (7) and (9), we have Integrating the inequality above from to on , we get, for , Continuing this progress, we conclude that, for each natural number , Since , it follows that for any , a contradiction. The proof is complete.

Let be a nonoscillatory solution to (1). It follows, from Lemma 3, that the function has to be eventually of constant sign. Hence, either or for sufficiently large .

Lemma 7. Let be a nonoscillatory solution to (1) on , and for . Then there exist with or and some , such that for

Proof. Putting for and , then and so we can pick . It is obvious that . Without loss of generality, we may assume that for . Note that for all . If , it is obvious that (18) holds. Next, we assume , then , for . We will show recursively that (18) holds.
We firstly show that for . Otherwise, for . Then, from (1) and (b), we have for . So, by (1) and (b), holds for all , which contradicts the condition (b). Hence, , for .
We now show that for . Otherwise, for . Then, from (1) and (b), we have for . So, by (1) and (b), holds for all , which contradicts the condition (b). This implies that , for .
Continuing this progress, we can prove that (18) holds. To complete the proof, it remains to show that . Suppose , then, from (1) and (d), we have on for . Note that and for all . It is easy to show that for all . But then, holds for all , which contradicts the condition (b). This completes the proof.

Lemma 8. Let be a nonoscillatory solution to (1) on , and for . Then there exist with or and some , such that for , either or

Proof. Without loss of generality, we may assume that for . Note that for all . We consider the following two cases.
(I) for . We will show recursively that (22) holds. We firstly show that for . Otherwise, for . Then, from (1) and (b), it is easy to see that for . So, by (1) and (b) again, holds for all , which contradicts the condition (b). Hence, and for . Continuing this progress similarly, we conclude that (22) holds.
(II) for . Analogically as in the proof of Lemma 7, we can prove that (23) holds. This completes the proof.

For the sake of convenience, we denote by , , and the set of all nonoscillatory solutions to the system (1) satisfying (18), (22), and (23) correspondingly. Denote by the set of all nonoscillatory solutions to the system (1). From Lemmas 7 and 8, we have the classification of nonoscillatory solutions to the system (1) as follows:(i) is odd and , (ii) is odd and , (iii) is even and , (iv) is even and ,

Remark 9. Assume that and for , where is a constant. Then for .

In fact, if for some . Let and suppose, without loss of generality, that for . Note that and for . It follows that for . On the other hand, for . By Lemma 5, which implies that , a contradiction.

Lemma 10. Let be a nonoscillatory solution to (1) on . Suppose that , and for . Then

Proof. Suppose that , , then there is a positive constant such that for . By (1) and (b), we have It follows from (b) that . Obviously there is a positive constant such that for . Similarly as above, we can prove that . Continuing this progress, we conclude that (29) holds. Using (29), it is easy to see that (30) holds. The proof is complete.

#### 3. Main Results and Proofs

We will now give the main results and their proofs. In the squeal, for the convenience of expressions, we define and by recursion formula as follows: where , , are continuous functions and .

Remark 11. From the definitions of and , it is easy to show the following two properties: (i)if for each , then for , ;(ii)if for each and for some , then for , .

Theorem 12. Assume that is odd and . If the following statements hold:
(1)there is a constant such that, for any , (2)for any , (3)there is a continuous function such that (4)for each even with , (5)then, for every nonoscillatory solution to (1), , .

Proof. Let be a nonoscillatory solution to (1). Without loss of generality, we may assume that for all . Because is odd and , the expression (26) holds. We consider the following five cases.
(I) on . In this case, we have By Lemma 4, . From (1) and (41), we have is a positive and nonincreasing function, which implies that . It follows, from Lemma 10, that , .
(II) on . In this case, we have From (1) and (42), there is a positive constant such that for . Integrating the first equation of (1) over , and using the inequality above, we get By the condition (b), . The conditions (35), (36), and (42) imply that is bounded on , which arrives a contradiction as for . Therefore, .
(III) on . In this case, we have Integrating the second equation of (1) from to , we obtain that, for and , Integrating the third equation of (1) from to , we get that, for and , Continuing this progress, we have that, for any and , , Combining inequalities above, we obtain that, for and , From (1), (b), (d), and (44), we have that, for and , Substituting (49) to (48), we have that, for and , Because of for , Using the monotonicity of , we have that, for and , Combining (50) and (52), we have that, for and , Multiplying the inequality above by and putting , we get that By Lemma 6, the inequality above has no eventually negative solution, a contradiction. Hence, .
(IV) on , . In this case, we have that, for , Integrating the first equation of (1) from to , we get that, for and , Integrating the equation of (1), we obtain, for and , , that Integrating the equation of (1), we obtain, for and , , that Combining inequalities above, we obtain that, for and , Note that (52) holds if , , , and for . So we have that, for and , ,
Putting , then Substituting (61) to (59), we have, on for some sufficiently large , that So, for This contradicts (39) and hence, .
(V) on . In this case, we have that, for ,
Similarly as in the case (IV) of the proof of this theorem, we can get, for some sufficiently large , that Using the monotonicity of , we have, for , that So, for which gives a contradiction with (40). Thus . The proof is complete.

Theorem 13. Assume that is even, in the system (1) and conditions (35)â€“(39) hold. Then, for every nonoscillatory solution to (1), we have , .

Proof. Let be a nonoscillatory solution to (1) and the expression (27) holds. Without loss of generality, we may assume that for all . We consider the following five cases.
(I) on . Analogically as the case (I) in the proof of Theorem 12, we can show that , .
(II) on . Also analogically as in the proof of Theorem 12(II), we can prove that ;
(III) on . In this case, we have We continue analogically as (III) in the proof of Theorem 12. We can get, for , that From (1), (b), (d), and (68), we have that, for and , We substitute (70) with (69), we have that, for and , By the same way as in the proof of Theorem 12(III), we can obtain that .
(IV) on , . In this case, we have that, for , By the same way as in the proof of Theorem 12(IV), we can obtain, for and , that