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Nonoscillatory solutions to third-order neutral dynamic equations on time scales
Advances in Difference Equations volume 2014, Article number: 309 (2014)
Abstract
In this paper, we establish the existence of nonoscillatory solutions to third-order nonlinear neutral dynamic equations on time scales of the form by employing Kranoselskii’s fixed point theorem. Three examples are included to illustrate the significance of the conclusions.
1 Introduction
In this paper, we study third-order nonlinear neutral dynamic equations of the form
on a time scale satisfying and .
Throughout this paper we shall assume that:
(C1) such that
(C2) and there exists a constant with such that
(C3) , , , and
where
If , there exists a sequence such that and .
(C4) , is nondecreasing in x and for and .
Hilger introduced the theory of time scales in his Ph.D. thesis [1] in 1988; see also [2]. More details of time scale calculus can be found in [3–6] and omitted here. In the last few years, there has been some research achievement as regards the existence of nonoscillatory solutions to neutral dynamic equations on time scales; see the papers [7–11] and the references therein.
Definition 1.1 By a solution of (1) we mean a continuous function which is defined on and satisfies (1) for . A solution of (1) is said to be eventually positive (or eventually negative) if there exists such that (or ) for all in . A solution x of (1) is said to be nonoscillatory if it is either eventually positive or eventually negative; otherwise, it is oscillatory.
In 1990s, some significative results for existence of nonoscillatory solutions to neutral functional differential equations were given in [7, 9]. In 2007, Zhu and Wang [11] discussed the existence of nonoscillatory solutions to first-order nonlinear neutral dynamic equations
on a time scale . In 2013, Gao and Wang [10] considered the second-order nonlinear neutral dynamic equations
under the condition , and established the existence of nonoscillatory solutions to (2) on a time scale. In 2014, Deng and Wang [8] studied the same problem of (2) under the condition .
In this paper, we shall establish the existence of nonoscillatory solutions to (1) by employing Kranoselskii’s fixed point theorem, and we give three examples to show the versatility of the results.
For simplicity, throughout this paper, we denote , where , and , , are denoted similarly.
2 Preliminary results
Let denote all continuous functions mapping into ℝ, and , . For , we define
Endowing with the norm , is a Banach space. Let , we say that X is uniformly Cauchy if for any given , there exists a such that, for any ,
X is said to be equi-continuous on if, for any given , there exists a such that, for any and with ,
We have the following lemma, which is an analog of the Arzela-Ascoli theorem on time scales.
Lemma 2.1 ([[11], Lemma 4])
Suppose that is bounded and uniformly Cauchy. Further, suppose that X is equi-continuous on for any . Then X is relatively compact.
In this section, our approach to the existence of nonoscillatory solutions to (1) is based largely on the application of Kranoselskii’s fixed point theorem (see [7]). For the sake of convenience, we state here this theorem as follows.
Lemma 2.2 (Kranoselskii’s fixed point theorem)
Suppose that X is a Banach space and Ω is a bounded, convex, and closed subset of X. Suppose further that there exist two operators such that
-
(i)
for all ;
-
(ii)
U is a contraction mapping;
-
(iii)
S is completely continuous.
Then has a fixed point in Ω.
If is an eventually negative solution of (1), then will satisfy
We may note that is nondecreasing in u and for and . Therefore, we will restrict our attention to eventually positive solutions of (1) in the following.
In the sequel, we use the notation
and have the following lemma.
Lemma 2.3 ([[8], Lemma 2.3])
Suppose that is an eventually positive solution of (1) and for and . Then we have:
-
(i)
If a is finite, then
-
(ii)
If a is infinite, then is unbounded, or
Let denote the set of all eventually positive solutions of (1) and
Now, we give the first theorem for a classification scheme of eventually positive solutions to (1).
Theorem 2.4 If is an eventually positive solution of (1), then belongs to , , , for some positive b, or .
Proof Suppose that is an eventually positive solution of (1). From (C2) and (C3), there exist and such that , , , and for . By (1) and (C4), it follows that, for ,
Hence, is strictly decreasing on . We claim that
Assume not; then there exists such that for . So there exist a constant and such that for , which means that
Integrating (6) from to , we obtain
Letting , by (C1) we have . Then there exists such that for , which implies that
Integrating (7) from to , we obtain
Letting , by (C1) we have . From (4), it follows that , then there exists such that or
By (C3), we can choose some positive integer such that for all . Then for any , we have
The inequality above implies that . It follows from (4) that and then contradicts . So (5) holds, and
where .
From (5), we have for , which means that is strictly increasing on . Hence, is either eventually positive or eventually negative. When is eventually negative, we have . Assume that there exists a constant such that
which means that
Integrating (9) from to , we obtain
Letting , by (C1) we have . Similarly, it will cause the contradiction as before. Hence, . When is eventually positive, we have for some positive b or . Therefore,
where .
When is eventually negative, which means that is eventually negative, then there exists such that for . It follows that is strictly decreasing on . Hence, is either eventually positive or eventually negative. If is eventually negative, we have or . Similarly, it will cause the contradiction as before. Therefore, is eventually positive, which means that for some positive b or .
When is eventually positive, it implies that is eventually positive. If is eventually negative, we have . Assume that . It will cause a similar contradiction to the one before. So . If is eventually positive, we have for some positive b or .
Therefore,
where .
It follows from L’Hôpital’s rule (see [[5], Theorem 1.120]) and (8), (10) that
and
When or for some positive b, we have . When , it implies that is eventually positive, which means that is eventually positive. It follows that for some positive b or . We have if for some positive b, and or for some positive b if . Then by Lemma 2.3, we see that must belong to , , , for some positive b, or . The proof is complete. □
3 Main results
In this section, by employing Kranoselskii’s fixed point theorem, we establish the existence criteria for each type of eventually positive solutions to (1).
Theorem 3.1 Equation (1) has an eventually positive solution in for some positive b if and only if there exists some constant such that
Proof Suppose that is an eventually positive solution of (1) in , i.e.,
Assume that (or ). By Lemma 2.3 we have (or ), which contradicts (12). Then we have
and there exists such that , , for . Integrating (1) from to , we obtain
Letting , we have
In view of (C4), it follows that
and
which means that (11) holds. The necessary condition is proved.
Conversely, suppose that there exists some constant such that (11) holds. There will be two cases to be considered: and .
Case 1: . Take such that , then .
When , since and (11) hold, we can choose a sufficiently large such that for , and
When , we can choose and the above such that
Furthermore, from (C3) there exists such that and for .
Define the Banach space as in (3) with , and let
It is easy to prove that is a bounded, convex, and closed subset of . By (C4), we have, for any ,
Now we define two operators and : as follows
Next, we will prove that and satisfy the conditions in Lemma 2.2.
(i) We prove that for any . Note that, for any , and . For any and , by (13), (14), and (16) we obtain
On the other hand, for and , we have
For , , and , we have and (15), and
Similarly, we can prove that for any and . Then we prove that for any and . In fact, for and , we have
For , , and , we have and (15), and
Therefore, we obtain for any .
(ii) We prove that is a contraction mapping. In fact, noting that and for , for we have
for , and
for . It follows that
for any . Therefore, is a contraction mapping.
(iii) We prove that is a completely continuous mapping.
Firstly, for , we have
and
That is, maps into .
Secondly, we prove the continuity of . For and , letting and as , we have
and
as . For , we have
For , we have . Then we obtain
Similar to Chen [7], by (18) and employing Lebesgue’s dominated convergence theorem [[5], Chapter 5], we conclude that
as . That is, is continuous.
Thirdly, we prove that is relatively compact. According to Lemma 2.1, it suffices to show that is bounded, uniformly Cauchy and equi-continuous. It is obvious that is bounded. Since and as , for any given there exists a sufficiently large such that and . Then, for any and , we have
Hence, is uniformly Cauchy.
Then, for , if with , we have
If with , we have
If , we always have
Therefore, there exists such that
whenever and . That is, is equi-continuous.
It follows from Lemma 2.1 that is relatively compact, and then is completely continuous.
By Lemma 2.2, there exists such that , which implies that is a solution of (1). In particular, for we have
Since
for and
we have
It is obvious that
The sufficiency holds when .
Case 2: . Take so that , then . Since and (11) hold, we can choose a sufficiently large such that
From (C3) there exists such that and for . Similarly, we introduce the Banach space and its subset as in (16). Define the operator as in (17) and the operator on as follows:
Next, we prove that for any . In fact, for any and , by (14) and (19) we obtain
and
That is, for any .
The remainder of the proof is similar to the case and we omit it here. By Lemma 2.2, there exists such that , which implies that is a solution of (1). In particular, for we have
Letting , we have
It is obvious that
The sufficiency holds when .
The proof is complete. □
Theorem 3.2 Equation (1) has an eventually positive solution in for some positive b if and only if there exists some constant such that
Proof Suppose that is an eventually positive solution of (1) in , i.e.,
Similarly, we have
and there exists such that , , for . Integrating (1) from to , we obtain
Letting , we have
or
Integrating (21) from to , we have
Letting , we have
In view of (C4), it follows that
and
which means that (20) holds. The necessary condition is proved.
Conversely, suppose that there exists some constant such that (20) holds. There will be two cases to be considered: and .
Case 1: . Take such that , then .
When , since and (20) hold, we can choose a sufficiently large such that for , and
When , we can choose and the above such that
Furthermore, from (C3) there exists such that and for .
Define the Banach space as in (3) with , and let
It is easy to prove that is a bounded, convex, and closed subset of . By (C4), we have, for any ,
Now we define two operators and as follows:
Next, we can prove that and satisfy the conditions in Lemma 2.2. The proof is similar to the case of Theorem 3.1 and omitted here.
By Lemma 2.2, there exists such that , which implies that is a solution of (1). In particular, for we have
Since
for and
we have
which implies that
The sufficiency holds when .
Case 2: . We introduce the Banach space and its subset as in (22). Define the operator as in (23) and the operator on as follows:
The following proof is similar to the case in Theorem 3.1 and we omit it here. By Lemma 2.2, there exists such that , which implies that is a solution of (1). In particular, for we have
Similarly, we have
which implies that
The sufficiency holds when .
The proof is complete. □
Theorem 3.3 Equation (1) has an eventually positive solution in for some positive b if and only if there exists some constant such that
Proof Suppose that is an eventually positive solution of (1) in , i.e.,
Then
and there exists such that , , for . Integrating (1) from to , we obtain
Letting , we have
or
Integrating (25) from to , we have
Letting , we have
or
Integrating (26) from to , we have
Letting , we have
In view of (C4), it follows that
and
which means that (24) holds. The necessary condition is proved.
Conversely, suppose that there exists some constant such that (24) holds. There will be two cases to be considered: and .
Case 1: . Take such that , then .
When , since and (24) hold, we can choose a sufficiently large such that for , and
When , we can choose and the above such that
Furthermore, from (C3) there exists such that and for .
Define the Banach space as in (3) with , and let
It is easy to prove that is a bounded, convex, and closed subset of . By (C4), we have, for any ,
Now we define two operators and as follows:
Next, we can prove that and satisfy the conditions in Lemma 2.2. The proof is similar to the case of Theorem 3.1 and omitted here.
By Lemma 2.2, there exists such that , which implies that is a solution of (1). In particular, for we have
Letting , we have
which implies that
The sufficiency holds when .
Case 2: . We introduce the Banach space and its subset as in (27). Define the operator as in (28) and the operator on as follows:
The following proof is similar to the case in Theorem 3.1 and we omit it here. By Lemma 2.2, there exists such that , which implies that is a solution of (1). In particular, for we have
Similarly, we have
which implies that
The sufficiency holds when .
The proof is complete. □
Theorem 3.4 Equation (1) has an eventually positive solution in , then
Conversely, if there exists a nonnegative constant M such that and
then (1) has an eventually positive solution in .
Proof Suppose that is an eventually positive solution of (1) in , i.e.,
Similarly, we have
and there exists such that for . From (C3) there exists such that and for . Integrating (1) from to , we obtain
Letting , we have
which implies that
by the monotonicity of f and for . Substituting s for in (31), we have
Integrating (32) from to , we have
Letting , we have
By the monotonicity of f and for , it follows that
and
which means that (29) holds. The necessary condition is proved.
Conversely, if there exists a positive constant M such that and (30) hold, then and we can choose a sufficiently large such that
From (C3) there exists such that and for .
Define the Banach space as in (3) with , and let
It is easy to prove that is a bounded, convex, and closed subset of . According to (C3) and (C4), we have, for any ,
Now we define two operators and as follows: