Abstract

The continuability, boundedness, monotonicity, and asymptotic properties of nonoscillatory solutions for a class of second-order nonlinear differential equations are discussed without monotonicity assumption for function g. It is proved that all solutions can be extended to infinity, are eventually monotonic, and can be classified into disjoint classes that are fully characterized in terms of several integral conditions. Moreover, necessary and sufficient conditions for the existence of solutions in each class and for the boundedness of all solutions are established.

1. Introduction

This paper studies the continuability, boundedness, monotonicity, and asymptotic properties of nonoscillatory solutions for a class of second-order nonlinear differential equations Some special cases of (1.1) such as half-linear equation where ,โ€‰โ€‰, is the so-called -Laplacian operator, Emden-Fowler equation and differential equation have been extensively discussed in the literature; see, for example, [1โ€“15] and references cited therein. Equation (1.1) with general nonlinear function is investigated in [16โ€“18]. It is worth to point out that is assumed to be monotonic in most cited papers, but [2, 6] explain that this assumption does not hold in some applications. The aim of this paper is to investigate the continuability, boundedness, monotonicity, and asymptotic properties of nonoscillatory solutions of (1.1) without monotonic assumption for . Some techniques and ideas have been used by the authors in [17].

By solution of (1.1), we mean a differentiable function such that is differentiable and satisfies (1.1) on the maximum existence interval , . A solution of (1.1) is said to be eventually monotonic if there exists a such that is monotonic on . In this paper, we consider only solutions that are not eventually identically equal to zero.

Throughout the paper, we always assume that

(H) ;

โ€ƒ;

โ€ƒ;

โ€ƒ.

(H1) There exists a constant such that

Remark 1.1. (H1) holds for -Laplacian operator; indeed, However, there are nonlinear functions that satisfy (H1) but not (1.6); see [17].

The paper is organized as follows: Section 1 briefly addresses the background and the motivation of the paper. Continuability, classification, and boundedness of solutions are discussed in Section 2. Sections 3 and 4 deal with the existence of class A and class B solutions, respectively. Finally, several remarks are provided in Section 5 to compare our results with existing ones.

2. Continuability, Classification, and Boundedness of Solutions

In this section we discuss continuability, classification, and boundedness of solutions of (1.1). First of all, we cite a result from [17] that will be used later on.

Lemma 2.1. If is a solution of (1.1) with maximal existence interval , , then is eventually monotonic. Moreover, if is bounded on all finite subinterval of , then .

Remark 2.2. From Lemma 2.1 all solutions of (1.1) except eventually trivial solutions can be classified into two classes

Next theorem establishes the continuability for all solutions of (1.1), in other words, all solutions can be extended to .

Theorem 2.3. Assume the following assmputions hold.(H2) There exists a real number and a continuous function such that is increasing and for , and for ; (H3) There exists a real number such that where . Then all solutions of (1.1) can be extended to .

Proof. The proof is similar to that of Theorem 2.3 [17]. We point out that as in the proof of Theorem 2.3 [17], for a class A solution , we have

Remark 2.4. The function is not monotonic. Clearly, , so is bounded by an increasing function . Therefore, the existing results which require the monotonic condition for would not apply, but Theorem 2.3 does.

From Remark 2.2 and Theorem 2.3, all solutions of (1.1) can be classified further into four disjoint classes

We will show that the existence of solutions in each class and the boundedness of all solutions are fully characterized by means of convergence or divergence of the following integrals:

Theorem 2.5. Let (H2) and (H3) hold. Then all positive (negative) solutions of (1.1) are bounded if and only if .

Proof. We consider positive solutions only since the case of negative solutions can be handled similarly.
Necessity. Let be a positive bounded class A solution. Then and for and . By the Extreme Value Theorem, we have . Hence Since is continuous and bounded and is continuous, then is bounded. Let for . Then By (H1), we have Integrating from to and letting , we have
Sufficiency. We will prove by contradiction. Let be a unbounded class A solution. Then and on , and there exists a real number such that for . Similar to the proof of Theorem 2.3, we have Letting and noting that , we have a contradiction to (H3). Therefore, is bounded.

Corollary 2.6. Let (H2) and (H3) hold. If (1.1) has a positive (negative) bounded class A solution, then all positive (negative) solutions are bounded. On the other hand, if (1.1) has an unbounded positive (negative) class A solution, then all positive (negative) solutions are unbounded.

3. Class A Solutions

In this section, we consider the existence of class and class solutions of (1.1). The necessary and sufficient conditions for the existence of class solutions and the sufficient conditions for the existence of class solutions are provided.

Theorem 3.1. Equation (1.1) has both positive and negative class A solutions.

Proof. Similar to the proof of Theorem 3.1 in [17].

Theorem 3.2. Equation (1.1) has a positive (negative) solution if and only if .

Proof. Necessity. Without loss of generality, we assume that is a positive solution. In this case, there exists a such that and for . Note that , we have Then and hence Taking on both sides and applying (H1) imply that Therefore
Sufficiency. Define Since , we may select a such that Let be the Banach space of all bounded and continuous functions defined on endowed with the supremum norm, and let . Clearly, is a bounded convex subset of . Define a mapping by In order to apply Schauder's fixed-point theorem to show that has a fixed point in , we need to prove that maps into and is continuous, and is precompact in .
Let . Considering (3.7), we have Hence, maps into .
Now, we show that if , and as , then . Indeed, for any fixed , since as , we have Note that and that By Lebesgueโ€™s dominated convergence theorem and considering (3.11) and (3.12) we have as . Therefore, is continuous in .
Finally, we show the precompactness of in , which means that for any sequence , has a convergent subsequence in . This can be proved by showing that has a convergent subsequence in for any compact subinterval of as well as the diagonal rule. In fact, is uniformly bounded on . Since By the Mean Value Theorem, we have Then is uniformly bounded and equicontinuous in . So has a convergent subsequence in by Arzelร -Ascoli Theorem.
Now all conditions of Schauder's fixed-point theorem are satisfied, so has a fixed point in , that is, It is easy to verify that . Hence, is a positive solution of (1.1). The proof is complete.

Theorem 3.3. Let (H2) and (H3) hold. Then(a) if and only if and .(b)Equation (1.1) has a positive (negative) solution if .

Proof. By Theorem 2.5 all solutions of (1.1) are bounded if and only if and , so part (a) follows.
If , there is no positive solution of (1.1) from Theorem 3.2. Therefore, Theorem 3.1 guarantees the existence of a positive solution of (1.1). Similarly, there exists a negative solution of (1.1) if .

4. Class B Solutions

In this section the existence of class , , and solutions are discussed. We assume that (1.1) has a unique solution for any initial conditions and .

Theorem 4.1. Assume the following assumptions hold.(H2a) There exists a continuous function such that G is increasing, for and ;(H4) There exists such that Then (1.1) has (a)both positive and negative solutions in class B; (b)no solution which is eventually identically equal to zero.

Proof. (a) We prove that class has a positive solution, the case of having a negative solution is similar. Assume . The solution of (1.1) with initial conditions and , denoted by , has the form Define two sets and as Then . Clearly, . We claim that is open. Indeed, if , there exists such that . For any , we have Since (1.1) has a unique solution for any initial conditions , , this solution is continuously dependent on initial data. If , we have uniformly for on . Hence, for all that are close to , this proves the openness of .
Next we show that . Define Let If there exists such that , then and . Otherwise, on . In this case, we claim on . If this is not true, since , there exists such that and for . Taking into account (4.6) we have This is a contradiction and hence on . Notice that we know . Clearly, is open, then . Take , is a nonincreasing nonnegative solution on . We will show that on . If not, there exists such that and for and . Note that for we have Dividing both sides by and integrating from to , we have That is a contradiction to (H4). Therefore, for and .
The proof of part (b) follows from the end part of the proof of part (a).

Theorem 4.2. Equation (1.1) has a positive (negative) solution if and only if .

Proof. Necessity. We assume that is a positive solution. The case of negative solution is similar. In this case, we have and for . Let and note that , . Then Integrating both sides of (1.1) from to implies that Hence, Again, integrating both sides of the above inequality we have
Sufficiency. Let Since we choose such that Let and as defined in Theorem 3.2. Define by For any , we have This proves that maps into . Similar to the proof of Theorem 3.2, we are able to show that is continuous in , and is precompact in . Then has a fixed-point in by Schauder's fixed-point theorem, that is, It is easy to verify that is a positive solution of (1.1). The proof is complete.

Theorem 4.3. Let (H2a) and (H4) hold and let . Then (1.1) has a positive (negative) solution if and only if ().

Proof. We prove the assertion for positive solutions without loss of generality.
Necessity. Assume is a positive solution. Then and for , , and . We claim . In fact, if , since is negative and increasing on , then and where .
Integrating both sides from to and noting that , we have a contradiction to and hence .
Integrating both sides of (1.1) from to we have Then Hence, Integrating both sides of the above inequality from to implies that Therefore, from (H4).
Sufficiency. By Theorem 4.1 (1.1) has a positive class B solution , either or . Note that implies that from Theorem 4.2. So . The proof is complete.

5. Remarks

In this section, we present several remarks about comparison of our results with the existing ones in the literature.

Theorems 2.3 and 2.5 improve [10, Theorem ] since (H3) reduces to (iii) of [10] if and the differentiability of and is not required. Theorems 2.3, 2.5, and 4.2 complement and generalize [2, Theorem 8]. Moreover, under (H2), Theorems 2.3, 2.5, and 4.2 improve [2, Theorem 8] since (H3) improves (22) of [2]; see the discussion in [16]. Theorem 2.5 generalizes [13, Theorem 3.9]. Theorems 2.3, 3.1, and 4.2 generalize [16, Theorem 1]. Theorem 3.2 generalizes [2, Theorem 3], [16, Theorem 3], and [18, Theorem 2.1]. Theorem 3.3 generalizes [18, Theorem 2.2]. Theorem 4.1 generalizes [13, Theorem 2.1] and improves [3, Theorem 6] under (H2a) since (hp) in [3] is replaced by a weaker condition (H4). Theorem 4.2 generalizes [2, Theorem 1], [16, Theorem 5], and [18, Theorem 3.1]. Theorem 4.3 generalize [16, Theorem 6] and [18, Theorem 3.2].