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Abstract and Applied Analysis
Volume 2011 (2011), Article ID 367541, 17 pages
http://dx.doi.org/10.1155/2011/367541
Research Article

Oscillation Criteria for Second-Order Superlinear Neutral Differential Equations

1School of Science, University of Jinan, Jinan, Shandong 250022, China
2School of Control Science and Engineering, Shandong University, Jinan, Shandong 250061, China

Received 5 September 2010; Accepted 20 January 2011

Academic Editor: Josef Diblík

Copyright © 2011 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.

Abstract

Some oscillation criteria are established for the second-order superlinear neutral differential equations (𝑟(𝑡)|𝑧(𝑡)|𝛼1𝑧(𝑡))+𝑓(𝑡,𝑥(𝜎(𝑡)))=0, 𝑡𝑡0, where 𝑧(𝑡)=𝑥(𝑡)+𝑝(𝑡)𝑥(𝜏(𝑡)), 𝜏(𝑡)𝑡, 𝜎(𝑡)𝑡, 𝑝𝐶([𝑡0,),[0,𝑝0]), and 𝛼1. Our results are based on the cases 𝑡01/𝑟1/𝛼(𝑡)d𝑡= or 𝑡01/𝑟1/𝛼(𝑡)d𝑡<. Two examples are also provided to illustrate these results.

1. Introduction

This paper is concerned with the oscillatory behavior of the second-order superlinear differential equation||𝑧𝑟(𝑡)||(𝑡)𝛼1𝑧(𝑡)+𝑓(𝑡,𝑥(𝜎(𝑡)))=0,𝑡𝑡0,(1.1) where 𝛼1 is a constant, 𝑧(𝑡)=𝑥(𝑡)+𝑝(𝑡)𝑥(𝜏(𝑡)).

Throughout this paper, we will assume the following hypotheses:(𝐴1)𝑟𝐶1([𝑡0,),),𝑟(𝑡)>0 for 𝑡𝑡0;(𝐴2)𝑝𝐶([𝑡0,),[0,𝑝0]), where 𝑝0 is a constant;(𝐴3)𝜏𝐶1([𝑡0,),), 𝜏(𝑡)𝜏0>0, 𝜏(𝑡)𝑡; (𝐴4)𝜎𝐶([𝑡0,),), 𝜎(𝑡)𝑡,𝜏𝜎=𝜎𝜏; (𝐴5)𝑓(𝑡,𝑢)𝐶([𝑡0,)×,), and there exists a function 𝑞𝐶([𝑡0,),[0,)) such that 𝑓(𝑡,𝑢)sign𝑢𝑞(𝑡)|𝑢|𝛼,for𝑢0,𝑡𝑡0.(1.2)

By a solution of (1.1), we mean a function 𝑥𝐶([𝑇𝑥,),) for some 𝑇𝑥𝑡0 which has the property that 𝑟(𝑡)|𝑧(𝑡)|𝛼1𝑧(𝑡)𝐶1([𝑇𝑥,),) and satisfies (1.1) on [𝑇𝑥,). We consider only those solutions 𝑥 which satisfy sup{|𝑥(𝑡)|𝑡𝑇}>0 for all 𝑇𝑇𝑥. As is customary, a solution of (1.1) is called oscillatory if it has arbitrarily large zeros on [𝑡0,), otherwise, it is called nonoscillatory. Equation (1.1) is said to be oscillatory if all its solutions are oscillatory.

We note that neutral differential equations find numerous applications in electric networks. For instance, they are frequently used for the study of distributed networks containing lossless transmission lines which rise in high-speed computers where the lossless transmission lines are used to interconnect switching circuits; see [1].

In the last few years, there are many studies that have been made on the oscillation and asymptotic behavior of solutions of discrete and continuous equations; see, for example, [228]. Agarwal et al. [5], Chern et al. [6], Džurina and Stavroulakis [7], Kusano and Yoshida [8], Kusano and Naito [9], Mirzov [10], and Sun and Meng [11] observed some similar properties between||𝑥𝑟(𝑡)||(𝑡)𝛼1𝑥(𝑡)||||+𝑞(𝑡)𝑥(𝜎(𝑡))𝛼1𝑥(𝜎(𝑡))=0(1.3) and the corresponding linear equations𝑟(𝑡)𝑥(𝑡)+𝑞(𝑡)𝑥(𝑡)=0.(1.4) Baculíková [12] established some new oscillation results for (1.3) when 𝛼=1. In 1989, Philos [13] obtained some Philos-type oscillation criteria for (1.4).

Recently, many results have been obtained on oscillation and nonoscillation of neutral differential equations, and we refer the reader to [1435] and the references cited therein. Liu and Bai [16], Xu and Meng [17, 18], Dong [19], Baculíková and Lacková [20], and Jiang and Li [21] established some oscillation criteria for (1.3) with neutral term under the assumptions 𝜏(𝑡)𝑡, 𝜎(𝑡)𝑡,𝑅(𝑡)=𝑡𝑡01𝑟1/𝛼(𝑠)d𝑠as𝑡,(1.5)𝑡01𝑟1/𝛼(𝑡)d𝑡<.(1.6)

Saker and O'Regan [24] studied the oscillatory behavior of (1.1) when 0𝑝(𝑡)<1, 𝜏(𝑡)𝑡 and 𝜎(𝑡)>𝑡.

Han et al. [26] examined the oscillation of second-order nonlinear neutral differential equation[]𝑟(𝑡)𝑥(𝑡)+𝑝(𝑡)𝑥(𝜏(𝑡))+𝑞(𝑡)𝑓(𝑥(𝜎(𝑡)))=0,𝑡𝑡0,(1.7) where 𝜏(𝑡)𝑡, 𝜎(𝑡)𝑡, 𝜏(𝑡)=𝜏0>0, 0𝑝(𝑡)𝑝0<, and the authors obtained some oscillation criteria for (1.7).

However, there are few results regarding the oscillatory problem of (1.1) when 𝜏(𝑡)𝑡 and 𝜎(𝑡)𝑡. Our aim in this paper is to establish some oscillation criteria for (1.1) under the case when 𝜏(𝑡)𝑡 and 𝜎(𝑡)𝑡.

The paper is organized as follows. In Section 2, we will establish an inequality to prove our results. In Section 3, some oscillation criteria are obtained for (1.1). In Section 4, we give two examples to show the importance of the main results.

All functional inequalities considered in this paper are assumed to hold eventually, that is, they are satisfied for all 𝑡 large enough.

2. Lemma

In this section, we give the following lemma, which we will use in the proofs of our main results.

Lemma 2.1. Assume that 𝛼1,  𝑎,𝑏. If 𝑎0,  𝑏0, then 𝑎𝛼+𝑏𝛼12𝛼1(𝑎+𝑏)𝛼(2.1) holds.

Proof. (i) Suppose that 𝑎=0 or 𝑏=0. Obviously, we have (2.1). (ii) Suppose that 𝑎>0, 𝑏>0. Define the function 𝑔 by 𝑔(𝑢)=𝑢𝛼,  𝑢(0,). Then 𝑔(𝑢)=𝛼(𝛼1)𝑢𝛼20 for 𝑢>0. Thus, 𝑔 is a convex function. By the definition of convex function, for 𝜆=1/2,𝑎,𝑏(0,), we have 𝑔𝑎+𝑏2𝑔(𝑎)+𝑔(𝑏)2,(2.2) that is, 𝑎𝛼+𝑏𝛼12𝛼1(𝑎+𝑏)𝛼.(2.3) This completes the proof.

3. Main Results

In this section, we will establish some oscillation criteria for (1.1).

First, we establish two comparison theorems which enable us to reduce the problem of the oscillation of (1.1) to the research of the first-order differential inequalities.

Theorem 3.1. Suppose that (1.5) holds. If the first-order neutral differential inequality 𝑝𝑢(𝑡)+0𝛼𝜏0𝑢(𝜏(𝑡))+12𝛼1𝑡𝑄(𝑡)𝑅(𝜎(𝑡))𝑅1𝛼𝑢(𝜎(𝑡))0(3.1) has no positive solution for all sufficiently large 𝑡1, where 𝑄(𝑡)=min{𝑞(𝑡),𝑞(𝜏(𝑡))}, then every solution of (1.1) oscillates.

Proof. Let 𝑥 be a nonoscillatory solution of (1.1). Without loss of generality, we assume that there exists 𝑡1𝑡0 such that 𝑥(𝑡)>0,  𝑥(𝜏(𝑡))>0, and 𝑥(𝜎(𝑡))>0 for all 𝑡𝑡1. Then 𝑧(𝑡)>0 for 𝑡𝑡1. In view of (1.1), we obtain ||𝑧𝑟(𝑡)||(𝑡)𝛼1𝑧(𝑡)𝑞(𝑡)𝑥𝛼(𝜎(𝑡))0,𝑡𝑡1.(3.2) Thus, 𝑟(𝑡)|𝑧(𝑡)|𝛼1𝑧(𝑡) is decreasing function. Now we have two possible cases for 𝑧(𝑡): (i) 𝑧(𝑡)<0 eventually, (ii) 𝑧(𝑡)>0 eventually.
Suppose that 𝑧(𝑡)<0 for 𝑡𝑡2𝑡1. Then, from (3.2), we get ||𝑧𝑟(𝑡)||(𝑡)𝛼1𝑧𝑡(𝑡)𝑟2||𝑧𝑡2||𝛼1𝑧𝑡2,𝑡𝑡2,(3.3) which implies that 𝑡𝑧(𝑡)𝑧2𝑟1/𝛼𝑡2||𝑧𝑡2||𝑡𝑡2𝑟1/𝛼(𝑠)d𝑠.(3.4) Letting 𝑡, by (1.5), we find 𝑧(𝑡), which is a contradiction. Thus 𝑧(𝑡)>0(3.5) for 𝑡𝑡2.
By applying (1.1), for all sufficiently large 𝑡, we obtain 𝑧𝑟(𝑡)(𝑡)𝛼+𝑞(𝑡)𝑥𝛼(𝑝𝜎(𝑡))+0𝛼𝑞(𝜏(𝑡))𝑥𝛼(𝑝𝜎(𝜏(𝑡)))+0𝛼𝜏𝑧(𝑡)𝑟(𝜏(𝑡))(𝜏(𝑡))𝛼0.(3.6) Using inequality (2.1), (3.2), (3.5), 𝜏𝜎=𝜎𝜏, and the definition of 𝑧, we conclude that 𝑧𝑟(𝑡)(𝑡)𝛼+𝑝0𝛼𝜏0𝑧𝑟(𝜏(𝑡))(𝜏(𝑡))𝛼+12𝛼1𝑄(𝑡)𝑧𝛼(𝜎(𝑡))0.(3.7) It follows from (3.2) and (3.5) that 𝑢(𝑡)=𝑟(𝑡)(𝑧(𝑡))𝛼>0 is decreasing and then 𝑧(𝑡)𝑡𝑡2𝑧𝑟(𝑠)(𝑠)𝛼1/𝛼𝑟1/𝛼(𝑠)d𝑠𝑢1/𝛼(𝑡)𝑡𝑡21𝑟1/𝛼(𝑠)d𝑠=𝑢1/𝛼𝑡(𝑡)𝑅(𝑡)𝑅2.(3.8) Thus, from (3.7) and the above inequality, we find 𝑝𝑢(𝑡)+0𝛼𝜏0𝑢(𝜏(𝑡))+12𝛼1𝑡𝑄(𝑡)𝑅(𝜎(𝑡))𝑅2𝛼𝑢(𝜎(𝑡))0.(3.9) That is, inequality (3.1) has a positive solution 𝑢; this is a contradiction. The proof is complete.

Theorem 3.2. Suppose that (1.5) holds. If the first-order differential inequality 𝜂𝜏(𝑡)+02𝛼1𝜏0+𝑝0𝛼𝑡𝑄(𝑡)𝑅(𝜎(𝑡))𝑅1𝛼𝜂(𝜎(𝑡))0(3.10) has no positive solution for all sufficiently large 𝑡1, where 𝑄 is defined as in Theorem 3.1, then every solution of (1.1) oscillates.

Proof. Let 𝑥 be a nonoscillatory solution of (1.1). Without loss of generality, we assume that there exists 𝑡1𝑡0 such that 𝑥(𝑡)>0,  𝑥(𝜏(𝑡))>0, and 𝑥(𝜎(𝑡))>0 for all 𝑡𝑡1. Then 𝑧(𝑡)>0 for 𝑡𝑡1. Proceeding as in the proof of Theorem 3.1, we obtain that 𝑢(𝑡)=𝑟(𝑡)(𝑧(𝑡))𝛼 is decreasing, and it satisfies inequality (3.1). Set 𝜂(𝑡)=𝑢(𝑡)+(𝑝0)𝛼𝑢(𝜏(𝑡))/𝜏0. From 𝜏(𝑡)𝑡, we get 𝑝𝜂(𝑡)=𝑢(𝑡)+0𝛼𝜏0𝑝𝑢(𝜏(𝑡))1+0𝛼𝜏0𝑢(𝑡).(3.11) In view of the above inequality and (3.1), we see that 𝜂𝜏(𝑡)+02𝛼1𝜏0+𝑝0𝛼𝑡𝑄(𝑡)𝑅(𝜎(𝑡))𝑅1𝛼𝜂(𝜎(𝑡))0.(3.12) That is, inequality (3.10) has a positive solution 𝜂; this is a contradiction. The proof is complete.

Next, using Riccati transformation technique, we obtain the following results.

Theorem 3.3. Suppose that (1.5) holds. Moreover, assume that there exists 𝜌𝐶1([𝑡0,),(0,)) such that limsup𝑡𝑡𝑡0𝜌(𝑠)𝑄(𝑠)2𝛼11(𝛼+1)𝛼+1𝑝1+0𝛼𝜏0𝜌𝑟(𝑠)+(𝑠)𝛼+1(𝜌(𝑠))𝛼d𝑠=(3.13) holds, where 𝑄 is defined as in Theorem 3.1, 𝜌+(𝑡)=max{0,𝜌(𝑡)}. Then every solution of (1.1) oscillates.

Proof. Let 𝑥 be a nonoscillatory solution of (1.1). Without loss of generality, we assume that there exists 𝑡1𝑡0 such that 𝑥(𝑡)>0, 𝑥(𝜏(𝑡))>0, and 𝑥(𝜎(𝑡))>0 for all 𝑡𝑡1. Then 𝑧(𝑡)>0 for 𝑡𝑡1. Proceeding as in the proof of Theorem 3.1, we obtain (3.2)–(3.7).
Define a Riccati substitution 𝑧𝜔(𝑡)=𝜌(𝑡)𝑟(𝑡)(𝑡)𝛼(𝑧(𝑡))𝛼,𝑡𝑡2.(3.14) Thus 𝜔(𝑡)>0 for 𝑡𝑡2. Differentiating (3.14) we find that 𝜔(𝑡)=𝜌𝑟𝑧(𝑡)(𝑡)(𝑡)𝛼(𝑧(𝑡))𝛼𝑟𝑧+𝜌(𝑡)(𝑡)(𝑡)𝛼(𝑧(𝑡))𝛼𝑟𝑧𝛼𝜌(𝑡)(𝑡)(𝑡)𝛼𝑧𝛼1(𝑡)𝑧(𝑡)(𝑧(𝑡))2𝛼.(3.15) Hence, by (3.14) and (3.15), we see that 𝜔𝜌(𝑡)=(𝑡)𝜌𝑟𝑧(𝑡)𝜔(𝑡)+𝜌(𝑡)(𝑡)(𝑡)𝛼(𝑧(𝑡))𝛼𝛼𝜌1/𝛼(𝑡)𝑟1/𝛼(𝜔𝑡)(𝛼+1)/𝛼(𝑡).(3.16)
Similarly, we introduce another Riccati substitution 𝑧𝜐(𝑡)=𝜌(𝑡)𝑟(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼,𝑡𝑡2.(3.17) Then 𝜐(𝑡)>0 for 𝑡𝑡2. From (3.2), (3.5), and 𝜏(𝑡)𝑡, we have 𝑧(𝑡)𝑟(𝜏(𝑡))𝑟(𝑡)1/𝛼𝑧(𝜏(𝑡)).(3.18) Differentiating (3.17), we find 𝜐(𝑡)=𝜌𝑟𝑧(𝑡)(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼𝑟𝑧+𝜌(𝑡)(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼𝑟𝑧𝛼𝜌(𝑡)(𝜏(𝑡))(𝜏(𝑡))𝛼𝑧𝛼1(𝑡)𝑧(𝑡)(𝑧(𝑡))2𝛼.(3.19) Therefore, by (3.17), (3.18), and (3.19), we see that 𝜐(𝑡)𝜌(𝑡)𝜌𝑟𝑧(𝑡)𝜐(𝑡)+𝜌(𝑡)(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼𝛼𝜌1/𝛼(𝑡)𝑟1/𝛼(𝜐𝑡)(𝛼+1)/𝛼(𝑡).(3.20) Thus, from (3.16) and (3.20), we have 𝜔𝑝(𝑡)+0𝛼𝜏0𝜐𝑟𝑧(𝑡)𝜌(𝑡)(𝑡)(𝑡)𝛼+𝑝0𝛼/𝜏0𝑟𝑧(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼+𝜌(𝑡)𝛼𝜌(𝑡)𝜔(𝑡)𝜌1/𝛼(𝑡)𝑟1/𝛼𝜔(𝑡)(𝛼+1)/𝛼(𝑝𝑡)+0𝛼𝜏0𝜌(𝑡)𝑝𝜌(𝑡)𝜐(𝑡)0𝛼𝜏0𝛼𝜌1/𝛼(𝑡)𝑟1/𝛼𝜐(𝑡)(𝛼+1)/𝛼(𝑡).(3.21) It follows from (3.5), (3.7), and 𝜎(𝑡)𝑡 that 𝜔(𝑝𝑡)+0𝛼𝜏0𝜐(1𝑡)2𝛼1𝜌𝜌(𝑡)𝑄(𝑡)++(𝑡)𝛼𝜌(𝑡)𝜔(𝑡)𝜌1/𝛼(𝑡)𝑟1/𝛼𝜔(𝑡)(𝛼+1)/𝛼(+𝑝𝑡)0𝛼𝜏0𝜌+(𝑡)𝑝𝜌(𝑡)𝜐(𝑡)0𝛼𝜏0𝛼𝜌1/𝛼(𝑡)𝑟1/𝛼𝜐(𝑡)(𝛼+1)/𝛼(𝑡).(3.22) Integrating the above inequality from 𝑡2 to 𝑡, we obtain 𝑡𝜔(𝑡)𝜔2+𝑝0𝛼𝜏0𝑝𝜐(𝑡)0𝛼𝜏0𝜐𝑡2𝑡𝑡212𝛼1𝜌(𝑠)𝑄(𝑠)d𝑠+𝑡𝑡2𝜌+(𝑠)𝛼𝜌(𝑠)𝜔(𝑠)𝜌1/𝛼(𝑠)𝑟1/𝛼𝜔(𝑠)(𝛼+1)/𝛼+(𝑠)d𝑠𝑡𝑡2𝑝0𝛼𝜏0𝜌+(𝑠)𝜌𝛼(𝑠)𝜐(𝑠)𝜌1/𝛼(𝑠)𝑟1/𝛼(𝜐𝑠)(𝛼+1)/𝛼(𝑠)d𝑠.(3.23) Define 𝛼𝐴=𝜌1/𝛼(𝑡)𝑟1/𝛼(𝑡)𝛼/(𝛼+1)𝜌𝜔(𝑡),𝐵=+(𝑡)𝛼𝜌(𝑡)𝛼𝛼+1𝜌1/𝛼(𝑡)𝑟1/𝛼(𝑡)𝛼/(𝛼+1)𝛼.(3.24) Using inequality 𝛼+1𝛼𝐴𝐵1/𝛼𝐴(𝛼+1)/𝛼1𝛼𝐵(𝛼+1)/𝛼,for𝐴0,𝐵0areconstants,(3.25) we have 𝜌+(𝑡)𝛼𝜌(𝑡)𝜔(𝑡)𝜌1/𝛼(𝑡)𝑟1/𝛼𝜔(𝑡)(𝛼+1)/𝛼1(𝑡)(𝛼+1)𝛼+1𝜌𝑟(𝑡)+(𝑡)𝛼+1𝜌(𝑡)𝛼.(3.26) Similarly, we obtain 𝜌+(𝑡)𝛼𝜌(𝑡)𝜐(𝑡)𝜌1/𝛼(𝑡)𝑟1/𝛼𝜐(𝑡)(𝛼+1)/𝛼1(𝑡)(𝛼+1)𝛼+1𝜌𝑟(𝑡)+(𝑡)𝛼+1𝜌(𝑡)𝛼.(3.27) Thus, from (3.23), we get 𝑡𝜔(𝑡)𝜔2+𝑝0𝛼𝜏0𝑝𝜐(𝑡)0𝛼𝜏0𝜐𝑡2𝑡𝑡2𝜌(𝑠)𝑄(𝑠)2𝛼11(𝛼+1)𝛼+1𝑝1+0𝛼𝜏0𝜌𝑟(𝑠)+(𝑠)𝛼+1𝜌(𝑠)𝛼d𝑠,(3.28) which contradicts (3.13). This completes the proof.

As an immediate consequence of Theorem 3.3 we get the following.

Corollary 3.4. Let assumption (3.13) in Theorem 3.3 be replaced by limsup𝑡𝑡𝑡0𝜌(𝑠)𝑄(𝑠)d𝑠=,limsup𝑡𝑡𝑡0𝜌𝑟(𝑠)+(𝑠)𝛼+1(𝜌(𝑠))𝛼d𝑠<.(3.29) Then every solution of (1.1) oscillates.

From Theorem 3.3 by choosing the function 𝜌, appropriately, we can obtain different sufficient conditions for oscillation of (1.1), and if we define a function 𝜌 by 𝜌(𝑡)=1, and 𝜌(𝑡)=𝑡, we have the following oscillation results.

Corollary 3.5. Suppose that (1.5) holds. If limsup𝑡𝑡𝑡0𝑄(𝑠)d𝑠=,(3.30) where 𝑄 is defined as in Theorem 3.1, then every solution of (1.1) oscillates.

Corollary 3.6. Suppose that (1.5) holds. If limsup𝑡𝑡𝑡0𝑠𝑄(𝑠)2𝛼11(𝛼+1)𝛼+1𝑝1+0𝛼𝜏0𝑟(𝑠)𝑠𝛼d𝑠=,(3.31) where 𝑄 is defined as in Theorem 3.1, then every solution of (1.1) oscillates.

In the following theorem, we present a Philos-type oscillation criterion for (1.1).

First, we introduce a class of functions . Let 𝔻0=(𝑡,𝑠)𝑡>𝑠𝑡0,𝔻=(𝑡,𝑠)𝑡𝑠𝑡0.(3.32) The function 𝐻𝐶(𝔻,) is said to belong to the class (defined by 𝐻, for short) if(i)𝐻(𝑡,𝑡)=0, for 𝑡𝑡0,  𝐻(𝑡,𝑠)>0, for (𝑡,𝑠)𝔻0;(ii)𝐻 has a continuous and nonpositive partial derivative 𝜕𝐻(𝑡,𝑠)/𝜕𝑠 on 𝐷0 with respect to 𝑠.

We assume that 𝜍(𝑡) and 𝜌(𝑡) for 𝑡𝑡0 are given continuous functions such that 𝜌(𝑡)>0 and differentiable and define 𝜌𝜃(𝑡)=(𝑡)𝜌(𝑡)+(𝛼+1)(𝜍(𝑡))1/𝛼[],𝜓(𝑡)=𝜌(𝑡)𝑟(𝑡)𝜍(𝑡)𝑟(𝑡)(𝜍(𝑡))(1+𝛼)/𝛼,𝜙(𝑡,𝑠)=𝑟(𝑠)𝜌(𝑠)(𝛼+1)𝛼+1𝜃(𝑠)+𝜕𝐻(𝑡,𝑠)/𝜕𝑠𝐻(𝑡,𝑠)𝛼+1.(3.33)

Now, we give the following result.

Theorem 3.7. Suppose that (1.5) holds and 𝛼 is a quotient of odd positive integers. Moreover, let 𝐻 be such that limsup𝑡1𝐻𝑡,𝑡0𝑡𝑡0𝐻(𝑡,𝑠)𝜌(𝑠)𝑄(𝑠)2𝛼1𝑝1+0𝛼𝜏0(𝜓(𝑠)+𝜙(𝑡,𝑠))d𝑠=(3.34) holds, where 𝑄 is defined as in Theorem 3.1. Then every solution of (1.1) oscillates.

Proof. Let 𝑥 be a nonoscillatory solution of (1.1). Without loss of generality, we assume that there exists 𝑡1𝑡0 such that 𝑥(𝑡)>0,  𝑥(𝜏(𝑡))>0, and 𝑥(𝜎(𝑡))>0 for all 𝑡𝑡1. Then 𝑧(𝑡)>0 for 𝑡𝑡1. Proceeding as in the proof of Theorem 3.1, we obtain (3.2)–(3.7). Define the Riccati substitution 𝜔 by 𝑧𝜔(𝑡)=𝜌(𝑡)𝑟(𝑡)(𝑡)𝛼(𝑧(𝑡))𝛼+𝑟(𝑡)𝜍(𝑡),𝑡𝑡2𝑡1.(3.35) Then, we have 𝜔(𝑡)=𝜌𝑧(𝑡)𝑟(𝑡)(𝑡)𝛼(𝑧(𝑡))𝛼𝑧+𝑟(𝑡)𝜍(𝑡)+𝜌(𝑡)𝑟(𝑡)(𝑡)𝛼(𝑧(𝑡))𝛼+𝑟(𝑡)𝜍(𝑡)=𝜌(𝑡)𝜌[](𝑡)𝜔(𝑡)+𝜌(𝑡)𝑟(𝑡)𝜍(𝑡)𝑟𝑧+𝜌(𝑡)(𝑡)(𝑡)𝛼(𝑧(𝑡))𝛼𝑟𝑧𝛼𝜌(𝑡)(𝑡)(𝑡)𝛼+1(𝑧(𝑡))𝛼+1.(3.36) Using (3.35), we get 𝜔𝜌(𝑡)=(𝑡)𝜌[](𝑡)𝜔(𝑡)+𝜌(𝑡)𝑟(𝑡)𝜍(𝑡)𝑟𝑧+𝜌(𝑡)(𝑡)(𝑡)𝛼(𝑧(𝑡))𝛼𝛼𝜌(𝑡)𝑟1/𝛼(𝑡)𝜔(𝑡)𝜌(𝑡)𝑟(𝑡)𝜍(𝑡)(1+𝛼)/𝛼.(3.37) Let 𝐴=𝜔(𝑡)𝜌(𝑡),𝐵=𝑟(𝑡)𝜍(𝑡).(3.38) By applying the inequality (see [21, 24]) 𝐴(1+𝛼)/𝛼(𝐴𝐵)1+𝛼/𝛼𝐵1/𝛼11+𝛼1𝐴𝛼𝐵,for𝛼=oddodd1,(3.39) we see that 𝜔(𝑡)𝜌(𝑡)𝑟(𝑡)𝜍(𝑡)(1+𝛼)/𝛼𝜔(1+𝛼)/𝛼(𝑡)𝜌(1+𝛼)/𝛼+1(𝑡)𝛼(𝑟(𝑡)𝜍(𝑡))(1+𝛼)/𝛼𝛼+1𝛼(𝑟(𝑡)𝜍(𝑡))1/𝛼𝜌(𝑡)𝜔(𝑡).(3.40) Substituting (3.40) into (3.37), we have 𝜔𝜌(𝑡)(𝑡)𝜌(𝑡)+(𝛼+1)(𝜍(𝑡))1/𝛼[]𝜔(𝑡)+𝜌(𝑡)𝑟(𝑡)𝜍(𝑡)𝑟(𝑡)(𝜍(𝑡))(1+𝛼)/𝛼𝑧+𝜌(𝑡)𝑟(𝑡)(𝑡)𝛼(𝑧(𝑡))𝛼𝛼𝑟1/𝛼(𝑡)𝜌1/𝛼𝜔(𝑡)(1+𝛼)/𝛼(𝑡).(3.41) That is, 𝜔𝑟𝑧(𝑡)𝜃(𝑡)𝜔(𝑡)+𝜓(𝑡)+𝜌(𝑡)(𝑡)(𝑡)𝛼(𝑧(𝑡))𝛼𝛼𝑟1/𝛼(𝑡)𝜌1/𝛼(𝜔𝑡)(1+𝛼)/𝛼(𝑡).(3.42)
Next, define another Riccati transformation 𝑢 by 𝑧𝑢(𝑡)=𝜌(𝑡)𝑟(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼+𝑟(𝑡)𝜍(𝑡),𝑡𝑡2𝑡1.(3.43) Then, we have 𝑢(𝑡)=𝜌𝑧(𝑡)𝑟(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼𝑧+𝑟(𝑡)𝜍(𝑡)+𝜌(𝑡)𝑟(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼+𝑟(𝑡)𝜍(𝑡)=𝜌(𝑡)[]𝜌(𝑡)𝑢(𝑡)+𝜌(𝑡)𝑟(𝑡)𝜍(𝑡)𝑧+𝜌(𝑡)𝑟(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼𝑧𝛼𝜌(𝑡)𝑟(𝜏(𝑡))(𝜏(𝑡))𝛼𝑧(𝑡)(𝑧(𝑡))𝛼+1.(3.44) From (3.2), (3.5), and 𝜏(𝑡)𝑡, we have that (3.18) holds. Hence, we obtain 𝑢𝜌(𝑡)(𝑡)[]𝜌(𝑡)𝑢(𝑡)+𝜌(𝑡)𝑟(𝑡)𝜍(𝑡)𝑧+𝜌(𝑡)𝑟(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼𝑧𝛼𝜌(𝑡)𝑟(𝜏(𝑡))(𝜏(𝑡))𝛼(1+𝛼)/𝛼𝑟1/𝛼(𝑡)(𝑧(𝑡))𝛼+1.(3.45) Using (3.43), we get 𝑢𝜌(𝑡)(𝑡)𝜌[](𝑡)𝑢(𝑡)+𝜌(𝑡)𝑟(𝑡)𝜍(𝑡)𝑟𝑧+𝜌(𝑡)(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼𝛼𝜌(𝑡)𝑟1/𝛼(𝑡)𝑢(𝑡)𝜌(𝑡)𝑟(𝑡)𝜍(𝑡)(1+𝛼)/𝛼.(3.46) Let 𝐴=𝑢(𝑡)𝜌(𝑡),𝐵=𝑟(𝑡)𝜍(𝑡).(3.47) By applying the inequality (3.39), we see that 𝑢(𝑡)𝜌(𝑡)𝑟(𝑡)𝜍(𝑡)(1+𝛼)/𝛼𝑢(1+𝛼)/𝛼(𝑡)𝜌(1+𝛼)/𝛼+1(𝑡)𝛼(𝑟(𝑡)𝜍(𝑡))(1+𝛼)/𝛼𝛼+1𝛼(𝑟(𝑡)𝜍(𝑡))1/𝛼𝜌(𝑡)𝑢(𝑡).(3.48) Substituting (3.48) into (3.46), we have 𝑢𝜌(𝑡)(𝑡)𝜌(𝑡)+(𝛼+1)(𝜍(𝑡))1/𝛼[]𝑢(𝑡)+𝜌(𝑡)𝑟(𝑡)𝜍(𝑡)𝑟(𝑡)(𝜍(𝑡))(1+𝛼)/𝛼𝑧+𝜌(𝑡)𝑟(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼𝛼𝑟1/𝛼(𝑡)𝜌1/𝛼𝑢(𝑡)(1+𝛼)/𝛼(𝑡).(3.49) That is, 𝑢𝑟𝑧(𝑡)𝜃(𝑡)𝑢(𝑡)+𝜓(𝑡)+𝜌(𝑡)(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼𝛼𝑟1/𝛼(𝑡)𝜌1/𝛼(𝑢𝑡)(1+𝛼)/𝛼(𝑡).(3.50) By (3.42) and (3.50), we find 𝜔𝑝(𝑡)+0𝛼𝜏0𝑢𝑝(𝑡)1+0𝛼𝜏0𝑟𝑧𝜓(𝑡)+𝜌(𝑡)(𝑡)(𝑡)𝛼+𝑝0𝛼/𝜏0𝑟𝑧(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝑡))𝛼𝛼+𝜃(𝑡)𝜔(𝑡)𝑟1/𝛼(𝑡)𝜌1/𝛼𝜔(𝑡)(1+𝛼)/𝛼𝑝(𝑡)+0𝛼𝜏0𝑝𝜃(𝑡)𝑢(𝑡)0𝛼𝜏0𝛼𝑟1/𝛼(𝑡)𝜌1/𝛼𝑢(𝑡)(1+𝛼)/𝛼(𝑡).(3.51) In view of the above inequality, (3.5), (3.7), and 𝜎(𝑡)𝑡, we get 𝜔𝑝(𝑡)+0𝛼𝜏0𝑢𝑝(𝑡)1+0𝛼𝜏0𝜓(𝑡)𝜌(𝑡)𝑄(𝑡)2𝛼1𝛼+𝜃(𝑡)𝜔(𝑡)𝑟1/𝛼(𝑡)𝜌1/𝛼𝜔(𝑡)(1+𝛼)/𝛼+𝑝(𝑡)0𝛼𝜏0𝑝𝜃(𝑡)𝑢(𝑡)0𝛼𝜏0𝛼𝑟1/𝛼(𝑡)𝜌1/𝛼(𝑢𝑡)(1+𝛼)/𝛼(𝑡),(3.52) which follows that 𝑡𝑡2𝐻(𝑡,𝑠)𝜌(𝑠)𝑄(𝑠)2𝛼1𝑝1+0𝛼𝜏0𝜓(𝑠)d𝑠𝑡𝑡2𝐻(𝑡,𝑠)𝜔(𝑠)d𝑠+𝑡𝑡2𝐻(𝑡,𝑠)𝜃(𝑠)𝜔(𝑠)d𝑠𝑡𝑡2𝐻(𝑡,𝑠)𝛼𝜔(1+𝛼)/𝛼(𝑠)𝑟1/𝛼(𝑠)𝜌1/𝛼𝑝(𝑠)d𝑠0𝛼𝜏0𝑡𝑡2𝐻(𝑡,𝑠)𝑢+𝑝(𝑠)d𝑠0𝛼𝜏0𝑡𝑡2𝑝𝐻(𝑡,𝑠)𝜃(𝑠)𝑢(𝑠)d𝑠0𝛼𝜏0𝑡𝑡2𝐻(𝑡,𝑠)𝛼𝑢(1+𝛼)/𝛼(𝑠)𝑟1/𝛼(𝑠)𝜌1/𝛼(𝑠)d𝑠.(3.53) Using the integration by parts formula and 𝐻(𝑡,𝑡)=0, we have 𝑡𝑡2𝐻(𝑡,𝑠)𝜔(𝑠)d𝑠=𝐻𝑡,𝑡2𝜔𝑡2𝑡𝑡2𝜕𝐻(𝑡,𝑠)𝜕𝑠𝜔(𝑠)d𝑠,𝑡𝑡2𝐻(𝑡,𝑠)𝑢(𝑠)d𝑠=𝐻𝑡,𝑡2𝑢𝑡2𝑡𝑡2𝜕𝐻(𝑡,𝑠)𝜕𝑠𝑢(𝑠)d𝑠.(3.54) So, by (3.53), we obtain 𝑡𝑡2𝐻(𝑡,𝑠)𝜌(𝑠)𝑄(𝑠)2𝛼1𝑝1+0𝛼𝜏0𝜓(𝑠)d𝑠𝐻𝑡,𝑡2𝜔𝑡2+𝑝0𝛼𝜏0𝐻𝑡,𝑡2𝑢𝑡2+𝑡𝑡2𝐻(𝑡,𝑠)𝜃(𝑠)+𝜕𝐻(𝑡,𝑠)/𝜕𝑠𝐻(𝑡,𝑠)𝜔(𝑠)d𝑠𝑡𝑡2𝐻(𝑡,𝑠)𝛼𝜔(1+𝛼)/𝛼(𝑠)𝑟1/𝛼(𝑠)𝜌1/𝛼+𝑝(𝑠)d𝑠0𝛼𝜏0𝑡𝑡2𝐻(𝑡,𝑠)𝜃(𝑠)+𝜕𝐻(𝑡,𝑠)/𝜕𝑠𝑝𝐻(𝑡,𝑠)𝑢(𝑠)d𝑠0𝛼𝜏0𝑡𝑡2𝐻(𝑡,𝑠)𝛼𝑢(1+𝛼)/𝛼(𝑠)𝑟1/𝛼(𝑠)𝜌1/𝛼(𝑠)d𝑠.(3.55) Using the inequality 𝐵𝑦𝐴𝑦(𝛼+1)/𝛼𝛼𝛼(𝛼+1)𝛼+1𝐵𝛼+1𝐴𝛼,(3.56) where 𝛼𝐴=𝑟1/𝛼(𝑠)𝜌1/𝛼(𝑠),𝐵=𝜃(𝑠)+𝜕𝐻(𝑡,𝑠)/𝜕𝑠𝐻(𝑡,𝑠),(3.57) we have 𝑡𝑡2𝐻(𝑡,𝑠)𝜌(𝑠)𝑄(𝑠)2𝛼1𝑝1+0𝛼𝜏0(𝜓(𝑠)+𝜙(𝑡,𝑠))d𝑠𝐻𝑡,𝑡2𝜔𝑡2+𝑝0𝛼𝜏0𝐻𝑡,𝑡2𝑢𝑡2(3.58) due to (3.55), which yields that 1𝐻𝑡,𝑡2𝑡𝑡2𝐻(𝑡,𝑠)𝜌(𝑠)𝑄(𝑠)2𝛼1𝑝1+0𝛼𝜏0𝑡(𝜓(𝑠)+𝜙(𝑡,𝑠))d𝑠𝜔2+𝑝0𝛼𝜏0𝑢𝑡2,(3.59) which contradicts (3.34). The proof is complete.

From Theorem 3.7, we can obtain different oscillation conditions for all solutions of (1.1) with different choices of 𝐻; the details are left to the reader.

Theorem 3.8. Assume that (1.6) and (3.30) hold. Furthermore, assume that 0𝑝(𝑡)𝑝1<1. If 𝑡01𝑟(𝑠)𝑠𝑡0𝑞(𝑢)d𝑢1/𝛼d𝑠=,(3.60) then every solution 𝑥 of (1.1) oscillates or lim𝑡𝑥(𝑡)=0.

Proof. Let 𝑥 be a nonoscillatory solution of (1.1). Without loss of generality, we assume that there exists 𝑡1𝑡0 such that 𝑥(𝑡)>0,  𝑥(𝜏(𝑡))>0, and 𝑥(𝜎(𝑡))>0 for all 𝑡𝑡1. Then 𝑧(𝑡)>0 for 𝑡𝑡1. Proceeding as in the proof of Theorem 3.1, we obtain (3.2). Thus 𝑟(𝑡)|𝑧(𝑡)|𝛼1𝑧(𝑡) is decreasing function, and there exists a 𝑡2𝑡1 such that 𝑧(𝑡)>0, 𝑡𝑡2 or 𝑧(𝑡)<0, 𝑡𝑡2.
Case 1. Assume that 𝑧(𝑡)>0, for 𝑡𝑡2. Proceeding as in the proof of Theorem 3.3 and setting 𝜌(𝑡)=𝑡, we can obtain a contradiction with (3.31).
Case 2. Assume that 𝑧(𝑡)<0, for 𝑡𝑡2. Then there exists a finite limit lim𝑡𝑧(𝑡)=𝑙,(3.61) where 𝑙0. Next, we claim that 𝑙=0. If not, then for any 𝜖>0, we have 𝑙<𝑧(𝑡)<𝑙+𝜖, eventually. Take 0<𝜖<𝑙(1𝑝1)/𝑝1. We calculate 𝑥(𝑡)=𝑧(𝑡)𝑝(𝑡)𝑥(𝜏(𝑡))>𝑙𝑝1𝑧(𝜏(𝑡))>𝑙𝑝1(𝑙+𝜖)=𝑚(𝑙+𝜖)>𝑚𝑧(𝑡),(3.62) where 𝑙𝑚=𝑙+𝜖𝑝1=𝑙1𝑝1𝜖𝑝1𝑙+𝜖>0.(3.63) From (3.2) and (3.62), we have 𝑟(𝑡)𝑧(𝑡)𝛼𝑞(𝑡)𝑥𝛼(𝜎(𝑡))(𝑚𝑙)𝛼𝑞(𝑡).(3.64) Integrating the above inequality from 𝑡2 to 𝑡, we get 𝑟(𝑡)𝑧(𝑡)𝛼𝑡𝑟2𝑧𝑡2𝛼(𝑚𝑙)𝛼𝑡𝑡2𝑞(𝑠)d𝑠,(3.65) which implies 𝑧1(𝑡)𝑚𝑙𝑟(𝑡)𝑡𝑡2𝑞(𝑠)d𝑠1/𝛼.(3.66) Integrating the above inequality from 𝑡2 to 𝑡, we have 𝑡𝑧(𝑡)𝑧2𝑚𝑙𝑡𝑡21𝑟(𝑠)𝑠𝑡2𝑞(𝑢)d𝑢1/𝛼d𝑠,(3.67) which yields 𝑧(𝑡); this is a contradiction. Hence, lim𝑡𝑧(𝑡)=0. Note that 0<𝑥(𝑡)𝑧(𝑡). Then, lim𝑡𝑥(𝑡)=0. The proof is complete.

4. Examples

In this section, we will give two examples to illustrate the main results.

Example 4.1. Consider the following linear neutral equation: (𝑥(𝑡)+2𝑥(𝑡+(2𝑛1)𝜋))+𝑥(𝑡+(2𝑚1)𝜋)=0,for𝑡𝑡0,(4.1) where 𝑛 and 𝑚 are positive integers.
Let 𝑟(𝑡)=1,𝑝(𝑡)=2,𝜏(𝑡)=𝑡+(2𝑛1)𝜋,𝑞(𝑡)=1,𝜎(𝑡)=𝑡+(2𝑚1)𝜋.(4.2) Hence, 𝑄(𝑡)=1. Obviously, all the conditions of Corollary 3.5 hold. Thus by Corollary 3.5, every solution of (4.1) is oscillatory. It is easy to verify that 𝑥(𝑡)=sin𝑡 is a solution of (4.1).

Example 4.2. Consider the following linear neutral equation: e2𝑡1𝑥(𝑡)+2𝑥(𝑡+3)+e2𝑡+1+12e2𝑡2𝑥(𝑡+1)=0,for𝑡𝑡0,(4.3) where 𝑛 and 𝑚 are positive integers.
Let 𝑟(𝑡)=e2𝑡1,𝑝(𝑡)=2,𝑞(𝑡)=e2𝑡+1+e2𝑡2/2,𝛼=1.(4.4) Clearly, all the conditions of Theorem 3.8 hold. Thus by Theorem 3.8, every solution of (4.3) is either oscillatory or lim𝑡𝑥(𝑡)=0. It is easy to verify that 𝑥(𝑡)=e𝑡 is a solution of (4.3).

Remark 4.3. Recent results cannot be applied to (4.1) and (4.3) since 𝜏(𝑡)𝑡 and 𝜎(𝑡)𝑡.

Remark 4.4. Using the method given in this paper, we can get other oscillation criteria for (1.1); the details are left to the reader.

Remark 4.5. It would be interesting to find another method to study (1.1) when 𝜏𝜎𝜎𝜏.

Acknowledgments

The authors sincerely thank the referees for their valuable suggestions and useful comments that have led to the present improved version of the original manuscript. This research is supported by the Natural Science Foundation of China (11071143, 60904024, 11026112), China Postdoctoral Science Foundation funded project (200902564), by Shandong Provincial Natural Science Foundation (ZR2010AL002, ZR2009AL003, Y2008A28), and also by University of Jinan Research Funds for Doctors (XBS0843).

References

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