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Abstract and Applied Analysis
Volume 2011 (2011), Article ID 927690, 15 pages
http://dx.doi.org/10.1155/2011/927690
Research Article

Oscillation of Second-Order Neutral Functional Differential Equations with Mixed Nonlinearities

1School of Science, University of Jinan, Jinan, Shandong 250022, China
2Department of Mathematics and Statistics, Missouri University of Science and Technology, Rolla, MO 65409-0020, USA
3School of Control Science and Engineering, Shandong University, Jinan, Shandong 250061, China

Received 2 September 2010; Revised 26 November 2010; Accepted 23 December 2010

Academic Editor: Miroslava Růžičková

Copyright © 2011 Shurong Sun 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

We study the following second-order neutral functional differential equation with mixed nonlinearities (𝑟(𝑡)|(𝑢(𝑡)+𝑝(𝑡)𝑢(𝑡𝜎))|𝛼1(𝑢(𝑡)+𝑝(𝑡)𝑢(𝑡𝜎)))+𝑞0(𝑡)|𝑢(𝜏0(𝑡))|𝛼1𝑢(𝜏0(𝑡))+𝑞1(𝑡)|𝑢(𝜏1(𝑡))|𝛽1𝑢(𝜏1(𝑡))+𝑞2(𝑡)|𝑢(𝜏2(𝑡))|𝛾1𝑢(𝜏2(𝑡))=0, where 𝛾>𝛼>𝛽>0, 𝑡0(1/𝑟1/𝛼(𝑡))d𝑡<. Oscillation results for the equation are established which improve the results obtained by Sun and Meng (2006), Xu and Meng (2006), Sun and Meng (2009), and Han et al. (2010).

1. Introduction

This paper is concerned with the oscillatory behavior of the second-order neutral functional differential equation with mixed nonlinearities ||𝑟(𝑡)(𝑢(𝑡)+𝑝(𝑡)𝑢(𝑡𝜎))||𝛼1(𝑢(𝑡)+𝑝(𝑡)𝑢(𝑡𝜎))+𝑞0||𝑢𝜏(𝑡)0||(𝑡)𝛼1𝑢𝜏0(𝑡)+𝑞1||𝑢𝜏(𝑡)1||(𝑡)𝛽1𝑢𝜏1(𝑡)+𝑞2||𝑢𝜏(𝑡)2||(𝑡)𝛾1𝑢𝜏2(𝑡)=0,𝑡𝑡0,(1.1) where 𝛾>𝛼>𝛽>0 are constants, 𝑟𝐶1([𝑡0,),(0,)), 𝑝𝐶([𝑡0,),[0,1)), 𝑞𝑖𝐶([𝑡0,),), 𝑖=0,1,2, are nonnegative, 𝜎0 is a constant. Here, we assume that there exists 𝜏𝐶1([𝑡0,),) such that 𝜏(𝑡)𝜏𝑖(𝑡), 𝜏(𝑡)𝑡, lim𝑡𝜏(𝑡)=, and 𝜏(𝑡)>0 for 𝑡𝑡0.

One of our motivations for studying (1.1) is the application of this type of equations in real word life problems. For instance, neutral delay equations appear in modeling of networks containing lossless transmission lines, in the study of vibrating masses attached to an elastic bar; see the Euler equation in some variational problems, in the theory of automatic control and in neuromechanical systems in which inertia plays an important role. We refer the reader to Hale [1] and Driver [2], and references cited therein.

Recently, there has been much research activity concerning the oscillation of second-order differential equations [38] and neutral delay differential equations [920]. For the particular case when 𝑝(𝑡)=0, (1.1) reduces to the following equation: ||||𝑟(𝑡)𝑢(𝑡)𝛼1𝑢(𝑡)+𝑞0||𝑢𝜏(𝑡)0||(𝑡)𝛼1𝑢𝜏0(𝑡)+𝑞1||𝑢𝜏(𝑡)1||(𝑡)𝛽1𝑢𝜏1(𝑡)+𝑞2||𝑢𝜏(𝑡)2||(𝑡)𝛾1𝑢𝜏2(𝑡)=0,𝑡𝑡0.(1.2)

Sun and Meng [6] established some oscillation criteria for (1.2), under the condition𝑡01𝑟1/𝛼(𝑡)d𝑡<,(1.3) they only obtained the sufficient condition [6, Theorem 5], which guarantees that every solution 𝑢 of (1.2) oscillates or tends to zero.

Sun and Meng [7] considered the oscillation of second-order nonlinear delay differential equation||𝑢𝑟(𝑡)||(𝑡)𝛼1𝑢(𝑡)+𝑞0||𝑢𝜏(𝑡)0||(𝑡)𝛼1𝑢𝜏0(𝑡)=0,𝑡𝑡0(1.4) and obtained some results for oscillation of (1.4), for example, under the case (1.3), they obtained some results which guarantee that every solution 𝑢 of (1.4) oscillates or tends to zero, see [7, Theorem 2.2].

Xu and Meng [10] discussed the oscillation of the second-order neutral delay differential equation||𝑟(𝑡)(𝑢(𝑡)+𝑝(𝑡)𝑢(𝑡𝜏))||𝛼1(𝑢(𝑡)+𝑝(𝑡)𝑢(𝑡𝜏))+𝑞(𝑡)𝑓(𝑢(𝜎(𝑡)))=0,𝑡𝑡0(1.5) and established the sufficient condition [10, Theorem 2.3], which guarantees that every solution 𝑢 of (1.5) oscillates or tends to zero.

Han et al. [11] examined the oscillation of second-order neutral delay differential equation||𝑟(𝑡)𝜓(𝑢(𝑡))(𝑢(𝑡)+𝑝(𝑡)𝑢(𝑡𝜏))||𝛼1(𝑢(𝑡)+𝑝(𝑡)𝑢(𝑡𝜏))+𝑞(𝑡)𝑓(𝑢(𝜎(𝑡)))=0,𝑡𝑡0(1.6) and established some sufficient conditions for oscillation of (1.6) under the conditions (1.3) and𝜎(𝑡)𝑡𝜏.(1.7) The condition (1.7) can be restrictive condition, since the results cannot be applied on the equation𝑒2𝑡1𝑢(𝑡)+2𝑢(𝑡2)𝑒+𝜆2𝑡+12𝑒2𝑡+2𝑢(𝑡1)=0,𝑡𝑡0.(1.8)

The aim of this paper is to derive some sufficient conditions for the oscillation of solutions of (1.1). The paper is organized as follows. In Section 2, we establish some oscillation criteria for (1.1) under the assumption (1.3). In Section 3, we will give three examples to illustrate the main results. In Section 4, we give some conclusions for this paper.

2. Main Results

In this section, we give some new oscillation criteria for (1.1).

Below, for the sake of convenience, we denote 𝑧(𝑡)=𝑢(𝑡)+𝑝(𝑡)𝑢(𝑡𝜎),𝑅(𝑡)=𝑡𝑡01𝑟1/𝛼(𝑠)d𝑠,(2.1)𝜉(𝑡)=𝑟1/𝛼(𝜏(𝑡))𝑡𝑡11𝑟(𝜏(𝑠))1/𝛼𝜏𝑄(𝑠)d𝑠,0𝜏(𝑡)=1𝑝0(𝑡)𝛼𝑞0(𝑡),𝑄1𝜏(𝑡)=1𝑝1(𝑡)𝛽𝑞1𝑄(𝑡),2𝜏(𝑡)=1𝑝2(𝑡)𝛾𝑞2𝜁(𝑡),0(𝑡)=𝑞0(1𝑡)1+𝑝(𝜌(𝑡))𝛼,𝜁1(𝑡)=𝑞1(1𝑡)1+𝑝(𝜌(𝑡))𝛽,𝜁2(𝑡)=𝑞21(𝑡)1+𝑝(𝜌(𝑡))𝛾,0(𝑡)=𝑞0(1𝑡)1+𝑝(𝑡)𝛼,1(𝑡)=𝑞1(1𝑡)1+𝑝(𝑡)𝛽,2(𝑡)=𝑞21(𝑡)1+𝑝(𝑡)𝛾,𝛿(𝑡)=𝜌(𝑡)1𝑟1/𝛼(𝑠)d𝑠,𝜋(𝑡)=𝑡1𝑟1/𝛼(𝑠)d𝑠,𝑘1=𝛾𝛽𝛾𝛼,𝑘2=𝛾𝛽,𝛼𝛽𝜑(𝑡)=𝑞0(𝑡)𝛿(𝑡)1+𝑝(𝜌(𝑡))𝛼+𝑞1(𝑡)𝛿(𝑡)1+𝑝(𝜌(𝑡))𝛽+𝑞2(𝑡)𝛿(𝑡)1+𝑝(𝜌(𝑡))𝛾.(1)

Theorem 2.1. Assume that (1.3) holds, 𝑝(𝑡)0, and there exists 𝜌𝐶1([𝑡0,),), such that 𝜌(𝑡)𝑡, 𝜌(𝑡)>0, 𝜏𝑖(𝑡)𝜌(𝑡)𝜎,𝑖=0,1,2. If for all sufficiently large 𝑡1, 𝑅𝛼𝑄(𝜏(𝑡))0𝑘(𝑡)+1𝑄1(𝑡)1/𝑘1𝑘2𝑄2(𝑡)1/𝑘2𝛼𝜏(𝑡)𝑅𝛼1(𝜏(𝑡))𝑟11/𝛼(𝜏(𝑡))𝜉𝛼(𝑡)d𝑡=,(2.2)𝜁0𝑘(𝑡)+1𝜁1(𝑡)1/𝑘1𝑘2𝜁2(𝑡)1/𝑘2𝛿𝛼𝛼(𝑡)𝛼+1𝛼+1𝜌(𝑡)𝛿(𝑡)𝑟1/𝛼(𝜌(𝑡))d𝑡=,(2.3) then (1.1) is oscillatory.

Proof. Suppose to the contrary that 𝑢 is a nonoscillatory solution of (1.1). Without loss of generality, we may assume that 𝑢(𝑡)>0 for all large 𝑡. The case of 𝑢(𝑡)<0 can be considered by the same method. From (1.1) and (1.3), we can easily obtain that there exists a 𝑡1𝑡0 such that 𝑧(𝑡)>0,𝑧||𝑧(𝑡)>0,𝑟(𝑡)||(𝑡)𝛼1𝑧(𝑡)0,(2.4) or 𝑧(𝑡)>0,𝑧||𝑧(𝑡)<0,𝑟(𝑡)||(𝑡)𝛼1𝑧(𝑡)0.(2.5) If (2.4) holds, we have 𝑧𝑟(𝑡)(𝑡)𝛼𝑧𝑟(𝜏(𝑡))(𝜏(𝑡))𝛼,𝑡𝑡1.(2.6) From the definition of 𝑧, we obtain 𝑢(𝑡)=𝑧(𝑡)𝑝(𝑡)𝑢(𝑡𝜎)𝑧(𝑡)𝑝(𝑡)𝑧(𝑡𝜎)(1𝑝(𝑡))𝑧(𝑡).(2.7) Define 𝜔(𝑡)=𝑅𝛼(𝑧𝜏(𝑡))𝑟(𝑡)(𝑡)𝛼(𝑧(𝜏(𝑡)))𝛼,𝑡𝑡1.(2.8) Then, 𝜔(𝑡)>0 for 𝑡𝑡1. Noting that 𝑧(𝑡)>0, we get 𝑧(𝜏𝑖(𝑡))𝑧(𝜏(𝑡)) for 𝑖=0,1,2. Thus, from (1.1), (2.7), and (2.8), it follows that 𝜔(𝑡)𝛼𝜏(𝑡)𝑅𝛼1(𝜏(𝑡))𝑟1/𝛼𝑧(𝜏(𝑡))𝑟(𝑡)(𝑡)𝛼(𝑧(𝜏(𝑡)))𝛼𝑅𝛼(𝜏𝜏(𝑡))1𝑝0(𝑡)𝛼𝑞0(𝑡)𝑅𝛼(𝜏𝜏(𝑡))1𝑝1(𝑡)𝛽𝑞1(𝑡)𝑧𝛽𝛼(𝜏𝜏(𝑡))+1𝑝2(𝑡)𝛾𝑞2(𝑡)𝑧𝛾𝛼(𝜏(𝑡))𝛼𝑅𝛼𝑧(𝜏(𝑡))𝑟(𝑡)(𝑡)𝛼(𝑧(𝜏(𝑡)))𝛼+1𝑧(𝜏(𝑡))𝜏(𝑡).(2.9) By (2.4), (2.9), and 𝜏(𝑡)>0, we get 𝜔(𝑡)𝛼𝜏(𝑡)𝑅𝛼1(𝜏(𝑡))𝑟1/𝛼𝑧(𝜏(𝑡))𝑟(𝑡)(𝑡)𝛼(𝑧(𝜏(𝑡)))𝛼𝑅𝛼(𝜏𝜏(𝑡))1𝑝0(𝑡)𝛼𝑞0(𝑡)𝑅𝛼(𝜏𝜏(𝑡))1𝑝1(𝑡)𝛽𝑞1(𝑡)𝑧𝛽𝛼(𝜏𝜏(𝑡))+1𝑝2(𝑡)𝛾𝑞2(𝑡)𝑧𝛾𝛼(.𝜏(𝑡))(2.10) In view of (2.4), (2.6), and (2.10), we have 𝜔(𝑡)𝛼𝜏(𝑡)𝑅𝛼1(𝜏(𝑡))𝑟1/𝛼𝑧(𝜏(𝑡))𝑟(𝜏(𝑡))(𝜏(𝑡))𝛼(𝑧(𝜏(𝑡)))𝛼𝑅𝛼(𝜏𝜏(𝑡))1𝑝0(𝑡)𝛼𝑞0(𝑡)𝑅𝛼(𝜏𝜏(𝑡))1𝑝1(𝑡)𝛽𝑞1(𝑡)𝑧𝛽𝛼(𝜏𝜏(𝑡))+1𝑝2(𝑡)𝛾𝑞2(𝑡)𝑧𝛾𝛼(.𝜏(𝑡))(2.11) By (2.4), we obtain 𝜏𝑡𝑧(𝜏(𝑡))=𝑧1+𝑡𝑡1𝑧(𝜏(𝑠))𝜏(𝜏𝑡𝑠)d𝑠=𝑧1+𝑡𝑡11𝑟(𝜏(𝑠))1/𝛼𝑧𝑟(𝜏(𝑠))(𝜏(𝑠))𝛼1/𝛼𝜏(𝑠)d𝑠𝑟1/𝛼(𝜏(𝑡))𝑧(𝜏(𝑡))𝑡𝑡11𝑟(𝜏(𝑠))1/𝛼𝜏(𝑠)d𝑠,(2.12) that is, 𝑧(𝜏(𝑡))𝜉(𝑡)𝑧(𝜏(𝑡)).(2.13) Set 𝑘𝑎=1𝑄1(𝑡)𝑧𝛽𝛼(𝜏(𝑡))1/𝑘1𝑘,𝑏=2𝑄2(𝑡)𝑧𝛾𝛼(𝜏(𝑡))1/𝑘2,𝑝=𝑘1,𝑞=𝑘2.(2.14) Using Young’s inequality ||||1𝑎𝑏𝑝|𝑎|𝑝+1𝑞||𝑏||𝑞1,𝑎,𝑏,𝑝>1,𝑞>1,𝑝+1𝑞=1,(2.15) we have 𝑄1(𝑡)𝑧𝛽𝛼(𝜏(𝑡))+𝑄2(𝑡)𝑧𝛾𝛼𝑘(𝜏(𝑡))1𝑄1(𝑡)1/𝑘1𝑘2𝑄2(𝑡)1/𝑘2.(2.16) Hence, by (2.11), (2.13), and (2.16), we obtain 𝜔(𝑡)𝛼𝜏(𝑡)𝑅𝛼1(𝜏(𝑡))𝑟11/𝛼(𝜏(𝑡))𝜉𝛼(𝑡)𝑅𝛼𝑄(𝜏(𝑡))0𝑘(𝑡)+1𝑄1(𝑡)1/𝑘1𝑘2𝑄2(𝑡)1/𝑘2.(2.17) Integrating (2.17) from 𝑡1 to 𝑡, we get 𝑡0<𝜔(𝑡)𝜔1,(2.18)𝑡𝑡1𝑅𝛼𝑄(𝜏(𝑠))0𝑘(𝑠)+1𝑄1(𝑠)1/𝑘1𝑘2𝑄2(𝑠)1/𝑘2𝛼𝜏(𝑠)𝑅𝛼1(𝜏(𝑠))𝑟11/𝛼(𝜏(𝑠))𝜉𝛼(𝑠)d𝑠.(2.19) Letting 𝑡 in (2.19), we get a contradiction to (2.2). If (2.5) holds, we define the function 𝜐 by 𝜐(𝑡)=𝑟(𝑡)𝑧(𝑡)𝛼1𝑧(𝑡)𝑧𝛼(𝜌(𝑡)),𝑡𝑡1.(2.20) Then, 𝜐(𝑡)<0 for 𝑡𝑡1. It follows from [𝑟(𝑡)|𝑧(𝑡)|𝛼1𝑧(𝑡)]0 that 𝑟(𝑡)|𝑧(𝑡)|𝛼1𝑧(𝑡) is nonincreasing. Thus, we have 𝑟1/𝛼(𝑠)𝑧(𝑠)𝑟1/𝛼(𝑡)𝑧(𝑡),𝑠𝑡.(2.21) Dividing (2.21) by 𝑟1/𝛼(𝑠) and integrating it from 𝜌(𝑡) to 𝑙, we obtain 𝑧(𝑙)𝑧(𝜌(𝑡))+𝑟1/𝛼(𝑡)𝑧(𝑡)𝑙𝜌(𝑡)d𝑠𝑟1/𝛼(𝑠),𝑙𝜌(𝑡).(2.22) Letting 𝑙 in the above inequality, we obtain 0𝑧(𝜌(𝑡))+𝑟1/𝛼(𝑡)𝑧(𝑡)𝛿(𝑡),𝑡𝑡1,(2.23) that is, 𝑟1/𝛼𝑧(𝑡)𝛿(𝑡)(𝑡)𝑧(𝜌(𝑡))1,𝑡𝑡1.(2.24) Hence, by (2.20), we have 1𝜐(𝑡)𝛿𝛼(𝑡)0,𝑡𝑡1.(2.25) Differentiating (2.20), we get 𝜐(𝑡)=𝑟(𝑡)𝑧(𝑡)𝛼1𝑧(𝑡)𝑧𝛼(𝜌(𝑡))𝛼𝑟(𝑡)𝑧(𝑡)𝛼1𝑧(𝑡)𝑧𝛼1(𝜌(𝑡))𝑧(𝜌(𝑡))𝜌(𝑡)𝑧2𝛼(𝜌(𝑡)),(2.26) by the above equality and (1.1), we obtain 𝜐(𝑡)=𝑞0𝑢(𝑡)𝛼𝜏0(𝑡)𝑧𝛼(𝜌(𝑡))𝑞1𝑢(𝑡)𝛽𝜏1(𝑡)𝑧𝛼(𝜌(𝑡))𝑞2𝑢(𝑡)𝛾𝜏2(𝑡)𝑧𝛼(𝜌(𝑡))𝛼𝑟(𝑡)𝑧(𝑡)𝛼1𝑧(𝑡)𝑧𝛼1(𝜌(𝑡))𝑧(𝜌(𝑡))𝜌(𝑡)𝑧2𝛼.(𝜌(𝑡))(2.27) Noticing that 𝑝(𝑡)0, from [10, Theorem 2.3], we see that 𝑢(𝑡)0 for 𝑡𝑡1, so by 𝜏𝑖(𝑡)𝜌(𝑡)𝜎, 𝑖=0,1,2, we have 𝑢𝛼𝜏0(𝑡)𝑧𝛼=𝑢𝜏(𝜌(𝑡))0(𝑡)𝑢(𝜌(𝑡))+𝑝(𝜌(𝑡))𝑢(𝜌(𝑡)𝜎)𝛼=1𝑢𝜏(𝜌(𝑡))/𝑢0𝜏(𝑡)+𝑝(𝜌(𝑡))𝑢(𝜌(𝑡)𝜎)/𝑢0(𝑡)𝛼11+𝑝(𝜌(𝑡))𝛼,𝑢𝛽𝜏1(𝑡)𝑧𝛼=𝑢𝜏(𝜌(𝑡))1(𝑡)𝑢(𝜌(𝑡))+𝑝(𝜌(𝑡))𝑢(𝜌(𝑡)𝜎)𝛽𝑧𝛽𝛼=1(𝜌(𝑡))𝜏𝑢(𝜌(𝑡))/𝑢1𝜏(𝑡)+𝑝(𝜌(𝑡))𝑢(𝜌(𝑡)𝜎)/𝑢1(𝑡)𝛽𝑧𝛽𝛼1(𝜌(𝑡))1+𝑝(𝜌(𝑡))𝛽𝑧𝛽𝛼𝑢(𝜌(𝑡)),𝛾𝜏2(𝑡)/𝑧𝛼=𝑢𝜏(𝜌(𝑡))2(𝑡)𝑢(𝜌(𝑡))+𝑝(𝜌(𝑡))𝑢(𝜌(𝑡)𝜎)𝛾𝑧𝛾𝛼=1(𝜌(𝑡))𝜏𝑢(𝜌(𝑡))/𝑢2𝜏(𝑡)+𝑝(𝜌(𝑡))𝑢(𝜌(𝑡)𝜎)/𝑢2(𝑡)𝛾𝑧𝛾𝛼1(𝜌(𝑡))1+𝑝(𝜌(𝑡))𝛾𝑧𝛾𝛼(𝜌(𝑡)).(2.28) On the other hand, from (𝑟(𝑡)(𝑧(𝑡))𝛼1𝑧(𝑡))0, 𝜌(𝑡)𝑡, we obtain 𝑧𝑟(𝜌(𝑡))1/𝛼(𝑡)𝑟1/𝛼𝑧(𝜌(𝑡))(𝑡).(2.29) Thus, by (2.20) and (2.27), we get 𝜐𝜁(𝑡)0(𝑡)+𝜁1(𝑡)𝑧𝛽𝛼(𝜌(𝑡))+𝜁2(𝑡)𝑧𝛾𝛼(𝜌(𝑡))𝛼𝜌(𝑡)𝑟1/𝛼(𝜌(𝑡))(𝜐(𝑡))(𝛼+1)/𝛼.(2.30) Set 𝑘𝑎=1𝜁1(𝑡)𝑧𝛽𝛼(𝜌(𝑡))1/𝑘1𝑘,𝑏=2𝜁2(𝑡)𝑧𝛾𝛼(𝜌(𝑡))1/𝑘2,𝑝=𝑘1,𝑞=𝑘2.(2.31) Using Young’s inequality (2.15), we obtain 𝜁1(𝑡)𝑧𝛽𝛼(𝜌(𝑡))+𝜁2(𝑡)𝑧𝛾𝛼𝑘(𝜌(𝑡))1𝜁1(𝑡)1/𝑘1𝑘2𝜁2(𝑡)1/𝑘2.(2.32) Hence, from (2.30), we have 𝜐𝜁(𝑡)0𝑘(𝑡)+1𝜁1(𝑡)1/𝑘1𝑘2𝜁2(𝑡)1/𝑘2𝛼𝜌(𝑡)𝑟1/𝛼(𝜌(𝑡))(𝜐(𝑡))(𝛼+1)/𝛼,(2.33) that is, 𝜐𝜁(𝑡)+0𝑘(𝑡)+1𝜁1(𝑡)1/𝑘1𝑘2𝜁2(𝑡)1/𝑘2+𝛼𝜌(𝑡)𝑟1/𝛼(𝜌(𝑡))(𝜐(𝑡))(𝛼+1)/𝛼0,𝑡𝑡1.(2.34) Multiplying (2.34) by 𝛿𝛼(𝑡) and integrating it from 𝑡1 to 𝑡 implies that 𝛿𝛼(𝑡)𝜐(𝑡)𝛿𝛼𝑡1𝜐𝑡1+𝛼𝑡𝑡1𝑟1/𝛼(𝜌(𝑠))𝜌(𝑠)𝛿𝛼1(+𝑠)𝜐(𝑠)d𝑠𝑡𝑡1𝜁0𝑘(𝑠)+1𝜁1(𝑠)1/𝑘1𝑘2𝜁2(𝑠)1/𝑘2𝛿𝛼(𝑠)d𝑠+𝛼𝑡𝑡1𝛿𝛼(𝑠)𝜌(𝑠)𝑟1/𝛼(𝜌(𝑠))(𝜐(𝑠))(𝛼+1)/𝛼d𝑠0.(2.35) Set 𝑝=(𝛼+1)/𝛼, 𝑞=𝛼+1, and 𝑎=(𝛼+1)𝛼/(𝛼+1)𝛿𝛼2/(𝛼+1)𝛼(𝑡)𝜐(𝑡),𝑏=(𝛼+1)𝛼/(𝛼+1)𝛿1/(𝛼+1)(𝑡).(2.36) Using Young's inequality (2.15), we get 𝛼𝛿𝛼1(𝑡)𝜐(𝑡)𝛼𝛿𝛼(𝑡)(𝜐(𝑡))(𝛼+1)/𝛼+𝛼𝛼+1𝛼+11.𝛿(𝑡)(2.37) Thus, 𝛼𝜌(𝑡)𝛿𝛼1(𝑡)𝜐(𝑡)𝑟1/𝛼(𝜌(𝑡))𝛼𝜌𝛿(𝑡)𝛼(𝑡)(𝜐(𝑡))(𝛼+1)/𝛼𝑟1/𝛼(𝜌(𝑡))+𝜌𝛼(𝑡)𝛼+1𝛼+11𝛿(𝑡)𝑟1/𝛼(𝜌(𝑡)).(2.38) Therefore, (2.35) yields 𝛿𝛼(𝑡)𝜐(𝑡)𝛿𝛼𝑡1𝜐𝑡1,𝑡𝑡1𝜁0𝑘(𝑠)+1𝜁1(𝑠)1/𝑘1𝑘2𝜁2(𝑠)1/𝑘2𝛿𝛼𝛼(𝑠)𝛼+1𝛼+1𝜌(𝑠)𝛿(𝑠)𝑟1/𝛼(𝜌(𝑠))d𝑠.(2.39) Letting 𝑡 in the above inequality, by (2.3), we get a contradiction with (2.25). This completes the proof of Theorem 2.1.

From Theorem 2.1, when 𝜌(𝑡)=𝑡, we have the following result.

Corollary 2.2. Assume that (1.3) holds, 𝑝(𝑡)0, and 𝜏𝑖(𝑡)𝑡𝜎, 𝑖=0,1,2. If for all sufficiently large 𝑡1 such that (2.2) holds and 0𝑘(𝑡)+11(𝑡)1/𝑘1𝑘22(𝑡)1/𝑘2𝜋𝛼𝛼(𝑡)𝛼+1𝛼+11𝜋(𝑡)𝑟1/𝛼(𝑡)d𝑡=,(2.40) then (1.1) is oscillatory.

Theorem 2.3. Assume that (1.3) holds, 𝑝(𝑡)0, and there exists 𝜌𝐶1([𝑡0,),), such that 𝜌(𝑡)𝑡, 𝜌(𝑡)>0, 𝜏𝑖(𝑡)𝜌(𝑡)𝜎, 𝑖=0,1,2. If for all sufficiently large 𝑡1 such that (2.2) holds and 𝜁0𝑘(𝑡)+1𝜁1(𝑡)1/𝑘1𝑘2𝜁2(𝑡)1/𝑘2𝛿𝛼+1(𝑡)d𝑡=,(2.41) then (1.1) is oscillatory.

Proof. Suppose to the contrary that 𝑢 is a nonoscillatory solution of (1.1). Without loss of generality, we may assume that 𝑢(𝑡)>0 for all large 𝑡. The case of 𝑢(𝑡)<0 can be considered by the same method. From (1.1) and (1.3), we can easily obtain that there exists a 𝑡1𝑡0 such that (2.4) or (2.5) holds.
If (2.4) holds, proceeding as in the proof of Theorem 2.1, we obtain a contradiction with (2.2).
If (2.5) holds, we proceed as in the proof of Theorem 2.1, then we get (2.25) and (2.34). Multiplying (2.34) by 𝛿𝛼+1(𝑡) and integrating it from 𝑡1 to 𝑡 implies that 𝛿𝛼+1(𝑡)𝜐(𝑡)𝛿𝛼+1𝑡1𝜐𝑡1+(𝛼+1)𝑡𝑡1𝑟1/𝛼(𝜌(𝑠))𝜌(𝑠)𝛿𝛼(+𝑠)𝜐(𝑠)d𝑠𝑡𝑡1𝜁0𝑘(𝑠)+1𝜁1(𝑠)1/𝑘1𝑘2𝜁2(𝑠)1/𝑘2𝛿𝛼+1(𝑠)d𝑠+𝛼𝑡𝑡1𝛿𝛼+1(𝑠)𝜌(𝑠)𝑟1/𝛼(𝜌(𝑠))(𝜐(𝑠))(𝛼+1)/𝛼d𝑠0.(2.42) In view of (2.25), we have 𝜐(𝑡)𝛿𝛼+1(𝑡)𝛿(𝑡)<, 𝑡. From (1.3), we get 𝑡𝑡1𝑟1/𝛼(𝜌(𝑠))𝜌(𝑠)𝛿𝛼(𝑠)𝜐(𝑠)d𝑠𝑡𝑡1𝑟1/𝛼(𝜌(𝑠))𝜌(𝑠)d𝑠=𝜌(𝑡)𝜌(𝑡1)𝑟1/𝛼(𝑢)d𝑢<,𝑡,𝑡𝑡1𝛿𝛼+1(𝑠)𝜌(𝑠)𝑟1/𝛼((𝜌(𝑠))𝜐(𝑠))(𝛼+1)/𝛼d𝑠𝜌𝑡𝜌(𝑡)1𝑟1/𝛼(𝑢)d𝑢<,𝑡.(2.43) Letting 𝑡 in (2.42) and using the last inequalities, we obtain 𝜁0𝑘(𝑡)+1𝜁1(𝑡)1/𝑘1𝑘2𝜁2(𝑡)1/𝑘2𝛿𝛼+1(𝑡)d𝑡<,(2.44) which contradicts (2.41). This completes the proof of Theorem 2.3.

From Theorem 2.3, when 𝜌(𝑡)=𝑡, we have the following result.

Corollary 2.4. Assume that (1.3) holds, 𝑝(𝑡)0, 𝜏𝑖(𝑡)𝑡𝜎, 𝑖=0,1,2. If for all sufficiently large 𝑡1 such that (2.2) holds and 0𝑘(𝑡)+11(𝑡)1/𝑘1𝑘22(𝑡)1/𝑘2𝜋𝛼+1(𝑡)d𝑡=,(2.45) then (1.1) is oscillatory.

Theorem 2.5. Assume that (1.3) holds, 𝑝(𝑡)0, and there exists 𝜌𝐶1([𝑡0,),), such that 𝜌(𝑡)𝑡, 𝜌(𝑡)>0, 𝜏𝑖(𝑡)𝜌(𝑡)𝜎, 𝑖=0,1,2. If for all sufficiently large 𝑡1 such that (2.2) holds and 𝑡1𝑟1/𝛼(𝑣)𝑣𝑡1𝜑(𝑢)d𝑢1/𝛼d𝑣=,(2.46) then (1.1) is oscillatory.

Proof. Suppose to the contrary that 𝑢 is a nonoscillatory solution of (1.1). Without loss of generality, we may assume that 𝑢(𝑡)>0 for all large 𝑡. The case of 𝑢(𝑡)<0 can be considered by the same method. From (1.1) and (1.3), we can easily obtain that there exists a 𝑡1𝑡0 such that (2.4) or (2.5) holds.
If (2.4) holds, proceeding as in the proof of Theorem 2.1, we obtain a contradiction with (2.2).
If (2.5) holds, we proceed as in the proof of Theorem 2.1, and we get (2.21). Dividing (2.21) by 𝑟1/𝛼(𝑠) and integrating it from 𝜌(𝑡) to 𝑙, letting 𝑙, yields 𝑧(𝜌(𝑡))𝑟1/𝛼(𝑡)𝑧(𝑡)𝜌(𝑡)𝑟1/𝛼(𝑠)d𝑠=𝑟1/𝛼(𝑡)𝑧(𝑡)𝛿(𝑡)𝑟1/𝛼𝑡1𝑧𝑡1𝛿(𝑡)=𝑎𝛿(𝑡).(2.47) By (1.1), we have 𝑟(𝑡)𝑧(𝑡)𝛼=𝑞0(𝑡)𝑢𝛼𝜏0(𝑡)+𝑞1(𝑡)𝑢𝛽𝜏1(𝑡)+𝑞2(𝑡)𝑢𝛾𝜏2(𝑡).(2.48) Noticing that 𝑝(𝑡)0, from [10, Theorem 2.3], we see that 𝑢(𝑡)0 for 𝑡𝑡1, so by 𝜏𝑖(𝑡)𝜌(𝑡)𝜎, 𝑖=0,1,2, we get 𝑢𝜏𝑖(𝑡)=𝑢𝜏𝑧(𝜌(𝑡))𝑖(𝑡)=1𝑢(𝜌(𝑡))+𝑝(𝜌(𝑡))𝑢(𝜌(𝑡)𝜎)𝜏𝑢(𝜌(𝑡))/𝑢𝑖(𝜏𝑡)+𝑝(𝜌(𝑡))𝑢(𝜌(𝑡)𝜎)/𝑢𝑖(1𝑡).1+𝑝(𝜌(𝑡))(2.49) Hence, we obtain 𝑟(𝑡)𝑧(𝑡)𝛼𝑏𝜑(𝑡),(2.50) where 𝑏=min{𝑎𝛼,𝑎𝛽,𝑎𝛾}. Integrating the above inequality from 𝑡1 to 𝑡, we have 𝑟(𝑡)𝑧(𝑡)𝛼𝑡𝑟1𝑧𝑡1𝛼+𝑏𝑡𝑡1𝜑(𝑢)d𝑢𝑏𝑡𝑡1𝜑(𝑢)d𝑢.(2.51) Integrating the above inequality from 𝑡1 to 𝑡, we obtain 𝑧𝑡1𝑧(𝑡)𝑏1/𝛼𝑡𝑡1𝑟1/𝛼(𝑣)𝑣𝑡1𝜑(𝑢)d𝑢1/𝛼d𝑣,(2.52) which contradicts (2.46). This completes the proof of Theorem 2.5.

3. Examples

In this section, three examples are worked out to illustrate the main results.

Example 3.1. Consider the second-order neutral delay differential equation (1.8), where 𝜆>0 is a constant.
Let 𝑟(𝑡)=e2𝑡, 𝑝(𝑡)=1/2, 𝜎=2, 𝑞0(𝑡)=𝜆(2e2𝑡+e2𝑡+2)/2, 𝛼=1, 𝜏0(𝑡)=𝑡1, 𝑞1(𝑡)=𝑞2(𝑡)=0, and 𝜏(𝑡)=𝜏0(𝑡), then 𝑅(𝑡)=𝑡𝑡01𝑟1/𝛼e(𝑠)d𝑠=2𝑡0e2𝑡2,𝜉(𝑡)=𝑟1/𝛼(𝜏(𝑡))𝑡𝑡11𝑟(𝜏(𝑠))1/𝛼𝜏e(𝑠)d𝑠=2(𝑡𝑡1)12,𝑄0𝑞(𝑡)=0(𝑡)2=𝜆2e2𝑡+e2𝑡+24,𝜁0(𝑡)=2𝑞0(𝑡)3=𝜆2e2𝑡+e2𝑡+23.(3.1)
Setting 𝜌(𝑡)=𝑡+1, we have 𝜏0(𝑡)=𝑡1𝜌(𝑡)𝜎, 𝛿(𝑡)=e2𝑡2/2. Therefore, for all sufficiently large 𝑡1, 𝑅𝛼𝑄(𝜏(𝑡))0𝑘(𝑡)+1𝑄1(𝑡)1/𝑘1𝑘2𝑄2(𝑡)1/𝑘2𝛼𝜏(𝑡)𝑅𝛼1(𝜏(𝑡))𝑟11/𝛼(𝜏(𝑡))𝜉𝛼(𝑡)d𝑡=,𝜁0𝑘(𝑡)+1𝜁1(𝑡)1/𝑘1𝑘2𝜁2(𝑡)1/𝑘2𝛿𝛼𝛼(𝑡)𝛼+1𝛼+1𝜌(𝑡)𝛿(𝑡)𝑟1/𝛼=(𝜌(𝑡))d𝑡𝜆2e2+136d𝑡=(3.2) if 𝜆>3/(2e2+1). Hence, by Theorem 2.1, (1.8) is oscillatory when 𝜆>3/(2e2+1).
Note that [11, Theorem 2.1] and [11, Theorem 2.2] cannot be applied in (1.8), since 𝜏0(𝑡)>𝑡2. On the other hand, applying [11, Theorem 3.2] to that (1.8), we obtain that (1.8) is oscillatory if 𝜆>3/(e2+2e4). So our results improve the results in [11].

Example 3.2. Consider the second-order neutral delay differential equation e𝑡1𝑢(𝑡)+2𝑢𝜋𝑡4+1265e𝑡𝑢1𝑡8arcsin6565=0,𝑡𝑡0.(3.3) Let 𝑟(𝑡)=e𝑡, 𝑝(𝑡)=1/2, 𝜎=𝜋/4, 𝑞0(𝑡)=1265e𝑡, 𝑞1(𝑡)=𝑞2(𝑡)=0,𝛼=1,𝜏0(𝑡)=𝑡(arcsin65/65)/8, 𝜌(𝑡)=𝑡+𝜋/4, and 𝜏(𝑡)=𝑡𝜋/4, then 𝑅(𝑡)=𝑡𝑡01𝑟1/𝛼(𝑠)d𝑠=e𝑡0e𝑡,𝜉(𝑡)=𝑟1/𝛼(𝜏(𝑡))𝑡𝑡11𝑟(𝜏(𝑠))1/𝛼𝜏(𝑠)d𝑠=e𝑡𝑡1𝑄1,0𝑞(𝑡)=0(𝑡)2=665e𝑡,𝜁0(𝑡)=2𝑞0(𝑡)3=865e𝑡,𝛿(𝑡)=e𝑡𝜋/4.(3.4) Therefore, for all sufficiently large 𝑡1, 𝑅𝛼𝑄(𝜏(𝑡))0𝑘(𝑡)+1𝑄1(𝑡)1/𝑘1𝑘2𝑄2(𝑡)1/𝑘2𝛼𝜏(𝑡)𝑅𝛼1(𝜏(𝑡))𝑟11/𝛼(𝜏(𝑡))𝜉𝛼(𝑡)d𝑡=,𝜁0𝑘(𝑡)+1𝜁1(𝑡)1/𝑘1𝑘2𝜁2(𝑡)1/𝑘2𝛿𝛼𝛼(𝑡)𝛼+1𝛼+1𝜌(𝑡)𝛿(𝑡)𝑟1/𝛼=(𝜌(𝑡))d𝑡865e𝜋/414d𝑡=.(3.5) Hence, by Theorem 2.1, (3.3) oscillates. For example, 𝑢(𝑡)=sin8𝑡 is a solution of (3.3).

Example 3.3. Consider the second-order neutral differential equation e𝑡𝑧(𝑡)+e2𝜆𝑡𝑢𝜆0𝑡+𝑞1(𝑡)𝑢1/3𝜆1𝑡+𝑞2(𝑡)𝑢5/3𝜆2𝑡=0,𝑡𝑡0,(3.6) where 𝑧(𝑡)=𝑢(𝑡)+𝑢(𝑡1)/2, 𝜆𝑖>0 for 𝑖=0,1,2, are constants, 𝑞1(𝑡)>0, 𝑞2(𝑡)>0 for 𝑡𝑡0.
Let 𝑟(𝑡)=e𝑡, 𝜎=1, 𝑞0(𝑡)=e2𝜆𝑡, 𝜆=max{𝜆0,𝜆1,𝜆2}, 𝜏𝑖(𝑡)=𝜆𝑖𝑡, 𝜏(𝑡)=𝜆𝑡, 0<𝜆<min{𝜆0,𝜆1,𝜆2,1}, 𝜌(𝑡)=𝜆𝑡+1, 𝛼=1, 𝛽=1/3, and 𝛾=5/3, then 𝑘1=𝑘2=2, 𝑅(𝑡)=𝑡𝑡01𝑟1/𝛼(𝑠)d𝑠=e𝑡0e𝑡,𝜉(𝑡)=𝑟1/𝛼(𝜏(𝑡))𝑡𝑡11𝑟(𝜏(𝑠))1/𝛼𝜏(𝑠)d𝑠=e𝜆(𝑡𝑡1)1,𝛿(𝑡)=e𝜆𝑡1.(3.7) It is easy to see that (2.2) and (2.41) hold for all sufficiently large 𝑡1. Hence, by Theorem 2.3, (3.6) is oscillatory.

4. Conclusions

In this paper, we consider the oscillatory behavior of second-order neutral functional differential equation (1.1). Our results can be applied to the case when 𝜏𝑖(𝑡)>𝑡, 𝑖=0,1,2; these results improve the results given in [6, 7, 10, 11].

Acknowledgments

The authors sincerely thank the reviewers for their valuable suggestions and useful comments that have led to the present improved version of the original paper. This research is supported by the Natural Science Foundation of China (nos. 11071143, 60904024, 11026112), China Postdoctoral Science Foundation funded Project (no. 200902564), the Natural Science Foundation of Shandong (nos. ZR2010AL002, ZR2009AL003, Y2008A28), and also the University of Jinan Research Funds for Doctors (no. XBS0843).

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