Advances in Difference Equations
Volume 2010 (2010), Article ID 586312, 23 pages
doi:10.1155/2010/586312
Research Article

Oscillation Behavior of Third-Order Neutral Emden-Fowler Delay Dynamic Equations on Time Scales

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

Received 14 September 2009; Revised 28 November 2009; Accepted 10 December 2009

Academic Editor: Leonid Berezansky

Copyright © 2010 Zhenlai Han 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 will establish some oscillation criteria for the third-order Emden-Fowler neutral delay dynamic equations ( 𝑟 ( 𝑡 ) ( 𝑥 ( 𝑡 ) 𝑎 ( 𝑡 ) 𝑥 ( 𝜏 ( 𝑡 ) ) ) Δ Δ ) Δ + 𝑝 ( 𝑡 ) 𝑥 𝛾 ( 𝛿 ( 𝑡 ) ) = 0 on a time scale 𝕋 , where 𝛾 > 0 is a quotient of odd positive integers with 𝑟 , 𝑎 , and 𝑝 real-valued positive rd-continuous functions defined on 𝕋 . To the best of our knowledge nothing is known regarding the qualitative behavior of these equations on time scales, so this paper initiates the study. Some examples are considered to illustrate the main results.

1. Introduction

The study of dynamic equations on time-scales, which goes back to its founder Hilger [1], is an area of mathematics that has recently received a lot of attention. It has been created in order to unify the study of differential and difference equations. Many results concerning differential equations carry over quite easily to corresponding results for difference equations, while other results seem to be completely different from their continuous counterparts. The study of dynamic equations on time-scales reveals such discrepancies, and helps avoid proving results twice—once for differential equations and once again for difference equations.

Several authors have expounded on various aspects of this new theory; see the survey paper by Agarwal et al. [2], Bohner and Guseinov [3], and references cited therein. A book on the subject of time-scales, by Bohner and Peterson [4], summarizes and organizes much of the time-scale calculus; see also the book by Bohner and Peterson [5] for advances in dynamic equations on time-scales.

In the recent years, there has been increasing interest in obtaining sufficient conditions for the oscillation and nonoscillation of solutions of various equations on time-scales; we refer the reader to the papers [638]. To the best of our knowledge, it seems to have few oscillation results for the oscillation of third-order dynamic equations; see, for example, [1416, 21, 35]. However, the paper which deals with the third-order delay dynamic equation is due to Hassan [21].

Hassan [21] considered the third-order nonlinear delay dynamic equations

𝑐 ( 𝑡 ) ( 𝑎 ( 𝑡 ) 𝑥 Δ ( 𝑡 ) ) Δ 𝛾 Δ + 𝑓 ( 𝑡 , 𝑥 ( 𝜏 ( 𝑡 ) ) ) = 0 , 𝑡 𝕋 , ( 1 . 1 ) where 𝜏 ( 𝜎 ( 𝑡 ) ) = 𝜎 ( 𝜏 ( 𝑡 ) ) is required, and the author established some oscillation criteria for (1.1) which extended the results given in [16].

To the best of our knowledge, there are no results regarding the oscillation of the solutions of the following third-order nonlinear neutral delay dynamic equations on time-scales up to now:

𝑟 ( 𝑡 ) ( 𝑥 ( 𝑡 ) 𝑎 ( 𝑡 ) 𝑥 ( 𝜏 ( 𝑡 ) ) ) Δ Δ Δ + 𝑝 ( 𝑡 ) 𝑥 𝛾 ( 𝛿 ( 𝑡 ) ) = 0 , 𝑡 𝕋 . ( 1 . 2 )

We assume that 𝛾 > 0 is a quotient of odd positive integers, 𝑟 , 𝑎 and 𝑝 are positive real-valued rd-continuous functions defined on 𝕋 such that 𝑟 Δ ( 𝑡 ) 0 , 0 < 𝑎 ( 𝑡 ) 𝑎 0 < 1 , l i m 𝑡 𝑎 ( 𝑡 ) = 𝑎 < 1 , the delay functions 𝜏 𝕋 𝕋 , 𝛿 𝕋 𝕋 are rd-continuous functions such that 𝜏 ( 𝑡 ) 𝑡 , 𝛿 ( 𝑡 ) 𝑡 , and l i m 𝑡 𝜏 ( 𝑡 ) = l i m 𝑡 𝛿 ( 𝑡 ) = .

As we are interested in oscillatory behavior, we assume throughout this paper that the given time-scale 𝕋 is unbounded above. We assume 𝑡 0 𝕋 and it is convenient to assume 𝑡 0 > 0 . We define the time-scale interval of the form [ 𝑡 0 , ) 𝕋 by [ 𝑡 0 , ) 𝕋 = [ 𝑡 0 , ) 𝕋 .

For the oscillation of neutral delay dynamic equations on time-scales, Mathsen et al. [26] considered the first-order neutral delay dynamic equations on time-scales

[ 𝑦 ] ( 𝑡 ) 𝑟 ( 𝑡 ) 𝑦 ( 𝜏 ( 𝑡 ) ) Δ + 𝑝 ( 𝑡 ) 𝑦 ( 𝛿 ( 𝑡 ) ) = 0 , 𝑡 𝕋 , ( 1 . 3 ) and established some new oscillation criteria of (1.3) which as a special case involve some well-known oscillation results for first-order neutral delay differential equations.

Agarwal et al. [7], Şahíner [28], Saker [31], Saker et al. [33], Wu et al. [34] studied the second-order nonlinear neutral delay dynamic equations on time-scales

𝑟 ( 𝑡 ) ( ( 𝑦 ( 𝑡 ) + 𝑝 ( 𝑡 ) 𝑦 ( 𝜏 ( 𝑡 ) ) ) Δ ) 𝛾 Δ + 𝑓 ( 𝑡 , 𝑦 ( 𝛿 ( 𝑡 ) ) ) = 0 , 𝑡 𝕋 , ( 1 . 4 ) by means of Riccati transformation technique, the authors established some oscillation criteria of (1.4).

Saker [32] investigated the second-order neutral Emden-Fowler delay dynamic equations on time-scales

𝑎 ( 𝑡 ) ( 𝑦 ( 𝑡 ) + 𝑟 ( 𝑡 ) 𝑦 ( 𝜏 ( 𝑡 ) ) ) Δ Δ + 𝑝 ( 𝑡 ) 𝑦 𝛾 ( 𝛿 ( 𝑡 ) ) = 0 , 𝑡 𝕋 , ( 1 . 5 ) and established some new oscillation for (1.5).

Our purpose in this paper is motivated by the question posed in [26]: What can be said about higher-order neutral dynamic equations on time-scales and the various generalizations? We refer the reader to the articles [23, 24] and we will consider the particular case when the order is 3, that is, (1.2). Set 𝑡 1 = m i n 𝑡 [ 𝑡 0 , ) 𝕋 { 𝜏 ( 𝑡 ) , 𝛿 ( 𝑡 ) } . By a solution of (1.2), we mean a nontrivial real-valued function 𝑥 𝐶 r d ( [ 𝑡 1 , ) 𝕋 , ) satisfying 𝑥 𝑎 𝑥 𝜏 𝐶 2 r d ( [ 𝑡 0 , ) 𝕋 , ) and 𝑟 ( 𝑥 𝑎 𝑥 𝜏 ) Δ Δ 𝐶 1 r d ( [ 𝑡 0 , ) 𝕋 , ) , and satisfying (1.2) for all 𝑡 [ 𝑡 0 , ) 𝕋 .

The paper is organized as follows. In Section 2, we apply a simple consequence of Keller’s chain rule, devoted to the proof of the sufficient conditions which guarantee that every solution of (1.2) oscillates or converges to zero. In Section 3, some examples are considered to illustrate the main results.

2. Main Results

In this section we give some new oscillation criteria for (1.2). In order to prove our main results, we will use the formula

( ( 𝑥 ( 𝑡 ) ) 𝛾 ) Δ = 𝛾 1 0 [ 𝑥 𝜎 ] ( 𝑡 ) + ( 1 ) 𝑥 ( 𝑡 ) 𝛾 1 𝑥 Δ ( 𝑡 ) d , ( 2 . 1 ) where 𝑥 is delta differentiable and eventually positive or eventually negative, which is a simple consequence of Keller’s chain rule (see Bohner and Peterson [4, Theorem 1 . 9 0 ]).

Before stating our main results, we begin with the following lemmas which are crucial in the proofs of the main results.

For the sake of convenience, we denote: 𝑧 ( 𝑡 ) = 𝑥 ( 𝑡 ) 𝑎 ( 𝑡 ) 𝑥 ( 𝜏 ( 𝑡 ) ) , for 𝑡 [ 𝑡 0 , ) 𝕋 . Also, we assume that

( 𝐻 ) there exists { 𝑐 𝑘 } 𝑘 0 𝕋 such that l i m 𝑘 𝑐 𝑘 = and 𝜏 ( 𝑐 𝑘 + 1 ) = 𝑐 𝑘 .

Lemma 2.1. Assume that ( 𝐻 ) holds. Further, assume that 𝑥 is an eventually positive solution of (1.2). If 𝑡 0 Δ 𝑡 𝑟 ( 𝑡 ) = , ( 2 . 2 ) then there are only the following three cases for 𝑡 𝑡 1 sufficiently large: ( i ) 𝑧 ( 𝑡 ) > 0 , 𝑧 Δ ( 𝑡 ) > 0 , 𝑧 Δ Δ ( 𝑡 ) > 0 , 𝑧 Δ Δ Δ ( 𝑡 ) < 0 , ( i i ) 𝑧 ( 𝑡 ) < 0 , 𝑧 Δ ( 𝑡 ) > 0 , 𝑧 Δ Δ ( 𝑡 ) > 0 , 𝑧 Δ Δ Δ ( 𝑡 ) < 0 , l i m 𝑡 𝑥 ( 𝑡 ) = 0 ,
or
( i i i ) 𝑧 ( 𝑡 ) > 0 , 𝑧 Δ ( 𝑡 ) < 0 , 𝑧 Δ Δ ( 𝑡 ) > 0 , 𝑧 Δ Δ Δ ( 𝑡 ) < 0 , l i m 𝑡 𝑧 ( 𝑡 ) = 𝑙 0 , l i m 𝑡 𝑥 ( 𝑡 ) = 𝑙 / ( 1 𝑎 ) 0 .

Proof. Let 𝑥 be an eventually positive solution of (1.2). Then there exists 𝑡 1 𝑡 0 such that 𝑥 ( 𝑡 ) > 0 , 𝑥 ( 𝜏 ( 𝑡 ) ) > 0 , and 𝑥 ( 𝛿 ( 𝑡 ) ) > 0 for all 𝑡 𝑡 1 . From (1.2) we have 𝑟 ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 ) Δ = 𝑝 ( 𝑡 ) 𝑥 𝛾 ( 𝛿 ( 𝑡 ) ) < 0 , 𝑡 𝑡 1 . ( 2 . 3 ) Hence 𝑟 ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 ) is strictly decreasing on [ 𝑡 1 , ) 𝕋 . We claim that 𝑧 Δ Δ ( 𝑡 ) > 0 eventually. Assume not, then there exists 𝑡 2 𝑡 1 such that 𝑟 ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 ) < 0 , 𝑡 𝑡 2 . ( 2 . 4 ) Then we can choose a negative 𝑐 and 𝑡 3 𝑡 2 such that 𝑟 ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 ) 𝑐 < 0 , 𝑡 𝑡 3 . ( 2 . 5 ) Dividing by 𝑟 ( 𝑡 ) and integrating from 𝑡 3 to 𝑡 , we have 𝑧 Δ ( 𝑡 ) 𝑧 Δ 𝑡 3 + 𝑐 𝑡 𝑡 3 Δ 𝑠 . 𝑟 ( 𝑠 ) ( 2 . 6 ) Letting 𝑡 , then 𝑧 Δ ( 𝑡 ) by (2.2). Thus, there is a 𝑡 4 𝑡 3 such that for 𝑡 𝑡 4 , 𝑧 Δ ( 𝑡 ) 𝑧 Δ 𝑡 4 < 0 . ( 2 . 7 ) Integrating the previous inequality from 𝑡 4 to 𝑡 , we obtain 𝑡 𝑧 ( 𝑡 ) 𝑧 4 𝑧 Δ 𝑡 4 𝑡 𝑡 4 . ( 2 . 8 ) Therefore, there exist 𝑑 > 0 and 𝑡 5 𝑡 4 such that 𝑥 ( 𝑡 ) 𝑑 + 𝑎 ( 𝑡 ) 𝑥 ( 𝜏 ( 𝑡 ) ) 𝑑 + 𝑎 0 𝑥 ( 𝜏 ( 𝑡 ) ) , 𝑡 𝑡 5 . ( 2 . 9 ) We can choose some positive integer 𝑘 0 such that 𝑐 𝑘 𝑡 5 , for 𝑘 𝑘 0 . Thus, we obtain 𝑥 𝑐 𝑘 𝑑 + 𝑎 0 𝑥 𝜏 𝑐 𝑘 = 𝑑 + 𝑎 0 𝑥 𝑐 𝑘 1 𝑑 𝑎 0 𝑑 + 𝑎 2 0 𝑥 𝜏 𝑐 𝑘 1 = 𝑑 𝑎 0 𝑑 + 𝑎 2 0 𝑥 𝑐 𝑘 2 𝑑 𝑎 0 𝑑 𝑎 𝑘 𝑘 0 0 1 𝑑 + 𝑎 𝑘 𝑘 0 0 𝑥 𝜏 𝑐 𝑘 0 + 1 = 𝑑 𝑎 0 𝑑 𝑎 𝑘 𝑘 0 0 1 𝑑 + 𝑎 𝑘 𝑘 0 0 𝑥 𝑐 𝑘 0 . ( 2 . 1 0 ) The above inequality implies that 𝑥 ( 𝑐 𝑘 ) < 0 for sufficiently large 𝑘 , which contradicts the fact that 𝑥 ( 𝑡 ) > 0 eventually. Hence we get 𝑧 Δ Δ ( 𝑡 ) > 0 . ( 2 . 1 1 ) It follows from this that either 𝑧 Δ ( 𝑡 ) > 0 or 𝑧 Δ ( 𝑡 ) < 0 . Since 𝑟 Δ ( 𝑡 ) 0 , 𝑟 ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 ) Δ = 𝑟 Δ ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 ) + 𝑟 𝜎 ( 𝑡 ) 𝑧 Δ Δ Δ ( 𝑡 ) < 0 , ( 2 . 1 2 ) which yields 𝑧 Δ Δ Δ ( 𝑡 ) < 0 . ( 2 . 1 3 ) If 𝑧 Δ ( 𝑡 ) > 0 , then there are two possible cases:(1) 𝑧 ( 𝑡 ) > 0 , eventually; or(2) 𝑧 ( 𝑡 ) < 0 , eventually.
If there exists a 𝑡 6 𝑡 1 such that case (2) holds, then l i m 𝑡 𝑧 ( 𝑡 ) exists, and l i m 𝑡 𝑧 ( 𝑡 ) = 𝑏 0 . We claim that l i m 𝑡 𝑧 ( 𝑡 ) = 0 . Otherwise, l i m 𝑡 𝑧 ( 𝑡 ) = 𝑏 < 0 . We can choose some positive integer 𝑘 0 such that 𝑐 𝑘 𝑡 6 , for 𝑘 𝑘 0 . Thus, we obtain
𝑥 𝑐 𝑘 𝑎 0 𝑥 𝜏 𝑐 𝑘 = 𝑎 0 𝑥 𝑐 𝑘 1 𝑎 2 0 𝑥 𝜏 𝑐 𝑘 1 = 𝑎 2 0 𝑥 𝑐 𝑘 2 𝑎 𝑘 𝑘 0 0 𝑥 𝜏 𝑐 𝑘 0 + 1 = 𝑎 𝑘 𝑘 0 0 𝑥 𝑐 𝑘 0 , ( 2 . 1 4 ) which implies that l i m 𝑘 𝑥 ( 𝑐 𝑘 ) = 0 , and from the definition of 𝑧 ( 𝑡 ) , we have l i m 𝑘 𝑧 ( 𝑐 𝑘 ) = 0 , which contradicts l i m 𝑡 𝑧 ( 𝑡 ) < 0 . Now, we assert that 𝑥 is bounded. If it is not true, there exists { 𝑠 𝑘 } 𝑘 [ 𝑡 6 , ) 𝕋 with 𝑠 𝑘 as 𝑘 such that 𝑥 𝑠 𝑘 = s u p 𝑡 0 𝑠 𝑠 𝑘 𝑥 ( 𝑠 ) , l i m 𝑘 𝑥 𝑠 𝑘 = . ( 2 . 1 5 ) From 𝜏 ( 𝑡 ) 𝑡 𝑧 𝑠 𝑘 𝑠 = 𝑥 𝑘 𝑠 𝑎 𝑘 𝑥 𝜏 𝑠 𝑘 1 𝑎 0 𝑥 𝑠 𝑘 , ( 2 . 1 6 ) which implies that l i m 𝑘 𝑧 ( 𝑠 𝑘 ) = , it contradicts that l i m 𝑡 𝑧 ( 𝑡 ) = 0 . Therefore, we can assume that l i m s u p 𝑡 𝑥 ( 𝑡 ) = 𝑥 1 , l i m i n f 𝑡 𝑥 ( 𝑡 ) = 𝑥 2 . ( 2 . 1 7 ) By 0 𝑎 < 1 , we get 𝑥 1 𝑎 𝑥 1 0 𝑥 2 𝑎 𝑥 2 , ( 2 . 1 8 ) which implies that 𝑥 1 𝑥 2 , so 𝑥 1 = 𝑥 2 , hence, l i m 𝑡 𝑥 ( 𝑡 ) = 0 .
Assume that 𝑧 Δ ( 𝑡 ) < 0 . We claim that 𝑧 ( 𝑡 ) 0 eventually. Otherwise, we have l i m 𝑡 𝑧 ( 𝑡 ) < 0 or l i m 𝑡 𝑧 ( 𝑡 ) = . By ( 𝐻 ) , there exists 𝑡 7 𝑡 1 , we can choose some positive integer 𝑘 0 such that 𝑐 𝑘 𝑡 7 for 𝑘 𝑘 0 , and we obtain
𝑥 𝑐 𝑘 𝑎 0 𝑥 𝜏 𝑐 𝑘 = 𝑎 0 𝑥 𝑐 𝑘 1 𝑎 2 0 𝑥 𝜏 𝑐 𝑘 1 = 𝑎 2 0 𝑥 𝑐 𝑘 2 𝑎 𝑘 𝑘 0 0 𝑥 𝜏 𝑐 𝑘 0 + 1 = 𝑎 𝑘 𝑘 0 0 𝑥 𝑐 𝑘 0 , ( 2 . 1 9 ) which implies that l i m 𝑘 𝑥 ( 𝑐 𝑘 ) = 0 , and from the definition of 𝑧 , we have l i m 𝑘 𝑧 ( 𝑐 𝑘 ) = 0 , which contradicts l i m 𝑡 𝑧 ( 𝑡 ) < 0 or l i m 𝑡 𝑧 ( 𝑡 ) = . Now, we have that l i m 𝑡 𝑧 ( 𝑡 ) = 𝑙 0 , here 𝑙 is finite. We assert that 𝑥 is bounded. If it is not true, there exists { 𝑠 𝑘 } 𝑘 [ 𝑡 6 , ) 𝕋 with 𝑠 𝑘 as 𝑘 such that 𝑥 𝑠 𝑘 = s u p 𝑡 0 𝑠 𝑠 𝑘 𝑥 ( 𝑠 ) , l i m 𝑘 𝑥 𝑠 𝑘 = . ( 2 . 2 0 ) From 𝜏 ( 𝑡 ) 𝑡 𝑧 𝑠 𝑘 𝑠 = 𝑥 𝑘 𝑠 𝑎 𝑘 𝑥 𝜏 𝑠 𝑘 1 𝑎 0 𝑥 𝑠 𝑘 , ( 2 . 2 1 ) which implies that l i m 𝑘 𝑧 ( 𝑠 𝑘 ) = , it contradicts that l i m 𝑡 𝑧 ( 𝑡 ) = 𝑙 0 . Therefore, we can assume that l i m s u p 𝑡 𝑥 ( 𝑡 ) = 𝑥 1 , l i m i n f 𝑡 𝑥 ( 𝑡 ) = 𝑥 2 . ( 2 . 2 2 ) By 0 𝑎 < 1 , we get 𝑥 1 𝑎 𝑥 1 𝑙 𝑥 2 𝑎 𝑥 2 , ( 2 . 2 3 ) which implies that 𝑥 1 𝑥 2 , so 𝑥 1 = 𝑥 2 , hence, l i m 𝑡 𝑥 ( 𝑡 ) = 𝑙 / ( 1 𝑎 ) 0 . This completes the proof.

In [4, Section 1 . 6 ] the Taylor monomials { 𝑛 ( 𝑡 , 𝑠 ) } 𝑛 = 0 are defined recursively by

0 ( 𝑡 , 𝑠 ) = 1 , 𝑛 + 1 ( 𝑡 , 𝑠 ) = 𝑡 𝑠 𝑛 ( 𝜏 , 𝑠 ) Δ 𝜏 , 𝑡 , 𝑠 𝕋 , 𝑛 1 . ( 2 . 2 4 ) It follows from [4, Section 1 . 6 ] that 1 ( 𝑡 , 𝑠 ) = 𝑡 𝑠 for any time-scale, but simple formulas in general do not hold for 𝑛 2 .

Lemma 2.2 (see [15, Lemma 4 ]). Assume that 𝑧 satisfies case (i) of Lemma 2.1. Then l i m i n f 𝑡 𝑡 𝑧 ( 𝑡 ) 2 𝑡 , 𝑡 0 𝑧 Δ ( 𝑡 ) 1 . ( 2 . 2 5 )

Lemma 2.3. Assume that 𝑥 is a solution of (1.2) satisfying case (i) of Lemma 2.1. If 𝑡 0 𝑝 ( 𝑡 ) 2 𝛿 ( 𝑡 ) , 𝑡 0 𝛾 Δ 𝑡 = , ( 2 . 2 6 ) then 𝑧 satisfies eventually 𝑧 Δ ( 𝑡 ) 𝑡 𝑧 Δ Δ ( 𝑧 𝑡 ) , Δ ( 𝑡 ) 𝑡 i s n o n i n c r e a s i n g . ( 2 . 2 7 )

Proof. Let 𝑥 be a solution of (1.2) such that case ( i ) of Lemma 2.1 holds for 𝑡 𝑡 1 . Define 𝑍 ( 𝑡 ) = 𝑧 Δ ( 𝑡 ) 𝑡 𝑧 Δ Δ ( 𝑡 ) . ( 2 . 2 8 ) Thus 𝑍 Δ ( 𝑡 ) = 𝜎 ( 𝑡 ) 𝑧 Δ Δ Δ ( 𝑡 ) > 0 . ( 2 . 2 9 ) We claim that 𝑍 ( 𝑡 ) > 0 eventually. Otherwise, there exists 𝑡 2 𝑡 1 such that 𝑍 ( 𝑡 ) < 0 for 𝑡 𝑡 2 . Therefore, 𝑧 Δ ( 𝑡 ) 𝑡 Δ = 𝑍 ( 𝑡 ) 𝑡 𝜎 ( 𝑡 ) > 0 , 𝑡 𝑡 2 , ( 2 . 3 0 ) which implies that 𝑧 Δ ( 𝑡 ) / 𝑡 is strictly increasing on [ 𝑡 2 , ) 𝕋 . Pick 𝑡 3 𝑡 2 such that 𝛿 ( 𝑡 ) 𝛿 ( 𝑡 3 ) 𝑡 2 , for 𝑡 𝑡 3 . Then we have 𝑧 Δ ( 𝛿 ( 𝑡 ) ) 𝑧 𝛿 ( 𝑡 ) Δ 𝛿 𝑡 3 𝛿 𝑡 3 = 𝑃 > 0 , ( 2 . 3 1 ) then 𝑧 Δ ( 𝛿 ( 𝑡 ) ) 𝑃 𝛿 ( 𝑡 ) for 𝑡 𝑡 3 . By Lemma 2.2, for any 0 < 𝑘 < 1 , there exists 𝑡 4 𝑡 3 such that 𝑧 ( 𝑡 ) 𝑧 Δ ( 𝑡 ) 𝑘 2 𝑡 , 𝑡 0 𝑡 , 𝑡 𝑡 4 . ( 2 . 3 2 ) Hence there exists 𝑡 5 𝑡 4 so that 𝑧 ( 𝛿 ( 𝑡 ) ) 𝑘 2 𝛿 ( 𝑡 ) , 𝑡 0 𝑧 𝛿 ( 𝑡 ) Δ ( 𝛿 ( 𝑡 ) ) 𝑃 𝑘 2 𝛿 ( 𝑡 ) , 𝑡 0 𝛿 ( 𝑡 ) 𝛿 ( 𝑡 ) = 𝑃 𝑘 2 𝛿 ( 𝑡 ) , 𝑡 0 , 𝑡 𝑡 5 . ( 2 . 3 3 ) By the definition of 𝑧 , we have that 𝑥 ( 𝑡 ) 𝑧 ( 𝑡 ) . ( 2 . 3 4 ) From (1.2), we obtain 𝑟 ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 ) Δ + 𝑝 ( 𝑡 ) 𝑧 𝛾 ( 𝛿 ( 𝑡 ) ) 0 . ( 2 . 3 5 ) Integrating both sides of (2.35) from 𝑡 5 to 𝑡 , we get 𝑟 ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 𝑡 ) 𝑟 5 𝑧 Δ Δ 𝑡 5 + ( 𝑃 𝑘 ) 𝛾 𝑡 𝑡 5 𝑝 ( 𝑠 ) 2 𝛿 ( 𝑠 ) , 𝑡 0 𝛾 Δ 𝑠 0 , ( 2 . 3 6 ) which yields that 𝑟 𝑡 5 𝑧 Δ Δ 𝑡 5 ( 𝑃 𝑘 ) 𝛾 𝑡 𝑡 5 𝑝 ( 𝑠 ) 2 𝛿 ( 𝑠 ) , 𝑡 0 𝛾 Δ 𝑠 , ( 2 . 3 7 ) which contradicts (2.26). Hence 𝑍 ( 𝑡 ) > 0 and 𝑧 Δ ( 𝑡 ) / 𝑡 is nonincreasing. The proof is complete.

Lemma 2.4. Assume that ( 𝐻 ) holds and 𝑥 is a solution of (1.2) which satisfies case (iii) of Lemma 2.1. If 𝑡 0 𝑝 ( 𝑠 ) 𝑅 𝜎 ( 𝑠 ) Δ 𝑠 = , ( 2 . 3 8 ) where 𝑅 ( 𝑡 ) = 𝑡 𝑡 0 ( 𝜎 ( 𝑢 ) / 𝑟 ( 𝑢 ) ) Δ 𝑢 for 𝑡 [ 𝑡 0 , ) 𝕋 , then l i m 𝑡 𝑥 ( 𝑡 ) = 0 .

Proof. Let 𝑥 be a solution of (1.2) such that case ( i i i ) of Lemma 2.1 holds for 𝑡 𝑡 1 . Then l i m 𝑡 𝑧 ( 𝑡 ) = 𝑙 0 , l i m 𝑡 𝑥 ( 𝑡 ) = 𝑙 / ( 1 𝑎 ) 0 . Next we claim that 𝑙 = 0 . Otherwise, there exists 𝑡 2 𝑡 1 such that 𝑧 ( 𝛿 ( 𝑡 ) ) 𝑙 > 0 for all 𝑡 𝑡 2 . By the definition of 𝑧 , we have that (2.35) holds. Integrating both sides of (2.35) from 𝑡 to , we get 𝑧 Δ Δ 1 ( 𝑡 ) 𝑟 ( 𝑡 ) 𝑡 𝑝 ( 𝑠 ) 𝑧 𝛾 ( 𝛿 ( 𝑠 ) ) Δ 𝑠 . ( 2 . 3 9 ) Integrating again from 𝑡 to , we have 𝑧 Δ ( 𝑡 ) 𝑡 1 𝑟 ( 𝑢 ) 𝑢 𝑝 ( 𝑠 ) 𝑧 𝛾 ( 𝛿 ( 𝑠 ) ) Δ 𝑠 Δ 𝑢 . ( 2 . 4 0 ) Integrating again from 𝑡 2 to , we obtain 𝑧 𝑡 1 𝑡 2 𝑣 1 𝑟 ( 𝑢 ) 𝑢 𝑝 ( 𝑠 ) 𝑧 𝛾 ( 𝛿 ( 𝑠 ) ) Δ 𝑠 Δ 𝑢 Δ 𝑣 𝑙 𝛾 𝑡 2 𝑣 1 𝑟 ( 𝑢 ) 𝑢 𝑝 ( 𝑠 ) Δ 𝑠 Δ 𝑢 Δ 𝑣 , ( 2 . 4 1 ) which contradicts (2.38), since by [23, Lemma 1 ] and [3, Remark 4 . 7 ], we get 𝑡 0 𝑣 1 𝑟 ( 𝑢 ) 𝑢 = 𝑝 ( 𝑠 ) Δ 𝑠 Δ 𝑢 Δ 𝑣 𝑡 0 𝑣 𝑢 1 = 𝑟 ( 𝑢 ) 𝑝 ( 𝑠 ) Δ 𝑠 Δ 𝑢 Δ 𝑣 𝑡 0 𝑣 𝑣 𝜎 ( 𝑠 ) 1 𝑝 𝑟 ( 𝑢 ) ( 𝑠 ) Δ 𝑢 Δ 𝑠 Δ 𝑣 = 𝑡 0 𝑡 𝜎 ( 𝑠 ) 0 𝑣 𝜎 ( 𝑠 ) 1 𝑝 = 𝑟 ( 𝑢 ) ( 𝑠 ) Δ 𝑢 Δ 𝑣 Δ 𝑠 𝑡 0 𝑝 ( 𝑠 ) 𝑡 𝜎 ( 𝑠 ) 0 𝑣 𝜎 ( 𝑠 ) 1 𝑟 ( 𝑢 ) Δ 𝑢 Δ 𝑣 Δ 𝑠 = 𝑡 0 𝑝 ( 𝑠 ) 𝑡 𝜎 ( 𝑠 ) 0 𝑡 𝜎 ( 𝑢 ) 0 1 = 𝑟 ( 𝑢 ) Δ 𝑣 Δ 𝑢 Δ 𝑠 𝑡 0 𝑝 ( 𝑠 ) 𝑡 𝜎 ( 𝑠 ) 0 1 𝑟 ( 𝑢 ) 𝑡 𝜎 ( 𝑢 ) 0 = Δ 𝑣 Δ 𝑢 Δ 𝑠 𝑡 0 𝑝 ( 𝑠 ) 𝑡 𝜎 ( 𝑠 ) 0 𝜎 ( 𝑢 ) 𝑡 0 𝑟 ( 𝑢 ) Δ 𝑢 Δ 𝑠 = 𝑡 0 𝑝 ( 𝑠 ) 𝑡 𝜎 ( 𝑠 ) 0 𝜎 ( 𝑢 ) 𝑟 ( 𝑢 ) Δ 𝑢 Δ 𝑠 = 𝑡 0 𝑝 ( 𝑠 ) 𝑅 𝜎 ( 𝑠 ) Δ 𝑠 . ( 2 . 4 2 ) Hence l i m 𝑡 𝑥 ( 𝑡 ) = 0 and completes the proof.

Theorem 2.5. Assume that ( 𝐻 ) , (2.2), (2.26), and (2.38) hold, 𝛾 1 . Furthermore, assume that there exists a positive function 𝜂 𝐶 1 r d ( [ 𝑡 0 , ) 𝕋 , ) such that for some 0 < 𝑘 < 1 and for all constants 𝑀 > 0 l i m s u p 𝑡 𝑡 𝑡 0 𝜂 𝜂 ( 𝑠 ) 𝑝 ( 𝑠 ) 𝜁 ( 𝑠 ) 𝑟 ( 𝑠 ) Δ ( 𝑠 ) 2 4 𝑘 𝛾 𝑀 𝛾 1 𝜂 ( 𝑠 ) Δ 𝑠 = , ( 2 . 4 3 ) where 𝜁 ( 𝑡 ) = ( 2 ( 𝛿 ( 𝑡 ) , 𝑡 0 ) / 𝑡 ) 𝛾 . Then every solution 𝑥 of (1.2) oscillates or l i m 𝑡 𝑥 ( 𝑡 ) = 0 .

Proof. Suppose that (1.2) has a nonoscillatory solution 𝑥 . We may assume without loss of generality that 𝑥 ( 𝑡 ) > 0 , 𝑥 ( 𝜏 ( 𝑡 ) ) > 0 , and 𝑥 ( 𝛿 ( 𝑡 ) ) > 0 for all 𝑡 [ 𝑡 1 , ) 𝕋 , 𝑡 1 [ 𝑡 0 , ) 𝕋 . Then by Lemma 2.1, 𝑧 satisfies three cases. Assume that 𝑧 satisfies case ( i ) . Define the function 𝜔 by 𝜔 ( 𝑡 ) = 𝜂 ( 𝑡 ) 𝑟 ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 ) 𝑧 Δ ( 𝑡 ) 𝛾 𝑡 , 𝑡 1 , 𝕋 . ( 2 . 4 4 ) Then 𝜔 ( 𝑡 ) > 0 . Using the product rule, we have 𝜔 Δ 𝑟 ( 𝑡 ) = ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 ) 𝜎 𝜂 ( 𝑡 ) 𝑧 Δ ( 𝑡 ) 𝛾 Δ + 𝑟 ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 ) Δ 𝜂 ( 𝑡 ) 𝑧 Δ ( 𝑡 ) 𝛾 . ( 2 . 4 5 ) By the quotient rule, we get 𝜔 Δ ( 𝑡 ) = 𝑟 ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 ) 𝜎 𝜂 Δ 𝑧 ( 𝑡 ) Δ ( 𝑡 ) 𝛾 𝑧 𝜂 ( 𝑡 ) Δ ( 𝑡 ) 𝛾 Δ 𝑧 Δ ( 𝑡 ) 𝛾 𝑧 Δ 𝜎 ( 𝑡 ) 𝛾 + 𝑟 ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 ) Δ 𝜂 ( 𝑡 ) 𝑧 Δ ( 𝑡 ) 𝛾 . ( 2 . 4 6 ) By the definition of 𝑧 and (1.2), we obtain (2.35). From (2.35) and (2.44), we have 𝜔 Δ 𝜂 ( 𝑡 ) Δ ( 𝑡 ) 𝜂 𝜎 𝜔 ( 𝑡 ) 𝜎 𝑧 ( 𝑡 ) 𝜂 ( 𝑡 ) 𝑝 ( 𝑡 ) 𝛾 ( 𝛿 ( 𝑡 ) ) 𝑧 Δ ( 𝑡 ) 𝛾 𝜂 ( 𝑡 ) 𝑟 ( 𝑡 ) 𝑧 Δ Δ ( 𝑡 ) 𝜎 𝑧 Δ ( 𝑡 ) 𝛾 Δ 𝑧 Δ ( 𝑡 ) 𝛾 𝑧 Δ 𝜎 ( 𝑡 ) 𝛾 , ( 2 . 4 7 ) from (2.25) and (2.27), for any 0 < 𝑘 < 1 , we obtain 𝑧 𝛾 ( 𝛿 ( 𝑡 ) ) 𝑧 Δ ( 𝑡 ) 𝛾 = 𝑧 𝛾 ( 𝛿 ( 𝑡 ) ) 𝑧 Δ ( 𝛿 ( 𝑡 ) ) 𝛾 𝑧 Δ ( 𝛿 ( 𝑡 ) ) 𝛾 𝑧 Δ ( 𝑡 ) 𝛾 𝑘 1 / 𝛾 2 ( 𝛿 ( 𝑡 ) , 𝑡 0 ) 𝛿 ( 𝑡 ) 𝛾 𝛿 ( 𝑡 ) 𝑡 𝛾 = 𝑘 2 ( 𝛿 ( 𝑡 ) , 𝑡 0 ) 𝑡 𝛾 , ( 2 . 4 8 ) hence by (2.48), we have 𝜔 Δ 𝜂 ( 𝑡 ) Δ ( 𝑡 ) 𝜂 𝜎 𝜔 ( 𝑡 ) 𝜎 ( 𝑡 ) 𝑘 𝜂 ( 𝑡 ) 𝑝 ( 𝑡 ) 𝜁 ( 𝑡 ) 𝜂 ( 𝑡