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Abstract and Applied Analysis
Volume 2011 (2011), Article ID 346745, 9 pages
http://dx.doi.org/10.1155/2011/346745
Research Article

Existence of Nonoscillatory Solutions of First-Order Neutral Differential Equations

Department of Mathematics, University of Žilina, 010 26 Žilina, Slovakia

Received 4 January 2011; Revised 4 March 2011; Accepted 27 April 2011

Academic Editor: Josef Diblík

Copyright © 2011 Božena Dorociaková 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

This paper contains some sufficient conditions for the existence of positive solutions which are bounded below and above by positive functions for the first-order nonlinear neutral differential equations. These equations can also support the existence of positive solutions approaching zero at infinity

1. Introduction

This paper is concerned with the existence of a positive solution of the neutral differential equations of the form𝑑[]𝑑𝑡𝑥(𝑡)𝑎(𝑡)𝑥(𝑡𝜏)=𝑝(𝑡)𝑓(𝑥(𝑡𝜎)),𝑡𝑡0,(1.1) where 𝜏>0,𝜎0,𝑎𝐶([𝑡0,),(0,)),𝑝𝐶(𝑅,(0,)),𝑓𝐶(𝑅,𝑅),𝑓 is nondecreasing function, and 𝑥𝑓(𝑥)>0,𝑥0.

By a solution of (1.1) we mean a function 𝑥𝐶([𝑡1𝑚,),𝑅),𝑚=max{𝜏,𝜎}, for some 𝑡1𝑡0, such that 𝑥(𝑡)𝑎(𝑡)𝑥(𝑡𝜏) is continuously differentiable on [𝑡1,) and such that (1.1) is satisfied for 𝑡𝑡1.

The problem of the existence of solutions of neutral differential equations has been studied by several authors in the recent years. For related results we refer the reader to [111] and the references cited therein. However there is no conception which guarantees the existence of positive solutions which are bounded below and above by positive functions. In this paper we have presented some conception. The method also supports the existence of positive solutions approaching zero at infinity.

As much as we know, for (1.1) in the literature, there is no result for the existence of solutions which are bounded by positive functions. Only the existence of solutions which are bounded by constants is treated, for example, in [6, 10, 11]. It seems that conditions of theorems are rather complicated, but cannot be simpler due to Corollaries 2.3, 2.6, and 3.2.

The following fixed point theorem will be used to prove the main results in the next section.

Lemma 1.1 ([see [6, 10] Krasnoselskii's fixed point theorem]). Let 𝑋 be a Banach space, let Ω be a bounded closed convex subset of 𝑋, and let 𝑆1,𝑆2 be maps of Ω into 𝑋 such that 𝑆1𝑥+𝑆2𝑦Ω for every pair 𝑥,𝑦Ω. If 𝑆1 is contractive and 𝑆2 is completely continuous, then the equation 𝑆1𝑥+𝑆2𝑥=𝑥(1.2) has a solution in Ω.

2. The Existence of Positive Solution

In this section we will consider the existence of a positive solution for (1.1). The next theorem gives us the sufficient conditions for the existence of a positive solution which is bounded by two positive functions.

Theorem 2.1. Suppose that there exist bounded functions 𝑢,𝑣𝐶1([𝑡0,),(0,)), constant 𝑐>0 and 𝑡1𝑡0+𝑚 such that 𝑢(𝑡)𝑣(𝑡),𝑡𝑡0,𝑣𝑡(2.1)(𝑡)𝑣1𝑡𝑢(𝑡)+𝑢10,𝑡0𝑡𝑡1,1(2.2)𝑢(𝑡𝜏)𝑢(𝑡)+𝑡1𝑝(𝑠)𝑓(𝑣(𝑠𝜎))𝑑𝑠𝑎(𝑡)𝑣𝑣(𝑡𝜏)(𝑡)+𝑡𝑝(𝑠)𝑓(𝑢(𝑠𝜎))𝑑𝑠𝑐<1,𝑡𝑡1.(2.3) Then (1.1) has a positive solution which is bounded by functions 𝑢,𝑣.

Proof. Let 𝐶([𝑡0,),𝑅) be the set of all continuous bounded functions with the norm 𝑥=sup𝑡𝑡0|𝑥(𝑡)|. Then 𝐶([𝑡0,),𝑅) is a Banach space. We define a closed, bounded, and convex subset Ω of 𝐶([𝑡0,),𝑅) as follows: 𝑡Ω=𝑥=𝑥(𝑡)𝐶0,,𝑅𝑢(𝑡)𝑥(𝑡)𝑣(𝑡),𝑡𝑡0.(2.4) We now define two maps 𝑆1 and 𝑆2Ω𝐶([𝑡0,),𝑅) as follows: 𝑆1𝑥(𝑡)=𝑎(𝑡)𝑥(𝑡𝜏),𝑡𝑡1,𝑆1𝑥𝑡1,𝑡0𝑡𝑡1,𝑆2𝑥(𝑡)=𝑡𝑝(𝑠)𝑓(𝑥(𝑠𝜎))𝑑𝑠,𝑡𝑡1,𝑆2𝑥𝑡1𝑡+𝑣(𝑡)𝑣1,𝑡0𝑡𝑡1.(2.5) We will show that for any 𝑥,𝑦Ω we have 𝑆1𝑥+𝑆2𝑦Ω. For every 𝑥,𝑦Ω and 𝑡𝑡1, we obtain 𝑆1𝑥𝑆(𝑡)+2𝑦(𝑡)𝑎(𝑡)𝑣(𝑡𝜏)𝑡𝑝(𝑠)𝑓(𝑢(𝑠𝜎))𝑑𝑠𝑣(𝑡).(2.6) For 𝑡[𝑡0,𝑡1], we have 𝑆1𝑥𝑆(𝑡)+2𝑦𝑆(𝑡)=1𝑥𝑡1+𝑆2𝑦𝑡1𝑡+𝑣(𝑡)𝑣1𝑡𝑣1𝑡+𝑣(𝑡)𝑣1=𝑣(𝑡).(2.7) Furthermore, for 𝑡𝑡1, we get 𝑆1𝑥𝑆(𝑡)+2𝑦(𝑡)𝑎(𝑡)𝑢(𝑡𝜏)𝑡𝑝(𝑠)𝑓(𝑣(𝑠𝜎))𝑑𝑠𝑢(𝑡).(2.8) Let 𝑡[𝑡0,𝑡1]. With regard to (2.2), we get 𝑣𝑡(𝑡)𝑣1𝑡+𝑢1𝑢(𝑡),𝑡0𝑡𝑡1.(2.9) Then for 𝑡[𝑡0,𝑡1] and any 𝑥,𝑦Ω, we obtain 𝑆1𝑥𝑆(𝑡)+2𝑦𝑆(𝑡)=1𝑥𝑡1+𝑆2𝑦𝑡1𝑡+𝑣(𝑡)𝑣1𝑡𝑢1𝑡+𝑣(𝑡)𝑣1𝑢(𝑡).(2.10) Thus we have proved that 𝑆1𝑥+𝑆2𝑦Ω for any 𝑥,𝑦Ω.
We will show that 𝑆1 is a contraction mapping on Ω. For 𝑥,𝑦Ω and 𝑡𝑡1 we have||𝑆1𝑥𝑆(𝑡)1𝑦||=||𝑎||(𝑡)(𝑡)𝑥(𝑡𝜏)𝑦(𝑡𝜏)𝑐𝑥𝑦.(2.11) This implies that 𝑆1𝑥𝑆1𝑦𝑐𝑥𝑦.(2.12) Also for 𝑡[𝑡0,𝑡1], the previous inequality is valid. We conclude that 𝑆1 is a contraction mapping on Ω.
We now show that 𝑆2 is completely continuous. First we will show that 𝑆2 is continuous. Let 𝑥𝑘=𝑥𝑘(𝑡)Ω be such that 𝑥𝑘(𝑡)𝑥(𝑡) as 𝑘. Because Ω is closed, 𝑥=𝑥(𝑡)Ω. For 𝑡𝑡1 we have||𝑆2𝑥𝑘𝑆(𝑡)2𝑥||||||(𝑡)𝑡𝑓𝑥𝑝(𝑠)𝑘||||(𝑠𝜎)𝑓(𝑥(𝑠𝜎))𝑑𝑠𝑡1||𝑓𝑥𝑝(𝑠)𝑘||(𝑠𝜎)𝑓(𝑥(𝑠𝜎))𝑑𝑠.(2.13) According to (2.8), we get 𝑡1𝑝(𝑠)𝑓(𝑣(𝑠𝜎))𝑑𝑠<.(2.14) Since |𝑓(𝑥𝑘(𝑠𝜎))𝑓(𝑥(𝑠𝜎))|0 as 𝑘, by applying the Lebesgue dominated convergence theorem, we obtain lim𝑘𝑆2𝑥𝑘𝑆(𝑡)2𝑥(𝑡)=0.(2.15) This means that 𝑆2 is continuous.
We now show that 𝑆2Ω is relatively compact. It is sufficient to show by the Arzela-Ascoli theorem that the family of functions {𝑆2𝑥𝑥Ω} is uniformly bounded and equicontinuous on [𝑡0,). The uniform boundedness follows from the definition of Ω. For the equicontinuity we only need to show, according to Levitans result [7], that for any given 𝜀>0 the interval [𝑡0,) can be decomposed into finite subintervals in such a way that on each subinterval all functions of the family have a change of amplitude less than 𝜀. Then with regard to condition (2.14), for 𝑥Ω and any 𝜀>0, we take 𝑡𝑡1 large enough so that𝑡𝜀𝑝(𝑠)𝑓(𝑥(𝑠𝜎))𝑑𝑠<2.(2.16) Then, for 𝑥Ω,𝑇2>𝑇1𝑡, we have ||𝑆2𝑥𝑇2𝑆2𝑥𝑇1||𝑇2+𝑝(𝑠)𝑓(𝑥(𝑠𝜎))𝑑𝑠𝑇1𝜀𝑝(𝑠)𝑓(𝑥(𝑠𝜎))𝑑𝑠<2+𝜀2=𝜀.(2.17) For 𝑥Ω and 𝑡1𝑇1<𝑇2𝑡, we get ||𝑆2𝑥𝑇2𝑆2𝑥𝑇1||𝑇2𝑇1𝑝(𝑠)𝑓(𝑥(𝑠𝜎))𝑑𝑠max𝑡1𝑠𝑡𝑇{𝑝(𝑠)𝑓(𝑥(𝑠𝜎))}2𝑇1.(2.18) Thus there exists 𝛿1=𝜀/𝑀, where 𝑀=max𝑡1𝑠𝑡{𝑝(𝑠)𝑓(𝑥(𝑠𝜎))}, such that ||𝑆2𝑥𝑇2𝑆2𝑥𝑇1||<𝜀if0<𝑇2𝑇1<𝛿1.(2.19) Finally for any 𝑥Ω,𝑡0𝑇1<𝑇2𝑡1, there exists a 𝛿2>0 such that ||𝑆2𝑥𝑇2𝑆2𝑥𝑇1||=||𝑣𝑇1𝑇𝑣2||=||||𝑇2𝑇1𝑣||||(𝑠)𝑑𝑠max𝑡0𝑠𝑡1||𝑣(||𝑇𝑠)2𝑇1<𝜀if0<𝑇2𝑇1<𝛿2.(2.20) Then {𝑆2𝑥𝑥Ω} is uniformly bounded and equicontinuous on [𝑡0,), and hence 𝑆2Ω is relatively compact subset of 𝐶([𝑡0,),𝑅). By Lemma 1.1 there is an 𝑥0Ω such that 𝑆1𝑥0+𝑆2𝑥0=𝑥0. We conclude that 𝑥0(𝑡) is a positive solution of (1.1). The proof is complete.

Corollary 2.2. Suppose that there exist functions 𝑢,𝑣𝐶1([𝑡0,),(0,)), constant 𝑐>0 and 𝑡1𝑡0+𝑚 such that (2.1), (2.3) hold and 𝑣(𝑡)𝑢(𝑡)0,𝑡0𝑡𝑡1.(2.21) Then (1.1) has a positive solution which is bounded by the functions 𝑢,𝑣.

Proof. We only need to prove that condition (2.21) implies (2.2). Let 𝑡[𝑡0,𝑡1] and set 𝐻𝑡(𝑡)=𝑣(𝑡)𝑣1𝑡𝑢(𝑡)+𝑢1.(2.22) Then with regard to (2.21), it follows that 𝐻(𝑡)=𝑣(𝑡)𝑢(𝑡)0,𝑡0𝑡𝑡1.(2.23) Since 𝐻(𝑡1)=0 and 𝐻(𝑡)0 for 𝑡[𝑡0,𝑡1], this implies that 𝐻𝑡(𝑡)=𝑣(𝑡)𝑣1𝑡𝑢(𝑡)+𝑢10,𝑡0𝑡𝑡1.(2.24) Thus all conditions of Theorem 2.1 are satisfied.

Corollary 2.3. Suppose that there exists a function 𝑣𝐶1([𝑡0,),(0,)), constant 𝑐>0 and 𝑡1𝑡0+𝑚 such that 1𝑎(𝑡)=𝑣(𝑡𝜏)𝑣(𝑡)+𝑡𝑝(𝑠)𝑓(𝑣(𝑠𝜎))𝑑𝑠𝑐<1,𝑡𝑡1.(2.25) Then (1.1) has a solution 𝑥(𝑡)=𝑣(𝑡),𝑡𝑡1.

Proof. We put 𝑢(𝑡)=𝑣(𝑡) and apply Theorem 2.1.

Theorem 2.4. Suppose that there exist functions 𝑢,𝑣𝐶1([𝑡0,),(0,)), constant 𝑐>0 and 𝑡1𝑡0+𝑚 such that (2.1), (2.2), and (2.3) hold and lim𝑡𝑣(𝑡)=0.(2.26) Then (1.1) has a positive solution which is bounded by the functions 𝑢,𝑣 and tends to zero.

Proof. The proof is similar to that of Theorem 2.1 and we omit it.

Corollary 2.5. Suppose that there exist functions 𝑢,𝑣𝐶1([𝑡0,),(0,)), constant 𝑐>0 and 𝑡1𝑡0+𝑚 such that (2.1), (2.3), (2.21), and (2.26) hold. Then (1.1) has a positive solution which is bounded by the functions 𝑢,𝑣 and tends to zero.

Proof. The proof is similar to that of Corollary 2.2, and we omitted it.

Corollary 2.6. Suppose that there exists a function 𝑣𝐶1([𝑡0,),(0,)), constant 𝑐>0 and 𝑡1𝑡0+𝑚 such that (2.25), (2.26) hold. Then (1.1) has a solution 𝑥(𝑡)=𝑣(𝑡),𝑡𝑡1 which tends to zero.

Proof. We put 𝑢(𝑡)=𝑣(𝑡) and apply Theorem 2.4.

3. Applications and Examples

In this section we give some applications of the theorems above.

Theorem 3.1. Suppose that 𝑡0𝑝(𝑡)𝑑𝑡=,(3.1)0<𝑘1𝑘2 and there exist constants 𝑐>0,𝛾0,𝑡1𝑡0+𝑚 such that 𝑘1𝑘2𝑘exp2𝑘1𝑡0𝑡0𝛾𝑝(𝑡)𝑑𝑡1,(3.2)exp𝑘2𝑡𝑡𝜏𝑘𝑝(𝑠)𝑑𝑠+exp2𝑡𝑡𝜏0𝛾×𝑝(𝑠)𝑑𝑠𝑡𝑝(𝑠)𝑓exp𝑘1𝑡𝑠𝜎0𝛾𝑝(𝜉)𝑑𝜉𝑑𝑠𝑎(𝑡)exp𝑘1𝑡𝑡𝜏𝑘𝑝(𝑠)𝑑𝑠+exp1𝑡𝑡𝜏0𝛾×𝑝(𝑠)𝑑𝑠𝑡𝑝(𝑠)𝑓exp𝑘2𝑡𝑠𝜎0𝛾𝑝(𝜉)𝑑𝜉𝑑𝑠𝑐<1,𝑡𝑡1.(3.3) Then (1.1) has a positive solution which tends to zero.

Proof. We set 𝑢(𝑡)=exp𝑘2𝑡𝑡0𝛾𝑝(𝑠)𝑑𝑠,𝑣(𝑡)=exp𝑘1𝑡𝑡0𝛾𝑝(𝑠)𝑑𝑠,𝑡𝑡0.(3.4) We will show that the conditions of Corollary 2.5 are satisfied. With regard to (2.21), for 𝑡[𝑡0,𝑡1], we get 𝑣(𝑡)𝑢(𝑡)=𝑘1𝑝(𝑡)𝑣(𝑡)+𝑘2𝑝(𝑡)𝑢(𝑡)=𝑝(𝑡)𝑣(𝑡)𝑘1+𝑘2𝑘𝑢(𝑡)exp1𝑡𝑡0𝛾𝑝(𝑠)𝑑𝑠=𝑝(𝑡)𝑣(𝑡)𝑘1+𝑘2𝑘exp1𝑘2𝑡𝑡0𝛾𝑝(𝑠)𝑑𝑠𝑝(𝑡)𝑣(𝑡)𝑘1+𝑘2𝑘exp1𝑘2𝑡0𝑡0𝛾𝑝(𝑠)𝑑𝑠0.(3.5) Other conditions of Corollary 2.5 are also satisfied. The proof is complete.

Corollary 3.2. Suppose that 𝑘>0,𝑐>0,𝑡1𝑡0+𝑚, (3.1) holds, and 𝑎(𝑡)=exp𝑘𝑡𝑡𝜏𝑘𝑝(𝑠)𝑑𝑠+exp𝑡𝑡𝜏0×𝑝(𝑠)𝑑𝑠𝑡𝑝(𝑠)𝑓exp𝑘𝑡𝑠𝜎0𝑝(𝜉)𝑑𝜉𝑑𝑠𝑐<1,𝑡𝑡1.(3.6) Then (1.1) has a solution 𝑥(𝑡)=exp𝑘𝑡𝑡0𝑝(𝑠)𝑑𝑠,𝑡𝑡1,(3.7) which tends to zero.

Proof. We put 𝑘1=𝑘2=𝑘,𝛾=0 and apply Theorem 3.1.

Example 3.3. Consider the nonlinear neutral differential equation []𝑥(𝑡)𝑎(𝑡)𝑥(𝑡2)=𝑝𝑥3(𝑡1),𝑡𝑡0,(3.8) where 𝑝(0,). We will show that the conditions of Theorem 3.1 are satisfied. Condition (3.1) obviously holds and (3.2) has a form 𝑘1𝑘2𝑘exp2𝑘1𝑝𝛾1,(3.9)0<𝑘1𝑘2,𝛾0. For function 𝑎(𝑡), we obtain exp2𝑝𝑘2+13𝑘1𝑝𝑘exp2𝛾𝑡023𝑘1𝛾𝑡0+𝑘123𝑘1𝑡𝑎(𝑡)exp2𝑝𝑘1+13𝑘2𝑝𝑘exp1𝛾𝑡023𝑘2𝛾𝑡0+𝑘113𝑘2𝑡,𝑡𝑡0.(3.10) For 𝑝=1,𝑘1=1,𝑘2=2,𝛾=1,𝑡0=1, condition (3.9) is satisfied and 𝑒4+1𝑒3𝑒𝑡𝑎(𝑡)𝑒2+𝑒46𝑒5𝑡,𝑡𝑡13.(3.11) If the function 𝑎(𝑡) satisfies (3.11), then (3.8) has a solution which is bounded by the functions 𝑢(𝑡)=exp(2𝑡),𝑣(𝑡)=exp(𝑡),𝑡3.
For example if 𝑝=1,𝑘1=𝑘2=1.5,𝛾=1,𝑡0=1, from (3.11) we obtain𝑎(𝑡)=𝑒3+𝑒1.5𝑒4.53𝑡,(3.12) and the equation 𝑒𝑥(𝑡)3+𝑒1.5𝑒4.53𝑡𝑥(𝑡2)=𝑥3(𝑡1),𝑡3,(3.13) has the solution 𝑥(𝑡)=exp(1.5𝑡) which is bounded by the function 𝑢(𝑡) and 𝑣(𝑡).

Acknowledgments

The research was supported by the Grant 1/0090/09 and the  Grant 1/1260/12 of the Scientific Grant Agency of the Ministry of Education of the Slovak Republic and Project APVV-0700-07 of the Slovak Research and Development Agency.

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