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Abstract and Applied Analysis
Volume 2011 (2011), Article ID 346745, 9 pages
doi: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 form 𝑑 [ ] 𝑑 𝑡 𝑥 ( 𝑡 ) 𝑎 ( 𝑡 ) 𝑥 ( 𝑡 𝜏 ) = 𝑝 ( 𝑡 ) 𝑓 ( 𝑥 ( 𝑡 𝜎 ) ) , 𝑡 𝑡 0 , ( 1 . 1 ) where 𝜏 > 0 , 𝜎 0 , 𝑎 𝐶 ( [ 𝑡 0 , ) , ( 0 , ) ) , 𝑝 𝐶 ( 𝑅 , ( 0 , ) ) , 𝑓 𝐶 ( 𝑅 , 𝑅 ) , 𝑓 is nondecreasing function, and 𝑥 𝑓 ( 𝑥 ) > 0 , 𝑥 0 .

By a solution of (1.1) we mean a function 𝑥 𝐶 ( [ 𝑡 1 𝑚 , ) , 𝑅 ) , 𝑚 = m a x { 𝜏 , 𝜎 } , for some 𝑡 1 𝑡 0 , such that 𝑥 ( 𝑡 ) 𝑎 ( 𝑡 ) 𝑥 ( 𝑡 𝜏 ) is continuously differentiable on [ 𝑡 1 , ) and such that (1.1) is satisfied for 𝑡 𝑡 1 .

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 𝑆 1 , 𝑆 2 be maps of Ω into 𝑋 such that 𝑆 1 𝑥 + 𝑆 2 𝑦 Ω for every pair 𝑥 , 𝑦 Ω . If 𝑆 1 is contractive and 𝑆 2 is completely continuous, then the equation 𝑆 1 𝑥 + 𝑆 2 𝑥 = 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 , 𝑣 𝐶 1 ( [ 𝑡 0 , ) , ( 0 , ) ) , constant 𝑐 > 0 and 𝑡 1 𝑡 0 + 𝑚 such that 𝑢 ( 𝑡 ) 𝑣 ( 𝑡 ) , 𝑡 𝑡 0 , 𝑣 𝑡 ( 2 . 1 ) ( 𝑡 ) 𝑣 1 𝑡 𝑢 ( 𝑡 ) + 𝑢 1 0 , 𝑡 0 𝑡 𝑡 1 , 1 ( 2 . 2 ) 𝑢 ( 𝑡 𝜏 ) 𝑢 ( 𝑡 ) + 𝑡 1 𝑝 ( 𝑠 ) 𝑓 ( 𝑣 ( 𝑠 𝜎 ) ) 𝑑 𝑠 𝑎 ( 𝑡 ) 𝑣 𝑣 ( 𝑡 𝜏 ) ( 𝑡 ) + 𝑡 𝑝 ( 𝑠 ) 𝑓 ( 𝑢 ( 𝑠 𝜎 ) ) 𝑑 𝑠 𝑐 < 1 , 𝑡 𝑡 1 . ( 2 . 3 ) Then (1.1) has a positive solution which is bounded by functions 𝑢 , 𝑣 .

Proof. Let 𝐶 ( [ 𝑡 0 , ) , 𝑅 ) be the set of all continuous bounded functions with the norm 𝑥 = s u p 𝑡 𝑡 0 | 𝑥 ( 𝑡 ) | . Then 𝐶 ( [ 𝑡 0 , ) , 𝑅 ) is a Banach space. We define a closed, bounded, and convex subset Ω of 𝐶 ( [ 𝑡 0 , ) , 𝑅 ) as follows: 𝑡 Ω = 𝑥 = 𝑥 ( 𝑡 ) 𝐶 0 , , 𝑅 𝑢 ( 𝑡 ) 𝑥 ( 𝑡 ) 𝑣 ( 𝑡 ) , 𝑡 𝑡 0 . ( 2 . 4 ) We now define two maps 𝑆 1 and 𝑆 2 Ω 𝐶 ( [ 𝑡 0 , ) , 𝑅 ) as follows: 𝑆 1 𝑥 ( 𝑡 ) = 𝑎 ( 𝑡 ) 𝑥 ( 𝑡 𝜏 ) , 𝑡 𝑡 1 , 𝑆 1 𝑥 𝑡 1 , 𝑡 0 𝑡 𝑡 1 , 𝑆 2 𝑥 ( 𝑡 ) = 𝑡 𝑝 ( 𝑠 ) 𝑓 ( 𝑥 ( 𝑠 𝜎 ) ) 𝑑 𝑠 , 𝑡 𝑡 1 , 𝑆 2 𝑥 𝑡 1 𝑡 + 𝑣 ( 𝑡 ) 𝑣 1 , 𝑡 0 𝑡 𝑡 1 . ( 2 . 5 ) We will show that for any 𝑥 , 𝑦 Ω we have 𝑆 1 𝑥 + 𝑆 2 𝑦 Ω . For every 𝑥 , 𝑦 Ω and 𝑡 𝑡 1 , we obtain 𝑆 1 𝑥 𝑆 ( 𝑡 ) + 2 𝑦 ( 𝑡 ) 𝑎 ( 𝑡 ) 𝑣 ( 𝑡 𝜏 ) 𝑡 𝑝 ( 𝑠 ) 𝑓 ( 𝑢 ( 𝑠 𝜎 ) ) 𝑑 𝑠 𝑣 ( 𝑡 ) . ( 2 . 6 ) For 𝑡 [ 𝑡 0 , 𝑡 1 ] , we have 𝑆 1 𝑥 𝑆 ( 𝑡 ) + 2 𝑦 𝑆 ( 𝑡 ) = 1 𝑥 𝑡 1 + 𝑆 2 𝑦 𝑡 1 𝑡 + 𝑣 ( 𝑡 ) 𝑣 1 𝑡 𝑣 1 𝑡 + 𝑣 ( 𝑡 ) 𝑣 1 = 𝑣 ( 𝑡 ) . ( 2 . 7 ) Furthermore, for 𝑡 𝑡 1 , we get 𝑆 1 𝑥 𝑆 ( 𝑡 ) + 2 𝑦 ( 𝑡 ) 𝑎 ( 𝑡 ) 𝑢 ( 𝑡 𝜏 ) 𝑡 𝑝 ( 𝑠 ) 𝑓 ( 𝑣 ( 𝑠 𝜎 ) ) 𝑑 𝑠 𝑢 ( 𝑡 ) . ( 2 . 8 ) Let 𝑡 [ 𝑡 0 , 𝑡 1 ] . With regard to (2.2), we get 𝑣 𝑡 ( 𝑡 ) 𝑣 1 𝑡 + 𝑢 1 𝑢 ( 𝑡 ) , 𝑡 0 𝑡 𝑡 1 . ( 2 . 9 ) Then for 𝑡 [ 𝑡 0 , 𝑡 1 ] and any 𝑥 , 𝑦 Ω , we obtain 𝑆 1 𝑥 𝑆 ( 𝑡 ) + 2 𝑦 𝑆 ( 𝑡 ) = 1 𝑥 𝑡 1 + 𝑆 2 𝑦 𝑡 1 𝑡 + 𝑣 ( 𝑡 ) 𝑣 1 𝑡 𝑢 1 𝑡 + 𝑣 ( 𝑡 ) 𝑣 1 𝑢 ( 𝑡 ) . ( 2 . 1 0 ) Thus we have proved that 𝑆 1 𝑥 + 𝑆 2 𝑦 Ω for any 𝑥 , 𝑦 Ω .
We will show that 𝑆 1 is a contraction mapping on Ω . For 𝑥 , 𝑦 Ω and 𝑡 𝑡 1 we have | | 𝑆 1 𝑥 𝑆 ( 𝑡 ) 1 𝑦 | | = | | 𝑎 | | ( 𝑡 ) ( 𝑡 ) 𝑥 ( 𝑡 𝜏 ) 𝑦 ( 𝑡 𝜏 ) 𝑐 𝑥 𝑦 . ( 2 . 1 1 ) This implies that 𝑆 1 𝑥 𝑆 1 𝑦 𝑐 𝑥 𝑦 . ( 2 . 1 2 ) Also for 𝑡 [ 𝑡 0 , 𝑡 1 ] , the previous inequality is valid. We conclude that 𝑆 1 is a contraction mapping on Ω .
We now show that 𝑆 2 is completely continuous. First we will show that 𝑆 2 is continuous. Let 𝑥 𝑘 = 𝑥 𝑘 ( 𝑡 ) Ω be such that 𝑥 𝑘 ( 𝑡 ) 𝑥 ( 𝑡 ) as 𝑘 . Because Ω is closed, 𝑥 = 𝑥 ( 𝑡 ) Ω . For 𝑡 𝑡 1 we have | | 𝑆 2 𝑥 𝑘 𝑆 ( 𝑡 ) 2 𝑥 | | | | | | ( 𝑡 ) 𝑡 𝑓 𝑥 𝑝 ( 𝑠 ) 𝑘 | | | | ( 𝑠 𝜎 ) 𝑓 ( 𝑥 ( 𝑠 𝜎 ) ) 𝑑 𝑠 𝑡 1 | | 𝑓 𝑥 𝑝 ( 𝑠 ) 𝑘 | | ( 𝑠 𝜎 ) 𝑓 ( 𝑥 ( 𝑠 𝜎 ) ) 𝑑 𝑠 . ( 2 . 1 3 ) According to (2.8), we get 𝑡 1 𝑝 ( 𝑠 ) 𝑓 ( 𝑣 ( 𝑠 𝜎 ) ) 𝑑 𝑠 < . ( 2 . 1 4 ) Since | 𝑓 ( 𝑥 𝑘 ( 𝑠 𝜎 ) ) 𝑓 ( 𝑥 ( 𝑠 𝜎 ) ) | 0 as 𝑘 , by applying the Lebesgue dominated convergence theorem, we obtain l i m 𝑘 𝑆 2 𝑥 𝑘 𝑆 ( 𝑡 ) 2 𝑥 ( 𝑡 ) = 0 . ( 2 . 1 5 ) This means that 𝑆 2 is continuous.
We now show that 𝑆 2 Ω is relatively compact. It is sufficient to show by the Arzela-Ascoli theorem that the family of functions { 𝑆 2 𝑥 𝑥 Ω } is uniformly bounded and equicontinuous on [ 𝑡 0 , ) . 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 [ 𝑡 0 , ) 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 𝑥 Ω and any 𝜀 > 0 , we take 𝑡 𝑡 1 large enough so that 𝑡 𝜀 𝑝 ( 𝑠 ) 𝑓 ( 𝑥 ( 𝑠 𝜎 ) ) 𝑑 𝑠 < 2 . ( 2 . 1 6 ) Then, for 𝑥 Ω , 𝑇 2 > 𝑇 1 𝑡 , we have | | 𝑆 2 𝑥 𝑇 2 𝑆 2 𝑥 𝑇 1 | | 𝑇 2 + 𝑝 ( 𝑠 ) 𝑓 ( 𝑥 ( 𝑠 𝜎 ) ) 𝑑 𝑠 𝑇 1 𝜀 𝑝 ( 𝑠 ) 𝑓 ( 𝑥 ( 𝑠 𝜎 ) ) 𝑑 𝑠 < 2 + 𝜀 2 = 𝜀 . ( 2 . 1 7 ) For 𝑥 Ω and 𝑡 1 𝑇 1 < 𝑇 2 𝑡 , we get | | 𝑆 2 𝑥 𝑇 2 𝑆 2 𝑥 𝑇 1 | | 𝑇 2 𝑇 1 𝑝 ( 𝑠 ) 𝑓 ( 𝑥 ( 𝑠 𝜎 ) ) 𝑑 𝑠 m a x 𝑡 1 𝑠 𝑡 𝑇 { 𝑝 ( 𝑠 ) 𝑓 ( 𝑥 ( 𝑠 𝜎 ) ) } 2 𝑇 1 . ( 2 . 1 8 ) Thus there exists 𝛿 1 = 𝜀 / 𝑀 , where 𝑀 = m a x 𝑡 1 𝑠 𝑡 { 𝑝 ( 𝑠 ) 𝑓 ( 𝑥 ( 𝑠 𝜎 ) ) } , such that | | 𝑆 2 𝑥 𝑇 2 𝑆 2 𝑥 𝑇 1 | | < 𝜀 i f 0 < 𝑇 2 𝑇 1 < 𝛿 1 . ( 2 . 1 9 ) Finally for any 𝑥 Ω , 𝑡 0 𝑇 1 < 𝑇 2 𝑡 1 , there exists a 𝛿 2 > 0 such that | | 𝑆 2 𝑥 𝑇 2 𝑆 2 𝑥 𝑇 1 | | = | | 𝑣 𝑇 1 𝑇 𝑣 2 | | = | | | | 𝑇 2 𝑇 1 𝑣 | | | | ( 𝑠 ) 𝑑 𝑠 m a x 𝑡 0 𝑠 𝑡 1 | | 𝑣 ( | | 𝑇 𝑠 ) 2 𝑇 1 < 𝜀 i f 0 < 𝑇 2 𝑇 1 < 𝛿 2 . ( 2 . 2 0 ) Then { 𝑆 2 𝑥 𝑥 Ω } is uniformly bounded and equicontinuous on [ 𝑡 0 , ) , and hence 𝑆 2 Ω is relatively compact subset of 𝐶 ( [ 𝑡 0 , ) , 𝑅 ) . By Lemma 1.1 there is an 𝑥 0 Ω such that 𝑆 1 𝑥 0 + 𝑆 2 𝑥 0 = 𝑥 0 . We conclude that 𝑥 0 ( 𝑡 ) is a positive solution of (1.1). The proof is complete.

Corollary 2.2. Suppose that there exist functions 𝑢 , 𝑣 𝐶 1 ( [ 𝑡 0 , ) , ( 0 , ) ) , constant 𝑐 > 0 and 𝑡 1 𝑡 0 + 𝑚 such that (2.1), (2.3) hold and 𝑣 ( 𝑡 ) 𝑢 ( 𝑡 ) 0 , 𝑡 0 𝑡 𝑡 1 . ( 2 . 2 1 ) 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 [ 𝑡 0 , 𝑡 1 ] and set 𝐻 𝑡 ( 𝑡 ) = 𝑣 ( 𝑡 ) 𝑣 1 𝑡 𝑢 ( 𝑡 ) + 𝑢 1 . ( 2 . 2 2 ) Then with regard to (2.21), it follows that 𝐻 ( 𝑡 ) = 𝑣 ( 𝑡 ) 𝑢 ( 𝑡 ) 0 , 𝑡 0 𝑡 𝑡 1 . ( 2 . 2 3 ) Since 𝐻 ( 𝑡 1 ) = 0 and 𝐻 ( 𝑡 ) 0 for 𝑡 [ 𝑡 0 , 𝑡 1 ] , this implies that 𝐻 𝑡 ( 𝑡 ) = 𝑣 ( 𝑡 ) 𝑣 1 𝑡 𝑢 ( 𝑡 ) + 𝑢 1 0 , 𝑡 0 𝑡 𝑡 1 . ( 2 . 2 4 ) Thus all conditions of Theorem 2.1 are satisfied.

Corollary 2.3. Suppose that there exists a function 𝑣 𝐶 1 ( [ 𝑡 0 , ) , ( 0 , ) ) , constant 𝑐 > 0 and 𝑡 1 𝑡 0 + 𝑚 such that 1 𝑎 ( 𝑡 ) = 𝑣 ( 𝑡 𝜏 ) 𝑣 ( 𝑡 ) + 𝑡 𝑝 ( 𝑠 ) 𝑓 ( 𝑣 ( 𝑠 𝜎 ) ) 𝑑 𝑠 𝑐 < 1 , 𝑡 𝑡 1 . ( 2 . 2 5 ) Then (1.1) has a solution 𝑥 ( 𝑡 ) = 𝑣 ( 𝑡 ) , 𝑡 𝑡 1 .

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

Theorem 2.4. Suppose that there exist functions 𝑢 , 𝑣 𝐶 1 ( [ 𝑡 0 , ) , ( 0 , ) ) , constant 𝑐 > 0 and 𝑡 1 𝑡 0 + 𝑚 such that (2.1), (2.2), and (2.3) hold and l i m 𝑡 𝑣 ( 𝑡 ) = 0 . ( 2 . 2 6 ) 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 , 𝑣 𝐶 1 ( [ 𝑡 0 , ) , ( 0 , ) ) , constant 𝑐 > 0 and 𝑡 1 𝑡 0 + 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 𝑣 𝐶 1 ( [ 𝑡 0 , ) , ( 0 , ) ) , constant 𝑐 > 0 and 𝑡 1 𝑡 0 + 𝑚 such that (2.25), (2.26) hold. Then (1.1) has a solution 𝑥 ( 𝑡 ) = 𝑣 ( 𝑡 ) , 𝑡 𝑡 1 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 𝑡 0 𝑝 ( 𝑡 ) 𝑑 𝑡 = , ( 3 . 1 ) 0 < 𝑘 1 𝑘 2 and there exist constants 𝑐 > 0 , 𝛾 0 , 𝑡 1 𝑡 0 + 𝑚 such that 𝑘 1 𝑘 2 𝑘 e x p 2 𝑘 1 𝑡 0 𝑡 0 𝛾 𝑝 ( 𝑡 ) 𝑑 𝑡 1 , ( 3 . 2 ) e x p 𝑘 2 𝑡 𝑡 𝜏 𝑘 𝑝 ( 𝑠 ) 𝑑 𝑠 + e x p 2 𝑡 𝑡 𝜏 0 𝛾 × 𝑝 ( 𝑠 ) 𝑑 𝑠 𝑡 𝑝 ( 𝑠 ) 𝑓 e x p 𝑘 1 𝑡 𝑠 𝜎 0 𝛾 𝑝 ( 𝜉 ) 𝑑 𝜉 𝑑 𝑠 𝑎 ( 𝑡 ) e x p 𝑘 1 𝑡 𝑡 𝜏 𝑘 𝑝 ( 𝑠 ) 𝑑 𝑠 + e x p 1 𝑡 𝑡 𝜏 0 𝛾 × 𝑝 ( 𝑠 ) 𝑑 𝑠 𝑡 𝑝 ( 𝑠 ) 𝑓 e x p 𝑘 2 𝑡 𝑠 𝜎 0 𝛾 𝑝 ( 𝜉 ) 𝑑 𝜉 𝑑 𝑠 𝑐 < 1 , 𝑡 𝑡 1 . ( 3 . 3 ) Then (1.1) has a positive solution which tends to zero.

Proof. We set 𝑢 ( 𝑡 ) = e x p 𝑘 2 𝑡 𝑡 0 𝛾 𝑝 ( 𝑠 ) 𝑑 𝑠 , 𝑣 ( 𝑡 ) = e x p 𝑘 1 𝑡 𝑡 0 𝛾 𝑝 ( 𝑠 ) 𝑑 𝑠 , 𝑡 𝑡 0 . ( 3 . 4 ) We will show that the conditions of Corollary 2.5 are satisfied. With regard to (2.21), for 𝑡 [ 𝑡 0 , 𝑡 1 ] , we get 𝑣 ( 𝑡 ) 𝑢 ( 𝑡 ) = 𝑘 1 𝑝 ( 𝑡 ) 𝑣 ( 𝑡 ) + 𝑘 2 𝑝 ( 𝑡 ) 𝑢 ( 𝑡 ) = 𝑝 ( 𝑡 ) 𝑣 ( 𝑡 ) 𝑘 1 + 𝑘 2 𝑘 𝑢 ( 𝑡 ) e x p 1 𝑡 𝑡 0 𝛾 𝑝 ( 𝑠 ) 𝑑 𝑠 = 𝑝 ( 𝑡 ) 𝑣 ( 𝑡 ) 𝑘 1 + 𝑘 2 𝑘 e x p 1 𝑘 2 𝑡 𝑡 0 𝛾 𝑝 ( 𝑠 ) 𝑑 𝑠 𝑝 ( 𝑡 ) 𝑣 ( 𝑡 ) 𝑘 1 + 𝑘 2 𝑘 e x p 1 𝑘 2 𝑡 0 𝑡 0 𝛾 𝑝 ( 𝑠 ) 𝑑 𝑠 0 . ( 3 . 5 ) Other conditions of Corollary 2.5 are also satisfied. The proof is complete.

Corollary 3.2. Suppose that 𝑘 > 0 , 𝑐 > 0 , 𝑡 1 𝑡 0 + 𝑚 , (3.1) holds, and 𝑎 ( 𝑡 ) = e x p 𝑘 𝑡 𝑡 𝜏 𝑘 𝑝 ( 𝑠 ) 𝑑 𝑠 + e x p 𝑡 𝑡 𝜏 0 × 𝑝 ( 𝑠 ) 𝑑 𝑠 𝑡 𝑝 ( 𝑠 ) 𝑓 e x p 𝑘 𝑡 𝑠 𝜎 0 𝑝 ( 𝜉 ) 𝑑 𝜉 𝑑 𝑠 𝑐 < 1 , 𝑡 𝑡 1 . ( 3 . 6 ) Then (1.1) has a solution 𝑥 ( 𝑡 ) = e x p 𝑘 𝑡 𝑡 0 𝑝 ( 𝑠 ) 𝑑 𝑠 , 𝑡 𝑡 1 , ( 3 . 7 ) which tends to zero.

Proof. We put 𝑘 1 = 𝑘 2 = 𝑘 , 𝛾 = 0 and apply Theorem 3.1.

Example 3.3. Consider the nonlinear neutral differential equation [ ] 𝑥 ( 𝑡 ) 𝑎 ( 𝑡 ) 𝑥 ( 𝑡 2 ) = 𝑝 𝑥 3 ( 𝑡 1 ) , 𝑡 𝑡 0 , ( 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 𝑘 1 𝑘 2 𝑘 e x p 2 𝑘 1 𝑝 𝛾 1 , ( 3 . 9 ) 0 < 𝑘 1 𝑘 2 , 𝛾 0 . For function 𝑎 ( 𝑡 ) , we obtain e x p 2 𝑝 𝑘 2 + 1 3 𝑘 1 𝑝 𝑘 e x p 2 𝛾 𝑡 0 2 3 𝑘 1 𝛾 𝑡 0 + 𝑘 1 2 3 𝑘 1 𝑡 𝑎 ( 𝑡 ) e x p 2 𝑝 𝑘 1 + 1 3 𝑘 2 𝑝 𝑘 e x p 1 𝛾 𝑡 0 2 3 𝑘 2 𝛾 𝑡 0 + 𝑘 1 1 3 𝑘 2 𝑡 , 𝑡 𝑡 0 . ( 3 . 1 0 ) For 𝑝 = 1 , 𝑘 1 = 1 , 𝑘 2 = 2 , 𝛾 = 1 , 𝑡 0 = 1 , condition (3.9) is satisfied and 𝑒 4 + 1 𝑒 3 𝑒 𝑡 𝑎 ( 𝑡 ) 𝑒 2 + 𝑒 4 6 𝑒 5 𝑡 , 𝑡 𝑡 1 3 . ( 3 . 1 1 ) If the function 𝑎 ( 𝑡 ) satisfies (3.11), then (3.8) has a solution which is bounded by the functions 𝑢 ( 𝑡 ) = e x p ( 2 𝑡 ) , 𝑣 ( 𝑡 ) = e x p ( 𝑡 ) , 𝑡 3 .
For example if 𝑝 = 1 , 𝑘 1 = 𝑘 2 = 1 . 5 , 𝛾 = 1 , 𝑡 0 = 1 , from (3.11) we obtain 𝑎 ( 𝑡 ) = 𝑒 3 + 𝑒 1 . 5 𝑒 4 . 5 3 𝑡 , ( 3 . 1 2 ) and the equation 𝑒 𝑥 ( 𝑡 ) 3 + 𝑒 1 . 5 𝑒 4 . 5 3 𝑡 𝑥 ( 𝑡 2 ) = 𝑥 3 ( 𝑡 1 ) , 𝑡 3 , ( 3 . 1 3 ) has the solution 𝑥 ( 𝑡 ) = e x p ( 1 . 5 𝑡 ) 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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