Existence of Nonoscillatory Solutions of First-Order Neutral Differential Equations

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 form𝑑𝑑𝑡[𝑥(𝑡)−𝑎(𝑡)𝑥(𝑡−𝜏)]=𝑝(𝑡)𝑓(𝑥(𝑡−𝜎)),𝑡≥𝑡⁰,(1.1) where 𝜏>0,𝜎≥0,𝑎∈𝐶([𝑡₀,∞),(0,∞)),𝑝∈𝐶(𝑅,(0,∞)),𝑓∈𝐶(𝑅,𝑅),𝑓 is nondecreasing function, and 𝑥𝑓(𝑥)>0,𝑥≠0.

By a solution of (1.1) we mean a function 𝑥∈𝐶([𝑡₁−𝑚,∞),𝑅),𝑚=max{𝜏,𝜎}, for some 𝑡₁≥𝑡₀, such that 𝑥(𝑡)−𝑎(𝑡)𝑥(𝑡−𝜏) is continuously differentiable on [𝑡₁,∞) and such that (1.1) is satisfied for 𝑡≥𝑡₁.

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 [1–11] 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+𝑆₂𝑦∈Ω for every pair 𝑥,𝑦∈Ω. 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,𝑣∈𝐶¹([𝑡₀,∞),(0,∞)), constant 𝑐>0 and 𝑡₁≥𝑡₀+𝑚 such that 𝑣(𝑡)−𝑣𝑡₁𝑢(𝑡)≤𝑣(𝑡),𝑡≥𝑡₀,(2.1)1−𝑢(𝑡)+𝑢𝑡₁≥0,𝑡₀≤𝑡≤𝑡₁,(2.2)𝑢(𝑡−𝜏)𝑢(𝑡)+^∞𝑡≤1𝑝(𝑠)𝑓(𝑣(𝑠−𝜎))𝑑𝑠≤𝑎(𝑡)𝑣(𝑡−𝜏)𝑣(𝑡)+^∞𝑡𝑝(𝑠)𝑓(𝑢(𝑠−𝜎))𝑑𝑠≤𝑐<1,𝑡≥𝑡₁.(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‖=sup_𝑡≥𝑡0|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: Ω=𝑥=𝑥(𝑡)∈𝐶𝑡₀,∞,𝑅∶𝑢(𝑡)≤𝑥(𝑡)≤𝑣(𝑡),𝑡≥𝑡₀.(2.4) We now define two maps 𝑆₁ and 𝑆₂∶Ω→𝐶([𝑡₀,∞),𝑅) as follows: 𝑆₁𝑥(𝑡)=𝑎(𝑡)𝑥(𝑡−𝜏),𝑡≥𝑡¹𝑆₁,𝑆₂𝑥𝑡₁,𝑡₀≤𝑡≤𝑡₁,𝑥(𝑡)=⎧⎪⎨⎪⎩−^∞𝑡𝑆₂𝑝(𝑠)𝑓(𝑥(𝑠−𝜎))𝑑𝑠,𝑡≥𝑡₁,𝑥𝑡₁+𝑣(𝑡)−𝑣𝑡₁,𝑡₀≤𝑡≤𝑡₁.(2.5) We will show that for any 𝑥,𝑦∈Ω we have 𝑆₁𝑥+𝑆₂𝑦∈Ω. For every 𝑥,𝑦∈Ω and 𝑡≥𝑡₁, we obtain 𝑆₁𝑥(𝑡)+𝑆₂𝑦(𝑡)≤𝑎(𝑡)𝑣(𝑡−𝜏)−^∞𝑡𝑝(𝑠)𝑓(𝑢(𝑠−𝜎))𝑑𝑠≤𝑣(𝑡).(2.6) For 𝑡∈[𝑡₀,𝑡₁], we have 𝑆₁𝑥(𝑡)+𝑆₂𝑦(𝑡)=≤𝑣𝑆₁𝑡₁𝑥𝑡₁+𝑆₂𝑦𝑡₁+𝑣(𝑡)−𝑣𝑡₁+𝑣(𝑡)−𝑣𝑡₁=𝑣(𝑡).(2.7) Furthermore, for 𝑡≥𝑡₁, we get 𝑆₁𝑥(𝑡)+𝑆₂𝑦(𝑡)≥𝑎(𝑡)𝑢(𝑡−𝜏)−^∞𝑡𝑝(𝑠)𝑓(𝑣(𝑠−𝜎))𝑑𝑠≥𝑢(𝑡).(2.8) Let 𝑡∈[𝑡₀,𝑡₁]. With regard to (2.2), we get 𝑣(𝑡)−𝑣𝑡₁+𝑢𝑡₁≥𝑢(𝑡),𝑡₀≤𝑡≤𝑡₁.(2.9) Then for 𝑡∈[𝑡₀,𝑡₁] and any 𝑥,𝑦∈Ω, we obtain 𝑆₁𝑥(𝑡)+𝑆₂𝑦(𝑡)=≥𝑢𝑆𝑡₁₁𝑥𝑡₁+𝑆₂𝑦𝑡₁+𝑣(𝑡)−𝑣𝑡₁+𝑣(𝑡)−𝑣𝑡₁≥𝑢(𝑡).(2.10) Thus we have proved that 𝑆₁𝑥+𝑆₂𝑦∈Ω for any 𝑥,𝑦∈Ω.
We will show that 𝑆₁ is a contraction mapping on Ω. For 𝑥,𝑦∈Ω and 𝑡≥𝑡₁ we have||𝑆₁𝑥(𝑡)−𝑆₁𝑦(𝑡)||=||𝑎(𝑡)‖𝑥(𝑡−𝜏)−𝑦(𝑡−𝜏)||≤𝑐‖𝑥−𝑦‖.(2.11) This implies that ‖‖𝑆₁𝑥−𝑆₁𝑦‖‖≤𝑐‖𝑥−𝑦‖.(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 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≥𝑡₁ we have||𝑆₂𝑥_𝑘(𝑡)−𝑆₂𝑥(𝑡)||≤≤||||^∞𝑡𝑝(𝑠)𝑓𝑥_𝑘(𝑠−𝜎)−𝑓(𝑥(𝑠−𝜎))𝑑𝑠_∞𝑡1||||𝑝(𝑠)||𝑓𝑥_𝑘(𝑠−𝜎)−𝑓(𝑥(𝑠−𝜎))||𝑑𝑠.(2.13) According to (2.8), we get ^∞𝑡₁𝑝(𝑠)𝑓(𝑣(𝑠−𝜎))𝑑𝑠<∞.(2.14) Since |𝑓(𝑥_𝑘(𝑠−𝜎))−𝑓(𝑥(𝑠−𝜎))|→0 as 𝑘→∞, by applying the Lebesgue dominated convergence theorem, we obtain lim_𝑘→∞‖‖𝑆₂𝑥_𝑘(𝑡)−𝑆₂𝑥(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 {𝑆₂𝑥∶𝑥∈Ω} 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 𝑥∈Ω and any 𝜀>0, we take 𝑡^∗≥𝑡₁ large enough so that^∞𝑡_∗𝑝(𝑠)𝑓(𝑥(𝑠−𝜎))𝑑𝑠<𝜀2.(2.16) Then, for 𝑥∈Ω,𝑇₂>𝑇₁≥𝑡^∗, we have ||𝑆₂𝑥𝑇₂−𝑆₂𝑥𝑇₁||≤+^∞𝑇₂𝑝(𝑠)𝑓(𝑥(𝑠−𝜎))𝑑𝑠_∞𝑇1𝑝(𝑠)𝑓(𝑥(𝑠−𝜎))𝑑𝑠<𝜀2+2𝜀=𝜀.(2.17) For 𝑥∈Ω and 𝑡₁≤𝑇₁<𝑇₂≤𝑡^∗, we get ||𝑆₂𝑥𝑇₂−𝑆₂𝑥𝑇₁||≤≤ma^𝑇2_𝑇1x𝑝_{𝑡1≤𝑠≤𝑡∗}(𝑠)𝑓(𝑥(𝑠−𝜎))𝑑𝑠{𝑝(𝑠)𝑓(𝑥(𝑠−𝜎))}𝑇₂−𝑇₁.(2.18) Thus there exists 𝛿₁=𝜀/𝑀, where 𝑀=max_{𝑡1≤𝑠≤𝑡∗}{𝑝(𝑠)𝑓(𝑥(𝑠−𝜎))}, such that ||𝑆₂𝑥𝑇₂−𝑆₂𝑥𝑇₁||<𝜀if0<𝑇₂−𝑇₁<𝛿₁.(2.19) Finally for any 𝑥∈Ω,𝑡₀≤𝑇₁<𝑇₂≤𝑡₁, there exists a 𝛿₂>0 such that ||𝑆₂𝑥𝑇₂−𝑆₂𝑥𝑇₁||≤=||m𝑣a𝑇x_{1𝑡0≤𝑠≤𝑡1}−𝑣𝑇₂||=||||^𝑇2_𝑇1𝑣^(𝑠)𝑑𝑠||||||𝑣^(𝑠)||𝑇₂−𝑇₁<𝜀if0<𝑇₂−𝑇₁<𝛿₂.(2.20) Then {𝑆₂𝑥∶𝑥∈Ω} 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 𝑥₀∈Ω 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,𝑣∈𝐶¹([𝑡₀,∞),(0,∞)), constant 𝑐>0 and 𝑡₁≥𝑡₀+𝑚 such that (2.1), (2.3) hold and 𝑣^(𝑡)−𝑢^(𝑡)≤0,𝑡₀≤𝑡≤𝑡₁.(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 𝐻(𝑡)=𝑣(𝑡)−𝑣𝑡₁−𝑢(𝑡)+𝑢𝑡₁.(2.22) Then with regard to (2.21), it follows that 𝐻′(𝑡)=𝑣′(𝑡)−𝑢′(𝑡)≤0,𝑡₀≤𝑡≤𝑡₁.(2.23) Since 𝐻(𝑡₁)=0 and 𝐻′(𝑡)≤0 for 𝑡∈[𝑡₀,𝑡₁], this implies that 𝐻(𝑡)=𝑣(𝑡)−𝑣𝑡₁−𝑢(𝑡)+𝑢𝑡₁≥0,𝑡₀≤𝑡≤𝑡₁.(2.24) Thus all conditions of Theorem 2.1 are satisfied.

Corollary 2.3. Suppose that there exists a function 𝑣∈𝐶¹([𝑡₀,∞),(0,∞)), constant 𝑐>0 and 𝑡₁≥𝑡₀+𝑚 such that 𝑎(𝑡)=1𝑣(𝑡−𝜏)𝑣(𝑡)+^∞𝑡𝑝(𝑠)𝑓(𝑣(𝑠−𝜎))𝑑𝑠≤𝑐<1,𝑡≥𝑡₁.(2.25) Then (1.1) has a solution 𝑥(𝑡)=𝑣(𝑡),𝑡≥𝑡₁.

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

Theorem 2.4. Suppose that there exist functions 𝑢,𝑣∈𝐶¹([𝑡₀,∞),(0,∞)), constant 𝑐>0 and 𝑡₁≥𝑡₀+𝑚 such that (2.1), (2.2), and (2.3) hold and lim_𝑡→∞𝑣(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,𝑣∈𝐶¹([𝑡₀,∞),(0,∞)), constant 𝑐>0 and 𝑡₁≥𝑡₀+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∈𝐶¹([𝑡₀,∞),(0,∞)), constant 𝑐>0 and 𝑡₁≥𝑡₀+𝑚 such that (2.25), (2.26) hold. Then (1.1) has a solution 𝑥(𝑡)=𝑣(𝑡),𝑡≥𝑡₁ 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<𝑘₁≤𝑘₂ and there exist constants 𝑐>0,𝛾≥0,𝑡₁≥𝑡₀+𝑚 such that 𝑘₁𝑘₂exp−𝑘e₂xp𝑘₂−𝑘₁^𝑡0_𝑡0−𝛾𝑝(𝑡)𝑑𝑡≥1,(3.2)^{𝑡𝑡−𝜏}𝑝(𝑠)𝑑𝑠+exp𝑘₂^{𝑡𝑡−𝜏}_0−𝛾×𝑝(𝑠)𝑑𝑠_∞𝑡𝑝(𝑠)𝑓exp−𝑘₁^{𝑡𝑠−𝜎}_0−𝛾≤exp−𝑘₁𝑝(𝜉)𝑑𝜉𝑑𝑠≤𝑎(𝑡)^{𝑡𝑡−𝜏}𝑝(𝑠)𝑑𝑠+exp𝑘₁^{𝑡𝑡−𝜏}_0−𝛾×𝑝(𝑠)𝑑𝑠_∞𝑡𝑝(𝑠)𝑓exp−𝑘₂^{𝑡𝑠−𝜎}_0−𝛾𝑝(𝜉)𝑑𝜉𝑑𝑠≤𝑐<1,𝑡≥𝑡₁.(3.3) Then (1.1) has a positive solution which tends to zero.

Proof. We set 𝑢(𝑡)=exp−𝑘₂^𝑡𝑡_0−𝛾𝑝(𝑠)𝑑𝑠,𝑣(𝑡)=exp−𝑘₁^𝑡𝑡_0−𝛾𝑝(𝑠)𝑑𝑠,𝑡≥𝑡₀.(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)==−𝑝(𝑘𝑡₁)𝑝𝑣((𝑡𝑡))𝑣(−𝑡)𝑘+₁𝑘₂𝑝(𝑡)𝑢(𝑡)+𝑘₂𝑢(𝑡)exp𝑘₁^𝑡𝑡_0−𝛾=𝑝(𝑡)𝑣(𝑡)−𝑘₁𝑝(𝑠)𝑑𝑠≤𝑝(𝑡)𝑣(𝑡)−𝑘₁+𝑘₂exp𝑘₁−𝑘₂^𝑡𝑡_0−𝛾𝑝(𝑠)𝑑𝑠+𝑘₂exp𝑘₁−𝑘₂^𝑡0_𝑡0−𝛾𝑝(𝑠)𝑑𝑠≤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,𝑡₁≥𝑡₀+𝑚, (3.1) holds, and 𝑎(𝑡)=×exp−𝑘^{𝑡𝑡−𝜏}𝑝(𝑠)𝑑𝑠+exp𝑘^{𝑡𝑡−𝜏}₀𝑝(𝑠)𝑑𝑠_∞𝑡𝑝(𝑠)𝑓exp−𝑘^{𝑡𝑠−𝜎}₀𝑝(𝜉)𝑑𝜉𝑑𝑠≤𝑐<1,𝑡≥𝑡₁.(3.6) Then (1.1) has a solution 𝑥(𝑡)=exp−𝑘^𝑡𝑡₀𝑝(𝑠)𝑑𝑠,𝑡≥𝑡₁,(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(𝑡−2)]^=𝑝𝑥³(𝑡−1),𝑡≥𝑡₀,(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 𝑘₁𝑘₂exp𝑘₂−𝑘₁𝑝𝛾≥1,(3.9)0<𝑘₁≤𝑘₂,𝛾≥0. For function 𝑎(𝑡), we obtain exp−2𝑝𝑘₂+13𝑘₁≤𝑎(𝑡)≤exp−e2x𝑝p𝑘₁𝑝𝑘₂𝛾−𝑡₀−2−3𝑘₁𝛾−𝑡₀−1+𝑘₂−3𝑘₁𝑡+13𝑘₂exp𝑝𝑘₁𝛾−𝑡₀−2−3𝑘₂𝛾−𝑡₀−1+𝑘₁−3𝑘₂𝑡,𝑡≥𝑡₀.(3.10) For 𝑝=1,𝑘₁=1,𝑘₂=2,𝛾=1,𝑡₀=1, condition (3.9) is satisfied and 𝑒⁻⁴+13𝑒𝑒^−𝑡≤𝑎(𝑡)≤𝑒⁻²+𝑒⁴6𝑒^−5𝑡,𝑡≥𝑡₁≥3.(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),𝑡≥3.
For example if 𝑝=1,𝑘₁=𝑘₂=1.5,𝛾=1,𝑡₀=1, from (3.11) we obtain𝑎(𝑡)=𝑒⁻³+𝑒^1.54.5𝑒^−3𝑡,(3.12) and the equation 𝑥(𝑡)−𝑒⁻³+𝑒^1.54.5𝑒^−3𝑡𝑥(𝑡−2)^=𝑥³(𝑡−1),𝑡≥3,(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.