Existence for Eventually Positive Solutions of High-Order Nonlinear Neutral Differential Equations with Distributed Delay

Abstract

We consider the existence for eventually positive solutions of high-order nonlinear neutral differential equations with distributed delay. We use Lebesgue's dominated convergence theorem to obtain new necessary and sufficient condition for the existence of eventually positive solutions.

1. Introduction and Preliminary

In this paper, we consider the high-order nonlinear neutral differential equation:𝑟(𝑡)𝑥(𝑡)−^𝑏_𝑎𝑝(𝑡,𝜏)𝑥(𝑡−𝜏)𝑑𝜏^(𝑛)+^𝑑_𝑐𝑞(𝑡,𝜎)𝑓(𝑥(𝑡−𝜎))𝑑𝜎=0,𝑡≥𝑡₀, (1.1) and the associated inequality:𝑟(𝑡)𝑥(𝑡)−^𝑏_𝑎𝑝(𝑡,𝜏)𝑥(𝑡−𝜏)𝑑𝜏^(𝑛)+^𝑑_𝑐𝑞(𝑡,𝜎)𝑓(𝑥(𝑡−𝜎))𝑑𝜎≤0,𝑡≥𝑡₀, (1.2)(1)where 𝑛 is a positive integer, 0<𝑎<𝑏,0<𝑐<𝑑,(2)𝑝∈𝐶([𝑡₀,∞)×[𝑎,𝑏],𝑅⁺), (3)𝑟∈𝐶([𝑡₀,∞),𝑅⁺),𝑟(𝑡)>0, 𝑞∈𝐶([𝑡₀,∞)×[𝑐,𝑑],𝑅⁺),(4)𝑓(𝑢) are continuously nondecreasing real function with respect to 𝑢 defined on R such that 𝑢𝑓(𝑢)>0, for 𝑢>0, for 𝑛 is positive odd integer; 𝑓(𝑢) are continuously decreasing real function with respect to 𝑢 defined on 𝑅 such that 𝑢𝑓(𝑢)>0, for 𝑢>0, for 𝑛 is positive even integer.

Recently, there has been a lot of activities concerning the existence of eventually positive solutions for nonlinear neutral differential equations. See [1–8]. In [1], Liu et al. have studied the even-order neutral differential equation:𝑎(𝑡)𝑥(𝑡)−^𝑚_𝑖=1𝑏_𝑖(𝑡)𝑥𝑡−𝜏_𝑖^(𝑛)−^𝑙_𝑗=1𝑝_𝑗(𝑡)𝑓_𝑗𝑡,𝑥𝑡−𝜎_𝑗=0,𝑡≥0, (1.3) and the associated differential inequality:𝑎(𝑡)𝑥(𝑡)−^𝑚_𝑖=1𝑏_𝑖(𝑡)𝑥𝑡−𝜏_𝑖^(𝑛)−^𝑙_𝑗=1𝑝_𝑗(𝑡)𝑓_𝑗𝑡,𝑥𝑡−𝜎_𝑗≥0,𝑡≥0. (1.4) They have obtained that the existences of eventually positive solutions of (1.3) and (1.4) are equivalent. In [2], Ouyang et al. has studied the odd-order neutral differential equation:𝑎(𝑡)𝑥(𝑡)−^𝑚_𝑖=1𝑏_𝑖(𝑡)𝑥𝑡−𝜏_𝑖^(𝑛)+^𝑙_𝑗=1𝑝_𝑗(𝑡)𝑓_𝑗𝑥𝑡−𝜎_𝑗=0,𝑡≥0, (1.5) and the associated differential inequality:𝑎(𝑡)𝑥(𝑡)−^𝑚_𝑖=1𝑏_𝑖(𝑡)𝑥𝑡−𝜏_𝑖^(𝑛)+^𝑙_𝑗=1𝑝_𝑗(𝑡)𝑓_𝑗𝑥𝑡−𝜎_𝑗≤0,𝑡≥0. (1.6) He has obtained that the existences of eventually positive solutions of (1.5) and (1.6) are equivalent.

As usual, a solution of (1.1) is a continuous function 𝑥(𝑡) defined on [−𝜇,∞) such that 𝑦(𝑡)∶=𝑟(𝑡)𝑥(𝑡)−∫^𝑏_𝑎𝑝(𝑡,𝜏)𝑥(𝑡−𝜏)𝑑𝜏 is 𝑛 times differentiable and (1.1) holds for all 𝑛≥0. Such a solution 𝑥(𝑡) is called an eventually positive solution if there is 𝑇≥𝑡₀, such that 𝑥(𝑡)>0, for 𝑡≥𝑇. Here, 𝜇=max{𝑏,𝑑}.

Lemma 1.1. Assume sup_𝑡≥𝑡0𝑟(𝑡)<∞, and ∫^𝑏_𝑎𝑝(𝑡,𝜏)𝑑𝜏/𝑟(𝑡)≤𝑀, let 𝑥(𝑡) is an eventually bounded positive solution of inequality (1.2), and set 𝑦(𝑡)=𝑟(𝑡)𝑥(𝑡)−^𝑏_𝑎𝑝(𝑡,𝜏)𝑥(𝑡−𝜏)𝑑𝜏. (1.7) If 𝑛 is a positive odd integer, then eventually, 𝑦^(𝑛)(𝑡)≤0,(−1)^𝑘+1𝑦^(𝑘)(𝑡)<0,(𝑘=0,1,…,𝑛−1),lim_𝑡→∞𝑦^(𝑘)(𝑡)=0,(𝑘=1,2,…,𝑛−1). (1.8) If 𝑛 is a positive even integer, then eventually, 𝑦^(𝑛)(𝑡)≤0,(−1)^𝑘𝑦^(𝑘)(𝑡)<0,(𝑘=0,1,…,𝑛−1),lim_𝑡→∞𝑦^(𝑘)(𝑡)=0,(𝑘=1,2,…,𝑛−1). (1.9)

Proof. We have the following cases.
Case 1 (If 𝑛 is a positive odd integer). Because 𝑥(𝑡) is bounded, and sup_𝑡≥𝑡0𝑟(𝑡)<∞,∫^𝑏_𝑎𝑝(𝑡,𝜏)𝑑𝜏/𝑟(𝑡)≤𝑀, thus 𝑦(𝑡) is bounded. From (1.2), we have 𝑦^(𝑛)(𝑡)≤−∫^𝑑_𝑐𝑞(𝑡,𝜎)𝑓(𝑥(𝑡−𝜎))𝑑𝜎≤0. Assume 𝑦^(𝑛−1)(𝑡)≤0, since 𝑦^(𝑛)(𝑡)≤0, 𝑦^(𝑛−1)(𝑡) decreases, set lim_𝑡→∞𝑦^(𝑛−1)(𝑡)=𝐿, then −∞≤𝐿<0.
Thus, 𝑦^(𝑛−2)(𝑡)−𝑦^(𝑛−2)𝑡₀=^𝑡_𝑡0𝑦^(𝑛−1)(𝑡)𝑑𝑡≤(𝐿+𝜀)𝑡−𝑡₀⟶−∞,(𝑡⟶∞), (1.10) that is, lim_𝑡→∞𝑦^(𝑛−2)(𝑡)=−∞. Simile, lim_𝑡→∞𝑦^(𝑘)(𝑡)=−∞,(𝑘=0,1,…,𝑛−2), this is a contradiction and 𝑦(𝑡) is bounded, therefore, 𝑦^(𝑛−1)(𝑡)>0.
Again, since it decreases, set lim_𝑡→∞𝑦^(𝑛−1)(𝑡)=𝐿, thus 𝐿≥0. Next, we proof 𝐿=0. Assume 𝐿>0, then, 𝑦^(𝑛−2)(𝑡)−𝑦^(𝑛−2)𝑡₀=^𝑡_𝑡0𝑦^(𝑛−1)(𝑡)𝑑𝑡≥(𝐿−𝜀)𝑡−𝑡₀⟶∞,(𝑡⟶∞), (1.11) then, lim_𝑡→∞𝑦^(𝑛−2)(𝑡)=∞. Simile, lim_𝑡→∞𝑦^(𝑘)(𝑡)=∞,(𝑘=0,1,…,𝑛−2), this is a contradiction and 𝑦(𝑡) is bounded, therefore, 𝐿=0. That is, lim_𝑡→∞𝑦^(𝑛−1)(𝑡)=0. Similarly, we obtain (−1)^𝑘+1𝑦^(𝑘)(𝑡)<0,(𝑘=0,1,…,𝑛−1),lim_𝑡→∞𝑦^(𝑘)(𝑡)=0,(𝑘=1,2,…,𝑛−1). (1.12)Case 2 (If 𝑛 is a positive even integer). The proof of Case 2 is similar to that of part Case 1, therefore, it is omitted. We obtain (−1)^𝑘𝑦^(𝑘)(𝑡)<0,(𝑘=0,1,…,𝑛−1),lim_𝑡→∞𝑦^(𝑘)(𝑡)=0,(𝑘=1,2,…,𝑛−1). (1.13)
The proof is complete.

Lemma 1.2 (see [3, page 21]). Let 𝜂∈(−∞,0),𝜏∈(0,∞),𝑡₀∈𝑅, and suppose that a function 𝑥(𝑡)∈𝐶([𝑡₀−𝜏,∞),𝑅) satisfies the inequality: 𝑥(𝑡)≤𝜂+max_{𝑡−𝜏≤𝑠≤𝑡}𝑥(𝑠),𝑡≥𝑡₀. (1.14) Then, 𝑥(𝑡) cannot be a nonnegative function.

Lemma 1.3. Suppose that sup_𝑡≥𝑡0𝑟(𝑡)<∞, and ∫^𝑏_𝑎𝑝(𝑡,𝜏)𝑑𝜏/𝑎(𝑡)≤1. Let 𝑥(𝑡) be an eventually positive solution of (1.2) and set 𝑦(𝑡)=𝑟(𝑡)𝑥(𝑡)−^𝑏_𝑎𝑝(𝑡,𝜏)𝑥(𝑡−𝜏)𝑑𝜏, (1.15) then eventually 𝑦(𝑡)>0. (1.16)

Proof. From (1.2) and (1.7), eventually, we have 𝑦^(𝑛)(𝑡)≤−^𝑑_𝑐𝑞(𝑡,𝜎)𝑓(𝑥(𝑡−𝜎))𝑑𝜎≤0, (1.17) and the hypotheses on 𝑞(𝑡,𝜎),𝑓(𝑥), and 𝑥(𝑡) yield that 𝑦^(𝑛) is not eventually zero. Thus, 𝑦^(𝑖)(𝑡)(𝑖=0,𝑙,…,𝑛−1) is eventually nonzero. Hence, if 𝑦(𝑡)>0 does not hold, then eventually 𝑦(𝑡)<0,𝑦^′(𝑡)<0, or 𝑦(𝑡)<0,𝑦^′(𝑡)>0.Case 1. 𝑦(𝑡)<0,𝑦^′(𝑡)<0, then there exists 𝑡₁>0 such that 𝑦(𝑡)≤𝑦(𝑡₁)<0 for 𝑡>𝑡₁. Then, 𝛼∶=−𝑦(𝑡₁)/sup_𝑡≥𝑡0𝑟(𝑡)>0. In view of (1.3) and ∫^𝑏_𝑎𝑝(𝑡,𝜏)𝑑𝜏/𝑎(𝑡)≤1, we obtain 𝑥(𝑡)=𝑦(𝑡)𝑟(𝑡)+1𝑟(𝑡)^𝑏_𝑎𝑝()𝑥()1𝑡,𝜏𝑡−𝜏𝑑𝜏≤−𝛼+𝑟(𝑡)^b_𝑎𝑝(𝑡,𝜏)𝑑𝜏max_{𝑡−𝑏≤𝑠≤𝑡}𝑥(𝑠)≤−𝛼+max_{𝑡−𝑏≤𝑠≤𝑡}𝑥(𝑠). (1.18) According to Lemma 1.2, we obtain 𝑥(𝑡)<0. This is a contradiction and so 𝑦(𝑡) is eventually positive.Case 2. 𝑦(𝑡)<0,𝑦^′(𝑡)>0, one argues that 𝑦^′′(𝑡)<0 and repeats the argument to obtain that 𝑦^(𝑛)(𝑡)>0 which contradicts the offset after 𝑦^(𝑛)(𝑡)<0.
The proof is complete.

2. Comparison Theory of Existence for Eventually Positive Solution

Theorem 2.1. Assume all conditions of Lemma 1.1 hold, 𝑛 is a positive odd integer. And(H₁)𝑝(𝑡,𝜏)+𝑞(𝑡,𝜎)>0, for sufficiently large 𝑡.Then (1.1) has an eventually bounded positive solution if and only if inequality (1.2) has an eventually bounded positive solution.

Proof. It is clear that an eventually bounded positive solution of (1.1) is also an eventually bounded positive solution of (1.2). So, it suffices to prove that if (1.2) has an eventually bounded positive solution 𝑥(𝑡), for 𝑡>𝑡₀, then so does (1.1). Set 𝑦(𝑡)=𝑟(𝑡)𝑥(𝑡)−^𝑏_𝑎𝑝(𝑡,𝜏)𝑥(𝑡−𝜏)𝑑𝜏. (2.1) It follows from Lemma 1.1 and (1.2) that eventually, 𝑦^(𝑛)(𝑡)≤0,(−1)^𝑘+1𝑦^(𝑘)(𝑡)>0,(𝑘=0,1,…,𝑛−1),lim_𝑡→∞𝑦^𝑘(𝑡)=0,(𝑘=1,2,…,𝑛−1). (2.2) By using (1.8) and integrating (1.2) from 𝑡 to ∞, we obtain 𝑦^(𝑛−1)(∞)−𝑦^(𝑛−1)(𝑡)≤−^∞_𝑡^𝑑_𝑐𝑞()𝑓(𝑥(𝑦𝑡,𝜎𝑡−𝜎))𝑑𝜎𝑑𝑠,_(𝑛−1)(𝑡)≥^∞_𝑡^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠. (2.3) By repeating the same procedure 𝑛 times and by using (1.8), we are led to the inequality: 𝑦(𝑡)≥^∞_𝑡𝑑𝑡_𝑛^∞_𝑡𝑛𝑑𝑡_𝑛−1^∞_𝑡𝑛−1𝑑𝑡_𝑛−2⋯^∞_𝑡2^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑡₀. (2.4) Using Tonelli′s theorem, we reverse the order of integration and obtain =^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑡₀. (2.5) That is, 𝑥(𝑡)≥1𝑟(𝑡)^𝑏_𝑎𝑝()𝑥()1𝑡,𝜏𝑡−𝜏𝑑𝜏+𝑟(𝑡)^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑡₀. (2.6) Let 𝑇≥𝑡₀ be such that (2.6) hold, and 𝑥(𝑡−𝜇)>0,𝑡≥𝑇. Now, we consider the set of functions Ω=𝑧∈𝐶𝑇−𝜇,∞,𝑅⁺∶0≤𝑧(𝑡)≤1,𝑡≥𝑇−𝜇 (2.7) and define an operator 𝑆 on Ω as follows: (𝑆𝑧)(𝑡)=𝑡−𝑇+𝜇𝜇(𝑆𝑧)(𝑇)+1−𝑡−𝑇+𝜇𝜇1,𝑇−𝜇≤𝑡<𝑇,𝑥(𝑡)1𝑟(𝑡)^𝑏_𝑎𝑝()𝑧()𝑥()+1𝑡,𝜏𝑡−𝜏𝑡−𝜏𝑑𝜏𝑟(𝑡)^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑧(𝑠−𝜎)𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑇. (2.8) Then, it follows from Lebesgue′s dominated convergence theorem that 𝑆 is continuous. By using (2.6), it is easy to see that 𝑆 maps Ω into itself, and for any 𝑧∈Ω, we have (𝑆𝑧)(𝑡)>0, for 𝑇−𝜇≤𝑡<𝑇. Next, we define the sequence 𝑧_𝑘(𝑡)∈Ω𝑧₀(𝑡)𝑧≡1,𝑡≥𝑇−𝜇,_𝑘+1(𝑡)=𝑆𝑧_𝑘(𝑡),𝑡≥𝑇−𝜇,𝑘=0,1,…. (2.9) Then, by using (2.6) and a simple induction, we can easily see that 0≤𝑧_𝑘+1(𝑡)≤𝑧_𝑘(𝑡)≤1,𝑡≥𝑇−𝜇. (2.10) Set lim_𝑘→∞𝑧_𝑘(𝑡)=𝑧(𝑡),𝑡≥𝑇−𝜇 then 𝑧(𝑡) satisfies 𝑧(𝑡)=𝑡−𝑇+𝜇𝜇(𝑆𝑧)(𝑇)+1−𝑡−𝑇+𝜇𝜇1,𝑇−𝜇≤𝑡<𝑇,𝑥(𝑡)1𝑟(𝑡)^𝑏_𝑎𝑝()𝑧()𝑥()+1𝑡,𝜏𝑡−𝜏𝑡−𝜏𝑑𝜏𝑟(𝑡)^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑧(𝑠−𝜎)𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑇. (2.11) Again, set 𝜔(𝑡)=𝑧(𝑡)𝑥(𝑡), then 𝜔(𝑡) satisfies 𝜔(𝑡)>0,𝑇−𝜇≤𝑡≤𝑇, and 𝜔(𝑡)=1𝑟(𝑡)^𝑏_𝑎𝑝()𝜔()1𝑡,𝜏𝑡−𝜏𝑑𝜏+𝑟(𝑡)^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝜔(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑇. (2.12) Thus, 𝜔(𝑡) is a positive solution of (1.1) for 𝑡≥𝑇.
The following: 𝜔(𝑡)>0,𝑡≥𝑇−𝜇. (2.13) Assume that there exists 𝑡^∗≥𝑇−𝜇, such that 𝜔(𝑡)>0, for 𝑇−𝜇≤𝑡≤𝑡^∗, and 𝜔(𝑡^∗)=0. Then, 0=𝜔𝑡^∗=1𝑟(𝑡^∗)^𝑏_𝑎𝑝𝑡^∗,𝜏𝜔𝑡^∗+1−𝜏𝑑𝜏𝑟(𝑡^∗)^∞_𝑡^∗𝑠−𝑡^∗𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝜔(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡^∗≥𝑇, (2.14) which implies 𝑝𝑡^∗𝑞()𝑓(𝜔(,𝜏=0,𝑠,𝜎𝑠−𝜎))≡0, (2.15) which contradicts (H₁). Thus, 𝜔(𝑡) is an eventually bounded positive solution of (1.1).
The proof is complete.

Theorem 2.2. Assume all conditions of Lemma 1.3 hold, 𝑛 is a positive odd integer. And(H₁)𝑝(𝑡,𝜏)+𝑞(𝑡,𝜎)>0, for sufficiently large 𝑡.Then, (1.1) has an eventually positive solution if and only if inequality (1.2) has an eventually positive solution.

Proof. It is clear that an eventually positive solution of (1.1) is also an eventually positive solution of (1.2). So, it suffices to prove that if (1.2) has an eventually positive solution 𝑥(𝑡), for 𝑡>𝑡₀, then so does (1.1). Set 𝑦(𝑡)=𝑟(𝑡)𝑥(𝑡)−^𝑏_𝑎𝑝(𝑡,𝜏)𝑥(𝑡−𝜏)𝑑𝜏. (2.16)
It follows from Lemma 1.3 and (1.2) that eventually, 𝑦^(𝑛)(𝑡)≤0,𝑦(𝑡)>0, which implies that there exists a nonnegative even integer 𝑛^∗≤𝑛−1, such that eventually 𝑦^(𝑘)(𝑡)>0,𝑘=0,1,…,𝑛^∗,(−1)^(𝑘)𝑦^(𝑘)(𝑡)>0,𝑘=𝑛^∗,…,𝑛−1. (2.17) We consider the following possible cases.
Case 1 (𝑛^∗=0). Since 𝑛 is a positive integer, 𝑛−1 is an even integer, we can easily see that there exists a 𝑇^′>0, such 𝑦^(𝑛−1)(𝑡)>0, and 𝑦^𝑛(𝑡)≤0,𝑡≥𝑇^′,𝑦^(𝑛−1)(∞)≥0.
By using (2.17) and integrating (1.2) from 𝑡 to ∞, we obtain 𝑦^(𝑛−1)(∞)−𝑦^(𝑛−1)(𝑡)≤−^∞_𝑡^𝑑_𝑐𝑞()𝑓(𝑥(𝑦𝑠,𝜎𝑠−𝜎))𝑑𝜎𝑑𝑠,_(𝑛−1)(𝑡)≥^∞_𝑡^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠. (2.18) By repeating the same procedure 𝑛 times and by using (2.31), we are led to the inequality: 𝑦(𝑡)≥^∞_𝑡𝑑𝑡_𝑛^∞_𝑡𝑛𝑑𝑡_𝑛−1^∞_𝑡𝑛−1𝑑𝑡_𝑛−2⋯^∞_𝑡2^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑡₀. (2.19) Using Tonelli′s theorem, we reverse the order of integration and obtain =^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑡₀. (2.20) That is, 𝑥(𝑡)≥1𝑟(𝑡)^𝑏_𝑎𝑝()𝑥()1𝑡,𝜏𝑡−𝜏𝑑𝜏+𝑟(𝑡)^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑡,𝜎)𝑓(𝑥(𝑡−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑡₀. (2.21) Let 𝑇≥𝑡₀ be such that (2.21) hold, and 𝑥(𝑡−𝜇)>0,𝑡≥𝑇. Now, we consider the set of functions Ω=𝑧∈𝐶𝑇−𝜇,∞,𝑅⁺∶0≤𝑧(𝑡)≤1,𝑡≥𝑇−𝜇, (2.22) and define an operator 𝑆 on Ω as follows: (𝑆𝑧)(𝑡)=𝑡−𝑇+𝜇𝜇(𝑆𝑧)(𝑇)+1−𝑡−𝑇+𝜇𝜇1,𝑇−𝜇≤𝑡<𝑇,𝑥(𝑡)1𝑟(𝑡)^𝑏_𝑎𝑝()𝑧()𝑥()+1𝑡,𝜏𝑡−𝜏𝑡−𝜏𝑑𝜏𝑟(𝑡)^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑧(𝑠−𝜎)𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑇. (2.23) Then, it follows from Lebesgue′s dominated convergence theorem that 𝑆 is continuous. By using (2.21), it is easy to see that 𝑆 maps Ω into itself, and for any 𝑧∈Ω, we have (𝑆𝑧)(𝑡)>0, for 𝑇−𝜇≤𝑡<𝑇. Next, we define the sequence 𝑧_𝑘(𝑡)∈Ω𝑧₀(𝑡)𝑧≡1,𝑡≥𝑇−𝜇,_𝑘+1(𝑡)=𝑆𝑧_𝑘(𝑡),𝑡≥𝑇−𝜇,𝑘=0,1,…. (2.24) Then, by using (2.21) and a simple induction, we can easily see that 0≤𝑧_𝑘+1(𝑡)≤𝑧_𝑘(𝑡)≤1,𝑡≥𝑇−𝜇. (2.25) Set lim_𝑘→∞𝑧_𝑘(𝑡)=𝑧(𝑡),𝑡≥𝑇−𝜇, then 𝑧(𝑡) satisfies: 𝑧(𝑡)=𝑡−𝑇+𝜇𝜇(𝑆𝑧)(𝑇)+1−𝑡−𝑇+𝜇𝜇1,𝑇−𝜇≤𝑡<𝑇,𝑥(𝑡)1𝑟(𝑡)^𝑏_𝑎𝑝()𝑧()𝑥()+1𝑡,𝜏𝑡−𝜏𝑡−𝜏𝑑𝜏𝑟(𝑡)^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑧(𝑠−𝜎)𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑇. (2.26) Again, set 𝜔(𝑡)=𝑧(𝑡)𝑥(𝑡), then 𝜔(𝑡) satisfies 𝜔(𝑡)>0,𝑇−𝜇≤𝑡≤𝑇, and 𝜔(𝑡)=1𝑟(𝑡)^𝑏_𝑎𝑝()𝜔()1𝑡,𝜏𝑡−𝜏𝑑𝜏+𝑟(𝑡)^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝜔(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑇. (2.27) Thus, 𝜔(𝑡) is a positive solution of (1.1) for 𝑡≥𝑇.
Consider the following: 𝜔(𝑡)>0,𝑡≥𝑇−𝜇. (2.28) Assume that there exists 𝑡^∗≥𝑇−𝜇, such that 𝜔(𝑡)>0, for 𝑇−𝜇≤𝑡≤𝑡^∗, and 𝜔(𝑡^∗)=0. Then, 0=𝜔𝑡^∗=1𝑟(𝑡^∗)^𝑏_𝑎𝑝𝑡^∗,𝜏𝜔𝑡^∗+1−𝜏𝑑𝜏𝑟(𝑡^∗)^∞_𝑡^∗𝑠−𝑡^∗𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝜔(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡^∗≥𝑇, (2.29) which implies 𝑝𝑡^∗𝑞()𝑓(𝜔(,𝜏=0,𝑠,𝜎𝑠−𝜎))≡0, (2.30) which contradicts (H₁). Thus, 𝜔(𝑡) is an eventually positive solution of equation.
Case 2 (2≤𝑛^∗≤𝑛−1). By using (2.17) and integrating (1.2) from 𝑡 to ∞, we obtain 𝑦^(𝑛∗)(𝑡)≥^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑡₀. (2.31) Let 𝑇≥0 be such that (2.17) and (H₁) hold. Integrating (2.31) from 𝑇 to 𝑡 and using (2.17), we have 𝑥(𝑡)≥1𝑟(𝑡)^𝑏_𝑎𝑝()𝑥()+1𝑡,𝜏𝑡−𝜏𝑑𝜏𝑟(𝑡)^𝑡_𝑇(𝑡−𝑠)^𝑛∗−1(𝑛^∗−1)!^∞_𝑡(𝑢−𝑠)^{𝑛−𝑛∗−1}(𝑛−𝑛^∗−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑢𝑑𝑠,𝑡≥𝑡₀. (2.32) Using a method similar to the proof of Case 1 yields that (1.1) also has an eventually positive solution.
The proof is complete.

Theorem 2.3. Assume all conditions of Lemma 1.1 hold, 𝑛 is a positive even integer, and(H₁)𝑝(𝑡,𝜏)+𝑞(𝑡,𝜎)>0, for sufficiently large 𝑡.Then, (1.1) has an eventually positive solution if and only if inequality (1.2) has an eventually positive solution.

Proof. It is clear that an eventually bounded positive solution of (1.1) is also an eventually bounded positive solution of (1.2). So, it suffices to prove that if (1.2) has an eventually bounded positive solution 𝑥(𝑡), for 𝑡>𝑡₀, then so does (1.1). Set 𝑦(𝑡)=𝑟(𝑡)𝑥(𝑡)−^𝑏_𝑎𝑝(𝑡,𝜏)𝑥(𝑡−𝜏)𝑑𝜏. (2.33) It follows from Lemma 1.1 and (1.2) that eventually, 𝑦^(𝑛)(𝑡)≤0,(−1)^𝑘𝑦^(𝑘)(𝑡)<0,(𝑘=0,1,…,𝑛−1),lim_𝑡→∞𝑦^𝑘(𝑡)=0,(𝑘=1,2,…,𝑛−1). (2.34) By using (1.9) and integrating (1.2) from 𝑡 to ∞, we obtain 𝑦^(𝑛−1)(∞)−𝑦^(𝑛−1)(𝑡)≤−^∞_𝑡^𝑑_𝑐𝑞()𝑓(𝑥(𝑦𝑠,𝜎𝑠−𝜎))𝑑𝜎𝑑𝑠,_(𝑛−1)(𝑡)≥^∞_𝑡^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠. (2.35) By repeating the same procedure 𝑛 times and by using (1.9), we are led to the inequality: 𝑦(𝑡)≤−^∞_𝑡𝑑𝑡_𝑛^∞_𝑡𝑛𝑑𝑡_𝑛−1^∞_𝑡𝑛−1𝑑𝑡_𝑛−2⋯^∞_𝑡2^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑡₀. (2.36) Using Tonelli′s theorem, we reverse the order of integration and obtain =−^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑡₀. (2.37) That is, 𝑥(𝑡)≤1𝑟(𝑡)^𝑏_𝑎𝑝(𝑡,𝜏)𝜔(𝑡−𝜏)𝑑𝜏−^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝑥(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑡₀. (2.38) Let 𝑇≥0 be such that (2.38) hold, and 𝑥(𝑡−𝜇)>0,𝑡≥𝑇. Now, we consider the set of functions: Ω=𝑧∈𝐶𝑇−𝜇,∞,𝑅∶0<𝑧(𝑡)≤1,𝑡≥𝑇−𝜇, (2.39) and define an operator 𝑆 on Ω as follows: (𝑆𝑧)(𝑡)=𝑡−𝑇+𝜇𝜇(𝑆𝑧)(𝑇)+1−𝑡−𝑇+𝜇𝜇,𝑥(𝑡)𝑇−𝜇≤𝑡<𝑇,1𝑟(𝑡)∫^𝑏_𝑎𝑝()𝑥()𝑡,𝜏𝑡−𝜏𝑧()1𝑡−𝜏𝑑𝜏−𝑟(𝑡)∫^∞_𝑡(𝑠−𝑡)^𝑛−1()!∫𝑛−1^𝑑𝑐𝑞()𝑓𝑥()𝑠,𝜎𝑠−𝜎𝑧(),𝑠−𝜎𝑑𝜎𝑑𝑠𝑡≥𝑇. (2.40) Then, it follows from Lebesgue′s dominated convergence theorem that 𝑆 is continuous. By using (2.38), it is easy to see that 𝑆 maps Ω into itself and, for any 𝑧∈Ω, we have (𝑆𝑧)(𝑡)<0, for 𝑇−𝜇≤𝑡<𝑇. Next, we define the sequence 𝑧_𝑘(𝑡)∈Ω: 𝑧₀(𝑡)𝑧≡1,𝑡≥𝑇−𝜇,_𝑘+1(𝑡)=𝑆𝑧_𝑘(𝑡),𝑡≥𝑇−𝜇,𝑘=0,1,…. (2.41) Then, by using (2.38) and a simple induction, we can easily see that 0<𝑧_𝑘+1(𝑡)≤𝑧_𝑘(𝑡)≤1,𝑡≥𝑇−𝜇. (2.42) Set lim_𝑘→∞𝑧_𝑘(𝑡)=𝑧(𝑡),𝑡≥𝑇−𝜇, then 𝑧(𝑡) satisfies 𝑧(𝑡)=𝑡−𝑇+𝜇𝜇(𝑆𝑧)(𝑇)+1−𝑡−𝑇+𝜇𝜇,𝑥(𝑡)𝑇−𝜇≤𝑡<𝑇,1𝑟(𝑡)∫^𝑏_𝑎𝑝()𝑥()𝑡,𝜏𝑡−𝜏𝑧()1𝑡−𝜏𝑑𝜏−𝑟(𝑡)∫^∞_𝑡(𝑠−𝑡)^𝑛−1()!∫𝑛−1^𝑑𝑐𝑞()𝑓𝑥()𝑠,𝜎𝑠−𝜎𝑧(),𝑠−𝜎𝑑𝜎𝑑𝑠𝑡≥𝑇. (2.43) Again, set 𝜔(𝑡)=𝑥(𝑡)/𝑧(𝑡), then 𝜔(𝑡) satisfies 𝜔(𝑡)>0,𝑇−𝜇≤𝑡≤𝑇, and 𝜔(𝑡)=1𝑟(𝑡)^𝑏_𝑎𝑝()𝜔()−1𝑡,𝜏𝑡−𝜏𝑑𝜏𝑟(𝑡)^∞_𝑡(𝑠−𝑡)^𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝜔(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡≥𝑇. (2.44) Thus, 𝜔(𝑡) is a positive solution of (1.1) for 𝑡≥𝑇.
Consider the following 𝜔(𝑡)>0,𝑡≥𝑇−𝜇. (2.45) Assume that there exists 𝑡^∗≥𝑇−𝜇, such that 𝜔(𝑡)>0, for 𝑇−𝜇≤𝑡≤𝑡^∗, and 𝜔(𝑡^∗)=0. Then, 0=𝜔𝑡^∗=1𝑟(𝑡^∗)^𝑏_𝑎𝑝𝑡^∗,𝜏𝜔𝑡^∗−1−𝜏𝑑𝜏𝑟(𝑡^∗)^∞_𝑡^∗𝑠−𝑡^∗𝑛−1(𝑛−1)!^𝑑_𝑐𝑞(𝑠,𝜎)𝑓(𝜔(𝑠−𝜎))𝑑𝜎𝑑𝑠,𝑡^∗≥𝑇, (2.46) which implies 𝑝𝑡^∗𝑞()𝑓(𝜔(,𝜏=0,𝑠,𝜎𝑠−𝜎))≡0, (2.47) which contradicts (H₁). Thus, 𝜔(𝑡) is an eventually positive solution of (1.1).
The proof is complete.

Example 2.4. Consider high-order neutral differential equation with distributed delay 𝑥(𝑡)−²₁𝑒^−𝑡𝑥(𝑡−𝜏)𝑑𝜏^(𝑛)+12𝑒⁻¹−3𝑒⁻²²₁𝜎𝑥(𝑡−𝜎)𝑑𝜎=0. (2.48) Here, 𝑟(𝑡)≡1,𝑝(𝑡−𝜏)=𝑒^−𝑡,𝑞(𝑡,𝜎)=𝜎/2𝑒⁻¹−3𝑒⁻²,𝑎=𝑐=1,𝑏=𝑑=2. It is easy to see that 𝑟(𝑡)∫≡1<∞,²¹𝑒^−𝑡𝑑𝜏1=𝑒^−𝑡<1,𝑡>0. (2.49) From Theorem 2.2, we have that (1.1) has an eventually positive solution if and only if inequality (1.2) has an eventually positive solution. In fact, 𝑥(𝑡)=𝑒^𝑡 is a positive solution of (1.1).

Acknowledgments

This research is supported by the Natural Sciences Foundation of Shanxi Province and Scientific Research Project of Shanxi Datong University. The authors are greatly indebted to the referees for their valuable suggestions and comments.