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# Nonoscillatory solutions to third-order neutral dynamic equations on time scales

## Abstract

In this paper, we establish the existence of nonoscillatory solutions to third-order nonlinear neutral dynamic equations on time scales of the form $( r 1 ( t ) ( r 2 ( t ) ( x ( t ) + p ( t ) x ( g ( t ) ) ) Δ ) Δ ) Δ +f(t,x(h(t)))=0$ by employing Kranoselskii’s fixed point theorem. Three examples are included to illustrate the significance of the conclusions.

## 1 Introduction

In this paper, we study third-order nonlinear neutral dynamic equations of the form

$( r 1 ( t ) ( r 2 ( t ) ( x ( t ) + p ( t ) x ( g ( t ) ) ) Δ ) Δ ) Δ +f ( t , x ( h ( t ) ) ) =0$
(1)

on a time scale $T$ satisfying $infT= t 0$ and $supT=∞$.

Throughout this paper we shall assume that:

(C1) $r 1 , r 2 ∈ C r d (T,(0,∞))$ such that

$∫ t 0 ∞ 1 r 1 ( t ) Δt=∞, ∫ t 0 ∞ 1 r 2 ( t ) Δt=∞.$

(C2) $p∈ C r d (T,R)$ and there exists a constant $p 0$ with $| p 0 |<1$ such that

$lim t → ∞ p(t)= p 0 .$

(C3) $g,h∈ C r d (T,T)$, $g(t)≤t$, $lim t → ∞ g(t)= lim t → ∞ h(t)=∞$, and

$lim t → ∞ R λ ( g ( t ) ) R λ ( t ) = η λ ∈(0,1],λ=1,2,$

where

$R 1 (t)=1+ ∫ t 0 t 1 r 2 ( s ) Δs, R 2 (t)=1+ ∫ t 0 t ∫ t 0 s 1 r 1 ( u ) r 2 ( s ) ΔuΔs.$

If $p 0 ∈(−1,0]$, there exists a sequence ${ c k } k ≥ 0$ such that $lim k → ∞ c k =∞$ and $g( c k + 1 )= c k$.

(C4) $f∈C(T×R,R)$, $f(t,x)$ is nondecreasing in x and $xf(t,x)>0$ for $t∈T$ and $x≠0$.

Hilger introduced the theory of time scales in his Ph.D. thesis  in 1988; see also . More details of time scale calculus can be found in  and omitted here. In the last few years, there has been some research achievement as regards the existence of nonoscillatory solutions to neutral dynamic equations on time scales; see the papers  and the references therein.

Definition 1.1 By a solution of (1) we mean a continuous function $x(t)$ which is defined on $T$ and satisfies (1) for $t≥ t 0$. A solution $x(t)$ of (1) is said to be eventually positive (or eventually negative) if there exists $c∈T$ such that $x(t)>0$ (or $x(t)<0$) for all $t≥c$ in $T$. A solution x of (1) is said to be nonoscillatory if it is either eventually positive or eventually negative; otherwise, it is oscillatory.

In 1990s, some significative results for existence of nonoscillatory solutions to neutral functional differential equations were given in [7, 9]. In 2007, Zhu and Wang  discussed the existence of nonoscillatory solutions to first-order nonlinear neutral dynamic equations

$[ x ( t ) + p ( t ) x ( g ( t ) ) ] Δ +f ( t , x ( h ( t ) ) ) =0$

on a time scale $T$. In 2013, Gao and Wang  considered the second-order nonlinear neutral dynamic equations

$[ r ( t ) ( x ( t ) + p ( t ) x ( g ( t ) ) ) Δ ] Δ +f ( t , x ( h ( t ) ) ) =0$
(2)

under the condition $∫ t 0 ∞ 1 r ( s ) Δs<∞$, and established the existence of nonoscillatory solutions to (2) on a time scale. In 2014, Deng and Wang  studied the same problem of (2) under the condition $∫ t 0 ∞ 1 r ( s ) Δs=∞$.

In this paper, we shall establish the existence of nonoscillatory solutions to (1) by employing Kranoselskii’s fixed point theorem, and we give three examples to show the versatility of the results.

For simplicity, throughout this paper, we denote $(a,b)∩T= ( a , b ) T$, where $a,b∈R$, and $[ a , b ] T$, $[ a , b ) T$, $( a , b ] T$ are denoted similarly.

## 2 Preliminary results

Let $C( [ T 0 , ∞ ) T ,R)$ denote all continuous functions mapping $[ T 0 , ∞ ) T$ into , and $R 0 (t)≡1$, $t∈ [ T 0 , ∞ ) T$. For $λ=0,1,2$, we define

(3)

Endowing $B C λ [ T 0 , ∞ ) T$ with the norm $∥ x ∥ λ = sup t ∈ [ T 0 , ∞ ) T | x ( t ) R λ 2 ( t ) |$, $(B C λ [ T 0 , ∞ ) T , ∥ ⋅ ∥ λ )$ is a Banach space. Let $X⊆B C λ [ T 0 , ∞ ) T$, we say that X is uniformly Cauchy if for any given $ϵ>0$, there exists a $T 1 ∈ [ T 0 , ∞ ) T$ such that, for any $x∈X$,

X is said to be equi-continuous on $[ a , b ] T$ if, for any given $ϵ>0$, there exists a $δ>0$ such that, for any $x∈X$ and $t 1 , t 2 ∈ [ a , b ] T$ with $| t 1 − t 2 |<δ$,

$| x ( t 1 ) R λ 2 ( t 1 ) − x ( t 2 ) R λ 2 ( t 2 ) |<ϵ.$

We have the following lemma, which is an analog of the Arzela-Ascoli theorem on time scales.

Lemma 2.1 ([, Lemma 4])

Suppose that $X⊆B C λ [ T 0 , ∞ ) T$ is bounded and uniformly Cauchy. Further, suppose that X is equi-continuous on $[ T 0 , T 1 ] T$ for any $T 1 ∈ [ T 0 , ∞ ) T$. Then X is relatively compact.

In this section, our approach to the existence of nonoscillatory solutions to (1) is based largely on the application of Kranoselskii’s fixed point theorem (see ). For the sake of convenience, we state here this theorem as follows.

Lemma 2.2 (Kranoselskii’s fixed point theorem)

Suppose that X is a Banach space and Ω is a bounded, convex, and closed subset of X. Suppose further that there exist two operators $U,S:Ω→X$ such that

1. (i)

$Ux+Sy∈Ω$ for all $x,y∈Ω$;

2. (ii)

U is a contraction mapping;

3. (iii)

S is completely continuous.

Then $U+S$ has a fixed point in Ω.

If $x(t)$ is an eventually negative solution of (1), then $y(t)=−x(t)$ will satisfy

$( r 1 ( t ) ( r 2 ( t ) ( y ( t ) + p ( t ) y ( g ( t ) ) ) Δ ) Δ ) Δ −f ( t , − y ( h ( t ) ) ) =0.$

We may note that $f ¯ (t,u):=−f(t,−u)$ is nondecreasing in u and $u f ¯ (t,u)>0$ for $t∈T$ and $u≠0$. Therefore, we will restrict our attention to eventually positive solutions of (1) in the following.

In the sequel, we use the notation

$z(t):=x(t)+p(t)x ( g ( t ) )$
(4)

and have the following lemma.

Lemma 2.3 ([, Lemma 2.3])

Suppose that $x(t)$ is an eventually positive solution of (1) and $lim t → ∞ z ( t ) R λ i ( t ) =a$ for $λ=1,2$ and $i=0,1$. Then we have:

1. (i)

If a is finite, then

$lim t → ∞ x ( t ) R λ i ( t ) = a 1 + p 0 η λ i .$
2. (ii)

If a is infinite, then $x ( t ) R λ i ( t )$ is unbounded, or

$lim sup t → ∞ x ( t ) R λ i ( t ) =+∞.$

Let $S +$ denote the set of all eventually positive solutions of (1) and

$A(α,β,γ)= { x ∈ S + : lim t → ∞ x ( t ) = α , lim t → ∞ x ( t ) R 1 ( t ) = β , lim t → ∞ x ( t ) R 2 ( t ) = γ } .$

Now, we give the first theorem for a classification scheme of eventually positive solutions to (1).

Theorem 2.4 If $x(t)$ is an eventually positive solution of (1), then $x(t)$ belongs to $A(0,0,0)$, $A(b,0,0)$, $A(∞,b,0)$, $A(∞,∞,b)$ for some positive b, or $A(∞,∞,0)$.

Proof Suppose that $x(t)$ is an eventually positive solution of (1). From (C2) and (C3), there exist $t 1 ∈ [ t 0 , ∞ ) T$ and $| p 0 |< p 1 <1$ such that $x(t)>0$, $x(g(t))>0$, $x(h(t))>0$, and $|p(t)|≤ p 1$ for $t∈ [ t 1 , ∞ ) T$. By (1) and (C4), it follows that, for $t∈ [ t 1 , ∞ ) T$,

$( r 1 ( t ) ( r 2 ( t ) z Δ ( t ) ) Δ ) Δ =−f ( t , x ( h ( t ) ) ) <0.$

Hence, $r 1 (t) ( r 2 ( t ) z Δ ( t ) ) Δ$ is strictly decreasing on $[ t 1 , ∞ ) T$. We claim that

$r 1 (t) ( r 2 ( t ) z Δ ( t ) ) Δ >0,t∈ [ t 1 , ∞ ) T .$
(5)

Assume not; then there exists $t 2 ∈ [ t 1 , ∞ ) T$ such that $r 1 (t) ( r 2 ( t ) z Δ ( t ) ) Δ <0$ for $t∈ [ t 2 , ∞ ) T$. So there exist a constant $c<0$ and $t 3 ∈ [ t 2 , ∞ ) T$ such that $r 1 (t) ( r 2 ( t ) z Δ ( t ) ) Δ ≤c$ for $t∈ [ t 3 , ∞ ) T$, which means that

$( r 2 ( t ) z Δ ( t ) ) Δ ≤ c r 1 ( t ) ,t∈ [ t 3 , ∞ ) T .$
(6)

Integrating (6) from $t 3$ to $t∈ [ σ ( t 3 ) , ∞ ) T$, we obtain

$r 2 (t) z Δ (t)≤ r 2 ( t 3 ) z Δ ( t 3 )+c ∫ t 3 t Δ s r 1 ( s ) .$

Letting $t→∞$, by (C1) we have $r 2 (t) z Δ (t)→−∞$. Then there exists $t 4 ∈ [ t 3 , ∞ ) T$ such that $r 2 (t) z Δ (t)≤ r 2 ( t 4 ) z Δ ( t 4 )<0$ for $t∈ [ t 4 , ∞ ) T$, which implies that

$z Δ (t)≤ r 2 ( t 4 ) z Δ ( t 4 )⋅ 1 r 2 ( t ) .$
(7)

Integrating (7) from $t 4$ to $t∈ [ σ ( t 4 ) , ∞ ) T$, we obtain

$z(t)−z( t 4 )≤ r 2 ( t 4 ) z Δ ( t 4 ) ∫ t 4 t Δ s r 2 ( s ) .$

Letting $t→∞$, by (C1) we have $z(t)→−∞$. From (4), it follows that $p 0 ∈(−1,0]$, then there exists $t 5 ∈ [ t 4 , ∞ ) T$ such that $z(t)<0$ or

$x(t)<−p(t)x ( g ( t ) ) < p 1 x ( g ( t ) ) ,t∈ [ t 5 , ∞ ) T .$

By (C3), we can choose some positive integer $k 0$ such that $c k ∈ [ t 5 , ∞ ) T$ for all $k≥ k 0$. Then for any $k≥ k 0 +1$, we have

$x ( c k ) < p 1 x ( g ( c k ) ) = p 1 x ( c k − 1 ) < p 1 2 x ( g ( c k − 1 ) ) = p 1 2 x ( c k − 2 ) < ⋯ < p 1 k − k 0 x ( g ( c k 0 + 1 ) ) = p 1 k − k 0 x ( c k 0 ) .$

The inequality above implies that $lim k → ∞ x( c k )=0$. It follows from (4) that $lim k → ∞ z( c k )=0$ and then contradicts $lim t → ∞ z(t)=−∞$. So (5) holds, and

$lim t → ∞ r 1 (t) ( r 2 ( t ) z Δ ( t ) ) Δ = L 2 ,$
(8)

where $0≤ L 2 <+∞$.

From (5), we have $( r 2 ( t ) z Δ ( t ) ) Δ >0$ for $t∈ [ t 1 , ∞ ) T$, which means that $r 2 (t) z Δ (t)$ is strictly increasing on $[ t 1 , ∞ ) T$. Hence, $r 2 (t) z Δ (t)$ is either eventually positive or eventually negative. When $r 2 (t) z Δ (t)$ is eventually negative, we have $lim t → ∞ r 2 (t) z Δ (t)≤0$. Assume that there exists a constant $d<0$ such that

$lim t → ∞ r 2 (t) z Δ (t)=d,$

which means that

$z Δ (t)≤ d r 2 ( t ) ,t∈ [ t 1 , ∞ ) T .$
(9)

Integrating (9) from $t 1$ to $t∈ [ σ ( t 1 ) , ∞ ) T$, we obtain

$z(t)≤z( t 1 )+d ∫ t 1 t Δ s r 2 ( s ) .$

Letting $t→∞$, by (C1) we have $z(t)→−∞$. Similarly, it will cause the contradiction as before. Hence, $lim t → ∞ r 2 (t) z Δ (t)=0$. When $r 2 (t) z Δ (t)$ is eventually positive, we have $lim t → ∞ r 2 (t) z Δ (t)=b$ for some positive b or $lim t → ∞ r 2 (t) z Δ (t)=+∞$. Therefore,

$lim t → ∞ r 2 (t) z Δ (t)= L 1 ,$
(10)

where $0≤ L 1 ≤+∞$.

When $r 2 (t) z Δ (t)$ is eventually negative, which means that $z Δ (t)$ is eventually negative, then there exists $t 6 ∈ [ t 1 , ∞ ) T$ such that $z Δ (t)<0$ for $t∈ [ t 6 , ∞ ) T$. It follows that $z(t)$ is strictly decreasing on $[ t 6 , ∞ ) T$. Hence, $z(t)$ is either eventually positive or eventually negative. If $z(t)$ is eventually negative, we have $lim t → ∞ z(t)=−∞$ or $lim t → ∞ z(t)<0$. Similarly, it will cause the contradiction as before. Therefore, $z(t)$ is eventually positive, which means that $lim t → ∞ z(t)=b$ for some positive b or $lim t → ∞ z(t)=0$.

When $r 2 (t) z Δ (t)$ is eventually positive, it implies that $z Δ (t)$ is eventually positive. If $z(t)$ is eventually negative, we have $lim t → ∞ z(t)≤0$. Assume that $lim t → ∞ z(t)<0$. It will cause a similar contradiction to the one before. So $lim t → ∞ z(t)=0$. If $z(t)$ is eventually positive, we have $lim t → ∞ z(t)=b$ for some positive b or $lim t → ∞ z(t)=+∞$.

Therefore,

$lim t → ∞ z(t)= L 0 ,$

where $0≤ L 0 ≤+∞$.

It follows from L’Hôpital’s rule (see [, Theorem 1.120]) and (8), (10) that

$lim t → ∞ r 2 (t) z Δ (t)= lim t → ∞ z ( t ) R 1 ( t ) = L 1$

and

$lim t → ∞ r 1 (t) ( r 2 ( t ) z Δ ( t ) ) Δ = lim t → ∞ z ( t ) R 2 ( t ) = L 2 .$

When $L 0 =0$ or $L 0 =b$ for some positive b, we have $L 1 = L 2 =0$. When $L 0 =+∞$, it implies that $z Δ (t)$ is eventually positive, which means that $r 2 (t) z Δ (t)$ is eventually positive. It follows that $L 1 =b$ for some positive b or $L 1 =+∞$. We have $L 2 =0$ if $L 1 =b$ for some positive b, and $L 2 =0$ or $L 2 =b$ for some positive b if $L 1 =+∞$. Then by Lemma 2.3, we see that $x(t)$ must belong to $A(0,0,0)$, $A(b,0,0)$, $A(∞,b,0)$, $A(∞,∞,b)$ for some positive b, or $A(∞,∞,0)$. The proof is complete. □

## 3 Main results

In this section, by employing Kranoselskii’s fixed point theorem, we establish the existence criteria for each type of eventually positive solutions to (1).

Theorem 3.1 Equation (1) has an eventually positive solution in $A(∞,∞,b)$ for some positive b if and only if there exists some constant $K>0$ such that

$∫ t 0 ∞ f ( t , K R 2 ( h ( t ) ) ) Δt<∞.$
(11)

Proof Suppose that $x(t)$ is an eventually positive solution of (1) in $A(∞,∞,b)$, i.e.,

$lim t → ∞ x(t)=∞, lim t → ∞ x ( t ) R 1 ( t ) =∞, lim t → ∞ x ( t ) R 2 ( t ) =b.$
(12)

Assume that $lim t → ∞ z(t)<∞$ (or $lim t → ∞ z ( t ) R 1 ( t ) <∞$). By Lemma 2.3 we have $lim t → ∞ x(t)<∞$ (or $lim t → ∞ x ( t ) R 1 ( t ) <∞$), which contradicts (12). Then we have

$lim t → ∞ z ( t ) = ∞ , lim t → ∞ z ( t ) R 1 ( t ) = ∞ , lim t → ∞ r 1 ( t ) ( r 2 ( t ) z Δ ( t ) ) Δ = lim t → ∞ z ( t ) R 2 ( t ) = ( 1 + p 0 η 2 ) b$

and there exists $T 1 ∈ [ t 0 , ∞ ) T$ such that $x(t)>0$, $x(g(t))>0$, $x(h(t))≥ b 2 R 2 (h(t))$ for $t∈ [ T 1 , ∞ ) T$. Integrating (1) from $T 1$ to $s∈ [ σ ( T 1 ) , ∞ ) T$, we obtain

$r 1 (s) ( r 2 ( s ) z Δ ( s ) ) Δ − r 1 ( T 1 ) ( r 2 ( T 1 ) z Δ ( T 1 ) ) Δ =− ∫ T 1 s f ( u , x ( h ( u ) ) ) Δu.$

Letting $s→∞$, we have

$∫ T 1 ∞ f ( u , x ( h ( u ) ) ) Δu<∞.$

In view of (C4), it follows that

$f ( u , b 2 R 2 ( h ( u ) ) ) ≤f ( u , x ( h ( u ) ) ) ,u∈ [ T 1 , ∞ ) T ,$

and

$∫ T 1 ∞ f ( u , b 2 R 2 ( h ( u ) ) ) Δu≤ ∫ T 1 ∞ f ( u , x ( h ( u ) ) ) Δu<∞,$

which means that (11) holds. The necessary condition is proved.

Conversely, suppose that there exists some constant $K>0$ such that (11) holds. There will be two cases to be considered: $0≤ p 0 <1$ and $−1< p 0 <0$.

Case 1: $0≤ p 0 <1$. Take $p 1$ such that $p 0 < p 1 <(1+4 p 0 )/5<1$, then $p 0 >(5 p 1 −1)/4$.

When $p 0 >0$, since $lim t → ∞ p(t)= p 0$ and (11) hold, we can choose a sufficiently large $T 0 ∈ [ t 0 , ∞ ) T$ such that $p(t)>0$ for $t∈ [ T 0 , ∞ ) T$, and

$5 p 1 − 1 4 ≤p(t)≤ p 1 <1,p(t) R 2 ( g ( t ) ) R 2 ( t ) ≥ 5 p 1 − 1 4 η 2 ,t∈ [ T 0 , ∞ ) T ,$
(13)
$∫ T 0 ∞ f ( t , K R 2 ( h ( t ) ) ) Δt≤ ( 1 − p 1 η 2 ) K 8 .$
(14)

When $p 0 =0$, we can choose $0< p 1 ≤1/13$ and the above $T 0$ such that

$| p ( t ) | ≤ p 1 ,t∈ [ T 0 , ∞ ) T .$
(15)

Furthermore, from (C3) there exists $T 1 ∈ ( T 0 , ∞ ) T$ such that $g(t)≥ T 0$ and $h(t)≥ T 0$ for $t∈ [ T 1 , ∞ ) T$.

Define the Banach space $B C 2 [ T 0 , ∞ ) T$ as in (3) with $λ=2$, and let

$Ω 1 = { x ( t ) ∈ B C 2 [ T 0 , ∞ ) T : K 2 R 2 ( t ) ≤ x ( t ) ≤ K R 2 ( t ) } .$
(16)

It is easy to prove that $Ω 1$ is a bounded, convex, and closed subset of $B C 2 [ T 0 , ∞ ) T$. By (C4), we have, for any $x∈ Ω 1$,

$f ( t , x ( h ( t ) ) ) ≤f ( t , K R 2 ( h ( t ) ) ) ,t∈ [ T 1 , ∞ ) T .$

Now we define two operators $U 1$ and $S 1$: $Ω 1 →B C 2 [ T 0 , ∞ ) T$ as follows

$( U 1 x ) ( t ) = { 3 4 K p 1 η 2 R 2 ( t ) − p ( T 1 ) x ( g ( T 1 ) ) R 2 ( T 1 ) R 2 ( t ) , t ∈ [ T 0 , T 1 ) T , 3 4 K p 1 η 2 R 2 ( t ) − p ( t ) x ( g ( t ) ) , t ∈ [ T 1 , ∞ ) T , ( S 1 x ) ( t ) = { 3 4 K R 2 ( t ) , t ∈ [ T 0 , T 1 ) T , 3 4 K R 2 ( t ) + ∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v , t ∈ [ T 1 , ∞ ) T .$
(17)

Next, we will prove that $U 1$ and $S 1$ satisfy the conditions in Lemma 2.2.

(i) We prove that $U 1 x+ S 1 y∈ Ω 1$ for any $x,y∈ Ω 1$. Note that, for any $x,y∈ Ω 1$, $K 2 R 2 (t)≤x(t)≤K R 2 (t)$ and $K 2 R 2 (t)≤y(t)≤K R 2 (t)$. For any $x,y∈ Ω 1$ and $t∈ [ T 1 , ∞ ) T$, by (13), (14), and (16) we obtain

$( U 1 x ) ( t ) + ( S 1 y ) ( t ) = 3 ( 1 + p 1 η 2 ) 4 K R 2 ( t ) − p ( t ) x ( g ( t ) ) + ∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , y ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v ≥ 3 ( 1 + p 1 η 2 ) 4 K R 2 ( t ) − p 1 η 2 K R 2 ( t ) = 3 − p 1 η 2 4 K R 2 ( t ) > K 2 R 2 ( t ) .$

On the other hand, for $t∈ [ T 1 , ∞ ) T$ and $p(t)≥0$, we have

$( U 1 x ) ( t ) + ( S 1 y ) ( t ) ≤ 3 ( 1 + p 1 η 2 ) 4 K R 2 ( t ) − K 2 p ( t ) R 2 ( g ( t ) ) + 1 − p 1 η 2 8 K R 2 ( t ) ≤ 3 ( 1 + p 1 η 2 ) 4 K R 2 ( t ) − K 2 5 p 1 − 1 4 η 2 R 2 ( t ) + 1 − p 1 η 2 8 K R 2 ( t ) = 7 + η 2 8 K R 2 ( t ) ≤ K R 2 ( t ) .$

For $t∈ [ T 1 , ∞ ) T$, $p(t)<0$, and $p 0 =0$, we have $0< p 1 ≤1/13$ and (15), and

$( U 1 x ) ( t ) + ( S 1 y ) ( t ) ≤ 3 ( 1 + p 1 η 2 ) 4 K R 2 ( t ) − K p ( t ) R 2 ( g ( t ) ) + 1 − p 1 η 2 8 K R 2 ( t ) ≤ 3 ( 1 + p 1 η 2 ) 4 K R 2 ( t ) + p 1 K R 2 ( t ) + 1 − p 1 η 2 8 K R 2 ( t ) = 7 + 8 p 1 + 5 p 1 η 2 8 K R 2 ( t ) ≤ 7 + 13 p 1 8 K R 2 ( t ) ≤ K R 2 ( t ) .$

Similarly, we can prove that $( U 1 x)(t)+( S 1 y)(t)≥K R 2 (t)/2$ for any $x,y∈ Ω 1$ and $t∈ [ T 0 , T 1 ] T$. Then we prove that $( U 1 x)(t)+( S 1 y)(t)≤K R 2 (t)$ for any $x,y∈ Ω 1$ and $t∈ [ T 0 , T 1 ] T$. In fact, for $t∈ [ T 0 , T 1 ] T$ and $p(t)≥0$, we have

$( U 1 x ) ( t ) + ( S 1 y ) ( t ) = 3 ( 1 + p 1 η 2 ) 4 K R 2 ( t ) − p ( T 1 ) x ( g ( T 1 ) ) R 2 ( T 1 ) R 2 ( t ) ≤ 3 ( 1 + p 1 η 2 ) 4 K R 2 ( t ) − K 2 5 p 1 − 1 4 η 2 R 2 ( t ) = 6 + p 1 η 2 + η 2 8 K R 2 ( t ) < K R 2 ( t ) .$

For $t∈ [ T 0 , T 1 ] T$, $p(t)<0$, and $p 0 =0$, we have $0< p 1 ≤1/13$ and (15), and

$( U 1 x ) ( t ) + ( S 1 y ) ( t ) ≤ 3 ( 1 + p 1 η 2 ) 4 K R 2 ( t ) + p 1 K R 2 ( t ) = 3 + 3 p 1 η 2 + 4 p 1 4 K R 2 ( t ) < K R 2 ( t ) .$

Therefore, we obtain $U 1 x+ S 1 y∈ Ω 1$ for any $x,y∈ Ω 1$.

(ii) We prove that $U 1$ is a contraction mapping. In fact, noting that $g(t)≤t$ and $R 2 (t)≥1$ for $t∈ [ T 0 , ∞ ) T$, for $x,y∈ Ω 1$ we have

$| ( U 1 x ) ( t ) R 2 2 ( t ) − ( U 1 y ) ( t ) R 2 2 ( t ) | = | p ( T 1 ) R 2 2 ( g ( T 1 ) ) R 2 ( t ) R 2 ( T 1 ) x ( g ( T 1 ) ) − y ( g ( T 1 ) ) R 2 2 ( g ( T 1 ) ) | ≤ p 1 sup t ∈ [ T 0 , ∞ ) T | x ( t ) R 2 2 ( t ) − y ( t ) R 2 2 ( t ) |$

for $t∈ [ T 0 , T 1 ] T$, and

$| ( U 1 x ) ( t ) R 2 2 ( t ) − ( U 1 y ) ( t ) R 2 2 ( t ) | = | p ( t ) R 2 2 ( g ( t ) ) R 2 2 ( t ) x ( g ( t ) ) − y ( g ( t ) ) R 2 2 ( g ( t ) ) | ≤ p 1 sup t ∈ [ T 0 , ∞ ) T | x ( t ) R 2 2 ( t ) − y ( t ) R 2 2 ( t ) |$

for $t∈ [ T 1 , ∞ ) T$. It follows that

$∥ U 1 x − U 1 y ∥ 2 ≤ p 1 ∥ x − y ∥ 2$

for any $x,y∈ Ω 1$. Therefore, $U 1$ is a contraction mapping.

(iii) We prove that $S 1$ is a completely continuous mapping.

Firstly, for $t∈ [ T 0 , ∞ ) T$, we have

$( S 1 x)(t)≥ 3 4 K R 2 (t)> K 2 R 2 (t)$

and

$( S 1 x)(t)≤ 3 4 K R 2 (t)+ 1 − p 1 η 2 8 K R 2 (t)= 7 − p 1 η 2 8 K R 2 (t)

That is, $S 1$ maps $Ω 1$ into $Ω 1$.

Secondly, we prove the continuity of $S 1$. For $x∈ Ω 1$ and $t∈ [ T 0 , ∞ ) T$, letting $x n ∈ Ω 1$ and $∥ x n − x ∥ 2 →0$ as $n→∞$, we have

$| f ( t , x n ( h ( t ) ) ) − f ( t , x ( h ( t ) ) ) | →0$
(18)

and

$| f ( t , x n ( h ( t ) ) ) − f ( t , x ( h ( t ) ) ) | ≤2f ( t , K R 2 ( h ( t ) ) )$

as $n→∞$. For $t∈ [ T 1 , ∞ ) T$, we have

$| ( S 1 x n ) ( t ) R 2 2 ( t ) − ( S 1 x ) ( t ) R 2 2 ( t ) | ≤ 1 R 2 2 ( t ) ∫ T 1 t ∫ T 1 v ∫ s ∞ | f ( u , x n ( h ( u ) ) ) − f ( u , x ( h ( u ) ) ) | r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v ≤ 1 R 2 ( t ) ∫ T 1 ∞ | f ( u , x n ( h ( u ) ) ) − f ( u , x ( h ( u ) ) ) | Δ u .$

For $t∈ [ T 0 , T 1 ] T$, we have $( S 1 x n )(t)−( S 1 x)(t)=0$. Then we obtain

$∥ S 1 x n − S 1 x ∥ 2 ≤ sup t ∈ [ t 0 , ∞ ) T 1 R 2 ( t ) ∫ T 1 ∞ | f ( u , x n ( h ( u ) ) ) − f ( u , x ( h ( u ) ) ) | Δu.$

Similar to Chen , by (18) and employing Lebesgue’s dominated convergence theorem [, Chapter 5], we conclude that

$∥ S 1 x n − S 1 x ∥ 2 →0$

as $n→∞$. That is, $S 1$ is continuous.

Thirdly, we prove that $S 1 Ω 1$ is relatively compact. According to Lemma 2.1, it suffices to show that $S 1 Ω 1$ is bounded, uniformly Cauchy and equi-continuous. It is obvious that $S 1 Ω 1$ is bounded. Since $∫ t 0 ∞ f(t,K R 2 (h(t)))Δt<∞$ and $R 2 (t)→∞$ as $t→∞$, for any given $ϵ>0$ there exists a sufficiently large $T 2 ∈ [ T 1 , ∞ ) T$ such that $R 2 ( T 2 )>3K/ϵ$ and $1 R 2 ( T 2 ) ∫ T 1 ∞ f(t,K R 2 (h(t)))Δt<ϵ/4$. Then, for any $x∈ Ω 1$ and $t 1 , t 2 ∈ [ T 2 , ∞ ) T$, we have

$| ( S 1 x ) ( t 1 ) R 2 2 ( t 1 ) − ( S 1 x ) ( t 2 ) R 2 2 ( t 2 ) | ≤ | 1 R 2 2 ( t 1 ) ∫ T 1 t 1 ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v − 1 R 2 2 ( t 2 ) ∫ T 1 t 2 ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v | + 3 4 K | 1 R 2 ( t 1 ) − 1 R 2 ( t 2 ) | ≤ 1 R 2 2 ( t 1 ) ∫ T 1 t 1 ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v + 1 R 2 2 ( t 2 ) ∫ T 1 t 2 ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v + 3 4 K ( 1 R 2 ( t 1 ) + 1 R 2 ( t 2 ) ) ≤ 1 R 2 ( T 2 ) ∫ T 1 ∞ f ( u , x ( h ( u ) ) ) Δ u + 1 R 2 ( T 2 ) ∫ T 1 ∞ f ( u , x ( h ( u ) ) ) Δ u + 3 K 2 R 2 ( T 2 ) < ϵ 4 + ϵ 4 + ϵ 2 = ϵ .$

Hence, $S 1 Ω 1$ is uniformly Cauchy.

Then, for $x∈ Ω 1$, if $t 1 , t 2 ∈T$ with $T 1 ≤ t 1 < t 2 < T 2 +1$, we have

$| ( S 1 x ) ( t 1 ) R 2 2 ( t 1 ) − ( S 1 x ) ( t 2 ) R 2 2 ( t 2 ) | ≤ | 1 R 2 2 ( t 1 ) ∫ T 1 t 1 ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v − 1 R 2 2 ( t 2 ) ∫ T 1 t 2 ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v | + 3 4 K | 1 R 2 ( t 1 ) − 1 R 2 ( t 2 ) | ≤ 1 R 2 2 ( T 1 ) ∫ t 1 t 2 ∫ T 1 v ∫ s ∞ f ( u , K R 2 ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v + 3 4 K | 1 R 2 ( t 1 ) − 1 R 2 ( t 2 ) | .$

If $t 1 , t 2 ∈T$ with $t 1 < T 1 ≤ t 2 < T 2 +1$, we have

$| ( S 1 x ) ( t 1 ) R 2 2 ( t 1 ) − ( S 1 x ) ( t 2 ) R 2 2 ( t 2 ) | ≤ 1 R 2 2 ( t 2 ) ∫ T 1 t 2 ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v + 3 4 K | 1 R 2 ( t 1 ) − 1 R 2 ( t 2 ) | ≤ 1 R 2 2 ( T 1 ) ∫ T 1 t 2 ∫ T 1 v ∫ s ∞ f ( u , K R 2 ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v + 3 4 K | 1 R 2 ( t 1 ) − 1 R 2 ( t 2 ) | .$

If $t 1 , t 2 ∈ [ T 0 , T 1 ] T$, we always have

$| ( S 1 x ) ( t 1 ) R 2 2 ( t 1 ) − ( S 1 x ) ( t 2 ) R 2 2 ( t 2 ) |= 3 4 K| 1 R 2 ( t 1 ) − 1 R 2 ( t 2 ) |.$

Therefore, there exists $0<δ<1$ such that

$| ( S 1 x ) ( t 1 ) R 2 2 ( t 1 ) − ( S 1 x ) ( t 2 ) R 2 2 ( t 2 ) |<ϵ$

whenever $t 1 , t 2 ∈ [ T 0 , T 2 + 1 ] T$ and $| t 2 − t 1 |<δ$. That is, $S 1 Ω 1$ is equi-continuous.

It follows from Lemma 2.1 that $S 1 Ω 1$ is relatively compact, and then $S 1$ is completely continuous.

By Lemma 2.2, there exists $x∈ Ω 1$ such that $( U 1 + S 1 )x=x$, which implies that $x(t)$ is a solution of (1). In particular, for $t∈ [ T 1 , ∞ ) T$ we have

$x(t)= 3 ( 1 + p 1 η 2 ) K 4 R 2 (t)−p(t)x ( g ( t ) ) + ∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv.$

Since

$∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv≤ ∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , K R 2 ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv$

for $t∈ [ T 1 , ∞ ) T$ and

$lim t → ∞ 1 R 2 ( t ) ∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , K R 2 ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v = lim t → ∞ ∫ t ∞ f ( u , K R 2 ( h ( u ) ) ) Δ u = 0 ,$

we have

$lim t → ∞ z ( t ) R 2 ( t ) = 3 ( 1 + p 1 η 2 ) K 4 and lim t → ∞ x ( t ) R 2 ( t ) = 3 ( 1 + p 1 η 2 ) K 4 ( 1 + p 0 η 2 ) >0.$

It is obvious that

$lim t → ∞ x(t)=∞, lim t → ∞ x ( t ) R 1 ( t ) =∞.$

The sufficiency holds when $0≤ p 0 <1$.

Case 2: $−1< p 0 <0$. Take $p 1$ so that $− p 0 < p 1 <(1−4 p 0 )/5<1$, then $p 0 <(1−5 p 1 )/4$. Since $lim t → ∞ p(t)= p 0$ and (11) hold, we can choose a sufficiently large $T 0 ∈ [ t 0 , ∞ ) T$ such that

$5 p 1 − 1 4 ≤−p(t)≤ p 1 <1,t∈ [ T 0 , ∞ ) T .$
(19)

From (C3) there exists $T 1 ∈ ( T 0 , ∞ ) T$ such that $g(t)≥ T 0$ and $h(t)≥ T 0$ for $t∈ [ T 1 , ∞ ) T$. Similarly, we introduce the Banach space $B C 2 [ T 0 , ∞ ) T$ and its subset $Ω 1$ as in (16). Define the operator $S 1$ as in (17) and the operator $U 1 ′$ on $Ω 1$ as follows:

$( U 1 ′ x ) (t)= { − 3 4 K p 1 η 2 R 2 ( t ) − p ( T 1 ) x ( g ( T 1 ) ) R 2 ( T 1 ) R 2 ( t ) , t ∈ [ T 0 , T 1 ) T , − 3 4 K p 1 η 2 R 2 ( t ) − p ( t ) x ( g ( t ) ) , t ∈ [ T 1 , ∞ ) T .$

Next, we prove that $U 1 ′ x+ S 1 y∈ Ω 1$ for any $x,y∈ Ω 1$. In fact, for any $x,y∈ Ω 1$ and $t∈ [ T 1 , ∞ ) T$, by (14) and (19) we obtain

$( U 1 ′ x ) ( t ) + ( S 1 y ) ( t ) = 3 ( 1 − p 1 η 2 ) 4 K R 2 ( t ) − p ( t ) x ( g ( t ) ) + ∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , y ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v ≥ 3 ( 1 − p 1 η 2 ) 4 K R 2 ( t ) + K 2 5 p 1 − 1 4 η 2 R 2 ( t ) = 6 − p 1 η 2 − η 2 8 K R 2 ( t ) > K 2 R 2 ( t )$

and

$( U 1 ′ x ) ( t ) + ( S 1 y ) ( t ) ≤ 3 ( 1 − p 1 η 2 ) 4 K R 2 ( t ) + p 1 η 2 K R 2 ( t ) + 1 − p 1 η 2 8 K R 2 ( t ) = 7 + p 1 η 2 8 K R 2 ( t ) < K R 2 ( t ) .$

That is, $U 1 ′ x+ S 1 y∈ Ω 1$ for any $x,y∈ Ω 1$.

The remainder of the proof is similar to the case $0≤ p 0 <1$ and we omit it here. By Lemma 2.2, there exists $x∈ Ω 1$ such that $( U 1 ′ + S 1 )x=x$, which implies that $x(t)$ is a solution of (1). In particular, for $t∈ [ T 1 , ∞ ) T$ we have

$x(t)= 3 ( 1 − p 1 η 2 ) K 4 R 2 (t)−p(t)x ( g ( t ) ) + ∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv.$

Letting $t→∞$, we have

$lim t → ∞ z ( t ) R 2 ( t ) = 3 ( 1 − p 1 η 2 ) K 4 and lim t → ∞ x ( t ) R 2 ( t ) = 3 ( 1 − p 1 η 2 ) K 4 ( 1 + p 0 η 2 ) >0.$

It is obvious that

$lim t → ∞ x(t)=∞, lim t → ∞ x ( t ) R 1 ( t ) =∞.$

The sufficiency holds when $−1< p 0 <0$.

The proof is complete. □

Theorem 3.2 Equation (1) has an eventually positive solution in $A(∞,b,0)$ for some positive b if and only if there exists some constant $K>0$ such that

$∫ t 0 ∞ ∫ s ∞ f ( u , K R 1 ( h ( u ) ) ) r 1 ( s ) ΔuΔs<∞.$
(20)

Proof Suppose that $x(t)$ is an eventually positive solution of (1) in $A(∞,b,0)$, i.e.,

$lim t → ∞ x(t)=∞, lim t → ∞ x ( t ) R 1 ( t ) =b, lim t → ∞ x ( t ) R 2 ( t ) =0.$

Similarly, we have

$lim t → ∞ z ( t ) = ∞ , lim t → ∞ r 2 ( t ) z Δ ( t ) = lim t → ∞ z ( t ) R 1 ( t ) = ( 1 + p 0 η 1 ) b , lim t → ∞ r 1 ( t ) ( r 2 ( t ) z Δ ( t ) ) Δ = lim t → ∞ z ( t ) R 2 ( t ) = 0$

and there exists $T 1 ∈ [ t 0 , ∞ ) T$ such that $x(t)≥b R 1 (t)/2$, $x(g(t))≥b R 1 (g(t))/2$, $x(h(t))≥b R 1 (h(t))/2$ for $t∈ [ T 1 , ∞ ) T$. Integrating (1) from $s∈ [ T 1 , ∞ ) T$ to $v∈ [ σ ( s ) , ∞ ) T$, we obtain

$r 1 (v) ( r 2 ( v ) z Δ ( v ) ) Δ − r 1 (s) ( r 2 ( s ) z Δ ( s ) ) Δ =− ∫ s v f ( u , x ( h ( u ) ) ) Δu.$

Letting $v→∞$, we have

$r 1 (s) ( r 2 ( s ) z Δ ( s ) ) Δ = ∫ s ∞ f ( u , x ( h ( u ) ) ) Δu,$

or

$( r 2 ( s ) z Δ ( s ) ) Δ = ∫ s ∞ f ( u , x ( h ( u ) ) ) Δ u r 1 ( s ) .$
(21)

Integrating (21) from $T 1$ to $t∈ [ σ ( T 1 ) , ∞ ) T$, we have

$r 2 (t) z Δ (t)− r 2 ( T 1 ) z Δ ( T 1 )= ∫ T 1 t ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) ΔuΔs.$

Letting $t→∞$, we have

$∫ T 1 ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) ΔuΔs<∞.$

In view of (C4), it follows that

$f ( u , b 2 R 1 ( h ( u ) ) ) ≤f ( u , x ( h ( u ) ) ) ,u∈ [ T 1 , ∞ ) T ,$

and

$∫ T 1 ∞ ∫ s ∞ f ( u , b R 1 ( h ( u ) ) / 2 ) r 1 ( s ) ΔuΔs≤ ∫ T 1 ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) ΔuΔs<∞,$

which means that (20) holds. The necessary condition is proved.

Conversely, suppose that there exists some constant $K>0$ such that (20) holds. There will be two cases to be considered: $0≤ p 0 <1$ and $−1< p 0 <0$.

Case 1: $0≤ p 0 <1$. Take $p 1$ such that $p 0 < p 1 <(1+4 p 0 )/5<1$, then $p 0 >(5 p 1 −1)/4$.

When $p 0 >0$, since $lim t → ∞ p(t)= p 0$ and (20) hold, we can choose a sufficiently large $T 0 ∈ [ t 0 , ∞ ) T$ such that $p(t)>0$ for $t∈ [ T 0 , ∞ ) T$, and

$5 p 1 − 1 4 ≤ p ( t ) ≤ p 1 < 1 , p ( t ) R 1 ( g ( t ) ) R 1 ( t ) ≥ 5 p 1 − 1 4 η 1 , t ∈ [ T 0 , ∞ ) T , ∫ T 0 ∞ ∫ s ∞ f ( u , K R 1 ( h ( u ) ) ) r 1 ( s ) Δ u Δ s ≤ ( 1 − p 1 η 1 ) K 8 .$

When $p 0 =0$, we can choose $0< p 1 ≤1/13$ and the above $T 0$ such that

$| p ( t ) | ≤ p 1 ,t∈ [ T 0 , ∞ ) T .$

Furthermore, from (C3) there exists $T 1 ∈ ( T 0 , ∞ ) T$ such that $g(t)≥ T 0$ and $h(t)≥ T 0$ for $t∈ [ T 1 , ∞ ) T$.

Define the Banach space $B C 1 [ T 0 , ∞ ) T$ as in (3) with $λ=1$, and let

$Ω 2 = { x ( t ) ∈ B C 1 [ T 0 , ∞ ) T : K 2 R 1 ( t ) ≤ x ( t ) ≤ K R 1 ( t ) } .$
(22)

It is easy to prove that $Ω 1$ is a bounded, convex, and closed subset of $B C 1 [ T 0 , ∞ ) T$. By (C4), we have, for any $x∈ Ω 2$,

$f ( t , x ( h ( t ) ) ) ≤f ( t , K R 1 ( h ( t ) ) ) ,t∈ [ T 1 , ∞ ) T .$

Now we define two operators $U 2$ and $S 2 : Ω 2 →B C 1 [ T 0 , ∞ ) T$ as follows:

$( U 2 x ) ( t ) = { 3 4 K p 1 η 1 R 1 ( t ) − p ( T 1 ) x ( g ( T 1 ) ) R 1 ( T 1 ) R 1 ( t ) , t ∈ [ T 0 , T 1 ) T , 3 4 K p 1 η 1 R 1 ( t ) − p ( t ) x ( g ( t ) ) , t ∈ [ T 1 , ∞ ) T , ( S 2 x ) ( t ) = { 3 4 K R 1 ( t ) , t ∈ [ T 0 , T 1 ) T , 3 4 K R 1 ( t ) + ∫ t ∞ ∫ v ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v , t ∈ [ T 1 , ∞ ) T .$
(23)

Next, we can prove that $U 2$ and $S 2$ satisfy the conditions in Lemma 2.2. The proof is similar to the case $0≤ p 0 <1$ of Theorem 3.1 and omitted here.

By Lemma 2.2, there exists $x∈ Ω 2$ such that $( U 2 + S 2 )x=x$, which implies that $x(t)$ is a solution of (1). In particular, for $t∈ [ T 1 , ∞ ) T$ we have

$x(t)= 3 ( 1 + p 1 η 1 ) K 4 R 1 (t)−p(t)x ( g ( t ) ) + ∫ t ∞ ∫ v ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv.$

Since

$∫ t ∞ ∫ v ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv≤ ∫ t ∞ ∫ v ∞ ∫ s ∞ f ( u , K R 1 ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv$

for $t∈ [ T 1 , ∞ ) T$ and

$lim t → ∞ 1 R 1 ( t ) ∫ t ∞ ∫ v ∞ ∫ s ∞ f ( u , K R 1 ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v = − lim t → ∞ ∫ t ∞ ∫ s ∞ f ( u , K R 1 ( h ( u ) ) ) r 1 ( s ) Δ u Δ s = 0 ,$

we have

$lim t → ∞ z ( t ) R 1 ( t ) = 3 ( 1 + p 1 η 1 ) K 4 and lim t → ∞ x ( t ) R 1 ( t ) = 3 ( 1 + p 1 η 1 ) K 4 ( 1 + p 0 η 1 ) >0,$

which implies that

$lim t → ∞ x(t)=∞, lim t → ∞ x ( t ) R 2 ( t ) =0.$

The sufficiency holds when $0≤ p 0 <1$.

Case 2: $−1< p 0 <0$. We introduce the Banach space $B C 1 [ T 0 , ∞ ) T$ and its subset $Ω 2$ as in (22). Define the operator $S 2$ as in (23) and the operator $U 2 ′$ on $Ω 2$ as follows:

$( U 2 ′ x ) (t)= { − 3 4 K p 1 η 1 R 1 ( t ) − p ( T 1 ) x ( g ( T 1 ) ) R 1 ( T 1 ) R 1 ( t ) , t ∈ [ T 0 , T 1 ) T , − 3 4 K p 1 η 1 R 1 ( t ) − p ( t ) x ( g ( t ) ) , t ∈ [ T 1 , ∞ ) T .$

The following proof is similar to the case $−1< p 0 <0$ in Theorem 3.1 and we omit it here. By Lemma 2.2, there exists $x∈ Ω 2$ such that $( U 2 ′ + S 2 )x=x$, which implies that $x(t)$ is a solution of (1). In particular, for $t∈ [ T 1 , ∞ ) T$ we have

$x(t)= 3 ( 1 − p 1 η 1 ) K 4 R 1 (t)−p(t)x ( g ( t ) ) + ∫ t ∞ ∫ v ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv.$

Similarly, we have

$lim t → ∞ z ( t ) R 1 ( t ) = 3 ( 1 − p 1 η 1 ) K 4 and lim t → ∞ x ( t ) R 1 ( t ) = 3 ( 1 − p 1 η 1 ) K 4 ( 1 + p 0 η 1 ) >0,$

which implies that

$lim t → ∞ x(t)=∞, lim t → ∞ x ( t ) R 2 ( t ) =0.$

The sufficiency holds when $−1< p 0 <0$.

The proof is complete. □

Theorem 3.3 Equation (1) has an eventually positive solution in $A(b,0,0)$ for some positive b if and only if there exists some constant $K>0$ such that

$∫ t 0 ∞ ∫ v ∞ ∫ s ∞ f ( u , K ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv<∞.$
(24)

Proof Suppose that $x(t)$ is an eventually positive solution of (1) in $A(b,0,0)$, i.e.,

$lim t → ∞ x(t)=b, lim t → ∞ x ( t ) R 1 ( t ) =0, lim t → ∞ x ( t ) R 2 ( t ) =0.$

Then

$lim t → ∞ z ( t ) = ( 1 + p 0 ) b , lim t → ∞ r 2 ( t ) z Δ ( t ) = lim t → ∞ z ( t ) R 1 ( t ) = 0 , lim t → ∞ r 1 ( t ) ( r 2 ( t ) z Δ ( t ) ) Δ = lim t → ∞ z ( t ) R 2 ( t ) = 0 ,$

and there exists $T 1 ∈ [ t 0 , ∞ ) T$ such that $x(t)≥b/2$, $x(g(t))≥b/2$, $x(h(t))≥b/2$ for $t∈ [ T 1 , ∞ ) T$. Integrating (1) from $s∈ [ T 1 , ∞ ) T$ to $v∈ [ σ ( s ) , ∞ ) T$, we obtain

$r 1 (v) ( r 2 ( v ) z Δ ( v ) ) Δ − r 1 (s) ( r 2 ( s ) z Δ ( s ) ) Δ =− ∫ s v f ( u , x ( h ( u ) ) ) Δu.$

Letting $v→∞$, we have

$r 1 (s) ( r 2 ( s ) z Δ ( s ) ) Δ = ∫ s ∞ f ( u , x ( h ( u ) ) ) Δu,$

or

$( r 2 ( s ) z Δ ( s ) ) Δ = ∫ s ∞ f ( u , x ( h ( u ) ) ) Δ u r 1 ( s ) .$
(25)

Integrating (25) from $v∈ [ T 1 , ∞ ) T$ to $w∈ [ σ ( v ) , ∞ ) T$, we have

$r 2 (w) z Δ (w)− r 2 (v) z Δ (v)= ∫ v w ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) ΔuΔs.$

Letting $w→∞$, we have

$r 2 (v) z Δ (v)=− ∫ v ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) ΔuΔs,$

or

$z Δ (v)=− ∫ v ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔs.$
(26)

Integrating (26) from $T 1$ to $t∈ [ σ ( T 1 ) , ∞ ) T$, we have

$z(t)−z( T 1 )=− ∫ T 1 t ∫ v ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv.$

Letting $t→∞$, we have

$∫ T 1 ∞ ∫ v ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv<∞.$

In view of (C4), it follows that

$f ( u , b 2 ) ≤f ( u , x ( h ( u ) ) ) ,u∈ [ T 1 , ∞ ) T ,$

and

$∫ T 1 ∞ ∫ v ∞ ∫ s ∞ f ( u , b / 2 ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv≤ ∫ T 1 ∞ ∫ v ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv<∞,$

which means that (24) holds. The necessary condition is proved.

Conversely, suppose that there exists some constant $K>0$ such that (24) holds. There will be two cases to be considered: $0≤ p 0 <1$ and $−1< p 0 <0$.

Case 1: $0≤ p 0 <1$. Take $p 1$ such that $p 0 < p 1 <(1+4 p 0 )/5<1$, then $p 0 >(5 p 1 −1)/4$.

When $p 0 >0$, since $lim t → ∞ p(t)= p 0$ and (24) hold, we can choose a sufficiently large $T 0 ∈ [ t 0 , ∞ ) T$ such that $p(t)>0$ for $t∈ [ T 0 , ∞ ) T$, and

$5 p 1 − 1 4 ≤ p ( t ) ≤ p 1 < 1 , t ∈ [ T 0 , ∞ ) T , ∫ T 0 ∞ ∫ v ∞ ∫ s ∞ f ( u , K ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v ≤ ( 1 − p 1 ) K 8 .$

When $p 0 =0$, we can choose $0< p 1 ≤1/13$ and the above $T 0$ such that

$| p ( t ) | ≤ p 1 ,t∈ [ T 0 , ∞ ) T .$

Furthermore, from (C3) there exists $T 1 ∈ ( T 0 , ∞ ) T$ such that $g(t)≥ T 0$ and $h(t)≥ T 0$ for $t∈ [ T 1 , ∞ ) T$.

Define the Banach space $B C 0 [ T 0 , ∞ ) T$ as in (3) with $λ=0$, and let

$Ω 3 = { x ( t ) ∈ B C 0 [ T 0 , ∞ ) T : K 2 ≤ x ( t ) ≤ K } .$
(27)

It is easy to prove that $Ω 3$ is a bounded, convex, and closed subset of $B C 0 [ T 0 , ∞ ) T$. By (C4), we have, for any $x∈ Ω 3$,

$f ( t , x ( h ( t ) ) ) ≤f(t,K),t∈ [ T 1 , ∞ ) T .$

Now we define two operators $U 3$ and $S 3 : Ω 3 →B C 0 [ T 0 , ∞ ) T$ as follows:

$( U 3 x ) ( t ) = { 3 4 K p 1 − p ( t ) x ( g ( t ) ) , t ∈ [ T 1 , ∞ ) T , ( U 3 x ) ( T 1 ) , t ∈ [ T 0 , T 1 ) T , ( S 3 x ) ( t ) = { 3 4 K + ∫ t ∞ ∫ v ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v , t ∈ [ T 1 , ∞ ) T , ( S 3 x ) ( T 1 ) , t ∈ [ T 0 , T 1 ) T .$
(28)

Next, we can prove that $U 3$ and $S 3$ satisfy the conditions in Lemma 2.2. The proof is similar to the case $0≤ p 0 <1$ of Theorem 3.1 and omitted here.

By Lemma 2.2, there exists $x∈ Ω 3$ such that $( U 3 + S 3 )x=x$, which implies that $x(t)$ is a solution of (1). In particular, for $t∈ [ T 1 , ∞ ) T$ we have

$x(t)= 3 ( 1 + p 1 ) K 4 −p(t)x ( g ( t ) ) + ∫ t ∞ ∫ v ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv.$

Letting $t→∞$, we have

$lim t → ∞ z(t)= 3 ( 1 + p 1 ) K 4 and lim t → ∞ x(t)= 3 ( 1 + p 1 ) K 4 ( 1 + p 0 ) >0,$

which implies that

$lim t → ∞ x ( t ) R 1 ( t ) = lim t → ∞ x ( t ) R 2 ( t ) =0.$

The sufficiency holds when $0≤ p 0 <1$.

Case 2: $−1< p 0 <0$. We introduce the Banach space $B C 0 [ T 0 , ∞ ) T$ and its subset $Ω 3$ as in (27). Define the operator $S 3$ as in (28) and the operator $U 3 ′$ on $Ω 3$ as follows:

$( U 3 ′ x ) (t)= { − 3 4 K p 1 − p ( t ) x ( g ( t ) ) , t ∈ [ T 1 , ∞ ) T , ( U 3 ′ x ) ( T 1 ) , t ∈ [ T 0 , T 1 ) T .$

The following proof is similar to the case $−1< p 0 <0$ in Theorem 3.1 and we omit it here. By Lemma 2.2, there exists $x∈ Ω 3$ such that $( U 3 ′ + S 3 )x=x$, which implies that $x(t)$ is a solution of (1). In particular, for $t∈ [ T 1 , ∞ ) T$ we have

$x(t)= 3 ( 1 − p 1 ) K 4 −p(t)x ( g ( t ) ) + ∫ t ∞ ∫ v ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv.$

Similarly, we have

$lim t → ∞ z(t)= 3 ( 1 − p 1 ) K 4 and lim t → ∞ x(t)= 3 ( 1 − p 1 ) K 4 ( 1 + p 0 ) >0,$

which implies that

$lim t → ∞ x ( t ) R 1 ( t ) = lim t → ∞ x ( t ) R 2 ( t ) =0.$

The sufficiency holds when $−1< p 0 <0$.

The proof is complete. □

Theorem 3.4 Equation (1) has an eventually positive solution in $A(∞,∞,0)$, then

$∫ t 0 ∞ f ( u , 3 4 R 1 ( h ( u ) ) ) Δu<∞, ∫ t 0 ∞ ∫ s ∞ f ( u , R 2 ( h ( u ) ) ) r 1 ( s ) ΔuΔs=∞.$
(29)

Conversely, if there exists a nonnegative constant M such that $|p(t) R 2 (t)|≤M$ and

$∫ t 0 ∞ f ( u , R 2 ( h ( u ) ) ) Δu<∞, ∫ t 0 ∞ ∫ s ∞ f ( u , ( M + 3 / 4 ) R 1 ( h ( u ) ) ) r 1 ( s ) ΔuΔs=∞,$
(30)

then (1) has an eventually positive solution in $A(∞,∞,0)$.

Proof Suppose that $x(t)$ is an eventually positive solution of (1) in $A(∞,∞,0)$, i.e.,

$lim t → ∞ x(t)=∞, lim t → ∞ x ( t ) R 1 ( t ) =∞, lim t → ∞ x ( t ) R 2 ( t ) =0.$

Similarly, we have

$lim t → ∞ z ( t ) = ∞ , lim t → ∞ r 2 ( t ) z Δ ( t ) = lim t → ∞ z ( t ) R 1 ( t ) = ∞ , lim t → ∞ r 1 ( t ) ( r 2 ( t ) z Δ ( t ) ) Δ = lim t → ∞ z ( t ) R 2 ( t ) = 0 ,$

and there exists $T 0 ∈ [ t 0 , ∞ ) T$ such that $3 R 1 (t)/4≤x(t)≤ R 2 (t)$ for $t∈ [ T 0 , ∞ ) T$. From (C3) there exists $T 1 ∈ ( T 0 , ∞ ) T$ such that $g(t)≥ T 0$ and $h(t)≥ T 0$ for $t∈ [ T 1 , ∞ ) T$. Integrating (1) from $T 1$ to $s∈ [ σ ( T 1 ) , ∞ ) T$, we obtain

$r 1 (s) ( r 2 ( s ) z Δ ( s ) ) Δ − r 1 ( T 1 ) ( r 2 ( T 1 ) z Δ ( T 1 ) ) Δ =− ∫ T 1 s f ( u , x ( h ( u ) ) ) Δu.$

Letting $s→∞$, we have

$r 1 ( T 1 ) ( r 2 ( T 1 ) z Δ ( T 1 ) ) Δ = ∫ T 1 ∞ f ( u , x ( h ( u ) ) ) Δu,$
(31)

which implies that

$∫ T 1 ∞ f ( u , 3 4 R 1 ( h ( u ) ) ) Δu<∞$

by the monotonicity of f and $3 R 1 (h(t))/4≤x(h(t))$ for $t∈ [ T 1 , ∞ ) T$. Substituting s for $T 1$ in (31), we have

$( r 2 ( s ) z Δ ( s ) ) Δ = ∫ s ∞ f ( u , x ( h ( u ) ) ) Δ u r 1 ( s ) .$
(32)

Integrating (32) from $T 1$ to $t∈ [ σ ( T 1 ) , ∞ ) T$, we have

$r 2 (t) z Δ (t)− r 2 ( T 1 ) z Δ ( T 1 )= ∫ T 1 t ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) ΔuΔs.$

Letting $t→∞$, we have

$∫ T 1 ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) ΔuΔs=∞.$

By the monotonicity of f and $x(h(t))≤ R 2 (h(t))$ for $t∈ [ T 1 , ∞ ) T$, it follows that

$f ( u , x ( h ( u ) ) ) ≤f ( u , R 2 ( h ( u ) ) ) ,u∈ [ T 1 , ∞ ) T ,$

and

$∫ T 1 ∞ ∫ s ∞ f ( u , R 2 ( h ( u ) ) ) r 1 ( s ) ΔuΔs≥ ∫ T 1 ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) ΔuΔs=∞,$

which means that (29) holds. The necessary condition is proved.

Conversely, if there exists a positive constant M such that $|p(t) R 2 (t)|≤M$ and (30) hold, then $lim t → ∞ p(t)=0$ and we can choose a sufficiently large $T 0 ∈ [ t 0 , ∞ ) T$ such that

$| p ( t ) | ≤ p 1 < 1 , | p ( t ) R 2 ( t ) | ≤ M , ( 2 M + 3 2 ) R 1 ( t ) ≤ 1 4 R 2 ( t ) , t ∈ [ T 0 , ∞ ) T , ∫ T 0 ∞ f ( u , R 2 ( h ( u ) ) ) Δ u ≤ 1 − p 1 8 .$

From (C3) there exists $T 1 ∈ ( T 0 , ∞ ) T$ such that $g(t)≥ T 0$ and $h(t)≥ T 0$ for $t∈ [ T 1 , ∞ ) T$.

Define the Banach space $B C 2 [ T 0 , ∞ ) T$ as in (3) with $λ=2$, and let

$Ω 4 = { x ( t ) ∈ B C 2 [ T 0 , ∞ ) T : ( M + 3 4 ) R 1 ( t ) ≤ x ( t ) ≤ R 2 ( t ) } .$

It is easy to prove that $Ω 4$ is a bounded, convex, and closed subset of $B C 2 [ T 0 , ∞ ) T$. According to (C3) and (C4), we have, for any $x∈ Ω 4$,

$x ( h ( t ) ) ≥ ( M + 3 4 ) R 1 ( h ( t ) ) ,f ( t , x ( h ( t ) ) ) ≤f ( t , R 2 ( h ( t ) ) ) ,t∈ [ T 1 , ∞ ) T .$

Now we define two operators $U 4$ and $S 4 : Ω 4 →B C 2 [ T 0 , ∞ ) T$ as follows:

$( U 4 x ) ( t ) = { ( M + 3 4 ) R 1 ( t ) − p ( T 1 ) x ( g ( T 1 ) ) R 2 ( T 1 ) R 2 ( t ) , t ∈ [ T 0 , T 1 ) T , ( M + 3 4 ) R 1 ( t ) − p ( t ) x ( g ( t ) ) , t ∈ [ T 1 , ∞ ) T , ( S 4 x ) ( t ) = { ( M + 3 4 ) R 1 ( t ) , t ∈ [ T 0 , T 1 ) T , ( M + 3 4 ) R 1 ( t ) + ∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v , t ∈ [ T 1 , ∞ ) T .$

Next, we can prove that $U 4$ and $S 4$ satisfy the conditions in Lemma 2.2. The proof is similar to Theorem 3.1 and omitted here. By Lemma 2.2, there exists $x∈ Ω 4$ such that $( U 4 + S 4 )x=x$, which implies that $x(t)$ is a solution of (1). In particular, for $t∈ [ T 1 , ∞ ) T$ we have

$x(t)= ( 2 M + 3 2 ) R 1 (t)−p(t)x ( g ( t ) ) + ∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv.$

Since $x(h(t))≥(M+3/4) R 1 (h(t))$ and

$∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv≤ ∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , R 2 ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv$

for $t∈ [ T 1 , ∞ ) T$, we have

$lim t → ∞ R 1 ( t ) R 2 ( t ) = lim t → ∞ 1 + ∫ t 0 t 1 r 2 ( s ) Δ s 1 + ∫ t 0 t ∫ t 0 s 1 r 1 ( u ) r 2 ( s ) Δ u Δ s = lim t → ∞ 1 ∫ t 0 t 1 r 1 ( u ) Δ u = 0 , lim t → ∞ 1 R 2 ( t ) ∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , R 2 ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v = lim t → ∞ ∫ t ∞ f ( u , R 2 ( h ( u ) ) ) Δ u = 0 , lim t → ∞ 1 R 1 ( t ) ∫ T 1 t ∫ T 1 v ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v = lim t → ∞ ∫ T 1 t ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) Δ u Δ s ≥ lim t → ∞ ∫ T 1 t ∫ s ∞ f ( u , ( M + 3 / 4 ) R 1 ( h ( u ) ) ) r 1 ( s ) Δ u Δ s = ∞ .$

It follows that

$lim t → ∞ z(t)=∞, lim t → ∞ z ( t ) R 1 ( t ) =∞, lim t → ∞ z ( t ) R 2 ( t ) =0.$

Since $|p(t)x(g(t))|≤|p(t) R 2 (t)|≤M$, by Lemma 2.3 we have

$lim t → ∞ x(t)=∞, lim t → ∞ x ( t ) R 1 ( t ) =∞, lim t → ∞ x ( t ) R 2 ( t ) =0.$

The proof is complete. □

When $p(t)≥0$ eventually, we have the following theorem.

Theorem 3.5 If there exist a constant $K>0$ and $T 0 ∈ [ t 0 , ∞ ) T$ with $T 0 >0$ such that, for $t∈ [ T 0 , ∞ ) T$,

$0≤p(t)≤Kg(t) e − t ,$
(33)
$∫ t ∞ ∫ v ∞ ∫ s ∞ f ( u , e − h ( u ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv≥(K+1) e − t$
(34)

and

$∫ t ∞ ∫ v ∞ ∫ s ∞ f ( u , 1 / h ( u ) ) r 1 ( s ) r 2 ( v ) ΔuΔsΔv≤ 1 t ,$
(35)

then (1) has an eventually positive solution in $A(0,0,0)$.

Proof From (C3) there exists $T 1 ∈ ( T 0 , ∞ ) T$ such that $g(t)≥ T 0$ and $h(t)≥ T 0$ for $t∈ [ T 1 , ∞ ) T$. Define the Banach space $B C 0 [ T 0 , ∞ ) T$ as in (3) with $λ=0$, and let

It is easy to prove that $Ω 5$ is a bounded, convex, and closed subset of $B C 0 [ T 0 , ∞ ) T$. Define an operator $S 5$ on $Ω 5$ as follows:

$( S 5 x)(t)= { − p ( t ) x ( g ( t ) ) + ∫ t ∞ ∫ v ∞ ∫ s ∞ f ( u , x ( h ( u ) ) ) r 1 ( s ) r 2 ( v ) Δ u Δ s Δ v , t ∈ [ T 1 , ∞ ) T , ( S 5 x ) ( T 1 ) , t ∈ [ T 0 , T 1 ) T .$

We prove that $S 5 x∈ Ω 5$ for any $x∈ Ω 5$. In fact, from (33)-(35), for $t∈ [ T 1 , ∞ ) T$ we have

$( S 5 x ) ( t ) = − p ( t ) x ( g ( t ) ) + ∫ t ∞ ∫ v ∞ ∫ s ∞ f ( u , x ( h$