Introduction

First and higher order boundary value problems (BVPs) have been studied in many papers, such as [3, 4, 10, 20, 22, 23] for multipoint BVP and [1, 2, 5, 9, 12, 15,16,17,18,19, 21] for infinite interval problem. Also there is a large bibliography on papers related to lower and upper solutions with nonlinear boundary value conditions for first and higher order equations. It is assumed in [11, 14] one pair of well-ordered upper and lower solutions and then applied some fixed point theorems or monotone iterative technique to obtain a solution. In the classical books of Bernfeld and Lakshmikantham [6] and Ladde et al. [13], the classical theory of the method of lower and upper solutions and the monotone iterative technique are given. This gives the solution as the limit of a monotone sequence formed by functions that solve linear problems related to the nonlinear equations considered. It is important to point out that to derive the existence of a solution a growth condition on the nonlinear part of the equation with respect to the dependence on the first derivative is imposed. The most usual condition is the so-called Nagumo condition imposes a quadratic growth in the dependence of the derivative.

In [19], Yan et al. established an upper and lower solution theory for the following second order boundary value problem

$$\begin{aligned} \left\{ \begin{array}{ll} y''(t) + \phi (t)f(t, y(t), y'(t))= 0, \, t \in (0, \infty ), \\ ay(0)- by'(0) = y_{0}\ge 0, \ \ \ \displaystyle \lim _{t\rightarrow +\infty }y'(t) = k,\\ \end{array}\right. \end{aligned}$$

where \(a, b, k > 0\). By using the upper and lower solutions method, the authors presented sufficient conditions for the existence of unbounded positive solutions.

In [15], Lian and Geng considered the second order Sturm–Liouville boundary value problem on the half line

$$\begin{aligned} \left\{ \begin{array}{ll} u''(t) + \phi (t)f(t, u(t), u'(t))= 0, \, t \in (0, \infty ), \\ u(0)- au'(0) = B, \ \ \ \ u'(+ \infty ) = C,\\ \end{array}\right. \end{aligned}$$

where \(a>0, B, C \in {\mathbb {R}}\). They concerned with the existence of multiple unbounded solutions.

We consider the following boundary value problem on an infinite time scale;

$$\begin{aligned} \left\{ \begin{array}{ll} -[p(t)u^{\triangle }(t)]^\nabla + \psi (t) f\left( t, u(t), u^{[\triangle ]}(t)\right) = 0, \, t \in (a, \infty ), \\ \alpha u(a)-\beta u^{[\triangle ]}(a) = \displaystyle \sum _{k=1}^{m-2} a_{k}u(\xi _{k}), \\ \displaystyle \gamma u(\infty )+ \delta u^{[\triangle ]}(\infty )=\sum _{k=1}^{m-2}b_{k}u(\xi _{k}), \\ \end{array}\right. \end{aligned}$$
(1.1)

where \(\alpha , \beta , \gamma , \delta , \xi _{k}, a_{k}, b_{k}\) (for \( k = 1, 2,\ldots , m-2\)) are complex constants such that \(|\alpha |+|\beta | \ne 0\), \(|\gamma |+|\delta | \ne 0\) and \(\xi _{k} \in {\mathbb {T}} \backslash \{a \}\), \(\psi :(a, \infty ) \rightarrow (0, \infty )\) is a continuous function, \(p: (a, \infty ) \rightarrow {\mathbb {C}}\) is \(\nabla -\) differentiable on \((a, \infty )_{\kappa }\), \(p(t)\ne 0\) for all \( t \in (a, \infty )\), \(p^{\nabla } : (a, \infty )_{\kappa } \rightarrow {\mathbb {C}}\), \(f :[a, \infty ) \times {\mathbb {R}} \times {\mathbb {R}} \rightarrow {\mathbb {R}}\) are continuous functions and \({\mathbb {T}}\) is any time scale such that \({\mathbb {T}}^{\kappa } = \infty \).

Throughout this paper, we mean \(u(\infty )= \displaystyle \lim _{t \rightarrow \infty } u(t)\) and \(u^{[\triangle ]}(t)=p(t)u^{\triangle }(t)\). Also, by an interval \((a, \infty )\), we mean the intersection of the real interval \((a, \infty )\) with the given time scale \({\mathbb {T}}\). Some preliminary definitions and theorems on time scales can be found in the books [7, 8].

This paper is organized as follows. In Section “The Preliminary Lemmas”, some definitions and lemmas are presented. We establish an upper and lower solution theory for problem (1.1) in Section “Main Result”. Sufficient conditions are given for the existence of at least one solution for this problem.

The Preliminary Lemmas

To prove the main results in this paper, we will employ several lemmas. We consider the following boundary value problem

$$\begin{aligned} \left\{ \begin{array}{ll} -[p(t)u^{\triangle }(t)]^\nabla =y(t), \, t \in (a, \infty ), \\ \alpha u(a)-\beta u^{[\triangle ]}(a) = \displaystyle \sum _{k=1}^{m-2} a_{k}u(\xi _{k}), \\ \displaystyle \gamma u(\infty )+ \delta u^{[\triangle ]}(\infty )=\sum _{k=1}^{m-2}b_{k}u(\xi _{k}). \\ \end{array}\right. \end{aligned}$$
(2.1)

Let \(\phi _{1}\) and \(\phi _{2}\) be the solutions of the linear problems

$$\begin{aligned} \left\{ \begin{array}{ll} -[p(t)\phi _{1}^{\triangle }(t)]^\nabla =0, \, t \in (a, \infty ), \\ \phi _{1}(a)= \beta , \ \ \ \phi _{1}^{[\triangle ]}(a)= \alpha \\ \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{array}{ll} -[p(t)\phi _{2}^{\triangle }(t)]^\nabla =0, \, t \in (a, \infty ), \\ \phi _{2}(\infty )= \delta , \ \ \ \phi _{2}^{[\triangle ]}(\infty )= -\gamma \\ \end{array}\right. \end{aligned}$$

respectively, so the functions \(\phi _{1}\) and \(\phi _{2}\) are given by \(\phi _{1}(t)=\beta + \alpha \displaystyle \int _{a}^{t} \frac{\Delta s}{p(s)} \) and \(\phi _{2}(t)=\delta + \gamma \displaystyle \int _{t}^{\infty } \frac{\Delta s}{p(s)} \).

Let we define \(d:=\gamma \beta + \alpha \delta + \alpha \gamma \displaystyle \int _{a}^{\infty } \frac{\Delta s}{p(s)} \)

and

$$\begin{aligned} \Omega :=\left| {\begin{array}{cc} -\sum \limits _{k=1}^{m-2} \alpha _{k}\phi _{1}(\xi _{k}) &{} d-\sum \limits _{k=1}^{m-2} \alpha _{k}\phi _{2}(\xi _{k})\\ d-\sum \limits _{k=1}^{m-2} \beta _{k}\phi _{1}(\xi _{k}) &{} -\sum \limits _{k=1}^{m-2} \beta _{k}\phi _{2}(\xi _{k}) \end{array} } \right| . \end{aligned}$$

The following lemmas are easily proved similar to Lemmas 2.3 and 2.4 in [20].

Lemma 2.1

Let \(\Omega , d \ne 0\). Then for \(y \in C[a, \infty )\), boundary value problem (2.1) has a unique solution

$$\begin{aligned} u(t)= \displaystyle \int _{a}^{\infty } G(t, s) y(s) \nabla s + A(y) \phi _{1}(t) + B(y) \phi _{2}(t), \end{aligned}$$

where

$$\begin{aligned} G(t, s) = \frac{1}{d} \left\{ \begin{array}{ll} \phi _{1}(s) \phi _{2}(t), &{} a \le s \le t< \infty ,\\ \phi _{1}(t) \phi _{2}(s), &{} a \le t \le s < \infty ,\\ \end{array}\right. \end{aligned}$$
$$\begin{aligned} A(y):=\frac{1}{\Omega }\left| {\begin{array}{cc} \sum \limits _{k=1}^{m-2} a_{k} \int _{a}^{\infty }G(\xi _{k}, s)y(s)\nabla s &{} d-\sum \limits _{k=1}^{m-2} a_{k}\phi _{2}(\xi _{k})\\ \sum \limits _{k=1}^{m-2} b_{k} \int _{a}^{\infty }G(\xi _{k}, s)y(s) \nabla s &{} -\sum \limits _{k=1}^{m-2} b_{k} \phi _{2}(\xi _{k}) \end{array} } \right| \end{aligned}$$

and

$$\begin{aligned} B(y):=\frac{1}{\Omega }\left| {\begin{array}{cc} -\sum \limits _{k=1}^{m-2} a_{k}\phi _{1}(\xi _{k}) &{} \sum \limits _{k=1}^{m-2} a_{k} \int _{a}^{\infty }G(\xi _{k}, s)y(s)\nabla s \\ d-\sum \limits _{k=1}^{m-2} b_{k} \phi _{1}(\xi _{k}) &{} \sum \limits _{k=1}^{m-2} b_{k} \int _{a}^{\infty }G(\xi _{k}, s)y(s)\nabla s \end{array} } \right| . \end{aligned}$$

We assume that the following conditions are satisfied:

\((H_{1})\):

\(\alpha , \gamma \ge 0\), \(\beta , \delta > 0\), \(a_{k}\), \(b_{k}\ge 0\) for \( k = 1, 2,\ldots , m-2\),

\((H_{2})\):

\(p: (a, \infty ) \rightarrow (0, \infty )\) is \(\nabla -\) differentiable on \((a, \infty )_{k}\) and \(p^{\nabla } : (a, \infty )_{k} \rightarrow {\mathbb {C}}\) is continuous function,

\((H_{3})\):

\(\Omega < 0\), \( d-\sum _{k=1}^{m-2}a_{k} \phi _{2}(\xi _{k}) > 0\), \( d-\sum _{k=1}^{m-2}b_{k} \phi _{1}(\xi _{k})> 0\).

Lemma 2.2

Assume that \((H_{1})-(H_{3})\) hold. Then,   \(0 < G(t, s) \le G(s, s)\) for \(t, s \in [a, \infty )\).

Lemma 2.3

If \(\int _{a}^{\infty } G(s, s)|y(s)| \nabla s<\infty \), then the following inequalities are satisfied:

$$\begin{aligned} |A(y)| \le A \int _{a}^{\infty }G(s, s)|y(s)| \nabla s,\quad |B(y)| \le B \int _{a}^{\infty }G(s,s)|y(s)| \nabla s, \end{aligned}$$

where

$$\begin{aligned} A=\frac{1}{\Omega }\left| {\begin{array}{cc} \sum \limits _{k=1}^{m-2} a_{k} &{} d-\sum \limits _{k=1}^{m-2} a_{k}\phi _{2}(\xi _{k})\\ \sum \limits _{k=1}^{m-2} b_{k} &{} -\sum \limits _{k=1}^{m-2} b_{k}\phi _{2}(\xi _{k}) \end{array} } \right| , \quad B=\frac{1}{\Omega }\left| {\begin{array}{cc} -\sum _{k=1}^{m-2} a_{k}\phi _{1}(\xi _{k}) &{} \sum \limits _{k=1}^{m-2} a_{k}\\ d-\sum \limits _{k=1}^{m-2} b_{k}\phi _{1}(\xi _{k}) &{} \sum \limits _{k=1}^{m-2} b_{k} \end{array} } \right| . \end{aligned}$$

Proof

It can be easily proven with simple calculation. \(\square \)

Now, we present some definitions and lemmas which are essential in the proof of our main results.

Definition 2.1

A function \(\underline{u}\in C^{1}[a, \infty ) \cap C^{2}(a, \infty )\) is called a lower solution of (1.1) if

$$\begin{aligned} \left\{ \begin{array}{ll} [p(t)\underline{u}^{\triangle }(t)]^\nabla + \psi (t)f(t, \underline{u}(t), \underline{u}^{[\triangle ]}(t))\ge 0, \, t \in (a, \infty ), \\ \alpha \underline{u}(a)-\beta \underline{u}^{[\triangle ]}(a) \le \displaystyle \sum _{k=1}^{m-2} a_{k}\underline{u}(\xi _{k}), \\ \displaystyle \gamma \underline{u}(\infty )+ \delta \underline{u}^{[\triangle ]}(\infty )\le \sum _{k=1}^{m-2}b_{k}\underline{u}(\xi _{k}). \\ \end{array}\right. \end{aligned}$$

Definition 2.2

A function \(\overline{u}\in C^{1}[a, \infty ) \cap C^{2}(a, \infty )\) is called an upper solution of (1.1) if

$$\begin{aligned} \left\{ \begin{array}{ll} [p(t)\overline{u}^{\triangle }(t)]^\nabla + \psi (t)f(t, \overline{u}(t), \overline{u}^{[\triangle ]}(t)) \le 0, \, t \in (a, \infty ), \\ \alpha \overline{u}(a)-\beta \overline{u}^{[\triangle ]}(a) \ge \displaystyle \sum _{k=1}^{m-2} a_{k}\overline{u}(\xi _{k}), \\ \displaystyle \gamma \overline{u}(\infty )+ \delta \overline{u}^{[\triangle ]}(\infty ) \ge \sum _{k=1}^{m-2}b_{k}\overline{u}(\xi _{k}). \\ \end{array}\right. \end{aligned}$$

Definition 2.3

Given a pair of functions \( \underline{u}, \overline{u} \in C^{1}[a, \infty )\) satisfying \(\underline{u} (t) \le \overline{u} (t)\), \(t \in [a, \infty ) \). A function f is said to satisfy the Nagumo condition with respect to the pair of functions \( \underline{u}, \overline{u} \), if there exist a non-negative function \(\varphi \in C[a, \infty )\) and a positive, increasing one \(h \in C[a, \infty )\) such that

$$\begin{aligned} |f(t, u, v) | \le \varphi (t) h(|v|), \end{aligned}$$

for all \(t \in [a, \infty )\), \(\underline{u}(t) \le u\le \overline{u}(t)\), \(v \in {\mathbb {R}}\) and

$$\begin{aligned} \displaystyle \int _{a}^{\infty } \varphi ( s) \psi (s) \nabla s<\infty , \, \, \displaystyle \int _{a}^{\infty } \frac{s}{h(|s|)} \nabla s= \infty . \end{aligned}$$

Lemma 2.4

Let \(g: {\mathbb {T}} \rightarrow {\mathbb {R}}\) be a strictly increasing function in \({\mathbb {T}}\) and \(\Delta -\)differentiable in \({\mathbb {T}}^{k}\), \(\tilde{\mathbb {T}}=g({\mathbb {T}})\) and \(f: \tilde{{\mathbb {T}}} \rightarrow {\mathbb {R}}\) be rd - continuous. Then for all \(a, t \in {\mathbb {T}} \), the following is satisfied:

$$\begin{aligned} \displaystyle \int _{a}^{t} f(g(s)) g^{\Delta } (s) \Delta s = \displaystyle \int _{g(a)}^{g(t)} f(r) \tilde{\Delta } r. \end{aligned}$$

Main Result

In this section, by using upper and lower solutions, we will prove the existence of the solution of problem (1.1). We shall work in the Banach space

$$\begin{aligned} E=\left\{ u \in C^{1}[a, \infty ) : \displaystyle \lim _{t \rightarrow \infty } \frac{u(t)}{1+ (t-a)}< \infty , \, \displaystyle \lim _{t \rightarrow \infty } u^{[\triangle ]} (t) < \infty \right\} \end{aligned}$$

with the norm

$$\begin{aligned} \Vert u\Vert = \max \{ \Vert u\Vert _{1}, \Vert u^{[\triangle ]}\Vert _{\infty } \}, \end{aligned}$$

where

$$\begin{aligned} \Vert u\Vert _{1}= \displaystyle \sup _{ t \in [a, \infty )} \frac{|u(t)|}{1+ (t-a)}, \quad \Vert u^{[\triangle ]}\Vert _{\infty }=\displaystyle \sup _{ t \in [a, \infty )} |u^{[\triangle ]} (t) |. \end{aligned}$$

We consider the following problem

$$\begin{aligned} \left\{ \begin{array}{ll} [p(t)u^{\triangle }(t)]^\nabla + \psi (t)f^{*}(t, u(t), u^{[\triangle ]}(t))=0, \, t \in (a, \infty ), \\ \alpha u(a)-\beta u^{[\triangle ]}(a) = \displaystyle \sum _{k=1}^{m-2} a_{k}u(\xi _{k}), \\ \displaystyle \gamma u(\infty )+ \delta u^{[\triangle ]}(\infty )=\sum _{k=1}^{m-2}b_{k}u(\xi _{k}),\\ \end{array}\right. \end{aligned}$$
(3.1)

where

$$\begin{aligned} f^{*}(t, u, v)= \left\{ \begin{array}{ll} f_{R}(t, \underline{u}(t), v)- \displaystyle \frac{u-\underline{u}(t)}{1+|u-\underline{u}(t)|}, &{} \underline{u}(t) > u,\\ f_{R}(t, u, v), &{} \underline{u}(t)\le u \le \overline{u}(t),\\ f_{R}(t, \overline{u}(t), v)- \displaystyle \frac{u-\overline{u}(t)}{1+|u-\overline{u}(t)|}, &{} \overline{u}(t)< u\\ \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} f_{R}(t, u, v)= \left\{ \begin{array}{ll} f(t,u, -R), &{} v < -R,\\ f(t, u, v), &{} -R \le v \le R ,\\ f(t, u, R), &{} v > R.\\ \end{array}\right. \end{aligned}$$

We define the operator \(F: E \rightarrow E\) by

$$\begin{aligned} Fu(t)=\displaystyle \int _{a}^{\infty } G(t, s) \psi (s) f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \nabla s +A( \psi f^{*}) \phi _{1}(t) + B( \psi f^{*} ) \phi _{2}(t). \end{aligned}$$

It is well known that the existence of the solution to the system (3.1) is equivalent to the existence of fixed point of the operator F. So we shall seek a fixed point of F .

In the rest of the paper, we assume that the following conditions are satisfied:

\((H_{4})\):

Problem (1.1) has a pair of upper and lower solutions \(\overline{u}\) , \(\underline{u}\) with \(\underline{u}(t) \le \overline{u}(t)\) for \(t \in (a, \infty )\),

\((H_{5})\):

\(f \in C([a,\infty ) \times {\mathbb {R}}^{2}, {\mathbb {R}})\) satisfies the Nagumo condition with respect to \(\underline{u}\) and \(\overline{u}\) also f(t, u, .) is non-decreasing in \({\mathbb {R}}\) and \(f(t, u, v)< 0\) for all \(t \in [a, \infty )\), \(\underline{u}(t) \le u\le \overline{u}(t)\) and \(v \in {\mathbb {R}}\),

\((H_{6})\):

\(\displaystyle \lim _{t \rightarrow \infty }\psi (t) \mu (t) (\varphi (t)+1)<\infty \), where \(\mu \) is graininess function.

Theorem 3.1

Assume that \((H_{1})-(H_{6})\) hold. Then problem (1.1) has at least one solution \(u \in E\) such that \(\underline{u}(t) \le u(t) \le \overline{u}(t)\) for \(t \in [a, \infty )\).

Proof

First, we show that the operator \(F: E \rightarrow E\) is well defined. For any fixed \(u \in E\), we have

$$\begin{aligned} \displaystyle \int _{a}^{\infty } \frac{G(t, s)}{1+ (t-a)} \psi (s) f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \nabla s \le \displaystyle \int _{a}^{\infty }G(s, s) \psi (s)(\varphi (s) H_{0} +1) \nabla s <\infty , \end{aligned}$$

where \(H_{0}= \displaystyle \max _{0\le s \le \Vert u^{[\triangle ]}\Vert _{\infty } }h(s)\). By Lebesgue dominated convergent theorem, we get

$$\begin{aligned} \displaystyle \lim _{t \rightarrow \infty }\frac{ F u(t)}{1+ (t-a)}= & {} \displaystyle \lim _{t \rightarrow \infty } \bigg \{ \displaystyle \int _{a}^{\infty } \frac{G(t, s)}{1+ (t-a)} \psi (s) f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \nabla s\\&+A( \psi f^{*})\frac{\phi _{1}(t)}{1+ (t-a)}\\&+ B( \psi f^{*} )\displaystyle \frac{\phi _{2}(t)}{1+ (t-a)} \bigg \}\\= & {} \displaystyle \int _{a}^{\infty } \lim _{t \rightarrow \infty } \frac{G(t, s)}{1+ (t-a)} \psi (s) f^{*}\left( s,u(s) , u^{[\triangle ]}(s)\right) \nabla s\\&+ A( \psi f^{*})\lim _{t \rightarrow \infty } \frac{\phi _{1}(t)}{1+ (t-a)}\\&+ B( \psi f^{*}) \displaystyle \lim _{t \rightarrow \infty } \frac{\phi _{2}(t)}{1+ (t-a)}\\< & {} \infty , \end{aligned}$$

and similarly, we find

$$\begin{aligned} \displaystyle \lim _{t \rightarrow \infty } (Fu)^{[\triangle ]} (t)= & {} \displaystyle \lim _{t \rightarrow \infty } \bigg \{\frac{\phi _{2}^{[\triangle ]} (t) }{d} \displaystyle \int _{a}^{t} \phi _{1} (s) \psi (s) f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \nabla s \\&+ \displaystyle \frac{ \phi _{1}^{[\triangle ]} (t) }{d} \displaystyle \int _{t}^{\infty } \phi _{2} (s) \psi (s) f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \nabla s +A( \psi f^{*}) \phi _{1}^{[\triangle ]} (t) \\&+B( \psi f^{*}) \phi _{2}^{[\triangle ]} (t) + \mu (t) \psi (\sigma (t)) f^{*}(\sigma (t), u(\sigma (t)) , u^{[\triangle ]}(\sigma (t))) \bigg \}\\= & {} \displaystyle \lim _{t \rightarrow \infty } \frac{\phi _{2}^{[\triangle ]} (t) }{d} \displaystyle \int _{a}^{t} \phi _{1} (s) \psi (s) f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \nabla s \\&+A( \psi f^{*}) \displaystyle \lim _{t \rightarrow \infty } \phi _{1}^{[\triangle ]} (t)+ B( \psi f^{*}) \displaystyle \lim _{t \rightarrow \infty } \phi _{2}^{[\triangle ]} (t)\\&+ \displaystyle \lim _{t \rightarrow \infty } \mu (t) \psi (\sigma (t)) f^{*}(\sigma (t), u(\sigma (t)) , u^{[\triangle ]}(\sigma (t)))\\= & {} \displaystyle \frac{-\gamma }{d} \displaystyle \int _{a}^{\infty } \phi _{1} (s) \psi (s) f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \nabla s +\alpha A( \psi f^{*}) -\gamma B( \psi f^{*})\\&+ \displaystyle \lim _{t \rightarrow \infty } \mu (t) \psi (\sigma (t)) f^{*}\left( \sigma (t), u(\sigma (t)) , u^{[\triangle ]}(\sigma (t))\right) \\< & {} \infty , \end{aligned}$$

so \(Fu \in E\).

Now, we shall prove that \(F: E \rightarrow E\) is continuous. For any convergent sequence \(u_{n} \rightarrow u \), there exists \(r_{1} > 0\) such that \(\Vert u_{n}\Vert , \Vert u\Vert \le r_{1}\). From then, we have

$$\begin{aligned}&\displaystyle \int _{a}^{\infty } \frac{G(t, s)}{1+ (t-a)} \psi (s)| f^{*}\left( s, u_{n}(s) , u_{n}^{[\triangle ]}(s)\right) \\&- f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) |\nabla s \le 2\displaystyle \int _{a}^{\infty }G(s, s) \psi (s)(\varphi (s) H_{r_{1}} +1) \nabla s <\infty , \end{aligned}$$

where \(H_{r_{1}}= \displaystyle \max _{0\le s \le r_{1} }h(s)\).

And then, for all \(t \in [a, \infty )\), we get

$$\begin{aligned} \displaystyle \frac{|Fu_{n}(t)- Fu(t)| }{1+(t-a)}= & {} \frac{1}{1+(t-a)}\bigg | \displaystyle \int _{a}^{\infty } G(t, s)\psi (s)\left[ f^{*}\left( s, u_{n}(s) , u_{n}^{[\triangle ]}(s)\right) \right. \\&\left. - f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \right] \nabla s \\&+\left[ A\left( \psi f^{*}\left( s, u_{n} , u_{n}^{[\triangle ]}\right) \right) - A\left( \psi f^{*}\left( s, u, u^{[\triangle ]}\right) \right) \right] \displaystyle \frac{ \phi _{1}(t)}{1+(t-a)} \\&+\left[ B\left( \psi f^{*}\left( s, u_{n} , u_{n}^{[\triangle ]}\right) \right) - B\left( \psi f^{*}\left( s, u, u^{[\triangle ]}\right) \right) \right] \displaystyle \frac{ \phi _{2}(t)}{1+(t-a)}\bigg |\\\le & {} \displaystyle \int _{a}^{\infty } \sup _{a\le t< \infty } \frac{G(t, s)}{1+(t-a)}\psi (s)\Bigg |f^{*}\left( s, u_{n}(s) , u_{n}^{[\triangle ]}(s)\right) \\&- f^{*}(s, u(s) , u^{[\triangle ]}(s))| \nabla s \\&+\Bigg |A\left( \psi \left[ f^{*}\left( s, u_{n} , u_{n}^{[\triangle ]}\right) - f^{*}\left( s, u, u^{[\triangle ]} \right) \right] \right) \Bigg |\displaystyle \sup _{a\le t< \infty }\frac{ \phi _{1}(t)}{1+(t-a)} \\&+\Bigg |B\left( \psi \left[ f^{*}\left( s, u_{n} , u_{n}^{[\triangle ]}\right) - f^{*}\left( s, u, u^{[\triangle ]} \right) \right] \right) \Bigg | \displaystyle \sup _{a\le t < \infty }\frac{ \phi _{2}(t)}{1+(t-a)}\\ \end{aligned}$$

from which, we find

$$\begin{aligned} \displaystyle \lim _{n \rightarrow \infty } \Vert Fu_{n} - Fu \Vert _{1} = 0. \end{aligned}$$

And similarly, for all \(t \in [a, \infty )\), we get

$$\begin{aligned} \left| (Fu_{n})^{[\triangle ]}(t)- (Fu)^{[\triangle ]}(t)\right|= & {} \bigg |\displaystyle \frac{\phi _{2}^{[\triangle ]} (t) }{d} \displaystyle \int _{a}^{t} \phi _{1} (s) \psi (s)\left[ f^{*}\left( s, u_{n}(s) , u_{n}^{[\triangle ]}(s)\right) \right. \\&\left. - f^{*}\left( s, u(s), u^{[\triangle ]}(s)\right) \right] \nabla s \\&+ \displaystyle \frac{ \phi _{1}^{[\triangle ]} (t) }{d} \displaystyle \int _{t}^{\infty } \phi _{2} (s) \psi (s)\left[ f^{*}\left( s, u_{n}(s) , u_{n}^{[\triangle ]}(s)\right) \right. \\&\left. -f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \right] \nabla s \\&+\left[ A\left( \psi f^{*}\left( s, u_{n} , u_{n}^{[\triangle ]}\right) \right) - A\left( \psi f^{*}\left( s, u, u^{[\triangle ]} \right) \right) \right] \phi _{1}^{[\triangle ]} (t) \\&+\left[ B\left( \psi f^{*}\left( s, u_{n} , u_{n}^{[\triangle ]}\right) \right) - B\left( \psi f^{*}\left( s, u, u^{[\triangle ]} \right) \right) \right] \phi _{2}^{[\triangle ]} (t) \\&+\mu (t) \psi (\sigma (t)) \left[ f^{*}\left( \sigma (t), u_{n}(\sigma (t)) , u_{n}^{[\triangle ]}(\sigma (t))\right) \right. \\&\left. -f^{*}\left( \sigma (t),u(\sigma (t)) , u^{[\triangle ]}(\sigma (t))\right) \right] \bigg |\\\le & {} \displaystyle \sup _{a\le t< \infty } \frac{\phi _{2}^{[\triangle ]} (t) }{d} \displaystyle \int _{a}^{t} \phi _{1} (s) \psi (s)\Bigg | f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \\&- f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \Bigg | \nabla s\\&+\displaystyle \sup _{a\le t< \infty } \frac{\phi _{1}^{[\triangle ]} (t) }{d} \displaystyle \int _{t}^{\infty } \phi _{2} (s) \psi (s)\Bigg | f^{*}\left( s, u_{n}(s) , u_{n}^{[\triangle ]}(s)\right) \\&- f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \Bigg |\nabla s \\&+\left| A\left( \psi \left[ f^{*}\left( s, u_{n} , u_{n}^{[\triangle ]}\right) - f^{*}\left( s, u, u^{[\triangle ]} \right) \right] \right) \Bigg |\displaystyle \sup _{a\le t< \infty } \phi _{1}^{[\triangle ]} (t)\right. \\&\left. +\left| B\left( \psi \left[ f^{*}\left( s, u_{n} , u_{n}^{[\triangle ]}\right) - f^{*}\left( s, u, u^{[\triangle ]}\right) \right] \right) \right| \displaystyle \sup _{a\le t < \infty } \phi _{2}^{[\triangle ]} (t)\right. \\&\left. +\mu (t) \psi (\sigma (t)) \Bigg |f^{*}\left( \sigma (t), u_{n}(\sigma (t)) , u_{n}^{[\triangle ]}(\sigma (t))\right) \right. \\&\left. -f^{*}\left( \sigma (t), u(\sigma (t)) ,u^{[\triangle ]}(\sigma (t))\right) \Bigg ] \right| , \end{aligned}$$

this implies

$$\begin{aligned} \displaystyle \lim _{n \rightarrow \infty } \left\| (Fu_{n})^{[\triangle ]}(t)- (Fu)^{[\triangle ]}(t) \right\| _{\infty } = 0. \end{aligned}$$

Thus \(F: E \rightarrow E\) is continuous.

Let \(D\subset E\) be any bounded set, then there exists \(r > 0\) such that for any \(u \in D\), it holds \(\Vert u\Vert \le r\). Then for any \(u \in D\) and \(t \in [a, \infty )\), we find

$$\begin{aligned} \displaystyle \frac{|Fu(t)|}{1+ (t-a)}= & {} \frac{1}{1+(t-a)}\bigg |\displaystyle \int _{a}^{\infty } G(t, s)\psi (s) f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \nabla s +A(\psi f^{*}) \phi _{1}(t)\\&+ B(\psi f^{*} )\phi _{2}(t)\bigg |\\\le & {} \displaystyle \int _{a}^{\infty } \sup _{a\le t< \infty } \frac{G(t, s)}{1+(t-a)}\psi (s)\left| f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \right| \nabla s \\&+ |A(\psi f^{*})| \displaystyle \sup _{a\le t< \infty }\frac{ \phi _{1}(t)}{1+(t-a)} + |B(\psi f^{*})| \displaystyle \sup _{a\le t< \infty }\frac{ \phi _{2}(t)}{1+(t-a)}\\\le & {} (1+A\phi _{1}(\infty )+B\phi _{2}(a))\displaystyle \int _{a}^{\infty } G(s, s)\psi (s)(\varphi (s)H_{r}+1) \nabla s < \infty ,\\ \end{aligned}$$

where \(H_{r}= \displaystyle \max _{0\le s \le r }h(s)\). And by the similar way, for all \(t \in [a, \infty )\), we get

$$\begin{aligned} \left| (Fu)^{[\triangle ]}(t)\right|= & {} \bigg |\displaystyle \frac{\phi _{2}^{[\triangle ]} (t) }{d} \displaystyle \int _{a}^{t} \phi _{1} (s) \psi (s) f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \nabla s\\&+ \displaystyle \frac{ \phi _{1}^{[\triangle ]} (t) }{d} \displaystyle \int _{t}^{\infty } \phi _{2} (s) \psi (s)f^{*}\left( s, u(s) , u^{[\triangle ]}(s)\right) \nabla s \\&+A( \psi f^{*} ) \phi _{1}^{[\triangle ]} (t) +B( \psi f^{*} ) \phi _{2}^{[\triangle ]} (t)\\&+\mu (t) \psi (\sigma (t)) f^{*}\left( \sigma (t),u(\sigma (t)) , u^{[\triangle ]}(\sigma (t))\right) \bigg |\\\le & {} \displaystyle \sup _{a\le t< \infty } \frac{ \phi _{2}^{[\triangle ]} (t) }{d} \displaystyle \int _{a}^{t} \phi _{1} (s) \psi (s) (\varphi (s)H_{r}+1) \nabla s\\&+\mu (t) \psi (\sigma (t))(\varphi (\sigma (t))H_{r}+1)\\&+\displaystyle \sup _{a\le t< \infty } \frac{ \phi _{1}^{[\triangle ]} (t) }{d} \displaystyle \int _{t}^{\infty } \phi _{2} (s) \psi (s) (\varphi (s)H_{r}+1) \nabla s \\&+\left[ A\displaystyle \sup _{a\le t< \infty } \left| \phi _{1}^{[\triangle ]} (t)\right| \right. \\&\left. +B\displaystyle \sup _{a\le t< \infty } \left| \phi _{2}^{[\triangle ]} (t)\right| \right] \displaystyle \int _{a}^{\infty } G(s, s)\psi (s)(\varphi (s)H_{r}+1) \nabla s<\infty ,\\ \end{aligned}$$

that is, F(D) is uniformly bounded. Moreover, we can easily show that F(D) is equicontinuous.

Finally, we show that F(D) is equiconvergent at infinity. In fact, we have

$$\begin{aligned}&\bigg |\displaystyle \frac{Fu(t)}{1+(t-a)}- \displaystyle \lim _{t \rightarrow \infty } \frac{Fu(t)}{1+(t-a)}\bigg |\\&\quad =\bigg |\displaystyle \int _{a}^{\infty } \bigg [\frac{G(t, s)}{1+(t-a)}- \displaystyle \lim _{t \rightarrow \infty } \frac{G(t, s)}{1+(t-a)} \bigg ] \psi (s) f^{*}(s, u(s) , u^{[\triangle ]}(s))\nabla s\\&\qquad + A(\psi f^{*}) \bigg [\displaystyle \frac{\phi _{1}(t)}{1+(t-a)}- \displaystyle \lim _{t \rightarrow \infty } \frac{\phi _{1}(t)}{1+(t-a)}\bigg ]\\&\qquad + B(\psi f^{*}) \bigg [\displaystyle \frac{\phi _{2}(t)}{1+(t-a)}- \displaystyle \lim _{t \rightarrow \infty } \frac{\phi _{2}(t)}{1+(t-a)}\bigg ] \bigg | \\&\quad \le \displaystyle \int _{a}^{\infty } \bigg | \frac{G(t, s)}{1+(t-a)}- \displaystyle \lim _{t \rightarrow \infty } \frac{G(t, s)}{1+(t-a)} \bigg | \psi (s) (\varphi (s)H_{r}+1) \nabla s\\&\qquad + A \bigg |\displaystyle \frac{\phi _{1}(t)}{1+(t-a)}- \displaystyle \lim _{t \rightarrow \infty } \frac{\phi _{1}(t)}{1+(t-a)}\bigg | \displaystyle \int _{a}^{\infty } G(s, s)\psi (s)(\varphi (s)H_{r}+1) \nabla s \\&\qquad + B \bigg |\displaystyle \frac{\phi _{2}(t)}{1+(t-a)}- \displaystyle \lim _{t \rightarrow \infty } \frac{\phi _{2}(t)}{1+(t-a)}\bigg | \displaystyle \int _{a}^{\infty } G(s, s)\psi (s)(\varphi (s)H_{r}+1) \nabla s \\&\qquad \rightarrow 0, \, as \, t \rightarrow \infty , \end{aligned}$$

and similarly, we get

$$\begin{aligned}&\left| (Fu)^{[\triangle ]}(t)-(Fu)^{[\triangle ]}(\infty )\right| \\&\le \quad \displaystyle \int _{a}^{\infty } \left| G^{[\triangle ]}(t, s)-\displaystyle \lim _{t \rightarrow \infty } G^{[\triangle ]}(t, s)\right| \psi (s)(\varphi (s)H_{r}+1) \nabla s\\&\qquad + A \left| \displaystyle \phi _{1}^{[\triangle ]}(t)- \displaystyle \lim _{t \rightarrow \infty } \phi _{1}^{[\triangle ]}(t)\right| \displaystyle \int _{a}^{\infty } G(s, s)\psi (s)(\varphi (s)H_{r}+1) \nabla s \\&\qquad +B \left| \displaystyle \phi _{2}^{[\triangle ]}(t)- \displaystyle \lim _{t \rightarrow \infty } \phi _{2}^{[\triangle ]}(t)\right| \displaystyle \int _{a}^{\infty } G(s, s)\psi (s)(\varphi (s)H_{r}+1) \nabla s \\&\qquad +|\mu (t) \psi (\sigma (t)) - \displaystyle \lim _{t \rightarrow \infty } \mu (t) \psi (\sigma (t)) | (\varphi (t)H_{r}+1)\\&\qquad \rightarrow 0, \, as \, t \rightarrow \infty . \end{aligned}$$

Then, we obtain that \(F: E \rightarrow E\) is completely continuous. By using the Schäuder fixed point theorem, we find that F has at least one fixed point \(u \in E\).

Now, we shall demonstrate that the function u satisfies \(\underline{u}(t) \le u(t) \le \overline{u}(t)\) and \(|u^{[\triangle ]}(t)| \le R\) for \(t \in [a, \infty )\) and positive real R. For this purpose, firstly, we show that \(\underline{u}(t) \le u(t)\), \(t \in [a, \infty )\). Therefore, let \(N > 0\) and choose

$$\begin{aligned}&\eta \ge \displaystyle \bigg (\sup _{t \in [a, \infty )}p(t) \bigg ) \max \bigg \{ \displaystyle \sup _{t \in [N, \infty ) } \frac{\overline{u}(t)-\underline{u}(a)}{t-a}, \displaystyle \sup _{t \in [N, \infty ) } \frac{\overline{u}(a)-\underline{u}(t)}{t-a} \bigg \} ,\\&\displaystyle \max \bigg \{\sup _{t \in [a, \infty )}\left| \underline{u}^{[\triangle ]}(t)\right| , \sup _{t \in [a, \infty )}\left| \overline{u}^{[\triangle ]}(t)\right| \bigg \} \le \eta , \end{aligned}$$

and \(R> \eta \) such that

$$\begin{aligned} \displaystyle \int _{\eta }^{R} \frac{s}{h(|s|)} \nabla s > \left( \sup _{t \in [a, \infty ) } p(t)\psi (t) \varphi (t)\right) \left( \sup _{t \in [a, \infty ) }\overline{u}(t)-\inf _{t \in [a, \infty ) }\underline{u}(t)\right) . \end{aligned}$$

Moreover, define \(g(t)= \underline{u}(t)-u(t)\). Suppose that \(g(t) > 0\) for some \(t \in [a, \infty ) \). Hence, there exists \(t_{0} \in [a, \infty )\) such that \(g(t_{0}) = \displaystyle \sup _{ t \in [a, \infty )} g(t)\). There are three cases to consider.

Case 1. If \(t_{0} \in (a, \infty )\), then, from [8, Lemma 6.17], we know

$$\begin{aligned} g^{\triangle }(t_{0}) \le 0 \text { and }[p(t_{0})g^{\triangle }(t_{0})]^\nabla \le 0, \end{aligned}$$

from which, the assumption \((H_{5})\) and \(R > \displaystyle \sup _{t \in [a, \infty )}| \underline{u}^{[\triangle ]}(t)|\), we obtain

$$\begin{aligned}{}[p(t_{0})g^{\triangle }(t_{0})]^\nabla= & {} [p(t_{0})\underline{u}^{\triangle }(t_{0})]^\nabla - [p(t_{0})u^{\triangle }(t_{0})]^\nabla \\\ge & {} -\psi (t)f\left( t_{0}, \underline{u}(t_{0}), \underline{u}^{[\triangle ]}(t_{0})\right) + \psi (t)f^{*}\left( t_{0}, u(t_{0}), u^{[\triangle ]}(t_{0})\right) \\= & {} -\psi (t) f\left( t_{0}, \underline{u}(t_{0}), \underline{u}^{[\triangle ]}(t_{0})\right) + \psi (t)f_{R}\left( t_{0}, \underline{u}(t_{0}), u^{[\triangle ]}(t_{0})\right) \\&- \displaystyle \psi (t)\frac{u(t_{0})-\underline{u}(t_{0})}{1+|u(t_{0})-\underline{u}(t_{0})|}\\> & {} \psi (t) \left[ f_{R}\left( t_{0}, \underline{u}(t_{0}), u^{[\triangle ]}(t_{0})\right) -f\left( t_{0}, \underline{u}(t_{0}), \underline{u}^{[\triangle ]}(t_{0})\right) \right] \\> & {} 0, \end{aligned}$$

which is a contradiction.

Case 2. If \(t_{0}=a\), then, by the boundary condition, we have

$$\begin{aligned} 0 \le \alpha (\underline{u}(a)- u(a) ) \le \beta p(a) \left( \underline{u}^{\triangle }(a)-u^{\triangle }(a) \right) , \end{aligned}$$

which is a contradiction.

Case 3. If \( g(\infty ) =\displaystyle \sup _{t \in [a, \infty ) } g(t) \), then, similarly, using the boundary condition, we get

$$\begin{aligned} 0 \le \displaystyle \gamma (\underline{u}(\infty ) - u(\infty ) ) \le \delta \left( u^{[\triangle ]}(\infty ) - \underline{u}^{[\triangle ]}(\infty )\right) , \end{aligned}$$

which is also a contradiction. Consequently, \(\underline{u}(t) \le u(t)\) holds for all \(t \in [a, \infty )\). Similarly, we can show that \(u(t) \le \overline{u}(t)\). Now, we demonstrate that \(|u^{[\triangle ]}(t)| \le R\) for \(t \in [a, \infty )\). There are three cases to be considered.

Case 1. If \(|u^{[\triangle ]}(t)| \le \eta \), \(\forall t \in [a, \infty )\), we take \(R=\eta \) and we complete the proof.

Case 2. If \(|u^{[\triangle ]}(t)| > \eta \), \(\forall t \in [a, \infty )\), without loss of generality, we suppose \(u^{[\triangle ]}(t) > \eta \) for \(t \in [a, \infty )\). While for any \(T >\, N\), we get

$$\begin{aligned} \displaystyle \sup _{t \in [a, \infty )}p(t) \displaystyle \bigg ( \frac{\overline{u}(T)-\underline{u}(a)}{T-a} \bigg )\ge & {} \displaystyle \sup _{t \in [a, \infty )}p(t) \displaystyle \bigg (\frac{u(T)-u(a)}{T-a}\bigg ) \\= & {} \displaystyle \frac{1}{T-a}\displaystyle \int _{a}^{T} \sup _{t \in [a, \infty )}p(t) u^{\triangle }(s)\Delta s \\> & {} \eta , \end{aligned}$$

which is a contradiction.

Case 3. There exist \(t_{0}, t_{1} \in [a, \infty )\) such that \(|u^{[\triangle ]}(t_{1})| > \eta \) and \(|u^{[\triangle ]}(t_{0})|\le \eta \).

In this case, firstly, suppose that \(u^{[\triangle ]}(t_{1}) > \eta \) and \(|u^{[\triangle ]}(t_{0})|\le \eta \), then by a convenient change of variable, we have

$$\begin{aligned} \displaystyle \int _{u^{[\triangle ]}(t_{0})}^{u^{[\triangle ]}(t_{1})} \frac{s}{h(|s|)} \nabla s= & {} \displaystyle \int _{t_{0}}^{t_{1}} \frac{u^{[\triangle ]}(t)}{h(|u^{[\triangle ]}(t)|)} \left[ p(t)u^{\triangle }(t)\right] ^\nabla \nabla t \\= & {} \displaystyle \int _{t_{0}}^{t_{1}} \frac{-u^{[\triangle ]}(t)}{h(|u^{[\triangle ]}(t)|)} \psi (t) f^{*}\left( t, u(t), u^{[\triangle ]}(t)\right) \nabla t\\= & {} \displaystyle \int _{t_{0}}^{c} \frac{-u^{[\triangle ]}(t)}{h(|u^{[\triangle ]}(t)|)} \psi (t) f^{*}\nabla t + \displaystyle \int _{c}^{t_{1}} \frac{-u^{[\triangle ]}(t)}{h(|u^{[\triangle ]}(t)|)} \psi (t) f^{*}\nabla t\\\le & {} \displaystyle \int _{c}^{t_{1}} \frac{-u^{[\triangle ]}(t)}{h(|u^{[\triangle ]}(t)|)} \psi (t) f^{*}\left( t, u(t), u^{[\triangle ]}(t\right) ) \nabla t\\\le & {} \displaystyle \int _{c}^{t_{1}} \frac{u^{[\triangle ]}(t)}{h(|u^{[\triangle ]}(t)|)} \psi (t) \varphi (t)h(|u^{[\triangle ]}(t)|) \nabla t\\= & {} \displaystyle \int _{c}^{t_{1}} p(t)u^{\triangle }(t)\psi (t) \varphi (t) \nabla t \\\le & {} \bigg (\displaystyle \sup _{t \in [a, \infty ) }\ p(t) \psi (t) \varphi (t)\bigg ) \int _{c}^{t_{1}}u^{\triangle }(t) \nabla t \\= & {} \bigg (\displaystyle \sup _{t \in [a, \infty ) }\ p(t) \psi (t) \varphi (t)\bigg ) \int _{c}^{t_{1}}u^{\nabla }(\sigma (t)) \nabla t \\= & {} \bigg (\displaystyle \sup _{t \in [a, \infty ) }\ p(t) \psi (t) \varphi (t)\bigg )(u(\sigma (t_{1})-u(\sigma (c))\\\le & {} \bigg (\displaystyle \sup _{t \in [a, \infty ) } p(t)\psi (t) \varphi (t)\bigg ) \bigg (\sup _{t \in [a, \infty ) }\overline{u}(t)-\inf _{t \in [a, \infty ) }\underline{u}(t)\bigg )\\< & {} \displaystyle \int _{\eta }^{R} \frac{s}{h(|s|)} \nabla s, \end{aligned}$$

this implies, \(u^{[\triangle ]}(t_{1}) < R\), where \(c =\inf \{t \in [t_{0}, t_{1}] : u^{[\triangle ]}(t)> 0 \}\).

Secondly, we suppose that \(-u^{[\triangle ]}(t_{1}) > \eta \) and \(|u^{[\triangle ]}(t_{0})|\le \eta \). Similarly, we get

$$\begin{aligned} \displaystyle \int _{-u^{[\triangle ]}(t_{0})}^{-u^{[\triangle ]}(t_{1})} \frac{s}{h(|s|)} \nabla s= & {} \displaystyle \int _{t_{0}}^{t_{1}} \frac{u^{[\triangle ]}(t)}{h(|u^{[\triangle ]}(t)|)} \left[ p(t)u^{\triangle }(t)\right] ^\nabla \nabla t \\= & {} \displaystyle \int _{t_{1}}^{t_{0}} \frac{u^{[\triangle ]}(t)}{h(|u^{[\triangle ]}(t)|)} \psi (t) f^{*}\left( t, u(t), u^{[\triangle ]}(t)\right) \nabla t\\= & {} \displaystyle \int _{t_{1}}^{k} \frac{u^{[\triangle ]}(t)}{h(|u^{[\triangle ]}(t)|)} \psi (t) f^{*}\nabla t + \displaystyle \int _{k}^{t_{0}} \frac{u^{[\triangle ]}(t)}{h(|u^{[\triangle ]}(t)|)} \psi (t) f^{*}\nabla t\\\le & {} \displaystyle \int _{t_{1}}^{k} \frac{u^{[\triangle ]}(t)}{h(|u^{[\triangle ]}(t)|)} \psi (t) f^{*}\left( t, u(t), u^{[\triangle ]}(t)\right) \nabla t\\\le & {} \displaystyle \int _{t_{1}}^{k} \frac{-u^{[\triangle ]}(t)}{h(|u^{[\triangle ]}(t)|)} \psi (t) \varphi (t)h(|u^{[\triangle ]}(t)|) \nabla t\\= & {} \displaystyle \int _{t_{1}}^{k} - p(t)u^{\triangle }(t)\psi (t) \varphi (t) \nabla t \\\le & {} \bigg (\displaystyle \sup _{t \in [a, \infty ) }\ p(t) \psi (t) \varphi (t)\bigg ) \displaystyle \int _{t_{1}}^{k} -u^{\triangle }(t) \nabla t \\= & {} \bigg (\displaystyle \sup _{t \in [a, \infty ) }\ p(t) \psi (t) \varphi (t)\bigg )\displaystyle \int _{t_{1}}^{k} -u^{\nabla }(\sigma (t)) \nabla t \\\le & {} \bigg (\displaystyle \sup _{t \in [a, \infty ) } p(t)\psi (t) \varphi (t)\bigg ) \bigg (\sup _{t \in [a, \infty ) }\overline{u}(t)-\inf _{t \in [a, \infty ) }\underline{u}(t)\bigg )\\< & {} \displaystyle \int _{\eta }^{R} \frac{s}{h(|s|)} \nabla s, \end{aligned}$$

hence, \(-u^{[\triangle ]}(t_{1}) < R\), where \(k =\sup \{t \in [t_{0}, t_{1}] : u^{[\triangle ]}(t)<0 \}\). Since \(t_{0}\) and \(t_{1}\) can be taken arbitrary, we conclude that if there exists a \(t \in [a, \infty ) \) with \(|u^{[\triangle ]}(t)| \ge \eta \), then \(|u^{[\triangle ]}(t)|\le R\). Therefore,

$$\begin{aligned}{}[p(t)u^{\triangle }(t)]^\nabla = - \psi (t)f^{*}(t, u(t), u^{[\triangle ]}(t))= -\psi (t)f (t, u(t), u^{[\triangle ]}(t)), \end{aligned}$$

that is, u is a solution of problem (1.1). This completes the proof. \(\square \)