Existence of an approximate solution for a class of fractional multi-point boundary value problems with the derivative term

Abstract

In this paper, we consider a class of fractional boundary value problems with the derivative term and nonlinear operator term. By establishing new mixed monotone fixed point theorems, we prove these problems to have a unique solution, and we construct the corresponding iterative sequences to approximate the unique solution.

1 Introduction

In this paper, we consider the existence and uniqueness of nontrivial solutions of the following fractional boundary value problem:

$$ \textstyle\begin{cases} {D}^{\beta }_{0^{+}}({D}^{\alpha }_{0^{+}} u(t)) = f(t,u(t),{D}^{\nu }_{0^{+}} u(t)) + g(t,u(t),(\mathcal{K} u)(t))-b , \quad t\in (0, 1), \\ u(0)=u'(0) =\cdots=u^{(n-2)}(0)=0, \\ {D}^{\alpha }_{0^{+}} u(0)= ({D}^{\alpha }_{0^{+}} u)^{\prime }(0) = \cdots =({D}^{\alpha }_{0^{+}} u)^{(n-2)}(0)=0, \\ {D}^{\gamma }_{0^{+}} u(1)=\sum _{i=1}^{m-2} \xi _{i}{D}^{ \gamma }_{0^{+}}u(\eta _{i}), \qquad {D}^{\alpha }_{0^{+}} u(1)=\sum _{i=1}^{m-2} \zeta _{i}{D}^{ \alpha }_{0^{+}}u(\eta _{i}), \end{cases} $$

(1.1)

where $b>0$, ${D}^{\alpha }_{0^{+}}$, ${D}^{\beta }_{0^{+}}$, ${D}^{\gamma }_{0^{+}}$, ${D}^{\nu }_{0^{+}}$ are the Riemann–Liouville fractional derivatives with $n-1<\alpha$, $\beta \leq n$, $n-2< \gamma \leq n-1$, $n\geq 2$ ($n\in \mathbb{N}$), $\alpha -\gamma -1>0$, $0< \nu \leq \gamma$, $0<\xi _{i}$, $\eta _{i},\zeta _{i}<1$, $i=1, 2, 3, \ldots , m-2$, $m\geq 2$, $\sum_{i=1}^{m-2} \xi _{i} \eta _{i}^{\alpha -\gamma -1}<1$, $\sum_{i=1}^{m-2} \zeta _{i}\eta _{i}^{\beta -1}<1$. $f,g:[0,1]\times (-\infty ,+\infty )\times (-\infty ,+\infty ) \longrightarrow (-\infty ,+\infty )$ are continuous.

In recent years, much attention has been paid to multi-point boundary value problems involving fractional order; see [1–29] and the references therein. We should mention related studies in [1–22], which motivated us to consider the problem (1.1). Lv [1] studied the existence of positive solutions of the following multi-point boundary value problem:

$$ \textstyle\begin{cases} {D}^{\alpha }_{0^{+}} u(t) + f(t,u(t)) =0,\quad t\in (0, 1), \\ u(0)=0, \qquad {D}^{\beta }_{0^{+}} u(1)=\sum _{i=1}^{m-2} \xi _{i}{D}^{ \beta }_{0^{+}}u(\eta _{i}), \end{cases} $$

(1.2)

where $1<\alpha \leq 2$, $0\leq \beta \leq 1$, $0\leq \alpha -\beta -1$, $0<\xi _{i}$, $\eta _{i}<1$ ($i=1,2,\ldots ,m-2$), and $\sum_{i=1}^{m-2} \xi _{i} \eta _{i}^{\alpha -\beta -1}<1$. $f:[0,1]\times [0,+\infty )\longrightarrow [0,+\infty )$ is a continuous function. Lv [1] first obtained the Green’s function of linear boundary value problem corresponding to the problem (1.2), which has been adopted in the proof of main theorems in [2–4]. Furthermore, Lv [2] studied the existence of solutions for nonlinear fractional m-point boundary value problems involving p-Laplacian operators by the fixed point index theorem. Li, Luo and Zhou [5] discussed Eq. (1.2) in the case of $m=3$ by using some fixed point theorems.

Very recently, Wang, Zhang and Wang [6] considered the following nonlinear fractional boundary value problem:

$$ \textstyle\begin{cases} -D_{0^{+}}^{\alpha }u(t)=f (t,u(t),u(t) )+g (t,u(t) ), \quad 0< t< 1, n-1< \alpha \leq n, \\ u^{(i)}(0)=0, \quad i=0,\ldots , n-2, \qquad D_{0^{+}}^{\nu }u(1)=b D_{0^{+}}^{\nu }u(\xi ), \quad n-2< \nu \leq n-1, \end{cases} $$

(1.3)

where $n-1< \alpha \leq n$ ($n>2,n\in \mathbb{N}$), $n-2<\nu \leq n-1$, $0\leq b\leq 1$, $0<\xi <1$ satisfying $\alpha -\nu -1\geq 0$ and $0\leq b \xi ^{\alpha -\nu -1}<1$. They established the existence and uniqueness of solutions of (1.3) by applying the properties of Green function and fixed point theorems for sum-type operator. On the other hand, they also gave the physical application of our system (1.1). The main feature of [6] is that the value of α is extended from $1<\alpha \leq 2$ in (1.2) to $n-1< \alpha \leq n$. Liang and Zhang [7] considered the existence of solutions of the problem (1.3) when $n=4$, $\nu =2$ and $f (t,u(t),u(t) )=0$. Moreover, Jleli and Samet [8] gave some sufficient conditions under which the problem (1.3) has a unique positive solution when $b=0$.

We note that Wang [9] studied the existence and uniqueness of positive solutions for singular fractional differential equations as follows:

$$ \textstyle\begin{cases} {D}^{\alpha }_{0^{+}} u(t) + p(t) f(t,u(t),{D}^{\beta }_{0^{+}} u(t)) + q(t) g(t,u(t),(H u)(t))=0, \quad 0< t< 1, \\ u(0)=u'(0) =\cdots=u^{(n-2)}(0)=0, \qquad [{D}^{\gamma }_{0^{+}} u(t)]_{t=1}=k(u(1)), \end{cases} $$

(1.4)

where $n-1<\alpha \leq n$, $n>3$, $1\leq \beta \leq \gamma \leq n-2$, p, $q\in C((0,1), [0, +\infty ))$, $p(t)$ and $q(t)$ are allowed to be singular at $t=0$ or $t=1$. $f, g: (0, 1)\times [0, +\infty )\times [0, +\infty )\longrightarrow [0, +\infty )$ are continuous, and $k: [0,1)\longrightarrow [0,+\infty )$ is also continuous. What attracts our attention is the nonlinear term contains not only the derivative term but also the operator term $(Hu)$. Similarly, Zhang and Tian [10] also studied the problem (1.4) with derivative term, but the difference is that the function g does not include the operator term. In [11], Ji et al. also investigated positive solutions for the nonlinear fractional differential equation with a derivative term. Goodrich [12] first obtained the Green function of the problem (1.4) when $k(u(1))=0$. In [13–15], they considered the fractional differential equations with integer order derivative, and they did not consider the boundary condition $[{D}^{\gamma }_{0^{+}} u(t)]_{t=1}$. Zhang [16] considered the singular fractional differential equations with multiple derivative terms, and obtained the existence of positive solutions.

We should mention the work of Jong [3], which directly is related to our problem (1.1). Jong investigated the following nonlinear fractional m-point boundary value problem with p-Laplacian operator:

$$ \textstyle\begin{cases} {D}^{\beta }_{0^{+}}(\varphi _{p}({D}^{\alpha }_{0^{+}} u))(t) = f(t,u(t)), \quad t\in (0, 1), \\ u(0)=0, \qquad {D}^{\alpha }_{0^{+}} u(0)=0, \\ {D}^{\gamma }_{0^{+}} u(1)=\sum _{i=1}^{m-2} \xi _{i}{D}^{ \gamma }_{0^{+}}u(\eta _{i}),\qquad \varphi _{p}({D}^{\alpha }_{0^{+}} u)(1)=\sum _{i=1}^{m-2} \zeta _{i}\varphi _{p}({D}^{\alpha }_{0^{+}}u)(\eta _{i}), \end{cases} $$

(1.5)

where $1<\alpha , \beta \leq 2$, $3<\alpha +\beta \leq 4$, $0<\gamma \leq 1$, $\alpha -\gamma -1>0$, $0<\xi _{i}$, $\eta _{i},\zeta _{i}<1$, $\sum_{i=1}^{m-2} \xi _{i} \eta _{i}^{\alpha -\gamma -1}<1$, $\sum_{i=1}^{m-2} \zeta _{i}\eta _{i}^{\beta -1}<1$, the p-Laplacian operator is defined as $\varphi _{p}(s)=|s|^{p-2}s$, $p>1$. Jong obtained that the problem (1.5) has a unique solution which is given by $u(t)=\int ^{1}_{0} G(t,s) \varphi _{p}^{-1} ( \int ^{1}_{0} H(s,\tau )f(\tau ,u(\tau ))\,d\tau )\,ds$. He first gave the Green function $H(s,\tau )$. The main tool of [3] is the Banach contraction mapping principle. Furthermore, he also showed the uniqueness of the problem (1.5) in [4] by the classic fixed point theorem of mixed monotone operators. Li and Qi [17] focused on p-Laplacian boundary value problems of higher order nonlinear differential equations. Tan and Li [18] used Kuratowski’s noncompactness measure and Sadovskii’s fixed point theorem to study the problem (1.5) when the boundary condition $\varphi _{p}({D}^{\alpha }_{0^{+}} u)(1)=\sum_{i=1}^{m-2} \zeta _{i}\varphi _{p}({D}^{\alpha }_{0^{+}}u)(\eta _{i})$ is removed. Wang and Xiang [19] considered the problem (1.5) when all boundary conditions are replaced by $u(0)=0$, ${D}^{\alpha }_{0^{+}} u(0)=0$, $u(1)=au(\xi )$ and $D_{0^{+}}^{\nu }u(1)=b D_{0^{+}}^{\alpha }u(\eta )$. Wang, Xiang and Liu [20] investigated the problem (1.5) when the boundary conditions are replaced by $u(0)=0$, ${D}^{\alpha }_{0^{+}} u(0)=0$ and $u(1)=au(\xi )$.

We should point out that the main tools and methods adopted in [1–20] are cone mapping theory. Therefore, nonlinearities in the problems (1.2)–(1.5) are usually required to be non-negative. But more and more authors are beginning to remove this restriction imposed on nonlinear terms. Very recently, Sang and Ren [21] dealt with the following fractional boundary value problem:

$$ \textstyle\begin{cases} -D_{0^{+}}^{\alpha }u(t)=f (t,u(t),u(t) )+g (t,u(t),u(t) )-b, \quad 0< t< 1, \\ u^{(i)}(0)=0,\quad 0\leq i\leq n-2, \qquad [D_{0^{+}}^{\beta }u(t) ]_{t=1}=0, \end{cases} $$

(1.6)

where $n-1< \alpha \leq n$, $1\leq \beta \leq n-2$, $n\geq 3$ $(n\in \mathbb{N})$, $b>0$ is a constant, $f,g:[0,1]\times (-\infty ,+\infty )\times (-\infty ,+\infty ) \longrightarrow (-\infty ,+\infty )$ are two continuous functions. In fact, Zhai and Wang [22] have introduced φ–$(h, e)$ operators, and established the existence and uniqueness of a nontrivial solution for a class of nonlinear fractional equations by using partial order method.

In this paper, the first goal is to obtain the fixed point of the solution of the operator equation $M(x,x)+ N(x,x)+ e = x$, where M and N are two mixed monotone operators. We will generalize the results of cone mapping to the non-cone case. Then we will provide some sufficient conditions under which the problem (1.1) has a unique solution and construct two iterative sequences of unique solution. Compared with [6, 9], we do not demand the assumption that nonlinearities are non-negative, and the more general boundary conditions are adopted.

Our paper is organized as follows. In Sect. 2, we will introduce some definitions and give some lemmas to prove the main conclusions. In Sect. 3, the existence of fixed point for the operator equation associated with the problem (1.1) is established. Then, based on our abstract results, the existence and uniqueness of the solution of the problem (1.1) are proved.

2 Preliminaries and related lemmas

In this section, we give some definitions and lemmas that are useful for the proof of our main results.

In this paper, $(E,\|\cdot \|)$ is a real Banach space. A partially ordered structure in E is induced by a cone $P \subset E$, i.e. $x\leq y$ if and only if $y-x\in P$. θ is the zero element. P is called normal if there exists $N>0$ such that $\theta \leq x \leq y \Rightarrow \|x\| \leq N\|y\|$. Given $h>\theta$, we denote by $P_{h}$

$$ P_{h}= \{ x\in E\mid \text{there exist } \lambda >0 , \mu >0 \text{ such that } \lambda h\leq x\leq \mu h \} . $$

Let $e\in P$ with $\theta \leq e\leq h$, denote

$$ P_{h,e}=\{x\in E\mid x+e\in P_{h}\}. $$

Definition 2.1

([30, 31])

If $A(x,y)$ is increasing in x, and decreasing in y, then $A:P_{h,e}\times P_{h,e}\rightarrow E$ is a mixed monotone operator. That is, for every $u_{i},v_{i}\in P_{h,e}$ ($i=1,2$) with $u_{1}\geq v_{1}$, $u_{2}\leq v_{2}$, we have $A(u_{1},u_{2})\geq A(v_{1},v_{2})$.

Definition 2.2

([32, 33])

The Riemann–Liouville fractional derivative of order $\alpha >0$ of a function $h\in C[0, 1]$ is defined by

$$ D_{0^{+}}^{\alpha }h(t)= \frac{1}{\Gamma (n-\alpha )} \biggl(\frac{d}{dt} \biggr)^{n} \int _{0}^{t} h(s) (t-s)^{n-\alpha -1}\,ds, $$

where $n=[\alpha ]+1$. The Riemann–Liouville fractional integral of order $\alpha >0$ of a function h is given by

$$ I^{\alpha }_{0^{+}} h(t)=\frac{1}{\Gamma (\alpha )} \int ^{t}_{0}(t-s)^{\alpha -1}h(s)\,ds . $$

Definition 2.3

([32])

Let $\alpha >-1$, $\nu >0$ and $t>0$. Then

$$ D_{0^{+}}^{\nu }t^{\alpha }= \frac{\Gamma (\alpha +1)}{\Gamma (\alpha -\nu +1)}t^{\alpha -\nu }. $$

Lemma 2.1

([34])

Let $u\in C[0,1]\cap L^{1}[0,1]$, $\alpha >0$, then

$$ I^{\alpha }_{0^{+}}D_{0^{+}}^{\alpha }u(t)=u(t)+c_{1}t^{\alpha -1}+c_{2}t^{ \alpha -2}+ \cdots +c_{n}t^{\alpha -n}, $$

where $c_{i}\in \mathbb{R}$, $i=1,2,\ldots ,n$ and $n=[\alpha ]+1$.

Lemma 2.2

Let $\sum_{i=1}^{m-2}\xi _{i}\eta _{i}^{\alpha - \gamma -1} \neq 1$. If $y(t)\in C[0,1]$, then the boundary value problem

$$ \textstyle\begin{cases} D_{0^{+}}^{\alpha }u(t)+y(t)=0,\quad 0< t< 1, n-1< \alpha \leq n, \\ u(0)=u'(0)=\cdots =u^{(n-2)}(0)=0, \\ {D}^{\gamma }_{0^{+}} u(1)=\sum _{i=1}^{m-2} \xi _{i}{D}^{ \gamma }_{0^{+}}u(\eta _{i}), \quad n-2< \gamma \leq n-1, \end{cases} $$

has a unique solution

$$ u(t)= \int _{0}^{1} G(t,s)y(s)\,ds, $$

where

$$\begin{aligned}& G(t,s)=G_{1}(t,s)+G_{2}(t,s), \\& G_{1}(t,s)=\frac{1}{\Gamma (\alpha )} \textstyle\begin{cases} t^{\alpha -1}(1-s)^{\alpha - \gamma -1}-(t-s)^{\alpha -1},& 0 \leq s\leq t\leq 1, \\ t^{\alpha -1}(1-s)^{\alpha -\gamma -1}, & 0\leq t\leq s\leq 1, \end{cases}\displaystyle \end{aligned}$$

and

$$ G_{2}(t,s)=\frac{1}{A\Gamma (\alpha )} \textstyle\begin{cases} t^{\alpha -1}\sum_{0\leq s\leq \eta _{i}}\xi _{i}[ \eta ^{\alpha -\gamma -1}_{i}(1-s)^{\alpha - \gamma -1}-(\eta _{i}-s)^{ \alpha - \gamma -1}],& 0\leq t\leq 1, \\ t^{\alpha -1}\sum_{\eta _{i}\leq s \leq 1}\xi _{i}\eta ^{\alpha -\gamma -1}_{i}(1-s)^{\alpha - \gamma -1}, & 0\leq t\leq 1, \end{cases} $$

with

$$ A=1-\sum_{i=1}^{m-2}\xi _{i} \eta _{i}^{\alpha - \gamma -1}. $$

Proof

Using Lemma 2.1, we get

$$ u(t)+c_{1}t^{\alpha -1}+c_{2}t^{\alpha -2}+ \cdots +c_{n}t^{\alpha -n}=-I^{ \alpha }_{0^{+}}y(t). $$

It follows from the condition $u(0)=u'(0) =\cdots=u^{(n-2)}(0)=0$ that $c_{n}=c_{n-1}=\cdots =c_{2}=0$. Thus

$$ u(t)=-I^{\alpha }_{0^{+}}y(t)-c_{1}t^{\alpha -1}. $$

The rest of our proof can be obtained from Lemma 2.1 in [1]. □

Lemma 2.3

Let $\sum_{i=1}^{m-2}\zeta _{i}\eta _{i}^{\beta -1} \neq 1$. If $f:[0,1]\times (-\infty ,+\infty )\times (-\infty ,+\infty )\to (- \infty ,+\infty )$ be a continuous function. Then the problem (1.1) has the following unique solution:

$$ u(t)= \int ^{1}_{0} G(t,s) \biggl( \int ^{1}_{0} H(s,\tau ) \bigl(f \bigl(\tau ,u( \tau ),{D}^{\nu }_{0^{+}} u(\tau ) \bigr) + g \bigl( \tau ,u( \tau ),(\mathcal{K} u) (\tau ) \bigr)-b \bigr)\,d\tau \biggr)\,ds, $$

where

$$ H(t,s)=H_{1}(t,s)+H_{2}(t,s), $$

in which

$$\begin{aligned}& H_{1}(t,s)=\frac{1}{\Gamma (\beta )} \textstyle\begin{cases} t^{\beta -1}(1-s)^{\beta -1}-(t-s)^{\beta -1}, & 0\leq s\leq t \leq 1, \\ t^{\beta -1}(1-s)^{\beta -1}, & 0\leq t\leq s\leq 1, \end{cases}\displaystyle \\& H_{2}(t,s)=\frac{1}{B\Gamma (\beta )} \textstyle\begin{cases} t^{\beta -1}\sum_{0\leq s\leq \eta _{i}}\zeta _{i}[ \eta ^{\beta -1}_{i}(1-s)^{\beta -1}-(\eta _{i}-s)^{\beta -1}], & 0 \leq t\leq 1, \\ t^{\beta -1}\sum_{\eta _{i}\leq s \leq 1}\zeta _{i}\eta ^{\beta -1}_{i}(1-s)^{\beta -1}, & 0\leq t\leq 1, \end{cases}\displaystyle \end{aligned}$$

where

$$ B=1-\sum_{i=1}^{m-2}\zeta _{i}\eta _{i}^{\beta -1}. $$

Proof

Let $h\in C[0,1]$, consider the boundary value problem:

$$ \textstyle\begin{cases} D_{0^{+}}^{\beta }v(t)+h(t)=0, \quad 0< t< 1, n-1< \beta \leq n, \\ v(0)=v'(0)=\cdots =v^{(n-2)}(0)=0, \qquad v(1)=\sum _{i=1}^{m-2} \zeta _{i}v(\eta _{i}). \end{cases} $$

Similarly, using Lemma 2.1, we deduce

$$ v(t)+c_{1}t^{\beta -1}+c_{2}t^{\beta -2}+ \cdots +c_{n}t^{\beta -n}=-I^{ \beta }_{0^{+}}h(t). $$

It follows from the condition $v(0)=v'(0) =\cdots=v^{(n-2)}(0)=0$ that $c_{n}=c_{n-1}=\cdots =c_{2}=0$. Thus

$$ v(t)=-I^{\beta }_{0^{+}}h(t)-c_{1}t^{\beta -1}. $$

The rest of our proof can be derived from Lemma 2.4 in [3]. □

Lemma 2.4

Let

$$\begin{aligned}& C(s)= \frac{1}{A}\sum_{0\leq s\leq \eta _{i}} \xi _{i} \bigl[\eta ^{\alpha -\gamma -1}_{i}(1-s)^{\alpha - \gamma -1}-( \eta _{i}-s)^{\alpha - \gamma -1} \bigr]+\sum _{ s\geq \eta _{i}} \xi _{i}\eta ^{\alpha -\gamma -1}_{i}(1-s)^{\alpha - \gamma -1}, \\& D= \frac{1}{A} \Biggl(1+\sum _{i=1}^{m-2}\xi _{i} \bigl(1- \eta _{i}^{\alpha - \gamma -1} \bigr) \Biggr). \end{aligned}$$

Then the function $G(t,s)$ defined in Lemma 2.2satisfies the following conditions:

$$ C(s)t^{\alpha -1}\leq \Gamma (\alpha )G(t,s)\leq D t^{\alpha -1}, $$

and

$$ C(s)t^{\alpha -\nu -1}\leq \Gamma (\alpha -\nu )D^{\nu }_{0^{+}}G(t,s) \leq D t^{\alpha -\nu -1}, $$

for every $t,s\in [0,1]$.

Proof

Since $G_{1}(t,s)\geq 0$ for $t\in [0,1]$, $s\in [0,1]$, it follows that

$$ G(t,s)\geq G_{2}(t,s)=\frac{C(s)}{\Gamma (\alpha )}t^{\alpha -1}. $$

At the same time, we have

$$ \begin{aligned} G(t,s)&=G_{1}(t,s)+G_{2}(t,s) \\ &\leq \frac{1}{\Gamma (\alpha )}t^{\alpha -1}+ \frac{1}{A\Gamma (\alpha )}\sum _{i=1}^{m-2}\xi _{i}t^{ \alpha -1} \\ &= \frac{1}{A\Gamma (\alpha )} \Biggl(1+\sum _{i=1}^{m-2}\xi _{i} \bigl(1-\eta ^{\alpha -\gamma -1}_{i} \bigr) \Biggr)t^{ \alpha -1} \\ &= \frac{D}{\Gamma (\alpha )}t^{\alpha -1}. \end{aligned} $$

Consequently

$$ C(s)t^{\alpha -1}\leq \Gamma (\alpha )G(t,s)\leq D t^{\alpha -1}. $$

Similarly

$$ {D}^{\nu }_{0^{+}}G(t,s)\geq {D}^{\nu }_{0^{+}}G_{2}(t,s)= \frac{C(s)}{\Gamma (\alpha -\nu )}t^{\alpha -\nu -1} $$

and

$$\begin{aligned}& \begin{aligned} {D}^{\nu }_{0^{+}}G(t,s)&= {D}^{\nu }_{0^{+}}G_{1}(t,s)+{D}^{ \nu }_{0^{+}}G_{2}(t,s) \\ &\leq \frac{1}{\Gamma (\alpha -\nu )}t^{ \alpha -\nu -1}+ \frac{1}{A\Gamma (\alpha -\nu )}\sum _{i=1}^{m-2} \xi _{i}t^{\alpha -\nu -1} \\ &= \frac{1}{A\Gamma (\alpha -\nu )} \Biggl(1+ \sum _{i=1}^{m-2}\xi _{i} \bigl(1-\eta ^{\alpha -\gamma -1}_{i} \bigr) \Biggr)t^{\alpha -\nu -1} \\ &= \frac{D}{\Gamma (\alpha -\nu )}t^{\alpha -\nu -1}. \end{aligned} \end{aligned}$$

Hence

$$ C(s)t^{\alpha -\nu -1}\leq \Gamma (\alpha -\nu )D^{\nu }_{0^{+}}G(t,s) \leq D t^{\alpha -\nu -1}. $$

 □

Lemma 2.5

([4])

Let

$$ E(s)= \frac{1}{B}\sum_{0\leq s\leq \eta _{i}} \zeta _{i} \bigl[\eta ^{\beta -1}_{i}(1-s)^{\beta -1}-( \eta _{i}-s)^{\beta -1} \bigr]+ \sum _{ s\geq \eta _{i}}\zeta _{i}\eta ^{\beta -1}_{i}(1-s)^{ \beta -1} $$

and

$$ F= \frac{1}{B} \Biggl(1+\sum _{i=1}^{m-2}\zeta _{i} \bigl(1- \eta _{i}^{\beta -1} \bigr) \Biggr). $$

Then

$$ E(s)t^{\beta -1}\leq \Gamma (\beta )H(t,s)\leq F t^{\beta -1}, $$

for every $t,s\in [0,1]$.

Lemma 2.6

Let P be a normal cone and $T: P_{h,e}\times P_{h,e}\longrightarrow E$ be a mixed monotone operator. Assume that the following conditions hold:

• (i) for every $\lambda \in (0, 1)$ and $u,v\in P_{h,e}$, there exists $\varphi (\lambda ,u,v)>\lambda$ such that

  $$ T \bigl(\lambda u+(\lambda -1)e,\lambda ^{-1}v+ \bigl(\lambda ^{-1}-1 \bigr)e \bigr)\geq \varphi (\lambda ,u,v)T(u,v)+ \bigl( \varphi (\lambda ,u,v)-1 \bigr)e; $$

• (ii) for fixed $t\in (0, 1)$ and $u\in P_{h,e}$, $\varphi (t,u,v)$ is decreasing in v, and for fixed $t\in (0, 1)$ and $v\in P_{h,e}$, $\varphi (t,u,v)$ is increasing in u;

• (iii) there exists $t_{0}\in (0,1)$ such that

  $$ \frac{t_{0}}{\varphi (t_{0},h,h)}h+ \biggl( \frac{t_{0}}{\varphi (t_{0},h,h)}-1 \biggr)e\leq T(h,h)\leq \frac{1}{t_{0}}h+ \biggl(\frac{1}{t_{0}}-1 \biggr)e. $$

Then:

• (1) T has a unique fixed point $x^{*}$ in $P_{h,e}$;

• (2) there exist initial values $u_{0},v_{0}\in P_{h,e}$, and $s\in (0,1)$ such that

  $$ sv_{0}\leq u_{0}< v_{0},\qquad u_{0}\leq T(u_{0},v_{0})\leq T(v_{0},u_{0}) \leq v_{0}; $$

• (3) for any $x_{0},y_{0}\in P_{h,e}$, taking the iterative sequences as follows:

  $$ x_{n}=T(x_{n-1},y_{n-1}),\qquad y_{n}=T(y_{n-1},x_{n-1}),\quad n=1,2,\ldots $$

  we have $x_{n}\rightarrow x^{*}$ and $y_{n}\rightarrow x^{*}$ as $n\rightarrow \infty$.

Proof

By (i), we have

$$ \begin{aligned} &T \bigl(\lambda ^{-1}u+ \bigl(\lambda ^{-1}-1 \bigr)e, \lambda v+(\lambda -1)e \bigr) \\ &\quad \leq \bigl[\varphi \bigl(\lambda ,\lambda ^{-1}u + \bigl(\lambda ^{-1}-1 \bigr)e, \lambda v+( \lambda -1)e \bigr) \bigr]^{-1}T(u,v) \\ &\qquad {}+ \bigl( \bigl[\varphi \bigl(\lambda ,\lambda ^{-1}u + \bigl( \lambda ^{-1}-1 \bigr)e, \lambda v+( \lambda -1)e \bigr) \bigr]^{-1} -1 \bigr)e, \end{aligned} $$

(2.1)

for every $\lambda \in (0, 1)$, $u, v\in P_{h,e}$. We can find a positive integer k with

$$ \biggl(\frac{\varphi (t_{0},h,h)}{t_{0}} \biggr)^{k} \geq \frac{1}{t_{0}} . $$

Let

$$\begin{aligned}& u_{n}=T(u_{n-1},v_{n-1}),\qquad v_{n}=T(v_{n-1},u_{n-1}), \quad n=1,2,\ldots . \\& x_{n}=t_{0}^{n}h+ \bigl(t_{0}^{n}-1 \bigr)e,\qquad y_{n}=t_{0}^{-n}h+ \bigl(t_{0}^{-n}-1 \bigr)e, \quad n=1,2,\ldots . \end{aligned}$$

Thus

$$ x_{n}=t_{0}x_{n-1}+(t_{0}-1)e , \qquad y_{n}=t_{0}^{-1}y_{n-1}+\bigl(t_{0}^{-1}-1\bigr)e , \quad n=1,2,\ldots . $$

Denote $u_{0}:=x_{k}$, $v_{0}:=y_{k}$, then $u_{0},v_{0}\in P_{h,e}$,

$$ u_{0}\leq v_{0} , \qquad u_{1}=T(u_{0},v_{0})\leq T(v_{0},u_{0})=v_{1} . $$

Since T is mixed monotone, we get

$$ u_{n}\leq v_{n}, \quad n=1,2,\ldots . $$

By the conditions (ii) and (iii), combining with (2.1), we have

$$\begin{aligned}& \begin{aligned} u_{1}&=T(u_{0},v_{0}) \\ &=T \bigl(t_{0}^{k}h+ \bigl(t_{0}^{k}-1 \bigr)e,t_{0}^{-k}h+ \bigl(t_{0}^{-k}-1 \bigr)e \bigr) \\ &=T \bigl(t_{0} \bigl(t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e \bigr)+(t_{0}-1)e,t_{0}^{-1} \bigl(t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e \bigr)+ \bigl(t_{0}^{-1}-1 \bigr)e \bigr) \\ &\geq \varphi \bigl(t_{0},t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e,t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e \bigr)T \bigl(t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e,t_{0}^{-k+1}h \\ &\quad {}+ \bigl(t_{0}^{-k+1}-1 \bigr)e \bigr)+ \bigl(\varphi \bigl(t_{0},t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e,t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e \bigr)-1 \bigr)e \\ &\geq \varphi \bigl(t_{0},t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e,t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e \bigr) \bigl[ \varphi \bigl(t_{0},t_{0}^{k-2}h+ \bigl(t_{0}^{k-2}-1 \bigr)e, \\ &\quad t_{0}^{-k+2}h+ \bigl(t_{0}^{-k+2}-1 \bigr)e \bigr)T \bigl(t_{0}^{k-2}h+ \bigl(t_{0}^{k-2}-1 \bigr)e,t_{0}^{-k+2}h+ \bigl(t_{0}^{-k+2}-1 \bigr)e \bigr) \\ &\quad {} + \bigl(\varphi \bigl(t_{0},t_{0}^{k-2}h+ \bigl(t_{0}^{k-2}-1 \bigr)e, t_{0}^{-k+2}h+ \bigl(t_{0}^{-k+2}-1 \bigr)e \bigr)-1 \bigr)e \bigr] \\ &\quad {} + \bigl(\varphi \bigl(t_{0},t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e,t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e \bigr)-1 \bigr)e \\ & =\varphi \bigl(t_{0},t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e,t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e \bigr) \varphi \bigl(t_{0},t_{0}^{k-2}h+ \bigl(t_{0}^{k-2}-1 \bigr)e,t_{0}^{-k+2}h \\ &\quad {}+ \bigl(t_{0}^{-k+2}-1 \bigr)e \bigr)T \bigl(t_{0}^{k-2}h+ \bigl(t_{0}^{k-2}-1 \bigr)e,t_{0}^{-k+2}h+ \bigl(t_{0}^{-k+2}-1 \bigr)e \bigr)+ \bigl[ \varphi \bigl(t_{0},t_{0}^{k-1}h \\ &\quad {}+ \bigl(t_{0}^{k-1}-1 \bigr)e, t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e \bigr)\varphi \bigl(t_{0},t_{0}^{k-2}h+ \bigl(t_{0}^{k-2}-1 \bigr)e,t_{0}^{-k+2}h \\ &\quad {}+ \bigl(t_{0}^{-k+2}-1 \bigr)e \bigr)-1 \bigr]e \\ & \geq \cdots \geq \varphi \bigl(t_{0},t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e,t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e \bigr) \varphi \bigl(t_{0},t_{0}^{k-2}h+ \bigl(t_{0}^{k-2}-1 \bigr)e, \\ &\quad t_{0}^{-k+2}h+ \bigl(t_{0}^{-k+2}-1 \bigr)e \bigr)\ldots \varphi (t_{0},h,h)T(h,h)+ \bigl[ \varphi \bigl(t_{0},t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e,t_{0}^{-k+1}h \\ &\quad {}+ \bigl(t_{0}^{-k+1}-1 \bigr)e \bigr)\varphi \bigl(t_{0}, t_{0}^{k-2}h+ \bigl(t_{0}^{k-2}-1 \bigr)e,t_{0}^{-k+2}h+ \bigl(t_{0}^{-k+2}-1 \bigr)e \bigr) \cdots \varphi (t_{0},h,h)-1 \bigr]e \\ & \geq t_{0}^{k-1}\varphi (t_{0},h,h)T(h,h)+ \bigl(t_{0}^{k-1}\varphi (t_{0},h,h)-1 \bigr)e \\ &\geq t_{0}^{k-1}\varphi (t_{0},h,h) \biggl( \frac{t_{0}}{\varphi (t_{0},h,h)}h+ \biggl( \frac{t_{0}}{\varphi (t_{0},h,h)}-1 \biggr)e \biggr) + \bigl(t_{0}^{k-1} \varphi (t_{0},h,h)-1 \bigr)e \\ & = t_{0}^{k}h+ \bigl(t_{0}^{k}-t_{0}^{k-1} \varphi (t_{0},h,h) \bigr)e+ \bigl(t_{0}^{k-1} \varphi (t_{0},h,h)-1 \bigr)e \\ & = t_{0}^{k}h+ \bigl(t_{0}^{k}-1 \bigr)e=x_{k}=u_{0} \end{aligned} \end{aligned}$$

and

$$\begin{aligned}& v_{1} =T(v_{0},u_{0}) \\& \hphantom{v_{1}} =T \bigl(t_{0}^{-k}h+ \bigl(t_{0}^{-k}-1 \bigr)e,t_{0}^{k}h+ \bigl(t_{0}^{k}-1 \bigr)e \bigr) \\& \hphantom{v_{1}} \leq \varphi \bigl(t_{0},t_{0}^{-k}h+ \bigl(t_{0}^{-k}-1 \bigr)e,t_{0}^{k}h+ \bigl(t_{0}^{k}-1 \bigr)e \bigr)^{-1}T \bigl(t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e,t_{0}^{k-1}h \\& \hphantom{v_{1}={}}{} + \bigl(t_{0}^{k-1}-1 \bigr)e \bigr)+ \bigl(\varphi \bigl(t_{0},t_{0}^{-k}h+ \bigl(t_{0}^{-k}-1 \bigr)e,t_{0}^{k}h+ \bigl(t_{0}^{k}-1 \bigr)e \bigr)^{-1}-1 \bigr)e \\& \hphantom{v_{1}} \leq \varphi \bigl(t_{0},t_{0}^{-k}h+ \bigl(t_{0}^{-k}-1 \bigr)e,t_{0}^{k}h+ \bigl(t_{0}^{k}-1 \bigr)e \bigr)^{-1} \varphi \bigl(t_{0},t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e,t_{0}^{k-1}h \\& \hphantom{v_{1}={}}{} + \bigl(t_{0}^{k-1}-1 \bigr)e \bigr)^{-1}T \bigl(t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e,t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e \bigr)+ \bigl[ \varphi \bigl(t_{0},t_{0}^{-k}h \\& \hphantom{v_{1}={}}{} + \bigl(t_{0}^{-k}-1 \bigr)e, t_{0}^{k}h+ \bigl(t_{0}^{k}-1 \bigr)e \bigr)^{-1} \varphi \bigl(t_{0},t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e,t_{0}^{k-1}h \\& \hphantom{v_{1}={}}{} + \bigl(t_{0}^{k-1}-1 \bigr)e \bigr)^{-1}-1 \bigr]e \\& \hphantom{v_{1}} \leq \cdots \leq \varphi \bigl(t_{0},t_{0}^{-k}h+ \bigl(t_{0}^{-k}-1 \bigr)e,t_{0}^{k}h+ \bigl(t_{0}^{k}-1 \bigr)e \bigr)^{-1} \varphi \bigl(t_{0},t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e, \\& \hphantom{v_{1}={}}{} t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e \bigr)^{-1}\cdots \varphi \bigl(t_{0}, t_{0} ^{-1} h+ \bigl(t_{0}^{-1}-1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr)^{-1}T(h,h) \\& \hphantom{v_{1}={}}{} + \bigl[\varphi \bigl(t_{0},t_{0}^{-k}h+ \bigl(t_{0}^{-k}-1 \bigr)e,t_{0}^{k}h+ \bigl(t_{0}^{k}-1 \bigr)e \bigr)^{-1} \varphi \bigl(t_{0},t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e,t_{0}^{k-1}h \\& \hphantom{v_{1}={}}{} + \bigl(t_{0}^{k-1}-1 \bigr)e \bigr)^{-1} \cdots \varphi \bigl(t_{0}, t_{0} ^{-1} h+ \bigl(t_{0}^{-1}-1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr)^{-1}-1 \bigr]e \\& \hphantom{v_{1}} \leq \varphi \bigl(t_{0}, t_{0} ^{-1} h+ \bigl(t_{0}^{-1}-1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr)^{-k}T(h,h) \\& \hphantom{v_{1}={}}{} + \bigl[\varphi \bigl(t_{0}, t_{0} ^{-1} h+ \bigl(t_{0}^{-1}-1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr)^{-k} -1 \bigr]e \\& \hphantom{v_{1}} \leq \varphi \bigl(t_{0}, t_{0}^{-1} h, t_{0} h \bigr)^{-k} T(h,h)+ \bigl[\varphi \bigl(t_{0}, t_{0}^{-1} h, t_{0} h \bigr)^{-k} -1 \bigr]e \\& \hphantom{v_{1}} \leq \varphi (t_{0}, h, h)^{-k} T(h,h)+ \bigl[\varphi (t_{0}, h, h)^{-k} -1 \bigr]e \\& \hphantom{v_{1}} \leq \varphi (t_{0}, h, h)^{-k} \bigl[t_{0}^{-1} h+ \bigl(t_{0}^{-1}-1 \bigr)e \bigr]+ \bigl[ \varphi (t_{0}, h, h)^{-k} -1 \bigr]e \\& \hphantom{v_{1}} =\varphi (t_{0}, h, h)^{-k} t_{0}^{-1} h +\varphi (t_{0}, h, h)^{-k} t_{0}^{-1}e -e \\& \hphantom{v_{1}} \leq t_{0}^{-k}h+ \bigl(t_{0}^{-k}-1 \bigr)e=y_{k}=v_{0}. \end{aligned}$$

Thus

$$ u_{0}\leq T(u_{0},v_{0})\leq T(v_{0},u_{0})\leq v_{0}. $$

We deduce for all $n\in \mathbb{N}$ that

$$ u_{0}\leq u_{1}\leq \cdots \leq u_{n}\leq \cdots \leq v_{n}\leq \cdots \leq v_{1}\leq v_{0}. $$

(2.2)

In addition

$$ u_{n}\geq u_{0}\geq sv_{0}+(s-1)e\geq sv_{n}+(s-1)e , \quad n=1,2, \ldots . $$

Let

$$ t_{n}=\sup \bigl\{ t>0\mid u_{n}\geq tv_{n}+(t-1)e\bigr\} . $$

Thus we have $u_{n}\geq t_{n}v_{n}+(t_{n}-1)e$, $n=1,2,\ldots$ . Consequently $\{t_{n}\}$ is increasing with $\{t_{n}\}\subset (0,1]$. Assume that $t_{n}\rightarrow t^{*}$ as $n\rightarrow \infty$, then $t^{*}=1$. If not, $0< t^{*}<1$.

Next, we need to prove that $t^{*}=1$. If $0< t^{*}<1$, we should discuss the following two cases.

Case 1: there is an integer N such that $t_{N}=t^{*}$. In this case, we have $t_{n}=t^{*}$ for all $n>N$. Then

$$ \begin{aligned} u_{n+1}&=T(u_{n},v_{n}) \geq T \bigl(t_{n}v_{n}+(t_{n}-1)e,t_{n}^{-1}u_{n}+ \bigl(t_{n}^{-1}-1 \bigr)e \bigr) \\ &=T \bigl(t^{*}v_{n}+ \bigl(t^{*}-1 \bigr)e, \bigl(t^{*} \bigr)^{-1}u_{n}+ \bigl( \bigl(t^{*} \bigr)^{-1}-1 \bigr)e \bigr) \\ &\geq \varphi \bigl(t^{*},v_{n},u_{n} \bigr)T(v_{n},u_{n})+ \bigl(\varphi \bigl(t^{*},v_{n},u_{n} \bigr)-1 \bigr)e \\ &\geq \varphi \bigl(t^{*},u_{0},v_{0} \bigr)T(u_{0},v_{0})+ \bigl(\varphi \bigl(t^{*},u_{0},v_{0} \bigr)-1 \bigr)e. \end{aligned} $$

We can get $t^{*}=t_{n+1}\geq \varphi (t^{*})>t^{*}$ from the definition of $t_{n+1}$, which is a contradiction.

Case 2: for all n, $t_{n}< t^{*}$, we have

$$\begin{aligned}& \begin{aligned} u_{n+1}&=T(u_{n},v_{n}) \geq T \bigl(t_{n}v_{n}+(t_{n}-1)e,t_{n}^{-1}u_{n}+ \bigl(t_{n}^{-1}-1 \bigr)e \bigr) \\ &= T \biggl(\frac{t_{n}}{t^{*}} \bigl(t^{*}v_{n}+ \bigl(t^{*}-1 \bigr)e \bigr)+ \biggl(\frac{t_{n}}{t^{*}}-1 \biggr)e, \biggl(\frac{t_{n}}{t^{*}} \biggr)^{-1} \bigl( \bigl(t^{*} \bigr)^{-1}u_{n}+ \bigl( \bigl(t^{*} \bigr)^{-1}-1 \bigr)e \bigr) \\ &\quad {}+ \biggl( \biggl(\frac{t_{n}}{t^{*}} \biggr)^{-1}-1 \biggr)e \biggr) \\ &\geq \varphi \biggl(\frac{t_{n}}{t^{*}},t^{*}v_{n}+ \bigl(t^{*}-1 \bigr)e, \bigl(t^{*} \bigr)^{-1}u_{n}+ \bigl( \bigl(t^{*} \bigr)^{-1}-1 \bigr)e \biggr)T \bigl(t^{*}v_{n}+ \bigl(t^{*}-1 \bigr)e, \bigl(t^{*} \bigr)^{-1}u_{n} \\ &\quad {} + \bigl( \bigl(t^{*} \bigr)^{-1}-1 \bigr)e \bigr)+ \biggl(\varphi \biggl( \frac{t_{n}}{t^{*}},t^{*}v_{n}+ \bigl(t^{*}-1 \bigr)e, \bigl(t^{*} \bigr)^{-1}u_{n}+ \bigl( \bigl(t^{*} \bigr)^{-1}-1 \bigr)e \biggr)-1 \biggr)e \\ &\geq \varphi \biggl(\frac{t_{n}}{t^{*}},t^{*}v_{n}+ \bigl(t^{*}-1 \bigr)e, \bigl(t^{*} \bigr)^{-1}u_{n}+ \bigl( \bigl(t^{*} \bigr)^{-1}-1 \bigr)e \biggr) \bigl[ \varphi \bigl(t^{*},v_{n},u_{n} \bigr)T(v_{n},u_{n}) \\ &\quad {} + \bigl(\varphi \bigl(t^{*},v_{n},u_{n} \bigr)-1 \bigr)e \bigr] \\ &\quad {}+ \biggl( \varphi \biggl(\frac{t_{n}}{t^{*}},t^{*}v_{n}+ \bigl(t^{*}-1 \bigr)e, \bigl(t^{*} \bigr)^{-1}u_{n}+ \bigl( \bigl(t^{*} \bigr)^{-1}-1 \bigr)e \biggr)-1 \biggr)e \\ &\geq \varphi \biggl(\frac{t_{n}}{t^{*}},t^{*}u_{0}+ \bigl(t^{*}-1 \bigr)e, \bigl(t^{*} \bigr)^{-1}v_{0}+ \bigl( \bigl(t^{*} \bigr)^{-1}-1 \bigr)e \biggr)\varphi \bigl(t^{*},u_{0},v_{0} \bigr)T(v_{n},u_{n}) \\ &\quad {}+ \biggl(\varphi \biggl(\frac{t_{n}}{t^{*}},t^{*}u_{0}+ \bigl(t^{*}-1 \bigr)e, \bigl(t^{*} \bigr)^{-1}v_{0}+ \bigl( \bigl(t^{*} \bigr)^{-1}-1 \bigr)e \biggr)\varphi \bigl(t^{*},u_{0},v_{0} \bigr)-1 \biggr)e. \end{aligned} \end{aligned}$$

By the definition of $t_{n+1}$, we have

$$ t_{n+1}\geq \varphi \biggl(\frac{t_{n}}{t^{*}},t^{*}u_{0}+\bigl(t^{*}-1\bigr)e,\bigl(t^{*}\bigr)^{-1}v_{0}+ \bigl(\bigl(t^{*}\bigr)^{-1}-1\bigr)e \biggr)\varphi \bigl(t^{*},u_{0},v_{0}\bigr) \geq \frac{t_{n}}{t^{*}}\varphi \bigl(t^{*},u_{0},v_{0}\bigr) . $$

Let $n\rightarrow \infty$, we have $t^{*}\geq \varphi (t^{*},u_{0},v_{0})>t^{*}$, which is a contradiction. Consequently $t^{*}=1$. Since P is normal, we have

$$ \|u_{n+p}-u_{n}\|\leq M(1-t_{n}) \Vert v_{0}+e \Vert , \qquad \Vert v_{n}-v_{n+p} \Vert \leq M(1-t_{n})\|v_{0}+e\|, $$

where M is the normality constant. Let $n\to \infty$, we get

$$ \|u_{n+p}-u_{n}\| \longrightarrow 0,\qquad \|v_{n}-v_{n+p}\| \longrightarrow 0. $$

Therefore ${u_{n}}$ and ${v_{n}}$ are Cauchy sequences. Repeating the proof of Lemma 2.3 in Sang and Ren [21], we derive that our conclusions hold. □

Lemma 2.7

Let P be a normal cone and $T: P_{h,e}\times P_{h,e}\longrightarrow E$ be a mixed monotone operator. Assume that the condition (i) in Lemma 2.6is satisfied. In addition, $\varphi (t,u,v)$ is decreasing in u and increasing in v for every $t\in (0, 1)$. Furthermore, there exists $t_{0}\in (0, 1)$ such that

$$ \begin{aligned} t_{0} h +(t_{0} -1) e&\leq T(h,h) \\ &\leq \frac{1}{t_{0}}\varphi \bigl(t_{0}, t_{0}^{-1}h + \bigl(t_{0}^{-1}-1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr)h \\ &\quad {}+ \biggl[\frac{1}{t_{0}}\varphi \bigl(t_{0}, t_{0}^{-1}h + \bigl(t_{0}^{-1}-1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr) -1 \biggr]e. \end{aligned} $$

(2.3)

Then the conclusions (1), (2), (3) in Lemma 2.6hold.

Proof

As in the proof of Lemma 2.6, we only need to check that $u_{1} =T(u_{0}, v_{0})\geq u_{0}$ and $v_{1} =T(v_{0}, u_{0})\leq v_{0}$ hold. For every $t\in (0,1)$, since $\varphi (t,u,v)$ is decreasing in u and increasing in v, by (2.3), we have

$$ \begin{aligned} u_{1}&=T(u_{0},v_{0}) \\ &=T \bigl(t_{0}^{k}h+ \bigl(t_{0}^{k}-1 \bigr)e,t_{0}^{-k}h+ \bigl(t_{0}^{-k}-1 \bigr)e \bigr) \\ &\geq \bigl[\varphi \bigl(t_{0},t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e,t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e \bigr) \varphi \bigl(t_{0},t_{0}^{k-2}h+ \bigl(t_{0}^{k-2}-1 \bigr)e,t_{0}^{-k+2}h \\ &\quad {}+ \bigl(t_{0}^{-k+2}-1 \bigr)e \bigr)\cdots \varphi (t_{0}, h, h) \bigr]T(h,h)+ \bigl[ \varphi \bigl(t_{0},t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e,t_{0}^{-k+1}h \\ &\quad {}+ \bigl(t_{0}^{-k+1}-1 \bigr)e \bigr)\varphi \bigl(t_{0}, t_{0}^{k-2}h+ \bigl(t_{0}^{k-2}-1 \bigr)e,t_{0}^{-k+2}h+ \bigl(t_{0}^{-k+2}-1 \bigr)e \bigr) \cdots \varphi (t_{0}, h, h)-1 \bigr]e \\ &\geq \bigl[\varphi (t_{0}, h, h) \bigr]^{k} T(h, h)+ \bigl( \bigl[\varphi (t_{0}, h, h) \bigr]^{k} -1 \bigr)e \\ &\geq \bigl[\varphi (t_{0}, h, h) \bigr]^{k} \bigl[t_{0} h +(t_{0} -1)e \bigr]+ \bigl( \bigl[\varphi (t_{0}, h, h) \bigr]^{k} -1 \bigr)e \\ &\geq t_{0}^{k} h+ \bigl(t_{0}^{k}-1 \bigr)e=x_{k}=u_{0} \end{aligned} $$

and

$$ \begin{aligned} v_{1}&=T(v_{0},u_{0}) \\ &=T \bigl(t_{0}^{-k}h+ \bigl(t_{0}^{-k}-1 \bigr)e,t_{0}^{k}h+ \bigl(t_{0}^{k}-1 \bigr)e \bigr) \\ &\leq \bigl[\varphi \bigl(t_{0},t_{0}^{-k}h+ \bigl(t_{0}^{-k}-1 \bigr)e,t_{0}^{k}h+ \bigl(t_{0}^{k}-1 \bigr)e \bigr)^{-1} \varphi \bigl(t_{0},t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e, \\ &\quad t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e \bigr)^{-1}\cdots \varphi \bigl(t_{0}, t_{0}^{-1} h+ \bigl(t_{0}^{-1} -1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr)^{-1} \bigr]T(h,h) \\ &\quad {}+ \bigl[\varphi \bigl(t_{0},t_{0}^{-k}h+ \bigl(t_{0}^{-k}-1 \bigr)e, t_{0}^{k}h+ \bigl(t_{0}^{k}-1 \bigr)e \bigr)^{-1} \varphi \bigl(t_{0},t_{0}^{-k+1}h+ \bigl(t_{0}^{-k+1}-1 \bigr)e, \\ &\quad t_{0}^{k-1}h+ \bigl(t_{0}^{k-1}-1 \bigr)e \bigr)^{-1}\cdots \varphi \bigl(t_{0}, t_{0}^{-1} h+ \bigl(t_{0}^{-1} -1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr)^{-1}-1 \bigr]e \\ &\leq t_{0}^{-k+1}\varphi \bigl(t_{0}, t_{0}^{-1} h+ \bigl(t_{0}^{-1} -1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr)^{-1} T(h,h) \\ &\quad {} + \bigl[t_{0}^{-k+1}\varphi \bigl(t_{0}, t_{0}^{-1} h+ \bigl(t_{0}^{-1} -1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr)^{-1}-1 \bigr]e \\ &\leq t_{0}^{-k}h+ \bigl(t_{0}^{-k}-1 \bigr)e=y_{k}=v_{0}.\end{aligned} $$

The rest of the proof is similar to that of Lemma 2.6, we omit it here. □

3 Main results

In this section, we will establish the existence and uniqueness of nontrivial solution for the problem (1.1). The main tools are fixed point theorems of an operator equation.

Theorem 3.1

Let P be a normal cone in E, and let M, $N: P_{h,e}\times P_{h,e}\longrightarrow E$ be two mixed monotone operators, and the following conditions are satisfied:

• (L1) for all $t \in (0, 1)$ and $x,y\in P_{h,e}$, there exists $\psi (t,x,y)>t$ such that

  $$ M \bigl(tx+(t-1)e,t^{-1}y+ \bigl(t^{-1}-1 \bigr)e \bigr) \geq \psi (t,x,y)M(x,y)+ \bigl(\psi (t,x,y)-1 \bigr)e; $$

• (L2) for fixed $t\in (0, 1)$ and $x\in P_{h,e}$, $\psi (t,x,y)$ is decreasing in y, and for fixed $t\in (0, 1)$ and $y\in P_{h,e}$, $\psi (t,x,y)$ is increasing in x;

• (L3) for all $t \in (0, 1)$ and $x,y\in P_{h,e}$,

  $$ N \bigl(tx+(t-1)e,t^{-1}y+ \bigl(t^{-1}-1 \bigr)e \bigr) \geq tN(x,y)+(t-1)e; $$

• (L4) there exists $t_{0}\in (0, 1)$ such that

  $$\begin{aligned}& \frac{t_{0}}{\psi (t_{0},h,h)}h+ \biggl(\frac{t_{0}}{\psi (t_{0},h,h)}-1 \biggr)e\leq M(h,h)\leq \frac{1}{t_{0}}h+ \biggl(\frac{1}{t_{0}}-1 \biggr)e, \\& N(h,h)\in P_{h,e}; \end{aligned}$$

• (L5) for all $x,y\in P_{h,e}$, there exists a constant $\delta >0$ such that

  $$ M(x,y)\geq \delta N(x,y)+(\delta -1)e. $$

Then the operator equation $M(x,x)+N(x,x)+e=x$ has a unique solution $x^{*}$ in $P_{h,e}$, and for any initial values $x_{0}, y_{0}\in P_{h,e}$, by setting the sequences $\{x_{n}\}$ and $\{y_{n}\}$ as follows:

$$\begin{aligned}& x_{n}=M(x_{n-1},y_{n-1})+N(x_{n-1},y_{n-1})+e,\quad n=1,2,\ldots , \\& y_{n}=M(y_{n-1},x_{n-1})+N(y_{n-1},x_{n-1})+e,\quad n=1,2,\ldots , \end{aligned}$$

we have $x_{n}\to x^{*}$ and $y_{n}\to x^{*}$ in E as $n\to \infty$.

Proof

For every $x_{i},y_{i}\in P_{h,e}$ ($i=1,2$) with $x_{1}\geq x_{2}$, $y_{1}\leq y_{2}$, the mixed monotone properties of $M(x,y)$ and $N(x,y)$ lead to

$$ M( x_{1},y_{1})\geq M( x_{2},y_{2}),\qquad N( x_{1},y_{1})\geq N( x_{2},y_{2}). $$

Now we define the operator $T: P_{h,e}\times P_{h,e}\to E$ by

$$ T(x,y)=M(x,y)+N(x,y)+e, \quad \text{for all } x,y\in P_{h,e}. $$

(3.1)

We have

$$ T(x_{1},y_{1})=M( x_{1},y_{1})+N( x_{1},y_{1})+e \geq M( x_{2},y_{2})+N( x_{2},y_{2})+e=T( x_{2},y_{2}). $$

Thus, T is a mixed monotone operator. Note that $N(h,h)\in P_{h,e}$, there exist $a_{1},a_{2}\in P_{h,e}$ such that

$$ a_{1}h+(a_{1}-1)e\leq N(h,h) \leq a_{2}h+(a_{2}-1)e. $$

From (3.1), we have

$$ T(h,h)=M(h,h)+N(h,h)+e. $$

By the condition (L4), we obtain

$$\begin{aligned}& \begin{aligned} T(h,h)&\geq \frac{t_{0}}{\psi (t_{0},h,h)}h+ \biggl( \frac{t_{0}}{\psi (t_{0},h,h)}-1 \biggr)e+ a_{1}h+(a_{1}-1)e \\ &= \biggl(\frac{t_{0}}{\psi (t_{0},h,h)}+a_{1} \biggr)h+ \biggl(\frac{t_{0}}{\psi (t_{0},h,h)}+a_{1}-1 \biggr)e \end{aligned} \end{aligned}$$

and

$$ \begin{aligned} T(h,h)&\leq \frac{1}{t_{0}}h+ \biggl(\frac{1}{t_{0}}-1 \biggr)e+a_{2}h+(a_{2}-1)e \\ &= \biggl(\frac{1}{t_{0}}+a_{2} \biggr)h+ \biggl(\frac{1}{t_{0}}+a_{2}-1 \biggr)e. \end{aligned} $$

Hence, the condition (iii) in Lemma 2.6 is proved. Next, by the condition (L4), we have

$$\begin{aligned}& M(x,y)+\delta M(x,y)\geq \delta N(x,y)+(\delta -1)e+\delta M(x,y) , \\& M(x,y)\geq \frac{\delta }{1+\delta }T(x,y)-\frac{e}{1+\delta }. \end{aligned}$$

(3.2)

By conditions (L1), (L3), (3.1) and (3.2), for every $x,y\in P_{h,e}$, we obtain

$$ \begin{aligned} &T \bigl(tx+(t-1)e,t^{-1}y+ \bigl(t^{-1}-1 \bigr)e \bigr)-tT(x,y) \\ &\quad =M \bigl(tx+(t-1)e,t^{-1}y+ \bigl(t^{-1}-1 \bigr)e \bigr)+N \bigl(tx+(t-1)e,t^{-1}y+ \bigl(t^{-1}-1 \bigr)e \bigr)+e \\ &\qquad {}-t \bigl(M(x,y)+N(x,y)+e \bigr) \\ &\quad \geq \psi (t,x,y)M(x,y)+ \bigl(\psi (t,x,y)-1 \bigr)e+tN(x,y)+(t-1)e+e-tM(x,y) \\ &\qquad {}-t N(x,y)-te \\ &\quad = \bigl(\psi (t,x,y)-t \bigr)M(x,y)+ \bigl(\psi (t,x,y)-1 \bigr)e \\ &\quad \geq \bigl(\psi (t,x,y)-t \bigr) \biggl( \frac{\delta }{1+\delta }T(x,y)- \frac{e}{1+\delta } \biggr)+ \bigl(\psi (t,x,y)-1 \bigr)e \\ &\quad = \frac{\delta (\psi (t,x,y)-t )}{1+\delta }T(x,y)+ \biggl( \psi (t,x,y)-1- \frac{\psi (t,x,y)-t}{1+\delta } \biggr)e. \end{aligned} $$

Therefore

$$ \begin{aligned} &T \bigl(tx+(t-1)e,t^{-1}y+ \bigl(t^{-1}-1 \bigr)e \bigr) \\ &\quad \geq \biggl( \frac{\delta (\psi (t,x,y)-t )}{1+\delta }+t \biggr)T(x,y)+ \biggl(\psi (t,x,y)-1-\frac{\psi (t,x,y)-t}{1+\delta } \biggr)e \\ &\quad = \frac{\delta \psi (t,x,y)+t}{1+\delta }T(x,y)+ \biggl(\frac{\delta \psi (t,x,y)+t}{1+\delta }-1 \biggr)e. \end{aligned} $$

(3.3)

Let $\varphi (t,x,y)=\frac{\delta \psi (t,x,y)+t}{1+\delta }$. Then $\varphi (t,x,y)>t$, $t\in (0, 1)$, together with (3.3), we obtain

$$ T \bigl(tx+(t-1)e,t^{-1}y+ \bigl(t^{-1}-1 \bigr)e \bigr) \geq \varphi (t,x,y)T(x,y)+ \bigl( \varphi (t,x,y)-1 \bigr)e , \quad \forall x,y\in P_{h,e} . $$

Thus condition (i) in Lemma 2.6 is proved. By (L2), for every $x_{1}\geq x_{2}$ and $y_{1}\leq y_{2}$ with $x_{i}, y_{i}\in P_{h,e}$, $i=1,2$, we have

$$ \psi (t,x_{1},y_{1})\geq \psi (t,x_{2},y_{2}). $$

Therefore

$$ \varphi (t,x_{1},y_{1})= \frac{\delta \psi (t,x_{1},y_{1})+t}{1+\delta }\geq \frac{\delta \psi (t,x_{2},y_{2})+t}{1+\delta }=\varphi (t,x_{2},y_{2}). $$

Thus we deduce condition (ii) in Lemma 2.6 to be met. According to Lemma 2.6, we get the conclusions of Theorem 3.1. □

In terms of Lemma 2.7, we can establish the following theorem, which is parallel with Theorem 3.1.

Theorem 3.1′

Let P be a normal cone in E, and M, $N: P_{h,e}\times P_{h,e}\longrightarrow E$ be two mixed monotone operators. The assumptions (L1), (L3) and (L5) in Theorem 3.1are satisfied. Furthermore, for fixed $t\in (0, 1)$ and $x\in P_{h,e}$, $\psi (t,x,y)$ is increasing in y, and for fixed $t\in (0, 1)$ and $y\in P_{h,e}$, $\psi (t,x,y)$ is decreasing in x. In addition, $N(h,h)\in P_{h,e}$, and there exists $t_{0}\in (0,1)$ such that

$$ \begin{aligned} t_{0} h +(t_{0} -1) e&\leq M(h,h) \\ &\leq \frac{1}{t_{0}}\psi \bigl(t_{0}, t_{0}^{-1}h + \bigl(t_{0}^{-1}-1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr)h \\ &\quad {}+ \biggl[\frac{1}{t_{0}}\psi \bigl(t_{0}, t_{0}^{-1}h + \bigl(t_{0}^{-1}-1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr) -1 \biggr]e. \end{aligned} $$

Then the conclusions of Theorem 3.1hold.

Define $E=\{x\mid x\in C[0,1],D_{0^{+}}^{\nu }x\in C[0,1]\}$. Then E is a Banach space with an order relation $u\leq v$ if $u(t)\leq v(t)$, $D_{0^{+}}^{\nu }u(t)\leq D_{0^{+}}^{\nu }v(t)$. Let $P\subset E$ be defined by $P=\{x\in E\mid x(t)\geq 0,D_{0^{+}}^{\nu }x(t)\geq 0\}$ for all $t\in [0,1]$. It is clear that P is a normal cone. Let

$$ e(t)=b \int ^{1}_{0}G(t,s) \biggl( \frac{s^{\beta -1}-s^{\beta }}{\Gamma (\beta +1)}+ \frac{\sum _{i=1}^{m-2} \zeta _{i}(\eta _{i}^{\beta -1}-\eta _{i}^{\beta }) s^{\beta -1}}{B\Gamma (\beta +1)} \biggr)\,ds, \quad t\in [0,1]. $$

Theorem 3.2

Assume that

• (H1) $f,g:[0,1]\times [-e^{*},+\infty )\times [-e^{*},+\infty )\to (- \infty ,+\infty )$ are continuous. For every $t\in [0,1]$, $g(t,0,\mathcal{K}(L))\geq 0$ with $g(t,0,\mathcal{K}(L))\not \equiv 0$ where $L\geq \frac{bDF}{\beta \Gamma (\alpha )\Gamma (\beta )}$ and $e^{*}=\max \{e(t):t\in [0,1]\}$;

• (H2) for fixed $t\in [0,1]$ and $y\in [-e^{*},+\infty )$, $f(t,x,y)$, $g(t,x,y)$ are increasing in $x\in [-e^{*},+\infty )$; for fixed $t\in [0,1]$ and $x\in [-e^{*},+\infty )$, $f(t,x,y)$, $g(t,x,y)$ are decreasing in $y\in [-e^{*},+\infty )$;

• (H3) for all $\lambda \in (0,1)$, there exists $\psi (\lambda ,x,y)\in (\lambda ,1)$ such that for all $t\in [0,1]$,

  • (a) $f (t,\lambda x+(\lambda -1)\rho _{1},\lambda ^{-1}D_{0^{+}}^{\nu }y+ (\lambda ^{-1}-1)\rho _{2} )\geq \psi (\lambda ,x,y)f(t,x,D_{0^{+}}^{\nu }y)$,

  • (b) $g (t,\lambda x+(\lambda -1)\rho _{3},\lambda ^{-1}y+(\lambda ^{-1}-1) \rho _{3} )\geq \lambda g(t,x,y)$, where $x,y\in [-e^{*},+\infty )$, $\rho _{1}, \rho _{3} \in [0,e^{*}]$, and $\rho _{2} \in [0, e_{*}]$ with $e_{*} =\max \{D_{0^{+}}^{\nu }e(t): t\in [0, 1]\}$,

  • (c) for fixed $t\in [0,1]$ and $y\in P_{h,e}$, $\psi (\lambda ,x,y)$ are increasing in $x\in P_{h,e}$ and for fixed $t\in [0,1]$ and $x\in P_{h,e}$, $\psi (\lambda ,x,y)$ are decreasing in $y\in P_{h,e}$;

• (H4) for all $t\in [0,1]$, $x,y\in [-e^{*},+\infty )$, there exists $\delta >0$ such that

  $$ f(t,x,y)\geq \delta g(t,x,0); $$

• (H5) $\mathcal{K}:C[0,1]\rightarrow C[0,1]$ and satisfies

  • (a) $\mathcal{K}u\geq 0$ for every $u\in P_{h,e}$,

  • (b) for $u, v\in P_{h,e}$, $u\leq v\Longrightarrow \mathcal{K}u\leq \mathcal{K}v$,

  • (c) for all $\lambda \in (0,1)$ and $u\in P_{h,e}$,

    $$ \mathcal{K} \bigl(\lambda u+(\lambda -1)\hat{e} \bigr)\geq \lambda \mathcal{K}(u)+( \lambda -1)\hat{e}, \quad \hat{e} \in \bigl[0, e^{*} \bigr]; $$

• (H6) there exists $t_{0}\in (0,1)$ such that

  $$ \begin{aligned} & \frac{t_{0}}{\psi (t_{0},h,h)}h(t)+ \frac{t_{0}}{\psi (t_{0},h,h)} e(t) \\ &\quad \leq \int ^{1}_{0} G(t,s) \biggl( \int ^{1}_{0} H(s,\tau )f \bigl(\tau ,h(\tau ),{D}^{\nu }_{0^{+}} h(\tau ) \bigr)\,d\tau \biggr)\,ds \\ &\quad \leq \frac{1}{t_{0}}h(t)+\frac{1}{t_{0}} e(t). \end{aligned} $$

Then the problem (1.1) has a unique solution $u^{*}$ in $P_{h,e}$, where $h(t)=Lt^{\alpha -1}$, for all $t\in [0,1]$. We can construct the following sequences:

$$\begin{aligned}& \begin{aligned} \omega _{n}(t)&= \int ^{1}_{0} G(t,s) \biggl( \int ^{1}_{0} H(s,\tau )f \bigl(\tau ,\omega _{n-1}(\tau ),{D}^{\nu }_{0^{+}} \sigma _{n-1}(\tau ) \bigr)\,d\tau \biggr)\,ds \\ &\quad {}+ \int ^{1}_{0} G(t,s) \biggl( \int ^{1}_{0} H(s, \tau )g \bigl(\tau ,\omega _{n-1}(\tau ),(\mathcal{K} \sigma _{n-1}) ( \tau ) \bigr)\,d\tau \biggr)\,ds-e(t), \quad n=1,2,\ldots , \end{aligned} \\& \begin{aligned} \sigma _{n}(t)&= \int ^{1}_{0} G(t,s) \biggl( \int ^{1}_{0} H(s,\tau )f \bigl(\tau ,\sigma _{n-1}(\tau ),{D}^{\nu }_{0^{+}} \omega _{n-1}(\tau ) \bigr)\,d\tau \biggr)\,ds \\ &\quad {}+ \int ^{1}_{0} G(t,s) \biggl( \int ^{1}_{0} H(s, \tau )g \bigl(\tau ,\sigma _{n-1},(\mathcal{K} \omega _{n-1}) (\tau ) \bigr)\,d\tau \biggr)\,ds-e(t), \quad n=1,2,\ldots , \end{aligned} \end{aligned}$$

for every initial value $\omega _{0}, \sigma _{0} \in P_{h,e}$,we have $\omega _{n}\rightarrow u^{*}$ and $\sigma _{n}\rightarrow u^{*}$ as $n\rightarrow \infty$.

Proof

By [3], we have

$$ \begin{aligned} \int ^{1}_{0}H(s,\tau )\,d\tau &= \int ^{1}_{0}H_{1}(s, \tau )\,d\tau + \int ^{1}_{0}H_{2}(s,\tau )\,d\tau \\ &= \frac{s^{\beta -1}-s^{\beta }}{\Gamma (\beta +1)}+ \frac{\sum _{i=1}^{m-2}\zeta _{i}(\eta _{i}^{\beta -1}-\eta _{i}^{\beta }) s^{\beta -1}}{B\Gamma (\beta +1)}. \end{aligned} $$

Furthermore, it follows from Lemmas 2.4 and 2.5 that

$$ \begin{aligned} 0&< e(t) \leq b \int ^{1}_{0} \frac{Dt^{\alpha -1}}{\Gamma (\alpha )} \biggl( \int ^{1}_{0} \frac{Fs^{\beta -1}}{\Gamma (\beta )}\,d\tau \biggr)\,ds \\ &= \frac{bDF}{\Gamma (\alpha )\Gamma (\beta )}t^{\alpha -1} \int ^{1}_{0}s^{ \beta -1}\,ds \\ &= \frac{bDF}{\beta \Gamma (\alpha )\Gamma (\beta )}t^{\alpha -1} \\ &\leq Lt^{\alpha -1}=h(t), \end{aligned} $$

where $L\geq \frac{bDF}{\beta \Gamma (\alpha )\Gamma (\beta )}$. Hence, $0< e(t)\leq h(t)$. By Lemma 2.3, we find that the problem (1.1) has the following expression:

$$ \begin{aligned} u(t)&= \int ^{1}_{0} G(t,s) \biggl( \int ^{1}_{0} H(s, \tau ) \bigl(f \bigl(\tau ,u(\tau ),{D}^{\nu }_{0^{+}} u(\tau ) \bigr) + g \bigl(\tau ,u( \tau ),(\mathcal{K} u) (\tau ) \bigr)-b \bigr)\,d\tau \biggr)\,ds \\ &= \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )f \bigl(\tau ,u(\tau ),{D}^{\nu }_{0^{+}} u(\tau ) \bigr)\,d\tau \,ds \\ &\quad {}+ \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )g \bigl(\tau ,u(\tau ),(\mathcal{K} u) (\tau ) \bigr)\,d\tau \,ds \\ &\quad {}- \int ^{1}_{0} G(t,s)b \int ^{1}_{0} H(s, \tau )\,d\tau \,ds \\ &= \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )f \bigl(\tau ,u(\tau ),{D}^{\nu }_{0^{+}} u(\tau ) \bigr)\,d\tau \,ds-e(t) \\ &\quad {}+ \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )g \bigl(\tau ,u(\tau ),(\mathcal{K} u) (\tau ) \bigr)\,d\tau \,ds-e(t)+e(t). \end{aligned}$$

For every $t\in [0,1]$ and $u,v\in P_{h,e}$, we consider the following operators:

$$ M(u,v) (t)= \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s,\tau )f \bigl(\tau ,u( \tau ),{D}^{\nu }_{0^{+}}v(\tau ) \bigr)\,d\tau \,ds-e(t) $$

(3.4)

and

$$ N(u,v) (t)= \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,u( \tau ),(\mathcal{K}v) (\tau ) \bigr)\,d\tau \,ds-e(t). $$

(3.5)

Clearly, $u(t)$ is the solution of problem (1.1) is equivalent to u is the fixed point of $M(u,v)(t)+N(u,v)(t)+e$. By (3.4) and (3.5), we get

$$\begin{aligned}& D_{0^{+}}^{\nu }M(u,v) (t)= \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )f \bigl(\tau ,u(\tau ),{D}^{\nu }_{0^{+}} v(\tau ) \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t), \\& D_{0^{+}}^{\nu }N(u,v) (t)= \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,u(\tau ),(\mathcal{K} v) (\tau ) \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t). \end{aligned}$$

(1) Firstly, we show that $M,N: P_{h.e}\times P_{h.e} \rightarrow E$ are two mixed monotone operators. By (H1) and (H2), for every $u_{i},v_{i}\in P_{h,e}$ ($i=1,2$) with $u_{1}\geq u_{2}$, $v_{1}\leq v_{2}$, we have

$$ \begin{aligned} M(u_{1},v_{1}) (t)&= \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )f \bigl(\tau ,u_{1}(\tau ),{D}^{\nu }_{0^{+}} v_{1}(\tau ) \bigr)\,d\tau \,ds-e(t) \\ & \geq \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )f \bigl(\tau ,u_{2}(\tau ),{D}^{\nu }_{0^{+}} v_{2}(\tau ) \bigr)\,d\tau \,ds-e(t)=M(u_{2},v_{2}) (t) \end{aligned} $$

and

$$ \begin{aligned} D_{0^{+}}^{\nu }M(u_{1},v_{1}) (t)&= \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )f \bigl(\tau ,u_{1}(\tau ),{D}^{\nu }_{0^{+}} v_{1}(\tau ) \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t) \\ & \geq \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )f \bigl(\tau ,u_{2}(\tau ),{D}^{\nu }_{0^{+}} v_{2}( \tau ) \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t) \\ &= D_{0^{+}}^{\nu }M(u_{2},v_{2}) (t). \end{aligned} $$

Hence, M is a mixed monotone operator. Similarly, we deduce

$$ \begin{aligned} N(u_{1},v_{1}) (t)&= \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )g \bigl(\tau ,u_{1}(\tau ),(\mathcal{K}v_{1}) (\tau ) \bigr)\,d\tau \,ds-e(t) \\ & \geq \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )g \bigl(\tau ,u_{2}(\tau ),(\mathcal{K}v_{2}) (\tau ) \bigr)\,d\tau \,ds-e(t)=N(u_{2},v_{2}) (t) \end{aligned} $$

and

$$ \begin{aligned} D_{0^{+}}^{\nu }N(u_{1},v_{1}) (t)&= \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,u_{1}(\tau ),( \mathcal{K}v_{1}) (\tau ) \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t) \\ & \geq \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,u_{2}(\tau ),(\mathcal{K}v_{2}) ( \tau ) \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t) \\ &=D_{0^{+}}^{\nu }N(u_{2},v_{2}) (t). \end{aligned} $$

Thus, N is also a mixed monotone operator.

(2) Next, by (H3), for every $t\in [0,1]$ and $\lambda \in (0,1)$, there exists $\psi (\lambda ,u,v)\in (\lambda ,1)$ such that, for every $u, v\in P_{h,e}$, we get

$$ \begin{aligned} &M \bigl(\lambda u+(\lambda -1)e,\lambda ^{-1}v+ \bigl(\lambda ^{-1}-1 \bigr)e \bigr) (t) \\ &\quad = \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s,\tau )f \bigl( \tau ,\lambda u+(\lambda -1)e,{D}^{\nu }_{0^{+}} \bigl(\lambda ^{-1}v+ \bigl( \lambda ^{-1}-1 \bigr)e \bigr) \bigr)\,d\tau \,ds-e(t) \\ &\quad = \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )f \bigl(\tau ,\lambda u+(\lambda -1)e,\lambda ^{-1}{D}^{\nu }_{0^{+}}v+ \bigl(\lambda ^{-1}-1 \bigr){D}^{\nu }_{0^{+}}e \bigr)\,d\tau \,ds-e(t) \\ &\quad \geq \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )\psi \bigl(\lambda ,u( \tau ),v(\tau ) \bigr)f \bigl(\tau ,u(\tau ),{D}^{ \nu }_{0^{+}}v( \tau ) \bigr)\,d\tau \,ds-e(t) \\ &\qquad {}+\psi (\lambda ,u,v)e(t)-\psi (\lambda ,u,v)e(t) \\ &\quad = \psi (\lambda ,u,v)M(u,v) (t)+ \bigl(\psi ( \lambda ,u,v)-1 \bigr)e(t) \end{aligned} $$

and

$$ \begin{aligned} &D_{0^{+}}^{\nu }M \bigl(\lambda u+(\lambda -1)e,\lambda ^{-1}v+ \bigl( \lambda ^{-1}-1 \bigr)e \bigr) (t) \\ &\quad = \int ^{1}_{0}D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )f \bigl(\tau ,\lambda u+(\lambda -1)e,{D}^{\nu }_{0^{+}} \bigl( \lambda ^{-1}v+ \bigl(\lambda ^{-1}-1 \bigr)e \bigr) \bigr)\,d\tau \,ds \\ &\qquad {}-D_{0^{+}}^{\nu }e(t) \\ &\quad = \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s, \tau )f \bigl(\tau ,\lambda u+(\lambda -1)e,\lambda ^{-1}{D}^{\nu }_{0^{+}}v+ \bigl(\lambda ^{-1}-1 \bigr){D}^{\nu }_{0^{+}}e \bigr)\,d\tau \,ds \\ &\qquad {}-D_{0^{+}}^{\nu }e(t) \\ &\quad \geq \int ^{1}_{0}D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s, \tau )\psi \bigl(\lambda ,u( \tau ),v(\tau ) \bigr)f \bigl(\tau ,u(\tau ),{D}^{ \nu }_{0^{+}}v( \tau ) \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t) \\ &\qquad {}+\psi (\lambda ,u,v)D_{0^{+}}^{\nu }e(t)-\psi (\lambda ,u,v)D_{0^{+}}^{\nu }e(t) \\ &\quad = \psi (\lambda ,u,v)D_{0^{+}}^{\nu }M(u,v) (t)+ \bigl( \psi (\lambda ,u,v)-1 \bigr)D_{0^{+}}^{\nu }e(t). \end{aligned} $$

Thus, $M (\lambda u+(\lambda -1)e,\lambda ^{-1}v+ (\lambda ^{-1}-1 )e )\geq \psi (\lambda ,u,v) M(u,v)+(\psi (\lambda ,u,v)-1)e$.

In view of (H3)(b) and (H5), we derive

$$ \begin{aligned} &\mathcal{K} \bigl(\lambda ^{-1} u+ \bigl( \lambda ^{-1}-1 \bigr)e \bigr)\leq \lambda ^{-1}( \mathcal{K}u)+ \bigl(\lambda ^{-1}-1 \bigr)e, \\ &N \bigl(\lambda u+(\lambda -1)e,\lambda ^{-1}v+ \bigl( \lambda ^{-1}-1 \bigr)e \bigr) (t) \\ &\quad = \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )g (\tau ,\lambda u+( \lambda -1)e, \bigl(\mathcal{K} \bigl(\lambda ^{-1}v+ \bigl(\lambda ^{-1}-1 \bigr)e \bigr) \bigr)\,d\tau \,ds-e(t) \\ &\quad \geq \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )g \bigl(\tau ,\lambda u+(\lambda -1)e,\lambda ^{-1}(\mathcal{K}v)+ \bigl(\lambda ^{-1}-1 \bigr)e \bigr)\,d\tau \,ds-e(t) \\ &\quad \geq \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )\lambda g \bigl(\tau ,u(\tau ),(\mathcal{K}v) (\tau ) \bigr)\,d\tau \,ds-e(t) \\ &\quad = \lambda \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )g \bigl(\tau ,u(\tau ),(\mathcal{K}v) (\tau ) \bigr)\,d\tau \,ds-e(t)+ \lambda e(t)-\lambda e(t) \\ &\quad = \lambda N(u,v) (t)+(\lambda -1)e(t), \end{aligned} $$

and

$$\begin{aligned} & D_{0^{+}}^{\nu }N \bigl(\lambda u+(\lambda -1)e, \lambda ^{-1}v+ \bigl(\lambda ^{-1}-1 \bigr)e \bigr) (t) \\ &\quad = \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )g (\tau ,\lambda u+( \lambda -1)e, \bigl(\mathcal{K} \bigl(\lambda ^{-1}v+ \bigl(\lambda ^{-1}-1 \bigr)e \bigr) \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t) \\ &\quad \geq \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,\lambda u+(\lambda -1)e, \lambda ^{-1}(\mathcal{K}v)+ \bigl(\lambda ^{-1}-1 \bigr)e \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t) \\ &\quad \geq \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )\lambda g \bigl(\tau ,u(\tau ),(\mathcal{K}v) ( \tau ) \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t) \\ &\quad = \lambda \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,u(\tau ),(\mathcal{K}v) (\tau ) \bigr)\,d\tau \,ds \\ &\qquad {}-D_{0^{+}}^{\nu }e(t)+ \lambda D_{0^{+}}^{\nu }e(t)- \lambda D_{0^{+}}^{\nu }e(t) \\ &\quad = \lambda D_{0^{+}}^{\nu }N(u,v) (t)+( \lambda -1)D_{0^{+}}^{\nu }e(t). \end{aligned}$$

Thus, $N (\lambda u+(\lambda -1)e,\lambda ^{-1}v+ (\lambda ^{-1}-1 )e )\geq \lambda N(u,v)+(\lambda -1)e$.

(3) In view of (H6), we have

$$\begin{aligned}& \begin{aligned} M(h,h) (t)&= \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s,\tau )f \bigl(\tau ,h(\tau ),{D}^{\nu }_{0^{+}}h(\tau ) \bigr)\,d\tau \,ds-e(t) \\ &\leq \frac{1}{t_{0}}h(t)+ \biggl( \frac{1}{t_{0}}-1 \biggr)e(t), \end{aligned} \\& M(h,h) (t)\geq \frac{t_{0}}{\psi (t_{0},h,h)}h(t)+ \biggl( \frac{t_{0}}{\psi (t_{0},h,h)}-1 \biggr)e(t), \end{aligned}$$

and

$$\begin{aligned}& \begin{aligned} D_{0^{+}}^{\nu }M(h,h) (t)&= \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )f \bigl(\tau ,h(\tau ),{D}^{\nu }_{0^{+}}h( \tau ) \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t) \\ &\leq D_{0^{+}}^{\nu }\frac{1}{t_{0}}h(t)+ \biggl(\frac{1}{t_{0}}-1 \biggr)D_{0^{+}}^{\nu }e(t), \end{aligned} \\& D_{0^{+}}^{\nu }M(h,h) (t)\geq D_{0^{+}}^{\nu }\frac{t_{0}}{\psi (t_{0},h,h)}h(t)+ \biggl( \frac{t_{0}}{\psi (t_{0},h,h)}-1 \biggr)D_{0^{+}}^{\nu }e(t). \end{aligned}$$

Thus,

$$ \frac{t_{0}}{\psi (t_{0},h,h)}h+ \biggl(\frac{t_{0}}{\psi (t_{0},h,h)}-1 \biggr)e\leq M(h,h)\leq \frac{\psi (t_{0},h,h)}{t_{0}}h+ \biggl( \frac{\psi (t_{0},h,h)}{t_{0}}-1 \biggr)e. $$

Next we show that $N(h,h)\in P_{h,e}$. It suffices to prove that $N(h,h)+e\in P_{h}$. From Lemma 2.4 and the condition (H2), we have

$$\begin{aligned}& \begin{aligned} N(h,h) (t)+e(t)&= \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,h(\tau ),(\mathcal{K}h) (\tau ) \bigr)\,d\tau \,ds \\ &\leq \int ^{1}_{0} \frac{Dt^{\alpha -1}}{\Gamma (\alpha )} \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,L\tau ^{\alpha -1},\mathcal{K} \bigl(L\tau ^{\alpha -1} \bigr) \bigr)\,d\tau \,ds \\ &\leq \frac{Dt^{\alpha -1}}{\Gamma (\alpha )} \int ^{1}_{0} \int ^{1}_{0} H(s,\tau )g (\tau ,L,0 )\,d\tau \,ds \\ &= \frac{D h(t)}{L\Gamma (\alpha )} \int ^{1}_{0} \int ^{1}_{0} H(s,\tau )g (\tau ,L,0 )\,d\tau \,ds, \end{aligned} \\& \begin{aligned} N(h,h) (t)+e(t)&= \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,h(\tau ),(\mathcal{K}h) (\tau ) \bigr)\,d\tau \,ds \\ &\geq \int ^{1}_{0} \frac{C(s)}{\Gamma (\alpha )}t^{\alpha -1} \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,L\tau ^{\alpha -1},\mathcal{K} \bigl(L\tau ^{\alpha -1} \bigr) \bigr)\,d\tau \,ds \\ &\geq \frac{t^{\alpha -1}}{\Gamma (\alpha )} \int ^{1}_{0}C(s) \int ^{1}_{0} H(s, \tau )g \bigl(\tau ,0, \mathcal{K}(L) \bigr)\,d\tau \,ds \\ &= \frac{h(t)}{L\Gamma (\alpha )} \int ^{1}_{0}C(s) \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,0, \mathcal{K}(L) \bigr)\,d\tau \,ds, \end{aligned} \end{aligned}$$

and

$$\begin{aligned}& \begin{aligned} D_{0^{+}}^{\nu }N(h,h) (t)+D_{0^{+}}^{\nu }e(t)&= \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s, \tau )g \bigl(\tau ,h(\tau ),(\mathcal{K}h) (\tau ) \bigr)\,d\tau \,ds \\ &\leq \int ^{1}_{0} \frac{Dt^{\alpha -\nu -1}}{\Gamma (\alpha -\nu )} \int ^{1}_{0} H(s, \tau )g \bigl(\tau ,L\tau ^{\alpha -1},\mathcal{K} \bigl(L\tau ^{\alpha -1} \bigr) \bigr)\,d\tau \,ds \\ &\leq \frac{Dt^{\alpha -\nu -1}}{\Gamma (\alpha -\nu )} \int ^{1}_{0} \int ^{1}_{0} H(s,\tau )g (\tau ,L,0 )\,d\tau \,ds \\ &= \frac{D D_{0^{+}}^{\nu }h(t)}{L\Gamma (\alpha )} \int ^{1}_{0} \int ^{1}_{0} H(s,\tau )g (\tau ,L,0 )\,d\tau \,ds, \end{aligned} \\& \begin{aligned} D_{0^{+}}^{\nu }N(h,h) (t)+D_{0^{+}}^{\nu }e(t)&= \int ^{1}_{0}D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,h(\tau ),(\mathcal{K}h) ( \tau ) \bigr)\,d\tau \,ds \\ &\geq \int ^{1}_{0} \frac{C(s)}{\Gamma (\alpha -\nu )}t^{\alpha -\nu -1} \int ^{1}_{0} H(s, \tau )g \bigl(\tau ,L\tau ^{\alpha -1},\mathcal{K} \bigl(L\tau ^{\alpha -1} \bigr) \bigr)\,d\tau \,ds \\ &\geq \frac{t^{\alpha -\nu -1}}{\Gamma (\alpha -\nu )} \int ^{1}_{0}C(s) \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,0, \mathcal{K}(L) \bigr)\,d\tau \,ds \\ &= \frac{D_{0^{+}}^{\nu }h(t)}{L\Gamma (\alpha )} \int ^{1}_{0}C(s) \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,0, \mathcal{K}(L) \bigr)\,d\tau \,ds. \end{aligned} \end{aligned}$$

Let

$$\begin{aligned}& l_{1}= \frac{D}{L\Gamma (\alpha )} \int ^{1}_{0} \int ^{1}_{0} H(s,\tau )g (\tau ,L,0 )\,d\tau \,ds, \\& l_{2}= \frac{1}{L\Gamma (\alpha )} \int ^{1}_{0}C(s) \int ^{1}_{0} H(s,\tau )g \bigl(\tau ,0, \mathcal{K}(L) \bigr)\,d\tau \,ds. \end{aligned}$$

Then $l_{2}h\leq N(h,h)+e\leq l_{1}h$, thus $N(h,h)\in P_{h,e}$. Therefore, the condition (L4) of Theorem 3.1 is proved.

(4) For every $u,v\in P_{h,e}$ and $t\in [0,1]$, we derive that

$$ \begin{aligned} M(u,v) (t)&= \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s,\tau )f \bigl(\tau ,u(\tau ),{D}^{\nu }_{0^{+}}v(\tau ) \bigr)\,d\tau \,ds-e(t) \\ &\geq \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )\delta g \bigl(\tau ,u( \tau ),0 \bigr)\,d\tau \,ds-e(t) \\ &\geq \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau )\delta g \bigl(\tau ,u( \tau ),(\mathcal{K}v) (\tau ) \bigr)\,d\tau \,ds-e(t) \\ &= \delta \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau ) g \bigl(\tau ,u(\tau ),(\mathcal{K}v) (\tau ) \bigr)\,d\tau \,ds-e(t)+\delta e(t)- \delta e(t) \\ &=\delta N(u,v) (t)+(\delta -1)e(t) \end{aligned} $$

and

$$\begin{aligned} D_{0^{+}}^{\nu }M(u,v) (t)&= \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )f \bigl(\tau ,u(\tau ),{D}^{\nu }_{0^{+}}v( \tau ) \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t) \\ &\geq \int ^{1}_{0} D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )\delta g \bigl(\tau ,u( \tau ),0 \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t) \\ &\geq \int ^{1}_{0}D_{0^{+}}^{\nu }G(t,s) \int ^{1}_{0} H(s,\tau )\delta g \bigl(\tau ,u( \tau ),(\mathcal{K}v) (\tau ) \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t) \\ &= \delta \int ^{1}_{0} G(t,s) \int ^{1}_{0} H(s, \tau ) g \bigl(\tau ,u(\tau ),(\mathcal{K}v) (\tau ) \bigr)\,d\tau \,ds-D_{0^{+}}^{\nu }e(t) \\ &+\delta D_{0^{+}}^{\nu }e(t)-\delta D_{0^{+}}^{\nu }e(t) \\ &=\delta D_{0^{+}}^{\nu }N(u,v) (t)+(\delta -1)D_{0^{+}}^{\nu }e(t). \end{aligned}$$

Therefore, $M(u,v)\geq \delta N(u,v)+(\delta -1)e$. That is, the condition (L5) of Theorem 3.1 is satisfied. Consequently, all the conditions of Theorem 3.1 are satisfied, the conclusions of Theorem 3.2 hold. □

By the proof of Theorem 3.2, combining with Theorem 3.1′, we can obtain the following result.

Theorem 3.2′

Assume that the conditions (H1), (H2), (H3)(a)(b), (H4) and (H5) in Theorem 3.2are satisfied. Moreover, for fixed $t\in [0,1]$ and $y\in P_{h,e}$, $\psi (\lambda ,x,y)$ are decreasing in $x\in P_{h,e}$ and for fixed $t\in [0,1]$ and $x\in P_{h,e}$, $\psi (\lambda ,x,y)$ are increasing in $y\in P_{h,e}$. In addition, there exists $t_{0}\in (0, 1)$ such that

$$ \begin{aligned} & t_{0} h(t)+(t_{0} -1) e(t) \\ &\quad \leq \int ^{1}_{0} G(t,s) \biggl( \int ^{1}_{0} H(s,\tau )f \bigl(\tau ,h(\tau ),{D}^{\nu }_{0^{+}} h(\tau ) \bigr)\,d\tau \biggr)\,ds -e(t) \\ &\quad \leq \frac{1}{t_{0}}\psi \bigl(t_{0}, t_{0}^{-1}h + \bigl(t_{0}^{-1}-1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr)h(t) \\ &\qquad {}+ \biggl[\frac{1}{t_{0}}\psi \bigl(t_{0}, t_{0}^{-1}h + \bigl(t_{0}^{-1}-1 \bigr)e, t_{0} h+(t_{0} -1)e \bigr) -1 \biggr] e(t). \end{aligned} $$

Then the conclusions of Theorem 3.2hold.

Lastly, let us give an example to illustrate our main results.

Example 3.1

Consider the following boundary value problem:

$$ \textstyle\begin{cases} D^{\frac{3}{2}}_{0^{+}}(D^{\frac{3}{2}}_{0^{+}}u)(t)=2t^{2}+1+ (u(t)+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 )^{\frac{1}{3}} + (u(t)+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 )^{\frac{1}{2}} \\ \hphantom{D^{\frac{3}{2}}_{0^{+}}(D^{\frac{3}{2}}_{0^{+}}u)(t)={}}{}+ (D^{\frac{1}{8}}_{0^{+}}u(t)+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 )^{\frac{-1}{5}} \\ \hphantom{D^{\frac{3}{2}}_{0^{+}}(D^{\frac{3}{2}}_{0^{+}}u)(t)={}}{}+ ( \int ^{t}_{0}(u(s)+\frac{15}{\Gamma ^{2} (\frac{3}{2})})\,ds+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 )^{-1}-10, \\ u(0)=0, \qquad D^{\frac{3}{2}}_{0^{+}}u(0)=0, \\ D^{\frac{1}{4}}_{0^{+}}u(1)=\frac{1}{10}D^{\frac{1}{4}}_{0^{+}}u( \frac{1}{4})+\frac{1}{10}D^{\frac{1}{4}}_{0^{+}}u(\frac{1}{2}) + \frac{1}{10}D^{\frac{1}{4}}_{0^{+}}u(\frac{3}{4}), \\ D^{\frac{3}{2}}_{0^{+}}u(1)=\frac{1}{10}D^{\frac{3}{2}}_{0^{+}}u( \frac{1}{4})+\frac{1}{10}D^{\frac{3}{2}}_{0^{+}}u(\frac{1}{2}) + \frac{1}{10}D^{\frac{3}{2}}_{0^{+}}u(\frac{3}{4}). \end{cases} $$

(3.6)

Then the problem (3.6) has a solution.

Proof

The problem (1.1) becomes the problem (3.6) when we choose $n=2$, $\alpha =\frac{3}{2}$, $\beta =\frac{3}{2}$, $\gamma =\frac{1}{4}$, $\nu =\frac{1}{8}$, $b=10$, $\eta _{1}=\frac{1}{4}$, $\eta _{2}=\frac{1}{2}$, $\eta _{3}=\frac{3}{4}$, $\xi _{1}=\xi _{2}=\xi _{3}=\frac{1}{10}$, and $\zeta _{1}=\zeta _{2}=\zeta _{3}=\frac{1}{10}$. Then we have

$$\begin{aligned}& A=1-\sum_{i=1}^{m-2}\xi _{i}\eta _{i}^{\alpha - \gamma -1}\approx 0.7521>0, \\& B=1-\sum_{i=1}^{m-2}\zeta _{i}\eta _{i}^{\beta -1}\approx 0.7927>0, \\& F=\frac{1}{B} \Biggl(1+\sum _{i=1}^{m-2} \zeta _{i} \bigl(1-\eta _{i}^{ \beta -1} \bigr) \Biggr)\approx 1.5571, \\& D=\frac{1}{A} \Biggl(1+\sum _{i=1}^{m-2}\xi _{i} \bigl(1-\eta _{i}^{ \alpha - \gamma -1} \bigr) \Biggr) \approx 1.3988. \end{aligned}$$

A direct computation leads to

$$\begin{aligned}& \begin{aligned} \int ^{1}_{0}H(s,\tau )\,d\tau &= \int ^{1}_{0}H_{1}(s, \tau )\,d\tau + \int ^{1}_{0}H_{2}(s,\tau )\,d\tau \\ &= \frac{s^{\frac{1}{2}}-s^{\frac{3}{2}}}{\Gamma (\frac{5}{2})}+ \frac{\frac{3}{10}[\frac{1}{4}^{\frac{1}{2}}-\frac{1}{4}^{\frac{3}{2}} +\frac{1}{2}^{\frac{1}{2}}-\frac{1}{2}^{\frac{3}{2}}+\frac{3}{4}^{\frac{1}{2}}- \frac{3}{4}^{\frac{3}{2}}]s^{\frac{1}{2}}}{B\Gamma (\frac{5}{2})}, \end{aligned} \\& \begin{aligned} e(t)&= 10 \int ^{1}_{0}G(t,s) \int ^{1}_{0}H(s,\tau )\,d\tau \,ds \\ &= 10 \int ^{1}_{0}G(t,s) \biggl[ \frac{s^{\frac{1}{2}}-s^{\frac{3}{2}}}{\Gamma (\frac{5}{2})}+ \frac{\frac{3}{10}[\frac{1}{4}^{\frac{1}{2}}- \frac{1}{4}^{\frac{3}{2}}+\frac{1}{2}^{\frac{1}{2}}-\frac{1}{2}^{\frac{3}{2}}+\frac{3}{4}^{\frac{1}{2}}-\frac{3}{4}^{\frac{3}{2}}]s^{\frac{1}{2}}}{B\Gamma (\frac{5}{2})} \biggr] \,ds \\ &\leq 10 \int ^{1}_{0} \frac{D t^{\frac{1}{2}}}{\Gamma (\frac{3}{2})} \biggl[ \frac{s^{\frac{1}{2}}-s^{\frac{3}{2}}}{\Gamma (\frac{5}{2})}+ \frac{\frac{3}{10}[\frac{1}{4}^{\frac{1}{2}}- \frac{1}{4}^{\frac{3}{2}}+\frac{1}{2}^{\frac{1}{2}}-\frac{1}{2}^{\frac{3}{2}}+\frac{3}{4}^{\frac{1}{2}}-\frac{3}{4}^{\frac{3}{2}}]s^{\frac{1}{2}}}{B\Gamma (\frac{5}{2})} \biggr] \,ds \\ &\leq \frac{15}{\Gamma ^{2} (\frac{3}{2})}t^{ \frac{1}{2}}= Lt^{\frac{1}{2}}=h(t), \end{aligned} \end{aligned}$$

and

$$ e^{\ast }\leq \frac{15}{\Gamma ^{2} (\frac{3}{2})},\qquad D_{0^{+}}^{ \frac{1}{8}} e(t) \leq \frac{15}{\Gamma (\frac{3}{2})\Gamma (\frac{11}{8})}. $$

Let

$$\begin{aligned}& f(t,u,v)=t^{2}+1+ \biggl(u(t)+\frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{\frac{1}{2}}+ \biggl(v(t)+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{-\frac{1}{5}}, \\& g(t,u,v)=t^{2}+ \biggl(u(t)+\frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{\frac{1}{3}}+ \biggl(v(t)+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{-1}, \\& (\mathcal{K}u) (t)= \int ^{t}_{0} \biggl(u(s)+ \frac{15}{\Gamma ^{2} (\frac{3}{2})} \biggr)\,ds. \end{aligned}$$

For $\lambda \in (0,1)$ and $\hat{e}\in [0,\frac{15}{\Gamma ^{2} (\frac{3}{2})}]$, we deduce

$$ \begin{aligned} \mathcal{K} \bigl(\lambda u+(\lambda -1)\hat{e} \bigr)&= \int ^{t}_{0} \biggl(\lambda u+(\lambda -1) \hat{e}+ \frac{15}{\Gamma ^{2} (\frac{3}{2})} \biggr)\,ds \\ &=\lambda \int ^{t}_{0}u \,ds+ \int ^{t}_{0}( \lambda -1)\hat{e}\,ds + \int _{0}^{t} \frac{15}{\Gamma ^{2} (\frac{3}{2})} \,ds \\ &\geq \lambda (\mathcal{K}u) (t)+(\lambda -1)\hat{e}, \end{aligned} $$

and $(\mathcal{K}u)$ is increasing in u, thus (H5) is satisfied. It is easy to check that $f,g:[0,1]\times [-\frac{15}{\Gamma ^{2} (\frac{3}{2})},+\infty ) \times [-\frac{15}{\Gamma ^{2} (\frac{3}{2})},+\infty )\to (-\infty ,+ \infty )$ are continuous, $f(t,u,v)$, $g(t,u,v)$ are both increasing in u and decreasing in v and $g(t,0,\mathcal{K}(L))=t^{2}+(\frac{15}{\Gamma ^{2} (\frac{3}{2})}+1)^{ \frac{1}{3}}+(\mathcal{K}(L)+\frac{15}{\Gamma ^{2} (\frac{3}{2})}+1)^{-1}>0$. Thus, (H1) and (H2) are satisfied.

For all $\lambda \in (0,1)$, $t\in [0, 1]$, $u,v\in P_{h,e}$, $\rho _{1}, \rho _{3} \in [0,\frac{15}{\Gamma ^{2} (\frac{3}{2})}]$ and $\rho _{2} \in [0, \frac{15}{\Gamma (\frac{3}{2})\Gamma (\frac{11}{8})}]$, there exists $\psi (\lambda ,u,v)=\lambda ^{\frac{1}{2}}$ such that

$$\begin{aligned}& f \bigl(t,\lambda u+(\lambda -1)\rho _{1},\lambda ^{-1}D_{0^{+}}^{ \frac{1}{8}} v+ \bigl(\lambda ^{-1}-1 \bigr)\rho _{2} \bigr) \\& \quad =t^{2}+1+ \biggl(\lambda u+(\lambda -1)\rho _{1}+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{\frac{1}{2}} \\& \qquad {}+ \biggl( \lambda ^{-1} D_{0^{+}}^{\frac{1}{8}} v+ \bigl(\lambda ^{-1}-1 \bigr) \rho _{2}+\frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{- \frac{1}{5}} \\& \quad =t^{2}+1+\lambda ^{\frac{1}{2}} \biggl(u+ \bigl(1-\lambda ^{-1} \bigr)\rho _{1}+ \lambda ^{-1} \frac{15}{\Gamma ^{2} (\frac{3}{2})}+\lambda ^{-1} \biggr)^{\frac{1}{2}} \\& \qquad {} +\lambda ^{\frac{1}{5}} \biggl(D_{0^{+}}^{\frac{1}{8}} v+(1- \lambda )\rho _{2}+\lambda \frac{15}{\Gamma ^{2} (\frac{3}{2})}+ \lambda \biggr)^{-\frac{1}{5}} \\& \quad \geq t^{2}+1+\lambda ^{\frac{1}{2}} \biggl(u+ \bigl(1-\lambda ^{-1} \bigr) \frac{15}{\Gamma ^{2} (\frac{3}{2})}+\lambda ^{-1} \frac{15}{\Gamma ^{2} (\frac{3}{2})}+\lambda ^{-1} \biggr)^{ \frac{1}{2}} \\& \qquad {} +\lambda ^{\frac{1}{5}} \biggl(D_{0^{+}}^{\frac{1}{8}} v+(1- \lambda )\frac{15}{\Gamma (\frac{3}{2})\Gamma (\frac{11}{8})}+ \lambda \frac{15}{\Gamma ^{2} (\frac{3}{2})}+\lambda \biggr)^{- \frac{1}{5}} \\& \quad \geq t^{2}+1+\lambda ^{\frac{1}{2}} \biggl(u+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{\frac{1}{2}} \\& \qquad {}+ \lambda ^{\frac{1}{5}} \biggl(D_{0^{+}}^{\frac{1}{8}} v+(1-\lambda ) \frac{15}{\Gamma ^{2} (\frac{3}{2})}+ \lambda \frac{15}{\Gamma ^{2} (\frac{3}{2})}+\lambda \biggr)^{-\frac{1}{5}} \\& \quad \geq t^{2}+1+\lambda ^{\frac{1}{2}} \biggl(u+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{\frac{1}{2}} + \lambda ^{\frac{1}{5}} \biggl(D_{0^{+}}^{\frac{1}{8}} v+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{-\frac{1}{5}} \\& \quad \geq \lambda ^{\frac{1}{2}} \biggl[ t^{2}+1+ \biggl(u+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{\frac{1}{2}} + \biggl(D_{0^{+}}^{ \frac{1}{8}} v+\frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{- \frac{1}{5}} \biggr] \\& \quad =\psi (\lambda ,u,v)f \bigl(t,u, D_{0^{+}}^{\frac{1}{8}} v \bigr). \end{aligned}$$

Moreover, we deduce

$$\begin{aligned}& g \bigl(t,\lambda u+(\lambda -1)\rho _{3},\lambda ^{-1}v+ \bigl( \lambda ^{-1}-1 \bigr)\rho _{3} \bigr) \\& \quad =t^{2}+ \biggl(\lambda u+(\lambda -1)\rho _{3}+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{\frac{1}{3}}+ \biggl( \lambda ^{-1}v+ \bigl(\lambda ^{-1}-1 \bigr)\rho _{3}+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{-1} \\& \quad =t^{2}+\lambda ^{\frac{1}{3}} \biggl(u+ \bigl(1-\lambda ^{-1} \bigr)\rho _{3}+ \frac{15\lambda ^{-1}}{\Gamma ^{2} (\frac{3}{2})}+\lambda ^{-1} \biggr)^{\frac{1}{3}} +\lambda \biggl(v+(1-\lambda )\rho _{3}+ \frac{15\lambda }{\Gamma ^{2} (\frac{3}{2})}+\lambda \biggr)^{-1} \\& \quad \geq \lambda t^{2}+\lambda \biggl(u+ \bigl(1-\lambda ^{-1} \bigr) \frac{15}{\Gamma ^{2} (\frac{3}{2})}+ \frac{15\lambda ^{-1}}{\Gamma ^{2} (\frac{3}{2})} +\lambda ^{-1} \biggr)^{\frac{1}{3}} \\& \qquad {}+\lambda \biggl(v+(1-\lambda ) \frac{15}{\Gamma ^{2} (\frac{3}{2})}+ \frac{15\lambda }{\Gamma ^{2} (\frac{3}{2})}+\lambda \biggr)^{-1} \\& \quad = \lambda t^{2}+\lambda \biggl(u+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+\lambda ^{-1} \biggr)^{ \frac{1}{3}} +\lambda \biggl(v+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+ \lambda \biggr)^{-1} \\& \quad \geq \lambda \biggl[ t^{2}+ \biggl(u+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{\frac{1}{3}} + \biggl(v+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{-1} \biggr] \\& \quad =\lambda g(t,u,v). \end{aligned}$$

Thus, (H3) is satisfied. Furthermore, for $u,v\in P_{h,e}$, we get

$$ \begin{aligned} f(t,u,v)&=t^{2}+1+ \biggl(u(t)+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{ \frac{1}{2}}+ \biggl(v(t)+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{- \frac{1}{5}} \\ &\geq t^{2}+1+ \biggl(u(t)+\frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{ \frac{1}{3}}+ \biggl(v(t)+\frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{-1} \\ &\geq t^{2}+ \biggl(u(t)+\frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{ \frac{1}{3}}+ \biggl(\frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{-1}=g(t,u,0), \end{aligned}$$

let $\delta =1$, we have $f(t,u,v)\geq \delta g(t,u,0)$. Thus (H4) is satisfied.

By Lemmas 2.4 and 2.5, we have

$$ \begin{aligned} & \int _{0}^{1} G(t,s) \biggl( \int _{0}^{1} H(s,\tau ) f \bigl( \tau , h(\tau ), D_{0^{+}}^{\nu }h(\tau ) \bigr)\,d\tau \biggr)\,ds \\ &\quad \leq \int _{0}^{1} G(t,s) \biggl( \int _{0}^{1} H(s,\tau ) \biggl(4+ h(\tau )+ \frac{15}{\Gamma ^{2}(\frac{3}{2})} \biggr)\,d\tau \biggr)\,ds \\ &\quad = \int _{0}^{1} G(t,s) \biggl( \int _{0}^{1} H(s,\tau ) h(\tau ) \,d\tau \biggr)\,ds + \biggl[ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+4 \biggr]\frac{e(t)}{10} \\ &\quad \leq \int _{0}^{1} \frac{D t^{\frac{1}{2}}}{\Gamma (\frac{3}{2})} \biggl( \int _{0}^{1} \frac{15 F s^{\frac{1}{2}}\tau ^{\frac{1}{2}}}{\Gamma ^{3} (\frac{3}{2})} \,d\tau \biggr)\,ds + \biggl[\frac{15}{\Gamma ^{2} (\frac{3}{2})}+4 \biggr] \frac{e(t)}{10} \\ &\quad =\frac{4 DF}{9 \Gamma ^{2} (\frac{3}{2})} h(t) + \biggl[ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+4 \biggr] \frac{e(t)}{10} \\ &\quad =1.2325 h(t) +2.3099 e(t) \end{aligned} $$

and

$$\begin{aligned}& \int _{0}^{1} G(t,s) \biggl( \int _{0}^{1} H(s,\tau ) f \bigl( \tau , h(\tau ), D_{0^{+}}^{\nu }h(\tau ) \bigr)\,d\tau \biggr)\,ds \\& \quad \geq \int _{0}^{1} \frac{c(s) t^{\frac{1}{2}}}{\Gamma (\frac{3}{2})} \biggl( \int _{s}^{1} \frac{1}{\Gamma (\frac{3}{2})} s^{\frac{1}{2}} (1-\tau )^{\frac{1}{2}} \tau ^{\frac{1}{4}} \,d\tau \biggr) \,ds +\frac{1}{10} e(t) \\& \quad \geq \frac{4 t^{\frac{1}{2}}}{5 \Gamma ^{2} (\frac{3}{2})} \int _{0}^{1} c(s) s^{\frac{1}{2}} \bigl(1-s^{\frac{5}{4}} \bigr) \,ds + \frac{1}{10} e(t) \\& \quad \geq \frac{4 t^{\frac{1}{2}}}{5 \Gamma ^{2} (\frac{3}{2})} \int _{\frac{1}{2}} ^{1} \frac{1}{10} \biggl( \frac{1}{2} \biggr)^{\frac{1}{4}}(1-s)^{ \frac{1}{4}} s^{\frac{1}{2}} \bigl(1-s^{\frac{5}{4}} \bigr) \,ds +\frac{1}{10} e(t) \\& \quad \geq 0.0001 h(t) +0.1 e(t). \end{aligned}$$

Choose $t_{0} =10^{-8}$, we deduce that the condition (H6) is satisfied. Therefore, all the assumptions of Theorem 3.2 are satisfied. We can construct the following iteration sequences:

$$ \begin{aligned} \omega _{n}(t)={}& \int ^{1}_{0} G(t,s) \biggl( \int ^{1}_{0} H(s,\tau ) \biggl(\tau ^{2}+ \biggl(\omega _{n-1}( \tau )+\frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{\frac{1}{3}} \\ &{}+ \biggl( \int ^{\tau }_{0} \biggl(\sigma _{n-1}(x)+ \frac{15}{\Gamma ^{2} (\frac{3}{2})} \biggr)\,dx + \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{-1} \biggr)\,d\tau \biggr)\,ds \\ &{}+ \int ^{1}_{0} G(t,s) \biggl( \int ^{1}_{0} H(s,\tau ) \biggl(\tau ^{2}+1+ \biggl(\omega _{n-1}(\tau )+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{\frac{1}{2}} \\ &{}+ \biggl({D}^{\nu }_{0^{+}} \sigma _{n-1}(\tau )+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{-\frac{1}{5}} \biggr)\,d\tau \biggr)\,ds-e(t), \quad n=1,2,\ldots , \end{aligned} $$

and

$$\begin{aligned}& \begin{aligned} \sigma _{n}(t)={}& \int ^{1}_{0} G(t,s) \biggl( \int ^{1}_{0} H(s,\tau ) \biggl(\tau ^{2}+ \biggl(\sigma _{n-1}( \tau )+\frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{\frac{1}{3}} \\ &{}+ \biggl( \int ^{\tau }_{0} \biggl(\omega _{n-1}(x)+ \frac{15}{\Gamma ^{2} (\frac{3}{2})} \biggr)\,dx + \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{-1} \biggr)\,d\tau \biggr)\,ds \\ &{}+ \int ^{1}_{0} G(t,s) \biggl( \int ^{1}_{0} H(s,\tau ) \biggl(\tau ^{2}+1+ \biggl(\sigma _{n-1}(\tau )+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{\frac{1}{2}} \\ &{} + \bigl({D}^{\nu }_{0^{+}} \omega _{n-1} \bigr) ( \tau )+ \frac{15}{\Gamma ^{2} (\frac{3}{2})}+1 \biggr)^{-\frac{1}{5}} \biggr)\,d\tau )\,ds-e(t),\quad n=1,2,\ldots , \end{aligned} \end{aligned}$$

for any initial values $\omega _{0}, \sigma _{0} \in P_{h,e}$,we have $\omega _{n}\rightarrow u^{*}$ and $\sigma _{n}\rightarrow u^{*}$ as $n\rightarrow \infty$. □

4 Conclusions

In this paper, we obtain two new mixed monotone fixed point theorems. By using our abstract results, we establish the existence and uniqueness theorems of the solution for a fractional m-point boundary value problem, which generalizes the well-known elastic beam equation. Furthermore, two iterative sequences to approximate the unique solution are also given.