# Fixed point theorems for sum operator with parameter

## Abstract

This paper develops some new existence and uniqueness theorems of a fixed point for a class of sum operator equations with parameter $$\lambda _{1}A(x,x)+\lambda _{2}B(x,x)+\lambda _{3}Cx+\lambda _{4}Dx=x$$, where A, B are two mixed monotone operators, C is an increasing operator, D is a decreasing operator. In the case of positive parameters, the results obtained in this paper extend many existing conclusions in the field of study. Furthermore, by using the properties of Green’s function and the above fixed point theorems of sum operator, the unique positive solution a class of fractional differential equations with integro-differential boundary value conditions is given. Application of the results to the study of fractional differential equations is also given in the article.

## Introduction

This paper discusses the existence and uniqueness of solution for a class of operator equation

$$\lambda _{1}A(x,x)+\lambda _{2}B(x,x)+\lambda _{3}Cx+\lambda _{4}Dx=x,$$

where A, B are two mixed monotone operators, C is an increasing operator, D is a decreasing operator, $$\lambda _{i}>0$$ ($$i=1,2,3,4$$) and satisfies the following conditions:

Situation 1:

1. 1.

$$A(\lambda x,\lambda ^{-1}y)\geq \varphi (\lambda )A(x,y)$$, $$\varphi (\lambda )\in (\lambda ,1]$$, $$\forall \lambda \in (0,1)$$, $$x, y \in P$$;

2. 2.

for any fixed $$y\in P$$, $$B(\cdot ,y): P\rightarrow P$$ is concave; for any $$x\in P$$, $$B(x,\cdot ): P\rightarrow P$$ is convex;

3. 3.

$$C: P \rightarrow P$$ is increasing sub-homogeneous.

4. 4.

$$D(\lambda ^{-1}y)\geq \lambda Dy$$, $$\forall \lambda \in (0,1)$$, $$y \in P$$.

Situation 2:

1. 1.

$$A(\lambda x,\lambda ^{-1}y)\geq \lambda A(x,y)$$, $$\forall \lambda \in (0,1)$$, $$x, y \in P$$;

2. 2.

for any fixed $$y\in P$$, $$B(\cdot ,y): P\rightarrow P$$ is concave; for any $$x\in P$$, $$B(x,\cdot ): P\rightarrow P$$ is convex;

3. 3.

$$C(\lambda x)\geq \varphi (\lambda )Cx$$, $$\varphi (\lambda )\in ( \lambda ,1]$$, $$\forall \lambda \in (0,1)$$, $$x,y \in P$$;

4. 4.

$$D(\lambda ^{-1}y)\geq \lambda Dy$$, $$\forall \lambda \in (0,1)$$, $$y \in P$$.

In fact, the operator equation $$\lambda _{1}A(x,x)+\lambda _{2}B(x,x)+\lambda _{3}Cx+\lambda _{4}Dx=x$$ generalizes and improves many articles.

When the parameter $$\lambda _{3}=1$$, $$\lambda _{i}=0$$ ($$i=1,2,4$$), situation 2 can be reduced to the case that an increasing operator C meets $$C(tx)\geq t^{\alpha (t)}Cx$$, $$\alpha (t) \in (0,1)$$, $$\forall t\in (0,1)$$, $$x \in P_{h}$$. Obviously, it is the result of the paper .

When the parameter $$\lambda _{1}=1$$, $$\lambda _{i}=0$$ ($$i=2,3,4$$), there is $$A(\lambda x,\lambda ^{-1}y)\geq \varphi (\lambda )A(x,y)$$, $$\varphi ( \lambda )\in (\lambda ,1]$$, $$\forall \lambda \in (0,1)$$, $$x, y \in P$$ in situation 1. In , we can see that the mixed monotone operator A meets the same properties.

When the parameter $$\lambda _{1}=1$$, $$\lambda _{2}=0$$, $$\lambda _{3}=1$$, $$\lambda _{4}=0$$, we deduce from situation 1 that Theorem 2.1 of  is established and from situation 2 that Theorem 2.4 of  holds.

When the parameter $$\lambda _{1}=1$$, $$\lambda _{2}=0$$, $$\lambda _{3}=1$$, $$\lambda _{4}=1$$, we derive Theorem 3.1 of  from situation 1 and Theorem 3.8 of  from situation 1.

When the parameter $$\lambda _{1}=1$$, $$\lambda _{2}=1$$, $$\lambda _{3}=1$$, $$\lambda _{4}=0$$, the conditions of situation 1 are the same as those of Theorem 3.1 in , thus they can get same conclusions.

The theory of nonlinear operators plays a significant role in modern mathematics and there are many excellent results (see ). Inspired by the these paper, we want to make some contributions to this field. As one of the applications of operator theory, fractional differential equations have attracted much attention by many researchers as a result of a myriad of their applications in many engineering and scientific disciplines, such as mechanics, biomedicine, physics, and so on, see  and the references therein. Besides, there are some methods such as comparison theorem, the monotone iterative technique, the method of lower and upper solutions, Leray-Schauder theory, Krasnoselskii’s fixed point theorems, and some other fixed point theorems in cones. They play an irreplaceable role in the existence, uniqueness, and multiplicity of positive solutions for fractional differential equations . In recent decades, more and more fractional differential equations are solved based on the nonlinear operators theory . In this paper, we make use of the new operator equation theory to investigate a class of new fractional differential equations.

The characteristic features of this paper are displayed as follows. Firstly, comparatively speaking, we generalize the results of the above article. Secondly, there are seldom investigated operator equations with parameters. A class of new operator equations with four operators $$\lambda _{1}A(x,x)+\lambda _{2}B(x,x)+\lambda _{3}Cx+\lambda _{4}Dx=x$$ is studied, and the fixed point theorem of this sum-type operator is obtained. We gain the existence and uniqueness solution of the operator equation, and construct two iterative sequences to uniformly approximate this solution in the fixed point theorem. Thirdly, by using the new results, we study a class of new fractional differential equations and get some great conclusions. Fourthly, some concrete examples are given to illustrate the main ideas.

The main body of this paper is organized as follows. In Sect. 2, we review the theory and results of fractional calculus and some definitions, notations in Banach space. In Sect. 3, a class of fixed point theorem is presented. In Sect. 4, by the theorem of Sect. 2, a kind of fractional differential equation is studied and some examples to illustrative our work are presented.

## Preliminaries

In this section, since all the work is in the Banach space, a brief review about the Banach space and relevant contents is given for the reader’s convenience; it includes some definitions, lemmas, and basic results. These will be used in the following proofs of our theorem. For more details, we refer the reader to .

Let $$(E,\|\cdot \|)$$ be a real Banach space. A nonempty closed convex set $$P\subset E$$ is called a cone if the following hold:

1. (i)

If $$x \in P$$, $$\lambda \geq 0$$, then $$\lambda x\in P$$;

2. (ii)

If $$x \in P$$ and $$-x \in P$$, then $$x = \theta$$,

where θ is the zero element of E.

There is partially ordered by a cone $$P\subset E$$, $$x\leq y$$, $$x,y \in E \Leftrightarrow y-x\in P$$, in which, if $$x\leq y$$ and $$x\neq y$$, we denote $$x< y$$ or $$y>x$$.

If $$\dot{P} = \{x \in P | x \mbox{ is an interior point of } P \}$$ is nonempty, cone P is said to be solid. Then, we call P normal if there exists a constant $$N>0$$ such that, for $$\forall x,y\in E$$, $$\theta \leq x\leq y$$, there is $$\|x\|\leq N\|y\|$$, where N is called the normality constant of P. For $$\forall x, y\in E$$, the denotation $$x\sim y$$ means that there exist $$\lambda , \mu >0$$ such that $$\lambda x\leq y\leq \mu x$$. Clearly, is an equivalence relation. Given $$h>\theta$$, and denoting the set $$P_{h}=\{x\in E\mid x\sim h\}$$, there is $$P_{h}\subset P$$.

### Definition 2.1

()

$$A:P\times P\rightarrow P$$ is called a mixed monotone operator if $$A(x,y)$$ is increasing in x and decreasing in y, i.e., $$u_{i},v_{i}\ (i=1,2)\in P$$, $$u_{1}\leq u_{2}$$, $$v_{1}\geq v_{2}$$ implies $$A(u_{1},v_{1})\leq A(u_{2},v_{2})$$. Element $$x\in P$$ is called a fixed point of A if $$A(x,x) = x$$.

### Definition 2.2

([9, 10])

An operator $$A : E\rightarrow E$$ is said to be homogeneous if it satisfies

$$A(tx) = tAx,\quad \forall t>0, x\in E.$$

An operator $$A : P\rightarrow P$$ is said to be sub-homogeneous if it satisfies

$$A(tx) \geq tAx, \quad \forall t\in (0,1), x\in P.$$

### Definition 2.3

([9, 10])

Let $$D = P$$ or $$D = \dot{P}$$ and a real number $$\alpha \in [0,1)$$. An operator $$A : D \rightarrow D$$ is said to be α-concave if it satisfies

$$A(tx)\geq t^{\alpha }Ax,\quad \forall t\in (0,1), x\in D.$$

### Definition 2.4

()

Let D be a convex subset in E. An operator $$A : D\rightarrow E$$ is called a convex operator if, for $$\forall x, y \in D$$ with $$y \leq x$$ and every $$t \in [0, 1]$$,

$$A\bigl(tx+(1-t)y\bigr)\leq tAx+(1-t)Ay,$$

$$A : D\rightarrow E$$ is called a concave operator if

$$A\bigl(tx+(1-t)y\bigr)\geq tAx+(1-t)Ay.$$

## Main results

In this section, we use the definitions, notations of section Preliminaries to investigate some new fixed point theorem. Furthermore, we can obtain the following sufficient conditions of existence and uniqueness of positive solutions for the operator equation $$\lambda _{1}A(x,x)+\lambda _{2}B(x,x)+\lambda _{3}Cx+\lambda _{4}Dx=x$$.

### Theorem 3.1

There are four operatorsA, B, C, D, where$$A, B: P\times P \rightarrow P$$are two mixed monotone operators, $$C: P \rightarrow P$$is an increasing sub-homogeneous operator, $$D: P \rightarrow P$$is a decreasing operator, and if the following conditions are satisfied:

($$A_{1}$$):

There exists$$\varphi (\lambda )\in (\lambda ,1]$$such that

\begin{aligned} A\bigl(\lambda x,\lambda ^{-1}y\bigr)\geq \varphi ( \lambda )A(x,y), \qquad D\bigl( \lambda ^{-1}y\bigr)\geq \lambda Dy, \quad \forall \lambda \in (0,1), x,y \in P; \end{aligned}
(1)
($$A_{2}$$):

For any fixed$$y\in P$$, $$B(\cdot ,y): P\rightarrow P$$is concave; for any$$x\in P$$, $$B(x,\cdot ): P\rightarrow P$$is convex;

($$A_{3}$$):

There exists$$\frac{1}{2}\leq \widetilde{c}\leq 1$$such that$$B(\theta ,lh)\geq \widetilde{c}B(lh,\theta )$$, $$l\geq 1$$;

($$A_{4}$$):

There exists$$h\in P$$with$$h\neq \theta$$such that$$A(h,h),B(h,h),Ch,Dh\in P_{h}$$;

($$A_{5}$$):

There exists$$\delta >0$$such that$$[\lambda _{2}B(x,y)+ \lambda _{3}Cx+\lambda _{4}Dy] \leq \delta \lambda _{1}A(x,y)$$, $$\forall x,y \in P_{h}$$.

Then the following conclusions hold:

$$(C1)$$:

$$A,B:P_{h}\times P_{h}\rightarrow P_{h}$$, $$C,D:P_{h}\rightarrow P_{h}$$;

$$(C2)$$:

There exist$$u_{0}, v_{0}\in P_{h}$$, $$r\in (0,1)$$such that$$rv_{0}\leq u_{0}< v_{0}$$,

\begin{aligned} u_{0} \leq & \lambda _{1}A(u_{0},v_{0})+ \lambda _{2}B(u_{0},v_{0})+ \lambda _{3}Cu_{0}+\lambda _{4}Dv_{0} \\ \leq &\lambda _{1}A(v_{0},u_{0})+\lambda _{2}B(v_{0},u_{0})+\lambda _{3}Cv_{0}+ \lambda _{4}Du_{0} \leq v_{0}; \end{aligned}
$$(C3)$$:

The operator equation$$\lambda _{1}A(x,x)+\lambda _{2}B(x,x)+\lambda _{3}Cx+\lambda _{4}Dx=x$$has a unique solution$$x^{*}\in P_{h}$$;

$$(C4)$$:

Constructing the iterative sequences as follows:

\begin{aligned}& x_{n}=\lambda _{1}A(x_{n-1},y_{n-1})+ \lambda _{2}B(x_{n-1},y_{n-1})+ \lambda _{3}Cx_{n-1}+\lambda _{4}Dy_{n-1}, \\& y_{n}=\lambda _{1}A(y_{n-1},x_{n-1})+ \lambda _{2}B(y_{n-1},x_{n-1})+ \lambda _{3}Cy_{n-1}+\lambda _{4}Dx_{n-1}, \quad n=1,2,\ldots , \end{aligned}

for any initial values$$x_{0},y_{0}\in P_{h}$$, we have

\begin{aligned} x_{n}\rightarrow x^{*}, \qquad y_{n} \rightarrow x^{*},\quad \textit{as } n\rightarrow \infty . \end{aligned}

### Proof

Define the operator $$T=\lambda _{1}A+\lambda _{2}B+\lambda _{3}C+\lambda _{4}D$$ by

$$T(x,y)=\lambda _{1}A(x,y)+\lambda _{2}B(x,y)+\lambda _{3}Cx+\lambda _{4}Dy,\quad \forall x,y\in P.$$

Firstly, we show that $$T:P_{h}\times P_{h}\rightarrow P_{h}$$. The proof of $$A,B:P_{h}\times P_{h}\rightarrow P_{h}$$, $$C,D:P_{h}\rightarrow P_{h}$$ is expanded. By $$A(\lambda x,\lambda ^{-1}y)\geq \varphi (\lambda )A(x,y)$$ of ($$A_{1}$$), we deduce

\begin{aligned} A\bigl(\lambda ^{-1}x,\lambda y\bigr)\leq \varphi ( \lambda )^{-1}A(x,y). \end{aligned}
(2)

Furthermore, the condition $$A(h,h)\in P_{h}$$ of ($$A_{4}$$) shows that there exist positive constants $$\overline{u_{1}}$$, $$\overline{v_{1}}$$ such that

\begin{aligned} \overline{u_{1}}h\leq A(h,h) \leq \overline{v_{1}}h. \end{aligned}
(3)

For $$\forall x, y \in P_{h}$$, we can find two sufficiently small numbers $$c_{1}, c_{2} \in (0, 1)$$ such that

\begin{aligned} c_{1}h\leq x \leq c^{-1}_{1}h, \qquad c_{2}h\leq y \leq c^{-1}_{2}h. \end{aligned}
(4)

Let $$c=\min \{c_{1},c_{2}\}$$, then by (1)–(4), we get

\begin{aligned}& A(x,y)\leq A\bigl(c^{-1}_{1}h,c_{2}h\bigr) \leq A\bigl(c^{-1}h,ch\bigr)\leq \varphi (c)^{-1}A(h,h) \leq \varphi (c)^{-1}\overline{v_{1}}h, \\& A(x,y)\geq A\bigl(c_{1}h,c^{-1}_{2}h\bigr) \geq A\bigl(ch,c^{-1}h\bigr)\geq \varphi (c)A(h,h) \geq \varphi (c) \overline{u_{1}}h. \end{aligned}

Thus, $$A:P_{h}\times P_{h}\rightarrow P_{h}$$ is proved. Then, due to ($$A_{2}$$), for any $$\lambda \in (0,1)$$, there is

$$B(x,y)=B\bigl(x,\lambda \lambda ^{-1} y+ (1-\lambda ) \theta \bigr) \leq \lambda B\bigl( x,\lambda ^{-1} y\bigr)+(1-\lambda ) B( x,\theta ),$$

thus, $$\lambda B( x,\lambda ^{-1} y)\geq B(x,y)-(1-\lambda ) B( x,\theta )$$. Subsequently, we can find a sufficiently large l such that $$x, y, \lambda ^{-1}y \leq lh$$. Combining with Definition 2.4 and from the condition of ($$A_{2}$$)–($$A_{3}$$), we know that

\begin{aligned} B\bigl(\lambda x,\lambda ^{-1}y\bigr) =&B\bigl(\lambda x+(1-\lambda ) \theta , \lambda ^{-1}y\bigr) \\ \geq & \lambda B\bigl(x, \lambda ^{-1}y\bigr)+(1-\lambda ) B\bigl( \theta ,\lambda ^{-1}y\bigr) \\ \geq & B(x,y)-(1-\lambda ) B( x,\theta )+(1-\lambda ) B\bigl(\theta , \lambda ^{-1}y\bigr) \\ \geq & B(x,y)+(1-\lambda ) \bigl(B( \theta ,lh)- B(lh,\theta )\bigr) \\ \geq & B(x,y)+(1-\lambda )\biggl[B( \theta ,lh)-\frac{1}{\widetilde{c}}B( \theta ,lh)\biggr] \\ \geq &\biggl[1+(1-\lambda ) \biggl(1-\frac{1}{\widetilde{c}}\biggr)\biggr]B(x,y) \\ =&\biggl[\biggl(2-\frac{1}{\widetilde{c}}\biggr)+\biggl(\frac{1}{\widetilde{c}}-1 \biggr)\lambda \biggr] B(x,y) \\ \geq & \lambda B(x,y), \end{aligned}

that is,

\begin{aligned} B\bigl(\lambda x,\lambda ^{-1}y\bigr)\geq \lambda B(x,y). \end{aligned}
(5)

Then we gain

\begin{aligned} B\bigl(\lambda ^{-1} x, \lambda y\bigr)\leq \lambda ^{-1}B(x,y), \quad \lambda \in (0,1). \end{aligned}

From the condition $$B(h,h)\in P_{h}$$, there exist two positive constants $$\overline{u_{2}}$$, $$\overline{v_{2}}$$ such that $$\overline{u_{2}}h\leq B(h,h)\leq \overline{v_{2}}h$$. For any $$x, y \in P_{h}$$,

\begin{aligned}& B(x,y)\leq B\bigl(c^{-1}_{1}h,c_{2}h\bigr) \leq B\bigl(c^{-1}h,ch\bigr)\leq c^{-1}B(h,h) \leq c^{-1}\overline{v_{2}}h, \\& B(x,y)\geq B\bigl(c_{1}h,c^{-1}_{2}h\bigr) \geq B\bigl(ch,c^{-1}h\bigr)\geq cB(h,h)\geq c \overline{u_{2}}h. \end{aligned}

Thus, $$B:P_{h}\times P_{h}\rightarrow P_{h}$$ holds. Since $$Ch, Dh\in P_{h}$$,

\begin{aligned} \overline{u_{3}}h\leq Ch \leq \overline{v_{3}}h, \qquad \overline{u_{4}}h \leq Dh\leq \overline{v_{4}}h, \end{aligned}
(6)

where $$\overline{u_{i}}$$, $$\overline{v_{i}}$$ ($$i=3,4$$) are positive constants. By the properties of operators C, D, for any $$\lambda \in (0,1)$$, $$x,y \in P$$, there is

$$C(tx)\geq tCx,\qquad C\bigl(t^{-1}x\bigr)\leq t^{-1}Cx, \qquad D\bigl(t^{-1}x\bigr)\geq tDx,\qquad D(tx)\leq t^{-1}Dx.$$

Using similar processes, we have

\begin{aligned}& Cx\leq C\bigl(c^{-1}_{1}h\bigr)\leq C \bigl(c^{-1}h\bigr)\leq c^{-1}Ch\leq c^{-1} \overline{v_{3}}h, \\& Cx\geq C(c_{1}h)\geq C(ch)\geq cCh\geq c\overline{u_{3}}h, \\& Dy\leq D(c_{2}h)\leq D(ch)\leq c^{-1}Dh\leq c^{-1}\overline{v_{4}}h, \\& Dy \geq D\bigl(c^{-1}_{2}h\bigr)\geq D \bigl(c^{-1}h\bigr)\geq cDh\geq c\overline{u_{4}}h, \end{aligned}

which shows that $$C,D:P_{h}\rightarrow P_{h}$$. From the above deduction, we have

\begin{aligned}& \bigl(\lambda _{1}\varphi (c)\overline{u_{1}}+\lambda _{2}c\overline{u_{2}}+ \lambda _{3}c \overline{u_{3}}+\lambda _{4}c\overline{u_{4}} \bigr)h \\& \quad \leq T(x,y) =\lambda _{1}A(x,y)+\lambda _{2}B(x,y)+ \lambda _{3}Cx+ \lambda _{4}Dy \\& \quad \leq \bigl(\lambda _{1}\varphi (c)^{-1} \overline{v_{1}}+\lambda _{2}c^{-1} \overline{v_{2}}+\lambda _{3}c^{-1} \overline{v_{3}}+\lambda _{4}c^{-1} \overline{v_{4}} \bigr)h, \end{aligned}

i.e., $$T:P_{h}\times P_{h}\rightarrow P_{h}$$, and then we can get $$T(h,h)\in P_{h}$$.

Secondly, we demonstrate that there exists $$\eta (t,x,y)\in (t,1]$$ such that

$$T\bigl(tx,t^{-1}y\bigr)\geq \eta (t,x,y)T(x,y), \quad \forall t \in (0,1), x,y \in P_{h}.$$

Combining with (1), (5) and the properties of operator C, D, we have

\begin{aligned} T\bigl(tx,t^{-1}y\bigr) =&\lambda _{1}A \bigl(tx,t^{-1}y\bigr)+\lambda _{2}B\bigl(tx,t^{-1}y \bigr)+ \lambda _{3}C(tx)+\lambda _{4}D \bigl(t^{-1}y\bigr) \\ \geq &\lambda _{1}\varphi (t)A(x,y)+\lambda _{2}tB(x,y)+ \lambda _{3}tCx+ \lambda _{4}tDy \\ =&\varphi (t)\lambda _{1}A(x,y)+t\bigl[\lambda _{2}B(x,y)+\lambda _{3}Cx+ \lambda _{4}Dy \bigr]. \end{aligned}

Owing to $$A,B:P_{h}\times P_{h}\rightarrow P_{h}$$, $$C,D:P_{h}\rightarrow P_{h}$$, there is $$\lambda _{1}A(x,y)\sim \lambda _{2}B(x,y)+\lambda _{3}Cx+\lambda _{4}Dy$$. Define $$K\{\frac{x}{y}\}=\inf \{k\in R|x\leq ky\}$$, and let

$$J(x,y)=K \biggl( \frac{\lambda _{2}B(x,y)+\lambda _{3}Cx+\lambda _{4}Dy}{\lambda _{1}A(x,y)} \biggr),$$

we can easily get $$J(x,y)\leq \delta$$ by ($$A_{5}$$). Next, considering a function $$\overline{h}(s)=\frac{\varphi (t)+J(x,y)t}{(J(x,y)+1)s}$$, it shows that h is continuous and strictly decreasing about s. Due to $$\varphi (t)>t$$, there is

$$\overline{h}\biggl(\frac{\delta t+\varphi (t)}{\delta +1}\biggr)= \frac{\varphi (t)+J(x,y)t}{(J(x,y)+1)\frac{\delta t+\varphi (t)}{\delta +1}}>1,\qquad \overline{h}\bigl(\varphi (t)\bigr)= \frac{\varphi (t)+J(x,y)t}{(J(x,y)+1)\varphi (t)}< 1.$$

Because of the monotone decreasing of , we can get that there exist $$\eta (t,x,y)\in (\frac{\delta t+\varphi (t)}{\delta +1}, \varphi (t)) \subset (t,1]$$,

$$\overline{h}\bigl(\eta (t,x,y)\bigr)= \frac{\varphi (t)+J(x,y)t}{(J(x,y)+1)\eta (t,x,y)}=1,$$

thus, $$J(x,y)=\frac{\varphi (t)-\eta (t,x,y)}{\eta (t,x,y)-t}$$. By the expression of K, J, there is

$$\lambda _{2}B(x,y)+\lambda _{3}Cx+\lambda _{4}Dy\leq \frac{\varphi (t)-\eta (t,x,y)}{\eta (t,x,y)-t}\lambda _{1}A(x,y), \quad \forall t\in (0,1), x,y \in P_{h}.$$

Thus, we obtain

\begin{aligned} T\bigl(tx,t^{-1}y\bigr) \geq &\varphi (t)\lambda _{1}A(x,y)+t \bigl[\lambda _{2}B(x,y)+ \lambda _{3}Cx+\lambda _{4}Dy\bigr] \\ =&\eta (t,x,y)\lambda _{1}A(x,y)+\bigl(\varphi (t)-\eta (t,x,y) \bigr)\lambda _{1}A(x,y) \\ &{}+t\bigl[\lambda _{2}B(x,y)+\lambda _{3}Cx+\lambda _{4}Dy\bigr] \\ \geq &\eta (t,x,y)\lambda _{1}A(x,y)+\bigl(\varphi (t)-\eta (t,x,y)\bigr) \frac{\eta (t,x,y)-t}{\varphi (t)-\eta (t,x,y)} \\ &{}\times \bigl[\lambda _{2}B(x,y)+\lambda _{3}Cx+ \lambda _{4}Dy\bigr] \\ &{}+t\bigl[\lambda _{2}B(x,y)+\lambda _{3}Cx+\lambda _{4}Dy\bigr] \\ =&\eta (t,x,y)\bigl[\lambda _{1}A(x,y)+\lambda _{2}B(x,y)+\lambda _{3}Cx+ \lambda _{4}Dy \bigr] \\ =&\eta (t,x,y)T(x,y),\quad \forall t\in (0,1), x,y \in P_{h}. \end{aligned}

Therefore, there exists $$\eta (t,x,y)\in (t,1]$$ such that

\begin{aligned} T\bigl(tx,t^{-1}y\bigr)\geq \eta (t,x,y)T(x,y), \quad \forall t \in (0,1), x,y \in P_{h}. \end{aligned}
(7)

Thirdly, we prove conclusion $$(C2)$$. Owing to $$T(h,h)\in P_{h}$$, we have $$t_{0}h\leq T(h,h)\leq t^{-1}_{0}h$$, where $$t_{0}\in (0,1)$$ is a small constant we choose. As a result of $$\eta (t,x,y)\in (t,1]$$, we have $$1< \frac{\eta (t_{0},x,y)}{t_{0}}\leq \frac{1}{t_{0}}$$. By the Archimedes principle, we can take a positive integer k such that $$(\frac{\eta (t_{0},x,y)}{t_{0}})^{k}\geq \frac{1}{t_{0}}$$. This inequality can be rewritten as

$$\eta (t_{0},x,y)\geq \biggl(\frac{1}{t_{0}} \biggr)^{\frac{1}{k}}\cdot t_{0},\qquad \frac{1}{\eta (t_{0},x,y)}\leq \frac{1}{t_{0}}\cdot t_{0}^{ \frac{1}{k}}.$$

Set $$u_{0}=t^{k}_{0}h$$, $$v_{0}=t^{-k}_{0}h$$. We can get $$u_{0}, v_{0}\in P_{h}$$, $$u_{0}=t^{2k}_{0}v_{0}< v_{0}$$. Consequently, there exists $$r\in (0,t^{2k}]$$ such that $$r\in (0,1)$$, $$u_{0}\geq rv_{0}$$. In addition, thanks to the mixed monotone properties of A, B, the increasing properties of C, and the decreasing properties of D, T is a mixed monotone operator, and then $$T(u_{0}, v_{0})\leq T(v_{0}, u_{0})$$. From (7), there is

\begin{aligned} T(u_{0}, v_{0}) =&T\bigl(t^{k}_{0}h,t^{-k}_{0}h \bigr) \\ =&T\bigl(t_{0}\cdot t^{k-1}_{0}h,t^{-1}_{0} \cdot t^{-k+1}_{0}h\bigr) \\ \geq & \eta \bigl(t_{0},t^{k-1}_{0}h,t^{-k+1}_{0}h \bigr)T\bigl(t^{k-1}_{0}h,t^{-k+1}_{0}h \bigr) \\ =&\eta \bigl(t_{0},t^{k-1}_{0}h,t^{-k+1}_{0}h \bigr)T\bigl(t_{0}\cdot t^{k-2}_{0}h,t^{-1}_{0} \cdot t^{-k+2}_{0}h\bigr) \\ \geq & \eta \bigl(t_{0},t^{k-1}_{0}h,t^{-k+1}_{0}h \bigr)\cdot \eta \bigl(t_{0},t^{k-2}_{0}h,t^{-k+2}_{0}h \bigr)T\bigl(t^{k-2}_{0}h, t^{-k+2}_{0}h \bigr) \geq \cdots \\ \geq & \biggl(\biggl(\frac{1}{t_{0}}\biggr)^{\frac{1}{k}}\cdot t_{0} \biggr)^{k}T(h,h) \geq \frac{1}{t_{0}}\cdot t_{0}^{k}t_{0}h=t_{0}^{k}h=u_{0}. \end{aligned}

By (7), we get $$T(t^{-1}x,ty)\leq \eta (t,t^{-1}x,ty)^{-1}T(x,y)$$, $$\forall t \in (0,1)$$, $$x,y \in P_{h}$$. Hence,

\begin{aligned} T(v_{0}, u_{0}) =&T\bigl(t^{-k}_{0}h,t^{k}_{0}h \bigr) \\ =&T\bigl(t^{-1}_{0}\cdot t^{-k+1}_{0}h,t_{0} \cdot t^{k-1}_{0}h\bigr) \\ \leq & \eta \bigl(t_{0},t^{-k}_{0}h,t^{k}_{0}h \bigr)^{-1}T\bigl(t^{-k+1}_{0}h,t^{k-1}_{0}h \bigr) \\ =&\eta \bigl(t_{0},t^{-k}_{0}h,t^{k}_{0}h \bigr)^{-1}T\bigl(t^{-1}_{0}\cdot t^{-k+2}_{0}h,t_{0} \cdot t^{k-2}_{0}h\bigr) \\ \leq & \eta \bigl(t_{0},t^{-k}_{0}h,t^{k}_{0}h \bigr)^{-1}\cdot \eta \bigl(t_{0},t^{-k+1}_{0}h,t^{k-1}_{0}h \bigr)^{-1}T\bigl( t^{-k+2}_{0}h, t^{k-2}_{0}h\bigr) \leq \cdots \\ \leq &\biggl(\frac{1}{t_{0}}\cdot t_{0}^{\frac{1}{k}} \biggr)^{k}T(h,h)\leq \frac{1}{t^{k}_{0}}\cdot t_{0}t^{-1}_{0}h= \frac{1}{t^{k}_{0}}h=v_{0}. \end{aligned}

Therefore, there is $$u_{0}\leq T(u_{0},v_{0})\leq T(v_{0},u_{0})\leq v_{0}$$. That is $$(C2)$$.

Fourthly, we show that the operator equation $$T(x,x)=x$$ has a unique solution $$x^{*}\in P_{h}$$. For $$u_{0}$$, $$v_{0}$$, construct successively the sequences as follows:

$$u_{n} = T(u_{n-1}, v_{n-1}),\qquad v_{n} = T(v_{n-1},u_{n-1}),\quad n = 1,2, \ldots ,n.$$

Thus, by (C2), we have $$u_{1}\leq v_{1}$$. And then due to the mixed monotone properties of operator T, there are $$u_{n}\leq v_{n}$$, $$n=1,2,3,\ldots$$ , and

\begin{aligned} u_{0}\leq u_{1}\leq \cdots \leq u_{n}\leq \cdots \leq v_{n}\leq \cdots \leq v_{1}\leq v_{0}. \end{aligned}
(8)

Considering $$x_{0}\geq ry_{0}$$ and (8), there is $$x_{n}\geq x_{0}\geq ry_{0}\geq ry_{n}$$ ($$n=1,2,3,\ldots$$). Put

$$t_{n}=\sup \{t>0|x_{n}\geq ty_{n}\},\quad n=1,2,3,\ldots .$$

It is clear that $$x_{n}\geq t_{n}y_{n}$$. Then from (8), there is $$x_{n+1}\geq x_{n}\geq t_{n}y_{n}\geq t_{n}y_{n+1}$$, $$n=1,2,3,\ldots$$ . Hence, $$t_{n+1}\geq t_{n}$$, i.e., $$\{t_{n}\}$$ is increasing about n, and $$\{t_{n}\}\subset (0,1]$$. Assume $$\lim_{n\rightarrow \infty }t_{n}\rightarrow t^{*}$$, so $$t^{*}=1$$. Otherwise, $$t\in (0,1)$$. From (7) and $$t_{n}\leq t^{*}$$,

\begin{aligned} x_{n+1} =&T(x_{n}, y_{n})\geq T \bigl(t_{n}y_{n},t_{n}^{-1}x_{n} \bigr)=T\biggl( \frac{t_{n}}{t^{*}}t^{*} y_{n}, \frac{t^{*}}{t_{n}}\frac{1}{t^{*}} x_{n}\biggr) \\ \geq &\frac{t_{n}}{t^{*}}T\biggl(t^{*}y_{n}, \frac{1}{t^{*}}x_{n}\biggr)\geq \frac{t_{n}}{t^{*}}\eta \bigl(t^{*},y_{n},x_{n}\bigr)T(y_{n},x_{n})= \frac{t_{n}}{t^{*}}\eta \bigl(t^{*},y_{n},x_{n} \bigr)y_{n+1}, \end{aligned}

combining with the definition of $$t_{n}$$, there is $$t_{n+1}\geq \frac{t_{n}}{t^{*}}\eta (t^{*},y_{n},x_{n})$$. Then $$\eta (t^{*},y_{n},x_{n})\leq \frac{t^{*}}{t_{n}}t_{n+1}$$. Owing to $$t^{*}<\frac{\delta t^{*}+\varphi (t^{*})}{\delta +1}< \eta (t^{*},y_{n},x_{n}) \leq \frac{t^{*}}{t_{n}}t_{n+1}$$, $$n=1,2,3,\ldots$$ , and $$\lim_{n\rightarrow \infty }\frac{t^{*}}{t_{n}}t_{n+1}=t^{*}$$, we know that

$$t^{*}< \frac{\delta t^{*}+\varphi (t^{*})}{\delta +1}\leq t^{*},$$

which is a contradiction. Thus, $$\lim_{n\rightarrow \infty } t_{n} = 1$$. For any natural number p, there is

\begin{aligned}& \theta \leq x_{n+p}-x_{n}\leq y_{n}-x_{n} \leq y_{n}-t_{n}y_{n}=(1-t_{n})v_{n} \leq (1-t_{n})v_{0}, \\& \theta \leq y_{n}-y_{n+p}\leq y_{n}-x_{n} \leq (1-t_{n})v_{0}. \end{aligned}

By the normality of cone P, there is

\begin{aligned}& \Vert x_{n+p}-x_{n} \Vert \leq N(1-t_{n}) \Vert v_{0} \Vert \rightarrow 0, \\& \Vert y_{n}-y_{n+p} \Vert \leq N(1-t_{n}) \Vert v_{0} \Vert \rightarrow 0\quad (n,p=1,2,3,\ldots ), \end{aligned}

where N is the normality constant. This shows that $$\{x_{n}\}$$ and $$\{y_{n}\}$$ are Cauchy sequences. Because of the complete continuity of A, there are $$x_{n}\rightarrow x^{*}$$, $$y_{n}\rightarrow y^{*}$$, when $$n\rightarrow \infty$$. And by (8), for $$\forall x^{*},y^{*}\in P_{h}$$, we obtain that $$x_{n} \leq x^{*}\leq y^{*}\leq y_{n}$$, $$\theta \leq y^{*}-x^{*}\leq y_{n}-x_{n}\leq (1-t_{n})v_{0}$$. Since cone P is normal, we have

$$\bigl\Vert y^{*}-x^{*} \bigr\Vert \leq N(1-t_{n}) \Vert v_{0} \Vert \rightarrow 0\quad (n \rightarrow \infty ),$$

hence, $$y^{*}=x^{*}$$. Set $$z^{*}:=y^{*}=x^{*}$$, there is

$$x_{n+1}=T(x_{n},y_{n})\leq T \bigl(z^{*},z^{*}\bigr)\leq T(y_{n},x_{n})=y_{n+1},$$

when $$n\rightarrow \infty$$, $$z^{*}=T(z^{*},z^{*})$$, i.e., $$z^{*}$$ is a fixed point of T in $$P_{h}$$.

Next, we show that $$z^{*}$$ is the unique fixed point of T. Assume that is another fixed point of T. Thanks to $$z^{*},\overline{z}\in P_{h}$$, there exist positive numbers $$a_{1}$$, $$a_{2}$$, $$b_{1}$$, $$b_{2}$$ such that

$$a_{1}h\leq z^{*} \leq a_{2}h, \qquad b_{1}h\leq \overline{z}\leq b_{2}h,$$

then $$\overline{z}\leq b_{2}h=\frac{b_{2}}{a_{1}}a_{1}h\leq \frac{b_{2}}{a_{1}}z^{*}$$, $$\overline{z}\geq b_{1}h=\frac{b_{1}}{a_{2}}a_{2}h \geq \frac{b_{1}}{a_{2}}z^{*}$$. We put

$$e_{1}=\sup \bigl\{ t>0, tz^{*}\leq \overline{z}\leq t^{-1}z^{*}\bigr\} .$$

Consequently, $$0< e_{1}\leq 1$$, $$e_{1}z^{*}\leq \overline{z}\leq e_{1}^{-1}z^{*}$$. Then $$e_{1}=1$$. Otherwise, $$0< e_{1}<1$$. There is

$$\overline{z}=T(\overline{z},\overline{z})\geq T\bigl(e_{1}z^{*},e_{1}^{-1}z^{*} \bigr) \geq \eta \bigl(e_{1},z^{*},z^{*} \bigr)T\bigl(z^{*},z^{*}\bigr)=\eta \bigl(e_{1},z^{*},z^{*} \bigr)z^{*}.$$

Hence, $$\eta (e_{1},z^{*},z^{*})\geq e_{1}$$, which contradicts the definition of $$e_{1}$$. So, $$e_{1}=1$$. Therefore, $$z^{*}=\overline{z}$$, i.e., T has a unique fixed point $$x^{*}$$ in $$P_{h}$$.

Eventually, we show that conclusion $$(C4)$$ holds. For any initial values $$x_{0},y_{0}\in P_{h}$$, we construct the iterative sequences:

$$x_{n}=T(x_{n-1},y_{n-1}),\qquad y_{n}=T(y_{n-1},x_{n-1}),\quad n=1,2,3,\ldots.$$

Due to $$x_{0},y_{0}\in P_{h}$$, we have $$e_{2}h\leq x_{0}\leq e_{2}^{-1}h$$, $$e_{3}h\leq y_{0}\leq e_{3}^{-1}h$$, where $$e_{2},e_{3}\in (0,1)$$ are two small numbers. Put $$e_{*}=\min \{e_{2},e_{3}\}$$, we deduce that $$e_{*}\in (0,1)$$ and $$e_{*}h\leq x_{0}$$, $$y_{0}\leq e_{*}^{-1}h$$. Then, by $$e_{*}<\eta (e_{*},x,y)\leq 1$$, there is $$1<\frac{\eta (e_{*},x,y)}{e_{*}}\leq \frac{1}{e_{*}}$$. By the Archimedes principle, there exists a sufficiently large positive integer m such that

$$\frac{\eta (e_{*},x,y)}{e_{*}}\geq \biggl(\frac{1}{e_{*}} \biggr)^{ \frac{1}{m}}.$$

Set $$\overline{u}_{0}=e^{m}_{*}h$$, $$\overline{v}_{0}=\frac{1}{e^{m}_{*}}h$$. Obviously, $$\overline{u}_{0}, \overline{v}_{0}\in P_{h}$$, and $$\overline{u}_{0}< x_{0}$$, $$y_{0}<\overline{v}_{0}$$. Put

$$\overline{u}_{n}=T(\overline{u}_{n-1}, \overline{v}_{n-1}), \qquad \overline{v}_{n}=T( \overline{v}_{n-1},\overline{u}_{n-1}),\quad n=1,2,3,\ldots .$$

Similarly, there exists $$y^{*}\in P_{h}$$ such that $$T(y^{*},y^{*})=y^{*}$$, $$\lim_{n\rightarrow \infty } \overline{u}_{n}= \lim_{n\rightarrow \infty } \overline{v}_{n}=y^{*}$$. Because T has the unique fixed points in $$P_{h}$$, we get $$x^{*}=y^{*}$$. And by induction, $$\overline{u}_{n}\leq x_{n}$$, $$y_{n}\leq \overline{v}_{n}$$, $$n=1,2,3, \ldots$$ . Thanks to cone P being normal, there is $$\lim_{n\rightarrow \infty }x_{n}= \lim_{n\rightarrow \infty } y_{n}=x^{*}$$, i.e., conclusion $$(C4)$$. The proof is complete. □

### Corollary 3.1

There are four operatorsA, B, C, D, where$$A, B: P\times P \rightarrow P$$are two mixed monotone operators, $$C:P \rightarrow P$$is an increasing homogeneous operator, and$$D: P \rightarrow P$$is a decreasing operator. Then, if ($$A_{2}$$)($$A_{5}$$) of Theorem 3.1are satisfied and meet the following:

($$A_{6}$$):

There exists$$\gamma \in (0,1)$$such that

\begin{aligned} A\bigl(\lambda x,\lambda ^{-1}y\bigr)\geq \lambda ^{\gamma }A(x,y), \qquad D\bigl(\lambda ^{-1}y\bigr) \geq \lambda Dy,\quad \forall \lambda \in (0,1), x,y \in P; \end{aligned}

then we can get conclusions$$(C1)$$$$(C4)$$of Theorem 3.1.

### Corollary 3.2

There are four operatorsA, B, C, D, where$$A, B: P_{h}\times P_{h} \rightarrow P_{h}$$are two mixed monotone operators, $$C:P_{h} \rightarrow P_{h}$$is an increasing sub-homogeneous operator, and$$D: P_{h} \rightarrow P_{h}$$is a decreasing operator. Then, if ($$A_{3}$$), ($$A_{5}$$) of Theorem 3.1are satisfied and have the following:

($$A_{7}$$):

There exists$$\varphi (\lambda )\in (\lambda ,1]$$such that

\begin{aligned} A\bigl(\lambda x,\lambda ^{-1}y\bigr)\geq \varphi (\lambda )A(x,y), \qquad D\bigl( \lambda ^{-1}y\bigr)\geq \lambda Dy, \quad \forall x,y \in P_{h}; \end{aligned}
($$A_{8}$$):

For any fixed$$y\in P_{h}$$, $$B(\cdot ,y): P_{h}\rightarrow P_{h}$$is concave; for any$$x\in P_{h}$$, $$B(x,\cdot ): P_{h}\rightarrow P_{h}$$is convex; then we can get conclusions$$(C2)$$$$(C4)$$of Theorem 3.1.

### Corollary 3.3

There are four operatorsA, B, C, D, where$$A, B: P_{h}\times P_{h} \rightarrow P_{h}$$are two mixed monotone operators, $$C:P_{h} \rightarrow P_{h}$$is an increasing homogeneous operator, and$$D: P_{h} \rightarrow P_{h}$$is a decreasing operator. Then, if ($$A_{3}$$), ($$A_{5}$$), ($$A_{7}$$), ($$A_{8}$$) are satisfied, then we can get conclusions$$(C2)$$$$(C4)$$of Theorem 3.1.

### Theorem 3.2

There are four operatorsA, B, C, D, where$$A, B: P\times P \rightarrow P$$are two mixed monotone operators, $$C: P \rightarrow P$$is an increasing operator, and$$D: P \rightarrow P$$is a decreasing operator. Then, if conditions ($$A_{2}$$)($$A_{4}$$) of Theorem 3.1and the following conditions are satisfied:

($$A_{9}$$):

There exists$$\varphi (\lambda )\in (\lambda ,1]$$such that

\begin{aligned}& \begin{aligned}&A\bigl(\lambda x,\lambda ^{-1}y\bigr)\geq \lambda A(x,y),\qquad C(\lambda x)\geq \varphi (\lambda )Cx, \\ & D\bigl(\lambda ^{-1}y\bigr)\geq \lambda Dy,\quad \forall \lambda \in (0,1), x,y \in P; \end{aligned} \end{aligned}
(9)
($$A_{ 10}$$):

There exists$$\delta >0$$such that$$\lambda _{1}A(x,y)+\lambda _{2}B(x,y)+\lambda _{4}Dy \leq \delta \lambda _{3}Cx$$, $$\forall x,y \in P_{h}$$;

then we can get conclusions$$(C1)$$$$(C4)$$of Theorem 3.1.

### Proof

Define the operator $$T=\lambda _{1}A+\lambda _{2}B+\lambda _{3}C+\lambda _{4}D$$ by

$$T(x,y)=\lambda _{1}A(x,y)+\lambda _{2}B(x,y)+\lambda _{3}Cx+\lambda _{4}Dy, \quad \forall x,y\in P.$$

Firstly, we show that $$T:P_{h}\times P_{h}\rightarrow P_{h}$$. According to Theorem 3.1, there is $$B:P_{h}\times P_{h}\rightarrow P_{h}$$, $$D:P_{h}\rightarrow P_{h}$$. Thus, we only prove that $$A:P_{h}\times P_{h}\rightarrow P_{h}$$, $$C:P_{h}\rightarrow P_{h}$$. Since (9), we obtain

\begin{aligned} A\bigl(\lambda ^{-1}x,\lambda y\bigr)\leq \lambda ^{-1}A(x,y),\qquad C\bigl(\lambda ^{-1}x\bigr) \leq \varphi (\lambda )^{-1} Cx, \quad \forall \lambda \in (0,1), x,y \in P. \end{aligned}
(10)

By $$A(h,h),Ch \in P_{h}$$, (3)–(4), (6) hold. By deduction, for any $$x, y \in P_{h}$$, we can get

\begin{aligned}& A(x,y)\leq A\bigl(c^{-1}_{1}h,c_{2}h\bigr) \leq A\bigl(c^{-1}h,ch\bigr)\leq c^{-1}A(h,h) \leq c^{-1}\overline{v_{1}}h, \\& A(x,y)\geq A\bigl(c_{1}h,c^{-1}_{2}h\bigr) \geq A\bigl(ch,c^{-1}h\bigr)\geq cA(h,h)\geq c \overline{u_{1}}h, \\& Cx\leq C\bigl(c^{-1}_{1}h\bigr)\leq C \bigl(c^{-1}h\bigr)\leq \varphi (c)^{-1}Ch\leq \varphi ( \lambda )^{-1}\overline{v_{3}}h, \\& Cx\geq C(c_{1}h)\geq C(ch)\geq \varphi (\lambda )Ch\geq \varphi ( \lambda )\overline{u_{3}}h, \end{aligned}

where $$c=\min \{c_{1},c_{2}\}$$. It shows that $$A:P_{h}\times P_{h}\rightarrow P_{h}$$, $$C:P_{h}\rightarrow P_{h}$$. With $$B:P_{h}\times P_{h}\rightarrow P_{h}$$, $$D:P_{h}\rightarrow P_{h}$$, $$T:P_{h}\times P_{h}\rightarrow P_{h}$$ holds. Clearly, $$T(h,h)\in P_{h}$$.

Secondly, we prove that there exists $$\eta (t,x,y)\in (t,1]$$ such that

\begin{aligned} T\bigl(tx,t^{-1}y\bigr)\geq \eta (t,x,y)T(x,y),\quad \forall t \in (0,1), x,y \in P_{h}. \end{aligned}
(11)

By $$A,B:P_{h}\times P_{h}\rightarrow P_{h}$$, $$C,D:P_{h}\rightarrow P_{h}$$, we get $$\lambda _{1}A(x,y)+\lambda _{2}B(x,y)+\lambda _{4}Dy\sim \lambda _{3}Cx$$. Set

$$J^{*}(x,y)=K \biggl( \frac{\lambda _{1}A(x,y)+\lambda _{2}B(x,y)+\lambda _{4}Dy}{\lambda _{3}Cx} \biggr).$$

Since ($$A_{10}$$), $$J^{*}(x,y)\leq \delta$$. Considering a similar function $$h^{*}(s)=\frac{\varphi (t)+J(x,y)t}{(J^{*}(x,y)+1)s}$$ of Theorem 3.1, $$J^{*}(x,y)=\frac{\varphi (t)-\eta (t,x,y)}{\eta (t,x,y)-t}$$ holds by Theorem 3.1. Thus, there is

$$\lambda _{1}A(x,y)+\lambda _{2}B(x,y)+\lambda _{4}Dy\leq \frac{\varphi (t)-\eta (t,x,y)}{\eta (t,x,y)-t}\lambda _{3}Cx,\quad \forall t\in (0,1), x,y \in P_{h}.$$

Combining with (5) and (9), for $$\forall t\in (0,1)$$, $$x,y \in P_{h}$$, we obtain

\begin{aligned} T\bigl(tx,t^{-1}y\bigr) =&\lambda _{1}A \bigl(tx,t^{-1}y\bigr)+\lambda _{2}B\bigl(tx,t^{-1}y \bigr)+ \lambda _{3}C(tx)+\lambda _{4}D \bigl(t^{-1}y\bigr) \\ \geq &t\lambda _{1}A(x,y)+t\lambda _{2}B(x,y)+\varphi (t)\lambda _{3}Cx+t \lambda _{4}Dy \\ =&\eta (t,x,y)\lambda _{3}Cx+\bigl(\varphi (t)-\eta (t,x,y)\bigr) \lambda _{3}Cx \\ &{}+t\bigl[\lambda _{1}A(x,y)+\lambda _{2}B(x,y)+ \lambda _{4}Dy\bigr] \\ =&\eta (t,x,y)\bigl[\lambda _{1}A(x,y)+\lambda _{2}B(x,y)+\lambda _{3}Cx+ \lambda _{4}Dy \bigr] \\ =&\eta (t,x,y)T(x,y). \end{aligned}

The next steps are the same as those of Theorem 3.1, then we can get conclusions $$(C1)$$$$(C4)$$ of Theorem 3.1. □

### Corollary 3.4

There are four operatorsA, B, C, D, where$$A, B: P\times P \rightarrow P$$are two mixed monotone operators, $$C: P \rightarrow P$$is an increasing operator, and$$D: P \rightarrow P$$is a decreasing operator. Then, if conditions ($$A_{2}$$)($$A_{4}$$) and ($$A_{10}$$) of Theorem 3.1and the following conditions are satisfied:

($$A_{11}$$):

There exists$$\gamma \in (0,1)$$such that

\begin{aligned}& A\bigl(\lambda x,\lambda ^{-1}y\bigr)\geq \lambda A(x,y), \qquad C( \lambda x)\geq \lambda ^{\gamma }Cx, \\& D\bigl(\lambda ^{-1}y\bigr)\geq \lambda Dy,\quad \forall \lambda \in (0,1), x,y \in P; \end{aligned}

then we can get conclusions$$(C1)$$$$(C4)$$of Theorem 3.1.

### Corollary 3.5

There are four operatorsA, B, C, D, where$$A, B: P_{h}\times P_{h} \rightarrow P_{h}$$are two mixed monotone operators, $$C:P_{h} \rightarrow P_{h}$$is an increasing sub-homogeneous operator, and$$D: P_{h} \rightarrow P_{h}$$is a decreasing operator. Then, if conditions ($$A_{3}$$), ($$A_{8}$$), ($$A_{10}$$) and the following conditions are satisfied:

($$A_{12}$$):

There exists$$\gamma \in (0,1)$$such that

\begin{aligned}& A\bigl(\lambda x,\lambda ^{-1}y\bigr)\geq \lambda A(x,y),\qquad C( \lambda x)\geq \varphi (\lambda )Cx, \\& D\bigl(\lambda ^{-1}y\bigr)\geq \lambda Dy, \quad \forall x,y \in P_{h}; \end{aligned}

then we can get conclusions (C2)(C4) of Theorem 3.1.

## Application

Nonlinear boundary value problems have attracted much attention for their applications in a variety of different areas such as mathematics, physics, biology, and so on. Thereafter there are a lot of interesting and important results, which include the uniqueness, existence, and multiplicity of positive solutions for the differential equation with the two-point, three-point, infinite-point boundary value problems or some integral boundary problems, etc. In the following, we study a class of integro-differential boundary problems for fractional differential equations:

\begin{aligned} \textstyle\begin{cases} D_{0^{+}}^{\alpha } x(t) +\lambda _{1}f(t,x(t),x(t))+\lambda _{2}g(t,x(t),x(t))+ \lambda _{3}\phi (t,x(t))+\lambda _{4}\psi (t,x(t))=0 , \\ x^{(i)}(0) =0,\quad i=0,1,\ldots ,n-2, \\ D_{0^{+}}^{\beta } x(1)= \int ^{\eta }_{0}x(s)\,\mathrm{d}s, \end{cases}\displaystyle \end{aligned}
(12)

where $$D_{0^{+}}^{\alpha }$$, $$D_{0^{+}}^{\beta }$$ is Riemann–Liouville fractional derivative, $$0<\beta <n-1<\alpha \leq n$$ ($$n>1$$), $$\alpha -\beta -1>0$$, $$t\in [0,1]$$, $$\eta \in (0,1]$$.

### Definition 4.1

()

Suppose that $$h\in C[0,1]$$, $$\alpha >0$$. Then the Riemann–Liouville fractional derivative of α order is defined to be

\begin{aligned} D^{\alpha }_{0^{+}}h(t)=\frac{1}{\varGamma (n-\alpha )} \frac{\mathrm{d^{n}}}{\mathrm{d}t^{n}} \int ^{t}_{0}(t-s)^{n-\alpha -1}h(s){ \, \mathrm{d}}s. \end{aligned}

The Riemann–Liouville fractional integral of α order is defined to be

\begin{aligned} I^{\alpha }_{0^{+}}h(t)=\frac{1}{\varGamma (\alpha )} \int ^{t}_{0}(t-s)^{ \alpha -1}h(s)\, \mathrm{d}s, \end{aligned}

where $$n=[\alpha ]+1$$, $$[\alpha ]$$ denotes the integer part of number $$[\alpha ]$$, provided that the right-hand side is pointwise defined on $$(0,1)$$.

### Lemma 4.1

Let$$h\in C[0,1]$$, the functionxis the solution of the following fractional differential equation:

\begin{aligned} \textstyle\begin{cases} D_{0^{+}}^{\alpha }x(t) +h(t)=0 ,\quad n-1< \alpha \leq n , \\ x^{(i)}(0) =0, \quad 0\leq i\leq n-2, \\ D_{0^{+}}^{\beta } x(1)= \int ^{\eta }_{0}x(s)\,\mathrm{d}s, \end{cases}\displaystyle \end{aligned}
(13)

if and only ifxsatisfies

$$x(t)= \int ^{1}_{0}G(t,s)h(s)\,\mathrm{d}s,$$

where

\begin{aligned} G(t, s)= \textstyle\begin{cases} \frac{\varGamma {(\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}-A(t-s)^{\alpha -1}-\varGamma {(\alpha -\beta )}t^{\alpha -1}(\eta -s)^{\alpha }}{A\varGamma (\alpha )}, \quad 0\leq s\leq t\leq 1, s\leq \eta , \\ \frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}-A(t-s)^{\alpha -1}}{A\varGamma (\alpha )}, \quad 0\leq \eta \leq s\leq t\leq 1, \\ \frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}-\varGamma { (\alpha -\beta )}t^{\alpha -1}(\eta -s)^{\alpha }}{A\varGamma (\alpha )}, \quad 0\leq t\leq s\leq \eta \leq 1, \\ \frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{A\varGamma (\alpha )}, \quad 0\leq t\leq s\leq 1, \eta \leq s, \end{cases}\displaystyle \end{aligned}
(14)

where$$A=\varGamma { (\alpha +1)}-\varGamma { (\alpha -\beta )}\eta ^{\alpha }$$.

### Proof

Integrating on both sides of the first formula of (13), we can obtain

\begin{aligned} x(t) =&-I^{\alpha }_{0^{+}}h(t)+c_{1}t^{\alpha -1}+c_{2}t^{\alpha -2}+ \cdots +c_{n}t^{\alpha -n} \\ =&- \int ^{t}_{0}\frac{(t-s)^{\alpha -1}}{\varGamma (\alpha )}h(s)\, \mathrm{d}s+c_{1}t^{ \alpha -1}+c_{2}t^{\alpha -2}+ \cdots +c_{n}t^{\alpha -n}. \end{aligned}

By the boundary condition $$x^{(i)}(0) =0$$ ($$2\leq i\leq n-2$$), we can easily get $$c_{n}=c_{n-1}=\cdots = c_{2}=0$$. Then we can get the solution of equation (13) of the following form:

$$x(t)=-I^{\alpha }_{0^{+}}h(t)+c_{1}t^{\alpha -1}=- \int ^{t}_{0} \frac{(t-s)^{\alpha -1}}{\varGamma (\alpha )}h(s)\, \mathrm{d}s+c_{1}t^{ \alpha -1}.$$
(15)

By the equality $$D_{0^{+}}^{\beta }t^{\alpha -1}= \frac{\varGamma { (\alpha )}}{\varGamma { (\alpha -\beta )}}t^{\alpha -\beta -1}$$, we get

$$D_{0^{+}}^{\beta } x(t)=-I^{\alpha -\beta }_{0^{+}}h(t)+D_{0^{+}}^{ \beta }c_{1}t^{\alpha -\beta -1}=- \int ^{t}_{0} \frac{(t-s)^{\alpha -\beta -1}}{\varGamma (\alpha -\beta )}h(s)\, \mathrm{d}s+c_{1} \frac{\varGamma { (\alpha )}}{\varGamma { (\alpha -\beta )}}t^{\alpha -\beta -1}.$$

Then integrating formula (15) from 0 to η, there is

\begin{aligned} \int ^{\eta }_{0}x(s)\,\mathrm{d}s =&- \int ^{\eta }_{0} \int ^{s}_{0} \frac{(s-\tau )^{\alpha -1}}{\varGamma (\alpha )}h(\tau )\, \mathrm{d}\tau { \,\mathrm{d}}s+ \int ^{\eta }_{0}c_{1}s^{\alpha -1} \,\mathrm{d}s \\ =&- \int ^{\eta }_{0}\frac{(\eta -s)^{\alpha }}{\varGamma (\alpha +1)}h(s){ \, \mathrm{d}}s+c_{1}\frac{\eta ^{\alpha }}{\alpha }. \end{aligned}

By the condition $$D_{0^{+}}^{\beta } x(1)= \int ^{\eta }_{0}x(s)\,\mathrm{d}s$$, we have

$$c_{1}=\frac{1}{A} \biggl\{ \int ^{1}_{0}\alpha (1-s)^{\alpha -\beta -1}h(s){ \,\mathrm{d}}s- \int ^{\eta }_{0} \frac{ \varGamma { (\alpha -\beta )}}{\varGamma { (\alpha )}} (\eta -s)^{ \alpha }h(s)\,\mathrm{d}s \biggr\} ,$$
(16)

where A is $$\varGamma { (\alpha +1)}-\varGamma { (\alpha -\beta )}\eta ^{\alpha }$$. Substituting (16) in (15), we conclude

\begin{aligned} x(t) =&- \int ^{t}_{0}\frac{(t-s)^{\alpha -1}}{\varGamma (\alpha )}h(s){ \, \mathrm{d}}s+c_{1}t^{\alpha -1} \\ =&- \int ^{t}_{0}\frac{(t-s)^{\alpha -1}}{\varGamma (\alpha )}h(s)\, \mathrm{d}s +\frac{\alpha t^{\alpha -1}}{A} \int ^{1}_{0}(1-s)^{\alpha -\beta -1}h(s){ \, \mathrm{d}}s \\ &{}- \frac{t^{\alpha -1}\varGamma { (\alpha -\beta )}}{A\varGamma { (\alpha )}} \int ^{\eta }_{0} (\eta -s)^{\alpha }h(s)\, \mathrm{d}s \\ =& \int ^{1}_{0}G(t,s)h(s)\,\mathrm{d}s, \end{aligned}

where $$G(t, s)$$ is defined as in (14). □

### Lemma 4.2

Let$$G(t, s)$$be as given in (14). If$$\varGamma (\alpha +1) \geq \varGamma (\alpha -\beta )\eta ^{\alpha }$$, then for any$$t, s\in [0,1]$$, we have

\begin{aligned} 0 \leq& \frac{\varGamma (\alpha -\beta )\eta ^{\alpha }[1-(1-s)^{\beta +1}](1-s)^{\alpha -\beta -1}t^{\alpha -1}}{A\varGamma (\alpha )} \\ \leq& G(t,s) \\ \leq& \frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{A\varGamma (\alpha )}. \end{aligned}
(17)

### Proof

When $$0\leq s\leq t\leq 1$$, $$s\leq \eta$$, by $$s\leq \eta$$, $$\alpha \geq 0$$, we observe that

$$(1-s)^{\alpha }>\biggl(1-\frac{s}{\eta }\biggr)^{\alpha }.$$

By $$s\leq t$$, $$\alpha -\beta -1 \geq 0$$, we have

\begin{aligned} t^{\alpha -1}(1-s)^{\alpha -\beta -1}>(t-s)^{\alpha -1}. \end{aligned}
(18)

Thus,

\begin{aligned} G(t, s) =&\frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}-A(t-s)^{\alpha -1}-\varGamma { (\alpha -\beta )}t^{\alpha -1}(\eta -s)^{\alpha }}{A\varGamma (\alpha )} \\ =&\frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}-\varGamma { (\alpha -\beta )}t^{\alpha -1}\eta ^{\alpha }(1-\frac{s}{\eta })^{\alpha }}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )}- \frac{(t-s)^{\alpha -1}}{\varGamma (\alpha )} \\ \geq & \frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}-\varGamma { (\alpha -\beta )}t^{\alpha -1}\eta ^{\alpha }(1-s)^{\alpha }}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )}- \frac{(t-s)^{\alpha -1}}{\varGamma (\alpha )} \\ =& \frac{[\varGamma { (\alpha +1)}-\varGamma { (\alpha -\beta )}\eta ^{\alpha }(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )}- \frac{(t-s)^{\alpha -1}}{\varGamma (\alpha )} \\ =&\frac{t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{\varGamma (\alpha )}+ \frac{\varGamma { (\alpha -\beta )}\eta ^{\alpha }[1-(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }] \varGamma (\alpha )} \\ &{}- \frac{(t-s)^{\alpha -1}}{\varGamma (\alpha )} \\ =&\frac{t^{\alpha -1}(1-s)^{\alpha -\beta -1}-(t-s)^{\alpha -1}}{\varGamma (\alpha )} + \frac{\varGamma { (\alpha -\beta )}\eta ^{\alpha }[1-(1-s)^{\beta +1}]t^{\alpha -1}(1-s) ^{\alpha -\beta -1}}{[\varGamma (\alpha +1) -\varGamma (\alpha -\beta )\eta ^{\alpha }] \varGamma (\alpha )} \\ =&\frac{t^{\alpha -1}(1-s)^{\alpha -\beta -1}-t^{\alpha -1}(1-\frac{s}{t})^{\alpha -1}}{ \varGamma (\alpha )} \\ &{}+ \frac{\varGamma { (\alpha -\beta )}\eta ^{\alpha }[1-(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )} \\ \geq &\frac{\varGamma { (\alpha -\beta )}\eta ^{\alpha }[1-(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )} \\ =&\frac{\varGamma { (\alpha -\beta )}\eta ^{\alpha }[1-(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{A\varGamma (\alpha )} \geq 0, \end{aligned}

and

\begin{aligned} G(t, s) =&\frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}-A(t-s)^{\alpha -1}-\varGamma { (\alpha -\beta )}t^{\alpha -1}(\eta -s)^{\alpha }}{A\varGamma (\alpha )} \\ \leq & \frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{A\varGamma (\alpha )}. \end{aligned}

When $$0\leq \eta \leq s\leq t\leq 1$$, from (18), there is

\begin{aligned} G(t, s) =&\frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}-A(t-s)^{\alpha -1}}{A\varGamma (\alpha )} \\ =&\frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )}- \frac{(t-s)^{\alpha -1}}{\varGamma (\alpha )} \\ \geq & \frac{[\varGamma { (\alpha +1)}-\varGamma { (\alpha -\beta )}\eta ^{\alpha }(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )}- \frac{(t-s)^{\alpha -1}}{\varGamma (\alpha )} \\ =&\frac{t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{\varGamma (\alpha )}+ \frac{\varGamma { (\alpha -\beta )}\eta ^{\alpha }[1-(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )}- \frac{(t-s)^{\alpha -1}}{\varGamma (\alpha )} \\ =&\frac{t^{\alpha -1}(1-s)^{\alpha -\beta -1}-(t-s)^{\alpha -1}}{\varGamma (\alpha )}+ \frac{\varGamma { (\alpha -\beta )}\eta ^{\alpha }[1-(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )} \\ \geq &\frac{\varGamma { (\alpha -\beta )}\eta ^{\alpha }[1-(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )} \\ =&\frac{\varGamma { (\alpha -\beta )}\eta ^{\alpha }[1-(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{A\varGamma (\alpha )} \geq 0, \end{aligned}

and

\begin{aligned} G(t, s) =&\frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}-A(t-s)^{\alpha -1}}{A\varGamma (\alpha )} \\ \leq & \frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{A\varGamma (\alpha )}. \end{aligned}

When $$0\leq t\leq s\leq \eta \leq 1$$,

\begin{aligned} G(t, s) =&\frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}-\varGamma { (\alpha -\beta )}t^{\alpha -1}(\eta -s)^{\alpha }}{A\varGamma (\alpha )} \\ \geq & \frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}-\varGamma { (\alpha -\beta )}t^{\alpha -1}\eta ^{\alpha }(1-s)^{\alpha }}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )} \\ =& \frac{[\varGamma { (\alpha +1)}-\varGamma { (\alpha -\beta )}\eta ^{\alpha }(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )} \\ =&\frac{t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{\varGamma (\alpha )}+ \frac{\varGamma { (\alpha -\beta )}\eta ^{\alpha }[1-(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )} \\ \geq &\frac{\varGamma { (\alpha -\beta )}\eta ^{\alpha }[1-(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{A\varGamma (\alpha )} \geq 0, \end{aligned}

and

\begin{aligned} G(t, s) =&\frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}-\varGamma { (\alpha -\beta )}t^{\alpha -1}(\eta -s)^{\alpha }}{A\varGamma (\alpha )} \\ \leq & \frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{A\varGamma (\alpha )}. \end{aligned}

When $$0\leq t\leq s\leq \eta \leq 1$$,

\begin{aligned} G(t, s) =&\frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{A\varGamma (\alpha )} \\ \geq & \frac{[\varGamma { (\alpha +1)}-\varGamma { (\alpha -\beta )}\eta ^{\alpha }(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{[\varGamma (\alpha +1)-\varGamma (\alpha -\beta )\eta ^{\alpha }]\varGamma (\alpha )} \\ \geq &\frac{\varGamma { (\alpha -\beta )}\eta ^{\alpha }[1-(1-s)^{\beta +1}]t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{A\varGamma (\alpha )} \geq 0. \end{aligned}

It is obvious that

\begin{aligned} G(t, s) \leq &\frac{\varGamma { (\alpha +1)}t^{\alpha -1}(1-s)^{\alpha -\beta -1}}{A\varGamma (\alpha )}. \end{aligned}

Then, for $$\forall t,s\in [0,1]$$, we obtain that $$G(t,s)$$ meets (17). □

### Theorem 4.1

Letf, $$g:[0,1]\times [0,+\infty )\times [0,+\infty )\rightarrow [0,+ \infty )$$, $$\phi ,\psi :[0,1]\times [0,+\infty )\rightarrow [0,+\infty )$$be continuous with$$f(t,0,1), g(t,0,1), \phi (t,0), \psi (t,1) \not \equiv 0$$and satisfy the following conditions:

($$N_{1}$$):

For any fixed$$t\in [0,1]$$, $$y\in [0,+\infty )$$, $$f(t,x,y)$$, $$g(t,x,y)$$, $$\phi (t,x)$$are increasing in$$x\in [0,+\infty )$$; for any fixed$$t\in [0,1]$$, $$x\in [0,+\infty )$$, $$f(t,x,y)$$, $$g(t,x,y)$$, $$\psi (t,y)$$are decreasing in$$y\in [0,+\infty )$$;

($$N_{2}$$):

For$$\forall \lambda \in (0,1)$$, $$t\in [0,1]$$, $$x,y\in [0,+\infty )$$, there exists$$\varphi (\lambda )\in (\lambda ,1]$$such that

\begin{aligned}& f\bigl(t,\lambda x,\lambda ^{-1}y\bigr)\geq \varphi (\lambda )f(t,x,y), \qquad \phi (t, \lambda x)\geq \lambda \phi (t,x), \\& \psi \bigl(t, \lambda ^{-1} y\bigr)\geq \lambda \psi (t,y); \end{aligned}

and for fixed$$t\in [0,1]$$, $$y\in [0,+\infty )$$, $$g(t,\cdot ,y)$$is concave; for fixed$$t\in [0,1]$$, $$x\in [0,+\infty )$$, $$g(t,x,\cdot )$$is convex;

($$N_{3}$$):

Let$$h(t)=t^{\alpha -1}$$, there exists$$c\geq 0$$such that$$g (s,\theta ,lh)\geq cg(s,lh,\theta )$$, $$l\geq 1$$;

($$N_{4}$$):

There exists a constant$$\delta >0$$such that$$[\lambda _{2}g(t,x,y)+\lambda _{3}\phi (t,x)+\lambda _{4}\psi (t,y) ] \leq \delta \lambda _{1}f(t,x,y)$$, $$\forall t\in [0,1]$$, $$x,y \in [0,+ \infty )$$.

Then

$$(S1)$$:

There exist$$u_{0}, v_{0}\in P_{h}$$, $$r\in (0,1)$$such that$$rv_{0}\leq u_{0}< v_{0}$$,

\begin{aligned} u_{0} \leq &\lambda _{1} \int ^{1}_{0}G(t,s)f\bigl(s,u_{0}(s),v_{0}(s) \bigr)\,\mathrm{d}s+ \lambda _{2} \int ^{1}_{0}G(t,s)g\bigl(s,u_{0}(s),v_{0}(s) \bigr)\,\mathrm{d}s \\ &{}+\lambda _{3} \int ^{1}_{0}G(t,s)\phi \bigl(s,u_{0}(s) \bigr)\,\mathrm{d}s +\lambda _{4} \int ^{1}_{0}G(t,s)\psi \bigl(s,v_{0}(s) \bigr)\,\mathrm{d}s \\ \leq &\lambda _{1} \int ^{1}_{0}G(t,s)f\bigl(s,v_{0}(s),u_{0}(s) \bigr)\,\mathrm{d}s + \lambda _{2} \int ^{1}_{0}G(t,s)g\bigl(s,v_{0}(s),u_{0}(s) \bigr)\,\mathrm{d}s \\ &{}+\lambda _{3} \int ^{1}_{0}G(t,s)\phi \bigl(s,v_{0}(s) \bigr)\,\mathrm{d}s +\lambda _{4} \int ^{1}_{0}G(t,s)\psi \bigl(s,u_{0}(s) \bigr)\,\mathrm{d}s\leq v_{0}; \end{aligned}
$$(S2)$$:

Equation (12) has a unique positive solution$$x^{*}$$in$$P_{h}$$, which meets$$\mu _{1} t^{\alpha -1}\leq x^{*}\leq \nu _{1} t^{\alpha -1}$$, $$\mu _{1}$$, $$\nu _{1} >0$$are two constants;

$$(S3)$$:

For any initial values$$x_{0}, y_{0}\in P_{h}$$, we construct successively the iterative sequences

\begin{aligned}& x_{n}(t)=\lambda _{1} \int ^{1}_{0}G(t,s)f\bigl(s,x_{n-1}(s),y_{n-1}(s) \bigr)\,\mathrm{d}s+ \lambda _{2} \int ^{1}_{0}G(t,s)g\bigl(s,x_{n-1}(s),y_{n-1}(s) \bigr)\,\mathrm{d}s \\& \hphantom{x_{n}(t)=}{}+\lambda _{3} \int ^{1}_{0}G(t,s)\phi \bigl(s,x_{n-1}(s) \bigr)\,\mathrm{d}s \\& \hphantom{x_{n}(t)=}{}+\lambda _{4} \int ^{1}_{0}G(t,s)\psi \bigl(s,y_{n-1}(s) \bigr)\,\mathrm{d}s, \quad n=1,2,\ldots , \\& y_{n}(t)=\lambda _{1} \int ^{1}_{0}G(t,s)f\bigl(s,y_{n-1}(s),x_{n-1}(s) \bigr)\,\mathrm{d}s+ \lambda _{2} \int ^{1}_{0}G(t,s)g\bigl(s,y_{n-1}(s),x_{n-1}(s) \bigr)\,\mathrm{d}s \\& \hphantom{y_{n}(t)=}{}+\lambda _{1} \int ^{1}_{0}G(t,s)\phi \bigl(s,y_{n-1}(s) \bigr)\,\mathrm{d}s \\& \hphantom{y_{n}(t)=}{}+\lambda _{4} \int ^{1}_{0}G(t,s)\phi \bigl(s,x_{n-1}(s) \bigr)\,\mathrm{d}s, \quad n=1,2,\ldots . \end{aligned}

Here, $$x_{n}\rightarrow x^{*}$$, $$y_{n}\rightarrow x^{*}$$, when$$n\rightarrow \infty$$.

### Proof

Let $$E = C[0,1]$$ and $$\|x\|=\sup_{0\leq t\leq 1}|x(t)|$$. It is obvious that $$(E,\|\cdot \|)$$ is a Banach space. Set $$P=\{x\in E|x(t)\geq 0, t \in [0,1]\}$$.

By Lemma 4.1, the unique solution of problem (12) has an integral formulation:

$$x(t)= \int ^{1}_{0}G(t,s) \bigl[\lambda _{1}f\bigl(s,x(s),x(s)\bigr)+\lambda _{2}g \bigl(s,x(s),x(s)\bigr)+ \lambda _{3}\phi \bigl(s,x(s)\bigr)+\lambda _{4}\psi \bigl(s,x(s)\bigr) \bigr]\,\mathrm{d}s.$$

Define four operators $$A,B,C,D:P\times P\rightarrow E$$ by

\begin{aligned}& A(x,y) (t)= \int ^{1}_{0}G(t,s)f\bigl(s,x(s),y(s)\bigr) \,\mathrm{d}s, \\& B(x,y) (t)= \int ^{1}_{0}G(t,s)g\bigl(s,x(s),y(s)\bigr){ \, \mathrm{d}}s, \\& Cx(t)= \int ^{1}_{0}G(t,s)\phi \bigl(s,x(s)\bigr)\, \mathrm{d}s, \\& Dy(t)= \int ^{1}_{0}G(t,s) \psi \bigl(s,y(s)\bigr)\, \mathrm{d}s. \end{aligned}

Therefore, x is the solution of equation (12) if and only if x is the solution of operator equation $$x = \lambda _{1}A(x,x)+\lambda _{2}B(x,x)+\lambda _{3}Cx+\lambda _{4}Dx$$.

Firstly, for $$\forall x, y \in P$$, in view of the definition of f, g, ϕ, ψ and Lemma 4.2, it is easy to obtain $$A(x,y)\geq 0$$, $$B(x,y)\geq 0$$, $$Cx\geq 0$$, $$Dy\geq 0$$, which shows that $$A, B:P\times P\rightarrow P$$, $$C,D:P\rightarrow P$$. In addition, due to ($$N_{1}$$), A, B are two monotone operators, C is an increasing operator, and D is a decreasing operator.

Secondly, we illustrate that C is a sub-homogeneous operator and A, D satisfy condition ($$A_{1}$$) of Theorem 3.1. From ($$N_{2}$$), we see that, for any $$\lambda \in (0,1)$$, $$x\in P$$,

\begin{aligned} C(\lambda x) =& \int ^{1}_{0}G(t,s)\phi \bigl(s,\lambda x(s) \bigr)\,\mathrm{d}s \\ \geq& \int ^{1}_{0}G(t,s)\lambda \phi \bigl(s,x(s) \bigr)\,\mathrm{d}s \\ =&\lambda \int ^{1}_{0}G(t,s)\phi \bigl(s,x(s)\bigr)\, \mathrm{d}s \\ =&\lambda Cx, \end{aligned}

which shows that C is a sub-homogeneous operator. Using ($$N_{2}$$) again, $$\forall \lambda \in (0,1)$$, $$x,y\in P$$, there exists $$\varphi (\lambda )\in (\lambda ,1]$$ such that

\begin{aligned} A\bigl(\lambda x,\lambda ^{-1}y\bigr) =& \int ^{1}_{0}G(t,s)f\bigl(s,\lambda x(s), \lambda ^{-1}y(s)\bigr)\,\mathrm{d}s \\ \geq & \int ^{1}_{0}G(t,s)\varphi (\lambda )f \bigl(s,x(s),y(s)\bigr)\,\mathrm{d}s \\ =&\varphi (\lambda ) \int ^{1}_{0}G(t,s)f\bigl(s,x(s),y(s)\bigr)\, \mathrm{d}s \\ =& \varphi (\lambda )A( x,y), \end{aligned}

and for any $$\lambda \in (0,1)$$, $$y\in P$$,

\begin{aligned} D\bigl(\lambda ^{-1} y\bigr) =& \int ^{1}_{0}G(t,s)\psi \bigl(s,\lambda ^{-1} y(s)\bigr){ \,\mathrm{d}}s \\ \geq& \int ^{1}_{0}G(t,s)\lambda \psi \bigl(s,y(s) \bigr)\,\mathrm{d}s \\ =&\lambda \int ^{1}_{0}G(t,s)\psi \bigl(s,y(s)\bigr)\, \mathrm{d}s \\ =&\lambda Dy. \end{aligned}

Thus, operator A, D satisfies ($$A_{1}$$) of Theorem 3.1.

Thirdly, we verified condition ($$A_{2}$$) of Theorem 3.1. From ($$N_{2}$$), for fixed $$t\in (0,1)$$, $$y\in P$$,

\begin{aligned}& B\bigl(\tau x_{1}+(1-\tau )x_{2},y\bigr) (t) \\& \quad = \int ^{1}_{0}G(t,s)g\bigl(s,\tau x_{1}(s)+(1- \tau )x_{2}(s),y(s)\bigr)\,\mathrm{d}s \\& \quad \geq \int ^{1}_{0}G(t,s)\bigl[\tau g\bigl(s, x_{1}(s),y(s)\bigr) +(1-\tau ) g\bigl(s, x_{2}(s),y(s) \bigr) \bigr]{\,\mathrm{d}}s \\& \quad = \tau \int ^{1}_{0}G(t,s) g\bigl(s, x_{1}(s),y(s)\bigr)\,\mathrm{d}s+ (1-\tau ) \int ^{1}_{0}G(t,s) g\bigl(s, x_{2}(s),y(s)\bigr)\,\mathrm{d}s \\& \quad = \tau B( x_{1},y)+ (1-\tau ) B( x_{2},y), \quad \forall \tau \in (0,1), x_{1},x_{2} \in P; \end{aligned}

for fixed $$t\in (0,1)$$, $$x\in P$$,

\begin{aligned}& B\bigl(x,\tau y_{1}+(1-\tau )y_{2}\bigr) (t) \\& \quad = \int ^{1}_{0}G(t,s)g\bigl(s,x(s),\tau y_{1}(s)+(1- \tau )y_{2}(s)\bigr)\,\mathrm{d}s \\& \quad \leq \int ^{1}_{0}G(t,s)\bigl[\tau g\bigl(s, x(s),y_{1}(s)\bigr) +(1-\tau ) g\bigl(s, x(s),y_{2}(s) \bigr)\bigr]{ \,\mathrm{d}}s \\& \quad = \tau \int ^{1}_{0}G(t,s) g\bigl(s, x(s),y_{1}(s)\bigr)\,\mathrm{d}s+ (1-\tau ) \int ^{1}_{0}G(t,s) g\bigl(s, x(s),y_{2}(s)\bigr)\,\mathrm{d}s \\& \quad = \tau B( x,y_{1})+ (1-\tau ) B( x,y_{2}), \quad \forall \tau \in (0,1), y_{1},y_{2} \in P, \end{aligned}

which indicate that, for fixed $$y\in P, B(\cdot ,y)$$ is concave; for fixed $$x\in P, B(x,\cdot )$$ is convex. That is condition ($$A_{2}$$) of Theorem 3.1.

Fourthly, we check condition ($$A_{3}$$) of Theorem 3.1. By ($$N_{3}$$), there exists $$\widetilde{c}\geq 0$$ such that

\begin{aligned} B(\theta ,lh) (t) =& \int ^{1}_{0}G(t,s)g\bigl(s,\theta ,lh(s)\bigr) \,\mathrm{d}s \\ \geq & \widetilde{c} \int ^{1}_{0}G(t,s)g\bigl(s,lh(s),\theta \bigr) \,\mathrm{d}s \\ =& \widetilde{c } B(lh,\theta ) (t). \end{aligned}

In the next step, we prove that $$A(h,h),B(h,h),Ch,Dh\in P_{h}$$. By Lemma 4.2, we have

\begin{aligned} A(h,h) (t) =& \int ^{1}_{0}G(t,s)f\bigl(s,s^{\alpha -1},s^{\alpha -1} \bigr)\,\mathrm{d}s \\ \geq & \frac{\varGamma (\alpha -\beta )\eta ^{\alpha }}{A\varGamma (\alpha )} \int ^{1}_{0}\bigl[1-(1-s)^{\beta +1} \bigr](1-s)^{\alpha -\beta -1}f\bigl(s,s^{ \alpha -1},s^{\alpha -1}\bigr)\, \mathrm{d}s \\ \geq & \biggl\{ \frac{\varGamma (\alpha -\beta )\eta ^{\alpha }}{A\varGamma (\alpha )} \int ^{1}_{0}\bigl[1-(1-s)^{\beta +1} \bigr](1-s)^{\alpha -\beta -1}f(s,0,1){ \,\mathrm{d}}s \biggr\} h(t), \end{aligned}

and

\begin{aligned} A(h,h) (t) =& \int ^{1}_{0}G(t,s)f\bigl(s,s^{\alpha -1},s^{\alpha -1} \bigr)\,\mathrm{d}s \\ \leq & \frac{\varGamma { (\alpha +1)}t^{\alpha -1}}{A\varGamma (\alpha )} \int ^{1}_{0}(1-s)^{\alpha -\beta -1}f \bigl(s,s^{\alpha -1},s^{\alpha -1}\bigr){ \,\mathrm{d}}s \\ \leq & \biggl\{ \frac{\varGamma { (\alpha +1)}}{A\varGamma (\alpha )} \int ^{1}_{0}(1-s)^{ \alpha -\beta -1}f(s,1,0)\, \mathrm{d}s \biggr\} h(t), \end{aligned}

where

\begin{aligned}& c_{1}= \frac{\varGamma (\alpha -\beta )\eta ^{\alpha }}{A\varGamma (\alpha )} \int ^{1}_{0}\bigl[1-(1-s)^{ \beta +1} \bigr](1-s)^{\alpha -\beta -1}f(s,0,1)\,\mathrm{d}s, \\& c_{2}= \frac{\varGamma { (\alpha +1)}}{A\varGamma (\alpha )} \int ^{1}_{0}(1-s)^{ \alpha -\beta -1}f(s,1,0)\, \mathrm{d}s. \end{aligned}

Thus, we can get $$c_{1}h(t) \leq A(h,h)(t) \leq c_{2}h(t)$$. Besides, due to $$f(s,1,0)\geq f(s,0,1)\geq 0$$ and $$f(s,0,1)\not \equiv 0$$, we derive that

$$\int ^{1}_{0}f(s,1,0)\,\mathrm{d}s \geq \int ^{1}_{0}f(s,0,1)\,\mathrm{d}s >0,$$

thus, $$A(h,h)\in P_{h}$$. Similarly, there are

\begin{aligned}& \biggl\{ \frac{\varGamma (\alpha -\beta )\eta ^{\alpha }}{A\varGamma (\alpha )} \int ^{1}_{0}\bigl[1-(1-s)^{\beta +1} \bigr](1-s)^{\alpha -\beta -1}g(s,0,1){ \,\mathrm{d}}s \biggr\} h(t) \\& \quad \leq B(h,h) (t)\leq \biggl\{ \frac{\varGamma { (\alpha +1)}}{A\varGamma (\alpha )} \int ^{1}_{0}(1-s)^{ \alpha -\beta -1}g(s,1,0)\, \mathrm{d}s \biggr\} h(t), \\& \biggl\{ \frac{\varGamma (\alpha -\beta )\eta ^{\alpha }}{A\varGamma (\alpha )} \int ^{1}_{0}\bigl[1-(1-s)^{ \beta +1} \bigr](1-s)^{\alpha -\beta -1}\phi (s,0)\,\mathrm{d}s \biggr\} h(t) \\& \quad \leq Ch(t)\leq \biggl\{ \frac{\varGamma { (\alpha +1)}}{A\varGamma (\alpha )} \int ^{1}_{0}(1-s)^{\alpha -\beta -1}\phi (s,1)\, \mathrm{d}s \biggr\} h(t), \end{aligned}

and

\begin{aligned}& \biggl\{ \frac{\varGamma (\alpha -\beta )\eta ^{\alpha }}{A\varGamma (\alpha )} \int ^{1}_{0}\bigl[1-(1-s)^{\beta +1} \bigr](1-s)^{\alpha -\beta -1}\psi (s,1){ \,\mathrm{d}}s \biggr\} h(t) \\& \quad \leq Dh(t)\leq \biggl\{ \frac{\varGamma { (\alpha +1)}}{A\varGamma (\alpha )} \int ^{1}_{0}(1-s)^{\alpha -\beta -1}\psi (s,0)\, \mathrm{d}s \biggr\} h(t). \end{aligned}

Then, by $$g(s,0,1),\phi (s,0),\psi (s,1)\not \equiv 0$$, we can obtain $$B(h,h),Ch,Dh\in P_{h}$$.

Eventually, we test condition $$(A_{5})$$ of Theorem 3.1. From $$(N_{5})$$, for any $$t\in [0, 1]$$, $$x, y \in P$$,

\begin{aligned}& \bigl[\lambda _{2}B(x,y) (t)+\lambda _{3}Cx(t)+\lambda _{4}Dy(t)\bigr] \\& \quad = \int ^{1}_{0}G(t,s)\bigl[\lambda _{2}g\bigl(s,x(s),y(s)\bigr)+\lambda _{3}\phi \bigl(s,x(s)\bigr)+ \lambda _{4}\psi \bigl(s,y(s)\bigr)\bigr]{\, \mathrm{d}}s \\& \quad \leq \delta \int ^{1}_{0}G(t,s)\lambda _{1}f \bigl(s,x(s),y(s)\bigr)\,\mathrm{d}s \\& \quad =\delta \lambda _{1}A(x,y) (t),\quad n=1,2,\ldots . \end{aligned}

From the above six steps, we verified all the conditions of Theorem 3.1, thus the conclusions of Theorem 4.1 hold with Theorem 3.1. □

### Theorem 4.2

Letf, $$g:[0,1]\times [0,+\infty )\times [0,+\infty )\rightarrow [0,+ \infty )$$, $$\phi :[0,1]\times [0,+\infty )\rightarrow [0,+\infty )$$, and$$\psi :[0,1]\times [0,+\infty )\rightarrow [0,+\infty )$$are continuous with$$f(t,0,1), g(t,0,1), \phi (t,0), \psi (t,1) \not \equiv 0$$. Then, if conditions ($$N_{1}$$), ($$N_{4}$$) and the following conditions are satisfied:

($$N_{5}$$):

For$$\forall \lambda \in (0,1)$$, $$t \in [0,1]$$, $$x,y\in [0,+\infty )$$, there exists$$\varphi (\lambda )\in (\lambda ,1]$$such that

\begin{aligned}& f\bigl(t,\lambda x,\lambda ^{-1}y\bigr)\geq \lambda f(t,x,y),\qquad \phi (t,\lambda x) \geq \varphi (\lambda ) \phi (t,x), \\& \psi \bigl(t,\lambda ^{-1} y\bigr)\geq \lambda ^{-1} \psi (t,y); \end{aligned}

and for fixed$$t\in [0,1]$$, $$y\in [0,+\infty )$$, $$g(t,\cdot ,y)$$is concave; for fixed$$t\in [0,1]$$, $$x\in [0,+\infty )$$, $$g(t,x,\cdot )$$is convex;

($$N_{6}$$):

There exists a constant$$\delta >0$$such that$$\lambda _{1}f(t,x,y)+\lambda _{2}g(t,x,y)+\lambda _{4}\psi (t,y) \leq \delta \lambda _{3}\phi (t,x)$$, $$\forall t\in [0,1]$$, $$x,y \in [0,+ \infty )$$;

then we can get conclusions$$(S1)$$$$(S3)$$of Theorem 4.1.

### Proof

The proof process is similar to that of Theorem 4.1. □

### Example 4.1

Consider the following problem:

\begin{aligned} \textstyle\begin{cases} D_{0^{+}}^{\frac{13}{2}}x(t) +2(x+3)^{\frac{1}{4}}+3(x+2)^{ \frac{1}{4}}+3(x+4)^{-\frac{1}{6}}+4(x+5)^{-\frac{1}{6}} \\ \quad {}-\frac{22}{5}e^{-x}+4e^{-y}+4t+2t^{2}+3t^{3}+4t^{4}+6=0,\quad 0< t< 1, \\ x^{i}(0) =0, \quad 0\leq i\leq 4, \qquad D_{0^{+}}^{\frac{5}{3}}x(1)=\int ^{ \frac{3}{4}}_{0}x(s)\,\mathrm{d}s. \end{cases}\displaystyle \end{aligned}

Here, $$\alpha =\frac{13}{2}\in (5,6)$$, $$\beta =\frac{5}{3}\in (1,2)$$, $$\alpha - \beta -1=\frac{23}{6}\geq 0$$, $$\eta =\frac{3}{4}$$, $$\lambda _{1}=1$$, $$\lambda _{2}=2$$, $$\lambda _{3}=3$$, $$\lambda _{4}=4$$, and

\begin{aligned}& f(t,x,y)=2(x+3)^{\frac{1}{4}}+3(y+4)^{-\frac{1}{6}}+6+4t, \\& g(t,x,y)=- \frac{11}{5}e^{-x}+2e^{-y}+3+t^{2}, \\& \phi (t,x)=(x+2)^{\frac{1}{4}}+t^{3}, \\& \psi (t,y)=(y+5)^{- \frac{1}{6}}+t^{4}. \end{aligned}

Thus, $$\varGamma (\alpha +1)- \varGamma (\alpha -\beta )\eta ^{\alpha }=\varGamma ( \frac{15}{2})-\varGamma (\frac{29}{6})\frac{3}{4}^{\frac{13}{2}}=1.868 \times 10^{-3}$$, $$f(t,0,1)=2\cdot 3^{\frac{1}{4}}+3\cdot 5^{-\frac{1}{5}}+6+4t\not \equiv 0$$, $$g(t,0,1)=-\frac{11}{5}+2e^{-1}+t^{2}\not \equiv 0$$, $$\phi (t,0)=2^{ \frac{1}{4}}+t^{3}\not \equiv 0$$, $$\psi (t,1)=3\times 6^{-\frac{1}{6}}+t^{4} \not \equiv 0$$.

For $$\forall \lambda \in (0,1)$$, $$t \in [0,1]$$, $$x,y\in [0,+\infty )$$, there exists $$\gamma =\frac{1}{4}\in (0,1)$$, we can get that

\begin{aligned} f\bigl(t,\lambda x,\lambda ^{-1} y\bigr) =&2(\lambda x+3)^{\frac{1}{4}}+3\bigl( \lambda ^{-1}y+4\bigr)^{-\frac{1}{6}}+6+4t \\ \geq & \lambda ^{\frac{1}{4}}2(x+3)^{\frac{1}{4}}+\lambda ^{ \frac{1}{6}}3(y+4)^{-\frac{1}{5}}+6+4t \\ \geq & \lambda ^{\frac{1}{4}}\bigl(2(x+3)^{\frac{1}{4}}+3(y+4)^{- \frac{1}{6}}+6+4t \bigr) \\ =&\lambda ^{\frac{1}{4}}f(t,x,y). \end{aligned}

For $$\forall t\in [0,1]$$, $$x,y\in [0,+\infty )$$, there is

$$g_{xx}^{\prime \prime }(t,x,y)=-\frac{11}{5}e^{-x} \leq 0, \qquad g_{yy}^{\prime \prime }(t,x,y)=2e^{-y} \geq 0,$$

we obtain that, for fixed $$t\in [0,1]$$, $$y\in [0,+\infty )$$, $$g(t,\cdot ,y)$$ is concave; for fixed $$t\in [0,1]$$, $$x\in [0,+\infty )$$, $$g(t,x,\cdot )$$ is convex. For $$\forall \lambda , t\in [0,1]$$, $$x,y\in [0,+\infty )$$, we have

\begin{aligned} \phi (t,\lambda x) =&(\lambda x+2)^{\frac{1}{4}}+t^{3}\geq \lambda ^{ \frac{1}{4}}( x+2)^{\frac{1}{4}}+t^{3} \\ \geq& \lambda \bigl((\lambda x+2)^{\frac{1}{4}}+t^{3}\bigr)= \lambda \phi (t, x) \end{aligned}

and

\begin{aligned} \psi \bigl(t,\lambda ^{-1} y\bigr)=\bigl(\lambda ^{-1}y+5 \bigr)^{-\frac{1}{6}}+t^{4}\geq \lambda \bigl((y+5)^{-\frac{1}{6}}+t^{4} \bigr)=\lambda \psi (t, y). \end{aligned}

Let $$l^{\prime }$$ be a sufficiently large constant and $$x, y\leq l^{\prime }$$, $$\widetilde{c}=\frac{1}{25}$$, there is

$$g\bigl(t,\theta ,l^{\prime }\bigr)=-\frac{11}{5}+2e^{-l^{\prime }}+3+t^{2} \geq \frac{1}{25}\biggl(- \frac{11}{5}e^{-l^{\prime }}+2+3+t^{2} \biggr)= \widetilde{c}g\bigl(t,l^{\prime },\theta \bigr).$$

Let $$\delta = 2$$, by calculation, we have

\begin{aligned}& \lambda _{2}g(t,x,y)+\lambda _{3}\phi (t,x)+\lambda _{4}\psi (t,y) \\& \quad =2\biggl[-\frac{11}{5}e^{-x}+2e^{-y}+3+t^{2} \biggr]+3\bigl[(x+2)^{\frac{1}{4}}+t^{3}\bigr]+4 \bigl[(y+5)^{- \frac{1}{6}}+t^{4}\bigr] \\& \quad \leq 3(x+2)^{\frac{1}{4}}+4(y+5)^{-\frac{1}{6}}+6- \frac{22}{5}e^{-x}+4e^{-y}+2t^{2}+3t^{3}+4t^{4} \\& \quad \leq 4(x+3)^{\frac{1}{4}}+6(y+4)^{-\frac{1}{6}}+12+8t \\& \quad \leq 2\bigl[2(x+2)^{\frac{1}{4}}+3(y+4)^{-\frac{1}{6}}+6+4t\bigr] \\& \quad = \delta \lambda _{1}f(t,x,y). \end{aligned}

Therefore, all the assumptions of Theorem 4.1 are satisfied, then Example 4.1 has a unique positive solution $$x^{*}\in P_{h}$$, where $$h(t)=t^{\frac{11}{2}}$$, $$t \in [0,1]$$. Besides, the other condition of Theorem 4.1 holds.

### Example 4.2

Consider the following problem:

\begin{aligned} \textstyle\begin{cases} D_{0^{+}}^{\frac{7}{3}}x(t) +(x+3)^{\frac{1}{4}}+2(x+2)^{\frac{1}{4}}+(y+4)^{- \frac{1}{5}}+3(y+5)^{-\frac{1}{6}} \\ \quad {}-\frac{11}{5}e^{-x}+2e^{-y}+4t+t^{2}+t^{3}+t^{4}+6=0,\quad 0< t< 1, \\ x^{i}(0) =0,\quad 0\leq i\leq 4,\qquad D_{0^{+}}^{\frac{1}{4}}x(1)=\int ^{ \frac{3}{4}}_{0}x(s)\,\mathrm{d}s. \end{cases}\displaystyle \end{aligned}

Here, $$\alpha =\frac{7}{3}\in (2,3)$$, $$\beta =\frac{1}{4}\in (0,1)$$, $$\alpha - \beta -1=\frac{13}{12}\geq 0$$, $$\eta =\frac{1}{2}$$, $$\lambda _{1}=2$$, $$\lambda _{2}=1$$, $$\lambda _{3}=4$$, $$\lambda _{4}=3$$, and

\begin{aligned}& f(t,x,y)=2(x+1)^{\frac{1}{2}}+\frac{1}{1+y}+\sin t, \\& g(t,x,y)=- \frac{3}{4}e^{-x}+\frac{12}{5}e^{-y}+3+ \cos t, \\& \phi (t,x)=(x+4)^{\frac{1}{2}}+2, \\& \psi (t,y)=\frac{1}{1+y}+t^{2}. \end{aligned}

Therefore, $$\varGamma (\alpha +1)- \varGamma (\alpha -\beta )\eta ^{\alpha }=\varGamma ( \frac{10}{3})-\varGamma (\frac{25}{12})\frac{1}{2}^{\frac{7}{3}}=2.5722$$, $$f(t,0,1)=2^{\frac{1}{2}}+\frac{1}{2}+\sin t$$, $$g(t,0,1)=-\frac{3}{4}+ \frac{12}{5}e^{-1}+\cos t$$, $$\phi (t,0)=4^{\frac{1}{2}}+2$$, $$\psi (t,1)= \frac{1}{2}+t^{2} \not \equiv 0$$.

For $$\forall \lambda \in(0,1)$$, $$t \in [0,1]$$, $$x,y\in [0,+\infty )$$, we can derive that

\begin{aligned} f\bigl(t,\lambda x,\lambda ^{-1} y\bigr) =&2(\lambda x+1)^{\frac{1}{2}}+ \frac{1}{1+y}+\sin t \\ \geq & \lambda ^{\frac{1}{2}}2(x+1)^{\frac{1}{2}}+\lambda \frac{1}{1+y}+\sin t \\ \geq & \lambda \biggl(2(x+1)^{\frac{1}{2}}+\frac{1}{1+y}+\sin t \biggr) \\ =&\lambda f(t,x,y). \end{aligned}

For expression of g, there is

$$g_{xx}^{\prime \prime }(t,x,y)=-\frac{3}{4}e^{-x} \leq 0,\qquad g_{yy}^{\prime \prime }(t,x,y)= \frac{12}{5}e^{-y} \geq 0,$$

i.e., $$g(t,\cdot ,y)$$ is concave, $$g(t,x,\cdot )$$ is convex. Further, there exist $$\gamma =\frac{1}{2}\in (0,1)$$,

\begin{aligned} \phi (t,\lambda x) =&(\lambda x+4)^{\frac{1}{2}}+2\geq \lambda ^{ \frac{1}{2}}\bigl(( x+2)^{\frac{1}{4}}+2\bigr)=\lambda ^{\frac{1}{2}} \phi (t, x), \end{aligned}

and

\begin{aligned} \psi \bigl(t,\lambda ^{-1} y\bigr) =&\frac{1}{1+\lambda ^{-1}y}+t^{2} \geq \lambda \biggl(\frac{1}{1+y}+t^{2}\biggr)=\lambda \psi (t, y). \end{aligned}

Let $$l^{\prime }$$ be a sufficiently large constant with $$x, y\leq l^{\prime }$$. $$c=\frac{1}{9}$$, we can obtain

$$g\bigl(t,\theta ,l^{\prime }\bigr)=-\frac{3}{4}+ \frac{12}{5}e^{-l^{\prime }}+3+\cos t\geq \frac{1}{9}\biggl(- \frac{3}{4}e^{-l^{\prime }}+\frac{12}{5}+3+\cos t\biggr)= cg \bigl(t,l^{\prime }, \theta \bigr).$$

Let $$\delta = 2.5$$, we get

\begin{aligned}& \lambda _{1}f(t,x,y)+\lambda _{2}g(t,x,y)+\lambda _{4}\psi (t,y) \\& \quad =2\biggl[2(x+1)^{\frac{1}{2}}+\frac{1}{1+y}+\sin t\biggr]+ \biggl[-\frac{3}{4}e^{-x}+ \frac{12}{5}e^{-y}+3+ \cos t\biggr] +3\biggl[\frac{1}{1+y}+t^{2}\biggr] \\& \quad =4(x+1)^{\frac{1}{2}}+\frac{5}{1+y}+2\sin t- \frac{3}{4}e^{-x}+ \frac{12}{5}e^{-y}+3+ \cos t+3t^{2} \\& \quad \leq 4(x+4)^{\frac{1}{2}}+17 \\& \quad \leq 2.5\times 4\bigl[(x+4)^{\frac{1}{2}}+2\bigr]=\delta \lambda _{3}\phi (t,x). \end{aligned}

Then Example 4.2 has a unique positive solution $$x^{*}\in P_{h}$$, where $$h(t)=t^{\frac{4}{3}}$$, $$t \in [0,1]$$.

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### Acknowledgements

First and foremost, we would like to express our deepest gratitude to our supervisor, Dr. Zhang Lingling, who has provided us with valuable help in the writing of this paper. Then we shall extend many thanks to Mr. Zhou Bibo, Miss Tian Huimin for all their kindness and help. Last but not least, we would like to thank the opening project. This paper is supported by the opening project of State Key Laboratory of Explosion Science and Technology. This paper is also supported by Key Research and Development program of Shanxi Province.

Not applicable.

## Funding

The project is the opening project of State Key Laboratory of Explosion Science and Technology (Beijing Institute of Technology). The opening project number is KFJJ19-06M. Another project is Key Research and Development program of Shanxi Province (International Cooperation). The project number is 201903D421042.

## Author information

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### Contributions

NZ and LZ carried out the concepts, design, definition of content, literature search, and manuscript preparation. NZ drafted the manuscript. BZ and HT performed manuscript review. All authors have read and approved the content of the manuscript.

### Corresponding author

Correspondence to Lingling Zhang.

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### Competing interests

The authors declare that they have no competing interests. 