Positive Solutions for a Three-Point Boundary Value Problem of Fractional Q-Difference Equations

Abstract

In this work, a three-point boundary value problem of fractional q-difference equations is discussed. By using fixed point theorems on mixed monotone operators, some sufficient conditions that guarantee the existence and uniqueness of positive solutions are given. In addition, an iterative scheme can be made to approximate the unique solution. Finally, some interesting examples are provided to illustrate the main results.

1. Introduction

We will deal with a fractional q-difference equation subject to three-point boundary conditions

$$\left\{\begin{array}{c}{D}_q^{\alpha }x\left(t\right)+f(t,x\left(t\right),x\left(t\right))+g(t,x\left(t\right))=0,0<t<1,2<\alpha <3,\hfill \\ x\left(0\right)=D_{q}x\left(0\right)=0,D_{q}x\left(1\right)=\beta D_{q}x\left(\eta \right),\hfill \end{array}\right.$$

(1)

where $0<\beta {\eta }^{\alpha -2}<1,0<q<1,$ $D_q^{\alpha }$ is the Riemann–Liouville fractional q-derivative of order $\alpha$.

Due to fast development in fractional calculus, many researchers studied q-difference calculus or quantum calculus. For this topic, the earlier results can be seen in Al-Salam [1] and Agarwal [2], and some recent results related to q-difference calculus in [3,4,5,6,7,8,9,10,11,12,13,14,15] and some references therein. Nowadays, fractional q-difference calculus has been given in wide applications of different science areas, which include basic hyper-geometric functions, mechanics, the theory of relativity, combinatorics and discrete mathematics. So many mathematical models have been abstracted out(see [16,17,18]) and problem (1) is one of the models. Therefore, fractional q-difference calculus has been of great interest and many good results can be found in [5,6,7,8] and references therein. Recently, the fruits about fractional q-difference equation boundary value problems emerge continuously. For different problems of fractional q-difference equations, the existence and the uniqueness of solutions have been always considered in literature. To solve these boundary value problems, some techniques have been applied, such as the monotone iterative technique, the lower-upper solution method, the Schauder fixed point theorem and the Krasnoselskii fixed point theorem. For details, one can see [13,14,15,19,20,21,22,23,24,25].

In [15], Liang and Zhang considered the existence and uniqueness of positive nondecreasing solutions for a fractional q-difference equation involving three-point boundary conditions

$$\left\{\begin{array}{c}{D}_q^{\alpha }x\left(t\right)+f(t,x\left(t\right))=0,2<\alpha <3,0<t<1,\hfill \\ x\left(0\right)=D_{q}x\left(0\right)=0,D_{q}x\left(1\right)=\beta D_{q}x\left(\eta \right),\hfill \end{array}\right.$$

(2)

where $0<\beta {\eta }^{\alpha -2}<1.$ They gave some sufficient conditions for Label (2), and their tool is a fixed point theorem in partially ordered sets.

In [19], Sriphanomwan et al. investigated the problem of fractional q-difference equations

$$\left\{\begin{array}{c}{D}_q^{\alpha }\left(D_q^{\beta }(1+p\left(t\right))\right)x\left(t\right)=f(t,x\left(t\right),D_{\theta }^{\mu }x\left(t\right),{\mathrm{\Psi }}_{\omega }^{v}x\left(t\right)),\hfill \\ x\left(0\right)=x\left(\eta \right),I_r^{\gamma }x\left(T\right)={\int }_0^T\frac{{(T-rs)}^{(\gamma -1)}}{{\mathrm{\Gamma }}_r\left(\gamma \right)}x\left(s\right)d_{r}s=g\left(x\right),\hfill \end{array}\right.$$

(3)

where $t\in I_{\chi }^T:=\{{\chi }^{k}T:k\in \mathbf{N}\cup \{0,T\}\},0<\alpha ,\beta ,\mu \le 1,1<\alpha +\beta \le 2,v,\gamma >0,\eta \in I_{\chi }^T-\{0,T\},$ and $p,q,r,\theta ,\omega$ are simple fractions. The existence and uniqueness of solutions for Label (3) was obtained. The used methods are the Banach contraction mapping principle and Krasnosel’skii fixed point theorem.

By using Schauder fixed point theorem and the Banach fixed point theorem, Yang [25] discussed a fractional q-difference equation with three-point boundary conditions:

$$\left\{\begin{array}{c}{D}_q^{\alpha }x\left(t\right)+f(t,x\left(t\right))=0,0\le t\le 1,1<\alpha \le 2,\hfill \\ x\left(0\right)=0,x\left(1\right)=\beta x\left(\xi \right),\hfill \end{array}\right.$$

(4)

where $0<\beta {\xi }^{\alpha -1}<1,0<\xi <1.$ The author gave the existence and uniqueness of positive solutions for Label (4).

In a very recent paper [24], the authors considered a special fractional q-difference equation with a three-point problem

$$\left\{\begin{array}{c}{D}_q^{\alpha }u\left(t\right)+f(t,u\left(t\right))=b,0<t<1,2<\alpha <3,\hfill \\ u\left(0\right)=D_{q}u\left(0\right)=0,D_{q}u\left(1\right)=\beta D_{q}u\left(\eta \right),\hfill \end{array}\right.$$

(5)

where $0<\beta {\eta }^{\alpha -2}<1,0<q<1,b\ge 0$ is a constant. The existence and uniqueness of solutions for Label (5) by using fixed point theorems for $\psi -(h,r)-$concave operators.

Motivated by [15,26], we consider the existence and uniqueness of positive solutions for Label (1). Different from the methods mentioned above, our tools are two fixed point theorems for mixed monotone operators. To the authors’ knowledge, Label (1) is a new form of fractional q-difference equations. We can give the existence and uniqueness of solutions for Label (1). Furthermore, we can make an iteration to approximate the unique solution.

2. Preliminaries

Here, we list some concepts and lemmas of fractional q-calculus. One can see [1,2,3,4,5,6,7,8], for example.

For $0<q<1$ and f defined on $[a,b],$ let

$$\left(I_{q}f\right)\left(t\right)={\int }_0^{t}f\left(s\right)d_{q}s=(1-q)\sum _{n=0}^{\infty }f\left(t{q}^n\right)t{q}^n,t\in [0,b].$$

Then,

$${\int }_a^{b}f\left(t\right)d_{q}t={\int }_a^{c}f\left(t\right)d_{q}t+{\int }_c^{b}f\left(t\right)d_{q}t,\forall c\in [a,b].$$

Definition 1.

(See [3]). $\alpha \ge 0$ and f is defined on $[0,1].$ The Riemann–Liouville fractional q-integral is $\left(I_q^{0}f\right)\left(t\right)=f\left(t\right)$ and

$$\left(I_q^{\alpha }f\right)\left(t\right)=\frac{1}{{\mathrm{\Gamma }}_q\left(\alpha \right)}{\int }_0^t{(t-qs)}^{(\alpha -1)}f\left(s\right)d_{q}s,\alpha >0.$$

Clearly, $\left(I_q^{\alpha }f\right)\left(t\right)=\left(I_{q}f\right)\left(t\right)$ when $\alpha =1.$

Lemma 1.

(See [22]). If $f,g$ are continuous on $[0,s]$ and $f\left(t\right)\le g\left(t\right)$ for $t\in [0,s],$ then

(i) 

${\int }_0^{s}f\left(t\right)d_{q}t\le {\int }_0^{s}g\left(t\right)d_{q}t.$ In addition, if $\alpha >1,$ then $I_q^{\alpha }f\left(s\right)\le I_q^{\alpha }g\left(s\right)$, $t\in [0,s],$

(ii) 

$\left|{\int }_0^{s}f\left(t\right)d_{q}t\right|\le {\int }_0^s\left|f\left(t\right)\right|d_{q}t$, $t\in [0,s].$

Definition 2.

(See [3]). The Riemann–Liouville fractional q-derivative of order $\alpha \ge 0$ is

$$\left(D_q^{\alpha }f\right)\left(t\right)=\left(D_q^n{I}_q^{n-\alpha }f\right)\left(t\right),\alpha >0,t\in [0,1],$$

where n denotes the smallest integer greater than or equal to α.

When $\alpha =1,$ $\left(D_q^{\alpha }f\right)\left(t\right)=D_{q}f\left(t\right).$ Furthermore,

$$\left(I_q^{\alpha }{D}_q^{p}f\right)\left(t\right)=\left(D_q^p{I}_q^{\alpha }f\right)\left(t\right)-\sum _{n=0}^{p-1}\frac{t^{\alpha -p+n}}{{\mathrm{\Gamma }}_q(\alpha -p+n+1)}\left(D_q^{n}f\right)\left(0\right),p\in \mathbf{N}.$$

Lemma 2.

If $f\left(t\right)$ is continuous with $f\left(t\right)\ge 0$ for $t\in [0,1],$ and there is $t_0\in (0,1)$ such that $f\left(t_0\right)\ne 0.$ Then,

$${\int }_0^{1}f\left(t\right)d_{q}t>0,t\in [0,1],$$

where

$${\int }_0^{1}f\left(t\right)d_{q}t=(1-q)\sum _{n=0}^{\infty }{q}^{n}f\left(q^n\right),q\in (0,1).$$

Proof. 

Because $f\left(t\right)\ge 0$ and $f\left(t_0\right)\ne 0,$ there is $n_0\in \mathbf{N}$ such that $t_0=q^{n_0},$ then

$$f\left(q^{n_0}\right)q^{n_0}>0,0<q<1,$$

and thus

$$(1-q)\sum _{n=0}^{\infty }{q}^{n}f\left(q^n\right)\ge (1-q)f\left(q^{n_0}\right)q^{n_0}=(1-q)f\left(t_0\right)t_0>0.$$

Hence, we have ${\int }_0^{1}f\left(t\right)d_{q}t>0.$ ☐

Here, we list other facts that are important in the sequel. See [26,27,28,29,30] for instance.

$(X,\parallel ·\parallel )$ is a real Banach space, its partial order induced by a cone K of $X,$ i.e., $x\le y$ if and only if $y-x\in K.$ If there is $N>0$ such that $\parallel x\parallel \le N\parallel y\parallel$ for $\theta \le x\le y,x,y\in X,$ then K is called normal, where $\theta$ denotes the zero element of X. The notation x–y denotes that there exist $\mu ,\nu >0$ such that $\mu x\le y\le \nu x,\forall x,y\in X$. For fixed $h>\theta ,$ define a set $K_h=\{x\in E\mid x\sim h\}.$ Then, $K_h\subset K$.

Definition 3.

(See [27]). Suppose $T:K\to K$ is a given operator. If

$$T\left(tx\right)\ge tTx,\forall t\in (0,1),x\in K,\left(6\right)$$

then T is said to be sub-homogeneous.

Definition 4.

(See [27]). Let $0\le \gamma <1.$ An operator $T:K\to K$ satisfies

$$T\left(tx\right)\ge t^{\gamma }Tx,\forall t\in (0,1),x\in K.$$

(7)

Then, T is said to be γ-concave.

Lemma 3.

(See [27]). Let $h>\theta ,0<\gamma <1$, $T_1:K\times K\to K$ be a mixed monotone operator and

$$T_1(tx,t^{-1}y)\ge t^{\gamma }{T}_1(x,y),\forall t\in (0,1),x,y\in K.$$

(8)

$T_2:K\to K$ is an increasing sub-homogeneous operator. Moreover,

(i) 

there exists $h_0\in K_h$ such that $T_1(h_0,h_0),T_2{h}_0\in K_h$;

(ii) 

there exists $\sigma >0$ such that $T_1(x,y)\ge \sigma T_{2}x,x,y\in K$.

Then:

(a) 

$T_1:K_h\times K_h\to K_h$ and $T_2:K_h\to K_h;$

(b) 

there are $u_0,v_0\in K_h$ and $\tau \in (0,1)$ satisfying

$$\tau v_0\le u_0<v_0,u_0\le T_1(u_0,v_0)+T_2{u}_0\le T_1(v_0,u_0)+T_2{v}_0\le v_0;$$

(c) 

$T_1(x,x)+T_{2}x=x$ exists a unique solution $x^{*}$ in $K_h;$

(d) 

for $x_0,y_0\in K_h,$ set

$$x_n=T_1(x_{n-1},y_{n-1})+T_2{x}_{n-1},y_n=T_1(y_{n-1},x_{n-1})+T_2{y}_{n-1},n=1,2,\dots ,$$

then $x_n\to x^{*},y_n\to x^{*}$ as $n\to \infty .$

Lemma 4.

(See [27]). Let $h>\theta ,0<\gamma <1$, $T_1:K\times K\to K$ be a mixed monotone operator and

$$T_1(tx,t^{-1}y)\ge t{T}_1(x,y),\forall t\in (0,1),x,y\in K.$$

(9)

$T_2:K\to K$ is an increasing γ-concave operator. Moreover,

(i) 

there exists $h_0\in K_h$ such that $T_1(h_0,h_0),T_2{h}_0\in K_h$;

(ii) 

there exists $\sigma >0$ such that $T_1(x,y)\le \sigma T_{2}x,x,y\in K$.

Then:

(a) 

$T_1:K_h\times K_h\to K_h$ and $T_2:K_h\to K_h;$

(b) 

there are $u_0,v_0\in K_h$ and $\tau \in (0,1)$ satisfying

$$\tau v_0\le u_0<v_0,u_0\le T_1(u_0,v_0)+T_2{u}_0\le T_1(v_0,u_0)+T_2{v}_0\le v_0;$$

(c) 

$T_1(x,x)+T_{2}x=x$ exists a unique solution $x^{*}$ in $K_h;$

(d) 

for $x_0,y_0\in K_h,$ set

$$x_n=T_1(x_{n-1},y_{n-1})+T_2{x}_{n-1},y_n=T_1(y_{n-1},x_{n-1})+T_2{y}_{n-1},n=1,2,\dots ,$$

then $x_n\to x^{*},y_n\to x^{*}$ as $n\to \infty .$

Remark 1.

From Lemmas 3 and 4, we have two special cases:

(i) 

Let $T_2=\theta$ in Lemma 3, we get the corresponding conclusion (see Corollary 2.2 in [27]);

(ii) 

Let $T_1=\theta$ in Lemma 4, we have the corresponding conclusion (see Theorem 2.7 in [31]).

3. Main Results

By using Lemmas 3 and 4, we will establish our main results for Label (1). Consider a Banach space $X=C[0,1]$, the norm is $\parallel u\parallel =\sup \left\{\right|u\left(t\right)|:t\in [0,1\left]\right\}.$ Set $K=\{x\in C[0,1\left]\right|x\left(t\right)\ge 0,t\in [0,1]\},$ a normal cone.

Lemma 5.

(See [15]). Let $g\in C[0,1],\beta {\eta }^{\alpha -2}\ne 1$ and $0<\eta <1$, then the unique solution of following three-point problem

$$\left\{\begin{array}{c}{D}_q^{\alpha }x\left(t\right)+g\left(t\right)=0,0<t<1,2<\alpha <3,\hfill \\ x\left(0\right)=D_{q}x\left(0\right)=0,D_{q}x\left(1\right)=\beta D_{q}x\left(\eta \right)\hfill \end{array}\right.$$

(10)

is

$$x\left(t\right)={\int }_0^{1}G(t,qs)g\left(s\right)d_{q}s+\frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)g\left(s\right)d_{q}s,$$

(11)

where

$$G(t,s)=\frac{1}{{\mathrm{\Gamma }}_q\left(\alpha \right)}\left\{\begin{array}{c}{(1-s)}^{(\alpha -2)}{t}^{\alpha -1}-{(t-s)}^{(\alpha -1)},0\le s\le t\le 1,\hfill \\ {(1-s)}^{(\alpha -2)}{t}^{\alpha -1},0\le t\le s\le 1,\hfill \end{array}\right.$$

(12)

$$\begin{array}{cc}\hfill H(t,s)& ={}_t{D}_{q}G(s,t)\hfill \\ & =\frac{{[\alpha -1]}_q}{{\mathrm{\Gamma }}_q\left(\alpha \right)}\left\{\begin{array}{c}{(1-s)}^{(\alpha -2)}{t}^{\alpha -2}-{(t-s)}^{(\alpha -2)},0\le s\le t\le 1,\hfill \\ {(1-s)}^{(\alpha -2)}{t}^{\alpha -2},0\le t\le s\le 1.\hfill \end{array}\right.\hfill \end{array}$$

Lemma 6.

(See [15]). For $G(t,qs)$ in (11), we obtain

(1) 

$G(t,qs)$ is continuous and $G(t,qs)\ge 0,t,s\in [0,1]\times [0,1];$

(2) 

$G(t,qs)$ is strictly increasing in $t\in [0,1]$.

Remark 2.

For $G(t,qs)$ in (11), we can easily get

$$G(t,qs)\le \frac{1}{{\mathrm{\Gamma }}_q\left(\alpha \right)}{(1-qs)}^{(\alpha -2)}{t}^{\alpha -1},t,s\in [0,1]\times [0,1].$$

By (2) in Lemma 6, we have ${}_t{D}_{q}G(qs,t)\ge 0,$ that is, $H(t,qs)\ge 0.$ Obviously,

$$H(t,qs)\le \frac{{[\alpha -1]}_q}{{\mathrm{\Gamma }}_q\left(\alpha \right)}{(1-qs)}^{(\alpha -2)}{t}^{\alpha -2}\le \frac{{[\alpha -1]}_q}{{\mathrm{\Gamma }}_q\left(\alpha \right)},t,s\in [0,1]\times [0,1].$$

Next, four assumptions are listed:

($H_1$) 

$f:[0,1]\times [0,+\infty )\times [0,+\infty )\to [0,+\infty )$ and $g:[0,1]\times [0,+\infty )\to [0,+\infty )$ are continuous;

($H_2$) 

$f(t,u,v)$ is increasing relative to u for fixed $t\in [0,1]$ and $v\in [0,+\infty ),$ decreasing relative to v for fixed $t\in [0,1]$ and $u\in [0,+\infty );$ $g(t,u)$ is increasing relative to u for fixed $t\in [0,1];$

($H_3$) 

for $\lambda \in (0,1),t\in [0,1],u\ge 0,$ $g(t,\lambda u)\ge \lambda g(t,u)$ is satisfied, and there is $\gamma \in (0,1)$ such that $f(t,\lambda u,{\lambda }^{-1}v)\ge {\lambda }^{\gamma }f(t,u,v)$ for $u,v\ge 0.$ In addition, $g(t,0)\not\equiv 0$;

($H_4$) 

there exists $\sigma >0$ such that $f(t,u,v)\ge \sigma g(t,u),\forall t\in [0,1],u,v\in [0,+\infty ).$

Theorem 1.

Let $\left(H_1\right)-\left(H_4\right)$ be satisfied, then

(a) 

there are $u_0,v_0\in K_h$ and $\tau \in (0,1)$ satisfying $\tau v_0\le u_0<v_0$ and

$$\begin{array}{ccc}\hfill u_0\left(t\right)& \le & \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,u_0\left(s\right),v_0\left(s\right))+g(s,u_0\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,u_0\left(s\right),v_0\left(s\right))+g(s,u_0\left(s\right))]d_{q}s,t\in [0,1],\hfill \end{array}$$

$$\begin{array}{ccc}\hfill v_0\left(t\right)& \ge & \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,v_0\left(s\right),u_0\left(s\right))+g(s,v_0\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,v_0\left(s\right),u_0\left(s\right))+g(s,v_0\left(s\right))]d_{q}s,t\in [0,1],\hfill \end{array}$$

where $h\left(t\right)=t^{\alpha -1}$ and $G(t,qs),H(t,qs)$ are defined as in Lemma 5;

(b) 

BVP (1) has a unique positive solution $u^{*}\in K_h;$

(c) 

for $x_0,y_0\in K_h,$ set

$$\begin{array}{ccc}\hfill x_{n+1}\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,x_n\left(s\right),y_n\left(s\right))+g(s,x_n\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,x_n\left(s\right),y_n\left(s\right))+g(s,x_n\left(s\right))]d_{q}s,n=1,2,\dots ,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill y_{n+1}\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,y_n\left(s\right),x_n\left(s\right))+g(s,y_n\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,y_n\left(s\right),x_n\left(s\right))+g(s,y_n\left(s\right))]d_{q}s,n=1,2,\dots ,\hfill \end{array}$$

then $\parallel x_n-u^{*}\parallel \to 0,\parallel y_n-u^{*}\parallel \to 0$ as $n\to \infty .$

Proof. 

By Lemma 5, the solution u of BVP (1) can be written by

$$\begin{array}{ccc}\hfill u\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,u\left(s\right),u\left(s\right))+g(s,u\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,u\left(s\right),u\left(s\right))+g(s,u\left(s\right))]d_{q}s.\hfill \end{array}$$

Now, we give two operators $T_1:K\times K\to X$ and $T_2:K\to X$ by

$$\begin{array}{ccc}\hfill T_1(u,v)\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,u\left(s\right),v\left(s\right))d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)f(s,u\left(s\right),v\left(s\right))d_{q}s,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \left(T_{2}u\right)\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)g(s,u\left(s\right))d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)g(s,u\left(s\right))d_{q}s.\hfill \end{array}$$

Obviously, u is a solution of Label (1) if and only if $u=T_1(u,u)+T_{2}u.$ By $\left(H_1\right),$ one has $T_1:K\times K\to K$ and $T_2:K\to K.$ We will prove that $T_1,T_2$ satisfy all the assumptions of Lemma 3. The proof consists of three steps.

Step 1. The aim of this step is to prove that $T_1$ is a mixed monotone operator.

For $u_i,v_i\in K,i=1,2$ with $u_1\ge u_2,v_1\le v_2,$ then $u_1\left(t\right)\ge u_2\left(t\right),v_1\left(t\right)\le v_2\left(t\right)$ for $t\in [0,1].$ From $\left(H_2\right)$ and Lemma 6,

$$\begin{array}{ccc}\hfill T_1(u_1,v_1)\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,u_1\left(s\right),v_1\left(s\right))d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)f(s,u_1\left(s\right),v_1\left(s\right))d_{q}s\hfill \\ & \ge & \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,u_2\left(s\right),v_2\left(s\right))d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)f(s,u_2\left(s\right),v_2\left(s\right))d_{q}s\hfill \\ & =& T_1(u_2,v_2)\left(t\right).\hfill \end{array}$$

Thus, $T_1(u_1,v_1)\ge T_1(u_2,v_2),$ that is, $T_1$ is mixed monotone.

Step 2. Our aim of this step is to show that $T_1$ satisfies the condition (8) and the operator $T_2$ is sub-homogeneous.

From $\left(H_2\right)$ and Lemma 6, $T_2$ is increasing. Furthermore, for $\lambda \in (0,1)$ and $u,v\in P,$ by $\left(H_3\right)$,

$$\begin{array}{ccc}\hfill T_1(\lambda u,{\lambda }^{-1}v)\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,\lambda u\left(s\right),{\lambda }^{-1}v\left(s\right))d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)f(s,\lambda u\left(s\right),{\lambda }^{-1}v\left(s\right))d_{q}s\hfill \\ & \ge & \frac{{\lambda }^{\gamma }\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,u\left(s\right),v\left(s\right))d_{q}s\hfill \\ & & +{\lambda }^{\gamma }{\int }_0^{1}G(t,qs)f(s,u_2\left(s\right),v_2\left(s\right))d_{q}s\hfill \\ & =& {\lambda }^{\gamma }{T}_1(u,v)\left(t\right),\hfill \end{array}$$

and thus $T_1(\lambda u,{\lambda }^{-1}v)\ge {\lambda }^{\gamma }{T}_1(u,v)$ for $\lambda \in (0,1),u,v\in K.$ Hence, the operator $T_1$ satisfies (8). In addition, for any $\lambda \in (0,1),u\in K,$ by $\left(H_3\right)$,

$$\begin{array}{ccc}\hfill T_2\left(\lambda u\right)\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)g(s,\lambda u\left(s\right))d_{q}s+{\int }_0^{1}G(t,qs)g(s,\lambda u\left(s\right))d_{q}s\hfill \\ & \ge & \frac{\lambda \beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)g(s,u\left(s\right))d_{q}s+\lambda {\int }_0^{1}G(t,qs)g(s,u\left(s\right))d_{q}s\hfill \\ & =& \lambda T_{2}u\left(t\right),\hfill \end{array}$$

that is, $T_2\left(\lambda u\right)\ge \lambda T_{2}u,u\in P.$ Thus, the operator $T_2$ is sub-homogeneous.

Step 3. The purpose of this step is to prove that $T_1(h,h),T_{2}h\in K_h.$ Furthermore, we also prove that there exists $\sigma >0$ such that $T_1(x,y)\ge \sigma T_{2}x,\forall x,y\in K.$

Firstly, in view of $\left(H_1\right),\left(H_2\right)$ and Lemma 6, for $t\in [0,1],$

$$\begin{array}{ccc}\hfill T_1(h,h)\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,h\left(s\right),h\left(s\right))d_{q}s+{\int }_0^{1}G(t,qs)f(s,h\left(s\right),h\left(s\right))d_{q}s\hfill \\ & =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,s^{\alpha -1},s^{\alpha -1})d_{q}s+{\int }_0^{1}G(t,qs)f(s,s^{\alpha -1},s^{\alpha -1})d_{q}s\hfill \\ & \le & \frac{\beta h\left(t\right)}{(1-\beta {\eta }^{\alpha -2}){\mathrm{\Gamma }}_q\left(\alpha \right)}{\int }_0^1{(1-qs)}^{(\alpha -2)}f(s,1,0)d_{q}s+\frac{h\left(t\right)}{{\mathrm{\Gamma }}_q\left(\alpha \right)}{\int }_0^1{(1-qs)}^{(\alpha -2)}f(s,1,0)d_{q}s.\hfill \end{array}$$

By the same arguments, for $t\in [0,1],$

$$\begin{array}{ccc}\hfill T_1(h,h)\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,s^{\alpha -1},s^{\alpha -1})d_{q}s+{\int }_0^{1}G(t,qs)f(s,s^{\alpha -1},s^{\alpha -1})d_{q}s\hfill \\ & \ge & \frac{h\left(t\right)}{{\mathrm{\Gamma }}_q\left(\alpha \right)}{\int }_0^1[{(1-qs)}^{(\alpha -2)}-{(1-qs)}^{(\alpha -1)}]f(s,0,1)d_{q}s.\hfill \end{array}$$

From $\left(H_2\right),\left(H_4\right),$

$${\int }_0^{1}f(s,1,0)d_{q}s\ge {\int }_0^{1}f(s,0,1)d_{q}s\ge \sigma {\int }_0^{1}g(s,0)d_{q}s>0.$$

Set

$$l_1=\left(\frac{1}{{\mathrm{\Gamma }}_q\left(\alpha \right)}+\frac{\beta }{(1-\beta {\eta }^{\alpha -2}){\mathrm{\Gamma }}_q\left(\alpha \right)}\right){\int }_0^1{(1-qs)}^{(\alpha -2)}f(s,1,0)d_{q}s,$$

$$l_2=\frac{1}{{\mathrm{\Gamma }}_q\left(\alpha \right)}{\int }_0^1\left[{(1-qs)}^{(\alpha -2)}-{(1-qs)}^{(\alpha -1)}\right]f(s,0,1)d_{q}s.$$

Then, $l_{2}h\left(t\right)\le T_1(h,h)\left(t\right)\le l_{1}h\left(t\right),t\in [0,1].$ It follows that $T_1(h,h)\in K_h.$ Similarly,

$$T_{2}h\left(t\right)\ge \frac{h\left(t\right)}{{\mathrm{\Gamma }}_q\left(\alpha \right)}{\int }_0^1\left[{(1-qs)}^{(\alpha -2)}-{(1-qs)}^{(\alpha -1)}\right]g(s,0)d_{q}s,$$

and

$$T_{2}h\left(t\right)\le \left(\frac{1}{{\mathrm{\Gamma }}_q\left(\alpha \right)}+\frac{\beta }{(1-\beta {\eta }^{\alpha -2}){\mathrm{\Gamma }}_q\left(\alpha \right)}\right)h\left(t\right){\int }_0^1{(1-qs)}^{(\alpha -2)}g(s,1)d_{q}s.$$

Since $g(t,0)\not\equiv 0,$ we also get $T_{2}h\in K_h.$ Thus, the condition $\left(i\right)$ of Lemma 3 holds. Next, we will indicate that $\left(ii\right)$ of Lemma 3 is still satisfied. For $t\in [0,1],u,v\in K,$ from $\left(H_4\right),$

$$\begin{array}{ccc}\hfill T_1(u,v)\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,u\left(s\right),v\left(s\right))d_{q}s+{\int }_0^{1}G(t,qs)f(s,u\left(s\right),v\left(s\right))d_{q}s\hfill \\ & \ge & \frac{\sigma \beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)g(s,u\left(s\right))d_{q}s+\sigma {\int }_0^{1}G(t,qs)g(s,u\left(s\right))d_{q}s\hfill \\ & =& \sigma T_{2}u\left(t\right).\hfill \end{array}$$

Then, $T_1(u,v)\ge \sigma T_{2}u$ for $u,v\in K.$ Therefore, by Lemma 3, we have: $u_0,v_0\in K_h$ and $\tau \in (0,1)$ satisfying $\tau v_0\le u_0<v_0,u_0\le T_1(u_0,v_0)+T_2{u}_0\le T_1(v_0,u_0)+T_2{v}_0\le v_0;$ the equation $T_1(u,u)+T_{2}u=u$ has a unique solution $u^{*}$ in $K_h;$ for $x_0,y_0\in K_h,$ set

$$x_n=T_1(x_{n-1},y_{n-1})+T_2{x}_{n-1},y_n=T_1(y_{n-1},x_{n-1})+T_2{y}_{n-1},n=1,2,\dots ,$$

one obtains $x_n\to u^{*},y_n\to u^{*}$ as $n\to \infty .$ Namely,

$$\begin{array}{ccc}\hfill u_0\left(t\right)& \le & \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,u_0\left(s\right),v_0\left(s\right))+g(s,u_0\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,u_0\left(s\right),v_0\left(s\right))+g(s,u_0\left(s\right))]d_{q}s,t\in [0,1],\hfill \end{array}$$

$$\begin{array}{ccc}\hfill v_0\left(t\right)& \ge & \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,v_0\left(s\right),u_0\left(s\right))+g(s,v_0\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,v_0\left(s\right),u_0\left(s\right))+g(s,v_0\left(s\right))]d_{q}s,t\in [0,1];\hfill \end{array}$$

Label (1) has a unique positive solution $u^{*}\in K_h;$ for $x_0,y_0\in K_h,$ the sequences

$$\begin{array}{ccc}\hfill x_{n+1}\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,x_n\left(s\right),y_n\left(s\right))+g(s,x_n\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,x_n\left(s\right),y_n\left(s\right))+g(s,x_n\left(s\right))]d_{q}s,n=1,2,\dots ,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill y_{n+1}\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,y_n\left(s\right),x_n\left(s\right))+g(s,y_n\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,y_n\left(s\right),x_n\left(s\right))+g(s,y_n\left(s\right))]d_{q}s,n=1,2,\dots \hfill \end{array}$$

satisfy $\parallel x_n-u^{*}\parallel \to 0,\parallel y_n-u^{*}\parallel \to 0$ as $n\to \infty .$ ☐

Theorem 2.

Let $\left(H_1\right),\left(H_2\right)$ and the following conditions be satisfied:

$\left(H_5\right)$

for $t\in [0,1],\lambda \in (0,1),u\ge 0$, there is $\gamma \in (0,1)$ such that $g(t,\lambda u)\ge {\lambda }^{\gamma }g(t,u)$ and $f(t,\lambda u,{\lambda }^{-1}v)\ge \lambda f(t,u,v)$ for $t\in [0,1],\lambda \in (0,1),u,v\ge 0;$

$\left(H_6\right)$

$f(t,0,1)\not\equiv 0$ for $t\in [0,1],$ and there is $\sigma >0$ satisfying $f(t,u,v)\le \sigma g(t,u),\forall t\in [0,1],u,v\ge 0.$

Then:

(a) 

there is $u_0,v_0\in P_h$ and $\tau \in (0,1)$ such that $\tau v_0\le u_0<v_0$ and

$$\begin{array}{ccc}\hfill u_0\left(t\right)& \le & \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,u_0\left(s\right),v_0\left(s\right))+g(s,u_0\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,u_0\left(s\right),v_0\left(s\right))+g(s,u_0\left(s\right))]d_{q}s,t\in [0,1],\hfill \end{array}$$

$$\begin{array}{ccc}\hfill v_0\left(t\right)& \ge & \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,v_0\left(s\right),u_0\left(s\right))+g(s,v_0\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,v_0\left(s\right),u_0\left(s\right))+g(s,v_0\left(s\right))]d_{q}s,t\in [0,1],\hfill \end{array}$$

where $h\left(t\right)=t^{\alpha -1}$ and $G(t,qs),H(t,qs)$ are defined as in Lemma 5;

(b) 

BVP (1) has a unique positive solution $u^{*}\in K_h;$

(c) 

for any $x_0,y_0\in K_h,$ set

$$\begin{array}{ccc}\hfill x_{n+1}\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,x_n\left(s\right),y_n\left(s\right))+g(s,x_n\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,x_n\left(s\right),y_n\left(s\right))+g(s,x_n\left(s\right))]d_{q}s,n=1,2,\dots ,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill y_{n+1}\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,y_n\left(s\right),x_n\left(s\right))+g(s,y_n\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,y_n\left(s\right),x_n\left(s\right))+g(s,y_n\left(s\right))]d_{q}s,n=1,2,\dots ,\hfill \end{array}$$

and we get $\parallel x_n-u^{*}\parallel \to 0,\parallel y_n-u^{*}\parallel \to 0$ as $n\to \infty .$

Proof. 

We also consider two operators $T_1,T_2.$ Given in the proof of Theorem 1, it has been shown that $T_1:K\times K\to K$ is mixed monotone and $T_2:K\to K$ is increasing. By $\left(H_5\right)$,

$$T_1(\lambda u,{\lambda }^{-1}v)\ge \lambda T_1(u,v),T_2\left(\lambda u\right)\ge {\lambda }^{\gamma }{T}_{2}u,\lambda \in (0,1),u,v\in K.$$

From $\left(H_2\right),\left(H_6\right),$

$$g(s,0)\ge \frac{1}{\sigma }f(s,0,1),f(s,1,0)\ge f(s,0,1),s\in [0,1].$$

Since $f(t,0,1)\not\equiv 0$, we obtain

$${\int }_0^{1}f(s,1,0)d_{q}s\ge {\int }_0^{1}f(s,0,1)d_{q}s>0,{\int }_0^{1}g(s,1)d_{q}s\ge {\int }_0^{1}g(s,0)d_{q}s\ge \frac{1}{\sigma }{\int }_0^{1}f(s,0,1)d_{q}s>0,$$

so

$$\begin{array}{c}\left(\frac{1}{{\mathrm{\Gamma }}_q\left(\alpha \right)}+\frac{\beta }{(1-\beta {\eta }^{\alpha -2}){\mathrm{\Gamma }}_q\left(\alpha \right)}\right){\int }_0^1{(1-qs)}^{(\alpha -2)}f(s,1,0)d_{q}s\\ \ge \frac{1}{{\mathrm{\Gamma }}_q\left(\alpha \right)}{\int }_0^1\left[{(1-qs)}^{(\alpha -2)}-{(1-qs)}^{(\alpha -1)}\right]f(s,0,1)d_{q}s>0,\end{array}$$

and

$$\begin{array}{c}\left(\frac{1}{{\mathrm{\Gamma }}_q\left(\alpha \right)}+\frac{\beta }{(1-\beta {\eta }^{\alpha -2}){\mathrm{\Gamma }}_q\left(\alpha \right)}\right){\int }_0^1{(1-qs)}^{(\alpha -2)}g(s,1)d_{q}s\\ \ge \frac{1}{{\mathrm{\Gamma }}_q\left(\alpha \right)}{\int }_0^1\left[{(1-qs)}^{(\alpha -2)}-{(1-qs)}^{(\alpha -1)}\right]g(s,0)d_{q}s>0.\end{array}$$

It can easily prove that $T_1(h,h),T_{2}h\in K_h.$ Furthermore, by $\left(H_6\right),$

$$\begin{array}{ccc}\hfill T_1(u,v)\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,u\left(s\right),v\left(s\right))d_{q}s+{\int }_0^{1}G(t,qs)f(s,u\left(s\right),v\left(s\right))d_{q}s\hfill \\ & \le & \frac{\sigma \beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)g(s,u\left(s\right))d_{q}s+\sigma {\int }_0^{1}G(t,qs)g(s,u\left(s\right))d_{q}s\hfill \\ & =& \sigma T_{2}u\left(t\right).\hfill \end{array}$$

Hence, $T_1(u,v)\le T_{2}u,$ for $u,v\in K.$ By Lemma 4, we can claim: there are $u_0,v_0\in P_h$ and $\tau \in (0,1)$ satisfying $\tau v_0\le u_0<v_0,u_0\le T_1(u_0,v_0)+T_2{u}_0\le T_1(v_0,u_0)+T_2{v}_0\le v_0;$ the equation $T_1(u,u)+T_{2}u=u$ has a unique solution $u^{*}$ in $K_h;$ for $x_0,y_0\in K_h,$ set

$$x_n=T_1(x_{n-1},y_{n-1})+T_2{x}_{n-1},y_n=T_1(y_{n-1},x_{n-1})+T_2{y}_{n-1},n=1,2,\dots ,$$

one has $x_n\to u^{*},y_n\to u^{*}$ as $n\to \infty .$ Namely,

$$\begin{array}{ccc}\hfill u_0\left(t\right)& \le & \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,u_0\left(s\right),v_0\left(s\right))+g(s,u_0\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,u_0\left(s\right),v_0\left(s\right))+g(s,u_0\left(s\right))]d_{q}s,t\in [0,1],\hfill \end{array}$$

$$\begin{array}{ccc}\hfill v_0\left(t\right)& \ge & \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,v_0\left(s\right),u_0\left(s\right))+g(s,v_0\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,v_0\left(s\right),u_0\left(s\right))+g(s,v_0\left(s\right))]d_{q}s,t\in [0,1];\hfill \end{array}$$

Label (1) has a unique positive solution $u^{*}\in K_h;$ for $x_0,y_0\in P_h,$ the sequences

$$\begin{array}{ccc}\hfill x_{n+1}\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,x_n\left(s\right),y_n\left(s\right))+g(s,x_n\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,x_n\left(s\right),y_n\left(s\right))+g(s,x_n\left(s\right))]d_{q}s,n=1,2,\dots ,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill y_{n+1}\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)[f(s,y_n\left(s\right),x_n\left(s\right))+g(s,y_n\left(s\right))]d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)[f(s,y_n\left(s\right),x_n\left(s\right))+g(s,y_n\left(s\right))]d_{q}s,n=1,2,\dots \hfill \end{array}$$

satisfy $\parallel x_n-u^{*}\parallel \to 0,\parallel y_n-u^{*}\parallel \to 0$ as $n\to \infty .$ ☐

In the sequel, we consider special cases of Label (1) with $g\equiv 0$ or $f\equiv 0.$ Similar to the proofs of Theorems 1 and 2 and according to Remark 1, we can draw the following conclusions:

Corollary 1.

Assume f satisfies $\left(H_1\right)-\left(H_4\right)$ and $f(t,0,1)\not\equiv 0,$ for $t\in [0,1].$ Then: $\left(a\right)$ there are $u_0,v_0\in K_h$ and $\tau \in (0,1)$ such that $\tau v_0\le u_0<v_0$ and

$$\begin{array}{ccc}\hfill u_0\left(t\right)& \le & \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,u_0\left(s\right),v_0\left(s\right))d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)f(s,u_0\left(s\right),v_0\left(s\right))d_{q}s,t\in [0,1],\hfill \end{array}$$

$$\begin{array}{ccc}\hfill v_0\left(t\right)& \ge & \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,v_0\left(s\right),u_0\left(s\right))d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)f(s,v_0\left(s\right),u_0\left(s\right))d_{q}s,t\in [0,1],\hfill \end{array}$$

where $h\left(t\right)=t^{\alpha -1}$ and $G(t,qs),H(t,qs)$ are given as in Lemma 5; $\left(b\right)$ the following BVP

$$\left\{\begin{array}{c}{D}_q^{\alpha }x\left(t\right)+f(t,x\left(t\right),x\left(t\right))=0,0<t<1,2<\alpha <3,\hfill \\ x\left(0\right)=D_{q}x\left(0\right)=0,D_{q}x\left(1\right)=\beta D_{q}x\left(\eta \right),\hfill \end{array}\right.$$

(13)

has a unique positive solution $u^{*}\in K_h;$ $\left(c\right)$ for $x_0,y_0\in K_h,$ set

$$\begin{array}{ccc}\hfill x_{n+1}\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,x_n\left(s\right),y_n\left(s\right))d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)f(s,x_n\left(s\right),y_n\left(s\right))d_{q}s,n=1,2,\dots ,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill y_{n+1}\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)f(s,y_n\left(s\right),x_n\left(s\right))d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)f(s,y_n\left(s\right),x_n\left(s\right))d_{q}s,n=1,2,\dots ,\hfill \end{array}$$

and we get $\parallel x_n-u^{*}\parallel \to 0,\parallel y_n-u^{*}\parallel \to 0$ as $n\to \infty .$

Corollary 2.

Assume g satisfies $\left(H_1\right),\left(H_2\right)$ and $\left(H_5\right),\left(H_6\right),$ $g(t,0)\not\equiv 0,$ for $t\in [0,1].$ Then:

$\left(a\right)$ there are $u_0,v_0\in K_h$ and $\tau \in (0,1)$ such that $\tau v_0\le u_0<v_0$ and

$$\begin{array}{ccc}\hfill u_0\left(t\right)& \le & \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)g(s,u_0\left(s\right))d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)g(s,u_0\left(s\right)),t\in [0,1],\hfill \end{array}$$

$$\begin{array}{ccc}\hfill v_0\left(t\right)& \ge & \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)g(s,v_0\left(s\right))d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)g(s,v_0\left(s\right))d_{q}s,t\in [0,1],\hfill \end{array}$$

where $h\left(t\right)=t^{\alpha -1}$ and $G(t,qs),H(t,qs)$ are given as in Lemma 5; $\left(b\right)$ the following problem

$$\left\{\begin{array}{c}{D}_q^{\alpha }x\left(t\right)+g(t,x\left(t\right))=0,0<t<1,2<\alpha <3,\hfill \\ x\left(0\right)=D_{q}x\left(0\right)=0,D_{q}x\left(1\right)=\beta D_{q}x\left(\eta \right),\hfill \end{array}\right.$$

(14)

has a unique positive solution $u^{*}\in K_h;$ $\left(c\right)$ for $x_0,y_0\in K_h,$ set

$$\begin{array}{ccc}\hfill x_{n+1}\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)g(s,x_n\left(s\right))d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)g(s,x_n\left(s\right))d_{q}s,n=1,2,\dots ,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill y_{n+1}\left(t\right)& =& \frac{\beta t^{\alpha -1}}{{[\alpha -1]}_q(1-\beta {\eta }^{\alpha -2})}{\int }_0^{1}H(\eta ,qs)g(s,y_n\left(s\right))d_{q}s\hfill \\ & & +{\int }_0^{1}G(t,qs)g(s,y_n\left(s\right))d_{q}s,n=1,2,\dots ,\hfill \end{array}$$

and we obtain $\parallel x_n-u^{*}\parallel \to 0,\parallel y_n-u^{*}\parallel \to 0$ as $n\to \infty .$

Remark 3.

In literature, we have not found such results as Theorems 1 and 2, and Corollaries 1 and 2 on fractional q-difference equation boundary value problems. The used methods in literature were not fixed point theorems for mixed monotone operators. Thus, our method is different from previous ones. We should point out that we can not only give the existence and uniqueness of solutions but also make an iteration to approximate the unique solution.

4. Examples

Example 1.

We consider a problem:

$$\left\{\begin{array}{c}{D}_q^{\frac{5}{2}}u\left(t\right)+u^{\frac{1}{5}}\left(t\right)+{\left[u\right(t)+4]}^{-\frac{1}{3}}+\frac{u\left(t\right)}{2+u\left(t\right)}{t}^3+3a=0,t\in (0,1),\hfill \\ u\left(0\right)=D_{q}u\left(0\right)=0,D_{q}u\left(1\right)=\frac{1}{2}{D}_{q}u\left(\frac{1}{2}\right),\hfill \end{array}\right.$$

(15)

where $q=\frac{1}{2},\alpha =\frac{5}{2},\beta =\eta =\frac{1}{2},a>0$. Take $0<b<a$ and let

$$f(t,u,v)=u^{\frac{1}{5}}+{[v+4]}^{-\frac{1}{3}}+b,g(t,u)=\frac{u}{2+u}{t}^3+3a-b,\gamma =\frac{1}{3}.$$

Then, $f:[0,1]\times [0,+\infty )\times [0,+\infty )\to [0,+\infty )$ and $g:[0,1]\times [0,+\infty )\to [0,+\infty )$ are continuous, $g(t,0)=3a-b>0.$ Furthermore, $f(t,u,v)$ is increasing relative to u for fixed $t\in [0,1]$ and $v\in [0,+\infty ),$ decreasing relative to v for fixed $t\in [0,1]$ and $u\in [0,+\infty ),$ $g(t,u)$ is increasing relative to u for fixed $t\in [0,1].$ On the other hand, for $\lambda \in (0,1),t\in [0,1],u,v\ge 0,$

$$g(t,\lambda u)=\frac{\lambda u\left(t\right)}{2+\lambda u\left(t\right)}{t}^3+3a-b\ge \frac{\lambda u\left(t\right)}{2+u\left(t\right)}{t}^3+\lambda (3a-b)=\lambda g(t,u),$$

and

$$f(t,\lambda u,{\lambda }^{-1}v)={\lambda }^{\frac{1}{5}}{u}^{\frac{1}{5}}+{\lambda }^{\frac{1}{3}}{[v+4\lambda ]}^{-\frac{1}{3}}+b\ge {\lambda }^{\frac{1}{3}}\left\{u^{\frac{1}{5}}+{[v+4]}^{-\frac{1}{3}}+b\right\}={\lambda }^{\gamma }f(t,u,v).$$

Then, $\left(H_1\right)–\left(H_3\right)$ holds. Moreover, taking $\sigma \in (0,\frac{b}{3a-b}],$ one has

$$f(t,u,v)=u^{\frac{1}{5}}+{[v+4]}^{-\frac{1}{3}}+b\ge b=\frac{b}{3a-b}·(3a-b)\ge \sigma \left[\frac{u}{2+u}{t}^3+3a-b\right]=\sigma g(t,u),$$

then $\left(H_4\right)$ holds. By means of Theorem 1, problem (15) has a unique positive solution $u^{*}\in K_h,$ where $h\left(t\right)=t^{\frac{3}{2}},t\in [0,1].$

Example 2.

In Example 4.1, we replace the nonlinear term $u^{\frac{1}{5}}\left(t\right)+{\left[u\right(t)+4]}^{-\frac{1}{3}}+\frac{u\left(t\right)}{2+u\left(t\right)}{t}^3+3a$ by

$${\sin }^{2}t+u^{\frac{1}{3}}\left(t\right)+\frac{1}{2+u\left(t\right)}+\frac{u\left(t\right)}{1+u\left(t\right)}+3.$$

By Theorem 2, we can also show that problem (4.1) has a unique positive solution $u^{*}\in K_h,$ where $h\left(t\right)=t^{\frac{3}{2}},t\in [0,1].$ In fact, let

$$f(t,u,v)={\sin }^{2}t+\frac{1}{2+v}+\frac{u}{1+u},g(t,u)=u^{\frac{1}{3}}+3,\gamma =\frac{1}{3}.$$

It is easy to check that $\left(H_1\right),\left(H_2\right)$ hold. We only show $\left(H_5\right),\left(H_6\right)$ are satisfied. For $\lambda \in (0,1),t\in [0,1],u,v\ge 0,$

$$g(t,\lambda u)={\lambda }^{\frac{1}{3}}{u}^{\frac{1}{3}}+3\ge {\lambda }^{\frac{1}{3}}[u^{\frac{1}{3}}+3]={\lambda }^{\gamma }g(t,u),$$

and

$$f(t,\lambda u,{\lambda }^{-1}v)={\sin }^{2}t+\frac{1}{2+{\lambda }^{-1}v}+\frac{\lambda u}{1+\lambda u}\ge {\sin }^{2}t+\frac{\lambda }{2+v}+\frac{\lambda u}{1+u}\ge \lambda f(t,u,v).$$

Furthermore, $f(t,0,1)={\sin }^{2}t+\frac{1}{3}\not\equiv 0$ and

$$f(t,u,v)\le 3\le u^{\frac{1}{3}}+3=g(t,u).$$

Take $\sigma \in [1,\infty )$ and then $\left(H_5\right),\left(H_6\right)$ hold.

Remark 4.

From Theorems 1 and 2 and Examples 1 and 2, we see that many boundary value problems can be studied by our methods under mixed monotone conditions. We can find that there are many functions that satisfy our conditions. In some works, the nonlinear terms required were super-linearity, sub-linearity or boundness, which guarantee existence of solutions, but the uniqueness has not been obtained.

5. Conclusions

In this article, we investigate a fractional q-difference equation with three-point boundary conditions (1). We obtain the existence and uniqueness of positive solutions in a special $K_h,$ where $h\left(t\right)=t^{\alpha -1}.$ The used methods here are some theorems for operator equation $T_1(x,x)+T_{2}x=x$, where $T_1$ is a mixed monotone operator and $T_2$ is an increasing operator. Our methods are new to fractional q-difference equation boundary value problems. Thus, we can claim that we give an alternative answer to fractional problems and our results are very limited in the literature. Finally, two interesting examples are presented to illustrate the main results. We should note that, to get the uniqueness, we must need the conditions of mixed monotonicity and monotonicity for nonlinear terms.