1 Introduction

The theory that fractional differential equations arise in the fields of science and engineering such as physics, chemistry, mechanics, economics, and biological sciences, etc.; see, for example, [1–6]. The q-difference calculus or quantum calculus is an old subject that was put forward by Jackson [7, 8]. The essential definitions and properties of q-difference calculus can be found in [9, 10]. Early development for q-fractional calculus can be seen in the papers by Al-Salam [11] and Agarwal [12] on the existence theory of fractional q-difference. These days the fractional q-difference equation have given fire to increasing scholars’ imaginations. Some works considered the existence of positive solutions for nonlinear q-fractional boundary value problem [13–32]. For example, Ferreira [13] studied the existence of positive solutions to the fractional q-difference equation

$$ \textstyle\begin{cases} D_{q}^{\alpha}u(t)+ f(t,u(t))=0, \quad 0< t< 1, 1< \alpha\leq2,\\ u(0)=u(1)=0. \end{cases} $$
(1.1)

Ferreira [14] also considered the existence of positive solutions to the nonlinear q-difference boundary value problem

$$ \textstyle\begin{cases} D_{q}^{\alpha}u(t)+ f(t,u(t))=0, \quad 0< t< 1, 1< \alpha\leq3,\\ u(0)=D_{q}u(0)=0, \qquad D_{q}u(1)=\beta\geq0. \end{cases} $$
(1.2)

EI-Shahed and AI-Askar [15] studied the existence of a positive solution to the fractional q-difference equation

$$ \textstyle\begin{cases} {}_{c} D_{q}^{\alpha}u(t)+a(t)f(t)=0, \quad 0\leq t \leq1, 2< \alpha\leq 3,\\ u(0)=D_{q}^{2}(0)=0, \qquad\gamma D_{q} u(1)+ \beta D_{q}^{2} u(1)=0, \end{cases} $$
(1.3)

where \(\gamma, \beta \leqslant0\), and \({}_{c} D_{q}^{\alpha}\) is the fractional q-derivative of Caputo type.

Darzi and Agheli [16] studied the existence of a positive solution to the fractional q-difference equation

$$ \textstyle\begin{cases} D_{q}^{\alpha}u(t)+a(t)f(t)=0, \quad 0\leq t \leq1, 3< \alpha\leq4,\\ u(0)=D_{q}u(0)=D_{q}^{2}u(0)=0, \qquad D_{q}^{2} u(1)= \beta D_{q}^{2} u(\eta), \end{cases} $$
(1.4)

where \(0 <\eta<1\) and \(1-\beta\eta^{\alpha-3}>0 \).

The methods used in the papers mentioned are mainly the Krasnoselskii fixed point theorem, the Schauder fixed point theorem, the Leggett–Williams fixed point theorem, and so on. Differently from methods used in the literature mentioned, on the basis of the enlightenment of the works [17, 18, 26], we will use fixed point theorems for mixed monotone operators to demonstrate the existence and uniqueness of positive solutions for integral boundary value problems of the form

$$ \textstyle\begin{cases} D_{q}^{\alpha}u(t)+f(t,u(t),u(t))+g(t,u(t))=0, \quad 0< t< 1,\\ u(0)=D_{q}u(0)=0, \qquad u(1)=\mu\int_{0}^{1}u(s)\,d_{q}s, \end{cases} $$
(1.5)

where \(D_{q}^{\alpha}\) is the fractional q-derivative of Riemann–Liouville type, \(0< q<1\), \(2<\alpha\leq3\), \(0<\mu<[\alpha]_{q}\). Our results ensure the existence of a unique positive solution. Moreover, an iterative scheme is constructed for approximating the solution. As far as we know, there are still very few works utilizing the fixed point results for mixed monotone operators to study the existence and uniqueness of a positive solution for fractional q-derivative integral boundary value problems.

The plan of the paper is as follows. In Sect. 2, we give not only basic definitions of q-fractional integral, but also some properties of certain Green’s functions, which play a fundamental role in the process of proofs. In Sect. 3, in light of some sufficient conditions, we obtained some results on the existence and uniqueness of positive solutions to problem (1.5). At the closing part, two examples are given to demonstrate the serviceability of our main results in Sect. 4.

2 Preliminaries

For convenience of the reader, on one hand, we recall some well-known facts on q-calculus and, on the other hand, some notations and lemmas that will be used in the proofs of our theorems.

A nonempty closed convex set \(P\subset E\) is a cone if (1) \(x\in P, r\geq0\Rightarrow r x\in P\) and (2) \(x\in P,-x\in P\Rightarrow x=\theta\) (θ is the zero element of E), where \((E, \|\cdot\| )\) is a real Banach space. For all \(x,y\in E\), if there exist \(\mu,\nu >0 \) such that \(\mu x\leq y\leq \nu x\), then we write \(x\sim y\). Obviously, ∼ is an equivalence relation. Let \(P_{h}=\{x\in E| x\sim h, h> \theta\}\).

Let \({q}\in(0,1)\). Then the q-number is given by

$$[a]_{q}= \frac{1-q^{a}}{1-q},\quad a\in R. $$

The q-analogue of the power function \((a-b)^{(n)}\) with \(n\in N_{0}\) is

$$(a-b)^{(0)}=1, \qquad(a-b)^{(n)}=\prod _{k=0} ^{n-1} \bigl(a-bq^{k}\bigr),\quad n\in N, a,b \in R. $$

More generally, if \(\alpha\in R\), then

$$(a-b)^{(\alpha)}=a^{\alpha} \prod_{k=0}^{\infty}\frac {a-bq^{k}}{a-bq^{\alpha+k}}, \quad\alpha\neq0. $$

Note that if \(b=0\), then \(a^{(\alpha)}=a^{\alpha}\). The q-gamma function is defined by

$$\Gamma_{q} (x)= \frac{(1-q)^{(x-1)}}{{1-q}^{x-1}}, \quad x \in R^{+}, $$

and satisfies \(\Gamma_{q} (x+1)=[x]_{q}\Gamma_{q} (x)\).

The q-derivative of a function f is defined by

$$(D_{q} f) (x)=\frac{f(qx)-f(x)}{(q-1)x}, \qquad(D_{q} f) (0)= \lim _{x \rightarrow0} (D_{q} f) (x), $$

and q-derivatives of higher order by

$$\bigl(D_{q}^{0} f\bigr) (x)=f(x), \qquad \bigl(D_{q}^{n} f\bigr) (x)=D_{q} \bigl(D_{q}^{n-1} f\bigr) (x), \quad n\in N. $$

The q-integral of a function f defined in the interval \([0,b]\) is given by

$$(I_{q} f) (x)= \int_{0} ^{x} f(s) \,d_{q}s = x(1-q)\sum _{k=0} ^{\infty}f\bigl(xq^{k} \bigr)q^{k},\quad x \in[0,b]. $$

If \(a \in[0,b]\) and f is defined in the interval \([0,b]\), then its integral from a to b is defined by

$$\int_{a} ^{b} f(s) \,d_{q}s= \int_{0} ^{b} f(s) \,d_{q}s - \int_{0} ^{a} f(s) \,d_{q}s. $$

Similarly to the derivatives, the operator \(I_{q}^{n}\) is given by

$$\bigl(I_{q}^{0} f\bigr) (x)=f(x),\qquad\bigl(I_{q}^{n} f\bigr) (x)=I_{q}\bigl(I_{q}^{n-1} f\bigr) (x), \quad n \in N. $$

The fundamental theorem of calculus applies to the operators \(I_{q}\) and \(D_{q}\), that is,

$$(D_{q} I_{q} f) (x)=f(x), $$

and if f is continuous at \(x=0\), then

$$(I_{q} D_{q} f) (x)=f(x)-f(0). $$

The following formulas will be used later (\({}_{t}D_{q}\) denotes the derivative with respect to variable t):

$$\begin{gathered} {}_{t}D_{q}(t-s)^{(\alpha)}=[\alpha]_{q} (t-s)^{(\alpha-1)}, \\ \biggl({}_{x} D_{q} \int_{0} ^{x} f(x,t)\,d_{q}t \biggr) (x) = \int_{0}^{x} {}_{x}D_{q} f(x,t)\,d_{q}t +f(qx,x). \end{gathered} $$

Definition 2.1

(see [4])

Let \(\alpha\geq0\), and let f be a function defined on \([0,1]\). The fractional q-integral of the Riemann–Liouville type is defined by \((I_{q}^{0} f)(x)=f(x)\) and

$$\bigl(I_{q}^{\alpha} f\bigr) (x)= \frac{1}{\Gamma_{q} (\alpha)} \int_{0} ^{x} (x-qt)^{(\alpha-1)} f(t) \,d_{q}t, \quad\alpha>0, x \in[0,1]. $$

Definition 2.2

(see [10])

The fractional q-derivative of the Riemann–Liouville type is defined by

$$\bigl(D_{q}^{0} f\bigr) (x)=f(x), \qquad \bigl(D_{q}^{\alpha} f\bigr) (x)= \bigl(D_{q}^{p} I_{q}^{p-\alpha }f\bigr) (x), \quad\alpha>0, $$

where p is the smallest integer greater than or equal to α.

Lemma 2.1

(see [10])

Let \(\alpha, \beta\geq0\), and let f be a function defined on \([0,1]\). Then the following formulas hold:

  1. (i)

    \((I_{q}^{\beta}I_{q}^{\alpha}f)(x)=(I_{q}^{\beta+\alpha} f)(x)\),

  2. (ii)

    \((D_{q}^{\alpha}I_{q}^{\alpha}f)(x)=f(x)\).

Lemma 2.2

(see [10])

Let \(\alpha>0\), and let be p be a positive integer. Then the following equality holds:

$$\bigl(I_{q}^{\alpha}D_{q}^{p} f\bigr) (x)= \bigl(D_{q}^{p} I_{q}^{\alpha}f \bigr) (x) - \sum_{k=0} ^{p-1} \frac{x^{\alpha-p+k}}{\Gamma_{q} (\alpha+k-p+1)} \bigl(D_{q}^{k} f\bigr) (0). $$

Lemma 2.3

(see [27])

Let \(2<\alpha\leq3\) and \(0<\mu<[\alpha]_{q}\). Let \(x\in C[0,1]\). Then the boundary value problem

$$\begin{aligned} &D_{q}^{\alpha}u(t)+ x(t)=0, \quad 0< t< 1, \\ \end{aligned}$$
(2.1)
$$\begin{aligned} &u(0)=D_{q}u(0)=0, \qquad u(1)=\mu \int_{0}^{1}u(s)\,d_{q}s, \end{aligned}$$
(2.2)

has a unique solution

$$u(t)= \int_{0} ^{1} G(t,qs)x(s)\,d_{q}s, $$

where

$$ G(t,s)= \textstyle\begin{cases} \frac{t^{\alpha-1}(1-s)^{(\alpha-1)} ([\alpha]_{q}-\mu+\mu q^{\alpha-1}s )- ([\alpha]_{q}-\mu )(t-s)^{\alpha -1}}{ ([\alpha]_{q}-\mu )\Gamma_{q} (\alpha)},& 0 \leq s\leq t \leq1,\\ \frac{t^{\alpha-1}(1-s)^{(\alpha-1)} ([\alpha]_{q}-\mu+\mu q^{\alpha-1}s )}{ ([\alpha]_{q}-\mu )\Gamma_{q} (\alpha)},& 0 \leq t\leq s\leq1. \end{cases} $$
(2.3)

Lemma 2.4

(see [27])

The function \(G(t,qs)\) defined by (2.3) has the following properties:

  1. (i)

    \(G(t,qs)\) is a continuous function and \(G(t,qs) \geq0\);

  2. (ii)

    \(\frac{\mu q^{\alpha}t^{\alpha-1}(1-qs)^{(\alpha-1)}s}{ ([\alpha]_{q}-\mu )\Gamma_{q} (\alpha)} \leq G(t,qs) \leq\frac {M_{0} t^{\alpha-1}}{ ([\alpha]_{q}-\mu )\Gamma_{q} (\alpha )}\), \(t, s \in[0,1]\),

where \(M_{0}=\max \{ [\alpha-1]_{q} ([\alpha]_{q}-\mu)+\mu q^{\alpha}, q^{\alpha-1} [\alpha]_{q} \}\).

Definition 2.3

(see [17])

An operator \(A: P \times P \rightarrow P\) is said to be a mixed monotone operator if \(A(x, y)\) is increasing in x and decreasing in y, that is, \(x_{i}, y_{i}\in P\) (\(i = 1, 2\)), \(x_{1} \leq x_{2}\), \(y_{1}\geq y_{2}\) imply \(A(x_{1}, y_{1})\leq A(x_{2}, y_{2})\). An element \(x\in P\) is called a fixed point of A if \(A(x, x) = x\).

Definition 2.4

(see [18])

An operator \(A: P\rightarrow P \) is said to be subhomogeneous if

$$\begin{aligned} A (tx)\geq t A(x)\quad \mbox{for any } t\in(0,1), x\in P. \end{aligned}$$
(2.4)

Definition 2.5

(see [18])

Let \(D = P \), and let γ be a real number with \(0\leq\gamma< 1 \). An operator \(A:D\rightarrow D\) is said to be γ-concave if it satisfies

$$\begin{aligned} A(tx)\geq t^{\gamma}A(x)\quad \mbox{for any } t\in(0,1),x\in D. \end{aligned}$$
(2.5)

Lemma 2.5

(see [17])

Let \(h > \theta\) and \(\gamma\in(0,1)\).

Let \(A: P \times P \rightarrow P\) be a mixed monotone operator satisfying

$$\begin{aligned} A\bigl(tx,t^{-1}y\bigr)\geq t^{\gamma}A(x,y) \quad \textit{for any } t \in(0,1), x,y \in P, \end{aligned}$$
(2.6)

and let \(B:P \rightarrow P \) be an increasing subhomogeneous operator. Assume that

  1. (i)

    there is \(h_{0} \in P_{h} \) such that \(A(h_{0},h_{0})\in P_{h}\) and \(B h_{0} \in P_{h} \);

  2. (ii)

    there exists a constant \(\delta_{0}\) such that \(A(x,y )\geq\delta_{0} B x\) for any \(x,y \in P \).

Then:

  1. (1)

    \(A: P_{h} \times P_{h} \rightarrow P_{h} \) and \(B:P_{h} \rightarrow P_{h}\);

  2. (2)

    there exist \(u_{0},v_{0} \in P_{h}\) and \(r \in(0,1)\) such that

    $$r v_{0} \leq u_{0} < v_{0}, \qquad u_{0} \leq A(u_{0},v_{0})+ B u_{0} \leq A ( v_{0},u_{0})+ B v_{0}\leq v_{0}; $$
  3. (3)

    the operator equation \(A(x,x )+ B x =x \) has a unique solution \(x^{*} \in P_{h}\);

  4. (4)

    for any initial values \(x_{0},y_{0} \in P_{h}\), constructing successively the sequences

    $$x_{n} = A(x_{n-1},y_{n-1})+ Bx_{n-1}, \qquad y_{n} = A(y_{n-1},x_{n-1})+ B y_{n-1}, \quad n=1,2,\ldots, $$

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

Remark 2.1

When \(B= \theta\) in Lemma 2.5, then the corresponding conclusion still holds.

Lemma 2.6

(see [18])

Let \(h > \theta\) and \(\gamma\in(0,1)\).

Let \(A:P \times P \rightarrow P\) be a mixed monotone operator satisfying

$$\begin{aligned} A\bigl(tx,t^{-1}y\bigr)\geq t A(x,y), \quad \textit{for any } t \in(0,1), x, y \in P, \end{aligned}$$
(2.7)

and let \(B:P \rightarrow P \) be an increasing γ-concave operator. Assume that

  1. (i)

    there is \(h_{0} \in P_{h} \) such that \(A(h_{0},h_{0})\in P_{h}\) and \(B h_{0} \in P_{h} \);

  2. (ii)

    there exists a constant \(\delta_{0}\) such that \(A(x,y )\leq\delta_{0} B x\) for any \(x,y \in P \).

Then:

  1. (1)

    \(A: P_{h} \times P_{h} \rightarrow P_{h} \) and \(B:P_{h} \rightarrow P_{h}\);

  2. (2)

    there exist \(u_{0},v_{0} \in P_{h}\) and \(r \in(0,1)\) such that

    $$r v_{0} \leq u_{0} < v_{0}, \qquad u_{0} \leq A(u_{0},v_{0})+ B u_{0} \leq A( v_{0},u_{0})+ B v_{0}\leq v_{0}; $$
  3. (3)

    the operator equation \(A (x,x )+ B x =x \) has a unique solution \(x^{*} \in P_{h}\);

  4. (4)

    for any initial values \(x_{0},y_{0} \in P_{h}\), constructing successively the sequences

    $$x_{n} = A (x_{n-1},y_{n-1})+ Bx_{n-1}, \qquad y_{n} =A(y_{n-1},x_{n-1})+B y_{n-1}, \quad n=1,2,\ldots, $$

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

Remark 2.2

When \(A= \theta\) in Lemma 2.6, then the corresponding conclusion still holds.

3 Main results

In this section, we give and prove our main results by applying Lemmas 2.5 and 2.6. We consider the Banach space \(X=C[0,1]\) endowed with standard norm \(\|x\|=\sup\{|x(t)|:{t\in[0,1]}\}\). Clearly, this space can be equipped with a partial order given by

$$x,y \in C[0,1], \quad x\leq y \quad \Leftrightarrow \quad x(t)\leq y(t) \quad \mbox{for }t \in[0,1]. $$

We define the cone \(P=\{x\in X:x(t) \geq 0,t\in[0,1]\}\). Notice that P is a normal cone in \(C[0,1]\) and the normality constant is 1.

Theorem 3.1

Suppose that

\((F_{1})\) :

a function \(f(t,x,y):[0,1]\times[0, +\infty)\times[0,+\infty)\rightarrow[0,+\infty)\) is continuous, increasing with respect to the second variable, and decreasing with respect to the third variable;

\((F_{2})\) :

a function \(g(t,x): [0,1]\times[0, +\infty)\rightarrow [0,+\infty) \) is continuous and increasing with respect to the second variable;

\((F_{3})\) :

there exists a constant \(\gamma\in(0,1)\) such that \(f(t,\lambda x,\lambda^{-1} y)\geq\lambda^{\gamma}f(t,x,y)\) for any \(t \in[0,1]\), \(\lambda\in(0,1)\), \(x, y \in [0,+\infty)\), and \(g(t,\lambda x)\geq\lambda g(t,x)\) for \(\lambda\in(0,1)\), \(t\in[0,1]\), \(u\in[0, +\infty)\), and \(g(t,0)\not\equiv0 \);

\((F_{4})\) :

there exists a constant \(\delta_{0} > 0\) such that \(f(t,x,y)\geq\delta_{0} g(t,x)\), \(t\in[0,1]\), \(x,y\geq0\).

Then:

  1. (1)

    there exist \(x_{0},y_{0}\in P_{h}\) and \(r\in(0,1)\) such that \(r y_{0}\leq x_{0} < y_{0}\) and

    $$\begin{gathered} x_{0}\leq \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,x_{0}(s),y_{0}(s) \bigr)+g\bigl(s,x_{0}(s)\bigr)\bigr]\,d_{q}s, \quad t\in[0,1], \\ y_{0}\geq \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,y_{0}(s),y_{0}(s) \bigr)+g\bigl(s,y_{0}(s)\bigr)\bigr]\,d_{q}s, \quad t\in[0,1], \end{gathered} $$

    where \(G(t,qs)\) is defined by (2.3), and \(h(t)= t^{\alpha-1}\), \(t \in[0,1]\);

  2. (2)

    the boundary value problem (1.5) has a unique positive solution \(u^{*} \) in \(P_{h}\), and for any \(x_{0},y_{0} \in P_{h}\), constructing successively the sequences

    $$\begin{gathered} x_{n+1}= \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,x_{n}(s),y_{n}(s) \bigr)+g\bigl(s,x_{n}(s)\bigr)\bigr]\,d_{q}s,\quad n=0,1,2,\ldots, \\ y_{n+1}= \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,y_{n}(s),x_{n}(s) \bigr)+g\bigl(s,y_{n}(s)\bigr)\bigr]\,d_{q}s,\quad n=0,1,2,\ldots, \end{gathered} $$

    we have \(\|x_{n}-u^{*}\|\rightarrow0\) and \(\|y_{n}-u^{*}\| \rightarrow0\) as \(n \rightarrow\infty\).

Proof

We note that if u is a solution of boundary value problem (1.5), then

$$\begin{aligned} u(t)= \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,u(s),u(s)\bigr)+g \bigl(s,u(s)\bigr)\bigr]\,d_{q}s, \quad0 \leq t \leq1. \end{aligned}$$
(3.1)

Define two operators \(T_{1}:P\times P \rightarrow E\) and \(T_{2}: P \rightarrow E \) by

$$\begin{aligned} \begin{aligned} &T_{1}(u,v) (t)= \int_{0}^{1} G(t,qs)f\bigl(s,u(s),v(s) \bigr)\,d_{q}s, \\ &(T_{2}u) (t)= \int _{0}^{1} G(t,qs)g\bigl(s,u(s) \bigr)\,d_{q}s. \end{aligned} \end{aligned}$$
(3.2)

We transform the boundary value problem (1.5) into a fixed point problem \(u = T_{1}(u,u)+ T_{2} u\). From \((F_{1})\), \((F_{2})\), and Lemma 2.4 it is easy to see that \(T_{1}: P\times P \rightarrow P\) and \(T_{2}: P \rightarrow P\). Next, we want to prove that \(T_{1}\) and \(T_{2}\) satisfy the conditions of Lemma 2.5.

To begin with, we prove that \(T_{1}\) is a mixed monotone operator. In fact, for \(u_{1},u_{2},v_{1},v_{2} \in P\) with \(u_{1}\geq u_{2}\) and \(v_{1}\leq v_{2}\), it is easy to see that \(u_{1}(t) \geq u_{2}(t)\), \(v_{1}(t)\leq v_{2}(t)\), \(t \in[0,1]\), and by Lemma 2.4 and \((F_{1})\),

$$\begin{aligned} T_{1}(u_{1},v_{1}) (t)&= \int_{0}^{1} G(t,qs)f\bigl(s,u_{1}(s),v_{1}(s) \bigr)\,d_{q}s \\ &\geq \int_{0}^{1} G(t,qs)f\bigl(s,u_{2}(s),v_{2}(s) \bigr)\,d_{q}s =T_{1}(u_{2},v_{2}) (t). \end{aligned}$$
(3.3)

For any \(\lambda\in(0,1)\) and \(u,v \in P \), by \((F_{3}) \) we have

$$ \begin{aligned}[b] T_{1}\bigl(\lambda u, \lambda^{-1}v\bigr) (t)&= \int_{0}^{1}G(t,qs)f\bigl(s,\lambda u(s), \lambda^{-1}v(s)\bigr)\,d_{q}s \\ &\geq\lambda^{\gamma}\int_{0}^{1}G(t,qs)f\bigl(s, u(s),v(s) \bigr)\,d_{q}s \geq\lambda^{\gamma}T_{1}(u,v) (t). \end{aligned} $$
(3.4)

So, the operator \(T_{1}\) satisfies (2.6).

For any \(u_{1}(t)\geq u_{2}(t)\), \(t \in[0,1]\), from \(G(t,qs) \geq0\) and \((F_{2})\) we know that

$$T_{2}u_{1}(t)= \int_{0}^{1} G(t,qs)g\bigl(s,u_{1}(s) \bigr)\,d_{q}s \geq \int_{0}^{1} G(t,qs)g\bigl(s,u_{2}(s) \bigr)\,d_{q}s=T_{2}u_{2}(t). $$

So \(T_{2}\) is increasing. Further, for any \(\lambda\in(0,1)\) and \(u \in P\), from hypothesis \((F_{3})\) we get

$$\begin{aligned} T_{2}(\lambda u) (t)= \int_{0}^{1}G(t,qs)g\bigl(s,\lambda u(s) \bigr)\,d_{q}s\geq\lambda \int_{0}^{1}G(t,qs)g\bigl(s,u(s)\bigr)\,d_{q}s =\lambda T_{2}u(t), \end{aligned}$$
(3.5)

that is, the operator \(T_{2}\) is subhomogeneous. By \((F_{1})\) and Lemma 2.4, for any \(t \in[0,1]\), we have

$$ \begin{aligned}[b] T_{1}(h,h) (t)&= \int_{0}^{1}G(t,qs)f\bigl(s,h(s),h(s) \bigr)\,d_{q}s \\ &= \int_{0}^{1}G(t,qs)f\bigl(s, s^{\alpha-1},s^{\alpha-1} \bigr)\,d_{q}s \\ &\leq\frac{M_{0}}{\Gamma_{q} (\alpha) ([\alpha]_{q}-\mu )}h(t) \int_{0}^{1} f(s,1,0)\,d_{q}s \end{aligned} $$
(3.6)

and

$$ \begin{aligned}[b] T_{1}(h,h) (t)&= \int_{0}^{1}G(t,qs)f\bigl(s,h(s),h(s) \bigr)\,d_{q}s \\ &= \int_{0}^{1}G(t,qs)f\bigl(s, s^{\alpha-1},s^{\alpha-1} \bigr)\,d_{q}s \\ &\geq\frac{\mu q^{\alpha}}{\Gamma_{q} (\alpha) ([\alpha]_{q}-\mu )}h(t) \int_{0}^{1}s (1-qs)^{(\alpha-1)}f(s,0,1)\,d_{q}s. \end{aligned} $$
(3.7)

From \((F_{2})\) and \((F_{4})\) we have the inequality

$$f(s,1,0)\geq f(s,0,1)\geq\delta_{0} g(s,0)\geq0. $$

Since \(g(t,0)\not\equiv0 \), we also obtain

$$ \int_{0}^{1} f(s,1,0)\,d_{q}s\geq \int_{0}^{1} f(s,0,1)\,d_{q}s \geq \delta_{0} \int _{0}^{1} g(s,0)\,d_{q}s > 0. $$
(3.8)

Let

$$\begin{aligned} &M_{1}=\frac{M_{0}}{\Gamma_{q} (\alpha) ([\alpha]_{q}-\mu )} \int_{0}^{1} f(s,1,0)\,d_{q}s, \\ &M_{2}=\frac{\mu q^{\alpha}}{\Gamma_{q} (\alpha) ([\alpha]_{q}-\mu )} \int_{0}^{1}s (1-qs)^{(\alpha-1)}f(s,0,1)\,d_{q}s, \\ &M_{3}=\frac{\mu q^{\alpha}}{\Gamma_{q} (\alpha) ([\alpha]_{q}-\mu )} \int_{0}^{1}s (1-qs)^{(\alpha-1)}g(s,0)\,d_{q}s, \\ &M_{4}=\frac{M_{0} }{\Gamma_{q} (\alpha) ([\alpha]_{q}-\mu )} \int_{0}^{1} g(s,1)\,d_{q}s. \end{aligned}$$

Thus we have \(M_{2}h(t)\leq T_{1}(h,h) \leq M_{1}h(t)\), \(M_{3}h(t)\leq T_{2}h\leq M_{4}h(t)\), \(t\in[0,1]\). So, \(T_{1}(h,h)\in P_{h}\). From \(g(t,0)\not\equiv0 \) it is easy to see that \(T_{2}h\in P_{h}\). So, there is \(h(t)=t^{\alpha-1}\in P_{h} \) such that \(T_{1}(h,h)\in P_{h}\) and \(T_{2}h \in P_{h} \).

Next, we prove that the operators \(T_{1}\) and \(T_{2}\) satisfy condition (ii) of Lemma 2.5. In fact, for \(u,v \in P \) and any \(t \in[0,1]\), by \((F_{4}) \) we have

$$ \begin{aligned}[b] T_{1}(u,v) (t)&= \int_{0}^{1}G(t,qs)f\bigl(s,u(s),v(s) \bigr)\,d_{q}s\\ &\geq\delta_{0} \int _{0}^{1}G(t,qs)g\bigl(s,u(s) \bigr)\,d_{q}s=\delta_{0} (T_{2}u) (t). \end{aligned} $$
(3.9)

Then we have \(T_{1}(u,v)\geq\delta_{0} T_{2} u\) for \(u,v\in P\). By Lemma 2.5 we can deduce: there exist \(u_{0},v_{0}\in P_{h}\) and \(r \in (0,1)\) such that \(rv_{0}\leq u_{0}\leq v_{0}\), \(u_{0}\leq T_{1}(u_{0},v_{0})+ T_{2} u_{0}\leq T_{1}(v_{0},u_{0})+T_{2}v_{0}\leq v_{0}\); the operator equation \(T_{1}(u,u)+T_{2}u=u\) has a unique solution \(u^{*} \in P_{h}\); and for any initial values \(x_{0}, y_{0} \in P_{h}\), constructing successively the sequences

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

we get \(x_{n}\rightarrow u^{*}\) and \(y_{n}\rightarrow u^{*} \) as \(n \rightarrow\infty\). We have the following two inequalities:

$$\begin{gathered} u_{0}(t)\leq \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,u_{0}(s),v_{0}(s) \bigr)+g\bigl(s,u_{0}(s)\bigr)\bigr]\,d_{q}s, \quad t\in[0,1], \\ v_{0}(t)\geq \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,v_{0}(s),u_{0}(s) \bigr)+g\bigl(s,v_{0}(s)\bigr)\bigr]\,d_{q}s, \quad t\in[0,1]. \end{gathered} $$

Thus problem (1.5) has a unique positive solution \(u^{*} \in P_{h}\); for any \(u_{0},v_{0} \in P_{h}\), constructing successively the sequences

$$\begin{gathered} x_{n+1}(t)= \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,x_{n}(s),y_{n}(s) \bigr)+g\bigl(s,x_{n}(s)\bigr)\bigr]\,d_{q}s,\quad n=0,1,2, \ldots, \\ y_{n+1}(t)= \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,y_{n}(s),x_{n}(s) \bigr)+g\bigl(s,y_{n}(s)\bigr)\bigr]\,d_{q}s,\quad n=0,1,2, \ldots, \end{gathered} $$

we have \(\|x_{n}-u^{*}\|\rightarrow0\) and \(\|y_{n}-u^{*}\|\rightarrow0 \) as \(n\rightarrow\infty\). □

Corollary 3.1

Suppose that f satisfies the conditions of Theorem 3.1 and \(g\equiv 0\), \(f(t,0,1)\not\equiv0\). Then:

  1. (i)

    there exist \(u_{0},v_{0} \in P_{h} \) and \(r\in(0,1)\) such that \(r v_{0} \leq u_{0} < v_{0}\), and

    $$\begin{gathered} u_{0}(t)\leq \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,u_{0}(s),v_{0}(s) \bigr)\bigr]\,d_{q}s, \quad t\in [0,1], \\ v_{0}(t)\geq \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,v_{0}(s),u_{0}(s) \bigr)\bigr]\,d_{q}s, \quad t\in[0,1], \end{gathered} $$

    where \(G(t,qs)\) is defined by (2.3), and \(h(t)= t^{\alpha-1}\), \(t \in[0,1]\);

  2. (ii)

    the BVP

    $$ \textstyle\begin{cases} D_{q}^{\alpha}u(t)+ f(t,u(t),u(t))=0, \quad 0< t< 1, 2< \alpha\leq3,\\ u(0)=D_{q}u(0)=0, \qquad u(1)=\mu\int_{0}^{1}u(s)\,d_{q}s, \end{cases} $$
    (3.10)

    has a unique positive solution \(u^{*} \) in \(P_{h}\);

  3. (iii)

    for any \(x_{0},y_{0} \in P_{h}\), the sequences

    $$\begin{gathered} x_{n+1}= \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,x_{n}(s),y_{n}(s) \bigr)\bigr]\,d_{q}s, \quad n=0,1,2,\ldots, \\ y_{n+1}= \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,y_{n}(s),x_{n}(s) \bigr)\bigr]\,d_{q}s, \quad n=0,1,2,\ldots, \end{gathered} $$

    satisfy \(\|x_{n}-u^{*}\|\rightarrow0\) and \(\|y_{n}-u^{*}\|\rightarrow0\) as \(n \rightarrow\infty\).

Theorem 3.2

Suppose that \((F_{1})\)–\((F_{2}) \) hold. In addition, suppose that f, g satisfy the following conditions:

\((F_{5})\) :

there exists a constant \(\gamma\in(0,1)\) such that \(g(t,\lambda u)\geq\lambda^{\gamma}g(t,u)\) for any \(t \in[0,1] \), \(\lambda\in(0,1)\), \(u \in[0, +\infty)\), and \(f(t,\lambda u,\lambda^{-1} v)\geq\lambda f(t,u,v)\) for \(\lambda\in (0,1)\), \(t\in[0,1]\), \(u,v\in[0,+\infty)\);

\((F_{6})\) :

\(f(t,0,1)\not\equiv0\) for \(t \in[0,1]\), and there exists a constant \(\delta_{0} > 0\) such that \(f(t,u,v)\leq\delta_{0} g(t,u)\), \(t\in[0,1]\), \(u,v \geq0\).

Then:

  1. (1)

    there exist \(u_{0},v_{0}\in P_{h}\) and \(r\in(0,1)\) such that \(r v_{0}\leq u_{0} < v_{0}\) and

    $$\begin{gathered} u_{0}\leq \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,u_{0}(s),v_{0}(s) \bigr)+g\bigl(s,u_{0}(s)\bigr)\bigr]\,d_{q}s, \quad t\in[0,1], \\ v_{0}\geq \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,v_{0}(s),u_{0}(s) \bigr)+g\bigl(s,v_{0}(s)\bigr)\bigr]\,d_{q}s, \quad t\in[0,1], \end{gathered} $$

    where \(G(t,qs)\) is defined by (2.3), and \(h(t)= t^{\alpha-1}\), \(t \in[0,1]\);

  2. (2)

    the boundary value problem (1.5) has a unique positive solution \(u^{*} \) in \(P_{h}\); and for any \(x_{0},y_{0} \in P_{h}\), the sequences

    $$\begin{gathered} x_{n+1}= \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,x_{n}(s),y_{n}(s) \bigr)+g\bigl(s,x_{n}(s)\bigr)\bigr]\,d_{q}s,\quad n=0,1,2, \ldots, \\ y_{n+1}= \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,y_{n}(s),x_{n}(s) \bigr)+g\bigl(s,y_{n}(s)\bigr)\bigr]\,d_{q}s,\quad n=0,1,2, \ldots, \end{gathered} $$

    satisfy \(\|x_{n}-u^{*}\|\rightarrow0\) and \(\|y_{n}-u^{*}\|\rightarrow0\) as \(n \rightarrow\infty\).

Proof

Similarly to the proof of Theorem 3.1, \(T_{1}\) and \(T_{2} \) are given in (3.2). From \((F_{1})\) and \((F_{2})\) we know that \(T_{1}: P\times P \rightarrow P \) is a mixed monotone operator and \(T_{2}: P\rightarrow P\) is increasing. By \((F_{5})\) we obtain

$$T_{1}\bigl(\lambda u,\lambda^{-1}v\bigr)\geq\lambda T_{1}(u,v), \qquad T_{2}(\lambda u)\geq\lambda^{\gamma}T_{2}u, \quad \mbox{for }\lambda\in(0,1), u,v \in P. $$

According to \((F_{2})\) and \((F_{6})\), we have

$$f(s,0,1)\leq\delta_{0} g(s,0), \qquad f(s,0,1)\leq f(s,1,0),\quad s \in[0,1]. $$

From \(f(t,0,1) \not\equiv0\) we get

$$\begin{aligned} &0< \int_{0}^{1}f(s,0,1)\,d_{q}s\leq \int_{0}^{1}f(s,1,0)\,d_{q}s, \\ &0< \frac{1}{\delta_{0}} \int_{0}^{1}f(s,0,1)\,d_{q}s\leq \int_{0}^{1}g(s,0)\,d_{q}s\leq \int_{0}^{1} g(s,1)\,d_{q}s, \end{aligned} $$

and the following inequalities hold:

$$\begin{aligned} & \begin{aligned}[b] 0&< \frac{\mu q^{\alpha}}{\Gamma_{q} (\alpha) ([\alpha]_{q}-\mu )} \int_{0}^{1}s (1-qs)^{(\alpha-1)}f(s,0,1)\,d_{q}s \\ &\leq\frac{M_{0} }{\Gamma_{q} (\alpha) ([\alpha]_{q}-\mu )} \int_{0}^{1} f(s,1,0)\,d_{q}s, \end{aligned} \end{aligned}$$
(3.11)
$$\begin{aligned} &\begin{aligned}[b] 0&< \frac{\mu q^{\alpha}}{\Gamma_{q} (\alpha) ([\alpha]_{q}-\mu )} \int_{0}^{1}s (1-qs)^{(\alpha-1)}g(s,0)\,d_{q}s \\ &\leq\frac{M_{0} }{\Gamma_{q} (\alpha) ([\alpha]_{q}-\mu )} \int_{0}^{1} g(s,1)\,d_{q}s. \end{aligned} \end{aligned}$$
(3.12)

Hence we can easily check that \(T_{1}(h,h) \in P\), \(T_{2}h \in P\), \(t\in [0,1]\), and, by using \((F_{6})\), we have

$$\begin{aligned} T_{1}(u,v) (t)&= \int_{0}^{1} G(t,s)f\bigl(s,u(s),v(s) \bigr)\,d_{q}s \\ &\leq\delta_{0} \int_{0}^{1} G(t,s)g\bigl(s,u(s)\bigr)\,d_{q}s= \delta_{0} T_{2}u(t). \end{aligned}$$
(3.13)

Then we have \(T_{1}(u,v)\leq\delta_{0} T_{2}u\) for \(u,v\in P\). Thus, from Lemma 2.6 we get that there exist \(u_{0},v_{0}\in P_{h}\) and \(r \in(0,1)\) such that \(rv_{0}\leq u_{0}\leq v_{0},u_{0}\leq T_{1}(u_{0},v_{0})+ T_{2} u_{0}\leq T_{1}(v_{0},u_{0})+T_{2}v_{0}\leq v_{0}\); the operator equation \(T_{1}(u,u)+T_{2}u=u\) has a unique solution \(u^{*} \in P_{h}\); and for any initial values \(x_{0}\), \(y_{0} \in P_{h}\), the sequences

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

satisfy \(x_{n}\rightarrow u^{*}\) and \(y_{n}\rightarrow u^{*} \) as \(n \rightarrow\infty\). That is,

$$\begin{gathered} u_{0}(t)\leq \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,u_{0}(s),v_{0}(s) \bigr)+g\bigl(s,u_{0}(s)\bigr)\bigr]\,d_{q}s, \quad t\in[0,1], \\ v_{0}(t)\geq \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,v_{0}(s),u_{0}(s) \bigr)+g\bigl(s,v_{0}(s)\bigr)\bigr]\,d_{q}s, \quad t\in[0,1], \end{gathered} $$

The boundary value problem (1.5) has a unique positive solution \(u^{*} \in P_{h}\); for \(u_{0},v_{0} \in P_{h}\), the sequences

$$\begin{gathered} x_{n+1}(t)= \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,x_{n}(s),y_{n}(s) \bigr)+g\bigl(s,x_{n}(s)\bigr)\bigr]\,d_{q}s,\quad n=0,1,2, \ldots, \\ y_{n+1}(t)= \int_{0}^{1} G(t,qs)\bigl[f\bigl(s,y_{n}(s),x_{n}(s) \bigr)+g\bigl(s,y_{n}(s)\bigr)\bigr]\,d_{q}s,\quad n=0,1,2, \ldots, \end{gathered} $$

satisfy \(\|x_{n}-u^{*}\|\rightarrow0\) and \(\|y_{n}-u^{*}\|\rightarrow0 \) as \(n\rightarrow\infty\). □

Corollary 3.2

Suppose that g satisfies the conditions of Theorem 3.2, \(f\equiv0\), and \(g(t,0)\not\equiv0\) for \(t \in[0,1]\). Then:

  1. (i)

    there exist \(u_{0},v_{0} \in P_{h} \) and \(r\in(0,1)\) such that \(r v_{0} \leq u_{0} < v_{0}\), and

    $$\begin{gathered} u_{0}(t)\leq \int_{0}^{1} G(t,qs)\bigl[g\bigl(s,u_{0}(s) \bigr)\bigr]\,d_{q}s, \\ v_{0}(t)\geq \int_{0}^{1} G(t,qs)\bigl[g\bigl(s,v_{0}(s) \bigr)\bigr]\,d_{q}s, \quad t\in[0,1], \end{gathered} $$

    where \(G(t,qs)\) is defined by (2.3), and \(h(t)= t^{\alpha-1}\), \(t \in[0,1]\);

  2. (ii)

    the BVP

    $$ \textstyle\begin{cases} D_{q}^{\alpha}u(t)+ g(t,u(t))=0, \quad 0< t< 1, 2< \alpha\leq3,\\ u(0)=D_{q}u(0)=0, \qquad u(1)=\mu\int_{0}^{1}u(s)\,d_{q}s, \end{cases} $$
    (3.14)

    has a unique positive solution \(u^{*} \) in \(P_{h}\); and for any \(x_{0},y_{0} \in P_{h}\), the sequences

    $$\begin{gathered} x_{n+1}= \int_{0}^{1} G(t,qs)g\bigl(s,x_{n}(s) \bigr)\,d_{q}s,\quad n=0,1,2,\ldots, \\ y_{n+1}= \int_{0}^{1} G(t,qs)g\bigl(s,y_{n}(s) \bigr)\,d_{q}s,\quad n=0,1,2,\ldots, \end{gathered} $$

    satisfy \(\|x_{n}-u^{*}\|\rightarrow0\) and \(\| y_{n}-u^{*}\|\rightarrow0\) as \(n \rightarrow\infty\).

4 Example

Now, we give two examples to illustrate our results.

Example 4.1

Consider the following boundary value problem:

$$ \textstyle\begin{cases} -D_{\frac{1}{2}}^{\frac{5}{2}} u(t)=u(t)^{\frac{1}{3}}+[u(t)+1]^{- \frac{1}{2}}+{\frac{u(t)}{1+u(t)}}t^{3}+t^{2}+4, \quad 0< t< 1,\\ u(0)=D_{\frac{1}{2}}u(0)=0, \qquad u(1)=\mu\int_{0}^{1}u(s)\,d_{\frac{1}{2}}s. \end{cases} $$
(4.1)

In this example, we let

$$\begin{aligned} &f(t,u,v )=u^{\frac{1}{3}}+[v+1]^{- \frac{1}{2}}+t^{2}+2, \qquad g(t,u)={\frac{u}{1+u}}t^{3}+2, \\ &\gamma={\frac{1}{2}}, \qquad\mu=\frac{1}{2}. \end{aligned}$$

It is not difficult to find that \(f(t,x,y):[0,1]\times[0, +\infty )\times[0,+\infty)\rightarrow[0,+\infty)\) is continuous, increasing with respect to the second variable, and decreasing with respect to the third variable and that \(g(t,x): [0,1]\times[0, +\infty)\rightarrow[0,+\infty) \) is continuous with \(g(t,0)=2>0\) and increasing with respect to the second variable. We also have

$$\begin{aligned}& g(t,\lambda u )={\frac{{\lambda}u}{1+\lambda{u}}}t^{3}+2 \geq{ \frac {{\lambda}u}{1+ {u}}}t^{3}+2 \lambda=\lambda g(t,u), \quad \lambda\in(0,1), \\& \begin{aligned} f\bigl(t,\lambda u,\lambda^{-1}v \bigr)&=\lambda^{\frac{1}{3}}u^{\frac {1}{3}}+ \lambda^{\frac{1}{2}}[v+\lambda]^{-\frac{1}{2}}+t^{2}+2 \\ &\geq\lambda^{\frac{1}{2}} \bigl\{ u^{\frac{1}{3}}+[v+1]^{-\frac {1}{2}}+t^{2}+2 \bigr\} \\ &=\lambda^{\gamma}f(t,u,v). \end{aligned} \end{aligned}$$

Further, if we take \(\delta_{0} \in(0,\frac{2}{3}]\), then we easily get

$$\begin{aligned} f(t,u,v )&=u^{\frac{1}{3}}+[v+1]^{- \frac{1}{2}}+t^{2}+2\geq2= \frac {2}{3} \cdot3 \\ &\geq\delta_{0}\biggl[{\frac{u}{1+u}}t^{3}+2\biggr]= \delta_{0} g(t,u). \end{aligned}$$

So f and g satisfy the conditions of Theorem 3.1. Thus by Theorem 3.1 the boundary value problem (4.1) has a unique positive solution in \(P_{h}\), where \(h(t)=t^{\alpha-1}=t^{\frac{3}{2}}\), \(t\in[0,1]\).

Example 4.2

Consider the following boundary value problem:

$$ \textstyle\begin{cases} -D_{\frac{1}{2}}^{\frac{5}{2}} u(t)={({\frac{u(t)}{1+u(t)}})}^{\frac {1}{4}}+[u(t)+1]^{- \frac{1}{3}}+t^{3}+u(t)^{\frac{1}{3}}+t^{2}+1, \quad 0< t< 1,\\ u(0)=D_{\frac{1}{2}}u(0)=0, \qquad u(1)=\mu\int_{0}^{1}u(s)\,d_{\frac{1}{2}}s. \end{cases} $$
(4.2)

We let

$$f(t,u,v )={\biggl({\frac{u}{1+u}}\biggr)}^{\frac{1}{4}}+[v+1]^{- \frac {1}{3}}+t^{3}, \qquad g(t,u)=u^{\frac{1}{3}}+t^{2}+1, \qquad \gamma={ \frac {1}{3}},\qquad \mu=\frac{1}{2}. $$

It is not difficult to find that \(f(t,x,y):[0,1]\times[0, +\infty )\times[0,+\infty)\rightarrow[0,+\infty)\) is continuous, increasing with respect to the second variable, and decreasing with respect to the third variable and that \(g(t,x): [0,1]\times[0, +\infty)\rightarrow[0,+\infty)\) and increasing with respect to the second variable. We also have

$$\begin{aligned} &g(t,\lambda u )=\lambda^{\frac{1}{3}}u^{\frac{1}{3}}+t^{2}+1 \geq \lambda^{\frac{1}{3}}\bigl[u^{\frac{1}{3}}+t^{2}+1\bigr] = \lambda^{\gamma}g(t,u), \quad \lambda\in(0,1), \\ & \begin{aligned} f\bigl(t,\lambda u,\lambda^{-1}v \bigr)&={\biggl({ \frac{\lambda u}{1+\lambda u}}\biggr)}^{\frac{1}{4}}+\bigl[\lambda^{-1}v+1 \bigr]^{- \frac{1}{3}}+t^{3} \\ &\geq\lambda^{\frac{1}{3}} \biggl\{ {\biggl({\frac{ u}{1+ u}} \biggr)}^{\frac {1}{4}}+[v+\lambda]^{- \frac{1}{3}}+t^{3} \biggr\} \\ &\geq\lambda \biggl\{ {\biggl({\frac{ u}{1+ u}}\biggr)}^{\frac{1}{4}}+[v+1]^{- \frac{1}{3}}+t^{3} \biggr\} \\ &=\lambda f(t,u,v). \end{aligned} \end{aligned}$$

If we take \(\delta_{0} =1>0\), then we have

$$f(t,u,v)={\biggl({\frac{u}{1+u}}\biggr)}^{\frac{1}{4}}+[v+1]^{- \frac {1}{3}}++t^{3} \leq u^{\frac{1}{4}}+t^{2}+1\leq u^{\frac{1}{3}}+t^{2}+1= \delta_{0} g(u,t). $$

So f and g satisfy the conditions of Theorem 3.2. Thus by Theorem 3.2 the boundary value problem (4.2) has a unique positive solution in \(P_{h}\), where \(h(t)=t^{\alpha-1}=t^{\frac{3}{2}}\), \(t\in[0,1]\).