1 Introduction

The differential equation arises in the modeling of different physical and natural phenomena, control system, nonlinear flow laws and many other branches of engineering. Fractional calculus is the extension of integer order calculus to arbitrary order calculus. With the development of fractional calculus, fractional differential equations have wide applications. In these years, there are many papers concerning integer order differential equations with p-Laplacian [1–7] and fractional differential equations with p-Laplacian [8–18].

By means of the monotone iterative technique, Sun et al. [1] investigated positive solutions for the following problems for the p-Laplacian operator:

$$\begin{aligned}& \bigl(\phi_{p} \bigl(u'(t) \bigr)\bigr)'+q(t) f\bigl(t,u(t),u'(t) \bigr)=0,\quad t\in (0,1), \\& u(0)-\alpha u'(\xi )=0,\quad u(1)+\beta u'(\eta )=0. \end{aligned}$$

Here \(\xi , \eta \in (0,1)\), \(\alpha \geq 0, \beta \geq 0\).

In [16], Liu et al. discussed the four-point problem for a class of fractional differential equation with mixed fractional derivatives and with p-Laplacian operator

$$\begin{aligned} &D_{0^{+}}^{\alpha } \bigl(\varphi_{p}\bigl(^{c}D_{0^{+}}^{\beta }u(t) \bigr) \bigr)= f\bigl(t,u(t),^{c}D_{0^{+}}^{\beta }u(t) \bigr),\quad t\in (0,1), \\ &^{c}D_{0^{+}}^{\beta }u(0)=u'(0)=0,\quad u(1)=r_{1}u(\eta ),\qquad {}^{c}D_{0^{+}}^{\beta }u(1)=r_{2}^{c}D_{0^{+}}^{\beta }u( \xi ). \end{aligned}$$

Based on the method of upper and lower solutions, they study the existence of positive solutions of the above boundary problem.

Motivated by the aforementioned work, this work discusses the existence of positive solutions for the following fractional differential equation:

$$\begin{aligned} \textstyle\begin{cases} D_{0^{+}}^{\beta }[\phi_{p}(D_{0^{+}}^{\alpha } u(t))]+f(t,u(t))=0, &t\in (0,1), \\ {}[\phi_{p}(D_{0^{+}}^{\alpha } u(0))]^{(i)}=0, &i=1,2,\ldots ,m-1, \\ u(0)-aD_{0^{+}}^{\alpha }u(\xi )=0, \\ (D_{0^{+}}^{\alpha } u(0))^{(j)}=0, &j=1,2,\ldots ,n-1, \\ u(1)+bD_{0^{+}}^{\alpha }u(\eta )=0, \end{cases}\displaystyle \end{aligned}$$
(1.1)

where \(0< n-1<\alpha \leq n, 0<m-1<\beta \leq m \) and \(m+n-1<\alpha +\beta \leq m+n, \phi_{p}(u)=\vert u \vert ^{p-2}u\), \(p>1\). \(D_{0^{+}}^{\alpha }\) and \(D_{0^{+}}^{\beta }\) are the Caputo fractional derivatives. We use the monotone iterative technique to obtain the existence of positive solutions for the boundary value problem and get iterative schemes for approximating the solutions. A function \(u(t)\) is a positive solution of the boundary value problem (1.1) if and only if \(u(t)\) satisfies the boundary value problem (1.1), and \(u(t)\geq 0\) for \(t\in [0,1]\). We will always suppose the following conditions are satisfied:

(\(H_{1}\)):

\(a,b\in (0,+\infty )\) are constants, \(0<\xi <\eta <1\);

(\(H_{2}\)):

\(f(t,u):[0,1]\times [0,\infty )\rightarrow [0,\infty )\) is continuous, and \(f(t,u)\not \equiv 0\) on any subinterval of \(t\in (0,1)\) for fixed \(u\in [0,\infty )\).

2 Preliminaries

To show the main result of this work, we give the following some basic definitions, which can be found in [19, 20].

Definition 2.1

The fractional integral of order \(\alpha >0\) of a function \(y:(0,+\infty )\rightarrow \mathbb{R}\) is given by

$$ I^{\alpha }_{0+}y(t)=\frac{1}{\Gamma (\alpha )} \int_{0}^{t}(t-s)^{ \alpha -1}y(s)\,\mathrm{d} s, $$

provided that the right side is pointwise defined on \((0,+\infty )\), where

$$ \Gamma (\alpha )= \int_{0}^{+\infty }e^{-x}x^{\alpha -1}\,{\mathrm{d}}x. $$

Definition 2.2

For a continuous function \(y:(0,+\infty ) \rightarrow \mathbb{R}\), the Caputo derivative of fractional order \(\alpha >0\) is defined as

$$ D^{\alpha }_{0+}y(t)=\frac{1}{\Gamma(n-\alpha )} \int_{0}^{t}(t-s)^{n- \alpha -1}y^{(n)}(s) \,\mathrm{d} s, $$

where \(n=[\alpha ]+1\), provided that the right side is pointwise defined on \((0,+\infty )\).

3 Main result

Lemma 3.1

The boundary value problem (1.1) is equivalent to the following equation:

$$\begin{aligned} u(t) =&a\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int _{0}^{\xi }(\xi -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{ \Gamma (\beta )} \biggr) \\ &{}+ \frac{1}{\Gamma (\alpha )} \int_{0}^{t}(t-s)^{\alpha -1}\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{s}(s-\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \end{aligned}$$

or

$$\begin{aligned} u(t) =&b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma ( \beta )} \biggr) \\ &{}+ \frac{1}{\Gamma (\alpha )} \int_{0}^{1}(1-s)^{\alpha -1}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &{}- \frac{1}{\Gamma (\alpha )} \int_{0}^{t}(t-s)^{\alpha -1}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s, \end{aligned}$$

where σ is the unique solution of the equation

$$\begin{aligned} &a\phi_{q} \biggl( \frac{\int_{0}^{t}(t-\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{\xi }(\xi - \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ &\quad =-\frac{1}{\Gamma (\alpha )} \int_{0}^{1}(1-s)^{\alpha -1}\phi_{q} \biggl( \frac{\int_{0}^{t}(t-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \\ &\quad\quad {}-\frac{\int_{0}^{s}(s-\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &\quad\quad {}-b\phi_{q} \biggl( \frac{\int_{0}^{t}(t-\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}- \frac{\int_{0}^{\eta }(\eta - \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr). \end{aligned}$$
(3.1)

Proof

From \(D_{0^{+}}^{\beta }[\phi_{p}(D_{0^{+}}^{\alpha } u(t))]+h(t) f(t,u(t))=0\), we get

$$\begin{aligned} \phi_{p} \bigl(D_{0^{+}}^{\alpha } u(t) \bigr)=c_{0}+c_{1}t+\cdots +c_{m-1}t^{m-1}- \frac{1}{ \Gamma (\beta )} \int_{0}^{t}(t-\tau )^{\beta -1}f \bigl(\tau ,u( \tau ) \bigr) \,\mathrm{d}\tau . \end{aligned}$$

In view of \([\phi_{p}(D_{0^{+}}^{\alpha } u(0))]^{(i)}=0,i=1,2,\ldots ,m-1\), we obtain

$$ c_{1}=c_{2}=\cdots =c_{m-1}=0, $$

that is,

$$\begin{aligned} \phi_{p} \bigl(D_{0^{+}}^{\alpha } u(t) \bigr)=c_{0}-\frac{1}{\Gamma (\beta )} \int _{0}^{t}(t-\tau )^{\beta -1}f \bigl(\tau ,u( \tau ) \bigr)\,\mathrm{d}\tau . \end{aligned}$$
(3.2)

By (3.2), we have

$$\begin{aligned} D_{0^{+}}^{\alpha } u(t)=\phi_{q} \biggl( c_{0}-\frac{1}{\Gamma (\beta )} \int_{0}^{t}(t-\tau )^{\beta -1}f \bigl(\tau ,u( \tau ) \bigr)\,\mathrm{d}\tau \biggr). \end{aligned}$$

For \(t\in [0,1]\), integrating from 0 to t, we get

$$\begin{aligned} u(t)={}&d_{0}+d_{1}t+\cdots +d_{n-1}t^{n-1} \\ &{}+\frac{1}{\Gamma (\alpha )} \int_{0}^{t}(t-s)^{\alpha -1}\phi_{q} \biggl( c_{0}-\frac{1}{\Gamma ( \beta )} \int_{0}^{s}(s-\tau )^{\beta -1}f \bigl(\tau ,u( \tau ) \bigr)\mathrm{d} \tau \biggr)\,\mathrm{d} s. \end{aligned}$$

In view of \((D_{0^{+}}^{\alpha } u(0))^{(j)}=0,j=1,2,\ldots ,n-1\), we obtain

$$ d_{1}=d_{2}=\cdots =d_{n-1}=0, $$

that is,

$$\begin{aligned} u(t)=d_{0}+\frac{1}{\Gamma (\alpha )} \int_{0}^{t}(t-s)^{\alpha -1}\phi _{q} \biggl( c_{0}-\frac{1}{\Gamma (\beta )} \int_{0}^{s}(s-\tau )^{ \beta -1}f \bigl(\tau ,u( \tau ) \bigr)\,\mathrm{d}\tau \biggr)\,\mathrm{d} s. \end{aligned}$$

By the use of \(u(0)-aD_{0^{+}}^{\alpha }u(\xi )=0\), we obtain

$$\begin{aligned} d_{0}=a\phi_{q} \biggl( c_{0}- \frac{1}{\Gamma (\beta )} \int_{0}^{\xi }( \xi -\tau )^{\beta -1}f \bigl( \tau ,u(\tau ) \bigr)\,\mathrm{d}\tau \biggr). \end{aligned}$$
(3.3)

By \(u(1)+bD_{0^{+}}^{\alpha }u(\eta )=0\), we obtain

$$\begin{aligned} d_{0}&=-\frac{1}{\Gamma (\alpha )} \int_{0}^{1}(1-s)^{\alpha -1}\phi _{q} \biggl( c_{0}-\frac{1}{\Gamma (\beta )} \int_{0}^{s}(s-\tau )^{ \beta -1}f \bigl(\tau ,u( \tau ) \bigr)\,\mathrm{d}\tau \biggr)\,\mathrm{d} s \\ &\quad {}-b\phi_{q} \biggl( c_{0}-\frac{1}{\Gamma (\beta )} \int_{0}^{\eta }( \eta -\tau )^{\beta -1}f \bigl( \tau ,u(\tau ) \bigr)\,\mathrm{d}\tau \biggr). \end{aligned}$$
(3.4)

By (3.3) and (3.4), we get

$$\begin{aligned} &a\phi_{q} \biggl( c_{0}-\frac{1}{\Gamma (\beta )} \int_{0}^{\xi }( \xi -\tau )^{\beta -1}f \bigl( \tau ,u(\tau ) \bigr)\,\mathrm{d}\tau \biggr) \\ &\quad =-\frac{1}{\Gamma (\alpha )} \int_{0}^{1}(1-s)^{\alpha -1}\phi_{q} \biggl( c_{0}-\frac{1}{\Gamma (\beta )} \int_{0}^{s}(s-\tau )^{\beta -1}f \bigl( \tau ,u(\tau ) \bigr)\,\mathrm{d}\tau \biggr)\,\mathrm{d} s \\ &\qquad {}-b\phi_{q} \biggl( c_{0}-\frac{1}{\Gamma (\beta )} \int_{0}^{\eta }( \eta -\tau )^{\beta -1}f \bigl( \tau ,u(\tau ) \bigr)\,\mathrm{d}\tau \biggr). \end{aligned}$$
(3.5)

Let

$$\begin{aligned} F(t) =&a\phi_{q} \biggl( \frac{\int_{0}^{t}(t-\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{\xi }(\xi - \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ &{}+\frac{1}{\Gamma (\alpha )} \int_{0}^{1}(1-s)^{\alpha -1}\phi_{q} \biggl( \frac{\int_{0}^{t}(t-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \\ &\quad\quad {}-\frac{\int_{0}^{s}(s-\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &\quad \quad {}+b\phi_{q} \biggl( \frac{\int_{0}^{t}(t-\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} - \frac{\int_{0}^{\eta }(\eta - \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr), \end{aligned}$$
(3.6)

obviously, \(F(t)\) is continuous and nondecreasing for \(t\in [0,1]\). From a direct calculation, we get

$$\begin{aligned} F(0) =&a\phi_{q} \biggl( -\frac{\int_{0}^{\xi }(\xi -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ &{}+\frac{1}{\Gamma (\alpha )} \int_{0}^{1}(1-s)^{\alpha -1}\phi_{q} \biggl( -\frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &{}+b\phi_{q} \biggl(-\frac{\int_{0}^{\eta }(\eta -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ \leq &a\phi_{q} \biggl( -\frac{\int_{0}^{\xi }(\xi -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ &{}+b \phi_{q} \biggl(-\frac{ \int_{0}^{\eta }(\eta -\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \biggr) \\ < &0 \end{aligned}$$

and

$$\begin{aligned} F(1) =&a\phi_{q} \biggl( \frac{\int_{0}^{1}(1-\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{\xi }(\xi - \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ &{}+\frac{1}{\Gamma (\alpha )} \int_{0}^{1}(1-s)^{\alpha -1}\phi_{q} \biggl( \frac{\int_{0}^{1}(1-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{s}(s-\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &{}+b\phi_{q} \biggl( \frac{\int_{0}^{1}(1-\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{\eta }(\eta - \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ \geq &a\phi_{q} \biggl( \frac{\int_{\xi }^{1}(1-\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ &{}+b \phi_{q} \biggl( \frac{ \int_{\eta }^{1}(1-\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{ \Gamma (\beta )} \biggr) \\ >& 0, \end{aligned}$$

so we see that the equation \(F(t)=0\) has a unique solution for \(t\in (0,1)\). Let σ be the unique solution of the equation \(F(t)=0\), then by (3.5) and (3.6), we obtain

$$ c_{0}=\frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}. $$

Consequently, we get

$$\begin{aligned} u(t) =&a\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int _{0}^{\xi }(\xi -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{ \Gamma (\beta )} \biggr) \\ &{}+ \frac{1}{\Gamma (\alpha )} \int_{0}^{t}(t-s)^{\alpha -1}\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \\ & {}-\frac{\int_{0}^{s}(s-\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \end{aligned}$$

or

$$\begin{aligned} u(t) =&b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma ( \beta )} \biggr) \\ &{}+ \frac{1}{\Gamma (\alpha )} \int_{0}^{1}(1-s)^{\alpha -1}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &{}- \frac{1}{\Gamma (\alpha )} \int_{0}^{t}(t-s)^{\alpha -1}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s. \end{aligned}$$

The proof is complete. □

Let the Banach space \(E=C[0,1]\) be endowed with the norm \(\Vert u \Vert = \max_{t\in [0,1]}\vert u(t) \vert \). Define the cone P by \(P=\{u\in E, u(t) \geq 0\}\). Then define the operator \(T:P\rightarrow E\), If \(0\leq t \leq \sigma \),

$$\begin{aligned} Tu(t) =&a\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int _{0}^{\xi }(\xi -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{ \Gamma (\beta )} \biggr) \\ &{}+ \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{s}(s-\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s, \end{aligned}$$

and if \(\sigma \leq t\leq 1\),

$$\begin{aligned} Tu(t) =&b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma ( \beta )} \biggr) \\ &{}+ \int_{0}^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &{}- \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s. \end{aligned}$$

Here σ is defined by (3.1). Obviously, \(u(t)\) is a solution of problem (1.1) if and only if \(u(t)\) is a fixed point of T.

The following theorem is the main result in this paper.

Theorem 3.1

Assume there exists \(r>0\) such that:

$$\begin{aligned} &\mathrm{(C1)}\quad f(t,u_{1})\leq f(t,u_{2}) \quad \textit{for any } 0\leq t\leq 1, 0\leq u_{1}\leq u_{2}\leq r ; \\ &\mathrm{(C2)} \quad \max_{t\in [0,\xi ]}f(t,r)\leq \biggl( \frac{b \Gamma (\alpha +1)}{\xi^{\alpha }} \biggr)^{p-1}\frac{(\eta -\xi )^{\beta }}{ \xi^{\beta }}\min _{t\in [\xi ,\eta ]}f(t,0) , \\ & \hphantom{\mathrm{(C2)} \quad }\max_{t\in [0,1]}f(t,r)\leq \bigl[a \Gamma (\alpha +1) \bigr]^{p-1}(\eta -\xi )^{\beta }\min_{t\in [\xi ,\eta ]}f(t,0) ; \\ &\mathrm{(C3)}\quad \max_{t\in [0,1]}f(t,r)\leq M=\min \biggl\{ \biggl( \frac{r}{L_{1}+1} \biggr)^{p-1}, \biggl(\frac{r}{L_{2}+1} \biggr)^{p-1} \biggr\} , \\ &\hphantom{\mathrm{(C3)}\quad}\textit{where} \\ &\hphantom{\mathrm{(C3)}\quad} L_{1}=a \biggl(\frac{\eta^{\beta }}{\Gamma (\beta +1)} \biggr)^{q-1}+ \frac{\eta^{\alpha }}{\Gamma (\alpha +1)} \biggl( \frac{\eta^{\beta }}{\Gamma ( \beta +1)} \biggr)^{q-1} , \\ & \hphantom{\mathrm{(C3)}\quad} L_{2}=b \biggl(\frac{\eta^{\beta }}{\Gamma (\beta +1)} \biggr)^{q-1} + \frac{1}{\Gamma ( \alpha +1)} \biggl(\frac{1}{\Gamma (\beta +1)} \biggr)^{q-1} . \end{aligned}$$

Then the problem (1.1) has two positive solutions \(w^{*}\) and \(v^{*}\) such that

$$\begin{aligned} &0\leq w^{*}< r, \quad 0\leq v^{*}< r, \\ &w^{*}=\lim_{k\rightarrow +\infty }w_{k},\qquad v^{*}=\lim_{k\rightarrow +\infty }v_{k}, \\ &w_{k}=Tw_{k-1}, \qquad v_{k}=Tv_{k-1}, \quad k=1,2,\ldots , \\ &w_{0}=\max \biggl\{ \frac{rL_{1}}{L_{1}+1}, \frac{rL_{2}}{L_{2}+1} \biggr\} , \quad v_{0}=0. \end{aligned}$$

Proof

We take four steps to prove the theorem.

Step 1. We prove \(\xi <\sigma <\eta \).

If \(0<\sigma <\xi \), then

$$\begin{aligned} Tu(\sigma ) =&a\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int _{0}^{\xi }(\xi -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{ \Gamma (\beta )} \biggr) \\ &{}+ \frac{1}{\Gamma (\alpha )} \int_{0}^{\sigma }(\sigma -s)^{\alpha -1} \phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{s}(s- \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \,\mathrm{d} s \\ < &\frac{1}{\Gamma (\alpha )} \int_{0}^{\sigma }(\sigma -s)^{\alpha -1} \phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{s}(s- \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \,\mathrm{d} s \\ \leq &\frac{1}{\Gamma (\alpha )} \int_{0}^{\xi }(\xi -s)^{\alpha -1} \phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \leq &\frac{\xi^{\alpha }}{\Gamma (\alpha +1)} \biggl( \frac{\xi^{ \beta }\max_{t\in [0,\xi ]}f(t,r)}{\Gamma (\beta +1)} \biggr)^{q-1} \end{aligned}$$

and

$$\begin{aligned} Tu(\sigma ) =&b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int _{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \biggr) \\ &{}+ \frac{1}{\Gamma (\alpha )} \int_{0}^{1}(1-s)^{\alpha -1}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &{}- \frac{1}{\Gamma (\alpha )} \int_{0}^{\sigma }(\sigma -s)^{\alpha -1} \phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau )) \,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma - \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \,\mathrm{d} s \\ >&b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma ( \beta )} \biggr) \\ \geq &b\phi_{q} \biggl(\frac{\int_{\xi }^{\eta }(\eta -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ \geq& b \biggl( \frac{(\eta -\xi )^{\beta }\min_{t\in [\xi ,\eta ]}f(t,0)}{\Gamma ( \beta +1)} \biggr)^{q-1}. \end{aligned}$$

In view of (C2), we get a contradiction. So we can obtain \(\sigma > \xi \).

If \(1>\sigma >\eta \), then

$$\begin{aligned} Tu(\sigma ) =&a\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int _{0}^{\xi }(\xi -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{ \Gamma (\beta )} \biggr) \\ &{}+ \frac{1}{\Gamma (\alpha )} \int_{0}^{\sigma }(\sigma -s)^{\alpha -1} \phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{s}(s- \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \,\mathrm{d} s \\ >&a\phi_{q} \biggl( \frac{\int_{\xi }^{\eta }(\eta -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ \geq& a \biggl( \frac{( \eta -\xi )^{\beta }\min_{t\in [\xi ,\eta ]}f(\tau ,0)}{\Gamma (\beta +1)} \biggr)^{q-1} \end{aligned}$$

and

$$\begin{aligned} Tu(\sigma ) =&b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int _{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \biggr) \\ &{}+ \frac{1}{\Gamma (\alpha )} \int_{0}^{1}(1-s)^{\alpha -1}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &{}- \frac{1}{\Gamma (\alpha )} \int_{0}^{\sigma }(\sigma -s)^{\alpha -1} \phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau )) \,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma - \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \,\mathrm{d} s \\ < & \frac{1}{\Gamma (\alpha )} \int_{0}^{1}(1-s)^{\alpha -1}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ < & \frac{1}{\Gamma (\alpha )} \int_{0}^{1}(1-s)^{\alpha -1}\phi_{q} \biggl( \frac{\int_{0}^{1}(1-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \leq & \frac{1}{\Gamma (\alpha +1)} \biggl(\frac{\max_{t\in [0,1]}f(t,r)}{ \Gamma (\beta +1)} \biggr)^{q-1}. \end{aligned}$$

In view of (C2), we get a contradiction. So we can obtain \(\sigma < \eta \).

Therefore we have \(\xi <\sigma <\eta \).

Step 2. We prove that \(T:P\rightarrow P\) is completely continuous.

For any \(u\in P\), if \(0\leq t\leq \sigma \),

$$\begin{aligned} Tu(t) =&a\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int _{0}^{\xi }(\xi -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{ \Gamma (\beta )} \biggr) \\ &{}+ \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{s}(s-\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \geq & \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi _{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{s}(s-\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \geq & 0. \end{aligned}$$

If \(\sigma \leq t\leq 1\),

$$\begin{aligned} Tu(t) =&b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma ( \beta )} \biggr) \\ &{}+ \int_{0}^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &{}- \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \geq & \int_{0}^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}\phi _{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau )) \,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma - \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &{}- \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \geq & \int_{t}^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}\phi _{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau )) \,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma - \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \geq & 0. \end{aligned}$$

So we get \(T:P\rightarrow P\). Obviously, T is continuous for the continuity of \(f(t,u)\).

Let \(\Omega \subset P\) be bounded, that is, there exists a positive constant l for any \(u\in \Omega \), and letting \(M_{1}= \max_{0\leq t\leq 1,0\leq u\leq l}f(t,u)+1\), then, for any \(u\in \Omega \), if \(0\leq t\leq \sigma \),

$$\begin{aligned} Tu(t) =&a\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int _{0}^{\xi }(\xi -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{ \Gamma (\beta )} \biggr) \\ &{}+ \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{s}(s-\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \leq &a\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ &{}+ \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \leq &a\phi_{q} \biggl( \frac{M_{1}\sigma^{\beta }}{\Gamma (\beta +1)} \biggr)+ \frac{\sigma^{\alpha }}{\Gamma (\alpha +1)}\phi_{q} \biggl( \frac{M _{1}\sigma^{\beta }}{\Gamma (\beta +1)} \biggr), \end{aligned}$$

if \(\sigma \leq t\leq 1\),

$$\begin{aligned} Tu(t) =&b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma ( \beta )} \biggr) \\ &{}+ \int_{0}^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &{}- \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \leq &b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ &{}+ \int_{0}^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \leq &b\phi_{q} \biggl(\frac{M_{1}\eta^{\beta }}{\Gamma (\beta +1)} \biggr)+\frac{1}{\Gamma (\alpha +1)} \phi_{q} \biggl( \frac{M_{1}}{\Gamma (\beta +1)} \biggr). \end{aligned}$$

Hence, \(T(\Omega )\) is uniformly bounded.

Now, we will prove that \(T(\Omega )\) is equicontinuous.

For each \(u\in \Omega \), if \(0\leq t_{1}< t_{2}\leq 1\),

$$\begin{aligned} & \bigl\vert (Tu) (t_{2})-(Tu) (t_{1}) \bigr\vert \\ &\quad = \biggl\vert \int_{0}^{t_{2}}\frac{(t_{2}-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{s}(s- \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &\qquad {}-\int_{0}^{t_{1}}\frac{(t_{1}-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{s}(s-\tau )^{\beta -1} f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d}s\biggr\vert \\ &\quad \leq \biggl\vert \int_{t_{1}}^{t_{2}}\frac{(t_{2}-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d}s \biggr\vert \\ &\qquad {}+ \int_{0}^{t_{1}} \biggl[ \frac{(t_{2}-s)^{\alpha -1}}{\Gamma (\alpha )}- \frac{(t_{1}-s)^{ \alpha -1}}{\Gamma (\alpha )} \biggr]\phi_{q} \biggl( \frac{\int_{0}^{ \sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{ \Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &\quad \leq \frac{(t_{2}-t_{1})^{\alpha }}{\Gamma (\alpha +1)}\phi_{q} \biggl( \frac{M_{1}\sigma^{\beta }}{\Gamma (\beta +1)} \biggr) +\frac{t_{2}^{\alpha }-t_{1}^{\alpha }}{\Gamma (\alpha +1)}\phi_{q} \biggl( \frac{M_{1}\sigma^{\beta }}{\Gamma (\beta +1)} \biggr) \end{aligned}$$

therefore, \(T(\Omega )\) is equicontinuous. Applying the Arzelá-Ascoli theorem, we conclude that T is a completely continuous operator.

Step 3. Let \(\overline{P}_{r}=\{u\in P: 0\leq \Vert u \Vert \leq r\}\), then we prove \(T:\overline{P}_{r}\rightarrow \overline{P}_{r}\).

For any \(u\in \overline{P}_{r}\), if \(0\leq t\leq \sigma \),

$$\begin{aligned} Tu(t) =&a\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int _{0}^{\xi }(\xi -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{ \Gamma (\beta )} \biggr) \\ &{}+ \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u( \tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{s}(s-\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \leq &a\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ &{}+ \int_{0}^{\sigma }\frac{(\sigma -s)^{\alpha -1}}{\Gamma (\alpha )} \phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \leq &a \biggl( \frac{\sigma^{\beta }M}{\Gamma (\beta +1)} \biggr)^{q-1}+ \frac{\sigma^{\alpha }}{\Gamma (\alpha +1)} \biggl( \frac{\sigma^{ \beta }M}{\Gamma (\beta +1)} \biggr)^{q-1} \\ \leq &a \biggl( \frac{\eta^{\beta }M}{\Gamma (\beta +1)} \biggr)^{q-1}+ \frac{ \eta^{\alpha }}{\Gamma (\alpha +1)} \biggl( \frac{\eta^{\beta }M}{ \Gamma (\beta +1)} \biggr)^{q-1} \\ =&M^{q-1}L_{1} \leq r. \end{aligned}$$

If \(t\leq \sigma \leq 1\),

$$\begin{aligned} Tu(t) =&b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma ( \beta )} \biggr) \\ &{}+ \int_{0}^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &{}- \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,u(\tau ))\mathrm{d} \tau }{\Gamma (\beta )}-\frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \leq &b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{\beta -1}f( \tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ &{}+ \int_{0}^{1}\frac{(1-s)^{ \alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{1}(1- \tau )^{\beta -1}f(\tau ,u(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \leq &b \biggl(\frac{\eta^{\beta }M}{\Gamma (\beta +1)} \biggr)^{q-1}+\frac{1}{ \Gamma (\alpha +1)} \biggl( \frac{M}{\Gamma (\beta +1)} \biggr)^{q-1} \\ =&M ^{q-1}L_{2} \leq r. \end{aligned}$$

Consequently, we get \(T:\overline{P}_{r}\rightarrow \overline{P}_{r}\).

Step 4. We prove \(w^{*}\) and \(v^{*}\) are two positive solutions of the problem (1.1).

Since \(w_{0}(t)=\max \{\frac{rL_{1}}{L_{1}+1}, \frac{rL_{2}}{L_{2}+1} \}\), obviously, \(w_{0}(t)\in \overline{P}_{r}\). Since \(w_{k+1}(t)=Tw _{k}(t), k=0,1,2,\ldots \) and \(T:\overline{P}_{r}\rightarrow \overline{P}_{r}\), we get \(w_{k}(t)\in \overline{P}_{r}\).

If \(0\leq t\leq \sigma \),

$$\begin{aligned} Tw_{0}(t) =&a\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,w_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}- \frac{ \int_{0}^{\xi }(\xi -\tau )^{\beta -1}f(\tau ,w_{0}(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \biggr) \\ &{}+ \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,w _{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{s}(s- \tau )^{\beta -1}f(\tau ,w_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \leq &a \biggl( \frac{\eta^{\beta }M}{\Gamma (\beta +1)} \biggr)^{q-1}+ \frac{ \eta^{\alpha }}{\Gamma (\alpha +1)} \biggl( \frac{\eta^{\beta }M}{ \Gamma (\beta +1)} \biggr)^{q-1} \\ =&M^{q-1}L_{1} \leq \frac{rL_{1}}{L_{1}+1}\leq w_{0}(t). \end{aligned}$$

If \(t\leq \sigma \leq 1\),

$$\begin{aligned} Tw_{0}(t) =&b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{ \beta -1}f(\tau ,w_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}- \frac{ \int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,w_{0}(\tau )) \,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ &{}+ \int_{0}^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,w_{0}(\tau )) \,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma - \tau )^{\beta -1}f(\tau ,w_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &{}- \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,w_{0}(\tau )) \,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma - \tau )^{\beta -1}f(\tau ,w_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \leq &b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{\beta -1}f( \tau ,w_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ &{}+ \int_{0}^{1}\frac{(1-s)^{ \alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{1}(1- \tau )^{\beta -1}f(\tau ,w_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \leq &b \biggl(\frac{\eta^{\beta }M}{\Gamma (\beta +1)} \biggr)^{q-1}+\frac{1}{ \Gamma (\alpha +1)} \biggl( \frac{M}{\Gamma (\beta +1)} \biggr)^{q-1} \\ =&M ^{q-1}L_{2} \leq \frac{L_{2}r}{L_{2}+1}\leq w_{0}(t). \end{aligned}$$

So we get \(w_{1}(t)\leq w_{0}(t)\), and on the basis of the definition of T, we obtain

$$ w_{2}(t)=Tw_{1}(t)\leq Tw_{0}(t)=w_{1}(t). $$

Hence by induction we obtain

$$ w_{k+1}(t)\leq w_{k}(t), \quad k=0,1,2,\ldots . $$

Thus, we get \(w^{*}\in \overline{P}_{r}\) such that \(w_{k}\rightarrow w^{*}\). Applying the continuity of T and \(w_{k+1}(t)=T w_{k}(t)\), we get \(w^{*}(t)=T w^{*}(t)\), hence \(w^{*}(t)\) is a positive solution of problem (1.1).

Since \(v_{0}(t)=0\), obviously, \(v_{0}(t)\in \overline{P}_{r}\). Since \(v_{k+1}(t)=Tv_{k}(t), k=0,1,2,\ldots \) and \(T:\overline{P}_{r}\rightarrow \overline{P}_{r}\), we get \(v_{k}(t)\in \overline{P}_{r}\).

If \(0\leq t\leq \sigma \),

$$\begin{aligned} Tv_{0}(t) =&a\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,v_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}- \frac{ \int_{0}^{\xi }(\xi -\tau )^{\beta -1}f(\tau ,v_{0}(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \biggr) \\ &{}+ \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,v _{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{s}(s- \tau )^{\beta -1}f(\tau ,v_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \geq &a\phi_{q} \biggl( \frac{\int_{0}^{\sigma }(\sigma -\tau )^{ \beta -1}f(\tau ,v_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}-\frac{ \int_{0}^{\xi }(\xi -\tau )^{\beta -1}f(\tau ,v_{0}(\tau ))\mathrm{d} \tau }{\Gamma (\beta )} \biggr) \\ \geq & 0=v_{0}(t). \end{aligned}$$

If \(t\leq \sigma \leq 1\),

$$\begin{aligned} Tv_{0}(t) =&b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{ \beta -1}f(\tau ,v_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )}- \frac{ \int_{0}^{\sigma }(\sigma -\tau )^{\beta -1}f(\tau ,v_{0}(\tau )) \,\mathrm{d}\tau }{\Gamma (\beta )} \biggr) \\ &{}+ \int_{0}^{1}\frac{(1-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,v_{0}(\tau )) \,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma - \tau )^{\beta -1}f(\tau ,v_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ &{}- \int_{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}\phi_{q} \biggl( \frac{\int_{0}^{s}(s-\tau )^{\beta -1}f(\tau ,v_{0}(\tau )) \,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{\sigma }(\sigma - \tau )^{\beta -1}f(\tau ,v_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \biggr)\,\mathrm{d} s \\ \geq &b\phi_{q} \biggl(\frac{\int_{0}^{\eta }(\eta -\tau )^{\beta -1}f(\tau ,v_{0}(\tau ))\,\mathrm{d}\tau }{\Gamma (\beta )} \\ &{}-\frac{\int_{0}^{ \sigma }(\sigma -\tau )^{\beta -1}f(\tau ,v_{0}(\tau ))\,\mathrm{d}\tau }{ \Gamma (\beta )} \biggr) \\ \geq &0=v_{0}(t). \end{aligned}$$

So we get \(v_{1}(t)\geq v_{0}(t)\), and on the basis of the definition of T, we obtain

$$ v_{2}(t)=Tv_{1}(t)\geq Tv_{0}(t)=v_{1}(t). $$

Hence by induction we obtain

$$ v_{k+1}(t)\geq v_{k}(t),\quad k=0,1,2,\ldots. $$

Thus, we get \(v^{*}\in \overline{P}_{r}\) such that \(v_{k}\rightarrow v^{*}\). Applying the continuity of T and \(v_{k+1}(t)=T v_{k}(t)\), we get \(v^{*}(t)=T v^{*}(t)\), hence \(v^{*}(t)\) is a positive solution of problem (1.1). □

Remark 3.1

The positive solutions \(w^{*}(t)\) and \(v^{*}(t)\) of problem (1.1) in Theorem 3.1 may coincide, and in this case the problem (1.1) has a positive solution in \(\overline{P}_{r}\).

4 Example

In this section, we give a simple example to explain the main theorem.

Example 4.1

For the problem (1.1), let \(\alpha =1.6, \beta =1.5, a=12, b=3, p=3, \xi =0.6, \eta =0.9, r=80\). \(f(t,u)=5t+\sin t+0.5u\).

From a direct calculation, we get \(q=1.5\),

$$\begin{aligned}& \max_{t\in [0,\xi ]}f(t,r) = 5\times 0.6+\sin 0.6+0.5\times 80=43+ \sin 0.6< 44, \\& \max_{t\in [0,1]}f(t,r) = 5\times 1+\sin 1+0.5\times 80=85+ \sin 1< 86, \\& \biggl( \frac{b \Gamma (\alpha +1)}{\xi^{\alpha }} \biggr)^{p-1}\frac{(\eta - \xi )^{\beta }}{\xi^{\beta }} \min _{t\in [\xi ,\eta ]}f(t,0) \approx 33.3471\min_{t\in [\xi ,\eta ]}f(t,0)>33 \times 3=99, \\& \bigl[a \Gamma (\alpha +1) \bigr]^{p-1}(\eta -\xi )^{\beta }\min _{t\in [\xi ,\eta ]}f(t,0)\approx 48.3603 \min_{t\in [\xi ,\eta ]}f(t,0)>48 \times 3=144, \\& L_{1} = a \biggl( \frac{\eta^{\beta }}{\Gamma (\beta +1)} \biggr)^{q-1}+ \frac{ \eta^{\alpha }}{\Gamma (\alpha +1)} \biggl( \frac{\eta^{\beta }}{\Gamma (\beta +1)} \biggr)^{q-1}\approx 10.0908, \\& L_{2} = b \biggl(\frac{\eta^{\beta }}{\Gamma (\beta +1)} \biggr)^{q-1} + \frac{1}{ \Gamma (\alpha +1)} \biggl(\frac{1}{\Gamma (\beta +1)} \biggr)^{q-1} \approx 3.0109, \\& M \approx 52.0366. \end{aligned}$$

So we get

$$\begin{aligned}& \max_{t\in [0,\xi ]}f(t,r)\leq \biggl( \frac{b \Gamma (\alpha +1)}{\xi^{\alpha }} \biggr)^{p-1}\frac{(\eta -\xi )^{\beta }}{\xi^{ \beta }}\min_{t\in [\xi ,\eta ]}f(t,0), \\& \max_{t\in [0,1]}f(t,r)\leq \bigl[a \Gamma (\alpha +1) \bigr]^{p-1} ( \eta -\xi )^{\beta }\min_{t\in [\xi ,\eta ]}f(t,0), \\& \max_{t\in [0,1]}f(t,r)\leq M. \end{aligned}$$

Then all the conditions of Theorem 3.1 are satisfied. Hence, by Theorem 3.1, we see that the aforementioned problem has two positive solutions \(w^{*}\) and \(v^{*}\).

5 Conclusions

The monotone iterative technique is used to solve the problem of a kind of nonlinear four-point Caputo fractional differential equation with p-Laplacian operator. Under certain nonlinear growth conditions of the nonlinearity, we acquire the existence of positive solutions for the boundary value problem and get iterative schemes for approximating the solutions.