1 Introduction

Fractional calculus has widespread applications in many fields of science and engineering, for example, physics, viscoelasticity, continuum mechanics, bioengineering, rheology, electrical networks, control theory of dynamical systems, optics and signal processing, and so on [1, 2].

Since the discussion of many problems can be summed up in the study of boundary value problems (BVPs for short) to nonlinear fractional differential equations, recently, the existence of solutions or positive solutions of BVPs for nonlinear fractional differential equations has received considerable attention from many authors, see [3–26] and the references therein.

In particular, in 2009, by using nonlinear alternative of Leray–Schauder type and Guo–Krasnoselskii’s fixed point theorem, Bai and Qiu [5] obtained the existence of a positive solution to the singular BVP

$$ \textstyle\begin{cases} ({}^{C}D_{0+}^{q}u)(t)+f(t,u(t))=0, \quad t\in (0,1), \\ u(0)=u^{\prime }(1)=u^{\prime \prime }(0)=0, \end{cases} $$
(1)

where \(2< q\leq 3 \) is a real number, \(f:(0,1]\times [0,+\infty )\rightarrow [0,+\infty )\) is continuous, and \(\lim _{t\rightarrow 0^{+}}f(t,\cdot )=+\infty \).

In 2012, Cabada and Wang [7] studied the existence of a positive solution for the BVP with integral boundary conditions

$$ \textstyle\begin{cases} ({}^{C}D_{0+}^{q}u)(t)+f(t,u(t))=0, \quad t\in (0,1), \\ u(0)=u^{\prime \prime }(0)=0,\quad\quad u(1)=\lambda \int ^{1}_{0}u(s)\,ds, \end{cases} $$
(2)

where \(2< q<3\), \(0<\lambda <2\), and \(f:[0,1] \times [0,+\infty )\rightarrow [0,+\infty )\) is continuous. Their analysis relied on Guo–Krasnoselskii’s fixed point theorem.

In 2014, Cabada and Hamdi [25] investigated the BVP with integral boundary conditions

$$ \textstyle\begin{cases} (D_{0+}^{q}u)(t)+f(t,u(t))=0,\quad t\in (0,1), \\ u(0)=u^{\prime }(0)=0,\quad\quad u(1)=\lambda \int _{0}^{1} u(s)\,ds, \end{cases} $$
(3)

where \(2< q\leq 3\), \(D_{0+}^{q}\) denotes the Riemann–Liouville fractional derivative, \(0<\lambda <q\), and \(f:[0,1] \times [0,+\infty )\rightarrow [0,+\infty )\) is a continuous function. The authors proved the existence of a positive solution to BVP (3) by employing Guo–Krasnoselskii’s fixed point theorem.

As it has been stated in [7], BVPs with integral boundary conditions have various applications in applied fields such as blood flow problems, chemical engineering, thermo-elasticity, underground water flow, population dynamics, and so forth. Motivated by the above-mentioned works, in this paper, we consider the existence of a positive solution for the following BVP of nonlinear fractional differential equation with integral boundary conditions:

$$ \textstyle\begin{cases} ({}^{C}D_{0+}^{q}u)(t)+f(t,u(t))=0, \quad t\in [0,1], \\ u^{\prime \prime }(0)=0, \\ \alpha u(0)-\beta u^{\prime }(0)=\int _{0}^{1}h_{1}(s)u(s)\,ds, \\ \gamma u(1)+\delta ({}^{C}D_{0+}^{\sigma }u)(1) =\int _{0}^{1}h_{2}(s)u(s)\,ds. \end{cases} $$
(4)

Throughout this paper, we always assume that \(2< q\leq 3\), \(0<\sigma \leq 1\), \(\alpha , \gamma , \delta \geq 0\), and \(\beta >0\) satisfying \(0<\rho :=(\alpha +\beta )\gamma + \frac{\alpha \delta }{\varGamma (2-\sigma )}<\beta [\gamma + \frac{\delta \varGamma (q)}{\varGamma (q-\sigma )} ]\), \(f :[0,1]\times [0,+\infty )\rightarrow [0,+\infty )\) and \(h_{i}\) (\(i=1,2\)): \([0,1] \rightarrow [0,+\infty )\) are continuous.

The main tool used is the following well-known Guo–Krasnoselskii’s fixed point theorem [27, 28].

Theorem 1.1

LetEbe a Banach space andKbe a cone in E. Assume that\(\varOmega _{1}\)and\(\varOmega _{2}\)are bounded open subsets ofEsuch that\(0 \in \varOmega _{1}\), \(\overline{\varOmega }_{1}\subset \varOmega _{2}\), and let\(T:K\cap (\overline{\varOmega }_{2}\backslash \varOmega _{1} ) \rightarrow K\)be a completely continuous operator such that either

  1. (1)

    \(\Vert Tu \Vert \leq \Vert u \Vert \)for\(u\in K\cap \partial \varOmega _{1}\)and\(\Vert Tu \Vert \geq \Vert u \Vert \)for\(u\in K\cap \partial \varOmega _{2}\), or

  2. (2)

    \(\Vert Tu \Vert \geq \Vert u \Vert \)for\(u\in K\cap \partial \varOmega _{1}\)and\(\Vert Tu \Vert \leq \Vert u \Vert \)for\(u\in K\cap \partial \varOmega _{2}\).

ThenThas a fixed point in\(K\cap (\overline{\varOmega }_{2}\setminus \varOmega _{1} )\).

2 Preliminaries

Let \([a,b] \) (\(-\infty < a< b<+\infty \)) be a finite interval on the real axis \(\mathbb{R}\), \(\mathbb{N}=\{1, 2, 3,\ldots\}\), \(\mu >0\) and \([\mu ]\) be the integer part of μ.

First, we present definitions of some spaces.

Let \(\operatorname{AC}[a,b]\) be the space of functions u which are absolutely continuous on \([a,b]\). For \(n\in \mathbb{N}\), we denote by \(\operatorname{AC}^{n}[a,b]\) the space of functions u which have continuous derivatives up to order \(n-1\) on \([a,b]\) such that \(u^{(n-1)}\in \operatorname{AC}[a,b]\). In particular, \(\operatorname{AC}^{1}[a,b]=\operatorname{AC}[a,b]\).

For \(m\in \mathbb{N}_{0}=\{0, 1, 2,\ldots\}\), we denote by \(C^{m}[a,b]\) the space of functions u which are m times continuously differentiable on \([a,b]\). In particular, for \(m=0\), \(C^{0}[a,b]=C[a,b]\) is the space of continuous functions u on \([a,b]\).

Next, we give the definitions of the Riemann–Liouville fractional integrals and fractional derivatives and the Caputo fractional derivatives on \([a,b]\), which may be found in [1].

Definition 2.1

The Riemann–Liouville fractional integrals \(I_{a+}^{\mu }u\) and \(I_{b-}^{\mu }u\) of order μ are defined by

$$ \bigl(I_{a+}^{\mu }u\bigr) (t):=\frac{1}{\varGamma (\mu )} \int _{a}^{t} \frac{u(s)\,ds}{(t-s)^{1-\mu }} \quad (t>a) $$

and

$$ \bigl(I_{b-}^{\mu }u\bigr) (t):=\frac{1}{\varGamma (\mu )} \int _{t}^{b} \frac{u(s)\,ds}{(s-t)^{1-\mu }} \quad (t< b), $$

respectively, where

$$ \varGamma (\mu )= \int _{0}^{+\infty }s^{\mu -1}e^{-s}\,ds. $$

Definition 2.2

The Riemann–Liouville fractional derivatives \(D_{a+}^{\mu }u\) and \(D_{b-}^{\mu }u\) of order μ are defined by

$$ \bigl(D_{a+}^{\mu }u\bigr) (t):= \biggl(\frac{d}{dt} \biggr)^{n}\bigl(I_{a+}^{n-\mu }u\bigr) (t)= \frac{1}{\varGamma (n-\mu )} \biggl(\frac{d}{dt} \biggr)^{n} \int _{a}^{t} \frac{u(s)\,ds}{(t-s)^{\mu -n+1}} \quad (t>a) $$

and

$$ \bigl(D_{b-}^{\mu }u\bigr) (t):= \biggl(-\frac{d}{dt} \biggr)^{n}\bigl(I_{b-}^{n-\mu }u\bigr) (t)= \frac{1}{\varGamma (n-\mu )} \biggl(-\frac{d}{dt} \biggr)^{n} \int _{t}^{b} \frac{u(s)\,ds}{(s-t)^{\mu -n+1}} \quad (t< b), $$

respectively, where \(n=[\mu ]+1\).

Definition 2.3

Let \(D_{a+}^{\mu }[u(s)](t)\equiv (D_{a+}^{\mu }u)(t)\) and \(D_{b-}^{\mu }[u(s)](t)\equiv (D_{b-}^{\mu }u)(t)\) be the Riemann–Liouville fractional derivatives of order μ, respectively. The Caputo fractional derivatives \({}^{C}D_{a+}^{\mu }u\) and \({}^{C}D_{b-}^{\mu }u\) of order μ on \([a,b]\) are defined by

$$ \bigl({}^{C}D_{a+}^{\mu }u\bigr) (t):= \Biggl(D_{a+}^{\mu } \Biggl[u(s)-\sum _{k=0}^{n-1} \frac{u^{(k)}(a)}{k!}(s-a)^{k} \Biggr] \Biggr) (t) $$

and

$$ \bigl({}^{C}D_{b-}^{\mu }u\bigr) (t):= \Biggl(D_{b-}^{\mu } \Biggl[u(s)-\sum _{k=0}^{n-1} \frac{u^{(k)}(b)}{k!}(b-s)^{k} \Biggr] \Biggr) (t), $$

respectively, where

$$ n= \textstyle\begin{cases} [\mu ]+1,&\mu \notin \mathbb{N}, \\ \mu , &\mu \in \mathbb{N}. \end{cases} $$
(5)

Lemma 2.1

(see [2])

Let\(\nu >\mu \). Then the equation\(({}^{C}D_{0+}^{\mu }I_{0+}^{\nu }u)(t)=(I_{0+}^{\nu -\mu }u)(t)\), \(t \in [0,1]\)is satisfied for\(u\in C [0,1 ]\).

Lemma 2.2

(see [1])

Letnbe given by (5). Then the following relations hold:

  1. (1)

    for\(k\in \{0,1,2,\ldots,n-1\}\), \({}^{C}D_{0+}^{\mu }t^{k}=0\);

  2. (2)

    if\(\nu >n\), \({}^{C}D_{0+}^{\mu }t^{\nu -1}= \frac{\varGamma (\nu )}{\varGamma (\nu -\mu )}t^{\nu -\mu -1}\).

Lemma 2.3

(see [1])

Letnbe given by (5). If\(u\in \operatorname{AC}^{n}[0,1]\)or\(u\in C^{n}[0,1]\), then

$$ \bigl(I_{0+}^{\mu }{{}^{C}D_{0+}^{\mu }}u \bigr) (t)=u(t)-\sum_{k=0}^{n-1} \frac{u^{k}(0)}{k!}t^{k}. $$

For convenience, we denote

$$\begin{aligned}& P_{i}=\frac{1}{\rho } \int _{0}^{1}(\alpha s+\beta )h_{i}(s) \,ds \end{aligned}$$

and

$$\begin{aligned}& Q_{i}=\frac{1}{\rho \varGamma (2-\sigma )} \int _{0}^{1}\bigl[\gamma \varGamma (2- \sigma ) (1-s)+\delta \bigr]h_{i}(s)\,ds,\quad i=1,2. \end{aligned}$$

Lemma 2.4

Let\((1-Q_{1})(1-P_{2})\neq P_{1}Q_{2}\). Then, for any\(y\in C [0,1 ]\), the BVP

$$ \textstyle\begin{cases} ({}^{C}D_{0+}^{q}u)(t)+y(t)=0, \quad t\in [0,1], \\ u^{\prime \prime }(0)=0, \\ \alpha u(0)-\beta u^{\prime }(0)=\int _{0}^{1}h_{1}(s)u(s)\,ds, \\ \gamma u(1)+\delta ({}^{C}D_{0+}^{\sigma }u)(1)=\int _{0}^{1}h_{2}(s)u(s)\,ds \end{cases} $$
(6)

has a unique solution

$$ u(t) = \int _{0}^{1} H(t,s) y(s) \,ds,\quad t \in [0,1], $$

here

$$\begin{aligned}& H(t,s)=G(t,s)+\sum_{i=1}^{2}\phi _{i}(t) \int _{0}^{1} G(\tau ,s)h_{i}( \tau )\,d \tau ,\quad (t,s)\in [0,1]\times [0,1], \end{aligned}$$

where

$$\begin{aligned}& G(t,s)=\frac{\alpha t+\beta }{\rho } \biggl[ \frac{\gamma (1-s)^{q-1}}{\varGamma (q)}+ \frac{\delta (1-s)^{q-\sigma -1}}{\varGamma (q-\sigma )} \biggr]- \textstyle\begin{cases} \frac{(t-s)^{q-1}}{\varGamma (q)},& 0\leq s\leq t\leq 1, \\ 0, &0\leq t\leq s\leq 1, \end{cases}\displaystyle \\& \phi _{1}(t)= \frac{\varGamma (2-\sigma )Q_{2}(\alpha t+\beta )+(1-P_{2}) [\gamma \varGamma (2-\sigma )(1-t)+\delta ]}{\rho \varGamma (2-\sigma ) [(1-Q_{1})(1-P_{2})-P_{1}Q_{2} ]}, \quad t\in [0,1], \end{aligned}$$

and

$$\begin{aligned}& \phi _{2}(t)= \frac{\varGamma (2-\sigma )(1-Q_{1})(\alpha t+\beta )+P_{1}[\gamma \varGamma (2-\sigma )(1-t)+\delta ]}{\rho \varGamma (2-\sigma )[(1-Q_{1})(1-P_{2})-P_{1}Q_{2}]}, \quad t\in [0,1]. \end{aligned}$$

Proof

In view of the equation in (6), Lemma 2.3, and \(u^{\prime \prime }(0)=0\), we have

$$ \begin{aligned} u(t)=-\bigl(I^{q}_{0+}y\bigr) (t)+u(0)+u^{\prime }(0)t,\quad t\in [0,1]. \end{aligned} $$
(7)

By (7), Lemma 2.1, and Lemma 2.2, we obtain

$$ \bigl({}^{C}D_{0+}^{\sigma }u\bigr) (t)=- \bigl(I_{0+}^{q-\sigma }y\bigr) (t)+ \frac{u^{\prime }(0)}{\varGamma (2-\sigma )}t^{1-\sigma }, \quad t\in [0,1]. $$
(8)

It follows from (7), (8), and the boundary conditions in (6) that

$$ \begin{aligned} u(0)={}&\frac{1}{\rho } \biggl[\beta \gamma \bigl(I_{0+}^{q}y \bigr) (1)+\beta \delta \bigl(I_{0+}^{q- \sigma }y\bigr) (1)+ \frac{\gamma \varGamma (2-\sigma )+\delta }{\varGamma (2-\sigma )} \int _{0}^{1}h_{1}(s)u(s)\,ds \\ &{}+ \beta \int _{0}^{1}h_{2}(s)u(s)\,ds \biggr] \end{aligned} $$

and

$$ u^{\prime }(0)=\frac{1}{\rho } \biggl[\alpha \gamma \bigl(I_{0+}^{q}y \bigr) (1)+ \alpha \delta \bigl(I_{0+}^{q-\sigma }y\bigr) (1)-\gamma \int _{0}^{1}h_{1}(s)u(s)\,ds+ \alpha \int _{0}^{1}h_{2}(s)u(s)\,ds \biggr], $$

which together with (7) shows that

$$\begin{aligned} u(t)={}& \int _{0}^{t} \biggl\{ -\frac{(t-s)^{q-1}}{\varGamma (q)}+ \frac{\alpha t+\beta }{\rho } \biggl[ \frac{\gamma (1-s)^{q-1}}{\varGamma (q)}+ \frac{\delta (1-s)^{q-\sigma -1}}{\varGamma (q-\sigma )} \biggr] \biggr\} y(s)\,ds \\ & {} + \int _{t}^{1} \biggl\{ \frac{\alpha t+\beta }{\rho } \biggl[ \frac{\gamma (1-s)^{q-1}}{\varGamma (q)}+ \frac{\delta (1-s)^{q-\sigma -1}}{\varGamma (q-\sigma )} \biggr] \biggr\} y(s)\,ds \\ & {} + \frac{\gamma \varGamma (2-\sigma )(1-t)+\delta }{\rho \varGamma (2-\sigma )} \int _{0}^{1}h_{1}(s)u(s)\,ds + \frac{\alpha t+\beta }{\rho } \int _{0}^{1}h_{2}(s)u(s)\,ds \\ ={}& \int _{0}^{1}G(t,s)y(s)\,ds+ \frac{\gamma \varGamma (2-\sigma )(1-t)+\delta }{\rho \varGamma (2-\sigma )} \int _{0}^{1}h_{1}(s)u(s)\,ds \\ & {} +\frac{\alpha t+\beta }{\rho } \int _{0}^{1}h_{2}(s)u(s)\,ds,\quad t\in [0,1]. \end{aligned}$$
(9)

From (9), we get

$$ (1-Q_{1}) \int _{0}^{1}h_{1}(s)u(s) \,ds-P_{1} \int _{0}^{1}h_{2}(s)u(s)\,ds= \int _{0}^{1}h_{1}(s) \int _{0}^{1}G(s,\tau )y(\tau )\,d\tau \,ds $$

and

$$ -Q_{2} \int _{0}^{1}h_{1}(s)u(s) \,ds+(1-P_{2}) \int _{0}^{1}h_{2}(s)u(s)\,ds= \int _{0}^{1}h_{2}(s) \int _{0}^{1}G(s,\tau )y(\tau )\,d\tau \,ds, $$

and so,

$$\begin{aligned}& \int _{0}^{1}h_{1}(s)u(s)\,ds \\& \quad = \frac{(1-P_{2})\int _{0}^{1}h_{1}(s)\int _{0}^{1}G(s,\tau )y(\tau )\,d\tau \,ds+P_{1}\int _{0}^{1}h_{2}(s)\int _{0}^{1}G(s,\tau )y(\tau )\,d\tau \,ds}{(1-Q_{1})(1-P_{2})- P_{1}Q_{2}} \end{aligned}$$

and

$$\begin{aligned}& \int _{0}^{1}h_{2}(s)u(s)\,ds \\& \quad = \frac{Q_{2}\int _{0}^{1}h_{1}(s)\int _{0}^{1}G(s,\tau )y(\tau )\,d\tau \,ds+(1-Q_{1})\int _{0}^{1}h_{2}(s)\int _{0}^{1}G(s,\tau )y(\tau )\,d\tau \,ds}{(1-Q_{1})(1-P_{2})-P_{1}Q_{2}}, \end{aligned}$$

which together with (9) implies that

$$ \begin{aligned} u(t)={}& \int _{0}^{1}G(t,s)y(s)\,ds+\sum _{i=1}^{2}\phi _{i}(t) \int _{0}^{1}h_{i}(s) \int _{0}^{1}G(s,\tau )y(\tau )\,d\tau \,ds \\ ={}& \int _{0}^{1}G(t,s)y(s)\,ds+\sum _{i=1}^{2}\phi _{i}(t) \int _{0}^{1}h_{i}( \tau ) \int _{0}^{1}G(\tau ,s)y(s)\,ds \,d\tau \\ ={}& \int _{0}^{1}G(t,s)y(s)\,ds+\sum _{i=1}^{2}\phi _{i}(t) \int _{0}^{1}y(s) \int _{0}^{1}G(\tau ,s)h_{i}(\tau )\,d \tau \,ds \\ ={}& \int _{0}^{1} \Biggl[G(t,s)+\sum _{i=1}^{2}\phi _{i}(t) \int _{0}^{1} G( \tau ,s)h_{i}(\tau )\,d \tau \Biggr]y(s)\,ds \\ ={}& \int _{0}^{1}H(t,s)y(s)\,ds,\quad t\in [0,1]. \end{aligned} $$

 □

In what follows, we let

$$ \begin{aligned} g(s)=\frac{\alpha +\beta }{\rho } \biggl[ \frac{\gamma (1-s)^{q-1}}{\varGamma (q)}+ \frac{\delta (1-s)^{q-\sigma -1}}{\varGamma (q-\sigma )} \biggr],\quad s\in [0,1] \end{aligned} $$

and

$$ \begin{aligned} \eta (s)= \frac{\beta \delta \varGamma (q)-\varGamma (q-\sigma )(\rho -\beta \gamma )(1-s)^{\sigma }}{(\alpha +\beta ) [\gamma \varGamma (q-\sigma )+\delta \varGamma (q) ]}, \quad s\in [0,1]. \end{aligned} $$

Lemma 2.5

\(G(t, s)\)satisfies the following properties:

  1. (1)

    \(G(t,s)\leq g(s)\), \((t,s)\in [0,1]\times [0,1]\);

  2. (2)

    \(G(t,s)\geq \eta (s)g(s)\), \((t,s)\in [0,1]\times [0,1]\).

Proof

Since (1) is obvious, we only need to prove that (2) holds.

First, it is clear that \(G(t,1)\geq \eta (1)g(1)\) for \(t\in [0,1]\).

Now, we verify that \(G(t,s)\geq \eta (s)g(s)\) for \((t,s)\in [0,1]\times [0,1)\). In fact, if \(s\leq t\), then

$$\begin{aligned}& \begin{aligned} \frac{G(t,s)}{g(s)}&= \frac{(\alpha t+\beta )[\gamma \varGamma (q-\sigma )(1-s)^{q-1}+\delta \varGamma (q)(1-s)^{q-\sigma -1}]-\rho \varGamma (q-\sigma )(t-s)^{q-1}}{(\alpha +\beta ) [\gamma \varGamma (q-\sigma )(1-s)^{q-1}+\delta \varGamma (q)(1-s)^{q-\sigma -1}]} \\ &\geq \frac{\beta \gamma \varGamma (q-\sigma )(1-s)^{\sigma } +\beta \delta \varGamma (q)-\rho \varGamma (q-\sigma )(1-s)^{\sigma }}{(\alpha +\beta )[\gamma \varGamma (q-\sigma )(1-s)^{\sigma }+\delta \varGamma (q)]} \\ &\geq \frac{\beta \delta \varGamma (q)-\varGamma (q-\sigma )(\rho -\beta \gamma )(1-s)^{\sigma }}{(\alpha +\beta ) [\gamma \varGamma (q-\sigma )+\delta \varGamma (q) ]} \\ &=\eta (s) \end{aligned} \end{aligned}$$

and if \(t\leq s\), then

$$\begin{aligned}& \begin{aligned} \frac{G(t,s)}{g(s)}&= \frac{(\alpha t+\beta )[\gamma \varGamma (q-\sigma )(1-s)^{q-1}+\delta \varGamma (q)(1-s)^{q-\sigma -1}]}{(\alpha +\beta ) [\gamma \varGamma (q-\sigma )(1-s)^{q-1}+\delta \varGamma (q)(1-s)^{q-\sigma -1}]} \\ &\geq \frac{\beta \gamma \varGamma (q-\sigma )(1-s)^{\sigma }+\beta \delta \varGamma (q)}{(\alpha +\beta )[\gamma \varGamma (q-\sigma )(1-s)^{\sigma }+\delta \varGamma (q)]} \\ &\geq \frac{\beta \gamma \varGamma (q-\sigma )(1-s)^{\sigma } +\beta \delta \varGamma (q)-\rho \varGamma (q-\sigma )(1-s)^{\sigma }}{(\alpha +\beta )[\gamma \varGamma (q-\sigma )(1-s)^{\sigma }+\delta \varGamma (q)]} \\ &\geq \frac{\beta \delta \varGamma (q)-\varGamma (q-\sigma )(\rho -\beta \gamma )(1-s)^{\sigma }}{(\alpha +\beta ) [\gamma \varGamma (q-\sigma )+\delta \varGamma (q) ]} \\ &=\eta (s). \end{aligned} \end{aligned}$$

 □

By the definition of η and the condition \(0<\rho <\beta [\gamma + \frac{\delta \varGamma (q)}{\varGamma (q-\sigma )} ]\), we may obtain the following remark.

Remark 2.1

η is increasing on \([0,1]\) and \(0<\eta (s)<1\) for \(s\in [0,1]\).

In the remainder of this paper, we always assume that the following conditions are satisfied:

$$ Q_{1}< 1,\qquad P_{2}< 1\quad \text{and}\quad (1-Q_{1}) (1-P_{2})> P_{1}Q_{2}. $$

Lemma 2.6

\(H(t,s)\)has the following property:

$$\begin{aligned}& \begin{aligned} m\eta (s)g(s)\leq H(t,s)\leq M g(s),\quad (t,s)\in [0,1]\times [0,1], \end{aligned} \end{aligned}$$

where

$$ \begin{aligned} m=1+\sum_{i=1}^{2} \min_{t\in [0,1]}\phi _{i}(t) \int _{0}^{1} h_{i}(\tau )\,d\tau \end{aligned} $$

and

$$ \begin{aligned} M=1+\sum_{i=1}^{2} \max_{t\in [0,1]}\phi _{i}(t) \int _{0}^{1} h_{i}(\tau )\,d\tau . \end{aligned} $$

Proof

On the one hand, in view of (1) of Lemma 2.5, we have

$$ \begin{aligned} H(t,s)&=G(t,s)+\sum_{i=1}^{2} \phi _{i}(t) \int _{0}^{1} G(\tau ,s)h_{i}( \tau )\,d \tau \\ &\leq \Biggl(1+\sum_{i=1}^{2}\phi _{i}(t) \int _{0}^{1}h_{i}(\tau )\,d \tau \Biggr)g(s) \\ &\leq Mg(s),\quad (t,s)\in [0,1]\times [0,1]. \end{aligned} $$

On the other hand, by (2) of Lemma 2.5, we get

$$ \begin{aligned} H(t,s)&=G(t,s)+\sum_{i=1}^{2} \phi _{i}(t) \int _{0}^{1} G(\tau ,s)h_{i}( \tau )\,d \tau \\ &\geq \Biggl(1+\sum_{i=1}^{2}\phi _{i}(t) \int _{0}^{1} h_{i}(\tau )\,d \tau \Biggr)\eta (s)g(s) \\ &\geq m\eta (s)g(s),\quad (t,s)\in [0,1]\times [0,1]. \end{aligned} $$

 □

Let \(E=C[0,1]\) be equipped with norm \(\Vert u \Vert =\max_{t\in [0,1]} \vert u(t) \vert \) and

$$ \begin{aligned} K=\bigl\{ u\in E:u(t)\geq \theta \Vert u \Vert , t\in [0,1]\bigr\} , \end{aligned} $$

where \(0<\theta =\frac{m\eta (0)}{M}<1\). Then it is easy to check that E is a Banach space and K is a cone in E.

Now, we define an operator T on K by

$$ (Tu) (t)= \int _{0}^{1}H(t,s)f\bigl(s,u(s)\bigr)\,ds,\quad u \in K, t\in [0,1]. $$

Obviously, if u is a fixed point of T, then u is a nonnegative solution of BVP (4).

Lemma 2.7

\(T:K\rightarrow K\)is completely continuous.

Proof

Let \(u\in K\). Then, in view of Lemma 2.6, we have

$$ \begin{aligned} & \Vert Tu \Vert \leq M \int _{0}^{1}g(s)f\bigl(s,u(s)\bigr)\,ds, \end{aligned} $$

which together with Lemma 2.6 and Remark 2.1 implies that

$$ \begin{aligned} (Tu) (t)&\geq m \int _{0}^{1}\eta (s)g(s)f\bigl(s,u(s)\bigr)\,ds \\ &\geq m\eta (0) \int _{0}^{1}g(s)f\bigl(s,u(s)\bigr)\,ds \\ &\geq \theta \Vert Tu \Vert ,\quad t\in [0,1]. \end{aligned} $$

This indicates that \(Tu\in K\). Furthermore, it is easy to prove that T is completely continuous by an application of Arzela–Ascoli theorem [29]. □

3 Main results

Define

$$\begin{aligned}& f^{0}=\limsup_{u\rightarrow 0^{+}}\max_{t\in [0,1]} \frac{f(t,u)}{u}, \quad\quad f^{\infty }=\limsup_{u\rightarrow +\infty } \max_{t\in [0,1]} \frac{f(t,u)}{u}, \\& f_{0}=\liminf_{u\rightarrow 0^{+}}\min_{t\in [0,1]} \frac{f(t,u)}{u}, \quad\quad f_{\infty }=\liminf_{u\rightarrow +\infty } \min_{t\in [0,1]} \frac{f(t,u)}{u}. \end{aligned}$$

Theorem 3.1

Suppose that one of the following conditions is satisfied:

  1. (i)

    \(f_{0}=+\infty \)and\(f^{\infty }=0\), or

  2. (ii)

    \(f^{0}=0\)and\(f_{\infty }=+\infty \).

Then BVP (4) has at least one positive solution.

Proof

First, we consider case (i): \(f_{0}=+\infty \) and \(f^{ \infty }=0\).

In view of \(f_{0}=+\infty \), there exists \(r_{1}>0\) such that

$$ \begin{aligned} f(t,u)\geq G_{1} u,\quad (t,u)\in [0,1] \times [0,r_{1}], \end{aligned} $$
(10)

where \(G_{1}\geq \frac{1}{m\theta \int _{0}^{1}\eta (s)g(s)\,ds}\).

Let \(\varOmega _{1}=\{u\in E: \Vert u \Vert < r_{1}\}\). Then, for any \(u\in K\cap \partial \varOmega _{1}\), by Lemma 2.6 and (10), we get

$$ \begin{aligned} (Tu) (t)&\geq m \int _{0}^{1}\eta (s)g(s)f\bigl(s,u(s)\bigr)\,ds \\ &\geq mG_{1} \int _{0}^{1}\eta (s)g(s)u(s)\,ds \\ &\geq mG_{1}\theta \Vert u \Vert \int _{0}^{1}\eta (s)g(s)\,ds \\ &\geq \Vert u \Vert ,\quad t\in [0,1], \end{aligned} $$

which shows that

$$ \Vert Tu \Vert \geq \Vert u \Vert \quad \text{for } u\in K\cap \partial \varOmega _{1}. $$
(11)

On the other hand, since \(f^{\infty }=0\), there exists \(U_{1}>0\) such that

$$ \begin{aligned} f(t,u)\leq \varepsilon _{1} u,\quad (t,u) \in [0,1]\times (U_{1},+\infty ), \end{aligned} $$

where \(\varepsilon _{1}>0\) satisfies \(\varepsilon _{1}\leq \frac{1}{2M\int _{0}^{1}g(s)\,ds}\).

Let \(M^{*}=\max_{(t,u)\in [0,1]\times [0,U_{1}]}f(t,u)\). Then we have

$$ \begin{aligned} f(t,u)\leq M^{*}+\varepsilon _{1}u,\quad (t,u)\in [0,1]\times [0,+\infty ). \end{aligned} $$
(12)

If we choose \(r_{2}=\max {\{2r_{1},2MM^{*}\int _{0}^{1}g(s)\,ds\}}\) and let \(\varOmega _{2}=\{u\in E: \Vert u \Vert < r_{2}\}\), then for any \(u\in K\cap \partial \varOmega _{2}\), from Lemma 2.6 and (12), we obtain

$$ \begin{aligned} (Tu) (t)&\leq M \int _{0}^{1}g(s)f\bigl(s,u(s)\bigr)\,ds \\ &\leq MM^{*} \int _{0}^{1}g(s)\,ds+M\varepsilon _{1} \Vert u \Vert \int _{0}^{1}g(s)\,ds \\ &\leq \frac{ \Vert u \Vert }{2}+\frac{ \Vert u \Vert }{2} \\ &= \Vert u \Vert ,\quad t\in [0,1], \end{aligned} $$

which indicates that

$$ \Vert Tu \Vert \leq \Vert u \Vert \quad \text{for } u\in K\cap \partial \varOmega _{2}. $$
(13)

Therefore, it follows from Theorem 1.1, Lemma 2.7, (11), and (13) that T has a fixed point \(u\in K\cap (\overline{\varOmega }_{2}\setminus \varOmega _{1} )\), which is a desired positive solution of BVP (4).

Next, we consider case (ii): \(f^{0}=0\) and \(f_{ \infty }=+\infty \).

In view of \(f^{0}=0\), there exists \(r_{3}>0\) such that

$$ \begin{aligned} f(t,u)\leq \varepsilon _{2}u,\quad (t,u) \in [0,1]\times [0,r_{3}], \end{aligned} $$
(14)

where \(\varepsilon _{2}>0\) satisfies \(\varepsilon _{2}\leq \frac{1}{M\int _{0}^{1}g(s)\,ds}\).

Let \(\varOmega _{3}=\{u\in E: \Vert u \Vert < r_{3}\}\). Then, for any \(u\in K\cap \partial \varOmega _{3}\), by Lemma 2.6 and (14), we get

$$ \begin{aligned} (Tu) (t)&\leq M \int _{0}^{1}g(s)f\bigl(s,u(s)\bigr)\,ds \\ &\leq M\varepsilon _{2} \int _{0}^{1}g(s)u(s)\,ds \\ &\leq M\varepsilon _{2} \Vert u \Vert \int _{0}^{1}g(s)\,ds \\ &\leq \Vert u \Vert ,\quad t\in [0,1], \end{aligned} $$

which shows that

$$ \begin{aligned} \Vert Tu \Vert \leq \Vert u \Vert \quad \text{for } u\in K\cap \partial \varOmega _{3}. \end{aligned} $$
(15)

On the other hand, since \(f_{\infty }=+\infty \), there exists \(U_{2}>0\) such that

$$ \begin{aligned} f(t,u)\geq G_{2}u,\quad (t,u)\in [0,1] \times [U_{2},+\infty ), \end{aligned} $$
(16)

where \(G_{2}\geq \frac{1}{m\theta \int _{0}^{1}\eta (s)g(s)\,ds}\).

If we choose \(r_{4}=\max \{\frac{U_{2}}{\theta },2r_{3}\}\) and let \(\varOmega _{4}=\{u\in E: \Vert u \Vert < r_{4}\}\), then for any \(u\in K\cap \partial \varOmega _{4}\), we know

$$ \begin{aligned} u(t)\geq \theta \Vert u \Vert =\theta r_{4}\geq U_{2},\quad t\in [0,1], \end{aligned} $$

which together with Lemma 2.6 and (16) implies that

$$ \begin{aligned} (Tu) (t)&\geq m \int _{0}^{1}\eta (s)g(s)f\bigl(s,u(s)\bigr)\,ds \\ &\geq mG_{2} \int _{0}^{1}\eta (s)g(s)u(s)\,ds \\ &\geq mG_{2}\theta \Vert u \Vert \int _{0}^{1}\eta (s)g(s)\,ds \\ &\geq \Vert u \Vert ,\quad t\in [0,1]. \end{aligned} $$

This indicates that

$$ \begin{aligned} \Vert Tu \Vert \geq \Vert u \Vert \quad \text{for } u\in K\cap \partial \varOmega _{4}. \end{aligned} $$
(17)

Therefore, it follows from Theorem 1.1, Lemma 2.7, (15), and (17) that T has a fixed point \(u\in K\cap (\overline{\varOmega }_{4}\setminus \varOmega _{3} )\), which is a desired positive solution of BVP (4). □

Example 3.1

Consider the following BVP:

$$ \textstyle\begin{cases} ({}^{C}D^{\frac{5}{2}}_{0+}u)(t)+ [\sin (\frac{\pi t}{2} )+1 ]u^{2}(t)=0,\quad t\in [0,1], \\ u^{\prime \prime }(0)=0, \\ u(0)-4u^{\prime }(0)=\int _{0}^{1}su(s)\,ds, \\ u(1)+({}^{C}D_{0+}^{\frac{1}{2}}u)(1)=\int _{0}^{1}(1-s)u(s)\,ds. \end{cases} $$
(18)

In view of \(q=\frac{5}{2}\), \(\sigma =\frac{1}{2}\), \(\alpha =\gamma =\delta =1\), \(\beta =4\), \(h_{1}(s)=s\), and \(h_{2}(s)=1-s\), \(s\in [0,1]\), a simple calculation shows that

$$ \begin{aligned} &0< \rho =5+\frac{2}{\pi }\sqrt{\pi }< \beta \biggl[\gamma + \frac{\delta \varGamma (q)}{\varGamma (q-\sigma )} \biggr]=4+3\sqrt{\pi } \end{aligned} $$

and

$$\begin{aligned}& P_{1}=\frac{7\sqrt{\pi }}{3(5\sqrt{\pi }+2)},\qquad P_{2}= \frac{13\sqrt{\pi }}{6(5\sqrt{\pi }+2)}, \\& Q_{1}= \frac{\sqrt{\pi }+6}{6(5\sqrt{\pi }+2)}, \qquad Q_{2}= \frac{\sqrt{\pi }+3}{3(5\sqrt{\pi }+2)}. \end{aligned}$$

Obviously, \(Q_{1}<1\), \(P_{2}<1\) and

$$ \begin{aligned} (1-Q_{1}) (1-P_{2})= \frac{(29\sqrt{\pi }+6)(17\sqrt{\pi }+12)}{36(5\sqrt{\pi }+2)^{2}}>P_{1}Q_{2}= \frac{7\sqrt{\pi }(\sqrt{\pi }+3)}{9(5\sqrt{\pi }+2)^{2}}. \end{aligned} $$

Moreover, since \(f(t,u)= [\sin (\frac{\pi t}{2})+1 ]u^{2}\), \((t,u)\in [0,1] \times [0,+\infty )\), it is easy to know that \(f:[0,1]\times [0,+\infty )\rightarrow [0,+\infty )\) is continuous and

$$ \begin{aligned} f^{0}=0,\quad\quad f_{\infty } =+ \infty . \end{aligned} $$

Therefore, it follows from Theorem 3.1 that BVP (18) has at least one positive solution.

4 Conclusion

In this paper, by applying Guo–Krasnoselskii’s fixed point theorem, we obtain the existence of at least one positive solution for a class of nonlinear boundary value problems involving fractional differential equation and integral boundary conditions. An illustrative example is also given to show the effectiveness of theoretical results.