Existence of positive solutions for a high order fractional differential equation integral boundary value problem with changing sign nonlinearity

Abstract

In this paper, we investigate the existence of positive solutions for a class of high order fractional differential equation integral boundary value problems with changing sign nonlinearity. By applying cone expansion and cone compression fixed point theorem, we have obtained and proved theorems related to the existence of positive solutions, which highlight the influences of the parameters in different ranges on the existence of positive solutions. Finally, we also give some examples to illustrate our main results.

1 Introduction

In this paper, we investigate the existence of positive solutions for a class of high order fractional differential equation integral boundary value problems with changing sign nonlinearity:

$$ \textstyle\begin{cases} D_{0^{+}}^{\alpha }u(t)+\lambda f(t,u(t))=0, \quad t\in (0,1),\\ u(0)= u'(0)=\cdots =u^{(n-2)}(0)=0,\\ D_{0^{+}}^{\beta }u(1)=\int _{0}^{1}D_{0^{+}}^{\beta }u(t)\,\mathrm{d}A(t), \end{cases} $$

(1.1)

where $D_{0{+}}^{\alpha }$ is the Riemann-Liouville fractional derivative of order α, $n-1<\alpha \leq n$, $n\geq 3$, $0<\beta \leq 1, \lambda >0$, and $\int _{0}^{1}D_{0^{+}}^{\beta }u(t)\,\mathrm{d}A(t)$ denotes the Riemann-Stieltjes integrals with respect to A, in which $A(t)$ is a monotone increasing function and $f:[0,1]\times \mathbb{R}^{+} \rightarrow \mathbb{R}$ may change sign, $\mathbb{R}^{+}=[0, +\infty )$.

In recent years, fractional differential equations arise in many engineering and scientific fields such as mathematical modeling of systems physics, chemistry, aerodynamics, electrodynamics of complex medium, polymer rheology, and so forth. Researchers have reached a significance in ordinary and partial differential equations involving fractional derivatives, see [1–3] and the references therein. Since the boundary value problems play an important role in fractional differential equations theory, more attention has been paid and plenty of meaningful results have been obtained, see [4–26].

Generally speaking, in order to guarantee the existence of positive solutions of boundary value problems, the nonlinearity is usually nonnegative, see [6, 7, 9, 10, 16, 17, 20, 21, 24] and the references therein. And with the nonlinearity changing sign, it will bring much more difficulties to the study of the problem, and the papers studying this are relatively few, see [11, 12, 15, 17, 27] and the references therein. Compared with Riemann integral, Riemann-Stieltjes integral is more general. For example, the Riemann-Stieltjes integral $\int _{0}^{1}u(t)\,\mathrm{d}A(t)$ will become the Riemann integral $\int _{0}^{1}A'(t)u(t)\,\mathrm{d}t$ when A has a continuous derivative. Moreover, Riemann-Stieltjes integral will become more important when A is not differentiable or A is discontinuous. Any finite or infinite sum can be expressed as a Riemann-Stieltjes integral by a suitable choice of discontinuous A, see [28]. Since the nonlocal boundary value problems include the multi-point boundary value problem (A is a step function) and the Riemann integral boundary value problem (A has a continuous derivative), it has become a more general case where we study the boundary value problem with integral boundary conditions of Riemann-Stieltjes type. Many researchers have done a lot of work on this class of boundary value problems, see [12, 22, 25, 29, 30] and the references therein.

In [12], Zhang studied the following nonlinear fractional boundary value problem:

$$ \textstyle\begin{cases} D_{0^{+}}^{\alpha }u(t)+f(t,u(t))=0, \quad t\in (0,1),\\ u(0)= u'(0)=0,\\ D_{0^{+}}^{\beta }u(1)=\int _{0}^{1}D_{0^{+}}^{\beta }u(t)\,\mathrm{d}A(t), \end{cases} $$

(1.2)

where $2<\alpha \leq 3$, $0<\beta \leq 1$, the nonlinearity $f(t,u)$ may change sign on some set. By means of the monotone iterative technique, the existence of nontrivial solutions or positive solutions is obtained. However, $f(t,u)$ is monotone increasing with respect to the fact that u in some interval is restricted, that is to say, $f(t,u)$ has local monotonicity.

In our work, it is not necessary to have monotonicity of the nonlinearity $f(t,u)$ since we derive the properties of the corresponding integral kernel function and get a more accurate inequality than the literature [12]. By a fixed point theorem, sufficient conditions for the existence of positive solutions of boundary value problem (1.1) are obtained. It is worth mentioning that the nonlinearity $f(t,u)$ does not need to be nonnegative and lower bounded. We also focus on studying the impact of the parameter λ on the existence of positive solutions, and we obtain sufficient conditions so that the problem has at least one positive solution when the parameter λ belongs to two intervals.

We say that f satisfies the $L^{1}$-Carathéodory conditions on $[0,1]\times \mathbb{R}^{+}$ if

• (1) $f(\cdot ,u)$ is measurable for all $u\in \mathbb{R}^{+}$;

• (2) $f(t,\cdot )$ is continuous for a.e. $t\in [0,1]$;

• (3) for each $r>0$, there exists $\varphi _{r}\in L^{1}[0,1]$, $\varphi _{r}(t)\geq 0$, such that $f(t,u)\leq \varphi _{r}(t)$ for all $u\in [0,r]$ and a.e. $t\in [0,1]$.

Denote $\Delta =\int _{0}^{1}t^{\alpha -\beta -1}\,\mathrm{d}A(t)$, in this paper, assume that $\int _{0}^{1}t^{\alpha -\beta -1}\,\mathrm{d}A(t)<1$ holds.

2 Preliminaries

The definitions of fractional integral and fractional derivative and the related lemmas can be found in [1–3].

Let the space $E= C[0,1]$, then E is a Banach space with the norm $\Vert u \Vert =\max _{t\in [0,1]}\vert u(t) \vert$.

Definition 2.1

If $u\in E$, $D_{0{+}}^{\alpha }u\in L^{1}[0,1]$ and satisfies (1.1), u is called a solution of fractional boundary value problem (1.1). Furthermore, if $u(t)>0, t\in (0,1)$, u is called a positive solution of fractional boundary value problem (1.1).

Lemma 2.1

(see[17])

If $D_{0^{+}}^{\alpha }u\in L^{1}[0,1]$, then $I_{0^{+}}^{n-\alpha }u\in AC^{n}[0,1]$.

Lemma 2.2

For any $y\in L^{1}[0,1]$, the unique solution of the boundary value problem

$$ \textstyle\begin{cases} D_{0^{+}}^{\alpha }u(t)+y(t)=0, \quad t\in (0,1),\\ u(0)= u'(0)=\cdots =u^{(n-2)}(0)=0,\\ D_{0^{+}}^{\beta }u(1)=\int _{0}^{1}D_{0^{+}}^{\beta }u(t)\,\mathrm{d}A(t) \end{cases} $$

(2.1)

is given by

$$\begin{aligned} u(t)= \int _{0}^{1}G(t,s)y(s)\,\mathrm{d}s, \end{aligned}$$

(2.2)

where

$$\begin{aligned}& G(t,s)=H(t,s)+\frac{t^{\alpha -1}}{1-\Delta } \int _{0}^{1}K(\tau ,s)\,\mathrm{d}A(\tau ), \end{aligned}$$

(2.3)

$$\begin{aligned}& H(t,s)=\frac{1}{\Gamma (\alpha )} \textstyle\begin{cases} t^{\mathrm{\alpha -1}}(1-s)^{\alpha -\beta -1}-(t-s)^{\mathrm{\alpha -1}},& 0\leq s< t\leq 1,\\ t^{\mathrm{\alpha -1}}(1-s)^{\alpha -\beta -1},& 0\leq t\leq s\leq 1, \end{cases}\displaystyle \end{aligned}$$

(2.4)

$$\begin{aligned}& K(\tau ,s)=\frac{1}{\Gamma (\alpha )} \textstyle\begin{cases} (\tau (1-s) )^{\mathrm{\alpha -\beta -1}}-(\tau -s)^{\alpha -\beta -1},& 0\leq s< \tau \leq 1,\\ (\tau (1-s) )^{\mathrm{\alpha -\beta -1}},& 0\leq \tau \leq s\leq 1. \end{cases}\displaystyle \end{aligned}$$

(2.5)

Proof

Suppose u is a solution to boundary value problem (2.1). Since $y\in L^{1}[0,1]$, then $D_{0^{+}}^{\alpha }u\in L^{1}[0,1]$. By Lemma 2.1, $I_{0^{+}}^{n-\alpha }u\in AC^{n}[0,1]$. Thus we have

$$ I_{0^{+}}^{\alpha } \bigl(D_{0^{+}}^{\alpha }u \bigr) (t)=u(t)-\sum _{k=1}^{n}c_{k}t^{\alpha -k}, $$

that is,

$$\begin{aligned} u(t)=-\frac{1}{\Gamma (\alpha )} \int _{0}^{t}(t-s)^{\alpha -1}y(s) \,\mathrm{d}s+c_{1}t^{\alpha -1}+c_{2}t^{\alpha -2}+ \cdots +c_{n}t^{\alpha -n}, \end{aligned}$$

(2.6)

where $c_{i}\in \mathbb{R}$, $i=1,2,\ldots ,n$.

From the boundary conditions $u(0)= u'(0)=\cdots =u^{(n-2)}(0)=0$, we can get $c_{n}=c_{n-1}=c_{n-2}=\cdots =c_{2}=0$. Thus,

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

(2.7)

and from the boundary condition $D_{0^{+}}^{\beta }u(1)=\int _{0}^{1}D_{0^{+}}^{\beta }u(t)\,\mathrm{d}A(t)$, we can get

$$ c_{1}=\frac{1}{\Gamma (\alpha )(1-\Delta )} \biggl( \int _{0}^{1}(1-s)^{\alpha -\beta -1}y(s) \,\mathrm{d}s- \int _{0}^{1} \biggl( \int _{0}^{\tau }(\tau -s)^{\alpha -\beta -1}y(s) \,\mathrm{d}s \biggr)\,\mathrm{d}A(\tau ) \biggr). $$

Therefore, the unique solution of boundary value problem (2.1) is

$$\begin{aligned} u(t)&= -\frac{1}{\Gamma (\alpha )} \int _{0}^{t}(t-s)^{\mathrm{\alpha -1}}y(s)\,\mathrm{d}s + \frac{t^{\alpha -1}}{\Gamma (\alpha )(1-\Delta )} \biggl( \int _{0}^{1}(1-s)^{\alpha -\beta -1}y(s) \,\mathrm{d}s \\ &\quad {}- \int _{0}^{1} \biggl( \int _{0}^{\tau }(\tau -s)^{\alpha -\beta -1}y(s)\,\mathrm{d}s \biggr)\,\mathrm{d} A(\tau ) \biggr) \\ &=-\frac{1}{\Gamma (\alpha )} \int _{0}^{t}(t-s)^{\mathrm{\alpha -1}}y(s)\,\mathrm{d}s + \frac{t^{\alpha -1}}{\Gamma (\alpha )} \int _{0}^{1}(1-s)^{\alpha -\beta -1}y(s)\,\mathrm{d}s \\ &\quad {}+\frac{t^{\alpha -1}}{\Gamma (\alpha )(1-\Delta )} \biggl( \int _{0}^{1}\biggl( \int _{0}^{1}\tau ^{\alpha -\beta -1}(1-s)^{\alpha -\beta -1}y(s) \,\mathrm{d}s\biggr)\,\mathrm{d}A(\tau ) \\ &\quad {}- \int _{0}^{1}\biggl( \int _{0}^{\tau }(\tau -s)^{\alpha -\beta -1}y(s)\,\mathrm{d}s \biggr)\,\mathrm{d} A(\tau ) \biggr) \\ &= \int _{0}^{1} \biggl(H(t,s)+\frac{t^{\alpha -1}}{\Gamma (\alpha )(1-\Delta )} \int _{0}^{1}K(\tau ,s)\,\mathrm{d}A(\tau ) \biggr)y(s)\,\mathrm{d}s \\ &= \int _{0}^{1}G(t,s)y(s)\,\mathrm{d}s, \end{aligned}$$

where $G(t,s)$, $H(t,s)$, and $K(t,s)$ are defined by (2.3), (2.4), and (2.5), respectively.

On the other hand, if u satisfies (2.2), then u will also satisfy (2.6). Thus, we have

$$ \bigl(D_{0^{+}}^{\alpha }u \bigr) (t) = \biggl(\frac{\,\mathrm{d}}{\,\mathrm{d}t} \biggr)^{n}I_{0^{+}}^{n-\alpha }u(t) =- \biggl( \frac{\,\mathrm{d}}{\,\mathrm{d}t} \biggr)^{n}\bigl(I_{0^{+}}^{n-\alpha } I_{0^{+}}^{\alpha }y\bigr) (t) =- \biggl(\frac{\,\mathrm{d}}{\,\mathrm{d}t} \biggr)^{n}I_{0^{+}}^{n}y(t)=-y(t), $$

which implies the equation of boundary value problem (2.1) is satisfied.

We can easily show that u satisfies the boundary condition of boundary value problem (2.1). □

Lemma 2.3

The function $H(t,s)$, which is defined by (2.4), satisfies the following conditions:

• (1) $H(t,s)\geq 0$ is continuous for $t,s\in [0,1]$ and $H(t,s)>0$ for $t,s\in (0,1)$;

• (2) For $t,s\in [0,1]$, we have

  $$ H(t,s)\leq \frac{1}{\Gamma (\alpha )}t^{\alpha -1}(1-s)^{\alpha -\beta -1}; $$

• (3) For $t,s\in [0,1]$, we have

  $$ \frac{\beta }{\Gamma (\alpha )}t^{\alpha -1}s(1-s)^{\alpha -\beta -1}\leq H(t,s)\leq \frac{1}{\Gamma (\alpha )}s(1-s)^{\alpha -\beta -1}. $$

Proof

(1) By (2.4), it is clear that $H(t,s)$ is continuous on $[0,1]\times [0,1]$ and $H(t,s)\geq 0$ when $0\leq t\leq s\leq 1$.

For $0\leq s\leq t\leq 1$, we have

$$\begin{aligned} t^{\mathrm{\alpha -1}}(1-s)^{\alpha -\beta -1}-(t-s)^{\mathrm{\alpha -1}} &\geq t^{\mathrm{\alpha -1}}(1-s)^{\alpha -\beta -1}-t^{\mathrm{\alpha -1}} (1-s)^{\mathrm{\alpha -1}} \\ &=t^{\mathrm{\alpha -1}}(1-s)^{\mathrm{\alpha -\beta -1}} \bigl(1-(1-s)^{\beta } \bigr) \geq 0. \end{aligned}$$

So, we have $H(t,s)\geq 0$ for any $t,s\in [0,1]$. Similarly, for $t,s\in (0,1)$, we have $H(t,s)>0$.

(2) It follows from (2.4) that $H(t,s)\leq \frac{1}{\Gamma (\alpha )}t^{\alpha -1}(1-s)^{\alpha -\beta -1}$.

(3) Since $n-1<\alpha \leq n$, $n\geq 3$, it is easy to show that $H(t,s)\leq \frac{1}{\Gamma (\alpha )}s(1-s)^{\alpha -\beta -1}$ for $0\leq t\leq s\leq 1$.

For $0\leq s\leq t\leq 1$, let

$$ h(t,s)=\frac{1}{\Gamma (\alpha )} \bigl(t^{\alpha -1}(1-s)^{\alpha -\beta -1} -(t-s)^{\alpha -1} \bigr)-\frac{1}{\Gamma (\alpha )}s(1-s)^{\alpha -\beta -1}. $$

We have

$$ h(1,s)\leq 0,\qquad h(s,s)\leq 0, $$

and

$$\begin{aligned} \frac{\partial h(t,s)}{\partial t} =\frac{\alpha -1}{\Gamma (\alpha )} \bigl(t^{\alpha -2}(1-s)^{\alpha -\beta -1} -(t-s)^{\alpha -2} \bigr). \end{aligned}$$

If there exists $t_{0}\in (s,1)$ such that $\frac{\partial h(t_{0},s)}{\partial t}=0$, then

$$ {t_{0}}^{\alpha -2}(1-s)^{\alpha -\beta -1}=(t_{0}-s)^{\alpha -2}, $$

which implies that

$$\begin{aligned} h(t_{0},s) &=\frac{1}{\Gamma (\alpha )} \bigl({t_{0}}^{\alpha -1}(1-s)^{\alpha -\beta -1} -(t_{0}-s)^{\alpha -1}-s(1-s)^{\alpha -\beta -1} \bigr) \\ &=\frac{1}{\Gamma (\alpha )} \bigl({t_{0}}^{\alpha -1}(1-s)^{\alpha -\beta -1} -(t_{0}-s)^{\alpha -2}(t_{0}-s)-s(1-s)^{\alpha -\beta -1} \bigr) \\ &=\frac{1}{\Gamma (\alpha )} \bigl({t_{0}}^{\alpha -1}(1-s)^{\alpha -\beta -1} -{t_{0}}^{\alpha -2}(1-s)^{\alpha -\beta -1}(t_{0}-s)-s(1-s)^{\alpha -\beta -1} \bigr) \\ &=\frac{1}{\Gamma (\alpha )}(1-s)^{\alpha -\beta -1} \bigl({t_{0}}^{\alpha -1} -{t_{0}}^{\alpha -2}(t_{0}-s)-s \bigr) \\ &=\frac{1}{\Gamma (\alpha )}(1-s)^{\alpha -\beta -1}s\bigl({t_{0}}^{\alpha -2}-1 \bigr)\leq 0. \end{aligned}$$

Hence, we have

$$ h(t,s)=\frac{1}{\Gamma (\alpha )} \bigl(t^{\alpha -1}(1-s)^{\alpha -\beta -1} -(t-s)^{\alpha -1} \bigr)-\frac{1}{\Gamma (\alpha )}s(1-s)^{\alpha -\beta -1}\leq 0, $$

which implies that $H(t,s)\leq \frac{1}{\Gamma (\alpha )}s(1-s)^{\alpha -\beta -1}$ for $0\leq s< t\leq 1$.

Therefore, we can get that

$$ H(t,s)\leq \frac{1}{\Gamma (\alpha )}s(1-s)^{\alpha -\beta -1} \quad \mbox{for any } t, s \in [0,1]. $$

On the other hand, for $0\leq s< t\leq 1$, we can show

$$\begin{aligned} H(t,s)&=\frac{1}{\Gamma (\alpha )} \bigl(t^{\mathrm{\alpha -1}}(1-s)^{\alpha -\beta -1}-(t-s)^{\mathrm{\alpha -1}} \bigr) \\ &\geq \frac{1}{\Gamma (\alpha )} \bigl(t^{\mathrm{\alpha -1}}(1-s)^{\alpha -\beta -1}-t^{\alpha -1}(1-s)^{\mathrm{\alpha -1}} \bigr) \\ &=\frac{1}{\Gamma (\alpha )}t^{\mathrm{\alpha -1}}(1-s)^{\alpha -\beta -1} \bigl(1-(1-s)^{\beta } \bigr) \\ &\geq \frac{\beta }{\Gamma (\alpha )}t^{\alpha -1}s(1-s)^{\alpha -\beta -1}. \end{aligned}$$

For $0\leq t\leq s\leq 1$, we have

$$\begin{aligned} H(t,s)=\frac{1}{\Gamma (\alpha )}t^{\mathrm{\alpha -1}}(1-s)^{\alpha -\beta -1}\geq \frac{\beta }{\Gamma (\alpha )}t^{\mathrm{\alpha -1}}(1-s)^{\alpha -\beta -1}s. \end{aligned}$$

Hence, (3) holds. □

Lemma 2.4

The function $K(\tau ,s)$, which is defined by (2.5), satisfies the following conditions:

$(1)$ $K(\tau ,s)\geq 0$ is continuous for $\tau , s\in [0,1]$ and $K(\tau ,s)>0$ for $\tau , s\in (0,1)$;

$(2)$ For $\tau \in [0,1],s\in [0,1]$, we have

$$ K(\tau ,s)\leq \frac{1}{\Gamma (\alpha )}(1-s)^{\alpha -\beta -1}; $$

$(3)$ For $\tau \in (0,1],s\in [0,1]$, we have

$$\begin{aligned} \frac{\min \{\alpha -\beta -1,1\}}{\Gamma (\alpha )}\tau ^{\alpha -\beta -1}(1-\tau ) s(1-s)^{\alpha -\beta -1} &\leq K(\tau,s) \\ &\leq \frac{\max \{\alpha -\beta -1,1\}}{\Gamma (\alpha )}\tau ^{\alpha -\beta -2} s(1-s)^{\alpha -\beta -1}. \end{aligned}$$

Proof

By the expression of $K(\tau ,s)$, it is easy to check that (1) and (2) hold.

(3) Similar to the proof of Lemma 2.8 in [12], we can prove

$$ K(\tau ,s)\geq \frac{\min \{\alpha -\beta -1,1\}}{\Gamma (\alpha )}\tau ^{\alpha -\beta -1} (1-\tau )s(1-s)^{\alpha -\beta -1}. $$

In the following we will prove

$$ K(\tau ,s) \leq \frac{\max \{\alpha -\beta -1,1\}}{\Gamma (\alpha )}\tau ^{\alpha -\beta -2} s(1-s)^{\alpha -\beta -1}. $$

We divide the proof into the following two cases for $\alpha -\beta -1\in (0,+\infty )$.

Case 1: $0<\alpha -\beta -1\leq 1$.

If $0\leq s<\tau \leq 1$, then

$$\begin{aligned} K(\tau ,s) &=\frac{1}{\Gamma (\alpha )} \bigl(\tau ^{\mathrm{\alpha -\beta -1}} (1-s)^{\mathrm{\alpha -\beta -1}}-( \tau -s)^{\mathrm{\alpha -\beta -1}} \bigr) \\ &=\frac{\tau ^{\mathrm{\alpha -\beta -2}}(1-s)^{\mathrm{\alpha -\beta -2}}}{\Gamma (\alpha )} \biggl(\tau (1-s)-(\tau -s)\frac{(1-\frac{s}{\tau })^{\mathrm{\alpha -\beta -2}}}{(1-s)^{\mathrm{\alpha -\beta -2}}} \biggr) \\ &\leq \frac{\tau ^{\mathrm{\alpha -\beta -2}}(1-s)^{\mathrm{\alpha -\beta -2}}}{\Gamma (\alpha )} \bigl(\tau (1-s)-(\tau -s) \bigr) \\ &=\frac{\tau ^{\mathrm{\alpha -\beta -2}}(1-s)^{\mathrm{\alpha -\beta -2}}}{\Gamma (\alpha )}s(1-\tau ) \\ &\leq \frac{1}{\Gamma (\alpha )}\tau ^{\mathrm{\alpha -\beta -2}}s(1-s)^{\mathrm{\alpha -\beta -1}}. \end{aligned}$$

If $0\leq \tau \leq s\leq 1$, then

$$ K(\tau ,s)=\frac{1}{\Gamma (\alpha )}\tau ^{\mathrm{\alpha -\beta -1}} (1-s)^{\mathrm{\alpha -\beta -1}} \leq \frac{1}{\Gamma (\alpha )}\tau ^{\mathrm{\alpha -\beta -2}}s(1-s) ^{\mathrm{\alpha -\beta -1}}. $$

Case 2: $1<\alpha -\beta -1$.

If $0\leq s<\tau \leq 1$, we have

$$\begin{aligned} K(\tau ,s) &=\frac{1}{\Gamma (\alpha )} \bigl(\tau ^{\mathrm{\alpha -\beta -1}} (1-s)^{\mathrm{\alpha -\beta -1}}-( \tau -s)^{\mathrm{\alpha -\beta -1}} \bigr) \\ &\leq \frac{\alpha -\beta -1}{\Gamma (\alpha )} \bigl(\tau (1-s) \bigr)^{\mathrm{\alpha -\beta -2}} \bigl(\tau (1-s)-(\tau -s) \bigr) \\ &\leq \frac{\alpha -\beta -1}{\Gamma (\alpha )} \bigl(\tau (1-s) \bigr)^{\mathrm{\alpha -\beta -2}} s(1-\tau ) \\ &\leq \frac{\alpha -\beta -1}{\Gamma (\alpha )}\tau ^{\mathrm{\alpha -\beta -2}} s(1-s)^{\mathrm{\alpha -\beta -1}}. \end{aligned}$$

If $0\leq \tau \leq s\leq 1$, we have

$$ K(\tau ,s)=\frac{1}{\Gamma (\alpha )}\tau ^{\mathrm{\alpha -\beta -1}} (1-s)^{\mathrm{\alpha -\beta -1}}\leq \frac{\alpha -\beta -1}{\Gamma (\alpha )} \tau ^{\mathrm{\alpha -\beta -2}}s(1-s)^{\mathrm{\alpha -\beta -1}}. $$

Therefore, we can get that (3) holds. □

Denote

$$\begin{aligned}& M_{1}=\frac{1}{\Gamma (\alpha )} \biggl(\beta +\frac{\min \{\alpha -\beta -1,1\} \int _{0}^{1}\tau ^{\mathrm{\alpha -\beta -1}}(1-\tau )\,\mathrm{d}A(\tau )}{1-\Delta } \biggr), \\& M_{2}=\frac{1}{\Gamma (\alpha )} \biggl(1+\frac{\max \{\alpha -\beta -1,1\} \int _{0}^{1}\tau ^{\mathrm{\alpha -\beta -2}}\,\mathrm{d}A(\tau )}{1-\Delta } \biggr), \\& M_{3}=\frac{1}{\Gamma (\alpha )} \biggl(1+\frac{A(1)-A(0)}{1-\Delta } \biggr). \end{aligned}$$

Lemma 2.5

The function $G(t,s)$, which is defined by (2.3), satisfies the following conditions:

• (1) $G(t,s)\geq 0$ is continuous for $t,s\in [0,1]$ and $G(t,s)>0$ for $t,s\in (0,1)$;

• (2) For $t,s\in [0,1]$, we have

  $$ G(t,s)\leq M_{3}t^{\alpha -1}; $$

• (3) For $t, s\in [0,1]$, we have

  $$ M_{1}t^{\alpha -1}s(1-s)^{\alpha -\beta -1}\leq G(t,s)\leq M_{2}s(1-s)^{\alpha -\beta -1}\leq M_{2}. $$

Proof

(1) From Lemmas 2.3 and 2.4, we obtain that $G(t,s)\geq 0$ is continuous for $t,s\in [0,1]$ and $G(t,s)>0$ for $t,s\in (0,1)$.

$(2)$ For any $t,s\in [0,1]$, from $(2)$ of Lemma 2.3 and $(2)$ of Lemma 2.4, we have

$$\begin{aligned} G(t,s) &=H(t,s)+\frac{t^{\alpha -1}}{1-\Delta } \int _{0}^{1}K(\tau ,s)\,\mathrm{d}A(\tau ) \\ &\leq \frac{1}{\Gamma (\alpha )}t^{\alpha -1}(1-s)^{\alpha -\beta -1}+ \frac{t^{\alpha -1} (A(1)-A(0))}{\Gamma (\alpha )(1-\Delta )}(1-s)^{\alpha -\beta -1} \\ &\leq \frac{1}{\Gamma (\alpha )} \biggl(1+\frac{A(1)-A(0)}{1-\Delta } \biggr)t^{\alpha -1} (1-s)^{\mathrm{\alpha -\beta -1}} \\ &\leq \frac{t^{\alpha -1}}{\Gamma (\alpha )} \biggl(1+\frac{A(1)-A(0)}{1-\Delta } \biggr) \\ &=M_{3}t^{\alpha -1}. \end{aligned}$$

$(3)$ For any $t,s\in [0,1]$, from $(3)$ of Lemma 2.3 and $(3)$ of Lemma 2.4, we have

$$\begin{aligned} G(t,s) &= H(t,s)+\frac{t^{\alpha -1}}{1-\Delta } \int _{0}^{1}K(\tau ,s)\,\mathrm{d}A(\tau ) \\ &\leq \frac{1}{\Gamma (\alpha )}s(1-s)^{\alpha -\beta -1}+\frac{t^{\alpha -1} \max \{\alpha -\beta -1,1\}\int _{0}^{1}\tau ^{\mathrm{\alpha -\beta -2}}\,\mathrm{d}A(\tau )}{\Gamma (\alpha )(1-\Delta )}s(1-s)^{\alpha -\beta -1} \\ &\leq \frac{s(1-s)^{\mathrm{\alpha -\beta -1}}}{\Gamma (\alpha )} \biggl(1+\frac{\max \{\alpha -\beta -1,1\} \int _{0}^{1}\tau ^{\mathrm{\alpha -\beta -2}}\,\mathrm{d}A(\tau )}{1-\Delta } \biggr) \\ &= M_{2}s(1-s)^{\mathrm{\alpha -\beta -1}} \\ &\leq M_{2}. \end{aligned}$$

On the other hand, for any $t,s\in [0,1]$, we have

$$\begin{aligned} G(t,s)&=H(t,s)+\frac{t^{\alpha -1}}{1-\Delta } \int _{0}^{1}K(\tau ,s)\,\mathrm{d}A(\tau ) \\ &\geq \frac{1}{\Gamma (\alpha )}t^{\alpha -1}s\beta (1-s)^{\alpha -\beta -1} \\ &\quad {}+\frac{t^{\alpha -1}\min \{\alpha -\beta -1,1\}\int _{0}^{1}\tau ^{\alpha -\beta -1} (1-\tau )\,\mathrm{d}A(\tau )}{\Gamma (\alpha )(1-\Delta )}s(1-s)^{\alpha -\beta -1} \\ &=\frac{t^{\alpha -1}s(1-s)^{\alpha -\beta -1}}{\Gamma (\alpha )} \biggl(\beta +\frac{\min \{\alpha -\beta -1,1\}\int _{0}^{1}\tau ^{\mathrm{\alpha -\beta -1}} (1-\tau )\,\mathrm{d}A(\tau )}{1-\Delta } \biggr) \\ &=M_{1}t^{\alpha -1}s(1-s)^{\alpha -\beta -1}. \end{aligned}$$

 □

3 Existence of positive solutions of the boundary value problem

We make the following assumption throughout this paper.

• (H1) There exists a nonnegative function $p\in L^{1}[0,1]$ and $\int _{0}^{1}p(s)\,\mathrm{d}s>0$ such that

  $$ f(t,u)\geq -p(t); $$

• (H2) There exists $[\theta _{1},\theta _{2}]\subset (0,1)$ such that

  $$ \liminf _{u\rightarrow +\infty }\inf _{t\in [\theta _{1},\theta _{2}]}\frac{f(t,u)}{u}=+\infty ; $$

• (H3) There exists $[\theta _{3},\theta _{4}]\subset (0,1)$ such that

  $$ \liminf _{u\rightarrow +\infty }\inf _{t\in [\theta _{3},\theta _{4}]}f(t,u)>\frac{2M_{2}M_{3}\int _{0}^{1}p(s)\,\mathrm{d}s}{M_{1}^{2}\theta _{3}^{\alpha -1}\int _{\theta _{3}}^{\theta _{4}}s(1-s)^{\alpha -\beta -1}\,\mathrm{d}s}, $$

  and

  $$ \limsup _{u\rightarrow +\infty }\sup _{t\in [0,1]}\frac{f(t,u)}{u}=0. $$

Define $P=\{u\in E: u(t)\geq \frac{M_{1}t^{\alpha -1}}{M_{2}}\Vert u \Vert , t\in [0,1]\}$. Obviously, $P\subset E$ is a cone of E.

We denote $B_{r}=\{u\in E:\Vert u \Vert < r\}$, $P_{r}=P\cap B_{r}$, and $\partial P_{r}=P\cap \partial B_{r}$.

Let $\omega (t)$ be the solution of the boundary value problem

$$ \textstyle\begin{cases} D_{0^{+}}^{\alpha }u(t)+\lambda p(t)=0, \quad t\in (0,1),\\ u(0)= u'(0)=\cdots =u^{(n-2)}(0)=0,\\ D_{0^{+}}^{\beta }u(1)=\int _{0}^{1}D_{0^{+}}^{\beta }u(t)\,\mathrm{d}A(t). \end{cases} $$

(3.1)

By Lemma 2.2, we have

$$ \omega (t)=\lambda \int _{0}^{1}G(t,s)p(s)\,\mathrm{d}s $$

is the unique solution of boundary value problem (3.1).

Denote $[u(t)-\omega (t)]^{+}=\max \{u(t)-\omega (t),0\}$, let

$$ f^{*} \bigl(t,u(t) \bigr)=f\bigl(t,\bigl[u(t)-\omega (t) \bigr]^{+}\bigr)+p(t), \quad t\in (0,1). $$

Next we will consider the following boundary value problem:

$$ \textstyle\begin{cases} D_{0^{+}}^{\alpha }u(t)+\lambda f^{*} (t,u(t) )=0, \quad t\in (0,1),\\ u(0)= u'(0)=\cdots =u^{(n-2)}(0)=0,\\ D_{0^{+}}^{\beta }u(1)=\int _{0}^{1}D_{0^{+}}^{\beta }u(t)\,\mathrm{d}A(t). \end{cases} $$

(3.2)

We define an operator $T:P\rightarrow E$ by

$$ Tu(t)=\lambda \int _{0}^{1}G(t,s)f^{*} \bigl(s,u(s) \bigr)\,\mathrm{d}s. $$

Lemma 3.1

Assume (H1) holds. Then $u^{*}$ is a positive solution of boundary value problem (1.1) if and only if $u=u^{*}+\omega$ is a positive solution of boundary value problem (3.2) and $u(t)\geq \omega (t)$ for $t\in [0,1]$.

Proof

If $u^{*}$ is a positive solution of boundary value problem (1.1), then

$$\begin{aligned} D_{0+}^{\alpha } \bigl(u^{*}(t)+\omega (t) \bigr)&= D_{0+}^{\alpha }u^{*}(t)+D_{0+}^{\alpha } \omega (t) \\ &= -\lambda f\bigl(t,u^{*}(t)\bigr)-\lambda p(t) \\ &= -\lambda \bigl(f\bigl(t,u^{*}(t)\bigr)+p(t) \bigr) \\ &= -\lambda f^{*} \bigl(t,\bigl(u^{*}(t)+\omega (t)\bigr) \bigr), \end{aligned}$$

which implies that

$$ D_{0+}^{\alpha }u(t)=-\lambda f^{*}\bigl(t,u(t)\bigr). $$

Since $u^{*}$ is a positive solution, then $u(t)\geq \omega (t)\geq 0$ for $t\in [0,1]$. It is easy to show that $u(t)$ satisfies the boundary conditions. Therefore, $u(t)$ is a positive solution of boundary value problem (3.2).

On the other hand, if $u=u^{*}+\omega$ is a positive solution of boundary value problem (3.2) and $u(t)\geq \omega (t)$ for $t\in [0,1]$, we can easily prove that $u^{*}$ is a positive solution of boundary value problem (1.1). □

Lemma 3.2

Assume that (H1) holds and f satisfies the $L^{1}$-Carathéodory conditions, then $T:P\rightarrow P$ is completely continuous.

Proof

(1) We can show that $T:P\rightarrow P$.

According to Lemmas 2.2 and 2.5, we have $Tu(t)\geq 0$ on $[0,1]$ for $u\in P$ and

$$\begin{aligned} Tu(t)&=\lambda \int _{0}^{1}G(t,s)f^{*} \bigl(s,u(s) \bigr)\,\mathrm{d}s \\ &\geq \lambda t^{\alpha -1}M_{1} \int _{0}^{1}s(1-s)^{\alpha -\beta -1}f^{*} \bigl(s,u(s) \bigr)\,\mathrm{d}s. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} Tu(t)&=\lambda \int _{0}^{1}G(t,s)f^{*} \bigl(s,u(s) \bigr)\,\mathrm{d}s \\ &\leq \lambda M_{2} \int _{0}^{1}s(1-s)^{\alpha -\beta -1}f^{*} \bigl(s,u(s) \bigr)\,\mathrm{d}s. \end{aligned}$$

Then $Tu(t)\geq \frac{M_{1}t^{\alpha -1}}{M_{2}}\Vert Tu \Vert$, which implies $T:P\rightarrow P$.

(2) We show that T is a continuous operator.

Let $\{u_{n}\}\subset P, u_{0}\in P$, and $\Vert u_{n}-u_{0} \Vert \rightarrow 0$ as $n\rightarrow \infty$, there exists a constant $r_{0}>0$ such that $\Vert u_{n} \Vert \leq r_{0}$ and $\Vert u_{0} \Vert \leq r_{0}$. Therefore, for a.e. $s\in [0,1]$, we have

$$ \bigl\vert f^{*} \bigl(s,u_{n}(s) \bigr)-f^{*}\bigl(s,u_{0}(s) \bigr)\bigr\vert \rightarrow 0, \quad n\rightarrow \infty , $$

and

$$ \bigl\vert f^{*} \bigl(s,u_{n}(s) \bigr)-f^{*}\bigl(s,u_{0}(s) \bigr)\bigr\vert \leq 2\varphi _{r_{0}}(s). $$

By the Lebesgue dominated convergence theorem, we can get

$$ \lim _{n\rightarrow \infty }\Vert Tu_{n}-Tu_{0} \Vert = 0, \quad n \rightarrow \infty . $$

Hence, $T:P\rightarrow P$ is continuous.

(3) $T:P\rightarrow P$ is relatively compact.

Let $\Omega \subset P$ be any bounded set, then there exists a constant $l>0$ such that $\Vert u \Vert \leq l$ for each $u\in \Omega$, we have

$$ \bigl\vert Tu(t) \bigr\vert =\biggl\vert \lambda \int _{0}^{1}G(t,s)f^{*} \bigl(s,u(s) \bigr)\,\mathrm{d}s \biggr\vert \leq \lambda M_{2} \int _{0}^{1} \bigl(\varphi _{l}(s)+p(s) \bigr)\,\mathrm{d}s< +\infty , $$

which implies that $\Vert Tu \Vert \leq \lambda M_{2}\int _{0}^{1} (\varphi _{l}(s)+p(s) )\,\mathrm{d}s$. Hence, $T(\Omega )$ is uniformly bounded.

In addition, for any given $u\in \Omega$, because $G(t,s)$ is continuous for $(t,s)\in [0,1]\times [0,1]$, then it must be uniformly continuous. So, for any $\varepsilon >0$, there exists a constant $\delta >0$ such that, for any $t_{1},t_{2},s\in [0,1]$, as $\vert t_{1}-t_{2} \vert <\delta$, we can get

$$ \bigl\vert G(t_{1},s)-G(t_{2},s) \bigr\vert < \frac{\varepsilon }{\lambda \int _{0}^{1} (\varphi _{l}(s)+p(s) )\,\mathrm{d}s+1}. $$

Then

$$ \bigl\vert Tu(t_{1})-Tu(t_{2}) \bigr\vert \leq \lambda \int _{0}^{1}\bigl\vert G(t_{1},s)-G(t_{2},s) \bigr\vert \bigl\vert f^{*}\bigl(s,u(s)\bigr) \bigr\vert \,\mathrm{d}s< \varepsilon . $$

Thus, we prove $T(\Omega )$ is equicontinuous.

According to the Arzela-Ascoli theorem, we conclude that $T(\Omega )$ is relatively compact.

Therefore, $T:P\rightarrow P$ is completely continuous. □

Theorem 3.3

Assume that (H1) and (H2) hold. Then there exists a constant $\lambda ^{*}>0$ such that boundary value problem (1.1) has at least one positive solution for any $\lambda \in (0,\lambda ^{*})$.

Proof

By assumption (H2), there exists a constant $N>0$ such that, for any $t\in [\theta _{1},\theta _{2}]$ and $u>N$, we have $f(t,u)>\tilde{L}u$, where $\tilde{L}=\frac{2M_{2}}{\lambda M_{1}^{2}\theta _{1}^{2(\alpha -1)}\int _{\theta _{1}}^{\theta _{2}} s(1-s)^{\alpha -\beta -1}\,\mathrm{d}s}$.

For any $\lambda >0$, let $r_{1}>\max \{\frac{2\lambda M_{2}M_{3}}{M_{1}}\int _{0}^{1}p(s)\,\mathrm{d}s,\frac{2M_{2}N}{M_{1}\theta _{1}^{\alpha -1}}\}$.

Then, for any $u\in \partial P_{r_{1}}, t\in [0,1]$, we have

$$\begin{aligned} u(t)-\omega (t)&\geq \frac{M_{1}t^{\alpha -1}}{M_{2}}r_{1}-\lambda \int _{0}^{1}G(t,s)p(s)\,\mathrm{d}s \\ &\geq \frac{M_{1}t^{\alpha -1}}{M_{2}}r_{1}-\lambda M_{3}t^{\alpha -1} \int _{0}^{1}p(s)\,\mathrm{d}s \\ &\geq t^{\alpha -1} \biggl(\frac{M_{1}}{M_{2}}r_{1}-\lambda M_{3} \int _{0}^{1}p(s)\,\mathrm{d}s \biggr) \\ &\geq t^{\alpha -1} \biggl(\frac{M_{1}}{M_{2}}r_{1}- \frac{M_{1}}{2M_{2}}r_{1} \biggr) \\ &= \frac{M_{1}}{2M_{2}}r_{1}t^{\alpha -1}\geq 0. \end{aligned}$$

Thus, for any $t\in [\theta _{1},\theta _{2}]$, we have

$$ u(t)-\omega (t)\geq \frac{M_{1}\theta _{1}^{\alpha -1}}{2M_{2}}r_{1}\geq N. $$

Hence, we get

$$\begin{aligned} \Vert Tu \Vert &=\max _{t\in [0,1]}\lambda \int _{0}^{1}G(t,s)f^{*}\bigl(s,u(s) \bigr)\,\mathrm{d}s \\ &=\max _{t\in [0,1]}\lambda \int _{0}^{1}G(t,s) \bigl(f \bigl(s,\bigl[u(s)- \omega (s)\bigr]^{+} \bigr)+p(s) \bigr)\,\mathrm{d}s \\ &=\max _{t\in [0,1]}\lambda \int _{0}^{1}G(t,s) \bigl(f \bigl(s,u(s)-\omega (s) \bigr)+p(s) \bigr)\,\mathrm{d}s \\ &\geq \max _{t\in [\theta _{1},\theta _{2}]}\lambda \int _{\theta _{1}}^{\theta _{2}}G(t,s)f \bigl(s,u(s)-\omega (s) \bigr) \,\mathrm{d}s \\ &\geq \max _{t\in [\theta _{1},\theta _{2}]}\lambda \tilde{L} \int _{\theta _{1}}^{\theta _{2}}G(t,s) \bigl(u(s)-\omega (s) \bigr) \,\mathrm{d}s \\ &\geq \lambda \tilde{L}\frac{M_{1}\theta _{1}^{\alpha -1}}{2M_{2}}r_{1} \int _{\theta _{1}}^{\theta _{2}}M_{1}\theta _{1}^{\alpha -1}s(1-s)^{\alpha -\beta -1}\,\mathrm{d}s \\ &=\frac{\lambda M_{1}^{2}\theta _{1}^{2(\alpha -1)}\int _{\theta _{1}}^{\theta _{2}}s(1-s)^{\alpha -\beta -1}\,\mathrm{d}s}{2M_{2}}\tilde{L}r_{1}=r_{1}. \end{aligned}$$

Thus

$$ \Vert Tu \Vert \geq \Vert u \Vert , \quad \mbox{for } u\in \partial P_{r_{1}}. $$

Take $0< r_{2}< r_{1}$. Choose $\lambda ^{*}=\min \{\frac{M_{1}r_{2}}{M_{2}M_{3}\int _{0}^{1}p(s)\,\mathrm{d}s}, \frac{r_{2}}{M_{2}\int _{0}^{1} (\varphi _{r_{2}}(s)+p(s) )\,\mathrm{d}s} \}$.

For any $\lambda \in (0,\lambda ^{*}), u\in \partial P_{r_{2}}, u(t)\geq \frac{M_{1}t^{\alpha -1}}{M_{2}}r_{2}$ and

$$ \omega (t)=\lambda \int _{0}^{1}G(t,s)p(s)\,\mathrm{d}s\leq \lambda M_{3}t^{\alpha -1} \int _{0}^{1}p(s)\,\mathrm{d}s. $$

We have

$$ 0\leq \frac{M_{1}t^{\alpha -1}}{M_{2}}r_{2}-\lambda M_{3}t^{\alpha -1} \int _{0}^{1}p(s)\,\mathrm{d}s\leq u(t)-\omega (t)\leq r_{2}. $$

Therefore,

$$\begin{aligned} Tu(t) &=\lambda \int _{0}^{1}G(t,s)f^{*}\bigl(s,u(s) \bigr)\,\mathrm{d}s \\ &=\lambda \int _{0}^{1}G(t,s) \bigl(f \bigl(s,u(s)-\omega (s) \bigr)+p(s) \bigr)\,\mathrm{d}s \\ &\leq \lambda M_{2} \int _{0}^{1} \bigl(\varphi _{r_{2}}(s)+p(s) \bigr)\,\mathrm{d}s \\ &< \lambda ^{*} M_{2} \int _{0}^{1} \bigl(\varphi _{r_{2}}(s)+p(s) \bigr)\,\mathrm{d}s \\ &\leq \frac{r_{2}}{M_{2}\int _{0}^{1} (\varphi _{r_{2}}(s)+p(s) )\,\mathrm{d}s}M_{2} \int _{0}^{1} \bigl(\varphi _{r_{2}}(s)+p(s) \bigr)\,\mathrm{d}s \\ &=r_{2}. \end{aligned}$$

Thus,

$$ \Vert Tu \Vert \leq \Vert u \Vert , \quad \mbox{for } u\in \partial P_{r_{2}}. $$

By Lemma 3.2, we know that T is a completely continuous operator.

According to cone expansion and cone compression fixed point theorem (see [31]), we can obtain that T has a fixed point u such that $r_{1}\leq \Vert u \Vert \leq r_{2}$ in $P\cap (\bar{B}_{r_{2}}\setminus B_{r_{1}})$.

Let $u^{*}=u-\omega$. For $t\in (0,1)$, we have

$$ u^{*}(t)=\frac{M_{1}t^{\alpha -1}}{M_{2}}\Vert u \Vert -\lambda M_{3}t^{\alpha -1} \int _{0}^{1}p(s)\,\mathrm{d}s> t^{\alpha -1} \biggl(\frac{M_{1}}{M_{2}}r_{2}-\lambda ^{*} M_{3} \int _{0}^{1}p(s)\,\mathrm{d}s \biggr)>0. $$

Then $u^{*}$ is the positive solution of (1.1).

Hence, there exists a constant $\lambda ^{*}>0$ such that boundary value problem (1.1) has at least one positive solution for any $\lambda \in (0,\lambda ^{*})$. □

Theorem 3.4

Assume that (H1) and (H3) hold. Then there exists a constant $\bar{\lambda }^{*}>0$ such that boundary value problem (1.1) has at least one positive solution for any $\lambda \in (\bar{\lambda }^{*},+\infty )$.

Proof

By the first limit of (H3), there exists a constant $N>0$ such that, for any $t\in [\theta _{3},\theta _{4}]$, $u>N$, we have

$$ f(t,u)\geq \frac{2M_{2}M_{3}\int _{0}^{1}p(s)\,\mathrm{d}s}{M_{1}^{2}\theta _{3}^{\alpha -1} \int _{\theta _{3}}^{\theta _{4}}s(1-s)^{\alpha -\beta -1}\,\mathrm{d}s}. $$

Choose $\bar{\lambda }^{*}=\frac{N\theta _{3}^{1-\alpha }}{M_{3}\int _{0}^{1}p(s)\,\mathrm{d}s}, R_{1}=\frac{2\lambda M_{2}M_{3}\int _{0}^{1}p(s)\,\mathrm{d}s}{M_{1}}$.

Then, for any $\lambda \in (\bar{\lambda }^{*},+\infty ), u\in \partial P_{R_{1}}$, we have

$$\begin{aligned} u(t)-\omega (t) &\geq \frac{M_{1}t^{\alpha -1}}{M_{2}}R_{1}-\lambda M_{3}t^{\alpha -1} \int _{0}^{1}p(s)\,\mathrm{d}s \\ &=\lambda M_{3}t^{\alpha -1} \int _{0}^{1}p(s)\,\mathrm{d}s. \end{aligned}$$

Hence, for any $t\in [\theta _{3},\theta _{4}],u(t)-\omega (t)\geq Nt^{\alpha -1}\theta _{3}^{1-\alpha }\geq N$, we get

$$\begin{aligned} Tu(t)&=\lambda \int _{0}^{1}G(t,s)f^{*} \bigl(s,u(s) \bigr)\,\mathrm{d}s \\ &\geq \lambda \int _{\theta _{3}}^{\theta _{4}}G(t,s)f \bigl(s,u(s)-\omega (s) \bigr) \,\mathrm{d}s \\ &\geq \frac{2\lambda M_{2}M_{3}\int _{0}^{1}p(s)\,\mathrm{d}s}{M_{1}^{2}\theta _{3}^{\alpha -1} \int _{\theta _{3}}^{\theta _{4}}s(1-s)^{\alpha -\beta -1}\,\mathrm{d}s} \int _{\theta _{3}}^{\theta _{4}} G(t,s)\,\mathrm{d}s \\ &\geq \frac{2\lambda M_{2}M_{3}\int _{0}^{1}p(s)\,\mathrm{d}s}{M_{1}^{2}\theta _{3}^{\alpha -1} \int _{\theta _{3}}^{\theta _{4}}s(1-s)^{\alpha -\beta -1}\,\mathrm{d}s} \int _{\theta _{3}}^{\theta _{4}} t^{\alpha -1}M_{1}s(1-s)^{\alpha -\beta -1} \,\mathrm{d}s \\ &\geq \frac{2\lambda M_{2}M_{3}\int _{0}^{1}p(s)\,\mathrm{d}s}{M_{1}}=R_{1}. \end{aligned}$$

Thus

$$ \Vert Tu \Vert \geq \Vert u \Vert , \quad \mbox{for } u\in \partial P_{R_{1}}. $$

On the other hand, let $\varepsilon =\frac{1}{3\lambda M_{2}}$, by (H3), $\limsup _{u\rightarrow +\infty }\sup _{t\in [0,1]}\frac{f(t,u)}{u}=0$, there exists a constant $M>R_{1}$ such that

$$ f(t,u)\leq \varepsilon u \quad \mbox{for } t\in [0,1] \mbox{ and } u>M. $$

Since f satisfies the $L^{1}$-Carathéodory conditions on $[0,1]\times [0,+\infty )$, we have

$$ f(t,u)\leq \varphi _{M}(t)\quad \mbox{for } t\in [0,1] \mbox{ and } 0\leq u\leq M. $$

Let

$$ R_{2}>\max \biggl\{ M, \frac{\lambda M_{2}M_{3}\int _{0}^{1}p(s)\,\mathrm{d}s}{M_{1}}, 3\lambda M_{2} \int _{0}^{1}p(s)\,\mathrm{d}s, 3\lambda M_{2} \int _{0}^{1}\varphi _{M}(s)\,\mathrm{d}s \biggr\} . $$

Then, for any $u\in \partial P_{R_{2}}$ and $t\in [0,1]$, we have

$$ u(t)-\omega (t)\geq \frac{M_{1}t^{\alpha -1}}{M_{2}}R_{2}-\lambda M_{3}t^{\alpha -1} \int _{0}^{1}p(s)\,\mathrm{d}s\geq 0. $$

Hence, for any $t\in [0,1]$, we get

$$\begin{aligned} Tu(t) &=\lambda \int _{0}^{1}G(t,s)f^{*} \bigl(s,u(s) \bigr)\,\mathrm{d}s \\ &=\lambda \int _{0}^{1}G(t,s) \bigl(f \bigl(s,u(s)-\omega (s) \bigr)+p(s) \bigr)\,\mathrm{d}s \\ &\leq \lambda M_{2} \int _{0}^{1} \bigl(f \bigl(s,u(s)-\omega (s) \bigr)+p(s) \bigr)\,\mathrm{d}s \\ &=\lambda M_{2} \int _{0}^{1}f \bigl(s,u(s)-\omega (s) \bigr) \,\mathrm{d}s+\lambda M_{2} \int _{0}^{1}p(s)\,\mathrm{d}s \\ &=\lambda M_{2} \biggl( \int _{u(s)-\omega (s)>M}f \bigl(s,u(s)-\omega (s) \bigr)\,\mathrm{d}s+ \int _{0\leq u(s)-\omega (s)\leq M}f \bigl(s,u(s)-\omega (s) \bigr)\,\mathrm{d}s \biggr) \\ &\quad {}+\lambda M_{2} \int _{0}^{1}p(s)\,\mathrm{d}s \\ &\leq \lambda M_{2}\varepsilon R_{2}+\lambda M_{2} \int _{0}^{1}\varphi _{M}(s)\,\mathrm{d}s+ \lambda M_{2} \int _{0}^{1}p(s)\,\mathrm{d}s \\ &\leq \frac{R_{2}}{3}+\frac{R_{2}}{3}+\frac{R_{2}}{3}=R_{2}. \end{aligned}$$

Thus,

$$ \Vert Tu \Vert \leq \Vert u \Vert \quad \mbox{for } u\in \partial P_{R_{2}}. $$

By Lemma 3.2, we know that T is a completely continuous operator.

According to cone expansion and cone compression fixed point theorem (see [31]), we can obtain that T has a fixed point u such that $R_{1}\leq \Vert u \Vert \leq R_{2}$ in $P\cap (\bar{B}_{R_{2}}\setminus B_{R_{1}})$.

Let $u^{*}(t)=u(t)-\omega (t), t\in [0,1]$. We have

$$ u^{*}(t)=\frac{M_{1}t^{\alpha -1}}{M_{2}}\Vert u \Vert -\lambda M_{3}t^{\alpha -1} \int _{0}^{1}p(s)\,\mathrm{d}s\geq \biggl(\lambda M_{3} \int _{0}^{1}p(s)\,\mathrm{d}s \biggr)t^{\alpha -1}. $$

Then $u^{*}$ is the positive solution of (1.1).

Therefore, there exists a constant $\bar{\lambda }^{*}>0$ such that boundary value problem (1.1) has at least one positive solution for any $\lambda \in (\bar{\lambda }^{*},+\infty )$. □

4 Illustration

To illustrate our main results, we present the following examples.

Example 4.1

We consider the boundary value problem

$$ \textstyle\begin{cases} D_{0^{+}}^{\frac{7}{2}}u(t)+\lambda (u^{\frac{3}{2}}(t)+\ln t)=0, \quad t\in (0,1),\\ u(0)= u'(0)=u''(0)=0,\\ D_{0^{+}}^{\frac{1}{2}}u(1)=\int _{0}^{1}D_{0^{+}}^{\frac{1}{2}}u(t)\,\mathrm{d}t, \end{cases} $$

(4.1)

where $\alpha =\frac{7}{2}$, $n=4$, $\beta =\frac{1}{2}$, $A(t)=t$. Choose $f(t,u)=u^{\frac{3}{2}}+\ln t$. Then

$$ -\vert \ln t \vert =\ln t\leq f(t,u)=u^{\frac{3}{2}}+\ln t,\quad t\in (0,1), $$

where

$$ p(t)=\vert \ln t \vert , \qquad q(t)=1, \qquad g(u)=u^{\frac{1}{2}}. $$

By direct calculation, we have

$$ M_{1}=0.2256758, \qquad M_{2}=M_{3}=0.7522528,\qquad \int _{0}^{1}p(s)\,\mathrm{d}s=1, $$

and

$$ \liminf _{u\rightarrow +\infty }\inf _{t\in [\theta _{1},\theta _{2}]}\frac{f(t,u)}{u}=+\infty . $$

So all the conditions of Theorem 3.3 are satisfied. By Theorem 3.3, there exists $\lambda ^{*}>0$ such that boundary value problem (4.1) has at least one positive solution provided $\lambda \in (0,\lambda ^{*})$.

Example 4.2

We consider the boundary value problem

$$ \textstyle\begin{cases} D_{0^{+}}^{\frac{8}{3}}u(t)+\lambda (-\frac{\sin (2\pi u(t))}{\sqrt{t(1-t)}}+u^{\frac{1}{2}}(t) )=0, \quad t\in (0,1),\\ u(0)= u'(0)=0,\\ D_{0^{+}}^{\frac{1}{8}}u(1)=D_{0^{+}}^{\frac{1}{8}}u(\frac{1}{4})+D_{0^{+}}^{\frac{1}{8}}u(\frac{3}{4}), \end{cases} $$

(4.2)

where $\alpha =\frac{8}{3}$, $n=3$, $\beta =\frac{1}{8}$, $f(t,u)=-\frac{\sin (2\pi u)}{\sqrt{t(1-t)}}+u^{\frac{1}{2}}$,

$$ A(t)= \textstyle\begin{cases} 0,& 0\leq t< \frac{1}{4},\\ 1,& \frac{1}{4}\leq t< \frac{3}{4},\\ 2,& \frac{3}{4}\leq t< 1. \end{cases} $$

Let $p(t)=\frac{2}{\sqrt{t(1-t)}}$. Clearly, for $t\in (0,1)$, we have

$$ f(t,u)>-p(t). $$

By direct calculation, we have

$$ M_{1}=2.461073,\qquad M_{2}=6.3273,\qquad M_{3}=6.197864,\qquad \int _{0}^{1}p(s)\,\mathrm{d}s=6.283185, $$

and

$$ \limsup _{u\rightarrow +\infty }\sup _{t\in [0,1]}\frac{f(t,u)}{u}=0. $$

So, all the conditions of Theorem 3.4 are satisfied. By Theorem 3.4, there exists $\bar{\lambda }^{*}>0$ such that boundary value problem (4.2) has at least one positive solution provided $\lambda \in (\bar{\lambda }^{*},+\infty )$.