Positive solutions of conformable fractional differential equations with integral boundary conditions

Abstract

In this paper, we discuss the existence of positive solutions of the conformable fractional differential equation $T_{\alpha }x(t)+f(t,x(t))=0$, $t\in [0,1]$, subject to the boundary conditions $x(0)=0$ and $x(1)= \lambda \int_{0}^{1}x(t)\,\mathrm{d}t$, where the order α belongs to $(1,2]$, $T_{\alpha }x(t)$ denotes the conformable fractional derivative of a function $x(t)$ of order α, and $f:[0,1]\times [0,\infty)\mapsto [0,\infty)$ is continuous. By use of the fixed point theorem in a cone, some criteria for the existence of at least one positive solution are established. The obtained conditions are generally weaker than those derived by using the classical norm-type expansion and compression theorem. A concrete example is given to illustrate the possible application of the obtained results.

1 Introduction

The fractional derivative is a generalization of the classical one to an arbitrary order, and the question of what is a fractional derivative was first raised by L’Hôpital in a letter to Leibniz in 1695. Since then, fractional calculus has been extensively studied, and it has been applied to almost every field of science, engineering, and mathematics in the last four decades [1–10]. It is worth emphasizing that there exist a number of definitions of fractional derivatives in the literature, and the different definitions are constructed to satisfy various constraints.

Recently, in [11] Khalil et al. introduced a new well-behaved definition of a fractional derivative termed the conformable fractional derivative. The new definition has drawn much interest from many researchers. And some results have been obtained on the properties of the conformable fractional derivative [11–13]. Several applications and generalizations of the definition were also discussed in [14–20], among which [14] indicated that several specific conformable fractional models are consistent with experimental date, and which [15] interpreted the physical and geometrical meaning of the conformable fractional derivative. Although the definite meaning indicates potential applications of the conformable fractional derivative in physics and engineering, it is worth noting that the investigation of the theory of conformable fractional differential equations has only entered an initial stage.

Initial value problems of conformable fractional differential equations were discussed in [21–23]; and analytical solutions to some specific conformable fractional partial differential equations were studied in [24–30]. For the discussion of boundary value problems (BVPs for short) of conformable fractional differential equations, some theoretical developments have also been achieved. In particular, Lyapunov type inequalities for some conformable boundary value problems were established in [31, 32]; a regular conformable fractional Sturm–Liouville eigenvalue problem was considered in [33]; solvability of some two-point fractional BVP was considered in [34–36] by using topological transversality theorem; a type of three-point fractional BVP was studied in [37] by means of fixed point theorems; and a class of periodic BVP was discussed in [38] by virtue of methods of lower and upper solutions. Applying approximation methods of operators and fixed point theorems, Xiaoyu Dong et al. [39] investigated the existence of positive solutions to a specific type of two-point BVP of p-Laplacian.

Motivated by the above-mentioned results and techniques in treating those BVPs of the conformable fractional differential equations, we then turn to investigating the existence of positive solutions for the BVP as follows:

$$ \textstyle\begin{cases} T_{\alpha }x(t)+f(t,x(t))=0, \quad t\in [0,1], \\ x(0)=0,\qquad x(1)=\mathcal{L}(x), \end{cases} $$

(1.1)

where α belongs to $(1,2]$, $T_{\alpha }$ denotes the conformable fractional derivative of order α, the function $f:[a,\infty)\times [0,\infty)\mapsto [0,\infty)$ is continuous, and $\mathcal{L}(x)=\lambda \int_{0}^{1}x(t)\,\mathrm{d}t$ for which the parameter λ is a positive number.

In the context of the conformable fractional derivatives, to the best of our knowledge, there have been very few results in the literature for the existence of positive solution to the conformable fractional differential equations with integral boundary conditions. It is worth pointing out that the obtained Green function in this work is singular, while the Green functions of BVPs of some new fractional derivatives with nonsingular kernels are nonsingular [40, 41].

The rest of paper is organized as follows. Section 2 preliminarily provides some definitions and lemmas which are crucial to the following discussion. In Sect. 3, we establish some criteria for the existence of at least one positive solution to the BVP (1.1) by means of the fixed point theorem in a cone. The obtained conditions are generally weaker than those derived by using the classical norm-type expansion and compression theorem [42]. Finally, an example is given to illustrate the possible application of the obtained results.

2 Preliminaries

In this section, we preliminarily provide some definitions and lemmas which play a key role in the following discussion.

Definition 2.1

([11, 12])

Let α be in $(0,1]$. The conformable fractional derivative of a function $f:[0,\infty)\mapsto \mathbb{R}$ of order α is defined by

$$ T_{\alpha }f(t)=\lim_{\epsilon \rightarrow 0}\frac{f(t+\epsilon t^{1-\alpha })-f(t)}{\epsilon }. $$

If $T_{\alpha }f(t)$ exists on $(0,b)$, then $T_{\alpha }f(0)=\lim_{t\rightarrow 0}T_{\alpha }f(t)$.

Definition 2.2

([11, 12])

Let α be in $(n,n+1]$. The conformable fractional derivative of a function $f:[0,\infty)\mapsto \mathbb{R}$ of order α is defined by

$$ T_{\alpha }f(t)=T_{\beta }f^{(n)}(t) \quad \text{for which } \beta = \alpha -n. $$

Definition 2.3

([12])

Let α be in $(n,n+1]$. The fractional integral of a function $f:[0,\infty)\mapsto \mathbb{R}$ of order α is defined by

$$ I_{\alpha }f(t)=\frac{1}{n!} \int_{0}^{t}(t-s)^{n}s^{\alpha -n-1}f(s) \,\mathrm{d}s. $$

Lemma 2.1

([11, 12])

Let α be in $(n,n+1]$. If f is a continuous function on $[0,\infty)$, then, for all $t>0$, $T_{\alpha }I_{\alpha }f(t)=f(t)$.

Lemma 2.2

([11, 12])

Let α be in $(n,n+1]$. Then $T_{\alpha }t^{k}=0$ for t in $[0,1]$ and $k=1,2,\dots,n$.

Lemma 2.3

([12])

Let α be in $(n,n+1]$. If $T_{\alpha }f(t)$ is continuous on $[0,\infty)$, then

$$ I_{\alpha }T_{\alpha }f(t)=f(t)+c_{0}+c_{1}t+ \cdots +c_{n}t^{n} $$

for some real numbers $c_{k}$, $k=1,2,\dots,n$.

By Lemma 2.3, we next present an integral presentation of the solution for the BVP of the linearized equation related to the BVP (1.1).

$$ \textstyle\begin{cases} T_{\alpha }x(t)+h(t)=0, \quad t\in [0,1], \alpha \in (1,2], \\ x(0)=0, \qquad x(1)=\mathcal{L}(x). \end{cases} $$

(2.1)

Lemma 2.4

Let h be in $C[0,1]$. If $\lambda \neq 2$, then the BVP (2.1) exists a unique solution defined on $[0,1]$ given by

$$ x(t)= \int_{0}^{1}\mathcal{K}(t,s)h(s)\,\mathrm{d}s, $$

(2.2)

where

$$\begin{aligned}& \mathcal{K}(t,s)=\mathcal{G}(t,s)+\mathcal{H}(t,s), \end{aligned}$$

(2.3)

$$\begin{aligned}& \mathcal{G}(t,s)= \textstyle\begin{cases} (1-t)s^{\alpha -1}, &0\leqslant s\leqslant t\leqslant 1, \\ ts^{\alpha -2}(1-s), &0< t\leqslant s\leqslant 1, \end{cases}\displaystyle \end{aligned}$$

(2.4)

and

$$ \mathcal{H}(t,s)=\frac{2\lambda t}{2-\lambda } \int_{0}^{1}\mathcal{G}(\tau,s)\,\mathrm{d}\tau. $$

(2.5)

Proof

By the continuity of h and Lemma 2.3, it follows from Eq. (2.1) that

$$ x(t)=c_{0}+c_{1}t-I_{\alpha }h(t). $$

This, together the boundary conditions, implies $c_{0}=0$ and

$$ c_{1}=I_{\alpha }h(1)+\mathcal{L}(x). $$

Hence

$$\begin{aligned} x(t) =&-I_{\alpha }h(t)+tI_{\alpha }h(1)+t\mathcal{L}(x) \\ =&- \int_{0}^{t}(t-s)s^{\alpha -2} h(s)\,\mathrm{d}s+ \int_{0}^{t}t(1-s)s^{\alpha -2} h(s)\,\mathrm{d}s \\ &{}+ \int_{t}^{1}t(1-s)s^{\alpha -2} h(s)\, \mathrm{d}s+t\mathcal{L}(x), \end{aligned}$$

which yields

$$ x(t)= \int_{0}^{1}\mathcal{G}(t,s)h(s)\,\mathrm{d}s+t \mathcal{L}(x). $$

(2.6)

Applying the transformation $\mathcal{L}$ to both sides of Eq. (2.6), we get

$$ \mathcal{L}(x)=\frac{2}{2-\lambda } \int_{0}^{1}\mathcal{L}\bigl(\mathcal{G}(t,s) \bigr)h(s)\,\mathrm{d}s. $$

(2.7)

Substituting the above expression into (2.6), we obtain the desired result. □

The functions $\mathcal{G}$, $\mathcal{H}$ and $\mathcal{K}$ have several important properties as follows.

Lemma 2.5

For any $(t,s)$ in $(0,1]\times (0,1]$,

$$ 0\leq q(t)\mathcal{G}(s,s)\leq \mathcal{G}(t,s)\leq \mathcal{G}(s,s). $$

(2.8)

Furthermore, if λ belongs to $[0,2)$, then

$$\begin{aligned}& 0\leq q(t)\mathcal{H}(1,s)\leq \mathcal{H}(t,s)\leq \mathcal{H}(1,s), \end{aligned}$$

(2.9)

$$\begin{aligned}& 0\leq q(t)\mathcal{M}(s)\leq \mathcal{K}(t,s)\leq \mathcal{M}(s), \end{aligned}$$

(2.10)

where $q(t)=t(1-t)$, and $\mathcal{M}(s)= \mathcal{G}(s,s)+ \mathcal{H}(1,s)$.

Proof

By the definition of $\mathcal{G}$, for $0\leqslant s\leqslant t \leqslant 1$,

$$ \mathcal{G}(t,s)=(1-t)s^{\alpha -1}\leq (1-s)s^{\alpha -1}= \mathcal{G}(s,s), $$

and for $0< t\leqslant s\leqslant 1$,

$$ \mathcal{G}(t,s)=ts^{\alpha -2}(1-s)=\frac{t}{s}(1-s)s^{\alpha -1} \leq (1-s)s^{\alpha -1}=\mathcal{G}(s,s). $$

Thus $\mathcal{G}(t,s)\leq \mathcal{G}(s,s)$ for $(t,s)$ in $(0,1]\times (0,1]$.

Moreover, observe that, for $0< s\leqslant t\leqslant 1$,

$$ \mathcal{G}(t,s)=(1-t)s^{\alpha -1}\geq (1-t)s^{\alpha -1}(1-s)\geq q(t) \mathcal{G}(s,s)\geq 0 $$

and that, for $0< t\leqslant s\leqslant 1$,

$$ \mathcal{G}(t,s)= ts^{\alpha -2}(1-s)\geq ts^{\alpha -1}(1-s)\geq q(t) \mathcal{G}(s,s)\geq 0. $$

Hence

$$ 0\leq q(t)\mathcal{G}(s,s)\leq \mathcal{G}(t,s)\leq \mathcal{G}(s,s) $$

for $(t,s)$ in $(0,1]\times (0,1]$. Furthermore, the above inequality and the definitions of $\mathcal{H}$ and $\mathcal{K}$ clearly yield the inequalities (2.9) and (2.10). The proof is complete. □

The key tool in our approach is the following well-known fixed point theorem in a cone [42, 43].

Lemma 2.6

Let $\mathfrak{B}$ be a Banach space, $\mathcal{P}\subseteq \mathfrak{B}$ a cone, and $\Omega_{1}$, $\Omega_{2}$ two bounded open balls of $\mathfrak{B}$ centered at the origin with $\bar{\Omega } _{1}\subset \Omega_{2}$. Suppose that $\Phi:\mathcal{P}\cap (\bar{ \Omega }_{2}\setminus \Omega_{1})\rightarrow \mathcal{P}$ is a completely continuous operator such that

• (C1) $\|\Phi x\|\leqslant \|x\|$, $x\in \mathcal{P}\cap \partial \Omega_{1}$.

• (C2) There exists $\psi \in \mathcal{P}\setminus \{0\}$ such that $x\neq \Phi x+\lambda \psi$ for $x\in \mathcal{P}\cap \partial \Omega_{2}$ and $\lambda >0$.

Then Φ has a fixed point in $\mathcal{P}\cap (\bar{\Omega }_{2}\setminus \Omega_{1})$. The same conclusion remains valid if (C1) holds on $\mathcal{P}\cap \partial \Omega_{2}$ and (C2) holds on $\mathcal{P}\cap \partial \Omega_{1}$.

3 Main results

In order to utilize the fixed point theorem to discuss the existence of solutions of the boundary value problem, we now make the basic assumption and define some sets of functions in $C[a,b]$ and operators.

• (H) The function f is nonnegative, and continuous on $[0,1]\times [0, \infty)$, and the parameter λ belongs to $[0,2)$.

Let $\mathfrak{B}=C[0,1]$ be the classical Banach space with the norm $\|x\|=\sup_{t\in [0,1]}|x(t)|$. Furthermore, define the cone $\mathcal{P}$ in $\mathfrak{B}$ by

$$ \mathcal{P}= \bigl\{ x\in \mathfrak{B} | x(t)\geqslant q(t) \Vert x \Vert \text{ for } t\in [0,1] \bigr\} . $$

Here the function $q(t)$ is defined as in Lemma 2.5.

Given a positive number r, define the subset $\Omega_{r}$ of $\mathfrak{B}$ by

$$ \Omega_{r}=\bigl\{ x\in \mathfrak{B}: \Vert x \Vert < r\bigr\} . $$

Also, define the operator from the space $\mathfrak{B}$ to itself by

$$ (\Phi x) (t)= \int_{0}^{1}\mathcal{K}(t,s)f\bigl(t,x(s)\bigr)\, \mathrm{d}s. $$

(3.1)

Under the hypothesis (H), the operator is well defined and has the following property.

Lemma 3.1

If the hypothesis (H) holds, then $\Phi (\mathcal{P})\subset \mathcal{P}$.

Proof

For any x in $\mathcal{P}$, the definition of Φ and the inequality (2.10) imply that

$$ (\Phi x) (t)= \int_{0}^{1}\mathcal{K}(t,s)f\bigl(t,x(s)\bigr)\, \mathrm{d}s\geq q(t) \int_{0}^{1}\mathcal{M}(s)f\bigl(t,x(s)\bigr)\, \mathrm{d}s $$

and that

$$ (\Phi x) (t))= \int_{0}^{1}\mathcal{K}(t,s)f\bigl(t,x(s)\bigr)\, \mathrm{d}s\leq \int_{0}^{1}\mathcal{M}(s)f\bigl(t,x(s)\bigr)\, \mathrm{d}s, $$

which yield

$$ (\Phi x) (t)\geq q(t) \Vert \Phi x \Vert . $$

Hence $\Phi x\in \mathcal{P}$. We thus complete the proof. □

We further discuss the complete continuity of the operator Φ. To this end, denote the operator Φ by

$$ \Phi =\Phi_{1}+\Phi_{2}, $$

(3.2)

where the operators $\Phi_{1}$ and $\Phi_{2}$ are defined, respectively, by

$$ (\Phi_{1}x) (t)= \int_{0}^{1}\mathcal{G}(t,s)f\bigl(s,x(s)\bigr)\, \mathrm{d}s $$

(3.3)

and

$$ (\Phi_{2}x) (t)= \int_{0}^{1}\mathcal{H}(t,s)f\bigl(s,x(s)\bigr)\, \mathrm{d}s. $$

(3.4)

And then we claim that $\Phi_{1}$ and $\Phi_{2}:\mathcal{P}\mapsto \mathcal{P}$ are completely continuous operators. Indeed, by an argument similar to the proof of Lemma 3.1, using the inequalities (2.8) and (2.9) we first infer that $\Phi_{1}( \mathcal{P})\subset \mathcal{P}$ and $\Phi_{2}(\mathcal{P})\subset \mathcal{P}$.

Furthermore, observe that the kernel $\mathcal{G}(t,s)$ of $\Phi_{1}$ is singular on $[0,1]\times [0,1]$, and that the complete continuity of the operator $\Phi_{1}$ was verified in [34, 39] by using approximations of the operator. As for the operator $\Phi_{2}$, its kernel $\mathcal{H}(t,s)$ is continuous on $[0,1]\times [0,1]$, and using the standard argument, we can easily check that it is also completely continuous. Thus we obtain the following lemma.

Lemma 3.2

If the hypothesis (H) holds, then the operator $\Phi:\mathcal{P} \mapsto \mathcal{P}$ completely continuous.

The next lemma transforms the BVP (1.1) into an equivalent fixed point problem.

Lemma 3.3

If the hypothesis (H) holds, then a function x in $C[0,1]$ is a positive solution of the BVP (1.1) if and only if it is a fixed point of Φ in $\mathcal{P}$.

Proof

Let x be a fixed point of Φ in $\mathcal{P}$, then

$$ x(t)= \int_{0}^{1}\mathcal{K}(t,s)f\bigl(s,x(s)\bigr)\, \mathrm{d}s =-I_{\alpha }f\bigl(t,x(t)\bigr)+tI _{\alpha }f\bigl(1,x(1) \bigr)+t\mathcal{L}(x), $$

(3.5)

and thus, by the continuity of f, Lemma 2.1, 2.2 and 2.3,

$$ T_{\alpha }x(t)+f\bigl(t,x(t)\bigr)=0. $$

Moreover, the equality (3.5) directly implies $x(0)=0$ and $x(1)=\mathcal{L}(x)$. Therefore x is a positive solution of the BVP (1.1).

On the other hand, if x is a positive solution of the BVP (1.1), then Lemma 2.4 implies $\Phi x=x$. Moreover, by the same type of argument as for the proof of Lemma 3.1, we also get $x(t)\geqslant q(t)\|x\|$ for $t\in [0,1]$. Hence x is a fixed point of Φ in $\mathcal{P}$. We consequently complete the proof. □

Before presenting the main results, we further introduce some notations as follows:

$$\begin{aligned}& f_{0}=\lim_{x\rightarrow 0}\min_{t\in [0,1]} \frac{f(t,x)}{x} \quad \text{and}\quad f^{\infty }=\lim_{x\rightarrow \infty } \max_{t\in [0,1]} \frac{f(t,x)}{x}; \\& f^{0}=\lim_{x\rightarrow 0}\max_{t\in [0,1]} \frac{f(t,x)}{x} \quad \text{and}\quad f_{\infty }=\lim_{x\rightarrow \infty } \min_{t\in [0,1]} \frac{f(t,x)}{x}; \\& \Lambda_{1}= \biggl(q(\delta) \int_{\delta }^{1-\delta }\mathcal{M}(s)\,\mathrm{d}s \biggr)^{-1} \quad \text{and}\quad \Lambda_{2}= \biggl( \int_{0}^{1}\mathcal{M}(s)\,\mathrm{d}s \biggr)^{-1}. \end{aligned}$$

Here δ is a positive number given in $(0,\frac{1}{2})$. The functions $\mathcal{M}(s)$ and $q(t)$ are defined as in Lemma 2.5.

Now we are in a position to give and show the main results.

Theorem 3.1

Assume that the hypothesis (H) holds. If $f_{0}>\Lambda_{1}$ and $f^{\infty }< \frac{\Lambda_{2}}{2}$, then the BVP (1.1) has at least one positive solution.

Proof

The assertion will be proven by Lemma 2.6. Observe that Lemma 3.2 ensures that the operator $\Phi:\mathcal{P}\rightarrow \mathcal{P}$ is completely continuous.

We first verify that the operator Φ satisfies the condition (C2) in Lemma 2.6. Since $f_{0}>\Lambda_{1}$, there exists a positive number $r_{1}$ such that

$$ f(t,x)\geqslant \Lambda_{1}x \quad \text{for } t\in [0,1] \text{ and } 0 \leqslant x\leqslant r_{1}. $$

Thus

$$ f\bigl(t,x(t)\bigr)\geqslant \Lambda_{1}x(t) \quad \text{for } t\in [0,1] \text{ and } x\in \mathcal{P}\cap \partial \Omega_{r_{1}}. $$

Now, choose the function $\psi \equiv 1$, and obviously, ψ belongs to $\mathcal{P}\setminus \{0\}$. We next show that, for the specified ψ,

$$ x\neq \Phi x+\lambda \psi $$

for $x\in \mathcal{P}\cap \partial \Omega_{r_{1}}$ and $\lambda >0$. If such were not the case, then there exist a function $x_{0}\in \mathcal{P}\cap \partial \Omega_{r_{1}}$ and a positive number $\lambda_{0}$ such that

$$ x_{0}=\Phi x_{0}+\lambda_{0}\psi. $$

Let $\bar{x}_{0}=\min_{t\in [\delta,1-\delta ]}x_{0}(t)$. Then, by the inequality (2.10), for each t in $[\delta,1-\delta ]$,

$$\begin{aligned} x_{0}(t) =& \int_{0}^{1}\mathcal{K}(t,s)f\bigl(s,x_{0}(s) \bigr)\,\mathrm{d}s+\lambda_{0} \\ \geqslant &\Lambda_{1} q(t) \int_{\delta }^{1-\delta }\mathcal{M}(s)x_{0}(s)\, \mathrm{d}s+\lambda_{0} \\ \geq &\Lambda_{1} q(\delta) \int_{\delta }^{1-\delta }\mathcal{M}(s)\,\mathrm{d}s\cdot \bar{x}_{0}+ \lambda_{0} \\ =&\bar{x}_{0}+\lambda_{0}. \end{aligned}$$

Thus, $\bar{x}_{0}\geq \bar{x}_{0}+\lambda_{0}$. This is a contradiction. Hence the operator Φ satisfies the condition (C2) in Lemma 2.6.

We now show that the operator Φ satisfies the condition (C1) in Lemma 2.6. From the assumption $f^{\infty }< \frac{\Lambda _{2}}{2}$, it follows that there exists a positive number $\gamma_{1}$ such that

$$ f(t,x)\leqslant \frac{\Lambda_{2}}{2} x \quad \text{for } t\in [0,1] \text{ and } x\geqslant \gamma_{1}, $$

(3.6)

Now let $\gamma_{2}=\max \{f(t,x):t\in [0,1], x\in [0,\gamma_{1}]\}$. Then the inequality (3.6) yields

$$ f(t,x)\leqslant \frac{\Lambda_{2}}{2} x+\gamma_{2} \quad \text{for } t\in [0,1] \text{ and } x\geqslant 0. $$

(3.7)

Set $r_{2}=\max \{2r_{1},2\gamma_{2} \int_{0}^{1}\mathcal{M}(s)\,\mathrm{d}s \}$ and let $x\in \mathcal{P} \cap \partial \Omega_{r_{2}}$. Then Lemma 2.5 and the inequality (3.7) imply

$$\begin{aligned} \Vert \Phi x \Vert =&\max_{t\in [0,1]} \int_{0}^{1}\mathcal{K}(t,s)f\bigl(s,x(s)\bigr)\, \mathrm{d}s \\ \leqslant & \int_{0}^{1}\mathcal{M}(s) \biggl(\frac{\Lambda_{2}}{2} x(s)+\gamma_{2} \biggr)\,\mathrm{d}s \\ \leqslant &\frac{\Lambda_{2}}{2} \int_{0}^{1}\mathcal{M}(s)\,\mathrm{d}s \Vert x \Vert +\gamma_{2} \int_{0}^{1}\mathcal{M}(s)\,\mathrm{d}s \\ \leqslant & \Vert x \Vert . \end{aligned}$$

Hence the operator Φ satisfies condition (C1) in Lemma 2.6. Consequently, the operator Φ has at least one fixed point $x\in \mathcal{P}\cap (\bar{\Omega }_{2}\setminus \Omega_{1})$, and Lemma 3.3 ensures that x is one positive solution of the BVP (1.1). The proof is complete. □

Theorem 3.2

Assume that the hypothesis (H) holds. If $f^{0}<\Lambda_{2}$ and $f_{\infty }>\Lambda_{1}$, then the BVP (1.1) has at least one positive solution.

Proof

The assertion will be shown by Lemma 2.6. Note that the complete continuity of the operator Φ is guaranteed by Lemma 3.2. We only need to prove that the operator Φ satisfies the conditions (C1) and (C2) in Lemma 2.6.

Since $f^{0}<\Lambda_{2}$ and $f_{\infty }>\Lambda_{1}$, there exist two positive numbers $r_{1}$ and $\gamma_{1}$ such that

$$\begin{aligned}& f(t,u)\leqslant \Lambda_{2}x \quad \text{for } t\in [0,1] \text{ and } 0\leqslant x\leqslant r_{1}, \end{aligned}$$

(3.8)

$$\begin{aligned}& f(t,x)\geqslant \Lambda_{1} x \quad \text{for } t\in [0,1] \text{ and } x\geqslant \gamma_{1}. \end{aligned}$$

(3.9)

It follows from Lemma 2.5 and the inequality (3.8) that, for $x\in \mathcal{P}\cap \partial \Omega_{r_{1}}$,

$$ \Vert \Phi x \Vert =\max_{t\in [0,1]} \int_{0}^{1}\mathcal{K}(t,s)f\bigl(s,x(s)\bigr)\, \mathrm{d}s\leqslant \Lambda_{2} \int_{0}^{1}\mathcal{M}(s)x(s)\,\mathrm{d}s\leqslant \Vert x \Vert . $$

Thus the operator Φ satisfies the condition (C1) in Lemma 2.6.

It remains to show that the operator Φ also satisfies the condition (C2) in Lemma 2.6. To this end, let $r_{2}=\max \{2r_{1},\gamma_{1}q^{-1}(\delta) \}$. If $x\in \mathcal{P} \cap \partial \Omega_{r_{2}}$, then

$$ x(t)\geqslant q(t) \Vert x \Vert \geqslant q(\delta)r_{2} \geqslant \gamma_{1} \quad \text{for } t\in [\delta,1-\delta ], $$

and hence, by the inequality (3.9),

$$ f\bigl(t,x(t)\bigr)\geqslant \Lambda_{1} x(t) \quad \text{for } t\in [ \delta,1- \delta ] \text{ and } x\in \mathcal{P}\cap \partial \Omega_{r_{2}}. $$

Now, choose the function $\psi \equiv 1$, and clearly, ψ belongs to $\mathcal{P}\setminus \{0\}$. We then claim that

$$ x\neq \Phi x+\lambda \psi $$

for $x\in \mathcal{P}\cap \partial \Omega_{r_{2}}$ and $\lambda >0$. Indeed, if the preceding assertion is not true, then there exist a function $x_{0}\in \mathcal{P}\cap \partial \Omega_{r_{2}}$ and a positive number $\lambda_{0}$ such that

$$ x_{0}=\Phi x_{0}+\lambda_{0}\psi. $$

Let $\bar{x}_{0}=\min_{t\in [\delta,1-\delta ]}x_{0}(t)$. Then, by the inequality (2.10), for each t in $[\delta,1-\delta ]$,

$$\begin{aligned} x_{0}(t) =& \int_{0}^{1}\mathcal{K}(t,s)f\bigl(s,x_{0}(s) \bigr)\,\mathrm{d}s+\lambda_{0} \\ \geqslant &\Lambda_{1} q(t) \int_{\delta }^{1-\delta }\mathcal{M}(s)x_{0}(s)\, \mathrm{d}s+\lambda_{0} \\ \geqslant &\Lambda_{1} q(\delta) \int_{\delta }^{1-\delta }\mathcal{M}(s)\,\mathrm{d}s\cdot \bar{x}_{0}+ \lambda_{0} \\ =&\bar{x}_{0}+\lambda_{0}. \end{aligned}$$

Therefore, $\bar{x}_{0}\geq \bar{x}_{0}+\lambda_{0}$. This contradiction ensures that the operator Φ satisfies the condition (C2) in Lemma 2.6. Therefore, in the light of Lemma 2.6, we conclude that the operator Φ has at least one fixed point $x\in \mathcal{P}\cap (\bar{\Omega }_{2}\setminus \Omega_{1})$, and by Lemma 3.3, the fixed point x is one positive solution of the BVP (1.1). The proof is complete. □

Remark 3.1

The conditions in Lemma 2.6 are weaker than those in the classical norm-type expansion and compression theorem [42], and accordingly, it is generally difficult to utilize the latter to prove Theorem 3.1 and 3.2.

By Theorem 3.1 and 3.2, we directly obtain the following corollary.

Corollary 3.1

If $f_{0}=\infty$ and $f^{\infty }=0$, or if $f^{0}=0$ and $f_{\infty }=\infty$, then the BVP (1.1) has at least one positive solution.

3.1 An illustrative example

Let $\mathfrak{D}=[0,1]\times [0,\infty)$, $f(t,x)=(t+1)(2+\sin x)$, and $\lambda \in [0,2)$. Then the function f is nonnegative, and continuous on $\mathfrak{D}$. Furthermore,

$$ f_{0}=\lim_{x\rightarrow 0}\min_{t\in [0,1]} \frac{f(t,x)}{x}= \lim_{x\rightarrow 0} \biggl(\frac{2}{x}+ \frac{\sin x}{x} \biggr)=\infty $$

and

$$ f^{\infty }=\lim_{x\rightarrow \infty }\max_{t\in [0,1]} \frac{f(t,x)}{x}=\lim_{x\rightarrow \infty } \biggl(\frac{4}{x}+ \frac{2 \sin x}{x} \biggr)=0. $$

Hence, the corresponding conditions in Corollary 3.1 are satisfied for the above specified function and parameters, which implies that to the boundary value problem (1.1) there exists at least one positive solution defined on $[0,1]$.

4 Conclusion

By using the fixed point theorem in a cone, we establish some criteria for the existence of at least one positive solution to the conformable fractional differential equations with integral boundary conditions. The obtained conditions are generally weaker than those derived by using the classical norm-type expansion and compression theorem and are easy to satisfy and check. We will further investigate boundary value problems of fractional differential equations with nonsingular kernel in the future.