Theory and Modern Applications

# Fractional boundary value problems with $$p(t)$$-Laplacian operator

## Abstract

The purpose of this paper is to discuss boundary value problems of fractional differential equations with $$p(t)$$-Laplacian operator. Some new existence, uniqueness, and multiplicity results were acquired by employing some fixed point theorems. Moreover, some examples are supplied to verify our main results.

## 1 Introduction

Fractional differential equations have been applied in many research fields in recent years (see ). Leszczynski and Blaszczyk  studied the following fractional mathematical model, which can be used to describe the height of granular material decreasing over time in a silo:

$${}^{C} D_{T^{-}}^{\alpha} D_{a^{+}}^{\alpha}h^{*}(t)+ \beta h^{*}(t) =0,\quad t\in[0,T],$$

where $${}^{C} D_{T^{-}}^{\alpha}$$ and $$D_{a^{+}}^{\alpha}$$ are respectively the right Caputo and left Riemann-Liouville fractional derivatives of order $$\alpha\in(0,1)$$. Moreover, many valuable results related to boundary value problems (BVPs) or initial problems for fractional differential equations have been achieved by some scholars (see ).

Bai and Lü  considered the fractional BVP

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} D_{0^{+} }^{\alpha}x(t) = f(t,x(t)),\quad t \in(0,1),\\ x(0)=x(1)=0, \end{array}\displaystyle \right . \end{aligned}
(1.1)

and obtained the existence and multiplicity of positive solutions by taking advantage of methods in cone. Here, $$D_{0^{+}}^{\alpha}$$ is the left Riemann-Liouville fractional derivative of order $$\alpha\in (1,2]$$, and $$f: [ {0,1} ] \times\mathbb{R}^{2} \to\mathbb{R}$$ is a continuous function.

Recently, many scholars have focused on fractional BVPs with p-Laplacian operator (see ). Chen and Liu  dealt with the following BVP involving p-Laplacian operator:

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} D_{0^{+} }^{\beta}\varphi_{p} (D_{0^{+} }^{\alpha}x(t)) = f(t,x(t)),\quad t \in[0,1],\\ x(0)=-x(1),\qquad D_{0^{+} }^{\alpha}x(0)=-D_{0^{+} }^{\alpha}x(1), \end{array}\displaystyle \right . \end{aligned}
(1.2)

where $$0 <\beta, \alpha \leq1$$, $$1<\alpha+\beta\leq2$$, $$D_{0^{+} }^{\alpha}$$ is the Caputo fractional derivative, $$\varphi_{p} (\cdot)$$ is the p-Laplacian operator, and $$f: [ {0,1} ] \times\mathbb{R}^{2} \to\mathbb{R}$$ is a continuous function.

However, many papers focused on the existence and multiplicity of solutions for fractional BVPs. The uniqueness of solution for fractional BVPs with $$p(t)$$-Laplacian operator has not been yet investigated. Thus, we deal with the following fractional BVP with $$p(t)$$-Laplacian operator and obtain the uniqueness of its solution by the method in cone (see Theorem 3.2):

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} D_{0^{+} }^{\beta}\varphi_{p(t)} (D_{0^{+} }^{\alpha}x(t)) +f(t,x(t))=0,\quad t \in[0,1], \\ x'(0)=x(1)=x''(0)=0,\qquad D_{0^{+} }^{\alpha}x(0)=0, \end{array}\displaystyle \right . \end{aligned}
(1.3)

where $$D_{0^{+} }^{\alpha}$$ is the Caputo fractional derivative, $$2< \alpha<3$$, $$0<\beta<1$$, $$\varphi_{p(t)} (\cdot)$$ is the $$p(t)$$-Laplacian operator with $$p(t)\in C^{1}[0,1]$$ such that $$p(t)>1$$. Moreover, f does not need to satisfy the Lipschitz condition, so the problem becomes more complicated. An iterative scheme is shown to approximate it.

Furthermore, we also discuss the existence and multiplicity of solutions for the BVP (1.3) (see Theorem 3.1 and Theorem 3.3). Noting that when $$p(t)=p$$, it becomes the well-known p-Laplacian operator, our results extend and enrich some existing papers. For the problems of integer differential equations with $$p(t)$$-Laplacian or p-Laplacian operator, we refer the readers to ).

## 2 Preliminaries

Let $$E=C[0,1]$$ with norm $$\Vert x \Vert _{\infty}= \max_{t \in [0,1]} \vert {x(t)} \vert$$, and P be a cone of E, where $$P=\{ x\in E| x(t)\geq0\}$$. Moreover, we define the partial ordering ≤ with respect to P by $$x \leq y$$ if only if $$y-x \in P$$. For $$u_{0} \in P$$ such that $$u_{0}> \theta$$ (i.e., $$u_{0}(t)$$ is unequal to zero identically), we denote

$$P_{u_{0}}=\bigl\{ x|x \in E, \exists\lambda(x)>0, \mu(x)>0\ \mathrm{s.t.}\ \lambda (x)u_{0}\leq x \leq\mu(x) u_{0}\bigr\} .$$

### Lemma 2.1

()

For any $$(t,x)\in[0,1]\times\mathbb {R}$$, $$\varphi_{p(t)}(x)=|x|^{p(t)-2}x$$ is a homeomorphism from $$\mathbb {R}$$ to $$\mathbb{R}$$ and strictly monotone increasing for any fixed t. Moreover, $$\varphi_{p(t)}^{-1}(\cdot)$$ is continuous, sends bounded sets to bounded sets, and is defined by

\begin{aligned} \varphi_{p(t)}^{-1}(x)=|x|^{\frac{2-p(t)}{p(t)-1}}x \quad \textit{for } x\in \mathbb{R}\setminus\{0\},\qquad\varphi_{p(t)}^{-1}(0)=0\quad \textit{for } x=0. \end{aligned}

### Definition 2.1

()

Let E be a Banach space. A cone $$P\subset E$$ is called normal if there is a constant $$N >0$$ such that $$\theta\leq x \leq y$$ and $$\|x\|\leq N\|y\|$$ for all $$x, y \in E$$.

### Lemma 2.2

([21, 22])

Let $$P\subset E$$ be a normal cone. Suppose that $$A:P_{u_{0}}\to P_{u_{0}}$$ is increasing and for any $$t\in(0,1)$$, there exists $$\eta(t)>0$$ such that

$$A(tx)\geq t\bigl(1+\eta(t)\bigr)Ax,\quad x\in P_{u_{0}}.$$

Then, A has a unique fixed point $$x^{*}$$ if and only if there exist $$w_{0}, v_{0}\in P_{u_{0}}$$ such that $$w_{0}\leq Aw_{0}\leq Av_{0}\leq v_{0}$$. Moreover, for any $$x_{0}\in[w_{0},v_{0}]$$, letting $$x_{n}=Ax_{n-1}$$ ($$n=1,2,\ldots$$), we have $$x_{n}\to x^{*}$$.

Define $$P(\theta,b,d) = \{ {x \in P|b \le\theta(x) \mbox{ and } \Vert x \Vert \le d} \}$$ and $$P_{c} = \{ {x \in P|\Vert x \Vert \le c} \}$$, where $$b,c,d>0$$.

### Lemma 2.3

()

Let P be a cone of Banach space E, and $$T:P_{c}\to P_{c}$$ be a completely continuous map. Suppose that there exists a nonnegative continuous concave functional θ such that $$\theta(x)\leq\|x\|$$ for $$x\in P$$ and numbers $$0 < a < b < d \le c$$ satisfying the following conditions:

1. (i)

$$\{ {x \in P(\theta,b,d)|\theta(x) > b} \} \ne \emptyset$$ and $$\theta(Tx) > b$$ for $$x \in P(\theta,b,d)$$.

2. (ii)

$$\Vert {Tx} \Vert < a$$ for $$x\in P_{a}$$.

3. (iii)

$$\theta(Tx) > b$$ for $$x \in P(\theta,b,c)$$ with $$\Vert {Tx} \Vert > d$$.

Then T has at least three fixed points $$x_{1}$$, $$x_{2}$$, $$x_{3}$$ in $$P_{c}$$.

### Definition 2.2

()

The Riemann-Liouville fractional integral operator of order α of a function x is given by

\begin{aligned} I_{0^{+}}^{\alpha}x(t)=\frac{1}{\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha-1}x(s)\,ds, \end{aligned}

provided that the right-hand side are pointwise defined on $$[0,1]$$.

### Definition 2.3

()

The Caputo fractional derivative of order α of a function x is given by

\begin{aligned} D_{0^{+}}^{\alpha}x(t)=I_{0^{+}}^{n-\alpha} \frac{d^{n}x(t)}{d t^{n}} =\frac{1}{\Gamma(n-\alpha)} \int_{0}^{t}(t-s)^{n-\alpha-1}x^{(n)}(s)\,ds, \end{aligned}

where $$n= [\alpha]+1$$, provided that the right-hand side is pointwise defined on $$[0,1]$$.

### Lemma 2.4

()

The general solution of the Caputo fractional differential equation

\begin{aligned} D_{0^{+}}^{\alpha}x(t)=0 \end{aligned}

is given by

\begin{aligned} x(t)=c_{0}+c_{1}t+c_{2}t^{2}+\cdots+c_{n-1}t^{n-1}, \end{aligned}

where $$c_{i}\in{\mathbb{R}}$$, $$i=0,1,2,\ldots,n-1$$, $$n= [\alpha]+1$$.

### Lemma 2.5

()

Assume that $$D_{0^{+}}^{\alpha}x(t)\in C[0,1]$$. Then

\begin{aligned} I_{0^{+}}^{\alpha} D_{0^{+}}^{\alpha}x(t)=x(t)+c_{0}+c_{1}t+c_{2}t^{2}+\cdots+c_{n-1}t^{n-1}, \end{aligned}

where $$c_{i}=-\frac{x^{(i)}(0)}{i!}$$, $$i=0,1,2,\ldots,n-1$$, $$n= [\alpha]+1$$.

### Lemma 2.6

If $$y(t)\in C[0,1]$$, then the unique solution of

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} D_{0^{+} }^{\beta}\varphi_{p(t)} (D_{0^{+} }^{\alpha}x(t)) +y(t)=0,\quad t \in[0,1], \\ x'(0)=x(1)=x''(0)=0,\qquad D_{0^{+} }^{\alpha}x(0)=0, \end{array}\displaystyle \right . \end{aligned}

can be expressed as the integral

\begin{aligned} x(t)= \int^{1}_{0}G(t,s)\varphi_{p(s)}^{-1} \bigl(I_{0^{+} }^{\beta}y(s)\bigr)\,ds, \end{aligned}
(2.1)

where

\begin{aligned}& G(t,s) = \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{(1 - s)^{\alpha- 1}-(t - s)^{\alpha - 1}}{{\Gamma(\alpha)}},& 0 \le s \le t \le1,\\ \frac{ (1 - s)^{\alpha - 1}}{{\Gamma(\alpha)}}, & 0 \le t \le s \le1. \end{array}\displaystyle \right . \end{aligned}

### Proof

Based on Lemma 2.5, we have

$$\varphi_{p(t)}\bigl(D_{0^{+} }^{\alpha}x(t)\bigr) = -I_{0^{+} }^{\beta}y(t)+c,\quad c\in \mathbb{R}.$$

Applying the operator $$\varphi^{-1} _{p(t)}$$ to both sides of this equality, we have

$$D_{0^{+} }^{\alpha}x(t) =\varphi^{-1} _{p(t)} \bigl(-I_{0^{+} }^{\beta}y(t)+c\bigr).$$

Combining this with $$D_{0^{+} }^{\alpha}x(0)=0$$, for fixed $$t=0$$, we have $$\varphi^{-1} _{p(0)}(c)=0$$. By Lemma 2.1 we have that $$c=0$$ and

$$x(t) =-\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1} \varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}y(s) \bigr)\,ds+c_{1}+c_{2}t+c_{3}t^{2},$$

where $$c_{i}\in\mathbb{R}$$, $$i=1,2,3$$. Since $$x'(0)=x(1)=x''(0)=0$$, we get that

$$c_{1}=\frac{1}{\Gamma(\alpha)} \int_{0}^{1}(1-s)^{\alpha-1} \varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}y(s) \bigr)\,ds$$

and $$c_{2}=c_{3}=0$$. Thus, we have

\begin{aligned} x(t)=-\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1} \varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}y(s) \bigr)\,ds+\frac{1}{\Gamma(\alpha)} \int _{0}^{1}(1-s)^{\alpha-1} \varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}y(s) \bigr)\,ds. \end{aligned}

Therefore, (2.1) holds. The proof is complete. □

### Lemma 2.7

$$G(t,s)$$ satisfies the following conditions:

(D1):

$$G(t,s)\in C([0,1]\times[0,1])$$, $$G(t,s)\geq0$$ for $$t, s \in [0,1]$$.

(D2):

$$\frac{1}{\Gamma(\alpha)}(1-t^{\alpha-1})(1 - s)^{\alpha- 1} \leq G(t,s)\leq\frac{1}{\Gamma(\alpha)}(1-t^{\alpha-1})$$ for $$t, s \in[0,1]$$.

### Proof

Clearly, (D1) is satisfied. For $$s\leq t$$ and $$2<\alpha<3$$, we have

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

For given t, we have

\begin{aligned} \frac{\partial G(t,s)}{\partial s}=\frac{1}{\Gamma(\alpha -1)}\bigl((t-s)^{\alpha-2}-(1-s)^{\alpha-2} \bigr)\leq0. \end{aligned}

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

For $$t\leq s$$, it is easy to get that

\begin{aligned}& (1 - s)^{\alpha - 1}\geq\bigl(1-t^{\alpha-1}\bigr) (1 - s)^{\alpha - 1},\\& (1 - s)^{\alpha - 1}\leq(1 - t)^{\alpha - 1}\leq1-t \leq 1-t^{\alpha-1}. \end{aligned}

Thus, (D2) is satisfied. □

## 3 Main result

Define the nonnegative continuous concave functional θ by

$$\theta({Ax}) =\min _{t \in[\tau,1-\tau]} Ax(t), \quad\tau \in\biggl(0,\frac{1}{2} \biggr), x\in P,$$

where

$$Ax(t)= \int^{1}_{0}G(t,s)\varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f\bigl(s,x(s)\bigr)\bigr)\,ds.$$

Let

\begin{aligned}& \Lambda_{1}:=\biggl(\frac{1}{\Gamma(\alpha)} \int^{1}_{0}(1-s)^{\alpha -1} \varphi^{-1} _{p(s)}\biggl(\frac{1}{\Gamma(\beta+1)}s^{\beta} \biggr)\,ds\biggr)^{-1},\\& \Lambda_{2}:= \biggl(\frac{1-(1-\tau)^{\alpha-1}}{\Gamma(\alpha)} \int ^{1}_{0}(1 - s)^{\alpha- 1} \varphi^{-1} _{p(s)}\biggl(\frac{1}{\Gamma(\beta +1)}s^{\beta} \biggr)\,ds\biggr)^{-1}. \end{aligned}

### Theorem 3.1

Assume that $$f:[0,1]\times[0,\infty)\to [0,\infty)$$ is continuous and there exist numbers $$0< a< b< d=c$$ satisfying the following conditions:

(I1):

$$f(t,x)<\varphi_{p(t)}(\Lambda_{1} a)$$ for $$[0,1]\times[0,a]$$.

(I2):

$$f(t,x)>\varphi_{p(t)}(\Lambda_{2} b)$$ for $$[\tau,1-\tau ]\times[b,c]$$.

(I3):

$$f(t,x)\leq\varphi_{p(t)}(\Lambda_{1} c)$$ for $$[0,1]\times [0,c]$$.

Then BVP (1.3) has three positive solutions.

### Proof

To begin with, for $$u\in P_{c}$$, we will prove that $$A:P_{c}\to P_{c}$$. By $$I_{3}$$ we have

\begin{aligned} \Vert {Ax} \Vert =&\max _{t \in [ {0,1} ]} \biggl| \int^{1}_{0}G(t,s)\varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f\bigl(s,x(s)\bigr)\bigr)\,ds\biggr| \\ \leq& \frac{1}{\Gamma(\alpha)} \int^{1}_{0}(1-s)^{\alpha-1}\varphi ^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f \bigl(s,x(s)\bigr)\bigr)\,ds \\ \leq& \Lambda_{1}c \frac{1}{\Gamma(\alpha)} \int^{1}_{0}(1-s)^{\alpha -1} \varphi^{-1} _{p(s)}\biggl(\frac{1}{\Gamma(\beta+1)}s^{\beta} \biggr)\,ds=c. \end{aligned}

Thus, $$AP_{c}\subset P_{c}$$. It is easy to get that A is continuous by the continuity of f. Let Ω be any bounded open subset of $$P_{c}$$. Since $$\varphi_{p(t)}^{-1}(\cdot)$$ and f are continuous, there exists a constant $$B>0$$ such that $$|\varphi^{-1} _{p(t)}(I_{0^{+} }^{\beta}f(t,x(t)))|\leq B$$ on $$[0,1]\times\overline{\Omega}$$. Thus, we have

\begin{aligned} \Vert {Ax} \Vert =&\mathop{\max}\limits _{t \in [ {0,1} ]} |Ax|\leq\frac{1}{\Gamma(\alpha)} \int^{1}_{0}(1-s)^{\alpha-1}B\,ds= \frac {B}{\Gamma(\alpha+1)}. \end{aligned}

Thus, AΩ̅ is uniformly bounded. On the other hand, for all $$t_{1}, t_{2}\in[0,1]$$ such that $$t_{1}\leq t_{2}$$ and for all $$x\in\overline{\Omega}$$, we have

\begin{aligned} \bigl\vert {Ax(t_{2} ) - Ax(t_{1} )} \bigr\vert =&\biggl| \int^{1}_{0}G(t_{2},s)\varphi ^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f \bigl(s,x(s)\bigr)\bigr)\,ds - \int ^{1}_{0}G(t_{1},s) \varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f \bigl(s,x(s)\bigr)\bigr)\,ds\biggr| \\ \leq& B \int_{0}^{1} {\bigl\vert {G(t_{2} ,s) - G(t_{1} ,s)} \bigr\vert \,ds} \\ \leq& \frac{B}{\Gamma(\alpha+1)}\bigl(t_{2}^{\alpha}-t_{1}^{\alpha} \bigr). \end{aligned}

Thus, we have

$$\bigl\vert {Ax(t_{2} ) - Ax(t_{1} )} \bigr\vert \to0\quad \mbox{uniformly as } t_{1}\to t_{2}.$$

Therefore, A is equicontinuous on Ω̅, so that $$A:P_{c}\to P_{c}$$ is completely continuous by the Arzelà-Ascoli theorem. Similarly, by $$I_{1}$$ we obtain that $$\|Ax\|< a$$ for $$x\in P_{a}$$. So, condition (ii) of Lemma 2.3 is satisfied. Let $$x_{0}(t) = \frac{{b + c}}{2}$$. Clearly, $$\|x_{0}\|\leq c$$ and $$\theta(x_{0})>b$$. Thus,

$$\bigl\{ {x \in P(\theta,b,d)|\theta(x) > b} \bigr\} \ne\emptyset.$$

For $$x\in P(\theta,b,d)$$, by (I2) we have $$b\leq x(t)\leq c$$, $$t\in[\tau,1-\tau]$$, and

\begin{aligned} \theta({Ax}) =&\mathop{\min}\limits _{t \in[\tau,1-\tau]} Ax(t) \\ \geq& \frac{1}{\Gamma(\alpha)} \mathop{\min}\limits _{t \in[\tau ,1-\tau]} \int^{1}_{0}\bigl(1-t^{\alpha-1}\bigr) (1 - s)^{\alpha- 1}\varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f\bigl(s,x(s)\bigr)\bigr)\,ds \\ >& b\Lambda_{2} \frac{1-(1-\tau)^{\alpha-1}}{\Gamma(\alpha)} \int ^{1}_{0}(1 - s)^{\alpha- 1} \varphi^{-1} _{p(s)}\biggl(\frac{1}{\Gamma(\beta +1)}s^{\beta} \biggr)\,ds=b. \end{aligned}

Thus, condition (i) of Lemma 2.3 holds. When $$d=c$$, condition (i) implies (iii) in Lemma 2.2. Then BVP (1.3) has three positive solutions. □

### Example 3.1

Consider the following BVP:

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} D_{0^{+} }^{\frac{1}{2}} \varphi_{3} (D_{0^{+} }^{\frac{8}{3}} x(t)) +f(t,x(t))=0,\quad t \in[0,1], \\ x'(0)=x(1)=x''(0)=0,\qquad D_{0^{+} }^{\frac{8}{3}} x(0)=0. \end{array}\displaystyle \right . \end{aligned}
(3.1)

Choose $$\alpha=\frac{8}{3}$$, $$\beta=\frac{1}{2}$$, $$p(t)=3$$, $$\tau=\frac {1}{4}$$,

$$f\bigl(t,x(t)\bigr)= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} t+45x^{3}, & x\leq10,\\ t+\frac{9}{200}x^{3}+44{,}955,& x>10, \end{array}\displaystyle \right .$$

$$a=\frac{1}{4}$$, $$b=10$$, $$c=50$$. By simple calculation we have

$$\Lambda_{1}=\frac{(\Gamma(\frac{3}{2}))^{\frac{1}{2}}\Gamma(\frac {47}{12})}{\Gamma(\frac{5}{4})}\approx5.29,\qquad \Lambda_{1}= \frac{(\Gamma (\frac{3}{2}))^{\frac{1}{2}}\Gamma(\frac{47}{12})}{\Gamma(\frac {5}{4})}\cdot\frac{\Gamma(\frac{8}{3})}{(1-(\frac{3}{4}))^{\frac {5}{3}}} \approx20.89.$$

So, we have

(I1):

$$f(t,x(t))\leq1.71<1.75\approx\varphi_{3}(\Lambda_{1} a)$$ for $$[0,1]\times[0,\frac{1}{4}]$$;

(I2):

$$f(t,x(t))\geq45{,}000.25>43{,}639.21=\varphi_{3}(\Lambda_{2} b)$$ for $$[\frac{1}{4},\frac{3}{4}]\times[10,50]$$;

(I3):

$$f(t,x(t))\leq50{,}580<69{,}960.25=\varphi_{3}(\Lambda_{1} c)$$ for $$[0,1]\times[0,50]$$.

Thus, Example 3.1 has three positive solutions.

### Theorem 3.2

Assume that $$f:[0,1]\times[0,\infty)\to (0,\infty)$$ is continuous and the following conditions hold:

(I4):

$$f(t,x)$$ is increasing in x.

(I5):

There exists $$r\in(0,1)$$ such that

$$f(t,rx)\geq r^{\frac{p(t)-1}{2}}f(t,x),\quad \textit{which implies } f\biggl(t, \frac {1}{r}x\biggr)\leq r^{-\frac{p(t)-1}{2}}f(t,x).$$

Then BVP (1.3) has a unique positive solution.

### Proof

Let $$\Theta(t)=1-t^{\alpha-1}$$, $$\lambda(x)=\frac {1}{\Gamma(\alpha)}\int^{1}_{0}(1 - s)^{\alpha- 1}\varphi^{-1} _{p(s)}(I_{0^{+} }^{\beta}f(s,x(s)))\,ds$$, $$\mu(x)=\frac{1}{\Gamma(\alpha )}\int^{1}_{0}\varphi^{-1} _{p(s)} (I_{0^{+} }^{\beta}f(s,x(s)))\,ds$$. For $$u\in P_{\Theta}$$, we will prove that $$A:P_{\Theta}\to P_{\Theta}$$. Indeed, by Lemma 2.7, we have

\begin{aligned}& Ax(t)\geq\Theta(t)\frac{1}{\Gamma(\alpha)} \int^{1}_{0}(1 - s)^{\alpha - 1} \varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f \bigl(s,x(s)\bigr)\bigr)\,ds,\\& Ax(t)\leq\Theta(t)\frac{1}{\Gamma(\alpha)} \int^{1}_{0}\varphi^{-1} _{p(s)} \bigl(I_{0^{+} }^{\beta}f\bigl(s,x(s)\bigr)\bigr)\,ds. \end{aligned}

Clearly, we have $$\lambda(x)\Theta(t)\leq Ax\leq\mu(x)\Theta(t)$$. Thus, we that $$A:P_{\Theta}\to P_{\Theta}$$, together with (I4) and the monotone increasing of $$\varphi^{-1} _{p(t)}(\cdot)$$, yields that A is an increasing operator. In view of (I5), we have

\begin{aligned} A(rx) \geq& r^{\frac{1}{2}} \int^{1}_{0}G(t,s)\varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f\bigl(s,x(s)\bigr)\bigr)\,ds \\ =& r\bigl(1+r^{-\frac{1}{2}}-1\bigr) \int^{1}_{0}G(t,s)\varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f\bigl(s,x(s)\bigr)\bigr)\,ds. \end{aligned}

We have $$\eta(r):=r^{-\frac{1}{2}}-1>0$$. Let

\begin{aligned}& b(t)= \int^{1}_{0}G(t,s)\varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f\bigl(s,\Theta(s)\bigr) \bigr)\,ds, \\& b_{1}=\min\biggl\{ 1,\frac{1}{\Gamma(\alpha)} \int ^{1}_{0}(1 - s)^{\alpha- 1} \varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f \bigl(s,\Theta(s)\bigr)\bigr)\,ds\biggr\} , \\& b_{2}=\max\biggl\{ 1,\frac{1}{\Gamma(\alpha)} \int ^{1}_{0}\varphi^{-1} _{p(s)} \bigl(I_{0^{+} }^{\beta}f\bigl(s,\Theta(s)\bigr)\bigr)\,ds\biggr\} . \end{aligned}

Thus, we have $$b_{1}\Theta(t)\leq b(t)\leq b_{2}\Theta(t)$$. Let $$a_{1}=\min\{\frac{1}{2},b_{1}\}$$, $$a_{2}>1$$, $$w_{0}(t)=a_{1}b(t)$$, $$v_{0}(t)=a_{2}b(t)$$. Since A is an increasing operator, we have $$Aw_{0}\leq A v_{0}$$. On one hand,

$$f(t,w_{0})=f\bigl(t,a_{1}b(t)\bigr)\geq f \bigl(t,a_{1}b_{1}\Theta(t)\bigr)\geq (a_{1}b_{1})^{\frac{p(t)-1}{2}}f \bigl(t,\Theta(t)\bigr).$$

For $$a_{1}\leq b_{1}$$, we have $$a_{1}\leq(a_{1}b_{1})^{\frac{1}{2}}$$. Therefore,

\begin{aligned} Aw_{0}(t) =& \int^{1}_{0}G(t,s) \int^{1}_{0}\varphi^{-1} _{p(s)} \bigl(I_{0^{+} }^{\beta}f\bigl(s,w_{0}(s)\bigr)\bigr)\,ds \\ \geq& (a_{1}b_{1})^{\frac{1}{2}} \int^{1}_{0}G(t,s)\varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f\bigl(s,\Theta(s)\bigr) \bigr)\,ds \\ \geq& a_{1} \int^{1}_{0}G(t,s)\varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f\bigl(s,\Theta(s)\bigr) \bigr)\,ds \\ =&w_{0}(t). \end{aligned}

On the other hand,

$$f(t,v_{0})=f\bigl(t,a_{2}b(t)\bigr)\leq f \bigl(t,a_{2}b_{2}\Theta(t)\bigr)\leq\biggl( \frac {1}{a_{2}b_{2}}\biggr)^{\frac{p(t)-1}{2}}f\bigl(t,\Theta(t)\bigr).$$

Hence,

\begin{aligned} Av_{0}(t) =& \int^{1}_{0}G(t,s)\varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f\bigl(s,v_{0}(s) \bigr)\bigr)\,ds \\ \leq& \biggl(\frac{1}{a_{2}b_{2}}\biggr)^{\frac{1}{2}} \int^{1}_{0}G(t,s)\varphi ^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f\bigl(s,\Theta(s)\bigr) \bigr)\,ds \\ \leq& a_{2} \int^{1}_{0}G(t,s)\varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f\bigl(s,\Theta(s)\bigr) \bigr)\,ds \\ =&v_{0}(t). \end{aligned}

Thus, we have $$w_{0}\leq Aw_{0}\leq Av_{0}\leq v_{0}$$. In view of Lemma 2.2, BVP (1.3) has a unique positive solution. Moreover, for any $$u_{0}\in[w_{0},v_{0}]$$, letting $$u_{n}=Au_{n-1}$$ ($$n=1,2,\ldots$$), we have $$u_{n}\to u^{*}$$. □

### Example 3.2

Consider the following BVP:

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} D_{0^{+} }^{\frac{4}{5}} \varphi_{4t^{2}+9} (D_{0^{+} }^{\frac{12}{5}} x(t)) +x^{2}(t)\sin t=0,\quad t \in[0,1], \\ x'(0)=x(1)=x''(0)=0,\qquad D_{0^{+} }^{\frac{12}{5}} x(0)=0, \end{array}\displaystyle \right . \end{aligned}
(3.2)

where $$\alpha=\frac{12}{5}$$, $$\beta=\frac{4}{5}$$, $$p(t)=4t^{2}+9$$, $$r=\frac{1}{2}$$, $$f(t,x(t))=x^{2}(t)\sin t$$. It is easy to verify that (I1) is satisfied. Moreover,

$$\biggl(\frac{1}{2}x\biggr)^{2}\sin t= \frac{1}{4}x^{2}\sin t\geq\biggl(\frac {1}{4} \biggr)^{t^{2}+2}x^{2}\sin t ,\qquad (2x)^{2} \sin t=4x^{2}\sin t\leq4^{t^{2}+2}x^{2} \sin t,$$

which yields that (I2) is satisfied. Thus, Example 3.2 has a unique positive solution.

In order to state Theorem 3.3, let $$P_{L}:=\min_{t\in[0,1]}p(t)$$, $$P_{M}:=\max_{t\in[0,1]}p(t)$$.

### Theorem 3.3

Assume that $$f:[0,1]\times\mathbb{R}\to \mathbb{R}$$ is continuous and the following conditions hold.

(I6):

There exist constants $$l_{1},l_{2}>0$$ such that

$$\bigl|f(t,x)\bigr|\leq l_{1}+l_{2}|x|^{r-1},\quad 1< r\leq P_{L}.$$
(I7):

$$\frac{K 2^{\frac{1}{P_{L}-1}}}{\Gamma(\alpha)(\Gamma(\beta +1))^{\frac{1}{P_{M}-1}}}<1$$, $$K=\max\{l_{2}^{\frac {1}{P_{L}-1}},l_{2}^{\frac{1}{P_{M}-1}}\}$$.

Then BVP (1.3) has a solution.

### Proof

In the same way as in the proof of Theorem 3.2, it is easy to prove that $$A:C[0,1]\to C[0,1]$$ is completely continuous. Denote

$$V=\bigl\{ x\in X|x=\lambda Ax, \lambda\in(0,1)\bigr\} .$$

According to Schaefer’s fixed point theorem, we just need to prove that V is bounded. For $$x\in V$$, we have

\begin{aligned} \bigl|I_{0^{+} }^{\beta}f\bigl(t,x(t)\bigr)\bigr| \leq&\frac{1}{\Gamma(\beta)} \int^{t}_{0}(t-s)^{\beta-1}\bigl|f\bigl(s, x(s) \bigr)\bigr|\,ds \\ \leq&\frac{1}{\Gamma(\beta)} \int^{1}_{0}(t-s)^{\beta -1} \bigl(l_{1}+l_{2}\bigl|x(s)\bigr|^{r-1}\bigr) \\ \leq&\frac{1}{\Gamma(\beta+1)}\bigl(l_{1}+l_{2}\|x \|_{\infty}^{r-1}\bigr). \end{aligned}

By the inequality $$(x+y)^{p}\leq2^{p}(x^{p}+y^{p})$$ for $$x,y,p>0$$ we have

\begin{aligned} \bigl|x(t)\bigr| =&\lambda\bigl|Ax(t)\bigr| \\ \leq& \int^{1}_{0}G(t,s)\varphi^{-1} _{p(s)}\bigl(I_{0^{+} }^{\beta}f\bigl(s,x(s)\bigr)\bigr)\,ds \\ \leq& \frac{1}{\Gamma(\alpha)} \int^{1}_{0}(1-s)^{\alpha-1} \biggl( \frac {2^{\frac{1}{p(s)-1}}}{(\Gamma(\beta+1))^{\frac {1}{p(s)-1}}}\bigl(l_{1}^{\frac{1}{p(s)-1}}+l_{2}^{\frac{1}{p(s)-1}} \|x\| _{\infty}^{\frac{r-1}{p(s)-1}}\bigr)\biggr)\,ds. \end{aligned}

Since $$\frac{r-1}{p(t)-1}\in(0,1]$$, by the inequality $$x^{\alpha}\leq x+1$$ for $$x>0$$ and $$\alpha\in(0,1]$$, we have

\begin{aligned} \|x\|_{\infty} \leq&\frac{2^{\frac{1}{P_{L}-1}}}{\Gamma(\alpha)(\Gamma(\beta +1))^{\frac{1}{P_{M}-1}}} \int^{1}_{0}(1-s)^{\alpha-1} \bigl(l_{1}^{\frac {1}{p(s)-1}}+l_{2}^{\frac{1}{p(s)-1}}\bigl(\|x \|_{\infty}+1\bigr)\bigr)\,ds. \end{aligned}

By (I7) there exists a constant $$\mu>0$$ such that $$\|x\|_{\infty }\leq\mu$$. Thus, the operator has a fixed point, which implies that BVP (1.3) has a solution. □

### Example 3.3

Consider the following BVP:

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} D_{0^{+} }^{\frac{3}{4}} \varphi_{t^{2}+2} (D_{0^{+} }^{\frac{5}{2}} x(t)) +1+\frac{1}{4}x^{2}(t)=0,\quad t \in[0,1], \\ x'(0)=x(1)=x''(0)=0,\qquad D_{0^{+} }^{\frac{5}{2}} x(0)=0, \end{array}\displaystyle \right . \end{aligned}
(3.3)

where $$\alpha=\frac{5}{2}$$, $$\beta=\frac{3}{4}$$, $$p(t)=t^{2}+2$$, $$r=2$$, $$f(t,x(t))=1+\frac{1}{4}x^{2}(t)$$, $$l_{1}=1$$, $$l_{2}=\frac{1}{4}$$. Clearly, (I6) holds. Moreover,

$$\frac{1}{\Gamma(\frac{5}{2})(\Gamma(\frac{7}{4}))^{\frac{1}{2}}}< 1,$$

which implies that (I7) is satisfied. Thus, BVP (3.3) has a solution.

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## Acknowledgements

The authors really appreciate the referee’s valuable suggestions and comments, which improved the former version of this paper. This research is supported by the National Natural Science Foundation of China (No. 11271364).

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Correspondence to Wenbin Liu.

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