Positive solutions to p-Laplacian fractional differential equations with infinite-point boundary value conditions

Abstract

This paper is devoted to an investigation of the multiple positive solutions to a class of infinite-point boundary value problems of nonlinear fractional differential equations coupled with the p-Laplacian operators and infinite-point boundary value conditions. By means of the properties of Green’s function and fixed point theorems, we establish the suitable criteria to guarantee the existence of positive solutions. Finally, an example is given in order to illustrate the main results.

1 Introduction

With the advance of technology, researchers are not satisfied with the limitations of the integer order calculus anymore; therefore, fractional calculus—the expansion and extension of the integer one—is brought into the public. There is no doubt the study on nonlinear fractional differential equations has become an issue of focus and is widely being used in many fields, such as physics [1, 2], biology [3], energy [4], chemical engineering [5], pharmacokinetics [6].

Based on practical application, researchers established lots of fractional differential equation models, so how to get the solutions to the equations is a big challenge to them. As is well known, it is difficult to give the exact solution, so giving the properties of the solutions is a significant point.

A great number of researchers are focused on the existence of solutions for linear or nonlinear FODE with different boundary conditions. Take Zhang [7] for example, one studied the existence of minimal non-negative solution for a class of nonlinear fractional integro-differential equations by applying the cone theory and the monotone iterative technique:

$$ \textstyle\begin{cases} D_{0^{+}}^{\alpha}x(t)+f(t,x(t),Tx(t),Sx(t))=0,\quad t\in[0,\infty),\\ x(0)=0,\qquad D_{0^{+}}^{\alpha-1} x(\infty)=x^{*}, \end{cases} $$

where

$$ (Tx) (t)= \int_{0}^{t}k(t,s)x(s)\,ds,\qquad (Sx) (t)= \int_{0}^{\infty}h(t,s)x(s)\,ds, $$

and $1<\alpha\le2, D_{0^{+}}^{\alpha}$ is the Riemann–Liouville derivative, $f, k, h$ are continuous functions.

Wang [8] considered the iterative positive solutions for a class of nonlocal fractional differential equations with nonlocal Hadamard integral and discrete boundary conditions by the monotone iterative method:

$$ \textstyle\begin{cases} ^{H}D^{\alpha}x(t)+\delta(t)f(t,x(t))=0,\quad t\in(1,\infty),\\ x(1)=x'(1)=0,\\ ^{H}D^{\alpha-1} x(\infty)=a^{H}I^{\beta}x(\xi)+b\sum_{i=1}^{m-2}c_{i}x(\eta_{i}), \end{cases} $$

where $2<\alpha\le3, a, b$ are real constants, $c_{i}\ (i=1, 2, \dots ,,m-2)$ are positive real constants, $1<\xi<\eta_{1}<\eta_{2}<\cdots<\eta _{m-2}<+\infty$, and $^{H}D^{\alpha}$ is the Hadamard fractional derivative.

As for Guo [9], they investigated the existence of solutions for the following infinite-point BVP with Caputo fractional derivative by Avery–Peterson’s fixed point theorem:

$$ \textstyle\begin{cases} ^{c}D^{\alpha}x(t)+f(t,x(t),x'(t))=0,\quad 0< t< 1,\\ x(0)=x^{''}(0)=0, \qquad x'(1)=\sum_{j=1}^{\infty}\zeta_{j}x(\xi_{j}), \end{cases} $$

where $2<\alpha\le3, \zeta_{j}\ge0, 0<\xi_{1}<\xi_{2}< \cdots<\xi_{i}< \cdots<1$, and $\sum_{i=1}^{\infty}\zeta_{i}\xi_{i}<1$.

Recently, some scientists have become interested in the study of the fractional differential equations with p-Laplacian operators [10,11,12,13]. For example, Xiping Liu [12, 13] obtained some new results on the existence of positive solutions for the four-point BVP with mixed fractional derivatives and p-Laplacian operator by a new method of lower and upper solutions which is based on the monotone iterative technique:

$$ \textstyle\begin{cases} D_{0^{+}}^{\beta}(\phi_{p}(^{C}D_{0^{+}}^{\alpha}x(t)))=f(t,x(t),^{C}D_{0^{+}}^{\alpha}x(t)),\quad t\in(0,1),\\ ^{c}D_{0^{+}}^{\alpha}x(0)=x'(0)=0,\\ x(1)=r_{1}x(\eta),\qquad ^{C}D_{0^{+}}^{\alpha}x(1)=r_{2}^{C}D_{0^{+}}^{\alpha}x(\xi), \end{cases} $$

where $1<\alpha, \beta\le2, r_{1}, r_{2}\ge0, f\in C([0,1]\times[0,+\infty )\times(-\infty,0], [0,+\infty))$, ${}^{c}D_{0^{+}}^{\alpha}$ is the Caputo derivative, $D_{0^{+}}^{\alpha}$ is the Riemann–Liouville derivative, $\phi _{p}$ is the p-Laplacian operator which is defined as $\phi_{p}(s)=\lvert s\lvert^{p-2}s (p>1),\phi_{q}$ is the inverse function of $\phi_{p}$ and $\frac{1}{p}+\frac{1}{q}=1$.

In this paper, we consider a class of nonlinear fractional differential equations coupled with the p-Laplacian operator and infinite-point boundary value conditions:

$$ \textstyle\begin{cases} -D_{0^{+}}^{\beta}(\phi_{p}(D_{0^{+}}^{\alpha}x(t)-\lambda x(t)))=f(t,x(t)),\quad t\in(0,1],\\ \lim_{t \to0^{+}}t^{1-\alpha}x(t)=\sum_{i=1}^{\infty}\mu_{i}x(\xi_{i}),\\ \lim_{t \to0^{+}}t^{2-\beta}(\phi_{p}(D_{0^{+}}^{\alpha}x(t)-\lambda x(t)))=\phi_{p}(D_{0^{+}}^{\alpha}x(1)-\lambda x(1))=0, \end{cases} $$

(1)

where $0<\alpha\le1, 1<\beta\le2, \lambda<0, \mu_{i} \ge0, 0<\xi _{i}<1, i\in\mathbb{R^{+}}, \sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}<1, f\in C ([0,1]\times[0,+\infty), [0,+\infty) ), \phi_{p}$ is the p-Laplacian operator and $D^{\alpha}, D^{\beta}$ are the Riemann–Liouville fractional derivatives.

To the best of our knowledge, few researchers studied fractional differential equations with p-Laplacian by the Riemann–Liouville derivative; this condition coupled with values at infinite number of points is not covered in the previous situations, and it is more general for the fractional differential equation models. By means of the properties of Green’s function, Krasnosel’skii fixed point theorem, the corollary of Krasnosel’skii fixed point theorem and Avery–Peterson fixed point theorem, we establish the suitable criteria to warrant the existence of the positive solutions for the BVP (1).

The rest of the paper is organized as follows. In Sect. 2, we list several basic definitions and lemmas to prepare for the further study in this paper. On top of that, the main result is showed in Sect. 3. In Sect. 4, we give an example to illustrate our main results.

2 Preliminaries

As is well known, $C[a,b]$ is the Banach space of all continuous real functions defined on $[a,b]$ with the norm $\lVert x\lVert=\sup_{t\in [a,b]}\lvert x(t)\lvert$ and $L^{1}[a,b]$ is for the Lebesgue integrable real functions defined on $[a,b]$. In this paper, we consider the following space:

$$ C_{1-\alpha}[0,1]= \bigl\{ x(t):t^{1-\alpha}x(t)\in C[0,1], 0< \alpha\le1 \bigr\} , $$

endowed with its norm $\lVert x\lVert_{1-\alpha}=\sup_{t\in [0,1]}t^{1-\alpha}\lvert x(t)\lvert$. Obviously, $C_{1-\alpha}[0,1]$ is also a Banach space; moreover, $C_{1-\alpha}[0,1]\subset L^{1}[0,1]$.

Definition 2.1

([14])

Let $\alpha>0$, the Riemann–Liouville fractional integral of the function f is defined by

$$ I_{0^{+}}^{\alpha}f(t)=\frac{1}{\varGamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}f(s)\,ds. $$

Definition 2.2

([14])

Let $\alpha>0$, the Riemann–Liouville fractional derivative of the function f is defined by

$$ D_{0^{+}}^{\alpha}f(t)=\frac{1}{\varGamma(n-\alpha)} \biggl( \frac{d}{dt} \biggr)^{n} \int_{0}^{t}(t-s)^{n-\alpha-1}f(s)\,ds, $$

where n is the smallest integer greater than or equal to α.

Definition 2.3

([15])

The two-parameter Mittag–Leffler function is defined by

$$ E_{\alpha,\beta}(\zeta)=\sum_{k=0}^{\infty}\frac{\zeta^{k}}{\varGamma(\alpha k+\beta)},\quad \alpha, \beta>0, \zeta\in\mathbb{R}. $$

Definition 2.4

([16])

If the map T satisfies

$$ T \bigl(ax+(1-a)y \bigr)\le(\ge)aT(x)+(1-a)T(y), \quad x,y\in P, a\in[0,1], $$

then T is called a non-negative convex (concave) operator on a cone P.

Lemma 2.1

([17])

Assume $\alpha>0, u \in C(0,1) \cap L(0,1)$, then

$$ I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}f(t)=u(t)+c_{1}t^{\alpha-1}+c_{2}t^{\alpha -2}+ \cdots+c_{n}t^{\alpha-n}, $$

where $c_{i}\in\mathbb {R}, i=1,2,\dots,n$ with n is the smallest integer greater than or equal to α.

Lemma 2.2

([18])

Assume $\alpha\in(0,1], \beta>0, \lambda$ is a negative constant, then the following conclusions hold:

• (i) $\lim_{t \to0^{+}}E_{\alpha,\beta}(\lambda t^{\alpha})=\frac {1}{\varGamma(\beta)}$,

• (ii) $E_{\alpha,\alpha}(\lambda t^{\alpha})>0$ for $\forall t\in [0,+\infty)$,

• (iii) $E_{\alpha,\alpha}(\lambda t^{\alpha})$ is decreasing in t for $\forall t\in(0,+\infty)$,

• (iv) $E_{\alpha,\alpha}(\lambda t^{\alpha})\le E_{\alpha,\alpha +1}(\lambda t^{\alpha})$ for $\forall t\in(0,+\infty)$.

Lemma 2.3

Assume $\sigma(t) \in C(0,1], 0<\alpha\le1$, then the problem

$$ \textstyle\begin{cases} D_{0+}^{\alpha}x(t)-\lambda x(t)=\sigma(t),\quad t\in(0,1],\\ \lim_{t \to0^{+}}t^{1-\alpha}x(t)=\sum_{i=1}^{\infty}\mu_{i}x(\xi_{i}), \end{cases} $$

(2)

has an unique solution

$$ x(t)= \int_{0}^{1} G(t,s)\sigma(s)\,ds, $$

(3)

where

$$ G(t,s)=\textstyle\begin{cases} \frac{\varGamma(\alpha)E_{\alpha,\alpha}(\lambda t^{\alpha})\sum_{s \le\xi _{i}}\mu_{i}(\xi_{i}-s)^{\alpha-1}E_{\alpha,\alpha}(\lambda(\xi_{i}-s)^{\alpha})}{t^{1-\alpha}(1-\varGamma(\alpha)\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha -1}E_{\alpha,\alpha}(\lambda\xi_{i}^{\alpha}))} +\frac{E_{\alpha,\alpha}(\lambda(t-\alpha)^{\alpha})}{(t-s)^{1-\alpha}},\\ \quad 0\le s< t \le1,\\ \frac{\varGamma(\alpha)E_{\alpha,\alpha}(\lambda t^{\alpha})\sum_{s \le\xi _{i}}\mu_{i}(\xi_{i}-s)^{\alpha-1}E_{\alpha,\alpha}(\lambda(\xi_{i}-s)^{\alpha})}{t^{1-\alpha}(1-\varGamma(\alpha)\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha -1}E_{\alpha,\alpha}(\lambda\xi_{i}^{\alpha}))},\\ \quad 0< t\le s\le1. \end{cases} $$

(4)

Proof

According to ([19]), if $\lim_{t \to0^{+}}t^{(1-\alpha )}x(t)=c_{0}$, we can easily get

$$ x(t)=c_{0}\varGamma(\alpha)t^{\alpha-1}E_{\alpha,\alpha} \bigl( \lambda t^{\alpha}\bigr)+ \int_{0}^{t}(t-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(\lambda(t-\alpha)^{\alpha}\bigr)\sigma(s)\,ds. $$

In particular, if we choose $c_{0}=\sum_{i=1}^{\infty}\mu_{i}x(\xi_{i})$, by simple calculation,

$$ c_{0}=\frac{\sum_{i=1}^{\infty}\mu_{i}\int_{0}^{\xi_{i}}(\xi_{i}-s)^{\alpha -1}E_{\alpha,\alpha}(\lambda(\xi_{i}-s)^{\alpha})\sigma(s)\,ds}{1-\varGamma (\alpha)\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}E_{\alpha,\alpha }(\lambda\xi_{i}^{\alpha})}. $$

Hence,

$$\begin{aligned} x(t)={}&\frac{\varGamma(\alpha)t^{\alpha-1}}{1-\varGamma(\alpha)\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}E_{\alpha,\alpha}(\lambda\xi_{i}^{\alpha})} \\ &{}\times\sum _{i=1}^{\infty}\mu_{i}E_{\alpha,\alpha} \bigl( \lambda t^{\alpha}\bigr) \int_{0}^{\xi _{i}}(\xi_{i}-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(\lambda(\xi_{i}-s)^{\alpha}\bigr)\sigma (s)\,ds \\ &{}+ \int_{0}^{t}(t-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(\lambda(t-\alpha)^{\alpha}\bigr)\sigma(s)\,ds \\ ={}& \int_{0}^{1}\frac{\varGamma(\alpha)E_{\alpha,\alpha}(\lambda t^{\alpha})\sum_{s \le\xi_{i}}\mu_{i}(\xi_{i}-s)^{\alpha-1}E_{\alpha,\alpha}(\lambda(\xi _{i}-s)^{\alpha})}{t^{1-\alpha}(1-\varGamma(\alpha)\sum_{i=1}^{\infty}\mu_{i}\xi _{i}^{\alpha-1}E_{\alpha,\alpha}(\lambda\xi_{i}^{\alpha}))}\sigma(s)\,ds \\ &{}+ \int _{0}^{t}\frac{E_{\alpha,\alpha}(\lambda(t-\alpha)^{\alpha})}{(t-s)^{1-\alpha }}\sigma(s)\,ds \\ ={}& \int_{0}^{1}G(t,s)\sigma(s)\,ds, \end{aligned}$$

(5)

where $G(t,s)$ is given by (4).

The proof is finished. □

Lemma 2.4

Assume $\alpha\in(0,1], \lambda$ is a negative constant, then

• (i) $G(t,s)>0$ for $t\in(0,1], s\in[0,1)$,

• (ii) $t^{1-\alpha}G(t,s)\ge\frac{\varGamma(\alpha)(E_{\alpha,\alpha }(\lambda))^{2}\sum_{s \le\xi_{i}}\mu_{i}\xi_{i}^{\alpha-1}}{ 1-\varGamma(\alpha)\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}E_{\alpha,\alpha }(\lambda)}\triangleq m$ for $t,s\in[0,1]$,

• (iii) $\min_{t\in[0,1]}t^{1-\alpha}G(t,s)\ge\rho\max_{t\in [0,1]}t^{1-\alpha}G(t,s)$ for $\rho\in(0,1)$,

• (iv) $t^{1-\alpha}\int_{0}^{1}G(t,s)\,ds\le(\varGamma(\alpha)(1-\sum_{i=0}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}))^{-1}\triangleq n$ for $t,s\in[0,1]$.

Proof

(i) By simple calculation,

$$ {1-\varGamma(\alpha)\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}E_{\alpha ,\alpha} \bigl(\lambda \xi_{i}^{\alpha}\bigr)}\ge1-\sum_{i=1}^{\infty}\mu_{i}\xi _{i}^{\alpha-1}>0. $$

In addition, according to Lemma 2.2, we can easily see that (i) holds.

(ii) Since $(t-s)^{\alpha-1}E_{\alpha,\alpha}(\lambda(t-s)^{\alpha})\ge0$ for all $0\le s< t\le1$,

$$\begin{aligned} t^{1-\alpha}G(t,s)&\ge\frac{\varGamma(\alpha)E_{\alpha,\alpha}(\lambda t^{\alpha})\sum_{s \le\xi_{i}}\mu_{i}(\xi_{i}-s)^{\alpha-1}E_{\alpha,\alpha }(\lambda(\xi_{i}-s)^{\alpha})}{1-\varGamma(\alpha)\sum_{i=1}^{\infty}\mu_{i}\xi _{i}^{\alpha-1}E_{\alpha,\alpha}(\lambda\xi_{i}^{\alpha})} \\ &\ge\frac{\varGamma(\alpha)(E_{\alpha,\alpha}(\lambda))^{2}\sum_{s \le\xi _{i}}\mu_{i}\xi_{i}^{\alpha-1}}{1-\varGamma(\alpha)E_{\alpha,\alpha}(\lambda )\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}} \\ &\triangleq m>0. \end{aligned}$$

(iii) As $t^{1-\alpha}G(t,s)$ is increasing in s and decreasing in t

$$ \max_{t\in[0,1]}t^{1-\alpha}G(t,s) = \frac{\sum_{s \le\xi_{i}}\mu_{i}(\xi_{i}-s)^{\alpha-1}E_{\alpha,\alpha }(\lambda(\xi_{i}-s)^{\alpha})}{1-\varGamma(\alpha)\sum_{i=1}^{\infty}\mu_{i}\xi _{i}^{\alpha-1}E_{\alpha,\alpha}(\lambda\xi_{i}^{\alpha})}. $$

According to Lemma 2.4(i), we have

$$\begin{aligned} \min_{t\in[0,1]}t^{1-\alpha}G(t,s)& \ge\varGamma(\alpha)E_{\alpha,\alpha }(\lambda)\max_{t\in[0,1]}t^{1-\alpha}G(t,s) \\ &\triangleq\rho \max_{t\in[0,1]}t^{1-\alpha}G(t,s), \end{aligned}$$

where $\rho=\varGamma(\alpha)E_{\alpha,\alpha}(\lambda)$, so $0<\rho<1$.

(iv) We divide this proof into two parts:

Part 1: For $0\le s \le t \le1$,

$$\begin{aligned} t^{1-\alpha} \int_{0}^{1}G(t,s)\,ds={}& \int_{0}^{1}\frac{\varGamma(\alpha)E_{\alpha ,\alpha}(\lambda t^{\alpha})\sum_{s \le\xi_{i}}\mu_{i}(\xi_{i}-s)^{\alpha -1}E_{\alpha,\alpha}(\lambda(\xi_{i}-s)^{\alpha})}{1-\varGamma(\alpha)\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}E_{\alpha,\alpha}(\lambda\xi_{i}^{\alpha})}\,ds \\ &{}+ \int_{0}^{t}t^{1-\alpha}(t-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(\lambda (t-\alpha)^{\alpha}\bigr)\,ds \\ \le{}&\frac{\varGamma(\alpha)E_{\alpha,\alpha}(\lambda t^{\alpha})\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}}{1-\varGamma(\alpha)\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}E_{\alpha,\alpha}(\lambda\xi_{i}^{\alpha})}+tE_{\alpha ,\alpha+1} \bigl(\lambda t^{\alpha}\bigr) \\ \le{}&\frac{\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}}{\varGamma(\alpha )(1-\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1})}+\frac{1}{\varGamma(\alpha )} \\ ={}&\frac{1}{\varGamma(\alpha)(1-\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha -1})}\triangleq n. \end{aligned}$$

Part 2: For $0\le t \le s \le1$,

$$\begin{aligned} t^{1-\alpha} \int_{0}^{1}G(t,s)\,ds={}& \int_{0}^{1}\frac{\varGamma(\alpha)E_{\alpha ,\alpha}(\lambda t^{\alpha})\sum_{s \le\xi_{i}}\mu_{i}(\xi_{i}-s)^{\alpha -1}E_{\alpha,\alpha}(\lambda(\xi_{i}-s)^{\alpha})}{1-\varGamma(\alpha)\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}E_{\alpha,\alpha}(\lambda\xi_{i}^{\alpha})}\,ds \\ \le{}&\frac{\varGamma(\alpha)E_{\alpha,\alpha}(\lambda t^{\alpha})\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}}{1-\varGamma(\alpha)\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}E_{\alpha,\alpha}(\lambda\xi_{i}^{\alpha})} \\ \le{}&\frac{\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1}}{\varGamma(\alpha )(1-\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha-1})} \\ < {}&\frac{1}{\varGamma(\alpha)(1-\sum_{i=1}^{\infty}\mu_{i}\xi_{i}^{\alpha -1})}\triangleq n. \end{aligned}$$

The proof is completed. □

Lemma 2.5

Assume $h(t) \in C(0,1], 1<\beta\le2$, then the problem

$$ \textstyle\begin{cases} -D_{0+}^{\beta}u(t)=h(t),\quad t\in(0,1],\\ \lim_{t \to0^{+}}t^{2-\beta}u(t)=u(1)=0, \end{cases} $$

(6)

has an unique solution

$$ u(t)= \int_{0}^{1} H(t,s)h(s)\,ds, $$

(7)

where

$$ H(t,s)=\frac{1}{\varGamma(\beta)}\textstyle\begin{cases} t^{\beta-1}(1-s)^{\beta-1}-(t-s)^{\beta-1}, &0\le s< t \le1,\\ t^{\beta-1}(1-s)^{\beta-1},& 0< t\le s\le1. \end{cases} $$

(8)

Proof

According to Lemma 2.1, we can easily get

$$ u(t)=-I_{0^{+}}^{\beta}h(t)+c_{1}t^{\beta-1}+c_{2}t^{\beta-2}, $$

by the condition (6), we obtain

$$ c_{1}=I_{0^{+}}^{\beta}h(1),\qquad c_{2}=0. $$

Hence,

$$\begin{aligned} u(t)={}&\frac{t^{\beta-1}}{\varGamma(\beta)} \int_{0}^{1}(1-s)^{\beta -1}h(s)\,ds- \frac{1}{\varGamma(\beta)} \int_{0}^{t}(t-s)^{\beta-1}h(s)\,ds = \int_{0}^{1}H(t,s)h(s)\,ds, \end{aligned}$$

where $H(t,s)$ is given by (8).

The proof is finished. □

Lemma 2.6

([20])

Assume $\beta\in(1,2]$, then

• (i) $H(t,s)>0$ for $t\in(0,1], s\in[0,1)$,

• (ii) $H(t,s)=H(1-s,1-t)$ for $t,s\in(0,1)$,

• (iii) $t^{\beta-1}(1-t)s(1-s)^{\beta-1}\le\varGamma(\beta)H(t,s)\le (\beta-1)s(1-s)^{\beta-1}$ for $t\in(0,1], s\in[0,1)$,

• (iv) $t^{\beta-1}(1-t)s(1-s)^{\beta-1}\le\varGamma(\beta)H(t,s)\le(\beta -1)(1-t)t^{\beta-1}$ for $t\in(0,1], s\in[0,1)$.

Lemma 2.7

Suppose $h(t)=f(t,x(t)), \sigma(t)=\phi_{q}(u(t))$ hold, then we obtain the unique solution of BVP (1):

$$ x(t)= \int_{0}^{1}G(t,s)\phi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds. $$

(9)

3 Main results

In this section, we establish the related condition to warrant the existence of the multiple positive solutions for BVP (1) by means of the properties of Green’s function we proved in the last section and the well-known Krasnosel’skii fixed point theorem, the corollary of Krasnosel’skii fixed point theorem and the Avery–Peterson fixed point theorem.

Let $E=\{x(t): x(t)\in C_{1-\alpha}[0,1]\}$. Define $P\subset E$,

$$ P= \Bigl\{ x(t)\in E: x(t)\ge0, \min_{t\in[0,1]}t^{1-\alpha}x(t) \ge\rho\max_{t\in[0,1]}t^{1-\alpha}x(t) \Bigr\} , $$

where ρ is defined in Lemma 2.4.

Suppose $T:P\to E$ is the operator defined by

$$ Tx(t):= \int_{0}^{1}G(t,s)\phi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds. $$

Throughout this section, suppose the following assumption holds:

$(\mathrm{H}_{0})$ $\lvert f(t,x) \lvert \le M$.

Lemma 3.1

$T: P\to P$ is a completely continuous operator.

Proof

Step 1: We show that $T(P)\subseteq P$.

For $\forall x\in P$, since the continuity of $G(t,s), H(t,s), f(t,x(t))$, it is easy to get $t^{1-\alpha}Tx(t)\in C[0,1]$. According to Lemma 2.4 and Lemma 2.6, we obtain $G(t,s), H(t,s)\ge0$, thanks to $f(t,x(t))$ being non-negative, therefore, $Tx(t)\ge0$. Furthermore,

$$\begin{aligned} \min_{t\in[0,1]}t^{1-\alpha}Tx(t)&= \min_{t\in[0,1]}t^{1-\alpha} \int _{0}^{1}G(t,s)\phi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds \\ &= \int_{0}^{1}\min_{t\in[0,1]}t^{1-\alpha}G(t,s) \phi_{q} \biggl( \int_{0}^{1}H(s,\tau )f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds \\ &\ge \int_{0}^{1}\rho\max_{t\in[0,1]}t^{1-\alpha}G(t,s) \phi_{q} \biggl( \int _{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds \\ &=\rho\max_{t\in[0,1]}Tx(t). \end{aligned}$$

Hence, $T(P)\subseteq P$.

Step 2: We will prove that T is uniformly bounded. We have

$$\begin{aligned} \bigl\lVert Tx(t)\bigr\lVert _{1-\alpha}&=\biggl\lvert t^{1-\alpha} \int_{0}^{1}G(t,s)\phi _{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds \biggr\lvert \\ &\le \int_{0}^{1} \bigl\lvert t^{1-\alpha}G(t,s)\bigr\lvert \phi_{q} \biggl( \int_{0}^{1}\bigl\lvert H(s,\tau)\bigr\lvert \bigl\lvert f \bigl( \tau,x(\tau) \bigr)\bigr\lvert \,d\tau \biggr)\,ds \\ &< n \phi_{q} \biggl(\frac{(\beta-1)M}{\varGamma(\beta)} \biggr) \\ &= n \phi_{q} \biggl(\frac{M}{\varGamma(\beta-1)} \biggr)\triangleq M_{1}. \end{aligned}$$

Therefore, $\lVert Tx(t)\lVert_{1-\alpha}< M_{1}$, which means T is uniformly bounded.

Step 3: We will give the proof of the equicontinuity of T.

Obviously, $t^{1-\alpha}G(t,s)$ is uniformly continuous in the interval $[0,1]\times[0,1]$. Therefore, for all $\varepsilon>0$, there exists $\delta>0$, without loss of generality, we choose $t_{1}, t_{2}\in[0,1]$, when $\lvert t_{1}-t_{2} \lvert<\delta$, we have

$$ \bigl\lvert G(t_{1},s)-G(t_{2},s) \bigr\lvert < \frac{\varepsilon}{\phi_{q}(M)\int _{0}^{1}\phi_{q} (\int_{0}^{1}H(s,\tau)\,d\tau )\,ds}. $$

Thus,

$$\begin{aligned} &\bigl\lVert Tx(t_{1})-Tx(t_{2}) \bigr\lVert _{1-\alpha}\\ &\quad = \biggl\lvert t_{1}^{1-\alpha} \int _{0}^{1}G(t_{1},s)\phi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds \\ &\qquad{}-t_{2}^{1-\alpha} \int_{0}^{1}G(t_{2},s)\phi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau ,x(\tau) \bigr)\,d \tau \biggr)\,ds \biggr\lvert \\ &\quad \le \int_{0}^{1} \bigl\lvert t_{1}^{1-\alpha}G(t_{1},s)-t_{2}^{1-\alpha }G(t_{2},s) \bigr\lvert \phi_{q} \biggl( \int_{0}^{1}\bigl\lvert H(s,\tau)\bigr\lvert \bigl\lvert f \bigl( \tau ,x(\tau) \bigr)\bigr\lvert \,d\tau \biggr)\,ds \\ &\quad < \frac{\varepsilon}{\phi_{q}(M)\int_{0}^{1}\phi_{q} (\int_{0}^{1}H(s,\tau)\,d\tau )\,ds} \phi_{q}(M) \int_{0}^{1}\phi_{q} \biggl( \int_{0}^{1}H(s,\tau)\,d\tau \biggr)\,ds \\ &\quad = \varepsilon. \end{aligned}$$

So, T is equicontinuous.

According to the Arzelà–Ascoli theorem, T is a completely continuous operator.

The proof is finished. □

Define $P_{k}=\{x\in P: \lVert x\lVert_{1-\alpha}< r_{k}\}$ and then we define the related boundary $\partial P_{k}=\{x\in P: \lVert x\lVert _{1-\alpha}=r_{k}\}$ where k is the positive integer.

Theorem 3.1

([21] Krasnosel’skii fixed point theorem)

Suppose E is a Banach space, and $P\subset E$ is a cone. Assume $P_{1}, P_{2}$ are two open and bounded subsets of E coupled with $0\in P_{1}, \overline{P_{1}}\subset P_{2}$ and suppose $T:P\cap (\overline {P_{2}}\setminus P_{1})\to P$ is a completely continuous operator such that one of the following conditions holds:

${(\mathrm{L}_{1})}$ :

$\lVert Tx\lVert\ge\lVert x\lVert$, if $x\in P\cap \partial P_{1}$, and $\lVert Tx\lVert\le\lVert x\lVert$, if $x\in P\cap \partial P_{2}$,

${(\mathrm{L}_{2})}$ :

$\lVert Tx\lVert\le\lVert x\lVert$, if $x\in P\cap \partial P_{1}$, and $\lVert Tx\lVert\ge\lVert x\lVert$, if $x\in P\cap \partial P_{2}$.

Then T has at least one fixed point in $P\cap (\overline {P_{2}}\setminus P_{1})$.

Theorem 3.2

If there exist $0< r_{1}< r_{2}, N_{1}\ge(\rho m\int_{0}^{1}\phi_{q}(\int_{0}^{1}H(s,\tau )\,d\tau)\,ds))^{-1}, 0\le N_{2}\le(n\int_{0}^{1}\phi_{q}(\int_{0}^{1}H(s,\tau)\,d\tau )\,ds))^{-1}$ and $N_{1} r_{1}< N_{2} r_{2}$ such that the following conditions hold:

${(\mathrm{L}_{3})}$ :

$f(t,x)\ge N_{1} r_{1}, (t,x)\in[0,1]\times[0,r_{1}]$,

${(\mathrm{L}_{4})}$ :

$f(t,x)\le N_{2} r_{2}, (t,x)\in[0,1]\times[0,r_{2}]$,

then T has at least one fixed point in $P\cap (\overline {P_{2}}\setminus P_{1})$. In addition, $r_{1}\le\lVert x\lVert_{1-\alpha}\le r_{2}$.

Proof

Step 1: Let $P_{1}=\{x\in P: \lVert x\lVert_{1-\alpha}< r_{1}\}$, by assumption ${(\mathrm{L}_{3})}$, for all $t\in[0,1]$,

$$\begin{aligned} \lVert Tx\lVert_{1-\alpha}&=t^{1-\alpha} \int_{0}^{1}G(t,s)\phi_{q} \biggl( \int _{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds \\ &\ge N_{1} r_{1} \int_{0}^{1}\rho m\phi_{q} \biggl( \int_{0}^{1}H(s,\tau)\,d\tau \biggr)\,ds \\ &\ge r_{1}=\lVert x\lVert_{1-\alpha}. \end{aligned}$$

Then we can get $\lVert Tx\lVert_{1-\alpha}\ge\lVert x\lVert_{1-\alpha }\ (\forall x \in P\cap \partial P_{1})$.

Step 2: Let $P_{2}=\{x\in P: \lVert x\lVert_{1-\alpha}< r_{2}\}$, by assumption ${(\mathrm{L}_{4})}$, for all $t\in[0,1]$,

$$\begin{aligned} \lVert Tx\lVert_{1-\alpha}&=t^{1-\alpha} \int_{0}^{1}G(t,s)\phi_{q} \biggl( \int _{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds \\ &\le N_{2} r_{2} \int_{0}^{1} n\phi_{q} \biggl( \int_{0}^{1}H(s,\tau)\,d\tau \biggr)\,ds \\ &\le r_{2}=\lVert x\lVert_{1-\alpha}. \end{aligned}$$

It is easy to obtain $\lVert Tx\lVert_{1-\alpha}\le\lVert x\lVert _{1-\alpha}\ (\forall x \in P\cap \partial P_{2})$.

To conclude, the above proof satisfies the condition ${(\mathrm{L}_{1})}$ of Theorem 3.1, that is to say, T has at least one fixed point in $P\cap (\overline{P_{2}}\setminus P_{1})$. In addition, $r_{1}\le\lVert x\lVert_{1-\alpha}\le r_{2}$.

The proof is finished. □

Theorem 3.3

([22])

Suppose E is a Banach space, and $P\subset E$ is a cone. Assume $P_{3}, P_{4}$ are two open and bounded subsets of E and $T:\overline{P_{3}}\to P$ is a completely continuous operator. If $\lVert Tx\lVert_{1-\alpha }< r_{3}$, for $x \in\partial P_{3}$ and $\lVert Tx\lVert_{1-\alpha}\ge r_{4}$, for $x \in\partial P_{4}$ hold, then T has at least two fixed points in $\overline{P_{3}}$, and

$$ 0\le\lVert x_{1}\lVert_{1-\alpha}< r_{3}< \lVert x_{2}\lVert_{1-\alpha}\le r_{4}. $$

Theorem 3.4

If there exist $0< r_{3}< r_{4}, 0\le N_{3}< (\rho m\int_{0}^{1}\phi_{q}(\int _{0}^{1}H(s,\tau)\,d\tau)\,ds))^{-1}, N_{4}\ge(n\int_{0}^{1}\phi_{q}(\int_{0}^{1}H(s,\tau )\,d\tau)\,ds))^{-1}$ such that the following conditions hold:

${(\mathrm{L}_{5})}$ :

$f(t,x)< N_{3} r_{3}, x \in\partial P_{3}$,

${(\mathrm{L}_{6})}$ :

$f(t,x)\ge N_{4} r_{4}, x \in\partial P_{4}$,

then T has at least two fixed points in $\overline{P_{3}}$, and

$$ 0\le\lVert x_{1}\lVert_{1-\alpha}< r_{3}< \lVert x_{2}\lVert_{1-\alpha}\le r_{4}. $$

Proof

On the one hand, $P_{3}=\{x\in P: \lVert x\lVert_{1-\alpha}< r_{3}\}$, by assumption ${(\mathrm{L}_{5})}$, for all $t\in[0,1]$,

$$\begin{aligned} \lVert Tx\lVert_{1-\alpha}&=t^{1-\alpha} \int_{0}^{1}G(t,s)\phi_{q} \biggl( \int _{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds \\ &\le N_{3} r_{3} \int_{0}^{1} n\phi_{q} \biggl( \int_{0}^{1}H(s,\tau)\,d\tau \biggr)\,ds \\ &< r_{3}. \end{aligned}$$

Then we can get $\lVert Tx\lVert_{1-\alpha}< r_{3}\ (\forall x \in\partial P_{3})$.

On the other hand, $P_{4}=\{x\in P: \lVert x\lVert_{1-\alpha}< r_{4}\}$, by assumption ${(\mathrm{L}_{6})}$, for all $t\in[0,1]$,

$$\begin{aligned} \lVert Tx\lVert_{1-\alpha}&=t^{1-\alpha} \int_{0}^{1}G(t,s)\phi_{q} \biggl( \int _{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds \\ &\ge N_{4} r_{4} \int_{0}^{1}\rho m\phi_{q} \biggl( \int_{0}^{1}H(s,\tau)\,d\tau \biggr)\,ds \\ &\ge r_{4}. \end{aligned}$$

Obviously, $\lVert Tx\lVert_{1-\alpha}\ge r_{4}\ (\forall x \in\partial P_{4})$ holds.

To conclude, the above proof satisfies the condition of Corollary 3.3, that is to say, T has at least two fixed points in $\overline {P_{3}}$, which satisfy

$$ 0\le\lVert x_{1}\lVert_{1-\alpha}< r_{3}< \lVert x_{2}\lVert_{1-\alpha}\le r_{4}. $$

The proof is finished. □

Suppose $\varphi,\psi$ are non-negative continuous convex functionals on $P, \theta$ is a non-negative continuous concave functional on P, and η is a non-negative continuous functional on P. Define the following convex set:

$$\begin{aligned} &P(\varphi; d)= \bigl\{ x\in P: \varphi(x)< d \bigr\} , \\ &P(\varphi, \theta; b, d)= \bigl\{ x\in P: b \le\theta(x), \varphi(x) \le d \bigr\} , \\ &P(\varphi, \psi, \theta; b, c, d)= \bigl\{ x\in P: b \le\theta(x), \psi(x) \le c, \varphi(x) \le d \bigr\} , \end{aligned}$$

and a closed set

$$ P(\varphi, \eta; a, d)= \bigl\{ x\in P: a \le\theta(x), \varphi(x) \le d \bigr\} , $$

where $a,b,c,d$ are positive constants.

Theorem 3.5

([23] Avery–Peterson fixed point theorem)

Suppose P is a cone on the real Banach space E. Suppose $\varphi ,\psi$ are non-negative continuous convex functionals on $P, \theta$ is a non-negative continuous concave functional on P, and η is a non-negative continuous functional on P satisfying $\eta(\rho x)\le \rho\eta(x)$ for $0\le\rho\le1$, for positive constants k and d, we have

$$ \theta(x)\le\eta(x), \qquad \lVert x\lVert\le k\varphi(x), \quad\forall x\in \overline{P(\varphi; d)}. $$

(10)

Suppose $T: \overline{P(\varphi; d)}\to\overline{P(\varphi; d)}$ is a completely continuous operator and there exist $a, b, c>0$ with $a< b$ such that the following conditions hold:

${(\mathrm{H}_{1})}$ :

$\{x\in P(\varphi, \psi, \theta; b, c, d):\theta(x)>b\}\ne \emptyset$ and $\theta(x)>b$ for $x\in P(\varphi, \psi, \theta; b, c, d)$,

${(\mathrm{H}_{2})}$ :

$\theta(Tx)>b$ for $x\in P(\varphi, \theta; b, d)$ with $\eta(Tx)>c$,

${(\mathrm{H}_{3})}$ :

$0 \notin P(\varphi, \eta; a, d)$ and $\eta(Tx)< a$, for $P(\varphi, \eta; a, d)$ with $\eta(x)=a$.

Then T has at least three fixed points $x_{1}, x_{2},x_{3}\in\overline {P(\varphi; d)}$ such that

$$ \varphi(x_{i})\le d,\qquad \theta(x_{1})>b,\qquad \theta(x_{2})>a,\qquad \eta(x_{2})< b,\qquad \eta(x_{3})< a, $$

where $i=1,2,3$.

Define the following functionals:

$$\begin{aligned} &\varphi(x)=\lVert x\lVert_{1-\alpha},\qquad \psi(x)= \eta(x)=\max_{t\in [0,1]}\bigl\lvert t^{1-\alpha}x(t)\bigr\lvert , \qquad \theta(x)= \min_{t\in[0,1]}\bigl\lvert t^{1-\alpha}x(t)\bigr\lvert , \\ &N=\rho m \int_{0}^{1}\phi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds, \end{aligned}$$

therefore, $\varphi, \psi, \eta, \theta$ satisfy the condition as mentioned above, and

$$ \rho\eta(x) \le\theta(x) \le\psi(x)=\eta(x), \qquad \lVert x\lVert_{1-\alpha }\le \varphi(x). $$

Theorem 3.6

Suppose ${(\mathrm{H}_{0})}$ holds, there exist positive constants $a, b, c, d$ with $a< b< c=b\rho^{-1}$,

${(\mathrm{H}_{4})}$ :

$f(t,x)\le\varGamma(\beta-1) \phi_{p}(dn^{-1})$ for $(t,x)\in [0,1]\times[0,d]$,

${(\mathrm{H}_{5})}$ :

$f(t,x)> \phi_{p}(bN^{-1})$ for $(t,x)\in[0,1]\times [b,c]$,

${(\mathrm{H}_{6})}$ :

$f(t,x)< \varGamma(\beta-1) \phi_{p}(an^{-1})$ for $(t,x)\in [0,1]\times[0,a]$,

Then T has at least three fixed points $x_{1}, x_{2},x_{3}\in\overline {P(\varphi; d)}$ such that

$$ \varphi(x_{i})\le d,\qquad \theta(x_{1})>b,\qquad \theta(x_{2})>a,\qquad \eta(x_{2})< b,\qquad \eta(x_{3})< a, $$

where $i=1,2,3$.

Proof

For $x(t)\in\overline{P(\varphi; d)}$, we have

$$ \varphi(x)=\lVert x\lVert_{1-\alpha}\le d. $$

According to ${(\mathrm{H}_{4})}$, we have

$$\begin{aligned} \bigl\lVert Tx(t)\bigr\lVert &=\biggl\lvert t^{1-\alpha} \int_{0}^{1}G(t,s)\phi_{q} \biggl( \int _{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds \biggr\lvert \\ &\le \int_{0}^{1} \bigl\lvert t^{1-\alpha}G(t,s)\bigr\lvert \phi_{q} \biggl( \int_{0}^{1}\bigl\lvert H(s,\tau)\bigr\lvert \bigl\lvert f \bigl( \tau,x(\tau) \bigr)\bigr\lvert \,d\tau \biggr)\,ds \\ &< n \phi_{q} \biggl(\frac{\varGamma(\beta-1) \phi_{q}(\frac{d}{n})}{\varGamma(\beta -1)} \biggr) \\ &=d. \end{aligned}$$

Hence, $\varphi(Tx)=\lVert Tx\lVert_{1-\alpha}\le d$, that is to say, $T: \overline{P(\varphi; d)}\to\overline{P(\varphi; d)}$ is a completely continuous operator.

Let $x(t)=\frac{b}{\rho}, x(t)\in P(\varphi, \psi, \theta; b, c, d)$ and $\theta(\frac{b}{\rho})>b$, and so $\{x(t)\in P(\varphi, \psi, \theta; b, c, d): \theta(\frac{b}{\rho})>b\}\ne\emptyset$.

For $x(t)\in P(\varphi, \psi, \theta; b, c, d)$, we obtain $b\le x(t) \le c=\frac{b}{\rho}$, then

$$\begin{aligned} \theta(Tx)&=\min_{t\in[0,1]}\bigl\lvert t^{1-\alpha}Tx(t)\bigr\lvert \\ &=\min_{t\in[0,1]}t^{1-\alpha} \int_{0}^{1}G(t,s)\phi_{q} \biggl( \int_{0}^{1}H(s,\tau )f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds \\ &\ge \int_{0}^{1} \rho\max_{t\in[0,1]} t^{1-\alpha}G(t,s) \phi_{q} \biggl( \int _{0}^{1} H(s,\tau) f \bigl(\tau,x(\tau) \bigr) \,d \tau \biggr)\,ds \\ &> \rho m \int_{0}^{1}\phi_{q} \biggl( \int_{0}^{1}H(s,\tau)f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds\frac{b}{N} \\ &=b. \end{aligned}$$

Hence, $\theta(Tx)>b$ for $x \in P(\varphi, \psi, \theta; b, c, d)$. That is to say, ${(\mathrm{H}_{1})}$ holds.

Let $x\in P(\varphi, \theta; b, d)$ with $\psi(Tx)>c=\frac{b}{\rho}$, we obtain

$$ \theta(Tx)\ge\rho\psi(Tx)>\rho c=\rho\frac{b}{\rho}=b. $$

Therefore, ${(\mathrm{H}_{2})}$ is satisfied.

By $\theta(0)=0< a$, we have $0 \notin P(\varphi, \eta; a, d)$.

Let $x\in P(\varphi, \eta; a, d)$ and $\eta(x)=a$, we obtain $\varphi (x)\le d$, according to ${(\mathrm{H}_{6})}$,

$$\begin{aligned} \eta(Tx)&=\max_{t\in[0,1]}\bigl\lvert t^{1-\alpha}Tx(t)\bigr\lvert \\ &=\max_{t\in[0,1]}t^{1-\alpha} \int_{0}^{1}G(t,s)\phi_{q} \biggl( \int_{0}^{1}H(s,\tau )f \bigl(\tau,x(\tau) \bigr)\,d \tau \biggr)\,ds \\ &\le n \int_{0}^{1} \phi_{q} \biggl( \int_{0}^{1} H(s,\tau) f \bigl(\tau,x(\tau) \bigr) \,d \tau \biggr)\,ds \\ &< n \phi_{q} \biggl(\frac{\varGamma(\beta-1) \phi_{q}(\frac{a}{n})}{\varGamma(\beta -1)} \biggr) \\ &=a. \end{aligned}$$

Hence, the condition ${(\mathrm{H}_{3})}$ holds.

The proof is finished. □

4 Illustrative example

Consider the following nonlinear fractional differential equations:

$$ \textstyle\begin{cases} -D_{0^{+}}^{1.7} (\phi_{3.5}(D_{0^{+}}^{0.5}x(t)-3x(t)) )=f(t,x(t)),\quad t\in(0,1], \\\lim_{t \to0^{+}}t^{0.5}x(t)=\sum_{k=1}^{\infty}\frac {1}{2^{2k}(k-1)!}x(\frac{1}{2^{2k}}),\\ \lim_{t \to0^{+}}t^{0.3}(\phi_{3.5}(D_{0^{+}}^{0.5} x(t)-3x(t)))=\phi _{3.5}(D_{0^{+}}^{0.5} x(1)-3x(1))=0. \end{cases} $$

(11)

Obviously, we choose $\alpha=0.5, \beta=1.7, p=3.5, \lambda=-3, \mu _{k}=\frac{1}{2^{2k}(k-1)!}, \xi_{k}=\frac{1}{2^{2k}}, \sum_{k=1}^{\infty}\mu _{k}\xi_{k}^{\alpha-1}=\frac{e^{0.5}}{2}<1$. By simple calculation, we obtain $\rho\approx0.04819, m\approx0.0011246, n\approx10.091$.

Part 1: In order to satisfy Theorem 3.1, by matlab, we get $N_{1}=3.1\times10^{4}, N_{2}=0.5, r_{1}=0.2, r_{2}=4\times10^{4}$, so we know that the BVP (11) has at least one positive solution $x_{1}$ satisfy $0.2\le\lVert x_{1}\lVert_{1-\alpha}\le4\times10^{4}$, where

$$ f(t,x)=7\times10^{3}e^{t}+\operatorname{sin}x,\quad (t,x)\in[0,1] \times \bigl[0,4\times10^{4} \bigr], $$

where $f(t,x)$ satisfies

• (i) $f(t,x)\ge N_{1} r_{1}=6.2\times10^{3}$ for $(t,x)\in[0,1]\times [0,0.2]$,

• (ii) $f(t,x)>N_{2} r_{2}=2\times10^{4}$ for $(t,x)\in[0,1]\times [0,4\times10^{4}]$.

Part 2: In order to make Theorem 3.3 hold, we choose $N_{3}=500, N_{4}=1, r_{3}=2, r_{4}=50$, then the BVP (11) has at least two positive solutions $x_{1}, x_{2}$, and

$$ 0\le\lVert x_{1}\lVert_{1-\alpha}< 2< \lVert x_{2} \lVert_{1-\alpha}\le50, $$

where

$$ f(t,x)=(t+5)^{3}+\frac{19}{4}x^{2},\quad (t,x)\in[0,1]\times[0,100], $$

where $f(t,x)$ satisfies

• (i) $f(t,x)< N_{3} r_{3}=1000$ for $(t,x)\in[0,1]\times[0,2]$,

• (ii) $f(t,x)>N_{4} r_{4}=50$ for $(t,x)\in[0,1]\times[0,50]$.

Part 3: In order to make Theorem 3.5 hold, we choose $a=0.75, b=2, c=b\rho^{-1}=41.51, d=0.5$, then we know the BVP (11) has at least three positive solutions $x_{1}, x_{2},x_{3}$ satisfying

$$\begin{aligned} &\lVert x\lVert_{0.5}(x_{i})\le 0.5\quad (i=1,2,3),\qquad \min_{t\in[0,1]}\bigl\lvert t^{0.5}x_{1} \bigr\lvert >2, \qquad \min_{t\in[0,1]}\bigl\lvert t^{0.5}x_{2} \bigr\lvert >0.75, \\ &\max_{t\in[0,1]}\bigl\lvert t^{0.5}x_{2}\bigr\lvert < 2,\qquad \max_{t\in[0,1]}\bigl\lvert t^{0.5}x_{3}\bigr\lvert < 0.75, \end{aligned}$$

where

$$ f(t,x)=\frac{1}{10^{3}}\sqrt {t}+x^{13},\quad (t,x)\in[0,1]\times[0,200], $$

where $f(t,x)$ satisfies

• (i) $f(t,x)<\varGamma(\beta-1) \phi_{p}(dn^{-1})=0.617$ for $(t,x)\in [0,1]\times[0,0.5]$,

• (ii) $f(t,x)> \phi_{p}(bN^{-1})=3189.505$ for $(t,x)\in[0,1]\times [2,41.51]$,

• (iii) $f(t,x)<\varGamma(\beta-1) \phi_{p}(an^{-1})=0.0342$ for $(t,x)\in [0,1]\times[0,0.75]$.

5 Conclusions

In this paper, we consider a class of nonlinear fractional differential equations coupled with the p-Laplacian operator and infinite-point boundary value conditions by means of the properties of Green’s function, Krasnosel’skii fixed point theorem, the corollary of Krasnosel’skii fixed point theorem and the Avery–Peterson fixed point theorem. On top of that, we establish the suitable criteria to warrant the existence of the positive solutions for the BVP we learned. To the best of our knowledge, few researchers studied fractional differential equations with p-Laplacian by the Riemann–Liouville derivative, and this condition coupled with values at an infinite number of points is not covered in the previous situations; last but not least, it is more general for the fractional differential equation models.