Multiple positive solutions for a coupled system of fractional multi-point BVP with p-Laplacian operator

Abstract

We investigate the existence of positive solutions for a system of fractional multi-point BVP with p-Laplacian operator. Our main tool is the fixed point theorem due to Leggett-Williams. The result obtained in this paper corrects some mistakes in (Al-Hossain in Differ. Equ. Dyn. Syst., 2016, doi:10.1007/s11590-013-0708-4) and essentially improves and extends some well-known results.

1 Introduction

In this paper, we are concerned with the existence of multiple positive solutions for the following p-Laplacian fractional operator multi-point BVP:

$$\begin{aligned}& D^{\beta_{1}}_{a^{+}} \bigl(\varphi_{p}\bigl(D^{\alpha_{1}}_{a^{+}}u(t) \bigr)\bigr)=f_{1}\bigl(t,u(t),v(t)\bigr), \quad t\in(a,b), \end{aligned}$$

(1.1)

$$\begin{aligned}& D^{\beta_{2}}_{a^{+}} \bigl(\varphi_{p}\bigl(D^{\alpha_{2}}_{a^{+}}v(t) \bigr)\bigr)=f_{2}\bigl(t,u(t),v(t)\bigr), \quad t\in(a,b), \end{aligned}$$

(1.2)

with the boundary conditions

$$\begin{aligned}& \left \{ \textstyle\begin{array} {l} u^{(j)}(a)=0,\quad j=0,1,2,\ldots,n-2,\qquad u^{(\mu_{1})}(b)=\sum_{i=1}^{\infty }\gamma_{i}u^{(\mu_{1})}(\xi_{i}) ,\\ \varphi_{p}(D^{\alpha_{1}}_{a^{+}}u(a))=0,\qquad D^{\mu_{2}}_{a^{+}}(\varphi_{p}(D^{\alpha_{1}}_{a^{+}}u(b)))=\sum_{i=1}^{\infty}\delta_{i} (D^{\mu_{2}}_{a^{+}}(\varphi_{p}(D^{\alpha _{1}}_{a^{+}}u(\eta_{i}))) ), \end{array}\displaystyle \right . \end{aligned}$$

(1.3)

$$\begin{aligned}& \left \{ \textstyle\begin{array} {l} v^{(k)}(a)=0,\quad k=0,1,2,\ldots,m-2,\qquad v^{(\mu_{3})}(b)=\sum_{l=1}^{\infty }\gamma_{l}v^{(\mu_{3})}(\xi_{l}),\\ \varphi_{p}(D^{\alpha_{2}}_{a^{+}}v(a))=0,\qquad D^{\mu_{4}}_{a^{+}}(\varphi_{p}(D^{\alpha_{2}}_{a^{+}}v(b)))=\sum_{l=1}^{\infty}\delta_{l} ( D^{\mu_{4}}_{a^{+}}(\varphi_{p}(D^{\alpha_{2}}_{a^{+}}v(\eta_{l}))) ), \end{array}\displaystyle \right . \end{aligned}$$

(1.4)

where $a,b\in R$ with $a< b$, $\varphi_{p}(s)=|s|^{p-2}s$, $p>1$, $\varphi^{-1}_{p}=\varphi_{q}$, $\frac{1}{p}+\frac{1}{q}=1$ and $D^{\alpha_{1}}_{a^{+}}$, $D^{\alpha_{2}}_{a^{+}}$, $D^{\beta_{1}}_{a^{+}}$, $D^{\beta_{2}}_{a^{+}}$, $D^{\mu_{2}}_{a^{+}}$, $D^{\mu_{4}}_{a^{+}}$ are the standard Riemann-Liouville fractional derivatives. We make the following assumptions:

($\mathrm{H}_{0}$):

$n-1<\alpha_{1} \le n$, $m-1<\alpha_{2} \le m$, for $n,m\ge3$, $\beta_{1},\beta_{2}\in(1,2]$, $\mu_{1}\in[1,\alpha_{1}-1)$, $\mu_{3}\in[1,\alpha_{2}-1)$, $\mu_{2},\mu_{4}\in(0,1]$ and $\mu_{2}\le\beta_{1}-1$, $\mu_{4}\le \beta_{2}-1$, $f_{i}(t,u,v)\in C([a,b] \times R^{2}, R^{+})$, for $i=1,2$,

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

$\gamma_{i}\ge0$ ($i=1,2,\ldots$) and $a<\xi_{1}<\xi_{2}<\cdots<\xi_{i-1}<\xi_{i}<\cdots<b$ satisfy that

$$\begin{aligned}& \Phi_{1}=(b-a)^{\alpha_{1}-\mu_{1}-1}-\sum_{i=1}^{\infty} \gamma_{i} (\xi_{i}-a)^{\alpha_{1}-\mu_{1}-1}>0, \\& \Phi_{2}=(b-a)^{\alpha_{2}-\mu_{3}-1}-\sum_{l=1}^{\infty} \gamma_{l} (\xi_{l}-a)^{\alpha_{2}-\mu_{3}-1}>0, \end{aligned}$$

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

$\delta_{i}\ge0$ ($i=1,2,\ldots$) and $a<\eta_{1}<\eta_{2}<\cdots<\eta_{i-1}<\eta_{i}<\cdots<b$ satisfy that

$$\begin{aligned}& \Delta_{1}=(b-a)^{\beta_{1}-\mu_{2}-1}-\sum_{i=1}^{\infty} \delta_{i} (\eta_{i}-a)^{\beta_{1}-\mu_{2}-1}>0, \\& \Delta_{2}=(b-a)^{\beta_{2}-\mu_{4}-1}-\sum_{l=1}^{\infty} \delta_{l} (\eta_{l}-a)^{\beta_{2}-\mu_{4}-1}>0. \end{aligned}$$

Recently, the existence of solutions or positive solutions of p-Laplacian fractional differential equations at nonresonance or resonance have been of great interest, readers can see [1–14] and the references cited therein. In particular, if $a=0$, $b=1$, $n=m=3$, $\delta_{i}=\gamma_{i}=0$ ($i=1,2,\ldots$), $\mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}=1$, $\alpha_{1}=\alpha_{2}=\alpha$, $\beta_{1}=\beta_{2}=\beta$, $f_{1}(t,u)=f_{2}(t,u)=\lambda f(u(t))$ with $u(t)=v(t)$, the problem becomes the problem studied in [14]. Han, Lu and Zhang investigated the existence of positive solutions for the eigenvalue problems of the fractional differential equation with generalized p-Laplacian

$$\left \{ \textstyle\begin{array} {l} D^{\beta}_{0^{+}} (\varphi_{p}(D^{\alpha}_{0^{+}}u(t)))=\lambda f(u(t)), \quad t\in(0,1), \\ u(0)=u'(0)=u'(1)=0,\qquad \varphi_{p}(D^{\alpha}_{0^{+}}u(0))=(\varphi_{p}(D^{\alpha}_{0^{+}}u(1)))'=0, \end{array}\displaystyle \right .$$

where $2<\alpha\le3$, $1<\beta\le2$, $\lambda>0$, $\varphi_{p}(s)=|s|^{p-2}s$, $\varphi^{-1}_{p}=\varphi_{q}$, $\frac{1}{p}+\frac{1}{q}=1$, $f:(0,+\infty)\to(0,+\infty)$ is continuous. By using the properties of Green’s function and Guo-Krasnosel’skii’s fixed point theorem, some existence results of at least one or two positive solutions in terms of a different eigenvalue interval are obtained.

If $\gamma_{1}=\xi$, $\xi_{1}=\eta$, $\gamma_{i}=0$ ($i=2,3,\ldots$), $\delta_{i}=0$ ($i=1,2,3,\ldots$), the problem becomes the problem considered in [1]. Al-Hossain investigated the existence of at least one positive solution for the following system of fractional order differential equations with p-Laplacian operators:

$$\begin{aligned}& D^{\beta_{1}}_{a^{+}} \bigl(\varphi_{p}\bigl(D^{\alpha_{1}}_{a^{+}}u(t) \bigr)\bigr)+f_{1}\bigl(t,u(t),v(t)\bigr)=0,\quad t\in(a,b), \end{aligned}$$

(1.5)

$$\begin{aligned}& D^{\beta_{2}}_{a^{+}} \bigl(\varphi_{p}\bigl(D^{\alpha_{2}}_{a^{+}}v(t) \bigr)\bigr)+f_{2}\bigl(t,u(t),v(t)\bigr)=0,\quad t\in(a,b), \end{aligned}$$

(1.6)

with the boundary conditions

$$\begin{aligned}& \left \{ \textstyle\begin{array} {l} u^{(j)}(a)=0,\quad j=0,1,2,\ldots,n-2,\qquad u^{(\mu_{1})}(b)=\xi u^{(\mu _{1})}(\eta) ,\\ \varphi_{p}(D^{\alpha_{1}}_{a^{+}}u(a))=0,\qquad D^{\mu_{2}}_{a^{+}}(\varphi_{p}(D^{\alpha_{2}}_{a^{+}}u(b)))=0, \end{array}\displaystyle \right . \end{aligned}$$

(1.7)

$$\begin{aligned}& \left \{ \textstyle\begin{array} {l} v^{(k)}(a)=0,\quad k=0,1,2,\ldots,m-2,\qquad v^{(\mu_{3})}(b)=\xi v^{(\mu _{3})}(\eta),\\ \varphi_{p}(D^{\alpha_{2}}_{a^{+}}v(a))=0,\qquad D^{\mu_{4}}_{a^{+}}(\varphi_{p}(D^{\alpha_{2}}_{a^{+}}v(b)))=0, \end{array}\displaystyle \right . \end{aligned}$$

(1.8)

where $\varphi_{p}(s)=|s|^{p-2}s$, $p>1$, $\varphi^{-1}_{p}=\varphi_{q}$, $\frac{1}{p}+\frac{1}{q}=1$, $\beta_{1},\beta_{2}\in(1,2]$, $n-1<\alpha_{1} \le n$, $m-1<\alpha_{2} \le m$ for $n,m\ge3$, $\mu_{2},\mu_{4}\in(0,1]$, $\mu_{1}\in[1,\alpha_{1}-2]$, $\mu_{3}\in[1,\alpha_{2}-2]$ are fixed integers, $\xi\in(0,\infty)$, $\eta\in(a,b)$ are constants with $0<\xi(\eta-a)^{\alpha_{1}-\mu_{1}-1}<(b-a)^{\alpha_{1}-\mu_{1}-1}$, $f_{i}(t,u,v)\in C([a,b] \times R^{2}, R^{+})$, for $i=1,2$, and $D^{\alpha_{1}}_{a^{+}}$, $D^{\alpha_{2}}_{a^{+}}$, $D^{\beta_{1}}_{a^{+}}$, $D^{\beta_{2}}_{a^{+}}$, $D^{\mu_{2}}_{a^{+}}$, $D^{\mu_{4}}_{a^{+}}$ are the standard Riemann-Liouville fractional derivatives. By using a fixed point theorem of the cone expansion and compression of functional type due to Avery, Henderson and O’Regan, the author established at least one existence result for positive solutions to the problem.

Unfortunately, there are some mistakes in paper [1]. First and foremost, the given Green’s function $H_{1}(t,s)$ is not to problem (1.5)-(1.8). In fact, the Green’s function $H_{1}(t,s)$ in [1] is the Green’s function of the following system of fractional order differential equations with p-Laplacian operators:

$$\begin{aligned}& D^{\beta_{1}}_{a^{+}} \bigl(\varphi_{p}\bigl(D^{\alpha_{1}}_{a^{+}}u(t) \bigr)\bigr)=f_{1}\bigl(t,u(t),v(t)\bigr),\quad t\in(a,b), \end{aligned}$$

(1.9)

$$\begin{aligned}& D^{\beta_{2}}_{a^{+}} \bigl(\varphi_{p}\bigl(D^{\alpha_{2}}_{a^{+}}v(t) \bigr)\bigr)=f_{2}\bigl(t,u(t),v(t)\bigr), \quad t\in(a,b), \end{aligned}$$

(1.10)

with the boundary conditions (1.7), (1.8). In [1], even if the author studied the above problem, there are still the following two mistakes in [1], which are very essential to the proof of the main results.

First, Lemma 3.3 in [1], the author gave the following result: if $s\le t$, then

$$\begin{aligned} G_{11}(t,s) =&\frac{1}{\Gamma(\alpha_{1})} \biggl[\frac{(t-a)^{\alpha _{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{(b-a)^{\alpha_{1}-\mu_{1}-1}}-(t-s)^{\alpha _{1}-1} \biggr] \\ =&\frac{(t-a)^{\alpha_{1}-1}}{\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-1}} \biggl[\frac{(b-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{ (b-a)^{\alpha_{1}-\mu_{1}-1}}-(b-s)^{\alpha_{1}-1} \biggr] \\ &{}+\frac{1}{\Gamma(\alpha_{1})} \biggl[\frac{(t-a)^{\alpha _{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{(b-a)^{\alpha_{1}-\mu_{1}-1}}-(t-s)^{\alpha _{1}-1} \biggr]. \end{aligned}$$

If $t\le s$, we have

$$\begin{aligned} G_{11}(t,s) =&\frac{1}{\Gamma(\alpha_{1})} \biggl[\frac{(t-a)^{\alpha _{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{(b-a)^{\alpha_{1}-\mu_{1}-1}} \biggr] \\ =&\frac{(t-a)^{\alpha_{1}-1}}{\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-1}} \biggl[\frac{(b-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{ (b-a)^{\alpha_{1}-\mu_{1}-1}}-(b-s)^{\alpha_{1}-1} \biggr] \\ &{}+\frac{1}{\Gamma(\alpha_{1})} \biggl[\frac{(t-a)^{\alpha _{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{(b-a)^{\alpha_{1}-\mu_{1}-1}} \biggr]. \end{aligned}$$

The above result is one of the mistakes. In fact, by

$$\begin{aligned}& \frac{(b-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{(b-a)^{\alpha_{1}-\mu _{1}-1}}-(b-s)^{\alpha_{1}-1} \\& \quad=\frac{(b-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}-(b-s)^{\alpha _{1}-1}(b-a)^{\alpha_{1}-\mu_{1}-1}}{(b-a)^{\alpha_{1}-\mu_{1}-1}} \\& \quad=\frac{(b-a)^{\alpha_{1}-\mu_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}[(b-a)^{\mu _{1}}-(b-s)^{\mu_{1}}]}{(b-a)^{\alpha_{1}-\mu_{1}-1}}\ge0, \end{aligned}$$

then, for $s\le t$, we have

$$\begin{aligned} G_{11}(t,s) =&\frac{1}{\Gamma(\alpha_{1})} \biggl[\frac{(t-a)^{\alpha _{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{(b-a)^{\alpha_{1}-\mu_{1}-1}}-(t-s)^{\alpha _{1}-1} \biggr] \\ \neq&\frac{(t-a)^{\alpha_{1}-1}}{\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-1}} \biggl[\frac{(b-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{ (b-a)^{\alpha_{1}-\mu_{1}-1}}-(b-s)^{\alpha_{1}-1} \biggr] \\ &{}+\frac{1}{\Gamma(\alpha_{1})} \biggl[\frac{(t-a)^{\alpha _{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{(b-a)^{\alpha_{1}-\mu_{1}-1}}-(t-s)^{\alpha _{1}-1} \biggr], \end{aligned}$$

and for $t\le s$, we have

$$\begin{aligned} G_{11}(t,s) =&\frac{1}{\Gamma(\alpha_{1})} \biggl[\frac{(t-a)^{\alpha _{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{(b-a)^{\alpha_{1}-\mu_{1}-1}} \biggr] \\ \neq&\frac{(t-a)^{\alpha_{1}-1}}{\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-1}} \biggl[\frac{(b-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{ (b-a)^{\alpha_{1}-\mu_{1}-1}}-(b-s)^{\alpha_{1}-1} \biggr] \\ &{}+\frac{1}{\Gamma(\alpha_{1})} \biggl[\frac{(t-a)^{\alpha _{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{(b-a)^{\alpha_{1}-\mu_{1}-1}} \biggr]. \end{aligned}$$

Thus, as the above wrong conclusion, the proof of the inequality of Lemma 3.3 is not correct. In this paper, we give a method to deal with this problem in the proof of Lemma 2.4.

Next, we want to point out, in [1], the following conclusion in Lemma 3.4(i) is also wrong: $H_{1}(t,s)\le H_{1}(b,s)$ for $(t,s)\in[a,b]\times[a,b]$. In fact, for $a\le s\le t \le b$,

Case 1: :

if $0<\mu_{2}\le\beta_{1}-1\le1$, we have

$$\begin{aligned} \frac{\partial}{\partial t} H_{1}(t,s) =&\frac{(\beta_{1}-1)[(t-a)^{\beta_{1}-2}(b-s)^{\beta_{1}-\mu _{2}-1}-(t-s)^{\beta_{1}-2}(b-a)^{\beta_{1}-\mu_{2}-1}]}{\Gamma(\beta _{1})(b-a)^{\beta_{1}-\mu_{2}-1}} \\ \le&\frac{(\beta_{1}-1)(t-s)^{\beta_{1}-2}[(b-s)^{\beta_{1}-\mu _{2}-1}-(b-a)^{\beta_{1}-\mu_{2}-1}]}{\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu _{2}-1}}\le 0. \end{aligned}$$

Case 2: :

if $0<\beta_{1}-1\le\mu_{2}\le1$, we have

$$\begin{aligned} \frac{\partial}{\partial t} H_{1}(t,s) =&\frac{(\beta_{1}-1)[(t-a)^{\beta_{1}-2}(b-s)^{\beta_{1}-\mu _{2}-1}-(t-s)^{\beta_{1}-2}(b-a)^{\beta_{1}-\mu_{2}-1}]}{\Gamma(\beta _{1})(b-a)^{\beta_{1}-\mu_{2}-1}} \\ =&\frac{(\beta_{1}-1)[(t-a)^{\mu_{2}-1}((t-a)(b-s))^{\beta_{1}-\mu _{2}-1}-(t-s)^{\mu_{2}-1}((t-s)(b-a))^{\beta_{1}-\mu_{2}-1}]}{\Gamma(\beta _{1})(b-a)^{\beta_{1}-\mu_{2}-1}}\\ \le&\frac{(\beta_{1}-1)[(t-s)(b-a)]^{\beta_{1}-\mu_{2}-1}[(t-a)^{\mu _{2}-1}-(t-s)^{\mu_{2}-1}]}{\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}}\le0. \end{aligned}$$

Thus, $H_{1}(t,s)$ in decreasing in t, which implies $H_{1}(t,s)\ge H_{1}(b,s)$ for $a\le s\le t \le b$.

Because of the wrong conclusion of Lemma 3.4 in [1], Theorems 4.2 and 4.3 in [1] cannot be established. The aim of this paper is to correct and improve the wrong results in [1]. By using the fixed point theorem due to Leggett-Williams, we will study the existence of triple positive solutions for the system of p-Laplacian fractional operator multi-point BVP (1.1)-(1.4). The associated Green’s function for the above problem is given at first, and some useful properties of the Green’s function are also obtained. The result obtained in this paper essentially improves and extends some known results. As application, an example is presented to illustrate the main result.

2 Preliminaries and lemmas

Lemma 2.1

[15]

Assume that $D_{a^{+}}^{\alpha}\in L^{1}(a,b)$ with a fractional derivative of order $\alpha>0$. Then

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

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

Lemma 2.2

If $y_{1}\in C[a,b]$ and ($\mathrm{H}_{0}$), ($\mathrm{H}_{1}$) hold, then the fractional differential equation BVPs

$$ \left \{ \textstyle\begin{array} {l} D^{\alpha_{1}}_{a^{+}} u(t)+y_{1}(t)=0, \quad t\in(a,b)\\ u^{(j)}(a)=0,\quad j=0,1,2,\ldots,n-2,\qquad u^{(\mu_{1})}(b)=\sum_{i=1}^{\infty}\gamma_{i}u^{(\mu_{1})}(\xi_{i}), \end{array}\displaystyle \right . $$

(2.1)

has a unique solution

$$u(t)= \int_{a}^{b} G_{1}(t,s)y_{1}(s) \,ds, $$

where

$$\begin{aligned}& \begin{aligned} G_{1}(t,s)&= \left \{ \textstyle\begin{array} {l} \frac{(t-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}-(t-s)^{\alpha _{1}-1}(b-a)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu _{1}-1}}\\ \quad{}+\frac{(t-a)^{\alpha_{1}-1}[\sum_{i=1}^{\infty}\gamma_{i}[(b-s)(\xi _{i}-a)]^{\alpha_{1}-\mu_{1}-1} -\sum_{i=1}^{\infty}\gamma_{i}[(\xi_{i}-s)(b-a)]^{\alpha_{1}-\mu_{1}-1}]}{\Phi _{1}\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu_{1}-1}},\\ \quad\quad a\le s\le\min\{t,\xi_{1}\},\\ \frac{(t-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}-(t-s)^{\alpha _{1}-1}(b-a)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu _{1}-1}}\\ \quad{}+\frac{(t-a)^{\alpha_{1}-1}[\sum_{i=1}^{\infty}\gamma_{i}[(b-s)(\xi _{i}-a)]^{\alpha_{1}-\mu_{1}-1} -\sum_{i=j}^{\infty}\gamma_{i}[(\xi_{i}-s)(b-a)]^{\alpha_{1}-\mu_{1}-1}]}{\Phi _{1}\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu_{1}-1}},\\ \quad\quad\xi_{j-1}< s\le\xi_{j}\le t, j=2,3,\ldots,\\ \frac{(t-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}-(t-s)^{\alpha _{1}-1}(b-a)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu _{1}-1}}\\ \quad{}+\frac{(t-a)^{\alpha_{1}-1}\sum_{i=1}^{\infty}\gamma_{i}[(b-s)(\xi _{i}-a)]^{\alpha_{1}-\mu_{1}-1}}{\Phi_{1}\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu _{1}-1}},\\ \quad\quad\lim_{i\rightarrow\infty}\xi_{i}\le s\le t,\\ \frac{(t-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha _{1})(b-a)^{\alpha_{1}-\mu_{1}-1}}\\ \quad{}+\frac{(t-a)^{\alpha_{1}-1}[\sum_{i=1}^{\infty}\gamma_{i}[(b-s)(\xi _{i}-a)]^{\alpha_{1}-\mu_{1}-1} -\sum_{i=1}^{\infty}\gamma_{i}[(\xi_{i}-s)(b-a)]^{\alpha_{1}-\mu_{1}-1}]}{\Phi _{1}\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu_{1}-1}},\\ \quad\quad t\le s\le\xi_{1},\\ \frac{(t-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha _{1})(b-a)^{\alpha_{1}-\mu_{1}-1}}\\ \quad{}+\frac{(t-a)^{\alpha_{1}-1}[\sum_{i=1}^{\infty}\gamma_{i}[(b-s)(\xi _{i}-a)]^{\alpha_{1}-\mu_{1}-1} -\sum_{i=j}^{\infty}\gamma_{i}[(\xi_{i}-s)(b-a)]^{\alpha_{1}-\mu_{1}-1}]}{\Phi _{1}\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu_{1}-1}},\\ \quad\quad t\le\xi_{j-1}< s\le\xi_{j}, j=2,3,\ldots,\\ \frac{(t-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha _{1})(b-a)^{\alpha_{1}-\mu_{1}-1}} +\frac{(t-a)^{\alpha_{1}-1}\sum_{i=1}^{\infty}\gamma_{i}[(b-s)(\xi _{i}-a)]^{\alpha_{1}-\mu_{1}-1}}{\Phi_{1}\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu _{1}-1}},\\ \quad\quad\max\{t,\lim_{i\rightarrow\infty}\xi_{i}\}\le s\le b, \end{array}\displaystyle \right .\\ &=G_{11}(t,s)+\frac{(t-a)^{\alpha_{1}-1}\sum_{i=1}^{\infty}\gamma_{i} G_{12}(\xi_{i},s)}{\Phi_{1}}, \end{aligned} \\& G_{11}(t,s)= \left \{ \textstyle\begin{array} {l@{\quad}l} \frac{(t-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}-(t-s)^{\alpha _{1}-1}(b-a)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu _{1}-1}},& s\le t,\\ \frac{(t-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha _{1})(b-a)^{\alpha_{1}-\mu_{1}-1}},& t\le s, \end{array}\displaystyle \right . \\& G_{12}(t,s)= \left \{ \textstyle\begin{array} {l@{\quad}l} \frac{[(b-s)(t-a)]^{\alpha_{1}-\mu_{1}-1}- [(t-s)(b-a)]^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu _{1}-1}},& s\le t,\\ \frac{ [(b-s)(t-a) ]^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha _{1})(b-a)^{\alpha_{1}-\mu_{1}-1}},& t\le s. \end{array}\displaystyle \right . \end{aligned}$$

Proof

Assume that $u\in C^{[\alpha_{1}]+1}[a,b]$ is a solution of fractional order BVPs (2.1) and is uniquely expressed as

$$I_{a^{+}}^{\alpha_{1}}D_{a^{+}}^{\alpha_{1}}u(t)=-I_{a^{+}}^{\alpha_{1}}y_{1}(t) $$

such that

$$u(t)=- \int_{a}^{t} \frac{(t-s)^{\alpha_{1}-1}}{\Gamma(\alpha_{1})}y_{1}(s) \,ds+c_{1} (t-a)^{\alpha_{1}-1}+c_{2} (t-a)^{\alpha_{1}-2}+ \cdots+c_{n} (t-a)^{\alpha_{1}-n}. $$

By $u^{(j)}(a)=0$, $j=0,1,2,\ldots,n-2$, we get that $c_{i}=0$, for $i=2,3,\ldots,n$, hence

$$u^{(\mu_{1})}(t)=-\prod_{j=1}^{\mu_{1}}( \alpha_{1}-j) \int_{a}^{t} \frac{(t-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})}y_{1}(s) \,ds+c_{1} \prod_{j=1}^{\mu_{1}}( \alpha_{1}-j) (t-a)^{\alpha_{1}-\mu_{1}-1}. $$

By $u^{(\mu_{1})}(b)=\sum_{i=1}^{\infty}\gamma_{i}u^{(\mu_{1})}(\xi_{i})$, we have

$$\begin{aligned}& -\prod_{j=1}^{\mu_{1}}(\alpha_{1}-j) \int_{a}^{b} \frac{(b-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})}y_{1}(s) \,ds+c_{1} \prod_{j=1}^{\mu_{1}}( \alpha_{1}-j) (b-a)^{\alpha_{1}-\mu_{1}-1} \\& \quad=\sum_{i=1}^{\infty}\gamma_{i} \Biggl[-\prod_{j=1}^{\mu_{1}}(\alpha _{1}-j) \int _{a}^{\xi_{i}} \frac{(\xi_{i}-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})}y_{1}(s) \,ds+c_{1} \prod_{j=1}^{\mu_{1}}( \alpha_{1}-j) (\xi_{i}-a)^{\alpha_{1}-\mu_{1}-1} \Biggr]. \end{aligned}$$

Then

$$\begin{aligned} c_{1} =&\frac{1}{\Phi_{1}} \Biggl[ \int_{a}^{b} \frac{(b-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})}y_{1}(s) \,ds-\sum_{i=1}^{\infty}\gamma_{i} \int _{a}^{\xi_{i}} \frac{(\xi_{i}-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})}y_{1}(s) \,ds \Biggr] \\ =& \biggl[\frac{1}{(b-a)^{\alpha_{1}-\mu_{1}-1}}+\frac{\sum_{i=1}^{\infty }\gamma_{i} (\xi_{i}-a)^{\alpha_{1}-\mu_{1}-1}}{\Phi_{1}(b-a)^{\alpha_{1}-\mu_{1}-1}} \biggr] \int_{a}^{b} \frac{(b-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})}y_{1}(s) \,ds \\ &{}-\frac{1}{\Phi_{1}}\sum_{i=1}^{\infty} \gamma_{i} \int _{a}^{\xi_{i}} \frac{(\xi_{i}-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})}y_{1}(s) \,ds \\ =& \int_{a}^{b} \frac{(b-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu _{1}-1}}y_{1}(s) \,ds +\frac{\sum_{i=1}^{\infty}\gamma_{i} \int_{a}^{b} \frac{(b-s)^{\alpha_{1}-\mu_{1}-1}(\xi_{i}-a)^{\alpha_{1}-\mu_{1}-1}}{\Gamma (\alpha_{1})(b-a)^{\alpha_{1}-\mu_{1}-1}}y_{1}(s)\,ds}{\Phi_{1}} \\ &{}-\frac{\sum_{i=1}^{\infty}\gamma_{i}\int _{a}^{\xi_{i}} \frac{(\xi_{i}-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha_{1})}y_{1}(s)\,ds}{\Phi_{1}}. \end{aligned}$$

Hence

$$\begin{aligned} u(t) =&- \int_{a}^{t} \frac{(t-s)^{\alpha_{1}-1}}{\Gamma(\alpha_{1})}y_{1}(s) \,ds+ \int_{a}^{b} \frac{(t-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{\Gamma(\alpha _{1})(b-a)^{\alpha_{1}-\mu_{1}-1}}y_{1}(s) \,ds \\ &{}+\frac{\sum_{i=1}^{\infty}\gamma_{i} \int_{a}^{b} \frac{(b-s)^{\alpha_{1}-\mu_{1}-1}(\xi_{i}-a)^{\alpha_{1}-\mu_{1}-1}}{\Gamma (\alpha_{1})(b-a)^{\alpha_{1}-\mu_{1}-1}}y_{1}(s)\,ds}{\Phi_{1}}+\frac{\sum_{i=1}^{\infty}\gamma_{i}\int_{a}^{b} G_{12}(\xi_{i},s)y_{1}(s)\,ds }{\Phi_{1}} \\ =& \int_{a}^{b} G_{11}(t,s)y_{1}(s) \,ds +\frac{(t-a)^{\alpha_{1}-1}\sum_{i=1}^{\infty}\gamma_{i} \int_{a}^{b} G_{12}(\xi_{i},s)y_{1}(s)\,ds}{\Phi_{1}} \\ =& \int_{a}^{b} G_{1}(t,s)y_{1}(s) \,ds. \end{aligned}$$

 □

Lemma 2.3

Suppose that ($\mathrm{H}_{0}$)-($\mathrm{H}_{2}$) are satisfied for $y_{2}\in C[a,b]$, then the fractional order BVP

$$\begin{aligned}& D^{\beta_{1}}_{a^{+}} \bigl(\varphi_{p}\bigl(D^{\alpha_{1}}_{a^{+}}u(t) \bigr)\bigr)=y_{2}(t), \quad t\in (a,b), \end{aligned}$$

(2.2)

$$\begin{aligned}& \left \{ \textstyle\begin{array} {l} u^{(j)}(a)=0,\quad j=0,1,2,\ldots,n-2,\qquad u^{(\mu_{1})}(b)=\sum_{i=1}^{\infty }\gamma_{i}u^{(\mu_{1})}(\xi_{i}),\\ \varphi_{p}(D^{\alpha_{1}}_{a^{+}}u(a))=0,\qquad D^{\mu_{2}}_{a^{+}}(\varphi_{p}(D^{\alpha_{1}}_{a^{+}}u(b)))=\sum_{i=1}^{\infty}\delta_{i} (D^{\mu_{2}}_{a^{+}}(\varphi_{p}(D^{\alpha _{1}}_{a^{+}}u(\eta_{i}))) ), \end{array}\displaystyle \right . \end{aligned}$$

(2.3)

has a unique solution

$$u(t)= \int_{a}^{b} G_{1}(t,s) \varphi_{q} \biggl( \int_{a}^{b} H_{1}(s,\tau)y_{2}( \tau)\,d\tau \biggr)\,ds, $$

where

$$\begin{aligned}& \begin{aligned} H_{1}(t,s)&= \left \{ \textstyle\begin{array} {l} \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(t-a)^{\beta_{1}-1}-(t-s)^{\beta _{1}-1}(b-a)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}}\\ \quad{}+\frac{(t-a)^{\beta_{1}-1}[\sum_{i=1}^{\infty}\delta_{i}[(b-s)(\eta _{i}-a)]^{\beta_{1}-\mu_{2}-1} -\sum_{i=1}^{\infty}\delta_{i}[(\eta_{i}-s)(b-a)]^{\beta_{1}-\mu_{2}-1}]}{\Delta _{1}\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}},\\ \quad\quad a\le s\le\min\{t,\eta_{1}\},\\ \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(t-a)^{\beta_{1}-1}-(t-s)^{\beta _{1}-1}(b-a)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}}\\ \quad{}+\frac{(t-a)^{\beta_{1}-1}[\sum_{i=1}^{\infty}\delta_{i}[(b-s)(\eta _{i}-a)]^{\beta_{1}-\mu_{2}-1} -\sum_{i=k}^{\infty}\delta_{i}[(\eta_{i}-s)(b-a)]^{\beta_{1}-\mu_{2}-1}]}{\Delta _{1}\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}},\\ \quad\quad\eta_{k-1}< s\le\eta_{k}\le t, k=2,3,\ldots,\\ \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(t-a)^{\beta_{1}-1}-(t-s)^{\beta _{1}-1}(b-a)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}}\\ \quad{}+\frac{(t-a)^{\beta_{1}-1}\sum_{i=1}^{\infty}\delta_{i}[(b-s)(\eta _{i}-a)]^{\beta_{1}-\mu_{2}-1} }{\Delta_{1}\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}},\\ \quad\quad\lim_{i\rightarrow\infty}\eta_{i}\le s\le t,\\ \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(t-a)^{\beta_{1}-1}}{\Gamma(\beta _{1})(b-a)^{\beta_{1}-\mu_{2}-1}}\\ \quad{}+\frac{(t-a)^{\beta_{1}-1}[\sum_{i=1}^{\infty}\delta_{i}[(b-s)(\eta _{i}-a)]^{\beta_{1}-\mu_{2}-1} -\sum_{i=1}^{\infty}\delta_{i}[(\eta_{i}-s)(b-a)]^{\beta_{1}-\mu_{2}-1}]}{\Delta _{1}\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}},\\ \quad\quad t\le s\le\eta_{1},\\ \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(t-a)^{\beta_{1}-1}}{\Gamma(\beta _{1})(b-a)^{\beta_{1}-\mu_{2}-1}}\\ \quad{}+\frac{(t-a)^{\beta_{1}-1}[\sum_{i=1}^{\infty}\delta_{i}[(b-s)(\eta _{i}-a)]^{\beta_{1}-\mu_{2}-1} -\sum_{i=k}^{\infty}\delta_{i}[(\eta_{i}-s)(b-a)]^{\beta_{1}-\mu_{2}-1}]}{\Delta _{1}\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}},\\ \quad\quad t\le\eta_{k-1}< s\le\eta_{k}, k=2,3,\ldots,\\ \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(t-a)^{\beta_{1}-1}}{\Gamma(\beta _{1})(b-a)^{\beta_{1}-\mu_{2}-1}}+\frac{(t-a)^{\beta_{1}-1}\sum_{i=1}^{\infty }\delta_{i}[(b-s)(\eta_{i}-a)]^{\beta_{1}-\mu_{2}-1} }{\Delta_{1}\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}},\\ \quad\quad\max\{t,\lim_{i\rightarrow+\infty}\eta_{i}\}\le s\le b, \end{array}\displaystyle \right . \\ &= H_{11}(t,s) +\frac{(t-a)^{\beta_{1}-1}\sum_{i=1}^{\infty}\delta_{i}H_{12}(\eta _{i},s)}{\Delta_{1}}, \end{aligned} \\& H_{11}(t,s)=\left \{ \textstyle\begin{array} {l@{\quad}l} \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(t-a)^{\beta_{1}-1}-(t-s)^{\beta _{1}-1}(b-a)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}},& s\le t,\\ \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(t-a)^{\beta_{1}-1}}{\Gamma(\beta _{1})(b-a)^{\beta_{1}-\mu_{2}-1}},& t\le s, \end{array}\displaystyle \right . \\& H_{12}(t,s)= \left \{ \textstyle\begin{array} {l@{\quad}l} \frac{[(b-s)(t-a)]^{\beta_{1}-\mu_{2}-1}- [(t-s)(b-a)]^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu _{2}-1}},& s\le t,\\ \frac{ [(b-s)(t-a) ]^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta _{1})(b-a)^{\beta_{1}-\mu_{2}-1}},& t\le s. \end{array}\displaystyle \right . \end{aligned}$$

Proof

It follows from Lemma 2.1 and $1<\beta_{1}\le2$ that

$$I_{a^{+}}^{\beta_{1}}D_{a^{+}}^{\beta_{1}}\bigl( \varphi_{p}\bigl(D_{a^{+}}^{\alpha _{1}}u(t)\bigr)\bigr)= \varphi_{p}\bigl(D_{a^{+}}^{\alpha_{1}}u(t) \bigr)+c_{1} (t-a)^{\beta_{1}-1}+c_{2} (t-a)^{\beta_{1}-2} \quad\text{for some }c_{1},c_{2}\in R. $$

Since (2.2), we get

$$I_{a^{+}}^{\beta_{1}}D_{a^{+}}^{\beta_{1}}\bigl( \varphi_{p}\bigl(D_{a^{+}}^{\alpha _{1}}u(t)\bigr) \bigr)=I_{a^{+}}^{\beta_{1}}y_{2}(t). $$

Then

$$\varphi_{p}\bigl(D_{a^{+}}^{\alpha_{1}}u(t) \bigr)=I_{a^{+}}^{\beta_{1}}y_{2}(t)+C_{1} (t-a)^{\beta_{1}-1}+C_{2} (t-a)^{\beta_{1}-2}\quad\text{for some }C_{1},C_{2}\in R. $$

Note that $\varphi_{p}(D^{\alpha_{1}}_{a^{+}}u(a))=0$, we have $C_{2}=0$, then

$$\begin{aligned} D^{\mu_{2}}_{a^{+}}\varphi_{p}\bigl(D_{a^{+}}^{\alpha_{1}}u(t) \bigr) =&D^{\mu _{2}}_{a^{+}}I_{a^{+}}^{\beta_{1}}y_{2}(t)+C_{1} D^{\mu_{2}}_{a^{+}}(t-a)^{\beta_{1}-1} \\ =&I_{a^{+}}^{\beta_{1}-\mu_{2}}y_{2}(t)+C_{1} \frac{\Gamma(\beta_{1})}{\Gamma(\beta _{1}-\mu_{2})} (t-a)^{\beta_{1}-\mu_{2}-1} \\ =& \int_{a}^{t} \frac{(t-\tau)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1}-\mu_{2})}y_{2}(\tau) \,d\tau +C_{1}\frac{\Gamma(\beta_{1})}{\Gamma(\beta_{1}-\mu_{2})} (t-a)^{\beta_{1}-\mu_{2}-1}. \end{aligned}$$

By $D^{\mu_{2}}_{a^{+}}(\varphi_{p}(D^{\alpha_{1}}_{a^{+}}u(b)))=\sum_{i=1}^{\infty}\delta_{i} (D^{\mu_{2}}_{a^{+}}(\varphi_{p}(D^{\alpha_{1}}_{a^{+}}u(\eta _{i}))) )$, we have

$$C_{1}=-\frac{1}{\Delta_{1}} \Biggl[ \int_{a}^{b} \frac{(b-\tau)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1})}y_{2}( \tau)\,d\tau -\sum_{i=1}^{\infty} \delta_{i} \int_{a}^{\eta_{i}} \frac{(\eta_{i}-\tau)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1})}y_{2}( \tau)\, d\tau \Biggr]. $$

Thus, the solution $u(t)$ of the fractional order BVP (2.2)-(2.3) satisfies

$$\begin{aligned} \varphi_{p}\bigl(D_{a^{+}}^{\alpha_{1}}u(t)\bigr) =& \int _{a}^{t}\frac{(t-\tau)^{\beta_{1}-1}}{\Gamma(\beta_{1})}y_{2}(\tau) \,d\tau\\ &{}-\frac {(t-a)^{\beta_{1}-1}}{\Delta_{1}} \\ &{}\times\Biggl[ \int _{a}^{b} \frac{(b-\tau)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1})}y_{2}( \tau)\,d\tau -\sum_{i=1}^{\infty} \delta_{i} \int _{a}^{\eta_{i}}\frac{ (\eta_{i}-\tau)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1})}y_{2}( \tau)\,d\tau \Biggr] \\ =& \int _{a}^{t}\frac{(t-\tau)^{\beta_{1}-1}}{\Gamma(\beta_{1})}y_{2}(\tau) \,d\tau\\ &{}- \biggl( \frac{1}{(b-a)^{\beta_{1}-\mu_{2}-1}}+\frac{\sum_{i=1}^{\infty}\delta_{i} (\eta_{i}-a)^{\beta_{1}-\mu_{2}-1}}{(b-a)^{\beta_{1}-\mu_{2}-1}\Delta_{1}} \biggr) \\ &{}\times(t-a)^{\beta_{1}-1} \int_{a}^{b} \frac{(b-\tau)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1})}y_{2}(\tau) \,d\tau \\ &{}+\frac{(t-a)^{\beta_{1}-1}\sum_{i=1}^{\infty}\delta_{i} \int_{a}^{\eta_{i}} \frac{(\eta_{i}-\tau)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1})}y_{2}(\tau)\,d\tau }{\Delta_{1}} \\ =&- \biggl( \int_{a}^{b} \frac{(b-\tau)^{\beta_{1}-\mu_{2}-1}(t-a)^{\beta_{1}-1}}{\Gamma(\beta _{1})(b-a)^{\beta_{1}-\mu_{2}-1}}y_{2}(\tau) \,d\tau- \int _{a}^{t}\frac{(t-\tau)^{\beta_{1}-1}}{\Gamma(\beta_{1})}y_{2}(\tau) \,d\tau \biggr) \\ &{}-\frac{(t-a)^{\beta_{1}-1}}{\Delta_{1}}\sum_{i=1}^{\infty} \delta_{i} \biggl( \int _{a}^{b}\frac{(b-\tau)^{\beta_{1}-\mu_{2}-1}(\eta_{i}-a)^{\beta_{1}-\mu_{2}-1}}{\Gamma (\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}} y_{2}( \tau)\,d\tau \\ &{}- \int_{a}^{\eta_{i}} \frac{(\eta_{i}-\tau)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1})}y_{2}( \tau)\,d\tau \biggr) \\ =&- \biggl( \int_{a}^{b} H_{11}(t, \tau)y_{2}(\tau)\,d\tau+\frac{(t-a)^{\beta_{1}-1}\sum_{i=1}^{\infty }\delta_{i}\int_{a}^{b} H_{12}(\eta_{i},\tau)y_{2}(\tau)\,d\tau}{\Delta_{1}} \biggr) \\ =&- \int_{a}^{b} H_{1}(t,\tau)y_{2}( \tau)\,d\tau. \end{aligned}$$

Then the fractional order BVP (2.2)-(2.3) is equivalent to the following problem:

$$\begin{aligned}& D_{a^{+}}^{\alpha_{1}}u(t)+\varphi_{q} \biggl( \int_{a}^{b} H_{1}(t,\tau)y_{2}( \tau)\,d\tau \biggr)=0\quad\text{for }t\in(a,b), \\& u^{(j)}(a)=0,\quad j=0,1,2,\ldots,n-2,\qquad u^{(\mu_{1})}(b)=\sum _{i=1}^{\infty}\gamma_{i}u^{(\mu_{1})}( \xi_{i}). \end{aligned}$$

In view of Lemma 2.2, we get

$$u(t)= \int_{a}^{b} G_{1}(t,s) \varphi_{q} \biggl( \int_{a}^{b} H_{1}(s,\tau)y_{2}( \tau)\,d\tau \biggr)\,ds. $$

 □

Lemma 2.4

Suppose ($\mathrm{H}_{0}$)-($\mathrm{H}_{2}$) hold. Then $G_{1}(t,s)$ has the following properties:

• (i) $0\le G_{1}(t,s)\le G_{1}(b,s)$ for all $(t,s)\in[a,b]\times[a,b]$,

• (ii) $G_{1}(t,s)\ge (\frac{1}{4} )^{\alpha_{1}-1}G_{1}(b,s)$ for all $(t,s)\in I\times(a,b)$, where $I= [\frac{3a+b}{4},\frac{a+3b}{4} ]$.

Proof

(i) For $t\le s$, we have

$$\frac{\partial}{\partial t} G_{11}(t,s)=\frac{(\alpha_{1}-1)(t-a)^{\alpha_{1}-2}(b-s)^{\alpha_{1}-\mu _{1}-1}}{\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu_{1}-1}}\ge 0. $$

For $t\ge s$, we get

$$\begin{aligned} \frac{\partial}{\partial t} G_{11}(t,s) =&\frac{(\alpha_{1}-1)[(t-a)^{\alpha_{1}-2}(b-s)^{\alpha_{1}-\mu _{1}-1}-(t-s)^{\alpha_{1}-2}(b-a)^{\alpha_{1}-\mu_{1}-1}]}{\Gamma(\alpha _{1})(b-a)^{\alpha_{1}-\mu_{1}-1}} \\ =&\frac{(\alpha_{1}-1)[(t-a)^{\mu_{1}-1}[(t-a)(b-s)]^{\alpha_{1}-\mu _{1}-1}-(t-s)^{\mu_{1}-1}[(t-s) (b-a)]^{\alpha_{1}-\mu_{1}-1}]}{\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu_{1}-1}} \\ \ge&\frac{(\alpha_{1}-1)(t-s)^{\mu_{1}-1}[((t-a)(b-s))^{\alpha_{1}-\mu_{1}-1}-((t-s) (b-a))^{\alpha_{1}-\mu_{1}-1}]}{\Gamma(\alpha_{1})(b-a)^{\alpha_{1}-\mu_{1}-1}}\ge0. \end{aligned}$$

Thus, for all $(t,s)\in[a,b]\times[a,b]$, we get

$$\frac{\partial}{\partial t} G_{1}(t,s)=\frac{\partial}{\partial t}G_{11}(t,s)+ \frac{(\alpha_{1}-1)(t-a)^{\alpha_{1}-2}\sum_{i=1}^{\infty}\gamma _{i}G_{12}(\xi_{i},s) }{\Phi_{1}}\ge0. $$

Then $0=G_{1}(a,s) \le G_{1}(t,s)\le G_{1}(b,s)$, $\forall(t,s)\in[a,b]\times[a,b]$.

(ii) For $t\le s$, we have

$$\frac{G_{11}(t,s)}{G_{11}(b,s)} = \frac{(t-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}}{(b-a)^{\alpha _{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}} = \biggl(\frac{t-a}{b-a} \biggr)^{\alpha_{1}-1}\quad \text{for }s\in(a,b). $$

For $t\ge s$, we get

$$\begin{aligned} \frac{G_{11}(t,s)}{G_{11}(b,s)} =& \frac{(t-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}-(t-s)^{\alpha _{1}-1}(b-a)^{\alpha_{1}-\mu_{1}-1}}{ (b-a)^{\alpha_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}-(b-s)^{\alpha _{1}-1}(b-a)^{\alpha_{1}-\mu_{1}-1}} \\ \ge& \frac{(t-a)^{\alpha_{1}-\mu_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}[(t-a)^{\mu _{1}}-(t-s)^{\mu_{1}}]}{ (b-a)^{\alpha_{1}-\mu_{1}-1}(b-s)^{\alpha_{1}-\mu_{1}-1}[(b-a)^{\mu _{1}}-(b-s)^{\mu_{1}}]} \\ \ge& \biggl(\frac{t-a}{b-a} \biggr)^{\alpha_{1}-1}\quad \text{for }s \in(a,b). \end{aligned}$$

Thus,

$$G_{11}(t,s)\ge \biggl(\frac{t-a}{b-a} \biggr)^{\alpha_{1}-1} G_{11}(b,s)\quad\text{for all } (t,s)\in[a,b]\times(a,b). $$

Then

$$\begin{aligned} G_{1}(t,s) =& G_{11}(t,s) +\frac{(t-a)^{\alpha_{1}-1}\sum_{i=1}^{\infty}\gamma_{i} G_{12}(\xi_{i},s)}{\Phi_{1}} \\ \ge& \biggl(\frac{t-a}{b-a} \biggr)^{\alpha_{1}-1} G_{11}(b,s)+ \biggl(\frac{t-a}{b-a} \biggr)^{\alpha_{1}-1}\frac{(b-a)^{\alpha _{1}-1}\sum_{i=1}^{\infty}\gamma_{i} G_{12}(\xi_{i},s)}{\Phi_{1}} \\ =& \biggl(\frac{t-a}{b-a} \biggr)^{\alpha_{1}-1} G_{1}(b,s)\ge \biggl(\frac{1}{4} \biggr)^{\alpha_{1}-1}G_{1}(b,s)\quad \text{for all }(t,s)\in I\times(a,b). \end{aligned}$$

 □

Lemma 2.5

Suppose ($\mathrm{H}_{0}$)-($\mathrm{H}_{2}$) hold. Then $H_{1}(t,s)$ has the following properties:

• (i) $0\le H_{1}(t,s)\le\omega_{1}(s)$ for all $(t,s)\in[a,b]\times[a,b]$, where

  $$\omega_{1}(s)= H_{11}(s,s) +\frac{(b-a)^{\beta_{1}-1}\sum_{i=1}^{\infty}\delta_{i}H_{12}(\eta _{i},s)}{\Delta_{1}}, $$

• (ii) $H_{1}(t,s)\ge h_{1}(s)H_{11}(s,s)$ for all $(t,s)\in I\times(a,b)$, where $I=[\frac{3a+b}{4},\frac{a+3b}{4}]$, and for $\zeta_{1}\in I$,

  $$h_{1}(s)=\left \{ \textstyle\begin{array} {l@{\quad}l} \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(\frac{3}{4})^{\beta_{1}-1}-(\frac {a+3b}{4}-s)^{\beta_{1}-1}(b-a)^{-\mu_{2}}}{(b-s)^{\beta_{1}-\mu_{2}-1}},& s\in (a,\zeta_{1}],\\ (\frac{1}{4})^{\beta_{1}-1},&s\in[\zeta_{1},b). \end{array}\displaystyle \right .$$

Proof

(i) For $t\le s$, it is easy to show that $\frac{\partial}{\partial t} H_{11}(t,s) \ge0$ for all $(t,s)\in[a,b]\times[a,b]$, then

$$0=H_{11}(a,s) \le H_{11}(t,s)\le H_{11}(s,s)\quad\text{for }t\le s. $$

For $t\ge s$, we have

$$\begin{aligned} \frac{\partial}{\partial t} H_{11}(t,s) =&\frac{(\beta_{1}-1)[(t-a)^{\beta_{1}-2}(b-s)^{\beta_{1}-\mu _{2}-1}-(t-s)^{\beta_{1}-2}(b-a)^{\beta_{1}-\mu_{2}-1}]}{\Gamma(\beta _{1})(b-a)^{\beta_{1}-\mu_{2}-1}} \\ \le&\frac{(\beta_{1}-1)(t-s)^{\beta_{1}-2}[(b-s)^{\beta_{1}-\mu _{2}-1}-(b-a)^{\beta_{1}-\mu_{2}-1}]}{\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu _{2}-1}}\le 0. \end{aligned}$$

Thus,

$$0=H_{11}(a,s)\le H_{11}(t,s)\le H_{11}(s,s)\quad \text{for }t\ge s. $$

Hence, for all $(t,s)\in[a,b]\times[a,b]$, by $\mu_{2}\le \beta_{1}-1$ and $(b-s)(t-a)\ge(t-s)(b-a)$, we can easily show that $H_{12}(t,s)\ge0$, then

$$\begin{aligned} 0 \le& H_{1}(t,s)\\ =& H_{11}(t,s) +\frac{(t-a)^{\beta_{1}-1}\sum_{i=1}^{\infty}\delta_{i}H_{12}(\eta _{i},s)}{\Delta_{1}} \\ \le& H_{11}(s,s) +\frac{(b-a)^{\beta_{1}-1}\sum_{i=1}^{\infty}\delta_{i}H_{12}(\eta _{i},s)}{\Delta_{1}} \\ =&\omega_{1}(s). \end{aligned}$$

(ii) Setting:

$$\begin{aligned}& h_{1}(t,s)=\frac{(b-s)^{\beta_{1}-\mu_{2}-1}(t-a)^{\beta_{1}-1}-(t-s)^{\beta _{1}-1}(b-a)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}}, \\& h_{2}(t,s)=\frac{(b-s)^{\beta_{1}-\mu_{2}-1}(t-a)^{\beta_{1}-1}}{\Gamma(\beta _{1})(b-a)^{\beta_{1}-\mu_{2}-1}}. \end{aligned}$$

Then

$$\begin{aligned} \min_{t\in I}H_{11}(t,s) =&\left \{ \textstyle\begin{array} {l@{\quad}l} h_{1}(\frac{a+3b}{4},s),& s\in(a,\frac{3a+b}{4}],\\ \min\{h_{1}(\frac{a+3b}{4},s),h_{2}(\frac{3a+b}{4},s)\}, & s\in[\frac {3a+b}{4},\frac{a+3b}{4}],\\ h_{2}(\frac{3a+b}{4},s),& s\in[\frac{a+3b}{4},b), \end{array}\displaystyle \right . \\ =& \left\{ \textstyle\begin{array} {l@{\quad}l} h_{1}(\frac{a+3b}{4},s),& s\in(a,\zeta_{1}],\\ h_{2}(\frac{3a+b}{4},s),& s\in[\zeta_{1},b), \end{array}\displaystyle \right . \\ =& \left\{ \textstyle\begin{array} {l@{\quad}l} \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(\frac{3}{4}(b-a))^{\beta_{1}-1}-(\frac {a+3b}{4}-s)^{\beta_{1}-1}(b-a)^{\beta_{1}-\mu_{2}-1}}{\Gamma(\beta _{1})(b-a)^{\beta_{1}-\mu_{2}-1}},& s\in(a,\zeta_{1}],\\ \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(\frac{1}{4}(b-a))^{\beta_{1}-1}}{\Gamma (\beta_{1})(b-a)^{\beta_{1}-\mu_{2}-1}},& s\in[\zeta_{1},b), \end{array}\displaystyle \right . \\ =& \left\{ \textstyle\begin{array} {l@{\quad}l} \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(\frac{3}{4}(b-a))^{\beta_{1}-1}-(\frac {a+3b}{4}-s)^{\beta_{1}-1}(b-a)^{\beta_{1}-\mu_{2}-1}}{(b-s)^{\beta_{1}-\mu _{2}-1}(s-a)^{\beta_{1}-1}} H_{11}(s,s),& s\in(a,\zeta_{1}],\\ \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(\frac{1}{4}(b-a))^{\beta _{1}-1}}{(b-s)^{\beta_{1}-\mu_{2}-1}(s-a)^{\beta_{1}-1}}H_{11}(s,s),&s\in[\zeta_{1},b), \end{array}\displaystyle \right . \\ \ge& \left\{ \textstyle\begin{array} {l@{\quad}l} \frac{(b-s)^{\beta_{1}-\mu_{2}-1}(\frac{3}{4})^{\beta_{1}-1}-(\frac {a+3b}{4}-s)^{\beta_{1}-1}(b-a)^{-\mu_{2}}}{(b-s)^{\beta_{1}-\mu _{2}-1}}H_{11}(s,s),& s\in(a,\zeta_{1}],\\ (\frac{1}{4})^{\beta_{1}-1}H_{11}(s,s),&s\in[\zeta_{1},b), \end{array}\displaystyle \right . \\ =&h_{1}(s)H_{11}(s,s)\quad\text{for }s\in(a,b). \end{aligned}$$

Therefore, for all $(t,s)\in I\times(a,b)$, we have

$$H_{1}(t,s)= H_{11}(t,s) +\frac{(t-a)^{\beta_{1}-1}\sum_{i=1}^{\infty}\delta_{i}H_{12}(\eta _{i},s)}{\Delta_{1}}\ge H_{11}(t,s)\ge h_{1}(s)H_{11}(s,s). $$

We complete the proof. □

Remark 2.6

In a similar manner, the results of the Green’s function $G_{2}(t,s)$ and $H_{2}(t,s)$ for the homogeneous BVP corresponding to the fractional differential equation (1.2) and (1.4) are obtained. Consider the following conditions:

• (i) $G_{i}(t,s)\ge\Lambda G_{i}(b,s)$ for all $(t,s)\in I\times(a,b)$, $i=1,2$,

• (ii) $H_{i}(t,s)\ge h(s) H_{i1}(s,s)$ for all $(t,s)\in I\times(a,b)$, $i=1,2$,

where $I= [\frac{3a+b}{4},\frac{a+3b}{4} ]$, $\Lambda=\min\{(\frac{1}{4})^{\alpha_{1}-1},(\frac{1}{4})^{\alpha_{2}-1}\}$, $h(s)=\min\{h_{1}(s), h_{2}(s)\}$.

Let c, d, e, r be positive real numbers, $K_{r}=\{x\in K: \|x\|< r\}$, $K(\psi,d,e)=\{x\in K:{ d\le\psi(x)},\|x\|\le e\}$.

Lemma 2.7

[16]

Let K be a cone in a real Banach space E, $K_{r}=\{x\in K: \|x\|< r\}$, ψ be a nonnegative continuous concave functional on K such that $\psi(x)\le\|x\|$, $\forall x\in\overline{K}_{r}$ and $K(\psi,d,e)=\{x\in K: d\le \psi(x),\|x\|\le e\}$. Suppose $T :\overline{K}_{r} \to\overline{K}_{r}$ is completely continuous and there exist constants $0 < c < d < e\le r$ such that

• (i) $\{x\in K(\psi,d,e)|\psi(x)>d\} \neq\emptyset$ and $\psi (Tx)>d$ for $x\in K(\psi,d,e)$;

• (ii) $\|Tx\|< c$ for $x \le c$;

• (iii) $\psi(Tx)>d$ for $x\in K(\psi,d,r)$ with $\|Tx\|>e$.

Then T has at least three fixed points $x_{1}$, $x_{2}$ and $x_{3}$ with $\|x_{1}\|< c$, $d<\psi(x_{2})$, $c<\|x_{3}\|$ with $\psi(x_{3})< d$.

3 The main result and proof

Let the Banach space $E = C[a, b]\times C[a, b]$ be endowed with the norm $\|(u, v)\|=\|u \|+\|v \|$ for $(u,v)\in E$ and $\|x\|=\max_{a\le t\le b} |x(t)|$. Define a cone $K\in E$ by

$$K= \Bigl\{ (u,v)\in E : \min_{a\le t\le b}u(t)\ge0, \min _{a\le t\le b}v(t)\ge0\text{ and }\min_{t\in I} \bigl\{ u(t)+v(t)\bigr\} \ge \Lambda\big\| (u,v)\big\| \Bigr\} , $$

where $I= [\frac{3a+b}{4},\frac{a+3b}{4} ]$.

It is well known that the system of fractional order BVP (1.1)-(1.4) is equivalent to

$$\begin{aligned}& u(t)= \int_{a}^{b} G_{1}(t,s) \varphi_{q} \biggl( \int_{a}^{b} H_{1}(s,\tau)f_{1} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr)\,ds, \\& v(t)= \int_{a}^{b} G_{2}(t,s) \varphi_{q} \biggl( \int_{a}^{b} H_{2}(s,\tau)f_{2} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr)\,ds. \end{aligned}$$

Define the operators $T_{1},T_{2} :K \to E$ by

$$\begin{aligned}& T_{1}(u,v) (t)= \int_{a}^{b} G_{1}(t,s) \varphi_{q} \biggl( \int_{a}^{b} H_{1}(s,\tau)f_{1} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr)\,ds, \\& T_{2}(u,v) (t)= \int_{a}^{b} G_{2}(t,s) \varphi_{q} \biggl( \int_{a}^{b} H_{2}(s,\tau)f_{2} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr)\,ds, \end{aligned}$$

and the operator $T:K\to E$ by

$$T(u,v)=\bigl(T_{1}(u,v),T_{2}(u,v)\bigr)\quad\text{for } (u,v)\in E. $$

It is clear that the existence of a positive solution to system (1.1)-(1.4) is equivalent to the existence of fixed points of the operator T.

Lemma 3.1

$T:K\to K$ is completely continuous.

Proof

The continuity of functions $G_{i}(t,s)$, $H_{i}(t,s)$ and $f_{i}(t,u(t),v(t))$ for $i=1,2$ implies that $T: K\to K$ is continuous. For all $(t,s)\in I\times[a,b]$ (where $I= [\frac{3a+b}{4},\frac{a+3b}{4} ]$), we have

$$\begin{aligned} \min_{t\in I} \bigl\{ T_{1}(u,v) (t)+T_{2}(u,v) (t) \bigr\} =&\min_{t\in I} \biggl\{ \int_{a}^{b} G_{1}(t,s) \varphi_{q} \biggl( \int_{a}^{b} H_{1}(s,\tau)f_{1} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr) \,ds \\ &{} + \int_{a}^{b} G_{2}(t,s) \varphi_{q} \biggl( \int_{a}^{b} H_{2}(s,\tau)f_{2} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr)\,ds \biggr\} \\ \ge& \Lambda \biggl\{ \int_{a}^{b} G_{1}(b,s) \varphi_{q} \biggl( \int_{a}^{b} H_{1}(s,\tau)f_{1} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr) \,ds \\ &{} + \int_{a}^{b} G_{2}(b,s) \varphi_{q} \biggl( \int_{a}^{b} H_{2}(s,\tau)f_{2} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr)\,ds \biggr\} \\ \ge&\Lambda\bigl(\big\| T_{1}(u,v)\big\| +\big\| T_{2}(u,v)\big\| \bigr) \\ =&\Lambda\big\| \bigl(T_{1}(u,v),T_{2}(u,v)\bigr)\big\| \\ =&\Lambda\big\| T(u,v)\big\| . \end{aligned}$$

Thus, $T(K)\subset K$. So, we can easily show that $T:K\to K$ is completely continuous by the Arzela-Ascoli theorem. The proof is completed. □

Let the nonnegative continuous concave functional ψ be defined on the cone K by

$$\psi(u,v)=\min_{t\in I}\big|u(t)+v(t)\big|,\quad\text{where } I= \biggl[ \frac{3a+b}{4},\frac{a+3b}{4} \biggr]. $$

For convenience, we denote

$$\begin{aligned} N =&\max \biggl\{ \biggl( \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} G_{1}(b,s) \varphi_{q} \biggl( \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} h(\tau)H_{11}(\tau,\tau)\,d \tau \biggr) \,ds \biggr)^{-1}, \\ & \biggl( \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} G_{2}(b,s) \varphi_{q} \biggl( \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} h(\tau)H_{21}(\tau,\tau)\,d \tau \biggr)\,ds \biggr)^{-1} \biggr\} , \end{aligned}$$

and

$$\begin{aligned} M =& \min \biggl\{ \biggl( \int_{a}^{b} G_{1}(b,s) \varphi_{q} \biggl( \int_{a}^{b} \omega_{1}(\tau)\,d\tau \biggr)\,ds \biggr)^{-1},\\ & \biggl( \int_{a}^{b} G_{2}(b,s) \varphi_{q} \biggl( \int_{a}^{b} \omega_{2}(\tau)\,d\tau \biggr)\,ds \biggr)^{-1} \biggr\} . \end{aligned}$$

Theorem 3.2

Suppose that ($\mathrm{H}_{0}$)-($\mathrm{H}_{2}$) hold. If there exist positive real numbers $0< c< d<\Lambda r$ such that the following conditions hold:

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

$f_{i}(t,u,v)< \varphi_{p}(\frac{r M}{2})$ for $i=1,2$, for all $t\in[a,b]$, $(u,v)\in[0,r]\times[0,r]$;

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

$f_{i}(t,u,v)> \varphi_{p}(\frac{d N}{2\Lambda})$ for $i=1,2$, for all $t\in I= [\frac{3a+b}{4},\frac{a+3b}{4} ]$, $(u,v)\in [d,\frac{d}{\Lambda}]\times[d,\frac{d}{\Lambda}]$;

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

$f_{i}(t,u,v)<\varphi_{p}(\frac{ c M}{2})$ for $i=1,2$, for all $t\in[a,b]$, $(u,v)\in[0,c]\times[0,c]$.

Then the system of fractional differential equations BVP (1.1)-(1.4) has at least three positive solutions $(u_{1},v_{1})$, $(u_{2},v_{2})$ and $(u_{3},v_{3})$ with

$$\begin{aligned}& \max_{a\le t\le b}\big|u_{1}(t)+v_{1}(t)\big|< c, \qquad d< \min_{t\in I}\big|u_{2}(t)+v_{2}(t)\big|< \max _{a\le t\le b}\big|u_{2}(t)+v_{2}(t)\big|< r,\\& c< \max _{a\le t\le b}\big|u_{3}(t)+v_{3}(t)\big|< r,\qquad\min _{t\in I}\big|u_{3}(t)+v_{3}(t)\big|< d. \end{aligned}$$

Proof

Firstly, if $(u,v)\in\overline{K}_{r}$, then we may assert that $T:\overline{K}_{r}\to\overline{K}_{r}$ is a completely continuous operator. To see this, suppose $(u,v)\in\overline{K}_{r}$, then $\|(u,v)\|\le r$. It follows from Lemma 2.4, Lemma 2.5 and ($\mathrm{H}_{3}$) that

$$\begin{aligned} \big\| T(u,v)\big\| =&\max_{a\le t\le b}\big|T(u,v) (t)\big| \\ =&\max_{a\le t\le b} \biggl( \int_{a}^{b} G_{1}(t,s) \varphi_{q} \biggl( \int_{a}^{b} H_{1}(s,\tau)f_{1} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr) \,ds \\ &{} + \int_{a}^{b} G_{2}(t,s) \varphi_{q} \biggl( \int_{a}^{b} H_{2}(s,\tau)f_{2} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr)\,ds \biggr) \\ \le& \frac{r M}{2} \biggl( \int_{a}^{b} G_{1}(b,s) \varphi_{q} \biggl( \int_{a}^{b} \omega_{1}(\tau)\,d\tau \biggr)\, ds+ \int_{a}^{b} G_{2}(b,s) \varphi_{q} \biggl( \int_{a}^{b} \omega_{2}(\tau)\,d\tau \biggr)\,ds \biggr) \\ \le& \frac{r }{2}+\frac{r }{2}=r. \end{aligned}$$

Therefore, $T:\overline{K}_{r}\to\overline{K}_{r}$. This together with Lemma 3.1 implies that $T:\overline{K}_{r}\to \overline{K}_{r}$ is a completely continuous operator. In the same way, if $(u,v)\in\overline{K}_{c}$, then assumption ($\mathrm{H}_{5}$) yields $\|T(u,v)\|< c$. Hence, condition (ii) of Lemma 2.7 is satisfied.

To check condition (i) of Lemma 2.7, we let $u(t)+v(t)=\frac{d}{\Lambda}$ for $t\in[a,b]$. It is easy to verify that $u(t)+v(t)=\frac{d}{\Lambda}\in K(\psi,d,\frac{d}{\Lambda})$ and $\psi(u,v)=\frac{d}{\Lambda}>d$, and so $\{(u,v)\in K(\psi,d,\frac{d}{\Lambda}):\psi(u,v)>d\}\neq\emptyset$. Thus, for all $(u,v)\in K(\psi,d,\frac{d}{\Lambda})$, we have that $d\le u(t)+v(t)\le\frac{d}{\Lambda}$ for $t\in I$ and $T(u,v)\in K$. From Lemma 2.4, Lemma 2.5 and ($\mathrm{H}_{4}$), one has

$$\begin{aligned} \psi\bigl( T(u,v) (t)\bigr) =&\min_{t\in I}\big|T(u,v) (t)\big| \\ =& \min_{t\in I} \biggl( \int_{a}^{b} G_{1}(t,s) \varphi_{q} \biggl( \int_{a}^{b} H_{1}(s,\tau)f_{1} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr) \,ds \\ &{} + \int_{a}^{b} G_{2}(t,s) \varphi_{q} \biggl( \int_{a}^{b} H_{2}(s,\tau)f_{2} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr)\,ds \biggr) \\ \ge& \biggl( \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} \Lambda G_{1}(b,s) \varphi_{q} \biggl( \int_{a}^{b} H_{1}(s,\tau)f_{1} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr) \,ds \\ &{} + \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} \Lambda G_{2}(b,s) \varphi_{q} \biggl( \int_{a}^{b} H_{2}(s,\tau)f_{2} \bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr)\,ds \biggr) \\ \ge& \biggl( \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} \Lambda G_{1}(b,s) \varphi_{q} \biggl( \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} h(\tau)H_{11}(\tau, \tau)f_{1}\bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr) \,ds \\ &{} + \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} \Lambda G_{2}(b,s) \varphi_{q} \biggl( \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} h(\tau) H_{21}(\tau, \tau)f_{2}\bigl(\tau,u(\tau),v(\tau)\bigr)\,d\tau \biggr)\,ds \biggr) \\ >&\frac{d N}{2\Lambda} \biggl( \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} \Lambda G_{1}(b,s) \varphi_{q} \biggl( \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} h(\tau)H_{11}(\tau,\tau)\,d \tau \biggr) \,ds \\ &{} + \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} \Lambda G_{2}(b,s) \varphi_{q} \biggl( \int_{\frac{3a+b}{4}}^{\frac{a+3b}{4}} h(\tau)H_{21}(\tau,\tau)\,d \tau \biggr)\,ds \biggr) \\ \ge&\frac{d}{2}+\frac{d}{2}= d. \end{aligned}$$

This shows that condition (i) of Lemma 2.7 holds.

Secondly, we verify that (iii) of Lemma 2.7 is satisfied. By Lemma 3.1, we have

$$\min_{{t\in I}}\big|T(u,v) (t)\big|\ge \Lambda\big\| T(u,v)\big\| >d\quad\text{for }(u,v)\in K(\psi,d,r)\text{ with } \big\| T(u,v)\big\| >\frac{d}{\Lambda}, $$

which shows that condition (iii) of Lemma 2.7 holds.

To sum up, all the conditions of Lemma 2.7 are satisfied, it follows from Lemma 2.7 that there exist three positive solutions $(u_{1},v_{1})$, $(u_{2},v_{2})$ and $(u_{3},v_{3})$ satisfying

$$\begin{aligned}& \max_{a\le t\le b}\big|u_{1}(t)+v_{1}(t)\big|< c,\qquad d< \min_{t\in I}\big|u_{2}(t)+v_{2}(t)\big|< \max _{a\le t\le b}\big|u_{2}(t)+v_{2}(t)\big|< r,\\& c< \max _{a\le t\le b}\big|u_{3}(t)+v_{3}(t)\big|< r,\qquad\min _{t\in I}\big|u_{3}(t)+v_{3}(t)\big|< d. \end{aligned}$$

 □

4 Example

Example 1

Consider the system of fractional differential equations BVP

$$\begin{aligned}& D^{\frac{3}{2}}_{0^{+}} \bigl(\varphi_{p} \bigl(D^{\frac{5}{2}}_{0^{+}}u(t)\bigr)\bigr)=f_{1} \bigl(t,u(t),v(t)\bigr),\quad t\in(0,1), \end{aligned}$$

(4.1)

$$\begin{aligned}& D^{\frac{3}{2}}_{0^{+}} \bigl(\varphi_{p} \bigl(D^{\frac{5}{2}}_{0^{+}}v(t)\bigr)\bigr)=f_{2} \bigl(t,u(t),v(t)\bigr),\quad t\in(0,1), \end{aligned}$$

(4.2)

with the boundary conditions

$$\begin{aligned}& \left\{ \textstyle\begin{array} {l} u(0)=u'(0)=0,\qquad u^{(\frac{5}{4})}(1)=\frac{1}{4} u^{(\frac {5}{4})}(\frac{1}{4}),\\ \varphi_{p}(D^{\frac{5}{2}}_{0^{+}}u(0))=0,\qquad D^{\frac{1}{4}}_{0^{+}}(\varphi_{p}(D^{\frac{5}{2}}_{0^{+}}u(1)))=\frac {1}{3}D^{\frac{1}{4}}_{0^{+}}(\varphi_{p}(D^{\frac{5}{2}}_{0^{+}}u(\frac {1}{4}))), \end{array}\displaystyle \right . \end{aligned}$$

(4.3)

$$\begin{aligned}& \left\{ \textstyle\begin{array} {l} v(0)=v'(0)=0,\qquad v^{(\frac{5}{4})}(1)=\frac{1}{4} v^{(\frac {5}{4})}(\frac{1}{4}),\\ \varphi_{p}(D^{\frac{5}{2}}_{0^{+}}v(0))=0,\qquad D^{\frac{1}{4}}_{0^{+}}(\varphi_{p}(D^{\frac{5}{2}}_{0^{+}}v(1)))=\frac {1}{3}D^{\frac{1}{4}}_{0^{+}}(\varphi_{p}(D^{\frac{5}{2}}_{0^{+}}v(\frac {1}{4}))), \end{array}\displaystyle \right . \end{aligned}$$

(4.4)

where

$$\begin{aligned}& f_{1}(t,u,v)=\left \{ \textstyle\begin{array} {l@{\quad}l} \frac{4(u+v)^{2}}{5(1+t^{2})},&0\le t\le1,0\le u+v \le\frac {1}{2}, \\ \frac{18{,}559.6(u+v)-9{,}279.6}{1+t^{2}},&0\le t\le1,\frac {1}{2}< u+v\le1, \\ \frac{ 145}{(1+t^{2})2^{u+v-7}},&0\le t\le1, u+v>1, \end{array}\displaystyle \right .\\& f_{2}(t,u,v)= \left\{ \textstyle\begin{array} {l@{\quad}l} \frac{1}{2}(1-t) (u+v),&0\le t\le1,0\le u+v \le\frac {1}{2}, \\ (1-t) (11{,}839.5(u+v)-5{,}919.5),&0\le t\le1,\frac{1}{2}< u+v\le1, \\ 92.5(1-t)2^{-(u+v)+7},&0\le t\le1, u+v>1. \end{array}\displaystyle \right . \end{aligned}$$

We note that $a=0$, $b=1$, $n=m=3$, $\alpha_{1}=\alpha_{2}=\frac{5}{2}$, $\beta_{1}=\beta_{2}=\frac{3}{2}$, $\mu_{1}=\mu_{3}=\frac{5}{4}$, $\gamma_{1}=\eta_{1}=\xi_{1}=\mu_{2}=\mu_{4}=\frac{1}{4}$, $\delta_{1}=\frac{1}{3}$, $\gamma_{i}=\delta_{i}=0$ ($i=2,3,4,\ldots$). Let $p=2$, an easy computation shows that $\Phi_{1}=\Phi_{2}\approx0.8232$, $\Delta_{1}=\Delta_{2}\approx0.7643$, $\Lambda=0.125$, $N\approx23.1092$, $M\approx2.6109$. Then, if we choose $c=\frac{1}{4}$, $d=\frac{1}{2}$, $r=7{,}109$, then $f_{i}(t,u,v)$ for $i=1,2$ satisfies the following conditions:

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

$f_{i}(t,u,v)< \varphi_{p}(\frac{r M}{2})\approx9{,}280.44$ for $i=1,2$, for all $t\in[0,1]$, $(u,v)\in [0,7{,}109]\times[0,7{,}109]$;

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

$f_{i}(t,u,v)> \varphi_{p}(\frac{d N}{2\Lambda})=46.2184$ for $i=1,2$, for all $t\in [\frac{1}{4},\frac{3}{4}]$, $(u,v)\in [\frac{1}{2},4]\times[\frac{1}{2},4]$;

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

$f_{i}(t,u,v)<\varphi_{p}(\frac{ c M}{2})\approx0.3264$ for $i=1,2$, for all $t\in[0,1]$, $(u,v)\in [0,\frac{1}{4}]\times[0,\frac{1}{4}]$.

Thus, all the hypotheses of Theorem 3.2 are satisfied. Hence, the system of fractional differential equations BVP (4.1)-(4.4) has at least three positive solutions.