The method of upper and lower solutions for a class of fractional differential coupled systems

Abstract

In this paper, we investigate a class of nonlocal boundary value problems of nonlinear fractional functional differential coupled systems with state dependent delays. The method of upper and lower solutions is established and some new results for the multiplicity of solutions of the boundary value problem are obtained. An example is also presented to illustrate our main results.

1 Introduction

It is well know that fractional calculus is the generalization of classical calculus from integer to real numbers, even in the complex field. In the past decades, fractional differential equations have been widely used in various research fields, such as chemical engineering, automatic control, and thermoelasticity. In consequence, the theoretical researches of fractional differential equations have been highly valued by more and more scholars; see [1–6].

When there are many state variables depending on each other in the system, the systems are often described as coupled systems; for details and examples, see [7–20] and the references therein. In recent years, the method of upper and lower solutions has played a more and more important role in the theoretical studies of differential equations. There are a large number of publications using the method of upper and lower solutions to study the existence and uniqueness of solutions of fractional differential equations; see [21–25] and the references therein.

In [10], the authors considered the three-point boundary value problems of nonlinear fractional coupled systems,

$$ \textstyle\begin{cases} D_{0^{+} }^{\alpha }u(t) = f(t,v(t),D_{0^{+} }^{\beta - 1} v(t)),\quad 0 < t < 1, \\ D_{0^{+} }^{\beta }v(t) = g(t,u(t),D_{0^{+} }^{\alpha - 1} u(t)),\quad 0 < t < 1, \\ u(0) = v(0) = 0,\qquad u(1) = \sigma _{1} u(\eta _{1} ),\qquad v(1) = \sigma _{2} v(\eta _{2} ), \end{cases} $$

where $D_{0^{+} }^{\alpha }$ is the Riemann–Liouville differentiation, $1 < \alpha$, $\beta \le 2$, $0 < \eta _{1}$, $\eta _{2} \le 1$, $\sigma _{1} ,\sigma _{2} > 0$, $\sigma _{1} \eta _{1} ^{\alpha - 1} = \sigma _{2} \eta _{2} ^{\beta - 1}$, $f,g \in C([0,1] \times \mathbb{R}^{2} ,\mathbb{R})$. The existence theorems of two solutions of boundary value problems at resonance are given by using the coincidence degree theory.

In scientific research, if the system is assumed to be controlled by an equation containing the current state and the rate of change of state, then we consider ordinary or partial differential equations. However, in some applications, the system may also include the past state of the system. In this case, it would be more accurate to describe them by functional differential equations; see [26–28]. In recent years, the boundary value problems of the fractional functional differential equations have attracted the attention of researchers and many research results have been obtained; see [29–31].

At the same time, some problems may have already occurred for some time before we begin to study them, such as infectious diseases. In this way, we have to consider the values of state variables for the previous period of time. The purpose of this paper is to study the existence of multiple solutions for nonlocal boundary value problems of fractional functional differential coupled systems with time delays,

$$ \textstyle\begin{cases} D_{0^{+} }^{\alpha }u(t) + f(t,v(t),v_{t} ) = 0,\quad t \in (0,1), \\ D_{0^{+} }^{\beta }v(t) + g(t,u(t),u_{t} ) = 0,\quad t \in (0,1), \\ u(t) = \phi (t),\qquad v(t) = \psi (t),\quad t \in [ - \tau ,0], \\ D_{0^{+} }^{\gamma _{1} } u(1) = aD_{0^{+} }^{\gamma _{1} } u(\xi ), \qquad D_{0^{+} }^{\gamma _{2} } v(1) = bD_{0^{+} }^{\gamma _{2} } v( \eta ), \end{cases} $$

(1.1)

where $0 < \gamma _{1}$, $\gamma _{2} \le 1$, $1 + \gamma _{1} \le \alpha \le 2$, $1 + \gamma _{2} \le \beta \le 2$, $\xi ,\eta \in (0,1)$, $a,b \in \mathbb{R}$. $D_{0^{+} }^{\alpha }$, $D_{0^{+} }^{\beta }$, $D_{0^{+} }^{\gamma _{1} }$, $D_{0^{+} }^{\gamma _{2} }$ are the Riemann–Liouville fractional derivative operators. The functions $f,g \in C([0,1] \times \mathbb{R}\times C[-\tau ,0])$. $u_{t}=u(t+\theta )$, $v_{t}=v(t+\theta )$, $\theta \in [ - \tau ,0]$, $\phi , \psi \in C([ - \tau ,0])$ and $\phi (0) = \psi (0) = 0$. Some new results for the existence of at least three solutions for the coupled system are established by using upper and lower solutions methods.

2 Preliminaries

In this section, we present some necessary definitions and lemmas which will be used in the proof of our main results.

Definition 2.1

(see [1])

The Riemann–Liouville fractional integral of a function $h:(0,\infty ) \to \mathbb{R}$ of order $\alpha > 0$ is given by

$$ I_{0^{+} }^{\alpha }h(t) = \frac{1}{{\Gamma (\alpha )}} \int _{0}^{t} {(t - s)^{\alpha - 1} h(s)\,ds}, $$

provided the right side is pointwise defined on $(0,\infty )$.

Definition 2.2

(see [1])

The Riemann–Liouville fractional derivative of order $\alpha > 0$ of a continuous function $h:(0,\infty ) \to \mathbb{R}$ is given by

$$ D_{0^{+} }^{\alpha }h(t) = D^{n} I_{0^{+} }^{n - \alpha } h(t) = \frac{1}{{\Gamma (n - \alpha )}}\biggl(\frac{d}{{dt}}\biggr)^{n} \int _{0}^{t} { \frac{{h(s)}}{{(t - s)^{\alpha - n + 1} }}\,ds}, $$

where $n=[\alpha ]+1$ and $[\alpha ]$ denote the integer parts of the real number α, provided the right side is pointwise defined on $(0,\infty )$.

Lemma 2.1

(see [1])

(1) If $h \in L(0,1)$, $\rho > \sigma > 0$, then

$$ D^{\sigma }I^{\rho }h(t) = I^{\rho - \sigma } h(t),D^{\sigma }I^{\sigma }h(t) = h(t). $$

(2) If $\rho > 0$, $\lambda > 0$, then

$$ D^{\rho }t^{\lambda - 1} = \frac{{\Gamma (\lambda )}}{{\Gamma (\lambda - \rho )}}t^{\lambda - \rho - 1}. $$

Lemma 2.2

(see [1])

Assume $\alpha > 0$, then the solution of the equation $D_{0^{+} }^{\alpha }h(t) = 0$ is given by

$$ h(t) = c_{1} t^{\alpha - 1} + c_{2} t^{\alpha - 2} + \cdots + c_{n} t^{ \alpha - n}, $$

where $c_{i} \in \mathbb{R}$, $i = 1,2, \ldots ,n$, and $n\in \mathbb{N}$ with $n - 1 < \alpha \le n$.

For convenience, we denote

$$ \rho _{\alpha }: = \Gamma (\alpha ) \bigl(1 - a\xi ^{\alpha - \gamma _{1} - 1} \bigr),\qquad \rho _{\beta }: = \Gamma (\beta ) \bigl(1 - b \eta ^{\beta - \gamma _{2} - 1} \bigr), $$

(2.1)

and we always assume that $\rho _{\alpha }>0$, $\rho _{\beta }>0$.

Lemma 2.3

Suppose $a_{0}$, $b_{0}$ are constants. Then for any given functions $x, y \in C[0,1]$, the boundary value problem of the linear fractional differential system

$$ \textstyle\begin{cases} D_{0^{+} }^{\alpha }u(t) + x(t) = 0,\quad t \in (0,1), \\ D_{0^{+} }^{\beta }v(t) + y(t) = 0,\quad t \in (0,1), \\ u(0) = v(0) = 0, \\ D_{0^{+} }^{\gamma _{1} } u(1) = aD_{0^{+} }^{\gamma _{1} } u(\xi ) - a_{0} , \qquad D_{0^{+} }^{\gamma _{2} } v(1) = bD_{0^{+} }^{\gamma _{2} } v(\eta ) - b_{0} \end{cases} $$

(2.2)

has a unique solution $(u,v) = (u(t),v(t))$ as

$$\begin{aligned}& u(t) = \int _{0}^{1} {G_{\alpha }(t,s)x(s)\,ds} - \frac{1}{{\rho _{\alpha }}}a_{0} \Gamma (\alpha - \gamma _{1} )t^{ \alpha - 1} , \end{aligned}$$

(2.3)

$$\begin{aligned}& v(t) = \int _{0}^{1} {G_{\beta }(t,s)y(s)\,ds} - \frac{1}{{\rho _{\beta }}}b_{0} \Gamma (\beta - \gamma _{2} )t^{\beta - 1}, \end{aligned}$$

(2.4)

where

$$\begin{aligned}& G_{\alpha }(t,s) = \frac{1}{{\rho _{\alpha }}} \textstyle\begin{cases} t^{\alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} - (1 - a\xi ^{ \alpha - \gamma _{1} - 1} )(t - s)^{\alpha - 1} - at^{\alpha - 1} ( \xi - s)^{\alpha - \gamma _{1} - 1} ,\\ \quad 0 \le s \le \min \{ t,\xi \} \le 1, \\ t^{\alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} - at^{\alpha - 1} ( \xi - s)^{\alpha - \gamma _{1} - 1} ,\\ \quad 0 \le t < s \le \xi \le 1, \\ t^{\alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} - (1 - a\xi ^{ \alpha - \gamma _{1} - 1} )(t - s)^{\alpha - 1} ,\\ \quad 0 \le \xi \le s < t \le 1, \\ t^{\alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} ,\\ \quad 0 \le \max \{ t, \xi \} \le s \le 1, \end{cases}\displaystyle \end{aligned}$$

(2.5)

$$\begin{aligned}& G_{\beta }(t,s) = \frac{1}{{\rho _{\beta }}} \textstyle\begin{cases} t^{\beta - 1} (1 - s)^{\beta - \gamma _{2} - 1} - (1 - b\eta ^{\beta - \gamma _{2} - 1} )(t - s)^{\beta - 1} - bt^{\beta - 1} (\eta - s)^{ \beta - \gamma _{2} - 1} ,\\ \quad 0 \le s \le \min \{ t,\eta \} \le 1, \\ t^{\beta - 1} (1 - s)^{\beta - \gamma _{2} - 1} - bt^{\beta - 1} ( \eta - s)^{\beta - \gamma _{2} - 1} ,\\ \quad 0 \le t < s \le \eta \le 1, \\ t^{\beta - 1} (1 - s)^{\beta - \gamma _{2} - 1} - (1 - b\eta ^{\beta - \gamma _{2} - 1} )(t - s)^{\beta - 1} ,\\ \quad 0 \le \eta \le s < t \le 1, \\ t^{\beta - 1} (1 - s)^{\beta - \gamma _{2} - 1} ,\\ \quad 0 \le \max \{ t, \eta \} \le s \le 1. \end{cases}\displaystyle \end{aligned}$$

(2.6)

Proof

Assume $(u,v) = (u(t),v(t))$ is a solution of the linear fractional system (2.2). By applying Lemma 2.2,

$$\begin{aligned}& u(t) = - I_{0^{+} }^{\alpha }x(t) + c_{1} t^{\alpha - 1} + c_{2} t^{ \alpha - 2}, \\& v(t) = - I_{0^{+} }^{\beta }y(t) + d_{1} t^{\beta - 1} + d_{2} t^{ \beta - 2} , \end{aligned}$$

where $c_{1} ,c_{2} ,d_{1} ,d_{2} \in \mathbb{R}$. In view of $u(0) = v(0) = 0$, we have $c_{2} = 0$, $d_{2} = 0$. Then

$$\begin{aligned}& u(t) = - I_{0^{+} }^{\alpha }x(t) + c_{1} t^{\alpha - 1} , \\& v(t) = - I_{0^{+} }^{\beta }y(t) + d_{1} t^{\beta - 1}. \end{aligned}$$

By Lemma 2.1, we have

$$\begin{aligned}& D_{0^{+} }^{\gamma _{1} } u(t) = - D_{0^{+} }^{\gamma _{1} } I_{0^{+} }^{\alpha }x(t) + c_{1} D_{0^{+} }^{\gamma _{1} } t^{\alpha - 1} = - I_{0^{+} }^{\alpha - \gamma _{1} }x(t) + c_{1} \frac{{\Gamma (\alpha )}}{{\Gamma (\alpha - \gamma _{1} )}}t^{\alpha - \gamma _{1} - 1} , \\& D_{0^{+} }^{\gamma _{2} } v(t) = - D_{0^{+} }^{\gamma _{2} } I_{0^{+} }^{\beta }y(t) + d_{1} D_{0^{+} }^{\gamma _{2} } t^{\beta - 1} = - I_{0^{+} }^{\beta - \gamma _{2} } y(t) + d_{1} \frac{{\Gamma (\beta )}}{{\Gamma (\beta - \gamma _{2} )}}t^{\beta - \gamma _{2} - 1} . \end{aligned}$$

And

$$\begin{aligned}& D_{0^{+} }^{\gamma _{1} } u(1) = - \frac{1}{{\Gamma (\alpha - \gamma _{1} )}} \int _{0}^{1} {(1 - s)^{ \alpha - \gamma _{1} - 1} x(s)\,ds} + c_{1} \frac{{\Gamma (\alpha )}}{{\Gamma (\alpha - \gamma _{1} )}}, \\& D_{0^{+} }^{\gamma _{1} } u(\xi ) = - \frac{1}{{\Gamma (\alpha - \gamma _{1} )}} \int _{0}^{\xi }{(\xi - s)^{ \alpha - \gamma _{1} - 1} x(s)\,ds} + c_{1} \frac{{\Gamma (\alpha )}}{{\Gamma (\alpha - \gamma _{1} )}}\xi ^{ \alpha - \gamma _{1} - 1} , \\& D_{0^{+} }^{\gamma _{2} } v(1) = - \frac{1}{{\Gamma (\beta - \gamma _{2} )}} \int _{0}^{1} {(1 - s)^{ \beta - \gamma _{2} - 1} y(s)\,ds} + d_{1} \frac{{\Gamma (\beta )}}{{\Gamma (\beta - \gamma _{2} )}}, \\& D_{0^{+} }^{\gamma _{2} } v(\eta ) = - \frac{1}{{\Gamma (\beta - \gamma _{2} )}} \int _{0}^{\eta }{(\eta - s)^{ \beta - \gamma _{2} - 1} y(s)\,ds} + d_{1} \frac{{\Gamma (\beta )}}{{\Gamma (\beta - \gamma _{2} )}}\eta ^{ \beta - \gamma _{2} - 1} . \end{aligned}$$

By the boundary conditions $D_{0^{+} }^{\gamma _{1} } u(1) = aD_{0^{+} }^{\gamma _{1} } u(\xi ) - a_{0}$, $D_{0^{+} }^{\gamma _{2} } v(1) = bD_{0^{+} }^{\gamma _{2} } v(\eta ) - b_{0}$, we have

$$\begin{aligned}& c_{1} = \frac{1}{{\rho _{\alpha }}} \biggl( \int _{0}^{1} {(1 - s)^{\alpha - \gamma _{1} - 1} x(s)\,ds} - a \int _{0}^{\xi }{(\xi - s)^{\alpha - \gamma _{1} - 1} x(s)\,ds} - a_{0} \Gamma (\alpha - \gamma _{1} ) \biggr), \\& d_{1} = \frac{1}{{\rho _{\beta }}} \biggl( \int _{0}^{1} {(1 - s)^{\beta - \gamma _{2} - 1} y(s)\,ds} - b \int _{0}^{\eta }{(\eta - s)^{\beta - \gamma _{2} - 1} y(s)\,ds} - b_{0} \Gamma (\beta - \gamma _{2} ) \biggr). \end{aligned}$$

So,

$$\begin{aligned} u(t) &= - \frac{1}{{\Gamma (\alpha )}} \int _{0}^{t} {(t - s)^{ \alpha - 1} x(s)\,ds} + \frac{{t^{\alpha - 1} }}{{\rho _{\alpha }}} \biggl( \int _{0}^{1} {(1 - s)^{\alpha - \gamma _{1} - 1} x(s)\,ds} \\ &\quad{}- a \int _{0}^{\xi }{(\xi - s)^{\alpha - \gamma _{1} - 1} x(s)\,ds} - a_{0} \Gamma (\alpha - \gamma _{1} ) \biggr) \\ &= \int _{0}^{1} {G_{\alpha }(t,s)x(s)\,ds} - \frac{1}{{\rho _{\alpha }}}a_{0} \Gamma (\alpha - \gamma _{1} )t^{\alpha - 1} . \end{aligned}$$

Similarly,

$$\begin{aligned} v(t) = \int _{0}^{1} {G_{\beta }(t,s)y(s)\,ds} - \frac{1}{{\rho _{\beta }}}b_{0} \Gamma (\beta - \gamma _{2} )t^{\beta - 1}. \end{aligned}$$

On the other hand, we can easily see that $(u,v) = (u(t),v(t))$ is a solution of the linear fractional system (2.2) if $u = u(t)$, $v=v(t)$ for $t \in [0,1]$ satisfy (2.3) and (2.4), respectively. □

Lemma 2.4

Assume $\rho _{\alpha }> 0$, $\rho _{\beta }> 0$, then $G_{\alpha }(t,s)$, $G_{\beta }(t,s)$ defined by (2.5) and (2.6), respectively, have the following properties:

(1) $G_{\alpha }(t,s)$, $G_{\beta }(t,s)$ are continuous on $(t,s) \in [0,1] \times [0,1]$;

(2) $0 < G_{\alpha }(t,s) \le \max_{0 \le t \le 1} G_{\alpha }(t,s) \le w_{1} (s)$, $t,s \in (0,1)$;

(3) $0 < G_{\beta }(t,s) \le \max_{0 \le t \le 1} G_{\beta }(t,s) \le w_{2} (s)$, $t,s \in (0,1)$, where $w_{1} (s)= \frac{{(1-s)^{\alpha - \gamma _{1} - 1} }}{{\Gamma (\alpha )}}(s^{ \alpha - 1} + \frac{{a\xi ^{\alpha - \gamma _{1} - 1} }}{{1 - a\xi ^{\alpha - \gamma _{1} - 1} }})$, $w_{2} (s)= \frac{{(1 - s)^{\beta - \gamma _{2} - 1} }}{{\Gamma (\beta )}}(s^{ \beta - 1} + \frac{{b\eta ^{\beta - \gamma _{2} - 1} }}{{1 - b\eta ^{\beta - \gamma _{2} - 1} }})$.

Proof

(1) The continuity of $G_{\alpha }(t,s)$, $G_{\beta }(t,s)$ is obvious by (2.5) and (2.6).

(2) Let

$$ g_{1} (t,s) = \frac{1}{{\Gamma (\alpha )}} \bigl(t^{\alpha - 1} (1 - s)^{ \alpha - \gamma _{1} - 1} - (t - s)^{\alpha - 1} \bigr),\quad t,s\in (0,1). $$

It is clear that

$$ \begin{aligned} g_{1} (t,s) &= \frac{1}{{\Gamma (\alpha )}} \bigl(t^{\alpha - 1} (1 - s)^{ \alpha - \gamma _{1} - 1} - (t - s)^{\alpha - 1} \bigr) \\ &> \frac{1}{{\Gamma (\alpha )}} \bigl(t^{\alpha - 1} (1 - s)^{\alpha - 1} - (t - s)^{\alpha - 1} \bigr) > 0, \quad t,s\in (0,1), \end{aligned} $$

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

$$ \begin{aligned} \frac{{\partial g_{1} (t,s)}}{{\partial t}} &= \frac{{\alpha - 1}}{{\Gamma (\alpha )}} \bigl(t^{\alpha - 2} (1 - s)^{ \alpha - \gamma _{1} - 1} - (t - s)^{\alpha - 2} \bigr) \\ &= \frac{1}{{\Gamma (\alpha - 1)}}t^{\alpha - 2} \biggl((1 - s)^{\alpha - \gamma _{1} - 1} - \biggl(1 - \frac{s}{t} \biggr)^{\alpha - 2} \biggr) < 0, \end{aligned} $$

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

$$ \begin{aligned} \frac{{\partial g_{1} (t,s)}}{{\partial t}} &= \frac{{\alpha - 1}}{{\Gamma (\alpha )}} \bigl(t^{\alpha - 2} (1 - s)^{ \alpha - \gamma _{1} - 1} - (t - s)^{\alpha - 2} \bigr) \\ &= \frac{1}{{\Gamma (\alpha - 1)}}t^{\alpha - 2} \biggl((1 - s)^{\alpha - \gamma _{1} - 1} - \biggl(1 - \frac{s}{t} \biggr)^{\alpha - 2} \biggr) > 0, \end{aligned} $$

which imply that $g_{1} (t,s) > 0$, and $g_{1} (t,s)$ is decreasing with respect to t as $s \le t$ and increasing with respect to t as $t \le s$.

So,

$$\begin{aligned}& \max_{0 \le t \le 1} g_{1} (t,s) = g_{1} (s,s) = \frac{1}{{\Gamma (\alpha )}}s^{\alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1}, \end{aligned}$$

(2.7)

$$\begin{aligned}& \min_{0 \le t \le 1} g_{1} (t,s) = g_{1} (1,s) \geq 0. \end{aligned}$$

(2.8)

For $0< s \le \min \{ t,\xi \}<1$,

$$\begin{aligned} G_{\alpha }(t,s) &= \frac{1}{{\rho _{\alpha }}} \bigl(t^{\alpha - 1} (1 - s)^{ \alpha - \gamma _{1} - 1} - \bigl(1 - a\xi ^{\alpha - \gamma _{1} - 1} \bigr) (t - s)^{\alpha - 1} - at^{\alpha - 1} (\xi - s)^{\alpha - \gamma _{1} - 1} \bigr) \\ &=\frac{1}{{\Gamma (\alpha )}}\biggl(1 + \frac{{a\xi ^{\alpha - \gamma _{1} - 1} }}{{1 - a\xi ^{\alpha - \gamma _{1} - 1} }}\biggr)t^{ \alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} \\ &\quad {}- \frac{1}{{\Gamma (\alpha )}}(t - s)^{\alpha - 1} - \frac{1}{{\rho _{\alpha }}}at^{\alpha - 1} (\xi - s)^{\alpha - \gamma _{1} - 1} \\ &= \frac{1}{{\Gamma (\alpha )}} \bigl(t^{\alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} - (t - s)^{\alpha - 1} \bigr)\\ &\quad {} + \frac{{at^{\alpha - 1} }}{{\rho _{\alpha }}} \bigl(\xi ^{\alpha - \gamma _{1} - 1} (1 - s)^{\alpha - \gamma _{1} - 1} - (\xi - s)^{\alpha - \gamma _{1} - 1} \bigr) \\ &= g_{1} (t,s) + \frac{{at^{\alpha - 1} }}{{\rho _{\alpha }}} \bigl(\xi ^{ \alpha - \gamma _{1} - 1} (1 - s)^{\alpha - \gamma _{1} - 1} - (\xi - s)^{ \alpha - \gamma _{1} - 1} \bigr) \\ &\ge g_{1} (t,s) + \frac{{at^{\alpha - 1} }}{{(1 - a\xi ^{\alpha - \gamma _{1} - 1} )}}g_{1} (\xi ,s) > 0. \end{aligned}$$

In a similar way we show $G_{\alpha }(t,s)>0$ for $0< t < s \le \xi <1$ or $0<\xi \le s<t<1$ or $0<\max \{ t,\xi \} \le s<1$. Hence, $G_{\alpha }(t,s)>0$ for $t,s \in (0,1)$, and it is obvious $G_{\alpha }(1,s)>0$ for $s \in (0,1)$.

Next, we will prove that $\max_{0 \le t \le 1} G_{\alpha }(t,s) \le w_{1} (s)$.

If $0\le s \le \min \{ t,\xi \}\le 1$, we have

$$\begin{aligned} \max_{s \le t \le 1} G_{\alpha }(t,s) &= \max_{s \le t \le 1} \biggl(g_{1} (t,s) + \frac{1}{{\rho _{\alpha }}} \bigl(at^{\alpha - 1} \bigl( \xi ^{\alpha - \gamma _{1} - 1} (1 - s)^{\alpha - \gamma _{1} - 1} - (\xi - s)^{\alpha - \gamma _{1} - 1} \bigr) \bigr) \biggr) \\ &\le g_{1} (s,s) + \frac{1}{{\rho _{\alpha }}}a\xi ^{\alpha - \gamma _{1} - 1} (1 - s)^{\alpha - \gamma _{1} - 1} = w_{1} (s). \end{aligned}$$

If $0<\xi \le s < t\le 1$, by (2.7), we get

$$\begin{aligned} & \max_{s \le t \le 1} G_{\alpha }(t,s) \\ &\quad = \max_{s\le t \le 1} \frac{1}{{\rho _{\alpha }}} \bigl(t^{ \alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} - \bigl(1 - a\xi ^{\alpha - \gamma _{1} - 1} \bigr) (t - s)^{\alpha - 1} \bigr) \\ &\quad = \max_{s \le t \le 1} \biggl( \frac{1}{{\Gamma (\alpha )}}\biggl(1 + \frac{{a\xi ^{\alpha - \gamma _{1} - 1} }}{{1 - a\xi ^{\alpha - \gamma _{1} - 1} }}\biggr)t^{ \alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} - \frac{1}{{\Gamma (\alpha )}}(t - s)^{\alpha - 1} \biggr) \\ &\quad = \max_{s \le t \le 1} \biggl( \frac{1}{{\Gamma (\alpha )}} \bigl(t^{\alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} - (t - s)^{\alpha - 1} \bigr) + \frac{1}{{\rho _{\alpha }}}a\xi ^{\alpha - \gamma _{1} - 1} t^{\alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} \biggr) \\ &\quad = \max_{s \le t \le 1} \biggl(g_{1} (t,s) + \frac{1}{{\rho _{\alpha }}}a\xi ^{\alpha - \gamma _{1} - 1} t^{\alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} \biggr) \\ &\quad \le \frac{1}{{\Gamma (\alpha )}}{s^{\alpha - 1}} {(1 - s)^{\alpha - { \gamma _{1}} - 1}} + \frac{1}{{{\rho _{\alpha }}}}a{\xi ^{\alpha - { \gamma _{1}} - 1}} {(1 - s)^{\alpha - {\gamma _{1}} - 1}} \\ &\quad \le w_{1} (s). \end{aligned}$$

If $0\le t < s \le \xi <1$, we can see that

$$\begin{aligned} \max_{0 \le t \le s} G_{\alpha }(t,s) &= \max_{0 \le t \le 1} \frac{1}{{\rho _{\alpha }}} \bigl(t^{ \alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} - at^{\alpha - 1} ( \xi - s)^{\alpha - \gamma _{1} - 1} \bigr) \\ &= \max_{0 \le t \le s} \biggl( \frac{{t^{\alpha - 1} (1 - s)^{\alpha - \gamma _{1}-1}}}{{\Gamma (\alpha )}}+ \frac{1}{{\rho _{\alpha }}}at^{\alpha -1} \bigl(\xi ^{\alpha -\gamma _{1}-1}(1-s)^{ \alpha -\gamma _{1}-1}-( \xi -s)^{\alpha -1} \bigr) \biggr) \\ &\le w_{1} (s). \end{aligned}$$

If $0\le \max \{ t,\xi \} \le s\le 1$, then

$$\begin{aligned} \max_{0 \le t \le s} G_{\alpha }(t,s) &= \max_{0 \le t \le s} \frac{1}{{\rho _{\alpha }}}t^{\alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} \\ &= \max_{0 \le t \le s} \frac{1}{{\Gamma (\alpha )}}\biggl(1 + \frac{{a\xi ^{\alpha - \gamma _{1} - 1} }}{{1 - a\xi ^{\alpha - \gamma _{1} - 1} }} \biggr)t^{ \alpha - 1} (1 - s)^{\alpha - \gamma _{1} - 1} \\ &\le w_{1} (s). \end{aligned}$$

Hence,

$$ 0< G_{\alpha }(t,s) \le \max_{0 \le t \le 1} G_{\alpha }(t,s) \le w_{1} (s),\quad s,t \in (0,1). $$

(3) Similarly, we can prove the inequality. □

By Lemma 2.3, let $a_{0}=b_{0}=0$, and we can get the following lemma.

Lemma 2.5

The fractional differential coupled system (1.1) is equivalent to the systems of integral systems

$$\begin{aligned}& u(t) = \textstyle\begin{cases} \int _{0}^{1} {G_{\alpha }(t,s)f(s,v(s),v_{s} )\,ds} ,&t \in (0,1], \\ \phi (t),& t \in [ - \tau ,0], \end{cases}\displaystyle \end{aligned}$$

(2.9)

$$\begin{aligned}& v(t) = \textstyle\begin{cases} \int _{0}^{1} {G_{\beta }(t,s)g(s,u(s),u_{s} )\,ds} ,&t \in (0,1], \\ \psi (t),& t \in [ - \tau ,0]. \end{cases}\displaystyle \end{aligned}$$

(2.10)

Let $E = \{ (u,v):u,v \in C[ - \tau ,1]\}$ and be endowed with norm

$$\bigl\| (u,v)\bigr\| _{E} = \max \Bigl\{ \max_{t \in [ - \tau ,1]} \bigl|u(t)\bigr|, \max_{t \in [ - \tau ,1]} \bigl|v(t)\bigr|\Bigr\} $$

and $C[ - \tau ,0]$ endowed with the norm $\|x\|_{\tau }=\max_{t \in [-\tau ,0]} |x(t)|$. Then $(E,\|\cdot \|_{E} )$ and $(C[-\tau ,0],\|\cdot \|_{\tau })$ are Banach spaces.

Let

$$ E_{0}=\bigl\{ (r,z)\in E:\bigl(r(t),z(t)\bigr)\equiv (0,0),t\in [-\tau ,0]\bigr\} $$

be endowed with norm

$$\bigl\| (r,z)\bigr\| _{E_{0}}= \max \Bigl\{ \max_{t\in [-\tau ,1]} \bigl|r(t)\bigr|, \max_{t \in [-\tau ,1]} \bigl|z(t)\bigr|\Bigr\} =\max \Bigl\{ \max_{t \in [0,1]} \bigl|r(t)\bigr|,\max_{t \in [0,1]} \bigl|z(t)\bigr|\Bigr\} $$

and $P=\{(r,z)\in E_{0}: r(t)\ge 0,z(t) \ge 0,t \in (0,1]\}$. Obviously, $E_{0}\subset E$ and $(E_{0} ,\| \cdot \|_{E_{0} } )$ is a Banach space, and $P\subset E_{0}$ is a normal solid cone.

For $(u_{1},v_{1} ),(u_{2},v_{2} ) \in E_{0}$, $(u_{1},v_{1})\preceq (u_{2},v_{2})$ if and only if $(u_{2}-u_{1},v_{2}-v_{1}) \in P$. Hence, $(E_{0},\preceq )$ is a partial order Banach space. We denote $(u_{1} ,v_{1} ) \prec (u_{2} ,v_{2} )$ if $(u_{1} ,v_{1} )\preceq (u_{2} ,v_{2} ) \in E_{0}$ and $(u_{1} ,v_{1} ) \ne (u_{2} ,v_{2} )$; we denote $(u_{1} ,v_{1} ) \prec \prec (u_{2} ,v_{2} )$ if $(u_{2} - u_{1} ,v_{2} - v_{1} ) \in \mathop{P}^{\circ }$.

Redefine functions $\phi (t)$ and $\psi (t)$ on $t \in [0,1]$. Let $\phi (t) = \psi (t) = 0$, $t \in [0,1]$. Obviously, $\phi ,\psi \in E$.

For any $(r,z) \in E_{0}$, let

$$ u(t)=\phi (t)+r(t)= \textstyle\begin{cases} r(t),& t \in [0,1], \\ \phi (t),& t \in [ - \tau ,0], \end{cases} $$

and

$$ v(t) = \psi (t) + z(t) = \textstyle\begin{cases} z(t),& t \in [0,1], \\ \psi (t),& t \in [ - \tau ,0]. \end{cases} $$

Then for $t \in [0,1]$, $u_{t}=\phi _{t}+r_{t}=\phi (t+\theta )+r(t+\theta )$, $v_{t}=\psi _{t}+z_{t}= \psi (t+\theta )+z(t+\theta )$, $\theta \in [-\tau ,0]$.

It is easy to see that the following lemma holds.

Lemma 2.6

$(u,v) \in E$ is a solution of systems (2.9) and (2.10) if and only if $(r,z) \in E_{0}$ is a solution of the following integral systems:

$$ r(t) = \textstyle\begin{cases} 0,& t \in [ - \tau ,0], \\ \int _{0}^{1} {G_{\alpha }(t,s)f(s,z(s),\psi _{s} + z_{s} )\,ds} ,& t \in (0,1], \end{cases} $$

(2.11)

and

$$ z(t) = \textstyle\begin{cases} 0,& t \in [ - \tau ,0], \\ \int _{0}^{1} {G_{\beta }(t,s)g(s,r(s),\phi _{s} + r_{s} )\,ds} ,& t \in (0,1], \end{cases} $$

(2.12)

which implies that $(u,v) \in E$ is a solution of system (1.1) if and only if $(r,z) \in E_{0}$ is a solution of the following coupled system:

$$ \textstyle\begin{cases} D_{0^{+} }^{\alpha }r(t) + f(t,z(t),\psi _{t} + z_{t} ) = 0,\quad t \in (0,1), \\ D_{0^{+} }^{\beta }z(t) + g(t,r(t),\phi _{t} + r_{t} ) = 0,\quad t \in (0,1), \\ D_{0^{+} }^{\gamma _{1} } r(1) = aD_{0^{+} }^{\gamma _{1} } r(\xi ), \qquad D_{0^{+} }^{\gamma _{2} } z(1) = bD_{0^{+} }^{\gamma _{2} } z( \eta ), \\ r(t) = 0,\qquad z(t) = 0,\quad t \in [ - \tau ,0]. \end{cases} $$

(2.13)

Define an operator $T:E_{0} \to E_{0}$ by

$$ T(r,z) = \bigl(A(r,z),B(r,z)\bigr), $$

where

$$\begin{aligned}& A(r,z) (t) = \textstyle\begin{cases} 0,& t \in [ - \tau ,0], \\ \int _{0}^{1} {G_{\alpha }(t,s)f(s,z(s),\psi _{s} + z_{s} )\,ds} ,& t \in (0,1], \end{cases}\displaystyle \end{aligned}$$

(2.14)

$$\begin{aligned}& B(r,z) (t) = \textstyle\begin{cases} 0,& t \in [ - \tau ,0], \\ \int _{0}^{1} {G_{\beta }(t,s)g(s,r(s),\phi _{s} + r_{s} )\,ds} ,& t \in (0,1]. \end{cases}\displaystyle \end{aligned}$$

(2.15)

It is clear that the Lemma 2.7 holds.

Lemma 2.7

A solution of the system (1.1) on E is equivalent to a fixed point of operator T on $E_{0}$.

Lemma 2.8

The operator $T:E_{0} \to E_{0}$ is completely continuous.

Proof

Firstly, we prove that operator T is continuous on $E_{0}$.

Let $\{ r_{n} ,z_{n} \} \subset E_{0}$, $(r,z) \in E_{0}$ such that $\|(r_{n} ,z_{n} ) - (r,z)\|_{E_{0} } \to 0$ as $n \to \infty$. Then, there exists a constant $M_{0} \ge 0$ such that $\|(r_{n} ,z_{n} )\|_{E_{0} } \le M_{0}$ for $n = 1,2, \ldots$ and $\|(r,z)\|_{E_{0}}\le M_{0}$, and then $\|r_{nt}\|_{\tau }\le M_{0}$, $\|z_{nt}\|_{\tau }\le M_{0}$, $\|r_{t}\|_{\tau }\le M_{0}$ and $\|z_{t}\|_{\tau }\le M_{0}$ for $t \in [0,1]$. Due to $\theta \in [ - \tau ,0]$, we have $\theta + t \in [ - \tau ,1]$ for $t \in [0,1]$. Therefore, we have $\|(r_{nt} ,z_{nt} ) - (r_{t},z_{t})\|_{E_{0} } \to 0$ as $n \to \infty$.

By the continuity of f, g,

$$\begin{aligned}& \lim_{n \to \infty } \bigl\vert f\bigl(t,z_{n} (t),\psi _{t} + z_{nt} \bigr) - f\bigl(t,z(t),\psi _{t} + z_{t} \bigr) \bigr\vert = 0\quad \mathrm{and}\\& \lim _{n \to \infty } \bigl\vert g\bigl(t,r_{n} (t),\phi _{t} + r_{nt} \bigr) - g\bigl(t,r(t), \phi _{t} + r_{t} \bigr) \bigr\vert = 0. \end{aligned}$$

And there exist constants $M_{1} ,M_{2} > 0$ such that

$$ \bigl\vert f\bigl(t,z_{n} (t),\psi _{t}+z_{nt} \bigr) \bigr\vert \le M_{1}\quad \mathrm{and}\quad \bigl\vert g \bigl(t,r_{n}(t), \phi _{t}+r_{nt}\bigr) \bigr\vert \le M_{2},\quad t \in [0,1],n = 1,2, \ldots , $$

and

$$ \bigl\vert f\bigl(t,z(t),\psi _{t}+z_{t} \bigr) \bigr\vert \le M_{1},\qquad \bigl\vert g\bigl(t,r(t),\phi _{t}+r_{t}\bigr) \bigr\vert \le M_{2}. $$

Then

$$\begin{aligned}& \bigl\vert f\bigl(t,z_{n} (t),\psi _{t} + z_{nt} \bigr) - f\bigl(t,z(t),\psi _{t} + z_{t} \bigr) \bigr\vert \le 2M_{1}\quad \mathrm{and}\\& \bigl\vert g \bigl(t,r_{n} (t),\phi _{t} + r_{nt} \bigr) - g \bigl(t,r(t), \phi _{t} + r_{t} \bigr) \bigr\vert \le 2M_{2}. \end{aligned}$$

It follows from Lemma 2.4 that for $t\in [-\tau ,1]$

$$\begin{aligned} \bigl\vert A (r_{n} ,z_{n} ) (t) - A (r,z) (t) \bigr\vert &\leq \biggl\vert \int _{0}^{1} G_{\alpha }(t,s) \bigl(f \bigl(t,z_{n} (s),\psi _{s} + z_{ns} \bigr) - f \bigl(s,z(s),\psi _{s} + z_{s} \bigr)\bigr) \biggr\vert \,ds \\ &\le \int _{0}^{1} w_{1} (s) \bigl\vert f \bigl(t,z_{n} (s),\psi _{s} + z_{ns} \bigr) - f \bigl(s,z(s), \psi _{s} + z_{s} \bigr) \bigr\vert \,ds \to 0 \end{aligned}$$

and

$$\begin{aligned} \bigl\vert B (r_{n} ,z_{n} ) (t) - B (r,z) (t) \bigr\vert &\leq \biggl\vert \int _{0}^{1} G_{\beta }(t,s) \bigl(g \bigl(t,r_{n} (s),\phi _{s} + r_{ns} \bigr) - f \bigl(s,r(s),\phi _{s} + r_{s} \bigr)\bigr) \biggr\vert \,ds \\ &\le \int _{0}^{1} w_{2} (s) \bigl\vert g \bigl(t,r_{n} (s),\phi _{s} + r_{ns} \bigr) - g \bigl(s,r(s), \phi _{s} + r_{s} \bigr) \bigr\vert \,ds \to 0. \end{aligned}$$

By Lebesgue’s dominated convergence theorem, as $n \to \infty$,

$$ A(r_{n} ,z_{n} )\to A(r,z),\qquad B(r_{n} ,z_{n} )\to B(r,z). $$

Consequently, the operator T is continuous.

Assume that $S \subset E_{0}$ is a bounded set, and there exists a constant $l_{1} > 0$ such that we have $\|(r,z)\|_{E_{0} } \le l_{1}$ for any $(r,z) \in S$. There exist constants $M_{3} > 0$ and $M_{4} > 0$ such that

$$ \bigl\vert f\bigl(t,z(t),\psi _{t} + z_{t} \bigr) \bigr\vert \le M_{3},\qquad \bigl\vert g\bigl(t,r(t),\phi _{t} + r_{t} \bigr) \bigr\vert \le M_{4}. $$

Therefore, by Lemma 2.4, we have

$$\begin{aligned} \bigl\vert A(r,z) (t) \bigr\vert &\le \int _{0}^{1} { \bigl\vert G_{\alpha }(t,s) \bigr|\bigl|f\bigl(s,z(s),\psi _{s} + z_{s} \bigr) \bigr\vert \,ds} \le {M_{3}} \int _{0}^{1} {{\omega _{1}}(s)\,ds} \\ &\le \frac{{M_{3} }}{{\Gamma (\alpha )}} \int _{0}^{1} {\biggl(s^{\alpha - 1} + \frac{{a\xi ^{\alpha - \gamma _{1} - 1} (1 - s)^{\alpha - \gamma _{1} - 1} }}{{1 - a\xi ^{\alpha - \gamma _{1} - 1} }}\biggr)\,ds} \\ &= \biggl(\frac{1}{{\Gamma (\alpha + 1)}} + \frac{{a{\xi ^{\alpha - {\gamma _{1}} - 1}}}}{{(\alpha - {\gamma _{1}}){\rho _{\alpha }}}}\biggr){M_{3}} . \end{aligned}$$

Similarly, we can prove that

$$ \bigl\vert B(r,z) (t) \bigr\vert \le \biggl(\frac{1}{{\Gamma (\beta + 1)}} + \frac{{b{\eta ^{\beta - {\gamma _{2}} - 1}}}}{{(\beta - {\gamma _{2}}){\rho _{\beta }}}}\biggr){M_{4}}. $$

Therefore, there exists a constant $l > 0$, such that $\|T(r,z)\|_{E_{0} } \le l$ for $(r,z) \in S$. So $T(S)$ is uniformly bounded.

By the continuity of $G_{\alpha }(t,s)$ and $G_{\beta }(t,s)$ on $[0,1] \times [0,1]$, we have $G_{\alpha }(t,s)$ and $G_{\beta }(t,s)$ are uniformly continuous on $[0,1] \times [0,1]$. Therefore, for any $\varepsilon > 0$, there exists $\delta > 0$ such that

$$ \bigl\vert G_{\alpha }(t_{1},s)-G_{\alpha }(t_{2},s) \bigr\vert < \frac{\varepsilon }{{M_{3}}}, \qquad \bigl\vert G_{\beta }(t_{1},s)-G_{\beta }(t_{2},s) \bigr\vert < \frac{\varepsilon }{{M_{4}}} $$

whenever $t_{1},t_{2},s\in [0,1]$ and $|t_{1}-t_{2}|<\delta$.

If ${t_{1}}, {t_{2}} \in [0,1]$ and $|t_{1} - t_{2} | < \delta$, we get

$$\begin{aligned} \bigl\vert A (r,z) (t_{1} ) - A (r,z) (t_{2} ) \bigr\vert &= \biggl\vert \int _{0}^{1} {\bigl(G_{\alpha }(t_{1} ,s) - G_{\alpha }(t_{2} ,s)\bigr)} f\bigl(s,z(s),\phi _{s} + z_{s} \bigr)\,ds \biggr\vert \\ &\le M_{3} \int _{0}^{1} { \bigl\vert G_{\alpha }(t_{1} ,s) - G_{\alpha }(t_{2} ,s) \bigr\vert } \,ds < \varepsilon . \end{aligned}$$

Similarly,

$$ \bigl\vert B (r,z) (t_{1} ) - B (r,z) (t_{2} ) \bigr\vert < \varepsilon . $$

And if ${t_{1}},{t_{2}} \in [ - \tau ,0]$ and $|t_{1} - t_{2} | < \delta$, we have

$$ \bigl\vert A(r,z) ({t_{1}}) - A(r,z) ({t_{2}}) \bigr\vert = 0,\qquad \bigl\vert B(r,z) ({t_{1}}) - B(r,z) ({t_{2}}) \bigr\vert = 0. $$

Hence, $T(S)$ is equicontinuous. According to the Arzela–Ascoli theorem, $T(S)$ is a relative compact set.

So the operator T is completely continuous on $E_{0}$. □

Lemma 2.9

(see [2])

Let E be a Banach space, and $P \subset E$ be a normal solid cone. Suppose that there exist $y_{1},z_{1},y_{2},z_{2} \in E$, with $y_{1} \prec z_{1} \prec y_{2} \prec z_{2}$, and $A:[y_{1} ,z_{2} ] \to E$ is a completely continuous strongly increasing operator such that

$$ y_{1} \preceq Ay_{1},\qquad Az_{1} \prec z_{1},\qquad y_{2} \preceq Ay_{2},\qquad Az_{2} \preceq z_{2}. $$

Then the operator A has at least three fixed points $x_{1}$, $x_{2}$, $x_{3}$ such that

$$ y_{1} \preceq x_{1} \prec \prec z_{1},\qquad y_{2} \prec \prec x_{2} \preceq z_{2}, \qquad y_{2} \npreceq x_{3} \npreceq z_{1}. $$

3 Multiple solutions of the coupled systems

Definition 3.1

$(u,v) \in E_{0}\cap (AC^{2}(0,1)\times AC^{2}(0,1))$ is called an upper solution of coupled system (2.13), if it satisfies

$$ \textstyle\begin{cases} D_{0^{+} }^{\alpha }u (t) + f(t,v (t),\psi _{t} + v_{t} ) \le 0,\quad t \in (0,1), \\ D_{0^{+} }^{\beta }v (t) + g(t,u (t),\phi _{t} + u_{t} ) \le 0,\quad t \in (0,1), \\ D_{0^{+} }^{\gamma _{1} } u (1) \le aD_{0^{+} }^{\gamma _{1} } u ( \xi ), \qquad D_{0^{+} }^{\gamma _{2} } v (1) \le bD_{0^{+} }^{ \gamma _{2} } v (\eta ), \\ u(t) = 0,\qquad v(t) = 0,\quad t \in [ - \tau ,0]. \end{cases} $$

(3.1)

Definition 3.2

$(x,y) \in E_{0}\cap (AC^{2}(0,1)\times AC^{2}(0,1))$ is called a lower solution of coupled system (2.13), if it satisfies

$$ \textstyle\begin{cases} D_{0^{+} }^{\alpha }x (t) + f(t,y (t),\psi _{t} + y_{t} ) \ge 0,\quad t \in (0,1), \\ D_{0^{+} }^{\beta }y (t) + g(t,x (t),\phi _{t} + x_{t} ) \ge 0,\quad t \in (0,1), \\ D_{0^{+} }^{\gamma _{1} } x (1) \ge aD_{0^{+} }^{\gamma _{1} } x ( \xi ), \qquad D_{0^{+} }^{\gamma _{2} } y (1) \ge bD_{0^{+} }^{ \gamma _{2} } y (\eta ), \\ x(t) = 0,\qquad y(t) = 0,\quad t \in [ - \tau ,0]. \end{cases} $$

(3.2)

Lemma 3.1

Let $(u,v) \in E_{0}\cap (AC^{2}(0,1)\times AC^{2}(0,1))$, if

$$ \textstyle\begin{cases} D_{0^{+} }^{\alpha }u(t) \le 0,\quad t \in (0,1), \\ D_{0^{+} }^{\beta }v(t) \le 0,\quad t \in (0,1), \\ D_{0^{+} }^{\gamma _{1} } u(1) \le aD_{0^{+} }^{\gamma _{1} } u( \xi ), \qquad D_{0^{+} }^{\gamma _{2} } v(1) \le bD_{0^{+} }^{\gamma _{2} } z(\eta ), \\ u(t) = 0,\qquad v(t) = 0,\quad t \in [ - \tau ,0], \end{cases} $$

(3.3)

then $u(t) \ge 0$ and $v(t) \ge 0$, $t \in [ 0 ,1]$.

Proof

Let $D_{0^{+} }^{\alpha }u(t) = - x(t) \le 0$, $D_{0^{+} }^{\alpha }v(t) = - y(t) \le 0$, $D_{0^{+} }^{\gamma _{1} } u(1) - aD_{0^{+} }^{\gamma _{1} } u( \xi ) = a_{0} \le 0$, $D_{0^{+} }^{\gamma _{2} } z(1) - bD_{0^{+} }^{ \gamma _{2} } z(\eta ) = b_{0} \le 0$. Then $x(t) \ge 0$, $y(t) \ge 0$, $a_{0} \le 0$, $b_{0} \le 0$. By Lemma 2.3, the coupled system

$$ \textstyle\begin{cases} D_{0^{+} }^{\alpha }u(t) + x(t) = 0,\quad t \in (0,1), \\ D_{0^{+} }^{\beta }v(t) + y(t) = 0,\quad t \in (0,1), \\ u(0) = v(0) = 0,\quad t \in [ - \tau ,0], \\ D_{0^{+} }^{\gamma _{1} } u(1) = aD_{0^{+} }^{\gamma _{1} } u(\xi ) - a_{0} , \qquad D_{0^{+} }^{\gamma _{2} } v(1) = bD_{0^{+} }^{\gamma _{2} } v(\eta ) - b_{0}, \end{cases} $$

has a unique solution

$$\begin{aligned}& u(t) = \textstyle\begin{cases} \int _{0}^{1} {G_{\alpha }(t,s)x(s)\,ds} - \frac{1}{{\rho _{\alpha }}}a_{0} \Gamma (\alpha - \gamma _{1} )t^{\alpha - 1} ,& t \in (0,1], \\ 0,& t \in [ - \tau ,0], \end{cases}\displaystyle \\& v(t) = \textstyle\begin{cases} \int _{0}^{1} {G_{\beta }(t,s)y(s)\,ds} - \frac{1}{{\rho _{\beta }}}b_{0} \Gamma (\beta - \gamma _{2} )t^{\beta - 1} ,& t \in (0,1], \\ 0,& t \in [ - \tau ,0]. \end{cases}\displaystyle \end{aligned}$$

By Lemma 2.4, $u(t) \ge 0$ and $v(t) \ge 0$, $t \in [ 0 ,1]$. □

For convenience, we assume that the functions f and g satisfy the following properties.

(H) For any $x_{1} ,x_{2} \in [0, + \infty )$ with $0 \le x_{1} \le x_{2}$ and any $p_{1} ,p_{2} \in C([-\tau ,0])$ with $p_{1}\leq p_{2}$,

$$ f(t,x_{1},p_{1})\le f(t,x_{2},p_{2}), \qquad g(t,x_{1},p_{1})\le g(t,x_{2},p_{2}), \quad t \in [0,1], $$

when at least $x_{1}< x_{2}$ and/or $p_{1}< p_{2}$ holds,

$$ f(t,x_{1},p_{1} ) < f(t,x_{2},p_{2}), \qquad g(t,x_{1},p_{1} )< g(t,x_{2},p_{2}), \quad t \in [0,1]. $$

Lemma 3.2

Suppose (H) holds, then T is a strongly increasing operator.

Proof

For any $(h_{1} ,k_{1} ),(h_{2} ,k_{2} ) \in E_{0}$ with $(h_{1} ,k_{1} ) \prec (h_{2} ,k_{2} )$, i.e., $h_{1} (t) \le h_{2} (t)$, $k_{1} (t) \le k_{2} (t)$ and $h_{1} (t)\not \equiv h_{2} (t)$ or $k_{1} (t)\not \equiv k_{2} (t)$ for $t\in [-\tau ,1]$.

By (H), we have

$$\begin{aligned}& f\bigl(t,k_{1}(t),\psi _{t}+k_{1t}\bigr) \le f \bigl(t,k_{2}(t),\psi _{t}+k_{2t}\bigr),\\& g \bigl(t,h_{1}(t),\phi _{t}+h_{1t}\bigr) \le g \bigl(t,h_{2}(t),\phi _{t}+h_{2t}\bigr), \quad t \in [0,1]. \end{aligned}$$

Since $h_{1} (t)\not \equiv h_{2} (t)$ and $k_{1} (t)\not \equiv k_{2} (t)$, there exist two intervals $[a_{1} ,b_{1} ],[a_{2} ,b_{2} ] \subset [0,1]$ such that $k_{1} (t) < k_{2} (t)$ for $t \in [a_{1} ,b_{1} ]$ or $h_{1} (t) < h_{2} (t)$ for $t \in [a_{2} ,b_{2} ]$. Then

$$ f\bigl(t,k_{1}(t),\psi _{t}+k_{1t} \bigr)< f\bigl(t,k_{2}(t),\psi _{t}+k_{2t}\bigr), \quad t \in [a_{1},b_{1}], $$

(3.4)

and

$$ g\bigl(t,h_{1} (t),\phi _{t}+h_{1t} \bigr)< g\bigl(t,h_{2}(t),\phi _{t}+h_{2t} \bigr), \quad t \in [a_{2},b_{2}]. $$

(3.5)

From (2.14), (2.15), (3.3), (3.4) and Lemma 2.4, for any $t \in (0,1]$,

$$\begin{aligned}& A(h_{2} ,k_{2} ) (t) - A(h_{1} ,k_{1} ) (t) \\& \quad = \int _{0}^{1} {G_{\alpha }(t,s)f \bigl(s,k_{2} (s),\psi _{s}+k_{2s}\bigr)\,ds}- \int _{0}^{1} {G_{\alpha }(t,s)f \bigl(s,k_{1} (s), \psi _{s}+k_{1s}\bigr)\,ds}> 0 , \\& B(h_{2},k_{2}) (t)-B(h_{1},k_{1}) (t) \\& \quad = \int _{0}^{1} {G_{\beta }(t,s)g \bigl(s,h_{2}(s), \phi _{s}+h_{2s}\bigr)\,ds} - \int _{0}^{1} {G_{\beta }(t,s)g \bigl(s,h_{1}(s), \phi _{s}+h_{1s}\bigr)\,ds}>0. \end{aligned}$$

For any $t \in [ - \tau ,0]$, we have

$$\begin{aligned}& A(h_{2} ,k_{2} ) (t) - A(h_{1} ,k_{1} ) (t)= 0, \end{aligned}$$

(3.6)

$$\begin{aligned}& B(h_{2} ,k_{2} ) (t) - B(h_{1} ,k_{1} ) (t)= 0. \end{aligned}$$

(3.7)

In conclusion, we have $T(h_{1},k_{1} )\prec \prec T(h_{2},k_{2} )$, T is a strongly increasing operator. □

Theorem 3.3

Suppose (H) holds, and there exist two lower solutions $(x_{1} ,y_{1} )$, $(x_{2} ,y_{2} )$ and two upper solutions $(u_{1} ,v_{1} )$, $(u_{2} ,v_{2} )$ of coupled system (2.13) such that $(x_{1},y_{1} )$, $(u_{2},v_{2} )$ are not solutions of the coupled system (2.13) with

$$ (x_{1},y_{1} ) \prec (u_{1},v_{1} ) \prec (x_{2},y_{2}) \prec (u_{2},v_{2}). $$

Then the coupled system (1.1) has at least three distinct solutions $(r_{1}+\phi ,z_{1}+\psi ),(r_{2}+\phi ,z_{2}+\psi ),(r_{3}+\phi ,z_{3}+ \psi )\in E$ and for $t \in [0,1]$,

Proof

By Lemma 2.8 and Lemma 3.2, we see that $T:E_{0}\to E_{0}$ is a completely continuous strongly increasing operator.

Let $A(x_{1} ,y_{1} ): = x_{1}^{(1)}$, $B(x_{1} ,y_{1} ): = y_{1}^{(1)}$, then from the definition of T,

$$ \textstyle\begin{cases} D_{0^{+} }^{\alpha }x_{1}^{(1)}(t) + f(t,y_{1} (t),\psi _{t} + y_{1t} ) = 0,\quad t \in (0,1), \\ D_{0^{+} }^{\beta }y_{1}^{(1)}(t) + g(t,x_{1} (t),\phi _{t} + x_{1t} ) = 0,\quad t \in (0,1), \\ D_{0^{+} }^{\gamma _{1} } x_{1}^{(1)}(1) = aD_{0^{+} }^{\gamma _{1} } x_{1} (\xi ), \qquad D_{0^{+} }^{\gamma _{2} } y_{1}^{(1)}(1) = bD_{0^{+} }^{\gamma _{2} } y_{1} (\eta ), \\ x_{1}^{(1)} (t) = 0,\qquad y_{1}^{(1)} (t) = 0,\quad t \in [ - \tau ,0]. \end{cases} $$

(3.8)

By (3.2) and (3.8),

$$\begin{aligned}& \begin{aligned} D_{0^{+} }^{\alpha }\bigl(x_{1} (t) - x_{1}^{(1)} (t)\bigr) &= D_{0^{+} }^{\alpha }x_{1} (t) - D_{0^{+} }^{\alpha }x_{{1}}^{(1)}(t) \\ &\ge - f\bigl(t,y_{1} (t),\psi _{t} + y_{1t} \bigr) + f\bigl(t,y_{1} (t),\psi _{t} + y_{1t} \bigr) = 0, \end{aligned} \\& \begin{aligned} D_{0^{+} }^{\alpha }\bigl(y_{1} (t) - y_{1}^{(1)}(t)\bigr) &= D_{0^{+} }^{\alpha }y_{1} (t) - D_{0^{+} }^{\alpha }y_{1}^{(1)}(t) \\ &\ge - g\bigl(t,x_{1} (t),\phi _{t} + x_{1t} \bigr) + g\bigl(t,x_{1} (t),\phi _{t} + x_{1t} \bigr) = 0. \end{aligned} \\& D_{0^{+} }^{\gamma _{1} } \bigl( x_{1} (1)-x_{1}^{(1)}(1) \bigr) \ge aD_{0^{+} }^{\gamma _{1} } x_{1} (\xi ) - aD_{0^{+} }^{\gamma _{1} } x_{1} ( \xi ), \\& D_{0^{+} }^{\gamma _{2} } \bigl( y_{1} (1)-y_{1}^{(1)}(1) \bigr) \ge bD_{0^{+} }^{ \gamma _{2} } y_{1} (\eta ) - bD_{0^{+} }^{\gamma _{2} } y_{1} ( \eta ). \end{aligned}$$

It is clear that

$$ x_{1}^{(1)}(t)-x_{1}(t)= 0,\qquad y_{1}^{(1)}(t)-y_{1}(t)= 0,\quad t \in [- \tau ,0]. $$

By Lemma 3.1,

$$ x_{1}^{(1)}(t)-x_{1}(t)\ge 0,\qquad y_{1}^{(1)}(t)-y_{1}(t)\ge 0,\quad t \in [0,1]. $$

Therefore,

$$ (x_{1},y_{1})\preceq T(x_{1} ,y_{1}). $$

Similarly, we can prove $T(x_{2} ,y_{2} )\preceq (x_{2} ,y_{2} )$. Because $(x_{2} ,y_{2} )$ is a lower solution of coupled system (2.13) and not a solution of (2.13), we have $T(x_{2} ,y_{2} ) \ne (x_{2} ,y_{2} )$. Thus

$$ T(x_{2} ,y_{2} ) \prec (x_{2} ,y_{2} ). $$

In the same way, we get

$$ T(u_{1} ,v_{1} ) \prec (u_{1} ,v_{1} ), \qquad T(u_{2} ,v_{2} ) \preceq (u_{2} ,v_{2} ). $$

It follows that T has at least three fixed points $(r_{1} ,z_{1} ),(r_{2} ,z_{2} ),(r_{3} ,z_{3} ) \in [(x_{1} ,y_{1} ),(u_{2} ,v_{2} )]$ from Lemma 2.9.

Hence, by Lemma 2.6, the coupled system (1.1) has at least three distinct solutions $(r_{1}+\phi ,z_{1}+\psi ),(r_{2}+\phi ,z_{2}+\psi ),(r_{3}+\phi ,z_{3}+ \psi )\in E$, and for $t \in [0,1]$,

 □

4 Illustration

To illustrate the applicability of the conclusion, we consider the following nonlinear differential fractional coupled system:

$$ \textstyle\begin{cases} D_{{0^{+} }}^{\frac{3}{2}}u(t) + \frac{1}{\pi }\arctan (\sqrt{t} v(t)) + 0.01 \Vert {v_{t}} \Vert _{\tau } = 0,\quad t \in [0,1], \\ D_{{0^{+} }}^{\frac{5}{4}}v(t) + \frac{1}{\pi }\arctan ({t^{ \frac{1}{4}}}u(t)) + 0.01 \Vert {u_{t}} \Vert _{\tau } = 0,\quad t \in [0,1], \\ u(t) = {t^{2}},\qquad v(t) = {t^{4}},\quad t \in [ - \frac{1}{2} ,0], \\ D_{{0^{+} }}^{\frac{3}{8}}u(1) = D_{{0^{+} }}^{\frac{3}{8}}u( \frac{1}{2}),\qquad D_{{0^{+} }}^{\frac{5}{8}}v(1) = \frac{{29}}{{50}}D_{{0^{+} }}^{\frac{5}{8}}v(\frac{1}{4}). \end{cases} $$

(4.1)

The coupled system (4.1) can be regarded as the form of (1.1), where $\alpha = \frac{3}{2}$, $\beta = \frac{5}{4}$, $\tau = \frac{1}{2}$, ${\gamma _{1}} = \frac{3}{8}$, ${\gamma _{2}} = \frac{5}{8}$, $\xi = \frac{1}{2}$, $\eta = \frac{1}{4}$, $a = 1$, $b = \frac{3}{5}$, $f(t,v(t),{v_{t}}) = \frac{1}{\pi }\arctan (\sqrt{t} v(t)) + 0.01\| {v_{t}}\|_{\tau }$, $g(t,u(t),{u_{t}}) = \frac{1}{\pi }\arctan ({t^{\frac{1}{4}}}u(t)) + 0.01\|{u_{t}}\|_{\tau }$, $\varphi (t) = {t^{2}}$, $\psi (t) = {t^{4}}$.

It is obvious that

$$ \begin{aligned} 1 - a{\xi ^{\alpha - {\gamma _{1}} - 1}} &= 1 - {\biggl(\frac{1}{2}\biggr)^{ \frac{3}{2} - \frac{3}{8} - 1}} \approx 0.082996 > 0, 1 - b{\eta ^{ \beta - {\gamma _{2}} - 1}} \\ &= 1 - \frac{{29}}{{50}}{ \biggl(\frac{1}{4}\biggr)^{ \frac{5}{4} - \frac{5}{8} - 1}} \approx 0.02456 > 0. \end{aligned} $$

Take

$$\begin{aligned}& {u_{1}}(t) = \textstyle\begin{cases} \frac{1}{{\sqrt{\pi }}}(16 - 2t)\sqrt{t} ,& t \in [0,1], \\ {t^{2}},& t \in [ - \frac{1}{2},0], \end{cases}\displaystyle \qquad {v_{1}}(t) = \textstyle\begin{cases} \frac{1}{{\Gamma (0.25)}}(10 - 2t){t^{\frac{1}{4}}},& t \in [0,1], \\ {t^{4}},& t \in [ - \frac{1}{2},0], \end{cases}\displaystyle \\& {u_{2}}(t) = \textstyle\begin{cases} \frac{1}{{\sqrt{\pi }}}(60 - 8t)\sqrt{t} ,& t \in [0,1], \\ {t^{2}},& t \in [ - \frac{1}{2},0], \end{cases}\displaystyle \qquad {v_{2}}(t) = \textstyle\begin{cases} \frac{1}{{\Gamma (0.25)}}(80 - 3t){t^{\frac{1}{4}}},& t \in [0,1], \\ {t^{4}},& t \in [ - \frac{1}{2},0], \end{cases}\displaystyle \\& {x_{1}}(t) = \textstyle\begin{cases} 0,& t \in [0,1], \\ {t^{2}},& t \in [ - \frac{1}{2},0], \end{cases}\displaystyle \qquad {y_{1}}(t) = \textstyle\begin{cases} 0,& t \in [0,1], \\ {t^{4}},& t \in [ - \frac{1}{2},0], \end{cases}\displaystyle \\& {x_{2}}(t) = \textstyle\begin{cases} \frac{{15\sqrt{\pi }}}{{\Gamma (\frac{{15}}{8})}}(5 - 4{t^{ \frac{1}{{16}}}}){t^{\frac{7}{8}}},& t \in [0,1], \\ {t^{2}},& t \in [ - \frac{1}{2},0], \end{cases}\displaystyle \qquad {y_{2}}(t) = \textstyle\begin{cases} \frac{{{t^{\frac{3}{8}}}}}{{\Gamma (\frac{5}{8})}}(23 - \frac{8}{5}{t^{ \frac{3}{4}}}),& t \in [0,1], \\ {t^{4}},& t \in [ - \frac{1}{2},0]. \end{cases}\displaystyle \end{aligned}$$

By simple computations, we have

$$\begin{aligned}& \textstyle\begin{cases} D_{{0^{+} }}^{\frac{3}{2}}{u_{1}}(t)+ f(t,{v_{1}}(t),{v_{1t}} + { \psi _{t}}) = - \frac{3}{2} + f(t,{v_{1}}(t),{v_{1t}} + {\psi _{t}}) \le 0,\quad t \in [0,1], \\ D_{{0^{+} }}^{\frac{5}{4}}{v_{1}}(t)+g(t,{u_{1}}(t),{u_{1t}} + { \phi _{t}}) \approx - 0.625 + g(t,{u_{1}}(t),{u_{1t}} + {\phi _{t}}) \le 0,\quad t \in [0,1], \\ {u_{1}}(t) = 0,\qquad {v_{1}}(t) = 0,\quad t \in [ - \frac{1}{2},0], \\ D_{{0^{+} }}^{\frac{3}{8}}{u_{1}}(1) \approx 7.07907 \le D_{{0^{+} }}^{ \frac{3}{8}}{u_{1}}(\frac{1}{2}) \approx 7.14069, \\ D_{{0^{+} }}^{\frac{5}{8}}{v_{1}}(1) \approx 1.04565 \le \frac{{29}}{{50}}D_{{0^{+} }}^{\frac{5}{8}}{v_{1}}(\frac{1}{4}) \approx 1.52995, \end{cases}\displaystyle \\& \textstyle\begin{cases} D_{{0^{+} }}^{\frac{3}{2}}{u_{2}}(t)+ f(t,{v_{2}}(t),{v_{2t}} + { \psi _{t}}) = - 6 + f(t,{v_{2}}(t),{v_{2t}} + {\psi _{t}}) \le 0,\quad t \in [0,1], \\ D_{{0^{+} }}^{\frac{5}{4}}{v_{2}}(t) + g(t,{u_{2}}(t),{u_{2t}} + { \phi _{t}})\approx - 0.9375 + g(t,{u_{2}}(t),{u_{2t}} + {\phi _{t}}) \le 0,\quad t \in [0,1], \\ {u_{2}}(t) = 0,{v_{2}}(t) = 0,\quad t \in [ - \frac{1}{2},0], \\ D_{{0^{+} }}^{\frac{3}{8}}{u_{2}}(1) \approx 26.19258 \le D_{{0^{+} }}^{ \frac{3}{8}}{u_{2}}(\frac{1}{2}) \approx 26.61531, \\ D_{{0^{+} }}^{\frac{5}{8}}{v_{2}}(1) \approx 12.89631 \le \frac{{29}}{{50}}D_{{0^{+} }}^{\frac{5}{8}}{v_{2}}(\frac{1}{4}) \approx 13.34455, \end{cases}\displaystyle \\& \textstyle\begin{cases} D_{{0^{+} }}^{\frac{3}{2}}{x_{1}}(t) + f(t,{y_{1}}(t),{y_{1t}} + { \psi _{t}})= 0 + f(t,{y_{1}}(t),{y_{1t}} + {\psi _{t}}) \ge 0,\quad t \in [0,1], \\ D_{{0^{+} }}^{\frac{5}{4}}{y_{1}}(t)+ g(t,{x_{1}}(t),{x_{1t}} + { \phi _{t}}) = 0 + g(t,{x_{1}}(t),{x_{1t}} + {\phi _{t}}) \ge 0,\quad t \in [0,1], \\ {x_{1}}(t) = 0,\qquad {y_{1}}(t) = 0,\quad t \in [ - \frac{1}{2},0], \\ D_{{0^{+} }}^{\frac{3}{8}}{x_{1}}(1) = 0 \ge D_{{0^{+} }}^{ \frac{3}{8}}{x_{1}}(\frac{1}{2}) = 0, \\ D_{{0^{+} }}^{\frac{5}{8}}{y_{1}}(1) = 0 \ge \frac{{29}}{{50}}D_{{0^{+} }}^{\frac{5}{8}}{y_{1}}(\frac{1}{4}) = 0, \end{cases}\displaystyle \\& \textstyle\begin{cases} D_{{0^{+} }}^{\frac{3}{2}}{x_{2}}(t)+ f(t,{y_{2}}(t),{y_{2t}} + { \psi _{t}}) \\ \quad =\frac{{5\pi }}{{64\Gamma (\frac{1}{2})}}(99{t^{ \frac{ - 5}{8}}}(\frac{5}{{\Gamma (\frac{{19}}{8})}} - \frac{{4{t^{\frac{1}{{16}}}}\Gamma (\frac{{31}}{{16}})}}{{\Gamma (\frac{{15}}{8})\Gamma (\frac{{39}}{{16}})}}) - \frac{{87\Gamma (\frac{{31}}{{16}}){t^{\frac{{ - 9}}{{16}}}}}}{{\Gamma (\frac{{15}}{8})\Gamma (\frac{{39}}{{16}})}}) +f(t,{y_{2}}(t),{y_{2t}} + { \psi _{t}}) \\ \quad \ge 0, \quad t \in [0,1], \\ D_{{0^{+} }}^{\frac{5}{4}}{y_{2}}(t)+ g(t,{x_{2}}(t),{x_{2t}} + { \phi _{t}})\\ \quad = ( \frac{{{{1842}^{\frac{1}{4}}}\Gamma (\frac{{11}}{8})}}{{\sqrt{\pi }\Gamma (\frac{5}{4})}} - \frac{{9{t^{\frac{3}{4}}}\Gamma (\frac{{17}}{8})}}{{\Gamma (\frac{{13}}{8})\Gamma (\frac{{23}}{8})}})64{t^{ \frac{{ - 7}}{8}}} - \frac{{3\Gamma (\frac{{17}}{8}){t^{\frac{{ - 1}}{8}}}}}{{2\Gamma (\frac{{13}}{8})\Gamma (\frac{{23}}{8})}} + g(t,{x_{2}}(t),{x_{2t}} + { \phi _{t}}) \\ \quad \ge 0,\quad t \in [0,1], \\ {x_{2}}(t) = 0,\qquad {y_{2}}(t) = 0, \quad t \in [ - \frac{1}{2},0], \\ D_{{0^{+} }}^{\frac{3}{8}}{x_{2}}(1) \approx 27.76526 \ge D_{{0^{+} }}^{ \frac{3}{8}}{x_{2}}(\frac{1}{2}) \approx 23.29748, \\ D_{{0^{+} }}^{\frac{5}{8}}{y_{2}}(1) \approx 10.29709 \ge \frac{{29}}{{50}}D_{{0^{+} }}^{\frac{5}{8}}{y_{2}}(\frac{1}{4}) \approx 9.15314, \end{cases}\displaystyle \end{aligned}$$

which show $({u_{1}},{v_{1}})$ and $({u_{2}},{u_{2}})$ are upper solutions, $({x_{1}},{y_{1}})$ and $({x_{2}},{y_{2}})$ are lower solutions of the coupled systems (4.1), and it is not hard to get $({x_{1}},{y_{1}}) \prec ({u_{1}},{v_{1}}) \prec ({x_{2}},{y_{2}}) \prec ({u_{2}},{v_{2}})$.

It is easy to obtain

$$ \begin{aligned} 0 &\le f\bigl(t,{v_{2}}(t),{v_{2t}} + {\psi _{t}} \bigr) - f\bigl(t,{v_{1}}(t),{v_{1t}} + {\phi _{t}}\bigr) \\ &\le \frac{1}{\pi } \Vert {v_{2}} - {v_{1}} \Vert + 0.01 \bigl\Vert ({v_{2t}} + {\psi _{t}}) - ({v_{1t}} + {\phi _{t}}) \bigr\Vert _{\tau } \le \biggl( \frac{1}{\pi } + 0.01\biggr) \Vert {v_{2}} - {v_{1}} \Vert \end{aligned} $$

and

$$ \begin{aligned} 0 &\le g\bigl(t,{u_{2}}(t),{u_{2t}} + {\psi _{t}} \bigr) - g\bigl(t,{u_{1}}(t),{u_{1t}} + {\phi _{t}}\bigr) \\ &\le \frac{1}{\pi } \Vert {u_{2}} - {u_{1}} \Vert + 0.01 \bigl\Vert {u_{2t}} + {\psi _{t}}) - ({u_{2t}} + {\phi _{t}}) \bigr\Vert _{\tau } \le \biggl( \frac{1}{\pi } + 0.01\biggr) \Vert {u_{2}} - {u_{1}} \Vert . \end{aligned} $$

Similarly,

$$ f\bigl(t,{y_{2}}(t),{y_{2t}} + {\psi _{t}}\bigr) \ge f\bigl(t,{y_{1}}(t),{y_{1t}} + { \psi _{t}} \bigr),\qquad g\bigl(t,{x_{2}}(t),{x_{2t}} + {\phi _{t}}\bigr) \ge g\bigl(t,{x_{1}}(t),{x_{1t}} + { \phi _{t}}\bigr). $$

Then condition (H) is satisfied. And all conditions of Theorem 3.3 are satisfied. In view of Theorem 3.3, the coupled system (4.1) has at least three distinct solutions $(r_{1}+\phi ,z_{1}+\psi ),(r_{2}+\phi ,z_{2}+\psi ),(r_{3}+\phi ,z_{3}+ \psi )\in [({x_{1}},{y_{1}}),({u_{2}},{v_{2}})]$ and, moreover,

5 Conclusion

In this paper, we present the method of upper and lower solutions for a class of fractional coupled systems including state dependent delays with nonlocal boundary conditions. By using the method of upper and lower solutions and fixed point theorems on the normal cone, the multiplicity results for the boundary value problem are established. The method and main results obtained in this paper can also be extended to the boundary value problems of function fractional differential high dimensional systems of the form

$$ \textstyle\begin{cases} D_{0^{+} }^{\alpha _{1}} u_{1}(t) + f_{1}(t,u_{2}(t),u_{2t} ) = 0,\quad t \in (0,1), \\ D_{0^{+} }^{\alpha _{2}} u_{2}(t) + f_{2}(t,u_{3}(t),u_{3t} ) = 0,\quad t \in (0,1), \\ \cdots \\ D_{0^{+} }^{\alpha _{m}} u_{m}(t) + f_{m}(t,u_{1}(t),u_{1t} ) = 0,\quad t \in (0,1), \\ u_{i}(t)=\varphi_{i}(t),\quad t\in [-\tau,0], \quad i=1,2,\cdots,m,\\ D_{0^{+} }^{\gamma _{i} } u_{i}(1) = a_{i} D_{0^{+} }^{\gamma _{i} } u_{i}(\xi _{i} ), \quad i=1,2,\cdots ,m. \end{cases} $$