Positive solutions for fractional differential systems with nonlocal Riemann–Liouville fractional integral boundary conditions

Abstract

In this paper, we study the positive solutions of fractional differential system with coupled nonlocal Riemann–Liouville fractional integral boundary conditions. Our analysis relies on Leggett–Williams and Guo–Krasnoselskii’s fixed point theorems. Two examples are worked out to illustrate our main results.

1 Introduction

The fractional order differential equations has been received much attention due to its various applications in science and engineering such as fluid dynamics, heat conduction, control theory, electroanalytical chemistry, economics, fractal theory, fractional biological neurons, etc. It is proved that the fractional order differential equation is a better tool for the description of hereditary properties of various materials and processes than the corresponding integer order differential equation. For a systematic development of the topic, we refer the books [1–7]. A variety of results on initial and boundary value problems of fractional differential equations and inclusions can easily be found in the literature on the topic. For some recent results, we can refer to [8–18] and references cited therein.

Recently in [19] the authors studied the existence of positive solutions to the boundary value problems of fractional differential equations of the form

$$\begin{aligned} D^{q}u(t)+ f(t, u(t))=0, \quad 1<q\le 2,\quad 0<t<1, \end{aligned}$$

(1.1)

subject to three point multi-term fractional integral boundary conditions

$$\begin{aligned} u(0)=0, \quad u(1)=\sum _{i=1}^{m}\alpha _{i}(I^{p_{i}}u)(\eta ), \quad 0<\eta <1, \end{aligned}$$

(1.2)

where $D^q$ is the standard Riemann–Liouville fractional derivative of order q,  $I^{p_{i}}$ is the Riemann–Liouville fractional integral of order $p_i>0,$ $i=1,2,\ldots , m,$  $f\in C([0,1]\times {\mathbb R})$ and $\alpha _i\ge 0$, $i=1,2,\ldots , m$, are real constants. The existence and multiplicity of positive solutions were obtained by using fixed point theorems. For some recent results on positive solutions of fractional differential equations we refer to [20–25] and references cited therein.

The main purpose in this paper is to investigate some sufficient conditions for existence of positive solutions to the following fractional system of differential equations subject to the nonlocal Riemann–Liouville fractional integral boundary conditions of the form

$$\begin{aligned} \left\{ \begin{array}{ll} D^{p}x(t)+f(t,x(t),y(t))=0,\quad 1<p\le 2, \,\,t\in (0,1),\\ D^{q}y(t)+g(t,x(t),y(t))=0,\quad 1<q\le 2,\,\,t\in (0,1),\\ x(0)=0,\quad x(1)=\sum \limits _{i=1}^{m} \alpha _{i}I^{\gamma _i}y(\eta ),\\ y(0)=0,\quad y(1)=\sum \limits _{j=1}^{n} \beta _{j}I^{\mu _j}x(\xi ),\\ \end{array}\right. \end{aligned}$$

(1.3)

where $D^{\phi }$ are Riemann–Liouville fractional derivatives of orders $\phi \in \{p,q\}$, $f,g\in C([0,1]\times \mathbb {R}_+^2,\mathbb {R}_+)$, $I^{\Phi }$ are Riemann–Liouville fractional integrals of order $\Phi \in \{\gamma _i,\mu _j\}$, $\alpha _i,\beta _j>0$, $i=1,\ldots ,m$, $j=1,\ldots ,n$ and the fixed constants $0<\eta<\xi <1$.

Many researchers have shown their interest in the study of systems of fractional differential equations. The motivation for those works stems from both the intensive development of the theory of fractional calculus itself and the applications. See for example [26–30] where systems for fractional differential equations were studied by using Banach contraction mapping principle and Schaefer’s fixed point theorem.

In this paper, we firstly derive the corresponding Green’s function and some of its properties are proved. Consequently problem (1.3) is deduced to a equivalent Fredholm integral equation of the second kind. Finally, by the means of some fixed-point theorems, the existence and multiplicity of positive solutions are obtained. Illustrative examples are also presented.

2 Preliminaries

In this section, we introduce some notations and definitions of Riemann–Liouville fractional calculus (see [4]) and present preliminary results needed in our proofs later.

Definition 2.1

The (left-sided) fractional integral of order $\alpha >0$ of a function $f:(0,\infty )\rightarrow \mathbb { R}$ is given by

$$\begin{aligned} (I^{\alpha }f)(t)=\frac{1}{\Gamma (\alpha )}\int _0^t (t-s)^{\alpha -1}f(s)ds,\quad t>0, \end{aligned}$$

(2.1)

provided the right-hand side is pointwise defined on $(0,\infty ),$ where $\Gamma (\alpha )$ is the Euler gamma function defined by $\Gamma (\alpha )=\int _0^\infty t^{\alpha -1}e^{-t}dt.$

Definition 2.2

The Riemann–Liouville fractional derivative of order $\alpha \ge 0$ for a function $f:(0,\infty )\rightarrow \mathbb {R}$ is given by

$$\begin{aligned} (D^{\alpha }f)(t)=\left( \frac{d}{dt}\right) ^{n}(I^{n-\alpha }f)(t)=\frac{1}{\Gamma (n-\alpha )}\left( \frac{d}{dt}\right) ^n\int _0^t\frac{f(s)}{(t-s)^{\alpha -n+1}}ds,\quad t>0,\nonumber \\ \end{aligned}$$

(2.2)

$n-1<\alpha <n$, provided that the right-hand side is pointwise defined on $(0,\infty )$.

We also denote the Riemann–Liouville fractional derivative of f by $D^\alpha f(t).$ If $\alpha =m\in \mathbb {N}$ then $D^m f(t)=f^{(m)}(t)$ for $t>0$, and if $\alpha =0$ then $D^0 f(t)=f(t)$ for $t>0.$

Lemma 2.1

Let $\alpha >0$ and $u\in C(0, 1)\cap L^1(0,1)$. Then the fractional differential equation $D^{\alpha }u(t)=0$ has a unique solution

$$\begin{aligned} u(t)=c_1t^{\alpha -1}+c_2t^{\alpha -2}+\cdots +c_nt^{\alpha -n},\quad 0<t<1, \end{aligned}$$

(2.3)

where $c_1,c_2,\ldots ,c_n$ are arbitrary real constants, and $n-1<\alpha <n.$

Lemma 2.2

Let $\alpha >0$, $n-1<\alpha \le n$ and $y\in AC(0, 1)$. (By AC we denote the space of absolutely continuous functions). The solution of the fractional differential equation $D^\alpha u(t)+y(t)=0$,    $0<t<1$, is

$$\begin{aligned} u(t)=-\frac{1}{\Gamma (\alpha )}\int _0^t (t-s)^{\alpha -1}y(s)ds+c_1t^{\alpha -1}+\cdots +c_nt^{\alpha -n},\quad 0<t<1,\nonumber \\ \end{aligned}$$

(2.4)

where $c_1,c_2,\ldots ,c_n$ are arbitrary real constants.

Lemma 2.3

Assume that $u,v\in AC([0,1],\mathbb {R}^+)$. Then the following system

$$\begin{aligned} \left\{ \begin{array}{ll} D^{p}x(t)+u(t)=0,\quad t\in (0,1),\\ D^{q}y(t)+v(t)=0,\quad t\in (0,1),\\ x(0)=0,\quad x(1)=\sum \limits _{i=1}^{n}\alpha _{i}I^{\gamma _i}y(\eta ),\\ y(0)=0,\quad y(1)=\sum \limits _{j=1}^{m}\beta _{j}I^{\mu _j}x(\xi ), \end{array}\right. \end{aligned}$$

(2.5)

can be written in the equivalent integral equations of the form

$$\begin{aligned} x(t)= & {} -\frac{1}{\Gamma (p)}\int _0^t (t-s)^{p-1}u(s)ds+t^{p-1}\bigg [\frac{1}{\Omega \Gamma (p)}\int _0^1 (1-s)^{p-1}u(s)ds \nonumber \\&-\frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _i}{\Gamma (q+{\gamma _i})}\int _0^{\eta } (\eta -s)^{q+{\gamma _i}-1}v(s)ds+\frac{\Lambda _1}{\Omega \Gamma (q)}\int _0^1 (1-s)^{q-1}v(s)ds \nonumber \\&-\frac{\Lambda _1}{\Omega }\sum _{j=1}^n\frac{\beta _j}{\Gamma (p+{\mu _j})}\int _0^\xi (\xi -s)^{p+{\mu _j}-1}u(s)ds\bigg ], \end{aligned}$$

(2.6)

and

$$\begin{aligned} y(t)= & {} -\frac{1}{\Gamma (q)}\int _0^t (t-s)^{q-1}v(s)ds+t^{p-1}\bigg [\frac{1}{\Omega \Gamma (q)}\int _0^1 (1-s)^{q-1}v(s)ds \nonumber \\&-\frac{1}{\Omega }\sum _{j=1}^n\frac{\beta _j}{\Gamma (p+{\mu _j})}\int _0^{\xi } (\xi -s)^{p+{\mu _j}-1}u(s)ds+\frac{\Lambda _2}{\Omega \Gamma (p)}\int _0^1 (1-s)^{p-1}u(s)ds \nonumber \\&-\frac{\Lambda _2}{\Omega }\sum _{i=1}^m\frac{\alpha _i}{\Gamma (q+{\gamma _i})}\int _0^\eta (\eta -s)^{q+{\gamma _i}-1}v(s)ds\bigg ], \end{aligned}$$

(2.7)

where

$$\begin{aligned} \Omega :=1-\Lambda _1\Lambda _2>0, \end{aligned}$$

with

$$\begin{aligned} \Lambda _1:= \sum _{i=1}^m\frac{\alpha _i\eta ^{q+\gamma _i-1}\Gamma (q)}{\Gamma (q+\gamma _i)},\quad \Lambda _2:= \sum _{j=1}^n\frac{\beta _j\xi ^{p+\mu _j-1}\Gamma (p)}{\Gamma (p+\mu _j)}. \end{aligned}$$

Proof

Applying Lemma 2.2, the first two equations of problem (2.5) can be expressed as

$$\begin{aligned} x(t)= & {} -\frac{1}{\Gamma (p)}\int _0^t (t-s)^{p-1}u(s)ds+c_1t^{p-1}+c_2t^{p-2},\\ y(t)= & {} -\frac{1}{\Gamma (q)}\int _0^t (t-s)^{q-1}v(s)ds+k_1t^{q-1}+k_2t^{q-2}, \end{aligned}$$

where $c_1,c_2,k_1,k_2\in \mathbb {R}$.

From the initial conditions of (2.5) that $x(0)=0$, $y(0)=0$, we have $c_2=k_2=0$. Therefore, we get the following equations

$$\begin{aligned} x(t) = -\frac{1}{\Gamma (p)}\int _0^t (t-s)^{p-1}u(s)ds+c_1t^{p-1}, \end{aligned}$$

(2.8)

and

$$\begin{aligned} y(t) = -\frac{1}{\Gamma (q)}\int _0^t (t-s)^{q-1}v(s)ds+k_1t^{q-1}. \end{aligned}$$

(2.9)

Taking the Riemann–Liouville fractional integral of orders $\mu _j$ and $\gamma _i$ to (2.8) and (2.9), and also substitution $t=\xi$ and $t=\eta$, respectively, we obtain

$$\begin{aligned} I^{\mu _j}x(\xi )= -\frac{1}{\Gamma (p+{\mu _j})}\int _0^{\xi } (\xi -s)^{p+{\mu _j}-1}u(s)ds+{c_1}\frac{\xi ^{p+{\mu _j}-1}\Gamma (p)}{\Gamma (p+{\mu _j})}, \end{aligned}$$

and

$$\begin{aligned} I^{\gamma _i}y(\eta )= -\frac{1}{\Gamma (q+{\gamma _i})}\int _0^{\eta } (\eta -s)^{q+{\gamma _i}-1}v(s)ds+{k_1}\frac{\eta ^{q+{\gamma _i-1}}\Gamma (q)}{\Gamma (q+{\gamma _i})}. \end{aligned}$$

Using the second nonlocal boundary conditions of (2.5), we deduce the following system

$$\begin{aligned} c_1-k_1\sum _{i=1}^m\frac{{\alpha _i}\eta ^{q+{\gamma _i}-1}\Gamma (q)}{\Gamma (q+{\gamma _i})}= & {} \frac{1}{\Gamma (p)}\int _0^1 (1-s)^{p-1}u(s)ds \\&-\sum _{i=1}^m \frac{\alpha _i}{\Gamma (q+{\gamma _i})}\int _0^\eta (\eta -s)^{q+{\gamma _i}-1}v(s)ds, \\ -c_1\sum _{j=1}^n\frac{{\beta _j}\xi ^{p+{\mu _j}-1}\Gamma (p)}{\Gamma (p+{\mu _j})}+k_1= & {} \frac{1}{\Gamma (q)}\int _0^1 (1-s)^{q-1}v(s)ds \\&-\sum _{j=1}^n \frac{\beta _j}{\Gamma (p+{\mu _j})}\int _0^\xi (\xi -s)^{p+{\mu _j}-1}u(s)ds. \end{aligned}$$

Solving the above system to find constants $c_1$ and $k_1$, we obtain

$$\begin{aligned} c_1= & {} \frac{1}{\Omega \Gamma (p)}\int _0^1 (1-s)^{p-1}u(s)ds-\frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _i}{\Gamma (q+{\gamma _i})}\int _0^{\eta } (\eta -s)^{q+{\gamma _i}-1}v(s)ds \nonumber \\&+\,\frac{\Lambda _1}{\Omega \Gamma (q)}\int _0^1 (1-s)^{q-1}v(s)ds-\frac{\Lambda _1}{\Omega }\sum _{j=1}^n\frac{\beta _j}{\Gamma (p+{\mu _j})}\int _0^\xi (\xi -s)^{p+{\mu _j}-1}u(s)ds, \end{aligned}$$

and

$$\begin{aligned} k_1= & {} \frac{1}{\Omega \Gamma (q)}\int _0^1 (1-s)^{q-1}v(s)ds-\frac{1}{\Omega }\sum _{j=1}^n\frac{\beta _j}{\Gamma (p+{\mu _j})}\int _0^{\xi } (\xi -s)^{p+{\mu _j}-1}u(s)ds \nonumber \\&+\,\frac{\Lambda _2}{\Omega \Gamma (p)}\int _0^1 (1-s)^{p-1}u(s)ds-\frac{\Lambda _2}{\Omega }\sum _{i=1}^m\frac{\alpha _i}{\Gamma (q+{\gamma _i})}\int _0^\eta (\eta -s)^{q+{\gamma _i}-1}v(s)ds. \end{aligned}$$

Substituting the values of $c_1$ and $k_1$ in (2.8) and (2.9), we deduce the integral equations (2.6) and (2.7), respectively, as desired. The converse follows by direct computation. This completes the proof. $\square$

Lemma 2.4

(Green’s function) The integral equations (2.6) and (2.7) in Lemma 2.3 can be expressed in the form of Green functions as

$$\begin{aligned} x(t)=\int _0^1 G_1(t,s)u(s)ds, \end{aligned}$$

(2.10)

$$\begin{aligned} y(t)=\int _0^1 G_2(t,s)v(s)ds, \end{aligned}$$

(2.11)

where $G_1$, $G_2$ are the Green’s functions given by

$$\begin{aligned} G_1(t,s)= & {} g_p(t,s)+\frac{\Lambda _1}{\Omega }\sum _{j=1}^n\frac{\beta _jt^{p-1}}{\Gamma (p+\mu _j)}g_{\mu _j}^p(\xi ,s)\nonumber \\&+\,\frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _it^{p-1}}{\Gamma (q+\gamma _i)}g_{\gamma _i}^q(\eta ,s), \end{aligned}$$

(2.12)

$$\begin{aligned} G_2(t,s)= & {} g_q(t,s)+\frac{\Lambda _2}{\Omega }\sum _{i=1}^m\frac{\alpha _it^{q-1}}{\Gamma (q+\gamma _i)}g_{\gamma _i}^q(\eta ,s)\nonumber \\&+\,\frac{1}{\Omega }\sum _{j=1}^n\frac{\beta _jt^{q-1}}{\Gamma (p+\mu _j)}g_{\mu _j}^p(\xi ,s), \end{aligned}$$

(2.13)

where

$$\begin{aligned} g_\phi (t,s)=\left\{ \begin{array}{ll} \displaystyle \frac{(1-s)^{\phi -1}t^{\phi -1}-(t-s)^{\phi -1}}{\Gamma (\phi )},&{}\quad 0\le s\le t\le 1,\\ \displaystyle \frac{(1-s)^{\phi -1}t^{\phi -1}}{\Gamma (\phi )}, &{} \quad 0\le t\le s\le 1, \end{array}\right. \end{aligned}$$

(2.14)

and

$$\begin{aligned} g_\psi ^{\phi }(\rho ,s)= \left\{ \begin{array}{ll}\displaystyle \rho ^{\phi +\psi -1}(1-s)^{\phi -1}-(\rho -s)^{\phi +\psi -1}, &{} \quad 0\le s\le \rho \le 1,\\ \displaystyle \rho ^{\phi +\psi -1}(1-s)^{\phi -1},&{} \quad 0\le \rho \le s\le 1, \end{array}\right. \end{aligned}$$

(2.15)

with $\phi \in \{p,q\}$, $\psi \in \{\mu _j,\gamma _i\}$, $\rho \in \{\xi ,\eta \}$.

Proof

From Lemma 2.3, by direct computation, we have

$$\begin{aligned} x(t)= & {} -\frac{1}{\Gamma (p)}\int _0^t (t-s)^{p-1}u(s)ds+t^{p-1}\bigg [\frac{1}{\Omega \Gamma (p)}\int _0^1 (1-s)^{p-1}u(s)ds\\&-\frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _i}{\Gamma (q+{\gamma _i})}\int _0^{\eta } (\eta -s)^{q+{\gamma _i}-1}v(s)ds+\frac{\Lambda _1}{\Omega \Gamma (q)}\int _0^1 (1-s)^{q-1}v(s)ds\\&-\frac{\Lambda _1}{\Omega }\sum _{j=1}^n\frac{\beta _j}{\Gamma (p+{\mu _j})}\int _0^\xi (\xi -s)^{p+{\mu _j}-1}u(s)ds\bigg ]\\= & {} \int _0^1 \frac{(1-s)^{p-1}t^{p-1}}{\Gamma (p)}u(s)ds-\frac{1}{\Gamma (p)}\int _0^t (t-s)^{p-1}u(s)ds\\&-\frac{(1-{\Lambda _1}{\Lambda _2})}{\Omega \Gamma (p)}\int _0^1 (1-s)^{p-1}t^{p-1}u(s)ds+\frac{1}{\Omega \Gamma (p)}\int _0^1 (1-s)^{p-1}t^{p-1}u(s)ds\\&-\frac{t^{p-1}}{\Omega }\sum _{i=1}^m\frac{\alpha _i}{\Gamma (q+{\gamma _i})}\int _0^{\eta } (\eta -s)^{q+{\gamma _i}-1}v(s)ds\\&+\frac{t^{p-1}\Lambda _1}{\Omega \Gamma (q)}\int _0^1 (1-s)^{q-1}v(s)ds-\frac{t^{p-1}\Lambda _1}{\Omega }\sum _{j=1}^n\frac{\beta _j}{\Gamma (p+{\mu _j})}\int _0^\xi (\xi -s)^{p+{\mu _j}-1}u(s)ds\\= & {} \int _0^1 g_p(t,s)u(s)ds+\frac{{\Lambda _1}}{\Omega }\sum _{j=1}^n\frac{{\beta _j}t^{p-1}}{\Gamma (p+{\mu _j})}\int _0^1 \xi ^{p+{\mu _j}-1}(1-s)^{p-1}u(s)ds\\&-\frac{\Lambda _1}{\Omega }\sum _{j=1}^n\frac{\beta _jt^{p-1}}{\Gamma (p+{\mu _j})}\int _0^\xi (\xi -s)^{p+{\mu _j}-1}u(s)ds\\&-\frac{t^{p-1}}{\Omega }\sum _{i=1}^m\frac{\alpha _i}{\Gamma (q+{\gamma _i})}\int _0^{\eta } (\eta -s)^{q+{\gamma _i}-1}v(s)ds+\frac{t^{p-1}\Lambda _1}{\Omega \Gamma (q)}\int _0^1 (1-s)^{q-1}v(s)ds\\= & {} \int _0^1 g_p(t,s)u(s)ds+\frac{{\Lambda _1}}{\Omega }\sum _{j=1}^n\frac{{\beta _j}t^{p-1}}{\Gamma (p+{\mu _j})}\int _0^1 g_{\mu _j}^p(\xi ,s)u(s)ds\\&+\frac{1}{\Omega }\sum _{i=1}^m\frac{{\alpha _i}t^{p-1}}{\Gamma (q+\gamma _i)}\int _0^1 g_{\gamma _i}^q(\eta ,s)v(s)ds\\= & {} \int _0^1 G_1(t,s)u(s)ds, \end{aligned}$$

which implies that (2.10) holds. In a similar way, we obtain

$$\begin{aligned} y(t)= & {} -\frac{1}{\Gamma (q)}\int _0^t (t-s)^{q-1}v(s)ds+\frac{t^{q-1}}{\Omega \Gamma (q)}\int _0^1 (1-s)^{q-1}v(s)ds \\&-\frac{t^{q-1}}{\Omega }\sum _{j=1}^n\frac{\beta _j}{\Gamma (p+{\mu _j})}\int _0^{\xi } (\xi -s)^{p+{\mu _j}-1}u(s)ds\\&\quad +\frac{t^{q-1}\Lambda _2}{\Omega \Gamma (p)}\int _0^1 (1-s)^{p-1}u(s)ds\\&-\frac{t^{q-1}\Lambda _2}{\Omega }\sum _{i=1}^m\frac{\alpha _i}{\Gamma (q+{\gamma _i})}\int _0^\eta (\eta -s)^{q+{\gamma _i}-1}v(s)ds\\&+\int _0^1 \frac{(1-s)^{q-1}t^{q-1}}{\Gamma (q)}v(s)ds-\int _0^1 \frac{(1-s)^{q-1}t^{q-1}}{\Gamma (q)}v(s)ds\\= & {} \int _0^1 \frac{(1-s)^{q-1}t^{q-1}}{\Gamma (q)}v(s)ds-\frac{1}{\Gamma (q)}\int _0^t (t-s)^{q-1}v(s)ds\\&+\frac{\Lambda _2}{\Omega }\sum _{i=1}^m \frac{{\alpha _i}t^{q-1}}{\Gamma (q+\gamma _i)}\int _0^1 (1-s)^{q-1}\eta ^{q+{\gamma _i}-1}v(s)ds\\&-\frac{\Lambda _2}{\Omega }\sum _{i=1}^m \frac{{\alpha _i}t^{q-1}}{\Gamma (q+\gamma _i)}\int _0^\eta (\eta -s)^{q+{\gamma _i}-1}v(s)ds\\&+\frac{1}{\Omega }\sum _{j=1}^n\frac{\beta _jt^{q-1}}{\Gamma (p+{\mu _j})}\int _0^1 (1-s)^{p-1}\xi ^{p+{\mu _j}-1}u(s)ds\\&-\frac{1}{\Omega }\sum _{j=1}^n\frac{\beta _jt^{q-1}}{\Gamma (p+{\mu _j})}\int _0^{\xi } (\xi -s)^{p+{\mu _j}-1}u(s)ds\\= & {} \int _0^1 {g_q}(t,s)v(s)ds+\frac{\Lambda _2}{\Omega }\sum _{i=1}^m \frac{{\alpha _i}t^{q-1}}{\Gamma (q+\gamma _i)}\int _0^1 {g_{\gamma _i}^q}(\eta ,s)v(s)ds\\&+\frac{1}{\Omega }\sum _{j=1}^n\frac{\beta _jt^{q-1}}{\Gamma (p+{\mu _j})}\int _0^1 g_{\mu _j}^p(\xi ,s)u(s)ds\\= & {} \int _0^1 G_2(t,s)v(s)ds, \end{aligned}$$

which proves that (2.11) is true. This completes the proof. $\square$

Before establishing some properties of the Green’s functions, we set the following constants

$$\begin{aligned} \Lambda _3= & {} \frac{\Gamma (p)}{\Gamma (2p)}+\frac{\Lambda _1}{\Omega }\displaystyle \sum _{j=1}^n\left( \frac{\mu _j+p(1-\xi )}{p\Gamma (p+\mu _j+1)}\right) \beta _j\xi ^{p+\mu _j-1} \\&+\frac{1}{\Omega }\sum _{i=1}^m\left( \frac{\gamma _i+q(1-\eta )}{q\Gamma (q+\gamma _i+1)}\right) \alpha _i\eta ^{q+\gamma _i-1},\\ \Lambda _4= & {} \frac{\Gamma (q)}{\Gamma (2q)}+\frac{\Lambda _2}{\Omega }\displaystyle \sum _{i=1}^m\left( \frac{\gamma _i+q(1-\eta )}{q\Gamma (q+\gamma _i+1)}\right) \alpha _i\eta ^{q+\gamma _i-1}\\&+\frac{1}{\Omega }\sum _{j=1}^n\left( \frac{\mu _j+p(1-\xi )}{p\Gamma (p+\mu _j+1)}\right) \beta _j\xi ^{p+\mu _j-1},\\ \Lambda _5= & {} \frac{\Lambda _1}{\Omega }\sum _{j=1}^n\frac{\beta _j\xi ^{2p+\mu _j-2}(1-\xi )^p}{p\Gamma (p+\mu _j)} +\frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _i\eta ^{p+q+\gamma _i-2}(1-\eta )^q}{q\Gamma (q+\gamma _i)},\\ \Lambda _6= & {} \frac{\Lambda _2}{\Omega }\sum _{i=1}^m\frac{\alpha _i\eta ^{2q+\gamma _i-2}(1-\eta )^q}{q\Gamma (q+\gamma _i)} +\frac{1}{\Omega }\sum _{j=1}^n\frac{\beta _j\xi ^{p+q+\mu _j-2}(1-\xi )^p}{p\Gamma (p+\mu _j)}. \end{aligned}$$

Lemma 2.5

The Green’s functions $G_1(t,s)$ and $G_2(t,s)$ in (2.12)–(2.13) satisfy the following properties:

• $(P_1)$ $G_1(t,s)$, $G_2(t,s)$ are continuous on $[0,1]\times [0,1]$;

• $(P_2)$ $G_1(t,s), G_2(t,s)\ge 0$ for all $0\le s$, $t\le 1$;

• $(P_3)$ $G_1(t,s) \le \displaystyle \sup \nolimits _{0\le t\le 1} G(t,s) \le g_p(s,s)+\frac{\Lambda _1}{\Omega }\sum \nolimits _{j=1}^{n}\frac{\beta _j g_{\mu _j}^p(\xi ,s)}{\Gamma (p+\mu _i)} +\frac{1}{\Omega }\sum \nolimits _{i=1}^m\frac{\alpha _i g_{\gamma _i}^q(\eta ,s)}{\Gamma (q+\gamma _i)}$, $G_2(t,s)\le \displaystyle \sup \nolimits _{0\le t\le 1} G(t,s) \le g_q(s,s)+\frac{\Lambda _2}{\Omega }\sum \nolimits _{i=1}^{m}\frac{\alpha _i g_{\gamma _i}^q(\eta ,s)}{\Gamma (q+\gamma _i)}+\frac{1}{\Omega }\sum \nolimits _{j=1}^n\frac{\beta _j g_{\mu _j}^p(\xi ,s)}{\Gamma (p+\mu _j)}$;

• $(P_4)$ $\displaystyle \int _0^1 \sup \nolimits _{0\le t \le 1}G_1(t,s)ds \le \Lambda _3$ and $\displaystyle \int _0^1\sup \nolimits _{0 \le t\le 1}G_2(t,s)ds \le \Lambda _4$;

• $(P_5)$ $\displaystyle \inf \nolimits _{\xi \le t\le 1}G_1(t,s)\ge \frac{\Lambda _1}{\Omega }\sum \nolimits _{j=1}^n\frac{\beta _j\xi ^{p-1}}{\Gamma (p+\mu _j)}g_{\mu _j}^p(\xi ,s) +\frac{1}{\Omega }\sum \nolimits _{i=1}^m\frac{\alpha _i\eta ^{p-1}}{\Gamma (q+\gamma _i)}g_{\gamma _i}^q(\eta ,s)$, $\displaystyle \inf \nolimits _{\xi \le t\le 1}G_2(t,s)\ge \frac{\Lambda _2}{\Omega }\sum \nolimits _{i=1}^m\frac{\alpha _i\eta ^{q-1}}{\Gamma (q+\gamma _i)}g_{\gamma _i}^q(\eta ,s) +\frac{1}{\Omega }\sum \nolimits _{j=1}^n\frac{\beta _j\xi ^{q-1}}{\Gamma (p+\mu _j)}g_{\mu _j}^p(\xi ,s)$;

• $(P_6)$ $\displaystyle \int _{\eta }^1\inf \nolimits _{\xi \le t\le 1}G_1(t,s)ds\ge \Lambda _5$ and $\displaystyle \int _{\eta }^1\inf \nolimits _{\xi \le t\le 1}G_2(t,s)ds\ge \Lambda _6.$

Proof

It is easy to prove that the condition $(P_1)$ holds. For $0\le s,t\le 1$, using the results in [19], we have $g_{\phi }(t,s)\ge 0$, $g_{\psi }^{\phi }(\rho ,s)\ge 0$, where $\phi \in \{p,q\}$, $\rho \in \{\eta , \xi \}$ and $\psi \in \{\mu _j,\gamma _i\}$, $i=1,2,\ldots ,m$, $j=1,2,\ldots ,n$, which leads to $G_1(t,s), G_2(t,s)\ge 0$. Therefore, the property $(P_2)$ is true.

From [19], we have $g_{\phi }(t,s)\le g_{\phi }(s,s)$ for $\phi \in \{p,q\}$, $(t,s)\in [0,1]$, which yields

$$\begin{aligned} G_1(t,s)\le & {} \displaystyle \sup _{0\le t\le 1} G_1(t,s) \le g_p(s,s)+\frac{\Lambda _1}{\Omega }\displaystyle \sum _{j=1}^{n}\frac{\beta _j g_{\mu _j}^p(\xi ,s)}{\Gamma (p+\mu _i)} \nonumber \\&+\frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _i g_{\gamma _i}^q(\eta ,s)}{\Gamma (q+\gamma _i)}, s\in [0,1], \end{aligned}$$

(2.16)

and also

$$\begin{aligned} G_2(t,s)\le & {} \sup _{0\le t\le 1} G_2(t,s) \le g_q(s,s)+\frac{\Lambda _2}{\Omega }\displaystyle \sum _{i=1}^{m}\frac{\alpha _i g_{\gamma _i}^q(\eta ,s)}{\Gamma (q+\gamma _i)}\nonumber \\&+\frac{1}{\Omega }\sum _{j=1}^n\frac{\beta _j g_{\mu _j}^p(\xi ,s)}{\Gamma (p+\mu _j)}, s\in [0,1]. \end{aligned}$$

(2.17)

Thus the condition $(P_3)$ is proved. Consequently, by direct integration, we get

$$\begin{aligned} \int _0^1\sup _{0\le t\le 1} G_1(t,s)ds\le & {} \int _0^1g_p(s,s)ds+\frac{\Lambda _1}{\Omega }\displaystyle \sum _{j=1}^{n}\frac{\beta _j}{\Gamma (p+\mu _i)} \int _0^1 g_{\mu _j}^p(\xi ,s)ds \\&\,+\frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _i}{\Gamma (q+\gamma _i)}\int _0^1 g_{\gamma _i}^q(\eta ,s)ds =\Lambda _3 \end{aligned}$$

and

$$\begin{aligned} \int _0^1\sup _{0\le t\le 1} G_2(t,s)ds\le & {} \int _0^1g_q(s,s)ds+\frac{\Lambda _2}{\Omega }\sum _{i=1}^{m}\frac{\alpha _i}{\Gamma (q+\gamma _i)} \int _0^1 g_{\gamma _i}^q(\eta ,s)ds\\&\,+\frac{1}{\Omega }\sum _{j=1}^n\frac{\beta _j}{\Gamma (p+\mu _j)}\int _0^1 g_{\mu _j}^p(\xi ,s)ds =\Lambda _4. \end{aligned}$$

Therefore, the condition $(P_4)$ holds.

From the positivity of the Green functions in $(P_2)$, we have

$$\begin{aligned} \inf _{\xi \le t\le 1}G_1(t,s)= & {} \displaystyle \inf _{\xi \le t\le 1} \Bigg ( g_p(t,s)+\frac{\Lambda _1}{\Omega }\sum _{j=1}^n\frac{\beta _jt^{p-1}}{\Gamma (p+\mu _j)}g_{\mu _j}^p(\xi ,s)\nonumber \\&+\,\frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _it^{p-1}}{\Gamma (q+\gamma _i)}g_{\gamma _i}^q(\eta ,s)\Bigg )\\\ge & {} \displaystyle \inf _{\xi \le t\le 1} g_p(t,s)+ \inf _{\xi \le t\le 1}\left( \frac{\Lambda _1}{\Omega }\sum _{j=1}^n\frac{\beta _jt^{p-1}}{\Gamma (p+\mu _j)}g_{\mu _j}^p(\xi ,s)\right) \nonumber \\&+\, \inf _{\xi \le t\le 1}\left( \frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _it^{p-1}}{\Gamma (q+\gamma _i)}g_{\gamma _i}^q(\eta ,s)\right) \\\ge & {} \frac{\Lambda _1}{\Omega }\sum _{j=1}^n\frac{\beta _j\xi ^{p-1}}{\Gamma (p+\mu _j)}g_{\mu _j}^p(\xi ,s) +\inf _{\eta \le t\le 1}\left( \frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _it^{p-1}}{\Gamma (q+\gamma _i)}g_{\gamma _i}^q(\eta ,s)\right) \\\ge & {} \frac{\Lambda _1}{\Omega }\sum _{j=1}^n\frac{\beta _j\xi ^{p-1}}{\Gamma (p+\mu _j)}g_{\mu _j}^p(\xi ,s) +\frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _i\eta ^{p-1}}{\Gamma (q+\gamma _i)}g_{\gamma _i}^q(\eta ,s). \end{aligned}$$

In the same method of the above inequalities, we obtain

$$\begin{aligned} \inf _{\xi \le t\le 1}G_2(t,s)\ge & {} \displaystyle \frac{\Lambda _2}{\Omega }\sum _{i=1}^m\frac{\alpha _i\eta ^{q-1}}{\Gamma (q+\gamma _i)}g_{\gamma _i}^q(\eta ,s) +\frac{1}{\Omega }\sum _{j=1}^n\frac{\beta _j\xi ^{q-1}}{\Gamma (p+\mu _j)}g_{\mu _j}^p(\xi ,s). \end{aligned}$$

Therefore, the inequalities in $(P_5)$ are satisfied.

To prove $(P_6)$, by directly integration, we have

$$\begin{aligned} \int _{\eta }^1\inf _{\xi \le t\le 1}G_1(t,s)ds\ge & {} \displaystyle \frac{\Lambda _1}{\Omega }\sum _{j=1}^n\frac{\beta _j\xi ^{p-1}}{\Gamma (p+\mu _j)}\int _{\eta }^1g_{\mu _j}^p(\xi ,s)ds\\&+\frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _i\eta ^{p-1}}{\Gamma (q+\gamma _i)}\int _{\eta }^1 g_{\gamma _i}^q(\eta ,s)ds\\\ge & {} \displaystyle \frac{\Lambda _1}{\Omega }\sum _{j=1}^n\frac{\beta _j\xi ^{p-1}}{\Gamma (p+\mu _j)}\int _{\xi }^1g_{\mu _j}^p(\xi ,s)ds\\&+\frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _i\eta ^{p-1}}{\Gamma (q+\gamma _i)}\int _{\eta }^1 g_{\gamma _i}^q(\eta ,s)ds = \Lambda _5 \end{aligned}$$

and

$$\begin{aligned} \int _{\eta }^1\inf _{\xi \le t\le 1}G_2(t,s)ds\ge & {} \displaystyle \frac{\Lambda _2}{\Omega }\sum _{i=1}^m\frac{\alpha _i\eta ^{q-1}}{\Gamma (q+\gamma _i)}\int _{\eta }^1g_{\gamma _i}^q(\eta ,s)ds\\&+\frac{1}{\Omega }\sum _{j=1}^n\frac{\beta _j\xi ^{q-1}}{\Gamma (p+\mu _j)}\int _{\xi }^1g_{\mu _j}^p(\xi ,s)ds =\Lambda _6. \end{aligned}$$

Therefore, we get the required inequality in $(P_6)$.

3 Main results

Let $E=C([0,1],\mathbb {R})\times C([0,1],\mathbb {R})$ be the Banach space with the norm $\Vert (x,y)\Vert :=\Vert x\Vert +\Vert y\Vert$, where $\Vert x\Vert =\displaystyle \sup \nolimits _{t\in [0,1]}|x(t)|,$ $\Vert y\Vert =\displaystyle \sup \nolimits _{t\in [0,1]}|y(t)|$. Then we define the positive cone $\mathcal {P}\subset E$ by

$$\begin{aligned} \mathcal {P}=\{(x,y)\in E:\,x(t)\ge 0\quad \text {and}\quad y(t)\ge 0, \quad 0\le t\le 1\}. \end{aligned}$$

Define an operator $\mathcal {Q}$ on E by

$$\begin{aligned} \mathcal {Q}(x,y)(t)=(A(x,y)(t), B(x,y)(t)),\quad \text {for all} \,\,t\in [0,1], \end{aligned}$$

(3.1)

where the operators $A:\mathcal {P}\rightarrow E$ and $B:\mathcal {P}\rightarrow E$ are defined by

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle A(x,y)(t):=\int _0^1 G_1(t,s)f(s,x(s),y(s))ds,\\ \displaystyle B(x,y)(t):=\int _0^1 G_2(t,s)g(s,x(s),y(s))ds. \end{array}\right. \end{aligned}$$

(3.2)

Lemma 3.1

The operator $\mathcal {Q}:\mathcal {P}\rightarrow \mathcal {P}$ is completely continuous.

Proof

Since $G_1(t,s)\ge 0, G_2(t,s)\ge 0$ for $s,t\in [0,1]$, we have $A(x,y)\ge 0, B(x,y)\ge 0$ for all $x, y\in \mathcal {P}$. Hence, $A, B:\mathcal {P}\rightarrow \mathcal {P}$.

For a constant $R>0$, we define $U=\{(x,y)\in \mathcal {P}$  :  $\Vert (x,y)\Vert <R\}$. Let

$$\begin{aligned} \displaystyle L=\max _{0\le t\le 1,\,\,0\le x\le R,\,\, 0\le y\le R}\left| f(t,x,y)\right| . \end{aligned}$$

Then for $(x,y)\in U$, from Lemma 2.5, one has

$$\begin{aligned} |A(x,y)(t)|= & {} \left| \int _0^1G_1(t,s)f(s,x(s), y(s))ds\right| \\\le & {} L\int _0^1G_1(t,s)ds\\\le & {} L\int _0^1\left( g_p(s,s)+\frac{\Lambda _1}{\Omega }\displaystyle \sum _{j=1}^{n}\frac{\beta _j g_{\mu _j}^p(\xi ,s)}{\Gamma (p+\mu _i)}+\frac{1}{\Omega }\sum _{i=1}^m\frac{\alpha _i g_{\gamma _i}^q(\eta ,s)}{\Gamma (q+\gamma _i)}\right) ds\\= & {} L\Bigg [\frac{\Gamma (p)}{\Gamma (2p)}+\frac{\Lambda _1}{\Omega }\displaystyle \sum _{j=1}^n\left( \frac{\mu _j+p(1-\xi )}{p\Gamma (p+\mu _j+1)}\right) \beta _j\xi ^{p+\mu _j-1}\\&+\frac{1}{\Omega }\sum _{i=1}^m\left( \frac{\gamma _i+q(1-\eta )}{q\Gamma (q+\gamma _i+1)}\right) \alpha _i\eta ^{q+\gamma _i-1}\Bigg ]:=M_1. \end{aligned}$$

Therefore, $\Vert A(x,y)\Vert \le M_1.$ Similarly we can prove that $\Vert B(x,y)\Vert \le M_2,$ and so $\mathcal {Q}(U)$ is uniformly bounded. In addition, it follows from the continuity of f, g, the uniform continuity of $G_1(t, s), G_2(t, s)$ on $[0, 1]\times [0, 1]$ that $\mathcal {Q}:E\rightarrow E$ is continuous.

Also as in Lemma 2.5 of [19] we can prove that $\mathcal {Q}(U)$ is equi-continuous. Applying the Arzelá-Ascoli Theorem, we have that $\overline{\mathcal {Q}(U)}$ is compact, i.e., $\mathcal {Q}:\mathcal {P}\rightarrow \mathcal {P}$ is a completely continuous operator. This completes the proof. $\square$

3.1 Existence result via Leggett–Williams fixed point theorem

In this subsection, the existence of at least three positive solutions will be proved using the Leggett–Williams fixed point theorem.

Definition 3.1

A continuous mapping $\omega :\mathcal {P}\rightarrow [0,\infty )$ is said to be a nonnegative continuous concave functional on the cone $\mathcal {P}$ of a real Banach space E provided that

$$\begin{aligned} \omega (\lambda {x}+(1-\lambda )y)\ge \lambda \omega (x)+(1-\lambda )\omega (y) \end{aligned}$$

for all $x,y\in \mathcal {P}$ and $\lambda \in [0,1]$.

Let $a,b,d>0$ be given constants and define $\mathcal {P}_d=\{(x,y)\in \mathcal {P}:\Vert (x,y)\Vert <d\}$, $\overline{\mathcal {P}}_d=\{(x,y)\in \mathcal {P}:\Vert (x,y)\Vert \le d\}$ and $\mathcal {P}(\omega ,a,b)=\{(x,y)\in \mathcal {P}:\omega ((x,y))\ge a,\,\Vert (x,y)\Vert \le b\}$.

Theorem 3.1

(Leggett–Williams fixed point theorem) Let $\mathcal {P}$ be a cone in the real Banach space E and $c>0$ be a constant. Assume that there exists a concave nonnegative continuous functional $\omega$ on $\mathcal {P}$ with $\omega (x)\le \Vert x\Vert$ for all $x\in \overline{\mathcal {P}}_c$. Let $Q:\overline{\mathcal {P}}_c\rightarrow \overline{\mathcal {P}}_c$ be a completely continuous operator. Suppose that there exist constants $0<a<b<d\le c$ such that the following conditions hold:

• (i) $\{x\in \mathcal {P}(\omega ,b,d):\omega (x)>b\}\ne \emptyset$ and $\omega (Qx)>b$ for $x\in \mathcal {P}(\omega ,b,d)$;

• (ii) $\Vert Qx\Vert <a$ for $x\le a$;

• (iii) $\omega (Qx)>b$ for $x\in \mathcal {P}(\omega ,b,c)$ with $\Vert Qx\Vert >d$.

Then Q has at least three fixed points $x_1,x_2$ and $x_3$ in $\overline{\mathcal {P}}_c$. In addition, $\Vert x_1\Vert <a$, $\omega (x_2)>b$, $\Vert x_3\Vert >a$ with $\omega (x_3)<b$.

Theorem 3.2

Let functions $f,g:[0,1]\times \mathbb {R}_+^2\rightarrow \mathbb {R}_+$ be continuous functions. Suppose that there exist constants $0<a<b<c$ such that the following assumptions hold:

• $(H_1)$ $\displaystyle f(t,x,y)<\frac{a}{2\Lambda _3}$ and $\displaystyle g(t,x,y)<\frac{a}{2\Lambda _4}$, for $(t,x,y)\in [0,1]\times [0,a]\times [0,a]$;

• $(H_2)$ $\displaystyle f(t,x,y)>\frac{b}{2\Lambda _5}$ and $\displaystyle g(t,x,y)>\frac{b}{2\Lambda _6}$, for $(t,x,y)\in [\eta ,1]\times [b,c]\times [b,c]$;

• $(H_3)$ $\displaystyle f(t,x,y)\le \frac{c}{2\Lambda _3}$ and $\displaystyle g(t,x,y)\le \frac{c}{2\Lambda _4}$, for $(t,x,y)\in [0,1]\times [0,c]\times [0,c]$.

Then, the problem (1.3) has at least three positive solutions $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ such that $\Vert (x_1,y_1)\Vert <a$, $\inf _{\xi \le t\le 1}(x_2,y_2)(t)>b$ and $\Vert (x_3,y_3)\Vert >a$ with $\inf _{\xi \le t\le 1}(x_3,y_3)(t)<b$.

Proof

Firstly, we will show that $\mathcal {Q}:\overline{\mathcal {P}}_c\rightarrow \overline{\mathcal {P}}_c$. For $(x,y)\in \overline{\mathcal {P}}_c$, it follows that $\Vert (x,y)\Vert \le c$. From the condition $(H_3)$ and Lemma 2.5, we have

$$\begin{aligned} \Vert \mathcal {Q}(x,y)\Vert= & {} \sup _{t\in [0,1]}|A(x,y)(t)|+\sup _{t\in [0,1]}|B(x,y)(t)|\\\le & {} \int _0^1 \sup _{t\in [0,1]}G_1(t,s)f(s,x(s),y(s))ds\\&\quad +\int _0^1 \sup _{t\in [0,1]}G_2(t,s)g(s,x(s),y(s))ds\\\le & {} \frac{c}{2\Lambda _3}\int _0^1\sup _{t\in [0,1]}G_1(t,s)ds+\frac{c}{2\Lambda _4}\int _0^1\sup _{t\in [0,1]}G_2(t,s)ds=c. \end{aligned}$$

This implies that $Q:\overline{\mathcal {P}}_c\rightarrow \overline{\mathcal {P}}_c$.

Let $(x,y)\in \overline{\mathcal {P}}_a$. The condition $(H_1)$ implies that

$$\begin{aligned} \Vert \mathcal {Q}(x,y)\Vert\le & {} \int _0^1\sup _{t\in [0,1]} G_1(t,s)f(s,x(s),y(s))ds\\&\quad +\int _0^1 \sup _{t\in [0,1]}G_2(t,s)g(s,x(s),y(s))ds\\< & {} \frac{a}{2\Lambda _3}\int _0^1\sup _{t\in [0,1]}G_1(t,s)ds+\frac{a}{2\Lambda _4}\int _0^1\sup _{t\in [0,1]}G_2(t,s)ds=a. \end{aligned}$$

Hence, the condition (ii) of Theorem 3.1 is fulfilled.

Now, we let a concave nonnegative continuous functional $\omega$ on $\mathcal {P}$ by $\omega (x,y)=\inf _{t\in [\xi ,1]}|x(t)|+\inf _{t\in [\xi ,1]}|y(t)|$. Choosing $(x,y)(t)=((b+c)/2,(b+c)/2)$ for all $t\in [0,1]$, we have that $(x,y)(t)\in \overline{\mathcal {P}}(\omega ,b,c)$ and $\omega ((x,y))=\omega ((b+c)/2, (b+c)/2))>b$. Then we obtain $\{(x,y)\in \mathcal {P}(\omega ,b,c)\,:\,\omega ((x,y))>b\}\ne \emptyset$. Thus, if $(x,y)\in \overline{P}(\omega ,b,c)$, then $b\le x(t)\le c$ and $b\le y(t)\le c$ for $t\in [\xi ,1]$. Using the condition $(H_2)$ and Lemma 2.5, we have

$$\begin{aligned} \omega (Q(x,y)(t))= & {} \inf _{\xi \le t\le 1}|A(x,y)(t)|+\inf _{\xi \le t\le 1}|B(x,y)(t)|\\\ge & {} \int _\eta ^1 \inf _{\xi \le t\le 1}G_1(t,s)f(s,x(s),y(s))ds\\&+\int _\eta ^1 \inf _{\xi \le t\le 1}G_2(t,s)g(s,x(s),y(s))ds\\> & {} \frac{b}{2\Lambda _5}\int _\eta ^1 \inf _{\xi \le t\le 1}G_1(t,s)ds+\frac{b}{2\Lambda _6}\int _\eta ^1 \inf _{\xi \le t\le 1}G_2(t,s)ds = b. \end{aligned}$$

Hence $\omega (\mathcal {Q}(x,y))>b$ for all $(x,y)\in \mathcal {P}(\omega ,b,c)$. This implies that the condition (i) of Theorem 3.1 is fulfilled.

Finally, we suppose that $(x,y)\in \mathcal {P}(\omega ,b,c)$ with $\Vert \mathcal {Q}(x,y)\Vert >d$, where $b<d\le c$. This implies that $b\le x(t)\le c$ and $b\le y(t)\le c$ for all $t\in [\xi ,1]$. By $(H_2)$ and Lemma 2.5, we obtain

$$\begin{aligned} \omega (Q(x,y)(t))= & {} \inf _{\xi \le t\le 1}|A(x,y)(t)|+\inf _{\xi \le t\le 1}|B(x,y)(t)|\\> & {} \frac{b}{2\Lambda _5}\int _\eta ^1 \inf _{\xi \le t\le 1}G_1(t,s)ds+\frac{b}{2\Lambda _6}\int _\eta ^1 \inf _{\xi \le t\le 1}G_2(t,s)ds =b, \end{aligned}$$

which leads to satisfy condition (iii) of Theorem 3.1. Therefore, by applying Theorem 3.1, we deduce that the problem (1.3) has at least three positive solutions $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ such that

$$\begin{aligned}&\Vert (x_1,y_1)\Vert<a,\, \displaystyle \inf \nolimits _{\xi \le t\le 1}(x_2,y_2)(t)>b \quad \text {and}\\&\quad \Vert (x_3,y_3)\Vert >a\,\,\text {with}\,\,\displaystyle \inf \nolimits _{\xi \le t\le 1}(x_3,y_3)(t)<b. \end{aligned}$$

This completes the proof. $\square$

3.2 Existence result via Guo–Krasnoselskii fixed point theorem

In this subsection, the existence theorems of at least one positive solution will be established using the Guo–Krasnoselskii fixed point theorem.

Theorem 3.3

(Guo–Krasnoselskii fixed point theorem) Let E be a Banach space, and let $\mathcal {P}\subset E$ be a cone. Assume that $\Phi _1,\Phi _2$ are bounded open subsets of E with $0\in \Phi _1$, $\overline{\Phi }_1\subset \Phi _2$, and let $Q:\mathcal {P}\cap (\overline{\Phi }_2{\setminus }\Phi _1)\rightarrow \mathcal {P}$ be a completely continuous operator such that:

• (i) $\Vert Qx\Vert \ge \Vert x\Vert$, $x\in \mathcal {P}\cap \partial \Phi _1$, and $\Vert Qx\Vert \le \Vert x\Vert$, $x\in \mathcal {P}\cap \partial \Phi _2$; or

• (ii) $\Vert Qx\Vert \le \Vert x\Vert$, $x\in \mathcal {P}\cap \partial \Phi _1$, and $\Vert Qx\Vert \ge \Vert x\Vert$, $x\in \mathcal {P}\cap \partial \Phi _2$.

Then Q has a fixed point in $\mathcal {P}\cap (\overline{\Phi }_2{\setminus }\Phi _1)$.

Theorem 3.4

Let $f,g:[0,1]\times \mathbb {R}_+^2\rightarrow \mathbb {R}_+$ be continuous functions. Suppose that there exist constants $\lambda _2>\lambda _1>0$, $\kappa _1\in (\Lambda _5^{-1},\infty )$, $\kappa _2\in (\Lambda _6^{-1},\infty )$, $\kappa _3\in (0,\Lambda _3^{-1})$ and $\kappa _4\in (0,\Lambda _4^{-1})$. In addition, assume the the following condition hold:

• $(H_4)$ $f(t,x,y)\ge \displaystyle \frac{\kappa _1\lambda _1}{2}$ for $(t,x,y)\in [0,1]\times [0,\lambda _1]\times [0,\lambda _1]$ and $g(t,x,y)\ge \displaystyle \frac{\kappa _2\lambda _1}{2}$ for $(t,x,y)\in [0,1]\times [0,\lambda _1]\times [0,\lambda _1]$;

• $(H_5)$ $f(t,x,y)\le \displaystyle \frac{\kappa _3\lambda _2}{2}$ for $(t,x,y)\in [0,1]\times [0,\lambda _2]\times [0,\lambda _2]$ and $g(t,x,y)\le \displaystyle \frac{\kappa _4\lambda _2}{2}$ for $(t,x,y)\in [0,1]\times [0,\lambda _2]\times [0,\lambda _2]$.

Then the problem (1.3) has at least one positive solution (x, y) such that

$$\begin{aligned} \lambda _1<\Vert (x,y)\Vert <\lambda _2. \end{aligned}$$

Proof

From Lemma 3.1, the operator $\mathcal {Q}:\mathcal {P}\rightarrow \mathcal {P}$ is completely continuous. Let $\Phi _1=\{(x,y)\in E:\Vert (x,y)\Vert <\lambda _1\}$. Hence, for any $(x,y)\in \mathcal {P}\cap \partial \Phi _1$, we get $0\le x(t)\le \lambda _1$ and $0\le y(t)\le \lambda _1$ for all $t\in [0,1]$. Using the condition $(H_4)$ and Lemma 2.5, one has

$$\begin{aligned} \Vert \mathcal {Q}(x,y)\Vert= & {} \sup _{t\in [0,1]}\int _0^1G_1(t,s)f(s, x(s),y(s))ds\\&\quad +\sup _{t\in [0,1]}\int _0^1G_2(t,s)g(s, x(s),y(s))ds\\\ge & {} \int _{\eta }^1\inf _{\xi \le t\le 1}G_1(t,s)f(s, x(s),y(s))ds\\&\quad +\int _{\eta }^1\inf _{\eta \le t\le 1}G_2(t,s)g(s, x(s),y(s))ds\\\ge & {} \frac{\kappa _1\lambda _1}{2}\int _{\eta }^1\inf _{\xi \le t\le 1}G_1(t,s)ds+\frac{\kappa _2\lambda _1}{2}\int _{\eta }^1\inf _{\eta \le t\le 1}G_2(t,s)ds \ge \lambda _1, \end{aligned}$$

which means that $\Vert \mathcal {Q}(x,y)\Vert \ge \Vert (x,y)\Vert$ for $(x,y)\in \mathcal {P}\cap \partial \Phi _1$.

Define $\Phi _2=\{(x,y)\in E:\Vert (x,y)\Vert <\lambda _2\}$. Therefore, for any $(x,y)\in \mathcal {P}\cap \partial \Phi _2$, we get $0\le x(t)\le \lambda _2$ and $0\le y(t)\le \lambda _2$ for all $t\in [0,1]$. From assumption $(H_5)$, we obtain

$$\begin{aligned} \Vert \mathcal {Q}(x,y)\Vert\le & {} \int _0^1\sup _{t\in [0,1]}G_1(t,s)f(s, x(s),y(s))ds\\&\quad +\int _0^1\sup _{t\in [0,1]}G_2(t,s)g(s, x(s),y(s))ds\\\le & {} \frac{\kappa _3\lambda _2}{2}\int _0^1\sup _{t\in [0,1]}G_1(t,s)ds+\frac{\kappa _4\lambda _2}{2}\int _0^1\sup _{t\in [0,1]}G_2(t,s)ds\le \lambda _2, \end{aligned}$$

which yields $\Vert \mathcal {Q}(x,y)\Vert \le \Vert (x,y)\Vert$ for $(x,y)\in \mathcal {P}\cap \partial \Phi _2$.

Therefore, the first part of Theorem 3.3 implies that the operator $\mathcal {Q}$ has a fixed point in $\mathcal {P}\cap (\overline{\Phi }_2{\setminus }\Phi _1)$ which is a positive solution of problem (1.3). Hence, the problem (1.3) has at least on positive solution (x, y) such that

$$\begin{aligned} \lambda _1<\Vert (x,y)\Vert <\lambda _2. \end{aligned}$$

The proof is complete. $\square$

Similarly to the previous theorem, we can prove the following result.

Theorem 3.5

Let $f,g:[0,1]\times \mathbb {R}_+^2\rightarrow \mathbb {R}_+$ be continuous functions. Assume that there exist constants $\lambda _2>\lambda _1>0$, $\kappa _1\in (\Lambda _5^{-1},\infty )$, $\kappa _2\in (\Lambda _6^{-1},\infty )$, $\kappa _3\in (0,\Lambda _3^{-1})$ and $\kappa _4\in (0,\Lambda _4^{-1})$. In addition, assume the the following condition hold.

• $(H_6)$ $f(t,x,y)\le \displaystyle \frac{\kappa _3\lambda _1}{2}$ for $(t,x,y)\in [0,1]\times [0,\lambda _1]\times [0,\lambda _1]$ and $g(t,x,y)\le \displaystyle \frac{\kappa _4\lambda _1}{2}$ for $(t,x,y)\in [0,1]\times [0,\lambda _1]\times [0,\lambda _1]$;

• $(H_7)$ $f(t,x,y)\ge \displaystyle \frac{\kappa _1\lambda _2}{2}$ for $(t,x,y)\in [0,1]\times [0,\lambda _2]\times [0,\lambda _2]$ and $g(t,x,y)\ge \displaystyle \frac{\kappa _2\lambda _2}{2}$ for $(t,x,y)\in [0,1]\times [0,\lambda _2]\times [0,\lambda _2]$.

Then the problem (1.3) has at least one positive solution (x, y) such that

$$\begin{aligned} \lambda _1<\Vert (x,y)\Vert <\lambda _2. \end{aligned}$$

4 Examples

In this section, we present two examples to illustrate our results.

Example 4.1

Consider following fractional system of differential equations subject to the nonlocal fractional integral boundary conditions of the form

$$\begin{aligned} \left\{ \begin{array}{ll} D^{\frac{3}{2}}x(t)+f(t,x(t),y(t))=0,\quad t\in (0,1),\\ D^{\frac{5}{3}}y(t)+g(t,x(t),y(t))=0,\quad t\in (0,1),\\ x(0)=0, \, x(1)=\displaystyle \frac{2}{3}I^{\frac{1}{2}}y\left( \frac{1}{4}\right) +\frac{\sqrt{3}}{5}I^{\frac{3}{2}}y\left( \frac{1}{4}\right) +\frac{\pi }{12}I^{\frac{5}{2}}y\left( \frac{1}{4}\right) ,\\ y(0)=0,\, y(1)=\displaystyle \frac{\sqrt{2}}{5}I^{\frac{1}{3}}x\left( \frac{4}{7}\right) +\frac{1}{\sqrt{7}}I^{\frac{2}{3}}x\left( \frac{4}{7}\right) +\frac{2}{\sqrt{e}}I^{\frac{4}{3}}x\left( \frac{4}{7}\right) +\frac{8}{13}I^{\frac{5}{3}}x\left( \frac{4}{7}\right) ,\\ \end{array}\right. \end{aligned}$$

(4.1)

where

$$\begin{aligned} f(t,x,y)=\left\{ \begin{array}{ll} \displaystyle x\left( \frac{2}{3}-x\right) +\frac{1}{2}y\left( \frac{2}{3}-y\right) +\frac{1}{5}(t+1);&{}0\le t\le 1;0\le x,y\le 2/3,\\ \displaystyle \frac{1}{5}(t+1)|\cos (x\pi )|+\frac{1}{5}(t+1)|\cos (y\pi )|\\ \displaystyle +\,45\left( x-\frac{2}{3}\right) \left( y-\frac{2}{3}\right) ;&{}0\le t\le 1;2/3\le x,y\le 4/3,\\ \displaystyle \frac{1}{5}(t+101)+\sin ^2\left( \left( x-\frac{4}{3}\right) \left( y-\frac{4}{3}\right) \right) ;&{}0\le t\le 1;4/3\le x,y< \infty , \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} g(t,x,y)=\left\{ \begin{array}{ll} \displaystyle xy\left( \frac{2}{3}-x\right) \left( \frac{2}{3}-y\right) +\frac{1}{8}(t+2);&{}0\le t\le 1;0\le x,y\le 2/3,\\ \displaystyle \frac{1}{8\sqrt{3}}(t+2)|\sin (x\pi )|+\frac{1}{8\sqrt{3}}(t+2)|\sin (y\pi )|\\ \displaystyle +\,27\left( x-\frac{2}{3}\right) \left( y-\frac{2}{3}\right) ;&{}0\le t\le 1;2/3\le x,y\le 4/3,\\ \displaystyle \frac{1}{8}(t+98)+\sin ^4\left( \left( x-\frac{4}{3}\right) \left( y-\frac{4}{3}\right) \right) ;&{}0\le t\le 1;4/3\le x,y< \infty . \end{array}\right. \end{aligned}$$

Here $p=3/2$, $q=5/3$, $m=3$, $\eta =1/4$, $\alpha _1=2/3$, $\gamma _1=1/2$, $\alpha _2=\sqrt{3}/5$, $\gamma _2=3/2$, $\alpha _3=\pi /12$, $\gamma _3=5/2$, $n=4$, $\xi =4/7$, $\beta _1=\sqrt{2}/5$, $\mu _1=1/3$, $\beta _2=1/\sqrt{7}$, $\mu _2=2/3$, $\beta _3=2/\sqrt{e}$, $\mu _3=4/3$, $\beta _4=8/13$, $\mu _4=5/3$. We find that $\Lambda _1=0.1173432604$ and $\Lambda _2=0.6208838470$ which leads to $\Omega =0.9271434651>0$. In addition, we can compute that $\Lambda _3=0.5487565277$, $\Lambda _4=0.6859288172$, $\Lambda _5=0.03857691941$ and $\Lambda _6=0.1101600049$.

Choosing $a=2/3$, $b=4/3$, $c=24$, we get

$$\begin{aligned} f(t,x,y)\le 0.5666666667 \quad \text {and}\quad g(t,x,y)\le 0.4120370370, \end{aligned}$$

which yields for $0\le t\le 1$ and $0\le x,y \le 2/3$,

$$\begin{aligned} f(t,x,y)<0.6074339286=\frac{a}{2\Lambda _3}\quad \text {and}\quad g(t,x,y)<0.4859590749=\frac{a}{2\Lambda _4}. \end{aligned}$$

In addition, we obtain

$$\begin{aligned} f(t,x,y)\ge 20.25000000 \quad \text {and}\quad g(t,x,y)\ge 12.28125000, \end{aligned}$$

which leads to

$$\begin{aligned} f(t,x,y)>17.28149051=\frac{b}{2\Lambda _5}\quad \text {and}\quad g(t,x,y)>6.051803167=\frac{b}{2\Lambda _6}, \end{aligned}$$

for $1/4\le t\le 1$ and $4/3\le x,y \le 24$. Also we have

$$\begin{aligned} f(t,x,y)\le 21.40000000 \quad \text {and}\quad g(t,x,y)\le 13.37500000, \end{aligned}$$

which gives

$$\begin{aligned} f(t,x,y)<21.86762143=\frac{c}{2\Lambda _3}\quad \text {and}\quad g(t,x,y)<17.49452670=\frac{c}{2\Lambda _4}, \end{aligned}$$

for $0\le t\le 1$ and $0\le x,y \le 24$.

Therefore, the conditions $(H_1$–$H_3)$ of Theorem 3.2 hold. Applying Theorem 3.2, we deduce that the problem (4.1) has at least three positive solutions $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ such that $\Vert (x_1,y_1)\Vert <2/3$, $\inf _{\frac{4}{7}\le t\le 1}(x_2,y_2)(t)>4/3$ and $\Vert (x_3,y_3)\Vert >2/3$ with $\inf _{\frac{4}{7}\le t\le 1}(x_3,y_3)(t)<4/3$.

Example 4.2

Consider following fractional system of differential equations subject to the nonlocal fractional integral boundary conditions of the form

$$\begin{aligned} \left\{ \begin{array}{ll} D^{\frac{7}{4}}x(t)+f(t,x(t),y(t))=0,\quad t\in (0,1),\\ D^{\frac{9}{5}}y(t)+g(t,x(t),y(t))=0,\quad t\in (0,1),\\ x(0)=0,\, x(1)=\displaystyle \frac{\sqrt{\pi }}{13}I^{\frac{1}{4}}y\left( \frac{3}{4}\right) +\frac{7}{12}I^{\frac{1}{2}}y\left( \frac{3}{4}\right) +\frac{\sqrt{2}}{15}I^{\frac{3}{4}}y\left( \frac{3}{4}\right) \\ \displaystyle \qquad \qquad \qquad \quad \quad \,\,\,\,\,\,\,+\frac{4}{\sqrt{11}}I^{\frac{5}{4}}y\left( \frac{3}{4}\right) ,\\ y(0)=0,\, y(1)=\displaystyle \frac{3}{16}I^{\frac{1}{5}}x\left( \frac{16}{19}\right) +\frac{2}{\sqrt{5}}I^{\frac{2}{5}}x\left( \frac{16}{19}\right) +\frac{1}{3e^2}I^{\frac{3}{5}}x\left( \frac{16}{19}\right) \\ \displaystyle \qquad \qquad \qquad \quad \quad \,\,\,\,\,\,\,+\frac{3}{8\pi }I^{\frac{4}{5}}x\left( \frac{16}{19}\right) +\frac{4}{13\sqrt{7}}I^{\frac{6}{5}}x\left( \frac{16}{19}\right) , \end{array}\right. \end{aligned}$$

(4.2)

where

$$\begin{aligned} f(t,x,y)=\left\{ \begin{array}{ll} \displaystyle x\left( 2-x\right) +y\left( 2-y\right) +8(1+t);&{}0\le t\le 1;0\le x,y\le 2,\\ \displaystyle 8(1+t)e^{2-x}+\sin ^2(\pi y);&{}0\le t\le 1;2\le x,y<\infty , \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} g(t,x,y)=\left\{ \begin{array}{ll} \displaystyle xy(2-x)^2e^{-y}+7(\sqrt{t}+1)+|\sin (\pi y)|;&{}0\le t\le 1;0\le x,y\le 2,\\ \displaystyle 7(\sqrt{t}+1)\cos ^2(2-x)+4\sin ^2(\pi x)\cos ^4(\pi y);&{}0\le t\le 1;2\le x,y<\infty . \end{array}\right. \end{aligned}$$

Here $p=7/4$, $q=9/5$, $m=4$, $\eta =3/4$, $\alpha _1=\sqrt{\pi }/13$, $\gamma _1=1/4$, $\alpha _2=7/12$, $\gamma _2=1/2$, $\alpha _3=\sqrt{2}/15$, $\gamma _3=3/4$, $\alpha _4=4/\sqrt{11}$, $\gamma _4=5/4$, $n=5$, $\xi =16/19$, $\beta _1=3/16$, $\mu _1=1/5$, $\beta _2=2/\sqrt{5}$, $\mu _2=2/5$, $\beta _3=1/3e^2$, $\mu _3=3/5$, $\beta _4=3/8\pi$, $\mu _4=4/5$, $\beta _5=4/13\sqrt{7}$, $\mu _5=6/5$. We find that $\Lambda _1=0.7502528482$ and $\Lambda _2=0.9064443536$ which yields $\Omega =0.3199375420>0$. Further, we can compute that $\Lambda _3=0.8012222566$, $\Lambda _4=0.9066096747$, $\Lambda _5=0.1389153084$ and $\Lambda _6=0.1437856901$.

Choosing $\lambda _1=2$, $\lambda _2=40$, $\kappa _1=8\in (\Lambda _5^{-1},\infty )=(7.198630673,\infty )$, $\kappa _2=7\in (\Lambda _6^{-1},\infty )=(6.954795010,\infty )$, $\kappa _3=1\in (0,\Lambda _3^{-1})=(0, 1.248093137)$ and $\kappa _4=1\in (0,\Lambda _4^{-1})=(0, 1.103010510)$, we have

$$\begin{aligned} f(t,x,y)\ge 8\ge \frac{\kappa _1\lambda _1}{2} \quad \text {and}\quad g(t,x,y)\ge 7\ge \frac{\kappa _2\lambda _1}{2}, \end{aligned}$$

for $0\le t\le 1$, $0\le x,y\le 2$. Also we have

$$\begin{aligned} f(t,x,y)\le 17\le \frac{\kappa _3\lambda _2}{2} \quad \text {and}\quad g(t,x,y)\le 18\le \frac{\kappa _4\lambda _2}{2}, \end{aligned}$$

for $0\le t\le 1$, $2\le x,y< \infty$.

Thus the conditions $(H_4$–$H_5)$ hold. By Theorem 3.4, we conclude that the problem (4.2) has at least one positive solution (x, y) such that $2<\Vert (x,y)\Vert <40$.