Positive solutions for Caputo fractional differential system with coupled boundary conditions

Abstract

This paper focuses on the Caputo fractional differential system involving coupled integral boundary conditions and parameters. Using the properties of the Green function, the Leray–Schauder’s alternative and the Banach contraction principle, the existence and uniqueness results of the system are established. An example is then given to demonstrate the validity of the result.

1 Introduction

Fractional differential systems have been of great interest recently. This paper mainly presents the existence and uniqueness solution of the following general fractional differential system involving coupled integral boundary conditions and parameters:

$$ \textstyle\begin{cases} {}^{\mathrm{c}}D^{\alpha_{1}}u(t)+\lambda_{1} f_{1}(t,u(t), v(t))=0, \\ {}^{\mathrm{c}}D^{\alpha_{2}}v(t)+\lambda_{2} f_{2}(t,u(t), v(t))=0,\quad 0< t< 1, \\ u'(0)=u''(0)=\cdots= u^{(n-1)}(0)=0,\qquad u(1)=\mu_{1}\int_{0}^{1}a(s)v(s)\,dA_{1}(s), \\ v'(0)=v''(0)=\cdots= v^{(m-1)}(0)=0,\qquad v(1)=\mu_{2}\int_{0}^{1}b(s)u(s)\,dA_{2}(s), \end{cases} $$

(1)

where $\lambda_{i}>0$ is a parameter, $n-1<\alpha_{1} \leq n$, $m-1<\alpha_{2} \leq m$, $n, m\geq2$, $D^{\alpha_{i}}_{0^{+}}$ is the standard Caputo derivative; $\mu_{i}>0$ is a constant, $\int_{0}^{1}a(s)v(s)\,dA_{1}(s)$, $\int _{0}^{1}b(s)u(s)\,dA_{2}(s)$ denote the Riemann–Stieltjes integral with a signed measure, that is, $A_{i}: [0, 1]\rightarrow[0, +\infty)$ is the function of bounded variation; $a, b : [0, 1]\rightarrow [0,+\infty)$ are continuous, $f_{i}: [0,1]\times [0,+\infty)\times[0,+\infty) \rightarrow[0, +\infty)$ is a continuous function, $i=1, 2$.

In the mathematical context, fractional differential equations involving different boundary value conditions have aroused the interest of many scholars, see references [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21] to name a few. Bai and Qiu [22] discussed the following nonlinear fractional differential equation with two-point boundary value conditions by using the Krasnoselskii’s fixed point theorem:

$$ \textstyle\begin{cases} D^{\alpha}_{0^{+}}u(t)+f(t, u(t))=0,\quad 0< t< 1, \\ u(0)=u'(1)=u''(0)=0, \end{cases} $$

where $2 < \alpha\leq3$, $D^{\alpha}_{0^{+}}$ is Caputo derivative.

Wang et al. [23] gave the existence and uniqueness results for the coupled fractional differential system

$$ \textstyle\begin{cases} D^{\alpha}_{0^{+}}u(t)+f(t, v(t))=0, \\ D^{\beta}_{0^{+}}v(t)+g(t, u(t))=0,\quad 0< t< 1, \\ u(0)=v{(0)}=0,\qquad u(1)=au(\xi),\qquad v(1)=bv(\xi), \end{cases} $$

where $1 < \alpha, \beta< 2$, $0\leq a, b < 1$, $0 < \xi< 1$, $D^{\alpha}_{0^{+}}$, $D^{\beta}_{0^{+}}$ are two standard Riemann–Liouville fractional derivatives, $f, g: [0, 1]\times[0, +\infty)\rightarrow[0, +\infty)$ are continuous. The whole discussion was based on the Banach fixed point theorem and the nonlinear alternative of Leray–Schauder type.

Recently, Henderson and Luca in [24] considered the system of fractional differential equations

$$ \textstyle\begin{cases} D^{\alpha_{1} }_{0^{+}}u(t)+\lambda_{1} f_{1}(t,u(t), v(t))=0, \\ D^{\alpha_{2} }_{0^{+}}v(t)+\lambda_{2} f_{2}(t,u(t), v(t))=0,\quad 0< t< 1, \end{cases} $$

(2)

with the multi-point boundary conditions

$$ \textstyle\begin{cases} u(0)=u'(0)=\cdots= u^{(n-2)}=0, \qquad u(1)=\sum_{i=1}^{p}a_{i}u(\xi_{i}), \\ v(0)=v'(0)=\cdots= v^{(m-2)}=0, \qquad v(1)=\sum_{i=1}^{q}b_{i}v(\eta_{i}), \end{cases} $$

where $n-1<\alpha_{1} \leq n$, $m-1<\alpha_{2} \leq m$, $n, m\geq2$, $\lambda_{i}>0$ is a parameter, $D^{\alpha_{i}}_{0^{+}}$, $D^{\beta_{i}}_{0^{+}}$ are Riemann–Liouville derivatives; $a_{i}>0$, $b_{i}>0$ are constants, $f_{i}: [0,1]\times [0,+\infty)\times[0,+\infty) \rightarrow[0, +\infty)$ is a continuous function. By the use of Krasnoselskii’s fixed point theorem, the authors in [24] got the existence of positive solutions for the above system. System (2) with coupled boundary value conditions

$$ \textstyle\begin{cases} u(0)=u'(0)=\cdots= u^{(n-2)}=0, \qquad u(1)=\mu_{1}\int _{0}^{1}v(s)\,dA_{1}(s), \\ v(0)=v'(0)=\cdots= v^{(m-2)}=0, \qquad v(1)=\mu_{2}\int _{0}^{1}u(s)\,dA_{2}(s) \end{cases} $$

has also been discussed in [25, 26], where $\mu_{i}>0$ is a constant.

Fractional differential systems involving derivatives with coupled boundary conditions have witnessed significant development, as shown by [27,28,29,30], but most of the authors considered the fractional equations with Riemann–Liouville derivatives. The equation discussed in this paper is exactly the Caputo fractional equation. The purpose of this paper is to investigate the existence and uniqueness of positive solutions for Caputo fractional differential systems with coupled integral boundary conditions. In this paper, the Caputo derivatives of orders $\alpha_{1}$ and $\alpha_{2}$ can be different, and in case $dA_{1}(s) = dA_{2}(s) = ds$ or $g(s)\,ds$, system (1) reduces to a multi-point boundary value problem as well.

2 Preliminaries and lemmas

Definition 2.1

([31, 32])

The Caputo fractional order derivative of order $\alpha>0$, $n-1 < \alpha< n$, $n\in \mathbb{N}$ is defined as

$${}^{\mathrm{c}}D^{\alpha}u(t)=\frac{1}{\varGamma(n-\alpha)} \int_{0}^{t}(t-s)^{n-\alpha-1}u^{(n)}(s) \,ds, $$

where $u \in C^{n}(J, \mathbb{R})$, $\mathbb{R}=(-\infty, +\infty)$, $\mathbb{N}$ denotes the natural number set, $n =[\alpha]+1$, and $[\alpha]$ denotes the integer part of α.

Definition 2.2

([31, 32])

Let $\alpha>0$ and let u be piecewise continuous on $(0, +\infty)$ and integrable on any finite subinterval of J. Then for $t>0$, we call

$$I^{\alpha}u(t)=\frac{1}{\varGamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}u(s)\,ds, $$

the Riemann–Liouville fractional integral of u of order α.

Lemma 2.1

([31, 32])

Let $n-1<\alpha\leq n$, $u \in C^{n}[0, 1]$. Then

$$I^{\alpha}\bigl({}^{\mathrm{c}}D^{\alpha}u\bigr) (t)=u(t)+c_{0}+c_{1} t+c_{2} t^{2}+ \cdots+c_{n-1} t^{n-1}, $$

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

Lemma 2.2

Assume the following condition ($\mathrm{H}_{0}$) holds:

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

$$k_{1}= \int_{0}^{1}a(t)\,dA_{1}(t)>0, \qquad k_{2}= \int_{0}^{1}b(t)\,dA_{2}(t)>0,\qquad 1-\mu _{1}\mu_{2}k_{1}k_{2}>0. $$

Let $h_{i}\in C(0, 1)\cap L(0, 1)$ ($i=1, 2$). Then the system with the coupled boundary conditions

$$ \textstyle\begin{cases} {}^{\mathrm{c}}D^{\alpha_{1}}u(t)+h_{1}(t)=0, \qquad {}^{\mathrm{c}}D^{\alpha_{2}}v(t)+h_{2}(t)=0,\quad 0< t< 1, \\ u'(0)=u''(0)=\cdots= u^{(n-1)}(0)=0, \qquad u(1)=\mu_{1}\int_{0}^{1}a(t)v(s)\,dA_{1}(s), \\ v'(0)=v''(0)=\cdots= v^{(m-1)}(0)=0, \qquad v(1)=\mu_{2}\int_{0}^{1}b(t)u(s)\,dA_{2}(s) \end{cases} $$

(3)

has a unique integral representation

$$ \textstyle\begin{cases} u(t)=\int_{0}^{1} K_{1}(t, s)h_{1}(s)\,ds+\int _{0}^{1} H_{1}(t, s)h_{2}(s)\,ds, \\ v(t)=\int_{0}^{1} K_{2}(t, s)h_{2}(s)\,ds+\int_{0}^{1} H_{2}(t, s)h_{1}(s)\,ds, \end{cases} $$

(4)

where

$$ \begin{aligned} &K_{1}(t, s)=\frac{\mu_{1}\mu_{2}k_{1} }{1-\mu _{1}\mu_{2}k_{1}k_{2}} \int_{0}^{1}G_{1}(t, s)b(t) \,dA_{2}(t)+ G_{1}(t, s), \\ &H_{1}(t, s)= \frac{\mu_{1} }{1-\mu_{1}\mu_{2}k_{1}k_{2}} \int_{0}^{1} G_{2}(t, s)a(t) \,dA_{1}(t), \\ &K_{2}(t, s)=\frac{\mu_{2}\mu_{1}k_{2} }{1-\mu_{1}\mu_{2}k_{1}k_{2}} \int_{0}^{1} G_{2}(t, s)a(t) \,dA_{1}(t)+ G_{2}(t, s), \\ &H_{2}(t, s)= \frac{\mu_{2} }{1-\mu_{1}\mu_{2}k_{1}k_{2}} \int_{0}^{1}G_{1}(t, s) b(t) \,dA_{2}(t), \end{aligned} $$

(5)

and

$$ G_{i}(t,s)= \textstyle\begin{cases} \frac{(1-s)^{\alpha_{i}-1}-(t-s)^{\alpha _{i}-1}}{\varGamma(\alpha_{i})},& 0\leq s\leq t\leq1, \\ \frac{(1-s)^{\alpha_{i}-1}}{\varGamma(\alpha_{i})},& 0\leq t\leq s\leq1, \end{cases}\displaystyle \quad i=1,2, $$

(6)

Proof

By Lemma 2.1, system (3) is equivalent to the following integral equations:

$$\begin{aligned}& u(t)=- \int_{0}^{t}\frac{(t-s)^{\alpha _{1}-1}}{\varGamma (\alpha_{1})} h_{1}(s) \,ds+c_{1}+c_{2}t+c_{3}t^{2}+ \cdots+c_{n}t^{n-1}, \end{aligned}$$

(7)

$$\begin{aligned}& v(t)=- \int_{0}^{t}\frac{(t-s)^{\alpha _{2}-1}}{\varGamma (\alpha_{2})} h_{2}(s) \,ds+\overline{c}_{1}+\overline{c}_{2}t+\overline {c}_{3}t^{2}+\cdots+\overline{c}_{m}t^{m-1}. \end{aligned}$$

(8)

Conditions $u'(0)=u''(0)=\cdots= u^{(n-1)}(0)=0$, $v'(0)=v''(0)=\cdots= v^{(m-1)}(0)=0$ imply that

$$c_{2}=c_{3}=\cdots=c_{n}=0,\qquad \overline{c}_{2}=\overline{c}_{3}=\cdots=\overline{c}_{m}=0. $$

That is,

$$\begin{aligned}& u(t)=- \int_{0}^{t}\frac{(t-s)^{\alpha _{1}-1}}{\varGamma (\alpha_{1})} h_{1}(s) \,ds+c_{1}, \\& v(t)=- \int_{0}^{t}\frac{(t-s)^{\alpha_{2}-1}}{\varGamma(\alpha_{2})} h_{2}(s) \,ds+\overline{c}_{1}. \end{aligned}$$

So, we get

$$\begin{aligned}& c_{1}=u(1)+ \int_{0}^{1}\frac{(1-s)^{\alpha _{1}-1}}{\varGamma (\alpha_{1})} h_{1}(s) \,ds, \\& \overline{c}_{1}=v(1)+ \int_{0}^{1}\frac{(1-s)^{\alpha_{2}-1}}{\varGamma (\alpha_{2})} h_{2}(s) \,ds. \end{aligned}$$

Together with (6), we have

$$\begin{aligned}& \begin{aligned}[b] u(t)&=u(1)+ \int_{0}^{1}\frac{(1-s)^{\alpha _{1}-1}}{\varGamma (\alpha_{1})} h_{1}(s) \,ds- \int_{0}^{t}\frac{(t-s)^{\alpha_{1}-1}}{\varGamma (\alpha_{1})} h_{1}(s) \,ds \\ &=u(1)+ \int_{0}^{1} G_{1}(t, s)h_{1}(s) \,ds, \end{aligned} \end{aligned}$$

(9)

$$\begin{aligned}& \begin{aligned}[b] v(t)&=v(1)+ \int_{0}^{1}\frac{(1-s)^{\alpha _{2}-1}}{\varGamma (\alpha_{2})} h_{2}(s) \,ds- \int_{0}^{t}\frac{(t-s)^{\alpha_{2}-1}}{\varGamma (\alpha_{2})} h_{2}(s) \,ds \\ &=v(1)+ \int_{0}^{1} G_{2}(t, s)h_{2}(s) \,ds. \end{aligned} \end{aligned}$$

(10)

Multiplying (9) and (10) by $b(t)$, $a(t)$, and integrating with respect to $dA_{2}(t)$, $dA_{1}(t)$, respectively, we have

$$ \begin{aligned} &\int_{0}^{1}b(t)u(t)\,dA_{2}(t)=u(1) \int _{0}^{1}b(t)\,dA_{2}(t)+ \int_{0}^{1}b(t) \int_{0}^{1}G_{1}(t, s) h_{1}(s) \,ds\,dA_{2}(t), \\ &\int_{0}^{1}a(t) v(t)\,dA_{1}(t)=v(1) \int_{0}^{1}a(t)\,dA_{1}(t)+ \int_{0}^{1}a(t) \int _{0}^{1} G_{2}(t, s)h_{2}(s)\,ds\,dA_{1}(t). \end{aligned} $$

(11)

Therefore, we obtain

$$\begin{aligned}& \frac{1}{\mu_{2}}v(1)-k_{2}u(1)= \int _{0}^{1}b(t) \int_{0}^{1}G_{1}(t, s) h_{1}(s) \,ds\,dA_{2}(t), \\& -k_{1} v(1)+\frac{1}{\mu_{1}}u(1)= \int_{0}^{1}a(t) \int_{0}^{1} G_{2}(t, s)h_{2}(s) \,ds\,dA_{1}(t). \end{aligned}$$

Note that

$$\left|\begin{array}{cc}\frac{1}{{\mu }_1}& -k_1\\ -k_2& \frac{1}{{\mu }_2}\end{array}\right|=\frac{1-{\mu }_1{\mu }_2{k}_1{k}_2}{{\mu }_1{\mu }_2}\ne 0.$$

Then, system (11) has a unique solution for $u(1)$ and $v(1)$. By Cramer’s rule, we get

$$\begin{aligned}& \begin{aligned}[b] u(1)&= \frac{\mu_{1}}{1-\mu_{1}\mu _{2}k_{1}k_{2}} \biggl( \int_{0}^{1}a(t) \int_{0}^{1} G_{2}(t, s)h_{2}(s) \,ds\,dA_{1}(t) \\ &\quad {}+\mu_{2}k_{1} \int_{0}^{1}b(t) \int_{0}^{1} G_{1}(t, s)h_{1}(s) \,ds\,dA_{2}(t)\biggr), \end{aligned} \end{aligned}$$

(12)

$$\begin{aligned}& \begin{aligned}[b] v(1)&= \frac{\mu_{2}}{1-\mu_{1}\mu _{2}k_{1}k_{2}}\biggl( \int _{0}^{1}b(t) \int_{0}^{1} G_{1}(t, s)h_{1}(s) \,ds\,dA_{2}(t) \\ &\quad {}+\mu_{1}k_{2} \int_{0}^{1}a(t) \int_{0}^{1} G_{2}(t, s)h_{2}(s) \,ds\,dA_{1}(t)\biggr). \end{aligned} \end{aligned}$$

(13)

Substituting (12) and (13) into (9) and (10), respectively, we can obtain (4). The proof is completed. □

Lemma 2.3

The Green function $G_{i}(t, s)$ ($i=1, 2$) defined by (6) has the following properties:

$$ \frac{(1-s)^{\alpha_{i}-1}(1-t^{\alpha _{i}-1})}{\varGamma(\alpha_{i})} \leq G_{i}(t, s)\leq \frac{(1-s)^{\alpha_{i}-1}}{\varGamma(\alpha _{i})}, \quad t, s\in[0, 1], i=1,2. $$

(14)

Proof

From the definition of $G_{i}(t, s)$ ($i=1, 2$), for $0\leq t\leq s\leq1$, it is obvious that (14) holds.

For $0\leq s\leq t\leq1$, we have $t-ts\geq t-s$, and then

$$\begin{aligned} (1-s)^{\alpha_{i}-1}-(t-s)^{\alpha_{i}-1} \geq& (1-s)^{\alpha_{i}-1}-(t-ts)^{\alpha_{i}-1} \\ \geq&(1-s)^{\alpha _{i}-1}-t^{\alpha_{i}-1}(1-s)^{\alpha_{i}-1} \\ =&(1-s)^{\alpha_{i}-1}\bigl(1-t^{\alpha_{i}-1}\bigr), \end{aligned}$$

so, we know $\frac{(1-s)^{\alpha_{i}-1}(1-t^{\alpha_{i}-1})}{\varGamma (\alpha_{i})} \leq G_{i}(t, s)$. From the definition of $G_{i}(t, s)$, we also obtain $G_{i}(t, s)\leq\frac{(1-s)^{\alpha_{i}}}{\varGamma(\alpha_{i})}$. Thus, (14) holds. The proof is completed. □

Lemma 2.4

For $t, s\in[0, 1]$, the functions $K_{i}(t, s)$ and $H_{i}(t, s)$ ($i=1, 2$) defined by (5) satisfy

$$\begin{aligned}& K_{1}(t, s), H_{2}(t, s)\leq\rho (1-s)^{\alpha_{1}-1},\qquad K_{2}(t, s), H_{1}(t, s)\leq \rho(1-s)^{\alpha_{2}-1}, \end{aligned}$$

(15)

$$\begin{aligned}& K_{1}(t, s), H_{2}(t, s)\geq \varrho(1-s)^{\alpha_{1}-1}, \qquad K_{2}(t, s), H_{1}(t, s) \geq\varrho(1-s)^{\alpha_{2}-1}, \end{aligned}$$

(16)

where

$$\begin{aligned}& \begin{aligned} \rho&=\max\biggl\{ \frac{\mu_{1}\mu_{2}k_{1} }{\varGamma(\alpha _{1})(1-\mu_{1}\mu_{2}k_{1}k_{2})} \int_{0}^{1} b(t) \,dA_{2}(t)+ \frac{1}{\varGamma(\alpha_{1})}, \\ &\quad \frac{\mu_{2} }{\varGamma(\alpha_{1})(1-\mu_{1}\mu_{2}k_{1}k_{2})} \int_{0}^{1}b(t)\,dA_{2}(t), \\ &\quad \frac{\mu_{1}\mu_{2}k_{2} }{\varGamma(\alpha_{2})(1-\mu_{1}\mu_{2}k_{1}k_{2})} \int_{0}^{1} a(t) \,dA_{1}(t)+ \frac{1}{\varGamma(\alpha_{2})}, \\ &\quad \frac{\mu_{1} }{\varGamma(\alpha_{2})(1-\mu_{1}\mu_{2}k_{1}k_{2})} \int_{0}^{1}a(t)\,dA_{1}(t)\biggr\} , \end{aligned} \\& \begin{aligned} \varrho&=\max \biggl\{ \frac{\mu_{1}\mu_{2}k_{1} }{\varGamma(\alpha _{1})(1-\mu_{1}\mu_{2}k_{1}k_{2})} \int_{0}^{1} b(t) \bigl(1-t^{\alpha_{1}-1}\bigr) \,dA_{2}(t), \\ &\quad \frac{\mu_{2} }{\varGamma (\alpha_{1})(1-\mu_{1}\mu_{2}k_{1}k_{2})} \int_{0}^{1}b(t) \bigl(1-t^{\alpha _{1}-1}\bigr) \,dA_{2}(t), \\ &\quad \frac{\mu_{1}\mu_{2}k_{2} }{\varGamma(\alpha_{2})(1-\mu_{1}\mu_{2}k_{1}k_{2})} \int_{0}^{1} a(t) \bigl(1-t^{\alpha_{2}-1}\bigr) \,dA_{1}(t), \\ &\quad \frac{\mu_{1} }{\varGamma (\alpha_{2})(1-\mu_{1}\mu_{2}k_{1}k_{2})} \int_{0}^{1}a(t) \bigl(1-t^{\alpha_{2}-1}\bigr) \,dA_{1}(t) \biggr\} . \end{aligned} \end{aligned}$$

Proof

By Lemma 2.3, together with the definitions of $K_{i}(t, s)$ and $H_{i}(t, s)$ in (5), for any $t, s\in[0,1]$, we have

$$\begin{aligned}& \begin{aligned}[b] K_{1}(t, s) &=\frac{\mu_{1}\mu_{2}k_{1} }{1-\mu _{1}\mu_{2}k_{1}k_{2}} \int_{0}^{1} G_{1}(t, s)b(t) \,dA_{2}(t)+ G_{1}(t, s) \\ &\leq\frac{\mu_{1}\mu _{2}k_{1} }{1-\mu_{1}\mu_{2}k_{1}k_{2}} \int_{0}^{1} \frac{(1-s)^{\alpha_{1}-1}b(t)}{\varGamma(\alpha _{1})}\,dA_{2}(t)+ \frac{(1-s)^{\alpha_{1}-1}}{\varGamma(\alpha_{1})} \\ &= \biggl(\frac{\mu_{1}\mu_{2}k_{1} }{\varGamma(\alpha_{1})(1-\mu_{1}\mu_{2}k_{1}k_{2})} \int_{0}^{1} b(t) \,dA_{2}(t)+ \frac{1}{\varGamma(\alpha_{1})} \biggr) (1-s)^{\alpha_{1}-1} \\ &\leq \rho(1-s)^{\alpha_{1}-1}, \end{aligned} \end{aligned}$$

(17)

$$\begin{aligned}& \begin{aligned}[b] H_{2}(t, s)&=\frac{\mu_{2} }{1-\mu_{1}\mu _{2}k_{1}k_{2}} \int _{0}^{1} G_{1}(t, s)b(t) \,dA_{2}(t) \\ &\leq\frac{\mu_{2} }{1-\mu_{1}\mu_{2}k_{1}k_{2}} \int_{0}^{1} \frac {(1-s)^{\alpha_{1}-1}b(t)}{\varGamma(\alpha_{1})}\,dA_{2}(t) \\ &= \biggl(\frac{\mu_{2} }{\varGamma(\alpha_{1})(1-\mu_{1}\mu_{2}k_{1}k_{2})} \int _{0}^{1}b(t)\,dA_{2}(t) \biggr) (1-s)^{\alpha_{1}-1} \\ &=\rho(1-s)^{\alpha_{1}-1}. \end{aligned} \end{aligned}$$

(18)

Similarly as in (17)–(18), we have $K_{2}(t, s), H_{1}(t, s)\leq\rho(1-s)^{\alpha_{2}-1}$, so the second inequality of (15) holds.

By Lemma 2.3, for any $t, s\in[0,1]$, we also have

$$\begin{aligned}& \begin{aligned}[b] K_{1}(t, s) &=\frac{\mu_{1}\mu_{2}k_{1} }{1-\mu _{1}\mu_{2}k_{1}k_{2}} \int_{0}^{1} G_{1}(t, s)b(t) \,dA_{2}(t)+ G_{1}(t, s) \\ &\geq\frac{\mu_{1}\mu _{2}k_{1} }{1-\mu_{1}\mu_{2}k_{1}k_{2}} \int_{0}^{1} \frac{b(t)(1-s)^{\alpha_{1}-1}(1-t^{\alpha_{1}-1})}{\varGamma (\alpha_{1})}\,dA_{2}(t) \\ & = \biggl(\frac{\mu_{1}\mu_{2}k_{1} }{\varGamma(\alpha_{1})(1-\mu_{1}\mu_{2}k_{1}k_{2})} \int_{0}^{1}b(t) \bigl(1-t^{\alpha_{1}-1}\bigr) \,dA_{2}(t) \biggr) (1-s)^{\alpha_{1}-1} \\ & \geq \varrho(1-s)^{\alpha_{1}-1} , \end{aligned} \end{aligned}$$

(19)

$$\begin{aligned}& \begin{aligned}[b] H_{2}(t, s)&=\frac{\mu_{2} }{1-\mu_{1}\mu _{2}k_{1}k_{2}} \int _{0}^{1} G_{1}(t, s)b(t) \,dA_{2}(t) \\ &\geq\frac{\mu_{2} }{1-\mu_{1}\mu_{2}k_{1}k_{2}} \int_{0}^{1} \frac {b(t)(1-s)^{\alpha_{1}-1}(1-t^{\alpha_{1}-1})}{\varGamma(\alpha _{1})}\,dA_{2}(t) \\ &= \biggl(\frac{\mu_{2} }{\varGamma(\alpha_{1})(1-\mu_{1}\mu_{2}k_{1}k_{2})} \int _{0}^{1}b(t) \bigl(1-t^{\alpha_{1}-1}\bigr) \,dA_{2}(t) \biggr) (1-s)^{\alpha_{1}-1} \\ &=\varrho(1-s)^{\alpha_{1}-1}. \end{aligned} \end{aligned}$$

(20)

Similarly as in (19)–(20), we have $K_{2}(t, s), H_{1}(t, s)\geq\varrho(1-s)^{\alpha_{2}-1}$, so the second inequality of (16) holds. The proof is completed. □

Let $X=C[0, 1]\times C[0, 1]$, then X is a Banach space with the norm

$$\bigl\Vert (u,v) \bigr\Vert =\|u\|+\|v\|, \qquad \|u\|=\max _{t\in[0, 1]} \bigl\vert u(t) \bigr\vert ,\qquad \|v\|=\max _{t\in[0, 1]} \bigl\vert v(t) \bigr\vert . $$

For any $(u,v)\in X$, we can define an integral operator $T: X\to X$ by

$$\begin{aligned}& T(u,v) (t)=\bigl(T_{1}(u,v) (t), T_{2}(u,v) (t)\bigr), \quad 0\leq t \leq 1, \\& T_{1}(u,v) (t)=\lambda_{1} \int_{0}^{1} K_{1}(t, s)f_{1} \bigl(s, u(s), v(s)\bigr)\,ds \\& \hphantom{T_{1}(u,v) (t)={}}{}+\lambda_{2} \int_{0}^{1} H_{1}(t, s)f_{2} \bigl(s, u(s), v(s)\bigr)\,ds,\quad 0\leq t \leq1, \\& T_{2}(u,v) (t)=\lambda _{2} \int_{0}^{1} K_{2}(t, s)f_{2} \bigl(s, u(s), v(s)\bigr)\,ds \\& \hphantom{T_{2}(u,v) (t)={}}{}+\lambda_{1} \int_{0}^{1} H_{2}(t, s)f_{1} \bigl(s, u(s), v(s)\bigr)\,ds, \quad 0\leq t \leq1. \end{aligned}$$

(21)

Then $(u, v)$ is a positive solutions of system (1) if and only if $(u, v)$ is a fixed point of T. It can be proved that the following Lemma 2.5 is correct.

Lemma 2.5

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

Lemma 2.6

([33])

Let E be a Banach space. Assume that $T : E\rightarrow E$ is a completely continuous operator. Let $V = \{x\in E|x = \mu Tx, 0< \mu< 1 \}$. Then either the set V is unbounded, or T has at least one fixed point.

3 Main results

Theorem 3.1

Assume that there exist real constants $m_{i} > 0$, and $n_{i}, l_{i} \geq0$, such that $\forall t\in[0, 1]$, $x, y \in[0, +\infty)$,

$$ f_{i}(t, x, y)\leq m_{i} + n_{i}|x| + l_{i}|y|, \quad i = 1, 2. $$

(22)

In addition, assume that

$$ 2M_{1}n_{1}+2M_{2}n_{2}< 1,\qquad 2M_{1}l_{1}+2M_{2}l_{2}< 1, $$

where

$$ M_{1}=\lambda_{1} \int_{0}^{1} \rho(1-s)^{\alpha _{1}-1}\,ds,\qquad M_{2}=\lambda_{2} \int_{0}^{1} \rho(1-s)^{\alpha_{2}-1}\,ds. $$

(23)

Then system (1) has at least one solution.

Proof

Let us confirm that the set $V = \{(u, v) \in X : (u, v) = \varsigma T(u, v), 0 \leq\varsigma\leq1\}$ is bounded. Let $(u, v) \in V$, then $(u, v) = \varsigma T(u, v)$. For any $t\in[0, 1]$, we have $u = \varsigma T_{1}(u, v)$, $v = \varsigma T_{2}(u, v)$. Then, by Lemma 2.4, we obtain

$$\begin{aligned}& \begin{aligned}[b] \bigl\vert u(t) \bigr\vert &\leq \biggl\vert \lambda_{1} \int_{0}^{1} K_{1}(t, s)f_{1} \bigl(s, u(s), v(s)\bigr)\,ds+\lambda_{2} \int_{0}^{1} H_{1}(t, s)f_{2} \bigl(s, u(s), v(s)\bigr)\,ds \biggr\vert \\ &\leq \lambda_{1} \int_{0}^{1} \rho(1-s)^{\alpha_{1}-1} f_{1}\bigl(s, u(s), v(s)\bigr)\,ds+\lambda_{2} \int_{0}^{1} \rho(1-s)^{\alpha_{2}-1}f_{2} \bigl(s, u(s), v(s)\bigr)\,ds \\ &\leq M_{1}\bigl(m_{1}+n_{1} \Vert u \Vert +l_{1} \Vert v \Vert \bigr)+M_{2}\bigl(m_{2}+n_{2} \Vert u \Vert +l_{2} \Vert v \Vert \bigr), \end{aligned} \end{aligned}$$

(24)

$$\begin{aligned}& \begin{aligned}[b] \bigl\vert v(t) \bigr\vert &\leq \biggl\vert \lambda_{2} \int_{0}^{1} K_{2}(t, s)f_{2} \bigl(s, u(s), v(s)\bigr)\,ds+\lambda_{1} \int_{0}^{1} H_{2}(t, s)f_{1} \bigl(s, u(s), v(s)\bigr)\,ds \biggr\vert \\ &\leq\lambda_{2} \int_{0}^{1} \rho(1-s)^{\alpha_{2}-1} f_{1}\bigl(s, u(s), v(s)\bigr)\,ds+\lambda_{1} \int_{0}^{1} \rho(1-s)^{\alpha_{1}-1}f_{1} \bigl(s, u(s), v(s)\bigr)\,ds \\ &\leq M_{2}\bigl(m_{2}+n_{2} \Vert u \Vert +l_{2} \Vert v \Vert \bigr)+M_{1}\bigl(m_{1}+n_{1} \Vert u \Vert +l_{1} \Vert v \Vert \bigr). \end{aligned} \end{aligned}$$

(25)

Combined with (24) and (25), we know

$$\begin{aligned} \Vert u \Vert + \Vert v \Vert \leq& 2M_{1} \bigl(m_{1}+n_{1} \Vert u \Vert +l_{1} \Vert v \Vert \bigr)+2M_{2}\bigl(m_{2}+n_{2} \Vert u \Vert +l_{2} \Vert v \Vert \bigr) \\ \leq& 2M_{1}m_{1}+2M_{2}m_{2}+(2M_{1}n_{1}+2M_{2}n_{2}) \Vert u \Vert +(2M_{1}l_{1}+2M_{2}l_{2}) \Vert v \Vert . \end{aligned}$$

Therefore

$$\begin{aligned} \bigl\Vert (u, v) \bigr\Vert =&\|u\|+\|v\| \\ \leq&\frac{2M_{1}m_{1} +2M_{2} m_{2}}{\min\{1-(2M_{1}n_{1}+2M_{2}n_{2}), 1-(2M_{1}l_{1}+2M_{2}l_{2})\}}. \end{aligned}$$

So we have proved that the set V is bounded. Thus, by Lemma 2.6, operator T has at least one fixed point. Hence system (1) has at least one solution. The proof is complete. □

Theorem 3.2

Assume that $f_{i}: [0, 1]\times[0, +\infty )\times[0, +\infty)\rightarrow[0, +\infty)$ is continuous and there exist real constants $\gamma_{i}, \delta_{i} \geq0$ such that $\forall t\in[0, 1]$, $x_{i}, y_{i}\in[0, +\infty)$,

$$ \bigl\vert f_{i}(t, x_{1}, y_{1})-f_{i}(t, x_{2}, y_{2}) \bigr\vert \leq \gamma _{i} \vert x_{1}-x_{2} \vert + \delta_{i} \vert y_{1}-y_{2} \vert , \quad i = 1, 2. $$

(26)

In addition, assume that $2M_{1}(\gamma_{1} + \delta_{1})+2M_{2}(\gamma_{2}+ \delta_{2})<1$, where $M_{1}$, $M_{2}$ are defined as (23). Then system (1) has a unique solution.

Proof

Denoting $\sup|f_{i}(t, 0, 0)|=\varTheta_{i}<+\infty$, by (26), we have

$$ \bigl\vert f_{i}(t, x, y) \bigr\vert \leq\varTheta_{i}+ \gamma_{i}|x| + \delta_{i}|y|, \quad i = 1, 2. $$

Let $r=\frac{2 M_{1}\varTheta_{1}+2M_{2}\varTheta_{2}}{1-2M_{1}(\gamma_{1} + \delta_{1})-2M_{2}(\gamma_{2}+ \delta_{2})}$, $K_{r} = \{(u, v)\in X:\|(u, v)\|< r\}$, we show that $TK_{r}\subset K_{r}$. For any $(u, v)\in K_{r}$, we have

$$\begin{aligned} \bigl\vert T_{1}(u, v) (t) \bigr\vert \leq&\max _{t\in[0, 1]} \biggl\vert \lambda_{1} \int_{0}^{1} K_{1}(t, s)f_{1} \bigl(s, u(s), v(s)\bigr)\,ds+\lambda_{2} \int_{0}^{1} H_{1}(t, s)f_{2} \bigl(s, u(s), v(s)\bigr)\,ds \biggr\vert \\ \leq&\lambda_{1} \int_{0}^{1} \rho(1-s)^{\alpha_{1}-1} f_{1}\bigl(s, u(s), v(s)\bigr)\,ds \\ &{}+\lambda_{2} \int_{0}^{1} \rho(1-s)^{\alpha_{2}-1}f_{2} \bigl(s, u(s), v(s)\bigr)\,ds \\ \leq& \lambda_{1} \int_{0}^{1} \rho(1-s)^{\alpha_{1}-1} \bigl( \varTheta_{1}+\gamma _{1} \Vert u \Vert + \delta_{1} \Vert v \Vert \bigr)\,ds \\ &{}+\lambda_{2} \int_{0}^{1} \rho(1-s)^{\alpha _{2}-1}\bigl( \varTheta_{2}+\gamma_{2} \Vert u \Vert + \delta_{2} \Vert v \Vert \bigr)\,ds \\ \leq& M_{1} \bigl(\varTheta_{1}+\gamma_{1} \Vert u \Vert + \delta_{1} \Vert v \Vert \bigr)+M_{2} \bigl(\varTheta_{2}+\gamma_{2} \Vert u \Vert + \delta_{2} \Vert v \Vert \bigr). \end{aligned}$$

Hence

$$ \bigl\Vert T_{1}(u, v) \bigr\Vert \leq M_{1} \bigl(\varTheta _{1}+(\gamma_{1} + \delta_{1})r \bigr)+M_{2} \bigl(\varTheta_{2}+( \gamma_{2}+ \delta_{2})r \bigr). $$

(27)

Similarly as in (27), for any $(u, v)\in K_{r}$, we can get

$$ \bigl\Vert T_{2}(u, v) \bigr\Vert \leq M_{1} \bigl(\varTheta _{1}+(\gamma_{1} + \delta_{1})r \bigr)+M_{2} \bigl(\varTheta_{2}+( \gamma_{2}+ \delta_{2})r \bigr). $$

(28)

By (27) and (28),

$$\begin{aligned} \bigl\Vert T(u, v) \bigr\Vert =& \bigl\Vert T_{1}(u, v) \bigr\Vert + \bigl\Vert T_{2}(u, v) \bigr\Vert \\ \leq&2 \bigl(M_{1} \bigl(\varTheta_{1}+( \gamma_{1} + \delta_{1})r \bigr)+M_{2} \bigl( \varTheta _{2}+(\gamma_{2}+ \delta_{2})r \bigr) \bigr) \\ \leq& r. \end{aligned}$$

Now for $(u_{1}, v_{1}), (u_{2}, v_{2})\in X$, and for any $t\in[0, 1]$, we have

$$\begin{aligned}& \bigl\vert T_{1}(u_{2}, v_{2}) (t)-T_{1}(u_{1}, v_{1}) (t) \bigr\vert \\& \quad \leq\lambda_{1} \int_{0}^{1} K_{1}(t, s) \bigl\vert f_{1}\bigl(s, u_{2}(s), v_{2}(s) \bigr)-f_{1}\bigl(s, u_{1}(s), v_{1}(s)\bigr) \bigr\vert \,ds \\& \qquad {}+\lambda_{2} \int_{0}^{1} H_{1}(t, s) \bigl\vert f_{2}\bigl(s, u_{2}(s), v_{2}(s) \bigr)-f_{2}\bigl(s, u_{1}(s), v_{1}(s)\bigr) \bigr\vert \,ds \\& \quad \leq\lambda_{1} \int_{0}^{1} \rho(1-s)^{\alpha_{1}-1} \bigl\vert f_{1}\bigl(s, u_{2}(s), v_{2}(s) \bigr)-f_{1}\bigl(s, u_{1}(s), v_{1}(s)\bigr) \bigr\vert \,ds \\& \qquad {}+\lambda_{2} \int_{0}^{1} \rho(1-s)^{\alpha_{2}-1} \bigl\vert f_{2}\bigl(s, u_{2}(s), v_{2}(s) \bigr)-f_{2}\bigl(s, u_{1}(s), v_{1}(s)\bigr) \bigr\vert \,ds \\& \quad \leq M_{1} \bigl(\gamma_{1} \Vert u_{2}-u_{1} \Vert + \delta_{1} \Vert v_{2}-v_{1} \Vert \bigr)+M_{2} \bigl( \gamma_{2} \Vert u_{2}-u_{1} \Vert + \delta_{2} \Vert v_{2}-v_{1} \Vert \bigr) \\& \quad \leq \bigl(M_{1}(\gamma_{1}+\delta_{1} )+M_{2}(\gamma_{2}+ \delta_{2}) \bigr) \bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr). \end{aligned}$$

Consequently, for $(u_{1}, v_{1}), (u_{2}, v_{2})\in X$, we obtain

$$ \bigl\Vert T_{1}(u_{2}, v_{2})-T_{1}(u_{1}, v_{1}) \bigr\Vert \leq \bigl(M_{1}(\gamma_{1}+ \delta_{1} )+M_{2}(\gamma_{2}+ \delta_{2}) \bigr) \bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr). $$

(29)

By a similar proof as for (29), for $(u_{1}, v_{1}), (u_{2}, v_{2})\in X$, we get

$$ \bigl\Vert T_{2}(u_{2}, v_{2})-T_{2}(u_{1}, v_{1}) \bigr\Vert \leq \bigl(M_{1}(\gamma_{1}+ \delta_{1} )+M_{2}(\gamma_{2}+ \delta_{2}) \bigr) \bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr). $$

(30)

It follows from (29) and (30) that

$$ \bigl\Vert T(u_{2}, v_{2})-T(u_{1}, v_{1}) \bigr\Vert \leq \bigl( 2 M_{1}(\gamma_{1}+ \delta_{1} )+2M_{2}(\gamma_{2}+ \delta_{2}) \bigr) \bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr). $$

Since $( 2 M_{1}(\gamma_{1}+\delta_{1} )+2M_{2}(\gamma_{2}+ \delta _{2}) )<1$, T is a contraction operator. By the contraction mapping principle, operator T has a unique fixed point, so system (1) has a unique solution. The proof is complete. □

4 Examples

An example is given to illustrate our main results in this paper. Consider the following problem:

$$ \textstyle\begin{cases} {}^{\mathrm{c}}D^{\frac{5}{2}}u(t)+ f_{1}(t,u(t), v(t))=0, \\ {}^{\mathrm{c}}D^{\frac{7}{3}}v(t)+ 2 f_{2}(t,u(t), v(t))=0, \quad 0< t< 1, \\ u'(0)=u''(0)=0, \qquad u(1)=v (\frac{1}{3} )+v (\frac{1}{2} ), \\ v'(0)=v''(0)=0,\qquad v(1)=\frac{1}{2}\int_{0}^{1}u(s)\,ds^{2}. \end{cases} $$

(31)

Let $\alpha_{1}=\frac{5}{2}$, $\alpha_{2}=\frac{7}{3}$, $\lambda_{1}=1$, $\lambda_{2}=2$, $\mu_{1}=1$, $\mu_{2}=\frac{1}{2}$, $a(t)= b(t)=1$,

$$ A(t)= \textstyle\begin{cases} 0,& 0\leq t< \frac{1}{3}, \\ 1,& \frac{1}{3}\leq t< \frac{1}{2}, \\ 2,& \frac{1}{2}\leq t\leq1, \end{cases}\displaystyle \qquad B(t)=t^{2}. $$

For $t\in[0, 1]$, $x, y\in[0, +\infty)$, take

$$\begin{aligned}& f_{1}(t, x, y)=\frac{t}{1+e^{t}} \biggl(1+\frac{1}{5} \sin^{2}x+\frac {1}{10}\cos y \biggr), \\& f_{2}(t, x, y)=\frac{t}{(1+t)^{3}} \biggl(1+3\cos x+\frac{1}{4} y \biggr). \end{aligned}$$

Notice that

$$\begin{aligned}& \bigl\vert f_{1}(t, x, y) \bigr\vert = \biggl\vert \frac{t}{1+e^{t}} \biggl(1+\frac{1}{5}\sin ^{2}x+ \frac{1}{10}\cos y \biggr) \biggr\vert \leq 1+\frac{1}{5}|x|+ \frac{1}{10}|y|, \\& \bigl\vert f_{2}(t, x, y) \bigr\vert = \biggl\vert \frac{t}{(2+t)^{3}} \biggl(\frac{2}{3}+3\cos x+2 y \biggr) \biggr\vert \leq \frac{1}{12}+\frac{3}{8}|x|+\frac{1}{2}|y|, \\& 2M_{1}n_{1}+2M_{2}n_{2}\doteq0.63467< 1, \qquad 2M_{1}l_{1}+2M_{2}l_{2} \doteq0.83802< 1. \end{aligned}$$

Therefore, all conditions of Theorem 3.1 are satisfied, and hence system (31) has at least one solution.