Existence and uniqueness results for a coupled fractional order systems with the multi-strip and multi-point mixed boundary conditions

Abstract

This paper is concerned with the existence and uniqueness of solutions for a coupled system of fractional differential equations supplemented with the multi-strip and multi-point mixed boundary conditions. The existence of solutions is derived by applying Leray-Schauder’s alternative, while the uniqueness of the solution is established via Banach’s contraction principle. We also show the existence and uniqueness results of a positive solution by applying the Krasnoselskii fixed point theorem.

1 Introduction

Fractional differential equations arise from the studies of complex problems in many engineering and scientific as the mathematical modeling of systems and processes in the fields of physics, chemistry, biology, economics, control theory, signal and image processing, biophysics, blood flow phenomena, aerodynamics, fitting of experimental data. Fractional differential equations are also an excellent tool for the description of hereditary properties of various materials and processes.

Therefore, fractional order boundary value problems, fractional evolution equations, impulsive fractional differential equations, fractional partial differential equations are found to be of significant interest for many researchers [1–4]. The fractional order boundary value problems are especially popular. Since recently, there are several kinds of boundary conditions in the study of fractional order boundary value problem such as two-point, multi-point, periodic/anti-periodic, nonlocal and integral boundary conditions. We can see the details of several recent works on the subject in [5–9].

Another interesting field of recent research is the coupled system of fractional differential equations. It have been shown to be more accurate and realistic that have many applications in real-world problems such as synchronization of chaotic systems [10–12], anomalous diffusion [13], disease models [14, 15], and ecological models [16].

In [17], the authors considered a nonlinear coupled system of Liouville-Caputo type fractional differential equations:

$$\begin{gathered} \textstyle\begin{cases} {}^{c}D^{q}x(t)=f (t,x(t),y(t), {}^{c}D^{\sigma_{1}}y(t) ),& t\in[0,1],\\ {}^{c}D^{p}y(t)=g (t,x(t), {}^{c}D^{\sigma_{2}}x(t),y(t) ),& t\in[0,1], \end{cases}\displaystyle \\ \textstyle\begin{cases} x(0)=\Psi_{1}(y),\qquad x'(0)=e_{1}y'(w_{1}),\qquad x(1)=a_{1}\int _{0}^{\xi}y(s)\,ds+b_{1}\sum_{i=1}^{m-2}\alpha_{i}y(\eta_{i}),\\ y(0)=\Psi_{2}(x),\qquad y'(0)=e_{2}x'(w_{2}),\qquad y(1)=a_{2}\int _{0}^{\xi}x(s)\,ds+b_{2}\sum_{i=1}^{m-2}\beta_{i}x(\eta_{i}), \end{cases}\displaystyle \end{gathered} $$

where $2< p,q\leq3$, $1<\sigma_{1},\sigma_{2}<2$, $0< w_{1}< w_{2}<\xi <\eta_{1}<\eta_{2}<\cdots<\eta_{m-2}<1$. The existence and uniqueness results were obtained by applying Banach’s fixed point theorem and Leray-Schauder’s alternative. We recommend the reader a series of papers studying on the coupled systems of fractional differential equations [18–27].

Motivated by the above mentioned work, we consider the existence and uniqueness results for the following fractional order differential systems:

$$ \textstyle\begin{cases} D_{0+}^{\alpha_{1}}u(t)+f_{1}(t,u(t),v(t))=0,& t\in(0,1),\\ D_{0+}^{\alpha_{2}}v(t)+f_{2}(t,u(t),v(t))=0,& t\in(0,1), \end{cases} $$

(1.1)

with the coupled integral and discrete mixed boundary conditions:

$$ \textstyle\begin{cases} u(0)=u'(0)=0,\qquad u(1)=\sum_{i=1}^{n}\int_{\xi_{i}}^{\eta _{i}}v(s)\,dA_{i}(s)+\sum_{i=1}^{m-2}b_{i}v(\sigma_{i}),\\ v(0)=v'(0)=0,\qquad v(1)=\sum_{i=1}^{n}\int_{\xi_{i}}^{\eta_{i}}u(s)\,dA_{i}(s)+\sum_{i=1}^{m-2}b_{i}u(\sigma_{i}), \end{cases} $$

(1.2)

where $2<\alpha_{k}\leq3$, $D^{\alpha_{k}}$ is the standard Riemann-Liouville fractional derivative of order $\alpha_{k}$, $k=1,2$; $0<\xi_{i}<\eta_{i}<1$, $A_{i}(s)$ is a nondecreasing function of bounded variation in $[0,1]$, $i=1,2,\ldots,n$; $0<\sigma_{i}<1$, $b_{i}\geq0$, $i=1,2,\ldots,m-2$.

We emphasize that the multi-strip and multi-point mixed boundary conditions in (1.2) state that the value of the unknown function at the right end point $t=1$ of the given interval is equal to the summation of the Riemann-Stieljes integral values of the unknown function on the sub-interval $[\xi_{i},\eta_{i}]$ ($i=1,2,\ldots,n$) plus the linear combination of discrete values of the unknown function on $\sigma_{i}$ ($i=1,2,\ldots,m-2$).

By applying Leray-Schauder’s alternative, Banach’s contraction principle and the fixed point theorems of cone expansion and compression of norm type, the sufficient conditions for the existence and uniqueness results to a general class of multi-strip and multi-point mixed boundary value problem for a coupled system of fractional differential equations are obtained.

2 Preliminaries

We present here the definitions, some lemmas from the theory of fractional calculus and some auxiliary results that will be used to prove our main theorems.

Definition 2.1

[1]

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

$$\bigl(I_{0+}^{\alpha}y\bigr) (t)=\frac{1}{\Gamma(\alpha)} \int _{0}^{t}(t-s)^{\alpha-1}y(s)\,ds,\quad t>0, $$

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

Definition 2.2

[1]

The Riemann-Liouville fractional derivative of order $\alpha\geq0$ for a function $y:(0,\infty)\rightarrow\mathbb{R}$ is given by

$$\bigl(D_{0+}^{\alpha}y\bigr) (t)=\frac{1}{\Gamma(n-\alpha)} \biggl( \frac {d}{dt} \biggr)^{n} \int_{0}^{t}y(s) (t-s)^{n-\alpha-1}\,ds,\quad t>0, $$

where $n=[\alpha]+1$, provided that the right-hand side is pointwise defined on $[0,\infty)$. The notation $[\alpha]$ stands for the largest integer not greater than α. We also denote the Riemann-Liouville fractional derivative of y by $D_{0+}^{\alpha}y(t)$. If $\alpha=m\in\mathbb {N}$ then $D_{0+}^{m}y(t)=y^{(m)}(t)$ for $t>0$, and if $\alpha=0$ then $D_{0+}^{0}y(t)=y(t)$ for $t>0$.

Lemma 2.1

Let $\alpha>0$ and $n=[\alpha]+1$ for $\alpha\notin\mathbb{N}$; that is, n is the smallest integer greater than or equal to α. Then the solutions of the fractional differential equation $D_{0+}^{\alpha}u(t)=0$, $0< t<1$, are

$$u(t)=c_{1}t^{\alpha-1}+c_{2}t^{\alpha-2}+ \cdots+c_{n}t^{\alpha-n}, \quad0< t< 1, $$

where $c_{1},c_{2},\ldots,c_{n}$ are arbitrary real constants.

Lemma 2.2

Let $\alpha>0$, n be the smallest integer greater than or equal to α ($n-1<\alpha\leq n$) and $y\in L^{1}(0,1)$. The solutions of the fractional equation $D_{0+}^{\alpha}u(t)+y(t)=0$, $0< t<1$, are

$$u(t)=-\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha -1}y(s)\,ds+c_{1}t^{\alpha-1}+ \cdots+c_{n}t^{\alpha-n},\quad0< t< 1, $$

where $c_{1},c_{2},\ldots,c_{n}$ are arbitrary real constants.

Now we derive the corresponding Green’s function for the boundary value problem (1.1)-(1.2) and obtain some properties of the Green’s function.

Lemma 2.3

For $x,y\in L^{1}(0,1)$, boundary value problem

$$ \textstyle\begin{cases} D_{0+}^{\alpha_{1}}u(t)+x(t)=0,\qquad D_{0+}^{\alpha _{2}}v(t)+y(t)=0,\quad t\in(0,1),\\ u(0)=u'(0)=0,\qquad u(1)=\sum_{i=1}^{n}\int_{\xi_{i}}^{\eta _{i}}v(s)\,dA_{i}(s)+\sum_{i=1}^{m-2}b_{i}v(\sigma_{i}),\\ v(0)=v'(0)=0,\qquad v(1)=\sum_{i=1}^{n}\int_{\xi_{i}}^{\eta_{i}}u(s)\,dA_{i}(s)+\sum_{i=1}^{m-2}b_{i}u(\sigma_{i}) \end{cases} $$

(2.1)

has an integral representation

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

(2.2)

where

$$\begin{aligned}& \begin{aligned} &K_{1}(t,s)=g_{1}(t,s)+ \frac{l_{2}t^{\alpha _{1}-1}}{1-l_{1}l_{2}} \Biggl[\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}}g_{1}(t,s)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i}g_{1}( \sigma_{i},s) \Biggr], \\ &H_{1}(t,s)=\frac{t^{\alpha_{1}-1}}{1-l_{1}l_{2}} \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}g_{2}(t,s)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i}g_{2}( \sigma_{i},s) \Biggr], \end{aligned} \end{aligned}$$

(2.3)

$$\begin{aligned}& \begin{aligned} &K_{2}(t,s)=g_{2}(t,s)+ \frac{l_{1}t^{\alpha _{2}-1}}{1-l_{1}l_{2}} \Biggl[\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}}g_{2}(t,s)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i}g_{2}( \sigma_{i},s) \Biggr], \\ &H_{2}(t,s)=\frac{t^{\alpha_{2}-1}}{1-l_{1}l_{2}} \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}g_{1}(t,s)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i}g_{1}( \sigma_{i},s) \Biggr], \end{aligned} \end{aligned}$$

(2.4)

$$\begin{aligned}& g_{k}(t,s)=\frac{1}{\Gamma(\alpha_{k})} \textstyle\begin{cases} t^{\alpha_{k}-1}(1-s)^{\alpha_{k}-1}-(t-s)^{\alpha_{k}-1}, &0\leq s\leq t\leq1,\\ t^{\alpha_{k}-1}(1-s)^{\alpha_{k}-1},&0\leq t\leq s\leq1, \end{cases}\displaystyle \end{aligned}$$

(2.5)

$$\begin{aligned}& l_{k}=\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}t^{\alpha _{k}-1}\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \sigma_{i}^{\alpha_{k}-1},\quad k=1,2. \end{aligned}$$

(2.6)

Proof

From Lemma 2.2, we can reduce (2.1) to the following equivalent integral equations:

$$ \begin{aligned} &u(t)=- \int_{0}^{t}\frac{(t-s)^{\alpha_{1}-1}}{\Gamma(\alpha _{1})}x(s)\,ds+c_{11}t^{\alpha_{1}-1}+c_{12}t^{\alpha _{1}-2}+c_{13}t^{\alpha_{1}-3}, \\ &v(t)=- \int_{0}^{t}\frac{(t-s)^{\alpha_{2}-1}}{\Gamma(\alpha _{2})}y(s)\,ds+c_{21}t^{\alpha_{2}-1}+c_{22}t^{\alpha _{2}-2}+c_{23}t^{\alpha_{2}-3}. \end{aligned} $$

(2.7)

From $u(0)=u'(0)=v(0)=v'(0)=0$, we have $c_{12}=c_{13}=c_{22}=c_{23}=0$. Thus (2.7) reduces to

$$ \begin{aligned} &u(t)=c_{11}t^{\alpha_{1}-1}- \int_{0}^{t}\frac{(t-s)^{\alpha _{1}-1}}{\Gamma(\alpha_{1})}x(s)\,ds, \\ &v(t)=c_{21}t^{\alpha_{2}-1}- \int_{0}^{t}\frac{(t-s)^{\alpha _{2}-1}}{\Gamma(\alpha_{2})}y(s)\,ds. \end{aligned} $$

(2.8)

Using the right side boundary conditions of (1.2) and combining (2.8), we have

$$ \begin{aligned} &u(t)=t^{\alpha_{1}-1} \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}}v(s)\,dA_{i}(s)+\sum _{i=1}^{m-2}b_{i}v(\sigma_{i}) \Biggr]+ \int _{0}^{1}g_{1}(t,s)x(s)\,ds, \\ &v(t)=t^{\alpha_{2}-1} \Biggl[\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}}u(s)\,dA_{i}(s)+\sum _{i=1}^{m-2}b_{i}u(\sigma_{i}) \Biggr]+ \int _{0}^{1}g_{2}(t,s)y(s)\,ds. \end{aligned} $$

(2.9)

Then we can get

$$\begin{aligned}& \begin{aligned}[b] &\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}v(t)\,dA_{i}(t)+\sum _{i=1}^{m-2}b_{i}v(\sigma_{i}) \\ &\quad=\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}} \int _{0}^{1}g_{2}(t,s)y(s)\,ds\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \int _{0}^{1}g_{2}(\sigma_{i},s)y(s)\,ds \\ &\qquad{}+ \Biggl[\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}t^{\alpha _{2}-1}\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \sigma_{i}^{\alpha_{2}-1} \Biggr] \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}u(t)\,dA_{i}(t)+\sum _{i=1}^{m-2}b_{i}u(\sigma_{i}) \Biggr],\hspace{-20pt} \end{aligned} \end{aligned}$$

(2.10)

$$\begin{aligned}& \begin{aligned}[b] &\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}u(t)\,dA_{i}(t)+\sum _{i=1}^{m-2}b_{i}u(\sigma_{i}) \\ &\quad=\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}} \int _{0}^{1}g_{1}(t,s)x(s)\,ds\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \int _{0}^{1}g_{1}(\sigma_{i},s)x(s)\,ds \\ &\qquad{}+ \Biggl[\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}t^{\alpha _{1}-1}\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \sigma_{i}^{\alpha_{1}-1} \Biggr] \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}v(t)\,dA_{i}(t)+\sum _{i=1}^{m-2}b_{i}v(\sigma_{i}) \Biggr]. \end{aligned} \end{aligned}$$

(2.11)

Combining (2.6), (2.10) and (2.11), we get

$$ \begin{aligned}[b] &\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}v(t)\,dA_{i}(t)+\sum _{i=1}^{m-2}b_{i}v(\sigma_{i}) \\ &\quad=\frac{1}{1-l_{1}l_{2}} \Biggl[ \Biggl(\sum_{i=1}^{n} \int_{\xi _{i}}^{\eta_{i}} \int_{0}^{1}g_{2}(t,s)y(s)\,ds\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \int_{0}^{1}g_{2}(\sigma_{i},s)y(s)\,ds \Biggr) \\ &\qquad{}+l_{2} \Biggl(\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}} \int _{0}^{1}g_{1}(t,s)x(s)\,ds\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \int _{0}^{1}g_{1}(\sigma_{i},s)x(s)\,ds \Biggr) \Biggr]. \end{aligned} $$

(2.12)

Similarly, we get

$$ \begin{aligned}[b] &\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}u(t)\,dA_{i}(t)+\sum _{i=1}^{m-2}b_{i}u(\sigma_{i}) \\ &\quad=\frac{1}{1-l_{1}l_{2}} \Biggl[ \Biggl(\sum_{i=1}^{n} \int_{\xi _{i}}^{\eta_{i}} \int_{0}^{1}g_{1}(t,s)x(s)\,ds\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \int_{0}^{1}g_{1}(\sigma_{i},s)x(s)\,ds \Biggr) \\ &\qquad{}+l_{1} \Biggl(\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}} \int _{0}^{1}g_{2}(t,s)y(s)\,ds\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \int _{0}^{1}g_{2}(\sigma_{i},s)y(s)\,ds \Biggr) \Biggr], \end{aligned} $$

(2.13)

where $l_{k}$ ($k=1,2$) is defined by (2.6). From (2.9), (2.12) and (2.13), we have

$$\begin{aligned}& \begin{aligned} u(t)&=\frac{t^{\alpha_{1}-1}}{1-l_{1}l_{2}} \Biggl(\sum _{i=1}^{n} \int _{\xi_{i}}^{\eta_{i}} \int_{0}^{1}g_{2}(t,s)y(s)\,ds\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \int_{0}^{1}g_{2}(\sigma_{i},s)y(s)\,ds \Biggr) \\ &\quad{} +\frac{l_{2}t^{\alpha_{1}-1}}{1-l_{1}l_{2}} \Biggl(\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}} \int _{0}^{1}g_{1}(t,s)x(s)\,ds\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \int _{0}^{1}g_{1}(\sigma_{i},s)x(s)\,ds \Biggr) \\ &\quad{}+ \int_{0}^{1}g_{1}(t,s)x(s)\,ds \\ &= \int_{0}^{1}g_{1}(t,s)x(s)\,ds+ \frac{l_{2}t^{\alpha _{1}-1}}{1-l_{1}l_{2}} \int_{0}^{1}x(s)\,ds \Biggl(\sum _{i=1}^{n} \int _{\xi_{i}}^{\eta_{i}}g_{1}(t,s)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i}g_{1}( \sigma_{i},s) \Biggr) \\ &\quad{}+\frac{t^{\alpha_{1}-1}}{1-l_{1}l_{2}} \int _{0}^{1}y(s)\,ds \Biggl(\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}}g_{2}(t,s)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i}g_{2}( \sigma_{i},s) \Biggr) \\ &= \int_{0}^{1}K_{1}(t,s)x(s)\,ds+ \int_{0}^{1}H_{1}(t,s)y(s)\,ds, \end{aligned} \\& \begin{aligned} v(t)&=\frac{t^{\alpha_{2}-1}}{1-l_{1}l_{2}} \Biggl(\sum _{i=1}^{n} \int _{\xi_{i}}^{\eta_{i}} \int_{0}^{1}g_{1}(t,s)x(s)\,ds\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \int_{0}^{1}g_{1}(\sigma_{i},s)x(s)\,ds \Biggr) \\ &\quad{}+\frac{l_{1}t^{\alpha_{2}-1}}{1-l_{1}l_{2}} \Biggl(\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}} \int _{0}^{1}g_{2}(t,s)y(s)\,ds\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \int _{0}^{1}g_{2}(\sigma_{i},s)y(s)\,ds \Biggr) \\ &\quad{}+ \int_{0}^{1}g_{2}(t,s)y(s)\,ds \\ &= \int_{0}^{1}g_{2}(t,s)y(s)\,ds+ \frac{l_{1}t^{\alpha _{2}-1}}{1-l_{1}l_{2}} \int_{0}^{1}y(s)\,ds \Biggl(\sum _{i=1}^{n} \int _{\xi_{i}}^{\eta_{i}}g_{2}(t,s)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i}g_{2}( \sigma_{i},s) \Biggr) \\ &\quad{}+\frac{t^{\alpha_{2}-1}}{1-l_{1}l_{2}} \int _{0}^{1}x(s)\,ds \Biggl(\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}}g_{1}(t,s)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i}g_{1}( \sigma_{i},s) \Biggr) \\ &= \int_{0}^{1}K_{2}(t,s)y(s)\,ds+ \int_{0}^{1}H_{2}(t,s)x(s)\,ds. \end{aligned} \end{aligned}$$

This completes the proof of the lemma. □

Lemma 2.4

Let $\psi_{k}(t)=t^{\alpha_{k}-1}(1-t)$, $\varphi_{k}(t)=(1-t)^{\alpha_{k}-1}t$, for $t\in[0,1]$, $k=1,2$. The function $g_{k}(t,s)$ defined by (2.5) has the following properties:

• (1) $g_{k}(t,s)$ is continuous for $(t,s)\in[0,1]\times[0,1]$ and $g_{k}(t,s)>0$ for $t,s\in(0,1)$;

• (2) $\psi_{k}(t)\varphi_{k}(s)\leq\Gamma(\alpha _{k})g_{k}(t,s)\leq(\alpha_{k}-1)\varphi_{k}(s)$ for $t,s\in[0,1]$;

• (3) $\psi_{k}(t)\varphi_{k}(s)\leq\Gamma(\alpha _{k})g_{k}(t,s)\leq(\alpha_{k}-1)\psi_{k}(t)$ for $t,s\in[0,1]$.

Proof

(1) The continuity of $g_{k}$ is easily checked.

(2) For $0\leq s\leq t\leq1$, we have

$$\begin{aligned}& \begin{aligned} \Gamma(\alpha_{k})g_{k}(t,s)&=t^{\alpha_{k}-1}(1-s)^{\alpha _{k}-1}-(t-s)^{\alpha_{k}-1} \\ &=(\alpha_{k}-1) \int_{t-s}^{t(1-s)}x^{\alpha_{k}-2}\,dx \\ &\leq(\alpha_{k}-1)t^{\alpha_{k}-2}(1-s)^{\alpha _{k}-2} \bigl[(t-ts)-(t-s)\bigr] \\ &=(\alpha_{k}-1)t^{\alpha_{k}-2}(1-s)^{\alpha_{k}-2}(1-t)s \\ &\leq(\alpha_{k}-1) (1-s)^{\alpha_{k}-2}(1-s)s \\ &=(\alpha_{k}-1)\varphi_{k}(s), \end{aligned} \\& \begin{aligned} \Gamma(\alpha_{k})g_{k}(t,s)&=t^{\alpha_{k}-1}(1-s)^{\alpha _{k}-1}-(t-s)^{\alpha_{k}-1} \\ &=t^{\alpha_{k}-2}(1-s)^{\alpha_{k}-2}(t-ts)-(t-s)^{\alpha _{k}-2}(t-s) \\ &\geq t^{\alpha_{k}-2}(1-s)^{\alpha_{k}-2}\bigl[(t-ts)-(t-s)\bigr] \\ &=t^{\alpha_{k}-2}(1-s)^{\alpha_{k}-2}(s-ts) \\ &\geq(1-s)^{\alpha_{k}-2}(s-ts)t^{\alpha_{k}-2}t(1-s) \\ &=\psi_{k}(t)\varphi_{k}(s). \end{aligned} \end{aligned}$$

For $0\leq t\leq s\leq1$, we have

$$\begin{aligned}& \begin{aligned} \Gamma(\alpha_{k})g_{k}(t,s)&=t^{\alpha_{k}-1}(1-s)^{\alpha _{k}-1}=t^{\alpha_{k}-2}(1-s)^{\alpha_{k}-1}t \\ &\leq(\alpha_{k}-1) (1-s)^{\alpha_{k}-1}s \\ &=(\alpha_{k}-1)\varphi_{k}(s), \end{aligned} \\& \begin{aligned} \Gamma(\alpha_{k})g_{k}(t,s)&=t^{\alpha_{k}-1}(1-s)^{\alpha _{k}-1} \geq t^{\alpha_{k}-1}(1-t) (1-s)^{\alpha_{k}-1}s \\ &=\psi_{k}(t)\varphi_{k}(s). \end{aligned} \end{aligned}$$

(3) We just need to prove the right end of the inequalities. For $0\leq s\leq t\leq1$, we get

$$\begin{aligned} \Gamma(\alpha_{k})g_{k}(t,s)&\leq(\alpha_{k}-1)t^{\alpha _{k}-2}(1-s)^{\alpha_{k}-2}(1-t)s \\ &\leq(\alpha_{k}-1)t^{\alpha_{k}-2}(1-t)t \\ &=(\alpha_{k}-1)\psi_{k}(t). \end{aligned}$$

For $0\leq t\leq s\leq1$, we get

$$\begin{aligned} \Gamma(\alpha_{k})g_{k}(t,s)&=t^{\alpha_{k}-1}(1-s)^{\alpha _{k}-1} \leq t^{\alpha_{k}-1}(1-s) \\ &\leq(\alpha_{k}-1)t^{\alpha_{k}-1}(1-s) \\ &\leq(\alpha_{k}-1)t^{\alpha_{k}-1}(1-t) \\ &=(\alpha_{k}-1)\psi_{k}(t). \end{aligned}$$

This completes the proof of the lemma. □

Denote

$$\begin{aligned} &\varrho_{1}=\frac{l_{2}}{\Gamma(\alpha_{1})(1-l_{1}l_{2})} \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}\psi_{1}(t)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \psi_{1}(\sigma_{i}) \Biggr], \\ &\varrho_{2}=\frac{1}{\Gamma(\alpha_{2})(1-l_{1}l_{2})} \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}\psi_{2}(t)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \psi_{2}(\sigma_{i}) \Biggr], \\ &\varrho_{3}=\frac{l_{1}}{\Gamma(\alpha_{2})(1-l_{1}l_{2})} \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}\psi_{2}(t)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \psi_{2}(\sigma_{i}) \Biggr], \\ &\varrho_{4}=\frac{1}{\Gamma(\alpha_{1})(1-l_{1}l_{2})} \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}\psi_{1}(t)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \psi_{1}(\sigma_{i}) \Biggr], \\ &\rho_{1}=\frac{\alpha_{1}-1}{\Gamma(\alpha_{1})} \Biggl[1+\frac {l_{2}}{1-l_{1}l_{2}} \Biggl(\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}} dA_{i}(t)+\sum _{i=1}^{m-2}b_{i} \Biggr) \Biggr], \\ &\rho_{2}=\frac{\alpha_{2}-1}{\Gamma(\alpha _{2})(1-l_{1}l_{2})} \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}} dA_{i}(t)+\sum _{i=1}^{m-2}b_{i} \Biggr], \\ &\rho_{3}=\frac{\alpha_{2}-1}{\Gamma(\alpha_{2})} \Biggl[1+\frac {l_{1}}{1-l_{1}l_{2}} \Biggl(\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}} dA_{i}(t)+\sum _{i=1}^{m-2}b_{i} \Biggr) \Biggr], \\ &\rho_{4}=\frac{\alpha_{1}-1}{\Gamma(\alpha _{1})(1-l_{1}l_{2})} \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}} dA_{i}(t)+\sum _{i=1}^{m-2}b_{i} \Biggr], \\ &\varrho=\min\{\varrho_{1},\varrho_{2}, \varrho_{3},\varrho_{4}\} ,\qquad \rho=\max\{ \rho_{1},\rho_{2},\rho_{3},\rho_{4}\}. \end{aligned}$$

Lemma 2.5

For $(t,s)\in[0,1]\times[0,1]$, the functions $K_{k}(t,s)$ and $H_{k}(t,s)$ ($k=1,2$) defined by (2.3) and (2.4) satisfy the following results:

• (1) $K_{k}(t,s)$ and $H_{k}(t,s)$ are continuous, $K_{k}(t,s)\geq0$ and $H_{k}(t,s)\geq0$;

• (2) $\varrho t^{\alpha_{k}-1}\varphi_{k}(s)\leq K_{k}(t,s)\leq\rho\varphi_{k}(s)$;

• (3) $\varrho t^{\alpha_{1}-1}\varphi_{2}(s)\leq H_{1}(t,s)\leq\rho\varphi_{2}(s)$, $\varrho t^{\alpha_{2}-1}\varphi_{1}(s)\leq H_{2}(t,s)\leq \rho\varphi_{1}(s)$;

• (4) $H_{k}(t,s)\leq\rho t^{\alpha_{k}-1}$, $K_{k}(t,s)\leq\rho t^{\alpha_{k}-1}$.

Proof

The continuity of $K_{k}$ and $H_{k}$ ($k=1,2$) is easily checked. According to the property (2) of Lemma 2.4 and (2.3), we have

$$\begin{aligned}& \begin{aligned} K_{1}(t,s)&=g_{1}(t,s)+ \frac{l_{2}t^{\alpha _{1}-1}}{1-l_{1}l_{2}} \Biggl[\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}}g_{1}(t,s)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i}g_{1}( \sigma_{i},s) \Biggr] \\ &\geq\frac{l_{2}t^{\alpha_{1}-1}}{1-l_{1}l_{2}} \Biggl[\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}\frac{\psi_{1}(t)\varphi _{1}(s)}{\Gamma(\alpha_{1})} \,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \frac {\psi_{1}(\sigma_{i})\varphi_{1}(s)}{\Gamma(\alpha_{1})} \Biggr] \\ &=\varrho_{1}t^{\alpha_{1}-1}\varphi_{1}(s), \end{aligned} \\& \begin{aligned} K_{1}(t,s)&\leq\frac{(\alpha_{1}-1)\varphi_{1}(s)}{\Gamma(\alpha _{1})}+ \frac{l_{2}t^{\alpha_{1}-1}}{1-l_{1}l_{2}} \Biggl[\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}\frac{(\alpha_{1}-1)\varphi _{1}(s)}{\Gamma(\alpha_{1})}\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \frac {(\alpha_{1}-1)\varphi_{1}(s)}{\Gamma(\alpha_{1})} \Biggr] \\ &\leq\frac{\alpha_{1}-1}{\Gamma(\alpha_{1})} \Biggl[1+\frac {l_{2}}{1-l_{1}l_{2}} \Biggl(\sum _{i=1}^{m-2} \int_{\xi_{i}}^{\eta _{i}} dA_{i}(t)+\sum _{i=1}^{m-2}b_{i} \Biggr) \Biggr] \varphi_{1}(s) \\ &=\rho_{1}\varphi_{1}(s), \end{aligned} \\& \begin{aligned} H_{1}(t,s)&=\frac{t^{\alpha_{1}-1}}{1-l_{1}l_{2}} \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}g_{2}(t,s)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i}g_{2}( \sigma_{i},s) \Biggr] \\ &\geq\frac{t^{\alpha_{1}-1}}{1-l_{1}l_{2}} \Biggl[\sum_{i=1}^{n} \int _{\xi_{i}}^{\eta_{i}}\frac{\psi_{2}(t)\varphi_{2}(s)}{\Gamma (\alpha_{2})} \,dA_{i}(t)+\sum_{i=1}^{m-2}b_{i} \frac{\psi_{2}(\sigma _{i})\varphi_{2}(s)}{\Gamma(\alpha_{2})} \Biggr] \\ &=\varrho_{2}t^{\alpha_{1}-1}\varphi_{2}(s), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} H_{1}(t,s)&\leq\frac{t^{\alpha_{1}-1}}{1-l_{1}l_{2}} \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}\frac{(\alpha_{2}-1)\varphi _{2}(s)}{\Gamma(\alpha_{2})} \,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \frac {(\alpha_{2}-1)\varphi_{2}(s)}{\Gamma(\alpha_{2})} \Biggr] \\ &\leq\frac{\alpha_{2}-1}{\Gamma(\alpha_{2})(1-l_{1}l_{2})} \Biggl[\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}} dA_{i}(t)+\sum _{i=1}^{m-2}b_{i} \Biggr] \varphi_{2}(s) \\ &=\rho_{2}\varphi_{2}(s). \end{aligned}$$

Similarly, from the property (2) of Lemma 2.4 and (2.4), we get

$$\begin{aligned} &\varrho_{3}t^{\alpha_{2}-1}\varphi_{2}(s)\leq K_{2}(t,s)\leq\rho _{3}\varphi_{2}(s), \\ &\varrho_{4}t^{\alpha_{2}-1}\varphi_{1}(s)\leq H_{2}(t,s)\leq\rho_{4}\varphi_{1}(s). \end{aligned} $$

On the other hand, according to the property (3) of Lemma 2.4 and (2.3), we obtain

$$\begin{aligned}& \begin{aligned} K_{1}(t,s)&=g_{1}(t,s)+ \frac{l_{2}t^{\alpha _{1}-1}}{1-l_{1}l_{2}} \Biggl[\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}}g_{1}(t,s)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i}g_{1}( \sigma_{i},s) \Biggr] \\ &\leq\frac{\alpha_{1}-1}{\Gamma(\alpha_{1})} \Biggl[\psi _{1}(t)+\frac{l_{2}t^{\alpha_{1}-1}}{1-l_{1}l_{2}} \Biggl(\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}\psi_{1}(t)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i} \psi_{1}(\sigma_{i}) \Biggr) \Biggr] \\ &\leq\frac{\alpha_{1}-1}{\Gamma(\alpha_{1})} \Biggl[t^{\alpha _{1}-1}+\frac{l_{2}t^{\alpha_{1}-1}}{1-l_{1}l_{2}} \Biggl(\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta_{1}}\,dA_{i}(t)+\sum _{i=1}^{m-2}b_{i} \Biggr) \Biggr] \\ &=\rho_{1}t^{\alpha_{1}-1}, \end{aligned} \\& \begin{aligned} H_{1}(t,s)&=\frac{t^{\alpha_{1}-1}}{1-l_{1}l_{2}} \Biggl[\sum _{i=1}^{n} \int_{\xi_{i}}^{\eta_{i}}g_{2}(t,s)\,dA_{i}(t)+ \sum_{i=1}^{m-2}b_{i}g_{2}( \sigma_{i},s) \Biggr] \\ &\leq\frac{(\alpha_{2}-1)t^{\alpha_{1}-1}}{\Gamma(\alpha _{2})(1-l_{1}l_{2})} \Biggl[\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}}t^{\alpha_{2}-1}(1-t) \,dA_{i}(t)+\sum_{i=1}^{m-2}b_{i} \sigma _{i}^{\alpha_{2}-1}(1-\sigma_{i}) \Biggr] \\ &\leq\frac{(\alpha_{2}-1)t^{\alpha_{1}-1}}{\Gamma(\alpha _{2})(1-l_{1}l_{2})} \Biggl[\sum_{i=1}^{n} \int_{\xi_{i}}^{\eta _{i}} dA_{i}(t)+\sum _{i=1}^{m-2}b_{i} \Biggr]= \rho_{2}t^{\alpha_{1}-1}. \end{aligned} \end{aligned}$$

Similarly, from the property (3) of Lemma 2.4 and (2.4), we get

$$K_{2}(t,s)\leq\rho_{3}t^{\alpha_{2}-1},\qquad H_{2}(t,s)\leq\rho _{4}t^{\alpha_{2}-1}. $$

This completes the proof of the lemma. □

Remark 2.1

From the Lemma 2.5, for $t,\tau,s\in[0,1]$, we have

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

where $\omega=\frac{\varrho}{\rho}$ and $0<\omega<1$.

Lemma 2.6

Leray-Schauder alternative

Let $F:E\rightarrow E$ be a completely continuous operator (i.e., a map that restricted to any bounded set in E is compact). Let

$$\varepsilon(F)= \bigl\{ x\in E:x=\lambda F(x), 0< \lambda< 1 \bigr\} . $$

Then either the set $\varepsilon(F)$ is unbounded, or F has at least one fixed point.

Lemma 2.7

Let X be a Banach space, and let $P\subset X$ be a cone in X. Assume $\Omega_{1}$, $\Omega_{2}$ are open subsets of X with $0\in\Omega_{1}\subset\overline{\Omega}_{1}\subset\Omega_{2}$, and let $S:P\rightarrow P$ be a completely continuous operator such that either

• (a) $\Vert Sw \Vert \leq \Vert w \Vert$, $w\in P\cap\partial\Omega_{1}$, $\Vert Sw \Vert \geq \Vert w \Vert$, $w\in P\cap\partial\Omega_{2}$, or

• (b) $\Vert Sw \Vert \geq \Vert w \Vert$, $w\in P\cap\partial\Omega_{1}$, $\Vert Sw \Vert \leq \Vert w \Vert$, $w\in P\cap\partial\Omega_{2}$.

Then S has a fixed point in $P\cap(\overline{\Omega}_{2}\setminus\Omega_{1})$.

Let us introduce the space $X= \{u(t)\mid u(t)\in C[0,1] \}$ endowed with the norm $\Vert u\Vert=\max_{\in[0,1]} \vert u(t) \vert$. Obviously $(X, \Vert \cdot \Vert )$ is a Banach space. Also let $Y= \{v(t)\mid v(t)\in C[0,1] \}$ endowed with the norm $\Vert v\Vert=\max_{t\in[0,1]} \vert v(t) \vert$. The product space $(X\times Y, \Vert (u,v) \Vert )$ is also a Banach space with the norm $\Vert (u,v) \Vert = \Vert u \Vert + \Vert v \Vert$.

In view of Lemma 2.3, we define the operator $T:X\times Y\rightarrow X\times Y$ by

$$ T(u,v) (t)= \left ( \begin{matrix} T_{1}(u,v)(t)\\ T_{2}(u,v)(t) \end{matrix} \right ), $$

where

$$\begin{aligned}& T_{1}(u,v) (t)= \int_{0}^{1}K_{1}(t,s)f_{1} \bigl(s,u(s),v(s)\bigr)\,ds+ \int _{0}^{1}H_{1}(t,s)f_{2} \bigl(s,u(s),v(s)\bigr)\,ds, \end{aligned}$$

(2.14)

$$\begin{aligned}& T_{2}(u,v) (t)= \int_{0}^{1}K_{2}(t,s)f_{2} \bigl(s,u(s),v(s)\bigr)\,ds+ \int _{0}^{1}H_{2}(t,s)f_{1} \bigl(s,u(s),v(s)\bigr)\,ds. \end{aligned}$$

(2.15)

Lemma 2.8

The operator $T:X\times Y\rightarrow X\times Y$ is completely continuous.

Proof

By continuity of the functions $f_{1}$ and $f_{2}$, the operator T is continuous.

Let $\Omega\subset X\times Y$ be bounded. Then there exist positive constants $A_{1}$ and $A_{2}$ such that

$$\bigl\vert f_{1}\bigl(t,u(t),v(t)\bigr) \bigr\vert \leq A_{1}, \qquad \bigl\vert f_{2}\bigl(t,u(t),v(t)\bigr) \bigr\vert \leq A_{2},\quad\forall (u,v)\in\Omega. $$

Then, for any $(u,v)\in\Omega$, we have

$$\begin{aligned} \bigl\vert T_{1}(u,v) (t) \bigr\vert &\leq \biggl\vert A_{1} \int _{0}^{1}K_{1}(t,s)\,ds+A_{2} \int_{0}^{1}H_{1}(t,s)\,ds \biggr\vert \\ &\leq A_{1}\rho t^{\alpha_{1}-1}+A_{2}\rho t^{\alpha_{1}-1} \\ &=\rho(A_{1}+A_{2})t^{\alpha_{1}-1}, \end{aligned}$$

which implies that

$$\bigl\Vert T_{1}(u,v) \bigr\Vert \leq\rho(A_{1}+A_{2}). $$

Similarly, we get

$$\bigl\Vert T_{2}(u,v) \bigr\Vert \leq\rho(A_{1}+A_{2}). $$

Thus, it follows from the above inequalities that the operator T is uniformly bounded.

Next, we show that T is equicontinuous. Let $t_{1},t_{2}\in[0,1]$ with $t_{1}\leq t_{2}$. Then we have

$$\begin{aligned} & \bigl\vert T_{1}\bigl(u(t_{2}),v(t_{2}) \bigr)-T_{1}\bigl(u(t_{1}),v(t_{1})\bigr) \bigr\vert \\ &\quad\leq A_{1} \biggl\vert \int_{0}^{1}\bigl(K_{1}(t_{2},s)-K_{1}(t_{1},s) \bigr)\,ds \biggr\vert +A_{2} \biggl\vert \int _{0}^{1}\bigl(H_{1}(t_{2},s)-H_{1}(t_{1},s) \bigr)\,ds \biggr\vert . \end{aligned}$$

Analogously, we can obtain

$$\begin{aligned} & \bigl\vert T_{2}\bigl(u(t_{2}),v(t_{2}) \bigr)-T_{2}\bigl(u(t_{1}),v(t_{1})\bigr) \bigr\vert \\ &\quad\leq A_{1} \biggl\vert \int_{0}^{1}\bigl(H_{2}(t_{2},s)-H_{2}(t_{1},s) \bigr)\,ds \biggr\vert +A_{2} \biggl\vert \int _{0}^{1}\bigl(K_{2}(t_{2},s)-K_{2}(t_{1},s) \bigr)\,ds \biggr\vert . \end{aligned}$$

Therefore, the operator $T(u,v)$ is equicontinuous, and thus the operator $T(u,v)$ is completely continuous. □

3 Existence and uniqueness results

Let us introduce the following hypotheses which are used hereafter.

(H₁) Assume there exist real constants $\mu_{i},\lambda_{i}\geq0$ ($i=1,2$) and $\mu_{0},\lambda_{0}>0$ such that $\forall u_{i}\in\mathbb{R}$, $i=1,2$, we have

$$\begin{aligned} & \bigl\vert f_{1}(t,u_{1},u_{2}) \bigr\vert \leq\mu_{0}+\mu _{1} \vert u_{1} \vert +\mu_{2} \vert u_{2} \vert , \\ & \bigl\vert f_{2}(t,u_{1},u_{2}) \bigr\vert \leq \lambda_{0}+\lambda_{1} \vert u_{1} \vert +\lambda _{2} \vert u_{2} \vert . \end{aligned}$$

(H₂) Assume that $f_{1},f_{2}:[0,1]\times\mathbb{R}^{2}\rightarrow\mathbb{R}$ are continuous functions and there exist constants $m_{i}$, $n_{i}$, $i=1,2$ such that, for all $t\in[0,1]$ and $u_{i},v_{i}\in\mathbb{R}$, $i=1,2$,

$$\begin{aligned}& \bigl\vert f_{1}(t,u_{1},u_{2})-f_{1}(t,v_{1},v_{2}) \bigr\vert \leq m_{1} \vert u_{1}-v_{1} \vert +m_{2} \vert u_{2}-v_{2} \vert , \\& \bigl\vert f_{2}(t,u_{1},u_{2})-f_{2}(t,v_{1},v_{2}) \bigr\vert \leq n_{1} \vert u_{1}-v_{1} \vert +n_{2} \vert u_{2}-v_{2} \vert . \end{aligned}$$

For the sake of convenience, we set

$$ M=\min \bigl\{ 1-2\rho(\mu_{1}+\lambda_{1}),1-2 \rho(\mu _{2}+\lambda_{2}) \bigr\} . $$

(3.1)

The first result is based on Leray-Schauder’s alternative.

Theorem 3.1

Assume that (H₁) holds. In addition it is assumed that

$$2\rho(\mu_{1}+\lambda_{1})< 1,\qquad 2\rho(\mu_{2}+ \lambda_{2})< 1. $$

Then the boundary value problem (1.1)-(1.2) has at least one solution.

Proof

It will be verified that the set $\varepsilon= \{(u,v)\in X\times Y\mid(u,v)=\lambda T(u,v), 0\leq\lambda\leq1 \}$ is bounded. Let $(u,v)\in\varepsilon$, then $(u,v)=\lambda T(u,v)$. For any $t\in[0,1]$, we have

$$u(t)=\lambda T_{1}(u,v) (t),\qquad v(t)=\lambda T_{2}(u,v) (t). $$

From (H₁), we have

$$\begin{aligned} \bigl\vert u(t) \bigr\vert &= \bigl\vert \lambda T_{1}(u,v) (t) \bigr\vert \leq \bigl\vert T_{1}(u,v) (t) \bigr\vert \\ &\leq\rho t^{\alpha_{1}-1}\bigl(\mu_{0}+\mu_{1} \Vert u \Vert +\mu_{2}\Vert v \Vert +\lambda_{0}+ \lambda_{1} \Vert u \Vert +\lambda_{2} \Vert v\Vert \bigr) \end{aligned}$$

and

$$\begin{aligned} \bigl\vert v(t) \bigr\vert &= \bigl\vert \lambda T_{2}(u,v) (t) \bigr\vert \leq \bigl\vert T_{2}(u,v) (t) \bigr\vert \\ &\leq\rho t^{\alpha_{2}-1}\bigl(\mu_{0}+\mu_{1} \Vert u \Vert +\mu_{2} \Vert v \Vert +\lambda_{0}+\lambda _{1} \Vert u \Vert +\lambda_{2} \Vert v \Vert \bigr). \end{aligned}$$

Hence we have

$$\begin{aligned} & \Vert u \Vert \leq\rho\bigl(\mu_{0}+\mu_{1} \Vert u \Vert +\mu_{2} \Vert v \Vert +\lambda_{0}+\lambda _{1} \Vert u \Vert +\lambda_{2} \Vert v \Vert \bigr), \\ & \Vert v \Vert \leq\rho\bigl(\mu_{0}+\mu_{1} \Vert u \Vert +\mu_{2} \Vert v \Vert +\lambda_{0}+\lambda _{1} \Vert u \Vert +\lambda_{2} \Vert v \Vert \bigr), \end{aligned}$$

which imply that

$$\Vert u \Vert + \Vert v \Vert \leq2\rho \bigl(\mu_{0}+ \lambda_{0}+(\mu_{1}+\lambda_{1}) \Vert u \Vert +(\mu_{2}+\lambda_{2}) \Vert v \Vert \bigr). $$

Consequently,

$$\bigl\Vert (u,v) \bigr\Vert \leq\frac{2\rho(\mu_{0}+\lambda_{0})}{M}, $$

for any $t\in[0,1]$, where M is defined by (3.1), which proves that ε is bounded. Thus, by Lemma 2.6, the operator T has at least one fixed point. Hence the boundary value problem (1.1)-(1.2) has at least one solution. The proof is complete. □

In the second result, we prove existence and uniqueness of solutions of the boundary value problem (1.1)-(1.2) via Banach’s contraction principle.

Theorem 3.2

Assume that (H₂) holds. In addition, assume that

$$2\rho(m_{1}+m_{2}+n_{1}+n_{2})< 1. $$

Then the boundary value problem (1.1)-(1.2) has a unique solution.

Proof

Define $\sup_{t\in[0,1]}f_{1}(t,0,0)=N_{1}<\infty$ and $\sup_{t\in [0,1]}f_{2}(t,0,0)=N_{2}<\infty$ such that

$$r\geq\frac{2\rho(N_{1}+N_{2})}{1-2\rho(m_{1}+m_{2}+n_{1}+n_{2})}. $$

We show that $TB_{r}\subset B_{r}$, where $B_{r}= \{(u,v)\in X\times Y: \Vert (u,v) \Vert \leq r \}$. For $(u,v)\in B_{r}$, we have

$$\begin{aligned} \bigl\vert T_{1}(u,v) (t) \bigr\vert &=\max_{0\leq t\leq1} \biggl\vert \int_{0}^{1}K_{1}(t,s) \bigl\vert f_{1}\bigl(s,u(s),v(s)\bigr)-f_{1}(s,0,0) \bigr\vert + \bigl\vert f_{1}(s,0,0) \bigr\vert \,ds \\ &\quad{}+ \int_{0}^{1}H_{1}(t,s) \bigl\vert f_{2}\bigl(s,u(s),v(s)\bigr)-f_{2}(s,0,0) \bigr\vert + \bigl\vert f_{2}(s,0,0) \bigr\vert \,ds \biggr\vert \\ &\leq\rho t^{\alpha_{1}-1}\bigl(m_{1} \Vert u \Vert +m_{2} \Vert v \Vert +N_{1}+n_{1} \Vert u \Vert +n_{2} \Vert v \Vert +N_{2}\bigr) \\ &\leq\rho \bigl[(m_{1}+m_{2})r+(n_{1}+n_{2})r+N_{1}+N_{2} \bigr]. \end{aligned}$$

Hence

$$\bigl\Vert T_{1}(u,v) \bigr\Vert \leq\rho \bigl[(m_{1}+m_{2}+n_{1}+n_{2})r+N_{1}+N_{2} \bigr]. $$

In the same way, we obtain

$$\bigl\Vert T_{2}(u,v) \bigr\Vert \leq\rho \bigl[(m_{1}+m_{2}+n_{1}+n_{2})r+N_{1}+N_{2} \bigr]. $$

Consequently, $\Vert T(u,v) \Vert \leq r$.

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

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

and consequently we obtain

$$ \bigl\Vert T_{1}(u_{2},v_{2})-T_{1}(u_{1},v_{1}) \bigr\Vert \leq\rho (m_{1}+m_{2}+n_{1}+n_{2}) \bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr). $$

(3.2)

Similarly,

$$ \bigl\Vert T_{2}(u_{2},v_{2})-T_{2}(u_{1},v_{1}) \bigr\Vert \leq\rho (m_{1}+m_{2}+n_{1}+n_{2}) \bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr). $$

(3.3)

It follows from (3.2) and(3.3) that

$$\bigl\Vert T(u_{2},v_{2})-T(u_{1},v_{1}) \bigr\Vert \leq2\rho (m_{1}+m_{2}+n_{1}+n_{2}) \bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr). $$

Since $2\rho(m_{1}+m_{2}+n_{1}+n_{2})<1$, T is a contraction operator. So, by Banach’s fixed point theorem, the operator T has a unique fixed point, which is the unique solution of problem (1.1)-(1.2). This completes the proof. □

Remark 3.1

In the case that the left parts of boundary conditions (1.2) are also the multi-strip and multi-point mixed boundary conditions, the existence and uniqueness conclusions could be obtained by using the definition of the Caputo type fractional q-derivative. This might be our forthcoming work.

Example 3.1

Consider the following fractional boundary value problem:

$$ \textstyle\begin{cases} D_{0+}^{2.5}u(t)=\frac{1}{8(t+2)^{2}}u(t)+\frac{1}{32}\sin v(t)+1,\\ D_{0+}^{2.5}v(t)=\frac{1}{64\pi}\sin(2\pi u(t))+\frac {1}{32}v(t)+\frac{1}{2},\\ u(0)=u'(0)=0,\qquad u(1)=\sum_{i=1}^{2}\int_{\xi_{i}}^{\eta _{i}}v(s)\,dA_{i}(s)+\sum_{i=1}^{2}b_{i}v(\sigma_{i}),\\ v(0)=v'(0)=0,\qquad v(1)=\sum_{i=1}^{2}\int_{\xi_{i}}^{\eta_{i}}u(s)\,dA_{i}(s)+\sum_{i=1}^{2}b_{i}u(\sigma_{i}), \end{cases} $$

(3.4)

where

$$\begin{aligned} &A_{1}(t)=0.5t,\qquad \xi_{1}=\frac{1}{8},\qquad \eta_{1}=\frac{3}{8},\qquad b_{1}=\frac{1}{3}, \qquad \sigma_{1}=\frac{1}{3}, \\ &A_{2}(t)=0.8t,\qquad \xi_{2}=\frac{5}{8},\qquad \eta_{2}=\frac{7}{8}, \qquad b_{2}= \frac{2}{3},\qquad \sigma_{2}=\frac{2}{3}. \end{aligned}$$

Then considering the condition (H₁)

$$\begin{aligned} & \bigl\vert f_{1}(t,x_{1},x_{2}) \bigr\vert \leq1+\frac{1}{32} \vert x_{1} \vert + \frac{1}{32} \vert x_{2} \vert , \\ & \bigl\vert f_{2}(t,x_{1},x_{2}) \bigr\vert \leq\frac{1}{2}+\frac {1}{32} \vert x_{1} \vert +\frac{1}{32} \vert x_{2} \vert . \end{aligned}$$

We get $\mu_{1}=\frac{1}{32}$, $\mu_{2}=\frac{1}{32}$, $\lambda_{1}=\frac {1}{32}$, $\lambda_{2}=\frac{1}{32}$. As for the condition (H₂)

$$\begin{aligned} & \bigl\vert f_{1}(t,u_{1},u_{2})-f_{1}(t,v_{1},v_{2}) \bigr\vert \leq \frac{1}{32} \vert u_{1}-u_{2} \vert +\frac{1}{32} \vert v_{1}-v_{2} \vert , \\ & \bigl\vert f_{2}(t,u_{1},u_{2})-f_{2}(t,v_{1},v_{2}) \bigr\vert \leq \frac{1}{32} \vert u_{1}-u_{2} \vert +\frac{1}{32} \vert v_{1}-v_{2} \vert . \end{aligned}$$

We get $m_{1}=\frac{1}{32}$, $m_{2}=\frac{1}{32}$, $n_{1}=\frac {1}{32}$, $n_{2}=\frac{1}{32}$. By simple computation, we have

$$\begin{aligned} &\rho_{1}=\rho_{3}=\frac{1.5}{\Gamma(2.5)} \Biggl[1+ \frac {1}{1-1_{1}l_{2}} \Biggl(\sum_{i=1}^{2} \int_{\xi_{i}}^{\eta _{i}} dA_{i}(t)+\sum _{i=1}^{2}b_{i} \Biggr) \Biggr] \approx3.3559, \\ &\rho_{2}=\rho_{4}=\frac{1.5}{\Gamma(2.5)(1-l_{1}l_{2})} \Biggl[\sum _{i=1}^{2} \int_{\xi_{i}}^{\eta_{i}} dA_{i}(t)+\sum _{i=1}^{2}b_{i} \Biggr]\approx 2.2276, \end{aligned}$$

with

$$l_{1}=l_{2}=\sum_{i=1}^{2} \int_{\xi_{i}}^{\eta _{i}}t^{1.5}\,dA_{i}(t)+ \sum_{i=1}^{2}b_{i} \sigma_{i}^{1.5}\approx 0.5734. $$

Then $\rho=\max \{\rho_{1},\rho_{2},\rho_{3},\rho_{4} \}\approx3.3559$. Therefore,

$$2\rho(\mu_{1}+\lambda_{1})\approx0.4194< 1,\qquad\rho(\mu _{2}+\lambda_{2})\approx0.4194< 1. $$

By Theorem 3.1, the coupled boundary value problem (3.4) has at least one solution. We also have

$$2\rho(m_{1}+m_{2}+n_{1}+n_{2}) \approx0.8390< 1. $$

By Theorem 3.2, the coupled boundary value problem (3.4) has a unique solution.

4 Existence and uniqueness results of a positive solution

In this section, X is the same Banach space as the space defined in Section 3. Denote

$$P= \bigl\{ (u,v)\in X:u(t)\geq\omega t^{\alpha_{1}-1} \bigl\Vert (u,v) \bigr\Vert ,v(t)\geq\omega t^{\alpha_{2}-1} \bigl\Vert (u,v) \bigr\Vert ,t \in[0,1] \bigr\} , $$

where ω is defined in Remark 2.1. It can easily be seen that P is a cone in X. For any real constants r and R with $0< r< R$, define

$$P_{r}= \bigl\{ (u,v)\in P: \bigl\Vert (u,v) \bigr\Vert < r \bigr\} ,\qquad P_{[r,R]}= \bigl\{ (u,v)\in P:r\leq \bigl\Vert (u,v) \bigr\Vert \leq R \bigr\} . $$

In what follows, we list the following assumptions for convenience.

(H₃) $f_{1}:[0,1]\times[0,\infty)\times[0,\infty)\rightarrow[0,\infty)$ is continuous, $f_{1}(t,u,v)$ is nondecreasing in u and nonincreasing in v, and there exist two constants $\theta_{1},\vartheta_{1}\in[0,1)$ such that

$$ \begin{aligned} &\kappa^{\theta_{1}}f_{1}(t,u,v) \leq f_{1}(t,\kappa u,v), \\ &f_{1}(t,u,\kappa v)\leq\kappa^{-\vartheta_{1}}f_{1}(t,u,v), \quad \forall u,v>0, \kappa\in(0,1). \end{aligned} $$

(4.1)

(H₄) $f_{2}:[0,1]\times[0,\infty)\times[0,\infty)\rightarrow[0,\infty)$ is continuous, $f_{2}(t,u,v)$ is nonincreasing in u and nondecreasing in v, and there exist two constants $\theta_{2},\vartheta_{2}\in[0,1)$ such that

$$ \begin{aligned} &\kappa^{\theta_{2}}f_{2}(t,u,v) \leq f_{2}(t,u,\kappa v), \\ &f_{2}(t,\kappa u,v)\leq\kappa^{-\vartheta_{2}}f_{2}(t,u,v), \quad \forall u,v>0, \kappa\in(0,1). \end{aligned} $$

(4.2)

(H₅) The following inequalities hold:

$$0< \int_{0}^{1}\varphi_{1}(s)f_{1} \bigl(s,1,s^{\alpha_{2}-1}\bigr)\,ds< +\infty ,\qquad0< \int_{0}^{1}\varphi_{2}(s)f_{2} \bigl(s,s^{\alpha _{1}-1},1\bigr)\,ds< +\infty, $$

where $\varphi_{1}$ and $\varphi_{2}$ are defined in Lemma 2.4.

Remark 4.1

From assumptions (H₃) and (H₄), we have

$$f_{1}\bigl(s,s^{\alpha_{2}-1},1\bigr)\leq f_{1} \bigl(s,1,s^{\alpha_{2}-1}\bigr),\qquad f_{2}\bigl(s,1,s^{\alpha_{1}-1} \bigr)\leq f_{2}\bigl(s,s^{\alpha_{1}-1},1\bigr). $$

This together with (H₅) yields

$$\begin{aligned} &0< \int_{0}^{1}\varphi_{1}(s)f_{1} \bigl(s,s^{\alpha_{2}-1},1\bigr)\,ds\leq \int _{0}^{1}\varphi_{1}(s)f_{1} \bigl(s,1,s^{\alpha_{2}-1}\bigr)\,ds< +\infty, \\ &0< \int_{0}^{1}\varphi_{2}(s)f_{2} \bigl(s,1,s^{\alpha_{1}-1}\bigr)\,ds\leq \int _{0}^{1}\varphi_{2}(s)f_{2} \bigl(s,s^{\alpha_{1}-1},1\bigr)\,ds< +\infty. \end{aligned}$$

Remark 4.2

The inequalities (4.1) and (4.2) imply that

$$\begin{aligned}& f_{1}(t,\kappa u,v)\leq\kappa^{\theta_{1}}f_{1}(t,u,v), \qquad f_{1}(t,u,v)\leq\kappa^{\vartheta_{1}}f_{1}(t,u,\kappa v),\quad \forall u,v>0, \kappa\in(1,\infty); \end{aligned}$$

(4.3)

$$\begin{aligned}& f_{2}(t,u,\kappa v)\leq\kappa^{\theta_{2}}f_{2}(t,u,v), \qquad f_{2}(t,u,v)\leq\kappa^{\vartheta_{2}}f_{2}(t,\kappa u,v),\quad \forall u,v>0, \kappa\in(1,\infty). \end{aligned}$$

(4.4)

From the above assumptions (H₃), (H₄) and (H₅), for any $(u,v)\in P\setminus \{(0,0) \}$, define the operator $T:P\setminus \{(0,0) \}\rightarrow P$ introduced by (2.14) and (2.15). Obviously, $(u,v)$ is a positive solutions of the boundary value problem (1.1)-(1.2) if and only if $(u,v)$ is a fixed point of T in $P\setminus \{(0,0) \}$.

Lemma 4.1

Assume that (H₃)-(H₅) hold. For any $0< r_{1}< r_{2}<+\infty, T:P_{[r_{1},r_{2}]}\rightarrow P$ is a completely continuous operator.

Proof

For any $(u,v)\in P\setminus \{(0,0) \}$, we can see that

$$ \begin{gathered}\omega t^{\alpha_{1}-1} \bigl\Vert (u,v) \bigr\Vert \leq u(t) \leq \bigl\Vert (u,v) \bigr\Vert ,\\ \omega t^{\alpha_{2}-1} \bigl\Vert (u,v) \bigr\Vert \leq v(t)\leq \bigl\Vert (u,v) \bigr\Vert ,\quad t \in[0,1]. \end{gathered}$$

(4.5)

Let $\kappa>1$ such that $\Vert (u,v) \Vert /\kappa<1$. From (H₃), (H₄), (4.3), (4.4) and (4.5), we have

$$ \begin{gathered} f_{1} \bigl(t,u(t),v(t) \bigr)\leq f_{1} \bigl(t,\kappa,\omega t^{\alpha_{2}-1} \bigl\Vert (u,v) \bigr\Vert \bigr)\leq\kappa ^{\theta_{1}}f_{1} \biggl(t,1, \frac{\omega \Vert (u,v) \Vert }{\kappa}t^{\alpha_{2}-1} \biggr) \\ \hphantom{f_{1} \bigl(t,u(t),v(t) \bigr)}\leq\kappa^{\theta_{1}+\vartheta_{1}} \bigl(\omega \bigl\Vert (u,v) \bigr\Vert \bigr)^{-\vartheta_{1}}f_{1} \bigl(t,1,t^{\alpha _{2}-1} \bigr), \\ f_{2} \bigl(t,u(t),v(t) \bigr)\leq f_{2} \bigl(t,\omega t^{\alpha _{1}-1} \bigl\Vert (u,v) \bigr\Vert ,\kappa \bigr)\leq\kappa ^{\theta_{2}}f_{2} \biggl(t,\frac{\omega \Vert (u,v) \Vert }{\kappa}t^{\alpha_{1}-1},1 \biggr) \\ \hphantom{f_{2} \bigl(t,u(t),v(t) \bigr)}\leq\kappa^{\theta_{2}+\vartheta_{2}} \bigl(\omega \bigl\Vert (u,v) \bigr\Vert \bigr)^{-\vartheta_{2}}f_{2} \bigl(t,t^{\alpha _{1}-1},1 \bigr). \end{gathered} $$

(4.6)

Hence, for any $t\in[0,1]$, by Lemma 2.5 and (4.6), we get

$$\begin{aligned}& \begin{aligned} T_{1}(u,v) (t)&= \int_{0}^{1}K_{1}(t,s)f_{1} \bigl(s,u(s),v(s) \bigr)\,ds+ \int_{0}^{1}H_{1}(t,s)f_{2} \bigl(s,u(s),v(s) \bigr)\,ds \\ &\leq \rho \biggl(\kappa^{\theta_{1}+\vartheta_{1}} \bigl(\omega \bigl\Vert (u,v) \bigr\Vert \bigr)^{-\vartheta_{1}} \int _{0}^{1}\varphi_{1}(s)f_{1} \bigl(s,1,s^{\alpha_{2}-1} \bigr)\,ds \\ &\quad {}+\kappa^{\theta_{2}+\vartheta_{2}} \bigl(\omega \bigl\Vert (u,v) \bigr\Vert \bigr)^{-\vartheta_{2}} \int_{0}^{1}\varphi _{2}(s)f_{2} \bigl(s,s^{\alpha_{1}-1},1 \bigr)\,ds \biggr) \\ &< +\infty, \end{aligned} \\& \begin{aligned} T_{2}(u,v) (t)&= \int_{0}^{1}K_{2}(t,s)f_{2} \bigl(s,u(s),v(s) \bigr)\,ds+ \int_{0}^{1}H_{2}(t,s)f_{1} \bigl(s,u(s),v(s) \bigr)\,ds \\ &\leq \rho \biggl(\kappa^{\theta_{1}+\vartheta_{1}} \bigl(\omega \bigl\Vert (u,v) \bigr\Vert \bigr)^{-\vartheta_{1}} \int _{0}^{1}\varphi_{1}(s)f_{1} \bigl(s,1,s^{\alpha_{2}-1} \bigr)\,ds \\ &\quad {}+\kappa^{\theta_{2}+\vartheta_{2}} \bigl(\omega \bigl\Vert (u,v) \bigr\Vert \bigr)^{-\vartheta_{2}} \int_{0}^{1}\varphi _{2}(s)f_{2} \bigl(s,s^{\alpha_{1}-1},1 \bigr)\,ds \biggr) \\ &< +\infty. \end{aligned} \end{aligned}$$

Together with the continuity of $K_{k}(t,s)$ and $H_{k}(t,s)$ ($k=1,2$), it is easy to see that $T_{k}\in C[0,1]$. Therefore, $T:P\setminus \{(0,0) \}\rightarrow P$ is well defined.

For any $(u,v)\in P_{[r_{1},r_{2}]}$ and $t,\tau\in[0,1]$, by Remark 2.1, we obtain

$$\begin{aligned}& \begin{aligned} T_{1}(u,v) (t)&= \int_{0}^{1}K_{1}(t,s)f_{1} \bigl(s,u(s),v(s) \bigr)\,ds+ \int_{0}^{1}H_{1}(t,s)f_{2} \bigl(s,u(s),v(s) \bigr)\,ds \\ &\geq \int_{0}^{1}\omega t^{\alpha_{1}-1}K_{1}( \tau,s)f_{1} \bigl(s,u(s),v(s) \bigr)\,ds+ \int_{0}^{1}\omega t^{\alpha_{1}-1}H_{1}( \tau ,s)f_{2} \bigl(s,u(s),v(s) \bigr)\,ds \\ &=\omega t^{\alpha_{1}-1}T_{1}(u,v) (\tau), \end{aligned} \\& \begin{aligned} T_{1}(u,v) (t)&= \int_{0}^{1}K_{1}(t,s)f_{1} \bigl(s,u(s),v(s) \bigr)\,ds+ \int_{0}^{1}H_{1}(t,s)f_{2} \bigl(s,u(s),v(s) \bigr)\,ds \\ &\geq \int_{0}^{1}\omega t^{\alpha_{1}-1}H_{2}( \tau,s)f_{1} \bigl(s,u(s),v(s) \bigr)\,ds+ \int_{0}^{1}\omega t^{\alpha_{1}-1}K_{2}( \tau ,s)f_{2} \bigl(s,u(s),v(s) \bigr)\,ds \\ &=\omega t^{\alpha_{1}-1}T_{2}(u,v) (\tau). \end{aligned} \end{aligned}$$

Then we have

$$\begin{aligned} &T_{1}(u,v) (t)\geq\omega t^{\alpha_{1}-1} \bigl\Vert T_{1}(u,v) \bigr\Vert ,\qquad T_{1}(u,v) (t)\geq\omega t^{\alpha_{1}-1} \bigl\Vert T_{2}(u,v) \bigr\Vert , \\ & T_{1}(u,v) (t)\geq\frac{1}{2}\omega t^{\alpha_{1}-1} \bigl\Vert T_{1}(u,v),T_{2}(u,v) \bigr\Vert . \end{aligned}$$

In the same way, we can prove that

$$\begin{aligned} &T_{2}(u,v) (t)\geq\omega t^{\alpha_{2}-1} \bigl\Vert T_{2}(u,v) \bigr\Vert ,\qquad T_{2}(u,v) (t)\geq\omega t^{\alpha_{2}-1} \bigl\Vert T_{1}(u,v) \bigr\Vert , \\ & T_{2}(u,v) (t)\geq\frac{1}{2}\omega t^{\alpha_{2}-1} \bigl\Vert T_{1}(u,v),T_{2}(u,v) \bigr\Vert . \end{aligned}$$

Therefore, we have $TP_{[r_{1},r_{2}]}\subseteq TP$. Further $T:P_{[r_{1},r_{2}]}\rightarrow P$ is completely continuous combining with Lemma 2.8. □

Theorem 4.2

Assume that (H₃)-(H₅) hold. Then the boundary value problem (1.1)-(1.2) has at least one positive solution $(u^{*},v^{*})$, and there exists a real number $0<\delta<1$ satisfying

$$\delta t^{\alpha_{1}-1}\leq u^{*}(t)\leq\delta^{-1}t^{\alpha _{1}-1}, \qquad\delta t^{\alpha_{2}-1}\leq v^{*}(t)\leq\delta ^{-1}t^{\alpha_{2}-1},\quad t\in[0,1]. $$

Proof

First, we show that the boundary value problem (1.1)-(1.2) has at least one positive solution.

There exists a positive constant c satisfying $0< c<1-c<1$. Choose r and R such that

$$\begin{aligned}& 0< r\leq\min \biggl\{ \biggl(\frac{1}{2}\varrho c^{\alpha _{1}-1} \omega^{\theta_{1}} \int_{0}^{1}\varphi_{1}(s)f_{1} \bigl(s,s^{\alpha_{1}-1},1 \bigr)\,ds \biggr)^{\frac{1}{1-\theta _{1}}},\frac{1}{4} \biggr\} , \\& \begin{aligned} R&\geq\max \biggl\{ \biggl(\frac{\rho}{2} \int_{0}^{1}\varphi _{1}(s)f_{1} \bigl(s,1,s^{\alpha_{2}-1} \bigr)\,ds \\ &\quad {}+\frac{\rho}{2} \int_{0}^{1}\varphi_{2}(s)f_{2} \bigl(s,s^{\alpha_{1}-1},1 \bigr)\,ds \biggr)^{\frac{1}{1-\max\{\theta _{1},\theta_{2}\}}},\frac{1}{\omega},1 \biggr\} . \end{aligned} \end{aligned}$$

For any $(u,v)\in\partial P_{r}$, we have

$$ r\omega t^{\alpha_{1}-1}\leq u(t)\leq r,\qquad r\omega t^{\alpha_{2}-1}\leq v(t)\leq r,\quad t\in[0,1]. $$

(4.7)

By Lemma 2.5, Remark 4.1, (H₃) and (H₄), for any $(u,v)\in\partial P_{r}$, we get

$$\begin{aligned} T_{1}(u,v) (t)&= \int_{0}^{1}K_{1}(t,s)f_{1} \bigl(s,u(s),v(s) \bigr)\,ds+ \int_{0}^{1}H_{1}(t,s)f_{2} \bigl(s,u(s),v(s) \bigr)\,ds \\ &\geq \int_{0}^{1}K_{1}(t,s)f_{1} \bigl(s,u(s),v(s) \bigr)\,ds\geq \varrho t^{\alpha_{1}-1} \int_{0}^{1}\varphi_{1}(s)f_{1} \bigl(s,r\omega s^{\alpha_{1}-1},r \bigr)\,ds \\ &\geq\varrho c^{\alpha_{1}-1}(r\omega)^{\theta_{1}} \int _{0}^{1}\varphi_{1}(s)f_{1} \bigl(s,s^{\alpha_{1}-1},1 \bigr)\,ds \\ & \geq2r= \bigl\Vert (u,v) \bigr\Vert ,\quad \text{for } t\in[c,1-c]. \end{aligned}$$

Similarly, we can get $T_{2}(u,v)(t)\geq \Vert u,v \Vert$, for $t\in [c,1-c]$. This guarantees that

$$ \bigl\Vert T(u,v) \bigr\Vert \geq \bigl\Vert (u,v) \bigr\Vert ,\quad\forall (u,v)\in\partial P_{r}. $$

(4.8)

On the other hand, for any $(u,v)\in\partial P_{R}$, we have

$$R\omega t^{\alpha_{1}-1}\leq u(t)\leq R,\qquad R\omega t^{\alpha _{2}-1}\leq v(t) \leq R,\quad t\in[0,1]. $$

By Lemma 2.5, ($H_{3}$)-($H_{5}$), (4.3) and (4.4), for any $(u,v)\in\partial P_{R}$, we get

$$\begin{aligned} T_{1}(u,v) (t)&= \int_{0}^{1}K_{1}(t,s)f_{1} \bigl(s,u(s),v(s) \bigr)\,ds+ \int_{0}^{1}H_{1}(t,s)f_{2} \bigl(s,u(s),v(s) \bigr)\,ds \\ &\leq\rho \int_{0}^{1}\varphi_{1}(s)f_{1} \bigl(s,R,R\omega s^{\alpha_{2}-1} \bigr)\,ds+\rho \int_{0}^{1}\varphi_{2}(s)f_{2} \bigl(s,R\omega s^{\alpha_{1}-1},R \bigr)\,ds \\ &\leq\rho R^{\theta_{1}} \int_{0}^{1}\varphi_{1}(s)f_{1} \bigl(s,1,s^{\alpha_{2}-1} \bigr)\,ds+\rho R^{\theta_{2}} \int _{0}^{1}\varphi_{2}(s)f_{2} \bigl(s,s^{\alpha_{1}-1},1 \bigr)\,ds \\ &\leq\rho R^{\max\{\theta_{1},\theta_{2}\}} \biggl( \int _{0}^{1}\varphi_{1}(s)f_{1} \bigl(s,1,s^{\alpha_{2}-1} \bigr)\,ds+ \int _{0}^{1}\varphi_{2}(s)f_{2} \bigl(s,s^{\alpha_{1}-1},1 \bigr)\,ds \biggr) \\ &\leq2R= \bigl\Vert (u,v) \bigr\Vert . \end{aligned}$$

In the same way, we have $T_{2}(u,v)(t)\leq2R= \Vert (u,v) \Vert$, for all $(u,v)\in\partial P_{R}$. So we have

$$ \bigl\Vert T(u,v) \bigr\Vert \leq \bigl\Vert (u,v) \bigr\Vert ,\quad\forall (u,v)\in\partial P_{R}. $$

(4.9)

By the complete continuity of T, (4.8), (4.9), and Lemma 2.7, we find that T has at least a fixed point $(u^{*},v^{*})\in P_{[r,R]}$. Consequently, boundary value problem (1.1)-(1.2) has a positive solution $(u^{*},v^{*})\in P_{[r,R]}$.

Next, we show there exists a real number $0<\delta<1$ satisfying

$$\delta t^{\alpha_{1}-1}\leq u^{*}(t)\leq\delta^{-1}t^{\alpha _{1}-1}, \qquad\delta t^{\alpha_{2}-1}\leq v^{*}(t)\leq\delta ^{-1}t^{\alpha_{2}-1},\quad t\in[0,1]. $$

From the Lemma 4.1, we know $(u^{*},v^{*})\in P\setminus \{(0,0) \}$. So, we have

$$\begin{aligned} &\omega t^{\alpha_{1}-1} \bigl\Vert \bigl(u^{*},v^{*} \bigr) \bigr\Vert \leq u^{*}(t)\leq \bigl\Vert \bigl(u^{*},v^{*} \bigr) \bigr\Vert ,\quad t\in[0,1], \\ &\omega t^{\alpha_{2}-1} \bigl\Vert \bigl(u^{*},v^{*} \bigr) \bigr\Vert \leq v^{*}(t)\leq \bigl\Vert \bigl(u^{*},v^{*} \bigr) \bigr\Vert ,\quad t\in[0,1]. \end{aligned}$$

Choose κ, such that $\Vert (u^{*},v^{*}) \Vert /\kappa<1$, $\kappa>1/\omega$. By Lemma 2.5, (H₃) and (H₄), for $t\in[0,1]$, we have

$$\begin{aligned} u^{*}(t)&= \int_{0}^{1}K_{1}(t,s)f_{1} \bigl(s,u^{*}(s),v^{*}(s) \bigr)\,ds+ \int_{0}^{1}H_{1}(t,s)f_{2} \bigl(s,u^{*}(s),v^{*}(s) \bigr)\,ds \\ &\leq \int_{0}^{1}\rho t^{\alpha_{1}-1}f_{1} \bigl(s,\kappa,\omega s^{\alpha_{2}-1} \bigl\Vert \bigl(u^{*},v^{*} \bigr) \bigr\Vert \bigr)\,ds+ \int_{0}^{1}\rho t^{\alpha_{1}-1}f_{2} \bigl(s,\omega s^{\alpha_{1}-1} \bigl\Vert \bigl(u^{*},v^{*} \bigr) \bigr\Vert ,\kappa \bigr)\,ds \\ &\leq\rho t^{\alpha_{1}-1} \biggl( \int_{0}^{1}f_{1} \biggl(s,\kappa , \frac{\omega \Vert (u^{*},v^{*}) \Vert }{\kappa }s^{\alpha_{2}-1} \biggr)\,ds+ \int_{0}^{1}f_{2} \biggl(s, \frac{\omega \Vert (u^{*},v^{*}) \Vert }{\kappa}s^{\alpha _{1}-1},\kappa \biggr)\,ds \biggr) \\ &\leq\rho t^{\alpha_{1}-1} \biggl(\kappa^{\theta_{1}+\vartheta _{1}} \bigl(\omega \bigl\Vert \bigl(u^{*},v^{*} \bigr) \bigr\Vert \bigr)^{-\vartheta_{1}} \int_{0}^{1}f_{1} \bigl(s,1,s^{\alpha _{2}-1} \bigr)\,ds \\ &\quad {}+\kappa^{\theta_{2}+\vartheta_{2}} \bigl(\omega \bigl\Vert \bigl(u^{*},v^{*} \bigr) \bigr\Vert \bigr)^{-\vartheta_{2}} \int _{0}^{1}f_{2} \bigl(s,s^{\alpha_{1}-1},1 \bigr)\,ds \biggr) \\ &\leq\rho t^{\alpha_{1}-1} \biggl(\kappa^{\theta_{1}+\vartheta _{1}}(2\omega R)^{-\vartheta_{1}} \int_{0}^{1}f_{1} \bigl(s,1,s^{\alpha_{2}-1} \bigr)\,ds \\ &\quad {}+\kappa^{\theta_{2}+\vartheta _{2}}(2\omega R)^{-\vartheta_{2}} \int_{0}^{1}f_{2} \bigl(s,s^{\alpha _{1}-1},1 \bigr)\,ds \biggr). \end{aligned}$$

In the same way, for $t\in[0,1]$, we also have

$$\begin{aligned} v^{*}(t)&\leq\rho t^{\alpha_{2}-1} \biggl(\kappa^{\theta _{1}+\vartheta_{1}} (2 \omega R )^{-\vartheta_{1}} \int _{0}^{1}f_{1} \bigl(s,1,s^{\alpha_{2}-1} \bigr)\,ds \\ &\quad {}+\kappa^{\theta_{2}+\vartheta_{2}} (2\omega R )^{-\vartheta_{2}} \int_{0}^{1}f_{2} \bigl(s,s^{\alpha_{1}-1},1 \bigr)\,ds \biggr). \end{aligned}$$

Choose

$$\begin{aligned} \delta&=\min \biggl\{ \omega r, \biggl(\rho\kappa^{\theta _{1}+\vartheta_{1}} (2\omega R )^{-\vartheta_{1}} \int _{0}^{1}f_{1} \bigl(s,1,s^{\alpha_{2}-1} \bigr)\,ds \\ &\quad {}+\rho\kappa^{\theta_{2}+\vartheta_{2}} (2\omega R )^{-\vartheta_{2}} \int_{0}^{1}f_{2} \bigl(s,s^{\alpha_{1}-1},1 \bigr)\,ds \biggr)^{-1},\frac{1}{2} \biggr\} , \end{aligned}$$

combining with (4.7), we have

$$\delta t^{\alpha_{1}-1}\leq u^{*}(t)\leq\delta^{-1}t^{\alpha _{1}-1}, \qquad\delta t^{\alpha_{2}-1}\leq v^{*}(t)\leq\delta ^{-1}t^{\alpha_{2}-1},\quad t\in[0,1]. $$

This completes the proof of Theorem 4.2. □

Theorem 4.3

Assume that (H₃)-(H₅) hold. Furthermore, assume $\theta_{1}+\vartheta_{1}<1$ and $\theta_{2}+\vartheta_{2}<1$. Then boundary value problem (1.1)-(1.2) has a unique positive solution on $[0,1]$.

Proof

Assume that the coupled boundary value problem (1.1)-(1.2) has two different positive solutions $(u_{1},v_{1})$ and $(u_{2},v_{2})$. By Theorem 4.2, there exist $0<\delta _{1}<1$ and $0<\delta_{2}<1$ such that

$$\begin{aligned} \begin{aligned} &\delta_{1}t^{\alpha_{1}-1}\leq u_{1}(t)\leq\delta _{1}^{-1}t^{\alpha_{1}-1}, \qquad\delta_{1}t^{\alpha_{2}-1}\leq v_{1}(t)\leq \delta_{1}^{-1}t^{\alpha_{2}-1},\quad t\in[0,1], \\ &\delta_{2}t^{\alpha_{1}-1}\leq u_{2}(t)\leq\delta _{2}^{-1}t^{\alpha_{1}-1},\qquad\delta_{2}t^{\alpha_{2}-1} \leq v_{2}(t)\leq\delta_{2}^{-1}t^{\alpha_{2}-1}, \quad t\in[0,1]. \end{aligned} \end{aligned}$$

Further, we have

$$\begin{aligned} &\delta_{1}\delta_{2}u_{2}(t)\leq u_{1}(t)\leq(\delta_{1}\delta _{2})^{-1}u_{2}(t), \quad t\in[0,1], \\ &\delta_{1}\delta_{2}v_{2}(t)\leq v_{1}(t)\leq(\delta_{1}\delta _{2})^{-1}v_{2}(t), \quad t\in[0,1]. \end{aligned}$$

Obviously, one has $\delta_{1}\delta_{2}\neq1$. Put

$$\Delta=\sup \bigl\{ \delta:\delta u_{2}(t)\leq u_{1}(t)\leq \delta ^{-1}u_{2}(t),\delta v_{2}(t)\leq v_{1}(t)\leq\delta ^{-1}v_{2}(t),t\in[0,1] \bigr\} . $$

It is easy to see that $0<\delta_{1}\delta_{2}<\Delta<1$, and

$$ \Delta u_{2}(t)\leq u_{1}(t)\leq \Delta^{-1}u_{2}(t),\qquad\Delta v_{2}(t)\leq v_{1}(t)\leq\Delta^{-1}v_{2}(t),\quad t\in[0,1]. $$

(4.10)

By (H₃), (H₄) and (4.10), we get

$$ \begin{gathered} f_{1} \bigl(t,u_{1}(t),v_{1}(t) \bigr) \geq f_{1} \bigl(t,\Delta u_{2}(t), \Delta^{-1}v_{2}(t) \bigr)\geq\Delta^{\theta _{1}+\vartheta_{1}}f_{1} \bigl(t,u_{2}(t),v_{2}(t) \bigr) \\ \hphantom{f_{1} \bigl(t,u_{1}(t),v_{1}(t) \bigr)}\geq\Delta^{\sigma}f_{1} \bigl(t,u_{2}(t),v_{2}(t) \bigr), \\ f_{2} \bigl(t,u_{1}(t),v_{1}(t) \bigr) \geq f_{2} \bigl(t,\Delta^{-1} u_{2}(t),\Delta v_{2}(t) \bigr)\geq\Delta^{\theta_{2}+\vartheta _{2}}f_{2} \bigl(t,u_{2}(t),v_{2}(t) \bigr) \\ \hphantom{f_{2} \bigl(t,u_{1}(t),v_{1}(t) \bigr)}\geq\Delta^{\sigma}f_{2} \bigl(t,u_{2}(t),v_{2}(t) \bigr), \end{gathered} $$

(4.11)

where $\sigma=\max \{\theta_{1}+\vartheta_{1},\theta_{2}+\vartheta _{2} \}$ such that $\sigma<1$. Similarly, by (H₃), (H₄) and (4.10), we have

$$ \begin{gathered} f_{1} \bigl(t,u_{2}(t),v_{2}(t) \bigr) \geq f_{1} \bigl(t,\Delta u_{1}(t), \Delta^{-1}v_{1}(t) \bigr)\geq\Delta^{\theta _{1}+\vartheta_{1}}f_{1} \bigl(t,u_{1}(t),v_{1}(t) \bigr) \\ \hphantom{f_{1} \bigl(t,u_{2}(t),v_{2}(t) \bigr)}\geq\Delta^{\sigma}f_{1} \bigl(t,u_{1}(t),v_{1}(t) \bigr), \\ f_{2} \bigl(t,u_{2}(t),v_{2}(t) \bigr) \geq f_{2} \bigl(t,\Delta^{-1} u_{1}(t),\Delta v_{1}(t) \bigr)\geq\Delta^{\theta_{2}+\vartheta _{2}}f_{2} \bigl(t,u_{1}(t),v_{1}(t) \bigr) \\ \hphantom{f_{2} \bigl(t,u_{2}(t),v_{2}(t) \bigr)}\geq\Delta^{\sigma}f_{2} \bigl(t,u_{1}(t),v_{1}(t) \bigr). \end{gathered} $$

(4.12)

From (4.11), for $t\in[0,1]$, we have

$$ \begin{gathered} u_{1}(t) =T_{1}(u_{1},v_{1}) (t) \\ \hphantom{ u_{1}(t)}= \int_{0}^{1}K_{1}(t,s)f_{1} \bigl(s,u_{1}(s),v_{1}(s) \bigr)\,ds+ \int _{0}^{1}H_{1}(t,s)f_{2} \bigl(s,u_{1}(s),v_{1}(s) \bigr)\,ds \\ \hphantom{ u_{1}(t)}\geq \int_{0}^{1}K_{1}(t,s) \Delta^{\sigma}f_{1} \bigl(s,u_{2}(s),v_{2}(s) \bigr)\,ds+ \int_{0}^{1}H_{1}(t,s) \Delta^{\sigma }f_{2} \bigl(s,u_{2}(s),v_{2}(s) \bigr)\,ds \\ \hphantom{ u_{1}(t)}=\Delta^{\sigma}T_{1}(u_{2},v_{2}) (t)= \Delta^{\sigma}u_{2}(t), \\ v_{1}(t) =T_{2}(u_{1},v_{1}) (t) \\ \hphantom{v_{1}(t)}= \int_{0}^{1}K_{2}(t,s)f_{2} \bigl(s,u_{1}(s),v_{1}(s) \bigr)\,ds+ \int _{0}^{1}H_{2}(t,s)f_{1} \bigl(s,u_{1}(s),v_{1}(s) \bigr)\,ds \\ \hphantom{v_{1}(t)}\geq \int_{0}^{1}K_{2}(t,s) \Delta^{\sigma}f_{2} \bigl(s,u_{2}(s),v_{2}(s) \bigr)\,ds+ \int_{0}^{1}H_{2}(t,s) \Delta^{\sigma }f_{1} \bigl(s,u_{2}(s),v_{2}(s) \bigr)\,ds \\ \hphantom{v_{1}(t)}=\Delta^{\sigma}T_{2}(u_{2},v_{2}) (t)= \Delta^{\sigma}v_{2}(t). \end{gathered} $$

(4.13)

Similarly, from (4.12), for $t\in[0,1]$, we have

$$ \begin{gathered} u_{2}(t) =T_{1}(u_{2},v_{2}) (t) \\ \hphantom{ u_{2}(t)}= \int_{0}^{1}K_{1}(t,s)f_{1} \bigl(s,u_{2}(s),v_{2}(s) \bigr)\,ds+ \int _{0}^{1}H_{1}(t,s)f_{2} \bigl(s,u_{2}(s),v_{2}(s) \bigr)\,ds \\ \hphantom{ u_{2}(t)}\geq \int_{0}^{1}K_{1}(t,s) \Delta^{\sigma}f_{1} \bigl(s,u_{1}(s),v_{1}(s) \bigr)\,ds+ \int_{0}^{1}H_{1}(t,s) \Delta^{\sigma }f_{2} \bigl(s,u_{1}(s),v_{1}(s) \bigr)\,ds \\ \hphantom{ u_{2}(t)}=\Delta^{\sigma}T_{1}(u_{1},v_{1}) (t)= \Delta^{\sigma}u_{1}(t), \\ v_{2}(t) =T_{2}(u_{2},v_{2}) (t) \\ \hphantom{v_{2}(t)}= \int_{0}^{1}K_{2}(t,s)f_{2} \bigl(s,u_{2}(s),v_{2}(s) \bigr)\,ds+ \int _{0}^{1}H_{2}(t,s)f_{1} \bigl(s,u_{2}(s),v_{2}(s) \bigr)\,ds \\ \hphantom{v_{2}(t)}\geq \int_{0}^{1}K_{2}(t,s) \Delta^{\sigma}f_{2} \bigl(s,u_{1}(s),v_{1}(s) \bigr)\,ds+ \int_{0}^{1}H_{2}(t,s) \Delta^{\sigma }f_{1} \bigl(s,u_{1}(s),v_{1}(s) \bigr)\,ds \\ \hphantom{v_{2}(t)}=\Delta^{\sigma}T_{2}(u_{1},v_{1}) (t)= \Delta^{\sigma}v_{1}(t). \end{gathered} $$

(4.14)

Combining (4.13) and (4.14), we can obtain

$$\Delta^{\sigma}u_{2}(t)\leq u_{1}(t)\leq \bigl( \Delta^{\sigma } \bigr)^{-1}u_{2}(t),\qquad \Delta^{\sigma}v_{2}(t)\leq v_{1}(t)\leq \bigl( \Delta^{\sigma} \bigr)^{-1}v_{2}(t),\quad t\in[0,1]. $$

Noticing that $0<\Delta, \sigma<1$, we get to a contradiction with the maximality of Δ. Thus, the boundary value problem (1.1)-(1.2) has a unique positive solution $(u^{*},v^{*})$. This completes the proof of Theorem 4.3. □

Example 4.1

Consider the following fractional boundary value problem:

$$ \textstyle\begin{cases} (D_{0+}^{2.5}u )(t)+\frac{\sqrt{u(t)}+1}{(2-t)(\sqrt [3]{v(t)}+1)}=0,\\ (D_{0+}^{2.5}v )(t)+\frac{\sqrt[3]{v(t)}+1}{(2-t)(\sqrt {u(t)}+1)}=0,\\ u(0)=u'(0)=0,\qquad u(1)=\sum_{i=1}^{2}\int_{\xi_{i}}^{\eta _{i}}v(s)\,dA_{i}(s)+\sum_{i=1}^{2}b_{i}v(\sigma_{i}),\\ v(0)=v'(0)=0,\qquad v(1)=\sum_{i=1}^{2}\int_{\xi_{i}}^{\eta_{i}}u(s)\,dA_{i}(s)+\sum_{i=1}^{2}b_{i}u(\sigma_{i}), \end{cases} $$

(4.15)

where

$$\begin{aligned} &A_{1}(t)=0.5t,\qquad \xi_{1}=\frac{1}{8},\qquad \eta_{1}=\frac{3}{8}, \qquad b_{1}= \frac{1}{3},\qquad \sigma_{1}=\frac{1}{3}, \\ &A_{2}(t)=0.8t,\qquad \xi_{2}=\frac{5}{8},\qquad \eta_{2}=\frac{7}{8},\qquad b_{2}=\frac{2}{3}, \qquad \sigma_{2}=\frac{2}{3}. \end{aligned}$$

Obviously, we have $\alpha_{1},\alpha_{2}=2.5$. We have

$$f_{1}(t,u,v)=\frac{\sqrt{u(t)}+1}{(2-t)(\sqrt[3]{v(t)}+1)},\qquad f_{2}(t,u,v)= \frac{\sqrt[3]{v(t)}+1}{(2-t)(\sqrt{u(t)}+1)}. $$

It is easy to see that $f_{1}:[0,1]\times[0,\infty)\times[0,\infty)$ is continuous, $f_{1}(t,u,v)$ is nondecreasing in u and nonincreasing in $v,f_{2}:[0,1]\times[0,\infty)\times[0,\infty)$ is continuous, $f_{2}(t,u,v)$ is nonincreasing in u and nondecreasing in v. Take

$$\theta_{1}=\frac{11}{20},\qquad\vartheta_{1}= \frac{2}{5},\qquad \theta_{2}=\frac{3}{5},\qquad \vartheta_{2}=\frac{1}{5}. $$

Then we know that condition (H₃) and (H₄) holds. As

$$\begin{aligned}& \begin{aligned} \int_{0}^{1}\varphi_{1}(s)f_{1} \bigl(s,1,s^{\alpha_{2}-1} \bigr)\,ds&= \int_{0}^{1}\frac{2(1-s)^{1.5}s}{(2-s)(s^{0.5}+1)}\,ds \\ &\leq \int_{0}^{1}\frac{2}{2-s}\,ds= 2\ln2 \approx1.3863< +\infty, \end{aligned} \\& \begin{aligned} \int_{0}^{1}\varphi_{2}(s)f_{2} \bigl(s,s^{\alpha_{1}-1},1 \bigr)\,ds&= \int_{0}^{1}\frac{2(1-s)^{1.5}s}{(2-s)(s^{0.75}+1)}\,ds \\ &\leq \int_{0}^{1}\frac{2}{2-s}\,ds= 2\ln2 \approx1.3863< +\infty. \end{aligned} \end{aligned}$$

The condition (H₅) is also satisfied. Therefore, by Theorem 4.2, we see that the coupled boundary value problem (4.15) has at least one positive solution $(u^{*},v^{*} )$. Furthermore,

$$\theta_{1}+\vartheta_{1}=\frac{19}{20}< 1,\qquad \theta_{2}+\vartheta_{2}=\frac{4}{5}< 1. $$

By Theorem 4.3, we see that $(u^{*},v^{*} )$ is the unique positive solution of the coupled boundary value problem (4.15).