Existence of solutions for a coupled system of fractional differential equations by means of topological degree theory

Abstract

This paper investigates the existence of solutions for a coupled system of fractional differential equations. The existence is proved by using the topological degree theory, and an example is given to show the applicability of our main result.

1 Introduction

In this manuscript, the following coupled system of fractional differential equations is discussed:

$$ \textstyle\begin{cases} D^{\theta }[x(t)-f(t,x(t))]=h(t,y(t),I^{\alpha }y(t)),\quad t\in [0,1], \\ D^{\theta }[y(t)-f(t,y(t))]=h(t,x(t),I^{\alpha }x(t)),\quad t\in [0,1], \\ x'(0)=y'(0)=0, \\ a_{1}x(0)-b_{1}x(\eta )-c_{1}x(1)=\frac{1}{\Gamma (\theta )}\int _{0}^{1} (1-s)^{\theta -1}\phi (s,x(s))\,ds, \\ a_{2}y(0)-b_{2}y(\xi )-c_{2}y(1)=\frac{1}{\Gamma (\theta )}\int _{0}^{1} (1-s)^{\theta -1}\psi (s,y(s))\,ds, \end{cases} $$

(1.1)

where $1<\theta \leq 2$, $\alpha >0$, $\eta ,\xi \in (0,1]$, $a_{j}$, $b_{j}$, $c_{j}$ ($j=1,2$) are real numbers with $a_{j}\neq b_{j}+ c_{j}$ ($j=1,2$). Further $f,\phi ,\psi : [0,1]\times \mathbb{R}\rightarrow [0,1]$, and $h:[0,1]\times \mathbb{R}\times \mathbb{R}\rightarrow [0,1]$ are continuous functions, $f(0,x(0))=0$, $\frac{\partial ^{i}f(t,x(t))}{\partial t^{i}}|_{t=0}=0$ for $i=1,2, \ldots , n-1$. $D^{\theta }$ represents the Caputo fractional derivative of order θ.

Fractional differential equations are widely used in many fields such as chemistry, physics, biology, and optimization theory [1–4]. In addition, coupled systems of fractional differential equations have attracted particular concern from scholars considering their appearance in the mathematical modeling of physical phenomena like chaos synchronization [5], anomalous diffusion [6], disease models [7], and so on. The existence theory to fractional differential equations with integral boundary conditions has widespread applications in optimization theory, many researchers have studied [8–13], and the existence of solutions is the basis of studying the stability and numerical solutions of differential equations [14]. For the existence of solutions of fractional differential equations, the authors use diverse methods, such as fixed point theory [15–19], upper and lower solutions method [20], monotone iterative technique and Mawhin’s continuation theorem [21], and topological degree theory [22]. When studying the existing literature, we find that fractional differential equations with integral boundary conditions are not properly tested via topological degree theory. Thus we investigate the existence result to a coupled system of fractional differential equations(1.1) through applying topological degree theory.

Bashiri et al. [23]investigated the existence of solutions for fractional differential equations by means of the coupled fixed point theorem of Krasnoselskii type

$$ \textstyle\begin{cases} D^{\theta }[x(t)-f(t,x(t))]=h(t,y(t),I^{\alpha }y(t)), \\ D^{\theta }[y(t)-f(t,y(t))]=h(t,x(t),I^{\alpha }x(t)), \\ x(0)=y(0)=0, \end{cases} $$

where $D^{\theta }$ denotes the Riemann–Liouville fractional derivative, $\theta \in (0,1)$, $\alpha >0$.

Ahmad et al. [24]established existence results as well as studied qualitative aspects of the proposed coupled system of fractional hybrid delay differential equations

$$ \textstyle\begin{cases} ^{C}D_{+0}^{\kappa }(r(t)-P_{1}(t,r(t),h(t)))=Q_{1}(t,r(\nu t),h(\nu t)),&t \in \mathcal{A}, \\ ^{C}D_{+0}^{\sigma }(h(t)-P_{2}(t,r(t),h(t)))=Q_{2}(t,r(\nu t),h(\nu t)),&t \in \mathcal{A}, \\ r(t)| _{t=0}=r_{0},\qquad h(t)| _{t=0}=h_{0}, \end{cases} $$

where $\mathcal{A}=[0,\tau ]$, $^{C}D_{+0}, \tau >0$ is Caputo’s derivative, and $r_{0}$, $h_{0}$ are real numbers, while the delay parameter is denoted by $\nu \in (0,1)$.

Muthaiah et al. [25]considered the existence and Hyers–Ulam type stability results for the nonlinear coupled system of Caputo–Hadamard type fractional differential equations

$$ \textstyle\begin{cases} ^{C}D^{\varrho }y(\tau )=f(\tau , y(\tau ),z(\tau )),\quad \tau \in [1,T]:= \mathcal{K}, \\ ^{C}D^{\varsigma }z(\tau )=g(\tau , y(\tau ),z(\tau )),\quad \tau \in [1,T]:= \mathcal{K}, \\ y(1)=0, y'(1)=0, y(T)=\alpha _{1}\sum _{j=1}^{k-2}\xi _{j}z( \zeta _{j})+\beta _{1}{^{H}I^{\varsigma _{1}}z(\vartheta )}, \\ z(1)=0, z'(1)=0, z(T)=\alpha _{2}\sum _{j=1}^{k-2}\upsilon _{j}z( \omega _{j})+\beta _{2}{^{H}I^{\varrho _{1}}y(\varphi )}, \\ 1< \vartheta < \varphi < \xi _{1}< \omega _{1}< \xi _{2}< \omega _{2}< \cdots < \xi _{k-2}< \omega _{k-2}< T, \end{cases} $$

where $^{C}D^{(\cdot )}$ denotes the Caputo–Hadamard fractional derivative, $^{H}I^{(\cdot )}$ denotes the Hadamard fractional integrals, $2<\varrho ,\varsigma \leq 3$, $0<\varrho _{1}, \varsigma _{1}<1$, $\alpha _{1}$, $\alpha _{2}$, $\beta _{1}$, $\beta _{2}$ are real constants and $\zeta _{j}$, $\upsilon _{j}$, $j=1,2,\ldots ,k-2$, are positive real constants. The consequence of existence is obtained by employing the alternative of Leray–Schauder and Krasnoselskii’s, whereas the uniqueness result is based on the principle of Banach contraction mapping.

Motivated especially by the aforementioned work, we consider the existence of solutions to a coupled system of fractional differential equations (1.1). According to our literature review, no scholars have studied equation (1.1), the results are entirely new. The remainder of this paper is as follows. In the second part, we display some definitions, facts, and results. We confirm the existence of solutions for system (1.1) in the third part. Finally, we provide an example to prove our results.

2 Preliminaries

In this part, we recollect a number of facts, definitions, and conclusions. Let $C([0,1]\times \mathbb{R}, [0,1])$ represent the space of all continuous functions $f,\phi ,\psi : [0,1]\times \mathbb{R}\rightarrow [0,1]$, and let $C([0,1]\times \mathbb{R}\times \mathbb{R}, [0,1])$ express the class of functions $h: [0,1]\times \mathbb{R}\times \mathbb{R}\rightarrow [0,1]$ such that

• (1) the map $t\rightarrow h(t,x,y)$ is measurable for each $x,y\in \mathbb{R}$,

• (2) the map $x\rightarrow h(t,x,y)$ is continuous for each $x\in \mathbb{R}]$,

• (3) the map $y\rightarrow h(t,x,y)$ is continuous for each $y\in \mathbb{R}$.

Let X be a Banach space and $\mathbb{B}\subset P(X)$, where $P(X)$ stands for the family of all bounded subsets of X. Next, we introduce some concepts.

Definition 2.1

([26])

The Kuratowski measure of noncompactness $\alpha : \mathbb{B}\rightarrow \mathbb{R_{+}}$ is defined as

$$\begin{aligned} \alpha (B)=\inf \{d>0, \text{where }B\in \mathbb{B}\text{ admits a finite cover by set of diameter }\leq d\}. \end{aligned}$$

Definition 2.2

([26])

Let $\mathcal{F}: \Omega \rightarrow X$ be a continuous bounded map, where $\Omega \subseteq X$. Then $\mathcal{F}$ is

$(1)$ α-Lipschitz if there exists $k\geq 0$, therefore $\alpha (\mathcal{F}(S))\leq k\alpha (S)$ for all bounded subsets $S\subseteq \Omega$;

$(2)$ strict α-contraction if there exists $0\leq k<1$ such that $\alpha (\mathcal{F}(S))\leq k\alpha (S)$ for all bounded subsets $S\subseteq \Omega$;

$(3)$ α-condensing if $\alpha (\mathcal{F}(S))< \alpha (S)$ for all bounded subsets $S\subseteq \Omega$ with $\alpha (S)>0$. In other words, $\alpha (\mathcal{F}(S))\geq \alpha (S)$ implies $\alpha (S)=0$.

All classes of strict α-contraction $\mathcal{F}: \Omega \rightarrow X$ and all classes of α-condensing maps $\mathcal{F}: \Omega \rightarrow X$ are represented by $\Lambda C_{\alpha }(\Omega )$ and $C_{\alpha }(\Omega )$, respectively. Then $\Lambda C_{\alpha }(\Omega )\subset C_{\alpha }(\Omega )$ and each $\mathcal{F}\in C_{\alpha }(\Omega )$ is α-Lipschitz with constant $k=1$. Moreover, $\mathcal{F}: \Omega \rightarrow X$ is Lipschitz whenever there is $k>0$, therefore

$$\bigl\Vert \mathcal{F}(x)-\mathcal{F}(y) \bigr\Vert \leq k \Vert x-y \Vert \quad \text{for all }x, y\in \Omega . $$

Further, $\mathcal{F}$ will be a strict contraction if $k<1$.

Proposition 2.3

([27])

If $\mathcal{F}, \mathcal{G} : \Omega \rightarrow X$ are α-Lipschitz with respective constants $k_{1}$ and $k_{2}$, then $\mathcal{F}+\mathcal{G}$ is α-Lipschitz with constant $k_{1}+k_{2}$.

Proposition 2.4

([27])

If $\mathcal{F }: \Omega \rightarrow X$ is Lipschitz with constant k, then $\mathcal{F}$ is α-Lipschitz with the equal constant k.

Proposition 2.5

([27])

If $\mathcal{F} : \Omega \rightarrow X$ is compact, then $\mathcal{F}$ is α-Lipschitz with constant $k=0$.

Theorem 2.6

([27])

If $\mathcal{F} : X\rightarrow X$ is α-condensing and

$$\begin{aligned} \Lambda =\{x\in X :\textit{there exists } 0\leq \nu \leq 1 \textit{ such that }x= \nu \mathcal{F}x\}. \end{aligned}$$

If Λ is a bounded set in X, so we have $r>0$ such that $\Lambda \subset B_{r}(0)$, then

$$\begin{aligned} D\bigl(I-\nu \mathcal{F}, B_{r}(0), 0\bigr))=1\quad \textit{for all }\nu \in [0,1]. \end{aligned}$$

Consequently, $\mathcal{F}$ has at least one fixed point, and the set of the fixed points of $\mathcal{F}$ lies in $B_{r}(0)$.

Definition 2.7

([28])

The fractional integral of order $\theta (\theta >0)$ of function $f: [0,\infty )\rightarrow R$ is defined as

$$\begin{aligned} I^{\theta }f(t)=\frac{1}{\Gamma (\theta )} \int _{0}^{t}(t-s)^{\theta -1}f(s)\,ds, \end{aligned}$$

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

Definition 2.8

([28])

The Caputo fractional derivative of order $\theta (\theta >0)$ of the function $f:[0,\infty )\rightarrow R$ is defined by

$$\begin{aligned} D^{\theta }f(t)=\frac{1}{\Gamma (n-\theta )} \int _{0}^{t}(t-s)^{n- \theta -1}f^{(n)}(s) \,ds, \end{aligned}$$

where $t>0$, $n=[\theta ]+1$.

Lemma 2.9

([28])

Let $\theta >0$, then the following result holds for fractional differential equations:

$$\begin{aligned} I^{\theta }\bigl[D^{\theta }f(t)\bigr]=f(t)+C_{0}+C_{1}t+C_{2}t^{2}+ \cdots +C_{n-1}t^{n-1} \end{aligned}$$

for arbitrary $n=[\theta ]+1$, $[\theta ]$ indicates the integer part of the real number $\theta >0$, $C_{i}\in \mathbb{R}$, $i=0,1,2,\ldots ,n-1$. $D^{\theta }$ is a Caputo fractional derivative.

3 Main results

In this part, we discuss the existence result for (1.1).

The space $X=C([0,1], \mathbb{R})$ of all continuous functions is a Banach space under the topological norm $\| x\| =\sup \{| x(t)| : t\in [0,1]\}$ and the product space $X\times X$ is a Banach space under the norm $\| (x,y)\| =\| x\| +\| y \|$ or $\| (x,y)\| =\max \{\| x\| ,\| y \| \}$.

In order to get the result of our result, we need the following hypotheses.

$(H_{1})$ For each $(t,x), (t,\bar{x}), (t,y), (t,\bar{y})\in [0,1]\times \mathbb{R}$, there exist constants $\lambda _{1}, \lambda _{2}\in [0,1)$ such that

$$\begin{aligned} \bigl\vert f(t,x)-f(t,\bar{x}) \bigr\vert \leq \lambda _{1} \Vert x-\bar{x} \Vert , \\ \bigl\vert f(t,y)-f(t,\bar{y}) \bigr\vert \leq \lambda _{2} \Vert y-\bar{y} \Vert . \end{aligned}$$

$(H_{2})$ For each $(t,x,y)\in \mathbb{R}$, there exist positive constants $l_{h}^{1}$, $l_{h}^{2}$, $M_{h}$ and $q_{1}\in [0,1)$ such that

$$\begin{aligned} \bigl\vert h(t,x,y) \bigr\vert \leq l_{h}^{1} \Vert x \Vert ^{q_{1}}+l_{h}^{2} \Vert y \Vert ^{q_{1}}+M_{h}. \end{aligned}$$

$(H_{3})$ For each $(t,x)\in [0,1]\times \mathbb{R}$, there exist positive constants $l_{f}$, $M_{f}$ and $q_{2}\in [0,1)$ such that

$$\begin{aligned} \bigl\vert f\bigl(t,x(t)\bigr) \bigr\vert \leq l_{f} \Vert x \Vert ^{q_{2}}+M_{f}. \end{aligned}$$

$(H_{4})$ For each $(t,x), (t,y)\in [0,1]\times \mathbb{R}$, there exist positive constants $c_{\phi }$, $c_{\psi }$, $M_{\phi }$, $M_{\psi }$, and $q_{2}\in [0,1)$ such that

$$\begin{aligned} \bigl\vert \phi (t,x) \bigr\vert \leq c_{\phi } \Vert x \Vert ^{q_{2}}+M_{ \phi }, \\ \bigl\vert \psi (t,y) \bigr\vert \leq c_{\psi } \Vert y \Vert ^{q_{2}}+M_{ \psi }. \end{aligned}$$

$(H_{5})$ For each $(t,x), (t,\bar{x}), (t,y), (t,\bar{y})\in [0,1]\times \mathbb{R}$, we have positive constants $b_{\phi }, b_{\psi }\in [0,1)$ such that

$$\begin{aligned} \bigl\vert \phi (t,x)-\phi (t,\bar{x}) \bigr\vert \leq b_{\phi } \Vert x-\bar{x} \Vert , \\ \bigl\vert \psi (t,y)-\psi (t,\bar{y}) \bigr\vert \leq b_{\psi } \Vert y-\bar{y} \Vert . \end{aligned}$$

Lemma 3.1

If $f(0,x(0))=0$, $\frac{\partial ^{i}f(t,x(t))}{\partial t^{i}}|_{t=0}=0$ for $i=1,2,\ldots , n-1$, then the consequence of fractional differential equations (1.1) is a conclusion of the following system of integral equations:

$$ \textstyle\begin{cases} x(t)=f(t,x(t))+\frac{1}{\Gamma (\theta )}\int _{0}^{t}(t-s)^{\theta -1}h(s,y(s),I^{ \alpha }y(s))\,ds \\ \hphantom{x(t)=}{} +\frac{1}{a_{1}-(b_{1}+c_{1})}\frac{1}{\Gamma (\theta )}\int _{0}^{1}(1-s)^{ \theta -1}\phi (s,x(s))\,ds \\ \hphantom{x(t)=}{} +\frac{b_{1}}{a_{1}-(b_{1}+c_{1})}f(\eta ,x(\eta ))+ \frac{c_{1}}{a_{1}-(b_{1}+c_{1})}f(1,x(1)) \\ \hphantom{x(t)=}{} +\frac{b_{1}}{a_{1}-(b_{1}+c_{1})}\frac{1}{\Gamma (\theta )} \int _{0}^{\eta }(\eta -s)^{\theta -1}h(s,y(s),I^{\alpha }y(s))\,ds \\ \hphantom{x(t)=}{} +\frac{c_{1}}{a_{1}-(b_{1}+c_{1})}\frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{\theta -1}h(s,y(s),I^{\alpha }y(s))\,ds, \\ y(t)=f(t,y(t))+\frac{1}{\Gamma (\theta )}\int _{0}^{t}(t-s)^{\theta -1}h(s,x(s),I^{ \alpha }x(s))\,ds \\ \hphantom{y(t)=}{} +\frac{1}{a_{2}-(b_{2}+c_{2})}\frac{1}{\Gamma (\theta )}\int _{0}^{1}(1-s)^{ \theta -1}\phi (s,y(s))\,ds \\ \hphantom{y(t)=}{} +\frac{b_{2}}{a_{2}-(b_{2}+c_{2})}f(\xi ,y(\xi ))+ \frac{c_{2}}{a_{2}-(b_{2}+c_{2})}f(1,y(1)) \\ \hphantom{y(t)=}{} +\frac{b_{2}}{a_{2}-(b_{2}+c_{2})}\frac{1}{\Gamma (\theta )} \int _{0}^{\xi }(\xi -s)^{\theta -1}h(s,x(s),I^{\alpha }x(s))\,ds \\ \hphantom{y(t)=}{} +\frac{c_{2}}{a_{2}-(b_{2}+c_{2})}\frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{\theta -1}h(s,x(s),I^{\alpha }x(s))\,ds. \end{cases} $$

(3.1)

Proof

Applying the fractional integrable operator $I^{\theta }$ on the equation of system (1.1) and through applying Lemma 2.9, we get

$$\begin{aligned} x(t)=f\bigl(t,x(t)\bigr)+\frac{1}{\Gamma (\theta )} \int _{0}^{t}(t-s)^{\theta -1}h \bigl(s,y(s),I^{ \alpha }y(s)\bigr)\,ds +C_{0}+C_{1}t. \end{aligned}$$

By applying the initial conditions $x'(0)=0$ and $\frac{\partial ^{i}f(t,x(t))}{\partial t^{i}}|_{t=0}=0$, we obtain $C_{1}=0$ and

$$\begin{aligned} x(t)=f\bigl(t,x(t)\bigr)+\frac{1}{\Gamma (\theta )} \int _{0}^{t}(t-s)^{\theta -1}h \bigl(s,y(s),I^{ \alpha }y(s)\bigr)\,ds +C_{0}. \end{aligned}$$

(3.2)

Now, applying the boundary conditions $a_{1}x(0)-b_{1}x(\eta )-c_{1}x(1)=\frac{1}{\Gamma (\theta )}\int _{0}^{1} (1-s)^{\theta -1}\phi (s, x(s))\,ds$ to (3.2), we have

$$\begin{aligned} (a_{1}-b_{1}-c_{1})C_{0} =&b_{1}f \bigl(\eta ,x(\eta )\bigr)+c_{1}f\bigl(1,x(1)\bigr)+ \frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{\theta -1}\phi \bigl(s,x(s)\bigr) \\ &{} +\frac{b_{1}}{\Gamma (\theta )} \int _{0}^{\eta }(\eta -s)^{ \theta -1}h \bigl(s,y(s),I^{\alpha }y(s)\bigr)\,ds \\ &{} +\frac{c_{1}}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{\theta -1}h \bigl(s,y(s),I^{ \alpha }y(s)\bigr)\,ds. \end{aligned}$$

By rearranging, we obtain

$$\begin{aligned} C_{0} =&\frac{b_{1}}{a_{1}-(b_{1}+c_{1})}f\bigl(\eta ,x(\eta )\bigr)+ \frac{c_{1}}{a_{1}-(b_{1}+c_{1})}f\bigl(1,x(1)\bigr) \\ &{} +\frac{1}{a_{1}-(b_{1}+c_{1})}\frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{ \theta -1}\phi \bigl(s,x(s)\bigr) \\ &{} +\frac{b_{1}}{a_{1}-(b_{1}+c_{1})}\frac{1}{\Gamma (\theta )} \int _{0}^{\eta }(\eta -s)^{\theta -1}h \bigl(s,y(s),I^{\alpha }y(s)\bigr)\,ds \\ &{} +\frac{c_{1}}{a_{1}-(b_{1}+c_{1})}\frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{\theta -1}h \bigl(s,y(s),I^{\alpha }y(s)\bigr)\,ds. \end{aligned}$$

Thus equation (3.2) becomes

$$\begin{aligned} x(t) =&f\bigl(t,x(t)\bigr)+\frac{1}{\Gamma (\theta )} \int _{0}^{t}(t-s)^{\theta -1}h \bigl(s,y(s),I^{ \alpha }y(s)\bigr)\,ds \\ &{} +\frac{1}{a_{1}-(b_{1}+c_{1})}\frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{ \theta -1}\phi \bigl(s,x(s)\bigr)\,ds \\ &{} +\frac{b_{1}}{a_{1}-(b_{1}+c_{1})}f\bigl(\eta ,x(\eta )\bigr)+ \frac{c_{1}}{a_{1}-(b_{1}+c_{1})}f \bigl(1,x(1)\bigr) \\ &{} +\frac{b_{1}}{a_{1}-(b_{1}+c_{1})}\frac{1}{\Gamma (\theta )} \int _{0}^{\eta }(\eta -s)^{\theta -1}h \bigl(s,y(s),I^{\alpha }y(s)\bigr)\,ds \\ &{} +\frac{c_{1}}{a_{1}-(b_{1}+c_{1})}\frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{\theta -1}h \bigl(s,y(s),I^{\alpha }y(s)\bigr)\,ds. \end{aligned}$$

Analogously, following the same steps in the process for the second equation of system (1.1), we get

$$\begin{aligned} y(t) =&f\bigl(t,y(t)\bigr)+\frac{1}{\Gamma (\theta )} \int _{0}^{t}(t-s)^{\theta -1}h \bigl(s,x(s),I^{ \alpha }x(s)\bigr)\,ds \\ &{} +\frac{1}{a_{2}-(b_{2}+c_{2})}\frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{ \theta -1}\psi \bigl(s,y(s)\bigr)\,ds \\ &{} +\frac{b_{2}}{a_{2}-(b_{2}+c_{2})}f\bigl(\xi ,y(\xi )\bigr)+ \frac{c_{2}}{a_{2}-(b_{2}+c_{2})}f \bigl(1,y(1)\bigr) \\ &{} +\frac{b_{2}}{a_{2}-(b_{2}+c_{2})}\frac{1}{\Gamma (\theta )} \int _{0}^{\xi }(\xi -s)^{\theta -1}h \bigl(s,x(s),I^{\alpha }x(s)\bigr)\,ds \\ &{} +\frac{c_{2}}{a_{2}-(b_{2}+c_{2})}\frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{\theta -1}h \bigl(s,x(s),I^{\alpha }x(s)\bigr)\,ds. \end{aligned}$$

 □

Define the operator $F, H, T : X\times X\rightarrow X\times X$ by

$$\begin{aligned}& F(x,y) (t)=\bigl(F_{1}x(t), F_{2}y(t)\bigr), \\& H(x,y) (t)=\bigl(H_{2}y(t), H_{1}x(t)\bigr), \\& T(x,y) (t)=F(x,y) (t)+H(x,y) (t), \end{aligned}$$

here $F_{1}, F_{2}, H_{1}, H_{2}: X\rightarrow X$ are

$$\begin{aligned}& \begin{aligned} F_{1}x(t)={}&f\bigl(t,x(t)\bigr)+\frac{b_{1}}{a_{1}-(b_{1}+c_{1})}f\bigl(\eta ,x( \eta )\bigr)+ \frac{c_{1}}{a_{1}-(b_{1}+c_{1})}f\bigl(1,x(1)\bigr) \\ &{} +\frac{1}{a_{1}-(b_{1}+c_{1})}\frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{ \theta -1}\phi \bigl(s,x(s)\bigr)\,ds, \end{aligned}\\& \begin{aligned} F_{2}y(t)={}&f\bigl(t,y(t)\bigr)+\frac{b_{2}}{a_{2}-(b_{2}+c_{2})}f\bigl(\xi ,y( \xi )\bigr)+ \frac{c_{2}}{a_{2}-(b_{2}+c_{2})}f\bigl(1,y(1)\bigr) \\ &{} +\frac{1}{a_{2}-(b_{2}+c_{2})}\frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{ \theta -1}\psi \bigl(s,y(s)\bigr)\,ds, \end{aligned}\\& \begin{aligned} H_{1}x(t)={}&\frac{1}{\Gamma (\theta )} \int _{0}^{t}(t-s)^{\theta -1}h \bigl(s,x(s),I^{ \alpha }x(s)\bigr)\,ds \\ &{} +\frac{b_{2}}{a_{2}-(b_{2}+c_{2})}\frac{1}{\Gamma (\theta )} \int _{0}^{\xi }(\xi -s)^{\theta -1}h \bigl(s,x(s),I^{\alpha }x(s)\bigr)\,ds \\ &{} +\frac{c_{2}}{a_{2}-(b_{2}+c_{2})}\frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{\theta -1}h \bigl(s,x(s),I^{\alpha }x(s)\bigr)\,ds, \end{aligned}\\& \begin{aligned} H_{2}y(t)={}&\frac{1}{\Gamma (\theta )} \int _{0}^{t}(t-s)^{\theta -1}h \bigl(s,y(s),I^{ \alpha }y(s)\bigr)\,ds \\ &{} +\frac{b_{1}}{a_{1}-(b_{1}+c_{1})}\frac{1}{\Gamma (\theta )} \int _{0}^{\eta }(\eta -s)^{\theta -1}h \bigl(s,y(s),I^{\alpha }y(s)\bigr)\,ds \\ &{} +\frac{c_{1}}{a_{1}-(b_{1}+c_{1})}\frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{\theta -1}h \bigl(s,y(s),I^{\alpha }y(s)\bigr)\,ds. \end{aligned} \end{aligned}$$

Then the system of integral equations (3.1) can be written as an operator equation

$$\begin{aligned} (x,y)=T(x,y)=F(x,y)+H(x,y), \end{aligned}$$

and fixed points of the operator equation are results of system (1.1).

Theorem 3.2

The operator F is Lipschitz with constant k. Therefore F is α-Lipschitz with the equal constant k and meets the following growth condition:

$$\begin{aligned} \bigl\Vert F\bigl(x(t),y(t)\bigr) \bigr\Vert \leq L_{F} \bigl\Vert (x,y) \bigr\Vert ^{q_{2}}+M_{F}. \end{aligned}$$

Proof

Now, we shall display that the operator F is Lipschitz with constant k. Let $x_{1}, x_{2}\in X$, then we get

$$\begin{aligned} &\bigl\vert F_{1}x_{1}(t)-F_{1}x_{2}(t) \bigr\vert \\ &\quad = \biggl\vert \bigl(f\bigl(t,x_{1}(t)\bigr)-f \bigl(t,x_{2}(t)\bigr)\bigr)+ \frac{b_{1}}{a_{1}-(b_{1}+c_{1})}\bigl(f\bigl(\eta ,x_{1}(\eta )\bigr)-f\bigl(\eta ,x_{2}( \eta )\bigr) \bigr) \\ &\qquad{} +\frac{c_{1}}{a_{1}-(b_{1}+c_{1})}\bigl(f\bigl(1,x_{1}(1)\bigr)-f \bigl(1,x_{2}(1)\bigr)\bigr) \\ &\qquad{} +\frac{1}{a_{1}-(b_{1}+c_{1})}\frac{1}{\Gamma (\theta )}\biggl( \int _{0}^{1}(1-s)^{ \theta -1}\phi \bigl(s,x_{1}(s)\bigr)\,ds- \int _{0}^{1}(1-s)^{\theta -1}\phi \bigl(s,x_{2}(s)\bigr)\,ds\biggr) \biggr\vert \\ &\quad \leq \bigl\vert f\bigl(t,x_{1}(t)\bigr)-f \bigl(t,x_{2}(t)\bigr) \bigr\vert \\ &\qquad{} +\frac{ \vert b_{1} \vert }{ \vert a_{1}-(b_{1}+c_{1}) \vert } \bigl\vert f\bigl( \eta ,x_{1}(\eta )\bigr)-f\bigl(\eta ,x_{2}(\eta )\bigr) \bigr\vert \\ &\qquad{} +\frac{ \vert c_{1} \vert }{ \vert a_{1}-(b_{1}+c_{1}) \vert } \bigl\vert f\bigl(1,x_{1}(1)\bigr)-f \bigl(1,x_{2}(1)\bigr) \bigr\vert \\ &\qquad{} +\frac{1}{ \vert a_{1}-(b_{1}+c_{1}) \vert \Gamma (\theta )} \int _{0}^{1}(1-s)^{ \theta -1} \bigl\vert \phi \bigl(s,x_{1}(s)\bigr)- \phi \bigl(s,x_{2}(s)\bigr) \bigr\vert \,ds. \end{aligned}$$

By using conditions ($H_{1}$) and ($H_{5}$), we can write

$$\begin{aligned} \bigl\Vert F_{1}x_{1}(t)-F_{1}x_{2}(t) \bigr\Vert & \leq \lambda _{1} \Vert x_{1}-x_{2} \Vert + \frac{ \vert b_{1} \vert \lambda _{1}}{ \vert a_{1}-(b_{1}+c_{1}) \vert } \Vert x_{1}-x_{2} \Vert \\ &\quad + \frac{ \vert c_{1} \vert \lambda _{1}}{ \vert a_{1}-(b_{1}+c_{1}) \vert } \Vert x_{1}-x_{2} \Vert + \frac{b_{\phi }}{ \vert a_{1}-(b_{1}+c_{1}) \vert \Gamma (\theta +1)} \Vert x_{1}-x_{2} \Vert \\ &= \biggl[\lambda _{1}+ \frac{( \vert b_{1} \vert + \vert c_{1} \vert ) \lambda _{1}}{ \vert a_{1}-(b_{1}+c_{1}) \vert }+ \frac{b_{\phi }}{ \vert a_{1}-(b_{1}+c_{1}) \vert \Gamma (\theta +1)} \biggr] \Vert x_{1}-x_{2} \Vert \\ &=k_{1} \Vert x_{1}-x_{2} \Vert , \end{aligned}$$

where $k_{1}=\lambda _{1}+ \frac{(| b_{1}| +| c_{1}| ) \lambda _{1}}{| a_{1}-(b_{1}+c_{1})| }+ \frac{b_{\phi }}{| a_{1}-(b_{1}+c_{1})| \Gamma (\theta +1)}$.

Similarly,

$$\begin{aligned} \bigl\Vert F_{2}y_{1}(t)-F_{2}y_{2}(t) \bigr\Vert & \leq \biggl[\lambda _{2}+ \frac{( \vert b_{2} \vert + \vert c_{2} \vert ) \lambda _{2}}{ \vert a_{2}-(b_{2}+c_{2}) \vert }+ \frac{b_{\psi }}{ \vert a_{2}-(b_{2}+c_{2}) \vert \Gamma (\theta +1)} \biggr] \Vert y_{1}-y_{2} \Vert \\ &=k_{2} \Vert y_{1}-y_{2} \Vert , \end{aligned}$$

where $k_{2}=\lambda _{2}+ \frac{(| b_{2}| +| c_{2}| ) \lambda _{2}}{| a_{2}-(b_{2}+c_{2})| }+ \frac{b_{\psi }}{| a_{2}-(b_{2}+c_{2})| \Gamma (\theta +1)}$. Thus

$$\begin{aligned} \bigl\Vert F(x_{1}, y_{1})-F(x_{2}, y_{2}) \bigr\Vert & = \bigl\Vert F_{1}x_{1}(t)-F_{1}x_{2}(t) \bigr\Vert + \bigl\Vert F_{2}y_{1}(t)-F_{2}y_{2}(t) \bigr\Vert \\ &\leq k_{1} \Vert x_{1}-x_{2} \Vert + k_{2} \Vert y_{1}-y_{2} \Vert \\ &\leq k \bigl\Vert (x_{1}, y_{1})-(x_{2}, y_{2}) \bigr\Vert , \end{aligned}$$

where $k=\max (\lambda _{1}+ \frac{(| b_{1}| +| c_{1}| ) \lambda _{1}}{| a_{1}-(b_{1}+c_{1})| }+ \frac{b_{\phi }}{| a_{1}-(b_{1}+c_{1})| \Gamma (\theta +1)}, \lambda _{2}+ \frac{(| b_{2}| +| c_{2}| ) \lambda _{2}}{| a_{2}-(b_{2}+c_{2})| }+ \frac{b_{\psi }}{| a_{2}-(b_{2}+c_{2})| \Gamma (\theta +1)} )$. Then F satisfies the Lipschitz condition, thus F is Lipschitz with constant k. According to Proposition 2.4, F is α-Lipschitz with constant k.

Moreover, we get

$$\begin{aligned} \bigl\vert F_{1}x(t) \bigr\vert \leq{}& \bigl\vert f\bigl(t,x(t) \bigr) \bigr\vert + \frac{ \vert b_{1} \vert }{ \vert a_{1}-(b_{1}+c_{1}) \vert } \bigl\vert f\bigl(\eta ,x( \eta ) \bigr) \bigr\vert +\frac{ \vert c_{1} \vert }{ \vert a_{1}-(b_{1}+c_{1}) \vert } \bigl\vert f\bigl(1,x(1)\bigr) \bigr\vert \\ &{} +\frac{1}{ \vert a_{1}-(b_{1}+c_{1}) \vert } \frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{\theta -1} \bigl\vert \phi \bigl(s,x(s)\bigr) \bigr\vert \,ds. \end{aligned}$$

By $(H_{3})$ and $(H_{4})$, we have

$$\begin{aligned} \bigl\vert F_{1}x(t) \bigr\vert \leq{}& l_{f} \Vert x \Vert ^{q_{2}}+M_{f}+ \frac{ \vert b_{1} \vert }{ \vert a_{1}-(b_{1}+c_{1}) \vert } \bigl(l_{f} \Vert x \Vert ^{q_{2}}+M_{f} \bigr)\\ &{} + \frac{ \vert c_{1} \vert }{ \vert a_{1}-(b_{1}+c_{1}) \vert }\bigl(l_{f} \Vert x \Vert ^{q_{2}}+M_{f}\bigr) \\ &{} +\frac{1}{ \vert a_{1}-(b_{1}+c_{1}) \vert } \frac{1}{\Gamma (\theta +1)}\bigl(c_{\phi } \Vert x \Vert ^{q_{2}}+M_{ \phi }\bigr) \\ ={}& \biggl[l_{f}+ \frac{( \vert b_{1} \vert + \vert c_{1} \vert )l_{f}}{ \vert a_{1}-(b_{1}+c_{1}) \vert }+ \frac{c_{\phi }}{ \vert a_{1}-(b_{1}+c_{1}) \vert \Gamma (\theta +1)} \biggr] \Vert x \Vert ^{q_{2}} \\ &{} + \biggl[M_{f}+ \frac{( \vert b_{1} \vert + \vert c_{1} \vert )M_{f}}{ \vert a_{1}-(b_{1}+c_{1}) \vert }+ \frac{M_{\phi }}{ \vert a_{1}-(b_{1}+c_{1}) \vert \Gamma (\theta +1)} \biggr]. \end{aligned}$$

Similarly,

$$\begin{aligned} \bigl\vert F_{2}y(t) \bigr\vert \leq{}& \biggl[l_{f}+ \frac{( \vert b_{2} \vert + \vert c_{2} \vert )l_{f}}{ \vert a_{2}-(b_{2}+c_{2}) \vert }+ \frac{c_{\psi }}{ \vert a_{2}-(b_{2}+c_{2}) \vert \Gamma (\theta +1)} \biggr] \Vert y \Vert ^{q_{2}} \\ &{} + \biggl[M_{f}+ \frac{( \vert b_{2} \vert + \vert c_{2} \vert )M_{f}}{ \vert a_{2}-(b_{2}+c_{2}) \vert }+ \frac{M_{\psi }}{ \vert a_{2}-(b_{2}+c_{2}) \vert \Gamma (\theta +1)} \biggr]. \end{aligned}$$

Hence it follows that

$$\begin{aligned} \bigl\Vert F\bigl(x(t),y(t)\bigr) \bigr\Vert ={}& \bigl\Vert \bigl(F_{1}(x), F_{2}(y)\bigr) \bigr\Vert \\ ={}& \bigl\Vert F_{1}(x) \bigr\Vert + \bigl\Vert F_{2}(y) \bigr\Vert \\ \leq{}& \biggl[l_{f}+ \frac{( \vert b_{1} \vert + \vert c_{1} \vert )l_{f}}{ \vert a_{1}-(b_{1}+c_{1}) \vert }+ \frac{c_{\phi }}{ \vert a_{1}-(b_{1}+c_{1}) \vert \Gamma (\theta +1)} \biggr] \Vert x \Vert ^{q_{2}} \\ &{}+ \biggl[l_{f}+ \frac{( \vert b_{2} \vert + \vert c_{2} \vert )l_{f}}{ \vert a_{2}-(b_{2}+c_{2}) \vert }+ \frac{c_{\psi }}{ \vert a_{2}-(b_{2}+c_{2}) \vert \Gamma (\theta +1)} \biggr] \Vert y \Vert ^{q_{2}} \\ &{} + \biggl[M_{f}+ \frac{( \vert b_{1} \vert + \vert c_{1} \vert )M_{f}}{ \vert a_{1}-(b_{1}+c_{1}) \vert }+ \frac{M_{\phi }}{ \vert a_{1}-(b_{1}+c_{1}) \vert \Gamma (\theta +1)} \biggr] \\ &{}+ \biggl[M_{f}+ \frac{( \vert b_{2} \vert + \vert c_{2} \vert )M_{f}}{ \vert a_{2}-(b_{2}+c_{2}) \vert }+ \frac{M_{\psi }}{ \vert a_{2}-(b_{2}+c_{2}) \vert \Gamma (\theta +1)} \biggr] \\ \leq{}& L_{F} \bigl\Vert (x,y) \bigr\Vert ^{q_{2}}+M_{F}, \end{aligned}$$

where

$$\begin{aligned} L_{F} =&\max \biggl(l_{f}+ \frac{(| b_{1}| +| c_{1}| )l_{f}}{| a_{1}-(b_{1}+c_{1})| }+ \frac{c_{\phi }}{| a_{1}-(b_{1}+c_{1})| \Gamma (\theta +1)},\\ &{}l_{f}+ \frac{(| b_{2}| +| c_{2}| )l_{f}}{| a_{2}-(b_{2}+c_{2})| }+ \frac{c_{\psi }}{| a_{2}-(b_{2}+c_{2})| \Gamma (\theta +1)} \biggr), \\ M_{F} =&2\max \biggl(M_{f}+ \frac{(| b_{1}| +| c_{1}| )M_{f}}{| a_{1}-(b_{1}+c_{1})| }+ \frac{M_{\phi }}{| a_{1}-(b_{1}+c_{1})| \Gamma (\theta +1)},\\ &{} M_{f}+ \frac{(| b_{2}| +| c_{2}| )M_{f}}{| a_{2}-(b_{2}+c_{2})| }+ \frac{M_{\psi }}{| a_{2}-(b_{2}+c_{2})| \Gamma (\theta +1)} \biggr). \end{aligned}$$

 □

Theorem 3.3

The operator $H: X\times X\rightarrow X\times X$ is continuous and meets the following growth condition:

$$\begin{aligned} \bigl\Vert H\bigl(x(t),y(t)\bigr) \bigr\Vert \leq L_{H} \bigl\Vert (x,y) \bigr\Vert ^{q_{1}}+M_{H}. \end{aligned}$$

Proof

Consider a bounded subset of $X \times X$ as

$$\begin{aligned} B_{r}=\bigl\{ \bigl\Vert (x,y) \bigr\Vert \leq r:(x,y)\in X \times X \bigr\} \subseteq X \times X. \end{aligned}$$

Let $\{(x_{n},y_{n})\}$ be a sequence in $B_{r}$ such that $(x_{n},y_{n})\rightarrow (x,y)$ as $n\rightarrow \infty$. To show that H is continuous, we consider

$$\begin{aligned} &\bigl\vert H_{1}(x_{n})-H_{1}(x) \bigr\vert \\ &\quad \leq \frac{1}{\Gamma (\theta )} \int _{0}^{t}(t-s)^{ \theta -1} \bigl\vert h\bigl(s,x_{n}(s),I^{\alpha }x_{n}(s) \bigr)-h\bigl(s,x(s),I^{\alpha }x(s)\bigr) \bigr\vert \,ds \\ &\qquad {}+\frac{ \vert b_{2} \vert }{ \vert a_{2}-(b_{2}+c_{2}) \vert } \frac{1}{\Gamma (\theta )} \int _{0}^{\xi }(\xi -s)^{\theta -1} \bigl\vert h\bigl(s,x_{n}(s),I^{ \alpha }x_{n}(s) \bigr)-h\bigl(s,x(s),I^{\alpha }x(s)\bigr) \bigr\vert \,ds \\ &\qquad{} +\frac{ \vert c_{2} \vert }{ \vert a_{2}-(b_{2}+c_{2}) \vert } \frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{\theta -1} \bigl\vert h\bigl(s,x_{n}(s),I^{ \alpha }x_{n}(s)\bigr)-h \bigl(s,x(s),I^{\alpha }x(s)\bigr) \bigr\vert \,ds. \end{aligned}$$

From the continuity of h, it follows that

$$\begin{aligned} h\bigl(s,x_{n}(s),I^{\alpha }x_{n}(s)\bigr) \rightarrow h\bigl(s,x(s),I^{\alpha }x(s)\bigr)\quad \text{as } n \rightarrow \infty . \end{aligned}$$

For every $t\in [0,1]$, by applying $(H_{2})$, and the Lebesgue dominated convergent theorem, we can get

$$\begin{aligned} \int _{0}^{t}\frac{(t-s)^{\theta -1}}{\Gamma (\theta )} \bigl\vert h\bigl(s,x_{n}(s),I^{ \alpha }x_{n}(s)\bigr)-h \bigl(s,x(s),I^{\alpha }x(s)\bigr) \bigr\vert \,ds\rightarrow 0\quad \text{as }n \rightarrow \infty . \end{aligned}$$

The same as the other terms approach 0 as $n\rightarrow \infty$, thus

$$\begin{aligned} \bigl\Vert H_{1}(x_{n})-H_{1}(x) \bigr\Vert \rightarrow 0\quad \text{as }n \rightarrow \infty . \end{aligned}$$

(3.3)

Then $H_{1}$ is continuous. By the same steps as above, one lightly gets that

$$\begin{aligned} \bigl\Vert H_{2}(y_{n})-H_{2}(y) \bigr\Vert \rightarrow 0\quad \text{as }n \rightarrow \infty . \end{aligned}$$

(3.4)

That is, $H_{2}$ is continuous. From (3.3) and (3.4), we have

$$\begin{aligned} \bigl\Vert H(x_{n},y_{n})-H(x,y) \bigr\Vert \rightarrow 0\quad \text{as }n \rightarrow \infty , \end{aligned}$$

which means that H is continuous.

Moreover, by $(H_{2})$, we have

$$\begin{aligned} \bigl\vert H_{1}x(t) \bigr\vert &= \biggl\vert \frac{1}{\Gamma (\theta )} \int _{0}^{t}(t-s)^{ \theta -1}h \bigl(s,x(s),I^{\alpha }x(s)\bigr)\,ds \\ &\quad +\frac{b_{2}}{a_{2}-(b_{2}+c_{2})}\frac{1}{\Gamma (\theta )} \int _{0}^{\xi }(\xi -s)^{\theta -1}h \bigl(s,x(s),I^{\alpha }x(s)\bigr)\,ds \\ &\quad +\frac{c_{2}}{a_{2}-(b_{2}+c_{2})}\frac{1}{\Gamma (\theta )} \int _{0}^{1}(1-s)^{\theta -1}h \bigl(s,x(s),I^{\alpha }x(s)\bigr)\,ds \biggr\vert \\ &\leq \frac{t^{\theta }}{\Gamma (\theta +1)}\bigl(l_{h}^{1} \Vert x \Vert ^{q_{1}}+l_{h}^{2} \bigl\Vert I^{\alpha }x \bigr\Vert ^{q_{1}}+M_{h}\bigr) \\ &\quad + \frac{ \vert b_{2} \vert \xi ^{\theta }}{ \vert a_{2}-(b_{2}+c_{2}) \vert \Gamma (\theta +1)}\bigl(l_{h}^{1} \Vert x \Vert ^{q_{1}}+l_{h}^{2} \bigl\Vert I^{\alpha }x \bigr\Vert ^{q_{1}}+M_{h}\bigr) \\ &\quad + \frac{ \vert c_{2} \vert }{ \vert a_{2}-(b_{2}+c_{2}) \vert \Gamma (\theta +1)}\bigl(l_{h}^{1} \Vert x \Vert ^{q_{1}}+l_{h}^{2} \bigl\Vert I^{\alpha }x \bigr\Vert ^{q_{1}}+M_{h}\bigr) \\ &\leq \biggl[\frac{1}{\Gamma (\theta +1)}+ \frac{ \vert b_{2} \vert + \vert c_{2} \vert }{ \vert a_{2}-(b_{2}+c_{2}) \vert \Gamma (\theta +1)} \biggr] \bigl(l_{h}^{1} \Vert x \Vert ^{q_{1}}+l_{h}^{2} \bigl\Vert I^{ \alpha }x \bigr\Vert ^{q_{1}}+M_{h} \bigr). \end{aligned}$$

Similarly,

$$\begin{aligned} \bigl\vert H_{2}y(t) \bigr\vert \leq \biggl[ \frac{1}{\Gamma (\theta +1)}+ \frac{ \vert b_{1} \vert + \vert c_{1} \vert }{ \vert a_{1}-(b_{1}+c_{1}) \vert \Gamma (\theta +1)} \biggr]\bigl(l_{h}^{1} \Vert y \Vert ^{q_{1}}+l_{h}^{2} \bigl\Vert I^{ \alpha }y \bigr\Vert ^{q_{1}}+M_{h} \bigr). \end{aligned}$$

(3.5)

Thus

$$\begin{aligned} \bigl\Vert H\bigl(x(t),y(t)\bigr) \bigr\Vert &= \bigl\Vert H_{1}x(t) \bigr\Vert + \bigl\Vert H_{2}y(t) \bigr\Vert \\ &\leq \biggl[\frac{1}{\Gamma (\theta +1)}+ \frac{ \vert b_{2} \vert + \vert c_{2} \vert }{ \vert a_{2}-(b_{2}+c_{2}) \vert \Gamma (\theta +1)} \biggr] \bigl(l_{h}^{1} \Vert x \Vert ^{q_{1}}+l_{h}^{2} \bigl\Vert I^{ \alpha }x \bigr\Vert ^{q_{1}}+M_{h} \bigr) \\ &\quad + \biggl[\frac{1}{\Gamma (\theta +1)}+ \frac{ \vert b_{1} \vert + \vert c_{1} \vert }{ \vert a_{1}-(b_{1}+c_{1}) \vert \Gamma (\theta +1)} \biggr] \bigl(l_{h}^{1} \Vert y \Vert ^{q_{1}}+l_{h}^{2} \bigl\Vert I^{ \alpha }y \bigr\Vert ^{q_{1}}+M_{h} \bigr) \\ &\leq \biggl[\frac{1}{\Gamma (\theta +1)}+ \frac{ \vert b_{2} \vert + \vert c_{2} \vert }{ \vert a_{2}-(b_{2}+c_{2}) \vert \Gamma (\theta +1)} \biggr] \bigl[l\bigl( \Vert x \Vert ^{q_{1}}+ \bigl\Vert I^{\alpha }x \bigr\Vert ^{q_{1}}\bigr)+M_{h}\bigr] \\ &\quad + \biggl[\frac{1}{\Gamma (\theta +1)}+ \frac{ \vert b_{1} \vert + \vert c_{1} \vert }{ \vert a_{1}-(b_{1}+c_{1}) \vert \Gamma (\theta +1)} \biggr] \bigl[l \bigl( \Vert y \Vert ^{q_{1}}+ \bigl\Vert I^{\alpha }y \bigr\Vert ^{q_{1}}\bigr)+M_{h}\bigr] \\ &\leq L_{h}\bigl[ \Vert x \Vert ^{q_{1}}+ \Vert y \Vert ^{q_{1}}+ \bigl\Vert I^{\alpha }x \bigr\Vert ^{q_{1}}+ \bigl\Vert I^{\alpha }y \bigr\Vert ^{q_{1}}+2M_{h}\bigr] \\ &\leq L_{H} \bigl\Vert (x,y) \bigr\Vert ^{q_{1}}+M_{H}, \end{aligned}$$

where $l=\max \{l_{h}^{1},l_{h}^{2}\}\in [0,1)$, $L_{h}=\max ( \frac{1}{\Gamma (\theta +1)}+ \frac{| b_{2}| +| c_{2}| }{| a_{2}-(b_{2}+c_{2})| \Gamma (\theta +1)}, \frac{1}{\Gamma (\theta +1)}+ \frac{| b_{1}| +| c_{1}| }{| a_{1}-(b_{1}+c_{1})| \Gamma (\theta +1)})$, $L_{H}=(1+(\frac{1}{\Gamma (\alpha +1)})^{q_{1}})L_{h}$, $M_{H}=2L_{h}M_{h}$. Hence H satisfies the growth condition. □

Theorem 3.4

The operator $H: X \times X\rightarrow X \times X$ is compact.

Proof

Let Ω be a bounded subset of $B_{r}\subseteq X \times X$ and $\{(x_{n},y_{n})\}$ be a sequence in Ω, through applying the growth condition of H, it is obvious that $H(\Omega )$ is uniformly bounded in $X \times X$. Now, we need to reveal that H is equicontinuous. Let $0\leq t\leq \tau \leq 1$, then we obtain

$$\begin{aligned} \bigl\vert H_{1}x_{n}(t)-H_{1}x_{n}( \tau ) \bigr\vert &= \biggl\vert \frac{1}{\Gamma (\theta )} \int _{0}^{t}(t-s)^{\theta -1}h \bigl(s,x_{n}(s),I^{ \alpha }x_{n}(s)\bigr)\,ds \\ &\quad -\frac{1}{\Gamma (\theta )} \int _{0}^{\tau }(\tau -s)^{\theta -1}h \bigl(s,x_{n}(s),I^{ \alpha }x_{n}(s)\bigr)\,ds \biggr\vert \\ &= \biggl\vert \frac{1}{\Gamma (\theta )} \int _{0}^{t}\bigl[(t-s)^{\theta -1}-( \tau -s)^{\theta -1}\bigr]h\bigl(s,x_{n}(s),I^{\alpha }x_{n}(s) \bigr)\,ds \\ &\quad -\frac{1}{\Gamma (\theta )} \int _{t}^{\tau }(\tau -s)^{\theta -1}h \bigl(s,x_{n}(s),I^{ \alpha }x_{n}(s)\bigr)\,ds \biggr\vert \\ &\leq \frac{1}{\Gamma (\theta )} \int _{0}^{t}\bigl[(t-s)^{\theta -1}-( \tau -s)^{\theta -1}\bigr] \bigl\vert h\bigl(s,x_{n}(s),I^{\alpha }x_{n}(s) \bigr) \bigr\vert \,ds \\ &\quad +\frac{1}{\Gamma (\theta )} \int _{t}^{\tau }(\tau -s)^{\theta -1} \bigl\vert h\bigl(s,x_{n}(s),I^{\alpha }x_{n}(s) \bigr) \bigr\vert \,ds \\ &\leq \frac{l_{h}^{1} \Vert x_{n} \Vert ^{q_{1}}+l_{h}^{2} \Vert I^{\alpha }x_{n} \Vert ^{q_{1}}+M_{h}}{\Gamma (\theta +1)}\bigl[t^{ \theta }-\tau ^{\theta }+2(\tau -t)^{\theta }\bigr]. \end{aligned}$$

Similarly,

$$\begin{aligned} \bigl\vert H_{2}y_{n}(t)-H_{2}y_{n}( \tau ) \bigr\vert \leq \frac{l_{h}^{1} \Vert y_{n} \Vert ^{q_{1}}+l_{h}^{2} \Vert I^{\alpha }y_{n} \Vert ^{q_{1}}+M_{h}}{\Gamma (\theta +1)}\bigl[t^{ \theta }-\tau ^{\theta }+2(\tau -t)^{\theta }\bigr]. \end{aligned}$$

Taking limit as $t\rightarrow \tau$, we get

$$\begin{aligned} \bigl\Vert H_{1}x_{n}(t)-H_{1}x_{n}( \tau ) \bigr\Vert \rightarrow 0 \end{aligned}$$

and

$$\begin{aligned} \bigl\Vert H_{2}y_{n}(t)-H_{2}y_{n}( \tau ) \bigr\Vert \rightarrow 0, \end{aligned}$$

which implies that

$$\begin{aligned} \bigl\Vert H(x_{n},y_{n}) (t)-H(x_{n},y_{n}) (\tau ) \bigr\Vert \rightarrow 0. \end{aligned}$$

This reveals that $H(x,y)$ is equicontinuous. $H(x,y)$ is compact by the Arzelá–Ascoli theorem. Hence, according to Proposition 2.5, H is α-Lipschitz with constant zero. □

Theorem 3.5

If $(H_{1})$–$(H_{5})$ hold and

$$\begin{aligned} k={}&\max \biggl(\lambda _{1}+ \frac{(| b_{1}| +| c_{1}| ) \lambda _{1}}{| a_{1}-(b_{1}+c_{1})| }+ \frac{b_{\phi }}{| a_{1}-(b_{1}+c_{1})| \Gamma (\theta +1)},\\ &\lambda _{2}+ \frac{(| b_{2}| +| c_{2}| ) \lambda _{2}}{| a_{2}-(b_{2}+c_{2})| }+ \frac{b_{\psi }}{| a_{2}-(b_{2}+c_{2})| \Gamma (\theta +1)}\biggr)\in [0,1), \end{aligned}$$

then coupled system (1.1) has at least one solution $(x,y)\in X\times X$. And the solution set of (1.1) is bounded in $X\times X$.

Proof

From Theorem 3.2 and $k\in [0,1)$, F is α-Lipschitz with constant $k\in [0,1)$, according to Theorem 3.4, H is α-Lipschitz with constant 0. By Proposition 2.3 and Definition 2.2, T is a strict α-contraction with constant k. Hence, T is α-condensing. Then, we think over the following set:

$$\begin{aligned} R=\bigl\{ (x,y)\in X\times X : \text{there exist }\zeta \in [0,1], (x,y)=\zeta T(x,y) \bigr\} . \end{aligned}$$

We have to reveal that R is bounded in $X\times X$. Let $(x,y)\in R$, then by applying the growth conditions of Theorem 3.2 and Theorem 3.3, we have

$$\begin{aligned} (x,y)=\zeta T(x,y)=\zeta \bigl(F(x,y)+H(x,y)\bigr), \end{aligned}$$

thus

$$\begin{aligned} \bigl\Vert (x,y) \bigr\Vert &=\zeta \bigl\Vert T(x,y) \bigr\Vert \\ &\leq \zeta \bigl( \bigl\Vert F(x,y) \bigr\Vert + \bigl\Vert H(x,y) \bigr\Vert \bigr) \\ &\leq \zeta \bigl[L_{F} \bigl\Vert (x,y) \bigr\Vert ^{q_{2}}+M_{F}+L_{H} \bigl\Vert (x,y) \bigr\Vert ^{q_{1}}+M_{H} \bigr] \\ &=\zeta \bigl(L_{F} \bigl\Vert (x,y) \bigr\Vert ^{q_{2}}+L_{H} \bigl\Vert (x,y) \bigr\Vert ^{q_{1}} \bigr)+\zeta (M_{F}+M_{H} ), \end{aligned}$$

where $q_{1},q_{2}\in [0,1)$. Thus R is bounded in $X \times X$. According to Theorem 2.6, there exists $r>0$ such that $R\subset B_{r}(0)$, then

$$\begin{aligned} D\bigl(I-\zeta T, B_{r}(0), 0\bigr))=1, \quad \text{for all }\zeta \in [0,1]. \end{aligned}$$

Therefore, T has at least one fixed point, then coupled system (1.1) has at least one solution. □

4 Examples

This part, we have the following example account for our main results.

Example 4.1

Give thought to the following equation:

$$ \textstyle\begin{cases} D^{\frac{3}{2}}[x(t)- \frac{\sin ^{2}(t) \vert x(t) \vert }{5(2+ \vert x(t) \vert )}]= \frac{e^{-\pi t}}{10}+ \frac{\cos \vert y(t) \vert +\sin \vert y(t) \vert }{51+t^{2}},\quad t\in [0,1], \\ D^{\frac{3}{2}}[y(t)- \frac{\sin^{2} (t) \vert y(t) \vert }{5(2+ \vert y(t) \vert )}]= \frac{e^{-\pi t}}{10}+ \frac{\cos \vert x(t) \vert +\sin \vert x(t) \vert }{51+t^{2}},\quad t\in [0,1], \\ x'(0)=y'(0)=0, \\ \frac{1}{4}x(0)-\frac{1}{2}x(\frac{1}{2})-6x(1)= \frac{1}{\Gamma (\frac{3}{2})}\int _{0}^{1} (1-s)^{\frac{1}{2}} \frac{\sin x(s)}{2}\,ds, \\ \frac{1}{5}y(0)-\frac{1}{7}y(\frac{1}{2})-8y(1)= \frac{1}{\Gamma (\frac{3}{2})}\int _{0}^{1} (1-s)^{\frac{1}{2}} \frac{\cos y(s)}{5}\,ds, \end{cases} $$

(4.1)

where $h=\frac{e^{-\pi t}}{10}+ \frac{\cos | y(t)| +\sin | y(t)| }{51+t^{2}}$, $\theta = \frac{3}{2}$, $a_{1}=\frac{1}{4}$, $b_{1}=\frac{1}{2}$, $c_{1}=6$, $a_{2}= \frac{1}{5}$, $b_{2}=\frac{1}{7}$, $c_{2}=8$, $\eta =\xi =\frac{1}{2}$. Let $\zeta =\frac{1}{5}$, then by routine calculation, we can have $c_{\phi }=b_{\phi }=\frac{1}{2}$, $c_{\psi }=b_{\psi }=\frac{1}{5}$, $M_{ \phi }=M_{\psi }=0$, $l_{h}^{1}=l_{h}^{2}=\frac{1}{51}$, $M_{h}= \frac{1}{10}$, $l_{f}=\frac{1}{10}$, $M_{f}=0$, $\lambda _{1}=\lambda _{2}= \frac{1}{10}$. Thus assumptions $(H_{1})-(H_{5})$ hold, and

$$\begin{aligned}& \lambda _{1}+ \frac{( \vert b_{1} \vert + \vert c_{1} \vert ) \lambda _{1}}{ \vert a_{1}-(b_{1}+c_{1}) \vert }+ \frac{b_{\phi }}{ \vert a_{1}-(b_{1}+c_{1}) \vert \Gamma (\theta +1)} \approx 0.389, \\& \lambda _{2}+ \frac{( \vert b_{2} \vert + \vert c_{2} \vert ) \lambda _{2}}{ \vert a_{2}-(b_{2}+c_{2}) \vert }+ \frac{b_{\psi }}{ \vert a_{2}-(b_{2}+c_{2}) \vert \Gamma (\theta +1)}) \approx 0.221. \end{aligned}$$

Next,

$$\begin{aligned} \bigl\vert F(x_{1}, y_{1}) (t)-F(x_{2}, y_{2}) (t) \bigr\vert \leq{}& \frac{1}{10}\bigl| x_{1}(t)-x_{2}(t)\bigr|+\frac{1}{10}\bigl|y_{1}(t)-y_{2}(t)\bigr|\\ &{} +\frac{1}{11.078} \int _{0}^{1} (1-s)^{\frac{1}{2}} \bigl\vert \sin (x_{1})- \sin (x_{2}) \bigr\vert \,ds \\ &{} +\frac{1}{35.196} \int _{0}^{1} (1-s)^{\frac{1}{2}} \bigl\vert \cos (y_{1})- \cos (y_{2}) \bigr\vert \,ds \\ \leq{}& 0.160 \Vert x_{1}-x_{2} \Vert + 0.119 \Vert y_{1}-y_{2} \Vert \\ \leq{}& 0.160 \bigl\Vert (x_{1}, y_{1})-(x_{2}, y_{2}) \bigr\Vert , \end{aligned}$$

thus F is α-Lipschitz with the constant 0.160, and thus H is α-Lipschitz with the constant zero, which means that T is α-Lipschitz with the constant 0.160. Since

$$\begin{aligned} R=\bigl\{ (x,y)\in X\times X : \text{there exist }\zeta \in [0,1], (x,y)=\zeta T(x,y) \bigr\} , \end{aligned}$$

then, by routine calculation, we obtain $L_{F}=0.216$, $L_{H}=2.380$, $M_{F}=0$, $M_{H}=0.238$.

Thus

$$\begin{aligned} \bigl\Vert (x,y) \bigr\Vert \cong 0.703\leq 1, \end{aligned}$$

then R is bounded, through Theorem 3.5, then problem (4.1) has at least one solution.