New analysis of fuzzy fractional Langevin differential equations in Caputo's derivative sense

Muhammad Akram; Ghulam Muhammad; Tofigh Allahviranloo; Ghada Ali; Muhammad Akram; Ghulam Muhammad; Tofigh Allahviranloo; Ghada Ali

doi:10.3934/math.20221016

AIMS Mathematics

2022, Volume 7, Issue 10: 18467-18496. doi: 10.3934/math.20221016

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Research article

New analysis of fuzzy fractional Langevin differential equations in Caputo's derivative sense

1.
Department of Mathematics, University of the Punjab, New Campus, Lahore- 54590, Pakistan
2.
Department of Mathematics, Lahore Garrison University, Lahore 54000, Pakistan
3.
Faculty of Engineering and Natural Sciences, Istinye University, Istanbul, Turkey
4.
Department of Mathematics, King Abdulaziz University Jeddah, Saudi Arabia

Received: 06 July 2022 Revised: 03 August 2022 Accepted: 11 August 2022 Published: 18 August 2022
MSC : 34A08, 34K37, 03E72

The extraction of analytical solution of uncertain fractional Langevin differential equations involving two independent fractional-order is frequently complex and difficult. As a result, developing a proper and comprehensive technique for the solution of this problem is very essential. In this article, we determine the explicit and analytical fuzzy solution for various classes of the fuzzy fractional Langevin differential equations (FFLDEs) with two independent fractional-orders both in homogeneous and non-homogeneous cases. The potential solution of FFLDEs is also extracted using the fuzzy Laplace transformation technique. Furthermore, the solution of FFLDEs is defined in terms of bivariate and trivariate Mittag-Leffler functions both in the general and special forms. FFLDEs are a new topic having many applications in science and engineering then to grasp the novelty of this work, we connect FFLDEs with RLC electrical circuit to visualize and support the theoretical results.

Keywords:

Citation: Muhammad Akram, Ghulam Muhammad, Tofigh Allahviranloo, Ghada Ali. New analysis of fuzzy fractional Langevin differential equations in Caputo's derivative sense[J]. AIMS Mathematics, 2022, 7(10): 18467-18496. doi: 10.3934/math.20221016

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Abstract

1. Introduction

Differential equations of arbitrary order have been shown to be useful in the study of models of many phenomena in various fields such as: Electrochemistry and material science, they are in fact described by differential equations of fractional order ^{[9,10,15,16,25,26,27,28,29]}. For more details, we refer the reader to the books of Hilfer ^[30], Podlubny ^[31], Kilbas et al. ^[34], Miller and Ross ^[2] and to the following research papers ^{[1,2,3,4,5,6,7,8,11,12,14,16,17,19,20,24,31,35,36,37,38,39,40,41,42]}. In this work, we discuss the existence and uniqueness of the solutions for multi-point boundary value problems of nonlinear fractional differential equations with two Riemann-Liouville fractionals:

$\begin{equation} \left\{\begin{array}{llll} D^{\alpha}x(t) = \sum^{m}_{i = 1}f_{i}(t, x(t), y(t), \varphi_{1}x(t), \phi_{1}y(t)) , \alpha\in]1, 2], t\in[0, T] &\\ D^{\beta }y(t) = \sum_{i = 1}^{m}g_{i}(t, x(t), y(t), \varphi_{2}x(t), \phi_{2}y(t)) , \beta\in]1, 2], t\in[0, T]& \\ I^{2-\alpha}x(0) = 0, \ D^{\alpha-2}x(T) = \theta I^{\alpha-1}(x(\eta)), \ 0 < \eta < T, &\\ I^{2-\beta}y(0) = 0, \ D^{\beta-2}x(T) = \omega I^{\beta-1}(x(\gamma)), \ 0 < \gamma < T, & \end{array}\right. \end{equation}$

(1.1)

where $D^{(.)}$ , $I^{(.)}$ denote the Riemann-Liouville derivative and integral of fractional order $(.)$ , respectively, $f_{i},$ $g_{i}:\left[0, T\right] \times \mathbb{R}^{4}\rightarrow \mathbb{R},$ $i = 1, \cdots, m$ are continuous functions on $[0, T]$ and

$(\varphi_{1}x)(t) = \int_{0}^{t}A_{1}'(t, s)x(s)ds, \ (\phi_{1}y)(t) = \int_{0}^{t}B_{1}'(t, s)y(s)ds,$

$(\varphi_{2}x)(t) = \int_{0}^{t}A_{2}'(t, s)x(s)ds, \ (\phi_{1}y)(t) = \int_{0}^{t}B_{2}'(t, s)y(s)ds,$

with $A_{i}$ and $B_{i}$ being continuous functions on $[0, 1]\times[0, 1].$ However, it is rare to find a work in nonlinear term $f_{i}$ depends on fractional derivative of unknown functions $x(t), y(t), \varphi_{1}x(t), \phi_{1}y(t)$ and solutions for multi-order fractional differential equations on the infinite interval $[0, T)$ . Motivated by ^{[8,11,12,13,14]} and the references therein, we consider the existence and unicity of solution for multi-order fractional differential equations on infinite interval $[0, T)$ .

The rest of this paper is organized as follow. In section 2, we present some preliminaries and lemmas. Section 3 is dedicated to showing the existence of a solution for problem (1.1). Finally, section 4 illustrated the proposed results with two examples.

Remark 1.1. This work generalizes the work of Houas and Benbachir ^[14] on different boundary conditions and for another type of integral.

2. Preliminaries

This section covers the basic concepts of Riemann-Liouville type fractional calculus that will be used throughout this paper.

Definition 2.1. ^[31,32] The Riemann-Liouville fractional integral operator of order $\alpha \geq 0,$ of a function $f:\left(0, \infty \right) \rightarrow \mathbb{R}$ is defined as

$\begin{eqnarray*} \left\{\begin{array}{ll} J^{\alpha }f\left( t\right) = \frac{1}{\Gamma \left( \alpha \right) }\int_{0}^{t}\left( t-\tau \right) ^{\alpha -1}f\left( \tau \right) d\tau , & \\ J^{0}f\left( t\right) = f\left( t\right) , & \end{array}\right. \end{eqnarray*}$

where $\Gamma \left(\alpha \right) : = \int_{0}^{\infty }e^{-u}u^{\alpha -1}du.$

Definition 2.2. ^[31,32] The Riemann-Liouville fractional derivative of order $\alpha > 0,$ of a continuous function $h:\left(0, \infty \right) \rightarrow \mathbb{R}$ is defined as

$\begin{equation*} D^{\alpha}h\left( t\right) = \frac{1}{\Gamma \left( n-\alpha \right) } \left( \frac{d}{dt}\right) ^{n}\int_{0}^{t}\left( t-\tau \right) ^{n-\alpha-1}h\left( \tau \right) d\tau = \left(\dfrac{d}{dt}\right)^{n}I^{n-\alpha}h(\tau), \end{equation*}$

where $n = \left[\alpha \right] +1.$

For $\alpha < 0$ , we use the convention that $D^{\alpha }h = J^{-\alpha}h.$ Also for $0\leq \rho < \alpha$ , it is valid that $D^{\rho }J^{\alpha }h = h^{\alpha -\rho }.$ We note that for $\varepsilon > -1$ and $\varepsilon \neq \alpha -1, \alpha-2, ..., \alpha-n,$ we have

$\begin{eqnarray*} D^{\alpha}t^{\varepsilon } & = &\frac{\Gamma \left( \varepsilon +1\right) }{\Gamma \left( \varepsilon -\alpha +1\right) }t^{\varepsilon -\alpha }, \\ D^{\alpha}t^{\alpha -i} & = &0, \text{ }i = 1, 2, ..., n. \end{eqnarray*}$

In particular, for the constant function $h(t) = 1$ , we obtain

$D^{\alpha}1 = \dfrac{1}{\Gamma(1-\alpha)}t^{-\alpha}, \alpha\notin\mathbb{N}.$

For $\alpha\in\mathbb{N}$ , we obtain, of course, $D^{\alpha}1 = 0$ because of the poles of the gamma function at the points $0, -1, -2, ...$ For $\alpha > 0$ , the general solution of the homgeneous equation $D^{\alpha}h(t) = 0$ in $C(0, T)\cap L(0, T)$ is

$h(t) = c_{0}t^{\alpha-n}+c_{1}t^{\alpha-n-1}+......+c_{n-2}t^{\alpha-2}+c_{n-1}t^{\alpha-1},$

where $c_{i}, i = 1, 2, ...., n-1$ , are arbitrary real constants. Further, we always have $D^{\alpha}I^{\alpha}h = h$ , and

$D^{\alpha}I^{\alpha}h(t) = h(t)+c_{0}t^{\alpha-n}+c_{1}t^{\alpha-n-1}+......+c_{n-2}t^{\alpha-2}+c_{n-1}t^{\alpha-1}.$

Lemma 2.1. ^[33] Let $E$ be Banach space. Assume that $T : E\longrightarrow E$ is a completely continuous operator. If the set $V = \{x \in E : x = \mu T x, \ 0 < \mu < 1\}$ is bounded, then $T$ has a fixed point in $E$ .

To define the solution for problem (1.1). We consider the following lemma.

Lemma 2.2. Suppose that $(H_{i})_{i = 1, \ldots, m}\subset C([0, 1], \mathbb{R})$ , and consider the problem

$\begin{equation} D^{\alpha}h(t)-\sum\limits_{i = 1}^{m}H_{i}(t) = 0, \ t\in j, \ 1 < \alpha < 2, \ m\in\mathbb{N}^{\ast}, \end{equation}$

(2.1)

with the conditions

$\begin{equation} I^{2-\alpha}h(0) = 0, \ D^{\alpha-2}h(T) = \theta I^{\alpha-1}(h(\eta)), \ 0 < \eta < T. \end{equation}$

(2.2)

Then we have

$\begin{eqnarray*} h(t)& = &\dfrac{1}{\Gamma(\alpha)}\sum\limits_{i = 1}^{m}\int^{t}_{0}(t-\tau)^{\alpha-1}H_{i}(\tau)d\tau\\ &&+\dfrac{t^{\alpha-1}}{\psi}\left(\sum\limits_{i = 1}^{m}\int^{T}_{0}(T-\tau)H_{i}(\tau)d\tau \right. \left. -\dfrac{\theta}{\Gamma(2\alpha)}\sum\limits_{i = 1}^{m}\int^{\eta}_{0}(\eta-\tau)^{2\alpha-2}H_{i}(\tau)d\tau\right) \end{eqnarray*}$

with $\psi = \theta\dfrac{\Gamma(\alpha)}{\Gamma(2\alpha-1)}\eta^{2\alpha-2}-\Gamma(\alpha)T.$

Proof. We have

$h(t) = \sum\limits_{i = 1}^{m}I^{\alpha}H_{i}(t)+c_{0}t^{\alpha-2}+c_{1}t^{\alpha-1},$

where $c_{i}\in\mathbb{R}, \ i = 0, 1.$

We obtain

$\begin{eqnarray*} I^{2-\alpha}h(\tau)& = &\sum\limits_{i = 1}^{m}I^{2}H_{i}(\tau)+c_{0}I^{2-\alpha}\tau^{\alpha-2}+c_{1}I^{2-\alpha}\tau^{\alpha-1}\\ & = &\sum\limits_{i = 1}^{m}I^{2}H_{i}(\tau)+c_{0}+c_{1}\tau , \\ I^{\alpha-1}h(\tau)& = &\sum\limits_{i = 1}^{m}I^{2\alpha-1}H_{i}(\tau)+c_{0}I^{\alpha-1}\tau^{\alpha-2}+c_{1}I^{\alpha-1}\tau^{\alpha-1}\\ & = &\sum\limits_{i = 1}^{m}I^{2\alpha-1}H_{i}(\tau)+c_{0}\dfrac{\Gamma(\alpha-1)}{\Gamma(2\alpha-2)}\tau^{2\alpha-3}+c_{1}\dfrac{\Gamma(\alpha)}{\Gamma(2\alpha-1)}\tau^{2\alpha-2}, \\ D^{\alpha-2}h(\tau)& = &\sum\limits_{i = 1}^{m}I^{2}H_{i}(\tau)+c_{0}\Gamma(\alpha-1)+c_{1}\Gamma(\alpha)\tau. \end{eqnarray*}$

Using the given conditions: $I^{2-\alpha}h(0) = 0$ , we find that $c_{0} = 0$ , and since $D^{\alpha-2}h(T)-\theta I^{\alpha-1}(h(\eta)) = 0$ , we have

$\begin{equation*} \sum\limits_{i = 1}^{m}I^{2}h_{i}(T)+c_{1}\Gamma(\alpha)T-\theta\left[\sum\limits_{i = 1}^{m}I^{2\alpha-1}h_{i}(\eta)+c_{1}\dfrac{\Gamma(\alpha)}{\Gamma(2\alpha-1)}\eta^{2\alpha-2}\right] = 0, \end{equation*}$

then

$\begin{equation*} c_{1}\left[\dfrac{\Gamma(\alpha)}{\Gamma(2\alpha-1)}\eta^{2\alpha-2}-\Gamma(\alpha)T\right] = \sum\limits_{i = 1}^{m}I^{2}h_{i}(T)-\theta\sum\limits_{i = 1}^{m}I^{2\alpha-1}h_{i}(\eta) \end{equation*}$

and

$\begin{eqnarray*} c_{1}& = &\dfrac{1}{\psi}\left(\sum\limits_{i = 1}^{m}I^{2}H_{i}(T)-\theta\sum\limits_{i = 1}^{m}I^{2\alpha-1}H_{i}(\eta)\right)\\ & = &\dfrac{1}{\psi}\left(\sum\limits_{i = 1}^{m}\int_{0}^{T}(T-\tau)H_{i}(\tau)d\tau-\dfrac{\theta}{\Gamma(2\alpha)}\sum\limits_{i = 1}^{m}\int_{0}^{\eta}(\eta-\tau)^{2\alpha-2}H_{i}(\tau)d\tau\right) \end{eqnarray*}$

with

$\begin{eqnarray*} \psi = \theta\dfrac{\Gamma(\alpha)}{\Gamma(2\alpha-1)}\eta^{2\alpha-2}-\Gamma(\alpha)T. \end{eqnarray*}$

Finally, the solution of (2.1) and (2.2) is

$\begin{eqnarray*} h(t)& = &\dfrac{1}{\Gamma(\alpha)}\sum\limits_{i = 1}^{m}\int^{t}_{0}(t-\tau)^{\alpha-1}H_{i}(\tau)d\tau+\dfrac{t^{\alpha-1}}{\psi}\left(\sum\limits_{i = 1}^{m}\int^{T}_{0}(T-\tau)H_{i}(\tau)d\tau \right. \\ &&\qquad \left. -\dfrac{\theta}{\Gamma(2\alpha)}\sum\limits_{i = 1}^{m}\int^{\eta}_{0}(\eta-\tau)^{2\alpha-2}H_{i}(\tau)d\tau\right). \end{eqnarray*}$

3. Main results

We denote by

$E = \{x, y\in C\left( \left[ 0, T\right], \mathbb{R} \right); \varphi_{i}x, \phi_{i}y\in C\left( \left[ 0, T\right], \mathbb{R} \right)\ \ i = 1, 2\},$

and the Banach space of all continuous functions from $\left[0, T \right]$ to $\mathbb{R}$ endowed with a topology of uniform convergence with the norm defined by

$||(x, y)||_{E} = \max\left(||x||, ||y||, ||\varphi_{1} x||, ||\phi_{1}y||, ||\varphi_{2}x||, ||\phi_{2}y||\right),$

where

$\begin{eqnarray*} &&||x|| = \sup\limits_{t\in j}|\varphi_{i}x(t)|, \; \; ||y|| = \sup\limits_{t \in j}|y(t)|, \\ && ||\phi_{i}x|| = \sup\limits_{t\in j}|\varphi_{i}x(t)|, \; \; ||\phi_{i}y|| = \sup\limits_{t \in j}|\phi_{i}y(t)|. \end{eqnarray*}$

In this section, we prove some existence and uniqueness results to the nonlinear fractional coupled system (1.1).

For the sake of convenience, we impose the following hypotheses:

$(H1)$ For each $i = 1, 2, \cdots, m$ , the functions $f_{i}$ and $g_{i}\ \ : [0, T]\times\mathbb{R}^4\longrightarrow\mathbb{R}$ are continuous.

$(H2)$ There exist nonnegative real numbers $\xi^{i}_{k}, \varphi^{i}_{k}, k = 1, 2, 3, 4, i = 1, 2, \cdots, m$ , such that for all $t\in \left[0, T\right]$ and all $(x_{1}, x_{2}, x_{3}, x_{4}),$ $(y_{1}, y_{2}, y_{3}, y_{4})\in \mathbb{R}^{4},$ we have

$\begin{equation*} |f_{i}(t, x_{1}, x_{2}, x_{3}, x_{4})-f_{i}(t, y_{1}, y_{2}, y_{3}, y_{4})|\leq\sum\limits_{k = 1}^{4}\ \xi^{i}_{k} |x_{k}-y_{k}|, \end{equation*}$

and

$\begin{equation*} |g_{i}(t, x_{1}, x_{2}, x_{3}, x_{4})-g_{i}(t, y_{1}, y_{2}, y_{3}, y_{4})|\leq\sum\limits_{k = 1}^{4}\ \chi^{i}_{k} |x_{k}-y_{k}|. \end{equation*}$

$(H3)$ There exist nonnegative constants $(L_{i}) \ \text{and}\ \ (K_{i}) \ i = 1, ..., m$ , such that: For each $t \in [0, T]$ and all $(x_{1}, x_{2}, x_{3}, x_{4}) \in \mathbb{R}^{4},$

$\begin{equation*} | f_{i} (t, x_{1}, x_{2}, x_{3}, x_{4})| \leq L_{i}, |g_{i} (t, x_{1}, x_{2}, x_{3}, x_{4})| \leq K_{i}, i = 1, ..., m. \end{equation*}$

We also consider the following quantities:

$\begin{eqnarray*} A_{1}& = &\dfrac{T^{\alpha}}{\Gamma(\alpha +1)}\sum\limits_{i = 1}^{m}(\xi_{1}^{i}+\xi_{2}^{i}+\xi_{3}^{i}+\xi_{4}^{i}), \\ A_{2}& = &\dfrac{T^{\beta}}{\Gamma(\beta +1)}\sum\limits_{i = 1}^{m}(\chi_{1}^{i}+\chi_{2}^{i}+\chi_{3}^{i}+\chi_{4}^{i}), \\ A_{3}& = &\max\limits_{t, s\in[0, 1]}||A'_{1}(t, s)||\times A_{1}, \\ A_{4}& = &\max\limits_{t, s\in[0, 1]}||A'_{2}(t, s)||\times A_{1}, \\ A_{5}& = &\max\limits_{t, s\in[0, 1]}||B'_{1}(t, s)||\times A_{2}, \\ A_{6}& = &\max\limits_{t, s\in[0, 1]}||B'_{2}(t, s)||\times A_{2}, \\ \nu_{1}& = &\left[\dfrac{T^{\alpha}}{\Gamma(\alpha+1)}+\dfrac{1}{\psi}\left(\dfrac{T^{\alpha+1}}{2}+\dfrac{\theta T^{3\alpha-2}}{(2\alpha-1)^{2}\Gamma(2\alpha-1)}\right)\right], \\ \nu_{2}& = &\left[\dfrac{T^{\beta}}{\Gamma(\beta+1)}+\dfrac{1}{\psi'}\left(\dfrac{T^{\beta+1}}{2}+\dfrac{\omega T^{3\beta-2}}{(2\beta-1)^{2}\Gamma(2\beta-1)}\right)\right], \\ \nu_{3}& = &\max\limits_{t, s\in[0, 1]}|A'_{1}(t, s)|\nu_{1}, \\ \nu_{4}& = &\max\limits_{t, s\in[0, 1]}|A'_{2}(t, s)|\nu_{1}, \\ \nu_{5}& = &\max\limits_{t, s\in[0, 1]}|B'_{1}(t, s)|\nu_{2}, \\ \nu_{6}& = &\max\limits_{t, s\in[0, 1]}|B'_{2}(t, s)|\nu_{2}. \end{eqnarray*}$

3.1. Existence of solutions

The first result is based on Banach contraction principle. We have

Theorem 3.1. Assume that $(H2)$ holds. If the inequality

$\begin{equation} \max(A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}) < 1, \end{equation}$

(3.1)

is valid, then the system (1.1) has a unique solution on $[0, T]$ .

Proof. We define the operator $T : E\longrightarrow E$ by

$T(x, y)(t) = (T_{1}(x, y)(t) , T_{2} (x, y)(t)) , t \in[0, T],$

such that

$\begin{eqnarray} T_{1}(x, y)(t)& = &\dfrac{1}{\Gamma(\alpha)}\sum\limits_{i = 1}^{m}\int^{t}_{0}(t-\tau)^{\alpha-1}H_{i}(\tau)d\tau+\dfrac{t^{\alpha-1}}{\psi}\left(\sum\limits_{i = 1}^{m}\int^{T}_{0}(T-\tau)H_{i}(\tau)d\tau \right.\\ &&\qquad \left. -\dfrac{\theta}{\Gamma(2\alpha)}\sum\limits_{i = 1}^{m}\int^{\eta}_{0}(\eta-\tau)^{2\alpha-2}H_{i}(\tau)d\tau\right) \end{eqnarray}$

(3.2)

and

$\begin{eqnarray} T_{2}(x, y)(t)& = &\dfrac{1}{\Gamma(\beta)}\sum\limits_{i = 1}^{m}\int^{t}_{0}(t-\tau)^{\beta-1}G_{i}(\tau)d\tau+\dfrac{t^{\beta-1}}{\psi'}\left(\sum\limits_{i = 1}^{m}\int^{T}_{0}(T-\tau)G_{i}(\tau)d\tau \right.\\ &&\qquad \left. -\dfrac{\omega}{\Gamma(2\beta)}\sum\limits_{i = 1}^{m}\int^{\gamma}_{0}(\gamma-\tau)^{2\beta-2}G_{i}(\tau)d\tau\right) \end{eqnarray}$

(3.3)

where

$H_{i}(\tau) = f_{i}(\tau, x(\tau), y(\tau), \varphi_{1}x(\tau), \phi_{1}y(\tau))$

and

$G_{i}(\tau) = g_{i}(\tau, x(\tau), y(\tau), \varphi_{2}x(\tau), \phi_{2}y(\tau)).$

We obtain

$\varphi_{i} T_{1}(x, y)(t) = \int^{t}_{0}A_{i}(t, s)T_{1}(x, y)(s)ds, \ \phi_{i}T_{2}(x, y)(t) = \int^{t}_{0}B_{i}(t, s)T_{2}(x, y)(s)ds$

where $i = 1, 2.$

We shall now prove that T is contractive.

Let $T_{1}(x_{1}, y_{1}), T_{2} (x_{2}, y_{2})\in E$ . Then, for each $t\in [0, T]$ , we have

$\begin{eqnarray} |T_{1}(x_{1}, y_{1})-T_{1}(x_{2}, y_{2})|&\leq &\left[\dfrac{1}{\Gamma(\alpha)}\sum\limits_{i = 1}^{m}\int^{t}_{0}(t-\tau)^{\alpha-1}d\tau+\dfrac{t^{\alpha-1}}{\psi}\left(\sum\limits_{i = 1}^{m}\int^{T}_{0}(T-\tau)d\tau \right.\right. \\ &&\qquad \left.\left. -\dfrac{\theta}{\Gamma(2\alpha)}\sum\limits_{i = 1}^{m}\int^{\eta}_{0}(\eta-\tau)^{2\alpha-2}d\tau\right)\right]\\ &&\times\max\limits_{\tau\in[0, T]}\sum\limits_{i = 1}^{m} \left|\left(\begin{array}{ll} f_{i}(\tau, x_{1}(\tau), y_{1}(\tau), \varphi_{1}x_{1}(\tau), \phi_{1}y_{1}(\tau))& \\ -f_{i}(\tau, x_{2}(\tau), y_{2}(\tau), \varphi_{1}x_{2}(\tau), \phi_{1}y_{2}(\tau)) & \end{array}\right)\right|\\ &\leq &\dfrac{T^{\alpha}}{\Gamma(\alpha +1)}\max\limits_{\tau\in[0, T]}\sum\limits_{i = 1}^{m} \left|\left(\begin{array}{ll} f_{i}(\tau, x_{1}(\tau), y_{1}(\tau), \varphi_{1}x_{1}(\tau), \phi_{1}y_{1}(\tau)) & \\ -f_{i}(\tau, x_{2}(\tau), y_{2}(\tau), \varphi_{1}x_{2}(\tau), \phi_{1}y_{2}(\tau)) \end{array}\right)\right|. \end{eqnarray}$

By $(H2)$ , it follows that

$\begin{eqnarray*} ||T_{1}(x_{1}, y_{1})-T_{1}(x_{2}, y_{2})||&\leq &\dfrac{T^{\alpha}}{\Gamma(\alpha +1)}\sum\limits_{i = 1}^{m}(\xi_{1}^{i}+\xi_{2}^{i}+\xi_{3}^{i}+\xi_{4}^{i})\times\max(||x_{1}-x_{2}||, ||y_{1}-y_{2}||, \\ &&||\varphi_{1}(x_{1}-x_{2})||, ||\varphi_{2}(x_{1}-x_{2})||, ||\phi_{1}(y_{1}-y_{2})||, ||\phi_{2}(y_{1}-y_{2})||). \end{eqnarray*}$

Hence,

$\begin{equation} ||T_{1}(x_{1}, y_{1})-T_{1}(x_{2}, y_{2})||\leq A_{1}||x_{1}-x_{2}, y_{1}-y_{2}||_{E}. \end{equation}$

(3.4)

With the same arguments as before, we can show that

$\begin{equation} ||T_{2}(x_{1}, y_{1})-T_{2}(x_{2}, y_{2})||\leq A_{2}||x_{1}-x_{2}, y_{1}-y_{2}||_{E}. \end{equation}$

(3.5)

On the other hand, we have

$\begin{eqnarray*} ||\varphi_{1}(T_{1}(x_{1}, y_{1})-T_{1}(x_{2}, y_{2}))||&\leq &\int_{0}^{t}||A'_{1}(t, s)||||T_{1}(x_{1}, y_{1})-T_{1}(x_{2}, y_{2})||ds\\ &\leq &\max\limits_{t, s\in[0, 1]}||A'_{1}(t, s)||\times A_{1}||x_{1}-x_{2}, y_{1}-y_{2}||_{E}. \end{eqnarray*}$

Hence,

$\begin{equation} ||\varphi_{1}(T_{1}(x_{1}, y_{1})-T_{1}(x_{2}, y_{2}))||\leq A_{3}||x_{1}-x_{2}, y_{1}-y_{2}||_{E} \end{equation}$

(3.6)

and

$\begin{equation} ||\varphi_{2}(T_{1}(x_{1}, y_{1})-T_{1}(x_{2}, y_{2}))||\leq A_{4}||x_{1}-x_{2}, y_{1}-y_{2}||_{E}. \end{equation}$

(3.7)

Also, we have

$\begin{equation} ||\phi_{1}(T_{2}(x_{1}, y_{1})-T_{2}(x_{2}, y_{2}))||\leq A_{5}||x_{1}-x_{2}, y_{1}-y_{2}||_{E} \end{equation}$

(3.8)

and

$\begin{equation} ||\phi_{2}(T_{2}(x_{1}, y_{1})-T_{2}(x_{2}, y_{2}))||\leq A_{6}||x_{1}-x_{2}, y_{1}-y_{2}||_{E}. \end{equation}$

(3.9)

Thanks to (3.4)–(3.9), we get

$\begin{eqnarray} ||T(x_{1}, y_{1})-T(x_{2}, y_{2})|| & \leq \max(A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6})\\ & \times ||(x_{1}-x_{2}, y_{1}-y_{2})||_{E}. \end{eqnarray}$

(3.10)

Thanks to (3.10), we conclude that T is a contractive operator. Therefore, by Banach fixed point theorem, $T$ has a unique fixed point which is the solution of the system (1.1).

3.2. Uniqueness of solutions

Our second main result is based on Lemma 2.1. We have

Theorem 3.2. Assume that the hypotheses $(H1)$ and $(H3)$ are satisfied. Then, system (1.1) has at least a solution on $[0, T]$ .

Proof. The operator $T$ is continuous on $E$ in view of the continuity of $f_{i}$ and $g_{i}$ (hypothesis ( $H1$ )).

Now, we show that $T$ is completely continuous:

(i) First, we prove that $T$ maps bounded sets of $E$ into bounded sets of $E$ . Taking $\lambda > 0$ , and $(x, y)\in \Omega_{\lambda} = \{(x, y) \in E; ||(x, y)||\leq\lambda\}$ , then for each $t\in [0, T]$ , we have:

$\begin{eqnarray*} |T_{1}(x, y)|&\leq &\left[\dfrac{1}{\Gamma(\alpha)}\int^{t}_{0}(t-\tau)^{\alpha-1}d\tau+\dfrac{t^{\alpha-1}}{\psi}\left(\int^{T}_{0}(T-\tau)d\tau \right.\right. \\ \nonumber &&\qquad \left.\left. -\dfrac{\theta}{\Gamma(2\alpha)}\int^{\eta}_{0}(\eta-\tau)^{2\alpha-2}d\tau\right)\right]\\ &&\times\sup\limits_{t\in[0, T]}\sum\limits_{i = 1}^{m} |f_{i}(t, x(t), y(t), \varphi_{1}x(t), \phi_{1}y(t))|\\ &\leq & \left[\dfrac{T^{\alpha}}{\Gamma(\alpha+1)}+\dfrac{1}{\psi}\left(\dfrac{T^{\alpha+1}}{2}+\dfrac{\theta T^{3\alpha-2}}{(2\alpha-1)^{2}\Gamma(2\alpha-1)}\right)\right] \\ &&\times\sup\limits_{t\in[0, T]}\sum\limits_{i = 1}^{m} |f_{i}(t, x(t), y(t), \varphi_{1}x(t), \phi_{1}y(t))|, \end{eqnarray*}$

Thanks to ( $H3$ ), we can write

$||T_{1}(x, y)||\leq\left[\dfrac{T^{\alpha}}{\Gamma(\alpha+1)}+\dfrac{1}{\psi}\left(\dfrac{T^{\alpha+1}}{2}+\dfrac{\theta T^{3\alpha-2}}{(2\alpha-1)^{2}\Gamma(2\alpha-1)}\right)\right]\sum\limits_{i = 1}^{m} L_{i}.$

Thus,

$\begin{equation} ||T_{1}(x, y)||\leq \nu_{1} \sum\limits_{i = 1}^{m} L_{i}. \end{equation}$

(3.11)

As before, we have

$\begin{equation} ||T_{2}(x, y)||\leq \nu_{2} \sum\limits_{i = 1}^{m} K_{i}. \end{equation}$

(3.12)

On the other hand, for all $j = 1, 2$ , we get

$\begin{eqnarray*} |\phi_{j}T_{1}(x, y)(t)| = \left|\int_{0}^{t}A'_{j}(t, s)T_{1}(x, y)(s)ds\right| \leq \max\limits_{t, s\in[0, 1]}|A'_{j}(t, s)|\nu_{1}\sum\limits_{i = 1}^{m} L_{i}. \end{eqnarray*}$

This implies that

$\begin{eqnarray} ||\phi_{1}T_{1}(x, y)(t)||&\leq &\nu_{3}\sum\limits_{i = 1}^{m} L_{i}, \end{eqnarray}$

(3.13)

$\begin{eqnarray} ||\phi_{2}T_{1}(x, y)(t)||&\leq &\nu_{4}\sum\limits_{i = 1}^{m} L_{i}. \end{eqnarray}$

(3.14)

Similarly, we have

$\begin{eqnarray} ||\varphi_{1}T_{2}(x, y)(t)||&\leq &\nu_{5}\sum\limits_{i = 1}^{m} K_{i}, \end{eqnarray}$

(3.15)

$\begin{eqnarray} ||\varphi_{2}T_{2}(x, y)(t)||&\leq &\nu_{6}\sum\limits_{i = 1}^{m} K_{i}. \end{eqnarray}$

(3.16)

It follows from (3.11)–(3.16) that:

$\begin{eqnarray*} ||T(x, y)||_{E}\leq \max \left(\begin{array}{ll} \nu_{1} \sum\limits_{i = 1}^{m} L_{i}, \nu_{2} \sum\limits_{i = 1}^{m} K_{i}, \nu_{3}\sum\limits_{i = 1}^{m} L_{i}, &\\ \nu_{4}\sum\limits_{i = 1}^{m} L_{i}, , \nu_{5}\sum\limits_{i = 1}^{m}, \nu_{6}\sum\limits_{i = 1}^{m}& \end{array}\right). \end{eqnarray*}$

Thus,

$||T(x, y)||_{E} < \infty.$

(ii) Second, we prove that $T$ is equi-continuous:

For any $0 \leq t_{1} < t_{2} \leq T$ and $(x, y) \in\Omega_{\lambda}$ , we have

$\begin{eqnarray*} &&|T_{1}(x, y)(t_{2})-T_{1}(x, y)(t_{1})|\\ &&\qquad\leq \left[\dfrac{1}{\Gamma(\alpha)}\int^{t_{1}}_{0}(t_{2}-\tau)^{\alpha-1}-(t_{1}-\tau)^{\alpha-1}d\tau+\dfrac{1}{\Gamma(\alpha)}\int^{t_{2}}_{t_{1}}(t_{2}-\tau)^{\alpha-1}d\tau\right. \\ &&\qquad\quad\left.+\dfrac{t_{2}^{\alpha-1}-t_{1}^{\alpha-1}}{\psi}\left(\dfrac{T^{2}}{2}-\dfrac{\theta\eta^{2\alpha-1}}{\Gamma(2\alpha-1)^{2}\Gamma(2\alpha-1)}\right)\right]\\ &&\qquad\times\sup\limits_{t\in[0, T]}\sum\limits_{i = 1}^{m} |f_{i}(t, x(t), y(t), \varphi_{1}x(t), \phi_{1}y(t))|\\ &&\qquad\leq\left[\dfrac{2}{\Gamma(\alpha+1)}(t_{2}-t_{1})^{\alpha-1}+(t_{2}^{\alpha-1}-t_{1}^{\alpha-1})\left[\dfrac{T^{2}}{2\psi}\right.\right.\\ &&\qquad\quad\left.\left.-\dfrac{\theta\eta^{2\alpha-1}}{\psi\Gamma(2\alpha-1)^{2}\Gamma(2\alpha-1)}+\dfrac{1}{\Gamma(\alpha+1)}\right]\right]\times\sum\limits_{i = 1}^{m}L_{i}. \end{eqnarray*}$

Therefore,

$\begin{eqnarray} &&||T_{1}(x, y)(t_{2})-T_{1}(x, y)(t_{1})||_{E} \\ &&\quad\left[\dfrac{2}{\Gamma(\alpha+1)}(t_{2}-t_{1})^{\alpha-1}+(t_{2}^{\alpha-1}-t_{1}^{\alpha-1})\left[\dfrac{T^{2}}{2\psi}+\dfrac{1}{\Gamma(\alpha+1)}\right]\right]\times\sum\limits_{i = 1}^{m}L_{i}. \end{eqnarray}$

(3.17)

We also have

$\begin{eqnarray} &&||T_{2}(x, y)(t_{2})-T_{2}(x, y)(t_{1})||_{E} \\ &&\quad\left[\dfrac{2}{\Gamma(\beta+1)}(t_{2}-t_{1})^{\beta-1}+(t_{2}^{\beta-1}-t_{1}^{\beta-1})\left[\dfrac{T^{2}}{2\psi'}+\dfrac{1}{\Gamma(\beta+1)}\right]\right]\times\sum\limits_{i = 1}^{m}K_{i}. \end{eqnarray}$

(3.18)

On the other hand,

$\begin{eqnarray*} &&|\phi_{i}T_{1}(x, y)(t_{2})-\phi_{i}T_{1}(x, y)(t_{1})|\\ &&\quad\leq\left[\max\limits_{s\in[0, 1]}|A_{i}'(t_{2}, s)-A_{i}'(t_{1}, s)|+(t_{2}-t_{1})\max\limits_{s\in[0, 1]}|A_{i}'(t_{1}, s)|\right] \times\sup\limits_{s\in[0, 1]}|T_{1}(x, y)(s)|. \end{eqnarray*}$

Consequently, for all $i = 1, 2,$ we obtain

$\begin{eqnarray} &&||\phi_{i}T_{1}(x, y)(t_{2})-\phi_{i}T_{1}(x, y)(t_{1})||\\ &&\quad\leq\left[\max\limits_{s\in[0, 1]}|A_{i}'(t_{2}, s)-A_{i}'(t_{1}, s)|+(t_{2}-t_{1})\max\limits_{s\in[0, 1]}|A_{i}'(t_{1}, s)|\right] \nu_{1}\sum\limits_{i = 1}^{m}L_{i}. \end{eqnarray}$

(3.19)

Similarly,

$\begin{eqnarray} &&||\varphi_{i}T_{1}(x, y)(t_{2})-\varphi_{i}T_{1}(x, y)(t_{1})||\\ &&\quad\leq\left[\max\limits_{s\in[0, 1]}|B_{i}'(t_{2}, s)-B_{i}'(t_{1}, s)|+(t_{2}-t_{1})\max\limits_{s\in[0, 1]}|B_{i}'(t_{1}, s)|\right] \nu_{2}\sum\limits_{i = 1}^{m}K_{i}. \end{eqnarray}$

(3.20)

where $i = 1, 2$ . Using (3.17)–(3.20), we deduce that

$||T(x, y)(t_{2})-T(x, y)(t_{1})||_{E}\longrightarrow 0$

as $t_{2}\rightarrow t_{1}.$

Combining (i) and (ii), we conclude that $T$ is completely continuous.

(iii) Finally, we shall prove that the set $F$ defined by

$F = \{(x, y) \in E, (x, y) = \rho T (x, y), \ 0 < \rho < 1 \}$

is bounded.

Let $(x, y)\in F$ , then $(x, y) = \rho T (x, y)$ , for some $0 < \rho < 1$ . Thus, for each $t \in [0, T]$ , we have:

$\begin{equation} x(t) = \rho T_{1}(x, y)(t), \ y (t) = \rho T_{2}(x, y)(t) . \end{equation}$

(3.21)

Thanks to ( $H3$ ) and using (3.11) and (3.12), we deduce that

$\begin{equation} ||x||\leq\rho\nu_{1}\sum\limits_{i = 1}^{m}L_{i}, \ ||y|| \leq\rho\nu_{2} \sum\limits_{i = 1}^{m}K_{i} . \end{equation}$

(3.22)

Using (3.13)–(3.16), it yields that

$\begin{eqnarray} \left\{\begin{array}{llll} ||\phi_{1}x||\leq \rho\nu_{3} \sum_{i = 1}^{m}L_{i} & \\ ||\phi_{2}x||\leq \rho\nu_{4} \sum_{i = 1}^{m}L_{i} & \\ ||\varphi_{1}y||\leq \rho\nu_{5} \sum_{i = 1}^{m}K_{i} & \\ ||\varphi_{2}y||\leq \rho\nu_{6} \sum_{i = 1}^{m}K_{i} & \end{array}\right.. \end{eqnarray}$

(3.23)

It follows from (3.22) and (3.23) that

$\begin{eqnarray*} ||T(x, y)||_{E}\leq \rho\max \left(\begin{array}{ll} \nu_{1} \sum_{i = 1}^{m} L_{i}, \nu_{2} \sum_{i = 1}^{m} K_{i}, \nu_{3}\sum_{i = 1}^{m} L_{i}, &\\ \nu_{4}\sum_{i = 1}^{m} L_{i}, , \nu_{5}\sum_{i = 1}^{m}, \nu_{6}\sum_{i = 1}^{m}& \end{array}\right). \end{eqnarray*}$

Consequently,

$||(x, y)||_{E} < \infty.$

This shows that $F$ is bounded. By Lemma (2.1), we deduce that $T$ has a fixed point, which is a solution of (1.1).

4. Related examples

To illustrate our main results, we treat the following examples.

Example 4.1. Consider the following system:

$\begin{eqnarray} \left\{\begin{array}{llllll} D^{\frac{3}{2}}x(t) = \dfrac{\cos(\pi t)(x+y+\varphi_{1}x(t)+\phi_{1}y(t))}{10\pi(x+y+\varphi_{1}x(t)+\phi_{1}y(t))}+\dfrac{1}{32\pi^{2}e}&\\ \\ \qquad\qquad\qquad \left(\cos x(t)+\cos y(t)+\dfrac{\varphi_{1}x(t)+\phi_{1}y(t)}{4\pi}\right), &\\ \\ D^{\frac{3}{2}}y(t) = \dfrac{1}{8\pi^{3}(t+1)}\left(\dfrac{x+y+\varphi_{2}x(t)+\phi_{2}y(t)}{3+x+y+\varphi_{2}x(t)+\phi_{2}y(t)}\right)+\dfrac{1}{(10\pi+e^{t})e^{(t+1)}}&\\ \\ \qquad\qquad\qquad \left(\dfrac{\sin x(t)+\sin y(t)+\cos\varphi_{2}x(t)+\cos\phi_{2}y(t) }{2+\sin x(t)+\sin y(t)+\cos\varphi_{2}x(t)+\cos\phi_{2}y(t)}\right), \\ \\ I^{\frac{1}{2}}x(0) = 0, \ D^{-\frac{1}{2}}x(T) = I^{\frac{1}{2}}(x(1)), &\\ I^{\frac{1}{2}}y(0) = 0, \ D^{-\frac{1}{2}}y(T) = I^{\frac{1}{2}}(y(1)).& \end{array}\right. \end{eqnarray}$

(4.1)

We have

$\alpha = \dfrac{3}{2}, \ \beta = \dfrac{3}{2}, \ T = 1, \ \theta = 1, \ \omega = 1, \ \gamma = 1, \ m = 2, \ \eta = 1.$

Also,

$\begin{eqnarray} f_{1}(t, x(t), y(t), \varphi_{1}x(t), \phi_{1}y(t))& = & \dfrac{\cos(\pi t)(x+y+\varphi_{1}x(t)+\phi_{1}y(t))}{10\pi(1+x+y+\varphi_{1}x(t)+\phi_{1}y(t))}, \end{eqnarray}$

(4.2)

$\begin{eqnarray} f_{2}(t, x(t), y(t), \varphi_{1}x(t), \phi_{1}y(t))& = &\dfrac{1}{32\pi^{2}e}\left(\cos x(t)+\cos y(t)+\dfrac{\varphi_{1}x(t)+\phi_{1}y(t)}{4\pi}\right).\\ \end{eqnarray}$

(4.3)

For $t \in [0, 1]$ and $(x_{1}, y_{1}, \varphi_{1}x_{1}, \phi_{1}y_{1}), (x_{2}, y_{2}, \varphi_{1}x_{2}, \phi_{1}y_{2})\in\mathbb{R}^{4}$ , we have

$\begin{eqnarray} &&|f_{1}(t, x_{1}, y_{1}, \varphi_{1}x_{1}, \phi_{1}y_{1})-f_{1}(t, x_{2}, y_{2}, \varphi_{1}x_{2}, \phi_{1}y_{2})|\\ &&\leq\dfrac{|\cos(\pi t)|}{10\pi}\left|\dfrac{x_{1}+y_{1}+\varphi_{1}x_{1}+\phi_{1}y_{1}}{1+x_{1}+y_{1}+\varphi_{1}x_{1}+\phi_{1}y_{1}}-\dfrac{x_{2}+y_{2}+\varphi_{1}x_{2}+\phi_{1}y_{2})}{1+x_{2}+y_{2}+\varphi_{1}x_{2}+\phi_{1}y_{2})}\right|\\ &&\leq \dfrac{1}{10\pi}(|x_{1}-x_{2}|+|y_{1}-y_{2}|+|\varphi_{1}x_{1}-\varphi_{1}x_{2}|+|\phi_{1}y_{1}-\phi_{1}y_{2}|) \end{eqnarray}$

(4.4)

and

$\begin{eqnarray} &&|f_{2}(t, x_{1}, y_{1}, \varphi_{1}x_{1}, \phi_{1}y_{1})-f_{2}(t, x_{2}, y_{2}, \varphi_{1}x_{2}, \phi_{1}y_{2})|\\ &&\quad\leq \dfrac{1}{32\pi e}(|x_{1}-x_{2}|+|y_{1}-y_{2}|+|\varphi_{1}x_{1}-\varphi_{1}x_{2}|+|\phi_{1}y_{1}-\phi_{1}y_{2}|). \end{eqnarray}$

(4.5)

So, we can take

$\xi_{1}^{1} = \xi_{2}^{1} = \xi_{3}^{1} = \xi_{4}^{1} = \dfrac{1}{10\pi},$

$\xi_{1}^{2} = \xi_{2}^{2} = \xi_{3}^{2} = \xi_{4}^{2} = \dfrac{1}{32\pi e}.$

We also have

$\begin{eqnarray} g_{1}(t, x(t), y(t), \varphi_{2}x(t), \phi_{2}y(t))& = &\dfrac{1}{8\pi^{3}(t+1)}\left(\dfrac{x+y+\varphi_{2}x(t)+\phi_{2}y(t)}{3+x+y+\varphi_{2}x(t)+\phi_{2}y(t)}\right) \end{eqnarray}$

and

$\begin{eqnarray} && g_{2}(t, x(t), y(t), \varphi_{2}x(t), \phi_{2}y(t))\\ && = \dfrac{1}{(10\pi +e^{t})e^{t+1}} \left(\dfrac{\sin x(t)+\sin y(t)+\cos\varphi_{2}x(t)+\cos\phi_{2}y(t) }{2+\sin x(t)+\sin y(t)+\cos\varphi_{2}x(t)+\cos\phi_{2}y(t)}\right) \end{eqnarray}$

(4.6)

For $t \in [0, 1]$ and $(x_{1}, y_{1}, \varphi_{2}x_{1}, \phi_{2}y_{1}), (x_{2}, y_{2}, \varphi_{2}x_{2}, \phi_{2}y_{2})\in\mathbb{R}^{4}$ , we can write

$\begin{eqnarray} &&|g_{1}(t, x_{1}, y_{1}, \varphi_{2}x_{1}, \phi_{2}y_{1})-g_{1}(t, x_{2}, y_{2}, \varphi_{2}x_{2}, \phi_{2}y_{2})|\\ &&\quad \leq \dfrac{1}{8\pi^{3}}(|x_{1}-x_{2}|+|y_{1}-y_{2}|+|\varphi_{2}x_{1}-\varphi_{2}x_{2}|+|\phi_{2}y_{1}-\phi_{2}y_{2}|), \end{eqnarray}$

(4.7)

and

$\begin{eqnarray} &&|g_{2}(t, x_{1}, y_{1}, \varphi_{2}x_{1}, \phi_{2}y_{1})-g_{2}(t, x_{2}, y_{2}, \varphi_{2}x_{2}, \phi_{2}y_{2})|\\ &&\quad \leq \dfrac{1}{10\pi e^{2}}(|x_{1}-x_{2}|+|y_{1}-y_{2}|+|\varphi_{2}x_{1}-\varphi_{2}x_{2}|+|\phi_{2}y_{1}-\phi_{2}y_{2}|). \end{eqnarray}$

(4.8)

Hence,

$\chi_{1}^{1} = \chi_{2}^{1} = \chi_{3}^{1} = \chi_{4}^{1} = \dfrac{1}{8\pi^{3}},$

$\chi_{1}^{2} = \chi_{2}^{2} = \chi_{3}^{2} = \chi_{4}^{2} = \dfrac{1}{10\pi e^{2}}.$

Therefore,

$A_{1} = 0.0589009676, \; \; A_{2} = 0.0250930393.$

Suppose

$A'_{i} = B'_{i} = 1, \ i = 1, 2,$

so,

$A_{1} = A_{3} = A_{4}, \; \; A_{2} = A_{5} = A_{6}.$

Thus,

$\begin{equation} \max(A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}) < 1, \end{equation}$

(4.9)

and by Theorem 3.1, we conclude that the system (4.1) has a unique solution on $[0, 1]$ .

Example 4.2.

$\begin{eqnarray} \left\{\begin{array}{llllll} D^{\frac{3}{2}}x(t) = \dfrac{\pi( t+1)\sin(\varphi_{1}x(t)+\phi_{1}y(t))}{2-\cos(x(t)+y(t))}&\\ \\ \qquad\qquad\qquad +\dfrac{e^{t}}{2\pi+\cos(x(t)+\varphi_{1}x(t))+\sin(\sin(y(t)+\phi_{1}y(t))}, \ t\in[0, 1], &\\ \\ D^{\frac{4}{3}}y(t) = \dfrac{e^{2}\sin(x(t)+y(t))}{2\pi+\cos(\varphi_{2}x(t)+\phi_{2}y(t))}&\\ \\ \qquad\qquad\qquad +\dfrac{3 t^{2}\cos y(t)}{e^{t^{3}+1}-\cos( x(t)+y(t)-\varphi_{2}x(t)-\phi_{2}y(t))}, \ t\in[0, 1], \\ \\ I^{\frac{1}{2}}x(0) = 0, \ D^{-\frac{1}{2}}x(T) = I^{\frac{1}{2}}(x(1)), &\\ I^{\frac{2}{3}}y(0) = 0, \ D^{-\frac{2}{3}}y(T) = I^{\frac{1}{3}}(y(1)).& \end{array}\right. \end{eqnarray}$

(4.10)

We have

$\alpha = \dfrac{3}{2}, \ \beta = \dfrac{4}{3}, \ T = 1, \ \theta = 1, \ \omega = 1, \ \gamma = 1, \ m = 2, \ \eta = 1.$

Since

$\begin{eqnarray} |f_{1}(t, x(t), y(t), \varphi_{1}x(t), \phi_{1}y(t))|& = &\left|\dfrac{\pi( t+1)\sin(\varphi_{1}x(t)+\phi_{1}y(t))}{2-\cos(x(t)+y(t))}\right| \leq 2\pi , \\ |f_{2}(t, x(t), y(t), \varphi_{1}x(t), \phi_{1}y(t))|& = &\left|\dfrac{e^{t}}{2\pi+\cos(x(t)+\varphi_{1}x(t))+\sin(\sin(y(t)+\phi_{1}y(t))}\right| \leq \dfrac{e}{2\pi +2}, \\ |g_{1}(t, x(t), y(t), \varphi_{2}x(t), \phi_{2}y(t))|& = &\left|\dfrac{e^{2}\sin(x(t)+y(t))}{2\pi+\cos(\varphi_{2}x(t)+\phi_{2}y(t))}\right| \leq \dfrac{e^{2}}{2\pi +1}, \\ |g_{2}(t, x(t), y(t), \varphi_{2}x(t), \phi_{2}y(t))|& = &\left|\dfrac{3 t^{2}\cos y(t)}{e^{t^{3}+1}-\cos( x(t)+y(t)-\varphi_{2}x(t)-\phi_{2}y(t))}\right| \leq \dfrac{3}{e-1}. \end{eqnarray}$

The functions $f_{1}, \ f_{2}, \ g_{1}\ \text{and}\ g_{2}$ are continuous and bounded on $[0, 1]\times\mathbb{R}^{4}$ . So, by Theorem 3.2, the system (4.10) has at least one solution on $[0, 1]$ .

5. Conclusions

We have proved the existence of solutions for fractional differential equations with integral and multi-point boundary conditions. The problem is solved by applying some fixed point theorems. We also provide examples to make our results clear.

Conflict of interest

The authors declare that they have no conflicts of interest in this paper.

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