Markov-switching threshold stochastic volatility models with regime changes

Ahmed Ghezal; Mohamed balegh; Imane Zemmouri; Ahmed Ghezal; Mohamed balegh; Imane Zemmouri

doi:10.3934/math.2024192

AIMS Mathematics

2024, Volume 9, Issue 2: 3895-3910. doi: 10.3934/math.2024192

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Research article Special Issues

Markov-switching threshold stochastic volatility models with regime changes

1.
Department of Mathematics, Abdelhafid Boussouf University Center of Mila, Algeria
2.
Department of Mathematics, College of Science and Arts, Muhayil, King Khalid University, Abha 61413, Saudi Arabia
3.
Department of Mathematics, University of Annaba, Elhadjar $23$ , Annaba, Algeria

Received: 16 November 2023 Revised: 18 December 2023 Accepted: 05 January 2024 Published: 11 January 2024
MSC : 60G10, 62F12

This paper introduces a comprehensive class of models known as Markov-Switching Threshold Stochastic Volatility (MS-TSV) models, specifically designed to address asymmetry and the leverage effect observed in the volatility of financial time series. Extending the classical threshold stochastic volatility model, our approach expresses the parameters governing log-volatility as a function of a homogeneous Markov chain with a finite state space. The primary goal of our proposed model is to capture the dynamic behavior of volatility driven by a Markov chain, enabling the accommodation of both gradual shifts due to economic forces and sudden changes caused by abnormal events. Following the model's definition, we derive several probabilistic properties of the MS-TSV models, including strict (or second-order) stationarity, causality, ergodicity, and the computation of higher-order moments. Additionally, we provide the expression for the covariance function of the squared (or powered) process. Furthermore, we establish the limit theory for the Quasi-Maximum Likelihood Estimator (QMLE) and demonstrate the strong consistency of this estimator. Finally, a simulation study is presented to assess the performance of the proposed estimation method.

Keywords:

Citation: Ahmed Ghezal, Mohamed balegh, Imane Zemmouri. Markov-switching threshold stochastic volatility models with regime changes[J]. AIMS Mathematics, 2024, 9(2): 3895-3910. doi: 10.3934/math.2024192

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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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Markov-switching threshold stochastic volatility models with regime changes

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Markov-switching threshold stochastic volatility models with regime changes

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4. Related examples

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