Semi-supervised multiscale dual-encoding method for faulty traffic data detection

Yongcan Huang; Jidong J. Yang; Yongcan Huang; Jidong J. Yang

doi:10.3934/aci.2022006

Applied Computing and Intelligence

2022, Volume 2, Issue 2: 99-114. doi: 10.3934/aci.2022006

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Research article

Semi-supervised multiscale dual-encoding method for faulty traffic data detection

Yongcan Huang ,
Jidong J. Yang ^,

Smart Mobility and Infrastructure Laboratory, College of Engineering, University of Georgia, Athens GA, USA

Received: 11 July 2022 Revised: 19 August 2022 Accepted: 12 September 2022 Published: 16 September 2022

Inspired by the recent success of deep learning in multiscale information encoding, we introduce a variational autoencoder (VAE) based semi-supervised method for detection of faulty traffic data, which is cast as a classification problem. Continuous wavelet transform (CWT) is applied to the time series of traffic volume data to obtain rich features embodied in time-frequency representation, followed by a twin of VAE models to separately encode normal data and faulty data. The resulting multiscale dual encodings are concatenated and fed to an attention-based classifier, consisting of a self-attention module and a multilayer perceptron. For comparison, the proposed architecture is evaluated against five different encoding schemes, including (1) VAE with only normal data encoding, (2) VAE with only faulty data encoding, (3) VAE with both normal and faulty data encodings, but without attention module in the classifier, (4) siamese encoding, and (5) cross-vision transformer (CViT) encoding. The first four encoding schemes adopt the same convolutional neural network (CNN) architecture while the fifth encoding scheme follows the transformer architecture of CViT. Our experiments show that the proposed architecture with the dual encoding scheme, coupled with attention module, outperforms other encoding schemes and results in classification accuracy of 96.4%, precision of 95.5%, and recall of 97.7%.

Keywords:

Citation: Yongcan Huang, Jidong J. Yang. Semi-supervised multiscale dual-encoding method for faulty traffic data detection[J]. Applied Computing and Intelligence, 2022, 2(2): 99-114. doi: 10.3934/aci.2022006

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Abstract

1. Introduction

The models of the differential and integral equations have been appeared in different applications (see ^{[1,2,3,4,6,7,9,12,13,14,15,16,17,18,19,20]}).

Boundary value problems involving fractional differential equations arise in physical sciences and applied mathematics. In some of these problems, subsidiary conditions are imposed locally. In some other cases, nonlocal conditions are imposed. It is sometimes better to impose nonlocal conditions since the measurements needed by a nonlocal condition may be more precise than the measurement given by a local condition. Consequently, a variety of excellent results on fractional boundary value problems (abbreviated BVPs) with resonant conditions have been achieved. For instance, Bai ^[4] studied a type of fractional differential equations with m-points boundary conditions. The existence of nontrivial solutions was established by using coincidence degree theory. Applying the same method, Kosmatov ^[17] investigated the fractional order three points BVP with resonant case.

Although the study of fractional BVPs at resonance has acquired fruitful achievements, it should be noted that such problems with Riemann-Stieltjes integrals are very scarce, so it is worthy of further explorations. Riemann-Stieltjes integral has been considered as both multipoint and integral in a single framework, which is more common, see the relevant works due to Ahmad et al. ^[1].

The boundary value problems with nonlocal, integral and infinite points boundary conditions have been studied by some authors (see, for example ^[8,10,11,12]).

Here, we discuss the boundary value problem of the nonlinear differential inclusions of arbitrary (fractional) orders

$\begin{equation} \frac{dx}{dt}\in F_{1}(t,x(t),I^{\gamma}f_{2}(t,D^{\alpha}x(t))),\; \; \alpha,\gamma\in(0,1] \; \; \; \; t\in(0,1) \end{equation}$

(1.1)

with the nonlocal boundary condition

$\begin{equation} {{\sum\limits^{m}_{k = 1} a_{k}}} x(\tau_{k} ) = x_{0},\; a_{k} > 0\; \; \tau_{k}\in[0,1], \end{equation}$

(1.2)

the integral condition

$\begin{equation} \int_{0}^{1}x(s)dg(s) = x_{0} \end{equation}$

(1.3)

and the infinite point boundary condition

$\begin{equation} {{\sum\limits^{\infty}_{k = 1} a_{k}}} x(\tau_{k} ) = x_{0},\; a_{k} > 0\; and\; \tau_{k}\in[0,1]. \end{equation}$

(1.4)

We study the existence of solutions $x\in C[0, 1]$ of the problems (1.1) and (1.2), and deduce the existence of solutions of the problem of (1.1) with the conditions (1.3) and (1.4). Then the existence of the maximal and minimal solutions will be proved. The sufficient condition for the uniqueness and continuous dependence of the solution will be studied.

This paper is organised as: In Section 2, we prove the existence of continuous solutions of the problems (1.1) and (1.2), and deduce the existence of solutions of the problem of (1.1) with the conditions (1.3) and (1.4). In Section 3, the existence of the maximal and minimal solutions is proved. In Section 4, the sufficient condition for the uniqueness and continuous dependence of the solution are studied. Next, in Section 5, we extend our results to the nonlocal problems (1.3) and (2.1). Finally, some existence results is proved for the nonlocal problems (1.4) and (2.1) in Section 6.

2. Main results

Consider the following assumptions:

(I) (i) The set $F_{1}(t, x, y)$ is nonempty, closed and convex for all $(t, x, y)\in [0, 1]\times R\times R$ .

(ii) $F_{1}(t, x, y)$ is measurable in $t\in [0, 1]$ for every $x, y\in R$ .

(iii) $F_{1}(t, x, y)$ is upper semicontinuous in $x$ and $y$ for every $t\in [0, 1]$ .

(iv) There exist a bounded measurable function $a_{1}: [0, 1]\longrightarrow R$ and a positive constant $K_{1}$ , such that

$\begin{eqnarray*} \|F_{1}(t,x,y)\|& = &\sup\{|f_{1}|:f_{1}\in F_{1}(t,x,y)\}\\ &\leq& |a_{1}(t)|+K_{1}(|x|+|y|). \end{eqnarray*}$

Remark 2.1. From the assumptions (i)–(iv) we can deduce that (see ^[3,6,7,13]) there exists $f_{1}\in F_{1}(t, x, y)$ , such that

(v) $f_{1}:[0, 1]\times R\times R\longrightarrow R$ is measurable in $t$ for every $x, y\in R$ and continuous in $x, y$ for $t\in [0, 1]$ , and there exist a bounded measurable function $a_{1}:[0, 1]\rightarrow R$ and a positive constant $K_{1} > 0$ such that

$|f_{1}(t,x,y)|\leq |a_{1}(t)|+K_{1}(|x|+|y|),$

and the functional $f_{1}$ satisfies the differential equation

$\begin{equation} \frac{dx}{dt} = f_{1}(t,x(t),I^{\gamma}f_{2}(t,D^{\alpha}x(t))),\; \; \alpha,\gamma\in(0,1]\; and\; t\in(0,1]. \end{equation}$

(2.1)

(II) $f_{2}:[0, 1]\times R\longrightarrow R$ is measurable in $t$ for any $x\in R$ and continuous in $x$ for $t\in [0, 1]$ , and there exist a bounded measurable function $a_{2}:[0, 1]\rightarrow R$ and a positive constant $K_{2} > 0$ such that

$|f_{2}(t,x)|\leq |a_{2}(t)|+K_{2}|x|,\; \forall t\in [0,1]\; \; \mbox {and}\; x\in R$

and

$\sup\limits_{t\in [0,1]}|a_{i}(t)|\leq a_{i}, i = 1,2.$

(III) $2K_{1}\gamma+K_{1}K_{2}\alpha < \alpha\gamma\Gamma(2-\alpha), \; \alpha, \gamma \in (0, 1].$

Remark 2.2. From (I) and (v) we can deduce that every solution of (1.1) is a solution of (2.1). Now, we shall prove the following lemma.

Lemma 2.1. If the solution of the problems (1.2)–(2.1) exists then it can be expressed by the integral equation

$\begin{eqnarray} y(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\limits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}} {\Gamma(\alpha)}y(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}} {\Gamma(\gamma)}f_{2}(\phi,y(\theta))d\theta)ds. \end{eqnarray}$

(2.2)

Proof. Consider the boundary value problems (1.2)–(2.1) be satisfied. Operating by $I^{1-\alpha}$ on both sides of (2.1) we can obtain

$\begin{equation} D^{\alpha}x(t) = I^{1-\alpha}\frac{dx}{dt} = I^{1-\alpha} f_{1}(t,x(t),I^{\gamma}f_{2}(t,D^{\alpha}x(t))). \end{equation}$

(2.3)

Taking

$\begin{equation} D^{\alpha}x(t) = y(t), \end{equation}$

(2.4)

then we obtain

$\begin{equation} x(t) = x(0)+I^{\alpha}y(t). \end{equation}$

(2.5)

Putting $t = \tau$ and multiplying (2.5) by $\Sigma_{k = 1}^{m}a_{k}$ , then we get

$\begin{equation} \Sigma_{k = 1}^{m}a_{k}x(\tau_{k}) = \Sigma_{k = 1}^{m}a_{k}x(0)+\Sigma_{k = 1}^{m}a_{k}I^{\alpha}y(\tau_{k}), \end{equation}$

(2.6)

$\begin{equation} x_{0} = \Sigma_{k = 1}^{m}a_{k}x(0)+\Sigma_{k = 1}^{m}a_{k}I^{\alpha}y(\tau_{k}) \end{equation}$

(2.7)

and

$\begin{equation} x(0) = \frac{1}{\Sigma_{k = 1}^{m}a_{k}}[x_{0}-\Sigma_{k = 1}^{m}a_{k}I^{\alpha}y(\tau_{k})]. \end{equation}$

(2.8)

Then

$\begin{equation} x(t) = \frac{1}{\Sigma_{k = 1}^{m}a_{k}}[x_{0}-\Sigma_{k = 1}^{m}a_{k}I^{\alpha}y(\tau_{k})]+I^{\alpha}y(t). \end{equation}$

(2.9)

Substituting (2.8) and (2.9) in (2.5), which completes the proof.

Theorem 2.1. Let assumptions (I)–(III) be satisfied. Then the integral equation (2.2) has at least one continuous solution.

Proof. Define a set $Q_{r}$ as

$Q_{r} = \{y\in C[0,1]:\|y\|\leq r\},\; \;$

$\begin{eqnarray*} r = \frac{\Gamma(\gamma+1)[a_{1}+K_{1}A|x_{0}|]+k_{1}a_{2}}{\alpha\gamma\Gamma(2-\alpha)-[2K_{1}\alpha+K_{1}K_{2}\gamma]},\; \; \; (\sum\limits^{m}_{k = 1}a_{k})^{-1} = A, \end{eqnarray*}$

and the operator $F$ by

$\begin{eqnarray*} Fy(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\limits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2}(\theta,y(\theta))d\theta)ds. \end{eqnarray*}$

For $y\in Q_{r}$ , then

$\begin{eqnarray*} |Fy(t)|& = &|\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\limits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}} {\Gamma(\alpha)}y(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}} {\Gamma(\gamma)}f_{2}(\theta,y(\theta))d\theta)ds|\\ &\leq& K_{1}\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)}[\frac{|x_{0}|}{\sum\limits^{m}_{k = 1}a_{k}}+\frac{\sum\limits^{m}_{k = 1} a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}|y(\theta)|d\theta}{\sum\limits^{m}_{k = 1}a_{k}}\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}|y(\theta)|d\theta+\int_{0}^{s}\frac{(s-\phi)^{\gamma-1}} {\Gamma(\gamma)}(K_{2}|y(\theta)|+|a_{2}(t)|)d\theta]+|a_{1}(t)|)ds\\ &\leq& \frac{1}{\Gamma(2-\alpha)}[K_{1}A|x_{0}|+\frac{K_{1}\|y\|}{\Gamma(\alpha+1)}+K_{1}\frac{\|y\|}{\Gamma(\alpha+1)} +\frac{K_{1}K_{2}\|y\|}{\Gamma(\gamma+1)}+\frac{K_{1}a_{2}}{\Gamma(\gamma+1)}+a_{1}]\\ &\leq& \frac{1}{\Gamma(2-\alpha)}[K_{1}A|x_{0}|+\frac{K_{1}r}{\Gamma(\alpha+1)}+K_{1}\frac{r}{\Gamma(\alpha+1)}+ \frac{K_{1}K_{2}r}{\Gamma(\gamma+1)}+\frac{K_{1}a_{2}}{\Gamma(\gamma+1)}+a_{1}]\\ &\leq& \frac{1}{\Gamma(2-\alpha)}[K_{1}A|x_{0}|+\frac{2K_{1}r}{\Gamma(\alpha+1)}+\frac{K_{1}K_{2}r}{\Gamma(\gamma+1)}+ \frac{K_{1}a_{2}}{\Gamma(\gamma+1)}+a_{1}]\; \leq r. \end{eqnarray*}$

Thus, the class of functions $\{Fy\}$ is uniformly bounded on $Q_{r}$ and $F:Q_{r}\rightarrow Q_{r}$ . Let $y\in Q_{r}$ and $t_{1}, t_{2}\in [0, 1]$ such that $|t_{2}-t_{1}| < \delta$ , then

$\begin{eqnarray*} &&|Fy(t_{2})-Fy(t_{1})|\\ & = &|\int_{0}^{t_{2}}\frac{(t_{2}-s)^{-\alpha}}{\Gamma(1-\alpha)} (f_{1}(s,x(s),I^{\gamma}f_{2}(s,y(s)))ds-\int_{0}^{t_{1}}\frac{(t_{1}-s)^{-\alpha}}{\Gamma(1-\alpha)} (f_{1}(s,x(s),I^{\gamma}f_{2}(s,y(s)))ds|\\ &\leq&|\int_{0}^{t_{2}}\frac{(t_{2}-s)^{-\alpha}}{\Gamma(1-\alpha)} (f_{1}(s,x(s),I^{\gamma}f_{2}(s,y(s)))ds-\int_{0}^{t_{1}}\frac{(t_{2}-s)^{-\alpha}}{\Gamma(1-\alpha)} (f_{1}(s,x(s),I^{\gamma}f_{2}(s,y(s)))ds\\ &&+\int_{0}^{t_{1}}\frac{(t_{2}-s)^{-\alpha}}{\Gamma(1-\alpha)} (f_{1}(s,x(s),I^{\gamma}f_{2}(s,y(s)))ds-\int_{0}^{t_{1}}\frac{(t_{1}-s)^{-\alpha}}{\Gamma(1-\alpha)} (f_{1}(s,x(s),I^{\gamma}f_{2}(s,y(s)))ds|\\ &\leq&\int_{t_{1}}^{t_{2}}\frac{(t_{2}-s)^{-\alpha}}{\Gamma(1-\alpha)} |(f_{1}(s,x(s),I^{\gamma}f_{2}(s,y(s)))|ds\\ &&+\int_{0}^{t_{1}}\frac{(t_{2}-s)^{-\alpha}-(t_{1}-s)^{-\alpha}}{\Gamma(1-\alpha)} |(f_{1}(s,x(s),I^{\gamma}f_{2}(s,y(s)))ds|\\ &\leq&\int_{t_{1}}^{t_{2}}\frac{(t_{2}-s)^{-\alpha}}{\Gamma(1-\alpha)} |(f_{1}(s,x(s),I^{\gamma}f_{2}(s,y(s)))|ds\\ &&+\int_{0}^{t_{1}}\frac{(t_{2}-s)^{\alpha}-(t_{1}-s)^{\alpha}}{\Gamma(1-\alpha)(t_{2}-s)^{\alpha}(t_{1}-s)^{\alpha}}. |(f_{1}(s,x(s),I^{\gamma}f_{2}(s,y(s)))|ds. \end{eqnarray*}$

Thus, the class of functions $\{Fy\}$ is equicontinuous on $Q_{r}$ and $\{Fy\}$ is compact operator by the Arzela-Ascoli Theorem ^[5].

Now we prove that $F$ is continuous operator. Let ${y_{n}}\subset Q_{r}$ be convergent sequence such that $y_{n}\rightarrow y$ , then

$\begin{eqnarray*} Fy_{n}(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{n}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{n}(\theta)d\theta, \int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2}(\theta,y_{n}(\theta)))ds. \end{eqnarray*}$

Using Lebesgue dominated convergence Theorem ^[5] and assumptions (iv)–(II) we have

$\begin{eqnarray*} \lim\limits_{n\to \infty} Fy_{n}(t)& = &\lim\limits_{n\to \infty}\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{n}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{n}(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2}(\theta,y_{n}(\theta)))ds\\ & = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}\lim\limits_{n\to \infty}y_{n}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}\lim\limits_{n\to \infty}y_{n}(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2}(\theta,\lim\limits_{n\to \infty}y_{n}(\theta)))ds\\ & = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2} (\theta,y(\theta)))ds\; = \; Fy(t). \end{eqnarray*}$

Then $F:Q_{r}\rightarrow Q_{r}$ is continuous, and by Schauder Fixed Point Theorem^[5] there exists at least one solution $y\in C[0, 1]$ of (2.2). Now

$\begin{eqnarray*} \frac{dx}{dt}& = &\frac{d}{dt}[x(0)+I^{\alpha}y(t)]\\ & = &\frac{d}{dt}I^{\alpha}I^{1-\alpha}f_{1}(t,x(t),I^{\gamma}f_{2}(t,y(t)))\\ & = &\frac{d}{dt}If_{1}(t,x(t),I^{\gamma}f_{2}(t,y(t)))\\ & = &f_{1}(t,x(t),I^{\gamma}f_{2}(t,y(t))). \end{eqnarray*}$

Putting $t = \tau$ and using (2.9), we obtain

$\begin{eqnarray*} &&x(t) = \frac{1}{\Sigma_{k = 1}^{m}a_{k}}[x_{0}-\Sigma_{k = 1}^{m}a_{k}I^{\alpha}y(\tau_{k})]+I^{\alpha}y(t),\\ &&\Sigma_{k = 1}^{m}a_{k}x(\tau_{k}) = \Sigma_{k = 1}^{m}a_{k}\frac{1}{\Sigma_{k = 1}^{m}a_{k}}[x_{0}-\Sigma_{k = 1}^{m}a_{k}I^{\alpha}y(\tau_{k})]+\Sigma_{k = 1}^{m}a_{k}I^{\alpha}y(\tau_{k}), \end{eqnarray*}$

then

$\begin{eqnarray*} &&\Sigma_{k = 1}^{m}a_{k}x(\tau_{k}) = x_{0}, \; a_{k} > 0\; and\; \tau_{k}\in[0,1]. \end{eqnarray*}$

This proves the equivalence between the problems (1.2)–(2.1) and the integral equation (2.2). Then there exists at least one solution $y\in C[0, 1]$ of the problems (1.2)–(2.1).

3. Maximal and minimal solutions

Here, we shall study the maximal and minimal solutions for the problems (1.2) and (2.1). Let $y(t)$ be any solution of (2.2), let $u(t)$ be a solution of (2.2), then $u(t)$ is said to be a maximal solution of (2.2) if it satisfies the inequality

$y(t)\leq u(t),\; \; \; t\in[0,1].$

A minimal solution can be defined by similar way by reversing the above inequality.

Lemma 3.1. Let the assumptions of Theorem 2.1 be satisfied. Assume that $x(t)$ and $y(t)$ are two continuous functions on $[0, 1]$ satisfying

$\begin{eqnarray*} y(t)&\leq&\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2}(\theta,y(\theta)))ds\; \; t\in[0,1],\\ x(t)&\geq&\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}x(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}x(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2} (\theta,x(\theta)))ds\; \; t\in[0,1], \end{eqnarray*}$

where one of them is strict.

Let functions $f_{1}$ and $f_{2}$ be monotonic nondecreasing in $y$ , then

$\begin{equation} y(t) < x(t),\; \; \; t > 0. \end{equation}$

(3.1)

Proof. Let the conclusion (3.1) be not true, then there exists $t_{1}$ with

$y(t_{1}) < x(t_{1}),\; \; \; t_{1} > 0\; \; \; \mbox{and}\; \; \; y(t) < x(t),\; \; \; 0 < t < t_{1}.$

Since $f_{1}$ and $f_{2}$ are monotonic functions in $y,$ then we have

$\begin{eqnarray*} y(t_{1})&\leq&\int_{0}^{t_{1}}\frac{(t_{1}-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2}(\theta,y(\theta)))ds\\ & < &\int_{0}^{t_{1}}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}x(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}x(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2} (\theta,x(\theta)))ds\\ & < &x(t_{1}),\; t_{1}\in[0,1]. \end{eqnarray*}$

This contradicts the fact that $y(t_{1}) = x(t_{1})$ , then $y(t) < x(t)$ . This completes the proof.

For the existence of the continuous maximal and minimal solutions for (2.1), we have the following theorem.

Theorem 3.1. Let the assumptions of Theorem 2.1 be hold. Moreover, if $f_{1}$ and $f_{2}$ are monotonic nondecreasing functions in $y$ for each $t\in [0, 1]$ , then Eq (2.1) has maximal and minimal solutions.

Proof. First, we should demonstrate the existence of the maximal solution of (2.1). Let $\epsilon > 0$ be given. Now consider the integral equation

$\begin{eqnarray} y_{\epsilon}(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1,\epsilon}\bigg(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}} {\Gamma(\alpha)}y_{\epsilon}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_\epsilon(\theta)d\theta,\int_{0}^{s} \frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2,\epsilon}(\theta,y_{\epsilon}(\theta))d\theta \bigg)ds,\; \; t\in[0,1], \end{eqnarray}$

where

$f_{1,\epsilon}(s,x_{\epsilon}(s),y_{\epsilon}(s) ) = f_{1}(s,x_{\epsilon}(s),y_{\epsilon}(s))+\epsilon,$

$f_{2,\epsilon}(s,x_{\epsilon}(s)) = f_{2}(s,x_{\epsilon}(s))+\epsilon.$

For $\epsilon_{2} > \epsilon_{1},$ we have

$\begin{eqnarray} y_{\epsilon_{2}}(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1,\epsilon_2}\bigg(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{\epsilon_{2}}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{\epsilon_{2}}(\theta)d\theta,\int_{0}^{s} \frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2,\epsilon_{2}}(\theta,y_{\epsilon_{2}}(\theta))d\theta\bigg )ds,\; \; t\in[0,1], \end{eqnarray}$

$\begin{eqnarray} y_{\epsilon_{2}}(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} \bigg[f_{1}\bigg(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{\epsilon_{2}}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{\epsilon_{2}}(\theta)d\theta,\int_{0}^{s} \frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}(f_{2}(\theta,y_{\epsilon_{2}}(\theta))+\epsilon_{2})d\theta \bigg)+\epsilon_2\bigg]ds,\; t\in[0,1]. \end{eqnarray}$

Also

$\begin{eqnarray*} y_{\epsilon_{1}}(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1,\epsilon_1}\bigg(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{\epsilon_{1}}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{\epsilon_{1}}(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^ {\gamma-1}}{\Gamma(\gamma)}f_{2,\epsilon_{1}}(\theta,y_{\epsilon_{1}}(\theta))d\theta\bigg)ds,\\ y_{\epsilon_{1}}(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} \bigg[f_{1}\bigg(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\phi)^{\alpha-1}} {\Gamma(\alpha)}y_{\epsilon_{1}}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{\epsilon_{1}}(\theta)d\theta,\int_{0}^{s} \frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}(f_{2}(\theta,y_{\epsilon_{1}}(\theta))+\epsilon_{1})d\theta \bigg)+\epsilon_1\bigg] ds \\ & > &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} \bigg[f_{1}\bigg(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}} {\Gamma(\alpha)}y_{\epsilon_{2}}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{\epsilon_{2}}(\theta)d\theta,\int_{0}^{s} \frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}(f_{2}(\theta,y_{\epsilon_{2}}(\tau))+\epsilon_{2})d\theta \bigg)+\epsilon_2\bigg]ds. \end{eqnarray*}$

Applying Lemma 3.1, we obtain

$y_{\epsilon_{2}} < y_{\epsilon_{1}},\; t\in[0,1].$

As shown before, the family of function $y_{\epsilon}(t)$ is equi-continuous and uniformly bounded, then by Arzela Theorem, there exist decreasing sequence $\epsilon_{n}$ , such that $\epsilon_{n}\rightarrow 0$ as $n\rightarrow \infty$ , and $u(t) = {\lim_{n\rightarrow \infty}}y_{\epsilon_{n}}(t)$ exists uniformly in $[0, 1]$ and denote this limit by $u(t)$ . From the continuity of the functions, $f_{2, \epsilon_n}(t, y_{\epsilon_n}(t))$ , we get $f_{2, \epsilon_n}(t, y_{\epsilon_n}(t))\longrightarrow f_{2}(t, y(t))$ as $n\rightarrow \infty$ and

$\begin{eqnarray*} &&u(t) = \lim\limits_{n\rightarrow \infty}y_{\epsilon_{n}}(t) = \int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} \bigg[f_{1}\bigg(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{\epsilon_{n}}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{\epsilon_{n}}(\theta)d\theta, \int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}(f_{2}(\theta,y_{\epsilon_{n}}(\theta))+\epsilon_{n})\bigg)+\epsilon_n \bigg]ds,\; \; t\in[0,1]. \end{eqnarray*}$

Now we prove that $u(t)$ is the maximal solution of (2.1). To do this, let $y(t)$ be any solution of (2.1), then

$\begin{eqnarray*} y(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}\bigg(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2}(\theta,y(\theta))d\theta \bigg)ds,\\ y_{\epsilon}(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} \bigg[f_{1}\bigg(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)} y_{\epsilon}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{\epsilon}(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}(f_{2} (\theta,y_{\epsilon}(\theta))+\epsilon)d\theta \bigg)+\epsilon \bigg]ds \end{eqnarray*}$

and

$\begin{eqnarray*} y_{\epsilon}(t)& > &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}\bigg(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{\epsilon}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y_{\epsilon}(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)} f_{2}(\theta,y_{\epsilon}(\theta))\bigg)ds. \end{eqnarray*}$

Applying Lemma 3.1, we obtain

$y(t) < y_{\epsilon}(t), t\in [0,1].$

From the uniqueness of the maximal solution it clear that $y_{\epsilon}(t)$ tends to $u(t)$ uniformly in $[0, 1]$ as $\epsilon\rightarrow 0.$

By a similar way as done above, we can prove the existence of the minimal solution.

4. Uniqueness of the solution

Here, we study the sufficient condition for the uniqueness of the solution $y\in C[0, 1]$ of problems (1.2) and (2.1). Consider the following assumptions:

( $I^{*}$ ) (i) The set $F_{1}(t, x, y)$ is nonempty, closed and convex for all $(t, x, y)\in [0, 1]\times R\times R$ .

(ii) $F_{1}(t, x, y)$ is measurable in $t\in [0, 1]$ for every $x, y\in R$ .

(iii) $F_{1}$ satisfies the Lipschitz condition with a positive constant $K_{1}$ such that

$H(F_{1}(t,x_{1},y_{1}),F_{1}(t,x_{2},y_{2})|\leq K_{1}(|x_{1}-x_{_{2}}|+|y_{1}-y_{2}|),$

where $H(A, B)$ is the Hausdorff metric between the two subsets $A, B\in [0, 1]\times E.\;$

Remark 4.1. From this assumptions we can deduce that there exists a function $f_{1}\in F_{1}(t, x, y)$ , such that

(iv) $f_{1}:[0, 1]\times R\times R \rightarrow R$ is measurable in $t\in [0, 1]$ for every $x, y\in R$ and satisfies Lipschitz condition with a positive constant $K_{1}$ such that (see ^[3,7])

$|f_{1}(t,x_{1},y_{1})-f_{1}(t,x_{2},y_{2})|\leq K_{1}(|x_{1}-x_{2}|+|y_{1}-y_{2}|).$

$(II^{*}) \; f_{2}:[0, 1]\times R\longrightarrow R$ is measurable in $t \in [0, T]$ and satisfies Lipschitz condition with positive constant $K_{2}$ , such that

$|f_{2}(t,x)-f_{2}(t,y)|\leq K_{2}|x-y|.$

From the assumption ( $I^{*}$ ), we have

$|f_{1}(t,x,y)|-|f_{1}(t,0,0)|\leq |f_{1}(t,x,y)-f_{1}(t,0,0)|\leq K_{1}(|x|+|y|).$

Then

$\begin{eqnarray*} |f_{1}(t,x,y)|&\leq& K_{1}(|x|+|y|)+|f_{1}(t,0,0)|\\ &\leq& K_{1}(|x|+|y|)+\mid a_{1}(t)\mid, \end{eqnarray*}$

where $|a_{1}(t)| = \sup\limits_{t\in I}|f_{1}(t, 0, 0)|.$

From the assumption ( $II^{*}$ ), we have

$|f_{2}(t,y)|-|f_{2}(t,0)|\leq|f_{2}(t,y)-f_{2}(t,0)|\leq K_{2}|y|.$

Then

$\begin{eqnarray*} |f_{2}(t,x)|&\leq& K_{2}|x|+|f_{2}(t,0)|\\ &\leq& K_{2}|x|+\mid a_{2}(t)\mid, \end{eqnarray*}$

where $|a_{2}(t)| = \sup\limits_{t\in I}|f_{1}(t, 0)|$ .

Theorem 4.1. Let the assumptions $(I^{*})$ and $(II^{*})$ be satisfied. Then the solution of the problems (1.2) and (2.1) is unique.

Proof. Let $y_{1}(t)$ and $y_{2}(t)$ be solutions of the problems (1.2) and (2.1), then

Then

$\|y_{1}-y_{2}\|[\alpha\gamma\Gamma(2-\alpha)-(2K_{1}\alpha+K_{1}K_{2}\gamma)] < 0.$

Since $(\alpha\gamma\Gamma(2-\alpha)-(2K_{1}\alpha+K_{1}K_{2}\gamma)) < 1$ , then $y_{1}(t) = y_{2}(t)$ and the solution of (1.2) and (2.1) is unique.

4.1. Continuous dependence of the solution

Definition 4.1. The unique solution of the problems (1.2) and (2.1) depends continuously on initial data $x_{0}$ , if $\epsilon > 0$ , $\exists \delta > 0$ , such that

$|x_{0}-x^{*}_{0}|\leq \delta\; \; \; \; \Rightarrow \|y-y^{*}\|\leq \epsilon,$

where $y^{*}$ is the unique solution of the integral equation

$\begin{eqnarray*} y^{*}(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x^{*}_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y^{*}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y^{*}(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}} {\Gamma(\gamma)}f_{2}(\theta,y^{*}(\theta))d\theta)ds|. \end{eqnarray*}$

Theorem 4.2. Let the assumptions $(I^{*})$ and $(II^{*})$ be satisfied, then the unique solution of (1.2) and (2.1) depends continuously on $x_{0}.$

Proof. Let $y(t)$ and $y^{*}(t)$ be the solutions of problems (1.2) and (2.1), then

then we obtain

$\begin{eqnarray*} &&|y(t)-y^{*}(t)|\leq(AK_{1}\delta)(\alpha\gamma\Gamma(2-\alpha)-(2K_{1}\alpha+K_{1}K_{2}\gamma))^{-1}\leq\epsilon \end{eqnarray*}$

and

$\begin{eqnarray*} &&\|y(t)-y^{*}(t)\|\leq\epsilon. \end{eqnarray*}$

Definition 4.2. The unique solution of the problems (1.2) and (2.1) depends continuously on initial data $a_{k}$ , if $\epsilon > 0$ , $\exists \delta > 0$ , such that

$\sum\limits^{m}_{k = 1}|a_{k}-a_{k}^{*}|\leq\delta\; \; \; \; \Rightarrow \|y-y^{*}\|\leq \epsilon,$

where $y^{*}$ is the unique solution of the integral equation

$\begin{eqnarray*} y^{*}(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}^{*}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y^{*}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y^{*}(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)} f_{2}(\theta,y^{*}(\theta))d \theta)ds|. \end{eqnarray*}$

Theorem 4.3. Let the assumptions $(I^{*})$ and $(II^{*})$ be satisfied, then the unique solution of problems (1.2) and (2.1) depends continuously on $a_{k}$ .

Proof. Let $y(t)$ and $y^{*}(t)$ be the solutions of problems (1.2) and (2.1) and $(\sum\nolimits^{m}_{k = 1}a^{*}_{k})^{-1} = A^{*}$ ,

$\begin{eqnarray*} |y(t)-y^{*}(t)|& = &|\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma (\alpha)}y(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^ {\gamma-1}}{\Gamma(\gamma)}f_{2}(\theta,y(\theta))d \theta)ds \\ &&-\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{m}_{k = 1}a^{*}_{k}}[x_{0}-\sum\limits^{m}_{k = 1}a^{*}_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y^{*}(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y^{*}(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}} {\Gamma(\gamma)}f_{2}(\theta,y^{*}(\theta))d \theta)ds|\\ &\leq&\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)}[\frac{|x_{0}|(\sum\nolimits^{m}_{k = 1}a^{*}_{k}-\sum\nolimits^{m}_{k = 1}a_{k})}{\sum\nolimits^{m}_{k = 1}a_{k}\sum\nolimits^{m}_{k = 1}a^{*}_{k}}\\ &&+\frac{\sum\nolimits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}|y(\theta)|d\theta}{\sum\nolimits^{m}_{k = 1}a_{k}} -\frac{\sum\nolimits^{m}_{k = 1}a^{*}_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}|y^{*}(\theta)|d\theta}{\sum\nolimits^{m}_{k = 1}a^{*}_{k}}\\ &&+K_{1}\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}|y(\theta)-y^{*}(\theta)|d\theta+K_{1}K_{2} \int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}|y(\phi)-y^{*}(\theta)|d\theta]ds\\ &\leq&\frac{1}{\Gamma(2-\alpha)}[\frac{K_{1}|x_{0}|\sum\nolimits^{m}_{k = 1}|a^{*}_{k}-a_{k}|}{\sum\nolimits^{m}_{k = 1}a_{k}\sum\nolimits^{m}_{k = 1}a^{*}_{k}}\\ &&+\frac{K_{1}\sum\nolimits^{m}_{k = 1}a_{k}(\sum\nolimits^{m}_{k = 1}a^{*}_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}|y(\theta)|d\theta- \sum\nolimits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}|y(\theta)|d\theta)}{\sum\nolimits^{m}_{k = 1}a_{k}\sum\nolimits^{m}_{k = 1}a^{*}_{k}}\\ &&+\frac{K_{1}\sum\nolimits^{m}_{k = 1}a^{*}_{k}(\sum\nolimits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}|y(\theta)|d\theta- \sum\nolimits^{m}_{k = 1}a^{*}_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}|y(\theta)|d\theta)}{\sum\nolimits^{m}_{k = 1}a_{k}\sum\nolimits^{m}_{k = 1}a^{*}_{k}}\\ &&+K_{1}\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}|y(\theta)-y^{*}(\theta)|d\theta+K_{1}K_{2} \int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}|y(\phi)-y^{*}(\theta)|d\theta]ds\\ &\leq&\frac{1}{\Gamma(2-\alpha)}[\frac{K_{1}|x_{0}|\delta}{\sum\nolimits^{m}_{k = 1}a_{k}\sum\nolimits^{m}_{k = 1}a^{*}_{k}}+ \frac{K_{1}(\sum\nolimits^{m}_{k = 1}a_{k}\delta\frac{r}{\Gamma(\alpha+1)}+\sum\nolimits^{m}_{k = 1}a^{*}_{k}\delta\frac{r}{\Gamma(\alpha+1)})}{\sum\nolimits^{m}_{k = 1}a_{k}\sum\nolimits^{m}_{k = 1}a^{*}_{k}}\\ &&+\frac{K_{1}\|y-y^{*}\|}{\Gamma(\alpha+1)}+\frac{K_{1}K_{2}\|y-y^{*}\|}{\Gamma(\gamma+1)}\\ &\leq&\frac{K_{1}\delta|x_{0}|}{\Gamma(2-\alpha)\sum\nolimits^{m}_{k = 1}a_{k}\sum\nolimits^{m}_{k = 1}a^{*}_{k}}+ \frac{K_{1}(\sum\nolimits^{m}_{k = 1}a_{k}\delta\frac{r}{\Gamma(\alpha+1)}+\sum\nolimits^{m}_{k = 1}a^{*}_{k}\delta\frac{r}{\Gamma(\alpha+1)})}{\Gamma(2-\alpha)\Gamma(\alpha+1)\sum\nolimits^{m}_{k = 1}a_{k} \sum\nolimits^{m}_{k = 1}a^{*}_{k}}\\ &&+\frac{K_{1}\|y-y^{*}\|}{\Gamma(2-\alpha)\Gamma(\alpha+1)}+\frac{K_{1}K_{2}\|y-y^{*}\|}{\Gamma(\gamma+1)\Gamma(2-\alpha)}, \end{eqnarray*}$

then we obtain

$\begin{eqnarray*} |y(t)-y^{*}(t)|\leq(AA^{*}K_{1}\delta [|x_{0}|+ r(\sum\limits^{m}_{k = 1}a_{k}+\sum\limits^{m}_{k = 1}a^{*}_{k})]).\; (\alpha\gamma\Gamma(2-\alpha)-(2K_{1}\alpha+K_{1}K_{2}\gamma))^{-1}\leq\epsilon \end{eqnarray*}$

and

$\begin{eqnarray*} &&\|y(t)-y^{*}(t)\|\leq\epsilon. \end{eqnarray*}$

5. Riemann-Stieltjes integral condition

Let $y\in C[0, 1]$ be the solution of the nonlocal boundary value problems (1.2) and (2.1). Let $a_{k} = (g(t_{k})-g(t_{k-1})$ , $g$ is increasing function, $\tau_{k}\in(t_{k-1}-t_{k})$ , $0 = t_{0} < t_{1} < t_{2}, ... < t_{m} = 1$ , then, as $m\longrightarrow \infty$ the nonlocal condition (1.2) will be

$\sum\limits^{m}_{k = 1}(g(t_{k})-g(t_{k-1})y(\tau_{k}) = x_{0}.$

As the limit $m\longrightarrow \infty$ , we obtain

$\lim\limits_{m\rightarrow \infty}\sum\limits^{m}_{k = 1}(g(t_{k})-g(t_{k-1})y(\tau_{k}) = \int_{0}^{1}y(s)dg(s) = x_{0}.$

Theorem 5.1. Let the assumptions (I)–(III) be satisfied. If $\sum\nolimits^{m}_{k = 1}a_{k}$ be convergent, then the nonlocal boundary value problems of (1.3) and (2.1) have at least one solution given by

$\begin{eqnarray*} y(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{g(1)-g(0)}[x_{0}-\int_{0}^{1}\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta dg(s)]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2} (\phi,y(\theta))d\theta)ds. \end{eqnarray*}$

Proof. As $m\rightarrow \infty$ , the solution of the nonlocal boundary value problem (2.1) will be

$\begin{eqnarray*} \lim\limits_{m\rightarrow \infty}y(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,(g(1)-g(0))^{-1}x_{0}\\ &&-(g(1)-g(0))^{-1}.\lim\limits_{m\rightarrow \infty}\sum\limits^{m}_{k = 1}[g(t_{k})-g(t_{k-1}]\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)} f_{2}(\phi,y(\theta))d\theta)ds\\ & = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{g(1)-g(0)}[x_{0}-\int_{0}^{1}\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta dg(s)]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)} f_{2}(\phi,y(\theta))d\theta)ds. \end{eqnarray*}$

6. Infinite-point boundary condition

Theorem 6.1. Let the assumptions (I)–(III) be satisfied, then the nonlocal boundary value problems of (1.4) and (2.1) have at least one solution given by

$\begin{eqnarray} y(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{\infty}_{k = 1}a_{k}}[x_{0}-\sum\limits^{\infty}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}} {\Gamma(\alpha)}y(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)} f_{2}(\phi,y(\theta))d\theta)ds. \end{eqnarray}$

Proof. Let the assumptions of Theorem 2.1 be satisfied. Let $\sum\nolimits^{m}_{k = 1} a_{k}$ be convergent, then take the limit to (1.4), we have

$\begin{eqnarray} \lim\limits_{m\rightarrow \infty}y(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\lim\limits_{m\rightarrow m}\frac{1}{\sum\nolimits^{m}_{k = 1}a_{k}}[x_{0}-\lim\limits_{m\rightarrow \infty}\sum\limits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta]\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2}(\phi,y(\theta))d\theta)ds. \end{eqnarray}$

Now

$\begin{eqnarray*} |a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta| \leq|a_{k}|\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}|y(\theta)|d\theta\leq\frac{|a_{k}|\|y\|}{\Gamma(\alpha+1)} \leq\frac{|a_{k}|r}{\Gamma(\alpha+1)} \end{eqnarray*}$

and by the comparison test $(\sum\nolimits^{m}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta)$ is convergent,

$\begin{eqnarray*} y(t)& = &\int_{0}^{t}\frac{(t-s)^{-\alpha}}{\Gamma(1-\alpha)} f_{1}(s,\frac{1}{\sum\nolimits^{\infty}_{k = 1}a_{k}}[x_{0}-\sum\limits^{\infty}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta]\nonumber\\ &&+\int_{0}^{s}\frac{(s-\theta)^{\alpha-1}}{\Gamma(\alpha)}y(\theta)d\theta,\int_{0}^{s}\frac{(s-\theta)^{\gamma-1}}{\Gamma(\gamma)}f_{2}(\phi,y(\theta))d\theta)ds.\nonumber \end{eqnarray*}$

Furthermore, from (2.9) we have

$\begin{eqnarray*} \sum\limits^{\infty}_{k = 1}a_{k}x(\tau_{k})& = &\sum\limits^{\infty}_{k = 1}a_{k}[\frac{1}{\sum\nolimits^{\infty}_{k = 1}a_{k}}(x_{0}-\sum\limits^{\infty}_{k = 1}a_{k}\int_{0}^{\tau_{k}}\frac{(\tau_{k}-s)^{\alpha-1}}{\Gamma(\alpha)}y(s)ds)]\\ &&+\int_{0}^{\tau_{k}}\frac{(\tau_{k}-s)^{\alpha-1}}{\Gamma(\alpha)}y(s)ds)]\\ & = &x_{0}. \end{eqnarray*}$

Example 6.1. Consider the following nonlinear integro-differential equation

$\begin{equation} \frac{dx}{dt} = t^{4}e^{-t}+\frac{x(t)}{\sqrt{t+2}}+\frac{1}{3}I^{\gamma}(\cos(5t+1)+\frac{1}{9}[t^{5}\sin D^{\frac{1}{3}}x(t)+e^{-3t}x(t)]) \end{equation}$

(6.1)

with boundary condition

$\begin{equation} {\sum\limits^{m}_{k = 1}\bigg[\frac{1}{k}-\frac{1}{k+1}\bigg]} x(\tau_k ) = x_{0},\; a_{k} > 0\; \; \tau_{k}\in[0,1]. \end{equation}$

(6.2)

Let

$\begin{eqnarray*} f_{1}(t,x(t),I^{\gamma}f_{2}(t,D^{\alpha}x(t))) = t^{4}e^{-t}+\frac{x(t)}{\sqrt{t+2}}+\frac{1}{3}I^{\gamma}(\cos(5t+1)+\frac{1}{9}[t^{5}\sin D^{\frac{1}{3}}x(t)+e^{-3t}x(t)]), \end{eqnarray*}$

then

$\begin{eqnarray*} |f_{1}(t,x(t),I^{\gamma}f_{2}(t,D^{\alpha}x(t))) = |t^{4}e^{-t}+\frac{x(t)}{\sqrt{2t+4}}+\\ \frac{1}{4}(\frac{4}{3}I^{\gamma}(\cos(5t+1)+\frac{1}{9}[t^{5}\sin D^{\frac{1}{3}}x(t)+e^{-3t}x(t)]))| \end{eqnarray*}$

and also

$\begin{eqnarray*} &&|I^{\gamma}f_{2}(t,D^{\alpha}x(t))|\leq \frac{1}{4}(\frac{4}{3}I^{\gamma}|\cos(5t+1)+\frac{1}{3}[t^{5}\sin D^{\frac{1}{3}}x(t)+e^{-3t}x(t)]|. \end{eqnarray*}$

It is clear that the assumptions (I) and (II) of Theorem 2.1 are satisfied with $a_{1}(t) = t^{4}e^{-t}\in L^{1}[0, 1]$ , $a_{2}(t) = \frac{4}{3}I^{\gamma}|(\cos(5t+1)|\in L^{1}[0, 1]$ , and let $\alpha = \frac{1}{3}$ , $\gamma = \frac{2}{3}$ , then $2K_{1}\gamma+K_{1}K_{2}\alpha = \frac{1}{2} < \alpha\gamma\Gamma(2-\alpha) = \frac{1}{2}\Gamma(2-\alpha)$ . Therefore, by applying Theorem 2.1, the nonlocal problems (6.1) and (6.2) has a continuous solution.

7. Conclusions

In this paper, we have studied a boundary value problem of fractional order differential inclusion with nonlocal, integral and infinite points boundary conditions. We have prove some existence results for that a single nonlocal boundary value problem, in of proving some existence results for a boundary value problem of fractional order differential inclusion with nonlocal, integral and infinite points boundary conditions. Next, we have proved the existence of maximal and minimal solutions. Then we have established the sufficient conditions for the uniqueness of solutions and continuous dependence of solution on some initial data and on the coefficients $a_k$ are studied. Finally, we have proved the existence of a nonlocal boundary value problem with Riemann-Stieltjes integral condition and with infinite-point boundary condition. An example is given to illustrate our results.

Conflict of interest

The authors declare no conflict of interest.

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