On fractional integro-differential inclusions via the extended fractional Caputo–Fabrizio derivation

Abstract

We first show that four fractional integro-differential inclusions have solutions. Also, we show that dimension of the set of solutions for the second fractional integro-differential inclusion problem is infinite dimensional under some different conditions.

1 Introduction

A lot of papers on fractional differential equations (see, for example, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18] and the references therein) have been published. As you know, most famous fractional derivations are the Caputo and Riemann–Liouville derivations. In 2015, Caputo and Fabrizio introduced a new fractional derivation without singular kernel [19]. Some researchers published some works about solving different equations including the new derivation (see, for example, [2, 3, 10, 20,21,22,23,24,25]). Some researchers investigated some results on dimension of the set of solutions for some fractional differential inclusions (see, for example, [26]).

Let $b>0$, $u\in H^{1}(0,b)$, and $\zeta \in (0,1)$. As you know, the Caputo–Fabrizio fractional derivative of order ζ is defined by

$$ {}^{\mathrm{CF}}{\mathcal{D}}^{\zeta }u(t)=\frac{(2-\zeta )M(\zeta )}{2(1-\zeta )} \int _{0}^{t}\exp \biggl(\frac{-\zeta }{1-\zeta }(t-s) \biggr)u^{\prime }(s)\,ds, $$

where $t\geq 0$ and $M(\zeta )$ is a normalization constant depending on ζ such that $M(0)= M(1)=1$ [19]. Losada and Nieto showed that ${}^{\mathrm{CF}}\mathcal{I}^{\zeta } u(t)=\frac{2(1-\zeta )}{(2- \zeta )M(\zeta )}u(t) +\frac{\zeta }{(2-\zeta )M(\zeta )}\int _{0}^{t} u(s) \,ds$ [27]. Also, they showed that $M(\zeta )=\frac{2}{2- \zeta }$ [27]. Hence, the fractional Caputo–Fabrizio derivative of order ζ is given by ${}^{\mathrm{CF}}{\mathcal{D}}^{\zeta }u(t)=\frac{1}{1- \zeta }\int _{0}^{t}\exp (-\frac{\zeta }{1-\zeta }(t-s))u^{\prime }(s)\,ds$, when $t\geq 0$ and $0<\zeta <1$ [27]. If $n\geq 1$ and $\zeta \in (0,1)$, then the fractional derivative ${}^{\mathrm{CF}}{\mathcal{D}} ^{\zeta +n}$ of order $n+\zeta$ is defined by ${}^{\mathrm{CF}}{\mathcal{D}} ^{\zeta +n}u:={}^{\mathrm{CF}}{\mathcal{D}}^{\zeta }({\mathcal{D}}^{n}u(t))$ [27]. Let $u,v\in H^{1}(0,1)$ and $\zeta \in (0,1)$. If $u^{(s)}(0)=0$ for all $s=1,2,\ldots ,n$, then ${}^{\mathrm{CF}}{\mathcal{D}} ^{\zeta }({}^{\mathrm{CF}}{\mathcal{D}}^{n}(u(t))={}^{\mathrm{CF}}{\mathcal{D}}^{n}({}^{\mathrm{CF}} {\mathcal{D}}^{\zeta }(u(t))$. Also, we have $\lim_{\zeta \to 0}{}^{\mathrm{CF}} {\mathcal{D}}^{\zeta }u(t)=u(t)-u(0)$, $\lim_{\zeta \to 1}{}^{\mathrm{CF}}{\mathcal{D}} ^{\zeta }u(t)=u(t)^{\prime }$, and ${}^{\mathrm{CF}}{\mathcal{D}}^{\zeta } ( \lambda u(t)+\gamma v(t) )=\lambda {}^{\mathrm{CF}}{\mathcal{D}}^{\zeta }u(t)+ \gamma ^{\mathrm{CF}}{\mathcal{D}}^{\zeta }v(t)$ [27]. It has been proved that the unique solution for the problem ${}^{\mathrm{CF}}{\mathcal{D}} ^{\zeta }u(t)=v(t)$ with boundary condition $u(0)=c$ is given by $u(t)=c+a_{\zeta }(v(t)-v(0))+b_{\zeta }\int _{0}^{t} v(s)\,ds$, where $a_{\zeta } = \frac{2(1-\zeta )}{(2-\zeta )M(\zeta )}=1-\zeta$ and $b_{\zeta } = \frac{2\zeta }{ (2-\zeta )M(\zeta )}=\zeta$ ([19] and [27]). Note that $v(0)=0$. Suppose that $u,v\in C_{\mathbb{R}}[0,1]$, $u(0)=0$, and there is a real constant L such that $|u(t)-v(t)|\leq L$ for all $t\in [0,1]$. Recently, Baleanu, Mousalou, and Rezapour proved that $|^{\mathrm{CF}}{\mathcal{D}}^{\zeta }u(t)-{^{\mathrm{CF}}}{\mathcal{D}}^{\zeta }v(t)| \leq \frac{1}{(1-\zeta )^{2}}L$ for all $t\in [0,1]$ [10]. This leads to $|^{\mathrm{CF}}{\mathcal{D}}^{\zeta }u(t)| \leq (\frac{1}{(1-\zeta )^{2}})L$ for all $t\in [0,1]$ whenever $u\in C_{\mathbb{R}}[0,1]$ and $|u(t)|\leq L$ for some $L\geq 0$ and all $t\in [0,1]$ with $u(0)=0$ [10]. Also, they showed that $|{}^{\mathrm{CF}}\mathcal{I}^{\zeta }u(t)-{}^{\mathrm{CF}}\mathcal{I}^{\zeta }v(t)| \leq L$ for all $t\in [0,1]$ [10] and so $|^{\mathrm{CF}}\mathcal{I}^{\zeta }u(t)| \leq L$ for all $t\in [0,1]$ whenever $u\in C_{\mathbb{R}}[0,1]$ with $|u(t)|\leq L$ for some $L\geq 0$ and all $t\in [0,1]$. For some more necessary definitions, see [1].

Let $u\in C_{\mathbb{R}}[0,d]$, $d>0$ and $\zeta \in (0,1)$. The extended fractional Caputo–Fabrizio derivation of order ζ is defined by [11]

$$\begin{aligned} {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }u(t) ={}& \frac{B(\zeta )}{1- \zeta } \bigl(u(t)-u (0)\bigr)\exp \biggl(\frac{-\zeta }{1-\zeta }t\biggr) \\ & {} +\frac{\zeta B(\zeta )}{(1-\zeta )^{2}} \int _{0}^{t} \bigl(u(t)-u (s)\bigr) \exp \biggl( \frac{-\zeta }{1-\zeta }(t-s)\biggr) \,ds. \end{aligned}$$

If $u(0)=0$, then we have ${}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }u(t)=\frac{B( \zeta )}{1-\zeta }u(t) - \frac{\zeta B(\zeta )}{(1-\zeta )^{2}}\int _{0}^{t} \exp (-\frac{\zeta }{1-\zeta }(t-s))u(s)\,ds$ [11].

Lemma 1

([11])

Let $u\in H^{1}(0,b)$, $b>0$, and $\zeta \in (0,1)$. Then ${}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }u(t)={}^{\mathrm{CF}}{\mathcal{D}}^{\zeta }u(t)$. If $u\in C_{\mathbb{R}}[0,b]$, then $\lim_{\zeta \to 0}~{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }u(t)=u(t)-u(0)$.

Lemma 2

([11])

Let $0<\zeta <1$. Then a solution for the problem ${}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \zeta }u(t)=v(t)$ with boundary condition $u(0)=0$ is given by $u(t)= a_{\zeta } v(t) + b_{\zeta }\int _{0}^{t} v(s)\,ds$.

Lemma 3

([11])

Let $u,v\in C_{\mathbb{R}}[0,1]$. If there is a real constant L such that $|u(t)-v(t)|\leq L$ for all $t\in [0,1]$, then $|{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }u(t)-{}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\zeta }v(t)| \leq \frac{(2-\zeta )B(\zeta ) }{(1-\zeta )^{2}}L$ for all $t\in [0,1]$. If $u(0)=v(0)$, then $|{}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\zeta }u(t)-{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }v(t)| \leq \frac{B( \zeta ) }{(1-\zeta )^{2}}L$.

This result implies that $|{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }u(t)| \leq \frac{(2-\zeta )B(\zeta ) }{(1-\zeta )^{2}}L$ for all $t\in [0,1]$ whenever $u\in C_{\mathbb{R}}[0,1]$ with $|u(t)|\leq L$ for some $L\geq 0$ and all $t\in [0,1]$.

We need the following results.

Lemma 4

([28])

Suppose that $\mathcal{Y}$ is a Banach space, ${\mathcal{F}}: I\times \mathcal{Y}\to \mathcal{P}_{cp,cv}( \mathcal{Y})$ is an $L^{1}$-Caratheodory multivalued and ϵ is a linear continuous mapping from $L^{1}(I,\mathcal{Y})$ to $C(I, \mathcal{Y})$. Then the mapping $\epsilon \circ S_{{\mathcal{F}}} : C(I, \mathcal{Y})\to \mathcal{P}_{cp,cv}C( I,\mathcal{Y})$ defined by $(\epsilon \circ S_{{\mathcal{F}}}) (y) = \epsilon (S_{{ \mathcal{F}},y})$ is a closed graph mapping in $C(I,\mathcal{Y})\times C(I,\mathcal{Y})$.

Theorem 5

([29])

Assume that Y is a Banach space, D is a closed and convex subset of Y, and W is an open subset of D with $0\in W$. If ${\mathcal{F}}: \bar{W}\to P_{cp,c}(D)$ is an upper semi-continuous compact map, then either ${\mathcal{F}}$ has a fixed point in W̄ or there is $x\in \partial W$ and $\delta \in (0,1)$ such that $x \in \delta {\mathcal{F}}(x)$.

Theorem 6

([30])

Suppose that $(\mathcal{Y},d)$ is a complete metric space. If ${\mathcal{G}}: \mathcal{Y} \to P_{cl}(\mathcal{Y})$ is a contraction, then ${\mathcal{G}}$ has a fixed point.

Theorem 7

([31])

Assume that $\mathcal{Y}$ is a Banach space, $\mathcal{E}\in P_{bd,cl,cv}(\mathcal{Y})$ and $\mathcal{F}, \mathcal{G}: \mathcal{E}\to P_{cp,cv}(\mathcal{Y})$ are two multivalued operators. If $\mathcal{F}y+\mathcal{G}y\subset \mathcal{E}$ for all $y\in \mathcal{E}$, $\mathcal{F}$ is a contraction and $\mathcal{G}$ is an upper semi-continuous compact map, then there is $y\in \mathcal{E}$ such that $y\in \mathcal{F}y+\mathcal{G}y$.

Theorem 8

([32])

Assume that $\mathcal{Y}$ is a Banach algebra, $D\in \mathcal{P}_{bd,cl,cv}(\mathcal{Y})$ and ${\mathcal{F}_{1}}: D \to \mathcal{P}_{cl,cv,bd} (\mathcal{Y})$ and ${\mathcal{F}_{2}}: D \to \mathcal{P}_{cp,cv} (\mathcal{Y})$ are two set-valued maps such that ${\mathcal{F}_{1}}$ is Lipschitz with a Lipschitz constant δ, ${\mathcal{F}}_{2}$ is upper semi-continuous and compact, ${\mathcal{F} _{1}} x{\mathcal{F}_{2}}x$ is a convex subset D for all $x\in D$ and $\mathcal{N}\delta <1$, where $\mathcal{N} =\|{\mathcal{F}_{2}}(D)\|= \sup \{\|{\mathcal{F}_{2}}x\|: x\in D\}$. Then there is $y\in D$ such that $y\in {\mathcal{F}_{1}}y{\mathcal{F}_{2}}y$.

Lemma 9

([26])

Let $\mathcal{A}$ mapping $[0,1]$ into $\mathcal{P}_{cp,cv} (\mathbb{R})$ be measurable such that the Lebesgue measure of the set $\{t: \dim \mathcal{A}(t)<1\}$ is zero. Then there are arbitrarily many linearly independent measurable selections $y_{1}(\cdot),\ldots , y_{m}(\cdot)$ of $\mathcal{A}$.

Theorem 10

([26])

Let $\mathcal{H}$ be a nonempty closed convex subset of a Banach space $\mathcal{Y}$ and $\mathcal{F}: \mathcal{H} \to \mathcal{P}_{cp,cv}(\mathcal{H})$ be a δ-contraction. If $\dim \mathcal{F}(x)\geq m$ for all $x\in \mathcal{H}$, then $\dim Fix (\mathcal{F})\geq m$.

2 Main results

Consider the Banach space $\mathcal{X}=C(I)$ of real-valued continuous functions on $I=[0,1]$ via the norm $\|x\|=\sup_{t\in I}|x(t)|$. Assume that $\zeta ,\iota :[0,1] \times [0,1]\to [0,\infty )$ are two continuous maps such that $\sup |\int _{0}^{t} \iota (t,s) \,ds|<\infty$ and $\sup |\int _{0}^{t} \zeta (t,s) \,ds|<\infty$. Consider the maps ϕ and φ defined by $(\phi w)(t)= \int _{0}^{t} \zeta (t,s)w(s)\,ds$ and $(\varphi w)(t)= \int _{0}^{t} \iota (t,s)w(s)\,ds$. Suppose that $\eta (t)\in L^{\infty }(I)$ with $\eta ^{\ast }= \sup_{t\in I} |\eta (t)|$. Put $\zeta _{0}=\sup |\int _{0}^{t} \zeta (t,s) \,ds|$ and $\iota _{0}=\sup |\int _{0}^{t} \iota (t,s) \,ds|$. First, we are going to investigate the fractional integro-differential inclusion

$$ {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta }x(t) \in \mathcal{F}\bigl(t, x(t),( \phi x) (t), (\varphi x) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{1}} x(t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{2}}x(t) , \ldots , {}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{ \beta _{m}}x(t)\bigr), $$

(1)

with boundary condition $x(0)=0$, where $\zeta ,\beta _{1},\dots , \beta _{m}\in (0,1)$.

We say that a function $x\in \mathcal{X}$ is a solution for problem (1) whenever there exists a function $f\in C(I)$ such that

$$ f(t)\in \mathcal{F}\bigl(t,x(t),(\phi x) (t),(\varphi x) (t),{}^{\mathrm{CF}}_{N} {\mathcal{D}}^{ \beta _{1}} x(t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{2}}x(t) ,\ldots ,{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{m}}x(t) \bigr) $$

for almost all $t\in I$ and $x(t)=a_{\zeta }f(t)+b_{\zeta }\int _{0} ^{t} f(s)\,ds$.

Theorem 11

Let $\mathcal{F}: I\times \mathbb{R}^{m+3}\to P_{cp,cv}(\mathbb{R})$ be a Caratheodory multivalued map such that

$$\begin{aligned} \bigl\Vert \mathcal{F}(t,x_{1},x_{2},x_{3},y_{1}, \ldots , y_{m}) \bigr\Vert _{p} & = \sup \bigl\{ \vert y \vert : y \in \mathcal{F} (t,x_{1}, x_{2}, x_{3}, y_{1}, \ldots , y_{m})\bigr\} \\ & \leq \eta (t) \Biggl( \vert x_{1} \vert + \vert x_{2} \vert + \vert x_{3} \vert + \sum _{i=1}^{m} \vert y_{i} \vert \Biggr) \end{aligned}$$

for all $t\in I$, $x_{i}, y_{j} \in \mathbb{R}$, $1\leq i\leq 3$ and $1\leq j \leq m$. If $\eta ^{*}(1+\zeta _{0} + \iota _{0} +\sum_{i=1} ^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} )\leq 1$, then inclusion (1) has one solution.

Proof

For $x\in \mathcal{X}$, define a selection set of $\mathcal{F}$ at $x\in \mathcal{X}$ by

$$\begin{aligned} \begin{aligned} S_{\mathcal{F},x} :={} & \bigl\{ f\in L^{1} (I,R): f(t) \in \mathcal{F} \bigl(t,x(t),( \phi x) (t),(\varphi x) (t), \\ & {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{1}} x(t),{}^{\mathrm{CF}} _{N}{\mathcal{D}}^{ \beta _{2}}x(t) , \ldots , {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{ \beta _{m}}x(t) \bigr) \text{ for all } t\in I\bigr\} . \end{aligned} \end{aligned}$$

Since $\mathcal{F}$ is a Caratheodory multifunction, by using Theorem 1.3.5 in [33], we get $S_{\mathcal{F},x}$ is nonempty. Define an operator $\varOmega \colon \mathcal{X} \to P( \mathcal{X})$ by $\varOmega (x)=\{g\in \mathcal{X}:\mbox{ there exists } f \in S_{\mathcal{F},x} \mbox{ such that } g(t)=a_{\zeta }f(t)+b_{\zeta }\int _{0}^{t} f(s)\,ds\mbox{ for all }t\in I\}$. We show that the operator Ω satisfies the hypothesis of Theorem 5. First, we show that $\varOmega (x)$ is convex for all $\in \mathcal{X}$.

Let $g_{1},g_{2}\in \varOmega (x)$ and $w\in [0,1]$. Choose $f_{1},f_{2} \in S_{\mathcal{F},x}$ such that $g_{i}(t)=a_{\zeta }f_{i}(t)+b_{ \zeta }\int _{0}^{t} f_{i}(s)\,ds$ for all $t\in I$. Then we have

$$ \bigl[wg_{1}+(1-w)g_{2}\bigr](t)=a_{\zeta } \bigl(wf_{1}+(1-w)f_{2}\bigr) (t)+b_{\zeta } \int _{0}^{t}\bigl(wf_{1}+(1-w)f_{2} \bigr) (s)\,ds $$

for all $t\in I$. Since $\mathcal{F}$ has convex values, it is easy to check that $S_{\mathcal{F},x}$ is convex, and so $wg_{1}+(1-w)g_{2} \in \varOmega (x)$. Now, we show that Ω maps bounded sets into bounded subsets. Let $\mathcal{B}_{r}=\{x\in \mathcal{X}:\|x\|\leq r \}$, $x\in \mathcal{B}_{r}$, and $g\in \varOmega (x)$. Choose $f\in S _{\mathcal{F},x}$ such that

$$\begin{aligned} \bigl\vert g(t) \bigr\vert \leq{} &a_{\zeta } \bigl\vert f(t) \bigr\vert + b_{\zeta } \int _{0}^{t} \bigl\vert f(s) \bigr\vert \,ds \leq a_{\zeta }\eta (t) \bigl( \vert x \vert + \bigl\vert \varphi (x) \bigr\vert + \bigl\vert \phi (x) \bigr\vert \\ & {} + \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{1}}x(t) \bigr\vert + \bigl\vert {}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{ \beta _{2}}x(t) \bigr\vert + \cdots + \bigl\vert {}^{\mathrm{CF}}_{N} {\mathcal{D}}^{ \beta _{m}}x(t) \bigr\vert \bigr) \\ & {} +b_{\zeta } \int _{0}^{t} \bigl( \vert x \vert + \bigl\vert \varphi (x) \bigr\vert + \bigl\vert \phi (x) \bigr\vert \\ & {} + \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{1}}x(s) \bigr\vert + \bigl\vert {}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{ \beta _{2}}x(s) \bigr\vert + \cdots + \bigl\vert {}^{\mathrm{CF}}_{N} {\mathcal{D}}^{ \beta _{m}}x(s) \bigr\vert \bigr)\eta (s)\,ds \\ \leq{}& a_{\zeta } \eta ^{\ast } \Biggl(r+\zeta _{0} r+ \iota _{0} r+ \sum_{i=1}^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}}r \Biggr) \\ & {} + b_{\zeta } \eta ^{\ast } \Biggl( r+\zeta _{0} r+\iota _{0} r+ \sum_{i=1}^{m} \frac{B( \beta _{i})}{ (1-\beta _{i})^{2}}r \Biggr) \\ ={}& \eta ^{\ast }\cdot r\cdot \Biggl(1+\zeta _{0}+\iota _{0}+ \sum_{i=1}^{m} \frac{B( \beta _{i})}{( 1-\beta _{i})^{2}} \Biggr) (a_{\zeta }+b_{\zeta })\leq r. \end{aligned}$$

Thus, $\|g\|=\max_{t\in I}| g(t) |\leq r$. This implies that Ω maps bounded sets into bounded sets in $\mathcal{X}$. Now, we show that Ω maps bounded sets of $\mathcal{X}$ into equi-continuous sets. Let $t_{1},t_{2}\in I$ with $t_{1}< t_{2}$, $x \in \mathcal{B}_{r}$ and $g\in \varOmega (x)$. Then we have

$$\begin{aligned} \bigl\vert g(t_{2})-g(t_{1}) \bigr\vert & = \biggl\vert a_{\zeta }f(t_{2}) + b_{\zeta } \int _{0}^{t _{2}}f(s) \,ds - a_{\zeta }f(t_{1}) -b_{\zeta } \int _{0}^{t_{1}}f(s) \,ds \biggr\vert \\ & \leq a_{\zeta } \bigl\vert f(t_{2}) - f(t_{1}) \bigr\vert +b_{\zeta } \int _{t_{1}}^{t _{2}} \bigl\vert f(s) \bigr\vert \,ds \\ & \leq r \Biggl( 1+\zeta _{0} + \iota _{0}+\sum _{i=1}^{m}\frac{B(\beta _{i})}{(1 -\beta _{i})^{2}} \Biggr) \bigl(\eta (t_{2}) - \eta (t_{1}) \bigr) (a_{\zeta }+b_{\zeta }). \end{aligned}$$

Hence, the right-hand side of the inequality tends to zero (independent on $x\in \mathcal{B}_{r}$) as $t_{2}\to t_{1}$. This implies that $\varOmega \colon \mathcal{X} \to P(\mathcal{X})$ is a compact multivalued map by using the Arzela–Ascoli theorem. We show that Ω has a closed graph. Let $x_{n} \to x_{\ast }$, $g_{n}\in \varOmega (x_{n})$ for all n and $g_{n} \to g_{\ast }$. It is sufficient to prove that $g_{\ast }\in \varOmega (x_{\ast })$. Since $g_{n}\in \varOmega (x_{n})$ for all n, there exist $f_{n}\in S_{\mathcal{F},x_{n}}$ such that $g_{n}(t)= a_{\zeta }f_{n}(t)+b_{\zeta } \int _{0}^{t} f_{n}(s)\,ds$ for all $t\in I$. Thus, we have to show that there exist $f_{\ast }\in S _{\mathcal{F},x_{\ast }}$ such that $g_{\ast }(t)=a_{\zeta }f_{\ast }(t)+b _{\zeta } \int _{0}^{t} f_{\ast }(s)\,ds$ for all $t\in I$. Consider the linear continuous operator $\theta \colon L^{1}(I,\mathbb{R})\to \mathcal{X}$ defined by $f\mapsto \theta (f)(t)$, where $\theta (f)(t)=a _{\zeta }f(t)+b_{\zeta } \int _{0}^{t} f(s)\,ds$ for all $t \in I$. Since θ is a linear continuous map, by using Lemma 4 we get $\theta \circ S_{\mathcal{F}}$ is a closed graph operator. Note that $g_{n} \in \theta \circ S_{\mathcal{F}}(x_{n})$ for all n. Since $x_{n} \to x_{\ast }$ and $g_{n} \to g_{\ast }$, there exists $f_{\ast }\in S_{\mathcal{F}}(x_{\ast })$ such that $g_{\ast }(t)=a _{\zeta }f_{\ast }(t)+b_{\zeta } \int _{0}^{t} f_{\ast }(s)\,ds$ for all $t\in I$. For $\lambda \in (0,1)$ and $x\in \lambda \varOmega (x)$, there exists $f\in S_{\mathcal{F},x}$ such that $x(t)= a_{\zeta }\lambda f(t)+b _{\zeta } \int _{0}^{t} \lambda f(s)\,ds$ for all $t\in I$. Hence,

$$\bigl\vert x(t) \bigr\vert \leq \lambda (a_{\zeta }+b_{\zeta }) \eta ^{\ast } \cdot \Biggl(1+\zeta _{0}+\iota _{0}+ \sum _{i=1}^{m} \frac{ B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \Vert x \Vert . $$

Thus, $\|x\|=\max_{t\in I}|x(t)|\leq \lambda \|x\|$. Put $\mathcal{W}= \{x\in \mathcal{X},\|x\|< r(1+\zeta _{0} +\iota _{0} + \sum_{i=1}^{m} \frac{B( \beta _{i})}{(1-\beta _{i})^{2}})\}$. Note that the operator $\varOmega \colon \overline{\mathcal{W}} \to P_{cp,cv}(\mathcal{X})$ is upper semi-continuous and compact. In view of the choice of $\mathcal{W}$, there is no $x\in \partial {\mathcal{W}}$ such that $x\in \lambda \varOmega (x)$ for some $\lambda \in (0,1)$. Hence, by using Theorem 5, Ω has a fixed point $x\in \overline{\mathcal{W}}$ which is a solution for problem (1). This completes the proof. □

Now consider the Banach space $\mathcal{X}=C(I)$ via the norm

$$ \Vert x \Vert =\max_{t\in I} \bigl\vert x(t) \bigr\vert + \sum_{i=1}^{m} \max_{t\in I} \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}}x(t) \bigr\vert + \sum_{j=1}^{n} \max _{t\in I} \bigl\vert {} ^{\mathrm{CF}} \mathcal{I}^{\gamma _{j}}x(t) \bigr\vert . $$

Here, we investigate the fractional integro-differential inclusion

$$ \begin{aligned}[b] {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\zeta }x(t) \in{} & \mathcal{F} \bigl( t, x(t), (\phi x) (t),( \varphi x) (t), \\ & {} {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{1}}x(t), {}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{\beta _{2}} x(t),\ldots , {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{m}} x(t), \\ & {} {}^{\mathrm{CF}}\mathcal{I}^{ \gamma _{1}}x(t),{}^{\mathrm{CF}} \mathcal{I}^{ \gamma _{2}}x(t), \ldots , {}^{\mathrm{CF}}\mathcal{I}^{ \gamma _{n}}x(t) \bigr), \end{aligned} $$

(2)

with boundary condition $x(0)=0$, where $\zeta ,\beta _{1},\dots ,\beta _{m},\gamma _{1},\dots ,\gamma _{n}\in (0,1)$. Similar to the last case, we say that a function $x\in C(I,\mathbb{R})$ is a solution for problem (2) whenever there exists a function $f\in L^{1}(I)$ such that

$$ \begin{aligned} f(t) & \in \mathcal{F}\bigl(t,x(t),(\phi x) (t),( \varphi x) (t),{}^{\mathrm{CF}}_{ N}{ \mathcal{D}}^{ \beta _{1}} x(t), {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{ \beta _{2}}x(t) , \ldots , \\ & \quad {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{m}}x(t) \bigr), ^{\mathrm{CF}} \mathcal{I}^{ \gamma _{1}}x(t), ^{\mathrm{CF}} \mathcal{I}^{ \gamma _{2}}x(t), \ldots , ^{\mathrm{CF}}\mathcal{I}^{ \gamma _{n}}x(t)) \end{aligned} $$

for almost all $t\in I$ and $x(t)=a_{\zeta }f(t)+b_{\zeta }\int _{0} ^{t} f(s)\,ds$ for all $t\in I$.

Theorem 12

Assume that $\mathcal{F} \colon I \times \mathbb{R}^{m+n+3 }\to P_{cv,cp}( \mathbb{R})$ is a multifunction such that the map $t \to \mathcal{F}(t,x_{1},x_{2},\ldots ,x_{3+m+n})$ is measurable for all $x_{1},x_{2},\ldots ,x_{m+n+3}\in \mathbb{R}$, the map $t \to d_{H}(0, \mathcal{F}(t,0,\ldots ,0))$ is integrably bounded for almost all $t\in I$ and

$$ \begin{aligned} & H_{d} \bigl( \mathcal{F}( t ,x_{1},x_{2},x_{3},y_{1},y_{2}, \ldots , y_{m},z_{1},z_{2},\ldots ,z_{n}), \\ & \qquad \mathcal{F}\bigl(t,x^{\prime }_{1},x^{\prime }_{2},x^{\prime } _{3},y^{\prime }_{1},y^{\prime }_{2}, \ldots ,y^{\prime }_{m},z^{ \prime }_{1},z^{\prime }_{2}, \ldots , z^{\prime }_{n}\bigr) \bigr) \\ &\quad \leq \eta (t) \Biggl( \bigl\vert x_{1}-x^{\prime }_{1} \bigr\vert + \bigl\vert x_{2}-x^{\prime }_{2} \bigr\vert + \bigl\vert x _{3}-x^{\prime }_{3} \bigr\vert +\sum_{i=1}^{m} \bigl\vert y_{i}-y^{\prime }_{i} \bigr\vert +\sum _{j=1} ^{n} \bigl\vert z_{j}-z^{\prime }_{j} \bigr\vert \Biggr) \end{aligned} $$

for all $t\in I$ and all $x_{1},x_{2},x_{3},x^{\prime }_{1},x^{\prime }_{2},x^{\prime }_{3},y_{1},\ldots ,y_{m},y^{\prime }_{1},\ldots ,y ^{\prime }_{m},z_{1},\ldots ,z_{n},z^{\prime }_{1},\ldots ,z^{\prime }_{n} \in \mathbb{R}$. If $\Delta \leq 1$, then the inclusion problem (2) has at least one solution, where

$$ \Delta =\eta ^{\ast } \Biggl(1+n+\zeta _{0}+\iota _{0}+\sum_{i=1}^{m} \frac{B( \beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \Biggl(1+n+\sum_{i=1}^{m} \frac{B( \beta _{i})}{(1-\beta _{i})^{2}} \Biggr). $$

Proof

By using the assumptions of Theorem III-6 in [34], we conclude that $\mathcal{F}$ admits a measurable selection $f\colon I \to \mathbb{R}$. Since $\mathcal{F}$ is integrable bounded, $f\in L^{1}(I,\mathbb{R})$ and so $S_{\mathcal{F},x}$ is nonempty for all $x\in \mathcal{X}$, where

$$ \begin{aligned} S_{\mathcal{F},x} ={} & \bigl\{ f\in L^{1} (I,R): f(t) \in \mathcal{F}\bigl(t,x(t),( \phi x) (t),(\varphi x) (t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{1}} x(t), \\ & \quad {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{2}}x(t) , \ldots ,{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{m}}x(t), {}^{\mathrm{CF}}\mathcal{I}^{ \gamma _{1}}x(t),{}^{\mathrm{CF}} \mathcal{I}^{ \gamma _{2}}x(t), \ldots , \\ & \quad {}^{\mathrm{CF}}\mathcal{I}^{ \gamma _{n}}x(t)\bigr) \text{ for all } t \in I\bigr\} . \end{aligned} $$

Define the operator $\varOmega \colon \mathcal{X} \to P(\mathcal{X})$ by

$$\begin{aligned} \varOmega (x)={}&\biggl\{ g\in \mathcal{X}: \mbox{ there exists } f\in S_{\mathcal{F},x} \mbox{ such that}\\ &g(t)=a_{\zeta }f(t)+b_{\zeta } \int _{0}^{t} f(s)\,ds\mbox{ for all }t\in I\biggr\} . \end{aligned}$$

First, we show that $\varOmega (x)\in P_{cl}(\mathcal{X})$ for all $x\in \mathcal{X}$. Let $g_{n}\in \varOmega (x)$ for all $n\geq 0$ and $g_{n}\rightarrow g_{\ast }$ for some $g\in \mathcal{X}$. For each n, choose $f_{n}\in S_{\mathcal{F},x}$ such that $g_{n}(t)= a_{\zeta }f _{n}(t)+b_{\zeta } \int _{0}^{t} f_{n}(s)\,ds$ for all $t\in I$. Since $\mathcal{F}$ has compact values, there is a subsequence of $f_{n}$ that converges to f in $L^{1}(I,\mathbb{R})$. Thus, $f\in S_{\mathcal{F},x}$ and $g_{n}(t)\rightarrow g_{\ast }(t)= a_{ \zeta }f(t)+b_{\zeta } \int _{0}^{t} f(s)\,ds$ for all $t\in I$. This implies that $g_{\ast }\in \varOmega$. Now, we show that there exists $\epsilon <1$ such that $H_{d}(\varOmega (x),\varOmega (y))\leq \epsilon \|x-y \|$ for all $x,y \in \mathcal{X}$. Let $x,y \in \mathcal{X}$ and $g_{1} \in \varOmega (x)$. Choose $f_{1} \in S_{\mathcal{F},x}$ such that $g_{1}(t)= a_{\zeta }f_{1}(t)+b_{\zeta } \int _{0}^{t} f_{1}(s)\,ds$ for all $t \in I$. Consider the multifunction $\tilde{\mathcal{F}}$ defined by

$$\begin{aligned} \tilde{\mathcal{F}}\bigl(t,x(t)\bigr)={}&\mathcal{F}\bigl(t,x(t),(\phi x) (t),( \varphi x) (t), {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{1}}x(t), {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{2}} x(t),\ldots ,{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{m}} x(t), \\ &{}{}^{\mathrm{CF}}\mathcal{I}^{ \gamma _{1}}x(t),{}^{\mathrm{CF}} \mathcal{I}^{ \gamma _{2}}x(t), \ldots ,{}^{\mathrm{CF}}\mathcal{I}^{ \gamma _{n}}x(t) \bigr). \end{aligned}$$

Then we have

$$ \begin{aligned} H_{d}( \tilde{ \mathcal{F}}\bigl(t,x(t) \bigr), \tilde{\mathcal{F}}\bigl(t,y(t)\bigr)\leq{} & \eta (t) \Biggl( \bigl\vert x(t)-y(t) \bigr\vert + \bigl\vert (\phi x) (t)-(\phi y) (t) \bigr\vert \\ & {} + \bigl\vert (\varphi x) (t)-(\varphi y) (t) \bigr\vert \\ & {} + \sum_{i=1}^{m} \bigl\vert {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}} x(t) - {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}} y(t) \bigr\vert \\ & {} + \sum_{j=1}^{n} \bigl\vert ^{\mathrm{CF}} \mathcal{I}^{ \gamma _{j}}x(t)-^{\mathrm{CF}} \mathcal{I}^{ \gamma _{j}}y(t) \bigr\vert \Biggr) \end{aligned} $$

for almost $t\in I$. Hence, there exists $w_{t} \in \tilde{\mathcal{F}}(t,y(t))$ such that

$$ \begin{aligned} \bigl\vert f_{1}(t)-w_{t} \bigr\vert \leq {}& \eta (t) \Biggl( \bigl\vert x(t) - y(t) \bigr\vert + \bigl\vert (\phi x) (t)-( \phi y) (t) \bigr\vert + \bigl\vert (\varphi x) (t)-( \varphi y) (t) \bigr\vert \\ & {} + \sum_{i=1}^{m} \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}} x(t) - {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}} y(t) \bigr\vert \\ & {} + \sum_{j=1}^{n} \bigl\vert ^{\mathrm{CF}}\mathcal{I}^{ \gamma _{j}}x(t)-^{\mathrm{CF}} \mathcal{I}^{ \gamma _{j}}y(t) \bigr\vert \Biggr):=M_{t} \end{aligned} $$

for almost $t\in I$. Define $V\colon I \to P(\mathbb{R})$ by $V(t)=\{u\in \mathbb{R}: |f_{1}(t)-u|\leq M_{t}\}$ for all $t\in I$. By using Theorem III-41 in [34], we get V is measurable. Since $t \mapsto V(t)\cap \tilde{\mathcal{F}}(t,y(t))$ is measurable (Proposition III-4 in [34]), we can choose $f_{2} \in S_{\mathcal{F},y}$ such that $|f_{1}(t)-f_{2}(t)|\leq M _{t}$ for almost all $t\in I$. Define $g_{2} \in \varOmega (y)$ by $g_{2}(t)= a_{\zeta }f_{2}(t)+b_{\zeta } \int _{0}^{t} f_{2}(s)\,ds$ for all $t \in I$. Then we have

$$\begin{aligned} \Vert g_{1}-g_{2} \Vert ={}&\max_{t\in I} \bigl\vert g_{1}(t)-g_{2}(t) \bigr\vert +\sum _{i=1}^{m} \max_{t\in I} \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}}g_{1}(t)-{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}}g_{2}(t) \bigr\vert \\ & {} + \sum_{i=1}^{n} \max _{t\in I} \bigl\vert {}^{\mathrm{CF}}\mathcal{I}^{\gamma _{i}}g _{1}(t)-{}^{\mathrm{CF}}\mathcal{I}^{\gamma _{i}}g_{2}(t) \bigr\vert \bigl\vert g_{1}(t)-g_{2}(t) \bigr\vert \\ \leq {}& a_{\zeta } \bigl\vert f_{1}(t)-f_{2}(t) \bigr\vert +b_{\zeta } \int _{0}^{t} \bigl\vert f_{1}(s)-f _{2}(s) \bigr\vert \,ds \\ \leq {}&\eta (t) \Biggl(1+n+\zeta _{0}+\iota _{0}+\sum _{i=1}^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) (a_{\zeta }+ b_{\zeta }) \Vert x-y \Vert , \end{aligned}$$

and so

$$\begin{aligned} &\bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}}g_{1}(t)-{}^{\mathrm{CF}}_{N} {\mathcal{D}}^{\beta _{i}}g_{2}(t) \bigr\vert \\ &\quad \leq \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \bigl\vert g_{1}(t)-g_{2}(t) \bigr\vert \\ &\quad \leq \eta (t)\frac{B( \beta _{i})}{(1-\beta _{i})^{2}} \Biggl(1+n+\zeta _{0}+\iota _{0} + \sum_{i=1}^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) (a _{\zeta }+ b_{\zeta }) \Vert x-y \Vert . \end{aligned}$$

Thus,

$$ \begin{aligned} \bigl\vert {}^{\mathrm{CF}} \mathcal{I}^{\gamma _{i}}g_{1}(t) - {}^{\mathrm{CF}}\mathcal{I}^{\gamma _{i}}g_{2}(t) \bigr\vert & \leq \bigl\vert g_{1}(t)-g_{2}(t) \bigr\vert \\ & \leq \eta (t) \Biggl(1 + n+\zeta _{0} + \iota _{0} +\sum_{i=1}^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) ( a _{\zeta }+ b_{\zeta }) \Vert x-y \Vert , \end{aligned} $$

and so

$$ \begin{aligned} \Vert g_{1}-g_{2} \Vert \leq{}& \eta ^{\ast } \Biggl(1 + n+ \zeta _{0}+\iota _{0} + \sum_{i=1}^{m} \frac{B( \beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \\ & {} \times \Biggl(1+n +\sum_{i=1}^{m} \frac{ B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \Vert x-y \Vert = \Delta \Vert x-y \Vert . \end{aligned} $$

Hence, $H_{d}(\varOmega (x), \varOmega (y))\leq \Delta \|x-y\|$. Since $\Delta <1$, Ω is a closed-valued contraction. By using Theorem 6, Ω has a fixed point which is a solution for the inclusion problem (2). □

Consider the Banach space ${\mathcal{X}}=\{x: x,{}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{i}}x \in C(I,\mathbb{R})\}$ endowed with the norm $\|x\|= \max_{t\in I} |x(t)|+\max_{t\in I} |{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{i}}x(t)|$. Here, we review the inclusion problem

$$ \begin{aligned}[b] {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\zeta }x(t) \in{}& \mathcal{F}\bigl(t,x(t),( \phi x) (t), {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{1}}x(t), {}^{\mathrm{CF}} _{N}{\mathcal{D}}^{\beta _{2}} x(t), \ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\beta _{n}} x(t)\bigr) \\ & {} +\mathcal{G}\bigl(t,x(t),(\varphi x) (t),{}^{\mathrm{CF}} \mathcal{I}^{ \beta _{1}}x(t),{}^{\mathrm{CF}}\mathcal{I}^{ \beta _{2}}x(t), \ldots , ^{\mathrm{CF}} \mathcal{I}^{ \beta _{n}}x(t)\bigr) \end{aligned} $$

(3)

with boundary condition $x(0)=0$, where $\zeta ,\beta _{1},\dots ,\beta _{n}\in (0,1)$. Define the set of the selections of $\mathcal{F}$ and $\mathcal{G}$ at x by

$$\begin{aligned} S_{\mathcal{F},x}={}&\bigl\{ v\in L^{1}[0,1]:v(t)\in \mathcal{F} \bigl(t,x(t),( \phi x) (t), \\ &{}{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{1}}x(t), {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{2}} x(t),\ldots ,{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{n}} x(t)\bigr)\text{ for almost all } t\in I\bigr\} \end{aligned}$$

and

$$\begin{aligned} S_{\mathcal{G},x}={}&\bigl\{ v\in L^{1}[0,1]:v(t)\in \mathcal{G} \bigl(t,x(t),( \varphi x) (t), \\ &{}{}^{\mathrm{CF}}\mathcal{I}^{ \beta _{1}}x(t),{}^{\mathrm{CF}} \mathcal{I}^{ \beta _{2}}x(t), \ldots ,{}^{\mathrm{CF}}\mathcal{I}^{ \beta _{n}}x(t) \bigr)\text{ for almost all }t \in I\bigr\} . \end{aligned}$$

We suppose that $S_{\mathcal{F},x}\neq \emptyset$ and $S_{ \mathcal{G},x}\neq \emptyset$ for all $x\in {\mathcal{X}}$. A function $x\in C(I,\mathbb{R})$ is a solution for problem (3) whenever there exist two functions $f\in H^{1}(I)$ and $f^{\prime }\in H^{1}(I)$ such that

$$ f(t) \in \mathcal{F}\bigl(t,x(t),(\phi x) (t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{ \beta _{1}} x(t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{2}}x(t) , \ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{n}}x(t) \bigr) $$

and $f^{\prime } \in \mathcal{G}(t,x(t),(\varphi x)(t),{}^{\mathrm{CF}} \mathcal{I}^{ \beta _{1}} x(t),{}^{\mathrm{CF}}\mathcal{I}^{ \beta _{2}}x(t), \ldots , ^{\mathrm{CF}}\mathcal{I}^{ \beta _{n}}x(t))$ for almost all $t\in I$ and

$$ x(t)=a_{\zeta }f(t)+b_{\zeta } \int _{0}^{t} f(s)\,ds + a_{\zeta }f^{ \prime }(t)+b_{\zeta } \int _{0}^{t} f^{\prime }(s)\,ds $$

for all $t\in I$.

Theorem 13

Let $\mathcal{F}: I\times \mathbb{R}^{n+2}\to P_{cp,cv}(\mathbb{R})$ be a multifunction and $\mathcal{G}: I\times \mathbb{R}^{n+2}\to P_{cp,cv}( \mathbb{R})$ be a Caratheodory set-valued map. Assume that there exist continuous functions $p, m:I\to (0,\infty )$ and $\eta (t) \in L^{ \infty }(I)$ such that $t\vdash \mathcal{F}(t,y_{1},\ldots ,y_{n+2})$ is measurable,

$$\begin{aligned}& \bigl\Vert \mathcal{F} \bigl(t,x(t),(\phi x) (t), {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{1}} x(t), {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{2}}x(t) , \ldots , {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{n}}x(t) \bigr) \bigr\Vert \leq m(t), \\& \bigl\Vert \mathcal{G}\bigl(t,x(t),(\varphi x) (t),{}^{\mathrm{CF}} \mathcal{I}^{ \beta _{1}}x(t),{}^{\mathrm{CF}} \mathcal{I}^{ \beta _{2}}x(t), \ldots , ^{\mathrm{CF}}\mathcal{I}^{ \beta _{n}}x(t)\bigr) \bigr\Vert \leq p(t), \end{aligned}$$

and

$$\begin{aligned}& H_{d}\bigl( \mathcal{F}(t,y_{1}, \ldots , y_{n+2}),\mathcal{F}\bigl(t,y^{\prime }_{1}, \ldots , y^{\prime }_{n+2}\bigr)\bigr) \leq \eta (t)\sum _{i=1}^{n+2} \bigl( \bigl\vert y _{i}-y^{\prime }_{i} \bigr\vert \bigr) \end{aligned}$$

for all $t\in I$, $x\in {\mathcal{X}}$ and $y_{1}, \ldots , y_{n+2}$, $y'_{1}, \ldots , y'_{n+2}\in \mathbb{R}$. If $L=\eta ^{\ast }(1+\sum_{i=1}^{n}\frac{B(\beta _{i})}{(1-\beta _{i})^{2}})(1+\zeta _{0}+\sum_{i=1}^{n}\frac{B(\beta _{i})}{(1-\beta _{i})^{2}})< 1$, then the inclusion problem (3) has at least one solution.

Proof

Put $\mathcal{Y}=\{x\in \mathcal{X}: \|x\|\leq M\}$, where $M=(1+ \sum_{i=1}^{n}\frac{B(\beta _{i})}{(1-\beta _{i})^{2}})(\|p\|_{\infty }+ \|m\|_{\infty })$. One can check that $\mathcal{Y}$ is a closed, bounded, and convex subset of $\mathcal{X}$. Define the multivalued operators $\mathcal{A},\mathcal{B}: \mathcal{Y}\to P(\mathcal{X})$ by

$$ \mathcal{A}x:=\biggl\{ x\in {\mathcal{X}}: ~\text{there is}~v\in S_{ \mathcal{F},x}~\text{such that}~x(t)=a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds~\text{for all}~t\in I\biggr\} $$

and $\mathcal{B}x:=\{x\in {\mathcal{X}}:~\text{there is}~v\in S_{ \mathcal{G},x}~\text{such that}~x(t)=a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds~\text{for all}~t\in I\}$. Note that problem (3) is equivalent to the inclusion fixed point problem $x\in \mathcal{A}x+\mathcal{B}x$. Also, the operator $\mathcal{A}$ is equivalent to the composition $\theta \circ S_{\mathcal{F}}$, where θ is the continuous linear operator on $L^{1}(0,1)$ into $\mathcal{X}$ defined by $\theta v(t)=a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds$. Let $x\in \mathcal{Y}$ and $\{v_{n}\}_{n\geq 1}$ be a sequence in $S_{\mathcal{F},x}$. Then $v_{n}(t)\in \mathcal{F}(t,x(t),( \phi x)(t),{}^{\mathrm{CF}}{\mathcal{D}}^{ \beta _{1}} x(t),{}^{\mathrm{CF}}{\mathcal{D}} ^{ \beta _{2}}x(t) ,\ldots ,{}^{\mathrm{CF}}{\mathcal{D}}^{ \beta _{n}}x(t) )$ for almost $t\in I$. Since

$$\mathcal{F}\bigl(t,x(t),(\phi x) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{1} } x(t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{2}} x(t) ,\ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{n}}x(t) \bigr) $$

is compact for all $t\in I$, there is a convergent subsequence of $\{v_{n}(t)\}$ (call it again $\{v_{n}(t)\}$) such that it converges in measure to some $v(t)\in S_{\mathcal{F},x}$ for almost all $t\in I$. Since θ is continuous, $\theta v_{n}(t)\to \theta v(t)$ pointwise on I. In order to show that the convergence is uniform, we show that $\{\theta v_{n}\}$ is an equi-continuous sequence. Let $\tau < t \in I$. Then we have

$$\bigl\vert \theta v_{n}(t)- \theta v_{n}(\tau ) \bigr\vert \leq a_{\zeta } \bigl\vert v_{n}(t)-v_{n}( \tau ) \bigr\vert +b_{\zeta } \int _{\tau }^{t} \bigl\vert v_{n}(s) \bigr\vert \,ds. $$

Since the right-hand of the above inequality tends to 0 as $t\to \tau$, the sequence $\{\theta v_{n}\}$ is equi-continuous. Now, by using the Arzela–Ascoli theorem, there is a uniformly convergent subsequence of $\{v_{n}\}$ (we show it again by $\{v_{n}\}$) such that $\theta v_{n}\to \theta v$. Note that $\theta v \in \theta (S_{ \mathcal{F},x})$. Hence, $\mathcal{A}x=\theta (S_{\mathcal{F},x})$ is compact for all $x\in \mathcal{Y}$. Now, we show that $\mathcal{A}x$ is convex for all $x\in \mathcal{Y}$. Let $u , u^{\prime }\in \mathcal{A}x$. Choose $v,v^{\prime }\in S_{\mathcal{F},x}$ such that $u(t)=a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds$ and $u'(t)=a_{ \zeta }v^{\prime }(t)+b_{\zeta } \int _{0}^{t}v^{\prime }(s)\,ds$ for almost all $t\in I$. Let $0\leq \lambda \leq 1$. Then we have

$$ \bigl(\lambda u + (1-\lambda ) u^{\prime } \bigr) (t) = a_{\zeta } \bigl(\lambda v(t)+(1-\lambda )v^{\prime }(t) \bigr) +b_{\zeta } \int _{0}^{t} \bigl(\lambda v(s)+(1-\lambda )v^{ \prime }(s) \bigr)\,ds. $$

Since $\mathcal{F}$ is convex-valued, $\lambda u+(1-\lambda )u^{ \prime }\in \mathcal{A}x$. Similarly, we can show that $\mathcal{B}$ is compact and convex-valued. Here, we show that $\mathcal{A}y+ \mathcal{B}y\subset \mathcal{Y}$ for all $y\in \mathcal{Y}$. Let $y\in \mathcal{Y}$, $u\in \mathcal{A}y$, and $u^{\prime }\in \mathcal{B}y$. Choose $v\in S_{\mathcal{F},y}$ and $v'\in S_{ \mathcal{G},y}$ such that $u(t)=a_{\zeta }v(t)+b_{\zeta } \int _{0} ^{t}v(s)\,ds$ and $u'(t)=a_{\zeta }v^{\prime }(t)+b_{\zeta } \int _{0} ^{t}v^{\prime }(s)\,ds$ for almost all $t\in I$. Hence,

$$\bigl\vert u(t)+u'(t) \bigr\vert \leq a_{\zeta } \bigl( \bigl\vert v(t) \bigr\vert + \bigl\vert v^{\prime }(t) \bigr\vert \bigr)+b_{\zeta } \int _{0}^{t} \bigl( \bigl\vert v(s) \bigr\vert + \bigl\vert v^{\prime }(s) \bigr\vert \bigr)\,ds, $$

and so

$$ \begin{aligned} \bigl\vert {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}}u(t)+{}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{i}}u'(t) \bigr\vert \leq{} & \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}}u(t) \bigr\vert + \bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}}u'(t) \bigr\vert \\ \leq {}&\frac{ a_{\zeta }B(\beta _{i})}{(1-\beta _{i})^{2}}\bigl(p(t)+m(t)\bigr) \\ & {} + \frac{ b_{\zeta }B(\beta _{i})}{(1-\beta _{i})^{2}} \bigl( \Vert p \Vert _{\infty }+ \Vert m \Vert _{\infty }\bigr) \end{aligned} $$

for $1\leq i \leq n$. This implies that

$$ \max_{t\in I} \bigl\vert u(t)+u'(t) \bigr\vert \leq a_{\zeta } \bigl( \Vert p \Vert _{\infty }+ \Vert m \Vert _{ \infty }\bigr)+b_{\zeta } \bigl( \Vert p \Vert _{\infty }+ \Vert m \Vert _{\infty }\bigr)= \Vert p \Vert _{\infty }+ \Vert m \Vert _{\infty } $$

and

$$ \begin{aligned} \max_{t \in I} \bigl\vert {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}}u(t)+{}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{\beta _{i}}u'(t) \bigr\vert \leq {}&\frac{ a_{\zeta }B( \beta _{i})}{(1-\beta _{i})^{2}}\bigl( \Vert p \Vert _{\infty }+ \Vert m \Vert _{\infty }\bigr) \\ & {} + \frac{ b_{\zeta }B(\beta _{i})}{(1-\beta _{i})^{2}} \bigl( \Vert p \Vert _{\infty }+ \Vert m \Vert _{\infty }\bigr) \\ = {}&\frac{ B(\beta _{i})( \Vert p \Vert _{\infty }+ \Vert m \Vert _{\infty })}{(1-\beta _{i})^{2}}. \end{aligned} $$

Thus, $\|u+u^{\prime }\|\leq (1+\sum_{i=1}^{n}(\frac{B(\beta _{i})}{(1- \beta _{i})^{2}})) (\|p\|_{\infty }+\|m\|_{\infty })=M$. Now, we show that the operator $\mathcal{B}$ is compact on $\mathcal{Y}$. To do this, we prove that $\mathcal{B}(\mathcal{Y})$ is uniformly bounded and equi-continuous in $\mathcal{X}$. Let $u\in \mathcal{B}(\mathcal{Y})$ be arbitrary. Choose $v\in S_{\mathcal{G},x}$ such that $u(t)=a_{ \zeta }v(t)+b_{\zeta }\int _{0}^{t} v(s)\,ds$ for some $x\in \mathcal{Y}$. Hence,

$$ \begin{aligned} \bigl\vert u(t) \bigr\vert & \leq a_{\zeta } \bigl\vert v(t) \bigr\vert +b_{\zeta } \int _{0}^{t} \bigl\vert v(s) \bigr\vert \,ds \bigl|{} ^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \beta _{i}}u(t))\bigr| \\ & \leq a_{\zeta } \bigl\vert {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}}v(t) \bigr\vert +b _{\zeta } \int _{0}^{t} \bigl\vert {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}}v(s) \bigr\vert \,ds \\ & \leq \frac{B(\beta _{i})(a_{\zeta }+ b_{\zeta })}{(1-\beta _{i})^{2}} p(t) \\ & =\frac{B(\beta _{i})}{(1-\beta _{i})^{2}} p(t). \end{aligned} $$

Thus, $\max_{t\in I}|u(t)|\leq (a_{\zeta }+b_{\zeta })\|p\|_{\infty }= \|p\|_{\infty }$ and $\max_{t\in I}|{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{i}}u_{i}(t)|\leq \frac{B(\beta _{i})}{(1-\beta _{i})^{2}}\|p\| _{\infty }$ for $i=1,\dots ,n$, and so $\|u\| \leq (1+\sum_{i=1}^{n} \frac{B( \beta _{i})}{(1-\beta _{i})^{2}})\|p\|_{\infty }$. Here, we show that $\mathcal{B}$ maps $\mathcal{Y}$ to equi-continuous subsets of $\mathcal{X}$. Let $t, \tau \in I$ with $\tau < t$, $x\in \mathcal{Y}$ and $u \in \mathcal{B}x$. Choose $v\in S_{\mathcal{G},x}$ such that $u(t)=a_{\zeta }v(t)+b_{\zeta }\int _{0}^{t} v(s)\,ds$ for all. Then we have

$$ \bigl\vert u(t)- u(\tau ) \bigr\vert \leq a_{\zeta }\bigl(v(t)-v(\tau ) \bigr)+b_{\zeta } \int _{\tau } ^{t}v(s)\,ds \leq a_{\zeta } \bigl(v(t)-v(\tau )\bigr)+b_{\zeta }(t-\tau ) \Vert p \Vert _{\infty } $$

and $|{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}}u(t)- {}^{\mathrm{CF}}_{N} {\mathcal{D}}^{\beta _{i}}u(\tau )|\leq \frac{B(\beta _{i})}{(1-\beta _{i})^{2}}|u(t)-u(\tau )|$. Since the right-hand of the inequality tends to 0 as $t\to \tau$, by using the Arzela–Ascoli theorem, we get $\mathcal{B}$ is compact. Now, we show that $\mathcal{B}$ has a closed graph. Let $x_{n}\in \mathcal{Y}$ and $u_{n}\in \mathcal{B}(x_{n})$ for all n with $x_{n}\to x_{0}$ and $u_{n}\to u_{0}$. We show that $u_{0}\in \mathcal{B}(x_{0})$. For each n, choose $v_{n}\in S_{ \mathcal{G},x_{n}}$ such that $u_{n}(t)= a_{\zeta }v_{n}(t)+b_{ \zeta }\int _{0}^{t}v_{n}(s) \,ds$ for all $t\in I$. Again, consider the continuous linear operator $\theta :L^{1}(0,1)\to {\mathcal{X}}$ defined by $\theta (v)(t)= a_{\zeta }v(t)+b_{\zeta }\int _{0}^{t}v(s) \,ds$. By using Lemma 4, $\theta o S_{\mathcal{G}}$ is a closed graph operator. Since $u_{n}\in \theta (S_{\mathcal{G},x_{n}})$ for all n and $x_{n}\to x_{0}$, there exists $v_{0}\in S_{\mathcal{G},x_{0}}$ such that $u_{0}(t)= a_{\zeta }v_{0}(t)+b_{\zeta }\int _{0}^{t}v_{0}(s) \,ds$. Hence, $u_{0}\in \mathcal{B}(x_{0})$. This implies that $\mathcal{B}$ has a closed graph, and so $\mathcal{B}$ is upper semi-continuous. Now, we show that $\mathcal{A}$ is a contraction multifunction. Let $x,y\in {\mathcal{X}}$ and $u\in \mathcal{A}y$. Choose $v\in S_{\mathcal{F},y}$ such that $u(t)= a_{\zeta }v(t)+b_{ \zeta }\int _{0}^{t}v(s) \,ds$ for all $t\in I$. Since

$$ \begin{aligned} &H_{d} \bigl( \mathcal{F}\bigl(t,x(t),(\phi x) (t),{}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{1}}x(t),\ldots ,{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{n}} x(t)\bigr), \\ & \qquad \mathcal{F}\bigl(t,y(t),(\phi y) (t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{1}}y(t), \ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{n}} y(t)\bigr) \bigr) \\ & \quad \leq \eta (t) \Biggl(1 + \zeta _{0} + \sum _{i=1}^{n}\frac{B(\beta _{i})}{(1- \beta _{i})^{2}} \Biggr) \Vert x-y \Vert \end{aligned} $$

for almost all $t\in I$, there exists $w\in \mathcal{F}(t,x(t),( \phi x)(t),{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{1}}x(t), \ldots , {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{n}} x(t))$ such that $|v(t)-w| \leq \eta (t) (1+\zeta _{0}+\sum_{i=1}^{n}\frac{B(\beta _{i})}{(1- \beta _{i})^{2}} )\|x-y\|$ for almost all $t\in I$. Consider the multifunction $U:I\to 2^{\mathbb{R}}$ defined by

$$ U(t)=\Biggl\{ w\in \mathbb{R}: \bigl\vert v(t)-w \bigr\vert \leq \eta (t) \Biggl(1+\zeta _{0}+\sum_{i=1}^{n} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \Vert x-y \Vert \text{ for almost all } t\in I \Biggr\} . $$

Since v and $\eta (1+\zeta _{0}+\sum_{i=1}^{n}\frac{B(\beta _{i})}{(1- \beta _{i})^{2}} )$ are measurable, we get

$$ U(\cdot)\cap \mathcal{F}\bigl(t,x(\cdot),(\phi x) (\cdot),{}^{\mathrm{CF}}{ \mathcal{D}}^{\beta _{1}}x(\cdot), \ldots ,{}^{\mathrm{CF}}{ \mathcal{D}}^{\beta _{n}} x(\cdot)\bigr) $$

is a measurable multifunction. Choose

$$ v^{\prime }(t)\in \mathcal{F}\bigl(t,x(t),(\phi x) (t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\beta _{1}}x(t), \ldots ,{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{n}} x(t) \bigr) $$

such that $|v(t)-v'(t)|\leq \eta (t) (1+\zeta _{0}+\sum_{i=1}^{n}\frac{B( \beta _{i})}{(1-\beta _{i})^{2}} )\|x-y\|$ and $u^{\prime }(t)=a_{ \zeta }v^{\prime }(t)+b_{\zeta }\int _{0}^{t}v^{\prime }(s)\,ds$ for all $t\in I$. Since $|u(t)-u'(t)|\leq a_{\zeta }(v(t)- v^{\prime }(t))+b _{\zeta }\int _{0}^{t}(v(s)-v^{\prime }(s))\,ds$ and

$$\bigl\vert {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{i}} u(t)-{}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\beta _{i}} u^{\prime }(t) \bigr\vert \leq \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \bigl\vert u(t)-u^{\prime }(t) \bigr\vert , $$

we get

$$ \begin{aligned} \max_{t\in I} \bigl\vert u(t)-u'(t) \bigr\vert \leq{}& a_{\zeta }\eta ^{\ast } \Biggl(1 + \zeta _{0}+\sum_{i=1}^{n} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \Vert x-y \Vert \\ & {}+b_{\zeta }\eta ^{\ast } \Biggl(1 +\zeta _{0} + \sum_{i=1}^{n}\frac{ B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggr) \Vert x-y \Vert \\ ={}&\eta ^{\ast } \Biggl(1+\zeta _{0} + \sum _{i=1}^{n}\frac{B(\beta _{i})}{(1- \beta _{i})^{2}} \Biggr) \Vert x-y \Vert \end{aligned} $$

and

$$ \max_{t\in I} \bigl\vert {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{i}}u(t)-{}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{\beta _{i}}u'(t) \bigr\vert \leq \eta ^{\ast }\frac{B(\beta _{i})}{(1-\beta _{i})^{2}} \Biggl(1+\zeta _{0}+\sum _{i=1}^{n}\frac{1}{(1- \beta _{i})^{2}} \Biggr) \Vert x-y \Vert | $$

for $1\leq i\leq n$. Hence, $\|u-u^{\prime }\| \leq \eta ^{\ast } (1+\sum_{i=1}^{n} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} ) (1+\zeta _{0}+\sum_{i=1}^{n}\frac{B(\beta _{i})}{(1-\beta _{i})^{2}} )\|x-y\|$. This implies that $H_{d}(\mathcal{A}x, \mathcal{A}y) \leq L \|x-y\|$. Now, by using Theorem 7, the inclusion fixed point problem $x\in \mathcal{A}x +\mathcal{B}x$ has a solution which is a solution for the inclusion problem (3). □

Now, we are ready to investigate the fractional integro-differential inclusion

$$\begin{aligned}& {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta } \biggl( \frac{x(t)}{g(t,x(t),( \phi x)(t),(\varphi x)(t),{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta _{1}}x(t), \dots , {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta _{n}}x(t))} \biggr) \\& \quad \in \mathcal{G}\bigl(t,x(t),(\phi x) (t),(\varphi x) (t),{}^{\mathrm{CF}}_{N} {\mathcal{D}}^{\beta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\beta _{k}}x(t)\bigr) \end{aligned}$$

(4)

with boundary condition $u(0)=0$, where $\zeta ,\zeta _{1},\dots ,\zeta _{n},\beta _{1},\dots ,\beta _{k}\in (0,1)$, $g: I\times \mathbb{R}^{n+3} \to \mathbb{R}\backslash \{0\}$ is continuous and $\mathcal{G}:I \times \mathbb{R}^{k+3}\to \mathcal{P}(\mathbb{R})$ is a multifunction. We say that $x\in \mathcal{X}$ is a solution for problem (4) whenever it satisfies the boundary conditions and there exists $v\in S_{ \mathcal{G},x}$ such that

$$ \begin{aligned} x(t) ={}& g\bigl(t,x(t),(\phi x) (t),(\varphi x) (t), {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\zeta _{1}}x(t),\ldots ,{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta _{n}}x(t)\bigr) \\ & {} \times \biggl(a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t} v(s)\,ds \biggr), \end{aligned} $$

where

$$ \begin{aligned} S_{\mathcal{G},x} ={}& \bigl\{ v\in L^{1}[0,1]:v(t) \in \mathcal{G}\bigl(t,x(t),( \phi x) (t),(\varphi x) (t), \\ & {} {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{k}}x(t) \bigr) \text{ for almost all } t \in I \bigr\} . \end{aligned} $$

Theorem 14

Suppose that $\mathcal{G}: I\times \mathbb{R}^{k+3}\to \mathcal{P} _{cp,cv}(\mathbb{R})$ is a Caratheodory set-valued map, $g: J\times \mathbb{R}^{n+3}\to \mathbb{R}\backslash \{0\}$ is a bounded continuous map with upper bound K and there are continuous functions $p, m:J\to (0,\infty )$ such that $\|\mathcal{G}(t, x_{1}, x_{2}, \ldots ,x_{k+3})\|\leq m(s)$ and

$$ \bigl\vert g(t,x_{1}, x_{2},\ldots ,x_{n+3})-g(t,y_{1}, y_{2},\ldots ,y_{n+3}) \bigr\vert \leq \eta (t)\sum _{i=1}^{n+3} \vert x_{i}-y_{i} \vert $$

for all $t\in I$. If $\eta ^{\ast }(1+\zeta _{0}+\iota _{0}+ \sum_{i=1} ^{n} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}})\cdot K\cdot \|m\|_{\infty }<1$, then the inclusion problem (4) has a solution.

Proof

Put $S=\{x\in {\mathcal{X}}: \|x\|\leq L\}$, where $L=K\|m\|_{\infty }$. It is clear that S is a convex, closed, and bounded subset of the Banach space $\mathcal{X}$. Define $\mathcal{A}, \mathcal{B} : S \to \mathcal{P}({\mathcal{X}})$ by

$$\mathcal{A}x(t)=g\bigl\{ t,x(t),(\phi x) (t),(\varphi x) (t), {}^{\mathrm{CF}}_{N} { \mathcal{D}}^{\zeta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{ \zeta _{n}}x(t)\bigr\} $$

and

$$ \mathcal{B}x(t)=\biggl\{ u\in {\mathcal{X}}: \text{ there is } v\in S_{ \mathcal{G},x} \text{ such that } u(t)= a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds \text{ for all } t\in I\biggr\} . $$

Thus, the problem of fractional differential inclusions is equivalent to the inclusion problem $x\in \mathcal{A}(x)\mathcal{B}(x)$. Consider the operator $\mathcal{B}=\theta \circ S_{\mathcal{G}}$, where θ is the continuous linear operator on $L^{1}(I)$ into $\mathcal{X}$ defined by $\theta v(s)=a_{\zeta }v(t)+b_{\zeta }\int _{0}^{t}v(s)\,ds$. Let $x\in S$ be arbitrary and $\{v_{n}\}$ be a sequence in $S_{\mathcal{G},x}$. Then $v_{n}(t)\in \mathcal{G}(t,x(t),(\phi x)(t),( \varphi x)(t),{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\beta _{k}}x(t) )$ for almost $t\in I$. Since

$$\mathcal{G}\bigl(t,x(t),(\phi x) (t),(\varphi x) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\beta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N} { \mathcal{D}}^{\beta _{k}}x(t) \bigr) $$

is compact for all $t\in I$, there is a convergent subsequence of $\{v_{n}(t)\}$ (show it by $\{v_{n}(t)\}$ again) to some $v\in S_{ \mathcal{G},x}$. Note that $\theta v_{n}(t)\to \theta v(t)$ pointwise on I because θ is continuous. Now, we show that $\{\theta v _{n}\}$ is an equi-continuous sequence. Let $\tau < t\in I$. Then we have $|\theta v_{n}(t)- \theta v_{n}(\tau )|\leq a_{\zeta }|v_{n}(t)-v _{n}(\tau )|+b_{\zeta }\int _{\tau }^{t} |v_{n}(s)| \,ds$. Thus, the sequence $\{\theta v_{n}\}$ is equi-continuous because the right-hand of the inequality tends to 0 as $t\to \tau$. Hence, it has a uniformly convergent subsequence by using the Arzela–Ascoli theorem. Choose a subsequence of $\{v_{n}\}$ (we show it again by $\{v_{n}\}$) such that $\theta v_{n}\to \theta v$. Hence, $\theta v \in \theta (S _{\mathcal{G} ,x})$ and so $\mathcal{B}=\theta (S_{\mathcal{G} ,x})$ is compact for all $x\in S$. Here, we prove that $\mathcal{B}x$ is convex for all $x\in S$. Let $x\in S$ and $u, u^{\prime }\in \mathcal{B}x$. Choose $v,v^{\prime }\in S_{\mathcal{G},x}$ such that $u(t)=a_{\zeta }v(t)+b_{\zeta }\int _{0}^{t}v(s)\,ds$ and $u'(t)=a_{\zeta }v^{\prime }(t)+b _{\zeta }\int _{0}^{t}v^{\prime }(s)\,ds$ for almost all $t\in I$. Let $0\leq \lambda \leq 1$. Then we have

$$ \lambda u(t)+(1-\lambda )u^{\prime }(t)=a_{\zeta } \bigl(\lambda v(t)+(1- \lambda ) v^{\prime }(t) \bigr)+b_{\zeta } \int _{0}^{t} \bigl(\lambda v(s)+(1- \lambda ) v^{\prime }(s) \bigr)\,ds. $$

Since $\mathcal{G}$ is convex-valued, $\lambda u+(1-\lambda )u^{ \prime }\in \mathcal{B}x$. It is clear that $\mathcal{A}$ is bounded, closed, and convex-valued. We show that $\mathcal{A}x \mathcal{B}x$ is a convex subset of S for all $x\in S$. Let $x\in S$ and $u,u'\in \mathcal{A}x \mathcal{B}x$. Choose $v,v'\in S_{\mathcal{G},x}$ such that

$$ u(t)=g\bigl(t,x(t),(\phi x) (t),(\varphi x) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\zeta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\zeta _{n}}x(t)\bigr) \times \biggl(a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds \biggr), $$

and

$$ \begin{aligned} u'(t) & =g\bigl(t,x(t),(\phi x) (t),( \varphi x) (t), {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{ \zeta _{1}}x(t), \ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\zeta _{n}}x(t)\bigr) \\ & \quad \times \biggl(a_{\zeta }v^{\prime }(t)+b_{\zeta } \int _{0}^{t}v ^{\prime }(s)\,ds \biggr) \end{aligned} $$

for almost all $t\in I$. Hence,

$$ \begin{aligned} \lambda u(t) + (1-\lambda )u^{\prime }(t) ={}& g \bigl(t,x(t),(\phi x) (t),( \varphi x) (t), \\ & {} {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta _{1}}x(t), \ldots , {}^{\mathrm{CF}}_{N}{\mathcal{D}}^{\zeta _{n}}x(t)\bigr) \\ & {}\times \biggl[a_{\zeta }\bigl(\lambda v(t)+(1-\lambda )v^{\prime }(t)\bigr) \\ & {} + b_{\zeta } \int _{0}^{t}\bigl(\lambda v(s)+(1-\lambda )v^{\prime }(s)\bigr)\,ds \biggr]. \end{aligned} $$

Note that $\lambda u+(1-\lambda )u^{\prime }\in \mathcal{A}x \mathcal{B}x$ because $\mathcal{G}$ is convex-valued. Hence, $\mathcal{A}x\mathcal{B}x$ is a convex subset of $\mathcal{X}$ for all $x\in \mathcal{X}$. However, we have

$$\begin{aligned} \bigl\vert u(t) \bigr\vert &= \biggl\vert g\bigl(t,x(t),(\phi x) (t),( \varphi x) (t), {}^{\mathrm{CF}}_{N} {\mathcal{D}}^{\zeta _{1}}x(t), \dots , {}^{\mathrm{CF}}_{N}{\mathcal{D}} ^{\zeta _{n}}x(t)\bigr) \times \biggl(a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds \biggr) \biggr\vert \\ &\leq K(a_{\zeta }+b_{\zeta }) \Vert m \Vert _{\infty }=L< 1 \end{aligned}$$

for all $t\in I$, and so $u\in S$ and $\mathcal{A}x\mathcal{B}x$ is a convex subset of S for all $x\in S$. Now, we show that the operator ${\mathcal{B}}$ is compact. It is enough to prove that $\mathcal{B}(S)$ is uniformly bounded and equi-continuous. Let $u\in \mathcal{B}(S)$. Choose $v\in S_{\mathcal{G},x}$ such that

$$ u(t)=g\bigl(t,x(t),(\phi x) (t),(\varphi x) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\zeta _{1}}x(t),\dots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\zeta _{n}}x(t)\bigr) \times \biggl(a_{\zeta }v(t)+b_{\zeta } \int _{0}^{t}v(s)\,ds \biggr) $$

for some $x\in S$. Since $|u(t)|\leq K(a_{\zeta }+b_{\zeta })\|m\| _{\infty }$, $\|u\|_{\infty }=\max_{t\in I}|u(t)|\leq K(a_{\zeta }+b _{\zeta })\|m\|_{\infty }$. Now, we prove that $\mathcal{B}$ maps S to equi-continuous subsets of $\mathcal{X}$. Let $t, \tau \in J$ with $\tau < t$, $x\in S$, and $u \in \mathcal{B}x$. Choose $v\in S _{\mathcal{G},x}$ such that $u(t)=a_{\zeta }v(t)+b_{\zeta }\int _{0} ^{t}v(s)\,ds$. Then we have

$$ \bigl\vert u(t)- u(\tau ) \bigr\vert \leq a_{\zeta } \bigl\vert v(t)-v(\tau ) \bigr\vert +b_{\zeta } \int _{\tau } ^{t} \bigl\vert v(s) \bigr\vert \,ds. $$

Note that the right-hand side of this inequality tends to 0 as $t\to \tau$. By using the Arzela–Ascoli theorem, we get $\mathcal{B}$ is compact. Here, we show that $\mathcal{B}$ has a closed graph. Let $x_{n}\in S$ and $u_{n}\in \mathcal{B}x_{n}$ for all n with $x_{n}\to x'$ and $u_{n}\to u'$. We show that $u'\in \mathcal{B}x'$. For each n, choose $v_{n}\in S_{\mathcal{G},x_{n}}$ such that $u_{n}(t)=a_{\zeta }v_{n}(t)+b_{\zeta }\int _{0}^{t}v_{n}(s)\,ds$ for all $t\in J$. Again, consider the continuous linear operator $\theta :L ^{1}(I)\to {\mathcal{X}}$ such that $\theta (v)(t)=u(t)=a_{\zeta }v(t)+b _{\zeta }\int _{0}^{t}v(s)\,ds$. By using Lemma 4, $\theta \circ S _{\mathcal{G}}$ is a closed graph operator. Since $x_{n}\to x'$ and $u_{n}\in \theta (S_{\mathcal{G},x_{n}})$ for all n, there is $v'\in S_{\mathcal{G},x'}$ such that $u'(s)=a_{\zeta }v^{\prime }(t)+b _{\zeta }\int _{0}^{t}v^{\prime }(s)\,ds$. Hence, $u'\in \mathcal{B}x'$. Thus, $\mathcal{B}$ has a closed graph and so $\mathcal{B}$ is upper semi-continuous. Finally note that

$$\begin{aligned} H(\mathcal{A} x, \mathcal{A}y) =& \Vert \mathcal{A}x-\mathcal{A}y \Vert \\ =&\max_{t\in I}\bigl| g\bigl(t,x(t),(\phi x) (t),(\varphi x) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \zeta _{1}}x(t),\ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{ \zeta _{n}}x(t)\bigr) \\ &{} -g\bigl(t,y(t),(\phi y) (t),(\varphi y) (t), {}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\zeta _{1}}y(t),\ldots , {}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\zeta _{n}}y(t)\bigr))\bigr| \\ \leq& \max_{t\in I} \bigl\vert \eta (t) \bigr\vert \Biggl( 1 + \zeta _{0}+\iota _{0}+\sum_{i=1}^{n} \frac{B(\zeta _{i})}{(1-\zeta _{i})^{2}} \Biggr) \bigl\vert x(t)-y(t) \bigr\vert \\ =&\eta ^{\ast } ( \Biggl(1+\zeta _{0}+\iota _{0}+ \sum_{i=1}^{n}\frac{B( \zeta _{i})}{(1-\zeta _{i})^{2}} \Biggr) \Vert x-y \Vert _{\infty } \end{aligned}$$

for all $x,y\in {\mathcal{X}}$. Now, by using Theorem 8, the inclusion problem $x\in \mathcal{A}x \mathcal{B}x$ has a solution which is a solution for problem (4). □

In this part, we show that the set of solutions for the second fractional integro-differential inclusion problem is infinite dimensional under some conditions. First we prove the next result.

Lemma 15

Suppose that $m\in L^{1}(I,\mathbb{R}^{+})$, $\mathcal{F}: I\times \mathbb{R}^{m+n+3}\to \mathcal{P}_{cv,cp}(\mathbb{R})$ is a multivalued map such that the map $t\vdash f(t,x_{1},x_{2},\ldots ,x_{3+m+n})$ is measurable and

$$\bigl\Vert \mathcal{F}(t,x_{1},x_{2},\ldots ,x_{m+n+3}) \bigr\Vert =\sup \bigl\{ \vert f \vert : f\in \mathcal{F}(t,x_{1},x_{2},\ldots ,x_{m+n+3})\bigr\} \leq m(t) $$

for almost all $t\in I$ and $\in x_{1},x_{2},\ldots ,x_{m+n+3}\in \mathbb{R}$. Define $\varPhi :{\mathcal{X}}\to \mathcal{P}({\mathcal{X}})$ by

$$ \varPhi (x)= \biggl\{ g\in {\mathcal{X}}: \textit{there is } f\in S_{ \mathcal{F},x} \textit{ such that } g(t)=a_{\zeta }f(t)+b_{\zeta } \int _{0} ^{t}f(s)\,ds \textit{ for all } t\in I \biggr\} . $$

Then $\varPhi (x)\in \mathcal{P}_{cp.cv}({\mathcal{X}})$ for all $x\in {\mathcal{X}}$.

Proof

Note that $\varPhi =\theta \circ S_{\mathcal{F}}$, where $\theta :L^{1}(I, \mathbb{R})\to \mathcal{X}$ is the continuous linear map defined by $\theta g(t)=a_{\zeta } f(t)+b_{\zeta } \int _{0}^{t}f(s)\,ds$. Let $x\in {\mathcal{X}}$ and $\{g_{n}\}$ be a sequence in $S_{\mathcal{F},x}$. Then we have

$$ g_{n}(t)\in \mathcal{F}\bigl(t,x(t),(\phi x) (t),(\psi x) (t),{}^{\mathrm{CF}}_{N} {\mathcal{D}}^{\beta _{1}}x(t),\dots ,{}^{\mathrm{CF}}_{N}{\mathcal{D}}^{ \beta _{m}}x(t), {}^{cF}\mathcal{I}^{\gamma _{1}}x(t),\dots ,{}^{cF} \mathcal{I}^{\gamma _{n}}x(t)\bigr) $$

for almost $t\in I$. Since

$$ \mathcal{F}\bigl(t,x(t),(\phi x) (t),(\psi x) (t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\beta _{1}}x(t),\dots ,{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{m}}x(t), {}^{cF} \mathcal{I}^{\gamma _{1}}x(t), \dots ,{}^{cF}\mathcal{I}^{\gamma _{n}}x(t)\bigr) $$

is compact for all $t\in I$, there is a convergent subsequence of $\{g_{n}(t)\}$ (show it by $\{g_{n}(t)\}$) which converges to some $g\in S_{\mathcal{F},x}$. Note that $\theta g_{n}(t)\to \theta g(t)$ pointwise on I because θ is continuous. Here, we prove that $\{\theta g_{n}\}$ is an equi-continuous sequence. Let $\tau < t \in I$. Then we have $|\theta g_{n}(t)- \theta g_{n}(\tau )|= a_{ \zeta } (f(t)-f(\tau ) )+b_{\zeta } \int _{\tau }^{t}f(s)\,ds$. Note that the sequence $\{\theta g_{n}\}$ is equi-continuous because the right-hand side of the inequality tends to zero when $\tau \to t$. Thus, there is a uniformly convergent subsequence of $\{g_{n}\}$ (show it by $\{g_{n}\}$ again) such that $\theta g_{n}\to \theta g$ (we use the Arzela–Ascoli theorem). This implies that $\theta g \in \theta (S _{\mathcal{F},x})$. Hence, $\varPhi x=\theta (S_{\mathcal{F},x})$ is compact for all $x\in {\mathcal{X}}$. Now, we show that Φx is convex for each $x\in {\mathcal{X}}$. Let $g, g^{\prime }\in \varPhi x$. Choose $f,f^{\prime }\in S_{\mathcal{F},x}$ such that $g(t)=a_{\zeta }f(t)+b_{\zeta } \int _{0}^{t}f(s)\,ds)$ and $g'(t)=a_{\zeta }f^{\prime }(t)+b_{\zeta } \int _{0}^{t}f^{\prime }(s)\,ds)$ for almost all $t\in I$. Let $0\leq \lambda \leq 1$. Then we have

$$ \lambda g(t)+(1-\lambda )g^{\prime }(t)=a_{\zeta } \bigl(\lambda f(t)+(1- \lambda )f^{\prime }(t) \bigr) +b_{\zeta } \int _{0}^{t} \bigl(\lambda f(s)+(1- \lambda )f^{\prime }(s) \bigr)\,ds. $$

Since $S_{\mathcal{F},x}$ is convex, $\lambda g+(1-\lambda )g^{\prime }\in \varPhi x$. This completes the proof. □

Note that the fixed point set of Φ is equal to the set of solutions for the inclusion problem (2). Now by using some different conditions, we show that the set of solutions for the fractional integro-differential inclusion problem could be infinite dimensional.

Theorem 16

Suppose that $\eta \in L^{1}(I,\mathbb{R}^{+})$, $\mathcal{F}: I \times \mathbb{R}^{m+n+3}\to \mathcal{P}_{cv,cp}(\mathbb{R})$ is a multivalued map such that the function $t\vdash \mathcal{F}(t,x_{1},x _{2},\ldots ,x_{m+n+3})$ is measurable,

$$H\bigl(\mathcal{F}(t,x_{1},x_{2},\ldots ,x_{m+n+3}), \mathcal{F}(t,y_{1},y _{2},\ldots ,y_{m+n+3})\bigr) \leq \eta (t)\sum_{i=1}^{m+n+3} \vert x_{i}-y_{i} \vert $$

and $\|\mathcal{F}(t,x_{1},x_{2},\ldots ,x_{m+n+3})\|=\sup \{|f| : f \in \mathcal{F}(t,x_{1},x_{2},\ldots ,x_{m+n+3})\}\leq \eta (t)$ for almost all $t\in I$ and $\in x_{1},x_{2},\ldots ,x_{m+n+3}, y_{1}, y _{2}, y_{m+n+3} \in \mathbb{R}$. If Lebesgue measure of the set

$$ \bigl\{ t: \dim \mathcal{F}(t,x_{1},x_{2},\ldots ,x_{m+n+3})< 1 \textit{ for some } x_{1},x_{2},\ldots ,x_{m+n+3}\in \mathbb{R}\bigr\} $$

is zero and $\Delta <1$, then the set of all solutions for problem (2) is infinite dimensional, where $\Delta =\eta ^{\ast } (1+n+\zeta _{0}+\iota _{0}+\sum_{i=1}^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} ) (1+n + \sum_{i=1}^{m} \frac{B(\beta _{i})}{(1-\beta _{i})^{2}} )$.

Proof

Similar to Lemma 15, define the multivalued map $\varPhi :{\mathcal{X}} \to \mathcal{P}({\mathcal{X}})$ by

$$ \varPhi (x)= \biggl\{ g\in {\mathcal{X}}: ~\text{there is} ~f\in S_{ \mathcal{F},x} ~\text{such that}~ g(t)=a_{\zeta }f(t)+b_{\zeta } \int _{0}^{t}f(s)\,ds ~\text{for all}~ t\in I \biggr\} . $$

By using Lemma 15, $\varPhi x\in \mathcal{P}_{cp,cv}({\mathcal{X}})$ for all $x\in \mathcal{X}$. By using a similar proof in Theorem 12, we can prove that Φ is a contractive multivalued map. Now, we show that $\dim \varPhi x>k$ for all $x\in \mathcal{X}$ and $k\geq 1$. Let $k\geq 1$, $x\in {\mathcal{X}}$, and

$$ \begin{aligned} \mathcal{G}(t)= {}& \mathcal{F}\bigl(t,x(t),(\phi x) (t),(\psi x) (t),{}^{\mathrm{CF}} _{N}{ \mathcal{D}}^{\beta _{1}}x(t),{}^{\mathrm{CF}}_{N}{ \mathcal{D}} ^{\beta _{2}}x(t), \dots ,{}^{\mathrm{CF}}_{N}{ \mathcal{D}}^{\beta _{m}}x(t), \\ & {} {}^{cF} \mathcal{I}^{\gamma _{1}}x(t), {}^{cF} \mathcal{I}^{ \gamma _{2}}x(t),\ldots ,{}^{cF}\mathcal{I}^{\gamma _{n}}x(t) \bigr) \end{aligned} $$

for all $t\in I$. By using Lemma 9, there are linearly independent measurable selections $g_{1},\dots ,g_{k}$ for $\mathcal{G}$. Consider the maps $h_{i}(t)=a_{\zeta }g_{i}(t)+b_{ \zeta }(t)\int _{0}^{t}g_{i}(s)\,ds$ for $i=1,\dots ,k$. Assume that $\sum_{i=1}^{k} a_{i} h_{i}(t)=0$ for almost $t\in I$. Since $a_{\zeta },b_{\zeta }\neq 0$, by using the Caputo–Fabrizio derivatives, we get $\sum_{i=1}^{k} a_{i} g_{i}(t)=0$ for almost $t\in I$. Hence, $a_{1}=\cdots =a_{k}=0$. This implies that $h_{1}, \dots ,h_{k}$ are linearly independent, and so $\dim \varPhi x\geq k$. Hence, we conclude that the set of fixed points of Φ is infinite dimensional by using Theorem 10. Thus, the set of all solutions for problem (2) is infinite dimensional. □

3 Conclusion

We guess that researchers will review different more fractional integro-differential inclusions in the near future. In this manuscript, we first investigate the existence of solutions for four fractional integro-differential inclusions including the new Caputo–Fabrizio derivation which has been introduced recently. Also, we show that dimension of the set of solutions for the second fractional integro-differential inclusion problem is infinite dimensional under some different conditions.