Theory and Modern Applications

# Existence results for multi-term fractional differential inclusions

## Abstract

In this paper, we study the existence of solutions for a new class of boundary value problems for nonlinear multi-term fractional differential inclusions. Our main result relies on the multi-valued form of Krasnoselskii’s fixed point theorem. An illustrative example is also presented.

## Introduction and preliminaries

In this paper we study the existence of solutions for the following multi-term fractional differential inclusions:

\begin{aligned} {}^{c}D^{\alpha}u(t) \in& F\bigl(t, u(t), u'(t), u''(t), {}^{c}D^{q_{1}}u(t), \ldots, {}^{c}D^{q_{k}}u(t) \bigr) \\ &{}+ G\bigl(t, u(t), u'(t), u''(t), {}^{c}D^{q_{1}}u(t), \ldots, {}^{c}D^{q_{k}}u(t) \bigr) \end{aligned}
(1.1)

supplemented with boundary conditions

$$u(0)=0, \qquad u'(0)=-u(1)- u'(1),\qquad u''( 0) = - u''(1)- {}^{c}D^{p}u(1),$$
(1.2)

where $${}^{c}D^{\alpha}$$, $${}^{c}D^{q_{i}}$$ denote the Caputo fractional derivatives, $$2 < \alpha\leq3$$, $$1 < q_{i} \leq2$$, $$i=1,2,\ldots, k$$, $$t\in J:=[0,1]$$, $$1< p\leq2$$, $$k\geq1$$, and $$F, G: J\times\mathbb {R}^{k+3} \to{\mathcal{P}}(\mathbb{R})$$ are multifunctions.

Many of published papers about fractional differential equations and inclusions apply the fixed point theory for proving the existence results. For instance, one can find a lot of papers in this field (see  and the references therein).

Let $$\alpha>0$$, $$n-1<\alpha<n$$, $$n=[\alpha]+1$$, and $$u\in C([a,b], \mathbb{R})$$. The Caputo derivative of fractional of order α for the function u is defined by $${}^{c}D^{\alpha} u(t)= \frac{1}{\Gamma (n-\alpha)}\int_{0}^{t} (t-\tau)^{n-\alpha-1}u^{(n)}(\tau)\, d\tau$$ (see for more details [11, 23, 2527]). Also, the Riemann-Liouville fractional order integral of the function u is defined by $$I^{\alpha}u(t)= \frac{1}{\Gamma(\alpha)} \int_{0}^{t} \frac{u(\tau)}{(t-\tau )^{1-\alpha}}\, d\tau$$ ($$t>0$$) whenever the integral exists [11, 23, 2527]. In , it has been proved that the general solution of the fractional differential equation $${}^{c}D^{\alpha}u( t)=0$$ is given by $$u( t)=c_{0}+c_{1}t+c_{2}t^{2}+\cdots+c_{n-1}t^{n-1}$$, where $$c_{0},\ldots,c_{n-1}$$ are real constants and $$n=[\alpha]+1$$. Also, for each $$T>0$$ and $$u\in C([0,T])$$ we have

$$I^{\alpha} {}^{c}D^{\alpha}u( t)=u(t)+c_{0}+c_{1}t+c_{2}t^{2}+ \cdots+c_{n-1}t^{n-1},$$

where $$c_{0},\ldots,c_{n-1}$$ are real constants and $$n=[\alpha]+1$$ .

Now, we review some definitions and notations as regards multifunctions [29, 30].

For a normed space $$(X, \|\cdot\|)$$, let $${\mathcal{P}}_{\mathrm{cl}}(X)=\{Y \in {\mathcal{P}}(X) : Y \mbox{ is closed}\}$$, $${\mathcal{P}}_{\mathrm{b}}(X)=\{Y \in{\mathcal{P}}(X) : Y \mbox{ is bounded}\}$$, $${\mathcal{P}}_{\mathrm{cp}}(X)=\{Y \in{\mathcal{P}}(X) : Y \mbox{ is compact}\}$$, and $${\mathcal{P}}_{\mathrm{cp}, \mathrm{cv}}(X)=\{Y \in{\mathcal{P}}(X) : Y \mbox{ is compact and convex}\}$$, $${\mathcal{P}}_{\mathrm{b}, \mathrm{cl}, \mathrm{cv}}(X)=\{Y \in{\mathcal{P}}(X) : Y \mbox{ is bounded, closed, and convex}\}$$. A multi-valued map $$G : X \to{\mathcal{P}}(X)$$ is convex (closed) valued if $$G(x)$$ is convex (closed) for all $$x \in X$$. The map G is bounded on bounded sets if $$G(\mathbb{B}) = \bigcup_{x \in\mathbb{B}}G(x)$$ is bounded in X for all $$\mathbb{B} \in{\mathcal{P}}_{\mathrm{b}}(X)$$ (i.e., $$\sup_{x \in\mathbb{B}}\{\sup\{|y| : y \in G(x)\}\} < \infty$$). G is called upper semi-continuous (u.s.c.) on X if for each $$x_{0} \in X$$, the set $$G(x_{0})$$ is a nonempty closed subset of X, and if for each open set N of X containing $$G(x_{0})$$, there exists an open neighborhood $$\mathcal{N}_{0}$$ of $$x_{0}$$ such that $$G(\mathcal{N}_{0}) \subseteq N$$. G is said to be completely continuous if $$G(\mathbb{B})$$ is relatively compact for every $$\mathbb{B} \in{\mathcal{P}}_{\mathrm{b}}(X)$$. If the multi-valued map G is completely continuous with nonempty compact values, then G is u.s.c. if and only if G has a closed graph, i.e., $$u_{n} \to u_{*}$$, $$y_{n} \to y_{*}$$, $$y_{n} \in G(u_{n})$$ imply $$y_{*} \in G(u_{*})$$. G has a fixed point if there is $$x \in X$$ such that $$x \in G(x)$$. The fixed point set of the multi-valued operator G will be denoted by FixG. A multi-valued map $$G : J \to {\mathcal{P}}_{\mathrm{cl}}(\mathbb{R})$$ is said to be measurable if for every $$y \in \mathbb{R}$$, the function $$t \mapsto d(y,G(t)) = \inf\{|y-z|: z \in G(t)\}$$ is measurable.

Consider the Pompeiu-Hausdorff metric $$H_{d} : {\mathcal{P}}(X) \times {\mathcal{P}}(X) \to\mathbb{R} \cup\{\infty\}$$ given by

$$H_{d}(A, B) = \max\Bigl\{ \sup_{a \in A}d(a,B), \sup _{b \in B}d(A,b)\Bigr\} ,$$

where $$d(A,b) = \inf_{a\in A}d(a;b)$$ and $$d(a,B) = \inf_{b\in B}d(a;b)$$. A multi-valued operator $$N : X \to{\mathcal{P}}_{\mathrm{cl}}(X)$$ is called contraction if there exists $$\gamma \in(0,1)$$ such that $$H_{d}(N(x),N(y)) \le\gamma d(x,y)$$ for each $$x, y \in X$$.

We say that $$F: J\times\mathbb{R}^{k+3} \rightarrow{\mathcal{P}}(\mathbb{R})$$ is a Carathéodory multifunction if $$t\mapsto F(t,u_{1}, \ldots, u_{k+3})$$ is measurable for all $$u_{i} \in\mathbb{R}$$ and $$(u_{1}, \ldots, u_{k+3})\mapsto F(t,u_{1}, \ldots, u_{k+3})$$ is upper semi-continuous for almost all $$t\in J$$ [29, 31]. Also, a Carathéodory multifunction $$F: J\times \mathbb{R}^{k+3} \rightarrow{\mathcal{P}}(\mathbb{R})$$ is called $$L^{1}$$-Carathéodory if for each $$\rho>0$$ there exists $$\phi_{\rho}\in L^{1}(J,\mathbb{R}^{+})$$ such that

$$\bigl\Vert F(t,u_{1}, \ldots, u_{k+3}) \bigr\Vert =\sup _{t\in J}\bigl\{ |s|:s\in F(t,u_{1}, \ldots, u_{k+3})\bigr\} \leq\phi_{\rho}(t)$$

for all $$|u_{1}|, \ldots, |u_{k+3}|\leq\rho$$ and for almost all $$t\in J$$ [29, 31].

Define the set of selections of F and G at $$u \in C(J,\mathbb{R})$$ by

$$S_{F,u}:=\bigl\{ v\in L^{1}(J,\mathbb{R}): v(t)\in F\bigl(t, u(t), u'(t), u''(t), {}^{c}D^{q_{1}}u(t), \ldots, {}^{c}D^{q_{k}}u(t) \bigr)\bigr\}$$

and

$$S_{G,u}:=\bigl\{ v_{1}\in L^{1}(J,\mathbb{R}): v_{1}(t)\in G\bigl(t, u(t), u'(t), u''(t), {}^{c}D^{q_{1}}u(t), \ldots, {}^{c}D^{q_{k}}u(t) \bigr)\bigr\}$$

for almost all $$t\in J$$. If F is an arbitrary multifunction, then it has been proved that $$S_{F}(u)\neq\emptyset$$ for all $$u\in C(J,X)$$ if $$\dim X<\infty$$ .

The graph of a function F is the set $$\operatorname{Gr}(F)=\{ (x,y)\in X\times Y: y\in F(x)\}$$ . The graph $$\operatorname{Gr}(F)$$ of $$F:X\to\mathcal {P}_{\mathrm{cl}}(Y)$$ is said to be a closed subset of $$X\times Y$$, if for every sequence $$\{u_{n}\}_{n \in\mathbb{N}} \subset X$$ and $$\{y_{n}\}_{n \in \mathbb{N}} \subset Y$$, when $$n \to\infty$$, $$u_{n} \to u_{0}$$, $$y_{n} \to y_{0}$$, and $$y_{n} \in F(u_{n})$$, then $$y_{0} \in F(u_{0})$$ .

We will use the following lemmas and theorem in our main result.

### Lemma 1.1

(, Proposition 1.2)

If $$F : X \to\mathcal{P}_{\mathrm{cl}}(Y)$$ is u.s.c., then $$\operatorname{Gr}(F)$$ is a closed subset of $$X \times Y$$. Conversely, if F is completely continuous and has a closed graph, then it is upper semi-continuous.

### Lemma 1.2

()

Let X be a separable Banach space. Let $$F : [0, 1] \times X^{k+3} \to{\mathcal{P}}_{\mathrm{cp},\mathrm{cv}}(X)$$ be an $$L^{1}$$-Carathéodory function. Then the operator

$$\Theta\circ S_{F} : C(J,X) \to{\mathcal{P}}_{\mathrm{cp},\mathrm{cv}} \bigl(C(J,X)\bigr),\qquad x \mapsto(\Theta\circ S_{F}) (x) = \Theta( S_{F,x})$$

is a closed graph operator.

### Theorem 1.3

(, Krasnoselskii’s fixed point theorem)

Let X be a Banach space, $$Y\in{\mathcal{P}}_{\mathrm{b}, \mathrm{cl}, \mathrm{cv}}(X)$$ and $$A, B: Y\to{\mathcal{P}}_{\mathrm{cp},\mathrm{cv}}(X)$$ two multi-valued operators. If the following conditions are satisfied:

1. (i)

$$Ay+By\subset Y$$ for all $$y\in Y$$;

2. (ii)

A is a contraction;

3. (iii)

B is u.s.c. and compact,

then there exists $$y\in Y$$ such that $$y\in Ay+By$$.

## Main results

Now, we are ready to prove our main result. Let $$X=\{u: u,u', u'', {}^{c}D^{q_{i}} u\in C(J,\mathbb{R}), i=1,2,\ldots, k\}$$. Then $$(X, \Vert \cdot\Vert)$$ endowed with the norm

$$\|u\|=\sup_{t\in J} \bigl\vert u(t)\bigr\vert +\sup _{t\in J}\bigl\vert u'(t)\bigr\vert + \sup _{t\in J}\bigl\vert u''(t)\bigr\vert + \sup_{t\in J}\bigl\vert {}^{c}D^{q_{i}} u(t) \bigr\vert \quad (i=1,\ldots, k)$$

is a Banach space .

### Lemma 2.1

Let $$y\in C(J,{\mathbb{R}})$$ and $$u\in C^{2}([0,1],\mathbb{R})$$ is a solution to the fractional boundary value problem

$$\left \{ \begin{array}{l} {}^{c}D^{\alpha}u(t)=y(t ), \\ u(0)=0,\qquad u'(0)=-u(1)- u'(1), \qquad u''( 0) = - u''(1)- {}^{c}D^{p}u(1), \end{array} \right .$$
(2.1)

then

\begin{aligned} u(t) =& \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha)}y(s)\,ds -\frac{t}{3}\int_{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha)}y(s)\,ds-\frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}y(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma(\alpha -2)} y(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}y(s)\,ds, \end{aligned}
(2.2)

and vice versa, where $$\Delta= \frac{\Gamma(3-p)}{4\Gamma(3-p) + 2} \neq0$$.

### Proof

It is well known that the solution of equation $${}^{c}D^{\alpha}u(t)=y(t)$$ can be written as

$$u( t)= \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha)} y(s )\,ds + c_{0} + c_{1}t+c_{2}t^{2},$$
(2.3)

where $$c_{0}, c_{1}, c_{2}\in\mathbb{R}$$. Then we get

\begin{aligned}& u'(t)=\int_{0}^{t} \frac{(t-s )^{\alpha -2}}{\Gamma(\alpha-1)}y(s)\,ds +c_{1}+2c_{2}t, \\& u''(t)=\int_{0}^{t} \frac{(t-s)^{\alpha -3}}{\Gamma(\alpha -2)}y(s)\,ds+2c_{2} \end{aligned}

and

$${}^{c}D^{p} u( t)=\int_{0}^{t} \frac{(t-s)^{\alpha-p-1 }}{\Gamma(\alpha -p)}y(s )\,ds + c_{2}\frac{2t^{2-p}}{\Gamma(3-p)} \quad (1< p\leq2).$$

By using the boundary value conditions, we obtain $$c_{0}=0$$ and

\begin{aligned} c_{1} =& -\frac{1}{3}\int_{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha )}y(s)\,ds -\frac{1}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha -1 )}y(s)\,ds \\ &{}+ \Delta\int_{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma(\alpha-2)} y(s) \,ds + \Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha-p)}y(s) \,ds \end{aligned}

and

$$c_{2} = - \Delta\int_{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma(\alpha-2)} y(s)\,ds - \Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha-p)}y(s)\,ds.$$

Substituting the values of $$c_{0}$$, $$c_{1}$$, and $$c_{2}$$ in (2.3) we get (2.2).

Conversely, applying the operator $${}^{c}D^{\alpha}$$ on (2.2) and taking into account (2.1), it follows that $${}^{c}D^{\alpha }u(t)=y(t)$$. From (2.2) it is easily to verify that the boundary conditions $$u(0)=0$$, $$u'(0)=-u(1)- u'(1)$$, $$u''( 0) = - u''(1)- {}^{c}D^{p}u(1)$$ are satisfied. This establishes the equivalence between (2.1) and (2.2). The proof is completed. □

### Definition 2.2

A function $$u\in C^{2}([0,1],\mathbb{R})$$ is called a solution for the problem (1.1)-(1.2) if it satisfies the boundary value conditions $$u(0)=0$$, $$u'(0)=-u(1)- u'(1)$$, and $$u''( 0) = - u''(1) - {}^{c}D^{p}u(1)$$, there exist functions $$v, v_{1}\in L^{1}(J, {\mathbb{R}})$$ such that $$v(t)\in F(t, u(t), u'(t), u''(t), {}^{c}D^{q_{1}}u(t), \ldots, {}^{c}D^{q_{k}}u(t))$$, $$v_{1}(t) \in G(t, u(t), u'(t), u''(t), {}^{c}D^{q_{1}}u(t), \ldots, {}^{c}D^{q_{k}}u(t))$$ for almost all $$t\in J$$ and

\begin{aligned} u(t) =& \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha)}v(s)\,ds -\frac{t}{3}\int_{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha)}v(s)\,ds \\ &{}-\frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma(\alpha -2)} v(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}v(s)\,ds \\ &{}+ \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha)}v_{1}(s) \,ds -\frac {t}{3}\int_{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha)}v_{1}(s)\,ds \\ &{}-\frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v_{1}(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma (\alpha-2)} v_{1}(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}v_{1}(s)\,ds. \end{aligned}
(2.4)

### Remark 2.3

For the sake of brevity, we set

\begin{aligned}& \Lambda_{1} = \frac{4}{3\Gamma(\alpha+1)}+\frac{1}{3\Gamma(\alpha )}+ \frac{\Delta}{4\Gamma(\alpha-1)}+\frac{\Delta}{4\Gamma(\alpha -p+1)} , \end{aligned}
(2.5)
\begin{aligned}& \Lambda_{2} = \frac{4}{3\Gamma(\alpha)}+\frac{1}{3\Gamma(\alpha +1)}+ \frac{\Delta}{\Gamma(\alpha-1)}+\frac{\Delta}{\Gamma(\alpha -p+1)}, \end{aligned}
(2.6)
\begin{aligned}& \Lambda_{3} = \frac{1+2\Delta}{\Gamma(\alpha-1 )}+\frac{2\Delta }{\Gamma(\alpha-p+1)}, \end{aligned}
(2.7)

and, for each $$i=1,\ldots, k$$,

$$\Lambda_{4}^{i} = \frac{1}{\Gamma(\alpha-q_{i} +1 )}+ \frac{2\Delta}{\Gamma (3-q_{i}) \Gamma(\alpha-1)}+\frac{2\Delta}{\Gamma(3-q_{i})\Gamma (\alpha-p+1)} .$$
(2.8)

Also in the following we use the notation $$\|x\|_{\infty}=\sup\{|x(t)|: t\in J\}$$.

### Theorem 2.4

Suppose that:

(H1):

$$F: J\times\mathbb{R}^{k+3} \to{\mathcal{P}}_{\mathrm{cp},c}(\mathbb{R})$$ is a multifunction and $$G:J\times\mathbb {R}^{k+3} \to{\mathcal{P}}_{\mathrm{cp},c}(\mathbb{R})$$ is a Carathéodory multifunction;

(H2):

there exist continuous functions $$p, m: J \to(0,\infty )$$ such that $$t\mapsto F(t,w_{1},w_{2},w_{3},z_{1}, \ldots, z_{k} )$$ is measurable and

$$\bigl\Vert F(t,w_{1},w_{2},w_{3},z_{1}, \ldots, z_{k} ) \bigr\Vert \leq m(t),\qquad \bigl\Vert G(t,w_{1},w_{2},w_{3},z_{1},\ldots, z_{k} ) \bigr\Vert \leq p(t);$$
(H3):

there exists a continuous function $$h:J\to(0,\infty)$$ such that

\begin{aligned}& H_{d} \bigl(F(t,w_{1},w_{2},w_{3},z_{1}, \ldots, z_{k}) , F\bigl(t,w'_{1},w'_{2},w'_{3},z'_{1}, \ldots, z'_{k}\bigr)\bigr) \\& \quad \leq h(t) \Biggl[ \sum_{i=1}^{3} \bigl\vert w_{i} -w'_{i} \bigr\vert + \sum _{i=1}^{k} \bigl\vert z_{i} -z'_{i} \bigr\vert \Biggr] \end{aligned}

for all $$t\in J$$ and for each $$w_{1},w_{2},w_{3},z_{1},\ldots, z_{k}, w'_{1},w'_{2},w'_{3},z'_{1},\ldots, z'_{k} \in\mathbb{R}$$.

If

$$L:= \Vert h\Vert_{\infty}\bigl(\Lambda_{1} + \Lambda_{2} + \Lambda_{3} +\Lambda_{4}^{i} \bigr)< 1$$

for $$i=1,2,\ldots,k$$, where the $$\Lambda_{j}$$ ($$j=1,\ldots,4$$) are defined in (2.5)-(2.8), then the inclusion problem (1.1)-(1.2) has at least one solution.

### Proof

We define the subset Y of X by $$Y=\{ u\in X: \Vert u\Vert\leq M \}$$, where

$$M=\bigl(\Vert p\Vert _{\infty}+ \Vert m\Vert_{\infty}\bigr) \bigl(\Lambda_{1} +\Lambda_{2} + \Lambda_{3} + \Lambda_{4}^{i} \bigr)\quad (i=1, \ldots, k).$$

It is clear that Y is closed, bounded, and convex subset of Banach space X. We define the multi-valued operators $$A,B:Y\to{\mathcal{P}}(X)$$ such that for some $$v\in S_{F,u}$$,

\begin{aligned} A(u) =&\biggl\{ u \in X: u(t)=\int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha)}v(s)\,ds -\frac {t}{3}\int_{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha)}v(s)\,ds \\ &{}-\frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma(\alpha -2)} v(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha-p)}v(s)\,ds \biggr\} , \end{aligned}

and for some $$v_{1}\in S_{G,u}$$,

\begin{aligned} B(u) =&\biggl\{ u \in X: u(t) = \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha)}v_{1}(s)\,ds -\frac {t}{3}\int _{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha)}v_{1}(s)\,ds \\ &{}-\frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v_{1}(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma (\alpha-2)} v_{1}(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}v_{1}(s)\,ds\biggr\} . \end{aligned}

In this way, the fractional differential inclusion (1.1)-(1.2) is equivalent to the inclusion problem $$u\in Au+Bu$$. We show that the multi-valued operators A and B satisfy the conditions of Theorem 1.3 on Y.

First, we show that the operators A and B define the multi-valued operators $$A,B: Y\to{\mathcal{P}}_{\mathrm{cp},\mathrm{cv}}(X)$$. First we prove that A is compact-valued on Y. Note that the operator A is equivalent to the composition $${\mathcal{L}} \circ S_{F}$$, where $${\mathcal{L}}$$ is the continuous linear operator on $$L^{1}(J, \mathbb{R})$$ into X, defined by

\begin{aligned} {\mathcal{L}} (v) (t) =& \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma (\alpha)}v(s)\,ds -\frac{t}{3}\int_{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma (\alpha)}v(s)\,ds \\ &{}- \frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma(\alpha -2)} v(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}v(s)\,ds . \end{aligned}

Suppose that $$u\in Y$$ is arbitrary and let $$\{v_{n}\}$$ be a sequence in $$S_{F,u}$$. Then, by definition of $$S_{F,u}$$, we have $$v_{n}(t)\in F(t, u(t), u'(t), u''(t), {}^{c}D^{q_{1}}u(t), \ldots, {}^{c}D^{q_{k}}u(t) )$$ for almost all $$t\in J$$. Since $$F(t, u(t), u'(t), u''(t), {}^{c}D^{q_{1}}u(t), \ldots, {}^{c}D^{q_{k}}u(t) )$$ is compact for all $$t\in J$$, there is a convergent subsequence of $$\{v_{n}(t)\}$$ (we denote it by $$\{v_{n}(t)\}$$ again) that converges in measure to some $$v(t)\in S_{F,u}$$ for almost all $$t\in J$$. On the other hand, $${\mathcal{L}}$$ is continuous, so $${\mathcal{L}} (v_{n})(t)\to{\mathcal{L}} (v)(t)$$ pointwise on J.

In order to show that the convergence is uniform, we have to show that $$\{{\mathcal{L}} (v_{n})\}$$ is an equi-continuous sequence. Let $$t_{1}, t_{2} \in J$$ with $$t_{1}< t_{2}$$. Then we have

\begin{aligned}& \bigl\vert {\mathcal{L}} (v_{n}) (t_{2}) - {\mathcal{L}} (v_{n}) (t_{1})\bigr\vert \\& \quad \leq \biggl\vert \frac{1}{\Gamma(\alpha)} \int_{0}^{t_{1}} \bigl[(t_{2}-s)^{\alpha -1}-(t_{1}-s)^{\alpha-1}\bigr] v_{n}(s)\,ds +\frac{1}{\Gamma(\alpha)} \int_{t_{1}}^{t_{2}} (t_{2}-s)^{\alpha-1}v_{n}(s)\,ds \biggr\vert \\& \qquad {} + \frac{\vert t_{2}-t_{1}\vert }{3 \Gamma(\alpha)}\int_{0}^{1} (1-s)^{\alpha-1}\bigl\vert v_{n}(s)\bigr\vert \,ds + \frac{\vert t_{2}-t_{1}\vert }{3 \Gamma(\alpha-1)}\int_{0}^{1} (1-s)^{\alpha-2} \bigl\vert v_{n}(s)\bigr\vert \,ds \\& \qquad {}+ \frac{\Delta \vert [(t_{2}-t_{1})-(t_{2}^{2} -t_{1}^{2} )]\vert }{\Gamma (\alpha-2)}\int_{0}^{1} (1-s)^{\alpha-3}\bigl\vert v_{n}(s)\bigr\vert \,ds \\& \qquad {}+ \frac{\Delta \vert [(t_{2}-t_{1})-(t_{2}^{2} -t_{1}^{2} )]\vert }{\Gamma (\alpha-p)}\int_{0}^{1} (1-s)^{\alpha-p-1}\bigl\vert v_{n}(s)\bigr\vert \,ds \\& \quad \leq \Vert m\Vert_{\infty}\biggl\{ \frac{\vert t_{2}^{\alpha}-t_{1}^{\alpha} \vert }{\Gamma(\alpha+1)} + \frac{\vert t_{2}-t_{1}\vert }{3\Gamma (\alpha+1)} + \frac{\vert t_{2}-t_{1} \vert }{3\Gamma(\alpha)}+ \frac {\Delta \vert [(t_{2}-t_{1})-(t_{2}^{2} -t_{1}^{2} )]\vert }{\Gamma(\alpha-1)} \\& \qquad {}+ \frac{\Delta \vert [(t_{2}-t_{1})-(t_{2}^{2} -t_{1}^{2} )]\vert }{\Gamma (\alpha-p+1)} \biggr\} . \end{aligned}

Continuing this process, we have

$$\bigl\vert \bigl({\mathcal{L}}' (v_{n}) (t_{2})\bigr)- \bigl({\mathcal{L}}' (v_{n}) (t_{1}) \bigr)\bigr\vert \leq \Vert m \Vert_{\infty}\biggl\{ \frac{\vert t_{2}^{\alpha-1} -t_{1}^{\alpha-1} \vert}{\Gamma(\alpha)} + \frac{\Delta\vert t_{2}-t_{1}\vert}{\Gamma(\alpha-1)} + \frac{\Delta\vert t_{2}-t_{1}\vert }{\Gamma(\alpha-p+1 )} \biggr\}$$

and

\begin{aligned} \bigl\vert \bigl({\mathcal{L}}'' (v_{n}) (t_{2})\bigr)- \bigl({\mathcal{L}}'' (v_{n}) (t_{1}) \bigr)\bigr\vert \leq& \biggl\vert \frac{1}{\Gamma(\alpha-2)}\int_{0}^{t_{1}} \bigl[(t_{2}-s)^{\alpha-3}-(t_{1}-s)^{\alpha-3}\bigr] v_{n}(s)\,ds \\ &{}+\frac{1}{\Gamma(\alpha-2)}\int_{t_{1}}^{t_{2}} (t_{2}-s)^{\alpha -3}v_{n}(s)\,ds \biggr\vert \\ \leq& \Vert m \Vert_{\infty}\frac{\vert t_{2}^{\alpha-2} -t_{1}^{\alpha -2} \vert}{\Gamma(\alpha-1 )}, \end{aligned}

and, finally, for every $$i=1,\ldots, k$$,

\begin{aligned}& \bigl\vert \bigl({}^{c}D^{q_{i}}{\mathcal{L}} (v_{n}) (t_{2})\bigr)- \bigl({}^{c}D^{q_{i}}{ \mathcal{L}} (v_{n}) (t_{1}) \bigr)\bigr\vert \\& \quad \leq \Vert m\Vert_{\infty}\biggl\{ \frac{\vert t_{2}^{\alpha- q_{i}} -t_{1}^{\alpha-q_{i}} \vert}{\Gamma(\alpha-q_{i}+1 )} + \frac{2\Delta\vert t_{2}^{2-q_{i}} -t_{1}^{2-q_{i}} \vert}{\Gamma(3-q_{i}) \Gamma(\alpha-1 )} + \frac{2\Delta\vert t_{2}^{2-q_{i}} -t_{1}^{2-q_{i}} \vert}{\Gamma(3-q_{i})\Gamma(\alpha-p+1 )} \biggr\} . \end{aligned}

We see that the right-hand sides of the above inequalities tend to zero as $$t_{2}\to t_{1}$$. Thus, the sequence $$\{{\mathcal{L}} (v_{n})\}$$ is equi-continuous and by using the Arzelá-Ascoli theorem, we see that there is a uniformly convergent subsequence. So, there is a subsequence of $$\{v_{n}\}$$ (we denote it again by $$\{v_{n}\}$$) such that $${\mathcal{L}} (v_{n})\to{\mathcal{L}} (v)$$. Note that $${\mathcal{L}} (v) \in {\mathcal{L}} (S_{F,u})$$. Hence, $$A(u) ={\mathcal{L}} (S_{F,u})$$ is compact for all $$u\in Y$$. So $$A(u)$$ is compact.

Now, we show that $$A(u)$$ is convex for all $$u\in X$$. Let $$z_{1},z_{2}\in A(u)$$. We select $$f_{1},f_{2}\in S_{F,u}$$ such that

\begin{aligned} z_{i}(t) =& \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha )}f_{i}(s)\,ds -\frac{t}{3}\int _{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma (\alpha)}f_{i}(s)\,ds \\ &{}- \frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}f_{i}(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma (\alpha-2)} f_{i}(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}f_{i}(s)\,ds, \quad i=1,2 \end{aligned}

for almost all $$t\in J$$. Let $$0\leq\lambda\leq1$$. Then we have

\begin{aligned} \bigl[\lambda z_{1}+(1-\lambda)z_{2}\bigr](t) =& \int _{0}^{t} \frac{(t-s )^{\alpha -1}}{\Gamma(\alpha)} \bigl[\lambda f_{1}(s) + (1-\lambda)f_{2}(s)\bigr] \,ds \\ &{}- \frac{t}{3}\int_{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha)} \bigl[\lambda f_{1}(s) + (1-\lambda)f_{2}(s) \bigr] \,ds \\ &{}- \frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )} \bigl[\lambda f_{1}(s) + (1-\lambda)f_{2}(s) \bigr]\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma(\alpha-2)} \bigl[\lambda f_{1}(s) + (1-\lambda)f_{2}(s) \bigr] \,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)} \bigl[\lambda f_{1}(s) + (1-\lambda)f_{2}(s) \bigr] \,ds. \end{aligned}

Since F has convex values, $$S_{F,u}$$ is convex and $$\lambda f_{1}(s) + (1-\lambda)f_{2}(s)\in S_{F,u}$$. Thus

$$\lambda z_{1}+(1-\lambda)z_{2}\in A(u) .$$

Consequently, A is convex-valued. Similarly, B is compact and convex-valued.

Here, we show that $$A(u)+ B(u)\subset Y$$ for all $$u\in Y$$. Suppose that $$u\in Y$$ and $$z_{1}\in A(u)$$, $$z_{2}\in B(u)$$ are arbitrary elements. Choose $$v_{1}\in S_{F,u}$$ and $$v_{2}\in S_{G,u}$$ such that

\begin{aligned} z_{1}(t) =& \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha )}v_{1}(s)\,ds -\frac{t}{3}\int _{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma (\alpha)}v_{1}(s)\,ds \\ &{}- \frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v_{1}(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma (\alpha-2)} v_{1}(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}v_{1}(s)\,ds \end{aligned}

and

\begin{aligned} z_{2}(t) =& \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha)}v_{2}(s)\,ds -\frac{t}{3}\int _{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha )}v_{2}(s)\,ds \\ &{}- \frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v_{2}(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma (\alpha-2)} v_{2}(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}v_{2}(s)\,ds \end{aligned}

for almost all $$t\in J$$. Hence, we get

\begin{aligned} \bigl\vert z_{1}(t) + z_{2}(t)\bigr\vert \leq& \frac{1}{\Gamma(\alpha)}\int_{0}^{t} (t-s )^{\alpha-1} \bigl(\bigl\vert v_{1}(s)\bigr\vert +\bigl\vert v_{2}(s)\bigr\vert \bigr) \,ds \\ &{}+ \frac{\vert t\vert }{3\Gamma(\alpha)}\int_{0}^{1} (1-s )^{\alpha -1}\bigl(\bigl\vert v_{1}(s)\bigr\vert +\bigl\vert v_{2}(s)\bigr\vert \bigr) \,ds \\ &{}+ \frac{\vert t\vert }{3\Gamma(\alpha-1)} \int_{0}^{1} (1-s )^{\alpha -2}\bigl(\bigl\vert v_{1}(s)\bigr\vert +\bigl\vert v_{2}(s)\bigr\vert \bigr) \,ds \\ &{}+ \frac{\Delta \vert t-t^{2}\vert }{\Gamma(\alpha-2)}\int_{0}^{1} (1-s)^{\alpha-3}\bigl(\bigl\vert v_{1}(s)\bigr\vert +\bigl\vert v_{2}(s)\bigr\vert \bigr) \,ds \\ &{}+ \frac{\Delta \vert t-t^{2}\vert }{\Gamma(\alpha-p)}\int_{0}^{1} (1-s)^{\alpha-p-1}\bigl(\bigl\vert v_{1}(s)\bigr\vert +\bigl\vert v_{2}(s)\bigr\vert \bigr) \,ds \\ \leq& \bigl(\Vert p\Vert _{\infty}+ \Vert m\Vert_{\infty}\bigr) \biggl\{ \frac {4}{3\Gamma(\alpha+1)}+\frac{1}{3\Gamma(\alpha)}+\frac{\Delta }{4\Gamma(\alpha-1)}+ \frac{\Delta}{4\Gamma(\alpha-p+1)} \biggr\} . \end{aligned}

Hence, $$\sup_{t\in J}\vert z_{1}(t)+z_{2}(t) \vert\leq(\Vert p\Vert_{\infty}+\Vert m\Vert_{\infty})\Lambda_{1}$$. Also we have

$$\bigl\vert z'_{1}(t)+z'_{2}(t) \bigr\vert \leq \bigl(\Vert p\Vert _{\infty}+ \Vert m\Vert _{\infty}\bigr) \biggl\{ \frac{4}{3\Gamma(\alpha)}+\frac{1}{3\Gamma(\alpha +1)}+ \frac{\Delta}{\Gamma(\alpha-1)}+\frac{\Delta}{\Gamma(\alpha -p+1)} \biggr\} ,$$

which implies that $$\sup_{t\in J}\vert z'_{1}(t)+z'_{2}(t) \vert\leq (\Vert p\Vert_{\infty}+\Vert m\Vert_{\infty})\Lambda_{2}$$ and

$$\bigl\vert z''_{1}(t)+z''_{2}(t) \bigr\vert \leq \bigl(\Vert p\Vert _{\infty}+ \Vert m\Vert _{\infty}\bigr) \biggl\{ \frac{1+2\Delta}{\Gamma(\alpha-1 )}+\frac{2\Delta }{\Gamma(\alpha-p+1)} \biggr\}$$

from which $$\sup_{t\in J}\vert z''_{1}(t)+z''_{2}(t) \vert\leq(\Vert p\Vert_{\infty}+\Vert m\Vert_{\infty})\Lambda_{3}$$. Finally, for all $$i=1, \ldots, k$$, we have

\begin{aligned}& \bigl\vert {}^{c}D^{q_{i}} z_{1}(t)+{}^{c}D^{q_{i}} z_{2}(t) \bigr\vert \\& \quad \leq \bigl(\Vert p\Vert _{\infty}+ \Vert m\Vert_{\infty}\bigr) \biggl\{ \frac{1}{\Gamma(\alpha -q_{i} +1 )}+ \frac{2\Delta}{\Gamma(3-q_{i}) \Gamma(\alpha-1)}+\frac{2\Delta }{\Gamma(3-q_{i})\Gamma(\alpha-p+1)} \biggr\} , \end{aligned}

and so $$\sup_{t\in J}\vert{}^{c}D^{q_{i}}z_{1}(t)+{}^{c}D^{q_{i}} z_{2}(t) \vert\leq(\Vert p\Vert_{\infty}+\Vert m\Vert_{\infty})\Lambda_{4}^{i}$$, $$i=1,2,\ldots, k$$. Hence, it follows that

$$\Vert z_{1}+z_{2} \Vert \leq \bigl(\Vert p\Vert _{\infty}+\Vert m\Vert_{\infty}\bigr) \bigl(\Lambda_{1} + \Lambda_{2} +\Lambda_{3} +\Lambda_{4}^{i} \bigr)=M,\quad i=1,2,\ldots,k.$$

Now, we show that the operator B is compact on Y. To do this, it is enough to prove that $$B(Y)$$ is uniformly bounded and equi-continuous in X. Let $$z\in B(Y)$$ be arbitrary. For some $$u\in Y$$, choose $$v_{1}\in S_{G,u}$$ such that

\begin{aligned} z(t) =& \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha)}v_{1}(s)\,ds -\frac{t}{3}\int _{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha )}v_{1}(s)\,ds \\ &{}- \frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v_{1}(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma (\alpha-2)} v_{1}(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}v_{1}(s)\,ds,\quad t\in J. \end{aligned}
(2.9)

Hence,

\begin{aligned}& \bigl\vert z(t)\bigr\vert \leq \Vert p\Vert_{\infty}\biggl\{ \frac{4}{3\Gamma (\alpha+1)}+\frac{1}{3\Gamma(\alpha)}+\frac{\Delta}{4\Gamma(\alpha -1)}+\frac{\Delta}{4\Gamma(\alpha-p+1)} \biggr\} , \\& \bigl\vert z'(t)\bigr\vert \leq \Vert p\Vert_{\infty}\biggl\{ \frac{4}{3\Gamma (\alpha)}+\frac{1}{3\Gamma(\alpha+1)}+\frac{\Delta}{\Gamma(\alpha -1)}+ \frac{\Delta}{\Gamma(\alpha-p+1)} \biggr\} , \\& \bigl\vert z''(t)\bigr\vert \leq \Vert p \Vert_{\infty}\biggl\{ \frac{1+2\Delta }{\Gamma(\alpha-1 )}+\frac{2\Delta}{\Gamma(\alpha-p+1)} \biggr\} , \\& \bigl\vert {}^{c}D^{q_{i}} z(t) \bigr\vert \leq \Vert p \Vert_{\infty}\biggl\{ \frac {1}{\Gamma(\alpha-q_{i} +1 )} + \frac{2\Delta}{\Gamma(3-q_{i}) \Gamma (\alpha-1)}+ \frac{2\Delta}{\Gamma(3-q_{i})\Gamma(\alpha-p+1)} \biggr\} \end{aligned}

for $$i=1,\ldots, k$$. Hence, $$\Vert z\Vert \leq \Vert p\Vert_{\infty}(\Lambda_{1}+\Lambda_{2}+\Lambda_{3} +\Lambda_{4}^{i} )$$, $$i=1,\ldots, k$$.

Now, we show that B maps Y to equi-continuous subsets of X. Let $$t_{1}, t_{2}\in J$$ with $$t_{1} < t_{2}$$, $$u\in Y$$, and $$z \in B(u)$$. Choose $$v_{1}\in S_{G,u}$$ such that $$z(t)$$ is given by (2.9). Then we have

\begin{aligned}& \bigl\vert z(t_{2})-z(t_{1})\bigr\vert \leq \Vert p \Vert_{\infty}\biggl\{ \frac {\vert t_{2}^{\alpha}-t_{1}^{\alpha}\vert}{\Gamma(\alpha+1)} + \frac{\vert t_{2}-t_{1}\vert}{3\Gamma(\alpha+1)} + \frac{\vert t_{2}-t_{1} \vert }{3\Gamma(\alpha)}+ \frac{\Delta\vert[(t_{2}-t_{1})-(t_{2}^{2} -t_{1}^{2} )]\vert}{\Gamma(\alpha-1)} \\& \hphantom{\bigl\vert z(t_{2})-z(t_{1})\bigr\vert \leq}{}+ \frac{\Delta\vert[(t_{2}-t_{1})-(t_{2}^{2} -t_{1}^{2} )]\vert}{\Gamma (\alpha-p+1)} \biggr\} , \\& \bigl\vert z'(t_{2})-z'(t_{1}) \bigr\vert \leq \Vert p\Vert_{\infty}\biggl\{ \frac {\vert t_{2}^{\alpha-1} -t_{1}^{\alpha-1} \vert}{\Gamma(\alpha)} + \frac{2\Delta\vert t_{2}-t_{1}\vert}{\Gamma(\alpha-1)} + \frac{2\Delta \vert t_{2}-t_{1}\vert}{\Gamma(\alpha-p+1 )} \biggr\} , \\& \bigl\vert z''(t_{2})-z''(t_{1}) \bigr\vert \leq \Vert p\Vert_{\infty}\frac{\vert t_{2}^{\alpha-2} -t_{1}^{\alpha-2} \vert}{\Gamma(\alpha-1 )} \end{aligned}

and

\begin{aligned}& \bigl\vert {}^{c}D^{q_{i}}z(t_{2})- {}^{c}D^{q_{i}}z(t_{1})\bigr\vert \\& \quad \leq \Vert p\Vert_{\infty}\biggl\{ \frac{\vert t_{2}^{\alpha- q_{i}} -t_{1}^{\alpha-q_{i}} \vert}{\Gamma(\alpha-q_{i}+1 )} + \frac{2\Delta\vert t_{2}^{2-q_{i}} -t_{1}^{2-q_{i}} \vert}{\Gamma(3-q_{i}) \Gamma(\alpha-1 )} + \frac{2\Delta\vert t_{2}^{2-q_{i}} -t_{1}^{2-q_{i}} \vert}{\Gamma(3-q_{i})\Gamma(\alpha-p+1 )} \biggr\} \end{aligned}

for each $$i=1,\ldots, k$$. It is seen that the right-hand sides of the above inequalities tend to zero as $$t_{2}\to t_{1}$$. Hence, by using the Arzelá-Ascoli theorem, B is compact.

Next, we prove that B has a closed graph. Let $$u_{n}\in Y$$ and $$z_{n}\in B(u_{n})$$ for all n such that $$u_{n}\to u_{0}$$ and $$z_{n}\to z_{0}$$. We show that $$z_{0}\in B(u_{0})$$. Associated with $$z_{n}\in B(u_{n})$$ for each $$n\in \mathbb{N}$$, there exists $$v_{n}\in S_{G,u_{n}}$$ such that

\begin{aligned} z_{n}(t) =& \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha)}v_{n}(s)\,ds -\frac{t}{3}\int _{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha )}v_{n}(s)\,ds \\ &{}- \frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v_{n}(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma (\alpha-2)} v_{n}(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}v_{n}(s)\,ds \end{aligned}

for all $$t\in J$$. It suffices to show that there exists $$v_{0}\in S_{G,u_{0}}$$ such that, for each $$t\in J$$,

\begin{aligned} z_{0}(t) =& \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha)}v_{0}(s)\,ds -\frac{t}{3}\int _{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha )}v_{0}(s)\,ds \\ &{}- \frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v_{0}(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma (\alpha-2)} v_{0}(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}v_{0}(s)\,ds . \end{aligned}

Consider the continuous linear operator $$\Theta:L^{1}(J,\mathbb{R})\to X$$ by

\begin{aligned} \begin{aligned} \Theta(v) (t) ={}& \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha )}v(s)\,ds -\frac{t}{3}\int_{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha )}v(s)\,ds \\ &{}- \frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma(\alpha -2)} v(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}v(s)\,ds . \end{aligned} \end{aligned}

Notice that

\begin{aligned} \bigl\Vert z_{n}(t)-z_{0}(t)\bigr\Vert =& \biggl\Vert \int_{0}^{t} \frac{(t-s )^{\alpha -1}}{\Gamma(\alpha)} \bigl(v_{n}(s)-v_{0}(s)\bigr)\,ds -\frac{t}{3}\int _{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha)}\bigl(v_{n}(s)-v_{0}(s) \bigr)\,ds \\ &{}- \frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}\bigl(v_{n}(s)-v_{0}(s)\bigr)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma(\alpha-2)} \bigl(v_{n}(s)-v_{0}(s)\bigr)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}\bigl(v_{n}(s)-v_{0}(s)\bigr)\,ds\biggr\Vert \to0\quad \mbox{as }n\to\infty. \end{aligned}

By using Lemma 1.2, $$\Theta\circ S_{G}$$ is a closed graph operator. Since $$z_{n}(t)\in\Theta(S_{G,u_{n}})$$ for all n, and $$u_{n}\to u_{0}$$, there is $$v_{0}\in S_{G,u_{0}}$$ such that

\begin{aligned} z_{0}(t) =& \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha)}v_{0}(s)\,ds -\frac{t}{3}\int _{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha )}v_{0}(s)\,ds \\ &{}- \frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v_{0}(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma (\alpha-2)} v_{0}(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}v_{0}(s)\,ds . \end{aligned}

Hence, $$z_{0}\in B(u_{0})$$. So, it follows that B has a closed graph and this implies that the operator B is upper semi-continuous.

Finally, we show that A is a contraction multifunction. Let $$u,w\in X$$ and $$z_{1}\in A(w)$$ is given. Then we can select $$v_{1}\in S_{F,w}$$ such that

\begin{aligned} z_{1}(t) =& \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha)}v_{1}(s)\,ds -\frac{t}{3}\int _{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha )}v_{1}(s)\,ds \\ &{}- \frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v_{1}(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma (\alpha-2)} v_{1}(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}v_{1}(s)\,ds \end{aligned}

for all $$t\in J$$. Since

\begin{aligned}& H_{d} \bigl(F\bigl(t, u(t), u'(t), u''(t), {}^{c}D^{q_{1}}u(t), \ldots, {}^{c}D^{q_{k}}u(t) \bigr) \\& \qquad {} - F\bigl(t, w(t), w'(t), w''(t), {}^{c}D^{q_{1}}w(t), \ldots, {}^{c}D^{q_{k}}w(t) \bigr) \bigr) \\& \quad \leq h(t) \Biggl[ \bigl\vert u(t) -w(t) \bigr\vert + \bigl\vert u'(t) -w'(t) \bigr\vert + \bigl\vert u''(t)-w''(t)\bigr\vert + \sum_{i=1}^{k} \bigl\vert {}^{c}D^{q_{i}} u(t)-{}^{c}D^{q_{i}}w(t) \bigr\vert \Biggr] \end{aligned}

for almost all $$t\in J$$, there exists $$y\in F(t, u(t), u'(t), u''(t), {}^{c}D^{q_{1}}u(t), \ldots, {}^{c}D^{q_{k}}u(t))$$ such that

\begin{aligned} \bigl\vert v_{1}(t)-y\bigr\vert \leq& m(t) \Biggl[ \bigl\vert u(t) -w(t) \bigr\vert + \bigl\vert u'(t) -w'(t) \bigr\vert + \bigl\vert u''(t)-w''(t) \bigr\vert \\ &{}+\sum_{i=1}^{k} \bigl\vert {}^{c}D^{q_{i}} u(t)-{}^{c}D^{q_{i}}w(t) \bigr\vert \Biggr] \end{aligned}

for almost all $$t\in J$$. Consider the multifunction $$U:J\to{\mathcal{P}}(\mathbb{R})$$ by

$$U(t)=\bigl\{ s\in\mathbb{R}: \bigl\vert v_{1}(t)-s\bigr\vert \leq m(t)g(t) \mbox { for almost all } t\in J\bigr\} ,$$

where

\begin{aligned} g(t) =& \Biggl[ \bigl\vert u(t) -w(t) \bigr\vert + \bigl\vert u'(t) -w'(t) \bigr\vert + \bigl\vert u''(t)-w''(t)\bigr\vert \\ &{}+ \sum_{i=1}^{k} \bigl\vert {}^{c}D^{q_{i}} u(t)-{}^{c}D^{q_{i}}w(t) \bigr\vert \Biggr] . \end{aligned}

Since $$v_{1}$$ and $$\varphi= mg$$ are measurable, $$U(\cdot)\cap F(\cdot, u(\cdot), u'(\cdot), u''(\cdot), {}^{c}D^{q_{1}}u(\cdot), \ldots, {}^{c}D^{q_{k}} u(\cdot) )$$ is a measurable multifunction. Thus, we can choose

$$v_{2}(t)\in F\bigl(t, u(t), u'(t), u''(t), {}^{c}D^{q_{1}}u(t), \ldots, {}^{c}D^{q_{k}}u(t) \bigr)$$

such that

\begin{aligned} \bigl\vert v_{1}(t)-v_{2}(t)\bigr\vert \leq& m(t) \Biggl[ \bigl\vert u(t) -w(t) \bigr\vert + \bigl\vert u'(t) -w'(t) \bigr\vert + \bigl\vert u''(t)-w''(t) \bigr\vert \\ &{}+ \sum_{i=1}^{k} \bigl\vert {}^{c}D^{q_{i}} u(t)-{}^{c}D^{q_{i}}w(t) \bigr\vert \Biggr] \end{aligned}

and

\begin{aligned} z_{2}(t) =& \int_{0}^{t} \frac{(t-s )^{\alpha-1}}{\Gamma(\alpha)}v_{2}(s)\,ds -\frac{t}{3}\int _{0}^{1} \frac{(1-s )^{\alpha-1}}{\Gamma(\alpha )}v_{2}(s)\,ds \\ &{}- \frac{t}{3} \int_{0}^{1} \frac{(1-s )^{\alpha-2}}{\Gamma(\alpha-1 )}v_{2}(s)\,ds+ \bigl(t-t^{2}\bigr)\Delta\int _{0}^{1} \frac{(1-s)^{\alpha-3}}{\Gamma (\alpha-2)} v_{2}(s)\,ds \\ &{}+ \bigl(t-t^{2}\bigr)\Delta\int_{0}^{1} \frac{(1-s)^{\alpha-p-1} }{\Gamma(\alpha -p)}v_{2}(s)\,ds \end{aligned}

for all $$t\in J$$. Now, we have

\begin{aligned} \bigl\vert z_{1}(t) - z_{2}(t)\bigr\vert \leq& \frac{1}{\Gamma(\alpha)}\int_{0}^{t} (t-s )^{\alpha-1} \bigl\vert v_{1}(s) - v_{2}(s)\bigr\vert \,ds \\ &{}+ \frac{\vert t\vert }{3\Gamma(\alpha)}\int_{0}^{1} (1-s )^{\alpha-1} \bigl\vert v_{1}(s) - v_{2}(s)\bigr\vert \,ds \\ &{}+ \frac{\vert t\vert }{3\Gamma(\alpha-1)} \int_{0}^{1} (1-s )^{\alpha -2} \bigl\vert v_{1}(s) - v_{2}(s)\bigr\vert \,ds \\ &{}+ \frac{\Delta \vert t-t^{2}\vert }{\Gamma(\alpha-2)}\int_{0}^{1} (1-s)^{\alpha-3} \bigl\vert v_{1}(s) - v_{2}(s)\bigr\vert \,ds \\ &{}+ \frac{\Delta \vert t-t^{2}\vert }{\Gamma(\alpha-p)}\int_{0}^{1} (1-s)^{\alpha-p-1} \bigl\vert v_{1}(s) - v_{2}(s)\bigr\vert \,ds \\ \leq& \Vert h\Vert_{\infty}\biggl\{ \frac{4}{3\Gamma(\alpha+1)}+ \frac {1}{3\Gamma(\alpha)}+\frac{\Delta}{4\Gamma(\alpha-1)}+\frac{\Delta }{4\Gamma(\alpha-p+1)} \biggr\} \| u-w\|. \end{aligned}

Similarly,

\begin{aligned}& \bigl\vert z'_{1}(t) - z'_{2}(t) \bigr\vert \leq \Vert h\Vert_{\infty}\biggl\{ \frac {4}{3\Gamma(\alpha)}+ \frac{1}{3\Gamma(\alpha+1)}+\frac{\Delta }{\Gamma(\alpha-1)}+\frac{\Delta}{\Gamma(\alpha-p+1)} \biggr\} \| u-w\|, \\& \bigl\vert z''_{1}(t) - z''_{2}(t)\bigr\vert \leq \Vert h \Vert_{\infty}\biggl\{ \frac {1+2\Delta}{\Gamma(\alpha-1 )}+\frac{2\Delta}{\Gamma(\alpha-p+1)} \biggr\} \|u-w \|, \\& \bigl\vert {}^{c}D^{q_{i}}z_{1}(t) - {}^{c}D^{q_{i}}z_{2}(t)\bigr\vert \\& \quad \leq \Vert h \Vert_{\infty}\biggl\{ \frac{1}{\Gamma(\alpha-q_{i} +1 )} + \frac{2\Delta }{\Gamma(3-q_{i}) \Gamma(\alpha-1)} + \frac{2\Delta}{\Gamma(3-q_{i})\Gamma(\alpha-p+1)} \biggr\} \|u-w\|. \end{aligned}

Hence,

\begin{aligned}& \sup_{t\in J}\bigl\vert z_{1}(t)-z_{2}(t) \bigr\vert \leq \Vert h\Vert_{\infty}\Lambda_{1} \|u-w\|, \\& \sup_{t\in J}\bigl\vert z'_{1}(t)- z'_{2}(t) \bigr\vert \leq \Vert h\Vert_{\infty}\Lambda_{2} \|u-w\|, \\& \sup_{t\in J}\bigl\vert z''_{1}(t)- z''_{2}(t) \bigr\vert \leq \Vert h \Vert_{\infty}\Lambda_{3} \|u-w\|, \\& \sup_{t\in J}\bigl\vert {}^{c}D^{q_{i}} z_{1}(t)- {}^{c}D^{q_{i}} z_{2}(t) \bigr\vert \leq \Vert h\Vert_{\infty}\Lambda_{4}^{i} \|u-w\| \end{aligned}

for each $$1\leq i\leq k$$. So

$$\Vert z_{1}-z_{2} \Vert \leq \Vert h\Vert_{\infty}\bigl(\Lambda_{1}+\Lambda _{2}+\Lambda_{3} + \Lambda_{4}^{i} \bigr)\Vert u-w\Vert,\quad i=1,2,\ldots,k .$$

This implies that $$H_{d}(A(u), A(w) )\leq L \Vert u-w\Vert$$. Thus A and B satisfy all the conditions of Theorem 1.3 and so the inclusion $$u\in A(u) +B(u)$$ has a solution in Y. Therefore the inclusion problem (1.1)-(1.2) has a solution in Y and the proof is completed. □

Finally, we give an example to illustrate the validity of our main result.

### Example 2.5

Consider the following fractional differential inclusion:

\begin{aligned} {}^{c}D^{\frac{5}{2}}u(t) \in& \biggl[ 0, \frac{ t | u(t)|^{3}}{100(1+| u(t) |^{3})} + \frac{t |2\sin(u'(t)) | }{200(| \sin(u'(t)) | + 1)} + \frac{0.01 t | u''(t) | }{| u''(t) | + 1 } \\ &{}+ \frac{ t | \cos({}^{c}D^{\frac{3}{2}}u(t)) | }{100(1+| \cos ({}^{c}D^{\frac{3}{2}}u(t) ) | ) } + \frac{t^{2} \vert\sin\frac{\pi }{2}t \vert\vert{}^{c}D^{\frac{3}{2}}u(t)\vert^{2}}{100t (\vert {}^{c}D^{\frac{3}{2}}u(t)\vert^{2} +1 )} \biggr] \\ &{}+ \biggl[ 0, \frac{e^{-t}\vert u(t)\vert}{(1+e^{t} )(1+\vert u(t)\vert )} + \frac{\vert\cos\pi t \vert\vert u'(t)\vert e^{-t}}{(1+e^{t})(1+\vert u'(t)\vert)} + \frac{e^{-t}\vert u''(t)\vert^{2} }{(1+\vert u''(t)\vert^{2} )(1+e^{t})} \\ &{}+ \frac{e^{-2t} \vert\sin({}^{c}D^{\frac{3}{2}}u(t))\vert }{(1+\vert\sin({}^{c}D^{\frac{3}{2}}u(t))\vert)e^{t} (1+e^{t})}+ \frac {e^{-3t}\vert{}^{c}D^{\frac{3}{2}}u(t)\vert^{3}}{(e^{2t}+e^{3t})(1+\vert {}^{c}D^{\frac{3}{2}}u(t)\vert^{3} )} \biggr], \end{aligned}
(2.10)

with the following boundary conditions:

$$u(0)=0, \qquad u'(0)=-u(1)- u'(1), \qquad u''( 0) = - u''(1)- {}^{c}D^{\frac{3}{2}}u(1),$$
(2.11)

where $$t\in[0,1]$$. In the above inclusion problem, we have $$\alpha= 5/2$$, $$p=3/2$$, $$k=2$$, and $$q_{1}=q_{2}=3/2$$. Also, we have $$\Delta=0.1597$$.

Now, we define $$F: [0,1] \times\mathbb{R} \times\mathbb{R} \times \mathbb{R} \times\mathbb{R} \times\mathbb{R} \to\mathcal{P}(\mathbb {R})$$ by

\begin{aligned} F(t, x,y,z,v,w) =& \biggl[ 0, \frac{ t | x|^{3}}{100(1+| x |^{3})} + \frac {t |2\sin y | }{200(| \sin y | + 1)} + \frac{0.01 t | z | }{| z | + 1 } \\ &{}+ \frac{ t | \cos v | }{100(1+| \cos v | ) } + \frac{t^{2} \vert\sin \frac{\pi}{2}t \vert w^{2}}{100t (w^{2} +1 )} \biggr], \end{aligned}

and $$G: [0,1] \times\mathbb{R} \times\mathbb{R} \times\mathbb{R} \times\mathbb{R} \times\mathbb{R} \to\mathcal{P}(\mathbb{R})$$ by

\begin{aligned} G(t, x,y,z,v,w) =& \biggl[ 0, \frac{e^{-t}\vert x\vert}{(1+e^{t} )(1+\vert x\vert)} + \frac{\vert\cos\pi t \vert\vert y\vert e^{-t}}{(1+e^{t})(1+\vert y\vert)} + \frac{e^{-t} z^{2} }{(1+ z^{2} )(1+e^{t})} \\ &{}+ \frac{e^{-2t} \vert\sin v\vert}{(1+\vert\sin v\vert)e^{t} (1+e^{t})}+ \frac{e^{-3t}\vert w\vert^{3}}{(e^{2t}+e^{3t})(1+\vert w\vert^{3} )} \biggr]. \end{aligned}

Then there exist continuous functions $$m, p:[0,1]\to(0, \infty)$$ given by

$$m(t)=5+ \frac{t}{100},\qquad p(t)= \frac{e^{-t}}{1+e^{t}} .$$

On the other hand, we can easily check that, for every $$x_{i}, y_{i} , z_{i} , v_{i}, w_{i} \in\mathbb{R}$$ ($$i=1,2$$),

\begin{aligned}& H_{d}\bigl(F(t, x_{1}, y_{1}, z_{1}, v_{1}, w_{1} ) -F(t,x_{2}, y_{2} , z_{2} , v_{2}, w_{2} ) \bigr) \\& \quad \leq h(t) \bigl( \vert x_{1}-x_{2} \vert + | y_{1}-y_{2} | + | z_{1}-z_{2} | + |v_{1}-v_{2}| + |w_{1}-w_{2} | \bigr) , \end{aligned}

where $$h: [0,1]\to(0, \infty)$$ is defined by $$h(t)= \frac{t}{100}$$. It can easily be found that $$\Lambda_{1} = 0.7369$$, $$\Lambda_{2} = 1.4434$$, $$\Lambda_{3} = 1.8102$$, $$\Lambda_{4}^{1} = 1.7687$$, and $$\Lambda_{4}^{2}=1.7687$$. Since $$\Vert h\Vert_{\infty}= \frac{1}{100}$$, we have $$L:= \Vert h\Vert _{\infty}(\Lambda_{1} +\Lambda_{2} + \Lambda_{3} +\Lambda_{4}^{1} + \Lambda_{4}^{2} ) = 0.01 \times7.5279=0.075279 < 1$$. Consequently all assumptions and conditions of Theorem 2.4 are satisfied. Hence, Theorem 2.4 implies that the fractional differential inclusion problem (2.10)-(2.11) has a solution.

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## Acknowledgements

We would like to thank the reviewers for their valuable comments and suggestions on the manuscript.

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