Open Access

Existence results for multi-term fractional differential inclusions

Advances in Difference Equations20152015:140

https://doi.org/10.1186/s13662-015-0481-z

Received: 20 January 2015

Accepted: 21 April 2015

Published: 6 May 2015

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.

Keywords

fractional differential inclusionboundary value problemexistencefixed point theorem

MSC

34A0834A6034B1034B15

1 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 [125] 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 [28], 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 \(n=[\alpha]+1\) [28].

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 \(\dim X<\infty\) [32].

The graph of a function F is the set \(\operatorname{Gr}(F)=\{ (x,y)\in X\times Y: y\in F(x)\} \) [29]. 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 \(y_{0} \in F(u_{0})\) [29].

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

Lemma 1.1

([29], 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

([32])

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

([33], 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 \(y\in Ay+By\).

2 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 [34].

We need the following auxiliary lemma. See also [35, 36].

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.

Declarations

Acknowledgements

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

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, University of Ioannina
(2)
Nonlinear Analysis and Applied Mathematics (NAAM) Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University
(3)
Young Researchers and Elite Club, Tabriz Branch, Islamic Azad University
(4)
Nonlinear Dynamic Analysis Research Center, Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok

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