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 [1–25] 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, 25–27]). 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, 25–27]. 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.