Abstract

We are concerned with the nonlinear (𝑛+1)-term fractional integrodifferential inclusions 𝐿(𝐷)𝑢𝐹(𝑡,𝑢,𝑇𝑢,𝑆𝑢), 𝑎.𝑒.𝑡[0,1], where 𝐿(𝐷)=𝐷𝛼𝑏𝑛𝐷𝛽𝑛𝑏𝑛1𝐷𝛽𝑛1𝑏1𝐷𝛽1, 0<𝛽1<𝛽2<<𝛽𝑛1<𝛽𝑛<1<𝛼<2, 𝑏1,𝑏2,,𝑏𝑛 are constant coefficients, and (𝑇𝑢)(𝑡)=10𝑘(𝑡,𝑠)𝑢(𝑠)𝑑𝑠, (𝑆𝑢)(𝑡)=𝑡0𝑙(𝑡,𝑠)𝑢(𝑠)𝑑𝑠, subject to the nonlocal conditions 𝑢(0)=0, 𝑢(1)=𝑚𝑖=1𝛾𝑖𝑢(𝜂𝑖). The existence results are obtained by using two fixed-point theorems due to Bohnenblust-Karlin and Covitz-Nadler, respectively. Our results partly generalize and improve the known ones.

1. Introduction

Fractional differential equations (FDEs) have received increasing interest for the last three decades. It is benefited by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in the modeling of many phenomena in various fields of science, engineering economics, and other fields, see for instance [112] and references therein.

Babakhani and Gejji [7] considered the existence of positive solutions for the nonlinear fractional differential equations𝐿(𝐷)𝑢=𝑓(𝑡,𝑢),𝑢(0)=0,0<𝑡<1,(1.1) where 𝐿(𝐷)=𝐷𝑠𝑛𝑎𝑛1𝐷𝑠𝑛1𝑎1𝐷𝑠1, 0<𝑠1<𝑠2<<𝑠𝑛<1, 𝑎𝑗>0, 𝑗=1,2,,𝑛1, and 𝐷𝑠𝑗 is the standard Riemann-Liouville fractional derivative. Some existence results of positive solutions are obtained by using some fixed-point theorems on a cone.

Stojanović [11] considered the existenceuniqueness of solutions for a nonlinear 𝑛-term fractional differential equation𝑏0𝐷𝛽0𝑢(𝑡)+𝑚1𝑖=1𝑏𝑖𝐷𝛽𝑖𝑢(𝑡)+𝑛1𝑖=𝑚𝑏𝑖𝐷𝛼𝑖𝑢(𝑡)+𝑏𝑛𝐷𝛼𝑛𝑢(𝑡)=𝑓(𝑡,𝑢(𝑡)),𝑡(0,1),(1.2) where 0<𝛽1<𝛽2<<𝛽𝑚1<1<𝛼𝑚<𝛼𝑚+1<<𝛼𝑛<2, with initial data 𝑢(0)=𝑓(0), 𝑢𝑡(0)=𝑔(0).

On the other hand, realistic problems arising from economics, optimal control, and so on can be modeled as differential inclusions. Recently, El-Sayed and Ibrahim [13] initiated the study of fractional differential equations inclusions. Differential inclusions have been widely investigated by many authors, see [1430] and references therein.

Very recently, in the survey paper [16], Agarwal et al. establish sufficient conditions for the existence and uniqueness of solutions for various classes of initial and boundary value problems for fractional differential equations and inclusions involving the Caputo fractional derivative.

Ouahab [26] studied the following boundary value problem of fractional differential inclusions:𝐷𝛼𝑦(𝑡)𝐹(𝑡,𝑦(𝑡)),a.e[].𝑡𝐽=0,1,1<𝛼2,𝑦(0)=𝑦(1)=0,(1.3) where 𝐷𝛼 is the standard Riemann-Liouville fractional derivative, 𝐹𝐽×𝒫() is a multivalued map with compact values.

Chang and Nieto [27] studied boundary value problem of fractional differential inclusions𝐶0𝐷𝛿[]𝑦(𝑡)𝐹(𝑡,𝑦(𝑡)),𝑡𝐽=0,1,𝛿(1,2),𝑦(0)=𝛼,𝑦(1)=𝛽,𝛼,𝛽0,(1.4) where 𝐶0𝐷𝛿𝑦(𝑡) is the Caputo's derivative, 𝐹𝐽×𝒫().

However, to the best of our knowledge, the existence of solutions for fractional integro-differential inclusions with multipoint boundary conditions has not been paid much attention. Our goal is to fill this gap in literature.

In the present work, we consider more general fractional integro-differential inclusions𝐿(𝐷)𝑢𝐹(𝑡,𝑢,𝑇𝑢,𝑆𝑢),a.e[].𝑡0,1(1.5) with multipoint boundary conditions𝑢(0)=0,𝑢(1)=𝑚𝑖=1𝛾𝑖𝑢𝜂𝑖,(1.6) where 𝐿(𝐷)=𝐷𝛼𝑏𝑛𝐷𝛽𝑛𝑏𝑛1𝐷𝛽𝑛1𝑏1𝐷𝛽1, 0<𝛽1<𝛽2<<𝛽𝑛1<𝛽𝑛<1<𝛼<2, 𝐷𝛼,𝐷𝛽𝑗, 𝑗=1,2,,𝑛, is the standard Riemann-Liouville fractional derivative, 0<𝜂1<𝜂2<𝜂𝑚<1, 0𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1<1, (𝑇𝑢)(𝑡)=10𝑘(𝑡,𝑠)𝑢(𝑠)d𝑠, (𝑆𝑢)(𝑡)=𝑡0𝑙(𝑡,𝑠)𝑢(𝑠)d𝑠, 𝑘,𝑙 are two continuous functions on [0,1]×[0,1], and 𝐹[0,1]×××𝒫() is a given multivalued function (𝒫() is the family of all nonempty subsets of ). We shall consider both the cases of convex and nonconvex valued right hand side and establish some sufficient conditions which admit that the integro-differential inclusions problem has at least one solution. These results obtained by applying two fixed point theorems due to Bohnenblust-Karlin and Covitz-Nadler are complement of previously known results.

The organization of this paper is as follows. In Section 2, we present some necessary definitions and preliminaries which are used throughout this paper. Main results and their proofs are given in Section 3.

2. Preliminaries and Several Lemmas

In this section, we recall some basic definitions and notations and give several lemmas which are useful in our discussion.

Definition 2.1. The Riemann-Liouville fractional integral of order 𝛼>0 of a function 𝑓+ is given by 𝐼𝛼1𝑓(𝑡)=Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑓(𝑠)d𝑡,(2.1) provided the right side is pointwise defined on +.

Definition 2.2. The Riemann-Liouville fractional derivative of function 𝑓+ is given by 𝐷𝛼𝑓(𝑡)=𝐷𝑛𝐼𝑛𝛼𝑓(𝑡),(2.2) where 𝑛=[𝛼]+1, 𝐷𝑛=d𝑛/d𝑡𝑛, provided the right side is pointwise defined on +.

Lemma 2.3. Let 𝛼>0, then 𝐼𝛼𝐷𝛼𝑥(𝑡)=𝑥(𝑡)+𝑐1𝑡𝛼1+𝑐2𝑡𝛼2++𝑐𝑛𝑡𝛼𝑛,(2.3) for some 𝑐𝑖, 𝑖=0,1,2,,𝑛1, 𝑛=[𝛼].

Remark 2.4. If the fractional derivative 𝐷𝛼𝑥(𝑡) is integrable, then 𝐼𝛼𝐷𝛽𝑥(𝑡)=𝐼𝛼𝛽𝐼𝑥(𝑡)1𝛽𝑥(𝑡)𝑡=0𝑡𝛼1Γ(𝛼),0<𝛽𝛼<1.(2.4) If 𝑥 is continuous on [0,1], then [𝐼1𝛽𝑥(𝑡)]𝑡=0=0, and (2.4) reduces to 𝐼𝛼𝐷𝛽𝑥(𝑡)=𝐼𝛼𝛽𝑥(𝑡),0<𝛽𝛼<1.(2.5)

The reader is referred to [4, 8, 9] for more details on fractional integrals and fractional derivatives.

Lemma 2.5. Let 𝑔𝐿([0,1],), 0<𝛽1<𝛽2<<𝛽𝑛1<𝛽𝑛<1<𝛼<2, and 0<𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1<1. Then 𝑢(𝑡) is a solution of the BVP 𝐷𝛼𝑏𝑛𝐷𝛽𝑛𝑏𝑛1𝐷𝛽𝑛1𝑏1𝐷𝛽1𝑢(𝑡)=𝑔(𝑡),a.e[],.𝑡0,1(2.6)𝑢(0)=0,𝑢(1)=𝑚𝑖=1𝛾𝑖𝑢𝜂𝑖.(2.7) if and only if 𝑢(𝑡) satisfies the integral equation 𝑢(𝑡)=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1𝑢(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1𝑢(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1𝑢(𝑠)d1𝑠+Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑔(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝑔(𝑠)d𝑠11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝑔(𝑠)d𝑠.(2.8)

Proof. In view of Lemma 2.3 and Remark 2.4, (2.6) is equivalent to the integral equation 𝑢(𝑡)=𝑐1𝑡𝛼1+𝑐2𝑡𝛼2+𝑛𝑖=1𝑏𝑖𝐼𝛼𝛽𝑖𝑢(𝑡)+𝐼𝛼𝑔(𝑡),(2.9) for some 𝑐1,𝑐2, that is, 𝑢(𝑡)=𝑐1𝑡𝛼1+𝑐2𝑡𝛼2+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1𝑢(𝑠)d1𝑠+Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑔(𝑠)d𝑠.(2.10) The boundary condition 𝑢(0)=0 implies 𝑐2=0. Thus, 𝑢(𝑡)=𝑐1𝑡𝛼1+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1𝑢(𝑠)d1𝑠+Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑔(𝑠)d𝑠.(2.11) In view of the boundary condition 𝑢(1)=𝑚𝑖=1𝛾𝑖𝑢(𝜂𝑖), we conclude that 𝑐1=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑢(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑢(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑔(𝑠)d𝑠11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑔(𝑠)d𝑠.(2.12) Therefore, the solution 𝑢(𝑡) of (2.6) and (2.7) satisfies (2.8).
Conversely, if 𝑢(𝑡) is a solution of (2.8), it is easy to verify that 𝑢(𝑡) satisfies (2.6) and (2.7). The proof is complete.

Now, we recall some facts from multivalued analysis.

Let (𝑋,𝑑) be a metric space, and 𝒫(𝑋)={𝑌𝑋𝑌}, 𝒫𝑏𝑑(𝑋)={𝑌𝒫(𝑋)𝑌bounded}, 𝒫𝑐𝑙(𝑋)={𝑌𝒫(𝑋)𝑌closed}, 𝒫𝑐𝑣(𝑋)={𝑌𝒫(𝑋)𝑌convex}, 𝒫𝑐𝑝(𝑋)={𝑌𝒫(𝑋)𝑌compact}, 𝒫𝑐𝑣,𝑐𝑝(𝑋)=𝒫𝑐𝑣(𝑋)𝒫𝑐𝑝(𝑋), and so forth.

Consider 𝐻𝑑𝒫(𝑋)×𝒫(𝑋)+{}, given by𝐻𝑑(𝐴,𝐵)=maxsup𝑥𝐴𝑑(𝑥,𝐵),sup𝑦𝐵,𝑑(𝐴,𝑦)(2.13) where 𝑑(𝑥,𝐵)=inf𝑦𝐵𝑑(𝑥,𝑦), 𝑑(𝐴,𝑦)=inf𝑥𝐴𝑑(𝑥,𝑦). Then (𝒫𝑏𝑑,𝑐𝑙(𝑋),𝐻𝑑) is a metric space and (𝒫𝑐𝑙(𝑋),𝐻𝑑) is a generalized metric space (see [31]).

A multivalued map 𝑁[0,1]𝒫𝑐𝑙(𝑋) is said to be measurable if for each 𝑢𝑋, the function 𝑌[0,1], defined by𝑌(𝑡)=𝑑(𝑢,𝑁(𝑡))=inf{𝑑(𝑢,𝑧)𝑧𝑁(𝑡)},(2.14) is measurable.

A multivalued map 𝐹𝑋𝒫(𝑋) is convex (closed) valued if 𝐹(𝑢) is convex (closed) for all 𝑢𝑋. 𝐹 is bounded on bounded sets if 𝐹(𝐵)=𝑢𝐵𝐹(𝑢) is bounded in 𝑋 for all 𝐵𝒫(𝑋). That is, sup𝑢𝐵{sup{|𝑦|𝑦𝐹(𝑢)}}<. 𝐹 is called upper semicontinuous (u.s.c for short) on 𝑋 if for each 𝑢0𝑋, the set 𝐹(𝑢0) is nonempty closed subset of 𝑋, and if for each open set 𝒰 of 𝑋 containing 𝐹(𝑢0), there exists an open neighborhood 𝒱 of 𝑢0 such that 𝐹(𝒱)𝒰. 𝐺 is said to be completely continuous if 𝐹(𝐵) is relatively compact for every 𝐵𝒫𝑏𝑑(𝑋). If the multivalued map 𝐹 is completely continuous with nonempty compact valued, then 𝐺 is u.s.c. if and only if 𝐹 has closed graph, that is, 𝑥𝑛𝑥, 𝑦𝑛𝑦, 𝑦𝑛𝐺(𝑥) imply 𝑦𝐺(𝑥).

More details on multivalued maps can be found in the books of Deimling [3], Górniewicz [32], Hu and Papageorgiou [33], and Tolstonogov [34].

Definition 2.6. The multivalued map 𝐹[0,1]×××𝒫() is 𝐿1-Carathéodory if (i)𝑡𝐹(𝑡,𝑢,𝑣,𝑤) is measurable for each (𝑢,𝑣,𝑤)××; (ii)(𝑢,𝑣,𝑤)𝐹(𝑡,𝑢,𝑣,𝑤) is upper semicontinuous for almost all 𝑡[0,1]; (iii)for each 𝑟>0, there exists 𝜑𝑟𝐿1([0,1],+) such that for all |𝑢|,|𝑣|,|𝑤|𝑟 and for almost all 𝑡[0,1],𝐹(𝑡,𝑢,𝑣,𝑤)𝒫||𝑓||=sup𝑓𝐹(𝑡,𝑢,𝑣,𝑤)𝜑𝑟(𝑡).(2.15)

For any 𝑢𝐶([0,1],), we define the set𝑆𝐹,𝑢=𝑓𝐿1([]0,1,)𝑓(𝑡)𝐹(𝑡,𝑢(𝑡),(𝑇𝑢)(𝑡),(𝑆𝑢)(𝑡))fora.e[],.𝑡0,1(2.16) which is known as the set of selection functions.

Lemma 2.7 (see [35]). Let 𝑋 be a Banach space. Let 𝐹[0,1]×𝑋𝒫𝑐𝑝,𝑐𝑣(𝑋) be an 𝐿1-Carathéodory multivalued map with 𝑆𝐹,𝑢, and let Γ be a linear continuous mapping from 𝐿1([0,1],𝑋) into 𝐶([0,1],𝑋). Then the operator Γ𝑆𝐹[]𝐶(0,1,𝑋)𝒫𝑐𝑝,𝑐𝑣[](𝐶(0,1,𝑋)),𝑢Γ𝑆𝐹(𝑆𝑦)=Γ𝐹,𝑢(2.17) is a closed graph operator in 𝐶([0,1],𝑋)×𝐶([0,1],𝑋).

The following Bohnenblust-Karlin fixed-point lemma and the fixed-point theorem for contractive multivalued operators given by Covitz and Nadler are of great importance in the proofs of our main results. The proofs of these results can be found in Bohnenblust and Karlin [30] and in Covitz and Nadler [36].

Lemma 2.8 (see [30]). Let 𝑋 be a Banach space, 𝐷 be a nonempty subset of 𝑋, which is bounded, closed, and convex. Suppose 𝐺𝐷𝒫(𝑋) is u.s.c. with closed, convex values, and such that 𝐺(𝐷)𝐷 and 𝐺(𝐷) is relatively compact. Then 𝐺 has a fixed point.

Lemma 2.9 (see [36]). Let (𝑋,𝑑) be a complete metric space. If 𝑁 is a contraction operator, then 𝐹𝑖𝑥𝑁.
For convenience, let us list some conditions. (H1)𝑚𝑖=1𝑛𝑗=1(𝛾𝑖𝑏𝑗𝜂𝛼𝛽𝑗𝑖/(1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1)Γ(𝛼𝛽𝑗+1))+𝑛𝑖=1(𝑏𝑖/(1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1)Γ(𝛼𝛽𝑖+1))+𝑛𝑖=1(𝑏𝑖/Γ(𝛼𝛽𝑖+1))<1;(H2)𝐹[0,1]×××𝒫𝑐𝑝,𝑐𝑣() is Carethéodory multivalued map.

Lemma 2.10. Assume that hypothesis (H1) is satisfied. For any 𝑔𝐶([0,1],), the integral equation 𝑢(𝑡)=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1𝑢(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1𝑢(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1𝑢(𝑠)d𝑠+𝑔(𝑡)(2.18) has a unique solution in 𝐶([0,1],).

Proof. We define the operator 𝐴 as follows: (𝐴𝑢)(𝑡)=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1𝑢(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1𝑢(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1𝑢(𝑠)d𝑠+𝑔(𝑡)(2.19) Obviously, 𝐴 is a map from 𝐶([0,1],) into 𝐶([0,1],). Also, we have ||𝐴𝑢1(𝑡)𝐴𝑢2||(𝑡)𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1||𝑢1(𝑠)𝑢2||(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1||𝑢1(𝑠)𝑢2||(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1||𝑢1(𝑠)𝑢2||(𝑠)d𝑠𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗𝜂𝛼𝛽𝑗𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗++1𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖++1𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑢+11𝑢2.(2.20) Therefore, 𝐴𝑢1𝐴𝑢2𝜆𝑢1𝑢2, where 𝜆=𝑚𝑖=1𝑛𝑗=1(𝛾𝑖𝑏𝑗𝜂𝛼𝛽𝑗𝑖/(1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1)Γ(𝛼𝛽𝑗+1))+𝑛𝑖=1(𝑏𝑖/(1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1)Γ(𝛼𝛽𝑖+1))+𝑛𝑖=1(𝑏𝑖/Γ(𝛼𝛽𝑖+1)). By (H1) and Banach contraction principle, the conclusion of lemma is true.

Definition 2.11. A function 𝑢𝐶[0,1] is said to be a solution of the BVP (1.5) and (1.6), if there exists a function 𝑓𝐿1([0,1],) such that 𝑓(𝑡)𝐹(𝑡,𝑢(𝑡),(𝑇𝑢)(𝑡),(𝑆𝑢)(𝑡)) a.e. on [0,1] and 𝑢(𝑡)=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1𝑢(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1𝑢(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1𝑢(𝑠)d1𝑠+Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑓(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝑓(𝑠)d𝑠11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝑓(𝑠)d𝑠.(2.21)

3. Main Results

In this section, we present our main results and prove them. Firstly, under convexity condition on the multivalued right-hand side, we are to establish the existence theorem of solutions for fractional differential inclusions (1.5) and (1.6), by employing the Bohnenblust-Karlin fixed-point theorem. Then, under nonconvexity condition on the multivalued right-hand side, the existence theorem of solutions are gotten, by employing the Covitz-Nadler fixed-point theorem.

Theorem 3.1. Assume that hypotheses (H1) and (H2) are satisfied. Then BVP (1.5) and (1.6) has at least one solution in 𝐶([0,1],), provided that (H3)𝜏/Γ(𝛼)+𝑚𝑖=1(𝛾𝑖𝜇𝑖/(1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1)Γ(𝛼))+(𝜔/(1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1)Γ(𝛼))<1𝜆,
where 𝜏=lim𝑟inf10𝜑max𝑟(𝑠),𝜑̃𝑘𝑟(𝑠),𝜑̃𝑙𝑟(𝑠)𝑟d𝜇𝑠,𝑖=lim𝑟inf𝜂𝑖0𝜂𝑖𝑠𝛼1𝜑max𝑟(𝑠),𝜑̃𝑘𝑟(𝑠),𝜑̃𝑙𝑟(𝑠)𝑟d𝑠,𝜔=lim𝑟inf10(1𝑠)𝛼1𝜑max𝑟(𝑠),𝜑̃𝑘𝑟(𝑠),𝜑̃𝑙𝑟(𝑠)𝑟d̃𝑠,𝑘=sup[]𝑡0,110𝑘(𝑡,𝑠)d̃𝑠,𝑙=sup[]𝑡0,1𝑡0𝑙(𝑡,𝑠)d𝑠,(3.1) and 𝜆 is defined in the proof of Lemma 2.10.

Proof. To transform the problem into a fixed-point problem, we consider the multivalued operator, 𝑁𝐶([0,1],)𝒫(𝐶([0,1],)), where for any 𝑢𝐶([0,1],),𝑁(𝑢) is defined by []𝐶(0,1,)(𝑡)=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1(𝑠)d1𝑠+Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑓(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝑓(𝑠)d𝑠11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝑓(𝑠)d𝑠,𝑓𝑆𝐹,𝑢.(3.2)
For any 𝑓𝐿1([0,1],), we have 𝑡0(𝑡𝑠)𝛼1𝑓(𝑠)d𝑠, 𝜂𝑖0(𝜂𝑖𝑠)𝛼1𝑡𝛼1𝑓(𝑠)d𝑠, 10(1𝑠)𝛼1𝑡𝛼1𝑓(𝑠)d𝑠𝐶([0,1],). In view of Lemma 2.10, 𝑁 is well defined. Moreover, it follows from the convexity of 𝑆𝐹,𝑢 (because 𝐹 has convex values) that 𝑁(𝑢) is convex for each 𝑢𝐶([0,1],). Clearly, the fixed points of 𝑁 are solutions of (1.5) and (1.6).
We shall show that 𝑁 has a fixed point in three steps.
Step 1. we claim that there exists a 𝑟>0, such that 𝑁(𝐵𝑟)𝐵𝑟, where 𝐵𝑟={𝑢𝐶([0,1],)𝑢𝑟}.
In fact, if it is not true, then for any 𝑟>0, there exists a function 𝑢𝑟𝐵𝑟, 𝑟𝑁(𝑢𝑟) but 𝑟𝐵𝑟, that is 𝑟>𝑟 and for some 𝑓𝑟𝑆𝐹,𝑢, 𝑟(𝑡)=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1𝑟(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1𝑟(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1𝑟(s)d1𝑠+Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑓𝑟(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝑓𝑟(𝑠)d𝑠11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝑓𝑟(𝑠)d𝑠.(3.3)
On the other hand, from (H2), we obtain ||𝑟(||𝑡)𝜆𝑟+1Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑓𝑟(𝑠)d𝑠+11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝑓𝑟(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝑓𝑟(𝑠)d𝑠𝜆𝑟+1Γ(𝛼)10(|𝑡𝑠|)𝛼1𝜑max{𝑟,̃𝑘𝑟,̃𝑙𝑟}(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝜑max{𝑟,̃𝑘𝑟,̃𝑙𝑟}(𝑠)d𝑠+11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝜑max{𝑟,̃𝑘𝑟,̃𝑙𝑟}(𝑠)d𝑠.(3.4) Thus, 𝑟<𝑟1(1𝜆)Γ(𝛼)10(|𝑡𝑠|)𝛼1𝜑max{𝑟,̃𝑘𝑟,̃𝑙𝑟}(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖(1𝜆)1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝜑max{𝑟,̃𝑘𝑟,̃𝑙𝑟}(𝑠)d𝑠+1(1𝜆)1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝜑max{𝑟,̃𝑘𝑟,̃𝑙𝑟}(𝑠)d𝑠.(3.5)
Dividing both sides by 𝑟 and taking the lower limit as 𝑟+, we get 𝜏1+(1𝜆)Γ(𝛼)𝑚𝑖=1𝛾𝑖𝜇𝑖(1𝜆)1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1+𝜔Γ(𝛼)(1𝜆)1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1.Γ(𝛼)(3.6)
This is a contraction to (H3). Hence there exists a 𝑟>0 such that 𝑁(𝐵𝑟)𝐵𝑟.
Step 2. 𝑁(𝐵𝑟) is equicontinuous.
Let 𝑡1,𝑡2[0,1], 𝑡1<𝑡2, 𝑢𝐵𝑟, and 𝑁(𝑢). Then there exists 𝑓𝑆𝐹,𝑢 such that for each 𝑡[0,1], we have (𝑡)=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1(𝑠)d1𝑠+Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑓(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝑓(𝑠)d𝑠11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝑓(𝑠)d𝑠.(3.7) Therefore, ||𝑡1𝑡2||𝑡𝑎2𝛼1𝑡1𝛼1+1Γ(𝛼)𝑡2𝑡1𝜑max{𝑟,̃𝑘𝑟,̃𝑙𝑟}(𝑠)d𝑠+1Γ(𝛼)𝑡10𝑡2𝑠𝛼1𝑡1𝑠𝛼1𝜑max{𝑟,̃𝑘𝑟,̃𝑙𝑟}(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖𝑟Γ𝛼𝛽𝑖𝑡+1𝛼𝛽𝑖2𝑡𝛼𝛽𝑖1𝑡+22𝑡1𝛼𝛽𝑖,(3.8) where 𝑎=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗𝜂𝛼𝛽𝑗𝑖𝑟1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗++1𝑛𝑖=1𝑏𝑖𝑟1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖++1𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝜑max{𝑟,̃𝑘𝑟,̃𝑙𝑟}(𝑠)d𝑠+11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝜑max{𝑟,̃𝑘𝑟,̃𝑙𝑟}(𝑠)d𝑠.(3.9)
The right-hand side of the above inequality tends to zeros independently of 𝐵𝑟, as 𝑡2𝑡10. This shows that 𝑁(𝐵𝑟) is equicontinuous. Step 3. 𝑁 has a closed graph.
Let 𝑢𝑛𝑢, 𝑛𝑁(𝑢𝑛) and 𝑛. We will prove that 𝑁(𝑢). Now 𝑛𝑁(𝑢𝑛) implies that there exists 𝑓𝑛𝑆𝐹,𝑢𝑛 such that for each 𝑡[0,1], 𝑛(𝑡)=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1𝑛(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1𝑛(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1𝑛(𝑠)d1𝑠+Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑓𝑛(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝑓𝑛(𝑠)d𝑠11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝑓𝑛(𝑠)d𝑠.(3.10)
We need to show that there exists 𝑓𝑆𝐹,𝑢 such that for each 𝑡[0,1], (𝑡)=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1(𝑠)d1𝑠+Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑓(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝑓(𝑠)d𝑠11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝑓(𝑠)d𝑠.(3.11) In (2.18), taking 1𝑔(𝑡)=Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑓(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝑓(𝑠)d𝑠11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝑓(𝑠)d𝑠.(3.12) In view of Lemma 2.10, we know that for any 𝑓𝐿([0,1],), the integral equation (2.18) has a unique solution, which we denote by 𝑓. Because of this, we can define the operator [][]Φ𝐿(0,1,)𝐶(0,1,),𝑓𝑓.(3.13) It is easy to verify that the operator Φ is linear. On the other hand, we can get 𝑓1(1𝜆)Γ(𝛼)10(|𝑡𝑠|)𝛼1||||𝑓(𝑠)d𝑠+1(1𝜆)1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1||𝑓||(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖(1𝜆)1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1||||𝑓(𝑠)d𝑠𝜇10||||𝑓(𝑠)d𝑠,(3.14) where 1𝜇=+(1𝜆)Γ(𝛼)𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1(1𝜆)1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1+1Γ(𝛼)(1𝜆)1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1,Γ(𝛼)(3.15) that is, Φ(𝑓)𝜇𝑓𝐿1, which implies Φ is continuous. From Lemma 2.7, it follows that Φ𝑆𝐹,𝑢 is a closed graph operator. Moreover, we know 𝑛Φ(𝑆𝐹,𝑢𝑛). Since 𝑢𝑛𝑢, it follows from Lemma 2.7 that (3.11) holds for some 𝑓𝑆𝐹,𝑢.
Therefore, 𝑁 is a compact multivalued map, u.s.c. with convex closed values. As a consequence of Lemma 2.8, we immediately conclude that 𝑁 has a fixed-point 𝑢 which is a solution of the problem (1.5) and (1.6). The proof is complete.

As a direct corollary of Theorem 3.1, we can immediately obtain the following result when the nonlinearity 𝐹 has sublinear growth in the state variable.

Corollary 3.2. Suppose that (H1) and (H2) are satisfied. Then the problem (1.5) and (1.6) has at least one solution in 𝐶([0,1],), provided that (H4)there exist functions 𝑝(𝑡),𝑞(𝑡),𝑟(𝑡),𝜃(𝑡)𝐿([0,1],), and 𝜎1,𝜎2,𝜎3[0,1) such that𝐹(𝑡,𝑢,𝑣,𝑤)𝒫𝑝(𝑡)|𝑢|𝜎1+𝑞(𝑡)|𝑣|𝜎2+𝑟(𝑡)|𝑢|𝜎3+𝜃(𝑡)(3.16) for each (𝑡,𝑢,𝑣,𝑤)[0,1]×××.

Proof. We only need to verify that (H4) implies (H3) in Theorem 3.1. Taking 𝜑(𝑡)=𝑝(𝑡)|𝑢|𝜎1+𝑞(𝑡)|𝑣|𝜎2+𝑟(𝑡)|𝑢|𝜎3+𝜃(𝑡), it is easy to prove 𝜏=lim𝑟inf10𝜑max𝑟(𝑠),𝜑̃𝑘𝑟(𝑠),𝜑̃𝑙𝑟(𝑠)𝑟d𝜇𝑠=0,𝑖=lim𝑟inf𝜂𝑖0𝜂𝑖𝑠𝛼1𝜑max𝑟(𝑠),𝜑̃𝑘𝑟(𝑠),𝜑̃𝑙𝑟(𝑠)𝑟d𝑠=0,𝜔=lim𝑟inf10(1𝑠)𝛼1𝜑max𝑟(𝑠),𝜑̃𝑘𝑟(𝑠),𝜑̃𝑙𝑟(𝑠)𝑟d𝑠=0,(3.17) because of 𝜎1,𝜎2,𝜎3[0,1). Then (H3) is satisfied. The proof is finished.

In the next part, we are concerned with the BVP (1.5) and (1.6) with nonconvex valued right-hand side. By using Covitz and Nadler fixed-point theorem, we obtain the following result.

Theorem 3.3. Assume that the following hypotheses hold: (A1) 𝐹[0,1]×××𝑃𝑐𝑙(); 𝑡𝐹(𝑡,𝑢,𝑣,𝑤) is measurable for each 𝑢,𝑣,𝑤;(A2)There exist three functions 𝑝,𝑞,𝑟𝐿1([0,1],) such that for a.e. 𝑡[0,1] and 𝑢1,𝑢2,𝑣1,𝑣2,𝑤1,𝑤2,𝐻𝑑𝐹𝑡,𝑢1,𝑣1,𝑤1,𝐹𝑡,𝑢2,𝑣2,𝑤2||𝑢𝑝(𝑡)1𝑢2||||𝑣+𝑞(𝑡)1𝑣2||||𝑤+𝑟(𝑡)1𝑤2||,𝑑(0,𝐹(𝑡,0,0,0))𝑚(𝑡).(3.18) Then the BVP (1.5) and (1.6) has at least one solution in 𝐶([0,1],) provided that ̃𝑎0+̃𝑏0+̃𝑐0+𝑚+1𝑖=1̃𝑎𝑖+̃𝑘̃𝑏𝑖+̃𝑙̃𝑐𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1<1𝜆,(3.19) where ̃𝑎0=1Γ(𝛼)10𝑝(𝑠)d𝑠,̃𝑎𝑖=1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑝(𝑠)d̃𝑏𝑠,𝑖=1,2,,𝑚+1,0=1Γ(𝛼)10𝑞(𝑠)d̃𝑏𝑠,𝑖=1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑞(𝑠)d𝑠,𝑖=1,2,,𝑚+1,̃𝑐0=1Γ(𝛼)10𝑝(𝑠)d𝑠,̃𝑐𝑖=1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑟(𝑠)d𝑠,𝑖=1,2,,𝑚+1,(3.20)𝜂𝑚+1=1 and 𝜆 defined as in Lemma 2.10, ̃̃𝑙𝑘, defined by (3.1) in Theorem 3.1.

Proof. Transform the problem into a fixed-point problem. Let 𝑁𝐶([0,1],)𝒫(𝐶([0,1],) be defined as in Theorem 3.1. We shall show that 𝑁 satisfies the assumptions of Lemma 2.9. The proof will be given in two steps.Step 1. 𝑁(𝑢)𝒫𝑐𝑙(𝐶([0,1],)) for each 𝑢𝐶([0,1],).
Indeed, let (𝑢𝑛)𝑛0𝑁(𝑢) such that 𝑢𝑛̃𝑢 in 𝐶([0,1],). Then there exists 𝑔𝑛𝑆𝐹,𝑢 such that for each 𝑡[0,1], 𝑢𝑛(𝑡)=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1𝑢𝑛(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1𝑢𝑛(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1𝑢𝑛(𝑠)d1𝑠+Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑔𝑛(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝑔𝑛(𝑠)d𝑠11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝑔𝑛(𝑠)d𝑠.(3.21)
For every 𝑤𝐹(𝑡,𝑢(𝑡),(𝑇𝑢)(𝑡),(𝑆𝑢)(𝑡)), from (A2), we have |𝑤|𝑑(0,𝐹(𝑡,0,0,0))+𝐻𝑑(𝐹(𝑡,0,0,0),𝐹(𝑡,𝑢(𝑡),(𝑇𝑢)(𝑡),(𝑆𝑢)(𝑡))).(3.22) Then, ||𝑔𝑛||||||||||||||(𝑡)𝑚(𝑡)+𝑝(𝑡)𝑢(𝑡)+𝑞(𝑡)(𝑇𝑢)(𝑡)+𝑟(𝑡)(𝑆𝑢)(𝑡)a.e[],.𝑡0,1(3.23) that is, 𝑔𝑛(𝑡)𝐵(0,𝑚(𝑡)+𝑝(𝑡)|𝑢(𝑡)|+𝑞(𝑡)|(𝑇𝑢)(𝑡)|+𝑟(𝑡)|(𝑆𝑢)(𝑡)|), where 𝐵||𝑢||||||||||=||||||(||||(||0,𝑚(𝑡)+𝑝(𝑡)(𝑡)+𝑞(𝑡)(𝑇𝑢)(𝑡)+𝑟(𝑡)(𝑆𝑢)(𝑡)𝑤|𝑤|𝑚(𝑡)+𝑝(𝑡)𝑢(𝑡)+𝑞(𝑡)𝑇𝑢)(𝑡)+𝑟(𝑡)𝑆𝑢)(𝑡)=Ψ(𝑡).(3.24)
It is clear that Ψ[0,1]𝒫𝑐𝑝,𝑐𝑣() is a multivalued integrable bounded map. Since 𝑔𝑛()Ψ(),𝑛1, we may pass to a subsequence if necessary to get that 𝑔𝑛 converges weakly to 𝑔 in 𝐿1𝑤([0,1],). From Mauz's lemma [37], there exists 𝑔conv{𝑔𝑛(𝑡)𝑛1}, then there exists a subsequence {𝑔𝑛𝑛1} in conv{𝑔𝑛𝑛1}, such that 𝑔𝑛 converges strongly to 𝑔 in 𝐿1([0,1],), which implies 𝑔𝐿1([0,1],). Then for each 𝑡[0,1], 𝑢𝑛(𝑡)̃𝑢(𝑡)=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1̃𝑢(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1̃𝑢(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1̃𝑢(𝑠)d1𝑠+Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑔(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝑔(𝑠)d𝑠11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝑔(𝑠)d𝑠.(3.25) So, ̃𝑢𝑁(𝑢)𝒫𝑐𝑙(𝐶([0,1],)). Step 2. There exists 𝛾<1, such that 𝐻𝑑𝑁𝑢(𝑢),𝑁𝛾𝑢𝑢foreach𝑢,𝑢([]𝐶0,1,).(3.26)
Let 𝑢,𝑢𝐶([0,1],) and 𝑁(𝑢). Then there exists 𝑓(𝑡)𝐹(𝑡,𝑢(t),(𝑇𝑢)(𝑡),(𝑆𝑢)(𝑡)) such that for each 𝑡[0,1](𝑡)=𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1(𝑠)d1𝑠+Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑓(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝑓(𝑠)d𝑠11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝑓(𝑠)d𝑠.(3.27) It follows from (A2) that 𝐻𝑑𝐹(𝑡,𝑢(𝑡),(𝑇𝑢)(𝑡),(𝑆𝑢)(𝑡)),𝐹𝑡,𝑢(𝑡),𝑇𝑢(𝑡),𝑆𝑢||(𝑡)𝑝(𝑡)𝑢(𝑡)𝑢(||||(𝑡)+𝑞(𝑡)𝑇𝑢)(𝑡)𝑇𝑢(||||(𝑡)+𝑟(𝑡)𝑆𝑢)(𝑡)𝑆𝑢(||[].𝑡),𝑡0,1(3.28) Hence, there is 𝑔𝐹(𝑡,𝑢(𝑡),(𝑇𝑢)(𝑡),(𝑆𝑢)(𝑡)) such that ||𝑓||||𝑢(𝑡)𝑔𝑝(𝑡)(𝑡)𝑢||||(𝑡)+𝑞(𝑡)(𝑇𝑢)(𝑡)𝑇𝑢||||(𝑡)+𝑟(𝑡)(𝑆𝑢)(𝑡)𝑆𝑢||[].(𝑡),𝑡0,1(3.29)
Consider 𝐺[0,1]𝒫(), given by =||||𝐺(𝑡)𝑔𝑓(𝑡)𝑔𝑝(t)||𝑢(𝑡)𝑢||||(𝑡)+𝑞(𝑡)(𝑇𝑢)(𝑡)𝑇𝑢||||(𝑡)+𝑟(𝑡)(𝑆𝑢)(𝑡)𝑆𝑢||.(𝑡)(3.30)
Since the multivalued operator 𝐺𝑡𝐺(𝑡)𝐹(𝑡,𝑢(𝑡),(𝑇𝑢)(𝑡),(𝑆𝑢)(𝑡)) is measurable (see in [38, Proposition  III.4]), there exists a function 𝑔, which is a measurable selection for 𝐺. So, 𝑔(𝑡)𝐹(𝑡,𝑢(𝑡),(𝑇𝑢)(𝑡),(𝑆𝑢)(𝑡)) and ||𝑓(𝑡)𝑔||||(𝑡)𝑝(𝑡)𝑢(𝑡)𝑢||||(𝑡)+𝑞(𝑡)(𝑇𝑢)(𝑡)𝑇𝑢||||(𝑡)+𝑟(𝑡)(𝑆𝑢)(𝑡)𝑆𝑢||[].(𝑡),𝑡0,1(3.31)
For each 𝑡[0,1], define (𝑡)=𝑚i𝑛=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1(𝑠)d𝑠𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1(𝑠)d1𝑠+Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑔(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1𝑔(𝑠)d𝑠11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1𝑔(𝑠)d𝑠.(3.32)
Then, for 𝑡[0,1]||(𝑡)||(𝑡)𝑚𝑛𝑖=1𝑗=1𝛾𝑖𝑏𝑗1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑗𝜂𝑖0𝜂𝑖𝑠𝛼𝛽𝑗1𝑡𝛼1||(𝑠)||(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ𝛼𝛽𝑖10(1𝑠)𝛼𝛽𝑖1𝑡𝛼1||(𝑠)||(𝑠)d𝑠+𝑛𝑖=1𝑏𝑖Γ𝛼𝛽𝑖𝑡0(𝑡𝑠)𝛼𝛽𝑖1||(𝑠)||(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1||𝑓(𝑠)𝑔||(𝑠)d𝑠+11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1||𝑓(𝑠)𝑔||(𝑠)d𝑠+1Γ(𝛼)𝑡0(𝑡𝑠)𝛼1||𝑓(𝑠)𝑔||(𝑠)d𝑠𝜆+𝑢𝑢×1Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑝̃̃𝑙(𝑠)+𝑞(𝑠)𝑘+𝑟(𝑠)d𝑠+𝑚𝑖=1𝛾𝑖1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)𝜂𝑖0𝜂𝑖𝑠𝛼1𝑡𝛼1̃̃𝑙𝑝(𝑠)+𝑞(𝑠)𝑘+𝑟(𝑠)d𝑠+11𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1Γ(𝛼)10(1𝑠)𝛼1𝑡𝛼1̃̃𝑙𝑝(𝑠)+𝑞(𝑠)𝑘+𝑟(𝑠)d𝑠.(3.33)
So, ̃𝑎0+̃𝑏0+̃𝑐0+1𝜆𝑚+1𝑖=1̃𝑎𝑖+̃𝑘̃𝑏𝑖+̃𝑙̃𝑐𝑖(1𝜆)1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1𝑢𝑢.(3.34)
By an analogous relation, obtained by interchanging the roles of 𝑢,𝑢, it follows that 𝐻𝑑𝑁𝑢(𝑢),𝑁𝐶𝑢𝑢.(3.35) where 𝐶=(̃𝑎0+̃𝑏0+̃𝑐0)/(1𝜆)+𝑚+1𝑖=1(̃𝑎𝑖+̃𝑘̃𝑏𝑖+̃𝑙̃𝑐𝑖/(1𝜆)(1𝑚𝑖=1𝛾𝑖𝜂𝑖𝛼1))<1. Therefore, 𝑁 is a contraction operator. With the help of Lemma 2.9, the problem (1.5) and (1.6) has at least one solution.

Remark 3.4. In (1.5) and (1.6), if 𝑏𝑖=0, 𝛾𝑗=0, 𝑖=1,,𝑛, 𝑗=1,,𝑚, 𝑘(𝑡,𝑠)=𝑙(𝑡,𝑠)0. Then we obtain the following fractional boundary value problem of differential inclusions 𝐷𝛼𝑦(𝑡)𝐹(𝑡,𝑦(𝑡)),a.e[].𝑡𝐽=0,1,0<𝛼2,𝑦(0)=𝑦(1)=0(3.36) which has been studied in [26]. It is easy to see that our results (Theorems 3.1 and 3.3) partly generalize and improve the results obtained in it.

Acknowledgments

This work is supported by the National Natural Science Foundation of PR China (no. 10701023, No. 10971221) and Shanghai Natural Science Foundation (no. 10ZR1400100).