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On a fractional differential inclusion via a new integral boundary condition

Abstract

In this paper, we investigate the existence of solution for two systems of fractional differential inclusions via some integral boundary value conditions. For this purpose, we use an endpoint result for multifunctions which has been proved in 2010 by Amini-Harandi (Nonlinear Anal. 72:132-134, 2010). Finally, we give an example for illustrating one of our results.

1 Introduction

As we know, diverse classes of fractional differential equations have been studied by researchers (see for example, [115] and the references therein). Much attention has been devoted to the fractional differential inclusions (see for example, [1632] and the references therein). Also, there have been provided many applications of this field (see for example, [33, 34] and [35]).

It is the aim of this paper to investigate the existence of solutions for two systems of fractional differential inclusions, subject to some integral boundary value conditions. In this respect, we use an endpoint result for multifunctions due to Amini-Harandi, [36]. We provide an example for illustrating one of our results.

2 Preliminaries

As is well known, the Riemann-Liouville fractional integral of order α>0 of a function f:(0,)R is given by I α f(t)= 1 Γ ( α ) 0 t ( t s ) α 1 f(s)ds, provided the right side is pointwise defined on (0,) (see [10, 13] and [14]). The Caputo fractional derivative of order α for a continuous function f is defined by D α c f(t)= 1 Γ ( n α ) 0 t f ( n ) ( s ) ( t s ) α n + 1 ds, where n=[α]+1 (see [10, 13] and [14]).

Recall that a multifunction G:J P c l (R) is said to be measurable whenever the function td(y,G(t)) is measurable for all yR, where J=[0,1] [37].

Let (X,d) be a metric space. We have the well-known Pompeiu-Hausdorff metric (see [38])

H d : 2 X × 2 X [0,), H d (A,B)=max { sup a A d ( a , B ) , sup b B d ( A , b ) } ,

where d(A,b)= inf a A d(a,b). Then (CB(X), H d ) is a metric space and (C(X), H d ) is a generalized metric space, where CB(X) is the set of closed and bounded subsets of X and C(X) is the set of closed subsets of X (see [27]).

Let T:X 2 X be a multifunction. An element xX is called an endpoint of T whenever Tx={x} [36]. Also, we say that T has the approximate endpoint property whenever inf x X sup y T x d(x,y)=0 [36]. A function g:RR is called upper semi-continuous whenever lim sup n g( λ n )g(λ) for all sequences { λ n } n 1 with λ n λ [36].

In 2010, Amini-Harandi proved the next result [36].

Lemma 2.1 Let ψ:[0,)[0,) be an upper semi-continuous function such that ψ(t)<t and lim inf t (tψ(t))>0, for all t>0, (X,d) a complete metric space and T:XCB(X) a multifunction such that H d (Tx,Ty)ψ(d(x,y)) for all x,yX. Then T has a unique endpoint if and only if T has approximate end point property.

In 2011, Ahmad et al. investigated the fractional inclusion problem D α c x(t)F(t,x(t)), via the integral boundary conditions x j (0) λ j x j (T)= μ j 0 1 g j (s,x(s))ds for j=0,1,2, where F is a multifunction (see for more details [20]).

In this paper, we are going to extend the problem in a sense. In this respect, we first investigate the existence of solution for the fractional differential inclusion problem

D α c x(t)F ( t , x ( t ) , x ( t ) , x ( t ) ) ,
(1)

via integral boundary value conditions

{ x ( 0 ) + x ( η ) + x ( 1 ) = 0 1 g 0 ( s , x ( s ) ) d s , D β c x ( 0 ) + c D β x ( η ) + c D β x ( 1 ) = 0 1 g 1 ( s , x ( s ) ) d s , D γ c x ( 0 ) + c D γ x ( η ) + c D γ x ( 1 ) = 0 1 g 2 ( s , x ( s ) ) d s ,
(2)

where tJ, 2<α3, 0<η,β<1, 1<γ<2, and F:J×R×R×R P c p (R) is a multifunction, g 1 , g 2 , g 3 :J×RR are continuous functions and D q c is the standard Caputo differentiation. Here, P c p (R) is the set of all compact subsets of .

Also, we investigate the existence of solution for the fractional differential inclusion problem

D α c x(t)F ( t , x ( t ) , c D γ 1 x ( t ) , , c D γ n x ( t ) ) ,
(3)

via integral boundary value conditions

{ x ( 0 ) + b x ( 1 ) = i = 1 n c D γ i x ( η ) , x ( 0 ) + a x ( 1 ) = i = 1 n I γ i x ( η ) ,
(4)

where tJ=[0,1], 1<α2, 0<η, γ i <1, α γ i 1 for all 1in, a> i = 1 n η γ i + 1 Γ ( γ i + 2 ) , b> i = 1 n η 1 γ i Γ ( 2 γ i ) , n1, and F:J× R n + 1 P(R) is a multifunction.

3 Main results

Now, we are ready to state and prove our main results. First, we give the following one.

Lemma 3.1 Let vC(J,R), α(2,3], β(0,1), γ(1,2) and g 0 , g 1 , g 2 :J×RR be continuous functions. The unique solution of the fractional differential problem

D α c x(t)=v(t)
(5)

via the boundary value conditions (2) is given by

x ( t ) = 1 Γ ( α ) 0 t ( t s ) ( α 1 ) v ( s ) d s + 1 3 0 1 g 0 ( s , x ( s ) ) d s 1 3 Γ ( α ) [ 0 1 ( 1 s ) α 1 v ( s ) d s + 0 η ( η s ) α 1 v ( s ) d s ] + 3 Γ ( 2 β ) t ( η + 1 ) Γ ( 2 β ) 3 ( η 1 β + 1 ) 0 1 g 1 ( s , x ( s ) ) d s + ( η + 1 ) Γ ( 2 β ) 3 Γ ( 2 β ) t 3 ( η 1 β + 1 ) Γ ( α β ) [ 0 1 ( 1 s ) α β 1 v ( s ) d s + 0 η ( η s ) α β 1 v ( s ) d s ] + ( 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t + 3 ( η 1 β + 1 ) Γ ( 3 γ ) Γ ( 3 β ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) ) × 0 1 g 2 ( s , x ( s ) ) d s + ( ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t 3 Γ ( 3 γ ) Γ ( 3 β ) ( η 1 β + 1 ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) ) × [ 0 1 ( 1 s ) α γ 1 v ( s ) d s 0 η ( η s ) α γ 1 v ( s ) d s ] .

Proof It is known that the general solution of (5) is

x(t)= I α v(t)+ c 0 + c 1 t+ c 2 t 2 ,

that is

x(t)= 1 Γ ( α ) 0 t ( t s ) α 1 v(s)ds+ c 0 + c 1 t+ c 2 t 2 ,
(6)

where c 0 , c 1 , c 2 are real arbitrary constants (see [10, 13] and [14]). Thus,

D β c x(t)= 1 Γ ( α β ) 0 t ( t s ) α β 1 v(s)ds+ c 1 t 1 β Γ ( 2 β ) + 2 c 2 t 2 β Γ ( 3 β )

and D γ c x(t)= 1 Γ ( α γ ) 0 t ( t s ) α γ 1 v(s)ds+ 2 c 2 t 2 γ Γ ( 3 γ ) . Hence,

x ( 0 ) + x ( η ) + x ( 1 ) = 3 c 0 + ( 1 + η ) c 1 + ( 1 + η 2 ) c 2 + 1 Γ ( α ) [ 0 1 ( 1 s ) α 1 v ( s ) d s + 0 η ( η s ) α 1 v ( s ) d s ] , D β c x ( 0 ) + c D β x ( η ) + c D β x ( 1 ) = c 1 η 1 β + 1 Γ ( 2 β ) + c 2 2 ( η 2 β + 1 ) Γ ( 3 β ) + 1 Γ ( α β ) [ 0 1 ( 1 s ) α β 1 v ( s ) d s + 0 η ( η s ) α β 1 v ( s ) d s ]

and

D γ c x ( 0 ) + c D γ x ( η ) + c D γ x ( 1 ) = c 2 2 ( η 2 γ + 1 ) Γ ( 3 γ ) + 1 Γ ( α γ ) [ 0 1 ( 1 s ) α γ 1 v ( s ) d s + 0 η ( η s ) α γ 1 v ( s ) d s ] .

By using the boundary conditions, we obtain

3 c 0 + ( 1 + η ) c 1 + ( 1 + η 2 ) c 2 = 0 1 g 0 ( s , x ( s ) ) d s 1 Γ ( α ) [ 0 1 ( 1 s ) α 1 v ( s ) d s + 0 η ( η s ) α 1 v ( s ) d s ] , c 1 η 1 β + 1 Γ ( 2 β ) + c 2 2 ( η 2 β + 1 ) Γ ( 3 β ) = 0 1 g 1 ( s , x ( s ) ) d s 1 Γ ( α β ) [ 0 1 ( 1 s ) α β 1 v ( s ) d s + 0 η ( η s ) α β 1 v ( s ) d s ]

and

c 2 2 ( η 2 γ + 1 ) Γ ( 3 γ ) = 0 1 g 2 ( s , x ( s ) ) d s 1 Γ ( α γ ) [ 0 1 ( 1 s ) α γ 1 v ( s ) d s + 0 η ( η s ) α β 1 v ( s ) d s ] .

This is a linear system of equations of triangular form, having c 0 , c 1 , and c 2 as unknowns. We solve by back substitution and find

c 0 = 1 3 0 1 g 0 ( s , x ( s ) ) d s 1 3 Γ ( α ) [ 0 1 ( 1 s ) α 1 v ( s ) d s + 0 η ( η s ) α 1 v ( s ) d s ] Γ ( 2 β ) ( η + 1 ) 3 ( η 1 β + 1 ) × 0 1 g 1 ( s , x ( s ) ) d s + ( η + 1 ) Γ ( 2 β ) 3 ( η 1 β + 1 ) Γ ( α β ) [ 0 1 ( 1 s ) α β 1 v ( s ) d s + 0 η ( η s ) α β 1 v ( s ) d s ] + 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) × 0 1 g 2 ( s , x ( s ) ) d s + ( ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) ) × [ 0 1 ( 1 s ) α γ 1 v ( s ) d s + 0 η ( η s ) α γ 1 v ( s ) d s ] , c 1 = Γ ( 2 β ) ( η 1 β + 1 ) 0 1 g 1 ( s , x ( s ) ) d s Γ ( 2 β ) ( η 1 β + 1 ) Γ ( α β ) [ 0 1 ( 1 s ) α β 1 v ( s ) d s + 0 η ( η s ) α β 1 v ( s ) d s ] ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) 0 1 g 2 ( s , x ( s ) ) d s + ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) × [ 0 1 ( 1 s ) α γ 1 v ( s ) d s + 0 η ( η s ) α γ 1 v ( s ) d s ] ,

and

c 2 = Γ ( 3 γ ) 2 ( η 2 γ + 1 ) 0 1 g 2 ( s , x ( s ) ) d s Γ ( 3 γ ) 2 ( η 2 γ + 1 ) Γ ( α γ ) × [ 0 1 ( 1 s ) α γ 1 v ( s ) d s + 0 η ( η s ) α γ 1 v ( s ) d s ] .

Now, we replace c 0 , c 1 , and c 2 in (6) and find the solution x(t) as we stated. This completes the proof. □

Let X= C 2 ([0,1]) endowed with the norm x= sup t J |x(t)|+ sup t J | x (t)|+ sup t J | x (t)|. Then (X,) is a Banach space. For xX, define

S F , x = { v L 1 [ 0 , 1 ] : v ( t ) F ( t , x ( t ) , x ( t ) , x ( t ) )  for almost all  t [ 0 , 1 ] } .

For the study of problem (1) and (2), we shall consider the following conditions.

(H1) F:J×R×R×R P c p (R) is an integrable bounded multifunction such that F(,x,y,z):[0,1] P c p (R) is measurable for all x,y,zR;

(H2) g 0 , g 1 , g 2 :J×RR be continuous functions, ψ:[0,)[0,) a nondecreasing upper semi-continuous map such that lim inf t (tψ(t))>0 and ψ(t)<t for all t>0;

(H3) There exist m, m 0 , m 1 , m 2 C(J,[0,)) such that

H d ( F ( t , x 1 , x 2 , x 3 ) , F ( t , y 1 , y 2 , y 3 ) ) 1 Λ 1 + Λ 2 + Λ 3 m(t)ψ ( | x 1 y 1 | + | x 2 y 2 | + | x 3 y 3 | )

and | g j (t,x) g j (t,y)| 1 Λ 1 + Λ 2 + Λ 3 m j (t)ψ(|xy|) for all tJ, x,y, x 1 , x 2 , x 3 , y 1 , y 2 , y 3 R and j=0,1,2, where

Λ 1 = [ m Γ ( α + 1 ) + m 0 3 + 2 m 3 Γ ( α + 1 ) + 5 Γ ( 2 β ) m 1 3 + 10 Γ ( 2 β ) m 3 Γ ( α β + 1 ) + 10 ( 2 Γ ( 2 β ) + Γ ( 3 β ) ) Γ ( 3 γ ) ( m 2 Γ ( α γ + 1 ) + 2 m ) 3 Γ ( 3 β ) Γ ( α γ + 1 ) ] , Λ 2 = [ m Γ ( α ) + 2 Γ ( 2 β ) m Γ ( α β + 1 ) + ( 2 Γ ( 2 β ) + Γ ( 3 β ) ) Γ ( 3 γ ) ( m 2 Γ ( α γ + 1 ) + 2 m ) Γ ( 3 β ) Γ ( α γ + 1 ) ] ,

and Λ 3 =[ m Γ ( α 1 ) + Γ ( 3 γ ) ( m 2 Γ ( α γ + 1 ) + 2 m ) Γ ( α γ + 1 ) ], and finally

(H4) N:X 2 X is given by

N(x)= { h X :  there exist  v S F , x  such that  h ( t ) = w ( t )  for all  t J } ,

where

w ( t ) = 1 Γ ( α ) 0 t ( t s ) ( α 1 ) v ( s ) d s + 1 3 0 1 g 0 ( s , x ( s ) ) d s 1 3 Γ ( α ) [ 0 1 ( 1 s ) α 1 v ( s ) d s + 0 η ( η s ) α 1 v ( s ) d s ] + 3 Γ ( 2 β ) t ( η + 1 ) Γ ( 2 β ) 3 ( η 1 β + 1 ) 0 1 g 1 ( s , x ( s ) ) d s + ( η + 1 ) Γ ( 2 β ) 3 Γ ( 2 β ) t 3 ( η 1 β + 1 ) Γ ( α β ) [ 0 1 ( 1 s ) α β 1 v ( s ) d s + 0 η ( η s ) α β 1 v ( s ) d s ] + ( 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t + 3 ( η 1 β + 1 ) Γ ( 3 γ ) Γ ( 3 β ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) ) × 0 1 g 2 ( s , x ( s ) ) d s + ( ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t 3 Γ ( 3 γ ) Γ ( 3 β ) ( η 1 β + 1 ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) ) × [ 0 1 ( 1 s ) α γ 1 v ( s ) d s + 0 η ( η s ) α γ 1 v ( s ) d s ] .

Theorem 3.1 Assume that (H1)-(H4) are satisfied. If the multifunction N has the approximate endpoint property, then the boundary value inclusion problem (1) and (2) has a solution.

Proof We show that the multifunction N:XP(X) has a endpoint which is a solution of the problem (1) and (2).

Note that the multivalued map tF(t,x(t), x (t), x (t)) is measurable and has closed values for all xX. Hence, it has measurable selection and so S F , x is nonempty for all xX. First, we show that N(x) is closed subset of X for all xX.

Let xX and { u n } n 1 be a sequence in N(x) with u n u. For each nN, choose v n S F , x such that

u n ( t ) = 1 Γ ( α ) 0 t ( t s ) ( α 1 ) v n ( s ) d s + 1 3 0 1 g 0 ( s , x ( s ) ) d s 1 3 Γ ( α ) [ 0 1 ( 1 s ) α 1 v n ( s ) d s + 0 η ( η s ) α 1 v n ( s ) d s ] + 3 Γ ( 2 β ) t ( η + 1 ) Γ ( 2 β ) 3 ( η 1 β + 1 ) 0 1 g 1 ( s , x ( s ) ) d s + ( η + 1 ) Γ ( 2 β ) 3 Γ ( 2 β ) t 3 ( η 1 β + 1 ) Γ ( α β ) [ 0 1 ( 1 s ) α β 1 v n ( s ) d s + 0 η ( η s ) α β 1 v n ( s ) d s ] + ( 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t + 3 ( η 1 β + 1 ) Γ ( 3 γ ) Γ ( 3 β ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) ) × 0 1 g 2 ( s , x ( s ) ) d s + ( ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t 3 Γ ( 3 γ ) Γ ( 3 β ) ( η 1 β + 1 ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) ) × [ 0 1 ( 1 s ) α γ 1 v n ( s ) d s + 0 η ( η s ) α γ 1 v n ( s ) d s ]

for all tJ.

Since F has compact values, { v n } n 1 has a subsequence which converges to some v L 1 [0,1]. We denote this subsequence again by { v n } n 1 .

It is easy to check that v S F , x and

u n ( t ) u ( t ) = 1 Γ ( α ) 0 t ( t s ) ( α 1 ) v ( s ) d s + 1 3 0 1 g 0 ( s , x ( s ) ) d s 1 3 Γ ( α ) [ 0 1 ( 1 s ) α 1 v ( s ) d s + 0 η ( η s ) α 1 v ( s ) d s ] + 3 Γ ( 2 β ) t ( η + 1 ) Γ ( 2 β ) 3 ( η 1 β + 1 ) 0 1 g 1 ( s , x ( s ) ) d s + ( η + 1 ) Γ ( 2 β ) 3 Γ ( 2 β ) t 3 ( η 1 β + 1 ) Γ ( α β ) [ 0 1 ( 1 s ) α β 1 v ( s ) d s + 0 η ( η s ) α β 1 v ( s ) d s ] + ( 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t + 3 ( η 1 β + 1 ) Γ ( 3 γ ) Γ ( 3 β ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) ) × 0 1 g 2 ( s , x ( s ) ) d s + ( ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t 3 Γ ( 3 γ ) Γ ( 3 β ) ( η 1 β + 1 ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) ) × [ 0 1 ( 1 s ) α γ 1 v ( s ) d s + 0 η ( η s ) α γ 1 v ( s ) d s ]

for all tJ. This implies that uN(x) and so N has closed values.

Since F is a compact multivalued map, it is easy to check that N(x) is a bounded set for all xX.

Now, we show that H d (N(x),N(y))ψ(xy).

Let x,yX and h 1 N(y). Choose v 1 S F , y such that

h 1 ( t ) = 1 Γ ( α ) 0 t ( t s ) ( α 1 ) v 1 ( s ) d s + 1 3 0 1 g 0 ( s , y ( s ) ) d s 1 3 Γ ( α ) [ 0 1 ( 1 s ) α 1 v 1 ( s ) d s + 0 η ( η s ) α 1 v 1 ( s ) d s ] + 3 Γ ( 2 β ) t ( η + 1 ) Γ ( 2 β ) 3 ( η 1 β + 1 ) 0 1 g 1 ( s , y ( s ) ) d s + ( η + 1 ) Γ ( 2 β ) 3 Γ ( 2 β ) t 3 ( η 1 β + 1 ) Γ ( α β ) [ 0 1 ( 1 s ) α β 1 v 1 ( s ) d s + 0 η ( η s ) α β 1 v 1 ( s ) d s ] + ( 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t + 3 ( η 1 β + 1 ) Γ ( 3 γ ) Γ ( 3 β ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) ) × 0 1 g 2 ( s , y ( s ) ) d s + ( ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t 3 Γ ( 3 γ ) Γ ( 3 β ) ( η 1 β + 1 ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) ) × [ 0 1 ( 1 s ) α γ 1 v 1 ( s ) d s + 0 η ( η s ) α γ 1 v 1 ( s ) d s ]

for almost all tJ.

Since

H d ( F ( t , x ( t ) , x ( t ) , x ( t ) ) , F ( t , y ( t ) , y ( t ) , y ( t ) ) ) 1 Λ 1 + Λ 2 + Λ 3 m ( t ) ψ ( | x ( t ) y ( t ) | + | x ( t ) y ( t ) | + | x ( t ) y ( t ) | )

for all tJ, there exists wF(t,x(t), x (t), x (t)) such that

| v 1 ( t ) w | 1 Λ 1 + Λ 2 + Λ 3 m(t)ψ ( | x ( t ) y ( t ) | + | x ( t ) y ( t ) | + | x ( t ) y ( t ) | )

for all tJ.

Consider the multivalued map U:JP(R) defined by

U ( t ) = { w R : | v 1 ( t ) w | 1 Λ 1 + Λ 2 + Λ 3 m ( t ) ψ ( | x ( t ) y ( t ) | + | x ( t ) y ( t ) | + | x ( t ) y ( t ) | ) } .

Since v 1 and φ=mψ(|xy|+| x y |+| x y |)( 1 Λ 1 + Λ 2 + Λ 3 ) are measurable, the multifunction U()F(,x(), x (), x ()) is measurable.

Choose v 2 (t)F(t,x(t), x (t), x (t)) such that

| v 1 ( t ) v 2 ( t ) | 1 Λ 1 + Λ 2 + Λ 3 m ( t ) ψ ( | x ( t ) y ( t ) | + | x ( t ) y ( t ) | + | x ( t ) y ( t ) | )

for all tJ.

Now, consider the element h 2 N(x), which is defined by

h 2 ( t ) = 1 Γ ( α ) 0 t ( t s ) ( α 1 ) v 2 ( s ) d s + 1 3 0 1 g 0 ( s , x ( s ) ) d s 1 3 Γ ( α ) [ 0 1 ( 1 s ) α 1 v 2 ( s ) d s + 0 η ( η s ) α 1 v 2 ( s ) d s ] + 3 Γ ( 2 β ) t ( η + 1 ) Γ ( 2 β ) 3 ( η 1 β + 1 ) 0 1 g 1 ( s , x ( s ) ) d s + ( η + 1 ) Γ ( 2 β ) 3 Γ ( 2 β ) t 3 ( η 1 β + 1 ) Γ ( α β ) [ 0 1 ( 1 s ) α β 1 v 2 ( s ) d s + 0 η ( η s ) α β 1 v 2 ( s ) d s ] + ( 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t + 3 ( η 1 β + 1 ) Γ ( 3 γ ) Γ ( 3 β ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) ) × 0 1 g 2 ( s , x ( s ) ) d s + ( ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t 3 Γ ( 3 γ ) Γ ( 3 β ) ( η 1 β + 1 ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) ) × [ 0 1 ( 1 s ) α γ 1 v 2 ( s ) d s + 0 η ( η s ) α γ 1 v 2 ( s ) d s ]

for all tJ. Thus,

| h 1 ( t ) h 2 ( t ) | 1 Γ ( α ) 0 t ( t s ) ( α 1 ) | v 1 ( s ) v 2 ( s ) | d s + 1 3 0 1 | g 0 ( s , y ( s ) ) g 0 ( s , x ( s ) ) | d s + 1 3 Γ ( α ) [ 0 1 ( 1 s ) α 1 | v 1 ( s ) v 2 ( s ) | d s + 0 η ( η s ) α 1 | v 1 ( s ) v 2 ( s ) | d s ] + | 3 Γ ( 2 β ) t ( η + 1 ) Γ ( 2 β ) 3 ( η 1 β + 1 ) | × 0 1 | g 1 ( s , y ( s ) ) g 1 ( s , x ( s ) ) | d s + | ( η + 1 ) Γ ( 2 β ) 3 Γ ( 2 β ) t 3 ( η 1 β + 1 ) Γ ( α β ) | × [ 0 1 ( 1 s ) α β 1 | v 1 ( s ) v 2 ( s ) | d s + 0 η ( η s ) α β 1 | v 1 ( s ) v 2 ( s ) | d s ] + | ( 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t + 3 ( η 1 β + 1 ) Γ ( 3 γ ) Γ ( 3 β ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) ) | × 0 1 | g 2 ( s , y ( s ) ) g 2 ( s , x ( s ) ) | d s + | ( ( η 2 + 1 ) Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) 2 ( η + 1 ) ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) + 6 ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) t 3 Γ ( 3 γ ) Γ ( 3 β ) ( η 1 β + 1 ) t 2 6 ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) ) | × [ 0 1 ( 1 s ) α γ 1 | v 1 ( s ) v 2 ( s ) | d s + 0 η ( η s ) α γ 1 | v 1 ( s ) v 2 ( s ) | d s ] 1 Λ 1 + Λ 2 + Λ 3 ψ ( x y ) [ m Γ ( α + 1 ) + m 0 3 + 2 m 3 Γ ( α + 1 ) + 5 Γ ( 2 β ) m 1 3 + 10 Γ ( 2 β ) m 3 Γ ( α β + 1 ) + 10 ( 2 Γ ( 2 β ) + Γ ( 3 β ) ) Γ ( 3 γ ) ( m 2 Γ ( α γ + 1 ) + 2 m ) 3 Γ ( 3 β ) Γ ( α γ + 1 ) ] = Λ 1 Λ 1 + Λ 2 + Λ 3 ψ ( x y ) , | h 1 ( t ) h 2 ( t ) | 1 Γ ( α 1 ) 0 t ( t s ) ( α 2 ) | v 1 ( s ) v 2 ( s ) | d s + Γ ( 2 β ) ( η 1 β + 1 ) Γ ( α β ) [ 0 1 ( 1 s ) α β 1 | v 1 ( s ) v 2 ( s ) | d s + 0 η ( η s ) α β 1 | v 1 ( s ) v 2 ( s ) | d s ] + | ( ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) + Γ ( 3 γ ) ( η 1 β + 1 ) Γ ( 3 β ) t ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) | × 0 1 | g 2 ( s , y ( s ) ) g 2 ( s , x ( s ) ) | d s + | ( η 2 β + 1 ) Γ ( 3 γ ) Γ ( 2 β ) Γ ( 3 γ ) Γ ( 3 β ) ( η 1 β + 1 ) t ( η 1 β + 1 ) ( η 2 γ + 1 ) Γ ( 3 β ) Γ ( α γ ) ) | × [ 0 1 ( 1 s ) α γ 1 | v 1 ( s ) v 2 ( s ) | d s + 0 η ( η s ) α γ 1 | v 1 ( s ) v 2 ( s ) | d s ] 1 Λ 1 + Λ 2 + Λ 3 ψ ( x y ) [ m Γ ( α ) + 2 Γ ( 2 β ) m Γ ( α β + 1 ) + ( 2 Γ ( 2 β ) + Γ ( 3 β ) ) Γ ( 3 γ ) ( m 2 Γ ( α γ + 1 ) + 2 m ) Γ ( 3 β ) Γ ( α γ + 1 ) ] = Λ 2 Λ 1 + Λ 2 + Λ 3 ψ ( x y ) ,

and

| h 1 ′′ ( t ) h 2 ′′ ( t ) | 1 Γ ( α 2 ) 0 t ( t s ) ( α 3 ) | v 1 ( s ) v 2 ( s ) | d s + Γ ( 3 γ ) ( η 2 γ + 1 ) 0 1 | g 2 ( s , y ( s ) ) g 2 ( s , x ( s ) ) | d s + Γ ( 3 γ ) ( η 2 γ + 1 ) Γ ( α γ ) [ 0 1 ( 1 s ) α γ 1 | v 1 ( s ) v 2 ( s ) | d s + 0 η ( η s ) α γ 1 | v 1 ( s ) v 2 ( s ) | d s ] 1 Λ 1 + Λ 2 + Λ 3 ψ ( x y ) [ m Γ ( α 1 ) + Γ ( 3 γ ) ( m 2 Γ ( α γ + 1 ) + 2 m ) Γ ( α γ + 1 ) ] = Λ 3 Λ 1 + Λ 2 + Λ 3 ψ ( x y ) .

Hence,

h 1 h 2 = sup t J | h 1 ( t ) h 2 ( t ) | + sup t J | h 1 ( t ) h 2 ( t ) | + sup t J | h 1 ′′ ( t ) h 2 ′′ ( t ) | 1 Λ 1 + Λ 2 + Λ 3 ψ ( x y ) ( Λ 1 + Λ 2 + Λ 3 ) = ψ ( x y ) .

Thus, it is easy to get H d (N(x),N(y))ψ(xy) for all x,yX.

Since the multifunction N has approximate endpoint property, by using Lemma 2.1 there exists x X such that N( x )={ x }. Hence by using Lemma 3.1, x is a solution of the problem (1) and (2). □

Now, we investigate the existence of solution for the fractional differential inclusion problem

D α c x(t)F ( t , x ( t ) , c D γ 1 x ( t ) , , c D γ n x ( t ) ) ,

via integral boundary value conditions

x (0)+b x (1)= i = 1 n c D γ i x(η),x(0)+ax(1)= i = 1 n I γ i x(η),

where tJ=[0,1], 1<α2, n2, 0<η, γ i <1, α γ i 1 for all 1in, a> i = 1 n η γ i + 1 Γ ( γ i + 2 ) , b> i = 1 n η 1 γ i Γ ( 2 γ i ) and F:J× R n + 1 P(R) is a multifunction.

Lemma 3.2 Let vC(J,R), α(1,2], η(0,1), n2 and β i (0,1) for i=1,,n. The unique solution of the fractional differential problem D α c x(t)=v(t) via the boundary value conditions

x(0)+ax(1)= i = 1 n I β i x(η), x (0)+b x (1)= i = 1 n c D β i x(η),

with a> i = 1 n η β i + 1 Γ ( β i + 2 ) , b> i = 1 n η 1 β i Γ ( 2 β i ) is

x(t)= 0 1 G(t,s)v(s)ds,

where G(t,s) is the Green function given by

G ( t , s ) = ( t s ) α 1 Γ ( α ) + 1 A i = 1 n ( η s ) α + β i 1 Γ ( α + β i ) a A Γ ( α ) ( 1 s ) α 1 1 A B ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) × i = 1 n ( η s ) α β i 1 Γ ( α β i ) b A B ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) ( 1 s ) α 2 Γ ( α 1 ) + t B i = 1 n ( η s ) α β i 1 Γ ( α β i ) b t B Γ ( α 1 ) ( 1 s ) α 2

whenever 0sηt1,

G ( t , s ) = ( t s ) α 1 Γ ( α ) a A Γ ( α ) ( 1 s ) α 1 b A B ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) ( 1 s ) α 2 Γ ( α 1 ) b t B Γ ( α 1 ) ( 1 s ) α 2

whenever 0ηst1,

G(t,s)= a A Γ ( α ) ( 1 s ) α 1 b A B ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) ( 1 s ) α 2 Γ ( α 1 ) b t B Γ ( α 1 ) ( 1 s ) α 2

whenever 0ηst1,

G ( t , s ) = ( t s ) α 1 Γ ( α ) + 1 A i = 1 n ( η s ) α + β i 1 Γ ( α + β i ) a A Γ ( α ) ( 1 s ) α 1 1 A B ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) × i = 1 n ( η s ) α β i 1 Γ ( α β i ) b A B ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) ( 1 s ) α 2 Γ ( α 1 ) + t B i = 1 n ( η s ) α β i 1 Γ ( α β i ) b t B Γ ( α 1 ) ( 1 s ) α 2

whenever 0stη1,

G ( t , s ) = 1 A i = 1 n ( η s ) α + β i 1 Γ ( α + β i ) a A Γ ( α ) ( 1 s ) α 1 1 A B ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) × i = 1 n ( η s ) α β i 1 Γ ( α β i ) b A B ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) ( 1 s ) α 2 Γ ( α 1 ) + t B i = 1 n ( η s ) α β i 1 Γ ( α β i ) b t B Γ ( α 1 ) ( 1 s ) α 2

whenever 0tsη1 and

G(t,s)= a A Γ ( α ) ( 1 s ) α 1 b A B ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) ( 1 s ) α 2 Γ ( α 1 ) b t B Γ ( α 1 ) ( 1 s ) α 2

whenever 0tηs1, where A=1+a i = 1 n η β i + 1 Γ ( β i + 2 ) and B=1+b i = 1 n η 1 β i Γ ( 2 β i ) .

Proof It is known that the general solution of the equation D α c x(t)=v(t) is

x(t)= I α v(t)+ c 0 + c 1 t= 1 Γ ( α ) 0 t ( t s ) ( α 1 ) v(s)ds+ c 0 + c 1 t,

where c 0 , c 1 R are arbitrary constants (see [10, 13] and [14]). Thus,

D β i c x ( t ) = 1 Γ ( α β i ) 0 t ( t s ) ( α β i 1 ) v ( s ) d s + c 1 t 1 β i Γ ( 2 β i ) , I β i x ( t ) = 1 Γ ( α + β i ) 0 t ( t s ) ( α + β i 1 ) v ( s ) d s + c 0 t 1 + β i Γ ( 2 + β i ) + c 1 t 2 + β i Γ ( 3 + β i ) ,

and x (t)= 1 Γ ( α 1 ) 0 t ( t s ) ( α 2 ) v(s)ds+ c 1 . Hence,

x(0)+ax(1)=(a+1) c 0 +a c 1 + a Γ ( α ) 0 1 ( 1 s ) ( α 1 ) v(s)ds

and

x (0)+b x (1)=(1+b) c 1 + b Γ ( α 1 ) 0 1 ( 1 s ) ( α 2 ) v(s)ds.

By using the boundary conditions, we obtain

c 0 ( 1 + a i = 1 n η β i + 1 Γ ( β i + 2 ) ) + c 1 ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) = i = 1 n 0 η ( η s ) α + β i 1 Γ ( β i + α ) v ( s ) d s a Γ ( α ) 0 1 ( 1 s ) α 1 v ( s ) d s

and

c 1 ( 1 + b i = 1 n η 1 β i Γ ( 2 β i ) ) = i = 1 n 0 η ( η s ) α β i 1 Γ ( α β i ) v(s)ds b Γ ( α 1 ) 0 1 ( 1 s ) α 2 v(s)ds.

Thus,

c 0 = 1 A i = 1 n 0 η ( η s ) α + β i 1 Γ ( α + β i ) v ( s ) d s a A Γ ( α ) 0 1 ( 1 s ) α 1 v ( s ) d s 1 A B ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) × i = 1 n 0 η ( η s ) α β i 1 Γ ( α β i ) v ( s ) d s b A B Γ ( α 1 ) ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) 0 1 ( 1 s ) α 2 v ( s ) d s , c 1 = 1 B i = 1 n 0 η ( η s ) α β i 1 Γ ( α β i ) v ( s ) d s b B Γ ( α 1 ) 0 1 ( 1 s ) α 2 v ( s ) d s .

Hence,

x ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 v ( s ) d s + 1 A i = 1 n 0 η ( η s ) α + β i 1 Γ ( α + β i ) v ( s ) d s a A Γ ( α ) 0 1 ( 1 s ) α 1 v ( s ) d s 1 A B ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) i = 1 n 0 η ( η s ) α β i 1 Γ ( α β i ) v ( s ) d s b A B Γ ( α 1 ) ( a i = 1 n η β i + 2 Γ ( β i + 3 ) ) × 0 1 ( 1 s ) α 2 v ( s ) d s + t B i = 1 n 0 η ( η s ) α β i 1 Γ ( α β i ) v ( s ) d s t b B Γ ( α 1 ) 0 1 ( 1 s ) α 2 v ( s ) d s = 0 1 G ( t , s ) v ( s ) d s .

This completes the proof. □

Suppose that X={x:x, c D γ i xC(J,R) for all i=1,,n} endowed with the norm x= sup t J |x(t)|+ i = 1 n sup t J | c D γ i x(t)|. Then (X,) is a Banach space [15]. For xX, define

S F , x = { v L 1 [ 0 , 1 ] : v ( t ) F ( t , x ( t ) , c D γ 1 x ( t ) , , c D γ n x ( t ) )  for almost all  t [ 0 , 1 ] } .

Now, put

L 1 = 1 Γ ( α + 1 ) + 1 A i = 1 n η α + γ i Γ ( α + γ i + 1 ) + a A Γ ( α + 1 ) + 1 A B ( | a i = 1 n η γ i + 2 Γ ( γ i + 3 ) | ) × i = 1 n η α γ i Γ ( α γ i + 1 ) + b A B Γ ( α ) ( | a i = 1 n η γ i + 2 Γ ( γ i + 3 ) | ) + 1 B i = 1 n η α γ i Γ ( α γ i + 1 ) + b B Γ ( α )

and L 2 j = 1 Γ ( α γ j + 1 ) + 1 B Γ ( 2 γ j ) i = 1 n η α γ i Γ ( α γ i + 1 ) + b B Γ ( 2 γ j ) Γ ( α ) for all 1jn.

Theorem 3.2 Let ψ:[0,)[0,) a nondecreasing upper semi-continuous map such that lim inf t (tψ(t))>0 and ψ(t)<t for all t>0, F:J× R n + 1 P c p (R) a multifunction such that F(, x 1 , x 2 ,, x n + 1 ):[0,1] P c p (R) is measurable and integrable bounded for all x 1 , x 2 ,, x n + 1 R. Assume that there exists mC(J,[0,)) such that

H d ( F ( t , x 1 , x 2 , , x n + 1 ) F ( t , y 1 , y 2 , , y n + 1 ) ) m ( t ) ψ ( | x 1 y 1 | + | x 2 y 2 | + + | x n + 1 y n + 1 | ) ( 1 m ( L 1 + j = 1 n L 2 j ) ) .

Define Ω:X 2 X by

Ω(x)= { h X :  there exist  v S F , x  such that  h ( t ) = 0 1 G ( t , s ) v ( s ) d s  for all  t J } .

If the multifunction Ω has the approximate endpoint property, then the boundary value inclusion problem (3) and (4) has a solution.

Proof We show that the multifunction Ω:XP(X) has a endpoint which is a solution of the problem (3) and (4).

First, we show that Ω(x) is closed subset of X for all xX.

Let xX and { u n } n 1 be a sequence in Ω(x) with u n u. For each nN, choose v n S F , x such that u n (t)= 0 1 G(t,s) v n (s)ds for all tJ. Since F has compact values, { v n } n 1 has a subsequence which converges to some v L 1 [0,1]. We denote this subsequence again by { v n } n 1 .

It is easy to check that v S F , x and u n (t)u(t)= 0 1 G(t,s)v(s)ds for all tJ. This implies that uΩ(x) and so Ω has closed values. Since F is a compact multivalued map, it is easy to check that Ω(x) is a bounded set for all xX.

Now, we show that for all x,yX, H d (Ω(x),Ω(y))ψ(xy).

Let x,yX and h 1 Ω(y). Choose v 1 S F , y such that h 1 (t)= 0 1 G(t,s) v 1 (s)ds for almost all tJ. Since

H d ( F ( t , x ( t ) , c D γ 1 x ( t ) , , c D γ n x ( t ) ) , F ( t , y ( t ) , c D γ 1 y ( t ) , , c D γ n y ( t ) ) ) m ( t ) ψ ( | x ( t ) y ( t ) | + | c D γ 1 x ( t ) c D γ 1 y ( t ) | + + | c D γ n x ( t ) c D γ n y ( t ) | ) × ( 1 m ( L 1 + j = 1 n L 2 j ) )

for all tJ, there exists wF(t,x(t), c D γ 1 x(t),, c D γ n x(t)) such that

| v 1 ( t ) w | m ( t ) ψ ( | x ( t ) y ( t ) | + | c D γ 1 x ( t ) c D γ 1 y ( t ) | + + | c D γ n x ( t ) c D γ n y ( t ) | ) ( 1 m ( L 1 + j = 1 n L 2 j ) )

for all tJ. Consider the multivalued map U:JP(R) defined by the rule

U ( t ) = { w R : | v 1 ( t ) w | m ( t ) ψ ( | x ( t ) y ( t ) | + | c D γ 1 x ( t ) c D γ 1 y ( t ) | + + | c D γ n x ( t ) c D γ n y ( t ) | ) ( 1 m ( L 1 + j = 1 n L 2 j ) ) } .

Since v 1 and

φ=mψ ( | x y | + | c D γ 1 x c D γ 1 y | + + | c D γ n x c D γ n y | ) ( 1 m ( L 1 + j = 1 n L 2 j ) )

are measurable, the multifunction

U()F ( t , x ( ) , c D γ 1 x ( ) , , c D γ n x ( ) )

is measurable.

Choose v 2 (t)F(t,x(t), c D γ 1 x(t),, c D γ n x(t)) such that

| v 1 ( t ) v 2 ( t ) | m ( t ) ψ ( | x ( t ) y ( t ) | + | c D γ 1 x ( t ) c D γ 1 y ( t ) | + + | c D γ n x ( t ) c D γ n y ( t ) | ) ( 1 m ( L 1 + j = 1 n L 2 j ) )

for all tJ. Now, consider the element h 2 Ω(x) which is defined by h 2 (t)= 0 1 G(t,s) v 2 (s)ds for all tJ.

Thus,

| h 1 ( t ) h 2 ( t ) | = | 0 1 G ( t , s ) v 1 ( s ) d s 0 1 G ( t , s ) v 2 ( s ) d s | = | 1 Γ ( α ) 0 t ( t s ) α 1 v 1 ( s ) d s + 1 A i = 1 n 0 η ( η s ) α + γ i 1 Γ ( α + γ i ) v 1 ( s ) d s a A Γ ( α ) 0 1 ( 1 s ) α 1 v 1 ( s ) d s 1 A B ( a i = 1 n η γ i + 2 Γ ( γ i + 3 ) ) × i = 1 n 0 η ( η s ) α γ i 1 Γ ( α γ i ) v 1 ( s ) d s b A B Γ ( α 1 ) ( a i = 1 n η γ i + 2 Γ ( γ i + 3 ) ) × 0 1 ( 1 s ) α 2 v 1 ( s ) d s + t B i = 1 n 0 η ( η s ) α γ i 1 Γ ( α γ i ) v 1 ( s ) d s t b B Γ ( α 1 ) 0 1 ( 1 s ) α 2 v 1 ( s ) d s 1 Γ ( α ) 0 t ( t s ) α 1 v 2 ( s ) d s 1 A i = 1 n 0 η ( η s ) α + γ i 1 Γ ( α + γ i ) v 2 ( s ) d s + a A Γ ( α ) 0 1 ( 1 s ) α 1 v 2 ( s ) d s + 1 A B ( a i = 1 n η γ i + 2 Γ ( γ i + 3 ) ) × i = 1 n 0 η ( η s ) α γ i 1 Γ ( α γ i ) v 2 ( s ) d s + b A B Γ ( α 1 ) ( a i = 1 n η γ i + 2 Γ ( γ i + 3 ) ) × 0 1 ( 1 s ) α 2 v 2 ( s ) d s t B i = 1 n 0 η ( η s ) α γ i 1 Γ ( α γ i ) v 2 ( s ) d s + t b B Γ ( α 1 ) 0 1 ( 1 s ) α 2 v 2 ( s ) d s | 1 Γ ( α ) 0 t ( t s ) α 1 | v 1 ( s ) v 2 ( s ) | d s + 1 A i = 1 n 0 η ( η s ) α + γ i 1 Γ ( α + γ i ) | v 1 ( s ) v 2 ( s ) | d s + a A Γ ( α ) 0 1 ( 1 s ) α 1 | v 1 ( s ) v 2 ( s ) | d s + 1 A B ( | a i = 1 n η γ i + 2 Γ ( γ i + 3 ) | ) i = 1 n 0 η ( η s ) α γ i 1 Γ ( α γ i ) | v 1 ( s ) v 2 ( s ) | d s + b A B Γ ( α 1 ) ( | a i = 1 n η γ i + 2 Γ