Open Access

On a fractional differential inclusion via a new integral boundary condition

  • R Ghorbanian1, 2,
  • Vahid Hedayati3,
  • Mihai Postolache4Email author and
  • Shahram Rezapour3
Journal of Inequalities and Applications20142014:319

https://doi.org/10.1186/1029-242X-2014-319

Received: 7 April 2014

Accepted: 31 July 2014

Published: 21 August 2014

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.

Keywords

Caputo fractional derivative endpoint fractional differential inclusion

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 ) d s , 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 d s , 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 t d ( y , G ( t ) ) is measurable for all y R , 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 ( C B ( X ) , H d ) is a metric space and ( C ( X ) , H d ) is a generalized metric space, where C B ( 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 x X is called an endpoint of T whenever T x = { 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 : R R 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 ( t ψ ( t ) ) > 0 , for all t > 0 , ( X , d ) a complete metric space and T : X C B ( X ) a multifunction such that H d ( T x , T y ) ψ ( d ( x , y ) ) for all x , y X . 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 ) ) d s 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 t J , 2 < α 3 , 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 × R R 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 t J = [ 0 , 1 ] , 1 < α 2 , 0 < η , γ i < 1 , α γ i 1 for all 1 i n , a > i = 1 n η γ i + 1 Γ ( γ i + 2 ) , b > i = 1 n η 1 γ i Γ ( 2 γ i ) , n 1 , 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 v C ( J , R ) , α ( 2 , 3 ] , β ( 0 , 1 ) , γ ( 1 , 2 ) and g 0 , g 1 , g 2 : J × R R 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 ) d s + 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 ) d s + 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 ) d s + 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 x X , 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 , z R ;

(H2) g 0 , g 1 , g 2 : J × R R be continuous functions, ψ : [ 0 , ) [ 0 , ) a nondecreasing upper semi-continuous map such that lim inf t ( t ψ ( 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 ( | x y | ) for all t J , 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 : X P ( X ) has a endpoint which is a solution of the problem (1) and (2).

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

Let x X and { u n } n 1 be a sequence in N ( x ) with u n u . For each n N , 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 t J .

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 t J . This implies that u N ( 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 x X .

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

Let x , y X 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 t J .

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 t J , there exists w F ( 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 t J .

Consider the multivalued map U : J P ( 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 = m ψ ( | x y | + | 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 t J .

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 t J . 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 ( x y ) for all x , y X .

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 ) + a x ( 1 ) = i = 1 n I γ i x ( η ) ,

where t J = [ 0 , 1 ] , 1 < α 2 , n 2 , 0 < η , γ i < 1 , α γ i 1 for all 1 i n , 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 v C ( J , R ) , α ( 1 , 2 ] , η ( 0 , 1 ) , n 2 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 ) + a x ( 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 ) d s ,
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 0 s η t 1 ,
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 0 η s t 1 ,
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 0 η s t 1 ,
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 0 s t η 1 ,
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 0 t s η 1 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 0 t η s 1 , 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 ) d s + 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 ) d s + c 1 . Hence,
x ( 0 ) + a x ( 1 ) = ( a + 1 ) c 0 + a c 1 + a Γ ( α ) 0 1 ( 1 s ) ( α 1 ) v ( s ) d s
and
x ( 0 ) + b x ( 1 ) = ( 1 + b ) c 1 + b Γ ( α 1 ) 0 1 ( 1 s ) ( α 2 ) v ( s ) d s .
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 ) d s b Γ ( α 1 ) 0 1 ( 1 s ) α 2 v ( s ) d s .
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 x C ( 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 x X , 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 1 j n .

Theorem 3.2 Let ψ : [ 0 , ) [ 0 , ) a nondecreasing upper semi-continuous map such that lim inf t ( t ψ ( 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 m C ( 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 : X P ( 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 x X .

Let x X and { u n } n 1 be a sequence in Ω ( x ) with u n u . For each n N , choose v n S F , x such that u n ( t ) = 0 1 G ( t , s ) v n ( s ) d s for all t J . 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 ) d s for all t J . This implies that u Ω ( 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 x X .

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

Let x , y X and h 1 Ω ( y ) . Choose v 1 S F , y such that h 1 ( t ) = 0 1 G ( t , s ) v 1 ( s ) d s for almost all t J . 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 t J , there exists w F ( 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 t J . Consider the multivalued map U : J P ( 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 = m ψ ( | 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 t J . Now, consider the element h 2 Ω ( x ) which is defined by h 2 ( t ) = 0 1 G ( t , s ) v 2 ( s ) d s for all t J .

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 ( γ i + 3 ) | ) × 0 1 ( 1 s ) α 2 | v 1 ( s ) v 2 ( s ) | d s + t B i = 1 n 0 η ( η s ) α γ i 1 Γ ( α γ i ) | v 1 ( s ) v 2 ( s ) | d s + t b B Γ ( α 1 ) 0 1 ( 1 s ) α 2 | v 1 ( s ) v 2 ( s ) | d s ( L 1 L 1 + j = 1 n L 2 j ) ψ ( x y ) ,
and
| c D γ j h 1 ( t ) c D γ j h 2 ( t ) | 1 Γ ( α γ j ) 0 t ( t s ) α γ j 1 | v 1 ( s ) v 2 ( s ) | d s + t 1 γ j B Γ ( 2 γ j ) × i = 1 n 0 η ( η s ) α γ i 1 Γ ( α γ i ) | v 1 ( s ) v 2 ( s ) | d s + b t 1 γ j B Γ ( 2 γ j ) Γ ( α 1 ) 0 1 ( 1 s ) α 2 | v 1 ( s ) v 2 ( s ) | d s ( L 2 j L 1 + j = 1 n Λ 2 j ) ψ ( x y )
for all 1 j n . Hence,
h 1 h 2 = sup t J | h 1 ( t ) h 2 ( t ) | + sup t J i = 1 n | c D γ i h 1 ( t ) c D γ i h 2 ( t ) | ψ ( x y )