Arab Journal of Mathematical Sciences

Arab J Math Sci 19(1) (2013), 73-83

On a multi point boundary value problem for a fractional order differential inclusion

Aurelian Cernea

Faculty of Mathematics and Informatics, University of Bucharest, Academiei 14, 010014 Bucharest, Romania

Abstract. The existence of solutions for a multi point boundary value problem of a fractional order differential inclusion is investigated. Several results are obtained by using suitable fixed point theorems when the right hand side has convex or non convex values.

Mathematics subject classification: 34A60; 34B18; 34B15

Keywords: Fractional derivative; Differential inclusion; Boundary value problem; Fixed point

Differential equations with fractional order have recently proved to be strong tools in the modeling of many physical phenomena; for a good bibliography on this topic we refer to [18]. As a consequence there was an intensive development of the theory of differential equations of fractional order [2,16,22] etc.. The study of fractional differential inclusions was initiated by El-Sayed and Ibrahim [13]. Very recently several qualitative results for fractional differential inclusions were obtained in [1,3,6-11,15,20] etc.

In this paper we study the following problem

Received 4 June 2012; accepted 12 July 2012 Available online 21 July 2012

1. Introduction

Dax(t) 2 F(t, x(t), x'{t)) a.e. [0, 1]

Tel.: +40 785810358.

E-mail address: acernea@fmi.unibuc.ro

Peer review under responsibility of King Saud University.

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ELSEVIER

1319-5166 © 2012 King Saud University. Production and hosting by Elsevier B.V. All rights reserved. http://dx.doi.Org/10.1016/j.ajmsc.2012.07.001

where Da is the standard Riemann-Liouville fractional derivative, a e(2,3], m P 1, 0 < 6 < ••• < ím < 1, Et^r1 < 1, k > 0, a, > 0, i = TTm and F: [0, 1] x R x R ! P(R) is a set-valued map.

The present paper is motivated by a recent paper of Nyamoradi [19], where it is considered problem (1.1) and (1.2) with F single valued and several existence results are provided.

The aim of our paper is to extend the study in [19] to the set-valued framework and to present some existence results for problem (1.1) and (1.2). Our results are essentially based on a nonlinear alternative of Leray-Schauder type, on Bressan-Colombo selection theorem for lower semicontinuous set-valued maps with decomposable values and on Covitz and Nadler set-valued contraction principle. The methods used are known ([1,8,9] etc.), however their exposition in the framework of problem (1.1) and (1.2) is new.

The paper is organized as follows: in Section 2 we recall some preliminary facts that we need in the sequel and in Section 3 we prove our main results.

2. Preliminaries

In this section we sum up some basic facts that we are going to use later.

Let (X, d) be a metric space with the corresponding norm I and let I c R be a compact interval. Denoted by L(I) the r-algebra of all Lebesgue measurable subsets of I, by P(X) the family of all nonempty subsets of X and by B(X) the family of all Borel subsets of X. If A c I then vA: I fi {0,1} denotes the characteristic function of A. For any subset A c X we denote by A the closure of A.

Recall that the Pompeiu-Hausdorff distance of the closed subsets A,B c X is defined

dH(A, B) = max{d*(A, B), d*(B, A)}, d*(A, B) = sup{d(a, B); a e A},

where d(x,B) = infyeBd(x,y).

As usual, we denote by C(I,X) the Banach space of all continuous functions x: I fi X endowed with the norm I x| C = supteI x(t)| and by L1(I,X) the Banach space of all (Bochner) integrable functions x: I fi X endowed with the norm |x|1 = J |x(t)|dt.

A subset D c L1(I,X) is said to be decomposable if for any m,v e D and any subset A e L(I) one has + vvB e D, where B = T\A.

Consider T: X ! P(X) a set-valued map. A point x e X is called a fixed point for T if x e T(x). T is said to be bounded on bounded sets if T(B):= UxeBT(x) is a bounded subset of X for all bounded sets B in X. T is said to be compact if T(B) is relatively compact for any bounded sets B in X. T is said to be totally compact if T(X) is a compact subset of X. T is said to be upper semicontinuous if for any open set D c X, the set {x e X: T(x) c D} is open in X. T is called completely continuous if it is upper semicon-tinuous and totally bounded on X.

It is well known that a compact set-valued map T with nonempty compact values is upper semicontinuous if and only if T has a closed graph.

We recall the following nonlinear alternative of Leray-Schauder type and its consequences.

Theorem 2.1 [21]. Let D and D be open and closed subsets in a normed linear space X such that 0 2 D and let T: D ! P(X) be a completely continuous set-valued map with compact convex values. Then either

(i) the inclusion x 2 T(x) has a solution, or

(ii) there exists x 2 OD (the boundary of D) such that kx 2 T(x) for some k > 1.

Corollary 2.2. Let Br(0) and Br(0) be the open and closed balls in a normed linear space X centered at the origin and of radius r and let T: Br(0) ! P(X) be a completely continuous set-valued map with compact convex values. Then either

(i) the inclusion x 2 T(x) has a solution, or

(ii) there exists x 2 X with I x| = r and kx 2 T(x) for some k > 1.

Corollary 2.3. Let Br(0) and Br(0) be the open and closed balls in a normed linear space X centered at the origin and of radius r and let T : Br(0)! X be a completely continuous single valued map with compact convex values. Then either

(i) the equation x = T(x) has a solution, or

(ii) there exists x 2 X with I x| = r and x = kT(x) for some k < 1.

We recall that a multifunction T : X ! P(X) is said to be lower semicontinuous if for any closed subset C c X, the subset {s 2 X: T(s) c C} is closed.

If F : I x R x R ! P(R) is a set-valued map with compact values and x 2 C(I,R) we define

Sf(x) = ff 2 L1(I, R) : f(t) 2 F(t, x(t), x'(t)) a.e. I}.

We say that F is of lower semicontinuous type if SF(.) is lower semicontinuous with closed and decomposable values.

Theorem 2.4 [4]. Let S be a separable metric space and G : S ! P(L1 (I, R)) be a lower semicontinuous set-valued map with closed decomposable values.

Then G has a continuous selection (i.e., there exists a continuous mapping g: S fi L1(I,R) such that g(s) 2 G(s) "s 2 S).

A set-valued map G : I! P(R) with nonempty compact convex values is said to be measurable if for any x 2 R the function t fi d(x,G(t)) is measurable.

A set-valued map F : I x R x R ! P(R) is said to be Caratheodory if t fi F(t,x,y) is measurable for all x,y 2 R and (x,y) fi F(t,x,y) is upper semicontinuous for almost all t 2 I.

F is said to be L1-Caratheodory if for any l > 0 there exists hi 2 L1(I,R) such that sup{ vl : v 2 F(t,x,y)} 6 h(t) a.e. I, 8x, y 2 Bl(0).

Theorem 2.5 [17]. Let Xbe a Banach space, let F : I x X ! P(X) be a L1-Caratheodory set-valued map with SF „ 0 and let f: L1 (I,X) fi C(I,X) be a linear continuous mapping.

Then the set-va/wed map C o SF : C(/, X) ! P(C(/, X)) defined by (r o Sf)(x) = T(Sf(x))

has compact convex va/wes and has a c/osed graph in C(/,X) X C(/,X).

Note that if dimX < 1 , and F is as in Theorem 2.5, then SF(x) „ 0 for any x € C(/,X)

Consider a set valued map T on X with nonempty values in X. T is said to be a k-contraction if there exists 0 < k <1 such that

dff(T(x), T(y)) 6 kd(x, y) 8x, y e X.

The set-valued contraction principle [12] states that if X is complete, and T: X ! P(X) is a set valued contraction with nonempty closed values, then T has a fixed point, i.e. a point z e X such that z e T(z).

Definition 2.6.

(a) The fractional integral of order a >0 of a Lebesgue integrable function f: (0, i) fi R is defined by

provided the right-hand side is pointwise defined on (0,1) and C is the (Euler's) Gamma function defined by C(a) = J0°° ta_1e~'dt. (b) The Riemann-LioMvi//e fractiona/ derivative of order a >0 of a continuous function f: (0,1) fi R is defined by

where n = [a] + 1, provided the right-hand side is pointwise defined on (0, i).By AC:([0,1],R) we denote the space of continuous real-valued functions whose first derivative exists and it is absolutely continuous on I. On AC1([0,1],R) we consider the norm

||x|| = max{ sup |x(t)|, sup |x'(t)|}.

Definition 2.7. A function x 2 AC1([0,1],R) is called a solution of problem (1.1) and (1.2) if there exists a function v G L1([0,1],R) with v(t) 2 F(t,x(t),x'(t)), a.e. [0,1] such that Dax(t) = v(t), a.e. [0,1] and conditions (1.2) are satisfied.

In what follows I = [0,1], a 2 (2,3], and A = E^nr1 2 (0, 1). Next we need the following technical result proved in [19].

Lemma 2.8 19. For any h 2 L1(I,R) the problem

Dax(t) = h(t) a.e. [0, 1],

(e.g., [17]).

í6[0,1

í6[0,1

x(0) = x'(0) = 0, x(1)^aix(ni) = k

has a unique solution given by

ta-i i-1 1 m r i

kt™ I t™ m i'I

x(t)=I—T + G(t,s)h{s)ds + -—^^ / G(£i,s)h(s)ds, t 2 [0, I], <j 0 i=i J 0

G{t s) ,= [t(I - s)]™-1 -(t - s)™-\ if 0 6 s < t 6 I,

1 ' ' r(ai)\ [t(I - s)]™-1, if 0 6 t < s 6 I.

Note that G(t,s) > 0 "t,s 2 I and G(t, s) 6 f^y, (e.g., Lemma 5 in [ 19]). If we denote Gi(t,s) = G{t,s) + Ph a-DG(t„s) one has \Gi{t,s)| 6 jfe (1 + and

|@Gi (t 6 2(az1l{1 + ZXiOi)

I @t (t; s)l 6 r(a) I 1 + 1-a j.

Let K1\=suptjs2^ Gi(t,s)\ and K2 := supt,s2/\ ^ (t, s)\. Finally, we denote z(t) = f^ and C1:=supt2I\\z(t)\\.

3. The main results

Now we are able to present the existence results for problem (1.1) and (1.2). We consider first the case when F is convex valued.

Hypothesis 3.1.

(i) F : I x R x R ! P(R) has nonempty compact convex values and is Caratheodory.

(ii) There exist u 2 L1(I,R) with u (t) > 0 a.e. I and there exists a nondecreasing function W:[0,1) fi (0,1) such that

sup{\v\, v 2 F(t, x,y)} 6 u(t)W(max{\x\, \y\}) a.e. I, 8x,y 2 R.

Theorem 3.2. Assume that Hypothesis 3.1 is satisfied and there exists r > 0 such that r > C1 + max{K1,K2}\u\1W(r). (3.1)

Then problem (1.1) and (1.2) has at least one solution x such that \\x\\ < r.

Proof. Let X = AC1(I,R) and consider r > 0 as in (3.1). It is obvious that the existence of solutions to problem (1.1) and (1.2) reduces to the existence of the solutions of the integral inclusion

x(t)2z(t)+[ G1 (t, s)F(s, x(s), x'(s))ds, t 2 I. (3.2)

Consider the set-valued map T: Br(0) ! P(AC1(/, R)) defined by

T(x) :=j v 2 AC1 (I, R); v(t)= z(t) +jf G(t, s)/(s)ds, / 2 SF(x)J. (3.3) We show that T satisfies the hypotheses of Corollary 2.2.

First, we show that T(x) c AC1(/,R) is convex for any x 2 AC1(/,R). If v1,v2 2 T(x) then there exist /1/2 2 SF(x) such that for any t 2 I one has

v,(0=z(t)+/ G1(t, s)/-(s)ds, i = 1,2. Jo

Let 0 6 a 6 1. Then for any t 2 I we have

(av1 + (1 - a)v,)(t)=z(t) + [ G1(t,s)[a/(s) + (1 - a)/,(s)]ds.

The values of F are convex, thus SFx) is a convex set and hence

av1 + (1 — a)v2 2 T(x).

Second, we show that T is bounded on bounded sets of AC1(/,R). Let B c AC1(/,R) be a bounded set. Then there exists m >0 such that ||x|| 6 m "x 2 B. If v 2 T(x) there exists /2 SF(x) such that v(t) = jJ G1(t, s)/(s)ds. One may write for any t 2 I

|v(t)| 6 |z(t)|+ f |G1(t,s)| • f(s)|ds Jo

6 |z(t)|+ f |G1(t,s)|u(s)W(max{|x(s)|, |x'(s)|})ds. Jo

On the other hand,

n1 , @G1

I (t, s)H

/»l -G

6 |z'(t)|+/ I"G(t,s)|u(s)W(max{|x(s)|, |x'(s)|})ds.

and therefore

|vk=niax{|v(t)|, |v'(t)|}

fi — g

6 maxmax{|z(t)|, |z'(t)|}+ / max{|Gi(t,s)|, |--1 teJ _/0 í,se/ — t

x(t,s)|}u(s)W(max{|x(s)|, |x'(s)|})ds 6 C1 + max{K1,K,}|u|1W(m) "v 2 T(x), i.e., T(B) is bounded.

We show next that T maps bounded sets into equi-continuous sets. Let B c AC1(I,R) be a bounded set as before and v 2 T(x) for some x 2 B. There exists f 2 SF(x) such that v(t) = z(t) + jJ G1 (t, s)f(s)ds. Then for any t,s 2 I we have

\v(t)- v(s)\ 6 \z(t)-z(s)\ + \ f1 G1(t, s)f(s)ds - f G1 (s, s)f(s)ds\

6 \z(t)-z(s)\+ f \G1(t, s) Jo

- G1(s,s)\u(s)W(max{\x(s)\, \x'(s)\})ds

6 \z(t)-z(s)\+ f \G1(t, s)-G1(s, s)\u(s)W(m)ds. J0

Similarly, we have

/1 —G —G

\ -G- (t, s)--G (s, s)\u(s)W(m)ds.

It follows that \ v(t) - v(s)\ fi 0 as t fi s . Therefore, T(B) is an equi-continuous set in AC1(I,R). We apply now Arzela-Ascoli's theorem we deduce that T is completely continuous on AC1(I,R).

In the next step of the proof we prove that T has a closed graph. Let xn 2 AC1(I,R) be a sequence such that xn fi x and vn 2 T(xn) "n 2 N such that vn fi v . We prove that v 2 T(x ). Since vn 2 T(xn), there exists fn 2 SF(xn) such that vn(t) = z(t)+ /1 G1(t,s)fn(s)ds. Define f L1(I,R) fi AC1(I,R) by (rf))(t) := /J G1(t,s)f(s)ds. One has

niax{\v„(t)-z(t)-(v*(t)-z(t))\,\v'n(t)-z' (t)-((v*)'(t)-z' (t))\ = niax{\v„(t)- v*(t)\,\v'n(t)-(v*)'(t)\} = kv„ - v*\\ ! 0 as n fi 1.

We apply Theorem 2.5 to find that roSF has closed graph and from the definition of r we get vn 2 ro SF(xn). Since xn fi x*, vn fi v* it follows the existence of f 2 SF(x*) such that v* (t)- z(t) = f! G1(t, s)f*(s)ds. Therefore, T is upper semicontinuous and compact on Br (0).

We apply Corollary 2.2 to deduce that either (i) the inclusion x 2 T(x) has a solution in Br(0), or (ii) there exists x 2 X with \\x\\ = r and kx 2 T(x) for some k >1.

Assume that (ii) is true. With the same arguments as in the second step of our proof we get r = \\x\\ 6 C1 + max{K1,K2}\ u\ 1W (r) which contradicts (3.1). Hence only (i) is valid and theorem is proved.

We consider now the case when F is not necessarily convex valued. Our first existence result in this case is based on the Leray-Schauder alternative for single valued maps and on Bressan Colombo selection theorem. □

Hypothesis 3.3.

(i) F : I x R x R ! P(R) has compact values, F is L(I)< B(R)< B(R) measurable and (x,y) fi F(t,x,y) is lower semicontinuous for almost all t 2 I.

(ii) There exist u 2 L1(I,R) with u (t) > 0 a.e. I and there exists a nondecreasing function W :[0,1) fi (0,1) such that

sup{|v|, v 2 F(t, x,y)} 6 u(t)W(max{|x|, |y|}) a.e. I, Vx,y 2 R.

Theorem 3.4. Assume that Hypothesis 3.3 is satisfied and there exists r > 0 sMch that conditio« (3.1) is satisfied. Then problem (1.1) and (1.2) has at least one solMtion on I.

Proof. We note first that if Hypothesis 3.3 is satisfied then F is of lower semicontinuous type (e.g., [14]). Therefore, we apply Theorem 2.4 to deduce that there exists f: AC1(I,R) fi L1(I,R) such that fx) 2 SF(x) "x 2 AC1(I,R).

We consider the corresponding problem

x(t)=z(t)+/ G1(t,s)f(x(s))ds, t 2 I (3.4)

in the space X = AC1(I,R). It is clear that if x 2 AC1(I,R) is a solution of the problem (3.4) then x is a solution to problem (1.1) and (1.2).

Let r >0 that satisfies the condition (3.1) and define the set-valued map T : Br(0)!p(AC!(I, R)) by

(T(x))(t) := z(t)+ / G1(t,s)f(x(s))ds. Jo

Obviously, the integral Eq. (3.4) is equivalent with the operator equation

x(t) = (T(x))(t), t 2 I. (3.5)

It remains to show that T satisfies the hypotheses of Corollary 2.3.

We show that T is continuous on Br(0). From Hypotheses 3.3. (ii) we have

|f(x(t))| 6 u(t)W(max{|x(t)|, |x'(t)|}) a.e. I

for all x 2 AC1(I,R). Let xn, x 2 Br(0) such that xn fi x. Then

f(x„(t))| 6 u(t)W(r) a.e. I.

From Lebesgue's dominated convergence theorem and the continuity off we obtain, for all t 2 I

lim (T(x„))(t)= z(t) + lim / G1(t,s)f(x„(s))ds = z(t)+ / G1(t,s)f(x(s))ds

n!1 o Jo

= (T(x))(t)

f i —G

lim (r(x„))'(t) = z'(t) + lim / —1 (t,s)/(x„(s))ds

n!1 J 0 —t

/»1 —g

= Z(t)+jf —G (t, s)f(x(s))ds =(T(x))'(t)

i.e., T is continuous on Br(0).

Repeating the arguments in the proof of Theorem 3.2 with corresponding modifications it follows that T is compact on Br(0). We apply Corollary 2.3 and we find that either (i) the equation x = T(x) has a solution in Br(0), or (ii) there exists x 2 X with ||x|| = r and x = kT(x) for some k <1.

As in the proof of Theorem 3.2 if the statement (ii) holds true, then we obtain a contradiction to (3.1). Thus only the statement (i) is true and problem (1.1) has a solution x 2 AC1(I,R) with ||x|| < r.

In order to obtain an existence result for problem (1.1) and (1.2) by using the set-valued contraction principle we introduce the following hypothesis on F. □

Hypothesis 3.5.

(i) F : I x R x R ! P(R) has nonempty compact values, is integrably bounded and for every x,y 2 R, F(.,x,y) is measurable.

(ii) There exists l1,l2 2 L1(I,R+) such that for almost all t 2 I,

dff(F(t,x1,yj,F(t,x2,y2)) 6 l1(t)|x1 - x2|+l2(t)|y! -y2| "x1,x2,y1,y2 2 R.

Theorem 3.6. Assume that Hypothesis 3.5. is satisfied and (I l7| 7 + I l2I 7)-max{K7,K2} < J. Then problem (1.1) and (1.2) has a solution.

Proof. We transform the problem (1.1) and (1.2) into a fixed point problem. Consider the set-valued map T: AC1 (I, R) ! p(aC1(I, R)) defined by

T(x) := {v 2 AC1 (I,R); v(t) = z(t)+ / G^t,s)f(s)ds, f 2 Sf(x)}.

Note that since the set-valued map F(.,x(.)) is measurable with the measurable selection theorem (e.g., Theorem III. 6 in [5]) it admits a measurable selection f: I fi R. Moreover, since F is integrably bounded, f 2 L1(I,R). Therefore, SF,x „ 0.

It is clear that the fixed points of T are solutions of problem (1.1) and (1.2). We shall prove that T fulfills the assumptions of Covitz Nadler contraction principle.

First, we note that since SF,x „ 0, T(x) „ 0 for any x 2 AC1(I,R).

Second, we prove that T(x) is closed for any x 2 AC1(I,R). Let {xn} np0 2 T(x) such that xn fi x* in AC1(I,R). Then x* 2 AC1(I,R) and there exists fn 2 SFx such that

xn(t)=z(t) + G1(t, s)fn(s)ds. J0

Since F has compact values and Hypothesis 3.5 is satisfied we may pass to a subsequence (if necessary) to get that fn converges to f 2 L1(I,R) in L1(I,R). In particular, f2 SF,x and for any t 2 I we have

xn(t) ! x*(t) = z(t)+ i G1(t, s)f(s)ds, J0

i.e., x* 2 T(x) and T(x) is closed.

Finally, we show that T is a contraction on AC1(I,R). Let x1,x2 2 AC1(I,R) and v1 2 T(x1). Then there exist f1 2 SF,x1 such that

v1 (t) = z(t)+ i G(t, s)f (s)ds, t 2 I. J0

Consider the set-valued map

H(t) := F(t, x2(t), x'2(t)) \{x 2 R; f (t)-x\

6 l1(t)\x1(t)- x2(t)\ + h(t)\x1(t)- x2(t)\}, t 2 I. From Hypothesis 3.5 one has

dH(F(t, x1(t), x1 (t)), F(t, x2(t), x2(t))) 6 l1(t)\x1(t)- x2(t)\+ l2(t)\x!1 (t)-x,2(t)\,

hence H has nonempty closed values. Moreover, since H is measurable, there exists f2 a measurable selection of H. It follows that f2 2 SF;x2 and for any t 2 I

\f1(t)-f2(t)\ 6 h(t)\x1(t)-x2(t)\ + l2(t)\x\(t)- x'2(t)\.

Define

v2(t)=z(t)+ ( G1 (t,s)f2(s)ds, t 2 I J0

and we have

\v 1 (t) v2(t)\ 6 f \G1(t, s)\-f1(s)-f2(s)\ds J0

6 f G1(t,s)[l1(s)\x1(s)-x2(s)\+ l2(s)\x[(s)- x2(s)\]ds J0

6 K1(\l1\1 + \l2\1)\\x1 - x2\. Similarly, we have

\v[(t)-v2 (t)\ 6 K2(\l1\1 + \l2\1)\x1 - x2\\.

So, |v1 — v2|| 6 (I lj 1 + I l2I 1)max{K1,K2}||x1 — x2||. From an analogous reasoning by interchanging the roles of x1 and x2 it follows

dH(T(x1), T(x2)) 6 (|l111 I 11211) max{K1, K2}kx1 — x2||.

Therefore, T admits a fixed point which is a solution to problem (1.1) and (1.2). □

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