Existence results for multi-term fractional differential inclusions

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1 Ntouyas et al. Advances in Difference Equations 215) 215:14 DOI /s z R E S E A R C H Open Access Existence results for multi-term fractional differential inclusions Sotiris K Ntouyas 12 SinaEtemad 3 Jessada Tariboon 4* * Correspondence: jessadat@kmutnb.ac.th 4 Nonlinear Dynamic Analysis Research Center Department of Mathematics Faculty of Applied Science King Mongkut s University of Technology North Bangkok Bangkok 18 Thail Full list of author information is available at the end of the article Abstract In this paper we study the existence of solutions for a new class of boundary value problems for nonlinear multi-term fractional differential inclusions. Our main result relies on the multi-valued form of Krasnoselskii s fixed point theorem. An illustrative example is also presented. MSC: 34A8; 34A6; 34B1; 34B15 Keywords: fractional differential inclusion; boundary value problem; existence; fixed point theorem 1 Introduction preliminaries In this paper we study the existence of solutions for the following multi-term fractional differential inclusions: c D α ut) F t ut) u t) u t) c D q 1 ut)... c D q k ut) ) G t ut) u t) u t) c D q 1 ut)... c D q k ut) ) 1.1) supplemented with boundary conditions u) = u ) = u1) u 1) u ) = u 1) c D p u1) 1.2) where c D α c D q i denote the Caputo fractional derivatives 2 < α 3 1 < q i 2 i = 12...k t J := [ 1] 1 < p 2 k 1 F G : J R k3 PR) are multifunctions. Many of published papers about fractional differential equations inclusions apply the fixed point theory for proving the existence results. For instance one can find a lot of papers in this field see [1 25] the references therein). Let α >n 1<α < n n =[α]1u C[a b] R). The Caputo derivative of fractional of order α for the function u is defined by c D α 1 t ut)= Ɣn α) t τ)n α 1 u n) τ) dτ see for more details [ ]). Also the Riemann-Liouville fractional order integral of the function u is defined by I α ut)= 1 uτ) dτ t > ) whenever the integral Ɣα) t τ) 1 α exists [ ]. In [28] it has been proved that the general solution of the fractional differential equation c D α ut) =isgivenbyut) =c c 1 t c 2 t 2 c n 1 t n 1 where c...c n 1 are real constants n =[α] 1.AlsoforeachT >u C[ T]) we 215 Ntouyas et al.; licensee Springer. This article is distributed under the terms of the Creative Commons Attribution 4. International License which permits unrestricted use distribution reproduction in any medium provided you give appropriate credit to the original authors) the source provide a link to the Creative Commons license indicate if changes were made.

2 Ntouyas etal. Advances in Difference Equations 215) 215:14 Page 2 of 15 have I αc D α ut)=ut)c c 1 t c 2 t 2 c n 1 t n 1 where c...c n 1 are real constants n =[α]1[28]. Now we review some definitions notations as regards multifunctions [29 3]. For a normed space X ) let P cl X) ={Y PX) :Y is closed P b X) ={Y PX) :Y is bounded P cp X) ={Y PX) :Y is compact P cpcv X) ={Y PX) : Y is compact convex P bclcv X) ={Y PX) :Y is bounded closed convex. A multi-valued map G : X PX) isconvexclosed)valuedifgx) isconvexclosed) for all x X. ThemapG is bounded on bounded sets if GB) = x B Gx) isbounded in X for all B P b X) i.e. sup x B {sup{ y : y Gx) < ). G is called upper semicontinuous u.s.c.) on X if for each x X thesetgx ) is a nonempty closed subset of X ifforeachopensetn of X containing Gx )thereexistsanopenneighborhood N of x such that GN ) N. G is said to be completely continuous if GB) is relatively compact for every B P b X). If the multi-valued map G is completely continuous with nonempty compact values then G is u.s.c. if only if G has a closed graph i.e. u n u y n y y n Gu n )implyy Gu ). G has a fixed point if there is x X such that x Gx). The fixed point set of the multi-valued operator G will be denoted by Fix G. A multi-valued map G : J P cl R) issaidtobemeasurableifforeveryy R the function t dy Gt)) = inf{ y z : z Gt) is measurable. Consider the Pompeiu-Hausdorff metric H d : PX) PX) R { given by { H d A B)=max sup a A da B) sup da b) b B where da b)=inf a A da; b)da B)=inf b B da; b). A multi-valued operator N : X P cl X) is called contraction if there exists γ 1) such that H d Nx) Ny)) γ dx y)for each x y X. We say that F : J R k3 PR) is a Carathéodory multifunction if t Ft u 1...u k3 ) is measurable for all u i R u 1...u k3 ) Ft u 1...u k3 ) is upper semi-continuous for almost all t J [29 31]. Also a Carathéodory multifunction F : J R k3 PR) is called L 1 -Carathéodory if for each ρ >thereexistsφ ρ L 1 J R )suchthat Ft u1...u k3 ) { = sup s : s Ft u1...u k3 ) φ ρ t) for all u 1... u k3 ρ for almost all t J [29 31]. Define the set of selections of F G at u CJ R)by S Fu := { v L 1 J R):vt) F t ut) u t) u t) c D q 1 ut)... c D q k ut) ) S Gu := { v 1 L 1 J R):v 1 t) G t ut) u t) u t) c D q 1 ut)... c D q k ut) ) for almost all t J.IfF is an arbitrary multifunction then it has been proved that S F u) for all u CJ X)ifdim X < [32].

3 Ntouyas etal. Advances in Difference Equations 215) 215:14 Page 3 of 15 The graph of a function F is the set GrF)={x y) X Y : y Fx) [29]. The graph GrF) off : X P cl Y ) is said to be a closed subset of X Y ifforeverysequence {u n n N X {y n n N Y whenn u n u y n y y n Fu n ) then y Fu )[29]. We will use the following lemmas theorem in our main result. Lemma 1.1 [29] Proposition 1.2) If F : X P cl Y ) is u.s.c. then GrF) is a closed subset of X Y. Conversely if F is completely continuous has a closed graph then it is upper semi-continuous. Lemma 1.2 [32]) Let X be a separable Banach space. Let F :[1] X k3 P cpcv X) be an L 1 -Carathéodory function. Then the operator S F : CJ X) P cpcv CJ X) ) x SF )x)=s Fx ) is a closed graph operator. Theorem 1.3 [33] Krasnoselskii s fixed point theorem) Let X be a Banach space Y P bclcv X) A B : Y P cpcv X) two multi-valued operators. If the following conditions are satisfied: i) Ay By Y for all y Y ; ii) A is a contraction; iii) B is u.s.c. compact then there exists y Ysuchthaty Ay By. 2 Main results Now we are ready to prove our main result. Let X = {u : u u u c D q iu CJ R) i = k. ThenX ) endowed with the norm u = sup ut) sup u t) sup u t) sup c D q i ut) i =1...k) is a Banach space [34]. We need the following auxiliary lemma. See also [35 36]. Lemma 2.1 Let y CJ R) u C 2 [ 1] R) is a solution to the fractional boundary value problem { c D α ut)=yt) u) = u ) = u1) u 1) u ) = u 1) c D p u1) 2.1) then ut)= ys) ds t Ɣα) 3 t t 2) vice versa where = ys) ds t Ɣα) 3 Ɣα 2) ys) ds t t 2) Ɣ3 p) 4Ɣ3 p)2. ys) ds Ɣα 1) ys) ds 2.2)

4 Ntouyas etal. Advances in Difference Equations 215) 215:14 Page 4 of 15 Proof It is well known that the solution of equation c D α ut)=yt)canbewrittenas ut)= ys) ds c c 1 t c 2 t 2 2.3) Ɣα) where c c 1 c 2 R.Thenweget u t)= u t)= t s) α 2 Ɣα 1) ys) ds c 1 2c 2 t t s) α 3 Ɣα 2) ys) ds 2c 2 c D p t s) α p 1 ut)= ys) ds c 2t 2 p 2 Ɣ3 p) 1 < p 2). By using the boundary value conditions we obtain c = c 1 = 1 3 c 2 = ys) ds 1 Ɣα) 3 ys) ds Ɣα 2) ys) ds Ɣα 2) ys) ds Ɣα 1) ys) ds ys) ds. Substituting the values of c c 1 c 2 in 2.3)weget2.2). Conversely applying the operator c D α on 2.2) taking into account 2.1) it follows that c D α ut)=yt). From 2.2) it is easily to verify that the boundary conditions u) = u ) = u1) u 1) u ) = u 1) c D p u1) are satisfied. This establishes the equivalence between 2.1)2.2). The proof is completed. Definition 2.2 Afunctionu C 2 [ 1] R) is called a solution for the problem 1.1)-1.2) if it satisfies the boundary value conditions u) = u ) = u1) u 1) u ) = u 1) c D p u1) there exist functions v v 1 L 1 J R) suchthatvt) Ft ut) u t) u t) c D q 1ut)... c D q k ut)) v 1 t) Gt ut) u t) u t) c D q 1ut)... c D q k ut)) for almost all t J ut)= vs) ds t vs) ds Ɣα) 3 Ɣα) t 3 Ɣα 1) vs) ds t t 2) vs) ds Ɣα 2) t t 2) Ɣα) vs) ds v 1 s) ds t v 1 s) ds 3 Ɣα)

5 Ntouyas etal. Advances in Difference Equations 215) 215:14 Page 5 of 15 t 3 t t 2) Ɣα 1) v 1s) ds t t 2) Remark 2.3 For the sake of brevity we set 1 = 2 = 4 3Ɣα 1) 1 3Ɣα) 4 3Ɣα) 1 3Ɣα 1) Ɣα 2) v 1s) ds v 1s) ds. 2.4) 4Ɣα 1) Ɣα 1) 4Ɣα p 1) 2.5) Ɣα p 1) 2.6) 3 = 12 Ɣα 1) 2 Ɣα p 1) 2.7) for each i =1...k i 4 = 1 Ɣα q i 1) 2 Ɣ3 q i )Ɣα 1) 2 Ɣ3 q i )Ɣα p 1). 2.8) Also in the following we use the notation x = sup{ xt) : t J. Theorem 2.4 Suppose that: H 1 ) F : J R k3 P cpc R) is a multifunction G : J R k3 P cpc R) is a Carathéodory multifunction; H 2 ) there exist continuous functions p m : J ) such that t Ft w 1 w 2 w 3 z 1...z k ) is measurable Ft w 1 w 2 w 3 z 1...z k ) mt) Gt w 1 w 2 w 3 z 1...z k ) pt); H 3 ) there exists a continuous function h : J ) such that If H d Ft w1 w 2 w 3 z 1...z k ) F t w 1 w 2 w 3 )) z 1...z k [ 3 ] k ht) w i w i z i z i i=1 i=1 for all t J for each w 1 w 2 w 3 z 1...z k w 1 w 2 w 3 z 1...z k R. L := h i 4) <1 for i =12...k where the j j =1...4)are defined in 2.5)-2.8) then the inclusion problem 1.1)-1.2) has at least one solution. Proof We define the subset Y of X by Y = {u X : u Mwhere M = ) p m i ) 4 i =1...k).

6 Ntouyas etal. Advances in Difference Equations 215) 215:14 Page 6 of 15 It is clear that Y is closed bounded convex subset of Banach space X.Wedefinethe multi-valued operators A B : Y PX)suchthatforsomev S Fu { Au)= u X : ut)= vs) ds t Ɣα) 3 t 3 Ɣα 1) vs) ds t t 2) t t 2) vs) ds for some v 1 S Gu { Bu)= u X : ut)= v 1 s) ds t Ɣα) 3 t 3 Ɣα 1) v 1s) ds t t 2) t t 2) v 1s) ds. vs) ds Ɣα) vs) ds Ɣα 2) v 1 s) ds Ɣα) Ɣα 2) v 1s) ds In this way the fractional differential inclusion 1.1)-1.2) is equivalent to the inclusion problem u Au Bu. We show that the multi-valued operators A B satisfy the conditions of Theorem 1.3 on Y. First we show that the operators A B define the multi-valued operators A B : Y P cpcv X). First we prove that A is compact-valued on Y. Note that the operator A is equivalent to the composition L S F wherelis the continuous linear operator on L 1 J R)into Xdefinedby Lv)t)= vs) ds t vs) ds Ɣα) 3 Ɣα) t 3 Ɣα 1) vs) ds t t 2) vs) ds Ɣα 2) t t 2) vs) ds. Suppose that u Y is arbitrary let {v n be a sequence in S Fu. Then by definition of S Fu wehavev n t) Ft ut) u t) u t) c D q 1ut)... c D q k ut)) for almost all t J. Since Ft ut) u t) u t) c D q 1ut)... c D q k ut)) is compact for all t J thereisaconvergent subsequence of {v n t) we denote it by {v n t) again) that converges in measure to some vt) S Fu for almost all t J. On the other h L is continuous so Lv n )t) Lv)t) pointwise on J. In order to show that the convergence is uniform we have to show that {Lv n ) is an equi-continuous sequence. Let t 1 t 2 J with t 1 < t 2.Thenwehave Lv n )t 2 ) Lv n )t 1 ) 1 Ɣα) 1 [ t2 s) α 1 t 1 s) α 1] v n s) ds 1 Ɣα) 2 t 1 t 2 s) α 1 v n s) ds

7 Ntouyas etal. Advances in Difference Equations 215) 215:14 Page 7 of 15 t 2 t 1 3Ɣα) [t 2 t 1 ) t 2 2 t2 1 )] Ɣα 2) [t 2 t 1 ) t 2 2 t2 1 )] vn s) t 2 t 1 ds 3Ɣα 1) vn s) ds v n s) ds vn s) ds { t α m 2 t1 α Ɣα 1) t 2 t 1 3Ɣα 1) t 2 t 1 3Ɣα) [t 2 t 1 ) t2 2 t2 1 )] Ɣα 1) [t 2 t 1 ) t 2 2 t2 1 )] Ɣα p 1) Continuing this process we have. L v n )t 2 ) ) L v n )t 1 ) ) { t α 1 2 t1 α 1 m t 2 t 1 Ɣα) Ɣα 1) t 2 t 1 Ɣα p 1) L v n )t 2 ) ) L v n )t 1 ) ) 1 Ɣα 2) finally for every i =1...k c D q i Lv n )t 2 ) ) c D q i Lv n )t 1 ) ) Ɣα 2) t 1 t2 α 2 t1 α 2 m Ɣα 1) [ t2 s) α 3 t 1 s) α 3] v n s) ds t 2 s) α 3 v n s) ds { α q t i 2 t α q i 1 m Ɣα q i 1) 2 t2 qi 2 t 2 q i 1 Ɣ3 q i )Ɣα 1) 2 t 2 qi 2 t 2 q i 1 Ɣ3 q i )Ɣα p 1) We see that the right-h sides of the above inequalities tend to zero as t 2 t 1.Thus the sequence {Lv n ) is equi-continuous by using the Arzelá-Ascoli theorem we see that there is a uniformly convergent subsequence. So there is a subsequence of {v n we denote it again by {v n )suchthatlv n ) Lv). Note that Lv) LS Fu ). Hence Au)= LS Fu )iscompactforallu Y.SoAu)iscompact. Now we show that Au) is convex for all u X. Letz 1 z 2 Au). We select f 1 f 2 S Fu such that z i t)= f i s) ds t f i s) ds Ɣα) 3 Ɣα) t 3 Ɣα 1) f is) ds t t 2) Ɣα 2) f is) ds t t 2) f is) ds i =12.

8 Ntouyas etal. Advances in Difference Equations 215) 215:14 Page 8 of 15 for almost all t J.Let λ 1. Then we have [ ] t [ λz1 1 λ)z 2 t)= λf1 s)1 λ)f 2 s) ] ds Ɣα) t [ λf1 s)1 λ)f 2 s) ] ds 3 Ɣα) t [ λf1 s)1 λ)f 2 s) ] ds 3 Ɣα 1) t t 2) t t 2) [ λf1 s)1 λ)f 2 s) ] ds Ɣα 2) [ λf1 s)1 λ)f 2 s) ] ds. Since F has convex values S Fu is convex λf 1 s)1 λ)f 2 s) S Fu.Thus λz 1 1 λ)z 2 Au). Consequently A is convex-valued. Similarly B is compact convex-valued. Here we show that Au) Bu) Y for all u Y.Supposethatu Y z 1 Au) z 2 Bu) are arbitrary elements. Choose v 1 S Fu v 2 S Gu such that z 1 t)= v 1 s) ds t v 1 s) ds Ɣα) 3 Ɣα) t 3 Ɣα 1) v 1s) ds t t 2) Ɣα 2) v 1s) ds t t 2) v 1s) ds z 2 t)= v 2 s) ds t v 2 s) ds Ɣα) 3 Ɣα) t 3 Ɣα 1) v 2s) ds t t 2) Ɣα 2) v 2s) ds t t 2) v 2s) ds for almost all t J.Henceweget z 1 t)z 2 t) 1 Ɣα) t 3Ɣα) v 1 s) v 2 s) ) ds t 3Ɣα 1) t t2 Ɣα 2) v1 s) v2 s) ) ds v1 s) v2 s) ) ds v 1 s) v 2 s) ) ds

9 Ntouyas etal. Advances in Difference Equations 215) 215:14 Page 9 of 15 t t2 v1 s) v2 s) ) ds p m ) { 4 3Ɣα 1) 1 3Ɣα) Hence sup z 1 t)z 2 t) p m ) 1.Alsowehave z 1 t)z 2 t) p m ) { 4 3Ɣα) 1 3Ɣα 1) which implies that sup z 1 t)z 2 t) p m ) 2 z 1 t)z 2 t) ) { 12 p m Ɣα 1) 2 Ɣα p 1) 4Ɣα 1) Ɣα 1). 4Ɣα p 1) Ɣα p 1) from which sup z 1 t)z 2 t) p m ) 3. Finally for all i =1...kwehave c D q i z 1 t) c D q i z 2 t) ) { 1 p m Ɣα q i 1) 2 Ɣ3 q i )Ɣα 1) 2 Ɣ3 q i )Ɣα p 1) so sup c D q iz 1 t) c D q iz 2 t) p m ) i 4 i = 12...k. Henceitfollows that z 1 z 2 p m ) i 4) = M i =12...k. Now we show that the operator B is compact on Y.Todothisitisenoughtoprovethat BY ) is uniformly bounded equi-continuous in X.Letz BY ) be arbitrary. For some u Y choosev 1 S Gu such that zt)= v 1 s) ds t v 1 s) ds Ɣα) 3 Ɣα) t 3 Ɣα 1) v 1s) ds t t 2) Ɣα 2) v 1s) ds t t 2) v 1s) ds t J. 2.9) Hence zt) { 4 p 3Ɣα 1) 1 3Ɣα) 4Ɣα 1) z t) p { 4 3Ɣα) 1 3Ɣα 1) z t) p { 12 Ɣα 1) c D q i zt) { p 2 Ɣα p 1) 1 Ɣα q i 1) Ɣα 1) 4Ɣα p 1) Ɣα p 1) 2 Ɣ3 q i )Ɣα 1) 2 Ɣ3 q i )Ɣα p 1) for i =1...k.Hence z p i 4 ) i =1...k.

10 Ntouyas etal.advances in Difference Equations 215) 215:14 Page 1 of 15 Now we show that B maps Y to equi-continuous subsets of X.Lett 1 t 2 J with t 1 < t 2 u Y z Bu). Choose v 1 S Gu such that zt)isgivenby2.9). Then we have zt2 ) zt 1 ) { t α 2 t1 α p Ɣα 1) t 2 t 1 3Ɣα 1) t 2 t 1 3Ɣα) [t 2 t 1 ) t2 2 t2 1 )] Ɣα 1) [t 2 t 1 ) t 2 2 t2 1 )] Ɣα p 1) z t 2 ) z t 1 ) { t α 1 2 t1 α 1 p 2 t 2 t 1 2 t 2 t 1 Ɣα) Ɣα 1) Ɣα p 1) z t 2 ) z t 1 ) p t α 2 2 t α 2 1 Ɣα 1) c D q i zt 2 ) c D q i zt 1 ) { α q t i 2 t α q i 1 p Ɣα q i 1) 2 t2 qi 2 t 2 q i 1 Ɣ3 q i )Ɣα 1) 2 t 2 qi 2 t 2 q i 1 Ɣ3 q i )Ɣα p 1) for each i =1...k. It is seen that the right-h sides of the above inequalities tend to zero as t 2 t 1. Hence by using the Arzelá-Ascoli theorem B is compact. Next we prove that B has a closed graph. Let u n Y z n Bu n ) for all n such that u n u z n z. We show that z Bu ). Associated with z n Bu n )foreachn N there exists v n S Gun such that z n t)= v n s) ds t v n s) ds Ɣα) 3 Ɣα) t 3 Ɣα 1) v ns) ds t t 2) Ɣα 2) v ns) ds t t 2) v ns) ds for all t J. It suffices to show that there exists v S Gu such that for each t J z t)= v s) ds t v s) ds Ɣα) 3 Ɣα) t 3 Ɣα 1) v s) ds t t 2) Ɣα 2) v s) ds t t 2) v s) ds. Consider the continuous linear operator : L 1 J R) X by v)t)= vs) ds t vs) ds Ɣα) 3 Ɣα) t 3 Ɣα 1) vs) ds t t 2) vs) ds Ɣα 2) t t 2) vs) ds.

11 Ntouyas etal.advances in Difference Equations 215) 215:14 Page 11 of 15 Notice that zn t) z t) t = vn s) v s) ) ds t vn s) v s) ) ds Ɣα) 3 Ɣα) t vn s) v s) ) ds 3 Ɣα 1) t t 2) t t 2) vn s) v s) ) ds Ɣα 2) vn s) v s) ) ds asn. By using Lemma 1.2 S G is a closed graph operator. Since z n t) S Gun ) for all n u n u thereisv S Gu such that z t)= v s) ds t v s) ds Ɣα) 3 Ɣα) t 3 Ɣα 1) v s) ds t t 2) Ɣα 2) v s) ds t t 2) v s) ds. Hence z Bu ). So it follows that B has a closed graph this implies that the operator B is upper semi-continuous. Finally we show that A is a contraction multifunction. Let u w X z 1 Aw) is given. Then we can select v 1 S Fw such that z 1 t)= v 1 s) ds t v 1 s) ds Ɣα) 3 Ɣα) t 3 Ɣα 1) v 1s) ds t t 2) Ɣα 2) v 1s) ds t t 2) v 1s) ds for all t J.Since H d F t ut) u t) u t) c D q 1 ut)... c D q k ut) ) F t wt) w t) w t) c D q 1 wt)... c D q k wt) )) ht)[ ut) wt) u t) w t) u t) w t) k c D q i ut) c D q i wt) ] i=1 for almost all t Jthereexistsy Ft ut) u t) u t) c D q 1ut)... c D q k ut)) such that v 1 t) y mt)[ ut) wt) u t) w t) u t) w t) k c D q i ut) c D q i wt) ] i=1

12 Ntouyas etal.advances in Difference Equations 215) 215:14 Page 12 of 15 for almost all t J. Consider the multifunction U : J PR)by Ut)= { s R : v 1 t) s mt)gt) for almost all t J where gt)=[ ut) wt) u t) w t) u t) w t) k c D q i ut) c D q i wt) ]. i=1 Since v 1 ϕ = mg are measurable U ) F u ) u ) u ) c D q 1u )... c D q k u )) is a measurable multifunction. Thus we can choose v 2 t) F t ut) u t) u t) c D q 1 ut)... c D q k ut) ) such that v1 t) v 2 t) mt) [ ut) wt) u t) w t) u t) w t) k c D q i ut) c D q i wt) ] i=1 z 2 t)= v 2 s) ds t v 2 s) ds Ɣα) 3 Ɣα) t 3 Ɣα 1) v 2s) ds t t 2) Ɣα 2) v 2s) ds t t 2) v 2s) ds for all t J.Nowwehave z 1 t) z 2 t) 1 Ɣα) t 3Ɣα) v 1 s) v 2 s) ds t 3Ɣα 1) t t2 Ɣα 2) t t2 v1 s) v 2 s) ds v 1 s) v 2 s) ds v1 s) v 2 s) ds v1 s) v 2 s) ds { 4 h 3Ɣα 1) 1 3Ɣα) 4Ɣα 1) u w. 4Ɣα p 1)

13 Ntouyas etal.advances in Difference Equations 215) 215:14 Page 13 of 15 Similarly z 1 t) z 2 t) h { 4 3Ɣα) 1 3Ɣα 1) { z 1 t) z 2 t) 12 h c D q i z 1 t) c D q i z 2 t) Ɣα 1) 2 Ɣα p 1) Ɣα 1) u w u w Ɣα p 1) { 1 h Ɣα q i 1) 2 Ɣ3 q i )Ɣα 1) 2 u w. Ɣ3 q i )Ɣα p 1) Hence supz 1 t) z 2 t) h 1 u w z 1 t) z 2 t) h 2 u w sup sup sup z 1 t) z 2 t) h 3 u w c D q i z 1 t) c D q i z 2 t) h i 4 u w for each 1 i k.so z 1 z 2 h i 4) u w i =12...k. This implies that H d Au) Aw)) L u w. ThusA B satisfy all the conditions of Theorem 1.3 so the inclusion u Au) Bu) hasasolutioniny. Therefore the inclusion problem 1.1)-1.2)hasasolutioninY the proofis completed. Finally we give an example to illustrate the validity of our main result. Example 2.5 Consider the following fractional differential inclusion: [ c D 2 5 t ut) 3 ut) 11 ut) 3 ) t 2 sinu t)) 2 sinu t)) 1).1t u t) u t) 1 t cos c D 2 3 ut)) 11 cos c D 2 3 ut)) ) t2 π sin 2 t c D 2 3 ut) 2 ] 1t c D 2 3 ut) 2 1) [ e t ut) 1 e t )1 ut) ) cos πt u t) e t 1 e t )1 u t) ) e t u t) 2 1 u t) 2 )1 e t ) e 2t sin c D 2 3 ut)) 1 sin c D 2 3 ut)) )e t 1 e t ) e 3t c 3 D 2 ut) 3 e 2t e 3t )1 c D 2 3 ut) 3 ) with the following boundary conditions: ] 2.1) u) = u ) = u1) u 1) u ) = u 1) c D 3 2 u1) 2.11) where t [ 1]. In the above inclusion problem we have α = 5/2 p = 3/2 k =2q 1 = q 2 =3/2.Alsowehave =.1597.

14 Ntouyas etal.advances in Difference Equations 215) 215:14 Page 14 of 15 Now we define F :[1] R R R R R PR)by [ t x 3 Ft x y z v w)= 11 x 3 ) t 2 sin y 2 sin y 1).1t z z 1 t cos v 11 cos v ) t2 sin π 2 t w2 1tw 2 1) G :[1] R R R R R PR)by [ e t x cos πt y e t Gt x y z v w)= 1 e t )1 x ) 1 e t )1 y ) e t z 2 ] e 2t sin v 1 sin v )e t 1 e t ) e 3t w 3 e 2t e 3t )1 w 3 ) 1 z 2 )1 e t ) ]. Then there exist continuous functions m p :[ 1] ) given by mt)=5 t e t pt)= 1 1e. t On the other h we can easily check that for every x i y i z i v i w i R i =12) H d Ft x1 y 1 z 1 v 1 w 1 ) Ft x 2 y 2 z 2 v 2 w 2 ) ) ht) x 1 x 2 y 1 y 2 z 1 z 2 v 1 v 2 w 1 w 2 ) where h :[1] ) isdefinedbyht)= t 1.Itcaneasilybefoundthat 1 = = = = = Since h = 1 1 wehave L := h )= = < 1. Consequently all assumptions conditions of Theorem 2.4 are satisfied. Hence Theorem 2.4 implies that the fractional differential inclusion problem 2.1)-2.11)hasasolution. Competing interests The authors declare that they have no competing interests. Authors contributions All authors contributed equally in this article. They read approved the final manuscript. Author details 1 Department of Mathematics University of Ioannina Ioannina Greece. 2 Nonlinear Analysis Applied Mathematics NAAM) Research Group Department of Mathematics Faculty of Science King Abdulaziz University P.O. Box 823 Jeddah Saudi Arabia. 3 Young Researchers Elite Club Tabriz Branch Islamic Azad University Tabriz Iran. 4 Nonlinear Dynamic Analysis Research Center Department of Mathematics Faculty of Applied Science King Mongkut s University of Technology North Bangkok Bangkok 18 Thail. Acknowledgements We would like to thank the reviewers for their valuable comments suggestions on the manuscript. Received: 2 January 215 Accepted: 21 April 215 References 1. Agarwal RP Ahmad B: Existence theory for anti-periodic boundary value problems of fractional differential equations inclusions. J. Appl. Math. Comput ) 2. Agarwal RP Belmekki M Benchohra M: A survey on semilinear differential equations inclusions involving Riemann-Liouville fractional derivative. Adv. Differ. Equ. 29 Article ID ) 3. Agarwal RP Benchohra M Hamani S: A survey on existence results for boundary value problems of nonlinear fractional differential equations inclusions. Acta Appl. Math )

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