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Differential inclusions of arbitrary fractional order with anti-periodic conditions in Banach spaces

JinRong Wang

1

, Ahmed G. Ibrahim

2

and Michal Feˇckan

B3, 4

1Department of Mathematics, Guizhou University, Guiyang, Guizhou 550025, P.R. China

2Department of Mathematics, Faculty of Science, King Faisal University, Al-Ahasa 31982, Saudi Arabia

3Department of Mathematical Analysis and Numerical Mathematics, Faculty of Mathematics, Physics and Informatics, Comenius University in Bratislava, Mlynská dolina, 842 48 Bratislava, Slovakia

4Mathematical Institute, Slovak Academy of Sciences, Štefánikova 49, 814 73 Bratislava, Slovakia

Received 30 December 2015, appeared 2 June 2016 Communicated by Paul Eloe

Abstract. In this paper, we establish various existence results of solutions for frac- tional differential equations and inclusions of arbitrary order q ∈ (m1,m), where mis an arbitrary natural number greater than or equal to two, in infinite dimensional Banach spaces, and involving the Caputo derivative in the generalized sense (via the Liouville–Riemann sense). We study the existence of solutions under both convexity and nonconvexity conditions on the multivalued side. Some examples of fractional differential inclusions on lattices are given to illustrate the obtained abstract results.

Keywords: fractional differential inclusions, anti-periodic solutions, Caputo derivative in the generalized sense, measure of noncompactness, fractional lattice inclusions.

2010 Mathematics Subject Classification: 26A33, 34K20, 34K45.

1 Introduction.

During the past two decades, fractional differential equations and fractional differential in- clusions have gained considerable importance due to their applications in various fields, such as physics, mechanics and engineering. For some of these applications, one can see [18,23]

and the references therein. For some recent developments on initial-value problems for dif- ferential equations and inclusions of fractional order, we refer the reader to the references [2,17,30,34,38–45].

Some applied problems in physics require fractional differential equations and inclusions with boundary conditions. Recently, many authors have been studied differential inclusions with various boundary conditions. Some of these works have been done in finite dimensional spaces and of positive integer order, for example, Ibrahim et al. [25] and Gomaa [19,20].

Several results have studied fractional differential equations and inclusions with various boundary value conditions in finite dimensional spaces. We refer, for example, to Agarwal

BCorresponding author. Email: Michal.Feckan@fmph.uniba.sk

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et al. [1] where conditions are established for the existence of solutions for various classes of initial and boundary value problems for fractional differential equations and inclusions involving the Caputo derivative. Next, Ouahab [34] studied a fractional differential inclusion with Dirichlet boundary conditions under both convexity and nonconvexity conditions on the multi-valued right-hand side and Ntouyas et al. [33] discussed the existence of solutions for fractional differential inclusions with three-point integral boundary conditions involving convex and non-convex multivalued maps.

For some recent works on boundary value problems for fractional differential equations and inclusions in infinite dimensional spaces, we refer to Benchohra et al. [7] where the ex- istence of solutions are established for nonlinear fractional differential inclusions with two point boundary conditions.

Anti-periodic boundary value problems occur in the mathematical modeling of a variety of physical problems and have received a considerable attention. Examples include anti-periodic trigonometric polynomials in the study of interpolation problems, anti-periodic wavelets, anti- periodic conditions in physics, and so forth (for details, see [3]).

Some recent works on anti-periodic boundary value problems of fractional orderqwhere m−1<q< mandm=2, 3, 4, 5 can be found in [1,3–5,11,12,26].

In this paper, we establish various existence results of solutions for fractional differential equations and inclusions of arbitrary order q ∈ (m−1,m), where m is a natural number greater than or equal to two, in infinite dimensional Banach spaces, and involving the Caputo derivative in the generalized sense (via the Riemann–Liouville sense). More precisely, let J = [0,T], T > 0, E be a real separable Banach space with a norm k · k. We study the following fractional differential equations and inclusions with anti-periodic conditions:

(cDqgx(t) = f(t,x(t)) a.e. on J,

x(k)(0) =−x(k)(T), k =0, 1, 2, . . . ,m−1, (1.1) and

(cDqgx(t)∈ F(t,x(t)) for a.e.t ∈ J,

x(k)(0) =−x(k)(T), k=0, 1, 2, . . . ,m−1, (1.2) respectively, where cDqgx(t) is the generalized Caputo derivative which is defined via the Riemann–Liouville fractional derivative of orderqwith the lower limit zero for the functionx at the pointt, f : J×E→EandF: J×E→2E is a multifunction.

We would like to point out that Agarwal et al. [1] considered the problems (1.1) and (1.2) when m = 4 and the dimension of E is finite. Thereafter, Ahmad et al. [3] considered the problem (1.1) when m = 2, Ahmad [4] proved the existence of solutions for (1.1) when m = 3 and the dimension of E is finite, Cernea [12] proved existence theorems of solutions for (1.2) when m = 3 and the dimension of E is finite, and Ibrahim [26] established various existence achievements in infinite dimensional Banach spaces for (1.1) and (1.2) whenm= 3.

Alsaedi et al. [5] considered (1.1) in finite dimensional spaces in the case whenm = 5. As a consequence, the obtained results in [1,3–5,11,26] are particular cases of our derived results.

We must mention that Kaslik and Sivasundaram [28] gave non-existence of periodic solutions of fractional order differential equations in the interval [0,) by using the Mellin transform approach. However, we consider fractional differential equations and inclusions with anti- periodic conditions on[0,T]in this paper, not on the interval[0,∞). In other words, we try to find solutions to fractional differential equations and inclusions with anti-periodic conditions

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on finite time intervals, not to find periodic solutions on the infinite interval. So our problem is much different from [28].

The present paper is organized as follows. In Section 2, we collect some background material and basic results about fractional calculus. Hence, we prove auxiliary lemmas which will be used later. In Section 3, we give existence results for (1.1). In Section 4, we prove various existence results for (1.2). We consider the case when the values of F are convex as well as nonconvex. In Section 5, we apply our abstract results to fractional differential inclusions on lattices continuing on our previous work [45].

The proofs rely on the methods and results for boundary value fractional differential in- clusions, the properties of noncompact measure and fixed point techniques.

2 Preliminaries and notations

Let C(J,E) be the space of E-valued continuous functions on J with the norm kxkC(J,E) = max{kx(t)k, t ∈ J}, ACn(J,E) be the space of E-valued functions f on J, which have con- tinuous derivatives up to the order n−1 on J such that fn1 is absolute continuous on J, L1(J,E) be the space of all E-valued Bochner integrable functions on J with the norm kfkL1(J,E) =Rb

0 kf(t)kdt,Pb(E) ={B⊆E: Bis nonempty and bounded},Pcl(E) ={B⊆E:B is nonempty and closed},Pk(E) ={B⊆ E:Bis nonempty and compact},Pck(E) ={B⊆E:B is nonempty, convex and compact}, Pcl,cv(E) = {B⊆ E : Bis nonempty, closed and convex}, conv(B)(respectively, conv(B)) be the convex hull (respectively, convex and closed hull inE) of a subset B.

Let G : J → 2E be a multifunction. By S1G we denote the set of integrable selections of G, i.e., S1G = {f ∈ L1(J,E) : f(t) ∈ G(t) a.e.}. This set may be empty. For Pcl(E)- valued measurable multifunction, S1G is nonempty and bounded in L1(J,E) if and only if t→sup{kxk: x∈G(t)} ∈L1(J,R+)(such a multifunction is said to be integrably bounded) (see [22, Theorem 3.2]). Note that S1G ⊆ L1(J,E)is closed if the values of Gare closed and it is convex if and only if for almost allt ∈ J,G(t)is convex set inE.

Definition 2.1. Let X and Y be two topological spaces. A multifunction G : X → P(Y) is said to be upper semicontinuous, if G1(V) ={x∈X :G(x)⊆V}is an open subset of Xfor every openV⊆Y.

For more information about multifunctions, see, [6,10,24,27]. Now, let us recall the fol- lowing definitions and facts about the integration and differentiation of fractional order.

Definition 2.2. [29, p. 69] The Riemann–Liouville fractional integral of order q > 0 with the lower limit zero for a function f ∈ Lp(J,E), p∈[1,∞)is defined as follows:

Iqf(t) = (gq∗ f)(t) =

Z t

0

(t−s)q1

Γ(q) f(s)ds, t ∈ J,

where the integration is in the sense of Bochner, Γis the Euler gamma function,gq(t) = Γtq(q1), for t >0, gq(t) =0, for t ≤ 0 and∗denotes the convolution of functions. For q= 0, we set I0f(t) = f(t).

It is known [29] that IqIβf(t) = Iq+βf(t), β, q ≥ 0. Moreover, by applying Young’s inequality, it follows that

kIqfkLp(J,E)= kgq∗fkLp(J,E)≤ kgqkL1(J,R)kfkLp(J,E) =gq+1(T)kfkLp(J,E).

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Then, Iq maps Lp(J,E) to Lp(J,E). Let dqe be the least integer greater than or equal to the numberq. The set of natural numbers is denoted byN.

Definition 2.3. [29, p. 70] Let q > 0, m = dqe and f ∈ L1(J,E) be such that gmq∗ f ∈ Wm,1(J,E). The Riemann–Liouville fractional derivative of orderqfor f is defined by

Dqf(J) = d

m

dtmImqf(t) = d

m

dtm(gmq∗ f)(t), where

Wm,1(J,E) = (

f(t) =

m1 k

=0

cktk

k!+Imϕ(t), t ∈ J : ϕ∈L1(J,E), ck ∈ E )

.

Note in the above definition that ϕ= f(m)andck = f(k)(0), k=0, 1, . . . ,m−1. In the fol- lowing lemma, we mention some elementary properties for the Riemann–Liouville fractional integrals and derivatives.

Lemma 2.4. Let q>0and m=dqe.

(i) If f ∈L1(J,E)then gmq∗(Iqf)∈Wm,1(J,E)and DqIqf(t) = f(t)a.e.

(ii) Ifγ> q and f ∈ L1(J,E), then DqIγf(t) = Iγqf(t)a.e. In particular, ifγ> k, k ∈N, then DkIγf(t) = Iγkf(t)a.e.

(iii) If p> 1q and f ∈ Lp(J,E)then Iqf is continuous on J.

Proof. Proofs of these properties are exactly as in the scalar case [36, Chapter 1].

Definition 2.5. [29, p. 91] Letq > 0 and m = dqe. The Caputo derivative in the generalized sense (via Riemann–Liouville fractional derivative) of orderqfor a given function f is defined by

cDqgf(t) =Dq

"

f(t)−

m1 k

=0

f(k)(0) k! tk

#

provided that the right side is well defined.

Remark 2.6. Let q>0,m=dqeand f : J →R, f(t) =tn,n=0, 1, 2, . . . ,(m−1). Then

cDqgtn= Dq

"

tn

m1 k

=0

f(k)(0) k! tk

#

=Dq[tn−tn] =0.

Remark 2.7. [29, Theorem 2.1] Let Ebe a reflexive Banach space, m∈N andq∈(m−1,m). If f ∈ ACm(J,E), thencDqgf(t)exists a.e. and

cDαgf(t) =Imq

f(m)(t)= 1 Γ(m−q)

Z t

0

(t−s)mq1f(m)(s)ds for a.e. t∈ J.

Moreover, if f ∈C(m)(J,E), then this equation is valid for all t∈ J.

To proceed, we state the following lemma as a simple consequence of Lemma 2.4 and formula [36, (3.13)].

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Lemma 2.8. Let m∈N, q∈ (m−1,m)and f ∈ L1γ(J,E), whereγ∈(0,q−(m−1)). Then (i) Ifσ >γthen the function t→ Iσf(t)is continuous on J.

(ii) For any t ∈ T and k ∈ {0, 1, 2, . . . ,m−1}, the function (Iqf)(k)(t)is continuous satisfying (Iqf)(k)(0) =0. Moreover,

cDqg(Iqf(t)) =Dq(Iqf(t)) for a.e. t∈ J, and hence,cDqg(Iqf(t)) = f(t)a.e. on J.

The following lemma is essential to derive existence results of solutions for (1.1) and (1.2).

Lemma 2.9. Let m∈ Nand q ∈ (m−1,m). If z ∈ Lγ1(J,E),γ ∈ (0,q−(m−1))and x: J →E is given by

x(t) =Iqz(t)−

m1 k

=0

bk+1tk, (2.1)

where

bm = 1

2(m−1)!Iq−(m1)z(T), (2.2) b1= 1

2

"

Iqz(T)−

m1 k

=1

bk+1Tk

#

, (2.3)

and for2≤n≤ m−1, bn= 1

2(n−1)!

"

Iq−(n1)z(T)−

m1 k

=n

k!bk+1

(k−(n−1))!Tk−(n1)

#

. (2.4)

Then, x(m1) is continuous on J,cDqgx(t)exists a.e. for t∈ J and (cDqgx(t) =z(t) for a.e. t∈ J,

x(k)(0) =−x(k)(T), k=0, 1, 2, . . . ,m−1. (2.5) Proof. First we note, since for any k = 0, . . . ,m−1, it holds q−k ≥ q−(m−1) > γ, by Lemma 2.8, the functions Iqkz, Dk(Iqz) are continuous on J. Hence bk+1 in (2.2), (2.3) and (2.4) are well-defined, andx(k)are continuous on Jfor all k=0, . . . ,m−1.

Now, in view of (ii) of Lemma 2.8, Dqgz(t)exists for a.e. t ∈ J and cDqgz(t) = Dqz(t)a.e.

Thus, for a.e.t ∈ J

cDqgx(t) =c Dqg(Iqz(t))−cDqg

m1 k

=0

bk+1tk

!

=z(t)−cDgq

m1 k

=0

bk+1tk

!

=z(t),

by Remark2.6. So (2.1) is a general solution of (2.5). Next, in view of (2.1), conditionx(0) =

−x(T)gives

0= x(0) +x(T) =

"

Iqz(T)−

m1 k

=1

bk+1Tk

#

−2b1,

which implies (2.3). Furthermore, by differentiating both sides of (2.1), we get from (ii) of Lemma2.4for 1≤n≤m−2

x(n)(t) =Iαnz(t)−(n!)bn+1

m1 k=

n+1

k!bk+1

(k−n))!tkn, t ∈ J.

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So conditionsx(n)(0) =−x(n)(T)give

0= x(n)(0) +x(n)(T) =Iqnz(T)−2(n!)bn+1

m1 k=

n+1

k!bk+1

(k−n))!Tkn, which implies (2.4). Similarly,

x(m1)(t) =Iq−(m1)z(t)−(m−1)!bm, t∈ J. (2.6) Then, conditionx(m1)(0) =−x(m1)(T)gives

0=x(m1)(0) +x(m1)(T) =Iq−(m1)z(T)−2(m−1)!bm, which implies (2.2). The proof is finished.

Now, by using Lemma2.9 to establish formulae of solutions for (2.5), when m= 2, 3, 4, 5, we get known results mentioned in Introduction. We list them one by one as follows.

(i) form=2, we get (see [3])

x(t) = Iqz(t)−1

2Iqz(T) + T−2t

4 Iq1z(T); (2.7) (ii) form=3, we get (see [4,12,26])

x(t) =Iqz(t)− 1

2Iqz(T) + T−2t

4 Iq1z(T) + t(T−t)

4 Iq2z(T); (2.8) (iii) form=4, we get (see [1])

x(t) =Iqz(t)− 1

2Iqz(T) + T−2t

4 Iq1z(T) + t(T−t)

4 Iq2z(T) +(6t2T−4t3−T3)

48 Iq3z(T);

(2.9)

(iv) form=5, we get (see [5]) x(t) =Iqz(t)−1

2Iqz(T) + T−2t

4 Iq1z(T) +t(T−t)

4 Iq2z(T) +(6t2T−4t3−T3)

48 Iq3z(T) + (2t3T−t4−tT3)

48 Iq4z(T).

(2.10)

Next, form=6, we get x(t) =Iqz(t)−1

2Iqz(T) + T−2t

4 Iq1z(T) +t(T−t)

4 Iq2z(T) +(6t2T−4t3−T3)

48 Iq3z(T) + (2t3T−t4−tT3)

48 Iq4z(T) +

T5 4(5!)− t

2T3 4(4!)+ t

4T 4(4!)− t

5

2(5!)

Iq5z(T).

(2.11)

From the above examples (2.7)–(2.10) and (2.11), it is worthwhile to observe that we expect that (2.1) has a form

x(t) =Iqz(t)−

m1 j

=0

θj(t)Iqjz(T) (2.12)

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for polynomialsθj(t)of degreej. Certainly by above arguments, such x(t)satisfiescDqx(t) = z(t)a.e. on J. The anti-periodic conditions of (2.5) imply

Iqkz(T) =

m1

j=k

θ(jk)(0) +θ(jk)(T)Iqjz(T), k=0, 1, 2, . . . ,m−1 (2.13) where we useθj(k)(t) =0 on J fork> j. Sincezis arbitrarily, we set

θ(kk)(t) = 1 2,

θ(jk)(0) +θ(jk)(T) =0, k =0, 1, 2, . . . ,j−1.

(2.14) It is easy to see that (2.14) determine uniquely θj(t). Since the solution of (2.5) is unique, we see that (2.12) with (2.14) give the solution (2.1), which coincides with the above computations (2.11). Moreover we see thatθj(t)are really independent ofm. Furthermore, set

ηj(s) = θj(sT)

Tj , s ∈[0, 1], j=0, 1, 2, . . . ,m−1.

Then

η(kk)(s) = 1 2,

ηj(k)(0) +η(jk)(1) =0, k =0, 1, 2, . . . ,j−1.

(2.15) Again, (2.15) determine uniquelyηj(t). But nowηj(t)are independent also ofT, so we have

ηj(s) = 1 2

j+1 k

=1

γk(j+1) (k−1)!s

k1, γ(kj)R, j,k =1, 2, 3, . . . ,m. (2.16) Then

θj(t) = 1 2

j+1 k

=1

γ(kj+1)Tj+1k

(k−1)! tk1, j=0, 1, 2, . . . ,m−1.

Hence

x(t) =Iqz(t)− 1 2

m1

j=0 j+1 k

=1

γ(kj+1)Tj+1k

(k−1)! tk1Iqjz(T)

= Iqz(t)− 1 2

m k=1

m1 j=

k1

γ(kj+1)Tj+1k

(k−1)! Iqjz(T)tk1.

(2.17)

By comparing (2.1) with (2.17), we obtain bk =

m1 j=

k1

γk(j+1)Tj+1k

2(k−1)! Iqjz(T), k=1, 2, 3, . . . ,m.

Consequently,γ(kj) is the coefficients of Tjk2Iq(−(kj11))!z(T) inbk for any j,k=1, 2, . . . ,mwith k≤ j.

Next, by inserting (2.16) into (2.15), after elementary calculations, we derive

γr(r)=1, r=1, 2, . . . ,m, γr(r)k =−

k n=1

γr(−(r)kn)

2(n)! , k=1, 2, . . . ,r−1. (2.18)

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Using (2.18), we can derive step by step, for instance γ(rr)1=−1

2, r≥2, γ(rr)2=0, r≥3, γr(r)3 = 1

24, r≥4 γ(rr)4=0, r≥5, γ(rr)5=− 1

2(5!), r≥6, γr(r)6 = 3

2(6!), r≥7.

(2.19)

Furthermore, let us set

˜ ηj(s) =

Z s

0 ηj1(z)dz−1 2

Z 1

0 ηj1(z)dz, j=1, 2, 3, . . . ,m−1.

We can easily check

˜

ηk(k)(s) = 1 2,

˜

η(jk)(0) +η˜(jk)(1) =0, k=0, 1, 2, . . . ,j−1.

So by uniqueness, ˜ηj =ηj, i.e., we get ηj(s) =

Z s

0 ηj1(z)dz−1 2

Z 1

0 ηj1(z)dz, j=1, 2, 3, . . . ,m−1. (2.20) On the other hand, (2.16) implies

γ(kj) =η(jk11)(0). Hence for anyr,k ∈N,r>k, by (2.20), we obtain

γr(r)k =ηr(r1k1)(0) =ηk(0) =γ(1k+1), which justifies (2.19).

Summarizing, we arrive at the following result.

Corollary 2.10. Let m ∈ Nand m−1 < q< m. If z ∈ L1σ(J,E), σ ∈ (0,q−(m−1)), then the function x given by(2.17)andγ(rr)k, r = 1, 2, 3, . . . ,m, k = 0, 1, 2, . . . ,r−1determined by(2.18), is a solution of (2.5).

Remark 2.11. One can verify that the expression (2.17) coincides with the known relations (2.7)–(2.10) and (2.11) form=2, 3, 4, 5, 6, respectively.

Remark 2.12. Let m ∈ N and m−1 < q < m. If z ∈ Lσ1(J,E), σ ∈ (0,q−(m−1)) then z∈ L1(J,E)and the functionx given by (2.17) satisfies the inequality

kxkC(J,E)3Tq

1

2Γ(q)kzkL1(J,E)+ T

q1

2

m2 j

=1

1 Γ(q−j)

j+1 k

=1

|γ(kj+1)|

(k−1)!kzkL1(J,E)

+ T

qσ

2Γ(q−m+1)(qm1+1σ

σ )1σ

m k=1

|γ(km)| (k−1)!kzk

L1σ(J,E),

(2.21)

where by Hölder’s inequality

kIq−(m1)z(T)k ≤ T

qm+1σ

Γ(q−m+1)(qm1+σ1σ)1σkzk

Lσ1(J,E).

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Our main tools are the Schaefer fixed point theorem for single-valued mappings and the O’Regan–Precup fixed point theorem [35, Theorem 3.1] for multivalued mappings, which is a generalization of the Mönch fixed point theorem.

Lemma 2.13. Let Z : E → E be continuous and map every bounded subset into relatively compact subset. If the set E(Z) ={x∈ E: x=λZ(x), λ∈[0, 1)}is bounded, then Z has a fixed point.

Lemma 2.14. Let D be a closed convex subset of E, and N : D→ Pc(D). Assume the graph of N is closed, N maps compact sets into relatively compact sets and that, for some x0∈U, one has

Z⊆ D, Z=conv({x0} ∪T(Z)), Z=C with C ⊆Z countable=⇒ Z is relatively compact.

(2.22) Then T has a fixed point.

For more about fixed point theorems see [16]. Finally, we give the concept of solutions for (1.1) and (1.2).

Definition 2.15. Let f : J ×E → E be a function. A function x ∈ C(m1)(J,E) is called a solution for (1.1), ifcDgqx(t)exists a.e. and

(cDgqx(t) = f(t,x(t)) a.e. on J = [0,T], x(k)(0) =−x(k)(T), k=0, 1, 2, . . . ,m−1.

Definition 2.16. Let F: J×E→2E be a multifunction. A functionx ∈C(m1)(J,E)is called a solution for (1.2), ifcDgqx(t)exists a.e. and

(cDgqx(t) =z(t) a.e. on J = [0,T], x(k)(0) =−x(k)(T), k=0, 1, 2, . . . ,m−1.

wherez ∈ L1(J,E)withz(t)∈F(t,x(t))for a.e. t ∈ J.

Of course, fandFin the above definitions are specified below with appropriate properties.

3 Existence results for (1.1)

In the following, we give the first existence result for (1.1).

Theorem 3.1. Let m∈N, m≥2and q∈ (m−1,m). Let f : J×E→E be a function such that the following conditions are satisfied.

(H1) f is continuous.

(H2) There exists a function ϕ∈ Lσ1(J,R+),σ∈ (0,q−(m−1))such that for any x∈E kf(t,x)k ≤ϕ(t)(1+kxk) for a.e. t∈ J.

(H3) For a.e. s∈ J, the function f(s,·)maps any bounded subset into relatively compact subset in E.

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Then, the problem(1.1)has a solution provided thatδ<1for δ=

"

3Tq1 2Γ(q) + T

q1

2

m2

j=1

1 Γ(q−j)

j+1 k

=1

|γk(j+1)| (k−1)!

#

kϕkL1(J,R+)

+ T

qσ

2Γ(q−m+1)(qm1+1σ

σ )1σ

m k=1

|γ(km)| (k−1)!kϕk

L1σ(J,R+).

(3.1)

Proof. Let us consider the operator N:C(J,E)→C(J,E)defined by (N(x))(t) = Iqz(t)−1

2

m k=1

m1 j=

k1

γ(kj+1)Tj+1k

(k−1)! Iqjz(T)tk1, t∈ J, (3.2) where z(t) = f(t,x(t)), t ∈ J. Note that, since f is continuous, then for every x ∈ C(J,E), the functionzis continuous. Therefore, the operatorNis well defined. We shall prove thatN satisfies the assumptions of the Schaefer fixed point theorem. We split the proof into several steps.

Step 1. Nis continuous.

Let{xn}nNbe a sequence such thatxn→x inC(J,E). For anyt ∈ J, k(N(xn))(t)−(N(x))(t)k

≤Tqmax

tJ kf(t,(xn(t))− f(t,(x(t))k

"

1

Γ(q+1)+1 2

m k=1

m1 j=

k1

|γ(kj+1)|

(k−1)(q+1−j)

# . Since f is continuous, we getkN(xn)−N(x)k →0, as,n→.

Step 2. Nmaps bounded sets into bounded sets.

Letr>0 andBr={x∈C(J,E):kxkC(J,E)≤r}. According to(H2), for anyx∈ Br kN(x)kC(J,E)≤ (1+r)kϕkL1(J,R+)

"

3Tq1 2Γ(q) + T

q1

2

m2 j

=1

1 Γ(q−j)

j+1 k

=1

|γ(kj+1)| (k−1)!

#

+ (1+r)Tqσ 2Γ(q−m+1)(qm1+1σ

σ )1σ

m k=1

|γ(km)| (k−1)!kϕk

Lσ1(J,R+)= (1+r)δ.

(3.3)

Then,N(Br)is bounded.

Step 3. Nmaps bounded sets into relatively compact subsets.

Let r > 0, x ∈ Br, t1,t2 ∈ J (t1 < t2). According to (H2), kf(t,x(t))k ≤ ϕ(t)(1+r) for a.e.

t∈ J. Then,

kN(x)(t2)−N(x)(t1)k

1+r Γ(q)

Z t1

0

h|(t2−s)q1−(t1−s)|q1iϕ(s)ds+

Z t2

t1

(t2−s)q1ϕ(s)ds

+r+1 2

m k=1

m1 j=

k1

|γ(kj+1)|Tj+1k

(k−1)! Iqjϕ(T)(tk21−tk11).

Sinceq>1, then clearly ast2→t1,k(N(x1)(t2))−(N(x1)(t1))k →0, independently ofxand uniformly fort1,t2. Therefore,N(Br)is equicontinuous.

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Now, lett ∈ J be fixed. We want to show that the subset G(t) = {(N(x))(t): x ∈ Br}is relatively compact inE. According to the definition ofN,(H2), the properties of the Hausdorff measure of noncompactness [32], one obtains

χ(G(t))≤ 1 Γ(q)

Z t

0

(t−s)q1χ({f(s,x(s)):x∈ Br})ds +1

2

m k=1

m1 j=

k1

|γk(j+1)|Tj (k−1)(q−j)

Z T

0

(T−s)qj1χ({f(s,x(s)): x∈ Br})ds.

(3.4)

Observe that in view of (H3), it followsχ({f(s,x(s)) : x ∈ Br}) = 0, for almosts ∈ J, Then, (3.4) impliesχ(G(t)) =0.

Step 4. The subset{x ∈C(J,E):x =λN(x), λ∈[0, 1)}is bounded.

Letλ∈ [0, 1)andx ∈C(J,E)be such that x=λN(x). IfkxkC(J,E)=r, then by (3.3) r= kxkC(J,E) ≤ kN(x)kC(J,E)≤(1+r)δ.

So

kxkC(J,E)δ 1−δ.

As a consequence of steps 1→4 and the Schaefer fixed point theorem, the functionNhas a fixed point. In view of(H2)and Corollary2.10, this fixed point is a solution for the problem (1.1). The proof is completed.

In the following theorem, we give another existence result for (1.1).

Theorem 3.2. Let m ∈ N, m ≥ 2and q ∈ (m−1,m). Let f : J×E → E be a function such that (H1),(H3)and the following condition is satisfied.

(H4) There exists ϕ∈ L1σ(J,R+),σ∈ (0,m−(q−1))such that for any x ∈E kf(t,x)k ≤ϕ(t) f or a.e.t ∈ J.

Then, the problem(1.1)has a solution.

Proof. Let us consider the operator N : C(J,E) → C(J,E)defined as (3.2). Arguing as in the proof of Theorem 3.1, we can show that N is continuous and maps any bounded subset to relatively compact subset. It remains to show that the set{x∈C(J,E): x=λN(x), λ∈ [0, 1)}

is bounded.

Let λ ∈ [0, 1)and x ∈ C(J,E) be such that x = λN(x). Then, by(H4)and arguments of (3.3), we have

kxkC(J,E)≤ kN(x)kC(J,E)δ.

This proves that the subset {x ∈ C(J,E) : x = λN(x),λ ∈ [0, 1)} is bounded. According to Schaefer’s fixed point theorem, the function N has a fixed point. In view of (H4) and Corollary2.10, this fixed point is a solution for the problem (1.1). The proof is completed.

Remark 3.3. The condition(H3)is satisfied, if the dimension ofEis finite.

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4 Existence results for (1.2)

4.1 Convex case

At first, we consider the case when values ofFare convex. In the following, we give the first existence result for (1.2).

Theorem 4.1. Let m∈ N, m ≥2and q ∈ (m−1,m). Let F : J×E→ Pck(E)be a multifunction.

Assume the following conditions.

(H5) For every x∈ E, t→ F(t,x)is measurable and for a.e. t∈ J, x→F(t,x)is upper semicontin- uous.

(H6) There exists a function ϕ∈ Lσ1(J,R+), σ ∈ (0,q−(m−1))and a nondecreasing continuous functionΩ:R+R+such that for any x ∈E

kF(t,x)k ≤ϕ(t)(kxk) for a.e. t∈ J. (H7) There exists a functionβ∈ L1ς(J,R+),ς∈(0,q−(m−1))such that

`:=

"

3Tq1 2Γ(q) + T

q1

2

m2 j

=1

1 Γ(q−j)

j+1 k

=1

|γ(kj+1)| (k−1)!

#

kβkL1(J,R+)

+ T

qς

2Γ(q−m+1)(qm1+1ς

ς )1ς

m k=1

|γ(km)| (k−1)!kβk

L1ς(J,R+)<1

(4.1)

and for every bounded subset D ⊆ E, χ(F(t,D)) ≤ β(t)χ(D)for a.e. t ∈ J, where χ is the Hausdorff measure of noncompactness in E.

(H8) There is a positive number r such that

δΩ(r)≤r (4.2)

forδ in(3.1).

Then the problem(1.2)has a solution.

Proof. In view of(H5)and [27, Theorems 1.3.1 and 1.3.5] for everyx ∈C(J,E), the multifunc- tiont → F(t,x(t))has a measurable selection and by (H6)this selection belongs to S1F,x(·)). So, we can define a multifunction R : C(J,E) → 2C(J,E) as follows: For any x ∈ C(J,E), a functiony∈ R(x)if and only if

y=N(f),

where f ∈ S1F,x(·)) and N : C(J,E) → C(J,E) is defined as (3.2) with z = f. According to (H6)and Corollary2.10, any fixed point forRis a solutions for (1.2). So, our aim is to prove that the multivalued function Rsatisfies the assumptions of Lemma2.14. Denote D= Br. It is clear since the values ofFare convex, the values of Rare convex also.

Step 1. R(D)⊆D.

Let x ∈ D and y ∈ R(x). Then, by using the same arguments in Step 2 of the proof of Theorem3.1, we can show that, for anyt∈ J

ky(t)kC(J,E)δΩ(r). (4.3)

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This inequality with (4.2) imply||y|| ≤r.

Step 2. R(D)is a equicontinuous set inC(J,E).

Let y∈ R(D)andt1,t2 ∈ J, (t1< t2). Then there is x ∈ Dwith y∈ R(x). By using the same arguments in Step 3 of the proof of Theorem3.1with (H6), and recalling the definition ofR, there is f ∈ S1F,x(·))such that

ky(t2)−y(t1)k=kN(f)(t2)− N(f)(t1)k

(r) Γ(q)

Z t

1

0

h|(t2−s)q1−(t1−s)|q1iϕ(s)ds+

Z t2

t1

(t2−s)q1ϕ(s)ds

+ r+1 2

m k=1

m1 j=

k1

|γk(j+1)|Tj+1k

(k−1)! Iqjϕ(T)(t2k1−tk11), t∈ J.

Then, the right hand side does not depend onx and uniformly tends to zero whent2→t1. Step 3. The implication (2.22) holds withx0=0.

Let Z ⊆ D, Z = conv({x0} ∪R(Z)), Z = C with C ⊆ Z countable. We claim that Z is relatively compact. Since Cis countable and sinceC⊆Z=conv({x0} ∪R(Z)), we can find a countable set H={yn: n≥1} ⊆ R(Z)withC⊆conv({x0} ∪H). Then, for anyn≥ 1, there exists xn ∈ Z withyn ∈ R(xn). This means that there is fn ∈ S1F,x

n(·)) such that yn = N(fn). From Z ⊆ C ⊆ conv ({x0} ∪H)we find that χ(Z(t)) ≤ χ(C(t)) ≤ χ(H(t)), t ∈ J. Observe that, by (H6), for every natural numbern, kfn(s)k ≤ ϕ(s)(r)a.e. Then, by using (H6)and the properties of the measure of noncompactness, one has fort ∈ J (see [9,21,32])

χ(Z(t))≤ 1 Γ(q)

Z t

0

(t−s)q1χ({fn(s):n≥1})ds +1

2

m k=1

m1 j=

k1

|γ(kj+1)|Tj (k−1)(q−j)

Z T

0

(T−s)qj1χ({fn(s):n≥1})ds.

(4.4)

Observe that, since Z ⊆ C ⊆conv ({x0} ∪H), then from Step 2, Z is equicontinuous. More- over, according to condition(H7)for a.e. t∈ J,

χ({fn(t):n≥1})≤ β(t)χ({xn(t):n≥1})≤ β(t)χ(Z(t))≤β(t)χC(J,E)(Z) So, we find from (4.4) that

χC(J,E)(Z) =max

tJ χ(Z(t))≤`χC(J,E)(Z). This inequality with (4.1) imply that Zis relatively compact.

Step 4. R maps compact sets into relatively compact sets.

Let G be a compact subset of D. From Step 2, R(G) is equicontinuous. Let {yn}nN, be a sequence in R(G). Then, there is a sequence {xn}nNin Gsuch thatyn ∈ R(xn). This means that

yn=N(fn) (4.5)

for fn∈S1F,x

n(·)). Arguing as above, one obtains from (4.5), fort ∈ J χ({yn(t):n≥1})≤ 1

Γ(q)

Z t

0

(t−s)q1χ({fn(s):n ≥1}ds) +1

2

m k=1

m1 j=

k1

|γ(kj+1)|Tj (k−1)(q−j)

Z T

0

(T−s)qj1χ({fn(s):n≥1})ds.

(4.6)

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