Nonlocal fractional differential inclusions with impulses ... · In the study of the topological structure of the solution sets of differential equations and inclusions, an important

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Surveys in Mathematics and its Applications

ISSN 1842-6298 (electronic), 1843-7265 (print)

Volume 14 (2019), 307 – 325

NONLOCAL FRACTIONAL DIFFERENTIALINCLUSIONS WITH IMPULSES AT VARIABLE

TIMES

Abdelghani Ouahab and Sarah Seghiri

Abstract. In this paper, we study the existence of mild solutions for a fractional semi-

linear differential inclusions posed in a Banach space with nonlocal conditions and impulses at

variable times. The main existence result is obtained by using fractional calculus, measure of

noncompactness, and multivalued fixed point theory. We study also the topological properties of

the solution set.

1 Introduction

Differential equations and inclusions of fractional order appear in many physicalphenomena of engineering science, such as problems in electro-chemistry, electro-magnetic, . . . (see, e.g., [24], [31], [35]). Many evolution processes in physics, chemicaltechnology, population dynamics, and natural sciences may change state abruptlyor be subject to short-term perturbations. These perturbations may be seen asimpulses. Differential equations with impulses were first considered by Milmanand Myshkis [34]. Since then, several research works have been published. Themonograph by Halanay and Wexler [22] presents the first impulsive problems.Particular attention has been given to differential equations and inclusions withimpulses at variable moments (see the papers of Bajo and Liz [8], Belarbi andBenchohra [10], Benchohra et al. [12], Agarwal et al [3] and Benchohra and Slimani[13] have considered impulsive fractional differential equations at variable moments.The results was extended to the multivalued case by Ait dads et al. [4]. Morerecently Cardinaly and Rubbioni [17] studied a nonlocal Cauchy problem in thepresent of impulses governed by a nonautonomous semi-linear differential inclusion.The study of semi-linear nonlocal initial value problem was initiated by Byszewski[15], and then followed by many works (see, e.g., [16], [21], [32], [18]).In this paper, we are concerned with the following fractional semi-linear differential

2010 Mathematics Subject Classification: 47H10; 26A33; 34A60; 34B37; 14F45.Keywords: Fixed point theorems; Fractional derivatives; Differential inclusions; Boundary value

problems with impulses; Topological properties.

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308 A. Ouahab and S. Seghiri

inclusion:

cDαy(t) ∈ Ay(t) + F (t, y(t)), t ∈ J, t = τk(y(t)), k = 1,m (1.1)

y(t+) = Ik(y(t)), t = τk(y(t)), k = 1,m (1.2)

y(0) = g(y), (1.3)

where J = (0, T ) and 1,m = 1, 2, . . . ,m. 0 < α < 1, cDα is the Caputo fractionalderivative, A : D(A) ⊂ X −→ X is the infinitesimal generator of a stronglycontinuous semigroup (C0−semigroup) T (t)t>0 on X with D(A) representing thedomain of the linear operator A. F : J×X −→ X is a Caratheodory multi-function,X is an ordered reflexive Banach space with norm ∥ · ∥, and the nonlocal term g isa given function.

Finally τk : X −→ R and Ik : X −→ X for k = 1,m. Since T (t)t>0 is stronglycontinuous, there exists a constant M such that M = sup

t∈J∥T (t)∥ < ∞. 1,m stands

for the set 1, 2, . . . , and y(t+) = lims→t+

y(s).

Differential inclusions of the form (1.1) were first considered by Aizicovici andGao [6] when g and T (t) are compact. In [7] and [33], the authors discussed (1.1)when A generates a compact semigroup. Finally, we mention Lian et al. [32] whostudied the existence of solutions to problem (1.1)-(1.3) without impulses.

In the study of the topological structure of the solution sets of differentialequations and inclusions, an important aspect is the Rδ− property, which includesacyclicity (in particular, compactness and connectedness). An Rδ-set may not bea singleton but, from the point of view of algebraic topology, it is equivalent to apoint, in the sense that it has the same cohomology groups as one point space. Thetopological structure of solution sets of differential inclusions on compact intervalshas been recently investigated by many authors, see Aronszajn [2], Deimling [18],Hu and Papageorgiou [25], and Peng and Zhou [40].

The aim of this paper is to extend the results of Lian et al.[32] when impulsesat variable times are involved. In section 2 we start with some backgrounds onmultivalued analysis, fractional derivatives, and measure of noncompactness. Insection 3, we define a generalized Cauchy operator and give some related properties.Section 4 is devoted to the existence of solutions for problem (1.1)-(1.3). Thetopological structure of the solution set is investigated in section 5, where someelements from algebraic topology are used.

2 Preliminaries

In this section we introduce some background material used throughout this paper.For more definitions and details about the multivalued mappings, we refer, e.g., to

******************************************************************************Surveys in Mathematics and its Applications 14 (2019), 307 – 325




Nonlocal fractional differential inclusions with impulses at variable times 309

[5], and [26]. In order to define the solution of problem (1.1)-(1.3) we shall considerthe space of functions

Σ = y : J −→ X : there exist 0 = t0 < t1 < . . . < tm < tm+1 = Tsuch that tk = τk(y(tk)), y(t

+k ) exists k = 1,m and yk ∈ C((tk−1, tk], X),

k = 1,m+ 1,

where X is an ordered reflexive Banach space, yk is the restriction of y over (tk−1, tk],for k = 1,m+ 1, and y(t+k ) = lim

t→t+k

y(t).

L1(J,X) will denote the Banach space of measurable functions from J into Xwhich are Bochner integrable and L(X) denote the space of bounded linear operatorfrom X into X. Consider the following subsets of X:

Pcl(X) = Y ∈ P (X) : Y is closed Pcp(X) = Y ∈ P (X) : Y is compact Pcv(X) = Y ∈ P (X) : Y is convex

Pcp,cv(X) = Pcp(X) ∩ Pcv(X).

Definition 1. Let X and Y be two topological spaces and F : X −→ P(Y ) amultivalued function.(1) F is said to be compact (convex) valued if F (x) is compact (convex) in Y for allx ∈ X.(2) F is said to be upper semi-continuous (u.s.c.) on X if F−1(V ) = x ∈ X/F (x) ⊂V is an open subset of X for every open subset V of Y .(3) F is said to be closed if its graph GF = (x, y) ∈ X × Y : y ∈ F (x) is a closedsubset of the topological space X×Y , that is xn → x, yn → y and yn ∈ F (xn) implyy ∈ F (x).(4)If Y = X, a point x of X is said to be fixed point of F if x ∈ F (x).(5) A function f : X −→ Y is said to be selection of F if f(x) ∈ F (x) for everyx ∈ X.

Definition 2. A sequence fn∞n=1 ⊂ L1(J,X) is said to be semi-compact if:(i) it is integrably bounded, that is, there exists ω ∈ L1(J,X) such that ∥fn(t)∥ ≤ω(t), for a.e. t ∈ J and every n ≥ 1,(ii) the set fn(t)∞n=1 is relatively compact in X for a.e. t ∈ J.

Lemma 3. [28] Every semi-compact sequence in L1(J,X) is weakly compact inL1(J,X).

Definition 4. A multivalued map F : J ×X −→ P(X) is said to be Caratheodoryif:(i) t ↦−→ F (t, y) is measurable for each y ∈ X,(ii) y ↦−→ F (t, y) is u.s.c. for almost all t ∈ J .






It is further an L1-Caratheodory if it is locally integrably bounded, i.e. for eachpositive r, there exists some hr ∈ L1(J,R+) such that

∥F (t, z)∥ ≤ hr(t) for a.e. t ∈ J and all ∥z∥ ≤ r.

For each y ∈ Σ, define the set of selections of F by

SF (y) = f ∈ L1(J,X) : f(t) ∈ F (t, y(t)), for a.e. t ∈ J.

When F is an L1-Caratheodory multi-valued mapping, we know from a resultdue to Lasota and Opial [30] that for each y ∈ C((tk−1, tk), the set SF (y) containsfunctions fk ∈ L1((tk−1, tk), k = 1,m.

Lemma 5. [30] Let F : J×X −→ Pcp,cv(X) be a Caratheodory multivalued map andlet G be a linear continuous mapping from L1(J,X) to C(J,X), then the operator

G SF : C(J,X) −→ Pcp,cv(X),

where (G SF )(y) = G(SF (y)), is a closed graph operator in C(J,X)× C(J,X).

Next some properties related to measure of non-compactness are recalled [28].

Definition 6. Let X be a Banach space and (A,≽) a partially ordered set. Afunction γ : P(X) −→ A is called a measure of non-compactness (for short M.N.C)in X if:

γ(coΩ) = γ(Ω), for every Ω ∈ P (X),

where coΩ is the convex hull of Ω. A measure of non-compactness γ is called:(a) monotone if for Ω0,Ω1 ∈ P(X), Ω0 ⊂ Ω1 =⇒ γ(Ω0) 6 γ(Ω1),(b) nonsingular if γ(a

⋃Ω) = γ(Ω), for every a ∈ X and Ω ∈ P(X),

(c) real if A = [0,∞] with natural ordering and Ω ∈ P(X),(d) regular if γ(Ω) = 0 is equivalent to the relative compactness of Ω.

We recall that for a bounded subset Ω of X, the Hausdorff M.N.C β is definedby

β(Ω) = ε > 0 : Ω has a finite ε− net in X.

We note that the Hausdorff MNC satisfies the above properties. Moreover, we have

Lemma 7. [9] If w ⊂ C(J,X) is bounded and equicontinuous, then β(w(t)) iscontinuous on J and

β(w) = supt∈J

β(w(t)).

Lemma 8. [23] If un∞n=1 ⊂ L1(J,X) is integrably bounded, then β(Un(t)∞n=1)is measurable and

β

(∫ t

0un(s)

∞

n=1

)6 2

∫ t

0β(un(s)∞n=1)ds.






Lemma 9. [36] If B ⊂ X is bounded, then for each ε > 0, there is a sequenceun∞n=1 in B such that

β(B) 6 2β(un∞n=1) + ε.

Definition 10. Let W be a closed subset of a Banach space X and γ a measureof non-compactness on X. A multi-mapping F : W −→ Pcp(X) is said to be γ-condensing if for every Ω ⊂ W , the relation γ(F (Ω)) > γ(Ω), implies the relativecompactness of Ω.

We will make use of the following fixed point theorem.

Theorem 11. [38] If M is a closed bounded and convex subset of a Banach spaceX and F : M −→ Pcp(M) is a closed γ−condensing multi-mapping, where γ is amonotone MNC defined on subsets of M . Then the fixed point set FixF = x ∈M : x ∈ F (x) is nonempty and compact.

Definition 12. Let α > 0 and f ∈ L1(J,X), then the fractional order integral of fof order α is defined by

Iαt f(t) =1

Γ(α)

∫ t

0(t− s)α−1f(s)ds, t > 0,

where Γ(.) is the Euler gamma function.

The basic definitions of fractional derivative and fractional integral are presentedbelow. For more details on the fractional calculus, we refer the reader to [29] and[37].

Definition 13. The Caputo derivative of order α > 0 of a function f : J −→ X isdefined as

cDαt f(t) =

1

Γ(n− α)

∫ t

0(t− s)n−α−1f (n)(s)ds, t > 0,

where n = [α] + 1. If 0 < α < 1, then

cDαt f(t) =

1

Γ(1− α)

∫ t

0(t− s)−αf ′(s)ds.

Before considering problem (1.1)-(1.3), let us start with the following problemwheach is already discussed in [32]

cDαy(t) ∈ Ay(t) + F (t, y(t)), t ∈ J = [0, T ], (2.1)

y(0) = g(y). (2.2)






Definition 14. A function y ∈ C(J,X) is said to be a mild solution of problem(2.1)-(2.2) if y(0) = g(y) and there exists f ∈ L1(J,X) such that f ∈ SF (y) and

y(t) = Sα(t)g(y) +

∫ t

0(t− s)α−1Pα(t− s)f(s)ds, for t ∈ J,

where

Sα(t) =

∫ ∞

0hα(θ)T (t

αθ)dθ, (2.3)

Pα(t) = α

∫ ∞

0θhα(θ)T (t

αθ)dθ. (2.4)

Here hα is the the probability density function on (0,∞) given by

hα(θ) =1

αθ−1− 1

αωα(θ− 1

α ), (2.5)

where

ωα(θ) =1

π

∞∑n=1

(−1)n−1θ−nα−1Γ(nα+ 1)

n!sin(πnα), θ ∈ (0,∞). (2.6)

Note that hα(θ) > 0 for θ ∈ (0,∞) and∫ ∞

0hα(θ)dθ = 1, (2.7)

∫ ∞

0θδhα(θ)dθ =

Γ(1 + δ)

Γ(1 + αδ), δ ∈ [0, 1]. (2.8)

Lemma 15. [27] The linear operators Sα(t) and Pα(t) have the following properties:

(1) For any fixed t > 0, Sα(t) and Pα(t) are bounded operators. More preciselyfor any x ∈ X, we have

∥Sα(t)x∥ 6M∥x∥, (2.9)

∥Pα(t)x∥ 6Mα

Γ(1 + α)∥x∥, (2.10)

where M = supt∈J

∥T (t)∥.

(2) Operators Sα(t) and Pα(t) are equicontinuous for t ∈ J if T (t)t>0 is equicontinuous.

The following result is easily checked.

Lemma 16. For θ ∈ (0, 1) and 0 < a 6 b, we have

|aθ − bθ| 6 (b− a)θ.






3 Main existence result

Definition 17. A function y of Σ is said to be a solution of problem (1.1)-(1.3) ifthere exists a function h ∈ L1(J,X) such that

h(t) ∈ Ay(t) + F (t, y(t)), for a.e. t ∈ J

and satisfies the equation

cDαtk(y(t)) = h(t), for a.e. t ∈ (tk, tk+1], t = τk(y(t)), k = 1,m

and the conditions y(t+) = Ik(y(t)), t = τk(y(t)), k = 1,m and y(0) = g(y) aresatisfied.

To prove the existence of mild solution for the problem (1.1)-(1.3), we list someassumptions:

(H1) The C0-semigroup T (t)t>0 generated by the linear infinitesimal operator Ais equicontinuous.

(H2) The operator g : Σ −→ X is continuous and compact.

(H3) The multivalued mapping F : J ×X −→ P (X) is Caratheodory, has compactand convex values, and satisfies:

(1) there exist a nondecreasing continuous function ψ : [0,∞) −→ [0,∞) andq ∈ L1(J,R+) such that

∥F (t, y)∥ 6 q(t)ψ(∥y∥), a.e. t ∈ J and all y ∈ X,

(2) there exists a function l ∈ L1(J,X) such that for every bounded D ⊂ X :β(F (t,D)) 6 l(t)β(D), t ∈ J.

(H4) The functions τk ∈ C(X,R) (k = 1,m) satisfy

0 < τ1(z) < τ2(z) < . . . < τm(z) < T, for all z ∈ X.

(H5) The functions Ik : X −→ X, k = 1,m are continuous nondecreasing and verify:

τk(Ik(z)) < τk(z) < τk+1(Ik(z)), for all z ∈ X.

(H6) For all y ∈ C(J,X) and k ∈ 1,m, the set Ek = Ek(y) = t ∈ [0, T ] : τk(y(t)) =t is finite, for all k = 1,m.






Theorem 18. [32] Assume that hypotheses (H1)-(H3) are satisfied and suppose that

limk→∞

supk

M

k

(µ(k) +

Ψ(k)

Γ(α+ 1)q0Tα

)) < 1, (3.1)

where µ(k) = sup∥g(y)∥ /∥y∥ 6 k, q0 = supq(t) : t ∈ J. Then the fractionaldifferential inclusion (2.1)-(2.2) has at least one mild solution on J and a compactsolution set.

Theorem 19. Assume that hypotheses of theorem 3.1 are satisfies. if the conditions(H4)-(H6) hold, then the problem (1.1)-(1.3) has a nonempty compact mild solutionset.

Proof. The proof will be given in several steps.

Step 1 . Using Theorem 3.1, the problem (2.1)-(2.2) has at least one mild solution.

Step 2 . Let y1 be a solution of problem (2.1)-(2.2). For k = 1,m, define the function

rk,1(t) = τk(y1(t))− t, for t > 0.

The condition (H4) implies that rk,1(0) = 0 for all k = 1,m. If rk,1(t) = 0 onJ for all k = 1,m, then y1 is a solution of problem (1.1)-(1.3).Consider the case when r1,1(t) = 0 for some t ∈ (0, T ]. Since r1,1(0) = 0 andr1,1 is continuous, then the set E1 = ti1, i ∈ I is nonempty and from (H6),E1 is finite. We distinguish between two cases:

Case 1. If E1 = t1 then r1,1(t1)) = 0 and r1,1(t) = 0, ∀ t ∈ (0, t1). By (H3),rk,1(t) = 0,∀ t ∈ (0, t1] and k = 1,m. Hence y1 is a solution of theproblem

cDαy(t) ∈ Ay(t) + F (t, y(t)), t ∈ [0, t1]y(0) = g(y).

Case 2. If E1 = ti1, i ∈ I is finite, take t1 = maxE1 and consider the problem

cDαt1y(t) ∈ Ay(t) + F (t, y(t)), for a.e. t ∈ [0, t1], t = ti1, i ∈ I,

with the impulsive conditions y(ti+1 ) = I1(y(ti1)), i ∈ I and the initial

condition y(0) = g(y). The solution for the above problem reads

y1(t) =

⎧⎪⎨⎪⎩y1(t), if t ∈ [0, t11]

Sα(t− ti1)I1(y(ti1)) +

∫ tti1(t− s)α−1Pα(t− s)f(s)ds,

if t ∈ (ti1, ti+11 ], i ∈ I.






We haver1,1(t1) = 0 and r1,1(t) = 0 for t ∈ (0, t1).

By (H4),rk,1(t) = 0 for all t ∈ [0, t1) and k = 1,m

and this y1 is a solution on [0, t1].

Step 3. Consider the problem

cDαt1y(t) ∈ Ay(t) + F (t, y(t)), for a.e. t ∈ [t1, T ], (3.2)

y(t+1 ) = I1(y1(t1)) (3.3)

and the operator R1 : C([t1, T ], X) −→ P (C([t1, T ], X)) defined by

R1(y) = h ∈ C([t1, T ], X) :

h(t) = Sα(t− t1)I1(y1(t1)) +∫ tt1(t− s)α−1Pα(t− s)f(s)ds,

f ∈ SF (y).Operator R1 is well defined. Indeed for t = t1, h(t1) = I1(y(t1)). As in Step 1,we can show that R1 satisfies the assumptions of Theorem 11 and deduce thatproblem (3.2)-(3.3) has a nonempty compact solution set. Denote a solutionof (3.2)-(3.3) by y2. and consider the map rk,2(t) = τk(y2(t))− t for t > t1. Ifrk,2(t) = 0 for t ∈ (t1, T ] and k = 1, k, then

y(t) =

y1(t), for t ∈ [0, t1]y2(t), for t ∈ (t1, T ].

is a solution of problem (1.1)-(1.3). Moreover, when r2,2(t) = 0 for somet ∈ (t1, T ], by (H5) we have

r2,2(t+1 ) = τ2(y2(t

+1 ))− t1

= τ2(I1(y1(t1)))− t1

> τ1(y1(t1))− t1

= r1,1(t1)

= 0.

Since r2,2 is continuous and r2,2(t1) > 0, the set E2 is nonempty and from(H6) E2 is finite. Let E2 = ti2, i ∈ I ′. Then we consider two cases:

Case 1. If E2 = t2 then r2,2(t2) = 0 and r2,2(t) = 0 for t ∈ (t1, t2). We haverk,2(t) == 0 for all t ∈ (t1, t2) and k = 2,m. For k = 1, we have

r1,2(t1) = τ1(y2(t1))− t1

= τ1(I1(y1(t1)))− t1

6 τ1(y1(t1))− t1

= r1,1(t1)

= 0,






i.e., r1,2(t1) < 0. Furthermore, from (H4) we have

r1,2(t2) = τ1(y2(t2))− t2

< τ2(y2(t2))− t2

= r2,2(t2) = 0,

i.e., r1,2(t2) < 0 and we know that ∀ t > t1, τ1(y(t)) = t. Then fort ∈ (t1, t2) τ1(y2(t)) = t i.e., r1,2(t) = 0. Moreover

r1,2(t) < 0, for t ∈ (t1, t2).

We conclude that rk,2(t) = 0, for t ∈ (t1, t2) and k = 1,m.

Case 2 If E2 = ti2, i ∈ I ′ is finite, let t2 = maxE2. Then the solution of problem(3.2)-(3.3) over (t1, t2] is

y2(t) =

⎧⎪⎨⎪⎩y2(t), if t ∈ [t1, t

12]

Sα(t− ti2)I2(y(ti2)) +

∫ tti2(t− s)α−1Pα(t− s)f(s)ds,

if t ∈ (ti2, ti+12 ], i ∈ I ′.

Step 4. We continue this process taking into account that ym+1 = y/[tm,T ] is a solutionto the problem

cDαtmy(t) ∈ Ay(t) + F (t, y(t)), for a.e. t ∈ (tm, T ], (3.4)

y(t+m) = Im(ym−1(t−m)). (3.5)

Finally the solution y of problem (1.1)-(1.3) is then defined by

y(t) =

⎧⎪⎪⎨⎪⎪⎩y1(t), if t ∈ [0, t1]y2(t), if t ∈ (t1, t2].. . .ym+1(t), if t ∈ (tm, T ].

In addition the solution set of problem (1.1)-(1.3) is compact.

Example 20. We consider the following fractional partial differential inclusion:⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩∂αt y(t, x) ∈ ∂2xy(t, x) +G(t, y(t, x)), if t ∈ [0, 1]and t = τk(y(t, x))for k = 1,m,

y(t+, x) = Ik(y(t, x)), if t = τk(y(t, x)) for k = 1,m,

y(t, 0) = y(t, π) = 0,

y(0, x) =∫ 10 h(s) sin(1+ | y(s, x) |))ds.

(3.6)






Where X = L2([0, π];R), ∂α is the Caputo fractional partial derivative of orderα with 0 < α < 1, h ∈ L1([0, 1];R), and G : [0, 1]×X −→ P(X).

We define the operator A by the Laplace operator, i.e. A = ∂2

∂x2 on the domain

D(A) = w ∈ X, w,w′are absolutly continuous and w

′′ ∈ X, w(0) = w(π) = 0.

Clearly, A generates a strongly continuous semigroup T (t), t ∈ [0, 1].Then the system above can be reformed as⎧⎪⎨⎪⎩

cDαy(t) ∈ Ay(t) + F (t, y(t)), t ∈ J = [0, 1], t = τk(y(t)), k = 1,m,

y(t+) = Ik(y(t)), t = τk(y(t)), k = 1,m,

y(0) = g(y),

(3.7)

where y(t)(x) = y(t, x), t ∈ [0, 1], x ∈ [0, π], τk(y(t, x)) = τk(y(t))(x), Ik(y(t, x)) =Ik(y(t))(x) and F (t, y(t))(x) = G(t, y(t, x)) Now we assume that F (t, y(t)) = f(t, y(t))such that f : [0, 1] ×X −→ X is a defined continuous function. Assume that thereare q ∈ L1([0, 1];R+) and ψ : [0,∞) −→ [0,∞) continuous and nondecreasing suchthat

∥ f(t, y) ∥6 q(t)ψ(∥ y ∥),and assume that there exist l ∈ L1([0, 1];X) such that, for every bounded D ⊂ X :β(f(t,D)) 6 l(t)β(D),the function g : [0, 1]×X −→ X is given by g(y)(x) =

∫ 10 h(s) sin(1+ | y(s, x) |))ds

is continuous and compact.Consider the functions

τk(y(t)) = t2 − 1ek(1+∥y∥L2 )

and Ik(y) = bky where bk ∈ [1e , 1], for k = 1,m.

Both τk and Ik are continuous for k = 1,m and we have

τk+1(y(t))− τk(y(t)) =e− 1

ek+1(1+ ∥ y ∥L2)> 0, for each k = 1,m,

τk(y(t))− τk(Ik(y)(t)) =(1− bk) ∥ y ∥L2

ek(1 + bk ∥ y ∥L2)(1+ ∥ y ∥L2)> 0, for each k = 1,m

and

τk+1(Ik(y)(t))− τk(y(t)) =e− 1 + (ebk − 1) ∥ y ∥L2

ek+1(1+ ∥ y ∥L2)(1 + bk ∥ y ∥L2)> 0 for each k = 1,m.

Suppose that there exists k0 > 0 such that

M

(µ(k0) +

Ψ(k0)

Γ(α+ 1)q0)< 1,

where µ(k0) = sup∥g(y)∥ /∥y∥ 6 k0, and q0 = supq(t), t ∈ J We can verifyeasly that F satisfies the hypothesis (H3). The equation τk(y(t)) = t is equivalent tot2 − t − 1

ek(1+∥y∥L2 )= 0 wich admis two solution at maximum (finite solution set),

then by theorem 4.1, the problem (3.6) admis at least one solution on [0, 1].






4 Topological structure of solution set

In this section, we prove that the solution set is in fact Rδ, hence acyclicity. First,we recall the general theory (see, e.g., for more details [19]). Let (X, d) and (Y, d′)be two metric spaces

Definition 21. A set A of X is called a contractible space if there exists a continuoushomotopy h : A× [0, 1] −→ Y and x0 ∈ A such that(a) h(x, 0) = x, for every x ∈ A,(b) h(x, 1) = x0, for every x ∈ A,i.e., if the identity map idA : A −→ Ais homotopic to a constant map.In particular any closed convex subset of X is contractible.

Definition 22. We say that a compact nonempty metric space X is an Rδ-set ifthere exists a decreasing sequence of compact nonempty contractible metric spaces(Xn)n∈N∗ such that X =

⋂∞n=1Xn.

Let Hn(X) denote the Cech cohomology in the space X with coefficients in agroup G.

Definition 23. A space A is called G-acyclic if Hn(A) = 0, for every n > 0.

Intuitively, acyclic set has no holes.

Proposition 24. If A is Rδ-set then it is acyclic.

An u.s.c map F : X −→ P (Y ) is called acyclic if for each x ∈ X, F (x) is acompact acyclic set.

Theorem 25. [19] Let ϕ : X −→ Pcp,cv(X) be an u.s.c multivalued map from metric

space X to a Banach space E. If ϕ(X) is a compact set, then there exists a sequenceof u.s.c mappings ϕn from X to co(ϕ(X)) which approximates ϕ in the sense that,for all x ∈ X, we have:

ϕ(x) ⊂ . . . ⊂ ϕn+1(x) ⊂ ϕn(x) ⊂ . . . ⊂ ϕ0(x), for all n > 0, (4.1)

for all ϵ > 0 there exists n0 = n0(ϵ, x) such that

ϕn(x) ⊂ Oϵ(ϕ(x)), for all n > n0. (4.2)

Theorem 26. Under conditions (H1)−(H6), the solution set of problem (1.1)-(1.3)is an Rδ-set.

Proof. First let us denote the set of mild solutions of problem (1.1)-(1.3) by S(g).By Theorem 5.1, there exists a sequence (Fn)n>0 that verifies (4.1) and (4.2). Forevery n > 0, consider the following semi-linear evolution inclusion:

cDαy(t) ∈ Ay(t) + Fn(t, y(t)), t ∈ J = [0, T ], t = τk(y(t)), k = 1,m (4.3)






y(t+) = Ik(y(t)), t = τk(y(t)), k = 1,m (4.4)

y(0) = g(y). (4.5)

Denote by Sn(g), the set of mild solutions of this problem. By Theorem 4.1, problem(4.3)-(4.5) has a nonempty compact solution set. Let y∗ = y∗[f∗], (f∗ ∈ SFn,y) bean element of Sn(g) and for λ ∈ [0, 1], let the problem

cDαy(t) ∈ Ay(t) + Fn(t, y(t)), t ∈ [λT, T ], t = τk(y(t)), k = 1,m (4.6)

y(t+) = Ik(y(t)), t = τk(y(t)), k = 1,m (4.7)

y(0) = y∗(t) t ∈ [0, λT ]. (4.8)

We know that problem (4.6)-(4.8) has a nonempty compact solution set forevery y∗ ∈ Sn(g). Moreover the solution depends continuously on (λ, y∗). Denotethis solution by y[y∗, λ](t). Consider h : Sn(g) × [0, 1] −→ Sn(g) be the mappinggiven by

h(y, λ)(t) =

y∗(t) if t ∈ [0, λT ]y[y∗, λ](t) if t ∈ [λT, T ].

Clearly h(y, λ)(.) is a solution of the problem (4.3)-(4.5). In fact, note that for anyy ∈ Sn(g), there exists f ∈ SF,y such that y = y[f ]. Let

f(t) = fχ[λT,T ](t) + f∗(t)χ[0,λT ](t),

for each t ∈ [0, T ] such that χ is the characteristic function. It is clear that f ∈ SFn,h

and it is checked that y[f ] = y[f ] is a solution to (4.3)-(4.5)for t ∈ [λT, T ] andy[f ] = y∗[f∗] is the solution for t ∈ [0, λT ]. Hence h(λ, y∗) ∈ Sn(g).

To show that h is continuous, let (yk, λk) ∈ Sn(g) × [0, 1] a sequence such that(yk, λk) → (y, λ), as k → ∞. Then

h(yk, λk)(t) =

yk(t), if t ∈ [0, λkT ],y[yk, λk](t), if t ∈ [λkT, T ].

We check that h(yk, λk) → h(y, λ), as k → ∞. Without loss of generality, assumethat λk 6 λ and distinguish between three cases.

• If t ∈ [λT, T ], then

∥h(yk, λk)− h(y∗, λ)∥[λT,T ] = |y[yk, λ](t)− y[y∗, λ](t)|= sup

t∈[λT,T ]|y[yk, λ](t)− y[y∗, λ](t)|,

which tends to 0, as k → ∞ for y[y∗, λ](t) depends continuously on (y∗, λ).






• If t ∈ [0, λkT ], then

∥h(ym, λm)− h(y∗, λ)∥ = |yk(t)− y∗(t)|,

which tends to 0, as k → ∞.

• If t ∈ [λkT, λT ], then

|h(yk, λk)(t)− h(y∗, λ)(t)| = |y[yk, λk](t)− y∗(t)|,6 |y[yk, λk](t)− y(t)|+ |yk(t)− y∗(t)|,−→ 0 as k → ∞.

Moreover, for all y ∈ Sn(g), we have thath(y, 0) = y∗,h(y, 1) = y[y∗, 1](t).

Hence the set Sn(g) is contractible for every n > 0. By Theorem 5.1, we have

F (t, y) ⊆ Fn+1(t, y) ⊆ Fn(t, y) ⊆ . . . ⊆ F1(t, y).

HenceS(g) ⊆ Sn+1(g) ⊆ Sn(g) ⊆ . . . ⊆ S1(g),

which implies that

S(g) ⊆⋂

n∈N∗

Sn(g).

To prove the converse, let y ∈⋂

n∈N∗Sn(g). Then there exists a sequence of selections

fnn∈N∗ ⊂ L1([0, T ], X) such that fn ∈ SFn,y and y = y[fn] for all n ∈ N∗. Letε > 0. From (4.2), there exists n0 = n0[ε, y] such that

Fn(x) ⊂ Oϵ(F (x)), for all n > n0.

Then∥Fn(t, y)∥ 6 ∥F (t, y)∥+ 2ε,

and without loss of generality

|fn(t)| 6 q(t)ψ(∥y∥) + 2ε, for a. e. t ∈ [0, T ] and n > n0.

Hence the sequence fnn∈N∗ is integrably bounded. From the reflexivity of X, thereexists a subsequence of fn still denoted by fn such that fn f weakly , f ∈L1([0, T ], X). By Mazur’s convexity theorem (see [14]), we obtain a sequence

fn ⊂ cofn : n > 1






such that fn → f. Moreover fn(t) → f(t), for a. e. t ∈ [0, T ] and fn(t) ∈ Fn(t, y).Dfine by N the subset of [0, T ]:

N = t ∈ [0, T ] : fn(t) → f(t).

For t ∈ N , we have

∥fn(t)∥ 6 ∥Fn(t, y(t))∥,6 ∥F (t, y(t))∥+ 2ε.

Since F has convex closed values, we conclude that f(t) ∈ F (t, y(t)) for t ∈ N .Moreover y(t) = yk(t) (for some k = 1,m) and y = yk[fn], then from Theorem 3.1we get Gfn → Gf, which implies that yk[fn] → yk[f ]. We deduce that y ∈ S(g) andthat S(g) =

⋂n∈N∗

Sn(g). Finally S(g) is Rδ-set.

Corollary 27. The solution set for problem (1.1)-(1.3) is acyclic.

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Abdelghani Ouahab

Department of Mathematics and Informatics,

The African University Ahmed Draia of Adrar,





























Adrar, Algeria.

e-mail: agh [email protected]

and

Laboratory of Mathematics,

Sidi-Bel-Abbes University,

PoBox 89, 22000 Sidi-Bel-Abbes, Algeria.

Sarah Seghiri

Departement of Mathematics,

Normal High School,

B.P N 92, Vieux Kouba, Algiers,

Algeria.

e-mail: [email protected]

and

Laboratory of Fixed Point Theory and Its Applications,

Normal High School,

B.P N 92, Vieux Kouba, Algiers,

Algeria.

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Nonlocal fractional differential inclusions with impulses ... · In the study of the topological structure of the solution sets of differential equations and inclusions, an important