Fractional order Pettis integral equations with multiple time delay in Banach spaces

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1 An. Ştiinţ. Univ. Al. I. Cuza Iaşi. Mat. (N.S. Tomul LXIII, 27, f. Fractional order Pettis integral equations with multiple time delay in Banach spaces Mouffak Benchohra Fatima-Zohra Mostefai Received: 29.IX.22 / Revised: 5.X.22 / Accepted: 26.X.22 Abstract This paper is devoted to study the existence of solutions under the Pettis integrability assumption for an integral equation of fractional order with multiple time delay in Banach space by using the technique of measure of weak noncompactness. Keywords Integral equation Measure of weak noncompactness Left sided mixed Pettis integral Weak solution Mathematics Subject Classification (2 26A33 34A8 Introduction In the last 2 years, the theory of differential and integral equations of fractional orders has become a new important branch and significant development has been done; see for instance the books by Abbas et al [], Kilbas et al. [9], and Lakshmikantham et al. []. Let us mention that this theory has many applications in describing numerous events and problems of the real world. For example, fractional differential equations are often applicable in engineering, physics, chemistry, and biology. See Baleanu et al. [2], Hilfer [8], Podlubny [6] and Tarasov [9]. Mouffak Benchohra Laboratoire de Mathématiques, Université de Sidi Bel-Abbès, B.P. 89, 22, Sidi Bel-Abbès, Algérie and Department of Mathematics, Faculty of Science, King Abdulaziz University, Saudi Arabia benchohra@univ-sba.dz Fatima-Zohra Mostefai Laboratoire de Mathématiques, Université de Saïda, B.P. 38, 2, Saïda, Algérie f.z.mostefai@gmail.com

2 2 Mouffak Benchohra, Fatima-Zohra Mostefai In this paper, we will investigate the existence of solutions for the following fractional integral equation u(x, y= g i (x, yu(x ξ i, y µ i Iθ α f(x, y, u(x, y; (x, y J a J b, (. u(x, y = Ψ(x, y; (x, y J = [ ξ, a] [ µ, b]\(, a] (, b], (.2 where J a = [, a], J b = [, b] for a, b >, θ = (,, ξ = max...m {ξ i }, µ = max...m {µ i }, Iθ α is the left sided mixed Pettis integral of order α, α = (α, α 2 (, (,, f : J a J b E E is a given function satisfying some assumptions that will be specified later, g i : J a J b R, i =,..., m are given functions, and Ψ : J E is a given continuous function such that Ψ(, y = g i (, yψ( ξ i, y µ i ; y [, b], Ψ(x, = g i (x, Ψ(x ξ i, µ i ; x [, a], E is a real Banach space with norm. In our investigation we apply the method associated with the technique of measures of weak noncompactness and a fixed point theorem of Mönch type. This technique was mainly initiated in the monograph of Bana`s and Goebel [3] and subsequently developed and used in many papers; see, for example, Bana`s et al. [4], Guo et al. [7], Krzyska and Kubiaczyk [], Mönch [2], O Regan [3, 4], Szufla [7], Szufla and Szukala [8], and the references therein. 2 Preliminaries In this section, we introduce notations, definitions, and preliminary facts that will be used in the remainder of this survey paper. Let R denote the real line and let J a = [, a] and J b = [, b] be two closed and bounded intervals in R for some real numbers a > and b >. Throughout the paper, E is a Banach space with norm. and dual E. Also (E, w = (E, σ(e, E denotes the space E with its weak topology. We take C(J a J b, E to be the Banach space of continuous functions u : J a J b E, with the usual supremum norm u = sup{ u(x, y, (x, y J a J b }. Definition 2. ([5] The function x : J a J b E is said to be Pettis integrable on J a J b if and only if there is an element x I J E corresponding to each I J J a J b (I and J are measurable, such that ϕ(x I J = I J ϕ(x(s, tdsdt for all ϕ E where the integral on the right is assumed to exist in the sense of Lebesgue (by definition, x I J = I J x(s, tdsdt. We let L (J a J b, E denote the Banach space of measurable functions u : J a J b E that are Pettis integrable, equipped with the norm u L = a b u(x, y dxdy.

3 Fractional order Pettis integral equations 3 Definition 2.2 A function h : E E is said to be weakly sequentially continuous if h takes each weakly convergent sequence in E to weakly convergent sequence in E (i.e., for any (x n n in E with x n x in E w, h(x n h(x in E w. Definition 2.3 ([6] Let E be a Banach space, Ω E be the family of all bounded subsets of E and B E the unit ball of E. The De Blasi measure of weak noncompactness is the map β : Ω E [, defined by β(x = inf{ɛ > : there exists a weakly compact subset Ω of E : X ɛb E Ω}. Properties: The De Blasi measure of noncompactness satisfies the following properties. (a A B β(a β(b, (b β(a = A is weakly relatively compact, (c β(a B = max{β(a, β(b}, (d β(a ω = β(a, (A ω denotes the weak closure of A, (e β(a B β(a β(b, (f β(λa = λ β(a, λ R, (g β(conv(a = β(a, (h β( λ h λa = hβ(a. The following result follows directly from the Hahn-Banach theorem. Proposition 2.4 Let E be a normed space with x then there exists ϕ E with ϕ = and ϕ(x = x. For completeness, we recall the definition of the fractional Pettis-integral of order α >. Let α, α 2 > and α = (α, α 2. For h L (J a J b, E, the expression (I α θ h(x, y = (x s α (y t α2 h(s, tdsdt, where the sign denotes the Pettis integral and Γ (. is the Euler gamma function, is called the left sided mixed Pettis integral of order α. For our purpose we will need the following fixed point theorem. Theorem 2.5 ([3] Let E be a Banach space with Q a nonempty, bounded, closed, convex and equicontinuous subset of metrizable locally convex vector space C(J a J b, E such that Q. Assume that T : Q Q is weakly-sequentially continuous. If the implication V = conv({} T (V V is relatively weakly compact, (2. holds for every subset V Q, then the operator T has a fixed point. 3 Existence of solutions First of all, we define what we mean by a solution of problem (.-(.2. Set J = [ ξ, a] [ µ, b]. Definition 3. A function u C(J, E is said to be a solution of (.-(.2 if u satisfies equation (. on J a J b and condition (.2 on J

4 4 Mouffak Benchohra, Fatima-Zohra Mostefai Set G = max...m {sup (x,y Ja J b g i (x, y }. We are now in the position to state and prove our existence result for the problem (.-(.2. We first list the following hypotheses. (H For each (x, y J a J b, f(x, y, is weakly sequentially continuous. (H2 For each u C(J a J b, E, f(,, u(, is Pettis integrable on J a J b. (H3 There exists p L (J a J b, R such that f(x, y, u p(x, y, for (x, y J a J b and each u E. (H4 Let r > be arbitrary (but fixed. For any ɛ > and for any subset X B r, there exists a closed subset I ɛ J a J b such that µ(j a J b \I ɛ < ɛ and β(f(t X sup (x,y T p(x, yβ(x, for each closed subset T of I ɛ, where µ denotes the Lebesgue measure in R 2. (H5 The functions g i : J a J b R, i =,..., m are continuous. Remark 3. (H4 is satisfied if the set f(x, y, B is weakly relatively compact in E for each (x, y J a J b and B a bounded set of E. Remark 3.2 If E is reflexive then (H4 is automatically satisfied since a subset of a reflexive Banach space is weakly compact if and only if it is closed in the weak topology and bounded in the norm topology. The main result in this paper reads as follows. Theorem 3.2 Assume that assumptions (H (H5 hold. If mg a α b α2 <, (3. Γ (α Γ (α 2 where = p, then problem (. (.2 has at least one solution on J. Proof. To transform the problem (.-(.2 into a fixed point problem, we define the operator T : C(J, E C(J, E as T (u(x, y Ψ(x, y; (x, y J, = g i (x, yu(x ξ i, y µ i Iθ αf(x, y, u(x, y; (x, y J a J b. (3.2 Now we prove that T satisfies all the assumptions of Theorem 2.5 and thus T has a fixed point which is a solution of problem (.-(.2. First notice that, for all u C(J a J b, E, f(x, y, u(x, y is Pettis integrable for a.e. (x, y J a J b (Assumption (H2 then ϕ(f(x, y, u(x, y L (J a J b for any ϕ E. From the definition of the integral of fractional order we have I α ϕ(f(x, y, u(x, y = = (x s α (y t α2 ϕ(f(s, t, u(s, tdsdt ( (x s α (y t α2 ϕ f(s, t, u(s, t dsdt

5 Fractional order Pettis integral equations 5 exists for almost every (x, y J a J b and is an element of L (J a J b, that is, for almost every (x, y J a J b, s (, x, t (, y the measurable function ( (x s α (y t α2 ϕ f(s, t, u(s, t = (x sα (y t α2 ϕ(f(s, t, u(s, t is Lebesgue integrable, hence the function (s, t (x sα (y t α2 f(s, t, u(s, t is Pettis integrable on J a J b, and thus the operator T is well defined. a α b α2 Let R >, be such that R, and consider the set Γ (α Γ (α 2 ( mg Q={u C(J, E : u R and u(x 2, y 2 u(x, y R g i (x 2, y 2 g i (x, y Γ (α Γ (α 2 [xα 2 yα2 2 x α yα2 ]; for (x, y, (x 2, y 2 J a J b }. Clearly, the subset Q is closed, convex and equicontinuous. The remainder of the proof will be given in three steps. Step : T maps Q into itself. To see this, take u Q, (x, y J a J b and assume that T u(x, y. Then there exists ϕ E with ϕ = such that T u(x, y = ϕ(t u(x, y. Thus, we obtain: ( m T u(x, y = ϕ(t u(x, y = ϕ g i (x, yu(x ξ i, y µ i (x s α (y t α2 f(s, t, u(s, tdsdt ( m = ϕ g i (x, yu(x ξ i, y µ i ( ϕ g i (x, y u(x ξ i, y µ i (x s α (y t α2 f(s, t, u(s, tdsdt (x s α (y t α2 f(s, t, u(s, t dsdt mgr (x s α (y t α2 dsdt a α b α2 mgr Γ (α Γ (α 2 R.

6 6 Mouffak Benchohra, Fatima-Zohra Mostefai On the other hand, for (x, y J, we have T u(x, y = ϕ(t u(x, y = ϕ(ψ(x, y R. Next, suppose that (x, y, (x 2, y 2 J a J b with x < x 2 and y < y 2, and let u Q, so T u(x, y T u(x 2, y 2. Then there exists ϕ E such that T u(x, y T u(x 2, y 2 = ϕ(t u(x, y T u(x 2, y 2 and ϕ =. Thus ( m T u(x 2, y 2 T u(x, y = ϕ g i (x 2, y 2 u(x 2 ξ i, y 2 µ i x2 (x 2 s α (y 2 t α2 f(s, t, u(s, tdsdt g i (x, y u(x ξ i, y µ i x y (x s α (y t α2 f(s, t, u(s, tdsdt ( m = ϕ g i (x 2, y 2 u(x 2 ξ i, y 2 µ i g i (x, y u(x ξ i, y µ i ( ϕ x2 x y x y f(s, t, u(s, tdsdt x y (x 2 s α (y 2 t α2 f(s, t, u(s, tdsdt [(x 2 s α (y 2 t α2 (x s α (y t α2 ] (x 2 s α (y 2 t α2 f(s, t, u(s, tdsdt x2 y (x 2 s α (y 2 t α2 f(s, t, u(s, tdsdt x g i (x 2, y 2 u(x 2 ξ i, y 2 µ i g i (x, y u(x ξ i, y µ i R x y x2 x y x y x2 y x [(x 2 s α (y 2 t α2 (x s α (y t α2 ]dsdt (x 2 s α (y 2 t α2 dsdt (x 2 s α (y 2 t α2 dsdt (x 2 s α (y 2 t α2 dsdt g i (x 2, y 2 g i (x, y Γ (α Γ (α 2 [xα 2 yα2 2 xα yα2 ].

7 Thus T (Q Q. Fractional order Pettis integral equations 7 Step 2: T is weakly sequentially continuous. Let (u n be a sequence in Q with u n (x, y u(x, y in (E, ω for each (x, y J. Obviously, T u n T u for any (x, y [ ξ, ] [ µ, ]. Fix (x, y J a J b, we have m g i(x, yu n (x ξ i, y µ i m g i(x, yu(x ξ i, y µ i and since f(x,y, is weakly sequentially continuous (Assumption (H we have immediately f(x, y, u n (x, y converging weakly uniformly to f(x, y, u(x, y. Then, Lebesgue Dominated Convergence Theorem for the Pettis integral implies that T u n (x, y converging weakly uniformly to T u(x, y in (E, ω. Since this holds, for each (x, y J a J b we have T u n T u, i.e T : Q Q is weakly sequentially continuous. Step 3: The implication (2. holds. Let V be a subset of Q such that V = conv(t (V {}. Obviously V (x, y conv(t (V (x, y {}, (x, y J. Further, as V is bounded and equicontinuous, by Ambrosetti Lemma (cf. [5], Lemma 3 the function (x, y υ(x, y = β(v (x, y is continuous on J. Since Ψ is continuous on [ ξ, ] [ µ, ], the set {Ψ(x, y, (x, y [ ξ, ] [ µ, ]} E is compact. By (H3 and the properties of the measure β, for any (x, y J a J b, we have υ(x, y β(t (V (x, y {} β(t (V (x, y ({ m β g i (x, yu(x ξ i, y µ i x y } (x s α (y t α2 f(s, t, u(s, tdsdt; u V ({ m } β g i (x, yu(x ξ i, y µ i ; u V ({ β β({g i (x, yu(x ξ i, y µ i ; u V } ({ x β g i (x, yβ(v (x, y y } (x s α (y t α2 f(s, t, u(s, tdsdt; u V } (x s α (y t α2 f(s, t, u(s, tdsdt; u V (x s α (y t α2 p(s, tβ(v (s, tdsdt a α b α2 mg υ Γ (α Γ (α 2 υ. a α b α 2 In particular, υ υ (mg Γ (α Γ (α. By (3. it follows that υ 2 =, that is υ(x, y = β(v (x, y =, for each (x, y J and then V is weakly relatively

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