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AIMSMathematics,6(8): 8654–8666. DOI:10.3934/math.2021503 Received: 15 January 2021 Accepted: 27 May 2021 http://www.aimspress.com/journal/Math Published: 07 June 2021 Research article Dynamicsandstability for Katugampola random fractional differential equations 1 ¨ 2,∗ 3 4 Fouzia Bekada , Saıd Abbas , Mouffak Benchohra and Juan J. Nieto 1 ¨ Laboratory of Mathematics, University of Saıda–Dr. Moulay Tahar, P. O. Box 138, EN-Nasr, 20000 ¨ Saıda, Algeria 2 ¨ Department of Mathematics, University of Saıda–Dr. Moulay Tahar, P. O. Box 138, EN-Nasr, ¨ 20000Saıda, Algeria 3 ` Laboratory of Mathematics, Djillali Liabes University of Sidi Bel-Abbes, P. O. Box 89, Sidi ` Bel-Abbes 22000, Algeria 4 ´ ´ ´ ´ Departamento de Estatistica, Analise Matematica e Optimizacion, Instituto de Matematicas, Universidade de Santiago de Compostela, Santiago de Compostela, Spain * Correspondence: Email: abbasmsaid@yahoo.fr, said.abbas@univ-saida.dz. Abstract: This paper deals with some existence of random solutions and the Ulam stability for a class of Katugampola random fractional differential equations in Banach spaces. A random fixed point theorem is used for the existence of random solutions, and we prove that our problem is generalized Ulam-Hyers-Rassias stable. An illustrative example is presented in the last section. Keywords: differential equation; Katugampola fractional integral; Katugampola fractional derivative; random solution; Banach space; Ulam stability; fixed point MathematicsSubjectClassification: 26A33, 34A37, 34G20 1. Introduction The history of fractional calculus dates back to the 17th century. So many mathematicians define the most used fractional derivatives, Riemann-Liouville in 1832, Hadamard in 1891 and Caputo in 1997 [24,28,34]. Fractional calculus plays a very important role in several fields such as physics, chemicaltechnology,economics,biology;see[2,24]andthereferencestherein. In2011,Katugampola introduced a derivative that is a generalization of the Riemann-Liouville fractional operators and the fractional integral of Hadamard in a single form [21,22]. There are several articles dealing with different types of fractional operators; see [1,3,9–13,16,32]. Variousresults about existence of solutions as well as Ulam stability are provided in [6–8,14,15,17,19, 8655 20,23,25–31,33]. In this article we investigate the following class of Katugampola random fractional differential equation ρ ς ( D0x)(ξ,w) = f(ξ, x(ξ,w),w); ξ ∈ I = [0,T], w ∈ Ω, (1.1) with the terminal condition x(T,w) = xT(w); w ∈ Ω, (1.2) ρ ς where x : Ω → E is a measurable function, ς ∈ (0,1], T > 0, f : I × E × Ω → E, D is the T 0 Katugampola operator of order ς, and Ω is the sample space in a probability space, and (E,k · k) is a Banach space. 2. Preliminaries By C(I) := C(I,E) we denote the Banach space of all continuous functions x : I → E with the norm kxk∞ = supkx(ξ)k, t∈I and L1(I,E) denotes the Banach space of measurable function x : I → E with are Bochner integrable, equipped with the norm Z kxkL1 = kx(ξ)kdξ. I Let Cς,ρ(I) be the weighted space of continuous functions defined by Cς,ρ(I) = {x : (0,T] → E : ξρ(1−ς)x(ξ) ∈ C(I)}, with the norm kxk := supkξρ(1−ς)x(ξ)k. C ξ∈I Definition2.1. [2]. The Riemann-Liouville fractional integral operator of the function h ∈ L1(I,E) of order ς ∈ R is defined by + Z 1 ξ RL ς r−1 I h(ξ) = (ξ − s) h(s)ds. 0 Γ(r) 0 Definition 2.2. [2]. The Riemann-Liouville fractional operator of order ς ∈ R is defined by + ! Z 1 d n ξ RL ς n−ς−1 D0h(ξ) = Γ(n−ς) dς 0 (ξ − s) h(s)ds. Definition 2.3. (Hadamard fractional integral) [4]. The Hadamard fractional integral of order r is defined as Z Iςh(ξ) = 1 ξ log ξς−1 h(s)ds, ς > 0, 0 Γ(ς) 1 s s provided that the left-hand side is well defined for almost every ξ ∈ (0,T). AIMSMathematics Volume6,Issue 8, 8654–8666. 8656 Definition 2.4. (Hadamard fractional derivative ) [4]. The Hadamard fractional derivative of order r is defined as ! Z n ξ Dςh(ξ) = 1 ξ d log ξ n−ς−1 h(s)ds, ς > 0, 0 Γ(n−ς) dξ 1 s s provided that the left-hand side is well defined for almost every ξ ∈ (0,T). Definition2.5. (Katugampolafractionalintegral)[21]. TheKatugampolafractionalintegralsoforder (ς > 0) is defined by Z 1−ς ξ ρ−1 ρIςx(ξ) = ρ s x(s)ds (2.1) 0 ρ ρ 1−ς Γ(ς) 0 (ξ − s ) for ρ > 0 and ξ ∈ I, provided that the left-hand side is well defined for almost every ξ ∈ (0,T). Definition 2.6. (Katugampola fractional derivative) [21]. The Katugampola fractional derivative of order ς > 0 is defined by: ! d n ρ r 1−ρ ρ n−r D0u(ξ) = ξ ( I u)(ξ) dξ 0 ! Z r−n+1 n ξ ρ−1 = ρ ξ1−ρ d s u(s)ds, ρ ρ r−n+1 Γ(n−r) dξ 0 (ξ − s ) provided that the left-hand side is well defined for almost every ξ ∈ (0,T). We present in the following theorem some properties of Katugampola fractional integrals and derivatives. Theorem2.7. [21]Let0 < Re(ς) < 1 and 0 < Re(η) < 1 and ρ > 0, for a > 0: • Index property: ρ ς ρ η ρ ς+η ( Da)( Dah)(t) = Da h(t) ρ r ρ η ρ r+η ( I )( I h)(t) = I h(t) a a a • Linearity property: ρ r ρ r ρ r D (h+g) = D h(t) + D g(t) a a a ρ r ρ r ρ r I (h + g) = I h(t)+ I g(t) a a a and we have d (t1−ρ )Ir(I1−r)u(s)ds. dt 0 0 Theorem2.8. [21]Letr be a complex number, Re(r) ≥ 0, n = [Re(r)] and ρ > 0. Then, for t > a; ρ r 1 R t r−1 (1) limρ→1( I h)(t) = (t − τ) h(τ)dτ. a Γ(r) a (2) lim +(ρIrh)(t) = 1 R t(log t)r−1h(τ)dτ. ρ→0 a Γ(r) a τ τ ρ r d n 1 R t h(τ) (3) limρ→1( Dah)(t) = (dt) Γ(n−r) r−n+1 dτ. a (t−τ) + ρ r 1 d n R t t n−r−1 dτ (4) limρ→0 ( Dah)(t) = Γ(n−r)(tdt) a (log τ) h(τ) τ . AIMSMathematics Volume6,Issue 8, 8654–8666. 8657 Remark2.9. ρ r RL r (1) limρ→1( I h)(t) = ( I h)(t). a a (2) lim +(ρIrh)(t) = (HIrh)(t). ρ→0 a a ρ r RL r (3) limρ→1( D h)(t) = ( D h)(t). a a (4) lim +(ρDrh)(t) = (HDrh)(t). ρ→0 a a ρ r Lemma2.10. Let0 < r < 1. The fractional equation ( D v)(t) = 0, has as solution 0 v(t) = ctρ(r−1), (2.2) with c ∈ R. Lemma2.11. Let0 < r < 1. Then ρ r ρ r ρ(r−1) I ( D0u)(t) = u(t) + ct . Proof. We have r r 1−p d ! r+1 r I D u(t) = t I D u(t) 0 0 dt 0 0 1−ρ d ρ−r Z t sρ−1 ρ r ! = (t ) ρ ρ −r( D0u(s))ds dt! Γ(r +1) 0 (t − s ) " ! # ! = t1−ρ d ρ−r Z t sρ−1 s1−ρ d (I1−ru)(s) ds ρ ρ −r 0 dt! Γ(r +1) 0 (t − s )" ds # ! 1−ρ d ρ−r Z t ρ ρ r d 1−r = t (t −s ) (I u)(s) ds . dt Γ(r + 1) ds 0 0 Thus, IrDru(t) = I + I , with 0 0 1 2 ! −r h i 1−ρ d ρ ρ ρ r 1−r t I = t (t −s ) I u(s) , 1 dt Γ(r +1) 0 0 and d ! ρ−r Z t 1−ρ ρ−1 ρ ρ r−1 1−r I2 = t rρs (t −s ) I u(s)ds. dt Γ(r +1) 0 0 Hence, we get I = ctρ(r−1) 1 and 1−ρ d ! ρ1−r Z t ρ−1 ρ ρ r−1 1−r I = t s (t −s ) I u(s)ds 2 dt Γ(r) 0 ! 0 = t1−ρ d Ir(I1−r)u(s)ds dt 0 0 = u(t). Finally we obtain r r ρ(r−1) (I )(D u)(t) = u(t) + ct . 0 0 AIMSMathematics Volume6,Issue 8, 8654–8666.
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