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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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