Available online at www.atnaa.org Research Article
Analysis of a fractional boundary value problem involving Riesz-Caputo fractional derivative.
Adjimi Naasa, Maamar Benbachirb, Mohamed S. Abdoc, Abdellatif Boutiaraa
aLaboratory of Mathematics And Applied Sciences, University of Ghardaia, Ghardaia 47000. Algeria.
bFaculty of Sciences, Saad Dahlab University, Blida, Algeria.
cDepartment of Mathematics, Hodeidah University, Al-Hodeidah, Yemen.
Abstract
In this paper, we investigate the existence and uniqueness of solutions for a class of fractional dierential equations with boundary conditions in the frame of Riesz-Caputo operators. We apply the methods of functional analysis such that the uniqueness result is established by using Banach's contraction principle, whereas Schaefer's and Krasnoslkii's xed point theorems are applied to obtain existence results. Some examples are given to illustrate our acquired results.
Keywords: Riesz-Caputo derivative Boundary value problem xed point theorem 2010 MSC: 26A33, 34A07, 93A30, 35R11
1. Introduction
Fractional calculus (FC) is a mathematical branch that investigates the properties of derivatives and integrals of non-integer order. The interested readers in the subject should refer to the books [34, 35, 36]. Fractional order models, which provide an excellent description of memory and genetic processes, are more accurate and appropriate than models with integer order. For the development of FC, there are sundry common denitions of fractional derivatives and integrals, such as Rimann-Liouville type, Caputo type, Hadamard type, Hilfer type, ψ-Caputo, ψ-Hilfer type, Caputo-Fabrizio type, Atangana-Baleanu type, conformable type, and Erdelyi-Kober type, etc, (see [11, 15, 16, 25, 32, 33, 37, 3]).
Email addresses: naasadjimi@gmail.com (Adjimi Naas), mbenbachir2001@gmail.com (Maamar Benbachir), msabdo1977@gmail.com (Mohamed S. Abdo), boutiara_a@yahoo.com (Abdellatif Boutiara)
Received April 26, 2021, Accepted September 18, 2021, Online September 21, 2021
Some recent contributions have been investigated the existence and uniqueness of solutions for dierent kinds of nonlinear fractional dierential equations (FDEs) and inclusion (FDIs) by using various types of xed point theorems, which can be found in [13, 6, 17, 7, 39, 38, 8, 19, 20, 21, 4, 5, 1, 2, 18], and the references cited therein. The study of FDEs or FDIs with anti-periodic boundary conditions, that are applied in numerous dierent elds, like chemical engineering, physics, economics, dynamics, etc., has received much attention recently, (see [23, 27, 40, 10, 26]) and the papers mentioned therein.
On the other hand, the authors in [31] investigated the existence results of the following FDEs
RC
0 DϑTκ(t) = g(t, κ(t)), 0 < γ ≤ 1, 0 ≤ t ≤ T , κ(0) = κ0, κ(T ) = κT,
where RC0 DϑT is the Riesz-Caputo derivative, g : [0, T ] × R → R is a continuous function, and κ0,κT are constants.
The positive solution of nonlinear FDEs with the Riesz space derivative
RC
0 Dϑ1κ(t) = h(t, κ(t)), 0 < ϑ ≤ 1, t ∈ [0, 1], κ(0) = κ0, κ(1) = κ1, κ0, κ1≥ 0,
has been studied by Yun Gu et al., in [24]. Also, Chen et al., in [27] discussed a class of FDEs with anti-periodic boundary condithions of the form
RC
0 DTϑκ(t) = g(t, κ(t)), 0 < γ ≤ 1, t ∈ [0, T ], κ(0) + κ(T ) = 0, κ0(0) + κ0(T ) = 0,
whereRC0 DϑT is the Riesz-Caputo derivative and g : [0, 1] × R → R is a continuous function.
Motivated by the above cited work, in this paper, we investigat the existence and uniqueness results of the following FDEs with the Riesz-Caputo derivative
RC
0 DTϑκ(t) + F(t, κ(t),RC0 DTςκ(t)) = 0, t ∈ J := [0, T ],
κ(0) + κ(T ) = 0, µκ0(0) + σκ0(T ) = 0, (1) where 1 < ϑ ≤ 2 and , 0 < ς ≤ 1 , RC0 DκT is the Riesz-Caputo fractional derivative of order κ ∈ {ϑ, ς}, F: J × R × R → R, is a continuous function, and µ,σ are nonnegative constants with µ > σ. We refer here to some very recent works that dealt with a similar analysis, see [28, 29, 30].
The paper is marshaled as follows. Section 2 has denitions and some of the most important basic concepts of the FC. In section 3, we prove the existence and uniqueness of solutions for the proposed problem with the Riesz-Caputo derivatives via Banach's, Schaefer's, and Krasnoselskii's xed point theorems. Some illustrative examples associated with our suggested problem are provided in Section 4.
2. Preliminaries
In this section ,we recall some basic concepts, and preliminary facts. By E = C(J , R) we denote the Banach space of all continuous functions from J into R as follows
E = n
κ : κ ∈ C([0, T ]), RCDδκ ∈ C([0, T ]) o
, endowed with the norm
kκkE = kκk + kRCDδκk, and
kκk = sup
t∈J|κ(t)|, kRCDδκk = sup
t∈J
|RCDδκ(t)|.
We start with denitions.
Denition 2.1. [9, 36] For 0 ≤ t ≤ T, the classical RieszCaputo fractional derivative is dened by
RC0 DTϑF(t) = 1 Γ(n − ϑ)
Z T 0
|t − ξ|n−ϑ−1F(n)(ξ)dξ
= 1
2(C0Dϑt + (−1)n Ct DTϑ)F(t),
whereC0Dtϑ andCt DTϑ are the left and right Caputo derivative, respectively
C
0DϑtF(t) = 1 Γ(n − ϑ)
Z t 0
(t − ξ)n−ϑ−1F(n)(ξ)dξ,
C
t DϑTF(t) = (−1)n Γ(n − ϑ)
Z T t
(ξ − t)n−ϑ−1F(n)(ξ)dξ.
Remark 2.2. ([27, 31]) In particular if F(t) ∈ C([0, T ]) and 0 < ϑ ≤ 1, then
RC
0 DϑTF(t) = 1
2(C0Dϑt − Ct DTϑ)F(t), if F(t) ∈ C2([0, T ]) and 1 < ϑ ≤ 2, then
RC
0 DϑTF(t) = 1
2(C0Dϑt + Ct DTϑ)F(t).
Denition 2.3. ([31]) The Riemann-Liouville fractional integrals concepts of order ϑ are dened as
0ItϑF(t) = 1 Γ(ϑ)
Z t 0
(t − ξ)ϑ−1F(ξ)dξ,
tITϑF(t) = 1 Γ(ϑ)
Z T t
(ξ − t)ϑ−1F(ξ)dξ,
0ITϑF(t) = 1 Γ(ϑ)
Z T 0
|ξ − t|ϑ−1F(ξ)dξ.
Lemma 2.4. ([27]) If F(t) ∈ Cn([0, T ]), then
0ItϑC0DtϑF(t) = F(t) −
n−1
X
k=0
F(k)(0)
k! (t − 0)k, and
tITϑCt DTϑF(t) = (−1)n
"
F(t) −
n−1
X
k=0
(−1)kF(k)(T )
k! (T − t)k
# . From the above denitions and lemmas, we have
0ITϑRC0 DϑTF(t) = 1 2
0ItϑC0Dtϑ+tITϑC0Dtϑ
F(t) + (−1)n1
2
0ItϑCt DϑT +tITϑCtDϑT
F(t)
= 1 2
0ItϑC0Dtϑ+ (−1)ntITϑCt DTϑ F(t).
In particular, if 1 < ϑ ≤ 2 and F(t) ∈ C2([0, T ]), then
0ITϑRC0 DTϑF(t) = F(t) − 1
2(F(0) + F(T )) −1
2F0(0)t + 1
2F0(T )(T − t). (2)
3. Main Results
Lemma 3.1. Assume that g ∈ C(J , R) and κ ∈ C2(J ). Then
RC0 DTϑκ(t) + g(t) = 0, t ∈ [0, T ], 1 < ϑ ≤ 2,
κ(0) + κ(T ) = 0, µκ0(0) + σκ0(T ) = 0, (3) is equivalent to the integral equation given by
κ(t) = (σ − µ)t + µT (µ + σ)Γ(ϑ − 1)
Z T 0
(T − s)ϑ−2g(s)ds − 1 Γ(ϑ)
Z t 0
(t − s)ϑ−1g(s)ds
− 1
Γ(ϑ) Z T
t
(s − t)ϑ−1g(s)ds. (2)
Proof. Applying Lemma (2.4) on equation (3), we obtain κ(t) = 1
2(κ(0) + κ(T )) +1
2κ0(0)t −1
2κ00RCITϑg(t)
= 1
2(κ(0) + κ(T )) +1
2κ0(0)t −1
2κ0(T )(T − t) − 1 Γ(ϑ)
Z t 0
(t − ξ)ϑ−1g(ξ)dξ
− 1
Γ(ϑ) Z T
t
(ξ − t)ϑ−1g(ξ)dξ. (3)
Then
κ0(t) = 1
2(κ0(0) + κ0(T )) − 1 Γ(ϑ − 1)
Z t 0
(t − ξ)ϑ−2g(ξ)dξ
+ 1
Γ(ϑ − 1) Z T
t
(ξ − t)ϑ−2g(ξ)dξ.
By the boundary conditions of (3), we nd that:
κ(0) = −σT
(µ + σ)Γ(ϑ − 1) Z T
0
(T − ξ)ϑ−2g(ξ)dξ + 1 Γ(ϑ)
Z T 0
(T − ξ)ϑ−1g(ξ)dξ,
κ(T ) = σT
(µ + σ)Γ(ϑ − 1) Z T
0
(T − ξ)ϑ−2g(ξ)dξ − 1 Γ(ϑ)
Z T 0
(T − ξ)ϑ−1g(ξ)dξ,
κ0(0) = 2σT (µ + σ)Γ(ϑ − 1)
Z T 0
(T − ξ)ϑ−2g(ξ)dξ,
κ0(T ) = −2µT (µ + σ)Γ(ϑ − 1)
Z T 0
(T − ξ)ϑ−2g(ξ)dξ.
Substituting the values of κ0(0)and κ0(T )into (3), we obtain (2).
Let us introduce the following notations:
Ω1= µTϑ
(µ + σ)Γ(ϑ) + 2Tϑ Γ(ϑ + 1), Ω2= Tϑ−ς(µ − σ) + 2µTϑΓ(2 − ς)
2(µ + σ)Γ(2 − ς)Γ(ϑ) + 2Tϑ−ς Γ(ϑ − ς + 1), k1 = Tϑ−ς
Γ(ϑ − ς + 1), k2 = Tϑ−ς
Γ(ϑ − ς + 1).
3.1. Uniqueness result via Banach's xed point theorem
Theorem 3.2. Let F : [0, T ] × R × R → R is a continuous function. Assume that
(H1) there exists nonnegative real numbers L1, L2 such that for all (ξ, v), (ξ0, v0) ∈ R2, we have
|F(t, ξ, v) − F(t, ξ0, v0)| ≤ L1|ξ − ξ0| + L2|v − v0|, if
(Ω1+ Ω2)(L1+ L2) < 1.
Then the problem (1) has a unique solution on J .
Proof. We transform BVP (1) into xed point problem. Then we dene the integral operator H : E → E by
Hκ(t) = (σ − µ)t + µT (µ + σ)Γ(ϑ − 1)
Z T 0
(T − ξ)ϑ−2F(ξ, κ(ξ),RCDςκ(ξ))dξ
− 1 (Γ)
Z t 0
(t − ξ)ϑ−1F(ξ, κ(ξ),RCDςκ(ξ))dξ
− 1 (Γ)
Z T t
(ξ − t)ϑ−1F(ξ, κ(ξ),RCDςκ(ξ))dξ.
Now, we prove that H is a contraction. For κ, $ ∈ E and for each t ∈ J , we have
|Hκ(t) − H$(t)|
≤ (σ − µ)t + µT | (µ + σ)|Γ(ϑ − 1)
Z T 0
(T − ξ)ϑ−2|F(ξ, κ(ξ),RCDςκ(ξ)) − F(ξ, $(ξ),RCDς$(ξ))|dξ
+ 1
Γ(ϑ) Z t
0
(t − ξ)ϑ−1|F(ξ, κ(ξ),RCDςκ(ξ)) − F(ξ, $(ξ),RCDς$(ξ))|dξ
+ 1
Γ(ϑ) Z T
t
(ξ − t)ϑ−1|F(ξ, κ(ξ),RCDςκ(ξ)) − F(ξ, $(ξ),RCDς$(ξ))|dξ
≤ µTϑ
(2µ + σ)Γ(ϑ) L1kκ − $k + L2kRCDςκ −RCDς$k + 2Tϑ
Γ(ϑ + 1) L1kκ − $k + L2kRCDςκ −RCDς$k
≤
µTϑ
(µ + σ)Γ(ϑ) + 2Tϑ Γ(ϑ + 1)
L1+ L2)(kκ − $k + kRCDςκ −RCDς$k . Consequently, we obtain
kHκ(t) − H$(t)k ≤ Ω1 L1+ L2)(kκ − $k + kRCDςκ −RCDς$k . (12)
On the other hand, we have
RC0 DςTHκ(t) −RC0 DTςHκ(t)
≤ 1
Γ(ϑ − ς) Z t
0
(t − ξ)ϑ−ς−1|F(ξ, κ(ξ),RCDςκ(ξ)) − F(ξ, $(ξ),RCDς$(ξ))|dξ
+ 1
Γ(ϑ − ς) Z T
t
(ξ − t)ϑ−ς−1|F(ξ, κ(ξ),RCDςκ(ξ)) − F(ξ, $(ξ),RCDς$(ξ))|dξ +
t1−ς− (T − t)1−ς (σ − µ) + 2µΓ(2 − ς)T 2(σ + µ)Γ(2 − ς)Γ(ϑ − 1)
Z T 0
(T − ξ)ϑ−2|F(ξ, (ξ),RCDςκ(ξ)) − F(ξ, $(ξ),RCDς$(ξ))|dξ
≤ Tϑ−ς(σ − µ) + 2µTϑΓ(2 − ς)
(σ + µ)Γ(2 − ς)Γ(ϑ) L1kκ − $k + L2kRCDςκ −RCDς$k + 2Tϑ−ς
Γ(ϑ − ς + 1) L1kκ − $k + L2kRCDςκ −RCDς$k . Thus,
kRC0 DςTHκ(t) −RC0 DςTHκ(t)k ≤ Ω2 L1+ L2)(kκ − $k + kRCDςκ −RCDς$k
(13) From (12) and (13), we get
kHκ(t) − H$(t)kE ≤ (Ω1+ Ω2) L1+ L2)(kκ − $k + kRCDςκ −RCDς$k .
Hence, H is a contraction. As a consequence of Banach contraction principle, the problem (1) has a unique solution on J .
3.2. Existence result via Shaefer xed point theorem
Lemma 3.3. Let E be a Banach space. Assume that H : E → E be a completely continuous operator, and the set
ω(H) = {$ ∈ E : $ = λH$, λ ∈ (0, 1)} . ω(F)is bounded. Then H has a xed point in E.
Theorem 3.4. Assume that there exists a positive M such that
|F(ξ, κ, $)| < M for t ∈ J , κ, $ ∈ R.
Then the problem (1) has at least one solution on J .
Proof. We will use the Sheafer's xed point theorem, to prove H has a xed point on E, we subdivided the proof into several steps :
Step 1. H is continuous on E: in view of continuity of F, we conclude that operator H is continuous.
step 2. H maps bounded sets into bounded sets in E.
For each κ ∈ Br= {κ ∈ E : kκkE ≤ r} and t ∈ J , we get
|Hκ(t)| ≤ |(σ − µ)t + µT | (σ + µ)Γ(ϑ − 1)
Z T 0
(T − ξ)ϑ−2|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
+ 1
Γ(ϑ) Z t
0
(t − ξ)ϑ−1|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
− 1
Γ(ϑ) Z T
t
(ξ − t)ϑ−1|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
≤ µTϑM
(σ + µ)Γ(ϑ) + 2MTϑ Γ(ϑ + 1).
Which implies that,
kHκ(t)k ≤ MΩ1, (14)
and,
|RC0 DςTHκ(t)| ≤ 1 Γ(ϑ − ς)
Z t 0
(t − ξ)ϑ−ς−1|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
+ 1
Γ(ϑ − ς) Z T
t
(t − ξ)ϑ−ς−1|F(ξ, κ(ξ),RCDςκ(ξ))|dξ +
t1−ς− (T − t)1−ς (σ − µ) + 2µΓ(2 − ς)T 2(σ + µ)|Γ(2 − ς)Γ(ϑ − 1)
× Z T
0
(T − ξ)ϑ−2|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
≤ Tϑ−ς(µ + σ) + 2µTϑΓ(2 − ς)
2|µ − σ|Γ(2 − ς)Γ(ϑ) M+ 2Tϑ−ς 2Γ(ϑ − ς + 1)M
≤ Tϑ−ς(σ − µ)) + 2µTϑΓ(2 − ς)
2(σ + µ)Γ(2 − ς)Γ(ϑ) M+ 2Tϑ−ς Γ(ϑ − ς + 1)M.
Which implies that,
kRC0 DTςHκ(t)k ≤ Ω2M. (15)
Adds side of inequality (14)and(15), we get
kRC0 DςTHκ(t)kE < M (Ω1+ Ω2) ∞.
Which implies that H maps bounded sets into bounded sets on E.
step 3. H maps bounded sets into equicontinuous sets in E.
Let Br be a bounded set of E as in step 2, and let κ ∈ Br. For each t1, t2 ∈ J , t1< t2, we have
|Hκ(t2) − Hκ(t1)| ≤ (σ − µ)(t2− t1) (σ + µ)Γ(ϑ − 1)
Z T 0
(T − ξ)ϑ−2|F(κ(ξ),RCDςκ(ξ))|dξ
+ 1
Γ(ϑ) Z t1
0
(t1− ξ)ϑ−1− (t2− ξ)ϑ−1
|F(κ(ξ),RCDςκ(ξ))|dξ
+ 1
Γ(ϑ) Z t2
t1
(ξ − t1)ϑ−1− (t2− ξ)ϑ−1
|F(κ(ξ),RCDςκ(ξ))|dξ
+ 1
Γ(ϑ) Z T
t2
(ξ − t1)ϑ−1− (ξ − t2)ϑ−1
|F(κ(ξ),RCDςκ(ξ))|dξ
≤
(tϑ1− tϑ2) + (t2− t1)ϑ
Γ(ϑ + 1) M+(σ − µ)|t2− t1|ϑTϑ (σ + µ)Γ(ϑ + 1) M +
(T − t1)ϑ− (T − t2)ϑ + (t2− t1)ϑ
Γ(ϑ + 1) M
≤
(tϑ1− tϑ2) + (t2− t1)ϑ
Γ(ϑ + 1) M+(σ − µ)|t2− t1|ϑTϑ (σ + µ)Γ(ϑ + 1) M +
(T − t1)ϑ− (T − t2)ϑ + (t2− t1)ϑ
Γ(ϑ + 1) M,
and,
|RC0 DTςHκ(t2) −RC0 DTςHκ(t1)| ≤ (tϑ−ς1 − tϑ−ς2 ) 2Γ(ϑ − ς + 1)M + (T − t2)ϑ−ς− (T − t1)ϑ−ς + (t2− t1)ϑ−ς
Γ(ϑ − ς + 1) M
+ (σ − µ)(t1−ς2 − t1−ς1 ) + ((T − t2)1−ς− (T − t1)1−ς) Tϑ−1M
2(σ + µ)Γ(2 − ς)Γ(ϑ) .
Hence,
kHκ(t2) − Hκ(t1)kκ≤
(tϑ1 − tϑ2) + (t2− t1)ϑ
Γ(ϑ + 1) M+(σ − µ)|t2− t1|ϑTϑ (σ + µ)Γ(ϑ + 1) M +
(T − t1)ϑ− (T − t2)ϑ + (t2− t1)ϑ
Γ(ϑ + 1) M
+ (tϑ−ς1 − tϑ−ς2 ) 2Γ(ϑ − ς + 1)M
+ (T − t2)ϑ−ς− (T − t1)ϑ−ς + (t2− t1)ϑ−ς
Γ(ϑ − ς + 1) M
+(σ − µ)(t1−ς2 − t1−ς1 ) + ((T − t2)1−ς− (T − t1)1−ς) Tϑ−1M
2(σ + µ)Γ(2 − ς)Γ(ϑ) .
Which implies that kHκ(t2) − Hκ(t1)kE → 0 as t2 → t1.By Arzela-Ascoli theorem, we conclude that H is completely continuous operator.
step 4. We show that the set ∆ dened by
∆ = {κ ∈ E, κ = ρH(κ), 0 < ρ < 1}
is bounded. Let κ ∈ ∆, for some ρ ∈ (0, 1). For each t ∈ J , we have 1
ρ|κ(t)| ≤ |(σ − µ)t + µT | (σ + µ)Γ(ϑ − 1)
Z T 0
(T − ξ)ϑ−2|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
+ 1
Γ(ϑ) Z t
0
(t − ξ)ϑ|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
+ 1
Γ(ϑ) Z T
t
(ξ − t)ϑ|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
≤ µTϑ
(σ + µ)Γ(ϑ)M+ 2Tϑ Γ(ϑ + 1)M.
Therefore,
kκk ≤ ρΩ1M. (16)
and,
1
ρ|RC0 DTςκ(t)| ≤ 1 Γ(ϑ − ς)
Z t 0
(t − ξ)ϑ−ς−1|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
+ 1
Γ(ϑ − ς) Z T
t
(ξ − t)ϑ−ς−1|F(ξ, κ(ξ),RCDςκ(ξ))|dξ +
t1−ς− (T − t)1−ς (σ − µ) + 2µΓ(2 − ς)T 2(σ + µ)Γ(2 − ς)Γ(ϑ − 1)
× Z T
0
(T − ξ)ϑ−2|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
≤ Tϑ−ς(µ + σ) + 2µTϑΓ(2 − ς)
2|µ − σ|Γ(2 − ς)Γ(ϑ) M+ 2Tϑ−ς 2Γ(ϑ − ς + 1)M
≤ Tϑ−ς(σ − µ) + 2µTϑΓ(2 − ς)
2(σ + µ)Γ(2 − ς)Γ(ϑ) M+ 2Tϑ−ς Γ(ϑ − ς + 1)M.
Therefore,
kRC0 DςTκ(t)k ≤ ρΩ2M. (17)
Adds side of inequality (16)and (17), we get
kκkE ≤ ρ(Ω1+ Ω2)M.
Hence,
kκkE < ∞.
This shows that ∆ is bounded.
As consequence of Schaefer's xed point theorem, the problem (1) has at least one solution in [0, T ].
3.3. Existence result via Karesnoslskii's xed point theorem
Lemma 3.5. (Karasnoselskii's xed point theorem) Let M a closed bounded, convex and nonempty subset of a Banach space E, let A, Bbe operator, such that
(a) Aκ + B$ ∈ M, whenever, κ, $ ∈ M, (b) A is compact and continuous,
(c) B is a contraction mapping, then there exist z ∈ M such that z = Az + Bz.
Theorem 3.6. Let F : J × R × R → R be a continuous function, and let the conditions (H1)-(H2) hold. In addition, the function F satisfying the assumptions :
(H3) There exists a nonnegative function Ω ∈ C(J ,R+) such that
|F(t, κ, $)| ≤ Ω(t) for any (t, κ, $) ∈ J × R × R, (H4) (k1+ k2)(L1+ L2) < 1 ,
Then the problem (1) has a least one solution in J . Proof. We dene two operators H1κ(t) and H2κ(t) as:
(H1κ)(t) = (σ − µ)t) + µT (σ + µ)Γ(ϑ − 1)
Z T 0
(T − ξ)ϑ−2F(ξ, κ(ξ),RCDςκ(ξ))dξ, (H2κ(t)) = − 1
Γ(ϑ) Z t
0
(t − ξ)ϑ−1F(ξ, κ(ξ),RCDςκ(ξ))dξ
− 1
Γ(ϑ) Z T
t
(ξ − t)ϑ−1F(ξ, κ(ξ),RCDςκ(ξ))dξ.
Choosing d ≥ (Ω1+ Ω2)(L1+ L2)kΩk, and we consider Bd= {κ ∈ E : kκkE ≤ d}. Step1 We shall prove that H1κ(t) + H2κ(t) ∈ Bd.
For any κ, $ ∈ Bd and for each then t ∈ J , we have
|H1κ(t) + H2$(t)| ≤ |(σ − µ)t + µT | (σ + µ)Γ(ϑ − 1)
Z T 0
(T − ξ)ϑ−2|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
+ 1
Γ(ϑ) Z t
0
(t − ξ)ϑ−1|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
− 1
Γ(ϑ) Z T
t
(ξ − t)ϑ−1|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
≤ µTϑ
(σ + µ)Γ(ϑ)(L1+ L2) kΩk + 2Tϑ
Γ(ϑ + 1)(L1+ L2)) kΩk.
Then
kH1κ(t) + H2$(t)k ≤ Ω1(L1+ L2)kΩk (18) On the other hand,
|RC0 DςTHκ(t) +RC0 DςTH$(t)| ≤ 1 Γ(ϑ − ς)
Z t 0
(t − ξ)ϑ−ς−1|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
+ 1
Γ(ϑ − ς) Z T
t
(ξ − t)ϑ−ς−1|F(ξ, κ(ξ),RCDςκ(ξ))|dξ +
t1−ς− (T − t)1−ς (σ − µ) + 2µΓ(2 − ς)T 2(σ + µ)Γ(2 − ς)Γ(ϑ − 1)
Z T 0
(T − ξ)ϑ−2|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
≤ Tϑ−ς(σ − µ) + 2µTϑΓ(2 − ς)
2(σ + µ)Γ(2 − ς)Γ(ϑ) (L1+ L2) kΩk + 2Tϑ−ς
Γ(ϑ − ς + 1)(L1+ L2) kΩk.
Hence
kRC0 DTςHκ(t) +RC0 DςTH$(t)k ≤ Ω1(L1+ L2)kΩk (19) It follows from (18) and (19) that
kH1κ(t) + H2$(t)kE ≤ (Ω1+ Ω2)(L1+ L2)kΩk ≤ d.
Hence, H1κ(t) + H2κ(t) ∈ Bd .
step2 We shall prove that H1is continuous and compact. The continuity of F implies that the operator H1
is continuous.
Now, we prove that H2 maps bounded sets into bounded sets of E.
For κ ∈ Bd,and for each t ∈ J , we have
|H1κ(t)| ≤
|(σ − µ) + µT | (σ + µ)Γ(ϑ − 1)
Z T 0
(T − ξ)ϑ−2|F(ξ, κ(ξ),RCDςκ(ξ))|dξ µTϑ
(σ + µ)Γ(ϑ − 1)(L1+ L2)kΩk.
Hence
kH1κ(t)k ≤ µTϑ
(σ + µ)Γ(ϑ − 1)(L1+ L2)kΩk, (20)
and
|RCDςH1κ(t)| ≤
((T − t)1−ς− t1−ς)(σ − µ) + 2µT Γ(2 − ς) 2(σ + µ)Γ(2 − ς)Γ(ϑ − 1)
× Z T
0
(T − ξ)ϑ−2|F(ξ, κ(ξ),RCDςκ(ξ))|dξ
≤ Tϑ−ς(σ − µ) + 2µTϑΓ(2 − ς)
2(σ + µ)Γ(2 − ς)Γ(ϑ) (L1+ L2)kΩk.
Hence
kRCDςH1κ(t)k ≤ Tϑ−ς(σ − µ) + 2µTϑΓ(2 − ς)
2(σ + µ)Γ(2 − ς)Γ(ϑ) (L1+ L2)kΩk. (21) Combining (20) and (21), we get
kH1κ(t)kE ≤
µTϑ
(σ − µ)Γ(ϑ − 1) +Tϑ−ς(σ − µ) + 2µTϑΓ(2 − ς) 2(σ + µ)Γ(2 − ς)Γ(ϑ)
(L1+ L2)kΩk, Consequently
kH1κ(t)kE ≤ ∞.
Thus, it follows the above inequality that operator H1 is uniformly bounded.
The operator H1 maps bounded sets into equicontinuous sets of E. Let t1, t2 ∈ J , t1 < t2, κ ∈ Bd, then we have :
|H1κ(t2) − H1κ(t1)| ≤ (σ − µ)(t2− t1) (σ + µ)Γ(ϑ − 1)
Z T 0
(T − ξ)ϑ−2|F(κ(ξ),RCDςκ(ξ))|dξ +(σ − µ)|t2− t1|ϑTϑ
(σ + µ)Γ(ϑ + 1) (L1+ L2)kκk.
On the other hand,
|RCDςH1κ(t2) −RCDςH1κ(t1)|
≤ (σ − µ)(t1−ς2 − t1−ς1 ) + ((T − t2)1−ς− (T − t1)1−ς) Tϑ−1(L1+ L2)kΩk 2(σ + µ)|Γ(2 − ς)Γ(ϑ)
It follows from (22) and (23) that
kH1κ(t2) − H1κ(t1)kE ≤ (σ − µ)|t2− t1|ϑTϑ
(σ + µ)Γ(ϑ + 1) (L1+ L2)kκk
+(σ − µ)(t1−ς2 − t1−ς1 ) + ((T − t2)1−ς− (T − t1)1−ς) Tϑ−1(L1+ L2)kΩk 2(σ + µ)|Γ(2 − ς)Γ(ϑ).
As t2→ t1, the right-hand side of this inequality tends to zeros.
Then as a consequence od steps, we can conclude that H1 is continuous and compact.
Step3 Now, we prove that H2 is contraction mapping . Let κ, $ ∈ E. Then, for each t ∈ J , we have
|H2κ(t) − H2$(t)| ≤ + 1 Γ(ϑ)
Z t 0
(t − ξ)ϑ−1|F(ξ, κ(ξ),RCDςκ(ξ)) − F(ξ, $(ξ),RCDς$(ξ))|dξ
+ 1
Γ(ϑ) Z T
t
(ξ − t)ϑ−1|F(ξ, κ(ξ),RCDςκ(ξ)) − F(ξ, $(ξ),RCDς$(ξ))|dξ
≤ 2Tϑ
Γ(ϑ + 1)(L1+ L2) kκ − $k + kRCDςκ −RCDς$k
Consequently we obtain
kH2κ(t) − H2$(t)k ≤ k1(L1+ L2)(kκ − $k + kRCDςκ −RCDς$k) (24) and
|RC0 DςTH2κ(t) −RC0 DςTH2κ(t)|
≤ 1
Γ(ϑ − ς) Z t
0
(t − ξ)ϑ−ς−1|F(ξ, κ(ξ),RCDςκ(ξ)) − F(ξ, $(ξ),RCDς$(ξ))|dξ
+ 1
Γ(ϑ − ς) Z T
t
(ξ − t)ϑ−ς−1|F(ξ, κ(ξ),RCDςκ(ξ)) − F(ξ, $(ξ),RCDς$(ξ))|dξ
≤ 2Tϑ−ς
Γ(ϑ − ς + 1)(L1+ L2) kκ − $k + kRCDςκ −RCDς$k , and
kRCDςH2κ(t) −RCDςH2$(t)k ≤ k2(L1+ L2)(kκ − $k + kRCDςκ −RCDς$k) (25) It follows from (24) and (25) that
kH2κ(t) − H2$(t)kE ≤ (k1+ k2)(L1+ L2) kκ − $k + kRCDςκ −RCDς$k
Using the condition (H4), we conclude that H2 is a contraction mapping. As consequence of a krasnosselski's
xed point theorem, we deduce that H has a xed point which as solution of (1).
Example 3.7. Consider following nonlinear FDE with Riesz-Caputo derivative:
RC0 D
3 2
Tκ(t) + (
√π+1)|κ(t)|
√
t2+144(1+|κ(t)|)+(1+e12π)2cos(RCD13κ(t)) = 0, t ∈ [0, 1],
$(0) + $(1) = 0, 2$0(0) + 12$0(1) = 0, (26)
Here, ϑ = 32, ς = 13, µ = 2, σ = 12, and,
F(t, κ(t),RCDςκ(t)) = (√
π + 1)|κ|
√
t2+ 144(1 + |κ|) + 1
(1 + e2π)2cos(RCD13κ(t)).
We have
|F(κ, $) − F(κ0, $0)| ≤
√π + 1
12 kκ − κ0k + 1
(1 + e2π)2k$ − $0k.
Then, the assumption (H1) is satised with L1 =
√π+1
12 , L2 = (1+e12π)2. Using the Matlab program, Ω1 = 2, 1493, Ω2 = 1, 9057.
Therefore, (L1+ L2)(Ω1+ Ω2) = 0, 9369 < 1. By using the theorem (3.2), the problem (26) has a unique solution on [0, 1].
Example 3.8. Consider following nonlinear FDE with Riesz-Caputo derivative:
RC0 D
5 3
Tκ(t) +
e−2tsin(1+|κ||κ| )
(e2π+2) + |RCD
1 2κ(t)|(
√π+1)
(|RCD12κ(t)|+1)(π+2)2
= 0, t ∈ [0, 1],
$(0) + $(1) = 0,35$0(0) + 23$0(1) = 0, (27)
Here, ϑ = 53, ς = 12, µ = 35, σ = 23, L1=
√π+1
12 , L2 = (1+e12π)2, and F(t, κ(t),RCDςκ(t)) =
e−2tsin(1+|κ||κ| )
(e2π+ 2) + |RCD12κ(t)|(
√π + 1) (|RCD12κ(t)| + 1)(π + 2)2
.
Moreover,
|F(κ, $) − F(κ0, $0)| ≤ 1
(e2π+ 2)kκ − κ0k +
√π + 1
(1 + π)2k$ − $0k.
Therefor,
|F(t, κ(t),RCDςκ(t))| ≤ e−2t e2π+2 +
√π + 1
(π + 2)2 = Ω(t).
Ω1= 2, 1493, Ω2= 1, 9057, K1 = 1, 5045, k2 = 0, 9239 .
Then (k1+ k2)(L1+ L2) = 0, 2406 < 1, (H4) is satised, by using the theorem (3.6), the problem (27) has at least one solution on [0, 1].
4. Conclusion
We have eectively achieved several necessary conditions describing the the existence and uniqueness of solutions for a class of fractional dierential equations with boundary conditions involving Riesz-Caputo fractional derivatives. Under some xed point theorems such as Banach, Schaefer, and Krasnoselskii, the necessary results have been investigated. Moreover, by giving appropriate examples, all the main results have been testied. In future such type of analysis can be established for more general type fractional dierential equations involving ψ-Riesz-Caputo fractional derivatives.
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