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Article

Existence and Uniqueness Results for Different Orders Coupled System of Fractional Integro-Differential Equations with Anti-Periodic Nonlocal Integral Boundary Conditions

1
Department of Mathematics, College of Science, Jouf University, P.O. Box 2014, Sakaka 72388, Saudi Arabia
2
Department of Mathematics and Statistics, Taif University, P.O. Box 11099, Taif 21944, Saudi Arabia
3
Department of Mathematics, College of Science, University of Jeddah, P.O. Box 80327, Jeddah 21589, Saudi Arabia
*
Author to whom correspondence should be addressed.
Symmetry 2023, 15(1), 182; https://doi.org/10.3390/sym15010182
Submission received: 6 December 2022 / Revised: 28 December 2022 / Accepted: 4 January 2023 / Published: 7 January 2023
(This article belongs to the Special Issue Applied Mathematics and Fractional Calculus II)

Abstract

:
This paper presents a new class of boundary value problems of integrodifferential fractional equations of different order equipped with coupled anti-periodic and nonlocal integral boundary conditions. We prove the existence and uniqueness criteria of the solutions by using the Leray-Schauder alternative and Banach contraction mapping principle. Examples are constructed for the illustration of our results.

1. Introduction

Fractional calculus has gained a rapid rise in popularity in the past few decades due to the nonlocal nature of the derivatives and integrals of fractional order [1]. As a matter of fact, this field incorporates the methods and concepts used to solve symmetrical differential equations with fractional derivatives. Thereby, it evolved in many theoretical and applications area. For application details in ecology, chaos and fractional dynamics, medical sciences, financial economics bio-engineering, immune system, etc., we refer the reader to the works [2,3,4,5,6,7,8,9]. For more theoretical aspects of fractional calculus, we refer the reader to the monographs [10,11,12,13,14,15,16,17,18].
During this development, nonlinear Fractional Differential Equations (FDEs) equipped with different kinds of Boundary Conditions (BCs) such as multi-point, periodic, anti-periodic, nonlocal, and integral conditions have also been widely studied and investigated. Many new results of variety boundary value problems were given in [19,20,21,22,23,24,25]. At the same time, fractional differential system subjects with different kinds of BCs also received the attention of such systems in the mathematical models with engineering and physical phenomena [26,27,28,29,30,31].
Recently, fractional Integro-Differential Equations (IDEs) with nonlocal conditions are considered a useful mathematical tool for the description of various real materials, for instance, see [32,33], and references therein. By side, several researchers have applied classical fixed point theorems to prove the existence and uniqueness results for such boundary value problems [19,31,34,35,36,37,38,39,40,41,42].
In addition, the authors in [43,44,45] investigated some coupled systems (CSs) of mixed-order FDEs with different kinds of BCs. To enrich the topic, we introduce and investigate a CS of fractional IDEs of Caputo type with different derivatives orders given by
c D q 1 [ κ 1 v ( t ) + λ 1 I x 1 θ 1 ϕ ( t , v ( t ) , u ( t ) ) ] = k ( t , v ( t ) , u ( t ) ) , 2 < q 1 3 , t [ x 1 , x 2 ] , c D q 2 [ κ 2 u ( t ) + λ 2 I x 1 θ 2 ψ ( t , v ( t ) , u ( t ) ) ] = p ( t , v ( t ) , u ( t ) ) , 1 < q 2 2 , t [ x 1 , x 2 ] ,
supplemented with coupled anti-periodic and nonlocal integral BCs:
v ( x 1 ) + v ( x 2 ) = 0 , v ( x 2 ) = 0 , v ( x 1 ) = h x 1 ξ u ( s ) d s , u ( x 1 ) + u ( x 2 ) = 0 , u ( x 2 ) = 0 ,
where c D Υ denotes the Caputo fractional differential operator of order Υ { q 1 , q 2 } , I x 1 Υ ¯ denotes the Riemann-Liouville fractional integral of order Υ ¯ { θ 1 , θ 2 } such that θ 1 , θ 2 > 1 , κ i , λ i , h , i = 1 , 2 are real constants with κ i , h 0 , ϕ , ψ , k , p : [ x 1 , x 2 ] × R × R R are given continuous functions and x 1 < ξ < x 2 .
For usefulness, we emphasize that the current study is novel, and contributes extensively to the existing results on the topic. Furthermore, new results follow as special cases of the present work.
The structure of this paper is as follows. In Section 2, we give some important definitions of fractional calculus and establish an auxiliary lemma that helps to transform the system (1) into equivalent integral equations. In Section 3, the existence and uniqueness results for the given system (1) are derived. Two examples are also presented to illustrate the obtained outcomes.

2. Preliminary Material

First, we outline some main definitions of fractional calculus.
Definition 1
([11]). Let U be an integrable function on x 1 z x 2 . The Riemann-Liouville fractional integral I x 1 ϑ of order ϑ R ( ϑ > 0 ) for U is given by
I x 1 ϑ U ( z ) = 1 Γ ( ϑ ) x 1 z ( z s ) ϑ 1 U ( s ) d s ,
where Γ is the Euler Gamma function.
Definition 2
([11]). The Caputo derivative for a function U A C r [ x 1 , x 2 ] of order ϑ ( r 1 , r ] , r N existing on [ x 1 , x 2 ] , is given by
c D ϑ U ( z ) = 1 Γ ( r ϑ ) x 1 z ( z l ) r ϑ 1 U ( r ) ( l ) d l , z [ x 1 , x 2 ] .
Lemma 1
([11]). The solution of the equation c D ϑ x ( z ) = 0 , r 1 < ϑ < r , z [ x 1 , x 2 ] , is
x ( z ) = m 0 + m 1 ( z x 1 ) + m 2 ( z x 1 ) 2 + . . . + m r 1 ( z x 1 ) r 1 ,
with m i R , i = 0 , 1 , . . . , r 1 . Moreover,
I x 1 ϑ c D ϑ x ( z ) = x ( z ) + i = 0 r 1 m i ( z x 1 ) i .
Next, we introduce an important lemma related to our new results.
Lemma 2.
For Φ , Ψ , K , P C ( [ x 1 , x 2 ] , R ) , the unique solution of the following linear system
c D q 1 [ κ 1 v ( t ) + λ 1 I x 1 θ 1 Φ ( t ) ] = K ( t ) , 2 < q 1 3 , t [ x 1 , x 2 ] , c D q 2 [ κ 2 u ( t ) + λ 2 I x 1 θ 2 Ψ ( t ) ] = P ( t ) , 1 < q 2 2 , t [ x 1 , x 2 ] ,
equipped with the BCs (2) is given by:
v ( t ) = 1 κ 1 x 1 t ( t s ) q 1 1 Γ ( q 1 ) K ( s ) d s λ 1 κ 1 x 1 t ( t s ) θ 1 1 Γ ( θ 1 ) Φ ( s ) d s + λ 1 2 κ 1 x 1 x 2 ( x 2 s ) θ 1 1 Γ ( θ 1 ) Φ ( s ) d s 1 2 κ 1 x 1 x 2 ( x 2 s ) q 1 1 Γ ( q 1 ) K ( s ) d s + ρ 1 ( t ) λ 2 x 1 x 2 ( x 2 s ) θ 2 1 Γ ( θ 2 ) Ψ ( s ) d s x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) P ( s ) d s + ρ 2 ( t ) λ 1 x 1 x 2 ( x 2 s ) θ 1 2 Γ ( θ 1 1 ) Φ ( s ) d s x 1 x 2 ( x 2 s ) q 1 2 Γ ( q 1 1 ) K ( s ) d s + ρ 3 ( t ) λ 2 x 1 x 2 ( x 2 s ) θ 2 2 Γ ( θ 2 1 ) Ψ ( s ) d s x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) P ( s ) d s + ρ 4 ( t ) x 1 ξ h λ 2 κ 2 x 1 s ( s τ ) θ 2 1 Γ ( θ 2 ) Ψ ( τ ) d τ h κ 2 x 1 s ( s τ ) q 2 1 Γ ( q 2 ) P ( τ ) d τ d s ,
u ( t ) = 1 κ 2 x 1 t ( t s ) q 2 1 Γ ( q 2 ) P ( s ) d s λ 2 κ 2 x 1 t ( t s ) θ 2 1 Γ ( θ 2 ) Ψ ( s ) d s + λ 2 2 κ 2 x 1 x 2 ( x 2 s ) θ 2 1 Γ ( θ 2 ) Ψ ( s ) d s 1 2 κ 2 x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) P ( s ) d s + ρ 5 ( t ) λ 2 x 1 x 2 ( x 2 s ) θ 2 2 Γ ( θ 2 1 ) Ψ ( s ) d s x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) P ( s ) d s ,
where
ρ 1 ( t ) = ϵ ( x 2 x 1 ) 4 κ 1 + ϵ ( t x 1 ) κ 1 ϵ ( t x 1 ) 2 2 κ 1 ( x 2 x 1 ) , ρ 2 ( t ) = ( x 2 x 1 ) 4 κ 1 + ( t x 1 ) 2 2 κ 1 ( x 2 x 1 ) , ρ 3 ( t ) = ϵ ( ξ x 2 ) ( x 2 x 1 ) 4 κ 1 + ϵ ( ξ x 2 ) ( t x 1 ) κ 1 ϵ ( ξ x 2 ) ( t x 1 ) 2 2 κ 1 ( x 2 x 1 ) , ρ 4 ( t ) = ( x 2 x 1 ) 4 ( t x 1 ) + ( t x 1 ) 2 2 ( x 2 x 1 ) , ρ 5 ( t ) = ( x 2 x 1 ) 2 κ 2 + ( t x 1 ) κ 2 ,
ϵ = h κ 1 ( ξ x 1 ) 2 κ 2 .
Proof. 
Using Lemma 1 and applying the integral operators I x 1 q 1 , I x 1 q 2 on both sides of the equations in (3), we get the general solution that can be written as
v ( t ) = 1 κ 1 x 1 t ( t s ) q 1 1 Γ ( q 1 ) K ( s ) d s λ 1 κ 1 x 1 t ( t s ) θ 1 1 Γ ( θ 1 ) Φ ( s ) d s + c 1 κ 1 + c 2 κ 1 ( t x 1 ) + c 3 κ 1 ( t x 1 ) 2 ,
v ( t ) = 1 κ 1 x 1 t ( t s ) q 1 2 Γ ( q 1 1 ) K ( s ) d s λ 1 κ 1 x 1 t ( t s ) θ 1 2 Γ ( θ 1 1 ) Φ ( s ) d s + c 2 κ 1 + 2 c 3 κ 1 ( t x 1 ) ,
u ( t ) = 1 κ 2 x 1 t ( t s ) q 2 1 Γ ( q 2 ) P ( s ) d s λ 2 κ 2 x 1 t ( t s ) θ 2 1 Γ ( θ 2 ) Ψ ( s ) d s + c 4 κ 2 + c 5 κ 2 ( t x 1 ) ,
u ( t ) = 1 κ 2 x 1 t ( t s ) q 2 2 Γ ( q 2 1 ) P ( s ) d s λ 2 κ 2 x 1 t ( t s ) θ 2 2 Γ ( θ 2 1 ) Ψ ( s ) d s + c 5 κ 2 ,
with c i R , i = 1 , . . . , 5 are unknown arbitrary constants.
Using the conditions (2) in Equations (8)–(11), we obtain a system of equations in c i ( i = 1 , . . . , 5 ) given by
2 c 1 + ( x 2 x 1 ) c 2 + ( x 2 x 1 ) 2 c 3 = I 1 , 2 c 4 + ( x 2 x 1 ) c 5 = I 2 , c 2 + 2 ( x 2 x 1 ) c 3 = I 3 , c 2 κ 1 + h ( ξ x 1 ) κ 2 c 4 + h ( ξ x 1 ) 2 2 κ 2 c 5 = I 4 , c 5 = I 5
where I i ; ( i = 1 , . . . , 5 ) are defined by
I 1 = λ 1 x 1 x 2 ( x 2 s ) θ 1 1 Γ ( θ 1 ) Φ ( s ) d s x 1 x 2 ( x 2 s ) q 1 1 Γ ( q 1 ) K ( s ) d s , I 2 = λ 2 x 1 x 2 ( x 2 s ) θ 2 1 Γ ( θ 2 ) Ψ ( s ) d s x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) P ( s ) d s , I 3 = λ 1 x 1 x 2 ( x 2 s ) θ 1 2 Γ ( θ 1 1 ) Φ ( s ) d s x 1 x 2 ( x 2 s ) q 1 2 Γ ( q 1 1 ) K ( s ) d s , I 4 = x 1 ξ h λ 2 κ 2 x 1 s ( s τ ) θ 2 1 Γ ( θ 2 ) Ψ ( τ ) d τ h κ 2 x 1 s ( s τ ) q 2 1 Γ ( q 2 ) G ( τ ) d τ d s , I 5 = λ 2 x 1 x 2 ( x 2 s ) θ 2 2 Γ ( θ 2 1 ) Ψ ( s ) d s x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) P ( s ) d s .
Solving the system (12) for c i ( i = 1 , . . . , 5 ) , we get that
c 1 = ϵ ( ξ x 2 ) ( x 2 x 1 ) 4 c 5 + 1 2 I 1 ϵ ( x 2 x 1 ) 4 I 2 ( x 2 x 1 ) 4 I 3 + κ 1 ( x 2 x 1 ) 4 I 4 , c 2 = ϵ ( ξ x 2 ) c 5 + ϵ I 2 κ 1 I 4 , c 3 = ϵ ( ξ x 2 ) 2 ( x 2 x 1 ) c 5 ϵ 2 ( x 2 x 1 ) I 2 + 1 2 ( x 2 x 1 ) I 3 + κ 1 2 ( x 2 x 1 ) I 4 , c 4 = ( x 2 x 1 ) 2 c 5 + 1 2 I 2 , c 5 = λ 2 x 1 x 2 ( x 2 s ) θ 2 2 Γ ( θ 2 1 ) Ψ ( s ) d s x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) P ( s ) d s ,
where ϵ is given by (7). Inserting the values of c i ( i = 1 , . . . 5 ) in (8) and (9) together with notations (6), we get (4) and (5). The converse follows by direct computation. This completes the proof. □

3. Existence and Uniqueness Results

Let V = { v | v C ( [ x 1 , x 2 ] , R ) } be a Banach space endowed with the norm
v = sup l [ x 1 , x 2 ] | v ( l ) | .
Obviously the product space ( V × V , . ) is also a Banach space with norm ( v , u ) = v + u for ( v , u ) V × V .
In view of Lemma 2, we define an operator J : V × V V × V as
J ( v , u ) ( t ) : = ( J 1 ( v , u ) ( t ) , J 2 ( v , u ) ( t ) ) ,
where
J 1 ( v , u ) ( t ) = 1 κ 1 x 1 t ( t s ) q 1 1 Γ ( q 1 ) k ( s , v ( s ) , u ( s ) ) d s λ 1 κ 1 x 1 t ( t s ) θ 1 1 Γ ( θ 1 ) ϕ ( s , v ( s ) , u ( s ) ) d s + λ 1 2 κ 1 x 1 x 2 ( x 2 s ) θ 1 1 Γ ( θ 1 ) ϕ ( s , v ( s ) , u ( s ) ) d s 1 2 κ 1 x 1 x 2 ( x 2 s ) q 1 1 Γ ( q 1 ) k ( s , v ( s ) , u ( s ) ) d s + ρ 1 ( t ) λ 2 x 1 x 2 ( x 2 s ) θ 2 1 Γ ( θ 2 ) ψ ( s , v ( s ) , u ( s ) ) d s x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) p ( s , v ( s ) , u ( s ) ) d s + ρ 2 ( t ) λ 1 x 1 x 2 ( x 2 s ) θ 1 2 Γ ( θ 1 1 ) ϕ ( s , v ( s ) , u ( s ) ) d s x 1 x 2 ( x 2 s ) q 1 2 Γ ( q 1 1 ) k ( s , v ( s ) , u ( s ) ) d s + ρ 3 ( t ) λ 2 x 1 x 2 ( x 2 s ) θ 2 2 Γ ( θ 2 1 ) ψ ( s , v ( s ) , u ( s ) ) d s x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) p ( s , v ( s ) , u ( s ) ) d s + ρ 4 ( t ) x 1 ξ h λ 2 κ 2 x 1 s ( s τ ) θ 2 1 Γ ( θ 2 ) ψ ( τ , v ( τ ) , u ( τ ) ) d τ h κ 2 x 1 s ( s τ ) q 2 1 Γ ( q 2 ) p ( τ , v ( τ ) , u ( τ ) ) d τ d s ,
J 2 ( v , u ) ( t ) = 1 κ 2 x 1 t ( t s ) q 2 1 Γ ( q 2 ) p ( s , v ( s ) , u ( s ) ) d s λ 2 κ 2 x 1 t ( t s ) θ 2 1 Γ ( θ 2 ) ψ ( s , v ( s ) , u ( s ) ) d s + λ 2 2 κ 2 x 1 x 2 ( x 2 s ) θ 2 1 Γ ( θ 2 ) ψ ( s , v ( s ) , u ( s ) ) d s 1 2 κ 2 x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) p ( s , v ( s ) , u ( s ) ) d s + ρ 5 ( t ) λ 2 x 1 x 2 ( x 2 s ) θ 2 2 Γ ( θ 2 1 ) ψ ( s , v ( s ) , u ( s ) ) d s x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) p ( s , v ( s ) , u ( s ) ) d s ,
where ρ i ( t ) , i = 1 , . . . . , 5 are given by (6). For brevity, we use the subsequent notations.
M 1 = 3 ( x 2 x 1 ) q 1 2 | κ 1 | | Γ ( q 1 + 1 ) + ρ ˜ 2 ( x 2 x 1 ) q 1 1 Γ ( q 1 ) ,
M 2 = ρ ˜ 1 ( x 2 x 1 ) q 2 Γ ( q 2 + 1 ) + ρ ˜ 3 ( x 2 x 1 ) q 2 1 Γ ( q 2 ) + ρ ˜ 4 | h | ( ξ x 1 ) q 2 + 1 | κ 2 | Γ ( q 2 + 2 ) ,
M 3 = 3 | λ 1 | ( x 2 x 1 ) θ 1 2 | κ 1 | | Γ ( θ 1 + 1 ) + ρ ˜ 2 | λ 1 | ( x 2 x 1 ) θ 1 1 Γ ( θ 1 ) ,
M 4 = ρ ˜ 1 | λ 2 | ( x 2 x 1 ) θ 2 Γ ( θ 2 + 1 ) + ρ ˜ 3 | λ 2 | ( x 2 x 1 ) θ 2 1 Γ ( θ 2 ) + ρ ˜ 4 | h | | λ 2 | ( ξ x 1 ) θ 2 + 1 | κ 2 | Γ ( θ 2 + 2 ) ,
M 5 = 3 ( x 2 x 1 ) q 2 2 | κ 2 | | Γ ( q 2 + 1 ) + ρ ˜ 5 ( x 2 x 1 ) q 2 1 Γ ( q 2 ) ,
M 6 = 3 | λ 2 | ( x 2 x 1 ) θ 2 2 | κ 2 | | Γ ( θ 2 + 1 ) + ρ ˜ 5 | λ 2 | ( x 2 x 1 ) θ 2 1 Γ ( θ 2 ) ,
where ρ ˜ i = sup t [ x 1 , x 2 ] | ρ i ( t ) | , i = 1 , , 5 .

3.1. Existence Result via Leray-Schauder Alternative

Lemma 3
([46]). (Leray-Schauder alternative) Let L : E E be a completely continuous operator. Let X ( L ) = { x E : x = δ L ( x ) f o r s o m e 0 < δ < 1 } . Then either the set X ( L ) is unbounded or L has at least one fixed point.
Theorem 1.
Assume the following assumption holds
( H 1 )
k , p , ϕ , ψ : [ x 1 , x 2 ] × R 2 R are continuous functions and there exist real constants γ i , ν i , μ i , ϖ i 0 ( i = 1 , 2 ) and γ 0 , ν 0 , μ 0 , ϖ 0 > 0 such that, for all t [ x 1 , x 2 ] and v , u R ,
| k ( t , v , u ) | γ 0 + γ 1 | v | + γ 2 | u | , | p ( t , v , u ) | ν 0 + ν 1 | v | + ν 2 | u | , | ϕ ( t , v , u ) | μ 0 + μ 1 | v | + μ 2 | u | , | ψ ( t , v , u ) | ϖ 0 + ϖ 1 | v | + ϖ 2 | u | ,
then the CS (1) and (2) has at least one solution on [ x 1 , x 2 ] if
N 1 = γ 1 M 1 + ν 1 ( M 2 + M 5 ) + μ 1 M 3 + ϖ 1 ( M 4 + M 6 ) < 1 ,
N 2 = γ 2 M 1 + ν 2 ( M 2 + M 5 ) + μ 2 M 3 + ϖ 2 ( M 4 + M 6 ) < 1 .
where M j , j = 1 , , 6 are given by (17)–(22) respectively.
Proof. 
First, we demonstrate that the operator J : V × V V × V is completely continuous. By continuity of the functions k , p , ϕ and ψ , it follows that the operators J 1 and J 2 are continuous. In consequence, the operator J is continuous. Let G V × V be a bounded set. Then ( v , u ) G , there exist positive constants L n , n = 1 , 2 , 3 , 4 such that:
| k ( t , v ( t ) , u ( t ) ) | L 1 , | p ( t , v ( t ) , u ( t ) ) | L 2 , | ϕ ( t , v ( t ) , u ( t ) ) | L 3 , | ψ ( t , v ( t ) , u ( t ) ) | L 4 .
Then, for any ( v , u ) G , we have
| J 1 ( v , u ) ( t ) | 1 | κ 1 | x 1 t ( t s ) q 1 1 Γ ( q 1 ) | k ( s , v ( s ) , u ( s ) ) | d s + | λ 1 | | κ 1 | x 1 t ( t s ) θ 1 1 Γ ( θ 1 ) | ϕ ( s , v ( s ) , u ( s ) ) | d s + | λ 1 | 2 | κ 1 | x 1 x 2 ( x 2 s ) θ 1 1 Γ ( θ 1 ) | ϕ ( s , v ( s ) , u ( s ) ) | d s + 1 2 | κ 1 | x 1 x 2 ( x 2 s ) q 1 1 Γ ( q 1 ) | k ( s , v ( s ) , u ( s ) ) | d s + | ρ 1 ( t ) | | λ 2 | x 1 x 2 ( x 2 s ) θ 2 1 Γ ( θ 2 ) | ψ ( s , v ( s ) , u ( s ) ) | d s + x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) | p ( s , v ( s ) , u ( s ) ) | d s + | ρ 2 ( t ) | | λ 1 | x 1 x 2 ( x 2 s ) θ 1 2 Γ ( θ 1 1 ) | ϕ ( s , v ( s ) , u ( s ) ) | d s + x 1 x 2 ( x 2 s ) q 1 2 Γ ( q 1 1 ) | k ( s , v ( s ) , u ( s ) ) | d s + | ρ 3 ( t ) | | λ 2 | x 1 x 2 ( x 2 s ) θ 2 2 Γ ( θ 2 1 ) | ψ ( s , v ( s ) , u ( s ) ) | d s + x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) | p ( s , v ( s ) , u ( s ) ) | d s + | ρ 4 ( t ) | [ x 1 ξ ( | h | | λ 2 | | κ 2 | x 1 s ( s τ ) θ 2 1 Γ ( θ 2 ) | ψ ( τ , v ( τ ) , u ( τ ) ) | d τ + | h | | κ 2 | x 1 s ( s τ ) q 2 1 Γ ( q 2 ) | p ( τ , v ( τ ) , u ( τ ) ) | d τ ) d s ] L 1 1 | κ 1 | x 1 t ( t s ) q 1 1 Γ ( q 1 ) d s + 1 2 | κ 1 | x 1 x 2 ( x 2 s ) q 1 1 Γ ( q 1 ) d s + | ρ 2 ( t ) | x 1 x 2 ( x 2 s ) q 1 2 Γ ( q 1 1 ) d s + L 2 | ρ 1 ( t ) | x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) d s + | ρ 3 ( t ) | x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) d s + | ρ 4 ( t ) | | h | | κ 2 | x 1 ξ x 1 s ( s τ ) q 2 1 Γ ( q 2 ) d τ d s + L 3 | λ 1 | | κ 1 | x 1 t ( t s ) θ 1 1 Γ ( θ 1 ) d s + | λ 1 | 2 | κ 1 | x 1 x 2 ( x 2 s ) θ 1 1 Γ ( θ 1 ) d s + | ρ 2 ( t ) | | λ 1 | x 1 x 2 ( x 2 s ) θ 1 2 Γ ( θ 1 1 ) d s + L 4 { | ρ 1 ( t ) | | λ 2 | x 1 x 2 ( x 2 s ) θ 2 1 Γ ( θ 2 ) d s + | ρ 3 ( t ) | | λ 2 | x 1 x 2 ( x 2 s ) θ 2 2 Γ ( θ 2 1 ) d s + | ρ 4 ( t ) | | h | | λ 2 | | κ 2 | x 1 ξ x 1 s ( s τ ) θ 2 1 Γ ( θ 2 ) d τ d s } ,
taking the norm for t [ x 1 , x 2 ] and using the notations (17)–(20) yields
J 1 ( v , u ) L 1 M 1 + L 2 M 2 + L 3 M 3 + L 4 M 4 .
Similarly, we have
J 2 ( v , u ) L 2 M 5 + L 4 M 6 .
From above inequalities (26) and (27), we deduce that J 1 and J 2 are uniformly bounded, which implies that
J ( v , u ) L 1 M 1 + L 2 ( M 2 + M 5 ) + L 3 M 3 + L 4 ( M 4 + M 6 ) .
Hence the operator J is uniformly bounded.
Next, we prove that J is equicontinuous. Let t 1 , t 2 [ x 1 , x 2 ] with t 1 < t 2 . Then we get
| J 1 ( v , u ) ( t 2 ) J 1 ( v , u ) ( t 1 ) | 1 | κ 1 | [ x 1 t 1 | ( t 2 s ) q 1 1 ( t 1 s ) q 1 1 | Γ ( q 1 ) | k ( s , v ( s ) , u ( s ) ) | d s + t 1 t 2 | ( t 2 s ) q 1 1 | Γ ( q 1 ) | k ( s , v ( s ) , u ( s ) ) | ] d s + | λ 1 | | κ 1 | [ x 1 t 1 | ( t 2 s ) θ 1 1 ( t 1 s ) θ 1 1 | Γ ( θ 1 ) | ϕ ( s , v ( s ) , u ( s ) ) | d s + t 1 t 2 | ( t 2 s ) θ 1 1 | Γ ( θ 1 ) | ϕ ( s , v ( s ) , u ( s ) ) | d s ] + | ρ 1 ( t 2 ) ρ 1 ( t 1 ) | | λ 2 | x 1 x 2 ( x 2 s ) θ 2 1 Γ ( θ 2 ) | ψ ( s , v ( s ) , u ( s ) ) | d s + x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) | p ( s , v ( s ) , u ( s ) ) | d s + | ρ 2 ( t 2 ) ρ 2 ( t 1 ) | | λ 1 | x 1 x 2 ( x 2 s ) θ 1 2 Γ ( θ 1 1 ) | ϕ ( s , v ( s ) , u ( s ) ) | d s + x 1 x 2 ( x 2 s ) q 1 2 Γ ( q 1 1 ) | k ( s , v ( s ) , u ( s ) ) | d s + | ρ 3 ( t 2 ) ρ 3 ( t 1 ) | | λ 2 | x 1 x 2 ( x 2 s ) θ 2 2 Γ ( θ 2 1 ) | ψ ( s , v ( s ) , u ( s ) ) | d s + x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) | p ( s , v ( s ) , u ( s ) ) | d s + | ρ 4 ( t 2 ) ρ 4 ( t 1 ) | [ x 1 ξ ( | h | | λ 2 | | κ 2 | x 1 s ( s τ ) θ 2 1 Γ ( θ 2 ) | ψ ( τ , v ( τ ) , u ( τ ) ) | d τ + | h | | κ 2 | x 1 s ( s τ ) q 2 1 Γ ( q 2 ) | p ( τ , v ( τ ) , u ( τ ) ) | d τ ) d s ] L 1 { 1 | κ 1 | Γ ( q 1 + 1 ) 2 ( t 2 t 1 ) q 1 + | ( t 2 x 1 ) q 1 ( t 1 x 1 ) q 1 | + | ρ 2 ( t 2 ) ρ 2 ( t 1 ) | x 1 x 2 ( x 2 s ) q 1 2 Γ ( q 1 1 ) d s } + L 2 { | ρ 1 ( t 2 ) ρ 1 ( t 1 ) | x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) d s + | ρ 3 ( t 2 ) ρ 3 ( t 1 ) | x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) d s + | ρ 4 ( t 2 ) ρ 4 ( t 1 ) | | h | | κ 2 | x 1 ξ x 1 s ( s τ ) q 2 1 Γ ( q 2 ) d τ d s } + L 3 { | λ 1 | | κ 1 | Γ ( θ 1 + 1 ) 2 ( t 2 t 1 ) θ 1 + | ( t 2 x 1 ) θ 1 ( t 1 x 1 ) θ 1 1 | + | ρ ¯ 2 ( t 2 ) ρ ¯ 2 ( t 1 ) | | λ 1 | x 1 x 2 ( x 2 s ) θ 1 2 Γ ( θ 1 1 ) d s } + L 4 { | ρ 1 ( t 2 ) ρ 1 ( t 1 ) | | λ 2 | x 1 x 2 ( x 2 s ) θ 2 1 Γ ( θ 2 ) d s + | ρ 3 ( t 2 ) ρ 3 ( t 1 ) | | λ 2 | x 1 x 2 ( x 2 s ) θ 2 2 Γ ( θ 2 1 ) d s + | ρ 4 ( t 2 ) ρ 4 ( t 1 ) | | h | | λ 2 | | κ 2 | x 1 ξ x 1 s ( s τ ) θ 2 1 Γ ( θ 2 ) d τ d s } ,
which imply that | J 1 ( v , u ) ( t 2 ) J 1 ( v , u ) ( t 1 ) | 0 independent of ( v , u ) G as t 2 t 1 . In a similar way, we get
| J 2 ( v , u ) ( t 2 ) J 2 ( v , u ) ( t 1 ) | 0
as t 2 t 1 . Thus J is equicontinuous. Therefore, by Arzela-Ascoli’s theorem, it follows that J is compact (completely continuous).
Finally, we ought to prove that Z ( J ) = { ( v , u ) V × V : ( v , u ) = δ J ( v , u ) ; 0 δ 1 } is bounded. Let ( v , u ) Z ( J ) . Then ( v , u ) = δ J ( v , u ) . For every t [ x 1 , x 2 ] , we have
v ( t ) = δ J 1 ( v , u ) ( t ) , u ( t ) = δ J 2 ( v , u ) ( t ) .
Using ( H 1 ) in (1), we get
| J 1 ( v , u ) ( t ) | 1 | κ 1 | x 1 t ( t s ) q 1 1 Γ ( q 1 ) γ 0 + γ 1 | v ( s ) | + γ 2 | u ( s ) | d s + | λ 1 | | κ 1 | x 1 t ( t s ) θ 1 1 Γ ( θ 1 ) μ 0 + μ 1 | v ( s ) | + μ 2 | u ( s ) | d s + | λ 1 | 2 | κ 1 | x 1 x 2 ( x 2 s ) θ 1 1 Γ ( θ 1 ) μ 0 + μ 1 | v ( s ) | + μ 2 | u ( s ) | d s + 1 2 | κ 1 | x 1 x 2 ( x 2 s ) q 1 1 Γ ( q 1 ) γ 0 + γ 1 | v ( s ) | + γ 2 | u ( s ) | d s + | ρ 1 ( t ) | { | λ 2 | x 1 x 2 ( x 2 s ) θ 2 1 Γ ( θ 2 ) ϖ 0 + ϖ 1 | v ( s ) | + ϖ 2 | u ( s ) | d s + x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) ν 0 + ν 1 | v ( s ) | + ν 2 | u ( s ) | d s } + | ρ 2 ( t ) | { | λ 1 | x 1 x 2 ( x 2 s ) θ 1 2 Γ ( θ 1 1 ) μ 0 + μ 1 | v ( s ) | + μ 2 | u ( s ) | d s + x 1 x 2 ( x 2 s ) q 1 2 Γ ( q 1 1 ) γ 0 + γ 1 | v ( s ) | + γ 2 | u ( s ) | d s } + | ρ 3 ( t ) | { | λ 2 | x 1 x 2 ( x 2 s ) θ 2 2 Γ ( θ 2 1 ) ϖ 0 + ϖ 1 | v ( s ) | + ϖ 2 | u ( s ) | d s + x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) ν 0 + ν 1 | v ( s ) | + ν 2 | u ( s ) | d s } + | ρ 4 ( t ) | { x 1 ξ ( | h | | λ 2 | | κ 2 | x 1 s ( s τ ) θ 2 1 Γ ( θ 2 ) ϖ 0 + ϖ 1 | v ( τ ) | + ϖ 2 | u ( τ ) | d τ + | h | | κ 2 | x 1 s ( s τ ) q 2 1 Γ ( q 2 ) ν 0 + ν 1 | v ( τ ) | + ν 2 | u ( τ ) | d τ ) d s } ,
which implies that
v γ 0 M 1 + ν 0 M 2 + μ 0 M 3 + ϖ 0 M 4 + γ 1 M 1 + ν 1 M 2 + μ 1 M 3 + ϖ 1 M 4 v + γ 2 M 1 + ν 2 M 2 + μ 2 M 3 + ϖ 2 M 4 u .
Similarly, we get
u ν 0 M 5 + ϖ 0 M 6 + ν 1 M 5 + ϖ 1 M 6 v + ν 2 M 5 + ϖ 2 M 6 u .
From inequalities (29) and (30), we have
( v , u ) 1 N γ 0 M 1 + ν 0 ( M 2 + M 5 ) + μ 0 M 3 + ϖ 0 ( M 4 + M 6 ) ,
with N = min 1 N 1 , 1 N 2 . The inequality (31) shows that Z ( J ) is bounded. Hence, J has at least one fixed point according to Lemma 3. Thus, there is at least one solution on [ x 1 , x 2 ] for the CS (1) and (2). □
Example 1.
Consider the CS of fractional differential equations given by
c D 7 3 1 3 v ( t ) + 2 110 I 23 5 ϕ ( t , v ( t ) , u ( t ) ) = k ( t , v ( t ) , u ( t ) ) , c D 5 4 7 9 u ( t ) + 3 70 I 11 3 ψ ( t , v ( t ) , u ( t ) ) = p ( t , v ( t ) , u ( t ) ) , t [ 0 , 1 ] ,
with the BCs
v ( 0 ) + v ( 1 ) = 0 , v ( 1 ) = 0 , v ( 0 ) = 3 125 0 2 / 5 u ( s ) d s , u ( 0 ) + u ( 1 ) = 0 , u ( 1 ) = 0 .
Here q 1 = 7 / 3 , q 2 = 5 / 4 , θ 1 = 23 / 5 , θ 2 = 11 / 3 , h = 3 / 125 , ξ = 2 / 5 with
k ( t , v ( t ) , u ( t ) ) = 3 t 2 6 + t 4 + sin v ( t ) 49 + t 2 + u ( t ) | v ( t ) | 70 ( 1 + | v ( t ) | ) , p ( t , v ( t ) , u ( t ) ) = 2 t 3 + 8 sin v ( t ) | tan 1 u ( t ) | 16 π ( t 3 + 1 ) + u ( t ) 19 , ϕ ( t , v ( t ) , u ( t ) ) = 3 11 + 20 v ( t ) ( t 2 + 8 ) 2 + 4 125 + t 2 3 u ( t ) , ψ ( t , v ( t ) , u ( t ) ) = ( t + 6 70 π ) tan 1 v ( t ) + v ( t ) | cos u ( t ) | t 5 + 22 + t 2 + 8 9 sin u ( t ) .
Clearly,
| k ( t , v ( t ) , u ( t ) ) | 1 2 + 1 7 v + 1 70 u , | p ( t , v ( t ) , u ( t ) ) | 2 3 + 1 4 v + 1 19 u , | ϕ ( t , v ( t ) , u ( t ) ) | 3 11 + 5 16 v + 4 5 u , | ψ ( t , v ( t ) , u ( t ) ) | 1 20 + 1 22 v + 1 3 u ,
and hence γ 0 = 1 2 , γ 1 = 1 7 , γ 2 = 1 70 , ν 0 = 2 3 , ν 1 = 1 4 , ν 2 = 1 19 , μ 0 = 3 11 , μ 1 = 5 16 , μ 2 = 4 5 , ϖ 0 = 1 20 , ϖ 1 = 1 22 and ϖ 2 = 1 3 . Using (23) and (24) with the given data we find that N 1 0.836430 < 1 , N 2 0.583054 < 1 . Therefore, by Theorem 1, the problem (32) and (33) has at least one solution on [ 0 , 1 ] .

3.2. Uniqueness Result via Banach’s Fixed Point Theorem

Theorem 2.
Assume the following assumption holds
( H 2 )
k , p , ϕ , ψ : [ x 1 , x 2 ] × R 2 R are continuous functions and there exist positive constants l m , m = 1 , , 4 such that t [ x 1 , x 2 ] , v i , u i , i = 1 , 2 R we have
| k ( t , v 1 , u 1 ) k ( t , v 2 , u 2 ) | l 1 | v 1 v 2 | + | u 1 u 2 | ,
| p ( t , v 1 , u 1 ) p ( t , v 2 , u 2 ) | l 2 | v 1 v 2 | + | u 1 u 2 | ,
| ϕ ( t , v 1 , u 1 ) ϕ ( t , v 2 , u 2 ) | l 3 | v 1 v 2 | + | u 1 u 2 | ,
| ψ ( t , v 1 , u 1 ) ψ ( t , v 2 , u 2 ) | l 4 | v 1 v 2 | + | u 1 u 2 | ,
then the CS (1) and (2) has a unique solution on [ x 1 , x 2 ] , provided that
M * = M 1 l 1 + ( M 2 + M 5 ) l 2 + M 3 l 3 + ( M 4 + M 6 ) l 4 < 1 ,
where M j , ( j = 1 , . . . , 6 ) are given by (17)–(22).
Proof. 
Define l 1 * = sup t [ x 1 , x 2 ] | k ( t , 0 , 0 ) | < , l 2 * = sup t [ x 1 , x 2 ] | p ( t , 0 , 0 ) | < , l 3 * = sup t [ x 1 , x 2 ] | ϕ ( t , 0 , 0 ) | < , l 4 * = sup t [ x 1 , x 2 ] | ψ ( t , 0 , 0 ) | < and K > 0 such that
K > M 1 l 1 * + ( M 2 + M 5 ) l 2 * + M 3 l 3 * + ( M 4 + M 6 ) l 4 * 1 ( M 1 l 1 + ( M 2 + M 5 ) l 2 + M 3 l 3 + ( M 4 + M 6 ) l 4 )
Firstly, we show that J B K B K , where
B K = { ( v , u ) V × V : ( v , u ) K }
For ( v , u ) B K , t [ x 1 , x 2 ] and by the assumption ( H 2 ) , we have
| k ( t , v ( t ) , u ( t ) | | k ( t , v ( t ) , u ( t ) ) k ( t , 0 , 0 ) | + | k ( t , 0 , 0 ) | l 1 | v ( t ) | + | u ( t ) | + l 1 * l 1 v V + u U + l 1 * l 1 K + l 1 * .
In the same manner, we can get,
| p ( t , v ( t ) , u ( t ) | l 2 K + l 2 * , | ϕ ( t , v ( t ) , u ( t ) | l 3 K + l 3 * , | ψ ( t , v ( t ) , u ( t ) | l 4 K + l 4 * .
Therefore, we have
| J 1 ( v , u ) ( t ) | l 1 K + l 1 * { 1 | κ 1 | x 1 t ( t s ) q 1 1 Γ ( q 1 ) d s + 1 2 | κ 1 | x 1 x 2 ( x 2 s ) q 1 1 Γ ( q 1 ) d s + | ρ 2 ( t ) | x 1 x 2 ( x 2 s ) q 1 2 Γ ( q 1 1 ) d s } + l 2 K + l 2 * { | ρ 1 ( t ) | x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) d s + | ρ 3 ( t ) | x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) d s + | ρ 4 ( t ) | | h | | κ 2 | x 1 ξ x 1 s ( s τ ) q 2 1 Γ ( q 2 ) d τ d s } + l 3 K + l 3 * { | λ 1 | | κ 1 | x 1 t ( t s ) θ 1 1 Γ ( θ 1 ) d s + | λ 1 | 2 | κ 1 | x 1 x 2 ( x 2 s ) θ 1 1 Γ ( θ 1 ) d s + | ρ 2 ( t ) | | λ 1 | x 1 x 2 ( x 2 s ) θ 1 2 Γ ( θ 1 1 ) d s } + l 4 K + l 4 * { | ρ 1 ( t ) | | λ 2 | x 1 x 2 ( x 2 s ) θ 2 1 Γ ( θ 2 ) d s + | ρ 3 ( t ) | | λ 2 | x 1 x 2 ( x 2 s ) θ 2 2 Γ ( θ 2 1 ) d s + | ρ 4 ( t ) | | h | | λ 2 | | κ 2 | x 1 ξ x 1 s ( s τ ) θ 2 1 Γ ( θ 2 ) d τ d s } , M 1 l 1 + M 2 l 2 + M 3 l 3 + M 4 l 4 K + M 1 l 1 * + M 2 l 2 * + M 3 l 3 * + M 4 l 4 * .
In consequence, we get
J 1 ( v , u ) M 1 l 1 + M 2 l 2 + M 3 l 3 + M 4 l 4 K + M 1 l 1 * + M 2 l 2 * + M 3 l 3 * + M 4 l 4 * .
Likewise, we can find that
J 2 ( v , u ) ( M 5 l 2 + M 6 l 4 ) K + M 5 l 2 * + M 6 l 4 * ,
and consequently, we get
J ( v , u ) M 1 l 1 + ( M 2 + M 5 ) l 2 + M 3 l 3 + ( M 4 + M 6 ) l 4 K + M 1 l 1 * + ( M 2 + M 5 ) l 2 * + M 3 l 3 * + ( M 4 + M 6 ) l 4 * K .
which implies that JB K B K .
Next, we show that the operator J is a contraction. For that, let v i , u i B K ; i = 1 , 2 and for each t [ x 1 , x 2 ] . Then we have
| J 1 ( v 1 , u 1 ) ( t ) J 1 ( v 2 , u 2 ) ( t ) | 1 | κ 1 | x 1 t ( t s ) q 1 1 Γ ( q 1 ) | k ( s , v 1 ( s ) , u 1 ( s ) ) k ( s , v 2 ( s ) , u 2 ( s ) ) | d s + | λ 1 | | κ 1 | x 1 t ( t s ) θ 1 1 Γ ( θ 1 ) | ϕ ( s , v 1 ( s ) , u 1 ( s ) ) ϕ ( s , v 2 ( s ) , u 2 ( s ) ) | d s + | λ 1 | 2 | κ 1 | x 1 x 2 ( x 2 s ) θ 1 1 Γ ( θ 1 ) | ϕ ( s , v 1 ( s ) , u 1 ( s ) ) ϕ ( s , v 2 ( s ) , u 2 ( s ) ) | d s + 1 2 | κ 1 | x 1 x 2 ( x 2 s ) q 1 1 Γ ( q 1 ) | k ( s , v 1 ( s ) , u 1 ( s ) ) k ( s , v 2 ( s ) , u 2 ( s ) ) | d s + | ρ 1 ( t ) | { | λ 2 | x 1 x 2 ( x 2 s ) θ 2 1 Γ ( θ 2 ) | ψ ( s , v 1 ( s ) , u 1 ( s ) ) ψ ( s , v 2 ( s ) , u 2 ( s ) ) | d s + x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) | p ( s , v 1 ( s ) , u 1 ( s ) ) p ( s , v 2 ( s ) , u 2 ( s ) ) | d s } + | ρ 2 ( t ) | { | λ 1 | x 1 x 2 ( x 2 s ) θ 1 2 Γ ( θ 1 1 ) | ϕ ( s , v 1 ( s ) , u 1 ( s ) ) ϕ ( s , v 2 ( s ) , u 2 ( s ) ) | d s + x 1 x 2 ( x 2 s ) q 1 2 Γ ( q 1 1 ) | k ( s , v 1 ( s ) , u 1 ( s ) ) k ( s , v 2 ( s ) , u 2 ( s ) ) | d s } + | ρ 3 ( t ) | { | λ 2 | x 1 x 2 ( x 2 s ) θ 2 2 Γ ( θ 2 1 ) | ψ ( s , v 1 ( s ) , u 1 ( s ) ) ψ ( s , v 2 ( s ) , u 2 ( s ) ) | d s + x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) | p ( s , v 1 ( s ) , u 1 ( s ) ) p ( s , v 2 ( s ) , u 2 ( s ) ) | d s } + | ρ 4 ( t ) | { x 1 ξ ( | h | | λ 2 | | κ 2 | x 1 s ( s τ ) θ 2 1 Γ ( θ 2 ) | ψ ( τ , v 1 ( τ ) , u 1 ( τ ) ) ψ ( τ , v 2 ( τ ) , u 2 ( τ ) ) | d τ + | h | | κ 2 | x 1 s ( s τ ) q 2 1 Γ ( q 2 ) | p ( τ , x 1 ( τ ) , y 1 ( τ ) ) p ( τ , v 2 ( τ ) , u 2 ( τ ) ) | d τ ) d s } .
By applying ( H 2 ) , we have
J 1 ( v 1 , u 1 ) J 1 ( v 2 , u 2 ) [ M 1 l 1 + M 2 l 2 + M 3 l 3 + M 4 l 4 ] ( v 1 v 2 + u 1 u 2 ) .
Similarly, we find
J 2 ( v 1 , u 1 ) J 1 ( v 2 , u 2 ) [ M 5 l 2 + M 6 l 4 ] ( v 1 v 2 + u 1 u 2 ) .
It follows from (39) and (40) that
J ( v 1 , u 1 ) J ( v 2 , u 2 ) [ M 1 l 1 + ( M 2 + M 5 ) l 2 + M 3 l 3 + ( M 4 + M 6 ) l 4 ] ( v 1 v 2 + u 1 u 2 ) .
The inequalities (38) and (41) shows that J is a contraction. Due to the Banach fixed point theorem, the operator J has a unique fixed point that corresponds to the unique solution of the system (1) and (2) on [ x 1 , x 2 ] . □
Example 2.
Consider the same system in Example (3.2) with
k ( t , v ( t ) , u ( t ) ) = 2 t 3 225 + t 8 | v ( t ) | 1 + | v ( t ) | + cos u ( t ) , p ( t , v ( t ) , u ( t ) ) = e 4 t 13 sin v ( t ) + u ( t ) + ln 7 , ϕ ( t , v ( t ) , u ( t ) ) = tan 1 ( t ) + 1 12 π sin 2 π v ( t ) + | u ( t ) | 6 ( 1 + | u ( t ) | ) , ψ ( t , v ( t ) , u ( t ) ) = 1 240 sin u ( t ) + 3 e t 720 v ( t ) .
Clearly,
| k ( t , v 1 , u 1 ) k ( t , v 2 , u 2 ) | l 1 ( v 1 v 2 + u 1 u 2 ) w i t h l 1 = 2 / 15 | p ( t , v 1 , u 1 ) p ( t , v 2 , u 2 ) | l 2 ( v 1 v 2 + u 1 u 2 ) w i t h l 2 = 1 / 13 | ϕ ( t , v 1 , u 1 ) ϕ ( t , v 2 , u 2 ) | l 3 ( v 1 v 2 + u 1 u 2 ) w i t h l 3 = 1 / 6 | ψ ( t , v 1 , u 1 ) ψ ( t , v 2 , u 2 ) | l 4 ( v 1 v 2 + u 1 u 2 ) w i t h l 4 = 1 / 240
Moreover, it is found that M * 0.402293 < 1 . So, the hypothesis of Theorem 2 is satisfied. Based on Theorem 2, there is a unique solution for the system (32) equipped with the conditions (33) on [ 0 , 1 ] .

4. Conclusions

In this work, we have successfully proved the existence and uniqueness results for a CS of nonlinear fractional IDEs of different orders type Caputo complemented with coupled anti-periodic and nonlocal integral BCs by using the Leray Schauder alternative and Banach fixed point theorem. As a special case, if we take λ 1 = λ 2 = 0 , consequently, our outcomes correspond to the solutions of the form:
J 1 * ( v , u ) ( t ) = 1 κ 1 x 1 t ( t s ) q 1 1 Γ ( q 1 ) k ( s , v ( s ) , u ( s ) ) d s 1 2 κ 1 x 1 x 2 ( x 2 s ) q 1 1 Γ ( q 1 ) k ( s , v ( s ) , u ( s ) ) d s ρ 1 ( t ) x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) p ( s , v ( s ) , u ( s ) ) d s ρ 2 ( t ) x 1 x 2 ( x 2 s ) q 1 2 Γ ( q 1 1 ) k ( s , v ( s ) , u ( s ) ) d s ρ 3 ( t ) x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) p ( s , v ( s ) , u ( s ) ) d s ρ 4 ( t ) x 1 ξ h κ 2 x 1 s ( s τ ) q 2 1 Γ ( q 2 ) p ( τ , v ( τ ) , u ( τ ) ) d τ d s ,
J 2 * ( v , u ) ( t ) = 1 κ 2 x 1 t ( t s ) q 2 1 Γ ( q 2 ) p ( s , v ( s ) , u ( s ) ) d s 1 2 κ 2 x 1 x 2 ( x 2 s ) q 2 1 Γ ( q 2 ) p ( s , v ( s ) , u ( s ) ) d s + ρ 5 ( t ) x 1 x 2 ( x 2 s ) q 2 2 Γ ( q 2 1 ) p ( s , v ( s ) , u ( s ) ) d s ,
and the values of M i , i = 1 , . . . , 6 given by (17)–(22) takes the following form in this situations:
M 1 * = 3 ( x 2 x 1 ) q 1 2 | κ 1 | | Γ ( q 1 + 1 ) + ρ ˜ 2 ( x 2 x 1 ) q 1 1 Γ ( q 1 ) , M 2 * = ρ ˜ 1 ( x 2 x 1 ) q 2 Γ ( q 2 + 1 ) + ρ ˜ 3 ( x 2 x 1 ) q 2 1 Γ ( q 2 ) + ρ ˜ 4 | h | ( ξ x 1 ) q 2 + 1 | κ 2 | Γ ( q 2 + 2 ) , M 5 * = 3 ( x 2 x 1 ) q 2 2 | κ 2 | | Γ ( q 2 + 1 ) + ρ ˜ 5 ( x 2 x 1 ) q 2 1 Γ ( q 2 ) , M 6 * = 3 | λ 2 | ( x 2 x 1 ) θ 2 2 | κ 2 | | Γ ( θ 2 + 1 ) + ρ ˜ 5 | λ 2 | ( x 2 x 1 ) θ 2 1 Γ ( θ 2 ) ,
In addition, the methods presented in this study can be utilized to solve the system of FDEs type Riemann-Liouville with the BCs (2).
The simulation results of such an equation are the goal of a numerical study which could be interesting for future work.

Author Contributions

Conceptualization, Y.A. and S.A.; writing—original draft preparation, Y.A. and S.A.; validation, L.A. and N.A.; investigation, L.A. and N.A.; writing—review and editing, A.B.M. All authors have read and agreed to the published version of the manuscript.

Funding

Taif University Researchers Supporting Project number (TURSP-2020/218), Taif University, Taif, Saudi Arabia.

Data Availability Statement

Not applicable.

Acknowledgments

The second author thanks Taif University Researchers Supporting Program (Project number: TURSP-2020/218), Taif University, Saudi Arabia for technical and financial support.

Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Alruwaily, Y.; Aljoudi, S.; Almaghamsi, L.; Ben Makhlouf, A.; Alghamdi, N. Existence and Uniqueness Results for Different Orders Coupled System of Fractional Integro-Differential Equations with Anti-Periodic Nonlocal Integral Boundary Conditions. Symmetry 2023, 15, 182. https://doi.org/10.3390/sym15010182

AMA Style

Alruwaily Y, Aljoudi S, Almaghamsi L, Ben Makhlouf A, Alghamdi N. Existence and Uniqueness Results for Different Orders Coupled System of Fractional Integro-Differential Equations with Anti-Periodic Nonlocal Integral Boundary Conditions. Symmetry. 2023; 15(1):182. https://doi.org/10.3390/sym15010182

Chicago/Turabian Style

Alruwaily, Ymnah, Shorog Aljoudi, Lamya Almaghamsi, Abdellatif Ben Makhlouf, and Najla Alghamdi. 2023. "Existence and Uniqueness Results for Different Orders Coupled System of Fractional Integro-Differential Equations with Anti-Periodic Nonlocal Integral Boundary Conditions" Symmetry 15, no. 1: 182. https://doi.org/10.3390/sym15010182

APA Style

Alruwaily, Y., Aljoudi, S., Almaghamsi, L., Ben Makhlouf, A., & Alghamdi, N. (2023). Existence and Uniqueness Results for Different Orders Coupled System of Fractional Integro-Differential Equations with Anti-Periodic Nonlocal Integral Boundary Conditions. Symmetry, 15(1), 182. https://doi.org/10.3390/sym15010182

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