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Article

On a Riemann–Liouville Type Implicit Coupled System via Generalized Boundary Conditions †

1
Department of Mathematics, University of Peshawar, Peshawar, Khyber Pakhtunkhwa 25000, Pakistan
2
School of Engineering, Monash University Malaysia, Selangor 47500, Malaysia
3
Department of Computing, Mathematics and Electronics, “1 Decembrie 1918” University of Alba Iulia, Alba Iulia 510009, Romania
4
Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz 51368, Iran
5
Department of Medical Research, China Medical University Hospital, China Medical University, Taichung 404, Taiwan
*
Authors to whom correspondence should be addressed.
These authors contributed equally to this work.
Mathematics 2021, 9(11), 1205; https://doi.org/10.3390/math9111205
Submission received: 16 April 2021 / Revised: 19 May 2021 / Accepted: 24 May 2021 / Published: 26 May 2021
(This article belongs to the Special Issue Dynamical Systems in Engineering)

Abstract

:
We study a coupled system of implicit differential equations with fractional-order differential boundary conditions and the Riemann–Liouville derivative. The existence, uniqueness, and at least one solution are established by applying the Banach contraction and Leray–Schauder fixed point theorem. Furthermore, Hyers–Ulam type stabilities are discussed. An example is presented to illustrate our main result. The suggested system is the generalization of fourth-order ordinary differential equations with anti-periodic, classical, and initial boundary conditions.

1. Introduction

The generalization of ordinary derivatives leads us to the theory of fractional derivatives. The concept of fractional derivatives was established in 1695, after the well-known conversation of Leibniz and L’Hospital [1]. Mathematicians like Riemann, Liouville, Caputo, Hadamard, Fourier, and Laplace contributed a lot and made the area more interesting for researchers. A fractional-order derivative is a global operator, which may act as a tool to modify or modernize different physical phenomena like control theory [2], dynamical process [3], electro-chemistry [4], mathematical biology [5], image and signal processing [6], etc. For more applications of the fractional differential equations ( F D E s ), we refer the reader to the works in [7,8,9,10,11]. Furthermore, the theory of coupled systems of differential equations is referred to as an important theory in the applied sciences envisaging different areas of biochemistry, ecology, biology, and classical fields of physical sciences and engineering. For details see in [12,13,14].
The theory regarding the existence of solutions of F D E s , drew significant attention of the researchers working on different boundary conditions, e.g., classical, integral, multi-point, non-local, periodic, and anti-periodic [15,16,17,18]. Among the qualitative properties of F D E s , the stability property of the solution is the central one, particularly the Hyers–Ulam ( H U ) stability [19,20,21,22,23,24,25,26]. Stability theory in the sense of H U was first discussed by Ulam [27] in the form of a question in 1940 and the following year, Hyers [28] answered his question in the context of Banach spaces. Recently, generalized H U stability was discussed by Alqifiary et al. [29] for linear differential equations. Razaei et al. [30] presented Laplace transform and H U stability of linear differential equations. Wang et al. [31] studied H U stability for two types of linear F D E s . Shen et al. [32] worked on the H U stability of linear F D E s with constant coefficients using Laplace transform method. Liu et al. [33] proved the H U stability of linear Caputo–Fabrizio F D E s . Liu et al. [34] studied the H U stability of linear Caputo–Fabrizio F D E s with the Mittag–Leffler kernel by Laplace transform method.
The above work motivate us to study the coupled implicit F D E s with fractional-order differential boundary conditions:
D α v ( t ) χ 1 ( t , u ( t ) , D α v ( t ) ) = 0 ; t J , D κ u ( t ) χ 2 ( t , v ( t ) , D κ u ( t ) ) = 0 ; t J , D α 4 v ( 0 ) = η 1 D α 4 v ( σ ) , D α 3 v ( 0 ) = η 2 D α 3 v ( σ ) , D α 2 v ( 0 ) = η 3 D α 2 v ( σ ) , D α 1 v ( 0 ) = η 4 D α 1 v ( σ ) , D κ 4 u ( 0 ) = η 5 D κ 4 u ( σ ) , D κ 3 u ( 0 ) = η 6 D κ 3 u ( σ ) , D κ 2 u ( 0 ) = η 7 D κ 2 u ( σ ) , D κ 1 u ( 0 ) = η 8 D κ 1 u ( σ ) ,
where 3 < α , κ 4 , J = [ 0 , σ ] , σ > 0 and η i 1 for i = 1 , 2 , , 8 . D α , D κ be the α , κ order denotes Riemann–Liouville fractional derivatives and χ 1 , χ 2 : J × R × R R be continuous functions.
Higher-order ordinary differential equations ( O D E s ) can be used to model problems arising from the field of applied sciences and engineering [35,36]. The generalization of fourth-order O D E s are F D E s (1) if α = κ = 4 . Fourth-order differential equations have important applications in mechanics, thus have attracted considerable attention over the last three decades. The problem of static deflection of a uniform beam, which can be modeled as a fourth-order initial value problem is a good example of a real problem in engineering [37,38].
This problem has been extensively analyzed, some new techniques were developed and numerous general and impressive results regarding the existence of solutions were established in [39,40,41,42]. Sometimes, mathematical modeling of the various physical phenomena may arise as a coupled system of the forgoing O D E s . Furthermore, for η i = 1 ( i = 1 , 2 , , 8 ) , we can obtain anti-periodic boundary conditions which are applicable in several mathematical models, some are given in [43,44].
The manuscript is categorized as follows. For our main results, we establish some basic notations, definitions, and lemma in Section 2. In Section 3, we present existence, uniqueness, and at least one solution of system (1) by applying the Banach contraction fixed point theorem and Leray–Schauder fixed point theorem. In Section 4, we discuss definitions of H U type stabilities, which help us to show that system (1) has H U type stabilities by two different approaches. In Section 5, by a particular example of the system (1), we show that our results are applicable.

2. Background Materials

In this fragment, we present basic notations with Banach spaces, definitions of the considered derivative and integral, and lemma, which will be utilized in the next sections.
Suppose C ( J ) is a Banach space with a norm defined as v = sup t J | v ( t ) | . For t J , we define v r ( t ) = t r v ( t ) , r 0 . Suppose that S 1 = C r ( J ) C ( J ) be the space of all functions v such that v r S 1 which yields to be a Banach space when endowed with the norm
v S 1 = max { sup t J t r | v ( t ) | , sup t J t r | D α v ( t ) | } .
Similarly, ( v , u ) S = v S 1 + u S 2 is the norm defined on the product space, where S = S 1 × S 2 . Obviously S , ( v , u ) S is a Banach space.
Definition 1.
[45] For a continuous function v : R + R , the Riemann–Liouville integral of order α > 0 is defined as
I α v ( t ) = 1 Γ ( α ) 0 t v ( τ ) ( t τ ) 1 α d τ ,
such that the integral is pointwise defined on R + .
Definition 2.
[45] For a continuous function v : R + R , the Riemann–Liouville derivative of order α > 0 is defined as
D α v ( t ) = 1 Γ ( n α ) d d t n 0 t v ( τ ) ( t τ ) α n + 1 d τ ,
where [ α ] represents the integer part of α and n = [ α ] + 1 . We note that for ϱ > 1 , ϱ α 1 , α 2 , , α n , we have
D α t ϱ = Γ ( ϱ + 1 ) Γ ( ϱ α + 1 ) t ϱ α
and
D α t α i = 0 , i = 1 , 2 , 3 , , n .
Lemma 1.
[45] Solution of the following Riemann–Liouville F D E of order n 1 < α n
D α v ( t ) = ϑ ( t ) ,
is
I α D α v ( t ) = I α ϑ ( t ) + k 0 t α n + k 1 t α n 1 + + k n 2 t α 2 + k n 1 t α 1 ,
where k i ( i = 1 , 2 , 3 , , n ) are unknowns.

3. Existence Theory

This section is devoted to the equivalent integral form of the proposed problem.
Lemma 2.
Let ϑ C ( J ) , the following α ( 3 , 4 ] order F D E with boundary conditions
D α v ( t ) = ϑ ( t ) ; t J , D α 4 v ( 0 ) = η 1 D α 4 v ( σ ) , D α 3 v ( 0 ) = η 2 D α 3 v ( σ ) , D α 2 v ( 0 ) = η 3 D α 2 v ( σ ) , D α 1 v ( 0 ) = η 4 D α 1 v ( σ ) ,
have the solution
v ( t ) = 0 σ G α ( t , τ ) ϑ ( τ ) d τ ,
where
G α ( t , τ ) = t τ α 1 Γ ( α ) + η 1 t α 4 σ τ 3 6 ( 1 η 1 ) Γ ( α 3 ) + ( 1 η 1 ) η 2 t α 3 + η 1 η 2 σ t α 4 ( α 3 ) σ τ 2 2 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 2 ) + η 3 t α 2 σ τ ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 σ t α 3 σ τ ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) η 3 σ 2 t α 4 σ τ 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) + η 4 t α 1 ( 1 η 4 ) Γ ( α ) + η 3 η 4 σ t α 2 ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 2 ( 1 + η 3 ) η 4 σ 2 t α 3 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 3 t α 4 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) , 0 τ < t σ , η 1 t α 4 σ τ 3 6 ( 1 η 1 ) Γ ( α 3 ) + ( 1 η 1 ) η 2 t α 3 + η 1 η 2 σ t α 4 ( α 3 ) σ τ 2 2 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 2 ) + η 3 t α 2 σ τ ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 σ t α 3 σ τ ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) η 3 σ 2 t α 4 σ τ 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) + η 4 t α 1 ( 1 η 4 ) Γ ( α ) + η 3 η 4 σ t α 2 ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 2 ( 1 + η 3 ) η 4 σ 2 t α 3 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 3 t α 4 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) , 0 t < τ σ .
Proof. 
Using Lemma 1 on F D E (2), we have
v ( t ) = 1 Γ ( α ) 0 t ( t τ ) α 1 ϑ ( τ ) d τ + k 3 t α 1 + k 2 t α 2 + k 1 t α 3 + k 0 t α 4 .
Applying boundary conditions of (2) on (4), we get unknowns
k 0 = η 1 ( 1 η 1 ) Γ ( α 3 ) [ 1 6 0 σ ( σ τ ) 3 ϑ ( τ ) d τ + η 2 σ 2 ( 1 η 2 ) 0 σ σ τ 2 ϑ ( τ ) d τ + ( 1 + η 2 ) η 3 σ 2 2 ( 1 η 2 ) ( 1 η 3 ) 0 σ σ τ ϑ ( τ ) d τ + ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 3 6 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) 0 σ ϑ ( τ ) d τ ] , k 1 = η 2 ( 1 η 2 ) Γ ( α 2 ) [ 1 2 0 σ σ τ 2 ϑ ( τ ) d τ + η 3 σ ( 1 η 3 ) 0 σ ( σ τ ) ϑ ( τ ) d τ + ( 1 + η 3 ) η 4 σ 2 2 ( 1 η 3 ) ( 1 η 4 ) 0 σ ϑ ( τ ) d τ ] , k 2 = η 3 ( 1 η 3 ) Γ ( α 1 ) 0 σ ( σ τ ) ϑ ( τ ) d τ + η 4 σ ( 1 η 4 ) 0 σ ϑ ( τ ) d τ , k 3 = η 4 ( 1 η 4 ) Γ ( α ) 0 σ ϑ ( τ ) d τ .
Put the values of k 0 , k 1 , k 2 and k 3 in Equation (4), we obtain
v ( t ) = 1 Γ ( α ) 0 t t τ α 1 ϑ ( τ ) d τ + η 1 t α 4 6 ( 1 η 1 ) Γ ( α 3 ) 0 σ σ τ 3 ϑ ( τ ) d τ + η 2 t α 3 2 ( 1 η 2 ) Γ ( α 2 ) + η 1 η 2 σ t α 4 2 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 3 ) 0 σ σ τ 2 ϑ ( τ ) d τ + η 3 t α 2 ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 σ t α 3 ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) η 3 σ 2 t α 4 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) × 0 σ σ τ ϑ ( τ ) d τ + [ η 4 t α 1 ( 1 η 4 ) Γ ( α ) + η 3 η 4 σ t α 2 ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 2 ( 1 + η 3 ) η 4 σ 2 t α 3 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) + + η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 3 t α 4 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) ] 0 σ ϑ ( τ ) d τ = 0 σ G α ( t , τ ) ϑ ( τ ) d τ ,
where G α ( t , τ ) is given by (3). □
Remark 1.
Let μ C ( J ) , the following κ ( 3 , 4 ] order F D E with boundary conditions
D κ u ( t ) = μ ( t ) ; t J , D κ 4 u ( 0 ) = η 5 D κ 4 u ( σ ) , D κ 3 u ( 0 ) = η 6 D κ 3 u ( σ ) , D κ 2 u ( 0 ) = η 7 D κ 2 u ( σ ) , D κ 1 u ( 0 ) = η 8 D κ 1 u ( σ )
has the solution
u ( t ) = 0 σ G κ ( t , τ ) μ ( τ ) D τ ,
where G κ ( t , τ ) is given by
G κ ( t , τ ) = t τ κ 1 Γ ( κ ) + η 5 t κ 4 σ τ 3 6 ( 1 η 5 ) Γ ( κ 3 ) + ( 1 η 5 ) η 6 t κ 3 + η 5 η 6 σ t κ 4 ( κ 3 ) σ τ 2 2 ( 1 η 5 ) ( 1 η 6 ) Γ ( κ 2 ) + η 7 t κ 2 σ τ ( 1 η 7 ) Γ ( κ 1 ) + η 6 η 7 σ t κ 3 σ τ ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 2 ) + η 5 ( 1 + η 6 ) η 7 σ 2 t κ 4 σ τ 2 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 3 ) + η 8 t κ 1 ( 1 η 8 ) Γ ( κ ) + η 7 η 8 σ t κ 2 ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 1 ) + η 6 ( 1 + η 7 ) η 8 σ 2 t κ 3 2 ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 2 ) + η 5 ( 1 + η 6 ) ( 1 + η 7 ) + η 6 + η 7 η 8 σ 3 t κ 4 6 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 3 ) , 0 τ < t σ , η 5 t κ 4 σ τ 3 6 ( 1 η 5 ) Γ ( κ 3 ) + ( 1 η 5 ) η 6 t κ 3 + η 5 η 6 σ t κ 4 ( κ 3 ) σ τ 2 2 ( 1 η 5 ) ( 1 η 6 ) Γ ( κ 2 ) + η 7 t κ 2 σ τ ( 1 η 7 ) Γ ( κ 1 ) + η 6 η 7 σ t κ 3 σ τ ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 2 ) + η 5 ( 1 + η 6 ) η 7 σ 2 t κ 4 σ τ 2 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 3 ) + η 8 t κ 1 ( 1 η 8 ) Γ ( κ ) + η 7 η 8 σ t κ 2 ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 1 ) + η 6 ( 1 + η 7 ) η 8 σ 2 t κ 3 2 ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 2 ) + η 5 ( 1 + η 6 ) ( 1 + η 7 ) + η 6 + η 7 η 8 σ 3 t κ 4 6 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 3 ) , 0 t < τ σ .
Remark 2.
Putting α = 4 and η 1 = η 2 = η 3 = η 4 = 1 in (3), gives Green’s function G α ( t , τ ) of fourth-order O D E with anti-periodic boundary conditions.
Remark 3.
Putting α = 4 and η 1 = η 2 = η 3 = η 4 = 0 in (5), gives the solution of fourth-order O D E having initial conditions.
For the reason of advantage, we set the following notations:
Q α = max { σ 4 Γ ( α + 1 ) + | η 1 σ 4 24 ( 1 η 1 ) Γ ( α 3 ) | + η 2 ( 1 η 1 ) σ 4 + η 1 η 2 σ 4 ( α 3 ) 6 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 2 ) | + | η 3 ( 1 η 2 ) σ 4 + η 2 η 3 σ 4 ( α 2 ) 2 ( 1 η 2 ) ( 1 η 3 ) Γ ( α 1 ) | + | η 1 ( 1 + η 2 ) η 3 σ 4 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) | + | ( 1 η 3 ) η 4 σ 4 + η 3 η 4 σ 4 ( α 1 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α ) | + | η 2 ( 1 + η 3 ) η 4 σ 4 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) | + | η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 4 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) | }
and
Q κ = max { σ 4 Γ ( κ + 1 ) + | η 5 σ 4 24 ( 1 η 5 ) Γ ( κ 3 ) | + η 6 ( 1 η 5 ) σ 4 + η 5 η 6 σ 4 ( κ 3 ) 6 ( 1 η 5 ) ( 1 η 6 ) Γ ( κ 2 ) | + | η 7 ( 1 η 6 ) σ 4 + η 6 η 7 σ 4 ( κ 2 ) 2 ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 1 ) | + | η 5 ( 1 + η 6 ) η 7 σ 4 2 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 3 ) | + | ( 1 η 7 ) η 8 σ 4 + η 7 η 8 σ 4 ( κ 1 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ ) | + | η 6 ( 1 + η 7 ) η 8 σ 4 2 ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 2 ) | + | η 5 ( 1 + η 6 ) ( 1 + η 7 ) + η 6 + η 7 η 8 σ 4 6 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 3 ) | }
If solution of system (1) is ( v , u ) and t J , then
v ( t ) = 1 Γ ( α ) 0 t t τ α 1 χ 1 ( τ , u ( τ ) , D α v ( τ ) ) d τ + η 1 t α 4 6 ( 1 η 1 ) Γ ( α 3 ) 0 σ σ τ 3 χ 1 ( τ , u ( τ ) , D α v ( τ ) ) d τ + η 2 t α 3 2 ( 1 η 2 ) Γ ( α 2 ) + η 1 η 2 σ t α 4 2 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 3 ) 0 σ σ τ 2 χ 1 ( τ , u ( τ ) , D α v ( τ ) ) d τ + η 3 t α 2 ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 σ t α 3 ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) η 3 σ 2 t α 4 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) × 0 σ σ τ χ 1 ( τ , u ( τ ) , D α v ( τ ) ) d τ + [ η 4 t α 1 ( 1 η 4 ) Γ ( α ) + η 3 η 4 σ t α 2 ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 2 ( 1 + η 3 ) η 4 σ 2 t α 3 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 3 t α 4 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) ] × 0 σ χ 1 ( τ , u ( τ ) , D α v ( τ ) ) d τ = 0 σ G α ( t , τ ) χ 1 ( τ , u ( τ ) , D α v ( τ ) ) d τ ,
and
u ( t ) = 1 Γ ( κ ) 0 t t τ κ 1 χ 2 ( τ , v ( τ ) , D κ u ( τ ) ) d τ + η 5 t κ 4 6 ( 1 η 5 ) Γ ( κ 3 ) 0 σ σ τ 3 χ 2 ( τ , v ( τ ) , D κ u ( τ ) ) d τ + η 6 t κ 3 2 ( 1 η 6 ) Γ ( κ 2 ) + η 5 η 6 σ t κ 4 2 ( 1 η 5 ) ( 1 η 6 ) Γ ( κ 3 ) 0 σ σ τ 2 χ 2 ( τ , v ( τ ) , D κ u ( τ ) ) d τ + η 7 t κ 2 ( 1 η 7 ) Γ ( κ 1 ) + η 6 η 7 σ t κ 3 ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 2 ) + η 5 ( 1 + η 6 ) η 7 σ 2 t κ 4 2 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 3 ) × 0 σ σ τ χ 2 ( τ , v ( τ ) , D κ u ( τ ) ) d τ + [ η 8 t κ 1 ( 1 η 8 ) Γ ( κ ) + η 7 η 8 σ t κ 2 ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 1 ) + η 6 ( 1 + η 7 ) η 8 σ 2 t κ 3 2 ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 2 ) + η 5 ( 1 + η 6 ) ( 1 + η 7 ) + η 6 + η 7 η 8 σ 3 t κ 4 6 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 3 ) ] × 0 σ χ 2 ( τ , v ( τ ) , D κ u ( τ ) ) d τ = 0 σ G κ ( t , τ ) χ 2 ( τ , v ( τ ) , D κ u ( τ ) ) d τ .
We use the following notations for convenience:
y ( t ) = χ 1 ( t , u ( t ) , D α v ( t ) ) = χ 1 ( t , u ( t ) , y ( t ) ) x ( t ) = χ 2 ( t , v ( t ) , D κ u ( t ) ) = χ 2 ( t , v ( t ) , x ( t ) ) .
Now, transform system (1) to the fixed point problem, let F : S S is an operator defined by
F ( v , u ) ( t ) = 0 t G α ( t , τ ) χ 1 ( τ , u ( τ ) , y ( τ ) ) d τ 0 t G κ ( t , τ ) χ 2 ( τ , v ( τ ) , x ( τ ) ) d τ = F α ( u , y ) ( t ) F κ ( v , x ) ( t ) = F α ( v ) ( t ) F κ ( u ) ( t ) .
Then, the fixed point of F and the solution of system (1) coincided, i.e.,
F α ( v ) ( t ) = 1 Γ ( α ) 0 t t τ α 1 y ( τ ) d τ + η 1 t α 4 6 ( 1 η 1 ) Γ ( α 3 ) 0 σ σ τ 3 v ( τ ) d τ + η 2 t α 3 2 ( 1 η 2 ) Γ ( α 2 ) + η 1 η 2 σ t α 4 2 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 3 ) 0 σ σ τ 2 y ( τ ) d τ + η 3 t α 2 ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 σ t α 3 ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) η 3 σ 2 t α 4 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) × 0 σ σ τ y ( τ ) d τ + [ η 4 t α 1 ( 1 η 4 ) Γ ( α ) + η 3 η 4 σ t α 2 ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 2 ( 1 + η 3 ) η 4 σ 2 t α 3 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 3 t α 4 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) ] 0 σ y ( τ ) d τ
and
F κ ( u ) ( t ) = 1 Γ ( κ ) 0 t t τ κ 1 x ( τ ) d τ + η 5 t κ 4 6 ( 1 η 5 ) Γ ( κ 3 ) 0 σ σ τ 3 x ( τ ) d τ + η 6 t κ 3 2 ( 1 η 6 ) Γ ( κ 2 ) + η 5 η 6 σ t κ 4 2 ( 1 η 5 ) ( 1 η 6 ) Γ ( κ 3 ) 0 σ σ τ 2 x ( τ ) d τ + η 7 t κ 2 ( 1 η 7 ) Γ ( κ 1 ) + η 6 η 7 σ t κ 3 ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 2 ) + η 5 ( 1 + η 6 ) η 7 σ 2 t κ 4 2 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 3 ) × 0 σ σ τ x ( τ ) d τ + [ η 8 t κ 1 ( 1 η 8 ) Γ ( κ ) + η 7 η 8 σ t κ 2 ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 1 ) + η 6 ( 1 + η 7 ) η 8 σ 2 t κ 3 2 ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 2 ) + η 5 ( 1 + η 6 ) ( 1 + η 7 ) + η 6 + η 7 η 8 σ 3 t κ 4 6 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 3 ) ] 0 σ x ( τ ) d τ .
Using Banach contraction theorem in the following, we prove the uniqueness of solution of system (1).
Theorem 1.
Let the functions χ 1 , χ 2 : J × R × R R are continuous and satisfy the hypothesis:
H 1 : For every t J and u , v , y , x , u ¯ , v ¯ , y ¯ , x ¯ : J R , there are L χ 1 , L χ 2 , L ¯ χ 1 , L ¯ χ 2 , such that
| χ 1 ( t , u ( t ) , y ( t ) ) χ 1 ( t , u ¯ ( t ) , y ¯ ( t ) ) | L χ 1 | u ( t ) u ¯ ( t ) | + L ¯ χ 1 | y ( t ) y ¯ ( t ) | ,
| χ 2 ( t , v ( t ) , x ( t ) ) χ 2 ( t , v ¯ ( t ) , x ¯ ( t ) ) | L χ 2 | v ( t ) v ¯ ( t ) | + L ¯ χ 2 | x ( t ) x ¯ ( t ) | .
In addition, suppose that
Q α L χ 1 ( 1 L ¯ χ 2 ) + Q κ L χ 2 ( 1 L ¯ χ 1 ) ( 1 L ¯ χ 2 ) ( 1 L ¯ χ 1 ) < 1 ,
where Q α and Q κ are defined by Equations (6) and (7), respectively. Furthermore, 0 L ¯ χ 1 , L ¯ χ 2 < 1 (through out the paper). Then, the solution of system (1) is unique.
Proof. 
Consider sup t J χ 1 ( t , 0 , 0 ) = Φ * < and sup t J χ 2 ( t , 0 , 0 ) = Ψ * < , such that
r 2 Q α Φ * ( 1 L ¯ χ 2 ) + 2 Q κ Ψ * ( 1 L ¯ χ 1 ) 2 ( 1 L ¯ χ 1 ) ( 1 L ¯ χ 2 ) Q α L χ 1 Q κ L χ 2 .
We show that F ( B r ) B r , where
B r = ( v , u ) S : ( v , u ) S r , v r 2 , u r 2 .
For ( v , u ) B r , we have
t 4 α | F α ( v ) ( t ) | t 4 α Γ ( α ) 0 t t τ α 1 ( | χ 1 ( τ , u ( τ ) , y ( τ ) ) χ 1 ( τ , 0 , 0 ) | + | χ 1 ( τ , 0 , 0 ) | ) d τ + | η 1 6 ( 1 η 1 ) Γ ( α 3 ) | × 0 σ σ τ 3 ( | χ 1 ( τ , u ( τ ) , y ( τ ) ) χ 1 ( τ , 0 , 0 ) | + | χ 1 ( τ , 0 , 0 ) | ) d τ + | η 2 t 2 ( 1 η 2 ) Γ ( α 2 ) + η 1 η 2 σ 2 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 3 ) | 0 T σ τ 2 ( | χ 1 ( τ , u ( τ ) , v ( τ ) ) χ 1 ( τ , 0 , 0 ) | + | χ 1 ( τ , 0 , 0 ) | ) d τ + | η 3 t 2 ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 t σ ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) η 3 σ 2 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) | × 0 σ σ τ | χ 1 ( τ , u ( τ ) , v ( τ ) ) χ 1 ( τ , 0 , 0 ) | + | χ 1 ( τ , 0 , 0 ) | d τ + | η 3 η 4 σ t 2 ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 4 t 3 ( 1 η 4 ) Γ ( α ) + η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 3 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) + η 2 ( 1 + η 3 ) η 4 σ 2 t 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) | 0 σ | χ 1 ( τ , u ( τ ) , v ( τ ) ) χ 1 ( τ , 0 , 0 ) | + | χ 1 ( τ , 0 , 0 ) | d τ .
Consider
| v ( t ) | | χ 1 ( t , u ( t ) , v ( t ) ) χ 1 ( t , 0 , 0 ) | + | χ 1 ( t , 0 , 0 ) | | χ 1 ( t , 0 , 0 ) | + L χ 1 | u ( t ) | + L ¯ χ 1 | v ( t ) | | χ 1 ( t , 0 , 0 ) | + L χ 1 | u ( t ) | 1 L ¯ χ 1 .
Substituting (10) in (9), we get
F α ( v ) [ σ 4 Γ ( α + 1 ) + | η 1 σ 4 24 ( 1 η 1 ) Γ ( α 3 ) | + η 2 ( 1 η 1 ) σ 4 + η 1 η 2 σ 4 ( α 3 ) 6 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 2 ) | + | η 3 ( 1 η 2 ) σ 4 + η 2 η 3 σ 4 ( α 2 ) 2 ( 1 η 2 ) ( 1 η 3 ) Γ ( α 1 ) | + | η 1 ( 1 + η 2 ) η 3 σ 4 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) | + | ( 1 η 3 ) η 4 σ 4 + η 3 η 4 σ 4 ( α 1 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α ) | + | η 2 ( 1 + η 3 ) η 4 σ 4 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) | + | η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 4 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) | ] 2 Φ * + L χ 1 r 2 ( 1 L ¯ χ 1 ) .
Therefore,
F α ( v ) Q α 2 Φ * + L χ 1 r 2 ( 1 L ¯ χ 1 ) .
On the same way, we can write
F κ ( u ) Q κ 2 Ψ * + L χ 2 r 2 ( 1 L ¯ χ 2 ) .
Inequalities (11) and (12) combined give
F ( v , u ) S r .
For any t J , and ( v 1 , u 1 ) , ( v 2 , u 2 ) S , we get
t 4 α | F α ( v 1 ) ( t ) F α ( v 2 ) ( t ) | t 4 α Γ ( α ) 0 t t τ α 1 | χ 1 ( τ , u 1 ( τ ) , y 1 ( τ ) ) χ 1 ( τ , u 2 ( τ ) , y 2 ( τ ) ) | d τ + | η 1 6 ( 1 η 1 ) Γ ( α 3 ) | × 0 σ σ τ 3 | χ 1 ( τ , u 1 ( τ ) , y 1 ( τ ) ) χ 1 ( τ , u 2 ( τ ) , y 2 ( τ ) ) | d τ + | η 2 t 2 ( 1 η 2 ) Γ ( α 2 ) + η 1 η 2 σ 2 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 3 ) | 0 σ σ τ 2 | χ 1 ( τ , u 1 ( τ ) , y 1 ( τ ) ) χ 1 ( τ , u 2 ( τ ) , y 2 ( τ ) ) | d τ + | η 3 t 2 ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 t σ ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) η 3 σ 2 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 c 0 ) Γ ( α 3 ) | × 0 σ σ τ | χ 1 ( τ , u 1 ( τ ) , y 1 ( τ ) ) χ 1 ( τ , u 2 ( τ ) , y 2 ( τ ) ) | d τ + | η 3 η 4 σ t 2 ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 4 t 3 ( 1 η 4 ) Γ ( α ) + η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 3 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) + η 2 ( 1 + η 3 ) η 4 σ 2 t 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) | 0 σ | χ 1 ( τ , u 1 ( τ ) , y 1 ( τ ) ) χ 1 ( τ , u 2 ( τ ) , y 2 ( τ ) ) | d τ [ σ 4 Γ ( α + 1 ) + | η 1 σ 4 24 ( 1 η 1 ) Γ ( α 3 ) | + η 2 ( 1 η 1 ) σ 4 + η 1 η 2 σ 4 ( α 3 ) 6 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 2 ) | + | η 3 ( 1 η 2 ) σ 4 + η 2 η 3 σ 4 ( α 2 ) 2 ( 1 η 2 ) ( 1 η 3 ) Γ ( α 1 ) | + | η 1 ( 1 + η 2 ) η 3 σ 4 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) | + | ( 1 η 3 ) η 4 σ 4 + η 3 η 4 σ 4 ( α 1 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α ) | + | η 2 ( 1 + η 3 ) η 4 σ 4 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) | + | η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 4 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) | ] L χ 1 1 L ¯ χ 1 u 1 u 2
and thus we get
F α ( v 1 ) F α ( v 2 ) Q α L χ 1 1 L ¯ χ 1 u 1 u 2 .
Similarly,
F κ ( u 1 ) F κ ( u 2 ) Q κ L χ 2 1 L ¯ χ 2 v 1 v 2 .
From the inequalities (13) and (14), we get that
F ( v 1 , u 1 ) F ( v 2 , u 2 ) S Q α L χ 1 ( 1 L ¯ χ 2 ) + Q κ L χ 2 ( 1 L ¯ χ 1 ) ( 1 L ¯ χ 2 ) ( 1 L ¯ χ 1 ) ( v 1 , u 1 ) ( v 2 , u 2 ) S .
Therefore, F is a contraction operator. Therefore, by Banach’s fixed point theorem, F has a unique fixed point, so the solution of the problem (1) is unique. □
The next result is based on the following Leray–Schauder alternative theorem.
Theorem 2.
[46] Let F : S S be an operator which is completely continuous (i.e., a map that restricted to any bounded set in S is compact). Suppose
B ( F ) = v S : v = λ F ( v ) , λ [ 0 , 1 ] .
Then, either the operator F has at least one fixed point or the set B ( F ) is unbounded.
Theorem 3.
Suppose the functions χ 1 , χ 2 : J × R × R R are continuous and satisfy the following hypothesis:
H2:
For every t J and u , v : J R , there are ϕ i ( i = 1 , 2 , 3 ) : J R + , such that
| χ 1 ( t , u ( t ) , y ( t ) ) | ϕ 1 ( t ) + ϕ 2 ( t ) | u ( t ) | + ϕ 3 ( t ) | y ( t ) | .
Similarly, for every t J and v , x : J R , there are φ i ( i = 1 , 2 , 3 ) : J R + , such that
| χ 2 ( t , v ( t ) , x ( t ) ) | φ 1 ( t ) + φ 2 ( t ) | u ( t ) | + φ 3 ( t ) | x ( t ) | ,
with sup t J ϕ i ( t ) = ϕ i * , sup t J φ i ( t ) = φ i * ( i = 1 , 2 , 3 ) .
In addition, it is assumed that
Q 0 = max Q κ φ 2 * 1 φ 3 * , Q α ϕ 2 * 1 ϕ 3 * < 1 and 0 ϕ 3 * , φ 3 * < 1 .
Then, the system (1) has at least one solution.
Proof. 
First, we prove that F is completely continuous. In view of continuity of χ 1 , χ 2 , the operator F is also continuous. For any ( v , u ) B r , we have
t 4 α | F α ( v ) ( t ) | t 4 α Γ ( α ) 0 t t τ α 1 | y ( τ ) | d τ + | η 1 6 ( 1 η 1 ) Γ ( α 3 ) | 0 σ σ τ 3 | y ( τ ) | d τ + | η 2 t 2 ( 1 η 2 ) Γ ( α 2 ) + η 1 η 2 σ 2 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 3 ) | 0 σ σ τ 2 | y ( τ ) | d τ + | η 3 t 2 ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 t σ ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) η 3 σ 2 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) | 0 σ σ τ | y ( τ ) | d τ + | η 4 t 3 ( 1 η 4 ) Γ ( α ) + η 3 η 4 σ t 2 ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 2 ( 1 + η 3 ) η 4 σ 2 t 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 3 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) | 0 σ | y ( τ ) | d τ .
Now by H 2 , we have
| y ( t ) | = | χ 1 ( t , u ( t ) , y ( t ) ) | ϕ 1 ( t ) + ϕ 2 ( t ) | u ( t ) | + ϕ 3 ( t ) | y ( t ) | ϕ 1 ( t ) + ϕ 2 ( t ) | u ( t ) | 1 ϕ 3 ( t ) .
Therefore, (16) implies
F α ( v ) [ σ 4 Γ ( α + 1 ) + | η 1 σ 4 24 ( 1 η 1 ) Γ ( α 3 ) | + η 2 ( 1 η 1 ) σ 4 + η 1 η 2 σ 4 ( α 3 ) 6 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 2 ) | + | η 3 ( 1 η 2 ) σ 4 + η 2 η 3 σ 4 ( α 2 ) 2 ( 1 η 2 ) ( 1 η 3 ) Γ ( α 1 ) | + | η 1 ( 1 + η 2 ) η 3 σ 4 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) | + | ( 1 η 3 ) η 4 σ 4 + η 3 η 4 σ 4 ( α 1 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α ) | + | η 2 ( 1 + η 3 ) η 4 σ 4 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) | + | η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 4 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) | ] 2 ϕ 1 * + ϕ 2 * r 2 ( 1 ϕ 3 * ) ,
which implies that
F α ( v ) Q α 2 ϕ 1 * + ϕ 2 * r 2 ( 1 ϕ 3 * ) .
Similarly, we get
F κ ( u ) Q κ 2 φ 1 * + φ 2 * r 2 ( 1 φ 3 * ) .
Thus, it follows from the inequalities (18) and (19) that F is uniformly bounded.
Now, we prove that F is equicontinuous. Let 0 t 2 t 1 t . Then, we have
| t 1 4 α F α ( v ) ( t 1 ) t 2 4 α F α ( v ) ( t 2 ) | = = | 1 Γ ( α ) 0 t 1 t 1 4 α ( t 1 τ ) α 1 t 2 4 α ( t 2 τ ) α 1 v ( τ ) d τ 1 Γ ( α ) t 1 t 2 t 2 4 α ( t 2 τ ) α 1 y ( τ ) d τ + η 4 ( t 1 3 t 2 3 ) ( 1 η 4 ) Γ ( α ) + η 3 η 4 σ ( t 1 2 t 2 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 2 ( 1 + η 3 ) η 4 σ 2 ( t 1 t 2 ) 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) 0 σ y ( τ ) d τ + η 3 ( t 1 2 t 2 2 ) ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 σ ( t 1 t 2 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) 0 σ σ τ y ( τ ) d τ + η 2 ( t 1 t 2 ) 2 ( 1 η 2 ) Γ ( α 2 ) 0 σ σ τ 2 y ( τ ) d τ | .
Therefore, we get
| t 1 4 α F α ( v ) ( t 1 ) t 2 4 α F α ( v ) ( t 2 ) | [ | 1 Γ ( α ) 0 t 1 t 1 4 α ( t 1 τ ) α 1 t 2 4 α ( t 2 τ ) α 1 d τ 1 Γ ( α ) t 1 t 2 t 2 4 α ( t 2 τ ) α 1 d τ | + | η 4 σ ( t 1 3 t 2 3 ) ( 1 η 4 ) Γ ( α ) + η 3 η 4 σ 2 ( t 1 2 t 2 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 2 ( 1 + η 3 ) η 4 σ 3 ( t 1 t 2 ) 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) | + | η 3 σ 2 ( t 1 2 t 2 2 ) 2 ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 σ 3 ( t 1 t 2 ) 2 ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) | + η 2 σ 3 ( t 1 t 2 ) 6 ( 1 η 2 ) Γ ( α 2 ) | ] ϕ 1 * + ϕ 2 * | u | 1 ϕ 3 * 0 as t 1 t 2 .
Similarly
| t 1 4 κ F κ ( u ) ( t 1 ) t 2 4 κ F κ ( u ) ( t 2 ) | [ | 1 Γ ( κ ) 0 t 1 t 1 4 κ ( t 1 τ ) κ 1 t 2 4 κ ( t 2 τ ) κ 1 d τ 1 Γ ( κ ) t 1 t 2 t 2 4 κ ( t 2 τ ) κ 1 d τ | + | η 4 σ ( t 1 3 t 2 3 ) ( 1 η 4 ) Γ ( κ ) + η 3 η 4 σ 2 ( t 1 2 t 2 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( κ 1 ) + η 2 ( 1 + η 3 ) η 4 σ 3 ( t 1 t 2 ) 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( κ 2 ) | + | η 3 σ 2 ( t 1 2 t 2 2 ) 2 ( 1 η 3 ) Γ ( κ 1 ) + η 2 η 3 σ 3 ( t 1 t 2 ) 2 ( 1 η 2 ) ( 1 η 3 ) Γ ( κ 2 ) | + η 2 σ 3 ( t 1 t 2 ) 6 ( 1 η 2 ) Γ ( κ 2 ) | ] φ 1 * + φ 2 * | v | 1 φ 3 * 0 as t 1 t 2 .
Therefore, F ( v , u ) is equicontinuous. Thus, we proved that the operator F ( v , u ) is continuous, uniformly bounded, and equicontinuous, concluding that F ( v , u ) is completely continuous. Now, by using Arzela–Ascoli theorem, the operator F ( v , u ) is compact.
Finally, we are going to check that B = ( v , u ) S | ( v , u ) = λ F ( v , u ) , λ [ 0 , 1 ] is bounded. Suppose ( v , u ) B , then ( v , u ) = λ F ( v , u ) . For t J , we have
v ( t ) = λ F α ( v ) ( t ) , u ( t ) = λ F κ ( u ) ( t ) .
Then,
t 4 α | v ( t ) | [ σ 4 Γ ( α + 1 ) + | η 1 σ 4 24 ( 1 η 1 ) Γ ( α 3 ) | + η 2 ( 1 η 1 ) σ 4 + η 1 η 2 σ 4 ( α 3 ) 6 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 2 ) | + | η 3 ( 1 η 2 ) σ 4 + η 2 η 3 σ 4 ( α 2 ) 2 ( 1 η 2 ) ( 1 η 3 ) Γ ( α 1 ) | + | η 1 ( 1 + η 2 ) η 3 σ 4 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) | + | ( 1 η 3 ) η 4 σ 4 + η 3 η 4 σ 4 ( α 1 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α ) | + | η 2 ( 1 + η 3 ) η 4 σ 4 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) | + | η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 4 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) | ] ϕ 1 ( t ) + ϕ 2 ( t ) | u ( t ) | 1 ϕ 3 ( t )
and
t 4 κ | u ( t ) | [ σ 4 Γ ( κ + 1 ) + | η 5 σ 4 24 ( 1 η 5 ) Γ ( κ 3 ) | + η 6 ( 1 η 5 ) σ 4 + η 5 η 6 σ 4 ( κ 3 ) 6 ( 1 η 5 ) ( 1 η 6 ) Γ ( κ 2 ) | + | η 7 ( 1 η 6 ) σ 4 + η 6 η 7 σ 4 ( κ 2 ) 2 ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 1 ) | + | η 5 ( 1 + η 6 ) η 7 σ 4 2 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 3 ) | + | ( 1 η 7 ) η 8 σ 4 + η 7 η 8 σ 4 ( κ 1 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ ) | + | η 6 ( 1 + η 7 ) η 8 σ 4 2 ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 2 ) | + | η 5 ( 1 + η 6 ) ( 1 + η 7 ) + η 6 + η 7 η 8 σ 4 6 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 3 ) | ] φ 1 ( t ) + φ 2 ( t ) | v ( t ) | 1 φ 3 ( t ) .
Therefore, from (20) and (21), we have
v Q α ϕ 1 * + ϕ 2 * u 1 ϕ 3 *
and
u Q κ φ 1 * + φ 2 * v 1 φ 3 * ,
which imply that
v + u = Q α ϕ 1 * 1 ϕ 3 * + Q κ φ 1 * 1 φ 3 * + Q κ φ 2 * v 1 φ 3 * + Q α ϕ 2 * u 1 ϕ 3 * .
Consequently, we get
( v , u ) S Q α ϕ 1 * + Q κ φ 1 * ( 1 ϕ 3 * ) ( 1 φ 3 * ) ( 1 Q 0 ) ,
for any t J , where Q 0 is defined by (15), which infer that B is bounded. Therefore, by Theorem 2, F has at least one fixed point. Thus, the system (1) has at least one solution. □

4. Stability Results

Let us recall some definitions related to H U stabilities:
Suppose the functions Θ α , Θ κ : J R + are nondecreasing and ϵ α , ϵ κ > 0 . Consider the inequalities given below.
| D α v ( t ) χ 1 ( t , u ( t ) , D α v ( t ) ) | ϵ α , t J , | D κ u ( t ) χ 2 ( t , v ( t ) , D κ u ( t ) ) | ϵ κ , t J ,
| D α v ( t ) χ 1 ( t , u ( t ) , D α v ( t ) ) | Θ α ( t ) ϵ α , t J , | D κ u ( t ) χ 2 ( t , v ( t ) , D κ u ( t ) ) | Θ κ ( t ) ϵ κ , t J ,
| D α v ( t ) χ 1 ( t , u ( t ) , D α v ( t ) ) | Θ α ( t ) , t J , | D κ u ( t ) χ 2 ( t , v ( t ) , D κ u ( t ) ) | Θ κ ( t ) , t J .
Definition 3.
[47] System (1) is H U stable, if there are C α , κ = max ( C α , C κ ) > 0 such that for some ϵ = max ( ϵ α , ϵ κ ) > 0 and for each solution ( v , u ) S of the inequality (22). There is a solution ( w , ζ ) S with
( v , u ) ( t ) ( w , ζ ) ( t ) C α , κ ϵ , t J .
Definition 4.
[47] System (1) is generalized H U stable, if there is Φ α , κ C ( R + , R + ) with Φ α , κ ( 0 ) = 0 , such that for each solution ( v , u ) S of (22), there is a solution ( w , ζ ) S of problem (1), which satisfies
( v , u ) ( t ) ( w , ζ ) ( t ) Φ α , κ ( ϵ ) , t J .
Definition 5.
[47] System (1) is H U –Rassias stable with respect to Θ α , κ = max ( Θ α , Θ κ ) C ( J , R ) , if there are constants C Θ α , Θ κ = max ( C Θ α , C Θ κ ) > 0 such that for some ϵ = ( ϵ α , ϵ κ ) > 0 and for each solution ( v , u ) S of the inequality (23). There is a solution ( w , ζ ) S with
( v , u ) ( t ) ( w , ζ ) ( t ) C Θ α , Θ κ Θ α , κ ( t ) ϵ , t J .
Definition 6.
[47] System (1) is generalized H U –Rassias stable with respect to Θ α , κ = max ( Θ α , Θ κ ) C ( J , R ) , if there is constant C Θ α , Θ κ = max ( C Θ α , C Θ κ ) > 0 , such that for each solution ( v , u ) S of the inequality (24). There is a solution ( w , ζ ) S of (1), which satisfies
( v , u ) ( t ) ( w , ζ ) ( t ) C Θ α , Θ κ Θ α , κ ( t ) , t J .
Remark 4.
We say that ( v , u ) S is a solution of the inequality (22), if there are Ψ χ 1 , Ψ χ 2 C ( J , R ) , which depends on v , u , respectively, such that ( A 1 ) | Ψ χ 1 ( t ) | ϵ α , | Ψ χ 2 ( t ) | ϵ κ , t J ;
( A 2 )
D α v ( t ) = χ 1 ( t , u ( t ) , D α v ( t ) ) + Ψ χ 1 ( t ) , t J , D κ u ( t ) = χ 2 ( t , v ( t ) , D κ u ( t ) ) + Ψ χ 2 ( t ) , t J .
Lemma 3.
Let ( v , u ) S be the solution of inequality (22), then we have
v m 1 Q α ϵ α , t J , u m 2 Q κ ϵ κ , t J .
Proof. 
By ( A 2 ) of Remark 4 and for t J , we have
D α v ( t ) = χ 1 ( t , u ( t ) , D α v ( t ) ) + Ψ χ 1 ( t ) , D κ u ( t ) = χ 2 ( t , v ( t ) , D κ u ( t ) ) + Ψ χ 2 ( t ) , D α 4 v ( 0 ) = η 1 D α 4 v ( σ ) , D α 3 v ( 0 ) = η 2 D α 3 v ( σ ) , D α 2 v ( 0 ) = η 3 D α 2 v ( σ ) , D α 1 v ( 0 ) = η 4 D α 1 v ( σ ) , D κ 4 u ( 0 ) = η 5 D κ 4 u ( σ ) , D κ 3 u ( 0 ) = η 6 D κ 3 u ( σ ) D κ 2 u ( 0 ) = η 7 D κ 2 u ( σ ) , D κ 1 u ( 0 ) = η 8 D κ 1 u ( σ ) .
By Lemma 1, the solution of (29) can be written as
v ( t ) = 1 Γ ( α ) 0 t t τ α 1 χ 1 ( τ , u ( τ ) , D α v ( τ ) ) + Ψ χ 1 ( τ ) d τ + η 1 t α 4 6 ( 1 η 1 ) Γ ( α 3 ) × 0 σ σ τ 3 χ 1 ( τ , u ( τ ) , D α v ( τ ) ) + Ψ χ 1 ( τ ) d τ + η 2 t α 3 2 ( 1 η 2 ) Γ ( α 2 ) + η 1 η 2 σ t α 4 2 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 3 ) × 0 σ σ τ 2 χ 1 ( τ , u ( τ ) , D α v ( τ ) ) + Ψ χ 1 ( τ ) d τ + [ η 3 t α 2 ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 σ t α 3 ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) η 3 σ 2 t α 4 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) ] 0 σ σ τ χ 1 ( τ , u ( τ ) , D α v ( τ ) ) + Ψ χ 1 ( τ ) d τ + [ η 4 t α 1 ( 1 η 4 ) Γ ( α ) + η 3 η 4 σ t α 2 ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 2 ( 1 + η 3 ) η 4 σ 2 t α 3 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 3 t α 4 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) ] 0 σ χ 1 ( τ , u ( τ ) , D α v ( τ ) ) + Ψ χ 1 ( τ ) d τ , u ( t ) = 1 Γ ( κ ) 0 t t τ κ 1 χ 2 ( τ , v ( τ ) , D κ u ( τ ) ) + Ψ χ 2 ( τ ) d τ + η 5 t κ 4 6 ( 1 η 5 ) Γ ( κ 3 ) × 0 σ σ τ 3 χ 2 ( τ , v ( τ ) , D κ u ( τ ) ) + Ψ χ 2 ( τ ) d τ + η 6 t κ 3 2 ( 1 η 6 ) Γ ( κ 2 ) + η 5 η 6 σ t κ 4 2 ( 1 η 5 ) ( 1 η 6 ) Γ ( κ 3 ) × 0 σ σ τ 2 χ 2 ( τ , v ( τ ) , D κ u ( τ ) ) + Ψ χ 2 ( τ ) d τ + [ η 7 t κ 2 ( 1 η 7 ) Γ ( κ 1 ) + η 6 η 7 σ t κ 3 ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 2 ) + η 5 ( 1 + η 6 ) η 7 σ 2 t κ 4 2 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 3 ) ] 0 σ σ τ χ 2 ( τ , v ( τ ) , D κ u ( τ ) ) + Ψ χ 2 ( τ ) d τ + [ η 8 t κ 1 ( 1 η 8 ) Γ ( κ ) + η 7 η 8 σ t κ 2 ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 1 ) + η 6 ( 1 + η 7 ) η 8 σ 2 t κ 3 2 ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 2 ) + η 5 ( 1 + η 6 ) ( 1 + η 7 ) + η 6 + η 7 η 8 σ 3 t κ 4 6 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 3 ) ] 0 σ χ 2 ( τ , v ( τ ) , D κ u ( τ ) ) + Ψ χ 2 ( τ ) d τ .
From first equation of (30), we have
t 4 α | v ( t ) m 1 ( t ) | t 4 α Γ ( α ) 0 t t τ α 1 | Ψ χ 1 ( τ ) | d τ + | η 1 6 ( 1 η 1 ) Γ ( α 3 ) | 0 σ σ τ 3 | Ψ χ 1 ( τ ) | d τ + | η 2 t 2 ( 1 η 2 ) Γ ( α 2 ) + η 1 η 2 σ 2 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 3 ) | 0 σ σ τ 2 | Ψ χ 1 ( τ ) | d τ + | η 3 t 2 ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 t σ ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) η 3 σ 2 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) | 0 σ σ τ | Ψ χ 1 ( τ ) | d τ + | η 4 t 3 ( 1 η 4 ) Γ ( α ) + η 3 η 4 σ t 2 ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 2 ( 1 + η 3 ) η 4 σ 2 t 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 3 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) | 0 σ | Ψ χ 1 ( τ ) | d τ .
where m 1 ( t ) are those terms which are free of Ψ χ 1 . Using (6) and ( A 1 ) of Remark 4, (31) becomes
v m 1 Q α ϵ α .
Similarly for second equation of (30), we obtain
u m 2 Q κ ϵ κ .

4.1. Method (I)

Theorem 4.
If hypothesis H 1 and
Λ = 1 Q α Q κ L χ 1 L χ 2 ( 1 Q α L ¯ χ 1 ) ( 1 Q κ L ¯ χ 2 ) > 0
hold, with 0 Q α L ¯ χ 1 , Q κ L ¯ χ 2 < 1 . Then system (1) is H U stable.
Proof. 
Let ( v , u ) S be the solution of (22) and ( w , ζ ) S be the solution of following system:
D α w ( t ) χ 1 ( t , ζ ( t ) , D α w ( t ) ) = 0 , t J , D κ ζ ( t ) χ 2 ( t , w ( t ) , D κ ζ ( t ) ) = 0 , t J , D α 4 w ( 0 ) = η 1 D α 4 w ( σ ) , D α 3 w ( 0 ) = η 2 D α 3 w ( σ ) , D α 2 w ( 0 ) = η 3 D α 2 w ( σ ) , D α 1 w ( 0 ) = η 4 D α 1 w ( σ ) , D κ 4 ζ ( 0 ) = η 5 D κ 4 ζ ( σ ) , D κ 3 ζ ( 0 ) = η 6 D κ 3 ζ ( σ ) , D κ 2 ζ ( 0 ) = η 7 D κ 2 ζ ( σ ) , D κ 1 w ( 0 ) = η 8 D κ 1 ζ ( σ ) .
Then in view of Lemma 1, for t J the solution of (33) is given by:
w ( t ) = 1 Γ ( α ) 0 t t τ α 1 χ 1 ( τ , ζ ( τ ) , D α w ( τ ) ) d τ + η 1 t α 4 6 ( 1 η 1 ) Γ ( α 3 ) × 0 σ σ τ 3 χ 1 ( τ , ζ ( τ ) , D α w ( τ ) ) d τ + η 2 t α 3 2 ( 1 η 2 ) Γ ( α 2 ) + η 1 η 2 σ t α 4 2 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 3 ) × 0 σ σ τ 2 χ 1 ( τ , ζ ( τ ) , D α w ( τ ) ) d τ + [ η 3 t α 2 ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 σ t α 3 ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) η 3 σ 2 t α 4 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) ] 0 σ σ τ χ 1 ( τ , ζ ( τ ) , D α w ( τ ) ) d τ + [ η 4 t α 1 ( 1 η 4 ) Γ ( α ) + η 3 η 4 σ t α 2 ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 2 ( 1 + η 3 ) η 4 σ 2 t α 3 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 3 t α 4 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) ] 0 σ χ 1 ( τ , ζ ( τ ) , D α w ( τ ) ) d τ , ζ ( t ) = 1 Γ ( κ ) 0 t t τ κ 1 χ 2 ( τ , w ( τ ) , D κ ζ ( τ ) ) d τ + η 5 t κ 4 6 ( 1 η 5 ) Γ ( κ 3 ) × 0 σ σ τ 3 χ 2 ( τ , w ( τ ) , D κ ζ ( τ ) ) d τ + η 6 t κ 3 2 ( 1 η 6 ) Γ ( κ 2 ) + η 5 η 6 σ t κ 4 2 ( 1 η 5 ) ( 1 η 6 ) Γ ( κ 3 ) × 0 σ σ τ 2 χ 2 ( τ , w ( τ ) , D κ ζ ( τ ) ) d τ + [ η 7 t κ 2 ( 1 η 7 ) Γ ( κ 1 ) + η 6 η 7 σ t κ 3 ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 2 ) + η 5 ( 1 + η 6 ) η 7 σ 2 t κ 4 2 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) Γ ( κ 3 ) ] 0 σ σ τ χ 2 ( τ , w ( τ ) , D κ ζ ( τ ) ) d τ + [ η 8 t κ 1 ( 1 η 8 ) Γ ( κ ) + η 7 η 8 σ t κ 2 ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 1 ) + η 6 ( 1 + η 7 ) η 8 σ 2 t κ 3 2 ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 2 ) + η 5 ( 1 + η 6 ) ( 1 + η 7 ) + η 6 + η 7 η 8 σ 3 t κ 4 6 ( 1 η 5 ) ( 1 η 6 ) ( 1 η 7 ) ( 1 η 8 ) Γ ( κ 3 ) ] 0 σ χ 2 ( τ , w ( τ ) , D κ ζ ( τ ) ) d τ .
Consider
t 4 α | v ( t ) w ( t ) | t 4 α | v ( t ) m 1 ( t ) | + t 4 α | m 1 ( t ) w ( t ) | .
Applying Lemma 3 in (35), we get
t 4 α | v ( t ) w ( t ) | Q α ϵ α + t 4 α Γ ( α ) 0 t t τ α 1 | χ 1 ( τ , u ( τ ) , D α v ( τ ) ) χ 1 ( τ , ζ ( τ ) , D α w ( τ ) ) | d τ + | η 1 6 ( 1 η 1 ) Γ ( α 3 ) | × 0 σ σ τ 3 | χ 1 ( τ , u ( τ ) , D α v ( τ ) ) χ 1 ( τ , ζ ( τ ) , D α w ( τ ) ) | d τ + | η 2 t 2 ( 1 η 2 ) Γ ( α 2 ) + η 1 η 2 σ 2 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 3 ) | 0 σ σ τ 2 | χ 1 ( τ , u ( τ ) , D α v ( τ ) ) χ 1 ( τ , ζ ( τ ) , D α w ( τ ) ) | d τ + | η 3 t 2 ( 1 η 3 ) Γ ( α 1 ) + η 2 η 3 t σ ( 1 η 2 ) ( 1 η 3 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) η 3 σ 2 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) | × 0 σ σ τ | χ 1 ( τ , u ( τ ) , D α v ( τ ) ) χ 1 ( τ , ζ ( τ ) , D α w ( τ ) ) | d τ + | η 4 t 3 ( 1 η 4 ) Γ ( α ) + η 3 η 4 σ t 2 ( 1 η 3 ) ( 1 η 4 ) Γ ( α 1 ) + η 2 ( 1 + η 3 ) η 4 σ 2 t 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) + η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 3 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) | × 0 σ | χ 1 ( τ , u ( τ ) , D α v ( τ ) ) χ 1 ( τ , ζ ( τ ) , D α w ( τ ) ) | d τ Q α ϵ α + [ σ 4 Γ ( α + 1 ) + | η 1 σ 4 24 ( 1 η 1 ) Γ ( α 3 ) | + η 2 ( 1 η 1 ) σ 4 + η 1 η 2 σ 4 ( α 3 ) 6 ( 1 η 1 ) ( 1 η 2 ) Γ ( α 2 ) | + | η 3 ( 1 η 2 ) σ 4 + η 2 η 3 σ 4 ( α 2 ) 2 ( 1 η 2 ) ( 1 η 3 ) Γ ( α 1 ) | + | η 1 ( 1 + η 2 ) η 3 σ 4 2 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) Γ ( α 3 ) | + | ( 1 η 3 ) η 4 σ 4 + η 3 η 4 σ 4 ( α 1 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α ) | + | η 2 ( 1 + η 3 ) η 4 σ 4 2 ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 2 ) | + | η 1 ( 1 + η 2 ) ( 1 + η 3 ) + η 2 + η 3 η 4 σ 4 6 ( 1 η 1 ) ( 1 η 2 ) ( 1 η 3 ) ( 1 η 4 ) Γ ( α 3 ) | ] L χ 1 u ζ + L ¯ χ 1 D α v D α w .
Using H 1 of Theorem 1 and (6) in (36), we have
v w Q α ϵ α 1 Q α L ¯ χ 1 + Q α L χ 1 1 Q α L ¯ χ 1 u ζ .
Similarly, we can get
u ζ Q κ ϵ κ 1 Q κ L ¯ χ 2 + Q κ L χ 2 1 Q κ L ¯ χ 2 v w .
We write (37) and (38) as
v w Q α L χ 1 1 Q α L ¯ χ 1 u ζ Q α ϵ α 1 Q α L ¯ χ 1 , u ζ Q κ L χ 2 1 Q κ L ¯ χ 2 v w Q κ ϵ κ 1 Q κ L ¯ χ 2 , 1 Q α L χ 1 1 Q α L ¯ χ 1 Q κ L χ 2 1 Q κ L ¯ χ 2 1 v w u ζ Q α ϵ α 1 Q α L ¯ χ 1 Q κ ϵ κ 1 Q κ L ¯ χ 2 .
From the above, we get
v w u ζ 1 Λ Q α L χ 1 Λ ( 1 Q α L ¯ χ 1 ) Q κ L χ 2 Λ ( 1 Q κ L ¯ χ 2 ) 1 Λ Q α ϵ α 1 Q α L ¯ χ 1 Q κ ϵ κ 1 Q κ L ¯ χ 2 ,
where
Λ = 1 Q α Q κ L χ 1 L χ 2 ( 1 Q α L ¯ χ 1 ) ( 1 Q κ L ¯ χ 2 ) > 0 .
Further simplification gives
v w Q α ϵ α Λ ( 1 Q α L ¯ χ 1 ) + Q α Q κ L χ 1 ϵ κ Λ ( 1 Q α L ¯ χ 1 ) ( 1 Q κ L ¯ χ 2 ) , u ζ Q κ ϵ κ Λ ( 1 Q κ L ¯ χ 2 ) + Q α Q κ L χ 2 ϵ α Λ ( 1 Q α L ¯ χ 1 ) ( 1 Q κ L ¯ χ 2 ) ,
from which we have
v w + u ζ Q α ϵ α Λ ( 1 Q α L ¯ χ 1 ) + Q κ ϵ κ Λ ( 1 Q κ L ¯ χ 2 ) + Q α Q κ L χ 1 ϵ κ Λ ( 1 Q α L ¯ χ 1 ) ( 1 Q κ L ¯ χ 2 ) + Q α Q κ L χ 2 ϵ α Λ ( 1 Q α L ¯ χ 1 ) ( 1 Q κ L ¯ χ 2 ) .
Let ϵ = max ϵ α , ϵ κ , then from (39) we have
( v , u ) ( w , ζ ) S C α , κ ϵ ,
where
C α , κ = Q α Λ ( 1 Q α L ¯ χ 1 ) + Q κ Λ ( 1 Q κ L ¯ χ 2 ) + Q α Q κ L χ 1 Λ ( 1 Q α L ¯ χ 1 ) ( 1 Q κ L ¯ χ 2 ) + Q α Q κ L χ 2 Λ ( 1 Q α L ¯ χ 1 ) ( 1 Q κ L ¯ χ 2 ) .
Remark 5.
By setting Φ α , κ ( ϵ ) = C α , κ ϵ , Φ α , κ ( 0 ) = 0 in (40), then by Definition 4 the problem (1) is generalized H U stable.
H3:
Let functions Θ α , Θ κ : J R + be nondecreasing. Then, there are ζ Θ α , ζ Θ κ > 0 , such that for every t J , the inequalities
I α Θ α ( t ) ζ Θ α Θ α ( t ) and I κ Θ κ ( t ) ζ Θ κ Θ κ ( t )
holds.
Remark 6.
Lemma 3 and Theorem 4 gives that the system (1) is H U –Rassias and generalized H U –Rassias stable, if ϵ α = Θ α ( t ) ϵ α and ϵ κ = Θ κ ( t ) ϵ κ with H 3 and Λ > 0 .

4.2. Method (II)

Theorem 5.
Under the hypothesis H 1 and if Λ * = 1 Q κ L χ 2 1 Q κ L ¯ χ 2 + Q α L χ 1 1 Q α L ¯ χ 1 > 0 . Then system (1) is H U stable.
Proof. 
From inequality (37) and (38), we have
v w + u ζ Q α ϵ α 1 Q α L ¯ χ 1 + Q κ ϵ κ 1 Q κ L ¯ χ 2 + Q κ L χ 2 1 Q κ L ¯ χ 2 v w + Q α L χ 1 1 Q α L ¯ χ 1 u ζ .
Let max ϵ α , ϵ κ = ϵ , then from (41) we obtain
( v , u ) ( w , ζ ) S C α , κ ϵ ,
where
C α , κ = Q α Λ * ( 1 Q α L ¯ χ 1 ) + Q κ Λ * ( 1 Q κ L ¯ χ 2 ) .
Remark 7.
With the help of Remark 5, we can obtain the generalized H U stability of system (1).
Remark 8.
Lemma 3 and Theorem 5 gives that the system (1) is H U –Rassias and generalized H U –Rassias stable, if ϵ α = Θ α ( t ) ϵ α and ϵ κ = Θ κ ( t ) ϵ κ with H 3 and Λ * > 0 .
Remark 9.
The results of coupled systems of fourth-order nonlinear F D E s gives the results of fourth-order nonlinear system of O D E s (If α , κ = 4 ) with anti-periodic and initial conditions, if η i = 1 ( i = 1 , 2 , , 8 ) and η i = 0 ( i = 1 , 2 , , 8 ) respectively.

5. Example

Example 1.
Consider the following coupled system of F D E s :
D α v ( t ) 1 4 ( t + 2 ) 2 | D α v ( t ) | 1 + | D α v ( t ) | + 1 16 sin 2 u ( t ) = 0 , t [ 0 , 1 ] , D κ u ( t ) 1 32 π sin ( 2 π v ( t ) ) + | D κ u ( t ) | 16 1 + | D κ u ( t ) | + 1 2 = 0 , t [ 0 , 1 ] , D α 4 v ( 0 ) = η 1 D α 4 v ( σ ) , D α 3 v ( 0 ) = η 2 D α 3 v ( σ ) , D α 2 v ( 0 ) = η 3 D α 2 v ( σ ) , D α 1 v ( 0 ) = η 4 D α 1 v ( σ ) , D κ 4 u ( 0 ) = η 5 D κ 4 u ( σ ) , D κ 3 u ( 0 ) = η 6 D κ 3 u ( σ ) , D κ 2 u ( 0 ) = η 7 D κ 2 u ( σ ) , D κ 1 u ( 0 ) = η 8 D κ 1 u ( σ ) .
From system (43), we can see α = κ = 10 3 , σ = 1 , η 1 = η 5 = 1 2 , η 2 = η 6 = 1 3 , η 3 = η 7 = 1 and η 4 = η 8 = 1 . Moreover, we have
| χ 1 ( t , u 1 ( t ) , D α v 1 ( t ) ) χ 1 ( t , u 2 ( t ) , D α v 2 ( t ) ) | 1 16 | u 1 ( t ) u 2 ( t ) | + 1 16 | D α v 1 ( t ) D α v 2 ( t ) | , | χ 2 ( t , v 1 ( t ) , D κ u 1 ( t ) ) χ 2 ( t , v 2 ( t ) , D κ u 2 ( t ) ) | 1 16 | v 1 ( t ) v 2 ( t ) | + 1 16 | D κ u 1 ( t ) D κ u 2 ( t ) | .
Therefore, we get L χ 1 = L ¯ χ 1 = L χ 2 = L ¯ χ 2 = 1 16 . Therefore,
Q α L χ 1 ( 1 L ¯ χ 2 ) + Q κ L χ 2 ( 1 L ¯ χ 1 ) ( 1 L ¯ χ 2 ) ( 1 L ¯ χ 1 ) 0.75141 < 1 ,
Thus, solution of (43) is unique. Moreover, system (43) is H U , generalized H U , H U –Rassias and generalized H U –Rassias stable by two different approaches under the conditions of Theorem 4 and Theorem 5, i.e., Λ > 0 and Λ * > 0 .

6. Conclusions

This paper concluded that the solution of coupled implicit F D E s s (1) is unique and exists by using the Banach contraction theorem and Leray–Schauder fixed point theorem. Under some assumptions, the aforesaid coupled system has at least one solution. Besides this, the considered coupled system is H U , generalized H U , H U –Rassias and generalized H U –Rassias stable. An example is presented to illustrate our obtained results. The proposed system (1) gives the following well-known system of O D E s , which has wide applications in applied sciences [5]
  • η i = 1 ( i = 1 , 2 , , 8 ) and α , κ = 4 , then we get fourth-order O D E s system with anti-periodic boundary conditions.
  • η i = 0 ( i = 1 , 2 , , 8 ) and α , κ = 4 , then we get fourth-order O D E s system with initial conditions.

Author Contributions

Conceptualization, U.R., A.Z., Z.A., I.-L.P., S.R., S.E.; investigation, U.R., A.Z., Z.A., I.-L.P., S.R., S.E.; writing—original draft preparation, U.R., A.Z., Z.A., I.-L.P., S.R., S.E.; writing—review and editing, U.R., A.Z., Z.A., I.-L.P., S.R., S.E. All authors have read and agreed to the published version of the manuscript.

Funding

Not applicable.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

The fifth author was supported by Azarbaijan Shahid Madani University.

Conflicts of Interest

The authors declare no conflict of interest.

Sample Availability

Data sharing not applicable to this article as no datasets were generated or analyzed during the current study.

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Riaz, U.; Zada, A.; Ali, Z.; Popa, I.-L.; Rezapour, S.; Etemad, S. On a Riemann–Liouville Type Implicit Coupled System via Generalized Boundary Conditions. Mathematics 2021, 9, 1205. https://doi.org/10.3390/math9111205

AMA Style

Riaz U, Zada A, Ali Z, Popa I-L, Rezapour S, Etemad S. On a Riemann–Liouville Type Implicit Coupled System via Generalized Boundary Conditions. Mathematics. 2021; 9(11):1205. https://doi.org/10.3390/math9111205

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Riaz, Usman, Akbar Zada, Zeeshan Ali, Ioan-Lucian Popa, Shahram Rezapour, and Sina Etemad. 2021. "On a Riemann–Liouville Type Implicit Coupled System via Generalized Boundary Conditions" Mathematics 9, no. 11: 1205. https://doi.org/10.3390/math9111205

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Riaz, U., Zada, A., Ali, Z., Popa, I. -L., Rezapour, S., & Etemad, S. (2021). On a Riemann–Liouville Type Implicit Coupled System via Generalized Boundary Conditions. Mathematics, 9(11), 1205. https://doi.org/10.3390/math9111205

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