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Article

Existence Results for Coupled Implicit \({\psi}\)-Riemann–Liouville Fractional Differential Equations with Nonlocal Conditions

1
Department of Mathematics, Huaiyin Normal University, Huaian 223300, China
2
Huaiyin High School in Jiangsu Province, Huaian 223002, China
*
Author to whom correspondence should be addressed.
Axioms 2022, 11(3), 103; https://doi.org/10.3390/axioms11030103
Submission received: 12 January 2022 / Revised: 18 February 2022 / Accepted: 21 February 2022 / Published: 25 February 2022

Abstract

:
In this paper, we study the existence and uniqueness of solutions for a coupled implicit system involving ψ -Riemann–Liouville fractional derivative with nonlocal conditions. We first transformed the coupled implicit problem into an integral system and then analyzed the uniqueness and existence of this integral system by means of Banach fixed-point theorem and Krasnoselskiis fixed-point theorem. Some known results in the literature are extended. Finally, an example is given to illustrate our theoretical result.

1. Introduction

The fractional calculus is an important branch of mathematics and its wide applications to many fields, such engineering, economics, physics, chemistry, finance, control of dynamical systems, and so on—see [1,2,3,4,5,6,7], and the references cited therein. One of the proposed generalizations of the fractional calculus operators is the ψ -fractional operator—see [8,9,10] and references therein for its wide applications. Some properties of this operator could be found in [11,12,13].
As we all know, the coupled system of fractional differential equations is becoming a more popular research field due to its vast applications in real-time problems, namely anomalous diffusion, chaotic systems, disease models, and ecological models [14,15,16]. Recently, the coupled system of fractional differential equations has been considered extensively in the literature. Alsaedi et al. [17] researched the uniqueness and existence of solutions for a nonlinear system of Riemann–Liouville fractional differential equations equipped with nonseparated semi-coupled integro-multipoint boundary conditions. Baleanu et al. [18] studied the uniqueness existence and Ulam stability for a coupled system involving generalized Sturm–Liouville problems and Langevin fractional differential equations described by Atangana–Baleanu–Caputo derivatives by virtue of the notable Mittag–Leffler kernel. Muthaiah et al. [19] presented the existence, uniqueness, and Hyers–Ulam stability of the coupled system of Caputo–Hadamard-type fractional differential equations with multipoint and nonlocal integral boundary conditions. Based on the features of the Hadamard fractional derivative, the implementation of fixed-point theorems, the employment of Urs’s stability approach, and the existence, uniqueness, and stability of the coupled system of nonlinear Langevin equations involving Caputo–Hadamard fractional derivative—subject to nonperiodic boundary conditions—are established by Matar et al. [20]. In [21], by using the coincidence degree theory, Zhang et al. established the existence and uniqueness theorems for the coupled systems of implicit fractional differential equations with periodic boundary conditions.
In recent years, the study of basic theories of initial and boundary value problems for implicit fractional differential equations and integral equations with Caputo fractional derivative and Riemann–Liouville fractional derivative has been paid to much attention. In [22], Benchohra and Souid obtained integrable solutions for initial value problem of implicit fractional differential equations. Nieto et al. [23] studied initial value problem for an implicit fractional differential equation using a fixed-point theory and approximation method. Furthermore, in [24] Benchohra and Bouriah established existence and various stability results for a class of boundary value problem for implicit fractional differential equation with Caputo fractional derivative. Implicit fractional differential equations play a key role in different problems, the readers are referred to see [25,26,27,28].
In [29], Benchohra et al. discussed the existence and Ulam stability analysis of the following nonlinear implicit fractional differential equation with initial value condition:
D 0 + α x ( t ) = f ( t , x ( t ) , D 0 + α x ( t ) ) , t ( 0 , T ] , t 1 α x ( t ) | t = 0 = x 0 , x 0 R ,
where D 0 + α is the standard Riemann–Liouville fractional derivative, f : ( 0 ; T ] × R × R R is a continuous function, and 0 < α < 1 .
Very recently, in [30], Lachouri et al. studied the existence and uniqueness of solutions for the following nonlinear implicit Riemann–Liouville fractional differential equation with nonlocal condition:
D 0 + α x ( t ) = f ( t , x ( t ) , D 0 + α x ( t ) ) , t ( 0 , T ] , t 1 α x ( t ) | t = 0 = x 0 g ( x ) , x 0 R ,
where D 0 + α and f are as in (1.1), g : C ( ( 0 , T ] , R ) R is a continuous nonlinear function.
Motivated by the above works, we consider the following coupled implicit ψ -Riemann–Liouville fractional differential equations with nonlocal conditions:
D 0 + α , ψ x ( t ) = f ( t , y ( t ) , D 0 + α , ψ x ( t ) ) , t ( 0 , T ] , D 0 + α , ψ y ( t ) = g ( t , x ( t ) , D 0 + α , ψ y ( t ) ) , t ( 0 , T ] , ( ψ ( t ) ψ ( 0 ) ) 1 α x ( t ) | t = 0 = y 0 r ( y ) , y 0 R , ( ψ ( t ) ψ ( 0 ) ) 1 α y ( t ) | t = 0 = x 0 h ( x ) , x 0 R ,
where D 0 + α , ψ x ( t ) is the Riemann–Liouville fractional derivative of a function x with respect to another function ψ , which is increasing, and ψ ( t ) 0 for all t [ 0 , T ] , f , g : ( 0 , T ] × R × R R are two continuous functions, and 0 < α < 1 , h , r : C ( ( 0 , T ] , R ) R are two continuous nonlinear functions.
To the best of our knowledge, there are no papers on coupled implicit fractional differential equations including fractional derivative of a function with respect to another function. We cover this gap in this paper.
In this paper, our aim is to present the sufficient conditions for the existence and uniqueness of solutions for coupled implicit system (3). First of all, we transform (3) into an integral system and then we study the existence and uniqueness of solutions by the Banach and Krasnoselskii fixed-point theorems. Finally, an example is given to illustrate our main results. Our results extend the main results of [30].
This paper will be organized as follows. In Section 2, we will briefly recall some notations, definitions and preliminaries. Section 3 is devoted to proving the existence and uniqueness of the solution for system (3). In Section 4, an example is given to illustrate our theoretical result. Finally, we present some conclusions in Section 5.

2. Preliminaries

In this section, we provided some basic definitions and lemmas which are used in the sequel.
Definition 1
([13]). Let α > 0 , f be an integrable function defined on [ a , b ] and ψ C 1 ( [ a , b ] ) be an increasing function with ψ ( t ) 0 for all t [ a , b ] . The left ψ-Riemann–Liouville fractional integral operator of order α of a function f is defined by:
I a + α , ψ f ( t ) = 1 Γ ( α ) a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) α 1 f ( s ) d s .
Definition 2
([13]). Let n 1 < α < n , f C n ( [ a , b ] ) and ψ C n ( [ a , b ] ) be an increasing function with ψ ( t ) 0 for all t [ a , b ] . The left ψ-Riemann–Liouville fractional derivative of order α of a function f is defined by:
D a + α , ψ f ( t ) = 1 ψ ( t ) d d t n I a + n α , ψ f ( t )
= 1 Γ ( n α ) 1 ψ ( t ) d d t n a t ψ ( s ) ( ψ ( t ) ψ ( s ) ) n α 1 f ( s ) d s ,
where n = [ α ] + 1 .
Lemma 1
([13]). Let α > 0 and β > 0 , then
(i) 
I a + α , ψ ( ψ ( s ) ψ ( a ) ) β 1 ( t ) = Γ ( β ) Γ ( β + α ) ( ψ ( t ) ψ ( a ) ) β + α 1 ,
(ii) 
D a + α , ψ ( ψ ( s ) ψ ( a ) ) β 1 ( t ) = Γ ( β ) Γ ( β α ) ( ψ ( t ) ψ ( a ) ) β α 1 .
In the following, we will give the combinations of the fractional integral and the fractional derivatives of a function with respect to another function.
Lemma 2
([11]). Let f C n ( [ a , b ] ) and n 1 < α < n . Then we have
(1) 
D a + α , ψ I a + α , ψ f ( t ) = f ( t ) ;
(2) 
I a + α , ψ D a + α , ψ f ( t ) = f ( t ) k = 1 n f [ k 1 ] ( a + ) Γ ( k α ) ( ψ ( t ) ψ ( a ) ) k α ,
where f [ k ] ( t ) : = 1 ψ ( t ) d d t k f ( t ) on [ a , b ] . In particular, given α ( 0 , 1 ) , one has
I a + α , ψ D a + α , ψ f ( t ) = f ( t ) c ( t a ) α 1 ,
where c is a constant.
Lemma 3.
( x , y ) solves (3) if—and only if—it is a solution of integral system.
x ( t ) = ( ψ ( t ) ψ ( 0 ) ) α 1 ( y 0 r ( y ) ) + 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 f ( s , y ( s ) , D 0 + α , ψ x ( s ) ) ψ ( s ) d s , y ( t ) = ( ψ ( t ) ψ ( 0 ) ) α 1 ( x 0 h ( x ) ) + 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 g ( s , x ( s ) , D 0 + α , ψ y ( s ) ) ψ ( s ) d s .
Proof. 
If ( x , y ) satisfies the problem (3), then applying I 0 + α , ψ to both sides of the first equation and second equation of (3), respectively, we have
I 0 + α , ψ D 0 + α , ψ x ( t ) = I 0 + α , ψ f ( t , y ( t ) , D 0 + α , ψ x ( t ) ) ,
and
I 0 + α , ϕ D 0 + α , ψ y ( t ) = I 0 + α , ψ g ( t , x ( t ) , D 0 + α , ψ y ( t ) ) .
By Lemma 2, we obtain
x ( t ) = c 1 ( ψ ( t ) ψ ( 0 ) ) α 1 + 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 f ( s , y ( s ) , D 0 + α , ψ x ( s ) ) ψ ( s ) d s , y ( t ) = c 2 ( ψ ( t ) ψ ( 0 ) ) α 1 + 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 g ( s , x ( s ) , D 0 + α , ψ y ( s ) ) ψ ( s ) d s ,
where t ( 0 , T ] . In view of the following conditions:
( ψ ( t ) ψ ( 0 ) ) 1 α x ( t ) | t = 0 = y 0 r ( y ) , ( ψ ( t ) ψ ( 0 ) ) 1 α y ( t ) | t = 0 = x 0 h ( x ) ,
we obtain
c 1 = y 0 r ( y ) , c 2 = x 0 h ( x ) .
Substituting (6) into (5) we obtain the integral system (4). □
Theorem 1 
((Krasnoselskll’s fixed point theorem) [31]). Let Ω be a non-empty closed bounded convex subset of a Banach space E. Suppose that F 1 and F 2 map Ω into E, such that
(i) 
F 1 x + F 2 y Ω for all x , y Ω ;
(ii) 
F 1 is continuous and compact;
(iii) 
F 2 is a contraction with constant k < 1 .
Then, there is a z Ω , with F 1 z + F 2 z = z .

3. Main Results

Let γ > 0 , T > 0 and J = [ 0 , T ] , we denote the weighted space of the following continuous functions:
C γ , ψ ( J , R ) = { x : ( 0 , T ] R | ( ψ ( t ) ψ ( 0 ) ) γ x C ( J , R ) } ,
with the norm
x C γ , ψ = sup t J | ( ψ ( t ) ψ ( 0 ) ) γ x ( t ) | .
In fact, we have (i) x C γ , ψ 0 , (ii) k x C γ , ψ = | k | x C γ , ψ , and
(iii) x + y C γ , ψ = sup t J | ( ψ ( t ) ψ ( 0 ) ) γ ( x ( t ) + y ( t ) ) |
sup t J | ( ψ ( t ) ψ ( 0 ) ) γ x ( t ) | + sup t J | ( ψ ( t ) ψ ( 0 ) ) γ y ( t ) | = x C γ , ψ + y C γ , ψ .
Thus, C γ , ψ ( J , R ) is a Banach space.
Define the operators A : C 1 α , ψ ( J , R ) × C 1 α , ψ ( J , R ) C 1 α , ψ ( J , R ) × C 1 α , ψ ( J , R ) by
A ( x , y ) ( t ) = ( A 1 ( x , y ) ( t ) , A 2 ( x , y ) ( t ) ) , t ( 0 , T ] ,
where
A 1 ( x , y ) ( t ) = ( ψ ( t ) ψ ( 0 ) ) α 1 ( y 0 r ( y ) ) + 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 p ( s ) ψ ( s ) d s ,
and
A 2 ( x , y ) ( t ) = ( ψ ( t ) ψ ( 0 ) ) α 1 ( x 0 h ( x ) ) + 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 q ( s ) ψ ( s ) d s ,
here, p , q : ( 0 , T ] R are two functions satisfying the functional equations
p ( t ) = f ( t , y ( t ) , p ( t ) ) , q ( t ) = g ( t , x ( t ) , q ( t ) ) .
The operator A is well-defined, i.e., for every ( x , y ) C 1 α , ψ ( J , R ) × C 1 α , ψ ( J , R ) and t > 0 , the following integrals
1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) p ( s ) d s ,
and
1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) q ( s ) d s
belong to C 1 α , ψ ( J , R ) .
For convenience, we allow the following hypothesis.
Hypothesis 1 (H1).
There exist constants L 1 , l 1 > 0 and L 2 , l 2 ( 0 , 1 ) such that
| f ( t , u 1 , v 1 ) f ( t , u 2 , v 2 ) | L 1 | u 1 u 2 | + L 2 | v 1 v 2 | ,
| g ( t , u 1 , v 1 ) g ( t , u 2 , v 2 ) | l 1 | u 1 u 2 | + l 2 | v 1 v 2 | ,
for t ( 0 , T ] , u 1 , u 2 , v 1 , v 2 R and f ( · , 0 , 0 ) , g ( · , 0 , 0 ) C 1 α , ψ ( J , R ) .
Hypothesis 2 (H2).
There exist two constant b 1 , b 2 ( 0 , 1 ) , such that
| h ( x ) h ( y ) | b 1 x y C 1 α , ψ , | r ( x ) r ( y ) | b 2 x y C 1 α , ψ ,
for x , y C 1 α , ψ ( J , R ) .
Hypothesis 3 (H3).
There exist m 1 , n 1 C 1 α , ψ ( J , R + ) , m 2 , n 2 , m 3 , n 3 C ( J , R + ) with m 3 * = sup t J m 3 ( t ) < 1 and n 3 * = sup t J n 3 ( t ) < 1 , such that
| f ( t , u , v ) | m 1 ( t ) + m 2 ( t ) | u | + m 3 ( t ) | v | ,
| g ( t , u , v ) | n 1 ( t ) + n 2 ( t ) | u | + n 3 ( t ) | v | ,
for t ( 0 , T ] and each u , v s . R .
Theorem 2.
Assume that (H1)–(H2) hold. If the following is true:
k = max b 2 + L 1 Γ ( α ) Γ ( 2 α ) ( 1 L 2 ) ( ψ ( T ) ψ ( 0 ) ) α , b 1 + l 1 Γ ( α ) Γ ( 2 α ) ( 1 l 2 ) ( ψ ( T ) ψ ( 0 ) ) α < 1 ,
then there exists a unique solution for the BVP (3) in the space C 1 α , ψ ( J , R ) × C 1 α , ψ ( J , R ) .
Proof. 
In the following, we will prove that the operator A has unique fixed point. From (8), one has by condition (H1) that
| p ( t ) | | f ( t , y ( t ) , p ( t ) ) f ( t , 0 , 0 ) | + | f ( t , 0 , 0 ) | L 1 | y ( t ) | + L 2 | p ( t ) | + | f ( t , 0 , 0 ) | ,
which implies that
| p ( t ) | L 1 1 L 2 | y ( t ) | + e 1 ( ψ ( t ) ψ ( 0 ) ) α 1 , t ( 0 , T ] ,
where e 1 = sup t J | ( ψ ( t ) ψ ( 0 ) ) 1 α f ( t , 0 , 0 ) | 1 L 2 < + . Similarly, we have
| q ( t ) | l 1 1 l 2 | x ( t ) | + e 2 ( ψ ( t ) ψ ( 0 ) ) α 1 , t ( 0 , T ] ,
where e 2 = sup t J | ( ψ ( t ) ψ ( 0 ) ) 1 α g ( t , 0 , 0 ) | 1 l 2 < + . For each x , y C 1 α , ψ ( J , R ) , by Lemma 1, we obtain
( ψ ( t ) ψ ( 0 ) ) 1 α Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) p ( s ) d s ( ψ ( t ) ψ ( 0 ) ) 1 α Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) | p ( s ) | d s ( ψ ( t ) ψ ( 0 ) ) 1 α Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) L 1 1 L 2 | y ( s ) | + e 1 ( ψ ( s ) ψ ( 0 ) ) α 1 d s ( ψ ( t ) ψ ( 0 ) ) 1 α Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) ( ψ ( s ) ψ ( 0 ) ) α 1 · L 1 1 L 2 | ( ψ ( s ) ψ ( 0 ) ) 1 α y ( s ) | + e 1 d s L 1 1 L 2 y C 1 α , ψ + e 1 ( ψ ( t ) ψ ( 0 ) ) 1 α I α , ψ ( ( ψ ( t ) ψ ( 0 ) ) α 1 ) L 1 1 L 2 y C 1 α , ψ + e 1 Γ ( α ) ( ψ ( t ) ψ ( 0 ) ) α Γ ( 2 α ) L 1 1 L 2 y C 1 α , ψ + e 1 Γ ( α ) Γ ( 2 α ) ( ψ ( T ) ψ ( 0 ) ) α .
Similarly, we have
( ψ ( t ) ψ ( 0 ) ) 1 α Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) q ( s ) d s
l 1 1 l 2 x C 1 α , ψ + e 2 Γ ( α ) Γ ( 2 α ) ( ψ ( T ) ψ ( 0 ) ) α .
Thus, the integrals exist and belong to C 1 α , ψ ( J , R ) . Let ( x , y ) , ( u , v ) C 1 α , ψ ( J , R ) × C 1 α , ψ ( J , R ) . Then, for each t ( 0 , T ] , we obtain
| A 1 ( x , y ) ( t ) A 1 ( u , v ) ) ( t ) | ( ψ ( t ) ψ ( 0 ) ) α 1 | r ( y ) r ( v ) | + 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) | p y ( s ) p v ( s ) | d s ,
and
| A 2 ( x , y ) ( t ) A 2 ( u , v ) ( t ) | ( ψ ( t ) ψ ( 0 ) ) α 1 | h ( x ) h ( u ) | + 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) | q x ( s ) q u ( s ) | d s ,
where p y , p v , q x , q u C 1 α , ψ ( J , R ) , such that
p y = f ( t , y ( t ) , p y ( t ) ) , p v = f ( t , v ( t ) , p v ( t ) ) , q x = g ( t , x ( t ) , q x ( t ) ) , q u = g ( t , u ( t ) , q u ( t ) ) .
In view of (H1), one has
| p y ( t ) p v ( t ) | = | f ( t , y ( t ) , p y ( t ) ) f ( t , v ( t ) , p v ( t ) ) | L 1 | y ( t ) v ( t ) | + L 2 | p y ( t ) p v ( t ) | ,
and
| q x ( t ) q u ( t ) | = | g ( t , x ( t ) , q x ( t ) ) g ( t , u ( t ) , q u ( t ) ) | l 1 | x ( t ) u ( t ) | + l 2 | q x ( t ) q u ( t ) | ,
which implies that
| p y ( t ) p v ( t ) | L 1 1 L 2 | y ( t ) v ( t ) | , | q x ( t ) q u ( t ) | l 1 1 l 2 | x ( t ) u ( t ) | .
Hence, for every t ( 0 , T ]
| A 1 ( x , y ) ( t ) A 1 ( u , v ) ( t ) | ( ψ ( t ) ψ ( 0 ) ) α 1 | r ( y ) r ( v ) | + 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) | p y ( s ) p v ( s ) | d s b 2 ( ψ ( t ) ψ ( 0 ) ) α 1 y v C 1 α , ψ + L 1 Γ ( α ) ( 1 L 2 ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ψ ( s ) | y ( s ) v ( s ) | d s b 2 ( ψ ( t ) ψ ( 0 ) ) α 1 y v C 1 α , ψ + L 1 Γ ( α ) ( 1 L 2 ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ( ψ ( s ) ψ ( 0 ) ) α 1 | ( ψ ( s ) ψ ( 0 ) ) 1 α ( y ( s ) v ( s ) ) | ψ ( s ) d s b 2 ( ψ ( t ) ψ ( 0 ) ) α 1 y v C 1 α , ψ + L 1 1 L 2 I α , ψ ( ( ψ ( t ) ψ ( 0 ) ) α 1 ) y v C 1 α , ψ b 2 ( ψ ( t ) ψ ( 0 ) ) α 1 y v C 1 α , ψ + L 1 Γ ( α ) Γ ( 2 α ) ( 1 L 2 ) ( ψ ( t ) ψ ( 0 ) ) 2 α 1 y v C 1 α , ψ .
So, we obtain the following:
( ψ ( t ) ψ ( 0 ) ) 1 α | A 1 ( x , y ) ( t ) A 1 ( u , v ) ( t ) | b 2 + L 1 Γ ( α ) Γ ( 2 α ) ( 1 L 2 ) ( ψ ( t ) ψ ( 0 ) ) α y v C 1 α , ψ
That is, as follows:
A 1 ( x , y ) A 1 ( u , v ) C 1 α , ψ b 2 + L 1 Γ ( α ) Γ ( 2 α ) ( 1 L 2 ) ( ψ ( T ) ψ ( 0 ) ) α y v C 1 α , ψ .
Similarly, we can obtain the following:
A 2 ( x , y ) A 2 ( u , v ) C 1 α , ψ b 1 + l 1 Γ ( α ) Γ ( 2 α ) ( 1 l 2 ) ( ψ ( T ) ψ ( 0 ) ) α x u C 1 α , ψ .
Thus, we have
A ( x , y ) A ( u , v ) C 1 α , ψ × C 1 α , ψ = A 1 ( x , y ) A 1 ( u , v ) C 1 α , ψ + A 2 ( x , y ) A 2 ( u , v ) C 1 α , ψ max b 2 + L 1 Γ ( α ) Γ ( 2 α ) ( 1 L 2 ) ( ψ ( T ) ψ ( 0 ) ) α , b 1 + l 1 Γ ( α ) Γ ( 2 α ) ( 1 l 2 ) ( ψ ( T ) ψ ( 0 ) ) α · ( x u C 1 α , ψ + y v C 1 α , ψ ) = k ( x , y ) ( u , v ) C 1 α , ψ × C 1 α , ψ .
where k = max b 2 + L 1 Γ ( α ) Γ ( 2 α ) ( 1 L 2 ) ( ψ ( T ) ψ ( 0 ) ) α , b 1 + l 1 Γ ( α ) Γ ( 2 α ) ( 1 l 2 ) ( ψ ( T ) ψ ( 0 ) ) α . From (7), we know that A is a contraction operator. By using of Banach’s fixed-point theorem, we obtain that A has a unique fixed point which is a unique solution of the problem (3). □
Theorem 3.
Suppose that (H1)–(H3) hold. If
μ = max b 2 + m 2 * Γ ( α ) ( ψ ( T ) ψ ( 0 ) ) α ( 1 m 3 * ) Γ ( 2 α ) , b 1 + n 2 * Γ ( α ) ( ψ ( T ) ψ ( 0 ) ) α ( 1 n 3 * ) Γ ( 2 α ) < 1 ,
where m 2 * = sup t J m 2 ( t ) and n 2 * = sup t J n 2 ( t ) . Then, the BVP (3) has at least one solution in Ω.
Proof. 
Let
R = 1 1 μ , Δ = | x 0 | + | y 0 | + Q 1 + Q 2 + m 1 Γ ( α ) ( ψ ( T ) ψ ( 0 ) ) α ( 1 m 3 * ) Γ ( 2 α ) + n 1 Γ ( α ) ( ψ ( T ) ψ ( 0 ) ) α ( 1 n 3 * ) Γ ( 2 α ) ,
where m 1 = sup t J { ( ψ ( t ) ψ ( 0 ) ) 1 α m 1 ( t ) } , n 1 = sup t J { ( ψ ( t ) ψ ( 0 ) ) 1 α n 1 ( t ) } , Q 1 = | h ( 0 ) | and Q 2 = | r ( 0 ) | .
Set the non-empty closed bounded convex subset as follows:
Ω = { ( x , y ) C 1 α , ψ ( J , R ) × C 1 α , ψ ( J , R ) : ( x , y ) C 1 α , ψ × C 1 α , ψ M } ,
where M R Δ is fixed. Define two operators F 1 , F 2 on Ω as follows:
F 1 ( x , y ) ( t ) = ( ( ψ ( t ) ψ ( 0 ) ) α 1 ( y 0 r ( y ) ) , ( ψ ( t ) ψ ( 0 ) ) α 1 ( x 0 h ( x ) ) ) , F 2 ( x , y ) ( t ) = 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 p x ( s ) ψ ( s ) d s , 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 q y ( s ) ψ ( s ) d s ,
where p x , q y : ( 0 , T ] R are two functions satisfying the following functional equations:
p x ( t ) = f ( t , y ( t ) , p x ( t ) ) , q y ( t ) = g ( t , x ( t ) , q y ( t ) ) .
In the following, we will prove that the operator F 1 + F 2 in Ω has at least one fixed point by using Krasnoselskii’s fixed-point theorem. The proof will be given in four steps.
Step 1. We prove that F 1 ( x , y ) + F 2 ( u , v ) Ω for all ( x , y ) , ( u , v ) Ω .
For any ( x , y ) , ( u , v ) Ω and t ( 0 , T ] , one has
F 1 ( x , y ) ( t ) + F 2 ( u , v ) ( t ) = ( H 1 ( x , y , u , v ) , H 2 ( x , y , u , v ) ) ,
where
H 1 ( x , y , u , v ) = ( ψ ( t ) ψ ( 0 ) ) α 1 ( y 0 r ( y ) ) + 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 p u ( s ) ψ ( s ) d s , H 2 ( x , y , u , v ) = ( ψ ( t ) ψ ( 0 ) ) α 1 ( x 0 h ( x ) ) + 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 q v ( s ) ψ ( s ) d s .
Here, p u , q v : ( 0 , T ] R are two functions satisfying the the functional equations:
p u ( t ) = f ( t , v ( t ) , p u ( t ) ) , q v ( t ) = g ( t , u ( t ) , q v ( t ) ) .
From (H3), for any t ( 0 , T ] , we obtain
| ( ψ ( t ) ψ ( 0 ) ) 1 α p u ( t ) | = | ( ψ ( t ) ψ ( 0 ) ) 1 α f ( t , v ( t ) , p u ( t ) ) | ( ψ ( t ) ψ ( 0 ) ) 1 α m 1 ( t ) + m 2 ( t ) | ( ψ ( t ) ψ ( 0 ) ) 1 α v ( t ) | + m 3 ( t ) | ( ψ ( t ) ψ ( 0 ) ) 1 α p u ( t ) | m 1 + m 2 * v C 1 α , ψ + m 3 * | ( ψ ( t ) ψ ( 0 ) ) 1 α p u ( t ) | ,
which implies that
| ( ψ ( t ) ψ ( 0 ) ) 1 α p u ( t ) | m 1 + m 2 * v C 1 α , ψ 1 m 3 * .
Similarly, we can obtain
| ( ψ ( t ) ψ ( 0 ) ) 1 α q v ( t ) | n 1 + n 2 * u C 1 α , ψ 1 n 3 * .
For any ( x , y ) , ( u , v ) Ω and t ( 0 , T ] , by (14), we obtain the following:
| H 1 ( x , y , u , v ) | ( ψ ( t ) ψ ( 0 ) ) α 1 | y 0 | + ( ψ ( t ) ψ ( 0 ) ) α 1 | r ( y ) r ( 0 ) | + ( ψ ( t ) ψ ( 0 ) ) α 1 | r ( 0 ) | + 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ( ψ ( s ) ψ ( 0 ) ) α 1 | ( ψ ( s ) ψ ( 0 ) ) 1 α p u ( s ) | ψ ( s ) d s ( ψ ( t ) ψ ( 0 ) ) α 1 | y 0 | + ( ψ ( t ) ψ ( 0 ) ) α 1 b 2 y C 1 α , ψ + ( ψ ( t ) ψ ( 0 ) ) α 1 Q 2 + m 1 + m 2 * v C 1 α , ψ 1 m 3 * 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ( ψ ( s ) ψ ( 0 ) ) α 1 ψ ( s ) d s = ( ψ ( t ) ψ ( 0 ) ) α 1 | y 0 | + ( ψ ( t ) ψ ( 0 ) ) α 1 b 2 y C 1 α , ψ + ( ψ ( t ) ψ ( 0 ) ) α 1 Q 2 + m 1 + m 2 * v C 1 α , ψ 1 m 3 * Γ ( α ) Γ ( 2 α ) ( ψ ( t ) ψ ( 0 ) ) 2 α 1 .
Thus,
( ψ ( t ) ψ ( 0 ) ) 1 α | H 1 ( x , y , u , v ) | | y 0 | + Q 2 + m 1 Γ ( α ) ( ψ ( T ) ψ ( 0 ) ) α ( 1 m 3 * ) Γ ( 2 α ) + b 2 y C 1 α , ψ + m 2 * Γ ( α ) ( ψ ( T ) ψ ( 0 ) ) α ( 1 m 3 * ) Γ ( 2 α ) v C 1 α , ψ .
Similarly, we have
( ψ ( t ) ψ ( 0 ) ) 1 α | H 2 ( x , y , u , v ) | | x 0 | + Q 1 + n 1 Γ ( α ) ( ψ ( T ) ψ ( 0 ) ) α ( 1 n 3 * ) Γ ( 2 α ) + b 1 x C 1 α , ψ + n 2 * Γ ( α ) ( ψ ( T ) ψ ( 0 ) ) α ( 1 n 3 * ) Γ ( 2 α ) u C 1 α , ψ .
Hence,
F 1 ( x , y ) + F 2 ( u , v ) C 1 α , ψ = H 1 ( x , y , u , v C 1 α , ψ + H 2 ( x , y , u , v C 1 α , ψ | x 0 | + | y 0 | + Q 1 + Q 2 + m 1 Γ ( α ) ( ψ ( T ) ψ ( 0 ) ) α ( 1 m 3 * ) Γ ( 2 α ) + n 1 Γ ( α ) ( ψ ( T ) ψ ( 0 ) ) α ( 1 n 3 * ) Γ ( 2 α ) + max b 2 + m 2 * Γ ( α ) ( ψ ( T ) ψ ( 0 ) ) α ( 1 m 3 * ) Γ ( 2 α ) , b 1 + n 2 * Γ ( α ) ( ψ ( T ) ψ ( 0 ) ) α ( 1 n 3 * ) Γ ( 2 α ) M = Δ + μ M M R + 1 1 R M = M .
Thus, F 1 ( x , y ) + F 2 ( u , v ) Ω for all x , y Ω .
Step 2. We show that F 1 is a contraction mapping.
For any ( x , y ) , ( u , v ) Ω × Ω , we obtain the following by (H2):
F 1 ( x , y ) ( t ) F 1 ( u , v ) ( t ) = ( F 11 , F 12 ) ,
where F 11 = ( ψ ( t ) ψ ( 0 ) ) α 1 ( r ( v ) r ( y ) ) and F 12 = ( ψ ( t ) ψ ( 0 ) ) α 1 ( h ( u ) h ( x ) ) .
So,
| F 11 | ( ψ ( t ) ψ ( 0 ) ) α 1 b 2 y v C 1 α , ψ , | F 12 | ( ψ ( t ) ψ ( 0 ) ) α 1 b 1 x u C 1 α , ψ . F 1 ( x , y ) F 1 ( u , v ) C 1 α , ψ × C 1 α , ψ = sup t J | ( ψ ( t ) ψ ( 0 ) ) 1 α F 11 | + sup t J | ( ψ ( t ) ψ ( 0 ) ) 1 α F 12 | b 1 x u C 1 α , ψ + b 2 y v C 1 α , ψ max { b 1 , b 2 } ( x u C 1 α , ψ + y v C 1 α , ψ ) = max { b 1 , b 2 } ( ( x , y ) ( u , v ) C 1 α , ψ × C 1 α , ψ .
Hence, the operator F 1 is a contraction.
Step 3. We show that F 2 is continuous.
Let { ( x n , y n ) } be a sequence such that ( x n , y n ) ( x , y ) in C 1 α , ψ ( J , R ) × C 1 α , ψ ( J , R ) , then, for each t ( 0 , T ] , we have
F 2 ( x n , y n ) ( t ) F 2 ( x , y ) ( t ) = ( F 21 , F 22 ) ,
where
F 21 = 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ( p x n ( s ) p x ( s ) ) ψ ( s ) d s ,
F 22 = 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ( q y n ( s ) q y ( s ) ) ψ ( s ) d s .
where p x n , p x , q y n , q y C 1 α , ψ ( J , R ) be such that
p x n ( t ) = f ( t , y n ( t ) , p x n ( t ) ) , p x ( t ) = f ( t , y ( t ) , p x ( t ) ) , q y n ( t ) = g ( t , x n ( t ) , q y n ( t ) ) , q y ( t ) = g ( t , x ( t ) , q y ( t ) ) .
By (H1), one has
| p x n ( t ) p x ( t ) | = | f ( t , y n ( t ) , p x n ( t ) ) f ( t , y ( t ) , p x ( t ) ) | L 1 | y n ( t ) y ( t ) | + L 2 | p x n ( t ) p x ( t ) | , | q y n ( t ) q y ( t ) | = | g ( t , x n ( t ) , q y n ( t ) ) g ( t , x ( t ) , q y ( t ) ) | l 1 | x n ( t ) x ( t ) | + l 2 | q y n ( t ) q y ( t ) | .
Then,
| p x n ( t ) p x ( t ) | L 1 1 L 2 | y n ( t ) y ( t ) | ,
and
| q y n ( t ) q y ( t ) | l 1 1 l 2 | x n ( t ) x ( t ) | .
By replacing (19) in Equation (17), we obtain the following:
| F 21 | 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 | p x n ( s ) p x ( s ) | ψ ( s ) d s L 1 ( 1 L 2 ) Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 | y n ( s ) y ( s ) | ψ ( s ) d s = L 1 ( 1 L 2 ) Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ( ψ ( s ) ψ ( 0 ) ) α 1 | ( ψ ( s ) ψ ( 0 ) ) 1 α ( y n ( s ) y ( s ) ) | ψ ( s ) d s = L 1 1 L 2 I α , ψ ( ( ψ ( t ) ψ ( 0 ) ) α 1 ) y n y C 1 α , ψ = L 1 Γ ( α ) ( 1 L 2 ) Γ ( 2 α ) ( ψ ( t ) ψ ( 0 ) ) 2 α 1 y n y C 1 α , ψ .
Similarly, we can obtain
| F 22 | l 1 Γ ( α ) ( 1 l 2 ) Γ ( 2 α ) ( ψ ( t ) ψ ( 0 ) ) 2 α 1 x n x C 1 α , ψ .
From (21), and (22), one has
F 2 ( x n , y n ) F 2 ( x , y ) C 1 α , ψ = ( ψ ( t ) ψ ( 0 ) ) 1 α F 21 C 1 α , ψ + ( ψ ( t ) ψ ( 0 ) ) 1 α F 22 C 1 α , ψ L 1 Γ ( α ) ( 1 L 2 ) Γ ( 2 α ) ( ψ ( t ) ψ ( 0 ) ) α y n y C 1 α , ψ + l 1 Γ ( α ) ( 1 l 2 ) Γ ( 2 α ) ( ψ ( t ) ψ ( 0 ) ) α x n x C 1 α , ψ max L 1 1 L 2 , l 1 1 l 2 Γ ( α ) Γ ( 2 α ) ( ψ ( T ) ψ ( 0 ) ) α ( x n x C 1 α , ψ + y n y C 1 α , ψ ) = max L 1 1 L 2 , l 1 1 l 2 Γ ( α ) Γ ( 2 α ) ( ψ ( T ) ψ ( 0 ) ) α ( x n , y n ) ( x , y ) C 1 α , ψ ,
which implies that
F 2 ( x n , y n ) F 2 ( x , y ) C 1 α , ψ 0 as n .
That is, F 2 is continuous.
Step 4. We prove that F 2 is compact.
For each ( x , y ) Ω × Ω and t ( 0 , T ] , one has
F 2 ( x , y ) ( t ) = 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 p x ( s ) ψ ( s ) d s , 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 q y ( s ) ψ ( s ) d s ,
where p x , q y are as in (13). Similar to the proof of (16), we obtain
1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 p x ( s ) ψ ( s ) d s 1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 ( ψ ( s ) ψ ( 0 ) ) α 1 | ( ψ ( s ) ψ ( 0 ) ) 1 α p x ( s ) | ψ ( s ) d s m 1 + m 2 * M 1 m 3 * Γ ( α ) Γ ( 2 α ) ( ψ ( t ) ψ ( 0 ) ) 2 α 1 ,
and
1 Γ ( α ) 0 t ( ψ ( t ) ψ ( s ) ) α 1 q y ( s ) ψ ( s ) d s n 1 + n 2 * M 1 n 3 * Γ ( α ) Γ ( 2 α ) ( ψ ( t ) ψ ( 0 ) ) 2 α 1 .
Hence,
F 2 ( x , y ) C 1 α , ψ max m 1 + m 2 * M 1 m 3 * , n 1 + n 2 * M 1 n 3 * Γ ( α ) Γ ( 2 α ) ( ψ ( T ) ψ ( 0 ) ) α .
Thus, F 2 ( Ω × Ω ) is uniformly bounded.
Finally, we show that F 2 ( Ω × Ω ) is equicontinuous, let 0 < t 1 < t 2 T and ( x , y ) Ω × Ω . Then,
( ψ ( t 2 ) ψ ( 0 ) ) 1 α F 2 ( x , y ) ( t 2 ) ( ψ ( t 1 ) ψ ( 0 ) ) 1 α F 2 ( x , y ) ( t 1 ) = ( K 21 , K 22 ) ,
where
K 21 = 1 Γ ( α ) 0 t 1 + t 1 t 2 ( ψ ( t 2 ) ψ ( 0 ) ) 1 α ( ψ ( t 2 ) ψ ( s ) ) α 1 p x ( s ) ψ ( s ) d s 0 t 1 ( ψ ( t 1 ) ψ ( 0 ) ) 1 α ( ψ ( t 1 ) ψ ( s ) ) α 1 p x ( s ) ψ ( s ) d s , K 22 = 1 Γ ( α ) 0 t 1 + t 1 t 2 ( ψ ( t 2 ) ψ ( 0 ) ) 1 α ( ψ ( t 2 ) ψ ( s ) ) α 1 q y ( s ) ψ ( s ) d s 0 t 1 ( ψ ( t 1 ) ψ ( 0 ) ) 1 α ( ψ ( t 1 ) ψ ( s ) ) α 1 q y ( s ) ψ ( s ) d s .
Then, by (14), we obtain the following:
| K 21 | 1 Γ ( α ) 0 t 1 | ( ψ ( t 2 ) ψ ( 0 ) ) 1 α ( ψ ( t 2 ) ψ ( s ) ) α 1 ( ψ ( s ) ψ ( 0 ) ) α 1 ( ψ ( t 1 ) ψ ( 0 ) ) 1 α ( ψ ( t 1 ) ψ ( s ) ) α 1 ( ψ ( s ) ψ ( 0 ) ) α 1 | | ( ψ ( s ) ψ ( 0 ) ) 1 α p x ( s ) | d s + 1 Γ ( α ) t 1 t 2 ( ψ ( t 2 ) ψ ( 0 ) ) 1 α ( ψ ( t 2 ) ψ ( s ) ) α 1 ( ψ ( s ) ψ ( 0 ) ) α 1 | ( ψ ( s ) ψ ( 0 ) ) 1 α p x ( s ) | d s m 1 + m 2 * M ( 1 m 3 * ) Γ ( α ) 0 t 1 | ( ψ ( t 2 ) ψ ( 0 ) ) 1 α ( ψ ( t 2 ) ψ ( s ) ) α 1 ( ψ ( t 1 ) ψ ( 0 ) ) 1 α ( ψ ( t 1 ) ψ ( s ) ) α 1 | ( ψ ( s ) ψ ( 0 ) ) α 1 d s + m 1 + m 2 * M ( 1 m 3 * ) Γ ( α ) t 1 t 2 ( ψ ( t 2 ) ψ ( 0 ) ) 1 α ( ψ ( t 2 ) ψ ( s ) ) α 1 ( ψ ( s ) ψ ( 0 ) ) α 1 d s 0 ,
as t 2 t 1 . Similarly, we obtain that K 22 0 as t 2 t 1 . Which yields that F 2 ( Ω × Ω ) is equicontinuous, Then, by the Ascoli–Arzela theorem, the operator F 2 is compact.
All the assume of the Theorem 1 are satisfied. Therefore, there exists a fixed point, ( x , y ) Ω × Ω , such that ( x , y ) = F 1 ( x , y ) + F 2 ( x , y ) , which is a solution of the problem (3). □

4. Example

Consider the following coupled implicit ψ -Riemann–Liouville fractional differential equations with nonlocal conditions
D 0 + 4 7 , ln ( 1 + t ) x ( t ) = f ( t , y ( t ) , D 0 + 4 7 , ln ( 1 + t ) x ( t ) ) , t ( 0 , 1 ] , D 0 + 4 7 , ln ( 1 + t ) y ( t ) = g ( t , x ( t ) , D 0 + 4 7 , ln ( 1 + t ) y ( t ) ) , t ( 0 , 1 ] , ln 3 7 ( 1 + t ) x ( t ) | t = 0 = 1 3 i = 1 n c i ln 3 7 ( 1 + t ) y ( t i ) , ln 3 7 ( 1 + t ) y ( t ) | t = 0 = 1 4 i = 1 n d i ln 3 7 ( 1 + t ) x ( t i ) ,
where 0 < t 1 < < t n < 1 , c i and d i are positive constants with i = 1 n c i 1 3 and i = 1 n d i 2 5 . Let
f ( t , u , v ) = 1 ( 2 + t ) 2 ( 2 + | u | + | v | ) + 2 sin t ln 3 7 ( 1 + t ) , t ( 0 , 1 ] , u , v s . R ,
g ( t , u , v ) = 1 3 exp ( 2 t ) arctan ( 1 + 3 | u | + | v | ) + cos t ln 3 7 ( 1 + t ) , t ( 0 , 1 ] , u , v s . R .
Thus, we have
C 1 α , ψ ( [ 0 , 1 ] , R ) = C 3 7 , ln ( 1 + t ) ( [ 0 , 1 ] , R ) = { h : ( 0 , 1 ] R : ln 3 7 ( 1 + t ) h C ( [ 0 , 1 ] , R ) } ,
where α = 4 7 . Obviously, the functions f and g are continuous, f ( · , 0 , 0 ) , g ( · , 0 , 0 ) C 3 7 , ln ( 1 + t ) ( [ 0 , 1 ] , R ) . For any u 1 , u 2 , v 1 , v 2 R and t ( 0 , 1 ] , one has
| f ( t , u 1 , v 1 ) f ( t , u 2 , v 2 ) | = 1 ( 2 + t ) 2 1 2 + | u 1 | + | v 1 | 1 2 + | u 2 | + | v 2 | | u 1 u 2 | + | v 1 v 2 | ( 2 + t ) 2 ( 2 + | u 1 | + | v 1 | ) ( 2 + | u 2 | + | v 2 | ) 1 4 ( | u 1 u 2 | + | v 1 v 2 | ) , | g ( t , u 1 , v 1 ) g ( t , u 2 , v 2 ) | = 1 3 exp ( 2 t ) arctan ( 1 + 3 | u 1 | + | v 1 | ) arctan ( 1 + 3 | u 2 | + | v 2 | ) 1 3 e ( 3 | u 1 u 2 | + | v 1 v 2 | ) .
Moreover, set the following:
r ( y ) = i = 1 n c i ln 3 7 ( 1 + t ) y ( t i ) , h ( x ) = i = 1 n d i ln 3 7 ( 1 + t ) x ( t i ) ,
For each x 1 , x 2 , y 1 , y 2 R , we have
| h ( x 1 ) h ( x 2 ) | i = 1 n d i ln 3 7 ( 1 + t ) | x 1 ( t i ) x 2 ( t i ) | i = 1 n d i x 1 x 2 C 3 7 , ln ( 1 + t ) 2 5 x 1 x 2 C 3 7 , ln ( 1 + t ) ,
and
| r ( y 1 ) r ( y 2 ) | i = 1 n c i ln 3 7 ( 1 + t ) | y 1 ( t i ) y 2 ( t i ) | i = 1 n c i y 1 y 2 C 3 7 , ln ( 1 + t ) 1 3 y 1 y 2 C 3 7 , ln ( 1 + t ) .
So, conditions (H1) and (H2) are satisfied with L 1 = 1 4 , L 2 = 1 4 , l 1 = 1 e , l 2 = 1 3 e , b 1 = 2 5 and b 2 = 1 3 . Moreover, the following condition:
max b 2 + L 1 Γ ( α ) Γ ( 2 α ) ( 1 L 2 ) ( ψ ( T ) ψ ( 0 ) ) α , b 1 + l 1 Γ ( α ) Γ ( 2 α ) ( 1 l 2 ) ( ψ ( T ) ψ ( 0 ) ) α = max 1 3 + 0.4504 , 2 5 + 0.5666 = 0.9666 < 1 ,
is satisfied with T = 1 . By Theorem 2, we have that the problem (23) has a unique coupled solution in the space C 3 7 , ln ( 1 + t ) ( [ 0 , 1 ] , R ) × C 3 7 , ln ( 1 + t ) ( [ 0 , 1 ] , R ) .

5. Conclusions

In this paper, we investigated a coupled implicit system that has ψ -Riemann–Liouville fractional derivative and nonlocal conditions. The interesting point is that two fractional implicit equations are coupled. By Banach fixed-point theorem and Krasnoselskii’s fixed-point theorem, the uniqueness and the existence results are proved. Our results obtained in this paper is new and complements the existing literature on this topic. We will study the corresponding problem in future research, and we hope to be able to make some progress.

Author Contributions

All the authors contributed equally and significantly in writing this article. All authors have read and agreed to the published version of the manuscript.

Funding

This work is supported by Natural Science Foundation of China (11571136).

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

We are really thankful to the reviewers for their careful reading of our manuscript and their many insightful comments and suggestions that have improved the quality of our manuscript.

Conflicts of Interest

The authors declare that there are no conflicts of interest.

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Jiang, D.; Bai, C. Existence Results for Coupled Implicit \({\psi}\)-Riemann–Liouville Fractional Differential Equations with Nonlocal Conditions. Axioms 2022, 11, 103. https://doi.org/10.3390/axioms11030103

AMA Style

Jiang D, Bai C. Existence Results for Coupled Implicit \({\psi}\)-Riemann–Liouville Fractional Differential Equations with Nonlocal Conditions. Axioms. 2022; 11(3):103. https://doi.org/10.3390/axioms11030103

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Jiang, Dinghong, and Chuanzhi Bai. 2022. "Existence Results for Coupled Implicit \({\psi}\)-Riemann–Liouville Fractional Differential Equations with Nonlocal Conditions" Axioms 11, no. 3: 103. https://doi.org/10.3390/axioms11030103

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Jiang, D., & Bai, C. (2022). Existence Results for Coupled Implicit \({\psi}\)-Riemann–Liouville Fractional Differential Equations with Nonlocal Conditions. Axioms, 11(3), 103. https://doi.org/10.3390/axioms11030103

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