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Article

Analytical Study of Two Nonlinear Coupled Hybrid Systems Involving Generalized Hilfer Fractional Operators

by
Mohammed A. Almalahi
1,2,†,
Omar Bazighifan
3,*,†,
Satish K. Panchal
2,†,
S. S. Askar
4,† and
Georgia Irina Oros
5,*,†
1
Department of Mathematics, Hajjah University, Hajjah 00967, Yemen
2
Department of Mathematics, Dr. Babasaheb Ambedkar Marathwada University, Aurangabad 431001, India
3
Section of Mathematics, International Telematic University Uninettuno, CorsoVittorio Emanuele II, 39, 00186 Rome, Italy
4
Department of Statistics and Operations Research, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia
5
Department of Mathematics and Computer Science, University of Oradea, 410087 Oradea, Romania
*
Authors to whom correspondence should be addressed.
These authors contributed equally to this work.
Fractal Fract. 2021, 5(4), 178; https://doi.org/10.3390/fractalfract5040178
Submission received: 29 September 2021 / Revised: 13 October 2021 / Accepted: 20 October 2021 / Published: 22 October 2021

Abstract

:
In this research paper, we dedicate our interest to an investigation of the sufficient conditions for the existence of solutions of two new types of a coupled systems of hybrid fractional differential equations involving ϕ -Hilfer fractional derivatives. The existence results are established in the weighted space of functions using Dhage’s hybrid fixed point theorem for three operators in a Banach algebra and Dhage’s helpful generalization of Krasnoselskii fixed- point theorem. Finally, simulated examples are provided to demonstrate the obtained results.

1. Introduction

Fractional differential equations (FDEs) have a profound physical background and rich theoretical connotations and have been particularly eye-catching in recent years. Several-order differential equations refer to equations that contain fractional derivatives or fractional integrals. Currently, fractional derivatives and fractions order integrals have a wide range of applications in many disciplines such as physics, biology, chemistry, etc. For more information see [1,2,3,4,5,6,7,8,9,10].
Several concepts about fractional derivatives such as Liouville, Caputo, Hadamard, Caputo–Fabrizio and Hilfer derivatives have been proposed in recent years. As a result, a wide range of arbitrary order differential equation designs based on these fractional operators have emerged. The qualitative properties of solutions (existence, uniqueness, and stability) have been studied by several researchers. For the applications and latest work regarding to these operators, we suggest readers to [11,12,13,14,15].
On the other hand, fractional hybrid differential equations emerge from a wide range of spaces of applied and physical sciences, this class of equations can be used to model and describe non-homogeneous physical events, e.g., in the deflections of a curved beam with a constant or varying cross-section, electromagnétic waves, or gravity-driven streams, etc. Many researchers have recently become interested in a novel class of mathematical modelings based on hybrid fractional differential equations with hybrid or non-hybrid boundary value conditions [16,17,18,19].
Coupled systems, including FDEs, are important to study because they appear in a wide range of practical applications. We refer to a collection of papers for some theoretical approaches on coupled systems [20,21,22,23]. Recent works related to our work were done by [24,25]. Sitho et. al. [24] studied the existence of hybrid fractional integrodifferential equations described as follows
D 0 + ϱ μ ( ı ) i = 1 m I 0 + q i , ϕ g i ı , μ ( ı ) f ı , μ ( ı ) = y ı , μ ( ı ) , ϱ ( 0 , 1 ) , μ ( 0 ) = 0 ,
and
D 0 + ϱ D 0 + ν μ ( ı ) i = 1 m I 0 + q i , ϕ g i ı , μ ( ı ) f ı , μ ( ı ) = y ı , μ ( ı ) , ϱ , ν ( 0 , 1 ) , μ ( 0 ) = 0 , D 0 + ν μ ( 0 ) = 0 ,
where D 0 + ϱ , D 0 + ν denotes the Caputo fractional derivatives of order ϱ , ν 0 , 1 and I 0 + q i denotes the Riemann-Liouville fractional integral of order q i > 0 , i = 1 , 2 , , m . The functions g i , y and f are continuous functions. Boutiara et. al. in [25] studied the existence of solutions for the following coupled system in the sense of ϕ -Caputo fractional operators
C D 0 + ϱ 1 , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) = y 1 ı , μ 1 ( ı ) , μ 2 ( ı ) , C D 0 + ϱ 2 , ϕ μ 2 ( ı ) i = 1 m I 0 + q i , ϕ g i 2 ı , μ 1 ( ı ) , μ 2 ( ı ) f 2 ı , μ 1 ( ı ) , μ 2 ( ı ) = y 2 ı , μ 1 ( ı ) , μ 2 ( ı ) μ k ( 0 ) = 0 , k = 1 , 2 .
Motivated by the novel advancements of hybrid fractional integrodifferential equations and their applications, also by the above argumentations, by means of Dhage’s hybrid fixed point theorem for three operators in a Banach algebra [26] and Dhage’s helpful generalization of Krasnoselskii’s fixed point theorem [27], we investigate the existence of a solution for two classes of coupled hybrid fractional integrodifferential equations. The first class described by
H D 0 + ϱ , β , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) = y 1 ı , μ 1 ( ı ) , μ 2 ( ı ) , ı J : = 0 , b , H D 0 + ϱ , β , ϕ μ 2 ( ı ) i = 1 m I 0 + q i , ϕ g i 2 ı , μ 1 ( ı ) , μ 2 ( ı ) f 2 ı , μ 1 ( ı ) , μ 2 ( ı ) = y 2 ı , μ 1 ( ı ) , μ 2 ( ı ) , I 0 + 1 γ , ϕ μ k ( ı ) i = 1 m I 0 + q i , ϕ g i k ı , μ 1 ( ı ) , μ 2 ( ı ) f k ı , μ 1 ( ı ) , μ 2 ( ı ) ı = 0 = 0 , k = 1 , 2 ,
where H D 0 + ϱ , β , ϕ is the ϕ -Hilfer fractional derivative of order ϱ 0 , 1 and type β 0 , 1 with respect to an increasing function ϕ C 1 J , R with ϕ ( ı ) 0 , for all ı J , I 0 + 1 γ , ϕ , I 0 + q i , ϕ are the ϕ -Riemann-Liouville fractional integral of order 1 γ > 0 , γ = ϱ + β ϱ β , q i > 0 , i = 1 , 2 , , m . The functions g i k , y k , f k C J × R 2 , R with f k 0 , 0 , 0 0 , g i k 0 , 0 , 0 = 0 , k = 1 , 2 , are continuous functions. We prove an existence result for the ϕ -Hilfer hybrid system (1) using Dhage’s hybrid fixed point theorem for three operators in a Banach algebra [26].
We will also look at the necessary conditions for the existence of a solution for ϕ -Hilfer hybrid system which described as follows
H D 0 + ϱ , β , ϕ H D 0 + ν , κ , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) = y 1 ı , μ 1 ( ı ) , μ 2 ( ı ) , ı J : = 0 , b , H D 0 + ϱ , β , ϕ H D 0 + ν , κ , ϕ μ 2 ( ı ) i = 1 m I 0 + q i , ϕ g i 2 ı , μ 1 ( ı ) , μ 2 ( ı ) f 2 ı , μ 1 ( ı ) , μ 2 ( ı ) = y 2 ı , μ 1 ( ı ) , μ 2 ( ı ) , I 0 + 1 γ , ϕ μ k ( 0 ) = 0 , H D 0 + ν , κ , ϕ μ k ( 0 ) = 0 , k = 1 , 2 ,
where H D 0 + A , B , ϕ is the ϕ -Hilfer fractional derivative of order A = ϱ , ν such that ϱ , ν 0 , 1 and types β , κ 0 , 1 with respect to an increasing function ϕ C 1 J , R with ϕ ( ı ) 0 , for all ı J , I 0 + 1 γ , ϕ , I 0 + q i , ϕ are the ϕ -Riemann-Liouville fractional integral of order 1 γ > 0 , γ = ν + κ ν κ , q i > 0 , i = 1 , 2 , , m . The functions g i k , y k , f k C J × R 2 , R with f k 0 , 0 , 0 0 , g i k 0 , 0 , 0 = 0 , k = 1 , 2 , are continuous functions that meet certain standards that will be mentioned later. We prove an existence result for the ϕ -Hilfer hybrid system (2) using a useful generalization of Krasnoselskii’s fixed point theory due to Dhage [27].
  • By adopting the same techniques used in [24], we derive the formula of solutions for ϕ -Hilfer hybrid system (1) and ϕ -Hilfer hybrid system (2).
  • We discuss two types of hybrid systems with generalized Hilfer fractional operator with respect to another increasing function ϕ with ϕ ( ı ) 0 , ı J .
  • The proposed problems (1) and (2) for different values of a function ϕ and parameters β encompasses the investigation of problems involving a variety of different fractional derivative operators.
  • The results obtained in this work includes the results of Sitho et al. [24], Boutiara et al. [25] and cover many problems which do not study yet.
The remainder of the paper is laid out as follows: We will go through some helpful preliminaries in Section 2. The existence of the solutions for ϕ -Hilfer hybrid system (1) has been investigated in Section 3, whereas the existence of the solutions for ϕ -Hilfer hybrid system (2) has been addressed in Section 4. We provide a relevant examples in Section 5 to demonstrate our findings. In the last section, we will provide some last observations about our findings.

2. Auxiliary Results

To achieve our main purposes, we present here some definitions and basic auxiliary results that are required throughout our paper. Let J : = 0 , b , let C J be the Banach space of continuous functions μ : J R equipped with the norm μ = sup μ ( ı ) : ı J . Consider the product Banach space C J × C J with the following norm
μ 1 , μ 2 = μ 1 + μ 2 ,
for each μ 1 , μ 2 C J × C J . Let γ = ϱ + β ϱ β such that ϱ 0 , 1 , β 0 , 1 and let ϕ C 1 J be an increasing function with ϕ ( ı ) 0 for each ı J . We define the weighted space C 1 γ ; ϕ J of continuous functions μ : J R by
C 1 γ ; ϕ J = μ : J R ; ϕ ( ı ) ϕ ( 0 ) 1 γ μ ( ı ) C J , 0 < γ 1 .
Obviously, C 1 γ ; ϕ J is a Banach space endowed with the norm
μ C 1 γ ; ϕ J = ϕ ( ı ) ϕ ( 0 ) 1 γ μ ( ı ) C J = max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ μ ( ı ) .
Let B : = C 1 γ ; ϕ J × C 1 γ ; ϕ J , be the product weighted space with the norm
μ 1 , μ 2 B = μ 1 C 1 γ ; ϕ J + μ 2 C 1 γ ; ϕ J ,
for each μ 1 , μ 2 B .
Definition 1 
([5]). The ϕ-RL fractional integral of f of order ϱ > 0 is defined by
I 0 + ϱ , ϕ f ( ı ) = 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ϱ 1 Γ ( ϱ ) f ( s ) d s ,
where Γ ( · ) is the gamma function.
Definition 2 
([28]). The ϕ-Hilfer fractional derivative of f C n J with order ϱ n 1 , n , n N , β 0 , 1 is defined by
H D 0 + ϱ , β , ϕ f ( ı ) = I 0 + β ( n ϱ ) ; ϕ D a + γ ; ϕ f ( ı ) , γ = ϱ + n β ϱ β ,
where
D 0 + γ ; ϕ f ( ı ) = f ϕ [ n ] I 0 + ( 1 β ) ( n ϱ ) ; ϕ f ( ı ) , a n d f ϕ [ n ] = 1 ϕ ( ı ) d d t n .
Lemma 1
([28]). Assume that γ = ϱ + β ϱ β , ϱ > 0 , β > 0 , and μ C 1 γ ; ϕ γ J . Then
I 0 + γ ; ϕ D 0 + γ ; ϕ μ = I 0 + ϱ ; ϕ H D 0 + ϱ , β ; ϕ μ   a n d   D 0 + γ ; ϕ I 0 + ϱ ; ϕ μ = D 0 + β ( 1 ϱ ) ; ϕ μ .
Theorem 1
([28]). Let 0 γ < ϱ and μ C 1 γ ; ϕ J . Then, I 0 + ϱ ; ϕ μ ( 0 ) = lim ı 0 + I 0 + ϱ ; ϕ μ ( ı ) = 0 .
Lemma 2
([2,28]). For ϱ , β , δ > 0 , we have
I 0 + ϱ , ϕ I 0 + β , ϕ f ( ı ) = I 0 + ϱ + β , ϕ f ( ı ) ,
I 0 + ϱ , ϕ ϕ ( ı ) ϕ ( 0 ) δ 1 = Γ ( δ ) ϕ ( ı ) ϕ ( 0 ) ϱ + δ 1 Γ ( ϱ + δ ) ,
and
H D 0 + ϱ , β , ϕ ϕ ( ı ) ϕ ( 0 ) δ 1 = 0 , δ = ϱ + n β ϱ β .
Lemma 3
([28]). Let f C n J , ϱ n 1 , n , n N , and β 0 , 1 . Then
I 0 + ϱ ; ϕ H D 0 + ϱ , β , ϕ f ( ı ) = f ( ı ) k = 1 n ϕ ( ı ) ϕ ( 0 ) γ k Γ ( γ k + 1 ) f ϕ n k I a + ( 1 β ) ( n ϱ ) ; ϕ f ( 0 ) ,
and
H D 0 + ϱ , β , ϕ I 0 + ϱ ; ϕ f ( ı ) = f ( ı ) .

3. Existence of Solution for ϕ -Hilfer Hybrid Systems (1)

In this section, we will study the existence of solutions for ϕ -Hilfer hybrid system (1) by means of Dhage’s hybrid fixed point theorem for three operators in a Banach algebra [26]. To achieve our main results, the following hypotheses must be satisfied
(H 1 ) The functions g i k , f k : J × R 2 R , f k ( 0 , 0 , 0 ) 0 , g i k ( 0 , 0 , 0 ) = 0 , i = 1 , 2 , , m , k = 1 , 2 , are continuous functions and there exist two positive functions Φ g i k ı , Φ f k ı , i = 1 , 2 , , m , k = 1 , 2 , with bound Φ ^ g i k and Φ ^ f k , i = 1 , 2 , , m , k = 1 , 2 , respectively, such that for each μ 1 , μ 2 , μ ¯ 1 , μ ¯ 2 B , we have
g i k ı , μ 1 ( ı ) , μ 2 ( ı ) y k ı , μ ¯ 1 ( ı ) , μ ¯ 2 ( ı ) Φ g i k ı μ 1 ( ı ) μ ¯ 1 ( ı ) + μ 2 ( ı ) μ ¯ 2 ( ı )
and
f k ı , μ 1 ( ı ) , μ 2 ( ı ) f k ı , μ ¯ 1 ( ı ) , μ ¯ 2 ( ı ) Φ f k ı μ 1 ( ı ) μ ¯ 1 ( ı ) + μ 2 ( ı ) μ ¯ 2 ( ı ) .
(H 2 ) The functions y k : J × R 2 R , k = 1 , 2 are continuous functions. For all ı , μ 1 , μ 2 J × B , there exists a functions ω y k : J R + with bound ω ^ y k , k = 1 , 2 and a continuous nondecreasing function Υ : R + R + such that
ϕ ( ı ) ϕ ( 0 ) 1 γ y k ı , μ 1 ( ı ) , μ 2 ( ı ) ω y k ı Υ μ 1 , μ 2 .
Lemma 4
([26]). Let X be a closed convex and bounded subset of the Banach algebra B and let A , K : B B and H : X be three operators such that
(a)
A and K are Lipschitzian with Lipschitz constant L A , L K , respectively,
(b)
H is compact and continuous,
(c)
μ = A μ H μ + K μ μ X , μ X ;
(d)
L A M H + L K < 1 , where M H = H ( X ) = sup { H μ : μ X } .
Then the operator equation μ = A μ H μ + K μ has a solution in X .
Lemma 5.
Let γ = ϱ + β ϱ β , ϱ 0 , 1 , β 0 , 1 , g i k , y k , f k : J × R 2 R , are continuous functions with f k 0 , 0 , 0 0 , g i k 0 , 0 , 0 = 0 , i = 1 , 2 , , m , k = 1 , 2 . If μ 1 , μ 2 B satisfies the ϕ-Hilfer hybrid system (1), then, μ 1 , μ 2 satisfies the following integral equations
μ 1 ( ı ) = f 1 ı , μ 1 , μ 2 I 0 + ϱ , ϕ y 1 ı , μ 1 , μ 2 + i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 , μ 2 , μ 2 ( ı ) = f 2 ı , μ 1 , μ 2 I 0 + ϱ , ϕ y 2 ı , μ 1 , μ 2 + i = 1 m I 0 + q i , ϕ g i 2 ı , μ 1 , μ 2 .
Proof. 
In the beginning, we assume that μ 1 , μ 2 B is a solution of ϕ -Hilfer hybrid system (1). We will prove that μ 1 , μ 2 satisfies the integral Equation (3). First, let
H D 0 + ϱ , β , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) = y 1 ı , μ 1 ( ı ) , μ 2 ( ı ) , I 0 + 1 γ , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) ı = 0 = 0 .
Taking the operator I 0 + ϱ , ϕ on both sides of the first Equation (4) and using Lemma 3, we have
μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) = ϕ ( ı ) ϕ ( 0 ) γ 1 Γ ( γ ) I 0 + 1 γ , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) ı = 0 + I 0 + ϱ , ϕ y 1 ı , μ 1 ( ı ) , μ 2 ( ı ) .
By the condition I 0 + 1 γ , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) ı = 0 = 0 , we obtain
μ 1 ( ı ) = f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) I 0 + ϱ , ϕ y 1 ı , μ 1 ( ı ) , μ 2 ( ı ) + i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) .
Next, let
H D 0 + ϱ , β , ϕ μ 2 ( ı ) i = 1 m I 0 + q i , ϕ g i 2 ı , μ 1 ( ı ) , μ 2 ( ı ) f 2 ı , μ 1 ( ı ) , μ 2 ( ı ) = y 2 ı , μ 1 ( ı ) , μ 2 ( ı ) , I 0 + 1 γ , ϕ μ 2 ( ı ) i = 1 m I 0 + q i , ϕ g i 2 ı , μ 1 ( ı ) , μ 2 ( ı ) f 2 ı , μ 1 ( ı ) , μ 2 ( ı ) ı = 0 = 0 .
By the same way, we obtain
μ 2 ( ı ) = f 2 ı , μ 1 ( ı ) , μ 2 ( ı ) I 0 + ϱ , ϕ y 2 ı , μ 1 ( ı ) , μ 2 ( ı ) + i = 1 m I 0 + q i , ϕ g i 2 ı , μ 1 ( ı ) , μ 2 ( ı ) .
From (6) and (7), we conclude that μ 1 , μ 2 satisfies the integral Equation (3).
Conversely, assume that μ 1 , μ 2 satisfies the integral Equation (3). Applying the operator I 0 + 1 γ , ϕ of the integral Equation (6), with replace ı by 0, we obtain
I 0 + 1 γ , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) ı = 0 = 0 .
Next, applying H D 0 + ϱ , β , ϕ on both sides of Equation (6), we have
H D 0 + ϱ , β , ϕ μ 1 ( ı ) = H D 0 + ϱ , β , ϕ f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) I 0 + ϱ , ϕ y 1 ı , μ 1 ( ı ) , μ 2 ( ı ) + i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) .
Using Lemma 2, we obtain
H D 0 + ϱ , β , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) = y 1 ı , μ 1 ( ı ) , μ 2 ( ı ) .
Similarly, we obtain
H D 0 + ϱ , β , ϕ μ 2 ( ı ) i = 1 m I 0 + q i , ϕ g i 2 ı , μ 1 ( ı ) , μ 2 ( ı ) f 2 ı , μ 1 ( ı ) , μ 2 ( ı ) = y 2 ı , μ 1 ( ı ) , μ 2 ( ı ) .
The proof is completed. □
Theorem 2.
Assume that (H 1 ), (H 2 ) hold. Then, ϕ-Hilfer hybrid system (1) has at least one solution in B , provided that
k = 1 2 Φ ^ f k ω ^ y k Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) + i = 1 m Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) Φ ^ g i k < 1 ,
where R is the radius of the closed ball set X defined in the following proof.
Proof. 
Let us consider a closed ball set
X = μ 1 , μ 2 B : μ 1 , μ 2 B R ,
with
R k = 1 2 F k ω ^ y k Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) + i = 1 m Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) G i k 1 k = 1 2 Φ ^ f k ω ^ y k Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) + i = 1 m Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) Φ ^ g i k ,
where F k = max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ f 1 ı , 0 , 0 , G i k = max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ g i k ı , 0 , 0 , k = 1 , 2 . By Lemma 5, the ϕ -Hilfer hybrid system (1) is equivalent to Equation (3). Define the operator V : B B as V = V 1 , V 2 , where
V 1 μ 1 , μ 2 ( ı ) = f 1 ı , μ 1 , μ 2 I 0 + ϱ , ϕ y 1 ı , μ 1 , μ 2 + i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 , μ 2 , V 2 μ 1 , μ 2 ( ı ) = f 2 ı , μ 1 , μ 2 I 0 + ϱ , ϕ y 2 ı , μ 1 , μ 2 + i = 1 m I 0 + q i , ϕ g i 2 ı , μ 1 , μ 2 .
To use Lemma 4, we define the following operators
A = A 1 , A 2 : B B by
A μ 1 , μ 2 ( ı ) = f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) , f 2 ı , μ 1 ( ı ) , μ 2 ( ı ) ,
H = H 1 , H 2 : X B by
H μ 1 , μ 2 ( ı ) = 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ϱ 1 Γ ( ϱ ) y 1 s , μ 1 ( s ) , μ 2 ( s ) d s , 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ϱ 1 Γ ( ϱ ) y 2 s , μ 1 ( s ) , μ 2 ( s ) d s ,
and K = K 1 , K 2 : B B by
K μ 1 , μ 2 ( ı ) = i = 1 m 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) q i 1 Γ ( q i ) g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) , i = 1 m 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) q i 1 Γ ( q i ) g i 2 ı , μ 1 ( ı ) , μ 2 ( ı ) .
Thus, the coupled system of the above hybrid integral Equation (9) can be written as a system of operator equations as
A μ 1 , μ 2 ( ı ) H μ 1 , μ 2 ( ı ) + K μ 1 , μ 2 ( ı ) = V μ 1 , μ 2 ( ı ) .
we will demonstrate that the operators A , H , and K meet all conditions in Lemma 4. This will be completed in the steps that follow.
Step (1): We begin by demonstrating that A and K are Lipschitz on B . For any μ 1 , μ 2 , μ 1 * , μ 2 * B . Then via (H1), we have
A 1 μ 1 , μ 2 A 1 μ 1 * , μ 2 * C 1 γ ; ϕ J = max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ A 1 μ 1 , μ 2 ( ı ) A 1 μ 1 * , μ 2 * ( ı ) = max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 * ( ı ) , μ 2 * ( ı ) Φ ^ f 1 μ 1 μ 1 * C 1 γ ; ϕ J + μ 2 μ 2 * C 1 γ ; ϕ J .
By the same way, one can obtain
A 2 μ 1 , μ 2 A 2 μ 1 * , μ 2 * C 1 γ ; ϕ J Φ ^ f 2 μ 1 μ 1 * C 1 γ ; ϕ J + μ 2 μ 2 * C 1 γ ; ϕ J .
Thus
A μ 1 , μ 2 A μ 1 * , μ 2 * B A 1 μ 1 , μ 2 A 1 μ 1 * , μ 2 * C 1 γ ; ϕ J + A 2 μ 1 , μ 2 A 2 μ 1 * , μ 2 * C 1 γ ; ϕ J k = 1 2 Φ ^ f k μ 1 , μ 2 μ 1 * , μ 2 * B .
Hence, A is a Lipschitzian on B with L A = k = 1 2 Φ ^ f k . Next, with the operator K , for any μ 1 , μ 2 , μ 1 * , μ 2 * B and via (H1), we have
K 1 μ 1 , μ 2 K 1 μ 1 * , μ 2 * C 1 γ ; ϕ J = max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ K 1 μ 1 , μ 2 ( ı ) K 1 μ 1 * , μ 2 * ( ı ) max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ i = 1 m 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) q i 1 Γ ( q i ) ϕ ( s ) ϕ ( 0 ) γ 1 ϕ ( s ) ϕ ( 0 ) 1 γ g i 1 s , μ 1 ( s ) , μ 2 ( s ) g i 1 s , μ 1 * ( s ) , μ 2 * ( s ) d s i = 1 m Γ γ ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) Φ ^ g i 1 μ 1 μ 1 * C 1 γ ; ϕ J + μ 2 μ 2 * C 1 γ ; ϕ J .
Similarly, we obtain
K 2 μ 1 , μ 2 K 2 μ 1 * , μ 2 * C 1 γ ; ϕ J i = 1 m Γ γ ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) Φ ^ g i 2 μ 1 μ 1 * C 1 γ ; ϕ J + μ 2 μ 2 * C 1 γ ; ϕ J .
Thus
K μ 1 , μ 2 K μ 1 * , μ 2 * B k = 1 2 i = 1 m Γ γ ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) Φ ^ g i k μ 1 , μ 2 μ 1 * , μ 2 * B .
Hence, K is a Lipschitzian on B with L K = k = 1 2 i = 1 m Γ γ ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) Φ ^ g i k .
Step (2): H is compact and continuous on X . For this purpose let μ 1 n , μ 2 n be a sequence in X such that μ 1 n , μ 2 n μ 1 , μ 2 X . Then by the Lebesgue dominated convergence theorem, for all ı J , we have
lim n H 1 μ 1 n , μ 2 n = lim n 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ϱ 1 Γ ( ϱ ) y 1 s , μ 1 n ( s ) , μ 2 n ( s ) d s = 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ϱ 1 Γ ( ϱ ) lim n y 1 s , μ 1 n ( s ) , μ 2 n ( s ) d s = 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ϱ 1 Γ ( ϱ ) y 1 s , μ 1 ( s ) , μ 2 ( s ) d s .
Similarly,
lim n H 2 μ 1 n , μ 2 n = 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ϱ 1 Γ ( ϱ ) y 2 s , μ 1 ( s ) , μ 2 ( s ) d s .
This means that H is continuous on X . Now, we prove that the set H X is uniformly bounded in X . For any μ 1 , μ 2 X , we have
H 1 μ 1 , μ 2 C 1 γ ; ϕ J ϕ ( ı ) ϕ ( 0 ) 1 γ 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ϱ 1 Γ ( ϱ ) ϕ ( s ) ϕ ( 0 ) γ 1 ϕ ( s ) ϕ ( 0 ) 1 γ y 1 s , μ 1 ( s ) , μ 2 ( s ) d s ω ^ y 1 Υ R Γ ( γ ) ϕ ( b ) ϕ ( s ) ϱ Γ ( ϱ + γ ) .
By the same way, we obtain
H 2 μ 1 , μ 2 C 1 γ ; ϕ J ϕ ( ı ) ϕ ( 0 ) 1 γ 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ϱ 1 Γ ( ϱ ) ϕ ( s ) ϕ ( 0 ) γ 1 ϕ ( s ) ϕ ( 0 ) 1 γ y 2 s , μ 1 ( s ) , μ 2 ( s ) d s ω ^ y 2 Υ R Γ ( γ ) ϕ ( b ) ϕ ( s ) ϱ Γ ( ϱ + γ ) .
Hence
H μ 1 , μ 2 B = H 1 μ 1 , μ 2 C 1 γ ; ϕ J + H 2 μ 1 , μ 2 C 1 γ ; ϕ J k = 1 2 ω ^ y k Υ R Γ ( γ ) ϕ ( b ) ϕ ( s ) ϱ Γ ( ϱ + γ ) = l .
This proves that H is uniformly bounded on X . Now, we shall show that H X is an equicontinuous set in X . Let 0 < ı 1 < ı 2 < b and μ 1 , μ 2 X . Then, we have
H 1 μ 1 , μ 2 ı 2 H 1 μ 1 , μ 2 ı 1 C 1 γ ; ϕ J max ı J ϕ ( ı 2 ) ϕ ( 0 ) 1 γ 0 ı 2 ϕ ( s ) ϕ ( ı 2 ) ϕ ( s ) ϱ 1 Γ ( ϱ ) y 1 s , μ 1 ( s ) , μ 2 ( s ) d s ϕ ( ı 1 ) ϕ ( 0 ) 1 γ 0 ı 1 ϕ ( s ) ϕ ( ı 1 ) ϕ ( s ) ϱ 1 Γ ( ϱ ) y 1 s , μ 1 ( s ) , μ 2 ( s ) d s 0 ı 1 ϕ ( s ) ϕ ( ı 2 ) ϕ ( s ) ϱ 1 ϕ ( ı 1 ) ϕ ( s ) ϱ 1 Γ ( ϱ ) ϕ ( s ) ϕ ( 0 ) γ 1 ϕ ( s ) ϕ ( 0 ) 1 γ y 1 s , μ 1 ( s ) , μ 2 ( s ) d s + ı 1 ı 2 ϕ ( s ) ϕ ( ı 2 ) ϕ ( s ) ϱ 1 Γ ( ϱ ) ϕ ( s ) ϕ ( 0 ) 1 γ y 1 s , μ 1 ( s ) , μ 2 ( s ) d s ω ^ y 1 Υ R Γ ( γ ) ϕ ( ı 2 ) ϕ ( 0 ) ϱ ϕ ( ı 1 ) ϕ ( 0 ) ϱ Γ ( ϱ + 1 ) 0   a s   ı 2 ı 1 .
On the other hand, by the same way, we obtain
H 2 μ 1 , μ 2 ı 2 H 2 μ 1 , μ 2 ı 1 1 γ ; ϕ = max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ H 2 μ 1 , μ 2 ı 2 H 2 μ 1 , μ 2 ı 1 0 a s ı 2 ı 1 .
This implies that
H μ 1 , μ 2 ı 2 H μ 1 , μ 2 ı 1 B 0 a s ı 2 ı 1 .
Thus, H X is an equicontinuous set in X .
Step (3): Let μ 1 , μ 2 B and μ 1 * , μ 2 * X such that
μ 1 ( ı ) = A 1 μ 1 , μ 2 ( ı ) H 1 μ 1 * , μ 2 * ( ı ) + K 1 μ 1 , μ 2 ( ı ) , μ 2 ( ı ) = A 2 μ 1 , μ 2 ( ı ) H 2 μ 1 * , μ 2 * ( ı ) + K 2 μ 1 , μ 2 ( ı ) .
Then, we have
μ 1 C 1 γ ; ϕ J A 1 μ 1 , μ 2 ( ı ) H 1 μ 1 * , μ 2 * ( ı ) + K 1 μ 1 , μ 2 ( ı ) ϕ ( ı ) ϕ ( 0 ) 1 γ f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ϱ 1 Γ ( ϱ ) ϕ ( s ) ϕ ( 0 ) γ 1 ϕ ( s ) ϕ ( 0 ) 1 γ y 1 s , μ 1 * ( s ) , μ 2 * ( s ) d s + ϕ ( ı ) ϕ ( 0 ) 1 γ i = 1 m 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) q i 1 Γ ( q i ) ϕ ( s ) ϕ ( 0 ) γ 1 ϕ ( s ) ϕ ( 0 ) 1 γ g i 1 s , μ 1 ( s ) , μ 2 ( s ) d s Φ ^ f 1 μ 1 , μ 2 B + F 1 ω ^ y 1 Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) + i = 1 m Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) Φ ^ g i 1 μ 1 , μ 2 B + G i 1 Φ ^ f 1 R + F 1 ω ^ y 1 Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) + i = 1 m Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) Φ ^ g i 1 R + G i 1 Φ ^ f 1 ω ^ y 1 Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) + i = 1 m Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) Φ ^ g i 1 R + F 1 ω ^ y 1 Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) + i = 1 m Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) G i 1 .
Similarly,
μ 2 C 1 γ ; ϕ J = A 2 μ 1 , μ 2 ( ı ) H 2 μ 1 * , μ 2 * ( ı ) + K 2 μ 1 , μ 2 ( ı ) Φ ^ f 2 ω ^ y 2 Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) + i = 1 m Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) Φ ^ g i 2 R + F 2 ω ^ y 2 Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) + i = 1 m Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) G i 2 .
Thus
μ 1 , μ 2 B k = 1 2 Φ ^ f k ω ^ y k Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) + i = 1 m Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) Φ ^ g i k R + k = 1 2 F k ω ^ y k Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) + i = 1 m Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) G i k R .
Therefore, μ 1 , μ 2 X .
Step (4): Finally, we will show that L A M H + L K < 1 . Since
M H = H X B = sup μ 1 , μ 2 X sup ı J H μ 1 , μ 2 ( ı ) k = 1 2 ω ^ y k Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) ,
then, by (8), one can obtain
L A k = 1 2 ω ^ y k Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) + L K < 1 ,
with L A = k = 1 2 Φ ^ f k and L K = k = 1 2 i = 1 m Γ γ ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) Φ ^ g i k . As a result of Lemma 4, we conclude that the operator equation
V μ 1 , μ 2 ( ı ) = A μ 1 , μ 2 ( ı ) H μ 1 , μ 2 ( ı ) + K μ 1 , μ 2 ( ı ) ,
has at least one solution in B . □

4. Existence of Solution for ϕ -Hilfer Hybrid System (2)

In this section, we will study the existence solution of ϕ -Hilfer hybrid system (2), according to Dhage’s helpful generalization of Krasnoselskii’s fixed point theorem [27]. We consider the following assumptions to obtain our main results:
(H 3 ) The functions y k , f k : J × R 2 R , f k 0 , 0 , 0 0 , k = 1 , 2 , are continuous and there exist two positive functions Φ y k , Φ f k with bound Φ ^ y k , Φ ^ f k , k = 1 , 2 , respectively, such that for each μ 1 , μ 2 , μ ¯ 1 , μ ¯ 2 B , we have
y k ı , μ 1 ( ı ) , μ 2 ( ı ) y k ı , μ ¯ 1 ( ı ) , μ ¯ 2 ( ı ) Φ y k ı μ 1 ( ı ) μ ¯ 1 ( ı ) + μ 2 ( ı ) μ ¯ 2 ( ı ) , f k ı , μ 1 ( ı ) , μ 2 ( ı ) f k ı , μ ¯ 1 ( ı ) , μ ¯ 2 ( ı ) Φ f k ı μ 1 ( ı ) μ ¯ 1 ( ı ) + μ 2 ( ı ) μ ¯ 2 ( ı ) .
(H 4 ) The functions y k , g i k , f k : J × R 2 R , f k 0 , 0 , 0 0 , g i k 0 , 0 , 0 = 0 , i = 1 , 2 , , m , k = 1 , 2 , are continuous. For all ı , μ 1 , μ 2 J × R 2 , there exists a functions ω y k , ω f k , ω g i k : J R + such that
ϕ ( ı ) ϕ ( 0 ) 1 γ y k ı , μ 1 ( ı ) , μ 2 ( ı ) ω y k ı , ϕ ( ı ) ϕ ( 0 ) 1 γ f k ı , μ 1 ( ı ) , μ 2 ( ı ) ω f k ı ,
and
ϕ ( ı ) ϕ ( 0 ) 1 γ g i k ı , μ 1 ( ı ) , μ 2 ( ı ) ω g i k ı .
Lemma 6
[27]). Let B be a Banach space and X B be a closed convex, bounded and nonempty subset of a Banach space B . Let Q : B B and K : X B be operators such that
(i)
K is completely continuous,
(ii)
Q is a contraction,
(iii)
μ = Q μ + K μ * for all μ * X μ X .
Then the operator equation μ = A μ + K μ * has a solution.
Definition 3.
A function μ 1 , μ 2 B is said to be a solution of ϕ-Hilfer hybrid system (2) if μ 1 H D 0 + ν , κ , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) and μ 2 H D 0 + ν , κ , ϕ μ 2 ( ı ) i = 1 m I 0 + q i , ϕ g i 2 ı , μ 1 ( ı ) , μ 2 ( ı ) f 2 ı , μ 1 ( ı ) , μ 2 ( ı ) are continuous for each μ 1 , μ 2 B and satisfies ϕ-Hilfer hybrid system and the conditions in (2).
Lemma 7.
Let γ = ν + κ ν κ , γ 1 = ϱ + β ϱ β such that ϱ , ν 0 , 1 , β , κ 0 , 1 , g i k , y k , f k : J × R 2 R , f k ( 0 , 0 , 0 ) 0 , g i k ( 0 , 0 , 0 ) = 0 , , i = 1 , 2 , , m , k = 1 , 2 are continuous functions. If μ 1 , μ 2 B satisfies the ϕ-Hilfer hybrid system (2), then, μ 1 , μ 2 satisfies the following integral equations
μ 1 ( ı ) = I 0 + ν , ϕ f 1 ı , μ 1 , μ 2 I 0 + ϱ , ϕ y 1 ı , μ 1 , μ 2 + i = 1 m I 0 + q i + ν , ϕ g i 1 ı , μ 1 , μ 2 , μ 2 ( ı ) = I 0 + ν , ϕ f 2 ı , μ 1 , μ 2 I 0 + ϱ , ϕ y 2 ı , μ 1 , μ 2 + i = 1 m I 0 + q i + ν , ϕ g i 2 ı , μ 1 , μ 2 .
Proof. 
In the beginning, we assume that μ 1 , μ 2 B is a solution of ϕ -Hilfer hybrid system (2). We will prove that μ 1 , μ 2 satisfies the integral Equation (10). First, let
H D 0 + ϱ , β , ϕ H D 0 + ν , κ , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) = y 1 ı , μ 1 ( ı ) , μ 2 ( ı ) , I 0 + 1 γ , ϕ μ 1 ( 0 ) = 0 , H D 0 + ν , κ , ϕ μ 1 ( 0 ) = 0 .
Taking the operator I 0 + ϱ , ϕ on both sides of the (11) and using Lemma 3, we have
H D 0 + ν , κ , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) = ϕ ( ı ) ϕ ( 0 ) γ 1 1 Γ ( γ ) I 0 + 1 γ 1 , ϕ H D 0 + ν , κ , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) ı = 0 + I 0 + ϱ , ϕ y 1 ı , μ 1 ( ı ) , μ 2 ( ı ) .
By the condition H D 0 + ν , κ , ϕ μ 1 ( 0 ) = 0 , g i 1 0 , 0 , 0 = 0 , we obtain
H D 0 + ν , κ , ϕ μ 1 ( ı ) = f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) I 0 + ϱ , ϕ y 1 ı , μ 1 ( ı ) , μ 2 ( ı ) + i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) .
Inserting the operator I 0 + ν , ϕ into both sides of Equation (12) and using Lemma 3, with semigroup property I 0 + q i , ϕ I 0 + ν , ϕ = I 0 + q i + ν , ϕ , we have
μ 1 ( ı ) = ϕ ( ı ) ϕ ( 0 ) γ 1 Γ ( γ ) I 0 + 1 γ , ϕ μ 1 ( 0 ) + I 0 + ν , ϕ f 1 ı , μ 1 , μ 2 I 0 + ϱ , ϕ y 1 ı , μ 1 , μ 2 + i = 1 m I 0 + q i + ν , ϕ g i 1 ı , μ 1 , μ 2 .
By the condition I 0 + 1 γ , ϕ μ 1 ( 0 ) = 0 , we obtain
μ 2 ( ı ) = I 0 + ν , ϕ f 2 ı , μ 1 , μ 2 I 0 + ϱ , ϕ y 2 ı , μ 1 , μ 2 + i = 1 m I 0 + q i + ν , ϕ g i 2 ı , μ 1 , μ 2 .
By the same way, we obtain
μ 2 ( ı ) = I 0 + ν , ϕ f 2 ı , μ 1 , μ 2 I 0 + ϱ , ϕ y 2 ı , μ 1 , μ 2 + i = 1 m I 0 + q i + ν , ϕ g i 2 ı , μ 1 , μ 2 .
From (13) and (14), we conclude that μ 1 , μ 2 satisfies the integral Equation (10).
Conversely, assume that μ 1 , μ 2 satisfies the integral Equation (10). Applying the operators H D 0 + ν , κ , ϕ and I 0 + 1 γ , ϕ of the integral Equation (13), with replace ı by 0, we obtain
H D 0 + ν , κ , ϕ μ 1 ( 0 ) = 0 a n d I 0 + 1 γ , ϕ μ 1 ( 0 ) = 0 .
Applying again the operators H D 0 + ν , κ , ϕ and I 0 + 1 γ , ϕ of the integral Equation (14), with replace ı by 0, we obtain
H D 0 + ν , κ , ϕ μ 2 ( 0 ) = 0 a n d I 0 + 1 γ , ϕ μ 2 ( 0 ) = 0 .
The proof is completed. □
In the following analyses, we use the following notations to keep things simple,
Z = k = 1 2 Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ν + ϱ + γ Γ ( ν + ϱ + γ ) ω ^ y k Φ ^ f k + ω ^ f k Φ ^ y k ,
and
P = k = 1 2 i = 1 m ω ^ g i k ϕ ( b ) ϕ ( 0 ) q i + ν Γ ( q i + ν + 1 ) ,
where ω ^ f k , ω ^ y k and ω ^ g i k , k = 1.2 , are bound of the functions ω y k , ω f k and ω g i k , respectively.
Theorem 3.
Assume that (H 3 ), (H 4 ) hold. If
Π = k = 1 2 Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ν + ϱ + γ Γ ( ν + ϱ + γ ) ω ^ y k Φ ^ f k + ω ^ f k Φ ^ y k < 1 ,
then, the ϕ-Hilfer hybrid system (2) has at least one solution on B .
Proof. 
Define a closed ball set
X = μ 1 , μ 2 B : μ 1 , μ 2 B R ,
with
R Z + P .
Define the operator V : B B as V = V 1 , V 2 , where
V 1 μ 1 , μ 2 ( ı ) = I 0 + ν , ϕ f 1 ı , μ 1 , μ 2 I 0 + ϱ , ϕ y 1 ı , μ 1 , μ 2 + i = 1 m I 0 + q i + ν , ϕ g i 1 ı , μ 1 , μ 2 , V 2 μ 1 , μ 2 ( ı ) = I 0 + ν , ϕ f 2 ı , μ 1 , μ 2 I 0 + ϱ , ϕ y 2 ı , μ 1 , μ 2 + i = 1 m I 0 + q i + ν , ϕ g i 2 ı , μ 1 , μ 2 .
To use Lemma 6, we define operators A = A 1 , A 2 : B B by
A μ 1 , μ 2 ( ı ) = f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) , f 2 ı , μ 1 ( ı ) , μ 2 ( ı ) ,
H = H 1 , H 2 : X B by
H μ 1 , μ 2 ( ı ) = 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ϱ 1 Γ ( ϱ ) y 1 s , μ 1 ( s ) , μ 2 ( s ) d s , 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ϱ 1 Γ ( ϱ ) y 2 s , μ 1 ( s ) , μ 2 ( s ) d s ,
and K = K 1 , K 2 : B B by
K μ 1 , μ 2 ( ı ) = i = 1 m 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) q i + ν 1 Γ ( q i + ν ) g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) , i = 1 m 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) q i + ν 1 Γ ( q i + ν ) g i 2 ı , μ 1 ( ı ) , μ 2 ( ı ) ,
and
Q μ 1 , μ 2 ( ı ) = 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ν 1 Γ ( ν ) A 1 μ 1 , μ 2 ( s ) H 1 μ 1 , μ 2 ( s ) d s , 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ν 1 Γ ( ν ) A 2 μ 1 , μ 2 ( s ) H 2 μ 1 , μ 2 ( s ) d s .
Thus, the coupled system of the above hybrid integral Equation (15) can be written as a system of operator equations as
V μ 1 , μ 2 ( ı ) = Q μ 1 , μ 2 ( ı ) + K μ 1 , μ 2 ( ı ) .
In the steps that follow, we will show that the operators Q and K obey the claims of Lemma 6.
Step (1): K is completely continuous. The operator K is obviously continuous. For μ 1 , μ 2 B , ı J , we have
K 1 μ 1 , μ 2 C 1 γ ; ϕ J i = 1 m 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) q i + ν 1 Γ ( q i + ν ) ϕ ( s ) ϕ ( 0 ) γ 1 ϕ ( s ) ϕ ( 0 ) 1 γ g i 1 s , μ 1 ( s ) , μ 2 ( s ) i = 1 m ω ^ g i 1 Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i + ν Γ ( q i + ν + γ ) .
Similarly, we obtain
K 2 μ 1 , μ 2 C 1 γ ; ϕ J i = 1 m 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) q i + ν 1 Γ ( q i + ν ) ϕ ( s ) ϕ ( 0 ) γ 1 ϕ ( s ) ϕ ( 0 ) 1 γ g i 2 s , μ 1 ( s ) , μ 2 ( s ) i = 1 m ω ^ g i 2 Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i + ν Γ ( q i + ν + γ ) .
Hence
K μ 1 , μ 2 B K 1 μ 1 , μ 2 C 1 γ ; ϕ J + K 2 μ 1 , μ 2 C 1 γ ; ϕ J k = 1 2 i = 1 m ω ^ g i k Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i + ν Γ ( q i + ν + γ ) = l .
Thus, K is bounded by
l = k = 1 2 i = 1 m ω ^ g i k Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i + ν Γ ( q i + ν + γ ) .
Let 0 < ı 1 < ı 2 < b and μ 1 , μ 2 X . Then, we have
K 1 μ 1 , μ 2 ( ı 2 ) K 1 μ 1 , μ 2 ( ı 1 ) C 1 γ ; ϕ J ϕ ( ı ) ϕ ( 0 ) 1 γ i = 1 m 0 ı 2 ϕ ( s ) ϕ ( ı 2 ) ϕ ( s ) q i + ν 1 Γ ( q i + ν ) ϕ ( s ) ϕ ( 0 ) γ 1 max ı , μ 1 , μ 2 J × X ϕ ( s ) ϕ ( 0 ) 1 γ g i 1 s , μ 1 ( s ) , μ 2 ( s ) d s ϕ ( ı ) ϕ ( 0 ) 1 γ i = 1 m 0 ı 1 ϕ ( s ) ϕ ( ı 1 ) ϕ ( s ) q i + ν 1 Γ ( q i + ν ) ϕ ( s ) ϕ ( 0 ) γ 1 max ı , μ 1 , μ 2 J × X ϕ ( s ) ϕ ( 0 ) 1 γ g i 1 s , μ 1 ( s ) , μ 2 ( s ) d s i = 1 m 0 ı 1 ϕ ( s ) ϕ ( ı 2 ) ϕ ( s ) q i + ν 1 ϕ ( ı 1 ) ϕ ( s ) q i + ν 1 Γ ( q i + ν ) ϕ ( s ) ϕ ( 0 ) γ 1 max ı , μ 1 , μ 2 J × X ϕ ( s ) ϕ ( 0 ) 1 γ g i 1 s , μ 1 ( s ) , μ 2 ( s ) d s + i = 1 m ı 1 ı 2 ϕ ( s ) ϕ ( ı 2 ) ϕ ( s ) q i + ν 1 Γ ( q i + ν ) ϕ ( s ) ϕ ( 0 ) γ 1 max ı , μ 1 , μ 2 J × X ϕ ( s ) ϕ ( 0 ) 1 γ g i 1 s , μ 1 ( s ) , μ 2 ( s ) d s i = 1 m ω ^ g i 1 Γ ( γ ) ϕ ( ı 2 ) ϕ ( 0 ) q i + ν ϕ ( ı 1 ) ϕ ( 0 ) q i + ν Γ ( q i + ν + γ ) .
By the same way, we obtain
K 2 μ 1 , μ 2 ( ı 2 ) K 2 μ 1 , μ 2 ( ı 1 ) C 1 γ ; ϕ J i = 1 m ω ^ g i 2 Γ ( γ ) ϕ ( ı 2 ) ϕ ( 0 ) q i + ν ϕ ( ı 1 ) ϕ ( 0 ) q i + ν Γ ( q i + ν + γ ) .
Thus
K μ 1 , μ 2 ( ı 2 ) K μ 1 , μ 2 ( ı 1 ) B i = 1 m ω ^ g i 1 Γ ( γ ) ϕ ( ı 2 ) ϕ ( 0 ) q i + ν ϕ ( ı 1 ) ϕ ( 0 ) q i + ν Γ ( q i + ν + γ ) + i = 1 m ω ^ g i 2 Γ ( γ ) ϕ ( ı 2 ) ϕ ( 0 ) q i + ν ϕ ( ı 1 ) ϕ ( 0 ) q i + ν Γ ( q i + ν + γ ) 0   a s   ı 2 ı 1 .
Thus, K is equicontinuous. Consequently, K is relatively compact on X . Hence, by the Arzelá-Ascoli theorem, we conclude that K is compact on X .
Step (2): Q is a contraction mapping. Let μ 1 , μ 2 , μ ¯ 1 , μ ¯ 2 X . Then for ı J , we have
Q 1 μ 1 , μ 2 Q 1 μ ¯ 1 , μ ¯ 2 C 1 γ ; ϕ J = max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ν 1 Γ ( ν ) A 1 μ 1 , μ 2 ( s ) H 1 μ 1 , μ 2 ( s ) A 1 μ ¯ 1 , μ ¯ 2 ( s ) H 1 μ ¯ 1 , μ ¯ 2 ( s ) d s = max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ν 1 Γ ( ν ) A 1 μ 1 , μ 2 ( s ) H 1 μ 1 , μ 2 ( s ) A 1 μ ¯ 1 , μ ¯ 2 ( s ) H 1 μ 1 , μ 2 ( s ) + A 1 μ ¯ 1 , μ ¯ 2 ( s ) H 1 μ 1 , μ 2 ( s ) + A 1 μ ¯ 1 , μ ¯ 2 ( s ) H 1 μ ¯ 1 , μ ¯ 2 ( s ) 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ν 1 Γ ( ν ) H 1 μ 1 , μ 2 ( s ) A 1 μ 1 , μ 2 ( s ) A 1 μ 1 , μ 2 ( s ) + 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ν 1 Γ ( ν ) A 1 μ 1 , μ 2 ( s ) H 1 μ 1 , μ 2 ( s ) H 1 μ 1 , μ 2 ( s ) Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ν + ϱ + γ Γ ( ν + ϱ + γ ) ω ^ y 1 Φ ^ f 1 + ω ^ f 1 Φ ^ y 1 μ 1 μ ¯ 1 C 1 γ ; ϕ J + μ 2 μ ¯ 2 C 1 γ ; ϕ J .
By same technique, one can obtain
Q 2 μ 1 , μ 2 Q 2 μ ¯ 1 , μ ¯ 2 C 1 γ ; ϕ J Γ ( γ ) ϕ ( ı ) ϕ ( 0 ) ν + ϱ + γ Γ ( ν + ϱ + γ ) ω ^ y 2 Φ ^ f 2 + ω ^ f 2 Φ ^ y 2 μ 1 μ ¯ 1 C 1 γ ; ϕ J + μ 2 μ ¯ 2 C 1 γ ; ϕ J .
Thus
Q μ 1 , μ 2 Q μ ¯ 1 , μ ¯ 2 B Π μ 1 , μ 2 μ ¯ 1 , μ ¯ 2 B .
Step (3): For any y 1 , y 2 , we have
V 1 μ 1 , μ 2 C 1 γ ; ϕ J = max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ V 1 μ 1 , μ 2 ( ı ) = max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ Q 1 μ 1 , μ 2 ( ı ) + K 1 y 1 , y 2 ( ı ) ω ^ f 1 ω ^ y 1 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) ν 1 Γ ( ν ) 0 s ϕ ( u ) ϕ ( s ) ϕ ( u ) ϱ 1 Γ ( ϱ ) d u d s + i = 1 m ω ^ g i 1 0 ı ϕ ( s ) ϕ ( ı ) ϕ ( s ) q i + ν 1 Γ ( q i + ν ) d s Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ν + ϱ + γ Γ ( ν + ϱ + γ ) ω ^ y 1 Φ ^ f 1 + ω ^ f 1 Φ ^ y 1 + i = 1 m ω ^ g i 1 Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i + ν Γ ( q i + ν + γ )
Similarly, one can obtain
V 2 μ 1 , μ 2 C 1 γ ; ϕ J = max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ V 2 μ 1 , μ 2 ( ı ) = max ı J ϕ ( ı ) ϕ ( 0 ) 1 γ Q 2 μ 1 , μ 2 ( ı ) + K 2 y 1 , y 2 ( ı ) Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ν + ϱ + γ Γ ( ν + ϱ + γ ) ω ^ y 2 Φ ^ f 2 + ω ^ f 2 Φ ^ y 2 + i = 1 m ω ^ g i 2 ϕ ( b ) ϕ ( 0 ) q i + ν Γ ( q i + ν + 1 ) .
It follows
V μ 1 , μ 2 B V 1 μ 1 , μ 2 C 1 γ ; ϕ J + V 2 μ 1 , μ 2 C 1 γ ; ϕ J Z + P R .
Thus, V μ 1 , μ 2 X . Hence, the last condition in Lemma 6 holds. According to above steps together with Lemma 6, we conclude that ϕ -Hilfer hybrid system (2) has at least one solution on B .  □

5. Examples

In this section, we cover our results with two examples that illustrate the applicability of the findings we have obtained.
Example 1.
Take the following coupled ϕ-Hilfer hybrid system
H D 0 + ϱ , β , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) = ı 1 2 + 3 35 13 ı 2 7 μ 1 ( ı ) + μ 2 ( ı ) + 15 , H D 0 + ϱ , β , ϕ μ 2 ( ı ) i = 1 m I 0 + q i , ϕ g i 2 ı , μ 1 ( ı ) , μ 2 ( ı ) f 2 ı , μ 1 ( ı ) , μ 2 ( ı ) = ı 1 2 + 3 5 12 ı 7 μ 1 ( ı ) + μ 2 ( ı ) + 15 , I 0 + 1 γ , ϕ μ k ( ı ) i = 1 m I 0 + q i , ϕ g i k ı , μ 1 ( ı ) , μ 2 ( ı ) f k ı , μ 1 ( ı ) , μ 2 ( ı ) ı = 0 = 0 , k = 1 , 2 ,
where ϱ = 1 2 , β = 1 2 , γ = 3 4 , ϕ ( ı ) = e ı , m = 3 , ı 0 , b : = 0 , 1 , q 1 = 1 2 , q 2 = 1 3 and q 3 = 1 4 . For k = 1 , 2 , we have
i = 1 3 I 0 + q i , ϕ g i k ı , μ 1 ( ı ) , μ 2 ( ı ) = I 0 + 1 2 , e ı 2 ı 5 4 + ı μ 1 ( ı ) + μ 2 ( ı ) μ 1 ( ı ) + μ 2 ( ı ) + 5 + I 0 + 1 3 , e ı ı sin ı 5 3 + e ı μ 1 ( ı ) + μ 2 ( ı ) μ 1 ( ı ) + μ 2 ( ı ) + 3 + I 0 + 1 4 , e ı ı sin ı 2 + 5 3 + e ı μ 1 ( ı ) + μ 2 ( ı ) μ 1 ( ı ) + μ 2 ( ı ) + 2 ,
and
f k ı , μ 1 ( ı ) , μ 2 ( ı ) = 1 100 μ 1 ( ı ) + μ 2 ( ı ) + 1 1 + ı .
In view of a given data, we noted that the functions g i k , f k : J × R 2 R with f k 0 , 0 , 0 = 1 100 0 , g i k 0 , 0 , 0 = 0 , i = 1 , 2 , 3 , k = 1 , 2 , are continuous functions. Moreover, for each μ 1 , μ 2 , μ ¯ 1 , μ ¯ 2 B , for k = 1 , 2 , we have
f k ı , μ 1 ( ı ) , μ 2 ( ı ) f k ı , μ ¯ 1 ( ı ) , μ ¯ 2 ( ı ) 1 100 μ 1 μ ¯ 1 + μ 2 μ ¯ 2 , g 1 k ı , μ 1 ( ı ) , μ 2 ( ı ) g 1 k ı , μ ¯ 1 ( ı ) , μ ¯ 2 ( ı ) 1 10 μ 1 μ ¯ 1 + μ 2 μ ¯ 2 , g 2 k ı , μ 1 ( ı ) , μ 2 ( ı ) g 2 k ı , μ ¯ 1 ( ı ) , μ ¯ 2 ( ı ) 2 5 3 + e μ 1 μ ¯ 1 + μ 2 μ ¯ 2 , g 3 k ı , μ 1 ( ı ) , μ 2 ( ı ) g 3 k ı , μ ¯ 1 ( ı ) , μ ¯ 2 ( ı ) 1 17 + 5 e μ 1 μ ¯ 1 + μ 2 μ ¯ 2 ,
with Φ ^ f 1 = Φ ^ f 2 = 1 100 , Φ ^ g 1 1 = Φ ^ g 1 2 = 1 10 , Φ ^ g 2 1 = Φ ^ g 2 2 = 2 5 3 + e and Φ ^ g 3 1 = Φ ^ g 3 2 = 1 17 + 5 e . By a given data, we conclude the condition (H 1 ) is satisfied. For all ı , μ 1 , μ 2 J × R 2 , there exists a functions ω y k : J R + and a continuous nondecreasing function Υ : R + R + such that
y k ı , μ 1 ( ı ) , μ 2 ( ı ) ω y k ı Υ μ 1 , μ 2 ,
where ω y k ı = ı 1 2 + 3 13 ı and Υ μ 1 , μ 2 = μ 1 , μ 2 5 + 3 7 . Then, one can find that ω ^ y k = 4 13 , k = 1 , 2 . Hence, the condition (H 2 ) is satisfied. Additionally
k = 1 2 Φ ^ f k ω ^ y k Υ R Γ ( γ ) ϕ ( b ) ϕ ( 0 ) ϱ Γ ( ϱ + γ ) + i = 1 m Γ ( γ ) ϕ ( b ) ϕ ( 0 ) q i Γ ( q i + γ ) Φ ^ g i k 0.65 < 1 .
Thus, by Theorem 2, we conclude that the ϕ-Hilfer hybrid system (16) has at least one solution in B .
Example 2.
Take the following coupled ϕ-Hilfer hybrid system
H D 0 + ϱ , β , ϕ H D 0 + ν , κ , ϕ μ 1 ( ı ) i = 1 m I 0 + q i , ϕ g i 1 ı , μ 1 ( ı ) , μ 2 ( ı ) f 1 ı , μ 1 ( ı ) , μ 2 ( ı ) = ı 1 2 + 3 35 13 ı 2 7 μ 1 ( ı ) + μ 2 ( ı ) + 15 , H D 0 + ϱ , β , ϕ H D 0 + ν , κ , ϕ μ 2 ( ı ) i = 1 m I 0 + q i , ϕ g i 2 ı , μ 1 ( ı ) , μ 2 ( ı ) f 2 ı , μ 1 ( ı ) , μ 2 ( ı ) = ı 1 2 + 3 5 12 ı 7 μ 1 ( ı ) + μ 2 ( ı ) + 15 , I 0 + 1 γ , ϕ μ k ( 0 ) = 0 , H D 0 + ν k , κ k , ϕ μ k ( 0 ) = 0 , k = 1 , 2 ,
where ϱ = 1 2 , β = 1 2 , ν = 3 4 , κ = 1 4 , γ = 13 16 , ı 0 , b : = 0 , 1 , ϕ ( ı ) = e ı , m = 3 , q 1 = 1 2 , q 2 = 1 3 and q 3 = 1 4 . For k = 1 , 2 , we have
i = 1 3 I 0 + q i , ϕ g i k ı , μ 1 ( ı ) , μ 2 ( ı ) = I 0 + 1 2 , e ı ı 3 + ı μ 1 ( ı ) + μ 2 ( ı ) μ 1 ( ı ) + μ 2 ( ı ) + 5 + I 0 + 1 3 , e ı sin ı 3 + e ı μ 1 ( ı ) + μ 2 ( ı ) μ 1 ( ı ) + μ 2 ( ı ) + 3 + I 0 + 1 4 , e ı ı 7 + e ı μ 1 ( ı ) + μ 2 ( ı ) μ 1 ( ı ) + μ 2 ( ı ) + 2 ,
and
f k ı , μ 1 ( ı ) , μ 2 ( ı ) = 1 50 μ 1 ( ı ) + μ 2 ( ı ) + 2 1 + ı .
In view of a given data, we noted that the functions g i k , f k : J × R 2 R with f k 0 , 0 , 0 = 1 25 0 , g i k 0 , 0 , 0 = 0 k = 1 , 2 , are continuous functions. Moreover, for each μ 1 , μ 2 , μ ¯ 1 , μ ¯ 2 B , there exist two positive functions Φ y k ( ı ) , Φ f k ( ı ) , k = 1 , 2 with bound Φ ^ y k , Φ ^ f k respectively, such that for each μ 1 , μ 2 , μ ¯ 1 , μ ¯ 2 B , we have
y k ı , μ 1 ( ı ) , μ 2 ( ı ) y k ı , μ ¯ 1 ( ı ) , μ ¯ 2 ( ı ) ı 1 2 + 3 13 ı μ 1 μ ¯ 1 + μ 2 μ ¯ 2 , f k ı , μ 1 ( ı ) , μ 2 ( ı ) f k ı , μ ¯ 1 ( ı ) , μ ¯ 2 ( ı ) 1 50 μ 1 μ ¯ 1 + μ 2 μ ¯ 2 ,
with Φ ^ f k = 1 50 and Φ ^ y k = 4 13 . By a given data, we conclude the condition (H 3 ) is satisfied. For all ı , μ 1 , μ 2 J × R 2 , there exists a functions ω y k , ω f k , ω g k : J R + such that
y k ı , μ 1 ( ı ) , μ 2 ( ı ) ı 1 2 + 3 13 ı , f k ı , μ 1 ( ı ) , μ 2 ( ı ) 1 50 ,
and
g 1 k ı , μ 1 ( ı ) , μ 2 ( ı ) ı 3 + ı , g 2 k ı , μ 1 ( ı ) , μ 2 ( ı ) sin ı 3 + e ı , g 3 k ı , μ 1 ( ı ) , μ 2 ( ı ) ı 7 + e ı
then, we obtain ω ^ y k = 4 13 , ω ^ f k = 1 50 , ω ^ g 1 k = 1 4 , ω ^ g 2 k = ω ^ g 2 2 = 1 3 + e and ω ^ g 3 k = 1 7 + e . Hence, the condition (H 4 ) is satisfied. Additionally,
Π = k = 1 2 ϕ ( b ) ϕ ( 0 ) ν + ϱ 1 Γ ( ν + ϱ + 1 ) ω ^ y k Φ ^ f k + ω ^ f k Φ ^ y k 0.25 < 1 .
Thus, by Theorem 3, we conclude that the ϕ-Hilfer hybrid system (17) has at least one solution on B .

6. Conclusions

Recently, the theory of fractional differential equations has attracted the interest of several researchers in different filed due to its various applications. In particular, those involving generalized fractional operators. It is important that we investigate the fractional systems with generalized Hilfer derivatives since these derivatives cover many systems in the literature and they contain a kernel with different values that generate many special cases.
The existence of solutions for two class ϕ -Hilfer hybrid fractional integrodifferential equations was investigated in this study. The first result was obtained by applying Dhage’s hybrid fixed point theorem for three operators in a Banach algebra [26], while the second result was reached by applying Dhage’s helpful generalization of Krasnoselskii’s fixed point theorem [27]. The main conclusions are well-illustrated with examples. The results obtained in this work includes the results of Sitho et al. [24], Boutiara et al. [25] and cover many problems which do not study yet.

Author Contributions

Conceptualization, M.A.A., O.B., S.K.P., S.S.A. and G.I.O.; Data curation, M.A.A., O.B., S.K.P., S.S.A. and G.I.O.; Formal analysis, M.A.A., O.B., S.K.P., S.S.A. and G.I.O.; Investigation, M.A.A., O.B., S.K.P., S.S.A. and G.I.O.; Methodology, M.A.A., O.B., S.K.P., S.S.A. and G.I.O. All authors read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

The authors thank the reviewers for their useful comments, which led to the improvement of the content of the paper. Research Supporting Project number (RSP-2021/167), King Saud University, Riyadh, Saudi Arabia.

Conflicts of Interest

The author declares no conflict of interest.

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Almalahi, M.A.; Bazighifan, O.; Panchal, S.K.; Askar, S.S.; Oros, G.I. Analytical Study of Two Nonlinear Coupled Hybrid Systems Involving Generalized Hilfer Fractional Operators. Fractal Fract. 2021, 5, 178. https://doi.org/10.3390/fractalfract5040178

AMA Style

Almalahi MA, Bazighifan O, Panchal SK, Askar SS, Oros GI. Analytical Study of Two Nonlinear Coupled Hybrid Systems Involving Generalized Hilfer Fractional Operators. Fractal and Fractional. 2021; 5(4):178. https://doi.org/10.3390/fractalfract5040178

Chicago/Turabian Style

Almalahi, Mohammed A., Omar Bazighifan, Satish K. Panchal, S. S. Askar, and Georgia Irina Oros. 2021. "Analytical Study of Two Nonlinear Coupled Hybrid Systems Involving Generalized Hilfer Fractional Operators" Fractal and Fractional 5, no. 4: 178. https://doi.org/10.3390/fractalfract5040178

APA Style

Almalahi, M. A., Bazighifan, O., Panchal, S. K., Askar, S. S., & Oros, G. I. (2021). Analytical Study of Two Nonlinear Coupled Hybrid Systems Involving Generalized Hilfer Fractional Operators. Fractal and Fractional, 5(4), 178. https://doi.org/10.3390/fractalfract5040178

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