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Article

Development on a Fractional Hybrid Differential Inclusion with a Nonlinear Nonlocal Fractional-Order Integral Inclusion

by
Ahmed M. A. El-Sayed
1,†,
Sheren A. Abd El-Salam
2,*,† and
Hind H. G. Hashem
1,†
1
Faculty of Science, Alexandria University, Alexandria 21544, Egypt
2
Faculty of Sciences, Damanhour University, Damanhour 22511, Egypt
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Mathematics 2022, 10(21), 4068; https://doi.org/10.3390/math10214068
Submission received: 18 September 2022 / Revised: 25 October 2022 / Accepted: 27 October 2022 / Published: 1 November 2022

Abstract

:
In this article, we consider a Riemann–Liouville fractional-order nonlinear hybrid delay differential inclusion with a nonlinear set-valued nonlocal integral condition of fractional order. We prove some existence and uniqueness results in C ( I , R ) . We also study the continuous dependence of the solutions on the two sets of selections of the two set-valued functions, considered in our problem, and on some other parameters. Finally, to validate our results, we present an example and some particular cases.

1. Preliminaries and Introduction

Great contributions dedicated to investigate the fractional-order initial and boundary value problems due to the various applications and real-world problems can be found in the literature (see [1,2,3,4,5,6,7,8]). For fractional-order differential inclusions and some existence results in particular, see [9,10,11,12] and references therein.
Dhage and Lakshmikantham [13] introduced and initiated study of a new category of nonlinear differential equation called ordinary hybrid differential equation.
Fractional hybrid differential equations can be employed in modeling and describing non-homogeneous physical phenomena that take place in their form. The importance of investigations of hybrid differential equations lies in the fact that it includes many dynamic systems as special cases. There have been many works on the theory of hybrid differential equations (see [2,3,4,13,14,15,16]).
Hybrid differential equations and inclusions with some multi-valued maps have been studied in many monographs (see [17,18]).
Kamenskii et al. [17] studied a semi-linear differential inclusion, involving a Caputo fractional derivative, in a separable Banach space. They proved the existence of a mild solution to this inclusion with a multi-valued condition by using the method of the generalized translation multi-valued operator and some fixed point theorems. Double phase implicit obstacle problems with convection and multi-valued mixed-boundary value conditions have been discussed in [18] by applying the Kakutani–Ky Fan fixed-point theorem for multi-valued operators along with the theory of nonsmooth analysis and variational methods for pseudo-monotone operators.
Motivated by the above results, we study the fractional-order hybrid delay differential inclusion
R D α x ( t ) x ( 0 ) g ( t , x ( ϕ 2 ( t ) ) ) F ( t , x ( ϕ 1 ( t ) ) ) , t I = [ 0 , T ] ,
subject to the multi-valued fractional-order nonlocal integral condition
x ( 0 ) x 0 0 τ ( τ s ) β 1 Γ ( β ) H ( s , x ( ϕ 3 ( s ) ) ) d s , β α , τ I ,
where R D α is the Riemann–Liouville derivative of order α ( 0 , 1 ) , and F , H : I × R + P ( R ) are two set-valued mappings where P ( R ) denote the family of nonempty subsets of R .
The set-valued functions F ( . , x ( . ) ) and H ( . , x ( . ) ) are assumed to have Carathéodory selections, as in [19], to prove the existence of solutions x C ( I , R ) of the Problem (1)–(2). Moreover, we discuss the continuous dependence of these solutions on the two sets of selections S F and S H , of the set-valued functions F and H , and on the data x 0 .
In order to achieve our task, we study first the single-valued problem corresponding to our considered Problem (1)–(2),
R D α x ( t ) x ( 0 ) g ( t , x ( ϕ 2 ( t ) ) ) = f ( t , x ( ϕ 1 ( t ) ) ) ,
subject to the fractional-order nonlocal integral condition
x ( 0 ) x 0 = 0 τ ( τ s ) β 1 Γ ( β ) h ( s , x ( ϕ 3 ( s ) ) ) d s ,
where f F and h H .
With the aim of proving the existence of the solutions x C ( I , R ) of the Problem
The paper is organized as follows: Section 2 contains main results for the single-valued Problem (3)–(4). In Section 3, as extensions of these results, we deduce similar results for the multi-valued Problem (1)–(2). In this case, we have two sets of selections correspond to the two multivalued functions. We prove the continuous dependence of the solutions of that problem on the two sets of selections. Finally, we present some special cases as examples in Section 4.
The following theorem will be needed.
Theorem 1
(Nonlinear alternative of Leray–Schauder type [20]). Let E be a Banach space and Ω be a bounded open subset of E, 0 Ω and T : Ω ¯ E be a completely continuous operator. Then, either there exists u Ω , λ > 1 such that T ( u ) = λ u , or there exists a fixed point u * Ω ¯ .

2. Single-Valued Problem

Let E = C ( I , R ) , with supremum norm | | x | | = sup t I | x ( t ) | for any x E . Consider now the single-valued Problem (3)–(4) with the following assumptions:
(i)
f , h : I × R R are measurable for almost all t for every x R and continuous in x for every t I .
(ii)
There exist two bounded measurable functions m i L 1 ( I ) and two positive constants b i , i = 1 , 2 such that
| f ( t , x ) | m 1 ( t ) + b 1 | x | , and | h ( t , x ) | m 2 ( t ) + b 2 | x | , t I , x R
with M = m a x { sup m 1 ( t ) , sup m 2 ( t ) , t I } and b = m a x { b 1 , b 2 } .
(iii)
g : I × R R \ { 0 } is continuous and there exists a positive constant a, such that
| g ( t , x ) g ( t , y ) | a | x y | and k 1 = sup t I | g ( t , 0 ) | .
(iv)
φ i : I I , i = 1 , 2 , 3 are continuous and φ i ( t ) t , t I .
(v)
There exists a real number r ( 0 , 1 ) that satisfies the quadratic algebraic equation
| x 0 | + A ( M ( 1 + k 1 ) + ( b k 1 + a M + b ) r + a b r 2 ) = r ,
where A = m a x { T β Γ ( β + 1 ) , T α Γ ( α + 1 ) } .
Definition 1.
By a solution of the Problem (3)–(4), we mean a function x E that satisfies the Problem (3)–(4).
Now, we have the following lemma.
Lemma 1.
If the solution of the Problem (3)–(4) exists, then it is given by the fractional-orders integral equation
x ( t ) = x 0 + 0 τ ( τ s ) β 1 Γ ( β ) h ( s , x ( ϕ 3 ( s ) ) ) d s + g ( t , x ( ϕ 2 ( t ) ) ) I α f ( t , x ( ϕ 1 ( t ) ) ) .
Equation (5) is a quadratic integral equation. Quadratic integral equations appear in several real problems and have been studied in many articles in the literature (for more results and investigations, see [21,22,23,24,25,26,27]).
Proof. 
Let x E be a solution of the Problem (3)–(4). Integrating (3), then using the properties of the fractional calculus [28] and assumption (i), we obtain
x ( t ) = x ( 0 ) + g ( t , x ( ϕ 2 ( t ) ) ) I α f ( t , x ( ϕ 1 ( t ) ) ) ,
and (5) can be obtained.
Secondly, let Let x E be a solution of the integral Equation (5). Then,
x ( t ) x ( 0 ) = g ( t , x ( ϕ 2 ( t ) ) ) I α f ( t , x ( ϕ 1 ( t ) ) ) , x ( t ) x ( 0 ) g ( t , x ( ϕ 2 ( t ) ) ) = I α f ( t , x ( ϕ 1 ( t ) ) )
and
R D α x ( t ) x ( 0 ) g ( t , x ( ϕ 2 ( t ) ) ) = d d t I 1 α x ( t ) x ( 0 ) g ( t , x ( ϕ 2 ( t ) ) ) = d d t I 1 α I α f ( t , x ( ϕ 1 ( t ) ) ) = f ( t , x ( ϕ 1 ( t ) ) ) .

2.1. Existence of Solutions

Now, we can prove the following existence theorem.
Theorem 2.
Let the assumptions (i)–(v) be satisfied. Then the Problem (3)–(4) has a solution x C ( I , R ) .
Proof. 
Let r be given by the assumptions (iv). Define the set Ω by
Ω = { x C ( I , R ) , | | x | | r }
and the operator F by
F x ( t ) = x 0 + 0 τ ( τ s ) β 1 Γ ( β ) h ( s , x ( ϕ 3 ( s ) ) ) d s + g ( t , x ( ϕ 2 ( t ) ) ) I α f ( t , x ( ϕ 1 ( t ) ) ) .
Let x Ω , λ > 1 such that F x = λ x . Now,
λ r = λ | | x | | = | | F x | | ,
and then we have
| F x | = | x 0 + 0 τ ( τ s ) β 1 Γ ( β ) h ( s , x ( ϕ 3 ( s ) ) ) d s + g ( t , x ( ϕ 2 ( t ) ) ) I α f ( t , x ( ϕ 1 ( t ) ) ) | | x 0 | + 0 τ ( τ s ) β 1 Γ ( β ) | h ( s , x ( ϕ 3 ( s ) ) ) | d s + g ( t , x ( ϕ 2 ( t ) ) ) | | I α f ( t , x ( ϕ 1 ( t ) ) ) | | x 0 | + 0 τ ( τ s ) β 1 Γ ( β ) m 2 ( s ) + b 2 | x ( ϕ 3 ( s ) ) | d s + | g ( t , 0 ) | + a | x ( ϕ 2 ( t ) ) | 0 t ( t s ) α 1 Γ ( α ) m 1 ( s ) + b 1 | x ( ϕ 1 ( s ) ) | d s | x 0 | + ( M + b | | x | | ) T β Γ ( β + 1 ) + ( k 1 + a | | x | | ) ( M + b | | x | | ) T α Γ ( α + 1 ) | x 0 | + ( M + b r ) T β Γ ( β + 1 ) + ( k 1 + a r ) ( M + b r ) T α Γ ( α + 1 )
| x 0 | + A ( M + b r + ( k 1 + a r ) ( M + b r ) )
= | x 0 | + A ( M ( 1 + k 1 ) + ( b k 1 + a M + b ) r + a b r 2 ) = r .
Then F : Ω Ω is uniformly bounded,
i . e . , | | F x | | r , and λ 1 .
This contradicts that λ > 1 .
In what follows, we show that F is an equicontinuous operator.
Now, let x Ω and define the two functions
θ 1 ( δ ) = { sup g | g ( t , x ) g ( s , x ) | : t , s I , | t s | < δ , | x | r }
and
θ 2 ( δ ) = { sup x Ω | x ( t ) x ( s ) | : t , s I , | t s | < δ , | x | r } .
Then from the uniform continuity of the functions g on I × [ r , r ] and x on I we deduce that x Ω and δ > 0 there exist ϵ 1 , ϵ 2 > 0 , such that
| g ( t , x ) g ( s , x ) | θ 1 ( δ ) < ϵ 1 , | x ( t ) x ( s ) | θ 2 ( δ ) < ϵ 2
independently of x Ω (see [29,30].)
Now, for any x Ω ¯ , let t 1 , t 2 I , t 1 < t 2 , then we have
| F x ( t 2 ) F x ( t 1 ) | = | g ( t 2 , x ( ϕ 2 ( t 2 ) ) ) I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) g ( t 1 , x ( ϕ 2 ( t 1 ) ) ) I α f ( t 1 , x ( ϕ 1 ( t 1 ) ) ) | = | g ( t 2 , x ( ϕ 2 ( t 2 ) ) ) I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) g ( t 1 , x ( ϕ 2 ( t 1 ) ) ) I α f ( t 1 , x ( ϕ 1 ( t 1 ) ) ) g ( t 1 , x ( ϕ 2 ( t 1 ) ) ) I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) + g ( t 1 , x ( ϕ 2 ( t 1 ) ) ) I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) | | g ( t 2 , x ( ϕ 2 ( t 2 ) ) ) g ( t 1 , x ( ϕ 2 ( t 1 ) ) ) | | I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) | + | g ( t 1 , x ( ϕ 2 ( t 1 ) ) ) | | I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) I α f ( t 1 , x ( ϕ 1 ( t 1 ) ) ) | = | g ( t 2 , x ( ϕ 2 ( t 2 ) ) ) g ( t 1 , x ( ϕ 2 ( t 1 ) ) ) g ( t 1 , x ( ϕ 2 ( t 2 ) ) ) + g ( t 1 , x ( ϕ 2 ( t 2 ) ) ) | | I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) | + | g ( t 1 , x ( ϕ 2 ( t 1 ) ) ) | | I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) I α f ( t 1 , x ( ϕ 1 ( t 1 ) ) ) | | g ( t 2 , x ( ϕ 2 ( t 2 ) ) ) g ( t 1 , x ( ϕ 2 ( t 2 ) ) ) | | I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) | + | g ( t 1 , x ( ϕ 2 ( t 2 ) ) ) g ( t 1 , x ( ϕ 2 ( t 1 ) ) ) | | I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) | + | g ( t 1 , x ( ϕ 2 ( t 1 ) ) ) | | I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) I α f ( t 1 , x ( ϕ 1 ( t 1 ) ) ) | θ 1 ( δ ) | I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) | + a | x ( ϕ 2 ( t 2 ) ) x ( ϕ 2 ( t 1 ) ) | | I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) | + | g ( t 1 , x ( ϕ 2 ( t 1 ) ) ) | | I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) I α f ( t 1 , x ( ϕ 1 ( t 1 ) ) ) | θ 1 ( δ ) + a | x ( ϕ 2 ( t 2 ) ) x ( ϕ 2 ( t 1 ) ) | I α m 1 ( t 2 ) + b 1 | x ( ϕ 1 ( t 2 ) ) | + | g ( t 1 , 0 ) | + a | x ( ϕ 2 ( t 1 ) ) | | I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) I α f ( t 1 , x ( ϕ 1 ( t 1 ) ) ) | θ 1 ( δ ) + a sup x Ω | x ( ϕ 2 ( t 2 ) ) x ( ϕ 2 ( t 1 ) ) | M T α Γ ( α + 1 ) + b | | x | | T α Γ ( 1 + α ) + ( k 1 + a | | x | | ) | I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) I α f ( t 1 , x ( ϕ 1 ( t 1 ) ) ) | θ 1 ( δ ) + a θ 2 ( δ ) M T α Γ ( α + 1 ) + b | | x | | T α Γ ( 1 + α ) + ( k 1 + a | | x | | ) | I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) I α f ( t 1 , x ( ϕ 1 ( t 1 ) ) ) | .
However, we have
| I α f ( t 2 , x ( ϕ 1 ( t 2 ) ) ) I α f ( t 1 , x ( ϕ 1 ( t 1 ) ) ) | = | 0 t 2 ( t 2 s ) α 1 Γ ( α ) f ( s , x ( ϕ 1 ( s ) ) ) d s 0 t 1 ( t 1 s ) α 1 Γ ( α ) f ( s , x ( ϕ 1 ( s ) ) ) d s | | 0 t 1 ( t 2 s ) α 1 Γ ( α ) f ( s , x ( ϕ 1 ( s ) ) ) d s + t 1 t 2 ( t 2 s ) α 1 Γ ( α ) f ( s , x ( ϕ 1 ( s ) ) ) d s 0 t 1 ( t 1 s ) α 1 Γ ( α ) f ( s , x ( ϕ 1 ( s ) ) ) d s | | 0 t 1 ( t 1 s ) α 1 Γ ( α ) f ( s , x ( ϕ 1 ( s ) ) ) d s + t 1 t 2 ( t 2 s ) α 1 Γ ( α ) f ( s , x ( ϕ 1 ( s ) ) ) d s 0 t 1 ( t 1 s ) α 1 Γ ( α ) f ( s , x ( ϕ 1 ( s ) ) ) d s | = | t 1 t 2 ( t 2 s ) α 1 Γ ( α ) f ( s , x ( ϕ 1 ( s ) ) ) d s | t 1 t 2 ( t 2 s ) α 1 Γ ( α ) | f ( s , x ( ϕ 1 ( s ) ) ) | d s ( M + b r ) t 1 t 2 ( t 2 s ) α 1 Γ ( α ) d s .
Consequently,
| F x ( t 2 ) F x ( t 1 ) | θ 1 ( δ ) + a θ 2 ( δ ) M T α Γ ( α + 1 ) + b | | x | | T α Γ ( 1 + α ) + ( M + b r ) t 1 t 2 ( t 2 s ) α 1 Γ ( α ) d s .
Then, F : Ω ¯ E is a completely continuous. Then, by Theorem 1.1 there exists a solution x C ( I , R ) of the Problem (3)–(4).
Thus F : Ω ¯ E is compact and has a fixed point x E . This proves the existence of solution x E of the Problem (3)–(4). □
Remark 1.
Several fixed-point problems involving product of operators have been investigated in many literature and monographs (for example, [21,23,27,31]). We can use a fixed-point theorem for the product of operators to prove Theorem 2.

2.2. Uniqueness of the Solution

Consider the following assumption:
( i i * ) there exist two positive constants b i , i = 1 , 2 such that
| f ( t , x ) f ( t , y ) | b 1 | x y | , | h ( t , x ) h ( t , y ) | b 2 | x y | ,
m 1 ( t ) = | f ( t , 0 ) | , m 2 ( t ) = | h ( t , 0 ) |
with M = m a x { sup m 1 ( t ) , sup m 2 ( t ) , t I } and b = m a x { b 1 , b 2 } .
Theorem 3.
Let the assumptions of Theorem 2 be satisfied and replace condition (ii) by ( i i * ) .
If
A 2 a b r + a M + b k 1 < 1 ,
then the Problem (3)–(4) has a unique solution x C ( I , R ) .
Proof. 
Let x 1 , x 2 be solutions of Equation (5), then
| x 1 ( t ) x 2 ( t ) | = | x 0 + 0 τ ( τ s ) β 1 Γ ( β ) h ( s , x 1 ( ϕ 3 ( s ) ) ) d s + g ( t , x 1 ( ϕ 2 ( t ) ) ) I α f ( t , x 1 ( ϕ 1 ( t ) ) ) x 0 0 τ ( τ s ) β 1 Γ ( β ) h ( s , x 2 ( ϕ 3 ( s ) ) ) d s g ( t , x 2 ( ϕ 2 ( t ) ) ) I α f ( t , x 2 ( ϕ 1 ( t ) ) ) | 0 τ ( τ s ) β 1 Γ ( β ) | h ( s , x 1 ( ϕ 3 ( s ) ) ) h ( s , x 2 ( ϕ 3 ( s ) ) ) | d s + | g ( t , x 1 ( ϕ 2 ( t ) ) ) I α f ( t , x 1 ( ϕ 1 ( t ) ) ) g ( t , x 2 ( ϕ 2 ( t ) ) ) I α f ( t , x 2 ( ϕ 1 ( t ) ) ) | b 0 τ ( τ s ) β 1 Γ ( β ) | x 1 ( ϕ 3 ( s ) ) x 2 ( ϕ 3 ( s ) ) | d s + | g ( t , x 1 ( ϕ 2 ( t ) ) ) I α f ( t , x 1 ( ϕ 1 ( t ) ) ) g ( t , x 2 ( ϕ 2 ( t ) ) ) I α f ( t , x 2 ( ϕ 1 ( t ) ) ) | ,
but
| g ( t , x 1 ( ϕ 2 ( t ) ) ) I α f ( t , x 1 ( ϕ 1 ( t ) ) ) g ( t , x 2 ( ϕ 2 ( t ) ) ) I α f ( t , x 2 ( ϕ 1 ( t ) ) ) |
= | g ( t , x 1 ( ϕ 2 ( t ) ) ) I α f ( t , x 1 ( ϕ 1 ( t ) ) ) g ( t , x 2 ( ϕ 2 ( t ) ) ) I α f ( t , x 2 ( ϕ 1 ( t ) ) ) + g ( t , x 2 ( ϕ 2 ( t ) ) ) I α f ( t , x 1 ( ϕ 1 ( t ) ) ) g ( t , x 2 ( ϕ 2 ( t ) ) ) I α f ( t , x 1 ( ϕ 1 ( t ) ) ) | | g ( t , x 1 ( ϕ 2 ( t ) ) ) I α f ( t , x 1 ( ϕ 1 ( t ) ) ) g ( t , x 2 ( ϕ 2 ( t ) ) ) I α f ( t , x 1 ( ϕ 1 ( t ) ) ) | + | g ( t , x 2 ( ϕ 2 ( t ) ) ) I α f ( t , x 1 ( ϕ 1 ( t ) ) ) g ( t , x 2 ( ϕ 2 ( t ) ) ) I α f ( t , x 2 ( ϕ 1 ( t ) ) ) | | g ( t , x 1 ( ϕ 2 ( t ) ) ) g ( t , x 2 ( ϕ 2 ( t ) ) ) | | I α f ( t , x 1 ( ϕ 1 ( t ) ) ) | + | g ( t , x 2 ( ϕ 2 ( t ) ) ) | | I α f ( t , x 1 ( ϕ 1 ( t ) ) ) I α f ( t , x 2 ( ϕ 1 ( t ) ) ) | a | x 1 ( ϕ 2 ( t ) ) x 2 ( ϕ 2 ( t ) ) | b | x 1 ( ϕ 1 ( t ) ) | + | f ( t , 0 ) | T α Γ ( 1 + α ) + a | x 2 ( ϕ 2 ( t ) ) | + | g ( t , 0 ) | b | x 1 ( ϕ 1 ( t ) ) x 2 ( ϕ 1 ( t ) ) | T α Γ ( 1 + α ) a | | x 1 x 2 | | b r + M T α Γ ( 1 + α ) + a r + k 1 b | | x 1 x 2 | | T α Γ ( 1 + α ) .
Then,
| | x 1 x 2 | | b T β Γ ( 1 + β ) | | x 1 x 2 | | + a T α Γ ( 1 + α ) | | x 1 x 2 | | b r + M + a r + k 1 b T α Γ ( 1 + α ) | | x 1 x 2 | | ,
and then
| | x 1 x 2 | | A 2 a b r + a M + b k 1 | | x 1 x 2 | | ,
1 A 2 a b r + a M + b k 1 | | x 1 x 2 | | 0 .
This implies that x 1 ( t ) = x 2 ( t ) . Hence, the solution of Problem (3)–(4) is unique. □

2.3. Continuous Dependency

In this subsection, we shall investigate the continuous dependence of the solution of the Problem (3)–(4) on x 0 and on the two functions f and h .
Definition 2.
The solution of the Problem (3)–(4) depends continuously on x 0 and on the two functions f and h , if ϵ > 0 ,   δ 1 , δ 2 , δ 3 > 0 be given such that
| x 0 x ˜ 0 | δ 1 , | f ( t , x ) f ˜ ( t , x ˜ ) | δ 2 and | h ( t , x ) h ˜ ( t , x ˜ ) | δ 3 . Then | x x ˜ | ϵ , where x ˜ be a solution of the quadratic integral equation
x ˜ ( t ) = x ˜ 0 + 0 τ ( τ s ) β 1 Γ ( β ) h ˜ ( s , x ˜ ( ϕ 3 ( s ) ) ) d s + g ( t , x ˜ ( ϕ 2 ( t ) ) ) I α f ˜ ( t , x ˜ ( ϕ 1 ( t ) ) ) .
Theorem 4.
Let the assumptions of Theorem 3 be satisfied, and then the solution of the integral Equation (5) (consequently the the Problem (3)–(4)) depends continuously on x 0 , f , and on h .
Proof. 
Let δ 1 , δ 2 , δ 3 > 0 be given such that | x 0 x ˜ 0 | δ 1 , | f ( t , x ) f ˜ ( t , x ˜ ) | δ 2 and | h ( t , x ) h ˜ ( t , x ˜ ) | δ 3 . Let x ˜ be a solution of the following integral Equation (9) Then,
| x ( t ) x ˜ ( t ) | = | x 0 + 0 τ ( τ s ) β 1 Γ ( β ) h ( s , x ( ϕ 3 ( s ) ) ) d s + g ( t , x ( ϕ 2 ( t ) ) ) I α f ( t , x ( ϕ 1 ( t ) ) ) x ˜ 0 0 τ ( τ s ) β 1 Γ ( β ) h ˜ ( s , x ˜ ( ϕ 3 ( s ) ) ) d s g ( t , x ˜ ( ϕ 2 ( t ) ) ) I α f ˜ ( t , x ˜ ( ϕ 1 ( t ) ) ) | | x 0 x ˜ 0 | + 0 τ ( τ s ) β 1 Γ ( β ) | h ( s , x ( ϕ 3 ( s ) ) ) h ˜ ( s , x ˜ ( ϕ 3 ( s ) ) ) | d s + | g ( t , x ( ϕ 2 ( t ) ) ) I α f ( t , x ( ϕ 1 ( t ) ) ) g ( t , x ˜ ( ϕ 2 ( t ) ) ) I α f ˜ ( t , x ˜ ( ϕ 1 ( t ) ) ) | | x 0 x ˜ 0 | + 0 τ ( τ s ) β 1 Γ ( β ) δ 3 d s + | g ( t , x ( ϕ 2 ( t ) ) ) I α f ( t , x ( ϕ 1 ( t ) ) ) g ( t , x ˜ ( ϕ 2 ( t ) ) ) I α f ( t , x ˜ ( ϕ 1 ( t ) ) ) | + | g ( t , x ˜ ( ϕ 2 ( t ) ) ) I α f ( t , x ˜ ( ϕ 1 ( t ) ) ) g ( t , x ˜ ( ϕ 2 ( t ) ) ) I α f ˜ ( t , x ˜ ( ϕ 1 ( t ) ) ) | δ 1 + δ 3 τ β Γ ( β + 1 ) + | g ( t , x ( ϕ 2 ( t ) ) ) I α f ( t , x ( ϕ 1 ( t ) ) ) g ( t , x ( ϕ 2 ( t ) ) ) I α f ( t , x ˜ ( ϕ 1 ( t ) ) ) | + | g ( t , x ( ϕ 2 ( t ) ) ) I α f ( t , x ˜ ( ϕ 1 ( t ) ) ) g ( t , x ˜ ( ϕ 2 ( t ) ) ) I α f ( t , x ˜ ( ϕ 1 ( t ) ) ) | + | g ( t , x ˜ ( ϕ 2 ( t ) ) ) | I α | f ( t , x ˜ ( ϕ 1 ( t ) ) ) f ˜ ( t , x ˜ ( ϕ 1 ( t ) ) ) | δ 1 + δ 3 τ β Γ ( β + 1 ) + | g ( t , x ( ϕ 2 ( t ) ) ) | I α | f ( t , x ( ϕ 1 ( t ) ) ) f ( t , x ˜ ( ϕ 1 ( t ) ) ) | + | g ( t , x ( ϕ 2 ( t ) ) ) g ( t , x ˜ ( ϕ 2 ( t ) ) ) | I α f ( t , x ˜ ( ϕ 1 ( t ) ) ) | + | g ( t , x ˜ ( ϕ 2 ( t ) ) ) | I α | f ( t , x ( ϕ 1 ( t ) ) ) f ˜ ( t , x ˜ ( ϕ 1 ( t ) ) ) | δ 1 + δ 3 T β Γ ( β + 1 ) + k 1 + a | x ( ϕ 2 ( t ) ) | b T α Γ ( α + 1 ) | | x x ˜ | | + a | | x x ˜ | | ( M + b r ) T α Γ ( α + 1 ) + ( k 1 + a r ) δ 2 T α Γ ( α + 1 ) δ 1 + δ 3 A + ( k 1 + a r ) A δ 2 + k 1 + a r b A + a A M + b r | | x x ˜ | | δ 1 + δ 3 A + ( k 1 + a r ) A δ 2 + b k 1 A + 2 b a r A + a A M | | x x ˜ | | δ 1 + δ 3 A + ( k 1 + a r ) A δ 2 + A b k 1 + 2 b a r + a M | | x x ˜ | | .
Then,
| | x x ˜ | | δ 1 + δ 3 A + ( k 1 + a r ) A δ 2 1 A b k 1 + 2 b a r + a M 1 ϵ .

3. Multi-Valued Problem

Now, consider the assumption:
( H 1 )
The set F ( t , x ) , H ( t , x ) are nonempty, closed and convex for all ( t , x ) I × R .
(I) F ( t , x ) , H ( t , x ) are measurable in t I for every x R .
(II) F ( t , x ) , H ( t , x ) are upper semicontinuous in x for every t I .
(III) There exist two bounded measurable functions m i L 1 ( I ) and two positive constants b i , i = 1 , 2 such that
| | F ( t , x ) | | = s u p { | f | : f F ( t , x ) } | m 1 ( t ) | + b 1 | x | , t I , x R ; | | H ( t , x ) | | = s u p { | h | : h F ( t , x ) } | m 2 ( t ) | + b 2 | x | , t I , x R .
Remark 2.
From the assumptions ( H 1 ) (which guarantee the existence of Carathéodory selections [32]) we can deduce that (see [11,19,33,34,35]) there exists f F ( t , x ) , h H ( t , x ) , such that f , h : I × R R are measurable for almost all t for every x R and continuous in x for every t I and there exist m 1 L 1 ( I ) , a bounded measurable function m 2 and b i 0 , i = 1 , 2 , such that
| f ( t , x ) | m 1 ( t ) + b 1 | x | , | h ( t , x ) | m 2 ( t ) + b 2 | x | , t I , x R ,
which satisfies the fractional-order hybrid differential Problem (3)–(4) which is equivalent to (5). Therefore, any solution of the nonlocal problem of the hybrid differential Equation (3) with any of the nonlocal boundary condition (4) is a solution of the nonlocal problem of the hybrid nonlinear differential inclusion with any one of the nonlocal conditions (1)–(2).

Existence Theorem

Now, from the main results obtained in Section 2, we deduce the following results for the fractional-order hybrid delay differential (1) inclusion with the multi-valued fractional-order nonlocal integral condition (2).
Theorem 5.
Let the assumptions ( H 1 ) and (iii)–(v) be satisfied. Then the Problem (1)–(2) has a solution x C ( I , R ) .
To present the continuous dependency of the unique solution on x 0 and on the two sets of selections S F and S H , replace the assumption ( H 1 ) instead of ( H 1 * ) .
  • The sets F and H are nonempty, closed, and convex for all ( t , x ) I × R and satisfy
    H ( F ( t , x 1 ) , F ( t , x 2 ) ) b 1 | x 1 x 2 | , x 1 , x 2 R ,
    H ( H ( t , x 1 ) , H ( t , x 2 ) ) b 2 | x 1 x 2 | , x 1 , x 2 R ,
    where H ( D , B ) is the Hausdorff metric [36] between the two subsets D , B I × R .
Remark 3.
From assumption ( H 1 * ) we can deduce that (see [34]) there exist functions f F and h H such that
f , h : I × R R are measurable for almost all t I for every x R and satisfy the Lipschitz condition with a positive constants b 1 and b 2 , such that
| f ( t , u 1 ) f ( t , u 2 ) | b 1 | u 1 u 2 | , u 1 , u 2 R a n d t I ,
| h ( t , u 1 ) h ( t , u 2 ) | b 2 | u 1 u 2 | , u 1 , u 2 R a n d t I .
Theorem 6.
Let the assumptions of Theorem 5 be satisfied and replace condition ( H 1 ) by ( H 1 * ) . If A b k 1 + 2 b a r + a M < 1 , then f F and h H the Problem (1)–(2) has a unique solution x C ( I , R ) .
Remark 4.
From Theorem 4, we can deduce some results for the continuous dependence of the solution of the Problem (1)–(2) on x 0 and on the sets of selections S F and S H .
  • The solutions of Problems (1)–(2) depend continuously on x 0 , if f F and h H , and the solution of Problems (3)–(4) depends continuously on x 0 .
  • The solutions of the Problem (1)–(2) depend continuously on the two sets S F and S H , if f F and h H , and the solution of Problems (3)–(4) depends continuously on the functions f and h .
Theorem 7.
Let the assumptions of Theorem 4 be satisfied, then the solutions of the Problem (1)–(2) depends continuously on the two sets of selections S F and S H (on the two functions f and h) and on data x 0 .
Remark 5.
The proofs of Theorems 5–7 can be deduced directly by the Kuratowski Selection Theorem (see [19,37]) and as done in [11,34].

4. Particular Cases and Example

  • Phanograph fractional differential inclusion with multi-valued condition Letting ϕ i ( t ) = η i t , t I and η i ( 0 , 1 ) , i = 1 , 2 , 3 , then the Problem (1)–(2) can be reduced to the fractional-order pantograph differential inclusion
    R D α x ( t ) x ( 0 ) g ( t , x ( η 2 t ) ) F ( t , x ( η 1 t ) ) , α ( 0 , 1 ) , t I ,
    subject to the multi-valued fractional-order nonlocal integral condition
    x ( 0 ) x 0 0 τ ( τ s ) β 1 Γ ( β ) H ( s , x ( η 3 s ) ) d s , β α , τ I .
  • Retarded fractional differential inclusion Letting ϕ i ( t ) = t r i , t r i > 0 , i = 1 , 2 and ϕ i ( t ) = 0 , t < r i , i = 1 , 2 . Then, in case h = 0 , the Problem (1)–(2) has the form
    R D α x ( t ) x ( 0 ) g ( t , x ( t r 2 ) ) F ( t , x ( t r 1 ) ) , α ( 0 , 1 ) , t r i > 0 , i = 1 , 2 ,
    subject to
    x ( 0 ) = x 0 , t r i , i = 1 , 2 ,
    which may be called a fractional-order retarded differential inclusion.
  • A conjugate order hybrid differential inclusion with multi-valued condition Letting β = 1 α , α 1 2 , then the Problem (1)–(2) yields the following particular case,
    R D α x ( t ) x ( 0 ) g ( t , x ( ϕ 2 ( t ) ) ) F ( t , x ( ϕ 1 ( t ) ) ) , α 1 2 , t I ,
    subject to the multi-valued fractional-order nonlocal integral condition
    x ( 0 ) x 0 0 τ ( τ s ) α Γ ( 1 α ) H ( s , x ( ϕ 3 ( s ) ) ) d s , α 1 2 , τ I .
    In particular, we consider the case when α = 1 2 :
    R D 1 2 x ( t ) x ( 0 ) g ( t , x ( ϕ 2 ( t ) ) ) F ( t , x ( ϕ 1 ( t ) ) ) , t I ,
    subject to the multi-valued fractional-order nonlocal integral condition
    x ( 0 ) x 0 0 τ ( τ s ) 1 2 Γ ( 1 2 ) H ( s , x ( ϕ 3 ( s ) ) ) d s , τ I .
  • Fractional differential inclusion with multi-valued condition Letting g ( t , x ) = 1 in (1), then we have the fractional-order differential inclusion
    R D α x ( t ) x ( 0 ) F ( t , x ( ϕ 1 ( t ) ) ) , t I ,
    subject to the multi-valued fractional-order nonlocal integral condition
    x ( 0 ) x 0 0 τ ( τ s ) β 1 Γ ( β ) H ( s , x ( ϕ 3 ( s ) ) ) d s , β α , τ I .
Example:
Consider the the fractional-order hybrid differential inclusion
R D 1 2 x ( t ) s i n ( t ) x ( t ) 1 + t 0 , 1 8 c o s ( t 2 + 2 ) + 1 1 + t x ( t ) , t [ 0 , 1 ] ,
subject to the multi-valued fractional-order nonlocal integral condition
x ( 0 ) 0 τ ( τ s ) 1 3 1 Γ ( 1 3 ) 0 , 1 4 s + 1 3 + s x ( s ) d s , τ [ 0 , 1 ] ,
where S F = { f : f ( t , x ) = 1 8 c o s ( t 2 + 2 ) + 1 1 + t x ( t ) } and S H = { h : h ( t , x ) = 1 4 t + 1 3 + t x ( t ) } .
Then we have x 0 = 0 , b = 1 4 , k 1 = 0 , M = 1 4 , a = 1 2 and A = 2 π .
Therefore, we can calculate r = 0.5678685902417353 ( 0 , 1 ) from the quadratic algebraic equation in assumption (v).
Moreover, we can verify the sufficient condition for the uniqueness of the solution of Problems (10)–(11),
A 2 a b r + a M + b k 1 = 0.30124 < 1 .

5. Conclusions

Many recent studies concerning the application of fractional differential equations to structural problems, including for example the modeling of spatial inclusions which lead to a spatial decay in delayed spatial long-range correlations, otherwise called nonlocal effects [7,38,39,40,41,42,43] have been conducted. Inspired by those results and other results of fractional delay differential inclusions (see references therein), we have acquainted a qualitative investigation for a hybrid differential inclusion with multivalued nonlocal integral condition of fractional order. Our analysis is located in the Banach space C ( I , R ) and based on the Leray–Schauder nonlinear alternative fixed point theorem. We have discussed some characteristics for the solution of that inclusion problem, such that the uniqueness of the solution and the continuous dependency on the initial data and the two sets of selections S F and S H of the set valued function F and H , respectively. In particular, taking β = 1 α , α 1 2 , we have a hybrid differential inclusion with multivalued nonlocal integral condition of conjugate order. Moreover, we have considered other special cases.

Author Contributions

These authors contributed equally to this work. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Not applicable.

Acknowledgments

The authors are thankful to the referees for remarks and suggestions for the improvement of this paper.

Conflicts of Interest

The authors declare no conflict of interest.

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El-Sayed, A.M.A.; El-Salam, S.A.A.; Hashem, H.H.G. Development on a Fractional Hybrid Differential Inclusion with a Nonlinear Nonlocal Fractional-Order Integral Inclusion. Mathematics 2022, 10, 4068. https://doi.org/10.3390/math10214068

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El-Sayed AMA, El-Salam SAA, Hashem HHG. Development on a Fractional Hybrid Differential Inclusion with a Nonlinear Nonlocal Fractional-Order Integral Inclusion. Mathematics. 2022; 10(21):4068. https://doi.org/10.3390/math10214068

Chicago/Turabian Style

El-Sayed, Ahmed M. A., Sheren A. Abd El-Salam, and Hind H. G. Hashem. 2022. "Development on a Fractional Hybrid Differential Inclusion with a Nonlinear Nonlocal Fractional-Order Integral Inclusion" Mathematics 10, no. 21: 4068. https://doi.org/10.3390/math10214068

APA Style

El-Sayed, A. M. A., El-Salam, S. A. A., & Hashem, H. H. G. (2022). Development on a Fractional Hybrid Differential Inclusion with a Nonlinear Nonlocal Fractional-Order Integral Inclusion. Mathematics, 10(21), 4068. https://doi.org/10.3390/math10214068

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