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Article

Solvability of a State–Dependence Functional Integro-Differential Inclusion with Delay Nonlocal Condition

by
Taher S. Hassan
1,2,3,
Reda Gamal Ahmed
4,
Ahmed M. A. El-Sayed
5,
Rami Ahmad El-Nabulsi
6,7,8,*,
Osama Moaaz
2,9 and
Mouataz Billah Mesmouli
1,2
1
Department of Mathematics, College of Science, University of Hail, Hail 2440, Saudi Arabia
2
Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt
3
Section of Mathematics, International Telematic University Uninettuno, Corso Vittorio Emanuele II, 39, 00186 Roma, Italy
4
Faculty of Science, Al-Azhar University, Cairo 11884, Egypt
5
Faculty of Science, Alexandria University, Alexandria 21522, Egypt
6
Research Center for Quantum Technology, Faculty of Science, Chiang Mai University, Chiang Mai 50200, Thailand
7
Department of Physics and Materials Science, Faculty of Science, Chiang Mai University, Chiang Mai 50200, Thailand
8
Mathematics and Physics Divisions, Athens Institute for Education and Research, 8 Valaoritou Street, Kolonaki, 10671 Athens, Greece
9
Department of Mathematics, College of Science, Qassim University, P.O. Box 6644, Buraydah 51452, Saudi Arabia
*
Author to whom correspondence should be addressed.
Mathematics 2022, 10(14), 2420; https://doi.org/10.3390/math10142420
Submission received: 31 March 2022 / Revised: 21 June 2022 / Accepted: 22 June 2022 / Published: 11 July 2022
(This article belongs to the Special Issue Orthogonal Polynomials and Special Functions)

Abstract

:
There is great focus on phenomena that depend on their past history or their past state. The mathematical models of these phenomena can be described by differential equations of a self-referred type. This paper is devoted to studying the solvability of a state-dependent integro-differential inclusion. The existence and uniqueness of solutions to a state-dependent functional integro-differential inclusion with delay nonlocal condition is studied. We, moreover, conclude the existence of solutions to the problem with the integral condition and the infinite-point boundary one. Some properties of the solutions are given. Finally, two examples illustrating the main result are presented.

1. Introduction

A functional equation is an equation involving an unknown function at more than an argument value. In functional equations, argument deviations are the variations between the argument values of an unknown function and an independent variable t. The functional differential equation, or differential equations with diverging arguments, is created by combining the concepts of differential and functional equations. Functional differential equations are used to describe many phenomena in different sciences, see [1]. Nonlocal problems in mathematical physics are problems in which, unlike traditional boundary value conditions, the desired function’s values at distinct places of the boundary (and/or its values at the boundary and outside it) are related. In fact, it is reasonable to treat the theory of functional differential equations and the theory of nonlocal problems as one indivisible theory.
For various reasons, many researchers have been interested in researching the nonlocal problem of functional differential equations with infinite point conditions; see for example [2,3,4,5,6,7,8,9,10]. The deviation of the argument in the current literature’s differential and integral equations with diverging arguments usually concerns only the time itself, see [11,12]. In theory and practice, however, another case exists in which the deviating arguments are dependent on both the state variable y and the time t. Several studies devoted to such differential equations have lately been published, for example [13,14,15,16,17]. The first papers studying this class of functional equations with self-reference, Eder [18], studied the existence of the unique solution for the differential equation
z ( t ) = z ( z ( t ) ) , z ( 0 ) = z 0 , t B R .
Wang [19] studied the strong solution and maximal strong solution for the equation
z ( t ) = h ( z ( z ( t ) ) ) ,
where h : R R is continuous and monotone, and h ( 0 ) = 0 .
Feckan [20] studied the existence solution
z ( t ) = h ( z ( z ( t ) ) ) , z ( 0 ) = 0 w h e r e h C 1 ( R , R ) .
Buica [13] studied the existence and continuous dependence of solutions, on z 0 of the functional differential equation
z ( t ) = h ( t , z ( z ( t ) ) ) , z ( t 0 ) = z 0 , t [ a , b ] ,
where t 0 , x 0 [ a , b ] and h C ( [ a , b ] , [ a , b ] ) .
Stanek [21] studied global properties of decreasing solution of the equation
z ( t ) = z ( z ( t ) ) + z ( t ) .
Stanek [16] studied global properties of solution of the functional differential equation
z ( t ) z ( t ) = k z ( z ( t ) ) 0 < | k | < 1 .
In this study, the initial value problem of the functional differential inclusion with self-dependence on a nonlinear delay integral operator
d y d t Ψ t , y 0 μ ( t ) ψ ( u , y ( u ) ) d u a . e . t ( 0 , B ] ,
with the nonlocal condition
y ( 0 ) + = 1 κ q y ( μ ( τ ) ) = y 0 , q > 0 , τ ( 0 , B ) ,
was investigated. We study the existence of the absolutely continuous solution y A C [ 0 , B ] and demonstrate the continuous dependence on y 0 and ψ . Moreover, as applications, we study the nonlocal problem of Equation (1) with the integral condition
y ( 0 ) + 0 B y ( μ ( u ) ) d h ( u ) = y 0 , h : [ 0 , B ] R ( increasing function )
and with the infinite-point boundary condition
y ( 0 ) + = 1 q y ( μ ( τ ) ) = y 0 ,
if = 1 q is convergent.
The paper is organized as follows: In Section 2, the equivalence of the functional differential inclusion with state-dependence on the nonlinear delay integral operator (1) with the nonlocal condition (2) is given. In Section 3, we study the existence of absolutely continuous solutions to problem (1) and (2), and conclude the existence of solutions to problem (1) with the integral condition (3) and the infinite-point boundary condition (4). In Section 4, we establish the existence of exactly one solution for (1) and (2). In Section 5, the continuous dependence of the solution is studied. Finally, in Section 6, two examples are given to corroborate the main existence result and a numerical example is given to demonstrate the difference between the exact solution and numerical solution.

2. Auxiliary Results

Consider the following assumptions:
(a)
Ψ ( t , y ) is nonempty, convex and closed ( t , y ) [ 0 , B ] × R .
(b)
Ψ ( t , y ) is measurable in t [ 0 , B ] for every y R .
(c)
Ψ ( t , y ) is upper semi continuous in y for every t [ 0 , B ] .
(d)
There exist a bounded measurable function c : [ 0 , B ] R and a positive constant b, such that
Ψ ( t , y ) = sup | r | : r Ψ ( t , y ) | c ( t ) | + b | y | , | c ( t ) | M .
From the assumptions (a)–(d), we can deduce that there exists a r Ψ ( t , y ) , such that the following is satisfied:
(1)
r : [ 0 , B ] × R R satisfies Carathéodory condition:
-
For each t [ 0 , B ] , r ( t , . ) is continuous;
-
For each ϱ R , r ( . , ϱ ) is measurable;
-
There exist a bounded measurable function c ( t ) and a positive constant b > 0 , such that
| r ( t , ϱ ) | | c ( t ) | + b | ϱ | , | c ( t ) | M .
and the functional r satisfies the integro-differential equation
d y d t = r t , y 0 μ ( t ) ψ ( u , y ( u ) ) d u a . e . t ( 0 , B ] ,
(2)
ψ : [ 0 , B ] × R R + satisfies Carathéodory condition:
-
For each t [ 0 , B ] , ψ ( t , . ) is continuous;
-
For each y R , ψ ( . , y ) is measurable;
-
| ψ ( t , y ) | 1 .
(3)
μ : [ 0 , B ] [ 0 , B ] is continuous and nondecreasing, μ ( t ) t , t [ 0 , B ] .
(4)
1 + 2 = 1 κ q b B < 1 .
Remark 1.
From ( a ) and ( 1 ) , we can deduce that every solution of (1) is also a solution of (5).
The equivalence of (5)–(2) and the integral equation is given in the following lemma.
Lemma 1.
If 1 4 hold, then the nonlocal problem (1) and (2) and the equation
y ( t ) = 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ϱ , y ( ϱ ) d ϱ d u + 0 t r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u for t [ 0 , B ]
are equivalent.
Proof. 
Assume that the solution of the nonlocal problem (1) and (2) exists. Integrating (5) from 0 to t, we get
y ( t ) = y ( 0 ) + 0 t r u , y 0 μ ( u ) ψ ϱ , y ( ϱ ) d ϱ d u .
Using the condition (2), we obtain
= 1 κ q y ( μ ( τ ) ) = y ( 0 ) = 1 κ q + = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u .
By = 1 κ q y ( μ ( τ ) ) = y 0 y ( 0 ) , we have
y 0 y ( 0 ) = y ( 0 ) = 1 κ q + = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u ,
which implies
y ( 0 ) = 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u .
Using (7) and (8), we obtain
y ( t ) = 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + 0 t r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u .
Now, suppose that y C [ 0 , B ] is a solution of (6). Now differentiate (6), we arrive at
d y d t = d d t 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + 0 t r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u = r t , y 0 μ ( t ) ψ ( ϱ , y ( ϱ ) ) d ϱ .
From (6), we find
y ( τ ) = 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + 0 τ r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u ,
y ( 0 ) = 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u ,
and
= 1 κ q y ( μ ( τ ) ) = 1 1 + = 1 κ q = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u .
From (9) and (10), we have
y ( 0 ) + = 1 κ q y ( μ ( τ ) ) = 1 1 + = 1 κ q 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u .
Then
y ( 0 ) + = 1 κ q y ( μ ( τ ) ) = y 0 .
This completes the proof. □

3. Existence of Solution

In the following theorem, using Schauder’s fixed point theorem, we establish the existence of at least one solution of (1) and (2).
Theorem 1.
If 1 4 hold, then the nonlocal problem (1) and (2) has at least one solution y A C [ 0 , B ] .
Proof. 
First, we define the operator A associated with Equation (6) and set P L by
A y ( t ) = 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + 0 t r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u .
Now P L = { y C [ 0 , B ] : | y ( t ) y ( u ) | L | t u | , t , u [ 0 , B ] } , where
L = b | y 0 | + ( 1 + = 1 κ q ) M ( 1 + = 1 κ q ) ( 1 + 2 = 1 κ q ) b B .
Then we have for y P ,
A y ( t ) = 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + 0 t r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u 1 1 + = 1 κ q | y 0 | + = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + 0 t r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u ,
and so
A y ( t ) 1 1 + = 1 κ q | y 0 | + = 1 κ q 0 μ ( τ ) | c ( u ) | + b y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + 0 t | c ( u ) | + b y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u 1 1 + = 1 κ q | y 0 | + = 1 κ q 0 μ ( τ ) ( | c ( u ) | + b y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y ( 0 ) + b | y ( 0 ) | ) d u + 0 t | c ( u ) | + b y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y ( 0 ) + b | y ( 0 ) | d u .
Thus,
1 1 + = 1 κ q | y 0 | + = 1 κ q 0 μ ( τ ) | c ( u ) | + b L 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ + b | y ( 0 ) | d u + 0 t | c ( u ) | + b L 0 μ ( u ) | ψ ( ϱ , y ( ϱ ) ) | d ϱ + b | y ( 0 ) | d u 1 1 + = 1 κ q | y 0 | + = 1 κ q 1 + = 1 κ q B M + b L B 2 + b B | y ( 0 ) | + B M + b L B 2 + b B | y ( 0 ) | = 1 1 + = 1 κ q | y 0 | + = 1 κ q 1 + = 1 κ q + 1 B M + b L B 2 + b B | y ( 0 ) | .
But
| y ( 0 ) | = 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u 1 1 + = 1 κ q y 0 + = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u
and so
| y ( 0 ) | 1 1 + = 1 κ q | y 0 | + = 1 κ q 0 μ ( τ ) | c ( u ) | + b y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y ( 0 ) + b | y ( 0 ) | d u 1 1 + = 1 κ q | y 0 | + = 1 κ q 0 μ ( τ ) | c ( u ) | + b L 0 μ ( u ) | ψ ( ϱ , y ( ϱ ) ) | d ϱ + b | y ( 0 ) | d u 1 1 + = 1 κ q | y 0 | + = 1 κ q B M + b L B 2 + b B | y ( 0 ) | .
Hence
| y ( 0 ) | | y 0 | + ( B M + b L B 2 ) = 1 κ q 1 + = 1 κ q b B = 1 κ q .
From (11) and (12), we obtain
| A y ( t ) | 1 1 + = 1 κ q | y 0 | + = 1 κ q 1 + = 1 κ q + 1 × ( B M + b L B 2 + b B | y ( 0 ) | ) 1 1 + = 1 κ q | y 0 | + = 1 κ q 1 + = 1 κ q + 1 × B M + b L B 2 + b B | y 0 | + ( B M + b L B 2 ) = 1 κ q 1 + = 1 κ q b B = 1 κ q = | y 0 | + ( B M + b L B 2 ) = 1 κ q 1 + = 1 κ q b B = 1 κ q + L B .
Now, let t 1 , t 2 ( 0 , B ] such that | t 2 t 1 | < δ , then
A y ( t 2 ) A y ( t 1 ) t 1 t 2 r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u t 1 t 2 | c ( u ) | + b y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y ( 0 ) + b | y ( 0 ) | d u t 1 t 2 | c ( u ) | + b L 0 μ ( u ) | ψ ( ϱ , y ( ϱ ) ) | d ϱ + b | y ( 0 ) | d u ( t 2 t 1 ) M + ( t 2 t 1 ) b L B + ( t 2 t 1 ) b | y ( 0 ) | = ( t 2 t 1 ) ( M + b L B + b | y ( 0 ) | ) .
From (12) and (13), we obtain
| A y ( t 2 ) A y ( t 1 ) | ( t 2 t 1 ) M + b L B + b | y 0 | + ( B M + b L B 2 ) = 1 κ q ( 1 + = 1 κ q ) b B = 1 κ q ( t 2 t 1 ) b | y 0 | + 1 + = 1 κ q ( M + b L B ) 1 + = 1 κ q b B = 1 κ q = ( t 2 t 1 ) L .
This proves that A : P L P L ; the class of functions { A y } is uniformly bounded and equi-continuous in P L .
Let y n P L , y n y ( n ) , then from assumptions 1 and 2 , we obtain r ( t , y n ( t ) , y n ( t ) ) r ( t , y ( t ) , y ( t ) ) and ψ ( t , y n ( t ) ) ψ ( t , y ( t ) ) as n . Also
lim n A y n ( t ) = lim n 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y n 0 μ ( u ) ψ ( ϱ , y n ( ϱ ) ) d ϱ d u + 0 t r ( u , y n ( 0 μ ( u ) ψ ( ϱ , y n ( ϱ ) ) d ϱ ) ) d u .
Now
y n 0 μ ( u ) ψ ( ϱ , y n ( ϱ ) ) d ϱ y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y n 0 μ ( u ) ψ ( ϱ , y n ( ϱ ) ) d ϱ y n 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ + y n 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ L 0 μ ( u ) ψ ( ϱ , y n ( ϱ ) ) ψ ( ϱ , y ( ϱ ) ) d ϱ + ϵ 2 ϵ 2 + ϵ 2 = ϵ .
Using (14) and (15) and Lebesgue’s dominated convergence theorem [22], we obtain
lim n A y n ( t ) = 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) lim n r u , y n 0 μ ( u ) ψ ( ϱ , y n ( ϱ ) ) d ϱ d u + 0 t lim n r u , y n 0 μ ( u ) ψ ( ϱ , y n ( ϱ ) ) d ϱ d u = A y ( t ) .
Then A y n A y as n , the operator A is continuous.
Hence by Schauder theorem [23] there exists at least solution y C [ 0 , B ] for the integral Equation (6), therefore there exists at least solution y A C [ 0 , B ] for the nonlocal problem (1) and (2). □
  • For the nonlocal integral condition, we present the following theorem.
Theorem 2.
If 1 4 hold, then the nonlocal problem of (1), (3) has at least one solution.
Proof. 
Let y A C [ 0 , B ] be the solution of the nonlocal problem (1) with (3). Let q = h ( t ) h ( t 1 ) , h is an increasing function, τ ( t 1 , t ) , 0 = t 0 < t 1 < t 2 , < t κ = B then, as κ the nonlocal condition (2) will be
y ( 0 ) + = 1 κ h ( t ) h ( t 1 ) y ( μ ( τ ) ) = y 0
and
y ( 0 ) + lim κ = 1 κ h ( t ) h ( t 1 ) y ( μ ( τ ) ) = y ( 0 ) + 0 B y ( μ ( u ) ) d h ( u ) = y 0 .
As κ , the solution of the nonlocal problem (1), (3) will be
y ( t ) = lim κ 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + 0 t r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u = 1 1 + h ( B ) h ( 0 ) y 0 lim κ = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u ( h ( t ) h ( t 1 ) ) + 0 t r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u = 1 1 + h ( B ) h ( 0 ) y 0 0 B r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d h ( u ) + 0 t r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) d ϱ d u .
This completes the proof. □
  • For the infinite-point boundary condition, we present the following theorem.
Theorem 3.
If 1 4 hold, then the nonlocal problem of (1), (4) has at least one solution.
Proof. 
Let = 1 κ q be convergent. Then
y κ ( t ) = 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + 0 t r u , y κ 0 μ ( u ) ψ ( ϱ , y κ ( ϱ ) ) d ϱ d u .
Take the limit to (16), as κ , we have
lim κ y κ ( t ) = lim κ 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + 0 t r u , y κ 0 μ ( u ) ψ ( ϱ , y κ ( ϱ ) ) d ϱ d u = lim κ 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + lim κ 0 t r u , y κ 0 μ ( u ) ψ ( ϱ , y κ ( ϱ ) ) d ϱ d u .
Now, | q y ( τ ) | | q | y , then by comparison test = 1 q y ( τ ) is convergent. Moreover,
0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u 0 μ ( τ ) | c ( u ) | + b y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u B M + b L B 2 + b B | y ( 0 ) | = M 1 .
Therefore,
q 0 τ r u , y 0 u ψ ( ϱ , y ( ϱ ) ) d ϱ d u M 1 q
and by the comparison test
= 1 q 0 τ r u , y 0 u ψ ( ϱ , y ( ϱ ) ) d ϱ d u
is convergent. Using assumptions 1 2 and Lebesgue’s dominated convergence theorem, see [22], from (17), we obtain
y ( t ) = 1 1 + = 1 q y 0 = 1 q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + 0 t r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u .
Then we have proved. □

4. Uniqueness of the Solution

Consider the following assumptions:
(â)
The set Ψ ( t , y ) is nonempty, convex and closed ( t , y ) [ 0 , B ] × R .
-
Ψ ( t , y ) is measurable in t [ 0 , B ] for every y R .
-
Ψ satisfies the Lipschitz condition with a positive constant b such that
H Ψ ( t , y ) Ψ ( t , q ) b | y q | ,
where H ( t , y ) is the Hausdorff metric between the two subsets A , B [ 0 , B ] × E .
Remark 2.
From this assumptions we can deduce that there exists a function r Ψ ( t , y ) , such that
(1*)
r : [ 0 , B ] × R R is measurable in t for any y R and satisfies the Lipschitz condition
| r ( t , y ) r ( t , q ) | b | y q | .
(2*)
ψ : [ 0 , B ] × R R is measurable in t for any y R and satisfies the Lipschitz condition
| ψ ( t , y ) ψ ( t , q ) | b 1 | y q | .
(3*)
1 + 2 = 1 κ q 1 + = 1 κ q b b 1 L B 2 + b B < 1 .
In the following theorem, we establish existence of exactly one solution of (1) and (2).
Theorem 4.
If 1 * 3 * hold, then the solution of the nonlocal problem (1) and (2) is unique.
Proof. 
Let x , y be two the solutions of (1) and (2). Then
x ( t ) y ( t ) 1 1 + = 1 κ q = 1 κ q 0 μ ( τ ) r t , x 0 μ ( u ) ψ ( ϱ , x ( ϱ ) ) d ϱ r t , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + 0 t r u , x 0 μ ( u ) ψ ( ϱ , x ( ϱ ) ) d ϱ r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u
and so
x ( t ) y ( t ) b 1 + = 1 κ q = 1 κ q 0 μ ( τ ) x 0 μ ( u ) ψ ( ϱ , x ( ϱ ) ) d ϱ y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + b 0 t x 0 μ ( u ) ψ ( ϱ , x ( ϱ ) ) d ϱ y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u .
Therefore, we obtain
x ( t ) y ( t ) b 1 + = 1 κ q = 1 κ q 0 μ ( τ ) x 0 μ ( u ) ψ ( ϱ , x ( ϱ ) ) d ϱ x 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + b 1 + = 1 κ q = 1 κ q 0 μ ( τ ) x 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + b 0 t x 0 μ ( u ) ψ ( ϱ , x ( ϱ ) ) d ϱ x 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + b 0 t x 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u
which reduces to
x ( t ) y ( t ) b L 1 + = 1 κ q = 1 κ q 0 μ ( τ ) 0 μ ( u ) ψ ( ϱ , x ( ϱ ) ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + B x y + b L 0 t 0 μ ( u ) ψ ( ϱ , x ( ϱ ) ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u + b B x y .
Thus, we arrive at
x ( t ) y ( t ) b L 1 + = 1 κ q = 1 κ q b 1 0 μ ( τ ) 0 μ ( u ) | x ( ϱ ) y ( ϱ ) | d ϱ d u + B x y + b b 1 L 0 t 0 μ ( u ) | x ( ϱ ) y ( ϱ ) | d ϱ d u + b B x y = 1 κ q 1 + = 1 κ q ( b b 1 L B 2 + b B ) x y + ( b b 1 L B 2 + b B ) x y = 1 + 2 = 1 κ q 1 + = 1 κ q ( b b 1 L B 2 + b B ) x y .
Hence,
1 1 + 2 = 1 κ q 1 + = 1 κ q ( b b 1 L B 2 + b B ) x y 0 .
Since
1 + 2 = 1 κ q 1 + = 1 κ q ( b b 1 L B 2 + b B ) < 1 .
which implies x ( t ) = y ( t ) and the solution of the nonlocal problem (1) and (2) is unique. □

5. Continuous Dependence

Theorem 5.
If 4 6 hold, then the solution of the nonlocal problem (1) and (2) depends continuously on y 0 .
Proof. 
Let y * be a solution of the integral equation
y ( t ) = 1 1 + = 1 κ q y 0 * = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ ϱ , y ( ϱ ) d ϱ d u + 0 t r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ d u ,
such that then | y 0 y 0 * | < δ . Then
y ( t ) y * ( t ) 1 1 + = 1 κ q | y 0 y 0 * | + 1 1 + = 1 κ q = 1 κ q 0 μ ( τ ) r t , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ r t , y * 0 μ ( u ) ψ ( ϱ , y * ( ϱ ) ) d ϱ d u + 0 t r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ r u , y * 0 μ ( u ) ψ ( ϱ , y * ( ϱ ) ) d ϱ d u
and so
y ( t ) y * ( t ) 1 1 + = 1 κ q δ + b 1 1 + = 1 κ q = 1 κ q 0 μ ( τ ) y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y * 0 μ ( u ) ψ ( ϱ , y * ( ϱ ) ) d ϱ d u + b 0 t y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y * 0 μ ( u ) ψ ( ϱ , y * ( ϱ ) ) d ϱ d u .
Thus, we arrive at
y ( t ) y * ( t ) 1 1 + = 1 κ q δ + b 1 1 + = 1 κ q = 1 κ q 0 μ ( τ ) y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y 0 μ ( u ) ψ ( ϱ , y * ( ϱ ) ) d ϱ d u + b 1 1 + = 1 κ q = 1 κ q 0 μ ( τ ) y 0 μ ( u ) ψ ( ϱ , y * ( ϱ ) ) d ϱ y * 0 μ ( u ) ψ ( ϱ , y * ( ϱ ) ) d ϱ d u + b 0 t y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y 0 μ ( u ) ψ ( ϱ , y * ( ϱ ) ) d ϱ d u + b 0 t y 0 μ ( u ) ψ ( ϱ , y * ( ϱ ) ) d ϱ y * 0 μ ( u ) ψ ( ϱ , y * ( ϱ ) ) d ϱ d u
which reduces to
y ( t ) y * ( t ) 1 1 + = 1 κ q δ + b L 1 1 + = 1 κ q = 1 κ q 0 μ ( τ ) 0 μ ( u ) | ψ ( ϱ , y ( ϱ ) ) ψ ( ϱ , y * ( ϱ ) ) | d ϱ d u + B y y * + b L 0 t 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) ψ ( ϱ , y * ( ϱ ) ) d ϱ d u + b B y y * .
Therefore, we obtain
y ( t ) y * ( t ) 1 1 + = 1 κ q δ + b L 1 1 + = 1 κ q = 1 κ q b 1 0 μ ( τ ) 0 μ ( u ) | y ( ϱ ) y * ( ϱ ) | d ϱ d u + B y y * + b b 1 L 0 t 0 μ ( u ) | y ( ϱ ) y * ( ϱ ) | d ϱ d u + b B y y * δ 1 + = 1 κ q + = 1 κ q 1 + = 1 κ q ( b b 1 L B 2 + b B ) y y * + ( b b 1 L B 2 + b B ) y y * = δ 1 + = 1 κ q + 1 + 2 = 1 κ q 1 + = 1 κ q ( b b 1 L B 2 + b B ) y y * .
Hence
y y * δ 1 + = 1 κ q 1 1 + 2 = 1 κ q 1 + = 1 κ q ( b b 1 L B 2 + b B ) = ϵ .
Then the solution of the nonlocal problem (1) and (2) depends continuously on y 0 . □
Theorem 6.
If 4 6 hold, then the solution of the nonlocal problem (1) and (2) continuously depends on the function ψ.
Proof. 
Let y * be a solution of the integral equation
y ( t ) = 1 1 + = 1 κ q y 0 = 1 κ q 0 μ ( τ ) r u , y 0 μ ( u ) ψ * ϱ , y ( ϱ ) d ϱ d u + 0 t r u , y 0 μ ( u ) ψ * ( ϱ , y ( ϱ ) ) d ϱ d u ,
such that | ψ ψ * | < δ . Then
y ( t ) y * ( t ) 1 1 + = 1 κ q = 1 κ q 0 μ ( τ ) r t , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ r t , y * 0 μ ( u ) ψ * ( ϱ , y * ( ϱ ) ) d ϱ d u + 0 t r u , y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ r u , y * 0 μ ( u ) ψ * ( ϱ , y * ( ϱ ) ) d ϱ d u ,
then
y ( t ) y * ( t ) b = 1 κ q 1 + = 1 κ q 0 μ ( τ ) y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y * 0 μ ( u ) ψ * ( ϱ , y * ( ϱ ) ) d ϱ d u + b 0 t y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y * 0 μ ( u ) ψ * ( ϱ , y * ( ϱ ) ) d ϱ d u .
Hence, we obtain
y ( t ) y * ( t ) b = 1 κ q 1 + = 1 κ q 0 μ ( τ ) y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y 0 μ ( u ) ψ * ( ϱ , y * ( ϱ ) ) d ϱ d u + b = 1 κ q 1 + = 1 κ q 0 μ ( τ ) y 0 μ ( u ) ψ * ( ϱ , y * ( ϱ ) ) d ϱ y * 0 μ ( u ) ψ * ( ϱ , y * ( ϱ ) ) d ϱ d u + b 0 t y 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) d ϱ y 0 μ ( u ) ψ * ( ϱ , y * ( ϱ ) ) d ϱ d u + b 0 t y 0 μ ( u ) ψ * ( ϱ , y * ( ϱ ) ) d ϱ y * 0 μ ( u ) ψ * ( ϱ , y * ( ϱ ) ) d ϱ d u ,
which yields
y ( t ) y * ( t ) b L = 1 κ q 1 + = 1 κ q 0 μ ( τ ) 0 μ ( u ) ψ ( ϱ , y ( ϱ ) ) ψ ( ϱ , y * ( ϱ ) ) + ψ ( ϱ , y * ( ϱ ) ) ψ * ( ϱ , y * ( ϱ ) ) | d ϱ d u + B y y * + b L 0 t 0 μ ( u ) | ψ ( ϱ , y ( ϱ ) ) ψ ( ϱ , y * ( ϱ ) ) + ψ ( ϱ , y * ( ϱ ) ) ψ * ( ϱ , y * ( ϱ ) ) | d ϱ d u + b B y y * .
Thus, we find
y ( t ) y * ( t ) b L B 2 δ = 1 κ q 1 + = 1 κ q + b L B 2 δ + b L = 1 κ q 1 + = 1 κ q b 1 0 μ ( τ ) 0 μ ( u ) | y ( ϱ ) y * ( ϱ ) | d ϱ d u + B y y * + b b 1 L 0 t 0 μ ( u ) | y ( ϱ ) y * ( ϱ ) | d ϱ d u + b B y y * b L B 2 δ = 1 κ q 1 + = 1 κ q + b L B 2 δ + 1 + 2 = 1 κ q 1 + = 1 κ q ( b b 1 L B 2 + b B ) y y * .
Therefore, we have
y y * b L B 2 δ = 1 κ q 1 + = 1 κ q + b L B 2 δ 1 1 + 2 = 1 κ q 1 + = 1 κ q ( b b 1 L B 2 + b B ) = ϵ .
Then the solution of the nonlocal problem (1) and (2) continuously depends on the function ψ . □

6. Examples

Example 1.
Consider the differential equation
d y d t = 1 3 t 3 + 1 t 2 + 5 y 0 β t cos 2 y 1 + e | y ( u ) | d u , t , β ( 0 , 1 ] ,
with condition
y ( 0 ) + = 1 1 4 y 1 = 1 .
y ( t ) = 1 1 + = 1 κ 4 1 = 1 κ 6 0 1 1 3 s 3 + 1 s 2 + 5 y 0 β s cos 2 y 1 + e | y ( u ) | d u d s + 0 t 1 3 s 3 + 1 s 2 + 5 y 0 β s cos 2 y 1 + e | y ( u ) | d u d s .
Set
r t , y 0 μ ( t ) ψ ( u , y ( u ) ) d u = 1 3 t 3 + 1 t 2 + 5 y 0 β t cos 2 y 1 + e | y ( u ) | d u .
Then
r t , y 0 μ ( t ) ψ ( u , y ( u ) ) d u 1 3 t 3 + 1 5 ( | y | ) ,
and also
| ψ ( u , y ( u ) ) | 1 .
It is clear that the assumptions 1–4 of Theorem 1 are satisfied with | c ( t ) | = | 1 3 t 3 | 1 3 is measurable bounded, b = 1 5 , L = ( 1 + 2 π 4 90 ) 1 5 0.63293 and the series: = 1 1 4 is convergent. Therefore, by applying to Theorem 1, the given nonlocal problem (21) and (22) has a solution given by the integral solution (23).
Example 2.
Consider the differential equation
d y d t = 1 4 ( t + 1 ) + 1 t + 16 y 0 t β sin 2 ( y ) 1 + u 2 d u t ( 0 , 1 ] , β 1 ,
with condition
y ( 0 ) + = 1 1 6 y ( 2 + 1 2 + ) = 1 .
The integral equation
y ( t ) = 1 1 + = 1 κ 6 1 = 1 κ 6 0 2 + 1 2 + 1 4 ( s + 1 ) + 1 s + 16 y 0 s β sin 2 ( y ) 1 + u 2 d u d s + 0 t 1 4 ( s + 1 ) + 1 s + 16 y 0 s β sin 2 ( y ) 1 + u 2 d u d s .
Set
r t , y 0 μ ( t ) ψ ( u , y ( u ) ) d u = t 4 + 1 t + 16 y 0 t β sin 2 ( y ) 1 + u 2 d u .
Then
r t , y 0 μ ( t ) ψ ( u , y ( u ) ) d u t 4 + 1 4 | y | ,
and also
| ψ ( u , y ( u ) ) | 1 .
It is clear that the assumptions 1 4 of Theorem 1 are satisfied with | c ( t ) | = | t 4 | 1 4 being measurably bounded, b = 1 3 , L = ( 1 + 2 π 6 945 ) 1 4 0.7586 and the series: = 1 1 6 being convergent. Therefore, by applying to Theorem 1, the given nonlocal problem (24) and (25) has a solution given by the integral solution (26).
Example 3.
Consider the differential equation
d y d t = 1 t 5 16 + ( 1 32 ( t + t 5 ) ) y 0 t 2 y ( u ) ( 1 + y 2 ( u ) ) 2 d u t ( 0 , 1 ] ,
with condition
y ( 0 ) + 0 1 y ( u ) d u = 1 2 .
The integral equation
y ( t ) = 1 2 1 2 0 1 0 θ 1 s 5 16 + ( 1 32 ( s + s 5 ) ) y 0 s 2 y ( u ) ( 1 + y 2 ( u ) ) 2 d u d s d θ + 0 t 1 s 5 16 t 2 + ( 1 32 ( s + s 5 ) ) y 0 s 2 y ( u ) ( 1 + y 2 ( u ) ) 2 d u d s .
Set
r t , y 0 μ ( t ) ψ ( u , y ( u ) ) d u = 1 t 5 16 t 2 + ( 1 32 ( t + t 5 ) ) y 0 t 2 1 1 + y 2 ( u ) d u .
Then
r t , y 0 μ ( t ) ψ ( u , y ( u ) ) d u 1 + 1 16 | y | ,
and also
| ψ ( u , y ( u ) ) | 1 .
It is clear that the assumptions 1 4 of Theorem 1 are satisfied with | c ( t ) | 1 being measurably bounded, b = 1 16 , L = M + 1 2 b | y 0 | 1 3 2 b 1 . Therefore, by applying to Theorem 1, the given nonlocal problem (27) and (28) has a solution given by the integral solution (29).
The exact solution of (27) and (28) is y ( t ) = t , we use Picard’s method to estimate the solution of (27) and (28).
y n ( t ) = 1 2 1 2 0 1 0 θ 1 s 5 16 + ( 1 32 ( s + s 5 ) ) y n 1 0 s 2 y n 1 ( u ) ( 1 + y n 1 2 ( u ) ) 2 d u d s d θ + 0 t 1 s 5 16 t 2 + ( 1 32 ( s + s 5 ) ) y n 1 0 s 2 y n 1 ( u ) ( 1 + y n 1 2 ( u ) ) 2 d u d s .
y 0 ( t ) = 1 4
and
y ( t ) = lim n y n ( t ) .
Figure 1 shows the valuesobtained from the exact solution, the first and the second iterations for different values of t, the three lines in Figure 1 semi coincide; at the top of Figure 1, the lines have been enlarged to show the difference between them.

7. Conclusions

In this work, the existence of an absolutely continuous solution using Schauder’s fixed point theorem, the uniqueness solution and the continuous dependence of the functional differential inclusion with self-dependence on a nonlinear delay integral operator were studied. Some examples were introduced to illustrate the benefits of our results. Lastly, the Picard method was used to estimate the solution of a given example and plot the solution.

Author Contributions

A.M.A.E.-S. and R.G.A. directed the study and helped with the inspection. All the authors carried out the main results of this article, drafted the manuscript. All authors have read and agreed to the published version of the manuscript.

Funding

This research has been funded by the Scientific Research Deanship at the University of Ha’il, Saudi Arabia, through project number RG-20 125.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

This research has been funded by the Scientific Research Deanship at the University of Ha’il, Saudi Arabia, through project number RG-20 125.

Conflicts of Interest

The authors declare that they have no competing interest. There are no non-financial competing interest (political, personal, religious, ideological, academic, intellectual, commercial or any other) to declare in relation to this manuscript.

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Figure 1. Exact solution, first approximation and second approximation.
Figure 1. Exact solution, first approximation and second approximation.
Mathematics 10 02420 g001
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Hassan, T.S.; Gamal Ahmed, R.; El-Sayed, A.M.A.; El-Nabulsi, R.A.; Moaaz, O.; Mesmouli, M.B. Solvability of a State–Dependence Functional Integro-Differential Inclusion with Delay Nonlocal Condition. Mathematics 2022, 10, 2420. https://doi.org/10.3390/math10142420

AMA Style

Hassan TS, Gamal Ahmed R, El-Sayed AMA, El-Nabulsi RA, Moaaz O, Mesmouli MB. Solvability of a State–Dependence Functional Integro-Differential Inclusion with Delay Nonlocal Condition. Mathematics. 2022; 10(14):2420. https://doi.org/10.3390/math10142420

Chicago/Turabian Style

Hassan, Taher S., Reda Gamal Ahmed, Ahmed M. A. El-Sayed, Rami Ahmad El-Nabulsi, Osama Moaaz, and Mouataz Billah Mesmouli. 2022. "Solvability of a State–Dependence Functional Integro-Differential Inclusion with Delay Nonlocal Condition" Mathematics 10, no. 14: 2420. https://doi.org/10.3390/math10142420

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