Next Article in Journal
On Approximation by an Absolutely Convergent Integral Related to the Mellin Transform
Next Article in Special Issue
Optimal Auxiliary Function Method for Analyzing Nonlinear System of Belousov–Zhabotinsky Equation with Caputo Operator
Previous Article in Journal
A Study on Existence and Controllability of Conformable Impulsive Equations
Previous Article in Special Issue
A Comprehensive Review on the Fejér-Type Inequality Pertaining to Fractional Integral Operators
 
 
Font Type:
Arial Georgia Verdana
Font Size:
Aa Aa Aa
Line Spacing:
Column Width:
Background:
Article

Analysis of a Fractional-Order Quadratic Functional Integro-Differential Equation with Nonlocal Fractional-Order Integro-Differential Condition

by
Ahmed M. A. El-Sayed
1,†,
Antisar A. A. Alhamali
2,*,† and
Eman M. A. Hamdallah
1,†
1
Faculty of Science, Alexandria University, Alexandria 21521, Egypt
2
Faculty of Science, Al Zaytona University, Misrata 51, Libya
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Axioms 2023, 12(8), 788; https://doi.org/10.3390/axioms12080788
Submission received: 26 June 2023 / Revised: 4 August 2023 / Accepted: 9 August 2023 / Published: 14 August 2023
(This article belongs to the Special Issue Recent Advances in Fractional Differential Equations and Inequalities)

Abstract

:
Here, we center on the solvability of a fractional-order quadratic functional integro-differential equation with a nonlocal fractional-order integro-differential condition in the class of continuous functions. The maximal and minimal solutions will be discussed. The continuous dependence of the solutions on a few parameters will be examined. Finally, the problems of conjugate orders and integer orders, and some other problems and remarks will be discussed and presented.

1. Introduction

The solvability of nonlocal problems of fractional orders has been discussed by many authors; see, for example, refs. [1,2,3,4,5,6,7,8,9].
Many existence and uniqueness results of differential equations and inclusions with nonlocal boundary conditions have been investigated by some researchers and using different techniques (see [4,5,6,7,8] and the references therein).
Some existence results under mixed Lipschitz and Caratheodory conditions have been discussed for functional integro-differential equations of fractional orders, and the existence of some properties of the solution has been proved in [6].
Subashini et al. [10] discussed the existence of a nonlocal fractional integro-differential problem of the Hilfer type by applying the Monch fixed-point theorem and techniques of measure of noncompactness.
An investigation described the applicability of the a priori estimate method on a nonlocal, nonlinear fractional differential equation for which the weak solution and the unique solution exist [11].
Motivated by the above results, we discuss the solvability of the nonlocal problem of the fractional-order quadratic functional integro-differential equation
d x d r ( r ) = f 1 r , D α x ( r ) × I β f 2 ( r , D γ x ( r ) ) , α , β , γ ( 0 , 1 ] , r ( 0 , 1 ]
with the nonlocal fractional-order integro-differential condition
x ( 0 ) + 0 1 h ( ς , x ( ς ) , D δ x ( ς ) ) d ς = x 0 , δ ( 0 , 1 ] ,
where I β is the fractional Riemann–Liouville integral of order β and D α is a Caputo fractional derivative of order α ( 0 , 1 ] .
We investigate the existence of a solution to problem (1) and (2) in C ( I ) of all continuous functions on I = [ 0 , 1 ] . Moreover, we establish some features and characteristics of these solutions and obtain some particular cases and remarks.
Now, we recall the following definitions.
Definition 1
([12]). Let h L 1 ( I ) , β R + . The Riemann–Liouville fractional integral of h of order β is given by
I β h ( r ) = 0 r ( r s ) β 1 Γ ( β ) h ( s ) d s ,
Definition 2.
Caputo, M. Fractional derivative D a α of order α ( 0 , 1 ] of h ( r ) A C ( I ) is given by (see [12])
D α h ( r ) = 0 r ( r s ) α Γ ( 1 α ) d d s h ( s ) d s = I 1 α d d r h ( r )

2. Existence of Solution

We take into account the following assumptions:
(i)
f 1 : I × R R is measurable in r I y R and continuous in y R r I . Furthermore, ∃ a bounded measurable function a 1 : I R and a constant b 1 > 0 where
| f 1 ( r , y ) | | a 1 ( r ) | + b 1 | y | a 1 * + b 1 | y | , a 1 * = sup r I I 1 α | a 1 ( r ) | .
(ii)
f 2 : I × R R is measurable in r I y R and continuous in y R r I . Furthermore, ∃ a bounded measurable function a 2 : I R and a constant b 2 > 0 where
| f 2 ( r , y ) | | a 2 ( r ) | + b 2 | y | a 2 * + b 2 | y | , a 2 * = sup r I I β | a 2 ( r ) | .
(iii)
r 1 is a positive root of the following equation:
a 1 * + b 1 a 2 * Γ ( 2 α ) 1 r 1 + b 1 b 2 r 1 2 Γ ( 2 α ) Γ ( α + β γ + 1 ) = 0 .
(iv)
h : I × R × R R is measurable in r for all x , y R and continuous in x , y for r I , and there exists a bounded measurable function a 3 : I R and a constant b 3 > 0 where
| h ( r , x , y ) | | a 3 ( r ) | + b 3 ( | x | + | y | )
where
sup r I 0 t | a 3 ( ς ) | d ς N .
(v)
x 0 > N + b 3 r 1 Γ ( α δ + 1 ) r 1 Γ ( α + 1 )
Now, we have to prove the following auxiliary result.
Lemma 1.
The solution to the problem (1) and (2) is given in the form, if it exists,
x ( r ) = x 0 0 1 h ( ς , x ( ς ) , I α δ y ( ς ) ) d ς + I α y ( r ) , δ , γ α .
where y satisfies the equation
y ( r ) = I 1 α f 1 r , y ( r ) × 0 r ( r θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ
and the two systems (1) and (2), and (4) and (5) are equivalent such that δ , γ α .
Proof. 
x is a solution of (1) and (2). Then, we obtain
D α x ( r ) = I 1 α d x d r = I 1 α f 1 ( r , D α x ( r ) × I β f 2 ( r , I α γ y ( r ) ) .
We take D α x ( r ) = y ( r ) ; thus
y ( r ) = I 1 α f 1 r , y ( r ) × 0 r ( r θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ
and
x ( r ) = x ( 0 ) + I α y ( r ) .
Now,
I α δ y ( r ) = I α δ D α x ( r ) = I α δ I 1 α d x d r = I 1 δ d x d r = D δ x ( r ) ,
then from (2) and (6), we obtain (4).
By letting y ( r ) = D α x ( r ) in (4), we obtain (2), and in (5), we obtain
I 1 α d d r x ( r ) = I 1 α f 1 r , D α x ( r ) × 0 r ( r θ ) β 1 Γ ( β ) f 2 ( θ , D γ x ( θ ) d θ .
Operating by I α , then by d d r , respectively, we obtain (1). This proves the equivalence of the two systems (1) and (2), and (4) and (5). □
Theorem 1.
Let us assume that the conditions (i)–(iv) hold; then, problem (1) and (2) has a solution in C ( I ) .
Proof. 
We characterize a closed ball Q r 1 and an operator F 1 as
Q r 1 = { y C ( I ) : y r 1 } , r 1 = a 1 * + 1 Γ ( 2 α ) ( r 1 b 1 a 2 * + b 1 b 2 r 1 2 Γ ( α + β γ + 1 ) )
and
F 1 y ( r ) = I 1 α f 1 r , y ( r ) × 0 r ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ d ς .
For y Q r 1 , we can obtain
| y ( r ) | = I 1 α | f 1 r , y ( r ) × I β f 2 ( r , I α γ y ( r ) ) | I 1 α | a 1 ( r ) | + b 1 | y ( r ) × I β f 2 ( r , I α γ y ( r ) ) | I 1 α | a 1 ( r ) | + b 1 y ( a 2 * + b 2 I α + β γ y ) I 1 α | a 1 ( r ) | + b 1 r 1 ( a 2 * + r 1 b 2 Γ ( α + β γ + 1 ) ) ,
I 1 α | f 1 ( r , y ( r ) × I β f 2 ( r , I α γ y ( r ) ) | a 1 * + 1 Γ ( 2 α ) ( r 1 b 1 a 2 * + r 1 2 b 1 b 2 Γ ( α + β γ + 1 ) )
and we deduce that
| F 1 y ( r ) | = | 0 r ( r ς ) α Γ ( 1 α ) f 1 ς , y ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ d ς | 0 r ( r ς ) α Γ ( 1 α ) a 1 + b 1 | y ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ | d ς 0 r ( r ς ) α Γ ( 1 α ) a 1 + b 1 y × ( a 2 * + b 2 y Γ ( α + β γ + 1 ) d ς a 1 * + 1 Γ ( 2 α ) ( r 1 b 1 a 2 * + b 1 b 2 r 1 2 Γ ( α + β γ + 1 ) ) = r 1 .
So, we have
F 1 y a 1 * + 1 Γ ( 2 α ) ( r 1 b 1 a 2 * + b 1 b 2 r 1 2 Γ ( α + β γ + 1 ) ) = r 1 .
This proves that F 1 : Q r 1 Q r 1 and the family { F 1 y } is uniformly bounded on Q r 1 .
Let y Q r 1 and r 1 , r 2 I , where r 2 > r 1 and r 1 r 2 δ ; therefore,
| F 1 y ( r 2 ) F 1 y ( r 1 ) | = | 0 r 2 ( r ς ) α Γ ( 1 α ) f 1 ς , y ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ d s 0 r 1 ( r ς ) α Γ ( 1 α ) f ς , y ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ d ς | | 0 r 1 ( r 2 ς ) α Γ ( 1 α ) f ς , y ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ d ς + r 1 r 2 ( r 2 ς ) α Γ ( 1 α ) f 1 ς , y ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ d ς 0 r 1 ( r 1 ς ) α Γ ( 1 α ) f 1 ς , y ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ d ς | | 0 r 1 ( r 2 ς ) α Γ ( 1 α ) ( r 1 ς ) α Γ ( 1 α ) ( a 1 * + r 1 b 1 a 2 * + b 1 b 2 r 1 2 Γ ( α + β γ + 1 ) ) d ς + r 1 r 2 ( r 2 ς ) α Γ ( 1 α ) ( a 1 + b 1 a 2 r 1 Γ ( 1 + β ) + b 1 b 2 r 1 2 Γ ( 1 + β ) Γ ( 1 + α γ ) ) d ς | 0 r 1 | ( r 2 ς ) α ( r 1 ς ) α Γ ( 1 α ) ( r 1 ς ) α ( r 2 ς ) α | ( a 1 * + r 1 b 1 a 2 * + b 1 b 2 r 1 2 Γ ( α + β γ + 1 ) ) d s + r 1 r 2 | ( r 2 ς ) α Γ ( 1 α ) ( r 2 ς ) α | ( a 1 * + r 1 b 1 a 2 * + b 1 b 2 r 1 2 Γ ( α + β γ + 1 ) ) d ς .
This shows that family { F 1 y } is equi-continuous on Q r 1 , and by the Arzela–Ascoli result [13], mapping F 1 is relatively compact.
Let { y n } Q r 1 be such that y n y ; then,
F 1 y n ( r ) = I 1 α f 1 r , y n ( r ) × 0 r ( r θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y n ( σ ) d σ ) d θ ,
lim n F 1 y n ( r ) = lim n I 1 α f 1 r , y n ( r ) × 0 r ( r θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y n ( σ ) d σ ) d θ = I 1 α f 1 r , y ( r ) × 0 r ( r θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) lim n y n ( σ ) d σ ) d θ d ς = 0 r ( r ς ) α Γ ( 1 α ) f 1 ς , y ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ d ς = F 1 y ( r ) .
This yields that F 1 y n ( r ) F 1 y ( r ) .
Thus, mapping F 1 is continuous, and by applying the Schauder fixed-point theorem [13], y Q r 1 C ( I ) is a solution of (5).
To prove that x C ( I ) exists and satisfies Equation (4), we prove the next result. □
Theorem 2.
Let us suppose that (i)–(iv) are satisfied; then, for a fixed function y (given by Theorem 1), x C ( I ) and satisfies (4).
Proof. 
Let Q r 2 be the closed ball
Q r 2 = { x C ( I ) : x r 2 } , r 2 = x 0 N b 3 r 1 Γ ( α δ + 1 ) + r 1 Γ ( α + 1 ) ( 1 + b 3 )
and define operator F 2
F 2 x ( r ) = x 0 0 1 h ( ς , x ( ς ) , I α δ y ( ς ) ) d ς + I α y ( r ) .
Let x Q r 2 ; then,
| F 2 x ( r ) | = | x 0 0 1 h ( ς , x ( ς ) , I α δ y ( ς ) ) d ς + I α y ( r ) | | x 0 | 0 1 | h ( ς , x ( ς ) , I α δ y ( ς ) ) | d ς + I α | y ( r ) | | x 0 | 0 1 | a 3 | + b 3 ( | x ( ς ) | + y Γ ( α δ + 1 ) ) d ς + y Γ ( α + 1 ) | x 0 | 0 1 | a 3 ( ς ) | d ς b 3 0 1 | x ( ς ) | d ς b 3 r 1 Γ ( α δ + 1 ) 0 1 d ς + r 1 Γ ( α + 1 ) x 0 N b 3 r 2 b 3 r 1 Γ ( α δ + 1 ) + r 1 Γ ( α + 1 ) = r 2
and
F 2 x | x 0 | N b 3 r 2 b 3 r 1 Γ ( α δ + 1 ) + r 1 Γ ( α + 1 ) = r 2
Similarly as performed above, F 2 : Q r 2 Q r 2 , and { F 2 x } is uniformly bounded on Q r 2 .
Let x Q r 2 and r 1 , r 2 I ; moreover, r 2 > r 1 and r 1 r 2 δ . Then,
| F 2 x ( r 2 ) F 2 x ( r 1 ) | = | x 0 0 1 h ( ς , x ( ς ) , I α δ y ( ς ) ) d ς + I α y ( r 2 ) x 0 + 0 1 h ( ς , x ( ς ) , I α δ y ( ς ) ) d ς I α y ( r 1 ) | 0 r 2 | f 1 ( ς , y ( ς ) ) × I β f 2 ( r , I α γ y ( ς ) ) | d ς 0 r 1 | f 1 ( ς , y ( ς ) ) × I β f 2 ( r , I α γ y ( ς ) ) | d ς r 1 r 2 | f 1 ( ς , y ( ς ) ) × I β f 2 ( r , I α γ y ( ς ) ) | d ς .
Thus, family { F 2 x } is equicontinuous on Q r 2 , and operator F 2 is relatively compact.
For { x n } Q r 2 , and x n y ; then,
F 2 x n ( r ) = x 0 0 1 h ( ς , x ( ς ) , I α δ y ( ς ) ) d ς + I α y ( r )
and
lim n F 2 x n ( r ) = lim n x 0 0 1 h ( ς , x ( ς ) , I α δ y ( ς ) ) d ς + I α y ( r )
Using the Lebesgue dominated convergence theorem, we have
lim n F 2 x n ( r ) = x 0 0 1 h ( ς , lim n x ( ς ) , I α δ y ( ς ) ) d ς + I α y ( r ) = x 0 0 1 h ( ς , x ( ς ) , I α δ y ( ς ) ) d ς + I α y ( r ) = F 2 x ( r )
This means that F 2 x n ( r ) F 2 x ( r ) . Hence, operator F 2 is continuous, and there exists x Q r 2 C ( I ) that satisfies (4).
Therefore, we prove the existence of a continuous solution x of Equation (4). □

2.1. Conjugate-Order Problem

For γ = 1 α , α [ 1 2 , 1 ) in problem (1) and (2), we have the nonlocal problem of conjugate-orders ( α , 1 α )
d x d r ( r ) = f 1 r , D α x ( r ) × I β f 2 ( r , D 1 α x ( r ) ) , α , β ( 0 , 1 ] , r ( 0 , 1 ]
with nonlocal fractional-order integro-differential condition (2).
By putting D α x ( r ) = y ( r ) , we can prove that nonlocal problem (7) with condition (2) is equivalent to the functional integral equation
y ( r ) = I 1 α f 1 r , y ( r ) × I β f 2 ( r , I 2 α 1 y ( r ) ) , α 1 2
with (4). Then, by applying Theorems 1 and 2, we can deduce the solvability of nonlocal problem (7) and (2) in C ( I ) .

2.2. Absolutely Continuous Solution

By taking y ( r ) = d x d r ( r ) , (1) and (2) reduce to the fractional-order quadratic integral equation
y ( r ) = f 1 r , I 1 α y ( r ) × I β f 2 ( r , I 1 γ y ( r ) ) , r ( 0 , 1 ]
and the nonlocal integro-differential condition
x ( 0 ) + 0 1 h ( ς , x 0 + 0 ς y ( s ) d s , I 1 δ y ( ς ) ) d ς = x 0 .
In the same way as in Lemma 1, we can prove the equivalence between (1) and (2), and system (9) and
x ( r ) = x 0 0 1 h ( ς , x ( ς ) , y ( ς ) ) d ς + 0 t y ( s ) d s
With the same technique as in [14], the existence of solution y L 1 ( I ) of (9) can be proved. So, the existence of an absolutely continuous solution
x ( r ) = x 0 0 1 h ( ς , x ( ς ) , I 1 δ y ( ς ) ) d ς + 0 t y ( s ) d s A C ( I )
of problem (1) and (2) can be proved.

2.3. Integer-Order Problem

By letting α , β , γ 1 in problem (1) and (2), we have the integer-order equation
d x d r ( r ) = f 1 r , d d r x ( r ) × 0 r f 2 ( ς , d d ς x ( ς ) ) d ς , r ( 0 , 1 ]
with the nonlocal integro-differential condition
x ( 0 ) + 0 1 h ( ς , x ( ς ) , d d ς x ( ς ) ) d ς = x 0 .
In the same way as in Lemma 1, we can prove the equivalence between (13) and (14), and the system
y ( r ) = f 1 r , y ( r ) × 0 r f 2 ( ς , y ( ς ) ) d ς , r ( 0 , 1 ]
and
x ( r ) = x 0 0 1 h ( ς , x ( ς ) , y ( ς ) ) d ς + 0 t y ( s ) d s
and as in [14], the existence of solution y L 1 ( I ) of (15) can be proved. Consequently, x A C ( I ) of integer-order problem (13) and (14) can be proved.

3. Some Characteristics of the Solution

3.1. Maximal and Minimal Solutions

Lemma 2.
Let us assume that the conditions of Theorem 1 hold. Let x , y C ( I ) satisfy
x ( r ) 0 r ( r ς ) α Γ ( 1 α ) f 1 ς , x ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) x ( σ ) d σ ) d θ d ς , y ( r ) 0 r ( r ς ) α Γ ( 1 α ) f 1 ς , y ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ d ς
such that one of them is strict. Let us assume that f 1 and f 2 are monotonic non-decreasing in the second argument; thus,
x ( r ) < y ( r ) , r > 0 .
Proof. 
Let us assume that result (17) is not true; then, t 1 where
x ( r 1 ) = y ( r 1 ) , r 1 > 0 , x ( r ) < y ( r ) 0 < r < r 1 .
From the monotonicity of f 1 and f 2 , we obtain
x ( r 1 ) 0 r 1 ( r 1 ς ) α Γ ( 1 α ) f 1 ς , x ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) x ( σ ) d σ ) d θ d ς < 0 r 1 ( r 1 ς ) α Γ ( 1 α ) f 1 ς , y ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ d ς = y ( r 1 ) .
Hence x ( r 1 ) < y ( r 1 ) , which contradicts that x ( r 1 ) = y ( t 1 ) ; then,
x ( r ) < y ( r ) , r I .  □
Theorem 3.
Let us assume that the conditions of Theorem 1 hold. For monotonic non-decreasing functions f 1 and f 2 in the second argument, problem (1) and (2) has maximal and minimal solutions.
Proof. 
To investigate the existence of the maximal solution of Equation (5), we take ϵ > 0 and consider the following integral equations:
y ϵ ( r ) = I 1 α f 1 ϵ r , y ϵ ( r ) × 0 r ( r θ ) β 1 Γ ( β ) f 2 ϵ ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ϵ ( σ ) d σ ) d θ ,
where
f 1 ϵ ( r , y ϵ ( r ) ) = f 1 ( r , y ϵ ( r ) ) + ϵ
f 2 ϵ ( r , y ϵ ( r ) ) = f 2 ( r , y ϵ ( r ) ) + ϵ .
Obviously, f 1 ϵ and f 2 ϵ satisfy the assumptions of Theorem 1; according to Theorem 1, Equation (18) has a continuous solution y ϵ and
y ϵ ( r ) = 0 r ( r ς ) α Γ ( 1 α ) × ϵ + f 1 ς , y ϵ ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) ϵ + f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ϵ ( σ ) d σ ) d θ d ς .
Next, we take ϵ 1 , ϵ 2 > 0 such that 0 < ϵ 2 < ϵ 1 < ϵ ; then,
y ϵ 1 ( r ) = 0 r ( r ς ) α Γ ( 1 α ) [ ϵ 1 + f 1 ( ς , y ϵ 1 ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) ϵ 1 + f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ϵ 1 ( σ ) d σ ) d θ } d ς ] > 0 r ( r ς ) α Γ ( 1 α ) [ ϵ 2 + f 1 ( ς , y ϵ 1 ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) ϵ 2 + f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ϵ 1 ( σ ) d σ ) d θ } d ς ]
y ϵ 2 ( r ) = 0 r ( r ς ) α Γ ( 1 α ) ϵ 2 + f 1 ς , y ϵ 2 ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) ϵ 2 + f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ϵ 2 ( σ ) d σ ) d θ d ς
By applying Lemma 2 on estimates (19) and (20), we obtain
y ϵ 2 ( r ) < y ϵ 1 ( r ) , r I .
As proved in Theorem 1, { y ϵ ( r ) } is equicontinuous and uniformly bounded on I; then, { y ϵ } is relatively compact, and there exists a decreasing sequence ϵ n such that ϵ n 0 ,   n and lim n y ϵ n ( r ) exists uniformly on I. Let
lim n y ϵ n ( r ) = q ( r ) .
Then,
0 r ( r ς ) α Γ ( 1 α ) f 1 ς , y ϵ n ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ϵ n ( σ ) d σ ) d θ d ς ) 0 r ( r ς ) α Γ ( 1 α ) f 1 ς , q ( ς ) × 0 ς ( ς ) θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) q ( σ ) d σ ) d θ d ς ,
then,
q ( r ) = lim n y ϵ n ( r ) = 0 r ( r ς ) α Γ ( 1 α ) f 1 ς , q ( ς ) × 0 ς ( ς ) θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) q ( σ ) d σ ) d θ d ς
which yields that q ( r ) is a solution of (5).
Now, if each y ( r ) is a solution of (5), then
y ϵ ( r ) = 0 r ( r ς ) α Γ ( 1 α ) [ ϵ + f 1 ( ς , y ϵ ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) ϵ + f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ϵ ( σ ) d σ ) d θ } d ς ] > 0 r ( r ς ) α Γ ( 1 α ) f 1 ς , y ϵ ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ϵ ( σ ) d σ ) d θ d ς
and
y ( r ) = 0 r ( r ς ) α Γ ( 1 α ) f 1 ς , y ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y ( σ ) d σ ) d θ d ς
Using Lemma 2 on estimates (21) and (22), we obtain
y ( r ) < y ϵ ( r ) , r I .
The uniqueness of the maximal solution yields that y ϵ ( r ) q ( r ) uniformly on I as ϵ 0 . Hence, q is the maximal solution of (5).
Similarly, we study the existence of the minimal solution of problem (1) and (2). □

3.2. Existence of Unique Solution

Let us assume the following conditions:
( i ) *
f 1 , f 2 : I × R R are measurable in r I x R and satisfy the Lipschitz condition,
| f 1 ( r , x ) f 1 ( r , y ) | L 1 | x y | , r I , x , y R .
| f 2 ( r , x ) f 2 ( r , y ) | L 2 | x y | , r I , x , y R .
Assumption ( i v ) implies that
| f 1 ( r , y ) | | f 1 ( r , 0 ) | + L 1 | y |
and
| f 1 ( r , y ) | a 1 + L 1 | y | , s u c h t h a t a 1 = sup r I | f 1 ( r , 0 ) | .
Furthermore,
| f 2 ( r , y ) | | f 2 ( r , 0 ) | + L 2 | y |
and
| f 2 ( r , y ) | a 2 + L 2 | y | , w h e r e a 2 = sup r I | f 2 ( r , 0 ) | .
( i i ) *
h : I × R × R R is measurable in r I x , y R and satisfies the Lipschitz condition,
| h ( r , x 1 , y 1 ) h ( r , x 2 , y 2 ) | L 3 ( | x 1 x 2 | + | y 1 y 2 | )
with Lipschitz condition L 3 > 0 .
Theorem 4.
Let us assume that (iii), (i*) are verified. If
2 b 1 b 2 r 1 Γ ( 2 α ) Γ ( α + β γ + 1 ) + b 1 a 2 * Γ ( 2 α ) < 1 .
then the solution of (5) is unique.
Proof. 
Let us consider that assumptions (iii), (i*) are satisfied; then, Theorem 1 is verified, and the solution of Equation (5) exists. Let y 1 , y 2 be two solutions of (5), where
| y 2 ( r ) y 1 ( r ) | = | 0 r ( r ς ) α Γ ( 1 α ) f ς , y 2 ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y 2 ( σ ) d σ ) d θ d ς 0 r ( r ς ) α Γ ( 1 α ) f ς , y 1 ( ς ) × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) y 1 ( σ ) d σ ) d θ d ς | b 1 0 r ( r ς ) α Γ ( 1 α ) ( | y 2 ( ς ) | × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) | y 2 ( σ ) | d σ ) d θ | y 1 ( ς ) | × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) | y 1 ( σ ) | d σ ) d θ ) d ς b 1 0 t ( t ς ) α Γ ( 1 α ) ( | y 2 ( ς ) | × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) | y 2 ( σ ) | d σ ) d θ | y 2 ( ς ) | × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) | y 1 ( σ ) | d σ ) d θ + | y 2 ( ς ) | × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) | y 1 ( σ ) | d σ ) d θ | y 1 ( ς ) | × 0 ς ( ς θ ) β 1 Γ ( β ) f 2 ( θ , 0 θ ( θ σ ) ( α γ ) 1 Γ ( α γ ) | y 1 ( σ ) | d σ ) d θ ) d ς b 1 0 r ( r ς ) α Γ ( 1 α ) ( y 2 × b 2 0 ς ( ς θ ) β 1 Γ ( β ) 0 θ ( θ σ ) α 1 Γ ( α ) | y 2 ( σ ) y 1 ( σ ) | d σ d θ + y 2 y 1 × ( a 2 * + b 2 r 1 Γ ( β + α γ + 1 ) ) ) d ς b 1 Γ ( 2 α ) ( y 2 y 1 b 2 r 1 Γ ( β + α γ + 1 ) + y 2 y 1 ( a 2 * + b 2 r 1 Γ ( β + α γ + 1 ) ) ) .
Hence,
y 2 y 1 1 ( 2 b 1 b 2 r 1 Γ ( 2 α ) Γ ( β + α γ + 1 ) + b 1 a 2 * Γ ( 2 α ) ) 0 .
which implies the uniqueness of the solution of (5). □
Corollary 1.
Let us suppose that Theorem (4) is verified. If L 3 < 1 , then for every solution y C ( I ) of Equation (5), there exists a unique solution x of (4).
Proof. 
Let y C ( I ) satisfy Equation (5) and x 1 , x 2 satisfy Equation (4); then, we have
| x 2 ( r ) x 1 ( r ) | = | x 0 0 1 h ( ς , x 2 ( ς ) , I α δ y ( ς ) ) d ς + I α y ( r ) x 0 + 0 1 h ( ς , x 1 ( ς ) , I α δ y ( ς ) ) d ς I α y ( r ) | 0 1 | h ( ς , x 2 ( ς ) , I α δ y ( ς ) ) | | h ( ς , x 1 ( ς ) , I α δ y ( ς ) ) | d ς L 3 | x 2 ( r ) x 1 ( r ) | ,
then,
x 2 x 1 ( 1 L 3 ) 0 x 2 = x 1 .
 □

3.3. Continuous Dependency Results

Definition 3.
The unique solution x C ( I ) of (4) depends continuously on x 0 , if ϵ > 0 , δ > 0 , where
| x 0 x 0 * | δ x x * ϵ
where x * satisfies
x * ( r ) = x 0 * 0 1 h ( ς , x * ( ς ) , I α δ y ( ς ) ) d ς + I α y ( r ) .
Theorem 5.
Let assumption (iv) hold; then, the unique solution of (4) depends continuously on x 0 .
Proof. 
Let us assume that the two functions x ( r ) and x * ( r ) satisfy (4); then,
| x ( r ) x * ( r ) | = | x 0 0 1 h ( ς , x ( ς ) , I α δ y ( ς ) ) d s + I α y ( r ) x 0 * + 0 1 h ( ς , x * ( ς ) , I α δ y ( ς ) ) d ς I α y ( r ) | | x 0 x 0 * | + 0 1 | h ( ς , x ( ς ) , I α δ y ( ς ) ) h ( ς , x * ( ς ) , I α δ y ( ς ) ) | d ς | x 0 x 0 * | + L 3 | x ( ς ) x * ( ς ) | δ + L 3 x x * .
Hence,
x x * ( 1 L 3 ) δ
and
x x * δ 1 L 3 = ϵ .
 □
Definition 4.
The unique solution x C ( I ) of Equation (4) depends continuously on y; if ϵ > 0 , δ > 0 , where
| y y * | δ x x * ϵ
where x * satisfies the integral equation
x * ( r ) = x 0 * 0 1 h ( ς , x * ( ς ) , I α δ y ( ς ) ) d ς + I α y ( r ) .
Theorem 6.
Let assumption (iv) hold; then, the unique solution of the integral Equation (4) depends continuously on y.
Proof. 
Let us assume that the two functions x ( r ) and x * ( r ) satisfy Equation (4); then,
| x ( r ) x * ( r ) | = | x 0 0 1 h ( ς , x ( ς ) , I α δ y ( ς ) ) d ς + I α y ( r ) x 0 * + 0 1 h ( ς , x * ( ς ) , I α δ y ( ς ) ) d ς I α y ( r ) | = | 0 1 ( h ( ς , x ( ς ) , I α δ y ( ς ) ) h ( ς , x * ( ς ) , I α δ y ( ς ) ) ) d ς + I α ( y ( r ) y * ( r ) ) | L 3 | x ( ς ) x * ( ς ) | + L 3 I α δ | y ( ς ) y * ( ς ) | + I α | y ( r ) y * ( r ) | L 3 x x * + L 3 δ Γ ( α δ + 1 ) + δ Γ ( α + 1 ) .
Hence,
x x * ( 1 L 3 ) L 3 δ Γ ( α δ + 1 ) + δ Γ ( α + 1 )
x x * L 3 δ Γ ( α δ + 1 ) + δ Γ ( α + 1 ) ( 1 L 3 ) = ϵ .
 □
Definition 5.
The unique solution x C ( I ) of Equation (4) depends continuously on h; if ϵ > 0 , δ > 0 , where
| h ( ς , x ( ς ) , I α δ y ( ς ) ) h * ( ς , x ( ς ) , I α δ y ( ς ) ) | δ x x * ϵ
where x * satisfies the integral equation
x * ( r ) = x 0 0 1 h * ( ς , x * ( ς ) , I α δ y ( ς ) ) d ς + I α y ( r ) .
Theorem 7.
Let assumption (iv) hold; then, the unique solution of (4) depends continuously on h.
Proof. 
Let us assume that the two functions x ( r ) and x * ( r ) satisfy Equation (4); then,
| x ( r ) x * ( r ) | = | x 0 0 1 h ( ς , x ( ς ) , I α δ y ( ς ) ) d s + I α y ( r ) x 0 + 0 1 h * ( ς , x * ( ς ) , I α δ y ( ς ) ) d ς I α y ( r ) | = | 0 1 h ( ς , x ( ς ) , I α δ y ( ς ) ) d ς 0 1 h * ( ς , x ( ς ) , I α δ y ( ς ) ) d ς + 0 1 h * ( ς , x ( ς ) , I α δ y ( ς ) ) d ς 0 1 h * ( ς , x * ( ς ) , I α δ y ( ς ) ) d ς | = 0 1 | h ( ς , x ( ς ) , I α δ y ( ς ) ) 0 1 h * ( ς , x ( ς ) , I α δ y ( ς ) ) | d ς + 0 1 | h * ( ς , x ( ς ) , I α δ y ( ς ) ) 0 1 h * ( ς , x * ( ς ) , I α δ y ( ς ) ) | d ς δ + L 3 | x ( ς ) x * ( ς ) | δ + L 3 x x * .
Hence,
x x * ( 1 L 3 ) δ
so,
x x * δ 1 L 3 = ϵ .
 □
Example 1.
We give the quadratic functional integro-differential equation
d x d r ( r ) = ( r 2 ) 2 + 1 2 ( D 3 4 x ( r ) × I 1 2 ( r 3 + 1 2 D 1 5 x ( ς ) ) ) . r ( 0 , 1 ]
with the nonlocal integro-differential condition
x ( 0 ) 1 = 0 1 ( r 2 ) 3 + 1 2 ( x ( r ) + D 1 2 x ( ς ) ) d ς .
Thus,
f 1 ( r , D α x ( r ) × I β f 2 ( r , D γ x ( r ) ) ) = ( r 2 ) 2 + 1 2 ( D 1 2 x ( r ) × I 1 2 ( r 3 + 1 2 D 1 5 x ( ς ) ) ) r I , f 2 ( r , D γ x ( r ) ) = r 3 + 1 2 D 1 5 x ( ς ) r I .
Obviously, Theorem 1 holds, when r = 1 ,
a 1 * = 0.196136 , a 2 * = 1 3 , a 3 * = 1 8 , b 1 = b 2 = b 3 = 1 2 , δ = 1 2 , α = 3 4 , β = 1 2 , γ = 1 5 a n d x 0 = 1 .
From (3), we have
41 64 r 1 2 19 24 r 1 + 0.19613558245703772 = 0 ,
then, r 1 = 0.342894 or r 1 = 0.892878 is a positive number. Then, problem (27) and (28) has at least one solution.
Example 2.
Given the problem of conjugate order
d x d r ( r ) = ( r 2 ) 2 + 1 2 ( D 1 2 x ( r ) × I 1 2 ( r 3 + 1 2 D 1 2 x ( ς ) ) ) . r ( 0 , 1 ]
with the nonlocal integro-differential condition
x ( 0 ) 1 = 0 1 ( r 2 ) 3 + 1 2 ( x ( r ) + D 1 2 x ( ς ) ) d ς .
we have
9 16 r 1 2 3 4 r 1 + 0.19613558245703772 = 0 ,
then, r 1 = 0.4211280954182888 or r 1 = 0.6899830156928223 .

4. Conclusions

Integral and differential equations contribute significantly to mathematical analysis and have numerous applications to issues in the real world. It has a wide range of uses in mechanics, population dynamics, mathematical biology, engineering, mathematical physics and other fields [1,2,3,8,15,16,17,18].
In this study, we establish the solvability of nonlocal problem (1) and (2). The existence of solution x C ( I ) of problem (1) and (2) is investigated, and the existence of the maximal and minimal solutions of problem (1) and (2) is proved. Furthermore, some continuous dependency results of solution x on fractional-order derivative y ( t ) , on parameter x 0 and on function h are also proved.

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.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

We thank the referee for their remarks and comments that help to improve our manuscript.

Conflicts of Interest

The authors declare no conflict of interest.

References

  1. Banaś, J.; Lecko, M.; El-Sayed, W.G. Existence theorems for some quadratic integral equations. J. Math. Anal. Appl. 1998, 222, 276–285. [Google Scholar] [CrossRef] [Green Version]
  2. Banaś, J.; Zajac, T. A new approach to the theory of functional integral equations of fractional order. J. Math. Anal. Appl. 2011, 375, 375–387. [Google Scholar] [CrossRef] [Green Version]
  3. Banaś, J.; Caballero, J.; Rocha, J.; Sadarangani, K. Monotonic solutions of a class of quadratic integral equations of Volterra type. Comput. Math. Appl. 2005, 49, 943–952. [Google Scholar] [CrossRef] [Green Version]
  4. Bader, R.; Papageorgiou, N.S. Nonlinear multivalued boundary value problems, Discussiones Mathematicae Differential Inclusions. Control Optim. 2001, 21, 127–148. [Google Scholar]
  5. Boucherif, A.; Precup, R. On the nonlocal initial value problem for first order differential equations. Fixed Point Theory 2003, 4, 205–212. [Google Scholar]
  6. El Borai, M.; Abbas, M.I. On Some Integro-Differential Equations of Fractional Orders Involving Caratheodory Nonlinearities. Int. J. Mod. Math. 2007, 2, 41–52. [Google Scholar]
  7. Kamenskii, M.; Obukhovskii, V.; Petrosyan, G.; Yao, J. Existence and approximation of solutions to nonlocal boundary value problems for fractional differential inclusions. Fixed Point Theory Appl. 2019, 2019, 2. [Google Scholar] [CrossRef] [Green Version]
  8. Zeng, S.; Rǎdulescu, V.D.; Winkert, P. Double phase implicit Obstacle problems with convection and multivalued mixed boundary value conditions. SIAM J. Math. Anal. 2022, 54, 1–25. [Google Scholar] [CrossRef]
  9. Sumelka, W. Fractional viscoplasticity. Mech. Res. Commun. 2014, 56, 31–36. [Google Scholar] [CrossRef]
  10. Subashini, R.; Jothimani, K.; Nisar, K.S.; Ravichandran, C. New results on nonlocal functional integrodifferential equations via Hilfer fractional derivative. Alex. Eng. J. 2020, 59, 2891–2899. [Google Scholar] [CrossRef]
  11. Aicha, S.; Merad, A. Solvability of nonlinear fractional integro-differential equation with nonlocal condition. Arab. J. Math. Sci. 2023, 29, 172–190. [Google Scholar] [CrossRef]
  12. Kilbas, A.A.; Srivastava, H.M.; Trujillo, J.J. Theory and Applications of Fractional Differential Equations; Elsevier: Amsterdam, The Netherlands, 2006. [Google Scholar]
  13. Curtain, R.F.; Pritchard, A.J. Functional Analysis in Modern Applied Mathematics; Academic Press: Cambridge, MA, USA, 1977. [Google Scholar]
  14. El-Sayed, A.M.A.; Hashem, H.G.; Al-Issa, S.M. An implicit hybrid delay functional integral equation: Existence of integrable solutions and continuous dependence. Mathematics 2021, 9, 3234. [Google Scholar] [CrossRef]
  15. El-Sayed, A.M.A.; Hashem, H.H.G.; Ziada, E.A.A. Picard and Adomian decomposition methods for a quadratic integral equation of fractional order. Comput. Appl. Math. 2014, 33, 95–109. [Google Scholar] [CrossRef]
  16. Failla, G.; Zingales, M. Advanced materials modeling via fractional calculus: Challenges and perspectives. Phil. Trans. R. Soc. 2020, 378, 20200050. [Google Scholar] [CrossRef] [PubMed]
  17. Fierro, R.; Martínez, C.; Morales, C.H. Caratheodory selections for multi-valued mappings. Nonlinear Anal. 2006, 64, 1229–1235. [Google Scholar] [CrossRef]
  18. Srivastava, H.M.; El-Sayed, A.M.A.; Gaafar, F.M. A Class of nonlinear boundary value problems for an arbitrary fractional-order differential equation with the Riemann-Stieltjes functional integral and infinite-point boundary conditions. Symmetry 2018, 10, 508. [Google Scholar] [CrossRef] [Green Version]
Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

Share and Cite

MDPI and ACS Style

El-Sayed, A.M.A.; Alhamali, A.A.A.; Hamdallah, E.M.A. Analysis of a Fractional-Order Quadratic Functional Integro-Differential Equation with Nonlocal Fractional-Order Integro-Differential Condition. Axioms 2023, 12, 788. https://doi.org/10.3390/axioms12080788

AMA Style

El-Sayed AMA, Alhamali AAA, Hamdallah EMA. Analysis of a Fractional-Order Quadratic Functional Integro-Differential Equation with Nonlocal Fractional-Order Integro-Differential Condition. Axioms. 2023; 12(8):788. https://doi.org/10.3390/axioms12080788

Chicago/Turabian Style

El-Sayed, Ahmed M. A., Antisar A. A. Alhamali, and Eman M. A. Hamdallah. 2023. "Analysis of a Fractional-Order Quadratic Functional Integro-Differential Equation with Nonlocal Fractional-Order Integro-Differential Condition" Axioms 12, no. 8: 788. https://doi.org/10.3390/axioms12080788

Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here.

Article Metrics

Back to TopTop