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Article

The Study of Fractional Quadratic Integral Equations Involves General Fractional Integrals

1
School of Mathematics and Physics, University of South China, Hengyang 421001, China
2
Faculty of Mathematics and Computational Science, Xiangtan University, Xiangtan 411105, China
*
Author to whom correspondence should be addressed.
Fractal Fract. 2025, 9(4), 249; https://doi.org/10.3390/fractalfract9040249
Submission received: 21 January 2025 / Revised: 19 February 2025 / Accepted: 20 March 2025 / Published: 15 April 2025

Abstract

:
This paper investigates the well-posedness of analytical solutions to fractional quadratic differential equations, which involve generalized fractional integrals with respect to other functions. The nonlinear components f and h depend on spatial variables and the general fractional integral, respectively. We use the operator equation T 1 ω T 2 ω + T 3 ω = ω to investigate the existence of solutions, marking the first study of its kind. Using an auxiliary function and Boyd and Wang’s fixed-point theorem, the uniqueness and continuous dependence of the solution are obtained. In particular, we applied nonlinear functional analysis to investigate Hyers-Ulam and Hyers-Ulam-Rassias stabilities for fractional quadratic integral equations. New results are provided for specific values of the parameter z, and a fundamental inequality is formulated to ensure the existence of maximal and minimal solutions. Some examples are given to illustrate our results.

1. Introduction

Fractional differential and integral equations, with their initial and boundary conditions, are widely applied in various scientific and engineering disciplines, including mechanics, physics, biology, economics, control systems, and medicine, see [1,2,3,4,5]. They are especially useful in modeling real-world processes that involve memory and hereditary effects, providing a more accurate description than classical differential equations. Over recent years, fractional differential and integral equations have garnered increasing attention due to their versatility in modeling nonlinear phenomena, as well as their growing importance in diverse fields of science and engineering [6,7,8]. In [9,10,11,12], fractional order integrals are proposed with general kernels, that is, with respect to some kernel functions z ( τ ) . The general type of fractional integrals (fractional integrals with respect to kernel function) are defined by
I a + α , z ω ( τ ) : = 1 Γ ( α ) a τ ( z ( τ ) z ( ς ) ) α 1 z ( ς ) ω ( ς ) d ς ,
where z ( τ ) is a strictly increasing function. The kinds of fractional integral can also be reduced to a well-known one, such as the following: if z ( τ ) = τ , z ( τ ) = ln τ , and z ( τ ) = τ r , then the above equation corresponds to the Riemann-Liouville fractional integral, Hadamard fractional integral, and Katugampola fractional integral, respectively. Recently, the physical meaning and constraint conditions of the function z ( τ ) were given by the continuous time random walk theory [13] and the boundedness theorem [14], respectively. We come across a long-tailed waiting time probability density function [13],
z ( τ ) 1 ( z ( τ ) z ( a ) ) α + 1
which is more general than the classical power law function
z ( τ ) 1 ( τ a ) α + 1 .
Quadratic integral equations have significant practical applications, such as modeling radiative transfer, the kinetic theory of gases, neutron transport, and traffic flow dynamics. Researchers have extensively studied the existence of solutions to various types of nonlinear quadratic integral equations. For example, numerical solutions [15,16,17,18,19], monotonic solutions [20,21,22], integrable and continuous solutions [23,24,25,26,27]. In fields such as queuing theory and biology, fractional quadratic integral equations are employed to model complex systems characterized by memory effects and non-local properties [3,28]. With the deepening application of fractional calculus theory, fractional quadratic integral equations have found increasing application in fields such as anomalous diffusion, complex viscoelasticity, mechatronic behaviors, biological systems, and rheology, garnering significant attention from researchers [29,30,31,32].
The stability of functional [33], differential [34,35], integral [36,37,38], and integro-differential equations [39] has been a major research focus in recent decades due to their numerous applications. In particular, Hyers-Ulam and Hyers-Ulam-Rassias stabilities have received significant attention. Hyers-Ulam and Hyers-Ulam-Rassias stabilities ensure that approximate solutions to fractional differential equations remain close to exact solutions, which is critical for modeling systems with inherent uncertainties. For instance, stochastic differential equations with Hyers-Ulam stability provide robust approximations for option pricing under market volatility. For the 1-D heat equation, Hyers-Ulam-Rassias stability guarantees that approximate solutions (e.g., from finite-difference methods) converge to exact temperature profiles, which is critical for thermal management in electronics. Hyers-Ulam and Hyers-Ulam-Rassias stabilities bridge theoretical rigor and practical reliability, ensuring models withstand real-world noise and approximations. However, to the best of the author’s knowledge, there are few results on the stability of solutions to fractional quadratic integral equations. This article aims to address this gap.
Recently, Ahood et al. [18] investigated an analytical and approximate solution via Banach’s fixed-point theorem, the Adomian decomposition method, and Picard’s method for the following fractional quadratic integral equation:
ω ( τ ) = f ( τ ) + g ( τ , ω ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , ω ( ς ) ) d ς ,
where τ I = [ 0 , 1 ] , z ( τ ) is a strictly increasing function. Under Equation (1), when z ( τ ) = τ , the equation was investigated in [5,40].
In (1), if we choose g ( τ , ω ( τ ) ) = 1 and z ( τ ) = τ , the equation simplifies to
ω ( τ ) = f ( τ ) + 0 τ ( τ ς ) α 1 Γ ( α ) h ( ς , ω ( ς ) ) d ς ,
which was analyzed in [41].
In (2), by selecting f ( τ ) = ω 0 , ω 0 is a constant, and α = 0 ; then, the integral equation reduces to
ω ( τ ) = ω 0 + 0 τ h ( ς , ω ( ς ) ) d ς ,
which was investigated in [42].
From the existing literature [5,18,40,41,42] and other studies on fractional quadratic integral equations [43,44], it is known that results are available when the nonlinear term f is either a constant or a function of time τ . In these studies, the nonlinear term, h, typically involves only a single spatial variable. This raises an important question: what happens when f depends on spatial variables, and h involves multiple spatial variables? Motivated by this idea, this paper focuses on the existence, uniqueness, and continuous dependence of solutions to nonlinear fractional quadratic integral equations of the form
ω ( τ ) = f ( τ , ω ( τ ) ) + g ( τ , ω ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) d ς ,
where τ I = [ 0 , 1 ] . The above integral equation can be alternatively expressed as
ω ( τ ) = f ( τ , ω ( τ ) ) + g ( τ , ω ( τ ) ) I 0 + α , z h ( τ , ω ( τ ) , I 0 + β , z ω ( τ ) ) ,
where α , β > 0 , and the functions are f C ( I × R , R ) , g C ( I × R , R { 0 } ) , and h C ( I × R × R , R ) . I 0 + * , z is a general fractional integral of order * = { α , β } , z ( τ ) and is a strictly increasing function. Specifically, when z ( τ ) = τ , our results correspond to the classic Riemann-Liouville fractional quadratic integral equation. When z ( τ ) = ln τ , our results correspond to the Hadamard fractional quadratic integral equation. When z ( τ ) = τ r ( r > 0 ) , our results correspond to the Katugampola fractional quadratic integral equation.
To our knowledge, no previous study has specifically addressed the nonlinear fractional quadratic integral regarding Equation (3). As a result, our aim is to fill this gap in the literature and contribute to enriching this academic area. From the literature [18,44,45] and the references therein, it is known that the Adomian decomposition method (ADM) and Picard’s method (PM) are powerful tools that can be used to study the existence and uniqueness of solutions to quadratic differential equations. However, since the nonlinear term f in our system (3) involves the spatial variable, ω , rather than f ( t ) or a constant, the main challenge arises from the fact that the ADM and PM are not directly applicable to the system (3). In this work, we approach the problem by using the operator equation T 1 ω T 2 ω + T 3 ω = ω to explore the existence of solutions to these equations. This marks the first work in this particular area. Recent studies on fractional quadratic integrals have focused on the existence and uniqueness of solutions. In this work, we extend these results to establish the well-posedness of (3), including the existence, uniqueness, continuous dependence, and Ulam-Hyers stability of the solutions. Notably, the nonlinear terms f and h in our study depend on the spatial variable and the general fractional integral, respectively, making our results more general and encompassing a broader range of known results.
The structure of the paper is as follows. In the next section, we review some key definitions and lemmas and outline the framework for our problem. Section 3 is dedicated to proving the existence of solutions to the nonlinear fractional quadratic integral in Equation (3). In Section 4, we establish the results on the uniqueness and continuous dependence of solutions using the Boyd and Wang fixed-point theorem. In Section 5, we apply nonlinear functional analysis to investigate the stability of Hyers-Ulam and Hyers-Ulam-Rassias for fractional quadratic integral equations. Section 6 investigates the existence of maximal and minimal solutions. Some examples are presented to support our main results in Section 7, and we conclude in Section 8.

2. Auxiliary Facts and Results

In this section, we present some fundamental definitions and lemmas that will be used in the subsequent analysis. We make use of X = C ( I , R ) to denote the set of all continuous functions of a real value mapped from I to R . We define the supremum norm   ·   on X as
ω = sup τ I | ω ( τ ) | ,
and the multiplication in X by
( u v ) ( τ ) = u ( τ ) v ( τ ) .
It is clear that X forms a Banach algebra with respect to the supremum norm and the multiplication defined above.
Very recently, the authors of [46] explained the function z ( τ ) by using the N-fold integral method. Much more importantly, the function space is A C δ n , and the algebraic structure of the function is obtained. Next, we provide some definitions for the fractional calculus in the general function space A C δ n , which depends on the general kernel function, z ( τ ) .
Definition 1 
([10,12,14] (General fractional integral)). Suppose ω L 1 [ a , b ] and z C 1 [ a , b ] are strictly increasing functions with z ( a ) 0 . The general fractional integral of order α > 0 is defined as
I a + α , z ω ( τ ) : = 1 Γ ( α ) a τ ( z ( τ ) z ( ς ) ) α 1 z ( ς ) ω ( ς ) d ς ,
where Γ ( α ) is the Euler gamma function, defined by Γ ( α ) = 0 τ α 1 e τ d τ .
The semi-group property of the general fractional integral can hold:
I a + α , z I a + β , z ω ( τ ) = I a + α + β ω ( τ ) , α , β > 0 .
The space A C δ n [ a , b ] is defined by the use of the δ derivative, namely
A C δ n [ a , b ] = ω : [ a , b ] C : ω [ n 1 ] A C [ a , b ] , ω [ n 1 ] = δ n 1 ω , δ : = 1 z ( τ ) d d t .
For further properties and additional definitions of fractional calculus operators, refer to [10,12,14].
To establish the existence of solutions to the nonlinear fractional quadratic integral in Equation (3), we apply the following hybrid fixed-point theorem for three operators in a Banach algebra, X, due to Dhage [4].
Lemma 1. 
Let E be a closed, convex, bounded, and nonempty subset of a Banach algebra, X, and let T 1 , T 3 : X X , and T 2 : E X be three operators satisfying the following conditions:
(i)
T 1 and T 3 are Lipschitz continuous, with Lipschitz constants L 1 and L 2 , respectively;
(ii)
T 2 is compact and continuous;
(iii)
ω = T 1 ω T 2 u + T 3 ω ω E for all u E ;
(iv)
L 1 K + L 2 < 1 , where K =   T 2 ( E ) .
Then, the operator equation T 1 ω T 2 ω + T 3 ω = ω has a solution in E.
Next, to prove the uniqueness of solutions for the nonlinear fractional quadratic integral in Equation (3), we apply the following definition and lemma:
Definition 2 
([1]). A mapping, T, acting in a Banach space, X, is said to be a nonlinear contraction if there exists a contraction, nondecreasing function, ψ, such that ψ ( 0 ) = 0 and ψ ( ω ) < ω for all ω > 0 and that T ω T u   ψ ( ω u ) for all ω , u X .
Lemma 2 
([1]). Let X be a Banach space, and let T : X X be a nonlinear contraction. Then, T has a unique fixed point in X.
To solve our problem, we define the following operator T : X X :
T ω ( τ ) = f ( τ , ω ( τ ) ) + g ( τ , ω ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) d ς .
Thus, solving Equation (3) is reduced to finding a fixed point of the operator T.

3. Existence of Solutions

In this section, we employ Lemma 1 to establish the existence of solutions to the nonlinear fractional quadratic integral in Equation (3). To achieve this, we introduce the following assumptions:
(H1)
The functions f : I × R R , g : I × R R { 0 } , h : I × R 2 R are continuous;
(H2)
There exist two positive functions, ϕ 1 and ϕ 2 , with norms ϕ 1 and ϕ 2 , respectively, such that
| f ( τ , ω ) f ( τ , u ) | ϕ 1 ( τ ) | ω u | , | g ( τ , ω ) g ( τ , u ) | ϕ 2 ( τ ) | ω u | ;
(H3)
There exists a function ψ L ( I , [ 0 , ) ) and a nondecreasing, sub-homogeneous (that is, Δ ( L ω ) L Δ ( ω ) for all L 1 and ω [ 0 , ) ) function, Δ : [ 0 , ) [ 0 , ) , such that
| h ( τ , ω , u ) | ψ ( τ ) Δ ( ω + u ) , ( τ , ω , u ) I × R 2 ;
(H4)
There exists a positive constant, R, such that
f 0 + g 0 Λ 1   ϕ 1     ϕ 2 Λ R ,
and
ϕ 2 Λ +   ϕ 1   < 1 ,
where f 0 = sup τ I f ( τ , 0 ) , g 0 = sup τ I g ( τ , 0 ) and
Λ = ψ Γ ( α + 1 ) Δ 1 + 1 Γ ( β + 1 ) R .
Now, we present the main results of this section as follows:
Theorem 1. 
If assumptions (H1)–(H4) are satisfied, then Equation (3) has at least one solution defined on I.
Proof. 
Let
E = { ω X | ω   R } .
Clearly, E is a closed, convex, and bounded subset of X. Define the three operators T 1 , T 3 : X X , and T 2 : E X as follows:
T 1 ω ( τ ) = g ( τ , ω ( τ ) ) , T 3 ω ( τ ) = f ( τ , ω ( τ ) ) , τ I ,
and
T 2 ω ( τ ) = 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) d ς , τ I .
This allows us to rewrite the integral in Equation (3) as
ω ( τ ) = T 1 ω ( τ ) T 2 ω ( τ ) + T 3 ω ( τ ) , τ I .
To prove the existence of solutions, we need to verify that the operators T i ( i = 1 , 2 , 3 ) satisfy all the conditions of Lemma 1. The proof is structured into four steps:
Step 1: Prove that the operators T 1 and T 3 satisfy the Lipschitz condition (Lemma 1 (i)). By employing condition (H2), we have
| T 1 ω ( τ ) T 1 u ( τ ) | = | g ( t , ω ( τ ) ) g ( τ , u ( τ ) ) | ϕ 2 ( τ ) | ω u |
for all ω , u X and τ I . By taking the supremum over τ for the above inequality, we obtain
T 1 ω T 1 u     ϕ 2 ω u ,
indicating that operator T 1 is Lipschitzian on X with the Lipschitz constant ϕ 2 .
Similarly, for T 3 , we have
T 3 ω T 3 u     ϕ 1 ω u ,
which means that operator T 3 is also Lipschitzian on X with the Lipschitz constant ϕ 1 .
Step 2: Prove that T 2 is completely continuous. By applying condition (H1), it is straightforward to see that T 2 is continuous.
Next, we prove that T 2 is uniformly bounded on E. For any ω E , when using Definition 1, we have
0 | ω ( τ ) |   ω   R , | I 0 + β , z ω ( ς ) |   ω 0 τ ( z ( τ ) z ( ς ) ) β 1 Γ ( β ) z ( ς ) d ς R Γ ( β + 1 ) ( z ( τ ) ) β R Γ ( β + 1 ) .
When using (H3), for ω E , we then find
| h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) | ψ ( ς ) Δ 1 + 1 Γ ( β + 1 ) R .
Therefore,
T 2 ω ( τ ) = 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) d ς ψ Δ 1 + 1 Γ ( β + 1 ) R 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) d ς ψ Γ ( α + 1 ) Δ 1 + 1 Γ ( β + 1 ) R ,
which means that T 2 ω   Λ for all ω E with Λ defined by (9). Thus, the operator T 2 is uniformly bounded on E.
Finally, we aim to demonstrate that T 2 ( E ) forms an equicontinuous set in X. Let τ 1 and τ 2 I with τ 1 < τ 2 and ω E ; we have
| T 2 ω ( τ 2 ) T 2 ω ( τ 1 ) | = 0 τ 1 ( z ( τ 2 ) z ( ς ) ) α 1 ( z ( τ 1 ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) | h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) | d ς + τ 1 τ 2 ( z ( τ 2 ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) d ς .
By using the bounds in condition (H3), this can be estimated as
| T 2 ω ( τ 2 ) T 2 ω ( τ 1 ) | ψ ( z ( τ 2 ) α z ( τ 1 ) α ) Γ ( α + 1 ) Δ 1 + 1 Γ ( β + 1 ) R
As τ 2 τ 1 , this expression tends to zero, showing that T 2 ( E ) is equicontinuous. By applying the Ascoli-Arzelà theorem, it follows that T 2 is completely continuous on E. Thus, condition (ii) of Lemma 1 is satisfied.
Step 3: Verify condition (iii) of Lemma 1. For any ω and u E such that ω = T 1 ω T 2 u + T 3 ω , according to (H2) and (H3), we then have
| ω ( τ ) | = | T 1 ω ( τ ) T 2 u ( τ ) + T 3 ω ( τ ) | | T 1 ω ( τ ) | | T 2 u ( τ ) | + | T 3 ω ( τ ) | ( | g ( τ , ω ( τ ) ) g ( τ , 0 ) | + | g ( τ , 0 ) | ) | T 2 u ( τ ) | + | f ( τ , ω ( t ) ) f ( τ , 0 ) | + | f ( τ , 0 ) | ( ϕ 2 | ω ( τ ) | + g 0 ) Λ +   ϕ 1 | ω ( τ ) | + f 0 .
This leads to
| ω ( τ ) | f 0 + g 0 Λ 1   ϕ 1     ϕ 2 Λ .
By taking the supremum over τ , we have
ω   f 0 + g 0 Λ 1   ϕ 1     ϕ 2 Λ R .
Step 4: Verify condition (iv) of Lemma 1. In view of
K =   T 2 ( E )   = sup ω E sup τ I | T 2 ω ( τ ) | Λ .
We have
ϕ 2 K + ϕ 1     ϕ 2 Λ +   ϕ 1   1 .
Here, L 1 = ϕ 2 and L 2 = ϕ 1 .
Since all conditions of Lemma 1 are satisfied, we conclude that operator equation ω = T 1 ω T 2 ω + T 3 ω has at least one solution in E. Thus, problem (3) has a solution on I. □
Corollary 1. 
Given the conditions of Theorem 1, when z ( τ ) = τ r ( r > 0 ) , Equation (3) simplifies to
ω ( τ ) = f ( τ , ω ( τ ) ) + g ( τ , ω ( τ ) ) 0 τ ( τ r ς r ) α 1 r α 1 ς r 1 Γ ( α ) h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) d ς .
Then, Equation (7) has a solution on I.
Proof. 
The proof process is similar to that of Theorem 1, and we omit it here. □

4. Uniqueness and Continuous Dependence of Solutions

In this section, we explore the uniqueness of the solution to Equation (3) and its continuous dependence using the auxiliary function ψ , as defined in (10).
For convenience, we define the following quantities
P = max τ I 0 τ ( z ( τ ) z ( ς ) ) α 1 z ( ς ) p ( ς ) d ς , Q = max τ I 0 τ ( z ( τ ) z ( ς ) ) α 1 z ( ς ) q ( ς ) d ς ,
where p ( ς ) , q ( ς ) , g m , and h m are defined in Theorem 2.
Additionally, the constant λ is determined by the equation
λ = ( λ + 1 ) ϕ 1   + g m Γ ( α ) P + Q Γ ( 1 + β ) + h m ϕ 2 ( 1 + λ ) Γ ( 1 + α ) .
Theorem 2. 
Assume that conditions (H1) and (H2) hold. Further, assume that functions g and h satisfy the following conditions:
(H5) 
There exists non-negative continuous functions p ( τ ) and q ( τ ) such that
| h ( τ , ω 1 , u 1 ) h ( τ , ω 2 , u 2 ) | p ( τ ) | ω 1 ω 2 | 1 + λ + q ( τ ) | u 1 u 2 | 1 + λ ,
where τ I , ω i and u i R , i = 1 , 2 ; the positive constant λ is defined in (9).
(H6) 
The functions g : I × R R and h : I × R 2 R are bounded with
g m = sup ( τ , ω ) I × R | g ( τ , ω ) | , h m = sup ( τ , ω , u ) I × R 2 | h ( τ , ω , u ) | .
Then, Equation (3) has a unique solution defined on I.
Proof. 
We define the operator T as in (5) and introduce the function
ψ ( u ) = λ u 1 + λ , u 0 .
It is clear that ψ is a nondecreasing function with ψ ( 0 ) = 0 and ψ ( u ) < u for all u > 0 .
For any ς I , ω , and u X , from (H5), we have
h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) p ( ς ) | ω ( ς ) u ( ς ) | 1 + λ + q ( ς ) | I 0 + β , z ω ( ς ) I 0 + β , z u ( ς ) | 1 + λ p ( ς ) 1 + λ + q ( ς ) Γ ( 1 + β ) ω u .
Thus, we can bound | T ω ( τ ) T u ( τ ) | as follows:
| T ω ( τ ) T u ( τ ) | | f ( τ , ω ( τ ) ) f ( τ , u ( τ ) ) | + | g ( τ , ω ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) d ς g ( τ , ω ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) d ς | + | g ( τ , ω ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) d ς g ( τ , u ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) d ς | | f ( τ , ω ( τ ) ) f ( τ , u ( τ ) ) | + | g ( τ , ω ( τ ) ) g ( τ , u ( τ ) ) | 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) | h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) | d ς + | g ( τ , ω ( τ ) ) | 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) | h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) | d ς
ϕ 1 ( τ ) | ω ( τ ) u ( τ ) | + g m Γ ( α ) 0 τ ( z ( τ ) z ( ς ) ) α 1 z ( ς ) ( p ( ς ) | ω ( ς ) u ( ς ) | 1 + λ + q ( ς ) | I 0 + β , z ω ( ς ) I 0 + β , z u ( ς ) | 1 + λ ) d ς + h m Γ ( α + 1 ) ϕ 2 ( τ ) | ω ( τ ) u ( τ ) | ( λ + 1 ) ϕ 1 λ + 1 + g m ( 1 + λ ) Γ ( α ) P + Q Γ ( 1 + β ) + h m ϕ 2 ( 1 + λ ) Γ ( 1 + α ) ( 1 + λ ) ω u = ( λ + 1 ) ϕ 1 λ + g m λ Γ ( α ) P + Q Γ ( 1 + β ) + h m ϕ 2 ( 1 + λ ) Γ ( 1 + α ) λ ψ ( ω u ) λ 1 ψ ( ω u ) λ = ψ ( ω u ) , ( by ( 9 ) )
i.e., T ω T u   ψ ( ω u ) . Therefore, T is a nonlinear contraction. By applying Lemma 2, we conclude that T has a unique fixed point in X, which corresponds to the unique solution of Equation (3). □
Corollary 2. 
Given the conditions of Theorem 2, then Equation (7) has a solution on I.
Proof. 
The proof process is similar to that of Theorem 2, and we omit it here. □
Theorem 3. 
The unique solution obtained in Theorem 2 depends continuously on the given functions f ( τ , ω ( τ ) ) , g ( τ , ω ( τ ) ) , and h ( τ , ω ( τ ) , I 0 + β , z ω ( τ ) ) .
Proof. 
Let functions f , f ˜ , g , g ˜ , and h , h ˜ satisfy all the conditions in Theorem 2. Let T and T ˜ denote the corresponding operators, as defined in (5). From Theorem 2, we know that there are unique corresponding solutions ω ( τ ) and ω ˜ ( τ ) such that
T ω = ω , T ˜ ω ˜ = ω ˜ .
Now, we estimate the difference, ω ω ˜ ,
ω ω ˜   =   T ω T ˜ ω ˜     T ω T ω ˜ + T ω ˜ T ˜ ω ˜   λ ω ω ˜ 1 + λ + T ω ˜ T ˜ ω ˜ .
Thus,
ω ω ˜   ( λ + 1 ) T ω ˜ T ˜ ω ˜ .
Next, we calculate the difference, | T ω ˜ ( τ ) T ˜ ω ˜ ( τ ) | ,
| T ω ˜ ( τ ) T ˜ ω ˜ ( τ ) | | f ( τ , ω ˜ ( τ ) ) f ˜ ( τ , ω ˜ ( τ ) ) | + | g ( τ , ω ˜ ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , ω ˜ ( ς ) , I 0 + β , z ω ˜ ( ς ) ) d ς g ˜ ( τ , ω ˜ ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , ω ˜ ( ς ) , I 0 + β , z ω ˜ ( ς ) ) d ς | + | g ˜ ( τ , ω ˜ ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , ω ˜ ( ς ) , I 0 + β , z ω ˜ ( ς ) ) d ς g ˜ ( τ , ω ˜ ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ˜ ( ς , ω ˜ ( ς ) , I 0 + β , z ω ˜ ( ς ) ) d ς | f f ˜   + h m g g ˜ Γ ( α + 1 ) + g ˜ m h h ˜ Γ ( α + 1 ) 1 + h m + g ˜ m Γ ( α + 1 ) max f f ˜ , g g ˜ , h h ˜ .
Hence, based on inequality (11), we derive the estimate
ω ω ˜   ( λ + 1 ) 1 + h m + g ˜ m Γ ( α + 1 ) max f f ˜ , g g ˜ , h h ˜ .
This concludes the proof. □
Corollary 3. 
The unique solution obtained in Corollary 2 depends continuously on the given functions f ( τ , ω ( τ ) ) , g ( τ , ω ( τ ) ) , and h ( τ , ω ( τ ) , I 0 + β , z ω ( τ ) ) .
Proof. 
The proof process is similar to that of Corollary 2, and we omit it here. □

5. Stability

In this section, we study the Hyers-Ulam-Rassias stability and the Hyers-Ulam stability of Equation (3) on the compact interval [ 0 , 1 ] . To achieve this, we first present the definitions of Hyers-Ulam-Rassias stability and Hyers-Ulam stability.
Definition 3. 
Let ω be a function expressed as
| ω ( τ ) f ( τ , ω ( τ ) ) g ( τ , ω ( τ ) ) I 0 + α , z h ( τ , ω ( τ ) , I 0 + β , z ω ( τ ) ) | ϵ η ( τ ) ,
where ϵ > 0 , and suppose there exists a solution, ω 0 , to Equation (3) and a positive constant c (independent of both ω and ω 0 ) such that
| ω ( τ ) ω 0 ( τ ) | c ϵ η ( τ ) ,
for τ I . Then, Equation (3) is called Hyers-Ulam-Rassias stable. If η ( τ ) is a constant function, Equation (3) is referred to as Hyers-Ulam stable.
To prove the stability of Equation (3), we introduce the following definitions and theorems.
Definition 4 
([47]). Let X be a nonempty set. A function, d : X × X [ 0 , + ] , is called a generalized metric on X if it satisfies the following conditions:
(i)
d ( ω , u ) = 0 if ω = u ;
(ii)
d ( ω , u ) = d ( u , ω ) for all ω , u X ;
(iii)
d ( ω , u ) d ( ω , v ) + d ( v , u ) for all ω , u , v X .
Unlike a standard metric space, a generalized metric space does not require all pairs of points in X to have a finite distance. Therefore, it is referred to as a generalized complete metric space.
Definition 5 
([48]). Let ( X , d ) be a metric space. A mapping, T : X X , satisfies a Lipschitz condition with a Lipschitz constant, L > 0 , if
d ( T ω , T u ) L d ( ω , u ) f o r   a l l ω , u X .
If the Lipschitz constant L < 1 is present, T is called a strictly contractive mapping.
Theorem 4 
([49]). Let ( X , d ) be a complete generalized metric space and T : X X be strictly contractive operator with the Lipschitz constant, L < 1 . If there exists a non-negative integral, k, such that d ( T k + 1 ω , T k ω ) < for some ω X , then the following are true:
(i)
The sequence { T k ω } converges to a fixed point ω * of T;
(ii)
ω * is the unique fixed point of T in the set X * = { u X | d ( T k ω * , u ) < } ;
(iii)
d ( u , ω * ) 1 1 L d ( T u , v ) for all u X * .
Theorem 5. 
Assume that conditions (H1), (H2), and (H6) hold. There exist non-negative continuous functions ϕ 3 ( · ) and ϕ 4 ( · ) such that
| h ( τ , ω 1 , u 1 ) h ( τ , ω 2 , u 2 ) | ϕ 3 ( τ ) | ω 1 ω 2 | + ϕ 4 ( τ ) | u 1 u 2 |
for all τ I , ω i , u i R , i = 1 , 2 . Furthermore, assume that
| ω ( τ ) f ( t , ω ( τ ) ) g ( τ , ω ( τ ) ) I 0 + α , z h ( τ , ω ( τ ) , I 0 + β , z ω ( τ ) ) | ϵ
and
0 < ϕ 1   + ϕ 2 h m Γ ( α + 1 ) + g m Γ ( α + 1 ) ϕ 3   +   ϕ 4 Γ ( β + 1 ) = L < 1 .
Then, the fractional quadratic integral in Equation (3) is Hyers-Ulam stable.
Proof. 
To prove stability, we introduce appropriate metrics in a Banach space, X, as follows:
d ( ω , u ) = inf { L [ 0 , ] : | ω ( τ ) u ( τ ) | L ϵ , τ I } .
According to Definition 4, we know that ( X , d ) is a generalized complete metric space. From (12) and metric d ( · ) , for any ς I , ω , u R , and we have
| h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) ϕ 3 ( ς ) | ω ( ς ) u ( ς ) | + ϕ 4 ( ς ) | I 0 + β , z ω ( ς ) I 0 + β , z u ( ς ) | ϕ 3 L ϵ +   ϕ 4 z β ( τ ) L ϵ Γ ( β + 1 ) ϕ 3   + ϕ 4 Γ ( β + 1 ) L ϵ .
Then, from the definition of T and conditions (H1), (H2), and (H6), we have
| T ω ( τ ) T u ( τ ) | = | f ( τ , ω ( τ ) ) f ( τ , u ( τ ) ) + g ( τ , ω ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) d ς g ( τ , u ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) d ς | | f ( τ , ω ( τ ) ) f ( τ , u ( τ ) ) | + | g ( τ , ω ( τ ) ) g ( τ , u ( τ ) ) | 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) | h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) | d ς + | g ( τ , ω ( τ ) ) | 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) | h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) | d ς
ϕ 1 ( ς ) | ω ( τ ) u ( τ ) | + ϕ 2 ( τ ) | ω ( τ ) u ( τ ) | h m Γ ( α ) 0 τ ( z ( τ ) z ( ς ) ) α 1 z ( ς ) d ς + g m Γ ( α ) ϕ 3   + ϕ 4 Γ ( β + 1 ) L ϵ 0 τ ( z ( τ ) z ( ς ) ) α 1 z ( ς ) d ς ϕ 1 L ϵ + ϕ 2 h m Γ ( α + 1 ) L ϵ + g m Γ ( α + 1 ) ϕ 3   + ϕ 4 Γ ( β + 1 ) L ϵ ϕ 1   + ϕ 2 h m Γ ( α + 1 ) + g m Γ ( α + 1 ) ϕ 3   + ϕ 4 Γ ( β + 1 ) L ϵ .
In view of 0 < L < 1 , we conclude that T is a contraction mapping. Let ω 0 X . Due to the continuous property of ω 0 X and T ω 0 X , there exists a positive constant, C 1 < , such that
| T ω 0 ( τ ) ω 0 ( τ ) | = | f ( t , ω 0 ( τ ) ) + g ( τ , ω 0 ( τ ) ) I 0 + α , z h ( τ , ω 0 ( τ ) , I 0 + β , z ω 0 ( τ ) ) ω 0 ( τ ) | C 1 ϵ
for any τ I . Thus, d ( T ω 0 , ω 0 ) < . Therefore, Theorem 4 (i) implies that there exists a continuous function, ω 0 : I R , such that T n ω 0 ω 0 in ( X , d ) as n , with ω 0 = T ω 0 , where ω 0 satisfies (3) for any τ I . If ω X , then both ω 0 and ω are continuous functions defined on a compact interval, I. Thus, there exists a constant, C 2 > 0 , such that
| ω ( τ ) ω 0 ( τ ) | C 2 ϵ
for any τ I . This implies that d ( ω , ω 0 ) < for every ω X , or equivalently, { ω X : d ( ω , ω 0 ) < } = X . Hence, according to Theorem 4 (ii), ω 0 is a unique continuous function with property (3). Furthermore, from (13), we have d ( T ω , ω ) ϵ for all τ I . Thus,
d ( ω , ω 0 ) 1 1 L d ( T ω , ω ) 1 1 L ϵ
This shows that the fractional quadratic integral in Equation (3) is Hyers-Ulam stable. □
Theorem 6. 
Assume that conditions (H1), (H2), (H6), and (12) hold. Furthermore, assume that
| ω ( τ ) f ( τ , ω ( τ ) ) g ( τ , ω ( τ ) ) I 0 + α , z h ( τ , ω ( τ ) , I 0 + β , z ω ( τ ) ) | ϵ η ( τ )
and
0 < ϕ 1   + ϕ 2 h m Γ ( α + 1 ) + g m Γ ( α ) ϕ 3   + ϕ 4 Γ ( β + 1 ) M C 3 = L < 1 ,
where η ( τ ) satisfies
0 τ ( η ( ς ) ) 1 p d ς p C 3 η ( τ ) a n d M = max τ I 0 τ ( z ( τ ) z ( ς ) ) 1 α z ( ς ) 1 q d ς q ,
with 1 p + 1 q = 1 , and C 3 is a positive constant. Then, the fractional quadratic integral in Equation (3) is Hyers-Ulam-Rassias stable.
Proof. 
For Hyers-Ulam-Rassias stability, one can introduce appropriate metrics in a Banach space, X, as follows:
d ( ω , u ) = inf { L [ 0 , ] : | ω ( τ ) u ( τ ) | L ϵ η ( τ ) , τ I } .
Clearly, ( X , d ) is a generalized complete metric space. From (12) and metric d ( · ) , for any ς I , ω , u R , we have
| h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) ϕ 3 ( ς ) | ω ( ς ) u ( ς ) | + ϕ 4 ( ς ) | I 0 + β , z ω ( ς ) I 0 + β , z u ( ς ) | ϕ 3 L ϵ η ( ς ) +   ϕ 4 z β ( τ ) L ϵ η ( ς ) Γ ( β + 1 ) ϕ 3   + ϕ 4 Γ ( β + 1 ) L ϵ η ( ς ) .
Then, from the definition of T and conditions (H1), (H2), and (H6), we have
| T ω ( τ ) T u ( τ ) | = | f ( τ , ω ( τ ) ) f ( τ , u ( τ ) ) + g ( τ , ω ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) d ς g ( τ , u ( τ ) ) 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) d ς | | f ( τ , ω ( τ ) ) f ( τ , u ( τ ) ) | + | g ( τ , ω ( τ ) ) g ( τ , u ( τ ) ) | 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) | h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) | d ς + | g ( τ , ω ( τ ) ) | 0 τ ( z ( τ ) z ( ς ) ) α 1 Γ ( α ) z ( ς ) | h ( ς , ω ( ς ) , I 0 + β , z ω ( ς ) ) h ( ς , u ( ς ) , I 0 + β , z u ( ς ) ) | d ς ϕ 1 ( ς ) | ω ( τ ) u ( τ ) | + ϕ 2 ( τ ) | ω ( τ ) u ( τ ) | h m Γ ( α ) 0 τ ( z ( τ ) z ( ς ) ) α 1 z ( ς ) d ς + g m Γ ( α ) ϕ 3   + ϕ 4 Γ ( β + 1 ) L ϵ 0 τ ( z ( τ ) z ( ς ) ) α 1 z ( ς ) η ( ς ) d ς ϕ 1 L ϵ η ( τ ) + ϕ 2 h m Γ ( α + 1 ) L ϵ η ( τ ) + g m Γ ( α ) ϕ 3   + ϕ 4 Γ ( β + 1 ) L ϵ 0 τ ( z ( τ ) z ( ς ) ) 1 α z ( ς ) 1 q d ς q × 0 τ ( η ( ς ) ) 1 p d ς p ϕ 1   + ϕ 2 h m Γ ( α + 1 ) + g m Γ ( α ) ϕ 3   + ϕ 4 Γ ( β + 1 ) M C 3 L ϵ η ( τ ) .
In view of 0 < L < 1 , we conclude that T is a contraction mapping. Let ω 0 X . Due to the continuous property of ω 0 X and T ω 0 X , there exists a positive constant, C 4 < , such that
| T ω 0 ( τ ) ω 0 ( τ ) | = | f ( τ , ω 0 ( τ ) ) + g ( τ , ω 0 ( τ ) ) I 0 + α , z h ( τ , ω 0 ( τ ) , I 0 + β , z ω 0 ( τ ) ) ω 0 ( τ ) | C 4 ϵ η ( τ )
for any τ I . Thus, d ( T ω 0 , ω 0 ) < . Therefore, according to Theorem 4 (i), there exists a continuous function, ω 0 : I R , such that T n ω 0 ω 0 in ( X , d ) , as n , and ω 0 = T ω 0 , where ω 0 satisfies (3) for any τ I . If ω X , then both ω 0 and ω are continuous functions defined on a compact interval, I. Thus, there exists a constant, C 5 > 0 , such that
| ω ( τ ) ω 0 ( τ ) | C 5 ϵ η ( τ )
for any τ I . This implies that d ( ω , ω 0 ) < for every ω X , or equivalently, { ω X : d ( ω , ω 0 ) < } = X . Hence, according to Theorem 4 (ii), ω 0 is a unique continuous function with property (3). Additionally, from Equation (14), we have
d ( T ω ( τ ) , ω ( τ ) ) ϵ η ( τ ) for   all τ I .
Thus,
d ( ω , ω 0 ) 1 1 L d ( T ω , ω ) 1 1 L ϵ η ( τ ) .
This shows that the fractional quadratic integral in Equation (3) is Hyers-Ulam stable. □
Remark 1. 
Similarly to Theorems 5 and 6, we can also prove that Equation (7) exhibits both Hyers-Ulam and Hyers-Ulam-Rassias stability.

6. Existence of Maximal and Minimal Solutions

This section demonstrates the existence of maximal and minimal solutions to Equation (3) in I, taking advantage of fractional integral inequalities.
Definition 6 
([50]). A solution, ω ¯ ( τ ) , of (3) is referred to as the maximal solution if all other solutions, ω ( τ ) , on I satisfy
ω ( τ ) ω ¯ ( τ ) , τ I .
Similarly, ω ̲ ( τ ) of (3) is the minimal solution if all other solutions, ω ( τ ) , on I satisfy
ω ( τ ) ω ̲ ( τ ) , τ I .
We begin by introducing a fundamental result for strict fractional integral inequalities related to Equation (3).
Lemma 3. 
Assume that f ( τ , ω ) and g ( τ , ω ) are nondecreasing functions in ω, and h ( τ , ω , u ) is a nondecreasing function in ω and u. If the assumptions of Theorem 1 hold, let ω ( τ ) and ϖ ( τ ) be continuous functions on I, satisfying
ω ( τ ) f ( τ , ω ( τ ) ) + g ( τ , ω ( τ ) ) I 0 + α , z h ( τ , ω ( τ ) , I 0 + β , z ω ( τ ) ) ,
and
ϖ ( τ ) f ( τ , ϖ ( τ ) ) + g ( τ , ϖ ( τ ) ) I 0 + α , z h ( τ , ϖ ( τ ) , I 0 + β , z ϖ ( τ ) ) ,
with at least one of the inequalities being strict. Then,
ω ( τ ) < ϖ ( τ ) .
Proof. 
Suppose that the inequalities (15) hold as strict inequalities, and suppose ω ( τ ) ϖ ( τ ) for some τ I . Then, there exists τ 1 I , τ 1 > 0 such that ω ( τ 1 ) = ϖ ( τ 1 ) and ω ( τ 1 ) < ϖ ( τ 1 ) for 0 < τ < τ 1 .
Given the monotonicity of f, g, and h, we have
ω ( τ 1 ) f ( τ 1 , ω ( τ 1 ) ) + g ( τ 1 , ω ( τ 1 ) ) I 0 + α , z h ( τ 1 , ω ( τ 1 ) , I 0 + β , z ω ( τ 1 ) ) < f ( τ 1 , ϖ ( τ 1 ) ) + g ( τ 1 , ϖ ( τ 1 ) ) I 0 + α , z h ( τ 1 , ϖ ( τ 1 ) , I 0 + β , z ϖ ( τ 1 ) ) < ϖ ( τ 1 ) .
This contradiction is ω ( τ 1 ) = ϖ ( τ 1 ) , establishing the truth of (17). □
For any small positive ϵ , consider the auxiliary equation
ω ( τ ) = f ϵ ( τ , ω ( τ ) ) + g ϵ ( τ , ω ( τ ) ) I 0 + α , z h ϵ ( τ , ω ( τ ) , I 0 + β , z ω ( τ ) ) ,
where
f ϵ ( τ , ω ( τ ) ) = f ( τ , ω ( τ ) ) + ϵ , g ϵ ( τ , ω ( τ ) ) = g ( τ , ω ( τ ) ) + ϵ ,
and
h ϵ ( τ , ω ( τ ) , I 0 + β , z ω ( τ ) ) = h ( τ , ω ( τ ) , I 0 + β , z ω ( τ ) ) + ϵ .
Theorem 7. 
Under the conditions of Theorem 1, for any small positive ϵ, Equation (18) admits a solution on I.
Proof. 
In view of ψ 2 Λ +   ψ 1   < 1 , there exists ϵ 0 > 0 such that
ψ 2 Λ + ϵ Γ ( α + 1 ) +   ψ 1 < 1 ,
for all ϵ < ϵ 0 . The remaining argument mirrors the proof of Theorem 1. □
Next, we will use Lemma 3 and Theorem 7 to prove the existence of maximal and minimal solutions to (3).
Theorem 8. 
Assume that f ( τ , ω ) and g ( τ , ω ) are nondecreasing functions in ω, and h ( τ , ω , u ) is a nondecreasing function in ω and u. Under the conditions of Theorem 1, Equation (3) has maximal and minimal solutions on I.
Proof. 
We focus on proving the existence of a maximal solution to Equation (3), as the argument for the minimal solution follows a similar process.
Consider a decreasing sequence, { ϵ } n , of positive real numbers such that
lim n ϵ n = 0 ,
and ϵ 0 > 0 satisfies
ψ 2 Λ + ϵ 0 Γ ( α + 1 ) +   ψ 1   < 1 .
Define
f ϵ n ( τ , ω ( τ ) ) = f ( τ , ω ( τ ) ) + ϵ n , g ϵ n ( τ , ω ( τ ) ) = g ( τ , ω ( τ ) ) + ϵ n ,
and
h ϵ n ( τ , ω ( τ ) , I 0 + β , z ω ( τ ) ) = h ( τ , ω ( τ ) , I 0 + β , z ω ( τ ) ) + ϵ n .
According to Theorem 7, the auxiliary equation
ω ( τ ) = f ϵ n ( τ , ω ( τ ) ) + g ϵ n ( τ , ω ( τ ) ) I 0 + α , z h ϵ n ( τ , ω ( τ ) , I 0 + β , z ω ( τ ) ) ,
admits a solution, denoted as r ( τ , ϵ n ) .
For any such solution, it holds that
r ( τ , ϵ n ) = f ϵ n ( τ , r ( τ , ϵ n ) ) + g ϵ n ( τ , r ( τ , ϵ n ) ) I 0 + α , z h ϵ n ( τ , r ( τ , ϵ n ) , I 0 + β , z r ( τ , ϵ n ) ) ,
where
f ϵ n ( τ , r ( τ , ϵ n ) ) = f ( τ , r ( τ , ϵ n ) ) + ϵ n , g ϵ n ( τ , r ( τ , ϵ n ) ) = g ( τ , r ( τ , ϵ n ) ) + ϵ n ,
and
h ϵ n ( τ , r ( τ , ϵ n ) , I 0 + β , z r ( τ , ϵ n ) ) = h ( τ , r ( τ , ϵ n ) , I 0 + β , z r ( τ , ϵ n ) ) + ϵ n .
By definition,
r ( τ , ϵ n ) f ( τ , r ( τ , ϵ n ) ) + g ( τ , r ( τ , ϵ n ) ) I 0 + α , z h ( τ , r ( τ , ϵ n ) , I 0 + β , z r ( τ , ϵ n ) ) .
For any solution, ω ( τ ) , to (3) that satisfies (15), Lemma 3 implies
ω ( τ ) < r ( τ , ϵ n ) .
Consider 0 < ϵ 2 < ϵ 1 < ϵ . From the definition,
r ( τ , ϵ 1 ) = f ϵ 1 ( τ , r ( τ , ϵ 1 ) ) + g ϵ 1 ( τ , r ( τ , ϵ 1 ) ) + I 0 + α , z h ϵ 1 ( τ , r ( τ , ϵ 1 ) , I 0 + β , z r ( τ , ϵ 1 ) ) = f ( τ , r ( τ , ϵ 1 ) + ϵ 1 + ( g ( τ , r ( τ , ϵ 1 ) ) + ϵ 1 ) I 0 + α , z ( h ( τ , r ( τ , ϵ 1 ) , I 0 + β , z r ( τ , ϵ 1 ) ) + ϵ 1 ) f ( τ , r ( τ , ϵ 1 ) + ϵ 2 + ( g ( τ , r ( τ , ϵ 1 ) ) + ϵ 2 ) I 0 + α , z ( h ( τ , r ( τ , ϵ 1 ) , I 0 + β , z r ( τ , ϵ 1 ) ) + ϵ 2 ) , r ( τ , ϵ 2 ) = f ( τ , r ( τ , ϵ 2 ) + ϵ 2 + ( g ( τ , r ( τ , ϵ 2 ) ) + ϵ 2 ) I 0 + α , z ( h ( τ , r ( τ , ϵ 2 ) , I 0 + β , z r ( τ , ϵ 2 ) ) + ϵ 2 ) .
By applying Lemma 3, from (22), we deduce
r ( τ , ϵ 1 ) r ( τ , ϵ 2 ) .
Thus, we infer that r ( τ , ϵ n ) is a decreasing sequence. Thus, its limit
lim n r ( τ , ϵ n ) = ω ¯ ( τ )
exists. Based on condition (H1), it can also be demonstrated that the convergence of r ( τ , ϵ n ) is uniform on I. Since r ( τ , ϵ n ) satisfies Equation (19), we have
r ( τ , ϵ n ) = f ϵ n ( τ , r ( τ , ϵ n ) ) + g ϵ n ( τ , r ( τ , ϵ n ) ) I 0 + α , z h ϵ n ( τ , r ( τ , ϵ n ) , I 0 + β , z r ( τ , ϵ n ) ) .
By taking the limit as n , it follows that
ω ¯ ( τ ) = f ( τ , ω ¯ ( τ ) ) + g ( τ , ω ¯ ( τ ) ) I 0 + α , z h ( τ , ω ¯ ( τ ) , I 0 + β , z ω ¯ ( τ ) )
which confirms that ω ¯ ( τ ) is, indeed, a solution to Equation (3). Thus, by combining this result with inequality (21) and the uniqueness of the maximal solution, as established in [50], we conclude that Equation (3) admits a maximal solution on I. □
Remark 2. 
Similarly to Theorem 8, we can also prove that Equation (7) has a maximal solution and a minimal solution on the interval I.

7. Examples

Example 1. 
In (3), let α = β = 0.5 , z ( τ ) = τ , and define the following functions:
f ( τ , ω ) = 3 τ 2 100 1 2 ( ω + 1 + ω 2 ) g ( τ , ω ) = 1 + τ 8 + π e 2 π τ cos ( π τ ) ( 6 π + 13 e τ ) 2 ω 1 + ω , h ( τ , ω , I 0 + 0.5 ω ) = 1 30 + τ 4 + 2 ω ( τ ) sin ( ω ) + π I 0 + 0.5 ω .
It is clear that condition (H1) is true. Upon computation, the following inequalities hold:
| f ( τ , ω ) f ( τ , u ) | 3 τ 2 100 | ω u | , | g ( τ , ω ) g ( τ , u ) | π e 2 π τ ( 6 π + 13 e τ ) 2 | ω u | , h ( τ , ω , I 0 + 0.5 ω ) 4 30 + τ ( 1 +   ω ) ,
where
ϕ 1 ( τ ) = 3 τ 2 100 , ϕ 2 ( τ ) = π e 2 π τ ( 6 π + 13 e τ ) 2 , ψ ( τ ) = 4 30 + τ , Δ ( ω ) = 1 + ω .
Further calculations yield
ϕ 1   = 3 100 , ϕ 2   = π ( 6 π + 13 ) 2 , ψ = 2 15 ,
and
f 0 = 3 200 , g 0 = 1 4 , Λ = 2 15 Γ ( 1.5 ) 1 + 2 + Γ ( 1.5 ) 1 + Γ ( 1.5 ) R .
Using computational tools, condition (H4) determines that the constant R satisfies the inequality 10 < R < 21,000. Therefore, all the conditions of Theorem 1 are satisfied, ensuring that Example 1 has at least one solution on I.
Example 2. 
In (3), let α = β = 0.5 , z ( τ ) = τ , and define the following functions:
f ( τ , ω ) = τ ω 4 + τ 2 , g ( τ , ω ) = sin ω + 1 200 ( 1 + τ 2 ) h ( τ , ω , u ) = 1 + τ 2 + sin ω 1 + λ + cos u 1 + λ .
Obviously, condition (H1) is true. By using computation, we have
| f ( τ , ω ) f ( τ , u ) | τ 4 + τ 2 | ω u | , | g ( τ , ω ) g ( τ , u ) | 1 200 ( 1 + τ 2 ) | ω u | , | h ( τ , ω 1 , u 1 ) h ( τ , ω 2 , u 2 ) | 1 1 + λ | ω 1 ω 2 | + | 1 1 + λ | u 1 u 2 |
and
g m = 1 100 , h m = 2 + 2 1 + λ , p ( τ ) = q ( τ ) = 1 , P = Q = 2 , ϕ 1   = 1 4 , ϕ 2   = 1 200 ,
where λ satisfies the equation
λ = ( λ + 1 ) ϕ 1   + g m Γ ( α ) P + Q Γ ( 1 + β ) + h m ϕ 2 ( 1 + λ ) Γ ( 1 + α ) .
By substituting all the parameters into the equation, it is found that λ is approximately equal to 0.31249. From Theorem 2, we know that Example 2 has a unique solution.
Example 3. 
In Example 2, the function h is replaced by
h ( τ , ω , u ) = h ( τ , ω , u ) = 1 + τ 2 + sin ω 1.31249 + cos u 1.31249 .
Take
ϕ 3   =   ϕ 3   = 1 1.31249 .
It is easy to verify that
( ϕ 1 + ϕ 2 h m Γ ( α + 1 ) + g m Γ ( α + 1 ) ϕ 3   + ϕ 4 Γ ( β + 1 ) ) 0.2895 < 1 .
Therefore, all the conditions of Theorem 5 are satisfied, and the equation is Hyers-Ulam stable.

8. Conclusions

This study establishes sufficient conditions for the existence, uniqueness, stability, and continuous dependence of solutions to, as well as for maximal and minimal solutions of, nonlinear fractional quadratic integral equations involving a general fractional integral. The operator equation T 1 ω T 2 ω + T 3 ω = ω , Boyd and Wang’s fixed-point theorem, and nonlinear functional analysis serve as key tools. Based on the results of this paper, it is evident that when the function z ( τ ) is equal to τ , log τ , or τ r , we can also obtain the corresponding results.
In future work, we will explore extending the results to stochastic fractional quadratic equations with random noise (e.g., Wiener processes) for applications in finance or biological systems.

Author Contributions

Formal analysis, M.Z.; Methodology, M.Z.; Writing-original draft, M.Z.; Writing-review and editing, H.W., Z.X. and L.Z. All authors have read and agreed to the published version of the manuscript.

Funding

This research was funded by the Basic Research Programs of Hengyang Bureau of Science and Technology (No. 202330046131).

Data Availability Statement

Data are contained within the article.

Acknowledgments

The authors express their gratitude to the reviewers for their meticulous evaluation and constructive feedback, which significantly improved this manuscript.

Conflicts of Interest

The authors declare no conflicts of interest.

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Zhou, M.; Wang, H.; Xu, Z.; Zhang, L. The Study of Fractional Quadratic Integral Equations Involves General Fractional Integrals. Fractal Fract. 2025, 9, 249. https://doi.org/10.3390/fractalfract9040249

AMA Style

Zhou M, Wang H, Xu Z, Zhang L. The Study of Fractional Quadratic Integral Equations Involves General Fractional Integrals. Fractal and Fractional. 2025; 9(4):249. https://doi.org/10.3390/fractalfract9040249

Chicago/Turabian Style

Zhou, Mi, Hengtai Wang, Zhenghua Xu, and Lu Zhang. 2025. "The Study of Fractional Quadratic Integral Equations Involves General Fractional Integrals" Fractal and Fractional 9, no. 4: 249. https://doi.org/10.3390/fractalfract9040249

APA Style

Zhou, M., Wang, H., Xu, Z., & Zhang, L. (2025). The Study of Fractional Quadratic Integral Equations Involves General Fractional Integrals. Fractal and Fractional, 9(4), 249. https://doi.org/10.3390/fractalfract9040249

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