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Article

q-Fractional Langevin Differential Equation with q-Fractional Integral Conditions

1
Bell Honors School, Nanjing University of Posts and Telecommunications, Nanjing 210023, China
2
Department of Mathematics, University of Peshawar, Peshawar 25120, Pakistan
3
Faculty of Science and Arts, Mohail Asser, King Khalid University, Abha 61421, Saudi Arabia
4
College of Science, Nanjing University of Posts and Telecommunications, Nanjing 210023, China
*
Author to whom correspondence should be addressed.
Mathematics 2023, 11(9), 2132; https://doi.org/10.3390/math11092132
Submission received: 1 April 2023 / Revised: 29 April 2023 / Accepted: 30 April 2023 / Published: 2 May 2023
(This article belongs to the Special Issue Fractional Calculus and Mathematical Applications)

Abstract

:
The major goal of this manuscript is to investigate the existence, uniqueness, and stability of a q-fractional Langevin differential equation with q-fractional integral conditions. We demonstrate the existence and uniqueness of the solution to the proposed q-fractional Langevin differential equation using the Banach contraction principle and Schaefer’s fixed-point theorem. We also elaborate on different kinds of Ulam stability. The theoretical outcomes are verified by examples.

1. Introduction

The most recently developed field of study is fractional calculus ( FC ). Many scientists from different areas of research have become interested in the topic of FC , which is based on integrals and derivatives of fractional order. It has numerous uses in a broad range of scientific and engineering disciplines, including optics, mathematical biology, plasma physics, electromagnetic theory, and many more. Currently, there are many significant results in the literature regarding the existence and uniqueness ( EU ) of solutions of fractional boundary value problems ( BVP s ) [1,2,3]. Fractional differential equations ( FDE s ) have been successfully used in recent years to model issues that arise in science and industry. The research on FDE s is interdisciplinary and occurs in many areas, including quantum mechanics, bio-engineering, control theory, and complex systems. As global operators, FDE s are also used to describe a variety of systems that have arisen in fields such as fluid mechanics, control theory, and probability theory [4,5,6,7,8,9,10,11]. This field has drawn the interest of many researchers because of its significance and broad variety of applications. Many scholars have developed numerous methods to demonstrate the presence of solutions to challenging fractional BVP s and benefited from various fixed-point theorems. For more on this topic, see [12,13,14,15] in detail.
Fractional q- DE s and q-calculus have recently received a lot of interest from mathematical engineers. q-calculus, also known as quantum calculus, dates back to Jackson’s study in 1908 [16]. The q- DE s are based on q-calculus and can describe some unique physical processes, e.g., in quantum mechanics, dynamical systems, stochastic processes, etc. For a basic introduction to the basic idea of q-calculus, the interested reader may consult [17,18,19]. The stability analysis of DE s in the Ulam sense and their various varieties is another field of research that has drawn significant attention from scholars. In fact, in 1940, Ulam [20] challenged the problem of functional equations, which Hyers [21] enhanced in 1941 using Banach spaces. Because of this reason, this problem led to the notion of stability and is popular as Ulam–Hyers stability ( UHS ). Rassias [22] then extended UHS to a generalized form and called it UH –Rassias stability. Additionally, many scientists have developed a widespread understanding of UHS for various classes of functional equations [23]. Ulam’s type of stability is important for a variety of applications, including optimization, numerical analysis, economics, biology, etc. Numerous mathematicians and technologists have considered Ulam-type stability. For more information on recent investigations of the Ulam-type stability of DE s , see, for the ordinary form, DE s [24], for the delayed form, DE s [25], and for the Caputo form, DE s [26].
Paul Langevin, a French physicist, introduced the Langevin equations ( LE s ) in 1908. The equation has the following form:
m d 2 z d ξ 2 = λ d z d ξ + Φ ( ξ ) .
The fractional Langevin equations ( FLE s ) have been a crucial topic in the field of physics, chemistry, and electrical engineering. In fact, the evolution of physical events in a changing environment is well described by LE . It is important to note that random variations are regarded as Gaussian noise in the LE s description of Brownian motion. In order to eliminate noise and staircase effects, many researchers used FDE to replace the ordinary DE s . The literature widely explores FLE s from both theoretical and numerical perspectives. Readers may consult [27,28] for more information. In generalizing the conventional one, FLE s provides a fractional Gaussian process parameterized by two indices that is more flexible for modeling fractal processes [29,30].
Etemad et al. [31] have studied the fractional q-integro-differential equation:
( λ D q α + ( 1 λ ) D q β ) z ( ξ ) = a f ( ξ , z ( ξ ) ) + b I q δ g ( ξ , z ( ξ ) ) , ξ [ 0 , 1 ] , a , b R +
with BC s
z ( 0 ) = 0 , μ 0 1 ( 1 q s ) ( γ 1 1 ) Γ q ( γ 1 ) z ( s ) d q s , ( 1 μ ) 0 1 ( 1 q s ) ( γ 2 1 ) Γ q ( γ 2 ) z ( s ) d q s = 0 , γ 1 , γ 2 > 0 ,
where q , δ ( 0 , 1 ) , λ ( 0 , 1 ] , μ ( 0 , 1 ] , 1 < α , β < 2 , α β > 1 , D q α denotes the Reimann–Liouville fractional q-derivative of order α , and f and g are appropriate functions.
Rizwan and Zada [32] studied the existence and stability analysis similar to the UH and UHRS of the FLE s by utilizing the generalized Diaz–Margoli’s fixed-point ( F P ) theorem:
C D ξ α ( C D ξ β + λ ) z ( ξ ) = F ( ξ , z ( ξ ) ) , 0 < α , β < 1 ,
where C D ξ α and C D ξ β represents the classical Caputo derivative ( CD ) [2] of order α and β , respectively, and λ > 0 .
Zhou and Liu [33] applied Monch’s FP theorem and a measure of weak non-compactness to analysis the fractional q-difference equation:
C D q α z ( ξ ) = f ( ξ , z ( ξ ) ) , ξ [ 0 , 1 ] , 0 < q < 1 , z ( 0 ) = D q 2 z ( 0 ) = 0 , γ ( D q z ) ( 0 ) + β ( D q 2 z ) ( 1 ) = 0 ,
where α ( 2 , 3 ) , γ , β 0 , and f is an appropriate function.
Alam et al. [34] demonstrated the implicit LE s involving mixed derivatives and Stieltjes IBC s :
C D 0 , ξ α ( D + λ ) z ( ξ ) = f ( ξ , z ( ξ ) ) , ξ ( 0 , 1 ] , z ( 0 ) = 0 , C D 0 , ξ β 0 z ( 1 ) = i = 1 p 0 1 C D 0 , ξ β i z ( ξ ) d μ i ξ ,
where C D 0 , ξ α represents the CD of order α with zero as a lower bound, α ( 0 , 1 ) , λ R \ { 0 } , p N , β i R (∀ i = 0 , , p ), 0 < β 1 , β 2 , , < β p < α , β 0 [ 0 , α ) , f is an appropriate function, and the Riemann–Stieltjes integrals are taken in the BC s with μ i ( i = 1 , , p ) being a function of the bounded variation.
Zada and Waheed [35] studied the class of implicit FDE with implicit IBC s :
C D α z ( ξ ) = f ( ξ , z ( ξ ) , C D α z ( ξ ) ) + 0 ξ ( ξ s ) σ 1 Γ ( δ ) g ( s , z ( s ) , C D α z ( s ) ) d s , ξ J = ( 0 , T ) , α ( 0 , 1 ] , z ( 0 ) = 0 ξ ( T η ) α 1 Γ ( α ) K ( η , z ( η ) , C D α z ( η ) ) d η ,
where C D α is the Caputo fractional derivative of order α on J, f , g , K are appropriate functions, δ > 0 , and σ > 0 .
From the above discussion and models, we are motivated to study the following implicit q- FDE s with FIBC s :
C D q α ( D + λ ) z ( ξ ) = f ( ξ , z ( ξ ) , C D q α z ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z ( s ) , C D q α z ( s ) ) d q s , ξ J = ( 0 , T ) , α ( 1 , 2 ] , z ( 0 ) = 0 T ( T q η ) α 1 Γ q ( α ) K ( η , z ( η ) , C D q α z ( η ) ) d q η , C D q z ( 0 ) = 0 , C D q β 0 z ( T ) = i = 1 p 0 T C D q β i z ( ξ ) d q ξ ,
where C D q α is the Caputo fractional q-derivative of order α on J, D denotes the ordinary differential operator, f , g , K are appropriate functions, λ R \ { 0 } , p N , β i R (∀ i = 0 , , p ) , 0 < β 1 , β 2 , , < β p < α , β 0 [ 0 , α ) , and δ > 0 and σ > 0 are constants. In comparison to the other studies on q-fractional Langevin DE s that hav ebeen published in the literature, we deal with the extended version of a new Caputo q–Langevin FDE with q-fractional integral conditions in which the uniqueness property of the solution is derived in terms of the Banach and Schefer FP theorems. These procedures on the proposed Langevin q- FDE   ( 1 ) have been implemented in research studies on q-fractional modelings. Furthermore, the Ulam stability of FDE is quite significant in the sense that it guarantees a bound between the exact and approximate solutions. Therefore, it can be required in a number of applications, such as optimization, approximation, and numerical analysis. Because of this importance of Ulam stability, we also investigate it for the proposed Langevin q- FDE   ( 1 ) .
The manuscript is organized as follows: In Section 2, we give some basic definitions and theorems associated with both q-fractional derivatives and q-fractional integrals. The existence and uniqueness of a solution for the proposed q-fractional Langevin differential equation is obtained in Section 3 utilizing the Banach contraction principle and Schaefer’s fixed-point theorem. Different kinds of Ulam stability for the considered system are investigated in Section 4. In Section 5, the results are verified with the help of examples.

2. Preliminary

Let J = [ 0 , T ] , X be the space of all continuous functions, i.e., X = { z : J R ; z C ( J , R ) } . Introducing the norm
z X = sup { | z ( ξ ) | : ξ J } ,
it is obvious that X is a Banach space.
Definition 1
([2]). The q - FI of order α from 0 to ξ for z is given by
I q α z ( ξ ) = 1 Γ q ( α ) 0 ξ ( ξ q s ) α 1 z ( s ) d q s , ξ > 0 , α > 0 ,
where Γ ( . ) is the gamma function.
Definition 2
([36]). The Caputo–q- FD of order α 0 for z on J is represented as
C D q α z ( ξ ) = I q α α D q α z ( ξ ) ,
where α is the smallest integer greater than α. Equivalently,
C D q α z ( ξ ) = 1 Γ q ( α ) 0 ξ z ( s ) ( ξ q s ) ( 1 + α ) d q s .
Lemma 1
([37]). Let α > 0 and p N . Then, the following holds:
I q α C D q p z ( ξ ) = C D q p I q α z ( ξ ) k = 0 p 1 ξ α p k Γ q ( α p k + 1 ) C D q k z ( 0 ) .
We recall the following from [19]. Let q ( 0 , 1 ) and define
[ a ] q = q a 1 q 1 = a a 1 + + 1 , a R .
The q-analogue of the power ( a b ) n is
( a b ) ( 0 ) = 1 , ( a b ) ( n ) = k = 0 n 1 ( a b q k ) , a , b R , n N .
If α is not a positive integer, then
( a b ) ( α ) = a α i = o ( 1 ( b a ) q i ) ( 1 ( b a ) q α + i ) .
Note that if b = 0 , then a ( α ) = a α . Using the above notations, we state the following definition.
Definition 3.
The q -Gamma function is defined as
Γ q ( α ) = ( 1 q ) ( α 1 ) ( 1 q ) α 1 , α R \ { 0 , 1 , 2 , } , 0 < q < 1
and satisfies
Γ q ( α + 1 ) = α q Γ q ( α ) .
We also point out the formulas, taken from [18], that will be used in our results.
[ a ( ξ s ) ] ( α ) = a α ( ξ s ) α , ξ D q ( ξ s ) α = α q ( ξ s ) ( α 1 ) .
Theorem 1
([38]). [Theorem of Banach about FP ] Consider a Banach space X and let a map ρ : X X be a contraction on X . Then, ρ has precisely one FP .
Theorem 2
([37]). [Theorem of Arzela–Ascoli] Let M C ( J , R ) ; then, it is relatively compact in C ( J , R )   i f the functions in M are uniformly bounded and equi-continuous on J.
Theorem 3
([37]). [Theorem of Schaefer about FP ] Let X be a Banach space and let the operator ρ : X X be a continuous and compact mapping. Moreover, we assume that
S = { z X | z = ϕ ρ z , 0 < ϕ < 1 }
is bounded. Then, S in X has at least one FP .
From [24], we state the following definition.
Definition 4.
(1) is UHS if there a real number C f , g > 0 in such a way that for each ϵ > 0 and x X satisfying
| C D q α x ( ξ ) f ( ξ , x ( ξ ) , C D q α x ( ξ ) ) 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , x ( s ) , C D q α x ( s ) ) d q s | ϵ , ξ J ,
there a solution z X of (1) with
| x ( ξ ) z ( ξ ) | C f , g ϵ , ξ J .
Definition 5.
(1) is generalized UHS   ( G U H S ) if there a function F f , g C ( R + , R + ) , F f , g ( 0 ) = 0 in such a way that for each solution x X of (3), there a solution z X of (1) with
| x ( ξ ) z ( ξ ) | F f , g ϵ , ξ J .
Definition 6.
(1) is UHRS corresponding to ψ C ( X , R + ) if C f , g , ψ > 0 in such a way that for each x X , satisfying
| C D q α x ( ξ ) f ( ξ , x ( ξ ) , C D q α x ( ξ ) ) 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , x ( s ) , C D q α x ( s ) ) d q s | ϵ ψ ( ξ ) , ξ J ,
there a solution z X of (1) with
| x ( ξ ) z ( ξ ) | C f , g ϵ ψ ( ξ ) , ξ J .
Definition 7.
(1) is GUHRS with respect to a function ψ C ( X , R + ) if there exists C f , g , ψ > 0 such that for each solution x X of
| C D q α x ( ξ ) f ( ξ , x ( ξ ) , C D q α x ( ξ ) ) 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , x ( s ) , C D q α x ( s ) ) d q s | ψ ( ξ ) , ξ J ,
there a solution z X of (1) with
| x ( ξ ) z ( ξ ) | C f , g ψ ( ξ ) , ξ J .
Remark 1.
It is obvious that
1. 
Definition 4 ⇒ Definition 5;
2. 
Definition 6 ⇒ Definition 7.
Remark 2.
x X is a solution of (3) if and only if there a function ψ X depending on x such that
1. 
| ψ ( ξ ) | ϵ , ξ J ;
2. 
C D q α ( D + λ ) x ( ξ ) = f ( ξ , x ( ξ ) , C D q α x ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , x ( s ) , C D q α x ( s ) ) d q s + ψ ( ξ ) , ξ J .
Remark 3.
x X is a solution of (5) if and only if there a functions ψ X depending on x such that
1. 
| ψ ( ξ ) | ϵ ψ ( ξ ) , ξ J ;
2. 
C D q α ( D + λ ) x ( ξ ) = f ( ξ , x ( ξ ) , C D q α x ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , x ( s ) , C D q α x ( s ) ) d q s + ψ ( ξ ) , ξ J .
Lemma 2.
Assume α ( 1 , 2 ] and ω C [ 0 , T ] ; then, the q- FDE
C D q α ( D + λ ) z ( ξ ) = ω ( ξ ) , ξ J = ( 0 , T ) , α ( 1 , 2 ] , z ( 0 ) = 0 T ( T q η ) α 1 Γ q ( α ) K ( η , z ( η ) , C D q α z ( η ) ) d q η , C D q z ( 0 ) = 0 , C D q β 0 z ( T ) = i = 1 p 0 T C D q β i z ( ξ ) d q ξ ,
has a solution given by
z ( ξ ) = 0 T G ω ( ξ , q s ) ω ( s ) d s + 0 T G χ ( ξ , q s ) χ ( s ) d q s ,
where ω X is given by
ω ( ξ ) = f ( ξ , z ( ξ ) , C D q α z ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z ( s ) , C D q α z ( s ) ) d q s
and
χ ( ξ ) = K ( ξ , z ( ξ ) , C D q α z ( ξ ) ) ,
G ( ξ , q s ) = G ω ( ξ , q s ) + G χ ( ξ , q s ) , G ω ( ξ , q s ) = G 1 ( ξ , q s ) + G 2 ( ξ , q s ) , G χ ( ξ , q s ) = G 3 ( ξ , q s ) , G 1 ( ξ , q s ) = ( λ ξ 1 + e λ ξ ) e λ ( T s ) I q α β 0 ( T ) Δ + e λ ( ξ s ) I q α ( T ) , 0 q s ξ T , ( λ ξ 1 + e λ ξ ) e λ ( T s ) I q α β 0 ( T ) Δ , 0 ξ q s T , G 2 ( ξ , q s ) = ( 1 λ ξ e λ ξ ) Δ i = 1 p 0 ξ e λ ( ξ s ) I q α β i ( T ) d q ξ , G 3 ( ξ , q s ) = ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 Δ Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) ,
and
Δ = i = 1 p 0 T ( λ ) β i e λ ξ d q ξ ( λ ) β 0 e λ T 0 .
Proof. 
Consider
C D q α ( D + λ ) z ( ξ ) = ω ( ξ ) , ξ J = ( 0 , T ) , α ( 1 , 2 ] .
Applying the operator I q α and using (2) with p = 2 , we obtain
( D + λ ) z ( ξ ) = I q α ω ( ξ ) + c 0 + c 1 ξ ,
where D denotes ordinary differential operator. This equation can also be rewritten as
D ( e λ ξ z ( ξ ) ) = e λ ξ [ I q α ω ( ξ ) + c 0 + c 1 ξ ] .
Integrating from 0 to ξ , we obtain
z ( ξ ) = 0 ξ e λ ( ξ s ) I q α ω ( s ) d s + c 0 λ ( 1 e λ ξ ) + c 1 λ 2 ( λ ξ 1 + e λ ξ ) + c 2 e λ ξ .
Using the boundary condition
z ( 0 ) = 0 T ( T q s ) α 1 Γ q ( α ) K ( s , z ( s ) , C D q α z ( s ) ) d q s ,
we obtain
c 2 = 0 T ( T q s ) α 1 Γ q ( α ) χ ( s ) d q s ,
where χ ( s ) = K ( s , z ( s ) , C D q α z ( s ) ) .
By applying C D q z ( 0 ) = 0 , we have
z ( ξ ) = 0 ξ e λ ( ξ s ) I q α ω ( s ) d s + c 0 λ ( 1 e λ ξ ) + c 1 λ 2 ( λ ξ 1 + e λ ξ ) 0 T ( T q s ) α 1 Γ q ( α ) χ ( s ) d q s ( e λ ξ ) .
Therefore,
C D q z ( ξ ) = c 0 e λ ξ + c 1 λ ( 1 e λ ξ ) + 0 ξ e λ ( ξ s ) I q α 1 ω ( s ) d s + λ e λ ξ 0 T ( T q s ) α 1 Γ q ( α ) χ ( s ) d q s .
At ξ = 0 , we obtain
C D q z ( 0 ) = c 0 + λ 0 T ( T q s ) α 1 Γ q ( α ) χ ( s ) d q s
and C D q z ( 0 ) = 0 , we obtain
c 0 = λ 0 T ( T q s ) α 1 Γ q ( α ) χ ( s ) d q s .
Using another boundary condition, we obtain
C D q β 0 z ( T ) = i = 1 p 0 T C D q β i z ( ξ ) d q ξ .
Fitting (9) in (10), we obtain
z ( T ) = 0 T e λ ( T s ) I q α ω ( s ) d s 0 T ( T q s ) α 1 Γ q ( α ) χ ( s ) d q s ( 1 e λ T ) + c 1 λ 2 ( λ T 1 + e λ T ) 0 T ( T q s ) α 1 Γ q ( α ) χ ( s ) d q s e λ T ,
i.e.,
z ( T ) = 0 T e λ ( T s ) I q α ω ( s ) d s 0 T ( T q s ) α 1 Γ q ( α ) χ ( s ) d q s + c 1 λ 2 ( λ T 1 + e λ T ) .
This implies
C D q β 0 z ( T ) = 0 T e λ ( T s ) I q α β 0 ω ( s ) d s 0 T ( T q s ) α β 0 1 Γ q ( α β 0 ) χ ( s ) d q s + c 1 λ 2 ( ( λ β 0 ) e λ T )
and
z ( ξ ) = 0 ξ e λ ( ξ s ) I q α ω ( s ) d s 0 T ( T q s ) α 1 Γ q ( α ) χ ( s ) d q s ( 1 e λ ξ ) + c 1 λ 2 ( λ ξ 1 + e λ ξ ) 0 T ( T q s ) α 1 Γ q ( α ) χ ( s ) d q s e λ ξ
z ( ξ ) = 0 ξ e λ ( ξ s ) I q α ω ( s ) d s 0 T ( T q s ) α 1 Γ q ( α ) χ ( s ) d q s + c 1 λ 2 ( λ ξ 1 + e λ ξ )
Therefore, (10) becomes
i = 1 p 0 T C D q β i z ( ξ ) d q ξ = i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i ω ( s ) d s d q ξ + c 1 λ 2 i = 1 p 0 T ( λ ) β i e λ ξ d q ξ .
From (11) and (12), we have
c 1 = λ 2 Δ 0 T e λ ( T s ) I q α β 0 ω ( s ) d s i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i ω ( s ) d s d q ξ 0 T ( T q s ) α β 0 1 Γ q ( α β 0 ) χ ( s ) d q s .
Putting all values of c 0 , c 1 , and c 2 in (9), we obtain
z ( ξ ) = 0 ξ e λ ( ξ s ) I q α ω ( s ) d s 0 T ( T q s ) α 1 Γ q ( α ) χ ( s ) d q s + ( λ ξ 1 + e λ ξ ) Δ ( 0 T e λ ( T s ) I q α β 0 ω ( s ) d s i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i ω ( s ) d s d q ξ 0 T ( T q s ) α β 0 1 Γ q ( α β 0 ) χ ( s ) d q s ) = 0 ξ e λ ( ξ s ) I q α ω ( s ) d s + ( λ ξ 1 + e λ ξ ) Δ 0 ξ e λ ( T s ) I q α β 0 ω ( s ) d s + ( λ ξ 1 + e λ ξ ) Δ ξ T e λ ( T s ) I q α β 0 ω ( s ) d s + ( λ ξ 1 + e λ ξ ) Δ i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i ω ( s ) d s d q ξ 0 T ( T q s ) α 1 Γ q ( α ) + ( λ ξ 1 + e λ ξ ) ( T q s ) α β 0 1 Δ Γ q ( α β 0 ) χ ( s ) d q s = 0 ξ e λ ( ξ s ) I q α + ( λ ξ 1 + e λ ξ ) e λ ( T s ) I q α β 0 Δ ω ( s ) d s + ( λ ξ 1 + e λ ξ ) Δ ξ T e λ ( T s ) I q α β 0 ω ( s ) d s + ( 1 λ ξ e λ ξ ) Δ i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i ω ( s ) d s d q ξ + 0 T ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 Δ Γ q ( α β 0 ) ( T q ) α 1 Γ q ( α ) χ ( s ) d q s = 0 T G 1 ( ξ , q s ) ω ( s ) d s + 0 T G 2 ( ξ , q s ) ω ( s ) d q s + 0 T G 3 ( ξ , q s ) χ ( s ) d s = 0 T ( G 1 ( ξ , q s ) + G 2 ( ξ , q s ) ) ω ( s ) d s + 0 T G 3 ( ξ , q s ) χ ( s ) d q s .
Thus
z ( ξ ) = 0 T G ω ( ξ , q s ) ω ( s ) d s + 0 T G χ ( ξ , q s ) χ ( s ) d q s .
   ☐

3. Existence of Solution

Here, we build up some sufficient assumptions for the EU of the solution to Problem (1).
In view of Lemma 2.15, Problem (6) has the following solution:
z ( ξ ) = 0 ξ e λ ( ξ s ) I q α ω ( s ) d s + ( λ ξ 1 + e λ ξ ) Δ 0 T e λ ( T s ) I q α β 0 ω ( s ) d s + ( 1 λ ξ e λ ξ ) Δ i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i ω ( s ) d s d q ξ + 0 T ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 Δ Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) χ ( s ) d q s ,
where
ω ( ξ ) = f ( ξ , z ( ξ ) , C D q α z ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z ( s ) , C D q α z ( s ) ) d q s ,
and
χ ( ξ ) = K ( ξ , z ( ξ ) , C D q α z ( ξ ) ) .
Assume
p ( ξ ) = f ( ξ , z ( ξ ) , C D q α z ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z ( s ) , C D q α z ( s ) ) d q s = f ( ξ , z ( ξ ) , p ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z ( s ) , p ( s ) ) d q s ,
and
v ( ξ ) = K ( ξ , z ( ξ ) , C D q α z ( ξ ) ) = K ( ξ , z ( ξ ) , v ( ξ ) ) .
Define an operator ρ : X X as
ρ z ( ξ ) = 0 ξ e λ ( ξ s ) I q α p ( s ) d s + ( λ ξ 1 + e λ ξ ) Δ 0 T e λ ( T s ) I q α β 0 p ( s ) d s + ( 1 λ ξ e λ ξ ) Δ i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i p ( s ) d s d q ξ + 0 T ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 Δ Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) v ( s ) d q s ,
where p and v X .
The following hypothesis will be used in further results.
( A 1 )
Δ = i = 1 p 0 T ( λ ) β i e λ ξ d q ξ ( λ ) β 0 e λ T 0 .
( A 2 )
f , g , K : J × R × R R are continuous.
( A 3 )
There ∃ N 1 > 0 and 0 < N 2 < 1 in such way that t J , and ∀ ζ , ζ ^ , θ , θ ^ R ; thus, we have
| f ( ξ , ζ , θ ) f ( ξ , ζ ^ , θ ^ ) | N 1 | ζ ζ ^ | + N 2 | θ θ ^ | .
( A 4 )
There ∃ N 3 > 0 and 0 < N 4 < 1 in such way that t J and ∀ ζ , ζ ^ , θ , θ ^ R ; thus, we have
| K ( ξ , ζ , θ ) K ( ξ , ζ ^ , θ ^ ) | N 3 | ζ ζ ^ | + N 4 | θ θ ^ | .
( A 5 )
There ∃ N 5 > 0 and 0 < N 6 < 1 in such a way that ξ J and ∀ ζ , ζ ^ , θ , θ ^ R ; we thus have
| g ( ξ , ζ , θ ) g ( ξ , ζ ^ , θ ^ ) | N 5 | ζ ζ ^ | + N 6 | θ θ ^ | .
( A 6 )
There ∃ l , m , n C ( J , R + ) , which is bounded and satisfies
| f ( ξ , ζ ( ξ ) , θ ( ξ ) ) | l ( ξ ) + m ( ξ ) | ζ ( ξ ) | + n ( ξ ) | θ ( ξ ) |
with n * = sup ξ J n ( ξ ) < 1 .
( A 7 )
There ∃ b , c , e C ( J , R + ) , which is bounded and satisfies
| K ( ξ , ζ ( ξ ) , θ ( ξ ) ) | b ( ξ ) + c ( ξ ) | ζ ( ξ ) | + e ( ξ ) | θ ( ξ ) |
with e * = sup ξ J e ( ξ ) < 1 .
( A 8 )
There ∃ i , j , k C ( J , R + ) , which is bounded and satisfies
| g ( ξ , ζ ( ξ ) , θ ( ξ ) ) | i ( ξ ) + j ( ξ ) | ζ ( ξ ) | + k ( ξ ) | θ ( ξ ) |
with k * = sup ξ J k ( ξ ) < 1 .
( A 9 )
There ∃ a non-decreasing ψ C ( J , R + ) and L ψ > 0 fulfilling
I q α ψ ( ξ ) L ψ ψ ( ξ ) , ξ J .
Lemma 3.
The Green’s functions G ω ( ξ , q s ) and G χ ( ξ , q s ) , which are obtained in Lemma 2, fulfill the following conditions
(1) 
G ω ( ξ , q s ) and G χ ( ξ , q s ) are continuous on J×J,
(2) 
max ξ J | 0 T G ω ( ξ , q s ) d s | Y ω    and     max ξ J | 0 T G χ ( ξ , q s ) d q s | Y χ ,
where
Y ω = T α ( 1 ( 1 q ) α ) λ q Γ q ( α + 1 ) + | ( λ T 1 + e λ T ) | T α β 0 ( 1 ( 1 q ) α β 0 ) λ q Γ q ( α β 0 + 1 ) + i = 1 p | ( 1 λ T e λ T ) | T α β i + 1 ( 1 ( 1 q ) α β i + 1 ) λ q Γ q ( α β i + 2 ) , Y χ = | 1 λ T e λ T | T α β 0 ( 1 ( 1 q ) α β 0 ) q Γ q ( α β 0 + 1 ) + T α ( 1 ( 1 q ) α ) q Γ q ( α ) a n d = i = 1 p ( λ ) β i e λ T λ ( ( λ ) β 0 e λ T ) .
Proof. (1) The continuities of G ω ( ξ , q s ) and G χ ( ξ , q s ) are obvious.
(2)
G ω ( ξ , q s ) and G χ ( ξ , q s ) have the following expression:
| 0 T G ω ( ξ , q s ) d s | = | 0 ξ e λ ( ξ s ) 0 ξ 1 Γ q ( α ) ( ξ q s ) α 1 d q s d s + ( λ ξ 1 + e λ ξ ) Δ 0 T e λ ( T s ) 0 ξ 1 Γ q ( α β 0 ) ( ξ q s ) α β 0 1 d q s d s + ( 1 λ ξ e λ ξ ) Δ i = 1 p 0 T 0 ξ e λ ( ξ s ) 0 ξ 1 Γ q ( α β i ) ( ξ q s ) α β i 1 d q s d s d q ξ | | 1 e λ ξ | ξ α ( 1 ( 1 q ) α ) λ q Γ q ( α + 1 ) + | ( λ ξ 1 + e λ ξ ) | | ( 1 e λ T ) | ξ α β 0 ( 1 ( 1 q ) α β 0 ) λ | Δ | q Γ q ( α β 0 + 1 ) + i = 1 p | ( 1 λ ξ e λ ξ ) | | ( 1 e λ ξ ) | T α β i + 1 ( 1 ( 1 q ) α β i + 1 ) λ | Δ | q Γ q ( α β i + 2 ) .
Now taking max ξ J , we obtain
max ξ J | 0 T G ω ( ξ , q s ) d s | | 1 e λ T | T α ( 1 ( 1 q ) α ) λ q Γ q ( α + 1 ) + | ( λ T 1 + e λ T ) | | ( 1 e λ T ) | T α β 0 ( 1 ( 1 q ) α β 0 ) λ q Γ q ( α β 0 + 1 ) + i = 1 p | ( 1 λ T e λ T ) | | ( 1 e λ T ) | T α β i + 1 ( 1 ( 1 q ) α β i + 1 ) λ q Γ q ( α β i + 2 ) T α ( 1 ( 1 q ) α ) λ q Γ q ( α + 1 ) + | ( λ T 1 + e λ T ) | T α β 0 ( 1 ( 1 q ) α β 0 ) λ q Γ q ( α β 0 + 1 ) + i = 1 p | ( 1 λ T e λ T ) | T α β i + 1 ( 1 ( 1 q ) α β i + 1 ) λ q Γ q ( α β i + 2 ) = Y ω .
Similarly,
| 0 T G χ ( ξ , q s ) d q s | | ( 1 λ ξ e λ ξ ) Δ 0 T ( T q s ) α β 0 1 Γ q ( α β 0 ) d q s 0 T ( T q s ) α 1 Γ q ( α ) d q s | | 1 λ ξ e λ ξ | T α β 0 ( 1 ( 1 q ) α β 0 ) | Δ | q Γ q ( α β 0 + 1 ) + T α ( 1 ( 1 q ) α ) q Γ q ( α ) .
Now taking max ξ J , we obtain
max ξ J | 0 T G χ ( ξ , q s ) d q s | | 1 λ T e λ T | T α β 0 ( 1 ( 1 q ) α β 0 ) q Γ q ( α β 0 + 1 ) + T α ( 1 ( 1 q ) α ) q Γ q ( α ) = Y χ .
   ☐
Theorem 4.
If the hypothesis ( A 2 ) ( A 5 ) and the inequality
Y ω N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + Y χ N 3 1 N 4 < 1
are satisfied, then (6) has a unique solution.
Proof. 
For z , x X , using (13), we have
ρ z ( ξ ) = 0 ξ e λ ( ξ s ) I q α | p ( s ) p ( s ) ^ | d s + ( λ ξ 1 + e λ ξ ) | Δ | 0 T e λ ( T s ) I q α β 0 | p ( s ) p ( s ) ^ | d s + ( 1 λ ξ e λ ξ ) | Δ | i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i | p ( s ) p ( s ) ^ | d s d q ξ + 0 T ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 | Δ | Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) | v ( s ) v ( s ) ^ | d q s ,
where p ^ and v ^ X are given by
p ( ξ ) ^ = f ( ξ , x ( ξ ) , p ( ξ ) ^ ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , x ( s ) , p ( s ) ^ ) d q s
and
v ( ξ ) ^ = K ( ξ , x ( ξ ) , v ( ξ ) ^ ) .
Using ( A 2 ) , ( A 3 ) and ( A 5 ) , we have
| p ( ξ ) p ( ξ ) ^ | = | f ( ξ , z ( ξ ) , p ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z ( s ) , p ( s ) ) d q s f ( ξ , x ( ξ ) , p ( ξ ) ^ ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , x ( s ) , p ( s ) ^ ) d q s | | f ( ξ , z ( ξ ) , p ( ξ ) ) f ( ξ , x ( ξ ) , p ( ξ ) ^ ) | + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) | g ( s , z ( s ) , p ( s ) ) g ( s , x ( s ) , p ( s ) ^ ) | d q s N 1 | z ( ξ ) x ( ξ ) | + N 2 | p ( ξ ) p ( ξ ) ^ | + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) N 5 | z ( s ) x ( s ) | + N 6 | p ( s ) p ( s ) ^ | d q s = N 1 | z ( ξ ) x ( ξ ) | + N 2 | p ( ξ ) p ( ξ ) ^ | + ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) N 5 | z ( ξ ) x ( ξ ) | + ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) N 6 | p ( ξ ) p ( ξ ) ^ | .
Thus,
| p ( ξ ) p ( ξ ) ^ | N 1 1 N 2 N 6 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) | z ( ξ ) x ( ξ ) | + N 5 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) | z ( ξ ) x ( ξ ) | .
Similarly,
| v ( ξ ) v ( ξ ) ^ | = | K ( ξ , z ( ξ ) , v ( ξ ) ) K ( ξ , x ( ξ ) , v ( ξ ) ^ ) | | K ( ξ , z ( ξ ) , v ( ξ ) ) | + | K ( ξ , x ( ξ ) , v ( ξ ) ^ ) | .
Using ( A 4 ) , we obtain
| v ( ξ ) v ( ξ ) ^ | N 3 | z ( ξ ) x ( ξ ) | + N 4 | v ( ξ ) v ( ξ ) ^ | .
Equivalently,
| v ( ξ ) v ( ξ ) ^ | N 3 1 N 4 | z ( ξ ) x ( ξ ) | .
Using (16) and (17) in (15), we have
| ρ z ( ξ ) ρ x ( ξ ) | [ | 1 e λ ξ | ξ α ( 1 ( 1 q ) α ) λ q Γ q ( α + 1 ) + | ( λ ξ 1 + e λ ξ ) | | ( 1 e λ T ) | ξ α β 0 ( 1 ( 1 q ) α β 0 ) λ | Δ | q Γ q ( α β 0 + 1 ) + i = 1 p | ( 1 λ ξ e λ ξ ) | | ( 1 e λ ξ ) | T α β i + 1 ( 1 ( 1 q ) α β i + 1 ) λ | Δ | q Γ q ( α β i + 2 ) [ N 1 1 N 2 N 6 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) ] + | 1 λ ξ e λ ξ | T α β 0 ( 1 ( 1 q ) α β 0 ) | Δ | q Γ q ( α + 1 ) + T α ( 1 ( 1 q ) α ) q Γ q ( α ) N 3 1 N 4 ] | z ( ξ ) x ( ξ ) | .
Since ξ [ 0 , T ] and using Lemma 3, we obtain
| ρ z ( ξ ) ρ x ( ξ ) | [ Y ω N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + Y χ N 3 1 N 4 ] | z ( ξ ) x ( ξ ) | .
Thus
| | ρ z ρ x | | X [ Y ω N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + Y χ N 3 1 N 4 ] | | z ( ξ ) x ( ξ ) | | X .
Moreover,
Y ω N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + Y χ N 3 1 N 4 < 1 .
Thus by the Banach theorem, ρ has a FP (unique), i.e., (6) has a unique, solution.    ☐
Theorem 5.
Using the hypothesis ( A 2 ) ( A 8 ) , (6) has at least one solution.
Proof. 
We use the predefined operator ρ and follow the steps of Schafer FP , as follows.
Step 1:
First, we prove that ρ is continuous. Consider a sequence z n X such that z n z X . For ξ J , we have
| ρ z n ( ξ ) ρ z ( ξ ) | 0 ξ e λ ( ξ s ) I q α | p n ( s ) p ( s ) | d s + ( λ ξ 1 + e λ ξ ) | Δ | 0 T e λ ( T s ) I q α β 0 | p n ( s ) p ( s ) | d s + ( 1 λ ξ e λ ξ ) | Δ | i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i | p n ( s ) p ( s ) | d s d q ξ + 0 T ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 | Δ | Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) | v n ( s ) v ( s ) | d q s ,
where p n , v n X
p n ( ξ ) = f ( ξ , z n ( ξ ) , p n ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z n ( s ) , p n ( s ) ) d q s
and
v n ( ξ ) = K ( ξ , z n ( ξ ) , v n ( ξ ) ) .
Hence, by ( A 3 ) ( A 5 ) , we obtain
| p n ( ξ ) p ( ξ ) | N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) | z n ( ξ ) z ( ξ ) | + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) | z n ( ξ ) z ( ξ ) | .
| v n ( ξ ) v ( ξ ) | N 3 1 N 4 | z n ( ξ ) z ( ξ ) |
| ρ z n ( ξ ) ρ z ( ξ ) | [ | 1 e λ ξ | ξ α ( 1 ( 1 q ) α ) λ q Γ q ( α + 1 ) + | ( λ ξ 1 + e λ ξ ) | | ( 1 e λ T ) | ξ α β 0 ( 1 ( 1 q ) α β 0 ) λ | Δ | q Γ q ( α β 0 + 1 ) + i = 1 p | ( 1 λ ξ e λ ξ ) | | ( 1 e λ ξ ) | T α β i + 1 ( 1 ( 1 q ) α β i + 1 ) λ | Δ | q Γ q ( α β i + 2 ) [ N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) ] + | 1 λ ξ e λ ξ | T α β 0 ( 1 ( 1 q ) α β 0 ) | Δ | q Γ q ( α + 1 ) + T α ( 1 ( 1 q ) α ) q Γ q ( α ) N 3 1 N 4 ] | z n ( ξ ) z ( ξ ) | .
For each ξ J , the sequence z n z as n ; thus, by the Lebesgue dominated convergence theorem,
| ρ z n ( ξ ) ρ z ( ξ ) | 0 n .
This implies that
| | ρ z n ρ z | | 0 n ;
i.e., ρ is continuous on J.
Step 2:
Next, for each z π k = { z X : | | z | | k } . We need to verify that | | ρ z | | L with some L > 0 . For ξ J , we obtain
| ρ z ( ξ ) | 0 ξ e λ ( ξ s ) I q α | p ( s ) | d s + ( λ ξ 1 + e λ ξ ) | Δ | 0 T e λ ( T s ) I q α β 0 | p ( s ) | d s + ( 1 λ ξ e λ ξ ) | Δ | i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i | p ( s ) | d s d q ξ + 0 T ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 | Δ | Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) | v ( s ) | d q s ,
where p , v X are given by
p ( ξ ) = f ( ξ , z ( ξ ) , p ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z ( s ) , p ( s ) ) d q s
and
v ( ξ ) = K ( ξ , z ( ξ ) , v ( ξ ) ) .
By ( A 5 ) and ( A 6 ) , we have
p ( ξ ) = f ( ξ , z ( ξ ) , p ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z ( s ) , p ( s ) ) d q s .
Thus,
| p ( ξ ) | = | f ( ξ , z ( ξ ) , p ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z ( s ) , p ( s ) ) d q s | | f ( ξ , z ( ξ ) , p ( ξ ) ) | + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) | g ( s , z ( s ) , p ( s ) ) | d q s l ( ξ ) + m ( ξ ) | z ( ξ ) | + n ( ξ ) | p ( ξ ) | + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) ( i ( s ) + j ( s ) | z ( s ) | + k ( s ) | p ( s ) | ) d q s l * + m * | | z | | X + n * | | p | | X + ( i * + j * | | z | | X + k * | | p | | X ) ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) ,
where l * = s u p ξ J l ( ξ ) , m * = s u p ξ J m ( ξ ) , n * = s u p ξ J n ( ξ ) < 1 , i * = s u p ξ J i ( ξ ) , j * = s u p ξ J j ( ξ ) , k * = s u p ξ J k ( ξ ) < 1
| p ( ξ ) | | | p | | X l * + m * | | z | | X 1 n * k * ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + ( i * + j * | | z | | X ) ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 n * k * ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) = h .
Similarly,
| v ( ξ ) | = | K ( ξ , z ( ξ ) , v ( ξ ) ) | .
By ( A 7 ) , we obtain
| v ( ξ ) | b ( ξ ) + c ( ξ ) | z ( ξ ) | + e ( ξ ) | v ( ξ ) | b * + c * | | z | | X + e * | | v | | X
| v ( ξ ) | b * + c * k 1 e * = h * .
Clearly, h and h * are constants. Using Lemma 3 and (18), we obtain
| | ρ ( z ) | | X Y ω ( h ) + Y χ ( h * ) = L .
Step 3:
Take ξ 1 , ξ 2 X in such way that ξ 1 < ξ 2 and assume that π k is bounded. Then, for z o k ,
| ρ z ( ξ 2 ) ρ z ( ξ 1 ) | = | 0 ξ 2 e λ ( ξ 2 s ) I q α p ( s ) d s 0 ξ 1 e λ ( ξ 1 s ) I q α p ( s ) d s + ( λ ξ 2 1 + e λ ξ 2 ) Δ 0 T e λ ( T s ) I q α β 0 p ( s ) d s ( λ ξ 1 1 + e λ ξ 1 ) Δ 0 T e λ ( T s ) I q α β 0 p ( s ) d s + ( 1 λ ξ 2 e λ ξ 2 ) Δ i = 1 p 0 T 0 ξ 2 e λ ( ξ 2 s ) I q α β i p ( s ) d s d q ξ 2 ( 1 λ ξ 1 e λ ξ 1 ) Δ i = 1 p 0 T 0 ξ 1 e λ ( ξ 1 s ) I q α β i p ( s ) d s d q ξ 1 |
In Step 2, we obtained that
| p ( ξ ) | | | p | | X l * + m * | | z | | X 1 n * k * ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + ( i * + j * | | z | | X ) ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 n * k * ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) = h ,
| ρ z ( ξ 2 ) ρ z ( ξ 1 ) | h | 0 ξ 2 e λ ( ξ 2 s ) I q α p ( s ) d s 0 ξ 1 e λ ( ξ 1 s ) I q α p ( s ) d s + ( λ ξ 2 1 + e λ ξ 2 ) Δ 0 T e λ ( T s ) I q α β 0 p ( s ) d s ( λ ξ 1 1 + e λ ξ 1 ) Δ 0 T e λ ( T s ) I q α β 0 p ( s ) d s + ( 1 λ ξ 2 e λ ξ 2 ) Δ i = 1 p 0 T 0 ξ 2 e λ ( ξ 2 s ) I q α β i p ( s ) d s d q ξ 2 ( 1 λ ξ 1 e λ ξ 1 ) Δ i = 1 p 0 T 0 ξ 1 e λ ( ξ 1 s ) I q α β i p ( s ) d s d q ξ 1 | .
It is obvious that the right-hand side of (19) tends to zero as ξ 1 ξ 2 . Therefore, from Step 1 to Step 3, the Arzela–Ascoli theorem ρ : X X is a completely continuous mapping.
Step 4:
Consider the set
S = { z X | z = ϕ ρ ( z ) , 0 < ϕ < 1 } .
We need to show that S is bounded. Let z S . Then, for some 0 < S < 1 with z = ϕ ρ ( z ) , we have
| z ( ξ ) | = | ϕ 0 ξ e λ ( ξ s ) I q α p ( s ) d s + ϕ ( λ ξ 1 + e λ ξ ) Δ 0 T e λ ( T s ) I q α β 0 p ( s ) d s + ϕ ( 1 λ ξ e λ ξ ) Δ i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i p ( s ) d s d q ξ + ϕ 0 T ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 Δ Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) v ( s ) d q s | .
This implies
| z ( ξ ) | 0 ξ e λ ( ξ s ) I q α | p ( s ) | d s + ( λ ξ 1 + e λ ξ ) | Δ | 0 T e λ ( T s ) I q α β 0 | p ( s ) | d s + ( 1 λ ξ e λ ξ ) | Δ | i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i | p ( s ) | d s d q ξ + 0 T ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 | Δ | Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) | v ( s ) | d q s .
By ( A 6 ) ( A 8 ) , we have
| p ( ξ ) | | | p | | X l * + m * | | z | | X 1 n * k * ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + ( i * + j * | | z | | X ) ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 n * k * ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) = h
and
| v ( ξ ) | b * + c * k 1 e * = h * .
Now, from Lemma 3 and (20),
| z ( ξ ) | h 0 ξ e λ ( ξ s ) I q α d s + h ( λ ξ 1 + e λ ξ ) | Δ | 0 T e λ ( T s ) I q α β 0 d s + h ( 1 λ ξ e λ ξ ) | Δ | i = 1 p 0 T 0 ξ e λ ( T s ) I q α β i d s d q ξ + h * 0 ξ ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 | Δ | Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) d q s Y ω ( h ) + Y χ ( h * ) = L .
i.e., | z ( ξ ) | L . This shows that S is bounded. Therefore, by Theorem 3, ρ has at least one FP , i.e., (6) has at least one solution.
   ☐

4. Stability

Here, we discuss different types of Ulam stability for (6).
Theorem 6.
If the hypothesis ( A 2 ) ( A 5 ) with (6) holds then (6) is UHS and GUHS .
Proof. 
Let x be an approximate solution of (3) and let z satisfy
C D q α ( D + λ ) z ( ξ ) = f ( ξ , z ( ξ ) , C D q α z ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z ( s ) , C D q α z ( s ) ) d q s , ξ J = ( 0 , T ) , α ( 1 , 2 ] , z ( 0 ) = 0 T ( T q η ) α 1 Γ q ( α ) K ( η , z ( η ) , C D q α z ( η ) ) d q η , C D q z ( 0 ) = 0 , C D q β 0 z ( T ) = i = 1 p 0 T C D q β i z ( ξ ) d q ξ .
By Lemma 2, we have
z ( ξ ) = 0 ξ e λ ( ξ s ) I q α p ( s ) d s + ( λ ξ 1 + e λ ξ ) Δ 0 T e λ ( T s ) I q α β 0 p ( s ) d s + ( 1 λ ξ e λ ξ ) Δ i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i p ( s ) d s d q ξ + 0 T ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 Δ Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) v ( s ) d q s ,
where p , v X are given by
p ( ξ ) = f ( ξ , z ( ξ ) , p ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z ( s ) , p ( s ) ) d q s
and
v ( ξ ) = K ( ξ , z ( ξ ) , v ( ξ ) ) .
As x is a solution of (3) and using Remark 2, we obtain
C D q α ( D + λ ) x ( ξ ) = f ( ξ , x ( ξ ) , C D q α x ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , x ( s ) , C D q α x ( s ) ) d q s + ψ ( ξ ) , ξ J = ( 0 , T ) , α ( 1 , 2 ] , x ( 0 ) = 0 T ( T q η ) α 1 Γ q ( α ) K ( η , x ( η ) , C D q α x ( η ) ) d q η , C D q x ( 0 ) = 0 , C D q β 0 x ( T ) = i = 1 p 0 T C D q β i x ( ξ ) d q ξ .
Clearly the solution of (21) will be
x ( ξ ) = 0 ξ e λ ( ξ s ) I q α p ( s ) ^ d s + ( λ ξ 1 + e λ ξ ) Δ 0 T e λ ( T s ) I q α β 0 p ( s ) ^ d s + ( 1 λ ξ e λ ξ ) Δ i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i p ( s ) ^ d s d q ξ 0 ξ e λ ( ξ s ) I q α ψ ( s ) d s + ( λ ξ 1 + e λ ξ ) Δ 0 T e λ ( T s ) I q α β 0 ψ ( s ) d s + ( 1 λ ξ e λ ξ ) Δ i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i ψ ( s ) d s d q ξ + 0 T ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 Δ Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) v ( s ) ^ d q s ,
where p ^ , v ^ X are given by
p ( ξ ) ^ = f ( ξ , x ( ξ ) , p ( ξ ) ^ ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , x ( s ) , p ( s ) ^ ) d q s
and
v ( ξ ) ^ = K ( ξ , x ( ξ ) , v ( ξ ) ^ ) .
For each ξ J , we have
| x ( ξ ) z ( ξ ) | 0 ξ e λ ( ξ s ) I q α | p ( s ) ^ p ( s ) | d s + ( λ ξ 1 + e λ ξ ) | Δ | 0 T e λ ( T s ) I q α β 0 | p ( s ) ^ p ( s ) | d s + ( 1 λ ξ e λ ξ ) | Δ | i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i | p ( s ) ^ p ( s ) | d s d q ξ + 0 ξ e λ ( ξ s ) I q α | ψ ( s ) | d s + ( λ ξ 1 + e λ ξ ) | Δ | 0 T e λ ( T s ) I q α β 0 | ψ ( s ) | d s + ( 1 λ ξ e λ ξ ) | Δ | i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i | ψ ( s ) | d s d q ξ + 0 T ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 | Δ | Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) | v ( s ) ^ v ( s ) | d q s .
By ( A 3 ) and ( A 5 ) , we obtain
| p ( ξ ) ^ p ( ξ ) | N 1 1 N 2 N 6 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) | x ( ξ ) z ( ξ ) | + N 5 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) | x ( ξ ) z ( ξ ) |
and
| v ( ξ ) ^ v ( ξ ) | N 3 1 N 4 | x ( ξ ) z ( ξ ) | .
Using (1) of Remark 2 in (22), we obtain
| x ( ξ ) z ( ξ ) | ( | 1 e λ ξ | ξ α ( 1 ( 1 q ) α ) λ q Γ q ( α + 1 ) + | ( λ ξ 1 + e λ ξ ) | | ( 1 e λ T ) | ξ α β 0 ( 1 ( 1 q ) α β 0 ) λ | Δ | q Γ q ( α β 0 + 1 ) + i = 1 p | ( 1 λ ξ e λ ξ ) | | ( 1 e λ ξ ) | T α β i + 1 ( 1 ( 1 q ) α β i + 1 ) λ | Δ | q Γ q ( α β i + 2 ) ) [ N 1 1 N 2 N 6 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) ] | x ( ξ ) z ( ξ ) | + ( | 1 e λ ξ | ξ α ( 1 ( 1 q ) α ) λ q Γ q ( α + 1 ) + | ( λ ξ 1 + e λ ξ ) | | ( 1 e λ T ) | ξ α β 0 ( 1 ( 1 q ) α β 0 ) λ | Δ | q Γ q ( α β 0 + 1 ) + i = 1 p | ( 1 λ ξ e λ ξ ) | | ( 1 e λ ξ ) | T α β i + 1 ( 1 ( 1 q ) α β i + 1 ) λ | Δ | q Γ q ( α β i + 2 ) ) | ψ ( ξ ) | + | 1 λ ξ e λ ξ | T α β 0 ( 1 ( 1 q ) α β 0 ) | Δ | q Γ q ( α + 1 ) + T α ( 1 ( 1 q ) α ) q Γ q ( α ) N 3 1 N 4 | x ( ξ ) z ( ξ ) | .
Since ξ [ 0 , T ] and using Lemma 3, we obtain
| x ( ξ ) z ( ξ ) | [ Y ω N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + Y χ N 3 1 N 4 ] | z ( ξ ) x ( ξ ) | + Y ω ϵ .
Thus,
| | x z | | X Y ω ϵ 1 Y ω N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) Y χ N 3 1 N 4 .
This implies
| | x z | | X ϵ C f , g ,
where
C f , g = Y ω ϵ 1 Y ω N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) Y χ N 3 1 N 4 .
Therefore, (6) is UHS . Furthermore, if we set F f ϵ = C f ϵ , F ( 0 ) = 0 , we see that (6) is GUHS .    ☐
Theorem 7.
If ( A 2 ) ( A 9 ) and (14) hold, then (6) is UHRS , and consequently, it is GUHRS .
Proof. 
Let x be an approximate solution of (5) and let z be the solution of
C D q α ( D + λ ) z ( ξ ) = f ( ξ , z ( ξ ) , C D q α z ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z ( s ) , C D q α z ( s ) ) d q s , ξ J = ( 0 , T ) , α ( 1 , 2 ] , z ( 0 ) = 0 T ( T q η ) α 1 Γ q ( α ) K ( η , z ( η ) , C D q α z ( η ) ) d q η , C D q z ( 0 ) = 0 , C D q β 0 z ( T ) = i = 1 p 0 T C D q β i z ( ξ ) d q ξ .
By Lemma 2, we have
z ( ξ ) = 0 ξ e λ ( ξ s ) I q α p ( s ) d s + ( λ ξ 1 + e λ ξ ) Δ 0 T e λ ( T s ) I q α β 0 p ( s ) d s + ( 1 λ ξ e λ ξ ) Δ i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i p ( s ) d s d q ξ + 0 T ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 Δ Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) v ( s ) d q s ,
where p , v X are given by
p ( ξ ) = f ( ξ , z ( ξ ) , p ( ξ ) ) + 0 ξ ( ξ q s ) σ 1 Γ q ( δ ) g ( s , z ( s ) , p ( s ) ) d q s
and
v ( ξ ) = K ( ξ , z ( ξ ) , v ( ξ ) ) .
Using the proof procedure of Theorem 6, for each ξ J , we obtain
| x ( ξ ) z ( ξ ) | 0 ξ e λ ( ξ s ) I q α | p ( s ) ^ p ( s ) | d s + ( λ ξ 1 + e λ ξ ) | Δ | 0 T e λ ( T s ) I q α β 0 | p ( s ) ^ p ( s ) | d s + ( 1 λ ξ e λ ξ ) | Δ | i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i | p ( s ) ^ p ( s ) | d s d q ξ 0 ξ e λ ( ξ s ) I q α | ψ ( s ) | d s + ( λ ξ 1 + e λ ξ ) | Δ | 0 T e λ ( T s ) I q α β 0 | ψ ( s ) | d s + ( 1 λ ξ e λ ξ ) | Δ | i = 1 p 0 T 0 ξ e λ ( ξ s ) I q α β i | ψ ( s ) | d s d q ξ + 0 T ( 1 λ ξ e λ ξ ) ( T q s ) α β 0 1 | Δ | Γ q ( α β 0 ) ( T q s ) α 1 Γ q ( α ) | v ( s ) ^ v ( s ) | d q s .
By ( A 3 ) and ( A 5 ) , we obtain
| p ( ξ ) ^ p ( ξ ) | N 1 1 N 2 N 6 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) | x ( ξ ) z ( ξ ) | + N 5 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) | x ( ξ ) z ( ξ ) |
and
| v ( ξ ) ^ v ( ξ ) | N 3 1 N 4 | x ( ξ ) z ( ξ ) | .
Using the last two inequalities and part (1) of Remark 3 in (23), we have
| x ( ξ ) z ( ξ ) | ( | 1 e λ ξ | ξ α ( 1 ( 1 q ) α ) λ q Γ q ( α + 1 ) + | ( λ ξ 1 + e λ ξ ) | | ( 1 e λ T ) | ξ α β 0 ( 1 ( 1 q ) α β 0 ) λ | Δ | q Γ q ( α β 0 + 1 ) + i = 1 p | ( 1 λ ξ e λ ξ ) | | ( 1 e λ ξ ) | T α β i + 1 ( 1 ( 1 q ) α β i + 1 ) λ | Δ | q Γ q ( α β i + 2 ) ) [ N 1 1 N 2 N 6 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 ξ σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) ] | x ( ξ ) z ( ξ ) | + ( | 1 e λ ξ | ξ α ( 1 ( 1 q ) α ) λ q Γ q ( α + 1 ) + | ( λ ξ 1 + e λ ξ ) | | ( 1 e λ T ) | ξ α β 0 ( 1 ( 1 q ) α β 0 ) λ | Δ | q Γ q ( α β 0 + 1 ) + i = 1 p | ( 1 λ ξ e λ ξ ) | | ( 1 e λ ξ ) | T α β i + 1 ( 1 ( 1 q ) α β i + 1 ) λ | Δ | q Γ q ( α β i + 2 ) ) | ψ ( ξ ) | + [ | 1 λ ξ e λ ξ | T α β 0 ( 1 ( 1 q ) α β 0 ) | Δ | q Γ q ( α + 1 ) + T α ( 1 ( 1 q ) α ) q Γ q ( α ) N 3 1 N 4 ] | x ( ξ ) z ( ξ ) | .
Since ξ [ 0 , T ] , using Lemma 3, we obtain
| x ( ξ ) z ( ξ ) | [ Y ω N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + Y χ N 3 1 N 4 ] | x ( ξ ) z ( ξ ) | + Y ω ϵ L ψ ψ ( ξ ) .
Thus,
| | x z | | X Y ω ϵ L ψ ψ ( ξ ) 1 Y ω N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) Y χ N 3 1 N 4 ;
i.e.,
| | x z | | X C f , g ϵ ,
where
C f , g = Y ω L ψ ψ ( ξ ) 1 Y ω N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) Y χ N 3 1 N 4 .
Therefore, (6) is UHRS . Similarly, we can obtain the GUHRS of the considered problem.    ☐

5. Examples

In this section, we investigate our theoretical results by two examples.
Example 1.
Take the q- FDE
C D 1 4 3 2 ( D 9 10 ) z ( ξ ) = 5 + | z ( ξ ) | + | C D 1 4 3 2 z ( ξ ) | 64 ( 4 + | z ( ξ ) | + | C D 1 4 3 2 z ( ξ ) | ) + 0 T ( ξ 1 4 s ) 5 2 1 Γ 1 4 ( 5 2 ) s sin | z ( s ) | + s i n | C D 1 4 3 2 z ( s ) | 50 d 1 4 s , ξ J = ( 0 , T ) , z ( 0 ) = 0 T ( T 1 4 s ) 3 2 1 Γ 1 4 ( 3 2 ) s sin | z ( s ) | + s i n | C D 1 4 3 2 z ( s ) | 50 d 1 4 s , C D 1 4 z ( 0 ) = 0 , C D 1 4 β 0 z ( T ) = i = 1 2 0 T C D 1 4 β i z ( ξ ) d 1 4 ξ ,
where q = 1 4 , α = 3 2 , δ = σ = 5 2 , β 1 = 7 6 , β 2 = 1 6 , β 0 = 1 3 , λ = 9 10 , T = π and p = 2 . Set
f ( ξ , ζ , θ ) = 5 + | ζ | + | θ | 64 ( 4 + | ζ | + | θ | ) , ζ C ( J , R )
g ( ξ , ζ , θ ) = ξ s i n | ζ | + s i n | θ | 50
and
K ( ξ , ζ , θ ) = ξ s i n | ζ | + s i n | θ | 50 .
Clearly, the functions f , g , K are continuous. For each ζ , ζ ^ X , θ , θ ^ R , ξ [ 0 , π ] , we have
| f ( ξ , ζ , θ ) f ( ξ , ζ ^ , θ ^ ) | 1 64 ( 5 + | ζ | + | θ | ) ,
which satisfies ( A 3 ) , N 1 = 1 64 , N 2 = 5 64 . We can observe that
| g ( ξ , ζ , θ ) g ( ξ , ζ ^ , θ ^ ) | | ζ ζ ^ | + | θ θ ^ | 50 ,
which satisfies ( A 4 ) and N 5 = N 6 = 1 50 , and
| K ( ξ , ζ , θ ) K ( ξ , ζ ^ , θ ^ ) | | ζ ζ ^ | + | θ θ ^ | 50
satisfies ( A 5 ) , N 3 = N 4 = 1 50 . We can see that
Y ω N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + Y χ N 3 1 N 4 0.976974 < 1 .
We can see that all the required conditions of Theorem 4 are satisfied; hence, (24) has at least one solution. By letting ψ ( ξ ) = ξ for all ξ J , we have
I 1 4 3 2 ψ ( ξ ) = 1 Γ q ( 3 2 ) 0 ξ ( ξ q s ) 3 2 1 s d q s = 8 π 5 2 15 π 2 ξ .
Hence ( A 9 ) is satisfied with L ψ = 2 . Therefore, by Theorem 4.2, the given problem is UHRS and consequently GUHRS .
Example 2.
Assume that q- FDE :
C D 2 3 3 2 ( D 4 5 ) z ( ξ ) = 3 + | z ( ξ ) | + | C D 2 3 3 2 z ( ξ ) | 21 e ( ξ + 1 ) ( 1 + | z ( ξ ) | + | C D 2 3 3 2 z ( ξ ) | ) + 0 T ( ξ 2 3 s ) 7 4 1 Γ 2 3 ( 7 4 ) 8 + | z ( ξ ) | + | C D 2 3 3 2 z ( ξ ) | 150 e ξ + 3 ( 1 + | z ( ξ ) | + | C D 2 3 3 2 z ( ξ ) | ) , ξ J = ( 0 , T ) , z ( 0 ) = 0 T ( T 2 3 s ) 3 2 1 Γ 2 3 ( 3 2 ) 8 + | z ( ξ ) | + | C D 2 3 3 2 z ( ξ ) | 150 e ( ξ + 3 ) ( 1 + | z ( ξ ) | + | C D 2 3 3 2 z ( ξ ) | ) d 2 3 s , C D 2 3 z ( 0 ) = 0 , C D 2 3 β 0 z ( T ) = i = 1 3 0 T C D 2 3 β i z ( ξ ) d 2 3 ξ ,
where q = 2 3 , α = 3 2 , δ = σ = 7 4 , β 1 = 1 8 , β 2 = 5 16 , β 3 = 3 16 , β 0 = 1 16 , λ = 4 5 , T = 1 and p = 3 .  Set
f ( ξ , ζ , θ ) = 3 + | ζ | + | θ | 21 e ( ξ + 1 ) ( 1 + | ζ | + | θ | ) , ζ C ( J , R )
g ( ξ , ζ , θ ) = 8 + | ζ | + | θ | 150 e ( ξ + 3 ) ( 1 + | ζ | + | θ | )
and
K ( ξ , ζ , θ ) = 8 + | ζ | + | θ | 150 e ( ξ + 3 ) ( 1 + | ζ | + | θ | ) .
Clearly, the functions f , g , K are continuous. For each ζ , ζ ^ X , θ , θ ^ R , ξ [ 0 , 1 ] , we have
| f ( ξ , ζ , θ ) f ( ξ , ζ ^ , θ ^ ) | | ζ ζ ^ | + | θ θ ^ | 21 e 1 ,
which satisfies ( A 3 ) and N 1 = N 2 = 1 21 e 1 . We can observe that
| g ( ξ , ζ , θ ) g ( ξ , ζ ^ , θ ^ ) | | ζ ζ ^ | + | θ θ ^ | 150 e 3
which satisfies ( A 4 ) and N 5 = N 6 = 1 150 e 3 , and
| K ( ξ , ζ , θ ) K ( ξ , ζ ^ , θ ^ ) | | ζ ζ ^ | + | θ θ ^ | 150 e 3
satisfies ( A 5 ) , N 3 = N 4 = 1 150 e 3 . Additionally, we check that
| Y ω N 1 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + N 5 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) 1 N 2 N 6 T σ ( 1 ( 1 q ) σ ) σ q Γ q ( δ ) + Y χ N 3 1 N 4 | 0.0315193 < 1 .
We see that all the required conditions of Theorem 4 are satisfied; hence, (25) has at least one solution. By letting ψ ( ξ ) = ξ 2 for all ξ J , we have
I 2 3 3 2 ψ ( ξ 2 ) = 1 Γ q ( 3 2 ) 0 ξ ( ξ q s ) 3 2 1 s 2 d q s = 32 7 2 105 π 2 π ξ 2 .
Hence, ( A 9 ) is satisfied with L ψ = 2 π . Therefore, by Theorem 4.2, the given problem is UHRS and consequently GUHRS .

6. Conclusions

In this article, we present some necessary conditions for obtaining the existence, uniqueness, and Ulam stabilities of a solution to System (1). Schaefer’s FP theorem, the Arzela–Ascoli theorem, and the Banach contraction principle are all used to show the desired results of (1). In this way, the Ulam stability results are shown to investigate the solution of our proposed system (1) under specific assumptions and conditions. We conclude by providing an interesting example to implement the theoretical results and demonstrate its validity. From the derived results, we also conclude that such a method is powerful for solving Langevin q- FDE . In our forthcoming work, we consider the inclusion version problem with R L derivatives.

Author Contributions

Conceptualization, J.Y.; Methodology, W.W., K.H.K., A.Z., S.B.M. and J.Y.; Validation, A.Z.; Formal analysis, W.W., K.H.K., A.Z., S.B.M. and J.Y.; Investigation, K.H.K. and A.Z. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Not Applicable.

Acknowledgments

The authors extend their appreciation to the Deanship of Scientific Research at King Khalid University for funding this work through large group Research Project under grant number RGP2/47/44.

Conflicts of Interest

The authors declare no conflict of interest.

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Wang, W.; Khalid, K.H.; Zada, A.; Ben Moussa, S.; Ye, J. q-Fractional Langevin Differential Equation with q-Fractional Integral Conditions. Mathematics 2023, 11, 2132. https://doi.org/10.3390/math11092132

AMA Style

Wang W, Khalid KH, Zada A, Ben Moussa S, Ye J. q-Fractional Langevin Differential Equation with q-Fractional Integral Conditions. Mathematics. 2023; 11(9):2132. https://doi.org/10.3390/math11092132

Chicago/Turabian Style

Wang, Wuyang, Khansa Hina Khalid, Akbar Zada, Sana Ben Moussa, and Jun Ye. 2023. "q-Fractional Langevin Differential Equation with q-Fractional Integral Conditions" Mathematics 11, no. 9: 2132. https://doi.org/10.3390/math11092132

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