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Article

# On Sequential Fractional q-Hahn Integrodifference Equations

1
Research Center in Mathematics and Applied Mathematics, Department of Mathematics, Faculty of Science, Chiang Mai University, Chiang Mai 50200, Thailand
2
Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok 10800, Thailand
3
Mathematics Department, Faculty of Science and Technology, Suan Dusit University, Bangkok 10300, Thailand
*
Authors to whom correspondence should be addressed.
Mathematics 2020, 8(5), 753; https://doi.org/10.3390/math8050753
Submission received: 10 April 2020 / Revised: 5 May 2020 / Accepted: 6 May 2020 / Published: 9 May 2020
(This article belongs to the Special Issue Nonlinear Equations: Theory, Methods, and Applications)

## Abstract

:
In this paper, we prove existence and uniqueness results for a fractional sequential fractional q-Hahn integrodifference equation with nonlocal mixed fractional q and fractional Hahn integral boundary condition, which is a new idea that studies q and Hahn calculus simultaneously.
MSC:
39A10; 39A13; 39A70

## 1. Introduction

A q-difference operator $D q$ is an important tool in areas of mathematics and applications [1,2,3,4] such as orthogonal polynomials problems and mathematical control theories. Basic definitions and properties for q-difference calculus were presented by Kac and Cheung [5], Al-Salam [6], Agarwal [7], and Annaby and Mansour [8]. There are many research works widely studying the q-difference operators (see [9,10,11,12,13,14,15,16,17,18,19,20,21,22,23]).
A Hahn difference operator $D q ,$ arose from the forward difference operator and the Jackson q-difference operator was introduced by Hahn [24] in 1949. Then, the right inverse of $D q , ω$ presented in terms of Jackson q-integral and Nörlund sum was proposed by Aldwoah [25,26] in 2009. The Hahn difference operator can be used in studied of families of orthogonal polynomials and approximation problems (see [27,28,29]). More research works about Hahn difference calculus can be found in [30,31,32,33,34,35,36,37,38,39].
The fractional Hahn difference operators was introduced by Brikshavana and Sitthiwirattham [40] in 2017, and Wang et al. [41] in 2018. The extension of this operator has been used in the study of existence results of solution of boundary value problems [42,43,44,45], a generalization of Minkowski’s inequality [46], and impulsive fractional quantum Hahn operator [47,48].
From the literature, we have found that the study of fractional q-difference and fractional Hahn difference operators simultaneously have not been studied. Therefore, in this article, we devote ourselves to study the boundary value problem for equations that contain both fractional q-difference and Hahn difference operators. Our problem is a nonlocal mixed fractional q and Hahn integral boundary value problem for sequential fractional q-Hahn integrodifference equation of the form
$D q α D q , ω β u ( t ) = F t , u ( t ) , Ψ q θ u ( t ) , Υ q , ω ϕ u ( t ) , t ∈ I q I q , ω [ ω 0 , T ] , u ( η ) = λ I q , ω γ u ( η ) , η ∈ ( ω 0 , T ) , u ( T ) = μ I q γ u ( T ) ,$
where $I q I q , ω [ ω 0 , T ] : = ⋃ k = 0 ∞ I q q k s + ω [ k ] q , s ∈ [ ω 0 , T ] ;$ $I q x : = { q n x : n ∈ N 0 } ∪ { 0 }$; $N 0 : = N ∪ { 0 }$; $0 < q < 1 ; ω > 0 ; T > ω 0$; $α , β , γ , θ , ϕ ∈ ( 0 , 1 ] ; α + β ∈ ( 1 , 2 ] ;$$λ , μ ∈ R +$; $F ∈ C [ 0 , T ] × R × R × R , R$ is given function; and for $ψ ∈ C [ 0 , T ] × [ ω 0 , T ] , [ 0 , ∞ ) , φ ∈ C [ ω 0 , T ] × [ ω 0 , T ] , [ 0 , ∞ )$, we define
$Ψ q θ u ( t ) : = I q θ ψ u ( t ) = 1 Γ q , ω ( θ ) ∫ 0 t t − σ q ( s ) q θ − 1 ̲ ψ ( t , s ) u ( s ) d q s , Ψ q , ω ϕ u ( t ) : = I q , ω ϕ φ u ( t ) = 1 Γ q , ω ( ϕ ) ∫ ω 0 t t − σ q , ω ( s ) q , ω ϕ − 1 ̲ φ ( t , s ) u ( s ) d q , ω s .$
This paper is organized as follows. In Section 2, we provide some definitions and lemmas for q-difference and Hahn difference operators. In Section 3, we prove the existence and uniqueness of a solution to problem (1) by using the Banach fixed point theorem. In the last section, we give an example to illustrate our results.

## 2. Preliminaries

In this section, we recall the notations, definitions, and lemmas for q and Hahn difference calculus. For $q ∈ ( 0 , 1 )$, $ω > 0$, we define
$[ n ] q : = 1 − q n 1 − q = q n − 1 + . . . + q + 1 and [ n ] q ! : = ∏ k = 1 n 1 − q k 1 − q , n ∈ N .$
The q-analogue of the power function $( a − b ) q n ̲$ with $n ∈ N 0$ is given by
$( a − b ) q 0 ̲ : = 1 , ( a − b ) q n ̲ : = ∏ k = 0 n − 1 ( a − b q k ) , a , b ∈ R .$
The $q , ω$-analogue of the power function $( a − b ) q , ω n ̲$ with $n ∈ N 0$ is given by
$( a − b ) q , ω 0 ̲ : = 1 , ( a − b ) q , ω n ̲ : = ∏ k = 0 n − 1 a − ( b q k + ω [ k ] q ) , a , b ∈ R .$
For $α ∈ R$, the power function is given by
$( a − b ) q α ̲ = a α ∏ n = 0 ∞ 1 − b a q n 1 − b a q α + n , a ≠ 0 ,$
$( a − b ) q , ω α ̲ = ( a − ω 0 ) α ∏ n = 0 ∞ 1 − b − ω 0 a − ω 0 q n 1 − b − ω 0 a − ω 0 q α + n = ( a − ω 0 ) − ( b − ω 0 ) q α ̲ , a ≠ ω 0 .$
We let the notations, $a q α ̲ = a α ,$$( a − ω 0 ) q , ω α ̲ = ( a − ω 0 ) α$, and $( 0 ) q α ̲ = ( ω 0 ) q , ω α ̲ = 0$ for $α > 0$.
The q-gamma and q-beta functions are defined by
$Γ q ( x ) : = ( 1 − q ) q x − 1 ̲ ( 1 − q ) x − 1 , x ∈ R \ { 0 , − 1 , − 2 , . . . } , B q ( x , s ) : = ∫ 0 1 t x − 1 ( 1 − q t ) q s − 1 ̲ d q t = Γ q ( x ) Γ q ( s ) Γ q ( x + s ) ,$
respectively.
For $k ∈ N$, the q-analogue and $q , ω$-analogue of forward jump operator are defined by
$σ q k ( t ) : = q k t and σ q , ω k ( t ) : = q k t + ω [ k ] q ,$
respectively. The q-analogue and $q , ω$-analogue of backward jump operator are defined by
$ρ q k ( t ) : = t q k , and ρ q , ω k ( t ) : = t − ω [ k ] q q k ,$
respectively.
Definition 1.
For $q ∈ ( 0 , 1 ) ,$ the q-differnce of a real function f is defined by
$D q f ( t ) = f ( t ) − f ( q t ) ( 1 − q ) t , t ≠ 0 and D q f ( 0 ) = lim t → 0 D q f ( t ) .$
Let f be a function defined on the interval $[ 0 , T ]$. q-integral is defined by
$I q f t = ∫ 0 t f s d q s = ( 1 − q ) t ∑ n = 0 ∞ q n f q n t$
where the infinite series is convergent.
Definition 2.
For $q ∈ ( 0 , 1 )$, $ω > 0$ and f defined on an interval $I ⊆ R$ which contains $ω 0 : = ω 1 − q$, the Hahn difference of f is defined by
$D q , ω f ( t ) = f ( q t + ω ) − f ( t ) t ( q − 1 ) + ω for t ≠ ω 0 ,$
and $D q , ω f ( ω 0 ) = f ′ ( ω 0 )$.
For $a , b ∈ I ⊆ R$ with $a < ω 0 < b$ and $[ k ] q = 1 − q k 1 − q , k ∈ N 0 : = N ∪ { 0 }$, the $q , ω$-interval is defined by
$[ a , b ] q , ω : = q k a + ω [ k ] q : k ∈ N 0 ∪ q k b + ω [ k ] q : k ∈ N 0 ∪ ω 0 = [ a , ω 0 ] q , ω ∪ [ ω 0 , b ] q , ω = ( a , b ) q , ω ∪ a , b = [ a , b ) q , ω ∪ b = ( a , b ] q , ω ∪ a .$
We note that, for each $s ∈ [ a , b ] q , ω$, the sequence $σ q , ω k ( s ) k = 0 ∞ = q k s + ω [ k ] q k = 0 ∞$ is uniformly convergent to $ω 0$.
Definition 3.
Let I be any closed interval of $R$ that contains $a , b$ and $ω 0$. Letting $f : I → R$ be a given function, $q , ω$-integral of f from a to b is defined by
$∫ a b f ( t ) d q , ω t : = ∫ ω 0 b f ( t ) d q , ω t − ∫ ω 0 a f ( t ) d q , ω t$
where $∫ ω 0 x f ( t ) d q , ω t : = x ( 1 − q ) − ω ∑ k = 0 ∞ q k f x q k + ω [ k ] q , x ∈ I ,$ and the series converges at $x = a$ and $x = b$ where the sum of the right-hand side is called the Jackson–Nörlund sum.
Note that the actual domain of function f is defined on $[ a , b ] q , ω ⊂ I .$
The following fractional q integral, fractional Hahn integral, fractional q difference, and fractional Hahn difference of Riemann–Liouville type are defined.
Definition 4.
Let f be defined on $[ 0 , T ]$ and $α ≥ 0$, the fractional q-integral of the Riemann–Liouville type is defined by
$( I q α f ) ( t ) : = 1 Γ q ( α ) ∫ 0 t ( t − q s ) q α − 1 ̲ f ( t ) d q s = t ( 1 − q ) Γ q ( α ) ∑ n = 0 ∞ q n t − q n + 1 t q α − 1 ̲ f ( q n t ) = t α ( 1 − q ) Γ q ( α ) ∑ n = 0 ∞ q n ( 1 − q n + 1 ) q α − 1 ̲ f ( q n t ) ,$
and $( I q 0 f ) ( x ) = f ( x )$.
Definition 5.
Let f be defined on $[ ω 0 , T ] q , ω$ and $α , ω > 0 , q ∈ ( 0 , 1 )$, and the fractional Hahn integral, is defined by
$I q , ω α f ( t ) : = 1 Γ q ( α ) ∫ ω 0 t t − σ q , ω ( s ) q , ω α − 1 ̲ f ( s ) d q , ω s = [ t ( 1 − q ) − ω ] Γ q ( α ) ∑ n = 0 ∞ q n t − σ q , ω n + 1 ( t ) q , ω α − 1 ̲ f σ q , ω n ( t ) = ( 1 − q ) ( t − ω 0 ) α Γ q ( α ) ∑ n = 0 ∞ q n 1 − q n + 1 q α − 1 ̲ f σ q , ω n ( t ) ,$
and $( I q , ω 0 f ) ( t ) = f ( t )$.
Definition 6.
Let f be defined on $[ 0 , T ]$ and $α ≥ 0$, the fractional q-derivative of the Riemann–Liouville type of order α, is defined by
$( D q α f ) ( t ) : = ( D q N I q N − α f ) ( t ) = 1 Γ q ( − α ) ∫ 0 t t − σ q ( s ) q − α − 1 ̲ f ( s ) d q s ,$
and $( D q 0 f ) ( x ) = f ( x )$, where N is the smallest integer that is greater than or equal to α.
Definition 7.
Let f be defined on $[ ω 0 , T ] q , ω$ and $α , ω > 0 , q ∈ ( 0 , 1 )$, the fractional Hahn difference of the Riemann–Liouville type of order α is defined by
$D q , ω α f ( t ) : = ( D q , ω N I q , ω N − α f ) ( t ) = 1 Γ q ( − α ) ∫ ω 0 t t − σ q , ω ( s ) q , ω − α − 1 ̲ f ( s ) d q , ω s ,$
and $D q , ω 0 f ( t ) = f ( t )$, where N is the smallest integer that is greater than or equal to α.
Lemma 1
([10]). Letting $α > 0 , q ∈ ( 0 , 1 )$ and $f : I q T → R$,
$I q α D q α f ( t ) = f ( t ) + C 1 t α − 1 + . . . + C N t α − N ,$
for some $C i ∈ R , i = { 1 , 2 , . . . , N }$ and $N − 1 < α ≤ N , N ∈ N .$
Lemma 2
([40]). Letting $α > 0 , q ∈ ( 0 , 1 ) , ω > 0$ and $f : I q , ω T → R$,
$I q , ω α D q , ω α f ( t ) = f ( t ) + C 1 ( t − ω 0 ) α − 1 + . . . + C N ( t − ω 0 ) α − N ,$
for some $C i ∈ R , i = { 1 , 2 , . . . , N }$ and $N − 1 < α ≤ N , N ∈ N .$
Some auxiliary lemmas used to investigate the solution of the linear variant of (1) are provided as follows.
Lemma 3
([16]). Let $α , β ≥ 0$ and $p , q ∈ ( 0 , 1 )$. Then, the following formulas hold:
$∫ 0 η ( η − q t ) q α − 1 ̲ t β d q t = η α + β B q ( β + 1 , α ) , ∫ 0 η ∫ 0 s ( η − p s ) p α − 1 ̲ ( s − q t ) q β − 1 ̲ d q t d p s = η α + β [ β ] q B q ( β + 1 , α ) .$
Lemma 4
([40]). Letting $α , β > 0 , p , q ∈ ( 0 , 1 )$ and $ω > 0$,
$∫ ω 0 t t − σ q , ω ( s ) q , ω α − 1 ̲ ( s − ω 0 ) β d q , ω s = ( t − ω 0 ) α + β B q ( β + 1 , α ) , ∫ ω 0 t ∫ ω 0 x t − σ p , ω ( x ) p , ω α − 1 ̲ x − σ q , ω ( s ) q , ω β − 1 ̲ d q , ω s d p , ω x = ( t − ω 0 ) α + β [ β ] q B q ( β + 1 , α ) .$
Employing Lemmas 3 and 4, we obtain the solution of the linear variant of problem (1) as shown in the following lemma.
Lemma 5.
Let $α , β , γ ∈ ( 0 , 1 ] , α + β ∈ ( 1 , 2 ]$; $0 < q < 1 ; ω > 0 ; T > ω 0$; $λ , μ ∈ R +$; $h ∈ C [ 0 , T ] , R$ be a given function. Then, the linear variant problem
$D q α D q , ω β u ( t ) = h ( t ) , t ∈ I q I q , ω [ ω 0 , T ] , u ( η ) = λ I q , ω γ u ( η ) , η ∈ ( ω 0 , T ) , u ( T ) = μ I q γ u ( T )$
has the unique solution which is in a form
$u ( t ) = 1 Γ q ( α ) Γ q ( β ) ∫ ω 0 t ∫ 0 x t − σ q , ω ( s ) q , ω β − 1 ̲ x − σ q ( s ) q α − 1 ̲ h ( s ) d q s d q , ω x + A T O η [ h ] − A η O T [ h ] 1 Ω Γ q ( β ) ∫ ω 0 t t − σ q , ω ( s ) q , ω β − 1 ̲ s α − 1 d q , ω s − B T O η [ h ] − B η O T [ h ] ( t − ω 0 ) β − 1 Ω$
for $t ∈ [ ω 0 , T ]$, where the functionals $O η [ h ]$ and $O T [ h ]$ are defined by
$O η [ h ] : = − 1 Γ q ( α ) Γ q ( β ) ∫ ω 0 η ∫ 0 x η − σ q , ω ( x ) q , ω β − 1 ̲ x − σ q ( s ) q α − 1 ̲ h ( s ) d q s d q , ω x$
$+ λ Γ q ( α ) Γ q ( β ) Γ q ( γ ) ∫ ω 0 η ∫ ω 0 r ∫ 0 x η − σ q , ω ( r ) q , ω γ − 1 ̲ r − σ q , ω ( x ) q , ω β − 1 ̲ × x − σ q ( s ) q α − 1 ̲ h ( s ) d q s d q , ω x d q , ω r , O T [ h ] : = − 1 Γ q ( α ) Γ q ( β ) ∫ ω 0 T ∫ 0 x T − σ q , ω ( x ) q , ω β − 1 ̲ x − σ q ( s ) q α − 1 ̲ h ( s ) d q s d q , ω x$
$+ μ Γ q ( α ) Γ q ( β ) Γ q ( γ ) ∫ 0 T ∫ ω 0 r ∫ 0 x T − σ q ( r ) q γ − 1 ̲ r − σ q , ω ( x ) q , ω β − 1 ̲ × x − σ q ( s ) q α − 1 ̲ h ( s ) d q s d q , ω x d q r ,$
and the constants $A η , A T , B η , B T , Ω$ are defined by
$A η : = ( η − ω 0 ) β − 1 − λ Γ q ( γ ) ∫ ω 0 T η − σ q , ω ( s ) q , ω γ − 1 ̲ ( s − ω 0 ) β − 1 d q , ω s ,$
$A T : = ( T − ω 0 ) β − 1 − μ Γ q ( γ ) ∫ 0 T T − σ q ( s ) q γ − 1 ̲ ( s − ω 0 ) β − 1 d q s ,$
$B η : = 1 Γ q ( β ) ∫ ω 0 η η − σ q , ω ( s ) q , ω β − 1 ̲ s α − 1 d q , ω s − λ Γ q ( β ) Γ q ( γ ) ∫ ω 0 η ∫ ω 0 x η − σ q , ω ( x ) q , ω γ − 1 ̲ x − σ q , ω ( s ) q , ω β − 1 ̲ s α − 1 d q , ω s d q , ω x ,$
$B T : = 1 Γ q ( β ) ∫ ω 0 T T − σ q , ω ( s ) q , ω β − 1 ̲ s α − 1 d q , ω s − μ Γ q ( β ) Γ q ( γ ) ∫ 0 T ∫ ω 0 x T − σ q ( x ) q γ − 1 ̲ x − σ q , ω ( s ) q , ω β − 1 ̲ s α − 1 d q , ω s d q x ,$
$Ω : = A T B η − A η B T ≠ 0 .$
Proof.
Firstly, we take fractional q-integral of order $α$ for (2). Then, we have
$D q , ω β u ( t ) = C 0 t α − 1 + ( 1 − q ) t α Γ q ( α ) ∑ k = 0 ∞ q k 1 − q k + 1 q α − 1 ̲ h σ q k ( t ) = C 0 t α − 1 + 1 Γ q ( α ) ∫ 0 t t − σ q ( s ) q α − 1 ̲ h ( x ) d q s ,$
for $t ∈ I q , ω [ ω 0 , T ] : = { q n s + ω [ n ] q : s ∈ [ ω 0 , T ] , n ∈ N 0 } ∪ { ω 0 } .$
Taking fractional Hahn integral of order $β$ for (11), we obtain
$u ( t ) = C 1 ( t − ω 0 ) β − 1 + C 0 Γ q ( β ) ( 1 − q ) ( t − ω 0 ) β ∑ k = 0 ∞ q k 1 − q k + 1 q β − 1 ̲ σ q , ω k ( t ) α − 1 + 1 Γ q ( α ) Γ q ( β ) ( 1 − q ) 2 ( t − ω 0 ) β ∑ h = 0 ∞ ∑ k = 0 ∞ q h + k 1 − q h + 1 q β − 1 ̲ × 1 − q k + 1 q α − 1 ̲ σ q , ω h ( t ) α h σ q k σ q , ω h ( t ) = C 1 ( t − ω 0 ) β − 1 + C 0 Γ q ( β ) ∫ ω 0 t t − σ q , ω ( s ) q , ω β − 1 ̲ s α − 1 d q , ω s + 1 Γ q ( α ) Γ q ( β ) ∫ ω 0 t ∫ 0 x t − σ q , ω ( x ) q , ω β − 1 ̲ x − σ q ( s ) q α − 1 ̲ h ( s ) d q s d q , ω x ,$
for $t ∈ [ ω 0 , T ] .$
Taking fractional q-integral of order $γ$ for (12), we have
$I q γ u ( t ) = C 1 Γ q ( γ ) ∫ 0 t t − σ q ( s ) q γ − 1 ̲ ( s − ω 0 ) β − 1 d q s + C 0 Γ q ( β ) Γ q ( γ ) ∫ 0 t ∫ ω 0 x t − σ q ( x ) q γ − 1 ̲ x − σ q , ω ( s ) q , ω β − 1 ̲ s α − 1 d q , ω s d q x , + 1 Γ q ( α ) Γ q ( β ) Γ q ( γ ) ∫ 0 t ∫ ω 0 r ∫ 0 x t − σ q ( r ) q γ − 1 ̲ r − σ q , ω ( x ) q , ω β − 1 ̲ × x − σ q ( s ) q α − 1 ̲ h ( s ) d q s d q , ω x d q r ,$
for $t ∈ [ 0 , T ]$.
In addition, we take fractional Hahn integral of order $γ$ for (12) to get
$I q , ω γ u ( t ) = C 1 Γ q ( γ ) ∫ ω 0 t t − σ q , ω ( s ) q , ω γ − 1 ̲ ( s − ω 0 ) β − 1 d q , ω s + C 0 Γ q ( β ) Γ q ( γ ) ∫ ω 0 t ∫ ω 0 x t − σ q , ω ( x ) q , ω γ − 1 ̲ x − σ q , ω ( s ) q , ω β − 1 ̲ s α − 1 d q , ω s d q , ω x , + 1 Γ q ( α ) Γ q ( β ) Γ q ( γ ) ∫ ω 0 t ∫ ω 0 r ∫ 0 x t − σ q , ω ( r ) q , ω γ − 1 ̲ r − σ q , ω ( x ) q , ω β − 1 ̲ × x − σ q ( s ) q α − 1 ̲ h ( s ) d q s d q , ω x d q , ω r ,$
for $t ∈ [ ω 0 , T ]$.
Substituting $t = η$ into (12) and (14), and employing the first condition of (2), we have
$A η C 1 + B η C 0 = O η [ h ] .$
Taking $t = T$ into (12) and (13), and employing the second condition of (2), we have
$A T C 1 + B T C 0 = O T [ h ] .$
Solving Equations (15) and (16), we obtain
$C 1 = B η O T [ h ] − B T O η [ h ] Ω and C 0 = A T O η [ h ] − A η O T [ h ] Ω .$
where $O η [ h ] , O T [ h ] , A η , A T , B η , B T$ and $Ω$ are defined by Equations (4)–(10).
Substituting $C 0$ and $C 1$ into (12), we obtain the solution (3). □

## 3. Existence Results

In this section, the existence and uniqueness result for the mixed q-Hahn problem (1) is studied. Let $C = C [ ω 0 , T ] , R$ be a Banach space of all function u with the norm defined by
$∥ u ∥ C = max t ∈ [ ω 0 , T ] | u ( t ) | .$
The operator $F : C → C$ is defined by
$( F u ) ( t ) : = 1 Γ q ( α ) Γ q ( β ) ∫ ω 0 t ∫ 0 x t − σ q , ω ( s ) q , ω β − 1 ̲ x − σ q ( s ) q α − 1 ̲ × F s , u ( s ) , Ψ q θ u ( s ) , Υ q , ω ϕ u ( s ) d q s d q , ω x + A T O η [ F u ] − A η O T [ F u ] 1 Ω Γ q ( β ) ∫ ω 0 t t − σ q , ω ( s ) q , ω β − 1 ̲ s α − 1 d q , ω s − B T O η [ F u ] − B η O T [ F u ] ( t − ω 0 ) β − 1 Ω$
where the functionals $O η [ F u ] , O T [ F u ]$ are defined by
$O η [ F u ] : = − 1 Γ q ( α ) Γ q ( β ) ∫ ω 0 η ∫ 0 x η − σ q , ω ( x ) q , ω β − 1 ̲ x − σ q ( s ) q α − 1 ̲ × F s , u ( s ) , Ψ q θ u ( s ) , Υ q , ω ϕ u ( s ) d q s d q , ω x + λ Γ q ( α ) Γ q ( β ) Γ q ( γ ) ∫ ω 0 η ∫ ω 0 r ∫ 0 x η − σ q , ω ( r ) q , ω γ − 1 ̲ r − σ q , ω ( x ) q , ω β − 1 ̲ × x − σ q ( s ) q α − 1 ̲ F s , u ( s ) , Ψ q θ u ( s ) , Υ q , ω ϕ u ( s ) d q s d q , ω x d q , ω r ,$
$O T [ F u ] : = − 1 Γ q ( α ) Γ q ( β ) ∫ ω 0 T ∫ 0 x T − σ q , ω ( x ) q , ω β − 1 ̲ x − σ q ( s ) q α − 1 ̲ × F s , u ( s ) , Ψ q θ u ( s ) , Υ q , ω ϕ u ( s ) d q s d q , ω x + μ Γ q ( α ) Γ q ( β ) Γ q ( γ ) ∫ 0 T ∫ ω 0 r ∫ 0 x T − σ q ( r ) q γ − 1 ̲ r − σ q , ω ( x ) q , ω β − 1 ̲ × x − σ q ( s ) q α − 1 ̲ F s , u ( s ) , Ψ q θ u ( s ) , Υ q , ω ϕ u ( s ) d q s d q , ω x d q r ,$
and the constants $A η , A T , B η , B T , Ω$ are defined by (6)–(10), respectively.
The problem (1) has solution if and only if the operator $F$ has fixed point. We show the proof in the following theorem.
Theorem 1.
Assume that $F : [ 0 , T ] × R × R × R → R$ is continuous, $ψ : [ 0 , T ] × [ ω 0 , T ] → [ 0 , ∞ )$ and $φ : [ ω 0 , T ] × [ ω 0 , T ] → [ 0 , ∞ )$ are continuous with $ψ 0 = max ψ ( t , s ) : ( t , s ) ∈ [ 0 , T ] × [ ω 0 , T ]$ and $φ 0 = max φ ( t , s ) : ( t , s ) ∈ [ ω 0 , T ] × [ ω 0 , T ]$. In addition, suppose that the following conditions hold:
$( H 1 )$
There exist constants $ℓ 1 , ℓ 2 , ℓ 3 > 0$ such that for each $t ∈ [ 0 , T ]$ and $u , v ∈ R ,$
$| F t , u , Ψ q θ u , Υ q , ω ϕ u − F t , v , Ψ q θ v , Υ q , ω ϕ v | ≤ ℓ 1 | u − v | + ℓ 2 | Ψ q θ u − Ψ q θ v | + ℓ 3 | Υ q , ω ϕ u − Υ q , ω ϕ v | .$
$( H 2 )$
$L Ξ < 1$,
where
$L : = ℓ 1 + ℓ 2 ψ 0 T θ Γ q ( θ + 1 ) + ℓ 3 φ 0 ( T − ω 0 ) ϕ Γ q ( ϕ + 1 ) ,$
$Ξ : = T α ( T − ω 0 ) β Γ q ( α + 1 ) Γ q ( β + 1 ) + Φ 1 Θ T + Φ 2 Θ η ,$
$Φ 1 : = η α ( η − ω 0 ) β Γ q ( α + 1 ) Γ q ( β + 1 ) 1 − λ ( η − ω 0 ) γ Γ q ( γ + 1 ) ,$
$Φ 2 : = T α ( T − ω 0 ) β Γ q ( α + 1 ) Γ q ( β + 1 ) 1 − μ T γ Γ q ( γ + 1 ) ,$
$Θ T : = 1 | Ω | | A T | T α − 1 ( T − ω 0 ) β Γ q ( β + 1 ) + | B T | ( T − ω 0 ) β − 1 ,$
$Θ η : = 1 | Ω | | A η | T α − 1 ( T − ω 0 ) β Γ q ( β + 1 ) + | B η | ( T − ω 0 ) β − 1 .$
Then, problem (1) has a unique solution.
Proof.
Firstly, we verify $F$ map bounded sets into bounded sets in $B L = { u ∈ C : ∥ u ∥ C ≤ L }$. Let $K = max t ∈ I q , ω T | F ( t , 0 , 0 , 0 ) |$, $L$ be a constant satisfied with
$L ≥ K Ξ 1 − L Ξ ,$
and the notation $| S ( t , u , 0 ) | = | F t , u , Ψ q θ u , Υ q , ω ϕ u − F t , 0 , 0 , 0 | + | F t , 0 , 0 , 0 | .$
For each $t ∈ [ 0 , T ]$ and $u ∈ B L$
$| O η [ F u ] | ≤ | 1 Γ q ( α ) Γ q ( β ) ∫ ω 0 η ∫ 0 x η − σ q , ω ( x ) q , ω β − 1 ̲ x − σ q ( s ) q α − 1 ̲ | S ( s , u , 0 ) | d q s d q , ω x − λ Γ q ( α ) Γ q ( β ) Γ q ( γ ) ∫ ω 0 η ∫ ω 0 r ∫ 0 x η − σ q , ω ( r ) q , ω γ − 1 ̲ r − σ q , ω ( x ) q , ω β − 1 ̲ × x − σ q ( s ) q α − 1 ̲ | S ( s , u , 0 ) | d q s d q , ω x d q , ω r | ≤ L ∥ u ∥ C + K Φ 1 ≤ L L + K Φ 1 .$
Similary,
$| O T [ F u ] | ≤ L L + K Φ 2 .$
From (27) and (28), we find that
$| ( F u ) ( t ) | ≤ 1 Γ q ( α ) Γ q ( β ) ∫ ω 0 T ∫ 0 x T − σ q , ω ( s ) q , ω β − 1 ̲ x − σ q ( s ) q α − 1 ̲ | S ( s , u , 0 ) | d q s d q , ω x + A T | O η [ F u ] | + A η | O T [ F u ] | 1 Ω Γ q ( β ) ∫ ω 0 T T − σ q , ω ( s ) q , ω β − 1 ̲ s α − 1 d q , ω s + B T | O η [ F u ] | + B η | O T [ F u ] | ( T − ω 0 ) β − 1 Ω ≤ Ξ L L + K ≤ L .$
Therefore, we obtain $∥ F u ∥ C ≤ L$, which implies that $F B L ⊂ B L$.
Next, we aim to prove that $F$ is contraction. Let the notation
$H | u − v | ( t ) = | F t , u ( t ) , Ψ q θ u ( t ) , Υ q , ω ϕ u ( t ) − F t , v ( t ) , Ψ q θ v ( t ) , Υ q , ω ϕ v ( t ) | ,$
for each $t ∈ [ 0 , T ]$ and $u , v ∈ C$. From (18), we find that
$| O η [ F u ] − O η [ F v ] | ≤ | 1 Γ q ( α ) Γ q ( β ) ∫ ω 0 η ∫ 0 x η − σ q , ω ( x ) q , ω β − 1 ̲ x − σ q ( s ) q α − 1 ̲ H | u − v | ( s ) d q s d q , ω x − λ Γ q ( α ) Γ q ( β ) Γ q ( γ ) ∫ ω 0 η ∫ ω 0 r ∫ 0 x η − σ q , ω ( r ) q , ω γ − 1 ̲ r − σ q , ω ( x ) q , ω β − 1 ̲ × x − σ q ( s ) q α − 1 ̲ H | u − v | ( s ) d q s d q , ω x d q , ω r | ≤ ℓ 1 | u − v | + ℓ 2 | Ψ q θ u − Ψ q θ v | + ℓ 3 | Υ q , ω ϕ u − Υ q , ω ϕ v | × η α ( η − ω 0 ) β Γ q ( α + 1 ) Γ q ( β + 1 ) − λ η α ( η − ω 0 ) β + γ Γ q ( α + 1 ) Γ q ( β + 1 ) Γ q ( γ + 1 ) ≤ ℓ 1 + ℓ 2 ψ 0 T θ Γ q ( θ + 1 ) + ℓ 3 φ 0 ( T − ω 0 ) ϕ Γ q ( ϕ + 1 ) | u − v | Φ 1 ≤ L Φ 1 ∥ u − v ∥ C .$
Similary, from (19), we have
$| O T [ F u ] − O T [ F v ] | ≤ L Φ 2 ∥ u − v ∥ C .$
Next, we find that
$| ( F u ) ( t ) − ( F v ) ( t ) | ≤ 1 Γ q ( α ) Γ q ( β ) ∫ ω 0 T ∫ 0 x T − σ q , ω ( s ) q , ω β − 1 ̲ x − σ q ( s ) q α − 1 ̲ H | u − v | ( s ) d q s d q , ω x + A T | O η [ F u ] − O η [ F v ] | + A η | O T [ F u ] − O T [ F v ] | 1 Ω Γ q ( β ) ∫ ω 0 T T − σ q , ω ( s ) q , ω β − 1 ̲ × s α − 1 d q , ω s + B T | O η [ F u ] − O η [ F v ] | + B η | O T [ F u ] − O T [ F v ] | ( T − ω 0 ) β − 1 Ω ≤ ∥ u − v ∥ C L [ T α ( T − ω 0 ) β Γ q ( α + 1 ) Γ q ( β + 1 ) + Φ 1 | Ω | | A T | T α − 1 ( T − ω 0 ) β Γ q ( β + 1 ) + | B T | ( T − ω 0 ) β − 1 + Φ 2 | Ω | | A η | T α − 1 ( T − ω 0 ) β Γ q ( β + 1 ) + | B η | ( T − ω 0 ) β − 1 ] ≤ L Ξ ∥ u − v ∥ C .$
By $( H 2 )$, we can conclude that $F$ is a contraction. From Banach fixed point theorem, $F$ has a fixed point. Therefore, problem (1) has a unique solution. □

## 4. Example

In this section, we give an example of nonlocal fractional q and Hahn integral boundary value problem for sequential fractional q-Hahn integrodifference equation:
$D 1 2 1 3 D 1 2 , 2 3 3 4 u ( t ) = 1 1000 e 2 + t 2 ( 1 + | u ( t ) | ) [ e − ( 4 t + π 3 ) u 2 + 2 | u | + e − ( e 3 + cos 2 π t ) Ψ 1 2 1 2 u ( t ) + e − ( 1 + sin 2 π t ) Ψ 1 2 , 2 3 2 5 u ( t ) ] , t ∈ I 1 2 I 1 2 , 2 3 [ 4 3 , 10 ] u 5 = 1 10 π I 1 2 , 2 3 1 5 u 5 , u 10 = 1 20 E I 1 2 1 5 u 10 ,$
where $ψ ( t , s ) = e − | s − t | ( t + 20 ) 3$ and $φ ( t , s ) = e − 2 | s − t | ( t + 30 ) 2$.
Here, $α = 1 3 , β = 3 4 , γ = 1 5 , θ = 1 2 , ϕ = 2 5 ,$$q = 1 2 , ω = 2 3 , ω 0 = ω 1 − q = 4 3 , T = 10 , η = 5$, $λ = 1 10 π , μ = 1 20 e ,$ and $F t , u ( t ) , Ψ q θ u ( t ) , Υ q , ω ϕ u ( t ) = 1 1000 e 2 + t 2 ( 1 + | u ( t ) | ) ×$ $e − ( 4 t + π 3 ) u 2 + 2 | u | + e − ( e 3 + cos 2 π t ) Ψ 1 2 1 2 u ( t ) + e − ( 1 + sin 2 π t ) Ψ 1 2 , 2 3 2 5 u ( t ) .$
After calculating, we get
$| A η | ≈ 0 . 7567 , | A T | ≈ 0 . 5984 , | B η | ≈ 0 . 9962 , | B T | ≈ 1 . 1816 ,$
$and | Ω | ≈ 0.2980 .$
For all $t ∈ 0 , 10$ and $u , v ∈ R$, we find that
$F t , u , Ψ q θ u , Υ q , ω ϕ u − F t , v , Ψ q θ v , Υ q , ω ϕ v ≤ 1 1000 e 2 + π 3 | u − v | + 1 1000 e 2 + e 3 Ψ q θ u − Ψ q θ v + 1 1000 e 3 Υ q , ω ϕ u − Υ q , ω ϕ v .$
Thus, $( H 1 )$ holds with $ℓ 1 = 0.0000475 , ℓ 2 = 0.0000547$, and $ℓ 3 = 0.0000498$.
Next, we find that
$ψ 0 = 0 . 00125 , φ 0 = 0 . 00111 , L = 0 . 000461 , Φ 1 = 4 . 9572 , Φ 2 = 12 . 1191 ,$
$Θ T = 4.6218 , Θ η = 4.8705 and Ξ = 92.4997 .$
Since
$L Ξ ≈ 0.0426 < 1 ,$
we see that the condition $( H 2 )$ holds.
Hence, by Theorem 1, problem (31) has a unique solution.

## 5. Conclusions

We have proved existence and uniqueness results of the sequential fractional q-Hahn integrodifference equation with nonlocal mixed fractional q and fractional Hahn integral boundary condition (1) by using the Banach fixed point theorem, and the existence of at least a solution by Schauder’s fixed point theorem. Our problem contains both fractional q-difference and fractional Hahn difference operators, which is a new idea.

## Author Contributions

Conceptualization, T.D., N.P. and T.S.; Formal analysis, T.D., N.P. and T.S.; Funding acquisition, N.P.; Investigation, T.D., N.P. and T.S.; Methodology, T.D., N.P. and T.S.; Writing—original draft, T.D., N.P. and T.S.; Writing—review and editing, T.D., N.P. and T.S. All authors contributed equally to this work. All authors have read and agreed to the published version of the manuscript.

## Funding

This research was funded by King Mongkut’s University of Technology North Bangkok. Contract No. KMUTNB-61-GOV-D-64.

## Acknowledgments

This research was supported by Chiang Mai University.

## Conflicts of Interest

The authors declare that they have no competing interests.

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Dumrongpokaphan, T.; Patanarapeelert, N.; Sitthiwirattham, T. On Sequential Fractional q-Hahn Integrodifference Equations. Mathematics 2020, 8, 753. https://doi.org/10.3390/math8050753

AMA Style

Dumrongpokaphan T, Patanarapeelert N, Sitthiwirattham T. On Sequential Fractional q-Hahn Integrodifference Equations. Mathematics. 2020; 8(5):753. https://doi.org/10.3390/math8050753

Chicago/Turabian Style

Dumrongpokaphan, Thongchai, Nichaphat Patanarapeelert, and Thanin Sitthiwirattham. 2020. "On Sequential Fractional q-Hahn Integrodifference Equations" Mathematics 8, no. 5: 753. https://doi.org/10.3390/math8050753

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