Existence results of nonlocal Robin boundary value problems for fractional (p,q)$(p,q)$-integrodifference equations

The existence results of a fractional $(p,q)$
(
p
,
q
)
-integrodifference equation with nonlocal Robin boundary condition are investigated by using Banach’s and Schauder’s fixed point theorems. Moreover, we study some properties of $(p,q)$
(
p
,
q
)
-integral that will be used as a tool for our calculations.

There are some recent papers studying the boundary value problem for (p, q)-difference equations [54][55][56]. However, the boundary value problem for fractional (p, q)-difference equations has not been studied since fractional (p, q)-operators have been defined lately. These motivate the authors for this research. This article investigates the existence results of a fractional (p, q)-integrodifference equation with nonlocal Robin boundary value conditions of the form is a given function; φ 1 , φ 2 : C(I T p,q , R) → R are given functionals; and for ϕ ∈ C(I T p,q × I T p,q , [0, ∞)), we define an operator of the (p, q)-integral of the product of functions ϕ and u as We aim to prove the existence and uniqueness of a solution for this problem by using Banach's fixed point theorem, and the existence of at least one solution by using Schauder's fixed point theorem. In addition, we provide an example to illustrate our results.

Preliminaries
In this section, we recall some basic definitions, notations, and lemmas. Letting 0 < q < p ≤ 1, we define the notations The (p, q)-forward jump and the (p, q)-backward jump operators are defined as The q-analogue of the power function (ab) n q with n ∈ N 0 := {0, 1, 2, . . .} is given by The (p, q)-analogue of the power function (ab) n p,q with n ∈ N 0 is given by For α ∈ R, we define a general form: Note that a α q = a α , a α p,q = ( a p ) α and (0) α q = (0) α p,q = 0 for α > 0. The (p, q)-gamma and (p, q)-beta functions are defined by respectively.

Definition 2.2
Let I be any closed interval of R containing a, b, and 0. Assuming that where provided that the series converges at x = a and Next, we define an operator I N p,q as The relations between (p, q)-difference and (p, q)-integral operators are given by ,q , and f , g be (p, q)-integrable on I T p,q . Then the following formulas hold: Lemma 2.6 ( [31], Fundamental theorem of (p, q)-calculus) Letting f : I → R be continuous at 0 and then F is continuous at 0 and D p,q F(x) exists for every x ∈ I where where t D p,q is (p, q)-difference with respect to t.
Next we introduce fractional (p, q)-integral and fractional (p, q)-difference of Riemann-Liouville type as follows.

Definition 2.3
For α > 0, 0 < q < p ≤ 1, and f defined on I T p,q , the fractional (p, q)-integral is defined by

Definition 2.4
For α > 0, 0 < q < p ≤ 1, and f defined on I T p,q , the fractional (p, q)difference operator of Riemann-Liouville type of order α is defined by Lemma 2.9 ([53]) Letting 0 < q < p ≤ 1 and f : I T p,q → R be continuous at 0,

Lemma 2.11
Let α, β > 0, 0 < q < p ≤ 1, and n ∈ Z. Then Proof From Lemma 2.10(a) and the definition of (p, q)-beta function, we have For n ∈ Z, we have t 0 (tqs) The proof is complete.
We next provide a lemma showing a result of the linear variant of problem (1.1).

4)
and the constants A η , A T , B η , B T , and Ω are defined by

Theorem 3.1 Assume that F
Suppose that the following conditions hold: (H 1 ) There exist constants 1 , 2 , 3 > 0 such that, for each t ∈ I T p,q and u i , v i ∈ R, i = 1, 2, 3, (H 2 ) There exist constants ω 1 , ω 2 > 0 such that, for each u, v ∈ C, . (3.9) Then problem (1.1) has a unique solution in I T p,q .
Proof For each t ∈ I T p,q and u, v ∈ C, we have Denote that Similarly, we have From (3.10) and (3.12), we obtain By (H 3 ) we can conclude that F is a contraction. Therefore, by using Banach's fixed point theorem, F has a fixed point which is a unique solution of problem (1.1) on I T p,q .

Existence of at least one solution
In this section, we present the existence of a solution to (1.1) by using Schauder's fixed point theorem.    Proof We organize the proof into three steps as follows.
Step II. The operator F is continuous on B L because of the continuity of F.