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Article

q1q2-Ostrowski-Type Integral Inequalities Involving Property of Generalized Higher-Order Strongly n-Polynomial Preinvexity

by
Humaira Kalsoom
1 and
Miguel Vivas-Cortez
2,*
1
Department of Mathematics, Zhejiang Normal University, Jinhua 321004, China
2
Escuela de Ciencias Físicas y Matemáticas, Facultad de Ciencias Naturales y Exactas, Pontificia Universidad Católica del Ecuador, Sede Quito 17-01-2184, Ecuador
*
Author to whom correspondence should be addressed.
Symmetry 2022, 14(4), 717; https://doi.org/10.3390/sym14040717
Submission received: 2 February 2022 / Revised: 20 February 2022 / Accepted: 23 February 2022 / Published: 1 April 2022
(This article belongs to the Special Issue Symmetry in Quantum Calculus)

Abstract

:
Quantum calculus has numerous applications in mathematics. This novel class of functions may be used to produce a variety of conclusions in convex analysis, special functions, quantum mechanics, related optimization theory, and mathematical inequalities. It can drive additional research in a variety of pure and applied fields. This article’s main objective is to introduce and study a new class of preinvex functions, which is called higher-order generalized strongly n-polynomial preinvex function. We derive a new q 1 q 2 -integral identity for mixed partial q 1 q 2 -differentiable functions. Because of the nature of generalized convexity theory, there is a strong link between preinvexity and symmetry. Utilizing this as an auxiliary result, we derive some estimates of upper bound for functions whose mixed partial q 1 q 2 -differentiable functions are higher-order generalized strongly n-polynomial preinvex functions on co-ordinates. Our results are the generalizations of the results in earlier papers. Quantum inequalities of this type and the techniques used to solve them have applications in a wide range of fields where symmetry is important.

1. Introduction

Calculus is a discipline of mathematics that assists us in the study of derivatives and integrals. The classical derivative was twisted with the strength regulation kind kernel, and this eventually gave rise to a new calculus known as the quantum calculus. The study of calculus without limitations is known as quantum calculus (abbreviated q-calculus). Quantum calculus, notably q-fractional calculus, q-integral calculus, and q-transform analysis, has become a valuable tool in many disciplines of mathematics and physics in recent decades. As one of the first few scholars to define the q-analogue of the derivative and integral operators, as well as to propose some of their applications, Jackson [1] was among the first few researchers to do so. To emphasize the importance of quantum integral inequalities, it is necessary to point out that they are more valuable and instructive than their classical equivalents. In its historical evolution, quantum calculus can be traced back to Euler and Jacobi; nevertheless, in recent decades, it has witnessed rapid progress, as illustrated in [2,3,4]. Because of this, numerous papers have been written in which new generalizations of classical ideas of quantum calculus have been proposed and discussed in detail. On finite intervals, Tariboon and Ntouyas [5,6] introduced quantum calculus ideas and obtained various q-analogues of conventional mathematical objects. As a result, numerous innovative conclusions relating to quantum analogs of classical mathematical achievements have already been published in the scientific literature due to this inspiration. It was discovered by Noor et al. [7] that they could generate a new q-analogues of inequality by using a first-order q-differentiable convex function. This work by Humaira et al. [8] provides a generalized q 1 q 2 -integral identity and numerous novel q 1 q 2 -differentiable convex functions over finite rectangles. Wu et al. [9] proposed a corrected q-analogue of the standard Simpson inequality for preinvex functions, for more details see in [10,11,12,13,14] and the references cited therein.
Integral inequalities play an essential role in understanding the universe, and there are many direct approaches to find the uniqueness and existence of the linear and nonlinear differential equations where symmetry plays an important role. Mathematical inequalities have streamlined the concept of classical convexity. Scientific observations and calculations rely on convex functions and their relationship to mathematical inequalities. The classical idea of convexity has recently been extended and generalized. According to Hanson [15], the introduction of invexity is one of the most important generalizations of convexity as it allows for the opening of a new type of convexity. The importance of invexity in optimization has been significantly increased due to this research, see in [16,17,18]. According to Mohan and Weir [19,20], they established a new class of functions known as preinvexity as a generalization of the concept of convexity.
Let Φ : J R R be convex on [ ϖ 1 , ϖ 2 ] , where ϖ 1 , ϖ 2 J with ϖ 1 < ϖ 2
Φ ϖ 1 + ϖ 2 2 1 ϖ 2 ϖ 1 ϖ 1 ϖ 2 Φ ( σ ) d σ Φ ( ϖ 1 ) + Φ ( ϖ 2 ) 2 .
Inequality (1) was introduced by C. Hermite [21] and investigated by J. Hadamard [22] in 1893. Both inequalities hold in the inverted direction if Φ is concave. Due to its role in various fields of modern mathematics, such as numerical analysis and functional and mathematical analyses, the validity of Hermit–Hadamard’s inequality is important in the literature.
Alexander Ostrowski introduced the Ostrowski’s inequality in [23], and with the passing of the years, generalizations on the same, involving derivatives of the function under study, have taken place.
Definition 1.
Let Φ : [ ϖ 1 , ϖ 2 ] R R be continuous on [ ϖ 1 , ϖ 2 ] and differentiable function on ϖ 1 , ϖ 2 , whose derivative Φ : ϖ 1 , ϖ 2 R is bounded on ϖ 1 , ϖ 2 , i.e., Φ ϱ M . Then, the following inequality holds:
Φ ( ϱ ) 1 ϖ 2 ϖ 1 ϖ 1 ϖ 2 Φ σ d σ 1 4 + ϱ ϖ 1 + ϖ 2 2 2 ϖ 2 ϖ 1 2 ϖ 2 ϖ 1 M ,
for all ϱ [ ϖ 1 , ϖ 2 ] . The constant 1 4 is the best possible constant.
The inequality (2) can be written in equivalent form as follows:
Φ ( ϱ ) 1 ϖ 2 ϖ 1 ϖ 1 ϖ 2 Φ σ d σ M ϖ 2 ϖ 1 ϱ ϖ 1 2 + ϖ 2 ϱ 2 2 .
In the field of mathematical inequalities, Ostrowski-type inequality has been given more attention by many mathematicians due to its applicability and usefulness. Recently, the generalizations, variants, and applications of the Ostrowski inequality have attracted the attention of many mathematicians, for more details see in [24,25,26,27,28,29,30,31,32] and the references cited therein.
Classical inequalities of the Ostrowski-type based on co-ordinated convex functions are determined in [33] by means of the following equality.
Theorem 1.
Let Φ : N R 2 R is a mixed partial differentiable function on N o and let [ ϖ 1 , ϖ 2 ] × [ ϖ 3 , ϖ 4 ] N o with ϖ 1 , ϖ 2 , ϖ 3 , ϖ 4 N o such that ϖ 1 < ϖ 2 , ϖ 3 < ϖ 4 . If 2 Φ z w L [ ϖ 1 , ϖ 2 ] × [ ϖ 3 , ϖ 4 ] , Then the following equality holds:
Φ ϱ , ρ + 1 ϖ 2 ϖ 1 ϖ 4 ϖ 3 ϖ 1 ϖ 2 ϖ 3 ϖ 4 Φ ( u , v ) d u d v Y = ( ϱ ϖ 1 ) 2 ( ρ ϖ 3 ) 2 ϖ 2 ϖ 1 ϖ 4 ϖ 3 0 1 0 1 z w 2 z w Φ ( z ϱ + ( 1 z ) ϖ 1 , w ρ + ( 1 w ) ϖ 3 ) d z d w ( ϱ ϖ 1 ) 2 ( ϖ 4 ρ ) 2 ϖ 2 ϖ 1 ϖ 4 ϖ 3 0 1 0 1 z w 2 z w Φ ( z ϱ + ( 1 z ) ϖ 1 , w ρ + ( 1 w ) ϖ 4 ) d z d w ( ϖ 2 ϱ ) 2 ( ρ ϖ 3 ) 2 ϖ 2 ϖ 1 ϖ 4 ϖ 3 0 1 0 1 z w 2 z w Φ ( z ϱ + ( 1 z ) ϖ 2 , w ρ + ( 1 w ) ϖ 3 ) d z d w + ( ϖ 2 ϱ ) 2 ( ϖ 4 ρ ) 2 ϖ 2 ϖ 1 ϖ 4 ϖ 3 0 1 0 1 z w 2 z w Φ ( z ϱ + ( 1 z ) ϖ 2 , w ρ + ( 1 w ) ϖ 4 ) d z d w
for all ϱ , ρ N and
Y = 1 ϖ 4 ϖ 3 ϖ 3 ϖ 4 Φ ( ϱ , v ) d v + 1 ϖ 2 ϖ 1 ϖ 1 ϖ 2 Φ ( u , ρ ) d u .
Noor et al. [7] utilized the quantum calculus to obtain some new q-analogues of the Ostrowski inequality.
Theorem 2.
Let Φ : N R be continuous with q ( 0 , 1 ) . If ϖ 1 D q Φ is an integrable function on N o , ϖ 1 , ϖ 2 N o such that ϖ 1 < ϖ 2 , Then, the following equality holds:
Φ ϱ 1 ϖ 2 ϖ 1 ϖ 1 ϖ 2 Φ ( u ) ϖ 1 d q u = q ( ϱ ϖ 1 ) 2 ϖ 2 ϖ 1 0 1 z ϖ 1 D q Φ 1 z ϖ 1 + z ϱ d q z + q ( ϖ 2 ϱ ) 2 ϖ 2 ϖ 1 0 1 z ϖ 1 D q Φ 1 z ϖ 2 + z ϱ d q z .
Alp et al. in [34] was proved quantum analogues of Hermite-Hadamard type by using class convexity of function on co-ordinates.
Theorem 3.
Let Φ : N = [ ϖ 1 , ϖ 2 ] × [ ϖ 3 , ϖ 4 ] R 2 R is a convex function on coordinates of N , q k ( 0 , 1 ) where k = 1 , 2 and ϖ 1 < ϖ 2 , ϖ 3 < ϖ 4 . Then, one has the inequalities
Φ q 1 ϖ 1 + ϖ 2 [ 2 ] q 1 , q 2 ϖ 3 + ϖ 4 [ 2 ] q 2 1 2 1 ϖ 2 ϖ 1 ϖ 1 ϖ 2 Φ z , q 2 ϖ 3 + ϖ 4 [ 2 ] q 2 d z + 1 ϖ 3 ϖ 4 ϖ 3 ϖ 4 Φ q 1 ϖ 1 + ϖ 2 [ 2 ] q 1 , w d w 1 ( ϖ 1 ϖ 2 ) ( ϖ 3 ϖ 4 ) ϖ 1 ϖ 2 ϖ 3 ϖ 4 Φ ( z , w ) d z d w 1 4 [ 1 ( [ 2 ] q 1 ) ( ϖ 2 ϖ 1 ) q 2 ϖ 1 ϖ 2 Φ ( z , ϖ 3 ) d z + ϖ 1 ϖ 2 Φ ( z , ϖ 4 ) d z + 1 ( [ 2 ] q 2 ) ( ϖ 3 ϖ 4 ) q 1 ϖ 3 ϖ 3 Φ ( ϖ 1 , w ) d w + ϖ 3 ϖ 4 Φ ( ϖ 2 , w ) d w ] q 1 q 2 Φ ( ϖ 1 , ϖ 3 ) + q 2 Φ ( ϖ 2 , ϖ 3 ) + q 1 Φ ( ϖ 1 , ϖ 4 ) + Φ ( ϖ 2 , ϖ 4 ) [ 2 ] q 1 [ 2 ] q 2 .
In [35], Latif et al. defined q ϖ 1 ϖ 3 -integral and partial q-derivatives for two variables functions as follows:
Definition 2.
Suppose that Φ : ϖ 1 , ϖ 2 × ϖ 3 , ϖ 4 R 2 R is continuous function. Then, the definite q ϖ 1 ϖ 3 -integral on ϖ 1 , ϖ 2 × ϖ 3 , ϖ 4 is defined by
ϖ 1 x ϖ 3 y Φ z , w ϖ 3 d q 2 w ϖ 1 d q 1 z = 1 q 1 1 q 2 x ϖ 1 y ϖ 3 × n = 0 m = 0 q 1 n q 2 m Φ q 1 n x + 1 q 1 n ϖ 1 , q 2 m y + 1 q 2 m ϖ 3
for x , y ϖ 1 , ϖ 2 × ϖ 3 , ϖ 4 .
Definition 3.
(See, e.g., Latif et al. [35]). Let Φ : ϖ 1 , ϖ 2 × ϖ 3 , ϖ 4 R 2 R be a continuous function of two variables. Then, the partial q 1 -derivatives, q 2 -derivatives and q 1 q 2 -derivatives at x , y ϖ 1 , ϖ 2 × ϖ 3 , ϖ 4 can be given as follows:
ϖ 1 q 1 Φ x , y ϖ 1 q 1 x = Φ q 1 x + 1 q 1 ϖ 1 , y Φ x , y 1 q 1 x ϖ 1 , x ϖ 1 ϖ 3 q 2 Φ x , y c q 2 y = Φ x , q 2 y + 1 q 2 ϖ 3 Φ x , y 1 q 2 y ϖ 3 , y ϖ 3 ϖ 1 , ϖ 3 q 1 , q 2 2 Φ x , y ϖ 1 q 1 x ϖ 3 q 2 y = 1 x ϖ 1 y ϖ 3 1 q 1 1 q 2 × Φ q 1 x + 1 q 1 ϖ 1 , q 2 y + 1 q 2 ϖ 3 Φ q 1 x + 1 q 1 ϖ 1 , y Φ x , q 2 y + 1 q 2 ϖ 3 + Φ x , y , x ϖ 1 , y ϖ 3 .
Recently, Budak et al. gave the following definitions of q ϖ 1 ϖ 4 , q ϖ 2 ϖ 3 and q ϖ 2 ϖ 4 integrals and related inequalities of Hermite-Hadamard type:
Definition 4.
(See, e.g., Budak et al. [36]). Suppose that Φ : ϖ 1 , ϖ 2 × ϖ 3 , ϖ 4 R 2 R is continuous function. Then, the following q ϖ 1 ϖ 4 , q ϖ 2 ϖ 3 and q ϖ 2 ϖ 4 integrals on ϖ 1 , ϖ 2 × ϖ 3 , ϖ 4 are defined by
ϖ 1 x y ϖ 4 Φ z , w ϖ 4 d q 2 w ϖ 1 d q 1 z = 1 q 1 1 q 2 x ϖ 1 ϖ 4 y × n = 0 m = 0 q 1 n q 2 m Φ q 1 n x + 1 q 1 n ϖ 1 , q 2 m y + 1 q 2 m d ,
x ϖ 2 ϖ 3 y Φ z , w ϖ 3 d q 2 w ϖ 2 d q 1 z = 1 q 1 1 q 2 ϖ 2 x y ϖ 3 × n = 0 m = 0 q 1 n q 2 m Φ q 1 n x + 1 q 1 n ϖ 2 , q 2 m y + 1 q 2 m ϖ 3
and
x ϖ 2 y ϖ 4 Φ z , w ϖ 4 d q 2 w ϖ 2 d q 1 z = 1 q 1 1 q 2 ϖ 2 x ϖ 4 y × n = 0 m = 0 q 1 n q 2 m Φ q 1 n x + 1 q 1 n ϖ 2 , q 2 m y + 1 q 2 m ϖ 4
respectively, for x , y ϖ 1 , ϖ 2 × ϖ 3 , ϖ 4 .
Definition 5.
(See, e.g., Budak et al. [36]). Let Φ : ϖ 1 , ϖ 2 × ϖ 3 , ϖ 4 R 2 R be a continuous function of two variables. Then, the partial q 1 -derivatives, q 2 -derivatives and q 1 q 2 -derivatives at x , y ϖ 1 , ϖ 2 × ϖ 3 , ϖ 4 can be given as follows:
ϖ 2 q 1 Φ x , y ϖ 2 q 1 x = Φ q 1 x + 1 q 1 ϖ 2 , y Φ x , y 1 q 1 ϖ 2 x , x ϖ 2 , ϖ 4 q 2 Φ x , y ϖ 2 q 2 y = Φ x , q 2 y + 1 q 2 ϖ 4 Φ x , y 1 q 2 ϖ 4 y , ϖ 4 y , ϖ 1 ϖ 4 q 1 , q 2 2 Φ x , y ϖ 1 q 1 x d q 2 y = 1 x ϖ 1 ϖ 4 y 1 q 1 1 q 2 × Φ q 1 x + 1 q 1 ϖ 1 , q 2 y + 1 q 2 ϖ 4 Φ q 1 x + 1 q 1 ϖ 1 , y Φ x , q 2 y + 1 q 2 ϖ 4 + Φ x , y , x ϖ 1 , y ϖ 4 , ϖ 3 ϖ 2 q 1 , q 2 2 Φ x , y ϖ 2 q 1 x ϖ 3 q 2 y = 1 ϖ 2 x y ϖ 3 1 q 1 1 q 2 × Φ q 1 x + 1 q 1 ϖ 2 , q 2 y + 1 q 2 ϖ 3 Φ q 1 x + 1 q 1 ϖ 2 , y Φ x , q 2 y + 1 q 2 ϖ 3 + Φ x , y , x ϖ 2 , y ϖ 3 , ϖ 2 , ϖ 4 q 1 , q 2 2 Φ x , y ϖ 2 q 1 x d q 2 y = 1 ϖ 2 x ϖ 4 y 1 q 1 1 q 2 × Φ q 1 x + 1 q 1 ϖ 2 , q 2 y + 1 q 2 ϖ 4 Φ q 1 x + 1 q 1 ϖ 2 , y Φ x , q 2 y + 1 q 2 ϖ 4 + Φ x , y , x ϖ 2 , y ϖ 4 .
Newly, Toplu et al. [37] presented and investigated a new fascinating type of convexity known as n-polynomial convexity and acquired numerous further improvements of Hermite–Hadamard’s inequality using the definition of n-polynomial convex functions.
The main objective of this paper is to introduce the notion of higher-order generalized strongly n-polynomial preinvex functions. To obtain further results of this paper, we also derive a new generalized quantum integral identity. Utilizing this new identity, we derive new quantum analogs of Ostrowski-type inequalities for higher-order generalized strongly n-polynomial preinvex functions on co-ordinates. This shows that the obtained results are quite a unifying one. The ideas and techniques of this paper are expected to open up a new place for further research in this area.

2. Discussions and Main Results

In this section, we introduce a new definition for higher-order generalized strongly n-polynomial preinvex function which is stated as follows.
Definition 6.
(See, e.g., Hudzik et al. [38]). If Ω ϑ R n and ϑ ( . , . ) : R n × R n R n be a continuous bifunction, then Ω ϑ R n is said to be invex set
ϖ 1 + σ ϑ ( ϖ 2 , ϖ 1 ) Ω ϑ , ϖ 1 , ϖ 2 Ω ϑ , σ [ 0 , 1 ] .
Bifunction ϑ having property ϑ ( ϖ 1 , ϖ 2 ) = ϖ 2 ϖ 1 .
Weir and Mond [19] presented the notation of preinvexity of functions:
Definition 7.
Let Φ : Ω ϑ R n R is called preinvexity of function if
Φ ϖ 1 + σ ϑ ( ϖ 2 , ϖ 1 ) ( 1 σ ) Φ ( ϖ 1 ) + σ Φ ( ϖ 2 )
for all ϖ 1 , ϖ 2 Ω ϑ , σ [ 0 , 1 ] .
This shows that the preinvexity plays an important and significant role in the development of various fields of pure and applied sciences. Now, we introduce a new concept, which is called higher-order generalized strongly n-polynomial preinvex function.
Definition 8.
Let n N . A non-negative function Φ : Ω ϑ R n R is called higher-order generalized strongly n-polynomial preinvex function for some μ 0 and θ > 0 , if
Φ ϖ 1 + σ ϑ ( ϖ 2 , ϖ 1 ) 1 n p = 1 n 1 ( 1 σ ) p Φ ( ϖ 1 ) + 1 n p = 1 n 1 σ p Φ ( ϖ 2 ) μ σ ( 1 σ ) ϑ ( ϖ 2 , ϖ 1 ) θ
for all ϖ 1 , ϖ 2 Ω ϑ , σ [ 0 , 1 ] .
Consider the fact that if θ = 2 , the class of higher-order generalized strongly n-polynomial preinvex functions reduces to the class of generalized strongly n-polynomial preinvex functions, which is itself a new class. Taking μ = 0 into account, we find that the class of higher-order generalized strongly n-polynomial preinvex functions reduce to the class of n-polynomial preinvex functions a novel class in the literature. We can also consider ϑ ( ϖ 2 , ϖ 1 ) = ϖ 2 ϖ 1 , which results in a new class of higher-order generalized strongly n-polynomial convex functions. Given an initial value of n = 1 , the above class simplifies to a simple higher-order generalized strongly preinvex function. If we consider n = 2 , we obtain a higher-order generalized strongly 2-polynomial preinvex function of the form
Φ ϖ 1 + σ ϑ ( ϖ 2 , ϖ 1 ) 3 σ σ 2 2 Φ ( ϖ 1 ) + 1 σ σ 2 2 Φ ( ϖ 2 ) μ σ ( 1 σ ) ϑ ( ϖ 2 , ϖ 1 ) θ ,
for all ϖ 1 , ϖ 2 Ω ϑ , σ [ 0 , 1 ] for some μ 0 and θ > 0 .
Proposition 1.
Let ϖ 2 > 0 and Φ α : [ ϖ 1 , ϖ 1 + ϑ ( ϖ 2 , ϖ 1 ) ] R be an arbitrary family of higher-order generalized strongly n-polynomial preinvex function and let Φ ( ϖ ) = sup α Φ α ( ϖ ) . If J ϑ = u [ ϖ 1 , ϖ 1 + ϑ ( ϖ 2 , ϖ 1 ) ] : Φ ( u ) < 1 is nonempty, then J ϑ is an interval and Φ is higher-order generalized strongly n-polynomial preinvex function on J ϑ .
Proof. 
Let σ [ 0 , 1 ] and ϖ 1 , ϖ 1 + ϑ ( ϖ 2 , ϖ 1 ) J ϑ be arbitrary. Then,
Φ ( ϖ 1 + σ ϑ ( ϖ 2 , ϖ 1 ) ) = sup α Φ α ( ϖ 1 + σ ϑ ( ϖ 2 , ϖ 1 ) ) sup α 1 n p = 1 n 1 ( 1 σ ) p Φ α ( ϖ 1 ) + 1 n p = 1 n 1 σ p Φ α ϖ 2 μ σ ( 1 σ ) ϑ ( ϖ 2 , ϖ 1 ) θ 1 n p = 1 n 1 ( 1 σ ) p sup α Φ α ( ϖ 1 ) + 1 n p = 1 n 1 σ p sup α Φ α ϖ 2 μ σ ( 1 σ ) ϑ ( ϖ 2 , ϖ 1 ) θ 1 n p = 1 n 1 ( 1 σ ) p Φ ( ϖ 1 ) + 1 n p = 1 n 1 σ p Φ ϖ 2 μ σ ( 1 σ ) ϑ ( ϖ 2 , ϖ 1 ) θ < ,
for some μ 0 and θ > 0 . This completes the proof. □
We now derive new quantum analogues of Ostrowski-type inequalities on co-ordinates using the concept of higher-order generalized strongly n-polynomial preinvex function for this, we need a key lemma.
Lemma 1.
Let Φ : N R 2 R be a mixed partial q 1 q 2 -differentiable function on N o (the interior of N ) with ϖ 1 , ϖ 3 q 1 q 2 2 Φ ( z , w ) ϖ 1 q 1 z ϖ 3 q 2 w , ϖ 1 ϖ 4 q 1 q 2 2 Φ ( z , w ) ϖ 1 q 1 z ϖ 4 q 2 w , ϖ 3 ϖ 2 q 1 q 2 2 Φ ( z , w ) ϖ 2 q 1 z ϖ 3 q 2 w and ϖ 2 , ϖ 4 q 1 q 2 2 Φ ( z , w ) ϖ 2 q 1 z ϖ 4 q 2 w are continuous and integrable on [ ϖ 1 , ϖ 1 + ϑ 1 ϖ 2 , ϖ 1 ] × [ ϖ 3 , ϖ 3 + ϑ 2 ϖ 4 , ϖ 3 ] N o for ϑ 1 ϖ 2 , ϖ 1 , ϑ 2 ϖ 4 , ϖ 3 > 0 and q 1 , q 2 ( 0 , 1 ) , then we have following equality:
Δ q 1 , q 2 ( , ϖ 1 , ϖ 2 , ϖ 3 , ϖ 4 ) = 1 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 ϖ 1 ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) ϖ 3 ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) Φ ( u , v ) ϖ 1 d q 1 u ϖ 3 d q 2 v + ϖ 1 ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ϖ 4 Φ ( u , v ) ϖ 1 d q 1 u ϖ 4 d q 2 v + ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) ϖ 2 ϖ 3 ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) Φ ( u , v ) ϖ 2 d q 1 u ϖ 3 d q 2 v + ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) ϖ 2 ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ϖ 4 Φ ( u , v ) ϖ 2 d q 1 u ϖ 4 d q 2 v A ϑ 1 ( ϱ , ϖ 1 ) ϑ 2 ( ρ , ϖ 3 ) Φ ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) + ϑ 2 ( ϖ 4 , ρ ) Φ ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ϑ 1 ( ϖ 2 , ϱ ) ϑ 2 ( ρ , ϖ 3 ) Φ ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) + ϑ 2 ( ϖ 4 , ρ ) Φ ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) , ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) = q 1 q 2 [ ϑ 1 ( ϱ , ϖ 1 ) ] 2 [ ϑ 2 ( ρ , ϖ 3 ) ] 2 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 0 1 0 1 z w ϖ 1 , ϖ 3 q 1 q 2 2 Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w d q 1 z d q 2 w + q 1 q 2 [ ϑ 1 ( ϱ , ϖ 1 ) ] 2 [ ϑ 2 ( ϖ 4 , ρ ) ] 2 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 0 1 0 1 z w ϖ 1 ϖ 4 q 1 q 2 2 Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 4 + w ϑ 2 ( ρ , ϖ 4 ) ) ϖ 1 q 1 z ϖ 4 q 2 w d q 1 z d q 2 w + q 1 q 2 [ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ϑ 2 ( ρ , ϖ 3 ) ] 2 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 0 1 0 1 z w ϖ 3 ϖ 2 q 1 q 2 2 Φ ( ϖ 2 + z ϑ 1 ( ϱ , ϖ 2 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 3 q 1 z ϖ 2 q 2 w d q 1 z d q 2 w + q 1 q 2 [ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ϑ 2 ( ϖ 4 , ρ ) ] 2 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 0 1 0 1 z w ϖ 2 , ϖ 4 q 1 q 2 2 Φ ( ϖ 2 + z ϑ 1 ( ϱ , ϖ 2 ) , ϖ 4 + w ϑ 2 ( ρ , ϖ 4 ) ) ϖ 2 q 1 z ϖ 4 q 2 w d q 1 z d q 2 w ,
where
A = ϑ 1 ( ϱ , ϖ 1 ) ϖ 3 ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) Φ ( ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , v ) ϖ 3 d q 2 v + ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ϖ 4 Φ ( ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , v ) ϖ 4 d q 2 v + ϑ 1 ( ϖ 2 , ϱ ) ϖ 3 ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) Φ ( ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) , v ) ϖ 3 d q 2 v + ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ϖ 4 Φ ( ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) , v ) ϖ 4 d q 2 v
+ ϑ 2 ( ρ , ϖ 3 ) ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) ϖ 2 Φ ( u , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) ) ϖ 2 d q 1 u + ϖ 1 ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) Φ ( u , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 d q 1 u + ϑ 2 ( ϖ 4 , ρ ) ϖ 2 + ϑ 1 ( ρ , ϖ 2 ) ϖ 2 Φ ( u , ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ) ϖ 2 d q 1 u + ϖ 1 ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) Φ ( u , ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ) ϖ 1 d q 1 u
for all ϱ , ρ N .
Proof. 
Now, we consider
0 1 0 1 z w ϖ 1 , ϖ 3 q 1 q 2 2 Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w d q 1 z d q 2 w .
By the Definition 2 and Definition 3 of partial q 1 q 2 -derivatives and definite q 1 q 2 ϖ 1 , ϖ 3 -integrals, respectively, we have
0 1 0 1 z w ϖ 1 , ϖ 3 q 1 q 2 2 Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w d q 1 z d q 2 w = 1 ( 1 q 1 ) ( 1 q 2 ) ϑ 1 ( ϱ , ϖ 1 ) ϑ 2 ( ρ , ϖ 3 ) × [ 0 1 0 1 Φ ( ϖ 1 + z q 1 ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w q 2 ϑ 2 ( ρ , ϖ 2 ) ) d q 1 z d q 2 w 0 1 0 1 Φ ( ϖ 1 + z q 1 ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) d q 1 z d q 2 w 0 1 0 1 Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w q 2 ϑ 2 ( ρ , ϖ 2 ) ) d q 1 z d q 2 w + 0 1 0 1 Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) d q 1 z d q 2 w ] = 1 ϑ 1 ( ϱ , ϖ 1 ) ϑ 2 ( ρ , ϖ 3 ) [ n = 0 m = 0 q 1 n q 2 m Φ ( ϖ 1 + z q 1 n + 1 ϑ 1 ( ϱ , ϖ 3 + q 2 m + 1 ϑ 2 ( ρ , ϖ 3 ) ) n = 0 m = 0 q 1 n q 2 m Φ ( ϖ 1 + z q 1 n + 1 ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) ) n = 0 m = 0 q 1 n q 2 m Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 3 + q 2 m + 1 ϑ 2 ( ρ , ϖ 3 ) ) + n = 0 m = 0 q 1 n q 2 m Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) ) ] = 1 ϑ 1 ( ϱ , ϖ 1 ) ϑ 2 ( ρ , ϖ 3 ) 1 q 1 q 2 n = 1 m = 1 q 1 n q 2 m Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) ) 1 q 1 n = 1 m = 0 q 1 n q 2 m Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) ) 1 q 2 n = 0 m = 1 q 1 n q 2 m Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) ) + n = 0 m = 0 q 1 n q 2 m Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) ) .
We observe that
1 q 1 q 2 n = 1 m = 1 q 1 n q 2 m Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) ) = Φ ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) q 1 q 2 1 q 1 q 2 n = 0 q n Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) ) 1 q 1 q 2 m = 0 q 2 m ( ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) ) + 1 q 1 q 2 n = 0 m = 0 q 1 n q 2 m Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) ) ,
1 q 1 n = 1 m = 0 q 1 n q 2 m Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) ) = 1 q 1 m = 0 q 2 m Φ ( ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) 1 q 1 n = 0 m = 0 q 1 n q 2 m Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) )
and
1 q 2 n = 0 m = 1 q 1 n q 2 m Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) ) = 1 q 2 n = 0 q 1 n Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + ϑ 3 ( ρ , ϖ 3 ) ) 1 q 2 n = 0 m = 0 q 1 n q 2 m Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) ) .
Utilizing (5)–(7) in (4), we get
0 1 0 1 z w ϖ 1 , ϖ 3 q 1 , q 2 2 ϖ 1 q 1 z ϖ 3 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) d q 1 z d q 2 w = 1 q 1 q 2 ϑ 1 ( ϱ , ϖ 1 ) ϑ 2 ( ρ , ϖ 3 ) Φ ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) 1 q 2 ϑ 2 ( ρ , ϖ 3 ) ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) m = 0 q 2 m ( ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) ) 1 q 1 ϑ 1 ( ϱ , ϖ 1 ) ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) n = 0 q 1 n Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) ) + 1 q 1 1 q 2 ϑ 1 ( ϱ , ϖ 1 ) ϑ 2 ( ρ , ϖ 3 ) ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) × n = 0 m = 0 q 1 n q 2 m Φ ( ϖ 1 + q 1 n ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + q 2 m ϑ 2 ( ρ , ϖ 3 ) )
and
0 1 0 1 z w ϖ 1 , ϖ 3 q 1 , q 2 2 ϖ 1 q 1 z ϖ 3 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) d q 1 z d q 2 w = 1 q 1 q 2 ϑ 1 ( ϱ , ϖ 1 ) ϑ 2 ( ρ , ϖ 3 ) Φ ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) 1 ϑ 2 ( ρ , ϖ 3 ) ϖ 3 ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) Φ ( ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , v ) ϖ 3 d q 2 v 1 ϑ 1 ( ϱ , ϖ 1 ) ϖ 1 ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) Φ ( u , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 d q 1 u + 1 ϑ 1 ( ϱ , ϖ 1 ) ϑ 2 ( ρ , ϖ 3 ) ϖ 1 ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) ϖ 3 ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) Φ ( u , v ) ϖ 1 d q 1 u ϖ 3 d q 2 v .
Multiplying both sides of equality (8) by q 1 q 2 ϑ 1 ( ϱ , ϖ 1 ) 2 ϑ 2 ( ρ , ϖ 3 ) 2 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 , then we acquire
q 1 q 2 ϑ 1 ( ϱ , ϖ 1 ) 2 ϑ 2 ( ρ , ϖ 3 ) 2 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 0 1 0 1 z w ϖ 1 , ϖ 3 q 1 , q 2 2 ϖ 1 q 1 z ϖ 3 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) d q 1 z d q 2 w = ϑ 1 ( ϱ , ϖ 1 ) ϑ 2 ( ρ , ϖ 3 ) ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 Φ ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) ϑ 1 ( ϱ , ϖ 1 ) ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 ϖ 3 ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) Φ ( ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , v ) ϖ 3 d q 2 v ϑ 2 ( ρ , ϖ 3 ) ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 ) ϖ 1 ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) Φ ( u , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 d q 1 u + 1 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 ϖ 1 ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) ϖ 3 ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) Φ ( u , v ) ϖ 1 d q 1 u ϖ 3 d q 2 v .
Similarly, by using Definition 4 and Definition 5, we calculate remaining integrals
q 1 q 2 [ ϑ 1 ( ϱ , ϖ 1 ) ] 2 [ ϑ 2 ( ϖ 4 , ρ ) ] 2 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 0 1 0 1 z w ϖ 1 ϖ 4 q 1 , q 2 2 ϖ 1 q 1 z ϖ 4 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 4 + w ϑ 2 ( ρ , ϖ 4 ) ) d q 1 z d q 2 w = ϑ 1 ( ϱ , ϖ 1 ) ϑ 2 ( ϖ 4 , ρ ) ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 Φ ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ϑ 1 ( ϱ , ϖ 1 ) ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ϖ 4 Φ ( ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) , v ) ϖ 4 d q 2 v ϑ 2 ( ϖ 4 , ρ ) ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 ) ϖ 1 ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) Φ ( u , ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ) ϖ 1 d q 1 u + 1 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 ϖ 1 ϖ 1 + ϑ 1 ( ϱ , ϖ 1 ) ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ϖ 4 Φ ( u , v ) ϖ 1 d q 1 u ϖ 4 d q 2 v ,
q 1 q 2 [ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ϑ 2 ( ρ , ϖ 3 ) ] 2 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 0 1 0 1 z w ϖ 3 ϖ 2 q 1 , q 2 2 ϖ 2 q 1 z ϖ 3 q 2 w Φ ( ϖ 2 + z ϑ 1 ( ϱ , ϖ 2 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) d q 1 z d q 2 w = ϑ 1 ( ϖ 2 , ϱ ) ϑ 2 ( ρ , ϖ 3 ) ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 Φ ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) ϑ 1 ( ϖ 2 , ϱ ) ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 ϖ 3 ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) Φ ( ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) , v ) ϖ 3 d q 2 v
ϑ 2 ( ρ , ϖ 3 ) ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 ) ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) ϖ 2 Φ ( u , ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) ) ϖ 2 d q 1 u + 1 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) ϖ 2 ϖ 3 ϖ 3 + ϑ 2 ( ρ , ϖ 3 ) Φ ( u , v ) ϖ 2 d q 1 u ϖ 3 d q 2 v
and
[ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ( ϖ 4 , ρ ) ] 2 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 0 1 0 1 z w ϖ 2 , ϖ 4 q 1 , q 2 2 ϖ 2 q 1 z ϖ 4 q 2 w Φ ( ϖ 2 + z ϑ 1 ( ϱ , ϖ 2 ) , ϖ 4 + w ϑ 2 ( ρ , ϖ 4 ) ) d q 1 z d q 2 w = ϑ 1 ( ϖ 2 , ϱ ) ϑ 2 ( ϖ 4 , ρ ) ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 Φ ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) , ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ϑ 2 ( ϖ 4 , ρ ) ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) ϖ 2 Φ ( u , ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ) ϖ 4 d q 1 u ϑ 1 ( ϖ 2 , ϱ ) ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 ) ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ϖ 4 Φ ( ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) , v ) ϖ 2 d q 1 v + 1 ϑ 1 ϖ 2 , ϖ 1 ϑ 2 ϖ 4 , ϖ 3 ϖ 2 + ϑ 1 ( ϱ , ϖ 2 ) ϖ 2 ϖ 4 + ϑ 2 ( ρ , ϖ 4 ) ϖ 4 Φ ( u , v ) ϖ 2 d q 1 u ϖ 4 d q 2 v .
From (9)–(12) in (3). Thus, Lemma 1 is proved. □
Theorem 4.
Suppose that n N and all the assumptions of Lemma 1 are true. If ϖ 1 , ϖ 3 q 1 q 2 2 Φ ( z , w ) ϖ 1 q 1 z ϖ 3 q 2 w σ 2 , ϖ 1 ϖ 4 q 1 q 2 2 Φ ( z , w ) ϖ 1 q 1 z ϖ 4 q 2 w σ 2 , ϖ 3 ϖ 2 q 1 q 2 2 Φ ( z , w ) ϖ 2 q 1 z ϖ 3 q 2 w σ 2 and ϖ 2 , ϖ 4 q 1 q 2 2 Φ ( z , w ) ϖ 2 q 1 z ϖ 4 q 2 w σ 2 are higher-order generalized strongly n-polynomial preinvex functions on co-ordinates on N for σ 1 , σ 2 > 1 , 1 σ 1 + 1 σ 2 = 1 with ϖ 1 , ϖ 3 q 1 q 2 2 Φ ( ϱ , ρ ) ϖ 1 q 1 z ϖ 3 q 2 w M , ϖ 1 ϖ 4 q 1 q 2 2 Φ ( ϱ , ρ ) ϖ 1 q 1 z ϖ 4 q 2 w M , ϖ 3 ϖ 2 q 1 q 2 2 Φ ( ϱ , ρ ) ϖ 2 q 1 z ϖ 3 q 2 w M and ϖ 2 , ϖ 4 q 1 q 2 2 Φ ( ϱ , ρ ) ϖ 2 q 1 z ϖ 4 q 2 w M , ϱ , ρ N , then the following inequality holds:
Δ q 1 , q 2 ( , ϖ 1 , ϖ 2 , ϖ 3 , ϖ 4 ) q 1 q 2 [ σ 1 + 1 ] q 1 [ σ 1 + 1 ] q 2 σ 1 ϑ 1 ( ϖ 2 , ϖ 1 ) ϑ 2 ( ϖ 4 , ϖ 3 ) × [ ϑ 1 ( ϱ , ϖ 1 ) ] 2 [ ϑ 2 ( ρ , ϖ 3 ) ] 2 M σ 2 ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) μ 1 X q 1 ϑ 1 θ ( ϱ , ϖ 1 ) 2 μ 2 X q 2 ϑ 2 θ ( ρ , ϖ 3 ) 1 σ 2 + [ ϑ 1 ( ϱ , ϖ 1 ) ] 2 [ ϑ 2 ( ϖ 4 , ρ ) ] 2 M σ 2 ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) μ 1 X q 1 ϑ 1 θ ( ϱ , ϖ 1 ) 2 μ 2 X q 2 ϑ 2 θ ( ρ , ϖ 4 ) 1 σ 2 + [ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ϑ 2 ( ρ , ϖ 3 ) ] 2 M σ 2 ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) μ 1 X q 1 ϑ 1 θ ( ϱ , ϖ 2 ) 2 μ 2 X q 2 ϑ 2 θ ( ρ , ϖ 3 ) 1 σ 2 + [ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ϑ 2 ( ϖ 4 , ρ ) ] 2 M σ 2 ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) μ 1 X q 1 ϑ 1 θ ( ϱ , ϖ 1 ) 2 μ 2 X q 2 ϑ 2 θ ( ρ , ϖ 4 ) 1 σ 2 .
Proof. 
Taking absolute value on both sides of (3), by applying power mean inequality for double integrals, we get the following inequality
Δ q 1 , q 2 ( , ϖ 1 , ϖ 2 , ϖ 3 , ϖ 4 ) 0 1 0 1 z σ 1 w σ 1 d q 1 z d q 2 w 1 σ 1 q 1 q 2 [ ϑ 1 ( ϱ , ϖ 1 ) ] 2 [ ϑ 2 ( ρ , ϖ 3 ) ] 2 ϑ 1 ( ϖ 2 , ϖ 1 ) ϑ 2 ( ϖ 4 , ϖ 3 )
× 0 1 0 1 ϖ 1 , ϖ 3 q 1 , q 2 2 ϖ 1 q 1 z ϖ 3 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) σ 2 d q 1 z d q 2 w 1 σ 2 + q 1 q 2 [ ϑ 1 ( ϱ , ϖ 1 ) ] 2 [ ϑ 2 ( ϖ 4 , ρ ) ] 2 ϑ 1 ( ϖ 2 , ϖ 1 ) ϑ 2 ( ϖ 4 , ϖ 3 ) × 0 1 0 1 ϖ 1 ϖ 4 q 1 , q 2 2 ϖ 1 q 1 z ϖ 4 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 4 + w ϑ 2 ( ρ , ϖ 4 ) ) σ 2 d q 1 z d q 2 w 1 σ 2 + q 1 q 2 [ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ϑ 2 ( ρ , ϖ 3 ) ] 2 ϑ 1 ( ϖ 2 , ϖ 1 ) ϑ 2 ( ϖ 4 , ϖ 3 ) × 0 1 0 1 ϖ 3 ϖ 2 q 1 , q 2 2 ϖ 2 q 1 z ϖ 3 q 2 w Φ ( ϖ 2 + z ϑ 1 ( ϱ , ϖ 2 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) σ 2 d q 1 z d q 2 w 1 σ 2 + q 1 q 2 [ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ϑ 2 ( ϖ 4 , ρ ) ] 2 ϑ 1 ( ϖ 2 , ϖ 1 ) ϑ 2 ( ϖ 4 , ϖ 3 ) × 0 1 0 1 ϖ 2 , ϖ 4 q 1 , q 2 2 ϖ 2 q 1 z ϖ 4 q 2 w Φ ( ϖ 2 + z ϑ 1 ( ϱ , ϖ 2 ) , ϖ 4 + w ϑ 2 ( ρ , ϖ 4 ) ) σ 2 d q 1 z d q 2 w 1 σ 2 ,
for all ϱ , ρ N .
Taking first integral together with the fact that ϖ 1 , ϖ 3 q 1 , q 2 2 ϖ 1 q 1 z ϖ 3 q 2 w Φ ( z , w ) σ 2 is higher-order generalized strongly n-polynomial preinvex functions, we get
0 1 0 1 ϖ 1 , ϖ 3 q 1 , q 2 2 ϖ 1 q 1 z ϖ 3 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) σ 2 d q 1 z d q 2 w 0 1 0 1 1 n p = 1 n 1 ( 1 z ) p ϖ 1 , ϖ 3 q 1 , q 2 2 Φ ( ϱ , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w σ 2 + 1 n p = 1 n 1 z p ϖ 1 , ϖ 3 q 1 , q 2 2 Φ ( ϖ 1 , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w σ 2 μ 1 z ( 1 z ) ϑ 1 θ ( ϖ 2 , ϖ 1 ) d q 1 z d q 2 w .
Computing the q 1 -integral on the right-hand side of (14), we have
0 1 1 n p = 1 n 1 ( 1 z ) p ϖ 1 , ϖ 3 q 1 , q 2 2 Φ ( ϱ , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w σ 2 d q 1 z + 0 1 1 n p = 1 n 1 z p ϖ 1 , ϖ 3 q 1 , q 2 2 Φ ( ϖ 1 , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w σ 2 d q 1 z 0 1 μ 1 z ( 1 z ) ϑ 1 θ ( ϱ , ϖ 1 ) d q 1 z .
In view of the Definition 2 and k = 1 , 2 , we get
Z q k = 1 n p = 1 n 0 1 1 ( 1 z ) p d q k z = 1 ( 1 q k ) n p = 1 n e = 0 q k e ( 1 q k e ) p , E q k = 1 n p = 1 n 0 1 1 z p d q k z
= 1 1 n p = 1 n 1 q k 1 q k p + 1 , X q k = 0 1 z ( 1 z ) d q k z = q k 2 ( 1 + q k ) ( 1 + q k + q k 2 ) .
Putting the above calculations into (14), we obtain
0 1 Z q 1 ϖ 1 , ϖ 3 q 1 , q 2 2 Φ ( ϱ , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w σ 2 + E q 1 ϖ 1 , ϖ 3 q 1 , q 2 2 Φ ( ϖ 1 , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w σ 2 μ 1 X q 1 ϑ 1 θ ( ϱ , ϖ 1 ) d q 2 w .
Similarly, by computing the q 2 -integral, utilizing the fact ϖ 1 , ϖ 3 q 1 , q 2 2 ϖ 1 q 1 z ϖ 3 q 2 w Φ ( ϱ , ρ ) M , ϱ , ρ N on the right-hand side of (15), we get
0 1 0 1 ϖ 1 , ϖ 3 q 1 , q 2 2 ϖ 1 q 1 z ϖ 1 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) σ 2 d q 1 z d q 2 w M σ 2 ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) μ 1 X q 1 ϑ 1 θ ( ϱ , ϖ 1 ) 2 μ 2 X q 2 ϑ 2 θ ( ρ , ϖ 3 ) .
Analogously, we also have
0 1 0 1 ϖ 1 ϖ 4 q 1 , q 2 2 ϖ 1 q 1 z ϖ 4 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 4 + w ϑ 2 ( ρ , ϖ 4 ) ) σ 2 d q 1 z d q 2 w M σ 2 ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) μ 1 X q 1 ϑ 1 θ ( ϱ , ϖ 1 ) 2 μ 2 X q 2 ϑ 2 θ ( ρ , ϖ 4 ) ,
0 1 0 1 ϖ 3 ϖ 2 q 1 , q 2 2 ϖ 2 q 1 z ϖ 3 q 2 w Φ ( ϖ 2 + z ϑ 1 ( ϱ , ϖ 2 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) σ 2 d q 1 z d q 2 w M σ 2 ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) μ 1 X q 1 ϑ 1 θ ( ϱ , ϖ 2 ) 2 μ 2 X q 2 ϑ 2 θ ( ρ , ϖ 3 )
and
0 1 0 1 ϖ 2 , ϖ 4 q 1 , q 2 2 ϖ 2 q 1 z ϖ 4 q 2 w Φ ( ϖ 2 + z ϑ 1 ( ϱ , ϖ 2 ) , ϖ 4 + w ϑ 2 ( ρ , ϖ 4 ) ) σ 2 d q 1 z d q 2 w M σ 2 ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) μ 1 X q 1 ϑ 1 θ ( ϱ , ϖ 2 ) 2 μ 2 X q 2 ϑ 2 θ ( ρ , ϖ 4 ) .
Now by making use of the inequalities (16)–(19) and the fact that
0 1 0 1 z σ 1 w σ 1 d q 1 z d q 2 w = 1 [ σ 1 + 1 ] q 1 [ σ 1 + 1 ] q 2 ,
we get the inequality (13). This completes the proof. □
Corollary 1.
I. By using the property of preinvex functions ϑ ( x , y ) = y x in Theorem 4, we get
| 1 ϖ 2 ϖ 1 ϖ 4 ϖ 3 ϖ 1 ϱ ϖ 3 ρ Φ ( u , v ) ϖ 1 d q 1 u ϖ 3 d q 2 v + ϖ 1 ϱ ρ ϖ 4 Φ ( u , v ) ϖ 1 d q 1 u ϖ 4 d q 2 v
+ ϱ ϖ 2 ϖ 3 ρ Φ ( u , v ) ϖ 2 d q 1 u ϖ 3 d q 2 v + ϱ ϖ 2 ρ ϖ 4 Φ ( u , v ) ϖ 2 d q 1 u ϖ 4 d q 2 v 1 ϖ 2 ϖ 1 ϖ 1 ϱ Φ ( u , ρ ) ϖ 1 d q 1 u + ϱ ϖ 2 Φ ( u , ρ ) ϖ 2 d q 1 u 1 ϖ 4 ϖ 3 ϖ 3 ρ Φ ( ϱ , v ) ϖ 3 d q 2 v + ρ ϖ 4 Φ ( ϱ , v ) ϖ 4 d q 2 v Φ ( ϱ , ρ ) | q 1 q 2 [ σ 1 + 1 ] q 1 [ σ 1 + 1 ] q 2 σ 1 ( ϖ 2 ϖ 1 ) ( ϖ 4 ϖ 3 ) × [ ϱ ϖ 1 ] 2 [ ρ ϖ 3 ] 2 M σ 2 ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) μ 1 X q 1 ( ϱ ϖ 1 ) θ 2 μ 2 X q 2 ( ρ ϖ 3 ) θ 1 σ 2 + [ ϱ ϖ 1 ] 2 [ ϖ 4 ρ ] 2 M σ 2 ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) μ 1 X q 1 ( ϱ ϖ 1 ) θ 2 μ 2 X q 2 ( ρ ϖ 4 ) θ 1 σ 2 + [ ϖ 2 ϱ ] 2 [ ρ ϖ 3 ] 2 M σ 2 ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) μ 1 X q 1 ( ϱ ϖ 2 ) θ 2 μ 2 X q 2 ( ρ ϖ 3 ) θ 1 σ 2 + [ ϖ 2 ϱ ] 2 [ ϖ 4 ρ ] 2 M σ 2 ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) μ 1 X q 1 ( ϱ ϖ 1 ) θ 2 μ 2 X q 2 ( ρ ϖ 4 ) θ 1 σ 2 .
II. Let μ k = 0 , for k = 1 , 2 in Theorem 4, we get
| Δ q 1 , q 2 ( , ϖ 1 , ϖ 2 , ϖ 3 , ϖ 4 ) | M q 1 q 2 [ σ 1 + 1 ] q 1 [ σ 1 + 1 ] q 2 σ 1 ϑ 1 ϱ , ϖ 1 2 + ϑ 1 ϖ 2 , ϱ 2 ϑ 1 ( ϖ 2 , ϖ 1 ) × ϑ 2 ρ , ϖ 3 2 + ϑ 2 ϖ 4 , ρ 2 ϑ 2 ( ϖ 4 , ϖ 3 ) ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) 1 σ 2 .
III. Let μ k = 0 , for k = 1 , 2 and property of preinvexity ϑ ( x , y ) = y x in Theorem 4, we get
| 1 ϖ 2 ϖ 1 ϖ 4 ϖ 3 ϖ 1 ϱ ϖ 3 ρ Φ ( u , v ) ϖ 1 d q 1 u ϖ 3 d q 2 v + ϖ 1 ϱ ρ ϖ 4 Φ ( u , v ) ϖ 1 d q 1 u ϖ 4 d q 2 v + ϱ ϖ 2 ϖ 3 ρ Φ ( u , v ) ϖ 2 d q 1 u ϖ 3 d q 2 v + ϱ ϖ 2 ρ ϖ 4 Φ ( u , v ) ϖ 2 d q 1 u ϖ 4 d q 2 v 1 ϖ 2 ϖ 1 ϖ 1 ϱ Φ ( u , ρ ) ϖ 1 d q 1 u + ϱ ϖ 2 Φ ( u , ρ ) ϖ 2 d q 1 u 1 ϖ 4 ϖ 3 ϖ 3 ρ Φ ( ϱ , v ) ϖ 3 d q 2 v + ρ ϖ 4 Φ ( ϱ , v ) ϖ 4 d q 2 v Φ ( ϱ , ρ ) | M q 1 q 2 [ σ 1 + 1 ] q 1 [ σ 1 + 1 ] q 2 σ 1 ϱ ϖ 1 2 + ϖ 2 ϱ 2 ϖ 2 ϖ 1 ρ ϖ 3 2 + ϖ 4 ρ 2 ϖ 4 ϖ 3 × ( Z q 1 + E q 1 ) ( Z q 2 + E q 2 ) 1 σ 2 .
IV. Let q k 1 , for k = 1 , 2 in part III, we have the following inequality, for more details, see in [33].
| K ϱ , ρ + 1 ( ϖ 2 ϖ 1 ) ( ϖ 4 ϖ 3 ) ϖ 1 ϖ 2 ϖ 3 ϖ 4 K ( u , v ) d v d u Q | M ( 1 + σ 1 ) 2 σ 1 ϱ ϖ 1 2 + ϖ 2 ϱ 2 ϖ 2 ϖ 1 ρ ϖ 3 2 + ϖ 4 ρ 2 ϖ 4 ϖ 3 , Q = 1 ϖ 2 ϖ 1 ϖ 1 ϖ 2 K ( u , ρ ) d u + 1 ϖ 4 ϖ 3 ϖ 3 ϖ 4 K ( ϱ , v ) d v .
Theorem 5.
Suppose that n N and all the assumptions of Lemma 1 are true. If ϖ 1 , ϖ 3 q 1 q 2 2 Φ ( z , w ) ϖ 1 q 1 z ϖ 3 q 2 w σ , ϖ 1 ϖ 4 q 1 q 2 2 Φ ( z , w ) ϖ 1 q 1 z ϖ 4 q 2 w σ , ϖ 3 ϖ 2 q 1 q 2 2 Φ ( z , w ) ϖ 2 q 1 z ϖ 3 q 2 w σ and ϖ 2 , ϖ 4 q 1 q 2 2 Φ ( z , w ) ϖ 2 q 1 z ϖ 4 q 2 w σ are higher-order generalized strongly n-polynomial preinvex functions on co-ordinates on N for σ 1 with ϖ 1 , ϖ 3 q 1 q 2 2 Φ ( ϱ , ρ ) ϖ 1 q 1 z ϖ 3 q 2 w M , ϖ 1 ϖ 4 q 1 q 2 2 Φ ( ϱ , ρ ) ϖ 1 q 1 z ϖ 4 q 2 w M , ϖ 3 ϖ 2 q 1 q 2 2 Φ ( ϱ , ρ ) ϖ 2 q 1 z ϖ 3 q 2 w M and ϖ 2 , ϖ 4 q 1 q 2 2 Φ ( ϱ , ρ ) ϖ 2 q 1 z ϖ 4 q 2 w M , ϱ , ρ N , then the following inequality holds:
Δ q 1 , q 2 ( , ϖ 1 , ϖ 2 , ϖ 3 , ϖ 4 ) q 1 q 2 ϑ 1 ( ϖ 2 , ϖ 1 ) ϑ 2 ( ϖ 4 , ϖ 3 ) [ 2 ] q 1 [ 2 ] q 2 1 1 σ × [ ϑ 1 ( ϱ , ϖ 1 ) ] 2 [ ϑ 2 ( ρ , ϖ 3 ) ] 2 M σ ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ϑ 1 θ ( ϱ , ϖ 1 ) 2 μ 2 U q 2 ϑ 2 θ ( ρ , ϖ 3 ) 1 σ + [ ϑ 1 ( ϱ , ϖ 1 ) ] 2 [ ϑ 2 ( ϖ 4 , ρ ) ] 2 M σ ( S q 1 + T q 1 ) ( Z q 2 + E q 2 ) μ 1 U q 1 ϑ 1 θ ( ϱ , ϖ 1 ) 2 μ 2 U q 2 ϑ 2 θ ( ρ , ϖ 4 ) 1 σ + [ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ϑ 2 ( ρ , ϖ 3 ) ] 2 M σ ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ϑ 1 θ ( ϱ , ϖ 2 ) 2 μ 2 U q 2 ϑ 2 θ ( ρ , ϖ 3 ) 1 σ + [ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ϑ 2 ( ϖ 4 , ρ ) ] 2 M σ ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ϑ 1 θ ( ϱ , ϖ 1 ) 2 μ 2 U q 2 ϑ 2 θ ( ρ , ϖ 4 ) 1 σ .
Proof. 
Taking absolute value on both sides of (3), by applying power mean inequality for double integrals, we get the following inequality:
Δ q 1 , q 2 ( , ϖ 1 , ϖ 2 , ϖ 3 , ϖ 4 ) 0 1 0 1 z w d q 1 z d q 2 w 1 1 σ q 1 q 2 [ ϑ 1 ( ϱ , ϖ 1 ) ] 2 [ ϑ 2 ( ρ , ϖ 3 ) ] 2 ϑ 1 ( ϖ 2 , ϖ 1 ) ϑ 2 ( ϖ 4 , ϖ 3 ) × 0 1 0 1 z w ϖ 1 , ϖ 3 q 1 , q 2 2 ϖ 1 q 1 z ϖ 3 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) σ d q 1 z d q 2 w 1 σ + q 1 q 2 [ ϑ 1 ( ϱ , ϖ 1 ) ] 2 [ ϑ 2 ( ϖ 4 , ρ ) ] 2 ϑ 1 ( ϖ 2 , ϖ 1 ) ϑ 2 ( ϖ 4 , ϖ 3 ) × 0 1 0 1 z w ϖ 1 ϖ 4 q 1 , q 2 2 ϖ 1 q 1 z ϖ 4 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 4 + w ϑ 2 ( ρ , ϖ 4 ) ) σ d q 1 z d q 2 w 1 σ + q 1 q 2 [ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ϑ 2 ( ρ , ϖ 3 ) ] 2 ϑ 1 ( ϖ 2 , ϖ 1 ) ϑ 2 ( ϖ 4 , ϖ 3 ) × 0 1 0 1 z w ϖ 3 ϖ 2 q 1 , q 2 2 ϖ 2 q 1 z ϖ 3 q 2 w Φ ( ϖ 2 + z ϑ 1 ( ϱ , ϖ 2 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) σ d q 1 z d q 2 w 1 σ + q 1 q 2 [ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ϑ 2 ( ϖ 4 , ρ ) ] 2 ϑ 1 ( ϖ 2 , ϖ 1 ) ϑ 2 ( ϖ 4 , ϖ 3 ) × 0 1 0 1 z w ϖ 2 , ϖ 4 q 1 , q 2 2 ϖ 2 q 1 z ϖ 4 q 2 w Φ ( ϖ 2 + z ϑ 1 ( ϱ , ϖ 2 ) , ϖ 4 + w ϑ 2 ( ρ , ϖ 4 ) ) σ d q 1 z d q 2 w 1 σ ,
for all ϱ , ρ N .
Taking first integral together with the fact that ϖ 1 , ϖ 3 q 1 , q 2 2 ϖ 1 q 1 z ϖ 3 q 2 w Φ ( z , w ) σ is higher-order generalized strongly n-polynomial preinvex functions, we get
0 1 0 1 z w ϖ 1 , ϖ 3 q 1 , q 2 2 ϖ 1 q 1 z ϖ 3 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) σ d q 1 z d q 2 w
0 1 w 0 1 z 1 n p = 1 n 1 ( 1 z ) p ϖ 1 , ϖ 3 q 1 , q 2 2 Φ ( ϱ , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w σ + 1 n p = 1 n 1 z p ϖ 1 , ϖ 3 q 1 , q 2 2 Φ ( ϖ 1 , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w σ μ 1 z ( 1 z ) ϑ 1 θ ( ϱ , ϖ 1 ) d q 1 z d q 2 w .
Computing the q 1 -integral on the right-hand side of (21), we have
0 1 z 1 n p = 1 n 1 ( 1 z ) p ϖ 1 , ϖ 3 q 1 , q 2 2 Φ ( ϱ , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w σ + 1 n p = 1 n 1 z p ϖ 1 , ϖ 3 q 1 , q 2 2 Φ ( ϖ 1 , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w σ μ 1 z ( 1 z ) ϑ 1 θ ( ϱ , ϖ 1 ) d q 1 z .
In view of the Definition 2 and k = 1 , 2 , we get
S q k = 1 n p = 1 n 0 1 z 1 ( 1 z ) p d q k z = 1 1 + q k 1 q k n p = 1 n e = 0 q k 2 e ( 1 q k e ) p , T q k = 1 n p = 1 n 0 1 z 1 z p d q k z = 1 1 + q k 1 n p = 1 n 1 q k 1 q k p + 2 , U q k = 0 1 z 2 ( 1 z ) d q k z = q k 3 ( 1 + q k + q k 2 ) ( 1 + q k + q k 2 + q k 3 ) .
Putting the above calculations into (21), we obtain
0 1 w Z q 1 ϖ 1 , ϖ 3 q 1 , q 2 2 Φ ( ϱ , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w σ + E q 1 ϖ 1 , ϖ 3 q 1 , q 2 2 Φ ( ϖ 1 , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) ϖ 1 q 1 z ϖ 3 q 2 w σ μ 1 U q 1 ϑ 1 θ ( ϱ , ϖ 1 ) d q 2 w .
Similarly, by computing the q 2 -integral, utilizing the fact ϖ 1 , ϖ 3 q 1 , q 2 2 ϖ 1 q 1 z ϖ 3 q 2 w Φ ( ϱ , ρ ) M , ϱ , ρ N on the right-hand side of (22), we get
0 1 0 1 z w ϖ 1 , ϖ 3 q 1 , q 2 2 ϖ 1 q 1 z ϖ 3 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) σ d q 1 z d q 2 w M σ ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ϑ 1 θ ( ϱ , ϖ 1 ) 2 μ 2 U q 2 ϑ 2 θ ( ρ , ϖ 3 ) .
Analogously, we also have
0 1 0 1 z w ϖ 1 ϖ 4 q 1 , q 2 2 ϖ 1 q 1 z ϖ 4 q 2 w Φ ( ϖ 1 + z ϑ 1 ( ϱ , ϖ 1 ) , ϖ 4 + w ϑ 2 ( ρ , ϖ 4 ) ) σ d q 1 z d q 2 w M σ ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ϑ 1 θ ( ϱ , ϖ 1 ) 2 μ 2 U q 2 ϑ 2 θ ( ρ , ϖ 4 ) ,
0 1 0 1 z w ϖ 3 ϖ 2 q 1 , q 2 2 ϖ 2 q 1 z ϖ 3 q 2 w Φ ( ϖ 2 + z ϑ 1 ( ϱ , ϖ 2 ) , ϖ 3 + w ϑ 2 ( ρ , ϖ 3 ) ) σ d q 1 z d q 2 w M σ ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ϑ 1 θ ( ϱ , ϖ 2 ) 2 μ 2 U q 2 ϑ 2 θ ( ρ , ϖ 3 )
and
0 1 0 1 z w ϖ 2 , ϖ 4 q 1 , q 2 2 ϖ 2 q 1 z ϖ 4 q 2 w Φ ( ϖ 2 + z ϑ 1 ( ϱ , ϖ 2 ) , ϖ 4 + w ϑ 2 ( ρ , ϖ 4 ) ) σ d q 1 z d q 2 w M σ ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ϑ 1 θ ( ϱ , ϖ 2 ) 2 μ 2 U q 2 ϑ 2 θ ( ρ , ϖ 4 ) .
Now, by making use of the inequalities (23)–(26) and the fact that
0 1 0 1 z w d q 1 z d q 2 w = 1 [ 2 ] q 1 [ 2 ] q 2 .
We get the inequality (20). This completes the proof. □
Corollary 2.
I. Let σ = 1 in Theorem 5, we have
Δ q 1 , q 2 ( , ϖ 1 , ϖ 2 , ϖ 3 , ϖ 4 ) q 1 q 2 ϑ 1 ( ϖ 2 , ϖ 1 ) ϑ 2 ( ϖ 4 , ϖ 3 ) × [ ϑ 1 ( ϱ , ϖ 1 ) ] 2 [ ϑ 2 ( ρ , ϖ 3 ) ] 2 M ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ϑ 1 θ ( ϱ , ϖ 1 ) 2 μ 2 U q 2 ϑ 2 θ ( ρ , ϖ 3 ) + [ ϑ 1 ( ϱ , ϖ 1 ) ] 2 [ ϑ 2 ( ϖ 4 , ρ ) ] 2 M ( S q 1 + T q 1 ) ( Z q 2 + E q 2 ) μ 1 U q 1 ϑ 1 θ ( ϱ , ϖ 1 ) 2 μ 2 U q 2 ϑ 2 θ ( ρ , ϖ 4 ) + [ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ϑ 2 ( ρ , ϖ 3 ) ] 2 M ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ϑ 1 θ ( ϱ , ϖ 2 ) 2 μ 2 U q 2 ϑ 2 θ ( ρ , ϖ 3 ) + [ ϑ 1 ( ϖ 2 , ϱ ) ] 2 [ ϑ 2 ( ϖ 4 , ρ ) ] 2 M ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ϑ 1 θ ( ϱ , ϖ 1 ) 2 μ 2 U q 2 ϑ 2 θ ( ρ , ϖ 4 ) .
II. By using the property of preinvex functions ϑ ( x , y ) = y x in Theorem 5, we get
| 1 ϖ 2 ϖ 1 ϖ 4 ϖ 3 ϖ 1 ϱ ϖ 3 ρ Φ ( u , v ) ϖ 1 d q 1 u ϖ 3 d q 2 v + ϖ 1 ϱ ρ ϖ 4 Φ ( u , v ) ϖ 1 d q 1 u ϖ 4 d q 2 v + ϱ ϖ 2 ϖ 3 ρ Φ ( u , v ) ϖ 2 d q 1 u ϖ 3 d q 2 v + ϱ ϖ 2 ρ ϖ 4 Φ ( u , v ) ϖ 2 d q 1 u ϖ 4 d q 2 v 1 ϖ 2 ϖ 1 ϖ 1 ϱ Φ ( u , ρ ) ϖ 1 d q 1 u + ϱ ϖ 2 Φ ( u , ρ ) ϖ 2 d q 1 u 1 ϖ 4 ϖ 3 ϖ 3 ρ Φ ( ϱ , v ) ϖ 3 d q 2 v + ρ ϖ 4 Φ ( ϱ , v ) ϖ 4 d q 2 v Φ ( ϱ , ρ ) |
q 1 q 2 ( ϖ 2 ϖ 1 ) ( ϖ 4 ϖ 3 ) [ 2 ] q 1 [ 2 ] q 2 1 1 σ × [ ϱ ϖ 1 ] 2 [ ρ ϖ 3 ] 2 M σ ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ( ϱ ϖ 1 ) θ 2 μ 2 U q 2 ( ρ ϖ 3 ) θ 1 σ + [ ϱ ϖ 1 ] 2 [ ϖ 4 ρ ] 2 M σ 2 ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ( ϱ ϖ 1 ) θ 2 μ 2 U q 2 ( ρ ϖ 4 ) θ 1 σ + [ ϖ 2 ϱ ] 2 [ ρ ϖ 3 ] 2 M σ ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ( ϱ ϖ 2 ) θ 2 μ 2 U q 2 ( ρ ϖ 3 ) θ 1 σ + [ ϖ 2 ϱ ] 2 [ ϖ 4 ρ ] 2 M σ ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) μ 1 U q 1 ( ϱ ϖ 1 ) θ 2 μ 2 U q 2 ( ρ ϖ 4 ) θ 1 σ .
III. Let μ k = 0 , for k = 1 , 2 in Theorem 5, we get
| Δ q 1 , q 2 ( , ϖ 1 , ϖ 2 , ϖ 3 , ϖ 4 ) | M q 1 q 2 [ 2 ] q 1 [ 2 ] q 2 1 1 σ ϑ 1 ϱ , ϖ 1 2 + ϑ 1 ϖ 2 , ϱ 2 ϑ 1 ( ϖ 2 , ϖ 1 ) × ϑ 2 ρ , ϖ 3 2 + ϑ 2 ϖ 4 , ρ 2 ϑ 2 ( ϖ 4 , ϖ 3 ) ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) 1 σ .
IV. Let μ k = 0 , for k = 1 , 2 and property of preinvexity ϑ ( x , y ) = y x in Theorem 5, we get
| 1 ϖ 2 ϖ 1 ϖ 4 ϖ 3 ϖ 1 ϱ ϖ 3 ρ Φ ( u , v ) ϖ 1 d q 1 u ϖ 3 d q 2 v + ϖ 1 ϱ ρ ϖ 4 Φ ( u , v ) ϖ 1 d q 1 u ϖ 4 d q 2 v + ϱ ϖ 2 ϖ 3 ρ Φ ( u , v ) ϖ 2 d q 1 u ϖ 3 d q 2 v + ϱ ϖ 2 ρ ϖ 4 Φ ( u , v ) ϖ 2 d q 1 u ϖ 4 d q 2 v 1 ϖ 2 ϖ 1 ϖ 1 ϱ Φ ( u , ρ ) ϖ 1 d q 1 u + ϱ ϖ 2 Φ ( u , ρ ) ϖ 2 d q 1 u 1 ϖ 4 ϖ 3 ϖ 3 ρ Φ ( ϱ , v ) ϖ 3 d q 2 v + ρ ϖ 4 Φ ( ϱ , v ) ϖ 4 d q 2 v Φ ( ϱ , ρ ) | M q 1 q 2 [ 2 ] q 1 [ 2 ] q 2 1 1 σ ϱ ϖ 1 2 + ϖ 2 ϱ 2 ϖ 2 ϖ 1 ρ ϖ 3 2 + ϖ 4 ρ 2 ϖ 4 ϖ 3 × ( S q 1 + T q 1 ) ( S q 2 + T q 2 ) 1 σ .
V. Let q k 1 , for k = 1 , 2 in part IV, we have the following inequality, for more details, see in [33].
| K ϱ , ρ + 1 ( ϖ 2 ϖ 1 ) ( ϖ 4 ϖ 3 ) ϖ 1 ϖ 2 ϖ 3 ϖ 4 K ( u , v ) d v d u Q | M 4 ϱ ϖ 1 2 + ϖ 2 ϱ 2 ϖ 2 ϖ 1 ρ ϖ 3 2 + ϖ 4 ρ 2 ϖ 4 ϖ 3 , Q = 1 ϖ 2 ϖ 1 ϖ 1 ϖ 2 K ( u , ρ ) d u + 1 ϖ 4 ϖ 3 ϖ 3 ϖ 4 K ( ϱ , v ) d v .

3. Conclusions

In this paper, a new class of preinvex functions, which is called higher-order generalized strongly n-polynomial preinvex function, is presented. Meanwhile, we derived a new quantum integral identity involving a second-order mixed partial differentiable function. Using this identity as an auxiliary result, we obtained q 1 q 2 -Ostrowski-type inequalities for higher-order generalized strongly n-polynomial preinvex functions on co-ordinates. We want to emphasize here that we can recapture some other new results from the main results of this article under some suitable conditions. For example, if we take θ = 2 , all the results reduce to the results for generalized strongly n-polynomial preinvex functions. We have results for higher-order generalized strongly n-polynomial convex functions if ϑ ( ϖ 2 , ϖ 1 ) = ϖ 2 ϖ 1 . Similarly, we have other new and known outcomes for other eligible options, and we have left the specifics to interested readers. This demonstrates that the conclusions reached in this article are pretty consistent. We hope that the ideas and strategies presented in this article may pique the interest of those who are interested.

Author Contributions

H.K. writing—original draft preparation, M.V.-C. review and editing. All authors have read and agreed to the published version of the manuscript.

Funding

Not Applicable.

Institutional Review Board Statement

Not Applicable.

Informed Consent Statement

Not Applicable.

Data Availability Statement

Not Applicable.

Acknowledgments

We want to give thanks to the Dirección de investigación from Pontificia Universidad Católica del Ecuador for technical support to our research project entitled: “Algunas desigualdades integrales para funciones convexas generalizadas y aplicaciones”.

Conflicts of Interest

The authors declare no conflict of interest.

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Kalsoom, H.; Vivas-Cortez, M. q1q2-Ostrowski-Type Integral Inequalities Involving Property of Generalized Higher-Order Strongly n-Polynomial Preinvexity. Symmetry 2022, 14, 717. https://doi.org/10.3390/sym14040717

AMA Style

Kalsoom H, Vivas-Cortez M. q1q2-Ostrowski-Type Integral Inequalities Involving Property of Generalized Higher-Order Strongly n-Polynomial Preinvexity. Symmetry. 2022; 14(4):717. https://doi.org/10.3390/sym14040717

Chicago/Turabian Style

Kalsoom, Humaira, and Miguel Vivas-Cortez. 2022. "q1q2-Ostrowski-Type Integral Inequalities Involving Property of Generalized Higher-Order Strongly n-Polynomial Preinvexity" Symmetry 14, no. 4: 717. https://doi.org/10.3390/sym14040717

APA Style

Kalsoom, H., & Vivas-Cortez, M. (2022). q1q2-Ostrowski-Type Integral Inequalities Involving Property of Generalized Higher-Order Strongly n-Polynomial Preinvexity. Symmetry, 14(4), 717. https://doi.org/10.3390/sym14040717

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