Second-Order Symmetric Duality for Multiple Objectives Nonlinear Programming Under Generalizations of Cone-Preinvexity Functions

In this article, we present generalizations of the cone-preinvexity functions and study a pair of second-order symmetric solutions for multiple objective nonlinear programming problems under these generalizations of the cone-preinvexity functions. In addition, we establish and prove the theorems of weak duality, strong duality, strict converse duality, and self-duality by assuming the skew-symmetric functions under these generalizations of the cone-preinvexity functions. Finally, we provide four nontrivial numerical examples to demonstrate that the results of the weak and strong duality theorems are true.


Study the Second-Order Symmetric Duality Theorems
Proof Assume the inverse. That is, Because the function f (., y) is a K − preinvexity w. r. t.
x f or f ixed y v, we get Using the mean-valued theorem, we get Since function − f (x, .) is a K − preinvexity w. r. t.y f or f ixed x, When we add the inequalities (1) and (2) together, we get a Because of assumption (iii), we have By adding the above inequalities, Then a relationship (3) is formed.
From the feasibility of the points (x, y, λ, p) and (u, v, λ, r ) for the (MONLP) and (MONLD) problems, respectively, we have Substituting (5) into (4), we get the following: That contradicts (*). Then the proof is complete.
In addition, the (MONLP) and (MONLD) problems become the forms: Assume the inverse of the weak duality theorem, that is, By using assumption (ii), we have Therefore, the relationship (I " ) gives From the feasibility points (x, y, λ, p), (u, v, λ, r ) for the (MONLP) and (MONLD) problems, respectively, we have We substitute into (3) to get This contradicts (* " ), then the proof is complete. [22]If there is x * a weak minimum for the (MONLP) problem, then there α * ∈ K * , β * ∈ Q * cannot be both zeros such that.  x, y, λ, p)is a weak minimum for the (MONLP) problem with a fix λ λ, r r and that:

Lemma 3.1 Yang
We claim that α 0, if a substitute is used y ∈ C 2 , x x ∈ C 1 , λ λ ∈ K * and p p ∈ K * the inequality (6) becomes If α 0 and belongs to K* and when y β + y ∈ C 2 we have.
We can conclude from assumption (i) that we obtained that β 0 this is not possible since,(α, β) 0 as a result,α 0.
When we substitute x x, y y and λ λ in (6), we get Furthermore, we know from assumption (ii) that.

β (α T e)y (9)
If we put y y, λ λ and p p in (6), we get That implies The following is obtained by differentiating the relationship (11) w. r. t. x This means it point (x, y, λ, r ) is feasible to solve the (MONLD) problem.
From the differential w. r. t.x, we get As a result of (7), and because β 0 we obtained We get the following result by differentiating the above inequality w. r. t.y When we put y 0 and y y in (13) and (14) respectively, we get From the differential equation w. r. t.y, we get the following when put y p. To show that the point (x, y, λ, p) is the weak maximum for the (MONLD) problem; otherwise, there exists a feasible point (u, v, λ, r ) such that Since This contradicts the weak duality theorem.
The following example illustrates the results of the strong duality theorem. the point (x, y, λ, p) be a weak minimum for the (MONLP)1 problem. Then, based on Lemma 3.1, there exists.

Example 3.2.1 Let
Equation (7) takes the form of If we put it α 0, x x ∈ C 1 , λ λ ∈ K * and p p ∈ K * this way ∀y ∈ C 2 , the inequality (15) becomes If α 0 ∈ K * and put y β + y ∈ C 2 ⇒. We get from (i) that We get the following by substituting x x, y y, λ λ in relation (13).
In addition, from (ii), we obtain In addition, when we substitute y y, λ λ and p p in relation (13), we get an For each x, x ∈ C 1 , x + x ∈ C 1 ⇒ x(3x 2 , 3x 2 ) ≥ 0. Then, by differentiating this inequality w. r. t. x, we obtain That shows the point (x, y, λ, r ) is feasible for the (MONLD) problem.
To demonstrate that point is the weak maximum for the (MONLD) problem; otherwise, there exists a feasible point (x, y, λ, r ) such that That contradicts the theorem.  x, y, λ, p) is a weak minimum for the (MONLP)2 problem; then, according to Lemma 3.1, there exists.
y(0, 2λ 2 y) ≥ 0 (14") By differentiable w. r. t. y, we get y(0, 2λ 2 ) + (0, 2λ 2 y) ≥ 0 ⇒ (0, 2λ 2 y) + p(0, 2λ 2 ) ∈ C * 2 When we put y 0, y y in the inequalities (13 " ) and (14 " ), respectively, we get y(0, 2λ 2 y) 0 Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder.