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

Abstract

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.

1 Introduction

Weir and Mond [21] introduced symmetric and self-dual for multiple objective nonlinear programming (MONLP) problems. Mond and Weir [17] introduced duality to the MONLP problems with pseudo-convexity and pseudo-concavity. Kim [13, 14] presented symmetric duality for these (MONLP) problems under pseudo-invexity functions. Devi [1] introduced symmetric programs involving $\eta$-bonvexity functions for these problems. Mishra [15] studied these programming problems under F-convexity. Yang [22] introduced converse duality for these programming problems under cone constraints. Seema [18] presented symmetric duality for these programming problems under cone-invex. Kassem [8,9,10,11] studied these programming problems under cone-invexity and (K, F)-pseudo-convexity functions. Suneja et al. [19] studied these programming problems with cone constraints. Dubey et al. [3,4,5,6,7] reported higher-order symmetric duality for non-differentiable fractional programming under cone constraints. Kassem [12] studied the second-order symmetric duality in vector optimization involving $(K,\eta )$-pseudobonvexity functions.

The article presents generalizations of the cone-preinvexity functions and studies a pair of second-order symmetric multiple objective nonlinear programming (MONLP) problems under these generalizations of the cone-preinvexity functions. In addition, we establish and prove duality theorems and a self-duality theorem for these (MONLP) problems under these generalizations of the cone-preinvexity functions. Finally, four nontrivial numerical examples are given to show that the results of the weak and strong duality theorems are true.

2 Notations and Preliminaries

Definition 2.1

[14] The set $X$ is called an invex at the point $u \in X$ with respect to (w. r. t.)$\eta :X \times X \to R^{n}$ if $\forall \,\,x \in X$ we have $u + \lambda \,\eta (x,u) \in X,\,0\mathop < \limits_{ = } \,\lambda \mathop < \limits_{ = } 1.$

If it $X$ is invex for all $u \in X$, the set $X$ is invex w. r. t. $\eta :X \times X \to R^{n}$.

Definition 2.2

[14] The invex set $X$ is called an invex cone, if $x \in X,\,\,\,\lambda \ge 0\,\, \Rightarrow \,\lambda \,x \in X.$

Definition 2.3

The function $f(x)$ is said to have a $K$-preinvexity w. r. t.$\eta :X \times X \to R^{n}$, if $\forall \,\,x,\,y \in R^{n} \,\,\,and\,\,\,\lambda \in (0,1)$ we have.

$\lambda \,f(x) + (1 - \lambda )\,f(y) - f(y + \lambda \,\eta (x,y)) \in K$.

Where $K$ is define as a closing invex cone.

Let us consider the following general (MONLP) problems:

$$ (MONLP)^{1} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,K - \min \,\,\,f_{i} (x),\,\,i = 1,2,..p $$

$$ {\text{subject}}\,\,\,{\text{to}}\,\,\,x \in X = \{ x \in C:\,\,\, - g_{j} (x) \in Q,\,\,j = 1,2,...m\} $$

where $C \subset R^{n} ,\,\,\,K\,\,\,and\,\,\,Q$ are closed invex cones and $f_{i} :R^{n} \to R^{p} ,\,\,\,g_{j} :R^{n} \to R^{m} .$

Definition 2.4

[15] The point $\overline{x} \in X$ is a weak minimum for the $(MONLP)^{1}$ problem, if $f(\overline{x}) - f(x) \notin {\text{int}} \,K\,\,\,\forall \,\,x \in X.$

Definition 2.5

[18] The positive cone $K^{*}$ is.

$$ K^{*} = \{ z \in R^{p} :\,\,x^{T} z \ge 0\,\,\,\,\forall \,\,\,x \in K\} . $$

Let us consider a pair of second-order symmetric duality $(MONLP)^{1}$ problems in the following form:

$$ ({\text{MONLP}}):\,\,\,\,\,\,\,K - \min \,\,f(x,y) - [y^{T} \nabla_{y} (\lambda^{T} f)(x,y)]e - [y^{T} \nabla_{yy} (\lambda^{T} f)(x,y)p]e $$

$$ {\text{subject}}\,\,\,{\text{to}}\,\,\,(x,y) \in C_{1} \times C_{2} , $$

$$ - \nabla_{y} (\lambda^{T} f)(x,y) - p^{T} \nabla_{yy} (\lambda^{t} f)(x,y) \in C_{2}^{*} , $$

$$ \lambda ,\,\,p \in K^{*} ,\,\,\,e \in {\text{int}} \,\,K,\,\,\,\lambda^{T} e = 1, $$

$$ (MONLD):\,\,\,\,\,\,\,\,K - \max \,\,\,f(u,v) - [u^{T} \nabla_{u} (\lambda^{T} f)(u,v)]e - [u^{T} \nabla_{uu} (\lambda^{T} f)(u,v)r]e $$

$$ {\text{subject}}\,\,{\text{to}}\,\,(u,v) \in C_{1} \times C_{2} , $$

$$ \nabla_{u} (\lambda^{T} f)(u,v) + r^{T} \nabla_{uu} (\lambda^{T} f)(u,v) \in C_{1}^{*} , $$

$$ \lambda ,\,\,r \in K^{*} ,\,\,\,e \in {\text{int}} \,\,K,\,\,\,\lambda^{T} e = 1, $$

where $f:R^{n} \times R^{m} \to R^{p}$ is a twice-differentiable function, these $C_{1}^{*} ,\,\,C_{2}^{*}$ are positive cones for closed invex cones $C_{1} \,,\,\,C_{2}$, respectively, $\nabla_{x} (\lambda^{T} f)(x,y)\,,\,\nabla_{y} (\lambda^{T} f)(x,y)$ and $\nabla_{xx} (\lambda^{T} f)(x,y),\,\,\nabla_{yy} (\lambda^{T} f)(x,y)$ the gradients and Hessian matrices for $(\lambda^{T} f)(x,y)$ w. r. t.$x\,,\,y$, respectively.

3 Study the Second-Order Symmetric Duality Theorems

Theorem 3.1

(Weak Duality) Assume the points $(x,y,\lambda ,p)\,\,\,\,and\,\,\,(u,v,\lambda ,r)$ are feasible for the (MONLP) and (MONLD) problems, respectively, and.

(i) The function $f(.,y)\,\,\,is\,\,a\,\,\,K -$ Preinvexity w. r. t. $x$ for a fixed $y$.

(ii) For a fixed $x$, the function $- f(x,.)\,\,\,is\,\,a\,\,\,\,K -$ Preinvexity w. r. t.$y$, and.

(iii) $\left( {\begin{array}{*{20}c} {\eta (x,u)} & {(v - y)} \\ { - (x - u)} & { - \eta (v,y)} \\ \end{array} } \right)\left( {\begin{array}{*{20}c} {\nabla_{u} f(u,v)} \\ {\nabla_{y} f(x,y)} \\ \end{array} } \right) \ge \left( {\begin{array}{*{20}c} 0 \\ 0 \\ \end{array} } \right)$.

Then

$$ \begin{aligned} & f(u,v) - [u^{T} \nabla_{u} (\lambda^{T} f)(u,v)]e - [u^{T} \nabla_{uu} (\lambda^{T} f)(u,v)r]e - f(x,y) + [y^{T} \nabla_{y} (\lambda^{T} f)(x,y)]e \\ & \quad + [y^{T} \nabla_{yy} (\lambda^{T} f)(x,y)p]e \notin \,{\text{int}} \,\,K. \\ \end{aligned} $$

Proof

Assume the inverse. That is,

$$ \begin{aligned} & f(u,v) - [u^{T} \nabla_{u} (\lambda^{T} f)(u,v)]e - [u^{T} \nabla_{uu} (\lambda^{T} f)(u,v)r]e - f(x,y) + [y^{T} \nabla_{y} (\lambda^{T} f)(x,y)]e \\ & \quad + [y^{T} \nabla_{yy} (\lambda^{T} f)(x,y)p]e \in \,{\text{int}} \,\,K \\ \end{aligned} $$

Because it $\lambda \in K^{*}$ implies

$$ \begin{aligned} & \lambda^{T} f(u,v) - \lambda^{T} [u^{T} \nabla_{u} (\lambda^{T} f)(u,v)]e - \lambda^{T} [u^{T} \nabla_{uu} (\lambda^{T} f)(u,v)r]e - \lambda^{T} f(x,y) + \lambda^{T} [y^{T} \nabla_{y} (\lambda^{T} f)(x,y)]e \\ & \quad + \lambda^{T} [y^{T} \nabla_{yy} (\lambda^{T} f)(x,y)p]e \ge 0\,\,\, \Rightarrow \\ \end{aligned} $$

$$ \begin{aligned} & \lambda^{T} f(u,v) - \lambda^{T} f(x,y) - [u^{T} \nabla_{u} (\lambda^{T} f)(u,v)] - [u^{T} \nabla_{uu} (\lambda^{T} f)(u,v)r] + [y^{T} \nabla_{y} (\lambda^{T} f)(x,y)] \\ & \quad + [y^{T} \nabla_{yy} (\lambda^{T} f)(x,y)p] \ge 0\,\,\,as\,\,\,\lambda^{T} e = 1. \\ \end{aligned} $$

(*)

Because the function $f(.,y)$ is a $K -$ preinvexity w. r. t.$x\,\,for\,\,fixed\,\,\,y = v$, we get

$$ \begin{aligned} & \lambda \,f(x,v) + (1 - \lambda )\,f(u,v) - f(u + \lambda \eta (x,u),v) \in K,\,\,\,\,\,\lambda \in (0,1)\,\,\,\,\, \Rightarrow \\ & \quad (f(x,v) - f(u,v)) - \left( {\frac{f(u + \lambda \eta (x,u),v) - f(u,v)}{\lambda }} \right) \in K \\ \end{aligned} $$

Using the mean-valued theorem, we get

$$ \begin{aligned} & f(x,v) - f(u,v) - \eta (x,u)\,\nabla_{u} f(u,v) \in K\,\,\, \Rightarrow \,\,for\,\,\,\lambda \in K^{*} \\ & \quad \lambda^{T} \,f(x,v) - \lambda^{\,T} f(u,v) \ge \lambda^{T} \eta (x,u)\,\nabla_{u} f(u,v), \\ \end{aligned} $$

(1)

Since function $- f(x,.)$ is a $K -$ preinvexity w. r. t.$y\,\,for\,\,fixed\,\,\,x,$

$$ \lambda^{T} \,f(x,y) - \lambda^{\,T} f(x,v) \ge - \lambda^{T} \eta (v,y)\,\nabla_{y} f(x,y) $$

(2)

When we add the inequalities (1) and (2) together, we get a

$$ \lambda^{T} \,f(x,y) - \lambda^{\,T} f(u,v) \ge \lambda^{T} \eta (x,u)\,\nabla_{u} f(u,v) - \lambda \eta (v,y)\nabla_{y} f(x,y) $$

(3)

Because of assumption (iii), we have

$$ \eta (x,u)\,\nabla_{u} f(u,v) \ge - (v - y)\nabla_{y} f(x,y);\,\,\,\,\,\,\,\,\,\,\, - (x - u)\nabla_{u} f(u,v) \ge \eta (v,y)\nabla_{y} f(x,y) $$

By adding the above inequalities,

$$ \eta (x,u)\,\nabla_{u} f(u,v) - \eta (v,y)\nabla_{y} f(x,y) \ge (x - u)\nabla_{u} f(u,v) - (v - y)\nabla_{y} f(x,y). $$

Then a relationship (3) is formed.

$$ \begin{aligned} & \lambda f(x,y) - \lambda f(u,v) - x^{T} \nabla_{u} (\lambda^{T} f)(u,v) + u^{T} \nabla_{u} (\lambda^{T} f)(u,v) \\ & \quad - y^{T} \nabla_{y} (\lambda^{T} f)(x,y) + v^{T} \nabla_{y} (\lambda^{T} f)(x,y) \ge 0. \\ \end{aligned} $$

(4)

From the feasibility of the points $(x,y,\lambda ,p)\,\,\,and\,\,\,(u,v,\lambda ,r)$ for the (MONLP) and (MONLD) problems, respectively, we have

$$ - \nabla_{y} (\lambda^{T} f)(x,y) \ge p^{T} \nabla_{yy} (\lambda^{T} f)(x,y),\,\,\,\,\,\nabla_{u} (\lambda^{T} f)(u,v) \ge - r^{T} \nabla_{uu} (\lambda^{T} f)(u,v) $$

(5)

Substituting (5) into (4), we get the following:

$$ \begin{aligned} & \lambda^{T} f(x,y) - y^{T} \nabla_{y} (\lambda^{T} f)(x,y) - y^{T} \nabla_{yy} (\lambda^{T} f)(x,y)p - \lambda^{T} f(u,v) \\ & \quad + u^{T} \nabla_{u} (\lambda^{T} f)(u,v) + u^{T} \nabla_{uu} (\lambda^{T} f)(u,v)r \ge 0, \\ \end{aligned} $$

That contradicts (*). Then the proof is complete.

To illustrate the results of this weak duality theorem, we introduce the following example:

Example 3.1.1

Consider the following:

$K = \{ (x,y) \in R^{2} :\,\,\left| y \right| \le x,\,\,\,x \ge 0\} ,$$f(x,y) = (x^{3} - y^{3} ,x^{3} ),$$C_{1} = C_{2} = R_{ + }$,

$$ \lambda f(x,v) + (1 - \lambda )f(u,v) - f(x + \lambda \eta (x,u),v) = (\lambda x^{3} - \lambda u^{3} + u^{3} - (x + \lambda \eta (x,u))^{3} , $$

$\lambda x^{3} - \lambda u^{3} + u^{3} - (x + \lambda \eta (x,u))^{3} )$.

There is $f(.,y)$ a $K -$ preinvexity function w. r. t.$x\,\,for\,\,fixed\,\,y.$ In addition, there is $- f(x,.)$ a $K -$ preinvexity function w. r. t.$y\,\,for\,\,fixed\,\,x$

Furthermore, the (MONLP) and (MONLD) problems take the form:

$$ (MONLP)\,1:\,\,\,\,\,\,\,\,\,\,\,\,K - \min \,\,\,(x^{3} - y^{3} ,x^{3} ) + 3\lambda_{1} y^{3} (e_{1} ,e_{2} ) + 6\lambda_{1} y^{2} p\,(e_{1} ,e_{2} ) $$

$$ subject\,\,\,to\,\,\,\,x \ge 0,\,\,\,y \ge 0, $$

$$3\lambda_{1} y^{2} + 6\lambda_{1} y\,p \ge 0$$

,

$$ \lambda = (\lambda_{1} ,\lambda_{2} ) \in K^{*} = K,\,\,\,(e_{1} ,e_{2} ) \in {\text{int}} \,K, $$

$$ \lambda_{1} e_{1} + \lambda_{2} e_{2} = 1, $$

and

$$ ({\text{MONLD}})\,1:\,\,\,\,\,\,\,\,\,\,\,K - \max \,\,(u^{3} - v^{3} ,u^{3} ) - 3(\lambda_{1} + \lambda_{2} )u^{3} (e_{1} ,e_{2} ) - 6(\lambda_{1} + \lambda_{2} )u^{2} r(e_{1} ,e_{2} ) $$

$$ subject\,\,\,to\,\,u \ge 0,\,\,\,v \ge 0, $$

$$ 3(\lambda_{1} + \lambda_{2} )u^{2} + 6(\lambda_{1} + \lambda_{2} )u\,r \ge 0, $$

$$ \lambda = (\lambda_{1} ,\lambda_{2} ) \in K^{\,*} = K,\,\,\,(e_{1} ,e_{2} ) \in {\text{int}} \,K, $$

$$ \lambda_{1} e_{1} + \lambda_{2} e_{2} = 1. $$

Assume the points $(x,y,\lambda ,p)\,\,and\,\,(u,v,\lambda ,r)$ are feasible for (MONLP)1 and (MONLD)1, problems, respectively.

Because $f(.,y)$ is a $K -$ preinvexity function w. r. t.$x$, we obtain

$$ f(x,v) - f(u,v) - \eta (x,u)\nabla_{u} f(u,v) \in K\,\,\, \Rightarrow \,\,\, $$

$(x^{3} - u^{3} - 3u^{2} \eta (x,u),\,x^{3} - u^{3} - 3u^{2} \eta (x,u)) \in K$.

Furthermore, because $- f(x,.)$ is a $K -$ preinvexity function w. r. t.$y$, we get

$$ (v^{3} - y^{3} - 3y^{2} \eta (v,y),0) \in K. $$

where $\lambda = (\lambda_{1} ,\lambda_{2} ) \in K^{*} = K,\,\,(e_{1} ,e_{2} ) \in {\text{int}} \,K$ we have

$$ (\lambda_{1} + \lambda_{2} )(x^{3} - u^{3} - 3u^{2} \eta (x,u)) \ge 0\,\,and\,\,\lambda_{1} (v^{3} - y^{3} - 3y^{2} \eta (v,y)) \ge 0. $$

Add the above two inequalities together, and we get

$$ (\lambda_{1} + \lambda_{2} )(x^{3} - u^{3} ) + \lambda_{1} (v^{3} - y^{3} ) \ge 3(\,\lambda_{1} + \lambda_{2} )u^{2} \eta (x,u) + 3\lambda_{1} y^{2} \eta (v,y)) \ge 0. $$

If we have the functions $\eta (x,u) = x - u,\,\,\,\,\eta (v,y) = v - y$,

$$ (\lambda_{1} + \lambda_{2} )(x^{3} - u^{3} ) + \lambda_{1} (v^{3} - y^{3} ) + 3(\,\lambda_{1} + \lambda_{2} )u^{3} + 3\lambda_{1} y^{3} - (\lambda_{1} + \lambda_{2} )u^{2} x - 3\lambda_{1} y^{2} v \ge 0 $$

Since the points $(x,y,\lambda ,p)\,\,and\,\,(u,v,\lambda ,r)$ are feasible for (MONLP)1 and (MONLD)1 problems, respectively, we get

$$ - 3(\lambda_{1} + \lambda_{2} )u^{2} \le 6(\lambda_{1} + \lambda_{2} )u\,r\,\,\,\,and\,\,\, - 3\lambda_{1} y^{2} \le 6\lambda_{1} y\,p \Rightarrow $$

$- 3(\lambda_{1} + \lambda_{2} )u^{2} x \le 6(\lambda_{1} + \lambda_{2} )u^{2} \,r\,\,\,\,and\,\,\, - 3\lambda_{1} y^{2} v \le 6\lambda_{1} y^{2} \,p$.

Then,

$$ \begin{aligned} & (\lambda_{1} + \lambda_{2} )(x^{\,3} - u^{\,3} ) + \lambda_{1} (v^{\,3} - y^{\,3} ) + 3(\,\lambda_{1} + \lambda_{2} )u^{\,3} + 3\lambda_{1} y^{\,3} \\ & \quad + 6(\lambda_{1} + \lambda_{2} )u^{2} r + 6\lambda_{1} y^{2} p \ge 0 \\ \end{aligned} $$

(**)

Because $e = (e_{1} ,e_{2} ) \in {\text{int}} \,K,\,\,\lambda_{1} e_{1} + \lambda_{2} e_{2} = 1,$ we get

$$ \begin{aligned} & (u^{\,3} - v^{\,3} ,u^{\,3} ) - 3(\lambda_{1} + \lambda_{2} )u^{\,3} (e_{1} ,e_{2} ) + 6(\lambda_{1} + \lambda_{2} )u^{\,2} r(e_{1} ,e_{2} ) - (x^{\,3} - y^{\,3} ,x^{\,3} ) \\ & \quad + 3\lambda_{1} y^{\,3} (e_{1} ,e_{2} ) + 6\lambda_{1} y^{2} p(e_{1} ,e_{2} ) \notin {\text{int}} \,K. \\ \end{aligned} $$

(***)

Because $\lambda = (\lambda_{1} ,\lambda_{2} )$ we get

$$ \begin{aligned} & \lambda_{1} (u^{3} - x^{3} + y^{3} - v^{3} - 3(\lambda_{1} + \lambda_{2} )u^{3} e_{1} - 6(\lambda_{1} + \lambda_{2} )u^{2} re_{1} - 3\lambda_{1} y^{3} e_{1} - 6\lambda_{1} y^{2} pe_{1} ) \\ & \quad + \lambda_{2} (u^{3} - x^{3} - 3(\lambda_{1} + \lambda_{2} )u^{3} e_{2} - 6(\lambda_{1} + \lambda_{2} )u^{2} re_{2} - 3\lambda_{1} y^{3} e_{2} - 6\lambda_{1} y^{2} pe_{2} ) \ge 0 \\ \end{aligned} $$

That is

$$ \begin{aligned} & (u^{3} - x^{3} )(\lambda_{1} + \lambda_{2} ) + \lambda_{1} (y^{3} - v^{3} ) - 3(\lambda_{1} + \lambda_{2} )u^{3} (\lambda_{1} e_{1} + \lambda_{2} e_{2} ) \\ & \quad - 6(\lambda_{1} + \lambda_{2} )u^{2} r(\lambda_{1} e_{1} + \lambda_{2} e_{2} ) - 3\lambda_{1} y^{3} (\lambda_{1} e_{1} + \lambda_{2} e_{2} ) - 6\lambda_{1} y^{2} p(\lambda_{1} e_{1} + \lambda_{2} e_{2} ) \ge 0\,\, \Rightarrow \,\, \\ & \quad (\lambda_{1} + \lambda_{2} )(u^{3} - x^{3} ) + \lambda_{1} (y^{3} - v^{3} ) - 3(\lambda_{1} + \lambda_{2} )u^{3} - 6(\lambda_{1} + \lambda_{2} )u^{2} r - 3\lambda_{1} y^{3} - 6\lambda_{1} y^{2} p \ge 0 \\ \end{aligned} $$

Or

$$ (\lambda_{1} + \lambda_{2} )(x^{3} - u^{3} ) + \lambda_{1} (v^{3} - y^{3} ) + 3(\lambda_{1} + \lambda_{2} )u^{3} + 6(\lambda_{1} + \lambda_{2} )u^{2} r + 3\lambda_{1} y^{3} + 6\lambda_{1} y^{2} p < 0 $$

That contradicts (**) and shows that the results of the duality theorem are weak.

Example 3.1.2

Let $\{ (x,y) \in R^{2} :\,\left| y \right| \ge x,\,\,x \ge 0\} ,\,\,C_{1} ,\,C_{2} = R_{ + }$, $f(x,y) = (x^{2} ,y^{2} - x^{2} )$.

In addition, the (MONLP) and (MONLD) problems become the forms:

$$ \begin{aligned}(MONLP)2{:}\qquad &K - \min \,(x^{2} ,y^{2} - x^{2} ) - 2\lambda_{2} y(e_{1} ,e_{2} ) - 2\lambda_{2} p(e_{1} ,e_{2} ) \\ &{\text{subject to}} \quad - 2\lambda_{2} y - 2\lambda_{2} p \ge 0 \\ &\qquad\qquad\qquad x \ge 0,\,\,y \ge 0 \end{aligned}$$

and

$$ \begin{aligned}(MONLD)2{:}\qquad &K - \max \,\,(u^{2} ,v^{2} - u^{2} ) - 2u^{2} (\lambda_{1} + \lambda_{2} )(e_{1} ,e_{2} ) - 2u(\lambda_{1} + \lambda_{2} )\,r\,(e_{1}^{{}} ,e_{2} ) \\ &{\text{subject to}} \quad 2(1 - 2\lambda_{2} )(u + r) \ge 0 \\ &\qquad\qquad\quad\,\, u \ge 0,\,\,v \ge 0 \end{aligned}$$

Assume the inverse of the weak duality theorem, that is,

$$ \begin{aligned} & \lambda^{T} f(u,v) - \lambda^{T} f(x,y) - [u^{T} \nabla_{u} (\lambda^{T} f)(u,v)] - [u^{T} \nabla_{uu} (\lambda^{T} f)(u,v)r] \\ & \quad + [y^{T} \nabla_{y} (\lambda^{T} f)(x,y)] + [y^{T} \nabla_{yy} (\lambda^{T} f)(x,y)p] \ge 0\,\, \Rightarrow \\ \end{aligned} $$

$$ \begin{aligned} & \lambda^{T} (u^{{^{2} }} ,v^{2} - u^{2} ) - \lambda^{T} (x^{2} ,y^{2} - x^{2} ) - \lambda^{T} u^{T} (2u, - 2u) - \lambda^{T} u^{T} (2, - 2)\,r + \lambda^{T} y^{T} (0,2y) \\ & \quad + \lambda^{T} y^{T} (0,2)p \ge 0 \\ \end{aligned} $$

(*")

Since $f(0,y)$ is a $K -$ preinvexity w. r. t. x for fixed y = v and $- f(x,0)$ is a $K -$ preinvexity w. r. t. y for fixed x, we get

$$ \lambda^{T} f(x,y) - \lambda^{T} f(u,v) \ge \lambda^{T} \eta (x,u)\nabla_{u} f(u,v) - \lambda^{T} \eta (v,y)\nabla_{y} f(x,y) $$

That indicates.

$$ \lambda^{T} (x^{2} ,y^{2} - x^{2} ) - \lambda^{T} (u^{2} ,v^{2} - u^{2} ) \ge \lambda^{T} \eta (x,u)(2u, - 2u) - \lambda^{T} \eta (v,y)(0,2y) $$

(I")

By using assumption (ii), we have

$$ \eta (x,u)\nabla_{u} f(u,v) - \eta (v,y)\nabla_{y} f(x,y) \ge (x - u)\nabla_{u} f(u,v) - (v - y)\nabla_{y} f(x,y) $$

Therefore, the relationship (I^") gives

$$ \begin{aligned} & \lambda^{T} (x^{2} ,y^{2} - x^{2} ) - \lambda^{T} (u^{2} ,v^{2} - u^{2} ) - \lambda^{T} x^{T} (2u, - 2u) + \lambda^{T} u^{T} (2u, - 2u) \\ & \quad - \lambda^{T} y^{T} (0,2y) + \lambda^{T} v^{T} (0,2y) \ge 0 \\ \end{aligned} $$

From the feasibility points $(x,y,\lambda ,p),\,\,(u,v,\lambda ,r)$ for the (MONLP) and (MONLD) problems, respectively, we have

$$ - \lambda^{T} (0,2y) \ge \lambda^{T} p^{T} (0,2),\,\,\,\,\,\,\,\lambda^{T} (2u, - 2u) \ge - \lambda^{T} r^{T} (2, - 2) $$

We substitute into (3) to get

$$ \begin{aligned} & \lambda^{T} (x^{2} ,y^{2} - x^{2} ) - \lambda^{T} y^{T} (0,2y) - \lambda^{T} y^{T} (0,2)p - \lambda^{T} (u^{2} ,v^{2} - u^{2} ) \\ & \quad + \lambda^{T} u^{T} (2u, - 2u) + \lambda^{T} u^{T} (2, - 2)\,r \ge 0 \\ \end{aligned} $$

This contradicts (*^"), then the proof is complete.

Lemma 3.1

Yang [22] If there is $x^{*}$ a weak minimum for the (MONLP) problem, then there $\alpha^{*} \in K^{*} ,\,\,\,\beta^{*} \in Q^{*}$ cannot be both zeros such that.

$(\alpha^{{*^{T} }} \nabla f(x^{*} )^{T} + \beta^{{*^{T} }} \nabla g(x^{*} )^{T} )(x - x^{*} ) \ge 0,\,\,\,\,\forall \,\,\,x \in X$ ,

$$ \beta^{{*^{T} }} g(x^{*} ) = 0. $$

Theorem 3.2

(Strong Duality) Assume the point $(\overline{x},\overline{y},\overline{\lambda },\overline{p})$ is a weak minimum for the (MONLP) problem with a fix $\lambda = \overline{\lambda },\,r = \overline{r}$ and that:

(i) $[\nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) + \nabla_{y} \{ \nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})\overline{p}\} ]$ is negative definite,

(ii) $\nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})$ is nonsingular, and

(iii) The assumptions of Theorem 3.1 are correct.

Then, with equal values of objective functions for the (MONLP) and (MONLD) problems, it point $(\overline{x},\overline{y},\overline{\lambda },\overline{r})$ is feasible to solve the (MONLD) problem.

Proof

Because the point $(\overline{x},\overline{y},\overline{\lambda },\overline{p})$ is a weak minimum for the (MONLP) problem, Lemma 3.1, states that there exists $\alpha \in K^{*} ,\,\,\,\beta \in (C_{2}^{*} ) = C_{2} ,\,\,\,(\alpha ,\beta ) \ne 0,$$(x,y) \in C_{1} \times C_{2} ,\,$ $\lambda ,\,\,p \in K^{*}$ such that.

$$ \begin{aligned} & [\alpha \nabla_{x} f(\overline{x},\overline{y}) + (\beta - (\alpha^{T} e)\overline{y})^{T} \nabla_{yx} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) \\ & \quad + (\beta - (\alpha^{T} e)\overline{y})^{T} \nabla_{x} \{ \nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})\overline{p}\} ](x - \overline{x}) \\ & \quad + [(\alpha - (\alpha^{T} e)\overline{\lambda })^{T} \nabla_{y} f(\overline{x},\overline{y}) + (\beta - (\alpha^{T} e)\overline{y} - (\alpha^{T} e)\overline{p})\nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) \\ & \quad + (\beta - (\alpha^{T} e)\overline{y})^{T} \nabla_{y} \{ \nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})\overline{p}\} ](y - \overline{y}) \\ & \quad + [(\beta - (\alpha^{T} e)\overline{\lambda })^{T} \nabla_{y} f(\overline{x},\overline{y}) + (\beta - (\alpha^{T} e)\overline{y})^{T} \nabla_{yy} f(\overline{x},\overline{y})\overline{p}](\lambda - \overline{\lambda }) \\ & \quad + [(\beta - (\alpha^{T} e)\overline{y})^{T} \nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})](p - \overline{p}) \ge 0 \\ \end{aligned} $$

(6)

$$ \beta [\nabla_{y} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) + \overline{p}^{T} \nabla_{yy} (\overline{\lambda }^{t} f)(\overline{x},\overline{y})] = 0 $$

(7)

We claim that $\alpha \ne 0$, if a substitute is used $y \in C_{2}$, $x = \overline{x} \in C_{1} ,\,\,\,\lambda = \overline{\lambda } \in K^{*} \,\,\,and\,\,\,p = \overline{p} \in K^{*}$ the inequality (6) becomes

$$ \begin{aligned} & [(\alpha - (\alpha^{T} e)\overline{\lambda })^{T} \nabla_{y} f(\overline{x},\overline{y}) + (\beta - (\alpha^{T} e)\overline{y} - (\alpha^{T} e)\overline{p})\nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) \\ & \quad + (\beta - (\alpha^{T} e)\overline{y})^{T} \nabla_{y} \{ \nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})\overline{p}\} ](y - \overline{y}) \ge 0 \\ \end{aligned} $$

(8)

If $\alpha =0$ and belongs to K* and when $y = \beta + \overline{y} \in C_{2}$ we have.

$\beta \,\,[\nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) + \nabla_{y} \{ \nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})\overline{p}\} ]\beta \ge 0$.

We can conclude from assumption (i) that we obtained that $\beta = 0$ this is not possible since,$(\alpha ,\beta ) \ne 0$ as a result,$\alpha \ne 0$.

When we substitute $x = \overline{x},\,\,y = \overline{y}\,\,\,and\,\,\,\lambda = \overline{\lambda }$ in (6), we get

$$ [(\beta - (\alpha^{T} e)\overline{y})^{T} \nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})](p - \overline{p}) \ge 0\,\,\,\forall \,\,p \in K^{*} \,\, \Rightarrow $$

$$ (\beta - (\alpha^{T} e)\overline{y})^{T} \nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) = 0\,\,\, $$

Furthermore, we know from assumption (ii) that.

$$ \beta = (\alpha^{T} e)\overline{y} $$

(9)

If we put $y = \overline{y},\,\,\,\lambda = \overline{\lambda }\,\,\,and\,\,\,\,p = \overline{p}$ in (6), we get

$$ \alpha \,\,\nabla_{x} f(\overline{x},\overline{y})(x - \overline{x}) \ge 0\,\, \Rightarrow \,\,\nabla_{x} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})(x - \overline{x}) = 0\,\,\,\,\forall \,\,x \in C_{1} . $$

(10)

That implies

$$ x,\,\,\overline{x} \in C_{1} ,\,\,\,x + \overline{x} \in C_{1} \,\, \Rightarrow \,\,x\nabla_{x} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) = 0 $$

(11)

The following is obtained by differentiating the relationship (11) w. r. t.$x$

$$ x\nabla_{xx} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) + \nabla_{x} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) = 0\,\, \Rightarrow \,\,\nabla_{x} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) + \overline{r}^{T} \nabla_{xx} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) \in C_{1}^{*} $$

This means it point $(\overline{x},\overline{y},\overline{\lambda },\overline{r})$ is feasible to solve the (MONLD) problem.

If we put $x = 0,\,\,x = \overline{x}$ in (10), we get $\overline{x}^{T} \nabla_{x} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) = 0$.

From the differential w. r. t.$x$, we get

$$ \overline{x}^{T} \nabla_{x} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) + x^{T} \nabla_{xx} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})\,\overline{r} = 0 $$

(I)

Substituting Eq. (9) into Eq. (8) yields

$$ \begin{aligned} & [(\alpha - (\alpha^{T} e)\overline{\lambda })^{T} \nabla_{y} f(\overline{x},\overline{y}) - (\alpha^{T} e)\overline{p}^{T} \nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})](y - \overline{y}) \ge 0\,\, \Rightarrow \\ & \quad \{ \nabla_{y} f(\overline{x},\overline{y}) - [\nabla_{y} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) + \overline{p}^{T} \nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})]\} (y - \overline{y}) \ge 0\,\, \\ \end{aligned} $$

(12)

As a result of (7), and because $\beta \ne 0$ we obtained

$$ \nabla_{y} f(\overline{x},\overline{y})(y - \overline{y}) \ge 0\,\,\,\,\forall \,\,\,y \in C_{2} \Rightarrow $$

(13)

$$ \forall \,\,y,\,\,\overline{y} \in C_{1} ,\,\,y + \overline{y} \in C_{1} \,\, \Rightarrow \,\,y\nabla_{y} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) \ge 0 $$

(14)

We get the following result by differentiating the above inequality w. r. t.$y$

$$ y\nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) + \nabla_{y} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) \ge 0\,\, \Rightarrow \,\,\nabla_{y} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) + \overline{p}^{T} \nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) \in C_{2}^{*} $$

When we put $y = 0\,\,\,and\,\,y = \overline{y}$ in (13) and (14) respectively, we get

$$ \overline{y}^{T} \nabla_{y} f(\overline{x},\overline{y}) = 0. $$

From the differential equation w. r. t.$y$, we get the following when put $y = p$.

$$ y^{T} \nabla_{y} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) + y^{T} \nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})\overline{p} = 0\,\, $$

(II)

Thus, from (I) and (II), the (MONLP) and (MONLD) problems are equal in the values of objective functions.

To show that the point $(\overline{x},\overline{y},\overline{\lambda },\overline{p})$ is the weak maximum for the (MONLD) problem; otherwise, there exists a feasible point $(\overline{u},\overline{v},\overline{\lambda },\overline{r})$ such that

$\begin{aligned} & f(\overline{u},\overline{v}) - \overline{u}^{T} \nabla_{u} (\overline{\lambda }^{T} f)(\overline{u},\overline{v})e - (\overline{u}^{T} \nabla_{uu} (\overline{\lambda }^{T} f)(\overline{u},\overline{v})\overline{r})e - f(\overline{x},\overline{y}) + \overline{x}^{T} \nabla_{x} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})e \\ & \quad + (\overline{x}^{T} \nabla_{xx} (\lambda^{T} f)(\overline{x},\overline{y})\overline{p})e \in \,{\text{int}} \,\,K \\ \end{aligned}$.

Since

$$ \nabla_{x} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) + \overline{x}^{T} \nabla_{xx} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) = \nabla_{y} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) + \overline{y}^{T} \nabla_{yy} (\overline{\lambda }^{T} f)(\overline{x},\overline{y}) $$

We got

$$ \begin{aligned} & f(\overline{u},\overline{v}) - \overline{u}^{T} \nabla_{u} (\overline{\lambda }^{T} f)(\overline{u},\overline{v})e - (\overline{u}^{T} \nabla_{uu} (\overline{\lambda }^{T} f)(\overline{u},\overline{v})\overline{r})e - f(\overline{x},\overline{y}) + \overline{y}^{T} \nabla_{y} (\overline{\lambda }^{T} f)(\overline{x},\overline{y})e \\ & \quad + (\overline{y}^{T} \nabla_{yy} (\lambda^{T} f)(\overline{x},\overline{y})\overline{p})e \in \,{\text{int}} \,\,K \\ \end{aligned} $$

(III)

This contradicts the weak duality theorem.

The following example illustrates the results of the strong duality theorem.

Example 3.2.1

Let the point $(\overline{x},\overline{y},\overline{\lambda },\overline{p})$ be a weak minimum for the (MONLP)1 problem. Then, based on Lemma 3.1, there exists.

$\alpha \in K^{*} ,\,\,\,\beta \in (C_{2}^{*} ) = C_{2} ,\,\,\,(\alpha ,\beta ) \ne 0,$ $(x,y) \in C_{1} \times C_{2} ,$ $\lambda ,\,\,p \in K^{*}$ such a thing as

$$ \begin{aligned} & [\alpha (3x^{2} ,3x^{2} )](x - \overline{x}) + [(\alpha - (\alpha^{T} e)\overline{\lambda })^{T} ( - 3y^{2} ,0) + (\beta - (\alpha^{T} e)\overline{y})^{T} - (\alpha^{T} e)\overline{p})( - 6\lambda_{1} y,0) \\ & \quad + (\beta - (\alpha^{T} e)\overline{y})^{T} ( - 6\lambda_{1} ,0)\overline{p}](y - \overline{y}) + [(\beta - (\alpha^{T} e)\overline{\lambda })^{T} ( - 3y^{2} ,0) \\ & \quad + (\beta - (\alpha^{T} e)\overline{y})^{T} ( - 6\lambda_{1} y,0)\overline{p}](\lambda - \overline{\lambda }) + [(\beta - (\alpha^{T} e)\overline{y})^{T} ) - (6\lambda_{1} y,0)](p - \overline{p}) \ge 0 \\ \end{aligned} $$

(15)

Equation (7) takes the form of

$$ \beta \,\,[( - 3\lambda_{1} y^{2} ,0) + ( - 6\lambda_{1} y,0)\overline{p}] = 0 $$

(16)

If we put it $\alpha \ne 0,$ $x = \overline{x} \in C_{1} ,\,\lambda = \overline{\lambda } \in K^{*} and\,p = \overline{p} \in K^{*}$ this way $\forall y \in C_{2} ,$ the inequality (15) becomes

$$ \begin{aligned} & [(\alpha - (\alpha^{T} e)\overline{\lambda })^{T} ( - 3y^{2} ,0) + (\beta - (\alpha^{T} e)\overline{y} - (\alpha^{T} e)\overline{p})( - 6\lambda_{1} y,0) \\ & \quad + (\beta - (\alpha^{T} e)\overline{y})^{T} ( - 6\lambda_{1} ,0)\overline{p}](y - \overline{y}) \ge 0 \\ \end{aligned} $$

(17)

If $\alpha = 0 \in K^{*} \,\,and\,\,put\,\,y = \beta + \overline{y} \in C_{2} \,\, \Rightarrow$.

We get from (i) that

$$ \beta = 0,\,\,\,(\alpha ,\beta ) \ne 0 \Rightarrow \,\,\alpha \ne 0 $$

We get the following by substituting $x = \vec{x},\,\,y = \overline{y},\,\,\lambda = \overline{\lambda }$ in relation (13).

$$ [(\beta - (\alpha^{T} e)\overline{y})^{T} ( - 6\lambda_{1} y,0)](p - \overline{p}) \ge 0\,\,\forall \,\,p \in K^{*} \,\, \Rightarrow $$

$$ (\beta - (\alpha^{T} e)\overline{y})^{T} ( - 6\lambda_{1} y,0) = 0 $$

In addition, from (ii), we obtain

$$ \beta = (\alpha^{T} e)\overline{y} $$

(18)

In addition, when we substitute $y = \overline{y},\,\,\lambda = \overline{\lambda }\,\,and\,\,p = \overline{p}$ in relation (13), we get an

$$ \alpha (3x^{2} ,3x^{2} )(x - \overline{x}) \ge 0\,\,\, \Rightarrow \,\,(3x^{2} ,3x^{2} )(x - \overline{x}) \ge 0\,\,\forall \,\,x \in C_{1} $$

For each $x,\,\,\overline{x} \in C_{1} ,\,\,x + \overline{x} \in C_{1} \,\, \Rightarrow \,\,x(3x^{2} ,3x^{2} ) \ge 0$.

Then, by differentiating this inequality w. r. t. x, we obtain

$$ x(6\lambda_{1} x,6\lambda_{2} x) + (3\lambda_{1} x^{2} ,2\lambda_{2} x^{2} ) \ge 0\,\, \Rightarrow \,\,(3\lambda_{1} x^{2} ,3\lambda_{2} x{}^{2}) + r(6\lambda_{1} x,6\lambda_{2} x) \in C_{1} $$

That shows the point $(\overline{x},\overline{y},\overline{\lambda },\overline{r})$ is feasible for the (MONLD) problem.

Furthermore, if we put $x = 0,\,\,x = \overline{x}$, we get $\overline{x}^{T} (3\lambda_{1} x^{3} ,3\lambda_{2} x^{3} ) + x(6\lambda_{1} x,6\lambda_{2} x)\overline{r} = 0$.

Substituting (16) into (15), we get the following:

$$ \begin{aligned} & [(\alpha - (\alpha^{T} e)\overline{\lambda })^{T} ( - 3y^{2} ,0) - (\alpha^{T} e)\overline{p}^{T} ( - 6\lambda_{1} y,0)](y - \overline{y}) \ge 0\,\, \Rightarrow \\ & \quad \{ ( - 3y^{2} ,0) - [( - 3\lambda_{1} y^{2} ,0) + \overline{p}^{T} ( - 6\lambda_{1} y,0)]\} (y - \overline{y}) \ge 0 \\ \end{aligned} $$

(19)

Because $\beta \ne 0$ (16) implies

$$ ( - 3y^{2} ,0)(y - \overline{y}) \ge 0\,\,\,\,\forall \,\,\,y \in C_{2} $$

(20)

For each $y,\,\overline{y} \in C_{1} ,\,\,y + \overline{y} \in C_{1} \,\, \Rightarrow$

$$ y( - 3\lambda_{1} y^{2} ,0) \ge 0 $$

(21)

By differentiating the above inequality w. r. t. y, we obtain the following:

$$ y( - 6\lambda_{1} y,0) + ( - 3\lambda_{1} y^{2} ,0) \ge 0\,\,\, \Rightarrow $$

$$ ( - 3\lambda_{1} y^{2} ,0) + \overline{p}^{T} ( - 6\lambda_{1} y,0) \in C_{2} $$

When we put $y = 0,\,\,y = \overline{y}$ in (17) and (19), respectively, we obtain.

$\overline{y}^{T} ( - 3y^{2} ,0) = 0$.

Then comes the differential w. r. t. y we get for $y = p$

$$ y^{T} ( - 3\lambda_{1} y^{2} ,0) + y^{T} ( - 6\lambda_{1} y,0)\overline{p} = 0 $$

(22)

Therefore, the (MONLP) and (MONLD) problems have equal objective function values.

To demonstrate that point is the weak maximum for the (MONLD) problem; otherwise, there exists a feasible point $(\overline{x},\overline{y},\overline{\lambda },\overline{r})$ such that

$$ \begin{aligned} & (u^{3} - v^{3} ,u^{3} ) - \overline{u}^{T} (3\lambda_{1} u^{2} ,3\lambda_{2} u^{2} )e - \overline{u}^{T} (6\lambda_{1} u,6\lambda_{2} u)re - (x^{3} - y^{3} ,x^{3} ) + x(3\lambda_{1} x^{2} ,3\lambda_{2} x^{2} )e \\ & \quad + \overline{x}^{t} (6\lambda_{1} x,6\lambda_{2} x)\overline{p})e \in \,{\text{int}} .\,K \\ \end{aligned} $$

If $p = r = 1$ we have

$$ (u^{3} - v^{3} - 3\lambda_{1} u^{3} - 6\lambda_{1} u^{2} - x^{3} + y^{3} - 3\lambda_{1} y^{3} - 6\lambda_{1} y^{2} ,\,\,u^{3} - 3\lambda_{2} u^{3} - 6\lambda_{2} u^{2} - x^{2} ) \in Int.\,K $$

That contradicts the theorem.

Example 3.2.2

Assume the point $(\overline{x},\overline{y},\overline{\lambda },\overline{p})$ is a weak minimum for the (MONLP)2 problem; then, according to Lemma 3.1, there exists.

$\alpha \in K^{*} ,\,\,\beta \in (C_{2}^{*} ),\,\,(\alpha ,\beta ) \ne 0,\,(x,y) \in C_{1} \times C_{2} ,\,\,\lambda ,p \in K^{*}$ such that

$$ \begin{aligned} & \beta^{T} [(2x, - 2x)](x - \overline{x}) + [(\alpha^{T} - (\alpha^{T} e)\overline{\lambda })^{T} (0,2y) + (\beta^{T} - \alpha^{T} e)\overline{y} - (\alpha^{T} e)\overline{p})(0,2\lambda_{2} )](y - \overline{y}) \\ & \quad + [(\beta - (\alpha^{T} e)\overline{\lambda })^{T} (0,2y) + (\beta - (\alpha^{T} e)\overline{y})^{T} (0,2)\overline{p}](\lambda - \overline{\lambda }) \\ & \quad + [(\beta - (\alpha^{T} e)\overline{y})^{T} (0,2\lambda_{2} )](p - \overline{p}) \ge 0 \\ \end{aligned} $$

(6")

Also, Eq. (7) takes the form:

$$ \beta [(0,2\lambda_{2} y) + \overline{p}^{T} (0,2\lambda_{2} )] = 0 $$

(7")

When we put $x = \overline{x} \in C_{1} ,\,\,\lambda = \overline{\lambda } \in K^{*} ,\,\,p = \overline{p}\,\,\forall \,y \in C_{2}$, then inequality (6)^" becomes.

$$ [(\alpha - (\alpha^{T} e)\overline{\lambda })^{T} (0,2y) + (\beta - (\alpha^{T} e)\overline{y} - (\alpha^{T} e)\overline{p})(0,2\lambda_{2} )] \ge 0 $$

(8")

If $\alpha = 0 \in K^{*} ,\,\,y = \beta + \overline{y} \in C_{2}$ we get

$$ \beta \,(0,2\lambda_{2} ) \ge 0 $$

Using assumption (i), we obtain $\beta = 0$.

This has not been possible since $(\alpha ,\beta ) \ne 0$ then $\alpha \ne 0$.

And if we put $x = \overline{x},\,\,y = \overline{y},\,\,\lambda = \overline{\lambda }$ it into Eq. (6^"), we get

$$ [(\beta - (\alpha^{T} e)\overline{y})^{T} (0,2\lambda_{2} )](p - \overline{p}) \ge 0\,\,\,\forall \,\,p \in K^{*} $$

Therefore, we obtained the following from assumption (ii),

$$ \beta = (\alpha^{T} e)\overline{y} $$

(9")

When we put $y = \overline{y},\,\,\lambda = \overline{\lambda },\,\,p = \overline{p}$ in inequality (6^"), we get

$$ \begin{aligned} & \alpha (2x, - 2x)(x - \overline{x}) \ge 0\,\, \Rightarrow \\ & \quad (2x, - 2x)(x - \overline{x}) = 0\,\,\,\forall \,\,x \in C_{1} \\ \end{aligned} $$

(10")

Furthermore, for each

$$ \begin{aligned} & x,\,\,\overline{x} \in C_{1} ,\,\,x + \overline{x} \in C_{1} \,\,\, \Rightarrow \\ & \quad x(2\lambda_{1} x, - 2\lambda_{2} x) = 0 \\ \end{aligned} $$

(11")

When we differentiate the above equation w. r. t. x, we get

$$ (2\lambda_{1} , - 2\lambda_{2} ) + (2\lambda_{1} x, - 2\lambda_{2} x) = 0 $$

This implies its point a feasible for the (MONLD)2 problem.

If we put $x = 0,\,\,x = \overline{x}$ in Eq. (10^"), we get

$$ \overline{x}(2\lambda_{1} x, - 2\lambda_{2} x) = 0 $$

Differentiating this equation w. r. t. x, we obtain.

$$ \overline{x}(2\lambda_{1} x, - 2\lambda_{2} x) + x(2\lambda_{1} , - 2\lambda_{2} )\,r = 0 $$

(I")

Substituting from (9^") into (8^") yields

$$ \begin{aligned} & [(\alpha - (\alpha^{T} e)\overline{\lambda })^{T} (0,2\overline{y}) - (\alpha^{T} e)\overline{p}^{T} (0,2\lambda_{2} )](y - \overline{y}) \ge 0\,\, \Rightarrow \\ & \quad \{ (0,2\overline{y}) - [(0,2\lambda_{2} \overline{y}) + \overline{p}^{T} (0,2\lambda_{2} )]\} (y - \overline{y}) \ge 0 \\ \end{aligned} $$

(12")

Since $\beta \ne 0$ Eq. (7^") takes the form.

$$ (0,2\overline{y})(y - \overline{y}) \ge 0\,\,\,\forall \,\,y \in C_{2} $$

(13")

For each $y,\,\,\overline{y} \in C_{1} ,\,\,\,y + \overline{y} \in C_{1}$, we have.

$$ \overline{y}(0,2\lambda_{2} \overline{y}) \ge 0 $$

(14")

By differentiable w. r. t. y, we get

$$ \begin{aligned} & \overline{y}(0,2\overline{\lambda }_{2} ) + (0,2\lambda_{2} \overline{y}) \ge 0\,\,\, \Rightarrow \\ & \quad (0,2\overline{\lambda }_{2} \overline{y}) + \overline{p}(0,2\lambda_{2} ) \in C_{2}^{*} \\ \end{aligned} $$

When we put $y = 0,\,\,y = \overline{y}$ in the inequalities (13^") and (14^"), respectively, we get

$$ \overline{y}(0,2\overline{\lambda }_{2} \overline{y}) = 0 $$

By differentiable w. r. t. y, we obtain for $y = p$.

$$ \overline{y}^{T} (0,2\overline{\lambda }_{2} \overline{y}) + \overline{y}^{T} (0,2\overline{\lambda }_{2} )\overline{p} = 0 $$

(II")

Therefore, from (I)^" and (II)^", the (MONLP)2 and (MONLD)2 problems are equal-value objective functions.

To demonstrate that this point is the weak maximum for the (MONLD)2 problem, otherwise, there exists a feasible point $(\overline{x},\overline{y},\overline{\lambda },\overline{r})$ such that

$$ \begin{aligned} & (u^{2} ,v^{2} - u^{2} ) - \overline{u}^{T} (2\lambda_{1} u, - 2\lambda_{2} u)e - \overline{u}^{T} (2\lambda_{1} , - 2\lambda_{2} )\,r\,e - (x^{2} ,y^{2} - x^{2} ) + \overline{x}^{T} (2\lambda_{1} x, - 2\lambda_{2} x)\,e \\ & \quad + \overline{x}^{T} (2\lambda_{1} , - 2\lambda_{2} )\,\,\overline{p}\,e\, \in {\text{int}} .\,K \\ \end{aligned} $$

This contradicts the theorem, so the proof is complete.

Theorem 3.3

(Strict Converse Duality). Assume that there is $(\overline{u},\overline{v},\overline{\lambda },\overline{r})$ a weak maximum for the (MONLD) problem with a fix $\lambda = \overline{\lambda },\,\,\,p = \overline{p}$ and that:

(i) $[\nabla_{uu} (\overline{\lambda }^{T} f)(\overline{u},\overline{v}) + \nabla_{u} \{ \nabla_{uu} (\overline{\lambda }^{T} f)(\overline{u},\overline{v})\overline{r}\} ]$ is positive definite,

(ii)$\nabla_{uu} (\overline{\lambda }^{T} f)(\overline{u},\overline{v})$ is nonsingular, in addition to.

(iii) Theorem 3. 1, assumptions are held.

Then, for the (MONLP) problem, the point $(\overline{u},\overline{v},\overline{\lambda },\overline{r})$ is feasible, and the value objective functions for the (MONLP) and (MONLD) problems have the same value.

The proof is similar to Theorem 3. 2.

4 Self-duality

Mathematical programming is called self-dual if it is formally identical to its dual (see Mishra [15, 16] and Kassem [12]), i.e., the dual can be recast in the form of a primal.

Definition 4.1

The function $f(x,y)$ is skew-symmetrical if $\forall \,x,\,y$ we have $f(x,y) = - f(y,x)$.

Assume the function $f$ is skew-symmetric, $m = n,\,\,p = r\,\,and\,\,\,C_{1} = C_{2}$.

We rewrite the (MONLD) problem as the following minimization problem:

$$ ({\text{MONLD}}^{\prime}):\,\,\,\,\,K - \min \,\,\, - f(u,v) + [u^{T} \nabla_{u} (\lambda^{T} f)(u,v)]e + [u^{T} \nabla_{uu} (\lambda^{T} f)(u,v)r]e $$

$subject\,\,\,to\,\,\,(u,v) \in C_{1} \times C_{2}$,

$\nabla_{u} (\lambda^{T} f)(u,v) + r^{T} \nabla_{uu} (\lambda^{T} f)(u,v) \in C_{1}^{*} ,$

$\lambda ,\,\,r \in K^{*} ,\,\,\,\,e \in {\text{int}} \,\,K,\,\,\,\,\lambda^{T} e = 1$.

Since $\nabla_{u} f(u,v) = - \nabla_{v} f(v,u)\,\,\,and\,\,\,\,\nabla_{uu} f(u,v) = - \nabla_{vv} f(v,u)$ then, the above (MONLD') problem has taken the form:

$$ ({\text{MONLD}}^{\prime})_{1} :\,\,\,\,\,K - \min \,\,\,f(v,u) - [u^{T} \nabla_{v} (\lambda^{T} f)(v,u)]e - [u^{T} \nabla_{vv} (\lambda^{T} f)(v,u)r]e $$

$subject\,\,\,to\,\,\,(v,u) \in C_{2} \times C_{1} ,$

$- \nabla_{v} (\lambda^{T} f)(v,u) - r^{T} \nabla_{vv} (\lambda^{T} f)(v,u) \in C_{2}^{*} ,$

$\lambda ,\,\,r \in K^{*} ,\,\,\,\,e \in {\text{int}} \,\,K,\,\,\,\,\lambda^{\,T} e = 1.$

This means the $(MONLD^{\prime})$ problem is formally identical to the (MONLP) problem, that is, the objective and constraint functions are identical. Therefore, the problem is self-dual.

Then, the point $(\overline{x},\overline{y},\overline{\lambda },\overline{p})$ of feasibility for the (MONLP) problem implies the point $(\overline{y},\overline{x},\overline{\lambda },\overline{r})$ of feasibility for the (MONLD) problem and vice versa.

Theorem 4.1

(Self-Duality) If the point $(x_{0} ,y_{0} ,\lambda_{0} ,r_{0} )$ is jointly optimal for the self-dual problem, then there exists a point $(y_{0} ,x_{0} ,\lambda_{0} ,r_{0} )$ such that.

$$ \begin{aligned} & f(y_{0} ,x_{0} ) - (y_{0}^{T} \nabla_{y} (\lambda_{0}^{T} f)(y_{0} ,x_{0} ))e - (y_{0}^{T} \nabla_{yy} (\lambda_{0}^{T} f)(y_{0} ,x_{0} )r_{0} )e \\ & \quad = f(x_{0} ,y_{0} ) - (x_{0}^{T} \nabla_{x} (\lambda_{0}^{T} f)(x_{0} ,y_{0} ))e - (x_{0}^{T} \nabla_{xx} (\lambda_{0}^{T} f)(x_{0} ,y_{0} )r_{0} )e \\ & \quad = 0 \\ \end{aligned} $$

Proof

As shown above, the problem $(MONLD^{\prime})$ is formally identical to the (MONLP) problem. Then the point $(x_{0} ,y_{0} ,\lambda_{0} ,r_{0} )$ is optimal for the $(MONLD^{\prime})$ problem, which implies the point $(y_{0} ,x_{0} ,\lambda_{0} ,r_{0} )$ is optimal for the (MONLP) problem. From the symmetric duality and the $(MONLD^{\prime})$ problem, we derive the following:

$$ \begin{aligned} & f(y_{0} ,x_{0} ) - (y_{0}^{T} \nabla_{y} (\lambda_{0}^{T} f)(y_{0} ,x_{0} ))e - (y_{0}^{T} \nabla_{yy} (\lambda_{0}^{T} f)(y_{0} ,x_{0} )r_{0} )e \\ & \quad = f(x_{0} ,y_{0} ) - (x_{0}^{T} \nabla_{x} (\lambda_{0}^{T} f)(x_{0} ,y_{0} ))e - (x_{0}^{T} \nabla_{xx} (\lambda_{0}^{T} f)(x_{0} ,y_{0} )r_{0} )e \\ & \quad = - f(y_{0} ,x_{0} ) + (y_{0}^{T} \nabla_{y} (\lambda_{0}^{T} f)(y_{0} ,x_{0} ))e + (y_{0}^{T} \nabla_{yy} (\lambda_{0}^{T} f)(y_{0} ,x_{0} )r_{0} )e \\ & \quad = 0 \\ \end{aligned} $$

5 Special Cases

• 1. Using $\eta (x,u) = x - u,\,\eta (v,y) = v - y\,\,and\,p = r = 0$ the (MONLP) and (MONLD) problems are related to the problems studied in [13].

• 2. If we put $p = r = 0$ and the objective functions do not contain the terms $y^{T} \nabla_{y} (\lambda^{T} f)(x,y)\,,\,u^{T} \nabla_{u} (\lambda^{T} f)(u,v)$, then the (MONLP) and (MONLD) problems are reduced to the problems studied in [12].

6 Conclusions

In this work, we presented generalizations of the cone-preinvexity functions and studied a pair of second-order symmetric dualities for the (MONLP) problems under these generalizations of the cone-preinvexity functions. In addition, we established and proved the weak duality, strong duality, strict converse duality, and self-duality theorems under these generalizations of the cone-preinvexity functions. Finally, four nontrivial numerical examples are presented to show that the results of the weak and strong duality theorems are true.

In the future, I will study this idea for higher-order fractional vector optimization problems.