Solutions for the Nonlinear Mixed Variational Inequality Problem in the System

Abstract

Our paper proposes a system of nonlinear mixed variational inequality problems (SNMVIPs) on Banach spaces. Under suitable assumptions, using the K-Fan fixed point theorem and Minty techniques, we demonstrate that the solution set to the SNMVIP is nonempty, weakly compact, and unique. Additionally, we suggest a stability result for the SNMVIPs by perturbing the duality mappings. Furthermore, we present an optimal control problem that is governed by the SNMVIPs and show that it can be solved.

1. Introduction

Lin [1] introduced the system of generalized quasi-variational inclusion problems. This system includes the set of problems suggested on a product set. It includes several well-known problems such as variational inequalities, equilibrium problems, vector equilibrium problems, and variational inclusions/disclusion problems. The system of variational inequalities has concealed symmetries in both variational inequalities and fixed point theory. However, the appearance of scale symmetry in this system creates a void of symmetric hiddenness and has a correlation effect.

Undoubtedly, in the realm of engineering, sciences, technology, chemical processes, and economics, several challenging and complex problems frequently result in inequalities instead of straightforward equations. In this scenario, variational inequalities have become a formidable mathematical resource. Variational inequalities (VIs) essentially arise from applied models with an underlying convex foundation and have been the subject of extensive research since the 1960s, encompassing mathematical theories, numerical techniques, and practical applications (among other significant sources, see [2,3,4,5,6,7]).

It should be noted that the results mentioned earlier cannot be applied to coupled systems that consist of two elliptical mixed variational inequalities. System of VIs are a mathematical tool used to analyze mixed boundary value problems, control problems, and similar problems. More details can be found in [8,9,10,11,12].

In this paper, we suggest SNMVIPs and using the K-Fan fixed point theorem, the Minty techniques, and inverse relaxed monotonicity to establish the existence, convergence, uniqueness, stability, and optimal control of the problems.

Before we proceed, let us define the problem that will be discussed in this article. Consider two reflexive Banach spaces, $(X,\parallel ·{\parallel }_X)$ and $(Y,\parallel ·{\parallel }_Y)$, with their dual spaces $(X^{*},\parallel ·{\parallel }_{X^{*}})$ and $(Y^{*},\parallel ·{\parallel }_{Y^{*}})$, respectively. We denote the duality pairing between $X^{*}$ and X by ${\langle ·,·\rangle }_X$ and between $Y^{*}$ and Y by ${\langle ·,·\rangle }_Y$. We use $\stackrel{w}{⟶}$ and → to denote the weak and the strong convergence in X, and $X_w$ denotes X with weak topology. The limits, lower limits, and upper limits are considered as n approaches infinity, even if not explicitly stated.

We can formulate SNMVIPs on Banach spaces:

Problem 1. 

Determine $(u,v)\in \Omega \times \mho$ such that

$${\langle \mathcal{A}(v,u),y-u\rangle }_X+\phi (u,y)-\phi (u,u)\ge {\langle \gamma ,y-u\rangle }_X,\forall y\in \Omega ,\gamma \in X^{*}$$

(1)

and

$${\langle \mathcal{B}(u,v),z-v\rangle }_Y+\varphi (v,z)-\varphi (v,v)\ge {\langle \zeta ,z-v\rangle }_Y,\forall z\in \mho ,\zeta \in Y^{*}.$$

(2)

We note that if $\phi (u,u)=\phi \left(u\right)$ and $\varphi (u,u)=\varphi \left(u\right)$, then Problem 1 reduces to the problem of [13] for finding $(u,v)\in \Omega \times \mho$ such that

$$\left\{\begin{array}{c}{\langle \mathcal{A}(v,u),y-u\rangle }_X+\phi \left(y\right)-\phi \left(u\right)\ge {\langle \gamma ,y-u\rangle }_X,\forall y\in \Omega \hfill \\ {\langle \mathcal{B}(u,v),z-v\rangle }_Y+\varphi \left(z\right)-\varphi \left(v\right)\ge {\langle \zeta ,z-v\rangle }_Y,\forall z\in \mho .\hfill \end{array}\right.$$

(3)

Definition 1.

Let $\Omega \ne Ø$ be a subset of a Banach space X. Let $\phi :\Omega \to \overline{\mathbb{R}}$ be a proper convex and lower semicontinuous function, and $\mathcal{Q}:\Omega \to X^{*}$. Then $\mathcal{Q}$ is called

(i) 

Monotone, if

$${\langle \mathcal{Q}u-\mathcal{Q}v,u-v\rangle }_X\ge 0,\forall u,v\in \Omega ;$$

(ii) 

Strictly monotone, if

$${\langle \mathcal{Q}u-\mathcal{Q}v,u-v\rangle }_X>0,\forall u,v\in \Omega andu\ne v;$$

(iii) 

Inverse relaxed monotone with constant ${\alpha }_{\mathcal{Q}}>0,$ if

$${\langle \mathcal{Q}u-\mathcal{Q}v,u-v\rangle }_X\ge -{\alpha }_{\mathcal{Q}}{\parallel Qu-Qv\parallel }_X^2,\forall u,v\in \Omega ;$$

(iv) 

Lipschitz continuous with constant ${\beta }_{\mathcal{Q}}>0,$ if

$${\parallel \mathcal{Q}u-\mathcal{Q}v\parallel }_X\le {\beta }_{\mathcal{Q}}{\parallel u-v\parallel }_X^2,\forall u,v\in \Omega ;$$

(v) 

φ-pseudomonotone, if

$${\langle \mathcal{Q}u,v-u\rangle }_X+\phi \left(v\right)-\phi \left(u\right)\ge 0,\forall u,v\in \Omega$$

then it implies that

$${\langle \mathcal{Q}v,v-u\rangle }_X+\phi \left(v\right)-\phi \left(u\right)\ge 0;$$

(vi) 

φ-stable-pseudomonotone with respect to the set $\mathcal{W}\subset X^{*}$, if $\mathcal{Q}$ and $u\rightarrowtail \mathcal{Q}u-w$ are φ-pseudomonotone for each $w\in \mathcal{W}.$

Let Z and Y be topological spaces and $Ø\ne V\subset Z$. We use the notation $2^V$ to represent the set of all possible subsets of the set V. Let $\mathcal{B}:Z\to 2^Y$ be a set-valued map. $Gr\left(\mathcal{B}\right)$ represents the graph of $\mathcal{B}$ and is defined as

$$Gr\left(\mathcal{B}\right)=\left\{\right(u,v)\in Z\times Y|v\in \mathcal{B}\left(u\right)\}\subset Z\times Y.$$

The graph of $\mathcal{B}$ is sequentially closed in $Z\times Y$ if any sequence $\left\{(u_n,v_n)\right\}\subset Gr\left(\mathcal{B}\right)$ converging to $(u,v)\in Z\times Y$ as $n\to \infty$. Then,

$$(u,v)\in Gr\left(\mathcal{B}\right)(\mathit{i}.\mathit{e}.,v\in \mathcal{B}(u\left)\right).$$

Theorem 1. 

[14] Let $D\ne Ø$ be a bounded, closed, and convex set of a subset of the reflexive Banach space Y. Let $\Lambda :D\to 2^D$ be a nonempty, closed, and convex set-values map whose graph is sequentially closed in the topology $Y_w\times Y_w$. Then, Λ has a fixed point.

2. Main Results

In this section, we focus on the uniqueness of solutions and their existence to Problem 1. We use the Minty methodology, Theorem 1, and the K-Fan fixed point theorem to establish the existence theorem for the solutions to Problem 1 under given modest assumptions. Additionally, we deploy the inverse relaxed monotonicity and Lipschitz continuity to prove two uniqueness results for Problem 1.

Furthermore, we propose that $\mathcal{S}:\mho \to 2^{\Omega }$ and $\mathcal{T}:\Omega \to 2^{\mho }$ be the set-valued maps described by

$$\mathcal{S}\left(v\right)=\{u\in \Omega |u\mathrm{solves}\left(1\right)\mathrm{which}\mathrm{corresponds}\mathrm{to}v\},\forall v\in \mho ,$$

and

$$\mathcal{T}\left(u\right)=\{v\in \mho |v\mathrm{solves}\left(2\right)\mathrm{which}\mathrm{corresponds}\mathrm{to}u\},\forall u\in \Omega ,$$

respectively.

The following assumptions must be made to solve Problem 1:

(A): 

$Ø\ne \Omega \subset X$ and $Ø\ne \mho \subset Y$ are closed and convex.

(B): 

$\gamma \in X^{*}$ and $\zeta \in Y^{*}$.

(C): 

$\phi :X\times X\to \overline{\mathbb{R}}$ is such that

(i)

$\phi (\tau ,·):X\to \overline{\mathbb{R}}$ is a proper, convex and lower semicontinuous function;

(ii)

There exists ${\varrho }_{\phi }\ge 0$ such that

$$\phi ({\tau }_1,v_2)-\phi ({\tau }_1,v_1)+\phi ({\tau }_2,v_1)-\phi ({\tau }_2,v_2)\le {\varrho }_{\phi }\parallel {\tau }_1-{\tau }_2{\parallel }_X{\parallel v_1-v_2\parallel }_X,\forall {\tau }_1,{\tau }_2,v_1,v_2\in X;$$

(iii)

For each $\tau \in X$, there exists ${\varrho }_{\phi }\left(\tau \right)>0$ such that [15]

$$\phi (\tau ,v_1)-\phi (\tau ,v_2)\le {\varrho }_{\phi }\left(\tau \right){\parallel v_1-v_2\parallel }_X,\forall v_1,v_2\in X$$

(D): 

$\mathcal{A}:Y\times X\to X^{*}$ is such that

(i)

$u\rightarrowtail \mathcal{A}(v,u)$ is $\phi$-stable-pseudomonotone with $\left\{\gamma \right\}$ and fulfills

$$\underset{\lambda \to 0}{lim}sup{\langle \mathcal{A}(v,\lambda y+(1-\lambda )u),y-u\rangle }_X\le {\langle \mathcal{A}(v,u),y-u\rangle }_X,\forall v\in Y\mathrm{and}u,y\in X.$$

(ii)

It possesses

$$\underset{n\to \infty }{lim}sup{\langle \mathcal{A}(v_n,y),y-u_n\rangle }_X\le {\langle \mathcal{A}(v,y),y-u\rangle }_X,$$

when $y\in X,(u,v)\in X\times Y,\left\{v_n\right\}\subset Y\mathrm{and}\left\{u_n\right\}\subset X$ are such that

$$v_n\stackrel{w}{\to }v\in Y\mathrm{and}{u}_n\stackrel{w}{⟶}u\in X\mathrm{as}n\to \infty ;$$

(iii)

$♭:{\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\to \mathbb{R}$ is a function such that

$${\langle \mathcal{A}(v,u),u\rangle }_X\ge {♭(\parallel u\parallel }_X{,\parallel v\parallel }_Y{)\parallel u\parallel }_X,\forall u\in X,v\in Y,$$

and

♠ every bounded set $Ø\ne D\subset {\mathbb{R}}^{+},$ we have

$$♭(t,s)\to +\infty \mathrm{as}t\to +\infty ,\forall s\in D,$$

♠ for any ${\varrho }_1,{\varrho }_2\ge 0,$ it holds that $♭(t,{\varrho }_{1}t+{\varrho }_2)\to +\infty$ as $t\to +\infty .$

(iv)

There exists ${\varrho }_{\mathcal{A}}>0$ such that

$${\parallel \mathcal{A}(v,u)\parallel }_{X^{*}}\le {\varrho }_{\mathcal{A}}{(1+\parallel u\parallel }_X+{\parallel v\parallel }_Y),\forall (u,v)\in X\times Y.$$

(E): 

$\varphi :Y\times Y\to \overline{\mathbb{R}}$ is such that

(i)

$\varphi (\tau ,·):Y\to \overline{\mathbb{R}}$ is a proper, convex and lower semicontinuous function;

(ii)

There exists ${\varrho }_{\varphi }\ge 0$ such that

$$\varphi ({\tau }_1,v_2)-\varphi ({\tau }_1,v_1)+\varphi ({\tau }_2,v_1)-\varphi ({\tau }_2,v_2)\le {\varrho }_{\varphi }\parallel {\tau }_1-{\tau }_2{\parallel }_Y{\parallel v_1-v_2\parallel }_Y,\forall {\tau }_1,{\tau }_2,v_1,v_2\in Y;$$

(iii)

For each $\tau \in X$, there exists ${\varrho }_{\varphi }\left(\tau \right)>0$ such that [15]

$$\varphi (\tau ,v_1)-\varphi (\tau ,v_2)\le {\varrho }_{\varphi }\left(\tau \right){\parallel v_1-v_2\parallel }_X,\forall v_1,v_2\in X.$$

(F): 

$\mathcal{B}:X\times Y\to Y^{*}$ is such that

(i)

$v\rightarrowtail \mathcal{B}(u,v)$ is $\varphi$-stable-pseudomonotone with $\left\{\zeta \right\}$ and fulfills

$$\underset{\lambda \to 0}{lim}sup{\langle \mathcal{B}(u,\lambda z+(1-\lambda )v),z-v\rangle }_Y\le {\langle \mathcal{B}(u,v),z-v\rangle }_Y,\forall z,v\in Y,u\in X.$$

(ii)

It possesses

$$\underset{n\to \infty }{lim}sup{\langle \mathcal{B}(u_n,z),z-v_n\rangle }_Y\le {\langle \mathcal{B}(u,z),z-v\rangle }_Y,$$

when $z\in Y,(u,v)\in X\times Y,\left\{v_n\right\}\subset Y$ and $\left\{u_n\right\}\subset X$ are such that

$$v_n\stackrel{w}{⟶}v\in Y\mathrm{and}{u}_n\stackrel{w}{⟶}u\in X\mathrm{as}n\to \infty ;$$

(iii)

$\ell :{\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\to \mathbb{R}$ is a function such that

$${\langle \mathcal{B}(u,v),v\rangle }_Y\ge {\ell (\parallel v\parallel }_Y{,\parallel u\parallel }_X{)\parallel v\parallel }_Y,\forall u\in X\mathrm{and}v\in Y,$$

and

♠ every bounded set $Ø\ne D\subset {\mathbb{R}}^{+}$, we have

$$\ell (t,s)\to +\infty \mathrm{as}t\to +\infty ,\mathrm{for}\mathrm{all}s\in D,$$

♠ any ${\varrho }_1,{\varrho }_2\ge 0$, it holds $\ell (t,{\varrho }_{1}t+{\varrho }_2)\to +\infty$ as $t\to +\infty .$

(iv)

∃ a constant ${\varrho }_{\mathcal{B}}>0$ such that

$${\parallel \mathcal{B}(u,v)\parallel }_{Y^{*}}\le {\varrho }_{\mathcal{B}}{(1+\parallel u\parallel }_X+{\parallel v\parallel }_Y),\forall (u,v)\in X\times Y.$$

Remark 1. 

When ♭ appears in (D)(iii) (or ℓ appears in (F)(iii)) but has no effect on the second variable, the condition (D)(iii) (or (F)(iii)) becomes a subsequent uniformly coercive condition:

(D)(iii)′: ∃$♭:{\mathbb{R}}^{+}\to \mathbb{R}$ with $♭\left(s\right)\to +\infty$ as $s\to +\infty$ such that

$${\langle \mathcal{A}(v,u),u\rangle }_X\ge {♭(\parallel u\parallel }_X{)\parallel u\parallel }_X,\forall u\in X\mathrm{and}v\in Y$$

(respectively,(F)(iii)′: ∃$\ell :{\mathbb{R}}^{+}\to \mathbb{R}$ with $\ell \left(s\right)\to +\infty$ as $s\to +\infty$ such that

$${\langle \mathcal{B}(u,v),v\rangle }_Y\ge {\ell (\parallel v\parallel }_Y{)\parallel v\parallel }_Y,\forall u\in X,andv\in Y).$$

The accessibility of solutions for Problem 1 is the main theorem of this article.

Theorem 2. 

Suppose that   (A), (B), (C), (D), (E), and  (F) are held. Then, the solution set denoted by $\Gamma (\gamma ,\zeta )$ of Problem 1 corresponding to $(\gamma ,\zeta )\in X^{*}\times Y^{*}$ is nonempty and weakly compact in $X\times Y$.

We require the following lemmas to prove this theorem:

Lemma 1. 

Assume that  (A), (B), (C),  and  (D) are satisfied. Then, the following statements hold:

(i) 

For fixed $v\in Y$, $u\in \Omega$ solves the (1), if and only if u solves the Minty inequality for determining $u\in \Omega$ such that

$${\langle \mathcal{A}(v,y),y-u\rangle }_X+\phi (u,y)-\phi (u,u)\ge {\langle \gamma ,y-u\rangle }_X,\forall y\in \Omega ;$$

(4)

(ii) 

For fixed $v\in Y$, the solution set $\mathcal{S}\left(v\right)$ of (1) is nonempty, bounded, closed and convex;

(iii) 

The graph of $\mathcal{S}:\mho \to 2^{\Omega }$ is sequentially closed in $Y_w\times X_w,$ implying that $\mathcal{S}$ is sequentially closed from a waek topology Y into the subsets of a weak topology X;

(iv) 

If the map $u\rightarrowtail \mathcal{A}(v,u)$ is strictly monotone for a fixed $v\in Y$, then $\mathcal{S}$ is a weakly continuous point-to-point mapping.

Proof. 

The assumptions (i) and (ii) are the straightforward consequences of ([16], Theorem 3.3). Now, we present the conclusion (iii).

Let $\left\{(v_n,u_n)\right\}\subset Gr\left(\mathcal{S}\right)$ be such that

$$v_n\stackrel{w}{⟶}v\in Y\mathrm{and}{u}_n\stackrel{w}{⟶}u\in X\mathrm{as}n\to \infty \mathrm{for}(u,v)\in X\times Y.$$

(5)

Then, for each $n\in \mathbb{N}$, we have $u_n\in \mathcal{S}\left(v_n\right),$ i.e.,

$${\langle \mathcal{A}(v_n,u_n),y-u_n\rangle }_X+\phi (u_n,y)-\phi (u_n,u_n)\ge {\langle \gamma ,y-u_n\rangle }_X,\forall y\in \Omega .$$

The (i) asserts that

$${\langle \mathcal{A}(v_n,y),y-u_n\rangle }_X+\phi (u_n,y)-\phi (u_n,u_n)\ge {\langle \gamma ,y-u_n\rangle }_X,\forall y\in \Omega .$$

(6)

To establish the upper limit as $n\to \infty$, we use assumption (D)(ii) and the weak lower semicontinuity of $\phi$ (because $\phi$ is convex and lower semicontinuous) to determine

$$\begin{array}{cc}\hfill {\langle \mathcal{A}(v,y),y-u\rangle }_X& +\phi (u,y)-\phi (u,u)\ge \underset{n\to \infty }{lim}sup{\langle \mathcal{A}(v_n,y),y-u_n\rangle }_X\hfill \\ \hfill & +\underset{n\to \infty }{lim}inf\left[\phi (u_n,y)\right]-\underset{n\to \infty }{lim}inf\left[\phi (u_n,u_n)\right]\hfill \\ \hfill & \ge \underset{n\to \infty }{lim}sup[{\langle \mathcal{A}(v_n,y),y-u_n\rangle }_X+\phi (u_n,y)-\phi (u_n,u_n)]\hfill \\ \hfill & \ge \underset{n\to \infty }{lim}sup{\langle \gamma ,y-u_n\rangle }_X\hfill \\ \hfill & ={\langle \gamma ,y-u\rangle }_X,\forall y\in \Omega .\hfill \end{array}$$

Using the assumption (i), we obtain

$$u\in \mathcal{S}\left(v\right).$$

Consequently, $(v,u)\in Gr\left(\mathcal{S}\right)$, which is the the graph of the mapping $\mathcal{S}:\mho \to 2^{\Omega }$, is sequentially closure in $Y_w\times X_w.$

Additionally, assume that $u\rightarrowtail \mathcal{A}(v,u)$ is strictly monotone. Let us consider $u_1,u_2\in \Omega$ be two solutions to (1). Then, we have

$${\langle \mathcal{A}(v,u_1),y-u_1\rangle }_X+\phi (u_1,y)-\phi (u_1,u_1)\ge {\langle \gamma ,y-u_1\rangle }_X,\forall y\in \Omega ,\gamma \in X^{*}$$

(7)

and

$${\langle \mathcal{A}(v,u_2),y-u_2\rangle }_X+\phi (u_2,y)-\phi (u_2,u_2)\ge {\langle \gamma ,y-u_2\rangle }_X,\forall y\in \Omega ,\zeta \in Y^{*}.$$

(8)

In putting $y=u_2$ into (7) and $y=u_1$ in (8), we have

$${\langle \mathcal{A}(v,u_1),u_2-u_1\rangle }_X+\phi (u_1,u_2)-\phi (u_1,u_1)\ge {\langle \gamma ,u_2-u_1\rangle }_X,$$

(9)

and

$${\langle \mathcal{A}(v,u_2),u_1-u_2\rangle }_X+\phi (u_2,u_1)-\phi (u_2,u_2)\ge {\langle \gamma ,u_1-u_2\rangle }_X.$$

(10)

Adding (9) and (10), we have

$${\langle \mathcal{A}(v,u_1)-\mathcal{A}(v,u_2),u_1-u_2\rangle }_X-\phi (u_1,u_2)+\phi (u_1,u_1)-\phi (u_2,u_1)+\phi (u_2,u_2)\le 0.$$

(11)

Given the assumption (C) and the strict monotonicity of $u\rightarrowtail \mathcal{A}(v,u)$, we obtain

$$u_1=u_2.$$

Therefore, $\mathcal{S}$ is a point-to-point mapping. However, assumption (iii) shows that $\mathcal{S}$ is weakly continuous. □

Similarly, Problem (2) has the following lemma.

Lemma 2. 

Assume that  (A), (B), (E), and (F) are satisfied. Then, the following statements hold:

(i) 

For each fixed $u\in X$, $v\in \mho$ solves (2) if and only if v solves the Minty inequality to determine $v\in \mho$ such that

$${\langle \mathcal{B}(u,z),z-v\rangle }_Y+\varphi (v,z)-\varphi (v,v)\ge {\langle \zeta ,z-v\rangle }_Y,\forall z\in \mho ;$$

(12)

(ii) 

For fixed $u\in X$, the solution set $\mathcal{T}\left(u\right)$ namely, of (2) is nonempty, bounded, closed, and convex;

(iii) 

The map $\mathcal{T}:\Omega \to 2^{\mho }$ has a sequentially closed graph in $X_w\times Y_w$;

(iv) 

If the map $v\rightarrowtail \mathcal{B}(u,v)$ is strictly monotone for a fixed $u\in X$, then $\mathcal{T}$ is a weakly continuous point-to-point mapping.

Furthermore, we provide an a priori appraisal of the solution to Problem 1.

Lemma 3. 

Let us assume that (A), (B), (C), (D), (E), and  (F) have been fulfilled satisfactorily. If the solution set $\Gamma (\gamma ,\zeta )$, namely, of Problem 1, is nonempty, then there exists $\mathcal{M}>0$ such that

$${\parallel u\parallel }_X\le \mathcal{M}and{\parallel v\parallel }_Y\le \mathcal{M},\forall (u,v)\in \Gamma (\gamma ,\zeta ).$$

(13)

Proof. 

Suppose that $\Gamma (\gamma ,\zeta )\ne Ø$. Let $(u,v)\in \Gamma (\gamma ,\zeta )$ be arbitrary and $(u_0,v_0)\in (D\left(\phi \right)\cap \Omega )\times (D\left(\varphi \right)\cap \mho ).$ By swapping $y=u_0$ and $z=v_0$ into (1) and (2), respectively, we obtain

$${\langle \mathcal{A}(v,u),u\rangle }_X\le {\langle \mathcal{A}(v,u),u_0\rangle }_X+\phi (u,u_0)-\phi (u,u)+{\langle \gamma ,u_0-u\rangle }_X$$

(14)

and

$${\langle \mathcal{B}(u,v),v\rangle }_Y\le {\langle \mathcal{B}(u,v),v_0\rangle }_Y+\varphi (v,v_0)-\varphi (v,v)+{\langle \zeta ,v_0-v\rangle }_Y.$$

(15)

Taking account of (14), we use hypotheses (C)(i), (iii), and (D)(iii)–(iv) to obtain

$$\begin{array}{cc}\hfill {♭(\parallel u\parallel }_X{,\parallel v\parallel }_Y{)\parallel u\parallel }_X& \le {\langle \mathcal{A}(v,u),u\rangle }_X\hfill \\ \hfill & \le {\langle \mathcal{A}(v,u),u_0\rangle }_X+\phi (u,u_0)-\phi (u,u)+{\langle \gamma ,u_0-u\rangle }_X\hfill \\ \hfill & \le {\parallel \mathcal{A}(v,u)\parallel }_{X^{*}}\parallel u_0{\parallel }_X+\phi (u,u_0)-\phi (u,u)+{\parallel \gamma \parallel }_{X^{*}}(\parallel u_0{\parallel }_X+{\parallel u\parallel }_X)\hfill \\ \hfill & \le {\varrho }_{\mathcal{A}}{(1+\parallel u\parallel }_X+{\parallel v\parallel }_Y)\parallel u_0{\parallel }_X+{\varrho }_{\phi }\left(u\right)(\parallel u_0{\parallel }_X+{\parallel u\parallel }_X{)+\parallel \gamma \parallel }_{X^{*}}(\parallel u_0{\parallel }_X+{\parallel u\parallel }_X).\hfill \end{array}$$

This implies that

$${♭(\parallel u\parallel }_X{,\parallel v\parallel }_Y)\le {\displaystyle \frac{{\varrho }_{\mathcal{A}}{(1+\parallel u\parallel }_X+{\parallel v\parallel }_Y)\parallel u_0{\parallel }_X}{{\parallel u\parallel }_X}}+{\displaystyle \frac{({\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}})\parallel u_0{\parallel }_X}{{\parallel u\parallel }_X}}+{\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}}.$$

(16)

Similarly, taking account of (15), we use hypotheses (E)(i), (iii) and (F)(iii)–(iv) to obtain

$$\begin{array}{cc}\hfill {\ell (\parallel v\parallel }_Y{,\parallel u\parallel }_X{)\parallel v\parallel }_Y& \le {\varrho }_{\mathcal{B}}{(1+\parallel u\parallel }_X+{\parallel v\parallel }_Y)\parallel v_0{\parallel }_Y+{\varrho }_{\varphi }\left(v\right)(\parallel v_0{\parallel }_Y+{\parallel v\parallel }_Y{)+\parallel \zeta \parallel }_{Y^{*}}(\parallel v_0{\parallel }_Y+{\parallel v\parallel }_Y)\hfill \\ \hfill ⟹& \hfill \\ \hfill \ell (\parallel v{\parallel }_Y,\parallel u{\parallel }_X)& \le {\displaystyle \frac{{\varrho }_{\mathcal{B}}{(1+\parallel u\parallel }_X+{\parallel v\parallel }_Y)\parallel v_0{\parallel }_Y}{{\parallel v\parallel }_Y}}+{\displaystyle \frac{({\varrho }_{\varphi }\left(v\right)+{\parallel \zeta \parallel }_{Y^{*}})\parallel v_0{\parallel }_Y}{{\parallel v\parallel }_Y}}+{\varrho }_{\varphi }\left(v\right)+{\parallel \zeta \parallel }_{Y^{*}}.\hfill \end{array}$$

(17)

Contrarily, suppose $\Gamma (\gamma ,\zeta )$ is unbounded. Then, taking a subsequence, if necessary, it is possible to suggest a sequence $\left\{(u_n,v_n)\right\}\subset \Omega \times \mho$ so that

$$\parallel u_n{\parallel }_X\uparrow +\infty \mathrm{as}n\to \infty ,$$

(18)

or

$$\parallel v_n{\parallel }_Y\uparrow +\infty \mathrm{as}n\to \infty .$$

(19)

Let us segregate the subsequent cases:

a

Assume (18) is satisfied and the sequence $\left\{v_n\right\}$ is bounded in Y;

b

Assume (19) is fulfilled and the sequence $\left\{u_n\right\}$ is bounded in X;

c

Assuming that both (18) and (19) are fulfilled.

Assuming a is valid, substitute $u=u_n$ and $v=v_n$ into (16) to obtain

$$\begin{array}{cc}\hfill ♭(\parallel u_n{\parallel }_X,\parallel v_n{\parallel }_Y)& \le {\displaystyle \frac{{\varrho }_{\mathcal{A}}(1+\parallel u_n{\parallel }_X+\parallel v_n{\parallel }_Y)\parallel u_0{\parallel }_X}{\parallel u_n{\parallel }_X}}+{\displaystyle \frac{({\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}})\parallel u_0{\parallel }_X}{\parallel u_n{\parallel }_X}}\hfill \\ \hfill & +{\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}}.\hfill \end{array}$$

(20)

When we let n approach infinity in the inequality (20) and make use of (18) along with property (D)(iii), we obtain the following:

$$\begin{array}{cc}\hfill +\infty & =\underset{n\to \infty }{lim}♭(\parallel u_n{\parallel }_X,\parallel v_n{\parallel }_Y)\hfill \\ \hfill & \le \underset{n\to \infty }{lim}\left[{\displaystyle \frac{{\varrho }_{\mathcal{A}}(1+\parallel u_n{\parallel }_X+\parallel v_n{\parallel }_Y)\parallel u_0{\parallel }_X}{\parallel u_n{\parallel }_X}}+{\displaystyle \frac{({\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}})\parallel u_0{\parallel }_X}{\parallel u_n{\parallel }_X}}+{\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}}\right]\hfill \\ \hfill & ={\varrho }_{\mathcal{A}}\parallel u_0{\parallel }_X+{\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}}.\hfill \end{array}$$

(21)

Consequently, (21) produces a contradiction. Similarly, for b, we could use (17) to obtain a contradiction. However, we assume that c holds, and we will proceed to discuss two additional situations:

(1)

${\displaystyle \frac{\parallel v_n{\parallel }_Y}{\parallel u_n{\parallel }_X}}\to +\infty \mathrm{as}n\to \infty ;$

(2)

There exist $n_0\in \mathbb{N}$ and $\widehat{{\varrho }_0}>0$ such that

$${\displaystyle \frac{\parallel v_n{\parallel }_Y}{\parallel u_n{\parallel }_X}}\le \widehat{{\varrho }_0},\forall n\ge n_0.$$

If item (1) is true, we enter $u=u_n$ and $v=v_n$ into (17) to yield

$$\ell (\parallel v_n{\parallel }_Y,\parallel u_n{\parallel }_X)\le {\displaystyle \frac{{\varrho }_{\mathcal{B}}(1+\parallel u_n{\parallel }_X+\parallel v_n{\parallel }_Y)\parallel v_0{\parallel }_Y}{\parallel v_n{\parallel }_Y}}+{\displaystyle \frac{({\varrho }_{\varphi }\left(v\right)+{\parallel \zeta \parallel }_{Y^{*}})\parallel v_0{\parallel }_Y}{\parallel v_n{\parallel }_Y}}+{\varrho }_{\varphi }\left(v\right)+{\parallel \zeta \parallel }_{Y^{*}}.$$

Taking the limit as n approaches infinity for the inequality mentioned above yields

$$\begin{array}{cc}\hfill +\infty & =\underset{n\to \infty }{lim}\ell (\parallel v_n{\parallel }_Y,\parallel u_n{\parallel }_X)\hfill \\ \hfill & \le \underset{n\to \infty }{lim}\left[{\displaystyle \frac{{\varrho }_{\mathcal{B}}(1+\parallel u_n{\parallel }_X+\parallel v_n{\parallel }_Y)\parallel v_0{\parallel }_Y}{\parallel v_n{\parallel }_Y}}+{\displaystyle \frac{({\varrho }_{\varphi }\left(v\right)+{\parallel \zeta \parallel }_{Y^{*}})\parallel v_0{\parallel }_Y}{\parallel v_n{\parallel }_Y}}+{\varrho }_{\varphi }\left(v\right)+{\parallel \zeta \parallel }_{Y^{*}}\right]\hfill \\ \hfill & ={\varrho }_{\mathcal{B}}\parallel v_0{\parallel }_Y+{\varrho }_{\varphi }\left(v\right)+{\parallel \zeta \parallel }_{Y^{*}}.\hfill \end{array}$$

(22)

It is obviously impossible; however, for a situation (2), we can deduce from (16) that

$$\begin{array}{cc}\hfill +\infty & \leftarrow ♭(\parallel u_n{\parallel }_X,\parallel v_n{\parallel }_Y)(\mathrm{as}n\to \infty )\hfill \\ \hfill & \le {\displaystyle \frac{{\varrho }_{\mathcal{A}}(1+\parallel u_n{\parallel }_X+\parallel v_n{\parallel }_Y)\parallel u_0{\parallel }_X}{\parallel u_n{\parallel }_X}}+{\displaystyle \frac{({\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}})\parallel u_0{\parallel }_X}{\parallel u_n{\parallel }_X}}+{\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}}\hfill \\ \hfill & ={\varrho }_{\mathcal{A}}(2+{\widehat{\varrho }}_0)\parallel u_0{\parallel }_X+{\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}}\parallel u_0{\parallel }_X+{\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}},\mathrm{for}n\ge n_1,\hfill \end{array}$$

(23)

where $n_1\ge n_0$ is such that

$$\parallel u_{n_1}{\parallel }_X>1.$$

This leads to a contradiction. Thus, $\Gamma (\gamma ,\zeta )$ is bounded in $X\times Y$, allowing us to determine $\mathcal{M}>0$ satisfying (13). □

Consider the set-valued mapping $\Lambda :\Omega \times \mho \to 2^{\Omega \times \mho }$ defined by

$$\Lambda (u,v)=\left(\mathcal{S}\right(v),\mathcal{T}(u\left)\right),\forall (u,v)\in \Omega \times \mho .$$

(24)

By invoking Lemma 1 and Lemma 2, it can be seen that $\Lambda$ is well defined. In addition, ∃ a bounded, closed, and convex set $\mathfrak{D}$ in $\Omega \times \mho$ such that $\Lambda$ maps $\mathfrak{D}$ into itself.

Lemma 4. 

Suppose (A), (B), (C), (D), (E), and  (F)  are met. Then, ∃ a constant $\widehat{\mathcal{M}}>0$ satisfy

$$\Lambda \overline{\left(\mathcal{B}(0,\widehat{\mathcal{M}})\right)}\subset \overline{\mathcal{B}(0,\widehat{\mathcal{M}})},$$

where $\overline{\mathcal{B}(0,\widehat{\mathcal{M}})}\subset X\times Y$ is defined as

$$\overline{\mathcal{B}(0,\widehat{\mathcal{M}})}={\{(u,v)\in \Omega \times \mho |\parallel u\parallel }_X\le \widehat{\mathcal{M}}{and\parallel v\parallel }_Y\le \widehat{\mathcal{M}}\}.$$

Proof. 

Our proof will be based on contradiction. Assume that

$$\Gamma \left(\overline{\mathcal{B}(0,n)}\right)\neg \subset \overline{\mathcal{B}(0,n)},\mathrm{for}n\in \mathbb{N}.$$

Then, for each $n\in \mathbb{N}$, we may determine $(u_n,v_n)\in \overline{\mathcal{B}(0,n)}$ and $(w_n,z_n)\in \Gamma (u_n,v_n)$ (i.e., $w_n\in \mathcal{S}(v_n$) and $z_n\in \mathcal{T}\left(u_n\right))$ so that

$$\parallel w_n{\parallel }_X>n\mathrm{or}{\parallel z_n\parallel }_Y>n.$$

(25)

Thus, assuming $\parallel w_n{\parallel }_X>n$ for each $n\in \mathbb{N}$ (similarly for $\parallel z_n{\parallel }_Y>n$ for each $n\in \mathbb{N}$). We employ (16); one has

$$♭(\parallel w_n{\parallel }_X,\parallel v_n{\parallel }_Y)\le {\displaystyle \frac{{\varrho }_{\mathcal{A}}(1+\parallel w_n{\parallel }_X+\parallel v_n{\parallel }_Y)\parallel u_0{\parallel }_X}{\parallel w_n{\parallel }_X}}+{\displaystyle \frac{({\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}})\parallel u_0{\parallel }_X}{\parallel w_n{\parallel }_X}}+{\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}}.$$

Since

$$\parallel v_n{\parallel }_Y\le n<{\parallel w_n\parallel }_X.$$

Therefore, passing to the limit as $n\to \infty$ for the inequality above, we have

$$\begin{array}{cc}\hfill +\infty & =\underset{n\to \infty }{lim}♭(\parallel w_n{\parallel }_X,\parallel v_n{\parallel }_Y)\hfill \\ \hfill & \le \underset{n\to \infty }{lim}\left[{\displaystyle \frac{{\varrho }_{\mathcal{A}}(1+\parallel w_n{\parallel }_X+\parallel v_n{\parallel }_Y)\parallel u_0{\parallel }_X}{\parallel w_n{\parallel }_X}}+{\displaystyle \frac{({\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}})\parallel u_0{\parallel }_X}{\parallel w_n{\parallel }_X}}+{\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}}\right]\hfill \\ \hfill & \le 2{\varrho }_{\mathcal{A}}\parallel u_0{\parallel }_X+{\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}}.\hfill \end{array}$$

This leads to a contradiction. Thus, ∃ a constant $\widehat{\mathcal{M}}>0$ satisfying

$$\Lambda \left(\overline{\mathcal{B}(0,\widehat{\mathcal{M}})}\right)\subset \overline{\mathcal{B}(0,\widehat{\mathcal{M}})}.$$

□

Proof. (Proof of Theorem 2)

Let us see that if $\Lambda$ has a fixed point $(u^{*},v^{*})$, then

$$u^{*}\in \mathcal{S}\left(v^{*}\right)\mathrm{and}{v}^{*}\in \mathcal{T}\left(u^{*}\right).$$

By employing the concepts of $\mathcal{S}$ and $\mathcal{T}$, it offers

$${\langle \mathcal{A}(v^{*},u^{*}),y-u^{*}\rangle }_X+\phi (u^{*},y)-\phi (u^{*},u^{*})\ge {\langle \gamma ,y-u^{*}\rangle }_X,\forall y\in \Omega ,$$

and

$${\langle \mathcal{B}(u^{*},v^{*}),z-v^{*}\rangle }_Y+\varphi (v^{*},z)-\varphi (v^{*},v^{*})\ge {\langle \zeta ,z-v^{*}\rangle }_Y,\forall z\in \mho .$$

Thus, it is clear that $(u^{*},v^{*})$ solves Problem 1. We will apply Theorem 1, the K-Fan fixed point theorem, to determine the existence of a fixed point for $\Lambda$.

Moreover, Lemmas 1, 2, and 4, in fact, infer that $\Lambda :\overline{\mathcal{B}(0,\widehat{\mathcal{M}})}\to 2^{\overline{\mathcal{B}(0,\widehat{\mathcal{M}})}}$ has nonempty, closed, and convex values; the graph of $\Lambda$ is sequentially closed in ${(X\times Y)}_w\times {(X\times Y)}_w.$ The conditions stated in Theorem 1 have been verified. From this theorem, It can be shown that Problem 1 has a solution $(u^{*},v^{*})\in \Omega \times \mho$, such that

$$(u^{*},v^{*})\in \Lambda (u^{*},v^{*}).$$

Hence,

$$\Gamma (\gamma ,\zeta )\ne Ø.$$

Lemma 3 clearly shows that $\Gamma (\gamma ,\zeta )$ is bounded in $X\times Y$. Therefore, we will demonstrate that $\Gamma (\gamma ,\zeta )$ is weakly closed. Consider $\left\{(u_n,v_n)\right\}\subset \Gamma (\gamma ,\zeta )$ such that

$$(u_n,v_n)\stackrel{w}{⟶}(u,v)\in X\times Y\mathrm{as}n\to \infty ,\mathrm{for}\mathrm{some}(u,v)\in \Omega \times \mho .$$

(26)

It is clear that for each natural number n, the pair $(u_n,v_n)\in \Lambda (u_n,v_n).$ Since $\Lambda$ is sequentially closed from ${(X\times Y)}_w$ to ${(X\times Y)}_w$ (see Lemma 1 and Lemma 2), we can conclude that

$$(u,v)\in \Lambda (u,v).$$

This signifies that

$$(u,v)\in \Gamma (\gamma ,\zeta ).$$

Thus, due to the boundedness of $\Gamma (\gamma ,\zeta )$, we can conclude that $\Gamma (\gamma ,\zeta )$ is weakly compact. □

Theorem 2 shows that the solution set of Problem 1 is both nonempty and weakly compact. However, it raises the question of whether it is possible to prove the uniqueness of the solution under certain assumptions. Fortunately, the theorems below provide a positive solution to this problem.

Theorem 3. 

Assume that  (A), (B), (C), (D), (E), and (F) are fulfilled. In addition, if the inequality below holds,

$$\begin{array}{cc}\hfill & {\langle \mathcal{A}(v_1,u_1)-\mathcal{A}(v_2,u_2),u_1-u_2\rangle }_X+{\langle \mathcal{B}(u_1,v_1)-\mathcal{B}(u_2,v_2),v_1-v_2\rangle }_Y+{\varrho }_{\phi }{\parallel u_1-u_2\parallel }_X^2\hfill \\ \hfill & +{\varrho }_{\varphi }{\parallel v_1-v_2\parallel }_Y^2>0,\forall (u_1,v_1),(u_2,v_2)\in X\times Ywith(u_1,v_1)\ne (u_2,v_2).\hfill \end{array}$$

(27)

Then Problem 1 has a unique solution.

Proof. 

Theorem 2 assures that

$$\Gamma (\gamma ,\zeta )\ne Ø.$$

We now demonstrate that Problem 1 has unique solution. Assume $(u_1,v_1),(u_2,v_2)\in \Gamma (\gamma ,\zeta )$ are the two different solutions. Then,

$${\langle \mathcal{A}(v_i,u_i),y-u_i\rangle }_X+\phi (u_i,y)-\phi (u_i,u_i)\ge {\langle \gamma ,y-u_i\rangle }_X,\forall y\in \Omega ,$$

(28)

and

$${\langle \mathcal{B}(u_i,v_i),z-v_i\rangle }_Y+\varphi (v_i,z)-\varphi (v_i,v_i)\ge {\langle \zeta ,z-v_i\rangle }_Y,\forall z\in \mho .$$

(29)

After setting $i=1$ to correspond to $y=u_2$ and $i=2$ to correspond to $y=u_1$ in Equation (28), we add the two equations to obtain

$${\langle \mathcal{A}(v_1,u_1)-\mathcal{A}(v_2,u_2),u_1-u_2\rangle }_X-\phi (u_1,u_2)+\phi (u_1,u_1)-\phi (u_2,u_1)+\phi (u_2,u_2)\le 0.$$

(30)

Similarly, assigning $i=1$ to correspond to $z=v_2$ and $i=2$ to correspond to $z=v_1$ in Equation (29), we add the two equations to obtain

$${\langle \mathcal{B}(u_1,v_1)-\mathcal{B}(u_2,v_2),v_1-v_2\rangle }_Y-\varphi (v_1,v_2)+\varphi (v_1,v_1)-\varphi (v_2,v_1)+\varphi (v_2,v_2)\le 0.$$

(31)

By using (30), (31), and assertions (C) and (E), we have

$${\langle \mathcal{A}(v_1,u_1)-\mathcal{A}(v_2,u_2),u_1-u_2\rangle }_X+{\langle \mathcal{B}(u_1,v_1)-\mathcal{B}(u_2,v_2),v_1-v_2\rangle }_Y+{\varrho }_{\phi }\parallel u_1-u_2{\parallel }_X^2+{\varrho }_{\varphi }{\parallel v_1-v_2\parallel }_Y^2\le 0.$$

This, combined with the condition (27), implies that $u_1=u_2$ and $v_1=v_2$. Thus, Problem 1 has a unique solution. □

By adding an additional condition to (27), the resulting theorem establishes a unique solution for Problem 1.

Theorem 4. 

Assume that  (A), (B), (C), (D), (E),  and (F) are fulfilled. If the following assumptions are met:

➀

The function $u\rightarrowtail \mathcal{A}(v,u)$ is inversely relaxed monotone and Lipschitz continuous for $v\in Y$, with constants ${\alpha }_{\mathcal{A}}>0$ and ${\beta }_{\mathcal{A}}>0$. Moreover, for each $u\in X$ the function $v\rightarrowtail \mathcal{A}(v,u)$ is Lipschitz continuous with ${\mathcal{L}}_{\mathcal{A}}>0$;

➁

The function $v\rightarrowtail \mathcal{B}(u,v)$ is inversely relaxed monotone and Lipschitz continuous for $u\in X$, with ${\alpha }_{\mathcal{B}}>0$ and ${\beta }_{\mathcal{B}}>0$. Moreover, for every $v\in Y$ the function $u\rightarrowtail \mathcal{B}(u,v)$ is Lipschitz continuous with ${\mathcal{L}}_{\mathcal{B}}>0$;

➂

${\displaystyle \frac{{\mathcal{L}}_{\mathcal{A}}{\mathcal{L}}_{\mathcal{B}}}{({\alpha }_{\mathcal{A}}{\beta }_{\mathcal{A}}+{\varrho }_{\phi })({\alpha }_{\mathcal{B}}{\beta }_{\mathcal{B}}+{\varrho }_{\varphi })}}<1.$

Then Problem 1 has a unique solution.

Proof. 

Let $(u_1,v_1)$ and $(u_2,v_2)$ be two solutions to Problem 1. Then, it has

$${\langle \mathcal{A}(v_1,u_1)-\mathcal{A}(v_2,u_2),u_1-u_2\rangle }_X-{\varrho }_{\phi }{\parallel u_1-u_2\parallel }_X^2\le 0,$$

(32)

and

$${\langle \mathcal{B}(u_1,v_1)-\mathcal{A}(u_2,v_2),v_1-v_2\rangle }_Y-{\varrho }_{\varphi }{\parallel v_1-v_2\parallel }_Y^2\le 0.$$

(33)

Again, from the inverse relaxed monotonicity and Lipschitz continuity of $\mathcal{A}$, we have

$$\begin{array}{cc}\hfill -{\alpha }_{\mathcal{A}}{\beta }_{\mathcal{A}}{\parallel u_1-u_2\parallel }_X^2& \le \langle \mathcal{A}{(v_1,u_1-\mathcal{A}(v_1,u_2),u_1-u_2\rangle }_X\hfill \\ \hfill & \le {\mathcal{L}}_{\mathcal{A}}\parallel v_1-v_2{\parallel }_Y{\parallel u_1-u_2\parallel }_X.\hfill \end{array}$$

(34)

Thus, from (32) and (34), we have

$$(-{\alpha }_{\mathcal{A}}{\beta }_{\mathcal{A}}-{\varrho }_{\phi })\parallel u_1-u_2{\parallel }_X^2\le {\mathcal{L}}_{\mathcal{A}}\parallel v_1-v_2{\parallel }_Y{\parallel u_1-u_2\parallel }_X.$$

(35)

Similarly, we obtain

$$\begin{array}{cc}\hfill -{\alpha }_{\mathcal{B}}{\beta }_{\mathcal{B}}{\parallel v_1-v_2\parallel }_Y^2& \le {\langle \mathcal{B}(u_1,v_1)-\mathcal{B}(u_1,v_2),v_1-v_2\rangle }_Y\hfill \\ \hfill & \le {\mathcal{L}}_{\mathcal{B}}\parallel v_1-v_2{\parallel }_Y{\parallel u_1-u_2\parallel }_X.\hfill \end{array}$$

(36)

Again, from (33) and (36), we have

$$(-{\alpha }_{\mathcal{B}}{\beta }_{\mathcal{B}}-{\varrho }_{\varphi })\parallel v_1-v_2{\parallel }_Y^2\le {\mathcal{L}}_{\mathcal{B}}\parallel v_1-v_2{\parallel }_Y{\parallel u_1-u_2\parallel }_X.$$

(37)

Combining Equations (35) and (37) yields

$$\parallel u_1-u_2{\parallel }_X\le {\displaystyle \frac{{\mathcal{L}}_{\mathcal{A}}{\mathcal{L}}_{\mathcal{B}}}{({\alpha }_{\mathcal{A}}{\beta }_{\mathcal{A}}+{\varrho }_{\phi })({\alpha }_{\mathcal{B}}{\beta }_{\mathcal{B}}+{\varrho }_{\varphi })}}{\parallel u_1-u_2\parallel }_X.$$

(38)

However, the inequality $\frac{{\mathcal{L}}_{\mathcal{A}}{\mathcal{L}}_{\mathcal{B}}}{({\alpha }_{\mathcal{A}}{\beta }_{\mathcal{A}}+{\varrho }_{\phi })({\alpha }_{\mathcal{B}}{\beta }_{\mathcal{B}}+{\varrho }_{\varphi })}<1$ implies that $u_1=u_2$ and $v_1=v_2.$

Therefore, Problem 1 has a unique solution. □

3. Stability Results

In this section, we delve into examining the stability of the system of nonlinear mixed variational inequality problems. Firstly, we present a set of regularized problems perturbed by duality mappings that correspond to Problem 1. Secondly, we arrive at a stability conclusion that demonstrates that every solution sequence to a regularized problem contains at least one subsequence that solves Problem 1.

Let X and Y be two reflexive Banach spaces, and let $X^{*}$ and $Y^{*}$ be their dual spaces. We assume that X and Y are strictly convex without losing the generality. Let $J_X:X\to X^{*}$ and $J_Y:Y\to Y^{*}$ be the duality mappings, so that

$$J_X\left(u\right)=\{u^{*}\in X^{*}|{\langle u^{*},u\rangle }_X={\parallel u\parallel }_X^2=\parallel u^{*}{\parallel }_{X^{*}}^2\},$$

$$J_Y\left(v\right)=\{v^{*}\in Y^{*}|{\langle v^{*},v\rangle }_Y={\parallel v\parallel }_Y^2=\parallel v^{*}{\parallel }_{Y^{*}}^2\}.$$

Let $\left\{{\delta }_n\right\}$ and $\left\{{\epsilon }_n\right\}$ be real sequences such that

$${\epsilon }_n>0,{\delta }_n>0,{\epsilon }_n\to 0\mathrm{and}{\delta }_n\to 0.$$

(39)

Consider the following perturbated problem for every $n\in \mathbb{N},$ which corresponds to Problem 1.

Problem 2.

Determine $(u_n,v_n)\in \Omega \times \mho$ so that

$${\langle \mathcal{A}(v_n,u_n)+{\epsilon }_n{J}_X\left(u_n\right),y-u_n\rangle }_X+\phi (u_n,y)-\phi (u_n,u_n)\ge {\langle \gamma ,y-u_n\rangle }_X,\forall y\in \Omega ,\gamma \in X^{*},$$

(40)

and

$${\langle \mathcal{B}(u_n,v_n)+{\delta }_n{J}_Y\left(v_n\right),z-v_n\rangle }_Y+\varphi (v_n,z)-\varphi (v_n,v_n)\ge {\langle \zeta ,z-v_n\rangle }_Y,\forall z\in \mho ,\zeta \in Y^{*}.$$

(41)

We assume the following.

(G): 

$u\rightarrowtail \mathcal{A}(v,u)$ and $v\rightarrowtail \mathcal{B}(u,v)$ are monotone, and meet

$$\underset{\lambda \to 0}{lim\; sup}{\langle \mathcal{A}(v,\lambda y+(1-\lambda )u),y-u\rangle }_X\le {\langle \mathcal{A}(v,u),y-u\rangle }_X,$$

$$\underset{\lambda \to 0}{lim\; sup}{\langle \mathcal{B}(u,\lambda z+(1-\lambda )v),z-v\rangle }_Y\le {\langle \mathcal{B}(u,v),z-v\rangle }_Y,\forall z,v\in Y,\forall y,u\in X.$$

(H): 

$u\rightarrowtail \mathcal{A}(v,u)$ is inverse relaxed monotone with ${\alpha }_{\mathcal{A}}>0$ and Lipschitz continuous with ${\beta }_{\mathcal{A}}>0$; similarly, $v\rightarrowtail \mathcal{B}(u,v)$ is inverse relaxed monotone with ${\alpha }_{\mathcal{B}}>0$ and Lipschitz continuous with ${\beta }_{\mathcal{B}}>0$, and fulfil

$$\underset{\lambda \to 0}{lim\; sup}{\langle \mathcal{A}(v,\lambda y+(1-\lambda )u),y-u\rangle }_X\le {\langle \mathcal{A}(v,u),y-u\rangle }_X,$$

$$\underset{\lambda \to 0}{lim\; sup}{\langle \mathcal{B}(u,\lambda z+(1-\lambda )v),z-v\rangle }_Y\le {\langle \mathcal{B}(u,v),z-v\rangle }_Y,\forall z,v\in Y,\forall y,u\in X.$$

The theory described below ensures that solutions to Problem 2 exist and converge.

Theorem 5. 

Assume that  (A), (B), (C), (D) (ii)–(iv), (E), and (F) (ii)–(iv) are satisfied. Then the following assertions hold:

(i) 

If, in addition to assumption(G), Problem 2 has at least one solution $(u_n,v_n)\in \Omega \times \mho$ for every $n\in \mathbb{N}$;

(ii) 

Furthermore, if(G)holds, there is a subsequence $\left\{(u_n,v_n)\right\}$ for every solution of the sequence $\left\{(u_n,v_n)\right\}$ to Problem 2, such that

$$(u_n,v_n)\stackrel{w}{⟶}(u,v)\in X\times Y\mathrm{as}n\to \infty ,$$

(42)

where $(u,v)\in \Omega \times \mho$ solves the Problem 2;

(iii) 

Under the conditions of(H), any sequence of solutions $\left\{(u_n,v_n)\right\}$ of Problem 2 has a subsequence $\left\{(u_n,v_n)\right\}$ such that

$$(u_n,v_n)\to (u,v)\in X\times Y\mathrm{as}n\to \infty ,$$

(43)

where $(x,y)\in \Omega \times \mho$ solves Problem 1.

Proof. 

(i)

Assign

$${\mathcal{A}}_n(v,u)=\mathcal{A}(v,u)+{\epsilon }_n{J}_X\left(u\right)$$

and

$${\mathcal{B}}_n(u,v)=\mathcal{B}(u,v)+{\delta }_n{J}_Y\left(v\right),\forall (u,v)\in X\times Y.$$

We shall confirm that ${\mathcal{A}}_n$ and ${\mathcal{B}}_n$ satisfy, respectively, (D) and (F). Observe that $J_X$ is demicontinuous and

$$0\le {(\parallel u\parallel }_X-{\parallel y\parallel }_X{)}^2\le {\langle J_X\left(u\right)-J_X\left(y\right),u-y\rangle }_X,\forall u,y\in X.$$

(44)

Using hypotheses (G), we determine that (D)(i) is satisfied for each $v\in Y$, $u\rightarrowtail {\mathcal{A}}_n(v,u)$. Utilizing the information that $\parallel J_X{\left(u\right)\parallel }_X={\parallel u\parallel }_X$ and

$${\langle J_X\left(u\right),u\rangle }_X={\parallel u\parallel }_X^2,\forall u\in X.$$

It is easy to show that ${\mathcal{A}}_n$ satisfies (D)(ii)–(iv). Similarly, ${\mathcal{B}}_n$ satisfies (F). Consequently, by using Theorem 2, we can argue that Problem 2 has a solution.

(ii)

Let $\left\{(u_n,v_n)\right\}$ be any arbitrary sequence that solves Problem 2. Next, a meticulous calculation yields

$$\begin{array}{cc}\hfill ♭(\parallel u_n{\parallel }_X,\parallel v_n{\parallel }_Y)& \le ♭(\parallel u_n{\parallel }_X,\parallel v_n{\parallel }_Y)+{\displaystyle \frac{{\epsilon }_n{\langle J_X\left(u_n\right),u_n\rangle }_X}{\parallel u_n{\parallel }_X}}\hfill \\ \hfill & \le {\displaystyle \frac{{\varrho }_{\mathcal{A}}(1+\parallel u_n{\parallel }_X+\parallel v_n{\parallel }_Y)\parallel u_0{\parallel }_X}{\parallel u_n{\parallel }_X}}+{\displaystyle \frac{({\epsilon }_n\parallel J_X\left(u_n\right){\parallel }_{X^{*}}+{\parallel \gamma \parallel }_{X^{*}})\parallel u_0{\parallel }_X+{\varrho }_{\phi }\left(u\right)}{\parallel u_n{\parallel }_X}}\hfill \\ \hfill & +{\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}}\hfill \\ \hfill & ={\displaystyle \frac{{\varrho }_{\mathcal{A}}(1+\parallel u_n{\parallel }_X+\parallel v_n{\parallel }_Y)\parallel u_0{\parallel }_X}{\parallel u_n{\parallel }_X}}+{\displaystyle \frac{({\epsilon }_n\parallel u_n{\parallel }_X+{\parallel \gamma \parallel }_{X^{*}})\parallel u_0{\parallel }_X+{\varrho }_{\phi }\left(u\right)}{\parallel u_n{\parallel }_X}}\hfill \\ \hfill & +{\varrho }_{\phi }\left(u\right)+{\parallel \gamma \parallel }_{X^{*}},\hfill \end{array}$$

(45)

and

$$\begin{array}{cc}\hfill \ell (\parallel v_n{\parallel }_Y,\parallel u_n{\parallel }_X)& \le {\displaystyle \frac{{\varrho }_{\mathcal{B}}(1+\parallel u_n{\parallel }_X+\parallel v_n{\parallel }_Y)\parallel v_0{\parallel }_Y}{\parallel v_n{\parallel }_Y}}+{\displaystyle \frac{({\delta }_n\parallel v_n{\parallel }_Y+{\parallel \zeta \parallel }_{Y^{*}})\parallel v_0{\parallel }_Y+{\varrho }_{\varphi }\left(v\right)}{\parallel v_n{\parallel }_Y}}\hfill \\ \hfill & +{\varrho }_{\varphi }\left(v\right)+{\parallel \zeta \parallel }_{Y^{*}}.\hfill \end{array}$$

(46)

The same argument that was employed in Lemma 3’s proof that $\left\{(u_n,v_n)\right\}$ is bounded in $X\times Y$.

If necessitated we can go to a relabeled subsequence and assume that

$$(u_n,v_n)\stackrel{w}{⟶}(u,v)\in X\times Y\mathrm{as}n\to \infty ,\mathrm{for}\mathrm{some}(u,v)\in \Omega \times \mho .$$

(47)

By using the monotonicity of $u\rightarrowtail \mathcal{A}(v,u)$ and $v\rightarrowtail \mathcal{B}(u,v)$, we can make the following deduction:

$${\langle \mathcal{A}(v_n,y)+{\epsilon }_n{J}_X\left(u_n\right),y-u_n\rangle }_X+\phi (u_n,y)-\phi (u_n,u_n)\ge {\langle \gamma ,y-u_n\rangle }_X,\forall y\in \Omega ,$$

(48)

and

$${\langle \mathcal{B}(u_n,z)+{\delta }_n{J}_Y\left(v_n\right),z-v_n\rangle }_Y+\varphi (v_n,z)-\varphi (v_n,v_n)\ge {\langle \zeta ,z-v_n\rangle }_Y,\forall z\in \mho .$$

(49)

By taking the upper limit as $n\to \infty$ and applying conditions (D)(ii) and (F)(ii), we infer that

$${\langle \mathcal{A}(v,y),y-u\rangle }_X+\phi (u,y)-\phi (u,u)\ge {\langle \gamma ,y-u\rangle }_X,\forall y\in \Omega ,$$

and

$${\langle \mathcal{B}(u,z),z-v\rangle }_Y+\varphi (v,z)-\varphi (v,v)\ge {\langle \zeta ,z-v\rangle }_Y,\forall z\in \mho ,$$

Here, we utilized the boundedness of $\left\{(u_n,v_n)\right\}\in X\times Y$. Using the Minty approach, we find $(u,v)\in \Omega \times \mho$ to solve Problem 1, i.e.,

$$(u,v)\in \Gamma (\gamma ,\zeta ).$$

(iii)

It can be deduced from (ii) that if we have a sequence of solutions denoted by $\left\{(u_n,v_n)\right\}$ for Problem 2, there will always exist a subsequence of $\left\{(u_n,v_n)\right\}$ that satisfies (42). We assert that the sequence $\left\{(u_n,v_n)\right\}$ has a strong convergence to $(u,v)$. It is simple to demonstrate that

$$\begin{array}{cc}\hfill -{\alpha }_{\mathcal{A}}{\beta }_{\mathcal{A}}{\parallel u_n-u\parallel }_X^2& \le {\langle \mathcal{A}(v_n,u_n)-\mathcal{A}(v_n,u),u_n-u\rangle }_X\hfill \\ \hfill & \le {\langle \mathcal{A}(v,u)-\mathcal{A}(v_n,u),u_n-u\rangle }_X+{\epsilon }_n{\langle J_X\left(u_n\right),u-u_n\rangle }_X\hfill \\ \hfill ⟹& \hfill \\ \hfill {\alpha }_{\mathcal{A}}{\beta }_{\mathcal{A}}{\parallel u_n-u\parallel }_X^2& \le {\langle \mathcal{A}(v,u)-\mathcal{A}(v_n,u),u-u_n\rangle }_X+{\epsilon }_n{\langle J_X\left(u_n\right),u_n-u\rangle }_X.\hfill \end{array}$$

(50)

By using hypothesis (D)(ii) and taking the upper limit as n approaches infinity on the above inequality, we obtain

$$\begin{array}{cc}\hfill 0& \le \underset{n\to \infty }{lim}inf{\alpha }_{\mathcal{A}}{\beta }_{\mathcal{A}}{\parallel u_n-u\parallel }_X^2\hfill \\ \hfill & \le \underset{n\to \infty }{lim}sup{\parallel u_n-u\parallel }_X^2\hfill \\ \hfill & \le {\langle \mathcal{A}(v,u)-\mathcal{A}(v_n,u),u-u_n\rangle }_X+\underset{n\to \infty }{lim}sup{\epsilon }_n\parallel u_n{\parallel }_X{\parallel u_n-u\parallel }_X\hfill \\ \hfill & \le 0.\hfill \end{array}$$

This implies that

$$u_n\to u\in X\mathrm{as}n\to \infty .$$

On the other hand, it has

$$v_n\to v\in Y\mathrm{as}n\to \infty .$$

□

4. Optimal Control

In this section, we explore optimal control for the SNMVIPs. Additionally, we examine and demonstrate the solveability of an optimal control problem that is influenced by the nonlinear mixed variational inequality system.

Consider two Banach spaces $Z_1$ and $Z_2$ with continuous embeddings from X to $Z_1$ and from Y to $Z_2$. Let $u_0\in Z_1$ and $v_0\in Z_2$ be two target profiles. We define subspaces $U\subset X^{*}$ and $V\subset Y^{*}$ such that the embeddings from U to $X^{*}$ and V to $Y^{*}$ are compact. We now examine the ensuing optimal control problem:

Problem 3.

Find $({\gamma }^{*},{\zeta }^{*})\in U\times V$ such that

$$\daleth ({\gamma }^{*},{\zeta }^{*})=\underset{(\gamma ,\zeta )\in U\times V}{inf}\daleth (\gamma ,\zeta ),$$

(51)

in which $\daleth :U\times V\to \mathbb{R}$ is defined as

$$\daleth (\gamma ,\zeta )=\underset{(u,v)\in \Gamma (\gamma ,\zeta )}{inf}\left({\displaystyle \frac{\rho }{2}}\parallel u-u_0{\parallel }_{Z_1}^2+{\displaystyle \frac{\theta }{2}}{\parallel v-v_0\parallel }_{Z_2}^2\right)+{\rm Y}(\gamma ,\zeta ).$$

(52)

Here, $\Gamma (\gamma ,\zeta )$ denotes the set of solution to Problem 1 for $(\gamma ,\zeta )\in X^{*}\times Y^{*}$, where the regularized parameters are $\rho >0$ and $\theta >0$.

We assume that the function ${\rm Y}$ satisfies the following conditions:

(K):${\rm Y}:U\times V\to \mathbb{R}$ is such that

(i)

${\rm Y}$ is bounded from beneath;

(ii)

${\rm Y}$ is coercive on $U\times V$, that is, it maintains

$$\underset{\begin{array}{c}(\gamma ,\zeta )\in U\times V\\ {\parallel \gamma \parallel }_U+{\parallel \zeta \parallel }_V\to \infty \end{array}}{lim}{\rm Y}(\gamma ,\zeta )\to +\infty ;$$

(iii)

${\rm Y}$ is weakly lower semicontinuous on $U\times V$, i.e.,

$$\underset{n\to \infty }{lim}inf{\rm Y}({\gamma }_n,{\zeta }_n)\ge {\rm Y}(\gamma ,\zeta ),$$

whenever $\left\{({\gamma }_n,{\zeta }_n)\right\}\subset U\times V$ and $(\gamma ,\zeta )\in U\times V$ are such that

$$({\gamma }_n,{\zeta }_n)\stackrel{w}{⟶}(\gamma ,\zeta )\in U\times V\mathrm{as}n\to \infty .$$

In this context, we are exploring the existence result for Problem 3.

Theorem 6.

Assume that  (A), (B), (C), (D)  (ii)–(iv), (E), and  (F) (ii)–(iv) hold. If (K) and  (G) are also satisfied, then Problem 3 has an optimal control pair.

Proof. 

For each fix $(\gamma ,\zeta )\in U\times V$, the closedness of $\Gamma (\gamma ,\zeta )$ ensures that $(\widehat{u},\widehat{v})\in \Gamma (\gamma ,\zeta )$ such that

$${\displaystyle \frac{\rho }{2}}\parallel \widehat{u}-u_0{\parallel }_{Z_1}^2+{\displaystyle \frac{\theta }{2}}{\parallel \widehat{v}-v_0\parallel }_{Z_2}^2=\underset{(u,v)\in \Gamma (\gamma ,\zeta )}{inf}\left({\displaystyle \frac{\rho }{2}}\parallel u-u_0{\parallel }_{Z_1}^2+{\displaystyle \frac{\theta }{2}}{\parallel v-v_0\parallel }_{Z_2}^2\right),$$

(53)

is attainable.

According to ℸ and (K)(i), there exists a minimizing sequence $\left\{({\gamma }_n,{\zeta }_n)\right\}\subset U\times V$ such that

$$\underset{n\to \infty }{lim}\daleth ({\gamma }_n,{\zeta }_n)=\underset{(\gamma ,\zeta )\in U\times V}{inf}\daleth (\gamma ,\zeta ).$$

(54)

We assume that the sequence $\left\{({\gamma }_n,{\zeta }_n)\right\}$ is bounded in $U\times V$. To arrive at a contradiction, we suppose that

$$\parallel {\gamma }_n{\parallel }_U+{\parallel {\zeta }_n\parallel }_V\to +\infty \mathrm{as}n\to \infty .$$

Using the latter with (K)(ii), we can conclude that

$$\begin{array}{cc}\hfill \underset{(\gamma ,\zeta )\in U\times V}{inf}\daleth (\gamma ,\zeta )& =\underset{n\to \infty }{lim}\daleth ({\gamma }_n,{\zeta }_n)\hfill \\ \hfill & \ge \underset{n\to \infty }{lim}{\rm Y}({\gamma }_n,{\zeta }_n)\hfill \\ \hfill & =+\infty .\hfill \end{array}$$

(55)

The result is a contradiction, which means that $\left\{({\gamma }_n,{\zeta }_n)\right\}$ is bounded in $U\to V$. Passing to a relabeled subsequence if necessary, we may assume that

$$({\gamma }_n,{\zeta }_n)\stackrel{w}{⟶}({\gamma }^{*},{\zeta }^{*})\in U\times V\mathrm{as}n\to \infty ,\mathrm{for}\mathrm{some}({\gamma }^{*},{\zeta }^{*})\in U\times V.$$

(56)

Let $\left\{(u_n,v_n)\right\}\subset \Omega \times \mho$ satisfy (53) by letting $\widehat{u}=u_n$, $\widehat{v}=v_n$, and $(\gamma ,\zeta )=({\gamma }_n,{\zeta }_n)$. We will now prove that $\left\{(u_n,v_n)\right\}\subset \Omega \times \mho$ is uniformly bounded in $X\times Y$. A simple computation reveals that

$$\begin{array}{cc}\hfill ♭(\parallel u_n{\parallel }_X,\parallel v_n{\parallel }_Y)& \le {\displaystyle \frac{{\varrho }_{\mathcal{A}}(1+\parallel u_n{\parallel }_X+\parallel v_n{\parallel }_Y)\parallel u_0{\parallel }_X}{\parallel u_n{\parallel }_X}}+{\displaystyle \frac{({\varrho }_{\phi }\left(u\right)+\parallel {\gamma }_n{\parallel }_{X^{*}})\parallel u_0{\parallel }_X}{\parallel u_n{\parallel }_X}}\hfill \\ \hfill & +{\varrho }_{\phi }\left(u\right)+{\parallel {\gamma }_n\parallel }_{X^{*}},\hfill \end{array}$$

(57)

and

$$\begin{array}{cc}\hfill \ell (\parallel v_n{\parallel }_Y,\parallel u_n{\parallel }_X)& \le {\displaystyle \frac{{\varrho }_{\mathcal{B}}(1+\parallel u_n{\parallel }_X+\parallel v_n{\parallel }_Y)\parallel v_0{\parallel }_Y}{\parallel v_n{\parallel }_Y}}+{\displaystyle \frac{({\varrho }_{\varphi }\left(v\right)+\parallel {\zeta }_n{\parallel }_{Y^{*}}\parallel v_0{\parallel }_Y}{\parallel v_n{\parallel }_Y}}\hfill \\ \hfill & +{\varrho }_{\varphi }\left(v\right)+{\parallel {\zeta }_n\parallel }_{Y^{*}}.\hfill \end{array}$$

(58)

Since there is continuity in the embeddings from U to $X^{*}$ and from V to $Y^{*}$, we use the same approach as in the proof of Lemma 3 to show that the sequence $\left\{(u_n,v_n)\right\}\subset \Omega \times \mho$ is uniformly bounded in $X\times Y$. Without loss of generality, we can assume that

$$(u_n,v_n)\stackrel{w}{⟶}(u^{*},v^{*})\in X\times Y,\mathrm{and}{Z}_1\times Z_2\mathrm{as}n\to \infty ,\mathrm{for}\mathrm{some}(u^{*},v^{*})\in \Omega \times \mho .$$

(59)

Using the Minty approach yields

$${\langle \mathcal{A}(v_n,y),y-u_n\rangle }_X+\phi (u_n,y)-\phi (u_n,u_n)\ge {\langle {\gamma }_n,y-u_n\rangle }_X,\forall y\in \Omega ,$$

(60)

and

$${\langle \mathcal{B}(u_n,z),z-v_n\rangle }_Y+\varphi (v_n,z)-\varphi (v_n,v_n)\ge {\langle {\zeta }_n,z-v_n\rangle }_Y,\forall z\in \mho .$$

(61)

The embedding from $(U,V)$ into $(X^{*},Y^{*})$ is compact, and (56) implies that

$$({\gamma }_n,{\zeta }_n)\to ({\gamma }^{*},{\zeta }^{*})\in X^{*}\times Y^{*}\mathrm{as}n\to \infty .$$

Taking the upper limit as $n\to \infty$ for (60) and (61), we obtain

$${\langle \mathcal{A}(v^{*},y),y-u^{*}\rangle }_X+\phi (u^{*},y)-\phi (u^{*},u^{*})\ge {\langle {\gamma }^{*},y-u^{*}\rangle }_X,\forall y\in \Omega ,$$

and

$${\langle \mathcal{B}(u^{*},z),z-v^{*}\rangle }_Y+\varphi (v^{*},z)-\varphi (v^{*},v^{*})\ge {\langle {\zeta }^{*},z-v^{*}\rangle }_Y,\forall z\in \mho ,$$

where we used (F)(ii) and (D)(ii). Using the Minty trick once more, we accomplish

$$(u^{*},v^{*})\in \Gamma ({\gamma }^{*},{\zeta }^{*}).$$

The weaker lower semicontinuity of ${\parallel ·\parallel }_{Z_1}$ and ${\parallel ·\parallel }_{Z_2}$, however, suggests that

$${\displaystyle \frac{\rho }{2}}\parallel u^{*}-u_0{\parallel }_{Z_1}^2+{\displaystyle \frac{\theta }{2}}{\parallel v^{*}-v_0\parallel }_{Z_2}^2\le \underset{n\to \infty }{lim\; inf}\left[{\displaystyle \frac{\rho }{2}}\parallel u_n-u_0{\parallel }_{Z_1}^2+{\displaystyle \frac{\theta }{2}}{\parallel v_n-v_0\parallel }_{Z_2}^2\right].$$

(62)

Note that ${\rm Y}$ is weakly lower semicontinuous on $U\times V$; it implies

$${\rm Y}({\gamma }^{*},{\zeta }^{*})\le \underset{n\to \infty }{lim\; inf}{\rm Y}({\gamma }_n,{\zeta }_n).$$

(63)

Referring to Equations (62) and (63), we have

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim\; inf}\daleth ({\gamma }_n,{\zeta }_n)& \ge \underset{n\to \infty }{lim\; inf}\underset{(u,v)\in \Gamma ({\gamma }_n,{\zeta }_n)}{inf}\left({\displaystyle \frac{\rho }{2}}\parallel u-u_0{\parallel }_{Z_1}^2+{\displaystyle \frac{\theta }{2}}{\parallel v-v_0\parallel }_{Z_2}^2\right)+\underset{n\to \infty }{lim\; inf}{\rm Y}({\gamma }_n,{\zeta }_n)\hfill \\ \hfill & =\underset{n\to \infty }{lim\; inf}\left({\displaystyle \frac{\rho }{2}}\parallel u_n-u_0{\parallel }_X^2+{\displaystyle \frac{\theta }{2}}{\parallel v_n-v_0\parallel }_Y^2\right)+\underset{n\to \infty }{lim\; inf}{\rm Y}({\gamma }_n,{\zeta }_n)\hfill \\ \hfill & \ge {\displaystyle \frac{\rho }{2}}\parallel u^{*}-u_0{\parallel }_{Z_1}^2+{\displaystyle \frac{\theta }{2}}{\parallel v^{*}-v_0\parallel }_{Z_2}^2+{\rm Y}({\gamma }^{*},{\zeta }^{*}),(\mathrm{where}(u^{*},v^{*})\in \Gamma ({\gamma }^{*},{\zeta }^{*}))\hfill \\ \hfill & \ge \underset{(u,v)\in \Gamma ({\gamma }^{*},{\zeta }^{*})}{inf}\left({\displaystyle \frac{\rho }{2}}\parallel u-u_0{\parallel }_{Z_1}^2+{\displaystyle \frac{\theta }{2}}{\parallel v-v_0\parallel }_{Z_2}^2\right)+{\rm Y}({\gamma }^{*},{\zeta }^{*})\hfill \\ \hfill & =\daleth ({\gamma }^{*},{\zeta }^{*}).\hfill \end{array}$$

(64)

We can use Equation (64) along with (54) to arrive at the following conclusion:

$$\daleth ({\gamma }^{*},{\zeta }^{*})\le \underset{(\gamma ,\zeta )\in U\times V}{inf}\daleth (\gamma ,\zeta ),$$

namely, $({\gamma }^{*},{\zeta }^{*})$ is an optimal control pair of Problem 3. □