1. Introduction
Lin [
1] introduced the system of generalized quasi-variational inclusion problems. This system consists of a set of problems defined on a product set. It includes several well-known problems such as variational inequalities, equilibrium problems, vector equilibrium problems, vector quasi-equilibrium problems, variational inclusions problems and variational disclusion problems.
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 emerged as a powerful mathematical resource. Variational inequalities 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. However, coupled variational inequalities can be a useful mathematical tool for examining various coupled mixed boundary value problems, control problems, and similar problems. More details can be found in [
8,
9,
10,
11,
12].
In this paper, we introduce a system of nonlinear mixed variational inequality problems and using the Kakutani-Ky 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 first define the problem that will play a crucial role in this paper. Consider two reflexive Banach spaces and , each with its dual spaces and , respectively. We use (resp., ) to denote the duality pairing between and X (resp., the duality pairing between and Y). We use the symbols and ⟶ to denote the weak and the strong convergence in the space X and employ the notation for the space X equipped with the weak topology. The limits, lower limits and upper limits are considered as even if we do not mention it explicitly. We can formulate a system of nonlinear mixed variational inequality problems on Banach spaces:
Problem 1.
Find such that
and
We note that if
and
, then Problem 1 reduces to the coupled system of variational inequality problems for finding
such that
studied in [
13].
Definition 2. Let Ω be a nonempty subset of a Banach space X. Let be a proper convex and lower semicontinuous function, and . Then is called
- (i)
- (ii)
strictly monotone, if it holds
- (iii)
strongly monotone with constant if it holds
- (iv)
relaxed strongly monotone with constant if it holds
- (v)
inverse relaxed monotone with constant if it holds
- (vi)
Lipschitz continuous with constant if it holds
- (vii)
-
pseudomonotone, if for any we have
- (viii)
stable pseudo monotone with respect to the set , if and are pseudo monotone for all
- (ix)
-
φ-pseudomonotone, if for any we have
- (x)
stable φ-pseudomonotone with respect to the set , if and are φ-pseudomonotone for each
Definition 3.
[14] Let be a proper, convex and lower semicontinuous function. The mapping defined by
for , is called the subdifferential of φ. An element is called a subgradient of φ in u.
Definition 4.
[14] Given a locally Lipschitz function , we denote by the Clarke generalized directional derivative of φ at the point in the direction defined by
The Clarke subdifferential or the generalized gradient of φ at , denoted by , is a subset of given by
Let
Z and
Y be topological spaces and
V be a nonempty subset of
Z. We denote by
the collection of subsets of
V. Given a set-valued mapping
, we use
to represent the graph of
. In other word,
is defined as follows:
We say that the graph of
is sequentially closed in
, or
is sequentially closed, if for any sequence
. If this sequence converges to
as
then we have
Theorem 1. [15], Let Y be a reflexive Banach space and D be a nonempty, bounded, closed and convex set subset of Y. Let be a set-valued map with nonempty, closed and convex values such that its graph is sequentially closed in topology. 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 Kakutani-Ky Fan fixed point theorem to establish the existence theorem for the solutions for the system of nonlinear mixed variational inequality problems under given modest assumptions. Additionally, we deploy the inverse relaxed monotonicity and Lipschitz continuity argument to present two uniqueness results for Problem 1.
Furthermore, we propose the set-valued mappings
and
, described by
and
respectively.
To investigate the existence of solutions to Problem 1, we now suggest the following appropriate assumptions.
-
(A):
and are both nonempty, closed and convex.
-
(B):
and .
-
(C):
-
is such that
- (i)
is a proper, convex and lower semicontinuous function,
- (ii)
there exists
such that
- (iii)
for each
, there exists
such that
[
16]
-
(D):
-
is such that
- (i)
for each
is stable
-pseudomonotone with
and fulfills
- (ii)
it holds
whenever
are such that
- (iii)
-
there exists a function
such that
and
♠ every nonempty and bounded set
we have
♠ for any constants it holds as
- (iv)
there exists a constant
such that
-
(E):
-
is such that
- (i)
is a proper, convex and lower semicontinuous function,
- (ii)
there exists
such that
- (iii)
for each
, there exists
such that
[
16]
-
(F):
-
is such that
- (i)
for each
is stable
-pseudomonotone with respect to
and satisfies
- (ii)
it holds
whenever
and
are such that
- (iii)
-
there exists a function
such that
and
♠ every nonempty and bounded set
, we have
♠ for any constants , it holds as
- (iv)
there exists a constant
such that
Remark 1.
If function r given in(D)(iii) (resp. ℓ given in(F)(iii)) is independent of its second variable, then condition(D)(iii) (resp.(F)(iii)) reduces to the following uniformly coercive condition:
(D)(iii)′: there exists a function with as such that
(resp.(F)(iii)′: there exists a function with as such that
The main result of this paper, related to the existence of solutions to Problem 1, is as follows.
Theorem 2.
Suppose that(A), (B), (C), (D), (E)and(F)are held. Then, the solution set denoted by of Problem 1 corresponding to is nonempty and weakly compact in .
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 each fix , is a solution of (1), if and only if, x solves the following Minty inequality for finding such that
- (ii)
for each fix , the solution set denoted by of (1) is nonempty, bounded, closed and convex;
- (iii)
the graph of the set-valued mapping is sequentially closed in i.e., is sequentially closed from Y endowed with the weak topology into the subsets of X with the weak topology;
- (iv)
for each fix , if the mapping is strictly monotone, then is a single-valued mapping and weakly continuous.
Proof. The assumptions (i) and (ii) are the straightforward consequences of [
17], Theorem 3.3] and [
18], Lemma 3.3]. Next, we present the conclusion (iii).
Let
be such that
Then, for each
, we have
i.e.,
The assertion (i) asserts that
To establish the upper limit as
, use assumption
(D)(ii) and weak lower semicontinuity of
(due to the convexity and lower semicontinuity of
) to determine
Using the assumption (i) again, we obtain
Consequently, , i.e., the graph of the set-valued mapping is sequentially closed in
Additionally, assume that
is strictly monotone. Let us consider
be two solutions to (
1). Then, we have
and
Putting
in (
7) and
in (
8), we have
and
Adding (
9) and (
10), we have
Hence, from the assumption (C) and the strict monotonicity of guarantees that . Therefore, is a single-valued mapping. But, by virtue of assumption (iii), we can observe that is weakly continuous. □
Similarly, for problem (
2), we have the following lemma.
Lemma 2.
Assume that(A), (B), (E)and(F)are satisfied. Then, the following statements hold
- (i)
for each fix , is a solution of (2), if and only if y solves the following Minty inequality for finding such that
- (ii)
for each fix , the solution set namely, of (2) is nonempty, bounded, closed and convex;
- (iii)
the graph of the set-valued mapping is sequentially closed in ;
- (iv)
for each fix , if the mapping is strictly monotone, then is a single-valued mapping and weakly continuous.
Furthermore, we focus on providing a priori estimates for the solutions of Problem 1.
Lemma 3.
Assume that(A), (B), (C), (D), (E)and(F)are satisfied. If the solution set namely, of Problem 1 is nonempty, then there exists a constant such that
Proof. Suppose that
. Let
be arbitrary and
By swapping
and
into (
1) and (
2),
respectively, we obtain
and
Taking account of (
14), we use hypotheses
(C)(i),(iii) and
(D)(iii)–(iv) to obtain
Similarly, taking account of (
15), we use hypotheses
(E)(i), (iii) and
(F)(iii)–(iv) to obtain
Contrary, suppose
is unbounded. Then, taking a subsequence if necessary, it is possible to find a sequence
such that it holds
or
Let us segregate the subsequent cases:
ⓐ Assuming that (
18) is fulfilled and the sequence
is bounded in the space
Y,
ⓑ Assuming that (
19) is fulfilled and the sequence
is bounded in the space
X,
ⓒ Assuming that both (
18) and (
19) are fulfilled.
Assume that ⓐ is valid, then we put
and
in (
16) to get
When we let
n approach infinity in the inequality (
20) and make use of (
18) along with property
(D)(iii), we obtain the following:
Hence, (
21) generates a contradiction. Similarly, for the case ⓑ we could also use (
17) to get a contradiction. However, we assume that ⓒ hold, we will proceed to discuss two additional situations:
- (1)
- (2)
there exist
and
such that
If the item (1) is true, then we put
and
into (
17) to yield
Taking the limit as
n approaches infinity for the inequality mentioned above yields:
It is obviously impossible, however for a situation (2), we can deduce from (
16) that
where
is such that
This leads to a contradiction. Thus, we conclude that
is bounded in
, allowing us to find a constant
satisfying (
13). □
Consider the set-valued mapping
given by
By invoking Lemma 1 and Lemma 2, it can be seen that is well-defined. Additionally, there exists a bounded, closed, and convex set in such that maps into itself.
Lemma 4.
Assume that conditions(A), (B), (C), (D), (E)and(F)are satisfied. Then, there exists a constant satisfying , where is defined by
Proof. Our proof will be based on contradiction. Suppose for each
, it holds
. Then, for every
, we can find
and
( i.e.,
) and
such that
Thus, passing to a relabeled subsequence if needed, we can assume that
for each
(since the proof for the case that
for each
is similar). Employing (
16), it determines
Therefore, passing to the limit as
for the inequality above, it gives
This leads to a contradiction. Thus, there exists a constant
satisfying
□
Proof. (Proof of Theorem 2). Let us see that if
is a fixed point of
, then we have
and
. Using the definitions of
and
, it provides
and
Consequently, it is clear that is also a solution to Problem 1. We will apply Theorem 1, the Kakutani-Ky Fan fixed point theorem, to determine the existence of a fixed point for .
Moreover, Lemma 1, Lemma 2, and Lemma 4, in fact, infer that
has nonempty, closed, and convex values, and that the graph of
is sequentially closed in
The conditions stated in Theorem 1 have been verified. By using this theorem, there exists a solution
to Problem 1 such that
and hence,
It is evident from Lemma 3 that
is bounded in
. Next, we will demonstrate that
is weakly closed. Let
be such that
It is easy to see that for each natural number
n, the pair
Since
is sequentially closed from
to
(
see Lemma 1 and Lemma 2), we can conclude that
Thus, due to the boundedness of , we can conclude that is weakly compact. □
Theorem 2 shows that the solution set of Problem 1 is both non-empty 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 following theorems provide a positive answer to this issue.
Theorem 3.
Assume that(A), (B), (C), (D), (E)and(F)are satisfied. If, in addition, the following inequality holds,
Then Problem 1 has a unique solution.
Proof. Theorem 2 assures that
. We now prove the uniqueness of Problem 1. Let
. Then, we have
and
After setting
to correspond to
and
to correspond to
in equation (
28), we add the two equations to obtain
Similarly, assigning
to correspond to
and
to correspond to
in equation (
29), we add the two equations to get
By using the (
30), (
31), and assertions
(C) and
(E), we have
This combined with the condition (
27) implies that
and
. 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 satisfied. If the following conditions also hold:
- ➀
for each , the function is inversely relaxed monotone and Lipschitz continuous with constants and , respectively, and for each the function is Lipschitz continuous with constant ,
- ➁
for each , the function is inversely relaxed monotone and Lipschitz continuous with constants and , respectively, and for each the function is Lipschitz continuous with constant ,
- ➂
Then Problem 1 has a unique solution.
Proof. Let
and
be two solutions to Problem 1. Then, it has
and
Again, from the inverse relaxed monotonicity and Lipschitz continuity of
, we have
Thus, from (
32) and (
34), we have
Again, from (
33) and (
36), we have
Combining equations (
35) and (
37) yields
However, the inequality implies that and 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 produce a stability result that shows that any solution sequence of a regularized problem has at least one subsequence that leads to a solution of the original problem, which is Problem 1.
Recall that
X and
Y are two reflexive Banach spaces that can be renormed to become strictly convex. Without loss of generality, we may assume that
X and
Y are strictly convex. Let
and
be the duality mappings of the spaces
X and
Y,
respectively, namely:
Let real sequences
and
be such that
For each consider the following perturbated problem corresponding to Problem 1.
Problem 5.
Find such that
and
We make the following assumptions.
-
(G):
and
are monotone, and satisfy
-
(H):
is inverse relaxed monotone with constant
and Lipschitz continuous with constant
; similarly,
is inverse relaxed monotone with constant
and Lipschitz continuous with constant
, and and satisfy
The theorem below guarantees the existence and convergence of solutions to Problem 5.
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,(G)holds, then for each , Problem 5 has at least one solution
- (ii)
-
furthermore, if(G)holds, then for any sequence of solutions of Problem 5, there exists a subsequence , such that
where is a solution of Problem 5;
- (iii)
-
if(H)holds, then for any sequence of solutions of Problem 5, there exists a subsequence such that
where is a solution of Problem 1.
We will verify that
and
satisfy
(D) and
(F),
respectively. Note that
is demicontinuous and
Using hypotheses
(G), we determine that
(D)(i) is satisfied for each
. By using the facts,
and
It is easy to show that satisfies (D)(ii)–(iv). Similarly, satisfies (F). Consequently, by using Theorem 2, we can argue that Problem 5 has a solution.
(ii) Let
be an arbitrary sequence of solutions of Problem 5. Next, a meticulous calculation yields
and
The same arguments used in the proof of Lemma 3 that is bounded in .
If necessitated we can proceed to a relabeled subsequence and presume that
By using the monotonicity of
and
, we can make the following deduction
and
Taking the upper limit as
and using hypotheses
(D)(ii) and
(F)(ii), we infer that
and
here we used the boundedness of
. Using Minty approach, we find
solve Problem 1,
i.e.,
(iii) It can be deduced from assertion (ii) that if we have a sequence of solutions denoted by
for Problem 5, there will always exist a subsequence of
which satisfies (
42). We claim that the sequence
converges strongly to
. It is simple to demonstrate that:
By using hypothesis
(D)(ii) and taking the upper limit as
n approaches infinity on the above inequality, we obtain
On the other hand, it has
□
4. Optimal Control
In this section, we explore optimal control for the system of nonlinear mixed variational inequality problems. Additionally, we consider an optimal control problem driven by the system of nonlinear mixed variational inequality problems and prove its solvability.
Consider two Banach spaces and with continuous embeddings from X to and from Y to . Let and be two target profiles. We define subspaces and such that the embeddings from U to and V to are compact. Now, we investigate the following optimal control problem:
Problem 6.
Find such that
where the cost function is defined by
Here, represents the solution set of Problem 1 for , with and as regularized parameters.
We assume that the function satisfies the following conditions:
(K): is such that
- (i)
is bounded from below;
- (ii)
is coercive on
, namely it holds
- (iii)
is weakly lower semicontinuous on
,
i.e.,
whenever
and
are such that
In this context, we are exploring the existence result for Problem 6.
Theorem 6.
Assume that(A), (B), (C), (D)(ii)–(iv),(E)and(F)(ii)–(iv) hold. If, in addition,(K)and(G)are fulfilled. Then Problem 6 has an optimal control pair.
Proof. For each fix
, the closedness of
(
see Theorem 2) guarantees that there exists
such that
is attainable.
From the definition of
ℸ and hypothesis
(K)(i), there exists a minimizing sequence
such that
We assert that the sequence
is bounded in
. Arguing by contradiction, we suppose that
The latter together with hypothesis
(K)(ii), leads to the conclusion:
This leads to a contradiction, so,
is bounded in
. Passing to a relabeled subsequence if necessary, we may assume that
Let
satisfy (
53) by taking
,
, and
. We will now prove that
is uniformly bounded in
. A direct computation shows that
and
Since the embeddings from
U to
and from
V to
are both continuous. We can apply the same arguments as in the proof of Lemma 3 to infer that
is uniformly bounded in
. Without loss of generality, we assume that
Using the Minty approach yields
and
The compactness of the embedding from
into
and (
56), implies that
Taking the upper limit as
for inequalities (
60)–(
61), we gain
and
where we used the conditions
(F)(ii) and
(D)(ii). Once more using the Minty trick, we get
However, the weak lower semicontinuity of
and
suggests
Note that
is weakly lower semicontinuous on
, it implies
Referring to equations (
62) and (
63), we have
We can use equation (
64) along with (
54) to arrive at the following conclusion:
namely
is an optimal control pair of Problem 6. □
Availability of data and material
The data sets used and/or analysed during the current study are available from the corresponding author upon reasonable request.
Competing interests
No potential conflict of interest was reported by the author(s).
Funding
The data sets used and/or analysed during the current study are available from the corresponding author upon reasonable request.
Author’s contributions
All authors read and approved the final manuscript.
Acknowledgements
The authors extend their appreciation to the Deputyship for Research & Innovation, Ministry of Education in Saudi Arabia for funding this research work through Project Number: ISP-2024.
Author’s information
Not applicable
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