The Optimal Control Problems for Generalized Elliptic Quasivariational Inequalities

In this article, we propose an optimal control problem for generalized elliptic quasi-variational inequality with unilateral constraints. Then, we discuss the sufficient assumptions that ensure the convergence of the solutions to the optimal control problem. The proofs depend on convergence results for generalized elliptic quasi-variational inequalities, obtained by the arguments of compactness, lower semi-continuity, monotonicity, penalty and different estimates. As an application, we addressed the abstract convergence results in the analysis of optimal control associated with boundary value problems.


Introduction
Many applications of optimal control theory can be found in physics, mechanics, automatics, systems theory, and financial management control theory. Although the control problems for linear systems are sufficiently well-studied, the situation is not so good for nonlinear systems. However, due to the complexity of nonlinear systems describing fluid motion and non-Newtonian fluid motion, such as polymers, various solutions, emulsions, blood, and many others, they have not been fully studied. In hydrodynamics, control (optimal) problems are often connected with fluid control by external forces. Usually, in solving such issues, a control is considered from a given finite set, see [1][2][3][4][5].
The concept of variational inequality was developed on the basis of monotonicity and convexity, including properties of the subdifferential of a convex function, see [6][7][8][9][10]. The study of optimal control problems for variational and hemivariational inequalities has been addressed in several works and is an expanding and vibrant branch of applied mathematics with numerous applications, see [11][12][13][14][15][16][17]. The theory and computational techniques for optimal control for equations and variational inequalities have been studied for quite some time now. In [18][19][20][21], the optimal control problem in the sense of boundary valued problems can be found in [22][23][24][25][26], and the computational issues have been addressed in [27][28][29][30][31]. Nonetheless, many important applied models have motivated the study of optimal control for more general quasi-variational problems.
In this paper, we consider an optimal control problem for a general class of elliptic quasivariational inequalities. Let X and Y be real Hilbert spaces endowed with the inner products ·, · X and ·, · Y respectively, K ⊂ X , N : X × X −→ X , J : X × X −→ R, γ : X −→ Y, f ∈ Y. Then, we consider the following inequality problem for finding x ∈ K such that and the admissible pairs set defined by and we consider a objective functional L : X × Y −→ R. Here and below, X × Y represents the product of the Hilbert spaces X and Y, equipped with the canonical inner product. Then, we study, in this paper, the following optimal control problem for finding (x * , f * ) ∈ V ad such that Next, consider a setK ⊂ X , an operatorN : X × X −→ X and an elementf ∈ Y. With these data, we suggest the following perturbation of (1) for findingx ∈K such that We associate to (4) the admissible pairs set given bŷ V ad = (x,f ) ∈K × Y such that (4) holds (5) and, for a objective functionalL : X × Y −→ R, we construct the following perturbation of the optimal control problem (3) for finding (x * ,f * ) ∈V ad such that The unique solvability of (1) and (4), and the solvability of (3) and (6) follow from well known results obtained in the literature, under sufficient assumptions on the data. Here, we shall use the existence and uniqueness results in [32][33][34][35], which will be resumed in the next section.
The first goal of this paper is to formulate adequate assumptions on the data which guarantee the convergence of the solutionx of (4) to the solution x of (1). The second goal is to demonstrate that, under the suitable circumstances, the solutions of (6) converge to a solution of (3). Finally, we investigate and describe the applications in contact mechanics and a heat transfer process.

Preliminaries
Throughout the text " " and " → " stand for the weak and the strong convergence, respectively. We denote by · X and · Y , the norms on the spaces X and Y, respectively. In our study of (1) we consider the following assumptions into the account: K is a nonempty, closed, convex subset of X .

Main Results
In this section, we state and prove a convergence result for the solution of (4), in the case where the problem has a dynamical structure. To the end, we consider two sequences {ζ n } ⊂ R, { f n } ⊂ Y and an operator G : X × X −→ X . For each n ∈ N let N n : X × X −→ X be the operator defined by Assume that ifN = N n andf = f n , then we have a problem for finding x n ∈K such that Remark 1.K = X then (28) denotes the penalty problem of (1), under the suitable assumptions of G , see [37,38].
To demonstrate the solvability of (28), we have the following observations: K is a nonempty, closed, convex subset of X .
G : X × X −→ X is a relaxed monotone, relaxed Lipschitz and Lipschitz continuous operator.
To investigate the behavior of the solution of (28) as n −→ ∞ we offer the following additional assumptions. and Remark 2. WhenK = X , condition (36)-(37) is satisfied for any penalty operator of the set K , see [40].  (37) and, for each n ∈ N, denote by x n the solution of (28). Then where x is the solution of (1).
Proof. There are several steps in the proof of the Theorem.
(i) The weak convergence. We assert that there is an elementx ∈K and a subsequence of {x n }, which is still denoted by {x n }, such that x n x ∈ X as n −→ ∞.
To prove the claim, we establish the boundedness of the sequence {x n } in X . Let n ∈ N. We make use of the assumption (35) and take y = x in (28) to see that Then, using the relaxed monotonicity and relaxed Lipschitz continuity of the operator N we have Next, assumption (37) implies that and assumptions (11)-(12), (22) yield On the other side, using (15) we find that Now adding inequalities (38)-(41), we have From (34) we see that the sequence { f n } is bounded in Y. Therefore, using inequality (42) and the smallness assumption (13), we deduce that there exists a constant ϑ > 0 independent of n such that This implies that the sequence {x n } is bounded in X . Thus, from the reflexivity of X , we deduce that Moreover, assumption (29) and the convergence (43) implies thatx ∈K and the proof of assertion is completed. (ii) The weak limit property.
Next, we show thatx is a solution to (1). Let y be a given element inK and let n ∈ N. We use (28) to obtain that 1 From the conditions (8)-(10), (34), (22), (15), and the boundedness of the sequence {x n }, we see that each term in the right-hand side of the inequality (44) is bounded. Therefore, there exists a constant h > 0 which does not depend on n, such that We now proceed to the upper limit in this inequality and use the convergence (33) Next, we take y =x in (45) and find that Therefore, using assumption (30) and a standard pseudomonotonicity argument (Proposition 1.23 in [38,41]) we obtain that We now combine the inequalities (47) and (45) to find that Using the assumption (37), we can deduce thatx ∈K . Now, consider an element y ∈K . We use (35) and (28) to get Therefore, using assumption (36) we find that Next, using (43) and assumption (25) we have On the other side, assumption (34), (26) and the convergence (43) yield Using the relations (48)-(50) to see that Now, taking y =x in (51) we get This inequality together with (43) and the pseudomonotonicity of N implies that Combining (53) and (51), we have Hence, it follows thatx ∈ K is a solution to (1), as claimed.
(iii) Result of weak convergence. Now, we prove that the whole sequence {x n } is weakly convergent. Since (1) has a unique solution x ∈ K , we deduce from the previous step thatx = x. Moreover, a comprehensive review of the proof in step (ii) specifies that every subsequence of {x n } that converges weakly in X has the weak limit point x. We should also note that the sequence {x n } is bounded in X . Therefore, using a standard argument we deduce that the whole sequence {x n } converges weakly in X to x, as n −→ ∞. (iv) Strong convergence.
In the final step of the proof, we prove that We take y =x ∈ K in (53) and use (52) to obtain Therefore, using equalityx = x, the relaxed monotonicity and relaxed Lipschitz continuity of N , and the convergence Hence, it follows that x n −→ x ∈ X , which completes the proof.

Optimal Control Analysis
In this section, we connect an optimal control problem with (28) for which we prove a convergence result. To this end, we hold the previous section's notations and assumptions and define the set of admissible pairs for (28) by Then, the optimal control problem associated to (28) is follows for In the study of (56), we assume that where B n and D n are functions which satisfy assumptions (17)- (19) and (20)- (21), for each n ∈ N, respectively. Note that, when we use these assumptions for the functions B n and D n , we refer to them as assumptions (17) 26) and (29)-(32) hold. Then, for each n ∈ N, there exists at least one solution (x n , f n ) ∈ V n ad of (56).
To study the behavior of the sequence of solutions of (56) as n −→ ∞ we consider the following additional hypotheses. (58) x n → x ∈ X as n −→ ∞, Proof. The proof is carried out in following manner.
We claim that the sequence { f n } is bounded in Y. Contrary we assume that { f n } is not bounded in Y, then passing to a subsequence still denoted { f n }, we have f n Y −→ +∞ as n −→ +∞.
By using the equality (57) and assumption (18) n , we have Therefore, passing to the limit as n −→ ∞ in this inequality and using (65) combined with assumption (60) we deduce that lim L n (x n , f n ) = +∞.
On the other side, since (x n , f n ) represents a solution to (56) for each n ∈ N we have We now denote by x 0 n the solution of (28) for f n = f . Then (x 0 n , f ) ∈ V n ad , and from (67) and (57) From (33) and (34) and Theorem 3 we have where x represents the solution of (1). Then, assumptions (58) and (61) imply that Relations (66), (68) and (70) lead to a contradiction, which concludes the claim.
In this step, we prove the convergence of the equations (62) and (63). First, since the sequence { f n } is bounded in Y, there exists a subsequence again denoted by { f n } and an element f ∈ Y such that (62) holds. Following that, x is a solution of (1) for f = f . Then we have Furthermore, assumption (33), the convergence (62) and Theorem 3 imply that (63) holds as well. (iii) The limit of optimality.
We now prove that (x , f ) is a solution to the optimal control (3). We use the convergences (62), (63) and assumptions (58), (59), to see that and, therefore, the structure (57) and (16) of the functionals L n and L shows that Next, we fix a solution (x 0 , f 0 ) of (3) and, moreover, for each n ∈ N we denote byx 0 n the solution of (28) for f n = f 0 . It follows from here that (x 0 n , f 0 ) ∈ V n ad and, by the optimality of the pair (x n , f n ), we have We proceed to the upper limit of this inequality to discover that Now, we know that x 0 is the solution of the inequality (1) for f = f 0 andx 0 n is the solution of the inequality (1) for f n = f 0 . As a result, the convergence (33) and Theorem 3 imply thatx 0 n −→ x 0 ∈ X as n −→ ∞ and, using assumptions (58) and (61), we find that We now use (57), (74) and (16) to get Therefore, (72), (73) and (75) imply that On the other side, since (x 0 , f 0 ) is a solution of (3), we have and, therefore, inclusion (71) implies that We now combine the inequalities (76) and (78) to see that Finally, relations (71), (79) and (77) imply that (64) holds and proof is completed.

Optimal Control Associated with Frictional Contact Problem
In this section, we will discuss the equilibrium of elastic bodies in a frictional contact problems, and in order to do so, we will require some notations and assumptions.
Let d ∈ {2, 3}. Consider S d to be a space of second order symmetric tensors on R d , and use the notation ·, · , · , 0 to represent the inner product, norm, and zero element of the spaces R d and S d , respectively. Let Ω ⊂ R d be a domain with a smooth boundary ∂Ω divided into three measurable disjoint parts 1 , 2 and 3 with meas( 1 ) > 0.
A generic point in Ω ∪ will be denoted by u = (u i ) and ν = ν i represents the unit outward normal to . We use the standard notation for Sobolev and Lebesgue spaces corresponding to Ω and . In particular, we use the spaces L 2 (Ω) d , L 2 ( 2 ) d , L 2 ( 3 ) and H 1 (Ω) d , endowed with their canonical inner products and associated norms. Furthermore, for an element ν ∈ H 1 (Ω) d we still write ν for the trace of ν to . We also considered the space which is a real Hilbert space with canonical inner product and the associated norm · V . Here and below ε represents the deformation operator, i.e., where an index that follows a comma denotes the partial derivative with respect to the corresponding component of u, e.g., The assumption that meas( 1 ) > 0 allows us to apply Korn's inequality which results in the completeness of the space V. We denote by 0 V the zero element of V, and for an element ν ∈ V, the normal and tangential components on are defined by respectively. Recall the trace inequality where d 0 denotes the positive constant. For the sake of convenience, we use the data F , p, f 0 , f 2 , µ and κ to satisfy the following conditions.
There exist ξ F > 0 and ς F such that There exist α F > 0 and β F ≥ 0 such that There exists ξ p > 0 such that Moreover, we use Y for the product space L 2 (Ω) d × L 2 ( 3 ) d equipped with the canonical inner product, and K for the set defined by So there is the inequality problem we consider in order to find x ∈ K such that Here, the elastic body Ω, which is subjected to external forces, is fixed on 1 and in frictional contact with 3 . The contact takes place with a layer of deformable material of thickness κ. The elasticity operator is denoted by F , and the density of applied body forces and traction acting on the body and the surface is denoted by f 0 and f 2 , 2 , respectively, p is a given function that defines the deformable material's reaction, and µ represents the coefficient of friction. Next, we consider the constants π 0 , π 2 , π 3 and a function θ such that We associate to (94) the set of admissible pairs V c ad and the cost functional L given by Furthermore, we consider the optimal control problem of finding ( Next, we take a look at a function q and a constantκ satisfy the following conditions.
There exists ξ q > 0 such that We introduce the setK = {y ∈ V : y ν ≤κ on 3 } and we assume that for each n ∈ N the functions f 0n , f 2n , θ n and the constant ζ n are given and satisfy the following conditions: Now, for each n ∈ N, we consider the following perturbation of (94) for finding x n ∈K such that The problem (107) is a variational formulation of the contact problem where the rigid body is covered by a layer of deformable material of thicknessκ. Here, the layer is divided into two parts: the first layer is located on the top of the rigid body with a thicknessκ − κ > 0 and the second layer is located above with a thickness κ. Since ζ n is the deformability coefficient of the first layer, therefore 1 ζ n denotes its stiffness coefficient, and q is a normal compliance function of the first layer.
With the problem (107), we associate the set of admissible pairs V cn ad and the cost function L n given by the solution of (107) converges to the solution of (94), i.e., (b) (98) has at least one solution and, for each n ∈ N, (109) has at least one solution. Moreover, if and {(x n , f n )} is a sequence of solutions of (109), there exists a subsequence of the sequence {(x n , f n )}, again denoted by {(x n , f n )}, and a solution (x , f ) of (98), such that Proof. First, we denote by γ : V −→ Y the operator y ιy, ζ 2 y , where ι : V −→ L 2 (Ω) d is the canonic embedding and ρ 2 : V −→ L 2 ( 2 ) d is the restriction to the trace map to 2 . Next, we take the operators N : V × V −→ V, G : V × V −→ V, the function  : V × V −→ R and the element f ∈ Y defined as follows: Then it is clear that x ∈ K is a solution of (94) if and only if For each n ∈ N, x n ∈K is a solution of (107) if and only if We can now continue with the proof of the two parts of the theorem.

A Stationary Heat Transfer Problem with Unilateral Constraints
Applying the abstract results of Sections 2 and 4, in this section we describes a heat transfer boundary value problem. The classical formulation of the following problem for finding a temperature field x : Ω −→ R such that x = 0, a.e. in 1 , (123) Here, Ω is a bounded domain in R d (d = 1, 2, 3) with smooth boundary ∂Ω = 1 ∪ 2 ∪ 3 and outer normal unit ν. Assume that 1 , 2 , 3 are disjoint measurable sets and, moreover, meas( 1 ) > 0. We do not mention the dependence of the different functions on the spatial variable x ∈ Ω ∪ ∂Ω. Let f be a internal energy function, b be the prescribed temperature field on 2 and q be the heat flux prescribed on 3 . Furthermore, ∂x ∂y denotes the normal derivative of x on 3 . For the variational analysis of (122)-(125) we consider the space there exists y 0 ∈ V such that 0 ≤ y 0 ∈ Ω and y 0 = b on 2 .
We introduce the set Then, the variational formulation of Equations (122)-(125), to obtained through standard arguments for finding x ∈ K such that Now, we introduce the set of admissible pairs for inequality (129) defined by Moreover, we consider two constants ω, and a function ψ such that Now we associate to (129) with above data, for finding optimal control problem (x , f ) ∈ V t ad such that Next, we introduce the setK = {y ∈ V : y ≥ 0 in Ω}.
For each n ∈ N, we assume that the functions f n , ψ n and the constants ζ n , ω n , n , are given and satisfy the following conditions: ζ n > 0, ω n > 0, n > 0, ψ n ∈ L 2 (Ω).
Then, for each n ∈ N, we consider the following perturbation of (129) for finding It is easy to see that (136) represents the variational formulation of the following boundary value problem for finding a temperature field x n : Ω −→ R such that x n = 0 a.e. on 1 , The set of admissible pairs for inequality (136) is defined by Furthermore, the associated optimal control problem for finding (x n , f n ) ∈ V tn ad such that the solution of problem (136) converges to the solution of problem (129), i.e., (b) problem (132) has at least one solution and, for each n ∈ N, problem (142) has at least one solution. Moreover, the solution of problem (132) is unique if and, for each n ∈ N, the solution of (142) is unique, if ψ n = 0 L 2 (Ω) .
(c) Assume that and let {(x n , f n )} be a sequence of solutions of (142). Then, there exists a subsequence of the sequence {(x n , f n )}, again denoted by {(x n , f n )}, and a solution (x , f ) of (132), such that f n f ∈ L 2 (Ω), x n → x ∈ V as n −→ ∞.
Proof. To begin, we will introduce some notation that will allow us to write the problem in an equivalent way. To this end, Let γ : V −→ L 2 (Ω) be the canonical inclusion of V in L 2 (Ω). Moreover, we consider the operators N : Then, x is a solution of (129) if and only if Moreover, for each n ∈ N, x n ∈K is a solution of (136) if and only if N (x, x), y − x V + 1 ζ n G (x n , x n ), y − x n V ≥ f n , y − x n L 2 (Ω) , ∀y ∈K .
Moreover, for each n ∈ N, (x n , f n ) is a solution of (132) if and only if (x n , f n ) ∈ V tn ad and L n (x n , f n ) = min (x , f )∈V tn ad L n (x, f ).
We now proceed with the proof of the two parts of the theorem. follows from a strict convexity argument. For any f ∈ L 2 (Ω), let x( f ) denote the solution of the variational inequality in (149). Then, in [32], it was demonstrated that the functional is strictly convex. Hence, the optimal control problem in (153) has a unique solution and the uniqueness of the solution of (142) in the case ψ n = 0 L 2 (Ω) follows from the same argument. Hence, combined with the equivalence results (153) and (154) allows us to conclude the proof of the (b) in Theorem 6. (c) The convergence (146) is a direct consequence of Theorem 4. The convergence (146) of the whole sequence {(x n , f n )} in the case ψ = 0 L 2 (Ω) follows from a standard argument, since in this case (132) has a unique solution.