Existence and uniqueness result for a class of mixed variational problems

Our aim in this paper is to investigate the existence and uniqueness result for a class of mixed variational problems. They are governed by two variational inequalities. By applying the saddle-point theory, we obtain the existence of solutions to mixed variational problems. Finally, some frictional contact problems are given to illustrate our main results.


Introduction
The abstract problem in this paper is a class of mixed variational problems governed by two variational inequalities, with a bilinear function and functional which is convex and lower semicontinuous. Considering such kinds of variational problems sets the functional background in the study of elastic contact problems with unilateral constraints and nonmonotone interface laws. For very recent work, see [1][2][3]. Our results improve the results in [4][5][6][7], which consider a bilinear functional.
Let X be a Hilbert space, Y a reflexive Banach space and be a nonempty, closed and convex subset Y . Let A : X → X, B : X × → R, φ : X → R be given maps to be specified later, and w, f ∈ X be fixed. We consider the mixed variational problems of following forms.
Recently, a lot of work was devoted to the modeling in contact mechanics. Weak formulations of contact problems involve the theory of variational inequalities and the theory of saddle-point problems; see e.g. [7][8][9][10][11][12][13]. The purpose of this paper is to investigate the weak solvability of a unilateral frictionless contact problem using a technique with dual Lagrange multipliers. The weak formulations with dual Lagrange multipliers allow one to write efficient algorithms in order to approximate the weak solutions. In the present work, the behavior of the materials is described by using the subdifferential of a proper, convex, lower semicontinuous functional and the contact is modeled with Signorini's condition with zero gap. The results extend and improve the results obtained in [5,6], where a unilateral frictionless contact model for nonlinearly elastic materials is analyzed.
In this paper, by applying the saddle-point theory, we obtain the existence of solutions to the mixed variational problems. The main results are that Problem 1 has a unique solution (u, λ) ∈ X × and the solution (u, λ) is Lipschitz continuous with respect to f and ω. Then we apply them to find that our model has a unique weak solution and the weak solution is Lipschitz continuous with respect to the data.
The rest of this paper is organized as follows. In Section 2, we will introduce some useful preliminaries and necessary materials. Section 3 is devoted to proving the existence and uniqueness results. In the last section, we give some examples of friction contact problems to illustrate our main results.

Preliminaries
In this section, we will introduce some basic preliminaries which are used throughout this paper. Denote by · X and (·, ·) X the norm and inner product of the Hilbert space X the inner product of X, respectively.
We now recall some elements of convex analysis. For more details, we refer to [10,14,15].

Definition 3 Let
A and B be nonempty sets. A pair (u, λ) ∈ C × D is said to be a saddle point of a functional L : In this paper, we apply the following result to prove our existence result. Theorem 4 Let X, Y be reflexive Banach spaces, and let A ⊂ X, B ⊂ Y be nonempty, closed, convex subsets. Assume that a functional L : C × D → R satisfies the following conditions: Then the functional L has at least one saddle point.
We also recall the following definition.
Definition 5 Let X be a Hilbert space and ϕ : X → R. ϕ is said to be Gâteaux differentiable at u ∈ X, if there exists an element ∇ϕ(u) ∈ X such that The element ∇ϕ(u) which satisfies the relation above is unique and is called the gradient of ϕ at u. ϕ : X → R is said to be Gâteaux differentiable if it is Gâteaux differentiable at every point of X. In this case, the operator ∇ϕ(u) : X → X that maps every element u ∈ X into the element ∇ϕ(u) is called the gradient operator of ϕ. The convexity of Gâteaux differentiable functions can be characterized as follows.
Lemma 6 ( [16,17]) Let X be a Hilbert space and ϕ : X → R be Gâteaux differentiable. Then ϕ is convex if and only if

Existence and uniqueness result
We will make the following hypotheses to obtain the existence and uniqueness result for Problem 1: We will prove the following existence and uniqueness result. Firstly, we give the following equivalence which can be deduced in the proof of Theorem 4 in [6].

Lemma 8 Problem 1 is equivalent to the following problem:
Find (u, λ) ∈ X × such that The proof of Theorem 7 is carried out in several steps. Let η ∈ X be given and consider the following auxiliary problem.
Define now the operator T : X → X by where u η ∈ X denotes the unique solution of Problem 9.

Lemma 10
The operator T has a unique fixed point.
Proof From Lemma 23 in the Appendix we know that Problem 9 has a unique solution (u η , λ η ) ∈ X × . Let η 1 , η 2 ∈ X and (u η 1 , λ η 1 ), (u η 2 , λ η 2 ) ∈ X × be the unique solutions of Problem 9 corresponding to η 1 , η 2 , respectively. From (7) we have Then and hence We use the hypothesis (3)(b) to obtain It is clear that m A ≤ L A and hence 0 < 1 - which implies that the operator T is a contraction. By applying the Banach contraction principle, we deduce that there exists a unique η * ∈ X such that η * = Tη * . This completes the proof of the lemma.
We now have all the ingredients to provide the proof of the main result in this section.
Next, we consider some stability results.

Theorem 12 Assume that
Proof From (1) and (2) we have and hence where k 1 , k 2 are strictly positive real constants. Choosing k 1 , k 2 such that and combining with (17) we get (16).

Contact problems
In this section, we consider some elastic frictional problems to illustrate our main results. The elastic body occupies an open bounded connected set ⊂ R d (d = 1, 2, 3) with boundary = ∂ , assume to be Lipschitz continuous. Assume that consists of three sets 1 , 2 and 3 , with mutually disjoint relatively open sets 1 , 2 and 3 , such that meas( 1 ) > 0.
We use the notation x = (x i ) for a generic point in ∪ and ν = (ν i ) for the outward unit normal at . The indices i, j run between 1 and d and, unless stated otherwise, the summation convention over repeated indices is used. The notation S d stands for the space of second order symmetric tensors on R d . On the spaces R d and S d we use the inner products and the Euclidean norms defined by respectively. For a vector field, notation u ν and u τ represent the normal and tangential components of u on given by u ν = u · ν and u τ = uu ν ν. Also, σ ν and σ τ represent the normal and tangential components of the stress field σ on the boundary, i.e. σ ν = (σ ν) · ν and σ τ = σ νσ ν ν. The classical model for the process is as follows.

Problem 13
Find a displacement field u : → R d and a stress field σ : → S d such that Div σ + f 0 = 0 in , σ ν = f 2 on 2 , Now, we describe (18)-(23) in the following. Equation (18) represents the elastic constitutive law. Equation (19) is the equation of equilibrium, where f 0 denotes the density of the body forces, (20) is the displacement homogeneous boundary condition which means that the body is fixed on 1 , and (21) is the traction boundary condition with surface tractions of density f 2 acting on 2 . Conditions (22) and (23), given on the contact surface 3 , represent the contact and the friction law, respectively. Here ζ is the friction bound.
In the rest of the paper we use standard notation for Lebesgue and Sobolev spaces and, in addition, we use the spaces V and H defined by Here and below we still denote by v the trace of an element v ∈ H 1 ( ; R d ). The space H will be endowed with the Hilbertian structure given by the inner product and the associated norm · H . On the space V we consider the inner product and the associated norm · V . Recall that, since meas( 1 ) > 0, it follows that V is a real Hilbert space. Moreover, by the Sobolev trace theorem, we have where γ is the norm of the trace operator γ : V → L 2 ( 3 ; R d ).
In the study of Problem 13 we assume that the following conditions hold: Finally, we assume that the densities of body forces and surface tractions have the regularity We define an operator A : for all u, v ∈ V . Moreover, we define an element f ∈ V * by for all v ∈ V . Next, we introduce the space of admissible displacement fields U defined by From Proposition 2.1 and Remark 2.2 in [5] we know that γ (U) can be organized as a Hilbert space. Let D T be the dual space of γ (U). We define λ ∈ D T by where ·, · T denotes the duality pairing between D T and γ (U). Furthermore, we define Let us introduce the following subset of D T : Obviously, λ ∈ . Moreover, from (23) it follows that Then, combining (18)-(23), from Problem 4.1 in [5], Problem 13 can be written as the following variational formulation: Now, we give the following existence, uniqueness and dependence results.

Problem 16
Find a displacement field u : → R d and a stress field σ : → S d such that Div σ + f 0 = 0 in , σ ν = f 2 on 2 , Let D S be the dual space of γ (V ). We define λ ∈ D S by where ·, · S denotes the duality pairing between D S and γ (V ). Furthermore, we define Let us introduce the following subset of D T : Obviously, λ ∈ . Moreover, from (31) it follows that Then Problem 16 can be written as the following variational formulation.
We have the following result which is analogous to Theorem 6.2 in [5] and Theorem 15.

Problem 19
Find a displacement field u : → R d and a stress field σ : → S d such that Div σ + f 0 = 0 in , σ ν = f 2 on 2 , Here g : 3 → R + denotes the gap between the deformable body and the foundation, measured along the outward normal τ .
We keep (32) and (33). We assume that there exists g ext : → R such that g ext ∈ H 1 ( ), γ g ext = 0 a.e. on 1 , γ g ext ≥ 0 a.e. on \ 1 , g = γ g ext a.e. on \ 3 , where γ : H 1 ( ) → L 2 ( ) is the well-known Sobolev trace operator. Moreover, we assume that the unit outward normal to 3  Then Problem 19 can be written as the following variational formulation: We have the following result which is analogous to Theorem 6.3 in [5] and Theorem 15.

Conclusion
In this paper, we study a class of mixed variational problems governed by two variational inequalities with dual Lagrange multipliers. We prove the existence and uniqueness of solution to the mixed variational problem and apply it to obtain the solutions to some frictional contact problems whose weak formulation can be transferred to the mixed variational problem. We point out that considering such kinds of mixed variational problems sets the functional background in the study of many frictional contact problems. To conclude, our results improve many results with bilinear cases and can be further studied. where Lemma 22 Problem 9 has a solution (u η , λ η ) ∈ X × if and only if (u η , λ η ) is a saddle point of the functional L η .
Proof For the necessity, let (u η , λ η ) ∈ X × be a solution of Problem 9. Inequality (8) implies that Moreover, by combining (7) and (37), we have Therefore, For the sufficiency, let (u η , λ η ) ∈ X × be a saddle point of functional L η . Note that which implies (8). Next, we will prove (7). Since we use (37) to see that, for all v ∈ X, Then, since ϕ is convex, we obtain Dividing by t and passing to the limit as t → 0, we get (7). This completes the proof of the lemma.
Proof By (5), it is easy to verify that the map v → L η (v, μ) is convex and l.s.c. for all μ ∈ , and μ → L η (v, μ) is concave and u.s.c. for all v ∈ X. Since φ is convex and l.s.c., it admits an affine minorant, see e.g. [18,Proposition 5.2.25], that is, there are l ∈ X * and b ∈ R such that Then we have Therefore, Next, we will prove that From Lemma 4.2 and Corollary 4.6 in [19], for every μ ∈ the following inequality: has a unique solution u μ ∈ X. It is easy to verify that Taking v = 0 X in (39), we obtain By Lemma 8 we deduce that inequality (39) is equivalent to the following variational equation: From (4)(b) we deduce that and combining with (41) we obtain It follows from (5) that Therefore, there exists c > 0 such that Then, combining (40) and (42), we deduce that there exists c > 0 such that Since μ ∈ is arbitrary, passing to the limit as μ Y → ∞ we get (38). By applying Theorem 4 we deduce that the functional L has at least one saddle point, and then we conclude Problem 9 has at least one solution by applying Lemma 22. Finally, we will show the uniqueness of the solution. In fact, let (u 1 η , λ 1 η ), (u 2 η , λ 2 η ) ∈ X × be two solutions of (41). Then we have Choosing v = u 2 ηu 1 η in the above equation, we get Combining (5)(c), (43) and (44), we conclude that u 1 η = u 2 η . Moreover, we have B v, λ 1 ηλ 2 η = 0, ∀v ∈ X.