On a class of generalized saddle-point problems arising from contact mechanics

In the present paper we consider a class of generalized saddle-point problems described by means of the following variational system: a(u,v−u)+b(v−u,λ)+j(v)−j(u)+J(u,v)−J(u,u)≥(f,v−u)X,b(u,μ−λ)−ψ(μ)+ψ(λ)≤0,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document} $$\begin{aligned} &a(u,v-u)+b(v-u,\lambda )+j(v)-j(u)+J(u,v)-J(u,u)\geq (f,v-u)_{X}, \\ &b(u,\mu -\lambda )-\psi (\mu )+\psi (\lambda )\leq 0, \end{aligned}$$ \end{document} (v∈K⊆X\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$v\in K\subseteq X$\end{document}, μ∈Λ⊂Y\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\mu \in \Lambda \subset Y$\end{document}), where (X,(⋅,⋅)X)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$(X,(\cdot,\cdot )_{X})$\end{document} and (Y,(⋅,⋅)Y)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$(Y,(\cdot,\cdot )_{Y})$\end{document} are Hilbert spaces. We use a fixed-point argument and a saddle-point technique in order to prove the existence of at least one solution. Then, we obtain uniqueness and stability results. Subsequently, we pay special attention to the case when our problem can be seen as a perturbed problem by setting ψ(⋅)=ϵψ¯(⋅)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\psi (\cdot )=\epsilon \bar{\psi}(\cdot )$\end{document}(ϵ>0)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$(\epsilon >0)$\end{document}. Then, we deliver a convergence result for ϵ→0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\epsilon \to 0$\end{document}, the case ψ≡0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\psi \equiv 0$\end{document} appearing like a limit case. The theory is illustrated by means of examples arising from contact mechanics, focusing on models with multicontact zones.


Introduction
In the present paper we bring the attention to the following mixed variational problem.
Problem 1 Given f ∈ X, find (u, λ) ∈ K × such that, for all v ∈ K and μ ∈ , We are going to study this problem under the following hypotheses: (h 1 ) (X, (·, ·) X , · X ) and (Y , (·, ·) Y , · Y ) are Hilbert spaces. (h 2 ) The form a : X × X → R is symmetric, bilinear continuous (of rank M a > 0) and X-elliptic (of rank m a > 0). (2) There exists α > 0 such that (h 4 ) The functional j : X → [0, ∞) is convex and Lipschitz continuous (of rank L j > 0). (h 5 ) The functional ψ : Y → [0, ∞) is convex and lower semicontinuous. When ψ ≡ 0, we assume that there exist c ψ > 0 and q > 1 such that ψ(μ) ≥ c ψ μ q Y for all μ ∈ Y . (h 6 ) • K ⊆ X is an unbounded closed convex subset such that 0 X ∈ K and ⊂ Y is a bounded closed convex subset such that 0 Y ∈ or • K ⊆ X is a linear subspace and ⊂ Y is an unbounded closed convex subset such that The functional J : X × X → [0, ∞) satisfies: (1) For each v ∈ X, J(v, ·) is convex and Lipschitz continuous of rank L J (v) > 0.
(2) There exists M J > 0 such that, for all v 1 , v 2 , w 1 , w 2 ∈ X, J(v 1 , w 2 ) -J(v 1 , w 1 ) + J(v 2 , w 1 ) - (3) There exists c J > 0 such that |J(u, v)| ≤ c J ( u X + 1) v X for all u, v ∈ X. Problem 1 can be seen as a generalization of the following mixed variational problem: Such a variational system appears in the weak formulations of some contact problems: see, e.g., [10] for unilateral contact problems or [11] for a class of bilateral contact problems. The unique solution of the problem (2) and (3) is the unique saddle point of the functional By analyzing contact models with two-contact zones, one can arrive at a variational system of the form below: given f ∈ X, find (u, λ) ∈ X × such that, for all v ∈ X and μ ∈ ⊂ Y , see, e.g., [15]. According to Theorem 2 in [15], the problem (4) and (5) has at least one solution (u, λ) that is unique in its first component; each solution is a saddle point of the functional By considering the perturbed problem one can associate the functional see the recent paper [3].
In the present paper, we focus on models with multicontact zones, arriving at variational formulations that can be cast in the more general form below.
Problem 2 Given f ∈ X, find (u, λ) ∈ K × such that, for all v ∈ K and μ ∈ , Problem 2 can be seen as a particular case of Problem 1, by setting ψ ≡ 0. Problem 1 is interesting in its own right and, even more complicated at first glance than Problem 2 due to the additional term "-ψ(μ) + ψ(λ)", actually, it brings us a significant advantage: assuming that ψ is strictly convex then its solution is unique in both components. In contrast, Problem 2 has at least one solution, unique only in its first component.
By setting ψ(·) = ψ (·) with > 0, Problem 1 can be seen as a perturbed problem. If ψ is strictly convex then the unique solution of the "perturbed" problem is the unique saddle point of a bifunctional that is strictly convex in its first argument and strictly concave in the second one. When goes to zero, then the case ψ ≡ 0 appears like a limit case. Thus, numerical reasons enhance our interest to study the case ψ ≡ 0.
In order to prove the well-posedness of Problem 1 under the hypotheses (h 1 )-(h 7 ), we use a fixed-point argument, a saddle-point technique, arguments in the theory of variational-quasivariational inequalities as well as a minimization argument.
The present paper brings a contribution to the literature devoted to the saddle-point problems and their applications in mechanics; references relevant to the matter are, for instance, [1,2,8,10,11,17].
To increase the clarity of the presentation, we provide below some tools we need. Then: • the set of the saddle points of L, For a proof, see Proposition 1.5 on page 169 in [6].
be two Hilbert spaces. Assume (6), (7), and (8). Moreover, we assume that Then, the functional L has at least one saddle point.
For the proof, see Proposition 2.4 on page 176 in [6]. The rest of the paper has the following structure. In Sect. 2 we deliver some auxiliary results. In Sect. 3 we study the well-posedness of Problem 1. In Sect. 4 we obtain a convergence result. In Sect. 5 we give examples of X, Y , a, b, j, J, K, , and ψ such that the working hypotheses are fulfilled; these examples are related to some weak formulations via Lagrange multipliers for a class of contact models with multicontact zones.

Auxiliary results
In this section we consider the following auxiliary problem.
Let us associate to Problem 3 the following functional: is concave and upper semicontinuous for all v ∈ X.
Since K ⊆ X is unbounded, we have to verify that Indeed, let μ * = 0 Y . We write, As j(·), J(·, ·) are nonnegative, Passing to the limit as v X → ∞, we obtain (9). If is unbounded, then we have to justify that keeping in mind that K ⊆ X is a linear subspace, according to (h 6 ). Indeed, let μ ∈ and let u μ ς ∈ K be the unique solution of the variational inequality of the second kind, or, equivalently, for all v ∈ K , where f μ ∈ X is defined by Riesz's representation theorem as follows, As u μ ς minimizes the functional Let us set v = 0 X in (11). Therefore, Then, we use the inf-sup property of b(·, ·) to write Therefore, As a result, there exists c = c(α, m a , M a , L j , L J (ς)) > 0 such that We observe that (10) is verified after we pass to the limit in the previous inequality. Thus, we conclude (i) by applying Proposition 2.
(ii) Let (u ς , λ ς ) ∈ K × be a solution of Problem 3. It is easy to observe that the second line of Problem 3 is equivalent with On the other hand, keeping in mind the symmetry of a(·, ·), for all v ∈ K , Therefore, (u ς , λ ς ) is a saddle point of L ς . Conversely, if (u ς , λ ς ) ∈ K × is a saddle point of L ς , keeping in mind (14), it is enough to prove that the saddle point (u ς , λ ς ) verifies the first line of Problem 3. To start, we write L ς (u ς , λ ς ) -L ς (w, λ ς ) ≤ 0 for all w ∈ K. Thus, for all w ∈ K , Subsequently, we divide by t > 0 and then we pass to the limit as t → 0. As a result, the pair (u ς , λ ς ) verifies the first line of Problem 3.

Proposition 4 Assume (h 1 )-(h 7 ).
If, in addition, ψ is strictly convex, then the functional L ς has a unique saddle point.
Proof According to Proposition 3, the functional L ς has at least one saddle point. How- We use now Proposition 1 to complete the proof.

Proposition 5
Assume (h 1 )-(h 7 ). Then, Problem 3 has at least one solution that is unique in its first argument. If, in addition, ψ is strictly convex then Problem 3 has a unique solution (u ς , λ ς ) ∈ K × .
Proof We apply Proposition 3. Furthermore, as v → L ς (v, μ) is strictly convex for all μ ∈ we deduce that Problem 3 has at least one solution that is unique in its first argument. If, in addition, ψ is strictly convex then we will apply Proposition 4.

Well-posedness results
To start, we deliver an existence, uniqueness, and stability result by using a Banach fixedpoint argument.
Theorem 1 Assume that (h 1 )-(h 7 ) hold true. If, in addition M J < m a , then Problem 1 has at least one solution that is unique and stable in its first component.
Proof Let ς ∈ K and let (u ς , λ ς ) ∈ K × be a solution of Problem 3. We define an operator According to Proposition 5 the operator T is well defined. Let us prove that T is a contraction. To start, we take ς 1 , ς 2 ∈ K and we denote by (u ς 1 , λ ς 1 ) and (u ς 2 , λ ς 2 ) two corresponding solutions of Problem 3. Thus, for every i ∈ {1, 2} we can write, for all v ∈ K and By setting v = u ς 2 and μ = λ ς 2 if i = 1 and v = u ς 1 and μ = λ ς 1 if i = 2, then, and from this we deduce that Therefore, As M J < m a , we conclude that T is a contraction. Thus, we can apply Banach's fixed-point theorem.
Let ς * be the unique fixed point of T. It is easy to observe that the pair (u ς * , λ ς * ) ∈ K × is a solution of Problem 1.
Let (u i , λ i ) ∈ K × (i ∈ {1, 2}) be two solutions of Problem 1. Thus, for all v ∈ K and μ ∈ , Setting v = u 2 and μ = λ 2 if i = 1 and v = u 1 and μ = λ 1 if i = 2, then we are led to As M J < m a , we obtain u 1 = u 2 .
Let f 1 , f 2 ∈ X be two data and let (u 1 , λ 1 ), (u 2 , λ 2 ) be two corresponding solutions. Then, for each i ∈ {1, 2} we can write, Setting v = u 2 , μ = λ 2 if i = 1 and v = u 1 , μ = λ 1 if i = 2 then, As M J < m a we deduce that As a result, the solution is stable in its first component.
Therefore, λ 1 and λ 2 are minimizers for the following functional Keeping in mind the properties of ψ(·) and b(·, ·), by using a standard minimization argument, see, e.g., Theorem 1.36 in [21], it is easy to observe that H u has a unique minimizer. Note that, when is unbounded, the property ψ(μ) ≥ c ψ μ q Y (q > 1) is crucial in order to obtain the coercivity of H u . In consequence, Problem 1 has a unique solution.
By passing to the limit as → 0 in the relation above, we obtain that u → u 0 . Due to the uniqueness of the limit, we conclude that u = u 0 . Because (λ ) is a weakly convergent sequence and 0 ≤ψ(λ ) ≤cψ λ p Y (p > 0), then ψ (λ ) → 0 as → 0. Therefore, keeping in mind the working hypotheses and passing to the limit as → 0 in Problem 4 we conclude that (u 0 , λ) ∈ K × verifies Problem 2. Thus, we can considerλ 0 = λ.
Remark 3 The entire sequence (u ) >0 is strongly convergent to u 0 when → 0 because the unique limit u 0 is independent of the subsequences (u ) >0 .

Examples
In this section we are going to illustrate the theory by means of examples of contact models involving multicontact zones.
Let us consider an elastic body that occupies a bounded domain ⊂ R 3 , with a Lipschitz continuous boundary. The boundary, denoted by , is partitioned in three measurable parts D , N , and C with positive measure. The body is clamped on D , body forces of density f 0 act on and surface traction of density f 2 acts on N . On the part C the body is or can be in contact (frictional or frictionless) with foundations or obstacles. The mathematical model can be described by means of the following boundary value problem.
As usual, u = (u i ) stands for the displacement field, ε = ε(u) = (ε ij (u)) denotes the infinitesimal strain tensor, E is the elastic tensor and σ = (σ ij ) is the Cauchy stress tensor in the linearized theory. By means of¯ we denote ∪ ∂ . Herein, S 3 denotes the space of second-order symmetric tensors on R 3 . Every field in R 3 or S 3 is typeset in boldface. Everywhere below, by · and : we will denote the inner product on R 3 and S 3 , respectively, and by · R 3 , · S 3 we will denote the Euclidean norm on R 3 and S 3 , respectively. The unit outward normal vector to the boundary is denoted by ν and is defined almost everywhere. The normal and the tangential components of the displacement field will be denoted by u ν = u · ν and u τ = uu ν ν, respectively; the normal and the tangential components of the Cauchy vector σ ν on the boundary will be given by the formulas σ ν = (σ ν) · ν, σ τ = σ νσ ν ν.