Variational Approaches for Contact Models with Multi-Contact Zones

Abstract

We consider a contact model with two contact zones, for linearly elastic materials, under the small deformation hypothesis. We pay attention to four possible variational formulations: one of them is a variational inequality of the second kind and the other three are mixed variational problems governed by variational inequalities on convex sets of Lagrange multipliers. We study the existence and the uniqueness of the solution for each of the four variational formulations. Some connections between these four weak formulations are also discussed. Our approach requires a background knowledge in the variational inequalities theory as well as in the saddle point theory.

1 Introduction

In the present paper, we draw attention to the possibility to deliver alternative variational formulations for contact models with multi-contact zones. Thus, depending on the motivation we have, one of several possible variational formulations can be picked. To exemplify, we consider a contact model with two contact zones, as follows.

Problem 1

Find ${\varvec{u}}:{\bar{\Omega }}\rightarrow {\mathbb {R}}^3$ and ${\varvec{\sigma }}:{\bar{\Omega }}\rightarrow {\mathbb {S}}^3$, such that

$$\begin{aligned}&{\text {Div}}{\varvec{\sigma }}+{\varvec{f}}_0=\mathbf{0}&\, \text{ in } \Omega ,\\&{\varvec{\sigma }}={{\mathcal {E}}}{\varvec{\varepsilon }}({\varvec{u}})&\, \text{ in } \Omega ,\\&{\varvec{u}}={\varvec{0}}&\, \text{ on } \Gamma _1,\\&{\varvec{\sigma }}\,{\varvec{\nu }}={\varvec{f}}_2&\, \text{ on } \Gamma _2,\\&{u}_{\nu }=0,\,\,\Vert {{\varvec{\sigma }}}_\tau \Vert _{{\mathbb {S}}^3}\le g,\,\,{{\varvec{\sigma }}}_\tau =-g\,\frac{{{\varvec{u}}}_\tau }{\Vert {{\varvec{u}}}_\tau \Vert _{{\mathbb {R}}^3}} \text{ if } {\varvec{u}}_{\tau }\ne {\varvec{0}}&\, \text{ on } \Gamma _3,\\&{{\varvec{\sigma }}}_{\tau }={\varvec{0}},\,\, {\sigma }_{\nu }\le 0,\,\, {u}_{\nu } \le 0,\,\, {\sigma }_{\nu }\, {u}_{\nu } =0&\, \text{ on } \Gamma _4, \end{aligned}$$

where ${\varvec{u}}=(u_i)$ is the displacement field and ${\varvec{\sigma }}=(\sigma _{ij})$ is the Cauchy stress tensor. Here, $\Omega \subset {\mathbb {R}}^{3}$ is a bounded domain with smooth boundary $\Gamma$ partitioned in four measurable parts, $\Gamma _{1},$ $\Gamma _2$, $\Gamma _{3},$ $\Gamma _4,$ such that all the parts have positive measure; ${\bar{\Omega }}=\Omega \cup \Gamma .$ We denote by ${\varvec{f}}_0:\Omega \rightarrow {\mathbb {R}}^3$ the density of the volume forces, by ${\varvec{f}}_2:\Gamma _2\rightarrow {\mathbb {R}}^3$ the density of the surface traction, by ${\varvec{\varepsilon }}={\varvec{\varepsilon }}({\varvec{u}})=(\varepsilon _{ij}({\varvec{u}}))$ the infinitesimal strain tensor, $\varepsilon _{ij}({\varvec{u}})=\frac{1}{2}\left( \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)$ for all $i, j\in \{1,2,3\}$ and by ${\mathcal {E}}$ the elastic tensor. As usual, ${\varvec{\nu }}$ is the outward unit normal vector to the boundary $\Gamma$. Also, by $\cdot$, we denote the inner product on ${\mathbb {R}}^3$, and by  : , we denote the inner product on ${\mathbb {S}}^3$, where ${\mathbb {S}}^3$ is the space of second-order symmetric tensors of ${\mathbb {R}}^3$ and by $\Vert \cdot \Vert _{{\mathbb {R}}^3}$ and $\Vert \cdot \Vert _{{\mathbb {S}}^3}$ we denote the Euclidean norm on ${\mathbb {R}}^3$ and ${\mathbb {S}}^3,$ respectively. The operator ${\text {Div}}$ is the divergence of a tensor, $Div \, {\varvec{\sigma }}=(\sum _{j=1}^{3}\frac{\partial \sigma _{ij}}{\partial x_j})$, $i\in \{1,2,3\}$.

Problem 1 models the deformation of a body in contact with two rigid foundations. On $\Gamma _3$, the contact is frictional bilateral with a positive friction bound g, and on $\Gamma _4$, we have a frictionless unilateral contact condition. It is worth to underline that on $\Gamma _3$, we do not know a priori in which points the friction force vanishes, while $\Gamma _4$ is a potential contact zone, because we do not know a priori in which points we have contact. Recall that $u_\nu ={\varvec{u}}\cdot {\varvec{\nu }}$, ${\varvec{u}}_{{\varvec{\tau }}}={\varvec{u}}-u_\nu \,{\varvec{\nu }}$, $\sigma _\nu =({\varvec{\sigma }}{\varvec{\nu }})\cdot {\varvec{\nu }}$, ${\varvec{\sigma }}_{\tau }={\varvec{\sigma }}{\varvec{\nu }}-\sigma _\nu \,{\varvec{\nu }}$, ${\varvec{\sigma }}{\varvec{\nu }}\cdot {\varvec{u}}=\sigma _\nu \,u_\nu +{\varvec{\sigma }}_{{\varvec{\tau }}}\cdot \,{\varvec{u}}_{{\varvec{\tau }}}$. For a background on the mathematical theory of contact mechanics models, see, e.g., [7, 13, 18].

If our interest consists only in computing the displacement field ${\varvec{u}}$, then we can choose the primal variational formulation consisting in a variational inequality of the second kind, or, for a more efficient approximation of the weak solution, we can choose a mixed variational formulation consisting of a variational equation and a variational inequality.

If, in addition to ${\varvec{u}}$, we are interested to compute the friction force ${\varvec{\sigma }}_{\tau },$ then a mixed variational formulation governed by a Lagrange multiplier related to ${\varvec{\sigma }}_{\tau }$ can be helpful. Or, if we are interested to compute the reaction of the foundation $-\sigma _{\nu }$, then a mixed variational formulation governed by a Lagrange multiplier related to $\sigma _{\nu }$ becomes convenient. In the last two cases, the weak formulations consist of systems of two variational inequalities.

The solvability of the weak formulations we deliver is based on the theory of variational inequalities and requires some elements of the saddle point theory. For a background on the variational inequalities of the second kind, the reader can consult, e.g., [18, 19], and for useful elements in the theory of the saddle point theory related to the solvability of mixed variational problems, we refer to, e.g., [1, 2, 6, 8]. Relevant to the matter are also the papers [4, 5, 11, 12].

The present paper is structured as follows. In Sect. 2, we review four abstract results related to a variational inequality of the second kind and three saddle point problems. In Sect. 3, we introduce the functional setting and the working hypotheses. In Sect. 4, we deliver four weak formulations. Section 5 is devoted to the weak solvability of these four variational formulations paying attention to the connection between them.

2 Preliminaries

Everywhere in this section $(X, (\cdot , \cdot )_X, \Vert \cdot \Vert _X)$ and $(Y, (\cdot , \cdot )_Y, \Vert \cdot \Vert _Y)$ are Hilbert spaces. We make the following assumptions:

A 1

$a:X\times X \rightarrow {\mathbb {R}}$ is a symmetric, bilinear form, such that

• (i) there exists $M_{a}> 0: |a(u,v)|\le M_{a}\,\Vert u\Vert _{X}\,\Vert v\Vert _{X}\,\,\,\,\,\,\,\text{ for } \text{ all }\,\,u,v\in X$;

• (ii) there exists $m_{a}> 0: a(v,v)\ge m_{a}\,\Vert v\Vert ^{2}_{X}\,\,\,\,\,\,\,\text{ for } \text{ all }\,\,v\in X$.

A 2

$b:X\times Y \rightarrow {\mathbb {R}}$ is a bilinear form, such that

• (a) there exists $M_{b}> 0: |b(v,\mu )|\le M_{b}\,\Vert v\Vert _{X}\,\Vert \mu \Vert _{Y}$ for all $v\in X,\,\mu \in Y$;

• (b) there exists $\alpha > 0:$ $\inf _{\mu \in Y, \mu \ne 0_{Y}}\,\sup _{v\in X, v\ne 0_{X}}\,\displaystyle \frac{b(v,\mu )}{\Vert v\Vert _{X}\,\Vert \mu \Vert _{Y}}\ge \alpha .$

A 3

$\phi :X\rightarrow {\mathbb {R}}_+$ is a convex functional. Moreover, $\phi$ is a Lipschitz continuous functional, i.e., there exists $L_\phi >0$:

$$\begin{aligned} |\phi (v)-\phi (w)|\le L_\phi \,\Vert v-w\Vert _X\,\,\, \text{ for } \text{ all }\,\,\, v,w\in X. \end{aligned}$$

A 4

$\Lambda$ is a closed, convex subset of Y that contains $0_{Y}$.

A 5

K is a closed, convex subset of X that contains $0_{X}$.

First, we focus on the following variational inequality of the second kind.

Problem 2

Given $f\in X$, find $u\in K$, such that

$$\begin{aligned} a(u, v-u)+\phi (v)-\phi (u)\ge (f, v-u)_X\quad \text{ for } \text{ all }\,\,\,v\in K\subseteq X. \end{aligned}$$

Theorem 1

The assumptions A 1, A 3 and A 5 hold true. Then, Problem 2 has a unique solution, $u\in K$.

For a proof of Theorem 1, the reader can consult, e.g., Theorem 3.1 in [19].

Notice that the unique solution of Problem 2 is the unique minimum of the functional

$$\begin{aligned}J:K\rightarrow {\mathbb {R}},\quad J(v)=\frac{1}{2}a(v,v)+\phi (v)-(f,v)_X.\end{aligned}$$

To proceed, we review three useful saddle point problems, Problems 3,4, 5, below.

Problem 3

Given $f\in X$, find $u\in X$ and $\lambda \in \Lambda \subseteq Y$, such that

$$\begin{aligned} a(u,v)+b(v,\lambda )= & {} (f,v)_X \quad \,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ for } \text{ all }\,\, v\in X\\ b(u,\mu -\lambda )\le & {} 0\,\,\,\,\,\quad \qquad \qquad \text{ for } \text{ all }\,\, \mu \in \Lambda . \end{aligned}$$

Theorem 2

The assumptions A 1, A 2 and A 4 hold true. Then, Problem 3 has a unique solution, $(u, \lambda )\in X\times \Lambda$.

For a proof, we send the reader to, e.g., Corollary 2 in [4].

Notice that a pair $(u, \lambda )\in X\times \Lambda$ is a solution of Problem 3 if and only if it is a saddle point of the following functional:

$$\begin{aligned}{\mathcal {L}}_1:X\times \Lambda \rightarrow {\mathbb {R}},\quad {\mathcal {L}}_1(v,\mu )=\frac{1}{2}a(v,v)-(f,v)_X+b(v,\mu );\end{aligned}$$

this motivates us to consider Problem 3 a saddle point problem.

Problem 4

Given $f \in X$, find $u\in K\subseteq X$ and $\lambda \in \Lambda \subseteq Y$, such that

$$\begin{aligned} a(u,v-u)+b (v-u,\lambda )\ge & {} (f,v-u)_X \,\,\,\,\,\,\,\,\,\,\text{ for } \text{ all }\,\, v\in K \\ b(u,\mu -\lambda )\le & {} 0\,\,\quad \,\,\,\quad \quad \quad \quad \,\text{ for } \text{ all }\,\, \mu \in \Lambda . \end{aligned}$$

Theorem 3

The assumptions A 1, A 2 (a), A 4 and A 5 hold true. If, in addition, $\Lambda \subseteq Y$ is a bounded subset, then Problem 4 has a solution $(u,\lambda )\in K\times \Lambda$, unique in its first component.

For details, see, e.g., Remark 3 in [4].

Notice that a pair $(u, \lambda )\in K\times \Lambda$ is a solution of Problem 4 if and only if it is a saddle point of the following functional:

$$\begin{aligned}{\mathcal {L}}_2:K\times \Lambda \rightarrow {\mathbb {R}},\quad {\mathcal {L}}_2(v,\mu )=\frac{1}{2}a(v,v)-(f,v)_X+b(v,\mu ).\end{aligned}$$

Thus, Problem 4 can be considered a saddle point problem.

Problem 5

Given $f \in X$, find $u\in X$ and $\lambda \in \Lambda \subseteq Y$, such that

$$\begin{aligned} a(u,v-u)+b (v-u,\lambda )+\phi (v)-\phi (u)\ge & {} (f,v-u)_X \qquad \text{ for } \text{ all }\,\, v\in X \\ b(u,\mu -\lambda )\le & {} 0\,\,\quad \,\,\,\quad \quad \quad \quad \,\text{ for } \text{ all }\,\, \mu \in \Lambda . \end{aligned}$$

Theorem 4

The Assumptions A 1–A 4 hold true. Then, Problem 5 has a solution $(u,\lambda )\in X\times \Lambda$, unique in its first component.

For a proof, we send the reader to, e.g., Theorem 3 in [4]. Similar techniques can be found in, e.g., [5, 11].

Notice that a pair $(u, \lambda )\in X\times \Lambda$ is a solution of Problem 5 if and only if it is a saddle point of the following functional:

$$\begin{aligned}{\mathcal {L}}_3:X\times \Lambda \rightarrow {\mathbb {R}},\quad {\mathcal {L}}_3(v,\mu )=\frac{1}{2}a(v,v)+\phi (v)-(f,v)_X+b(v,\mu ).\end{aligned}$$

Hence, Problem 5 is a saddle point problem, too.

3 Functional Setting and Working Hypotheses

To start, we remind some useful Hilbert Lebesgue spaces.

• $L^2(\Omega )^3=\{{\varvec{v}}=(v_i)\,|\,\,v_i\in L^2(\Omega ),\,\,\,1\le i\le 3\}$ endowed with the inner product $({\varvec{u}},{\varvec{v}})_{L^2(\Omega )^3}=\sum _{i=1}^{3}\int _\Omega u_i\,v_i\,\mathrm{d}x=\int _\Omega {\varvec{u}}\cdot {\varvec{v}}\,\mathrm{d}x$ and the associated norm $\Vert {\varvec{v}}\Vert _{L^2(\Omega )^3}=\left( \sum _{i=1}^{3}\int _\Omega v_i\,v_i\,\mathrm{d}x\right) ^{1/2}.$

• $L^2(\Omega )^{3\times 3}=\{{\varvec{\tau }}=({\varvec{\tau }}_{ij})\,|\,\,{\varvec{\tau }}_{ij}\in L^2(\Omega ),\,\,\,1\le i, j\le 3\}$ endowed with $({\varvec{\sigma }}, {\varvec{\tau }})_{L^2(\Omega )^{3\times 3}}=\sum _{i,j=1}^{3}\int _\Omega \sigma _{ij} \,\tau _{ij}\,\mathrm{d}x=\int _\Omega {\varvec{\sigma }}:{\varvec{\tau }}\,\mathrm{d}x$ and the corresponding norm $\Vert {\varvec{\tau }}\Vert _{L^2(\Omega )^{3\times 3}}=\left( \sum _{i,j=1}^{3}\int _\Omega \tau _{ij}\,\tau _{ij}\,\mathrm{d}x\right) ^{1/2}.$

• $L_s^2(\Omega )^{3\times 3}=\{{\varvec{\tau }}=({\varvec{\tau }}_{ij})\,|\,\,{\varvec{\tau }}_{ij}={\varvec{\tau }}_{j\,i}\in L^2(\Omega ),\,\,\,1\le i, j\le 3\},$ endowed with $({\varvec{\sigma }}, {\varvec{\tau }})_{L_s^2(\Omega )^{3\times 3}}=({\varvec{\sigma }}, {\varvec{\tau }})_{L^2(\Omega )^{3\times 3}},$ and the norm $\Vert {\varvec{\sigma }}\Vert _{L_s^2(\Omega )^{3\times 3}}=\Vert {\varvec{\sigma }}\Vert _{L^2(\Omega )^{3\times 3}}.$

Subsequently, we introduce some useful Hilbert Sobolev spaces.

• $H^1(\Omega )^3=\{{\varvec{v}}=(v_i)\,|\,\,\,v_i\in H^1(\Omega ),\,\,\, 1\le i\le 3\}$ endowed with the canonical inner product

  $$\begin{aligned} ({\varvec{u}}, {\varvec{v}})_{H^1(\Omega )^3}= & {} \sum _{i=1}^3 (u_i, v_i)_{L^2(\Omega )}+\sum _{i=1}^3(\nabla u_i,\nabla v_i)_{L^2(\Omega )^3}, \end{aligned}$$

  and the associated norm

  $$\begin{aligned} \Vert {\varvec{v}}\Vert _{H^1(\Omega )^3}= & {} \sqrt{\sum _{i=1}^3 \Vert v_i\Vert ^2_{L^2(\Omega )}+\sum _{i=1}^3 \Vert \nabla v_i\Vert ^2_{L^2(\Omega )^3}}. \end{aligned}$$

  Furthermore, $H^1(\Omega )^3$ can be endowed with the following particular inner product:

  $$\begin{aligned} (({\varvec{u}}, {\varvec{v}}))_{H^1(\Omega )^3}=({\varvec{u}}, {\varvec{v}})_{L^2(\Omega )^3}+({\varvec{\varepsilon }}({\varvec{u}}), {\varvec{\varepsilon }}({\varvec{v}}))_{L^2(\Omega )^{3\times 3}}\quad \text{ for } \text{ all }\,\,{\varvec{u}}, {\varvec{v}}\in H^1(\Omega )^3 \end{aligned}$$

  and the associated norm

  $$\begin{aligned} |||{\varvec{v}}|||_{H^1(\Omega )^3}=(\Vert {\varvec{v}}\Vert _{L^2(\Omega )^3}^2+\Vert {\varvec{\varepsilon }}({\varvec{v}})\Vert _{L^2(\Omega )^{3\times 3}}^2)^{1/2}\quad \text{ for } \text{ all }\,\, {\varvec{v}}\in H^1(\Omega )^3, \end{aligned}$$

  where ${\varvec{\varepsilon }}:H^1(\Omega )^3\rightarrow L_s^2(\Omega )^{3\times 3}$, ${\varvec{\varepsilon }}({\varvec{u}})=\frac{1}{2}(\nabla {\varvec{u}}+\nabla {\varvec{u}}^T)$ is a linear and continuous tensor; see, e.g., [18].

  Recall that there exists a constant $c>0$, such that, for all ${\varvec{v}}\in H^1(\Omega )^3$, we have

  $$\begin{aligned} \sum _{i=1}^3\!\int _\Omega v_i \,v_i\, \mathrm{d}x+\!\sum _{i=1}^3\int _\Omega \nabla v_i\cdot \nabla v_i\, \mathrm{d}x\le \! c\left( \sum _{i=1}^3\int _\Omega v_i\,v_i\,\mathrm{d}x+\int _\Omega {\varvec{\varepsilon }}({\varvec{v}}):{\varvec{\varepsilon }}({\varvec{v}})\,\mathrm{d}x\right) ;\quad \end{aligned}$$

  (1)

  see for instance [9]. Thus, we deduce that $\Vert \cdot \Vert _{H^1(\Omega )^3}$ is equivalent with $|||\cdot |||_{H^1(\Omega )^3}$. Therefore, the space $(H^1(\Omega )^3, ((\cdot ,\cdot ))_{H^1(\Omega )^3}, |||\cdot |||_{H^1(\Omega )^3})$ is a Hilbert space.

• $$\begin{aligned} H^{1/2}(\Gamma )^3= & {} {\varvec{\gamma }}(H^1(\Omega )^3)\\= & {} \{{\varvec{w}}=(w_1, w_2, w_3)^T\,\,|\,\,w_i\in H^{1/2}(\Gamma ),\,\,\,1\le i\le 3\}. \end{aligned}$$

  This space is endowed with the following inner product:

  $$\begin{aligned}&({\varvec{\chi }}, {\varvec{w}})_{H^{1/2}(\Gamma )^3}=\sum _{i=1}^3(\chi _i, w_i)_{H^{1/2}(\Gamma )}\\&\quad =\sum _{i=1}^3(\chi _i,w_i)_{L^2(\Gamma )} +\sum _{i=1}^3\int _{\Gamma }\int _{\Gamma }\frac{(\chi _i({\varvec{x}})-\chi _i({\varvec{y}}))(w_i({\varvec{x}})-w_i({\varvec{y}}))}{\Vert {\varvec{x}}-{\varvec{y}}\Vert ^{3}}ds({\varvec{x}})ds({\varvec{y}}) \end{aligned}$$

  and the corresponding norm

  $$\begin{aligned} \Vert {\varvec{w}}\Vert _{H^{1/2}(\Gamma )^3}=\left( \Vert {\varvec{w}}\Vert ^2_{{L^2}\,({\Gamma )}^3}+ \int _{\Gamma }\int _{\Gamma }\frac{({\varvec{w}}({\varvec{x}})-{\varvec{w}}({\varvec{y}}))^2}{\Vert {\varvec{x}}-{\varvec{y}}\Vert ^{3}}ds({\varvec{x}})ds({\varvec{y}})\right) ^{1/2}. \end{aligned}$$

  For more details about the Hilbert spaces $H^{1/2}(\Gamma )$ and $H^{1/2}(\Gamma )^3$, the reader can consult, e.g., [7, 14]. Relevant to the matter are also, e.g., [10, 15].

Recall that ${\varvec{\gamma }}:H^1(\Omega )^3\rightarrow L^2(\Gamma )^3$ is a linear, continuous and compact operator and ${\varvec{\gamma }}:H^1(\Omega )^3\rightarrow H^{1/2}(\Gamma )^3$ is a linear, continuous and surjective operator. Furthermore, there exists a linear, continuous operator ${\varvec{l}}:H^{1/2}(\Gamma )^3\rightarrow H^1(\Omega )^3$, such that ${\varvec{\gamma }}({\varvec{l}}({\varvec{\xi }}))={\varvec{\xi }}\quad \text{ for } \text{ all }\,\,\,{\varvec{\xi }}\in H^{1/2}(\Gamma )^3.$ The operator ${\varvec{l}}$ is called the right inverse of the trace operator ${\varvec{\gamma }}$.

To proceed, we introduce useful closed subspaces of the space $H^1(\Omega )^3.$

• $X_0=\{{\varvec{v}}\in H^1(\Omega )^3\,\, |\,\,{\varvec{\gamma }}\, {\varvec{v}}={\varvec{0}} \text{ a.e. } \text{ on } \Gamma _{1}\},$ $meas(\Gamma _1)>0$; see, e.g., [18]. Recall that there exists $c_K=c_K(\Omega ,\Gamma _1)>0$, such that

  $$\begin{aligned} \Vert {\varvec{\varepsilon }}({\varvec{v}})\Vert _{L^2(\Omega )^{3\times 3}}\ge c_K|||{\varvec{v}}|||_{H^1(\Omega )^3}\quad \text{ for } \text{ all }\,\,\,{\varvec{v}}\in X_0. \end{aligned}$$

  (2)

  This is the Korn’s inequality and for a proof of it the reader can consult, e.g., [16]. Let us introduce the following inner product:

  $$\begin{aligned} ({\varvec{u}}, {\varvec{v}})_{X_0}=\int _\Omega {\varvec{\varepsilon }}({\varvec{u}}):{\varvec{\varepsilon }}({\varvec{v}})\,\mathrm{d}x=({\varvec{\varepsilon }}({\varvec{u}}), {\varvec{\varepsilon }}({\varvec{v}}))_{L^2(\Omega )^{3\times 3}} \end{aligned}$$

  with the corresponding norm

  $$\begin{aligned} \Vert {\varvec{v}}\Vert _{X_0}=\left( \int _\Omega {\varvec{\varepsilon }}({\varvec{v}}):{\varvec{\varepsilon }}({\varvec{v}})\,\mathrm{d}x\right) ^{1/2}=\Vert {\varvec{\varepsilon }}({\varvec{v}})\Vert _{L^2(\Omega )^{3\times 3}}. \end{aligned}$$

  Using Korn’s inequality, we get the equivalence between $|||\cdot |||_{H^1(\Omega )^3}$ and $\Vert \cdot \Vert _{X_0}$. Therefore, $(X_0, (\cdot ,\cdot )_{X_0}, \Vert \cdot \Vert _{X_0})$ is a Hilbert space. Furthermore, there exists a positive constant $c_0$, such that

  $$\begin{aligned} \Vert {\varvec{\gamma }}{\varvec{v}}\Vert _{L^2(\Gamma _3)^3}\le c_0\Vert {\varvec{v}}\Vert _{X_0}\quad \text{ for } \text{ all }\,\,\,{\varvec{v}}\in X_0. \end{aligned}$$

  (3)

• Let us introduce a subspace of the space $X_0$ as follows:

  $$\begin{aligned} X=\{\,{\varvec{v}}\in X_0\,\, |\,\, v_\nu =0 \text{ a.e. } \text{ on } \Gamma _3 \}, \end{aligned}$$

  (4)

  where $v_\nu ={\varvec{\gamma }}\,{\varvec{v}}\cdot {\varvec{\nu }}$. As it is known, the space $(X, (\cdot ,\cdot )_{X_0}, \Vert \cdot \Vert _{X_0})$ is a Hilbert space; see, e.g., [18].

• $S={\varvec{\gamma }}(X)=\{{\varvec{w}}= {\varvec{\gamma }}{\varvec{v}}\quad \text{ a.e. } \text{ on }\,\,\,\Gamma ,\,\,\, {\varvec{v}}\in X\}.$ We sent the reader to, e.g., Proposition 2.1 in [12] for a proof of the fact that ${\varvec{\gamma }}(X_0)$ and ${\varvec{\gamma }}(X)$ are closed subspaces of the Hilbert space $H^{1/2}(\Gamma )^3$. Therefore, $(S, (\cdot ,\cdot )_{H^{1/2}(\Gamma )^3}, \Vert \cdot \Vert _{H^{1/2}(\Gamma )^3})$ is a Hilbert space.

• $Y=S\,',$ Y being the dual of S.

• We also need a convex subset of X as follows:

  $$\begin{aligned} K=\{{\varvec{v}}\in X\,|\,\,\,v_\nu \le 0 \,\,\text { a.e. on}\,\,\Gamma _4\}. \end{aligned}$$

  (5)

  The set K is a nonempty, unbounded, closed, convex subset of X containing $0_X$.

In the study of Problem 1, we admit the following hypotheses:

H 1

${\mathcal {E}}=({\mathcal {E}}_{ijkl}):{\mathbb {S}}^3\rightarrow {\mathbb {S}}^3$ is a fourth-order tensor, such that

(a):

${\mathcal {E}}{\varvec{\sigma }}:{\varvec{\tau }}={\varvec{\sigma }}:{\mathcal {E}}{\varvec{\tau }}\quad \text{ for } \text{ all }\,\,\,{\varvec{\sigma }}, {\varvec{\tau }}\in {\mathbb {S}}^3;$

(b):

$\text{ there } \text{ exists } \, m_{{\mathcal {E}}}>0$ such that ${{\mathcal {E}}}{\varvec{\tau }}:{\varvec{\tau }}\ge m_{{{\mathcal {E}}}} \Vert {\varvec{\tau }}\Vert _{{\mathbb {S}}^3}^2\quad \text{ for } \text{ all }\,\,\,{\varvec{\tau }}\in {\mathbb {S}}^3.$

H 2

${\varvec{f}}_{0}\in L^2(\Omega )^3$, ${\varvec{f}}_{2}\in L^{2}(\Gamma _{2})^{3},$ $g>0$.

4 Weak Formulations

Let ${\varvec{u}}$ and ${\varvec{\sigma }}$ be smooth enough functions which verify Problem 1. For all ${\varvec{v}}\in X$, we have

$$\begin{aligned} \int _\Omega {\text {Div}} {\varvec{\sigma }}\cdot {\varvec{v}}\,\mathrm{d}x+\int _\Omega {\varvec{f}}_0\cdot {\varvec{v}}\, \mathrm{d}x=0. \end{aligned}$$

Using a Green formula for tensors, see, e.g., page 89 in [18] and taking into account the boundary conditions, we get

$$\begin{aligned} \int _\Gamma {\varvec{\sigma }}{\varvec{\nu }}\cdot {\varvec{\gamma }}{\varvec{v}}\,\mathrm{d}\Gamma -\int _\Omega {\varvec{\sigma }}:{\varvec{\varepsilon }}({\varvec{v}})\,\mathrm{d}x+\int _\Omega {\varvec{f}}_0\cdot {\varvec{v}}\,\mathrm{d}x=0\quad \text{ for } \text{ all } {\varvec{v}}\in X. \end{aligned}$$

Notice that ${\varvec{\sigma }}{\varvec{\nu }}\cdot {\varvec{\gamma }}{\varvec{v}}=\sigma _\nu \,v_\nu +{\varvec{\sigma }}_{{\varvec{\tau }}}\cdot {\varvec{v}}_{{\varvec{\tau }}}={\varvec{\sigma }}_{{\varvec{\tau }}}\cdot {\varvec{v}}_{{\varvec{\tau }}}$ on $\Gamma _3$. Furthermore, since ${\varvec{v}}_{{\varvec{\tau }}}={\varvec{\gamma }}{\varvec{v}}-v_\nu {\varvec{\nu }}={\varvec{\gamma }}{\varvec{v}}$ on $\Gamma _3$, we obtain that ${\varvec{\sigma }}{\varvec{\nu }}\cdot {\varvec{\gamma }}{\varvec{v}}={\varvec{\sigma }}_{{\varvec{\tau }}}\cdot {\varvec{\gamma }}{\varvec{v}}$ on $\Gamma _3$. Consequently

$$\begin{aligned} \int _{\Gamma _3} {\varvec{\sigma }}{\varvec{\nu }}\cdot {\varvec{\gamma }}{\varvec{v}}\,\mathrm{d}\Gamma =\int _{\Gamma _3} {\varvec{\sigma }}_{{\varvec{\tau }}}\cdot {\varvec{\gamma }}{\varvec{v}}\,\mathrm{d}\Gamma \quad \text{ for } \text{ all }\,\,\,{\varvec{v}}\in X. \end{aligned}$$

On the other hand, ${\varvec{\sigma }}{\varvec{\nu }}\cdot {\varvec{\gamma }}{\varvec{v}}=\sigma _\nu \,v_{\nu }$ on $\Gamma _4$. Therefore

$$\begin{aligned} \int _{\Gamma _4} {\varvec{\sigma }}{\varvec{\nu }}\cdot {\varvec{\gamma }}{\varvec{v}}\,\mathrm{d}\Gamma =\int _{\Gamma _4} \sigma _\nu {\varvec{\nu }}\cdot {\varvec{\gamma }}{\varvec{v}}\,\mathrm{d}\Gamma \quad \text{ for } \text{ all }\,\,\,{\varvec{v}}\in X. \end{aligned}$$

Then, for all ${\varvec{v}}\in X$, we obtain

$$\begin{aligned}&\int _\Omega {\mathcal {E}}{\varvec{\varepsilon }}({\varvec{u}}):{\varvec{\varepsilon }}({\varvec{v}}) \mathrm{d}x-\int _{\Gamma _3} {\varvec{\sigma }}_{{\varvec{\tau }}}\cdot {\varvec{\gamma }}{\varvec{v}}\, \mathrm{d}\Gamma -\int _{\Gamma _4} \sigma _\nu {\varvec{\nu }}\cdot {\varvec{\gamma }}{\varvec{v}}\,\, \mathrm{d}\Gamma \nonumber \\&\quad =\int _\Omega {\varvec{f}}_0\cdot {\varvec{v}}\,\mathrm{d}x+\int _{\Gamma _2} {\varvec{f}}_2\cdot {\varvec{\gamma }}{\varvec{v}}\, \mathrm{d}\Gamma . \end{aligned}$$

(6)

By (6), for all ${\varvec{v}}\in X$, we can write

$$\begin{aligned}&\int _\Omega {\mathcal {E}}{\varvec{\varepsilon }}({\varvec{u}}):({\varvec{\varepsilon }}({\varvec{v}})-{\varvec{\varepsilon }}({\varvec{u}})) \mathrm{d}x-\int _{\Gamma _3} {\varvec{\sigma }}_{{\varvec{\tau }}}\cdot ({\varvec{\gamma }}{\varvec{v}}-{\varvec{\gamma }}{\varvec{u}}) \mathrm{d}\Gamma \nonumber \\&\quad -\int _\Omega {\varvec{f}}_0\cdot ({\varvec{v}}-{\varvec{u}}) \mathrm{d}x-\int _{\Gamma _4} \sigma _\nu {\varvec{\nu }}\cdot ({\varvec{\gamma }}{\varvec{v}}-{\varvec{\gamma }}{\varvec{u}}) \mathrm{d}\Gamma =\int _{\Gamma _2} {\varvec{f}}_2\cdot ({\varvec{\gamma }}{\varvec{v}}-{\varvec{\gamma }}{\varvec{u}}) \mathrm{d}\Gamma . \end{aligned}$$

(7)

Let us define a bilinear form as follows:

$$\begin{aligned} a:X\times X\rightarrow {\mathbb {R}},\quad a({\varvec{u}},\,{\varvec{v}})=\int _{\Omega }{{\mathcal {E}}}\,{\varvec{\varepsilon }}({\varvec{u}}): {\varvec{\varepsilon }}({\varvec{v}})\,\mathrm{d}x\quad \text{ for } \text{ all } \,\,\, {\varvec{u}},{\varvec{v}}\in X. \end{aligned}$$

(8)

The form a is well defined. Indeed

$$\begin{aligned} \int _\Omega {\mathcal {E}}\,{\varvec{\varepsilon }}({\varvec{u}})({\varvec{x}}): {\varvec{\varepsilon }}({\varvec{v}})({\varvec{x}})\,\mathrm{d}x=({\mathcal {N}}_{{\mathcal {E}}}({\varvec{\varepsilon }}({\varvec{u}})), {\varvec{\varepsilon }}({\varvec{v}}))_{L^2(\Omega )^{3\times 3}}, \end{aligned}$$

where ${\mathcal {N}}_{{\mathcal {E}}}:L^2(\Omega )^{3\times 3}\rightarrow L^2(\Omega )^{3\times 3},\quad {\mathcal {N}}_{{\mathcal {E}}}({\varvec{\varepsilon }}({\varvec{u}}))({\varvec{x}})={\mathcal {E}}({\varvec{\varepsilon }}({\varvec{u}})({\varvec{x}}))$ for a.e. ${\varvec{x}}\in \Omega$ is the Nemytskii operator; see, e.g., page 370 in [17].

In addition, by the Riesz’s representation theorem, we define ${\varvec{f}}\in X$ as follows:

$$\begin{aligned} ({\varvec{f}},{\varvec{v}})_{X}=\int _{\Omega }\,{\varvec{f}}_0\cdot {\varvec{v}}\,\mathrm{d}x+\int _{\Gamma _2}\,{\varvec{f}}_2\cdot {\varvec{\gamma }}{\varvec{v}}\,\mathrm{d}\Gamma \,\,\,\text{ for } \text{ all }\,\,\,{\varvec{v}}\in X. \end{aligned}$$

(9)

4.1 The First Weak Formulation

Due to the boundary conditions on $\Gamma _4,$ by taking ${\varvec{v}}\in K$, we can write

$$\begin{aligned} -\int _{\Gamma _4} \sigma _\nu {\varvec{\nu }}\cdot ({\varvec{\gamma }}{\varvec{v}}-{\varvec{\gamma }}{\varvec{u}}) \,\mathrm{d}\Gamma =-\int _{\Gamma _4} \sigma _\nu {\varvec{\nu }}\cdot {\varvec{\gamma }}{\varvec{v}}\,\mathrm{d}\Gamma \le 0. \end{aligned}$$

Based on relations (7), (8) and (9), for all ${\varvec{v}}\in K$, we get

$$\begin{aligned} a({\varvec{u}},{\varvec{v}}-{\varvec{u}})-\int _{\Gamma _3} {\varvec{\sigma }}_{{\varvec{\tau }}}\cdot ({\varvec{\gamma }}{\varvec{v}}-{\varvec{\gamma }}{\varvec{u}}) \,\mathrm{d}\Gamma\ge & {} ({\varvec{f}},{\varvec{v}}-{\varvec{u}})_{ X}. \end{aligned}$$

(10)

Let us define a functional as follows:

$$\begin{aligned} \phi :X\rightarrow {\mathbb {R}}_+,\,\, \quad \phi ({\varvec{v}})=\int _{\Gamma _3}\,g\,\Vert {\varvec{v}}_{\tau }\Vert _{{\mathbb {R}}^3}\,\mathrm{d}\Gamma . \end{aligned}$$

(11)

We observe that for all ${\varvec{v}}\in K$

$$\begin{aligned}&-\int _{\Gamma _3} {\varvec{\sigma }}_{{\varvec{\tau }}}\cdot {\varvec{\gamma }}{\varvec{v}}\mathrm{d}\Gamma \le \int _{\Gamma _3}\Vert {\varvec{\sigma }}_{{\varvec{\tau }}}\Vert _{{\mathbb {S}}^3}\Vert {\varvec{\gamma }}{\varvec{v}}\Vert _{{\mathbb {R}}^3} \mathrm{d}\Gamma \le \int _{\Gamma _3} g\Vert {\varvec{v}}_{{\varvec{\tau }}}\Vert _{{\mathbb {R}}^3} \mathrm{d}\Gamma =\phi ({\varvec{v}}),\\&-\int _{\Gamma _3} {\varvec{\sigma }}_{{\varvec{\tau }}}\cdot {\varvec{\gamma }}{\varvec{u}}\mathrm{d}\Gamma = \int _{\Gamma _3} g\Vert {\varvec{u}}_{{\varvec{\tau }}}\Vert _{{\mathbb {R}}^3} \mathrm{d}\Gamma =\phi ({\varvec{u}}). \end{aligned}$$

As a result

$$\begin{aligned} \phi ({\varvec{v}})-\phi ({\varvec{u}})\ge -\int _{\Gamma _3} {\varvec{\sigma }}_{{\varvec{\tau }}}\cdot ({\varvec{\gamma }}{\varvec{v}}-{\varvec{\gamma }}{\varvec{u}}) \mathrm{d}\Gamma . \end{aligned}$$

Therefore, we arrive at the following weak formulation.

Problem 6

Given ${\varvec{f}}\in X$, find ${\varvec{u}}\in K\subset X$, such that

$$\begin{aligned} a({\varvec{u}},{\varvec{v}}-{\varvec{u}})+\phi ({\varvec{v}})-\phi ({\varvec{u}})\ge ({\varvec{f}},{\varvec{v}}-{\varvec{u}})_{X} \,\,\,\,\,\text{ for } \text{ all }\,\, {\varvec{v}}\in K\subset X. \end{aligned}$$

4.2 The Second Weak Formulation

Let us introduce the Lagrange multiplier ${\varvec{\lambda }}_2\in X'$ as follows:

$$\begin{aligned} ({\varvec{\lambda }}_2, {\varvec{v}})_{X',X} =-\int _{\Gamma _3} {\varvec{\sigma }}_{{\varvec{\tau }}}\cdot {\varvec{\gamma }}{\varvec{v}}\,\mathrm{d}\Gamma -\int _{\Gamma _4} \sigma _\nu v_{\nu }\, \mathrm{d}\Gamma \quad \text{ for } \text{ all }\,\,\,{\varvec{v}}\in X, \end{aligned}$$

where $(\cdot , \cdot )_{X',X}$ denotes the duality pairing between $X'$ and X;  herein and everywhere below, $X'$ stands for the dual of X.

Let us introduce the following set of the Lagrange multipliers:

$$\begin{aligned} \Lambda _2=\{{\varvec{\mu }}\in X'\,\,| \,\,\,({\varvec{\mu }},{\varvec{v}})_{X',X}\le \int _{\Gamma _3} g\Vert {\varvec{v}}_{\tau }\Vert _{{\mathbb {R}}^3} \,\mathrm{d}\Gamma \,\,\,\text{ for } \text{ all }\,\,\,{\varvec{v}}\in K\}. \end{aligned}$$

Keeping in mind the boundary conditions on $\Gamma _3$ and $\Gamma _4,$ we immediately deduce that ${\varvec{\lambda }}_2\in \Lambda _2.$ Afterwards, we define a bilinear form

$$\begin{aligned} {\widetilde{b}}:X\times X'\rightarrow {\mathbb {R}},\quad \quad {\widetilde{b}}({\varvec{v}},{\varvec{\mu }})=({\varvec{\mu }}, {\varvec{v}})_{X',X}\quad \text{ for } \text{ all } \,\,\,{\varvec{v}}\in X,\, {\varvec{\mu }}\in X'. \end{aligned}$$

(12)

According to (6), we can write

$$\begin{aligned} a({\varvec{u}},{\varvec{v}})+{\widetilde{b}}({\varvec{v}},{\varvec{\lambda }}_2)=({\varvec{f}},{\varvec{v}})_{X} \quad \text{ for } \text{ all } {\varvec{v}}\in X, \end{aligned}$$

where the forms a and ${\widetilde{b}}$ were defined in (8) and (12), respectively, and ${\varvec{f}}$ was introduced in (9). On the other hand

$$\begin{aligned} {\widetilde{b}}({\varvec{u}},{\varvec{\mu }})\le \int _{\Gamma _3} g\Vert {\varvec{u}}_{{\varvec{\tau }}}\Vert _{{\mathbb {R}}^3} \,\mathrm{d}\Gamma \quad \text{ for } \text{ all } {\varvec{\mu }}\in \Lambda _2, \end{aligned}$$

and

$$\begin{aligned} {\widetilde{b}}({\varvec{u}}, {\varvec{\lambda }}_2)= & {} ({\varvec{\lambda }}_2, {\varvec{u}})_{X',X}\\= & {} -\int _{\Gamma _3} {\varvec{\sigma }}_{{\varvec{\tau }}}\cdot {\varvec{\gamma }}{\varvec{u}}\, \mathrm{d}\Gamma -\int _{\Gamma _4}\sigma _{\nu }u_{\nu }\,\mathrm{d}\Gamma \\= & {} \int _{\Gamma _3} g\Vert {\varvec{u}}_{{\varvec{\tau }}}\Vert _{{\mathbb {R}}^3} \mathrm{d}\Gamma . \end{aligned}$$

As a consequence

$$\begin{aligned} {\widetilde{b}}({\varvec{u}},{\varvec{\mu }}-{\varvec{\lambda }}_2)\le 0\quad \text{ for } \text{ all }\,\,\,{\varvec{\mu }}\in \Lambda _2. \end{aligned}$$

(13)

Therefore, we can state the following variational formulation.

Problem 7

Given ${\varvec{f}}\in X$, find ${\varvec{u}}\in X$ and ${\varvec{\lambda }}_2\in \Lambda _2\subset X'$, such that

$$\begin{aligned} a({\varvec{u}},{\varvec{v}})+{\widetilde{b}}({\varvec{v}},{\varvec{\lambda }}_2)= & {} ({\varvec{f}},{\varvec{v}})_{X} \quad \qquad \qquad \text{ for } \text{ all }\,\, {\varvec{v}}\in X \\ {\widetilde{b}}({\varvec{u}},{\varvec{\mu }}-{\varvec{\lambda }}_2)\le & {} {\varvec{0}}\qquad \,\,\quad \qquad \qquad \text{ for } \text{ all }\,\, {\varvec{\mu }}\in \Lambda _2. \end{aligned}$$

4.3 The Third Weak Formulation

Let us define the Lagrange multiplier ${\varvec{\lambda }}_3\in Y$

$$\begin{aligned} \langle {\varvec{\lambda }}_3, {\varvec{w}}\rangle =-\int _{\Gamma _3}\,{\varvec{\sigma }}_\tau \,\cdot {\varvec{w}}\,\mathrm{d}\Gamma \,\,\text{ for } \text{ all }\,\,\,{\varvec{w}}\in S; \end{aligned}$$

herein and everywhere below, $\langle \cdot ,\cdot \rangle$ denotes the duality product between Y and S.

We also define a set of Lagrange multipliers as follows:

$$\begin{aligned} \Lambda _3=\{{\varvec{\mu }}\in Y\,\,| \,\,\langle {\varvec{\mu }}, {\varvec{w}}\rangle \le \int _{\Gamma _3}\,g\,\Vert {\varvec{w}}\Vert _{{\mathbb {R}}^3}\,\mathrm{d}\Gamma \,\,\,\text{ for } \text{ all }\,\,\,{\varvec{w}}\in S\}. \end{aligned}$$

(14)

Due to the boundary conditions on $\Gamma _3$, we deduce that ${\varvec{\lambda }}_3\in \Lambda _3$.

We define now a bilinear form as follows:

$$\begin{aligned} b:X\times Y\rightarrow {\mathbb {R}},\quad \quad b({\varvec{v}},{\varvec{\mu }})=\langle {\varvec{\mu }}, {\varvec{\gamma }}{\varvec{v}}\rangle \quad \text{ for } \text{ all } \,\,\,{\varvec{v}}\in X, {\varvec{\mu }}\in Y. \end{aligned}$$

(15)

According to (10), we can write

$$\begin{aligned} a({\varvec{u}},{\varvec{v}}-{\varvec{u}})+b({\varvec{v}}-{\varvec{u}},{\varvec{\lambda }}_3)\ge ({\varvec{f}},{\varvec{v}}-{\varvec{u}})_{X}\quad \text{ for } \text{ all } \,\, {\varvec{v}}\in K, \end{aligned}$$

(16)

where a and b are the forms introduced in (8) and (15), ${\varvec{f}}\in X$ is the element defined in (9), and K is the set defined in (5).

Furthermore

$$\begin{aligned} b({\varvec{u}},{\varvec{\mu }})\le \int _{\Gamma _3} g\Vert {\varvec{u}}_{{\varvec{\tau }}}\Vert _{{\mathbb {R}}^3}\, \mathrm{d}\Gamma \quad \text{ for } \text{ all } {\varvec{\mu }}\in \Lambda _3. \end{aligned}$$

On the other hand

$$\begin{aligned} b({\varvec{u}}, {\varvec{\lambda }}_3)=\langle {\varvec{\lambda }}_3, {\varvec{\gamma }}{\varvec{u}}\rangle =-\int _{\Gamma _3} {\varvec{\sigma }}_{{\varvec{\tau }}}\cdot {\varvec{\gamma }}{\varvec{u}}\,\mathrm{d}\Gamma =\int _{\Gamma _3} g\Vert {\varvec{u}}_{{\varvec{\tau }}}\Vert _{{\mathbb {R}}^3} \,\mathrm{d}\Gamma . \end{aligned}$$

As a consequence

$$\begin{aligned} b({\varvec{u}},{\varvec{\mu }}-{\varvec{\lambda }}_3)\le 0\quad \text{ for } \text{ all }\,\,\,{\varvec{\mu }}\in \Lambda _3. \end{aligned}$$

(17)

By (16) and (13), we arrive at the following weak formulation.

Problem 8

Given ${\varvec{f}}\in X$, find ${\varvec{u}}\in K\subset X$ and ${\varvec{\lambda }}_3\in \Lambda _3\subset Y$, such that

$$\begin{aligned} a({\varvec{u}},{\varvec{v}}-{\varvec{u}})+b({\varvec{v}}-{\varvec{u}},{\varvec{\lambda }}_3)\ge & {} ({\varvec{f}},{\varvec{v}}-{\varvec{u}})_{X}\qquad \quad \text{ for } \text{ all }\,\, {\varvec{v}}\in K \\ b({\varvec{u}},{\varvec{\mu }}-{\varvec{\lambda }}_3)\le & {} {\varvec{0}}\quad \quad \qquad \qquad \quad \,\, \text{ for } \text{ all }\,\, {\varvec{\mu }}\in \Lambda _3. \end{aligned}$$

4.4 The Fourth Weak Formulation

Let us consider a Lagrange multiplier ${\varvec{\lambda }}_4\in Y$, such that

$$\begin{aligned} \langle {\varvec{\lambda }}_4, {{\varvec{w}}}\rangle =-\int _{\Gamma _4}\,\sigma _{\nu }\,{\varvec{w}}\cdot {\varvec{\nu }}\,\mathrm{d}\Gamma \quad \text{ for } \text{ all } {\varvec{w}}\in S \end{aligned}$$

and the set of the Lagrange multipliers as follows:

$$\begin{aligned} \Lambda _4=\{{\varvec{\mu }}\in Y\,\,|\,\,\langle {\varvec{\mu }},\,{\varvec{w}}\rangle \le 0\quad \text{ for } \text{ all } {\varvec{w}}\in S \text{ such } \text{ that } {\varvec{w}}\cdot {\varvec{\nu }}\le 0 \text{ a.e. } \text{ on } \Gamma _4\}. \end{aligned}$$

Since $\sigma _\nu \le 0$ a.e. on $\Gamma _4$, it results that ${\varvec{\lambda }}_4\in \Lambda _4.$

Keeping in mind (7), (8), (9) (11) and (15), we get

$$\begin{aligned} a({\varvec{u}},{\varvec{v}}-{\varvec{u}})+b({\varvec{v}}-{\varvec{u}},{\varvec{\lambda }}_4)+\phi ({\varvec{v}})-\phi ({\varvec{u}})\ge ({\varvec{f}},{\varvec{v}}-{\varvec{u}})_{X} \,\,\,\,\,\text{ for } \text{ all }\,\, {\varvec{v}}\in X.\nonumber \\ \end{aligned}$$

(18)

Furthermore

$$\begin{aligned} b({\varvec{u}},{\varvec{\mu }})\le 0\quad \text{ for } \text{ all } {\varvec{\mu }}\in \Lambda _4 \end{aligned}$$

and

$$\begin{aligned} b({\varvec{u}},{\varvec{\lambda }}_4)=\langle {\varvec{\lambda }}_4,{\varvec{\gamma }}{\varvec{u}}\rangle =-\int _{\Gamma _4}\sigma _\nu {\varvec{\gamma }}{\varvec{u}}\cdot {\varvec{\nu }}\, \mathrm{d}\Gamma =0. \end{aligned}$$

As a result

$$\begin{aligned} b({\varvec{u}},{\varvec{\mu }}-{\varvec{\lambda }}_4)\le 0\quad \text{ for } \text{ all }\,\,\, {\varvec{\mu }}\in \Lambda _4.\ \end{aligned}$$

(19)

By (18) and (19), we arrive at the following variational problem.

Problem 9

Given ${\varvec{f}}\in X$, find ${\varvec{u}}\in X$ and ${\varvec{\lambda }}_4\in \Lambda _4\subset Y$, such that

$$\begin{aligned} a({\varvec{u}},{\varvec{v}}-{\varvec{u}})+b({\varvec{v}}-{\varvec{u}},{\varvec{\lambda }}_4)+\phi ({\varvec{v}})-\phi ({\varvec{u}})\ge & {} ({\varvec{f}},{\varvec{v}}-{\varvec{u}})_{X} \,\,\,\,\,\text{ for } \text{ all }\,\, {\varvec{v}}\in X \\ b({\varvec{u}},{\varvec{\mu }}-{\varvec{\lambda }}_4)\le & {} 0\,\,\quad \,\,\,\quad \quad \quad \,\,\,\text{ for } \text{ all }\,\, {\varvec{\mu }}\in \Lambda _4, \end{aligned}$$

where a, $\phi$, ${\varvec{f}}$, and b are those introduced in (8), (11), (9), and (15).

5 Main Results

In this section, we focus on the weak solvability of Problem 1 using successively the four variational formulations delivered in Sect. 4. Our first result is the following existence and uniqueness result regarding Problem 6.

Theorem 5

We admit hypotheses H 1 and H 2. Then, Problem 6 has a unique solution, ${\varvec{u}}_1\in K.$

Proof

We note that X defined in (4) is a Hilbert space. Obviously, K defined in (5) is a closed, convex subset of X, such that it contains $0_X$. Therefore, the assumption A 5 holds true.

The form a in (8) is a symmetric and bilinear form. Moreover, (i) in the assumption A 1 holds true. Indeed

$$\begin{aligned} |a({\varvec{u}},{\varvec{v}})|= & {} \left| \int _\Omega {\mathcal {E}} {\varvec{\varepsilon }}({\varvec{u}})({\varvec{x}}):{\varvec{\varepsilon }}({\varvec{v}})({\varvec{x}})\,\mathrm{d}x\right| \\\le & {} \int _\Omega \Vert {\mathcal {E}} {\varvec{\varepsilon }}({\varvec{u}})({\varvec{x}})\Vert _{{\mathbb {S}}^3}\Vert {\varvec{\varepsilon }}({\varvec{v}})({\varvec{x}})\Vert _{{\mathbb {S}}^3}\,\mathrm{d}x\\\le & {} \max _{i,j,k,l}|{\mathcal {E}}_{ijkl}|\int _\Omega \Vert {\varvec{\varepsilon }}({\varvec{u}})({\varvec{x}})\Vert _{{\mathbb {S}}^3}\Vert {\varvec{\varepsilon }}({\varvec{v}})({\varvec{x}})\Vert _{{\mathbb {S}}^3}\,\mathrm{d}x\\\le & {} \max _{i,j,k,l}| {\mathcal {E}}_{ijkl}|\left( \int _\Omega \Vert {\varvec{\varepsilon }}({\varvec{u}})({\varvec{x}})\Vert ^2_{{\mathbb {S}}^3}\right) ^{1/2}\left( \int _\Omega \Vert {\varvec{\varepsilon }}({\varvec{v}})({\varvec{x}})\Vert ^2_{{\mathbb {S}}^3}\,\mathrm{d}x\right) ^{1/2}\\= & {} \max _{i,j,k,l}|{\mathcal {E}}_{ijkl}|\,\Vert {\varvec{u}}\Vert _{X_0}\,\Vert {\varvec{v}}\Vert _{X_0}. \end{aligned}$$

Thus, we can choose $M_a=\max _{i,j,k,l}|{\mathcal {E}}_{ijkl}|.$

To prove (ii) in the assumption A 1, we evaluate

$$\begin{aligned} a({\varvec{v}}, {\varvec{v}})= & {} \int _\Omega {\mathcal {E}} {\varvec{\varepsilon }}({\varvec{v}})({\varvec{x}}):{\varvec{\varepsilon }}({\varvec{v}})({\varvec{x}})\,\mathrm{d}x\ge \int _\Omega m_{\mathcal {E}} \Vert {\varvec{\varepsilon }}({\varvec{v}})({\varvec{x}})\Vert _{{\mathbb {S}}^3}^2\,\mathrm{d}x\\= & {} m_{\mathcal {E}} \Vert {\varvec{\varepsilon }}({\varvec{v}})\Vert _{L^2(\Omega )^{3\times 3}}^2=m_{\mathcal {E}} \Vert {\varvec{v}}\Vert ^2_{X_0}. \end{aligned}$$

We can take $m_a=m_{\mathcal {E}}.$ Therefore, the assumption A 1 holds true.

Furthermore, the assumption A 3 is fulfilled. It is obviously that the functional $\phi$ in (11) is convex. Let us show that $\phi$ is a Lipschitz continuous functional. We write

$$\begin{aligned} |\phi ({\varvec{v}})-\phi ({\varvec{u}})|= & {} \left| \int _{\Gamma _3} (g\Vert {\varvec{v}}_{{\varvec{\tau }}}({\varvec{x}})\Vert _{{\mathbb {R}}^3}-g\Vert {\varvec{u}}_{{\varvec{\tau }}}({\varvec{x}})\Vert _{{\mathbb {R}}^3})\,\mathrm{d}\Gamma \right| \\\le & {} g\int _{\Gamma _3}\Vert {\varvec{v}}_{{\varvec{\tau }}}({\varvec{x}})-{\varvec{u}}_{{\varvec{\tau }}}({\varvec{x}})\Vert _{{\mathbb {R}}^3}\,\mathrm{d}\Gamma \\\le & {} g\sqrt{meas(\Gamma _3)}\Vert {\varvec{\gamma }}{\varvec{v}}-{\varvec{\gamma }}{\varvec{u}}\Vert _{L^2(\Gamma _3)^3}\\\le & {} c_0\,g\sqrt{meas(\Gamma _3)}\Vert {\varvec{v}}-{\varvec{u}}\Vert _{X_0}, \end{aligned}$$

where $c_0>0$ is the constant in (3). Therefore, we can take $L_\phi =c_0\,g\sqrt{meas(\Gamma _3)}.$

Therefore, we can apply Theorem 1. $\square$

To proceed, we focus on the solvability of Problem 7.

Theorem 6

We admit the hypotheses H 1 and H 2. Then, Problem 7 has a unique solution, $({\varvec{u}}_2,{\varvec{\lambda }}_2)\in X\times \Lambda _2.$

Proof

The assumptions A 1, A 2 and A4 are fulfilled. Hence, we can apply Theorem 2. $\square$

Subsequently, we deliver a characterization of the solution $({\varvec{u}}_2,{\varvec{\lambda }}_2)$ in terms of the unique solution of Problem 6, ${\varvec{u}}_1$.

Let ${\varvec{u}}_1\in K$ be the unique solution of Problem 6. Next, we define ${\varvec{\lambda }}_1\in X'$ as follows:

$$\begin{aligned} ({\varvec{\lambda }}_1,{\varvec{v}})_{X',X}=({\varvec{f}},{\varvec{v}})_{X}-a({\varvec{u}}_1,{\varvec{v}})\quad \text{ for } \text{ all } {\varvec{v}}\in X. \end{aligned}$$

(20)

Proposition 1

If ${\varvec{u}}_1\in K$ is the unique solution of Problem 6 and ${\varvec{\lambda }}_1$ is the element of $X'$ defined in (20), then $({\varvec{u}}_1,{\varvec{\lambda }}_1)$ is the unique solution of Problem 7.

Proof

Setting ${\varvec{v}}={\varvec{0}}$ and ${\varvec{v}}=2{\varvec{u}}_1$, respectively, in Problem 6, we deduce that

$$\begin{aligned} a({\varvec{u}}_1,{\varvec{u}}_1)+\phi ({\varvec{u}}_1)=({\varvec{f}},{\varvec{u}}_1)_{X}. \end{aligned}$$

(21)

And from this, keeping in mind Problem 6

$$\begin{aligned}a({\varvec{u}}_1,{\varvec{v}})+\phi ({\varvec{v}})\ge ({\varvec{f}},{\varvec{v}})_{X} \text{ for } \text{ all } {\varvec{v}}\in K.\end{aligned}$$

As a result

$$\begin{aligned} \phi ({\varvec{v}})\ge ({\varvec{f}},{\varvec{v}})_{X}-a({\varvec{u}}_1,{\varvec{v}}) \text{ for } \text{ all } {\varvec{v}}\in K. \end{aligned}$$

Hence, keeping in mind (20), we immediately observe that ${\varvec{\lambda }}_1\in \Lambda _2.$

On the other hand, by (20), it is straightforward to see that

$$\begin{aligned} a({\varvec{u}}_1,{\varvec{v}})+b({\varvec{v}},{\varvec{\lambda }}_1)= ({\varvec{f}},{\varvec{v}})_{X} \text{ for } \text{ all } {\varvec{v}}\in X. \end{aligned}$$

(22)

Moreover, for all ${\varvec{\mu }}\in \Lambda _2,$ keeping in mind the definition of $\Lambda _2,$ we have

$$\begin{aligned} b({\varvec{u}}_1,{\varvec{\mu }})\le \phi ({\varvec{u}}_1). \end{aligned}$$

(23)

Due to (21), we can write

$$\begin{aligned} \phi ({\varvec{u}}_1)=({\varvec{f}},{\varvec{u}}_1)_{X}-a({\varvec{u}}_1,{\varvec{u}}_1). \end{aligned}$$

Using now (20), we get

$$\begin{aligned} b({\varvec{u}}_1,{\varvec{\lambda }}_1)=\phi ({\varvec{u}}_1). \end{aligned}$$

(24)

By (23) and (24), we get

$$\begin{aligned} b({\varvec{u}}_1,{\varvec{\mu }}-{\varvec{\lambda }}_1)\le 0\quad \text{ for } \text{ all } {\varvec{\mu }}\in \Lambda _2. \end{aligned}$$

(25)

As a consequence, due to (22) and (25), the pair $({\varvec{u}}_1,{\varvec{\lambda }}_1)\in K\times \Lambda _2$ where ${\varvec{u}}_1$ is the unique solution of Problem 6 and ${\varvec{\lambda }}_1$ is defined in (20) is a solution of Problem 7. However, Problem 7 has a unique solution $({\varvec{u}}_2,{\varvec{\lambda }}_2)\in X\times \Lambda _2.$ We conclude that ${\varvec{u}}_2={\varvec{u}}_1$ and ${\varvec{\lambda }}_2={\varvec{\lambda }}_1.$ $\square$

Remark 1

According to Proposition 1, the first component of the unique pair solution of Problem 7 is the unique solution of Problem 6. Also, it is worth to underline that ${\varvec{u}}_2\in K.$

Remark 2

Proposition 1 allows us to give a new characterization of the unique solution of Problem 6. Indeed, the unique solution ${\varvec{u}}_1$ of Problem 6 is the first component of the unique saddle point $({\varvec{u}}_2,{\varvec{\lambda }}_2)$ of the functional

$$\begin{aligned}{\mathcal {L}}:X\times \Lambda _2\rightarrow {\mathbb {R}},\quad {\mathcal {L}}({\varvec{v}},{\varvec{\mu }})=\frac{1}{2}a({\varvec{v}},{\varvec{v}})-({\varvec{f}},{\varvec{v}})_X+b({\varvec{v}},{\varvec{\mu }}). \end{aligned}$$

Afterwards, we pay attention to Problem 8.

Theorem 7

We admit the hypotheses H 1 and H 2. Then, Problem 8 has a solution $({\varvec{u}}_3, {\varvec{\lambda }}_3)\in K\times \Lambda _3,$ unique in its first component.

Proof

The form b defined in (15) is bilinear. To verify (a) in the assumption A 2, we can write

$$\begin{aligned} |b({\varvec{v}}, {\varvec{\mu }})|= & {} |\langle {\varvec{\mu }}, {\varvec{\gamma }}{\varvec{v}}\rangle |\le \Vert {\varvec{\mu }}\Vert _Y\Vert {\varvec{\gamma }}{\varvec{v}}\Vert _{H^{1/2}(\Gamma )^3}\le c_\gamma \Vert {\varvec{\mu }}\Vert _Y\Vert {\varvec{v}}\Vert _{H^1(\Omega )^3}\\\le & {} c_\gamma \,c\,\Vert {\varvec{\mu }}\Vert _Y|||{\varvec{v}}|||_{H^1(\Omega )^3}\le c_K^{-1}\,c_\gamma \,c\,\Vert {\varvec{\mu }}\Vert _Y\Vert {\varvec{v}}\Vert _{X_0}, \end{aligned}$$

where $c_K>0$ is the constant in the Korn’s inequality (2), $c>0$ is the constant in (1) and $c_{\gamma }>0$ in the constant in the trace theorem. We can choose $M_b=c_K^{-1}\,c_\gamma \,c.$

The assumptions A 1, A 2 (a), A 4 and A 5 hold true. We claim that $\Lambda _3$ in (14) is a bounded set. Indeed, let ${\varvec{\mu }}\in \Lambda _3$. Then

$$\begin{aligned} \frac{\langle {\varvec{\mu }}, {\varvec{\gamma }}{\varvec{v}}\rangle }{\Vert {\varvec{v}}\Vert _{X_0}}\le & {} \frac{\int _{\Gamma _3}g\Vert {\varvec{v}}_{{\varvec{\tau }}}\Vert _{{\mathbb {R}}^3}\,\mathrm{d}\Gamma }{\Vert {\varvec{v}}\Vert _{X_0}}\le \frac{g\,\sqrt{meas(\Gamma _3)}\,\Vert {\varvec{\gamma }}{\varvec{v}}\Vert _{L^2(\Gamma _3)^3}}{\Vert {\varvec{v}}\Vert _{X_0}}\\\le & {} c_0\,g\,\sqrt{meas(\Gamma _3)}, \end{aligned}$$

where $c_0>0$ is the constant in (3). Consequently

$$\begin{aligned} \Vert {\varvec{\mu }}\Vert _Y\le c_0\,g\sqrt{meas(\Gamma _3)}\quad \text{ for } \text{ all }\,\,\,{\varvec{\mu }}\in \Lambda _3. \end{aligned}$$

As $\Lambda _3$ is a bounded subset of the Hilbert space Y, we can apply Theorem 3 to obtain the existence of a solution $({\varvec{u}}_3,{\varvec{\lambda }}_3)\in K\times \Lambda _3$ which is unique in its first component. $\square$

Remark 3

We observe that ${\varvec{u}}_3\in K.$ However, the equality ${\varvec{u}}_3={\varvec{u}}_1$ is left open.

Finally, we address Problem 9.

Theorem 8

We admit the hypotheses H 1 and H 2. Then, Problem 9 has a solution $({\varvec{u}}_4,{\varvec{\lambda }}_4)\in X\times \Lambda _4$, unique in its first component.

Proof

Clearly, the set of the Lagrange multipliers $\Lambda _4$ is a closed, convex subset of Y which contains $0_Y$.

In addition, $\Lambda _4$ is an unbounded subset of Y, since there exists a sequence $(\mu _n)_n\subset \Lambda _4$, such that $\Vert \mu _n\Vert _Y\rightarrow \infty$ as $n\rightarrow \infty$. Indeed, we can construct a sequence $(\mu _n)_n$, such that for each positive integer n, $\mu _n=n\,\mu _0$, where $\mu _0\in \Lambda _4$ is defined as follows:

$$\begin{aligned} \langle \mu _0,{\varvec{w}}\rangle =\int _{\Gamma _4}{\varvec{w}}({\varvec{x}})\cdot {\varvec{\nu }}({\varvec{x}})\,\mathrm{d}\Gamma \quad \text{ for } \text{ all }\,\,\,\,{\varvec{w}}\in S. \end{aligned}$$

If ${\varvec{w}}\in S$, such that ${\varvec{w}}\cdot {\varvec{\nu }}\le 0$ a.e. on $\Gamma _4$, then

$$\begin{aligned} \int _{\Gamma _4}{\varvec{w}}({\varvec{x}})\cdot {\varvec{\nu }}({\varvec{x}})\,\mathrm{d}\Gamma \le 0. \end{aligned}$$

Thus, $\mu _0\in \Lambda _4.$ As a result, $\mu _n=n\,\mu _0\in \Lambda _4\,\,\,\,\text{ for } \text{ all }\,\,\,\,n\in {\mathbb {N}}.$

Furthermore, for all positive integers n

$$\begin{aligned} \Vert \mu _n\Vert _Y= n\Vert \mu _0\Vert _Y. \end{aligned}$$

Passing to the limit as $n\rightarrow \infty$ in this last relation, we are lead to $\Vert \mu _n\Vert _Y\rightarrow \infty$.

Let us prove the inf-sup property of the form b, i.e., (b) in the assumption A 2. Using similar arguments with those used in [3, 12], we can write

$$\begin{aligned} \Vert {\varvec{\mu }}\Vert _Y= & {} \sup _{{\varvec{w}}\in S,\,{\varvec{w}}\ne 0_S}\frac{\langle {\varvec{\mu }}, {\varvec{w}}\rangle }{\Vert {\varvec{w}}\Vert _{H^{1/2}(\Gamma )^3}}\le \sup _{{\varvec{z}}\in X,\,{\varvec{\gamma }}{\varvec{z}}\ne 0_S}\frac{b({\varvec{z}}, {\varvec{\mu }})}{\Vert {\varvec{\gamma }}{\varvec{z}}\Vert _{H^{1/2}(\Gamma )^3}}\\\le & {} \sup _{{\varvec{z}}\in X,\,{\varvec{\gamma }}{\varvec{z}}\ne 0_S}\frac{b({\varvec{l}}({\varvec{\gamma }}{\varvec{z}}), {\varvec{\mu }})}{\Vert {\varvec{\gamma }}{\varvec{z}}\Vert _{H^{1/2}(\Gamma )^3}}\le c_l\,\sup _{{\varvec{z}}\in X,\,{\varvec{l}}({\varvec{\gamma }}{\varvec{z}})\ne 0_X}\frac{b({\varvec{l}}({\varvec{\gamma }}{\varvec{z}}), {\varvec{\mu }})}{|||{\varvec{l}}({\varvec{\gamma }}{\varvec{z}})|||_{H^1(\Omega )^3}}\\\le & {} {\widetilde{c}}_l\,\sup _{{\varvec{v}}\in X,\,{\varvec{v}}\ne 0_X} \frac{b({\varvec{v}}, {\varvec{\mu }})}{\Vert {\varvec{v}}\Vert _{X_0}}. \end{aligned}$$

Thus, we can take $\alpha ={\widetilde{c}}_l^{-1}$ to conclude that the assumption A 2 (b) is fulfilled.

Since the assumptions A 1- A 4 hold true, we are going to apply Theorem 4 to get the conclusion. $\square$

Remark 4

The equality ${\varvec{u}}_4={\varvec{u}}_1$ is an open question.