On a nonlinear elasticity problem with friction and Sobolev spaces with variable exponents

We consider a nonlinear elasticity problem in a bounded domain, its boundary is decomposed in three parts: lower, upper, and lateral. The displacement of the substance, which is the unknown of the problem, is assumed to satisfy the homogeneous Dirichlet boundary conditions on the upper part, and not homogeneous one on the lateral part, while on the lower part, friction conditions are considered. In addition, the problem is governed by a particular constitutive law of elasticity system with a strongly nonlinear strain tensor. The functional framework leads to using Sobolev spaces with variable exponents. The formulation of the problem leads to a variational inequality, for which we prove the existence and uniqueness of the solution of the associated variational problem. 74B20; 70F40; 46E30; 35R35


Introduction
The study of partial differential equation problems with variable exponents comes from the theory of nonlinear elasticity, elastic mechanics, fluid dynamics, electrorheological fluids, image processing, etc. (see [2,15,19] The corresponding nonlinear constraints tensor σ (u) = (σ ij (u(x))) 1≤i,j≤3 is then given by which describes a nonlinear relation between the stress tensor (σ ij ) i,j=1,2,3 and the deformation tensor (E ij ) i,j=1,2,3 . The coefficients of elasticity a ijkh (see [3]) satisfy the following symmetry properties: a ijkh = a jikh = a ijhk for all 1 ≤ i, j, k, h ≤ 3. (1. 3) The aim of this paper is to prove the existence and uniqueness of weak solution for the following nonlinear problem, encountered in the theory of nonlinear elasticity [3]: Let w be a bounded domain in R 2 situated in the plane of equation x 3 = 0. We suppose that w represents the lower surface of the domain occupied by the substance. The upper surface 1 is defined by where h is a function defined and bounded on w, that is to say, there exist h * and h * in R such that We study the displacement of a substance in = x , x 3 ∈ R 3 : x , 0 ∈ w and 0 < x 3 < h x the boundary ∂ = = w ∪ 1 ∪ L , where L is the lateral surface of . The outer normal vector unitary on is denoted by n = (n 1 , n 2 , n 3 ). The outer normal vector unitary on w is the vector (0, 0, -1).
Einstein's convention, which consists of making the sum on the repeated indices, will be used unless otherwise stated.
We define the normal and the tangential components u n and u t = (u t 1 , u t 2 , u t 3 ), of the displacement variable u by For normal and tangential components σ n and σ t = (σ t 1 , σ t 2 , σ t 3 ) of the strain tensor, the definition is as follows: σ n = (σ .n).n = σ ij n i n j , σ t i = σ ij n jσ n n i i and j = 1, 2, 3. (1.5) In this section, we are interested in the following equation: where f = (f 1 , f 2 , f 3 ) represents a mass density of the external forces. For boundary conditions, it is assumed that Condition (1.9) means that there is a tangential force exerted by the surface w on the substance. This tangential effort cannot exceed a certain threshold. The Tresca law assumes that this threshold is fixed and known where K is a given positive function called coefficient of friction and |σ t | is the modulus of the tangential constraint defined on w by (1.5).
As long as the tangential constraint σ t has not reached the threshold K , the substance moves with a given displacement s, which is the displacement of the lower surface w (adhesion). When the threshold is reached, the substance and the surface move tangentially relative to each other and there is proportional sliding. What can be summarized as follows [8]: on w, (1.11) where the positive real λ is unknown. This problem models the behavior of a heterogeneous material with the above Tresca friction free boundary condition. The consideration of this general material is in no way restrictive. Indeed, we can apply this study to the most particular elastic materials, but this particular case makes it easy to describe the different stages of this work. The tensor of the constraints considered here is nonlinear and grouped, as special cases, some models used in Ciarlet [3], Lions [12], and Dautray and Lions [4]. Let us cite by way of example (see [3,12]).
The complete problem (P 0 ) is therefore to find the displacement field u, satisfying the following equation and boundary conditions: We consider the functional framework of the considered problem (P 0 ) using Lebesgue and Sobolev spaces with variable exponents, see for example [6]. However, it is not necessary to use this notion of Lebesgue and Sobolev spaces with variable exponents to study this problem. But we see it as a good generalization to the same study with Lebesgue and Sobolev spaces with fixed exponents. Several authors studied the system of elasticity with laws of particular behavior and using various techniques in Sobolev spaces with constant exponents. For example, in [3] Ciarlet used the implicit function theorem to show the existence and uniqueness of a solution; in [4] Dautray and Lions studied the linear problem in a regular boundary domain; in [21] Zoubai and Merouani studied the existence and uniqueness of the solutions of the nonlinear elasticity system by topological degree; and in [13,20] Zoubai and Merouani studied the existence and uniqueness of the solution of Dirichlet's and Neumann's problems in Sobolev spaces with variable exponents.
In Sect. 2, we recall some definitions and properties of Lebesgue and Sobolev spaces with variable exponents (see for example [5-7, 10, 11] for the proofs and more details). This notion of Sobolev spaces with variable exponents is also used in many works (see for example [1,9,14]).
The need to work with the concept of Sobolev spaces with variable exponents is motivated by the appearance of these spaces when modeling electrorheological and thermorheological fluids (see [16]) and in image restoration (see [2]).
In Sect. 3, using this notion of Sobolev spaces with variable exponents, we give the convenient functional framework for the considered problem (P 0 ) to lead to variational problem 3.1. Then we prove in Theorem 3.2 the existence part by checking all hypotheses of Theorem 8.1 page 251 in [12]. And finally, in Sect. 4, we obtain also the uniqueness of the solution to variational problem (3.1).

Generalized Lebesgue and Sobolev spaces
Let ⊂ R N , let p(·) : − → [1, +∞] be a measurable function, called the variable exponent. In the following, we adopt the following notations: We define the generalized Lebesgue space L p(·) ( ), also called Lebesgue space with variable exponent, as the set of measurable functions u : → R for which the convex modulus is convex, so also the function u → ρ p(·) (u).
In the writing of variational formulations, the convex modulus ρ p(·) appears, which leads us to state the following results.

Variational problem and existence result
We introduce now the following functional space: G ∈ W 2,p(·) ( ) 3 such that G| 1 ∪ L = g and G · n = 0 on w. (3.1) The variational formulation of problem (P 0 ), see for example [8], leads to the following variational problem. Problem 3.1 Let f ∈ (L p (·) ( )) 3 and G satisfying (3.1) be given.
Find u such that u -G ∈ V ( ) and satisfying the following variational inequality hold To prove the existence of a solution to Problem 3.1, let us assume the following assumptions: (H 1 ) 3 < p(x) < +∞ for x ∈ , (H 2 ) ∃α 0 > 0; ∃β > 0 such that α 0 ≤ a ijkh (x) ≤ β a.e. in , ∀i, j, k, h = 1 to 3 we need the three properties of the operator E kh in the following theorem.
Thus, for v ∈ W p(·) ( ), we have ∂v i ∂x j ∈ L p(·) ( ) for i, j = 1 to 3. Hence, by the Hölder inequality, we obtain that 2. holds. The third 3. property comes from [17]. Proof For the existence part, we apply Theorem 8.5 page 251 in [12] and the first three properties of E kh cited in Theorem 3.1. First, we rewrite variational inequality (3.2) in the form of this Theorem 8.5 page 251 in [12]. 1 ∪ L ( )) 3 ∩ (W 2,p(·) ( )) 3 is a separable and reflexive Banach space, then its closed subspace V ( ) is also a separable and reflexive Banach space.
is linear and continuous, so it is an element of V ( ). We note by T(u -G) this application, so we have • Similarly, we have the application V ( ) → R, which associates is linear and continuous, so is an element V ( ). We note by f this application, so we have • We check now that the operator T is pseudo-monotonic. a) Let u be bounded in V ( ), we have Let u ∈ V ( ), by Remark 2.3 we get E kh (∇u) ∈ L p(·) ( ), and as p(x) > p (x), as soon as p(x) > 3 and bounded, we have E kh (∇u) ∈ L p (·) ( ) ∀1 ≤ k, h, ≤ 3. Using now hypothesis (H 2 ) and the Hölder inequality, with (3.3) we obtain

From (3.3) and (2.1) we get
and from (ii)-(iii) of Proposition 2.1 we have For all h and all k ∈ {1, 2, 3}, we get that E kh L p (·) ( ) is bounded for all h and all k ∈ {1, 2, 3}, consequently T(u) V ( ) is bounded. b) Let u, v, w ∈ V ( ) and λ ∈ R, we check that the application of R in R: λ → T(u + λv), w is continuous. For this, let us consider {λ n } to be a sequence of R that converges to λ. Let us denote The E kh being continuous, we therefore have, for all h and all k ∈ {1, 2, 3}, and we have also with (H 2 ) and the definition of E kh : As the sequence (λ n ) n∈N is convergent in R, then ∃m ∈ R : |λ n | ≤ m, so We define now the function L by We obtain that L ∈ L p(·) ( ). Thus from Lebesgue's dominated convergence theorem, we deduce that which shows that T is hemi-continuous. c) By hypothesis (H 2 ) and the monotony of E kh , we have so T is monotonous. As it is also bounded and hemi-continuous, then T is pseudo-monotonic. • The functional J is proper, convex, and lower semi-continuous on V ( ). Indeed, let u and v be two elements of V ( ), and λ ∈ [0, 1], we have where C is the constant of the continuous injection from V ( ) on (L p (·) (w)) 2 . Thus, J is Lipschitzian, so it is fortiori lower semi-continuous on V ( ).
We can now apply Theorem 8.5 page 251 in [12] to obtain the existence of u such that u -G in V ( ) satisfying variational inequality (3.2).

On the uniqueness of the result
then the operators are monotonous.
Proof We have the following result in [13]. We give also the proof for the convenience of the readers. Using the rule 1 2 (a 2 + b 2 ) ≥ -ab, with a = ∂u m ∂x i and b = ∂u m ∂x j , we have and consequently, ∀i, j = 1 to 3, To conclude, we must prove that the second member of (4.1) is positive. For that, we separate the second member of (4.1) in linear and nonlinear parts.
Let the linear function , i, j = 1 to 3, and the nonlinear function The functions A ij and B ij are continuous for p(·) > 3. It remains to show that, ∀i, j = 1 to 3, the A ij are increasing on R, the B ij are increasing on R -, and the A ij + B ij are increasing on ] -∞, 1 3 ]. The function t − → 1 2 t of R − → R, being increasing on R, we have 1 2 Therefore, the A ij are increasing.