Source term model for elasticity system with nonlinear dissipative term in a thin domain

Mohamed Dilmi; Mourad Dilmi; Salah Boulaaras; Hamid Benseridi

doi:10.1515/dema-2022-0033

Article Open Access

Source term model for elasticity system with nonlinear dissipative term in a thin domain

Mohamed Dilmi , Mourad Dilmi , Salah Boulaaras and Hamid Benseridi

Published/Copyright: August 26, 2022

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From the journal Demonstratio Mathematica Volume 55 Issue 1

Abstract

This article establishes an asymptotic behavior for the elasticity systems with nonlinear source and dissipative terms in a three-dimensional thin domain, which generalizes some previous works. We consider the limit when the thickness tends to zero, and we prove that the limit solution u ∗ is a solution of a two-dimensional boundary value problem with lower Tresca’s free-boundary conditions. Moreover, we obtain the weak Reynolds-type equation.

Keywords: asymptotic behavior; dissipative term; source term; Tresca friction law; weak solution

MSC 2010: 35R35; 76F10; 78M35; 35B40; 35J85; 49J40

1 Introduction

Many contemporary problems of the technology of materials require the study of what happens in the limiting case of thin films. The aim of the present article is to study the asymptotic behavior of the deformation of isotropic and homogeneous elastic body with a nonlinear source and dissipative terms, in a thin domain Ω ε ⊂ R 3 , where 0 < ε < 1 is a small parameter that tends to zero. The boundary of the domain Γ ε is assumed to be composed of three portions, where part of its boundary is subjected to conditions of Tresca and another part is subjected to the conditions of Dirichlet. In recent years, many mathematical articles were devoted to the asymptotic analyses of problems in thin domains. For instance, in the case of linearly thin elastic structures, Saadallah et al. [1] carried out the asymptotic convergence of the dynamical problem of nonisothermal linear elasticity with friction but without including the dissipative and source terms. Bayada and Lhalouani [2] carried out the asymptotic and numerical analyses for a unilateral contact problem with Coulomb’s friction between an elastic body and a thin elastic soft layer. In the context of the linear thin elasticity, the authors in [3] proved the asymptotic analysis of an interface problem with a dissipative term in a dynamic regime. Paumier [4] studied the asymptotic modeling of a thin elastic plate in unilateral contact with friction against a rigid obstacle, where he also proved that any family of solutions of the three-dimensional problem of Signorini with friction strongly converges toward a unique solution of a two-dimensional Signorini problem without friction. Similar problems for an incompressible fluid in a thin domain, with friction boundary conditions studied only for the stationary case, have been considered in [5,6, 7,8]. Otherwise, the existence, uniqueness, and the stabilization of the elasticity systems with nonlinear source terms were studied by several authors. Lions in [9,10] studied the existence and the uniqueness of the solution for some problems of linear elasticity with frictions of the Tresca and Coulomb types. In addition, Lagnese [11,12] proved some uniform stability results of elasticity systems with a linear dissipative term. For the linear damping case ∂ u ∂ t , the source terms ∣ u ∣ p − 2 u were considered by Levine [13,14]. Nakao [15] considered the dissipative term u 1 + u 2 , which has finite limits at infinity.

The aim of this article is first to generalize what has been considered in [1,6,7,16]. Compared to previous works, several additional difficulties appear. We have considered a dynamical problem for elasticity systems with non linear source ρ ( u ε ) and dissipative terms g ∂ u ε ∂ t , where the nonlinear functions ρ ( ⋅ ) and g ( ⋅ ) satisfy certain structural conditions.

The article is structured as follows. In Section 2, we give a strong formulation of the problem. In Section 3, the related weak formulation of the problem is given, and we discuss the existence and uniqueness theorem of the weak solution; then, we introduce a new scaling taking into account the small parameter ε . Some estimates and the convergence theorem are given in Section 4. Finally, in Section 5, the limit problem and a specific weak form of the Reynolds equation are obtained.

2 Formulation of the problem

To formulate the system of elasticity with a nonlinear source and dissipative terms, we begin with the following notations. Let Ω ε be a bounded domain of R 3 . The boundary of Ω ε will be denoted Γ ε = ω ¯ ∪ Γ ¯ 1 ε ∪ Γ ¯ L ε , where ω is a fixed bounded domain of R 2 for the equation x 3 = 0 , which is the bottom of the domain Ω ε and Γ ¯ 1 ε is the upper surface for the equation x 3 = ε h ( x 1 , x 2 ) , where h ( . ) is a function of class C 1 defined on ω such that

0 < h ̲ = h min ≤ h ( x 1 , x 2 ) ≤ h max = h ¯ , ∀ ( x 1 , x 2 , 0 ) ∈ ω .

Γ L ε is the lateral surface.

We denote by x = ( x ′ , x 3 ) ∈ R 3 , x ′ = ( x 1 , x 2 ) ∈ R 2 . The thin domain Ω ε is given by

Ω ε = { ( x ′ , x 3 ) ∈ R 3 : ( x 1 , x 2 ) ∈ ω , 0 < x 3 < ε h ( x ′ ) } .

Let u ε ( x , t ) be the displacement at a point x and time t in [0,T]. We denote by d i j ( u ε ) , 1 ≤ i , j ≤ 3 , the strain tensor, which is given by

d i j ( u ε ) = 1 2 ∂ u i ε ∂ x j + ∂ u j ε ∂ x i , 1 ≤ i , j ≤ 3 .

The stress–strain relation can be expressed as

σ i j ε ( u ε ) = 2 μ d i j ( u ε ) + λ ∑ k = 1 3 d k k ( u ε ) δ i j , i , j = 1 , 2 , 3 ,

where λ and μ are the Lamé’s coefficients and δ i j is the Kronecker symbol.

The normal and the tangential components of u ε on the boundary are given by

u n ε = u ε . n , u τ ε = u ε − u n ε . n .

Similarly, we denote by σ n ε and σ τ ε the normal and tangential components of σ ε , with

σ n ε = ( σ ε . n ) . n = ∑ i , j = 1 3 ( σ i j ε . n i ) . n j , σ τ ε = ( σ τ i ε ) 1 ≤ i ≤ 3 with σ τ i ε = ∑ j = 1 3 σ i j ε . n j − σ n ε . n i , 1 ≤ i ≤ 3 ,

where n is the unit external normal vector.

The elasticity system with dissipative and source terms in a dynamic regime is described by

(1) ∂ 2 u i ε ∂ t 2 − ∂ σ i j ε ( u ε ) ∂ x j + α ε ρ i ( u i ε ) + γ ε g i ∂ u i ε ∂ t = f i ε ( x , t ) , in Ω ε × ] 0 , T [ ,

where f ε represents the body forces and α ε , γ ε are positive constants.

The actual displacements on the boundary are as follows:

Dirichlet boundary conditions are imposed on ( Γ 1 ε ∪ Γ L ε ) × ] 0 , T [
(2) u ε = 0 .
On ω × ] 0 , T [ the velocity satisfies the following condition:
(3) ∂ u ε ∂ t . n = 0 .
The tangential velocity on ω × ] 0 , T [ is unknown and satisfies the Tresca friction law
(4) ∣ σ τ ε ∣ < k ε ⇒ ∂ u ε ∂ t τ = 0 ∣ σ τ ε ∣ = k ε ⇒ ∃ β > 0 such that ∂ u ε ∂ t τ = − β σ τ ε ,
where ∣ . ∣ denotes the Euclidean norm of R 2 and k ε is a given function.

Let us define the initial condition

(5) u ε ( x , 0 ) = u 0 ( x ) , ∂ u ε ∂ t = u 1 ( x ) , ∀ x ∈ Ω ε .

Finally, the dissipative and the source terms satisfy the following hypothesis:

1. For the source term

ρ i : R → R is a continuous increasing function, with
0 ≤ ρ i ′ ( u ) ≤ c ρ ∗ , i = 1 , 2 , 3 , ∀ u ∈ R ,
where c ρ ∗ is a positive constant independent of u .
ρ i ( 0 ) = 0 .

2. For the dissipative term
g i : R → R is a continuous increasing function.
g i ( 0 ) = 0 .
∀ u , v ∈ R , there exists a positive constant c g ∗ independent of u and v , such that
∣ g i ( u ) − g i ( v ) ∣ ≤ c g ∗ ∣ u − v ∣ , i = 1 , 2 , 3 .

A typical example of these functions is

ρ i ( u ) = sin ( u ) + 2 u e − 1 u 2 if u ≠ 0 0 if u = 0 , g i ( u ) = u 1 + u 2 , ∀ u ∈ R .

Remark 1

If ( a ρ ) and ( b ρ ) are satisfied, then

∣ ρ i ( u ) ∣ ≤ c ρ ∗ u , ∀ u ∈ R + .

3 Weak formulation

We multiply the equation (1) with φ − ∂ u ε ∂ t ; using integration by parts and the boundary conditions (2)–(4), we obtain the following variational problem:

Problem ( PK ε ) . Find u ε ∈ K ε where ∂ u ε ∂ t ∈ K ε , ∀ t ∈ ] 0 , T [ , such that

(6) ∂ 2 u ε ∂ t 2 , φ − ∂ u ε ∂ t + a u ε , φ − ∂ u ε ∂ t + α ε ρ ( u ε ) , φ − ∂ u ε ∂ t + γ ε g ∂ u ε ∂ t , φ − ∂ u ε ∂ t + j ε ( φ ) − j ε ∂ u ε ∂ t ≥ f ε , φ − ∂ u ε ∂ t , ∀ φ ∈ K ε ,

u ε ( x , 0 ) = u 0 ( x ) , ∂ u ε ∂ t ( x , 0 ) = u 1 ( x )

with

K ε = { v ∈ H 1 ( Ω ε ) 3 : v = 0 on Γ 1 ε ∪ Γ L ε , v . n = 0 on ω } ,

a ( u , v ) = 2 μ ∫ Ω ε d ( u ) d ( v ) d x + λ ∫ Ω ε div ( u ) div ( v ) d x ,

j ε ( v ) = ∫ ω k ε ∣ v ∣ d x ′ and ( f , v ) = ∫ Ω ε f ⋅ v ⋅ d x = ∑ i = 1 3 ∫ Ω ε f i v i d x , ∀ v ∈ H 1 ( Ω ε ) 3 .

The existence and unique results of the weak solution to the problem (6) are obtained in the following theorem.

Theorem 1

If the following assumptions are realized

f ε , ∂ f ε ∂ t , ∂ 2 f ε ∂ t 2 ∈ L 2 ( 0 , T ; L 2 ( Ω ε ) 3 ) ,

k ε ∈ C 0 ∞ ( ω ) , k ε > 0 i s i n d e p e n d e n t o f t ,

(7) u 0 ∈ H 1 ( Ω ε ) 3 , u 1 ∈ H 1 ( Ω ε ) 3 , ( u 1 ) τ = 0 .

Then, there exists a unique solution u ε of (6) with

u ε , ∂ u ε ∂ t ∈ L ∞ ( 0 , T ; H 1 ( Ω ε ) 3 ) ,

∂ 2 u ε ∂ t 2 , ρ ( u ε ) , g ∂ u ε ∂ t ∈ L ∞ ( 0 , T ; L 2 ( Ω ε ) 3 ) .

For the proof of this theorem, we use the same technique in [9,10].

3.1 The problem in a fixed domain

According to the change of variables z = x 3 ∕ ε , we define the fixed domain, in which the ε -convergence will be proved by

Ω = { ( x ′ , z ) ∈ R 3 : ( x ′ , 0 ) ∈ ω , 0 < z < h ( x ′ ) } ,

with Γ = ω ¯ ∪ Γ ¯ 1 ∪ Γ ¯ L being the boundary of Ω . We define the new unknowns

u ˆ i ε ( x ′ , z , t ) = u i ε ( x ′ , x 3 , t ) , i = 1 , 2 ; u ˆ 3 ε ( x ′ , z , t ) = ε − 1 u 3 ε ( x ′ , x 3 , t ) .

For the data of problem (6), it is assumed that they depend on ε as follows:

f ˆ ( x ′ , z , t ) = ε 2 f ε ( x ′ , x 3 , t ) ; k ˆ = ε k ε ; α ˆ = ε α ε ; γ ˆ = ε 2 γ ε ,

with f ˆ , k ˆ , α ˆ , and γ ˆ not depending on ε .

We introduce the following spaces:

K = { φ ∈ H 1 ( Ω ) 3 : φ = 0 on Γ 1 ∪ Γ L and φ . n = 0 on ω } ,

Π ( K ) = { φ ∈ H 1 ( Ω ) 2 : φ = ( φ 1 , φ 2 ) , φ i = 0 on Γ 1 for i = 1 , 2 } .

The variational problem (6) is reformulated on the fixed domain as follows:

Problem (PK). Find u ˆ ε ∈ K , with ∂ u ˆ ε ∂ t ∈ K , ∀ t ∈ ] 0 , T [ such that

(8) ∑ i = 1 2 ε 2 ∂ 2 u ˆ i ε ∂ t 2 , φ ˆ i − ∂ u ˆ i ε ∂ t + ε 4 ∂ 2 u ˆ 3 ε ∂ t 2 , φ ˆ 3 − ∂ u ˆ 3 ε ∂ t + a ˆ u ˆ ε , φ ˆ − ∂ u ˆ ε ∂ t + ε α ˆ ∑ i = 1 2 ρ i ( u ˆ i ε ) , φ ˆ i − ∂ u ˆ i ε ∂ t + ε 2 α ˆ ρ 3 ( ε u ˆ 3 ε ) , φ ˆ 3 − ∂ u ˆ 3 ε ∂ t + γ ˆ ∑ i = 1 2 g i ∂ u ˆ i ε ∂ t , φ ˆ i − ∂ u ˆ i ε ∂ t + ε γ ˆ g 3 ε ∂ u ˆ 3 ε ∂ t , φ ˆ 3 − ∂ u ˆ 3 ε ∂ t + J ˆ ( φ ˆ ) − J ˆ ∂ u ˆ ε ∂ t ≥ ∑ i = 1 2 f ˆ i , φ ˆ i − ∂ u ˆ i ε ∂ t + ε f ˆ 3 , φ ˆ 3 − ∂ u ˆ 3 ε ∂ t , ∀ φ ∈ K ,

u ˆ ε ( x , 0 ) = u ˆ 0 ( x ) , ∂ u ˆ ε ∂ t ( x , 0 ) = u ˆ 1 ( x ) ,

where

J ˆ ( φ ˆ ) = ∫ ω k ˆ ∣ φ ˆ ∣ d x ′

and

a ˆ u ˆ ε , φ ˆ − ∂ u ˆ ε ∂ t = μ ε 2 ∑ i , j = 1 2 ∫ Ω ∂ u ˆ i ε ∂ x j + ∂ u ˆ j ε ∂ x i ∂ ∂ x j φ ˆ i − ∂ u ˆ i ε ∂ t d x ′ d z + μ ∑ i = 1 2 ∫ Ω ∂ u ˆ i ε ∂ z + ε 2 ∂ u ˆ 3 ε ∂ x i ∂ ∂ z φ ˆ i − ∂ u ˆ i ε ∂ t + ε 2 ∂ ∂ x i φ ˆ 3 − ∂ u ˆ 3 ε ∂ t d x ′ d z + 2 μ ε 2 ∫ Ω ∂ u ˆ 3 ε ∂ z . ∂ ∂ z φ ˆ 3 − ∂ u ˆ 3 ε ∂ t d x ′ d z + λ ε 2 ∫ Ω div ( u ˆ ε ) div φ ˆ − ∂ u ˆ ε ∂ t d x ′ d z .

4 Some estimates and convergence theorem

To establish some estimates and convergence results for the solutions of the variational problem (8), we use the following inequalities [2].

The Korn inequality

(9) μ C K ∥ ∇ u ε ∥ L 2 ( Ω ε ) 3 × 3 ≤ a ( u ε , u ε ) ( C K independent of ε ) , ∀ u ε ∈ K ε .

The Poincaré inequality

(10) ∥ u ε ∥ L 2 ( Ω ε ) 3 ≤ ε h ¯ ∂ u ε ∂ x 3 L 2 ( Ω ε ) 3 , ∀ u ε ∈ K ε .

Theorem 2

Under the assumptions of Theorem 1, there exists a constant c independent of ε , such that the following estimates hold

(11) ∑ i = 1 2 ∂ u ˆ i ε ∂ z ∂ u ˆ i ε ∂ t L 2 ( Ω ) 2 + ε ∂ 2 u ˆ i ε ∂ t 2 L 2 ( Ω ) 2 + ε 2 ∂ ∂ x i ∂ u ˆ 3 ε ∂ t L 2 ( Ω ) 2 + ∑ i , j = 1 2 ε ∂ ∂ x j ∂ u ˆ i ε ∂ t L 2 ( Ω ) 2 + ε ∂ ∂ z ∂ u ˆ 3 ε ∂ t L 2 ( Ω ) 2 + ε 2 ∂ 2 u ˆ 3 ε ∂ t 2 L 2 ( Ω ) 2 ≤ c ,

(12) ∑ i = 1 2 ∂ u ˆ i ε ∂ z L 2 ( Ω ) 2 + ε ∂ u ˆ i ε ∂ t L 2 ( Ω ) 2 + ε 2 ∂ u ˆ 3 ε ∂ x i L 2 ( Ω ) 2 + ∑ i , j = 1 2 ε ∂ u ˆ i ε ∂ x j L 2 ( Ω ) 2 + ε ∂ u ˆ 3 ε ∂ z L 2 ( Ω ) 2 + ε 2 ∂ u ˆ 3 ε ∂ t L 2 ( Ω ) 2 ≤ c .

Proof

The functional j ε ( . ) is convex but nondifferentiable. To overcome this difficulty, we regularize it by j ζ ε ( . ) , where

j ζ ε ( v ) = ∫ ω k ε ( x ′ ) ϕ ζ ( ∣ v τ ∣ 2 ) d x ′ where ϕ ζ ( λ ) = 1 1 + ζ ∣ λ ∣ ( 1 + ζ ) , ζ > 0 .

Clearly, j ζ ( . ) is convex and Gateaux differentiable with Gateaux derivative ( j ζ ) ′ defined on K ε and given by

( ( j ζ ) ′ ( v ) , w ) = ∫ ω k ε ( x ′ ) ϕ ζ ′ ( ∣ v ∣ 2 ) v . w d x ′ .

We consider the approximate equation

(13) ∂ 2 u ζ ε ∂ t 2 , φ + a ( u ζ ε , φ ) + α ε ( ρ ( u ζ ε ) , φ ) + γ ε g ∂ u ζ ε ∂ t , φ + ( j ζ ε ) ′ ∂ u ζ ε ∂ t , φ = ( f ε , φ ) , ∀ φ ∈ K ε ,

with

u ζ ε ( x , 0 ) = u 0 ( x ) , ∂ u ζ ε ∂ t ( x , 0 ) = u 1 ( x ) .

To show (11), we derive (13) for t and take φ = ∂ 2 u ζ ε ∂ t 2 , and we obtain

∂ 3 u ζ ε ∂ t 3 , ∂ 2 u ζ ε ∂ t 2 + a ∂ u ζ ε ∂ t , ∂ 2 u ζ ε ∂ t 2 + α ε ∂ u ζ ε ∂ t ρ ′ ( u ζ ε ) , ∂ 2 u ζ ε ∂ t 2 + γ ε ∂ 2 u ζ ε ∂ t 2 g ′ ∂ u ζ ε ∂ t , ∂ 2 u ζ ε ∂ t 2 + ∂ ∂ t ( j ζ ε ) ′ ∂ u ζ ε ∂ t , ∂ 2 u ζ ε ∂ t 2 = ∂ f ε ∂ t , ∂ 2 u ζ ε ∂ t 2 ,

as ∂ ∂ t ( j ζ ε ) ′ ∂ u ζ ε ∂ t , ∂ 2 u ζ ε ∂ t 2 ≥ 0 and g i ( . ) , i = 1 , 2 , 3 is increasing, we obtain

1 2 d d t ∂ 2 u ζ ε ∂ t 2 L 2 ( Ω ε ) 3 2 + a ∂ u ζ ε ∂ t , ∂ u ζ ε ∂ t + α ε ∫ Ω ε ρ ′ ( u ζ ε ) ∂ u ζ ε ∂ t . ∂ 2 u ζ ε ∂ t 2 d x ′ d x 3 ≤ ∂ f ε ∂ t , ∂ 2 u ζ ε ∂ t 2 .

Integrating this last inequality from 0 to t . Then, we use the Korn inequality (9) and the hypotheses ( a ρ ) to obtain

(14) ∂ 2 u ζ ε ∂ t 2 L 2 ( Ω ε ) 3 2 + 2 μ C K ∇ ∂ u ζ ε ∂ t L 2 ( Ω ε ) 3 × 3 2 ≤ 2 ∫ 0 t ∂ f ε ( s ) ∂ t , ∂ 2 u ζ ε ( s ) ∂ t 2 d s + α ε c ρ ∗ ∫ Ω ε ∂ u ζ ε ∂ t ∂ 2 u ζ ε ∂ t 2 d x ′ d x 3 + ∂ 2 u ζ ε ( 0 ) ∂ t 2 L 2 ( Ω ε ) 3 2 + ( 2 μ + λ ) ∥ ∇ u 1 ∥ L 2 ( Ω ε ) 3 × 3 2 .

Using Cauchy-Schwartz’s, Poincaré’s (10), and Young’s inequalities, we obtain

(15) 2 ∫ 0 t ∂ f ε ( s ) ∂ t , ∂ 2 u ζ ε ( s ) ∂ t 2 d s ≤ μ C K ∇ ∂ u ζ ε ∂ t L 2 ( Ω ε ) 3 × 3 2 + 4 ε 2 h ¯ 2 μ C K ∂ f ε ∂ t L 2 ( Ω ε ) 3 2 + 4 ε 2 h ¯ 2 μ C K ∂ f ε ( 0 ) ∂ t L 2 ( Ω ε ) 3 2 + μ C K ∥ ∇ u 1 ∥ L 2 ( Ω ε ) 3 × 3 2 + 4 ( ε h ¯ ) 2 μ C K ∫ 0 t ∂ 2 f ε ( s ) ∂ t 2 L 2 ( Ω ε ) 3 2 d s + ∫ 0 t μ C K ∇ ∂ u ζ ε ∂ t ( s ) L 2 ( Ω ε ) 3 × 3 2 d s .

Also, using Poincaré’s inequality (10), we obtain

(16) α ε c ρ ∗ ∫ Ω ε ∂ u ζ ε ∂ t ∂ 2 u ζ ε ∂ t 2 d x ′ d x 3 ≤ ( α ε c ρ ∗ ε h ¯ ) 2 ∫ 0 t ∇ ∂ u ζ ε ∂ t ( s ) L 2 ( Ω ε ) 3 × 3 2 d s + ∫ 0 t ∂ 2 u ζ ε ∂ t 2 ( s ) L 2 ( Ω ε ) 3 2 d s .

From (15), (16), and (14), we obtain

(17) ∂ 2 u ζ ε ∂ t 2 L 2 ( Ω ε ) 3 2 + μ C K ∇ ∂ u ζ ε ∂ t L 2 ( Ω ε ) 3 × 3 2 ≤ ∫ 0 t ∂ 2 u ζ ε ( s ) ∂ t 2 L 2 ( Ω ε ) 3 2 + ( μ C K + α ˆ c ρ ∗ h ¯ ) ∇ ∂ u ζ ε ∂ t ( s ) L 2 ( Ω ε ) 3 × 3 2 d s + ∂ 2 u ζ ε ∂ t 2 ( 0 ) L 2 ( Ω ε ) 3 2 + 4 ε 2 h ¯ 2 μ C K ∂ f ε ∂ t L 2 ( Ω ε ) 3 2 + 4 ε 2 h ¯ 2 μ C K ∂ f ε ( 0 ) ∂ t L 2 ( Ω ε ) 3 2 + μ C K ∥ ∇ u 1 ∥ L 2 ( Ω ε ) 3 × 3 2 + 4 ( ε h ¯ ) 2 μ C K ∫ 0 t ∂ 2 f ε ( s ) ∂ t 2 L 2 ( Ω ε ) 3 2 d s .

It is necessary to estimate ∂ 2 u ζ ε ∂ t 2 ( 0 ) . We deduce from (13) and (7) that

∂ 2 u ζ ε ∂ t 2 ( 0 ) , φ = ( f ε ( 0 ) , φ ) − a ( u 0 , φ ) − γ ε ( g ( u 1 ) , φ ) − α ε ( ρ ( u 0 ) , φ ) , ∀ φ ∈ K ε .

Let us apply Cauchy-Schwartz’s inequality and then Poincaré’s formula to obtain

∂ 2 u ζ ε ∂ t 2 ( 0 ) , φ ≤ ( ε h ¯ ∥ f ε ( 0 ) ∥ L 2 ( Ω ε ) 3 + ( 2 μ + ε α ε h ¯ 2 c ρ ∗ ) ∥ u 0 ∥ H 1 ( Ω ε ) 3 + ε 2 γ ε h ¯ 2 c g ∗ ∥ u 1 ∥ H 1 ( Ω ε ) 3 ) ∥ φ ∥ H 1 ( Ω ε ) 3 .

By multiplying this inequality by ε , we obtain

(18) ε ∂ 2 u ζ ε ∂ t 2 ( 0 ) L 2 ( Ω ε ) 3 ≤ C ′ ,

where C ′ = h ¯ ∥ f ˆ ( 0 ) ∥ L 2 ( Ω ) 3 + ( 2 μ + α ˆ h ¯ 2 c ρ ∗ ) ∥ u ˆ 0 ∥ H 1 ( Ω ) 3 + γ ˆ h ¯ 2 c g ∗ ∥ u ˆ 1 ∥ H 1 ( Ω ) 3 does not depend on ε .

We pass to the limit in (17) as ζ tends to zero and find

(19) ∂ 2 u ε ∂ t 2 L 2 ( Ω ε ) 3 2 + 2 μ C K ∇ ∂ u ε ∂ t L 2 ( Ω ε ) 3 × 3 2

≤ ∫ 0 t ∂ 2 u ε ( s ) ∂ t 2 L 2 ( Ω ε ) 3 2 + ( μ C K + α ˆ c ρ ∗ h ¯ ) ∇ ∂ u ε ∂ t ( s ) L 2 ( Ω ε ) 3 × 3 2 d s + ∂ 2 u ε ∂ t 2 ( 0 ) L 2 ( Ω ε ) 3 2 + 4 ε 2 h ¯ 2 μ C K ∂ f ε ∂ t L 2 ( Ω ε ) 3 2 + 4 ε 2 h ¯ 2 μ C K ∂ f ε ( 0 ) ∂ t L 2 ( Ω ε ) 3 2 + 4 ( ε h ¯ ) 2 μ C K ∫ 0 t ∂ 2 f ε ( s ) ∂ t 2 L 2 ( Ω ε ) 3 2 d s + μ C K ∥ ∇ u 1 ∥ L 2 ( Ω ε ) 3 × 3 2 .

ε 2 ∥ f ε ∥ L 2 ( Ω ε ) 3 2 = ε − 1 ∥ f ˆ ∥ L 2 ( Ω ) 3 2 ,

multiplying now the inequality (19) by ε , we obtain

(20) ε ∂ 2 u ε ∂ t 2 L 2 ( Ω ε ) 3 2 + 2 μ C K ∇ ∂ u ε ∂ t L 2 ( Ω ε ) 3 × 3 2 ≤ A + ∫ 0 t ε ∂ 2 u ε ( s ) ∂ t 2 L 2 ( Ω ε ) 3 2 + μ C K ∇ ∂ u ε ∂ t ( s ) L 2 ( Ω ε ) 3 × 3 2 d s ,

where A is a constant that does not depend on ε with

A = 4 h ¯ 2 μ C K ∂ f ˆ ∂ t L 2 ( Ω ) 3 2 + 4 h ¯ 2 μ C K ∂ f ˆ ( 0 ) ∂ t L 2 ( Ω ) 3 2 + C ′ + μ C K ∥ ∇ u ˆ 1 ∥ L 2 ( Ω ) 3 × 3 2 + 4 h ¯ 2 μ C K ∂ 2 f ˆ ( s ) ∂ t 2 L 2 ( 0 , T ; L 2 ( Ω ) 3 ) 2 .

Using Gronwall’s lemma for (20), we obtain

(21) ε ∂ 2 u ε ∂ t 2 L 2 ( Ω ε ) 3 2 + ε ∇ ∂ u ε ∂ t L 2 ( Ω ε ) 3 × 3 2 ≤ c ,

where c is a positive constant independent of ε . So we deduce (11) from (21).

Let u ε be the solution of the problem (6); we choose φ = 0 and obtain

∂ 2 u ε ∂ t 2 , ∂ u ε ∂ t + a u ε , ∂ u ε ∂ t + α ε ρ ( u ε ) , ∂ u ε ∂ t + γ ε g ∂ u ε ∂ t , ∂ u ε ∂ t ≤ f ε , ∂ u ε ∂ t .

As g i ( ⋅ ) is monotone and g i ( 0 ) = 0 , i = 1 , 2 , 3 ; this implies that

1 2 d d t ∂ u ε ∂ t L 2 ( Ω ε ) 3 2 + a ( u ε , u ε ) ≤ f ε , ∂ u ε ∂ t − α ε ρ ( u ε ) , ∂ u ε ∂ t ,

integrating over ( 0 , t ) and using Korn’s inequality (9), we find

(22) ∂ u ε ∂ t L 2 ( Ω ε ) 3 2 + 2 μ C K ∥ ∇ u ε ∥ L 2 ( Ω ε ) 3 × 3 2 ≤ − 2 α ε ρ ( u ε ) , ∂ u ε ∂ t + [ ∥ u 1 ∥ L 2 ( Ω ε ) 3 2 + ( 2 μ + 3 λ ) ∥ ∇ u 0 ∥ L 2 ( Ω ε ) 3 × 3 2 ] + μ C K ∥ ∇ u ε ∥ L 2 ( Ω ε ) 3 × 3 2 + 4 ε 2 h ¯ 2 μ C K ∥ f ε ∥ L 2 ( Ω ε ) 3 2 + 4 ε 2 h ¯ 2 μ C K ∥ f ε ( 0 ) ∥ L 2 ( Ω ε ) 3 2 + μ C K ∥ ∇ u 0 ∥ L 2 ( Ω ε ) 3 × 3 2 + 4 ( ε h ¯ ) 2 μ C K ∫ 0 t ∂ f ε ( s ) ∂ t L 2 ( Ω ε ) 3 2 d s + ∫ 0 t μ C K ∥ ∇ u ε ( s ) ∥ L 2 ( Ω ε ) 3 × 3 2 d s .

Using Cauchy-Schwartz, Poincaré’s (10), and Young’s inequalities, we obtain

(23) − 2 α ε ∫ 0 t ρ ( u ε ( s ) ) , ∂ u ε ( s ) ∂ t d s ≤ 2 α ε c ρ ∗ ∫ 0 t ∥ u ε ∥ L 2 ( Ω ε ) 3 ∂ u ε ∂ t L 2 ( Ω ε ) 3 ≤ ( 2 α ˆ h ¯ 2 c ρ ∗ ) 2 ∫ 0 t ∥ ∇ u ε ( s ) ∥ L 2 ( Ω ε ) 3 × 3 2 d s + ∫ 0 t ∂ u ε ( s ) ∂ t L 2 ( Ω ε ) 3 × 3 2 d s .

According to (22) and (23), we find

∂ u ε ∂ t L 2 ( Ω ε ) 3 2 + 2 μ C K ∥ ∇ u ε ∥ L 2 ( Ω ε ) 3 × 3 2 ≤ ∫ 0 t ∂ u ε ( s ) ∂ t L 2 ( Ω ε ) 3 2 + ( μ C K + ( 2 α ˆ h ¯ 2 c ρ ∗ ) 2 ) ∥ ∇ u ε ( s ) ∥ L 2 ( Ω ε ) 3 × 3 2 d s + μ C K ∥ ∇ u ε ∥ L 2 ( Ω ε ) 3 × 3 2 + μ C K ∥ ∇ u 0 ∥ L 2 ( Ω ε ) 3 × 3 2 + 4 ( ε h ¯ ) 2 μ C K ∫ 0 t ∂ f ε ( s ) ∂ t L 2 ( Ω ε ) 3 2 d s + ∥ u 1 ∥ L 2 ( Ω ε ) 3 2 + ( μ C K + 2 μ + 3 λ ) ∥ ∇ u 0 ∥ L 2 ( Ω ε ) 3 × 3 2 + 4 ε 2 h ¯ 2 μ C K ∥ f ε ( 0 ) ∥ L 2 ( Ω ε ) 3 2 + 4 ε 2 h ¯ 2 μ C K ∥ f ε ∥ L 2 ( Ω ε ) 3 2 ,

and we multiply this inequality by ε and deduce

ε ∂ u ε ∂ t L 2 ( Ω ε ) 3 2 + μ C K ∥ ∇ u ε ∥ L 2 ( Ω ε ) 3 × 3 2 ≤ B + ∫ 0 t ε ∂ u ε ∂ t L 2 ( Ω ε ) 3 2 + ( μ C K + ( α ˆ h ¯ 2 c ρ ∗ ) 2 ) ∥ ∇ u ε ∥ L 2 ( Ω ε ) 3 × 3 2 d s ,

where B is a constant that does not depend on ε with

B = ∥ u ˆ 1 ∥ L 2 ( Ω ) 3 2 + ( μ C K + 2 μ + 3 λ ) ∥ ∇ u ˆ 0 ∥ L 2 ( Ω ) 3 × 3 2 + 4 h ¯ 2 μ C K ∥ f ˆ ( 0 ) ∥ L 2 ( Ω ) 3 2 + 4 h ¯ 2 μ C K ∥ f ˆ ∥ L 2 ( Ω ) 3 2 + 4 h ¯ 2 μ C K ∂ f ˆ ∂ t L 2 ( 0 , T ; L 2 ( Ω ) 3 ) 2 .

Using Gronwall’s lemma, there exists a constant c that does not depend on ε such that

ε ∂ u ε ∂ t L 2 ( Ω ε ) 3 2 + ε ∥ ∇ u ε ∥ L 2 ( Ω ε ) 3 × 3 2 ≤ c ,

thus we obtain the second estimate (12).□

Now, we will try to obtain the theorem of convergence when ε → 0 ; for this purpose, we introduce the Banach space

V z = v = ( v 1 , v 2 ) ∈ L 2 ( Ω ) 2 : ∂ v i ∂ z ∈ L 2 ( Ω ) , i = 1 , 2 and v = 0 on Γ 1 ,

with norm

∥ v ∥ V z = ∑ i = 1 2 ∥ v i ∥ L 2 ( Ω ) 2 + ∂ v i ∂ z L 2 ( Ω ) 2 1 2 .

Theorem 3

Under the hypotheses of Theorem 2, there exists u i ∗ ∈ L 2 ( 0 , T ; V z ) ∩ L ∞ ( 0 , T ; V z ) , i = 1 , 2 such that

(24) u ˆ i ε ⇀ u i ∗ , i = 1 , 2 ∂ u ˆ i ε ∂ t ⇀ ∂ u i ∗ ∂ t , i = 1 , 2 w e a k l y i n L 2 ( 0 , T ; V z ) a n d w e a k l y ∗ i n L ∞ ( 0 , T ; V z ) ,

(25) ε ρ i ( u ˆ i ε ) ⇀ 0 , i = 1 , 2 ε ρ 3 ( ε u ˆ 3 ε ) ⇀ 0 w e a k l y i n L 2 ( 0 , T ; L 2 ( Ω ) ) a n d w e a k l y ∗ i n L ∞ ( 0 , T ; L 2 ( Ω ) ) ,

(26) g i ∂ u ˆ i ε ∂ t ⇀ g i ∂ u i ∗ ∂ t , i = 1 , 2 ε g 3 ε ∂ u ˆ 3 ε ∂ t ⇀ 0 w e a k l y i n L 2 ( 0 , T ; L 2 ( Ω ) ) a n d w e a k l y ∗ i n L ∞ ( 0 , T ; L 2 ( Ω ) ) ,

(27) ε ∂ u ˆ i ε ∂ t ⇀ 0 , i = 1 , 2 ε 2 ∂ u ˆ 3 ε ∂ t ⇀ 0 ε 2 ∂ u ˆ 3 ε ∂ x i ⇀ 0 , i = 1 , 2 ε 2 ∂ u ˆ 3 ε ∂ z ⇀ 0 ε ∂ u ˆ i ε ∂ x j ⇀ 0 , i , j = 1 , 2 w e a k l y i n L 2 ( 0 , T ; L 2 ( Ω ) ) a n d w e a k l y ∗ in L ∞ ( 0 , T ; L 2 ( Ω ) ) ,

(28) ε ∂ 2 u ˆ i ε ∂ t 2 ⇀ 0 , i = 1 , 2 ε ∂ 2 u ˆ i ε ∂ x j ∂ t ⇀ 0 , i , j = 1 , 2 ε 2 ∂ 2 u ˆ 3 ε ∂ x i ∂ t ⇀ 0 , i = 1 , 2 ε 2 ∂ 2 u ˆ 3 ε ∂ z ∂ t ⇀ 0 ε 2 ∂ 2 u ˆ 3 ε ∂ t 2 ⇀ 0 w e a k l y i n L 2 ( 0 , T ; L 2 ( Ω ) ) a n d w e a k l y ∗ i n L ∞ ( 0 , T ; L 2 ( Ω ) ) .

Proof

From Theorem (2), there exists a constant c independent of ε such that

∂ u ˆ i ε ∂ z L 2 ( Ω ) 2 ≤ c , i = 1 , 2 .

Using this estimate with the Poincaré inequality in the domain Ω , we obtain

∥ u ˆ i ε ∥ L 2 ( Ω ) 2 ≤ ∂ u ˆ i ε ∂ z L 2 ( Ω ) 2 ≤ c , i = 1 , 2 .

So ( u ˆ 1 ε , u ˆ 2 ε ) ε is bounded in L 2 ( 0 , T , V z ) ∩ L ∞ ( 0 , T , V z ) , which implies the existence of an element ( u 1 ∗ , u 2 ∗ ) in L 2 ( 0 , T , V z ) ∩ L ∞ ( 0 , T , V z ) such that ( u ˆ 1 ε , u ˆ 2 ε ) ε converges weakly to ( u 1 ∗ , u 2 ∗ ) in L 2 ( 0 , T , V z ) ∩ L ∞ ( 0 , T , V z ) , the same for ∂ u ˆ 1 ε ∂ t , ∂ u ˆ 2 ε ∂ t ε , and thus we obtain (24). From (25) and (26), we have ∣ ρ ( u ˆ ε ) ∣ ≤ c ρ ∗ ∣ u ˆ ε ∣ , which implies that

∑ i = 1 2 ρ i ( u ˆ i ε ) 2 + ρ 3 ( ε u ˆ 3 ε ) 2 ≤ ( c ρ ∗ ) 2 ∑ i = 1 2 ( u ˆ i ε ) 2 + ( ε u ˆ 3 ε ) 2 ,

integrating this inequality on Ω , we obtain

∑ i = 1 2 ∥ ρ i ( u ˆ i ε ) ∥ L 2 ( Ω ) 2 + ∥ ρ 3 ( ε u ˆ 3 ε ) ∥ L 2 ( Ω ) 2 ≤ ( c ρ ∗ ) 2 ∑ i = 1 2 ∥ u ˆ i ε ∥ L 2 ( Ω ) 2 + ∥ ε u ˆ 3 ε ∥ L 2 ( Ω ) 2 .

On the other hand, by estimate (12) and the inequality

∥ ε u ˆ 3 ε ∥ L 2 ( Ω ) ≤ h ¯ ε ∂ u ˆ 3 ε ∂ z L 2 ( Ω ) ,

we find

∑ i = 1 2 ∥ u ˆ i ε ∥ L 2 ( Ω ) 2 + ∥ ε u ˆ 3 ε ∥ L 2 ( Ω ) 2 ≤ c ,

then

∑ i = 1 2 ∥ ρ i ( u ˆ i ε ) ∥ L 2 ( Ω ) 2 + ∥ ρ 3 ( ε u ˆ 3 ε ) ∥ L 2 ( Ω ) 2 ≤ c ,

from this we conclude that ε ρ i ( u ˆ i ε ) , i = 1 , 2 , converges weakly to 0, and ε ρ 3 ( ε u ˆ 3 ε ) converges weakly to 0.

Using hypotheses ( b g ) , ( c g ) , and estimate (11), we find

∑ i = 1 2 g i ∂ u ˆ i ε ∂ t L 2 ( Ω ) 2 + g 3 ε ∂ u ˆ 3 ε ∂ t L 2 ( Ω ) 2 ≤ c ,

so, we obtain g i ∂ u ˆ i ε ∂ t converges weakly to g i ∂ u i ∗ ∂ t i = 1 , 2 , and ε g 3 ε ∂ u ˆ 3 ε ∂ t converges weakly to 0.

5 Study the limit problem

In this section, we seek the limit system that satisfies by u ˆ ε when ε → 0 .

Theorem 4

With the same assumptions of Theorem 3, u ∗ satisfies the limit variational problem

(29) μ ∑ i = 1 2 ∫ Ω ∂ u i ∗ ∂ z . ∂ ∂ z φ ˆ i − ∂ u i ∗ ∂ t d x ′ d z + γ ˆ ∑ i = 1 2 ∫ Ω g i ∂ u i ∗ ∂ t φ ˆ i − ∂ u i ∗ ∂ t d x ′ d z + J ˆ ( φ ˆ ) − J ˆ ∂ u ∗ ∂ t ≥ ∑ i = 1 2 f ˆ i , φ ˆ i − ∂ u i ∗ ∂ t , ∀ φ ˆ ∈ Π ( K ) ,

and the limit problem

(30) − μ ∂ 2 u i ∗ ∂ z 2 + γ ˆ g i ∂ u i ∗ ∂ t = f ˆ i ( t ) , f o r i = 1 , 2 i n L 2 ( Ω ) ,

u i ∗ ( x , 0 ) = u ˆ 0 , i ( x ) , i = 1 , 2 .

Proof

By passage to the limit when ε tends to zero in the variational inequality (8), using the convergence results of Theorem 3 and the fact that J ˆ ( ⋅ ) is convex and lower semi-continuous, we deduce

(31) μ ∑ i = 1 2 ∫ Ω ∂ u i ∗ ∂ z . ∂ ∂ z φ ˆ i − ∂ u i ∗ ∂ t d x ′ d z + γ ˆ ∑ i = 1 2 ∫ Ω g i ∂ u i ∗ ∂ t φ ˆ i − ∂ u i ∗ ∂ t d x ′ d z + J ˆ ( φ ˆ ) − J ˆ ∂ u ∗ ∂ t ≥ ∑ i = 1 2 f ˆ i , φ ˆ i − ∂ u i ∗ ∂ t .

We choose now in (31)

φ ˆ i = ∂ u i ∗ ∂ t ± ψ i , ψ i ∈ H 0 1 ( Ω ) , i = 1 , 2 ,

and using Green’s formula, then choosing ψ 1 = 0 and ψ 2 ∈ H 0 1 ( Ω ) , then ψ 2 = 0 and ψ 1 ∈ H 0 1 ( Ω ) , we obtain

− μ ∫ Ω ∂ ∂ z ∂ u i ∗ ∂ z ψ i d x ′ d z = ∫ Ω f ˆ i ψ i d x ′ d z − γ ˆ ∑ i = 1 2 ∫ Ω g i ∂ u i ∗ ∂ t ψ i d x ′ d z .

Thus,

(32) − μ ∂ 2 u i ∗ ∂ z 2 + γ ˆ g i ∂ u i ∗ ∂ t = f ˆ i for i = 1 , 2 in H − 1 ( Ω ) ,

and as f ˆ i ∈ L 2 ( Ω ) , i = 1 , 2 , then (32), is valid in L 2 ( Ω ) .□

Theorem 5

Let

τ ∗ ( x ′ , t ) = ∂ u ∗ ∂ z ( x ′ , 0 , t ) , s ∗ ( x ′ , t ) = u ∗ ( x ′ , 0 , t ) ,

the trace of the displacement. Then, the traces τ ∗ , s ∗ satisfy the following inequality:

(33) ∫ ω k ˆ ψ + ∂ s ∗ ∂ t − ∂ s ∗ ∂ t d x ′ − ∫ ω μ τ ∗ ψ d x ′ ≥ 0 , ∀ ψ ∈ L 2 ( ω ) 2 ,

and the following limit form of the Tresca boundary conditions

(34) μ ∣ τ ∗ ∣ < k ˆ ⇒ ∂ s ∗ ∂ t = 0 , μ ∣ τ ∗ ∣ = k ˆ ⇒ ∃ β > 0 : ∂ s ∗ ∂ t = β τ ∗ , a . e o n ω × ] 0 , T [ .

Also, u ∗ satisfies the weak generalized equation

(35) ∫ ω F ˜ − μ h 2 s ∗ + μ ∫ 0 h u ∗ ( x ′ , z , t ) d z + G ˜ ∇ ψ ( x ′ ) d x = 0 , ∀ ψ ∈ H 1 ( ω ) ,

where

F ˜ ( x ′ , h , t ) = ∫ 0 h F ( x ′ , z , t ) d z − h 2 F ( x ′ , h , t ) , F ( x ′ , z , t ) = ∫ 0 z ∫ 0 ζ f ˆ ( x ′ , η , t ) d η d ζ ,

G ˜ ( x ′ , h , t ) = − γ ˆ ∫ 0 h G ( x ′ , z , t ) d z + γ ˆ h 2 G ( x ′ , h , t ) , G ( x ′ , z , t ) = ∫ 0 z ∫ 0 ζ g ∂ u ∗ ∂ t ( x ′ , η , t ) d η d ζ .

Proof

For the proof of (33), choosing in (29) φ = ( ∂ u 1 ∗ ∂ t + ψ 1 , ∂ u 2 ∗ ∂ t + ψ 2 ) , where ψ ∈ Π ( K ) , we obtain

∫ ω k ˆ ψ + ∂ s ∗ ∂ t − ∂ s ∗ ∂ t d x ′ ≥ − ∑ i = 1 2 ∫ Ω μ ∂ u i ∗ ∂ z ∂ ψ i ∂ z d x ′ d z − ∑ i = 1 2 ∫ Ω γ ˆ g ˆ i ∂ u i ∗ ∂ t ψ i d x ′ d z + ∑ i = 1 2 ∫ Ω f ˆ i . ψ i d x ′ d z ,

using now the Green formula, equality (30), and the fact that ψ i = 0 on Γ 1 ∪ Γ L , we obtain

∫ ω k ˆ ψ + ∂ s ∗ ∂ t − ∂ s ∗ ∂ t d x ′ − ∫ ω μ τ ∗ . ψ ≥ 0 , ∀ ψ ∈ Π ( K ) .

This inequality remains valid for any ψ ∈ D ( ω ) 2 , and by density of D ( ω ) 2 in L 2 ( ω ) 2 , it also remains valid for any ψ ∈ L 2 ( ω ) 2 . For (34), we choose ψ = ± ∂ s ∗ ∂ t in (33), and then we follow the same techniques of [5].

To prove (35) by integrating (30) from 0 to z , we see that

− μ ∂ u i ∗ ∂ z ( x ′ , z , t ) + μ ∂ u i ∗ ∂ z ( x ′ , 0 , t ) = ∫ 0 z f ˆ i ( x ′ , η , t ) d η − γ ˆ ∫ 0 z g i ∂ u i ∗ ∂ t ( x ′ , η , t ) d η .

By integrating for the second time between 0 and z , we obtain

(36) μ u i ∗ ( x ′ , z , t ) = μ s i ∗ + z μ τ i ∗ + γ ˆ ∫ 0 z ∫ 0 ζ g i ∂ u i ∗ ∂ t ( x ′ , η , t ) d η d ζ − ∫ 0 z ∫ 0 ζ f ˆ i ( x ′ , η , t ) d η d ζ ,

we replace z by h ( x ′ ) and hence

(37) μ s i ∗ + h μ τ i ∗ = − γ ˆ ∫ 0 h ∫ 0 ζ g i ∂ u i ∗ ∂ t ( x ′ , η , t ) d η d ζ + ∫ 0 h ∫ 0 ζ f i ( x ′ , η , t ) d η d ζ .

Integrating (36) from 0 to h , we obtain

(38) μ ∫ 0 h u i ∗ ( x ′ , z , t ) d z = h μ s i ∗ + 1 2 h 2 μ τ i ∗ + γ ˆ ∫ 0 h ∫ 0 z ∫ 0 ζ g i ∂ u i ∗ ∂ t ( x ′ , η , t ) d η d ζ d z − ∫ 0 h ∫ 0 z ∫ 0 ζ f ˆ i ( x ′ , η , t ) d η d ζ d z ,

and from (37) and (38), we deduce that

F ˜ − h 2 μ s ∗ + μ ∫ 0 h u ∗ ( x ′ , z , t ) d z + G ˜ = 0 .

Then

∫ ω F ˜ − h 2 μ s ∗ + μ ∫ 0 h u ∗ ( x ′ , z , t ) d z + G ˜ ∇ ψ ( x ′ ) d x ′ = 0 . □

Theorem 6

The solution u i ∗ , i = 1 , 2 , of the limit problem (29) and (30) is unique in L 2 ( 0 , T ; V z ) ∩ L ∞ ( 0 , T ; V z ) .

Proof

Suppose that there exist two solutions u i ∗ and u i ∗ ∗ , i = 1 , 2 , of the variational inequality (29), we have

(39) μ ∑ i = 1 2 ∫ Ω ∂ u i ∗ ∂ z . ∂ ∂ z φ ˆ i − ∂ u i ∗ ∂ t d x ′ d z + γ ˆ ∑ i = 1 2 ∫ Ω g i ∂ u i ∗ ∂ t φ ˆ i − ∂ u i ∗ ∂ t d x ′ d z + J ˆ ( φ ˆ ) − J ˆ ∂ u ∗ ∂ t ≥ ∑ i = 1 2 f ˆ i , φ ˆ i − ∂ u i ∗ ∂ t

and

(40) μ ∑ i = 1 2 ∫ Ω ∂ u i ∗ ∗ ∂ z . ∂ ∂ z φ ˆ i − ∂ u i ∗ ∗ ∂ t d x ′ d z + γ ˆ ∑ i = 1 2 ∫ Ω g i ∂ u i ∗ ∗ ∂ t φ ˆ i − ∂ u i ∗ ∗ ∂ t d x ′ d z + J ˆ ( φ ˆ ) − J ˆ ∂ u i ∗ ∗ ∂ t ≥ ∑ i = 1 2 f ˆ i , φ ˆ i − ∂ u i ∗ ∗ ∂ t .

We take φ ˆ = ∂ u ∗ ∗ ∂ t in (39), then φ ˆ = ∂ u ∗ ∂ t in (40), and by summing the two inequalities, we obtain

μ ∑ i = 1 2 ∫ Ω ∂ ∂ z ( u i ∗ − u i ∗ ∗ ) . ∂ ∂ z ∂ u i ∗ ∂ t − ∂ u i ∗ ∗ ∂ t d x d z + γ ˆ ∑ i = 1 2 ∫ Ω g i ∂ u i ∗ ∂ t − g i ∂ u i ∗ ∗ ∂ t . ∂ u i ∗ ∂ t − ∂ u i ∗ ∗ ∂ t d x d z ≤ 0 ,

this implies that

μ d d t ∑ i = 1 2 ∂ ∂ z ( u i ∗ − u i ∗ ∗ ) L 2 ( Ω ) 2 ≤ 0 .

Then, we conclude

∥ u ∗ − u ∗ ∗ ∥ L 2 ( 0 , T ; V z ) = ∥ u ∗ − u ∗ ∗ ∥ L ∞ ( 0 , T ; V z ) = 0 . □

6 Conclusion

In this article, we studied the theoretical analysis of the elasticity system with nonlinear source and dissipative terms in a dynamic regime in a three-dimensional thin domain of Ω ε ⊂ R 3 with the Tresca friction law. First, the problem statement and variational formulation of the problem are formulated. We then obtained the estimates on displacement independently of the parameter ε . Finally, we obtained the main results concerning the limit of weak problem and its uniqueness.

S.Boularas@qu.edu.sa

Acknowledgments

The authors would like to thank the Deanship of Scientific Research, Qassim University for funding the publication of this project.

Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.
Conflict of interest: Authors state no conflict of interest.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2021-10-25

Revised: 2022-06-16

Accepted: 2022-06-29

Published Online: 2022-08-26

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/dema-2022-0033

Keywords for this article

asymptotic behavior; dissipative term; source term; Tresca friction law; weak solution

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