Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 3 (2025) 

  

964 
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Variational Analysis of a Dynamic Frictional Contact 

Problem with Adhesion and Long Memory in 

Viscoelastic Materials 

 

H. Hammar 1, S. Boutechebak 2 

1 Applied Mathematics Laboratory, Department of Mathematics, Faculty of Sciences, University of Setif 1, 

19000, Algeria.  

E-mail:hena.hammar@univ-setif.dz 

2 Applied Mathematics Laboratory, Department of Mathematics, Faculty of Sciences, University of Setif 1, 

19000, Algeria.  

E-mail: souraya.boutechbak@univ-setif.dz 

 

Article History: 

Received: 26-01-2025 

Revised: 15-03-2025 

Accepted: 28-05-2025 

Abstract: We consider a mathematical model that describes a dynamic frictional 

contact between a foundation and a viscoelastic body with long memory. The contact 

is modelled with a normal compliance condition in that the penetration is limited and 

restricted to a unilateral constraint and associated to the nonlocal friction law with 

adhesion, where the coefficient of friction is an independent solution. The adhesion 

of the contact surfaces is considered and modelled with a surface variable. We derive 

a variational formulation written as the coupling between a variational inequality and 

a differential equation. The existence and uniqueness result of the weak solution 

under a smallness assumption on the coefficient of friction is established. The proof 

is based on arguments of nonlinear evolution equation with monotone operators, a 

classical existence, differential equations, and the Banach fixed point theorem. 

Keywords: viscoelastic; normal compliance; dynamic process; adhesion; differential 

equations; friction; weak solution.. 

Mathematics subject classification 2010: 74M15, 74M10, 74F15, 49J40. 

 

1. Introduction 

Contact issues with deformable bodies are prevalent in industrial applications and daily life, 

significantly impacting structural and mechanical systems. The last fifty years, variational 

inequalities have become a formidable tool in the mathematical study of many non-linear 

problems in physics and mechanics and the analysis of mathematical models in contact mechanics 

has expanded quickly over the past few decades, the complexity of the boundary conditions and 

the diversity of the constituent equations leading to variational formulations of the inequation 

type. The state of the art in mathematics, mechanics and numerical analysis is contained in [18]. 

This reference finds numerical investigations and a thorough analysis of the adhesive contact 

problem. General models for unilateral and frictional contact problems with adhesion can be found 

in [5, 7, 9, 13, 19, 20]. In [21], a quasistatic viscoelastic unilateral and frictional contact problem 

with adhesion and long memory was studied. This paper aims to model and establish the 

mailto:hena.hammar@univ-setif.dz
mailto:souraya.boutechbak@univ-setif.dz


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variational analysis for a dynamic process of unilateral and frictional contact with adhesion and 

long memory. Remember that [2, 3, 4, 6, 12, 14, 15, 16] has studied models for dynamic or 

quasistatic processes of frictionless adhesive contact between a deformable body and foundation. 

Following [10,11], the bondind field is used as an extra variable β wich satisfies the restriction 0 ≤

𝛽 ≤ 1, at a point of the contact surface of the boundary, when 𝛽 = 1 all of the bonds are active 

and the adhesion is complete and for 𝛽 = 0 the bonds are inactive, severed and there is no 

adhesion; when 0 < 𝛽 < 1 the adhesion is partial and only a fraction 𝛽 of the bonds is active. 

We direct the reader’s attention to the comprehensive bibgraphy on the topic in [1, 10, 17, 18, 19]. 

In this paper, we deal with the study of a dynamic problem, we derive a variational formulation 

of the problem which is set a system coupling a variational second order evolution inequality. We 

establish the existence and the uniqueness of a weak solution of the model. The idea is to reduce 

the second order evolution inequality of the system to first order evolution inequality. After this, 

we use classical results on first order evolution inequalities and differential equations and the fixed 

point arguments. The rest of the paper is structured as follows. In section 2 and 3 we present some 

notations and preliminaires, and the viscoelastic unilateral and frictional contact model with 

adhesion and long memory, and provide comments on the contact boundary conditions. In section 

4, we list the assumptions on the data and derive the variational formulation. In section 5, we 

present our main results on existence and uniqueness which state the unique weak solvability. 

2. Notations and preliminaries 

Throughout this paper, 𝕊𝑑  represents the space of second order symmetric tensors on ℝ𝑑(d =2,3) 

while |. | represents the Euclidean norm on ℝ𝑑and 𝕊𝑑 . 

Thus, for every 𝑢, 𝑣 ∈ ℝ𝑑 , 𝑢. 𝜐 = 𝑢𝑖𝜐𝑖 and |𝜐| = (𝜐, 𝜐)
1

2, and for every  𝜎, 𝜏 ∈ 𝕊𝑑 , 𝜎. 𝜏 = 𝜎𝑖𝑗𝜏𝑖𝑗 ,   

|𝜏| = (𝜏. 𝜏)
1
2. 

The summing convention over represented indices is used here and below, where the indices 

𝑖 and 𝑗 range from 1 to d. Let Ω ⊂ ℝ𝑑  be a bounded domain with a Lipschitz boundary Γand let 

𝜈 denote the unit outer normal on Γ. For Lebesgue and Sololev spaces associated to Ω and Γand 

introduce the spaces, 

𝐻 = 𝕃2(Ω)𝑑 = {𝑢 = (𝑢𝑖)/𝑢𝑖 ∈ 𝕃
2(Ω)}, 

ℋ= {𝜎 = (𝜎𝑖𝑗) /𝜎𝑖𝑗 = 𝜎𝑗𝑖 ∈ 𝕃
2(Ω)},  

 H1 = {u = (ui)/ε(u) ∈ ℋ}, 

ℋ1 = {σ ∈ ℋ/Di𝜐σ ∈ H }. 

Here The deformation ε and divergence Di𝜐 are operators defined by 

𝜀(𝑢) = (𝜀𝑖𝑗(𝑢)),    𝜀𝑖𝑗(𝑢) =
1

2
(𝑢𝑖,𝑗 + 𝑢𝑗,𝑖),    𝐷𝑖𝜐𝜎 = (𝜎𝑖𝑗,𝑗). 

The spaces H, ℋ, H1 and ℋ1 are real Hilbert spaces endowed with the canon- ical inner products 

given by 



Communications on Applied Nonlinear Analysis 

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(𝑢, 𝜐)𝐻 = ∫ 𝑢𝑖𝜐𝑖  𝑑𝑥
Ω

          ∀ 𝑢, 𝜐 ∈ 𝐻, 

(𝜎, 𝜏)ℋ = ∫ 𝜎𝑖𝑗𝜏𝑖𝑗 𝑑𝑥
Ω

           ∀𝜎, 𝜏 ∈ ℋ, 

  (u, 𝜐)H1 = (u, 𝜐)H +  (𝜀(𝑢), 𝜀(𝜐))ℋ  ∀u, 𝜐 ∈ H1, 

(𝜎, 𝜏 )ℋ1
 =  (𝜎, 𝜏 )ℋ + (Di𝜐σ, Di𝜐τ )H ∀σ, τ ∈ ℋ1. 

The associated norms on the spaces H, ℋ, H1 and ℋ1 are denoted by |.|H,|. |ℋ, |.|H1 and  

|. |ℋ1
, respectively. For every element 𝜐 ∈ H1 we also use the    notation 𝜈 for the trace of 𝜈 

on Γ and we denote by 𝜐𝜈 and 𝜐𝜏 the normal and the tangential components of 𝜐 on Γ given by 

𝜐𝜈 = 𝜐. 𝜈,   𝜐𝜏 = 𝜐 − 𝜐𝜈𝜈.            (2.1)  

We also denote by 𝜐𝜈 and 𝜐𝜏 the normal and tangential traces of a function 𝜎 ∈ ℋ1 and we 

recall that when σ is a regular function then 

𝜎𝜈 = (𝜎𝜈). 𝜈,       𝜎𝜏 = 𝜎𝜈 − 𝜎𝜈𝜈.                                         (2.2) 

 And the following Green’s formula holds:  

(𝜎, 𝜀(𝜐))
ℋ
+ (𝐷𝑖𝜐𝜎, 𝜐)𝐻 = ∫ 𝜎𝜈. 𝜐 𝑑𝑎    ∀𝜐 ∈ 𝐻1,Γ

      (2.3) 

where the surface measure element is da. Let 𝑇 > 0, for every real Hilbert space 𝑋 we employ 

the usual notation for the spaces 𝕃𝑝(0,𝑇;𝑋), 1 ≤ 𝑝 ≤∞ and 𝑊1,∞ (0, 𝑇;𝑋). Recall that the 

norm on the space 𝑊1,∞ (0,𝑇; 𝑋) is given by 

‖𝑢‖𝑊1,∞(0.𝑇;𝑋) = ‖𝑢‖𝕃∞(0.𝑇;𝑋) + ‖�̇�‖𝕃∞(0.𝑇,𝑋), 

where �̇� denote the first derivative of u with respect to time. Fanilly, the space of continuous 

functions from [0, T ] to X  is denoted by C ([0,T ] ; X) with the norm 

 ‖𝑥‖𝐶([0,𝑇];𝑋) = max
𝑡∈[0,𝑇]

‖𝑥(𝑡)‖𝑋. 

Moreover, for a real number r, r+ is used to represent its positive part, that is r+ = max {r,0}. 

3. Problem statement 

The physical setting is the following. A viscoelastic body with long memory occupies a bounded 

domain Ω ⊂ ℝ𝑑(d = 2, 3) with a regular boundary Γ that is partitioned into three disjoint 

measurable parts Γ1, Γ2 and Γ3such that meas Γ1> 0. The body is acted upon by a volume 

force of density 𝜑1on Ω and a surface traction of density 𝜑2on Γ2 and it is in unilate contact 

with adhesion following the nonlocal friction law with a foundation, over the potential contact 

surfaceΓ3. Therefore, the mechanical problem’s classical formulation is written as follows. 

Problem P1.  

Find a displacement field u: Ω × [0,T] →ℝ𝑑 , a stress field 𝜎: Ω × [0,T ] → 𝕊𝑑and a bonding 

field β: Γ3× [0,T ] → [0,1] such that for all 𝑡 ∈ [0,T ]: 



Communications on Applied Nonlinear Analysis 

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   𝜎(𝑡) = 𝒜𝜀(𝑢(𝑡)) + 𝒢𝜀(�̇�(𝑡)) + ∫ ℱ(𝑡 − 𝑠)𝜀(𝑢(𝑠))𝑑𝑠
𝑡

0
  in Ω,                  (3.1) 

𝐷𝑖𝜐𝜎(𝑡) + 𝜑1(𝑡) = 𝜌�̈�   in Ω,                                                                         (3.2) 

  𝑢(𝑡) = 0  on Γ1,                                                                                               (3.3)    

   𝜎(𝑡)𝜈 = 𝜑2  on Γ2,                                                                                          (3.4) 

  
𝑢𝜈 ≤ 𝑔; 𝜎𝜈(𝑡) + 𝑝(𝑢𝜈(𝑡)) − 𝑐𝜈𝛽

2(𝑡)𝑅𝜈(𝑢𝜈(𝑡)) ≤ 0

(𝜎𝜈(𝑡) + 𝑝(𝑢𝜈(𝑡)) − 𝑐𝜈𝛽
2(𝑡)𝑅𝜈(𝑢𝜈(𝑡))) (𝑢𝜈(𝑡) − 𝑔) = 0

}  on Γ3,              (3.5) 

      

|𝜎𝜏(𝑡) + 𝑐𝜏𝛽
2(𝑡)𝑅𝜏(𝑢𝜏(𝑡))| ≤ 𝜇|𝑅𝜎𝜏(𝑢(𝑡))|

|𝜎𝜏(𝑡) + 𝑐𝜏𝛽
2(𝑡)𝑅𝜏(𝑢𝜏(𝑡))| < 𝜇|𝑅𝜎𝜏(𝑢(𝑡))| ⇒ 𝑢𝜏(𝑡) = 0

|𝜎𝜏(𝑡) + 𝑐𝜏𝛽
2(𝑡)𝑅𝜏(𝑢𝜏(𝑡))| = 𝜇|𝑅𝜎𝜏(𝑢(𝑡))| ⟹

∃𝜆 ≥ 0 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑢𝜏(𝑡) = −𝜆 (𝜎𝜏(𝑡) + 𝑐𝜏𝛽
2(𝑡)𝑅𝜏(𝑢𝜏(𝑡)))}

 
 

 
 

  on Γ3,        (3.6) 

     �̇�(𝑡) = − [𝛽(𝑡) (𝑐𝜈 (𝑅𝜈(𝑢𝜈(𝑡)))
2

+ 𝑐𝜏|𝑅𝜈𝑢𝜏(𝑡)|
2 − 𝜀𝑎)]

+
       on Γ3,         (3.7) 

β (0) = β0 on Γ3,         (3.8) 

𝑢(0) = 𝑢0,   𝑢(𝑡) = 𝑢1̇  in Ω.         (3.9) 

The viscoelastic constitutive law with long memory of the material is represented by the equation 

(3.1). Here 𝒢 and 𝒜 are nonlinear operators describing the purely viscous and the elastic 

properties of the material, respectively and  ∫ ℱ(𝑡 − 𝑠)𝜀(𝑢(𝑠))𝑑𝑠
𝑡

0
 is the memory term in which ℱ 

denotes the tensor of relaxation, the stress σ (t) at current instant t depends on the whole history  

of strains up to this moment of time. Equation (3.2) represents the equation of motion where ρ 

denotes the material mass density, while (3.3) and (3.4) are the displacement and traction 

boundary conditions, respectively, in which 𝜎𝜈 represents the Cauchy stress vector. The 

conditions (3.5) represents the unilateral contact with adhesion in which 𝑐𝜈 is a given a 

adhesion coefficient which may dependent on 𝑥 ∈ Γ3 and 𝑅𝜈  𝑎𝑛𝑑 𝑅𝜏 are truncation operators 

defined by  

  𝑅𝜈(𝑠) = {

𝐿  𝑖𝑓 𝑠 < −𝐿
−𝑠  𝑖𝑓 − 𝐿 ≤ 𝑠 ≤ 0

0  𝑖𝑓  𝑠 > 0
,       

   𝑅𝜏(𝑠) = {
𝜐     𝑖𝑓    |𝜐| ≤ 𝐿,

𝐿
𝜐

|𝜐|
   𝑖𝑓  |𝜐| > 𝐿.

 

Here 𝐿 > 0 is the characteristic length of the bond, beyond which the latter has no additional 

traction (see [18]) and p is a normal compliance function which satisfies the assumption (4.14), 

𝑔 denotes the maximum value of the penetration which satisfies 𝑔 ≥ 0. When 𝑢𝜈  < 0 i.e. when 

there is separation between the body and the foundation then the condition (3.5) combined with 

hypothese (3.23) and definition of 𝑅𝜈 shows that 𝜎𝜈 = 𝑐𝜈𝛽
2𝑅𝜈(𝑢𝜈) and does not exeed the 

value 𝐿 cν  L∞(Γ3). When g > 0, the body may interpenetrate into the fondation, but the 



Communications on Applied Nonlinear Analysis 

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penetration is limited that is uν ≤ 𝑔. In this case of penetration (i.e. uν ≥ 0), when 0 ≤ uν < 𝑔 

then −σν = p(uν) which means that the reaction of the foundation is uniquely determined by 

the normal displacement and σν ≤ 0. Since p is an increasing function then the reaction is 

increasing with the penetration. When uν = 𝑔 then −σν ≥ p (𝑔) and σν is not uniquely 

determined. When 𝑔 > 0 and p = 0, condition (3.5) become the Signorini’s contact conditions 

with a gap and adhesion 

uν ≤ 𝑔,  σν − cνβ
2𝑅ν(uν) ≤ 0, (σν − cνβ

2𝑅ν(uν))(uν − 𝑔) = 0. 

When 𝑔 = 0, the conditions (3.5) combined with hypothese (3.23) lead to the Signorini contact 

conditions with adhesion, with zero gap, given by 

uν ≤ 0,  σν − cνβ
2𝑅ν(uν) ≤ 0, (σν − cνβ

2𝑅ν(uν))uν = 0. 

These contact conditions were used in [20]. It follows from (3.5) that there is no penetration 

between the body and the foundation, since uν ≤ 0 during the process. Also, note that when the 

bonding field vanishes, then the contact conditions (3.5) become the classical Signorini contact 

conditions with zero gap, that is, 

uν ≤ 0,  σν ≤ 0,  σνuν = 0. 

Condition (3.6) represent Couloub’s law of dry friction with adhesion where 𝜇 denotes the 

coefficient of friction. Equation (3.7) represents the ordinary differentail equation which 

describes the evolution of the bonding field and it was already used in [20, 21] . Since β ≤ 0 on 

𝛤3×[0, T ] , once debonding occurs bonding cannot be reestablished, indeed, the adhesion process 

is irreversible. Also from [19] it must be pointed out clearly that condition (3.7) does not allow for 

complete debonding in finite time. In Equation (3.8) β0 denotes the initial bonding. Finally, in 

equation (3.9) u0 is the initial displacement and u1 the initial velocity. 

4. Variational formulation 

For a weak formulation of problem P1, let V be the closed subspace of H1 defined by 

V = {𝜐 ∈ H1 : 𝜐 = 0 on Γ1} . 

And the convex subset of admissible displacement given by 

K = {𝜐 ∈ V : 𝜐ν ≤ 𝑔 a.e on Γ3} . 

Since meas Γ1 > 0, the following Korn’s inequality holds [8] 

 ε (𝜐)  ℋ ≥ 𝑐Ω 𝜐  H1  ∀𝜐 ∈ V.       (4.1)  

Where 𝑐Ω > 0 is a constant which depends only on Ω and Γ1. We equip 𝑉 with the inner 

product 

(𝑢, 𝜐)𝑉 = 〈𝜀(𝑢), 𝜀(𝜐)〉ℋ . 

And  .  V is the associated norm. It follows from Korn’s inequality (4.1) that the norms  .  

H1 and  .  V are equivalent on V. Then (V, .  V ) is a real Hilbert space. Moreover by Sobolev’s 

trace theorem, there exists 𝑑Ω> 0 which only depends on the domain Ω, Γ1 and Γ3 such that 



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 𝜐  (𝕃2(Γ3))d ≤ 𝑑Ω 𝜐  V  ∀𝜐 ∈ V.                                                (4.2)  

The body forces and surface tractions have the regularity 

   𝜑1 ∈ 𝐶([0, 𝑇]; 𝐻),       𝜑2 ∈ 𝐶 ([0, 𝑇]; (𝕃
2(Γ3))

𝑑
)                                (4.3) 

The function 𝑓 : [0.T ] → V  defined by 

(𝑓(𝑡), 𝜐)𝑉 = ∫ 𝜑1(𝑡)𝜐𝑑𝑥Ω
+ ∫ 𝜑2(𝑡)𝑑𝑎     ∀𝜐 ∈ 𝑉,Γ2

    𝑡 ∈ [0, 𝑇],          (4.4) 

And that (4.3) and (4.4) imply 

𝑓 ∈ C ([0.T ] ; V ) 

In the study of the mechanical problem P1 (3.1)-(3.9) let the following assumptions:  

The elasticity operator 𝒜 satisfies 

{
 
 
 
 

 
 
 
 
(𝑎) 𝒜: Ω × 𝕊𝑑 ⟶ 𝕊𝑑 ,                                                                                                
(𝑏) 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑀 > 0 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 ∶  |𝒜(𝑥, 𝜀1) − 𝒜(𝑥, 𝜀2)| ≤ 𝑀|𝜀1 − 𝜀2|

∀𝜀1, 𝜀2 ∈ 𝕊
𝑑 , 𝑎. 𝑒.  𝑥 ∈ Ω

(𝑐) 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑚 > 0 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 ∶                                                                    

(𝒜(𝑥, 𝜀1) − 𝒜(𝑥, 𝜀2)). (𝜀1 − 𝜀2) ≥ 𝑚|𝜀1 − 𝜀2|
2

∀ 𝜀1, 𝜀2 ∈ 𝕊
𝑑 , 𝑎. 𝑒.  𝑥 ∈ Ω,

(𝑑) 𝑡ℎ𝑒 𝑚𝑎𝑝𝑝𝑖𝑛𝑔 𝑥 ⟶ 𝒜(𝑥, 𝜀) 𝑖𝑠 𝑙𝑒𝑏𝑒𝑠𝑔𝑢𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑎𝑏𝑙𝑒 𝑖𝑛 Ω  𝑓𝑜𝑟 𝑎𝑙𝑙  

𝜀 ∈ 𝕊𝑑

(𝑒) 𝑡ℎ𝑒 𝑚𝑎𝑝𝑝𝑖𝑛𝑔 𝑥 ⟶ 𝒜(𝑥, 0) ∈ ℋ.                                                                 

       (4.5) 

The viscosity operator 𝒢 satisfies 

   

{
  
 

  
 
(𝑎) 𝒢: Ω × 𝕊𝑑 ⟶ 𝕊𝑑 ,                                                                                             
(𝑏) 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝐿𝒢 > 0 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡: |𝒢(𝑥, 𝜉1) − 𝒢(𝑥, 𝜉2)| ≤ 𝐿𝒢|𝜉1 − 𝜉2| 

∀𝜉1, 𝜉2 ∈ 𝕊
𝑑 ,

(𝑐) 𝑡ℎ𝑒 𝑚𝑎𝑝𝑝𝑖𝑛𝑔 𝑥 ⟶ 𝒢(𝑥, 𝜉) 𝑖𝑠 𝑙𝑒𝑏𝑒𝑠𝑔𝑢𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑎𝑏𝑙𝑒 𝑖𝑛 Ω 𝑓𝑜𝑟 𝑎𝑙𝑙

𝜉 ∈ 𝕊𝑑 ,
(𝑑) 𝑡ℎ𝑒 𝑚𝑎𝑝𝑝𝑖𝑛𝑔 𝑥 ⟶ 𝒢(𝑥, 0) ∈ ℋ.                                                             

         (4.6) 

The mass density satisfies 

ρ ∈ 𝕃∞ (Ω) , there exists ρ∗ > 0 such that ρ(x) ≥ ρ∗, a.e. x ∈ Ω.                                  (4.7)  

The space of the tensors of fourth order defined by 

ℋ∞ = {𝜀 = (𝜀𝑖𝑗𝑘𝑙): 𝜀𝑖𝑗𝑘𝑙 =𝜀𝑗𝑖𝑘𝑙 = 𝜀𝑘𝑙𝑖𝑗 ∈ 𝕃∞ (Ω), 1 ≤ i, j, k, l ≤ d}. 

Which is the real Banach space with the norm 

‖𝜀‖ℋ∞
= max

1≤𝑖,𝑗,𝑘,𝑙≤𝑑
‖𝜀𝑖𝑗𝑘𝑙‖𝕃∞(Ω). 

The tensor of relaxation ℱ satisfies 

ℱ ∈𝐶 ([0.T ] ; ℋ∞)                                                       (4.8) 



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The adhesion coefficients cν, cτ and εa satisfy 

cν, cτ ∈ 𝕃∞ (Γ3) , εa ∈ 𝕃2 (Γ3) and cν, cτ , εa > 0  a.e. on Γ3 .                                      (4.9) 

That the initial bonding field satisfies 

β0 ∈ 𝕃2 (Γ3) , 0 ≤ β0 ≤ 1 a.e on Γ3.                                                      (4.10) 

Finally, the initial data satisfy 

u0 ∈ V and u1 ∈ H.                                                                        (4.11) 

A modified inner product on the Hilbert space H = 𝕃2 (Ω)d given by 

((u, 𝜐))H = (ρu, 𝜐)H ∀u, 𝜐 ∈ H, 

that, it is weighted with ρ and let  .  H be the associated norm i.e. 

‖𝜐‖𝐻 = (𝜌𝜐, 𝜐)𝐻

1
2     ∀𝜐 ∈ 𝐻. 

It follows from assumptions (4.7) that  .  H and |.|H are equivalent norms on 

H and also the inclusion mapping of (V, |.|V ) into (H, |.|H ) is continuons and 

dense. Let 𝑉 ́ be the dual space of 𝑉. Identifying H with its own dual, it can 

write the Gelfand triple 

𝑉  ⊂ H ⊂ 𝑉 ́; 

The notation (. , . )𝑉′×𝑉 represents the duality pairing between 𝑉′and V recall that 

(𝑢, 𝜐)𝑉′×𝑉 = ((𝑢, 𝜐))𝐻       ∀𝑢 ∈ 𝐻, ∀𝜐 ∈ 𝑉. 

Next, the subset 𝑊 of H1 are defined as 

𝑊 = {𝜐 ∈ H1 : div (𝜐) ∈ H}, 

and let 𝑗c : V×V → R, 𝑗f : (V ∩ W ) × V → ℝ be the functionals given by 

    𝑗𝑐(𝑢, 𝜐) = ∫ 𝑝(𝑢𝜈)𝜐𝜈𝑑𝑎      ∀(𝑢, 𝜐) ∈ 𝑉 × 𝑉,Γ3
  

  𝑗𝑓(𝑢, 𝜐) = ∫ 𝜇|𝑅𝜎𝜈(𝑢)||𝜐𝜏|𝑑𝑎    ∀(𝑢, 𝜐) ∈ (𝑉 ∩𝑊) × 𝑉,Γ3
 

Where 

𝑅: 𝐻
1

2(Γ) ⟶ 𝕃2(Γ3) is a linear and continuous mapping (𝑠𝑒𝑒 [7]).                        (4.12) 

The coefficient of friction 𝜇 is assumed to satisfy 

𝜇 ∈ 𝕃∞ (Γ3) and 𝜇 ≥ 0 a.e on Γ3                                                                         (4.13) 

Next, let 

𝑗= 𝑗c + 𝑗f 



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The functional ℎ is defined by 

h : 𝕃2 (Γ3) × V × V→ ℝ, 

  ℎ(𝛽, 𝑢, 𝜐) = ∫ (−𝑐𝜈𝛽
2𝑅𝜈(𝑢𝜈)𝜐𝜈 + 𝑐𝜏𝛽

2𝑅𝜏(𝑢𝜏)𝜐𝜏)𝑑𝑎, ∀(𝛽, 𝑢, 𝜐) ∈ 𝕃
2(Γ3) × 𝑉 × 𝑉   Γ3

 

where the normal compliance function 𝑝 satisfies: 

  

{
 
 
 
 

 
 
 
 
(𝑎) 𝑝: Γ3 × ℝ⟶ ℝ+,                                                                            
(𝑏) ∃𝐿𝑝 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 ∶  |𝑝(𝑥 − 𝑟1) − 𝑝(𝑥 − 𝑟2)| ≤ 𝐿𝑝|𝑟1 − 𝑟2|       

∀ 𝑟1, 𝑟2 ∈ ℝ,   𝑎. 𝑒.  𝑥 ∈ Γ3,
(𝑐) (𝑝(𝑥, 𝑟1) − 𝑝(𝑥, 𝑟2))(𝑟1 − 𝑟2) ≥ 0,                                               

∀ 𝑟1, 𝑟2 ∈ ℝ,   𝑎. 𝑒.  𝑥 ∈ Γ3,
(𝑑) 𝑡ℎ𝑒 𝑚𝑎𝑝𝑝𝑖𝑛𝑔 𝑥 ⟶ 𝑝(𝑥, 𝑟) 𝑖𝑠 𝐿𝑒𝑏𝑒𝑠𝑔𝑢𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑛 Γ3,

𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑟 ∈ ℝ,
(𝑒) 𝑝(𝑥, 0) = 0, 𝑎. 𝑒.  𝑥 ∈ Γ3.                                                                 

                              (4.14) 

Finaly, the following set of the bonding field: 

𝐵 = {𝜃: [0, 𝑇] ⟶ 𝕃2(Γ3): 0 ≤ 𝜃(𝑡) ≤ 1, ∀∈ [0, 𝑇], 𝑎. 𝑒. 𝑜𝑛 Γ3 }. 

By a standard procedure based on Green’s formula the following variational formulation of 

problem P1 is derived in terms of displacement and bonding field. 

Problem PV. Find a displacement field u: Ω ×[0.T] → ℝ𝑑 , a stress field  

σ: Ω × [0.T] → 𝕊𝑑 and a bonding field β: Γ3 × [0.T ] →[0.1] such that u (t) ∈ 𝐾 ∩  𝑊, 

𝜎(𝑡) = 𝒜𝜀(𝑢(𝑡)) + 𝒢𝜀(�̇�(𝑡)) + ∫ ℱ(𝑡 − 𝑠)𝜀(𝑢(𝑠))𝑑𝑠    𝑎. 𝑒. 𝑡 ∈ [0, 𝑇]
1

0
  

((𝑢,̈ 𝜔 − �̇�))
𝐻
+ (𝒜𝜀(𝑢), 𝜀(𝜔) − 𝜀(�̇�))

ℋ
+ (𝒢𝜀(�̇�), 𝜀(𝜔) − 𝜀(�̇�))

ℋ
  

+(∫ ℱ(𝑡 − 𝑠)𝜀(𝑢(𝑠))𝑑𝑠, 𝜀(𝜔) − 𝜀(�̇�)
𝑡

0

)
ℋ

+ ℎ(𝛽(𝑡), 𝑢(𝑡), 𝜔 − �̇�) + 𝑗𝑐(𝑢(𝑡), 𝜔) 

−𝑗𝑐(𝑢(𝑡), �̇�) + 𝑗𝑓(𝑢(𝑡), 𝜔) − 𝑗𝑓(𝑢(𝑡), �̇�) ≥ (𝑓(𝑡), 𝜔(𝑡) − �̇�(𝑡))
𝑉

                                   (4.15) 

 

�̇�(𝑡) = − [𝛽(𝑡) (𝑐𝜈 (𝑅𝜈(𝑢𝜈(𝑡)))
2

+ 𝑐𝜏|𝑅𝜏(𝑢𝜏(𝑡))|
2
− 𝜀𝑎)]

+
  𝑎. 𝑒. 𝑡 ∈ [0, 𝑇]                   (4.16)  

  𝑢(0) = 𝑢0,   �̇�(0) = 𝑢1 = 𝜐0,   𝛽(0) = 𝛽0.                                                                         (4.17)  

5. Existence and uniqueness result 

The main result in this section is the following existence and uniqueness result. 

Theorem 5.1: Let the assumptions (4.3)-(4.13) hold. Then, there exists a constant 𝜇0 > 0 

such that problem PV has a unique solution (u, σ, β) which satisfies if 

    ‖𝜇‖𝕃∞(Γ3) < 𝜇0. 



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The proof of theorem 5.1 is carried out several steps. In the first step, the closed subset Z of 

the space 𝐶([0.T ] ; 𝕃2(Γ3)) is defined as 

  𝑍 = {𝜃 ∈ 𝐶([0, 𝑇]; 𝕃2(Γ3)) ∩ 𝐵: 𝜃(0) = 𝛽0}, 

where the Banach space 𝐶([0.T ] ; 𝕃2(Γ3)) is endowed with the norm 

  ‖𝐵‖𝑘 = max
𝑡∈[0,𝑇]

[𝑒𝑥𝑝(−𝑘𝑡)‖𝛽(𝑡)‖𝕃2(Γ3)], 𝑘 > 0. 

Next for a given ξ ∈ Z, let the following variational problem. 

Problem P1ξ. Find 𝑢𝜉  ∈ 𝐶 ([0.T ] ; 𝑉 ) such that 𝑢𝜉  ∈ K ∩ W 

  ((�̈�𝜉 , 𝜔 − �̇�𝜉)) + (𝒜𝜀(𝑢𝜉), 𝜀(𝜔) − 𝜀(�̇�𝜉))
ℋ
+ (𝒢𝜀(�̇�𝜉), 𝜀(𝜔) − 𝜀(�̇�𝜉))

ℋ
  

+(∫ ℱ(𝑡 − 𝑠)
𝑡

0
𝜀 (𝑢𝜉(𝑠)) 𝑑𝑠, 𝜀(𝜔) − 𝜀(�̇�𝜉))

ℋ
+ ℎ(𝛽(𝑡), 𝑢𝜉(𝑡), 𝜔 − �̇�𝜉) + 𝑗𝑐(𝑢𝜉(𝑡), 𝜔) −

𝑗𝑐(𝑢𝜉(𝑡), �̇�𝜉) + 𝑗𝑓(𝑢𝜉(𝑡), 𝜔) − 𝑗𝑓(𝑢𝜉(𝑡), �̇�𝜉) ≥ (𝑓(𝑡), 𝜔(𝑡) − �̇�𝜉(𝑡))
𝑉
∀  𝜔 ∈ 𝐾, 𝑡 ∈ [0, 𝑇]      (5.1)                       

We have the following results  

Theorem 5.2. There exists a constant µ1 > 0 such that problem Pξ has a 

unique solution if 

 µ  𝕃∞(Γ3) < µ1 

Let 𝜂 ∈𝐶 ([0.T ] ; V ) be given the following intermediate problem is introduced by: 

  Problem 𝑃𝜉𝜂. Find 𝑢𝜉𝜂 ∈ 𝐶([0, 𝑇]; 𝑉) such that 𝑢𝜉𝜂 ∈ 𝐾 ∩𝑊 

((�̈�𝜉𝜂 , 𝜔 − �̇�𝜉𝜂)) + (𝒢𝜀(�̇�𝜉𝜂), 𝜀(𝜔) − 𝜀(�̇�𝜉𝜂))
ℋ
+ (𝜂(𝑡), 𝜀(𝜔) − 𝜀(�̇�𝜉𝜂))

ℋ
 

+𝑗𝑓(𝑢𝜉𝜂(𝑡), 𝜔) − 𝑗𝑓(𝑢𝜉𝜂(𝑡), �̇�𝜉𝜂) ≥ (𝑓(𝑡), 𝜔(𝑡) − �̇�𝜉𝜂(𝑡))
𝑉

   

𝑢𝜉𝜂(0) = 𝑢0,   �̇�𝜉𝜂(0) = 𝑢1  ∀ 𝜔∈ 𝐾, 𝑡 ∈ [0, 𝑇]                        (5.2)    

 Since Riesz’s representation theorem representation theorem implies that there exists an element 

𝑓𝜂∈ 𝐶 ([0.T ] ; V ) such that 

(𝑓𝜂(𝑡), 𝜔)𝑉 =
(𝑓(𝑡), 𝜔)𝑉 − (𝜂(𝑡), 𝜀(𝜔))ℋ, 

the problem 𝑃𝜉𝜂 is equivalent to the following problem. 

Problem  𝑃2𝜉𝜂 . Find 𝑢𝜉𝜂  ∈ 𝐶 ([0.T ] ; 𝑉) such that 

   𝑢𝜉𝜂 ∈ 𝐾 ∩𝑊, 

((�̈�𝜉𝜂 , 𝜔 − �̇�𝜉𝜂)) + (𝒢𝜀(�̇�𝜉𝜂), 𝜀(𝜔) − 𝜀(�̇�𝜉𝜂))
ℋ

 +𝑗𝑓(𝑢𝜉𝜂(𝑡), 𝜔) − 𝑗𝑓(𝑢𝜉𝜂(𝑡), �̇�𝜉𝜂)    

≥ (𝑓𝜂(𝑡),𝜔 − 𝑢𝜉𝜂)𝑉          ∀𝜔 ∈ 𝐾, 𝑡 ∈
[0, 𝑇]                                                              (5.3) 



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Lemma 5.3 There exists a constant 𝜇1 > 0 such that problem P2ξη has a unique solution 

if    µ  𝕃
∞(Γ ) < µ1. 

The proof is based on several step by using arguments on Banach fixed point theorem. Indeed, 

let q ∈ C+ where C+ is a non-empty closed subset of 𝕃2(Γ3) 

defined as    𝐶+ = {𝑠 ∈ 𝕃
2(Γ3); 𝑠 ≥ 0 𝑎. 𝑒. 𝑜𝑛Γ3 }

  

  And  let the functional     𝑗𝑞: 𝑉 ⟶ ℝ given by  

𝑗𝑞(𝜐) = ∫ 𝜇𝑞|𝜐𝜏|𝑑𝑎    ∀𝜐 ∈ 𝑉.Γ3
    

We consider the following auxiliary problem. 

Problem 𝑷𝝃𝜼𝒒. Find 𝑢𝜉𝜂𝑞  ∈ C ([0.T ] ; V) such that 

  𝑢𝜉𝜂𝑞 ∈ 𝐾, ((�̈�𝜉𝜂𝑞 , 𝜔 − �̇�𝜉𝜂𝑞)) + (𝒢𝜀(�̇�𝜉𝜂𝑞), 𝜀(𝜔) − 𝜀(�̇�𝜉𝜂𝑞))
ℋ
+ 𝑗𝑞(𝜔) 

  −𝑗𝑞(𝑢𝜉𝜂𝑞(𝑡), �̇�𝜉𝜂𝑞) ≥ (𝑓𝜂(𝑡),𝜔 − 𝑢𝜉𝜂𝑞)𝑉;  ∀𝜔 ∈ 𝐾, 𝑡 ∈
[0, 𝑇]                                     (5.4) 

Lemma 5.4. Problem 𝑷𝝃𝜼𝒒 has a unique solution with the regularity 𝜐𝜉𝜂𝑞 ∈ C(0, T ; H) ∩ 

𝕃2(0, T ; V ) ∩ W 1,2(0, T ; 𝑉′). 

Proof: The continuous injection of V into 𝕃2(Γ3)d implies that j is continuous  

and convex. We define the sequence:          

   𝑗𝜀(𝜐) = ∫ 𝑢𝑞√|𝜐𝜏|2 + 𝜀2 𝑑𝑠  Γ3
 

  ∀ 𝜐 ∈ 𝑉 , ∀𝜀 > 0 

 Its derivative of F ré chet is given by : 

  𝑗𝜀
′(𝜐). 𝜔 = ∫ 𝑢𝑞

(𝜐𝜏,𝜔𝜏)

√|𝜐𝜏|
2+𝜀2

𝑑𝑠,   ∀𝜐 ∈ 𝑉, ∀𝜀 > 0.
Γ3

 

Then 𝑗𝜀 is of class C1. Direct algebraic calculations show that ∀ α ≥ 0, 

 β ≥ 0 such that α + β = 1 and for any real x and y, n ≥ 1 : 

√(𝛼𝑥 + 𝛽𝑦)2 +
1

𝑛
≤ 𝛼√𝑥2 +

1

𝑛
+ 𝛽√𝑦2 +

1

𝑛
 

So jε is convex ∀ε > 0. also: 

∃C > 0, ∀𝜔 ∈ V, |𝑗𝜀
′(𝜔)|V ′ ≤ C|𝑔|𝕃2(Γ3)                                                              (5.5) 

The hypothesis (4.5) (a) implies that G : V → 𝑉′is a continuous Lipschitz operator. Since 

𝑗𝜀
′  is continuous then G +𝑗𝜀

′  is a continuous and therefore 

hemicontinuous operator. Now, according to (4.5)(b) and the monotony of 𝑗𝜀
′ we find : 

〈(𝐺 + 𝑗𝜀
′)𝑢 − (𝐺 + 𝑗𝜀

′)𝜐, 𝑢 − 𝜐〉𝑉′×𝑉 ≥ 𝑚𝒢|𝑢 − 𝜐|𝑉
2         ∀𝑢, 𝜐 ∈ 𝑉                              (5.6) 



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Then G +𝑗𝜀
′ is a monotone operator. By taking 𝜐 = 0𝑉 on (5.6) and using the Inequality 𝑎𝑏 ≤

𝑚𝒢

2
𝑎2 +

1

2𝑚𝒢
𝑏2, 

 it results ∀u, 𝜐 ∈ V : 

   〈(𝐺 + 𝑗𝜀
′)𝑢, 𝑢〉𝑉′×𝑉 ≥ 𝑚𝒢|𝑢|𝑉

2 − |𝐺0𝑉|𝑉′ . 

And |𝑢|𝑉 ≥
1

2
𝑚𝒢|𝑢|𝑉

2 −
1

2𝑚𝒢
|𝐺0𝑉|𝑉′

2 ,    For  𝜔 =
1

2
𝑚𝒢 , 𝐶 =

1

2𝑚𝒢
|𝐺0𝑉|𝑉′

2 ∈ ℝ.    

Next, by using (4.5)(a) and (5.5) then : 

  |(𝐺 + 𝑗𝜀
′)𝑢 − (𝐺 + 𝑗𝜀

′)𝜐|𝑉′ ≤ 𝐿𝒢|𝑢 − 𝜐|𝑉 + 𝐶. 

Choosing 𝜐 = 0𝑉  it result: 

|(𝐺 + 𝑗𝜀
′)𝑢|𝑉′ ≤ 𝐶(|𝑢 − 𝜐|𝑉 + 1), ∀𝑢 ∈ 𝑉        

Finally, by using (4.11) that there exists 𝜐𝜂
𝜀∈ 𝕃2(0, T ; V ) ∩ C([0, T ]; H) and 

 �̇�𝜂
𝜀 ∈ 𝕃2([0, 𝑇]; 𝑉′),  such that :

         
 

   {
�̇�𝜂
𝜀(𝑡) + 𝐺𝜐𝜂

𝜀(𝑡) + 𝑗𝜂
𝜀(𝜐𝜂

𝜀) = 𝑓𝜂(𝑡)   𝑖𝑛 𝑉
′𝑎. 𝑒. 𝑡 ∈ [0, 𝑇],   

𝜐𝜂
𝜀(0) = 𝑢1.

                                (5.7)
                                               

( )
                       

 

Then 𝜐𝜂
𝜀 ∈ 𝕃2([0, 𝑇]; 𝑉) ∩ 𝑊1,2(0, 𝑇; 𝑉′)whic h sa t i s f i e s :  

(�̇�𝜂
𝜀(𝑡), 𝜔 − 𝜐𝜂

𝜀(𝑡))
𝑉′×𝑉

+ (𝒢𝜐𝜂
𝜀(𝑡), 𝜔 − 𝜐𝜂

𝜀(𝑡))
𝑉′×𝑉

+ 𝑗𝜀(𝜔) − 𝑗𝜀(𝜐𝜂
𝜀) ≥ (𝑓𝜂(𝑡), 𝜔 − 𝜐𝜂

𝜀(𝑡))
𝑉′×𝑉

                                                           

  𝜐𝜂(0) = 𝑢1                                                                                                                                 (5.8) 

 Using (5.7) to obtain: 

(�̇�𝜂
𝜀(𝑡), 𝜐𝜂

𝜀(𝑡))
𝑉′×𝑉

+ (𝒢𝜐𝜂
𝜀  (𝑡), 𝜐𝜂

𝜀  (𝑡))𝑉′×𝑉 + (𝑗𝜀
′(𝜐𝜂

𝜀), 𝜐𝜂
𝜀)
𝑉′×𝑉

 

= (𝑓𝜂(𝑡), 𝜐𝜂
𝜀(𝑡))

𝑉′×𝑉
 

   𝜐𝜂(0) = 𝑢1                                                                                                                          (5.9) 

By using (4.8), the monotony of 𝑗𝜀
′ and (5.3) it comes that : 

∃ 𝐶 > 0, ∀𝑡 ∈ [0, 𝑇], |𝜐𝜂
𝜀(𝑡)|

𝐻
≤ 𝐶, ∫ |𝜐𝜂

𝜀(𝑡)|
𝐻

2
𝑑𝑡 ≤ 𝐶, ∫ |�̇�𝜂

𝜀(𝑡)|
𝐻

2
𝑑𝑡 ≤ 𝐶.

𝑇

0

𝑇

0
                               (5.10) 

 So there is a sub-sequence (vη) such that : 

𝜐𝜂
𝜀 ⟶ 𝜐𝜂weakly in 𝕃2(0, T ; V ) and weakly star in 𝕃∞(0, T ; H). 

 �̇�𝜂
𝜀 ⟶ �̇�η and weakly star in 𝕃2(0, T ; V)                                                  (5.11) 

It comes that :       

𝜐𝜂 ∈ 𝐶(0, 𝑇; 𝐻) and 𝜐𝜂
𝜀(𝑡) ⟶ 𝜐𝜂(𝑡) weakly in 𝐻, ∀𝑡 ∈ [0, 𝑇].                               (5.12)  



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By integration of (5.8),then ∀𝜔 ∈ 𝕃2(0, T ; V ) : 

∫ (�̇�𝜂
𝜀(𝑡), 𝜔)

𝑉′×𝑉
𝑑𝑡 + ∫ (𝒢𝜐𝜂

𝜀(𝑡), 𝜔)
𝑉′×𝑉

𝑑𝑡 + ∫ 𝑗𝜀(𝜔)𝑑𝑡
𝑇

0

𝑇

0

𝑇

0

 

                              

   ≥ ∫ (�̇�𝜂
𝜀(𝑡), 𝜐𝜂

𝜀(𝑡))
𝑉′×𝑉

𝑑𝑡 + ∫ (𝒢𝜐𝜂
𝜀(𝑡), 𝜐𝜂

𝜀(𝑡))
𝑉′×𝑉

𝑑𝑡
𝑇

0

𝑇

0
 

  +∫ 𝑗𝜀(𝜐𝜂
𝜀)𝑑𝑡 + ∫ (𝑓𝜂(𝑡), 𝜔 − 𝜐𝜂

𝜀(𝑡))
𝑉′×𝑉

𝑑𝑡
𝑇

0

𝑇

0
 

    ≥
1

2
|𝜐𝜂
𝜀(𝑇)|

𝐻

2
−
1

2
|𝜐𝜂
𝜀(0)|

𝐻

2
+ ∫ (𝒢𝜐𝜂

𝜀(𝑡), 𝜐𝜂
𝜀(𝑡))

𝑉′×𝑉

𝑇

0
 

  +∫ 𝑗𝜀(𝜐𝜂
𝜀)𝑑𝑡 + ∫ (𝑓𝜂(𝑡), 𝜔 − 𝜐𝜂

𝜀(𝑡))
𝑉′×𝑉

𝑑𝑡.
𝑇

0

𝑇

0
                                                                                 (5.13) 

By (5.11), (5.12) and the weak semi-continuity below it result that: 

  ∀𝜔 ∈ 𝕃2(0, 𝑇; 𝑉), 

     ∫ (�̇�𝜂(𝑡), 𝜔 − 𝜐𝜂(𝑡))
𝑉′×𝑉

𝑑𝑡 + ∫ (𝒢𝜐𝜂(𝑡), 𝜔 − 𝜐𝜂(𝑡))
𝑉′×𝑉

𝑑𝑡
𝑇

0

𝑇

0
+ ∫ (𝑗𝑞(𝜔) − 𝑗𝑞(𝜐𝜂)) ≥

𝑇

0

∫ (𝑓𝜂(𝑡), 𝜔 − 𝜐𝜂(𝑡))
𝑉′×𝑉

𝑑𝑡
𝑇

0
 

Which implies that: 

(�̇�𝜂(𝑡), 𝜔 − 𝜐𝜂(𝑡))
𝑉′×𝑉

+ (𝒢𝜐𝜂(𝑡),𝜔 − 𝜐𝜂(𝑡))
𝑉′×𝑉

+ 𝑗𝑞(𝜔) − 𝑗𝑞(𝜐𝜂) 

   ≥ (𝑓𝜂(𝑡),𝜔 − 𝜐𝜂(𝑡))
𝑉′×𝑉

 , ∀𝜔 ∈ 𝑉, ∀𝑡 ∈ [0, 𝑇]. 

So the problem 𝑃𝜂𝑞 has a solution 𝜐𝜂 ∈ 𝐶(0, 𝑇;𝐻) ∩ 𝕃
2(0, 𝑇; 𝑉) ∩ 𝑊1,2(0, 𝑇; 𝑉′).  For uniqueness, 

let 𝜐𝜂
1, 𝜐𝜂

2 be two solutions of 𝑃𝜂𝑞 . Then for all 𝑡 ∈ [0, 𝑇], 

(�̇�𝜂
2(𝑡) − �̇�𝜂

1(𝑡), 𝜐𝜂
2(𝑡) − 𝜐𝜂

1(𝑡))
𝑉′×𝑉

+ (𝒢𝜐𝜂
2(𝑡) − 𝒢𝜐𝜂

1(𝑡), 𝜐𝜂
2(𝑡) − 𝜐𝜂

1(𝑡))
𝑉′×𝑉

≤ 0.  

by integrating the previous inequation and using (4.5) then : 

1

2
|𝜐𝜂
2(𝑡) − 𝜐𝜂

1(𝑡)|
𝑉

2
+𝑚𝒢 ∫ |𝜐𝜂

2(𝑠) − 𝜐𝜂
1(𝑠)|

𝑉

2𝑇

0
𝑑𝑠 ≤ 0, ∀𝑡 ∈ [0, 𝑇].    

It implies 𝜐𝜂
2=𝜐𝜂

1. 

In the study of the problem 𝑃𝜂𝑞 we have the following result : 

Lemma 5.5 : The problem 𝑃𝜐𝜂𝑞 has a unique solution 𝑢𝜂𝑞  ∈ 𝑊 1,2(0,T; V) ∩ C1(0, T ; 𝐻) ∩ 

𝑊2,2(0, T ; 𝑉′). Moreover, if  𝑢1, 𝑢2 two solutions of the problem 𝑃𝜐𝜂𝑞 corresponding to the data 

𝜂1, 𝜂2 ∈ 𝕃2(0, T ; 𝑉′) and 𝑞1, 𝑞2∈ C+ then there exists c > 0 such that : 

|�̇�𝜂1𝑞1(𝑡) − �̇�𝜂2𝑞2(𝑡)|𝑉
2
≤ 𝑐 ∫ |𝜂1(𝑠) − 𝜂2(𝑠)|𝑉′

2 𝑑𝑠 + ∫ |𝑞1(𝑠) − 𝑞2(𝑠)|𝑉′
2 𝑑𝑠

𝑡

0

𝑡

0
   

|𝑢𝜂1𝑞1(𝑡) − 𝑢𝜂2𝑞2(𝑡)|𝑉
2
≤ 𝑐 ∫ |𝜂1(𝑠) − 𝜂2(𝑠)|𝑉′

2 𝑑𝑠 + ∫ |𝑞1(𝑠) − 𝑞2(𝑠)|𝑉′
2 𝑑𝑠.

𝑡

0

𝑡

0
               (5.14)  



Communications on Applied Nonlinear Analysis 

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Proof: The proof is a consequence of the lemma (5.4) and the relation (4.5). For proving the 

inequality (5.14), let 𝑢𝜂1,𝑢𝜂2 be two solutions of problems 𝑃𝜐𝜂1 and 𝑃𝜐𝜂2 respectively, then: 

(�̈�𝜂𝑖𝑞𝑖, �̇�𝜂𝑗𝑞𝑗 − �̇�𝜂𝑖𝑞𝑖) + (𝒢�̇�𝜂𝑖𝑞𝑖 , �̇�𝜂𝑗𝑞𝑗 − �̇�𝜂𝑖𝑞𝑖) + 𝑗𝑞 (�̇�𝜂𝑗𝑞𝑗) 

−𝑗𝑞(�̇�𝜂𝑖𝑞𝑖) ≥ (𝑓 − 𝜂𝑖 , �̇�𝜂𝑗𝑞𝑗 − �̇�𝜂𝑖𝑞𝑖). 

  where 𝑖 = 1 if 𝑗 = 2 and 𝑖 = 2 if 𝑗 = 1.  

Doing the addition so we have   

(�̈�𝜂1𝑞1 − �̈�𝜂2𝑞2 , �̇�𝜂2𝑞2 − �̇�𝜂1𝑞1) + (𝒢�̇�𝜂1𝑞1 − 𝒢�̇�𝜂2𝑞2 , �̇�𝜂2𝑞2 − �̇�𝜂1𝑞1)   

+𝑗𝑞1(�̇�𝜂2𝑞2) − 𝑗𝑞1(�̇�𝜂1𝑞1) + 𝑗𝑞2(�̇�𝜂1𝑞1) − 𝑗𝑞2(�̇�𝜂2𝑞2) ≥ (𝜂1 − 𝜂2, �̇�𝜂1𝑞1 − �̇�𝜂2𝑞2)                  (5.15) 

Using (4.6) and integring, the  inequation (5.15) becomes : 

∫ (�̈�𝜂1𝑞1(𝑠) − �̈�𝜂2𝑞2(𝑠), �̇�𝜂2𝑞2(𝑠) − �̇�𝜂1𝑞1(𝑠))𝑑𝑠
𝑡

0

 

+∫ (𝒢�̇�𝜂1𝑞1(𝑠) − 𝒢�̇�𝜂2𝑞2(𝑠), �̇�𝜂2𝑞2(𝑠) − �̇�𝜂1𝑞1(𝑠)) 𝑑𝑠 + ∫ 𝑞1|�̇�𝜂2𝑞2(𝑠)|𝑑𝑠
𝑡

0

𝑡

0

 

 +∫ 𝑞1|�̇�𝜂1𝑞1(𝑠)|𝑑𝑠 + ∫ 𝑞2|�̇�𝜂1𝑞1(𝑠)|𝑑𝑠 − ∫ 𝑞2|�̇�𝜂2𝑞2(𝑠)|𝑑𝑠
𝑡

0

𝑡

0

𝑡

0
−≥ ∫ |𝜂1(𝑠) −

𝑡

0

𝜂2(𝑠)|𝑉′|�̇�𝜂1𝑞1(𝑠) − �̇�𝜂2𝑞2(𝑠)|𝑉𝑑𝑠.                                                                                                  

And we have 〈�̈�, �̇�〉 =
1

2
〈�̇�, �̇�〉′ =

1

2

𝑑

𝑑𝑡
|�̇�|′ 

According to this equation we find 

1

2

𝑑

𝑑𝑡
|�̇�𝜂1𝑞1 − �̇�𝜂2𝑞2|𝑉

2
+𝑚𝒢|�̇�𝜂1𝑞1 − �̇�𝜂2𝑞2|𝑉

2
 

≤ |𝜂1 − 𝜂2||�̇�𝜂1𝑞1 − �̇�𝜂2𝑞2| + 𝐴. 

With 𝐴 = 𝑗𝑞1(�̇�𝜂2𝑞2) − 𝑗𝑞1(�̇�𝜂1𝑞1) + 𝑗𝑞2(�̇�𝜂1𝑞1) − 𝑗𝑞2(�̇�𝜂2𝑞2). 

𝐴 = ∫ 𝜇[(𝑞2 − 𝑞1)|�̇�𝜂1𝑞1(𝑠)| + (𝑞1 − 𝑞2)|�̇�𝜂2𝑞2(𝑠)|]𝑑𝑠.
Γ3

 

𝐴 ≤ 𝜇1|𝑞2 − 𝑞1|𝕃2(Γ3)|�̇�𝜂1𝑞1 − �̇�𝜂2𝑞2|𝕃2(Ω). 

Then: 
1

2
|�̇�𝜂1𝑞1 − �̇�𝜂2𝑞2|𝑉

2
+
𝑚𝒢

2
∫ |�̇�𝜂1𝑞1(𝑠) − �̇�𝜂2𝑞2(𝑠)|𝑉

2
𝑑𝑠

𝑡

0
 

≤ ∫ |𝜂1 − 𝜂2||�̇�𝜂1𝑞1(𝑠) − �̇�𝜂2𝑞2(𝑠)|𝑑𝑠 + ∫ 𝐴𝑑𝑠.
𝑡

0

𝑡

0

 

So 



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1

2
|�̇�𝜂1𝑞1 − �̇�𝜂2𝑞2|𝑉

2
+𝑚𝒢∫ |�̇�𝜂1𝑞1(𝑠) − �̇�𝜂2𝑞2(𝑠)|𝑉

2
𝑑𝑠

𝑡

0

 

≤ ∫ (
𝑚𝒢

2
|𝜂1(𝑠) − 𝜂2(𝑠)|

2 +
1

2𝑚𝒢
|�̇�𝜂1𝑞1(𝑠) − �̇�𝜂2𝑞2(𝑠)|

2
)𝑑𝑠

𝑡

0

 

+∫ 𝜇1|𝑞2 − 𝑞1|𝕃2(Γ3)|�̇�𝜂1𝑞1(𝑠) − �̇�𝜂2𝑞2(𝑠)|𝕃2(Ω)𝑑𝑠.
𝑡

0
                                                     (5.16)                                                                                                                            

But (𝑎𝑏 ≤
𝑚𝒢

2
𝑎2 +

1

2𝑚𝒢
𝑏2) then:  

1

2
|�̇�𝜂1𝑞1 − �̇�𝜂2𝑞2|𝑉

2
+𝑚𝒢∫ |�̇�𝜂1𝑞1(𝑠) − �̇�𝜂2𝑞2(𝑠)|𝑉

2
𝑑𝑠

1

0

 

≤ ∫ (
𝑚𝒢

2
|�̇�𝜂1𝑞2(𝑠) − �̇�𝜂2𝑞2|𝑉

2
+

1

2𝑚𝒢
|𝜂1(𝑠) − 𝜂2(𝑠)|𝑉′

2 )𝑑𝑠.
𝑡

0
                                              (5.17) 

From which:  

1

2
|�̇�𝜂1𝑞1 − �̇�𝜂2𝑞2|𝑉

2
+
𝑚𝒢

2
∫ |�̇�𝜂1𝑞1(𝑠) − �̇�𝜂2𝑞2(𝑠)|𝑉

2
𝑑𝑠

𝑡

0
  ≤

1

2𝑚𝒢
∫ |𝜂1(𝑠) − 𝜂2(𝑠)|𝑉′

2 𝑑𝑠
𝑡

0
                                                                                                                                                  

(5.18) 

It comes 

{
|�̇�𝜂1𝑞1 − �̇�𝜂2𝑞2|𝑉

2
≤ 𝑐 ∫ |𝜂1(𝑠) − 𝜂2(𝑠)|𝑉′

2 𝑑𝑠
1

0

∫ |�̇�𝜂1𝑞1(𝑠) − �̇�𝜂2𝑞2(𝑠)|𝑉
2𝑡

0
𝑑𝑠 ≤ 𝑐 ∫ |𝜂1(𝑠) − 𝜂2(𝑠)|𝑉′

2 𝑑𝑠
𝑡

0

                                               (5.19) 

On the other part of 𝑢1(0) = 𝑢2(0) = 𝑢0 then:  

|𝑢𝜂1𝑞1(𝑠) − 𝑢𝜂2𝑞2(𝑠)|𝑉
2
≤ ∫ |�̇�𝜂1𝑞1(𝑠) − �̇�𝜂2𝑞2(𝑠)|𝑉𝑑𝑠

𝑡

0
                                                    (5.20)          

And: 

|𝑢𝜂1𝑞1(𝑠) − 𝑢𝜂2𝑞2(𝑠)|𝑉
2
≤ ∫ |�̇�𝜂1𝑞1(𝑠) − �̇�𝜂2𝑞2(𝑠)|𝑉

2
𝑑𝑠

𝑡

0
                                                     (5.21) 

And so: 

|𝑢𝜂1𝑞1(𝑠) − 𝑢𝜂2𝑞2(𝑠)|𝑉
2
≤ 𝑐 ∫ |𝜂1(𝑠) − 𝜂2(𝑠)|𝑉′

2 𝑑𝑠
𝑡

0
                                                            (5.22 ) 

Hence it  result (5.14). 

Now, let the map ψt : C+ →C+ be defined by 

𝜓𝑡(𝑞) = |𝑅𝜎𝜈 (𝑢𝜉𝜂𝑞(𝑡))|. 

Lemma 5.6 There exists a constant µ1 > 0 such that the mapping ψt has a unique fixed point 

q∗ and 𝑢𝜉𝜂𝑞∗ (t) is a unique solution of the inequality (5.3) if ‖𝜇‖𝕃∞(Γ3) < 𝜇1 

Proof. Let q1, q2 ∈ C+. Using (4.16), it follows that there exists a constant c0 >0 such that 



Communications on Applied Nonlinear Analysis 

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     ‖𝜓𝑡(𝑞1) − 𝜓𝑡(𝑞2)‖𝕃2(Γ3) ≤ 𝑐0‖𝜎𝜈 (𝑢𝜉𝜂𝑞1(𝑡)) − 𝜎𝜈 (𝑢𝜉𝜂𝑞2(𝑡))‖𝐻−
1
2(Γ)

.                           (5.23) 

Moreover using (4.5) (b) yields 

  ‖𝜎𝜈 (𝑢𝜉𝜂𝑞1(𝑡)) − 𝜎𝜈 (𝑢𝜉𝜂𝑞2(𝑡))‖𝐻−
1
2(Γ)

≤ 𝑀‖𝑢𝜉𝜂𝑞1(𝑡) − 𝑢𝜉𝜂𝑞2(𝑡)‖𝑉
 .                                   (5.24) 

Using (4.2), (4.5) (c), (4.13) (c) and the properties of Rν and Rτ to find after some calculus 

algebra that 

‖𝑢𝜉𝜂𝑞1(𝑡) − 𝑢𝜉𝜂𝑞2(𝑡)‖𝑉 ≤
‖𝜇‖𝕃∞(Γ3)

𝑑Ω

𝑚
‖𝑞1 − 𝑞2‖𝕃2(Γ3).                                                       (5.25) 

Hence, taking into account (4.12) ,combining (5.23) , (5.24) and (5.25) to deduce that 

‖𝜓𝑡(𝑞1) − 𝜓𝑡(𝑞2)‖𝕃2(Γ3) ≤ ‖𝜇‖𝕃∞(Γ3)
𝑐0𝑀𝑑Ω

𝑚
‖𝑞1 − 𝑞2‖𝕃2(Γ3). 

Take 𝜇1 = 𝑚 𝑐0𝑀𝑑Ω⁄ , then this inequality shows that if ‖𝜇‖𝕃∞(Γ3) < 𝜇1,  

ψ is a contraction; thus it has unique fixed point q∗ and uηq∗ (t) is a unique solution of (5.3). 

Denote uξηq∗ = uη.Now shall see that uξη ∈ C ([0.T]; V). 

Indeed, let t1, t2 ∈ [0, T] . Taking 𝜐 = 𝑢𝜉𝜂(𝑡2) in (5.3) written for 𝑡 = 𝑡1 and then 

 𝜐 = 𝑢𝜉𝜂(𝑡1) in the same inequality written for 𝑡 = 𝑡2 Using (4.5) (c) , (4.11) , (4.13) (c) and 

the properties of 𝑅𝜈 and 𝑅𝜏 , and adding the resulting inequalities, it follows that there exists 

a constant c1 > 0 such that 

‖𝑢𝜉𝜂(𝑡2) − 𝑢𝜉𝜂(𝑡1)‖𝑉 ≤
𝑐1

𝑚−‖𝜇‖𝕃∞(Γ3)
(‖𝜉(𝑡2) − 𝜉(𝑡1)‖𝕃2(Γ3) +‖𝜂(𝑡2) − 𝜂(𝑡1)‖ℋ 

+‖𝑓(𝑡2) − 𝑓(𝑡1)‖𝑉). 

Then, as 𝜉 ∈ 𝐶([0, 𝑇]; 𝕃2(Γ3)), 𝜂 ∈ 𝐶([0, 𝑇];ℋ) and𝑓 ∈ ([0, 𝑇]; 𝑉),it immediately concludes 

that  𝑢𝜉𝜂(𝑡) ∈ 𝑊, ∀𝑡 ∈ [0, 𝑇]. Indeed, for each t ∈ [0, T ] , 

denote 𝜎 (𝑢𝜉𝜂(𝑡)) = 𝒜𝜀 (𝑢𝜉𝜂(𝑡)) + 𝜂(𝑡), take 𝜐 = 𝑢𝜉𝜂(𝑡) ± 𝜑 in inequality (5.3) where 

𝜑 ∈ (𝐶0
∞(Ω))

𝑑
and use Green’s formula with regularity 𝜑1(𝑡) ∈ 𝐻 

leads to div𝜎 (𝑢𝜉𝜂(𝑡)) ∈ 𝐻 and then 𝑢𝜉𝜂(𝑡) ∈ 𝑊. 

Now introducing the operator Λ𝜉 : 𝐶([0,T];ℋ)⟶ 𝐶([0, 𝑇];ℋ) with 𝜂 ⟶ Λ𝜉 

defined by                                                                                                                        ( 5.26) 

〈Λ𝜉𝜂, 𝜔〉 = 〈𝒜𝜀(𝑢𝜉𝜂), 𝜀(𝜔)〉ℋ + ℎ(𝑢𝛽𝜂 , 𝜔) + 𝑗𝑐(𝑢𝜉𝜂) +∫ ℱ(𝑡 − 𝑠)𝜀(𝑢𝜉𝜂)𝑑𝑠
𝑡

0

 

Lemma 5.7. The operator Λ𝜉 has a unique fixed point 𝜂𝜉. 

Proof. Let 𝜂1, 𝜂2 ∈ 𝐶([0, 𝑇];ℋ). Using (5.4) , (5.26) and (4.8) we obtain for  



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‖𝜇‖𝕃∞(Γ3) < 𝜇1at follow that 

‖Λ𝜉𝜂1(𝑡) − Λ𝜉𝜂2(𝑡)‖ℋ ≤ 𝑐2 ∫ ‖𝜂1(𝑡) − 𝜂2(𝑡)‖ℋ𝑑𝑠.  𝑡 ∈ [0, 𝑇]
𝑡

0
  

where c2 > 0. Reiterating this inequality 𝑛 times, yields 

‖Λ𝜉
𝑛𝜂1(𝑡) − Λ𝜉

𝑛𝜂2(𝑡)‖𝐶([𝑂,𝑇];ℋ)
≤

(𝑐2𝑇)
𝑛

𝑛!
‖𝜂1 − 𝜂2‖𝐶([0,𝑇];ℋ)  

As lim
𝑛→+∞

(𝑐2𝑇)
𝑛

𝑛!
= 0, it follows that for a positive integer n sufficiently large, Λ𝜉

𝑛 is a 

contraction; then, by using the Banach fixed point theorem, it has a unique fixed point 𝜂𝜉  

which is also a unique fixed of Λ𝜉  i.e., 

Λ𝜉𝜂𝜉 = 𝜂𝜉(𝑡), ∀𝑡 ∈ [0, 𝑇]                                                                                       (5.27) 

Then by (4.3) and (4.27) we conclude that 𝑢𝜉𝜂𝜉  is the unique solution of  

Problem 𝑃1𝜉 . In the second step stating the following problem.  

Problem 𝑃𝑎𝑑. Find 𝛽∗ : [0, T ] → 𝕃2(Γ3)such that   

�̇�∗(𝑡) = − [𝛽∗(𝑡) (𝑐𝜈 (𝑅𝜈 (𝑢𝛽∗𝜈(𝑡)))
2

+ 𝑐𝜏 |𝑅𝜏 (𝑢𝛽∗𝜏(𝑡))|
2

− 𝜀𝑎)]
+

                                  (5.28)                                                                                                                  

𝛽∗(0) = 𝛽0                                                                                                                                     (5.29) 

Let obtain the following result be given 

Proposition 5.8. Problem Pad has a unique solution 𝛽∗which satisfies 

𝛽∗ ∈ 𝑊1,∞(0, 𝑇; 𝕃2(Γ3)) ∩ 𝐵. 

Proof. Let t ∈ [0, T ] and consider the mapping ϕ:Z → Z defined by 

𝜙𝛽(𝑡) = 𝛽0 −∫ [𝛽(𝑠) (𝑐𝜈 (𝑅𝜈 (𝑢𝛽(𝑠))))

2

+ 𝑐𝜏 |𝑅𝜏 (𝑢𝛽𝜏(𝑠))|
2

− 𝜀𝑎]

+

𝑑𝑠
𝑡

0

, 

where 𝑢𝛽   is the solution of Problem 𝑃1𝛽. For𝛽1, 𝛽2 ∈ 𝐵, there exists a constant c3 > 0 

 such that 

‖𝜙𝛽1(𝑡) − 𝜙𝛽2(𝑡)‖𝕃2(Γ3) 

≤ 𝑐3 ∫ ‖𝛽1(𝑠) (𝑅𝜈 (𝑢𝛽1𝜈(𝑠)))
2

− 𝛽2(𝑠) (𝑅𝜈 (𝑢𝛽2𝜈(𝑠)))
2

‖
𝕃2(Γ3)

𝑑𝑠
𝑡

0
  

+𝑐3 ∫ ‖𝛽1(𝑠) (𝑅𝜏 (𝑢𝛽1𝜏(𝑠)))
2

− 𝛽2(𝑠) (𝑅𝜏 (𝑢𝛽2(𝑠)))
2

‖
𝕃2(Γ3)

𝑑𝑠
𝑡

0
  

As in [20, 21] it deduces 

‖𝜙𝛽1(𝑡) − 𝜙𝛽2(𝑡)‖𝕃2(Γ3) ≤ 𝑐4 (∫ ‖𝛽1(𝑠) − 𝛽2(𝑠)‖𝕃2(Γ3)
𝑡

0
𝑑𝑠 + ∫ ‖𝑢𝛽1(𝑠) − 𝑢𝛽2(𝑠)‖𝑉

𝑡

0
𝑑𝑠).  



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(5.30) 

For some constant 𝑐4 > 0. Now to continue the proof it has needed to prove  

the following lemma. 

Lemma 5.9. There exists a constant 𝜇0> 0 such that: 

‖𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡)‖𝑉 ≤ 𝑐‖𝛽1(𝑡) − 𝛽2(𝑡)‖𝕃2(Γ3) ∀t ∈ [0, T ] , 

Proof. Let t ∈ [0, T ] . Take 𝑢𝛽2 (t) in (4.1) satisfied by 𝑢𝛽1 (t), then take 

𝑢𝛽1 (t) in the same inequality satisfied by 𝑢𝛽2 (t) ; by addding the resulting inequalities 

〈�̈�1 − �̈�2, �̇�1 − �̇�2〉 + 〈𝒜𝜀 (𝑢𝛽1(𝑡)) − 𝒜𝜀 (𝑢𝛽2(𝑡)) , 𝜀 (�̇�𝛽1(𝑡)) − 𝜀 (�̇�𝛽2(𝑡))〉ℋ 

≤ 〈∫ ℱ(𝑡 − 𝑠) (𝜀 (𝑢𝛽1(𝑠)) − 𝜀 (𝑢𝛽2(𝑠))) 𝑑𝑠, 𝜀 (𝑢𝛽2(𝑡)) − 𝜀 (𝑢𝛽1(𝑡))
𝑡

0

〉ℋ 

+ 〈𝒢𝜀 (�̇�𝛽1(𝑡)) − 𝒢𝜀 (�̇�𝛽2(𝑡)) , 𝜀 (�̇�𝛽2(𝑡)) − 𝜀 (�̇�𝛽1(𝑡))〉ℋ 

+ℎ (𝛽1(𝑡), 𝑢𝛽1(𝑡), 𝑢𝛽2(𝑡) − 𝑢𝛽1(𝑡)) + 𝑗𝑐 (𝑢𝛽1(𝑡), 𝑢𝛽2(𝑡)) 

+ℎ (𝛽2(𝑡), 𝑢𝛽2(𝑡), 𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡)) − 𝑗𝑐 (𝑢𝛽2(𝑡), 𝑢𝛽2(𝑡)) 

+𝑗𝑐 (𝑢𝛽2(𝑡), 𝑢𝛽1(𝑡)) − 𝑗𝑐 (𝑢𝛽2(𝑡), 𝑢𝛽2(𝑡)) + 𝑗𝑓 (𝑢𝛽1(𝑡), 𝑢𝛽2(𝑡)) 

−𝑗𝑓 (𝑢𝛽1(𝑡), 𝑢𝛽1(𝑡)) + 𝑗𝑓 (𝑢𝛽2(𝑡), 𝑢𝛽1(𝑡)) − 𝑗𝑓 (𝑢𝛽2(𝑡), 𝑢𝛽2(𝑡)). 

  and using (4.4) (b) then we obtain 

 1
2

𝑑

𝑑𝑡
|�̇�1 − �̇�2|

2 +𝑚‖𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡)‖𝑉
2

 

≤ 〈∫ ℱ(𝑡 − 𝑠) (𝜀 (𝑢𝛽1(𝑠)) − 𝜀 (𝑢𝛽2(𝑠))) 𝑑𝑠, 𝜀 (𝑢𝛽2(𝑡) − 𝑢𝛽1(𝑡))
𝑡

0
〉ℋ  

+ 〈𝒢𝜀(�̇�𝛽1) − 𝒢𝜀(�̇�𝛽2), 𝜀 (�̇�𝛽2(𝑡)) − 𝜀 (�̇�𝛽1(𝑡))〉ℋ  

+ℎ (𝛽1(𝑡), 𝑢𝛽1(𝑡), 𝑢𝛽2(𝑡) − 𝑢𝛽1(𝑡)) + ℎ (𝛽2(𝑡), 𝑢𝛽2(𝑡), 𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡))  

+𝑗𝑐 (𝑢𝛽1(𝑡), 𝑢𝛽2(𝑡)) − 𝑗𝑐 (𝑢𝛽1(𝑡), 𝑢𝛽1(𝑡)) + 𝑗𝑐 (𝑢𝛽2(𝑡), 𝑢𝛽1(𝑡))  

−𝑗𝑐 (𝑢𝛽2(𝑡), 𝑢𝛽2(𝑡)) + 𝑗𝑓 (𝑢𝛽1(𝑡), 𝑢𝛽2(𝑡)) − 𝑗𝑓 (𝑢𝛽1(𝑡), 𝑢𝛽1(𝑡))  

+𝑗𝑓 (𝑢𝛽2(𝑡), 𝑢𝛽1(𝑡)) − 𝑗𝑓 (𝑢𝛽2(𝑡), 𝑢𝛽2(𝑡))                                                                          (5.31) 

We  have 

〈∫ ℱ(𝑡 − 𝑠) (𝜀 (𝑢𝛽1(𝑠) − 𝜀(𝑢𝛽2(𝑠)) 𝑑𝑠, 𝜀 (�̇�𝛽2(𝑡)) − 𝜀 (�̇�𝛽1(𝑡)))
𝑡

0
〉ℋ  



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   ≤ 𝑐5 (∫ ‖𝑢𝛽1(𝑠) − 𝑢𝛽2(𝑠)‖𝑉
𝑑𝑠

𝑡

0
) ‖𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡)‖𝑉

 

for some positive constant c5.Using Young’s inequality, it finds  

〈∫ ℱ(𝑡 − 𝑠) (𝜀 (𝑢𝛽1(𝑠)) − 𝜀(𝑢𝛽2(𝑠)))
𝑡

0
𝑑𝑠, 𝜀 (�̇�𝛽1(𝑡)) − 𝜀(�̇�𝛽2(𝑡))〉ℋ  

≤
𝑐5
2

2𝑚
(∫ ‖𝑢𝛽1(𝑠) − 𝑢𝛽2(𝑠)‖𝑉𝑑𝑠

𝑡

0
)
2

+
𝑚

2
‖𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡)‖𝑉

2
                                               (5.32) 

Using the properties of Rν and Rτ (see [1, 20,21]), we have 

ℎ (𝛽1(𝑡), 𝑢𝛽1(𝑡), 𝑢𝛽2(𝑡) − 𝑢𝛽1(𝑡)) + ℎ (𝛽2(𝑡), 𝑢𝛽2(𝑡), 𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡)) 

≤ 𝑐6‖𝛽1(𝑡) − 𝛽2(𝑡)‖𝕃2(Γ3)‖𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡)‖𝑉
 

where c6 > 0. Using also (4.2) , (4.13) and (4.14) (c) yields  

  And using Young’s inequality it results: 

𝑐6‖𝛽1(𝑡) − 𝛽2(𝑡)‖𝕃2(Γ3)‖𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡)‖𝑉  

 ≤ 𝑐7‖𝛽1(𝑡) − 𝛽2(𝑡)‖𝕃2(Γ3)
2 +

𝑚

4
‖𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡)‖𝑉

2
                                                      (5.33) 

for some contant c7 > 0. Then (5.33) implies that 

1

2

𝑑

𝑑𝑡
|�̇�1 − �̇�2|

2 +
𝑚

4
‖𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡)‖𝑉

2
 

≤ 𝑐0𝑀𝑑Ω‖𝜇‖𝕃∞(Γ3)‖𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡)‖𝑉
2
+
𝑐5
2

2𝑚
(∫ ‖𝑢𝛽1(𝑠) − 𝑢𝛽2(𝑠)‖𝑉𝑑𝑠

𝑡

0

)

2

 

+𝑐7‖𝛽1(𝑡) − 𝛽2(𝑡)‖𝕃2(Γ3)
2 +

𝑚𝒢

2
|�̇�1 − �̇�2|

2   

Let now 𝜇0 =
𝜇1

4
, then if ‖𝜇‖𝕃2(Γ3) < 𝜇0,  it deduces that there exists a constant 

𝑐8 > 0 such that  

∫
1

2

𝑑

𝑑𝑡
|�̇�1 − �̇�2|

2𝑑𝑡
𝑠

0
+ ∫ ‖𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡)‖𝑉

2
𝑑𝑡

𝑠

0
  

≤ 𝑐8 ∫ (∫ ‖𝑢𝛽1(𝑠) − 𝑢𝛽2(𝑠)‖𝑉
2
𝑑𝑠 + ‖𝛽1(𝑡) − 𝛽2(𝑡)‖𝕃2(Γ3)

2𝑡

0
)

𝑠

0
𝑑𝑡  +

𝑀𝒢

2
|�̇�1 − �̇�2|

2. 

Then using Gronwall’s argument, it follows that there exists a constant 𝑐 > 0 

Such that  

1

2
|�̇�1 − �̇�2|

2 +∫ ‖𝑢𝛽1(𝑡) − 𝑢𝛽2(𝑡)‖𝑉
2
𝑑𝑡

𝑠

0

 

∫ (∫ ‖𝑢𝛽1(𝑠) − 𝑢𝛽2(𝑠)‖𝑉
2
𝑑𝑠 + ‖𝛽1(𝑠) − 𝛽2(𝑠)‖𝕃2(Γ3)

2
𝑡

0

)𝑑𝑡
𝑠

0

+
𝑀𝒢

2
|�̇�1 − �̇�2|

2. 

                                                                                                           (5.34) 



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Now to end the proof of Proposition 5.8 using (5.30) and (5.34) to deduce 

‖𝜙𝛽1(𝑡) − 𝜙𝛽2(𝑡)‖𝕃2(Γ3) ≤ 𝑐9 ∫ ‖𝛽1(𝑠) − 𝛽2(𝑠)‖𝕃2(Γ3)𝑑𝑠  ∀𝑡 ∈ [0, 𝑇]
𝑡

0
,  

where c9 > 0 and then we obtain  

And reiterating this inequality n times, yield

‖𝜙𝛽1 − 𝜙𝛽2‖𝕃2(Γ3) ≤
𝑐9
𝑘
‖𝛽1 − 𝛽2‖𝕃2(Γ3). 

‖𝜙𝑛𝛽1(𝑡) − 𝜙
𝑛𝛽2(𝑡)‖𝕃2(Γ3) ≤ (

𝑐9𝑇

𝑘
)
𝑛 1

𝑛!
‖𝛽1 − 𝛽2‖𝕃2(Γ3). 

As lim
𝑛→+∞

(
𝑐9𝑇

𝑘
)
𝑛 1

𝑛!
= 0,     

It follows that for a position integer 𝑛 sufficiently large, 𝜙𝑛  is a contraction; then, by using the 

Banach fixed point theorem, it has a unique fixed point 𝛽∗ which satisfies (5.28) and (5.29). Now 

we have all ingredients to prove Theorem 5.1. 

Proof of theorem 5.1. Existence.  Let 𝛽 = 𝜉∗ and let 𝑢𝜉∗ the solution 

of problem P1. We conclude by (5.1), (5.28) and (5.29) that (𝑢𝜉∗ , 𝜉
∗) is a solution of problem PV. 

Uniqueness. Suppose that (𝑢, 𝛽) is a solution of problem PV which satisfies (4.15), (4.16) and 

(4.17) . It follows from (4.15) that 𝑢 is a solution to problem 𝑃1𝜉  and from Theorem 5.2 

that 𝑢 = 𝑢𝛽.Take 𝑢 = 𝑢𝛽in (4.15) and use the initial condition (4.17), we deduce that 𝛽 is a 

solution to problem𝑃𝑎𝑑.. 

Therefor, we obtain from Proposition 5.8 that 𝛽 = 𝛽∗ and then we conclude that (𝑢𝛽∗ , 𝛽
∗) is a 

unique solution to problem PV. 

Let now  𝜎∗ be the function defined by (3.1) which corresponds to the function uβ∗ . Then, it results 

from (4.5), (4.6) and (4.8) that𝜎∗ ∈ 𝐶([0, 𝑇];ℋ).Using also a standard argument, it follows 

from the inequality (4.15) that 

𝐷𝑖𝜐𝜎∗(𝑡) + 𝜑1(𝑡) = 𝜌𝑢 ̈  𝑖𝑛 Ω, for all t ∈ [0, 𝑇].     

Therefor, using the regularity  𝜑1 ∈ C ([0, T ] ; H) , we deduce that div σ∗ ∈ C ([0, T ] ; H) which 

implies that σ∗ ∈ C ([0, T ] ; ℋ1) . The triple (uβ∗ , σ∗, β∗) which satisfies (3.1) and (4.15)−(4.17) is 

called a weak solution of problem P1.Moreover, the regularity of the weak solution is uβ∗ ∈ C ([0, 

T ] ; V ) , σ∗ ∈ C ([0, T ] ; ℋ1) and β∗ ∈ W 1,∞([0, T ] ; 𝕃2(Γ3)) ∩ B. 

6. CONCLUDING REMARK 

Scientific study and contemporary publication in mechanics focus on two primary components: 

one pertaining to the laws of behaviour and other concerning the boundary conditions imposed in 

the body. 

Numerous publications have employed constitutive laws incorporating internal variables to 

represent the influence of internal variable on the behaviour of materials such as metals, rocks and 



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polymers, wherein the rate of deformation is contingent upon these internal variables. Our model 

is obtained by comnining the viscoelastic constitutive law with friction, long memory and internal 

state variable β, which describes the pointwise fractional density of active bonds on the contact 

surface and is sometimes referred to as the intensity of adhesion. 

Mathematically, the idea is to reduce the second order nonlinear evolution inequality of the system 

to the first order evolution inequality. After this, we use classical results on first order evolution 

nonlinear inequalities, differential equations and the fixed point arguments. 

References 

[1] L. Cangemi. Frottement et adh´erence: mod`ele, traitement num´erique application a` 

l’interface fibre/ matrice, PH. D. Thesis, Univ. M´editerran´ee, Aix Marseille I, 1997. 

[2] O. Chau, J. Fernandez, W. Han and M. Sofonea, Variational and numerical analysis of 

a dynamic frictionless contact problem with adhesion, Journal of Computational and 

Applied Mathematics, 156 (2003), 127-157. 

[3] O. Chau, J.R. Fernandes, M. Shillor and M. Sofonea. Variational and numerical analysis 

of a quasistatic viscoelastic contact problem with adhesion, Journal of Computational 

and Applied Mathematics, 159(2003), 431-465. 

[4] O. Chau, M. Shillor and M. Sofonea. Dynamic frictionless contact with adhesion, J. 

Appl. Math. Phys. (ZAMP), 55 (2004), 32-47. 

[5] M. Cocou and R. Rocca. Existence results for unilateral quasistatic contact problems with 

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