analysis of a viscoelastic frictionless contact problem with adhesion

ANALYSIS OF A VISCOELASTIC FRICTIONLESS
CONTACT PROBLEM WITH ADHESION
AREZKI TOUZALINE
We consider a quasistatic frictionless contact problem for viscoelastic bodies with
long memory. The contact is modelled with normal compliance in such a way that
the penetration is limited and restricted to unilateral contraints. The adhesion
between contact surfaces is taken into account and the evolution of the bonding
field is described by a first order differential equation. We derive a variational formulation of the mechanical problem and we establish an existence and uniqueness
result by using arguments of time-dependent variational inequalities, differential
equations and Banach fixed point theorem. Moreover, using compactness properties we study a regularized problem which has a unique solution and we obtain
the solution of the original model by passing to the limit as the regularization
parameter converges to zero.
AMS 2010 Subject Classification: 74H20, 74M15, 47J20, 49J40.
Key words: viscoelastic, normal compliance, adhesion, frictionless, contact, weak
solution, fixed point.
1. INTRODUCTION
Contact problems involving deformable bodies are quite frequent in the
industry as well as in daily life and play an important role in structural and mechanical systems. Contact processes involve complicated surface phenomena,
and are modelled by highly nonlinear initial boundary value problems. Taking
into account various contact conditions associated with more and more complex behavior laws lead to the introduction of new and non standard models,
expressed by the aid of evolution variational inequalities. An early attempt to
study contact problems within the framework of variational inequalities was
made in [6]. The mathematical, mechanical and numerical state of the art
can be found in [19]. In this reference we find results on the mathematical
analysis and numerical studies of various adhesive contact problems. Recently
a new book [21] introduces the reader into the theory of variational inequalities with emphasis on the study of contact mechanics and more specifically,
on antiplane frictional contact problems. Also, recently existence results for
the continuous case were established in [1, 5, 7, 16] in the study of unilateral
REV. ROUMAINE MATH. PURES APPL., 55 (2010), 5, 411–430
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and frictional contact problems for linear elastic materials. In this paper as
in [13] we study a mathematical model which describes a frictionless and adhesive contact problem between a viscoelastic body with long memory and a
foundation. The elasticity operator is assumed to vanish for zero strain, to be
Lipschitz continuous and strictly monotone. As in [12] the contact is modelled with normal compliance in such a way that the penetration is limited
and restricted to unilateral constraints. We recall that models for dynamic
or quasistatic processes of frictionless adhesive contact between a deformable
body and a foundation have been studied in [3, 4, 8, 18, 19, 20, 22]. Following [9, 10] we use the bonding field as an additional state variable β, defined
on the contact surface of the boundary. The variable satisfies the restrictions
0 ≤ β ≤ 1. At a point on the boundary contact surface, when β = 1 the
adhesion is complete and all the bonds are active; when β = 0 all 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 refer the reader to the
extensive bibliography on the subject in [2, 9, 10, 11, 15, 17, 18, 19]. According to [12], the method presented here considers a compliance model in which
the compliance term doesn’t represent necessarily a compact perturbation of
the original problem without contact. This leads to study such models, where
a strictly limited penetration is permitted to perform the limit procedure to
the Signorini contact problem. In this work as in [23] we extend the result
established in [22] to the unilateral contact problem with a modified normal
compliance when the penetration is finite and the adhesion between contact
surfaces is taken into account. We derive a variational formulation of the mechanical problem for which we prove the existence of a unique weak solution,
and obtain a partial regularity result for the solution. Moreover, we study a
regularized problem which we consider as a frictionless contact problem and
adhesion with unlimited penetration. We prove its unique weak solvability
and show that the solution of the original model is obtained by passing to the
limit as the regularization parameter converges to zero.
The paper is structured as follows. In Section 2 we present some notations
and give the variational formulation. In Section 3 we state and prove our main
existence and uniqueness result, Theorem 3.1. Finally, in Section 4, we prove
a convergence result of a regularized problem, Theorem 4.2.
2. PROBLEM STATEMENT
AND VARIATIONAL FORMULATION
Let Ω ⊂ Rd (d = 2, 3), be a domain initially occupied by a viscoelastic
body with long memory. Ω is supposed to be open, bounded, with a sufficiently
regular boundary Γ. We assume that Γ is composed of three sets Γ̄1 , Γ̄2 , and
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Analysis of a viscoelastic frictionless contact problem with adhesion
413
Γ̄3 , with the mutually disjoint relatively open sets Γ1 , Γ2 and Γ3 , such that
meas(Γ1 ) > 0. The body is acted upon by a volume force of density ϕ1 on
Ω and a surface traction of density ϕ2 on Γ2 . On Γ3 the body is in adhesive
frictionless contact with a foundation.
Thus, the classical formulation of the mechanical problem is written as
follows.
Problem P1 . Find a displacement u : Ω × [0, T ] → Rd and a bonding
field β : Γ3 × [0, T ] → [0, 1] such that
Z t
F (t − s) ε (u (s)) ds in Ω,
(2.1)
σ(t) = Aε (u(t)) +
0
(2.2)
div σ(t) + ϕ1 (t) = 0 in Ω,
(2.3)
u(t) = 0 on Γ1 ,
(2.4)
σ(t)ν = ϕ2 (t) on Γ2 ,
(2.5)
(2.6)
(2.7)
(2.8)
uν (t) ≤ g, σν (t) + p (uν (t)) − cν β 2 (t)Rν (uν (t)) ≤ 0
σν (t) + p (uν (t)) − cν β 2 (t)Rν (uν (t)) (uν (t) − g) = 0
)
on Γ3 ,
στ (t) = 0 on Γ3 ,
β̇(t) = − cν β(t) (Rν (uν (t)))2 − εa
+
on Γ3 ,
β (0) = β0 on Γ3 .
Equation (2.1) represents the viscoelastic constitutive law of the material
in which A and F denote the elasticity operator and the relaxation fourth-order
respectively. Equation (2.2) represents the equilibrium equation while (2.3)
and (2.4) are the displacement and traction boundary conditions, respectively,
in which ν denotes the unit outward normal vector on Γ and σν represents the
Cauchy stress vector. Condition (2.5) represents the unilateral contact with
adhesion in which cν is a given adhesion coefficient which may dependent on
x ∈ Γ3 and Rν is the truncation operator defined by

 L if s < −L,
−s if − L ≤ s ≤ 0,
Rν (s) =

0
if s > 0.
Here L > 0 is the characteristic length of the bond, beyond which it does not
offer any additional traction (see [20]) and p is a normal compliance function
which satisfies the assumption (2.17); g denotes the maximum value of the
penetration which satisfies g ≥ 0. When uν < 0, i.e., when there is separation
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between the body and the foundation, condition (2.5) combined with assumption (2.17) and definition of Rν implies that σν = cν β 2 Rν (uν ) and does not
exeed the value L kcν kL∞ (Γ3 ) . When g > 0, the body may interpenetrate into
the foundation, but the 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, from (2.5) we recover the
contact conditions
uν ≤ g,
σν − cν β 2 Rν (uν ) ≤ 0,
(σν − cν β 2 Rν (uν ))(uν − g) = 0.
When g = 0, conditions (2.5) combined with hypothese (2.16) lead to the
contact conditions
uν ≤ 0,
σν − cν β 2 Rν (uν ) ≤ 0,
(σν − cν β 2 Rν (uν ))uν = 0.
These contact conditions were used in [20, 22] to model the unilateral contact
with adhesion. It follows from (2.5) 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 (2.5) become the
classical Signorini contact conditions with zero gap function, that is,
uν ≤ 0,
σν ≤ 0,
σν uν = 0.
Equation (2.6) represents a frictionless contact conditon and shows that
the tangential stress vanishes on the contact surface during the process. Also it
means that the glue does not provide any resistance to the tangential motion
of the body on the foundation. Equation (2.7) represents the ordinary differential equation which describes the evolution of the bonding field, in which
r+ = max {r, 0}, and it was already used in [3]. Since β̇ ≤ 0 on Γ3 × (0, T ),
once debonding occurs bonding cannot be reestablished and, indeed, the adhesive process is irreversible. Also from [14] it must be pointed out clearly that
condition (2.7) does not allow for complete debonding in finite time. Finally,
(2.8) is the initial condition, in which β0 denotes the initial bonding field. In
(2.7) a dot above a variable represents its derivative with respect to time. We
denote by Sd the space of second order symmetric tensors on Rd (d = 2, 3)
while k · k represents the Euclidean norm on Rd and Sd . Thus, for every
1
u, v ∈ Rd , u.v = ui vi , kvk = (v.v) 2 , and for every σ, τ ∈ Sd , σ.τ = σij τij ,
1
kτ k = (τ.τ ) 2 . Here and below, the indices i and j run between 1 and d and
the summation convention over repeated indices is adopted. Now, to proceed
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Analysis of a viscoelastic frictionless contact problem with adhesion
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with the variational formulation, we need the function spaces
d
d
H = L2 (Ω) , H1 = H 1 (Ω) ,
Q = τ = (τij ) ; τij = τji ∈ L2 (Ω) , Q1 = {σ ∈ Q; div σ ∈ H} .
Note that H and Q are real Hilbert spaces endowed with the respective canonical inner products
Z
Z
σij τij dx.
ui vi dx, hσ, τ iQ =
(u, v)H =
Ω
Ω
The strain tensor is
1
(ui,j + uj,i ) ;
2
div σ = (σij,j ) is the divergence of σ. For every v ∈ H1 we denote by vν and
vτ the normal and tangential components of v on the boundary Γ given by
ε (u) = (εij (u)) =
vν = v.ν,
vτ = v − vν ν.
Also, we denote by σν and στ the normal and the tangential traces of a function
σ ∈ Q1 , and when σ is a regular function then
σν = (σν) .ν, στ = σν − σν ν,
and the following Green’s formula holds
Z
hσ, ε (v)iQ + (div σ, v)H =
σν.vda ∀v ∈ H1 ,
Γ
where da is the surface measure element. Now, let V be the closed subspace
of H1 defined by
V = {v ∈ H1 ; v = 0 on Γ1 } ,
and let the convex subset of admissible displacements given by
K = {v ∈ V ; vν ≤ g a.e. on Γ3 } .
Since meas(Γ1 ) > 0, Korn’s inequality [6], holds
(2.9)
kε (v)kQ ≥ cΩ kvkH1
∀v ∈ V,
where cΩ > 0 is a constant which depends only on Ω and Γ1 . We equip V with
the inner product
(u, v)V = hε (u) , ε (v)iQ
and k · kV is the associated norm. It follows from Korn’s inequality (2.7) that
the norms k·kH1 and k·kV are equivalent on V. Then (V, k · kV ) 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
(2.10)
kvk(L2 (Γ3 ))d ≤ dΩ kvkV
∀v ∈ V.
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For p ∈ [1, ∞], we use the standard norm of Lp (0, T ; V ). We also use the
Sobolev space W 1,∞ (0, T ; V ) equipped with the norm
kvkW 1,∞ (0,T ;V ) = kvkL∞ (0,T ;V ) + kv̇kL∞ (0,T ;V ) .
For every real Banach space (X, k·kX) and T > 0 we use the notation C([0, T ]; X)
for the space of continuous functions from [0, T ] to X; recall that C ([0, T ]; X)
is a real Banach space with the norm
kxkC([0,T ];X) = max kx(t)kX .
t∈[0,T ]
We suppose that the body forces and surface tractions have the regularity
d (2.11)
ϕ1 ∈ C ([0, T ]; H) , ϕ2 ∈ C [0, T ]; L2 (Γ2 )
.
Next, we denote by f : [0, T ] → V the function defined by
Z
Z
ϕ1 (t) .vdx +
ϕ2 (t).vda ∀v ∈ V , t ∈ [0, T ],
(2.12)
(f (t), v)V =
Γ2
Ω
and we note that (2.11) and (2.12) imply
f ∈ C ([0, T ]; V ) .
In the study of the mechanical problem P1 we assume that the elasticity
operator A : Ω × Sd → Sd , satisfies
(2.13)

(a) there exists M > 0 such that kA (x, ε1 ) − A (x, ε2 )k ≤ M kε1 − ε2 k




for all ε1 , ε2 in Sd , a.e. x in Ω;




(b) there exists m > 0 such that (A (x, ε1 )−A (x, ε2 ))·(ε1 −ε2 ) ≥ m kε1 −ε2 k2 ,
for allε1 , ε2 in Sd , a.e. x in Ω;






(c) the mapping x → A (x, ε) is Lebesgue measurable on Ω for any ε in Sd ;



(d) A (x, 0) = 0 for a.e. x in Ω.
Also, we need to introduce the space of the tensors of fourth order defined by
Q∞ = {E = (Eijkl ) ; Eijkl = Ejikl = Eklij ∈ L∞ (Ω)} ,
which is the real Banach space with the norm
kEkQ∞ =
max
0≤i,j,k,l≤d
kEijkl kL∞ (Ω) .
We assume that the tensor of relaxation F satisfies
(2.14)
F ∈ C ([0, T ]; Q∞ ) .
The adhesion coefficients satisfy
(2.15)
cν , εa ∈ L∞ (Γ3 ) and cν , εa ≥ 0 a.e. on Γ3 ,
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Analysis of a viscoelastic frictionless contact problem with adhesion
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and we assume that the initial bonding field satisfies
β0 ∈ L2 (Γ3 ) ; 0 ≤ β0 ≤ 1 a.e. on Γ3 .
(2.16)
Next, we define the functional j : L2 (Γ3 ) × V × V → R by
Z
p (uν ) − cν β 2 Rν (uν ) vν da ∀β ∈ L2 (Γ3 ) , ∀u, v ∈ V.
j (β, u, v) =
Γ3
As in [12] we assume that the normal compliance function p satisfies

(a) p : ] − ∞, g] → R;





 (b) there exists Lp > 0 such that
|p (r1 ) − p (r2 )| ≤ Lp |r1 − r2 | for all r1 , r2 ≤ g;
(2.17)



(c) (p (r1 ) − p (r2 )) (r1 − r2 ) ≥ 0 for all r1 , r2 ≤ g;



(d) p (r) = 0 for all r < 0.
Following [19], the properties satisfied by the functional j are
(2.18)
j (β1 , u1 , u2 −u1 )+j (β2 , u2 , u1 −u2 ) ≤ c kβ1 −β2 kL2 (Γ2 ) ku1 −u2 kV ,
(2.19)
j (β, u1 , u2 − u1 ) + j (β, u2 , u1 − u2 ) ≤ 0,
(2.20)
j (β, u1 , v) − j (β, u2 , v) ≤ c ku1 − u2 kV kvkV ,
where c is a positive constant,
j (β, v, v) ≥ 0.
(2.21)
Finally, we need to introduce the set
B = θ : [0, T ] → L2 (Γ3 ) ; 0 ≤ θ(t) ≤ 1, ∀t ∈ [0, T ] , a.e. on Γ3 .
We now use Green’s formula and techniques similar to those presented
in [20] to obtain the following variational formulation of the problem P1 .
Problem P2 . Find a displacement
field u ∈ C ([0, T ]; V ) and a bonding
field β ∈ W 1,∞ 0, T ; L2 (Γ3 ) ∩ B such that
(2.22)
u(t) ∈ K, hAε (u (t)) , ε (v) − ε (u (t))iQ +
Z t
+
F (t − s) ε (u(s)) ds, ε (v) − ε (u(t))
+
0
Q
+j (β(t), u(t), v − u(t)) ≥ (f (t), v − u(t))V
(2.23)
(2.24)
β̇(t) = − cν β(t)(Rν (uν (t)))2 − εa
β (0) = β0 .
+
∀v ∈ K, t ∈ [0, T ],
a.e. t ∈ (0, T ) ,
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3. EXISTENCE AND UNIQUENESS RESULT
The main result in this section is the following existence and uniqueness
theorem.
Theorem 3.1. Let (2.11), (2.13), (2.14), (2.15), (2.16), and (2.17) hold.
Then Problem P2 has a unique solution.
The proof of Theorem 2.1 is carried out in several steps. In the first step,
let k > 0 and consider the space X defined as
h
i
X = β ∈ C [0, T ]; L2 (Γ3 ) ; sup exp (−kt) kβ(t)kL2 (Γ3 ) < +∞ .
t∈[0,T ]
It is well known that X is a Banach space with the norm
h
i
kβkX = sup exp (−kt) kβ(t)kL2 (Γ3 ) .
t∈[0,T ]
Next for a given β ∈ X, we consider the following variational problem.
Problem P1β . Find uβ ∈ C ([0, T ]; V ) such that
Z
(3.1)
+
t
uβ (t) ∈ K, hAε (uβ (t)) , ε (v) − ε (uβ (t))iQ +
+ j (uβ (t), v) −
F (t − s) ε (uβ (s)) ds, ε (v) − ε (uβ (t))
0
Q
−j (uβ (t) , uβ (t)) ≥ (f (t), v − uβ (t))V
∀v ∈ K, t ∈ [0, T ].
We have the following result.
Proposition 3.2. Problem P1β has a unique solution.
For the proof of this proposition we consider the following intermediate
problem.
Problem P1βη . For η ∈ C ([0, T ]; Q), find uβη ∈ C ([0, T ]; V ) such that
(3.2)
uβη (t) ∈ K, hAε (uβη (t)) , ε (v − uβη (t))iQ + hη, ε (v − uβη (t))iQ +
+j (β(t), uβη (t), v − uβη (t)) ≥ (f (t), v − uβη (t))V
∀v ∈ K, t ∈ [0, T ] .
We have the following result.
Lemma 3.3. Problem P1βη has a unique solution.
Proof. Riesz’s representation theorem leads to the existence of an element
fη ∈ C ([0, T ]; V ) such that
(fη (t), v)V = (f (t), v)V − hη, ε(v)iQ .
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Analysis of a viscoelastic frictionless contact problem with adhesion
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Let t ∈ [0, T ] and let At : V → V be the operator defined by
(At u, v)V = hAε (u) , ε(v)iQ + j (β(t), u, v)
∀u, v ∈ V.
Using the hypotheses on A and (2.18)–(2.21) we see that At is strongly monotone and Lipschitz continuous. Then from [21], since K is a nonempty closed
convex subset of V , using the standard results for elliptic variational inequalities, we deduce that there exists a unique element uβη (t) ∈ K which satisfies
the inequality
hAε (uβη (t)) , ε (v − uβη (t))iQ + j (β(t), uβη (t), v − uβη (t)) ≥
(3.3)
≥ (fη (t), v − uβη (t))V
∀v ∈ K, t ∈ [0, T ],
and then it satisfies (3.2). To show that uβη ∈ C ([0, T ]; V ), it suffices to
see [19] . Now, to end the proof we need to introduce the operator
Λβ : C ([0, T ]; Q) → C ([0, T ]; Q)
defined by
Z
(3.4)
t
F (t − s) ε (uβη (s)) ds ∀η ∈ C ([0, T ]; Q) , t ∈ [0, T ] .
Λβ η(t) =
0
Lemma 3.4. The operator Λβ has a unique fixed point ηβ .
Proof. Let η1 , η2 ∈ C ([0, T ] ; Q). Relations (3.3), (3.4) and assumption
(2.14) on F imply that there exists a constant c > 0 such that
Z t
kΛβ η1 (t) − Λβ η2 (t)kQ ≤ c
kη1 (s) − η2 (s)kQ ds ∀t ∈ [0, T ].
0
We deduce that for a positive integer m, Λm
β is a contraction; then it admits
a unique fixed point ηβ which is a unique fixed point of Λβ , i.e.,
(3.5)
Λβ ηβ (t) = ηβ (t)
∀t ∈ [0, T ].
Then by (3.3) and (3.5) we conclude that uβηβ is the unique solution of (3.1)
and Proposition 3.2 is proved. Denote uβ = uβηβ . In the next step we consider the following problem.
Problem P2β . Find β ∗ : [0, T ] → L∞ (Γ3 ) such that
(3.6)
β̇ ∗ (t) = − cν β ∗ (t) (Rν (uβ ∗ ν (t)))2 − εa
a.e. t ∈ (0, T ) ,
+
(3.7)
β ∗ (0) = β0 .
We obtain the following result.
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Lemma 3.5. Problem P2β has a unique solution β ∗ which satisfies
β ∗ ∈ W 1,∞ 0, T ; L2 (Γ3 ) ∩ B.
Proof. Consider the mapping T : X → X given by
Z t
T β(t) = β0 −
cν β (s) (Rν (uβν (s)))2 − εa + ds,
0
where uβ is the solution of Problem P1β . Then for β1 , β2 ∈ X, we have
kT β1 (t) − T β2 (t)kL2 (Γ3 ) ≤
Z t
2
≤ (cν β1 (s) Rν (uβ1 ν (s)) −εa )+ −(cν β2 (s)(Rν (uβ2 ν (s)))2 −εa )+
L2 (Γ3 )
0
ds.
Using the definition of Rν and writing β1 = β1 − β2 + β2 , we deduce as in [19]
that there exists a constant c > 0 such that
Z t
kT β1 (t) − T β2 (t)kL2 (Γ3 ) ≤ c
kβ1 (s) − β2 (s)kL2 (Γ3 ) ds.
0
On the other hand, we have
Z t
exp (kt)
kβ1 (s) − β2 (s)kL2 (Γ3 ) ds ≤ kβ1 − β2 kX
.
k
0
Therefore,
kT β1 (t) − T β2 (t)kL2 (Γ3 ) ≤ c kβ1 − β2 kX
exp (kt)
k
∀t ∈ [0, T ],
which yields
exp (−kt) kT β1 (t) − T β2 (t)kL2 (Γ3 ) ≤
c
kβ1 − β2 kX
k
∀t ∈ [0, T ].
Hence
(3.8)
kT β1 − T β2 kX ≤
c
kβ1 − β2 kX .
k
Inequality (3.8) shows that for k > c, T is a contraction. Then it has a
unique fixed point β ∗ which satisfies (3.6) and (3.7) . To prove that β ∗ ∈ B, it
suffices to see [20, Remark 3.1]. Finally, as in [20, 22], we conclude that
(uβ ∗ , β ∗ ) ∈ C ([0, T ]; K) × (W 1,∞ 0, T ; L2 (Γ3 ) ∩ B)
is the unique solution to Problem P2 and Theorem 3.1 is proved.
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Analysis of a viscoelastic frictionless contact problem with adhesion
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4. THE REGULARIZED PROBLEM
In this section we study a frictionless contact problem with normal compliance and adhesion with unlimited penetration. The contact condition (2.5)
is replaced by the contact condition
−σδν = pδ (uδν ) − cν β 2 Rν (uδν ) on Γ3 × (0, T ) ,
where as in [12] the regularized functional pδ : R → R is defined by

if r ≤ g,
 p(r)
pδ (r) =
r−g

+ p(g) if r > g,
δ
and δ > 0 denotes the regularization parameter. We study the behavior of
the solution as δ → 0 and prove that in the limit we obtain the solution
of adhesive frictionless contact problem with normal compliance and finite
penetration. Next, we define the functional jδ : L2 (Γ3 ) × V × V → R by
Z
jδ (β, u, v) =
pδ (uν ) − cν β 2 Rν (uν ) vν da ∀β ∈ L2 (Γ3 ) , ∀u, v ∈ V.
Γ3
With this notation, the formulation of the regularized problem with frictionless
contact and adhesion is the following.
Problem P1δ . Find a displacement field uδ : Ω × [0, T ] → Rd and a
bonding field βδ : Γ3 × [0, T ] → [0, 1] such that
Z t
F (t − s) ε (uδ (s)) ds in Ω,
(4.1)
σδ (t) = Aε (uδ (t)) +
0
(4.2)
div σδ (t) + ϕ1 (t) = 0 in Ω,
(4.3)
uδ (t) = 0 on Γ1 ,
(4.5)
σδ (t)ν = ϕ2 (t) on Γ2 ,
(4.6)
−σδν (t) = pδ (uδν (t)) − cν β 2 (t)Rν (uδν (t)) on Γ3 ,
(4.7)
β̇δ (t) = − cν βδ (t) (Rν (uδν (t)))2 − εa
(4.8)
+
on Γ3 ,
βδ (0) = β0 on Γ3 .
Problem P1δ has the following variational formulation.
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Problem P2δ . Find (uδ , βδ ) ∈ C([0, T ]; V ) × (W 1,∞ 0, T ; L2 (Γ3 ) ∩ B)
such that
Z t
hAε (uδ (t)) , ε(v)iQ +
F (t − s) ε (uδ (s)) ds, ε(v)
(4.9)
+
0
Q
+jδ (βδ (t), uδ (t) , v) = (f (t), v)V
(4.10)
(4.11)
∀v ∈ V , t ∈ [0, T ],
β̇δ (t) = − cν βδ (t) (Rν (uδν (t)))2 − εa
+
a.e. t ∈ (0, T ) ,
βδ (0) = β0 .
We have the following result.
Theorem 4.1. Problem P2δ has a unique solution.
Proof. The proof of Theorem 4.1 is similar to the proof of Theorem 3.1
and it is carried out in several steps. So, we omit the details of the proof. We
recall that the steps are as follows.
(i) For any β ∈ X, we prove that there exists a unique uδ ∈ C ([0, T ]; V )
such that
Z t
(4.12)
hAε (uδ (t)) , ε(v)iQ +
F (t − s) ε (uδ (s)) ds, ε(v)
+
0
+ jδ (β(t), uδ (t), v) = (f (t), v)V
Q
∀v ∈ V , t ∈ [0, T ].
To provide this step for all t ∈ [0, T ] we consider the operator Tt : V → V
defined by
(Tt u, v)V = hAε (u) , ε(v)iQ + jδ (β(t), u, v)
∀u, v ∈ V.
We use the hypotheses on A, the properties (2.18)–(2.21) satisfied by the
functional j and the definition of the function pδ to see that the operator Tt
is strongly monotone and Lipschitz continuous. Then we conclude by using
arguments similar to those used in the proof of Lemma 3.3.
(ii) There exists a unique βδ such that
(4.13)
βδ ∈ W 1,∞ 0, T ; L2 (Γ3 ) ,
(4.14)
(4.15)
β̇δ (t) = − cν βδ (t)(Rν (uδν (t)))2 − εa
+
a.e. t ∈ (0, T ) ,
βδ (0) = β0 .
(iii) Let βδ defined in (ii) and denote again by uδ the function obtained in
step (i) for β = βδ . Then, by using (4.13)–(4.15) it is easy to see that (uδ , βδ )
is the unique solution to Problem P2δ and it satisfies
(uδ , βδ ) ∈ C ([0, T ]; V ) × (W 1,∞ 0, T ; L2 (Γ3 ) ∩ B). 13
Analysis of a viscoelastic frictionless contact problem with adhesion
423
The behavior of the solution (uδ , βδ ) as δ → 0 is given by the following
theorem.
Theorem 4.2. Assume that (2.11), (2.13), (2.14), (2.15), (2.16), and
(2.17) hold. Then the solution (uδ , βδ ) of Problem P2δ converges to the solution
(u, β) of Problem P2 , that is,
lim kuδ (t) − u(t)kV = 0 for all t ∈ [0, T ] ,
(4.16)
(4.17)
δ→0
lim kβδ (t) − β(t)kL2 (Γ3 ) = 0 for all t ∈ [0, T ].
δ→0
The proof is carried out in several steps. In the first step, we show the
following lemma.
Lemma 4.3. For each t ∈ [0, T ], there exists ū(t) ∈ K such that after
passing to a subsequence still denoted (uδ (t)) we have
uδ (t) → ū(t) weakly in V.
(4.18)
Proof. Let t ∈ [0, T ]. Take v = uδ (t) in (4.9) then
(4.19)
hAε (uδ (t)) , ε (uδ (t))iQ + j (βδ (t), uδ (t), uδ (t)) +
Z t
+
F (t − s) ε (uδ (s)) ds, ε (uδ (t))
= (f (t), uδ (t))V .
0
Q
Since
jδ (βδ (t), uδ (t) , uδ (t)) ≥ 0,
from (4.19) we deduce that
Z
(4.20)
hAε (uδ (t)) , ε (uδ (t))iQ +
t
F (t − s) ε (uδ (s)) ds, ε (uδ (t))
≤
0
Q
≤ (f (t), uδ (t))V .
Now, we use Ries’z representation theorem to define the operator D :
[0, T ] → L (V ) by
Z t
(D(t)u, v)V =
F (t − s) ε (u(s)) ds, ε(v)
∀u, v ∈ V.
0
Q
As in [21] it is clear that
D ∈ C ([0, T ]; L (V )) .
Moreover from (4.20) it follows that
m kuδ (t)k2V − kDkC([0,T ];L(V )) kuδ (t)k2V ≤ kf (t)kV kuδ (t)kV .
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Arezki Touzaline
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This inequality implies that if
kDkC([0,T ];L(V )) < m,
there exists a constant c > 0 such that
kuδ (t)kV ≤ c kf (t)kV .
The sequence (uδ (t)) is bounded in V . Then there exists ū(t) ∈ V and a
subsequence again denoted (uδ (t)) such that (4.18) holds. Also from (4.19)
we have
Z t
jδ (βδ (t), uδ (t), uδ (t)) ≤ (f (t), uδ (t))V −
F(t − s)ε (uδ (s)) ds, ε (uδ (t)) .
0
Q
On the other hand,
Z
pδ (uδν (t)) − cν βδ2 (t)Rν (uδν (t)) uδν (t)da.
jδ (βδ (t), uδ (t) , uδ (t)) =
Γ3
Since
Z
cν βδ2 (t)Rν (uδν (t)) uδν (t)da ≤ 0,
Γ3
we have
Z t
Z
pδ (uδν (t))uδν (t)da ≤ (f (t), uδ (t))V −
F(t − s)ε(uδ (s))ds, ε(uδ (t)) .
Γ3
0
Q
Now, according to [12],
Z
Z
pδ (uδν (t)) uδν (t)da =
pδ (uδν (t)) uδν (t)da+
Γ3
Γ3 ∩{uδν (t)≤g}
Z
+
pδ (uδν (t)) uδν (t)da.
Γ3 ∩{uδν (t)>g}
As
Z
Z
p (uδν (t)) uδν (t)da ≥ 0,
pδ (uδν (t)) uδν (t)da =
Γ3 ∩{uδν (t)≤g}
Γ3 ∩{uδν (t)≤g}
we find that
Z
pδ (uδν (t)) uδν (t)da ≤
Γ3 ∩{uδν (t)>g}
Z t
(f (t), uδ (t))V −
F (t − s) ε (uδ (s)) ds, ε (uδ (t))
.
0
Q
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Analysis of a viscoelastic frictionless contact problem with adhesion
425
The left hand side of the previous inequaliy can be written as
Z
Z
uδν (t)−g
pδ (uδν (t))uδν (t)da =
(uδν (t)−g) da+
δ
Γ3 ∩{uδν (t)>g}
Γ3 ∩{uδν (t)>g}
Z
Z
g
p (g) uδν (t)da,
+
(uδν (t) − g)da +
Γ3 ∩{uδν (t)>g} δ
Γ3 ∩{uδν (t)>g}
from which we deduce
Z
Γ3 ∩{uδν (t)>g}
t
Z
≤ (f (t), uδ (t))V −
uδν (t) − g
δ
(uδν (t) − g) da ≤
F (t − s) ε (uδ (s)) ds, ε (uδ (t))
.
0
Q
This inequality implies that
(uδν (t) − g) 2
+
L2 (Γ3 )
≤ δc,
where c > 0. Hence, using (4.18) , we deduce that
(ūν (t) − g) 2
(4.21)
+ L (Γ ) ≤ lim inf (uδν (t) − g)+ L2 (Γ
3
3)
δ→0
= 0.
Therefore, it follows from (4.21) that (ūν (t) − g)+ = 0 a.e. on Γ3 , i.e., ūν (t) ≤ g
a.e. on Γ3 and then ū(t) ∈ K. Now, we state the following auxilliary problem.
Problem P3 . Find βa : [0, T ] → L∞ (Γ3 ) such that
β̇a (t) = − cν βa (t) (Rν (ūν (t)))2 − εa
a.e. t ∈ (0, T ) ,
+
βa (0) = β0 .
As in [22, Lemma 3.2] we have the following result.
Lemma 4.4. Problem P3 has a unique solution
βa ∈ W 1,∞ 0, T ; L2 (Γ3 ) ∩ B.
Also as in [22, Lemma 4.5] we have the following convergence result.
Lemma 4.5. We have
(4.22)
lim kβδ (t) − βa (t)kL2 (Γ3 ) = 0 for all t ∈ [0, T ].
δ→0
Next, we prove the following lemma.
Lemma 4.6. We have ū(t) = u(t) for all t ∈ [0, T ].
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Proof. Let v ∈ K and take v − uδ (t) in (4.9) , we have
hAε (uδ (t)) , ε (v − uδ (t))iQ + jδ (βδ (t), uδ (t) , v − uδ (t)) +
F (t − s) ε (uδ (s)) ds, ε (v − uδ (t))
= (f (t), v − uδ (t))V .
(4.23)
t
Z
+
0
Q
Since
jδ (βδ (t), uδ (t) , v − uδ (t)) =
Z
=
Γ3 ∩{uδν (t)≤g}
Z
+
Γ3 ∩{uδν (t)>g}
p (uδν (t)) − cν βδ2 (t)Rν (uδν (t)) (vν − uδν (t)) da+
pδ (uδν (t)) − cν βδ2 (t)Rν (uδν (t)) (vν − uδν (t)) da,
we use the definition of the operator Rν to see that
Z
pδ (uδν (t)) − cν βδ2 (t)Rν (uδν (t)) (vν − uδν (t)) da =
Γ3 ∩{uδν (t)>g}
Z
=
pδ (uδν (t)) (vν − uδν (t)) da.
Γ3 ∩{uδν (t)>g}
Therefore, using the definition of the function pδ , the right hand side of the
previous equality is written as
Z
pδ (uδν (t)) (vν − uδν (t)) da =
Γ3 ∩{uδν (t)>g}
Z
=
Γ3 ∩{uδν (t)>g}
uδν (t) − g
δ
((vν − g) − (uδν (t) − g))da+
Z
(p (g)) ((vν − g) − (uδν (t) − g))da,
+
Γ3 ∩{uδν (t)>g}
which implies that
Z
pδ (uδν (t)) (vν − uδν (t)) da ≤ 0.
Γ3 ∩{uδν (t)>g}
Then we deduce that
(4.24)
Z
≤
Γ3 ∩{uδν (t)≤g}
jδ (βδ (t), uδ (t) , v − uδ (t)) ≤
p (uδν (t)) − cν βδ2 (t)Rν (uδν (t)) (vν − uδν (t)) da. 17
Analysis of a viscoelastic frictionless contact problem with adhesion
427
Now, take v = ū(t) in (4.23) to obtain by using the assumption (2.13) (b) on
A that
m kuδ (t) − ū(t)k2V ≤
(4.25)
t
Z
≤
F (t − s) (ε (uδ (s)) − ε (ū (s))) ds, ε (ū(t)) − ε (uδ (t))
+
0
Q
Z
p (uδν (t)) − cν βδ2 (t)Rν (uδν (t)) (ūν (t) − uδν (t)) da +
+
Z
Γ3 ∩{uδν (t)≤g}
t
+
0
F (t − s) ε (ū(s)) ds, ε (uδ (t)) − ε(ū(t))
+ (f (t), uδ (t) − ū (t))V .
Q
Next, we denote
Z
cδ (t) =
p (uδν (t)) − cν βδ2 (t)Rν (uδν (t)) (ūν (t) − uδν (t)) da +
Γ3 ∩{uδν (t)≤g}
Z
+
0
+ hAε (ū(t)) , ε (ū(t)) − ε(uδ (t))iQ +
t
+ (f (t), uδ (t) − ū(t))V .
F (t − s) ε (ū(s)) ds, ε (uδ (t)) − ε(ū(t))
Q
As in [21] we have
Z t
≤
F (t − s) (ε (ū(s)) − ε (uδ (s))) ds, ε (uδ (t) − ū(t))
0
Q
Z
≤m
0
t
kuδ (s) − ū(s)kV ds kuδ (t) − ū(t)kV .
2
2
Using the elementary inequality ab ≤ a2 + b2 , we find that
Z t
(4.26)
≤
F (t − s) (ε (u(s)) − ε (uδ (s))) ds, ε (uδ (t) − u(t))
0
≤
m
2
Q
Z
0
t
2
m
kuδ (s) − u(s)kV ds + kuδ (t) − u(t)k2V .
2
Now, we combine inequalities (4.25) and (4.26) to obtain
Z t
2
m
m
2
kuδ (s) − ū(s)kV ds + |cδ (t)| ,
kuδ (t) − ū(t)kV ≤
2
2
0
which yields
kuδ (t) −
ū(t)k2V
Z
≤c
t
kuδ (s) −
0
ū(s)k2V
ds + |cδ (t)| .
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Arezki Touzaline
18
where c > 0. The Gronwall argument implies that there exists a constant
c1 > 0 such that
p
(4.27)
kuδ (t) − ū(t)kV ≤ c1 |cδ (t)|.
Now, using Lemma 4.3, we get
Z
p (uδν (t)) − cν βδ2 (t)Rν (uδν (t)) (ūν (t) − uδν (t)) da → 0,
Γ3 ∩{uδν (t)≤g}
hAε (ū(t)) , ε (ū(t) − uδ (t))iQ +
Z
+
t
F (t − s) ε (ū(s)) ds, ε (uδ (t)) − ε(ū(t))
→ 0,
0
Q
(f (t), uδ (t) − ū(t))V → 0,
which imply
cδ (t) → 0 as δ → 0.
Hence
kuδ (t) − ū(t)kV → 0 for all t ∈ [0, T ].
(4.28)
Now, from (4.23) and (4.24) we deduce the inequality
(4.29)
Z
+
Γ3 ∩{uδν (t)≤g}
Z
+
0
t
hAε (uδ (t)) , ε (v − uδ (t))iQ +
p (uδν (t)) − cν βδ2 (t)Rν (uδν (t)) (vν − uδν (t)) da+
≥ (f (t), v − uδ (t))V ∀v ∈ K.
F (t − s) ε (uδ (s)) ds, ε (v − uδ (t))
Q
Therefore, using (4.22), (4.28), (2.17) (b) and the properties of Rν , we obtain
by passing to the limit as δ → 0 in inequality (4.29) that
(4.30)
Z
hAε (ū(t)) , ε(v) − ε (ū(t))iQ +
t
+
F (t − s) ε (ū(s)) ds, ε(v) − ε (ū(t))
0
+j (βa (t), ū(t), v − ū (t)) ≥ (f (t), v − ū(t))V
Q
∀v ∈ K, t ∈ [0, T ].
Hence we combine Lemma 4.3, (4.30) and Problem P3 to deduce that (ū, βa )
satisfies (2.22)–(2.24) and so it is a solution of Problem P2 . As this problem
has a unique solution, Lemma 4.6 is proved. Moreover from (4.28), Lemmas 4.5 and 4.6, we deduce (4.16) and (4.17)
and then Theorem 4.2 is proved. Acknowledgements. I want to express my gratitude to the anonymous referee for
his/her helpful suggestions and comments.
19
Analysis of a viscoelastic frictionless contact problem with adhesion
429
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Received 24 November 2009
Faculté de Mathématiques, U.S.T.H.B
Laboratoire de Systèmes Dynamiques
BP 32 EL ALIA
Bab-Ezzouar, 16111, Algérie
[email protected]