Akram Ben
email: akram.benaissa@fsm.rnu.tn

Mohamed Ferhat
email: ferhat22@hotmail.fr

STABILITY RESULT FOR VISCOELASTIC WAVE EQUATION WITH DYNAMIC BOUNDARY CONDITIONS

Keywords: 2010 Mathematics Subject Classification. 35L05, 35L15, 35L70, 93D15 Energy decay, infinite memory, dynamic boundary condition

In this paper we consider wave viscoelastic equation with dynamic boundary condition in a bounded domain, we establish a general decay result of energy by exploiting the frequency domain method which consists in combining a contradiction argument and a special analysis for the resolvent of the operator of interest with assumptions on past history relaxation function. Contents 1. Introduction 2 2. Preliminaries 4 3. Well-posedness of the problem 5 4. Stability result 11

Introduction

We omit the space variable x of u(x, t), u t (x, t) and for simplicity reason denote u(x, t) = u, u t (x, t) = u t , when no confusion arises also the functions considered are all real valued, here u t = ∂u(t)/∂t, u tt = ∂ 2 u(t)/∂t 2 . Our main interest lies in the following system of viscoelastic equation :

                                 u tt -∆u + ∞ 0 g(s)∆u(x, t -s)ds = 0,
x ∈ Ω, t > 0

u tt = - ∂u ∂ν (x, t) - ∞ 0 g(s) ∂u ∂ν (x, t -s)ds , x ∈ Γ 1 , t > 0 u(x, t) = 0, x ∈ Γ 0 , t > 0 u(x, -t) = u 0 (x, t), x ∈ Ω, t > 0 u t (x, 0) = u 1 (x), x ∈ Ω, u(x, 0) = u 0 (x), x ∈ Ω, (1.1)
The main difficulty of the problem considered is related to the non ordinary boundary conditions defined on Γ 1 . Very little attention has been paid to this type of boundary conditions. From the mathematical point of view, these problems do not neglect acceleration terms on the boundary. Such types of boundary conditions are usually called dynamic boundary conditions. They are not only important from the theoretical point of view but also arise in several physical applications. For instance in one space dimension, problem (1.1) can modelize the dynamic evolution of a viscoelastic rod that is fixed at one end and has a tip mass attached to its free end. The dynamic boundary conditions represent the Newton's law for the attached mass, (see [START_REF] Andrews | Second order evolution equations with dynamic boundary conditions[END_REF][START_REF] Conrad | stabilization of a flexible beam with a tip mass[END_REF] for more details) which arise when we consider the transverse motion of a flexible membrane whose boundary may be affected by the vibrations only in a region. Also some of them as in problem (1.1) appear when we assume that is an exterior domain of R 3 in which homogeneous fluid is at rest except for sound waves. Each point of the boundary is subjected to small normal displacements into the obstacle (see [START_REF] Conrad | stabilization of a flexible beam with a tip mass[END_REF] for more details). Among the early results dealing with the dynamic boundary conditions are those of Grobbelaar-Van Dalsen [START_REF] Caroll | Singular and Degenerate Cauchy Problems[END_REF][START_REF] Grobbelaar-Van Dalsen | On the initial-boundary-value problem for the extensible beam with attached load[END_REF] in which the authors have made contributions to this field and in [START_REF] Grobbelaar-Van Dalsen | On the solvability of the boundary-value problem for the elastic beam with attached load[END_REF] the authors have studied the following problem:

                         u tt -∆u + δ∆u t = |u| p-1 u, x ∈ Ω, t > u tt = - ∂u(x, t) ∂ν (x, t) + δ ∂u(x, t) ∂ν (x, t) + α|u t | m-1 u(x, t) , x ∈ Γ 1 , t > u(x, t) = 0, x ∈ Γ 0 , t > u t (x, 0) = u 1 (x), x ∈ Ω, u(x, 0) = u 0 (x), x ∈ Ω, (1.
2) and they have obtained several results concerning local existence which extended to the global existence by using the concept of stable sets, the authors have obtained also the energy decay and the blow up of the solutions for positive initial energy.

The same problem has traited by [START_REF] Gerbi | Global existence and exponential growth for visceolastic wave equation with dynamic boundary conditions[END_REF], they showed the existence and uniqueness of a local in time solution and under some restrictions on the initial data, the solution continues to exist globally in time. On the other hand, if the interior source dominates the boundary damping, they proved that the solution is unbounded and grows as an exponential function. In addition, in the absence of the strong damping, they proved also the solution ceases to exist and blows up in finite time. Related problem as [START_REF] Gerbi | Asymptotic stability and blow up for a semilinear damped wave equation with dynamic boundary conditions[END_REF], M. M. Cavalcanti, A. Khemmoudj and M. Medjden [START_REF] Cavalcanti | Uniform stabilisation of the damped-Ventcel problem with variable coefficients and dynamic boundary conditions[END_REF] studied the following system:

                               u tt + Au + a(x)g 1 (u t ) = 0, x ∈ Ω, t > 0, u tt + ∂u(x, t) ∂ν A + A T υ + g 2 (υ t ) = 0, x ∈ Γ 1 , t > 0, u(x, t) = 0 x ∈ Γ 0 , t > 0, u(x, t) = υ, x ∈ Γ 1 , t > 0, (u t (x, 0), υ t (x, 0)) = (u 1 , υ 1 ), x ∈ (Ω, Γ 1 ), (u(x, 0), υ(0)) = (u 0 (x), υ 1 ), x ∈ (Ω, Γ 1 ).
They supposed that the second-order differential operators A and A T satisfy certain uniform ellipticity conditions, and they obtained uniform stabilization by using Riemannian geometry methods. Motivated by the previous works, it is interesting to show more general decay result to that in [START_REF] Gerbi | Asymptotic stability and blow up for a semilinear damped wave equation with dynamic boundary conditions[END_REF] and [START_REF] Gerbi | Local existence and exponential growth for a semilinear damped wave equation with dynamic boundary conditions[END_REF], we analyze the influence of the viscoelastic, on the solutions to (1.1). Under suitable assumption on function g(.), the initial data and the parameters in the equations. The content of this paper is organized as follows: In Section 2, we provide assumptions that will be used later. In Section 3, we state and prove the local existence result. In Section 4, by exploiting the frequency domain method we prove the stability result.

Preliminaries

In this section, we present some material and assumptions for the proof of our results. Denote

H 1 Γ 0 (Ω) = u ∈ H 1 (Ω) : u Γ 0 = 0 , H 1 Γ 0 (Γ) = u ∈ H 1 (Γ) : u Γ 0 = 0 , we set γ 1 the trace operator from H 1 Γ 0 (Ω) on L 2 (Γ 1 ) and H 1 2 (Γ 1 ) = γ 1 (H 1 Γ 0 (Ω))
. We denote by B the norm of γ 1 namely:

∀u ∈ H 1 Γ 0 (Ω), u 2.Γ 1 ≤ B ∇u 2 .
We will use the following embeddings

H 1 Γ 0 (Ω) → L q (Ω) for 2 ≤ q ≤ 2n n -2 , if n ≥ 3 and q ≥ 2, if n = 1, 2 L r (Ω) → L q (Ω), for q < r.
Then for some c s > 0,

ν q ≤ c s ∇ν 2 , ν q ≤ c s ν r for ν ∈ H 1 Γ 0 (Ω). We recall that H 1 2 (Γ 1 ) is dense in L 2 (Γ 1 ). We denote E(∆, L 2 (Ω)) = u ∈ H 1 (Ω) such that ∆u ∈ L 2 (Ω) , recall that for a function u ∈ E(∆, L 2 (Ω)), ∂u ∂ν ∈ H -1 2 (Γ 1
). We will usually use the following Green's formula

Ω ∇u(x)∇ω(x)dx = - Ω ∆u(x)ω(x)dx + Γ 1 ∂u ∂ν (x)ω(x)dΓ 1 , ∀ω ∈ H 1 Γ 0 (Ω). (2.3)
For studying the problem (1.1) we will need the following assumptions (A1).

• The relaxation function g is differentiable function such that, for s ≥ 0

g(s) ≥ 0, 1 - ∞ 0 g(s)ds = > 0, (2.4) • ∃ ζ 0 , ζ 1 > 0 : -ζ 1 g(t) ≤ g (t) ≤ -ζ 0 g(t), ∀ t ∈ R.
(2.5)

Well-posedness of the problem

In order to prove the existence of solutions of problem (1.1), we follow the approach of Dafermos [START_REF] Dafermos | Asymptotic stability in viscoelasticity[END_REF], by considering a new auxiliary variable the relative history of u as follows:

η := η t (x, s) = u(x, t) -u(x, t -s) in Ω × (0, ∞) × (0, ∞).
and the weighted L 2 -spaces

M = L 2 g (R + ; H 1 Γ 0 (Ω)) = ξ : R + → H 1 Γ 0 (Ω)) : ∞ 0 g(s) ∇ξ(s) 2 2 ds < ∞ ,
which is a Hilbert space endowed with inner product and norm consecutively

ξ, ζ M = ∞ 0 g(s) Ω ∇ξ(s)∇ζ(s)dx ds,
and

ξ 2 M = ∞ 0 g(s) ∇ξ(s) 2 2 ds.
Our analysis is given on the phase space

H = H 1 Γ 0 (Ω) × ×L 2 (Ω) × L 2 (Γ 1 ) × M. (3.6)
If we denote V := (u, u t , γ 1 (u t ), η), clearly, H is a Hilbert space with respect to the inner product

V 1 , V 2 H = (1 -g 0 ) Ω ∇u 1 .∇u 2 dx + Ω υ 1 .υ 2 dx + Γ 1 w 1 .w 2 dσ + ∞ 0 g(s) Ω ∇η 1 (s).∇η 2 (s)dx ds, (3.7) for V 1 = (u 1 , υ 1 , w 1 , η 1 ) T and V 2 = (u 2 , υ 2 , w 2 , η 2 ) T . Therefore, problem (1.1) is equivalent to                                          u tt -∆u - ∞ 0 g(s)∆η t (x, s)ds = 0, x ∈ Ω, t > 0, u tt = - ∂u ∂ν (x, t) + ∞ 0 g(s) ∂u ∂ν (x, t -s)ds , x ∈ Γ 1 , t > 0, η t t (x, t) + η t s (x, s) = u t (x, t), x ∈ Ω, t > 0, s > 0, u(x, t) = η t (x, 0) = 0, x ∈ Γ 0 , t > 0, u(x, -t) = u 0 (x, t), x ∈ Ω, t > 0, u t (x, 0) = u 1 (x), x ∈ Ω, u(x, 0) = u 0 (x), x ∈ Ω. (3.8) If V 0 ∈ H and V ∈ H , the problem (3.8
) is formally equivalent to the following abstract evolution equation in the Hilbert space

H      V (t) = AV (t), t > 0 V (0 = V 0 , .
(3.9) such that V 0 = (u 0 , u 1 , γ 1 (u 1 ), η 0 ) T and the operator A is defined by

A      u υ ω η      =           υ (1 -g 0 )∆u + ∞ 0 g(s)∆η(s)ds - ∂u ∂ν - ∞ 0 g(s) ∂ω ∂ν (x, t -s)ds - ∂η ∂s + υ           (3.10)
The domain of A is the set of V = (u, υ, ω, η) T such that the domain of A is defined by

D(A) =      (u, υ, ω, η) ∈ H 2 (Ω) ∩ H 1 Γ 0 (Ω) × H 1 Γ 0 (Ω) × L 2 (Γ 1 ) × M, (1 -g 0 )u + ∞ 0 g(s)η(s)ds ∈ L 2 (Ω), ω = γ 1 (u) = u 0 (., 0), η(0) = 0 on Γ 1     
Now, our main result is stated as follows:

Theorem 3.1. Let V 0 ∈ H. Then, system (3.8) has a unique weak solution V ∈ C(R + ; H) Moreover, if V 0 ∈ D(A)
, then the solution of (3.9) satisfies

V ∈ C 1 (R + ; H) ∩ C(R + ; H)
Proof. By Lumer-Phillips'Theorem, it suffices to show that A is mdissipative. We first prove that A is dissipative. Indeed, for any V = (u, υ, ω, η) T ∈ D(A), we have

AV, V H =           υ (1 -g 0 )∆u + ∞ 0 g(s)∆η(s)ds - ∂u ∂ν - ∞ 0 g(s) ∂u ∂ν (x, t -s)ds - ∂η ∂s + υ           ,      u υ ω η      = (1 -g 0 ) Ω ∇υ.∇udx + (1 -g 0 ) Ω ∆u.υdx + Ω ∞ 0 g(s)∆η(s)υ(s)ds + Γ 1 -∂u ∂ν - ∞ 0 g(s) ∂u ∂ν (x, t -s)ds ωdσ + -∂∂η ∂s + υ, η L 2 g (3.11) Noting that Γ 1 -∂u ∂ν - ∞ 0 g(s) ∂u ∂ν (x, t -s)ds ωdσ = 0 (3.12)
By exploiting Green's formula and integrating by parts and using the fact that η(0) = 0 (definition of D(A)), we obtain

-∂η ∂s , η L 2 g = 1 2 ∞ 0 g (s) ∇η(s) 2 .

ds

Inserting the previous inequalities into (3.11) , we obtain

AV, V H = 1 2 ∞ 0 g (s) ∇η(s) 2 ds,
which implies that AV, V H ≤ 0, since g is nonincreasing. This means that A is dissipative. Note that, thanks to (A1) and the fact that η ∈ L 2 g (R;

H 1 Γ 0 (Ω)), ∞ 0 g (s) ∇η(s) 2 ds = - ∞ 0 g (s) ∇η(s) 2 ds ≤ ζ 1 ∞ 0 g(s) ∇η(s) 2 ds < +∞, (3.13)
Next, we shall prove that Iλ-A is surjective for λ > 0. Indeed, let

F = (f 1 , f 2 , f 3 , f 4 ) T ∈ H, we show that there exists W = (ω 1 , ω 2 , ω 3 , ω 4 ) T ∈ D(A) satisfying (Iλ -A)W = F (3.14)
As previously , we have

A =           0 I 0 0 (1 -g 0 )∆ 0 0 ∞ 0 g(s)∆ds -∂ ∂ν 0 - ∞ 0 g(s) ∂ ∂ν ds 0 0 I 0 - ∂ ∂s          
which gives us

                   λω 1 -ω 2 = f 1 -(1 -g 0 )∆ω 1 + λω 2 - ∞ 0 g(s)∆ω 4 (s)ds = f 2 λω 3 + ∂ω 1 ∂ν + ∞ 0 g(s) ∂ω 3 (s) ∂ν ds = f 3 -ω 2 + λω 4 + ∂ ∂s ω 4 = f 4 (3.15)
we note that the first in (3.15) equation gives (3.17)

ω 2 = λω 1 -f 1 (3.
From the first and the second equation in (3.15) we can deduce the following

λ 2 ω 1 -(1 -g 0 )∆ω 1 = (f 2 + λf 1 ) + ∞ 0 g(s)ω 4 (s)ds. (3.18) Putting ū = ω 1 + ∞ 0 g(s)ω 3 (s)ds.
Then from equation (3.18), ū must satisfy

λ 2 ū -(1 -g 0 )∆ū = λ 2 ∞ 0 g(s)ω 3 (s)ds -(1 -g 0 ) ∞ 0 g(s)∆ω 3 (s)ds + (f 2 + λf 1 ) + ∞ 0 g(s)ω 4 (s)ds (3.19)
with the boundary conditions

ū = 0 on Γ 0 (3.20) ∂ ū ∂ν = f 3 -λū + λu 0 (x)(1 -l) on Γ 1 . ( 3
a(ū, ϕ) = l(ϕ) ∀ ϕ ∈ H 1 Γ 0 (Ω) (3.22) where a(ū, ϕ) = Ω λ 2 ū.ϕ + (1 -g 0) )∇ū.∇ϕ dx + λ Γ 1 ū(σ)ϕ(σ)dσ (3.23) and l(ϕ) = Ω λ 2 ∞ 0 g(s)ω 3 (s)dsϕdx + (1 -g 0) ∞ 0 g(s)∇ω 3 (s)ds∇ϕdx + (f 2 + λf 1 )ϕdx + Ω ∞ 0 g(s)ω 4 (s)dsϕdx + λ Γ 1 u 0 (σ)ϕ(σ)dσ (3.24)
It is clear that a is a bilinear and continuous form on H 1 Γ 0 (Ω) and l is linear and continuous form on H 1 Γ 0 (Ω). On the other hand, (3.23) implies that there exists a positive constant a 0 such that

a(ū, ū) = Ω λ 2 |ū| 2 dx + (1 -g 0) ) Ω |∇ū| 2 dx + λ Γ 1 |ū(σ)| 2 dσ ≥ a 0 ū 2 2 ∀ū ∈ H 1 Γ 0 (Ω), (3.25)
which implies that a is coercive. Therefore, using the Lax-Milgram Theorem, we conclude that (3.19) has a unique solution ū in H 1 Γ 0 (Ω). By classical regularity arguments, we conclude that the solution ū of (3.19) belongs into H 2 (Ω) ∩ H 1 Γ 0 (Ω) and satisfies (3.19). Consequently, using (3.16) and (3.17), we deduce that (3.8) has a unique solution V ∈ D(A). This proves that (λI -A) is surjective and hence A is an infinitesimal generator of a linear C 0 semigroup of contractions on H.

The energy associated with (3.8) is defined by Since

E(t) = 1 2 u t (t) 2 2 + ∇u(t) 2 2 + η 2 L 2 g , ( 3
u t (x, t) = η t (x, s) + η s (x, s), (x, s) ∈ Ω × R + , t ≥ 0, we have ∞ 0 g(s) Ω ∇η(s)∇u t (t)dxds = ∞ 0 g(s) Ω ∇η(s)∇η t (t)dxds + ∞ 0 g(s) Ω ∇η(s)∇η s (t)dxds = 1 2 ∞ 0 g(s) d dt ∇η(s) 2 2 ds - 1 2 ∞ 0 g (s) ∇η(s) 2 2 ds + ∞ 0 g(s) Ω ∇η(s)∇η t (t)dxds, (3.29)
Due to Young's inequality, we have for any δ > 0

∞ 0 g(s) Ω ∇η(s)∇η t (t)dxds ≤ ∞ 0 g(s) 1 4δ ∇η(s) 2 2 + δ ∇η t 2 2 ds ≤ δ ∞ 0 g(s)ds ∇η t 2 2 + 1 4δ ∞ 0 g(s) ∇η(s) 2 2 ds = δg 0 ∇η t 2 2 + 1 4δ η 2 L 2 g , ( 3

Stability result

The necessary and sufficient conditions for the exponential stability of the C 0 -semigroup of contractions on a Hilbert space were obtained by Gearhart [START_REF] Gearhart | Spectral theory for contraction semigroups on Hilbert spaces[END_REF] and Huang [START_REF] Huang | Characteristic condition for exponential stability of linear dynamical systems in Hilbert spaces[END_REF] independently, see also Prüss [START_REF] Prüss | On the spectrum of C0-semigroups[END_REF]. We will use the following result due to Gearhart.

Lemma 4.1. A Semigroup {e tA } t≥0 of contractions on a Hilbert space X is exponentially stable if and only if

iR ≡ {iβ; β ∈ R} ⊂ ρ(A) (4.31) and lim sup |β|→∞ (iβI -A) -1 X < ∞ (4.32)
Our main result reads as follows:

Theorem 4.2. The semigroup of system (3.8) decays exponentially as

e tA V 0 H ≤ Ce -γt V 0 D(A) , ∀ V 0 ∈ D(A), t > 0 (4.33)
Proof. The proof is splinted into two parts the first part consists to prove (4.31) which is equivalent to prove the following two assertions

(1) If β is a real number, then (iβI -A) is injectif and (2) If β is a real number, then (iβI -A) is surjectif.

It is the objective of the two following lemmas.

Lemma 4.3. If β is a real number, then iβ is not an eigenvalue of A Proof. We will show that the equation

AZ = iβZ (4.34)
with Z = (u, υ, ω, η) T ∈ D(A) and β ∈ R has only the trivial solution. Equation (4.34) can be written as

iβu -υ = 0 (4.35) iβυ -(1 -g 0 )∆u -∞ 0 g(s)∆η(s)ds = 0 (4.36) iβω + ∂u ∂ν + ∞ 0 g(s) ∂ω(s) ∂ν ds = 0 (4.37) iβη + ∂η ∂s -υ = 0 (4.38)
By taking the inner product of (4.34) with Z ∈ D(A) and using (3.27), we get:

( AZ, Z H ) ≤ ∞ 0 g (s) ∇η(s) 2 ds ≤ - ∞ 0 g(s) ∇η(s) 2 ds = -η 2 M ≤ 0 (4.39)
Thus we obtain that: η = 0, moreover as η satisfies (4.38) by integration, we obtain

η(s) = s 0
e iβy υ(y))dy e -iβs since η = 0 we deduce that υ = 0 and from (4.35) we have u = 0. since ω = γ 1 (u) = u 0 (., 0), we obtain also ω = 0. Thus the only solution of (4.34) is the trivial one. Hence the proof is completed.

Next, we show that A has no continuous spectrum on the imaginary axis. Proof. In view of Lemma it is enough to show that A is surjective.

For F = (f 1 , f 2 , f 3 , f 4 ) T ∈ H, let V = (u, υ, ω, η) T ∈ D(A) solution of (iβI -A)V = F (4.40) which is                    iβu -υ = f 1 -(1 -g 0 )∆u + iβυ - ∞ 0 g(s)∆η(s)ds = f 2 iβω + ∂u ∂ν + ∞ 0 g(s) ∂ω(s) ∂ν ds = f 3 -υ + iβη + ∂η ∂s = f 4
(iβ) 2 ū -(1 -g 0 )∆ū = (iβ) 2 ∞ 0 g(s)ω 3 (s)ds -(1 -g 0 ) ∞ 0 g(s)∆ω 3 (s)ds + (f 2 + iβf 1 ) + ∞ 0 g(s)ω 4 (s)ds
ū = 0 on Γ 0 (4.46) ∂ ū ∂ν = f 3 -iβ ū + iβu 0 (x)(1 -l) on Γ 1 . ( 4
(ū, ϕ) = l(ϕ) ∀ ϕ ∈ H 1 Γ 0 (Ω) (4.48) where b(ū, ϕ) = Ω (iβ) 2 ū.ϕ + (1 -g 0) )∇ū.∇ϕ dx + iβ Γ 1 ū(σ)ϕ(σ)dσ (4.49) and l(ϕ) = Ω (iβ) 2 ∞ 0 g(s)ω 3 (s)dsϕdx + (1 -g 0) ∞ 0 g(s)∇ω 3 (s)ds∇ϕdx + (f 2 + iβf 1 )ϕdx + Ω ∞ 0 g(s)ω 4 (s)dsϕdx + iβ Γ 1 u 0 (σ)ϕ(σ)dσ (4.
50) It is clear that b is a bilinear and continuous form on H 1 Γ 0 (Ω) and l is linear and continuous form on H 1 Γ 0 (Ω). On the other hand (4.49) implies that there exists a positive constant a 0 such that

a(ū, ū) = Ω (iβ) 2 |ū| 2 dx + (1 -g 0) ) Ω |∇ū| 2 dx + iβ Γ 1 |ū(σ)| 2 dσ ≥ a 0 ū 2 2 ∀ū ∈ H 1 Γ 0 (Ω), ( 4

  16) and the last equation in (3.15) with η(0) = 0 has unique solution ω 4 (s) = s 0 e y (f 4 (y) + ω 2 (y))dy e -s .

  .30) by replacing (3.29) and (3.30) into (3.28) we get the desired result.

Lemma 4 . 4 .

 44 If β is a real number, then iβ to the resolvent set ρ(A) of A

( 4 .

 4 41) we note that the first equation in (4.41) gives υ = iβω 1 -f 1 . (4.42) The last equation in (4.41) with η(0) = 0 has unique solution ω 4 (s) = s 0 e iβy (f 4 (y) + ω 2 (y))dy e -iβs (4.43) Another time, from the first and the second equation in (4.41) we can deduce the following (iβ) 2 ω 1 -(1 -g 0 )∆ω 1 = (f 2 + iβf 1 ) + )ω 3 (s)ds = ū, then from equation (4.44) ū must satisfy

  (4.61) Now, multiplying the equation (4.54) by υ n and (4.55) by u n , adding them and taking the real parts , we obtain -)∇η n (s)∇u n (t)ds → 0 in L 2 (Ω).

( 4 .

 4 62) Due to Young's inequality, we have for any δ > 0 ∞ 0 g(s) Ω ∇η n (s)∇u n (t)dxds ≤ Replacing the last inequality in (4.62), for δ sufficiently small we get ∇u n → 0 in L 2 (Ω).

  u n → 0 in H 1 Γ 0 (Ω).(4.65) By using (4.56) and trace theorem we getω n → 0 in L 2 (Γ 1 )(4.66) which contradicts (4.52) . Thus (4.32) is proved.

  .26)

	Lemma 3.2. The functional defined in (3.26) satisfies the following
	inequality
			E (t) ≤	1 2	0	∞	g (s) ∇η(s) 2 2 ds, ∀ t ≥ 0,	(3.27)
	Proof. By multiplying the first equation in (3.8) by u t (t), and integrat-
	ing over Ω we get
	1 2	d dt	u t (t) 2 2 +	1 2	∇u(t) 2 2 +	0	∞	g(s)

R n ∇η(s)∇u t (t)dsdx.

(3.28)

  .51) which implies that b is coercive. Therefore, using the Lax-Milgram Theorem, we conclude that (4.45) has a unique solution ū in H 1 Γ 0 (Ω). By classical regularity arguments, we conclude that the solution ū of (4.42) belongs into H 2 (Ω) ∩ H 1 Γ 0 (Ω). Consequently, using (4.42) and (4.43), we deduce that (4.34) has a unique solution V ∈ D(A). This proves that (iβ -A) is surjective.Lemma 4.5. The resolvent operator of A satisfies (4.32). that condition (4.32) is false. By Banach-steinhaus Theorem ([START_REF] Dunford | Linear Operators, Part I[END_REF]), there exists a sequence of real numbers β n → +∞ and a sequence of vectorsZ n = (u n , υ n , ω n , η n ) T ∈ D(A) with Z n H = 1 (4.52) such that (iβ n I -A)Z n H → 0 as n → ∞. (iβ n u n -υ n ) ≡ f n → 0, in H 1 ΓOur aim is to derive from (4.53) that Z n H converges to zero, thus there is a contradiction.| (iβ n I -A)Z n , Z n H | ≤ (iβ n I -A)Z n H (4.58) Using the hypotheses on g , we find that By exploiting the convergence (4.59) and (4.60), we can deduce from (4.54) that υ n → 0 in L 2 (Ω) and u n → 0 in L 2 (Ω).

	Suppose (4.53)
	That's
	iβ n ω n +	∂u n ∂ν	+	0	∞	g(s)	∂ω n (s) ∂ν	ds ≡ h n → 0, in L 2 (Γ 1 )	(4.56)
	iβ n η n +	∂η n ∂s	-υ n ≡ k n → 0, in M.	(4.57)
					η n → 0 in L 2 g (R + ; H 1 Γ 0 (Ω))	(4.59)
	and
	η n (s) =

0 (Ω) (4.54)

iβ n υ n -(1 -g 0 )∆u n -∞ 0 g(s)∆η n (s)ds ≡ g n → 0, in L 2 (Ω) (4.

55) s 0 e iβy k n (y) e -iβs + s 0 e iβy υ n (y)dy e -iβs . (4.60)