Asymptotic behaviour of the energy to partially ...rivera/Art_Pub/Tesepere.pdf · Asymptotic...

Asymptotic behaviour of the energy to partially

viscoelastic materials∗

Jaime E. Munoz Rivera Alfonso Peres Salvatierra

Abstract

In this paper we study models of materials consisting of an elastic part (without memory)and a viscoelastic part, where the dissipation given by the memory is effective. We showthat the solutions of the corresponding partial viscoelastic model decay exponentially tozero, provided the relaxation function also decays exponentially, no matter how small is theviscoelastic part of the material.

AMS classification code: 35B40, 35L05, 35L70

Keywords and phrases: viscoelasticity, exponential decay, localized damping, materials with memory.

1 Introduction

Let us consider a n-dimensional body which in its reference configuration is homogeneous and

occupies the open bounded set Ω ⊂ IRn with smooth boundary Γ. Let x 7→ u(x, t) be the

position of the material particle x at time t. Then the viscoelastic equation of motion is given

by

ρutt − κ∆u +

∫ ∞

0g(s) div a(x)∇u(·, t − s) ds = f in Ω×]0,∞[

u(x, t) = 0 on ∂Ω×]0,∞[= Γ×]0,∞[

u(x, 0) = u0(x), ut(x, 0) = ut(x), in Ω.

where ρ is the mass density function, g is the relaxation function and f denotes the body force.

Here we are mainly interested on the asymptotic behaviour of the solution u when t tends to

infinity. Note that the above model is dissipative, and the dissipation is given by the memory

term, where a ≥ 0. The memory is effective only in a part of the body Ω where a > 0.

Concerning the exponential stability of dissipative systems, it is well known by now that the

solution of the wave equation with frictional damping

∗Supported by a grant 305406/88-4 of CNPq-BRASIL (To appear in Quarterly of Applied Mathematics)

1

utt − ∆u + a(x)ut = 0 in Ω×]0,∞[,

u(x, t) = 0, on Γ×]0,∞[,

decays exponentially to zero as time goes to infinity provided a(x) ≥ a0 > 0 a.e. in Ω. In this

case, since a(x) is a positive function, the dissipative effect is working in the whole domain Ω.

In this direction we may ask, whether the damping term a(x)ut continues to be effective when

the function a satifies only a(x) ≥ 0. That is, suppose that the function a(x) vanishes outside

an open subset ω of Ω. Is the dissipative term a(x)ut strong enough to produce the exponential

decay of the solution?. Under what conditions on ω we may expect an uniform rate of decay

for the solution? A first answer to these questions was given in [6], [14], [18], [27]. In those

works the authors proved that the solution of the wave equation with ”local” damping decays

exponentially to zero provided ω is a neighborhood of the boundary Γ. This means that a

dissipation mechanism which is effective over a strategic part of the material is strong enough to

produce exponential rate of decay of the total energy. Such results are connected with Control

Theory. It was proved by Lions [13] that the solution of the wave equation can be controlled

acting only in a neighborhood of the boundary. Roughly speaking, any time that we can control

the solution acting on a part of the material, it is possible to introduce a dissipative mechanism

at the same place producing uniform decay in time of the energy. The methods in Bardos et al.

[1] allow to give a necessary and sufficient condition for the uniform decay of the solutions of the

dissipative wave equation. Namely, this condition requests that every ray of geometric optics

that propagates in Ω and is reflected on the boundary enters the region a > 0 in a uniform

time. In fact, the necessity of this condition is due to the construction of Ralston [22, 23] of

the gaussian beam solutions of the wave equation. Other examples of localized damping are

given by introducing dissipative boundary conditions acting in a part of the boundary. In this

direction there is an extensive literature, see for example [4], [5], [6], [7], [8], [10], [11], [12], [17],

[19], [20], [21], [24], [26], [28] among others.

In this paper we also consider locally distributed dissipation; but this dissipation does not

appear by the introduction of any artificial mechanism. On the contrary, it arises because of

the mixed structure of the material. That is, we consider a body consisting of an elastic and

a viscoelastic part. So, the dissipation is due to the memory effect which works only over a

portion of the material. The one dimensional case is represented in the next picture, but our

analysis applies without restriction on the dimension

2

@ @

Viscoelastic PartElastic Part

0 LL1

Denoting by σ the stress and by ∗ the convolution product g ∗ f =∫ t0 g(t − τ)f(τ) dτ , the

constitutive law we use in this paper is given by

σ = κ∇u + a(x)g ∗ ∇u,

so that, the corresponding motion equation for κ = 1 may be written as

utt − ∆u +

∫ t

0g(t − τ)div a(x)∇u dτ = 0, in Ω×]0,∞[ (1.1)

with initial data

u(x, 0) = u0(x), ut(x, 0) = u1(x) in Ω. (1.2)

and Dirichlet’s boundary condition

u(x, t) = 0, on Γ×]0,∞[. (1.3)

For materials with memory the stress depends not only on the present values but also on the

entire temporal history of the motion. Therefore, we have also to prescribe the history of u prior

to 0 as the initial data. Here we assume that u vanishes identically for t < 0, that is

u(x, t) = 0, for t < 0.

We do not assume that u(·, 0+) = u(·, 0−) or ut(·, 0+) = ut(·, 0−).

Let x 7→ a(x) be a non negative C2-function defined over Ω and let us denote by ωε the set

ωε = (∪x∈Γ0Bε(x)) ∩ Ω,

where Bε(x) = x ∈ IRn; ‖x‖ ≤ ε and Γ0 is given by

Γ0 = x ∈ Γ; (x − x0) · ν ≥ 0 ,

where ν is the unitary external normal defined over Γ and x0 any point of IRn. For the one

dimensional case we have that ωε =]L − ε, L[. The hypotheses we use on a are the following

3

x 7→ a(x) ∈ C2(Ω); a(x) =

0 on Ω \ ωε

1 on ωε/2.(1.4)

|∇a(x)|2 ≤ c|a(x)|. (1.5)

We can built a function a satisfying condition (1.5), by taking a = b2, with b as in (1.4). In

the next picture Ω is a rectangle, the set ωε denotes the viscoelastic part of the body. Note that

ωε is behind the part that an observer can see when situated on x0. This is a particular case in

which the geometric requeriments of Bardos et al. [1] and Ralston [22, 23] are satisfied

ZZ

ZZ

ZZ

ZZ

ZZ

s x0

Ω

s@@I

ωε

-ε

6?

ε

6?

ε

Elastic Part

Let us mention some other papers related to the problems we address. Dafermos in [2] proved

that the solution to viscoelastic system goes to zero as time goes to infinity; but without giving

explicit rate of decay. Lagnese in [9] considered the linear viscoelastic plate equation obtaining

uniform rates of decay but introducing additional damping terms acting on the boundary. Uni-

form rates of decay for the solutions of linear viscoelastic system with memory were obtained

recently by M. Rivera et al [15]. Unfortunately the method used to achieve uniform rates of

decay in those works are based on second order estimates, which are time depending in our

problem. Thus, the methods that have been used for establishing uniform rates of decay fail

in the case of partially viscoelastic equation. Therefore a new asymptotic technique has to be

devised. For nonlinear models see [16].

The aim of this paper is to show that in the geometrial setting above, the energy decays

exponential provided the kernel g also decays exponentially. More especifically, we assume that

g satisfies

g ∈ C3(]0,∞[), g(t) > 0, |g′′(t)| ≤ cg(t), |g′′′(t)| ≤ cg(t) (1.6)

4

−κ0g(t) ≤ g′(t) ≤ κ1g(t) (1.7)

α := 1 −∫ ∞

0g(τ) dτ < 1 (1.8)

To facilitate our analysis, we introduce the following binary operators

g <= ∇u =

∫ t

0g(t − τ)

∫

Ωa(x)|∇u(x, t) −∇u(x, τ)|2 dxdτ

g <= u =

∫ t

0g(t − τ)

∫

Ωa(x)|u(x, t) − u(x, τ)|2 dxdτ

Under the above conditions the main result of this paper is given by

Theorem 1.1 Under the above assumptions on Ω, ω and a and with the kernel g satisfying

(1.6)–(1.8), the weak solution of the viscoelastic equation (1.1)–(1.3), decays exponentially as

time goes to infinity. That is, there exist positive constants c and γ that do not depend on the

initial data, such that

E(t) ≤ cE(0)e−γt

where by E(t) we are denoting the first order energy

E(t) =1

2

∫

Ω|ut|2 +

1 − a(x)

∫ t

0g(τ) dτ

|∇u|2 dx +1

2g <= ∇u.

The method we use here is based on the construction of a functional L for which an inequality

of the form

d

dtL(t) ≤ −cL(t)

holds, with c > 0. To construct such functional L we start from the energy identity. Then,

we look for other functions whose derivatives introduce negative terms such as: −∫

a(x)|ut|2,−∫

a(x)|∇u|2, etc. until we are able to construct the whole energy in the right hand side of the

energy identity. Finally, we take L as the summatory of such functions. Unfortunately, the above

process also introduces terms without definite sign. To overcome this difficulty, we introduce a

new multiplier which allows us to get the appropiate estimates. Finally, we choose carefully the

coefficients of each term of L, such that the resulting summatory satisfies the required inequality.

The remaining part of this article is organized as follows. In the next section 2, we establish

the existence and regularity result to equation (1.1)–(1.3) as well as the energy identity. Section

3 deals with the regularity of the convolution term. Finally in section 4 we show the exponential

decay of the solution of the equation (1.1)–(1.3).

5

2 Existence results and preliminaries

Our starting point is given by the following Lemma

Lemma 2.1 For any v ∈ C1(0, T ;H1(0, L)) we get

∫

Ω

∫ t

0g(t − τ)a(x)∇v dτ · ∇vt dx = −1

2g(t)

∫

Ωa(x)|∇v|2 dx +

1

2g′ <= ∇v

−1

2

d

dt

g <= ∇v −(∫ t

0g dτ

) ∫

Ωa(x)|∇v|2 dx

∫

Ω

∫ t

0g(t − τ)a(x)v dτ · vt dx = −1

2g(t)

∫

Ωa(x)|v|2 dx +

1

2g′ <= v

−1

2

d

dt

g <= v −(∫ t

0g dτ

)∫

Ωa(x)|v|2 dx

Proof.- It is easy to see that

d

dtg <= ∇v = g′ <= ∇v − 2

∫

Ω

∫ t

0g(t − τ)a(x)∇v(τ)dτ · ∇vt(x, t) dτdx

+2

∫ t

0g(t − τ) dτ

∫

Ωa(x)∇v∇vt dx

= g′ <= ∇v − 2

∫

Ω

∫ t

0g(t − τ)a(x)∇v(τ) dτ · ∇vt dx

+d

dt

∫ t

0g(τ) dτ

∫

Ωa(x)|∇v|2 dx

− g(t)

∫

Ωa(x)|∇v|2 dx

This shows our result, the proof of the other identity is similar. <=

It is not difficult to show that there exists only one solution to equation (1.1)–(1.3). We sum-

marize the existence result in the following theorem

Theorem 2.1 Let us suppose that g is a C0-function and that the initial data satisfies

(u0, u1) ∈ H10 (Ω) × L2(Ω),

then, there exists only one weak solution u to equation (1.1)–(1.3) with the following regularity,

u ∈ L∞(0,∞;H10 (Ω)), ut ∈ L∞(0,∞;L2(Ω)).

In addition, if g ∈ C1 and

(u0, u1) ∈ H2(Ω) ∩ H10 (Ω) × H1

0 (Ω);

6

then, there exist only one strong solution u of equation (1.1)–(1.3) satisfying

u ∈ C2−i(0,∞;H10 (Ω) ∩ H i(Ω)), i = 1, 2. u ∈ C2(0,∞;L2(Ω)).

<=

The dissipative property of the viscoelastic equation is summarized in the following Lemma:

Lemma 2.2 Any strong solution of (1.1)–(1.3) satisfies

d

dtE(t) =

1

2g′ <= ∇u − 1

2g(t)

∫

Ωa(x)|∇u|2 dx.

Proof.-Multiplying equation (1.1) by ut and integration over Ω yields

d

dt

∫

Ω

(

|ut|2 + |∇u|2)

dx =

∫

Ωa(x)g ∗ ∇u · ∇ut dx.

From Lemma 2.1 our conclusion follows <=

Lemma 2.2 tells us that the dissipation given by the memory term is effective only on the

support of the function a. We will show in section 4 that such dissipation is enough to produce

the exponential decay of the energy, as time goes to infinity.

3 Regularity of the convolution

Let us denote by ‖ · ‖C0 the norm in C0(Ω). The following Lemma will play an important role

in the sequel.

Lemma 3.1 Let us suppose that g is a positive function satisfying condition (1.8), a ∈ C 0(Ω)

is such that ‖a‖C0 ≤ 1 and finally let us take f ∈ Lp(0, T ;L2(Ω)) with 1 ≤ p < ∞. In this

conditions we have that there exists only one solution v of the Volterra’s equation

v(x, t) −∫ t

0g(t − τ)a(x)v(·, τ) dτ = f(x, t), a.e. (x, t) ∈ Ω×]0, T [

satisfying

v ∈ Lp(0, T ;L2(Ω)).

Besides, there exists a positive constant c independent of T , such that

‖v‖Lp(0,T ;L2) ≤ c‖f‖Lp(0,T ;L2).

7

Proof.-We will use the Picard’s method. Let us denote by

v0 := f, v−1 = 0,

and consider the iterative equation

vµ(x, t) =

∫ t

0g(t − τ)a(x)vµ−1(·, τ) dτ + f(x, t).

Denoting by

wµ = vµ − vµ−1,

we have that

wµ(x, t) =

∫ t

0g(t − τ)a(x)wµ−1(·, τ) dτ,

from where it follows that

‖wµ(t)‖L2 ≤∫ t

0g(t − τ)‖a‖C0‖wµ−1(·, τ)‖L2 dτ.

Using Young’s inequality we get

‖wµ‖Lp(0,T ;L2) ≤(∫ T

0g(τ) dτ

)

‖a‖C0‖wµ−1‖Lp(0,T ;L2)

From where it follows that:

‖wµ‖Lp(0,T ;L2) ≤[

(

∫ T

0g dτ)‖a‖C0

]2

‖wµ−2‖Lp(0,T ;L2)

≤[

(

∫ T

0g dτ)‖a‖C0

]µ

‖w0‖Lp(0,T ;L2)

≤[

(

∫ T

0g dτ)‖a‖C0

]µ

‖f‖Lp(0,T ;L2)

Since (∫ T0 g dτ)‖a‖C0 < 1, then it follows that

∞∑

µ=1

‖wµ‖Lp(0,T ;L2) ≤‖f‖Lp(0,T ;L2)

1 −∫ T0 g dτ‖a‖C0

≤‖f‖Lp(0,T ;L2)

1 − (1 − α)‖a‖C0

.

Where α was defined in (1.8). Recalling that

vµ − v0 =µ∑

i=1

wi,

we conclude that the sequence vµ is convergent. So, there exists a function v ∈ Lp(0, T ;L2(Ω))

for which we have

vµ → v strong in Lp(0, T ;L2(Ω)).

8

Besides, we have that

‖v‖Lp(0,T ;L2) ≤

1

1 − (1 − α)‖a‖C0

+ 1

︸︷︷︸

:=C

‖f‖Lp(0,T ;L2).

Where C is a positive constant which does not depend on T . To show the uniqueness, let us

suppose that there exists another solution v. Denoting by V = v − v, we have that

V =

∫ t

0g(t − τ)a(x)V (x, τ) dτ.

In particular,

|V | ≤∫ t

0g(t − τ)|a(x)V (x, τ)| dτ.

From Gronwall’s inequality we conclude that V = 0. The proof is now complete <=

In the next Lemma we will show that integration in time is equivalent to remove spatial

derivatives.

Lemma 3.2 Let us suppose that 0 ≤ a(x) ≤ 1 satisfies conditions (1.6)–(1.8) and that g is a

positive function satisfying (1.8). If u is a weak solution of (1.1)–(1.3) satisfying

ut ∈ L∞(0, T ;L2(Ω)), u ∈ L∞(0, T ;H10 (Ω)),

then we have that

g ∗ u ∈ L2(0, T ;H2(Ω))

and

‖g ∗ u‖L2(0,T ;H2) ≤ C

∫ T

0E(t) dt + CE(0). (3.1)

where C is a positive constant independent of T .

Proof.-Applying convolution to equation (1.1) we have

−∆g ∗ u + g ∗ g ∗ div a(x)∇u = −g ∗ utt.

Performing an integration by parts over ]0, t[ we get

g ∗ utt = g(0)ut − g(t)ut(·, 0) +

∫ t

0g′(t − τ)ut dτ := −F.

From the hypotheses we conclude that F ∈ L2(0, T ;L2(Ω)). Denoting by v = g ∗ ∆u, we have

that

−v + a(x)g ∗ v = G (3.2)

9

where

G = F − g ∗ ∇a(x) · ∇u. (3.3)

Applying Lemma 3.1 for p = 2 and since ‖a‖C0 ≤ 1 we conclude that

v ∈ L2(0, T ;L2(Ω)).

Since

∫ T

0

∫

Ω|F |2 dx ≤ 2g(0)2

∫ T

0

∫

Ω|ut|2 dxdt +

∫ T

0g(t)

∫

Ω|u1|2 dxdt +

∫ T

0

∫

Ω|g′ ∗ ut|2 dxdt

︸︷︷︸

≤∫ T

0|g′| dt

∫ T

0

∫

Ω|ut|2 dx

we have that

‖v‖L2(0,T ;L2(Ω)) ≤ c‖G‖L2(0,T ;L2(Ω))

≤ C

∫ T

0E(t) dt + C

∫ T

0g(t) dtE(0),

which implies that g ∗ ∆u ∈ L2(0, T ;L2(Ω)) from where we obtain

g ∗ u ∈ L2(0, T ;H2(Ω))

Finally, using elliptic regularity, inequality (3.1) follows. The proof is now complete <=

4 Exponential Decay

To show the exponential decay of the solution let us introduce the following functional

I(t) :=

∫

Ωa(x)

ut(g ∗ u)t −1

2g(0)|u|2 −

∫ t

0g dτ |u|2 dx

dx− 1

2g′′ <= u+

1

2

∫

Ωa2(x)|g∗∇u|2 dx.

In these conditions we have:

Lemma 4.1 Under the above conditions and for g ∈ C 3, satisfying conditions (1.6)–(1.8), we

have that for any δ > 0 there exists Cδ satisfying

d

dtI(t) ≤ −g(0)

∫

Ωa(x)|ut|2 dx + δ

∫

Ωa(x)|∇u|2 dx + Cδg <= ∇u

+Cδg(t)

∫

Ωa(x)|∇u|2 dx + Cδ

∫

ωε

∫ t

0g(t − τ)|u(x, t) − u(x, τ)|2 dxdt + Cδ

∫

ωε

|u(x, t)|2 dx.

Here Cδ → ∞ when δ → 0.

10

Proof.-Multiplying equation (1.1) by a(x)(g ∗ u)t we get:

∫

Ωutta(x)(g ∗ u)t dx

︸︷︷︸

:=I1

−∫

Ω∆ua(x)(g ∗ u)t dx

︸︷︷︸

:=I2

−∫

Ωa(x)g ∗ ∇u · ∇ a(x)(g ∗ u)t dx

︸︷︷︸

:=I3

= 0,

from where we have:

I1 =d

dt

∫

Ωuta(x)(g ∗ u)t dx −

∫

Ωuta(x)(g ∗ u)tt dx

=d

dt

∫

Ωuta(x)(g ∗ u)t dx −

∫

Ωuta(x)(g(0)u + g′ ∗ u)t dx

=d

dt

∫

Ωuta(x)(g ∗ u)t dx − g(0)

∫

Ωa(x)|ut|2 dx −

∫

Ωa(x)ut

g′(0)u + g′′ ∗ u

dx

=d

dt

∫

Ωuta(x)(g ∗ u)t dx − g(0)

∫

Ωa(x)|ut|2 dx − g′(0)

2

d

dt

∫

Ωa(x)|u|2 dx

−∫

Ωa(x)utg

′′ ∗ u dx.

Using Lemma 2.1 we get that

∫

Ωa(x)utg

′′ ∗ u dx =1

2g′′′ <= u − 1

2

∫

Ωa(x)|u|2 dx − 1

2

d

dt

g′′ <= u −∫ t

0g dτ

∫

Ωa(x)|u|2 dx

,


I1(t) =d

dtI0(t) − g(0)

∫

Ωa(x)|ut|2 dx +

1

2g′′(t)

∫

Ωa(x)|u|2 dx − 1

2g′′′ <= u,

where by I0 we are denoting

I0 =

∫

Ωa(x)

ut(g ∗ u)t −g′(0)

2|u|2

dx − 1

2g′′ <= u −

∫ t

0g dτ

∫

Ωa(x)|u|2 dx.

On the other hand:

I2 = −∫

Ωa(x)∆u

g(0)u + g′ ∗ u

dx

=

∫

Ω∇u · ∇

a(x)g(0)u + a(x)g′ ∗ u

dx

=

∫

Ω[∇u · ∇a(x)]g(0)u + g(0)

∫

Ωa(x)|∇u|2 dx +

∫

Ω[∇u · ∇a]g′ ∗ u dx

+

∫

Ωa(x)∇u · g′ ∗ ∇u dx.

Note that

11

∫

Ω[∇u · ∇a]g(0)u +

∫

Ω[∇u · ∇a]g′ ∗ u dx =

g(t)

∫

Ω[∇u · ∇a]u +

∫

Ω∇u · ∇a

∫ t

0g′(t − τ) u(·, τ) − u(·, t) dx.

Similarly, we have

g(0)

∫

Ωa(x)|∇u|2 dx +

∫

Ωa(x)∇u · ∇g′ ∗ u dx =

g(t)

∫

Ωa(x)|∇u|2 dx +

∫

Ωa(x)∇u ·

∫ t

0g′(t − τ) ∇u(·, τ) −∇u(·, t) dx.

Using the above formulas we conclude that I2 may be written as

I2(t) = g(t)

∫

Ω[∇u · ∇a]u +

∫

Ω[∇u · ∇a]

∫ t

0g′(t − τ) u(·, τ) − u(·, t) dx

+g(t)

∫

Ωa(x)|∇u|2 dx +

∫

Ωa(x)∇u ·

∫ t

0g′(t − τ) ∇u(·, τ) −∇u(·, t) dx.

From hypotheses (1.5), (1.7)-(1.8) we have that

∫

Ω[∇u · ∇a]

∫ t

0g′(t − τ) u(·, τ) − u(·, t) dτdx ≤

c

∫

Ω

√a|∇u|

∫ t

0g dτ

1/2 ∫ t

0g(t − τ)|u(·, t) − u(·, τ)|2 dτ

1/2

dx

≤ δ

∫


∫ t

0

∫

ωε

|u(·, t) − u(·, τ)|2 dxdt.

Using similar ideas to estimate the term

∫

Ωa(x)∇u ·

∫ t

0g′(t − τ) ∇u(·, τ) −∇u(·, t) dxdt

we arrive at

I2(t) ≤ Cδg(t)

∫

Ωa(x)|∇u|2 dx + Cδg <= ∇u + δ

∫

Ωa(x)|∇u|2 dx

+Cδ

∫ t

0

∫

ωε

g′(t − τ)|u(x, t) − u(x, τ)|2 dxdτ,

where δ is a small parameter to be fixed later and Cδ → ∞ as δ → 0. Finally,

12

I3(t) = −∫

Ωag ∗ ∇u · ∇(ag ∗ u)t dx

= −∫

Ωa(x)

∫ t

0g(t − τ)[∇u(·, τ) −∇u(·, t) · ∇a](g ∗ u)t

−∫ t

0g dτ

∫

Ωa(x)[∇u · ∇a](g ∗ u)t dx − 1

2

d

dt

∫

Ωa2|g ∗ ∇u|2 dx

= −∫

Ωa(x)

∫ t

0g(t − τ) ∇u(·, τ) −∇u(·, t) · ∇a(x)

g(0)u + g′ ∗ u

dx

−∫ t

0g dτ

∫

Ωa(x)∇u · ∇a(x)

g(0)u + g′ ∗ u

− 1

2

d

dt

∫

Ωa2|g ∗ ∇u|2 dx.

Using the identity

g(0)u + g′ ∗ u = g(t) +

∫ t

0g′(t − τ) u(·, τ) − u(·, t) dτ

and since a(x) ≤ 1, with the same arguments we used to estimate I2 we have

−∫

Ωa(x)

∫ t

0g(t − τ)[∇u(·, τ) −∇u(·, t) · ∇a] dτ

g(0)u + g′ ∗ u

dx ≤

C

g <= ∇u + g(t)

∫

Ωa(x)|∇u|2 dx

+ δ

∫


∫

ωε

|u|2 dx.

+Cδ

∫ t

0

∫

ωε

g(t − τ)|u(·, t) − u(·, τ)|2 dxdt,

and

∫ t

0g dτ

∫

Ωa(x)∇u · ∇a(x)

g(0)u + g′ ∗ u

≤

δ

∫


∫

ωε

|u|2 dx + Cδ

∫ t

0

∫

ωε

g(t − τ)|u(·, t) − u(·, τ)|2 dxdt,


I3(t) ≤ −1

2

d

dt

∫

Ωa2|g ∗ ∇u|2 dx + Cg <= ∇u + δ

∫

Ωa(x)|∇u|2 dx+

+Cδ

∫

ωε

|u|2 dx + Cδ

∫ t

0

∫

ωε

g(t − τ)|u(·, t) − u(·, τ)|2 dxdt

Since

−I1 = I2 + I3

our conclusion follows by substitution of the relations for Ii, i = 1, 2, 3 into the above identity.

The proof is now complete. <=

13

Lemma 4.2 With the same hypotheses as Lemma 4.1 we have that the solution of equation

(1.1)–(1.3) satisfies,

d

dt

∫

Ωauut dx ≤

∫

Ωa(x)|ut|2 dx − c0

∫

Ωa(x)|∇u|2 dx + c

∫

ωε

|u|2 dx

+Cδ

∫ t

0

∫

ωε

g(t − τ)|u(·, t) − u(·, τ)|2 dxdt.

Proof.-Let us multiply equation (1.1) by a(x)u(x, t) to get

∫

Ωauttu dx

︸︷︷︸

:=I1

−∫

Ωa(x)u∆u dx

︸︷︷︸

:=I2

+

∫

Ωdiv ag ∗ ∇u au dx

︸︷︷︸

:=I3

= 0.

Note that

I1(t) =d

dt

∫

Ωautu dx −

∫

Ωa(x)|ut|2 dx.

On the other hand

I2(t) =

∫

Ω∇u∇au dx

=

∫

Ωu∇u∇a dx +

∫

Ωa(x)|∇u|2 dx

= −1

2

∫

Ω∆a(x)|u|2 dx +

∫

Ωa(x)|∇u|2 dx.

Finally,

I3(t) = −∫

Ωa(x)g ∗ ∇u · ∇ au dx

= −∫

Ωa(x)[g ∗ ∇u · ∇a]u dx −

∫

Ωa2g ∗ ∇u · ∇u dx

= −∫

Ωa(x)[g ∗ ∇u · ∇a]u dx +

∫ t

0g dτ

∫

Ωa2|∇u|2 dx

−∫

Ωa2∫ t

0g(t − τ) ∇u(·, τ) − u(·, t) · ∇u dx.

Summing up I1, I2, and I3 we get

d

dt

∫

Ωauut dx =

∫

Ωa(x)|ut|2 dx −

∫

Ωa(1 − a(x)

∫ t

0g dτ)|∇u|2 dx +

∫

Ω(∆a)|u|2 dx

+

∫

Ωa(x)

∫ t

0g(t − τ) ∇u(·, τ) −∇u(·, t)∇a(x)u dτ dx

+

∫

Ωa2∫ t

0g(t − τ) ∇u(·, τ) −∇u(·, t) dτ∇u dx.

14

Using |∆a(x)| ≤ C our conclusion follows <=

The following Lemma is proved by Lions [13], for convenience we rewrite it here.

Lemma 4.3 Let us denote by qk a C1-function, then any strong solution (u ∈ C i(0, T ;H2−i(Ω))

for i = 0, 1, 2) of the wave equation

utt − ∆u = f, (4.1)

u(x, t) = 0, on Γ×]0,∞[,

satisfies the following identity

d

dt

∫

Ωutqk

∂u

∂xkdx =

∫

Ωfqk

∂u

∂xkdx +

1

2

∫

Γqkνk|

∂u

∂ν|2 dΓ +

∫

Γ

∂u

∂νqk

∂u

∂xkdΓ

+1

2

∫

Ω

∂qk

∂xk

|ut|2 + |∇u|2

dx +

∫

Ω∇u · ∇qk

∂u

∂xkdx.

Proof.-Let us multiply equation (4.1) by qk∂u∂xk

to get

∫

Ωutt − ∆u qk

∂u

∂xkdx =

∫

Ωfqk

∂u

∂xkdx. (4.2)

It is easy to see that

∫

Ωuttqk

∂u

∂xkdx =

d

dt

∫

Ωutqk

∂u

∂xkdx −

∫

Ωutqk

∂ut

∂xkdx

=d

dt

∫

Ωutqk

∂u

∂xkdx − 1

2

∫

Ωqk

∂|ut|2∂xk

dx

=d

dt

∫

Ωutqk

∂u

∂xkdx +

1

2

∫

Ω

∂qk

∂xk|ut|2 dx.

On the other hand

−∫

Ω∆uqk

∂u

∂xkdx = −

∫

Γ

∂u

∂νqk

∂u

∂xkdΓ +

∫

Ω∇u · ∇q

∂u

∂xkdx +

∫

Ω∇u · q ∂∇u

∂xkdx

= −∫

Γ

∂u

∂νqk

∂u

∂xkdΓ +

∫

Ω∇u · ∇q

∂u

∂xkdx +

1

2

∫

Ωqk

∂|∇u|2∂xk

dx

= −∫

Γ

∂u

∂νqk

∂u

∂xkdΓ +

∫

Ω∇u · ∇q

∂u

∂xkdx +

1

2

∫

Γqkνk|∇u|2 dΓ

−1

2

∫

Ω

∂qk

∂xk|∇u|2 dx.

Since u(x, t) = 0 on Γ then we have that

∂u

∂xk= νk

∂u

∂ν,

from where our conclusion follows <=

15

Lemma 4.4 Let us take qk = a2(x)hk, where hk ∈ C2(Ω) is such that hk = νk on Γ. Then we

have that

− d

dt

∫

Ωa2(x)uthk

∂u

∂xkdx ≤ −1

2

∫

Γ0

|∂u

∂ν|2 dΓ + C

∫

Ωa(x)

|ut|2 + |∇u|2

dx

+C

∫

Ωa(x)|h(x) div ag ∗ ∇u |2 dx,

for any solution of equation (1.1)–(1.3)

Proof.-From Lemma 4.3 applied to qk = a2(x)hk we have

− d

dt

∫

Ωa2hkut

∂u

∂xkdx = −

∫

Ωfa2(x)hk

∂u

∂xkdx =

∫

Γa2(x)hkνk|

∂u

∂ν|2 dx

+1

2

∫

Ω

∂a2(x)hk

∂xk

|ut|2 − |∇u|2

dx +

∫

Ω∇u · ∇a2hk

∂u

∂xkdx.

Since a = 1 on Γ0 and hk = νk on Γ, using the Cauchy-Schwarz inequality, our conclusion

follows. <=

Let us denote by LN (t) the functional

LN (t) = NE(t) + I(t) +g(0)

2

∫

Ωa(x)uut dx − δ0

∫

Ωa2(x)uthk

∂u

∂xkdx.

In these conditions we get

Lemma 4.5 Under the above notations we have

d

dtLN (t) ≤ −κ0

∫

Ωa(x)

|ut|2 + |∇u|2

dx − N

2

g <= ∇u + g(t)

∫

Ωa(x)|∇u|2 dx

(4.3)

−δ0

2

∫

Γ|∂u

∂ν|2 dΓ + Cδ0

∫

Ωa(x)|h(x) div ag ∗ ∇u |2 dx + C

∫

ωε

|u|2 dx

+C

∫ t

0

∫

ωε

g(t − τ)|u(x, t) − u(x, τ)|2 dxdτ.

Proof.-From Lemma 4.1 and Lemma 4.2 we get that

d

dt

I(t) +g(0)

2

∫

Ωa(x)uut dx

≤ −g(0)

2

∫

Ωa(x)|ut|2 dx − (C − δ)

∫

Ωa(x)|∇u|2 dx

+C

∫

ωε

|u|2 dx + Cδ

g <= ∇u + g(t)

∫

Ωa(x)|∇u|2 dx

+C

∫ t

0

∫

ωε

g(t − τ)|u(x, t) − u(x, τ)|2 dxdτ.

16

where G is given by (3.3). Using similar arguments, we conclude that

‖v‖2L2(0,T ;L2) ≤

∫ T

0

∫

Ωa(x)|G|2 dxdt

≤∫ T

0

∫

Ωa(x)

|ut|2 + |∇u|2

dxdt + C

∫ T

0g dtE(0).

Therefore it follows that

∫ T

0

∫

Ωa(x)| div ag ∗ ∇u |2 dxdt ≤ C

∫ T

0

∫

Ωa(x)

|ut|2 + |∇u|2

dxdt + C

∫ T

0g(t) dtE(0),

for a positive constant C. The proof is now complete <=

Lemma 4.7 Let us suppose that ϕ is a weak solution of the wave equation

ϕtt − ∆ϕ = 0

ϕ(x, 0) = ϕ0, ϕt = ϕ1

ϕ(x, t) = 0, on Σ = Γ×]0,∞[.

Then, for x0 ∈ IRn and T > 2R(x0), there exists a positive constant C > 0 for which we have

E(0) ≤ C

∫ T

0

∫

ω|ϕt|2 + |∇ϕ|2 dxdt

for any (ϕ0, ϕ1) ∈ H10 (Ω) × L2(Ω) where

R(x0) = maxx∈Ω

|n∑

k=1

(xk − x0k)

2|1/2

Proof.-See [13] Lemma 2.3, Chapter VIII, pag 411. <=

Our next step is to estimate the term∫

ω |u|2 dx. To do this we will use the following Lemma

Lemma 4.8 Let us suppose that u is a weak solution of (1.1)–(1.3), then for any ε > 0 there

exist a positive constant Cε for which we have

∫ T

0

∫

Ω|u|2 dxdt ≤ Cε

∫ T

0g(t)

∫

Ωa(x)|∇u|2 dxdt +

∫ T

0g <= ∇u dt

+ε

∫ T

0

∫

Ωa(x)

|∇u|2 + |ut|2 + | div ag ∗ ∇u|2

dxdt,

∫ T

0

∫

Ω|g ∗ ∇u|2 dxdt ≤ Cε

∫ T

0g(t)

∫


∫ T

0g <= ∇u dt

+ε

∫ T

0

∫

Ωa(x)

|∇u|2 + |ut|2 + | div ag ∗ ∇u|2

dxdt,

18

and

∫ T

0

∫ t

0

∫

Ωg(σ − t)|u(x, σ) − u(x, t)|2 dxdσdt ≤

Cε

∫ T

0g(t)

∫


∫ T

0g <= ∇u dt

+ε

∫ T

0

∫

Ωa(x)

|∇u|2 + |ut|2 + | div ag ∗ ∇u|2

dxdt,

provided T is large enough.

Proof.-We argue by contradiction. Suppose that there exists ε0 > 0 and a sequence of

functions such that

∫ T

0

∫

Ω|uν |2 dxdt ≥ ν

∫ T

0g(t)

∫

Ωa(x)|∇uν |2 dxdt +

∫ T

0g <= ∇uν dt

+ε0

∫ T

0

∫

Ωa(x)

|∇uν |2 + |uνt |2 + | div ag ∗ ∇u|2

dxdt, (4.6)

for ν → ∞. By the linearity of the problem we may suppose that

∫ T

0

∫

Ω|uν |2 dxdt = 1, ∀ν ∈ IN. (4.7)

So, we get that

g(t)a(x)|∇uν |2+

∫ t

0a(·)g(t−τ)|uν(·, τ)−uν(·, t)|2 dτ → 0 strongly in L1(]0,∞[×Ω).

(4.8)

Let us decompose uν into:

uν = wν + vν ,

where

wνtt − ∆wν = −div ag ∗ ∇uν (bounded in L2(0, T ;L2(Ω))).

wν(x, 0) = 0, wνt (x, 0) = 0, in Ω

wν(x, t) = 0, on Γ×]0,∞[,

and

vνtt − ∆vν = 0,

vν(x, 0) = uν(x, 0), vνt (x, 0) = uν

t (x, 0), in Ω

19

vν(x, t) = 0, on Γ×]0,∞[

From (4.6) and (4.7) it follows that uν is bounded in

W 1,∞(0, T ;L2(ω)) ∩ L∞(0, T ;H1(ω)).

Note that wν is also bounded in

W 1,∞(0, T ;L2(Ω)) ∩ L∞(0, T ;H10 (Ω)).

Thereby, we conclude that vν = uν − wν satisfies

vνt is bounded in L2(0, T ;L2(ω)),

vν is bounded in L2(0, T ;H1(ω)).

Using Lemma 4.7 we have

(uν(·, 0), uνt (·, 0)) = (vν(·, 0), vν

t (·, 0)), is bounded in H10 (Ω) × L2(Ω).

which implies that

vν is bounded in W 1,∞(0, T ;L2(Ω)) ∩ L∞(0, T ;H10 (Ω)).

Hence

uν = wν + vν is bounded in W 1,∞(0, T ;L2(Ω)) ∩ L∞(0, T ;H10 (Ω)).

Therefore there exists a subsequence (which we still denote in the same way) and a function

u ∈ W 1,∞(0, T ;L2(Ω)) such that

uν → u weak * in W 1,∞(0, T ;L2(Ω))

and satisfying

utt − ∆u = 0,

u(x, 0) = u0(x), ut(x, 0) = u1(x), in Ω

u(x, t) = 0, on Γ×]0, T [

From (4.8) we conclude that

u = 0 on ωε×]0, T [

Using the Holmgren’s Theorem for T > 2diam (Ω \ ωε) we get that u = 0 on Ω×]0, T [. But

this is contradictory with (4.7) since due to the compactness of the embedding H 1(Ω×]0, T [) ⊂

20

L2(Ω×]0, T [), the sequence uν converges stromgly in L2(Ω×]0, T [). This contradiction proves

the first inequality. To prove the other we use similar arguments. Thereby, our conclusion

follows. <=

Using the inequalities (4.3), (4.5), Lemma 4.8 and taking ε > 0 small enough we arrive at

LN (T ) −LN (0) ≤ −κ0

∫ T

0M(t) dt + CεE(0) + Cε

∫ T

0E(t) dt (4.9)

for N > 2C; where by M we are denoting

M(t) =

∫

Ωa(x)

|ut|2 + |∇u|2

dx + g <= ∇u +

∫

Γ0

|∂u

∂ν|2 dΓ.

Now we are in conditions to prove the main result of this paper.

Proof of Theorem 1.1 We will suppose that the initial data belongs to H 2(Ω) ∩H10 (Ω)×

H10 (Ω). Our conclusion will follow using standard density arguments. Using Lemma 4.3 for

q = x − x0 we conclude that

− d

dt

∫

Ωutqk

∂u

∂xkdx = −

∫

Ωfqk

∂u

∂xkdx +

1

2

∫

Ω

∂qk

∂xk

|ut|2 − |∇u|2

dx

+

∫

Ω∇u · ∇qk

∂u

∂xkdx − 1

2

∫

Γqkνk|

∂u

∂ν|2,

from where it follows

− d

dt

∫

Ωutqk

∂u

∂xkdx = −

∫

Ωfqk

∂u

∂xkdx +

n

2

∫

Ω

|ut|2 − |∇u|2

dx

+

∫

Ω|∇u|2 dx − 1

2

∫

Γqkνk|

∂u

∂ν|2,

which implies that

− d

dt

∫

Ωutqk

∂u

∂xkdx = −

∫

Ωfqk

∂u

∂xkdx +

n − 1

2

∫

Ω

|ut|2 − |∇u|2

dx (4.10)

+1

2

∫

Ω|ut|2 + |∇u|2 dx − 1

2

∫

Γqkνk|

∂u

∂ν|2.

Multiplying by u equation (1.1) we get

d

dt

∫

Ωuut dx =

∫

Ω

|ut|2 − |∇u|2

dx +

∫

Ωag ∗ ∇u · ∇u dx.

Inserting this identity into (4.10) we have

21

− d

dt

∫

Ωutqk

∂u

∂xkdx = −

∫

Ωfqk

∂u

∂xkdx +

n − 1

2

d

dt

∫

Ωuut dx

−n − 1

2

∫

Ωag ∗ ∇u · ∇u dx +

1

2

∫

Ω|ut|2 + |∇u|2 dx

−1

2

∫

Γqkνk|

∂u

∂ν|2,

from where we have

d

dt

−∫

Ωutqk

∂u

∂xkdx − n − 1

2

∫

Ωuut dx

︸︷︷︸

:=X(t)

= −∫

Ωfqk

∂u

∂xkdx − n − 1

2

∫

Ωag ∗ ∇u · ∇u dx

+1

2

∫

Ω|ut|2 + |∇u|2 dx − 1

2

∫

Γqkνk|

∂u

∂ν|2.

Integrating over [0, T ] we get

X(T ) − X(0) = −∫ T

0

∫

Ωfqk

∂u

∂xkdx dt − n − 1

2

∫ T

0

∫

Ωag ∗ ∇u · ∇u dx dt (4.11)

+1

2

∫ T

0

∫

Ω|ut|2 + |∇u|2 dx dt − 1

2

∫ T

0

∫

Γqkνk|

∂u

∂ν|2 dt.

Since

X(T ) ≤ CE(T ), X(0) ≤ CE(0),

and using∫ T

0E(t) dt ≤ C

∫ T

0

∫

Ω|ut|2 + |∇u|2 dx dt +

∫ T

0g <= ∇u dt

,

together with inequality (4.11) we conclude that

∫ T

0E(t) dt ≤ C

∫ T

0M(t) dt + C E(T ) + E(0) .

From the energy identity we get

E(0) ≤ E(T ) +

∫ T

0M(t) dt. (4.12)

Therefore, there exist a positive constant C1 such that

∫ T

0E(t) dt ≤ C1

∫ T

0M(t) dt + C1E(T ). (4.13)

22

Since E(t) is a decreasing function we have that

E(T ) ≤ 1

T

∫ T

0E(t) dt.

Inserting the above inequality into (4.13) we get

(1 − C

T)

∫ T

0E(t) dt ≤ C1

∫ T

0M(t) dt. (4.14)

On the other hand, it is not difficult to see that

c0E(t) ≤ L(t) ≤ c1E(t). (4.15)

Therefore using (4.9), (4.14), and (4.15) we conclude that

L(T ) −L(0) ≤ −κ0

∫ T

0M(t) dt + CεE(0) + Cε

∫ T

0E(t) dt (using (4.12))

≤ −κ0

∫ T

0M(t) dt + Cε

E(T ) +

∫ T

0M(t) dt

+ Cε

∫ T

0E(t) dt

≤ −κ1

∫ T

0L(t) dt.

provided ε small enough. Using inequality (4.15) we can establish that

∫ T

0L(t) dt ≥ c

∫ T

0E(t) dt ≥ cTE(T ) ≥ c2TL(T ),

which implies that

L(T ) −L(0) ≤ −κ1cTL(T ).

This is equivalent to

L(T ) ≤ 1

1 + CTL(0).

Repeating the above process from T to 2T we get that

L(2T ) ≤ 1

1 + CTL(T ) ≤ 1

(1 + CT )2L(0).

In general we have that

L(nT ) ≤ 1

(1 + CT )nL(0).

Since any number t can be written as t = nT + r where r < T and E(t) is a decreasing function,

from (4.15) we arrive at

L(t) ≤ cL(t − r) ≤ C

(1 + CT )(t−r)/TL(0) ≤ c0e

−γtL(0),

where γ = ln(1+CT )T , from where the exponential decay follows.

23

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Jaime E. Munoz Rivera, National Laboratory for Scientific Computation, Department of Re-

search and Development, Rua Getulio Vargas 333, Quitandinha CEP 25651-070. Petropollis,

RJ, Brasil, and IM, Federal University of Rio de Janeiro.

Alfonso Peres Salvatierra, Universidad Nacional Mayor de San Marcos. Av. Venezuela s/n,

Lima - Peru.

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