General decay for a porous thermoelastic system with thermo-viscoelastic damping

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General decay for a porous thermoelastic system with

thermo-viscoelastic damping

Kilani Brahim, Djebabla Abdelhak

To cite this version:

General decay for a porous thermoelastic

system with thermo-viscoelastic damping

Kilani Brahim

and Djebabla Abdelhak

(1)

_{Mathematics, Dynamics and Modelization Laboratory,}

University Badji Mokhtar

Department of Mathematics

PO BOX 12 Annaba 23000, Algeria

E-mail: kilbra2000@yahoo.fr

and

(2)

_{Laboratory of Applied Mathematics}

University Badji Mokhtar

Department of Mathematics

PO BOX 12 Annaba 23000, Algeria

E-mail: abdelhak.djebabla@univ-annaba.dz

Abstract

In this article, we consider the one-dimensional viscoelastic porous thermoelastic system. We introduce two new numbers χ0 and χ1 that

characterizes the decay. We establish a general decay result for the case χ0 = χ1 = 0. To better observe the effect of the weak

dissipa-tion (thermal effect or thermo-viscoelastic damping) on changing the nature of the stability numbers, we refer the reader to [1, 2, 3] .

Keywords: General decay; Porous; Relaxation function; Vis-coelastic damping; Thermal effect.

1

Introduction

The one-dimensional porous-elastic model has the form 

ρ0utt= µuxx+ βϕx

ρ0κϕtt = αϕxx− βux− τ ϕt− ξϕ

where u is the longitudinal displacement, ϕ is the volume fraction, ρ0 > 0 is

the mass density, κ > 0 is the equilibrated inertia, and µ, α, β, τ , and ξ are the constitutive constants which are positive and satisfy µξ > β2_{. The first}

contribution in this direction has been investigated in 2003 by Quintanilla [6]. The authors analyzed this model in a bounded domain with initial conditions and mixed boundary conditions and showed that the damping in the porous equation (−τ ϕt) is not strong enough to obtain an exponential decay. Only

the slow decay has been proved. The case τ = 0 and with viscoelastic damping term of the form γutxx acting on the right- hand side of the first

equation has been investigated by Rivera et al. [1], they proved that the decay rate of the solution is polynomial and cannot be exponential. Casas and Quintanilla [7] considered a system of the form

               ρ0utt− µuxx− bϕx+ βθx = 0 in (0, ∞) × (0, π) ρ0κϕtt− αϕxx+ bux+ ζϕt− mθ = 0 in (0, ∞) × (0, π) cθt− κθxx+ βutx+ mϕt = 0 in (0, ∞) × (0, π), u (x, 0) = u0(x), ϕ(x, 0) = ϕ0(x), θ(x, 0) = θ0(x) in (0, π) ut(x, 0) = u1(x), ϕt(x, 0) = ϕ1(x), in (0, π) u (x, t) = ϕx(x, t) = θx(x, t) = 0, x = 0, π, t ≥ 0

where θ denotes the temperature difference, they showed that the presence of the macrotemperature and the porous dissipations acting together stabilize the system exponentially. Soufyane [15] introduced a viscoelastic damping in the porous equation together with a macrotemperature effect and showed that the decay is exponential (resp. polynomial) when the relaxation function is of exponential (resp.polynomial) decay. A similar result was also obtained by Soufyane et al. [16,17], for the same system, with the frictional damping −τ ϕt replaced by two boundary viscoelastic dissipations of the form

u(L, t) = −R₀tg1(t − s) [µux(L, s) + bϕ(L, s)] ds,

v(L, t) = −αR₀tg2(t − s) ϕ (L, s) ds.

where g1 and g2 are positive nonincreasing functions. Recently, Messaoudi

system of type III with the presence of a viscoelastic damping                ρ1ϕtt− k (ϕx+ ψ)_x+ θx= 0, ρ2ψtt− αψxx+ k (ϕx+ ψ) − θ + Rt 0 g (t − s) ψxx(x, s) ds = 0, ρ3θtt− κθxx− δθtxx+ βϕttx+ βψtt= 0, ϕ (x, 0) = ϕ0(x) , ϕt(x, 0) = ϕ1(x) , ψ (x, 0) = ψ0(x) , ψt(x, 0) = ψ1(x) , θ (x, 0) = θ0(x) , θt(x, 0) = θ1(x) , ϕ (0, t) = ϕ (1, t) = ψ (0, t) = ψ (1, t) = θ (0, t) = θ (1, t) = 0

where ϕ is the longitudinal displacement, ψ is the volume fraction, θ is the difference in temperature and the relaxation function g : R+ → R is a

non-increasing function. They established a general decay result for the case of equal speeds _ρk

1 =

α ρ2



. Notice that the term δθtxx in the third equation

represents a strong damping. The term βθtprovides a weaker damping

(fric-tional damping) and so we consider with initial data and boundary conditions the following system

   ρ1ϕtt− k (ϕx+ ψ)_x+ γθx = 0, ρ2ψtt− bψxx + k (ϕx+ ψ) − γθ + Rt 0 g (t − s) ψxx(x, s) ds = 0, ρ3θtt− lθxx+ βθt+ γϕttx+ γψtt = 0,

in Ω × (0, +∞), where Ω = [0, L] and the coefficients ρ1, ρ2, ρ3, k, b, γ, β

and l are positive constants.

Our aim in this work is to prove the general decay of solutions in the energy norm. If we make the change of functions

 ϕ_t = χ ∈ H₀1(Ω) , ψt = ξ ∈ H01(Ω) , we find    ρ1χtt− k (χx+ ξ)x+ γθtx = 0, ρ2ξtt− bψxx+ k (χx+ ψ) − γθt+ Rt 0 g (t − s) ξxx(x, s) ds = 0, ρ3θtt− lθxx+ βθt+ γχtx+ γξt= 0, (1)

to be considered with the following initial data and boundary conditions

χ (0, .) = χ0, χt(0, .) = χ1, ψ (0, .) = ψ0,

ψt(0, .) = ψ1, θ (0, .) = θ0, θt(0, .) = θ1,

The plan of the paper is as follows: In the next section we prepare some material needed to prove our result. We present the different functionals by which we modify the classical energy to get an equivalent one. Section 3 is devoted mainly to the statement and proof the general decay result.

2

Preliminaries

In this section, we present our hypotheses, and state without proof a global existence result. First, we denote by ∗ the usual convolution term

(f ∗ h) (t) = R₀tf (t − s) h (s) ds (3) and the binary operators ♦ and o respectively, by

(f ♦h) (t) = R0tf (t − s) (h (t) − h (s)) ds (4)

and

(f oh) (t) = R₀tf (t − s) (h (t) − h (s))2ds. (5) For the relaxation function g we assume:

(H1) g : R+→ R+ is differentiable function such that

g (0) > 0, λ := b −R₀∞g (s) ds := b − ¯g > 0 (6) (H2) There exists a nonincreasing differentiable function ζ : R+→ R+

g0(t) ≤ −ζ (t) g (t) , ∀ t ≥ 0. (7) Remark. There are many functions satisfying (H1) and (H2) (see [21]).

Proposition 1 Let ((χ0, χ1) , (ξ0, ξ1) , (θ0, θ1)) ∈ (H01(0, L) × L2(0, L)) 3

be given and assume that g satisfied (H1) and (H2). Then, problem (1.2) has a unique global solution:

then the solution satisfies

(χ, ξ, θ) ∈ (C (R+; H2(0, L)) ∩ C1(R+; H01(0, L)) ∩ C2(R+; L2(0, L))) 3

.

Remark 1 Proposition 1 can be established using standard methods such as the Galerkin method (see [4] for example).

The first-order energy associated with problem (1-2) is given as

E(t) = 1 2 R Ωρ 2 1χ2t + ρ22ξt2+ ρ3θt2+ k (χx+ ξ)2+ lθx2 +  b −R₀tg (s) ds  ξ_x2+ goξx o dx, t ≥ 0. (8)

Lemma 1 Under assumptions (H1) and (H2), we have

R Ωξt Rt 0g(t − s)ξxx(s) dsdx = 1 2 d dt h (goξx) −  Rt 0 g (s) ds  R Ωξ 2 xdx i −1 2(g 0_oξ x) + 1₂g(t) R Ωξ 2 xdx. t ≥ 0

Proof. Integrating by parts and using the boundary conditions, we get

R Ωξt Rt 0 g(t − s)ξxx(s) dsdx = −  Rt 0 g(s)ds  R Ωξ 2 x(t) dsdx −R Ω Rt 0 g(t − s)ξtx(t) [ξx(s) − ξx(t)] dsdx = 1_2dtd hR_Ω(goξx) dx −  Rt 0 g(s)ds  R Ωξ 2 x(t) dx i −1 2 R Ω(g 0_oξ x) dx +1₂g(t) R Ωξ 2 x(t) dx.

Lemma 2 For any function g ∈ C ([0 ∞) , R) and for any h ∈ L2_{(Ω) we}

have that

[(g♦h) (t)]2 ≤Rt

0 g(τ )dτ (goh) (t), t ≥ 0.

Proof. By Cauchy-Schwarz inequality, we have

[(g♦h) (t)]2 ≤hRt 0 g (t − s) (h (t) − h (s)) ds i2 = h Rt 0g 1 2 (t − s) g 1 2 (t − s) (h (t) − h (s)) ds i2 ≤   Rt 0g (t − s) ds 1₂  Rt 0 g (t − s) (h (t) − h (s)) 2 ds 1 2 2 ,

Lemma 3 There exists a positive constant C0 such that R Ω(bξx− (g ∗ ξx)) 2 dx ≤ C0 R Ωξ 2 xdx + 4g R Ω(goξx) dx, t ≥ 0, with C0 = (2b2+ 4g2) .

Proof. Using the fact that (a2_{+ b}2_{) ≤ 2a}2_{+ 2b}2 _{and lemma 3, we get}

R Ω(bξx− (g ∗ ξx)) 2 dx ≤ 2b2R_Ωξ_x2dx + 2R_Ω(g ∗ ξx) 2 dx ≤ 2b2R Ωξ 2 xdx + 2 R Ω Rt 0g(t − s) (ξx(t) − ξx(s) − ξx(t)) 2 dsdx ≤  2b2+ 4R₀tg(t − s)ds 2 R Ωξ 2 xdx + 4 R Ω(g♦ξx) 2 dx ≤ (2b2_{+ 4g}2₎R Ωξ 2 xdx + 4g R Ω(goξx) dx.

Our result reads as follows.

Theorem 1 Let ((χ0, χ1) , (ξ0, ξ1) , (u0, u1)) ∈ (H01(0, 1) × L2(0, 1)) 3

be given, assume that g satisfies (H1) and (H2) and the coefficients of the system sat-isfy the condition

κ0 = γ + bρ_k1 − ρ2 = 0 and κ1 = ρ_ρ3₂b + bγ_k − l = 0. (9)

Then, there exist two positive constants c0 and c1such that

E (t) ≤ c0e−c1 Rt1

t0 ζ(s)ds_, ∀t ≥ t_0.

3

Proof of the main result

The proof of our main result will be established through several lemmas.

Lemma 4 Let (ϕ, ψ, θ) be the solution of (1-2). Then the energy functional E, defined by (8) satisfies d dtE(t) = −β R Ωθ 2 tdx − 1 2g(t) R Ωξ 2 xdx + 1 2 R Ω(g 0_oξ x) dx ≤ 0, t ≥ 0. (10)

Proof : Multiplying the first equation of (1-2) by χt, the second by ξt

and the third by θt then we integrate over (0, L) and using lemma 2, we get

(10) for any regular solution.

Next, we introduce the multiplier w given by the solution of the Dirichlet problem

Lemma 5 Let (ϕ, ψ, θ) be the solution of (1-2). Then the functional I1(t) = ρ1 R Ωχtwdx + ρ2 R Ωξtξdx, t ≥ 0. (11)

satisfies, for any positive constant ε1, the estimate d dtI1(t) ≤ − λ 2 R Ωξ 2 xdx + ε1 R Ωχ 2 tdx +  ρ2+ ρ2 1Cp 4ε1  R Ωξ 2 tdx 3γ2_C p λb R Ωθ 2 tdx + 3g 2λ R Ω(goξx) (t)dx, t ≥ 0. . (12)

where Cp is the Poincar´e constant and λ is defined in (H1).

Proof : Differentiating I1(t) in (11) and using the first and second

equa-tions of (1-2), we obtain d dtI1(t) = ρ1 R Ωχtwtdx + k R Ωw 2 xdx + γ R Ωθtwxdx + ρ2 R Ωξ 2 tdx −bR Ωξ 2 xdx − k R Ωξ 2_{dx + γ}R Ωθtξdx + R Ω(g ∗ ξx) ξx(t) dx. Hence, thanks to R Ωw 2 tdx ≤ Cp R Ωw 2 txdx ≤ Cp R Ωψ 2 tdx,

and the Young inequality, we obtain, for all δ > 0,

d dtI1(t) ≤ ε1 R Ωχ 2 tdx + ρ2 1Cp 4ε1 R Ωξ 2 tdx + δ R Ωξ 2 xdx + γ2Cp 4δ R Ωθ 2 tdx +ρ2 R Ωξ 2 tdx − b R Ωξ 2 xdx + δ R Ωξ 2 xdx + γ2_C p 4δ R Ωθ 2 tdx +R₀tg (τ ) dτR_Ωξ2 xdx + R Ω(g♦ξx) ξx(t)dx. (13)

Next, exploiting the Young’s inequality for the last term in (13), and lemma 3, we obtain, for all δ > 0,

R Ω(g♦ξx) ξx(t)dx ≤ δ R Ωξ 2 xdx +4δ1 Rt 0 g(τ )dτ R Ω(goξx) (t)dx.

The choice of δ = λ₆ gives the result.

Lemma 6 Let (ϕ, ψ, θ) be the solution of (1-2). Then the functional

I2(t) := −ρ1

RL

0 χtχdx − ρ2

RL

0 ξtξdx, t ≥ 0, (14)

satisfies, the estimate

Proof. A Straightforward computation, using the .rst and the second equations in (1-2), yields d dtI2(t) = −ρ1 R Ωχ 2 tdx − ρ2 R Ωξ 2 tdx + k R Ω(χx+ ξ) 2 dx +bR_Ωξ_x2dx − γR_Ωθt(χx+ ξ) dx − Rt 0 g(τ )dτ R Ωξ 2 xdx −R Ω(g♦ξx) ξx(t) dx.

Thanks to Young.s inequality, we get

−γR Ωθt(χx+ ξ) dx ≤ γ2 2k R Ωθ 2 tdx + k2 R Ω(χx+ ξ) 2 dx −R Ω(g♦ξx) ξx(t) dx ≤ b 2 R Ωξ 2 xdx +2b1 R Ω(g♦ξx) 2 dx.

Finally, it remains to conclude by lemma 3.

Lemma 7 Let (ϕ, ψ, θ) be the solution of (1-2). Then the functional

I3(t) := −ρ2

R

Ωξt(g♦ξ) (t) dx, t ≥ 0. (16)

satisfies, for any positive constant ε2 and δ1 > 0, the estimate

I₃0 (t) ≤ −ρ2 Rt 0 g (τ ) dτ − δ1  RL 0 ξ 2 tdx + ε2(1 + g) R Ωξ 2 xdx +γ₂ R_Ωθ2 xdx + kε2 R Ω(χx+ ξ) 2 dx + C (ε2) R Ω(goξx) dx −ρ22 4δ1g(0)Cp R Ω(g 0_oξ x) (t)dx, t ≥ 0. (17) where C(ε2) = h ε2+_2ε1₂ + b 2 4ε2 + kCp 4ε2 + γCp 2  gi.

Proof. Differentiating the functional I3(t) and using the second equation

We now estimate the terms in this last identity, using Young’s inequality and lemma 3, we obtain, for all δ1 > 0,

−ρ2 R Ωξt(g 0 ♦ξ) (t)dx ≤ δ1 R Ωξ 2 tdx + ρ2 2 4δ1 Rt 0(−g 0_{(τ ) dτ )}R Ω(−g 0_{oξ) (t)dx} ≤ δ1 R Ωξ 2 tdx + ρ2 2 4δ1g(0)Cp R Ω(−g 0_oξ x) (t)dx

Similary, we have for ε2 > 0,

bR_Ωξx(g♦ξx) (t)dx ≤ ε2 R Ωξ 2 xdx + b2_g 4ε2 R Ω(goξx) (t)dx, kR Ω(χx+ ξ) (g♦ξ) (t)dx ≤ kε2 R Ω(χx+ ξ) 2 dx +kCpg 4ε2 R Ω(goξx) (t)dx, −γR Ωθt(g♦ξ) (t)dx ≤ γ 2 R Ωθ 2 xdx + γCpg 2 R Ω(goξx) (t)dx, finally −R Ω  Rt 0 g (t − s) (ξx(s) ds)  (g♦ξx) (t)dx ≤ _2ε1₂ R Ω(g♦ξx) (t)dx +ε2 2 R Ω  Rt 0 g (t − s) (ξx(t) − ξx(s) − ξx(t) ds) 2 dx ≤ ε2  Rt 0 g(τ )dτ  R Ωξ 2 x(t) dx +  ε2 +_2ε1₂  R Ω(g♦ξx) (t)dx ≤ ε2  Rt 0 g(τ )dτ  R Ωξ 2 x(t) dx +  ε2+_2ε1 2  gR Ω(goξx) (t)dx.

Combining all the above estimates, we prove the assertion of the lemma.

Lemma 8 Let (ϕ, ψ, θ) be the solution of (1-2). Then the functional

I4(t) := ρ3 R Ωθtθdx + γ R Ω(χx+ ξ) θdx + β 2 R Ωθ 2_{dx, (18)}

satisfies, for any positive constant ε3, the estimate

I₄0 (t) ≤ −lR₀Lθ2 xdx +  γ2 4kε3 + ρ3  RL 0 θ 2 tdx + kε3 RL 0 (χx+ ξ) 2 dx. (19)

Proof. Differentiating I4(t) and using the third equation in (1-2), we

By using Young’s inequality, we obtain for any ε3 > 0, I₄0 (t) ≤ −lR_Ωθ2_xdx +  γ2 4kε3 + ρ3  R Ωθ 2 tdx + kε3 R Ω(χx+ ξ) 2 dx, which is exactly (19).

Lemma 9 Assume that the coefficients ρ1, ρ2, ρ3, b, k and γ satisfy the

rela-tion (9). Then the funcrela-tional I5(t) := ρ2 R Ωξt(χx+ ξ) dx + (ρ2− γ) R Ωχtξxdx −γ_k +ρ3 ρ2  R Ωθx(g ∗ ξx) (t) dx − ρ1 k R Ωχt(g ∗ ξx) (t) dx +ρ3 R Ωξtθtdx + l R Ωθxξxdx, (20)

satisfies, for any positive constant ε3, the estimate

I₅0 (t) ≤ −k₂R_Ω(χx+ ξ) 2 dx + ε3 R Ωχ 2 tdx + ε3 R Ωθ 2 xdx + ρ2 R Ωξ 2 tdx + [χx(bξx− (g ∗ ξx) (t))]x=L_x=0 + C1(ε3) R Ωθ 2 tdx − g 2ε3C2 R Ω(g 0_oξ x) (t) dx +g 2₍₀₎ 2ε3 C2 R Ωξ 2 xdx, ∀t ≥ 0, (21) where C1(ε3) = γ 2 k + β2 4γ + ρ3γ ρ2 + ρ23k ρ2 2 and C2 =   γ k + ρ3 ρ2 2 + ρ1 k
2  .

Proof. A straifhtfoward computation, using the second, the third equa-tions in (1-2) and integrating by parts, yields

I₅0(t) = −bR_Ωξx(χx+ ξ)xdx − k R Ω(χx+ ξ) 2 dx +γR Ωθt(χx+ ξ) dx + R Ω(g ∗ ξx) (t) (χx+ ξ)xdx + [χx(bξx− (g ∗ ξx) (t))] x=L x=0 + ρ2 R Ωξ 2 tdx +ρ2 R Ωξtχtxdx + (ρ2− γ) R Ωχttξxdx + (ρ2− γ) R Ωχtξtxdx −γ_k +ρ3 ρ2  R Ωθtx(g ∗ ξx) (t) dx −  γ k + ρ3 ρ2  R Ωθx(g ∗ ξx)t(t) dx −ρ1 k R Ωχtt(g ∗ ξx) (t) dx − ρ1 k R Ωχt(g ∗ ξx)t(t) dx −ρ3b ρ2 R Ωξxθtxdx − ρ3k ρ2 R Ωθt(χx+ ξ) dx + ρ3γ ρ3 R Ωθ 2 tdx +ρ3 ρ2 R Ωθtx(g ∗ ξx) (t) dx − l R Ωθxξtxdx − β R Ωξtθtdx −γR Ωξtχtxdx − γ R Ωξ 2 tdx + l R Ωθtxξxdx + l R Ωθxξtxdx,

By using Young.s inequality, relation (9) and the properties of g, the result follows.

As in [24], to deal with the boundary terms appearing in (21), we define the function

q(x) = 2 − 4x_L, x ∈ Ω

Lemma 10 Let (ϕ, Ψ, u) be the solution of (1-2). The functionals J1 and J2

defined by J1(t) := ε_k3  ρ₁R Ωχtq (x) χxdx + ρ3 R Ωθtq (x) θxdx +γR Ωχxq (x) θxdx,  (22) J2(t) := _4ερ2 3 R Ωξtq (x)  bξx− Rt 0 g (t − s) ξx(s) ds  dx, (23) satisfy, for any ε3 > 0, the estimates

J₁0(t) ≤ −ε3[χ2x(L) + χ2x(0)] + 2ε3ρ1 kL R Ωχ 2 tdx +ε3 _L2 + 1
R Ωχ 2 xdx + ε3 k 2l L + β + γ
R Ωθ 2 xdx ε3 R Ωξ 2 xdx + ε3 k 2ρ3 L + β
R Ωθ 2 tdx + ε3γ k R Ωξ 2 tdx (24) and J₂0 (t) ≤ −_4ε1 3   bξx(L) − Rt 0g (t − s) ξx(L, s) ds 2 − 1 4ε3   bξx(0) − Rt 0 g (t − s) ξx(0, s) ds 2 C3(ε3) R Ωξ 2 xdx + kε3 R Ω(χx+ ξ) 2 dx +ρ2 2ε3 1 + 1 L
R Ωξ 2 tdx + γ2_L 4ε3 R Ωθ 2 tdx +_Lε3g 3 + kg 4ε3 3  R Ω(goξx) dx − ρ2g(0) 4ε3 R Ω(g 0_oξ x) dx, (25) where C3(ε3) =  1 2Lε3 + 1 Lε3 + k 8ε3 3  (b2_{+ 2g}2_{) +} ρ2 4ε3g 2_(0).

Proof. Differentiating J1(t) and using the first and third equations in

Thanks to Young’s inequality, we obtain the desired result.

Next, from the second equation in (1-2) and integrating by parts, we get

J₂0 (t) := _2Lε1 3 R Ω  bξx− Rt 0g (t − s) ξx(s) ds 2 dx − 1 4ε3   bξx(L) − Rt 0 g (t − s) ξx(L, s) ds 2 − 1 4ε3   bξx(0) − Rt 0 g (t − s) ξx(0, s) ds 2 + γ 4ε3 R Ωθtq (x)  bξx− Rt 0 g (t − s) ξx(s) ds  dx − k 4ε3 R Ω(ϕx+ ξ) q (x)  bξx− Rt 0 g (t − s) ξx(s) ds  dx +ρ2 4ε3 R Ωξtq (x)  bξx− Rt 0 g (t − s) ξx(s) ds  tdx. (26)

Next, by using Young’s inequality, lemma 3, lemma 4 and the fact that q2(x) ≤ 4, ∀x ∈ [0, L] , we estimate the terms in (26) as follows,

By combining all the above estimates, we estabilish the assertion of the lemma.

Proof of Theorem. For some positive constants; N , N1, N2, to be chosen

appropriately later. We define the Lyapunov functional by

£ (t) = N E + N1I1(t) +1₄I2(t) + N2I3(t)

+N3I4(t) + I5(t) + J1(t) + J2(t) , ∀t > 0.

(27)

Next, taking into account (10), (12), (15), (17), (19), (21), (24), (25) and the following relations RL 0 χ 2 xdx ≤ 2 RL 0 (χx+ ξ) 2 dx + 2Cp RL 0 ξ 2 xdx, h χx  bξx− Rt 0g (t − s) ξx(x, s) ds ix=L x=0 ≤ ε3[χ 2 x(L) + χ2x(0)] +_4ε1 3   bξx(L) − Rt 0g (t − s) ξx(L, s) ds 2 +_4ε1 3   bξx(0) − Rt 0 g (t − s) ξx(0, s) ds 2 , we obtain £ (t) ≤ −hλN1 2 − 3b 8 − N2ε2(1 + g) − g2(0) 2ε3 C2− ε3− C3(ε3) − 2ε3 2 L+ 1
Cp i R Ωξ 2 xdx −hN2  ρ2 Rt 0 g (τ ) dτ − δ1  − 5ρ2 4 − N1  ρ2+ ρ2 1Cp 4ε1  − ε3γ k − ρ2 2ε3 1 + 1 L
iRL 0 ξ 2 tdx −hN β −γ_8k2 − N3  γ2 4kε3 + ρ3  −3N1γ2Cp λb − C1(ε3) − ε3 k 2ρ3 L + β
− γ2_L 4ε3 i R Ωθ 2 tdx −_k 8 − kε2N2− ε3 N3k + k + 2 2 L + 1

 R Ω(χx+ ξ) 2 dx −ρ1 4 − N1ε1− ε3 1 + 2ρ1 kL
 R Ωχ 2 tdx −N3l −γ₂N2− ε_k3 2l_L + β + γ + 1
 RL 0 θ 2 xdx +h3gN1 2λ + g 8b + N2C (ε2) +  3g Lε3 + kg 4ε3 3 i R Ω(goξx) (t)dx +hN 2 − N2 ρ2 2 4δ1g(0)Cp− ρ2g(0) 4ε3 − g 2ε3C2 i R Ω(g 0_oξ x) dx (28) Since g is continuous, positive and g (0) > 0, then for any t0 > 0, we

have Z t 0 g (s) ds ≥ Z t0 0 g (s) ds = g0, ∀t ≥ t0.

Now all the terms in the right-hand side of (28) become negative if we select carefully our constants. First, let us take δ1 = _4Nk

enough so that

ε1 ≤

ρ1

8N1

,

Second, by we select N1 large enough so that

λN1 4 − 3b 8 − g2₍₀₎ 2ε3 C2− C3(ε3) − 2ε3  2 L+ 1 + 1 2Cp  Cp > 0,

then select ε3 so small so that

ε3 ≤ min " ρ1 8  1 + 2ρ1 kL −1 , k 16  N3k + k + 2  2 L+ 1 −1# .

Next, we pick N3 large enough so that

N3l − γ 2N2− ε3 k  2l L + β + γ + 1  > 0,

and we choose N2 large enough so that

N2g0− 1 4 − 5ρ2 4 − N1  ρ2+ ρ2 1Cp 4ε1  − ε3γ k − ρ2 2ε3  1 + 1 L  > 0

and ε2 so small so that

ε2 < min  1 8N2 , λN1 4N2(1 + g) 

Finally, we choose N large enough so that

N β −γ 2 8k− N3  γ2 4kε3 + ρ3  −3N1γ 2_C p λb − C1(ε3) − ε3 k  2ρ₃ L + β  −γ 2_L 4ε3 > 0, and N 2 − N2 ρ2₂ 4δ1 g(0)Cp− ρ2g(0) 4ε3 − g 2ε3 C2 > 0

Therefore, (28) takes the form

£0(t) ≤ −k0E (t) + c (goξx) (t) , ∀t ≥ t0, (29)

for two positive constants k0 and c.

ζ (t) £0(t) ≤ −k0ζ (t) E (t) + cζ (t) (goψx) (t)

≤ −k0ζ (t) E (t) − c1(g0oξx) (t)

≤ −k0ζ (t) E (t) − c1E0(t)

[ζ (t) £ (t) + cE (t)]0− ζ0(t) E (t) ≤ −k0ζ (t) E (t) , ∀t ≥ t0.

Using the fact that ζ0(t) ≤ 0, we have

(ζ (t) £ (t) + cE (t))0 ≤ −k0ζ (t) E (t) , ∀t ≥ t0.

Again, by noting that

R (t) = ζ (t) £ (t) + cE (t) ∼ E (t) . we obtain for some positive constant α,

R0(t) ≤ −αζ (t) R (t) , ∀t ≥ t0. (30)

A simple integration of (30) over (t0, t) leads to

R (t) ≤ R (0) e−c1

Rt1

t0 ζ(s)ds_, ∀t ≥ t_{0. (31)}

Finally, the assertion of Theorem 11 is then obtained by virtue of the bound-edness of E and ζ and the fact that R ∼ E.

Acknowledgment

The authors are grateful to Prof. Said Mazouzi from the University of Annaba, Algeria, for his helpful comments.

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