Rapid Decay of Solutions for a Coupled System of Wave Equations with Class of Relaxation Functions in any Space Dimension

Rapid Decay of Solutions for a Coupled System of Wave Equations with Class of Relaxation Functions in any Space Dimension

Khaled ZENNIR Qassim University Department of Mathematics

Colle. Sci. Arts, Al-Ras Kingdom of Saudi Arabia

[email protected]

Saleh BOULAARES Qassim University Department of Mathematics

Colle. Sci. Arts, Al-Ras Kingdom of Saudi Arabia

[email protected]

Ali ALLAHEM Qassim University Department of Mathematics College of Sciences and Arts

Kingdom of Saudi Arabia [email protected]

Abstract:We consider a coupled system of viscoelastic wave equations. In weighted spaces, we shall prove a fast decay of energy associated to a coupled system with class of relaxation functions, as T → ∞ in IRⁿ.

Key–Words:Lyapunov function, viscoelastic, density, decay rate, weighted spaces, coupled system

1 Introduction

A coupled system of viscoelastic wave equations in any space dimension is given by



















|u⁰|^q−2u^0
0− φ(x)^∆xu −^R₀^tg1(t − s)∆xu(s)ds^

= αv,

|v⁰|^q−2v^0
0− φ(x)^∆_xv −^R₀^tg₂(t − s)∆_xv(s)ds^

= αu.

(1)

Here x ∈ IRⁿ, α 6= 0, t > 0, q ≥ 2, n > 2 and the scalar functions gi(s), i = 1, 2 are assumed to satisfy (A1).

Our problem (1) is supplemented with the next initial data.

u (0, x) = u0(x) ∈ H(IRⁿ),

u⁰(0, x) = u1(x) ∈ L^q_ρ(IRⁿ), (2) v (0, x) = v₀(x) ∈ H(IRⁿ),

v⁰(0, x) = v₁(x) ∈ L^q_ρ(IRⁿ). (3) We introduce the weighted spaces in Definition 1, where (φ(x))⁻¹= ρ(x) satisfies

ρ : IRⁿ→ IR^∗₊, ρ(x) ∈ C^0,˜γ(IRⁿ) (4) with ˜γ ∈ (0, 1) and ρ ∈ L^s(IRⁿ) ∩ L^∞(IRⁿ), where s = _2n−qn+2q²ⁿ .

There are many results about the existence by standard Galerkin method, (see [9], [10], [15], [16], [19], [21]), it is well known that, for any initial data u0, v0 ∈ H(IRⁿ) and u₁, v₁ ∈ L^q_ρ(IRⁿ), the problem (1)-(3) has a unique weak solution

(u, v) ∈ C([0, ∞), H(IRⁿ)) × C([0, ∞), H(IRⁿ),

(u⁰, v⁰) ∈ C([0, ∞), L^q_ρ(IRⁿ)) × C([0, ∞), L^q_ρ(IRⁿ)), under hypotheses (A1) − (A2). For completeness, if u₀, v₀ ∈ D(IRⁿ) ∩ H(IRⁿ) and u₁, v₁ ∈ H(IRⁿ), we state without proof, the regularity result as

(u, v) ∈ C([0, ∞), D(IRⁿ) ∩ H(IRⁿ)) × C([0, ∞), D(IRⁿ) ∩ H(IRⁿ)),

(u⁰, v⁰) ∈ C([0, ∞), H(IRⁿ)) × C([0, ∞), H(IRⁿ)).

This kind of problem with viscoelasticity was first introduced in [8], where a many qualitative results are obtained (see in this direction [1], [2], [6], [10], [11], [12], [13], [16], [18], [20],[21], [22],[23] [24]).

2 Definitions of Function Spaces and Assumption

We first, state without proof some useful results. Let us make use of the following assumption

(A1) The functions gi : IR⁺ −→ IR⁺are of class C¹ satisfying:

1 − gi= li> 0, gi(0) = g0i> 0, (5) where gi =^R₀^∞gi(t)dt.

(A2) There exists a positive function H ∈ C¹(IR⁺) such that

g⁰_i(t) + H(gi(t)) ≤ 0, t ≥ 0, H(0) = 0 (6) and H is defined below.

A- With t1> 0 such that for i = 1, 2:

KHALED SALEH ALI

SAUDI ARABIA SAUDI ARABIA SAUDI ARABIA

1) ∀t ≥ t1: We have

s→+∞lim g_i(s) = 0, which gives that

s→+∞lim

− g⁰_i(s)^

cannot be positive, so

s→+∞lim

− g_i⁰(s)^= 0.

Then gi(t1) > 0 and

max{g₁(s), g₂(s), −g₁⁰(s), −g₂⁰(s)}

< min{r, H(r), H0(r)},

where H0(t) = H(D(t)) provided that D is a positive C¹ function, with D(0) = 0, for which H₀ is strictly increasing and strictly convex C² function on (0, r]

and

Z +∞

0

g_i(s)H₀(−g⁰_i(s))ds < +∞.

2) ∀t ∈ [0, t1]: As g_iare nonincreasing, gi(0) > 0 and g_i(t₁) > 0 then g_i(t) > 0 and

gi(0) ≥ gi(t) ≥ gi(t1) > 0.

Then

e ≤ H(g₁(t)) ≤ f

e⁰ ≤ H(g₂(t)) ≤ f⁰

for some positive constants e, f, e⁰ and f⁰. Conse- quently,

g⁰_i(t) ≤ −H(g_i(t)) ≤ −kg_i(t), k > 0 gives

g_i⁰(t) + kg_i(t) ≤ 0, k > 0 (7) B- Let H₀^∗be the convex conjugate of H₀in the sense of Young (see [3], pages 61-64), then

H₀^∗(s) = s(H₀⁰)⁻¹(s) − H₀[(H₀⁰)⁻¹(s)], s ∈ (0, H₀⁰(r)) and satisfies the following Young’s inequality

AB ≤ H₀^∗(A) + H0(B), A ∈ (0, H₀⁰(r)), B ∈ (0, r].

Definition 1 ([10], [18]) The function spaces of our problem and their norm is defined as follows:

H(IRⁿ) =ⁿu ∈ L^2n/(n−2)(IRⁿ) : ∇_xu ∈ (L²(IRⁿ))^no, (8) with the normkuk_H(IRn)=^R_IRn|∇u|²dx^1/2. D(IRⁿ) =ⁿu ∈ L^2n/(n−2)(IRⁿ) : ∆xu ∈ L²(IRⁿ)^o.

(9) We define the norm

kuk_D(IRn) =^ Z

IRⁿ

|∆u|²dx^1/2,

where D(IRⁿ) can be embedded continuously in L^2n/(n−2)(IRⁿ), i.e there exists k > 0 such that

kuk_L2n/(n−2) ≤ kkuk_D. (10) and the spaceL²_ρ(IRⁿ) to be the closure of C₀^∞(IRⁿ) functions with respect to the inner product

(f, h)_L²

ρ(IRⁿ)= Z

IRⁿ

ρf hdx. (11) For1 < p < ∞, if f is a measurable function on IRⁿ, we define

kf k_L^p

ρ(IRⁿ)=

Z

IRⁿ

ρ|f |^qdx

1/p

. (12) The space L²_ρ(IRⁿ) is a separable Hilbert space.

Lemma 2 ([7]) Let h, w ∈ C¹(IR) be any two func- tions andθ ∈ [0, 1], then we have

w⁰(t) Zt

0

h(t − s)w(s)ds

= −1 2

d dt

t

Z

0

h(t − s)|w(t) − w(s)|²ds

+1 2

d dt





t

Z

0

h(s)ds



|w(t)|²

+1 2 Zt

0

h⁰(t − s)|w(t) − w(s)|²ds −1

2h(t)|w(t)|². and

   

Z t 0

h(t − s)(w(t) − w(s))ds    

2

≤

Z t 0

|h(s)|^2(1−θ)ds







t

Z

0

|h(t − s)|^2θ|w(t) − w(s)|²ds





The modified energy associate to (u, v) at time t is given as

E(t) = (q − 1) q

hku⁰k^q_Lq

ρ(IRⁿ)+ kv⁰k^q_Lq ρ(IRⁿ)

i

+α Z

IRⁿ

ρuvdx +1 2

1 − Z t

0

g₁(s)ds^k∇_xuk²₂

+1 2

 1 −

Z t 0

g2(s)ds^k∇_xvk²₂

+1

2(g₁◦ ∇_xu) + 1

2(g₂◦ ∇_xv) (13) We easily deduce for c > 0, that

E(t) ≥ (1 − c|α|kρk⁻¹_Ln/2) (q − 1)

q

hku⁰k^q_Lq

ρ(IRⁿ)+ kv⁰k^q_Lq ρ(IRⁿ)

i

+1 2

1 − Z t

0

g₁(s)ds^k∇_xuk²₂

+1 2

 1 −

Z t 0

g2(s)ds^k∇_xvk²₂

+1

2(g₁◦ ∇_xu) +1

2(g₂◦ ∇_xv) (14) for α small enough and by using Lemma 3.

The first derivative of the energy functional for all t ≥ 0 is given by

E⁰(t) ≤ 1

2(g₁⁰ ◦ ∇_xu)(t) +1

2(g₂⁰ ◦ ∇_xv)(t), (15) Noting by

(g_i◦∇_xh)(t) = Z t

0

g_i(t−τ ) k∇_xh(t) − ∇_xh(τ )k²₂dτ, (16) for ψ(t) ∈ H(IRⁿ), t ≥ 0, i = 1, 2.

3 Main results and Proofs

Lemma 3 [13] Let ρ satisfy (4), then for any u ∈ H(IRⁿ)

kuk_L^p

ρ(IRⁿ)≤ kρk_Ls(IRⁿ)k∇_xuk_L²_(IRⁿ₎ (17) withs = _2n−pn+2p²ⁿ , 2 ≤ p ≤ _n−2²ⁿ

Our main result reads as follows.

Theorem 4 Let (u₀, v₀) ∈ (H(IRⁿ))², (u₁, v₁) ∈ (L^q_ρ(IRⁿ))². Then there exists a positive constants a, b, c, d such that the energy of solution of problem (1)-(3) satisfies,

E(t) ≤ dH₁⁻¹(bt + c), for all t ≥ 0, where(A1) − (A2) hold

H₁(t) = Z 1

t

1

sH₀⁰(as)ds (18) To prove Theorem 4, let us define

L(t) = ξ1E(t) + ψ1(t) + ξ2ψ2(t) (19) for ξ1, ξ₂ > 1. Let

ψ₁(t) = Z

IRⁿ

ρ(x)^hu|u⁰|^q−2u⁰+ v|v⁰|^q−2v⁰ⁱdx, (20) and

ψ₂(t) = (21)

− Z

IRⁿ

ρ(x)|u⁰|^q−2u⁰ Z t

0

g1(t − s)(u(t) − u(s))dsdx

− Z

IRⁿ

ρ(x)|v⁰|^q−2v⁰ Z t

0

g₂(t − s)(v(t) − v(s))dsdx.

Lemma 5 Suppose that (A1) and (A2) hold. Then, the functionalψ1 satisfies, along the solution of (1)-(3)

ψ⁰₁(t) ≤ ku⁰k^q

L^qρ(IRⁿ)+ kv⁰k^q

L^qρ(IRⁿ)

+ (c_σαkρk²_Ln/2− l)^hk∇_xuk²₂+ k∇_xvk²₂ⁱ

+ (1 − l)

4σ [(g1◦ ∇_xu) + (g2◦ ∇_xv)] , wherel = min{l1, l2}.

Proof. From (20), integrating over IRⁿ, we have ψ⁰₁(t) =

Z

IRⁿ

ρ(x)|u⁰|^qdx

+ Z

IRⁿ

ρ(x)u^|u⁰|^q−2u^00dx

+ Z

IRⁿ

ρ(x)|v⁰|^qdx + Z

IRⁿ

ρ(x)v^|v⁰|^q−2v^00dx

= Z

IRⁿ

ρ(x)|u⁰|^q+ u∆_xu

−u Z t

0

g1(t − s)∆xu(s, x)ds − αρ(x)uv^dx

+ Z

IRⁿ

ρ(x)|v⁰|^q+ v∆_xv

−v Z t

0

g2(t − s)∆xv(s, x)ds − αρ(x)uv^dx

= Z

IRⁿ

ρ(x)|u⁰|^q− ∇_xu∇_xu

+∇xu Z t

0

g1(t − s)∇xu(s, x)ds − αρ(x)uv^dx

+ Z

IRⁿ



ρ(x)|v⁰|^q− ∇_xv∇xv

+∇_xv Z t

0

g₂(t − s)∇_xv(s, x)ds − αρ(x)uv^dx

= Z

IRⁿ

ρ(x)|u⁰|^q− (∇_xu)²

+∇_xu Z t

0

g₁(t − s)(∇_xu(s) − ∇_xu(t))ds^dx

+ Z

IRⁿ

ρ(x)|v⁰|^q− (∇_xv)²

+∇xv Z t

0

g2(t − s)(∇xv(s) − ∇xv(t))ds^dx

+ Z

IRⁿ

(∇_xu)² Z t

0

g₁(s)dsdx

+ Z

IRⁿ

(∇xv)² Z t

0

g2(s)dsdx − 2α Z

IRⁿ

ρ(x)uvdx Thanks to Young’s inequality and Lemma 2 and for θ = 1/2, we get for small enough positive constant σ

∇_xu Z t

0

g1(t − s)(∇xu(s) − ∇xu(t))dsdx

≤ σk∇_xuk²₂

+ 1 4σ

Z

IRⁿ

Z _t

0

g₁(t − s)|∇_xu(s) − ∇_xu(t)|ds

2

dx

≤ σk∇_xuk²₂+1 − l1

4σ (g₁◦ ∇_xu)(t) and

∇_xv Z t

0

g₂(t − s)(∇_xv(s) − ∇_xv(t))dsdx

≤ σk∇_xvk²₂

+ 1 4σ

Z

IRⁿ

Z t 0

g₂(t − s)|∇_xv(s) − ∇_xv(t)|ds

2

dx

≤ σk∇_xvk²₂+ 1 − l2

4σ (g2◦ ∇_xv)(t)

By (A1) and the fact that^R₀^tg(s)ds <^R₀^∞g(s)ds ψ⁰₁(t) ≤ ku⁰k^q_Lq

ρ(IRⁿ)

+kv⁰k^q

L^qρ(IRⁿ)+ (σ − l₁)k∇_xuk²₂ +(σ − l2)k∇xvk²₂− 2α

Z

IRⁿ

ρ(x)uvdx

+1 − l₁

4σ (g1◦ ∇_xu) +1 − l₂

4σ (g2◦ ∇_xv) Using Holder’s, Young’s inequalities and Lemma 3 to get

Z

IRⁿ

|ρ(x)uv|dx

= Z

IRⁿ

|(ρ(x)^1/2u)(ρ(x)^1/2v)|dx

≤

Z

IRⁿ

ρ(x)|u|²dx

1/2Z

IRⁿ

ρ(x)|v|²dx

1/2

≤ σ Z

IRⁿ

ρ(x)|u|²dx + 1 4σ

Z

IRⁿ

ρ(x)|v|²dx

≤ kρk²_Ln/2

 σ

Z

IRⁿ

|∇_xu|²dx + 1 4σ

Z

IRⁿ

|∇_xv|²dx

 . Then, we obtain for l = min{l1, l₂}, c_σ > 0

ψ⁰₁(t) ≤ ku⁰k^q

L^qρ(IRⁿ)+ kv⁰k^q

L^qρ(IRⁿ)

+ (cσαkρk²_Ln/2− l)^k∇_xuk²₂+ k∇xvk²₂^

+ (1 − l)

4σ ((g₁◦ ∇_xu) + (g₂◦ ∇_xv)) . Lemma 6 Suppose that (A1) and (A2) hold. Then, the functionalψ₂ satisfies, along the solution of (1)- (3), for anyσ ∈ (0, 1)

ψ⁰₂(t) ≤ σ^1 + αkρk²_Ln/2(IRⁿ)

 hk∇_xuk²₂+ k∇_xvk²₂ⁱ

+ cσ



1 + αkρk²_Ln/2



[(g1◦ ∇_xu) + (g2◦ ∇_xv)]

− c_σkρk^q_Ls

h(g⁰₁◦ ∇_xu)^q/2+ (g₂⁰ ◦ ∇_xv)^q/2i

+

 σ −

Z t 0

g(s)ds

h ku⁰k^q/2

L^qρ(IRⁿ)+ kv⁰k^q/2

L^qρ(IRⁿ)

i.

where Z t

0

g(s)ds ≤ min

Z _t

0

g₁(s)ds, Z t

0

g₂(s)ds

 (22) Proof. Exploiting Eqs. (1) to get

ψ⁰₂(t) = − Z

IRⁿ

ρ(x)^|u⁰|^q−2u^00× Z t

0

g₁(t − s)(u(t) − u(s))dsdx

− Z

IRⁿ

ρ(x)|u⁰|^q−2u⁰× Z t

0

g⁰₁(t − s)(u(t) − u(s))dsdx − Z t

0

g₁(s)dsku⁰k^q_Lq ρ(IRⁿ)

− Z

IRⁿ

ρ(x)^|v⁰|^q−2v^00 Z t

0

g2(t − s)(v(t) − v(s))dsdx

− Z

IRⁿ

ρ(x)|v⁰|^q−2v⁰ Z _t

0

g₂⁰(t − s)(v(t) − v(s))dsdx

− Z t

0

g2(s)dskv⁰k^q_Lq ρ(IRⁿ)

= Z

IRⁿ

∇_xu Z t

0

g₁(t − s)(∇_xu(t) − ∇_xu(s))dsdx

− Z

IRⁿ

Z _t

0

g1(t − s)∇xu(s, x)ds



×

Z t 0

g₁(t − s)(∇_xu(t) − ∇_xu(s))ds

 dx

− Z

IRⁿ

ρ(x)|u⁰|^q−2u⁰ Z t

0

g₁⁰(t − s)(u(t) − u(s))dsdx

− Z t

0

g₁(s)dsku⁰k^q

L^qρ(IRⁿ)

+α Z

IRⁿ

ρ(x)v Z t

0

g1(t − s)(u(t) − u(s))dsdx

+ Z

IRⁿ

∇_xv Z t

0

g₂(t − s)(∇_xv(t) − ∇_xv(s))dsdx

− Z

IRⁿ

Z _t

0

g2(t − s)∇xv(s, x)ds



×

Z _t

0

g2(t − s)(∇xv(t) − ∇xv(s))ds

 dx

− Z

IRⁿ

ρ(x)|v⁰|^q−2v⁰ Z t

0

g₂⁰(t − s)(v(t) − v(s))dsdx

− Z t

0

g2(s)dskv⁰k^q_Lq ρ(IRⁿ)

+α Z

IRⁿ

ρ(x)u Z t

0

g2(t − s)(v(t) − v(s))dsdx, then

ψ⁰₂(t) =

 1 −

Z t 0

g₁(s)ds



× Z

IRⁿ

∇_xu Z t

0

g1(t − s)(∇xu(t) − ∇xu(s))dsdx

+ Z

IRⁿ

Z _t

0

g₁(t − s)(∇_xu(t) − ∇_xu(s))ds

2

dx

− Z

IRⁿ

ρ(x)|u⁰|^q−2u⁰ Z t

0

g₁⁰(t − s)(u(t) − u(s))dsdx

− Z t

0

g1(s)dsku⁰k^q

L^qρ(IRⁿ)+ c(g1◦ ∇_xu)(t) +

 1 −

Z _t

0

g₂(s)ds



× Z

IRⁿ

∇_xv Z t

0

g2(t − s)(∇xv(t) − ∇xv(s))dsdx

+ Z

IRⁿ

Z t 0

g₂(t − s)(∇_xv(t) − ∇_xv(s))ds

2

dx

− Z

IRⁿ

ρ(x)|v⁰|^q−2v⁰ Z t

0

g₂⁰(t − s)(v(t) − v(s))dsdx

− Z t

0

g₂(s)dskv⁰k^q

L^qρ(IRⁿ)+ c(g₂◦ ∇_xv)(t) +α

Z

IRⁿ

ρ(x)^v Z t

0

g1(t − s)(u(t) − u(s))ds

+u Z t

0

g₂(t − s)(v(t) − v(s))ds^dx

Thanks to Holder’s and Young’s inequalities and with Lemma 3, we estimate

Z

IRⁿ

ρ(x)v Z t

0

g₁(t − s)(u(t) − u(s))dsdx

≤

Z

IRⁿ

ρ(x)|v|²dx

1/2

×

Z

IRⁿ

ρ(x) Z t

0

g1(t − s)(u(t) − u(s))ds²

1/2

≤ σkvk²_L2 ρ(IRⁿ)

+c_σ Z t

0

g₁(t − s)(u(t) − u(s))ds ²

L²_ρ(IRⁿ)

≤ σkρk²_Ln/2(IRn)k∇_xvk²₂+ c_σkρk²_Ln/2(IRn)(g₁◦ ∇_xu)(t).

and Z

IRⁿ

ρ(x)u Z t

0

g2(t − s)(v(t) − v(s))dsdx

≤

Z

IRⁿ

ρ(x)|u|²dx

1/2

×

Z

IRⁿ

ρ(x) Z t

0

g2(t − s)(v(t) − v(s))ds²

1/2

≤ σkuk²_L2 ρ(IRⁿ)

+cσ

Z t 0

g2(t − s)(v(t) − v(s))ds ²

L²_ρ(IRⁿ)

≤ σkρk²_Ln/2(IRn)k∇_xuk²₂

+c_σkρk²_Ln/2(IRn)(g₂◦ ∇_xv)(t).

and for the exponents _q−1^q , q

− Z

IRⁿ

ρ(x)|u⁰|^q−2u⁰ Z t

0

g₁⁰(t − s)(u(t) − u(s))dsdx

≤

Z

IRⁿ

ρ(x)|u⁰|^qdx

(q−1)/q

×

Z

IRⁿ

ρ(x) Z _t

0

g₁⁰(t − s)(u(t) − u(s))ds^q

1/q

≤ σku⁰k^q_Lq ρ(IRⁿ)

+c_σ Z t

0

−g⁰₁(t − s)(u(t) − u(s))ds ^q

L^q_ρ(IRⁿ)

≤ σku⁰k^q_Lq

ρ(IRⁿ)− c_σkρk^q_Ls(IRⁿ)(g₁⁰ ◦ ∇_xu)^q/2(t).

and

− Z

IRⁿ

ρ(x)|v⁰|^q−2v⁰ Z t

0

g₂⁰(t − s)(v(t) − v(s))dsdx

≤

Z

IRⁿ

ρ(x)|v⁰|^qdx

(q−1)/q

×

Z

IRⁿ

ρ(x) Z t

0

g₂⁰(t − s)(v(t) − v(s))ds^q

1/q

≤ σkv⁰k^q

L^qρ(IRⁿ)

+cσ

Z t 0

−g⁰₂(t − s)(v(t) − v(s))ds ^q

L^qρ(IRⁿ)

≤ σkv⁰k^q

L^qρ(IRⁿ)− c_σkρk^q_Ls(IRn)(g₂⁰◦ ∇_xv)^q/2(t).

Thanks to Young’s and Poincare’s inequalities and us- ing Lemma 2 for θ = 1/2, we obtain

ψ⁰₂(t) ≤ σ^1 + αkρk²_Ln/2(IRⁿ)

 k∇_xuk²₂+ k∇_xvk²₂^

+c_σ^1 + αkρk²_Ln/2(IRⁿ)

((g₁◦ ∇_xu) + (g₂◦ ∇_xv))

−c_σkρk^q_Ls(IRⁿ)

(g⁰₁◦ ∇_xu)^q/2+ (g₂⁰ ◦ ∇_xv)^q/2

+

 σ −

Z t 0

g₁(s)ds



ku⁰k^q/2

L^qρ(IRⁿ)

+

 σ −

Z t 0

g2(s)ds



kv⁰k^q/2_Lq ρ(IRⁿ). For ξ1, ξ2 > 1, we have

β₁L(t) ≤ E(t) ≤ β₂L(t) (23) which holds for two positive constants β1and β2. Lemma 7 For ξ1, ξ2 > 1, we have

L(t) ∼ E(t). (24)

Proof. We have by (19)

|L(t) − ξ₁E(t)|

≤ |ψ₁(t)| + ξ2|ψ₂(t)|

≤ Z

IRⁿ

 

ρ(x)u|u⁰|^q−2u⁰dx + Z

IRⁿ

 

ρ(x)v|v⁰|^q−2v⁰dx +ξ₂

Z

IRⁿ

   

ρ(x)|u⁰|^q−2u⁰ Z t

0

g₁(t − s)(u(t) − u(s))ds    

dx

+ξ₂ Z

IRⁿ

   

ρ(x)|v⁰|^q−2v⁰ Z t

0

g₂(t − s)(v(t) − v(s))ds    

dx.

Using Holder’s and Young’s inequalities with _q−1^q , q, since q ≥ 2, we have

Z

IRⁿ

 

ρ(x)u|u⁰|^q−2u⁰dx

≤

Z

IRⁿ

ρ(x)|u|^qdx

1/qZ

IRⁿ

ρ(x)|u⁰|^qdx

(q−1)/q

≤ 1 q

Z

IRⁿ

ρ(x)|u|^qdx



+ q − 1 q

Z

IRⁿ

ρ(x)|u⁰|^qdx



≤ cku⁰k^q

L^qρ(IRⁿ)+ ckρk^q_Ls(IRⁿ)k∇_xuk^q₂

and Z

IRⁿ

 

ρ(x)v|v⁰|^q−2v⁰dx

≤

Z

IRⁿ

ρ(x)|v|^qdx

1/qZ

IRⁿ

ρ(x)|v⁰|^qdx

(q−1)/q

≤ 1 q

Z

IRⁿ

ρ(x)|v|^qdx



+ q − 1 q

Z

IRⁿ

ρ(x)|v⁰|^qdx



≤ ckv⁰k^q

L^qρ(IRⁿ)+ ckρk^q_Ls(IRⁿ)k∇_xvk^q₂ and

Z

IRⁿ

  

 ρ(x)

q−1

q |u⁰|^q−2u⁰



×

 ρ(x)^1q

Z t 0

g1(t − s)(u(t) − u(s))ds

  dx

≤

Z

IRⁿ

ρ(x)|u⁰|^qdx

(q−1)/q

×

Z

IRⁿ

ρ(x) Z t

0

g₁(t − s)(u(t) − u(s))ds^qdx

1/q

≤ q − 1 q ku⁰k^q_Lq

ρ(IRⁿ)

+1 q

Z t 0

g₁(t − s)(u(t) − u(s))ds ^q

L^q_ρ(IRⁿ)

≤ q − 1 q ku⁰k^q

L^qρ(IRⁿ)+1

qkρk^q_Ls(IRⁿ)(g₁◦ ∇_xu)^q/2(t).

and Z

IRⁿ

  



ρ(x)^q−1q |v⁰|^q−2v⁰



×

 ρ(x)^1q

Z t 0

g2(t − s)(v(t) − v(s))ds

  dx

≤

Z

IRⁿ

ρ(x)|v⁰|^qdx

(q−1)/q

×

Z

IRⁿ

ρ(x) Z t

0

g2(t − s)(v(t) − v(s))ds^qdx

1/q

≤ q − 1 q kv⁰k^q_Lq

ρ(IRⁿ)

+1 q

Z t 0

g₂(t − s)(v(t) − v(s))ds ^q

L^qρ(IRⁿ)

≤ q − 1 q kv⁰k^q_Lq

ρ(IRⁿ)+1

qkρk^q_Ls(IRⁿ)(g2◦ ∇_xv)^q/2(t).

Then, since q ≥ 2, we have

|L(t) − ξ₁E(t)| ≤ c(E(t) + E^q/2(t))

≤ c(E(t) + E(t)E^(q/2)−1(t))

≤ c(E(t) + E(t)E^(q/2)−1(0))

≤ cE(t).

Therefore, we can choose ξ1so that

L(t) ∼ E(t). (25)

Proof of Theorem 4 By (15), Lemma 5 and Lemma 6, we have

L⁰(t) = ξ1E⁰(t) + ψ⁰₁(t) + ξ2ψ₂⁰(t)

≤ (ξ₁

2 − c_σξ₂kρk^q_Lsb) × h

(g₁⁰ ◦ ∇_xu)^q/2+ (g₂⁰ ◦ ∇_xv)^q/2i +M0[(g2◦ ∇_xu) + (g2◦ ∇_xv)]

−M₁^hku⁰k^q

L^qρ+ kv⁰k^q

L^qρ

i

−M₂^hk∇_xuk²₂+ k∇xvk²₂ⁱ where

M₀= (4ξ2c^1 + αkρk²_Ln/2

+ (1 − l)

4σ ),

M1 =

 ξ2

Z _t1

0

g(s)ds − σ



− 1

 , M2 =^−ξ₂σ^1 + αkρk²_Ln/2

+ (l − cσαkρk²_Ln/2)^, and t₁was given in A.

We now choose σ so small that ξ1> 2cσkρk^q_Ls(IRⁿ)ξ2. Whence σ is fixed, we can choose ξ1, ξ₂ large enough so that M₁, M₂> 0, which yield, for all t ≥ t₁

L⁰(t) ≤ M₀[(g₁◦ ∇_xu) + (g₂◦ ∇_xv)] − cE(t).(26) Let us introduce a new functional F (t) = L(t) + cE(t). Then by (26), we get for some positive con- stant c and t ≥ t1

F⁰(t) = L⁰(t) + cE⁰(t) (27)

≤ −cE(t)

+c Z

IRⁿ

Z t t1

g1(t − s)|∇xu(t) − ∇xu(s)|²dsdx

+c Z

IRⁿ

Z t t1

g₂(t − s)|∇_xv(t) − ∇_xv(s)|²dsdx.

By (7) and (15), we have for all t ≥ t1

Z

IRⁿ

Z t1

0

g1(t − s)|∇xu(t) − ∇xu(s)|²dsdx

+ Z

IRⁿ

Z t1

0

g₂(t − s)|∇_xv(t) − ∇_xv(s)|²dsdx

≤ −1 k

Z

IRⁿ

Z t1

0

g₁⁰(t − s)|∇_xu(t) − ∇_xu(s)|²dsdx

+ Z

IRⁿ

Z t1

0

g₂⁰(t − s)|∇xv(t) − ∇xv(s)|²dsdx^

≤ −cE⁰(t).

At this point, we define I(t) =

Z t t1

H0(−g⁰₁(s))(g1◦ ∇_xu)(t)ds

+ Z t

t1

H₀(−g₂⁰(s))(g₂◦ ∇_xv)(t)ds. (28) Since^R₀^+∞H₀(−g_i⁰(s))g(s)ds < +∞, i = 1, 2, from (15) we have

I(t) = Z t

t1

H₀(−g₁⁰(s)) × Z

IRⁿ

g1(s)|∇xu(t) − ∇xu(t − s)|²dxds

+ Z t

t1

H₀(−g⁰₂(s)) × Z

IRⁿ

g₂(s)|∇_xv(t) − ∇_xv(t − s)|²dxds

≤ 2 Z t

t1

H0(−g₁⁰(s))g1(s) × Z

IRⁿ

|∇_xu(t)|²+ |∇_xu(t − s)|²dxds

+2 Z t

t1

H0(−g₂⁰(s))g2(s) × Z

IRⁿ

|∇_xv(t)|²+ |∇xv(t − s)|²dxds

≤ cE(0)^h Z t

t1

H₀(−g⁰₁(s))g₁(s)ds

+ Z t

t1

H0(−g₂⁰(s))g2(s)dsⁱ. (29) We have I(t) < 1 (see[16], Eq. (3.11)). We now define the functional λ(t) related to I(t) as

λ(t) = − Z t

t1

H₀(−g⁰₁(s))g₁⁰(s) × Z

IRⁿ

g₁(s)|∇_xu(t) − ∇_xu(t − s)|²dxds

− Z t

t1

H0(−g⁰₂(s))g₂⁰(s) × (30) Z

IRⁿ

g₂(s)|∇_xv(t) − ∇_xv(t − s)|²dxds.

By (A1)-(A2), we get

H₀(−g_i⁰(s))g_i(s) ≤ H₀(H(g_i(s)))g_i(s)

= D(gi(s))gi(s) ≤ k0. for some positive constant k₀. Then, for all t ≥ t₁

λ(t)

≤ −k₀ Z t

t1

g₁⁰(s) Z

IRⁿ

|∇_xu(t) − ∇xu(t − s)|²dxds

−k₀ Z t

t1

g₂⁰(s) Z

IRⁿ

|∇_xv(t) − ∇_xv(t − s)|²dxds

≤ −k₀ Z t

t1

g₁⁰(s) Z

IRⁿ

|∇_xu(t)|²+ |∇xu(t − s)|²dxds

−k₀ Z t

t1

g₂⁰(s) Z

IRⁿ

|∇_xv(t)|²+ |∇_xv(t − s)|²dxds

≤ −cE(0)

Z _t

t1

g⁰₁(s)ds + Z t

t1

g₂⁰(s)ds



≤ cE(0) max {g₁(t₁), g₂(t₁)}

< min{r, H(r), H₀(r)}. (31) By the definition of H0, and for x ∈ (0, r], θ ∈ [0, 1]

we have

H0(θx) ≤ θH0(x).

Using (29), (31)leads to λ(t) = I⁻¹(t)ⁿ

Z t t1

I(t)H₀[H₀⁻¹(−g₁⁰(s))] ×