Weighted integral inequality and applications in general energy decay estimate for a variable density wave equation with memory

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R E S E A R C H

Open Access

Weighted integral inequality and

applications in general energy decay

estimate for a variable density wave equation

with memory

Fushan Li

1*

_{and Fengying Hu}

1

*_{Correspondence:} [email protected]

1_{School of Mathematical Sciences,}

Qufu Normal University, Qufu, P.R. China

Abstract

This paper develops a weighted integral inequality to derive decay estimates for the quasilinear viscoelastic wave equation with variable density

|ut|ρutt–

u

–

u

tt+

t

0

g(t–s)u(s)ds= 0 in

Ω

×(0,∞)

with initial conditions and boundary condition, wheregis a memory kernel function and

ρ

is a positive constant. Depending on the properties of convolution kernelgat inﬁnity, we establish a general decay rate of the solution such that the exponential and polynomial decay results in some literature are special cases of this paper, and we improve the integral method used in the literature.

MSC: 35B35; 35G05; 35G16; 35L30

Keywords: Weighted integral inequality; Viscoelastic equation; Memory kernel; General decay estimate

1 Introduction

Using Lyapunov technique for some perturbed energy, Messaoudiet and Khulaiﬁ [24] studied the following problem:

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

|ut|ρutt–u–utt+ t

0g(t–s)u(s)ds= 0 inΩ×(0,∞),

u= 0 on∂Ω×(0,∞),

u|t=0=u0, ut|t=0=u1 inΩ,

(1.1)

whereΩ⊂Rn ₍_n_≥_{1) is a bounded domain with smooth boundary}_∂Ω _{such that the}

divergence theorem can be applied.

Cavalcanti et al. [3] proved that the ﬁnite energy solutions of nonlinear abstract PDE with a memory term exhibit exponential decay rates when strong damping –utis active,

this uniform decay is no longer valid (by spectral analysis arguments) for dynamics sub-jected to frictional damping only, sayg= 0. Viscoelastic equations with variable density

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have been studied by many authors, and several stability results have been established in [2, 9,10,23,25,26]. As we know, all the stability results were obtained by establishing differential inequalities on the functional equivalent to the original energy. Our approach is based on integral inequalities and multiplier techniques. Indeed, instead of using Lya-punov technique for some perturbed energy, we rather concentrate on the original energy, showing that it satisfies a weighted integral inequality which, in turn, yields the final decay estimate. We prove a general decay rate from which the exponential decay and the poly-nomial decay are only special cases. Due to the assumption ong, the weighted inequality established in this paper improves the integral inequality in [1].

We mention here some related works concerning the energy for the evolution equations. For the nonlinear damped wave equations and Marguerre–von Karman system, some en-ergy decay rate estimates were obtained in [11–15,19,20] and the references therein. Li and his coauthors [7,16–18] studied the blow up phenomena of the solutions for evolu-tion equaevolu-tions. This research laid a good foundaevolu-tion for our further study. For the stability and convergence results of evolution equations, the readers can refer to [6,21,22].

The outline of this paper is as follows. In Sect.2, we present the preliminaries and our im-portant result. In Sect.3, we construct an energy inequality, prove the main Theorem2.2

and give applications to various functionsξ(t).

To simplify calculations in our analysis, we introduce the following notations:

G(t) =

t

0

g(s)ds– strength of memory, g◦h=

t

0

g(t–s)h(s) –h(t)2ds,

g∗h=

t

0

g(t–s)h(s)ds, (u,v) =

Ω

uv dx, R+:= [0, +∞),

vρ+2:=vLρ+2₍_Ω₎, v:=v_L2₍_Ω₎.

2 Preliminaries and main result

In this section we prepare some material needed in the proof of our result and state our main result. Throughout this paper,Cdenotes a generic positive constant. We impose the following assumptions onρandg.

Assumption 2.1 SetG(t) =₀tg(s)ds. We assume that 1.

0 <ρ≤ 2

n– 2 ifn≥3; ρ> 0 ifn= 1, 2,

which implies that

H₀1(Ω)→L2(ρ+1)_(Ω).

2. g: [0,∞)→[0,∞)is a locally absolutely continuous function such that

G(∞) < 1, g(0) > 0, g(t)≤0, for a.e.t≥0.

3. There exists a non-increasing functionξ∈C1_{[0, +}_∞₎_with+∞

0 ξ(τ)dτ = +∞ satisfying

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Theorem 2.1([24]) Suppose that(u0,u1)∈H01(Ω)×H01(Ω).Under Assumption2.1,there

exists a unique solution u of(1.1)satisfying

u∈L∞ 0,∞;H₀1(Ω), ut∈L∞ 0,∞;H01(Ω)

, utt∈L∞ 0,∞;L2(Ω)

,

and

u(x,t)→u0(x) in H01(Ω), ut(x,t)→u1(x) in H01(Ω),as t→0+.

Lemma 2.1([4,5,8]) H01(Ω)→Lr(Ω)with

r:

⎧ ⎨ ⎩

0 <r≤ 2n

n–2, n≥3,

≥2, n= 1, 2,

which implies

ϕr≤B∇ϕ2, ∀ϕ∈H01(Ω).

Lemma 2.2 Let u be the global solution of the problem(1.1),then for any suitable function w one has

|ut|ρutt,w

= 1

ρ+ 1

d dt |ut|

ρ

ut,w

– 1

ρ+ 1 |ut|

ρ

ut,wt

.

Proof From

d dt |ut|

ρ_u t,w

= d

dt

u2

t ρ

2_u t,w

= |ut|ρutt,w

+ρ 2

u2_t

ρ–2 2 ₂_u

tuttut,w

+ |ut|ρut,wt

= (ρ+ 1) |ut|ρutt,w

+ |ut|ρut,wt

,

we have

|ut|ρutt,w

= 1

ρ+ 1

d dt |ut|

ρ

ut,w

– 1

ρ+ 1 |ut|

ρ

ut,wt

.

Lemma 2.3 Let u be the global solution of the problem(1.1),then

d dtE(t) =

1 2g

_{◦ ∇}_u₍_t_{) –}1

2g(t)∇u 2_,

where

E(t) = 1 ρ+ 2ut

ρ+2

ρ+2+ 1

2 1 –G(t)∇u(t) 2

+1 2∇ut

2₊1

2(g◦ ∇u)(t).

Proof Multiplying equation (1.1) byut, integrating by parts overΩand using Lemma2.2,

we obtain the conclusion.

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Theorem 2.2 Let u be the global solution of problem(1.1)with Assumption2.1.We deﬁne the energy functional as

E(t) = 1 ρ+ 2ut

ρ+2

ρ+2+ 1

2 1 –G(t)∇u(t) 2

+1 2∇ut

2₊1

2(g◦ ∇u)(t).

Then,for some t0> 0there exist positive constants C0andωsuch that

E(t)≤C0e–

ω_tt 0ξ(s)ds_.

3 The proof of main result

In order to derive the desired result of Theorem2.2by the integral method, we establish the following weighted integral inequality.

Lemma 3.1 Let u be the solution of(1.1)under Assumption2.1,then

T

S

ξ(t)E(t)dt≤CE(S)

for some constant C> 0.

To prove the above inequality, we need the following two lemmas.

T

S

ξ(t)E(t)dt≤C

T

S

ξ(t)ut2dt+C T

S

ξ(t)∇ut2dt+CE(S).

Proof Multiplying byξ(t)u(t) both sides of equation (1.1), integrating the resulting equa-tion overΩ×[S,T] (0≤S≤T), then using the boundary conditions and Lemma2.2, we have

0 =

T

S

ξ(t)

u,|ut|ρutt–u–utt+ t

0

g(t–s)u(s)ds

dt

=

T

S

ξ(t) u,|ut|ρutt

dt+

T

S

ξ(t)∇u2dt+

T

S

ξ(t)(∇u,∇utt)dt

–

T

S

ξ(t)

∇u,

t

0

g(t–s)∇u(s)ds

dt

=

T

S

ξ(t) u,|ut|ρutt

dt+

T

S

ξ(t)1 –G(t)∇u2dt+

T

S

ξ(t)(∇u,∇utt)dt

–

T

S

ξ(t)

∇u,

t

0

g(t–s) ∇u(s) –∇u(t)ds

dt. (3.1)

According to the deﬁnition of the energy functionalE(t), we get

1 –G(t)∇u2= 2E(t) – 2 ρ+ 2ut

ρ+2

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Combining (3.1) with (3.2), we deduce that

T

S

ξ(t)E(t)dt= 1 ρ+ 2

T

S

ξ(t)utρρ+2+2dt+ 1 2

T

S

ξ(t)∇ut2dt

+1 2

T

S

ξ(t)(g◦ ∇u)(t)dt

–1 2

T

S

ξ(t)u,|ut|ρutt

dt–1 2

T

S

ξ(t)(∇u,∇utt)dt

+1 2

T

S

ξ(t)

∇u,

t

0

g(t–s)∇u(s) –∇(t)ds

dt. (3.3)

From Lemma2.3, we see that

d dtE(t) =

1 2g

_{◦ ∇}_u₍_t_{) –}1

2g(t)∇u 2_≤1

2g

_{◦ ∇}_u₍_t_),

that is,

–g◦ ∇u(t)≤–2E(t),

which, together with Assumption2.1, implies

T

S

ξ(t)(g◦ ∇u)(t)dt≤–

T

S

g◦ ∇u(t)dt≤–2

T

S

E(t)dt. (3.4)

For the fourth term on the right-hand side of (3.3), integrating by parts and using Lemma2.2, we have

–

T

S

ξ(t)u,|ut|ρutt

dt= – 1

ρ+ 1 ξ(t)u,|ut|

ρ

ut T S +

1 ρ+ 1

T

S

ξ(t)u_t,|ut|ρut

dt

= – 1

ρ+ 1 ξ(t)u,|ut|

ρ

utT_S +

1 ρ+ 1

T

S

ξ(t)u,|ut|ρut

dt

+ 1

ρ+ 1

T

S

ξ(t)utρρ+2+2dt. (3.5)

By Young inequality, Lemma2.1and the deﬁnition ofE(t), we have

– 1

ρ+ 1 ξ(t)u,|ut|

ρ

ut≤

1 ρ+ 1

ξ(t)

1 ρ+ 2u

ρ+2

ρ+2+ ρ+ 1 ρ+ 2ut

ρ+2

≤ξ(t)

1 (ρ+ 1)(ρ+ 2)

2 1 –G(∞)

ρ

2+1

Bρ+2_Eρ₂_{(0) + 1}

E(t)

≤k1ξ(t)E(t),

with some positive constantk1. Hence,

– 1

ρ+ 1 ξ(t)u,|ut|

ρ

ut T S

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Similarly,

_ρ1_{+ 1}_ST ξ(t)u,|ut|ρut

dt≤k1

T

S

ξ(t)E(t)dt= –k1

T

S

ξ(t)E(t)dt. (3.7)

For the ﬁfth term on the right-hand side of (3.3), integrating by parts, we have

–

T

S

ξ(t)(∇u,∇utt)dt

= –ξ(t)∇u,∇utT_S + T

S

ξ(t)∇u_t,∇ut

dt

≤ξ(t)

1 2∇u

2₊1 2∇ut

2

T

S

+

T

S

ξ(t)∇u,∇ut

dt+

T

S

ξ(t)∇ut2dt

≤

1 1 –G(∞)+ 1

2ξ(0)E(S) +

T

S

ξ(t)E(t)dt

+

T

S

ξ(t)∇ut2dt

≤2k2ξ(0)E(S) –k2

T

S

ξ(t)E(t)dt+

T

S

ξ(t)∇ut2dt, (3.8)

with some positive constantk2.

For the sixth term on the right-hand side of (3.3), we have

∇u,

t

0

g(t–s)∇u(s) –∇u(t)ds

≤ε∇u2+ 1 4ε

Ω t

0

2

dx

≤ε∇u2+ 1 4ε

t

0

g(s)ds

Ω t

0

g(t–s)∇u(s) –∇u(t)2ds dx

≤ 2ε

1 –G(∞)E(t) +

G(∞)

2ε (g◦ ∇u)(t).

Combining with (3.4), we obtain

T

S

ξ(t)

∇u(t),

t

0

dt

≤ 2ε

1 –G(∞)

T

S

ξ(t)E(t)dt–G(∞) ε

T

S

E(t)dt. (3.9)

By (3.3)–(3.9), we get

T

S

ξ(t)E(t)dt≤

1 ρ+ 2+

1 2(ρ+ 1)

T

S

ξ(t)ut ρ+2

ρ+2dt+ 3 2

T

S

ξ(t)∇ut2dt

+ ε

1 –G(∞)

T

S

ξ(t)E(t)dt–1

2(k1+k2)

T

S

ξ(t)E(t)dt

–

1 +G(∞) 2ε

T

S

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Integrating by parts (and notingE(t)≤0), we have

–

T

S

ξ(t)E(t)dt= –ξ(t)E(t)|T_S +

T

S

ξ(t)E(t)dt

= –ξ(T)E(T) +ξ(S)E(S) +

T

S

ξ(t)E(t)dt

≤ξ(0)E(S), (3.11)

and

–

T

S

E(t)dt=E(S) –E(T)≤E(S). (3.12)

Owing to (3.10)–(3.12), we get

T

S

ξ(t)E(t)dt

≤

1 ρ+ 2+

1 2(ρ+ 1)

T

S

ξ(t)ut ρ+2

ρ+2dt+ 3 2

T

S

ξ(t)∇ut2dt

+ ε

1 –G(∞)

T

S

ξ(t)E(t)dt+

1 +G(∞)

2ε +

3

2(k1+k2)ξ(0)

E(S). (3.13)

Choosingεsmall enough, we obtain from (3.13) that

T

S

ξ(t)E(t)dt≤C

T

S

ξ(t)ut2dt+C T

S

ξ(t)∇ut2dt+CE(S).

The proof of Lemma3.2is completed.

T

S

ξ(t)utρρ+2+2dt+

T

S

ξ(t)∇ut2dt≤εC T

S

ξ(t)E(t)dt+C(ε)E(S).

Proof Multiplying byξ(t)₀tg(t–s)(u(s) –u(t))dsboth sides of equation (1.1) and then integrating the resulting equation overΩ×[S,T] (0≤S≤T) gives

T

S

ξ(t)

t

0

g(t–s) u(s) –u(t)ds,

|ut|ρutt–u–utt+ t

0

g(t–s)u(s)ds

dt= 0. (3.14)

Integrating by parts and using Lemma2.2, we obtain

T

S

|ut|ρutt,ξ(t) t

0

g(t–s) u(s) –u(t)ds

dt

= 1

ρ+ 1

|ut|ρut,ξ(t) t

0

T

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– 1 ρ+ 1

T

S

|ut|ρut,ξ(t) t

0

dt

– 1

ρ+ 1

T

S

|ut|ρut,ξ(t) t

0

dt

+ 1

ρ+ 1

T

S

ξ(t)G(t)ut(t) ρ+2

ρ+2dt.

Moreover, we have

T

S

ξ(t)

t

0

g(t–s) u(s) –u(t)ds, –utt dt = T S

ξ(t)

t

0

g(t–s)∇u(s) –∇u(t)ds,∇utt

dt

=

∇ut,ξ(t) t

0

T S – T S

∇ut,ξ(t) t

0

dt – T S

∇ut,ξ(t) t

0

dt+

T

S

ξ(t)G(t)∇ut2dt

and

T

S

ξ(t)

t

0

g(t–s) u(s) –u(t)ds, –u+

t

0

g(t–s)u(s)ds

dt = T S

ξ(t)

t

0

g(t–s)∇u(s) –∇u(t)ds,∇u(t)dx

dt – T S

ξ(t)

t

0

g(t–s) ∇u(s) –∇u(t)ds,

t

0

g(t–s)∇u(s)ds

dt

= –

T

S

ξ(t)

t

0

2

dt

+

T

S

ξ(t)1 –G(t)

t

0

g(t–s) ∇u(s) –∇u(t)ds,∇u(t)

dt.

Therefore, plugging the above three identities into (3.14), we get

1 ρ+ 1

T

S

ξ(t)G(t)ut(t) ρ+2

ρ+2dt+

T

S

ξ(t)G(t)∇ut2dt

= – 1 ρ+ 1

|ut|ρut,ξ(t) t

0

T

S

+ 1

ρ+ 1

T

S

|ut|ρut,ξ(t) t

0

g(t–s)u(s) –u(t)ds

dt

+ 1

ρ+ 1

T

S

|ut|ρut,ξ(t) t

0

dt

–

∇ut,ξ(t) t

0

T S + T S

∇ut,ξ(t) t

0

(9)

+

T

S

∇ut,ξ(t) t

0

+

T

S

ξ(t)

t

0

2

dt

–

T

S

ξ(t)1 –G(t)

t

0

g(t–s)∇u(s) –∇u(t),∇u(t)dx

dt. (3.15)

Using Cauchy and Hölder inequalities as well as Lemma2.1, we have

|ut|ρut, t

0

≤ρ+ 1 ρ+ 2ut

ρ+2

ρ+2+ 1 ρ+ 2

Ω t

0

ρ+2

dx

=ρ+ 1 ρ+ 2ut

ρ+2

ρ+2+ 1 ρ+ 2

Ω t

0

gρρ+1+2₍_t_–_s₎_gρ1+2₍_t_–_s₎_u₍_s_{) –}_u₍_t₎_ds ρ+2

dx

≤ρ+ 1 ρ+ 2ut

ρ+2

ρ+2+ 1 ρ+ 2

t

0

g(t–s)ds

ρ+1

Ω t

0

g(t–s)u(s) –u(t)ρ+2ds

dx

≤ρ+ 1 ρ+ 2ut

ρ+2

ρ+2+ 1 ρ+ 2G

ρ+1₍_t₎

t

0

g(t–s)u(s) –u(t)ρ_ρ+2₊₂ds

≤ρ+ 1 ρ+ 2ut

ρ+2

ρ+2+ 1 ρ+ 2G

ρ+1₍_t₎_Bρ+2

t

0

g(t–s)∇u(s) –∇u(t)ρ+2ds

≤ρ+ 1 ρ+ 2ut

ρ+2

ρ+2+ 1 ρ+ 2G

ρ+1₍_t₎

×Bρ+2

t

0

g(t–s) ∇u(s)+∇u(t)ρ∇u(s) –∇u(t)2ds

≤ρ+ 1 ρ+ 2ut

ρ+2

ρ+2+ 1 ρ+ 2G

ρ+1₍_t₎

×Bρ+2

t

0

g(t–s)

2

2E(0) 1 –G(∞)

1 2ρ

∇u(s) –∇u(t)2ds

=ρ+ 1 ρ+ 2ut

ρ+2

ρ+2+ 2ρ

ρ+ 2G

ρ+1₍_t₎_Bρ+2

2E(0) 1 –G(∞)

ρ

2 t

0

g(t–s)∇u(s) –∇u(t)2ds

≤(ρ+ 1)E(t) + 2

ρ

ρ+ 2G

ρ+1₍_t₎_Bρ+2

2E(0) 1 –G(∞)

ρ

2

g◦ ∇u

≤

(ρ+ 1) + 2

ρ+1

ρ+ 2G

ρ+1₍_t₎_Bρ+2

2E(0) 1 –G(∞)

ρ

2

E(t), (3.16)

which implies that

– 1

ρ+ 1

|ut|ρut,ξ(t) t

0

T

S

≤2ξ(0) ρ+ 1

(ρ+ 1) + 2

ρ+1

ρ+ 2G

ρ+1₍_t₎_Bρ+2

2E(0) 1 –G(∞)

ρ

2

E(S)

≤2ξ(0) ρ+ 1

(ρ+ 1) + 2

ρ+1

ρ+ 2G

ρ+1₍_t₎_Bρ+2

2E(0) 1 –G(∞)

ρ

2

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In order to estimate the second term on the right-hand side of (3.15), we apply (3.16) and (3.11) to get

1 ρ+ 1

T

S

|ut|ρut,ξ(t) t

0

dt

≤ 1

ρ+ 1

(ρ+ 1) + 2

ρ+1

ρ+ 2G

ρ+1₍_t₎_Bρ+2

2E(0) 1 –G(∞)

ρ

2 T

S

ξ(t)E(t)dt

= – 1 ρ+ 1

(ρ+ 1) + 2

ρ+1

ρ+ 2G

ρ+1₍_t₎_Bρ+2

2E(0) 1 –G(∞)

ρ

2 T

S

ξ(t)E(t)dt

≤ ξ(0) ρ+ 1

(ρ+ 1) + 2

ρ+1

ρ+ 2G

ρ+1₍_t₎_Bρ+2

2E(0) 1 –G(∞)

ρ

2

E(S). (3.18)

In addition, for anyδ> 0, using Young inequality, we have

T

S

|ut|ρut,ξ(t) t

0

dt = T S

ξρρ+1+2₍_t₎|_u t|ρut,ξ

1

ρ+2₍_t₎ t

0

dt

≤δ

T

S

ξ(t)utρρ+2+2dt+C(δ)

T

S

ξ(t)

Ω t

0

ρ+2

dx

dt.

Using Hölder inequality and Lemmas2.1and2.3, we have

T

S

ξ(t)

Ω t 0 g ρ+1

ρ+2₍_t_–_s₎_gρ1+2₍_t_–_s₎_u₍_s_{) –}_u₍_t₎_ds ρ+2 dx dt ≤ T S

ξ(t)

t

0

–g(t–s)dt

ρ+1

Ω t

0

g(t–s)u(s) –u(t)ρ+2ds

dx

dt

≤gρ+1(0)ξ(0)

T

S t

0

g(t–s)u(s) –u(t)ρ_ρ+2₊₂ds

dt

≤gρ+1_(0)ξ₍₀₎_Bρ+2

T

S t

0

g(t–s)∇u(s) –∇u(t)ρ_ρ+2₊₂ds

dt

≤2ρgρ+1(0)Bρ+2

2E(0) 1 –G(∞)

ρ 2 ξ(0) T S t 0

dt

= –2ρgρ+1(0)Bρ+2

2E(0) 1 –G(∞)

ρ 2 ξ(0) T S t 0

dt

≤–2ρ+1gρ+1(0)Bρ+2

2E(0) 1 –G(∞)

ρ

2

ξ(0)

T

S

E(t)dt

≤2ρ+1gρ+1(0)Bρ+2

2E(0) 1 –G(∞)

ρ

2

ξ(0)E(S).

Therefore,

T

S

|ut|ρut,ξ(t) t

0

dt

≤δ

T

S

ξ(t)ut ρ+2

ρ+2dt+ 2ρ+1C(δ)gρ+1(0)Bρ+2

2E(0) 1 –G(∞)

ρ

2

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Since

∇ut,ξ(t) t

0

≤1

2ξ(t)∇ut 2₊1

2ξ(t)

Ω t

0

2

dx

=1

2ξ(t)∇ut 2₊1

2ξ(t)G(t)

t

0

g(t–s)∇u(s) –∇u(t)2ds dx

≤ξ(0)E(t) +1

2ξ(t)G(t)g◦ ∇u≤2ξ(0)E(t), (3.20)

one obtains

–

∇ut,ξ(t) t

0

T

S

≤4ξ(0)E(S). (3.21)

Using (3.20) and (3.11), we have

T

S

∇ut,ξ(t) t

0

≤2

T

S

ξ(t)E(t)≤2ξ(0)E(S). (3.22)

Similarly,

T

S

∇ut,ξ(t) t

0

dt

≤δ 2

T

S

ξ(t)∇ut2dt

+ 1 2δ

T

S

ξ(t)

–

t

0

g(t–s)ds

·

t

0

dt

=δ 2

T

S

ξ(t)∇ut2dt–

1 2δ

T

S

ξ(t)

t

0

g(s)ds

t

0

g(t–s)∇u(s) –∇u(t)2ds dt

≤δ 2

T

S

ξ(t)∇ut2dt+

g(0) 2δ

T

S

ξ(t)

t

0

g(t–s)∇u(s) –∇u(t)2ds dt

≤δ 2

T

S

ξ(t)∇ut2dt–

g(0) δ

T

S

ξ(t)E(t)dt

≤δ 2

T

S

ξ(t)∇ut2dt+

g(0)ξ(0)

δ E(S). (3.23)

Now, sinceg(t)≥0 andG(∞) < 1, we get

T

S

ξ(t)

t

0

2

dt

≤

T

S

ξ(t)

t

0

g(s)ds

t

0

dt

≤G(∞)

T

S

ξ(t)

t

0

(12)

≤G(∞)

T

S t

0 ξ(t)

ξ(t–s)ξ(t–s)g(t–s)∇u(s) –∇u(t) 2

ds

dt

≤G(∞)

T

S t

0

ξ(t–s)g(t–s)∇u(s) –∇u(t)2ds

dt

≤–G(∞)

T

S t

0

dt

≤–2G(∞)

T

S

E(t)dt≤2G(∞)E(S)≤2E(S) (3.24)

and

–

T

S

ξ(t)1 –G(t)

t

0

g(t–s)∇u(s) –∇u(t),∇u(t)

dt

≤ε

T

S

ξ(t)∇u2dt+G(∞) 4ε

T

S

ξ(t)

t

0

dt

=ε

T

S

ξ(t)∇u2dt

+G(∞) 4ε

T

S t

0 ξ(t)

ξ(t–s)ξ(t–s)g(t–s)∇u(s) –∇u(t) 2

ds

dt

≤ε

T

S

ξ(t)∇u2dt+G(∞) 4ε

T

S t

0

ξ(t–s)g(t–s)∇u(s) –∇u(t)2ds

dt

≤ε

T

S

ξ(t)∇u2dt–G(∞) 4ε

T

S t

0

dt

≤ε

T

S

ξ(t)∇u2dt–G(∞) 2ε

T

S

E(t)dt

≤ε

T

S

ξ(t)∇u2dt+ 1

2εE(S)≤ε

T

S

ξ(t)E(t)dt+ 1

2εE(S). (3.25)

Combining (3.15)–(3.25), we obtain

1 ρ+ 1

T

S

ξ(t)G(t)utρρ+2+2dt+

T

S

ξ(t)G(t)∇ut2dt

≤δ

T

S

ξ(t)ut ρ+2

ρ+2dt+ δ 2

T

S

ξ(t)∇ut2dt+ε T

S

ξ(t)E(t)dt

+

C1+

g(0)ξ(0)

δ +

1 2ε

E(S). (3.26)

Sincegis continuous andg(0) > 0, for anyt0> 0, we have

G(t) =

t

0

g(s)ds≥

t0

0

g(s)dx> 0, ∀t≥t0.

Now, if we ﬁxδ> 0 small enough such that

δ<

t0

0

(13)

then by (3.26), forT>S≥t0, we have

T

S

ξ(t)ut ρ+2

ρ+2dt+

T

S

ξ(t)∇ut2dt≤εC T

S

ξ(t)E(t)dt+C(ε)E(S).

Proof of Lemma3.1 Plugging the estimate of Lemma3.3into the inequality of Lemma3.2, and ﬁxingεsmall enough, we obtain

T

S

ξ(t)E(t)dt≤CE(S)

for some constantC> 0. LettingT→+∞, we have

+∞

t

ψ(s)E(s)ds≤cE(t), ∀t≥t0, (3.27)

with

ψ(t) :=

t

t0

ξ(s)ds.

Proof of Theorem2.2 From Assumption2.1and Lemma2.3, we know thatE(t) is a non-increasing function andψ: [t0, +∞)→R+is a strictly increasingC2 function such that ψ(t0) = 0 andlimt→+∞ψ(t) = +∞. Firstly, we deﬁne a new functionf : [t0,∞)→R+as follows:

f(τ) =E ψ–1(τ),

thenf is a non-increasing function such that

ψ(T)

ψ(S)

f(τ)dτ =

ψ(T)

ψ(S)

E ψ–1(τ)dτ=

T

S

E(t)ψ(t)dt

≤

+∞

S

E(t)ψ(t)dt≤cE(S) =cf ψ(S), ∀t0≤S<T<∞.

Sett=ψ(S). SincelimT→+∞ψ(T) = +∞, we get

+∞

t

f(τ)dτ≤cf(t), ∀t≥t0. (3.28)

Next, we deﬁne the following function:

h(x) =e1cx

+∞

x

f(s)ds, (3.29)

wherec> 0 is a constant. Noting (3.28), we obtain

h(x) = 1

ce

1

cx

+∞

x

f(s)ds–cf(x)

(14)

Integrating (3.30) over [t0,t] and noting (3.28), we have

h(t)≤h(t0) =e

1

ct0 +∞

t0

f(s)ds≤c1f(t0). (3.31)

Furthermore, using (3.29) and (3.31), we obtain

+∞

t

f(s)ds≤c1f(t0)e–

1

ct_. _(3.32)

On the other hand, noting thatf is a positive non-increasing function, it is easy to see that

+∞

t

f(s)ds≥

t+c

t

f(s)ds≥cf(t+c). (3.33)

Combining (3.32) with (3.33), we get

f(t+c)≤c2f(t0)e–

1

ct_. _(3.34)

Lettings=t+cin (3.34), we have

f(s)≤c2f(t0)e1–

1

cs_,

that is,

E ψ–1(s)≤c2f(t0)e1–

1

cs_. _(3.35)

Moreover, lettingt=ψ–1(s) in (3.35), we get

E(t)≤c2f(t0)e1–

1

cψ(t)_,

that is,

E(t)≤C0e–ωψ(t)=C0e–ω t

t0ξ(s)ds_,

for some constantsC0andω=1_c> 0.

The proof of Theorem2.2is completed.

Remark3.1 From Theorem 2.2, if we choose diﬀerentξ(t), we can get diﬀerent decay results. Choosingξ(t)≡a, we get the exponential decay result

E(t)≤Ce–ωt, ∀t≥t0.

Now considerξ(t) =₍₁₊1_t₎γ (0 <γ ≤1). Ifγ = 1, we get the polynomial decay result

E(t)≤C(1 +t)–ω_, _ω_{> 0,}_∀_t_≥_t

0.

If 0 <γ< 1, we get a decay result of the form

(15)

4 Conclusions

In this paper, we present a weighted integral inequality to derive decay estimates for a quasilinear viscoelastic wave equation with variable density and memory. Due to the as-sumption on the memory kernel function, the weighted inequality established in this pa-per improves the integral inequality in [1]. We establish a general decay rate of the solution such that the exponential and polynomial decay results are special cases of this paper.

Acknowledgements

The authors would like to thank the referee for his/her careful reading and kind suggestions.

Funding

This work was supported by National Natural Science Foundation of China (No. 11201258).

Availability of data and materials

The data used to support the ﬁndings of this study are available from the corresponding author upon request.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

The authors contributed equally to the writing of this paper. The authors read and approved the ﬁnal manuscript.

Authors’ information

Fushan Li currently works at the School of Mathematical Sciences, Qufu Normal University, P.R. China. He does research in Applied Mathematics and Analysis. He and his group are engaged in the research on the well-posedness and longtime dynamics for some nonlinear evolution equations. Fengying Hu is studying for a Master’s degree in Qufu Normal University. She is a member of Li’s group.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations.

Received: 16 August 2018 Accepted: 21 October 2018 References

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