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

Abstract

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

$$\begin{aligned} \vert u_{t} \vert ^{\rho }u_{tt}-\Delta u-\Delta u_{tt}+ \int^{t}_{0}g(t-s)\Delta u(s)\,ds=0 \quad \text{in } \varOmega \times (0, \infty ) \end{aligned}$$

with initial conditions and boundary condition, where g is a memory kernel function and ρ is a positive constant. Depending on the properties of convolution kernel g at 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.

Introduction

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

$$\begin{aligned} \textstyle\begin{cases} \vert u_{t} \vert ^{\rho }u_{tt}-\Delta u-\Delta u_{tt}+\int^{t}_{0}g(t-s) \Delta u(s)\,ds=0 \quad \text{in } \varOmega \times (0, \infty ),\\ u=0 \quad \text{on } \partial \varOmega \times (0, \infty ), \\ u| _{t=0}=u_{0},\quad\quad u_{t}| _{t=0}=u_{1} \quad \text{in } \varOmega , \end{cases}\displaystyle \end{aligned}$$

(1.1)

where $\varOmega \subset \mathbb{R}^{n}$ ($n\geq 1$) 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with a memory term exhibit exponential decay rates when strong damping $-\Delta u_{t}$ is active, this uniform decay is no longer valid (by spectral analysis arguments) for dynamics subjected to frictional damping only, say ${g=0}$. Viscoelastic equations with variable density 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 Lyapunov 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 polynomial decay are only special cases. Due to the assumption on g, 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 energy 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 evolution equations. This research laid a good foundation 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 important result. In Sect. 3, we construct an energy inequality, prove the main Theorem 2.2 and give applications to various functions $\xi (t)$.

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

$$\begin{aligned}& G(t) = \int_{0}^{t} g(s)\,ds-\text{strength of memory}, \quad\quad g \circ h= \int_{0}^{t} g(t-s) \bigl\Vert h(s)-h(t) \bigr\Vert ^{2}\,ds, \\& g\ast h = \int_{0}^{t} g(t-s)h(s)\,ds,\quad\quad ( u, v ) = \int_{\varOmega }uv\,dx,\quad\quad \mathbb{R}^{+}:=[0,+\infty ), \\& \Vert v \Vert _{\rho +2}:= \Vert v \Vert _{L^{\rho +2}(\varOmega )},\quad\quad \Vert v \Vert := \Vert v \Vert _{L ^{2}(\varOmega )}. \end{aligned}$$

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, C denotes a generic positive constant. We impose the following assumptions on ρ and g.

Assumption 2.1

Set $G(t) =\int_{0}^{t} g(s)\,ds$. We assume that

• 1.
  $$\begin{aligned} 0< \rho \leqslant \frac{2}{n-2} \quad \text{if } n\geq 3; \quad\quad \rho >0 \quad \text{if } n=1,2, \end{aligned}$$

  which implies that

  $$\begin{aligned} H_{0}^{1}(\varOmega )\hookrightarrow L^{2(\rho +1)}( \varOmega ). \end{aligned}$$

• 2. $g: [0, \infty ) \rightarrow [0, \infty )$ is a locally absolutely continuous function such that

  $$\begin{aligned} G(\infty ) < 1,\quad\quad g(0) > 0, \quad\quad g'(t) \leqslant 0, \quad \text{for a.e. } t \geqslant 0. \end{aligned}$$

• 3. There exists a non-increasing function $\xi \in C^{1}[0,+\infty )$ with $\int_{0}^{+\infty }\xi (\tau )\,d\tau =+\infty$ satisfying

  $$\begin{aligned} g'(t)\leqslant -\xi (t)g(t),\quad\quad \xi (t)>0, \quad \forall t \geqslant 0. \end{aligned}$$

Theorem 2.1

([24])

Suppose that $( u_{0},u _{1} ) \in H^{1}_{0}(\varOmega )\times H^{1}_{0}(\varOmega )$. Under Assumption 2.1, there exists a unique solution u of (1.1) satisfying

$$\begin{aligned} u \in L^{\infty }\bigl(0, \infty ; H_{0}^{1}(\varOmega )\bigr),\quad\quad u_{t}\in L^{ \infty }\bigl(0, \infty ; H_{0}^{1}(\varOmega )\bigr),\quad\quad u_{tt}\in L^{\infty }\bigl(0, \infty ; L^{2}(\varOmega )\bigr), \end{aligned}$$

and

$$\begin{aligned} u(x,t)\rightarrow u_{0}(x)\quad \textit{in } H_{0}^{1}( \varOmega ),\quad \quad u_{t}(x,t)\rightarrow u_{1}(x) \quad \textit{in } H_{0}^{1}(\varOmega ), \textit{as } t\rightarrow 0^{+}. \end{aligned}$$

Lemma 2.1

([4, 5, 8])

$H_{0}^{1}(\varOmega )\hookrightarrow L^{r}(\varOmega )$ with

$$\begin{aligned} r: \textstyle\begin{cases} 0< r\leqslant \frac{2n}{n-2}, & n\geqslant 3, \\ \geqslant 2, & n=1,2, \end{cases}\displaystyle \end{aligned}$$

which implies

$$\begin{aligned} \Vert \varphi \Vert _{r}\leqslant B \Vert \nabla \varphi \Vert _{2},\quad \forall \varphi \in H_{0}^{1}( \varOmega ). \end{aligned}$$

Lemma 2.2

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

$$\begin{aligned} \bigl( \vert u_{t} \vert ^{\rho }u_{tt}, w \bigr) =\frac{1}{\rho +1}\frac{d}{dt} \bigl( \vert u_{t} \vert ^{\rho }u_{t},w \bigr) -\frac{1}{\rho +1} \bigl( \vert u_{t} \vert ^{ \rho }u_{t},w_{t} \bigr) . \end{aligned}$$

Proof

From

$$\begin{aligned} \frac{d}{dt} \bigl( \vert u_{t} \vert ^{\rho }u_{t},w \bigr) &=\frac{d}{dt} \bigl( \bigl( u_{t}^{2} \bigr) ^{\frac{\rho }{2}}u_{t},w \bigr) \\ &= \bigl( \vert u_{t} \vert ^{\rho }u_{tt}, w \bigr) +\frac{\rho }{2} \bigl( \bigl( u_{t}^{2} \bigr) ^{\frac{\rho -2}{2}}2u_{t}u_{tt}u_{t},w \bigr) + \bigl( \vert u_{t} \vert ^{\rho }u_{t},w_{t} \bigr) \\ &=(\rho +1) \bigl( \vert u_{t} \vert ^{\rho }u_{tt}, w \bigr) + \bigl( \vert u_{t} \vert ^{ \rho }u_{t},w_{t} \bigr) , \end{aligned}$$

we have

$$\begin{aligned} \bigl( \vert u_{t} \vert ^{\rho }u_{tt}, w \bigr) =\frac{1}{\rho +1}\frac{d}{dt} \bigl( \vert u_{t} \vert ^{\rho }u_{t},w \bigr) -\frac{1}{\rho +1} \bigl( \vert u_{t} \vert ^{ \rho }u_{t},w_{t} \bigr) . \end{aligned}$$

 □

Lemma 2.3

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

$$\begin{aligned} \frac{d}{dt}E(t) =\frac{1}{2}g'\circ \nabla u(t)- \frac{1}{2}g(t) \Vert \nabla u \Vert ^{2}, \end{aligned}$$

where

$$\begin{aligned} E(t)=\frac{1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+ \frac{1}{2}\bigl(1-G(t)\bigr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}+\frac{1}{2} \Vert \nabla u_{t} \Vert ^{2} +\frac{1}{2}(g \circ \nabla u) (t). \end{aligned}$$

Proof

Multiplying equation (1.1) by $u_{t}$, integrating by parts over Ω and using Lemma 2.2, we obtain the conclusion. □

Our main result is the following decay theorem.

Theorem 2.2

Let u be the global solution of problem (1.1) with Assumption 2.1. We define the energy functional as

$$\begin{aligned} E(t)=\frac{1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+ \frac{1}{2}\bigl(1-G(t)\bigr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}+\frac{1}{2} \Vert \nabla u_{t} \Vert ^{2} +\frac{1}{2}(g \circ \nabla u) (t). \end{aligned}$$

Then, for some $t_{0}>0$ there exist positive constants $C_{0}$ and ω such that

$$\begin{aligned} E(t)\leqslant C_{0}e^{-\omega \int_{t_{0}}^{t}\xi (s)\,ds}. \end{aligned}$$

The proof of main result

In order to derive the desired result of Theorem 2.2 by the integral method, we establish the following weighted integral inequality.

Lemma 3.1

Let u be the solution of (1.1) under Assumption 2.1, then

$$\begin{aligned} \int_{S}^{T}\xi (t)E(t)\,dt \leqslant C E(S) \end{aligned}$$

for some constant $C>0$.

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

Lemma 3.2

Let u be the solution of (1.1) under Assumption 2.1, then

$$\begin{aligned} \int_{S}^{T}\xi (t)E(t)\,dt \leqslant & C \int_{S}^{T}\xi (t) \Vert u_{t} \Vert ^{2}\,dt+ C \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt+CE(S). \end{aligned}$$

Proof

Multiplying by $\xi (t)u(t)$ both sides of equation (1.1), integrating the resulting equation over $\varOmega \times [S,T]$ ($0\leqslant S\leqslant T$), then using the boundary conditions and Lemma 2.2, we have

$$\begin{aligned} 0={}& \int_{S}^{T}\xi (t) \biggl( u, \vert u_{t} \vert ^{\rho }u_{tt}-\Delta u-\Delta u_{tt}+ \int^{t}_{0}g(t-s)\Delta u(s)\,ds \biggr) \,dt \\ ={}& \int_{S}^{T}\xi (t) \bigl( u, \vert u_{t} \vert ^{\rho }u_{tt} \bigr) \,dt+ \int _{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt+ \int_{S}^{T}\xi (t) ( \nabla u, \nabla u_{tt} ) \,dt \\ & {} - \int_{S}^{T}\xi (t) \biggl( \nabla u, \int^{t}_{0}g(t-s)\nabla u(s)\,ds \biggr) \,dt \\ ={}& \int_{S}^{T}\xi (t) \bigl( u, \vert u_{t} \vert ^{\rho }u_{tt} \bigr) \,dt + \int _{S}^{T}\xi (t) \bigl( 1-G(t) \bigr) \Vert \nabla u \Vert ^{2}\,dt+ \int_{S}^{T} \xi (t) ( \nabla u,\nabla u_{tt} ) \,dt \\ & {} - \int_{S}^{T}\xi (t) \biggl( \nabla u, \int^{t}_{0}g(t-s) \bigl( \nabla u(s)-\nabla u(t) \bigr) \,ds \biggr) \,dt. \end{aligned}$$

(3.1)

According to the definition of the energy functional $E(t)$, we get

$$\begin{aligned} \bigl( 1-G(t) \bigr) \Vert \nabla u \Vert ^{2}=2E(t)-\frac{2}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}- \Vert \nabla u_{t} \Vert ^{2}-g\circ \nabla u(t). \end{aligned}$$

(3.2)

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

$$\begin{aligned} \int_{S}^{T}\xi (t)E(t)\,dt ={}& \frac{1}{\rho +2} \int_{S}^{T}\xi (t) \Vert u _{t} \Vert ^{\rho +2}_{\rho +2}\,dt+\frac{1}{2} \int_{S}^{T}\xi (t) \Vert \nabla u _{t} \Vert ^{2}\,dt \\ & {}+\frac{1}{2} \int_{S}^{T}\xi (t) (g\circ \nabla u) (t)\,dt \\ & {} -\frac{1}{2} \int_{S}^{T}\xi (t) \bigl( u, \vert u_{t} \vert ^{\rho }u_{tt} \bigr) \,dt - \frac{1}{2} \int_{S}^{T}\xi (t) ( \nabla u,\nabla u_{tt} ) \,dt \\ & {} +\frac{1}{2} \int_{S}^{T}\xi (t) \biggl( \nabla u, \int^{t}_{0}g(t-s) \bigl( \nabla u(s)-\nabla (t) \bigr) \,ds \biggr) \,dt. \end{aligned}$$

(3.3)

From Lemma 2.3, we see that

$$\begin{aligned} \frac{d}{dt}E(t) =\frac{1}{2}g'\circ \nabla u(t)- \frac{1}{2}g(t) \Vert \nabla u \Vert ^{2} \leqslant \frac{1}{2}g'\circ \nabla u(t), \end{aligned}$$

that is,

$$\begin{aligned} -g'\circ \nabla u(t)\leqslant -2E'(t), \end{aligned}$$

which, together with Assumption 2.1, implies

$$\begin{aligned} \int_{S}^{T}\xi (t) ( g\circ \nabla u ) (t)\,dt \leqslant - \int _{S}^{T} \bigl( g'\circ \nabla u \bigr) (t)\,dt\leqslant -2 \int_{S}^{T}E'(t)\,dt. \end{aligned}$$

(3.4)

For the fourth term on the right-hand side of (3.3), integrating by parts and using Lemma 2.2, we have

$$\begin{aligned} - \int_{S}^{T}\xi (t) \bigl( u, \vert u_{t} \vert ^{\rho }u_{tt} \bigr) \,dt ={} & - \frac{1}{\rho +1} \bigl( \xi (t)u, \vert u_{t} \vert ^{\rho }u_{t} \bigr)\big| _{S}^{T}+ \frac{1}{\rho +1} \int_{S}^{T} \bigl( \bigl(\xi (t)u \bigr)_{t}, \vert u _{t} \vert ^{\rho }u_{t} \bigr) \,dt \\ ={}& -\frac{1}{\rho +1} \bigl( \xi (t)u, \vert u_{t} \vert ^{\rho }u_{t} \bigr)\big| _{S}^{T}+ \frac{1}{\rho +1} \int_{S}^{T} \bigl( \xi '(t)u, \vert u_{t} \vert ^{\rho }u_{t} \bigr) \,dt \\ & {} +\frac{1}{\rho +1} \int_{S}^{T}\xi (t) \Vert u_{t} \Vert _{\rho +2}^{\rho +2}\,dt. \end{aligned}$$

(3.5)

By Young inequality, Lemma 2.1 and the definition of $E(t)$, we have

$$\begin{aligned} \biggl\vert -\frac{1}{\rho +1} \bigl( \xi (t)u, \vert u_{t} \vert ^{\rho }u_{t} \bigr) \biggr\vert \leqslant{} & \frac{1}{\rho +1} \biggl[ \xi (t) \biggl( \frac{1}{ \rho +2} \Vert u \Vert _{\rho +2}^{\rho +2}+\frac{\rho +1}{\rho +2} \Vert u_{t} \Vert _{\rho +2}^{\rho +2} \biggr) \biggr] \\ \leqslant {}& \xi (t) \biggl( \frac{1}{(\rho +1)(\rho +2)} \biggl( \frac{2}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}+1}B^{\rho +2}E^{\frac{\rho }{2}}(0)+1 \biggr) E(t) \\ \leqslant {}& k_{1}\xi (t)E(t), \end{aligned}$$

with some positive constant $k_{1}$. Hence,

$$\begin{aligned} \biggl\vert -\frac{1}{\rho +1} \bigl( \xi (t)u, \vert u_{t} \vert ^{\rho }u_{t} \bigr)\big| _{S}^{T} \biggr\vert \leqslant 2k_{1}\xi (0)E(S). \end{aligned}$$

(3.6)

Similarly,

$$\begin{aligned} \biggl\vert \frac{1}{\rho +1} \int_{S}^{T} \bigl( \xi '(t) u, \vert u_{t} \vert ^{\rho }u _{t} \bigr) \,dt \biggr\vert \leqslant {}& k_{1} \int_{S}^{T} \bigl\vert \xi '(t) \bigr\vert E(t)\,dt=-k _{1} \int_{S}^{T}\xi '(t)E(t)\,dt. \end{aligned}$$

(3.7)

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

$$\begin{aligned} &- \int_{S}^{T}\xi (t) ( \nabla u,\nabla u_{tt} ) \,dt \\ &\quad =- \bigl( \xi (t)\nabla u,\nabla u_{t} \bigr)\big| _{S}^{T}+ \int_{S} ^{T} \bigl( \bigl( \xi (t)\nabla u \bigr) _{t},\nabla u_{t} \bigr) \,dt \\ & \quad \leqslant \xi (t) \biggl( \frac{1}{2} \Vert \nabla u \Vert ^{2}+ \frac{1}{2} \Vert \nabla u_{t} \Vert ^{2} \biggr)\bigg| _{S}^{T}+ \int_{S}^{T} \bigl( \xi '(t)\nabla u, \nabla u_{t} \bigr) \,dt + \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt \\ &\quad \leqslant \biggl( \frac{1}{1-G(\infty )}+1 \biggr) \biggl[ 2\xi (0)E(S)+ \int_{S}^{T} \bigl\vert \xi '(t) \bigr\vert E(t)\,dt \biggr] + \int_{S}^{T}\xi (t) \Vert \nabla u _{t} \Vert ^{2}\,dt \\ &\quad \leqslant 2 k_{2}\xi (0)E(S)-k_{2} \int_{S}^{T}\xi '(t)E(t)\,dt+ \int _{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt, \end{aligned}$$

(3.8)

with some positive constant $k_{2}$.

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

$$\begin{aligned} & \biggl( \nabla u, \int_{0}^{t}g(t-s) \bigl(\nabla u(s)-\nabla u(t) \bigr)\,ds \biggr) \\ &\quad \leqslant \varepsilon \Vert \nabla u \Vert ^{2}+ \frac{1}{4\varepsilon } \int_{\varOmega } \biggl( \int_{0}^{t} g(t-s) \bigl(\nabla u(s)-\nabla u(t) \bigr)\,ds \biggr) ^{2}\,dx \\ &\quad \leqslant \varepsilon \Vert \nabla u \Vert ^{2}+ \frac{1}{4\varepsilon } \int_{0}^{t}g(s)\,ds \int_{\varOmega } \int_{0}^{t}g(t-s) \bigl\vert \nabla u(s)- \nabla u(t) \bigr\vert ^{2}\,ds\,dx \\ &\quad \leqslant \frac{2\varepsilon }{1-G(\infty )} E(t)+\frac{G( \infty )}{2\varepsilon }(g\circ \nabla u) (t). \end{aligned}$$

Combining with (3.4), we obtain

$$\begin{aligned} & \int_{S}^{T}\xi (t) \biggl( \nabla u(t), \int_{0}^{t}g(t-s) \bigl( \nabla u(s)-\nabla u(t) \bigr) \,ds \biggr) \,dt \\ &\quad \leqslant \frac{2\varepsilon }{1-G(\infty )} \int_{S}^{T} \xi (t)E(t)\,dt -\frac{G(\infty )}{\varepsilon } \int_{S}^{T}E'(t)\,dt. \end{aligned}$$

(3.9)

By (3.3)–(3.9), we get

$$\begin{aligned} \int_{S}^{T}\xi (t)E(t)\,dt \leqslant {}& \biggl( \frac{1}{\rho +2}+\frac{1}{2( \rho +1)} \biggr) \int_{S}^{T}\xi (t) \Vert u_{t} \Vert _{\rho +2}^{\rho +2}\,dt+ \frac{3}{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt \\ & {} +\frac{\varepsilon }{1-G(\infty )} \int_{S}^{T}\xi (t)E(t)\,dt- \frac{1}{2} ( k_{1}+k_{2} ) \int_{S}^{T}\xi '(t)E(t)\,dt \\ & {} - \biggl( 1+\frac{G(\infty )}{2\varepsilon } \biggr) \int_{S}^{T}E'(t)\,dt+ ( k_{1}+k_{2} ) \xi (0)E(S). \end{aligned}$$

(3.10)

Integrating by parts (and noting $E'(t)\leqslant 0$), we have

$$\begin{aligned} - \int_{S}^{T}\xi '(t)E(t)\,dt &= -\xi (t)E(t)| _{S}^{T}+ \int _{S}^{T}\xi (t)E'(t)\,dt \\ &= -\xi (T)E(T)+\xi (S)E(S)+ \int_{S}^{T}\xi (t)E'(t)\,dt \\ &\leqslant \xi (0)E(S), \end{aligned}$$

(3.11)

and

$$\begin{aligned} - \int_{S}^{T}E'(t)\,dt=E(S)-E(T) \leqslant E(S). \end{aligned}$$

(3.12)

Owing to (3.10)–(3.12), we get

$$\begin{aligned} & \int_{S}^{T}\xi (t)E(t)\,dt \\ &\quad \leqslant \biggl( \frac{1}{\rho +2}+\frac{1}{2( \rho +1)} \biggr) \int_{S}^{T}\xi (t) \Vert u_{t} \Vert _{\rho +2}^{\rho +2}\,dt+ \frac{3}{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt \\ & \quad\quad {} +\frac{\varepsilon }{1-G(\infty )} \int_{S}^{T}\xi (t)E(t)\,dt + \biggl( 1+ \frac{G(\infty )}{2\varepsilon }+\frac{3}{2}(k_{1}+k_{2})\xi (0) \biggr) E(S). \end{aligned}$$

(3.13)

Choosing ε small enough, we obtain from (3.13) that

$$\begin{aligned} \int_{S}^{T}\xi (t)E(t)\,dt \leqslant & C \int_{S}^{T}\xi (t) \Vert u_{t} \Vert ^{2}\,dt+ C \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt+CE(S). \end{aligned}$$

The proof of Lemma 3.2 is completed. □

Lemma 3.3

Let u be the solution of (1.1) under Assumption 2.1, then

$$\begin{aligned} \int_{S}^{T}\xi (t) \Vert u_{t} \Vert _{\rho +2}^{\rho +2}\,dt+ \int_{S}^{T} \xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt\leqslant \varepsilon C \int_{S}^{T} \xi (t)E(t)\,dt+ C(\varepsilon )E(S). \end{aligned}$$

Proof

Multiplying by $\xi (t)\int_{0}^{t}g(t-s) ( u(s)-u(t) ) \,ds$ both sides of equation (1.1) and then integrating the resulting equation over $\varOmega \times [S,T]$ ($0\leqslant S\leqslant T$) gives

$$\begin{aligned} \begin{aligned}[b] & \int_{S}^{T} \biggl( \xi (t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds, \\ &\quad \vert u _{t} \vert ^{\rho }u_{tt}-\Delta u-\Delta u_{tt}+ \int^{t}_{0}g(t-s)\Delta u(s)\,ds \biggr) \,dt=0.\end{aligned} \end{aligned}$$

(3.14)

Integrating by parts and using Lemma 2.2, we obtain

$$\begin{aligned} &\int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho }u_{tt},\xi (t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad =\frac{1}{\rho +1} \biggl( \vert u_{t} \vert ^{\rho }u_{t}, \xi (t) \int _{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr)\bigg| _{S}^{T} \\ &\quad \quad {} -\frac{1}{\rho +1} \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho } u_{t}, \xi '(t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad \quad {} -\frac{1}{\rho +1} \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho }u_{t}, \xi (t) \int_{0}^{t}g'(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad \quad {} +\frac{1}{\rho +1} \int_{S}^{T} \bigl( \xi (t)G(t) \bigr) \bigl\Vert u _{t}(t) \bigr\Vert _{\rho +2}^{\rho +2}\,dt. \end{aligned}$$

Moreover, we have

$$\begin{aligned} &\int_{S}^{T} \biggl( \xi (t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds,- \Delta u_{tt} \biggr) \,dt \\ &\quad = \int_{S}^{T} \biggl( \xi (t) \int_{0}^{t}g(t-s) \bigl( \nabla u(s)- \nabla u(t) \bigr) \,ds,\nabla u_{tt} \biggr) \,dt \\ &\quad = \biggl( \nabla u_{t},\xi (t) \int^{t}_{0}g(t-s) \bigl(\nabla u(s)- \nabla u(t) \bigr)\,ds \biggr)\bigg| _{S}^{T} \\ & \quad\quad {} - \int_{S}^{T} \biggl( \nabla u_{t},\xi '(t) \int^{t}_{0}g(t-s) \bigl(\nabla u(s)-\nabla u(t) \bigr)\,ds \biggr) \,dt \\ &\quad\quad {} - \int_{S}^{T} \biggl( \nabla u_{t},\xi (t) \int^{t}_{0}g'(t-s) \bigl(\nabla u(s)- \nabla u(t)\bigr)\,ds \biggr) \,dt + \int_{S}^{T} \xi (t)G(t) \Vert \nabla u_{t} \Vert ^{2}\,dt \end{aligned}$$

and

$$\begin{aligned} & \int_{S}^{T} \biggl( \xi (t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds,- \Delta u+ \int_{0}^{t} g(t-s)\Delta u(s)\,ds \biggr) \,dt \\ &\quad = \int_{S}^{T} \biggl( \xi (t) \int_{0}^{t}g(t-s) \bigl( \nabla u(s)- \nabla u(t) \bigr) \,ds,\nabla u(t)\,dx \biggr) \,dt \\ &\quad \quad {} - \int_{S}^{T} \biggl( \xi (t) \int_{0}^{t}g(t-s) \bigl( \nabla u(s)- \nabla u(t) \bigr) \,ds, \int_{0}^{t} g(t-s)\nabla u(s)\,ds \biggr) \,dt \\ &\quad =- \int_{S}^{T}\xi (t) \biggl\Vert \int_{0}^{t}g(t-s) \bigl( \nabla u(s)- \nabla u(t) \bigr) \,ds \biggr\Vert ^{2}\,dt \\ &\quad \quad {} + \int_{S}^{T}\xi (t) \bigl( 1-G(t) \bigr) \biggl( \int_{0}^{t}g(t-s) \bigl( \nabla u(s)-\nabla u(t) \bigr) \,ds,\nabla u(t) \biggr) \,dt. \end{aligned}$$

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

$$\begin{aligned} &\frac{1}{\rho +1} \int_{S}^{T} \bigl( \xi (t)G(t) \bigr) \bigl\Vert u_{t}(t) \bigr\Vert _{\rho +2}^{\rho +2}\,dt+ \int_{S}^{T} \xi (t)G(t) \Vert \nabla u_{t} \Vert ^{2}\,dt \\ &\quad = -\frac{1}{\rho +1} \biggl( \vert u_{t} \vert ^{\rho }u_{t},\xi (t) \int _{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr)\bigg| _{S}^{T} \\ & \quad\quad {} +\frac{1}{\rho +1} \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho }u_{t},\xi '(t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ & \quad\quad {} +\frac{1}{\rho +1} \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho } u_{t},\xi (t) \int_{0}^{t}g'(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ & \quad\quad {} - \biggl( \nabla u_{t},\xi (t) \int^{t}_{0}g(t-s) \bigl(\nabla u(s)- \nabla u(t) \bigr)\,ds \biggr)\bigg| _{S}^{T} \\ & \quad\quad {} + \int_{S}^{T} \biggl( \nabla u_{t},\xi '(t) \int^{t}_{0}g(t-s) \bigl(\nabla u(s)-\nabla u(t) \bigr)\,ds \biggr) \\ & \quad\quad {} + \int_{S}^{T} \biggl( \nabla u_{t},\xi (t) \int^{t}_{0}g'(t-s) \bigl(\nabla u(s)- \nabla u(t)\bigr)\,ds \biggr) \\ & \quad\quad {} + \int_{S}^{T}\xi (t) \biggl\Vert \int_{0}^{t}g(t-s) \bigl( \nabla u(s)- \nabla u(t) \bigr) \,ds \biggr\Vert ^{2}\,dt \\ &\quad\quad {} - \int_{S}^{T}\xi (t) \bigl( 1-G(t) \bigr) \biggl( \int_{0}^{t}g(t-s) \bigl( \nabla u(s)-\nabla u(t) \bigr) ,\nabla u(t)\,dx \biggr) \,dt. \end{aligned}$$

(3.15)

Using Cauchy and Hölder inequalities as well as Lemma 2.1, we have

$$\begin{aligned} & \biggl\vert \biggl( \vert u_{t} \vert ^{\rho }u_{t}, \int_{0}^{t} g(t-s) \bigl(u(s)-u(t)\bigr) \,ds \biggr) \biggr\vert \\ &\quad \leqslant \frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+\frac{1}{ \rho +2} \int_{\varOmega } \biggl( \int_{0}^{t} g(t-s) \bigl\vert u(s)-u(t) \bigr\vert \,ds \biggr) ^{\rho +2}\,dx \\ &\quad =\frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+ \frac{1}{ \rho +2} \int_{\varOmega } \biggl( \int_{0}^{t} g^{\frac{\rho +1}{\rho +2}}(t-s)g ^{\frac{1}{\rho +2}}(t-s) \bigl\vert u(s)-u(t) \bigr\vert \,ds \biggr) ^{\rho +2}\,dx \\ &\quad \leqslant \frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+\frac{1}{ \rho +2} \biggl( \int_{0}^{t} g(t-s)\,ds \biggr) ^{\rho +1} \int_{\varOmega } \biggl( \int_{0}^{t} g(t-s) \bigl\vert u(s)-u(t) \bigr\vert ^{\rho +2} \,ds \biggr) \,dx \\ &\quad \leqslant \frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+\frac{1}{ \rho +2}G^{\rho +1}(t) \int_{0}^{t} g(t-s) \bigl\Vert u(s)- u(t) \bigr\Vert ^{\rho +2}_{ \rho +2} \,ds \\ &\quad \leqslant \frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+\frac{1}{ \rho +2}G^{\rho +1}(t)B^{\rho +2} \int_{0}^{t} g(t-s) \bigl\Vert \nabla u(s)- \nabla u(t) \bigr\Vert ^{\rho +2} \,ds \\ &\quad \leqslant \frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+\frac{1}{ \rho +2}G^{\rho +1}(t) \\ &\quad \quad{}\times B^{\rho +2} \int_{0}^{t} g(t-s) \bigl( \bigl\Vert \nabla u(s) \bigr\Vert + \bigl\Vert \nabla u(t) \bigr\Vert \bigr) ^{\rho } \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2} \,ds \\ &\quad \leqslant \frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+\frac{1}{ \rho +2}G^{\rho +1}(t) \\ &\quad \quad{}\times B^{\rho +2} \int_{0}^{t} g(t-s) \biggl[ 2 \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{1}{2}} \biggr] ^{\rho } \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \\ &\quad =\frac{\rho +1}{\rho +2} \Vert u_{t} \Vert ^{\rho +2}_{\rho +2}+ \frac{2^{ \rho }}{\rho +2}G^{\rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}} \int_{0}^{t} g(t-s) \bigl\Vert \nabla u(s)- \nabla u(t) \bigr\Vert ^{2} \,ds \\ &\quad \leqslant (\rho +1)E(t)+\frac{2^{\rho }}{\rho +2}G^{\rho +1}(t)B ^{\rho +2} \biggl( \frac{2E(0)}{1-G(\infty )} \biggr) ^{\frac{\rho }{2}} g\circ \nabla u \\ &\quad \leqslant \biggl( (\rho +1)+\frac{{2^{\rho +1}}}{\rho +2}G^{ \rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G(\infty )} \biggr) ^{\frac{ \rho }{2}} \biggr) E(t), \end{aligned}$$

(3.16)

which implies that

$$\begin{aligned} & {-} \frac{1}{\rho +1} \biggl( \vert u_{t} \vert ^{\rho }u_{t},\xi (t) \int _{0}^{t} g(t-s) \bigl(u(s)-u(t)\bigr) \,ds \biggr)\bigg| _{S}^{T} \\ &\quad \leqslant \frac{2\xi (0)}{\rho +1} \biggl( (\rho +1)+\frac{ {2^{\rho +1}}}{\rho +2}G^{\rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}} \biggr) E(S) \\ &\quad \leqslant \frac{2\xi (0)}{\rho +1} \biggl( (\rho +1)+\frac{ {2^{\rho +1}}}{\rho +2}G^{\rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}} \biggr) E(S). \end{aligned}$$

(3.17)

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

$$\begin{aligned} &\frac{1}{\rho +1} \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho } u_{t},\xi '(t) \int_{0}^{t}g(t-s) \bigl( u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad \leqslant \frac{1}{\rho +1} \biggl( (\rho +1)+\frac{{2^{\rho +1}}}{ \rho +2}G^{\rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G(\infty )} \biggr) ^{\frac{\rho }{2}} \biggr) \int_{S} ^{T} \bigl\vert \xi '(t) \bigr\vert E(t)\,dt \\ &\quad = -\frac{1}{\rho +1} \biggl( (\rho +1)+\frac{{2^{\rho +1}}}{ \rho +2}G^{\rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G(\infty )} \biggr) ^{\frac{\rho }{2}} \biggr) \int_{S} ^{T}\xi '(t)E(t)\,dt \\ &\quad \leqslant \frac{\xi (0)}{\rho +1} \biggl( (\rho +1)+\frac{{2^{ \rho +1}}}{\rho +2}G^{\rho +1}(t)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}} \biggr) E(S). \end{aligned}$$

(3.18)

In addition, for any $\delta >0$, using Young inequality, we have

$$\begin{aligned} & \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho }u_{t}, \xi (t) \int_{0}^{t} g'(t-s) \bigl(u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad = \int_{S}^{T} \biggl( \xi^{\frac{\rho +1}{\rho +2}}(t) \vert u_{t} \vert ^{ \rho }u_{t}, \xi^{\frac{1}{\rho +2}}(t) \int_{0}^{t} g'(t-s) \bigl(u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad \leqslant \delta \int_{S}^{T}\xi (t) \Vert u_{t} \Vert ^{\rho +2}_{ \rho +2} \,dt+C(\delta ) \int_{S}^{T}\xi (t) \biggl[ \int_{\varOmega } \biggl( \int_{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\vert u(s)-u(t) \bigr\vert \,ds \biggr) ^{\rho +2}\,dx \biggr] \,dt. \end{aligned}$$

Using Hölder inequality and Lemmas 2.1 and 2.3, we have

$$\begin{aligned} & \int_{S}^{T}\xi (t) \biggl[ \int_{\varOmega } \biggl( \int_{0}^{t} \bigl\vert g' \bigr\vert ^{\frac{ \rho +1}{\rho +2}}(t-s) \bigl\vert g' \bigr\vert ^{\frac{1}{\rho +2}}(t-s) \bigl\vert u(s)-u(t) \bigr\vert \,ds \biggr) ^{\rho +2}\,dx \biggr] \,dt \\ &\quad \leqslant \int_{S}^{T} \xi (t) \biggl( \int_{0}^{t} -g'(t-s)\,dt \biggr) ^{\rho +1} \biggl[ \int_{\varOmega } \biggl( \int_{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\vert u(s)-u(t) \bigr\vert ^{\rho +2}\,ds \biggr) \,dx \biggr] \,dt \\ &\quad \leqslant g^{\rho +1}(0)\xi (0) \int_{S}^{T} \biggl( \int_{0} ^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\Vert u(s)-u(t) \bigr\Vert ^{\rho +2}_{\rho +2} \,ds \biggr) \,dt \\ &\quad \leqslant g^{\rho +1}(0)\xi (0)B^{\rho +2} \int_{S}^{T} \biggl( \int_{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{\rho +2}_{\rho +2} \,ds \biggr) \,dt \\ &\quad \leqslant 2^{\rho }g^{\rho +1}(0)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}}\xi (0) \int_{S}^{T} \biggl( \int _{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad =-2^{\rho }g^{\rho +1}(0)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}}\xi (0) \int_{S}^{T} \biggl( \int _{0}^{t} g'(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2} \,ds \biggr) \,dt \\ &\quad \leqslant -2^{\rho +1}g^{\rho +1}(0)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}}\xi (0) \int_{S}^{T}E'(t)\,dt \\ &\quad \leqslant 2^{\rho +1}g^{\rho +1}(0)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}}\xi (0) E(S). \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned}[b] & \int_{S}^{T} \biggl( \vert u_{t} \vert ^{\rho }u_{t}, \xi (t) \int_{0}^{t} g'(t-s) \bigl(u(s)-u(t) \bigr) \,ds \biggr) \,dt \\ &\quad \leqslant \delta \int_{S}^{T}\xi (t) \Vert u_{t} \Vert ^{\rho +2}_{ \rho +2} \,dt+2^{\rho +1}C(\delta )g^{\rho +1}(0)B^{\rho +2} \biggl( \frac{2E(0)}{1-G( \infty )} \biggr) ^{\frac{\rho }{2}}\xi (0) E(S). \end{aligned} \end{aligned}$$

(3.19)

Since

$$\begin{aligned} & \biggl\vert \biggl( \nabla u_{t}, \xi (t) \int_{0}^{t} g(t-s) \bigl(\nabla u(s)- \nabla u(t) \bigr) \,ds \biggr) \biggr\vert \\ &\quad \leqslant \frac{1}{2}\xi (t) \Vert \nabla u_{t} \Vert ^{2}+\frac{1}{2} \xi (t) \int_{\varOmega } \biggl( \int_{0}^{t} g(t-s) \bigl\vert \nabla u(s)-\nabla u(t) \bigr\vert \,ds \biggr) ^{2}\,dx \\ &\quad =\frac{1}{2}\xi (t) \Vert \nabla u_{t} \Vert ^{2}+\frac{1}{2}\xi (t)G(t) \int_{0}^{t} g(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2} \,ds\,dx \\ &\quad \leqslant \xi (0)E(t)+\frac{1}{2}\xi (t)G(t)g\circ \nabla u \leqslant 2\xi (0)E(t), \end{aligned}$$

(3.20)

one obtains

$$\begin{aligned} - \biggl( \nabla u_{t},\xi (t) \int_{0}^{t} g(t-s) \bigl(\nabla u(s)- \nabla u(t) \bigr) \,ds \biggr)\bigg| _{S}^{T}\leqslant 4\xi (0)E(S). \end{aligned}$$

(3.21)

Using (3.20) and (3.11), we have

$$\begin{aligned} \int_{S}^{T} \biggl( \nabla u_{t},\xi '(t) \int^{t}_{0}g(t-s) \bigl(\nabla u(s)- \nabla u(t) \bigr)\,ds \biggr) \leqslant 2 \int_{S}^{T} \bigl\vert \xi '(t) \bigr\vert E(t)\leqslant 2\xi (0)E(S). \end{aligned}$$

(3.22)

Similarly,

$$\begin{aligned} & \int_{S}^{T} \biggl( \nabla u_{t}, \xi (t) \int_{0}^{t} g'(t-s) \bigl(\nabla u(s)- \nabla u(t)\bigr) \,ds \biggr) \,dt \\ &\quad \leqslant \frac{\delta }{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2} \,dt \\ & \quad \quad {} +\frac{1}{2\delta } \int_{S}^{T} \xi (t) \biggl( - \int_{0}^{t}g'(t-s)\,ds \biggr) \cdot \biggl( \int_{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2} \,ds \biggr) \,dt \\ &\quad =\frac{\delta }{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2} \,dt-\frac{1}{2 \delta } \int_{S}^{T}\xi (t) \int_{0}^{t} g'(s) \,ds \int_{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2} \,ds\,dt \\ &\quad \leqslant \frac{\delta }{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2} \,dt+\frac{g(0)}{2\delta } \int_{S}^{T}\xi (t) \int_{0}^{t} \bigl\vert g'(t-s) \bigr\vert \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2} \,ds\,dt \\ &\quad \leqslant \frac{\delta }{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2} \,dt-\frac{g(0)}{\delta } \int_{S}^{T}\xi (t)E'(t)\,dt \\ &\quad \leqslant \frac{\delta }{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2} \,dt+\frac{g(0)\xi (0)}{\delta }E(S). \end{aligned}$$

(3.23)

Now, since $g(t)\geqslant 0$ and $G(\infty )<1$, we get

$$\begin{aligned} & \int_{S}^{T}\xi (t) \biggl\Vert \int_{0}^{t}g(t-s) \bigl( \nabla u(s)- \nabla u(t) \bigr) \,ds \biggr\Vert ^{2}\,dt \\ &\quad \leqslant \int_{S}^{T}\xi (t) \biggl( \int_{0}^{t}g(s)\,ds \biggr) \biggl( \int_{0}^{t}g(t-s) \bigl\Vert \nabla u(s)- \nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant G(\infty ) \int_{S}^{T}\xi (t) \biggl( \int_{0}^{t}g(t-s) \bigl\Vert \nabla u(s)- \nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant G(\infty ) \int_{S}^{T} \biggl( \int_{0}^{t}\frac{ \xi (t)}{\xi (t-s)}\xi (t-s)g(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant G(\infty ) \int_{S}^{T} \biggl( \int_{0}^{t}\xi (t-s)g(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant -G(\infty ) \int_{S}^{T} \biggl( \int_{0}^{t}g'(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant - 2G(\infty ) \int_{S}^{T}E'(t)\,dt\leqslant 2G( \infty )E(S)\leqslant 2E(S) \end{aligned}$$

(3.24)

and

$$\begin{aligned} & {-} \int_{S}^{T}\xi (t) \bigl( 1-G(t) \bigr) \biggl( \int_{0}^{t}g(t-s) \bigl( \nabla u(s)-\nabla u(t) \bigr) ,\nabla u(t) \biggr) \,dt \\ &\quad \leqslant \varepsilon \int_{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt+\frac{G( \infty )}{4\varepsilon } \int_{S}^{T}\xi (t) \biggl( \int_{0}^{t}g(t-s) \bigl\Vert \nabla u(s)- \nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad =\varepsilon \int_{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt \\ &\quad\quad{} +\frac{G( \infty )}{4\varepsilon } \int_{S}^{T} \biggl( \int_{0}^{t}\frac{\xi (t)}{ \xi (t-s)}\xi (t-s)g(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant \varepsilon \int_{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt+\frac{G( \infty )}{4\varepsilon } \int_{S}^{T} \biggl( \int_{0}^{t}\xi (t-s)g(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant \varepsilon \int_{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt-\frac{G( \infty )}{4\varepsilon } \int_{S}^{T} \biggl( \int_{0}^{t}g'(t-s) \bigl\Vert \nabla u(s)-\nabla u(t) \bigr\Vert ^{2}\,ds \biggr) \,dt \\ &\quad \leqslant \varepsilon \int_{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt-\frac{G( \infty )}{2\varepsilon } \int_{S}^{T}E'(t)\,dt \\ &\quad \leqslant \varepsilon \int_{S}^{T}\xi (t) \Vert \nabla u \Vert ^{2}\,dt+\frac{1}{2 \varepsilon }E(S)\leqslant \varepsilon \int_{S}^{T}\xi (t)E(t)\,dt+\frac{1}{2 \varepsilon }E(S). \end{aligned}$$

(3.25)

Combining (3.15)–(3.25), we obtain

$$\begin{aligned} &\frac{1}{\rho +1} \int_{S}^{T} \bigl( \xi (t)G(t) \bigr) \Vert u_{t} \Vert _{ \rho +2}^{\rho +2}\,dt+ \int_{S}^{T} \xi (t)G(t) \Vert \nabla u_{t} \Vert ^{2}\,dt \\ &\quad \leqslant \delta \int_{S}^{T}\xi (t) \Vert u_{t} \Vert ^{\rho +2}_{\rho +2} \,dt+\frac{ \delta }{2} \int_{S}^{T}\xi (t) \Vert \nabla u_{t} \Vert ^{2} \,dt+\varepsilon \int _{S}^{T}\xi (t)E(t)\,dt \\ &\quad \quad {} + \biggl[ C_{1}+\frac{g(0)\xi (0)}{\delta }+\frac{1}{2\varepsilon } \biggr] E(S). \end{aligned}$$

(3.26)

Since g is continuous and $g(0)>0$, for any $t_{0}>0$, we have

$$\begin{aligned} G(t)= \int_{0}^{t}g(s)\,ds\geqslant \int_{0}^{t_{0}}g(s)\,dx>0,\quad \forall t\geqslant t_{0}. \end{aligned}$$

Now, if we fix $\delta >0$ small enough such that

$$\begin{aligned} \delta < \int_{0}^{t_{0}}g(s)\,ds, \end{aligned}$$

then by (3.26), for $T>S\geqslant t_{0}$, we have

$$\begin{aligned} \int_{S}^{T}\xi (t) \Vert u_{t} \Vert _{\rho +2}^{\rho +2}\,dt+ \int_{S}^{T} \xi (t) \Vert \nabla u_{t} \Vert ^{2}\,dt\leqslant \varepsilon C \int_{S}^{T} \xi (t)E(t)\,dt+ C(\varepsilon )E(S). \end{aligned}$$

 □

Proof of Lemma 3.1

Plugging the estimate of Lemma 3.3 into the inequality of Lemma 3.2, and fixing ε small enough, we obtain

$$\begin{aligned} \int_{S}^{T}\xi (t)E(t)\,dt \leqslant C E(S) \end{aligned}$$

for some constant $C>0$. Letting $T\rightarrow +\infty$, we have

$$\begin{aligned} \int_{t}^{+\infty }\psi '(s)E(s)\,ds\leqslant {cE(t)},\quad \forall t\geqslant t_{0}, \end{aligned}$$

(3.27)

with

$$\begin{aligned} \psi (t):= \int_{t_{0}}^{t}\xi (s)\,ds. \end{aligned}$$

 □

Proof of Theorem 2.2

From Assumption 2.1 and Lemma 2.3, we know that $E(t)$ is a non-increasing function and $\psi :[t_{0},+ \infty )\rightarrow \mathbb{R}^{+}$ is a strictly increasing $C^{2}$ function such that $\psi (t_{0})=0$ and $\lim_{t\rightarrow +\infty } \psi (t)=+\infty$. Firstly, we define a new function $f:[t_{0},\infty )\rightarrow \mathbb{R}^{+}$ as follows:

$$\begin{aligned} f(\tau )=E\bigl(\psi^{-1}(\tau )\bigr), \end{aligned}$$

then f is a non-increasing function such that

$$\begin{aligned} \int_{\psi (S)}^{\psi (T)}f(\tau )\,d\tau &= \int_{\psi (S)}^{\psi (T)}E\bigl( \psi^{-1}(\tau ) \bigr)\,d\tau = \int_{S}^{T}E(t)\psi '(t)\,dt \\ &\leqslant \int_{S}^{+\infty }E(t)\psi '(t)\,dt\leqslant cE(S)=cf\bigl( \psi (S)\bigr),\quad \forall t_{0}\leqslant S< T< \infty . \end{aligned}$$

Set $t=\psi (S)$. Since $\lim_{T\rightarrow +\infty }\psi (T)=+ \infty$, we get

$$\begin{aligned} \int_{t}^{+\infty }f(\tau )\,d\tau \leqslant cf(t),\quad \forall t \geqslant t_{0}. \end{aligned}$$

(3.28)

Next, we define the following function:

$$\begin{aligned} h(x)=e^{\frac{1}{c}x} \int_{x}^{+\infty }f(s)\,ds, \end{aligned}$$

(3.29)

where $c>0$ is a constant. Noting (3.28), we obtain

$$\begin{aligned} h'(x)=\frac{1}{c}e^{\frac{1}{c}x} \biggl( \int_{x}^{+\infty }f(s)\,ds-cf(x) \biggr) \leqslant 0. \end{aligned}$$

(3.30)

Integrating (3.30) over $[t_{0},t]$ and noting (3.28), we have

$$\begin{aligned} h(t)\leqslant h(t_{0})=e^{\frac{1}{c}t_{0}} \int_{t_{0}}^{+\infty }f(s)\,ds \leqslant c_{1}f(t_{0}). \end{aligned}$$

(3.31)

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

$$\begin{aligned} \int_{t}^{+\infty }f(s)\,ds\leqslant c_{1}f(t_{0})e^{-\frac{1}{c}t}. \end{aligned}$$

(3.32)

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

$$\begin{aligned} \int_{t}^{+\infty }f(s)\,ds\geqslant \int_{t}^{t+c}f(s)\,ds\geqslant cf(t+c). \end{aligned}$$

(3.33)

Combining (3.32) with (3.33), we get

$$\begin{aligned} f(t+c)\leqslant c_{2}f(t_{0})e^{-\frac{1}{c}t}. \end{aligned}$$

(3.34)

Letting $s=t+c$ in (3.34), we have

$$\begin{aligned} f(s)\leqslant c_{2}f(t_{0})e^{1-\frac{1}{c}s}, \end{aligned}$$

that is,

$$\begin{aligned} E\bigl(\psi^{-1}(s)\bigr)\leqslant c_{2}f(t_{0})e^{1-\frac{1}{c}s}. \end{aligned}$$

(3.35)

Moreover, letting $t=\psi^{-1}(s)$ in (3.35), we get

$$\begin{aligned} E(t)\leqslant c_{2}f(t_{0})e^{1-\frac{1}{c}\psi (t)}, \end{aligned}$$

that is,

$$\begin{aligned} E(t)\leqslant C_{0}e^{-\omega \psi (t)}=C_{0}e^{-\omega \int_{t_{0}} ^{t}\xi (s)\,ds}, \end{aligned}$$

for some constants $C_{0}$ and $\omega =\frac{1}{c}>0$.

The proof of Theorem 2.2 is completed. □

Remark 3.1

From Theorem 2.2, if we choose different $\xi (t)$, we can get different decay results. Choosing $\xi (t) \equiv {a}$, we get the exponential decay result

$$\begin{aligned} E(t)\leqslant Ce^{-\omega t},\quad \forall t\geqslant t_{0}. \end{aligned}$$

Now consider $\xi (t)=\frac{1}{(1+t)^{\gamma }}$ ($0<\gamma \leqslant 1$). If $\gamma =1$, we get the polynomial decay result

$$\begin{aligned} E(t)\leqslant C (1+t)^{-\omega },\quad \omega >0, \forall t\geqslant t_{0}. \end{aligned}$$

If $0<\gamma <1$, we get a decay result of the form

$$\begin{aligned} E(t)\leqslant Ce^{-\frac{\omega }{1-\gamma }(1+t)^{1-\gamma }},\quad \omega >0, \forall t\geqslant t_{0}. \end{aligned}$$

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 assumption on the memory kernel function, the weighted inequality established in this paper 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.