A general decay result for a semilinear heat equation with past and finite history memories

Abstract

In this paper, we consider the initial-boundary value problem of the following semilinear heat equation with past and finite history memories:

$$\begin{aligned} &u_{t}-\Delta u + \int _{0}^{t} {{g_{1}}(t - s) \operatorname{div}\bigl({a_{1}}(x) \nabla u(s)\bigr)\,ds} \\ &\quad{} + \int _{0}^{ + \infty } {{g_{2}}(s) \operatorname{div}\bigl({a_{2}}(x)\nabla u(t - s)\bigr)\,ds}+ f(u)=0, \quad(x,t)\in \varOmega \times [0,+\infty ), \end{aligned}$$

where Ω is a bounded domain. Under suitable conditions on $a_{1}$ and $a_{2}$, for a large class of relation functions $g_{1}$ and $g_{2}$, we establish a general decay estimate, including the usual exponential and polynomial decay cases.

1 Introduction

In this paper, we aim at studying a generalized decay result of the following problem:

$$\begin{aligned} \textstyle\begin{cases} u_{t}-\Delta u + \int _{0}^{t} {{g_{1}}(t - s)\operatorname{div}({a_{1}}(x) \nabla u(s))\,ds} \\ \quad{}+ \int _{0}^{ + \infty } {{g_{2}}(s)\operatorname{div}({a_{2}}(x)\nabla u(t - s))\,ds}+ f(u)=0,\quad (x,t)\in \varOmega \times [0,+\infty ), \\ u(x,t) = 0,\quad (x,t)\in \partial \varOmega \times [0,+\infty ), \\ u(x,-t) = u_{0}(x,t),\quad (x,t)\in \varOmega \times [0,+\infty ), \end{cases}\displaystyle \end{aligned}$$

(1.1)

where $\varOmega \subset \mathbb{R}^{N}$ ($N \ge 1$) is a bounded domain with smooth boundary ∂Ω, $g_{1}$ and $g_{2}$ are positive nonincreasing functions defined on $\mathbb{R}^{+}:=[0,+ \infty )$, $a_{1}$ and $a_{2}$ are essentially bounded nonnegative functions defined on Ω, f is the nonlinear term, and $u_{0}$ is a given initial condition, respectively, satisfying assumptions (H1)–(H5) shown in Sect. 2.

This type of equations describes many mathematical models in engineering and physical sciences; we refer to [1]. For example, in the study of heat conduction in materials with memory, the classical Fourier law of the heat flux is replaced by the following form (see [2]):

$$ q=-d\nabla u- \int _{-\infty }^{t}g(t-s)\nabla \bigl[a(x)u(x,s)\bigr]\,ds, $$

where q, u, d, and the integral term represent the heat flux, temperature, diffusion coefficient, and memory effect in the material, respectively. From the mathematical point of view, we would expect that the leading term $-d\nabla u$ dominates the integral term in the equation. Hence theory of parabolic equations can be applied to this type of equations.

The study on the global existence, blow-up, and energy decay of solutions for this type of problems involving a finite history memory term has attracted much attention; see [3,4,5,6,7,8,9,10,11,12] and the references therein. More precisely, Messaoudi and Tellab [9] studied the quasilinear parabolic system of the form

$$ A(t) \vert u_{t} \vert ^{m-2}u_{t}-\Delta u + \int _{0}^{t} {{g}(t - s)\Delta u(s))\,ds}=0, \quad(x,t) \in \varOmega \times [0,+\infty ), $$

for $m\ge 2$, where $A(t)$ a bounded and positive definite matrix, and proved a general property of energy decay result, with usual exponential and polynomial decays as particular cases. Fang and Qiu [11] considered the mixed boundary problem of the equation

$$ A(x,t)u_{t}-\Delta u + \int _{0}^{t} {{g}(t - s)\operatorname{div} \bigl(a(x) \nabla u(s)\bigr)\,ds}=0,\quad(x,t)\in \varOmega \times [0,+\infty ), $$

where $A(x,t)$ is a positive function such that $A_{t}(x,t)\le 0$. By the technique of Lyapunov functional they proved the existence and uniqueness of a global solution and that the energy functional decays exponentially or polynomially to zero as time tends to infinity. Later, Li et al. [12] were concerned with a mixed boundary value problem of the semilinear parabolic equation

$$ u_{t}-\Delta u + \int _{0}^{t} {{g}(t - s)\operatorname{div} \bigl(a(x)\nabla u(s)\bigr)\,ds}=0,\quad(x,t) \in \varOmega \times [0,+\infty ). $$

Under suitable conditions, a generalized property of energy decay was proved, in which the exponential and polynomial decay results are only particular cases. However, to the best of the authors’ knowledge, the decay results for semilinear heat equations with past (infinite) history memory have not been discussed yet. Motivated by this observation, we intend to study the generalized property of energy decay for problem (1.1) in the presence of past and finite history memories, which allows a wide class of memory kernel functions, where the exponential and polynomial decay results are only particular cases (see Example 3.1). Our result is also valid for the problem with past memory or finite memory term case; see Remark 3.2. It is necessary to point out that the argument in [9, 11, 12] cannot be extended to problem (1.1) due to the past memory term. In this paper, we adopt a new approach introduced by Guesmia [13], who investigated a class of hyperbolic problems.

This paper is organized as follows. In Sect. 2, we present preliminaries and some lemmas needed for later work. In Sect. 3, we establish the general decay result and give some examples to illustrate its wide application.

2 Preliminaries and some lemmas

In this section, we give some assumptions, definitions, and lemmas, which will be used to establish our main result.

We first state the following assumptions.

$\mathrm{(H1)}$ :

${g_{i}}:\mathbb{R}^{+} \to \mathbb{R}^{+}$ are differentiable nonincreasing functions satisfying $g_{i}(0) > 0$, $i=1,2$, and $1 - \Vert a_{1} \Vert _{\infty }\int _{0}^{ + \infty } g_{1}(s)\,ds - \Vert a_{2} \Vert _{\infty }\int _{0}^{ + \infty } g_{2}(s)\,ds = l > 0$.

$\mathrm{(H2)}$ :

There exists a positive differentiable nonincreasing function $\xi :\mathbb{R}^{+} \to \mathbb{R}^{+}$ such that

$$\begin{aligned} {g_{1}}'(t) \le - \xi (t){g_{1}}(t), \quad t \ge 0. \end{aligned}$$

(2.1)

$\mathrm{(H3)}$ :

There exist a positive constant σ and an increasing strictly convex function $G:\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ of class ${C^{1}} ( \mathbb{R^{+}} ) \cap {C^{2}}(0, + \infty )$, satisfying $G(0) = G'(0) = 0$ and $\lim_{t \to + \infty } G'(t) = + \infty$, such that

$$\begin{aligned} {g_{2}}'(t) \le - \sigma {g_{2}}(t),\quad t \ge 0, \end{aligned}$$

(2.2)

or

$$\begin{aligned} \int _{0}^{ + \infty } {\frac{{{g_{2}}(t)}}{{{G^{ - 1}}( - {g_{2}}'(t))}}} \,dt + \sup_{t \in {\mathbb{R}^{+} }} \frac{ {{g_{2}}(t)}}{{{G^{ - 1}}( - {g_{2}}'(t))}} < + \infty . \end{aligned}$$

(2.3)

$\mathrm{(H4)}$ :

${a_{i}}:\overline{\varOmega } \to \mathbb{R}^{+}$ ($i=1,2$) are in $C^{1}(\overline{\varOmega })$, and there exist two positive constants $b_{0}$ and $b_{1}$ such that $|\nabla a_{1}(x)| \le b_{1}a_{1}(x)$ for $x\in \overline{\varOmega }$ and

$$\begin{aligned} a_{1}(x)+a_{2}(x)\ge b_{0},\quad x\in \overline{ \varOmega }. \end{aligned}$$

$\mathrm{(H5)}$ :

$f:\mathbb{R}\rightarrow \mathbb{R}$ is Lipschitz continuous and satisfies

$$ f(s)s \ge 2F(s) \ge 0, \quad s\in \mathbb{R}, $$

where $F(s) = \int _{0}^{s} {f(z)\,dz}$.

To get an equivalent problem of (1.1) for convenience, we define

$$\begin{aligned} {\eta ^{t}}(x,s) = u(x,t) - u(x,t - s),\quad \forall x\in \varOmega , s,t\geq 0. \end{aligned}$$

Obviously, this gives

$$\begin{aligned} \eta _{t}^{t}(x,s) + \eta _{s}^{t}(x,s) = {u_{t}}(x,t). \end{aligned}$$

Together with (1.1), we deduce the problem

$$\begin{aligned} \textstyle\begin{cases} {u_{t}} - \operatorname{div}[(1 - {a_{2}}(x)\int _{0}^{ + \infty } {{g_{2}}(s)\,ds} )\nabla u(t)] + \int _{0}^{t} {{g_{1}}(t - s) \operatorname{div}({a_{1}}(x)\nabla u(s))\,ds} \\ \quad {}-\int _{0}^{ + \infty } {{g_{2}}(s)\operatorname{div}({a_{2}}(x)\nabla {\eta ^{t}}(s))\,ds} + f(u) =0,\quad(x,t)\in \varOmega \times [0,\infty ), \\ \eta _{t}^{t}(x,s) + \eta _{s}^{t}(x,s) - {u_{t}}(x,t) = 0,\quad x \in \varOmega , s,t \ge 0, \end{cases}\displaystyle \end{aligned}$$

(2.4)

and the following initial and boundary conditions:

$$\begin{aligned} \textstyle\begin{cases} u(x,t) = {\eta ^{t}}(x,s) = 0,\quad \forall x \in \partial \varOmega , s,t \ge 0, \\ u(x, - t) = {u_{0}}(x,t),\quad {\eta ^{t}}(x,0) = 0, (x,t)\in \varOmega \times [0,\infty ), \\ {\eta ^{0}}(x,s) = {u_{0}}(x,0) - {u_{0}}(x,s),\quad x \in \varOmega , s,t \ge 0. \end{cases}\displaystyle \end{aligned}$$

(2.5)

Now we define the Hilbert space

$$ H_{0}^{1}(\varOmega ) = \bigl\{ u \in {H^{1}}(\varOmega )| u = 0\text{ on } \partial \varOmega \bigr\} $$

and the weight space with respect to $g_{2}$

$$ {\mathrm{M}} = \biggl\{ \xi :{\mathbb{R}^{+} } \to H_{0}^{1}(\varOmega )\Big| { \int _{0}^{ + \infty } {{g_{2}}(s)} \bigl\Vert { \sqrt{{a_{2}}(x)} \nabla \xi (s)} \bigr\Vert _{2}^{2}\,ds < + \infty } \biggr\} $$

endowed with scalar product and norm

$$\begin{aligned}& {(\xi ,\varsigma )_{\mathrm{M}}} = \int _{0}^{ + \infty } {{g_{2}}(s)} \int _{\varOmega }{{a_{2}}(x)\nabla \xi (s)\nabla \varsigma (s)\,dx\,ds}, \\& \Vert \xi \Vert _{\mathrm{M}}^{2} = \int _{0}^{ + \infty } {{g _{2}}(s)} \bigl\Vert {\sqrt{{a_{2}}(x)} \nabla \xi (s)} \bigr\Vert _{2} ^{2}\,ds < + \infty , \end{aligned}$$

where ${ \Vert \cdot \Vert _{q}}$, $1\leq q\leq \infty$, is the norm of $L^{q}(\varOmega )$.

We now state the definition of a solution of problem (2.4)–(2.5).

Definition 2.1

A solution of problem (2.4)–(2.5) is a function $u \in C([0,T];H_{0}^{1}(\varOmega )) \cap {C^{1}}([0,T];{L^{2}}(\varOmega ))$, $T>0$, that satisfies

$$\begin{aligned} & \int _{\varOmega }{u_{t}}(t)\omega (x,t)\,dx + \int _{\varOmega }{\biggl(1 - {a_{2}}(x) \int _{0}^{ + \infty } {{g_{2}}(s)\,ds} \biggr) \nabla u(t)\nabla \omega (x,t)\,dx} \\ &\quad{}- \int _{\varOmega }{ \int _{0}^{t} {{g_{1}}(t - s ){a_{1}}(x)} \nabla u(s) \nabla \omega (x,t)\,ds \,dx} + \int _{\varOmega }{{a_{2}}(x) \int _{0}^{ + \infty } {{g_{2}}(s)\nabla {\eta ^{t}}(s)\nabla \omega (x,t)\,ds} \,dx} \\ &\quad{} + \int _{\varOmega }{f(u)\omega (x,t)\,dx} = 0, \\ &{\bigl({\eta _{t} ^{t}}(s),\xi \bigr)_{\mathrm{M}}} = - { \bigl({\eta _{s} ^{t}}(s), \xi \bigr)_{\mathrm{M}}} + { \bigl({u_{t}}(t),\xi \bigr)_{\mathrm{M}}}, \end{aligned}$$

for all $\omega \in C([0,T];H_{0}^{1}(\varOmega ))$ and $\xi \in {\mathrm{M}}$.

Remark 2.2

The existence and uniqueness of a global solution for problem (2.4)–(2.5) can be established by using the Galerkin method, the contraction mapping principle, and a continuation argument. The process is similar to that of [11].

Define the modified energy functional of problem (2.4)–(2.5) by

$$\begin{aligned} E(t)&: = E[u](t) \\ &= \frac{1}{2} \int _{\varOmega }{k(x,t){{ \bigl\vert {\nabla u}(t) \bigr\vert }^{2}}\,dx} + \frac{1}{2}({g_{1}} \circ \nabla u) (t) + \frac{1}{2}\bigl({g_{2}} \circ \nabla {\eta ^{t}}\bigr) (t) + \int _{\varOmega } {F\bigl(u(t)\bigr)\,dx}, \end{aligned}$$

(2.6)

where

$$\begin{aligned}& k(x,t) = 1 - {a_{1}}(x) \int _{0}^{t} {{g_{1}}(s)\,ds} - {a_{2}}(x) \int _{0}^{ + \infty } {{g_{2}}(s)\,ds}, \\& ({g_{1}} \circ \nabla u) (t) = \int _{0}^{t} {{g_{1}}(t - s)} \bigl\Vert {\sqrt{{a_{1}}(x)} \bigl(\nabla u(t) - \nabla u(s)\bigr)} \bigr\Vert _{2}^{2}\,ds, \quad \forall u \in H_{0}^{1}( \varOmega ), \\& \bigl({g_{2}} \circ \nabla {\eta ^{t}}\bigr) (t) = \int _{0}^{ + \infty } {{g_{2}}(s)} \bigl\Vert { \sqrt{{a_{2}}(x)} \nabla {\eta ^{t}}(s)} \bigr\Vert _{2}^{2}\,ds. \end{aligned}$$

We show the following lemma, which is useful to get $\frac{dE(t)}{dt}$.

Lemma 2.1

For all ${\eta ^{t}}(s) \in {\mathrm{M}}$, we have the equality

$$ \frac{1}{2} \int _{\varOmega }{{a_{2}}(x){{ \int _{0}^{ + \infty } {{g_{2}}'(s) \bigl\vert {\nabla {\eta ^{t}}(s)} \bigr\vert } }^{2}}\,ds \,dx} = - \int _{\varOmega }{{a_{2}}(x) \int _{0}^{ + \infty } {{g_{2}}(s)\nabla {\eta ^{t}}(s) \nabla \eta _{s}^{t}(s)} \,ds \,dx}. $$

Proof

The result can be obtained directly from the calculation

$$\begin{aligned} 0&= \frac{1}{2} \int _{\varOmega }a_{2}(x)\frac{d}{ds} \biggl( \int _{0}^{ + \infty } g_{2}(s) \bigl\vert \nabla {\eta ^{t}}(s) \bigr\vert ^{2}\,ds \biggr)\,dx \\ &= \frac{1}{2} \int _{\varOmega }a_{2}(x) \int _{0}^{ + \infty } g_{2}'(s) \bigl\vert {\nabla {\eta ^{t}}(s)} \bigr\vert ^{2}\,ds \,dx \\ & \quad{}+ \int _{\varOmega }a_{2}(x) \int _{0}^{ + \infty } {{g_{2}}(s)\nabla {\eta ^{t}}(s)\nabla \eta _{s}^{t}(s)} \,ds \,dx. \end{aligned}$$

 □

Lemma 2.2

The energy functional $E(t)$ of problem (2.4)–(2.5) satisfies

$$\begin{aligned} \frac{d}{{dt}}E(t) &= - \frac{1}{2}{g_{1}}(t) \bigl\Vert { \sqrt{{a_{1}}(x)} \nabla u(t)} \bigr\Vert _{2}^{2} + \frac{1}{2}\bigl( {g_{1}}' \circ \nabla u\bigr) (t) + \frac{1}{2}\bigl({g_{2}}' \circ \nabla {\eta ^{t}}\bigr) (t) - \bigl\Vert u_{t}(t) \bigr\Vert _{2}^{2} \\ &\le 0. \end{aligned}$$

(2.7)

Proof

Differentiating $E(t)$ in (2.6), we obtain

$$\begin{aligned} \frac{d}{{dt}}E(t) &= - \frac{1}{2}{g_{1}}(t) \bigl\Vert { \sqrt{{a_{1}}(x)} \nabla u(t)} \bigr\Vert _{2}^{2}+ \int _{\varOmega }k(x,t) \nabla u(t)\nabla u_{t}(t)\,dx+ \frac{1}{2}\bigl(g_{1}'\circ \nabla u\bigr) (t) \\ &\quad{} +\frac{1}{2}\bigl(g_{2}'\circ \nabla \eta ^{t}\bigr) (t)+ \int _{\varOmega }a_{1}(x) \int _{0}^{t}g_{1}(t-s) \bigl(\nabla u(t)- \nabla u(s)\bigr)\nabla u_{t}(t)\,ds \,dx \\ &\quad{} + \int _{\varOmega }a_{2}(x) \int _{0}^{+\infty } g_{2}(s)\nabla \eta ^{t}(s) \nabla \eta ^{t}_{t}(s)\,ds \,dx+ \int _{\varOmega }f(u)u_{t}(t)\,dx. \end{aligned}$$

Multiplying the first equality in (2.4) by $u_{t}$ and then integrating the result over Ω, we obtain

$$\begin{aligned} & \int _{\varOmega }u_{t}^{2}(t)\,dx+ \int _{\varOmega }\nabla u(t) \nabla u_{t}(t)\,dx- \int _{\varOmega } \int _{0}^{t}g_{1}(t-s)a_{1}(x) \nabla u(s)\,ds\nabla u_{t}(t)\,dx \\ &\quad{}- \int _{\varOmega } \int _{0}^{+\infty } g_{2}(s)a_{2}(x) \nabla u(t-s)\,ds \nabla u_{t}(t)\,dx+ \int _{\varOmega }f(u)u_{t}(t)\,dx=0. \end{aligned}$$

Subtracting the above two equalities and using Lemma 2.1, we get (2.7). □

Note that here the positive constant c or C denotes different constants in different places.

3 General decay

In this section, we establish the estimate of general energy decay, which is the main result of this paper. For this purpose, we introduce the perturbed energy functional

$$\begin{aligned} \varPsi (t) = E(t) + \varepsilon _{1}\chi (t)+\varepsilon _{2} \phi (t), \end{aligned}$$

where $\varepsilon _{1}$ and $\varepsilon _{2}$ are small positive constants, and

$$\begin{aligned}& \chi (t) = \frac{1}{2} \int _{\varOmega }u^{2}(t)\,dx, \\& \phi (t) = \int _{\varOmega }u(t) \int _{0}^{t}g_{1}(t-s)a_{1}(x)u(s)\,ds \,dx. \end{aligned}$$

We prove the following three lemmas for later use.

Lemma 3.1

Assume that $u(x,t)$ is a solution of problem (2.4)–(2.5). Then there exist two constants $c_{1},c_{2}>0$ such that

$$\begin{aligned} {c_{1}}E(t) \le \varPsi (t) \le {c_{2}}E(t). \end{aligned}$$

(3.1)

Proof

Let $c_{*}$ be the best constant for the Poincaré inequality $\|u\|_{2}\le c_{*}\|\nabla u\|_{2}$. It follows directly that

$$\begin{aligned} \bigl\vert \chi (t) \bigr\vert = \frac{1}{2} \biggl\vert \int _{\varOmega }{u^{2}(t)}\,dx \biggr\vert \le \frac{c_{*}^{2}}{2} \Vert \nabla u \Vert _{2}^{2} \le c_{01}E(t). \end{aligned}$$

Applying the Hölder inequality, we obtain

$$\begin{aligned} \bigl\vert \phi (t) \bigr\vert &\le \biggl\vert \int _{\varOmega }u(t) \int _{0}^{t}g_{1}(t-s)a_{1}(x) \bigl(u(t)-u(s)\bigr)\,ds \,dx \biggr\vert \\&\quad{} + \biggl\vert \int _{\varOmega }u(t) \int _{0}^{t}g_{1}(t-s)a_{1}(x)u(t)\,ds \,dx \biggr\vert \\ &\le \biggl(\frac{\delta _{1}}{2}+ \Vert a_{1} \Vert _{\infty } \int _{0}^{t}g _{1}(s)\,ds \biggr)c_{*}^{2} \Vert \nabla u \Vert _{2}^{2}+ \frac{ \Vert a_{1} \Vert _{ \infty }\int _{0}^{t}g_{1}(s)\,ds}{2\delta _{1}}c_{*}^{2}(g_{1} \circ \nabla u) (t)\\ &\le c_{02}E(t). \end{aligned}$$

Then, selecting ${c_{1}} = 1 - {\varepsilon _{1}}c_{01}-{\varepsilon _{2}}c_{02}$ and ${c_{2}} = 1 + {\varepsilon _{1}}c_{01}+{\varepsilon _{2}}c_{02}$, we get(3.1). □

Lemma 3.2

Assume that (H1)–(H5) hold and ${u_{0}} \in H_{0}^{1}(\varOmega ) \cap {H^{2}}(\varOmega )$. Let $u(x,t)$ be a solution of problem (2.4)–(2.5). Then there exists a small constant $\mu _{1}$ such that

$$\begin{aligned} \chi '(t)&\le -(l-\mu _{1}) \Vert \nabla u \Vert _{2}^{2}+\frac{1-l}{2\mu _{1}}\bigl[(g _{1}\circ \nabla u) (t)+\bigl(g_{2}\circ \nabla \eta ^{t}\bigr) (t)\bigr]- \int _{\varOmega }uf(u)\,dx. \end{aligned}$$

(3.2)

Proof

Differentiating $\chi (t)$ and using Green’s formula and Hölder’s inequality, we deduce

$$\begin{aligned} \chi '(t)&= \int _{\varOmega }u{u_{t}}\,dx \\ &= - \int _{\varOmega }\nabla u\biggl[\biggl(1 - {a_{2}}(x) \int _{0}^{ + \infty } {{g _{2}}(s)\,ds} \biggr) \nabla u(t)\biggr]\,dx\\&\quad{} + \int _{\varOmega }\nabla u \int _{0}^{t} g_{1}(t - s) \bigl(a_{1}(x)\nabla u(s)\bigr)\,ds \\ &\quad{}- \int _{\varOmega }\nabla u \int _{0}^{ + \infty } g_{2}(s)a_{2}(x) \nabla \eta ^{t}(s)\,ds \,dx - \int _{\varOmega }uf(u)\,dx \\ &= - \int _{\varOmega }k(x,t) \bigl\vert \nabla u(t) \bigr\vert ^{2}\,dx + \int _{\varOmega }\nabla u \int _{0}^{t} g_{1}(t - s)a_{1}(x) \bigl(\nabla u(s)-\nabla u(t)\bigr)\,ds \\ &\quad{}- \int _{\varOmega }\nabla u \int _{0}^{ + \infty } g_{2}(s)a_{2}(x) \nabla \eta ^{t}(s)\,ds \,dx - \int _{\varOmega }uf(u)\,dx \\ &\le -l \Vert \nabla u \Vert _{2}^{2}+ \frac{\mu _{1}+\mu _{2}}{2} \Vert \nabla u \Vert _{2}^{2}+ \frac{ \Vert a_{1} \Vert _{\infty }\int _{0}^{t}g_{1}(s)\,ds}{2\mu _{1}}(g _{1}\circ \nabla u) (t) \\ &\quad{} +\frac{ \Vert a_{2} \Vert _{\infty }\int _{0}^{\infty }g_{2}(s)\,ds}{2\mu _{2}}\bigl(g _{2}\circ \nabla \eta ^{t}\bigr) (t)- \int _{\varOmega }uf(u)\,dx \\ &\le -(l-\mu _{1}) \Vert \nabla u \Vert _{2}^{2}+ \frac{1-l}{2\mu _{1}}\bigl[(g_{1} \circ \nabla u) (t)+\bigl(g_{2} \circ \nabla \eta ^{t}\bigr) (t)\bigr]- \int _{\varOmega }uf(u)\,dx. \end{aligned}$$

Setting $\mu _{1}=\mu _{2}$ in the last inequality, we get (3.2). □

Lemma 3.3

Assume that (H1)–(H5) hold and ${u_{0}} \in H_{0}^{1}(\varOmega ) \cap {H^{2}}(\varOmega )$. Let $u(x,t)$ be a solution of problem (2.4)–(2.5). Then

$$\begin{aligned} \bigl\vert \phi '(t) \bigr\vert &\le \biggl[1+ \frac{c_{*}^{2}}{2}+2g_{1}(0) \Vert a_{1} \Vert _{ \infty }c_{*}^{2}+\bigl(3b_{1}^{2}c_{*}^{2}+3+c_{*}^{2} \bigr) (1-l)^{2} \biggr] \Vert \nabla u \Vert _{2}^{2} \\ &\quad{} +\bigl(3b_{1}^{2}c_{*}^{2}+4+c_{*}^{2} \bigr) (1-l) (g_{1}\circ \nabla u) (t)+ (1-l) \bigl(g _{2} \circ \nabla {\eta ^{t}}\bigr) (t) \\ &\quad{} +\frac{1}{2} \int _{\varOmega } \bigl\vert f(u) \bigr\vert ^{2}\,dx- \frac{c_{*}^{2}}{2} \Vert a_{1} \Vert _{\infty }g_{1}(0) \bigl(g_{1}'\circ \nabla u\bigr) (t). \end{aligned}$$

(3.3)

Proof

Differentiating $\phi (t)$ and using Green’s formula and Hölder’s inequality, we deduce

$$\begin{aligned} \phi '(t) &= \int _{\varOmega }u_{t}(t) \int _{0}^{t}g_{1}(t-s)a_{1}(x)u(s)\,ds \,dx \\ &\quad{} + \int _{\varOmega }u(t) \int _{0}^{t}g_{1}'(t-s)a_{1}(x)u(s)\,ds \,dx+g_{1}(0) \int _{\varOmega } \bigl\vert u(t) \bigr\vert ^{2}a_{1}(x)\,dx \\ &= \int _{\varOmega }{\operatorname{div}} \biggl[ \biggl(1 - {a_{2}}(x) \int _{0} ^{ + \infty } {{g_{2}}(s)\,ds} \biggr) \nabla u(t) \biggr] \int _{0}^{t}g _{1}(t-s)a_{1}(x)u(s)\,ds \,dx \\ &\quad{}- \int _{\varOmega } \int _{0}^{t} {{g_{1}}(t - s) \operatorname{div}\bigl({a_{1}}(x) \nabla u(s)\bigr)\,ds} \int _{0}^{t}g_{1}(t-s)a_{1}(x)u(s)\,ds \,dx \\ &\quad{} + \int _{\varOmega } \int _{0}^{ + \infty } {{g_{2}}(s) \operatorname{div}\bigl({a_{2}}(x) \nabla {\eta ^{t}}(s) \bigr)\,ds} \int _{0}^{t}g_{1}(t-s)a_{1}(x)u(s)\,ds \,dx \\ &\quad{} - \int _{\varOmega }f(u) \int _{0}^{t}g_{1}(t-s)a_{1}(x)u(s)\,ds \,dx \\ &\quad{} + \int _{\varOmega }u(t) \int _{0}^{t}g_{1}'(t-s)a_{1}(x)u(s)\,ds \,dx+g_{1}(0) \int _{\varOmega } \bigl\vert u(t) \bigr\vert ^{2}a_{1}(x)\,dx \\ &\left . \begin{aligned} =- \int _{\varOmega }k(x,t)\nabla u(t) \int _{0}^{t}g_{1}(t-s)\nabla a_{1}(x)u(s)\,ds \,dx \\ \quad {}- \int _{\varOmega }k(x,t)\nabla u(t) \int _{0}^{t}g_{1}(t-s)a_{1}(x) \nabla u(s)\,ds \,dx \end{aligned} \right \} :=I_{1} \\ &\left . \begin{aligned} \quad {}+ \int _{\varOmega } \int _{0}^{t} {{g_{1}}(t - s){a_{1}}(x) \bigl(\nabla u(s)- \nabla u(t)\bigr)\,ds} \int _{0}^{t}g_{1}(t-s)\nabla a_{1}(x)u(s)\,ds \,dx \\ \quad {}+ \int _{\varOmega } \int _{0}^{t} {{g_{1}}(t - s){a_{1}}(x) \bigl(\nabla u(s)- \nabla u(t)\bigr)\,ds} \int _{0}^{t}g_{1}(t-s)a_{1}(x) \nabla u(s)\,ds \,dx \end{aligned} \right \} :=I_{2} \\ &\left . \begin{aligned} \quad {}- \int _{\varOmega } \int _{0}^{ + \infty } g_{2}(s) \bigl({a_{2}}(x)\nabla {\eta ^{t}}(s)\bigr)\,ds \int _{0}^{t}g_{1}(t-s)\nabla a_{1}(x)u(s)\,ds \,dx \\ \quad {}- \int _{\varOmega } \int _{0}^{ + \infty } g_{2}(s) \bigl({a_{2}}(x)\nabla {\eta ^{t}}(s)\bigr)\,ds \int _{0}^{t}g_{1}(t-s)a_{1}(x) \nabla u(s)\,ds \,dx \end{aligned} \right \} :=I_{3} \\ &\quad {} - \int _{\varOmega }f(u) \int _{0}^{t}g_{1}(t-s)a_{1}(x)u(s)\,ds \,dx \\ &\quad {}+ \int _{\varOmega }u(t) \int _{0}^{t}g_{1}'(t-s)a_{1}(x)u(s)\,ds \,dx+g_{1}(0) \int _{\varOmega } \bigl\vert u(t) \bigr\vert ^{2}a_{1}(x)\,dx. \end{aligned}$$

(3.4)

Using (H4), we estimate the first two terms in the right-hand side of this equality:

$$\begin{aligned} \vert I_{1} \vert &\le \biggl\vert \int _{\varOmega }k(x,t)\nabla u(t) \int _{0}^{t}g_{1}(t-s) \nabla a_{1}(x)u(s)\,ds \,dx \biggr\vert \\ &\quad{} + \biggl\vert \int _{\varOmega }k(x,t)\nabla u(t) \int _{0}^{t}g_{1}(t-s)a_{1}(x) \nabla u(s)\,ds \,dx \biggr\vert \\ &\le \frac{\mu _{3}+\mu _{4}}{2} \int _{\varOmega } \bigl\vert k(x,t)\nabla u(t) \bigr\vert ^{2}\,dx \\&\quad{}+ \biggl(\frac{b_{1}^{2}c_{*}^{2}}{2\mu _{3}}+\frac{1}{2\mu _{4}} \biggr) \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t-s) a_{1}(x) \nabla u(s) \biggr) ^{2}\,ds \,dx \\ &\le \frac{\mu _{3}+\mu _{4}}{2} \Vert \nabla u \Vert _{2}^{2} \\&\quad{}+ \biggl(\frac{b _{1}^{2}c_{*}^{2}}{2\mu _{3}}+\frac{1}{2\mu _{4}} \biggr) \Vert a_{1} \Vert _{ \infty } \int _{0}^{t}g_{1}(s)\,ds \int _{\varOmega } \int _{0}^{t}g_{1}(t-s) a _{1}(x) \bigl\vert \nabla u(s) \bigr\vert ^{2}\,ds \,dx \\ &\le \biggl[\frac{\mu _{3}+\mu _{4}}{2}+ \biggl(\frac{b_{1}^{2}c_{*} ^{2}}{2\mu _{3}}+\frac{1}{2\mu _{4}} \biggr) (1+\mu _{5}) \biggl( \Vert a_{1} \Vert _{\infty } \int _{0}^{t}g_{1}(s)\,ds \biggr)^{2} \biggr] \Vert \nabla u \Vert _{2} ^{2} \\ &\quad{} + \biggl(\frac{b_{1}^{2}c_{*}^{2}}{2\mu _{3}}+\frac{1}{2\mu _{4}} \biggr) \biggl(1+ \frac{1}{ \mu _{5}}\biggr) \Vert a_{1} \Vert _{\infty } \int _{0}^{t}g_{1}(s)\,ds(g_{1} \circ \nabla u) (t). \end{aligned}$$

(3.5)

We estimate the second two terms in the right-hand side of (3.4):

$$\begin{aligned} \vert I_{2} \vert &\le \biggl\vert \int _{\varOmega } \int _{0}^{t} {{g_{1}}(t - s){a_{1}}(x) \bigl( \nabla u(s)-\nabla u(t)\bigr)\,ds} \int _{0}^{t}g_{1}(t-s)\nabla a_{1}(x)u(s)\,ds \,dx \biggr\vert \\ &\quad{} + \biggl\vert \int _{\varOmega } \int _{0}^{t} {{g_{1}}(t - s){a_{1}}(x) \bigl(\nabla u(s)-\nabla u(t)\bigr)\,ds} \int _{0}^{t}g_{1}(t-s)a_{1}(x) \nabla u(s)\,ds \,dx \biggr\vert \\ &\le \frac{\mu _{6}+\mu _{7}}{2} \int _{\varOmega } \biggl( \int _{0}^{t} {g_{1}}(t - s){a_{1}}(x) \bigl(\nabla u(s)-\nabla u(t)\bigr)\,ds \biggr)^{2}\,dx \\ &\quad{} + \biggl(\frac{b_{1}^{2}c_{*}^{2}}{2\mu _{6}}+\frac{1}{2\mu _{7}} \biggr) \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t-s) a_{1}(x) \nabla u(s)\,ds \biggr) ^{2}\,dx \\ &\le \biggl(\frac{b_{1}^{2}c_{*}^{2}}{2\mu _{6}}+\frac{1}{2\mu _{7}} \biggr) (1+ \mu _{5}) \biggl( \Vert a_{1} \Vert _{\infty } \int _{0}^{t}g_{1}(s)\,ds \biggr)^{2} \Vert \nabla u \Vert _{2}^{2} \\ &\quad{} +\biggl[\frac{\mu _{6}+\mu _{7}}{2}+ \biggl(\frac{b_{1}^{2}c_{*}^{2}}{2\mu _{6}}+\frac{1}{2\mu _{7}} \biggr) \biggl(1+\frac{1}{\mu _{5}}\biggr)\biggr] \Vert a_{1} \Vert _{ \infty } \int _{0}^{t}g_{1}(s)\,ds(g_{1} \circ \nabla u) (t). \end{aligned}$$

(3.6)

Similarly, we estimate the third two terms in the right-hand side of (3.4):

$$\begin{aligned} \vert I_{3} \vert &\le \biggl| \int _{\varOmega } \int _{0}^{ + \infty } g_{2}(s) \bigl({a _{2}}(x)\nabla {\eta ^{t}}(s)\bigr)\,ds \int _{0}^{t}g_{1}(t-s)\nabla a_{1}(x)u(s)\,ds \,dx \biggr\vert \\ &\quad{} + \biggl\vert \int _{\varOmega } \int _{0}^{ + \infty } g_{2}(s) \bigl({a_{2}}(x) \nabla {\eta ^{t}}(s)\bigr)\,ds \int _{0}^{t}g_{1}(t-s)a_{1}(x) \nabla u(s)\,ds \,dx\biggr| \\ &\le \frac{\mu _{8}+\mu _{9}}{2} \Vert a_{2} \Vert _{\infty } \int _{0}^{ + \infty } g_{2}(s)\,ds \bigl(g_{2}\circ \nabla {\eta ^{t}}\bigr) \\ &\quad{} + \biggl(\frac{b_{1}^{2}c_{*}^{2}}{2\mu _{8}}+\frac{1}{2\mu _{9}} \biggr) (1+ \mu _{5}) \biggl( \Vert a_{1} \Vert _{\infty } \int _{0}^{t}g_{1}(s)\,ds \biggr)^{2} \Vert \nabla u \Vert _{2}^{2} \\ &\quad{} + \biggl(\frac{b_{1}^{2}c_{*}^{2}}{2\mu _{8}}+\frac{1}{2\mu _{9}} \biggr) \biggl(1+ \frac{1}{ \mu _{5}}\biggr) \Vert a_{1} \Vert _{\infty } \int _{0}^{t}g_{1}(s)\,ds(g_{1} \circ \nabla u) (t). \end{aligned}$$

(3.7)

The seventh term in the right-hand side of (3.4) gives

$$\begin{aligned} & \biggl\vert \int _{\varOmega }f(u) \int _{0}^{t}g_{1}(t-s)a_{1}(x)u(s)\,ds \,dx \biggr\vert \\ &\quad \le \biggl\vert \int _{\varOmega }f(u) \int _{0}^{t}g_{1}(t-s)a_{1}(x) \bigl(u(s)-u(t)\bigr)\,ds \,dx \biggr\vert \\ &\qquad{} + \biggl\vert \int _{\varOmega }f(u) \int _{0}^{t}g_{1}(t-s)a_{1}(x)u(t)\,ds \,dx \biggr\vert \\ &\quad \le \frac{\mu _{10}}{2} \int _{\varOmega } \bigl\vert f(u) \bigr\vert ^{2}\,dx+ \frac{1}{2\mu _{10}} \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t-s)a_{1}(x)u(s)\,ds \biggr) ^{2}\,dx \\ &\quad \le \frac{\mu _{10}}{2} \int _{\varOmega } \bigl\vert f(u) \bigr\vert ^{2}\,dx+ \frac{c_{*}^{2}}{2 \mu _{10}}(1+\mu _{5}) \biggl( \Vert a_{1} \Vert _{\infty } \int _{0}^{t}g_{1}(s)\,ds \biggr) ^{2} \Vert \nabla u \Vert _{2}^{2} \\ &\qquad{} +\frac{c_{*}^{2}}{2\mu _{10}}\biggl(1+\frac{1}{\mu _{5}}\biggr) \Vert a_{1} \Vert _{\infty } \int _{0}^{t}g_{1}(s)\,ds(g_{1} \circ \nabla u) (t). \end{aligned}$$

(3.8)

From the last two terms in the right-hand side of (3.4) we get

$$\begin{aligned} & \biggl\vert \int _{\varOmega }u(t) \int _{0}^{t}g_{1}'(t-s)a_{1}(x)u(s)\,ds \,dx \biggr\vert \\ &\quad \le \biggl\vert \int _{\varOmega }u(t) \int _{0}^{t}g_{1}'(t-s)a_{1}(x) \bigl(u(t)-u(s)\bigr)\,ds \,dx \biggr\vert \\ &\qquad{} + \biggl\vert \int _{\varOmega }\bigl(u(t)\bigr)^{2} \int _{0}^{t}g_{1}'(t-s)a_{1}(x)\,ds \,dx \biggr\vert \\ &\quad \le \frac{\mu _{11}c_{*}^{2}}{2} \Vert \nabla u \Vert _{2}^{2}+ \frac{1}{2\mu _{11}} \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}'(t-s)a_{1}(x) \bigl(u(t)-u(s)\bigr)\,ds \biggr) ^{2}\,dx \\ &\qquad{} + \Vert a_{1} \Vert _{\infty } \bigl\vert g_{1}(t)-g_{1}(0) \bigr\vert c_{*}^{2} \Vert \nabla u \Vert _{2}^{2} \\ &\quad \le \biggl(\frac{\mu _{11}c_{*}^{2}}{2}+ \Vert a_{1} \Vert _{\infty } \bigl\vert g_{1}(t)-g _{1}(0) \bigr\vert c_{*}^{2} \biggr) \Vert \nabla u \Vert _{2}^{2} \\ &\quad {}-\frac{c_{*}^{2}}{2\mu _{11}} \Vert a_{1} \Vert _{\infty } \bigl\vert g_{1}(t)-g_{1}(0) \bigr\vert \bigl(g_{1}'\circ \nabla u\bigr) (t) \end{aligned}$$

(3.9)

and

$$\begin{aligned} g_{1}(0) \int _{\varOmega } \bigl\vert u(t) \bigr\vert ^{2}a_{1}(x)\,dx \le g_{1}(0) \Vert a_{1} \Vert _{ \infty }c_{*}^{2} \Vert \nabla u \Vert _{2}^{2}. \end{aligned}$$

(3.10)

Substituting inequalities (3.5)–(3.10) into (3.4), we obtain

$$\begin{aligned} \bigl\vert \phi '(t) \bigr\vert &\le \biggl[ \frac{\mu _{3}+\mu _{4}}{2}+\frac{\mu _{11}c_{*} ^{2}}{2}+ \Vert a_{1} \Vert _{\infty } \bigl\vert g_{1}(t)-g_{1}(0) \bigr\vert c_{*}^{2}+g_{1}(0) \Vert a _{1} \Vert _{\infty }c_{*}^{2} \\ &\quad {}+ \biggl(\frac{b_{1}^{2}c_{*}^{2}}{2\mu _{3}}+\frac{1}{2\mu _{4}}+\frac{b _{1}^{2}c_{*}^{2}}{2\mu _{6}}+ \frac{1}{2\mu _{7}}+\frac{b_{1}^{2}c_{*} ^{2}}{2\mu _{8}}+\frac{1}{2\mu _{9}}+\frac{c_{*}^{2}}{2\mu _{10}} \biggr)\\ &\quad {}\times (1+ \mu _{5}) \biggl( \Vert a_{1} \Vert _{\infty } \int _{0}^{t}g_{1}(s)\,ds \biggr)^{2} \biggr]\Vert \nabla u \Vert _{2}^{2} \\ &\quad {}+ \biggl[ \frac{\mu _{6}+\mu _{7}}{2}+ \biggl(\frac{b _{1}^{2}c_{*}^{2}}{2\mu _{3}}+\frac{1}{2\mu _{4}}+ \frac{b_{1}^{2}c_{*} ^{2}}{2\mu _{6}}+\frac{1}{2\mu _{7}}+ \frac{b_{1}^{2}c_{*}^{2}}{2\mu _{8}}+\frac{1}{2\mu _{9}}+ \frac{c_{*} ^{2}}{2\mu _{10}} \biggr) \biggl(1+\frac{1}{\mu _{5}}\biggr) \biggr] \\ &\quad {}\times \Vert a_{1} \Vert _{\infty } \int _{0}^{t}g_{1}(s)\,ds(g_{1} \circ \nabla u) (t) \\ &\quad{} + \frac{\mu _{8}+\mu _{9}}{2} \Vert a_{2} \Vert _{\infty } \int _{0}^{ + \infty } g _{2}(s)\,ds \bigl(g_{2}\circ \nabla {\eta ^{t}}\bigr) \\ &\quad{} +\frac{\mu _{10}}{2} \int _{\varOmega } \bigl\vert f(u) \bigr\vert ^{2}\,dx \\ &\quad{}-\frac{c_{*}^{2}}{2\mu _{11}} \Vert a_{1} \Vert _{\infty } \bigl\vert g_{1}(t)-g_{1}(0) \bigr\vert \bigl(g _{1}' \circ \nabla u\bigr) (t). \end{aligned}$$

Set

$$ \mu _{3}=\mu _{4}=\mu _{5}=\mu _{6}= \mu _{7}=\mu _{8}=\mu _{9}=\mu _{10}= \mu _{11}=1. $$

Then we obtain

$$\begin{aligned} \bigl\vert \phi '(t) \bigr\vert &\le \biggl[1+ \frac{c_{*}^{2}}{2}+2g_{1}(0) \Vert a_{1} \Vert _{ \infty }c_{*}^{2}+\bigl(3b_{1}^{2}c_{*}^{2}+3+c_{*}^{2} \bigr) (1-l)^{2} \biggr] \Vert \nabla u \Vert _{2}^{2} \\ &\quad{} +\bigl(3b_{1}^{2}c_{*}^{2}+4+c_{*}^{2} \bigr) (1-l) (g_{1}\circ \nabla u) (t)+ (1-l) \bigl(g _{2} \circ \nabla {\eta ^{t}}\bigr) (t) \\ &\quad{} +\frac{1}{2} \int _{\varOmega } \bigl\vert f(u) \bigr\vert ^{2}\,dx- \frac{c_{*}^{2}}{2} \Vert a_{1} \Vert _{\infty }g_{1}(0) \bigl(g_{1}'\circ \nabla u\bigr) (t). \end{aligned}$$

 □

We now state and prove our main theorem.

Theorem 3.1

Assume that (H1)–(H5) hold and ${u_{0}} \in H_{0}^{1}(\varOmega ) \cap {H^{2}}(\varOmega )$. Assume further that in case of (2.3), there exists ${M_{0}} > 0$ satisfying

$$\begin{aligned} \int _{\varOmega }{{{ \bigl\vert {\nabla {u_{0}}(x,t)} \bigr\vert }^{2}}\,dx} \le {M_{0}},\quad t > 0. \end{aligned}$$

(3.11)

Then there exist positive constants $\varepsilon _{0}$, $k_{1}$, and $k_{2}$ such that the energy functional of problem (2.4)–(2.5) satisfies

$$\begin{aligned} E(t) \le {k_{1}}G_{1}^{ - 1} \biggl({k_{2}} \int _{0}^{t} {\xi (s)\,ds} \biggr),\quad t > 0, \end{aligned}$$

(3.12)

where $G_{1}(t) = \int _{t}^{1} {\frac{1}{{{G_{0}}(s)}}} \,ds$, and

$$\begin{aligned} G_{0}(t) = \textstyle\begin{cases} t &\textit{if }(2.2) \textit{ holds}, \\ tG'({\varepsilon _{0}}t) &\textit{if }(2.3) \textit{ holds}. \end{cases}\displaystyle \end{aligned}$$

Remark 3.2

This result contains two particular cases: (i) $a_{1}=0$, where we can take $\xi \equiv 1$, and (ii) $a_{2}=0$, where we can take ${G_{0}}(t) = t$. Hence our result is more general; for example, case (ii) contains Theorem 1 in [12].

Proof

Applying (2.7), (3.2), and (3.3), for some $\varepsilon _{1}>0$ and $\varepsilon _{2}>0$, we obtain

$$\begin{aligned} \varPsi '(t)&= E'(t)+\varepsilon _{1}\chi '(t)+\varepsilon _{2}\phi '(t) \\ &\le - \frac{1}{2}{g_{1}}(t) \bigl\Vert {\sqrt{{a_{1}}(x)} \nabla u(t)} \bigr\Vert _{2}^{2} + \frac{1}{2} \bigl({g_{2}}' \circ \nabla {\eta ^{t}}\bigr) (t) \\ &\quad{} - \bigl\Vert u_{t}(t) \bigr\Vert _{2}^{2}+ \frac{1-l}{2\mu _{1}}\varepsilon _{1}\bigl[(g_{1}\circ \nabla u) (t)+\bigl(g_{2}\circ \nabla \eta ^{t}\bigr) (t)\bigr]- \varepsilon _{1} \int _{\varOmega }uf(u)\,dx \\ &\quad{} + \biggl\{ \biggl[1+\frac{c_{*}^{2}}{2}+2g_{1}(0) \Vert a_{1} \Vert _{\infty }c _{*}^{2}+ \bigl(3b_{1}^{2}c_{*}^{2}+3+c_{*}^{2} \bigr) (1-l)^{2} \biggr]\varepsilon _{2} -(l-\mu _{1})\varepsilon _{1} \biggr\} \Vert \nabla u \Vert _{2}^{2} \\ &\quad{} +\bigl(3b_{1}^{2}c_{*}^{2}+4+c_{*}^{2} \bigr) (1-l)\varepsilon _{2}(g_{1}\circ \nabla u) (t)+ (1-l) \varepsilon _{2}\bigl(g_{2}\circ \nabla {\eta ^{t}}\bigr) (t) \\ &\quad{} +\frac{1}{2}\varepsilon _{2} \int _{\varOmega } \bigl\vert f(u) \bigr\vert ^{2}\,dx+ \frac{1}{2}\bigl[1-c _{*}^{2}\varepsilon _{2} \Vert a_{1} \Vert _{\infty }g_{1}(0) \bigr]\bigl(g_{1}'\circ \nabla u\bigr) (t). \end{aligned}$$

Choosing $\varepsilon _{1}$ and $\varepsilon _{2}$ such that

$$\begin{aligned} & \biggl[1+\frac{c_{*}^{2}}{2}+2g_{1}(0) \Vert a_{1} \Vert _{\infty }c_{*}^{2}+\bigl(3b _{1}^{2}c_{*}^{2}+3+c_{*}^{2} \bigr) (1-l)^{2} \biggr]\varepsilon _{2} -(l-\mu _{1})\varepsilon _{1}< 0 \end{aligned}$$

and

$$\begin{aligned} &1-c_{*}^{2}\varepsilon _{2} \Vert a_{1} \Vert _{\infty }g_{1}(0)>0, \end{aligned}$$

we deduce

$$\begin{aligned} \varPsi '(t)&\le -cE(t)+C\bigl(g_{1}\circ \nabla u+g_{2}\circ \nabla \eta ^{t}\bigr),\quad t > 0. \end{aligned}$$

(3.13)

To get our conclusion, we deal with two cases to estimate $({g_{2}} \circ \nabla {\eta ^{t}})(t)$:

Case 1. Condition (2.2) holds. Using (2.7), we get

$$\begin{aligned} \bigl({g_{2}} \circ \nabla {\eta ^{t}}\bigr) (t) \le - \frac{1}{\sigma }\bigl({g_{2}}' \circ \nabla { \eta ^{t}}\bigr) (t) \le - \frac{2}{\sigma }E'(t),\quad t> 0. \end{aligned}$$

(3.14)

Case 2. Condition (2.3) holds. We follow the idea of [13]. Let $G^{*}(t) = \sup _{s \ge 0} \{ ts - G(s) \}$ be the dual function of the convex function G and set $K(s) = \frac{s}{{{G^{ - 1}}(s)}}$, $s \in \mathbb{R}^{+}$. Noting that ${G^{ - 1}}$ is a positive concave function and ${G^{ - 1}}(0) = 0$, for any $0 \le {s_{1}} < {s_{2}}$, we derive

$$\begin{aligned} K({s_{1}})&= \frac{{{s_{1}}}}{{{G^{ - 1}}( \frac{{{s_{1}}}}{{{s_{2}}}}{s_{2}} + (1 - \frac{{{s_{1}}}}{{{s_{2}}}})0)}} \\ &\le \frac{{{s_{1}}}}{{\frac{{{s_{1}}}}{{{s_{2}}}}{G^{ - 1}}({s_{2}})+ (1 - \frac{{{s_{1}}}}{{{s_{2}}}}){G^{ - 1}}(0)}} = \frac{{{s_{2}}}}{ {{G^{ - 1}}({s_{2}})}} = K({s_{2}}), \end{aligned}$$

which implies that $K(s)$ is nondecreasing.

From (2.6) and (3.11) we arrive at

$$\begin{aligned} \bigl\Vert {\sqrt{{a_{2}}(x)} \nabla {\eta ^{t}}} \bigr\Vert _{2}^{2}& \le 2 \bigl\Vert {\sqrt{{a_{2}}(x)} \nabla u(t)} \bigr\Vert _{2}^{2} + 2 \bigl\Vert { \sqrt{{a_{2}}(x)} \nabla u(t-s)} \bigr\Vert _{2}^{2} \le cE(t) \le c. \end{aligned}$$

Choosing $\tau _{1},{\tau _{2}} > 0$ and applying the Young inequality $ts \le G(t) + G^{*}(s)$ and the fact that $K(s)$ and $G^{*}$ are nondecreasing, we have

$$\begin{aligned} \bigl({g_{2}} \circ \nabla {\eta ^{t}}\bigr) (t)&= \frac{1}{{{\tau _{1}}G'({\varepsilon _{0}}E(t))}} \int _{0}^{ + \infty } {G^{ - 1}} \bigl( - {\tau _{2}} {g_{2}}'(s) \bigl\Vert { \sqrt{{a_{2}}(x)} \nabla {\eta ^{t}}} \bigr\Vert _{2}^{2} \bigr)\\&\quad{}\times \frac{{{\tau _{1}}G'({\varepsilon _{0}}E(t)) {g_{2}}(s) \Vert {\sqrt{{a_{2}}(x)} \nabla {\eta ^{t}}} \Vert _{2}^{2}}}{{{G^{ - 1}} ( - {\tau _{2}}{g_{2}}'(s) \Vert {\sqrt{ {a_{2}}(x)} \nabla {\eta ^{t}}} \Vert _{2}^{2} )}} \,ds \\ &\le - \frac{{{\tau _{2}}}}{{{\tau _{1}}G'({\varepsilon _{0}}E(t))}}\bigl( {g_{2}}' \circ \nabla { \eta ^{t}}\bigr) (t)\\&\quad{} + \frac{1}{{{\tau _{1}}G'({\varepsilon _{0}}E(t))}} \int _{0}^{ + \infty } {G^{*} \biggl( \frac{{c{\tau _{1}}G'({\varepsilon _{0}}E(t)){g_{2}}(s)}}{ {{G^{ - 1}}( - c{\tau _{2}}{g_{2}}'(s))}} \biggr)\,ds}. \end{aligned}$$

Using (2.7) and $G^{*}(s) = s{(G')^{ - 1}}(s) - G({(G')^{ - 1}}(s)) \le s{(G')^{ - 1}}(s)$, we get

$$\begin{aligned} \bigl({g_{2}} \circ \nabla {\eta ^{t}}\bigr) (t)& \le - \frac{{2{\tau _{2}}}}{{{\tau _{1}}G'({\varepsilon _{0}}E(t))}}E'(t) \\&\quad{}+ c \int _{0}^{ + \infty } \frac{ {{g_{2}}(s)}}{{{G^{ - 1}}( - c{\tau _{2}}{g_{2}}'(s))}}{{ \bigl(G'\bigr)}^{ - 1}} \biggl(\frac{{c{\tau _{1}}G'({\varepsilon _{0}}E(t)){g_{2}}(s)}}{{{G ^{ - 1}}( - c{\tau _{2}}{g_{2}}'(s))}} \biggr)\,ds. \end{aligned}$$

Setting ${\tau _{2}} = \frac{1}{c}$, by (2.3) we know that

$$ \bigl({g_{2}} \circ \nabla {\eta ^{t}}\bigr) (t) \le - \frac{2}{{c{\tau _{1}}G'( {\varepsilon _{0}}E(t))}}E'(t) + c{\bigl(G' \bigr)^{ - 1}}\bigl(c{\tau _{1}}G'\bigl({ \varepsilon _{0}}E(t)\bigr)\bigr) \int _{0}^{ + \infty } {\frac{{{g_{2}}(s)}}{{{G^{ - 1}}( - {g_{2}}'(s))}}\,ds}. $$

Then, setting ${\tau _{1}} = \frac{1}{c}$ and using (2.3) again, we obtain

$$\begin{aligned} G'\bigl({\varepsilon _{0}}E(t)\bigr) \bigl({g_{2}} \circ \nabla {\eta ^{t}}\bigr) (t) \le - cE'(t) + c{\varepsilon _{0}}E(t)G'\bigl({ \varepsilon _{0}}E(t)\bigr),\quad t > 0. \end{aligned}$$

(3.15)

From (3.14) and (3.15) it is easy to see that

$$\begin{aligned} \frac{{{G_{0}}(E(t))}}{{E(t)}}\bigl({g_{2}} \circ \nabla {\eta ^{t}}\bigr) (t) \le - cE'(t) + c{\varepsilon _{0}} {G_{0}}\bigl(E(t)\bigr),\quad t > 0, \end{aligned}$$

(3.16)

with $G_{0}$ given in Theorem 3.1.

Now, multiplying (3.13) by $\frac{{G_{0}}(E(t))}{E(t)}$, then using (3.16), and selecting $\varepsilon _{0}$ sufficiently small, we get

$$\begin{aligned} \frac{{{G_{0}}(E(t))}}{{E(t)}}\varPsi '(t) + cE'(t) \le - c{G_{0}}\bigl(E(t)\bigr)+ C\frac{{{G_{0}}(E(t))}}{{E(t)}}({g_{1}} \circ \nabla u) (t),\quad t > 0. \end{aligned}$$

(3.17)

Define

$$\begin{aligned} L(t) = \frac{{G_{0}}(E(t))}{E(t)}{\varPsi }(t) + cE(t),\quad t > 0. \end{aligned}$$

From the definition of $G_{0}$ we have that $t \mapsto \frac{{{G_{0}}(E(t))}}{ {E(t)}}$ is nonincreasing and nonnegative. Applying Lemma 3.1, we have $L(t) \sim E(t)$ for $t > 0$.

Differentiating $L(t)$, we get

$$\begin{aligned} L'(t)& = \biggl(\frac{{G_{0}}(E(t))}{E(t)} \biggr)'{\varPsi }(t)+ \frac{ {{G_{0}}(E(t))}}{{E(t)}}\varPsi '(t) + cE'(t) \\ &\le - c{G_{0}}\bigl(E(t)\bigr) + C\frac{{{G_{0}}(E(t))}}{{E(t)}}({g_{1}} \circ \nabla u) (t),\quad t > 0. \end{aligned}$$

(3.18)

To handle the last term in (3.18), multiplying it by $\xi (t)$ in both sides, and using (H2), (2.7), the fact that $t \mapsto \frac{{{G_{0}}(E(t))}}{{E(t)}}$ is nonnegative, we obtain

$$\begin{aligned} \xi (t)L'(t) &\le - c\xi (t){G_{0}}\bigl(E(t)\bigr) + C \frac{{{G_{0}}(E(t))}}{ {E(t)}}\xi (t) ({g_{1}} \circ \nabla u) (t) \\ &\le - c\xi (t){G_{0}}\bigl(E(t)\bigr) - C\bigl({g_{1}}' \circ \nabla u\bigr) (t) \\ &\le - c\xi (t){G_{0}}\bigl(E(t)\bigr) - CE'(t),\quad t > 0. \end{aligned}$$

(3.19)

Finally, set $I(t) = \xi (t)L(t) + CE(t)$. Obviously, $I(t) \sim E(t)$. Since $\xi (t)$ is nonincreasing, it follows form (3.19) that

$$\begin{aligned} I'(t) \le - c \xi (t){G_{0}}\bigl(E(t) \bigr),\quad t > 0. \end{aligned}$$

(3.20)

Since $G_{0}(s)>0$ for $s>0$ and $I(t) \sim E(t)$, we can see that

$$\begin{aligned} - \frac{{I'(t)}}{{{G_{0}}(I(t))}} \ge c\xi (t),\quad t > 0, \end{aligned}$$

and thus

$$\begin{aligned} \bigl( {{G_{1}}\bigl(I(t)\bigr)} \bigr)' \ge c\xi (t), \quad t > 0. \end{aligned}$$

Integrating this inequality over $(0,t)$, we obtain

$$\begin{aligned} {G_{1}}\bigl(I(t)\bigr) \ge c \int _{0}^{t} {\xi (s)\,ds} + {G_{1}} \bigl(I(0)\bigr)\ge c \int _{0}^{t} {\xi (s)\,ds},\quad t > 0. \end{aligned}$$

Since $G_{1}$ is nonincreasing, it is easy to get that

$$\begin{aligned} I(t) \le {G_{1}}^{ - 1} \biggl(c \int _{0}^{t} {\xi (s)\,ds} \biggr),\quad t >0. \end{aligned}$$

Since $I(t) \sim E(t)$, we obtain (3.12). The proof is completed. □

Now we give some examples to illustrate the result in Theorem 3.1, in which the exponential and polynomial decay estimates are only particular cases.

Example 3.1

• (i) Let $g_{1}(t)= \lambda e^{-(t+1)^{p}}$, $\lambda > 0$, $p\ge 1$, and $g_{2}(t)=e^{-(t+1)^{q}}$, $q\ge 1$. We can see that (2.1) and (2.2) hold for $\xi =p$ and $\sigma =q$, respectively. Then (3.12) gives the exponential decay estimate

  $$\begin{aligned} E(t) \le {k_{1}}e^{-k_{2} p t}. \end{aligned}$$

• (ii) Let $g_{1}(t)=(1+t)^{v}$, $v<-1$, and $g_{2}(t)= \lambda e^{-(t+1)^{p}}$, $\lambda > 0$, $p\ge 1$. Similarly, we can check that (2.1) and (2.2) hold for $\xi (t)= \frac{-v}{1+t}$ and $\sigma =p$, respectively. Then (3.12) gives the polynomial decay estimate

  $$\begin{aligned} E(t) \le {k_{1}}(1+t)^{k_{2}v}. \end{aligned}$$

• (iii) Let $g_{1}(t)=\frac{1}{(2+t)^{v}(\ln (2+t))^{\lambda }}$, $v,\lambda >0$, and $g_{2}(t)=\frac{1}{(1+t)^{q}}$, $q>1$. We can see that (2.1) holds for $\xi (t)=\frac{v}{2+t}+\frac{\lambda }{ \ln (2+t)}$ and (2.3) holds for $G(t)=t^{\frac{1}{p}+1}$, $p\in (0,\frac{q-1}{2})$. Then we obtain the decay estimate

  $$\begin{aligned} &E(t) \le \frac{k_{1}}{(k_{2}\int ^{t}_{0}\xi (s)\,ds+1)^{p}} = \frac{k _{1}}{(k_{2}\ln [(2+t)^{v}(\ln (2+t))^{\lambda }]-k_{2}\ln [2^{v}( \ln 2)^{\lambda }]+1)^{p}}. \end{aligned}$$