Asymptotic behaviour of solution for multidimensional viscoelasticity equation with nonlinear source term

Abstract

In this paper we study the initial-boundary value problem of the multidimensional viscoelasticity equation with nonlinear source term $u_{tt}-\mathrm{\Delta }{u}_t-{\sum }_{i=1}^N\frac{\partial }{\partial x_i}{\sigma }_i(u_{x_i})=f(u)$. By using the potential well method, we first prove the global existence. Then we prove that when time $t\to +\mathrm{\infty }$, the solution decays to zero exponentially under some assumptions on nonlinear functions and the initial data.

1 Introduction

This paper considers the initial-boundary value problem (IBVP) of the multidimensional viscoelasticity equation with nonlinear source term

(1.1)

(1.2)

(1.3)

where $u(x,t)$ is the unknown function with respect to the spacial variable $x\in \mathrm{\Omega }$ and the time variable $t,\mathrm{\Omega }\subset {\mathbb{R}}^N$ is a bounded domain.

The viscoelasticity equation

$$u_{tt}-u_{xxt}=\sigma {(u_x)}_x$$

(1.4)

was suggested and studied by Greenberg et al. [1, 2] from viscoelasticity mechanics in 1968. Under the condition ${\sigma }^{\prime }(s)>0$ and higher smooth conditions on $\sigma (s)$ and the initial data, they obtained the global existence of classical solutions for the initial-boundary value problem of Eq. (1.4).

After that many authors [3–11] studied the global well-posedness of IBVP for Eq. (1.4). In [3–10] the global existence, uniqueness and stability of solution were studied thoroughly. And in [11] the blow up of solution was discussed. Furthermore, in [12–14] the global existence of solution for IBVP of some multidimensional viscoelasticity equation was considered. And in [11] the blow up of solution for IBVP of the multidimensional generalisation of Eq. (1.4) was proved. Recently, in [15] and [16], the IBVP of the multidimensional viscoelasticity equation with nonlinear damping and source terms

(1.5)

(1.6)

(1.7)

was studied, and by using the potential well method, the global existence of weak solution was proved under some assumptions on nonlinear functions ${\sigma }_i(s)$, $f(s)$, $g(s)$ and the initial data. But we do not know how the global solution behaves as the time goes to infinity, namely the asymptotic behaviour of problem (1.1)-(1.3) is still open up to now. In the present paper, we try to study this problem by the multiplier method [17–22].

The main purpose of present paper is to consider the asymptotic behaviour of solution for problem (1.1)-(1.3). Since in the proof of the asymptotic behaviour of solution the global existence theory is required, it is necessary to give the proof of global existence of solution for problem (1.1)-(1.3).

In this paper, suppose that $\sigma (s)=({\sigma }_1(s),\dots ,{\sigma }_N(s))$ and $f(s)$ satisfy the following assumptions:

where the constants in (H₁) and (H₂) are all positive and satisfy

In this paper, we first give some definitions and lemmas (Section 2). Then we prove the global existence of solution (Section 3). Finally, we prove the asymptotic behaviour of solution (Section 4).

In this paper, we denote ${\parallel \cdot \parallel }_{L^p(\mathrm{\Omega })}$ by ${\parallel \cdot \parallel }_p$, $\parallel \cdot \parallel ={\parallel \cdot \parallel }_{L^2(\mathrm{\Omega })}$ and $(u,v)={\int }_{\mathrm{\Omega }}uv\mathrm{d}x$.

2 Preliminaries

In this section, we will give some definitions and prove some lemmas for problem (1.1)-(1.3).

For problem (1.1)-(1.3), we define

Remark 2.1 Note that the definitions of $J(u)$ and $I(u)$ in the present paper are different from those in [11] and [15]. The definitions given in this paper will be shown more natural and rational because they are a part of the total energy $E(t)$.

Lemma 2.2 Let (H₁) and (H₂) hold. Set

$$\overline{{\sigma }_i}(s)={\sigma }_i(s)-as,\overline{G_i}(s)={\int }_0^s\overline{{\sigma }_i}(s)\mathrm{d}\tau .$$

Then the following hold:

• (i) $\overline{{\sigma }_i}(s)$ is increasing and $s\overline{{\sigma }_i}(s)\ge 0$ $\mathrm{\forall }s\in \mathbb{R}$;

• (ii) $0\le \overline{G_i}(s)\le s\overline{{\sigma }_i}(s)$ $\mathrm{\forall }s\in \mathbb{R}$.

Proof This lemma follows from $\overline{{\sigma }_i}(0)=0$ and ${\overline{\sigma }}_i^{\prime }(s)\ge 0$. □

Lemma 2.3 Let (H₁) and (H₂) hold, $u\in W_0^{1,m+1}(\mathrm{\Omega })$. Then the following hold:

• (i) If $0<{\parallel \mathrm{\nabla }u\parallel }_{m+1}<r(\delta )$, then $I_{\delta }(u)>0$;

• (ii) If $I_{\delta }(u)<0$, then ${\parallel \mathrm{\nabla }u\parallel }_{m+1}>r(\delta )$;

• (iii) If $I_{\delta }(u)=0$, then ${\parallel \mathrm{\nabla }u\parallel }_{m+1}\ge r(\delta )$,

where

$$r(\delta )={\left(\frac{A\delta }{b{C}_{\ast }^{q+1}}\right)}^{\frac{1}{q-m}},C_{\ast }=\underset{u\in W_0^{1,m+1}(\mathrm{\Omega })/0}{sup}\frac{{\parallel u\parallel }_{q+1}}{{\parallel \mathrm{\nabla }u\parallel }_{m+1}}.$$

Proof

• (i) If $0<{\parallel \mathrm{\nabla }u\parallel }_{m+1}<r(\delta )$, then we have

  $$\begin{array}{rl}{\int }_{\mathrm{\Omega }}uf(u)\mathrm{d}x& \le {\int }_{\mathrm{\Omega }}|uf(u)|\mathrm{d}x\le b{\int }_{\mathrm{\Omega }}{|u|}^{q+1}\mathrm{d}x\\ =b{\parallel u\parallel }_{q+1}^{q+1}\le b{C}_{\ast }^{q+1}{\parallel \mathrm{\nabla }u\parallel }_{m+1}^{q+1}\\ =\frac{b{C}_{\ast }^{q+1}}{A}{\parallel \mathrm{\nabla }u\parallel }_{m+1}^{q-m}A{\parallel \mathrm{\nabla }u\parallel }_{m+1}^{m+1}\\ <\delta \sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{x_i}{\sigma }_i(u_{x_i})\mathrm{d}x,\end{array}$$

which gives $I_{\delta }(u)>0$.

• (ii) If $I_{\delta }(u)<0$, then we have

  $$\delta A{\parallel \mathrm{\nabla }u\parallel }_{m+1}^{m+1}\le \delta \sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{x_i}{\sigma }_i(u_{x_i})\mathrm{d}x<{\int }_{\mathrm{\Omega }}uf(u)\mathrm{d}x\le b{C}_{\ast }^{q+1}{\parallel \mathrm{\nabla }u\parallel }_{q-m}^{m+1}{\parallel \mathrm{\nabla }u\parallel }_{m+1}^{m+1},$$

which gives

$${\parallel \mathrm{\nabla }u\parallel }_{m+1}>r(\delta ).$$

• (iii) If $I_{\delta }(u)=0$ and $u\ne 0$, then by

  $$\delta A{\parallel \mathrm{\nabla }u\parallel }_{m+1}^{m+1}\le \delta \sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{x_i}{\sigma }_i(u_{x_i})\mathrm{d}x<{\int }_{\mathrm{\Omega }}uf(u)\mathrm{d}x\le b{C}_{\ast }^{q+1}{\parallel \mathrm{\nabla }u\parallel }_{m+1}^{q-m}{\parallel \mathrm{\nabla }u\parallel }_{m+1}^{m+1},$$

we get

$${\parallel \mathrm{\nabla }u\parallel }_{m+1}\ge r(\delta ).$$

 □

Lemma 2.4 Let (H₁) and (H₂) hold. Then the following holds:

$$d\ge d_0=\frac{(p-l)A}{(p+1)(l+1)}{\left(\frac{A}{b{C}_{\ast }^{q+1}}\right)}^{\frac{m+1}{q-m}}.$$

(2.1)

Proof For any $u\in \mathcal{N}$, by Lemma 2.3, we have ${\parallel \mathrm{\nabla }u\parallel }_{m+1}\ge r(1)$ and

$$\begin{array}{rl}J(u)& =\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{G}_i(u_{x_i})\mathrm{d}x-{\int }_{\mathrm{\Omega }}F(u)\mathrm{d}x\\ \ge \frac{1}{l+1}\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{x_i}{\sigma }_i(u_{x_i})\mathrm{d}x-\frac{1}{p+1}{\int }_{\mathrm{\Omega }}uf(u)\mathrm{d}x\\ =(\frac{1}{l+1}-\frac{1}{p+1})\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{x_i}{\sigma }_i(u_{x_i})\mathrm{d}x+\frac{1}{p+1}I(u)\\ =\frac{p-l}{(p+1)(l+1)}\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{x_i}{\sigma }_i(u_{x_i})\mathrm{d}x\\ \ge \frac{p-l}{(p+1)(l+1)}A{\parallel \mathrm{\nabla }u\parallel }_{m+1}^{m+1}\\ \ge \frac{p-l}{(p+1)(l+1)}A{r}^{m+1}(1)=\frac{(p-l)A}{(p+1)(l+1)}{\left(\frac{A}{b{C}_{\ast }^{q+1}}\right)}^{\frac{m+1}{q-m}},\end{array}$$

which gives (2.1). □

Now, for problem (1.1)-(1.3), we define

$$W=\{u\in W_0^{1,m+1}(\mathrm{\Omega })\mid I(u)>0\}\cup \{0\}.$$

3 Global existence of solution

In this section, we prove the global existence of weak solution for problem (1.1)-(1.3).

Definition 3.1 We call $u=u(x,t)$ a weak solution of problem (1.1)-(1.3) on $\mathrm{\Omega }\times [0,T)$ if $u\in L^{\mathrm{\infty }}(0,T;W_0^{1,m+1}(\mathrm{\Omega }))$, $u_t\in L^{\mathrm{\infty }}(0,T;L^2(\mathrm{\Omega }))\cap L^2(0,T;H_0^1(\mathrm{\Omega }))$ satisfying

• (i)
  
• (ii)
  $$u(x,0)=u_0(x)\text{in }{W}_0^{1,m+1}(\mathrm{\Omega });u_t(x,0)=u_1(x)\text{in }{L}^2(\mathrm{\Omega }).$$

Theorem 3.2 Let (H₁) and (H₂) hold, $u_0(x)\in W_0^{1,m+1}(\mathrm{\Omega })$, $u_1(x)\in L^2(\mathrm{\Omega })$. Assume that $E(0)<d$, $u_0(x)\in W$. Then problem (1.1)-(1.3) admits a global weak solution $u\in L^{\mathrm{\infty }}(0,\mathrm{\infty };W_0^{1,m+1}(\mathrm{\Omega }))$ and $u_t\in L^{\mathrm{\infty }}(0,\mathrm{\infty };L^2(\mathrm{\Omega }))\cap L^2(0,\mathrm{\infty };H_0^1(\mathrm{\Omega }))$.

Proof Let ${\{w_j(x)\}}_{j=1}^{\mathrm{\infty }}$ be a system of base functions in $W_0^{1,m+1}(\mathrm{\Omega })$. Construct the approximate solutions of problem (1.1)-(1.3)

$$u_n(x,t)=\sum _{j=1}^n{g}_{jn}(t)w_j(x),n=1,2,\dots ,$$

satisfying

(3.1)

(3.2)

(3.3)

Multiplying (3.1) by $g_{sn}^{\prime }(t)$ and summing for s, we get

$$\frac{\mathrm{d}}{\mathrm{d}t}{E}_n(t)+{\parallel \mathrm{\nabla }{u}_{nt}\parallel }^2=0,0\le t<\mathrm{\infty }$$

(3.4)

and

$$E_n(t)+{\int }_0^t{\parallel \mathrm{\nabla }{u}_{n\tau }\parallel }^2\mathrm{d}\tau =E_n(0),0\le t<\mathrm{\infty },$$

(3.5)

where

$$E_n(t)=\frac{1}{2}{\parallel u_{nt}\parallel }^2+J(u_n).$$

From (3.2) and (3.3), we have $E_n(0)\to E(0)$ as $n\to \mathrm{\infty }$. Hence, for sufficiently large n, we have $E_n(0)<d$ and

$$\frac{1}{2}{\parallel u_{nt}\parallel }^2+J(u_n)+{\int }_0^t{\parallel \mathrm{\nabla }{u}_{n\tau }\parallel }^2\mathrm{d}\tau <d,0\le t<\mathrm{\infty }.$$

(3.6)

On the other hand, since W is an open set in $W_0^{1,m+1}(\mathrm{\Omega })$, Eq. (3.2) implies that for sufficiently large n, we have $u_n(0)\in W$. Next, we prove that $u_n(t)\in W$ for $0<t<\mathrm{\infty }$ and sufficiently large n. If it is false, then there exists a $t_0>0$ such that $u_n(t_0)\in \partial W$, i.e. $I(u_n(t_0))=0$ and $u_n(t_0)\ne 0$, i.e. $u_n(t_0)\in \mathcal{N}$. So, by the definition of d, we get $J(u_n(t_0))\ge d$, which contradicts (3.6).

From (3.6) we have

$$\frac{1}{2}{\parallel u_{nt}\parallel }^2+\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{G}_i(u_{n{x}_i})\mathrm{d}x-{\int }_{\mathrm{\Omega }}F(u_n)\mathrm{d}x+{\int }_0^t{\parallel \mathrm{\nabla }{u}_{n\tau }\parallel }^2\mathrm{d}\tau <d,0\le t<\mathrm{\infty },$$

which gives

$$\frac{1}{2}{\parallel u_{nt}\parallel }^2+\frac{1}{l+1}\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{n{x}_i}{\sigma }_i(u_{n{x}_i})\mathrm{d}x-\frac{1}{p+1}{\int }_{\mathrm{\Omega }}{u}_{n}f(u_n)\mathrm{d}x+{\int }_0^t{\parallel \mathrm{\nabla }{u}_{n\tau }\parallel }^2\mathrm{d}\tau <d$$

and

which together with $u_n(t)\in W$ gives

$$\frac{1}{2}{\parallel u_{nt}\parallel }^2+\frac{p-l}{(p+1)(l+1)}\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{n{x}_i}{\sigma }_i(u_{n{x}_i})\mathrm{d}x+{\int }_0^t{\parallel \mathrm{\nabla }{u}_{n\tau }\parallel }^2\mathrm{d}\tau <d,$$

and

$$\frac{1}{2}{\parallel u_{nt}\parallel }^2+\frac{p-l}{(p+1)(l+1)}A{\parallel \mathrm{\nabla }{u}_n\parallel }_{m+1}^{m+1}+{\int }_0^t{\parallel \mathrm{\nabla }{u}_{n\tau }\parallel }^2\mathrm{d}\tau <d,0\le t<\mathrm{\infty }.$$

(3.7)

From (3.7) we can get

(3.8)

(3.9)

(3.10)

(3.11)

(3.12)

Hence there exist u, $\chi =({\chi }_1,{\chi }_2,\dots ,{\chi }_N)$, η and a subsequence $\{u_{\nu }\}$ of $\{u_n\}$ such that as $\nu \to \mathrm{\infty }$, $u_{\nu }\to u$ in $u\in L^{\mathrm{\infty }}(0,\mathrm{\infty };W_0^{1,m+1}(\mathrm{\Omega }))$ weak-star, and a.e. in $Q=\mathrm{\Omega }\times [0,\mathrm{\infty })$, $u_{\nu t}\to u_t$ in $L^{\mathrm{\infty }}(0,\mathrm{\infty };L^2(\mathrm{\Omega }))$ weak-star and in $L^2(0,\mathrm{\infty };H_0^1(\mathrm{\Omega }))$ weakly, ${\sigma }_i(u_{\nu x_i})\to {\chi }_i={\sigma }_i(u_{x_i})$ in $L^{\mathrm{\infty }}(0,\mathrm{\infty };L^{{(m+1)}^{\prime }}(\mathrm{\Omega }))$ weak-star, ${(m+1)}^{\prime }=\frac{m+1}{m}$, $f(u_v)\to \eta =f(u)$ in $L^{\mathrm{\infty }}(0,\mathrm{\infty };L^{{(q+1)}^{\prime }}(\mathrm{\Omega }))$ weak-star, ${(q+1)}^{\prime }=\frac{q+1}{q}$.

Integrating (3.1) with respect to t, we have

(3.13)

Letting $n=\nu \to \mathrm{\infty }$ in (3.13), we get

and

On the other hand, from (3.2) and (3.3), we get $u(x,0)=u_0(x)$ in $W_0^{1,m+1}(\mathrm{\Omega })$, $u_t(x,0)=u_1(x)$ in $L^2(\mathrm{\Omega })$. Therefore u is a global weak solution of problem (1.1)-(1.3). □

4 Asymptotic behaviour of solution

In this section, we prove the main conclusion of this paper - the asymptotic behaviour of solution for problem (1.1)-(1.3).

Lemma 4.1 Let (H₁) and (H₂) hold, $u_0(x)\in W_0^{1,m+1}(\mathrm{\Omega })$, $u_1(x)\in L^2(\mathrm{\Omega })$. Then, for the approximate solutions $u_n(x,t)$ of problem (1.1)-(1.3) constructed in the proof of Theorem 3.2, the following hold:

• (i)
  $$I(u_n)={\parallel u_{nt}\parallel }^2-\frac{\mathrm{d}}{\mathrm{d}t}((u_{nt},u_n)+\frac{1}{2}{\parallel \mathrm{\nabla }{u}_n\parallel }^2);$$

  (4.1)

• (ii) Furthermore, if $E(0)<d_0$ and $u_0(x)\in W$, then for sufficiently large n, there exists a ${\delta }_1\in (0,1)$ such that

  $$I(u_n)\ge (1-{\delta }_1)\sum _{i=1}^N({\sigma }_i(u_{n{x}_i}),u_{n{x}_i}).$$

  (4.2)

Proof (i) Multiplying (3.1) by $g_{sn}(t)$ and summing for s, we get (4.1).

• (ii) From

  $$E(0)<d_0=\frac{p-l}{(p+1)(l+1)}A{\left(\frac{A}{b{C}_{\ast }^{q+1}}\right)}^{\frac{m+1}{q-m}}$$

it follows that there exists a ${\delta }_1\in (0,1)$ such that

$$E(0)<\frac{p-l}{(p+1)(l+1)}A{\left(\frac{A{\delta }_1}{b{C}_{\ast }^{q+1}}\right)}^{\frac{m+1}{q-m}}\equiv d({\delta }_1).$$

(4.3)

From (3.2), (3.3) and (4.3), it follows that $E_m(0)<d({\delta }_1)$ for sufficiently large n. Hence from (3.5) we have

and

$$\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{G}_i(u_{n{x}_i})\mathrm{d}x-{\int }_{\mathrm{\Omega }}F(u_n)\mathrm{d}x<d({\delta }_1),$$

which gives

$$\frac{1}{l+1}\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{n{x}_i}{\sigma }_i(u_{n{x}_i})\mathrm{d}x-\frac{1}{p+1}{\int }_{\mathrm{\Omega }}{u}_{n}f(u_n)\mathrm{d}x<d({\delta }_1)$$

and

$$\frac{p-l}{(p+1)(l+1)}\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{n{x}_i}{\sigma }_i(u_{n{x}_i})\mathrm{d}x+\frac{1}{p+1}I(u_n)<d({\delta }_1),$$

which together with $u_n\in W$ for sufficiently large n gives

$$\frac{p-l}{(p+1)(l+1)}A{\parallel \mathrm{\nabla }{u}_n\parallel }_{m+1}^{m+1}<\frac{p-l}{(p+1)(l+1)}A{\left(\frac{A{\delta }_1}{b{C}_{\ast }^{q+1}}\right)}^{\frac{m+1}{q-m}}$$

and

$${\parallel \mathrm{\nabla }{u}_n\parallel }_{m+1}<{\left(\frac{A{\delta }_1}{b{C}_{\ast }^{q+1}}\right)}^{\frac{1}{q-m}}=r({\delta }_1).$$

Hence, by Lemma 2.3, we have $I_{{\delta }_1}(u_n)>0$ or $u_n=0$. So, we have

$$\begin{array}{rl}I(u_n)& =\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{n{x}_i}{\sigma }_i(u_{n{x}_i})\mathrm{d}x-{\int }_{\mathrm{\Omega }}{u}_{n}f(u_n)\mathrm{d}x\\ =(1-{\delta }_1)\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{n{x}_i}{\sigma }_i(u_{n{x}_i})\mathrm{d}x+I_{{\delta }_1}(u_n)\ge (1-{\delta }_1)\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{n{x}_i}{\sigma }_i(u_{n{x}_i})\mathrm{d}x.\end{array}$$

 □

Theorem 4.2 Let (H₁) and (H₂) hold, $u_0(x)\in W_0^{1,m+1}(\mathrm{\Omega })$, $u_1(x)\in L^2(\mathrm{\Omega })$. Assume that $E(0)<d_0$, $u_0(x)\in W$. Then, for the global weak solution u given in Theorem 3.2, there exist positive constants C and λ such that

$${\parallel u_t\parallel }^2+{\parallel \mathrm{\nabla }u\parallel }_{m+1}^{m+1}\le C{e}^{-\lambda t},0\le t<\mathrm{\infty }.$$

(4.4)

Proof Let $\{u_n\}$ be the approximate solutions of problem (1.1)-(1.3) in the proof of Theorem 3.2, then (3.4) holds. Multiplying (3.4) by $e^{\alpha t}$ ($\alpha >0$), we get

$$\frac{\mathrm{d}}{\mathrm{d}t}(e^{\alpha t}{E}_n(t))+e^{\alpha t}{\parallel \mathrm{\nabla }{u}_{nt}\parallel }^2=\alpha e^{\alpha t}{E}_n(t)$$

and

$$e^{\alpha t}{E}_n(t)+{\int }_0^t{e}^{\alpha \tau }{\parallel \mathrm{\nabla }{u}_{n\tau }\parallel }^2\mathrm{d}\tau =E_n(0)+\alpha {\int }_0^t{e}^{\alpha \tau }{E}_n(\tau )\mathrm{d}\tau ,0\le t<\mathrm{\infty }.$$

(4.5)

From (H₂), Lemma 2.2 and Lemma 4.1, we get

where

$$C(r,{\delta }_1)=\frac{1}{1-{\delta }_1}\frac{r}{r+1}+\frac{1}{r+1}.$$

Hence we have

(4.6)

and

(4.7)

From

$$\frac{1}{2}{\parallel u_{nt}\parallel }^2+\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{G}_i(u_{n{x}_i})\mathrm{d}x-{\int }_{\mathrm{\Omega }}F(u_n)\mathrm{d}x=E_n(t)$$

we get

$$\frac{1}{2}{\parallel u_{nt}\parallel }^2+\frac{1}{l+1}\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{n{x}_i}{\sigma }_i(u_{n{x}_i})\mathrm{d}x-\frac{1}{p+1}{\int }_{\mathrm{\Omega }}{u}_{n}f(u_n)\mathrm{d}x\le E_n(t)$$

and

$$\frac{1}{2}{\parallel u_{nt}\parallel }^2+\frac{p-l}{(p+1)(l+1)}\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{n{x}_i}{\sigma }_i(u_{n{x}_i})\mathrm{d}x+\frac{1}{p+1}I(u_n)\le E_n(t),$$

which together with $u_n\in W$ for sufficiently large n gives

$$\begin{array}{r}\frac{1}{2}{\parallel u_{nt}\parallel }^2+\frac{p-l}{(p+1)(l+1)}\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{n{x}_i}{\sigma }_i(u_{n{x}_i})\mathrm{d}x\le E_n(t),\\ \frac{1}{2}{\parallel u_{nt}\parallel }^2+\frac{p-l}{(p+1)(l+1)}(a{\parallel \mathrm{\nabla }{u}_n\parallel }^2+\sum _{i=1}^N{\int }_{\mathrm{\Omega }}{u}_{n{x}_i}{\overline{\sigma }}_i(u_{n{x}_i})\mathrm{d}x)\le E_n(t),\end{array}$$

(4.8)

and

$$\frac{1}{2}{\parallel u_{nt}\parallel }^2+\frac{p-l}{(p+1)(l+1)}a{\parallel \mathrm{\nabla }{u}_n\parallel }^2\le E_n(t),0\le t<\mathrm{\infty }.$$

(4.9)

From (4.8) and the Poincaré inequality ${\parallel \mathrm{\nabla }u\parallel }^2\ge {\lambda }_1{\parallel u\parallel }^2$, it follows that there exists a constant $C_0=C_0(p,l,a,{\lambda }_1)>0$ such that

$$({\parallel u_{nt}\parallel }^2+{\parallel u_n\parallel }^2+{\parallel \mathrm{\nabla }{u}_n\parallel }^2)\le C_0{E}_n(t),0\le t<\mathrm{\infty }.$$

(4.10)

From (4.5)-(4.10) it follows that there exists a $C_0$ such that

(4.11)

Choose α such that

$$0<\alpha <min\{\frac{1}{2C_0},\frac{{\lambda }_1}{\frac{1}{2}+C(r,{\delta }_1)}\}.$$

Then from (4.11) we get

$$\begin{array}{rl}{e}^{\alpha t}{E}_n(t)& \le 2(C_0\alpha +1)E_n(0)+2{\alpha }^2{C}_0{\int }_0^t{e}^{\alpha \tau }{E}_n(\tau )\mathrm{d}\tau \\ \le 2(C_0\alpha +1)d_0+2{\alpha }^2{C}_0{\int }_0^t{e}^{\alpha \tau }{E}_n(\tau )\mathrm{d}\tau \\ \le 3d_0+2\alpha C_0{\int }_0^t{e}^{\alpha \tau }{E}_n(\tau )\mathrm{d}\tau .\end{array}$$

From this and the Gronwall inequality, we get

$$e^{\alpha t}{E}_n(t)\le 3d_0{e}^{-2{\alpha }^2{C}_{0}t}$$

and

$$E_n(t)\le 3d_0{e}^{-\lambda t},0\le t<\mathrm{\infty },\lambda =\alpha (1-2C_0\alpha )>0.$$

(4.12)

On the other hand, from (4.8) we get

$$\frac{1}{2}{\parallel u_{nt}\parallel }^2+\frac{p-l}{(p+1)(l+1)}A{\parallel \mathrm{\nabla }{u}_n\parallel }_{m+1}^{m+1}\le E_n(t),0\le t<\mathrm{\infty }.$$

Hence, there exists a $C_1=C_1(p,l,A)$ such that

$${\parallel u_{nt}\parallel }^2+{\parallel \mathrm{\nabla }{u}_n\parallel }_{m+1}^{m+1}\le C_1{E}_n(t),0\le t<\mathrm{\infty }.$$

(4.13)

Let $\{u_{\nu }\}$ be the subsequence of $\{u_n\}$ in the proof of Theorem 3.2. Then from (4.13) and(4.12), we obtain

which gives (4.4), where $C=3d_0{C}_1$, $\lambda =\alpha (1-2C_0\alpha )$. □