一类具有局部记忆阻尼的弱耦合系统的能量衰减估计


扩展功能
	加入收藏夹

	复制引文信息

	加入引用管理器

	Email Alert

	RSS
本文作者相关文章
	章春国

	张海燕

	谷尚武

章春国, 张海燕,谷尚武

(杭州电子科技大学数学系杭州 310018)

收稿日期: 2013-11-07;修订日期: 2014-10-04

基金项目:国家自然科学基金(61374096, 11271104) 资助

E-mail:作者简介：章春国, cgzhang@hdu.edu.cn

摘要：研究具有局部记忆阻尼弱耦合梁-弦系统.首先在合适的假设条件下, 应用线性算子半群理论证明了系统的适定性; 进而运用线性算子半群的频域定理证明了具有局部记忆阻尼弱耦合梁-弦系统的能量是一致指数衰减的.

关键词：耦合梁-弦系统线性算子半群局部记忆阻尼一致指数衰减

Energy Decay Estimates for the Weakly Coupled Systems with Local Memory Damping

Zhang Chunguo, Zhang Haiyan,Gu Shangwu

epartment of Mathematics, Hangzhou Dianzi University, Hangzhou Zhejiang 310018

Abstract : This paper studies the weakly coupled beam-string systems with local memory damping. First, under the appropriate hypothesis, we proved that the well-posedness of the system by using the theory of linear operator semigroup. And then, we show that the energy of the weakly coupled beam-string system with local memory damping is uniform exponential decay by applying the frequence domain result on Hilbert space.

Key words: Coupled beam-string system Linear operator semigroup Local memory damping Uniform exponential decay

1 引言

在实际力学系统中,柔性梁(柔性结构)的镇定和稳定是一个十分重要的控制问题. 基于梁、弦、板等系统在空间科学及机器人学中的广泛应用, 由智能材料制成的补钉黏贴在或嵌入到基底结构中作为主动或被动的阻尼器, 为了获得最优配置(最佳控制)结果,往往需要知道系统能量衰减与系统参数之间的定量关系, 许多实际系统的控制作用往往取决于系统的这一指标. 为了满足科技日益发展的需要, 因此对各类系统模型进行稳定性分析是一件十分有意义的事情.

近二十年来,随着应用的发展和技术要求的不断提高, 驱使国内外许多数学和力学工作者研究各种具有不同类型阻尼的Euler-Bernoulli 梁, Timoshenko梁,Rayleigh梁以及Petrovsky板等系统(耦合系统)的稳定性. 例如: 文献^{[1, 2]}研究的是非线性阻尼耦合振动的Petrovsky系统, 文献^{[3, 4, 5, 6, 7, 8, 9, 10]}考虑的是各种具有不同阻尼的弦、梁及波系统的稳定性. 然而局部阻尼在阻尼器的边缘往往会发生骤变(不连续)现象,在数学上, 这种不连续现象会对系统的稳定性分析产生许多困难. 这些困难往往会激发学者对这些问题的研究兴趣. 文献^[11] 研究的是具有全局阻尼Petrovsky 系统稳定性, 本文将文献^[11]的结果推广到更一般的局部记忆阻尼情形. 我们研究的是一类具有局部记忆阻尼弱耦合梁-弦系统的能量估计. 更精确地说,考虑如下的初边值问题

\begin{equation}\left\{\begin{array}{lll} (u_{1})_{tt}+u_{1}^{(4)}+a u_{2}- \int_{0}^{+\infty}g_{1s}[b_{1}(u''_{1}(x,t)-u''_{1}(x,t-s))]''{\rm d}s =0,& 0\leq x\leq L,t>0,\\[3mm] (u_{2})_{tt}-u''_{2}+a u_{1}+\int_{0}^{+\infty}g_{2s}[b_{2}(u'_{2}(x,t)-u'_{2}(x,t-s))]'{\rm d}s =0,& 0\leq x\leq L,t>0,\\[2mm] u_{1}(0)=u'_{1}(0)=u_{1}(L)=u'_{1}(L)=u_{2}(0)=u_{2}(L)=0, &t>0,\\ u_{i}(x,0)=u_{i}^{0}(x),(u_{i})_{t}(x,0)=u_{i}^{1}(x), &0\leq x\leq L,i=1,2, \end{array}\right. \label{1.1} \end{equation}

(1)

其中“$'$”表示对空间变量$x$的导数.

本文首先假设

(H1) 弱耦合系数$a(x)\in L^{\infty}(0,L)$.

(H2) 局部阻尼系数$b_{1}(x)\in C^{2}(0,L),b_{2}(x)\in C^{1}(0,L)$. 且满足: 对于 $i=1,2$, 当$x\in(0,\alpha)$ 时,$b_{i}(x)=0$; 当$x\in(\alpha,L)$时, $b_{i}(x)\geq C>0$,其中$C$为正常数.

(H3) 松弛函数$g_{i}(s),i=1,2$ 满足如下条件

$(g1)$~ $ g_{i}(s)\in C^{2}(0,+\infty)\cap C[0,+\infty)$和 $ g_{i}(s)\in L^{1}(0,+\infty)$;

$(g2)$~ 在$(0,+\infty)$上,$g_{i}>0,g_{is}<0,g_{iss}>0$;

$(g3)$~ 在$(0,+\infty)$上,存在$k_{2}>k_{1}>0$使得$-k_{1}g_{is} $(g4)$~ $g_{i}(\infty)=0$.

本文的方法源于文献^[5]和^[9]. 在合适的假设下,应用经典结果^[12]和$C_{0}$-半群生成元的预解式在虚轴上的有界性频域结果^[13],并且运用线性算子半群理论、分片乘子技巧以及矛盾的讨论得到了系统(1)的适定性和能量是一致指数衰减性质.

2 预备知识与主要结果

为了研究系统(1)的能量衰减性质,我们进一步假设

(H4)

\begin{equation}\|a\|_{L^{\infty}(0,L)}<\frac{1}{2\sqrt{c_{1}c_{2}}}, \label{2}\end{equation}

(2)

这里$c_{1},c_{2}$是两个正常数,同时 $$\|u\|_{H^{2}(0,L)}^{2}\leq c_{1}\int_{0}^{L}|u''|^{2}{\rm d}x,\quad \forall u\in H_{0}^{2}(0,L),$$ $$\|u\|_{H^{1}(0,L)}^{2}\leq c_{2}\int_{0}^{L}|u'|^{2}{\rm d}x,\quad \forall u\in H_{0}^{1}(0,L).$$ 为了方便起见,记$\|a\|_{L^{\infty}(0,L)}$为$\|a\|_{\infty}$.

(H5)

\begin{equation}\gamma_{i}=1-\frac{3}{2}\sqrt{c_{1}c_{2}}\|a\|_{\infty} -\frac{1}{2}\|b_{i}\|\int_{0}^{+\infty}|g_{is}(s)|{\rm d}s>0,\quad i=1,2, \label{3}\end{equation}

(3)

这里$\|b_{i}\|=\sup\limits_{x\in[0,L]}\left|b_{i}(x)\right|,i=1,2$. 由(H4)得

\begin{eqnarray} \int_{0}^{L}a(u_{1}\overline{u}_{2}+\overline{u}_{1}u_{2}){\rm d}x &\geq&-\|a\|_{\infty} \int_{0}^{L}\left[\sqrt{\frac{c_{2}}{c_{1}}}|u_{1}|^{2} +\sqrt{\frac{c_{1}}{c_{2}}}|u_{2}|^{2}\right]{\rm d}x \nonumber\\ &\geq &-\sqrt{c_{1}c_{2}}\|a\|_{\infty} \int_{0}^{L}\left[|u''_{1}|^{2} +|u'_{2}|^{2}\right]{\rm d}x.\end{eqnarray}

(4)

定义系统(1)在时刻$t$的能量为

\begin{eqnarray} E(t)&=&\frac{1}{2}\int_{0}^{L}\left(|u_{1t}|^{2}+|u_{2t}|^{2}+|u''_{1}|^{2}+|u'_{2}|^{2}\right){\rm d}x +\frac{1}{2}\int_{0}^{L}a(u_{1}\overline{u}_{2}+\overline{u}_{1}u_{2}){\rm d}x \nonumber\\ &&+\frac{1}{2}\int_{0}^{+\infty}|g_{1s}|\int_{0}^{L}b_{1}|y''_{1}|^{2}{\rm d}x{\rm d}s +\frac{1}{2}\int_{0}^{+\infty}|g_{2s}|\int_{0}^{L}b_{2} |y'_{2}|^{2}{\rm d}x{\rm d}s, \end{eqnarray}

(5)

其中$y_{1}(x,t,s)=u_{1}(x,t)-u_{1}(x,t-s), y_{2}(x,t,s)=u_{2}(x,t)-u_{2}(x,t-s)$.

引入函数空间 $$H_{0}^{2}(0,L)=\{u|u\in H^{2}(0,L),u(0)=u'(0)=u(L)=u'(L)=0\},$$ $$ H_{0}^{1}(0,L)=\{u|u\in H^{1}(0,L),u(0)=u(L)=0\},$$ 其中$H^{k}(0,L)$是$k$阶Sobolev空间(参见文献^[14]). $$V=H_{0}^{2}(0,L)\times H_{0}^{1}(0,L),\quad H=L^{2}(0,L)\times L^{2}(0,L),$$ 赋予范数 $$\|(u_{1},u_{2})\|_{V}^{2}=\int_{0}^{L}(|u''_{1}|^{2}+|u'_{2}|^{2}){\rm d}x +\int_{0}^{L}a(u_{1}\overline{u}_{2}+\overline{u}_{1}u_{2}){\rm d}x, $$ $$\|(v_{1},v_{2})\|_{H}^{2}=\int_{0}^{L}(|v_{1}|^{2}+|v_{2}|^{2}){\rm d}x.$$ 同时,记 $$W_{1}=\left\{v|v,v',v''\in L^{2}(\delta,L),\forall\delta\in(\alpha,L), \sqrt{b_{1}}v''\in L^{2}(\alpha,L),v(L)=0\right\},$$ $$W_{2}= \left\{v|v,v'\in L^{2}(\delta,L),\forall\delta\in(\alpha,L), \sqrt{b_{2}}v'\in L^{2}(\alpha,L),v(L)=0\right\},$$ $$ Y=L^{2}(0,+\infty; W_{1})\times L^{2}(0,+\infty; W_{2}),$$ $$\|(y_{1},y_{2})\|_{Y}^{2}=\int_{0}^{+\infty}|g_{1s}|\int_{0}^{L}b_{1}|y''_{1}|^{2}{\rm d}x{\rm d}s+ \int_{0}^{+\infty}|g_{2s}|\int_{0}^{L}b_{2}|y'_{2}|^{2}{\rm d}x{\rm d}s.$$ 因此$V,H,Y$均为实(复) Hilbert空间,且具有范数$\|(\cdot,\cdot)\|_{V},\|(\cdot,\cdot)\|_{H},\|(\cdot,\cdot)\|_{Y}$.

设${{\cal H}}=V\times H\times Y$,赋予范数 $$\|Z\|_{{\cal H}}^{2}=\|(u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})\|_{{\cal H}}^{2} =\|(u_{1},u_{2})\|_{V}^{2}+\|(v_{1},v_{2})\|_{H}^{2}+\|(y_{1},y_{2})\|_{Y}^{2},$$ 那么${\cal H}$也是实(复) Hilbert空间.

接下来,设$Z=(u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})$,在 ${\cal H}$中定义线性算子${\cal A}$如下

\begin{equation}D({\cal A})=\left\{\begin{array}{lll} Z\in{{\cal H}}|(v_{1},v_{2})\in V ,\quad (y_{1s},y_{2s})\in Y, \quad y_{1}(\cdot,0)=y_{2}(\cdot,0)=0,\\[2mm] u''_{1}-\int_{0}^{+\infty}g_{1s}b_{1}y''_{1}{\rm d}s\in H^{2}(0,L),\quad -u'_{2}+\int_{0}^{+\infty}g_{2s}b_{2}y'_{2}{\rm d}s\in H^{1}(0,L). \end{array}\right\}\label{6}\end{equation}

(6)

\begin{equation}{\cal A}Z=\left(\begin{array}{lll} v_{1},v_{2},-au_{2}-u_{1}^{(4)} +\int_{0}^{+\infty}g_{1s}(b_{1}y''_{1})''{\rm d}s,\\[3mm] -au_{1}+u''_{2}-\int_{0}^{+\infty}g_{2s}(b_{2}y'_{2})'{\rm d}s, v_{1}-y_{1s},v_{2}-y_{2s}\end{array}\right).\label{7}\end{equation}

(7)

因此,可以将系统(1)改写成${\cal H}$上的抽象Cauchy问题 $$\frac{{\rm d}Z(t)}{{\rm d}t}={\cal A}Z(t),$$ 其中$Z=(u_{1},u_{2},v_{1},v_{2},y_{1},y_{2}),$ $ Z_{0}=Z(0)=(u_{1}^{0},u_{2}^{0},v_{1}^{0},v_{2}^{0},0,0)$.

现在我们叙述本文的主要结果.

$bf 定理 2.1$ 如果假设(H1)-(H5)成立, 那么${\cal A}$生成${\cal H}$上压缩$C_{0}$ -半群${\rm e}^{{\cal A}t}$. 进一步,\\ 若$(u_{1}^{0},u_{2}^{0},v_{1}^{0},v_{2}^{0},0,0)\in D({\cal A})$, 则系统(1)存在唯一的强解; 若$(u_{1}^{0},u_{2}^{0},v_{1}^{0},v_{2}^{0},0,0)\in {{\cal H}}$, 则系统(1)存在唯一的弱解.

$bf 定理 2.2$ 如果假设(H1)-(H5)成立, 那么系统(1)的能量是一致指数衰减的.

3 适定性和谱性质

本节我们先给出系统(1)的适定性,进一步给出线性算子${\cal A}$的谱性质.

类似文献^[5]的定理2.1,我们先证明一个有用的引理.

$\bf 引理 3.1$ 假设$g_{1}(\cdot),g_{2}(\cdot)$满足条件(g1)-(g4), $(l,\tau)\in Y,{\rm Re}\lambda>-\frac{k_{1}}{2}$, 若记 $$y_{1}(x,s)=\int_{0}^{s}{\rm e}^{-\lambda(s-t)}l(x,t){\rm d}t,\quad y_{2}(x,s)=\int_{0}^{s}{\rm e}^{-\lambda(s-t)}\tau(x,t){\rm d}t, $$ 则 $$(y_{1},y_{2})\in Y\cap{C([0,+\infty); W_{1})\times C([0,+\infty); W_{2})}, \quad (y_{1s},y_{2s})\in Y.$$ 且对某个$\delta\in(0,2{\rm Re}\lambda+k_{1})$,有

\begin{equation}\|(y_{1},y_{2})\|_{Y}^{2}\leq\frac{1}{\delta(2{\rm Re}\lambda+k_{1}-\delta)} \|(l,\tau)\|_{Y}^{2}\label{8}\end{equation}

(8)

和

\begin{equation}{\rm Re}\langle(y_{1s},y_{2s}),(y_{1},y_{2})\rangle_{Y} \geq\frac{k_{1}}{2}\|(y_{1},y_{2})\|_{Y}^{2}.\label{9}\end{equation}

(9)

$\bf 证$ 首先由文献[5,定理2.1]知: 当$s\downarrow0$或者$s\uparrow \infty$时,

\begin{equation}|sg_{is}(s)|\rightarrow0,\quad i=1,2. \label{10}\end{equation}

(10)

其次,对于任意的$T>0$,有$y_{i}\in L^{2}(0,T; W_{i}),i=1,2$. 因此,由$y_{i}~(i=1,2)$的定义,我们有 $$y_{1}\in C(0,+\infty; W_{1})\bigcap H^{2}(0,T; W_{1}); \quad y_{2}\in C(0,+\infty; W_{2})\bigcap H^{1}(0,T; W_{2}).$$ 且当$s\in(0,+\infty)$时有

\begin{equation}y_{1s}+\lambda y_{1}=l,\quad y_{2s}+\lambda y_{2} =\tau.\label{11}\end{equation}

(11)

注意到$\left|\int_{0}^{s}y''_{1s}(x,r){\rm d}r\right|^{2}\leq s\int_{0}^{s}\left|y''_{1s}(x,r)\right|^{2}{\rm d}r$和 $\left|\int_{0}^{s}y'_{2s}(x,r){\rm d}r\right|^{2}\leq s\int_{0}^{s}\left|y'_{2s}(x,r)\right|^{2}{\rm d}r$. 于是

\begin{eqnarray} &&|g_{1s}(s)| \left[\int_{\alpha}^{L}b_{1}\left|y''_{1}(x,s)\right|^{2}{\rm d}x\right]^{\frac{1}{2}}+ |g_{2s}(s)|\left[\int_{\alpha}^{L}b_{2}\left|y'_{2}(x,s)\right|^{2}{\rm d}x\right]^{\frac{1}{2}} \nonumber\\ &=&|g_{1s}(s)|\left[\int_{\alpha}^{L}b_{1}\left|\int_{0}^{s}y''_{1s}(x,r){\rm d}r\right|^{2}{\rm d}x\right]^{\frac{1}{2}}+ |g_{2s}(s)|\left[\int_{\alpha}^{L}b_{2}\left|\int_{0}^{s}y'_{2s}(x,s){\rm d}r\right|^{2}{\rm d}x\right]^{\frac{1}{2}} \nonumber\\ &\leq& s^{\frac{1}{2}}|g_{1s}(s)|\left[\int_{\alpha}^{L}b_{1}\int_{0}^{s}\left|y''_{1s}(x,r) \right|^{2}{\rm d}r {\rm d}x\right]^{\frac{1}{2}}+ s^{\frac{1}{2}}|g_{2s}(s)|\left[\int_{\alpha}^{L}b_{2} \int_{0}^{s}\left|y'_{2s}(x,s)\right|^{2}{\rm d}r{\rm d}x\right]^{\frac{1}{2}} \nonumber\\ &\leq &\left|sg_{1s}\right|^{\frac{1}{2}}\left[\int_{0}^{s}|g_{1s}(s)| \int_{\alpha}^{L}b_{1}\left|y''_{1s}\right|^{2} {\rm d}x{\rm d}r\right]^{\frac{1}{2}}+ \left|sg_{1s}\right|^{\frac{1}{2}}\left[\int_{0}^{s}|g_{2s}(s)|\int_{\alpha}^{L}b_{2} \left|y'_{2s}(x,s)\right|^{2}{\rm d}x{\rm d}r\right]^{\frac{1}{2}} \nonumber\\ &\leq &\left|s\widetilde{g}_{s}\right|^{\frac{1}{2}} \left\|(y_{1s},y_{2s})\right\|_{Y}\rightarrow0\quad (s\downarrow0), \label{12} \end{eqnarray}

(12)

其中$\left|\widetilde{g}_{s}\right|=\max \left\{\left|g_{1s}\right|,\left|g_{2s}\right|\right\}$.

由条件(g2)和(g3)得,对$T>\varepsilon>0$,

\begin{eqnarray} &&-k_{1}\int_{\varepsilon}^{T}g_{1s}\int_{\alpha}^{L}b_{1}\left|y''_{1}(x,s)\right|^{2}{\rm d}x{\rm d}s -k_{1}\int_{\varepsilon}^{T}g_{2s}\int_{\alpha}^{L}b_{2}\left|y'_{2}(x,s)\right|^{2}{\rm d}x{\rm d}s \nonumber\\ &\leq & \int_{\varepsilon}^{T}g_{1ss}\int_{\alpha}^{L}b_{1}\left|y''_{1}(x,s)\right|^{2}{\rm d}x{\rm d}s +\int_{\varepsilon}^{T}g_{2ss}\int_{\alpha}^{L}b_{2}\left|y'_{2}(x,s)\right|^{2}{\rm d}x{\rm d}s \nonumber\\ &=&g_{1s}(T)\int_{\alpha}^{L}b_{1}\left|y''_{1s}(x,T)\right|^{2} {\rm d}x -g_{1s}(\varepsilon)\int_{\alpha}^{L}b_{1}\left|y''_{1s}(x,\varepsilon)\right|^{2}{\rm d}x \nonumber\\ &&+g_{2s}(T)\int_{\alpha}^{L}b_{2}\left|y'_{2s}(x,T)\right|^{2} {\rm d}x -g_{2s}(\varepsilon)\int_{\alpha}^{L}b_{2}\left|y'_{2s}(x,\varepsilon)\right|^{2}{\rm d}x \nonumber\\ &&-2{\rm Re}\left(\int_{\varepsilon}^{T}g_{1s}\int_{\alpha}^{L}b_{1}y''_{1s}\overline{y}''_{1}{\rm d}x{\rm d}s\right) -2{\rm Re}\left(\int_{\varepsilon}^{T}g_{2s}\int_{\alpha}^{L}b_{2}y'_{2s}\overline{y}'_{2}{\rm d}x{\rm d}s\right) \nonumber\\ &\leq& -2{\rm Re}\left(\int_{\varepsilon}^{T}g_{1s}\int_{\alpha}^{L}b_{1}y''_{1s}\overline{y}''_{1}{\rm d}x{\rm d}s\right) -2{\rm Re}\left(\int_{\varepsilon}^{T}g_{2s}\int_{\alpha}^{L}b_{2}y'_{2s}\overline{y}'_{2}{\rm d}x{\rm d}s\right)+o(1). \label{13}\end{eqnarray}

(13)

若当$s\downarrow0$时,有$o(1)\rightarrow0$. 由(11),(13)式和带权的Cauchy不等式得到 \begin{eqnarray*} && -k_{1}\int_{\varepsilon}^{T}g_{1s}\int_{\alpha}^{L}b_{1}\left|y''_{1}(x,s)\right|^{2}{\rm d}x{\rm d}s -k_{1}\int_{\varepsilon}^{T}g_{2s}\int_{\alpha}^{L}b_{2}\left|y'_{2}(x,s)\right|^{2}{\rm d}x{\rm d}s\\ &\leq&-2{\rm Re}\left(\int_{\varepsilon}^{T}g_{1s}\int_{\alpha}^{L}b_{1}y''_{1s}\overline{y}''_{1}{\rm d}x{\rm d}s\right) -2{\rm Re}\left(\int_{\varepsilon}^{T}g_{2s}\int_{\alpha}^{L}b_{2}y'_{2s}\overline{y}'_{2}{\rm d}x{\rm d}s\right)+o(1)\\ &=&-2{\rm Re}\left(\int_{\varepsilon}^{T}g_{1s}\int_{\alpha}^{L}b_{1} \left(l''-\lambda y''_{1}\right)\overline{y}''_{1}{\rm d}x{\rm d}s\right)\\ &&-2{\rm Re}\left(\int_{\varepsilon}^{T}g_{2s}\int_{\alpha}^{L}b_{2} \left(\tau'-\lambda y'_{2}\right)\overline{y}'_{2}{\rm d}x{\rm d}s\right)+o(1)\\ &\leq&2{\rm Re}\lambda\int_{\varepsilon}^{T}g_{1s}\int_{\alpha}^{L}b_{1}\left|y''_{1}\right|^{2}{\rm d}x{\rm d}s +\delta\int_{\varepsilon}^{T}\left|g_{1s}\right|\int_{\alpha}^{L}b_{1}\left|y''_{1}\right|^{2}{\rm d}x{\rm d}s\\ &&+\frac{1}{\delta}\int_{\varepsilon}^{T}\left|g_{1s}\right|\int_{\alpha}^{L}b_{1}\left|l''\right|^{2}{\rm d}x{\rm d}s +2{\rm Re}\lambda\int_{\varepsilon}^{T}g_{2s}\int_{\alpha}^{L}b_{2}\left|y'_{2}\right|^{2}{\rm d}x{\rm d}s\\ &&+\delta\int_{\varepsilon}^{T}\left|g_{2s}\right|\int_{\alpha}^{L}b_{2}\left|y'_{2}\right|^{2}{\rm d}x{\rm d}s+ \frac{1}{\delta}\int_{\varepsilon}^{T}\left|g_{2s}\right|\int_{\alpha}^{L}b_{2}\left|\tau'\right|^{2}{\rm d}x{\rm d}s+o(1)\\ &=&-2{\rm Re}\lambda\left(\int_{\varepsilon}^{T}\left|g_{1s}\right| \int_{\alpha}^{L}b_{1}\left|y''_{1}\right|^{2}{\rm d}x{\rm d}s+ \int_{\varepsilon}^{T}\left|g_{2s}\right|\int_{\alpha}^{L}b_{2}\left|y'_{2}\right|^{2}{\rm d}x{\rm d}s\right)\\ &&+\delta\left(\int_{\varepsilon}^{T}\left|g_{1s}\right|\int_{\alpha}^{L}b_{1}\left|y''_{1}\right|^{2}{\rm d}x{\rm d}s+ \int_{\varepsilon}^{T}\left|g_{2s}\right|\int_{\alpha}^{L}b_{2}\left|y'_{2}\right|^{2}{\rm d}x{\rm d}s\right)\\ &&+\frac{1}{\delta}\left(\int_{\varepsilon}^{T}\left|g_{1s}\right| \int_{\alpha}^{L}b_{1}\left|l''\right|^{2}{\rm d}x{\rm d}s +\int_{\varepsilon}^{T}\left|g_{2s}\right|\int_{\alpha}^{L}b_{2}\left|\tau'\right|^{2}{\rm d}x{\rm d}s\right)+o(1). \end{eqnarray*} 从而有

\begin{eqnarray} &&(2{\rm Re}\lambda+k_{1}-\delta) \left(\int_{\varepsilon}^{T}\left|g_{1s}\right|\int_{\alpha}^{L}b_{1}\left|y''_{1}\right|^{2}{\rm d}x{\rm d}s+ \int_{\varepsilon}^{T}\left|g_{2s}\right|\int_{\alpha}^{L}b_{2}\left|y'_{2}\right|^{2}{\rm d}x{\rm d}s\right) \nonumber\\ &\leq&\frac{1}{\delta}\left\|(l,\tau)\right\|_{Y}^{2}+o(1).\label{14} \end{eqnarray}

(14)

对于上式,选取足够小的$\delta$使得 $\delta\in(0,2{\rm Re}\lambda+k_{1})$ 并令$\varepsilon\downarrow0$,则有$o(1)\rightarrow0$, 且令$T\uparrow\infty$,即得(8)式. 同时,当$(y_{1},y_{2})\in Y$时,由(11)式得$(y_{1s},y_{2s})\in Y$, 再由(13)式得到(9)式.

$\bf 定理2.1的证明$ 对于任意的$Z=(u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})\in D({\cal A})$. 由分部积分并应用引理3.1得到

\begin{eqnarray} &&{\rm Re}\langle {\cal A}Z,Z\rangle_{{\cal H}} \nonumber\\ &=&{\rm Re}\langle {\cal A}(u_{1},u_{2},v_{1},v_{2},y_{1},y_{2}), (u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})\rangle_{{\cal H}} \nonumber\\ &=&{\rm Re}\left\langle \left(\begin{array}{lll} v_{1},v_{2},-au_{2}-u_{1}^{(4)}+ \int_{0}^{+\infty}g_{1s}(s)(b_{1}y''_{1})''{\rm d}s, \\[3mm] -au_{1}+u''_{2}-\int_{0}^{+\infty}g_{2s}(s)(b_{2}y'_{2})'{\rm d}s, v_{1}-y_{1s},v_{2}-y_{2s}\end{array}\right), (u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})\right\rangle_{{\cal H}} \nonumber\\ &=&{\rm Re}\left\{\int_{0}^{L}[v''_{1}\overline{u}''_{1}+v'_{2}\overline{u}'_{2} +av_{1}\overline{u}_{2}+av_{2}\overline{u}_{1}+(-au_{2}-u_{1}^{(4)}+ \int_{0}^{+\infty}g_{1s}(s)(b_{1}y''_{1})''{\rm d}s)\overline{v}_{1}]{\rm d}x\right\} \nonumber\\ &&+{\rm Re}\bigg\{\int_{0}^{L}(-au_{1} +u''_{2}-\int_{0}^{+\infty}g_{2s}(s)(b_{2}y'_{2})'{\rm d}s)\overline{v}_{2}{\rm d}x \nonumber\\ &&+\int_{0}^{+\infty}|g_{1s}(s)|\int_{0}^{L}b_{1}(v''_{1}-y''_{1s})\overline{y}''_{1}{\rm d}x{\rm d}s\bigg\} \nonumber\\ &&+{\rm Re}\left\{\int_{0}^{+\infty}|g_{2s}(s)|\int_{0}^{L}b_{2}(v'_{2}-y'_{2s}) \overline{y}'_{2}{\rm d}x{\rm d}s\right\} \nonumber\\ &=&{\rm Re}\left\{\int_{0}^{+\infty}g_{1s}(s)\int_{0}^{L}(b_{1}y''_{1})''\overline{v}_{1}{\rm d}x{\rm d}s- \int_{0}^{+\infty}g_{2s}(s)\int_{0}^{L}(b_{2}y'_{2})'\overline{v}_{2}{\rm d}x{\rm d}s\right\} \nonumber\\ &&+{\rm Re}\left\{-\int_{0}^{+\infty}g_{1s}(s)\int_{0}^{L}b_{1}(v_{1}-y_{1s})''\overline{y}''_{1}{\rm d}x{\rm d}s- \int_{0}^{+\infty}g_{2s}(s)\int_{0}^{L}b_{2}(v_{2}-y_{2s})'\overline{y}'_{2}{\rm d}x{\rm d}s\right\} \nonumber\\ &=&{\rm Re}\left\{\int_{0}^{+\infty}g_{1s}(s)\int_{0}^{L}b_{1}y''_{1s}\overline{y}''_{1}{\rm d}x{\rm d}s+ \int_{0}^{+\infty}g_{2s}(s)\int_{0}^{L}b_{2}y'_{2s}\overline{y}'_{2}{\rm d}x{\rm d}s\right\} \nonumber\\ &=&-{\rm Re}\left\langle(y_{1s},y_{2s}),(y_{1},y_{2})\right\rangle_{Y}\leq-\frac{k_{1}}{2} \|(y_{1},y_{2})\|_{Y}^{2}\leq0.\label{15} \end{eqnarray}

(15)

因此${\cal A}$在${\cal H}$中是耗散的.

对$\forall Z_{1}=(f_{1},f_{2},f_{3},f_{4},l,\tau)\in {\cal H}$,我们考虑方程

\begin{equation}{\cal A}(u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})=(f_{1},f_{2},f_{3},f_{4},l,\tau), \label{16}\end{equation}

(16)

解此方程得

\begin{equation}\left\{\begin{array}{lll} v_{1}=f_{1},\quad v_{2}=f_{2},\\[2mm] y_{1}=\int_{0}^{s}(f_{1}(x)-l(x,s)){\rm d}s,\quad y_{2}=\int_{0}^{s}(f_{2}(x)-\tau(x,s)){\rm d}s,\\ -au_{2}-u_{1}^{(4)}+\int_{0}^{+\infty}g_{1s}(s)(b_{1}y''_{1})''{\rm d}s=f_{3},\\ -au_{1}+u''_{2}-\int_{0}^{+\infty}g_{2s}(s)(b_{2}y'_{2})'{\rm d}s=f_{4}. \end{array}\right. \label{17} \end{equation}

(17)

对(17)式的后两式分别乘以$\overline{u}_{1},\overline{u}_{2}$, 并从$0$到$L$积分,分部积分并应用边界条件得

\begin{equation}\int_{0}^{L}|u''_{1}|^{2}{\rm d}x+\int_{0}^{L}a\overline{u}_{1}u_{2}{\rm d}x -\int_{0}^{L}\overline{u}_{1}\int_{0}^{+\infty}g_{1s}(b_{1}y''_{1})''{\rm d}s{\rm d}x=-\int_{0}^{L}\overline{u}_{1}f_{3}{\rm d}x. \label{18}\end{equation}

(18)

\begin{equation}\int_{0}^{L}|u'_{2}|^{2}{\rm d}x+\int_{0}^{L}au_{1}\overline{u}_{2}{\rm d}x +\int_{0}^{L}\overline{u}_{2}\int_{0}^{+\infty}g_{2s}(b_{2}y'_{2})'{\rm d}s{\rm d}x=-\int_{0}^{L}\overline{u}_{2}f_{4}{\rm d}x. \label{19}\end{equation}

(19)

将(18),(19)式两式相加得

\begin{eqnarray} &&\int_{0}^{L}|u''_{1}|^{2}{\rm d}x +\int_{0}^{L}|u'_{2}|^{2}{\rm d}x+\int_{0}^{L}a(\overline{u}_{1}u_{2}+u_{1}\overline{u}_{2}){\rm d}x \nonumber\\ &=&-\int_{0}^{L}\overline{u}_{1}f_{3}{\rm d}x-\int_{0}^{L}\overline{u}_{2}f_{4}{\rm d}x +\int_{0}^{L}\overline{u}_{1}\int_{0}^{+\infty}g_{1s}(b_{1}y''_{1})''{\rm d}s{\rm d}x \nonumber\\ &&-\int_{0}^{L}\overline{u}_{2}\int_{0}^{+\infty}g_{2s}(b_{2}y'_{2})'{\rm d}s{\rm d}x. \label{20} \end{eqnarray}

(20)

由上式经分部积分,并应用带权的Cauchy不等式和(H4)得到

\begin{eqnarray} \left\|(u_{1},u_{2})\right\|_{V}^{2} &=&\int_{0}^{L}(|u''_{1}|^{2}+|u'_{2}|^{2}){\rm d}x+\int_{0}^{L}a(\overline{u}_{1}u_{2}+u_{1}\overline{u}_{2}){\rm d}x \nonumber\\ &=&\bigg|-\int_{0}^{L}\overline{u}_{1}f_{3}{\rm d}x-\int_{0}^{L}\overline{u}_{2}f_{4}{\rm d}x +\int_{0}^{L}\overline{u}_{1}\int_{0}^{+\infty}g_{1s}(b_{1}y''_{1})''{\rm d}s{\rm d}x \nonumber\\ &&-\int_{0}^{L}\overline{u}_{2}\int_{0}^{+\infty}g_{2s}(b_{2}y'_{2})'{\rm d}s{\rm d}x\bigg| \nonumber\\ &=&\bigg|-\int_{0}^{L}\overline{u}_{1}f_{3}{\rm d}x-\int_{0}^{L}\overline{u}_{2}f_{4}{\rm d}x +\int_{0}^{+\infty}g_{1s}\int_{0}^{L}b_{1}y''_{1}\overline{u}''_{1}{\rm d}x{\rm d}s \nonumber\\ &&+\int_{0}^{+\infty}g_{2s}\int_{0}^{L}b_{2}y'_{2}\overline{u}'_{2}{\rm d}x{\rm d}s\bigg| \nonumber\\ &\leq& \frac{\varepsilon_{1}}{2}\int_{0}^{L}\left|u_{1}\right|^{2}{\rm d}x +\frac{1}{2\varepsilon_{1}}\int_{0}^{L}\left|f_{3}\right|^{2}{\rm d}x +\frac{\varepsilon_{2}}{2}\int_{0}^{L}\left|u_{2}\right|^{2}{\rm d}x +\frac{1}{2\varepsilon_{2}}\int_{0}^{L}\left|f_{4}\right|^{2}{\rm d}x \nonumber\\ &&+ \frac{1}{2}\int_{0}^{+\infty}\left|g_{1s}\right|\int_{0}^{L}b_{1}\left|u''_{1}\right|^{2}{\rm d}x{\rm d}s +\frac{1}{2}\int_{0}^{+\infty}\left|g_{1s}\right|\int_{\alpha}^{L}b_{1}\left|y''_{1}\right|^{2}{\rm d}x \nonumber\\ &&+ \frac{1}{2}\int_{0}^{+\infty}\left|g_{2s}\right|\int_{0}^{L}b_{2}\left|u'_{2}\right|^{2}{\rm d}x{\rm d}s +\frac{1}{2}\int_{0}^{+\infty}\left|g_{2s}\right|\int_{\alpha}^{L}b_{2}\left|y'_{2}\right|^{2}{\rm d}x \nonumber\\ &\leq& \frac{1}{2}\sqrt{c_{1}c_{2}}\left\|a\right\|_{\infty}\int_{0}^{L}(\left|u''_{1}\right|^{2} +\left|u'_{2}\right|^{2}){\rm d}x+\frac{\widetilde{\varepsilon}}{2}\left\|(f_{3},f_{4})\right\|_{H}^{2} \nonumber\\ &&+ \frac{1}{2}\int_{0}^{+\infty}\left|g_{1s}\right|\int_{0}^{L}b_{1}\left|u''_{1}\right|^{2}{\rm d}x{\rm d}s + \frac{1}{2}\int_{0}^{+\infty}\left|g_{2s}\right|\int_{0}^{L}b_{2}\left|u'_{2}\right|^{2}{\rm d}x{\rm d}s \nonumber\\ &&+\frac{1}{2}\left\|(y_{1},y_{2})\right\|_{Y}^{2} \nonumber\\ &\leq& \frac{1}{2}(\sqrt{c_{1}c_{2}}\left\|a\right\|_{\infty} +\left\|b_{1}\right\|\int_{0}^{+\infty}\left|g_{1s}\right|{\rm d}s)\int_{0}^{L}\left|u''_{1}\right|^{2}{\rm d}x \nonumber\\ &&+\frac{1}{2}(\sqrt{c_{1}c_{2}}\left\|a\right\|_{\infty} +\left\|b_{2}\right\|\int_{0}^{+\infty}\left|g_{2s}\right|{\rm d}s)\int_{0}^{L}\left|u'_{2}\right|^{2}{\rm d}x \nonumber\\ &&+\frac{\widetilde{\varepsilon}}{2}\left\|(f_{3},f_{4})\right\|_{H}^{2} +\frac{1}{2}\left\|(y_{1},y_{2})\right\|_{Y}^{2}, \label{21} \end{eqnarray}

(21)

其中$\varepsilon_{1}=\sqrt{\frac{c_{2}}{c_{1}}} \left\|a\right\|_{\infty},\varepsilon_{2} =\sqrt{\frac{c_{1}}{c_{2}}}\left\|a\right\|_{\infty}$ 及 $\widetilde{\varepsilon}={\rm max}\{\varepsilon_{1}^{-1}, \varepsilon_{2}^{-1}\}$.

因为

\begin{eqnarray} \left\|(u_{1},u_{2})\right\|_{V}^{2} &=&\int_{0}^{L}(|u''_{1}|^{2}+|u'_{2}|^{2}){\rm d}x +\int_{0}^{L}a(\overline{u}_{1}u_{2}+u_{1}\overline{u}_{2}){\rm d}x \nonumber\\ &\geq& \int_{0}^{L}(\left|u''_{1}\right|^{2} +\left|u'_{2}\right|^{2}){\rm d}x-\sqrt{c_{1}c_{2}}\left\|a\right\|_{\infty}\int_{0}^{L}(\left|u''_{1}\right|^{2} +\left|u'_{2}\right|^{2}){\rm d}x \nonumber\\ &=&(1-\sqrt{c_{1}c_{2}}\left\|a\right\|_{\infty})\int_{0}^{L}(\left|u''_{1}\right|^{2} +\left|u'_{2}\right|^{2}){\rm d}x. \label{22} \end{eqnarray}

(22)

所以结合(21)和(22)式得

\begin{eqnarray} &&(1-\sqrt{c_{1}c_{2}}\left\|a\right\|_{\infty})\int_{0}^{L}(\left|u''_{1}\right|^{2} +\left|u'_{2}\right|^{2}){\rm d}x \nonumber\\ &\leq& \frac{1}{2}(\sqrt{c_{1}c_{2}}\left\|a\right\|_{\infty} +\left\|b_{1}\right\|\int_{0}^{+\infty}\left|g_{1s}\right|{\rm d}s)\int_{0}^{L}\left|u''_{1}\right|^{2}{\rm d}x +\frac{\widetilde{\varepsilon}}{2}\left\|(f_{3},f_{4})\right\|_{H}^{2} \nonumber\\ &&+\frac{1}{2}(\sqrt{c_{1}c_{2}}\left\|a\right\|_{\infty} +\left\|b_{2}\right\|\int_{0}^{+\infty}\left|g_{2s}\right|{\rm d}s)\int_{0}^{L}\left|u'_{2}\right|^{2}{\rm d}x +\frac{1}{2}\left\|(y_{1},y_{2})\right\|_{Y}^{2}. \label{23} \end{eqnarray}

(23)

于是有

\begin{eqnarray} &&(1-\frac{3}{2}\sqrt{c_{1}c_{2}}\left\|a\right\|_{\infty} -\frac{1}{2}\left\|b_{1}\right\|\int_{0}^{+\infty}\left|g_{1s}\right|{\rm d}s)\int_{0}^{L}\left|u''_{1}\right|^{2}{\rm d}x \nonumber\\ &&+(1-\frac{3}{2}\sqrt{c_{1}c_{2}}\left\|a\right\|_{\infty} -\frac{1}{2}\left\|b_{2}\right\|\int_{0}^{+\infty}\left|g_{2s}\right|{\rm d}s)\int_{0}^{L}\left|u'_{2}\right|^{2}{\rm d}x \nonumber\\ &\leq& \frac{\widetilde{\varepsilon}}{2}\left\|(f_{3},f_{4})\right\|_{H}^{2} +\frac{1}{2}\left\|(y_{1},y_{2})\right\|_{Y}^{2}. \label{24} \end{eqnarray}

(24)

从而由假设(H5)知,若记$\gamma={\rm min}\{\gamma_{1},\gamma_{2}\}$, 则(24)式可化为

\begin{equation}\int_{0}^{L}(\left|u''_{1}\right|^{2}+\left|u'_{2}\right|^{2}){\rm d}x \leq\frac{1}{2\gamma}(\widetilde{\varepsilon}\left\|(f_{3},f_{4})\right\|_{H}^{2} +\left\|(y_{1},y_{2})\right\|_{Y}^{2}). \label{25}\end{equation}

(25)

因此我们有如下估计

\begin{eqnarray} \left\|(u_{1},u_{2})\right\|_{V}^{2} &=&\int_{0}^{L}(|u''_{1}|^{2}+|u'_{2}|^{2}){\rm d}x +\int_{0}^{L}a(\overline{u}_{1}u_{2}+u_{1}\overline{u}_{2}){\rm d}x \nonumber\\ &\leq&\int_{0}^{L}(\left|u''_{1}\right|^{2} +\left|u'_{2}\right|^{2}){\rm d}x+\sqrt{c_{1}c_{2}}\left\|a\right\|_{\infty}\int_{0}^{L}(\left|u''_{1}\right|^{2} +\left|u'_{2}\right|^{2}){\rm d}x \nonumber\\ &=&(1+\sqrt{c_{1}c_{2}}\left\|a\right\|_{\infty})\int_{0}^{L}(\left|u''_{1}\right|^{2} +\left|u'_{2}\right|^{2}){\rm d}x \nonumber\\ &\leq& \frac{m}{2\gamma}(\widetilde{\varepsilon}\left\|(f_{3},f_{4})\right\|_{H}^{2} +\left\|(y_{1},y_{2})\right\|_{Y}^{2}), \label{26} \end{eqnarray}

(26)

这里$m=1+\sqrt{c_{1}c_{2}}\left\|a\right\|_{\infty}$.

注意到

\begin{equation}\|(f_{1},f_{2})\|_{Y}^{2} =\int_{0}^{+\infty}|g_{1s}|\int_{\alpha}^{L}b_{1}|f''_{1}|^{2}{\rm d}x{\rm d}s +\int_{0}^{+\infty}|g_{2s}|\int_{\alpha}^{L}b_{2}|f'_{2}|^{2}{\rm d}x{\rm d}s \leq C_{1}\|(f_{1},f_{2})\|_{V}^{2}, \label{27}\end{equation}

(27)

其中$C_{1}$是一个确定的正常数.

由(H4)得

\begin{equation} \|(v_{1},v_{2})\|_{H}^{2}=\|(f_{1},f_{2})\|_{H}^{2}=\int_{0}^{L}(|f_{1}|^{2}+|f_{2}|^{2}){\rm d}x \leq C_{2}\|(f_{1},f_{2})\|_{V}^{2}, \label{28}\end{equation}

(28)

其中$C_{2}$是一个确定的正常数. 并且

\begin{eqnarray} \|(y_{1},y_{2})\|_{Y}^{2} &=& \bigg\|\bigg(\int_{0}^{s}(f_{1}-l){\rm d}s,\int_{0}^{s}(f_{2}-\tau) {\rm d}s\bigg)\bigg\|_{Y}^{2} \nonumber\\ &\leq& \frac{4}{k_{1}^{2}}\|(f_{1}-l,f_{2}-\tau)\|_{Y}^{2} \leq\frac{8C_{1}}{k_{1}^{2}}(\|(f_{1},f_{2})\|_{V}^{2}+\|(l,\tau)\|_{Y}^{2}). \label{29} \end{eqnarray}

(29)

综合上面的推理过程知: 存在某个常数$M>0$, 使得$\|Z\|_{{\cal H}}\leq M\|Z_{1}\|_{{\cal H}}$.

因此${\cal A}^{-1}\in {\cal L}({\cal H}),0\in \rho({\cal A})$(${\cal A}$的预解集), 且${\cal A}$是闭的. 同时,容易验证${\rm ker}{\cal A}=\{0\}$. 且对足够小的$\lambda>0$,有 $R(\lambda I-{\cal A})={\cal H}$($R({\cal A})$表示${\cal A}$的值域), 故由文献^[12]中定理1.4.6知$\overline{D({\cal A})}={\cal H}$. 那么应用Lumer-Phillips定理, 这就证明了${\cal A}$生成${\cal H}$上压缩$C_{0}$ -半群${\rm e}^{{\cal A}t}$.

进一步,若$(u_{1}^{0},u_{2}^{0},v_{1}^{0},v_{2}^{0},0,0)\in D({\cal A})$,则系统 (1) 存在唯一的强解 $$(u_{1}(\cdot),u_{2}(\cdot))\in C^{1}((0,+\infty);V)\cap C^{2}((0,+\infty);H);$$ 若$(u_{1}^{0},u_{2}^{0},v_{1}^{0},v_{2}^{0},0,0)\in {\cal H}$,则系统(1)存在唯一的弱解 $$(u_{1}(\cdot),u_{2}(\cdot))\in C((0,+\infty);V)\cap C^{1}((0,+\infty);H).$$ 这就证明了定理2.1.

$\bf 性质 3.1$ 假设(H1)-(H3)成立, 那么i${\Bbb R}\subset \rho({\cal A})$.

$\bf 证$ 易证${\cal A}$具有紧的预解式,因此用文献[15引理4.1]类似的方法可得 $$\{\lambda\in\sigma({\cal A})| {\rm Im}\lambda\neq0\} \subset\sigma_{{\rm P}}({\cal A}).$$ 这里$\sigma_{{\rm P}}({\cal A})$意指算子${\cal A}$的点谱. 接下来要证i${\Bbb R}\subset \rho({\cal A})$, 只需要证明i$\omega\overline{\in}\sigma_{{\rm P}}({\cal A})$ ($\forall\omega\in{\Bbb R}$). 采用反证法, 假设结论不成立,则存在$\omega\in {\Bbb R},\omega\neq0$满足 i$\omega\in\sigma_{{\rm P}}({\cal A})$, 故存在$(u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})\in D({\cal A}), (u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})\neq0$使得

\begin{equation} ({\rm i}\omega-{\cal A})(u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})=0. \label{30}\end{equation}

(30)

则有

\begin{eqnarray}0 &=&{\rm Re}\left\langle({\rm i}\omega-{\cal A})(u_{1},u_{2},v_{1},v_{2},y_{1},y_{2}), (u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})\right\rangle_{{\cal H}} \nonumber\\ &=&-{\rm Re}\left\langle(y_{1s},y_{2s}), (y_{1},y_{2})\right\rangle_{Y}\label{31}\end{eqnarray}

(31)

由引理3.1的(9)式知: 在$(\alpha,L)$中,有

\begin{equation} y_{1}=0,\quad y_{2}=0. \label{32}\end{equation}

(32)

又由线性算子${\cal A}$的定义和(30)式得到

\begin{equation}\left\{\begin{array}{lll} v_{1}={\rm i}\omega u_{1},\quad v_{2}={\rm i}\omega u_{2},\\[2mm] -au_{2}-u_{1}^{(4)}+\int_{0}^{+\infty}g_{1s}(s)(b_{1}y''_{1})''{\rm d}s={\rm i}\omega v_{1},\\ [3mm] -au_{1}+u''_{2}-\int_{0}^{+\infty}g_{2s}(s)(b_{2}y'_{2})'{\rm d}s={\rm i}\omega v_{2},\\ [2mm] v_{1}-y_{1s}={\rm i}\omega y_{1},\quad v_{2}-y_{2s}={\rm i}\omega y_{2}. \end{array}\right.\label{33}\end{equation}

(33)

由(32)和(33)式得到: 在$(\alpha,L)$中,有

\begin{equation}(u_{1},u_{2},v_{1},v_{2})=0.\label{34}\end{equation}

(34)

上式结合(32)式得到: 在$(\alpha,L)$中,有 $$(u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})=0.$$ 再由假设(H2)和(6),(33)式得 $$\left\{\begin{array}{lll} &\omega^{2}u_{1}-au_{2}-u_{1}^{(4)}=0,\\ &\omega^{2}u_{2}-au_{1}+u''_{2}=0,\\ &u_{1}|_{x=0}=u'_{1}|_{x=0}=u_{1}|_{x=L}=u'_{1}|_{x=L} =u_{2}|_{x=0}=u_{2}|_{x=L}=0. \end{array}\right.$$ 因此,对于$\forall x_{0}\in(\alpha,L)$, 有$W|_{x=x_{0}}=(u_{1},u'_{1},u''_{1},u'''_{1},u_{2},u'_{2})^{{\rm T}}|_{x=x_{0}}=0$. 则由上式得 $$\frac{{\rm d}W}{{\rm d}x}=A_{0}W,\quad W|_{x=x_{0}}=0,$$ 其中 $$A_{0}=\left(\begin{array}{cccccc} 0&~~1~~&0&~~0~~&0&~~0\\ 0&0&1&0&0&~~0\\ 0&0&0&1&0&~~0\\ \omega^{2}&0&0&0&-a&~~0\\ 0&0&0&0&0&~~1\\ a&0&0&0&-\omega^{2}&~~0 \end{array}\right).$$ 于是,在$(0,L)$ 中,由常微分方程的初值问题的唯一性定理^[16] 知$(u_{1},u_{2})=0$, 由此结合(33)式得: $(u_{1},u_{2},v_{1},v_{2})=0$, 又在$(0,\alpha)$中,再次由假设(H2)和(32),(33)式得到: 当 $x\in(0,\alpha)$时,$y_{1}=y_{2}=0$.

综上得: 在$(0,L)$上,恒有$(u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})=0$ 与假设矛盾. 性质3.1得证.

4 定理2.2的证明

本节我们证明本文的主要结果.

$\bf 定理2.2的证明$ 假设(H1)-(H3)成立,如果$u_{1}(x,t),u_{2}(x,t)$是系统(1)的解. 那么有 $$(u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})={\rm e}^{At}(u_{1}^{0},u_{2}^{0},v_{1}^{0},v_{2}^{0},0,0),$$ 和 $$\|(u_{1},u_{2},v_{1},v_{2},y_{1},y_{2})\|_{{\cal H}}^{2}=2E(t),\quad t>0.$$ 众所周知: 系统(1)的能量$E(t)$的一致指数衰减性等价于$C_{0}$ -半群${\rm e}^{{\cal A}t}$的一致指数稳定性. 由性质3.1和频域结果^[13]知,要证明$C_{0}$ -半群${\rm e}^{{\cal A}t}$的一致指数稳定性,只需证明

\begin{equation} \sup\{\|({\rm i}\beta-{\cal A})^{-1}\||\beta\in {\Bbb R}\}<+\infty. \label{35}\end{equation}

(35)

假设(35)式不成立,即$\sup\{\|({\rm i}\beta-{\cal A})^{-1}\||\beta\in {\Bbb R}\}=+\infty$, 由一致有界性定理和预解式的连续性知,存在$\{\beta_{n}\}\subset {\Bbb R}$和 $\left\{Z_{n}=(u_{1n},u_{2n},v_{1n},v_{2n},y_{1n},y_{2n})\right\}\subset D({\cal A})$使得

\begin{equation} \|Z_{n}\|_{{\cal H}}=1,|\beta_{n}|\longrightarrow+\infty. \label{36}\end{equation}

(36)

且在${\cal H}$中,有

\begin{equation} ({\rm i}\beta_{n}-{\cal A})Z_{n}=(f_{1n},f_{2n},f_{3n},f_{4n},l_{n},\tau_{n})\longrightarrow0. \label{37}\end{equation}

(37)

即在$V$中,

\begin{equation} ({\rm i}\beta_{n}u_{1n}-v_{1n},\quad {\rm i}\beta_{n}u_{2n}-v_{2n})=(f_{1n},f_{2n})\longrightarrow0, \label{38}\end{equation}

(38)

在$H$中,

\begin{eqnarray} & &\left({\rm i}\beta_{n}v_{1n}+au_{2n}+u_{1n}^{(4)} -\int_{0}^{+\infty}g_{1s}(b_{1}y''_{1n})''{\rm d}s,\ {\rm i}\beta_{n}v_{2n}+au_{1n}-u''_{2n}+\int_{0}^{+\infty}g_{2s}(b_{2}y'_{2n})'{\rm d}s\right) \nonumber\\ &=&(f_{3n},f_{4n})\longrightarrow0, \label{39}\end{eqnarray}

(39)

在$Y$中,

\begin{equation} ({\rm i}\beta_{n}y_{1n}-v_{1n}+y_{1ns},\ {\rm i}\beta_{n}y_{2n}-v_{2n}+y_{2ns})=(l_{n},\tau_{n})\longrightarrow0. \label{40}\end{equation}

(40)

为了证明定理2.2,我们需要下面的引理.

$\bf 引理 4.1$ 由(36)-(37)式所定义的序列 $(u_{1n},u_{2n},v_{1n},v_{2n},y_{1n},y_{2n})$有如下性质

\begin{equation} \|(y_{1n},y_{2n})\|_{Y}\longrightarrow0, \label{41}\end{equation}

(41)

\begin{equation} \int_{\alpha}^{L}(|u''_{1n}|^{2}+|u'_{2n}|^{2}){\rm d}x\longrightarrow0. \label{42}\end{equation}

(42)

$\bf 证$ 注意到$y_{1n}(\cdot,0)=0,$ $ y_{2n}(\cdot,0)=0$, 由(40)式得到

\begin{eqnarray} y_{1n}&=&\int_{0}^{s}{\rm e}^{-{\rm i}\beta_{n}(s-\lambda)}(v_{1n}(x)+l_{n}(x,\lambda)){\rm d}\lambda \nonumber\\ &=&\frac{1}{{\rm i}\beta_{n}}(1-{\rm e}^{-{\rm i}\beta_{n}s})v_{1n}(x) +\int_{0}^{s}{\rm e}^{-{\rm i}\beta_{n}(s-\lambda)}l_{n}(x,\lambda){\rm d}\lambda. \label{43}\end{eqnarray}

(43)

\begin{eqnarray} y_{2n}&=&\int_{0}^{s}{\rm e}^{-{\rm i}\beta_{n}(s-\lambda)}(v_{2n}(x)+\tau_{n}(x,\lambda)){\rm d}\lambda \nonumber\\ &=&\frac{1}{{\rm i}\beta_{n}}(1-{\rm e}^{-{\rm i}\beta_{n}s})v_{2n}(x) +\int_{0}^{s}{\rm e}^{-{\rm i}\beta_{n}(s-\lambda)}\tau_{n}(x,\lambda){\rm d}\lambda. \label{44} \end{eqnarray}

(44)

因此,由引理3.1和(36)-(37)式可得 $$\frac{k_{1}}{2}\|(y_{1n},y_{2n})\|_{Y}^{2} \leq {\rm Re}\left\langle(y_{1ns},y_{2ns}),(y_{1n},y_{2n})\right\rangle_{Y}= {\rm Re}\left\langle({\rm i}\beta_{n}-{\cal A})Z_{n},Z_{n}\right\rangle_{Y}\longrightarrow0.$$ 上式即蕴含(41)式.

接下来,由(40),(41),(43)和(44)式可以推出 $$\left\|\left(\frac{1}{{\rm i}\beta_{n}}(1-{\rm e}^{-{\rm i}\beta_{n}s})v_{1n}, \frac{1}{{\rm i}\beta_{n}}(1-{\rm e}^{-{\rm i}\beta_{n}s})v_{2n}\right)\right\|_{Y}\longrightarrow0,$$ 即

\begin{equation} \int_{0}^{+\infty}(\cos\beta_{n}s-1)g_{1s}{\rm d}s \int_{\alpha}^{L}b_{1}\left|\frac{1}{{\rm i}\beta_{n}}v''_{1n}\right|^{2}{\rm d}x +\int_{0}^{+\infty}(\cos\beta_{n}s-1)g_{2s}{\rm d}s \int_{\alpha}^{L}b_{2}\left|\frac{1}{{\rm i}\beta_{n}}v'_{2n}\right|^{2}{\rm d}x\longrightarrow0. \label{45}\end{equation}

(45)

由(g1)-(g3),对 $\forall\beta\neq0$,有 $$\xi_{j}(\beta):=\int_{0}^{+\infty}(\cos\beta s-1)g_{js}(s){\rm d}s>0,\quad j=1,2.$$ 并由文献^[17]中的Riemann-Lebesgue引理得 $$\lim_{\beta\rightarrow\infty}\xi_{j}(\beta)=g_{j}(0)>0,\quad j=1,2.$$ 由于$\xi_{j}(\beta)~(j=1,2)$在${\Bbb R}$上连续,则存在正常数$\delta_{0},\widetilde{C}$使得 $$\inf_{|\beta|\geq\delta_{0}}\xi_{j}(\beta)\geq\widetilde{C}>0,\quad j=1,2.$$ 故由(38),(45)式和假设(H2)立即得到(42)式.

接下来继续证明定理2.2. 由假设(H4)和(38)式得 \begin{eqnarray*} \|(f_{1n},f_{2n})\|_{V}^{2} &=&\int_{0}^{L}(|f''_{1n}|^{2}+|f'_{2n}|^{2}){\rm d}x +\int_{0}^{L}a(\overline{f}_{1n}f_{2n}+f_{1n}\overline{f}_{2n}){\rm d}x\\ &\geq &\int_{0}^{L}(|f''_{1n}|^{2}+|f'_{2n}|^{2}){\rm d}x -\sqrt{c_{1}c_{2}}\|a\|_{\infty}\int_{0}^{L}(|f''_{1n}|^{2}+|f'_{2n}|^{2}){\rm d}x\\ &=&(1-\sqrt{c_{1}c_{2}}\|a\|_{\infty}) \int_{0}^{L}(|f''_{1n}|^{2}+|f'_{2n}|^{2}){\rm d}x\longrightarrow0. \end{eqnarray*} 上式结合(39)式得: 在$L^{2}(0,L)$中有

\begin{equation} f''_{1n}\rightarrow0,\quad f'_{2n}\rightarrow0,\quad f_{3n}\rightarrow0,\quad f_{4n}\rightarrow0. \label{46}\end{equation}

(46)

由于$\left\langle(f_{1n},f_{2n}),(v_{1n},v_{2n})\right\rangle_{H}\rightarrow0$, 对(38)式用$(v_{1n},v_{2n})$在$H$中作内积有

\begin{equation} {\rm i}\beta_{n}\int_{0}^{L}u_{1n}\overline{v}_{1n}{\rm d}x+{\rm i}\beta_{n}\int_{0}^{L}u_{2n}\overline{v}_{2n}{\rm d}x -\int_{0}^{L}(|v_{1n}|^{2}+|v_{2n}|^{2}){\rm d}x\rightarrow0. \label{47}\end{equation}

(47)

又由于$\left\langle(f_{3n},f_{4n}),(u_{1n},u_{2n})\right\rangle_{H}\rightarrow0$, 对(39)式用$(u_{1n},u_{2n})$在$H$中作内积,并通过分部积分,由引理4.1得

\begin{equation} {\rm i}\beta_{n}\int_{0}^{L}v_{1n}\overline{u}_{1n}{\rm d}x+{\rm i}\beta_{n}\int_{0}^{L}v_{2n}\overline{u}_{2n}{\rm d}x +\int_{0}^{L}(|u''_{1n}|^{2}+|u'_{2n}|^{2}){\rm d}x\rightarrow0. \label{48}\end{equation}

(48)

将(47)和(48)式两式相加,并取实部得

\begin{equation} \int_{0}^{L}(|u''_{1n}|^{2}+|u'_{2n}|^{2}){\rm d}x-\int_{0}^{L}(|v_{1n}|^{2}+|v_{2n}|^{2}){\rm d}x\rightarrow0. \label{49}\end{equation}

(49)

注意到(38)式: 当$|\beta_{n}|\rightarrow+\infty$时, 有$\big|\int_{0}^{L}au_{1n}u_{2n}{\rm d}x\big|\rightarrow0$. 从而由(49)式结合(36)式得

\begin{equation} \left\{\begin{array}{lll} \int_{0}^{L}(|u''_{1n}|^{2}+|u'_{2n}|^{2}){\rm d}x\rightarrow\frac{1}{2},\\ \int_{0}^{L}(|v_{1n}|^{2}+|v_{2n}|^{2}){\rm d}x\rightarrow\frac{1}{2}.\end{array}\right. \label{50}\end{equation}

(50)

为了得到矛盾的讨论,由引理4.1知,接下来只需运用分片乘子法证明

\begin{equation} \int_{0}^{\alpha}(|u''_{1n}|^{2}+|u'_{2n}|^{2}){\rm d}x\rightarrow0. \label{51}\end{equation}

(51)

结合(36),(38)式可得: 在$L^{2}(0,L)$中有

\begin{equation} u_{1n}\rightarrow0\quad u_{2n}\rightarrow0. \label{52}\end{equation}

(52)

由(50)式知: $u''_{1n}$ 在$L^{2}(0,L)$中有界,由(52)式知 $$\int_{0}^{L}|u'_{1n}|{\rm d}x=u'_{1n}\overline{u}_{1n}|_{0}^{L} -\int_{0}^{L}u''_{1n}\overline{u}_{1n}{\rm d}x\rightarrow0 $$ 从而,在$L^{2}(0,L)$中,有

\begin{equation} u'_{1n}\rightarrow0. \label{53}\end{equation}

(53)

由(38)和(39)式得: 在$L^{2}(0,L)$中有

\begin{equation} \beta_{n}^{2}u_{1n}-au_{2n}-u_{1n}^{(4)} +\int_{0}^{+\infty}g_{1s}(b_{1}y''_{1n})''{\rm d}s=-({\rm i}\beta_{n}f_{1n}+f_{3n})\rightarrow0. \label{54}\end{equation}

(54)

\begin{equation} \beta_{n}^{2}u_{2n}-au_{1n}+u''_{2n} -\int_{0}^{+\infty}g_{2s}(b_{2}y'_{2n})'{\rm d}s=-({\rm i}\beta_{n}f_{2n}+f_{4n})\rightarrow0. \label{55}\end{equation}

(55)

现取$q=\gamma({\rm e}^{\eta x}-1)$,这里的$\eta$是一个给定的正常数, 对于某个足够小的$\varepsilon_{0}$,选择$\gamma$使得 $$\left\{\begin{array}{lll} \gamma=1,\quad x\in [0,\alpha],\\ \gamma=0,\quad x\in [\alpha+\varepsilon_{0},L], \end{array}\right.$$ 这里$0\leq\gamma\leq1$,且$\gamma\in C^{2}[0,L]$.

记$\alpha_{1}=\alpha+\varepsilon_{0}$, 并用$q\overline{u}'_{1n}$乘(54)式, 在$[0,L]$上积分,且分部积分并取实部得

\begin{equation} {\rm Re}\left\{\int_{0}^{L}q(\beta_{n}^{2}u_{1n}-au_{2n}-u_{1n}^{(4)})\overline{u}'_{1n}{\rm d}x +\int_{0}^{+\infty}g_{1s}\int_{0}^{L}q(b_{1}y''_{1n})''\overline{u}'_{1n}{\rm d}x{\rm d}s\right\}\rightarrow0. \label{56}\end{equation}

(56)

注意到$q(0)=q(L)=0$和 \begin{eqnarray*} &&{\rm Re}\left\{\int_{0}^{L}qu_{1n}^{(4)}\overline{u}'_{1n}{\rm d}x\right\}\\ &=&{\rm Re}\left\{qu'''_{1n}\overline{u}'_{1n}|_{0}^{L}-\int_{0}^{L}u'''_{1n}(q\overline{u}'_{1n})'{\rm d}x\right\}\\ &=&-{\rm Re}\left\{u''_{1n}(q'\overline{u}'_{1n}+q\overline{u}''_{1n})|_{0}^{L} -\int_{0}^{L}u''_{1n}(q\overline{u}'_{1n})''{\rm d}x\right\}\\ &=&{\rm Re}\left\{\int_{0}^{L}u''_{1n}(q''\overline{u}'_{1n} +2q'\overline{u}''_{1n}+q\overline{u}'''_{1n}){\rm d}x\right\}\\ &=&{\rm Re}\left\{\int_{0}^{L}u''_{1n}q''\overline{u}'_{1n}{\rm d}x\right\} +{\rm Re}\left\{\int_{0}^{L}qu''_{1n}\overline{u}'''_{1n}{\rm d}x\right\}+2\int_{0}^{L}q'|u''_{1n}|^{2}{\rm d}x.\\ \end{eqnarray*} 而且 $$\int_{0}^{L}qu''_{1n}\overline{u}'''_{1n}{\rm d}x=q|u''_{1n}|^{2}|_{0}^{L} -\int_{0}^{L}q'|u''_{1n}|^{2}{\rm d}x-\int_{0}^{L}qu'''_{1n}\overline{u}''_{1n}{\rm d}x.$$ 于是 $${\rm Re}\left\{\int_{0}^{L}qu''_{1n}\overline{u}'''_{1n}{\rm d}x\right\} =-\frac{1}{2}\int_{0}^{L}q'|u''_{1n}|^{2}{\rm d}x.$$ 因此由(56)式得

\begin{eqnarray} && {\rm Re}\left\{\int_{0}^{L}q(\beta_{n}^{2}u_{1n}-au_{2n}-u_{1n}^{(4)})\overline{u}'_{1n}{\rm d}x +\int_{0}^{+\infty}g_{1s}\int_{0}^{L}q(b_{1}y''_{1n})''\overline{u}'_{1n}{\rm d}x{\rm d}s\right\} \nonumber\\ &=&-\frac{1}{2}\int_{0}^{L}q'|\beta_{n}u_{1n}|^{2}{\rm d}x-\frac{3}{2}\int_{0}^{L}q'|u''_{1n}|^{2}{\rm d}x -{\rm Re}\left\{\int_{0}^{L}u''_{1n}q''\overline{u}'_{1n}{\rm d}x\right\} \nonumber\\ &&-{\rm Re}\left\{\int_{0}^{L}qau_{2n}\overline{u}'_{1n}{\rm d}x\right\} +{\rm Re}\left\{\int_{0}^{+\infty}g_{1s} \int_{0}^{L}q(b_{1}y''_{1n})''\overline{u}'_{1n}{\rm d}x{\rm d}s\right\}\rightarrow0. \label{57} \end{eqnarray}

(57)

由(50)式及$q$的取法知: $u''_{1n}q''$与$qa\overline{u}'_{1n}$都在$L^{2}(0,L)$上有界,当$x\in(0,\alpha)$时, $b_{1}(x)=0$,从而由(57)式结合(52)-(53)式可得

\begin{equation}\int_{0}^{\alpha}q'|\beta_{n}u_{1n}|^{2}{\rm d}x+3\int_{0}^{\alpha}q'|u''_{1n}|^{2}{\rm d}x \rightarrow0. \label{58}\end{equation}

(58)

由于在$[0,\alpha]$上,$q'=\eta {\rm e}^{\eta x}>0$,因此,在$[0,\alpha]$上有

\begin{equation}\beta_{n}u_{1n}\rightarrow0,\quad u''_{1n}\rightarrow0. \label{59}\end{equation}

(59)

同理,用$q\overline{u}'_{2n}$乘(55)式, 在$[0,L]$上积分,且分部积分并取实部得

\begin{equation} {\rm Re}\left\{\int_{0}^{L}q(\beta_{n}^{2}u_{2n}-au_{1n}+u''_{2n})\overline{u}'_{2n}{\rm d}x -\int_{0}^{+\infty}g_{2s}\int_{0}^{L}q(b_{2}y'_{2n})'\overline{u}'_{2n}{\rm d}x{\rm d}s\right\}\rightarrow0. \label{60}\end{equation}

(60)

注意到$q(0)=q(L)=0$和 \begin{eqnarray*} {\rm Re}\left\{\int_{0}^{L}qu''_{2n}\overline{u}'_{2n}{\rm d}x\right\} &=&{\rm Re}\left\{q|u'_{2n}|^{2}|_{0}^{L}-\int_{0}^{L}u'_{2n}(q\overline{u}'_{2n})'{\rm d}x\right\} \\ &=&-{\rm Re}\left\{\int_{0}^{L}u'_{2n}(q'\overline{u}'_{2n} +q\overline{u}''_{2n}){\rm d}x\right\}. \end{eqnarray*} 于是 $${\rm Re}\left\{\int_{0}^{L}qu''_{2n}\overline{u}'_{2n}{\rm d}x\right\} =-\frac{1}{2}\int_{0}^{L}q'|u'_{2n}|^{2}{\rm d}x.$$ 因此由(60)式得

\begin{eqnarray} && {\rm Re}\left\{\int_{0}^{L}q(\beta_{n}^{2}u_{2n}-au_{2n}+u''_{2n})\overline{u}'_{2n}{\rm d}x +\int_{0}^{+\infty}g_{2s}\int_{0}^{L}q(b_{2}y'_{2n})'\overline{u}'_{2n}{\rm d}x{\rm d}s\right\} \nonumber\\ &=&-\frac{1}{2}\int_{0}^{L}q'|\beta_{n}u_{2n}|^{2}{\rm d}x-\frac{1}{2}\int_{0}^{L}q'|u'_{2n}|^{2}{\rm d}x \nonumber\\ &&-{\rm Re}\left\{\int_{0}^{L}qau_{1n}\overline{u}'_{2n}{\rm d}x\right\} +{\rm Re}\left\{\int_{0}^{+\infty}g_{2s} \int_{0}^{L}q(b_{2}y'_{2n})'\overline{u}'_{2n}{\rm d}x{\rm d}s\right\}\rightarrow0. \label{61} \end{eqnarray}

(61)

由(50)式及$q$的取法知: $qa\overline{u}'_{2n}$在$L^{2}(0,L)$上有界, 当$x\in(0,\alpha)$时, $b_{2}(x)=0$,从而由(61)式结合(52)式可得

\begin{equation}\int_{0}^{\alpha}q'|\beta_{n}u_{1n}|^{2}{\rm d}x+\int_{0}^{\alpha}q'|u'_{2n}|^{2}{\rm d}x \rightarrow0. \label{62}\end{equation}

(62)

由于在$[0,\alpha]$上,$q'=\eta {\rm e}^{\eta x}>0$,因此,在$[0,\alpha]$上有

\begin{equation}\beta_{n}u_{2n}\rightarrow0,\quad u'_{2n}\rightarrow0. \label{63}\end{equation}

(63)

由(59)和(63)式立即得到(51)式. 所以,(42)式结合(51)式得到

\begin{equation} \int_{0}^{L}(|u''_{1n}|^{2}+|u'_{2n}|^{2}){\rm d}x\rightarrow0. \label{64}\end{equation}

(64)

显然,上式与(50)式矛盾. 这样就证明了定理2.2.

参考文献

[1]	Aissa Guesmia. Energy decay for a damped nonlinear coupled system. Journal of Mathematical Analysis and Applications, 1999, 239: 38-48
[2]	Han X S, Wang M X. Energy decay rate for a coupled hyperbolic system with nonlinear damping. Nonlinear Analysis, 2009, 70: 3264-3272
[3]	Aissa Guesmia, Salim A Messaoudi. General energy decay estimates of Timoshenko systems with frictional versus viscoelastic damping. Math Meth Appl Sci, 2009, 32: 2102-2122
[4]	Han X S, Wang M X. General decay of energy for a viscoelastic equation with nonlinear damping. Math Meth Appl Sci, 2009, 32: 346-358
[5]	Liu K S, Liu Z Y. On the type of $C_{0}$-semigroup associated with the abstract linear viscoelastic system. Z Angew Math Phys, 1996, 47(2): 1-15
[6]	Liu K S, Liu Z Y. Exponential decay of energy of vibrating strings with local viscoelasticity. Z Angew Math Phys, 2002, 53(2): 265-280
[7]	Rivera J E M, Salverra A P. Asymptotic behavior of energy in partially viscoelastic material. Quart Appl Math, 2001, 1(3): 557-578
[8]	Mauro de Lima Santos. Decay rates for solutions of a system of wave equations with memory. Electronic Journal of Differential Equations, 2002, 38(1): 1-17
[9]	章春国. 具有局部记忆阻尼的非均质Timoshenko梁的稳定性. 数学物理学报, 2012, 32(1): 186-200
[10]	Jong Yeoul Park, Sun Hye Park. General decay for a quasilinear system of viscoelastic equations with nonlinear damping. Acta Mathematica Scientia, 2012, 32B(4): 1321-1332
[11]	章春国, 谷尚武, 姜敬华. 具有Boltzmann阻尼Petrovsky系统的稳定性. 系统科学与数学, 2013, 33(7): 807-817
[12]	Pazy A. Semigroups of Linear Operators and Applications to Partical Differential Euqations. New York: Springer-Verlag, 1983
[13]	Huang F L. Characteristic condition for exponential stability of linear dynamical systems in Hilbert spaces. Ann of Diff Eqs, 1985, 1(1): 43-56
[14]	Adams R A. Sobolev Spaces. New York: Acadamic Press, 1975
[15]	Chen S P, Liu K S, Liu Z Y. Spectrum and stability for elastic systems with global or local Kelvin-Voigt damping. SIAM J Appl Math, 1998, 2(59): 651-668
[16]	Hartman P. Ordinary Differential Equations (2nd ed). Boston, Basel, Stuttgart: Birkhäuser, 1982
[17]	Hille E, Hillips R S. Functional Analysis and Semi-Groups. American Mathematical Society Colloquium Publications. 1957, 31: 141-150