0 引言

本文考虑如下带有线性记忆的阻尼耦合吊桥方程解的渐近性:

$ \begin{align} \left\{\!\!\begin{array}{l} u_{tt}+\alpha u_{t}+\Delta^{2} u-\int_{0}^{\infty }\mu_{1}(s) \Delta^{2}u(t-s){\text{d}}s+k(u-v)^{+}+f_1(u)=g_{1}(x), \\ ~(x, t)\in[0, L] \times{\mathbb{R}}^{+}, \\ v_{tt}+\beta v_{t}- \Delta v-\int_{0}^{\infty }\mu_{2}(s)\Delta v(t-s){\text{d}}s-k(u-v)^{+}+f_2(v)=g_{2}(x), \\ ~(x, t)\in[0, L]\times{\mathbb{R}}^{+}, \\ u(0)=\Delta u(0)=u(L)=\Delta u(L)=0, v(0)=v(L)=0, t\geq0, \\ u(x, \tau)=u_{0}(x, \tau), u_{t}(x, \tau)=\partial_{t}u_{0}(x, \tau), ~(x, \tau)\in[0, L]\times(-\infty, 0], \\ v(x, \tau)=v_{0}(x, \tau), v_{t}(x, \tau)=\partial_{t}v_{0}(x, \tau), ~(x, \tau)\in[0, L]\times(-\infty, 0], \end{array} \right. \end{align} $

(0.1)

其中$\Delta^{2}u=u_{xxxx}, -\Delta u=-u_{xx}, $ $\alpha>0, \beta>0$是粘性阻尼, $k>0$是弹性系数, $g_{1}(x)\in L^{2}[0, L], g_{2}(x)\in L^{2}[0, L]$, 函数$(u-v)^{+}=\max\{u-v, 0\}$.

耦合吊桥方程描述了吊桥路面及其主链在竖直平面的运动情况, 由Lazer和McKenna在文献[1]中作为非线性分析中的一个新问题提出, Ahmed和Harbi在文献[2]中研究了这一问题弱解的存在性.马巧珍等人在文献[3-4]中得到了耦合吊桥方程强解和强全局吸引子的存在性, 在文献[5]中则继续讨论了其拉回吸引子的存在性.文献[6]得到了一致吸引子的存在性. Kang等人在文献[7]中研究了强拓扑空间中非自治耦合吊桥方程拉回吸引子的存在性.关于带有记忆项的动力系统, 在文献[8-9]中作者证明了耦合吊桥方程的长时间行为, Pata在文献[10]中研究了带线性记忆的阻尼双曲方程的全局吸引子.进一步, Park和Kang在文献[11]中得到了带非线性阻尼吊桥方程的全局吸引子.最近, 在文献[12]中作者利用收缩函数的方法在强拓扑空间中证明了单个吊桥方程全局吸引子的存在性.受上述文献的启发, 我们研究带双线性记忆的耦合吊桥方程全局吸引子的存在性.与文献[10, 12-13, 17]一样, 我们引入下面表示历史位移的变量

$ \begin{align} \begin{array}{l} \eta=\eta^{t}(x, s)=u(x, t)-u(x, t-s), ~(x, s)\in[0, L]\times{\mathbb{R}}^{+}, ~t\geqslant0, \\ \phi=\phi^{t}(x, s)=v(x, t)-v(x, t-s), ~(x, s)\in[0, L]\times{\mathbb{R}}^{+}, ~t\geqslant0. \end{array} \end{align} $

(0.2)

则

$ \begin{align} \begin{array}{l} \eta^{t}_{t}(x, s)=-\eta^{t}_{s}(x, s)+u_{t}(x, t), ~(x, s)\in[0, L] \times{\mathbb{R}}^{+}, ~t\geqslant0, \\ \phi^{t}_{t}(x, s)=-\phi^{t}_{s}(x, s)+v_{t}(x, t), ~(x, s)\in[0, L] \times{\mathbb{R}}^{+}, ~t\geqslant0. \end{array} \end{align} $

(0.3)

令$Ɫ_{1}=1-\int_{0}^{\infty }\mu_{1}(s){\text{d}}s>0$, $Ɫ_{2}=1-\int_{0}^{\infty }\mu_{2}(s){\text{d}}s>0$, 且$\mu_{1}\in L^{1}({\mathbb{R}}^{+})$, $\mu_{2}\in L^{1}({\mathbb{R}}^{+})$, 则$(0.1)$转化为

$ \begin{align} \left\{\begin{array}{l} u_{tt}+\alpha u_{t}+L_{1}\Delta^ {2}u+\int_{0}^{\infty }\mu_{1}(s)\Delta^{2}\eta^{t}(s){\text{d}}s+k(u-v)^{+}+f_{1}(u)=g_{1}(x), \\~(x, t)\in[0, L]\times{\mathbb{R}}^{+}, \\ v_{tt}+\beta v_{t}- L_{2}\Delta v+\int_{0}^{\infty }\mu_{2}(s)\Delta\phi^{t}(s) {\text{d}}s-k(u-v)^{+}+f_{2}(v)=g_{2}(x), \\ ~(x, t)\in[0, L]\times{\mathbb{R}}^{+}, \\ \eta_{t}=-\eta_{s}+u_{t}, \phi_{t}=-\phi_{s}+v_{t}, ~ (x, t, s)\in[0, L]\times{\mathbb{R}}^{+}\times{\mathbb{R}}^{+}. \end{array} \right. \end{align} $

(0.4)

相应的边值条件为

$ \begin{align} \begin{array}{l} u(0)=\Delta u(0)=u(L)=\Delta u(L)=0, v(0)=v(L)=0, \\ \eta(0)=\Delta \eta(0)=\eta(L)=\Delta \eta(L)=0, \phi(0)=\phi(L)=0 . \end{array} \end{align} $

(0.5)

初值条件为

$ \begin{align} \begin{array}{l} u(x, 0)=u_{0}(x), ~u_{t}(x, 0)=u_{1}(x), ~\eta^{t}(x, 0)=0, ~\eta^{0}(x, s)=\eta_{0}(x, s), \\ v(x, 0)=v_{0}(x), ~v_{t}(x, 0)=v_{1}(x), ~\phi^{t}(x, 0)=0, ~\phi^{0}(x, s)=\phi_{0}(x, s), \end{array} \end{align} $

(6)

其中

$ \begin{align*} \left\{\begin{array}{ll} u_{0}(x)=u_{0}(x, 0), v_{0}(x)=v_{0}(x, 0), x\in[0, L], \\ u_{1}(x)=\partial_{t}u_{0}(x, t)\mid_{t=0}, v_{1}(x)= \partial_{t}v_{0}(x, t)\mid_{t=0}, ~x\in[0, L], \\ \eta_{0}(x, s)=u_{0}(x, 0)-u_{0}(x, -s), \phi_{0}(x, s)=v_{0}(x, 0)-v_{0} (x, -s), \\ (x, s)\in[0, L]\times{\mathbb{R}}^{+}. \end{array} \right. \end{align*} $

1 准备工作

为方便起见, 记$V_{0}=L^{2}[0, L], V_{1}=H^{2}[0, L]\cap H^{1}_{0}[0, L], $$V_{2}= H^{1}_{0}[0, L]$, 并记

$ \begin{align*} H=V_{0}\times V_{0}, V=V_{1}\times V_{2}, \end{align*} $

用$(\cdot)$和$|\cdot|$分别表示空间$H$的内积与范数, 因此对任意的$u=(u_{1}, u_{2})$, $v=(v_{1}, v_{2})$, 定义

$ \begin{align*} (u, v)=\int_{[0, L]}(u_{1}v_{1}+u_{2}v_{2}){\text{d}}x, ~|u|^{2}_{2}=|u_{1}|^{2}_{2}+|u_{2}|^{2}_{2}. \end{align*} $

同理, $((\cdot, \cdot))$和$||\cdot||$分别表示空间$V$的内积与范数, 即

$ \begin{align*} ((u, v))=\int_{[0, L]}(\Delta u_{1}\Delta v_{1}+\nabla u_{2}\nabla v_{2}){\text{d}}x, ~||u||^{2}_{2}=|\Delta u_{1}|^{2}_{2}+|\nabla u_{2}|^{2}_{2}. \end{align*} $

下面, 我们考虑特征值问题

$ \begin{align} \left\{\!\!\begin{array}{l} -\nabla v=\lambda v, ~x\in[0, L], \\ v(0)=v(L)=0, \end{array} \right. \end{align} $

(1.1)

其中$\lambda_{1}$是$-\nabla$的第一特征值, 那么

$ \begin{align} \left\{\!\!\begin{array}{l} \Delta^{2} u=\lambda^{2} u, ~x\in[0, L], \\ u(0)=\Delta u(0)=u(L)=\Delta u(L)=0, \end{array} \right. \end{align} $

(1.2)

则$\Delta^2$的第一特征值为$\lambda_{1}^{2}$.取$\lambda=\min\{\lambda_{1}, \lambda_{1}^{2}\}$, 利用Poincare不等式, 有$||u||^{2}\geq\lambda|u|^{2}$, $\forall u=(u_{1}, u_{2})\in V_{1}\times V_{2}.$此外, 假设记忆核函数$\mu_{i}(\cdot), i=1, 2$, 满足如下条件:

$(H_{1})~\mu_{i}\in C^{1}({\mathbb{R}}^{+})\cap L^{1}({\mathbb{R}}^{+}), $$\mu_{i}'(s)\leqslant0\leqslant \mu_{i}(s), ~\forall s\in{\mathbb{R}}^{+}$;

$(H_{2})~\int_{0}^{\infty }\mu_{i}(s){\text{d}}s =\mu_{0}>0, ~\forall s\in{\mathbb{R}}^{+}$;

$(H_{3})~\mu_{i}'(s)+\delta\mu_{i}(s)\leqslant0, ~\forall s\in{\mathbb{R}}^{+}, ~\delta>0$.

非线性函数$f_{i}\in C^2({{\mathbb{R}}, {\mathbb{R}}}), i=1, 2, $满足如下假设条件:

$(F_{1})~\liminf\limits_{|s|\rightarrow\infty}\frac{f_{i}(s)}{s} \geqslant\delta, ~\forall ~s\in{\mathbb{R}}$, 其中$\delta>0$;

$(F_{2})~f_i(0)=0, |f_i(u)-f_i(v)|\leqslant k_{0}(1+|u|^{p}+|v|^{p})|u-v|, ~\forall u, v\in{\mathbb{R}}$, 其中$k_{0}>0$且$p>0$.

根据记忆核函数$\mu_{i}(\cdot), i=1, 2, $满足的条件, 设$L^{2}_{\mu_{i}}({\mathbb{R}}^{+};V_{i}), i=1, 2, $为定义于${\mathbb{R}}^{+}$上取值于$V_{i}$的一个Hilbert空间, 并且赋予相应的内积和范数:

$ \begin{align*} (u, v)_{\mu, V_{i}}=\int_{0}^{\infty }\mu(r)(u(r), v(r))_{V_{i}} {\text{d}}r, \|u\|_{\mu, V_{i}}^{2}=(u, u)_{\mu, V_{i}}=\int_{0}^{\infty } \mu(r)\|u(r)\|_{V_{i}}^{2}{\text{d}}r. \end{align*} $

定义如下Hilbert空间: ${\mathcal{H}}=V\times H\times L^{2}_{\mu}({\mathbb{R}}^{+}\times{\mathbb{R}}^{+};V)$, 其中$L^{2}_{\mu}({\mathbb{R}}^{+}\times{\mathbb{R}}^{+};V)=L^{2}_ {\mu_{1}}({\mathbb{R}}^{+};H^{2}[0, L]\cap H^{1}_{0}[0, L])$$\times L^{2}_{\mu_{2}}({\mathbb{R}}^{+}; H^{1}_{0}[0, L])$.为了得到问题$(0.4)$--$(0.6)$的全局吸引子, 我们还需要以下结果.

引理1.1^[8]  设记忆项$\mu_{i}(s)$满足$(H_{1})$--$(H_{3})$, 非线性项$f_{i}$满足$(F_{1})$--$(F_{2})$, $g_{i}\in L^{2}(0, L), i=1, 2$, 那么

(ⅰ) 若初值$(u_{0}, v_{0}, , u_{1}, v_{1}, \eta_{0}, \phi_{0})\in\mathcal{H}$, 则问题$(0.4)$--$(0.6)$有一个弱解

$ \begin{align*} (u, v, u_{t}, v_{t}, \eta, \phi)\in C([0, T], \mathcal{H}), ~\forall T>0, \end{align*} $

满足

$ \begin{align*} u\in L^{\infty}(0, T;V_{1}), ~u_{t}\in L^{\infty}(0, T;V_{0}), ~\eta\in L^{\infty}(0, T;L^{2}_{\mu_{1}}({\mathbb{R}}^{+}\times{\mathbb{R}}^ {+}, V_{1})), \\[1mm] v\in L^{\infty}(0, T;V_{2}), ~v_{t}\in L^{\infty}(0, T;V_{0}), ~\phi \in L^{\infty}(0, T;L^{2}_{\mu_{2}}({\mathbb{R}}^{+}\times{\mathbb{R}}^{+}, V_{2})). \end{align*} $

(ⅱ) 令$z_{i}=(u^{i}, v^{i}, u_{t}^{i}, v_{t}^{i}, \eta^{i}, \phi^{i})$是问题$(0.4)$--$(0.6)$对应于初值$z_{i}(0)=(u_{0}^{i}, v_{0}^{i}, u_{1}^{i}, v_{1}^{i}, \eta_{0}^{i}, \phi_{0}^{i})$的一个弱解, 则对常数$c>0$, 有

$ \begin{align*} |z_{1}(t)-z_{2}(t)|_{\mathcal{H}}\leqslant {\text{e}}^{ct}|z_{1}(0)-z_{2}(0)|_{\mathcal{H}}, ~t\in[0, T]. \end{align*} $

问题$(0.4)$--$(0.6)$的适定性结果表明, 算子$S(t):\mathcal{H}\rightarrow\mathcal{H}$可定义为

$ \begin{align*} S(t)(u_{0}, v_{0}, u_{1}, v_{1}, \eta_{0}, \phi_{0})=(u(t), v(t), u_{t}(t), v_{t}(t), \eta^{t}, \phi^{t}), ~t\geqslant0, \end{align*} $

其中$(u(t), v(t), u_{t}(t), v_{t}(t), \eta^{t}, \phi(t))$是问题$(0.4)$--$(0.6)$唯一的弱解, 算子$S(t)$满足半群的性质, 且可定义一个在$\mathcal{H}$上局部Lipschitz连续的非线性$C_{0}$-半群.

定义1.2^[12]  设$X$为Banach空间, $B$为 $X$中的有界集, 定义于$X\times X$上的函数$\phi(\cdot, \cdot)$称为$B\times B$上的收缩函数, 如果对于任意的序列$\{x_{n}\}^{\infty}_{n=1}\subset B$, 存在子列$\{x_{n_{k}}\}^{\infty}_{k=1}\subset \{x_{n}\}^{\infty}_{n=1}$, 使得

$ \begin{align} \lim\limits_{k\rightarrow\infty}\lim\limits_{l\rightarrow\infty}\phi(x_{n_{k}}, x_{n_{l}})=0. \end{align} $

(1.3)

令${\mathfrak{C}}$表示定义于$B\times B$上的收缩函数的集合.

引理1.3^[12]  设$\{S(t)\}_{t\geqslant0}$为Banach空间$(X, \|\cdot\|)$上的半群, 并存在有界吸收集$B_{0}$.进一步, 对任意的$\epsilon>0$, 存在$T=T(B_{0}, \epsilon)$以及$\phi_{T}(\cdot, \cdot)\in{\mathfrak{C}}(B_{0})$, 使得

$ \begin{align*} \|S(T)x-S(T)y\|\leqslant\epsilon+\phi_{T}(x, y), ~\forall(x, y)\in B_{0}, \end{align*} $

其中$\phi_{T}$依赖于$T$.那么$\{S(t)\}_{t\geqslant0}$在$X$中渐近紧, 即对于任意的有界序列$\{y_{n}\}^{\infty}_{n=1}\subset X$和$\{t_{n}\}$, 当$n\rightarrow\infty, t_{n}\rightarrow\infty$时, $\{S(t_{n})y_{n}\}_{n=1}^{\infty}$在$X$中相对紧.

注: 一个半群称之为耗散的是指它拥有一个紧的吸收集.

定理1.4^[17]  一个耗散的动力系统$(\mathcal{H}, S(t))$有一个紧的全局吸引子当且仅当它是渐近紧的.

本文的主要结果是以下定理.

定理1.5  设记忆项$\mu_{i}(s)$满足 $(H_{1})$--$(H_{3})$, 非线性项$f_{i}$满足$(F_{1})$--$(F_{2})$, $g_{i}\in L^{2}(\Omega), i=1, 2$, 则由问题$(1.4)$--$(1.6)$产生的动力系统$(\mathcal{H}, S(t))$有一个紧的全局吸引子$\mathcal{A}_{\alpha}\subset\mathcal{H}, \alpha\geqslant0$, 且以范数$\|\cdot\|$吸引$\mathcal{H}$中的任意有界集.

2 $\mathcal{H}$空间的全局吸引子

为了证明定理$1.5$, 我们将用到第$1$节中的一些结果.首先需要证明动力系统$(\mathcal{H}, S(t))$是耗散的, 即它有一个有界吸收集; 其次需要证明它的渐近紧性, 从而由定理$1.4$可得其全局吸引子的存在性.

2.1 $\mathcal{H}$中的有界吸收集

首先, 由$(F_{1})$可知, 存在常数$K_{1}, K_{2}, K_{3}, K_{4}>0, \eta=\eta(\delta)>0, \phi=\phi(\delta)>0$, 使得

$ f_{1}(s)s+\eta s^{2}+K_{1}\geqslant0, ~\forall s\in{\mathbb{R}}; $

(2.1)

$ F_{1}(s)+\eta s^{2}+K_{2}\geqslant0, ~\forall s\in{\mathbb{R}}; $

(2.2)

$ f_{2}(s)s+\phi s^{2}+K_{3}\geqslant0, ~\forall s\in{\mathbb{R}}; $

(2.3)

$ F_{2}(s)+\phi s^{2}+K_{4}\geqslant0, ~\forall s\in{\mathbb{R}}, $

(2.4)

其中$F_{i}(u)=\int_{0}^{u}f_{i}(s){\text{d}}s, i=1, 2$.取$0<\sigma<1$, 用$\varphi=u_{t}+\sigma u, ~\psi=v_{t}+\sigma v$与$(0.4)$在$H$中作内积后得

$ \begin{align} \begin{array}{l} ~~~\dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t}(L_{1}|\Delta u|^{2}_{2}+|\varphi|^{2}_{2}+|\psi|^{2}_{2}+ L_{2}|\nabla v|^{2}_{2})+\sigma L_{1}|\Delta u|^{2}_{2}\\ +(\alpha-\sigma)|\varphi|^{2}_{2}-\sigma(\alpha-\sigma) (u, \varphi) +\sigma L_{2}\|\nabla v\|^{2}_{2}+(\beta-\sigma)|\psi|^{2}_{2}-\sigma(\beta-\sigma)(v, \psi)\\ +(\eta^{t}, u_{t})_{\mu_{1}, V_{1}}+ \sigma(\eta^{t}, u)_{\mu_{1}, V_{1}}+(\phi^{t}, v_{t})_{\mu_{2}, V_{2}}+\sigma(\phi^{t}, v)_{\mu_{2}, V_{2}}+k((u-v)^{+}, \varphi-\psi)\\ +(f_{1}(u), \varphi)+(f_{2}(v), \psi)=(g_{1}, \varphi)+(g_{2}, \psi). \end{array} \end{align} $

(2.5)

结合$(0.4), (H_{2}), (H_{3})$以及Hölder不等式, 有

$ (\eta^{t}, u_{t})_{\mu_{1}, V_{1}}+(\phi^{t}, v_{t})_{\mu_{2}, V_{2}}\\ = (\eta^{t}, \eta^{t}_{t}+\eta^{t}_{s})_{\mu_{1}, V_{1}}+(\phi^{t}, \phi^{t}_{t}+\phi^{t}_{s})_{\mu_{2}, V_{2}}\\ = \dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t}| \eta^{t}|^{2}_{\mu_{1}, V_{1}}+\dfrac{1}{2}\dfrac{{\text{d}}} {{\text{d}}t}\|\phi^{t}\|^{2}_{\mu_{2}, V_{2}} +\int_{0}^{\infty }\mu_{1}(s)(\eta^{t}(s), \eta^{t}_{s}(s))_{V_{1}}{\text{d}}s+\\ \int_{0}^{\infty }\mu_{2}(s)(\phi^{t}(s), \phi^{t}_{s} (s))_{V_{2}}{\text{d}}s\\ =\dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t}| \eta^{t}|^{2}_{\mu_{1}, V_{1}}+\dfrac{1}{2}\int_{0}^{\infty } \mu_{1}(s){\text{d}}|\eta^{t}(s)|^{2}_{V_{1}}+ \dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t} \|\phi^{t}\|^{2}_{\mu_{2}, V_{2}}+ \dfrac{1}{2}\int_{0}^{\infty }\mu_{1}(s){\text{d}} \|\phi^{t}(s)\|^{2}_{V_{2}}\\ = \dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t}|\eta^{t}|^ {2}_{\mu_{1}, V_{1}}- \dfrac{1}{2}\int_{0}^{\infty }\mu_{1}'(s)|\eta^{t}(s) |^{2}_{V_{1}}{\text{d}}s +\dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t} \|\phi^{t}\|^{2}_{\mu_{2}, V_{2}}-\dfrac{1}{2}\int_{0}^{\infty }\mu_{2}'(s)|\phi^{t}(s)|^{2}_{V_{2}}{\text{d}}s \\ \geqslant \dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t}| \eta^{t}|^{2}_{\mu_{1}, V_{1}}+\dfrac{\delta}{2}\int_{0}^{\infty }\mu_{1}(s)|\eta^{t}(s)|^{2}_{V_{1}}{\text{d}}s+ \dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t} \|\phi^{t}\|^{2}_{\mu_{2}, V_{2}}+\dfrac{\delta}{2} \int_{0}^{\infty } \mu_{2}(s)\|\phi^{t}(s)\|^{2}_{V_{2}} {\text{d}}s \\ =\dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t}|\eta^{t}| ^{2}_{\mu_{1}, V_{1}}+\dfrac{\delta}{2}|\eta^{t}|^{2}_{\mu_{1}, V_{1}}+ \dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t} \|\phi^{t}\|^{2}_{\mu_{2}, V_{2}}+\dfrac{\delta}{2} \|\phi^{t}\|^{2}_{\mu_{2}, V_{2}}, $

及

$ \begin{align*} \sigma(\eta^{t}, u)_{\mu_{1}, V_{1}}\geqslant- \dfrac{\delta}{4}|\eta^{t}|^{2}_{\mu_{1}, V_{1}}- \dfrac{\mu_{0}\sigma^{2}}{\delta}|\Delta u|^{2}_{2}, \\ \sigma(\phi^{t}, v)_{\mu_{2}, V_{2}}\geqslant- \dfrac{\delta}{4}\|\phi^{t}\|^{2}_{\mu_{2}, V_{2}}- \dfrac{\mu_{0}\sigma^{2}}{\delta}\|\Delta v\|^{2}_{2}. \end{align*} $

从而由式$(2.5)$可得

$ \begin{align} \begin{array}{l} \dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t}(|L_{1}\Delta u|^{2}_{2}+|\varphi|^{2}_{2}+|\psi|^{2}_{2}+\|L_{2}\nabla v\|^{2}_{2} +|\eta^{t}|^{2}_{\mu_{1}, V_{1}}+\|\phi^{t}\|^{2}_{\mu_{2}, V_{2}})\\ ~~~~~~+\sigma\Big(L_{1}-\dfrac{\mu_{0}\sigma}{\delta}\Big)|\Delta u|^{2}_{2} +(\varepsilon-\sigma)\|\phi\|^{2}_{2}-\sigma (\varepsilon-\sigma)(u, \phi)+\dfrac{\delta}{4} |\eta^{t}|^{2}_{\mu_{1}, V_{1}}\\ ~~~~~~+\sigma\Big(L_{2}-\dfrac{\mu_{0}\sigma}{\delta}\Big)\|\nabla v\|^{2}_{2} +(\beta-\sigma)|\psi|^{2}_{2}-\sigma(\beta-\sigma)(u, \psi)+ \dfrac{\delta}{4}|\psi^{t}|^{2}_{\mu_{2}, V_{2}}\\ ~~~~~~+k((u-v)^{+}, \varphi-\psi) +(f_{1}(u), \varphi)+(f_{2}(v), \psi)\leq(g_{1}, \varphi)+(g_{2}, \psi). \end{array} \end{align} $

(2.6)

取充分小的$\sigma$和充分大的$L_{1}, L_{2}$, 使得

$ \begin{align} L_{1}-\dfrac{\mu_{0}\sigma}{\delta}-\dfrac{\varepsilon \sigma}{2\lambda}\geqslant1-\sigma, ~\dfrac{\alpha}{2}-\sigma\geqslant\dfrac{\alpha}{4}, ~L_{2}-\dfrac{\mu_{0}\sigma}{\delta}- \dfrac{\beta\sigma}{2\lambda}\geqslant1-\sigma, ~\dfrac{\beta}{2}-\sigma\geqslant\dfrac{\beta}{4} . \end{align} $

(2.7)

结合Hölder、Young和Poincare不等式, 有

$ \begin{align} \begin{array}{l} \sigma\Big(L_{1}-\dfrac{\mu_{0}\sigma}{\delta}\Big)| \Delta u|^{2}_{2}+(\alpha-\sigma)|\varphi|^{2}_{2}- \sigma(\alpha-\sigma)(u, \varphi)\\ ~~~+\sigma\Big(L_{2}-\dfrac{\mu_{0}\sigma}{\delta}\Big) \|\nabla v\|^{2}_{2}+(\beta-\sigma)|\psi|^{2}_{2}- \sigma(\beta-\sigma)(v, \psi)\\ \geqslant \sigma\Big(L_{1}-\dfrac{\mu_{0}\sigma}{\delta}\Big) |\Delta u|^{2}_{2}+(\alpha-\sigma)|\varphi|^{2}_{2}- \dfrac{\alpha\sigma}{\sqrt{\lambda}}|\Delta u|_{2}|\varphi|_{2}\\ ~~~+\sigma\Big(L_{2}-\dfrac{\mu_{0}\sigma}{\delta}\Big)\|\nabla v\|^{2}_{2}+(\beta-\sigma)|\psi|^{2}_{2}-\dfrac{\beta\sigma} {\sqrt{\lambda}}\|\nabla v\|_{2}|\psi|_{2} \\ \geqslant \sigma\Big(L_{1}-\dfrac{\mu_{0}\sigma}{\delta}\Big) |\Delta u|^{2}_{2}+(\alpha-\sigma)|\varphi|^{2}_{2}-\Big(\dfrac {\alpha\sigma^{2}}{2\lambda}|\Delta u|^{2}_{2}+ \dfrac{\alpha}{2}|\varphi|^{2}_{2}\Big)\\ ~~~+\sigma\Big(L_{2}-\dfrac{\mu_{0}\sigma}{\delta}\Big) \|\nabla v\|^{2}_{2}+(\beta-\sigma)|\psi|^{2}_{2}-\Big( \dfrac{\beta\sigma^{2}}{2\lambda}\|\nabla v\|^{2}_{2}+ \dfrac{\beta}{2}|\psi|^{2}_{2}\Big)\\ =\sigma\Big(L_{1}-\dfrac{\mu_{0}\sigma}{\delta}- \dfrac{\alpha\sigma}{2\lambda}\Big)|\Delta u|^{2}_{2}+\Big(\dfrac{\alpha}{2}-\sigma\Big)| \varphi|^{2}_{2}+\sigma\Big(L_{2}-\dfrac{\mu_{0}\sigma}{\delta}- \dfrac{\beta\sigma}{2\lambda}\Big)\\ \|\nabla v\|^{2}_{2}+\Big(\dfrac{\beta}{2}-\sigma\Big)\psi|^{2}_{2}\\ \geqslant \sigma(1-\sigma)|\Delta u|^{2}_{2}+\dfrac{\alpha}{4}|\varphi|^{2}_{2}+\sigma(1-\sigma) \|\nabla v\|^{2}_{2}+\dfrac{\beta}{4}|\psi|^{2}_{2}. \end{array} \end{align} $

(2.8)

此外,

$ \begin{align} \begin{array}{l} \displaystyle(f_{1}(u), \varphi)=(f_{1}(u), u_{t}+\sigma u)=\dfrac{{\text{d}}} {{\text{d}}t}\int_{[0, L]}F_{1}(u){\text{d}}x+ \sigma\int_{[0, L]}f_{1}(u)u{\text{d}}x, \\ \displaystyle(f_{2}(v), \psi)=(f_{2}(v), v_{t}+\sigma v)=\frac{{\text{d}}}{{\text{d}}t}\int_{[0, L]} F_{2}(v){\text{d}}x+\sigma\int_{[0, L]}f_{2}(v)v {\text{d}}x. \end{array} \end{align} $

(2.9)

且

$ \begin{align} (g_{1}, \varphi)=\dfrac{{\text{d}}}{{\text{d}}t}\int_{[0, L]} gu{\text{d}}x+\sigma\int_{[0, L]}g_{1}u{\text{d}}x, ~(g_{2}, \psi)=\dfrac{{\text{d}}}{{\text{d}}t}\int_{[0, L]}g_{2}v{\text{d}}x+\sigma\int_{[0, L]}g_{2}{\text{d}}x, \end{align} $

(2.10)

$ \begin{align} k((u-v)^{+}, \varphi-\psi)=&k((u-v)^{+}, (u-v)_{t})+\sigma k((u-v)^{+}, (u-v))\notag\\ =&\dfrac{1}{2}\cdot\dfrac{{\text{d}}}{dt}(k|(u-v)^{+}|^{2})+\sigma k|(u-v)^{+}|^{2}. \end{align} $

(2.11)

将式$(2.8)$--$(2.11)$代入式$(2.6)$, 整理可得

$ \dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t}(L_{1}|\Delta u|^{2}_{2}+|\varphi|^{2}_{2}+|\eta^{t}|^{2}_{\mu_{1}, V_{1}} +k|(u-v)^{+}|^{2}_{2}+L_{2}\|\nabla v\|^{2}_{2}+|\psi|^{2}_{2}+ \|\phi^{t}\|^{2}_{\mu_{2}, V_{2}}\notag\\ \ \ \ \ \ \ \ \ \ \ +\!2\int_{[0, L]}F_{1}(u){\text{d}}x\!-\!2 \int_{[0, L]}g _{1}u{\text{d}}x\!+\!2\int_{[0, L]}F_{2}(v){\text{d}}x \!-\!2\int_{[0, L]}g_{2}v{\text{d}}x)\!+\!\sigma(1\!-\!\sigma)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \|\Delta u\|^{2}_{2} \!+\!\dfrac{\alpha}{4}|\varphi|^{2}_{2}\notag\\ \ \ \ \ \ \ \ \ \ \ +\dfrac{\delta}{4}|\eta^{t}|^{2}_{\mu_{1}, V_{1}}+ \sigma(1-\sigma)\|\nabla v\|^{2}_{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\dfrac{\beta}{4}|\psi|^{2}_{2}+\dfrac{\delta}{4}\|\phi^{t}\| ^{2}_{\mu_{2}, V_{2}}+\sigma k|(u-v)^{+}|^{2}_{2}+\sigma\int_{[0, L]}f_{1}(u)u{\text{d}}x\notag\\ \ \ \ \ \ \ \ \ \ \ -\sigma\int_{[0, L]}g_{1}u{\text{d}}x+\sigma\int_{[0, L]} f_{2}(v)v{\text{d}}x-\sigma\int_{[0, L]}g_{2} v{\text{d}}x\leqslant0. $

(2.12)

取$\sigma_{0}=\min\{\sigma, \frac{\alpha}{2}, \frac{\delta}{2}, \frac{\beta}{2}, \sigma(1-\sigma)\}$, 令

$ \begin{align} \begin{array}{l} E(t)=L_{1}|\Delta u|^{2}_{2}+|\varphi|^{2}_{2}+|\eta^{t}|^{2}_{\mu_{1}, V_{1}} +k|(u-v)^{+}|^{2}_{2}+L_{2}\|\nabla v\|^{2}_{2}+|\psi|^{2} _{2}+\|\phi^{t}\|^{2}_{\mu_{2}, V_{2}}\\[2mm] \qquad\quad ~ \displaystyle+2\int_{[0, L]}F_{1}(u){\text{d}}x-2\int_{[0, L]} g_{1}u{\text{d}}x+2\int_{[0, L]}F_{2}(v){\text{d}}x -2\int_{[0, L]}g_{2}v{\text{d}}x, \end{array} \end{align} $

(2.13)

和

$ \begin{align} \begin{array}{l} I(t)=|\Delta u|^{2}_{2}+|\varphi|^{2}_{2} +|\eta^{t}|^{2}_{\mu_{1}, V_{1}}+\|\nabla v\|^{2}_{2}+ |\psi|^{2}_{2}+ \|\phi^{t}\|^{2}_{\mu_{2}, V_{2}}+ k|(u-v)^{+}|^{2}_{2}\\[2mm] \qquad \quad \displaystyle+\int_{[0, L]}f_{1}(u)u{\text{d}}x-\int_{[0, L]} g_{1}u{\text{d}}x +\int_{[0, L]}f_{2}(v)v{\text{d}}x-\int_{[0, L]} g_{2}v{\text{d}}x, \end{array} \end{align} $

(2.14)

从而有

$ \begin{align*} \dfrac{\text{d}}{{\text{d}}t}E(t)+\sigma_{0}I(t)\leqslant0, \end{align*} $

即

$ \begin{align} \displaystyle E(t)\leqslant-\sigma_{0}\int_{0}^{t}I(\tau) {\text{d}}\tau+E(0), \end{align} $

(2.15)

其中

$ \begin{align} \begin{array}{l} E(0)=Ɫ_{1}|\Delta u_{0}|^{2}_{2}+|u_{1}+\sigma u_{0}|^{2} _{2}+|\eta_{0}|^{2}_{\mu_{1}, V_{1}}\\[2mm] \qquad \quad ~\, \displaystyle+k|(u-v)^{+}_{0}|^{2}_{2}+2\int_{[0, L]} F_{1}(u_{0}){\text{d}}x-2\int_{[0, L]}g_{1}u_{0} {\text{d}}x+L_{2}|\nabla v_{0}\|^{2}_{2}\\[2mm] \qquad \quad ~\, \displaystyle+\|v_{1}+\sigma v_{0}\|^{2}_{2}+\|\phi_{0}\|^{2}_{\mu_{2}, V_{2}}+2 \int_{[0, L]} F_{2}(v_{0}){\text{d}}x-2\int_{[0, L]}g_{2}v_{0} {\text{d}}x. \end{array} \end{align} $

(2.16)

根据$(2.1)$--$(2.4)$和$(2.13)$--$(2.14)$, 利用Sobolev紧嵌入定理可得

$ \begin{align} E(t)\geqslant &L_{1}|\Delta u|^{2}_{2}+|\varphi|^{2}_{2}+|\eta^{t}|^{2}_{\mu_{1}, V_{1}}+k|(u-v)^{+}| ^{2}_{2}\\ &-2\int_{[0, L]}(\eta u^{2}+K_{3}){\text{d}}x-\sigma_{0}|u|^{2}_{2}-\dfrac{1}{\sigma_{0}}\|g_{1}\|^{2}_{2}\notag\\ &+L_{2}\|\nabla v\|^{2}_{2} +|\psi|^{2}_{2}+\|\phi^{t}\|^{2}_{\mu_{2}, V_{2}}-2\int_{[0, L]}(\phi v^{2}+K_{4}){\text{d}}x\\ &-\sigma_{0}\|v\|^{2}_{2}-\dfrac{1} {\sigma_{0}}\|g_{2}\|^{2}_{2}\notag\\ \geqslant&\Big(1-\dfrac{k+2\eta+\sigma_{0}}{\lambda}\Big) |\Delta u|^{2}_{2}+|\varphi|^{2}_{2}+|\eta^{t}|^{2}_{\mu_{1}, V_{1}}-M_{1}\notag\\ &+\Big(1-\dfrac{k+2\eta+\sigma_{0}}{\lambda}\Big)\|\nabla v\|^{2}_{2}+|\psi|^{2}_{2}+\|\phi^{t}\|^{2}_{\mu_{2}, V_{2}} -M_{2}, \end{align} $

(2.17)

其中$M_{1}=2K_{3}|[0, L]|+\frac{|g_{1}|^{2}_{2}}{\sigma_{0}}, M_{2}=2K_{4}|[0, L]|+\frac{|g_{2}|^{2}_{2}}{\sigma_{0}}$.类似地,

$ \begin{align} I(t)\geqslant&|\Delta u|^{2}_{2}+|\varphi|^{2}_{2}+|\eta^{t}| ^{2}_{\mu_{1}, V_{1}}+k|(u-v)^{+}|^{2}_{2}+2\int_{[0, L]} f_{1}(u)u{\text{d}}x-2\int_{[0, L]}g_{1}u{\text{d}}x\notag\\ &+\|\nabla v\|^{2}_{2}+|\psi|^{2}_{2}+\|\phi^{t}\|^{2}_{\mu_{2}, V_{2}}+2 \int_{[0, L]}f_{2}(v)v{\text{d}}x-2\int_{[0, L]} g_{2}v{\text{d}}x \notag\\\geqslant & \Big(1-\dfrac{k+2\eta+\sigma_{0}}{\lambda}\Big)| \Delta u|^{2}_{2}+|\varphi|^{2}_{2}+|\eta^{t}|^{2}_ {\mu_{1}, V_{1}}-M_{3}\notag\\ &+\Big(1-\dfrac{k+2\eta+\sigma_{0}}{\lambda}\Big)\|\nabla v\|^{2}_{2}+|\psi|^{2}_{2}+\|\phi^{t}\|^{2}_{\mu_{2}, V_{2}}-M_{4}. \end{align} $

(2.18)

其中$M_{3}=2K_{1}|[0, L]|+\frac{|g_{1}|^{2}_{2}}{\sigma_{0}}, M_{4}=2K_{2}|[0, L]|+\frac{|g_{2}|^{2}_{2}}{\sigma_{0}}$.令$\frac{k+2\eta}{\lambda}<1, \frac{k+2\phi}{\lambda}<1$, 且$0<\sigma_{0}<\lambda-k-2\eta, 0<\sigma_{0}<\lambda-k-2\phi$, 则有

$ \begin{align} 1-\dfrac{k+2\eta+\sigma_{0}}{\lambda}>0, 1-\dfrac{k+2\phi+\sigma_{0}}{\lambda}>0. \end{align} $

(2.19)

结合式$(2.17)$--$(2.19)$, 存在一个常数$C_{1}$使得

$ E(t)\geqslant C_{1}(|\Delta u|^{2}_{2}+|\varphi|^{2}_{2}+|\eta^{t}|^{2}_{\mu_{1}, V_{1}}+\|\nabla v\|^{2}_{2}+|\psi|^{2}_{2}+\|\phi^{t}\|^{2}_{\mu_{2}, V_{2}})-M_{1} -M_{2}, $

(2.20)

$ I(t)\geqslant C_{1}(|\Delta u|^{2}_{2}+|\varphi|^{2}_{2}+|\eta^{t}|^{2}_{\mu_{1}, V_{1}}+\|\nabla v\|^{2}_{2}+|\psi|^{2}_{2}+\|\phi^{t}\|^{2}_{\mu_{2}, V_{2}})-M_{3}-M_{4}, $

(2.21)

故由式$(2.20)$--$(2.21)$和式$(2.15)$可知

$ \begin{align} \begin{array}{l} C_{1}(|\Delta u|^{2}_{2}+|\varphi|^{2}_{2}+|\eta^{t}|^{2}_{\mu_{1}, V_{1}}+\|\nabla v\|^{2}_{2}+|\psi|^{2}_{2}+\|\phi^{t}\|^{2}_{\mu_{2}, V_{2}})-M_{1}-M_{2}\\ \leqslant \!-\!\sigma_{0}\int_{0}^{t}[C_{1}(|\Delta u|^{2}_{2}\!+\!|\varphi|^{2}_{2}\!+\!|\eta^{t}|^{2}_{\mu_{1}, V_{1}}\!+\!\|\nabla v\|^{2}_{2}\!+\!|\psi|^{2}_{2}\!+\!\|\phi^{t}\|^{2}_{\mu_{2}, V_{2}})\!-\!M_{3}\!-\!M_{4}]{\text{d}}t\!+\!E(0). \end{array} \end{align} $

(2.22)

因此, 对$\forall K>\frac{M_{3}+M_{4}}{C_{1}}$, 存在$t_{0}=t_{0}(B)$, 使得

$ \begin{align} |\Delta u(t_{0})|^{2}_{2}+|\varphi(t_{0})|^{2}_{2}+|\eta^{t_{0}}|^{2}_{\mu_{1}, V_{1}}+\|\nabla v(t_{0})\|^{2}_{2}+|\psi(t_{0})|^{2}_{2}+\|\phi^{t_{0}}\|^{2}_{\mu_{2}, V_{2}}\leqslant K. \end{align} $

(2.23)

所以, 如果$u, v$是系统$(0.4)$--$(0.6)$的解, 令$B_{0}=\bigcup_{t\geqslant0}S(t)B$, 其中

$ B\!=\!\{(u_{0}, v_{0}, u_{1}, v_{1}, \eta_{0}, \phi_{0})^{\text{T}}\in\mathcal{H}_{0}:|\Delta u_{0}|^{2}_{2}\!+\!|\varphi_{0}|^{2}_{2}\!+\!|\eta_{0}|^{2}_ {\mu_{1}, V_{1}}\!+\\ \!||\nabla v_{0}\|^{2}_{2}\!+\!|\psi_{0}|^{2} _{2}+ \|\phi_{0}\|^{2}_{\mu_{2}, V_{2}}\leqslant K\}, $

则$B_{0}$是半群$\{S(t)\}_{t\geqslant0}$的一个有界吸收集.

定理2.1  假设记忆项$\mu_{i}(s)$满足$(H_{1})$--$(H_{3})$, 非线性项$f_{i}$满足$(F_{1})$--$(F_{2})$, $g_{i}\in L^{2}[0, L], i=1, 2$, 则空间$\mathcal{H}$中以0为中心, $\rho_{1}$为半径的球$B_{0}=B_{\mathcal{H}}(0, \rho_{1})$是问题$(0.4)$--$(0.6)$生成的解半群$\{S(t)\}_{t\geqslant0}$在$\mathcal{H}$中的有界吸收集, 即对$\mathcal{H}$中的任意有界集$B$, 存在$t=t_{0}(B)$, 使得当$t\geqslant t_{0}(B)$时, 有$S(t)B\subset B_{0}$.

有界吸收集的存在性意味着对于依赖于有界集$B\subset\mathcal{H}$的初值, 问题$(0.4)$--$(0.6)$的解全局有界, 即, 若$(u, v, u_{t}, v_{t}, \eta, \phi)$是问题$(0.4)$--$(0.6)$在有界集$B$上对应于初值$(u_{0}, v_{0}, u_{1}, v_{1}, \eta_{0}, \phi_{0})$的解, 则

$ \begin{align} \|(u(t), v(t), u_{t}(t), v_{t}(t), \eta^{t}, \phi^{t})\|_{\mathcal{H}}\leqslant C_{B}, ~\forall t\geqslant0, \end{align} $

(2.24)

其中$ C_{B}>0$是依赖于$B$的常数.

2.2 全局吸引子的存在性

引理2.2  在定理$1.5$的假设条件下, 给定一个有界集$B\subset\mathcal{H}$, 令$z_{1}=(u, v, u_{t}, v_{t}, \eta^{t}, \phi^{t}), z_{2}=(x, y, x_{t}, y_{t}, \xi^{t}, \gamma^{t})$是问题$(0.4)$--$(0.6)$在$B$上分别满足初值$z_{1}(0)=(u_{0}, v_{0}, u_{1}, v_{1}, \eta_{0}^{t}, \phi_{0}^{t}), z_{2}(0)=$$(x_{0}, y_{0}, x_{1}, y_{1}, \xi_{0}^{t}, \gamma_{0}^{t})$的两个弱解, 则

$ \begin{align} \begin{array}{l} |(z_{1}(t)-z_{2}(t)|^{2}_{\mathcal{H}}\leqslant {\text{e}}^{-\alpha_{2}t}|(z_{1}(0)-z_{2}(0))|^{2}_{\mathcal{H}}\\ \displaystyle+C_{3}\int_{0}^{t}{\text{e}}^{-\alpha_{2}(t-s)} |(u(s)-x(s))+(v(s)-y(s))|^{2}_{2(p+1)}{\text{d}}s, ~\forall t\geqslant0, \end{array} \end{align} $

(2.25)

其中, $\alpha_{2}, C_{3}$是正常数.

证明  给定一个有界集$B\subset\mathcal{H}$.令$w=u-x, \omega=v-y, \zeta=\eta-\xi, $$\vartheta=\phi-\gamma$, 根据$(0.4)$, 可得

$ \begin{align} \left\{\!\!\begin{array}{l} w_{tt}+\alpha w_{t}+L_{1}\Delta^{2}w+\int_{0}^{\infty }\mu_{1}(s)\Delta^{2}\zeta^{t}(s){\text{d}}s+\\ k(u-v)^{+}-k(x-y)^{+}+f_{1}(u)-f_{1}(x)=0, \\ \omega_{tt}+\beta \omega_{t}-L_{2}\nabla^{2}\omega+\int_{0}^{\infty }\mu_{2}(s)\Delta^{2}\vartheta^{t}(s){\text{d}} s\\ -k(u-v)^{+}+k(x-y)^{+}+f_{2}(v)-f_{2}(y)=0, \\ \zeta_{t}=-\zeta_{s}+w_{t}, \vartheta_{t}=- \vartheta_{s}+\omega_{t}. \end{array} \right. \end{align} $

(2.26)

相应的初值条件为

$ w(0)=u_{0}-x_{0}, ~w_{t}(0)=u_{1}-x_{1}, ~\zeta^{0}=\eta_{0}-\xi_{0}, \omega(0)=v_{0}-y_{0}, ~\omega_{t}(0)=v_{1}-y_{1}, \\ ~\vartheta^{0}=\phi_{0}-\gamma_{0}. $

用$\varsigma_{1}=w_{t}+\sigma w, \varsigma_{2}=\omega_{t}+\sigma \omega$分别与$(3.26)$在$H$上作内积, 可得

$ \begin{align} \dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t}& (L_{1}|\Delta w|^{2}_{2}+L_{2}\|\nabla \omega\|^{2}_{2}+ |\varsigma_{1}|^{2}_{2}+\|\varsigma_{2}\|^{2}_{2})+\\ &\sigma L_{1}|\Delta w|^{2}_{2}+\sigma L_{2}\|\nabla\omega\|^{2}_{2}+ (\alpha-\sigma)(w_{t}, \varsigma_{1})\notag\\ &+(\beta-\sigma)(\omega_{t}, \varsigma_{2}) +(\zeta^{t}, w_{t})_{\mu_{1}, V_{1}}\\ &+\sigma(\zeta^{t}, w)_{\mu_{1}, V_{1}}+(\vartheta^{t}, \omega_{t})_{\mu_{2}, V_{2}} +\sigma(\vartheta^{t}, \omega)_{\mu_{2}, V_{2}}\notag\\ &+(k(u-v)^{+}-k(x-y)^{+}, \varsigma_{1})-(k(u-v)^{+}-k(x-y)^{+}, \varsigma_{2})\notag\\ &+(f_{1}(u)-f_{1}(x), \varsigma_{1})+(f_{2}(v)-f_{2}(y), \varsigma_{2})=0. \end{align} $

(2.27)

类似前面的讨论, 我们有

$ \begin{align*} (\alpha-\sigma)(w_{t}, \varsigma_{1})\!=\!(\alpha\!-\!\sigma) |\varsigma_{1}|^{2}_{2}\!-\!\sigma(\alpha\!-\!\sigma)(w, \varsigma_{1}), (\beta\!-\!\sigma)(\omega_{t}, \varsigma_{2})\!=\!(\beta\!-\!\sigma) \|\varsigma_{2}\|^{2}_{2}\!-\!\sigma(\beta\!-\!\sigma)(\omega, \varsigma_{2}) \end{align*} $

和

$ \begin{align*} (\zeta^{t}, w_{t})_{\mu_{1}, V_{1}}\geqslant\dfrac{1}{2}\cdot \dfrac{{\text{d}}}{{\text{d}}t}|\zeta^{t}|^{2}_{\mu_{1}, V_{1}} +\dfrac{\delta}{2}|\zeta^{t}|^{2}_{\mu_{1}, V_{1}}, (\vartheta^{t}, \omega_{t})_{\mu_{2}, V_{2}}\geqslant\dfrac{1}{2}\cdot \dfrac{{\text{d}}}{{\text{d}}t}\|\vartheta^{t}\|^{2}_{\mu_{2}, V_{2}}+\dfrac{\delta}{2}\|\vartheta^{t}\|^{2}_{\mu_{2}, V_{2}} \end{align*} $

及

$ \begin{align*} \sigma(\zeta^{t}, w)_{\mu_{1}, V_{1}}\geqslant-\dfrac {\delta}{4}|\zeta^{t}|^{2}_{\mu_{1}, V_{1}}-\dfrac{\mu_{0} \sigma^{2}}{\delta}|\Delta w|^{2}_{2}, \sigma(\vartheta^{t}, \omega)_{\mu_{2}, V_{2}} \geqslant-\dfrac{\delta}{4}\|\vartheta^{t}\|^{2}_{\mu_{2}, V_{2}}-\dfrac{\mu_{0}\sigma^{2}}{\delta}\|\nabla \omega\|^{2}_{2}. \end{align*} $

进一步, 类似式(2.8)的估计, 可得

$ \begin{align} &\sigma\Big(1-\dfrac{\mu_{0}\sigma}{\delta}\Big)|\Delta w|^{2}_ {2}+(\alpha-\sigma)|\varsigma_{1}|^{2}_{2}-\sigma(\alpha- \sigma)(w, \varsigma_{1})+\sigma\Big(1-\dfrac{\mu_{0}\sigma} {\delta}\Big)\|\nabla \omega\|^{2}_{2}\notag\\ &+\!(\beta\!-\!\sigma)\|\varsigma_{2}\|^{2}_{2}\!-\!\sigma(\beta\!-\! \sigma)(\omega, \varsigma_{2}) \geqslant\sigma(1\!-\!\sigma)|\triangle w|^{2}_{2}\!+\!\dfrac{\alpha}{4}|\varsigma_{1}|^{2}_{2}\!+\! \sigma(1\!-\!\sigma)\|\nabla \omega\|^{2}_{2}\!+\!\dfrac{\beta}{4}\|\varsigma_{2}\|^{2}_{2}. \end{align} $

(2.28)

从而由式$(2.27)$得

$ \begin{align} \dfrac{1}{2}\cdot\dfrac{{\text{d}}}{{\text{d}}t}& (|\Delta w|^{2}_{2}+|\varsigma_{1}|^{2}_{2}+|\zeta^{t}|^ {2}_{\mu_{1}, V_{1}}+\|\nabla \omega\|^{2}_{2}+\|\varsigma_{2} \|^{2}_{2}+\|\vartheta^{t}\|^{2}_{\mu_{2}, V_{2}})\notag\\ &\quad \, +\sigma(1-\sigma)|\Delta w|^{2}_{2}+\dfrac{\alpha}{4}| \varsigma_{1}|^{2}_{2}+\dfrac{\delta}{4}|\zeta^{t}|^{2}_ {\mu_{1}, V_{1}}+\sigma(1-\sigma)\\ &\|\nabla \omega\|^{2}_{2}+ \dfrac{\beta}{4}\|\varsigma_{2}\|^{2}_{2}+\dfrac{\delta}{4} \|\vartheta^{t}\|^{2}_{\mu_{2}, V_{2}}\notag\\ &=-(k(u-v)^{+}-k(x-y)^{+}, \varsigma_{1})+(k(u-v)^{+}-k(x-y)^{+}, \varsigma_{2})\notag\\& \quad \, -(f_{1}(u)-f_{1}(x), \varsigma_{1})-(f_{2}(v)-f_{2}(y), \varsigma _{2}).\end{align} $

(2.29)

根据Young不等式和Poincare不等式, 有

$ \begin{align} |\!-\!(k(u-v)^{\!+\!}\!-\!k(x-y)^{\!+\!}, \varsigma_{1})|\!=\! &\Big|-\int_ {\Omega} (k(u-v)^{+}-k(x-y)^{+})(w_{t}+\sigma w){\text{d}}x\Big|\notag\\ \leqslant\!&\int_{\Omega} (k(u\!-\!v)^{+}\!-\!k(x\!-\!y)^{+})w_{t}{\text{d}}x\!+\!\\ & \int_{\Omega} (k(u\!-\!v)^{+}\!-\!k(x\!-\!y)^{+})\\ &\sigma w{\text{d}}x\notag\\ \leqslant & \dfrac{k^{2}}{\delta_{0}}|((u-v)^{+}-(x-y)^{+})|^{2}_{2}+ \dfrac{\delta_{0}}{4}|w_{t}|^{2}_{2}+\sigma k|w|^{2}_{2}\notag\\ \leqslant&\dfrac{k^{2}c_{0}}{\delta_{0}}|w|^{2}_{2(p+1)}+ \dfrac{\delta_{0}}{4}|\varsigma_{1}|_{2}^{2}+\Big(\dfrac{ \delta_{0}}{4\lambda}+\dfrac{\sigma k}{\lambda}\Big)|\Delta w|^{2}_{2}. \end{align} $

(2.30)

同理可得

$ \begin{align} |(k(u-v)^{+}-k(x-y)^{+}), \varsigma_{2})|\leqslant\dfrac {k^{2}c_{0}}{\delta_{0}}\|\omega\|^{2}_{2(p+1)}+ \dfrac{\delta_{0}}{4}\|\varsigma_{2}\|_{2}^{2}+\Big(\dfrac {\delta_{0}}{4\lambda}+\dfrac{\sigma k}{\lambda}\Big)\|\nabla \omega\|^{2}_{2}. \end{align} $

(2.31)

事实上, $|(u-v)^{+}-(x-y)^{+}|\leqslant C|(u-v)-(x-y)|, $ $|w_{t}|_{2}^{2}=\varsigma_{1}-\sigma w|_{2}^{2}\leq|\varsigma_{1}|_{2}^{2}+\sigma^{2}|w|_{2}^{2}, \|\omega_{t}\|_{2}^{2}=$ $\|\varsigma_{2}-\sigma \omega\|_{2}^{2}\leq\|\varsigma_{2}\|_{2}^{2}+\sigma^{2} \|\omega\|_{2}^{2}$且$c_{0}>0$是由嵌入$L^{2(p+1)}(\Omega)\hookrightarrow L^{2}(\Omega)$产生的常数.

利用Hölder不等式$(\frac{p}{2(p+1)}+\frac{1}{2(p+1)}+ \frac{1}{2}=1)$, 结合条件$(F_{2})$, 估计式$(2.24)$和Yöung不等式可得

$ \begin{align} |-(f_{1}(u)-f_{1}(x), \varsigma_{1})|=&\Big|-\int_{\Omega} (f_{1}(u)-f_{1}(x))(w_{t}+\sigma w){\text{d}}x\Big|\notag\\ \leqslant & k_{0}\int_{\Omega}(1+|u|^{p}+|x|^{p})|w||w_{t}+\sigma w|{\text{d}}x\notag\\ \leqslant&k_{0}\Big(|\Omega|^{\frac{p}{2(p+1)}}+|u|^{p}_{2(p+1)}+ |x|^{p}_{2(p+1)}\Big)|w|_{2(p+1)}(|w_{t}|_{2}+|\sigma w|_{2})\notag\\ \leqslant&C_{B}|w|_{2(p+1)}|w_{t}|_{2}+C_{B}|w|_{2(p+1)}| \sigma w|_{2}\notag\\ \leqslant &\dfrac{2C_{B}^{2}}{\delta_{0}}|w|_{2(p+1)}^{2}+ \dfrac{\delta_{0}}{4}|w_{t}|^{2}_{2}+ \dfrac{\delta_{0}\sigma^{2}}{4}|w|^{2}_{2}\notag\\ \leqslant &\dfrac{2C_{B}^{2}}{\delta_{0}}|w|_{2(p+1)}^{2}+ \dfrac{\delta_{0}}{4}|\varsigma_{1}|_{2}^{2}+ \dfrac{\delta_{0}\sigma^{2}}{2\lambda} |\triangle w|^{2}_{2}. \end{align} $

(2.32)

同理可得

$ \begin{align} |-(f_{2}(v)-f_{2}(y), \varsigma_{2})|\leqslant\dfrac{2C_{B} ^{2}}{\delta_{0}}\|\omega\|_{2(p+1)}^{2}+ \dfrac{\delta_{0}}{4}\|\varsigma_{2}||_{2}^{2}+ \dfrac{\delta_{0}\sigma^{2}}{2\lambda} \|\nabla \omega\|^{2}_{2}. \end{align} $

(2.33)

将式$(2.30)$--$(2.33)$代入式$(2.29)$, 整理得

$ \begin{align} \dfrac{{\text{d}}}{{\text{d}}t}&(|\Delta w|^{2}_{2}+ |\varsigma_{1}|^{2}_{2}+|\zeta^{t}|^{2}_{\mu_{1}, V_{1}}+\|\nabla \omega\|^{2}_{2}+\|\varsigma_{2}\| ^{2}_{2}+\|\vartheta^{t}\|^{2}_{\mu_{2}, V_{2}})\notag\\ &+\Big(2\sigma(1-\sigma)-\dfrac{\delta_{0}}{\sqrt{\lambda}} -\dfrac{2\sigma}{\lambda}\Big(k+\dfrac{\delta_{0}\sigma}{4}\Big) \Big)(|\Delta w|^{2}_{2}+\|\nabla\omega\|_{2}^{2})\notag\\ &\!+\!\dfrac{\alpha}{2}|\varsigma_{1}|^{2}_{2}\!+\! \dfrac{\delta}{2}|\zeta^{t}|^{2}_{\mu_{1}, V_{1}} \!+\!\dfrac{\beta}{2}\|\varsigma_{2}\|^{2}_{2}\!+\! \dfrac{\delta}{2}\|\vartheta^{t}\|^{2}_{\mu_{2}, V_{2}} \leqslant \Big(\dfrac{2k^{2}c_{0}}{\delta_{0}}\!+\!\dfrac{4C_{B}^{2}} {\delta_{0}}\Big)\\ &(|w|^{2}_{2(p\!+\!1)}\!+\!\|\omega\|^{2}_{2(p\!+\!1)}). \end{align} $

(2.34)

取足够小的$\delta_{0}$, 使

$ \begin{align*} 2\sigma(1-\sigma)-\dfrac{\delta_{0}}{\sqrt{\lambda}} -\dfrac{2\sigma}{\lambda}\Big(k+\dfrac{\delta_{0}\sigma}{4}\Big) >0. \end{align*} $

令$\alpha_{2}=\min\{2\sigma(1-\sigma)-\frac{\delta_{0}} {\sqrt{\lambda}} -\frac{2\sigma}{\lambda}(k+\frac{\delta_{0}\sigma}{4}), \frac{\alpha}{2}, \frac{\delta}{2}\}, $$~C_{3}=\frac{2k^{2}c_{0}}{\delta_{0}}+\frac{4C_{B}^{2}} {\delta_{0}}.$

定义如下泛函:

$ \begin{align*} E(t)=|\Delta w|^{2}_{2}+|\varsigma_{1}|^{2}_{2}+|\zeta^{t}|^{2}_{\mu, V}+\|\nabla \omega\|^{2}_{2}+\|\varsigma_{2}\|^{2}_{2}+\|\vartheta^{t}\|^{2}_{\mu_{2}, V}, \end{align*} $

则

$ \begin{align*} \dfrac{{\text{d}}}{{\text{d}}t}E(t)+\alpha_{2}E(t)\leqslant C_{3}(|w|^{2}_{2(p+1)}+\|\omega\|^{2}_{2(p+1)}). \end{align*} $

根据Gronwall引理, 可知

$ \begin{align*} E(t)\leqslant {\text{e}}^{-\alpha_{2}t}E(0)+C_{3}\int_{0}^{t}{\text{e}}^{-\alpha_{2}(t-s)} (|w|^{2}_{2(p+1)}+\|\omega\|^{2}_{2(p+1)}){\text{d}}s. \end{align*} $

由于$E(t)\geqslant|(z_{1}(t)-z_{2}(t)+z_{3}(t)- z_{4}(t))|^{2}_{\mathcal{H}}$, 故可得式$(2.25)$.

引理2.3  在定理$2.6$的假设条件下, 问题(0.4)--(0.6)生成的动力系统$(\mathcal{H}, S(t))$渐近紧.

证明  令$B$是$\mathcal{H}$上关于$S(t)$的一个正不变有界子集, 记$C_{B}$为仅依赖于$B$的常数.对于$z_{1}, z_{2}\in B$, $S(t)z_{1}=(u(t), v(t), u_{t}(t), v_{t}(t), \eta^{t}, \phi^{t}), $$S(t)z_{2}=(x(t), y(t), x_{t}(t), y_{t}(t), \xi^{t}, \gamma^{t})$是$(0.4)$--$(0.6)$的解.则给定一个$\epsilon>0$, 根据式$(2.25)$, 存在一个$T>0$使得

$ \begin{align} |(S(t)z_{1}-S(t)z_{2}|_{\mathcal{H}} \leqslant\epsilon+C_{B}\Big(\int_{0}^{T}|(u(s)-x(s)) +(v(s)-y(s))|^{2}_{2(p+1)}{\text{d}}s\Big)^{\frac{1}{2}}. \end{align} $

(2.35)

由于$p>0$, 从而$2<2(p+1)<\infty$.取$\theta=\frac{1}{2}(1-\frac{1}{p+1})$, 利用Gagliardo-Nirenberg不等式可得

$ \begin{align*} |(u(t)-x(t))+&(v(t)-y(t))|_{2(p+1)}\\[1mm] &\leqslant C|\Delta(u(t)-x(t))+\nabla(v(t)-y(t))|_{2}^ {\theta}\cdot|(u(t)-x(t))+(v(t)-y(t))|^{1-\theta}_{2}\\[1mm]& \leqslant C_{B}|(u(t)-x(t))+(v(t)-y(t))|^{1-\theta}_{2}. \end{align*} $

由于$|\Delta u(t)|_{2}^{2}, |\triangle x(t)|_{2}^{2}$和$\|\nabla v(t)\|_{2}^{2}, \|\nabla y(t)\|_{2}^{2}$一致有界, 故存在一个常数$C_{B}>0$使得

$ \begin{align} |(u(t)-x(t))+(v(t)-y(t))|_{2(p+1)}^{2}\leqslant C_{B}|(u(t)-x(t))+(v(t)-y(t))|^{2(1-\theta)}_{2}. \end{align} $

(2.36)

根据式$(2.35)$和式$(2.36)$, 有

$ \begin{align*} |(S(t)z_{1}-S(t)z_{2})|_{\mathcal{H}} \leqslant\epsilon+\rho_{T}(z_{1}, z_{2}), \end{align*} $

其中

$ \begin{align*} \rho_{T}(z_{1}, z_{2})=C_{B}\Big(\int_{0}^{T}|(u(s)- x(s))+(v(s)-y(s))|^{2(1-\theta)}_{2}{\text{d}}s\Big) ^{\frac{1}{2}}. \end{align*} $

下证$\rho_{T}\in\mathfrak{C}$, 即$\rho_{T}$满足$(1.1)$.

给定一个序列$(z^{n}_{0})=(u^{n}_{0}, v^{n}_{0}, u^{n}_{1}, v^{n}_{1}\eta^{n}_{0}, \phi^{n}_{0})\in B$, 令$S(t)(z^{n}_{0})=(u^{n}(t), v^{n}(t), u^{n}_{t}(t), v^{n}_{t}(t), $ $\eta^{n, t}, \phi^{n, t})$.由于当$t\geqslant0$时, $B$关于$S(t)$正不变, 从而序列$(u^{n}(t), v^{n}(t), u^{n}_{t}(t), v^{n}_{t}(t), \eta^{n, t}, \phi^{n, t})$在$\mathcal{H}$上一致有界.另一方面, $(u^{n}, v^{n}, u^{n}_{t}, v^{n}_{t})$在$C([0, T], V\times H)$中有界.

由紧嵌入$V\subset H$, 存在一个子列$(u^{n_{k}})$在$C([0, T], H)$中强收敛.因此

$ \begin{align*} \lim\limits_{k\rightarrow\infty}\lim\limits_ {l\rightarrow\infty}\int_{0}^{T}|u^{n_{k}}(s) -u^{n_{l}}(s)|^{2(1-\theta)}_{2}{\text{d}}s=0. \end{align*} $

即$(1.1)$成立.结论得证.

定理1.5的证明  由定理$2.1$和引理$2.3$可知, $(\mathcal{H}, S(t))$是耗散的动力系统且渐近紧, 从而根据定理$1.4$, 它有一个紧的全局吸引子$\mathcal{A}_{\alpha}\subset\mathcal{H}, \alpha\geqslant0$, 且以范数$|\cdot|$吸引$\mathcal{H}$中的任意有界集.