在本文中, 我们考虑边界充分光滑的有界域
$ \begin{equation} \left\{\begin{array}{ll} \displaystyle \varepsilon(t)u_{tt}+k(0)A^{\theta}u+ \int_{0}^{\infty }{{}}k^{\prime}(s)A^{\theta} u(t-s)\text{d}s+g(u)=f, & (x, t) \in\Omega\times\mathbb{R}, \\[8pt] u(x, t)=0, & x\in\partial\Omega, ~t\in\mathbb{R}, \\[8pt] u(x, t)=u^{\tau}(x, t), ~u_{t}(x, t)=u^{\tau}_{t}(x, t), & x\in\Omega, ~t\leqslant\tau, \end{array} \right. \end{equation} $ | (0.1) |
其中
(1)
$ \begin{align} \lim\limits_{t\to +\infty}\varepsilon(t)=0; \end{align} $ | (0.2) |
特别地, 存在常数
$ \begin{align} \sup\limits_{t\in \mathbb{R}}(|\varepsilon(t)|+\left|\varepsilon'(t)\right|) \leqslant L. \end{align} $ | (0.3) |
(2) 函数
$ \begin{align} \left|g''(y)\right| &\leqslant C(1+|y|), \quad \forall y \in \mathbb{R}; \end{align} $ | (0.4) |
$ \begin{align} &\liminf\limits_{|s| \to +\infty} \frac{g(s)}{s}>-\lambda_{1}^{\theta}, \end{align} $ | (0.5) |
其中
$ \begin{align*} G(u)=\int_{0}^{u}{{}}g(y)\text{d}y. \end{align*} $ |
此外, 我们假设
$ \begin{eqnarray} 2\langle g(u), u\rangle \geqslant 2\langle G(u), 1\rangle-(1-\nu)\|u\|^{2}_{\theta}-C. \end{eqnarray} $ | (0.6) |
关于时间依赖吸引子, 前人在不同的方程模型上已获得一些成果.在文献[1-2]中, Plinio和Conti修正了拉回吸引子的经典定义, 建立了验证拉回吸引性的新方法.在文献[3]中, Conti等人研究了波方程时间依赖吸引子的渐近结构.在文献[4-5]中, 刘婷婷、马巧珍等人分别关于Plate方程和非经典反应扩散方程得到了时间依赖全局吸引子的存在性和正则性结果.
方程(0.1)来源于文献[6-8]中建立的等温黏弹性理论, 描述了一种各向同性的黏弹性物质的能量耗散过程.方程(0.1)不含时间依赖项时, 张玉宝等人在文献[9]中证明了弱耗散条件下强全局吸引子的存在性.据我们所知, 方程(0.1)时间依赖吸引子的渐近性态尚未有人研究.在研究过程中, 发现存在一些难以克服的本质性困难:首先, 由于系统中正的递减函数
本文结构如下:第1节, 介绍所研究问题的一些预备知识, 包括空间定义, 一些符号和一般的抽象结果; 第2节, 用先验估计和算子分解的方法证明了方程(0.1)时间依赖全局吸引子的存在性和正则性.
为了方便估计, 本文中出现的
设
$ \begin{equation*} \left\{\!\!\begin{array}{ll} A \omega_{j}=\lambda_{j} \omega_{j}, \\[5pt] 0 < \lambda_{1}\leqslant\lambda_{2}\leqslant\cdots\leqslant \lambda_{j}, \lambda_{j} \rightarrow \infty (j\rightarrow \infty). \end{array} \right. \end{equation*} $ |
利用这组基定义与
$ \begin{align*} \langle u, v\rangle _{\theta}=\sum \limits^{\infty}_{j=1}\lambda^{\theta}_{j}(u, \omega_{j})(v, \omega_{j}), \quad \|u\|^{2}_{\theta} =\langle u, u\rangle_{\theta}, \quad \forall u, v \in V_{\theta}, \end{align*} $ |
则其构成Hilbert空间, 且易知算子
$ \begin{align} V_{\theta}\hookrightarrow &L^{\frac{2n}{n-2\theta}}, \end{align} $ | (1.1) |
并且有嵌入不等式
$ \begin{eqnarray} \lambda_{1}^{\theta}\int_{\Omega }{{}}|v|^{2}\text{d}x\leqslant\int_{\Omega }{{}}|A^{\frac{\theta}{2}}v|^{2}\text{d}x, \quad \forall v\in V_{\theta}. \end{eqnarray} $ | (1.2) |
为便于估计, 空间
$ \begin{align*} \langle u, v\rangle &=\int_{\Omega }{{}} u(x)v(x)\text{d}x, \quad \|u\|^{2}=\int_{\Omega }{{}} |u(x)|^{2}\text{d}x, \quad \forall u, v \in H; \end{align*} $ |
$ \begin{align*} \langle u, v\rangle _{\theta}&=\int_{\Omega }{{}} A^{\frac{\theta}{2}}u(x)A^{\frac{\theta}{2}}v(x)\text{d}x, \quad \|u\|^{2}_{\theta}=\int_{\Omega }{{}} |A^{\frac{\theta}{2}}u(x)|^{2}\text{d}x, \quad \forall u, v \in V_{\theta}. \end{align*} $ |
$ \begin{align*} \langle \varphi, \psi\rangle_{\mu, \theta}=\int_{0}^{\infty }{{}}\mu(s)\int_{\Omega }{{}}A^{\frac{\theta}{2}}\varphi A^{\frac{\theta}{2}} \psi \text{d}x\text{d}s, \quad \|\varphi\|^{2}_{\mu, \theta}=\int_{0}^{\infty }{{}}\mu(s)\int_{\Omega }{{}}|A^{\frac{\theta}{2}}\varphi|^{2} \text{d}x\text{d}s. \end{align*} $ |
定义时间依赖空间族
$ \begin{align*} \mathcal{H}_{t}^{\sigma}=V_{\theta+\sigma} \times V_{\sigma} \times L_{\mu}^{2}(\mathbb{R}^{+};V_{\theta+\sigma}), \end{align*} $ |
并且赋予相应的范数:
$ \begin{align*} \|z\|^{2}_{\mathcal{H}_{t}^{\sigma}}=\left\|(u, u_{t}, \eta^{t})\right\|^{2}_{\mathcal{H}_{t}^{\sigma}} =\|u\|^{2}_{\theta+\sigma}+\varepsilon(t)\|u_{t}\|^{2} _{\sigma} +\|\eta^{t}\|^{2}_{\mu, \theta+\sigma}. \end{align*} $ |
当
$ \begin{align*} \|z\|^{2}_{\mathcal{H}_{t}}=\left\|(u, u_{t}, \eta^{t})\right\|^{2}_{\mathcal{H}_{t}}=\|u\|^{2}_{\theta} +\varepsilon(t)\|u_{t}\|^{2}+\|\eta^{t}\|^{2}_{\mu, \theta}. \end{align*} $ |
定义变量
$ \begin{align*} \eta^{t}(s)=\eta^{t}(x, s)=u(x, t)-u(x, t-s). \end{align*} $ |
设
$ \begin{equation} \left\{\begin{array}{ll} \displaystyle \varepsilon(t)u_{tt}+A^{\theta}u+\int_{0}^{\infty }{{}}\mu(s)A^{\theta}\eta^{t}(s)\text{d}s+g(u)=f, \\[8pt] \eta^{t}_{t}=-\eta^{t}_{s}+u_{t}, \end{array} \right. \end{equation} $ | (1.3) |
相应的初-边值条件为:
$ \begin{equation} \left\{\begin{array}{ll} u(x, t)=0, \eta^{t}(x, s)=0, & x\in\partial\Omega, \\[8pt] u(x, \tau)= u_{\tau}(x), u_{t}(x, \tau)=u_{t\tau}(x), & x\in\Omega, \\[8pt] \eta^{\tau}(x, s)=\eta_{\tau}(x, s)=u(x, \tau)-u(x, \tau-s), & (x, s)\in\Omega\times\mathbb{R^{+}}. \end{array} \right. \end{equation} $ | (1.4) |
对记忆核函数
$ \begin{align} \mu(s)\in{{\rm{C}}^{\rm{1}}}(\mathbb{R^{+}})\cap L^{1}(\mathbb{R^{+}}), &\quad \mu(s)\geqslant0, \mu^{\prime}(s)\leqslant0, \quad \forall s\in\mathbb{R^{+}}; \end{align} $ | (1.5) |
$ \int_{0}^{\infty }{{}}\mu (s)\text{d}s={{k}_{0}}; $ | (1.6) |
$ \begin{align} \mu^{\prime}(s)+\delta&\mu(s)\leqslant0, \quad \forall s\in \mathbb{R^{+}}, \end{align} $ | (1.7) |
其中
引理1.1[10] 设记忆核函数
引理1.2(Gronwall-型引理)[11] 设
首先, 关于方程(1.3)
定义2.1 记
$ \begin{align*} \left\{\!\!\begin{array}{ll} \displaystyle \left\langle\varepsilon(t)u_{tt}, v\right\rangle+\langle u, v\rangle_{\theta}+\left\langle\eta^{t}(s), v\right\rangle_{\mu, \theta}+\langle g(u), v\rangle=\langle f, v\rangle, \\[8pt] \left\langle\eta^{t}_{t}(s)+\eta^{t}_{s}(s), \varphi(s)\right\rangle_{\mu, \theta}=\left\langle u_{t}, \varphi(s)\right\rangle_{\mu, \theta}, \end{array} \right. \end{align*} $ |
对于任意的
应用文献[12-13]中的方法, 可得到方程(1.3)
定理2.1 假设条件(0.2)
根据定理2.1, 可以定义下面的过程族
引理2.1 假设条件(0.2)
$ \begin{align} \|U(t, \tau)z_{1}(\tau)- U(t, \tau)z_{2}(\tau)\|_{\mathcal{H}_{t}}\leqslant \text{e}^{C(t-\tau)}\|z_{1}(\tau)-z_{2}(\tau)\|_{\mathcal{H}_{\tau}}, \quad \forall t\geqslant\tau. \end{align} $ | (2.8) |
为了证明引理2.1, 我们先证明下面的结论.
2.2 时间依赖吸收集引理2.2 在引理2.1的假设条件下, 设
$ \begin{align} \|U(t, \tau)z(\tau)\|_{\mathcal{H}_{t}}\leqslant R_{0}, \quad \forall t\geqslant \tau. \end{align} $ | (2.9) |
证明 设
$ \begin{align} \left\langle \varepsilon(t)u_{tt}, 2u_{t}+2\rho u\right\rangle+&\left\langle A^{\theta} u, 2u_{t}+2\rho u\right\rangle+ \left\langle\int_{0}^{\infty }{{}}\mu(s)A^{\theta}\eta^{t}(s)\text{d}s, 2u_{t}+2\rho u\right \rangle\notag\\ +\left\langle g(u), 2u_{t}+2\rho u\right\rangle&=\left\langle f, 2u_{t}+2\rho u\right\rangle. \end{align} $ | (2.10) |
对于记忆项, 由方程(1.3)的第二个式子, 结合引理1.1, 并利用Hölder不等式和Young不等式, 有
$ \begin{align} \left\langle\int_{0}^{\infty }{{}}\mu(s)A^{\theta}\eta^{t}(s)\text{d}s, 2u_{t}\right\rangle= & \int_{\Omega }{{}}\int_{0}^{\infty }{{}}2(\eta^{t}_{t}+\eta^{t}_{s})\mu(s)A^{\theta}\eta^{t}(s)\text{d}s\text{d}x \\[1mm] \geqslant&\frac{\rm{d}}{\text{d}t}\|\eta^{t}\|_{\mu, \theta}^{2}+\delta \|\eta^{t}\|_{\mu, \theta}^{2}, \end{align} $ | (2.11) |
$ \begin{align} \left\langle\int_{0}^{\infty }{{}}\mu(s)A^{\theta}\eta^{t}(s)\text{d}s, 2\rho u\right\rangle \geqslant&-2\rho\int_{\Omega }{{}}|A^{\frac{\theta}{2}}u|\int_{0}^{\infty }{{}} \mu(s)|A^{\frac{\theta}{2}}\eta^{t}(s)|\text{d}s\text{d}x\notag \\ \geqslant&-\frac{\rho\nu}{2}\|u\|_{\theta}^{2}-\frac{2k_{0}\rho}{\nu}\|\eta^{t}\|_{\mu, \theta}^{2}. \end{align} $ | (2.12) |
对合适的常数
$ \begin{align} \mathcal{E}(t)=& \|u\|^{2}_{\theta}+\varepsilon(t)\|u_{t}\|^{2}+\|\eta^{t}\|^{2}_{\mu, \theta}+2\rho\varepsilon(t)\left\langle u, u_{t}\right\rangle+2\langle G(u), 1\rangle-2\langle f, u\rangle+\breve{C}\notag \\ =& E(t)+2\rho\varepsilon(t)\left\langle u, u_{t}\right\rangle+2\langle G(u), 1\rangle-2\langle f, u\rangle+\breve{C}, \end{align} $ | (2.13) |
则对足够小的
$ \begin{eqnarray} \nu_{0} E(t)-C\leqslant \mathcal{E}(t)\leqslant C(E(t)). \end{eqnarray} $ | (2.14) |
事实上, 由Hölder不等式和Young不等式, 并结合条件(0.3)和(1.2)式, 有
$ \begin{align*} 2\rho\varepsilon(t)|\left\langle u, u_{t}\right\rangle|&\leqslant \rho\varepsilon(t)\|u\|_{\theta}^{2}+\frac{\rho\varepsilon(t)}{\lambda_{1}^{\theta}}\|u_{t}\|^{2}. \end{align*} $ |
因为
$ \begin{align} \langle G(u), 1\rangle\leqslant \int_{\Omega }{{}}\int_{0}^{u}{{}} C(1+|s|^{3}){\rm d}s{\rm d}x\leqslant C\|u\|_{L^{4}}^{4}+C\leqslant C\|u\|_{\theta}^{4}+C. \end{align} $ | (2.15) |
由条件(0.5), 存在足够小的常数
$ \begin{align*} 2\langle G(u), 1\rangle\geqslant-(1-\nu)\|u\|_{\theta}^{2}-C. \end{align*} $ |
对于(2.10)式, 由(2.11)
$ \begin{align} \dfrac{\rm{d}}{\text{d}t}&\mathcal{E}(t)+\rho\mathcal{E}(t)+ \dfrac{\rho\nu}{2}\|u\|_{\theta}^{2}-(\varepsilon'(t)+3\rho\varepsilon(t)) \|u_{t}\|^{2}+\Big(\delta-\dfrac{2k_{0}\rho}{\nu}-\rho\Big)\|\eta^{t} \|_{\mu, \theta}^{2}\notag \\ &-2\rho(\varepsilon'(t)+\rho\varepsilon(t))\langle u, u_{t}\rangle\leqslant \rho C. \end{align} $ | (2.16) |
由Hölder不等式和Young不等式, 并结合条件(0.3)和(1.2)式, 有
$ \begin{align} -2\rho(\varepsilon'(t)+\rho\varepsilon(t))\left\langle u, u_{t}\right\rangle \geqslant-\dfrac{\rho\nu}{2}\|u\|_{\theta}^{2}-\dfrac{2\rho}{\nu\lambda_{1}^{\theta}}L^{2}\|u_{t}\|^{2}. \end{align} $ | (2.17) |
将(2.17)式带入到(2.16)式, 有
$ \begin{align} \dfrac{\rm{d}}{\text{d}t}&\mathcal{E}(t)+\rho\mathcal{E}(t)- \Big(\varepsilon'(t)+3\rho\varepsilon(t)+\dfrac{2\rho}{\nu\lambda_{1}^{\theta}}L^{2}\Big) \|u_{t}\|^{2}+ \Big(\delta-\dfrac{2k_{0}\rho}{\nu}-\rho\Big)\|\eta^{t}\|_{\mu, \theta}^{2}\leqslant \rho C. \end{align} $ | (2.18) |
并取
$ \begin{align*} \dfrac{\rm{d}}{\text{d}t}\mathcal{E}(t)+\rho \mathcal{E}(t) \leqslant \rho C. \end{align*} $ |
应用Gronwall引理并结合(2.14)式, 证得(2.9)式成立.证毕.
引理2.1的证明 对于给定的初值
$ \begin{eqnarray} \|U(t, \tau)z_{i}(\tau)\|_{\mathcal{H}_{t}}\leqslant R_{0}. \end{eqnarray} $ | (2.19) |
定义
$ \begin{eqnarray} \varepsilon(t)\overline{u}_{tt}+A^{\theta}\overline{u}+\int_{0}^{\infty }{{}}\mu(s)A^{\theta}\overline{\eta}^{t}(s)\text{d}s +g(u_{1})-g(u_{2})=0. \end{eqnarray} $ | (2.20) |
用
$ \begin{eqnarray} \dfrac{\rm{d}}{\text{d}t}\|\overline{z}\|_{\mathcal{H}_{t}}^{2}-\varepsilon'(t)\|\overline{u}_{t}\|^{2}+\delta \|\overline{\eta}^{t}\|_{\mu, \theta}^{2}\leqslant-2\left\langle g(u_{1})-g(u_{2}), \overline{u}_{t}\right\rangle. \end{eqnarray} $ | (2.21) |
因为
$ \begin{align} -2\left\langle g(u_{1})-g(u_{2}), \overline{u}_{t}\right\rangle \leqslant&-2\int_{\Omega }{{}} |g'(\xi)||\overline{u}||\overline{u}_{t}|\text{d}x\notag \\ \leqslant&2C\Big(\int_{\Omega }{{}}(1+|u_{2}|^{2}+ |u_{1}|^{2})^{\frac{n}{\theta}}\text{d}x\Big)^{\frac{\theta}{n}}\|\overline{u}\|_{L^{\frac{2n} {n-2\theta}}} \|\overline{u}_{t}\|\notag \\ \leqslant&2C(1+\|u_{2}\|^{2}_{\theta}+\|u_{1}\|^{2}_{\theta})\|\overline{u}\|_{\theta}\|\overline{u}_{t}\|\notag \\ \leqslant&C(\|\overline{u}\|_{\theta}^{2}+\|\overline{u}_{t}\|^{2}). \end{align} $ | (2.22) |
因此将(2.22)式代入到(2.21)式, 有
$ \begin{align*} \dfrac{\rm{d}}{\text{d}t}\|\overline{z}(t)\|_{\mathcal{H}_{t}}^{2}\leqslant C(\|\overline{u}\|_{\theta}^{2}+\|\overline{u_{t}}\|^{2})\leqslant C\|\overline{z}(t)\|_{\mathcal{H}_{t}}^{2}. \end{align*} $ |
在区间
记
引理2.3 假设条件(0.2)
$ \begin{align} \sup\limits_{z_{\tau}\in \mathbb{B}_{\tau}(R_{0})}\left\{\|U(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}}^{2}+\int_{\tau }^{\infty }{{}}\|u_{t}(y)\|^{2}\text{d}y\right\}\leqslant M_{0}, \quad \forall t\in \mathbb{R}. \end{align} $ | (2.23) |
证明 由引理2.2直接可以得到时间依赖吸收集的存在性.为了证明(2.23)式, 只需令(2.18)式中的
定理2.2 关于方程(1.3)
为了证明过程的渐近紧性, 需要给出紧集的一个拉回吸引集族.为此, 可以将过程分解为衰减部分和紧性部分的和.
在条件(0.4)
$ \begin{align} g_{1}'(s)&\leqslant k, \quad \forall s\in \mathbb{R}, \end{align} $ | (2.24) |
$ \begin{align} |g_{0}''(s)|&\leqslant k(1+|s|), \quad \forall s\in \mathbb{R}, \end{align} $ | (2.25) |
$ \begin{align} g_{0}(0)&=g_{0}'(0)=0, \end{align} $ | (2.26) |
$ \begin{align} g_{0}(s)s&\geqslant0, \quad \forall s\in \mathbb{R}. \end{align} $ | (2.27) |
设
$ \begin{align*} U(t, \tau)z_{\tau}=z(t)=(u(t), u_{t}(t), \eta^{t}(s))=U_{0}(t, \tau)z_{\tau}+U_{1}(t, \tau)z_{\tau}, \end{align*} $ |
其中
$ \begin{align*} U_{0}(t, \tau)z_{\tau}=(v(t), v_{t}(t), \zeta^{t}(s)), \quad U_{1}(t, \tau)z_{\tau}=(w(t), w_{t}(t), \xi^{t}(s)), \end{align*} $ |
分别满足
$ \begin{eqnarray} \left\{\!\!\begin{array}{ll} \displaystyle \varepsilon(t) v_{tt}+A^{\theta}v+\int_{0}^{\infty }{{}}\mu(s)A^{\theta}\zeta^{t}(s)\text{d}s+g_{0}(v)=0, \\[8pt] \zeta_{t}^{t}=-\zeta_{s}^{t}+v, \\[8pt] v(x, t)|_{\partial\Omega}=0, \quad \zeta^{t}(x, t)|_{\partial\Omega}=0, \\[8pt] v(x, \tau)=u^{\tau}(x), \quad v_{t}(x, \tau)=u^{\tau}_{t}(x), \quad \zeta^{\tau}(x, s)=\eta^{\tau}(x, s), \end{array} \right. \end{eqnarray} $ | (2.28) |
和
$ \begin{align} \left\{\!\!\begin{array}{ll} \displaystyle \varepsilon(t) w_{tt}+A^{\theta}w+\int_{0}^{\infty }{{}}\mu(s)A^{\theta}\xi^{t}(s)\text{d}s+g(u)-g_{0}(v)=f, \\[3mm] \xi_{t}^{t}=-\xi_{s}^{t}+w, \\[3mm] w(x, t)|_{\partial\Omega}=0, \quad \xi^{t}(x, t)|_{\partial\Omega}=0, \\[3mm] w(x, \tau)=0, \quad w_{t}(x, \tau)=0, \quad \xi^{\tau}(x, s)=0. \end{array} \right. \end{align} $ | (2.29) |
引理2.4 存在常数
$ \|{{U}_{0}}(t,\tau ){{z}_{\tau }}{{\|}_{{{\mathcal{H}}_{t}}}}\le C{{\text{e}}^{-\alpha (t-\tau )}},\quad \forall t\ge \tau . $ | (2.30) |
证明 用
$ \begin{eqnarray} \|U_{0}(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}}\leqslant C. \end{eqnarray} $ | (2.31) |
设
$ \begin{align} \dfrac{\rm{d}}{\text{d}t}&(\|v\|^{2}_{\theta}+\varepsilon(t)\|v_{t}\|^{2}+\|\zeta^{t}\|^{2}_{\mu, \theta} \\ +2\rho\varepsilon(t)\left\langle v, v_{t}\right\rangle+2\langle G_{0}(v), 1\rangle)+ \\\dfrac{3\rho}{2}\|v\|_{\theta}^{2}-(\varepsilon'(t)+2\rho\varepsilon(t))\|v_{t}\|^{2}\notag \\ &+(\delta-2\rho k_{0})\|\zeta^{t}\|_{\mu, \theta}^{2} \\-2\rho\varepsilon'(t)\langle v, v_{t}\rangle+2\rho\left\langle g_{0}(v), v\right\rangle\leqslant0. \end{align} $ | (2.32) |
由Hölder不等式和Young不等式, 并结合条件(0.3)和(1.2)式, 得到
$ \begin{eqnarray} -2\rho\varepsilon'(t)\left\langle v, v_{t}\right\rangle\geqslant -\dfrac{\rho}{2}\|v\|_{\theta}^{2}-\dfrac{2\rho L^{2}}{\lambda_{1}^{\theta}}\|v_{t}\|^{2}. \end{eqnarray} $ | (2.33) |
定义泛函
$ \begin{align} \mathcal{E}_{0}(t)=&\|v\|^{2}_{\theta}+\varepsilon(t)\|v_{t}\|^{2}+\|\zeta^{t}\|^{2}_{\mu, \theta}+2\rho\varepsilon(t)\left\langle v, v_{t}\right\rangle+2\langle G_{0}(v), 1\rangle\notag \\ = &\|U_{0}(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}}^{2}+2\rho\varepsilon(t)\left\langle v, v_{t}\right\rangle+2\langle G_{0}(v), 1\rangle, \end{align} $ | (2.34) |
其中
$ \begin{align} \dfrac{1}{2}\|U_{0}(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}}^{2}\leqslant\mathcal{E}_{0}(t)\leqslant C\|U_{0}(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}}^{2}. \end{align} $ | (2.35) |
事实上, 由条件(2.27), 有
$ \begin{align*} \langle G_{0}(v), 1\rangle\leqslant C\|v\|_{\theta}^{4}+C\leqslant C. \end{align*} $ |
由Hölder不等式和Young不等式, 并结合条件(0.3)和(1.2)式, 得到
$ \begin{align*} 2\rho\varepsilon(t)|\left\langle v, v_{t}\right\rangle|\leqslant \dfrac{1}{2}\|v\|_{\theta}^{2}+\dfrac{2\rho^{2} L}{\lambda_{1}^{\theta}}\varepsilon(t)\|v_{t}\|^{2}. \end{align*} $ |
将(2.33)
$ \begin{align*} \dfrac{\rm{d}}{\text{d}t}\mathcal{E}_{0}(t)+ \rho\|U_{0}(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}}^{2}-\Big(\varepsilon'(t)+3\rho\varepsilon(t)+\dfrac{2\rho L^{2}}{\lambda_{1}^{\theta}}\Big)\|v_{t}\|^{2}+ (\delta-\rho-2\rho k_{0})\|\zeta^{t}\|_{\mu, \theta}^{2}\leqslant C. \end{align*} $ |
取
$ \begin{align} \varepsilon'(t)+3\rho\varepsilon(t)+\dfrac{2\rho L^{2}}{\lambda_{1}^{\theta}}\leqslant0, \quad \delta-\rho-2\rho k_{0}\geqslant0, \end{align} $ | (2.36) |
则有
$ \begin{align*} \dfrac{\rm{d}}{\text{d}t}\mathcal{E}_{0}(t)+\rho\|U_{0}(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}}^{2}\leqslant C. \end{align*} $ |
利用Gronwall引理, 结合(2.35)式, 即得(2.30)式成立.证毕.
由以上证明可知, 下面的估计式成立.
$ \begin{eqnarray} \sup\limits_{t\geqslant\tau}\left\{\|U(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}}+\|U_{0}(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}}+ \|U_{1}(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}}\right\}\leqslant C. \end{eqnarray} $ | (2.37) |
引理2.5 存在
$ \begin{eqnarray} \sup\limits_{t\geqslant\tau}\|U_{1}(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}^{1/3}}\leqslant M. \end{eqnarray} $ | (2.38) |
证明 设
$ \begin{gathered} \frac{{\text{d}}}{{{\text{d}}t}}\left( {\left\| w \right\|_{\theta + 1/3}^2 + \varepsilon \left( t \right)\left\| {{w_t}} \right\|_{1/3}^2 + \left\| {{\xi ^t}} \right\|_{\mu ,\theta + 1/3}^2 + 2\rho \varepsilon \left( t \right)} \right. \hfill \\ \left. {\left\langle {{w_t},{A^{1/3}}w} \right\rangle + 2\left\langle {g\left( u \right) - {g_0}\left( v \right) - f,{A^{1/3}}w} \right\rangle } \right) + \hfill \\ \frac{{3\rho }}{2}\left\| w \right\|_{\theta + 1/3}^2 - (\varepsilon '\left( t \right) + 2\rho \varepsilon \left( t \right))\left\| {{w_t}} \right\|_{1/3}^2 + \hfill \\ (\delta - 2\rho {k_0})\left\| {{\xi ^t}} \right\|_{\mu ,\theta + \frac{1}{3}}^2 - 2\rho \varepsilon '\left( t \right)\left\langle {{w_t},{A^{1/3}}w} \right\rangle \hfill \\ + 2\rho \left\langle {g\left( u \right) - {g_0}\left( v \right) - f,{A^{1/3}}w} \right\rangle {I_1} + {I_2} + {I_3}, \hfill \\ \end{gathered} $ | (2.39) |
其中
$ \begin{align*} I_{1}=&2\left\langle (g'_{0}(u)-g'_{0}(v))u_{t}, A^{1/3}w \right\rangle, \\ I_{2}=&2\left\langle g'_{0}(v)w_{t}, A^{1/3}w \right\rangle, \\ I_{3}=&2\left\langle g'_{1}(u)u_{t}, A^{1/3}w \right\rangle. \end{align*} $ |
对足够小的
$ \begin{align} \Lambda(t)=&\|w\|^{2}_{\theta+1/3}+\varepsilon(t)\|w_{t}\|^{2}_{1/3} +\|\xi^{t}\|^{2}_{\mu, \theta+1/3} +2\rho\varepsilon(t)\left\langle w_{t}, A^{1/3}w \right\rangle \\ &+2\left\langle g(u)-g_{0}(v)-f, A^{1/3}w \right\rangle+ C^{*}\notag \\ =&\|U_{1}(t, \tau)z_{\tau}\|^{2}_{\mathcal{H}^{1/3}_{t}} +2\rho\varepsilon(t)\left\langle w_{t}, A^{1/3}w \right\rangle+2\left\langle g(u)-g_{0}(v)-f, A^{1/3}w \right\rangle+ C^{*}, \end{align} $ | (2.40) |
我们有
$ \begin{eqnarray} \dfrac{1}{2}\|U_{1}(t , \tau)z_{\tau}\|^{2}_{\mathcal{H}^{1/3}_{t}}\leqslant\Lambda(t)\leqslant 2\|U_{1}(t, \tau)z_{\tau}\|^{2}_{\mathcal{H}^{1/3}_{t}}+C. \end{eqnarray} $ | (2.41) |
事实上, 由Hölder不等式和Young不等式, 并结合条件(0.3)和(1.2)式, 有
$ \begin{align*} 2\rho\varepsilon(t)\left|\left\langle w_{t}, A^{1/3}w \right\rangle\right| \leqslant \rho\|w\|_{\theta+1/3}^{2}+\dfrac{\rho L}{\lambda_{1}^{\theta}}\varepsilon(t)\|w_{t}\|_{1/3}^{2}. \end{align*} $ |
因为
$ \begin{align*} 2\left|\left\langle g(u)-g_{0}(v), A^{1/3}w\right\rangle\right|\leqslant & 2\left|\left\langle g(u)-g(v), A^{1/3}w \right\rangle\right|+2\left|\left\langle g_{1}(v), A^{1/3}w\right\rangle\right|\notag \\[1mm] \leqslant&2\int_{\Omega }{{}} |g'(\xi)||w||A^{1/3}w|\text{d}x+2\int_{\Omega }{{}} |g'_{1}(\xi)||v||A^{1/3}w|\text{d}x \notag \\[1mm] \leqslant&C(\int_{\Omega }{{}}(1+|u|^{2}+|v|^{2})^{\frac{n}{\theta}} \text{d}x)^{\frac{\theta}{n}} \|w\|_{L^{\frac{2n}{n-2\theta}}}\|A^{1/3}w\|+ 2k\|v\|\|A^{1/3}w\| \notag \\ \leqslant& C\|w\|_{\theta}\|w\|_{2/3} +C\|v\|_{\theta}\|w\|_{2/3}\notag \\ \leqslant&\dfrac{1}{4}\|w\|^{2}_{\theta+1/3}+C. \end{align*} $ |
将(2.40)式代入到(2.39)式, 有
$ \begin{align} \dfrac{\rm{d}}{\text{d}t}&\Lambda(t)+ \rho\Lambda(t)+\dfrac{\rho}{2}\|w\|^{2}_{\theta+1/3}-(\varepsilon'(t)+3\rho\varepsilon(t))\|w_{t}\|^{2}_{1/3}+(\delta-\rho-2\rho k_{0})\|\xi^{t}\|_{\mu, \theta+1/3}^{2}\notag\\ &-2\rho(\varepsilon'(t)+\rho\varepsilon(t))\left\langle w_{t}, A^{1/3}w\right\rangle\leqslant I_{1}+I_{2}+I_{3}+\rho C^{*}. \end{align} $ | (2.42) |
由Hölder不等式和Young不等式, 并结合条件(0.3)和(1.2)式, 得到
$ \begin{eqnarray} -2\rho(\varepsilon'(t)+\rho\varepsilon(t))\left\langle w_{t}, A^{1/3}w\right\rangle \geqslant -\dfrac{\rho}{2}\|w\|_{\theta+1/3}^{2}-\dfrac{2\rho L^{2}}{\lambda_{1}^{\theta}}\|w_{t}\|^{2}. \end{eqnarray} $ | (2.43) |
将(2.43)式代入到(2.42)式, 并取
$ \begin{eqnarray} \dfrac{\rm{d}}{\text{d}t}\Lambda(t)+\rho\Lambda(t)\leqslant I_{1}+I_{2}+I_{3}+\rho C^{*}. \end{eqnarray} $ | (2.44) |
因为
$ \begin{align*} I_{1}\leqslant&C (1+\|u\|_{L^{\frac{2n}{4\theta-n}}}+\|v\| _{L^{\frac{2n}{4\theta-n}}})\|u_{t}\|\|w\|_{L^{\frac{2n}{n-2 (\theta+1/3)}}}\|A^{1/3}w\|_{L^{\frac{2n}{n-2(\theta-1/3)}}}\\ \leqslant&C(1+\|u\|_{\theta}+\|v\|_{\theta})\|u_{t}\|\|w\|_{\theta+1/3}\|A^{1/3}w\|_{\theta-1/3}\\ \leqslant&\dfrac{\rho}{2}\|w\|^{2}_{\theta+1/3} +C \|u_{t}\|^{2} \|w\|^{2}_{\theta+1/3}\\ \leqslant&\dfrac{\rho}{2}\Lambda(t) +C \|u_{t}\|^{2} \|w\|^{2}_{\theta+1/3}, \\ I_{2}\leqslant & C\Big(\int_{\Omega }{{}}(|v|+|v|^{2})^{\frac{n}{\theta}} \text{d}x\Big)^{\frac{\theta}{n}}\|w_{t}\|_{1/3}\|A^{1/3}w\|_{L^{\frac{2n}{n-2(\theta-1/3)}}}\\ \leqslant& C (1+\|v\|_{\theta}+\|v\|_{\theta}^{2})\|w_{t}\|_{1/3}\|w\|_{\theta+1/3}\\ \leqslant&C\|w_{t}\|_{1/3}^{2} +C\|v\|_{\theta}^{2}\|w\|^{2}_{\theta+1/3}. \end{align*} $ |
此外, 由条件(2.24), 有
$ \begin{align*} I_{3}\leqslant k \|u_{t}\| \|A^{1/3}w\| \leqslant \lambda_{1}^{2(\theta-1/3)}\|u_{t}\|^{2} \|w\|^{2}_{2/3}+C \leqslant \|u_{t}\|^{2} \|w\|^{2}_{\theta+1/3}+C. \end{align*} $ |
因此, 由不等式(2.44), 可得
$ \begin{align*} \dfrac{\rm{d}}{\text{d}t}\Lambda(t) +\dfrac{\rho}{2}\Lambda(t) \leqslant q (t)\Lambda(t) + C, \end{align*} $ |
其中
$ \begin{align*} \int_{\tau }^{\infty }{{}}q(y)\text{d}y\leqslant C. \end{align*} $ |
并且由引理1.2, 引理2.3和引理2.4, 得到
$ \Lambda (t)\le C\Lambda (\tau ){{\text{e}}^{-\frac{\rho }{4}(t-\tau )}}+C,\text{ } $ | (2.45) |
再结合(2.41)式, 证明了
为了构造紧集的一个拉回吸引集, 我们还需要记忆项的紧性.
引理2.6[12-14] 假定
(ⅰ)
(ⅱ)
那么
另外, 对任意的
$ \begin{equation} \left\{\!\!\begin{array}{ll} \xi_{t}^{t}=-\xi_{s}^{t}+w_{t}, \quad t\geqslant\tau, \\ \xi^{\tau}=\xi_{\tau} \end{array} \right. \end{equation} $ | (2.46) |
有唯一解
$ \begin{equation} \xi^{t}(x, s)=\left\{\!\!\begin{array}{ll} w(x, t)-w(x, t-s), &\tau < s < t, \\ w(x, t)-w(\tau), &s\geqslant t.\end{array} \right. \end{equation} $ | (2.47) |
设
引理2.7[15] 假定非线性项
$ \begin{align*} \mathcal{K}_{T}=\Pi U_{1}(T, \tau)\mathfrak{B}, \end{align*} $ |
则存在一个正常数
(ⅰ)
(ⅱ)
其中
引理2.8 在引理2.6的条件下, 令
根据引理2.5和引理2.8, 可以考虑
$ \begin{align*} K_{t}=\left\{z(t)\in \mathcal{H}_{t}^{1/3}:\|z(t)\|_{\mathcal{H}_{t}^{1/3}}\leqslant M\right\}. \end{align*} $ |
由此可知
$ \begin{align*} \text{dist}(U(t, \tau)\mathbb{B}_{\tau}(R_{0}), K_{t})\leqslant C{{\text{e}}}^{-\alpha(t-\tau)}, \quad \forall t\geqslant\tau, \end{align*} $ |
其中
在
定理2.3
为了证明
$ \begin{align*} U(t, \tau)z_{\tau}=z(t)=(u(t), u_{t}(t), \eta^{t}(s))=U_{3}(t, \tau)z_{\tau}+U_{4}(t, \tau)z_{\tau}, \end{align*} $ |
其中
$ \begin{align*} U_{3}(t, \tau)z_{\tau}=(v(t), v_{t}(t), \zeta^{t}(s)), \quad U_{4}(t, \tau)z_{\tau}=(w(t), w_{t}(t), \xi^{t}(s)), \end{align*} $ |
分别满足
$ \begin{align} \left\{\!\!\begin{array}{ll} \varepsilon(t) v_{tt}+A^{\theta}v+\int_{0}^{\infty }{{}}\mu(s)A^{\theta}\zeta^{t}(s)\text{d}s=0, \\[1mm] \zeta_{t}^{t}=-\zeta_{s}^{t}+v, \\[1mm] v(x, t)|_{\partial\Omega}=0, \quad \zeta^{t}(x, t)|_{\partial\Omega}=0, \\[1mm] v(x, \tau)=u^{\tau}(x), \quad v_{t}(x, \tau)=u^{\tau}_{t}(x), \quad \zeta^{\tau}(x, s)=\eta^{\tau}(x, s), \end{array} \right. \end{align} $ | (2.48) |
和
$ \begin{align} \left\{\!\!\begin{array}{ll} \displaystyle \varepsilon(t) w_{tt}+A^{\theta}w+\int_{0}^{\infty }{{}}\mu(s)A^{\theta}\xi^{t}(s)\text{d}s+g(u)=f, \\[1mm] \xi_{t}^{t}=-\xi_{s}^{t}+w, \\[1mm] w(x, t)|_{\partial\Omega=0}, \quad \xi^{t}(x, t)|_{\partial\Omega}=0, \\[1mm] w(x, \tau)=0, \quad w_{t}(x, \tau)=0, \quad \xi^{\tau}(x, s)=0. \end{array} \right. \end{align} $ | (2.49) |
作为引理2.4的一个特例, 我们可以得到
$ \begin{eqnarray} \|U_{3}(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}}\leqslant C{{\text{e}}}^{-\alpha(t-\tau)}, \quad \forall t\geqslant\tau. \end{eqnarray} $ | (2.50) |
引理2.9 存在常数
$ \begin{eqnarray} \sup\limits_{t\geqslant\tau}\|U_{4}(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}^{1}}\leqslant M_{1}. \end{eqnarray} $ | (2.51) |
证明 设
$ \begin{align} \dfrac{\rm{d}}{\text{d}t}&(\|w\|_{\theta+1}^{2}+\varepsilon(t)\|w_{t}\|_{1}^{2} +\|\xi\|_{\mu, \theta+1}^{2} +2\rho\varepsilon(t)\left\langle w_{t} , Aw \right\rangle-2\langle f, Aw \rangle) \\ &+\dfrac{3}{2}\rho\|w\|^{2}_{\theta+1}-(\varepsilon'(t)+2\rho\varepsilon(t))\|w_{t}\|^{2}_{1}\notag +(\delta-2\rho k_{0})\|\xi^{t}\|^{2}_{\mu, \theta+1} \\ & -2\rho\varepsilon'(t)\left\langle w_{t}, Aw\right\rangle-2\rho\langle f, Aw \rangle\leqslant-2\left\langle g(u), A(w_{t}+\rho w)\right\rangle. \end{align} $ | (2.52) |
对足够小的
$ \begin{align} \mathcal{E}_{1}(t)=&\|w\|_{\theta+1}^{2}+\varepsilon(t)\|w_{t}\|_{1}^{2}+\|\xi\|_{\mu, \theta+1}^{2}+2\rho\varepsilon(t)\left\langle w_{t}, Aw \right\rangle-2\langle f, Aw \rangle+\widetilde{C}\notag \\ =&\|U_{4}(t, \tau)z_{\tau}\|^{2}_{\mathcal{H}^{1}_{t}}+2\rho\varepsilon(t)\left\langle w_{t}, Aw \right\rangle-2\langle f, Aw \rangle+\widetilde{C}, \end{align} $ | (2.53) |
我们有
$ \begin{eqnarray} \dfrac{1}{4}\|U_{4}(t, \tau)z_{\tau}\|^{2}_{\mathcal{H}^{1}_{t}} \leqslant \mathcal{E}_{1}(t)\leqslant 2\|U_{4}(t, \tau)z\|^{2}_{\mathcal{H}^{1}_{t}}+C. \end{eqnarray} $ | (2.54) |
将(2.53)式代入到(2.52)式, 有
$ \begin{align} \dfrac{\rm{d}}{\text{d}t}&\mathcal{E}_{1}(t)+ \rho\mathcal{E}_{1}(t)+\dfrac{1}{2}\rho\|w\|^{2}_{\theta+1}-(\varepsilon'(t)+3\rho\varepsilon(t))\|w_{t}\|^{2}_{1}+(\delta-\rho-2\rho k_{0})\|\xi\|_{\mu, \theta+1}^{2}\notag \\ &-2\rho(\varepsilon'(t)+\rho\varepsilon(t))\langle w_{t}, A w\rangle\leqslant-2\left\langle g(u), A(w_{t}+\rho w)\right\rangle+\rho \widetilde{C}, \end{align} $ | (2.55) |
由Hölder不等式和Young不等式, 并结合条件(0.3)和(1.2)式, 得到
$ \begin{eqnarray} -2\rho(\varepsilon'(t)+\rho\varepsilon(t))\left\langle w_{t}, A w\right\rangle \geqslant -\dfrac{\rho}{2}\|w\|_{\theta+1}^{2}-\dfrac{2\rho L^{2}}{\lambda_{1}^{\theta}}\|w_{t}\|_{1}^{2}. \end{eqnarray} $ | (2.56) |
将(2.56)式代入(2.55)式, 并取
$ \begin{align} \dfrac{\rm{d}}{\text{d}t}\mathcal{E}_{1}(t)+\rho \mathcal{E}_{1}(t)\leqslant-2\left\langle g(u), A(w_{t}+\rho w)\right\rangle+\rho \widetilde{C}. \end{align} $ | (2.57) |
由吸引子的不变性, 可得
$ \begin{align*} \|U(t, \tau)z_{\tau}\|_{\mathcal{H}_{t}^{1/3}}\leqslant C, \end{align*} $ |
其中
$ \begin{align} -2\langle g(u), Aw_{t}\rangle-2\rho \langle g(u), Aw\rangle \leqslant&2\int_{\Omega }{{}}|g'(u)||A^{1/2}u||A^{1/2}w_{t}|\text{d}x+2\int_{\Omega }{{}}|g'(u)||A^{1/2}u||A^{1/2}w|\text{d}x\notag \\[2mm] \leqslant&C\Big(\int_{\Omega }{{}}(1+|u|^{2})^{\frac{n}{n- 2(\theta+1/3)}}\text{d}x\Big)^{\frac{{n-2(\theta+1/3)}}{n}} \\ &\times\|A^{1/2}u \|_{L^{{\frac{2n}{4(\theta+1/3)-n}}}}(\|w_{t}\|_{1}+\|w\|_{1})\notag\\[2mm] \leqslant&C\Big(1+\|u\|_{L^{\frac{2n}{n-2(\theta+1/3)}}}^{2}\Big) \|A^{1/2}u\|_{\theta-2/3}(\|w_{t}\|_{1}+\|w\|_{1}) \notag\\ \leqslant&C(1+\|u\|^{2}_{\theta+1/3})\|u\|_{\theta+1/3}(\|w_{t}\|_{1}+\|w\|_{1})\notag\\ \leqslant&\dfrac{\rho}{2}\mathcal{E}_{1}(t)+C. \end{align} $ | (2.58) |
将(2.58)式代入(2.57)式, 有
$ \begin{align*} \dfrac{\rm{d}}{\text{d}t}\mathcal{E}_{1}(t)+\dfrac{\rho}{2}\mathcal{E}_{1}(t)\leqslant C. \end{align*} $ |
应用Gronwall引理, 结合(2.53)式, 可得
定理2.3的证明 令
$ \begin{align*} K^{1}_{t}=\left\{z(t)\in \mathcal{H}^{1}_{t}: \|z(t)\|_{\mathcal{H}^{1}_{t}} \leqslant M_{1} \right\}. \end{align*} $ |
由不等式(2.50)和引理2.9, 对
$ \begin{align*} \lim\limits_{\tau \rightarrow -\infty} \text{dist}(U(t, \tau)A_{\tau}, K^{1}_{t})=0. \end{align*} $ |
从而由
$ \begin{align*} \text{dist}(A_{t}, K^{1}_{t})=0. \end{align*} $ |
因此,
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