设$ \Omega\subset R^N $是具有光滑边界的有界区域, 本文主要研究$ \Omega $上具有变系数非局部高阶Kirchhoff方程的渐近行为:

$ \begin{eqnarray} \left\{ \begin{array}{c} u_{tt}+a(x)(-\Delta)^mu_t+b(x)M(\|\nabla ^mu\|^2)(-\Delta)^m u+g(x, u) = f(x), \\ u = 0, \frac{\partial^i u}{\partial v^i} = 0, i = 1, 2, \cdots, m-1, x\in\Gamma, t\ge0, \\ u(x, 0) = u_0(x), u_t(x, 0) = u_{1}(x), x\in\Omega. \end{array} \right. \end{eqnarray} $

(1.1)

其中$ \Gamma $为$ \Omega $的光滑边界, $ v $是边界$ \Gamma $上的外法向量, $ m>1 $, $ a(x), b(x) $为关于空间变量$ x $的变系数函数, $ f(x) $是外力项, $ g(x, u) $是满足一定增长条件和耗散条件的关于未知函数$ u $的非线性源项.

一般而言, 整体吸引子是一个具有不变性的非空紧集, 并且能够吸引任何有界集. 对于耗散动力系统, 整体吸引子是解释系统解的长时间动力学的核心概念, 对于自治动力系统, 通常用整体吸引子来刻画其渐近行为.

吸引子包含了系统几乎所有的长时间信息, 因此关于吸引子存在性及结构的研究对整个系统的长时间演变过程有着非常重要的意义. Kirchhoff方程作为一重要的数学物理方程, 研究其吸引子就显得相当重要, 近年来, 得到了许多有关Kirchhoff型方程的长时间动力学行为, I.Chueshov^[1]研究了带非线性强阻尼Kirchhoff方程$ u_{tt}-\sigma(\|\nabla u\|^2)\Delta u_t-\phi(\|\nabla u\|^2)\Delta u+f(u) = h(x) $的长时间动力行为. Guoguang Lin, Penghui Lv和Ruijin Lou^[2]研究了具阻尼项的广义非线性Kirchhoff-Boussinesq方程的整体动力学, 研究得到了该类方程的整体吸引子和指数吸引子.更多有关低阶波动方程或Kirchhoff方程的研究可见文献[3-5].

随着研究的深入, 学者开始研究高阶波动方程的长时间动力学行为. Salim A. Messaoudia, Belkacem Said Houari^[6]研究了具有Dirichlet边界条件的多维高阶Kirchhoff方程, 估计了在正初始能量下解爆破. 林国广, 李卓茜^[7]研究了一类非线性非局部且具强阻尼项的高阶Kirchhoff方程的初边值问题, 得到了该类高阶方程的整体解的存在唯一性, 并且得到了该类Kirchhoff方程的整体吸引子族, 且整体吸引子族具有有限的Hausdorff维数和Fractal维数. Fucai Li^[8]研究了有界域上具有非线性耗散的高阶Kirchhoff型方程

$ \begin{eqnarray*} u_{tt}+(\int_{\Omega}|D^mu|^2\mathrm{d}x)^q(-\Delta)^mu+u_t|u_t|^r = |u|^pu, x\in\Omega, t>0, \end{eqnarray*} $

其中$ m>1, q, p, r>0 $. 当$ p\le r $时, 得到方程的整体解, 同时当$ p>\max\{r, 2q\} $时, 对于所有具负初始能量的初始数据, 解在$ L^{p+2} $范数下将有限时间内爆破.详细的关于更多高阶Kirchhoff方程的相关研究可见文献[9-11].

目前带变系数的高阶Kirchhoff方程的渐近行为见文较少.本文在得到问题(1.1)的渐近行为时, 主要的问题是如何处理变系数, 在变系数情况下如何得到渐近紧性.本文运用合理的假设和莱布尼兹公式克服了变系数带来的困难, 进而得到有界吸收集和渐近紧性.

本文第2部分介绍有关概念和动力系统的相关定义及理论; 第3部分得到问题的先验估计和整体解; 第4部分得到问题(1.1)的整体吸引子族.

本节主要给出动力系统和整体吸引子(族)的相关理论.

首先引进本文需用到的相关记号:

定义$ H = L^2(\Omega) $上的内积和范数分别为$ (\cdot, \cdot) $和$ \|\cdot\| $, $ L^p = L^p(\Omega), \|\cdot\|_p = \|\cdot\|_{L^p} $, 其中$ p\ge1 $. 现令

$ \begin{eqnarray*} V_{m} = H^{m}_0(\Omega) = H^{m}(\Omega)\cap H^1_0(\Omega), V_{m+k} = H^{m+k}_0(\Omega) = H^{m+k}(\Omega)\cap H^1_0(\Omega), \quad k = 0, 1, \cdots, m. \end{eqnarray*} $

其相应的内积和范数分别为

$ \begin{eqnarray*} (u, v)_{V_{m+k}} = (\nabla^{m+k}u, \nabla^{m+k}v), \|u\|_{V_{m+k}} = \|\nabla^{m+k}u\|_H. \end{eqnarray*} $

同时, 有一般形式的Poincare不等式: $ \lambda_1\|\nabla^ru\|^2\le\|\nabla^{r+1}u\|^2 $, 其中$ \lambda_1 $是$ -\Delta $的第一特征值.文中$ C_i $为正常数, $ C(\cdot) $表示依赖于括号内参数的正常数, $ C^n_m $为对应的组合数.

定义 2.1^[7]  设$ X $是完备度量空间, $ S(t):X\to X, t\in R^+ $是连续算子, 若对任意$ s, t\in R^+, S(t) $满足$ S(0) = I, S(t+s) = S(t)S(t) $.则称$ \{S(t)\}_{t\ge0} $是$ X $上的算子半群, 称$ (S(t), X) $构成连续动力系统, $ X $称为该动力系统的相空间.

定义 2.2(整体吸引子)^[11]  设$ X $是Banach空间, $ \{S(t)\}_{t\ge0} $为连续的算子半群, 如果紧集$ A_0\subset X $满足:

(1)  不变性  在半群$ \{S(t)\}_{t\ge0} $作用下为不变集, 即$ S(t)A_0 = A_0(\forall t\ge0) $;

(2)  吸引性  $ A_0 $吸引$ X $中一切有界集, 即$ \forall B\subset X $为$ X $中的有界集, 有

$ \begin{eqnarray*} {\rm{dist}}(S(t)B, A_0) = \sup\limits_{x\in B}\inf\limits_{y\in A_0}\|S(t)x-y\|_X\to0(t\in\infty). \end{eqnarray*} $

特别地, 当$ t\to\infty $时, 从$ u_0 $出发的一切轨道$ S(t)u_0 $收敛于$ A_0 $内, 即有dist$ (S(t)u_0, A_0)\to0(t\to\infty) $, 则称紧集$ A_0 $为半群$ \{S(t)\}_{t\ge0} $的整体吸引子.

引理 2.3^[7]  设$ X $是Banach空间, 连续的算子半群$ \{S(t)\}_{t\ge0} $满足

(1) 半群$ \{S(t)\}_{t\ge0} $在$ X $中一致有界, 即$ \forall R>0 $存在正常数$ C(R) $使得$ \|u\|_X\le R $, 有

$ \|S(t)u\|_X\le C(R), (\forall t\in[0, +\infty)); $

(2) 存在$ X $中有界吸收集$ B_0 $, 则任意一个有界集$ B\subset X $, 存在一个时刻$ t_0 $, 使得

$ S(t)B\subset B_0(t\ge t_0); $

(3) 对$ t>0, S(t) $是全连续算子;

则半群$ \{S(t)\}_{t\ge0} $具有紧的整体吸引子$ A_0 $.

定义 2.4 (整体吸引子族)  设$ X_k, k = 1, 2, \cdots, m $, 是Banach空间, $ \{S(t)\}_{t\ge0} $为连续的算子半群, 如果紧集$ A_k\subset X_k $满足

(1)  不变性  在半群$ \{S(t)\}_{t\ge0} $作用下为不变集, 即$ S(t)A_k = A_k(\forall t\ge0) $;

(2)  吸引性  $ A_k $吸引$ X_k $中一切有界集, 即$ \forall B_k\subset X_k $为$ X_k $中的有界集, 有

$ \begin{eqnarray*} {\rm {dist}}(S(t)B_k, A_k) = \sup\limits_{x\in B_k}\inf\limits_{y\in A_k}\|S(t)x-y\|_{X_k}\to0(t\in\infty). \end{eqnarray*} $

特别地, 当$ t\to\infty $时, 从$ u_0 $出发的一切轨道$ S(t)u_0 $收敛于$ A_k $内, 即有dist$ (S(t)u_0, A_k)\to0(t\to\infty) $, 则称紧集$ A_k $为半群$ \{S(t)\}_{t\ge0} $的整体吸引子族.

引理 2.5  设$ X_k, k = 1, 2, \cdots, m $, 是Banach空间, 连续的算子半群$ \{S(t)\}_{t\ge0} $满足

(1) 半群$ \{S(t)\}_{t\ge0} $在$ X_k $中一致有界, 即$ \forall R>0 $存在正常数$ C_k(R_k) $使得$ \|u\|_{X_k}\le R_k $, 有

$ \begin{eqnarray*} \|S(t)u\|_{X_k}\le C_k(R_k), (\forall t\in[0, +\infty)); \end{eqnarray*} $

(2) 存在$ X_k $中有界吸收集$ B_{0k} $, 则任意一个有界集$ B\subset X_k $, 存在一个时刻$ t_{0k} $, 使得

$ \begin{eqnarray*} S(t)B\subset B_{0k}(t\ge t_{0k}); \end{eqnarray*} $

(3) 对$ t>0, S(t) $是全连续算子;

则半群$ \{S(t)\}_{t\ge0} $具有紧的整体吸引子族$ A_k $.

首先给出问题(1.1)的假设条件:

$ (H) $变系数$ a(x), b(x), a(x) = c(x)\cdot b(x) $, 满足$ a\in C_0^{\infty}(\Omega), a(x)\ge a_{00}>0, \frac{\partial^ia}{\partial^iv}|_{\Gamma} = 0, a_0 = \|a(x)\|_{\infty} $, $ b\in C_0^{\infty}(\Omega), b(x)\ge b_{00}>0, \frac{\partial^ib}{\partial^iv}|_{\Gamma} = 0, b_0 = \|b(x)\|_{\infty} $, $ c\in C_0^{\infty}(\Omega), c(x)\ge c_{00}>0, \frac{\partial^ic}{\partial^iv}|_{\Gamma} = 0, c_0 = \|c(x)\|_{\infty}, c_i = \|\nabla^ic(x)\|_{\infty} $, 且$ \mu(x) = b(x)^{-1}\ge \mu_{00}>0, \mu\in C_0^{\infty}(\Omega), \mu_i = \|\nabla^i\mu(x)\|_{\infty} $;

$ (M)M\in C^1(R^+) $, 且$ 0<M_0\le M(s)\le M_1, \forall s\in R^+, x\in\Omega, k = 0, 1, \cdots, m $, 非线性项$ g(x, u) $满足下列条件: 存在正常数$ \beta_1, \beta_2, \beta_3, \beta_4, \beta_5>0 $, 对$ \forall u\in R, x\in\Omega $, 满足

$ \begin{eqnarray} &&|g(x, u)|\le \beta_1|p|^p+\phi_1(x), \phi_1\in L^2(\Omega), \end{eqnarray} $

(3.1)

$ \begin{eqnarray} &&ug(x, u)-\beta_2G(x, u)\ge\phi_2(x), \phi_2\in L^1(\Omega), \end{eqnarray} $

(3.2)

$ \begin{eqnarray} &&G(x, u)\ge \beta_3|u|^{p+1}-\phi_3(x), \phi_3\in L^1(\Omega), \end{eqnarray} $

(3.3)

$ \begin{eqnarray} &&|g_u(x, u)|\le \beta_4|u|^{p-1}+\phi_4(x), \phi_4\in V_m(\Omega), \end{eqnarray} $

(3.4)

$ \begin{eqnarray} &&|\nabla^k_xg(x, u)|\le \beta_5|u|^p+\phi_5(x), \phi_5\in V_k(\Omega), \end{eqnarray} $

(3.5)

当$ N = 1, 2 $时, $ 1\le p<+\infty $, 当$ N = 3, 4 $时, $ 1\le p<\frac N{N-2} $, 其中$ G(x, u) = \int^u_0g(x, s)\mathrm{d}s $.由方程(3.1)和(3.2)可得

$ \begin{eqnarray} G(x, u)\le \beta_6(|u|^2+|u|^{p+1}+\phi^2_1+\phi_2). \end{eqnarray} $

(3.6)

定义相空间$ X_k = V_{m+k}\times V_k, k = 0, 1, \cdots, m $, 当$ k = 0 $时, $ V_0 = L^2 $, 则定义范数为$ \|(u, v)\|_{X_k}^2 = \|u\|_{V_{m+k}}^2+\|v\|_{V_k}^2 $.

设$ \varepsilon>0 $足够小, 且$ c_{00} $及$ \varepsilon $满足

$ \begin{eqnarray} &&\big(\frac{7c_{00}}4\lambda^m_1-2\sum^m_{i = 1}C^i_mC_{0i}c_i\lambda^{\frac{m(3-\alpha_{0i})}2}_1\big)\\ &&-\varepsilon\Big(\frac{c_{00}}4\lambda^m_1+ 3\|\mu(x)\|_{\infty}+ 2\sum^m_{i = 1}(1-\alpha_{0i})(\frac{c_{00}}{8\alpha_{0i}m})^{\frac{\alpha_{0i}}{\alpha_{0i}-1}} (\frac{2m}{M_0}(C^i_mc_iC_{0i})^2)^{\frac1{1-\alpha_{0i}}}\Big)>0, \\ &&\frac{3M_0}4-\varepsilon\|c(x)\|_{\infty}>0, \frac{4-\varepsilon}4a_{00}\lambda^m_1-2\varepsilon>0, \frac{3b_{00}M_0}4-\varepsilon\|a(x)\|_{\infty}>0. \end{eqnarray} $

(3.7)

引理 3.1  设$ (H) $成立, $ M $满足$ (M) $, $ f\in H $, (3.1)–(3.5)成立, $ (u_0, u_1)\in X_0 $, 由问题(1.1)确定的$ (u, v) $满足

$ \begin{eqnarray} &&(\mu(x)v, v)+M_0(M_1)\|\nabla^mu\|^2-\varepsilon(c(x)\nabla^mu, \nabla^mu)+\varepsilon^2(\mu(x)u, u)+2\int_{\Omega}G(x, u)\mathrm{d}x\\ &\le &\mathrm{e}^{-\sigma_1t}\big[(\mu(x)v_0, v_0)+M_0(M_1)\|\nabla^mu_0\|^2-\varepsilon(c(x)\nabla^mu_0, \nabla^mu_0)+\varepsilon^2(\mu(x)u_0, u_0)\\ &&+2\int_{\Omega}G(x, u_0)\mathrm{d}x]+\frac{4\lambda^{-m}_1\|\mu(x)\|^2_{\infty}\|f(x)\|^2}{c_{00}\sigma_1}, \end{eqnarray} $

(3.8)

其中$ v = u_t+\varepsilon u $.且存在一个正常数$ R_0 $和$ t_0>0 $, 使得

$ \begin{eqnarray*} \|(u, v)\|^2_{X_0} = \|v\|^2+\|\nabla^{m}u\|^2\le R_0, t>t_0. \end{eqnarray*} $

证  将$ v $与方程组(1.1)在$ L^2(\Omega) $中作内积, 得

$ \begin{eqnarray} &&\frac12\frac{\mathrm{d}}{\mathrm{d}t}(\mu(x)v, v)-\varepsilon(\mu(x)v, v)+(c(x)(-\Delta)^mv, v)-\varepsilon(c(x)(-\Delta)^mu, v)+\varepsilon^2(\mu(x)u, v)\\ &&+(M(\|\nabla^mu\|^2)(-\Delta)^mu, v)+(\mu(x)g(x, u), v) = (\mu(x)f(x), v), \end{eqnarray} $

(3.9)

分别处理(3.9)中各项:

$ \begin{eqnarray} &&(c(x)(-\Delta)^mv, v) = (c(x)\nabla^mv, \nabla^mv)+(\sum^m_{i = 1}C^i_m\nabla^{m-i}v\nabla^ic(x), \nabla^mv), \end{eqnarray} $

(3.10)

$ \begin{eqnarray} &&\varepsilon(c(x)(-\Delta)^mu, v) \end{eqnarray} $

(3.11)

$ \begin{eqnarray} & = &\varepsilon(c(x)\nabla^mu, \nabla^mv)+\varepsilon(\sum^m_{i = 1}C^i_m\nabla^{m-i}v\nabla^ic(x), \nabla^mu)\\ & = &\varepsilon\frac12\frac{\mathrm{d}}{\mathrm{d}t}(c(x)\nabla^mu, \nabla^mu)+\varepsilon^2(c(x)\nabla^mu, \nabla^mu)+ \varepsilon(\sum^m_{i = 1}C^i_m\nabla^{m-i}v\nabla^ic(x), \nabla^mu), \\ &&\varepsilon^2(\mu(x)u, v) = \varepsilon^2(\mu(x)u, (u_t+\varepsilon u)) = \frac{\varepsilon^2}2\frac{\mathrm{d}}{\mathrm{d}t}(\mu(x)u, u)+\varepsilon^3(\mu(x)u, u), \end{eqnarray} $

(3.12)

$ \begin{array}{l} \;\;\;\;\;\;(M(\|\nabla^mu\|^2)(-\Delta)^mu, v) = M(\|\nabla^mu\|^2)(\nabla^{m}u, \nabla^{m}u_t+\varepsilon\nabla^{m}u) \\ = \frac12M(\|\nabla^mu\|^2)\frac{\mathrm{d}}{\mathrm{d}t}\|\nabla^{m}u\|^2+\varepsilon M(\|\nabla^mu\|^2)\|\nabla^{m}u\|^2, \nonumber \end{array}$

(3.13)

当$ \frac{\mathrm{d}}{\mathrm{d}t}\|\nabla^{m}u\|^2+2\varepsilon\|\nabla^{m}u\|^2\ge0 $时,

$ \begin{eqnarray} M(\|\nabla^mu\|^2)((-\Delta)^mu, v)\ge\frac12\frac{\mathrm{d}}{\mathrm{d}t}(M_0\|\nabla^{m}u\|^2)+\varepsilon M_0\|\nabla^{m}u\|^2, \end{eqnarray} $

(3.14)

当$ \frac{\mathrm{d}}{\mathrm{d}t}\|\nabla^{m}u\|^2+2\varepsilon\|\nabla^{m}u\|^2<0 $时,

$ \begin{eqnarray} &M(\|\nabla^mu\|^2)((-\Delta)^mu, v)\ge\frac12\frac{\mathrm{d}}{\mathrm{d}t}(M_1\|\nabla^{m}u\|^2)+\varepsilon M_1\|\nabla^{m}u\|^2, \end{eqnarray} $

(3.15)

$ \begin{eqnarray} &(\mu(x)g(x, u), v) = (\mu(x)g(x, u), u_t+\varepsilon u) = \frac{\mathrm{d}}{\mathrm{d}t} \int_{\Omega}\mu(x)G(x, u)\mathrm{d}x+\varepsilon(\mu(x)g(x, u), u), \qquad \end{eqnarray} $

(3.16)

$ \begin{eqnarray} &(\mu(x)f(x), v)\le\|\mu(x)\|_{\infty}\|f(x)\|\|v\| \le\frac{c_{00}}8\|\nabla^mv\|^2+\frac{2\lambda^{-m}_1\|\mu(x)\|^2_{\infty}}{c_{00}}\|f(x)\|^2, \end{eqnarray} $

(3.17)

又有

$ \begin{eqnarray*} (\nabla^{m-i}v\nabla^ic(x), \nabla^mv)\le c_i\|\nabla^{m-i}v\|\|\nabla^mv\|, i = 1, 2, \cdots, m, c_i = \|\nabla^ic(x)\|_{\infty}, \end{eqnarray*} $

又根据内插不等式, 得

$ \begin{eqnarray*} \|\nabla^{m-i}v\|\le C_{0i}\|\nabla^mv\|^{\alpha_{0i}}\|v\|^{1-\alpha_{0i}}, \alpha_{ki} = \frac{m-i}m, \end{eqnarray*} $

则

$ \begin{eqnarray} \sum^m_{i = 1}C^i_m(\nabla^{m-i}v\nabla^ic(x), \nabla^mv)&\le& \sum^m_{i = 1}C^i_mC_{0i}c_i\|v\|^{1-\alpha_{0i}}\|\nabla^mv\|^{1+\alpha_{0i}}\\ &\le& \sum^m_{i = 1}C^i_mC_{0i}c_i\lambda^{\frac{m(1-\alpha_{0i})}2}_1\|\nabla^mv\|^2, \end{eqnarray} $

(3.18)

$ \begin{eqnarray} &&(C^i_m\nabla^{m-i}v\nabla^ic(x), \nabla^mu)\le\frac{M_0}{8m}\|\nabla^mu\|^2+\frac{2m}{M_0}(C^i_mc_i)^2\|\nabla^{m-i}v\|^2\\ &\le&\frac{M_0}{8m}\|\nabla^mu\|^2+ \frac{c_{00}}{8m}\|\nabla^mv\|^2 +(1-\alpha_{0i})(\frac{c_{00}}{8\alpha_{0i}m})^{\frac{\alpha_{0i}}{\alpha_{0i}-1}}(\frac{2m}{M_0}(C^i_mC_{0i}c_i)^2)^{\frac1{1-\alpha_{0i}}}\|v\|^2, \qquad \end{eqnarray} $

(3.19)

将(3.10)–(3.19)代入(3.9), 得到

$ \begin{eqnarray*} &&\frac{\mathrm{d}}{\mathrm{d}t}\big[(\mu(x)v, v)+M_0(M_1)\|\nabla^mu\|^2-\varepsilon(c(x)\nabla ^mu, \nabla^mu)+\varepsilon^2(\mu(x)u, u) + 2\int_{\Omega}\mu G(x, u)\mathrm{d}x\big]\nonumber\\ &&+2(c(x)\nabla^mv, \nabla^mv)-\frac{c_{00}}4\|\nabla v\|^2-2\sum^m_{i = 1}C^i_mC_{0i}c_i\lambda^{\frac{m(1-\alpha_{0i})}2}_1\|\nabla^mv\|^2 - 2\varepsilon(\mu(x)v, v)\nonumber\\ &&+2\varepsilon M_0(M_1)\|\nabla^mu\|^2-2\varepsilon^2(c(x)\nabla^mu, \nabla^mu)+2\varepsilon^3(\mu(x)u, u)- \frac{\varepsilon M_0}4\|\nabla^mu\|^2\nonumber\\ &&-\frac{\varepsilon c_{00}}4\|\nabla^mv\|^2 -2\varepsilon^2(c(x)\nabla^mu, \nabla^mu)-\frac{\varepsilon M_0}4\|\nabla^mu\|^2-\frac{\varepsilon c_{00}}4\|\nabla^mv\|^2+ 2\varepsilon^3(\mu(x)u, u)\nonumber\\ &&-2\varepsilon\sum^m_{i = 1}(1-\alpha_{0i})(\frac{c_{00}}{8\alpha_{0i}m})^{\frac{\alpha_{0i}}{\alpha_{0i}-1}}(\frac{2m}{M_0}(C^i_mC_{0i}c_i)^2)^{\frac1{1-\alpha_{0i}}}\|v\|^2+ 2\varepsilon(\mu(x)g(x, u), u)\nonumber\\ \end{eqnarray*} $

$ \begin{equation} \le\frac{4\lambda^{-m}_1\|\mu(x)\|^2_{\infty}}{c_{00}}\|f(x)\|^2, \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{equation} $

(3.20)

根据(3.2), 得

$ \begin{eqnarray} 2\varepsilon(\mu(x)g(x, u), u)\ge2\varepsilon(\beta_2\int_{\Omega}\mu G(x, u)\mathrm{d}x+\int_{\Omega}\mu\phi_2(x)\mathrm{d}x), \end{eqnarray} $

(3.21)

由$ (c(x)\nabla^mu, \nabla^mu)\le\|c(x)\|_{\infty}\|\nabla^mu\|^2, 2(c(x)\nabla^mv, \nabla^mv)\le2c_{0}\|\nabla^mv\|^2 $, (3.7), (3.21), 可得适当小的$ \sigma_1 = \frac{\varepsilon}2 $, 则

$ \begin{eqnarray*} &&\frac{\mathrm{d}}{\mathrm{d}t}\big[(\mu(x)v, v)+M_0(M_1)\|\nabla^mu\|^2-\varepsilon(c(x)\nabla ^mu, \nabla^mu)+\varepsilon^2(\mu(x)u, u) + 2\int_{\Omega}\mu G(x, u)\mathrm{d}x\big]\nonumber\\&&+ \sigma_1\big[(\mu(x)v, v)+M_0(M_1)\|\nabla^mu\|^2-\varepsilon(c(x)\nabla^mu, \nabla^mu)+ \varepsilon^2(\mu(x)u, u)+2\int_{\Omega}\mu G(x, u)\mathrm{d}x\big] \end{eqnarray*} $

$ \begin{equation} + 2\varepsilon\int_{\Omega}\mu\phi_2(x)\mathrm{d}x \le\frac{4\lambda^{-m}_1\|\mu(x)\|^2_{\infty}}{c_{00}}\|f(x)\|^2, \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \end{equation} $

(3.22)

由Gronwall不等式, (3.22)可得

$ \begin{eqnarray} &&(\mu(x)v, v)+M_0(M_1)\|\nabla^mu\|^2-\varepsilon(c(x)\nabla ^mu, \nabla^mu)+\varepsilon^2(\mu(x)u, u) +2\int_{\Omega}\mu G(x, u)\mathrm{d}x\\ &\le&\mathrm{e}^{-\sigma_1t}\big[(\mu(x)v_0, v_0)+M_0(M_1)\|\nabla^mu_0\|^2-\varepsilon(c(x)\nabla ^mu_0, \nabla^mu_0)+\varepsilon^2(\mu(x)u_0, u_0) \\ &&+2\int_{\Omega}\mu G(x, u_0)\mathrm{d}x\big]+\frac{4\lambda^{-m}_1\|\mu(x)\|^2_{\infty}}{c_{00}\sigma_1}\|f(x)\|^2, \end{eqnarray} $

(3.23)

由$ (\mu(x)v, v)\le\mu_{0}\|v\|^2 $, (3.3), 则

$ \begin{eqnarray} \|v\|^2+\|\nabla^mu\|^2&\le& C_0\mathrm{e}^{-\sigma_1t}\big[(\mu(x)v_0, v_0)+M_0(M_1)\|\nabla^mu_0\|^2-\varepsilon(c(x)\nabla ^mu_0, \nabla^mu_0)\\ &&+\varepsilon^2(\mu(x)u_0, u_0) +2\int_{\Omega}\mu G(x, u_0)\mathrm{d}x\big]+\frac{4C_0\lambda^{-m}_1\|\mu(x)\|^2_{\infty}}{c_{00}\sigma_1}\|f(x)\|^2, \qquad\, \; \end{eqnarray} $

(3.24)

并且

$ \begin{eqnarray} \overline{\lim\limits_{t\to\infty}}\|(u, v)\|^2_{X_0}\le\frac{4C_0\lambda^{-m}_1\|\mu(x)\|^2_{\infty}}{c_{00}\sigma_1}\|f(x)\|^2, \end{eqnarray} $

(3.25)

因此, 存在一个正常数$ R_0 $和$ t_0>0 $, 使得

$ \begin{eqnarray} \|(u, v)\|^2_{X_0} = \|v\|^2+\|\nabla^mu\|^2\le R_0, t>t_0. \end{eqnarray} $

(3.26)

引理3.1证毕.

引理 3.2  设$ (H) $成立, $ M $满足$ (M) $, $ f\in V_k $, (3.1)-(3.5)成立, $ (u_0, u_1)\in X_k, k = 1, 2, \cdots, m $, 由问题(1.1)确定的$ (u, v) $满足

$ \begin{eqnarray} \|\nabla^kv\|^2+\|\Delta^{m+k}u\|^2 &\le& C_k\mathrm{e}^{-\sigma_2t}\big[\|\nabla^kv_0\|^2+M_0(M_1)(b(x)\nabla^{m+k}u_0, \nabla^{m+k}u_0)\\ &&-\varepsilon(a(x)\nabla^{m+k}u_0, \nabla^{m+k}u_0)+ \varepsilon^2\|\nabla^ku_0\|^2\big]\\ &&+C_k\frac{C(R_0)+\frac{4\lambda^{-m}_1}{a_{00}}\|\nabla^kf(x)\|^2}{\sigma_2}, \end{eqnarray} $

(3.27)

其中$ v = u_t+\varepsilon u $.且存在一个正常数$ R_k $和$ t_k>0 $, 使得

$ \begin{eqnarray*} \|(u, v)\|^2_{X_k} = \|\nabla^kv\|^2+\|\nabla^{m+k}u\|^2\le R_k, t>t_k. \end{eqnarray*} $

证  将$ (-\Delta)^kv, k = 1, 2, \cdots, m-1 $与方程组(1.1)在$ L^2(\Omega) $中作内积, 得

$ \begin{eqnarray} &&\frac12\frac{\mathrm{d}}{\mathrm{d}t}\|\nabla^kv\|^2-\varepsilon\|\nabla^kv\|^2+(a(x)(-\Delta)^mv, (-\Delta)^kv)-\varepsilon(a(x)(-\Delta)^mu, (-\Delta)^kv)\\ &&+\varepsilon^2(u, (-\Delta)^kv)+(b(x)M(\|\nabla^mu\|^2)(-\Delta)^mu, (-\Delta)^kv)+(g(x, u), (-\Delta)^kv)\\ & = &(f(x), (-\Delta)^kv), \end{eqnarray} $

(3.28)

分别处理(3.28)中各项:

$ \begin{eqnarray} &&(a(x)(-\Delta)^mv, (-\Delta)^kv)\\ & = &(a(x)\nabla^{m+k}v, \nabla^{m+k}v)+ (\sum^{m-k}_{i = 1}C^i_{m-k}\nabla^{m+k-i}v\nabla^ia(x), \nabla^{m+k}v), \end{eqnarray} $

(3.29)

$ \begin{eqnarray} &&\varepsilon(a(x)(-\Delta)^mu, (-\Delta)^kv)\\& = &\varepsilon(a(x)\nabla^{m+k}u, \nabla^{m+k}v)+ \varepsilon(\sum^{m-k}_{i = 1}C^i_{m-k}\nabla^{m+k-i}v\nabla^ia(x), \nabla^{m+k}u)\\ & = &\varepsilon\frac12\frac{\mathrm{d}}{\mathrm{d}t}(a(x)\nabla^{m+k}u, \nabla^{m+k}u)+\varepsilon^2(a(x)\nabla^{m+k}u, \nabla^{m+k}u)\\ &&+\varepsilon(\sum^{m-k}_{i = 1}C^i_{m-k}\nabla^{m+k-i}v\nabla^ia(x), \nabla^{m+k}u), \end{eqnarray} $

(3.30)

$ \begin{eqnarray} \varepsilon^2(u, (-\Delta)^kv) = \varepsilon^2(u, (-\Delta)^k(u_t+\varepsilon u)) = \frac{\varepsilon^2}2\frac{\mathrm{d}}{\mathrm{d}t}\|\nabla^ku\|^2+\varepsilon^3\|\nabla^ku\|^2, \end{eqnarray} $

(3.31)

$ \begin{eqnarray} &&M(\|\nabla^mu\|^2)(b(x)(-\Delta)^mu, (-\Delta)^kv)\\ & = &M(\|\nabla^mu\|^2)(b(x)\nabla^{m+k}u, \nabla^{m+k}v)+ M(\|\nabla^mu\|^2)(\sum^{m-k}_{i = 1}C^i_{m-k}\nabla^{m+k-i}v\nabla^ib(x), \nabla^{m+k}u)\\ & = &M(\|\nabla^mu\|^2)(b(x)\nabla^{m+k}u, \nabla^{m+k}(u_t+\varepsilon u))\\&&+ M(\|\nabla^mu\|^2)(\sum^{m-k}_{i = 1}C^i_{m-k}\nabla^{m+k-i}v\nabla^ib(x), \nabla^{m+k}u)\\ & = &\frac12M(\|\nabla^mu\|^2)\frac{\mathrm{d}}{\mathrm{d}t}(b(x)\nabla^{m+k}u, \nabla^{m+k}u)+ \varepsilon M(\|\nabla^mu\|^2)(b(x)\nabla^{m+k}u, \nabla^{m+k}u)\\ &&+M(\|\nabla^mu\|^2)(\sum^{m-k}_{i = 1}C^i_{m-k}\nabla^{m+k-i}v\nabla^ib(x), \nabla^{m+k}u), \end{eqnarray} $

(3.32)

$ \begin{eqnarray} &&(g(x, u), (-\Delta)^kv) = (\nabla^k_xg(x, u), \nabla^kv)\\ & = &|\int_{\Omega}(\beta_5|u|^p+\phi_5(x))\nabla^kv\mathrm{d}x|\le \beta_5|\int_{\Omega}|u|^p\nabla^kv\mathrm{d}x|+|\int_{\Omega}\phi_5(x)\nabla^kv\mathrm{d}x|\\ &\le&\beta_5\|u\|^p_{L^{2p}}\|\nabla^kv\|+\|\phi_5(x)\|\|\nabla^kv\|\le \frac{a_{00}}8\|\nabla^kv\|^2+C_{k-1}(R^{\frac p2}_0+\|\phi_5(x)\|^2), \end{eqnarray} $

(3.33)

$ \begin{eqnarray} (f(x), (-\Delta)^kv)& = &(\nabla^kf(x), \nabla^kv)\le\|\nabla^kf(x)\|\|\nabla^kv\|\\ &\le&\frac{a_{00}}8\|\nabla^{m+k}v\|^2+ \frac{2\lambda^{-m}_1\|\nabla^kf(x)\|^2}{a_{00}}, \end{eqnarray} $

(3.34)

又有

$ \begin{eqnarray} (\nabla^{m+k-i}v\nabla^ia(x), \nabla^{m+k}v)&\le& a_i\|\nabla^{m+k-i}v\|\|\nabla^{m+k}v\|, \\ &&i = 1, 2, \cdots, m-k, a_i = \|\nabla^ia(x)\|_{\infty}, \end{eqnarray} $

(3.35)

又根据内插不等式, 得

$ \begin{eqnarray*} \|\nabla^{m+k-i}v\|\le C_{ki}\|\nabla^{m+k}v\|^{\alpha_{ki}}\|v\|^{1-\alpha_{ki}}, \alpha_{ki} = \frac{m+k-i}{m+k}, \end{eqnarray*} $

则

$ \begin{eqnarray} &&C^i_{m-k}(\nabla^{m+k-i}v\nabla^ia(x), \nabla^{m+k}v)\le C^i_{m-k}C_{ki}a_i\|v\|^{1-\alpha_{ki}}\|\nabla^{m+k}v\|^{1+\alpha_{ki}}\\ &\le&\frac{a_{00}}{8(m-k)}\|\nabla^{m+k}v\|^2+\frac{1-\alpha_{ki}}2(\frac{a_{00}}{4(1-\alpha_{ki})(m-k)})^{\frac{1+\alpha_{ki}}{1-\alpha_{ki}}}(C^i_{m-k}C_{ki}a_i)^{\frac2{1-\alpha_{ki}}}\|v\|^2, \qquad \end{eqnarray} $

(3.36)

$ \begin{eqnarray} &&(C^i_k\nabla^{m+k-i}v\nabla^ia(x), \nabla^{m+k}u)\le\frac{b_{00}M_0}{8(m-k)}\|\nabla^{m+k}u\|^2+ \frac{2(m-k)}{b_{00}M_0}(C^i_{m-k}a_i)^2\|\nabla^{m+k-i}v\|^2\\ &\le&\frac{b_{00}M_0}{8(m-k)}\|\nabla^{m+k}u\|^2+\frac{a_{00}}{8(m-k)}\|\nabla^{m+k}v\|^2\\ &&+(1-\alpha_{ki})(\frac{a_{00}}{8\alpha_{ki}(m-k)})^{\frac{\alpha_{ki}}{\alpha_{ki}-1}}(\frac{2(m-k)}{b_{00}M_0}(C^i_{m-k}C_{ki}a_i)^2)^{\frac1{1-\alpha_{ki}}}\|v\|^2, \end{eqnarray} $

(3.37)

$ \begin{eqnarray} &&M(\|\nabla^mu\|^2)(\sum^{m-k}_{i = 1}C^i_{m-k}\nabla^{m+k-i}v\nabla^ib(x), \nabla^{m+k}u) \le M_1\sum^{m-k}_{i = 1}C^i_{m-k}b_i\|\nabla^{m+k-i}v\|\|\nabla^{m+k}u\|\\ &\le&\frac{\varepsilon b_{00}M_{0}}8\|\nabla^{m+k}u\|^2+ \varepsilon^{-1}b^{-1}_{00}M^{-1}_0\sum^{m-k}_{i = 1}(C^i_{m-k}M_1B_i)^2\|\nabla^{m+k-i}v\|^2\\ &\le&\frac{\varepsilon b_{00}M_{0}}8\|\nabla^{m+k}u\|^2+\frac{a_{00}}8\|\nabla^{m+k}v\|^2\\ &&+(\varepsilon^{-1}b^{-1}_{00}M^{-1}_0)^{\frac1{1-\alpha_{ki}}}\sum^{m-k}_{i = 1}(1-\alpha_{ki}) (\frac{a_{00}}{8\alpha_{ki}(m-k)})^{\frac{\alpha_{ki}}{\alpha_{ki}-1}}(C^i_{m-k}b_iC_{ki}M_1)^{\frac2{1-\alpha_{ki}}}\|v\|^2, \end{eqnarray} $

(3.38)

结合(3.29)–(3.38), (3.28)得到

$ \begin{eqnarray} &&\frac{\mathrm{d}}{\mathrm{d}t}\big[\|\nabla^kv\|^2-\varepsilon(a(x)\nabla^{m+k}u, \nabla^{m+k}u)+\varepsilon^2\|\nabla^ku\|^2\big]+ M(\|\nabla^mu\|^2)\frac {\mathrm{d}}{\mathrm{d}t}(b(x)\nabla^{m+k}u, \nabla^{m+k}u)\\ &&+2(a(x)\nabla^{m+k}v, \nabla^{m+k}v)- \frac{4+\varepsilon}4a_{00}\|\nabla^{m+k}v\|^2-2\varepsilon\|\nabla^kv\|^2\\ &&+2\varepsilon M(\|\nabla^mu\|^2)(b(x)\nabla^{m+k}u, \nabla^{m+k}u) -\\ &&\frac{\varepsilon b_{00}M_0}2\|\nabla^{m+k}u\|^2-2\varepsilon^2(a(x)\nabla^{m+k}u, \nabla^{m+k}u)+2\varepsilon^3\|\nabla^ku\|^2\\ &\le& \sum^{m-k}_{i = 1}(1-\alpha_{ki})(\frac{a_{00}}{4(1-\alpha_{ki})(m-k)})^{\frac{1+\alpha_{ki}}{1-\alpha_{ki}}}(C^i_{m-k}C_{ki}a_i)^{\frac2{1-\alpha_{ki}}}\|v\|^2\\ &&+2\varepsilon\sum^{m-k}_{i = k}(1-\alpha_{ki})(\frac{a_{00}}{8\alpha_{ki}(m-k)})^{\frac{\alpha_{ki}}{\alpha_{ki}-1}}(\frac{2(m-k)}{b_{00}M_0}(C^i_{m-k}C_{ki}a_i)^2)^{\frac1{1-\alpha_{ki}}}\|v\|^2\\ &&+2(\varepsilon^{-1}b^{-1}_{00}M^{-1}_0)^{\frac1{1-\alpha_{ki}}}\sum^{m-k}_{i = 1}(1-\alpha_{ki}) (\frac{a_{00}}{8\alpha_{ki}(m-k)})^{\frac{\alpha_{ki}}{\alpha_{ki}-1}}(C^i_{m-k}b_iC_{ki}M_1)^{\frac2{1-\alpha_{ki}}}\|v\|^2\\ &&+\frac{4\lambda^{-m}_1\|\nabla^kf(x)\|^2}{a_{00}}+2C_{k-1}(R^{\frac p2}_0+\|\phi_5(x)\|^2)\\ &\le& C(R_0)+\frac{4\lambda^{-m}_1\|\nabla^kf(x)\|^2}{a_{00}}, \end{eqnarray} $

(3.39)

当$ \frac{\mathrm{d}}{\mathrm{d}t}(b(x)\nabla^{m+k}u, \nabla^{m+k}u)+2\varepsilon(b(x)\nabla^{m+k}u, \nabla^{m+k}u)\ge0 $时,

$ \begin{eqnarray*} &&M(\|\nabla^mu\|^2)(\frac{\mathrm{d}}{\mathrm{d}t}(b(x)\nabla^{m+k}u, \nabla^{m+k}u)+2\varepsilon(b(x)\nabla^{m+k}u, \nabla^{m+k}u))\nonumber\\ &\ge&\frac{\mathrm{d}}{\mathrm{d}t}(M_0(b(x)\nabla^{m+k}u, \nabla^{m+k}u))+2\varepsilon M_0(b(x)\nabla^{m+k}u, \nabla^{m+k}u), \end{eqnarray*} $

当$ \frac{\mathrm{d}}{\mathrm{d}t}(b(x)\nabla^{m+k}u, \nabla^{m+k}u)+2\varepsilon(b(x)\nabla^{m+k}u, \nabla^{m+k}u)<0 $时,

$ \begin{eqnarray*} &&M(\|\nabla^mu\|^2)(\frac{\mathrm{d}}{\mathrm{d}t}(b(x)\nabla^{m+k}u, \nabla^{m+k}u)+2\varepsilon(b(x)\nabla^{m+k}u, \nabla^{m+k}u)\nonumber\\ &\ge&\frac{\mathrm{d}}{\mathrm{d}t}(M_1(b(x)\nabla^{m+k}u, \nabla^{m+k}u))+2\varepsilon M_1(b(x)\nabla^{m+k}u, \nabla^{m+k}u), \end{eqnarray*} $

则(3.39)变换为

$ \begin{eqnarray} &&\frac {\mathrm{d}}{\mathrm{d}t}\big[\|\nabla^kv\|^2+M_0(M_1)(b(x)\nabla^{m+k}u, \nabla^{m+k}u)-\varepsilon(a(x)\nabla^{m+k}u, \nabla^{m+k}u)+\varepsilon^2\|\nabla^ku\|^2\big]\\ &&+2(a(x)\nabla^{m+k}v, \nabla^{m+k}v)-\frac{4+\varepsilon}4a_{00}\|\nabla^{m+k}v\|^2\\ &&-2\varepsilon\|\nabla^kv\|^2 +2\varepsilon M_0(M_1)(b(x)\nabla^{m+k}u, \nabla^{m+k}u)\\ &&-\frac{\varepsilon b_{00}M_{0}}2\|\nabla^{m+k}u\|^2-2\varepsilon^2(a(x)\nabla^{m+k}u, \nabla^{m+k}u) +2\varepsilon^3\|\nabla^ku\|^2\\ &\le& C(R_0)+\frac{4\lambda^{-m}_1}{a_{00}}\|\nabla^kf(x)\|^2, \end{eqnarray} $

(3.40)

由$ 2(a(x)\nabla^{m+k}v, \nabla^{m+k}v)\ge2a_{00}\|\nabla^{m+k}v\|^2, (b(x)\nabla^{m+k}u, \nabla^{m+k}u)\ge b_{00}\|\nabla^{m+k}u\|^2 $, (3.7),

令$ \sigma_2 = \min\{\frac{\varepsilon}2, \frac{4-\varepsilon}8a_{00}\lambda^m_1-\varepsilon\} $, 则有

$ \begin{eqnarray} &&\frac {\mathrm{d}}{\mathrm{d}t}\big[\|\nabla^kv\|^2+M_0(M_1)(b(x)\nabla^{m+k}u, \nabla^{m+k}u)-\varepsilon(a(x)\nabla^{m+k}u, \nabla^{m+k}u)+\varepsilon^2\|\nabla^ku\|^2\big]\\ &&+\sigma_2\big[\|\nabla^kv\|^2+M_0(M_1)(b(x)\nabla^{m+k}u, \nabla^{m+k}u)-\varepsilon(a(x)\nabla^{m+k}u, \nabla^{m+k}u)+\varepsilon^2\|\nabla^ku\|^2\big]\\ &\le& C(R_0)+\frac{4\lambda^{-m}_1}{a_{00}}\|\nabla^kf(x)\|^2, \end{eqnarray} $

(3.41)

当时$ k = m $时, (3.41)亦成立, 此处不再详述.

利用Gronwall不等式, 可得

$ \begin{eqnarray} &&\|\nabla^kv\|^2+M_0(M_1)(b(x)\nabla^{m+k}u, \nabla^{m+k}u)-\varepsilon(a(x)\nabla^{m+k}u, \nabla^{m+k}u)+\varepsilon^2\|\nabla^ku\|^2 \\ &\le&\mathrm{e}^{-\sigma_2t}\big[\|\nabla^kv_0\|^2+M_0(M_1)(b(x)\nabla^{m+k}u_0, \nabla^{m+k}u_0)-\varepsilon(a(x)\nabla^{m+k}u_0, \nabla^{m+k}u_0) \\&&+\varepsilon^2\|\nabla^ku_0\|^2\big] +\frac{C(R_0)+\frac4{a_{00}}\lambda^{-m}_1\|\nabla^kf(x)\|^2}{\sigma_2}, \end{eqnarray} $

(3.42)

由假设$ (H) $, 得

$ \begin{eqnarray} &&\|\nabla^kv\|^2+\|\nabla^{m+k}u\|^2 \\ &\le& C_k\mathrm{e}^{-\sigma_2t}\big[\|\nabla^kv_0\|^2+M_0(M_1)(b(x)\nabla^{m+k}u_0, \nabla^{m+k}u_0)-\varepsilon(a(x)\nabla^{m+k}u_0, \nabla^{m+k}u_0)\\ &&+ \varepsilon^2\|\nabla^ku_0\|^2\big] +C_k\frac{C(R_0)+\frac4{a_{00}}\lambda^{-m}_1\|\nabla^kf(x)\|^2}{\sigma_2}, \end{eqnarray} $

(3.43)

并且

$ \begin{eqnarray*} \overline{\lim\limits_{t\to\infty}}\|(u, v)\|^2_{X_k}\le C_k\frac{C(R_0)+\frac4{a_{00}}\lambda^{-m}_1\|\nabla^kf(x)\|^2}{\sigma_2}. \end{eqnarray*} $

因此, 存在一个正常数$ R_k $和$ t_k>0 $, 使得

$ \begin{eqnarray} \|(u, v)\|^2_{X_k} = \|\nabla^kv\|^2+\|\nabla^{m+k}u\|^2\le R_k, t>t_k. \end{eqnarray} $

(3.44)

引理3.2证毕.

引理 3.3 (整体解的存在唯一性)  在引理3.1和引理3.2假设条件下, $ (u_0, u_1)\in X_k, k = 0, 1, \cdots, m $, 则初边值问题(1.1)存在唯一的整体解$ (u, v)\in L^{\infty}([0, +\infty), X_k) $.

证 存在性  利用Galerkin方法证明整体解的存在性.

第一步, 近似解构造

设$ (-\Delta)^{m+k}w_j = \lambda^{m+k}_jw_j, k = 0, 1, 2, \cdots, m $, 其中$ \lambda_j $是$ -\Delta $在$ \Omega $上带有齐次Dirichlet边界的特征值, $ w_j $为特征值$ \lambda_j $的特征函数, 由特征值理论知$ w_1, w_2, \cdots, w_n $构成$ H $的标准正交基.

设问题(1.1)的近似解是$ u_n(t) = \sum\limits^n_{j = 1}h_j(t)w_j $, 其中$ h_j(t) $由下面非线性常微分方程组确定

$ \begin{eqnarray} (u_{tt}+a(x)(-\Delta)^{m}u_t+b(x)M(\|\nabla^mu\|^2)(-\Delta)^mu+g(x, u), w_j)& = & (f(x), w_j), \\ j& = &1, 2, \cdots, n, \end{eqnarray} $

(3.45)

满足初始条件$ u_n(0) = u_{n0}, u_{nt}(0) = u_{n1} $, 当$ n\to+\infty $时, 在$ X_k $中$ (u_{n0}, u_{n1})\to(u_0, u_1) $, 由常微分方程的基本理论可知近似解$ u_n(t) $在$ (0, t_n) $存在.

第二步, 先验估计

现需证明$ X_k(k = 0, 1, \cdots, m) $空间解的存在在性, 故在(3.45)两端同时乘以$ \lambda^k_j(h'_j(t)+\varepsilon h_j(t)) $, 并对$ j $求和, 令$ v_n(t) = u_{nt}(t)+\varepsilon u_n(t) $.

由引理3.1和引理3.2得

当$ k = 0 $时, 得到$ X_0 $空间中解的先验估计

$ \begin{eqnarray} \|(u_n, v_n)\|^2_{X_0} = \|v_n\|^2+\|\nabla^{m}u_n\|^2\le R_0, t>t_0. \end{eqnarray} $

(3.46)

当$ k = 1, 2, \cdots, m $时, 得到$ X_k $空间中解的先验估计

$ \begin{eqnarray} \|(u_n, v_n)\|^2_{X_k} = \|\nabla^kv_n\|^2+\|\nabla^{m+k}u_n\|^2\le R_k, t>t_k. \end{eqnarray} $

(3.47)

由此可知, $ (u_n, v_n) $在$ L^{\infty}([0, +\infty);X_0) $中有界, $ (u_n, v_n) $在$ L^{\infty}([0, +\infty);X_k) $中有界.

第三步, 极限过程

在$ X_k, k = 0, 1, \cdots, m $空间中, 从序列$ u_n $中选取子列, 仍用$ u_n $表示, 则

$ \begin{eqnarray} (u_n, v_n)\to(u, v) \end{eqnarray} $

(3.48)

在$ L^{\infty}([0, +\infty);X_k) $中弱*收敛.

由Rellich-Kohdrachov紧嵌入定理知$ X_k(k = 1, 2, \cdots, m) $紧嵌入$ X_0 $, 有$ (u_n, v_n)\to(u, v) $在$ X_0 $中几乎处处强收敛.

由(3.48)得

$ \begin{eqnarray} (u_{nt}, (-\Delta)^kw_j) = (v_n, \lambda^k_jw_j)-(\varepsilon u_n, \lambda^k_jw_j), \end{eqnarray} $

(3.49)

故$ (u_{nt}, (-\Delta)^kw_j)\to(v, \lambda^k_jw_j)-(\varepsilon u, \lambda^k_jw_j) $在$ L^{\infty}[0, +\infty) $中弱*收敛.

由

$ \begin{eqnarray} (u_{ntt}, (-\Delta)^kw_j) = \frac{\mathrm{d}}{\mathrm{d}t}(u_{nt}, (-\Delta)^kw_j), \end{eqnarray} $

(3.50)

故$ (u_{ntt}, (-\Delta)^kw_j)\to(u_{tt}, \lambda^k_jw_j) $在$ D'[0, +\infty) $中收敛, $ D'[0, +\infty) $是$ D[0, +\infty) $无穷可微空间的共轭空间.

由于

$ \begin{eqnarray*} &&M(\|\nabla^mu_n\|^2)(b(x)(-\Delta)^mu_{n}, (-\Delta)^kw_j) = M(\|\nabla^mu_n\|^2)(b(x)\nabla^{m+k}u_{n}, \lambda^{m+k}_jw_j)\nonumber\\ &&+ M(\|\nabla^mu_n\|^2)(\sum^{m-k}_{i = 1}C^i_{m-k}\lambda^{m+k-i}_jw_j\nabla^ib(x), \nabla^{m+k}u_n), \end{eqnarray*} $

故

$ \begin{eqnarray} &&M(\|\nabla^mu_n\|^2)(b(x)(-\Delta)^mu_{n}, (-\Delta)^kw_j)\\ & = &M(\|\nabla^mu_n\|^2)(b(x)\nabla^{m+k}u_{n}, \lambda^{m+k}_jw_j)\\ &&+ M(\|\nabla^mu_n\|^2)(\sum^{m-k}_{i = 1}C^i_{m-k}\lambda^{m+k-i}_jw_j\nabla^ib(x), \nabla^{m+k}u_n)\\ &\to& M(\|\nabla^mu\|^2)(b(x)\nabla^{m+k}u, \lambda^{m+k}_jw_j)+ M(\|\nabla^mu\|^2)(\sum^{m-k}_{i = 1}C^i_{m-k}\lambda^{m+k-i}_jw_j\nabla^ib(x), \nabla^{m+k}u), \end{eqnarray} $

(3.51)

在$ L^{\infty}[0, +\infty) $中弱*收敛.

由

$ \begin{eqnarray*} &&(a(x)(-\Delta)^mu_{nt}, (-\Delta)^kw_j)\nonumber\\ & = &(a(x)\nabla^{m+k}u_{nt}, \lambda^{m+k}_jw_j)+ (\sum^{m-k}_{i = 1}C^i_{m-k}\lambda^{m+k-i}_jw_j\nabla^ia(x), \nabla^{m+k}u_{nt})\nonumber\\ & = &(a(x)\nabla^{m+k}v_{n}, \lambda^{m+k}_jw_j)- (\varepsilon a(x)\nabla^{m+k}u_{n}, \lambda^{m+k}_jw_j)\nonumber\\ &&+(\sum^{m-k}_{i = 1}C^i_{m-k}\lambda^{m+k-i}_jw_j\nabla^ia(x), \nabla^{m+k}v_{n})- (\sum^{m-k}_{i = 1}C^i_{m-k}\lambda^{m+k-i}_jw_j\nabla^ia(x), \varepsilon\nabla^{m+k}u_{n}), \end{eqnarray*} $

则

$ \begin{eqnarray} &&(a(x)(-\Delta)^mu_{nt}, (-\Delta)^kw_j)\\ & = &(a(x)\nabla^{m+k}u_{nt}, \lambda^{m+k}_jw_j)+ (\sum^{m-k}_{i = 1}C^i_{m-k}\lambda^{m+k-i}_jw_j\nabla^ia(x), \nabla^{m+k}u_{nt})\\ & = &(a(x)\nabla^{m+k}v_{n}, \lambda^{m+k}_jw_j)- (\varepsilon a(x)\nabla^{m+k}u_{n}, \lambda^{m+k}_jw_j)\\ &&+(\sum^{m-k}_{i = 1}C^i_{m-k}\lambda^{m+k-i}_jw_j\nabla^ia(x), \nabla^{m+k}v_{n})- (\sum^{m-k}_{i = 1}C^i_{m-k}\lambda^{m+k-i}_jw_j\nabla^ia(x), \varepsilon\nabla^{m+k}u_{n})\\ &\to&(a(x)\nabla^{m+k}v, \lambda^{m+k}_jw_j)- (\varepsilon a(x)\nabla^{m+k}u, \lambda^{m+k}_jw_j)\\ &&+(\sum^{m-k}_{i = 1}C^i_{m-k}\lambda^{m+k-i}_jw_j\nabla^ia(x), \nabla^{m+k}v)- (\sum^{m-k}_{i = 1}C^i_{m-k}\lambda^{m+k-i}_jw_j\nabla^ia(x), \varepsilon\nabla^{m+k}u), \end{eqnarray} $

(3.52)

在$ L^{\infty}[0, +\infty) $中弱*收敛.

根据假设可知$ (g(x, u_n), (-\Delta)^kw_j)\to(g(x, u), \lambda^k_jw_j) $在$ L^{\infty}[0, +\infty) $中弱*收敛.

易得$ (u_n(0), u_{nt}(0))\to(u_0, u_1) $在$ X_k $中弱收敛.当$ j\to+\infty $及$ n\to+\infty $时, 可得

$ \begin{eqnarray} (u_{tt}+a(x)(-\Delta)^mu_t+b(x)M(\|\nabla^mu\|^2)(-\Delta)^mu+g(x, u), (-\Delta)^kw_j)& = &(f(x), (-\Delta)^kw_j), \\ j& = &1, 2, \cdots, n, \end{eqnarray} $

(3.53)

因此得到问题(1.1)解的存在性.

下证解的唯一性

设$ u_1 $和$ u_2 $是方程的两个解, 令$ w = u_1-u_2 $, 则将$ w_t $与问题(1.1)在$ L^2(\Omega) $中作内积, 得

$ \begin{eqnarray} &&\frac12\frac{\mathrm{d}}{\mathrm{d}t}\|w_t\|^2+(a(x)(-\Delta)^mw_t, w_t)\\ &&+ (b(x)M(\|\nabla^mu_1\|^2)(-\Delta)^mu_1-b(x)M(\|\nabla^mu_2\|^2)(-\Delta)^mu_2, w_t)\\ &&+(g(x, u_1)-g(x, u_2), w_t) = 0, \end{eqnarray} $

(3.54)

类似于引理3.1处理方法, 得

$ \begin{eqnarray} &&(a(x)(-\Delta)^mw_t, w_t) = (a(x)\nabla^mw_t, \nabla^mw_t)+(\sum^m_{i = 1}C^i_m\nabla^{m-i}w_t\nabla^ia(x), \nabla^mw_t), \end{eqnarray} $

(3.55)

$ \begin{eqnarray} &&(b(x)M(\|\nabla^mu_1\|^2)(-\Delta)^mu_1-b(x)M(\|\nabla^mu_2\|^2)(-\Delta)^mu_2, w_t) \\ & = &(M(\|\nabla^mu_1\|^2)(b(x)(-\Delta)^mw, w_t)+(M(\|\nabla^mu_1\|^2)-M(\|\nabla^mu_2\|^2))(b(x)(-\Delta)^mu_2, w_t) \\ & = &\frac12M(\|\nabla^mu_1\|^2)\frac{\mathrm{d}}{\mathrm{d}t}(b(x)\nabla^mw, \nabla^mw)+ M(\|\nabla^mu_1\|^2)(\sum^{m}_{i = 1}C^i_m\nabla^{m-i}w_t\nabla^ib(x), \nabla^mw)\\ &&+(M(\|\nabla^mu_1\|^2)-M(\|\nabla^mu_2\|^2))(b(x)\nabla^mu_2, \nabla^mw_t)\\ &&+(M(\|\nabla^mu_1\|^2)-M(\|\nabla^mu_2\|^2))(\sum^{m}_{i = 1}C^i_m\nabla^{m-i}w_t\nabla^ib(x), \nabla^mu_2), \end{eqnarray} $

(3.56)

又有

$ \begin{eqnarray} (\nabla^{k-i}w_t\nabla^ia(x), \nabla^mw_t)\le a_i\|\nabla^{m-i}w_t\|\|\nabla^{m}w_t\|, i = 1, 2, \cdots, m, a_i = \|\nabla^ia(x)\|_{\infty}, \end{eqnarray} $

(3.57)

又根据内插不等式, 得

$ \begin{eqnarray*} \|\nabla^{m-i}w_t\|\le C_{0i}\|\nabla^mw_t\|^{\alpha_{0i}}\|w_t\|^{1-\alpha_{0i}}, \alpha_{0i} = \frac{m-i}{m}, \end{eqnarray*} $

则

$ \begin{eqnarray} &&C^i_m(\nabla^{m-i}w_t\nabla^ia(x), \nabla^mw_t)\le C^i_{m}C_{0i}a_i\|w_t\|^{1-\alpha_{0i}}\|\nabla^{m}w_t\|^{1+\alpha_{ki}}\\ &\le&\frac{a_{00}}{8m}\|\nabla^{m}w_t\|^2+\frac{1-\alpha_{i}}2(\frac{a_{00}}{4(1-\alpha_{0i})m})^{\frac{1+\alpha_{0i}}{1-\alpha_{0i}}}(C^i_mC_{0i}a_i)^{\frac2{1-\alpha_{0i}}}\|w_t\|^2, \end{eqnarray} $

(3.58)

$ \begin{eqnarray} &&M(\|\nabla^mu_1\|^2)(\sum^{m}_{i = 1}C^i_{m}\nabla^{m-i}w_t\nabla^ib(x), \nabla^{m}w)\le M_1\sum^{m}_{i = 1}C^i_{m}b_i\|\nabla^{m-i}w_t\|\|\nabla^{m}w\| \end{eqnarray} $

(3.59)

$ \begin{eqnarray} &\le& M_1\|\nabla^mw\|^2+M_1\sum^{m}_{i = 1}(C^i_{m}b_i)^2\|\nabla^{m-i}w_t\|^2\le M_1\|\nabla^mw\|^2+\frac{a_{00}}8\|\nabla^{m}w_t\|^2\\ &&+(M_1)^{\frac1{1-\alpha_{0i}}}\sum^m_{i = 1}(1-\alpha_{0i})(\frac{a_{00}}{8\alpha_{0i}m})^{\frac{\alpha_{0i}}{\alpha_{0i}-1}}(C^i_mb_iC_{0i})^{\frac2{1-\alpha_{0i}}}\|w_t\|^2, \\ &&(M(\|\nabla^mu_1\|^2)-M(\|\nabla^mu_2\|^2))(\sum^{m}_{i = 1}C^i_m\nabla^{m-i}w_t\nabla^ib(x), \nabla^mu_2) \end{eqnarray} $

(3.60)

$ \begin{eqnarray} &\le&C(M_0, M_1)\sum^m_{i = 1}|(C^i_m\nabla^{m-i}w_t\nabla^ib(x), \nabla^mu_2)|\le C(R_0, M_0, M_1)\sum^m_{i = 1}C^i_mb_i\|\nabla^{m-i}w_t\|\\ &\le&\frac{a_{00}}8\|\nabla^mw_t\|^2 +(C(R_0, M_0, M_1))^{\frac1{1-\alpha_{0i}}}\sum^m_{i = 1}(1-\alpha_{0i})(\frac{a_{00}}{8\alpha_{0i}m})^{\frac{\alpha_{0i}}{\alpha_{0i}-1}}(C^i_mb_iC_{0i})^{\frac2{1-\alpha_{0i}}}\|w_t\|^2, \\ &&(M(\|\nabla^mu_1\|^2)-M(\|\nabla^mu_2\|^2))(b(x)\nabla^mu_2, \nabla^mw_t)\\ &\le&C_{0-1}b_0M'(\xi_1)(\|\nabla^mu_1\|+\|\nabla^mu_2\|)\|\nabla^mu_2\|\|\nabla^mw\|\|\nabla^mw_t\|\\ &\le&\frac{a_{00}}8\|\nabla^mw_t\|^2+C(R_0)\|\nabla^mw\|^2, \end{eqnarray} $

(3.61)

其中$ \xi_1 = \theta\|\nabla^mu_1\|^2+(1-\theta)\|\nabla^mu_2\|^2, \theta\in(0, 1) $,

$ \begin{eqnarray} (g(x, u_1)-g(x, u_2), w_t)\le g'(x, \xi_2)\|w\|\|w_t\|\le\frac{C_{0-2}\lambda^m_1}2\|\nabla^mw\|^2+\frac{C_{0-2}}2\|w_t\|^2, \end{eqnarray} $

(3.62)

其中$ \xi_2 = \theta u_1+(1-\theta)u_2, \theta\in(0, 1) $,

当$ \frac{\mathrm{d}}{\mathrm{d}t}(b(x)\nabla^{m}w, \nabla^{m}w)\ge0 $时,

$ \begin{eqnarray} &&M(\|\nabla^mu_1\|^2)(\frac{\mathrm{d}}{\mathrm{d}t}(b(x)\nabla^{m}w, \nabla^{m}w)\ge \frac{\mathrm{d}}{\mathrm{d}t}(M_0(b(x)\nabla^{m}w, \nabla^{m}w)), \end{eqnarray} $

(3.63)

当$ \frac{\mathrm{d}}{\mathrm{d}t}(b(x)\nabla^{m}w, \nabla^{m}w)<0 $时,

$ \begin{eqnarray} &&M(\|\nabla^mu_1\|^2)(\frac{\mathrm{d}}{\mathrm{d}t}(b(x)\nabla^{m}w, \nabla^{m}w)\ge \frac{\mathrm{d}}{\mathrm{d}t}(M_1(b(x)\nabla^{m}w, \nabla^{m}w)), \end{eqnarray} $

(3.64)

结合(3.55)–(3.65), (3.54)化为

$ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}t}(\|w_t\|^2+M_0(M_1)(b(x)\nabla^{m}w, \nabla^{m}w))\le C_{0-3}(\|w_t\|^2+M_0(M_1)(b(x)\nabla^{m}w, \nabla^{m}w)). \end{eqnarray} $

(3.65)

由Gronwall不等式得

$ \begin{eqnarray*} \|w_t\|^2+M_0(M_1)(b(x)\nabla^{m}w, \nabla^{m}w)\le \mathrm{e}^{C_{0-3}t}(\|w_1\|^2+M_0(M_1)(b(x)\nabla^{m}w_0, \nabla^{m}w_0)), \end{eqnarray*} $

由此可得

$ \begin{eqnarray*} &&\|w_t\|^2+\|\nabla^mw\|^2\le C_{0-4}\mathrm{e}^{C_{0-3}t}(\|w_1\|^2+M_0(M_1)(b(x)\nabla^{m}w_0, \nabla^{m}w_0)). \end{eqnarray*} $

从而解的唯一性得证.

定理 4.1  在引理3.1和3.2的假设条件下及引理3.3, 问题(1.1)存在整体吸引子族:

$ \begin{eqnarray*} A_k = \omega(B_{0k}) = \bigcap\limits_{\tau\ge0}\overline{\bigcup\limits_{t\ge\tau}S(t)B_{0k}}, k = 1, 2, \cdots, m, \end{eqnarray*} $

其中$ B_{0k} = \{(u, v)\in X_k:\|(u, v)\|^2_{X_k} = \|\nabla^{m+k}u\|^2+\|\nabla^kv\|^2\le R_k\} $是$ X_k $中的有界吸收集,

(1)  $ S(t)A_k = A_k, (\forall t\ge0) $,

(2)  $ A_k $吸引$ X_k $中一切有界集, 即$ \forall B_k\subset X_k $为$ X_k $中的有界集, 有

$ \begin{eqnarray*} dist(S(t)B_k, A_k) = \sup\limits_{x\in B_k}\inf\limits_{y\in A_k}\|S(t)x-y\|_{X_k}\to0(t\to\infty), \end{eqnarray*} $

$ \{S(t)\}_{t\ge0} $是问题(1.1)生成的解半群.

证  根据引理2.5, 由引理3.3, 可知问题(1.1)生成半群$ S(t):X_k\to X_k $.

由引理3.1和引理3.2可得, $ \forall B_k\subset X_k $且是包含在球$ \{\|(u, v)\|^2_{X_k}\le R_k\} $的有界集, 则

$ \begin{eqnarray*} \|S(t)(u_0, v_0)\|^2_{X_k} = \|u\|^2_{V_{m+k}}+\|v\|^2_{V_k}\le\|u_0\|^2_{V_{m+k}}+\|v_0\|^2_{V_k}+C\le R_k+C, \end{eqnarray*} $

表明$ \{S(t)\}_{t\ge0} $在$ X_k $上一致有界.

进一步, 对$ \forall(u_0, v_0)\in X_k $, 有

$ \begin{eqnarray*} \|S(t)(u_0, v_0)\|^2_{X_k} = \|u\|^2_{V_{m+k}}+\|v\|^2_{V_k}\le R_k, \end{eqnarray*} $

因此, $ B_{0k} = \{(u, v)\in X_k:\|(u, v)\|^2_{X_k} = \|\nabla^{m+k}u\|^2+\|\nabla^kv\|^2\le R_k\} $是半群$ \{S(t)\}_{t\ge0} $中的有界吸收集. 因为$ X_k\hookrightarrow\hookrightarrow X_0 $是紧嵌入, 即$ X_k $中的有界集是$ X_0 $中的紧集, 因此解半群$ \{S(t)\}_{t\ge0} $是全连续算子. 综上, 得到解半群$ \{S(t)\}_{t\ge0} $的整体吸引子族:

$ \begin{eqnarray*} A_k = \omega(B_{0k}) = \bigcap\limits_{\tau\ge0}\overline{\bigcup\limits_{t\ge\tau}S(t)B_{0k}}, k = 1, 2, \cdots, m, \end{eqnarray*} $

定理4.1证毕.