该文研究下列具阻尼的Boussinesq型方程初边值问题

$ \begin{equation} \label{eq:1.1} \left\{\begin{array}{lll} u_{tt}-\Delta u+\Delta^{2} u-\Delta u_{t}-\Delta g(u)=f(x), \\ u|_{\partial\Omega}=\Delta u |_{\partial\Omega}=0, t>0, \\ u(x, 0)=u_{0}(x), u_{t}(x, 0)=u_{1}(x), x\in\Omega \end{array}\right. \end{equation} $

(1.1)

解的长时间行为, 其中$\Omega$是${\Bbb R}^{N}$中具有光滑边界$\partial\Omega$的有界域, $\nu$是$\partial\Omega$的单位外法向量.

1872年, Bussinesq^[1]建立了描述浅水波水面长波传播的方程

$ \begin{equation} \label{eq:1.2} u_{tt}-u_{xx}-\alpha u_{xxxx}=\beta(u^{2})_{xx}, \end{equation} $

(1.2)

其中$u$表示流体自由表面的运动, 常系数$\alpha, \beta$依赖于流体的深度和水波的特征速度.当$\alpha<0$时, 方程(1.2)被称为``好"的Boussinesq方程. Bona和Sachs<sup>[2]</sup>研究了``好"的Boussinesq方程的初值问题的局部解的适定性. Sachs^[3]研究了方程(1.2)的初值问题整体解的不存在性.当$\alpha>0$时, 方程(1.2)被称为``坏"的Boussinesq方程. 1982年, Deift等<sup>[4]</sup>将反散射理论应用于``坏"的Boussinesq方程的研究, 首次证明在初始函数呈负指数阶一致衰减的条件下, 下面的Boussinesq方程

$ \begin{equation}\label{eq:1.3} u_{tt}-u_{xx}-3 u_{xxxx}=-12(u^{2})_{xx} \end{equation} $

(1.3)

的初值问题是可解的. 1985年Levine和Sleeman^[5]进一步指出在一定条件下, 方程(1.3)的初边值问题不可能存在整体解. 1996年陈国旺和杨志坚^[6]用不同的方法讨论了更一般的``坏"的Boussinesq方程的初边值问题的解的Blowup问题.

2008年, 宋长明等^[18]证明了一维情况下具强阻尼``坏"的Boussinesq方程

$ \begin{equation} \label{eq:1.4} \left\{ \begin{array}{ll} u_{tt}-u_{xx}+2ku_{xxt}-\alpha u_{xxxx}=\beta (u^{n})_{xx}, ~~&(x, t)\in {\Bbb R}\times(0, +\infty), \\ u(x, 0)=u_{0}(x), \;\;\;u_{t}(x, 0)=u_{1}(x), &x\in {\Bbb R} \end{array} \right.\end{equation} $

(1.4)

存在一个整体光滑解$u\in C^{\infty}((0, T];H^{\infty}({\Bbb R}))\cap C([0, T];H^{1}({\Bbb R}))\cap C([0, T];H^{-1}({\Bbb R})), \forall T>0$, 其中$k, \alpha$是正实数, $\beta$是实数, $n$是大于2的整数.

2008年, 杨志坚和郭柏灵^[7]证明了多维Boussinesq方程的初值问题

$ \begin{equation}\label{eq:1.5}\left\{ \begin{array}{ll} u_{tt}-\Delta u+\mu \Delta^{2}u=\Delta\sigma(u), (x, t)\in {\Bbb R}^N\times(0, +\infty), \\ u(x, 0)=u_{0}(x), \;\;\;u_{t}(x, 0)=u_{1}(x), \;\;\;x\in {\Bbb R}^N \end{array} \right.\end{equation} $

(1.5)

整体弱解的存在性, 其中:(ⅰ) $\mu>0, \sigma(u)\in C({\Bbb R})$, $|\sigma(s)|\leq b|s|^{p}, s\in {\Bbb R}$; (ⅱ)当增长阶$1<p<\frac{N+2}{(N-2)^{+}}$时, $0\leq\sigma(s)s\leq\beta\int_{0}^{s}\sigma(\tau){\rm d}\tau, $ $s\in {\Bbb R}, $ $\beta>0$.

Lai等^[10]研究了更一般的具阻尼Boussinesq方程

$ \begin{equation}\label{eq:1.6} u_{tt}-au_{ttxx}-2bu_{txx}=-cu_{xxxx}+u_{xx}-p^{2}u+\beta(u^{2})_{xx} \end{equation} $

(1.6)

的Cauchy问题的整体适定性, 给出一个长时间的渐近解.

整体吸引子是研究具有耗散项非线性发展方程的长时间行为的一个基本概念, 已有很多研究(参见文献[11-12]). Boussinesq型方程的整体吸引子问题受到广泛关注(参见文献[9, 21-23]).

2012年, 杨志坚^[20]研究了具阻尼项的Boussinesq方程

$ \begin{equation}\label{eq:1.7} u_{tt}+\Delta^{2}u-\Delta u_{t}-\Delta g(u)=f(x) \end{equation} $

(1.7)

解的长时间行为, 在$g(u)$非超临界情况下得到方程对应解算子半群整体吸引子及指数吸引子的存在性.

最近, 杨志坚^[16]研究了在多维情况下弹性波导管模型方程的初边值问题

$ \left\{ {\begin{array}{*{20}{l}} {{u_{tt}} - \Delta u - \Delta {u_{tt}} + {\Delta ^2}u - \Delta {u_t} - \Delta g(u) = f(x),(x,t) \in \Omega \times {\mathbb{R}^ + },} \\ {u{|_{\partial \Omega }} = 0,\frac{{\partial u}}{{\partial \nu }}{|_{\partial \Omega }} = 0,} \\ {u(x,0) = {u_0}(x),\;\;\;{u_t}(x,0) = {u_1}(x),\;\;\;x \in \Omega } \end{array}} \right. $

(1.8)

的解的长时间行为, 其中$\Omega$是${\Bbb R}^N$中具有光滑边界$\partial\Omega$的有界域, $\nu$是$\partial\Omega$的单位外法向量.陈国旺^[24]、王书彬^[25]分别讨论了在一维情况下, 上述方程的初边值问题和初值问题的整体解的存在性.

杨志坚^[17]证明了问题

$ \begin{equation}\label{eq:1.9} u_{tt}-{\rm div}\{\sigma(|\Delta u|^{2})\Delta u\}-\Delta u_{t}+\Delta^{2}u+h(u_{t})+g(u)=f(x) \end{equation} $

(1.9)

在相空间$E=V_{2}\times H$中整体吸引子的存在性并讨论了其Hausdorff维数和分形维数的有限性, 其中$\sigma(s)=s^{\frac{m-1}{2}}, s\geq 0, m\geq 1$.

受文献[13-17]的启发, 对于问题(1.1), 我们考虑下列问题:当$1< p\leq\frac{N+2}{(N-2)^{+}}, N\leq5 $时, 具弱阻尼项的Boussinesq型方程的解的长时间行为是否有类似的结果呢?它的整体吸引子是否具有有限的Hausdorff维数?这些问题都有待解决.

本文不同于文献[16-17], 作者进一步证明:

(ⅰ) 在非线性项$g(u)$限制条件弱的情况下, 具弱阻尼项的Boussinesq型方程初边值问题(1.1)解的存在性唯一性, 见定理2.1;

(ⅱ) 当增长阶$1< p\leq\frac{N+2}{(N-2)^{+}}, N\leq5 $时, 用半群分解理论证明问题(1.1)对应的半群$S(t)$在能量相空间$E=V_{2}\times H$中整体吸引子的存在性和吸引子Hausdorff维数的有限性, 见定理3.1和定理3.2;

(ⅲ) 对非线性项$g(u)$的抽象条件给出具体实例, 见注记3.1.

该文的安排如下:第一章引言; 第二章中讨论了问题(1.1)整体解的存在唯一性; 在第三章第一节中讨论了问题(1.1)对应的无穷维动力系统整体吸引子的存在性, 第二节中证明了整体吸引子有有限的Hausdorff维数, 对非线性项$g(u)$给出具体实例.

为叙述方便, 我们引入符号

$ L^{p}=L^{p}(\Omega), ~~ H^{k}_{0}=W^{k, 2}_{0}(\Omega), ~~ V_2=H^{2}\cap H^{1}_{0}, $

$ H=L^{2}, ~~ V'_{2}=V_{-2}, ~~ \|\cdot\|_{p}=\|\cdot\|_{L^{p}}, ~~ \|\cdot\|=\|\cdot\|_{L^{2}}, $

其中$p\geq1, (\cdot, \cdot)$表示$H$的内积, 也表示两个对偶空间元素的对偶积, 为方便其见, 用$C$表示不同的正常数$, C(\cdot, \cdot)$表示依赖与参数的正常数.

易知${V_{2}}\hookrightarrow{H}={H{'}}\hookrightarrow{V'_{2}}.$定义算子$A:{V_{2}}\rightarrow{V'_{2}}, $

$ \begin{equation}\label{eq:2.1} (Au, v)=(\Delta u, \Delta v), \ \ \ \forall u, v\in V_{2}. \end{equation} $

(2.1)

$D(A)=\{u\in H^{4}|u|_{\partial\Omega}=\Delta u |_{\partial\Omega}=0\}, \ \ \forall u \in D(A), Au=\Delta^{2}u.$则$A\in {\cal L}(V_{2}, V'_{2})$是自伴的严格正算子, 并且$A$是${V_{2}}\rightarrow{V'_{2}}$和${D(A)}\rightarrow{H}$的同构, 因此可以定义算子方幂$A^{s}(s\in {\Bbb R})$和空间$V_{s}=D(A^{\frac{s}{4}})(s\in {\Bbb R})$且在下面范数意义下, $V_{s}$组成Hilbert空间.

$ \begin{equation}\label{eq:2.2} (u, v)_{s}=(A^{\frac{s}{4}} u, A^{\frac{s}{4}} v), \ \ \ \|{u}\|_{V_{s}}= \|{A^{\frac{s}{4}}u}\|. \end{equation} $

(2.2)

定义Hilbert空间$E_{s}={V_{s+2}}\times {V_{s}}$, 特别地, $E=E_{0}={V_{2}}\times{H}$.在$E_{s}$上装备范数

$ \begin{equation}\label{eq:2.3} \|(u, v)\|_{E_{s}}=\|u\|_{V_{s+2}}+\|v\|_{V_{s}}, \end{equation} $

(2.3)

则问题(1.1)等价于下面算子方程的Cauchy问题

$ \begin{equation}\label{eq:2.4} \left \{ \begin{array}{ll} A^{-\frac{1}{2}}u_{tt}+(I+A^{\frac{1}{2}})u+u_{t}+g(u)=A^{-\frac{1}{2}}f, \\ u(0)=u_{0}, u_{t}(0)=u_{1}, \end{array} \right. \end{equation} $

(2.4)

定理2.1(解的适定性) 假定

$(H_1)$ $g\in{C^{1}({\Bbb R})}, $

$ \begin{equation}\label{eq:2.5}\begin{array}{ll} |g(s)|\leq K_{1}(|s|^{p}+1), ~~|g'(s)|\leq K_{2}(|s|^{p-1}+1), \\ |g'(s_{1})-g'(s_{2})|\leq K_{3}(|s_{1}|^{p-1-\gamma}+|s_{2}|^{p-1-\gamma}+1)|s_{1}-s_{2}|^{\gamma}, \\ \end{array}\end{equation} $

(2.5)

其中

$ \begin{equation}\label{eq:2.6} \left\{\begin{array}{ll} 0<\gamma<p-1, ~~&1<p<2, \\ \gamma=1, &p\geq 2, \end{array}\right. \end{equation} $

(2.6)

其中$1< p<\frac{N+2}{(N-2)^{+}}, (a)^{+}=\max\{0, a\}, K_{i}>0, i=1, 2, 3$;

$ (H_2)\\\ \ \ \ \ \ \ \ \ \begin{equation}\label{eq:2.7} \liminf\limits_{|s|\rightarrow + \infty}{\frac{G(s)}{|s|^{2}}\geq 0}, ~~\liminf\limits_{|s|\rightarrow + \infty}{\frac {sg(s)-\rho G(s)}{|s|^{2}}\geq 0}, \end{equation} $

(2.7)

其中$G(s)=\int_{0}^{s}g(\tau){\rm d}\tau, 0<\rho<2$;

$ (H_3)\ \ f\in V_{-1}, \;\;\;(u_{0}, u_{1})\in V_{2}\times H. $

则问题(1.1)存在唯一的解$u, (u, u_{t})\in C_{b}({\Bbb R}^{+}, V_{2}\times H)$, 并且$\|(u, u_{t})\|_{E}\leq C_{1}{\rm e}^{-\delta t}+C_{0}, $其中$C_{1}=C(\|(u_{0}, u_{1})\|_{E}), C_{0}=C(\|f\|_{V_{-1}}).$同时, $N\leq 5$时, $(u, u_{t})$连续依赖$E$中的初值$(u_{0}, u_{1}).$

注2.1 (2.7)式蕴含$\forall \eta >0, \exists C_{\eta}>0, \tilde{C}_{\eta}>0, $使得

$ \begin{equation}\label{eq:2.8} G(s)+\eta |s^{2}|\geq -C_{\eta}, sg(s)-\rho G(s)+\eta |s^{2}|\geq -\tilde{C}_{\eta}, \end{equation} $

(2.8)

其中$G(s)=\int_{0}^{s}g(\tau){\rm d}\tau .$

引理2.1^[12] 设$X, Y$为两个Banach空间, 且$X \hookrightarrow Y$.若$\varphi \in L^{\infty}((0, T);X)\bigcap C_{w}([0, T];$ $Y)$则$\varphi \in C_{w}([0, T];X).$

引理2.2 当$1< p< \frac{N+2}{(N-2)^{+}}(N\leq5), $ $\|g(u)-g(v)\|_{V_{1}}\leq C( R)\|u-v\|_{V_{2}}, $ $\forall u, v \in V_{2}, $ $\|u\|_{V_{2}}+\|v\|_{V_{2}}\leq R.$

定理2.1的证明 令$v=u_{t}+\varepsilon u, $则(2.4)式变为

$ \begin{equation}\label{eq:2.9} \left\{ \begin{array}{ll} A^{-\frac{1}{2}}(v_{t}-\epsilon v+\epsilon ^{2}u)+(I+A^{\frac{1}{2}})u+v-\epsilon u+g(u)=A^{-\frac{1}{2}}f, \\ v(0)=u_{1}+\epsilon u_{0}=v_{0}. \end{array} \right. \end{equation} $

(2.9)

对上式用$v$作$H$内积并作先验估计使用由Gronwall引理可得

$ \begin{equation}\label{eq:2.10} \|A^{\frac{1}{4}}u\|^{2}+\|A^{-\frac{1}{4}}u_{t}\|^{2} \leq C_{1}{\rm e}^{-\delta t}+C_{0}, \end{equation} $

(2.10)

其中$C_{1}=C_{1}(\|(u_{0}, u_{1})\|_{E}), C_{0}=C_{0}(\|f\|_{V_{-1}}), \delta =\epsilon \in (0, 1).$

用$A^{\frac{1}{2}}u_{t}+\epsilon A^{\frac{1}{2}}u$与(2.4)式中方程作$H$内积, 可得

$ \begin{equation}\label{eq:2.11} \frac{\rm d}{{\rm d}t}H_{2}(u)+K_{2}(u)=0, t>0. \end{equation} $

(2.11)

显然

$ \begin{eqnarray}\label{eq:2.12} H_{2}(u)&=&\frac{1}{2}[\|u_{t}\|^{2}+(1+\epsilon)\|A^{\frac{1}{4}}u\|^{2}+\|A^{\frac{1}{2}}u\|^{2}]+\epsilon(u, u_{t})-(u, f) \\&\geq &\frac{1}{4}(\|A^{\frac{1}{4}}u\|^{2}+\|u_{t}\|^{2}+\|A^{\frac{1}{2}}u\|^{2})-C_{0}, \end{eqnarray} $

(2.12)

$ \begin{equation}\label{eq:2.13} H_{2}(u)\leq C(\|A^{\frac{1}{4}}u\|^{2}+\|u_{t}\|^{2}+\|A^{\frac{1}{2}}u\|^{2})-C_{0}, \end{equation} $

(2.13)

其中$C_{0}=C_{0}(\|f\|_{V_{-1}}).$

$ \begin{equation}\label{eq:2.14} K_{2}(u)\geq C(\|A^{\frac{1}{4}}u_{t}\|^{2}+\|A^{\frac{1}{2}}u\|^{2})-C_{1}{\rm e}^{-\delta t}-C_{0}. \end{equation} $

(2.14)

事实上, 当$1\leq p<\frac{N+2}{(N-2)^{+}}$时,

$ \begin{equation}\label{eq:2.15} |(A^{\frac{1}{4}}g(u), A^{\frac{1}{4}}u_{t})| \leq \frac {1}{4}\|A^{\frac{1}{4}}u_{t}\|^{2}+\frac {\varepsilon}{4}\|A^{\frac{1}{2}}u\|^{2}+C, \end{equation} $

(2.15)

$ \begin{equation}\label{eq:2.16} |(A^{\frac{1}{2}}u, g(u))|\leq \frac {1}{4}\|A^{\frac{1}{2}}u\|^{2}+C, \end{equation} $

(2.16)

结合(2.11)-(2.16)式可得

$ \begin{equation}\label{eq:2.17} \frac{\rm d}{{\rm d}t}H_{2}(u)+\delta H_{2}(u)+\delta\|A^{\frac{1}{4}}u_{t}(t)\|^{2}\leq C_{1}{\rm e}^{-\delta t}+C_{0}, \end{equation} $

(2.17)

由Gronwall引理可得

$ \begin{equation}\label{eq:2.18} H_{2}(u)\leq H_{2}(u_{0}){\rm e}^{-\delta t}+ C_{0}, \end{equation} $

(2.18)

结合(2.18)式可知

$ \begin{equation}\label{eq:2.19} \|A^{\frac{1}{2}}u\|^{2}+\|A^{\frac{1}{4}}u\|^{2}+\|u_{t}\|^{2}\leq C_{2}{\rm e}^{-\delta t}+C_{0}, t>0, \\ \end{equation} $

(2.19)

故

$ ||(u,{u_t})|{|_E} + \int_t^{t + 1} {||{A^{\frac{1}{4}}}{u_t}(} \tau )|{|^2}{\text{d}}\tau \leqslant {C_2}{{\text{e}}^{ - \delta t}} + {C_0},t > 0. $

(2.20)

下面设方程(1.1)的近似解$u^{n}$为

$ \begin{equation} u^{n}(t)=\sum \limits^{n}\limits_{j=1}T_{jn}(t)w_{j}, \end{equation} $

(2.21)

其中$Aw_{j}=\lambda_{j}w_{j}, T_{jn}(t)=(u^{n}, w_{j}), \lambda_{j}$是算子$A$所对应的特征值, $w_{j}$是算子$A$对应于特征值$\lambda_{j}$的特征元, 由此$\{{w_{j}}\}_{j=1}^{n}$构成$H$中的一个正交规范基, 并且同时是$V_{2}$中的正交基.

由Galerkin方法可知, $u^{n}(t)$满足

$ \begin{equation}\label{eq:2.22} \left\{ \begin{array}{ll} (A^{-\frac{1}{2}}u^{n}_{tt}, w_{j})+((I+A^{\frac{1}{2}})u^{n}, w_{j})+(u^{n}_{t}, w_{j})+(g(u^{n}), w_{j})=(A^{-\frac{1}{2}}f, w_{j}), \\ u^{n}(0)=u_{0n}, u^{n}_{t}(0)=u_{1n}, \end{array} \right. \end{equation} $

(2.22)

其中$j=1, \cdots, n, u_{0n}\rightarrow u_{0}$在$V_{2}$中, $u_{1n}\rightarrow u_{1}$在$H$中.并且(2.19)式对于$u^{n}(t)$也成立, 即对于任意的$t\in[0, T]$, 有

$ \begin{equation}\label{eq:2.23} \|A^{\frac{1}{2}}u^{n}\|^{2}+\|u^{n}_{t}\|^{2}+\int^{t+1}_{t}\|A^{\frac{1}{4}}u_{t}(\tau)\|^{2}{\rm d}\tau\leq C_{2}{\rm e}^{-\delta t}+C_{0}, t>0, \\ \end{equation} $

(2.23)

则可抽取子列仍记为$u^{n}$, 满足

$ \mbox{$u^{n}\rightarrow u$在$L^{\infty}({\Bbb R}^+, V_{2})$中弱$^{*}$;} $

$ \mbox{$u_{t}^{n}\rightarrow u_{t}$在$L^{\infty}({\Bbb R}^+;H)\bigcap L^{2}(0, T;V_{1})$ 中弱$^{*}$;} $

$ \mbox{$u^{n}\rightarrow u$在$C([0, T], V_{1})$中.} $

对(2.22)式在$(0, t)$上积分可得

$ \begin{eqnarray}\label{eq:2.24} &&(A^{-\frac{1}{2}}u^{n}_{t}, w_{j})+\int^{t}_{0}((I+A^{\frac{1}{2}})u^{n}, w_{j}){\rm d}\tau+\int^{t}_{0}(u_{t}^{n}, w_{j}){\rm d}\tau +\int^{t}_{0}(g(u^{n}), w_{j}){\rm d}\tau\\ &=&\int^{t}_{0}(A^{-\frac{1}{2}}f, w_{j}){\rm d}\tau+(A^{-\frac{1}{2}}u_{1n}, w_{j}), \end{eqnarray} $

(2.24)

(1) 当$1\leq p\leq\frac{N}{(N-2)^{+}}$时,

$ \begin{equation}\label{eq:2.25} \int^{t}_{0}|(g(u^{n})-g(u), w_{j})|{\rm d}\tau\leq C\int^{t}_{0}(\|u^{n}-u\|+\|A^{\frac{1}{4}}u^{n}-A^{\frac{1}{4}}u\|){\rm d}\tau\rightarrow 0. \end{equation} $

(2.25)

(2) 当$\frac{N}{(N-2)^{+}}< p<\frac{N+2}{(N-2)^{+}}$时,

$ \begin{eqnarray}\label{eq:2.26} && \int^{t}_{0}|(g(u^{n})-g(u), w_{j})|{\rm d}\tau\\ &\leq& C\int^{t}_{0}(\|u^{n}-u\|+\|A^{\frac{1}{4}}(u^{n}-u)\|^{\theta}\|A^{\frac{1}{2}}(u^{n}-u)\|^{1-\theta}){\rm d}\tau\rightarrow 0, \end{eqnarray} $

(2.26)

$ \begin{equation}\label{eq:2.27} \int^{t}_{0}(g(u^{n}), w_{j}){\rm d}\tau\rightarrow\int^{t}_{0}(g(u), w_{j}){\rm d}\tau, n\rightarrow +\infty. \end{equation} $

(2.27)

在(2.24)式中令$n\rightarrow +\infty$, 所得方程两边对$t$求导, 由$w_{j}$在$H$中稠密性可知

$ \begin{equation}\label{eq:2.28} (u, u_{t})\in L^{\infty}({\Bbb R}^+, V_{2}\times H) \end{equation} $

(2.28)

是问题(1.1)的解.

由$u_{tt}=f-A^{\frac{1}{2}}u-A^{\frac{1}{2}}u_{t}-Au-A^{\frac{1}{2}}g(u)\in L^{\infty}({\Bbb R}^{+}, V_{-2})$, 对$\forall T>0$,

$ \begin{equation}\label{eq:2.29} \begin{array}{ll} u\in L^{\infty}({\Bbb R}^{+}, V_{2})\cap C_{w}([0, T], V_{1}), \\ u_{t}\in L^{\infty}({\Bbb R}^{+}, H)\cap C_{w}([0, T], V_{-2}), \end{array} \end{equation} $

(2.29)

由引理2.1可知$(u, u_{t})\in C_{w}([0, T], V_{2}\times H).$

$A^{\frac{1}{2}}u_{t}$与(2.4)式中方程作$H$内积式子在$(t_{0}, t)$上进行积分, 当$t\rightarrow t_{0}$时,

$ \begin{eqnarray}\label{eq:2.30} &&(\|u_{t}(t)\|^{2}+\|A^{\frac{1}{4}}u(t)\|^{2}+ \|A^{\frac{1}{2}}u(t)\|^{2})-(\|u_{t}(t_{0})\|^{2}+\|A^{\frac{1}{4}}u(t_{0})\|^{2}+ \|A^{\frac{1}{2}}u(t_{0})\|^{2}) \\ &=&2\int^{t}_{t_{0}}(u, f){\rm d}t+2\int^{t}_{t_{0}}\|A^{\frac{1}{4}}u_{t}\|^{2}{\rm d}t +2\int^{t}_{t_{0}}(A^{\frac{1}{4}}g(u), A^{\frac{1}{4}}u_{t}){\rm d}t \\ &\rightarrow &0, \end{eqnarray} $

(2.30)

即当$t\rightarrow t_{0}$时,

$ \begin{equation}\label{eq:2.31} \|(u, u_{t})(t)\|_{V_{2}\times H}-\|(u, u_{t})(t_{0})\|_{V_{2}\times H}\rightarrow 0, \end{equation} $

(2.31)

故

$ \begin{equation}\label{eq:2.32} (u, u_{t})\in C([0, T], V_{2}\times H). \end{equation} $

(2.32)

现在我们证明$V_{2}\times H$中解对初值的连续依赖性.

设$u(t), v(t)$是方程(2.4)在$C_{b}({\Bbb R}^{+}, V_{2}\times H)$中的两个解, $(u_{0}, u_{1}), (v_{0}, v_{1})$分别是其对应的初值.令$w=u-v, $则$w$满足

$ \begin{equation}\label{eq:2.33} \left\{ \begin{array}{ll} A^{-\frac{1}{2}}w_{tt}+(I+A^{\frac{1}{2}})w+w_{t}+g(u)-g(v)=0, \\ w(0)=u_{0}-v_{0}=w_{0}, w_{t}(0)=u_{1}-v_{1}=w_{1}, \end{array} \right. \end{equation} $

(2.33)

对上面方程用$A^{\frac{1}{2}}w_{t}$做内积, 并利用引理2.2, 可得

$ \begin{eqnarray}\label{eq:2.34} &&\frac {1}{2}\frac {\rm d}{{\rm d}t}( \|w_{t}\|^{2}+\|A^{\frac{1}{4}}w\|^{2}+ \|A^{\frac{1}{2}}w\|^{2})+\|A^{\frac{1}{4}} w_{t}\|^{2}\\ &=&-(A^{\frac{1}{2}}w_{t}, g(u)-g(v))\\ &\leq& \frac{1}{2}\|A^{\frac{1}{4}}w_{t}\|^{2}+ C(R)\|A^{\frac{1}{2}}w\|^{2}, \end{eqnarray} $

(2.34)

由Gronwall不等式

$ \begin{equation}\label{eq:2.35} \|(w, w_{t})\|^{2}_{E}\leq C(T)\|(w_{0}, w_{1})\|^{2}_{E}, 0\leq t \leq T. \end{equation} $

(2.35)

定理2.1得证.

引理3.1 对于下面的方程

$ \begin{equation}\label{eq:3.1} \left\{ \begin{array}{ll} A^{-\frac{1}{2}}u_{tt}+(I+A^{\frac{1}{2}})u+u_{t}=h, \\ u(0)=u_{0}, u_{t}(0)=u_{1}, \end{array} \right. \end{equation} $

(3.1)

(1) 若$(u_{0}, u_{1})\in V_{2}\times H, h\in C_{b}({\Bbb R}^+, H), $则方程(3.1)有唯一解$u, (u, u_{t})\in C_{b}({\Bbb R}^+, V_{2}\times H)$, 并且下式成立

$ \begin{equation}\label{eq:3.2} \|A^{\frac{1}{2}}u\|^{2}+\|u_{t}\|^{2}\leq C {\rm e}^{-\delta t}+C_{0}, \;\;\;t>0, \end{equation} $

(3.2)

其中$C=C(\|(u_{0}, u_{1})\|_{E}), C_{0}=C_{0}(\|h\|_{ C_{b}({\Bbb R}^+, H)}).$

(2) 若$(u_{0}, u_{1})\in V_{s+2}\times V_{s}, h\in C_{b}({\Bbb R}^+, V_{s}), $则方程(3.1)有唯一解$u, (u, u_{t})\in C_{b}({\Bbb R}^+, $ $V_{s+2}\times V_{s})$, 并且下式成立

$ \begin{equation}\label{eq:3.3} \|A^{\frac{s+2}{4}}u\|^{2}+\|A^{\frac{s}{4}}u_{t}\|^{2}\leq C {\rm e}^{-\delta t}+C_{0}, \;\;\;t>0, \end{equation} $

(3.3)

其中$C=C(\|(u_{0}, u_{1})\|_{E_{s}}), C_{0}=C_{0}(\|h\|_{ C_{b}({\Bbb R}^+, V_{s})}).$

(3) 若$(u, u_{t})\in C_{b}({\Bbb R}^+, V_{s+2}\times V_{s}), $ $s\in {\Bbb R}, $是方程(3.1)的解且$u_{0}=u_{1}=0, $ $h\in C_{b}({\Bbb R}^+;V_{\sigma }), $ $\sigma >s(\sigma>0), $则$(u, u_{t})\in C_{b}({\Bbb R}^+, $ $V_{\sigma+2}\times V_{\sigma})$.

证  (1) 类似于定理1的证明及结论, 易得解得存在唯一性及(3.1)式成立.

(2) 用$A^{\frac{s}{4}}$作用于方程(3.1), 并令$v=A^{\frac{s}{4}}u, \tilde{h}=A^{\frac{s}{4}}h$可得

$ \begin{equation}\label{eq:3.4} \left\{ \begin{array}{ll} A^{-\frac{1}{2}}v_{tt}+(I+A^{\frac{1}{2}})v+v_{t}=\tilde{h}, \\ v(0)=v_{0}=A^{\frac{1}{4}}u_{0}, \;\;\;v_{t}(0)=v_{1}=A^{\frac{1}{4}}u_{1}, \end{array} \right. \end{equation} $

(3.4)

当$(u_{0}, u_{1})\in V_{s+2}\times V_{s}, h\in C_{b}({\Bbb R}^+, V_{s})$时,

$ \begin{equation}\label{eq:3.5} (v_{0}, v_{1})\in V_{2}\times H, \;\;\;\tilde{h}\in C_{b}({\Bbb R}^+, H), \end{equation} $

(3.5)

由(1)可知方程(3.4)有唯一解$v, (v, v_{t})\in C_{b}({\Bbb R}^+, V_{2}\times H)$, 即方程(3.1)有唯一解$u, (u, u_{t})\in C_{b}({\Bbb R}^+, V_{s+2}\times V_{s})$和(3.3)式成立.

(3) 用$A^{\frac{\sigma}{4}}$作用于方程(3.1), 并令$v=A^{\frac{\sigma}{4}}u, \tilde{h}=A^{\frac{\sigma}{4}}h$, 可得

$ \begin{equation}\label{eq:3.6} \left\{ \begin{array}{ll} A^{-\frac{1}{2}}v_{tt}+(I+A^{\frac{1}{2}})v+v_{t}=\tilde{h}, \\ v(0)=0, v_{t}(0)=0. \end{array} \right. \end{equation} $

(3.6)

当$h\in C_{b}({\Bbb R}^+, V_{\sigma})$时,

$ \begin{equation}\label{eq:3.7} \tilde{h}\in C_{b}({\Bbb R}^+, H), \end{equation} $

(3.7)

由(1)可知方程(3.6)有唯一解$v, (v, v_{t})\in C_{b}({\Bbb R}^+, V_{2}\times H)$, 即方程(3.1)有唯一解$u, (u, u_{t})\in C_{b}({\Bbb R}^+, V_{\sigma+2}\times V_{\sigma})$.

引理3.2 若$u$是问题(1.1)在定理1中的解, 且$g$满足$( H_{1})$中假设条件$(N\leq5)$, 则存在$\sigma:0<\sigma<1$, 使$g'(u)u_{t}\in C_{b}({\Bbb R}^+, V_{\sigma -2}).$

证  $\forall \psi \in V_{2-\sigma}$有

$ \begin{equation}\label{eq:3.8} |(g'(u)u_{t}, \psi)|\leq C\|\psi\|_{V_{2-\sigma}}, \end{equation} $

(3.8)

其中$N=5$时,

$ \begin{equation}\label{eq:3.9} \|g'(u)\|_{\frac{5}{2-\sigma}}\leq C(1+\|u\|^{p-1}_{\frac{5(p-1)}{2-\sigma}})\leq C.\end{equation} $

(3.9)

证毕.

引理3.3^[12] 若度量空间$X$上的半群$S(t)$有一有界吸收集$B_{0}$, 且存在$X$中的一紧集$K$, 对于X中的任意有界集$B$, 使得当$t\rightarrow\infty$时,

$ \begin{equation}\label{eq:3.10} {\rm dist}\{S(t)B, K\}\rightarrow 0, \end{equation} $

(3.10)

则${\cal A}=\omega(B_{0})(B_{0}$的$\omega$极限集)是$S(t)$的整体吸引子.

引理3.4^[17] 设$E=V_{2}\times H, E_{\sigma}=V_{\sigma+2}\times V_{\sigma}, 0<\sigma<1, K_{1}$是$E_{\sigma}$中的有界集, $K=\overline{K_{1}^{E}}$, 则K是$E_{\sigma}$中的有界集.

在定理1的假定下, 定义算子$S(t):E\rightarrow E, S(t)(u_{0}, u_{1})=(u(t), u_{t}(t))$, 其中$u$是问题(2.4)的解.则$N\leq 5$时, $S(t)$是$E$上的连续算子半群.

定理3.1(整体吸引子的存在性) 在定理1的假定下, $S(t)$在$E$存在一个连通的整体吸引子${\cal A}\subset K$, 其中$K$是$E_{\sigma}$中的有界集, $0<\sigma<1$.

证 由(2.19)式结论可知集合$B_{0}=\{(u, v)\in E\mid\|(u, v)\|^{2}_{E}\leq C_{0}\}$是算子半群$S(t)$在$E$中的吸收集, 对于$E$中任意的有界集$B$, 下式成立

$ \begin{equation}\label{eq:3.11} \begin{array}{ l} {\rm dist}(S(t)B, B_{0})\leq C(B){\rm e}^{-\delta t}, \;\;\;\forall B\subset E. \end{array}\end{equation} $

(3.11)

将$S(t)$分解为$:S(t)=S_{1}(t)+S_{2}(t)$, 其中$S_{2}(t):V_{2}\times H\rightarrow V_{2}\times H, S_{2}(t)(u_{0}, u_{1})=(\bar{u}, \bar{u}_{t})$且满足

$ \begin{equation}\label{eq:3.12} \left\{ \begin{array}{ll} A^{-\frac{1}{2}}\bar{u}_{tt}+(I+A^{\frac{1}{2}})\bar{u}+\bar{u}_{t}=0, t>0, \\ \bar{u}(0)=u_{0}, \bar{u}_{t}(0)=u_{1}, \end{array} \right. \end{equation} $

(3.12)

由引理3.1可知

$ \begin{equation}\label{eq:3.13} \|S_{2}(t)\|_{{\cal L}(E)}\leq C {\rm e}^{-\delta t}, \end{equation} $

(3.13)

$S_{1}(t)=S(t)-S_{2}(t), S_{1}(t):E\rightarrow E, S_{1}(t)(u_{0}, u_{1})=(\hat{u}, \hat{u}_{t})$且满足

$ \begin{equation}\label{eq:3.14} \left\{ \begin{array}{ll} A^{-\frac{1}{2}}\hat{u_{tt}}+(I+A^{\frac{1}{2}})\hat{u}+\hat{u}_{t}=A^{-\frac{1}{2}}f-g(u), t>0, \\ \hat{u}(0)=0, \hat{u}_{t}(0)=0, \end{array} \right. \end{equation} $

(3.14)

其中$u\in C_{b}({\Bbb R}^+, V_{2}).$

将(3.14)式对$t$求导, 并令$w=\hat {u}_{t}, $可得

$ \begin{equation}\label{eq:3.15} \left\{ \begin{array}{ll} A^{-\frac{1}{2}}w_{tt}+(I+A^{\frac{1}{2}})w+w_{t}=-g'(u)u_{t}, \\ w(0)=0=w_{0}, w_{t}(0)=f-A^{\frac{1}{2}}g(u_{0})=w_{1}, \end{array} \right. \end{equation} $

(3.15)

因为$f\in V_{\sigma -2}, A^{\frac{1}{2}}g(u_{0})\in V_{\sigma -2}$, 所以$w_{1}\in V_{\sigma -2}, g'(u)u_{t}\in C_{b}({\Bbb R}^+, V_{\sigma -2})$, 由引理3.1可知

$ \begin{equation}\label{eq:3.16} (\hat{u}_{t}, \hat{u}_{tt})\in C_{b}({\Bbb R}^+, V_{\sigma} \times V_{\sigma -2}), \end{equation} $

(3.16)

又$(I+A^{\frac{1}{2}})\hat{u}=A^{-\frac{1}{2}}f-g(u)-A^{-\frac{1}{2}}\hat{u}_{tt}-\hat{u}_{t}\in C_{b}({\Bbb R}^+, V_{\sigma})$, 所以$\hat{u} \in C_{b}({\Bbb R}^+, V_{\sigma+2})$.

由此,

$ \begin{equation}\label{eq:3.17} (\hat{u}, \hat{u}_{t}) \in C_{b}({\Bbb R}^+, V_{\sigma+2}\times V_{\sigma}), \end{equation} $

(3.17)

则$\bigcup\limits_{t\geq 0\\, \|(\hat{u}, \hat{u}_{t})\|_{E_{\sigma}}\leq R}(\hat{u}, \hat{u}_{t})$是$V_{\sigma+2}\times V_{\sigma}$中的有界集.又${V_{\sigma+2}\times V_{\sigma}}\hookrightarrow\hookrightarrow{V_{2}\times H}$, 由引理3.4知

$ K=\overline{\bigcup\limits_{t\geq 0\\, \|(\hat{u}, \hat{u}_{t})\|_{E_{\sigma}}\leq R}(\hat{u}, \hat{u}_{t})^{E}} $

是$E=V_{2}\times H$中的紧集.对任意有界集$B\subset E$, 由(3.11)式可得

$ \begin{eqnarray}\label{eq:3.18} {\rm dist} \{S(t)B, K \}&=&\sup\limits_{(u_{0}, u_{1})\in B} \inf\limits_{(\hat{u}, \hat{u}_{t})\in K} {\rm dist} \{ S(t)(u_{0}, u_{1}), (\hat{u}, \hat{u}_{t})\} \\ &\leq& \sup\limits_{(u_{0}, u_{1})\in B}\|(\bar{u}, \bar{u}_{t})\|_{E}\rightarrow 0, \;\;\;t\rightarrow \infty. \end{eqnarray} $

(3.18)

因此, 由引理3.3可知, $S(t)$由一个整体吸引子${\cal A}=\omega(B_{0})$且

$ \begin{equation}\label{eq:3.19} {\rm dist}\{{\cal A}, K\}={\rm dist}\{S(t){\cal A}, K\}\rightarrow 0, \;\;\;t\rightarrow \infty, \end{equation} $

(3.19)

即${\cal A}\subset K$.

引理3.5^[19] 设$X$为一Hilbert空间, $B\subset X$是一个紧集, $S:B\rightarrow X$是一个连续映射, 并且$SB=B$, 如果

(1)  $S$在$B$上是一致可微的, 即存在线性算子$L(u)\in {\cal L}(X), $使得

$ \begin{equation}\label{eq:3.20} \sup\limits_{0<\|v-u\|_{X}\leq\epsilon}\frac{\|Sv-Su-L(u)(v-u)\|_{X}}{\|v-u\|_{X}}\rightarrow 0\ (\epsilon\rightarrow 0); \end{equation} $

(3.20)

(2)  $\sup\limits_{u \in B}\|L(u)\|_{{\cal L}(X)}<\infty;$

(3) 对某一个$d>0, $成立$\bar{\omega_{\rm d}}\leq\sup\limits_{u \in B}\omega_{\rm d}(L(u))<1;$则$B$的Hausdorff维数不超过$d$, 即$\dim_{H}\{B, X\}\leq d<\infty.$

定理3.2 设定理1的假定成立, 并且存在$\sigma_{1}:0<\sigma_{1}<1, \gamma:0<\gamma\ll 1$使得

$(H_{4})$

$ \begin{equation}\label{eq:3.21} \|g'(u)\|_{{\cal L}_{(V_{2-\sigma_{1}}, V_{1})}}\leq C(R), \;\;\;\forall u \in V_{2+ \sigma }, \|u\|_{V_{2+\sigma}}\leq R, \end{equation} $

(3.21)

$(H_{5})$

$ \begin{equation}\label{eq:3.22}\begin{array}{ll} \|g'(u_{1})-g'(u_{2})\|_{{\cal L}(V_{2}, V_{1})}\leq C(R)\|u_{1}-u_{2}\|_{V_{2}}^{\gamma}, \\ \forall u_{1}, u_{2} \in V_{2+ \sigma}, \|u_{1}\|_{V_{2+\sigma}}+\|u_{2}\|_{V_{2+\sigma}}\leq R. \end{array}\end{equation} $

(3.22)

则上述吸引子${\cal A}$有有限的Hausdorff维数.

证 下面分三步证明.

第一步 变分方程解的存在性.

考虑方程(1.1)的变分方程

$ \begin{equation}\label{eq:3.23} \left\{ \begin{array}{ll} A^{-\frac{1}{2}}v_{tt}+(I+A^{\frac{1}{2}})v+v_{t}+g'(u)v=0, \\ v(0)=\xi, v_{t}(0)=\eta, \end{array} \right. \end{equation} $

(3.23)

其中$(u, u_{t})=S(t)(u_{0}, u_{1}), (u_{0}, u_{1})\in{\cal A}\subset K$.即

$ \begin{equation}\label{eq:3.24} \|(u, u_{t})(t)\|_{E_{\sigma}}\leq R, \;\;\;t\geq 0. \end{equation} $

(3.24)

用$A^{\frac{1}{2}}v_{t}$和(3.23)式中方程作内积

$ \begin{equation}\label{eq:3.25} \frac {1}{2}\frac {\rm d}{{\rm d}t}( \|v_{t}\|^{2}+\|A^{\frac{1}{4}}v\|^{2}+ \|A^{\frac{1}{2}}v\|^{2})+\|A^{\frac{1}{4}} v_{t}\|^{2}+(A^{\frac{1}{2}}v_{t}, g'(u)v)=0, t>0. \end{equation} $

(3.25)

因为

$ \begin{equation}\label{eq:3.26} |(A^{\frac{1}{2}}v_{t}, g'(u)v)|\leq\frac{1}{4}\|A^{\frac{1}{4}}v_{t}\|^{2}+C(R)\|v\|^{2}_{V_{2-\sigma_{1}}}, \end{equation} $

(3.26)

可知

$ \begin{equation}\label{eq:3.27} \frac {\rm d}{{\rm d}t}( \|v_{t}\|^{2}+\|A^{\frac{1}{4}}v\|^{2}+ \|A^{\frac{1}{2}}v\|^{2})+\frac {1}{2}\|A^{\frac{1}{4}}v_{t}\|^{2}\leq C(R)\|v\|^{2}_{V_{2-\sigma_{1}}}, \end{equation} $

(3.27)

从而

$ \begin{equation}\label{eq:3.28} \|v_{t}\|^{2}+\|A^{\frac{1}{4}}v\|^{2}+ \|A^{\frac{1}{2}}v\|^{2}+\int^{t}_{0}\|A^{\frac{1}{4}} v_{t}\|^{2}{\rm d}\tau \leq C(R, T), t\in [0, T]. \end{equation} $

(3.28)

利用经典Faedo-Galerkin的近似解方法得

$ v\in L^{\infty}(0, T;V_{2}), v_{t}\in L^{\infty}(0, T;H)\bigcap L^{2}(0, T;V_{1}) $

是方程(3.23)的解.则由引理2.1.可知

$ (v, v_{t})\in C_{w}(0, T;V_{2}\times H). $

对(3.23)式中方程用$A^{\frac{1}{2}}v_{t}$作内积, 可得

$ \begin{equation}\label{eq:3.29} \frac {1}{2}\frac {\rm d}{{\rm d}t}( \|v_{t}\|^{2}+\|A^{\frac{1}{4}}v\|^{2}+ \|A^{\frac{1}{2}}v\|^{2}) +\|A^{\frac{1}{4}}v_{t}\|^{2}+(A^{\frac{1}{2}}v_{t}, g'(u)v)=0, t>0, \end{equation} $

(3.29)

对上式在$(t_{0}, t)$上积分可得$t\rightarrow t_{0}$时, 有

$ \begin{eqnarray}\label{eq:3.30} &&\frac {1}{2}(\|v_{t}\|^{2}+\|A^{\frac{1}{4}}v\|^{2}+ \|A^{\frac{1}{2}}v\|^{2}-\|\eta\|^{2}+\|A^{\frac{1}{4}}\xi\|^{2}+ \|A^{\frac{1}{2}}\xi\|^{2} )\\ &=&-\int_{t_{0}}^{t}\|A^{\frac{1}{4}}v_{t}\|^{2}{\rm d}\tau-\int_{t_{0}}^{t}(A^{\frac{1}{2}}v_{t}, g'(u)v){\rm d}\tau\\ &\rightarrow& 0, \end{eqnarray} $

(3.30)

因此

$ \begin{equation}\label{eq:3.31} (v, v_{t})\in C([0, T], V_{2}\times H). \end{equation} $

(3.31)

第二步  $S(t)$在${\cal A}$上的可微性.

令$\varphi_{0}=(u_{0}, u_{1}), \tilde{\varphi}_{0}=(u_{0}+\xi, u_{1}+\eta)\in {\cal A}, $以及

$ \varphi(t)=S(t)\varphi_{0}=(u(t), u_{t}(t)), \tilde{\varphi}(t)=S(t)\tilde{\varphi}_{0}=(\tilde{u}(t), \tilde{u}_{t}(t)), $

令$w=\tilde{u}-u$, 则$w$满足

$ \begin{equation}\label{eq:3.32} \left\{ \begin{array}{ll} A^{-\frac{1}{2}}w_{tt}+(I+A^{\frac{1}{2}})w+w_{t}+g(\tilde{u})-g(u)=0, \\ w(0)= \xi, w_{t}(0)=\eta, \end{array} \right. \end{equation} $

(3.32)

且据(2.35)式可知

$ \begin{equation}\label{eq:3.33} \|(w, w_{t})\|^{2}_{E}\leq C(T)\|(\xi, \eta)\|^{2}_{E}, t\in [0, T]. \end{equation} $

(3.33)

定义算子:${\cal L}(t, \varphi_{0}):E\rightarrow E, {\cal L}(t, \varphi_{0})(\xi, \eta)=(v(t), v_{t}(t)) $, 其中$\varphi_{0}\in {\cal A}\subset E_{\sigma}, $ $v(t)$是变分方程(2.35)的解, 下面只需证明

$ \begin{equation}\label{eq:3.34} {\cal L}(t, \varphi_{0})=DS(t)\varphi_{0}. \end{equation} $

(3.34)

事实上, 令$\theta=w-v=\tilde{u}-u-v, \theta$满足

$ \begin{equation}\label{eq:3.35} A^{-\frac{1}{2}}\theta_{tt}+(I+A^{\frac{1}{2}})\theta+\theta_{t}+g'(u)\theta =-\bigg\{\int^{1}_{0}[g' (\lambda\tilde{u}+(1-\lambda)u)-g'(u)]{\rm d}\lambda\bigg\}(\tilde{u}-u), \end{equation} $

(3.35)

用$A^{\frac{1}{2}}\theta_{t}$作$H$内积

$ \begin{eqnarray}\label{eq:3.36} & &\frac {1}{2} \frac {\rm d}{{\rm d}t}( \|\theta_{t}\|^{2}+\|A^{\frac{1}{4}}\theta\|^{2}+ \|A^{\frac{1}{2}}\theta\|^{2})+\|A^{\frac{1}{4}} \theta_{t}\|^{2}+(g'(u)\theta, A^{\frac{1}{2}}\theta_{t})\\ &=&-\bigg(\int^{1}_{0}[g'(\lambda\tilde{u}+(1-\lambda)u)-g'(u)]{\rm d}\lambda(\tilde{u}-u), A^{\frac{1}{2}}\theta_{t}\bigg), t>0, \end{eqnarray} $

(3.36)

其中

$ \begin{equation}\label{eq:3.37} |(g'(u)\theta, A^{\frac{1}{2}}\theta_{t})|\leq\frac{1}{8}\|A^{\frac{1}{4}}\theta_{t}\|^{2}+C(R)\|\theta\|^{2}_{V_{2}}, \end{equation} $

(3.37)

$ \begin{eqnarray}\label{eq:3.38} &&\bigg|\bigg(\int^{1}_{0}[g'(\lambda\tilde{u}+(1-\lambda)u)-g'(u)]{\rm d}\lambda(\tilde{u}-u), A^{\frac{1}{2}}\theta_{t} \bigg)\bigg| \\ &\leq&\frac{1}{8}\|A^{\frac{1}{4}}\theta_{t}\|^{2}+C(R)\|A^{\frac{1}{2}}(\tilde{u}-u)\|^{2(1+\delta)}. \end{eqnarray} $

(3.38)

综合(3.36)-(3.38)式, 并利用Gronwall引理可得

$ \begin{equation}\label{eq:3.39} \|\theta_{t}\|^{2}+\|A^{\frac{1}{4}}\theta\|^{2}+ \|A^{\frac{1}{2}}\theta\|^{2}\leq C(R, T)\|(\xi, \eta)\|^{2(1+\delta)}_{E}, \end{equation} $

(3.39)

由此可得S (t)在${\cal A}$上的可微性.

第三步  ${\cal A}$的Hausdorff维数.

引理3.6^[8] 设$X$为一Hilbert空间, $q(t, \cdot), \gamma(t, \cdot)$是$X$中的两个双线性形式, $t\in[0, T], $ $0<T\leq\infty, \{L(t), 0\leq t<T\}$是$X$中一族线性连续算子, 并且$L(0)=I, $如果

(1)  $\forall \eta_{0}\in X, q(t, L(t) \eta_{0}), \gamma(t, L(t) \eta_{0})$是绝对连续的, 且$\forall t\in [0, T)$, 满足

$ \begin{equation}\label{eq:3.40} \frac{\rm d}{{\rm d}t}q(t, L(t) \eta_{0})=\gamma(t, L(t)\eta_{0}); \end{equation} $

(3.40)

(2) 存在两个正实数$\alpha=\alpha(T), \beta=\beta(T)$, 使得$\forall t\in [0, T)$, 满足

$ \begin{equation}\label{eq:3.41} \alpha\|\eta\|^{2}_{E}\leq q(t, L(t)\eta)\leq \beta\|\eta\|^{2}_{E}\ \ \, t\in [0, T); \end{equation} $

(3.41)

(3) 存在$X$上非负自伴紧算子$K, $及$C_{0}>0, \sigma\in (0, 1]$使得对几乎处处的$t\in[0, T)$,

$ \begin{equation}\label{eq:3.42} \|\gamma(t, \eta)\|\leq C_{0}\|\eta\|^{2(1-\sigma)}_{X}(K\eta, \eta)^{\sigma}\ \ \, \forall \eta \in X. \end{equation} $

(3.42)

则

$ \begin{equation}\label{eq:3.43} \det\limits_{1\leq i, j\leq m}(L(t) \eta^{i}_{0}, L(t) \eta^{j}_{0}) \leq\Big(\frac{\beta}{\alpha}\Big)^{m}{\rm exp} \bigg\{\bigg(\frac{C_{0}}{\alpha}\sum^{m}\limits_{l=1}k_{l}^{\sigma}\bigg)t\bigg\}\det \limits_{1\leq i, j\leq m}(\eta^{i}_{0}, \eta^{j}_{0}), \end{equation} $

(3.43)

其中$\eta^{1}_{0}, \cdots, \eta^{m}_{0}$是$X$中任意元素, $\{k_{i}\}_{i=1}^{\infty}$是$K$特征值的非增序列.

下面通过引理3.6证明$\bar{\omega_{\rm d}}<1$.记$w=v_{t}+\epsilon v$, 则(3.23)式中方程可化为

$ \begin{equation}\label{eq:3.44} A^{\frac{1}{2}}(w_{t}-\epsilon w+\epsilon^{2}v)+\big(I+A^{\frac{1}{2}}\big)v+w-\epsilon v+g'(u)v=0, \end{equation} $

(3.44)

对(3.44)式两边用$A^{\frac{1}{2}}w$作内积得

$ \begin{equation}\label{eq:3.45} \frac{1}{2}\frac{\rm d}{{\rm d}t}q_{\epsilon}(v, v_{t})-2\epsilon\|w\|^{2}+\|A^{\frac{1}{4}}w\|^{2}+\epsilon q_{\epsilon}(v, v_{t}) =-\big(g'(u)v, A^{\frac{1}{2}}w\big), \end{equation} $

(3.45)

其中

$ \begin{equation}\label{eq:3.46} q_{\epsilon}(v, v_{t})=\|v_{t}+\epsilon v\|^{2}+\epsilon^{2}\|v\|^{2}+(1-\epsilon)\|A^{\frac{1}{4}}v\|^{2}+\|A^{\frac{1}{2}}v\|^{2}, \end{equation} $

(3.46)

因为

$ \begin{equation}\label{eq:3.47} \frac{3}{4}\|A^{\frac{1}{4}}w\|^{2}-2\epsilon\|w\|^{2} \geq \Big(\frac{1}{2}-\frac{2\epsilon}{\lambda_{1}}\Big)\|A^{\frac{1}{4}}w\|^{2}+\frac{1}{4}\|A^{\frac{1}{4}}w\|^{2}, \end{equation} $

(3.47)

$ \begin{eqnarray}\label{eq:3.48} |(g'(u)v, A^{\frac{1}{2}}w)|&\leq& \|g'(u)\|_{{\cal L}(V_{2-\sigma_{1}}, V_{1})}\|v\|_{V_{2-\sigma_{1}}}\|A^{\frac{1}{2}}w\|_{V_{-1}} \\&\leq& C(R)\|v\|_{V_{2-\sigma_{1}}}\|A^{\frac{1}{4}}w\| \\&\leq &\frac{1}{4}\|A^{\frac{1}{4}}w\|^{2}+C(R)|v\|^{2}_{V_{2-\sigma_{1}}}, \end{eqnarray} $

(3.48)

其中$\lambda_{1}$是$A$的第一个特征值, (3.48)式用到了$(H_{4})$.

将(3.47)和(3.48)式应用于(3.45)式, 可得

$ \begin{equation}\label{eq:3.49} \frac {\rm d}{{\rm d}t}q_{\epsilon}(v, v_{t})+k q_{\epsilon}(v, v_{t})+\frac{1}{2}\|A^{\frac{1}{4}}w\|^{2}\leq C(R)\|v\|^{2}_{V_{2-\sigma_{1}}}. \end{equation} $

(3.49)

定义算子$K:E\rightarrow E, K\{u, v\}=A^{-\frac{\sigma_{1}}{2}}\{u, v\}, \forall u, v\in E.$令

则由(3.49)式可以得到

$ \begin{equation}\label{eq:3.50} \frac {\rm d}{{\rm d}t}q_{\epsilon}(v, v_{t})+k q_{\epsilon}(v, v_{t})\leq C(R)(K\{u, v\}, \{u, v\})_{E}, \end{equation} $

(3.50)

因为$K\in {\cal L}(E)$且$K:E=V_{2}\times H\rightarrow V_{2+2\sigma_{1}}\times V_{2\sigma_{1}}\hookrightarrow\hookrightarrow V_{2}\times H$, 故K是紧算子.

对(3.45)式两边同时乘${\rm e}^{kt}$, 得

$ \begin{equation}\label{eq:3.51} \frac {\rm d}{{\rm d}t}\tilde{q}_{\epsilon}(\tilde{v}, \tilde{v}_{t})\leq C(R)(\|\tilde{v}\|^{2}_{V_{2-\sigma_{1}}}+\|\tilde{v}_{t}\|^{2}_{V_{-\sigma_{1}}}), \end{equation} $

(3.51)

其中$\tilde{v}={\rm e}^{\frac{k}{2}t}v, \tilde{v}_{t}=\frac{k}{2}{\rm e}^{\frac{k}{2}t}v+{\rm e}^{\frac{k}{2}t}v_{t}.$当$\epsilon, k$适当小时,

$ \begin{eqnarray}\label{eq:3.52} \tilde{q_{\epsilon}}(\tilde{v}, \tilde{v}_{t}) &=&{\rm e}^{kt}q_{\epsilon}(v, v_{t})\\ &=&\Big\|\tilde{v}_{t}+\Big(\epsilon-\frac{k}{2}\Big)\tilde{v}\Big\|^{2}+\epsilon^{2}\|\tilde{v}\|^{2}+(1- \epsilon)\|A^{\frac{1}{4}}\tilde{v}\|^{2}+\|A^{\frac{1}{2}}\tilde{v}\|^{2}\\ &\sim& \|(\tilde{v}, \tilde{v}_{t})\|^{2}_{E}, \end{eqnarray} $

(3.52)

即存在$ \alpha=\alpha(T)>0, \beta=\beta(T)>0, $使得

$ \begin{equation}\label{eq:3.53} \alpha\|(\tilde{v}, \tilde{v}_{t})\|_{E}\leq \tilde{q_{\epsilon}}(\tilde{v}, \tilde{v_{t}})\leq \beta\|(\tilde{v}, \tilde{v_{t}})\|_{E}, t\in [0, T]. \end{equation} $

(3.53)

令

$ {R_\kappa } = \left( {\begin{array}{*{20}{c}} 1&0 \\ {\frac{k}{2}}&1 \end{array}} \right), $

(3.54)

显然

$ \begin{equation}\label{eq:3.55} \left(\begin{array}{ll} \tilde{v}\\\tilde{v}_{t} \end{array}\right)={\rm e}^{\frac{kt}{2}}R_{k} \left(\begin{array}{cc} v\\v_{t} \end{array}\right). \end{equation} $

(3.55)

定义算子$L_{1}(t, \varphi_{0}):E\rightarrow E, L_{1}(t, \varphi_{0})\psi_{0}=\psi(t).$其中

$ {\psi _0} = \left( {\begin{array}{*{20}{c}} {{{\tilde v}_0}} \\ {{{\tilde v}_1}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} \xi \\ {\eta + \frac{{k\xi }}{2}} \end{array}} \right),\qquad \psi (t) = \left( {\begin{array}{*{20}{c}} {\tilde v} \\ {\widetilde {{v_t}}} \end{array}} \right), $

则

$ \begin{equation}\label{eq:3.56} L_{1}(t, \varphi_{0})={\rm e}^{\frac{k t}{2}}R_{k}{\cal L}(t, \varphi_{0})R_{-k}, \end{equation} $

(3.56)

则(3.51)式可写成

$ \begin{equation}\label{eq:3.57} \frac {\rm d}{{\rm d}t}\tilde{q}_{\epsilon}(t, L_{1}(t, \varphi_{0})\psi_{0}) \leq C(R)(K\{\tilde{v}, \tilde{v}_{t}\}, \{\tilde{v}, \tilde{v}_{t}\})_{E}, \end{equation} $

(3.57)

由引理3.6得

$ \begin{eqnarray}\label{eq:3.58} \det\limits_{1\leq i, j\leq m}({\cal L}(t, \varphi_{0})\eta^{i}_{0}, ({\cal L}(t, \varphi_{0})\eta^{j}_{0})) &=&\|\wedge^{m}L_{1}(t, \varphi_{0})(\eta^{1}_{0}\wedge\cdots \wedge\eta^{m}_{0})\|^{2}_{\wedge^{m}E}\\ &\leq&\Big(\frac{\beta}{\alpha}\Big)^{m}\exp \bigg\{\Big(\frac{C}{\alpha}\sum^{m} \limits_{l=1}k_{l}\Big)t\bigg\}\|(\eta^{1}_{0}\wedge\cdots \wedge\eta^{m}_{0})\|^{2}_{\wedge^{m}E}, \qquad \end{eqnarray} $

(3.58)

并且

$ \begin{equation}\label{eq:3.59} \sup\limits_{\varphi_{0}\in{\cal A}}\|\wedge^{m}L_{1}(t, \varphi_{0})\|^{2}_{{\cal L}(\wedge^{m}E)} \leq \bigg[\Big(\frac{\beta}{\alpha}\Big)^{\frac{m}{2}}\exp\bigg\{ \Big(\frac{C}{2\alpha}\sum^{m}\limits_{l=1}k_{l}^{\sigma}\Big)t\bigg\}\bigg]^{2}, \end{equation} $

(3.59)

故存在$ m_{0}$, 当$m\geq m_{0}$时,

$ \begin{eqnarray}\label{eq:3.60} \bar{\omega_{m}}(t)&=&\sup\limits_{\varphi_{0}\in{\cal A}}\omega_{m}({\cal L}(t, \varphi_{0})) \\ &=&{\rm e}^{-\frac{mkt}{2}}\sup\limits_{\varphi_{0}\in{\cal A}}\|\wedge^{m}(R_{-k}L_{1}(t, \varphi_{0})R_{k}) \|_{{\cal L}(\wedge^{m}E)}\\ &\leq& {\rm e}^{-\frac{mkt}{2}}\sup\limits_{\varphi_{0}\in{\cal A}}\|\wedge^{m}R_{-k}\|_{{\cal L}(\wedge^{m}E)} \|\wedge^{m}L_{1}(t, \varphi_{0})\|_{{\cal L}(\wedge^{m}E)}\|\wedge^{m} R_{k})\|_{{\cal L}(\wedge^{m}E)} \\ &\leq&\bigg[\Big(\frac{\beta}{\alpha}\Big)^{\frac{m}{2}}\exp\bigg\{-\frac{m}{2} \Big(k-\frac{C}{m\alpha}\sum^{m}\limits_{l=1}k_{l}\Big)t\bigg\}\bigg]^{2} \\ &<&1. \end{eqnarray} $

(3.60)

特别地,

$ \begin{equation}\label{eq:3.61} \sup\limits_{\varphi_{0}\in{\cal A}}\|{\cal L}(t, \varphi_{0})\|_{{\cal L}(E)} \leq\Big(\frac{\beta}{\alpha}\Big)^{\frac{1}{2}}\exp\bigg\{-\frac{m}{2} \Big(\frac{k}{2}-\frac{Ck_{l}}{2\alpha}\Big)t\bigg\}<\infty, \end{equation} $

(3.61)

由引理3.6可知, $\dim_{H}\{{\cal A}, E\}\leq m_{0}<\infty.$定理3.2得证.

注3.1 设$g(s)=a|s|^{p-1}s$, 其中$a\in {\Bbb R}^+, 2\leq p<\frac{N+2}{(N-2)^{+}}$.易验证$g(s)$满足条件$(H_{1})-(H_{4})$.从而$g(s)=a|s|^{p-1}s$使得定理1-3中所有结论成立.