带有非局部 Laplace 算子的饱和 Schrödinger-Klein-Gordon 方程的概自守动力学

带有非局部 Laplace 算子的饱和 Schrödinger-Klein-Gordon 方程的概自守动力学

张天伟^,, 李永昆^,^*

云南大学数学与统计学院昆明 650500

Almost Automorphic Dynamics of Nonlocal Laplacian Saturating Schrödinger-Klein-Gordon Equations

Zhang Tianwei^,, Li Yongkun^,^*

School of Mathematics and Statistics, Yunnan University, Kunming 650500

通讯作者: * 李永昆,Email: yklie@ynu.edu.cn

收稿日期: 2022-06-8 修回日期: 2023-10-7

基金资助:

国家自然科学基金(12261098)
国家自然科学基金(11861072)

Received: 2022-06-8 Revised: 2023-10-7

Fund supported:

NSFC(12261098)
NSFC(11861072)

作者简介 About authors

张天伟,Email:yntwzhang@outlook.com

摘要

迄今为止, 几乎没有学者研究 Schrödinger 或 Klein-Gordon 方程的概自守动力学. 该文结合 Galerkin 方法、 Laplace 变换、Fourier 级数和 Picard 迭代研究了带有非局部 Laplace 算子饱和 Schrödinger-Klein-Gordon 方程的概自守弱解的一些结果. 此外, 还考虑了该方程的全局指数收敛性.

关键词： Schrödinger; Klein-Gordon; Galerkin 方法; Fourier 级数; Picard 迭代

Abstract

To the best of the authors' knowledge, almost no literature focuses on the almost automorphic dynamics to Schrödinger or Klein-Gordon equations. This paper gives some results on almost automorphic weak solutions to a nonlocal Laplacian saturating Schrödinger-Klein-Gordon equations by employing a mix of Galerkin method, Laplace transform, Fourier series and Picard iteration. Beyond that, global exponential convergence of the equations is investigated.

Keywords： Schrödinger; Klein-Gordon; Galerkin method; Fourier series; Picard iteration

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本文引用格式

张天伟, 李永昆. 带有非局部 Laplace 算子的饱和 Schrödinger-Klein-Gordon 方程的概自守动力学[J]. 数学物理学报, 2024, 44(2): 326-353

Zhang Tianwei, Li Yongkun. Almost Automorphic Dynamics of Nonlocal Laplacian Saturating Schrödinger-Klein-Gordon Equations[J]. Acta Mathematica Scientia, 2024, 44(2): 326-353

1 引言

本文用 $\mathbf{R}^{d}$ 表示 $d$ 维欧式空间, $\mathbf{R}_0=[0,+\infty)$, $\mathbf{R}_+=(0,+\infty)$; $\mathbf{Z}_+=\{1,2,\cdots \}$; 用 $\mathbf{C}$ 表示复数集. 设 $\Omega$ 是 $\mathbf{R}^d$ 中一个具有光滑边界 $\partial \Omega$ 的有界开域, $\bar{\Omega}:=\Omega\cup\partial \Omega$. 本文致力于研究如下带有非局部 Laplace 算子的饱和 Schrödinger-Klein-Gordon 方程

(1.1) $ \begin{cases} \mathrm{i}\hbar\frac{\partial}{\partial t}\mathbf{z}-\hbar^2(-\Delta)^{\alpha} \mathbf{z}+\mathrm{i}a\mathbf{z}+\frac{\mu\mathbf{z} \mathbf{w} }{1+\rho|\mathbf{z}|+\varrho| \mathbf{w} |}=f(x,t), &\quad \mbox{in} \Omega \times \mathbf{R}_+, \\ \frac{\partial^2}{\partial t^2} \mathbf{w} +(-\Delta)^{\beta} \mathbf{w} +b\frac{\partial}{\partial t} \mathbf{w} +c \mathbf{w} =\frac{\nu|\mathbf{z}|^{2}}{1+\varepsilon|\mathbf{z}|}+g(x,t), &\quad \mbox{in} \Omega \times \mathbf{R}_+, \\\mathbf{z}= \mathbf{w} =0,&\quad \mbox{in} \partial \Omega \times\mathbf{R}_0,\end{cases}$

其中 $\mathbf{z}=\mathbf{z}(x,t)$, $ \mathbf{w} = \mathbf{w} (x,t)$ 分别表示复标量核子场和实介子场; $\mathrm{i}=\sqrt{-1}$ 是虚数单位; $\alpha,\beta\in(0,1)$ 满足 $2\max\{\alpha,\beta\}<d$; $\hbar>0$ 为普朗克常数; $a>0$ 为系统的耗散机制; $b>0$ 是阻尼系数; $c>0$ 为介子的质量; $\mu, \nu>0$ 是耦合常数; $f\in\mathbf{C}$, $g\in\mathbf{R}$ 表示外力; 饱和函数 $ \mathbf{Sat} _1$ 和 $ \mathbf{Sat} _2$ 定义为

$ \mathbf{Sat} _1(\mathbf{z} \mathbf{w} )=\frac{\mathbf{z} \mathbf{w} }{1+\rho|\mathbf{z}|+\varrho| \mathbf{w} |},\quad \mathbf{Sat} _2(|\mathbf{z}|^2)=\frac{|\mathbf{z}|^{2}}{1+\varepsilon|\mathbf{z}|}, $

其中 $\mathbf{z}=\hat{\mathbf{z}}+\mathrm{i}\breve{\mathbf{z}}\in\mathbf{C}$, $|\mathbf{z}|=\sqrt{\hat{\mathbf{z}}^2+\check{\mathbf{z}}^2}$, $\rho, \varrho$ 和 $\varepsilon$ 为正饱和系数. 令 $\rho, \varrho, \varepsilon\rightarrow0$, 则方程 (1.1) 转化为非饱和方程, 这类方程在过去的几十年中被广泛研究, 参见文献 [1],[2],[3],[4],[5].

方程 (1.1) 的初值条件为

$\mathbf{z}(x,0)=\varphi(x)=\hat{\varphi}(x)+\mathrm{i}\breve{\varphi}(x),\quad \mathbf{w} (x,0)=\phi(x),\quad \lim\limits_{t\rightarrow0}\frac{\partial}{\partial t} \mathbf{w} (x,t)=\psi(x),\quad \mbox{在 $\Omega$ 中}. $

著名的 Schrödinger-Klein-Gordon 方程是量子场论中描述守恒的复中子场与实介子场相互作用的经典模型. 耦合的 Schrödinger-Klein-Gordon 方程, 因其相当精确地描述了物理系统关于时间的量子态, 而且这种耦合方程的发展为科学和工程领域开辟了新的前景, 因此这类方程吸引了许多学者的关注. 近年来 Schrödinger-Klein-Gordon 系统成为描述量子力学的重要工具, 被众多学者所研究, 参见文献 [3],[4],[5],[6],[7]. 另一方面, 问题 (1.1) 中的非线性项 $ \mathbf{Sat} _1$ 和 $ \mathbf{Sat} _2$ 可以看作一种特殊的"饱和非 Kerr 律", 出现在许多应用中. 饱和定律精确地描述了经过激光束的气体蒸气介电常数的变化, 这种现象也发生在光纤的孤子传播中. 在半导体掺杂光纤中, 孤子输运模型采用饱和非线性而不是通常的 Kerr 非线性. 这么做的主要动机是在半导体掺杂玻璃和其他复合材料可以观察到较弱的非线性. 饱和函数已经被广泛研究, 参见文献 [8],[9],[10],[11],[12].

Bochner 在其微分几何著作^[13]中首次提出了概自守的概念, 它是概周期的推广. 概自守的概念在研究一类常微分方程、抛物方程和广义微分方程的概自守动力学中是必不可少的, 这一事实进一步说明了概自守在动力学定性研究中的重要性. 首先, 就从系数空间到解空间的升力性质而言, 动力学在概周期的范畴内通常不是封闭的, 但是在概自守的范畴内是封闭的. 其次, 概自守动力学的出现说明了周期系统与概周期系统的主要区别. 例如, 在单调动力系统中, 周期系数的"升力"永远不可能是概自守的. 研究概自守动力学的另一个重要意义是它与 Levitan $N$-概周期性的关系. 因为概自守函数在本质上是 $N$-概周期的, 所以目前对概自守动力学的研究与 $N$-概周期的情况密切相关, 参见文献 [14],[15],[16],[17],[18],[19],[20].

纵观历史, 几乎全部现有的文章都集中在 Schrödinger 或 Klein-Gordon 方程的概周期动力学的研究上, 参见文献 [21],[22],[23],[24],[25],[26]. 例如, 利用 KAM 定理, 作者在文献 [21],[22],[23],[24] 中研究了一类特殊的 Schrödinger 方程的概周期解. Signing^[25] 考虑了具有概周期集的 Klein-Gordon 型方程的均匀化问题. Abdallah 等^[26]研究了具有非线性概周期项的 Klein-Gordon-Schrödinger 无限维格系统的一致全局吸引子的存在性. 然而, 据我们所知, 目前几乎没有文献涉及到 Schrödinger 或 Klein-Gordon 方程的概自守动力学研究. 受此启发, 本文将结合 Galerkin 方法、Fourier 级数、Laplace 变换和 Picard 迭代, 研究带有非局部 Laplace 算子的饱和 Schrödinger-Klein-Gordon 方程 (1.1) 在合适的函数空间中概自守弱解的存在性和唯一性.

1.1 非局部 Laplace 算子

设 $s\in(0,1)$, $\mathscr{S}$ 是速降函数组成的 Schwartz 空间, 分数阶 Laplace 算子^[27] $(-\Delta)^{s}:\mathscr{S}\rightarrow \mathrm{L}^2(\mathbf{R}^d)$ 表示为

$ \label{dy1} (-\Delta)^{s}{ \mathbf{u} }(x)=C(d,s)\lim\limits_{\epsilon\rightarrow 0^+}\int_{\mathbf{R}^d\setminus \mathbf{B}(x,\epsilon)} \frac{{ \mathbf{u} }(x)-{ \mathbf{u} }(y)}{|x-y|^{d+2s}} \mathrm{d}y,\quad\forall { \mathbf{u} }\in\mathscr{S}, x\in\mathbf{R}^d, $

其中 $\mathbf{B}(x,\epsilon)$ 表示球心为 $x\in\mathbf{R}^d$, 半径为 $\epsilon$ 的球, $C(d,s)$ 是标准化常数, 其表达式为

$ \frac{1}{C(d,s)}=\int_{\mathbf{R}^d}\frac{1-\cos\zeta_1}{|\zeta|^{d+2s}} \mathrm{d}\zeta, \quad \zeta=(\zeta_1,\zeta'), \zeta'\in\mathbf{R}^{d-1}. $

对 $s\in(0,1)$, $(-\Delta)^{s}$ 的一个等价形式为

$ \label{dy2} (-\Delta)^{s} \mathbf{u} (x)=-\frac{1}{2}C(d,s)\int_{\mathbf{R}^d} \frac{ \mathbf{u} (x+y)+ \mathbf{u} (x-y)-2 \mathbf{u} (x)}{|y|^{d+2s}} \mathrm{d}y,\quad \forall \mathbf{u} \in\mathscr{S}, x\in\mathbf{R}^d. $

设 $\mathrm{X}^s(\Omega)$ 表示所有 Lebesgue 可测函数 ${ \mathbf{u} }:\mathbf{R}^d\rightarrow\mathbf{R}$ 组成的线性空间, 使得在 $\mathrm{X}^s(\Omega)$ 中任意函数 ${ \mathbf{u} }$ 对 $\Omega$ 的限制属于 $\mathrm{L}^2(\Omega)$, 定义 $\mathrm{X}^s(\Omega)$ 的范数为

$ | \mathbf{u} |_{\mathrm{X}^s}=| \mathbf{u} |_{2} +\bigg(\int_{Q\times Q}\frac{| \mathbf{u} (x)- \mathbf{u} (y)|^2}{|x-y|^{d+2s}} \mathrm{d}x\mathrm{d}y\bigg)^{\frac{1}{2}}, \quad\forall \mathbf{u} \in\mathrm{X}^s(\Omega), $

其中 $|\cdot|_2$ 表示 $\mathrm{L}^2(\Omega)$ 的范数, $Q:=(\mathbf{R}^d\times \mathbf{R}^d)\backslash(\eth\Omega\times\eth\Omega)$, $\eth\Omega:=\mathbf{R}^d\backslash\Omega$. 定义

$\mathrm{X}_0^s(\Omega):=\{u\in\mathrm{X}^s(\Omega): \mathbf{u} =0\ \text{在}\ \mathbf{R}^d\backslash\Omega\ \text{中几乎处处成立}\},$

其范数为

$ | \mathbf{u} |_{\mathrm{X}_0^s}=\bigg(\int_{\mathbf{R}^d\times\mathbf{R}^d}\frac{| \mathbf{u} (x)- \mathbf{u} (y)|^2}{|x-y|^{d+2s}} \mathrm{d}x\mathrm{d}y\bigg)^{\frac{1}{2}},\quad \forall \mathbf{u} \in\mathrm{X}_0^s(\Omega), x\in\mathbf{R}^d.$

$\left(\mathrm{X}_0^s(\Omega), |\cdot|_{\mathrm{X}_0^s}\right)$ 是 Hilbert 空间, 其内积为

$ \big\langle \mathbf{u}, \mathbf{v} \big\rangle_{\mathrm{X}_0^s}= \int_{\mathbf{R}^d\times\mathbf{R}^d}\frac{[ \mathbf{u} (x)- \mathbf{u} (y)][ \mathbf{v} (x)- \mathbf{v} (y)]}{|x-y|^{d+2s}} \mathrm{d}x\mathrm{d}y, \quad \forall \mathbf{u}, \mathbf{v} \in\mathrm{X}_0^s(\Omega). $

定义空间 $\mathrm{L}_{s}^2(\Omega)=\{ \mathbf{u} :(-\Delta)^s \mathbf{u} \in\mathrm{L}^2(\Omega)\}$, $| \mathbf{u} |_{s,2}:=|(-\Delta)^s \mathbf{u} |_2$, $\forall{ \mathbf{u} }\in\mathrm{L}_{s}^2(\Omega)$.

考虑如下特征值问题

(1.2) $ \begin{cases} (-\Delta)^s{ \mathbf{u} }=\lambda { \mathbf{u} }, &\quad \mbox{在}\ \Omega\ \mbox{中}, \\[0.25cm]{ \mathbf{u} }=0,&\quad \mbox{在}\ \mathbf{R}^d\backslash\Omega\ \mbox{中},\end{cases}$

其中 $s\in(0,1)$, $d>2s$. 问题 (1.2) 的弱解表示为

$ \label{45.10} \int_{\mathbf{R}^d\times\mathbf{R}^d}\frac{[{ \mathbf{u} }(x)-{ \mathbf{u} }(y)][ \mathbf{v} (x)- \mathbf{v} (y)]}{|x-y|^{d+2s}} \mathrm{d}x\mathrm{d}y=\lambda\int_{\Omega}{ \mathbf{u} }(x) \mathbf{v} (x) \mathrm{d}x, \quad \forall \mathbf{v} \in \mathrm{X}_0^s(\Omega), $

其中 ${ \mathbf{u} }\in \mathrm{X}_0^s(\Omega)$. 根据文献 [28] 可知, 在 $\mathrm{L}^2(\Omega,\mathbf{R})$ 中存在对应于非局部 Laplace 算子 $(-\Delta)^{s}$ 的一个完备的规范正交基 $\{e_k\}_{k=1}^{\infty}\subseteq\mathrm{X}_0^s(\Omega)$. 同时, 它也是 $\mathrm{X}_0^s(\Omega)$ 的正交基. 设 $\{\lambda_k\}_{k=1}^{\infty}$ 为 $\{e_k\}_{k=1}^{\infty}$ 对应的特征值序列, 则 $0<\lambda_1\leq \lambda_2\leq \cdots\leq \lambda_k\leq \cdots$, $\lim\limits_{k\rightarrow\infty}\lambda_k=\infty$.

1.2 主要结果

设

$ \mathbb{E}_{\mathbf{z}}=\left\{ \mathbf{z}: \mathbf{z}\in\mathrm{L}_{\mathrm{loc}}^2(\mathbf{R}_+,\mathrm{X}_0^\alpha(\Omega)), \dot{\mathbf{z}}_t\in \mathrm{L}_{\mathrm{loc}}^2(\mathbf{R}_+,\mathrm{L}^2(\Omega)) \right\}, $

$ \mathbb{E}_{ \mathbf{w} }=\left\{ \mathbf{w} : \mathbf{w} \in\mathrm{L}_{\mathrm{loc}}^2(\mathbf{R}_+,\mathrm{X}_0^\beta(\Omega)), \dot{ \mathbf{w} }_t\in\mathrm{L}_{\mathrm{loc}}^2(\mathbf{R}_+,\mathrm{X}_0^\beta(\Omega)), \ddot{ \mathbf{w} }_t\in \mathrm{L}_{\mathrm{loc}}^2(\mathbf{R}_+,\mathrm{L}^2(\Omega)) \right\}, $

其中 $\dot{\mathbf{z}}_t:=\frac{\partial\mathbf{z}}{\partial t}$, $\dot{ \mathbf{w} }_t:=\frac{\partial \mathbf{w} }{\partial t}$, $\ddot{ \mathbf{w} }_t:=\frac{\partial^2 \mathbf{w} }{\partial t^2}$, $\forall t\in\mathbf{R}_0$.

定义1.1 若 $(\mathbf{z}, \mathbf{w} )\in \mathbb{E}_{\mathbf{z}}\times\mathbb{E}_{ \mathbf{w} }$ 满足

$\begin{eqnarray*} \left\{ \begin{array}{lll} \Big\langle\mathrm{i}\hbar\dot{\mathbf{z}}_t+ \mathrm{i}a\mathbf{z} + \mu \mathbf{Sat} _1(\mathbf{z} \mathbf{w} ),\xi\Big\rangle_{\mathrm{L}^2} -\frac{1}{2}c_{\alpha}\hbar^2\langle\mathbf{z},\xi\rangle_{\mathrm{X}_0^\alpha}=\langle f,\xi\rangle_{\mathrm{L}^2},\\ \Big\langle\ddot{ \mathbf{w} }_t+b\dot{ \mathbf{w} }_t+c \mathbf{w},\zeta\Big\rangle_{\mathrm{L}^2} +\frac{c_{\beta}}{2}\langle \mathbf{w},\zeta\rangle_{\mathrm{X}_0^\beta} =\big\langle \nu \mathbf{Sat} _2(|\mathbf{z}|^2)+g,\zeta\big\rangle_{\mathrm{L}^2},\\ \mathbf{z}(0)=\varphi,\quad \mathbf{w} (0)=\phi,\quad \dot{ \mathbf{w} }_t\big|_{t=0}=\psi, \end{array} \right. \end{eqnarray*}$

其中 $t\in\mathbf{R}_+$, $c_\alpha:=C(d,\alpha)$, $c_\beta:=C(d,\beta)$, $\xi, \zeta\in \mathrm{L}^2(\Omega)$. 则 $(\mathbf{z}, \mathbf{w} )\in \mathbb{E}_{\mathbf{z}}\times\mathbb{E}_{ \mathbf{w} }$ 是方程 (1.1) 的一个全局弱解.

利用 Galerkin 方法, 本章的第一个主要结果如下所示

定理1.1 假设如下条件成立.

$(\mathcal{K}_1)$ $ (\varphi,\phi,\psi)=(\hat{\varphi},\breve{\varphi},\phi,\psi) \in \{\mathrm{C}^2(\Omega)\cap\mathrm{X}_0^\alpha(\Omega)\cap\mathrm{L}_\alpha^2(\Omega)\}^2\times \{\mathrm{C}^2(\Omega)\cap\mathrm{X}_0^\beta(\Omega)\cap\mathrm{L}_{\beta}^2(\Omega)\}\times\mathrm{L}^2(\Omega).$

$(\mathcal{K}_2)$ $f, g\in \mathrm{L}^\infty(\mathbf{R}_0,\mathrm{L}^{2}(\Omega))\cap C^1(\mathbf{R}_0,\mathrm{L}^{2}(\Omega))$, $\dot{f}_t:=\frac{\partial f}{\partial t}, \dot{g}_t:=\frac{\partial g}{\partial t} \in \mathrm{L}_{\mathrm{loc}}^\infty(\mathbf{R}_0,\mathrm{L}^{2}(\Omega))$.

则方程 (1.1) 在 $\mathbb{E}_{\mathbf{z}}\times\mathbb{E}_{ \mathbf{w} }$ 中至少存在一个全局弱解 $(\mathbf{z}_0, \mathbf{w} _0)$.

设 $\{e_k^\alpha\}_{k=1}^{\infty}\subseteq\mathrm{X}_0^\alpha(\Omega)$ 和 $\{e_k^\beta\}_{k=1}^{\infty}\subseteq\mathrm{X}_0^\beta(\Omega)$ 为 $\mathrm{L}^2(\Omega,\mathbf{R})$ 中分别对应于非局部 Laplace 算子 $(-\Delta)^{\alpha}$ 和 $(-\Delta)^{\beta}$ 的两个完备的规范正交基. 同时, 它们也分别是 $\mathrm{X}_0^\alpha(\Omega)$ 和 $\mathrm{X}_0^\beta(\Omega)$ 的正交基. 设$\{\lambda_k^\alpha\}_{k=1}^{\infty}$ 和 $\{\lambda_k^\beta\}_{k=1}^{\infty}$ 分别为 $\{e_k^\alpha\}_{k=1}^{\infty}$ 和 $\{e_k^\beta\}_{k=1}^{\infty}$ 对应的特征值序列. 则 $\lambda_k^\alpha$ 和 $\lambda_k^\beta$ 具有有限的多重性且满足

$ 0<\lambda_1^\alpha\leq \lambda_2^\alpha\leq \cdots\leq \lambda_k^\alpha\leq \cdots,\quad \lim\limits_{k\rightarrow\infty}\lambda_k^\alpha=\infty, $

$ 0<\lambda_1^\beta\leq \lambda_2^\beta\leq \cdots\leq \lambda_k^\beta\leq \cdots,\quad \lim\limits_{k\rightarrow\infty}\lambda_k^\beta=\infty. $

定义 $(\cdot)_k:=\langle \cdot,e_k\rangle_{\mathrm{L}^2}$, $k\in\mathbf{Z}_+$, $\langle\cdot,\cdot\rangle_{\mathrm{L}^2}$ 表示 $\mathrm{L}^2(\Omega)$ 的内积. 根据状态估计、矩阵理论和 Laplace 变换, 可知 $(\mathbf{z}_0, \mathbf{w} _0)=(\hat{\mathbf{z}}_0,\breve{\mathbf{z}}_0, \mathbf{w} _0)$ 可以表示为如下形式的 Fourier 级数

(1.3) $ \begin{array}{l} {\left[\begin{array}{c} \hat{\mathbf{z}}_{0} \\ \breve{\mathbf{z}}_{0} \end{array}\right](t)=\sum_{k=1}^{\infty}\left\{e^{-\frac{a}{\hbar} t}\left[\begin{array}{cc} \cos \left(\dot{\lambda}_{k}^{\alpha} t\right) & \sin \left(\dot{\lambda}_{k}^{\alpha} t\right) \\ -\sin \left(\dot{\lambda}_{k}^{\alpha} t\right) & \cos \left(\dot{\lambda}_{k}^{\alpha} t\right) \end{array}\right]\left[\begin{array}{c} \hat{\varphi}_{k} \\ \breve{\varphi}_{k} \end{array}\right]\right.} \\ +\frac{1}{\hbar} \int_{0}^{t} e^{-\frac{a}{\hbar}(t-s)}\left[\begin{array}{cc} \cos \left(\dot{\lambda}_{k}^{\alpha}(t-s)\right) & \sin \left(\dot{\lambda}_{k}^{\alpha}(t-s)\right) \\ \sin \left(\dot{\lambda}_{k}^{\alpha}(s-t)\right) & \cos \left(\dot{\lambda}_{k}^{\alpha}(t-s)\right) \end{array}\right] \\ \left.\times\left[\begin{array}{c} -\mu \widehat{\mathbf{S a t}}_{1, k}\left(\mathbf{z}_{0} \mathbf{w}_{0}\right)+\operatorname{Im} f_{k} \\ \mu \widetilde{\mathbf{S a t}}_{1, k}\left(\mathbf{z}_{0} \mathbf{w}_{0}\right)-\operatorname{Re} f_{k} \end{array}\right](s) \mathrm{d} s\right\} e_{k}^{\alpha}, \\ \end{array} $

(1.4) $ \mathbf{w} _{0}(t)=\sum\limits_{k=1}^\infty\Bigg\{e^{-\frac{b}{2}t}\left[ \cos(\grave{\lambda}_k^\beta t) +\frac{b}{2\grave{\lambda}_k^\beta}\sin(\grave{\lambda}_k^\beta t) \right]\phi_k + \frac{1}{\grave{\lambda}_k^\beta}e^{-\frac{b}{2}t}\sin(\grave{\lambda}_k^\beta t)\psi_k \\ +\frac{1}{\grave{\lambda}_k^\beta}\int_{0}^te^{-\frac{b}{2}(t-s)}\sin(\grave{\lambda}_k^\beta(t-s)) \left[\nu \mathbf{Sat} _{2,k}(|\mathbf{z}_0(t)|^2)+g_k(s)\right]\mathrm{d}s\Bigg\}e_k^\beta, $

其中 $\mathbf{z}_0=\hat{\mathbf{z}}_0+\mathrm{i}\breve{\mathbf{z}}_0$, $t\in \mathbf{R}_+$,

$ \widehat{ \mathbf{Sat} }_1(\mathbf{z}_0 \mathbf{w} _0)=\frac{\hat{\mathbf{z}}_0 \mathbf{w} _0}{1+\rho|\mathbf{z}_0|+\varrho| \mathbf{w} _0|},\quad \widetilde{ \mathbf{Sat} }_1(\mathbf{z}_0 \mathbf{w} _0)=\frac{\breve{\mathbf{z}}_0 \mathbf{w} _0}{1+\rho|\mathbf{z}_0|+\varrho| \mathbf{w} _0|}, $

$ \acute{\lambda}_k^\alpha=\frac{1}{2}c_{\alpha}\hbar\lambda_k^\alpha,\quad \grave{\lambda}_k^\beta=\sqrt{\frac{1}{2}c_{\beta}\lambda_k^\beta+c-\frac{b^2}{4}}, \quad k=1,2,\cdots. $

基于式 (1.3)-(1.4), 可知方程 (1.1) 的全局弱解 $({z}_0,{w}_0)$ 具有如下的唯一性、能量估计和全局指数收敛性.

定理1.2 设 $(\mathcal{K}_1)$-$(\mathcal{K}_2)$ 及以下条件成立, 则 $(\mathbf{z}_0, \mathbf{w} _0)=(\hat{\mathbf{z}}_0,\breve{\mathbf{z}}_0, \mathbf{w} _0)$ 是方程 (1.1) 在 $\mathrm{C}(\mathbf{R}_0,\{\mathrm{L}^2(\Omega)\}^3) \cap\{\mathbb{E}_{\mathbf{z}}\times\mathbb{E}_{ \mathbf{w} }\}$ 中的唯一全局弱解.

$ \left(\mathcal{K}_{3}\right) b^{2}<2 c_{\beta} \lambda_{1}^{\beta}+4 c.$

而且, 若 $a\varrho>2\mu$, 那么我们有以下的能量估计

$ \max\limits_{t\in\mathbf{R}_0}|\mathbf{z}_0(t)|_2\leq \frac{2\varrho(a|\varphi|_{2}+f_{\infty})}{a\varrho-2\mu}:=\mathbf{z}_{\infty}, $

$ \max\limits_{t\in\mathbf{R}_0}| \mathbf{w} _0(t)|_{2} \leq\frac{2\grave{\lambda}_1^\beta+b}{2\grave{\lambda}_1^\beta}|\phi|_{2} +\frac{1}{\grave{\lambda}_1^\beta}|\psi|_{2} +\frac{2}{\grave{\lambda}_1^\beta b}\Big[\frac{\nu}{\varepsilon}\mathbf{z}_\infty+g_{\infty}\Big]:= \mathbf{w} _{\infty}, $

其中 $f_\infty:=\sup\limits_{t\in\mathbf{R}_0}|f(\cdot,t)|_{2}$, $g_\infty:=\sup\limits_{t\in\mathbf{R}_0}|g(\cdot,t)|_{2}$.

定理1.3 设 $(\mathcal{K}_1)$-$(\mathcal{K}_3)$ 以及如下假设成立.

$(\mathcal{K}_4)$ $ \min\left\{\frac{a}{\hbar},\frac{b}{2}\right\}>\max\left\{ \frac{64\mu^2}{a \hbar\rho_0^2},\frac{9\nu^2}{2b(\grave{\lambda}_1^\beta)^2\varepsilon^2}\right\}$, 其中 $\rho_0:=\min\{\rho,\varrho\}. $

则方程 (1.1) 在弱解意义下是全局指数收敛的.

最后, 为了研究方程 (1.1) 的渐近概自守弱解, 我们利用 (1.3)-(1.4) 构造了一个 Picard 迭代序列并有如下结论

定理1.4 设 $(\mathcal{K}_1)$-$(\mathcal{K}_4)$ 及如下假设成立.

$(\mathcal{K}_5)$ $(f,g)=(\mathrm{Re}f,\mathrm{Im}f,g)\in\mathbb{AAA}(\mathbf{R}_0,\{\mathrm{L}^2(\Omega)\}^3)$.

$(\mathcal{K}_6)$ $ \gamma:=\max\bigg\{\frac{8\sqrt{2}\mu}{a\rho_0}, \frac{3\nu}{b\grave{\lambda}_1^\beta\varepsilon}\bigg\}<1$.

则方程 (1.1) 存在唯一的全局指数稳定的渐近概自守弱解, 其中 $\mathbb{AAA}(\mathbf{R}_0,\{\mathrm{L}^2(\Omega)\}^3)$ 表示从 $\mathbf{R}_0$ 映到 $\{\mathrm{L}^2(\Omega)\}^3$ 的所有渐近概自守函数组成的空间, 其定义将在第 5 节中给出.

本文的结构安排如下. 在第 3 节中, 引入了一些基本函数空间, 并用 Galerkin 方法证明了方程 (1.1) 全局弱解的存在性. 第 4 节利用矩阵理论和 Laplace 变换构造了全局弱解的 Fourier 级数, 并证明了该解的唯一性和收敛性. 并且, 给出了定理 1.2 和 1.3 的证明. 在第 5 节中, 利用 Picard 迭代方法研究了方程 (1.1) 的唯一全局渐近概自守弱解的存在性.

2 能量估计和全局弱解

令 $I_T=(0,T)$, $\bar{I}_T=[T]$, $\forall T>0$, $\mathbb{K}$ 是某个有限维度量空间的子集, $\mathbb{H}$ 表示距离为 $\|\cdot\|_{\mathbb{H}}$ 的某个度量空间.

$\bullet$ $\mathrm{C}^m(\mathbb{K}):=\mathrm{C}^m(\mathbb{K},\mathbb{H})$ 表示从 $\mathbb{K}$ 映到 $\mathbb{H}$ 的所有 $m$ 阶连续可微函数组成的空间, 其范数为 $\|\cdot\|_{\mathrm{C}^m}$, $\forall m\in\mathbf{Z}_+$ 或 $m=\infty$. $\mathrm{C}_0^m(\mathbb{K}):=\{\eta\in\mathrm{C}^m(\mathbb{K}): \mathrm{supp}\eta\subset\subset\Omega\}$.

$\bullet$ $\mathrm{L}^p(\Omega):=\mathrm{L}^p(\Omega,\mathbf{C})$ 表示所有可测复值映射 $\eta:\Omega\rightarrow \mathbf{C}$ 组成的集合, 其范数为

$ |\eta|_{p}=\left[\int_{\Omega}|\eta(x)|^p \mathrm{d}x\right]^{\frac{1}{p}}. $

特别地, $\mathrm{L}^2(\Omega)$ 是 Hilbert 空间, 其内积为

$ \langle\eta,\eta\rangle_{\mathrm{L}^2}=\int_{\Omega}\eta(x)\bar{\eta}(x) \mathrm{d}x, $

相应的范数为 $|\eta|_2=\sqrt{\langle\eta,\eta\rangle_{\mathrm{L}^2}}$. 此外, $\mathrm{L}^\infty(\Omega)$ 表示 $\Omega$ 上所有本性有界映射组成的集合, 其范数为 $|\cdot|_{\infty}$.

$\bullet$ $\mathrm{L}^p(I_T):=\mathrm{L}^p(I_T,\mathbb{H})$ 表示所有的可测向量值映射 $\eta:I_T\rightarrow \mathbb{H}$ 组成的集合, 其范数为 $\|\cdot\|_{\mathrm{L}^p}$, 其中

$ \| \eta\|_{\mathrm{L}^p}=\left\{ \begin{array}{ll} \displaystyle\bigg[\int_{0}^T\| \eta(t)\|_{\mathbb{H}}^p \mathrm{d}t\bigg]^{\frac{1}{p}}, \quad& \hbox{if $p\in[1,\infty)$;} \\ \mathrm{ess}\sup\limits_{t\in I_T}\| \eta(t)\|_{\mathbb{H}}, & \hbox{if $p=\infty$.} \end{array} \right. $

此外, $\mathrm{L}_{\mathrm{loc}}^p(I_T,\mathbb{H})$ 为由 $\mathrm{L}_{\mathrm{loc}}^p(I_T,\mathbb{H})=\big\{ \eta\in \mathrm{L}^p(I_T): \forall T>0\big\}$ 定义的局部 Lebesgue 空间.

定义 $\mathbb{K}_n^\alpha=\mathrm{span}\{e_1^\alpha,\cdots,e_n^\alpha\},$ $\mathbb{K}_n^\beta=\mathrm{span}\{e_1^\beta,\cdots,e_n^\beta\}$, $n\in\mathbf{Z}_+$,

$ \hat{\mathbf{z}}_n(t)=\sum\limits_{k=1}^n\hat{\mathbf{z}}_{k,n}(t)e_k^\alpha,\quad \breve{\mathbf{z}}_n(t)=\sum\limits_{k=1}^n\breve{\mathbf{z}}_{k,n}(t)e_k^\alpha, $

$ \mathbf{z}_n(t)=\hat{\mathbf{z}}_n(t)+\mathrm{i}\breve{\mathbf{z}}_n(t)=\sum\limits_{k=1}^n\Big[\hat{\mathbf{z}}_{k,n}(t) +\mathrm{i}\breve{\mathbf{z}}_{k,n}(t)\Big]e_k^\alpha,\quad \mathbf{w} _n(t)=\sum\limits_{k=1}^n \mathbf{w} _{k,n}(t)e_k^\beta, $

其中 $(\mathbf{z}_n, \mathbf{w} _n)$ 是如下方程的解

(2.1) $ \begin{cases} \Big\langle\mathrm{i}\hbar\frac{\partial\mathbf{z}_n}{\partial t}+ \mathrm{i}a\mathbf{z}_n+ \mu \mathbf{Sat} _1(\mathbf{z}_n \mathbf{w} _n),\xi\Big\rangle_{\mathrm{L}^2}-\frac{1}{2}c_{\alpha}\hbar^2\langle\mathbf{z}_n,\xi\rangle_{\mathrm{X}_0^\alpha}=\langle f,\xi\rangle_{\mathrm{L}^2},\\ \Big\langle\frac{\partial^2 \mathbf{w} _n}{\partial t^2}+b\frac{\partial \mathbf{w} _n}{\partial t}+c \mathbf{w} _n,\zeta\Big\rangle_{\mathrm{L}^2}+\frac{1}{2}c_{\beta}\langle \mathbf{w} _n,\zeta\rangle_{\mathrm{X}_0^\beta}=\big\langle \nu \mathbf{Sat} _2(|\mathbf{z}_n|^2)+g,\zeta\big\rangle_{\mathrm{L}^2},\\ \mathbf{z}_n(0)=\varphi+\frac{\varphi}{n},\quad \mathbf{w} _n(0)=\phi+\frac{\phi}{n},\quad \frac{\partial \mathbf{w} _n(t)}{\partial t}\bigg|_{t=0}=\psi+\frac{\psi}{n},\end{cases}$

$\forall\xi\in \mathbb{K}_n^\alpha$, $\zeta\in \mathbb{K}_n^\beta$, $t\in\bar{I}_T$, $n\in\mathbf{Z}_+$. 根据常微分方程基本理论, 可知若 $(f,g)\in\mathrm{C}^1\times\mathrm{C}^1$, 则方程 (2.1) 存在唯一的解 $(\mathbf{z}_n, \mathbf{w} _n)\in \mathrm{C}^2\times\mathrm{C}^3$, $\forall t\in\bar{I}_T$, $n\in\mathbf{Z}_+$.

2.1 状态估计

定义 $f_{T_\infty}:=\sup\limits_{t\in\bar{I}_T}\{|f(\cdot,t)|_{2},|\dot{f}_t(\cdot,t)|_{2}\},$ $g_{T_\infty}:=\sup\limits_{t\in\bar{I}_T}\{|g(\cdot,t)|_{2},|\dot{g}_t(\cdot,t)|_{2}\}.$ 接着, 我们将给出方程 (2.1) 的一些状态估计.

引理2.1 若 $(\mathcal{K}_1)$-$(\mathcal{K}_2)$ 成立. 则存在某些正常数 $M_{\mathbf{z}}^T$, $M_{ \mathbf{w} }^T$, $M_{\mathbf{z}_\alpha}^T$, $M_{ \mathbf{w} _\beta}^T$, $M_{\dot{\mathbf{z}}}^T$, $M_{\dot{ \mathbf{w} }}^T$, $M_{\dot{ \mathbf{w} }_\beta}^T$, $M_{\ddot{ \mathbf{w} }}^T$ 使得

$\begin{cases} \left|\mathbf{z}_n(t)\right|_2^2\leq M_{\mathbf{z}}^T,\quad\left| \mathbf{w} _n(t)\right|_2^2\leq M_{ \mathbf{w} }^T,\quad \left|\frac{\partial}{\partial t}\mathbf{z}_n(t)\right|_2^2\leq M_{\dot{\mathbf{z}}}^T,\quad \left|\frac{\partial}{\partial t} \mathbf{w} _n(t)\right|_2^2\leq M_{\dot{ \mathbf{w} }}^T,\\ \left|\mathbf{z}_n(t)\right|_{\mathrm{X}_0^\alpha}^2\leq M_{\mathbf{z}_{\alpha}}^T,\quad\left| \mathbf{w} _n(t)\right|_{\mathrm{X}_0^\beta}^2\leq M_{ \mathbf{w} _{\beta}}^T,\quad \left|\frac{\partial}{\partial t} \mathbf{w} _n(t)\right|_{\mathrm{X}_0^\beta}^2\leq M_{\dot{ \mathbf{w} }_{\beta}}^T,\quad \left|\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\right|_2^2\leq M_{\ddot{ \mathbf{w} }}^T,\end{cases}$

$\forall t\in \bar{I}_T\backslash\{0\}$, $n\in\mathbf{Z}_+.$

证 1) $\{\mathbf{z}_n\}$, $\{ \mathbf{w} _n\}$, $\left\{\frac{\partial}{\partial t} \mathbf{w} _n\right\}$ 的 $\mathrm{L}^2$-估计和 $\{ \mathbf{w} _n\}$ 的 $\mathrm{X}_0^\beta$-估计.

在 (2.1) 式的第一个方程中令 $\xi={z}_n$, 取对应的虚部, 由插值不等式, 可得

(2.2) $ \frac{\hbar}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left|\mathbf{z}_{n}(t)\right|_{2}^{2}+a\left|\mathbf{z}_{n}(t)\right|_{2}^{2} \leq|f|_{2}\left|\mathbf{z}_{n}(t)\right|_{2} \leq \frac{1}{2}\left|\mathbf{z}_{n}(t)\right|_{2}^{2}+\frac{1}{2}|f|_{2}^{2},$

即

$\frac{\mathrm{d}}{\mathrm{d} t}\left|\mathbf{z}_{n}(t)\right|_{2}^{2} \leq \frac{1}{\hbar}\left|\mathbf{z}_{n}(t)\right|_{2}^{2}+\frac{1}{\hbar}|f|_{2}^{2} \Rightarrow\left|\mathbf{z}_{n}(t)\right|_{2}^{2} \leq 4|\varphi|_{2}^{2}+\frac{t}{\hbar} f_{T_{\infty}}^{2}+\frac{1}{\hbar} \int_{0}^{t}\left|\mathbf{z}_{n}(s)\right|_{2}^{2} \mathrm{~d} s, $

其中 $t\in \bar{I}_T\backslash\{0\}$, $n\in\mathbf{Z}_+$. 利用 Gronwall 不等式, 可得

$\left|\mathbf{z}_{n}(t)\right|_{2}^{2} \leq\left[4|\varphi|_{2}^{2}+\frac{T}{\hbar} f_{T_{\infty}}^{2}\right] e^{\frac{T}{\hbar}}:=M_{\mathbf{z}}^{T}, \quad \forall t \in \bar{I}_{T}, n \in \mathbf{Z}_{+} $

同理, 在 (2.1) 式的第二方程中取 $\zeta=\frac{\partial}{\partial t}{w}_n$, 有

$\begin{aligned} & \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left|\frac{\partial}{\partial t} \mathbf{w}_{n}(t)\right|_{2}^{2}+b\left|\frac{\partial}{\partial t} \mathbf{w}_{n}(t)\right|_{2}^{2}+\frac{c_{\beta}}{4} \frac{\mathrm{d}}{\mathrm{d} t}\left|\mathbf{w}_{n}(t)\right|_{\mathrm{X}_{0}^{\beta}}^{2}+\frac{c}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left|\mathbf{w}_{n}(t)\right|_{2}^{2} \\ \leq & {\left[\frac{\nu}{\varepsilon}\left|\mathbf{z}_{n}(t)\right|_{2}+g_{T_{\infty}}\right]\left|\frac{\partial}{\partial t} \mathbf{w}_{n}(t)\right|_{2} } \end{aligned}$

进而

(2.3) $ \frac{\mathrm{d}}{\mathrm{d}t}\left|\frac{\partial}{\partial t} \mathbf{w} _n(t)\right|_2^2 +\frac{c_{\beta}}{2}\frac{\mathrm{d}}{\mathrm{d}t}\left| \mathbf{w} _n(t)\right|_{\mathrm{X}_0^\beta}^2 +c\frac{\mathrm{d}}{\mathrm{d}t}\left| \mathbf{w} _n(t)\right|_2^2 \leq\frac{2}{b}\Big[\frac{\nu}{\varepsilon}|\mathbf{z}_n(t)|_{2}+g_{T_\infty}\Big]^2, $

其中 $t\in \bar{I}_T\backslash\{0\}$, $n\in\mathbf{Z}_+$. 因此,

(2.4) $ \begin{aligned} & \left|\frac{\partial}{\partial t} \mathbf{w}_{n}(t)\right|_{2}^{2}+\frac{c_{\beta}}{2}\left|\mathbf{w}_{n}(t)\right|_{\mathbf{X}_{0}^{\beta}}^{2}+c\left|\mathbf{w}_{n}(t)\right|_{2}^{2} \\ \leq & \frac{2 T}{b}\left[\frac{\nu}{\varepsilon} \sqrt{M_{\mathbf{z}}^{T}}+g_{T_{\infty}}\right]^{2}+4|\psi|_{2}^{2}+2 c_{\beta}|\phi|_{\mathbf{X}_{0}^{\beta}}^{2}+4 c|\phi|_{2}^{2}:=a_{1}, \end{aligned}$

其中 $t\in \bar{I}_T\backslash\{0\}$, $n\in\mathbf{Z}_+.$

由不等式 (2.4), 可知

$ \left|\frac{\partial}{\partial t} \mathbf{w} _n(t)\right|_2^2\leq a_1:=M_{\dot{ \mathbf{w} }}^T,\quad\left| \mathbf{w} _n(t)\right|_{\mathrm{X}_0^\beta}^2\leq \frac{2a_1}{c_{\beta}}:=M_{ \mathbf{w} _{\beta}}^T,\quad \left| \mathbf{w} _n(t)\right|_2^2\leq \frac{a_1}{c}:=M_{ \mathbf{w} }^T,$

其中 $t\in \bar{I}_T\backslash\{0\}$, $n\in\mathbf{Z}_+$.

2) $\left\{\frac{\partial}{\partial t}\mathbf{z}_n\right\}$ 的 $\mathrm{L}^2$-估计和 $\{\mathbf{z}_n\}$ 的 $\mathrm{X}_0^\alpha$-估计.

对 (2.1) 式的第一个方程两边同时取微分, 令 $\xi=\frac{\partial}{\partial t}{z}_n$, 取其虚部, 可得

(2.5) $ \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\left|\frac{\partial}{\partial t}\mathbf{z}_n(t)\right|_2^2 +a\left|\frac{\partial}{\partial t}\mathbf{z}_n(t)\right|_2^2 =\mathrm{Im}\left\langle-\mu\frac{\partial}{\partial t} \mathbf{Sat} _1(\mathbf{z}_n(t) \mathbf{w} _n(t))+\dot{f}_t, \frac{\partial\mathbf{z}_n(t)}{\partial t}\right\rangle_{\mathrm{L}^2}\qquad\quad $

对 $t\in \bar{I}_T\backslash\{0\}$ 几乎处处成立, 其中 $n\in\mathbf{Z}_+$. 通过直接计算可得

$\begin{aligned} & \left|\operatorname{Im}\left\langle\mu \frac{\partial}{\partial t} \mathbf{S a t}_{1}\left(\mathbf{z}_{n}(t) \mathbf{w}_{n}(t)\right), \frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right\rangle_{\mathrm{L}^{2}}\right| \\ \leq & \left|\mu \operatorname{Im}\left\langle\frac{\mathbf{z}_{n}(t) \frac{\partial \mathbf{w}_{n}(t)}{\partial t}}{1+\rho\left|\mathbf{z}_{n}(t)\right|+\varrho\left|\mathbf{w}_{n}(t)\right|}, \frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right\rangle_{\mathrm{L}^{2}}\right| \\ & +\left|\mu \operatorname{Im}\left\langle\frac{\rho \mathbf{z}_{n}^{2}(t) \mathbf{w}_{n}(t) \frac{\partial \overline{\mathbf{z}}_{n}(t)}{\partial t}}{2\left|\mathbf{z}_{n}(t)\right|\left(1+\rho\left|\mathbf{z}_{n}(t)\right|+\varrho\left|\mathbf{w}_{n}(t)\right|\right)^{2}}, \frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right\rangle_{\mathrm{L}^{2}}\right| \\ & +\left|\mu \operatorname{Im}\left\langle\frac{\varrho \mathbf{z}_{n}(t) \mathbf{w}_{n}^{2}(t) \frac{\partial \mathbf{w}_{n}(t)}{\partial t}}{\left|\mathbf{w}_{n}(t)\right|\left(1+\rho\left|\mathbf{z}_{n}(t)\right|+\varrho\left|\mathbf{w}_{n}(t)\right|\right)^{2}}, \frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right\rangle_{\mathrm{L}^{2}}\right| \\ \leq & \frac{2 \mu}{\rho}\left|\frac{\partial}{\partial t} \mathbf{w}_{n}(t)\right|_{2}\left|\frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right|_{2}+\frac{\mu}{2 \varrho}\left|\frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right|_{2}^{2} \\ \leq & \frac{(\rho+2 \varrho) \mu}{2 \rho \varrho}\left|\frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right|_{2}^{2}+\frac{\mu}{\rho} M_{\dot{\mathbf{w}}}^{T}, \quad \forall t \in \bar{I}_{T} \backslash\{0\} \text { a.e., } \end{aligned} $

对 $t\in \bar{I}_T\backslash\{0\}$ 几乎处处成立, 其中 $n\in\mathbf{Z}_+$. 结合 (2.5) 式可知

$\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left|\frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right|_{2}^{2} \leq\left[\frac{(\rho+2 \varrho) \mu}{2 \rho \varrho}+\frac{1}{2}\right]\left|\frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right|_{2}^{2}+\frac{\mu}{\rho} M_{\dot{\mathbf{w}}}^{T}+\frac{1}{2} f_{T_{\infty}}^{2} $

对 $t\in \bar{I}_T\backslash\{0\}$ 几乎处处成立, 其中 $n\in\mathbf{Z}_+$. 由 Gronwall 不等式可得

$\left|\frac{\partial}{\partial t}\mathbf{z}_n(t)\right|_2^2 \leq \left[\left|\dot{\mathbf{z}}_n(0)\right|_2^2 + \frac{2\mu}{\rho}M_{\dot{ \mathbf{w} }}^T+f_{T_\infty}^2\right] \exp\left\{\left[ \frac{(\rho+2\varrho)\mu}{\rho\varrho}+1\right]T\right\}, $

对 $t\in \bar{I}_T\backslash\{0\}$ 几乎处处成立, 其中 $n\in\mathbf{Z}_+$. 其中 $\dot{\mathbf{z}}_n(0):=\frac{\partial}{\partial t}\mathbf{z}_n(t)\Big|_{t=0}$, $n\in\mathbf{Z}_+$.

在 (2.1) 式的第一个方程中, 依次取 $t\rightarrow0$, $\xi=\dot{{z}}_n(0)$, 由分部积分公式可得

$\Big\langle\mathrm{i}\hbar\dot{\mathbf{z}}_n(0)+ \mathrm{i}a\varphi_n + \mu \mathbf{Sat} _1(\varphi_n\phi_n),\dot{\mathbf{z}}_n(0)\Big\rangle_{\mathrm{L}^2} -\hbar^2\langle(-\Delta)^\alpha\varphi_n,\dot{\mathbf{z}}_n(0)\rangle_{\mathrm{L}^2}=\langle f(\cdot,0),\dot{\mathbf{z}}_n(0)\rangle_{\mathrm{L}^2}, $

其中 $\varphi_n=\varphi+\frac{\varphi}{n}$, $\phi_n=\phi+\frac{\phi}{n}$, $n\in\mathbf{Z}_+$. 取相应的虚部可得

$\left|\dot{\mathbf{z}}_n(0)\right|_2 \leq \frac{1}{\hbar}\left[2 a|\varphi|_2+\frac{2 \mu}{\varrho}|\varphi|_2+2 \hbar^2|\varphi|_{\alpha, 2}+f_{T_{\infty}}\right]:=a_2, \quad \forall n \in \mathbf{Z}_{+}$

即

$|\dot{\mathbf{z}}_n(0)|_2\leq \frac{1}{\hbar}\left[ 2a|\varphi|_2 +\frac{2\mu}{\varrho}|\varphi|_2 +2\hbar^2|\varphi|_{\alpha,2}+f_{T_\infty}\right]:=a_2,\quad \forall n\in\mathbf{Z}_+. $

利用不等式 (2.6) 可知

$\left|\frac{\partial}{\partial t}\mathbf{z}_n(t)\right|_2^2 \leq \left[a_2^2 + \frac{2\mu}{\rho}M_{\dot{ \mathbf{w} }}^T+f_{T_\infty}^2\right] \exp\left\{\left[ \frac{(\rho+2\varrho)\mu}{\rho\varrho}+1\right]T\right\}:=M_{\dot{\mathbf{z}}}^T, $

其中 $t\in \bar{I}_T\backslash\{0\}$, $n\in\mathbf{Z}_+$.

在 (2.1) 式的第一个方程中, 令 $\xi=\hbar\frac{\partial}{\partial t}{z}_n+a{z}_n$, 并取其实部可得

(2.7) $ \begin{aligned} \frac{1}{2} c_{\alpha} \hbar^{2} \operatorname{Re}\left\langle\mathbf{z}_{n}(t), \hbar \frac{\partial}{\partial t} \mathbf{z}_{n}(t)+a \mathbf{z}_{n}(t)\right\rangle_{\mathbf{X}_{0}^{\alpha}}= & \operatorname{Re}\left\langle\mu \operatorname{Sat}_{1}\left(\mathbf{z}_{n}(t) \mathbf{w}_{n}(t)\right), \hbar \frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right\rangle_{\mathrm{L}^{2}} \\ & +\operatorname{Re}\left\langle\mu \mathbf{S a t}_{1}\left(\mathbf{z}_{n}(t) \mathbf{w}_{n}(t)\right), a \mathbf{z}_{n}(t)\right\rangle_{\mathrm{L}^{2}} \\ & -\operatorname{Re}\left\langle f, \hbar \frac{\partial}{\partial t} \mathbf{z}_{n}(t)+a \mathbf{z}_{n}(t)\right\rangle_{\mathrm{L}^{2}}, \end{aligned} $

其中

$\begin{aligned} \operatorname{Re}\left\langle\mu \operatorname{Sat}_{1}\left(\mathbf{z}_{n}(t) \mathbf{w}_{n}(t)\right), \hbar \frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right\rangle_{\mathrm{L}^{2}} & \leq \mu \hbar\left|\mathbf{S a t}_{1}\left(\mathbf{z}_{n}(t) \mathbf{w}_{n}(t)\right)\right|_{2}\left|\frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right|_{2} \\ & \leq \frac{\mu \hbar}{\varrho}\left|\mathbf{z}_{n}(t)\right|_{2}\left|\frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right|_{2} \leq \frac{\mu \hbar}{\varrho} \sqrt{M_{\mathbf{z}}^{T}} \sqrt{M_{\dot{\mathbf{z}}}^{T}}, \\ \operatorname{Re}\left\langle\mu \mathbf{S a t}_{1}\left(\mathbf{z}_{n}(t) \mathbf{w}_{n}(t)\right), a \mathbf{z}_{n}(t)\right\rangle_{\mathrm{L}^{2}} & \leq a \mu\left|\mathbf{S a t}_{1}\left(\mathbf{z}_{n}(t) \mathbf{w}_{n}(t)\right)\right|_{2}\left|\mathbf{z}_{n}(t)\right|_{2} \\ & \leq \frac{a \mu}{\varrho}\left|\mathbf{z}_{n}(t)\right|_{2}^{2} \leq \frac{a \mu}{\varrho} M_{\mathbf{z}}^{T}, \\ -\operatorname{Re}\left\langle f, \hbar \frac{\partial}{\partial t} \mathbf{z}_{n}(t)+a \mathbf{z}_{n}(t)\right\rangle_{\mathrm{L}^{2}} & \leq \hbar|f|_{2}\left|\frac{\partial}{\partial t} \mathbf{z}_{n}(t)\right|_{2}+a|f|_{2}\left|\mathbf{z}_{n}(t)\right|_{2} \\ & \leq \hbar f_{T_{\infty}} \sqrt{M_{\dot{\mathbf{z}}}^{T}}+a f_{T_{\infty}} \sqrt{M_{\mathbf{z}}^{T}}, \end{aligned} $

其中 $t\in \bar{I}_T\backslash\{0\}$, $n\in\mathbf{Z}_+.$

将上述结果代入式 (2.7) 可得

(2.8) $ \frac{c_{\alpha}\hbar^3}{4}\frac{\mathrm{d}}{\mathrm{d}t}|\mathbf{z}_n(t)|_{\mathrm{X}_0^\alpha}^2 +\frac{ac_{\alpha}\hbar^2}{2}|\mathbf{z}_n(t)|_{\mathrm{X}_0^\alpha}^2\leq a_3,\quad \forall t\in \bar{I}_T\backslash\{0\}, n\in\mathbf{Z}_+, $

其中

$ a_3:=\frac{\mu\hbar}{\varrho}\sqrt{M_{\mathbf{z}}^T}\sqrt{M_{\dot{\mathbf{z}}}^T}+ \frac{a\mu}{\varrho}M_{\mathbf{z}}^T+ \hbar f_{T_\infty}\sqrt{M_{\dot{\mathbf{z}}}^T}+af_{T_\infty}\sqrt{M_{\mathbf{z}}^T}. $

利用不等式 (2.8) 可得

$|\mathbf{z}_n(t)|_{\mathrm{X}_0^\alpha}^2\leq 4|\varphi|_{\mathrm{X}_0^\alpha}^2+ \frac{4a_3T}{c_{\alpha}\hbar^3}:=M_{\mathbf{z}_\alpha}^T,\quad \forall t\in \bar{I}_T\backslash\{0\}, n\in\mathbf{Z}_+. $

3) $\left\{\frac{\partial^2}{\partial t^2} \mathbf{w} _n\right\}$ 的 $\mathrm{L}^2$-估计和 $\left\{\frac{\partial}{\partial t} \mathbf{w} _n\right\}$ 的 $\mathrm{X}_0^\beta$-估计.

对 (2.1) 式的第二个方程两边同时微分, 取 $\zeta=\frac{\partial^2}{\partial t^2}{w}_n$, 有

(2.9) $ \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\left|\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\right|_2^2 +b\left|\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\right|_2^2 +\frac{c}{2}\frac{\mathrm{d}}{\mathrm{d}t}\left|\frac{\partial}{\partial t} \mathbf{w} _n(t)\right|_2^2 +\frac{c_{\beta}}{4}\frac{\mathrm{d}}{\mathrm{d}t} \left|\frac{\partial}{\partial t} \mathbf{w} _n(t)\right|_{\mathrm{X}_0^\beta}^2 \\ =\left\langle\nu\frac{\partial}{\partial t} \mathbf{Sat} _2(|\mathbf{z}_n(t)|^2) +\dot{g}_t,\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\right\rangle_{\mathrm{L}^2} \\ \leq \nu\left|\left\langle\frac{(2+\varepsilon|\mathbf{z}_n(t)|)\mathbf{z}_n(t)}{2(1+\varepsilon|\mathbf{z}_n(t)|)^2} \frac{\partial}{\partial t}\bar{\mathbf{z}}_n(t),\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\right\rangle_{\mathrm{L}^2} \right| \\ +\nu\left|\left\langle\frac{(2+\varepsilon|\mathbf{z}_n(t)|)\bar{\mathbf{z}}_n(t)}{2(1+\varepsilon|\mathbf{z}_n(t)|)^2} \frac{\partial}{\partial t}\mathbf{z}_n(t),\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\right\rangle_{\mathrm{L}^2} \right|+g_{T_\infty}\left|\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\right|_2 \\ \leq \frac{2\nu}{\varepsilon}\left|\frac{\partial}{\partial t}\mathbf{z}_n(t)\right|_2 \left|\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\right| +g_{T_\infty}\left|\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\right| \\ \leq \frac{8\nu^2}{b\varepsilon^2}\left|\frac{\partial}{\partial t}\mathbf{z}_n(t)\right|_2^2 +b\left|\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\right|^2+\frac{2g_{T_\infty}^2}{b} \\ \leq \frac{8\nu^2}{b\varepsilon^2}M_{\dot{\mathbf{z}}}^T +\frac{2g_{T_\infty}^2}{b}+b\left|\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\right|^2 \\ :=a_4+b\left|\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\right|^2, $

对 $t\in \bar{I}_T\backslash\{0\}$ 几乎处处成立, 其中 $n\in\mathbf{Z}_+.$

定义 $\ddot{ \mathbf{w} }_n(0):=\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\Big|_{t=0}$, $\forall n\in\mathbf{Z}_+$. 在 (2.1) 式的第二个方程中依次令 $t\rightarrow0$, $\zeta=\ddot{{w}}_n(0)$ 可得

$\begin{aligned} \left|\ddot{\mathbf{w}}_{n}(0)\right|_{2}^{2} & \leq\left[b\left|\psi_{n}\right|_{2}+c\left|\phi_{n}\right|_{2}+\left|(-\Delta)^{\beta} \phi_{n}\right|_{2}+\nu\left|\mathbf{S a t}_{2}\left(\left|\varphi_{n}\right|^{2}\right)\right|_{2}+g_{T_{\infty}}\right]\left|\ddot{\mathbf{w}}_{n}(0)\right|_{2} \\ & \leq\left[2 b|\psi|_{2}+2 c|\phi|_{2}+2|\phi|_{\beta, 2}+\frac{2 \nu}{\varepsilon}|\varphi|_{2}+g_{T_{\infty}}\right]\left|\ddot{\mathbf{w}}_{n}(0)\right|_{2}, \end{aligned}$

其中

$\varphi_n=\varphi+\frac{\varphi}{n},\quad \phi_n=\phi+\frac{\phi}{n},\quad \psi_n=\psi+\frac{\psi}{n}, n\in\mathbf{Z}_+.$

因此,

$|\ddot{ \mathbf{w} }_n(0)|_2\leq 2b|\psi|_2+2c|\phi|_2 +2|\phi|_{\beta,2} +\frac{2\nu}{\varepsilon}|\varphi|_2 +g_{T_\infty}:=a_5,\quad \forall n\in\mathbf{Z}_+. $

由 (2.9) 式可知

$ \left|\frac{\partial^2}{\partial t^2} \mathbf{w} _n(t)\right|_2^2\leq 2a_4T+a_5^2+4c\psi^2+2c_\beta|\psi|_{\mathrm{X}_0^{\beta}}^2:=M_{\ddot{ \mathbf{w} }}^T,\quad \left|\frac{\partial}{\partial t} \mathbf{w} _n(t)\right|_{\mathrm{X}_0^\beta}^2\leq \frac{2M_{\ddot{ \mathbf{w} }}^T}{c_\beta}:=M_{\dot{ \mathbf{w} }_{\beta}}^T,$

其中 $t\in \bar{I}_T\backslash\{0\}$, $n\in\mathbf{Z}_+.$ 证毕.

2.2 全局弱解

定理1.1 的证明 由引理 2.1 可得

$\begin{cases} \{{\mathbf{z}}_n\} \text{ 在 }\mathrm{L}^\infty(I_T,\mathrm{L}^2(\Omega)\cap\mathrm{X}_0^\alpha(\Omega))\text{ 中有界};\\[2mm] \left\{\frac{\partial}{\partial t}{\mathbf{z}}_n\right\} \text{ 在 }\mathrm{L}^\infty(I_T,\mathrm{L}^2(\Omega))\text{ 中有界};\\[2mm] \{{ \mathbf{w} }_n\} \text{ 在 }\mathrm{L}^\infty(I_T,\mathrm{L}^2(\Omega)\cap\mathrm{X}_0^\beta(\Omega))\text{ 中有界};\\[2mm] \left\{\frac{\partial}{\partial t}{ \mathbf{w} }_n\right\} \text{ 在 }\mathrm{L}^\infty(I_T,\mathrm{L}^2(\Omega)\cap\mathrm{X}_0^\beta(\Omega))\text{ 中有界};\\ \left\{\frac{\partial^2}{\partial t^2}{ \mathbf{w} }_n\right\} \text{ 在 }\mathrm{L}^\infty(I_T,\mathrm{L}^2(\Omega))\text{ 中有界}.\end{cases}$

因此, 存在 $\{({\mathbf{z}}_n,{ \mathbf{w} }_n)\}_{n\in\mathbf{Z}_+}$ 的一个子列 (仍记为 $\{({\mathbf{z}}_n,{ \mathbf{w} }_n)\}_{n\in\mathbf{Z}_+}$) 使得当 $n\rightarrow\infty$ 时, 有

$\begin{cases} \text{在$\mathrm{L}_{\mathrm{loc}}^2(\mathbf{R}_+,\mathrm{L}^2(\Omega)\cap\mathrm{X}_0^\alpha(\Omega))$ 中 }{\mathbf{z}}_n\stackrel{}{\rightharpoonup}{\mathbf{z}}_0;\\[2mm] \text{在 $\mathrm{L}_{\mathrm{loc}}^2(\mathbf{R}_+,\mathrm{L}^2(\Omega))$ 中 }\frac{\partial}{\partial t}{\mathbf{z}}_n\stackrel{}{\rightharpoonup}\frac{\partial}{\partial t}{\mathbf{z}}_0;\\[2mm] \text{在$\mathrm{L}_{\mathrm{loc}}^2(\mathbf{R}_+,\mathrm{L}^2(\Omega)\cap\mathrm{X}_0^\beta(\Omega))$ 中 }{ \mathbf{w} }_n\stackrel{}{\rightharpoonup}{ \mathbf{w} }_0;\\[2mm] \text{在 $\mathrm{L}_{\mathrm{loc}}^2(\mathbf{R}_+,\mathrm{L}^2(\Omega)\cap\mathrm{X}_0^\beta(\Omega))$ 中 }\frac{\partial}{\partial t}{ \mathbf{w} }_n\stackrel{}{\rightharpoonup}\frac{\partial}{\partial t}{ \mathbf{w} }_0;\\ \text{在$\mathrm{L}_{\mathrm{loc}}^2(\mathbf{R}_+,\mathrm{L}^2(\Omega))$ 中 }\frac{\partial^2}{\partial t^2}{ \mathbf{w} }_n \stackrel{}{\rightharpoonup}\frac{\partial^2}{\partial t^2}{ \mathbf{w} }_0.\end{cases}$

因为 $\mathrm{X}_0^\alpha(\Omega)\hookrightarrow\mathrm{L}^2(\Omega)$ 和 $\mathrm{X}_0^\beta(\Omega)\hookrightarrow\mathrm{L}^2(\Omega)$ 是紧的, 所以可以选取 $\{({\mathbf{z}}_n,{ \mathbf{w} }_n)\}_{n\in\mathbf{Z}_+}$ 的一个子列 (仍记为 $\{({\mathbf{z}}_n,{ \mathbf{w} }_n)\}_{n\in\mathbf{Z}_+}$) 使得当 $n\rightarrow\infty$ 时, 在 $\mathrm{L}^2(\Omega)$ 中对每个 $t\in\mathbf{R}_+$ 有

${\mathbf{z}}_n\rightarrow{\mathbf{z}}_0,\quad { \mathbf{w} }_n\rightarrow{ \mathbf{w} }_0. $

由于在 $\mathrm{L}^2(\Omega)$ 中, 当 $n\rightarrow\infty$ 时, 有${z}_n\rightarrow{z}_0$. 因此当 $n\rightarrow\infty$ 时, 有

$ \|{\mathbf{z}}_0(t)\|_{2}\leq\|{\mathbf{z}}_n(t)\|_{2}+\|{\mathbf{z}}_n(t)-{\mathbf{z}}_0(t)\|_{2} \leq \sqrt{M_{{\mathbf{z}}}^T}+\|{\mathbf{z}}_n(t)-{\mathbf{z}}_0(t)\|_{2}\rightarrow \sqrt{M_{{\mathbf{z}}}^T}, $

即 $\|{\mathbf{z}}_0(t)\|_{2}^2\leq M_{{\mathbf{z}}}^T$, $\forall t\in\bar{I}_T$.

注意到

$ |\langle {\mathbf{z}}_n(t),{\mathbf{z}}_0(t)\rangle_{\mathrm{X}_0^\alpha}| \leq \|{\mathbf{z}}_n(t)\|_{\mathrm{X}_0^\alpha}\|{\mathbf{z}}_0(t)\|_{\mathrm{X}_0^\alpha}\leq \sqrt{M_{{\mathbf{z}}_\alpha}^T}\|{\mathbf{z}}_0(t)\|_{\mathrm{X}_0^\alpha},\quad n\in\mathbf{Z}_+, t\in\bar{I}_T\backslash\{0\}. $

在上述不等式中令 $n\rightarrow\infty$ 可得

$\|{\mathbf{z}}_0(t)\|_{\mathrm{X}_0^\alpha}^2\leq M_{{\mathbf{z}}_\alpha}^T,\quad \forall t\in\bar{I}_T.$

同理, 有

$\|{ \mathbf{w} }_0(t)\|_{2}^2\leq M_{{ \mathbf{w} }}^T,\quad \|{ \mathbf{w} }_0(t)\|_{\mathrm{X}_0^\beta}^2\leq M_{{ \mathbf{w} }_\beta}^T,\quad \forall t\in\bar{I}_T.$

由于

$\left| \mathbf{Sat} _1(\mathbf{z}_n \mathbf{w} _n)- \mathbf{Sat} _1(\mathbf{z}_0 \mathbf{w} _0)\right| =\bigg|\frac{(\mathbf{z}_n-\mathbf{z}_0) \mathbf{w} _n}{1+\rho|\mathbf{z}_n|+\varrho| \mathbf{w} _n|}+\frac{\mathbf{z}_0( \mathbf{w} _n- \mathbf{w} _0)}{1+\rho|\mathbf{z}_0|+\varrho| \mathbf{w} _0|} \\ +\frac{\mathbf{z}_0 \mathbf{w} _n}{1+\rho|\mathbf{z}_n|+\varrho| \mathbf{w} _n|}-\frac{\mathbf{z}_0 \mathbf{w} _n}{1+\rho|\mathbf{z}_0|+\varrho| \mathbf{w} _0|}\bigg| \\ \leq \frac{2}{\varrho}|\mathbf{z}_n-\mathbf{z}_0|+\frac{2}{\rho}| \mathbf{w} _n- \mathbf{w} _0|, $

从而,

$ \begin{aligned} \Big|\left\langle \mathbf{Sat} _1(\mathbf{z}_n \mathbf{w} _n)- \mathbf{Sat} _1(\mathbf{z}_0 \mathbf{w} _0),\xi\right\rangle_{\mathrm{L}^2}\Big| &\leq&\left|\frac{2}{\varrho} \left\langle|\mathbf{z}_n-\mathbf{z}_0|,|\xi|\right\rangle_{\mathrm{L}^2} \right| +\left|\frac{2}{\rho} \left\langle| \mathbf{w} _n- \mathbf{w} _0|,|\xi|\right\rangle_{\mathrm{L}^2} \right| \\ &\leq&\frac{2}{\varrho} |\mathbf{z}_n-\mathbf{z}_0|_2|\xi|_2 +\frac{2}{\rho} | \mathbf{w} _n- \mathbf{w} _0|_2|\xi|_2, \end{aligned} $

即

$ \lim\limits_{n\rightarrow\infty}\left\langle \mathbf{Sat} _1(\mathbf{z}_n \mathbf{w} _n)- \mathbf{Sat} _1(\mathbf{z}_0 \mathbf{w} _0),\xi\right\rangle_{\mathrm{L}^2}=0,\quad \forall\xi \in \mathbb{K}_m^\alpha, m\in\mathbf{Z}_+. $

注意到 $\bigcup\limits_{m=1}^{\infty}\mathbb{K}_m^\alpha $ 在 $\mathrm{L}^2(\Omega)$ 中稠密, 所以, 当 $n\rightarrow\infty$ 时, 在 $\mathrm{L}^2(\Omega)$ 中有

$ \mathbf{Sat} _1(\mathbf{z}_n \mathbf{w} _n)\rightharpoonup \mathbf{Sat} _1(\mathbf{z}_0 \mathbf{w} _0). $

同理, 当 $n\rightarrow\infty$ 时, 在 $\mathrm{L}^2(\Omega)$ 中有 $ \mathbf{Sat} _2(|\mathbf{z}_n|^2)\rightharpoonup \mathbf{Sat} _2(|\mathbf{z}_0|^2)$.

对 (2.1) 式的两边同时取极限, 利用 Lebesgue 控制收敛定理以及 $\bigcup\limits_{n=1}^{\infty}\mathbb{K}_n^{\alpha} $ 和 $\bigcup\limits_{n=1}^{\infty}\mathbb{K}_n^{\beta}$ 在 $\mathrm{L}^2(\Omega)$ 中稠密, 可得

$\begin{matrix} \left\{ \begin{array}{lll} \left\langle\mathrm{i}\hbar\frac{\partial}{\partial t}\mathbf{z}_0+ \mathrm{i}a\mathbf{z}_0 + \mu \mathbf{Sat} _1(\mathbf{z}_0 \mathbf{w} _0)-f,\xi\right\rangle_{\mathrm{L}^2} =\frac{1}{2}c_{\alpha}\hbar^2\langle\mathbf{z}_0,\xi\rangle_{\mathrm{X}_0^\alpha},\\ \left\langle\left(\frac{\partial^2}{\partial t^2}+b\frac{\partial}{\partial t}+c\right) \mathbf{w} _0 - \nu \mathbf{Sat} _2(|\mathbf{z}_0|^2)-g,\zeta\right\rangle_{\mathrm{L}^2} +\frac{1}{2}c_{\beta}\langle \mathbf{w} _0,\zeta\rangle_{\mathrm{X}_0^\beta} =0,\\ \mathbf{z}_0(0)=\varphi,\quad \mathbf{w} _0(0)=\phi,\quad \frac{\partial \mathbf{w} _0(t)}{\partial t}\bigg|_{t=0}=\psi, \end{array} \right. \end{matrix}$

其中 $\xi, \zeta\in\mathrm{L}^2(\Omega)$, $t\in \mathbf{R}_+$. 显然, $(\mathbf{z}_0, \mathbf{w} _0)\in \mathbb{E}_{\mathbf{z}}\times\mathbb{E}_{ \mathbf{w} }$ 是方程 (2.1) 的一个全局弱解 (见定义 1.1). 证毕.

3 Fourier 级数, 唯一性和收敛性

3.1 $(\mathbf{z}_0, \mathbf{w} _0)$ 的 Fourier 级数

定义 $\mathbf{z}_0(t)=\hat{\mathbf{z}}_0(t)+\mathrm{i}\breve{\mathbf{z}}_0(t)$, $ \mathbf{w} _0(t)=\sum\limits_{k=1}^\infty \mathbf{w} _{k,0}(t)e_k^\beta$, 其中

$ \hat{\mathbf{z}}_0(t)=\sum\limits_{k=1}^\infty\hat{\mathbf{z}}_{k,0}(t)e_k^\alpha,\quad \breve{\mathbf{z}}_0(t)=\sum\limits_{k=1}^n\breve{\mathbf{z}}_{k,0}(t)e_k^\alpha,\quad\forall t\in \mathbf{R}_+. $

因为在 $\mathrm{X}_0^\alpha(\Omega)$ 中, 当 $n\rightarrow\infty$ 时有 ${z}_n\stackrel{}{\rightharpoonup}{z}_0$, 所以

$\langle \mathbf{z}_0, e_k^\alpha\rangle_{\mathrm{X}_0^\alpha} =\lim\limits_{n\rightarrow\infty}\langle \mathbf{z}_n, e_k^\alpha\rangle_{\mathrm{X}_0^\alpha} =\lambda_k^\alpha\lim\limits_{n\rightarrow\infty}\langle\mathbf{z}_n, e_k^\alpha\rangle_{\mathrm{L}^2} =\lambda_k^\alpha\langle\mathbf{z}_0, e_k^\alpha\rangle_{\mathrm{L}^2},\quad k=1,2,\cdots. $

同理, $\langle \mathbf{w} _0, e_k^\beta\rangle_{\mathrm{X}_0^\beta} =\lambda_k^\beta\langle \mathbf{w} _0, e_k^\beta\rangle_{\mathrm{L}^2}$, $\forall k=1,2,\cdots $. 因此, 模型 (2.10) 的第一个方程可转化为

$\begin{eqnarray*} \left\{ \begin{array}{lll} \left\langle\hbar\frac{\partial}{\partial t}\hat{\mathbf{z}}_0(t),e_k^\alpha\right\rangle_{\mathrm{L}^2}&=& \frac{1}{2}c_{\alpha}\hbar^2\lambda_k^\alpha\langle\breve{\mathbf{z}}_0(t),e_k^\alpha\rangle_{\mathrm{L}^2} -a\langle\hat{\mathbf{z}}_0(t),e_k^\alpha\rangle_{\mathrm{L}^2}\\ && -\mu\langle \widehat{ \mathbf{Sat} }_1(\mathbf{z}_0(t) \mathbf{w} _0(t)),e_k^\alpha\rangle_{\mathrm{L}^2}+\langle\mathrm{Im} f(x,t),e_k^\alpha\rangle_{\mathrm{L}^2},\\ \left\langle\hbar\frac{\partial}{\partial t}\breve{\mathbf{z}}_0(t),e_k^\alpha\right\rangle_{\mathrm{L}^2}&=& -\frac{1}{2}c_{\alpha}\hbar^2\lambda_k^\alpha\langle\hat{\mathbf{z}}_0(t),e_k^\alpha\rangle_{\mathrm{L}^2} -a\langle\breve{\mathbf{z}}_0(t),e_k^\alpha\rangle_{\mathrm{L}^2}\\ && +\mu\langle\widetilde{ \mathbf{Sat} }_1(\mathbf{z}_0(t) \mathbf{w} _0(t)),e_k^\alpha\rangle_{\mathrm{L}^2}-\langle\mathrm{Re} f(x,t),e_k^\alpha\rangle_{\mathrm{L}^2} \end{array} \right. \end{eqnarray*}$

等价于

(3.1) $ \frac{\mathrm{d}}{\mathrm{d}t}\left[\begin{array}{c}\hat{\mathbf{z}}_{k,0} \\\breve{\mathbf{z}}_{k,0} \\\end{array}\right](t)=\frac{1}{\hbar}\left[\begin{array}{cc}-a & \frac{1}{2}c_{\alpha}\hbar^2\lambda_k^\alpha \\ -\frac{1}{2}c_{\alpha}\hbar^2\lambda_k^\alpha & -a \\\end{array}\right]\left[\begin{array}{c}\hat{\mathbf{z}}_{k,0}\\\breve{\mathbf{z}}_{k,0} \\\end{array}\right](t)+\frac{1}{\hbar}\left[\begin{array}{c}-\mu\widehat{ \mathbf{Sat} }_{1,k}(\mathbf{z}_0 \mathbf{w} _0)+\mathrm{Im}f_k \\\mu\widetilde{ \mathbf{Sat} }_{1,k}(\mathbf{z}_0 \mathbf{w} _0)-\mathrm{Re}f_k \\\end{array}\right](t),$

其中 $\hat{\mathbf{z}}_{k,0}(0)=\hat{\varphi}_k,$ $\breve{\mathbf{z}}_{k,0}(0)=\breve{\varphi}_k$, $t\in \mathbf{R}_+$, $k=1,2,\cdots $.

结合方程 (3.1) 可得

(3.2) $ \begin{aligned} {\left[\begin{array}{l} \hat{\mathbf{z}}_{k, 0} \\ \breve{\mathbf{z}}_{k, 0} \end{array}\right](t)=} & e^{-\frac{a}{\hbar} t}\left[\begin{array}{cc} \cos \left(\dot{\lambda}_{k}^{\alpha} t\right) & \sin \left(\dot{\lambda}_{k}^{\alpha} t\right) \\ -\sin \left(\dot{\lambda}_{k}^{\alpha} t\right) & \cos \left(\dot{\lambda}_{k}^{\alpha} t\right) \end{array}\right]\left[\begin{array}{l} \hat{\varphi}_{k} \\ \breve{\varphi}_{k} \end{array}\right]+\frac{1}{\hbar} \int_{0}^{t} e^{-\frac{a}{\hbar}(t-s)} \\ & \times\left[\begin{array}{ll} \cos \left(\dot{\lambda}_{k}^{\alpha}(t-s)\right) & \sin \left(\dot{\lambda}_{k}^{\alpha}(t-s)\right) \\ \sin \left(\dot{\lambda}_{k}^{\alpha}(s-t)\right) & \cos \left(\dot{\lambda}_{k}^{\alpha}(t-s)\right) \end{array}\right]\left[\begin{array}{c} -\mu \widehat{\operatorname{Sat}}_{1, k}\left(\mathbf{z}_{0} \mathbf{w}_{0}\right)+\operatorname{Im} f_{k} \\ \mu \widetilde{\mathbf{S a t}}_{1, k}\left(\mathbf{z}_{0} \mathbf{w}_{0}\right)-\operatorname{Re} f_{k} \end{array}\right](s) \mathrm{d} s, \end{aligned} $

其中 $t\in \mathbf{R}_0$, $k=1,2,\cdots $. 由式 (3.2), 可知 $\mathbf{z}_0=(\hat{\mathbf{z}}_0,\breve{\mathbf{z}}_0)$ 可以表示为 Fourier 级数 (1.3).

引理3.1 若定理 1.2 的所有假设均成立. 则 $\max\limits_{t\in\mathbf{R}_0}|\mathbf{z}_0(t)|_2\leq\mathbf{z}_{\infty}.$

证考虑部分和

$\hat{\mathbf{z}}_0^{n}(t)=\sum\limits_{k=1}^{n}\hat{\mathbf{z}}_{k,0}(t)e_k^\alpha,\quad \breve{\mathbf{z}}_0^{n}(t)=\sum\limits_{k=1}^{n}\breve{\mathbf{z}}_{k,0}(t)e_k^\alpha,\quad t\in\mathbf{R}_+, n=1,2,\cdots.$

显然,

$\max\limits_{t\in\bar{I}_T}\{|\hat{\mathbf{z}}_0^{n}(t)|_{2}^2,|\breve{\mathbf{z}}_0^{n}(t)|_{2}^2\}\leq M_{\mathbf{z}}^T,$

并且在 $\{\mathrm{L}^2(\Omega)\}^2$ 中, $\{(\hat{\mathbf{z}}_0^{n}(t),\breve{\mathbf{z}}_0^{n}(t) \}_{n\in\mathbf{Z}_+}$ 收敛到 $(\hat{\mathbf{z}}_0(t),\breve{\mathbf{z}}_0(t) )$, $\forall t\in \bar{I}_T$.

根据式 (1.3)、Minkowski 不等式、Hölder 不等式和 Bessel 不等式, 可得

$ \begin{aligned} \max\limits_{t\in\bar{I}_T}|\hat{\mathbf{z}}_0^n(t)|_{2}&=&\max\limits_{t\in\bar{I}_T}\Bigg\{\sum\limits_{k=1}^{n}\Bigg\{ e^{-\frac{a}{\hbar}t}\left[ \begin{array}{c} \cos(\acute{\lambda}_k^\alpha t) \\ \sin(\acute{\lambda}_k^\alpha t) \end{array} \right]^\top \left[ \begin{array}{c} \hat{\varphi}_k\\ \breve{\varphi}_k \\ \end{array} \right] + \frac{1}{\hbar}\int_{0}^t e^{-\frac{a}{\hbar}(t-s)} \\ &&\times\left[ \begin{array}{c} \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (t-s)) \end{array} \right]^\top \bigg[ \begin{array}{c} -\mu\widehat{ \mathbf{Sat} }_{1,k}(\mathbf{z}_0 \mathbf{w} _0)+\mathrm{Im}f_k \\ \mu\widetilde{ \mathbf{Sat} }_{1,k}(\mathbf{z}_0 \mathbf{w} _0)-\mathrm{Re}f_k \\ \end{array} \bigg](s)\mathrm{d}s\Bigg\}^2\Bigg\}^{\frac{1}{2}} \\ &\leq&|\varphi|_{2}+\max\limits_{t\in\bar{I}_T}\bigg\{\sum\limits_{k=1}^{n}\bigg\{ \frac{1}{\hbar}\int_{0}^t e^{-\frac{a}{\hbar}(t-s)} \\ &&\times\left[ \begin{array}{c} \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (t-s)) \end{array} \right]^\top \bigg[ \begin{array}{c} -\mu\widehat{ \mathbf{Sat} }_{1,k}(\mathbf{z}_0 \mathbf{w} _0)+\mathrm{Im}f_k \\ \mu\widetilde{ \mathbf{Sat} }_{1,k}(\mathbf{z}_0 \mathbf{w} _0)-\mathrm{Re}f_k \\ \end{array} \bigg](s)\mathrm{d}s \bigg\}^2\bigg\}^{\frac{1}{2}} \\ &\leq&|\varphi|_{2} +\frac{1}{\hbar}\max\limits_{t\in\bar{I}_T}\bigg\{\sum\limits_{k=1}^{n}\bigg[\int_{0}^te^{-\frac{a}{\hbar}(t-s)}\mathrm{d}s \int_{0}^te^{-\frac{a}{\hbar}(t-s)} \\ &&\times \bigg( \left[ \begin{array}{c} \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (t-s)) \end{array} \right]^\top \bigg[ \begin{array}{c} -\mu\widehat{ \mathbf{Sat} }_{1,k}(\mathbf{z}_0 \mathbf{w} _0)+\mathrm{Im}f_k \\ \mu\widetilde{ \mathbf{Sat} }_{1,k}(\mathbf{z}_0 \mathbf{w} _0)-\mathrm{Re}f_k \\ \end{array} \bigg](s) \bigg)^2\mathrm{d}s\bigg]\bigg\}^{\frac{1}{2}} \\ &\leq&|\varphi|_{2} +\frac{1}{a}\Big[\frac{\mu}{\varrho}\max\limits_{t\in\bar{I}_T}|\mathbf{z}_0(t)|_{2}+f_{\infty}\Big]. \end{aligned} $

同理,

$ \max\limits_{t\in\bar{I}_T}|\breve{\mathbf{z}}_0^n(t)|_{2}\leq|\varphi|_{2} +\frac{1}{a}\Big[\frac{\mu}{\varrho}\max\limits_{t\in\bar{I}_T}|\mathbf{z}_0(t)|_{2}+f_{\infty}\Big], \quad \forall n\in\mathbf{Z}_+. $

在上述两个不等式中令 $n\rightarrow\infty$, 则

$ \max\limits_{t\in\bar{I}_T}|\mathbf{z}_0(t)|_2 \leq \frac{2\varrho(a|\varphi|_{2}+f_{\infty})}{a\varrho-2\mu} =\mathbf{z}_{\infty}. $

进而, 令 $T\rightarrow\infty$, 则 $\max\limits_{t\in\mathbf{R}_0}|\mathbf{z}_0(t)|_2\leq\mathbf{z}_{\infty}.$ 证毕.

由 (2.10) 式的第二个方程可知

(3.3) $ \frac{\mathrm{d}^2}{\mathrm{d}t^2} \mathbf{w} _{k,0}(t) +b\frac{\mathrm{d}}{\mathrm{d}t} \mathbf{w} _{k,0}(t) +\frac{1}{2}c_{\beta}\lambda_k^\beta \mathbf{w} _{k,0}(t) +c \mathbf{w} _{k,0}(t)-\nu \mathbf{Sat} _{2,k}(|\mathbf{z}_0(t)|^2)=g_k(t), $

其中 $ \mathbf{w} _{k,0}(0)=\phi_k,$ $\frac{\mathrm{d}}{\mathrm{d}t} \mathbf{w} _{k,0}(t)\Big|_{t=0}=\psi_k,$ $t\in \mathbf{R}_+$, $k=1,2,\cdots.$

结合不等式 (2.2) 计算得到,

$ \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}|\mathbf{z}_n(t)|_2^2&\leq&\frac{2f_\infty^2}{a\hbar} \Rightarrow |\mathbf{z}_n(t)|_2^2\leq 4|\varphi|_2^2+\frac{2f_\infty^2}{a\hbar}t\leq h_0e^{t}, \end{aligned} $

其中 $h_0:=4|\varphi|_2^2+\frac{2f_\infty^2}{a\hbar}$, $t\in \bar{I}_T\backslash\{0\}$, $n\in\mathbf{Z}_+$. 由不等式 (2.3) 可得

$\frac{\mathrm{d}}{\mathrm{d}t}\left|\frac{\partial}{\partial t} \mathbf{w} _n(t)\right|_2^2 + \frac{c_{\beta}}{2}\frac{\mathrm{d}}{\mathrm{d}t}\left| \mathbf{w} _n(t)\right|_{\mathrm{X}_0^\beta}^2 +c\frac{\mathrm{d}}{\mathrm{d}t}\left| \mathbf{w} _n(t)\right|_2^2 \leq\frac{4}{b}\Big[\frac{\nu^2}{\varepsilon^2}|\mathbf{z}_n(t)|_{2}^2+g_{\infty}^2\Big], $

因此,

$ \begin{aligned} \left|\frac{\partial}{\partial t} \mathbf{w} _n(t)\right|_2^2 +c\left| \mathbf{w} _n(t)\right|_2^2 &\leq&\frac{4\nu^2h_0}{b\varepsilon^2}\int_{0}^te^{s}\mathrm{d}s+\frac{4g_\infty^2 t}{b} +h_1 \\ &\leq&\left[\frac{4\nu^2h_0}{b\varepsilon^2}+\frac{4g_\infty^2}{b} +h_1\right]e^{t} :=h_2e^{t}, \end{aligned} $

其中 $h_1:=4\left|\psi\right|_2^2+2c_{\beta}|\phi|_{\mathrm{X}_0^\beta}^2+4c\left|\phi\right|_2^2$, $h_2:=\frac{4\nu^2h_0}{b\varepsilon^2}+\frac{4g_\infty^2}{b} +h_1$, $t\in \bar{I}_T\backslash\{0\}$, $n\in\mathbf{Z}_+$. 由于在 $\mathrm{L}^2(\Omega)$ 中, 当 $n\rightarrow\infty$ 时对每个 $t\in\mathbf{R}_+$ 都有 $ \mathbf{w} _n\rightarrow \mathbf{w} _0,$ $\frac{\partial}{\partial t} \mathbf{w} _n\rightarrow\frac{\partial}{\partial t} \mathbf{w} _0$, 所以

(3.4) $ \begin{aligned} \left|\frac{\partial}{\partial t} \mathbf{w} _0(t)\right|_2^2 +c\left| \mathbf{w} _0(t)\right|_2^2 &\leq&h_2e^{t}, \quad \forall t\in\mathbf{R}_+. \end{aligned} $

由不等式 (3.4) 可知, $ \mathbf{w} _{k,0}$, $\frac{\mathrm{d}}{\mathrm{d}t} \mathbf{w} _{k,0}$ 是有限增长的. 若 $(\mathcal{K}_3)$ 成立, 对式 (3.3) 两边同时进行 Laplace 变换可得

$ \begin{aligned} \mathbf{w} _{k,0}(t)&=&e^{-\frac{b}{2}t}\bigg[\cos(\grave{\lambda}_k^\beta t) +\frac{b}{2\grave{\lambda}_k^\beta}\sin(\grave{\lambda}_k^\beta t)\bigg]\phi_k +\frac{1}{\grave{\lambda}_k^\beta}e^{-\frac{b}{2}t}\sin(\grave{\lambda}_k^\beta t)\psi_k \\ &&+\frac{1}{\grave{\lambda}_k^\beta}\int_{0}^te^{-\frac{b}{2}(t-s)}\sin(\grave{\lambda}_k^\beta(t-s)) \left[\nu \mathbf{Sat} _{2,k}(|\mathbf{z}_0(t)|^2)+g_k(s)\right]\mathrm{d}s, \end{aligned} $

其中 $t\in \mathbf{R}_0$, $k=1,2,\cdots $. 因此, $ \mathbf{w} _0$ 可以表示成 Fourier 级数 (1.4).

引理3.2 设定理 1.2 的所有条件均成立. 则 $\max\limits_{t\in\mathbf{R}_0}| \mathbf{w} _0(t)|_{2}\leq \mathbf{w} _{\infty}.$

证考虑部分和

$ \mathbf{w} _0^{n}(t)=\sum\limits_{k=1}^{n} \mathbf{w} _{k,0}(t)e_k^\beta,\quad t\in\mathbf{R}_+, n=1,2,\cdots.$

显然,

$\max\limits_{t\in\bar{I}_T}\{| \mathbf{w} _0^{n}(t)|_{2}^2\}_{n\in\mathbf{Z}_+}\leq M_{\mathbf{z}}^T,$

而且在 $\mathrm{L}^2(\Omega)$ 中, $\{ \mathbf{w} _0^{n}(t))\}_{n\in\mathbf{Z}_+}$ 收敛于 $ \mathbf{w} _0(t)$, $\forall t\in \bar{I}_T$. 由方程 (1.4) 可知,

$ \begin{aligned} \max\limits_{t\in\bar{I}_T}| \mathbf{w} _0^n(t)|_{2} &\leq&\max\limits_{t\in\bar{I}_T}\bigg\{\sum\limits_{k=1}^{n}\bigg\{e^{-\frac{b}{2}t}\bigg[ \cos(\grave{\lambda}_k^\beta t) +\frac{b}{2\grave{\lambda}_k^\beta}\sin(\grave{\lambda}_k^\beta t) \bigg]\phi_k\bigg\}^2\bigg\}^{\frac{1}{2}} \\ &&+\max\limits_{t\in\bar{I}_T}\bigg\{\sum\limits_{k=1}^{n}\bigg[\frac{1}{\grave{\lambda}_k^\beta} e^{-\frac{b}{2}t}\sin(\grave{\lambda}_k^\beta t)\psi_k\bigg]^2\bigg\}^{\frac{1}{2}} +\max\limits_{t\in\bar{I}_T}\bigg\{\sum\limits_{k=1}^{n}\bigg[ \frac{1}{\grave{\lambda}_k^\beta} \\ &&\times\int_{0}^te^{-\frac{b}{2}(t-s)}\sin(\grave{\lambda}_k^\beta(t-s)) \left[\nu \mathbf{Sat} _{2,k}(|\mathbf{z}_0(t)|^2)+g_k(s)\right]\mathrm{d}s \bigg]^2\bigg\}^{\frac{1}{2}} \\ &\leq&\frac{2\grave{\lambda}_1^\beta+b}{2\grave{\lambda}_1^\beta}|\phi|_{2} +\frac{1}{\grave{\lambda}_1^\beta}|\psi|_{2} +\frac{2}{\grave{\lambda}_1^\beta b}\Big[\frac{\nu}{\varepsilon}\max\limits_{t\in\bar{I}_T}|\mathbf{z}_0(t)|_2+g_{\infty}\Big] \\ &\leq&\frac{2\grave{\lambda}_1^\beta+b}{2\grave{\lambda}_1^\beta}|\phi|_{2} +\frac{1}{\grave{\lambda}_1^\beta}|\psi|_{2} +\frac{2}{\grave{\lambda}_1^\beta b}\Big[\frac{\nu}{\varepsilon}\mathbf{z}_\infty+g_{\infty}\Big],\quad \forall n\in\mathbf{Z}_+. \end{aligned} $

由 $T$ 的任意性可知, 结论成立. 证毕.

3.2 定理 1.2 的证明

证我们运用反证法来证明唯一性. 设 $(\mathbf{z}_*, \mathbf{w} _*)=(\hat{\mathbf{z}}_*,\breve{\mathbf{z}}_*, \mathbf{w} _*)\in\mathrm{C}(\mathbf{R}_0,\{\mathrm{L}^2(\Omega)\}^3) \cap\{\mathbb{E}_{\mathbf{z}}\times\mathbb{E}_{ \mathbf{w} }\}$ 也是方程 (1.1) 的一个全局弱解. 定义

$ \hat{\mathbf{z}}_*(t)=\sum\limits_{k=1}^{\infty}\hat{\mathbf{z}}_{k,*}(t)e_k^\alpha,\quad \breve{\mathbf{z}}_*(t)=\sum\limits_{k=1}^{\infty}\breve{\mathbf{z}}_{k,*}(t)e_k^\alpha,\quad \mathbf{w} _*(t)=\sum\limits_{k=1}^{\infty} \mathbf{w} _{k,*}(t)e_k^\beta, $

其中 $\hat{\mathbf{z}}_{k,*}(t)=\langle\hat{\mathbf{z}}_{*}(t), e_k^\alpha\rangle_{\mathrm{L}^2}$, $\breve{\mathbf{z}}_{k,*}(t)=\langle\breve{\mathbf{z}}_{*}(t), e_k^\alpha\rangle_{\mathrm{L}^2}$, $ \mathbf{w} _{k,*}(t)=\langle \mathbf{w} _{*}(t), e_k^\beta\rangle_{\mathrm{L}^2}$, $t\in\mathbf{R}_0$, $k=1,2,\cdots $. 令 $\hat{\mathbf{z}}_*^n(t)$, $\breve{\mathbf{z}}_*^n(t)$, $ \mathbf{w} _*^n(t)$ 分别表示 $\hat{\mathbf{z}}_*(t)$, $\breve{\mathbf{z}}_*(t)$, $ \mathbf{w} _*(t)$ 的部分和, 其表达式为

$ \hat{\mathbf{z}}_*^n(t)=\sum\limits_{k=1}^{n}\hat{\mathbf{z}}_{k,*}(t)e_k^\alpha,\quad \breve{\mathbf{z}}_*^n(t)=\sum\limits_{k=1}^{n}\breve{\mathbf{z}}_{k,*}(t)e_k^\alpha,\quad \mathbf{w} _*^n(t)=\sum\limits_{k=1}^{n} \mathbf{w} _{k,*}(t)e_k^\beta, $

其中 $t\in\mathbf{R}_0$, $n=1,2,\cdots $. 显然, $\{(\hat{\mathbf{z}}_*^n,\breve{\mathbf{z}}_*^n, \mathbf{w} _*^n)\}_{n\in\mathbf{Z}_+}$ 在 $\{\mathrm{L}^2(\Omega)\}^3$ 中收敛到 $(\hat{\mathbf{z}}_*,\breve{\mathbf{z}}_*, \mathbf{w} _*)$, $\forall t\in \mathbf{R}_0$. 进而, $(\hat{\mathbf{z}}_*,\breve{\mathbf{z}}_*, \mathbf{w} _*)$ 可以表示成相应的 Fourier 形式 (1.3) 和 (1.4). 因此,

$ \begin{aligned} |\hat{\mathbf{z}}_0^{n}(t)-\hat{\mathbf{z}}_*^{n}(t)|_{2}^2&=&\sum\limits_{k=1}^{n}\big[\hat{\mathbf{z}}_{k,0}^{n}(t)-\hat{\mathbf{z}}_{k,*}^{n}(t)\big]^2 \\ &=&\sum\limits_{k=1}^{n}\bigg\{\frac{\mu}{\hbar}\int_{0}^t e^{-\frac{a}{\hbar}(t-s)} \left[ \begin{array}{c} \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (t-s)) \\ \end{array} \right]^\top \\ &&\times \bigg[ \begin{array}{c} \widehat{ \mathbf{Sat} }_{1,k}(\mathbf{z}_0 \mathbf{w} _0)-\widehat{ \mathbf{Sat} }_{1,k}(\mathbf{z}_* \mathbf{w} _*) \\ \widetilde{ \mathbf{Sat} }_{1,k}(\mathbf{z}_0 \mathbf{w} _0)-\widetilde{ \mathbf{Sat} }_{1,k}(\mathbf{z}_* \mathbf{w} _*) \\ \end{array} \bigg](s)\mathrm{d}s \bigg\}^2 \\ &\leq&\frac{\mu^2}{\hbar^2} \int_{0}^te^{-\frac{a}{\hbar}(t-s)}\mathrm{d}s \int_{0}^te^{-\frac{a}{\hbar}(t-s)} \sum\limits_{k=1}^{n}\bigg\{ \left|\widehat{ \mathbf{Sat} }_{1,k}(\mathbf{z}_0 \mathbf{w} _0)-\widehat{ \mathbf{Sat} }_{1,k}(\mathbf{z}_* \mathbf{w} _*)\right| \\ && + \left|\widetilde{ \mathbf{Sat} }_{1,k}(\mathbf{z}_0 \mathbf{w} _0)-\widetilde{ \mathbf{Sat} }_{1,k}(\mathbf{z}_* \mathbf{w} _*)\right| \bigg\}^2(s)\mathrm{d}s \\ &\leq&\frac{32\mu^2}{a\hbar\rho_0^2} \int_{0}^t \bigg[ |\mathbf{z}_0(s)-\mathbf{z}_*(s)|_{2}^2+ | \mathbf{w} _0(s)- \mathbf{w} _*(s)|_{2}^2 \bigg]\mathrm{d}s, \end{aligned} $

其中 $t\in\mathbf{R}_0$, $n=1,2,\cdots.$ 注意到 $\langle\cdot\rangle_{\mathrm{L}^2}$ 关于 $|\cdot|_{2}$ 是连续的, 在上述不等式中令 $n\rightarrow\infty$, 则

$ |\hat{\mathbf{z}}_0(t)-\hat{\mathbf{z}}_*(t)|_{2}^2 \leq\frac{32\mu^2}{a\hbar\rho_0^2} \int_{0}^t \bigg[ |\mathbf{z}_0(s)-\mathbf{z}_*(s)|_{2}^2+ | \mathbf{w} _0(s)- \mathbf{w} _*(s)|_{2}^2 \bigg]\mathrm{d}s,\quad \forall t\in\mathbf{R}_0. $

同理,

$ |\breve{\mathbf{z}}_0(t)-\breve{\mathbf{z}}_*(t)|_{2}^2 \leq\frac{32\mu^2}{a\hbar\rho_0^2} \int_{0}^t \bigg[ |\mathbf{z}_0(s)-\mathbf{z}_*(s)|_{2}^2+ | \mathbf{w} _0(s)- \mathbf{w} _*(s)|_{2}^2 \bigg]\mathrm{d}s,\quad \forall t\in\mathbf{R}_0. $

所以,

$ |\mathbf{z}_0(t)-\mathbf{z}_*(t)|_{2}^2 \leq\frac{64\mu^2}{a\hbar\rho_0^2} \int_{0}^t \bigg[ |\mathbf{z}_0(s)-\mathbf{z}_*(s)|_{2}^2+ | \mathbf{w} _0(s)- \mathbf{w} _*(s)|_{2}^2 \bigg]\mathrm{d}s,\quad \forall t\in\mathbf{R}_0. $

利用方程 (1.4) 可得,

$ \begin{aligned} | \mathbf{w} _0^{n}(t)- \mathbf{w} _*^{n}(t)|_{2}^2&=&\sum\limits_{k=1}^{n}\big[ \mathbf{w} _{k,0}^{n}(t)- \mathbf{w} _{k,*}^{n}(t)\big]^2 \\ &=&\sum\limits_{k=1}^{n}\bigg\{\frac{1}{\grave{\lambda}_k^\beta}\int_{0}^te^{-\frac{b}{2}(t-s)}\sin(\grave{\lambda}_k^\beta(t-s)) \\&& \times\left[\nu \mathbf{Sat} _{2,k}(|\mathbf{z}_0|^2)-\nu \mathbf{Sat} _{2,k}(|\mathbf{z}_*|^2)\right](s)\mathrm{d}s \bigg\}^2 \\ &\leq&\frac{2\nu^2}{b(\grave{\lambda}_1^\beta)^2} \int_{0}^te^{-\frac{b}{2}(t-s)} \left| \mathbf{Sat} _2(|\mathbf{z}_0|^2)- \mathbf{Sat} _2(|\mathbf{z}_*|^2)\right|^2(s)\mathrm{d}s \\ &\leq&\frac{9\nu^2}{2\varepsilon^2b(\grave{\lambda}_1^\beta)^2} \int_{0}^t |\mathbf{z}_0(s)-\mathbf{z}_*(s)|_{2}^2\mathrm{d}s,\quad \forall t\in\mathbf{R}_0, n=1,2,\cdots. \end{aligned} $

在上述不等式中令 $n\rightarrow\infty$, 有

$ | \mathbf{w} _0(t)- \mathbf{w} _*(t)|_{2}^2 \leq\frac{9\nu^2}{2\varepsilon^2b(\grave{\lambda}_1^\beta)^2} \int_{0}^t \bigg[ |\mathbf{z}_0(s)-\mathbf{z}_*(s)|_{2}^2+ | \mathbf{w} _0(s)- \mathbf{w} _*(s)|_{2}^2 \bigg]\mathrm{d}s,\quad \forall t\in\mathbf{R}_0. $

定义 $\mathcal{U}(t)=|\mathbf{z}_0(t)-\mathbf{z}_*(t)|_{2}^2+| \mathbf{w} _0(t)- \mathbf{w} _*(t)|_{2}^2$, $\forall t\in\mathbf{R}_0$. 则

$ \mathcal{U}(t) \leq \left\{ \frac{64\mu^2}{a\hbar\rho_0}^2+\frac{9\nu^2}{2\varepsilon^2b(\grave{\lambda}_1^\beta)^2} \right\} \int_{0}^t \mathcal{U}(s)\mathrm{d}s,\quad \forall t\in\mathbf{R}_0. $

利用 Gronwall 不等式可知, $\mathcal{U}(t)\equiv0$, $\forall t\in\mathbf{R}_0$. 证毕.

3.3 定理 1.3 的证明

证由 $(\mathcal{K}_4)$ 可知, 存在足够小的 $\epsilon>0$ 使得

$ \min\left\{\frac{a}{\hbar},\frac{b}{2}\right\}>\max\left\{ \frac{64\mu^2(1+\epsilon)}{a \hbar\rho_0^2},\frac{9(1+\epsilon)\nu^2}{2b(\grave{\lambda}_1^\beta)^2\varepsilon^2}\right\}. $

设 $(\mathbf{z}_*, \mathbf{w} _*)=(\hat{\mathbf{z}}_*,\breve{\mathbf{z}}_*, \mathbf{w} _*)\in\mathrm{C}(\mathbf{R}_0,\{\mathrm{L}^2(\Omega)\}^3) \cap\{\mathbb{E}_{\mathbf{z}}\times\mathbb{E}_{ \mathbf{w} }\}$ 也是方程 (1.1) 的一个全局弱解且其初始条件为

$\hat{\mathbf{z}}_*(x,0)=\hat{\varphi}_*(x),\quad \breve{\mathbf{z}}_*(x,0)=\breve{\varphi}_*(x),\quad \mathbf{w} _*(x,0)=\phi_*(x),\quad \dot{ \mathbf{w} }_*(x,0)=\psi_*(x),\quad \forall x\in\Omega.$ 令 $\varphi_*=(\hat{\varphi}_*,\breve{\varphi}_*)$, $\hat{\mathbf{z}}_*^n(t)$, $\breve{\mathbf{z}}_*^n(t)$, $ \mathbf{w} _*^n(t)$ 分别表示如上节所述的 $\hat{\mathbf{z}}_*(t)$, $\breve{\mathbf{z}}_*(t)$, $ \mathbf{w} _*(t)$ 的部分和, 其中 $t\in\mathbf{R}_0$, $n=1,2,\cdots $. 而且, $\{(\hat{\mathbf{z}}_*^n,\breve{\mathbf{z}}_*^n, \mathbf{w} _*^n)\}_{n\in\mathbf{Z}_+}$ 在 $\{\mathrm{L}^2(\Omega)\}^3$ 中收敛于 $(\hat{\mathbf{z}}_*,\breve{\mathbf{z}}_*, \mathbf{w} _*)$, $\forall t\in \mathbf{R}_0$, $(\hat{\mathbf{z}}_*,\breve{\mathbf{z}}_*, \mathbf{w} _*)$ 可以用相应的 Fourier 级数 (1.3) 和 (1.4) 来表示. 因此,

$ \begin{aligned} |\hat{\mathbf{z}}_0^{n}(t)-\hat{\mathbf{z}}_*^{n}(t)|_{2}^2 &\leq&(1+\epsilon^{-1})e^{-\frac{2a}{\hbar}t} |\varphi-\varphi_*|_2^2 \\ &&+ \sum\limits_{k=1}^{n}(1+\epsilon)\bigg\{\frac{\mu}{\hbar}\int_{0}^t e^{-\frac{a}{\hbar}(t-s)} \left[ \begin{array}{cc} \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (t-s)) \end{array} \right]^\top \\ && \times\bigg[ \begin{array}{c} -\widehat{ \mathbf{Sat} }_{1,k}(\mathbf{z}_{0} \mathbf{w} _{0})+\widehat{ \mathbf{Sat} }_{1,k}(\mathbf{z}_{*} \mathbf{w} _{*}) \\ \widetilde{ \mathbf{Sat} }_{1,k}(\mathbf{z}_{0} \mathbf{w} _{0})-\widetilde{ \mathbf{Sat} }_{1,k}(\mathbf{z}_{*} \mathbf{w} _{*}) \\ \end{array} \bigg](s)\mathrm{d}s\bigg\}^2 \\ &\leq&(1+\epsilon^{-1})e^{-\frac{2a}{\hbar}t} |\varphi-\varphi_*|_2^2 \\ &&+ \frac{(1+\epsilon)\mu^2}{a\hbar} \int_{0}^te^{-\frac{a}{\hbar}(t-s)}\Big| \widehat{ \mathbf{Sat} }_{1}(\mathbf{z}_{0} \mathbf{w} _{0})-\widehat{ \mathbf{Sat} }_{1}(\mathbf{z}_{*} \mathbf{w} _{*}) \Big|_2^2\mathrm{d}s \\ &&+\frac{(1+\epsilon)\mu^2}{a\hbar} \int_{0}^te^{-\frac{a}{\hbar}(t-s)}\Big| \widetilde{ \mathbf{Sat} }_{1}(\mathbf{z}_{0} \mathbf{w} _{0})-\widetilde{ \mathbf{Sat} }_{1}(\mathbf{z}_{*} \mathbf{w} _{*}) \Big|_2^2\mathrm{d}s \\ &\leq&(1+\epsilon^{-1})e^{-\frac{2a}{\hbar}t} |\varphi-\varphi_*|_2^2+\frac{16\mu^2(1+\epsilon)}{a \hbar\rho_0^2} \\ &&\times\int_{0}^te^{-\frac{a}{\hbar}(t-s)}\left[\big|\mathbf{z}_{0}(s)-\mathbf{z}_{*}(s)\big|_{2}^2 +\big| \mathbf{w} _{0}(s)- \mathbf{w} _{*}(s)\big|_{2}^2\right]\mathrm{d}s. \end{aligned}$

在上述不等式中令 $n\rightarrow\infty$, 结合 Bohr 不等式

$(a+b)^2\leq (1+\epsilon^{-1})a^2+(1+\epsilon)b^2,\quad \forall\epsilon>0, a,b\in\mathbf{R}$

可得

$ \begin{aligned} |\hat{\mathbf{z}}_0(t)-\hat{\mathbf{z}}_*(t)|_{2}^2 &\leq&(1+\epsilon^{-1})e^{-\frac{2a}{\hbar}t} |\varphi-\varphi_*|_2^2+\frac{16\mu^2(1+\epsilon)}{a \hbar\rho_0^2} \\ &&\times\int_{0}^te^{-\frac{a}{\hbar}(t-s)}\left[\big|\mathbf{z}_{0}(s)-\mathbf{z}_{*}(s)\big|_{2}^2 +\big| \mathbf{w} _{0}(s)- \mathbf{w} _{*}(s)\big|_{2}^2\right]\mathrm{d}s,\quad \forall t\in\mathbf{R}_0. \end{aligned} $

同理,

$ \begin{aligned} |\breve{\mathbf{z}}_0(t)-\breve{\mathbf{z}}_*(t)|_{2}^2 &\leq&(1+\epsilon^{-1})e^{-\frac{2a}{\hbar}t} |\varphi-\varphi_*|_2^2+\frac{16\mu^2(1+\epsilon)}{a \hbar\rho_0^2} \\ &&\times\int_{0}^te^{-\frac{a}{\hbar}(t-s)}\left[ \big|\mathbf{z}_{0}(s)-\mathbf{z}_{*}(s)\big|_{2}^2 +\big| \mathbf{w} _{0}(s)- \mathbf{w} _{*}(s)\big|_{2}^2\right]\mathrm{d}s,\quad \forall t\in\mathbf{R}_0. \end{aligned}$

所以,

$ \begin{aligned} |\mathbf{z}_0(t)-\mathbf{z}_*(t)|_{2}^2 &\leq&2(1+\epsilon^{-1})e^{-\frac{2a}{\hbar}t} |\varphi-\varphi_*|_2^2+\frac{32\mu^2(1+\epsilon)}{a \hbar\rho_0^2} \\ &&\times\int_{0}^te^{-\frac{a}{\hbar}(t-s)}\left[\big|\mathbf{z}_{0}(s)-\mathbf{z}_{*}(s)\big|_{2}^2 +\big| \mathbf{w} _{0}(s)- \mathbf{w} _{*}(s)\big|_{2}^2\right]\mathrm{d}s,\ \forall t\in\mathbf{R}_0. \end{aligned}$

另一方面, 由方程 (1.4) 可知,

$ \begin{aligned} | \mathbf{w} _0^{n}(t)- \mathbf{w} _*^{n}(t)|_{2}^2 &\leq&2(1+\epsilon^{-1})\left(1+\frac{b}{2\grave{\lambda}_1^\beta}\right)^2 |\phi-\phi_*|_2^2e^{-bt} +2(1+\epsilon^{-1})\frac{1}{(\grave{\lambda}_1^\beta)^2} |\psi-\psi_*|_2^2e^{-bt} \\ &&+ \sum\limits_{k=1}^{n}(1+\epsilon)\bigg[ \frac{\nu}{\grave{\lambda}_k^\beta}\int_{0}^te^{-\frac{b}{2}(t-s)}\sin(\grave{\lambda}_k^\beta(t-s)) \\ &&\times \left[ \mathbf{Sat} _{2,k}(|\mathbf{z}_{0}(s)|^2)- \mathbf{Sat} _{2,k}(|\mathbf{z}_{*}(s)|^2)\right]\mathrm{d}s\bigg]^2 \\ &\leq&2(1+\epsilon^{-1})\left(1+\frac{b}{2\grave{\lambda}_1^\beta}\right)^2 |\phi-\phi_*|_2^2e^{-bt} +2(1+\epsilon^{-1})\frac{1}{(\grave{\lambda}_1^\beta)^2} |\psi-\psi_*|_2^2e^{-bt} \\ &&+ \frac{2(1+\epsilon)\nu^2}{b(\grave{\lambda}_1^\beta)^2} \int_{0}^te^{-\frac{b}{2}(t-s)}\Big| \mathbf{Sat} _{2}(|\mathbf{z}_{0}(s)|^2)- \mathbf{Sat} _{2}(|\mathbf{z}_{*}(s)|^2) \Big|_2^2\mathrm{d}s \\ &\leq&2(1+\epsilon^{-1})\left(1+\frac{b}{2\grave{\lambda}_1^\beta}\right)^2 |\phi-\phi_*|_2^2e^{-bt} +2(1+\epsilon^{-1})\frac{1}{(\grave{\lambda}_1^\beta)^2} |\psi-\psi_*|_2^2e^{-bt} \\ &&+\frac{9(1+\epsilon)\nu^2}{2b(\grave{\lambda}_1^\beta)^2\varepsilon^2} \int_{0}^te^{-\frac{b}{2}(t-s)}\big|\mathbf{z}_{0}(s)-\mathbf{z}_{*}(s)\big|_{2}^2\mathrm{d}s. \end{aligned} $

在上述不等式在令 $n\rightarrow\infty$ 可得

$ \begin{aligned} | \mathbf{w} _0(t)- \mathbf{w} _*(t)|_{2}^2 &\leq&2(1+\epsilon^{-1})\left(1+\frac{b}{2\grave{\lambda}_1^\beta}\right)^2 |\phi-\phi_*|_2^2e^{-bt} +2(1+\epsilon^{-1})\frac{1}{(\grave{\lambda}_1^\beta)^2} |\psi-\psi_*|_2^2e^{-bt} \\ &&+\frac{9(1+\epsilon)\nu^2}{2b(\grave{\lambda}_1^\beta)^2\varepsilon^2} \int_{0}^te^{-\frac{b}{2}(t-s)}|\mathbf{z}_{0}(s)-\mathbf{z}_{*}(s)|_{2}^2\mathrm{d}s,\quad \forall t\in\mathbf{R}_0. \end{aligned} $

定义

$\mathcal{V}(t)=\max\left\{\big|\mathbf{z}_{0}(t)-\mathbf{z}_{*}(t)\big|_{2}^2, \big| \mathbf{w} _{0}(t)- \mathbf{w} _{*}(t)\big|_{2}^2\right\},\quad \forall t\in \mathbf{R}_0.$

结合不等式 (3.5) 和 (3.6) 可得,

$ \mathcal{V}(t)\leq \sigma_1 e^{-\varpi t}+\sigma_2\int_{0}^te^{-\varpi(t-s)}\mathcal{V}(s)\mathrm{d}s, \quad\forall t\in \mathbf{R}_0, $

其中

$ \sigma_1:=\max\bigg\{2(1+\epsilon^{-1}) |\varphi-\varphi_*|_2^2, 2(1+\epsilon^{-1})\bigg(1+\frac{b}{2\grave{\lambda}_1^\beta}\bigg)^2 |\phi\!-\!\phi_*|_2^2\!+\! 2(1+\epsilon^{-1})\frac{1}{(\grave{\lambda}_1^\beta)^2} |\psi\!-\!\psi_*|_2^2\bigg\},$

$ \sigma_2:=\max\left\{ \frac{64\mu^2(1+\epsilon)}{a \hbar\rho_0^2},\frac{9(1+\epsilon)\nu^2}{2b(\grave{\lambda}_1^\beta)^2\varepsilon^2}\right\},\quad \varpi=\min\left\{\frac{a}{\hbar},\frac{b}{2}\right\}.$

由文献 [引理 13] 可得

$ \mathcal{V}(t)\leq \sigma_1e^{-(\varpi-\sigma_2)t}\rightarrow0 (t\rightarrow\infty). $

证毕.

4 概自守弱解

这里我们介绍概自守向量值函数的概念, 详情请参阅文献 [30].

假设 $\mathbb{H}$ 表示某个 Hilbert 空间. 如果对于每一个序列 $\{\bar{t}_p\}_{p\in\mathbf{Z}_+}$, 都存在一个子序列 $\{t_p\}_{p\in\mathbf{Z}_+}\subseteq\{\bar{t}_p\}_{p\in\mathbf{Z}_+}$ 和一个函数 $\tilde{f}:\mathbf{R}\rightarrow\mathbb{H}$ 使得对任意的 $t\in\mathbf{R}$, 在 $\mathbb{H}$ 中有

$ \tilde{f}(t)=\lim\limits_{p\rightarrow\infty}f(t+t_p),\quad \lim\limits_{p\rightarrow\infty}\tilde{f}(t-t_p)=f(t), $

则称 $\mathbb{H}$-值函数 $f\in\mathrm{C}(\mathbf{R},\mathbb{H})$ 为概自守函数. 所有这些函数的集合用 $\mathbb{AA}(\mathbf{R},\mathbb{H})$ 表示. 设 $\mathbb{B}\subseteq\mathbb{H}$ 为任意有界集. 如果映射 $f=f(\eta,t):\mathbb{H}\times\mathbf{R}\rightarrow \mathbb{H}$, $\forall(\eta,t)\in \mathbb{H}\times\mathbf{R}$ 关于变量 $t$ 是概自守的, 对 $\eta\in\mathbb{B}$ 是一致的, 则称映射 $f=f(\eta,t)$ 关于变量 $t$ 是概自守的. 这类映射的集合记为 $\mathbb{AA}(\mathbb{H}\times\mathbf{R},\mathbb{H})$.

令 $\aleph(\mathbf{R}_0,\mathbb{H})=\{\mathbf{h}\in\mathrm{C}(\mathbf{R}_0,\mathbb{H}): \lim\limits_{t\rightarrow\infty}\|\mathbf{h}(t)\|_{\mathbb{H}}=0\}$. 若存在 $\mathbf{f}\in \mathbb{AA}(\mathbf{R},\mathbb{H})$ 且 $\mathbf{h}\in \aleph(\mathbf{R}_0, \mathbb{H})$ 满足 $f=\mathbf{f}+\mathbf{h}$, 则称 $\mathbb{H}$-值连续函数 $f:\mathbf{R}_0\rightarrow \mathbb{H}$ 为渐近概自守函数. 所有这些函数组成的集合用 $\mathbb{AAA}(\mathbf{R}_0,\mathbb{H})$表示. 特别地, 若 $\mathbb{H}$ 是 Banach 空间, 则 $\mathbb{AAA}(\mathbf{R}_0,\mathbb{H})$ 也是 Banach 空间, 其范数为 $ \|f\|_{\mathbb{AAA}}=\sup\limits_{t\in\mathbf{R}}\|\mathbf{f}\|_{\mathbb{H}}+\sup\limits_{t\in\mathbf{R}_0}\|\mathbf{h}\|_{\mathbb{H}}, $ 其中 $f=\mathbf{f}+\mathbf{h}\in \mathbb{AAA}(\mathbf{R}_0,\mathbb{H})$, $\mathbf{f}\in \mathbb{AA}(\mathbf{R},\mathbb{H})$, $\mathbf{h}\in \aleph(\mathbf{R}_0, \mathbb{H})$.

设连续函数 $\mathbf{h}:\mathbb{H}\times\mathbf{R}_0\rightarrow\mathbb{H}$ 对所有的 $\eta\in\mathbb{B}$ 满足 $\lim\limits_{t\rightarrow\infty}\|\mathbf{h}(\eta,t)\|_{\mathbb{H}}=0$. 所有这类函数组成的集合用 $\aleph(\mathbb{H}\times\mathbf{R}_0, \mathbb{H})$ 表示. 若函数 $f=f(\eta,t):\mathbb{H}\times\mathbf{R}_0\rightarrow \mathbb{H}$, $\forall(\eta,t)\in \mathbb{H}\times\mathbf{R}_0$ 对任意的 $\eta\in\mathbb{B}$ 关于变量 $t$ 是渐近概自守的, 则称函数 $f=f(\eta,t)$ 关于变量 $t$ 是渐近概自守的. 即 $f=\mathbf{f}+\mathbf{h}$, 其中主项 $\mathbf{f}\in\mathbb{AA}(\mathbb{H}\times\mathbf{R},\mathbb{H})$, $\mathbf{h}\in \aleph(\mathbb{H}\times\mathbf{R}_0, \mathbb{H})$. 这类函数用 $\mathbb{AAA}(\mathbb{X}\times\mathbf{R}_0,\mathbb{X})$ 表示.

4.1 Picard 序列

本节我们将使用 Picard 迭代方法. 基于式 (1.3) 和 (1.4), 构造一个 Picard 迭代序列 $(\delta_m,\eta_m) =(\hat{\delta}_m,\breve{\delta}_m,\eta_m)\in\{\mathrm{L}^2(\Omega)\}^3$, 如下所示

(4.1) $ \begin{matrix}\label{px1} \left\{ \begin{array}{lll} \left[ \begin{array}{c} \hat{\delta}_{0} \\ \breve{\delta}_{0} \\ \end{array} \right](t)&=&\left[ \begin{array}{c} \hat{\varphi} \\ \breve{\varphi} \\ \end{array} \right]\in \left[\{\mathrm{L}^2(\Omega)\}^2\right]^\top,\\[6mm] \left[ \begin{array}{c} \hat{\delta}_{m} \\ \breve{\delta}_{m} \\ \end{array} \right](t) &=& \sum\limits_{k=1}^\infty\Bigg\{ e^{-\frac{a}{\hbar}t}\left[ \begin{array}{cc} \cos(\acute{\lambda}_k^\alpha t) & \sin(\acute{\lambda}_k^\alpha t) \\ -\sin(\acute{\lambda}_k^\alpha t) & \cos(\acute{\lambda}_k^\alpha t) \\ \end{array} \right] \left[ \begin{array}{c} \hat{\varphi}_k\\ \breve{\varphi}_k \\ \end{array} \right] \\[5mm] && + \frac{1}{\hbar}\int_{0}^t e^{-\frac{a}{\hbar}(t-s)} \left[ \begin{array}{cc} \cos(\acute{\lambda}_k^\alpha (t-s)) & \sin(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (s-t)) & \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \end{array} \right] \\[5mm] && \times\left[ \begin{array}{c} -\mu\widehat{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1})+\mathrm{Im}f_k \\ \mu\widetilde{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1})-\mathrm{Re}f_k \\ \end{array} \right](s)\mathrm{d}s\Bigg\}e_k^\alpha, \end{array} \right. \end{matrix} $

(4.2) $ \begin{matrix}\label{px2} \left\{ \begin{array}{lll} \eta_{0}(t)&=&\phi\in \mathrm{L}^2(\Omega), \\[2mm] \eta_{m}(t)&=& \sum\limits_{k=1}^\infty\Bigg\{e^{-\frac{b}{2}t}\left[ \cos(\grave{\lambda}_k^\beta t) +\frac{b}{2\grave{\lambda}_k^\beta}\sin(\grave{\lambda}_k^\beta t) \right]\phi_k + \frac{1}{\grave{\lambda}_k^\beta}e^{-\frac{b}{2}t}\sin(\grave{\lambda}_k^\beta t)\psi_k\\[5mm] && +\frac{1}{\grave{\lambda}_k^\beta}\int_{0}^te^{-\frac{b}{2}(t-s)}\sin(\grave{\lambda}_k^\beta(t-s)) \left[\nu \mathbf{Sat} _{2,k}(|\delta_{m-1}(s)|^2)+g_k(s)\right]\mathrm{d}s\Bigg\}e_k^\beta, \end{array} \right. \end{matrix}$

其中 $k=1,2,\cdots $, $m=1,2,\cdots,$ $t\in\mathbf{R}_0$.

定义 $|\delta_m|_2=\sqrt{|\hat{\delta}_m|_2^2+|\breve{\delta}_m|_2^2}$, $\forall m=0,1,\cdots $.

命题4.1 设定理 1.4 的所有条件都成立. 则 $\{(\delta_m,\eta_m)=(\hat{\delta}_m,\breve{\delta}_m,\eta_m): m=1,2,\cdots \}$ 在 $\{\mathrm{L}^2(\Omega)\}^3$ 中是有定义的并且是 $\mathrm{L}^2$-连续的.

证结合方程 (4.1), 利用 Minkowski 不等式和 Hölder 不等式可得,

$ \begin{aligned} |\hat{\delta}_1(t)|_2^2 &=&\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}\Bigg\{ e^{-\frac{a}{\hbar}t}\left[ \begin{array}{c} \cos(\acute{\lambda}_k^\alpha t) \\ \sin(\acute{\lambda}_k^\alpha t) \end{array} \right]^\top \left[ \begin{array}{c} \hat{\varphi}_k\\ \breve{\varphi}_k \\ \end{array} \right] \\ &&+ \frac{1}{\hbar}\int_{0}^t e^{-\frac{a}{\hbar}(t-s)} \left[ \begin{array}{c} \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (t-s)) \end{array} \right]^\top \left[ \begin{array}{c} -\mu\widehat{ \mathbf{Sat} }_{1,k}(\delta_{0}\eta_{0})+\mathrm{Im}f_k \\ \mu\widetilde{ \mathbf{Sat} }_{1,k}(\delta_{0}\eta_{0})-\mathrm{Re}f_k \\ \end{array} \right](s)\mathrm{d}s\Bigg\}^2 \\ &\leq&\lim\limits_{n\rightarrow\infty}\bigg\{e^{-\frac{a}{\hbar}t}|\varphi|_{2} + \frac{\mu}{\hbar}\bigg[ \int_{0}^{t}e^{-\frac{a}{\hbar}(t-s)}\mathrm{d}s \\ &&\times\int_{0}^{t}e^{-\frac{a}{\hbar}(t-s)} \sum\limits_{k=1}^{n}\Big( \widehat{ \mathbf{Sat} }_{1,k}^2(\delta_{0}\eta_{0}) + \widetilde{ \mathbf{Sat} }_{1,k}^2(\delta_{0}\eta_{0}) \Big)\mathrm{d}s\bigg]^{\frac{1}{2}} \\ &&+ \frac{1}{\hbar}\bigg[ \int_{0}^{t}e^{-\frac{a}{\hbar}(t-s)}\mathrm{d}s\int_{0}^{t}e^{-\frac{a}{\hbar}(t-s)}\sum\limits_{k=1}^{n}\Big( \mathrm{Im}f_k^2 + \mathrm{Re}f_k^2 \Big)(s)\mathrm{d}s\bigg]^{\frac{1}{2}} \bigg\}^2 \\ &\leq&\bigg( |\varphi|_{2}+\frac{\mu}{a\varrho}|\varphi|_{2}+\frac{1}{a}f_\infty \bigg)^2,\quad \forall t\in\mathbf{R}_0. \end{aligned} $

同理,

$ |\breve{\delta}_1(t)|_2^2\leq\bigg( |\varphi|_{2}+\frac{\mu}{a\varrho}|\varphi|_{2}+\frac{1}{a}f_\infty \bigg)^2,\quad \forall t\in\mathbf{R}_0. $

所以,

$ |\delta_1(t)|_2^2\leq2\bigg( |\varphi|_{2}+\frac{\mu}{a\varrho}|\varphi|_{2}+\frac{1}{a}f_\infty \bigg)^2,\quad \forall t\in\mathbf{R}_0. $

另一方面, 有

$ \begin{aligned} |\hat{\delta}_2(t)|_2^2 &=&\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}\Bigg\{ e^{-\frac{a}{\hbar}t}\left[ \begin{array}{c} \cos(\acute{\lambda}_k^\alpha t) \\ \sin(\acute{\lambda}_k^\alpha t) \end{array} \right]^\top \left[ \begin{array}{c} \hat{\varphi}_k\\ \breve{\varphi}_k \\ \end{array} \right] \\ &&+ \frac{1}{\hbar}\int_{0}^t e^{-\frac{a}{\hbar}(t-s)} \left[ \begin{array}{c} \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (t-s)) \end{array} \right]^\top \left[ \begin{array}{c} -\mu\widehat{ \mathbf{Sat} }_{1,k}(\delta_{1}\eta_{1})+\mathrm{Im}f_k \\ \mu\widetilde{ \mathbf{Sat} }_{1,k}(\delta_{1}\eta_{1})-\mathrm{Re}f_k \\ \end{array} \right](s)\mathrm{d}s\Bigg\}^2 \\ &\leq&\bigg[|\varphi|_{2}+ \frac{\mu\sup\limits_{t\in\mathbf{R}_0}|\delta_1(t)|_{2}}{a\varrho} +\frac{f_\infty}{a} \bigg]^2 \\ &\leq&\bigg\{ |\varphi|_{2}\bigg[ 1+\frac{\sqrt{2}\mu}{a\varrho} +\bigg(\frac{\sqrt{2}\mu}{a\varrho}\bigg)^2 \bigg]+\frac{f_\infty}{a}\bigg( 1+\frac{\sqrt{2}\mu}{a\varrho} \bigg) \bigg\}^2,\quad \forall t\in\mathbf{R}_0. \end{aligned} $

同理,

$ |\breve{\delta}_2(t)|_2^2\leq \bigg\{ |\varphi|_{2}\bigg[ 1+\frac{\sqrt{2}\mu}{a\varrho} +\bigg(\frac{\sqrt{2}\mu}{a\varrho}\bigg)^2 \bigg]+\frac{f_\infty}{a}\bigg( 1+\frac{\sqrt{2}\mu}{a\varrho}\bigg)\bigg\}^2, $

$ |\delta_2(t)|_2^2\leq 2\bigg\{ |\varphi|_{2}\bigg[ 1+\frac{\sqrt{2}\mu}{a\varrho} +\bigg(\frac{\sqrt{2}\mu}{a\varrho}\bigg)^2 \bigg]+\frac{f_\infty}{a}\bigg( 1+\frac{\sqrt{2}\mu}{a\varrho}\bigg)\bigg\}^2, \quad\forall t\in\mathbf{R}_0. $

通过数学归纳法可得,

$ |\delta_m(t)|_2^2\leq 2\bigg\{ |\varphi|_{2}\sum\limits_{k=0}^{m}\bigg(\frac{\sqrt{2}\mu}{a\varrho}\bigg)^{k} +\frac{f_\infty}{a}\sum\limits_{k=0}^{m-1}\bigg(\frac{\sqrt{2}\mu}{a\varrho}\bigg)^{k} \bigg\}^2\leq2\bigg[ \frac{a\varrho|\varphi|_{2}+\varrho f_\infty}{a\varrho-\sqrt{2}\mu} \bigg]^2:=\delta_\infty^2, $

其中 $t\in\mathbf{R}_0$, $m=3,4,\cdots.$ 因此, $\delta_m=(\hat{\delta}_m,\breve{\delta}_m)\in\{\mathrm{L}^2(\Omega)\}^2(m=1,2,\cdots )$ 是有定义的.

另一方面, 基于方程 (4.2) 可知

$ \begin{aligned} |\eta_m(t)|_2^2 &=&\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}\Bigg\{ e^{-\frac{b}{2}t}\bigg[\cos(\grave{\lambda}_k^\beta t) +\frac{b}{2\grave{\lambda}_k^\beta}\sin(\grave{\lambda}_k^\beta t)\bigg]\phi_k +\frac{e^{-\frac{b}{2}t}}{\grave{\lambda}_k^\beta}\sin(\grave{\lambda}_k^\beta t)\psi_k\\ &&+\frac{1}{\grave{\lambda}_k^\beta}\int_{0}^te^{-\frac{b}{2}(t-s)}\sin(\grave{\lambda}_k^\beta(t-s)) \left[\nu \mathbf{Sat} _{2,k}(|\delta_{m-1}(s)|^2)+g_k(s)\right]\mathrm{d}s\Bigg\}^2 \\ &\leq&\lim\limits_{n\rightarrow\infty}\bigg\{e^{-\frac{b}{2}t}\bigg[\sum\limits_{k=1}^{n}\bigg[ \cos(\grave{\lambda}_k^\beta t) +\frac{b}{2\grave{\lambda}_k^\beta}\sin(\grave{\lambda}_k^\beta t) \bigg]^2\phi_k^2\bigg]^{\frac{1}{2}} \\&&+\frac{e^{-\frac{b}{2}t}}{\grave{\lambda}_1^\beta}\bigg[\sum\limits_{k=1}^{n}\psi_k^2\bigg]^{\frac{1}{2}} +\bigg[\sum\limits_{k=1}^{n}\bigg[\frac{1}{\grave{\lambda}_k^\beta}\int_{0}^{t}e^{-\frac{b}{2}(t-s)} \nu \mathbf{Sat} _{2,k}(|\delta_{m-1}(s)|^2)\mathrm{d}s\bigg]^2\bigg]^{\frac{1}{2}} \\&&+\bigg[\sum\limits_{k=1}^{n} \bigg[\frac{1}{\grave{\lambda}_k^\beta}\int_{0}^{t}e^{-\frac{b}{2}(t-s)} g_k(s)\mathrm{d}s\bigg]^2\bigg]^{\frac{1}{2}}\bigg\}^2 \\ &\leq&\lim\limits_{n\rightarrow\infty}\bigg\{\frac{e^{-\frac{b}{2}t}(2\grave{\lambda}_1^\beta+b)}{2\grave{\lambda}_1^\beta}|\phi|_2 +\frac{1}{\grave{\lambda}_1^\beta}|\psi|_2 \\ &&+\frac{1}{\grave{\lambda}_1^\beta}\bigg[ \int_{0}^{t}e^{-\frac{b}{2}(t-s)}\mathrm{d}s\int_{0}^{t}e^{-\frac{b}{2}(t-s)} \sum\limits_{k=1}^{n}\nu^2 \mathbf{Sat} _{2,k}^2(|\delta_{m-1}(s)|^2)\mathrm{d}s\bigg]^{\frac{1}{2}} \\ &&+\frac{1}{\grave{\lambda}_1^\beta}\bigg[ \int_{0}^{t}e^{-\frac{b}{2}(t-s)}\mathrm{d}s\int_{0}^{t}e^{-\frac{b}{2}(t-s)}\sum\limits_{k=1}^{n} g_k^2(s)\mathrm{d}s\bigg]^{\frac{1}{2}}\bigg\}^2 \\ &\leq&\bigg[\frac{(2\grave{\lambda}_1^\beta+b)}{2\grave{\lambda}_1^\beta}|\phi|_2 +\frac{1}{\grave{\lambda}_1^\beta}|\psi|_2 +\frac{2\nu}{\grave{\lambda}_1^\beta b\varepsilon}\delta_\infty +\frac{2}{\grave{\lambda}_1^\beta b}g_\infty \bigg]^2:=\eta_\infty, \end{aligned} $

其中 $t\in\mathbf{R}_0$, $m=1,2,\cdots $. 因此, $\{\eta_m:m=1,2,\cdots \}\subseteq\mathrm{L}^2(\Omega)$ 是有定义的.

对任意的 $\epsilon\in(0,1)$, 取任意的 $t_0, t_1\in\mathbf{R}_0$ 满足 $t_1\geq t_0$ 以及

$ \max\bigg\{\bigg[1-e^{-\frac{a}{\hbar}(t_1-t_0)}\bigg]^2, \bigg[1-e^{-\frac{a}{\hbar}(t_1-t_0)} +e^{-\frac{a}{\hbar}t_1}-e^{-\frac{a}{\hbar}t_0}\bigg]^2\bigg\} <\frac{a^2\varrho^2\epsilon^2}{4\mu^2\delta_\infty^2}, $

$ \max\bigg\{\bigg[1-e^{-\frac{b}{2}(t_1-t_0)}\bigg]^2, \bigg[1-e^{-\frac{b}{2}(t_1-t_0)} +e^{-\frac{b}{2}t_1}-e^{-\frac{b}{2}t_0}\bigg]^2\bigg\} <\frac{(\grave{\lambda}_k^\beta)^2\varepsilon^2b^2\epsilon^2}{16\nu^2\delta_\infty^2}. $

利用迭代序列 (4.1) 和 Hölder 不等式, 直接计算得到

$ \begin{aligned} |\hat{\delta}_m(t_1)-\hat{\delta}_m(t_0)|_2^2 &=&\frac{1}{\hbar^2}\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}\bigg\{ \int_{0}^{t_1} e^{-\frac{a}{\hbar}(t_1-s)} \bigg[ \begin{array}{c} \cos(\acute{\lambda}_k^\alpha (t_1-s)) \\ \sin(\acute{\lambda}_k^\alpha (t_1-s)) \end{array} \bigg]^\top \\&&\times \bigg[ \begin{array}{c} -\mu\widehat{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1})\\ \mu\widetilde{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1})\\ \end{array} \bigg](s)\mathrm{d}s-\int_{0}^{t_0} e^{-\frac{a}{\hbar}(t_0-s)} \\&&\times\bigg[ \begin{array}{c} \cos(\acute{\lambda}_k^\alpha (t_0-s)) \\ \sin(\acute{\lambda}_k^\alpha (t_0-s)) \end{array} \bigg]^\top \bigg[ \begin{array}{c} -\mu\widehat{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1})\\ \mu\widetilde{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1})\\ \end{array} \bigg](s)\mathrm{d}s\bigg\}^2 \\ &\leq&\frac{2}{\hbar^2}\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}\bigg[ \int_{t_0}^{t_1} e^{-\frac{a}{\hbar}(t_1-s)} \bigg[ \begin{array}{c} 1 \\ 1 \end{array} \bigg]^\top \bigg[ \begin{array}{c} \mu|\widehat{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1})|\\ \mu|\widetilde{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1})|\\ \end{array} \bigg](s) \mathrm{d}s\bigg]^2 \\ &&+\frac{2}{\hbar^2}\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}\bigg[\int_{0}^{t_0}\bigg[e^{-\frac{a}{\hbar}(t_0-s)} -e^{-\frac{a}{\hbar}(t_1-s)}\bigg] \bigg[ \begin{array}{c} 1 \\ 1 \end{array} \bigg]^\top \\&&\times \bigg[ \begin{array}{c} \mu|\widehat{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1})|\\ \mu|\widetilde{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1})|\\ \end{array} \bigg](s) \mathrm{d}s\bigg]^2 \\ &\leq&\frac{2\mu^2}{\hbar^2\varrho^2} \int_{t_0}^{t_1}e^{-\frac{a}{\hbar}(t_1-s)}\mathrm{d}s \int_{t_0}^{t_1}e^{-\frac{a}{\hbar}(t_1-s)}|\delta_{m-1}(s)|_2^2\mathrm{d}s +\frac{2\mu^2}{\hbar^2\varrho^2}\int_{0}^{t_0} \\ &&\times\bigg[e^{-\frac{a}{\hbar}(t_0-s)} -e^{-\frac{a}{\hbar}(t_1-s)}\bigg]\mathrm{d}s\int_{0}^{t_0} \bigg[e^{-\frac{a}{\hbar}(t_0-s)} -e^{-\frac{a}{\hbar}(t_1-s)}\bigg]|\delta_{m-1}(s)|_2^2\mathrm{d}s \\ &\leq&\frac{2\mu^2\delta_\infty^2}{a^2\varrho^2} \bigg[1-e^{-\frac{a}{\hbar}(t_1-t_0)}\bigg]^2 +\frac{2\mu^2\delta_\infty^2}{a^2\varrho^2}\bigg[1-e^{-\frac{a}{\hbar}(t_1-t_0)} +e^{-\frac{a}{\hbar}t_1}-e^{-\frac{a}{\hbar}t_0}\bigg]^2 \\ &\leq&\epsilon^2,\quad \forall m=1,2,\cdots. \end{aligned} $

同理, $|\breve{\delta}_m(t_1)-\breve{\delta}_m(t_0)|_2^2\leq \epsilon^2$, $\forall m=1,2,\cdots $.

类似地, 结合方程 (4.2) 可得

$ \begin{aligned} |\eta_m(t_1)-\eta_m(t_0)|_2^2 &=&\frac{\nu^2}{(\grave{\lambda}_1^\beta)^2}\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}\bigg\{ \int_{0}^{t_1}e^{-\frac{b}{2}(t_1-s)}\sin(\grave{\lambda}_k^\beta(t_1-s)) \mathbf{Sat} _{2,k}(|\delta_m(s)|^2)\mathrm{d}s \\&&-\int_{0}^{t_0} e^{-\frac{b}{2}(t_0-s)}\sin(\grave{\lambda}_k^\beta(t_0-s)) \mathbf{Sat} _{2,k}(|\delta_m(s)|^2)\mathrm{d}s\bigg\}^2 \\ &\leq&\frac{8\nu^2\delta_\infty^2}{(\grave{\lambda}_1^\beta)^2\varepsilon^2b^2}\bigg\{ \bigg[1-e^{-\frac{b}{2}(t_1-t_0)}\bigg]^2 +\bigg[1-e^{-\frac{b}{2}(t_1-t_0)} +e^{-\frac{b}{2}t_1}-e^{-\frac{b}{2}t_0}\bigg]^2\bigg\} \\ &\leq&\epsilon^2,\quad \forall m=1,2,\cdots. \end{aligned}$

综上所述, $\{(\hat{\delta}_m,\breve{\delta}_m,\eta_m): m=0,1,\cdots \}$ 是 $\mathrm{L}^2$-连续的. 证毕.

4.2 概自守

考虑 $(\hat{\delta}_m,\breve{\delta}_m,\eta_m)$ 的部分和, 其表达式为

(4.3) $ \begin{aligned} \label{415.1} \left[ \begin{array}{c} \hat{\delta}_{m}^n\\ \breve{\delta}_{m}^n\\ \end{array} \right](t)&=&\sum\limits_{k=1}^{n}\Bigg\{ e^{-\frac{a}{\hbar}t}\left[ \begin{array}{cc} \cos(\acute{\lambda}_k^\alpha t) & \sin(\acute{\lambda}_k^\alpha t) \\ -\sin(\acute{\lambda}_k^\alpha t) & \cos(\acute{\lambda}_k^\alpha t) \\ \end{array} \right] \left[ \begin{array}{c} \hat{\varphi}_k\\ \breve{\varphi}_k \\ \end{array} \right] \\ && + \frac{1}{\hbar}\int_{0}^t e^{-\frac{a}{\hbar}(t-s)} \left[ \begin{array}{cc} \cos(\acute{\lambda}_k^\alpha (t-s)) & \sin(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (s-t)) & \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \end{array} \right] \\ && \times\left[ \begin{array}{c} -\mu\widehat{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1})+\mathrm{Im}f_k \\ \mu\widetilde{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1})-\mathrm{Re}f_k \\ \end{array} \right](s)\mathrm{d}s\Bigg\}e_k^\alpha, \end{aligned}$

(4.4) $ \begin{aligned} \label{415.2} \eta_{m}^n(t)&=& \sum\limits_{k=1}^n\Bigg\{e^{-\frac{b}{2}t}\left[ \cos(\grave{\lambda}_k^\beta t) +\frac{b}{2\grave{\lambda}_k^\beta}\sin(\grave{\lambda}_k^\beta t)\right]\phi_k +\frac{1}{\grave{\lambda}_k^\beta}e^{-\frac{b}{2}t}\sin(\grave{\lambda}_k^\beta t)\psi_k \\ &&+\frac{1}{\grave{\lambda}_k^\beta}\int_{0}^te^{-\frac{b}{2}(t-s)}\sin(\grave{\lambda}_k^\beta(t-s)) \left[\nu \mathbf{Sat} _{2,k}(|\delta_{m-1}(s)|^2)+g_k(s)\right]\mathrm{d}s\Bigg\}e_k^\beta, \end{aligned}$

其中 $t\in\mathbf{R}_0$, $m,n=1,2,\cdots.$ 通过命题 4.1 的证明可知 $\{(\hat{\delta}_m^n,\breve{\delta}_m^n,\eta_m^n): n=1,2,\cdots \}$ 在 $\{\mathrm{L}^2(\Omega)\}^3$ 中对任意的 $t\in\bar{I}_T$ 一致收敛于 $(\hat{\delta}_m,\breve{\delta}_m,\eta_m)$, $m=1,2,\cdots $.

命题4.2 若定理 1.4 所有假设均成立. 则 $\{(\delta_m,\eta_m)=(\hat{\delta}_m,\breve{\delta}_m,\eta_m): m=0,1,\cdots \} \subseteq\mathbb{AAA}(\mathbf{R}_0,\{\mathrm{L}^2(\Omega)\}^3)$.

证简记

$\hat{\mathcal{F}}(\varphi,\phi,t)=-\mu\widehat{ \mathbf{Sat} }_{1}(\varphi\phi)+\mathrm{Im}f(\cdot,t),\quad \breve{\mathcal{F}}(\varphi,\phi,t)=\mu\widetilde{ \mathbf{Sat} }_{1}(\varphi\phi)-\mathrm{Re}f(\cdot,t), $

$ \mathcal{G}(\varphi,t)=\nu \mathbf{Sat} _{2}(|\varphi|^2)+g(\cdot,t),\quad \forall t\in\mathbf{R}_0.$

由于 $(\varphi,\phi)=(\hat{\varphi},\breve{\varphi},\phi)\in\mathbb{AAA}(\mathbf{R}_0,\{\mathrm{L}^2(\Omega)\}^3)$, 利用文献 [30,定理 3.6] 可知,

$ \hat{\mathcal{F}}(\varphi,\phi,t), \breve{\mathcal{F}}(\varphi,\phi,t) \in\mathbb{AAA}(\{\mathrm{L}^2(\Omega)\}^3\times\mathbf{R}_0,\mathrm{L}^2(\Omega)), $

$ \mathcal{G}(\varphi,t)\in\mathbb{AAA}(\{\mathrm{L}^2(\Omega)\}^2\times\mathbf{R}_0,\mathrm{L}^2(\Omega)),\quad \forall t\in\mathbf{R}_0.$

因此,

$ \hat{\mathcal{F}}(\varphi,\phi,\cdot)=\hat{\mathcal{F}}_{\mathbb{A}}(\varphi,\phi,\cdot) +\hat{\mathcal{F}}_{\mathbb{N}}(\varphi,\phi,\cdot),\quad \breve{\mathcal{F}}(\varphi,\phi,\cdot)=\breve{\mathcal{F}}_{\mathbb{A}}(\varphi,\phi,\cdot) +\breve{\mathcal{F}}_{\mathbb{N}}(\varphi,\phi,\cdot), $

$\mathcal{G}(\varphi,\cdot)=\mathcal{G}_{\mathbb{A}}(\varphi,\cdot)+\mathcal{G}_{\mathbb{N}}(\varphi,\cdot),$

其中 $\hat{\mathcal{F}}_{\mathbb{A}}, \breve{\mathcal{F}}_{\mathbb{A}} \in\mathbb{AA}(\{\mathrm{L}^2(\Omega)\}^3\times\mathbf{R}_0,\mathrm{L}^2(\Omega))$, $\hat{\mathcal{F}}_{\mathbb{N}}, \breve{\mathcal{F}}_{\mathbb{N}} \in\aleph(\{\mathrm{L}^2(\Omega)\}^3\times\mathbf{R}_0,\mathrm{L}^2(\Omega))$, $\mathcal{G}_{\mathbb{A}}\in\mathbb{AA}(\{\mathrm{L}^2(\Omega)\}^2\times\mathbf{R}_0,\mathrm{L}^2(\Omega)),$ $\mathcal{G}_{\mathbb{N}}\in \aleph(\{\mathrm{L}^2(\Omega)\}^2\times\mathbf{R}_0,\mathrm{L}^2(\Omega))$.

若 $m=1$, 则 $\hat{\delta}_{1}^n(t)=\sum\limits_{i=1}^4\hat{\delta}_{1,i}^n(t)$, 其中

$ \left[ \begin{array}{c} \hat{\delta}_{1,1}^n\\ \breve{\delta}_{1,1}^n\\ \end{array} \right](t)=\sum\limits_{k=1}^{n} e^{-\frac{a}{\hbar}t}\left[ \begin{array}{cc} \cos(\acute{\lambda}_k^\alpha t) & \sin(\acute{\lambda}_k^\alpha t) \\ -\sin(\acute{\lambda}_k^\alpha t) & \cos(\acute{\lambda}_k^\alpha t) \\ \end{array} \right] \left[ \begin{array}{c} \hat{\varphi}_k\\ \breve{\varphi}_k \\ \end{array} \right]e_k^\alpha, $

$ \left[ \begin{array}{c} \hat{\delta}_{1,2}^n\\ \breve{\delta}_{1,2}^n\\ \end{array} \right](t)=\frac{1}{\hbar}\sum\limits_{k=1}^{n} \int_{0}^t e^{-\frac{a}{\hbar}(t-s)} \left[ \begin{array}{cc} \cos(\acute{\lambda}_k^\alpha (t-s)) & \sin(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (s-t)) & \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \end{array} \right] \left[ \begin{array}{c} \hat{\mathcal{F}}_{\mathbb{N},k}(\varphi,\phi,s) \\ \breve{\mathcal{F}}_{\mathbb{N},k}(\varphi,\phi,s) \\ \end{array} \right]\mathrm{d}s e_k^\alpha, $

$ \left[ \begin{array}{c} \hat{\delta}_{1,3}^n\\ \breve{\delta}_{1,3}^n\\ \end{array} \right](t)=\sum\limits_{k=1}^{n} \int_{-\infty}^0 \frac{-e^{-\frac{a}{\hbar}(t-s)}}{\hbar} \left[ \begin{array}{cc} \cos(\acute{\lambda}_k^\alpha (t-s)) & \sin(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (s-t)) & \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \end{array} \right] \left[ \begin{array}{c} \hat{\mathcal{F}}_{\mathbb{A},k}(\varphi,\phi,s) \\ \breve{\mathcal{F}}_{\mathbb{A},k}(\varphi,\phi,s) \\ \end{array} \right]\mathrm{d}s e_k^\alpha, $

$ \left[ \begin{array}{c} \hat{\delta}_{1,4}^n\\ \breve{\delta}_{1,4}^n\\ \end{array} \right](t)=\frac{1}{\hbar}\sum\limits_{k=1}^{n} \int_{-\infty}^t e^{-\frac{a}{\hbar}(t-s)} \left[ \begin{array}{cc} \cos(\acute{\lambda}_k^\alpha (t-s)) & \sin(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (s-t)) & \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \end{array} \right] \left[ \begin{array}{c} \hat{\mathcal{F}}_{\mathbb{A},k}(\varphi,\phi,s) \\ \breve{\mathcal{F}}_{\mathbb{A},k}(\varphi,\phi,s) \\ \end{array} \right]\mathrm{d}s e_k^\alpha, $

$ \eta_{1,1}^n(t)=\sum\limits_{k=1}^{n}\Bigg\{e^{-\frac{b}{2}t}\left[ \cos(\grave{\lambda}_k^\beta t) +\frac{b}{2\grave{\lambda}_k^\beta}\sin(\grave{\lambda}_k^\beta t) \right]\phi_k+ \frac{1}{\grave{\lambda}_k^\beta}e^{-\frac{b}{2}t}\sin(\grave{\lambda}_k^\beta t)\psi_k\Bigg\} e_k^\beta, $

$ \eta_{1,2}^n(t)=\frac{1}{\grave{\lambda}_k^\beta}\sum\limits_{k=1}^{n}\int_{0}^t e^{-\frac{b}{2}(t-s)}\sin(\grave{\lambda}_k^\beta(t-s)) \mathcal{G}_{\mathbb{N},k}(\varphi,s)\mathrm{d}s e_k^\beta, $

$ \eta_{1,3}^n(t)=\frac{-1}{\grave{\lambda}_k^\beta}\sum\limits_{k=1}^{n}\int_{-\infty}^0 e^{-\frac{b}{2}(t-s)}\sin(\grave{\lambda}_k^\beta(t-s)) \mathcal{G}_{\mathbb{A},k}(\varphi,s)\mathrm{d}s e_k^\beta, $

$ \eta_{1,4}^n(t)=\frac{1}{\grave{\lambda}_k^\beta}\sum\limits_{k=1}^{n}\int_{-\infty}^t e^{-\frac{b}{2}(t-s)}\sin(\grave{\lambda}_k^\beta(t-s)) \mathcal{G}_{\mathbb{A},k}(\varphi,s)\mathrm{d}s e_k^\beta,\quad n\in\mathbf{Z}_+\cup\{\infty\}, t\in\mathbf{R}_0. $

与命题 4.1 类似, $\{(\hat{\delta}_{1,i}^n(\cdot),\breve{\delta}_{1,i}^n(\cdot),\eta_{1,i}^n(\cdot))\}$ 在 $\{\mathrm{L}^2(\Omega)\}^3$ 中收敛于 $(\hat{\delta}_{1,i}^\infty(\cdot),\breve{\delta}_{1,i}^\infty(\cdot),\eta_{1,i}^\infty(\cdot))$, $i=1,2,3$. 此外,

$ \begin{aligned} \lim\limits_{t\rightarrow\infty}|\hat{\delta}_{1,1}^\infty(t)|_2^2&\leq& \lim\limits_{t\rightarrow\infty}\lim\limits_{n\rightarrow\infty}e^{-\frac{2a}{\hbar}t}\sum\limits_{k=1}^{n}(\hat{\varphi}_k^2+\breve{\varphi}_k^2) \leq \lim\limits_{t\rightarrow\infty}e^{-\frac{2a}{\hbar}t}|\varphi|_2^2=0, \\ \lim\limits_{t\rightarrow\infty}|\hat{\delta}_{1,2}^\infty(t)|_2^2 &\leq& \lim\limits_{t\rightarrow\infty}\lim\limits_{n\rightarrow\infty}\frac{1}{a\hbar} \int_{0}^te^{-\frac{a}{\hbar}(t-s)}\sum\limits_{k=1}^{n}\left[\hat{\mathcal{F}}_{\mathbb{N},k}^2(\varphi,\phi,s) +\breve{\mathcal{F}}_{\mathbb{N},k}^2(\varphi,\phi,s)\right]\mathrm{d}s \\ &\leq& \frac{1}{a\hbar}\lim\limits_{t\rightarrow\infty} \int_{0}^{\frac{t}{2}}e^{-\frac{a}{\hbar}(t-s)}\left[\big|\hat{\mathcal{F}}_{\mathbb{N}}(\varphi,\phi,s)\big|_2^2 +\big|\breve{\mathcal{F}}_{\mathbb{N}}(\varphi,\phi,s)\big|_2^2\right]\mathrm{d}s \\ &&+\frac{1}{a\hbar}\lim\limits_{t\rightarrow\infty} \int_{\frac{t}{2}}^te^{-\frac{a}{\hbar}(t-s)}\left[\big|\hat{\mathcal{F}}_{\mathbb{N}}(\varphi,\phi,s)\big|_2^2 +\big|\breve{\mathcal{F}}_{\mathbb{N}}(\varphi,\phi,s)\big|_2^2\right]\mathrm{d}s \\ &\leq&\frac{1}{a^2}\lim\limits_{t\rightarrow\infty}(e^{-\frac{at}{2\hbar}}-e^{-\frac{at}{\hbar}}) \sup\limits_{s\in\mathbf{R}_0}\left[\big|\hat{\mathcal{F}}_{\mathbb{N}}(\varphi,\phi,s)\big|_2^2 +\big|\breve{\mathcal{F}}_{\mathbb{N}}(\varphi,\phi,s)\big|_2^2\right] \\ &&+\frac{1}{a^2}\lim\limits_{t\rightarrow\infty} \sup\limits_{s\in[\frac{t}{2},\infty)}\left[\big|\hat{\mathcal{F}}_{\mathbb{N}}(\varphi,\phi,s)\big|_2^2 +\big|\breve{\mathcal{F}}_{\mathbb{N}}(\varphi,\phi,s)\big|_2^2\right]=0, \\ \lim\limits_{t\rightarrow\infty}|\hat{\delta}_{1,3}^\infty(t)|_2^2 &\leq&\frac{1}{a\hbar} \lim\limits_{t\rightarrow\infty}\lim\limits_{n\rightarrow\infty} \int_{-\infty}^0e^{-\frac{a}{\hbar}(t-s)}\sum\limits_{k=1}^{n}\left[\hat{\mathcal{F}}_{\mathbb{A},k}^2(\varphi,\phi,s) +\breve{\mathcal{F}}_{\mathbb{A},k}^2(\varphi,\phi,s)\right]\mathrm{d}s \\ &\leq&\frac{1}{a^2} \lim\limits_{t\rightarrow\infty}e^{-\frac{at}{\hbar}} \sup\limits_{s\in\mathbf{R}_0}\left[\big|\hat{\mathcal{F}}_{\mathbb{A}}(\varphi,\phi,s)\big|_2^2 +\big|\breve{\mathcal{F}}_{\mathbb{A}}(\varphi,\phi,s)\big|_2^2\right]=0, \end{aligned}$

因此, $\hat{\delta}_{1,i}^\infty\in\aleph(\mathbf{R}_0, \mathrm{L}^2(\Omega))$, $i=1,2,3$.

由于 $\hat{\mathcal{F}}_{\mathbb{A}}, \breve{\mathcal{F}}_{\mathbb{A}} \in\mathbb{AA}(\{\mathrm{L}^2(\Omega)\}^3\times\mathbf{R},\mathrm{L}^2(\Omega))$, 因而对于每个序列 $\{\bar{t}_p\}_{p\in\mathbf{Z}_+}$, 都存在一个子列 $\{t_p\}_{p\in\mathbf{Z}_+}\subseteq\{\bar{t}_p\}_{p\in\mathbf{Z}_+}$ 和函数 $\hat{\mathcal{F}}_{\mathbb{A}}^\circ, \breve{\mathcal{F}}_{\mathbb{A}}^\circ: \{\mathrm{L}^2(\Omega)\}^3\times\mathbf{R}\rightarrow\mathrm{L}^2(\Omega)$ 使得在 $\mathrm{L}^2(\Omega)$ 中, 对于每个 $t\in\mathbf{R}$

有

$ \hat{\mathcal{F}}_{\mathbb{A}}^\circ(\delta,\eta,t) =\lim\limits_{p\rightarrow\infty}\hat{\mathcal{F}}_{\mathbb{A}}(\delta,\eta,t+t_p),\quad \breve{\mathcal{F}}_{\mathbb{A}}^\circ(\delta,\eta,t) =\lim\limits_{p\rightarrow\infty}\breve{\mathcal{F}}_{\mathbb{A}}(\delta,\eta,t+t_p), $

$ \lim\limits_{p\rightarrow\infty}\hat{\mathcal{F}}_{\mathbb{A}}^\circ(\delta,\eta,t-t_p) =\hat{\mathcal{F}}_{\mathbb{A}}(\delta,\eta,t),\quad \lim\limits_{p\rightarrow\infty}\breve{\mathcal{F}}_{\mathbb{A}}^\circ(\delta,\eta,t-t_p) =\breve{\mathcal{F}}_{\mathbb{A}}(\delta,\eta,t) $

令

$\hat{\delta}_{1,4}^{\circ,n}(t)= \frac{1}{\hbar}\sum\limits_{k=1}^{n} \int_{-\infty}^t e^{-\frac{a}{\hbar}(t-s)} \left[ \begin{array}{c} \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (t-s)) \end{array} \right]^\top \left[ \begin{array}{c} \hat{\mathcal{F}}_{\mathbb{A},k}^\circ(\varphi,\phi,s) \\ \breve{\mathcal{F}}_{\mathbb{A},k}^\circ(\varphi,\phi,s) \\ \end{array} \right]\mathrm{d}s e_k^\alpha, $

其中 $n\in\mathbf{Z}_+\cup\{\infty\}$, $t\in\mathbf{R}$. 与方程 (4.3) 和 (4.4) 的方法类似, 有 $\{\hat{\delta}_{1,4}^{\circ,n}(t):n=1,2,\cdots \}$ 在 $\mathrm{L}^2(\Omega)$ 中关于 $t\in\bar{I}_T$ 一致收敛于 $\hat{\delta}_{1,4}^{\circ,\infty}(t)$. 因此,

$ \begin{aligned} \Big|\hat{\delta}_{1,4}^{n}(t+t_p)-\hat{\delta}_{1,4}^{\circ,n}(t)\Big|_2^2 &=&\frac{1}{\hbar^2}\int_{-\infty}^te^{-\frac{a}{\hbar}(t-s)}\mathrm{d}s\int_{-\infty}^te^{-\frac{a}{\hbar}(t-s)} \\ &&\times\bigg[\sum\limits_{k=1}^{n}\Big(\hat{\mathcal{F}}_{\mathbb{A},k}(\varphi,\phi,s+t_p) -\hat{\mathcal{F}}_{\mathbb{A},k}^\circ(\varphi,\phi,s)\Big)^2 \\&&+ \sum\limits_{k=1}^{n}\Big(\breve{\mathcal{F}}_{\mathbb{A},k}(\varphi,\phi,s+t_p) -\breve{\mathcal{F}}_{\mathbb{A},k}^\circ(\varphi,\phi,s)\Big)^2 \bigg]\mathrm{d}s \\ &\leq&\frac{1}{a\hbar}\int_{-\infty}^te^{-\frac{a}{\hbar}(t-s)} \Big|\hat{\mathcal{F}}_{\mathbb{A}}(\varphi,\phi,s+t_p) -\hat{\mathcal{F}}_{\mathbb{A}}^\circ(\varphi,\phi,s)\Big|_2^2\mathrm{d}s \\ &&+\frac{1}{a\hbar}\int_{-\infty}^te^{-\frac{a}{\hbar}(t-s)} \Big|\breve{\mathcal{F}}_{\mathbb{A}}(\varphi,\phi,s+t_p) -\breve{\mathcal{F}}_{\mathbb{A}}^\circ(\varphi,\phi,s)\Big|_2^2\mathrm{d}s. \end{aligned}$

在上述不等式中令 $n\rightarrow\infty$ 可得

$ \begin{aligned} \Big|\hat{\delta}_{1,4}^{\infty}(t+t_p)-\hat{\delta}_{1,4}^{\circ,\infty}(t)\Big|_2^2 &\leq&\frac{1}{a\hbar}\int_{-\infty}^te^{-\frac{a}{\hbar}(t-s)} \Big|\hat{\mathcal{F}}_{\mathbb{A}}(\varphi,\phi,s+t_p) -\hat{\mathcal{F}}_{\mathbb{A}}^\circ(\varphi,\phi,s)\Big|_2^2\mathrm{d}s \\ &&+\frac{1}{a\hbar}\int_{-\infty}^te^{-\frac{a}{\hbar}(t-s)} \Big|\breve{\mathcal{F}}_{\mathbb{A}}(\varphi,\phi,s+t_p) -\breve{\mathcal{F}}_{\mathbb{A}}^\circ(\varphi,\phi,s)\Big|_2^2\mathrm{d}s. \end{aligned} $

由 Lebesgue 控制收敛定理得到

$ \lim\limits_{p\rightarrow\infty}\Big|\hat{\delta}_{1,4}^{\infty}(t+t_p)-\hat{\delta}_{1,4}^{\circ,\infty}(t)\Big|_2^2=0, \quad \forall t\in\mathbf{R}. $

同理,

$\lim\limits_{p\rightarrow\infty}\Big|\hat{\delta}_{1,4}^{\circ,\infty}(t-t_p)-\hat{\delta}_{1,4}^{\infty}(t)\Big|_2^2=0, \quad \forall t\in\mathbf{R}. $

因此, $\hat{\delta}_{1,4}^\infty(\cdot)\in\mathbb{AA}(\mathbf{R},\mathrm{L}^2(\Omega))$ 并且当 $n\rightarrow\infty$ 时, 有

$ \hat{\delta}_{1}^n(\cdot)=\hat{\delta}_{1,4}^n(\cdot) +\sum\limits_{i=1}^3\hat{\delta}_{1,i}^n(\cdot)\xrightarrow[]{\mathrm{L}^2(\Omega)} \hat{\delta}_{1}(\cdot)=\hat{\delta}_{1,4}^\infty(\cdot) +\sum\limits_{i=1}^3\hat{\delta}_{1,i}^\infty(\cdot)\in\mathbb{AAA}(\mathbf{R}_0,\mathrm{L}^2(\Omega)) $

同理, 当 $n\rightarrow\infty$ 时, 有

$ \breve{\delta}_{1}^n(\cdot)=\breve{\delta}_{1,4}^n(\cdot) +\sum\limits_{i=1}^3\breve{\delta}_{1,i}^n(\cdot)\xrightarrow[]{\mathrm{L}^2(\Omega)} \breve{\delta}_{1}(\cdot)=\breve{\delta}_{1,4}^\infty(\cdot) +\sum\limits_{i=1}^3\breve{\delta}_{1,i}^\infty(\cdot)\in\mathbb{AAA}(\mathbf{R}_0,\mathrm{L}^2(\Omega)), $

$ \eta_{1}^n(\cdot)=\eta_{1,4}^n(\cdot)+\sum\limits_{i=1}^3\eta_{1,i}^n(\cdot)\xrightarrow[]{\mathrm{L}^2(\Omega)} \eta_{1}(\cdot)=\eta_{1,4}^\infty(\cdot) +\sum\limits_{i=1}^3\eta_{1,i}^\infty(\cdot)\in\mathbb{AAA}(\mathbf{R}_0,\mathrm{L}^2(\Omega)). $

由数学归纳法可知 $\{(\delta_m,\eta_m)=(\hat{\delta}_m,\breve{\delta}_m,\eta_m): m=2,3,\cdots \} \subseteq\mathbb{AAA}(\mathbf{R}_0,\{\mathrm{L}^2(\Omega)\}^3)$. 证毕.

定义范数 $|\xi|_{\sup}=\sup\limits_{t\in\mathbf{R}_0}|\xi(t)|_2$, $\forall \xi\in\mathbb{AAA}(\mathbf{R}_0,\mathrm{L}^2(\Omega))$. 则根据文献 [30] 可知 $(\mathbb{AAA}(\mathbf{R}_0,$ $\mathrm{L}^2(\Omega)), |\cdot|_{\sup})$ 是 Banach 空间.

命题4.3 若定理 1.4 的所有假设都成立. 则 $\{(\delta_m,\eta_m)=(\hat{\delta}_m,\breve{\delta}_m,\eta_m): m=0,1,\cdots \}$ 是 $\mathbb{AAA}(\mathbf{R}_0,\{\mathrm{L}^2(\Omega)\}^3)$ 中的 Cauchy 序列.

证由 $\{\hat{\delta}_m: m=1,2,\cdots \}$ 的部分和得,

$ \begin{aligned} \big|\hat{\delta}_{m+1}^n(t)-\hat{\delta}_{m}^n(t)\big|_{2}^2&=&\sum\limits_{k=1}^{n}\bigg\{ \frac{\mu}{\hbar}\int_{0}^t e^{-\frac{a}{\hbar}(t-s)} \left[ \begin{array}{cc} \cos(\acute{\lambda}_k^\alpha (t-s)) \\ \sin(\acute{\lambda}_k^\alpha (t-s)) \end{array} \right]^\top \\ && \times\bigg[ \begin{array}{c} -\widehat{ \mathbf{Sat} }_{1,k}(\delta_{m}\eta_{m})+\widehat{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1}) \\ \widetilde{ \mathbf{Sat} }_{1,k}(\delta_{m}\eta_{m})-\widetilde{ \mathbf{Sat} }_{1,k}(\delta_{m-1}\eta_{m-1}) \\ \end{array} \bigg](s)\mathrm{d}s\bigg\}^2 \\ &\leq&\frac{\mu^2}{a\hbar} \int_{0}^te^{-\frac{a}{\hbar}(t-s)}\Big| \widehat{ \mathbf{Sat} }_{1}(\delta_{m}\eta_{m})-\widehat{ \mathbf{Sat} }_{1}(\delta_{m-1}\eta_{m-1}) \Big|_2^2\mathrm{d}s \\ &&+\frac{\mu^2}{a\hbar} \int_{0}^te^{-\frac{a}{\hbar}(t-s)}\Big| \widetilde{ \mathbf{Sat} }_{1}(\delta_{m}\eta_{m})-\widetilde{ \mathbf{Sat} }_{1}(\delta_{m-1}\eta_{m-1}) \Big|_2^2\mathrm{d}s \\ &\leq&\frac{16\mu^2}{a^2\rho_0^2} \left(\big|\delta_{m}-\delta_{m-1}\big|_{\sup}^2+\big|\eta_{m}-\eta_{m-1}\big|_{\sup}^2\right), \end{aligned} $

其中

$\big|\delta_{m}-\delta_{m-1}\big|_{\sup}^2:=\big|\hat{\delta}_{m}-\hat{\delta}_{m-1}\big|_{\sup}^2 +\big|\breve{\delta}_{m}-\breve{\delta}_{m-1}\big|_{\sup}^2,\quad t\in\mathbf{R}_0, m,n=1,2,\cdots.$

在上述不等式中令 $n\rightarrow\infty$, 可得

$ \big|\hat{\delta}_{m+1}-\hat{\delta}_{m}\big|_{\sup}^2\leq\frac{16\mu^2}{a^2\rho_0^2} \left(\big|\delta_{m}-\delta_{m-1}\big|_{\sup}^2+\big|\eta_{m}-\eta_{m-1}\big|_{\sup}^2\right), \quad m=1,2,\cdots. $

同理,

$ \big|\breve{\delta}_{m+1}-\breve{\delta}_{m}\big|_{\sup}^2\leq\frac{16\mu^2}{a^2\rho_0^2} \left(\big|\delta_{m}-\delta_{m-1}\big|_{\sup}^2+\big|\eta_{m}-\eta_{m-1}\big|_{\sup}^2\right), \quad m=1,2,\cdots. $

综上所述,

$\big|\delta_{m+1}-\delta_{m}\big|_{\sup}\leq\frac{4\sqrt{2}\mu}{a\rho_0} \left(\big|\delta_{m}-\delta_{m-1}\big|_{\sup}+\big|\eta_{m}-\eta_{m-1}\big|_{\sup}\right), \quad m=1,2,\cdots. $

同理, 由方程 (4.4) 可得

$ \begin{aligned} \big|\eta_{m+1}^n(t)-\eta_{m}^n(t)\big|_{2}^2&=&\sum\limits_{k=1}^{n}\Bigg\{ \frac{\nu}{\grave{\lambda}_k^\beta}\int_{0}^te^{-\frac{b}{2}(t-s)}\sin(\grave{\lambda}_k^\beta(t-s)) \\ &&\times \left[ \mathbf{Sat} _{2,k}(|\delta_{m}(s)|^2)- \mathbf{Sat} _{2,k}(|\delta_{m-1}(s)|^2)\right]\mathrm{d}s\Bigg\}^2 \\ &\leq&\frac{2\nu^2}{b(\grave{\lambda}_1^\beta)^2} \int_{0}^te^{-\frac{b}{2}(t-s)}\Big| \mathbf{Sat} _{2}(|\delta_{m}(s)|^2)- \mathbf{Sat} _{2}(|\delta_{m-1}(s)|^2) \Big|_2^2\mathrm{d}s \\ &\leq&\frac{9\nu^2}{b^2(\grave{\lambda}_1^\beta)^2\varepsilon^2} \big|\delta_{m}-\delta_{m-1}\big|_{\sup}^2,\quad \forall t\in\mathbf{R}_0. \end{aligned} $

所以,

(4.6) $ \big|\eta_{m+1}-\eta_{m}\big|_{\sup}\leq\frac{3\nu}{b\grave{\lambda}_1^\beta\varepsilon}\big|\delta_{m}-\delta_{m-1}\big|_{\sup},\quad m=1,2,\cdots.$

定义

$\Pi_0=0,\quad \Pi_m:=\Pi_{m-1}+\max\{\big|\delta_{m}-\delta_{m-1}\big|_{0}, \big|\eta_{m}-\eta_{m-1}\big|_{0}\},\quad m=1,2,\cdots.$

结合不等式 (4.5) 和 (4.6) 可知,

$ \Pi_{m+1}-\Pi_{m}\leq \max\bigg\{\frac{8\sqrt{2}\mu}{a\rho_0}, \frac{3\nu}{b\grave{\lambda}_1^\beta\varepsilon} \bigg\}(\Pi_{m}-\Pi_{m-1})\leq\gamma^m\Pi_{1},\quad m=1,2,\cdots. $

对任意的 $p\in\mathbf{Z}_+$, 当 $m\rightarrow\infty$ 时, 有

$ \max\left\{\big|\delta_{m+p}-\delta_{m}\big|_{0}, \big|\eta_{m+p}-\eta_{m}\big|_{0} \right\}\leq\sum\limits_{k=m}^{m+p-1}\left(\Pi_{k+1}-\Pi_{k} \right)\leq\sum\limits_{k=m}^{m+p-1}\gamma^k \Pi_{1}=\frac{\gamma^m\Pi_{1}}{1-\gamma} \rightarrow0. $

因此, $\{\hat{\delta}_m\}$, $\{\breve{\delta}_m\}$和$\{\eta_m\}$ 是 $\mathbb{AAA}(\mathbf{R}_0,\mathrm{L}^2(\Omega))$ 中的 Cauchy 序列. 证毕.

由命题 4.3 和 $\mathbb{AAA}(\mathbf{R}_0,\{\mathrm{L}^2(\Omega)\}^3)$ 的完备性可知, 存在 $(\delta_*,\eta_*)=(\hat{\delta}_*,\breve{\delta}_*,\eta_*)$ 使得 $\{(\hat{\delta}_m,\breve{\delta}_m,\eta_m)\}$ 在 $\mathbb{AAA}(\mathbf{R}_0,\{\mathrm{L}^2(\Omega)\}^3)$ 中收敛于 $(\hat{\delta}_*,\breve{\delta}_*,\eta_*)$. 取 $(\tilde{\delta}_*,\tilde{\eta}_*)=(\hat{\tilde{\delta}}_*,\breve{\tilde{\delta}}_*,\tilde{\eta}_*)$ 为

$\begin{array}{l} {\left[\begin{array}{c} \hat{\tilde{\delta}}_{*} \\ \breve{\tilde{\delta}}_{*} \end{array}\right](t)=\sum_{k=1}^{\infty}\left\{e^{-\frac{a}{\hbar} t}\left[\begin{array}{cc} \cos \left(\dot{\lambda}_{k}^{\alpha} t\right) & \sin \left(\dot{\lambda}_{k}^{\alpha} t\right) \\ -\sin \left(\dot{\lambda}_{k}^{\alpha} t\right) & \cos \left(\dot{\lambda}_{k}^{\alpha} t\right) \end{array}\right]\left[\begin{array}{c} \hat{\varphi}_{k} \\ \breve{\varphi}_{k} \end{array}\right]\right.} \\ +\frac{1}{\hbar} \int_{0}^{t} e^{-\frac{a}{\hbar}(t-s)}\left[\begin{array}{ll} \cos \left(\dot{\lambda}_{k}^{\alpha}(t-s)\right) & \sin \left(\dot{\lambda}_{k}^{\alpha}(t-s)\right) \\ \sin \left(\hat{\lambda}_{k}^{\alpha}(s-t)\right) & \cos \left(\hat{\lambda}_{k}^{\alpha}(t-s)\right) \end{array}\right] \\ \left.\times\left[\begin{array}{c} -\mu \widehat{\boldsymbol{S a t}}_{1, k}\left(\delta_{*} \eta_{*}\right)+\operatorname{Im} f_{k} \\ \mu \widetilde{\mathbf{S a t}}_{1, k}\left(\delta_{*} \eta_{*}\right)-\operatorname{Re} f_{k} \end{array}\right](s) \mathrm{d} s\right\} e_{k}^{\alpha}, \quad \forall t \in \mathbf{R}_{0}, \\ \tilde{\eta}_{*}(t)=\sum_{k=1}^{\infty}\left\{e^{-\frac{b}{2} t}\left[\cos \left(\grave{\lambda}_{k}^{\beta} t\right)+\frac{b}{2 \grave{\lambda}_{k}^{\beta}} \sin \left(\grave{\lambda}_{k}^{\beta} t\right)\right] \phi_{k}+\frac{1}{\grave{\lambda}_{k}^{\beta}} e^{-\frac{b}{2} t} \sin \left(\grave{\lambda}_{k}^{\beta} t\right) \psi_{k}\right. \\ \left.+\frac{1}{\grave{\lambda}_{k}^{\beta}} \int_{0}^{t} e^{-\frac{b}{2}(t-s)} \sin \left(\grave{\lambda}_{k}^{\beta}(t-s)\right)\left[\nu \mathbf{S a t}_{2, k}\left(\left|\delta_{*}(s)\right|^{2}\right)+g_{k}(s)\right] \mathrm{d} s\right\} e_{k}^{\beta}, \quad \forall t \in \mathbf{R}_{0}. \\ \end{array}$

与命题 4.1 类似, $(\hat{\tilde{\delta}}_*,\breve{\tilde{\delta}}_*,\tilde{\eta}_*)$ 在 $\{\mathrm{L}^2(\Omega)\}^3$ 中是有定义的.

定义 $(\tilde{\delta}_*^n,\tilde{\eta}_*^n)=(\hat{\tilde{\delta}}_*^n,\breve{\tilde{\delta}}_*^n,\tilde{\eta}_*^n)$ 为 $(\hat{\tilde{\delta}}_*,\breve{\tilde{\delta}}_*,\tilde{\eta}_*)$ 的前 $n$ 项的部分和, $n=1,2,\cdots $. 利用与方程 (4.3) 和 (4.4) 类似的方法可知, $\{(\hat{\tilde{\delta}}_*^n(t),\breve{\tilde{\delta}}_*^n(t),\tilde{\eta}_*^n(t))\}$ 在 $\{\mathrm{L}^2(\Omega)\}^3$ 中收敛于 $(\hat{\tilde{\delta}}_*(t),\breve{\tilde{\delta}}_*(t),\tilde{\eta}_*(t))$, $\forall t\in\mathbf{R}_0$. 类似于不等式 (4.5) 和 (4.6) 的讨论可得,

$ \big|\delta_{m}^n(t)-\tilde{\delta}_{*}^n(t)\big|_{2}\leq\frac{4\sqrt{2}\mu}{a\rho_0} \left[\big|\delta_{m}-\delta_{*}\big|_{\sup}+\big|\eta_{m}-\eta_{*}\big|_{\sup}\right], $

$ \big|\eta_{m}^n(t)-\tilde{\eta}_{*}^n(t)\big|_2\leq\frac{\sqrt{3}\nu}{b\grave{\lambda}_1^\beta\varepsilon} \big|\delta_{m}-\delta_{*}\big|_{\sup}, $

其中

$\big|\delta_{m}^n(t)-\tilde{\delta}_{*}^n(t)\big|_{2}:= \sqrt{\big|\hat{\delta}_{m}^n(t)-\hat{\tilde{\delta}}_{*}^n(t)\big|_{2}^2 +\big|\breve{\delta}_{m}^n(t)-\breve{\tilde{\delta}}_{*}^n(t)\big|_{2}^2},\quad t\in\mathbf{R}_0, m, n=1,2,\cdots.$

在上述最后两个不等式中依次取 $n\rightarrow\infty$, $m\rightarrow\infty$, 得

$ \big|\hat{\delta}_{*}(t)-\hat{\tilde{\delta}}_{*}(t)\big|_{2}=0,\quad \big|\breve{\delta}_{*}(t)-\breve{\tilde{\delta}}_{*}(t)\big|_{2}=0,\quad \big|\eta_{*}(t)-\tilde{\eta}_{*}(t)\big|_{2}=0,\quad\forall t\in\mathbf{R}_0. $

定理 1.4 的证明 由定理 1.2 知, $(\mathbf{z}_0, \mathbf{w} _0)=(\hat{\mathbf{z}}_0,\breve{\mathbf{z}}_0, \mathbf{w} _0)$ 是方程 (1.1) 在 $\mathrm{C}(\mathbf{R}_0,\{\mathrm{L}^2(\Omega)\}^3)$ 中的唯一解, 并且该解满足 Fourier 级数 (1.3) 和 (1.4). 类似与定理 1.2 的证明, 有$(\hat{\delta}_*,\breve{\delta}_*,\eta_*)=(\hat{\mathbf{z}}_0,\breve{\mathbf{z}}_0, \mathbf{w} _0)$. 证毕.

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原文顺序

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... 其中 $\mathbf{z}=\hat{\mathbf{z}}+\mathrm{i}\breve{\mathbf{z}}\in\mathbf{C}$, $|\mathbf{z}|=\sqrt{\hat{\mathbf{z}}^2+\check{\mathbf{z}}^2}$, $\rho, \varrho$ 和 $\varepsilon$ 为正饱和系数. 令 $\rho, \varrho, \varepsilon\rightarrow0$, 则方程 (1.1) 转化为非饱和方程, 这类方程在过去的几十年中被广泛研究, 参见文献 [1],[2],[3],[4],[5]. ...

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2022

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2020

... 著名的 Schrödinger-Klein-Gordon 方程是量子场论中描述守恒的复中子场与实介子场相互作用的经典模型. 耦合的 Schrödinger-Klein-Gordon 方程, 因其相当精确地描述了物理系统关于时间的量子态, 而且这种耦合方程的发展为科学和工程领域开辟了新的前景, 因此这类方程吸引了许多学者的关注. 近年来 Schrödinger-Klein-Gordon 系统成为描述量子力学的重要工具, 被众多学者所研究, 参见文献 [3],[4],[5],[6],[7]. 另一方面, 问题 (1.1) 中的非线性项 $ \mathbf{Sat} _1$ 和 $ \mathbf{Sat} _2$ 可以看作一种特殊的"饱和非 Kerr 律", 出现在许多应用中. 饱和定律精确地描述了经过激光束的气体蒸气介电常数的变化, 这种现象也发生在光纤的孤子传播中. 在半导体掺杂光纤中, 孤子输运模型采用饱和非线性而不是通常的 Kerr 非线性. 这么做的主要动机是在半导体掺杂玻璃和其他复合材料可以观察到较弱的非线性. 饱和函数已经被广泛研究, 参见文献 [8],[9],[10],[11],[12]. ...

Structure-preserving algorithms for the two-dimensional fractional Klein-Gordon-Schr?dinger equation

2020

Global attractor of a dissipative fractional Klein Gordon Schr?dinger System

2022

Energy-preserving scheme for the nonlinear fractional Klein-Gordon Schr?dinger equation

2021

Fractional Klein-Gordon-Schr?dinger equations with Mittag-Leffler memory

2020

Finite time extinction for a class of damped Schr?dinger equations with a singular saturated nonlinearity

2022

Existence and concentration of positive solutions for a system of coupled saturable Schr?dinger equations

2020

Bound state for a strongly coupled nonlinear Schr?dinger system with saturation

2019

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2017

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2012

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1955

... Bochner 在其微分几何著作^[13]中首次提出了概自守的概念, 它是概周期的推广. 概自守的概念在研究一类常微分方程、抛物方程和广义微分方程的概自守动力学中是必不可少的, 这一事实进一步说明了概自守在动力学定性研究中的重要性. 首先, 就从系数空间到解空间的升力性质而言, 动力学在概周期的范畴内通常不是封闭的, 但是在概自守的范畴内是封闭的. 其次, 概自守动力学的出现说明了周期系统与概周期系统的主要区别. 例如, 在单调动力系统中, 周期系数的"升力"永远不可能是概自守的. 研究概自守动力学的另一个重要意义是它与 Levitan $N$-概周期性的关系. 因为概自守函数在本质上是 $N$-概周期的, 所以目前对概自守动力学的研究与 $N$-概周期的情况密切相关, 参见文献 [14],[15],[16],[17],[18],[19],[20]. ...

1998

Almost automorphically and almost periodically forced circle flows of almost periodic parabolic equations on $S^1$

2019

Compact almost automorphic function on time scales and its application

2021

2009

2013

Global exponential stability of discrete-time almost automorphic Caputo-Fabrizio BAM fuzzy neural networks via exponential Euler technique

2022

On the construction of almost periodic solutions for a nonlinear Schr?dinger equation

2002

... 纵观历史, 几乎全部现有的文章都集中在 Schrödinger 或 Klein-Gordon 方程的概周期动力学的研究上, 参见文献 [21],[22],[23],[24],[25],[26]. 例如, 利用 KAM 定理, 作者在文献 [21],[22],[23],[24] 中研究了一类特殊的 Schrödinger 方程的概周期解. Signing^[25] 考虑了具有概周期集的 Klein-Gordon 型方程的均匀化问题. Abdallah 等^[26]研究了具有非线性概周期项的 Klein-Gordon-Schrödinger 无限维格系统的一致全局吸引子的存在性. 然而, 据我们所知, 目前几乎没有文献涉及到 Schrödinger 或 Klein-Gordon 方程的概自守动力学研究. 受此启发, 本文将结合 Galerkin 方法、Fourier 级数、Laplace 变换和 Picard 迭代, 研究带有非局部 Laplace 算子的饱和 Schrödinger-Klein-Gordon 方程 (1.1) 在合适的函数空间中概自守弱解的存在性和唯一性. ...

... ]. 例如, 利用 KAM 定理, 作者在文献 [21],[22],[23],[24] 中研究了一类特殊的 Schrödinger 方程的概周期解. Signing^[25] 考虑了具有概周期集的 Klein-Gordon 型方程的均匀化问题. Abdallah 等^[26]研究了具有非线性概周期项的 Klein-Gordon-Schrödinger 无限维格系统的一致全局吸引子的存在性. 然而, 据我们所知, 目前几乎没有文献涉及到 Schrödinger 或 Klein-Gordon 方程的概自守动力学研究. 受此启发, 本文将结合 Galerkin 方法、Fourier 级数、Laplace 变换和 Picard 迭代, 研究带有非局部 Laplace 算子的饱和 Schrödinger-Klein-Gordon 方程 (1.1) 在合适的函数空间中概自守弱解的存在性和唯一性. ...

The existence of almost-periodic solutions for 1-dimensional nonlinear Schr?dinger equation with quasi-periodic forcing

2020

... ],[22],[23],[24] 中研究了一类特殊的 Schrödinger 方程的概周期解. Signing^[25] 考虑了具有概周期集的 Klein-Gordon 型方程的均匀化问题. Abdallah 等^[26]研究了具有非线性概周期项的 Klein-Gordon-Schrödinger 无限维格系统的一致全局吸引子的存在性. 然而, 据我们所知, 目前几乎没有文献涉及到 Schrödinger 或 Klein-Gordon 方程的概自守动力学研究. 受此启发, 本文将结合 Galerkin 方法、Fourier 级数、Laplace 变换和 Picard 迭代, 研究带有非局部 Laplace 算子的饱和 Schrödinger-Klein-Gordon 方程 (1.1) 在合适的函数空间中概自守弱解的存在性和唯一性. ...

Almost periodic solutions for a class of higher dimensional Schr?dinger equations

2009

... ],[23],[24] 中研究了一类特殊的 Schrödinger 方程的概周期解. Signing^[25] 考虑了具有概周期集的 Klein-Gordon 型方程的均匀化问题. Abdallah 等^[26]研究了具有非线性概周期项的 Klein-Gordon-Schrödinger 无限维格系统的一致全局吸引子的存在性. 然而, 据我们所知, 目前几乎没有文献涉及到 Schrödinger 或 Klein-Gordon 方程的概自守动力学研究. 受此启发, 本文将结合 Galerkin 方法、Fourier 级数、Laplace 变换和 Picard 迭代, 研究带有非局部 Laplace 算子的饱和 Schrödinger-Klein-Gordon 方程 (1.1) 在合适的函数空间中概自守弱解的存在性和唯一性. ...

Almost-periodic solutions for a quasi-periodically forced nonlinear Schr?dinger equation

2021

... ],[24] 中研究了一类特殊的 Schrödinger 方程的概周期解. Signing^[25] 考虑了具有概周期集的 Klein-Gordon 型方程的均匀化问题. Abdallah 等^[26]研究了具有非线性概周期项的 Klein-Gordon-Schrödinger 无限维格系统的一致全局吸引子的存在性. 然而, 据我们所知, 目前几乎没有文献涉及到 Schrödinger 或 Klein-Gordon 方程的概自守动力学研究. 受此启发, 本文将结合 Galerkin 方法、Fourier 级数、Laplace 变换和 Picard 迭代, 研究带有非局部 Laplace 算子的饱和 Schrödinger-Klein-Gordon 方程 (1.1) 在合适的函数空间中概自守弱解的存在性和唯一性. ...

Almost periodic homogenization of the Klein-Gordon type equation

2020

... [25] 考虑了具有概周期集的 Klein-Gordon 型方程的均匀化问题. Abdallah 等^[26]研究了具有非线性概周期项的 Klein-Gordon-Schrödinger 无限维格系统的一致全局吸引子的存在性. 然而, 据我们所知, 目前几乎没有文献涉及到 Schrödinger 或 Klein-Gordon 方程的概自守动力学研究. 受此启发, 本文将结合 Galerkin 方法、Fourier 级数、Laplace 变换和 Picard 迭代, 研究带有非局部 Laplace 算子的饱和 Schrödinger-Klein-Gordon 方程 (1.1) 在合适的函数空间中概自守弱解的存在性和唯一性. ...

Attractors of the Klein-Gordon-Schr?dinger lattice systems with almost periodic nonlinear part

2022

... [26]研究了具有非线性概周期项的 Klein-Gordon-Schrödinger 无限维格系统的一致全局吸引子的存在性. 然而, 据我们所知, 目前几乎没有文献涉及到 Schrödinger 或 Klein-Gordon 方程的概自守动力学研究. 受此启发, 本文将结合 Galerkin 方法、Fourier 级数、Laplace 变换和 Picard 迭代, 研究带有非局部 Laplace 算子的饱和 Schrödinger-Klein-Gordon 方程 (1.1) 在合适的函数空间中概自守弱解的存在性和唯一性. ...

Hitchhiker's guide to the fractional Sobolev spaces

2012

... 设 $s\in(0,1)$, $\mathscr{S}$ 是速降函数组成的 Schwartz 空间, 分数阶 Laplace 算子^[27] $(-\Delta)^{s}:\mathscr{S}\rightarrow \mathrm{L}^2(\mathbf{R}^d)$ 表示为 ...

Variational methods for non-local operators of elliptic type

2013

... 其中 ${ \mathbf{u} }\in \mathrm{X}_0^s(\Omega)$. 根据文献 [28] 可知, 在 $\mathrm{L}^2(\Omega,\mathbf{R})$ 中存在对应于非局部 Laplace 算子 $(-\Delta)^{s}$ 的一个完备的规范正交基 $\{e_k\}_{k=1}^{\infty}\subseteq\mathrm{X}_0^s(\Omega)$. 同时, 它也是 $\mathrm{X}_0^s(\Omega)$ 的正交基. 设 $\{\lambda_k\}_{k=1}^{\infty}$ 为 $\{e_k\}_{k=1}^{\infty}$ 对应的特征值序列, 则 $0<\lambda_1\leq \lambda_2\leq \cdots\leq \lambda_k\leq \cdots$, $\lim\limits_{k\rightarrow\infty}\lambda_k=\infty$. ...

Boundary value problems for semilinear differential inclusions of fractional order in a Banach space

2018

2021

... 这里我们介绍概自守向量值函数的概念, 详情请参阅文献 [30]. ...

... 由于 $(\varphi,\phi)=(\hat{\varphi},\breve{\varphi},\phi)\in\mathbb{AAA}(\mathbf{R}_0,\{\mathrm{L}^2(\Omega)\}^3)$, 利用文献 [30,定理 3.6] 可知, ...

... 定义范数 $|\xi|_{\sup}=\sup\limits_{t\in\mathbf{R}_0}|\xi(t)|_2$, $\forall \xi\in\mathbb{AAA}(\mathbf{R}_0,\mathrm{L}^2(\Omega))$. 则根据文献 [30] 可知 $(\mathbb{AAA}(\mathbf{R}_0,$ $\mathrm{L}^2(\Omega)), |\cdot|_{\sup})$ 是 Banach 空间. ...

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