各向异性Sobolev空间中分数阶拟地转方程的稳定性

Stability Results of Fractional Surface Quasi-Geostrophic Equation in Anisotropic Sobolev Space

摘要: 本文针对二维分数阶表面准地转SQG方程的存在全局解，在初始数据${\theta }_0$ 属于非齐次各向异性Sobolev空间$H^{0,\alpha }\left(\frac{1}{2}<\alpha <1\right)$ 条件下，利用Bony分解理论和Littlewood-Paley分解技术，证明了该方程解的稳定性。

Abstract: We prove the stability of a global solution for the fractional SQG equation under the conditions that the initial data ${\theta }_0$ belongs to the nonhomogeneous anisotropic Sobolev space $H^{0,\alpha }$ with $\frac{1}{2}<\alpha <1$ . The main tools employed in our analysis are the Bony decomposition theory and Littlewood-Paley decomposition technique.

1. 引言

二维分数阶表面准地转SQG方程是由三维不可压缩欧拉方程推导出的二维模型。本文重点研究由下式给出的耗散SQG方程：

$\{\begin{array}{l}\frac{\partial \theta }{\partial t}+u\cdot \nabla \theta +v{\left(-\Delta \right)}^{\alpha }\theta =0,\hfill \\ \theta \left(x,0\right)={\theta }_0\left(x\right),\hfill \end{array}$ (1)

其中$\alpha \in \left(1/2,1\right),v>0$ 是耗散系数，而$\theta =\theta \left(x,t\right)$ 是两个空间变量$x$ 和时间$t$ 的实标量函数。速度场$u$ 通过Riesz变换确定：

$u=\left(-\frac{{\partial }_2}{\sqrt{-\Delta }}\theta ,\frac{{\partial }_1}{\sqrt{-\Delta }}\theta \right)=\left(-{ℛ}_2\theta ,{ℛ}_1\theta \right)={ℛ}^{\perp }\theta .$ (2)

显然，

$\text{div}u={\partial }_1{u}_1+{\partial }_2{u}_2=0.$ (3)

${\left(-\Delta \right)}^{\alpha }\left(0<\alpha <1\right)$ 通过Fourier变换定义，

$\widehat{{\left(-\Delta \right)}^{\alpha }f}\left(\xi \right)={\left|\xi \right|}^{2\alpha }\widehat{f}\left(\xi \right),$

其中$\widehat{f}$ 表示[1]的Fourier变换。为便于记号，我们写为${\left(-\Delta \right)}^{\frac{1}{2}}=\Lambda$ 。

分数阶SQG方程是对经典SQG方程的扩展，不同于经典拟地转方程，分数阶SQG方程能更精确地描述具有非局部性和依赖性的动力学过程。因此，各向异性分数阶方程成为捕捉复杂方向系统中的强大工具，例如大气动力学、多孔介质流动及生物流体力学领域。

分数阶SQG方程的过往研究主要集中于各向同性的情形。Kiselev与Nazarov在[2]中取得突破性进展，证明了临界SQG方程的全局正则性。对于次临界分数阶SQG方程$\left(\alpha <1\right)$ ，Constantin与Vicol在[3]中证明了其全局存在性与唯一性。

除解的全局正则性外，弱解及其对初始数据的依赖性研究亦取得进展。Resnick探索了经典SQG方程的弱解及其长期行为，该研究至今仍对理解非正则解具有重要影响[4]。

Danchin与Paicu通过推进微积分方法，在临界空间中深入剖析解的精细特性。这些方法论在研究分数阶SQG方程及相关模型时发挥了关键作用。分数阶SQG方程的各向异性情形比各向同性情形更为复杂，这源于其对称性的缺失及额外的正则性要求。近期研究通过用分数阶算子${\left(-\Delta \right)}^{\alpha }$ 替代标准扩散项$k{\partial }_{11}\theta$ 和$k^{\prime }{\partial }_{22}\theta$ [5] [6]。

在(1)式中，${\left(-\Delta \right)}^{\alpha }\theta$ 取代了经典SQG方程中的扩散项，构建了更一般的理论框架。文章定理以及证明借鉴了文献[7]的主要技术思想和方程中能量估计的方法，文献[8]提供了处理分数阶算子的分析技巧，将该思想融入方程中耗散项与非线性项的进一步估计，因此本文将方法结合在一起，将其推广应用于各向异性Sobolev空间中分数阶拟地转方程的稳定性问题。即如下定理，

定理1 存在常数$C$ ，使得对于所有初始数据${\theta }_0\in H^{0,\alpha }$ ，且$\frac{1}{2}<\alpha <1$ ,满足${\Vert {\theta }_0\Vert }_{H^{0,\alpha }}<\frac{v}{C}$ ，方程(1)和(2)的解在$L^{\infty }\left(\left[0,\infty \right];H^{0,\alpha }\right)\cap L^2\left(\left[0,\infty \right];H^{1,\alpha }\right)$ 中具有一致有界性。因此，我们得到

${\Vert \theta \left(t\right)\Vert }_{H^{0,\alpha }}^2+v{\displaystyle {\int }_0^t{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,\alpha }}^2}\text{d}t\le {\left(\frac{v}{C}\right)}^2.$ (4)

2. 预备知识

让我们首先回顾：$S$ 表示${ℝ}^N$ 上光滑函数构成的Schwartz空间，其中函数的各阶导数在无穷远处衰减；$S\prime$ 为缓增分布空间，是$S$ 在通常配对下的对偶空间。本文利用Littlewood-Paley分解详细刻画各向异

性Sobolev空间$H^{0,\alpha }$ ，分析中主要采用Bony分解[9]。设$\phi \left(\xi \right)$ 为支架包含于环域$\left\{\xi \in ℝ|\frac{3}{4}\le \left|\xi \right|\le \frac{8}{3}\right\}$ 的光滑函数，$\chi$ 为支集包含于球$\left\{\xi \in ℝ|\left|\xi \right|\le \frac{4}{3}\right\}$ 的光滑函数，则有

$\forall \xi \in ℝ,\chi \left(\xi \right)+{\displaystyle \sum _{q\in \mathbb{N}}\phi }\left(2^{-q}\xi \right)=1.$

为方便起见，我们用${\Delta }_j$ 表示非齐次各向异性二进块，在$x_2$ 方向上的定义如下：

$\begin{array}{ll}{\Delta }_{j}a=0\hfill & �j\le -2,\hfill \\ {\Delta }_{j}a={ℱ}^{-1}\left(\chi \left({\xi }_2\right)\widehat{a}\right)\hfill & �j=-1,\hfill \\ {\Delta }_{j}a={ℱ}^{-1}\left(\phi \left(2^{-j}{\xi }_2\right)\widehat{a}\right)\hfill & �j\ge 0.\hfill \end{array}$ (5)

此外，我们引入以下低频截断算子：

$S_{q}u:={\displaystyle \sum _{p\le q-1}{\Delta }_p}u,\begin{array}{c}�所有\end{array}q\in N.$ (6)

命题1 对于$f\in S^{\prime }$ ，可直接验证如下恒等式成立：

$\begin{array}{l}{\Delta }_p{\Delta }_{q}u=0,\begin{array}{cccc}& & & 若\end{array}\left|p-q\right|\ge 2.\hfill \\ {\Delta }_p\left(S_{q-1}u{\Delta }_{q}u\right)=0,\begin{array}{cc}& 若\left|p-q\right|\ge 5\end{array}.\hfill \end{array}$ (7)

命题2 (Bernstein不等式[9])设$k\in N$ 。设$\left(R_1,R_2\right)$ 满足$0<R_1<R_2$ 。则存在仅依赖于$R_1,R_2$ 的常数$C$ ，使得对于任意$1\le a\le b\le \infty$ 及任意$u\in L^a\left({ℝ}^2\right)$ ，下列不等式成立：

$\begin{array}{l}\text{supp}\widehat{u}\subset ℬ\left(0,R_1\lambda \right)\Rightarrow \underset{\left|\alpha \right|=k}{\sup }{\Vert {\partial }^{\alpha }u\Vert }_{L^b}\le C^{k+1}{\lambda }^{k+2\left(\frac{1}{a}-\frac{1}{b}\right)}{\Vert u\Vert }_{L^a},\\ \text{supp}\widehat{u}\subset \mathcal{C}\left(0,R_1\lambda ,R_2\lambda \right)\Rightarrow C^{-k-1}{\lambda }^k{\Vert u\Vert }_{L^a}\le \underset{\left|\alpha \right|=k}{\sup }{\Vert {\partial }^{\alpha }u\Vert }_{L^a}\le C^{k+1}{\lambda }^k{\Vert u\Vert }_{L^a}.\end{array}$

定义1非齐次Sobolev空间$H^{\alpha }$ 由以下范数定义：

${\Vert a\Vert }_{H^{\alpha }(ℝ)}={\left({\displaystyle {\int }_{ℝ}{\left(1+{\left|\xi \right|}^2\right)}^{\alpha }}{\left|\widehat{a}\left(\xi \right)\right|}^2\text{d}\xi \right)}^{\frac{1}{2}}$

其中$\widehat{a}\left(\xi \right)$ 表示$a$ 的Fourier变换。

命题3 (一维非齐次Sobolev空间中的乘积法则)设$a\in H^{{\alpha }_0}\left(ℝ\right)$ 与$b\in H^{{\alpha }_1}\left(ℝ\right)$ ，$-\frac{1}{2}<{\alpha }_0,{\alpha }_1<\frac{1}{2}$ ，${\alpha }_0+{\alpha }_1>0$ ，则在一维非齐次Sobolev空间中，$ab\in H^{{\alpha }_0+{\alpha }_1-\frac{1}{2}}\left(ℝ\right)$ 与乘积法则成立如下：

${\Vert ab\Vert }_{H^{{\alpha }_0+{\alpha }_1-\frac{1}{2}}\left(ℝ\right)}\le C{\Vert a\Vert }_{H^{{\alpha }_0}\left(ℝ\right)}{\Vert b\Vert }_{H^{{\alpha }_1}\left(ℝ\right)}.$ (8)

定义2 Sobolev空间$H^{\alpha ,{\alpha }^{\prime }}\left({ℝ}^2\right)\left(\alpha ,{\alpha }^{\prime }\in ℝ\right)$ 定义为

${\Vert a\Vert }_{H^{\alpha ,{\alpha }^{\prime }}\left({ℝ}^2\right)}={\left({\displaystyle {\int }_{{ℝ}^2}{\left(1+{\left|{\xi }_1\right|}^2\right)}^{\alpha }}{\left(1+{\left|{\xi }_2\right|}^2\right)}^{{\alpha }^{\prime }}{\left|\widehat{a}\left(\xi \right)\right|}^2\text{d}{\xi }_1\text{d}{\xi }_2\right)}^{\frac{1}{2}}$

命题4对于$a\in H^{\alpha ,{\alpha }^{\prime }}\left({ℝ}^2\right)$ ，Sobolev范数可等价表示为

${\Vert a\Vert }_{H^{\alpha ,{\alpha }^{\prime }}\left({ℝ}^2\right)}={\left({\displaystyle \sum _{j,k\in \mathbb{Z}}{2}^{2\left(j{\alpha }^{\prime }+k\alpha \right)}}{\displaystyle {\int }_{{ℝ}^2}{\left|{\Delta }_j{\Delta }_{k}a\left(x_1,x_2\right)\right|}^2}\text{d}{x}_1\text{d}{x}_2\right)}^{\frac{1}{2}}.$

命题5 Sobolev范数也可表示为$L^2$ 空间与Sobolev空间的混合范数，有

${\Vert a\Vert }_{H^{\alpha ,{\alpha }^{\prime }}\left({ℝ}^2\right)}={\Vert 2^{j{\alpha }^{\prime }}{\Vert {\Delta }_{j}a\Vert }_{L^2\left({ℝ}_{x_2};H^{\alpha }\left({ℝ}_{x_1}\right)\right)}\Vert }_{l^2\left(\mathbb{Z}\right)},$

其中${\Vert \cdot \Vert }_{L^2\left({ℝ}_{x_2};H^{\alpha }\left({ℝ}_{x_1}\right)\right)}$ 表示混合范数，是将$x_2$ 方向上的$L^2$ 范数与$x_1$ 方向上的Sobolev范数相结合。

引理1 [7]对于${\alpha }^{\prime }>\frac{1}{2}$ 和$\alpha \in ℝ$ ，有

${\Vert a\Vert }_{L^{\infty }\left(ℝ;H^{\alpha }\left(ℝ\right)\right)}\le C{\Vert a\Vert }_{H^{\alpha ,{\alpha }^{\prime }}}.$ (9)

引理2 [10]设$a\left(t\right)$ 是定义在$ℝ$ 上的函数，设$a_k\left(t\right)={\Vert {\Delta }_{k}a\Vert }_{L^2}\cdot 2^{\frac{k\alpha }{q^{\prime }}}$ 是$l^2$ 空间中随时间变化的函数序列，满足$\alpha >\frac{1}{2},2<q<\infty ,1<q^{\prime }<2$ ，则有

${\Vert a_k\left(t\right)\Vert }_{l^2}\le C{\Vert a\left(t\right)\Vert }_{L^2}^{\frac{1}{q}}{\Vert {\Lambda }^{\alpha }a\left(t\right)\Vert }_{L^2}^{\frac{1}{q^{\prime }}},\text{where}\frac{1}{q}+\frac{1}{q^{\prime }}=1.$

引理3对满足$\frac{1}{2}<a^{\prime }\le \alpha <1$ 的实数$\alpha$ 和${\alpha }^{\prime }$ ，当$\theta \in H^{\frac{1}{4},\alpha }\cap H^{\frac{1}{4},{\alpha }^{\prime }}\cap H^{0,{\alpha }^{\prime }}$ 和${\Lambda }^{\alpha }\theta \in H^{\frac{1}{4},\alpha }\cap H^{0,\alpha }$ ，则存在常数$C$ 使得

${\left|\langle {\text{Δ}}_j\left(u\cdot \nabla \theta \right)|{\text{Δ}}_j\theta \rangle \right|}_{L^2}\le C\cdot d_j\cdot 2^{-j\alpha }{\Vert \theta \Vert }_{H^{\frac{1}{4},\alpha }}\left({\Vert \theta \Vert }_{H^{0,{\alpha }^{\prime }}}{\Vert {\text{Λ}}^{\alpha }\theta \Vert }_{H^{\frac{1}{4},\alpha }}+{\Vert \theta \Vert }_{H^{\frac{1}{4},{\alpha }^{\prime }}}{\Vert {\text{Λ}}^{\alpha }\theta \Vert }_{H^{0,\alpha }}\right),$ (10)

其中$u=\left(-{ℛ}_2\theta ,{ℛ}_1\theta \right)$ 和$d_j\in l^1\left(\mathbb{Z}\right)$ 。

3. 引理3证明

为消除各向异性的影响，我们将估计分为两个部分：

$F_j^1={\Delta }_j\left(u_1{\partial }_1\theta \right),F_j^2={\Delta }_j\left(u_2{\partial }_2\theta \right).$

首先，我们估计$F_j^1$ 项

$\begin{array}{c}{\Vert F_j^1\Vert }_{L^2\left({ℝ}_{x_2};H^{-\frac{1}{4}}\right)}={\Vert {\text{Δ}}_j\left(u_1{\partial }_1\theta \right)\Vert }_{L^2\left({ℝ}_{x_2};H^{-\frac{1}{4}}\right)}\\ ={\Vert {\Delta }_j\left(T_{u_1}{\partial }_1\theta +T_{{\partial }_1\theta }{u}_1+R\left(u_1,{\partial }_1\theta \right)\right)\Vert }_{L^2\left({ℝ}_{x_2};H^{-\frac{1}{4}}\right)}.\end{array}$

由(6)式可得，

${\Vert {\Vert {\Delta }_j{T}_{u_1}{\partial }_1\theta \Vert }_{H^{-\frac{1}{4}}\left({ℝ}_{x_1}\right)}\Vert }_{L^2\left({ℝ}_{x_2}\right)}\le {\Vert {\displaystyle \sum _{\left|k-j\right|\le 2}{\Vert {\Delta }_j\left(S_{k-1}{u}_1{\Delta }_k{\partial }_1\theta \right)\Vert }_{H^{-\frac{1}{4}}\left({ℝ}_{x_1}\right)}}\Vert }_{L^2\left({ℝ}_{x_2}\right)}$ ，

应用Sobolev空间的乘积规则，(8)式可得

${\Vert {\Vert {\Delta }_j{T}_{u_1}{\partial }_1\theta \Vert }_{H^{-\frac{1}{4}}\left({ℝ}_{x_1}\right)}\Vert }_{L^2\left({ℝ}_{x_2}\right)}\le {\displaystyle \sum _{\left|k-j\right|\le 2}{\Vert {\Vert S_{k-1}{u}_1\Vert }_{H^{\frac{1}{4}}}{\Vert {\Delta }_k{\partial }_1\theta \Vert }_{H^0}\Vert }_{L^2\left({ℝ}_{x_2}\right)}}$ ，

利用引理1和Bernstein不等式，我们得到

$\begin{array}{c}{\Vert {\Vert {\Delta }_j{T}_{u_1}{\partial }_1\theta \Vert }_{H^{-\frac{1}{4}}\left({ℝ}_{x_1}\right)}\Vert }_{L^2\left({ℝ}_{x_2}\right)}\le {\displaystyle \sum _{\left|k-j\right|\le 2}{\Vert S_{k-1}{u}_1\Vert }_{L^{\infty }\left({ℝ}_{x_2};H^{\frac{1}{4}}\right)}{\Vert {\Delta }_k{\partial }_1\theta \Vert }_{L^2}}\\ \le {\displaystyle \sum _{\left|k-j\right|\le 2}C}\cdot 2^k{\Vert u_1\Vert }_{H^{\frac{1}{4},{\alpha }^{\prime }}}{\Vert {\Delta }_k\theta \Vert }_{L^2},\end{array}$

利用引理2，令$q=\infty ,q^{\prime }=1$ ，则

(11)

同样地，我们可以在Sobolev空间中应用引理1、引理2和Sobolev空间的乘积规则(8)式，我们还得到：

$\begin{array}{c}{\Vert {\Vert {\Delta }_j{T}_{{\partial }_1\theta }{u}_1\Vert }_{H^{-\frac{1}{4}}\left({ℝ}_{x_1}\right)}\Vert }_{{}_{L^2\left({ℝ}_{x_2}\right)}}\le {\Vert {\displaystyle \sum _{\left|k-j\right|\le 2}{\Vert {\Delta }_j\left(S_{k-1}{\partial }_1\theta {\Delta }_k{u}_1\right)\Vert }_{H^{-\frac{1}{4}}\left({ℝ}_{x_1}\right)}}\Vert }_{L^2\left({ℝ}_{x_2}\right)}\\ \le {\displaystyle \sum _{\left|k-j\right|\le 2}{\Vert {\Vert S_{k-1}{\partial }_1\theta \Vert }_{H^0}{\Vert {\Delta }_k{u}_1\Vert }_{H^{\frac{1}{4}}}\Vert }_{L^2\left({ℝ}_{x_2}\right)}}\\ \le {\displaystyle \sum _{\left|k-j\right|\le 2}{\Vert {\Vert S_{k-1}{\partial }_1\theta \Vert }_{H^0\left({ℝ}_{x_1}\right)}\Vert }_{L^{\infty }\left({ℝ}_{x_2}\right)}{\Vert {\Vert {\Delta }_k{u}_1\Vert }_{H^{\frac{1}{4}}\left({ℝ}_{x_1}\right)}\Vert }_{L^2\left({ℝ}_{x_2}\right)}}\\ \le {\displaystyle \sum _{\left|k-j\right|\le 2}{2}^k}{\Vert {\Delta }_k\theta \Vert }_{L^2}{\Vert {\Delta }_k{u}_1\Vert }_{L^2\left({ℝ}_{x_2};H^{\frac{1}{4}}\right)}\\ \le C\cdot c_j{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,{\alpha }^{\prime }}}{\Vert u_1\Vert }_{H^{\frac{1}{4},\alpha }}.\end{array}$ (12)

类似地，我们应用引理1、引理2和Bernstein不等式得到

$\begin{array}{c}{\Vert {\Vert {\Delta }_{j}R\left(u_1,{\partial }_1\theta \right)\Vert }_{H^{-\frac{1}{4}}\left({ℝ}_{x_1}\right)}\Vert }_{L^2\left({ℝ}_{x_2}\right)}\le {\Vert {\displaystyle \sum _{\begin{array}{c}\left|j-k\right|\le 2\\ \left|k-k^{\prime }\right|\le 1\end{array}}{\Vert {\Delta }_j\left({\Delta }_k{u}_1{{\Delta }^{\prime }}_k{\partial }_1\theta \right)\Vert }_{H^{-\frac{1}{4}}\left({ℝ}_{x_1}\right)}}\Vert }_{L^2\left({ℝ}_{x_2}\right)}\\ \le {\Vert {\displaystyle \sum _{\begin{array}{c}\left|j-k\right|\le 2\\ \left|k-k^{\prime }\right|\le 1\end{array}}{\Vert {\Delta }_k{u}_1\Vert }_{H^{\frac{1}{4}}}}{\Vert {\Delta }_{k^{\prime }}{\partial }_1\theta \Vert }_{H^0}\Vert }_{L^2\left({ℝ}_{x_2}\right)}\\ \le {\displaystyle \sum _{\begin{array}{c}\left|j-k\right|\le 2\\ \left|k-k^{\prime }\right|\le 1\end{array}}C}\cdot 2^k{\Vert {\Delta }_k{u}_1\Vert }_{L^{\infty }\left({ℝ}_{x_2};H^{\frac{1}{4}}\right)}{\Vert {\Delta }_{k^{\prime }}\theta \Vert }_{L^2\left({ℝ}_{x_2};H^0\right)}\end{array}$ (13)

$\begin{array}{l}\le {\displaystyle \sum _{\left|j-k\right|\le 2}C}\cdot 2^k{\Vert u_1\Vert }_{H^{\frac{1}{4},{\alpha }^{\prime }}}{\Vert {\Lambda }^{\alpha }\theta \Vert }_{L^2}\\ \le C\cdot c_j{\Vert u_1\Vert }_{H^{\frac{1}{4},{\alpha }^{\prime }}}{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,\alpha }}.\end{array}$

因此，由(11)、(12)和(13)式可得

$\begin{array}{c}{\Vert F_j^1\Vert }_{L^2\left({ℝ}_{X_2};H^{-\frac{1}{4}}\right)}={\Vert {\Delta }_j\left(T_{u_1}{\partial }_1\theta +T_{{\partial }_1\theta }{u}_1+R\left(u_1,{\partial }_1\theta \right)\right)\Vert }_{L^2\left({ℝ}_{x_2};H^{-\frac{1}{4}}\right)}\\ \le C\cdot c_j{\Vert u_1\Vert }_{H^{\frac{1}{4},{\alpha }^{\prime }}}{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,\alpha }}+C\cdot c_j{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,{\alpha }^{\prime }}}{\Vert u_1\Vert }_{H^{\frac{1}{4},\alpha }}+C\cdot c_j{\Vert u_1\Vert }_{H^{\frac{1}{4},{\alpha }^{\prime }}}{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,\alpha }}\\ \le C\cdot c_j\left({\Vert u_1\Vert }_{H^{\frac{1}{4},{\alpha }^{\prime }}}{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,\alpha }}+{\Vert u_1\Vert }_{H^{\frac{1}{4},\alpha }}{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,{\alpha }^{\prime }}}\right),\end{array}$

其中$c_j$ 是$l^2\left(\mathbb{Z}\right)$ 中的一般元素，且对于$j\ge -1,2^{-j\alpha }\le c_j$ ，且$c_j^2=d_j$ 。

$\begin{array}{c}\left|{\langle {\Delta }_j\left(u_1{\partial }_1\theta \right)|{\Delta }_j\theta \rangle }_{L^2}\right|\le {\Vert {\Delta }_j\left(u_1{\partial }_1\theta \right)\Vert }_{L^2\left({ℝ}_{x_2};H^{-\frac{1}{4}}\right)}{\Vert {\Delta }_j\theta \Vert }_{L^2\left({ℝ}_{x_2};H^{\frac{1}{4}}\right)}\\ \le {\Vert {\Delta }_j\left(u_1{\partial }_1\theta \right)\Vert }_{L^2\left({ℝ}_{x_2};H^{-\frac{1}{4}}\right)}\cdot c_j\cdot 2^{-j\alpha }{\Vert \theta \Vert }_{H^{\frac{1}{4},\alpha }}\\ \le C{d}_j\cdot 2^{-j\alpha }{\Vert \theta \Vert }_{H^{\frac{1}{4},\alpha }}\left({\Vert u_1\Vert }_{H^{\frac{1}{4},{\alpha }^{\prime }}}{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,\alpha }}+{\Vert u_1\Vert }_{H^{\frac{1}{4},\alpha }}{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,\alpha }}\right).\end{array}$

基于Riesz变换${ℛ}_k\left(k=1,2\right)$ ，其中${ℛ}_k$ 是满足$\frac{{\xi }_k}{\sqrt{{\xi }_1^2+{\xi }_2^2}}<1$ 的Fourier乘子，我们有：

$\begin{array}{c}{\Vert {ℛ}_k\theta \Vert }_{H^{\alpha ,{\alpha }^{\prime }}}={\Vert 2^{j{\alpha }^{\prime }}{\Vert {\left(1+{\left|\xi \right|}^2\right)}^{\frac{\alpha }{2}}\widehat{{\Delta }_j{R}_k\theta }\Vert }_{L^2}\Vert }_{l^2}\\ ={\Vert 2^{j{\alpha }^{\prime }}{\Vert {\left(1+{\left|\xi \right|}^2\right)}^{\frac{\alpha }{2}}{\phi }_j\cdot \frac{{\xi }_k}{\sqrt{{\xi }_1^2+{\xi }_2^2}}\widehat{\theta }\Vert }_{L^2}\Vert }_{l^2}\\ \le {\Vert 2^{j{\alpha }^{\prime }}{\Vert {\left(1+{\left|\xi \right|}^2\right)}^{\frac{\alpha }{2}}{\phi }_j\cdot \widehat{\theta }\Vert }_{L^2}\Vert }_{l^2}={\Vert \theta \Vert }_{H^{\alpha ,{\alpha }^{\prime }}}.\end{array}$ (14)

接下来，我们通过Bony分解对$u_2{\partial }_2\theta$ 进行估计，得到：

$\begin{array}{c}{\Delta }_j\left(u_2{\partial }_2\theta \right)={\Delta }_j\left({\displaystyle \sum _{k=-\infty }^{+\infty }{\displaystyle \sum _{k^{\prime }=-\infty }^{+\infty }{\Delta }_{k^{\prime }}}}{u}_2{\Delta }_k{\partial }_2\theta \right)\\ ={\Delta }_j\left({\displaystyle \sum _{k=-\infty }^{+\infty }\left({\displaystyle \sum _{k^{\prime }=-\infty }^{k+1}{\Delta }_{k^{\prime }}}{u}_2{\Delta }_k{\partial }_2\theta +{\displaystyle \sum _{k^{\prime }=k+2}^{+\infty }{\Delta }_{k^{\prime }}}{u}_2{\Delta }_k{\partial }_2\theta \right)}\right)\\ ={\Delta }_j\left({\displaystyle \sum _{k=-\infty }^{+\infty }{S}_{k+2}}{u}_2{\Delta }_k{\partial }_2\theta +{\displaystyle \sum _{k=-\infty }^{+\infty }{\displaystyle \sum _{k^{\prime }=k+2}^{+\infty }{\Delta }_{k^{\prime }}}}{u}_2{\Delta }_k{\partial }_2\theta \right)\\ ={\Delta }_j(\left({\displaystyle \sum _{k\ge j-3}+{\displaystyle \sum _{k<j-3}}}\right)S_{k+2}{u}_2{\Delta }_k{\partial }_2\theta +{\displaystyle \sum _{k=-\infty }{S}_{k-1}}{\partial }_2\theta {\Delta }_k{u}_2)\\ ={\Delta }_j(\left({\displaystyle \sum _{k\ge j-3}+}{\displaystyle \sum _{k<j-3}}\right)S_{k+2}{u}_2{\Delta }_k{\partial }_2\theta +\left({\displaystyle \sum _{\left|j-k\right|\le 4}+}{\displaystyle \sum _{k<j-4}+}{\displaystyle \sum _{k>j+4}}\right)S_{k-1}{\partial }_2\theta {\Delta }_k{u}_2)\\ ={\Delta }_j{\displaystyle \sum _{k\ge j-3}{S}_{k+2}}{u}_2{\Delta }_k{\partial }_2\theta +{\Delta }_j{\displaystyle \sum _{k<j-3}{S}_{k+2}}{u}_2{\Delta }_k{\partial }_2\theta +{\Delta }_j{\displaystyle \sum _{\left|j-k\right|\le 4}{S}_{k-1}}{\partial }_2\theta {\Delta }_k{u}_2\\ +{\Delta }_j{\displaystyle \sum _{k<j-4}{S}_{k-1}}{\partial }_2\theta {\Delta }_k{u}_2+{\Delta }_j{\displaystyle \sum _{k>j+4}{S}_{k-1}}{\partial }_2\theta {\Delta }_k{u}_2\end{array}$ (15)

$=I_1+I_2+I_3+I_4+I_5.$

对于$I_2,I_4,I_5$ ，由(7)可得：

$I_2={\Delta }_j{\displaystyle \sum _{k<j-3}{S}_{k+2}}{u}_2{\Delta }_k{\partial }_2\theta ={\Delta }_j{\displaystyle \sum _{k<j-3}{\displaystyle \sum _{l\le k+1}{\Delta }_l}}{u}_2{\Delta }_k{\partial }_2\theta$ ，

这里$l\le k+1<j-2$ ，那么$\left|j-l\right|>2$ ，即$I_2=0$

$I_4={\Delta }_j{\displaystyle \sum _{k\le j-4}{S}_{k-1}}{\partial }_2\theta {\Delta }_k{u}_2={\Delta }_j{\displaystyle \sum _{k\le j-4}{\displaystyle \sum _{l\le k-2}{\Delta }_l{\partial }_2\theta }}{\Delta }_k{u}_2$ ，

这里$l\le k-2\le j-6$ ，那么$\left|j-l\right|\ge 6$ ，即$I_4=0$

$I_5={\Delta }_j{\displaystyle \sum _{k>j+4}{S}_{k-1}}{\partial }_2\theta {\Delta }_k{u}_2={\Delta }_j{\displaystyle \sum _{k>j+4}{\displaystyle \sum _{l\le k-2}{\Delta }_l{\partial }_2\theta {\Delta }_k{u}_2}}$ ，

这里$\left|j-l\right|>2$ ，即$I_5=0$

所以$I_2$ ，$I_4$ 和$I_5$ 均为0，这意味着

$F_j^2={\Delta }_j\left(u_2{\partial }_2\theta \right)={\Delta }_j{\displaystyle \sum _{k\ge j-3}{S}_{k+2}}\left({\partial }_2\theta \right){\Delta }_k{u}_2+{\Delta }_j{\displaystyle \sum _{\left|j-k\right|\le 4}{S}_{k-1}}\left(u_2\right){\Delta }_k{\partial }_2\theta =F_j^{2,1}+F_j^{2,2}.$ (16)

因此，利用(8)式、Höder\不等式、Bernstein不等式以及散度为0条件${\partial }_2{u}_2=-{\partial }_1{u}_1$ ，

$\begin{array}{c}{\Vert {\Vert F_j^{2,1}\Vert }_{H^{-\frac{1}{4}}\left({ℝ}_{x_1}\right)}\Vert }_{L^2\left({ℝ}_{x_2}\right)}={\Vert {\displaystyle \sum _{k\ge j-3}{\Vert {\Delta }_j{S}_{k+2}\left({\partial }_2\theta \right){\Delta }_k{u}_2\Vert }_{H^{-\frac{1}{4}}}}\Vert }_{L^2\left({ℝ}_{x_2}\right)}\\ \le {\displaystyle \sum _{k\ge j-3}{\Vert {\Vert S_{k+2}\left({\partial }_2\theta \right)\Vert }_{H^{\frac{1}{4}}}{\Vert {\Delta }_k{u}_2\Vert }_{H^0}\Vert }_{L^2\left({ℝ}_{x_2}\right)}}\\ \le C\cdot 2^{\frac{j}{2}}{\displaystyle \sum _{k\ge j-3}{\Vert {\Vert S_{k+2}\left({\partial }_2\theta \right)\Vert }_{H^{\frac{1}{4}}}{\Vert {\Delta }_k{u}_2\Vert }_{H^0}\Vert }_{L^2\left({ℝ}_{x_2}\right)}}\\ \le C{\displaystyle \sum _{k\ge j-3}{2}^{\frac{j}{2}}}{\Vert {\Vert S_{k+2}\left({\partial }_2\theta \right)\Vert }_{H^{\frac{1}{4}}}\Vert {}_{L^2}\Vert {\Vert {\Delta }_k{u}_2\Vert }_{H^0}\Vert }_{L^2}\\ \le C{\displaystyle \sum _{k\ge j-3}{\displaystyle \sum _{l\le k+1}{2}^{\frac{j}{2}}}}{2}^l{\Vert {\Delta }_l\theta \Vert }_{\left({ℝ}_{x_2},H^{\frac{1}{4}}\right)}{\Vert {\Delta }_k{u}_1\Vert }_{L^2}\\ \le C{\displaystyle \sum _{k\ge j-3}{\displaystyle \sum _{l\le k+1}{2}^{\frac{j}{2}}}}{2}^{k\left(l-\alpha \right)}{2}^{l\alpha }{2}^{-2k{\alpha }^{\prime }}{2}^{2k{\alpha }^{\prime }}{\Vert {\Delta }_l\theta \Vert }_{L^2\left({ℝ}_{x_2};H^{\frac{1}{4}}\right)}{\Vert {\Lambda }^{\alpha }{u}_1\Vert }_{L^2}\\ \le C{\displaystyle \sum _{k\ge j-3}{2}^{\frac{j}{2}}}{2}^{k\left(l-\alpha \right)}{2}^{-2k{\alpha }^{\prime }}{\Vert \theta \Vert }_{H^{\frac{1}{4},\alpha }}{\Vert {\Lambda }^{\alpha }{u}_1\Vert }_{H^{0,{\alpha }^{\prime }}},\end{array}$

其中${\Vert {\Delta }_k{u}_2\Vert }_{L^2\left({ℝ}^2\right)}\le C\cdot {\Vert {\Delta }_k{u}_1\Vert }_{L^2\left({ℝ}^2\right)}$ .

由于${\alpha }^{\prime }>\frac{1}{2}$ ，我们得出结论：

${\Vert F_j^{2,1}\Vert }_{L^2\left({ℝ}_2;H^{-\frac{1}{4}}\right)}\le C\cdot d_j{\Vert \theta \Vert }_{H^{\frac{1}{4},\alpha }}{\Vert {\Lambda }^{\alpha }{u}_1\Vert }_{H^{0,{\alpha }^{\prime }}}.$ (17)

随后，我们估计$F_j^{2,2}$ ，其中${\theta }_j={\Delta }_j\theta$ 。利用分解${\Delta }_j=\left({\Delta }_{j-1}+{\Delta }_j+{\Delta }_{j+1}\right){\Delta }_j$ ，即$F_j^{2,2}={\Delta }_j{\displaystyle \sum _{\left|j-k\right|\le 4}{S}_{k-1}}{u}_2{\Delta }_k{\partial }_2\theta$ 。

由于

$\left[{\Delta }_j,S_{k-1}{u}_2\right]{\Delta }_k\left({\partial }_2\theta \right)={\Delta }_j\left(S_{k-1}\left(u_2\right){\Delta }_k\left({\partial }_2\theta \right)\right)-S_{k-1}\left(u_2\right){\Delta }_j\left({\Delta }_k\left({\partial }_2\theta \right)\right).$ (18)

因此

$\begin{array}{l}{\Delta }_j{\displaystyle \sum _{\left|j-k\right|\le 4}{S}_{k-1}}\left(u_2\right){\Delta }_k\left({\partial }_2\theta \right)\\ ={\Delta }_j{\displaystyle \sum _{\left|j-k\right|\le 4}\left[{\Delta }_j,S_{k-1}{u}_2\right]{\Delta }_k{\partial }_2\theta }+{\Delta }_j{\displaystyle \sum _{\left|j-k\right|\le 4}{S}_{k-1}}\left(\left(u_2\right){\Delta }_k\left({\partial }_2\theta \right)\right)\\ ={\Delta }_j{\displaystyle \sum _{\left|j-k\right|\le 4}\left[{\Delta }_j,S_{k-1}{u}_2\right]{\Delta }_k{\partial }_2\theta }+{\Delta }_j{\displaystyle \sum _{\left|j-k\right|\le 4}\left(\left(S_{k-1}-S_j\right)u_2\left({\Delta }_{j-1}+{\Delta }_j+{\Delta }_j\right){\Delta }_j\left({\Delta }_k\left({\partial }_2\theta \right)\right)\right)}\\ ={\Delta }_j{\displaystyle \sum _{\left|j-k\right|\le 4}\left[{\Delta }_j,S_{k-1}{u}_2\right]{\Delta }_k{\partial }_2\theta }+S_j\left(u_2\right){\partial }_2{\theta }_j+{\displaystyle \sum _{\left|j-k\right|\le 1}\left(S_{k-1}-S_j\right)u_2{\partial }_2{\Delta }_k{\theta }_j}.\end{array}$ (19)

则

$\begin{array}{c}{\langle F_j^{2,2}|{\theta }_j\rangle }_{L^2}={\langle S_j\left(u_2\right){\partial }_2{\theta }_j|{\theta }_j\rangle }_{L^2}+{\displaystyle \sum _{\left|j-k\right|\le 4}{\langle \left[{\Delta }_j,S_{k-1}{u}_2\right]{\Delta }_k{u}_2{\partial }_2\theta |{\theta }_j\rangle }_{L^2}}\\ +{\displaystyle \sum _{\left|j-k\right|\le 1}{\langle \left(S_{k-1}-S_j\right)u_2{\partial }_2{\Delta }_k{\theta }_j|{\theta }_j\rangle }_{L^2}}\\ =\text{I}+\text{II}+\text{III},\end{array}$ (20)

其中，利用分部积分法

$\begin{array}{c}{\langle S_j\left(u_2\right){\partial }_2{\theta }_j|{\theta }_j\rangle }_{L^2}={\displaystyle {\int }_{ℝ}{S}_j}\left(u_2\right){\partial }_2{\theta }_j\cdot {\theta }_j\text{d}{x}_2\\ ={\displaystyle {\int }_{ℝ}\left({\partial }_2{S}_j\left(u_2\right)\cdot {\theta }_j+S_j\left(u_2\right)\cdot {\partial }_2{\theta }_j\right){\theta }_j\text{d}{x}_2}\\ =-{\displaystyle {\int }_{ℝ}{\theta }_j}{\partial }_2{S}_j\left(u_2\right){\theta }_j\text{d}{x}_2-{\langle S_j\left(u_2\right){\partial }_2{\theta }_j|{\theta }_j\rangle }_{L^2}.\end{array}$

所以，

${\langle S_j\left(u_2\right){\partial }_2{\theta }_j|{\theta }_j\rangle }_{L^2}=-\frac{1}{2}{\langle S_j\left({\partial }_2{u}_2\right){\theta }_j|{\theta }_j\rangle }_{L^2}=\frac{1}{2}{\langle S_j\left({\partial }_1{u}_1\right){\theta }_j|{\theta }_j\rangle }_{L^2}.$

对于$\text{I}={\langle S_j\left(u_2\right){\partial }_2{\theta }_j|{\theta }_j\rangle }_{L^2}$ ，我们可以用类似于$F_j^1$ 进行估计，从而得到

${\langle {\theta }_j{S}_j\left({\partial }_1{u}_1\right)|{\text{Δ}}_j\theta \rangle }_{L^2}\le C\cdot d_j{\Vert \theta \Vert }_{H^{\frac{1}{4},{\alpha }^{\prime }}}\left({\Vert \theta \Vert }_{H^{\frac{1}{4},{\alpha }^{\prime }}}{\Vert {\text{Λ}}^{\alpha }{u}_1\Vert }_{H^{0,\alpha }}+{\Vert \theta \Vert }_{H^{\frac{1}{4},\alpha }}{\Vert {\text{Λ}}^{\alpha }{u}_1\Vert }_{H^{0,{\alpha }^{\prime }}}\right).$ (21)

对于$\text{II}={\displaystyle \sum _{\left|j-k\right|\le 4}{\langle \left[{\Delta }_j,S_{k-1}{u}_2\right]{\Delta }_k{\partial }_2\theta |{\theta }_j\rangle }_{L^2}}$ ，令$h={ℱ}^{-1}\phi$ 和$h_1\left(x\right)=xh\left(x\right)$ 。那么，

$\begin{array}{l}\left[{\Delta }_j,S_{k-1}{u}_2\right]{\tilde{\theta }}_K\left(x_1,x_2\right)\\ ={\Delta }_j{S}_{k-1}{u}_2\left(x_1,x_2\right)-S_{k-1}{u}_2{\Delta }_j{\tilde{\theta }}_k\left(x_1,x_2\right)\\ ={ℱ}^{-1}\left[\phi \left(2^{-j}{y}_2\right)\left(\xi \right)\widehat{S_{k-1}{u}_2}\right]\left(x_1,x_2\right)-S_{k-1}{u}_2\left(x_1-x_2\right){ℱ}^{-1}\left[\phi \left(2^{-j}{y}_2\right)\left(\xi \right){\tilde{\theta }}_k\widehat{\left(x_1,x_2\right)}\right]\\ =2^j{\displaystyle {\int }_{ℝ}h}\left(2^j{y}_2\right)\left(S_{k-1}{u}_2\left(x_1,x_2-y_2\right)-S_{k-1}{u}_2\left(x_1,x_2\right)\right){\tilde{\theta }}_k\left(x_1,x_2-y_2\right)\text{d}{y}_2\\ =-2^j{\displaystyle {\int }_{ℝ\times \left[0,1\right]}{y}_2}h\left(2^j{y}_2\right)S_{k-1}{\partial }_2{u}_2\left(x_1,x_2-t{y}_2\right){\tilde{\theta }}_k\left(x_1,x_2-y_2\right)\text{d}{y}_2\text{d}t\\ ={\displaystyle {\int }_{ℝ\times \left[0,1\right]}{h}_1}\left(2^j{y}_2\right)S_{k-1}{\partial }_1{u}_1\left(x_1,x_2-t{y}_2\right){\tilde{\theta }}_k\left(x_1,x_2-y_2\right)\text{d}{y}_2\text{d}t.\end{array}$

通过(8)，我们推导出

${\Vert \left[{\Delta }_j,S_{k-1}{u}_2\right]{\tilde{\theta }}_k\left(\cdot ,x_2\right)\Vert }_{H-\frac{1}{4}}\le C{\displaystyle {\int }_{ℝ}\left|h_1\left(2^j{y}_2\right)\right|}{\Vert S_{k-1}{\partial }_1{u}_1\Vert }_{L^{\infty }\left({ℝ}_{x_2};L^2\right)}{\Vert {\tilde{\theta }}_k\left(\cdot ,x_2-y_2\right)\Vert }_{H^{\frac{1}{4}}}\text{d}{y}_2.$

通过对两边取$L^2$ 范数，并利用范数的积分被积分的范数所限制的性质，结合卷积不等式，我们得到：

${\Vert {\Vert \left[{\Delta }_j,S_{k-1}{u}_2\right]{\tilde{\theta }}_k\left(\cdot ,x_2\right)\Vert }_{H^{-\frac{1}{4}}}\Vert }_{L^2}\le C\cdot 2^{-j}{\Vert S_{k-1}{\partial }_1{u}_1\Vert }_{L^{\infty }\left(R_{x_2},L^2\right)}{\Vert {\tilde{\theta }}_k\Vert }_{L^2\left(R_{x_2};H^{\frac{1}{4}}\right)}.$

此外，应用引理1、引理2、Bernstein不等式，并利用$\left|j-k\right|\le 4$ 这一条件，我们得到：

$\begin{array}{l}{\displaystyle \sum _{\left|j-k\right|\le 4}{\langle \left[{\Delta }_j,S_{k-1}{u}_2\right]{\Delta }_k{\partial }_2\theta |{\theta }_j\rangle }_{L^2}}\\ \le {\displaystyle \sum _{\left|j-k\right|\le 4}C}\cdot 2^{-j}{\Vert S_{k-1}{\partial }_1{u}_1\Vert }_{L^{\infty }\left({ℝ}_{x_2};L^2\right)}{\Vert {\theta }_k\Vert }_{L^2\left({ℝ}_{x_2};H^{\frac{1}{4}}\right)}{\Vert {\theta }_j\Vert }_{L^2\left({ℝ}_{x_2};H^{\frac{1}{4}}\right)}\\ \le {\displaystyle \sum _{\left|j-k\right|\le 4}C}\cdot 2^{k-j}{\Vert {\Delta }_k{u}_1\Vert }_{L^2\left({ℝ}^2\right)}{\displaystyle \sum _{k=-\infty }^{+\infty }{2}^{ks}}{2}^{-ks}{\Vert {\theta }_k\Vert }_{L^2\left({ℝ}_{x_2};H^{\frac{1}{4}}\right)}\cdot {\chi }_{\left|j-k\right|\le 4}\\ \le {\displaystyle \sum _{\left|j-k\right|\le 4}C}\cdot 2^{k-j}{\Vert {\Lambda }^{\alpha }{u}_1\Vert }_{L^2\left({ℝ}^2\right)}\cdot 2^{-j\alpha }{\Vert {\theta }_k\Vert }_{H^{\frac{1}{4},\alpha }}{\Vert {\theta }_j\Vert }_{H^{\frac{1}{4},\alpha }}\\ \le C\cdot d_j\cdot 2^{-j\alpha }{\Vert {\Lambda }^{\alpha }{u}_1\Vert }_{H^{0,{\alpha }^{\prime }}}{\Vert \theta \Vert }_{H^{\frac{1}{4},\alpha }}^2.\end{array}$

对于$\text{III}={\displaystyle \sum _{\left|j-k\right|\le 1}{\langle \left(S_{k-1}-S_j\right)u_2{\partial }_2{\Delta }_k{\theta }_j|{\theta }_j\rangle }_{L^2}}$ 中$\left(S_{k-1}-S_j\right)u_2{\partial }_2{\Delta }_k{\theta }_j$ 我们可以估计为$F_j^{2,1}$ 。

此外，通过应用Riesz定理，我们得到

$\left|{\langle {\Delta }_j\left(u\cdot \nabla \theta \right)|{\Delta }_j\theta \rangle }_{L^2}\right|\le C{d}_j\cdot 2^{-j\alpha }{\Vert \theta \Vert }_{H^{\frac{1}{4},\alpha }}\left({\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,{\alpha }^{\prime }}}{\Vert \theta \Vert }_{H^{\frac{1}{4},\alpha }}+{\Vert \theta \Vert }_{H^{\frac{1}{4},{\alpha }^{\prime }}}{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,\alpha }}\right).$

引理3已证明。

4. 定理1的证明

局部光滑解可简单得到。若存在全局先验界，则全局解的存在性可通过标准紧性方法推导。首先，将算子${\Delta }_j$ 作用于方程(1)，

${\Delta }_j\left({\partial }_t\theta \right)+{\Delta }_j\left(u\cdot \nabla \theta \right)+v{\Delta }_j{\left(-\Delta \right)}^{\alpha }\theta =0,$ (22)

并与${\Delta }_j\theta$ 进行内积运算，我们得到能量估计式

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert {\Delta }_j\theta \Vert }_{L^2}^2+v{\Vert {\Lambda }^{\alpha }{\Delta }_j\theta \Vert }_{L^2}^2\le {\langle {\Delta }_j\left(u\cdot \nabla \theta \right)|{\Delta }_j\theta \rangle }_{L^2},$ (">)">

利用${\theta }_j=\Delta j\theta$ ，并应用引理3 (其中$\alpha ={\alpha }^{\prime }$ )，可得：

$\frac{\text{d}}{\text{d}t}{\Vert {\Delta }_j\theta \Vert }_{L^2}^2+2v{\Vert {\Lambda }^{\alpha }{\Delta }_j\theta \Vert }_{L^2}^2\le C{d}_j\cdot 2^{-j\alpha }{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,\alpha }}{\Vert \theta \Vert }_{H^{\frac{1}{4},\alpha }}^2.$ (24)

两边同时乘以$2^{j\alpha }$ 并取$l^1$ 范数，可得出结论：

$\frac{\text{d}}{\text{d}t}{\Vert \theta \left(t\right)\Vert }_{H^{0,\alpha }}^2+2v{\Vert {\Lambda }^{\alpha }\theta \left(t\right)\Vert }_{H^{0,\alpha }}^2\le C\cdot {\Vert \theta \Vert }_{H^{\frac{1}{4},\alpha }}^2{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,\alpha }}.$ (25)

根据Sobolev嵌入$H^{\frac{1}{2},\alpha }\looparrowright H^{\frac{1}{4},\alpha }$ ，我们可以推导出

$\frac{\text{d}}{\text{d}t}{\Vert \theta \left(t\right)\Vert }_{H^{0,\alpha }}^2+2v{\Vert {\Lambda }^{\alpha }\theta \left(t\right)\Vert }_{H^{0,\alpha }}^2\le C{\Vert \theta \Vert }_{H^{0,\alpha }}{\Vert \theta \Vert }_{H^{\frac{1}{2},\alpha }}^2{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,\alpha }}.$ (26)

应用插值不等式

${\Vert \theta \Vert }_{H^{\frac{1}{2},\alpha }}^2\le C\cdot {\Vert \theta \Vert }_{H^{0,\alpha }}{\Vert \theta \Vert }_{H^{1,\alpha }}\le C\cdot {\Vert \theta \Vert }_{H^{0,\alpha }}{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,\alpha }},$ (27)

由此可知

$\frac{\text{d}}{\text{d}t}{\Vert \theta \left(t\right)\Vert }_{H^{0,\alpha }}^2+2v{\Vert {\Lambda }^{\alpha }\theta \left(t\right)\Vert }_{H^{0,\alpha }}^2\le C{\Vert \theta \Vert }_{H^{0,\alpha }}{\Vert {\Lambda }^{\alpha }\theta \Vert }_{H^{0,\alpha }}^2.$

对于所有$t\in \left[0,\infty \right]$ ，${\Vert \theta \left(t\right)\Vert }_{H^{0,\alpha }}$ 满足以下全局界限与小性条件：

${\Vert \theta \left(t\right)\Vert }_{H^{0,\alpha }}<\frac{v}{C}.$

因此，

$\frac{\text{d}}{\text{d}t}{\Vert \theta \left(t\right)\Vert }_{H^{0,\alpha }}^2+v{\Vert {\Lambda }^{\alpha }\theta \left(t\right)\Vert }_{H^{0,\alpha }}^2\le 0.$

假设$t=t^{\ast }$ ，我们得到${\Vert \theta \left(t^{*}\right)\Vert }_{H^{0,\alpha }}=\frac{v}{C}$ ，并且

${\displaystyle {\int }_0^{t^{*}}\frac{\text{d}}{\text{d}t}}{\Vert \theta \left(t\right)\Vert }_{H^{0,\alpha }}^2\text{d}t+v{\displaystyle {\int }_0^{t^{*}}{\Vert {\Lambda }^{\alpha }\theta \left(t\right)\Vert }_{H^{0,\alpha }}^2\text{d}t}\le 0,t\in \left[0,t^{*}\right].$ (28)

我们得到${\Vert \theta \left(t^{*}\right)\Vert }_{H^{0,\alpha }}^2+v{\displaystyle {\int }_0^{t^{*}}{\Vert {\Lambda }^{\alpha }\theta \left(t\right)\Vert }_{H^{0,\alpha }}^2}\text{d}t\le {\Vert \theta \left(0\right)\Vert }_{H^{0,\alpha }}^2$ ，那么${\Vert \theta \left(t^{*}\right)\Vert }_{H^{0,\alpha }}\le {\Vert \theta \left(0\right)\Vert }_{H^{0,\alpha }}<\frac{v}{C}$ ，这与假设相矛盾。

因此对于$t\in \left[0,\infty \right)$

${\Vert \theta \left(t\right)\Vert }_{H^{0,\alpha }}^2+v{\displaystyle {\int }_0^t{\Vert {\Lambda }^{\alpha }\theta \left(t\right)\Vert }_{H^{0,\alpha }}^2\text{d}t}\le {\left(\frac{v}{C}\right)}^2.$

因此，定理1的证明至此完成。

基金项目

本研究得到国家自然科学基金(12461020)的支持。