分数阶Navier-Stokes方程在齐次Sobolev空间中解的爆破准则

On the Blow-Up Criterion for Solutions of 3D Fractional Navier-Stokes Equations in Homogeneous Sobolev Spaces

摘要: 本文在最大时间T_v^*有限时，利用Fourier变换的性质，齐次Sobolev空间中的插值结果以及乘积定理，研究了分数阶三维不可压缩Navier-Stokes方程在齐次Sobolev空间H^s中解的爆破性和L²范数的衰减性，以及解关于H^2-a范数、范数和范数的有界性，是对Benameur J的经典Navier-Stokes方程结论的推广。

Abstract: In this paper, when the maximum time T_v^* is finite, the blow-up of the solutions to the fractional 3D incompressible Navier-Stokes equations in H^s spaces and the decay in L² norm and the boundedness of the solution with respect to H^2-a norm, norm and norm are stud-ied, via using the property of Fourier transform, interpolation results and product law in the homo-geneous Sobolev spaces. It’s a generalization of the classical Navier-Stokes equations conclusion of Benameur J.

1. 引言

本文研究分数阶不可压缩Navier-Stokes方程的初值问题

$\{\begin{array}{l}{\partial }_{t}u+\nu {\left(-\Delta \right)}^{\alpha }u+\left(u\cdot \nabla \right)u=-\nabla p,\\ \text{div}u=0,\\ u\left(0\right)=u^0,\end{array}$ (1.1)

在最大存在时间附近解的性质以及在齐次Sobolev空间的爆破性。其中 $\nu$ 为流体的粘性系数， $\alpha >0$ 表示“耗散强度”。向量函数 $u\left(t,x\right)=\left(u_1\left(t,x\right),u_2\left(t,x\right),u_3\left(t,x\right)\right)$ 表示流体在 $\left(t,x\right)\in R^{+}\times R^3$ 的未知速度， $u^0=\left(u_1^0\left(t,x\right),u_2^0\left(t,x\right),u_3^0\left(t,x\right)\right)$ 是给定的初始速度，标量函数 $p=p\left(t,x\right)$ 表示流体在 $\left(t,x\right)\in R^{+}\times R^3$ 所受的未知压力， $\Delta ={\displaystyle {\sum }_{j=1}^n{\partial }_{x_j}^2}$ 是关于空间变量 $x=\left(x_1,x_2,x_3\right)$ 的Laplacian微分算子。

对于经典不可压缩Navier-Stokes方程，Kato T在文献 [1] 和Leray J在文献 [2] 中研究了解的局部存在性和唯一性。Benameur J在文献 [3] 中研究了非齐次Sobolev空间中解的爆破准则，即 ${\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}\ge {\left(T_{\nu }^{\text{*}}-t\right)}^{-\frac{s}{3}}\left(s>\frac{5}{2}\right)$，其中 $T_{\nu }^{\text{*}}$ 是最大存在时间。Robinson JC，Sadowski W，Silva RP在文献 [4] 中证明了 ${\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}\ge {\left(T_{\nu }^{\text{*}}-t\right)}^{-\frac{2s-1}{4}}\left(\frac{1}{2}<s<\frac{5}{2},s\ne \frac{3}{2}\right)$， ${\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}\ge {\left(T_{\nu }^{\text{*}}-t\right)}^{-\frac{2s}{5}}\left(s>\frac{5}{2}\right)$。Cortissoz C，Montero JA，Pinilla CE在文献 [5] 中研究了解的 ${\stackrel{\dot{}}{H}}^{\frac{3}{2}}$ 范数和 ${\stackrel{\dot{}}{H}}^{\frac{5}{2}}$ 范数的下界。Cheskidov A，Zaya K在文献 [6] 中证明了解在 ${\stackrel{\dot{}}{H}}^{\frac{3}{2}}$ 空间的强爆破估计。Benameur J在文献 [7] 中研究了解在 ${\stackrel{\dot{}}{H}}^{\frac{5}{2}}$ 空间的爆破准则。

本文研究分数阶不可压缩Navier-Stokes方程(1.1)的解在 ${\stackrel{\dot{}}{H}}^s$ 空间中的爆破和 $L^2$ 范数衰减，以及解关于 ${\stackrel{\dot{}}{H}}^{2-\alpha }$ 范数和 ${\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }$ 范数和 ${\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }$ 范数的有界性，是对Benameur J文献 [3] [7] 中结论的推广。主要结果如下：

定理1 设 $\alpha >1$， $s>\max \left\{\alpha -1,\frac{3}{2}\right\}$， $\lim {\sup }_{t\to T_{\nu }^{*}}{\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}^2=\infty$ 且 $u\in C\left(\left[0,T_{\nu }^{*}\right);{\stackrel{\dot{}}{H}}^s\right)$ 是方程(1.1)的极大解，若 $T_{\nu }^{*}<\infty$，则有

$\frac{c{\nu }^{\frac{s}{3}}}{{\left(T_{\nu }^{*}-t\right)}^{\frac{s}{3}}}\le {\Vert u\left(t\right)\Vert }_{L^2}^{\frac{2s}{3}-1}{\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}.$

定理2 设 $1<\alpha <2$， $\lim {\sup }_{t\to T_{\nu }^{*}}{\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2=\infty$ 且 $u\in C\left(\left[0,T_{\nu }^{*}\right);{\stackrel{\dot{}}{H}}^{2-\alpha }\right)$ 是方程(1.1)的极大解，若 $T_{\nu }^{*}<\infty$，则有

$\frac{c\sqrt{\nu }}{\sqrt{T_{\nu }^{*}-t}}\le {\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{H}}^{1-\alpha }}\text{.}$

定理3 设 $1<\alpha <\frac{5}{2}$， $u\in C\left(\left[0,T_{\nu }^{*}\right);{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }\right)$ 是方程(1.1)的极大解，若 $T_{\nu }^{*}<\infty$，则有

$\frac{c\sqrt{\nu }}{\sqrt{T_{\nu }^{*}-t}}\le {\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}.$

定理4 设 $1<\alpha <\frac{7}{2}$， $u\in C\left(\left[0,T_{\nu }^{*}\right);{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }\right)$ 是方程(1.1)的极大解，若 $T_{\nu }^{*}<\infty$，则有

$\frac{c{\left(\alpha -1\right)}^{\frac{1}{\alpha +1}}}{{\left(T_{\nu }^{*}-t\right)}^{\frac{2}{\alpha +1}}}\le {\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}.$

2. 预备知识

1) [7] Fourier变换定义为：

$\mathcal{F}\left(f\right)\left(\xi \right)=\hat{f}\left(\xi \right)={\displaystyle {\int }_{R^3}\exp }\left(-ix\cdot \xi \right)f\left(x\right)\text{d}x,\xi =\left({\xi }_1,{\xi }_2,{\xi }_3\right)\in R^3,f\in L^1\left(R^3\right).$

2) 分数阶Laplacian微分算子通过Fourier变换定义为：

$\mathcal{F}\left({\left(-\Delta \right)}^{\alpha }u\right)\left(t,\xi \right)={\left|\xi \right|}^{2\alpha }\mathcal{F}u\left(t,\xi \right).$

关于 ${\left(-\Delta \right)}^{\alpha }$ 的更多详细描述见文献 [8]。

3) [7] $R^3$ 上函数 $f\left(x\right),g\left(x\right)$ 定义其卷积为：

$f\ast g\left(x\right)={\displaystyle {\int }_{R^3}f\left(x-y\right)g\left(y\right)\text{d}y}={\displaystyle {\int }_{R^3}f}\left(y\right)g\left(x-y\right)\text{d}y.$

4) [7] 若 $f=\left(f_1,f_2,f_3\right)$ 和 $g=\left(g_1,g_2,g_3\right)$ 是两个向量场，则

$f\otimes g:=\left(g_{1}f,g_{2}f,g_{3}f\right),$

$\text{div}\left(f\otimes g\right):=\left(\text{div}\left(g_{1}f\right),\text{div}\left(g_{2}f\right),\text{div}\left(g_{3}f\right)\right).$

5) [7] 齐次Sobolev空间定义为： ${\stackrel{\dot{}}{H}}^s\left(R^3\right)=\left\{f\in S^{\prime }\left(R^3\right),\hat{f}\in L_{loc}^1,{\left|\xi \right|}^s\hat{f}\in L^2\left(R^3\right)\right\}$，其范数为

${\Vert f\Vert }_{{\stackrel{\dot{}}{H}}^s}={\left({\displaystyle {\int }_{R^3}{\left|\xi \right|}^{2s}}{\left|\hat{f}\left(\xi \right)\right|}^2\text{d}\xi \right)}^{\frac{1}{2}}.$

6) [7] Lei-Lin空间定义为： ${\chi }^{\sigma }\left(R^3\right)=\left\{f\in S^{\prime }\left(R^3\right),\hat{f}\in L_{loc}^1,{\left|\xi \right|}^{\sigma }\hat{f}\in L^1\left(R^3\right)\right\}$，其范数为

${\Vert f\Vert }_{{\chi }^{\sigma }}={\displaystyle {\int }_{R^3}{\left|\xi \right|}^{\sigma }}\left|\hat{f}\left(\xi \right)\right|\text{d}\xi .$

引理1 [9] 对于 $s>\frac{3}{2}$，有

${\Vert \hat{u}\Vert }_{L^1}\le C{\Vert u\Vert }_{L^2}^{1-\frac{3}{2s}}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^s}^{\frac{3}{2s}}\text{.}$

其中 $C=2\sqrt{\frac{\pi }{3}}\left({\left(\frac{2s}{3}-1\right)}^{\frac{3}{4s}}+{\left(\frac{2s}{3}-1\right)}^{-1+\frac{3}{4s}}\right)$。

引理2 [10] 设 $s,s^{\prime }\in R$ 且 $s<1$， $s+s^{\prime }>0$，则存在常数 $C=C\left(s,s^{\prime }\right)$，使得 $f,g\in {\stackrel{\dot{}}{H}}^s\cap {\stackrel{\dot{}}{H}}^{s^{\prime }}$，有

${\Vert fg\Vert }_{{\stackrel{\dot{}}{H}}^{s+s^{\prime }-1}}\le C\left({\Vert f\Vert }_{{\stackrel{\dot{}}{H}}^s}{\Vert g\Vert }_{{\stackrel{\dot{}}{H}}^{s^{\prime }}}+{\Vert f\Vert }_{{\stackrel{\dot{}}{H}}^{s^{\prime }}}{\Vert g\Vert }_{{\stackrel{\dot{}}{H}}^s}\right).$

若 $s,s^{\prime }<1$， $s+s^{\prime }>0$，则存在常数 $C=C\left(s,s^{\prime }\right)$，使得 $f\in {\stackrel{\dot{}}{H}}^s,g\in {\stackrel{\dot{}}{H}}^{s^{\prime }}$，有

${\Vert fg\Vert }_{{\stackrel{\dot{}}{H}}^{s+s^{\prime }-1}}\le C{\Vert f\Vert }_{{\stackrel{\dot{}}{H}}^s}{\Vert g\Vert }_{{\stackrel{\dot{}}{H}}^{s^{\prime }}}.$

引理3 [11] 设 $s,s^{\prime }\in R$ 且 $s<\frac{3}{2}$， $s+s^{\prime }>0$，则存在常数 $C=C\left(s,s^{\prime }\right)$，使得 $f,g\in {\stackrel{\dot{}}{H}}^s\cap {\stackrel{\dot{}}{H}}^{s^{\prime }}$，

${\Vert fg\Vert }_{{\stackrel{\dot{}}{H}}^{s+s^{\prime }-\frac{3}{2}}}\le C\left({\Vert f\Vert }_{{\stackrel{\dot{}}{H}}^s}{\Vert g\Vert }_{{\stackrel{\dot{}}{H}}^{s^{\prime }}}+{\Vert f\Vert }_{{\stackrel{\dot{}}{H}}^s}{\Vert g\Vert }_{{\stackrel{\dot{}}{H}}^{s^{\prime }}}\right).$

若 $s,s^{\prime }<\frac{3}{2}$， $s+s^{\prime }>0$，则存在常数 $C=C\left(s,s^{\prime }\right)$，

${\Vert fg\Vert }_{{\stackrel{\dot{}}{H}}^{s+s^{\prime }-\frac{3}{2}}}\le C{\Vert f\Vert }_{{\stackrel{\dot{}}{H}}^s}{\Vert g\Vert }_{{\stackrel{\dot{}}{H}}^{s^{\prime }}}.$

引理4 [7] 对于 $s\ge 0$，若 $f\in {\stackrel{\dot{}}{H}}^s\cap {\chi }^1$，则存在常数 $C>0$，

$\left|{\langle f\cdot \nabla f\rangle }_{{\stackrel{\dot{}}{H}}^s}\right|\le C{\Vert f\Vert }_{{\chi }^1}{\Vert f\Vert }_{{\stackrel{\dot{}}{H}}^s}^2\text{.}$

引理5 [12] Gronwall不等式(微分形式)设 $y\left(t\right)$ 是 $\left[0,T\right]$ 上的非负绝对连续函数， $h\left(t\right)$， $g\left(t\right)$ 是 $\left[0,T\right]$ 上的非负可积函数，且满足

$y^{\prime }\left(t\right)\le h\left(t\right)y\left(t\right)+g\left(t\right),\text{a}\text{.e}.t\in \left[0,T\right].$

那么

$y\left(t\right)\le {\text{e}}^{{\displaystyle {\int }_0^{t}h}\left(\tau \right)\text{d}\tau }\left[y\left(0\right)+{\displaystyle {\int }_0^{t}g}\left(\tau \right)\text{d}\tau \right].$

引理6 对于 $f\in {\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }$，存在常数 $C>0$，使得 $0<a<b<\infty$，有

${\Vert f\Vert }_{{\chi }^1}\le C\sqrt{4\pi }\left[a^{\frac{5}{2}}{\Vert u\Vert }_{L^2}+\sqrt{\frac{b^{2\alpha -2}}{\alpha -1}}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}+\frac{1}{b}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}}}\right]\text{.}$

证明：设 $0<a<b<\infty$

${\Vert f\Vert }_{{\chi }^1}={\displaystyle \int \left|\xi \right|\left|\hat{f}\left(\xi \right)\right|\text{d}\xi }=I_a+J_{a,b}+K_b,$

$I_a={\displaystyle {\int }_{\left|\xi \right|<a}\left|\xi \right|\left|\hat{f}\left(\xi \right)\right|\text{d}\xi }\le {\left({\displaystyle {\int }_{\left|\xi \right|<a}{\left|\xi \right|}^2\text{d}\xi }\right)}^{\frac{1}{2}}{\Vert f\Vert }_{L^2}\le \sqrt{\frac{4\pi }{5}}{a}^{\frac{5}{2}}{\Vert f\Vert }_{L^2}.$

$\begin{array}{c}{J}_{a,b}={\displaystyle {\int }_{a<\left|\xi \right|<b}\left|\xi \right|\left|\hat{f}\left(\xi \right)\right|\text{d}\xi }\le {\left({\displaystyle {\int }_{a<\left|\xi \right|<b}{\left|\xi \right|}^{2\alpha -5}\text{d}\xi }\right)}^{\frac{1}{2}}{\Vert f\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}\\ \le \sqrt{\frac{2\pi }{\alpha -1}}\sqrt{b^{2\alpha -2}-a^{2\alpha -2}}{\Vert f\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}\le \sqrt{\frac{2\pi }{\alpha -1}}\sqrt{b^{2\alpha -2}}{\Vert f\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}\text{.}\end{array}$

$K_b={\displaystyle {\int }_{\left|\xi \right|>b}\left|\xi \right|\left|\hat{f}\left(\xi \right)\right|\text{d}\xi }\le {\left({\displaystyle {\int }_{\left|\xi \right|>b}{\left|\xi \right|}^{-5}\text{d}\xi }\right)}^{\frac{1}{2}}{\Vert f\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}}}\le \sqrt{2\pi }\frac{1}{b}{\Vert f\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}}}\text{.}$

3. 定理1的证明

证明：方程(1.1)在 ${\stackrel{\dot{}}{H}}^s$ 空间下与u取内积

${\langle {\partial }_{t}u,u\rangle }_{{\stackrel{\dot{}}{H}}^s}+\nu {\langle {\left(-\Delta \right)}^{\alpha }u,u\rangle }_{{\stackrel{\dot{}}{H}}^s}+{\langle u\cdot \nabla u,u\rangle }_{{\stackrel{\dot{}}{H}}^s}={\langle -\nabla p,u\rangle }_{{\stackrel{\dot{}}{H}}^s},$

则有

$\begin{array}{c}\frac{1}{2}{\partial }_t{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^s}^2+\nu {\Vert {\Lambda }^{\alpha }u\Vert }_{{\stackrel{\dot{}}{H}}^s}^2\le \left|{\langle u\cdot \nabla u,u\rangle }_{{\stackrel{\dot{}}{H}}^s}\right|\\ ={\displaystyle {\int }_{R^3}{\left|\xi \right|}^{2s}\widehat{u\cdot \nabla u}\cdot \hat{u}\text{d}\xi }\\ ={\displaystyle {\int }_{R^3}{\left|\xi \right|}^{2s+1}\widehat{u\otimes u}\cdot \hat{u}\text{d}\xi }\\ ={\displaystyle {\int }_{R^3}{\left|\xi \right|}^{s-\alpha +1}\widehat{u\otimes u}{\left|\xi \right|}^s\widehat{{\Lambda }^{\alpha }}u\text{d}\xi }\\ \le {\left({\displaystyle {\int }_{R^3}{\left|\xi \right|}^{2\left(s-\alpha +1\right)}\widehat{u\otimes u}{}^2\text{d}\xi }\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_{R^3}{\left|\xi \right|}^{2s}\widehat{{\Lambda }^{\alpha }}{u}^2\text{d}\xi }\right)}^{\frac{1}{2}}\\ ={\Vert u\otimes u\Vert }_{{\stackrel{\dot{}}{H}}^{s-\alpha +1}}{\Vert {\Lambda }^{\alpha }u\Vert }_{{\stackrel{\dot{}}{H}}^s}.\end{array}$

又

$\begin{array}{c}{\Vert u\otimes u\Vert }_{{\stackrel{\dot{}}{H}}^{s-\alpha +1}}^2={\displaystyle {\int }_{R^3}{\left|\xi \right|}^{2\left(s-\alpha +1\right)}{\left|\hat{u}\ast \hat{u}\right|}^2\text{d}\xi }\\ ={\displaystyle {\int }_{R^3}{\left|\xi \right|}^{2\left(s-\alpha +1\right)}{\left|{\displaystyle {\int }_{R^3}\hat{u}\left(\eta \right)\hat{u}\left(\xi -\eta \right)\text{d}\eta }\right|}^2\text{d}\xi }\\ \le {\displaystyle {\int }_{R^3}{\left|\xi \right|}^{2\left(s-\alpha +1\right)}{\left({\displaystyle {\int }_{R^3}\left|\hat{u}\left(\eta \right)\right|\left|\hat{u}\left(\xi -\eta \right)\right|\text{d}\eta }\right)}^2\text{d}\xi }\\ \le {\displaystyle {\int }_{R^3}{\left({\displaystyle {\int }_{R^3}{\left|\xi \right|}^{s-\alpha +1}\left|\hat{u}\left(\eta \right)\right|\left|\hat{u}\left(\xi -\eta \right)\right|\text{d}\eta }\right)}^2\text{d}\xi }\text{.}\end{array}$

由

$\begin{array}{c}{\left|\xi \right|}^{s-\alpha +1}={\left|\xi -\eta +\eta \right|}^{s-\alpha +1}\le {\left(\left|\xi -\eta \right|+\left|\eta \right|\right)}^{s-\alpha +1}\le {\left(2\max \left(\left|\xi -\eta \right|,\left|\eta \right|\right)\right)}^{s-\alpha +1}\\ \le 2^{s-\alpha +1}\left({\left|\xi -\eta \right|}^{s-\alpha +1}+{\left|\eta \right|}^{s-\alpha +1}\right)\text{.}\end{array}$

$\begin{array}{c}{\Vert u\otimes u\Vert }_{{\stackrel{\dot{}}{H}}^{s-\alpha +1}}^2\le {\displaystyle {\int }_{R^3}{\left({\displaystyle {\int }_{R^3}{2}^{s-\alpha +1}}{\left|\xi -\eta \right|}^{s-\alpha +1}\left|\hat{u}\left(\eta \right)\right|\left|\hat{u}\left(\xi -\eta \right)\right|\text{d}\eta \right)}^2\text{d}\xi }\\ +{\displaystyle {\int }_{R^3}{\left({\displaystyle {\int }_{R^3}{2}^{s-\alpha +1}}{\left|\eta \right|}^{s-\alpha +1}\left|\hat{u}\left(\eta \right)\right|\left|\hat{u}\left(\xi -\eta \right)\right|\text{d}\eta \right)}^2\text{d}\xi }\\ =2^{2\left(s-\alpha +1\right)+1}{\displaystyle {\int }_{R^3}{\left({\displaystyle {\int }_{R^3}{\left|\eta \right|}^{s-\alpha +1}\left|\hat{u}\left(\eta \right)\right|\left|\hat{u}\left(\xi -\eta \right)\right|\text{d}\eta }\right)}^2\text{d}\xi }\\ =2^{2\left(s-\alpha +1\right)+1}{\displaystyle {\int }_{R^3}{\left(\left({\left|\cdot \right|}^{s-\alpha +1}\left|\hat{u}\right|\right)\ast \left|\hat{u}\right|\right)}^2\text{d}\xi }\end{array}$

$\begin{array}{l}=2^{2\left(s-\alpha +1\right)+1}{\Vert \left({\left|\cdot \right|}^{s-\alpha +1}\left|\hat{u}\right|\right)\ast \left|\hat{u}\right|\Vert }_{L^2}^2\\ \le 2^{2\left(s-\alpha +1\right)+1}{\Vert \left({\left|\cdot \right|}^{s-\alpha +1}\left|\hat{u}\right|\right)\Vert }_{L^2}^2{\Vert \hat{u}\Vert }_{L^1}^2\\ =2^{2\left(s-\alpha +1\right)+1}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{s-\alpha +1}}^2{\Vert \hat{u}\Vert }_{L^1}^2\text{.}\end{array}$

则有

$\begin{array}{c}\frac{1}{2}{\partial }_t{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^s}^2+\nu {\Vert {\Lambda }^{\alpha }u\Vert }_{{\stackrel{\dot{}}{H}}^s}^2\le 2^{s-\alpha +\frac{3}{2}}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{s-\alpha +1}}{\Vert \hat{u}\Vert }_{L^1}{\Vert {\Lambda }^{\alpha }u\Vert }_{{\stackrel{\dot{}}{H}}^s}\\ \le 2^{2s-2\alpha +3}{\nu }^{-1}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{s-\alpha +1}}^2{\Vert \hat{u}\Vert }_{L^1}^2+\frac{\nu }{2}{\Vert {\Lambda }^{\alpha }u\Vert }_{{\stackrel{\dot{}}{H}}^s}^2\text{.}\end{array}$

则

${\partial }_t{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^s}^2+\nu {\Vert {\Lambda }^{\alpha }u\Vert }_{{\stackrel{\dot{}}{H}}^s}^2\le C{\nu }^{-1}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{s-\alpha +1}}^2{\Vert \hat{u}\Vert }_{L^1}^2\le C{\nu }^{-1}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^s}^2{\Vert \hat{u}\Vert }_{L^1}^2\text{.}$

由引理5，对 $0\le a\le t<T_{\nu }^{*}$

${\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}^2\le {\Vert u\left(a\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}^2{\text{e}}^{c{\nu }^{-1}{\displaystyle {\int }_a^t{\Vert \hat{u}\Vert }_{L^1}^2\text{d}\tau }}\text{.}$

因为 $\lim {\sup }_{t\to T_{\nu }^{*}}{\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}^2=\infty$，所以 ${\displaystyle {\int }_a^{T_{\nu }^{*}}{\Vert \hat{u}\Vert }_{L^1}^2\text{d}\tau }=\infty$。

由引理1， ${\Vert \hat{u}\Vert }_{L^1}\le C{\Vert u\Vert }_{L^2}^{1-\frac{3}{2s}}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^s}^{\frac{3}{2s}}$

$\frac{{\Vert \hat{u}\Vert }_{L^1}^{\frac{4s}{3}}}{C{\Vert u\Vert }_{L^2}^{\frac{4s}{3}-2}}\le {\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^s}^2\le {\Vert u\left(a\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}^2{\text{e}}^{c{\nu }^{-1}{\displaystyle {\int }_a^t{\Vert \hat{u}\Vert }_{L^1}^2\text{d}\tau }},$

${\Vert \hat{u}\Vert }_{L^1}^{\frac{4s}{3}}\le C{\Vert u\Vert }_{L^2}^{\frac{4s}{3}-2}{\Vert u\left(a\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}^2{\text{e}}^{c{\nu }^{-1}{\displaystyle {\int }_a^t{\Vert \hat{u}\Vert }_{L^1}^2\text{d}\tau }},$

${\Vert \hat{u}\Vert }_{L^1}^2{\text{e}}^{-c{\nu }^{-1}{\displaystyle {\int }_a^t{\Vert \hat{u}\Vert }_{L^1}^2\text{d}\tau }}\le C{\Vert u\Vert }_{L^2}^{2-\frac{3}{s}}{\Vert u\left(a\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}^{\frac{3}{s}}\text{.}$

在 $\left[a,T\right]$ 上对上式积分有

$1-{\text{e}}^{-c{\nu }^{-1}{\displaystyle {\int }_a^t{\Vert \hat{u}\Vert }_{L^1}^2\text{d}\tau }}\le C{\nu }^{-1}\left(T-a\right){\Vert u\Vert }_{L^2}^{2-\frac{3}{s}}{\Vert u\left(a\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}^{\frac{3}{s}}.$

当 $T\to T_{\nu }^{*}$ 时，有

$1\le C{\nu }^{-1}\left(T_{\nu }^{*}-a\right){\Vert u\Vert }_{L^2}^{2-\frac{3}{s}}{\Vert u\left(a\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}^{\frac{3}{s}}$

$\frac{C{\nu }^{\frac{s}{3}}}{{\left(T_{\nu }^{*}-t\right)}^{\frac{s}{3}}}\le {\Vert u\left(t\right)\Vert }_{L^2}^{\frac{2s}{3}-1}{\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^s}.$

从而结论得证。

4. 定理2的证明

证明：方程(1.1)在 ${\stackrel{\dot{}}{H}}^{2-\alpha }$ 空间下与u取内积

${\langle {\partial }_{t}u,u\rangle }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}+\nu {\langle {\left(-\Delta \right)}^{\alpha }u,u\rangle }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}+{\langle u\cdot \nabla u,u\rangle }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}={\langle -\nabla p,u\rangle }_{{\stackrel{\dot{}}{H}}^{2-\alpha }},$

则有

$\begin{array}{c}\frac{1}{2}{\partial }_t{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2+\nu {\Vert {\Lambda }^{\alpha }u\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2\le \left|{\langle u\cdot \nabla u,u\rangle }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}\right|\\ ={\displaystyle {\int }_{R^3}{\left|\xi \right|}^{5-2\alpha }\widehat{u\otimes u}\cdot \hat{u}\text{d}\xi }\\ ={\displaystyle {\int }_{R^3}{\left|\xi \right|}^{3-2\alpha }\widehat{u\otimes u}{\left|\xi \right|}^{2-\alpha }{\left|\xi \right|}^{\alpha }\hat{u}\text{d}\xi }\\ \le {\Vert u\otimes u\Vert }_{{\stackrel{\dot{}}{H}}^{3-2\alpha }}{\Vert {\Lambda }^{\alpha }u\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}.\end{array}$

由引理2及不等式 $ab\le \frac{a^2}{2}+\frac{b^2}{2}$ 有

$\begin{array}{c}\frac{1}{2}{\partial }_t{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2+\nu {\Vert {\Lambda }^{\alpha }u\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2\le C{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}{\Vert {\Lambda }^{\alpha }u\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}\\ \le C{\nu }^{-1}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^4+\frac{\nu }{2}{\Vert {\Lambda }^{\alpha }u\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2.\end{array}$

则

${\partial }_t{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^s}^2+\nu {\Vert {\Lambda }^{\alpha }u\Vert }_{{\stackrel{\dot{}}{H}}^s}^2\le C{\nu }^{-1}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^4.$

由引理5，对于 $0\le a\le t<T_{\nu }^{*}$，

${\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2\le {\Vert u\left(a\right)\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2{\text{e}}^{C{\nu }^{-1}{\displaystyle {\int }_a^t{\Vert u\left(\tau \right)\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2\text{d}\tau }}\text{.}$

因为 $\lim {\sup }_{t\to T_{\nu }^{*}}{\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2=\infty$，所以 ${\displaystyle {\int }_a^{T_{\nu }^{*}}{\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2}=\infty$。

${\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2{\text{e}}^{-C{\nu }^{-1}{\displaystyle {\int }_a^t{\Vert u\left(\tau \right)\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2\text{d}\tau }}\le {\Vert u\left(a\right)\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2.$

在 $\left[a,T\right]$ 上对上式积分有

$1-{\text{e}}^{-C{\nu }^{-1}{\displaystyle {\int }_a^T{\Vert u\left(\tau \right)\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2\text{d}\tau }}\le C{\nu }^{-1}\left(T-a\right){\Vert u\left(a\right)\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2.$

当 $T\to T_{\nu }^{*}$ 时，有

$1\le C{\nu }^{-1}\left(T_{\nu }^{*}-a\right){\Vert u\left(a\right)\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}^2,$

$\frac{c\sqrt{\nu }}{\sqrt{T_{\nu }^{*}-t}}\le {\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^{2-\alpha }}\text{.}$

$\frac{c\sqrt{\nu }}{\sqrt{T_{\nu }^{*}-t}}\le {\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{H}}^{1-\alpha }}.$

从而结论得证。

5. 定理3的证明

证明：方程(1.1)在 ${\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }$ 空间下与u取内积

${\langle {\partial }_{t}u,u\rangle }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}+\nu {\langle {\left(-\Delta \right)}^{\alpha }u,u\rangle }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}+{\langle u\cdot \nabla u,u\rangle }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}={\langle -\nabla p,u\rangle }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }},$

则有

$\begin{array}{c}\frac{1}{2}{\partial }_t{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}^2+\nu {\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}}}^2\le \left|{\langle u\cdot \nabla u,u\rangle }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}\right|\\ ={\displaystyle {\int }_{R^3}{\left|\xi \right|}^{6-2\alpha }\widehat{u\otimes u}\cdot \hat{u}\text{d}\xi }\\ ={\displaystyle {\int }_{R^3}{\left|\xi \right|}^{\frac{7}{2}-2\alpha }\widehat{u\otimes u}{\left|\xi \right|}^{\frac{5}{2}}\hat{u}\text{d}\xi }\\ \le {\Vert u\otimes u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-2\alpha }}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}}}.\end{array}$

由引理3及不等式 $ab\le \frac{a^2}{2}+\frac{b^2}{2}$ 有

$\frac{1}{2}{\partial }_t{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}^2+\nu {\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}}}^2\le C{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}}}\le C{\nu }^{-1}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}^4+\frac{\nu }{2}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}}}^2.$

${\partial }_t{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}^2\le C{\nu }^{-1}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}^4\text{.}$

在 $\left[t,T_{\nu }^{*}\right)\subset \left[0,T_{\nu }^{*}\right)$ 上对上式积分有

${\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}^2\le C{\nu }^{-1}\left(T_{\nu }^{*}-t\right){\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}^4,$

$\frac{c\sqrt{\nu }}{\sqrt{T_{\nu }^{*}-t}}\le {\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{5}{2}-\alpha }}\text{.}$

从而结论得证。

6. 定理4的证明

证明：方程(1.1)在 ${\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }$ 空间下与u取内积

${\langle {\partial }_{t}u,u\rangle }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}+\nu {\langle {\left(-\Delta \right)}^{\alpha }u,u\rangle }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}+{\langle u\cdot \nabla u,u\rangle }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}={\langle -\nabla p,u\rangle }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }},$

则有

$\frac{1}{2}{\partial }_t{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2+\nu {\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}}}^2\le \left|{\langle u\cdot \nabla u,u\rangle }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}\right|\text{.}$

由引理4，

$\frac{1}{2}{\partial }_t{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2+\nu {\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}}}^2\le c{\Vert u\Vert }_{{\chi }^1}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2\text{.}$

由引理6，

$\begin{array}{c}\frac{1}{2}{\partial }_t{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2+\nu {\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}}}^2\le c{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2\left[a^{\frac{5}{2}}{\Vert u\Vert }_{L^2}+\sqrt{\frac{b^{2\alpha -2}}{\alpha -1}}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}+\frac{1}{b}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}}}\right]\\ \le c{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2\left[a^{\frac{5}{2}}{\Vert u\Vert }_{L^2}+\sqrt{\frac{b^{2\alpha -2}}{\alpha -1}}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}\right]+\frac{c}{\nu b^2}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^4+\frac{\nu }{2}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}}}^2\text{.}\end{array}$

则有

${\partial }_t{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2\le c{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2\left[a^{\frac{5}{2}}{\Vert u\Vert }_{L^2}+\sqrt{\frac{b^{2\alpha -2}}{\alpha -1}}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}+\frac{1}{\nu b^2}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2\right].$

取 $t_0=\inf \left\{t\in \left[0,T_{\nu }^{*}\right),{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}=2{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}\right\}$，对上式积分有

${\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2-{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2\le c{T}_{\nu }^{*}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2\left[a^{\frac{5}{2}}{\Vert u\Vert }_{L^2}+\sqrt{\frac{b^{2\alpha -2}}{\alpha -1}}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}+\frac{1}{\nu b^2}{\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2\right],$

$3{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2\le c{T}_{\nu }^{*}{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2\left[a^{\frac{5}{2}}{\Vert u\Vert }_{L^2}+\sqrt{\frac{b^{2\alpha -2}}{\alpha -1}}{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}+\frac{1}{\nu b^2}{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2\right],$

$1\le c{T}_{\nu }^{*}\left[a^{\frac{5}{2}}{\Vert u\Vert }_{L^2}+\sqrt{\frac{b^{2\alpha -2}}{\alpha -1}}{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}+\frac{1}{\nu b^2}{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^2\right]\text{.}$

取 $a={\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^{\frac{2}{5}}<b=\sqrt{{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}}$，

$1\le c{T}_{\nu }^{*}\left[{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}{\Vert u\Vert }_{L^2}+\frac{1}{\sqrt{\alpha -1}}\sqrt{{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^{\alpha -1}}{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}+\frac{1}{\nu }{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}\right],$

$1\le c{T}_{\nu }^{*}{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}\left[{\Vert u\Vert }_{L^2}+\frac{1}{\sqrt{\alpha -1}}\sqrt{{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^{\alpha -1}}+\frac{1}{\nu }\right].$

设 $t_1\in \left[0,T_{\nu }^{*}\right)$ 有

$\frac{1}{\sqrt{\alpha -1}}\sqrt{{\Vert u^0\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^{\alpha -1}}\ge 2\left({\Vert u\Vert }_{L^2}+\frac{1}{\nu }\right).$

则

$1\le c\left(T_{\nu }^{*}-t_1\right){\Vert u\left(t_1\right)\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}\frac{1}{\sqrt{\alpha -1}}\sqrt{{\Vert u\left(t_1\right)\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}^{\alpha -1}}.$

用任意的 $t\in \left[t_1,T_{\nu }^{*}\right)$ 替换 $t_1$ 有

$\frac{c{\left(\alpha -1\right)}^{\frac{1}{\alpha +1}}}{{\left(T_{\nu }^{*}-t\right)}^{\frac{2}{\alpha +1}}}\le {\Vert u\left(t\right)\Vert }_{{\stackrel{\dot{}}{H}}^{\frac{7}{2}-\alpha }}.$

因此结论得证。

NOTES

^*通讯作者。