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

doi:10.12677/AAM.2023.121026

Advances in Applied Mathematics
Vol. 12 No. 01 ( 2023 ), Article ID: 60844 , 9 pages
10.12677/AAM.2023.121026

分数阶Navier-Stokes方程在齐次Sobolev空间中解的爆破准则

徐郜婷，孙小春^*，吴育联

●How to Cite this Article

西北师范大学，甘肃兰州

收稿日期：2022年12月28日；录用日期：2023年1月24日；发布日期：2023年1月31日

摘要

本文在最大时间 $T_{ν}^{*}$ 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 ${\dot{H}}^{s}$ 中解的爆破性和 $L^{2}$ 范数的衰减性，以及解关于 ${\dot{H}}^{2 - α}$ 范数、 ${\dot{H}}^{\frac{5}{2} - α}$ 范数和 ${\dot{H}}^{\frac{7}{2} - α}$ 范数的有界性，是对Benameur J的经典Navier-Stokes方程结论的推广。

关键词

分数阶Navier-Stokes方程，衰减性，爆破准则

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

Gaoting Xu, Xiaochun Sun^*, Yulian Wu

Northwest Normal University, Lanzhou Gansu

Received: Dec. 28^th, 2022; accepted: Jan. 24^th, 2023; published: Jan. 31^st, 2023

ABSTRACT

In this paper, when the maximum time $T_{ν}^{*}$ is finite, the blow-up of the solutions to the fractional 3D incompressible Navier-Stokes equations in ${\dot{H}}^{s}$ spaces and the decay in $L^{2}$ norm and the boundedness of the solution with respect to ${\dot{H}}^{2 - α}$ norm, ${\dot{H}}^{\frac{5}{2} - α}$ norm and ${\dot{H}}^{\frac{7}{2} - α}$ norm are studied, via using the property of Fourier transform, interpolation results and product law in the homogeneous Sobolev spaces. It’s a generalization of the classical Navier-Stokes equations conclusion of Benameur J.

Keywords:Fractional Navier-Stokes Equation, Decay, Blow-Up Criterion

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

本文研究分数阶不可压缩Navier-Stokes方程的初值问题

${\begin{cases} \partial_{t} u + ν {(- Δ)}^{α} u + (u \cdot \nabla) u = - \nabla p, \\ div u = 0, \\ u (0) = u^{0}, \end{cases}$ (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 $ν$ 为流体的粘性系数， $α > 0$ 表示“耗散强度”。向量函数 $u (t, x) = (u_{1} (t, x), u_{2} (t, x), u_{3} (t, x))$ 表示流体在 $(t, x) \in R^{+} \times R^{3}$ 的未知速度， $u^{0} = (u_{1}^{0} (t, x), u_{2}^{0} (t, x), u_{3}^{0} (t, x))$ 是给定的初始速度，标量函数 $p = p (t, x)$ 表示流体在 $(t, x) \in R^{+} \times R^{3}$ 所受的未知压力， $Δ = \sum_{j = 1}^{n} \partial_{x_{j}}^{2}$ 是关于空间变量 $x = (x_{1}, x_{2}, x_{3})$ 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 ${‖ u (t) ‖}_{{\dot{H}}^{s}} \geq {(T_{ν}^{*} - t)}^{- \frac{s}{3}} (s > \frac{5}{2})$ ，其中 $T_{ν}^{*}$ 是最大存在时间。Robinson JC，Sadowski W，Silva RP在文献 [4] 中证明了 ${‖ u (t) ‖}_{{\dot{H}}^{s}} \geq {(T_{ν}^{*} - t)}^{- \frac{2 s - 1}{4}} (\frac{1}{2} < s < \frac{5}{2}, s \neq \frac{3}{2})$ ， ${‖ u (t) ‖}_{{\dot{H}}^{s}} \geq {(T_{ν}^{*} - t)}^{- \frac{2 s}{5}} (s > \frac{5}{2})$ 。Cortissoz C，Montero JA，Pinilla CE在文献 [5] 中研究了解的 ${\dot{H}}^{\frac{3}{2}}$ 范数和 ${\dot{H}}^{\frac{5}{2}}$ 范数的下界。Cheskidov A，Zaya K在文献 [6] 中证明了解在 ${\dot{H}}^{\frac{3}{2}}$ 空间的强爆破估计。Benameur J在文献 [7] 中研究了解在 ${\dot{H}}^{\frac{5}{2}}$ 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 ${\dot{H}}^{s}$ 空间中的爆破和 $L^{2}$ 范数衰减，以及解关于 ${\dot{H}}^{2 - α}$ 范数和 ${\dot{H}}^{\frac{5}{2} - α}$ 范数和 ${\dot{H}}^{\frac{7}{2} - α}$ 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 $α > 1$ ， $s > \max {α - 1, \frac{3}{2}}$ ， $\lim \sup_{t \to T_{ν}^{*}} {‖ u (t) ‖}_{{\dot{H}}^{s}}^{2} = \infty$ 且 $u \in C ([0, T_{ν}^{*}); {\dot{H}}^{s})$ 是方程(1.1)的极大解，若 $T_{ν}^{*} < \infty$ ，则有

$\frac{c ν^{\frac{s}{3}}}{{(T_{ν}^{*} - t)}^{\frac{s}{3}}} \leq {‖ u (t) ‖}_{L^{2}}^{\frac{2 s}{3} - 1} {‖ u (t) ‖}_{{\dot{H}}^{s}} .$

定理2 设 $1 < α < 2$ ， $\lim \sup_{t \to T_{ν}^{*}} {‖ u (t) ‖}_{{\dot{H}}^{2 - α}}^{2} = \infty$ 且 $u \in C ([0, T_{ν}^{*}); {\dot{H}}^{2 - α})$ 是方程(1.1)的极大解，若 $T_{ν}^{*} < \infty$ ，则有

$\frac{c \sqrt{ν}}{\sqrt{T_{ν}^{*} - t}} \leq {‖ \nabla u ‖}_{{\dot{H}}^{1 - α}} .$

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

$\frac{c \sqrt{ν}}{\sqrt{T_{ν}^{*} - t}} \leq {‖ u (t) ‖}_{{\dot{H}}^{\frac{5}{2} - α}} .$

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

$\frac{c {(α - 1)}^{\frac{1}{α + 1}}}{{(T_{ν}^{*} - t)}^{\frac{2}{α + 1}}} \leq {‖ u (t) ‖}_{{\dot{H}}^{\frac{7}{2} - α}} .$

2. 预备知识

1) [7] Fourier变换定义为：

$F (f) (ξ) = \hat{f} (ξ) = \int_{R^{3}} \exp (- i x \cdot ξ) f (x) d x, ξ = (ξ_{1}, ξ_{2}, ξ_{3}) \in R^{3}, f \in L^{1} (R^{3}) .$

2) 分数阶Laplacian微分算子通过Fourier变换定义为：

$F ({(- Δ)}^{α} u) (t, ξ) = {| ξ |}^{2 α} F u (t, ξ) .$

关于 ${(- Δ)}^{α}$ 的更多详细描述见文献 [8]。

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

$f * g (x) = \int_{R^{3}} f (x - y) g (y) d y = \int_{R^{3}} f (y) g (x - y) d y .$

4) [7] 若 $f = (f_{1}, f_{2}, f_{3})$ 和 $g = (g_{1}, g_{2}, g_{3})$ 是两个向量场，则

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

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

5) [7] 齐次Sobolev空间定义为： ${\dot{H}}^{s} (R^{3}) = {f \in S^{'} (R^{3}), \hat{f} \in L_{l o c}^{1}, {| ξ |}^{s} \hat{f} \in L^{2} (R^{3})}$ ，其范数为

${‖ f ‖}_{{\dot{H}}^{s}} = {(\int_{R^{3}} {| ξ |}^{2 s} {| \hat{f} (ξ) |}^{2} d ξ)}^{\frac{1}{2}} .$

6) [7] Lei-Lin空间定义为： $χ^{σ} (R^{3}) = {f \in S^{'} (R^{3}), \hat{f} \in L_{l o c}^{1}, {| ξ |}^{σ} \hat{f} \in L^{1} (R^{3})}$ ，其范数为

${‖ f ‖}_{χ^{σ}} = \int_{R^{3}} {| ξ |}^{σ} | \hat{f} (ξ) | d ξ .$

引理1 [9] 对于 $s > \frac{3}{2}$ ，有

${‖ \hat{u} ‖}_{L^{1}} \leq C {‖ u ‖}_{L^{2}}^{1 - \frac{3}{2 s}} {‖ u ‖}_{{\dot{H}}^{s}}^{\frac{3}{2 s}} .$

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

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

${‖ f g ‖}_{{\dot{H}}^{s + s^{'} - 1}} \leq C ({‖ f ‖}_{{\dot{H}}^{s}} {‖ g ‖}_{{\dot{H}}^{s^{'}}} + {‖ f ‖}_{{\dot{H}}^{s^{'}}} {‖ g ‖}_{{\dot{H}}^{s}}) .$

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

${‖ f g ‖}_{{\dot{H}}^{s + s^{'} - 1}} \leq C {‖ f ‖}_{{\dot{H}}^{s}} {‖ g ‖}_{{\dot{H}}^{s^{'}}} .$

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

${‖ f g ‖}_{{\dot{H}}^{s + s^{'} - \frac{3}{2}}} \leq C ({‖ f ‖}_{{\dot{H}}^{s}} {‖ g ‖}_{{\dot{H}}^{s^{'}}} + {‖ f ‖}_{{\dot{H}}^{s}} {‖ g ‖}_{{\dot{H}}^{s^{'}}}) .$

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

${‖ f g ‖}_{{\dot{H}}^{s + s^{'} - \frac{3}{2}}} \leq C {‖ f ‖}_{{\dot{H}}^{s}} {‖ g ‖}_{{\dot{H}}^{s^{'}}} .$

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

$| {〈 f \cdot \nabla f 〉}_{{\dot{H}}^{s}} | \leq C {‖ f ‖}_{χ^{1}} {‖ f ‖}_{{\dot{H}}^{s}}^{2} .$

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

$y^{'} (t) \leq h (t) y (t) + g (t), a .e . t \in [0, T] .$

那么

$y (t) \leq e^{\int_{0}^{t} h (τ) d τ} [y (0) + \int_{0}^{t} g (τ) d τ] .$

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

${‖ f ‖}_{χ^{1}} \leq C \sqrt{4 π} [a^{\frac{5}{2}} {‖ u ‖}_{L^{2}} + \sqrt{\frac{b^{2 α - 2}}{α - 1}} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}} + \frac{1}{b} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2}}}] .$

证明：设 $0 < a < b < \infty$

${‖ f ‖}_{χ^{1}} = \int | ξ | | \hat{f} (ξ) | d ξ = I_{a} + J_{a, b} + K_{b},$

$I_{a} = \int_{| ξ | < a} | ξ | | \hat{f} (ξ) | d ξ \leq {(\int_{| ξ | < a} {| ξ |}^{2} d ξ)}^{\frac{1}{2}} {‖ f ‖}_{L^{2}} \leq \sqrt{\frac{4 π}{5}} a^{\frac{5}{2}} {‖ f ‖}_{L^{2}} .$

$\begin{matrix} J_{a, b} = \int_{a < | ξ | < b} | ξ | | \hat{f} (ξ) | d ξ \leq {(\int_{a < | ξ | < b} {| ξ |}^{2 α - 5} d ξ)}^{\frac{1}{2}} {‖ f ‖}_{{\dot{H}}^{\frac{7}{2} - α}} \\ \leq \sqrt{\frac{2 π}{α - 1}} \sqrt{b^{2 α - 2} - a^{2 α - 2}} {‖ f ‖}_{{\dot{H}}^{\frac{7}{2} - α}} \leq \sqrt{\frac{2 π}{α - 1}} \sqrt{b^{2 α - 2}} {‖ f ‖}_{{\dot{H}}^{\frac{7}{2} - α}} . \end{matrix}$

$K_{b} = \int_{| ξ | > b} | ξ | | \hat{f} (ξ) | d ξ \leq {(\int_{| ξ | > b} {| ξ |}^{- 5} d ξ)}^{\frac{1}{2}} {‖ f ‖}_{{\dot{H}}^{\frac{7}{2}}} \leq \sqrt{2 π} \frac{1}{b} {‖ f ‖}_{{\dot{H}}^{\frac{7}{2}}} .$

3. 定理1的证明

证明：方程(1.1)在 ${\dot{H}}^{s}$ 空间下与u取内积

${〈 \partial_{t} u, u 〉}_{{\dot{H}}^{s}} + ν {〈 {(- Δ)}^{α} u, u 〉}_{{\dot{H}}^{s}} + {〈 u \cdot \nabla u, u 〉}_{{\dot{H}}^{s}} = {〈 - \nabla p, u 〉}_{{\dot{H}}^{s}},$

则有

$\begin{matrix} \frac{1}{2} \partial_{t} {‖ u ‖}_{{\dot{H}}^{s}}^{2} + ν {‖ Λ^{α} u ‖}_{{\dot{H}}^{s}}^{2} \leq | {〈 u \cdot \nabla u, u 〉}_{{\dot{H}}^{s}} | \\ = \int_{R^{3}} {| ξ |}^{2 s} \hat{u \cdot \nabla u} \cdot \hat{u} d ξ \\ = \int_{R^{3}} {| ξ |}^{2 s + 1} \hat{u \otimes u} \cdot \hat{u} d ξ \\ = \int_{R^{3}} {| ξ |}^{s - α + 1} \hat{u \otimes u} {| ξ |}^{s} \hat{Λ^{α}} u d ξ \\ \leq {(\int_{R^{3}} {| ξ |}^{2 (s - α + 1)} \hat{u \otimes u}^{2} d ξ)}^{\frac{1}{2}} {(\int_{R^{3}} {| ξ |}^{2 s} \hat{Λ^{α}} u^{2} d ξ)}^{\frac{1}{2}} \\ = {‖ u \otimes u ‖}_{{\dot{H}}^{s - α + 1}} {‖ Λ^{α} u ‖}_{{\dot{H}}^{s}} . \end{matrix}$

又

$\begin{matrix} {‖ u \otimes u ‖}_{{\dot{H}}^{s - α + 1}}^{2} = \int_{R^{3}} {| ξ |}^{2 (s - α + 1)} {| \hat{u} * \hat{u} |}^{2} d ξ \\ = \int_{R^{3}} {| ξ |}^{2 (s - α + 1)} {| \int_{R^{3}} \hat{u} (η) \hat{u} (ξ - η) d η |}^{2} d ξ \\ \leq \int_{R^{3}} {| ξ |}^{2 (s - α + 1)} {(\int_{R^{3}} | \hat{u} (η) | | \hat{u} (ξ - η) | d η)}^{2} d ξ \\ \leq \int_{R^{3}} {(\int_{R^{3}} {| ξ |}^{s - α + 1} | \hat{u} (η) | | \hat{u} (ξ - η) | d η)}^{2} d ξ . \end{matrix}$

由

$\begin{matrix} {| ξ |}^{s - α + 1} = {| ξ - η + η |}^{s - α + 1} \leq {(| ξ - η | + | η |)}^{s - α + 1} \leq {(2 \max (| ξ - η |, | η |))}^{s - α + 1} \\ \leq 2^{s - α + 1} ({| ξ - η |}^{s - α + 1} + {| η |}^{s - α + 1}) . \end{matrix}$

$\begin{matrix} {‖ u \otimes u ‖}_{{\dot{H}}^{s - α + 1}}^{2} \leq \int_{R^{3}} {(\int_{R^{3}} 2^{s - α + 1} {| ξ - η |}^{s - α + 1} | \hat{u} (η) | | \hat{u} (ξ - η) | d η)}^{2} d ξ \\ + \int_{R^{3}} {(\int_{R^{3}} 2^{s - α + 1} {| η |}^{s - α + 1} | \hat{u} (η) | | \hat{u} (ξ - η) | d η)}^{2} d ξ \\ = 2^{2 (s - α + 1) + 1} \int_{R^{3}} {(\int_{R^{3}} {| η |}^{s - α + 1} | \hat{u} (η) | | \hat{u} (ξ - η) | d η)}^{2} d ξ \\ = 2^{2 (s - α + 1) + 1} \int_{R^{3}} {(({| \cdot |}^{s - α + 1} | \hat{u} |) * | \hat{u} |)}^{2} d ξ \end{matrix}$

$\begin{array}{l} = 2^{2 (s - α + 1) + 1} {‖ ({| \cdot |}^{s - α + 1} | \hat{u} |) * | \hat{u} | ‖}_{L^{2}}^{2} \\ \leq 2^{2 (s - α + 1) + 1} {‖ ({| \cdot |}^{s - α + 1} | \hat{u} |) ‖}_{L^{2}}^{2} {‖ \hat{u} ‖}_{L^{1}}^{2} \\ = 2^{2 (s - α + 1) + 1} {‖ u ‖}_{{\dot{H}}^{s - α + 1}}^{2} {‖ \hat{u} ‖}_{L^{1}}^{2} . \end{array}$

则有

$\begin{matrix} \frac{1}{2} \partial_{t} {‖ u ‖}_{{\dot{H}}^{s}}^{2} + ν {‖ Λ^{α} u ‖}_{{\dot{H}}^{s}}^{2} \leq 2^{s - α + \frac{3}{2}} {‖ u ‖}_{{\dot{H}}^{s - α + 1}} {‖ \hat{u} ‖}_{L^{1}} {‖ Λ^{α} u ‖}_{{\dot{H}}^{s}} \\ \leq 2^{2 s - 2 α + 3} ν^{- 1} {‖ u ‖}_{{\dot{H}}^{s - α + 1}}^{2} {‖ \hat{u} ‖}_{L^{1}}^{2} + \frac{ν}{2} {‖ Λ^{α} u ‖}_{{\dot{H}}^{s}}^{2} . \end{matrix}$

则

$\partial_{t} {‖ u ‖}_{{\dot{H}}^{s}}^{2} + ν {‖ Λ^{α} u ‖}_{{\dot{H}}^{s}}^{2} \leq C ν^{- 1} {‖ u ‖}_{{\dot{H}}^{s - α + 1}}^{2} {‖ \hat{u} ‖}_{L^{1}}^{2} \leq C ν^{- 1} {‖ u ‖}_{{\dot{H}}^{s}}^{2} {‖ \hat{u} ‖}_{L^{1}}^{2} .$

由引理5，对 $0 \leq a \leq t < T_{ν}^{*}$

${‖ u (t) ‖}_{{\dot{H}}^{s}}^{2} \leq {‖ u (a) ‖}_{{\dot{H}}^{s}}^{2} e^{c ν^{- 1} \int_{a}^{t} {‖ \hat{u} ‖}_{L^{1}}^{2} d τ} .$

因为 $\lim \sup_{t \to T_{ν}^{*}} {‖ u (t) ‖}_{{\dot{H}}^{s}}^{2} = \infty$ ，所以 $\int_{a}^{T_{ν}^{*}} {‖ \hat{u} ‖}_{L^{1}}^{2} d τ = \infty$ 。

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

$\frac{{‖ \hat{u} ‖}_{L^{1}}^{\frac{4 s}{3}}}{C {‖ u ‖}_{L^{2}}^{\frac{4 s}{3} - 2}} \leq {‖ u ‖}_{{\dot{H}}^{s}}^{2} \leq {‖ u (a) ‖}_{{\dot{H}}^{s}}^{2} e^{c ν^{- 1} \int_{a}^{t} {‖ \hat{u} ‖}_{L^{1}}^{2} d τ},$

${‖ \hat{u} ‖}_{L^{1}}^{\frac{4 s}{3}} \leq C {‖ u ‖}_{L^{2}}^{\frac{4 s}{3} - 2} {‖ u (a) ‖}_{{\dot{H}}^{s}}^{2} e^{c ν^{- 1} \int_{a}^{t} {‖ \hat{u} ‖}_{L^{1}}^{2} d τ},$

${‖ \hat{u} ‖}_{L^{1}}^{2} e^{- c ν^{- 1} \int_{a}^{t} {‖ \hat{u} ‖}_{L^{1}}^{2} d τ} \leq C {‖ u ‖}_{L^{2}}^{2 - \frac{3}{s}} {‖ u (a) ‖}_{{\dot{H}}^{s}}^{\frac{3}{s}} .$

在 $[a, T]$ 上对上式积分有

$1 - e^{- c ν^{- 1} \int_{a}^{t} {‖ \hat{u} ‖}_{L^{1}}^{2} d τ} \leq C ν^{- 1} (T - a) {‖ u ‖}_{L^{2}}^{2 - \frac{3}{s}} {‖ u (a) ‖}_{{\dot{H}}^{s}}^{\frac{3}{s}} .$

当 $T \to T_{ν}^{*}$ 时，有

$1 \leq C ν^{- 1} (T_{ν}^{*} - a) {‖ u ‖}_{L^{2}}^{2 - \frac{3}{s}} {‖ u (a) ‖}_{{\dot{H}}^{s}}^{\frac{3}{s}}$

$\frac{C ν^{\frac{s}{3}}}{{(T_{ν}^{*} - t)}^{\frac{s}{3}}} \leq {‖ u (t) ‖}_{L^{2}}^{\frac{2 s}{3} - 1} {‖ u (t) ‖}_{{\dot{H}}^{s}} .$

从而结论得证。

4. 定理2的证明

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

${〈 \partial_{t} u, u 〉}_{{\dot{H}}^{2 - α}} + ν {〈 {(- Δ)}^{α} u, u 〉}_{{\dot{H}}^{2 - α}} + {〈 u \cdot \nabla u, u 〉}_{{\dot{H}}^{2 - α}} = {〈 - \nabla p, u 〉}_{{\dot{H}}^{2 - α}},$

则有

$\begin{matrix} \frac{1}{2} \partial_{t} {‖ u ‖}_{{\dot{H}}^{2 - α}}^{2} + ν {‖ Λ^{α} u ‖}_{{\dot{H}}^{2 - α}}^{2} \leq | {〈 u \cdot \nabla u, u 〉}_{{\dot{H}}^{2 - α}} | \\ = \int_{R^{3}} {| ξ |}^{5 - 2 α} \hat{u \otimes u} \cdot \hat{u} d ξ \\ = \int_{R^{3}} {| ξ |}^{3 - 2 α} \hat{u \otimes u} {| ξ |}^{2 - α} {| ξ |}^{α} \hat{u} d ξ \\ \leq {‖ u \otimes u ‖}_{{\dot{H}}^{3 - 2 α}} {‖ Λ^{α} u ‖}_{{\dot{H}}^{2 - α}} . \end{matrix}$

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

$\begin{matrix} \frac{1}{2} \partial_{t} {‖ u ‖}_{{\dot{H}}^{2 - α}}^{2} + ν {‖ Λ^{α} u ‖}_{{\dot{H}}^{2 - α}}^{2} \leq C {‖ u ‖}_{{\dot{H}}^{2 - α}} {‖ u ‖}_{{\dot{H}}^{2 - α}} {‖ Λ^{α} u ‖}_{{\dot{H}}^{2 - α}} \\ \leq C ν^{- 1} {‖ u ‖}_{{\dot{H}}^{2 - α}}^{4} + \frac{ν}{2} {‖ Λ^{α} u ‖}_{{\dot{H}}^{2 - α}}^{2} . \end{matrix}$

则

$\partial_{t} {‖ u ‖}_{{\dot{H}}^{s}}^{2} + ν {‖ Λ^{α} u ‖}_{{\dot{H}}^{s}}^{2} \leq C ν^{- 1} {‖ u ‖}_{{\dot{H}}^{2 - α}}^{4} .$

由引理5，对于 $0 \leq a \leq t < T_{ν}^{*}$ ，

${‖ u (t) ‖}_{{\dot{H}}^{2 - α}}^{2} \leq {‖ u (a) ‖}_{{\dot{H}}^{2 - α}}^{2} e^{C ν^{- 1} \int_{a}^{t} {‖ u (τ) ‖}_{{\dot{H}}^{2 - α}}^{2} d τ} .$

因为 $\lim \sup_{t \to T_{ν}^{*}} {‖ u (t) ‖}_{{\dot{H}}^{2 - α}}^{2} = \infty$ ，所以 $\int_{a}^{T_{ν}^{*}} {‖ u (t) ‖}_{{\dot{H}}^{2 - α}}^{2} = \infty$ 。

${‖ u ‖}_{{\dot{H}}^{2 - α}}^{2} e^{- C ν^{- 1} \int_{a}^{t} {‖ u (τ) ‖}_{{\dot{H}}^{2 - α}}^{2} d τ} \leq {‖ u (a) ‖}_{{\dot{H}}^{2 - α}}^{2} .$

在 $[a, T]$ 上对上式积分有

$1 - e^{- C ν^{- 1} \int_{a}^{T} {‖ u (τ) ‖}_{{\dot{H}}^{2 - α}}^{2} d τ} \leq C ν^{- 1} (T - a) {‖ u (a) ‖}_{{\dot{H}}^{2 - α}}^{2} .$

当 $T \to T_{ν}^{*}$ 时，有

$1 \leq C ν^{- 1} (T_{ν}^{*} - a) {‖ u (a) ‖}_{{\dot{H}}^{2 - α}}^{2},$

$\frac{c \sqrt{ν}}{\sqrt{T_{ν}^{*} - t}} \leq {‖ u (t) ‖}_{{\dot{H}}^{2 - α}} .$

$\frac{c \sqrt{ν}}{\sqrt{T_{ν}^{*} - t}} \leq {‖ \nabla u ‖}_{{\dot{H}}^{1 - α}} .$

从而结论得证。

5. 定理3的证明

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

${〈 \partial_{t} u, u 〉}_{{\dot{H}}^{\frac{5}{2} - α}} + ν {〈 {(- Δ)}^{α} u, u 〉}_{{\dot{H}}^{\frac{5}{2} - α}} + {〈 u \cdot \nabla u, u 〉}_{{\dot{H}}^{\frac{5}{2} - α}} = {〈 - \nabla p, u 〉}_{{\dot{H}}^{\frac{5}{2} - α}},$

则有

$\begin{matrix} \frac{1}{2} \partial_{t} {‖ u ‖}_{{\dot{H}}^{\frac{5}{2} - α}}^{2} + ν {‖ u ‖}_{{\dot{H}}^{\frac{5}{2}}}^{2} \leq | {〈 u \cdot \nabla u, u 〉}_{{\dot{H}}^{\frac{5}{2} - α}} | \\ = \int_{R^{3}} {| ξ |}^{6 - 2 α} \hat{u \otimes u} \cdot \hat{u} d ξ \\ = \int_{R^{3}} {| ξ |}^{\frac{7}{2} - 2 α} \hat{u \otimes u} {| ξ |}^{\frac{5}{2}} \hat{u} d ξ \\ \leq {‖ u \otimes u ‖}_{{\dot{H}}^{\frac{7}{2} - 2 α}} {‖ u ‖}_{{\dot{H}}^{\frac{5}{2}}} . \end{matrix}$

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

$\frac{1}{2} \partial_{t} {‖ u ‖}_{{\dot{H}}^{\frac{5}{2} - α}}^{2} + ν {‖ u ‖}_{{\dot{H}}^{\frac{5}{2}}}^{2} \leq C {‖ u ‖}_{{\dot{H}}^{\frac{5}{2} - α}} {‖ u ‖}_{{\dot{H}}^{\frac{5}{2} - α}} {‖ u ‖}_{{\dot{H}}^{\frac{5}{2}}} \leq C ν^{- 1} {‖ u ‖}_{{\dot{H}}^{\frac{5}{2} - α}}^{4} + \frac{ν}{2} {‖ u ‖}_{{\dot{H}}^{\frac{5}{2}}}^{2} .$

$\partial_{t} {‖ u ‖}_{{\dot{H}}^{\frac{5}{2} - α}}^{2} \leq C ν^{- 1} {‖ u ‖}_{{\dot{H}}^{\frac{5}{2} - α}}^{4} .$

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

${‖ u ‖}_{{\dot{H}}^{\frac{5}{2} - α}}^{2} \leq C ν^{- 1} (T_{ν}^{*} - t) {‖ u ‖}_{{\dot{H}}^{\frac{5}{2} - α}}^{4},$

$\frac{c \sqrt{ν}}{\sqrt{T_{ν}^{*} - t}} \leq {‖ u (t) ‖}_{{\dot{H}}^{\frac{5}{2} - α}} .$

从而结论得证。

6. 定理4的证明

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

${〈 \partial_{t} u, u 〉}_{{\dot{H}}^{\frac{7}{2} - α}} + ν {〈 {(- Δ)}^{α} u, u 〉}_{{\dot{H}}^{\frac{7}{2} - α}} + {〈 u \cdot \nabla u, u 〉}_{{\dot{H}}^{\frac{7}{2} - α}} = {〈 - \nabla p, u 〉}_{{\dot{H}}^{\frac{7}{2} - α}},$

则有

$\frac{1}{2} \partial_{t} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2} + ν {‖ u ‖}_{{\dot{H}}^{\frac{7}{2}}}^{2} \leq | {〈 u \cdot \nabla u, u 〉}_{{\dot{H}}^{\frac{7}{2} - α}} | .$

由引理4，

$\frac{1}{2} \partial_{t} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2} + ν {‖ u ‖}_{{\dot{H}}^{\frac{7}{2}}}^{2} \leq c {‖ u ‖}_{χ^{1}} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2} .$

由引理6，

$\begin{matrix} \frac{1}{2} \partial_{t} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2} + ν {‖ u ‖}_{{\dot{H}}^{\frac{7}{2}}}^{2} \leq c {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2} [a^{\frac{5}{2}} {‖ u ‖}_{L^{2}} + \sqrt{\frac{b^{2 α - 2}}{α - 1}} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}} + \frac{1}{b} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2}}}] \\ \leq c {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2} [a^{\frac{5}{2}} {‖ u ‖}_{L^{2}} + \sqrt{\frac{b^{2 α - 2}}{α - 1}} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}] + \frac{c}{ν b^{2}} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{4} + \frac{ν}{2} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2}}}^{2} . \end{matrix}$

则有

$\partial_{t} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2} \leq c {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2} [a^{\frac{5}{2}} {‖ u ‖}_{L^{2}} + \sqrt{\frac{b^{2 α - 2}}{α - 1}} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}} + \frac{1}{ν b^{2}} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2}] .$

取 $t_{0} = \inf {t \in [0, T_{ν}^{*}), {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}} = 2 {‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}}}$ ，对上式积分有

${‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2} - {‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2} \leq c T_{ν}^{*} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2} [a^{\frac{5}{2}} {‖ u ‖}_{L^{2}} + \sqrt{\frac{b^{2 α - 2}}{α - 1}} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}} + \frac{1}{ν b^{2}} {‖ u ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2}],$

$3 {‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2} \leq c T_{ν}^{*} {‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2} [a^{\frac{5}{2}} {‖ u ‖}_{L^{2}} + \sqrt{\frac{b^{2 α - 2}}{α - 1}} {‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}} + \frac{1}{ν b^{2}} {‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2}],$

$1 \leq c T_{ν}^{*} [a^{\frac{5}{2}} {‖ u ‖}_{L^{2}} + \sqrt{\frac{b^{2 α - 2}}{α - 1}} {‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}} + \frac{1}{ν b^{2}} {‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{2}] .$

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

$1 \leq c T_{ν}^{*} [{‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}} {‖ u ‖}_{L^{2}} + \frac{1}{\sqrt{α - 1}} \sqrt{{‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{α - 1}} {‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}} + \frac{1}{ν} {‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}}],$

$1 \leq c T_{ν}^{*} {‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}} [{‖ u ‖}_{L^{2}} + \frac{1}{\sqrt{α - 1}} \sqrt{{‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{α - 1}} + \frac{1}{ν}] .$

设 $t_{1} \in [0, T_{ν}^{*})$ 有

$\frac{1}{\sqrt{α - 1}} \sqrt{{‖ u^{0} ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{α - 1}} \geq 2 ({‖ u ‖}_{L^{2}} + \frac{1}{ν}) .$

则

$1 \leq c (T_{ν}^{*} - t_{1}) {‖ u (t_{1}) ‖}_{{\dot{H}}^{\frac{7}{2} - α}} \frac{1}{\sqrt{α - 1}} \sqrt{{‖ u (t_{1}) ‖}_{{\dot{H}}^{\frac{7}{2} - α}}^{α - 1}} .$

用任意的 $t \in [t_{1}, T_{ν}^{*})$ 替换 $t_{1}$ 有

$\frac{c {(α - 1)}^{\frac{1}{α + 1}}}{{(T_{ν}^{*} - t)}^{\frac{2}{α + 1}}} \leq {‖ u (t) ‖}_{{\dot{H}}^{\frac{7}{2} - α}} .$

因此结论得证。

文章引用

徐郜婷,孙小春,吴育联. 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[J]. 应用数学进展, 2023, 12(01): 231-239. https://doi.org/10.12677/AAM.2023.121026

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NOTES

^*通讯作者。

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