理论数学  >> Vol. 11 No. 4 (April 2021)

三维不可压磁流体力学方程弱解正则准则
Regularity Criterion of the Weak Solution for the 3D Incompressible MHD Equations

DOI: 10.12677/PM.2021.114077, PDF, HTML, XML, 下载: 20  浏览: 68  科研立项经费支持

作者: 李天理:安徽职业技术学院基础教学部,安徽 合肥;王 力:安徽职业技术学院铁道学院,安徽 合肥

关键词: MHD方程正则准则Sobolev嵌入不等式Young不等式MHD Equation Regularity Criterion Sobolev Embedding Inequality Young Inequality

摘要: 文章考虑在三维情况下不可压的磁流体力学(MHD)方程解的正则性。使用Young不等式,Hölder不等式及Sobolev嵌入技术等,扩大了弱解正则的函数空间,证明了当∂3u,∂3b∈Lp(0,T;Lq(R3)),2/p+3/q=46/25+3/25q, 31/8≤q≤∞时,或者当∂3u,∂3b∈Lp(0,T;Lq(R3)),2/p+3/q=22/13+3/13q,19/8≤q≤∞时,且都有,则三维不可压MHD方程弱解(u, b)在(0,T]上是正则的。
Abstract: This paper considers the regularity of weak solutions for incompressible MHD equations in 3D cases. Here, Yuong inequalities, Hölder inequalities and Sobolev embedding techniques are used to expand the integral space to which the weak solution belongs. Here, it is proved that the weak solution (u, b) is regular on (0,T], if ∂3u,∂3b∈Lp(0,T;Lq(R3)) and 2/p+3/q=46/25+3/25q, 31/8≤q≤∞ or ∂3u,∂3b∈Lp(0,T;Lq(R3)), 2/p+3/q=22/13+3/13q, 19/8≤q≤∞ ,together with .

文章引用: 李天理, 王力. 三维不可压磁流体力学方程弱解正则准则[J]. 理论数学, 2021, 11(4): 640-646. https://doi.org/10.12677/PM.2021.114077

1. 引言及主要结论

本文主要考虑下列三维不可压MHD方程弱解的正则性

{ t u + ( u ) u + p = Δ u + ( b ) b , t b + ( u ) b = Δ b + ( b ) u , u = b = 0 , u ( x , 0 ) = u 0 ( x ) , b ( x , 0 ) = b 0 ( x ) , ( x , t ) R 3 × ( 0 , T ] (1)

这里 u = ( u 1 , u 2 , u 3 ) b = ( b 1 , b 2 , b 3 ) p = p ( x , t ) 分别表示未知流体速度向量场,磁流体速度向量场和标量压力场, u 0 ( x ) b 0 ( x ) 是初始条件,且满足在广义下有 u 0 = b 0 = 0

众所周知在文献 [1] 中,指出满足初始条件时,方程(1)存在弱解是局部正则解。然而,是否存在全局正则性的弱解仍然是一个充满挑战的公开问题。但是,文献中有大量结果表明,如果对弱解施加一些附加条件,则对这个问题的回答是肯定的(参见 [2] - [14] )。对压力p附加一些条件同样也可得到弱解具有正则性(参见 [15] [16] )。在这些结果中,我们对仅涉及速度u,磁场b更感兴趣。

最近,Sadek Gala,Maria Alessandra Ragusa在 [7] 中证明了

3 ( u ± b ) L p ( ( 0 , T ) ; L q ( R 3 ) ) , 2 p + 3 q = 8 5 + 3 5 q , 4 q

或者 3 ( u ± b ) L p ( ( 0 , T ) ; L q ( R 3 ) ) , 2 p + 3 q = 4 11 + 9 11 q , 5 2 q

则弱解在 ( 0 , T ] 上是正则的。

重要结论之一是著名的Ni-Guo-Zhou准则(参见 [8] 定理1.3,类似的结果也可参考 [17] 定理 1.2)。如下

{ b , u 3 L p 1 ( 0 , T ; L q 1 ( R 3 ) ) , 2 / p 1 + 3 / q 1 1 , 3 < q 1 , b , 3 u L p 2 ( 0 , T ; L q 2 ( R 3 ) ) , 2 / p 2 + 3 / q 2 2 , 3 / 2 < q 2 . (2)

受文献 [7] 和 [8] 的启发,我们得到比文献 [7] 如下更优的结果如下定理1。

这里 w ± = u ± b , w 0 ± = u 0 ± b 0 .

定理1 假设 u 0 , b 0 L 2 ( R 3 ) ,且在广义积分条件下有 u 0 = b 0 = 0 , T > 0 ,设 ( u , b ) 是MHD方程的1在 ( 0 , T ] 上满足初始条件的弱解,且 b 1 , b 2 L 2 s s 3 ( 0 , T ; L s ( R 3 ) ) , s > 3 ,如果有

3 w L p ( ( 0 , T ) ; L q ( R 3 ) ) , 2 p + 3 q = 46 25 + 3 25 q , 31 8 q (3)

或者

3 w L p ( ( 0 , T ) ; L q ( R 3 ) ) , 2 p + 3 q = 22 13 + 3 13 q , 19 8 q (4)

则弱解在 ( 0 , T ] 上是正则的,

注:上式的q值比文献 [7] 的值相应减小,等式值相应增大,所以此结果比文献 [7] 结果更优。

2. 主要结果的证明

我们把方程(1)的第一个方程和第二个方程加减运算,把方程1转化为如下方程

{ t w + + w w + + p = Δ w + , t w + w + w + p = Δ w , w + = w = 0 , w + ( x , 0 ) = w 0 + ( x ) , w ( x , 0 ) = w 0 ( x ) , ( x , t ) R 3 × ( 0 , T ] (5)

为了证明定理1,让我们介绍一些辅助结论。我们给出三维乘法的Sobolev不等式及压力估计如下:

引理1 存在一个常数 C > 0 ,使得下式成立

u L 3 q C 1 u L 2 1 3 2 u L 2 1 3 3 u L q 1 3 , 1 q . (6)

引理2 假设 ( w + , w , P ) 是方程2.1组的光滑解,那么有下式成立

i) P L r C w + L 2 r w L 2 r , (7)

ii) 3 P L r C w 3 w + L r , (8)

iii) 3 P L r C w + 3 w L r . (9)

注:这里 1 < r < ,引理1的详细证明参见Wu和Cao文献 [3],引理2的证明可参见文献 [16]。

定理1的证明:

因为 当 31 8 q 时, L 25 q 23 q 36 ( 0 , T ; L q ( R 3 ) ) L 2 q 2 q 3 ( 0 , T ; L q ( R 3 ) )

19 8 q 时, L 13 q 11 q 18 ( 0 , T ; L q ( R 3 ) ) L 2 q 2 q 3 ( 0 , T ; L q ( R 3 ) )

根据定理1和(2)式的假设,定理1的证明只要证明下式成立

u 3 , b 3 L 3 ( 0 , T ; L 9 ( R 3 ) ) .

在方程(5)的第一个和第二个方程两边分别用 w + , w 作内积,利用 w + = w = 0 ,分部积分相加得:

w + ( , t ) L 2 2 + w ( , t ) L 2 2 + 2 0 t ( w + ( , τ ) L 2 2 + w ( , τ ) L 2 2 ) d τ w + ( , 0 ) L 2 2 + w ( , 0 ) L 2 2 C . (10)

| w 3 + | w 3 + , | w 3 | w 3 分别与方程(5)的第一个和第二个方程作内积,利用 w + = w = 0 ,然后相加得:

1 3 d d t ( w 3 + L 3 3 + w 3 L 3 3 ) + 4 9 R 3 ( | | w 3 + | 3 / 2 | 2 + | | w 3 | 3 / 2 | 2 ) d x = R 3 3 p w 3 + | w 3 + | d x R 3 3 p w 3 | w 3 | d x = I 1 + I 2 . (11)

针对定理1第一种情况的证明,利用Hölder不等式,Yuong不等式和式(8),对式(11)右边第一项进行估计得

| I 1 | C 3 p L 3 w 3 + L 3 2 (由Hölder不等式)

C w + 3 w L 3 w 3 + L 3 2 (由式(8))

C w + L 3 q / ( q 3 ) 3 w L q w 3 + L 3 2 C w + L 2 ( 8 q 31 ) / 3 ( 4 q 3 ) w + L 3 q ( q + 18 ) / 2 ( 4 q 3 ) 3 w L q w 3 + L 3 2 C 1 w + L 2 ( q + 18 ) / 6 ( 4 q 3 ) 2 w + L 2 ( q + 18 ) / 6 ( 4 q 3 ) 3 w + L q ( q + 18 ) / 6 ( 4 q 3 ) 3 w L q w 3 + L 3 2 C w + L 2 ( q + 18 ) / 3 ( 4 q 3 ) 3 w + L q ( q + 18 ) / 6 ( 4 q 3 ) 3 w L q w 3 + L 3 2 C ( w + L 2 2 + 3 w + L q ( q + 18 ) / ( 23 q 36 ) 3 w L q 6 ( 4 q 3 ) / ( 23 q 36 ) ) w 3 + L 3 2 (12)

C ( w + L 2 2 + 3 w + L q 25 q / ( 23 q 36 ) + 3 w L q 25 q / ( 23 q 36 ) ) w 3 + L 3 2

类似 | I 1 | 的估计方法,得到式(11)右边第二项 | I 2 | 的估计如下:

| I 2 | C ( w L 2 2 + 3 w L q 25 q / ( 23 q 36 ) + 3 w + L q 25 q / ( 23 q 36 ) ) w 3 + L 3 2 . (13)

| I 1 | | I 2 | 的估计式(12)和(13)代入式(11)得

d d t ( w 3 + L 3 3 + w 3 L 3 3 ) d d t ( w 3 + L 3 3 + w 3 L 3 3 ) + 4 9 ( | w 3 + | 3 2 L 2 2 + | w 3 | 3 2 L 2 2 ) C ( w + L 2 2 + w L 2 2 + 3 w + L q 25 q / ( 23 q 36 ) + 3 w L q 25 q / ( 23 q 36 ) ) ( w 3 + L 3 2 + w 3 L 3 2 ) . (14)

在上式两边同时除以 ( w 3 + L 3 2 + w 3 L 3 2 ) ,在时间空间上积分,利用能量不等式(10)和式(3)得到

w 3 + L ( 0 , T ; L 3 ( R 3 ) ) + w 3 L ( 0 , T ; L 3 ( R 3 ) ) C .

由式(14),上式和能量不等式(10)可得

4 9 ( | w 3 + | 3 2 L 2 2 + | w 3 | 3 2 L 2 2 ) C ( w + L 2 2 + w L 2 2 + 3 w + L q 25 q / ( 23 q 36 ) + 3 w L q 25 q / ( 23 q 36 ) ) ( w 3 + L 3 2 + w 3 L 3 2 ) C .

所以由嵌入不等式得

w 3 + L 3 ( 0 , T ; L 9 ( R 3 ) ) + w 3 L 3 ( 0 , T ; L 9 ( R 3 ) ) = | w 3 + | 3 2 L 2 ( 0 , T ; L 6 ( R 3 ) ) 2 3 + | w 3 | 3 2 L 2 ( 0 , T ; L 6 ( R 3 ) ) 2 3 | w 3 + | 3 2 L 2 ( 0 , T ; L 2 ( R 3 ) ) 2 3 + | w 3 | 3 2 L 2 ( 0 , T ; L 2 ( R 3 ) ) 2 3 C .

所以得

u 3 L 3 ( 0 , T ; L 9 ( R 3 ) ) = 1 2 ( u 3 + b 3 ) + ( u 3 b 3 ) L 3 ( 0 , T ; L 9 ( R 3 ) ) 1 2 ( w 3 + L 3 ( 0 , T ; L 9 ( R 3 ) ) + w 3 L 3 ( 0 , T ; L 9 ( R 3 ) ) ) C .

同理可证

b 3 L 3 ( 0 , T ; L 9 ( R 3 ) ) = 1 2 ( u 3 + b 3 ) ( u 3 b 3 ) L 3 ( 0 , T ; L 9 ( R 3 ) ) C .

所以定理1第一种情况证明完成。

定理1第二种情况的证明。利用Hölder不等式,Yuong不等式和式(7),对式(11)右边第一项进行另一种估计得

| I 1 | = | R 3 3 p w 3 + | w 3 + | d x | C R 3 | p | | 3 w 3 + | | w 3 + | d x C p L 3 q / ( 2 q 3 ) 3 w 3 + L q w 3 + L 3 C ( w + L 6 q / ( 2 q 3 ) 2 + w L 6 q / ( 2 q 3 ) 2 ) 3 w 3 + L q w 3 + L 3 C w + L 2 2 ( 8 q 19 ) / 3 ( 4 q 3 ) w + L 3 q 2 ( q + 9 ) / 2 ( 4 q 3 ) 3 w 3 + L q w 3 + L 3

+ C w L 2 2 ( 8 q 19 ) / 3 ( 4 q 3 ) w L 3 q 2 ( q + 9 ) / 2 ( 4 q 3 ) 3 w 3 + L q w 3 + L 3 C 1 w + L 2 ( q + 9 ) / 3 ( 4 q 3 ) 2 w + L 2 ( q + 9 ) / 3 ( 4 q 3 ) 3 w + L q ( q + 9 ) / 3 ( 4 q 3 ) 3 w 3 + L q w 3 + L 3 + C 1 w L 2 ( q + 9 ) / 3 ( 4 q 3 ) 2 w L 2 ( q + 9 ) / 3 ( 4 q 3 ) 3 w L q ( q + 9 ) / 3 ( 4 q 3 ) 3 w 3 + L q w 3 + L 3 C w + L 2 2 ( q + 9 ) / 3 ( 4 q 3 ) 3 w + L q 13 q / 3 ( 4 q 3 ) w 3 + L 3 + C w L 2 2 ( q + 9 ) / 3 ( 4 q 3 ) 3 w L q ( q + 9 ) / 3 ( 4 q 3 ) 3 w + L q w 3 + L 3 (15)

C ( w + L 2 2 + 3 w + L q 13 q / ( 11 q 18 ) ) w 3 + L 3 + C ( w L 2 2 + 3 w L q ( q + 9 ) / ( 11 q 18 ) 3 w + L q 3 ( 4 q 3 ) / ( 11 q 18 ) ) w 3 + L 3 C ( w + L 2 2 + 3 w + L q 13 q / ( 11 q 18 ) ) w 3 + L 3 + C ( w L 2 2 + 3 w L q ( q + 9 ) / ( 11 q 18 ) 3 w + L q 3 ( 4 q 3 ) / ( 11 q 18 ) ) w 3 + L 3 C ( w + L 2 2 + w + L 2 2 + 3 w + L q 13 q / ( 11 q 18 ) + 3 w L q 13 q / ( 11 q 18 ) ) w 3 + L 3

同理可证

| I 2 | C ( w + L 2 2 + w + L 2 2 + 3 w + L q 13 q / ( 11 q 18 ) + 3 w L q 13 q / ( 11 q 18 ) ) w 3 + L 3 . (16)

| I 1 | | I 2 | 的估计式(15)和(16)代入式(11)得

d d t ( w 3 + L 3 3 + w 3 L 3 3 ) d d t ( w 3 + L 3 3 + w 3 L 3 3 ) + 4 9 ( | w 3 + | 3 2 L 2 2 + | w 3 | 3 2 L 2 2 ) C ( w + L 2 2 + w + L 2 2 + 3 w + L q 13 q / ( 11 q 18 ) + 3 w L q 13 q / ( 11 q 18 ) ) ( w 3 + L 3 + w 3 L 3 )

由上式两边同时除以 ( w 3 + L 3 + w 3 L 3 ) ,并利用能量不等式(10)和式(4)得

w 3 + L ( 0 , T ; L 3 ( R 3 ) ) + w 3 L ( 0 , T ; L 3 ( R 3 ) ) C .

同理定理1第一种情况的证明,可证得 u 3 , b 3 L 3 ( 0 , T ; L 9 ( R 3 ) ) ,所以定理第二种情况得证。

基金项目

安徽省省级自然科学基金(KJ20180002)。

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