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数学物理学报, 2019, 39(5): 1125-1135 doi:

论文

带有分数阶耗散的三维广义Oldroyd-B模型的整体正则性

张秋月,

Global Regularity for 3D Generalized Oldroyd-B Type Models with Fractional Dissipation

Zhang Qiuyue,

通讯作者: 张秋月, E-mail: qyzhang0722@163.com

收稿日期: 2018-07-20  

Received: 2018-07-20  

摘要

该文考虑了三维共旋情形下,带有分数阶耗散(-△)η1u和(-△)η2τ的广义不可压缩Oldroyd-B模型.假设初始数据(u0τ0)充分光滑,若耗散指标η1 ≥ 5/4,η2 ≥ 5/4,利用能量方法,得到了方程组经典解的整体正则性.

关键词: Oldroyd-B模型 ; 分数阶Laplace耗散 ; 整体正则性

Abstract

In this paper, we consider the 3D generalized Oldroyd-B type models with fractional Laplacian dissipation (-△)η1u and (-△)η2τ in the corotational case. By using energy method, for η1 ≥ 5/4 and η2 ≥ 5/4, we obtain the global regularity of classical solutions when the initial data (u0, τ0) are sufficiently smooth.

Keywords: Oldroyd-B type models ; Fractional Laplacian dissipation ; Global regularity

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本文引用格式

张秋月. 带有分数阶耗散的三维广义Oldroyd-B模型的整体正则性. 数学物理学报[J], 2019, 39(5): 1125-1135 doi:

Zhang Qiuyue. Global Regularity for 3D Generalized Oldroyd-B Type Models with Fractional Dissipation. Acta Mathematica Scientia[J], 2019, 39(5): 1125-1135 doi:

1 引言

为了从数学的角度来描述一些诸如血液和高分子聚合物等粘弹性流体的运动, Oldroyd在文献[1]中首次引入了Oldroyd-B模型,该模型后来又被Bird等[2]广泛讨论.这组方程给出了一种典型的本构关系,即流体的应力张量和流体的速度梯度之间满足线性关系,这不符合牛顿粘性定律.有关Oldroyd-B模型更详细的推导可在文献[3]中找到.

d维广义不可压缩Oldroyd-B模型的具体形式如下

{tu+uu+p+ν(Δ)η1u=Kdiv(τ),tτ+uτ+βτ+μ(Δ)η2τ=αDu+Q(u,τ),u=0,u(x,0)=u0(x),τ(x,0)=τ0(x),
(1.1)

这里xRd,t>0,参数ν,μ,β0, K,α>0,耗散指标η1,η20. uRd表示流体的速度, τSd(R)是一个d阶对称方阵,表示流体应力张量的非牛顿部分, pR表示流体的压强. Du是流体速度梯度的对称部分,称为流体的应变张量,即

Du=u+(u)2.

Q(u,τ)是一个给定的双线性项,一般的有

Q(u,τ)=ΩττΩ+b(Duτ+τDu),

其中Ω=u(u)2是流体速度梯度的非对称部分,称为流体的涡度张量,常数b[1,1].b=0,我们称方程组(1.1)是共旋情形下的Oldroyd-B模型. u0(x)τ0(x)是给定的初始数据并且满足u0=0.这里,分数阶Laplace算子(Δ)η可通过傅里叶变换定义如下

^(Δ)η1f(ξ)=|ξ|2η1ˆf(ξ),ˆf(ξ)=(2π)d2Rdeixξf(x)dx.

在该文中,我们记i=xi, (u)ij=jui以及(τ)i=jτij, i,j=1,2,,d,这里用到了Einstein求和法则,即重复指标表示对该指标从1到d的求和.

近年来,对Oldroyd-B模型的数学研究吸引了大量学者的关注,并得到了许多有意义的成果(例如强解的局部或整体适定性,光滑解的爆破准则,经典解的整体正则性以及小初值整体解等结论).在此,我们做以下简要回顾:当Oldroyd-B模型只带有速度耗散时,即方程组(1.1)中ν>0, μ=0, η1=1, Guillopé和Saut在文献[4-5]中验证了方程组的局部解在Hilbert空间Hs(Rd)(s>d2)中的存在唯一性.在临界Besov空间中, Chemin和Masmoudi[6]证明了方程组存在小初值整体解.更多有关小初值整体解的结论可参考文献[7-14].另外, Lions和Masmoudi[15]建立了一般初值条件下,方程组在共旋情形下弱解的整体存在性,但唯一性还未得.对只带有应力张量扩散的Oldroyd-B模型,即方程组(1.1)中ν=0, μ>0, η2=1, Elgindi和Rousset[16]得到了双线性项Q(u,τ)=0时,方程组在二维情形下经典解的整体正则性;进一步地,对任意的双线性项Q(u,τ),他们还验证了方程组小初值光滑解的整体适定性.后来, Elgindi和Liu在文献[17]中将文献[16]中的小初值光滑解的整体适定性推广到了三维情形下,并得到了一个光滑解的爆破准则.当Oldroyd-B模型既带有速度耗散又带有应力张量扩散时,即方程组(1.1)中ν>0, μ>0, η1=η2=1, Constantin和Kliegl在文献[18]中证明了方程组在二维情形下强解的整体适定性.更多有关Oldroyd-B模型及其相关模型弱解的整体存在性以及光滑解的整体适定性结论,可参见文献[19-22].

当粘弹性流体应力张量的非牛顿部分τ=0时,方程组(1.1)就退化成了广义的Navier-Stokes (GNS)方程组.众所周知,当η154时,三维GNS方程组经典解的整体正则性已得到. Wu在文献[23]中研究了当η154, η254时,三维广义磁流体(GMHD)方程组经典解的整体正则性.若η154, η274, Wan在文献[24]中证明了三维广义Hall-MHD方程组存在唯一整体光滑解.受以上文章的启发,本文考虑三维共旋情形下的广义Oldroyd-B模型经典解的整体正则性问题,即方程组(1.1)中d=3b=0.

下面我们给出本文的主要研究结果.

定理1.1  令T>0,考虑三维方程组(1.1)中ν>0, μ>0以及b=0.假设初始数据(u0,τ0)Hs(R3)×Hs(R3)(s>52),并且满足u0=0.η154, η254,则方程组存在唯一整体经典解(u,τ)满足

uL([0,T];Hs(R3))L2([0,T];Hs+η1(R3)),

τL([0,T];Hs(R3))L2([0,T];Hs+η2(R3)).

注1.1  若双线性项Q(u,τ)=0, Wan在文献[25]中证明了二维方程组(1.1)在以下两类次临界情形下经典解的整体适定性: (1) ν>0, μ=0, η1>1, η2=0; (2) ν>0, μ>0, η1=1, η2>0.于是,本文结果推广了Wan的结论,得到了当Q(u,τ)0时,三维广义Oldroyd-B模型在共旋情形下经典解的整体正则性.

注1.2  证明定理1.1的关键在于得到对非线性项uu, uτ以及双线性项Q(u,τ)H1估计.为了解决这些困难,我们需要利用几个Gagliardo-Nirenberg型三插值不等式,即2.2小节中的不等式(3.7)–(3.10),其详细证明在附录中给出.

该文其余部分的结构如下:第二部分给出证明该文结果所需的两个引理;在第三部分中,我们具体给出定理1.1的证明;附录给出不等式(3.7)–(3.10)的详细证明.不失一般性,我们令α=K=1,且C是一个与被估计函数无关的正的常数,其大小随具体情况而定.

2 预备知识

我们首先给出下述Gagliardo-Nirenberg型插值不等式(详见文献[26,引理2.1]),其证明可通过Littlewood-Paley分解理论得到.

引理2.1  令1p, η>d(121p),则存在常数C=C(d,p,η)使得,对任意的d维向量函数fHη(Rd)

(2.1)

p\neq\infty 时,上述不等式对 \eta = d(\frac{1}{2}-\frac{1}{p}) 也成立.

其次,我们给出如下的交换子估计以及乘积的高阶导数估计(读者可参见文献[27]).

引理2.2  令 0<p<\infty , \eta>0 ,则存在一个不依赖于函数 f g 的常数 C 使得,对任意的 f\in\dot{W}^{1, p_{1}}\cap\dot{W}^{\eta, p_{3}} , g\in\dot{W}^{\eta-1, p_{2}}\cap L^{p_{4}} ,有

\begin{eqnarray*} \|[\Lambda^{\eta}, f]g\|_{L^{p}}\leq C(\|\nabla f\|_{L^{p_{1}}}\|\Lambda^{\eta-1}g\|_{L^{p_{2}}}+\|\Lambda^{\eta}f\|_{L^{p_{3}}}\|g\|_{L^{p_{4}}}), \end{eqnarray*}

以及

\|\Lambda^{\eta}(fg)\|_{L^{p}}\leq C(\|f\|_{L^{p_{1}}}\|\Lambda^{\eta}g\|_{L^{p_{2}}}+\|\Lambda^{\eta}f\|_{L^{p_{3}}}\| g\|_{L^{p_{4}}}),

其中 [\Lambda^{\eta}, f]g = \Lambda^{\eta}(fg)-f\Lambda^{\eta}g , \Lambda = (-\Delta)^{\frac{1}{2}} ,且 1\leq p_{1}, p_{4}\leq\infty , 1< p_{2}, p_{3}<\infty 满足

\frac{1}{p} = \frac{1}{p_{1}}+\frac{1}{p_{2}} = \frac{1}{p_{3}}+\frac{1}{p_{4}}.

3 定理1.1的证明

本节致力于证明定理1.1.首先,对于初始值 (u_{0}, \tau_{0})\in H^{s}({\Bbb R} ^{3}) \times H^{s}({\Bbb R} ^{3}) , s>\frac{5}{2} ,方程组(1.1)局部解的存在唯一性可由文献[28,第三章]中的标准方法得到.因此,为了证明本文结论,我们只需得到局部解的先验估计使得 \|(u, \tau)\|_{H^{s}} 一致有界.采用能量方法和几个Gagliardo-Nirenberg型三插值不等式,局部解的正则性可一步一步地提高,进而用迭代方法可得到方程组局部解的高阶导数估计.具体过程如下:

3.1 (u, \tau) L^2 -范数估计

在方程组 (1.1) 前两个方程的两端分别点乘 u \tau ,并在全空间 {\Bbb R} ^3 上对空间变量 x 进行分部积分,再将两式相加,我们得到

\begin{eqnarray} \frac{1}{2}\frac{{\rm d}}{{\rm d}t}(\|u(t)\|_{L^{2}}^{2}+\|\tau(t)\|_{L^{2}}^{2}) +\nu\|\Lambda^{\eta_{1}}u\|_{L^{2}}^{2}+\mu \|\Lambda^{\eta_{2}}\tau\|_{L^{2}}^{2}+\beta \|\tau\|_{L^{2}}^{2} = 0. \end{eqnarray}
(3.1)

由于 \nabla\cdot u = 0 以及 \tau 是一个3阶对称方阵,这里我们用到了以下结论:

\int_{{\Bbb R} ^{3}}(u\cdot\nabla u)\cdot u{\rm d}x = 0, \quad\int_{{\Bbb R} ^{3}}\nabla p\cdot u{\rm d}x = 0, \quad\int_{{\Bbb R} ^{3}}(u\cdot\nabla \tau): \tau {\rm d}x = 0,

\begin{eqnarray*} \int_{{\Bbb R} ^{3}}Du:\tau {\rm d}x & = &\frac{1}{2}\int_{{\Bbb R} ^{3}}(\partial_{i}u_{k}\tau_{ik}+\partial_{k}u_{i}\tau_{ik}){\rm d}x\nonumber\\ & = &-\frac{1}{2}\int_{{\Bbb R} ^{3}}(u_{k}\partial_{i}\tau_{ki}+u_{i}\partial_{k}\tau_{ik}){\rm d}x\nonumber\\ & = &-\int_{{\Bbb R} ^{3}}{\rm div}(\tau)\cdot u{\rm d}x, \end{eqnarray*}

\begin{eqnarray*} \int_{{\Bbb R} ^{3}}Q(\nabla u, \tau): \tau {\rm d}x & = &\frac{1}{2}\int_{{\Bbb R} ^{3}}[(\partial_{i}u_{k}-\partial_{k}u_{i})\tau_{il}\tau_{kl} -\tau_{lk}(\partial_{i}u_{k}-\partial_{k}u_{i})\tau_{li}]{\rm d}x\nonumber\\ & = &\frac{1}{2}\int_{{\Bbb R} ^{3}}[\partial_{i}u_{k}(\tau_{il}\tau_{kl}-\tau_{lk}\tau_{li}) -\partial_{k}u_{i}(\tau_{il}\tau_{kl}-\tau_{lk}\tau_{li})]{\rm d}x\nonumber\\ & = &0, \end{eqnarray*}

其中符号":"表示对任意两个方阵 A = (a_{ij})_{3\times3} B = (b_{ij})_{3\times3} 作内积,即 A:B = a_{ij}b_{ij} .对(3.1)式两端同时关于时间 t [0, T] 上进行积分,可推得

\begin{eqnarray} &&\|u(T)\|_{L^{2}}^{2}+\|\tau(T)\|_{L^{2}}^{2} +2\nu\int_{0}^{T}\|\Lambda^{\eta_{1}}u(t)\|_{L^{2}}^{2}{\rm d}t+2\mu \int_{0}^{T}\|\Lambda^{\eta_{2}}\tau(t)\|_{L^{2}}^{2}{\rm d}t\\ &&+2\beta\int_{0}^{T}\|\tau(t)\|_{L^{2}}^{2}{\rm d}t = \|u_{0}\|_{L^{2}}^{2}+\|\tau_{0}\|_{L^{2}}^{2}. \end{eqnarray}
(3.2)

3.2 (u, \tau) H^1 -范数估计

在方程 (1.1)_1 和方程 (1.1)_2 的两端分别作用 \Lambda ,并分别关于 \Lambda u \Lambda \tau 在全空间 {\Bbb R} ^3 上作内积,对 x 进行分部积分,再将两式相加,我们有

\begin{eqnarray} &&\frac{1}{2}\frac{{\rm d}}{{\rm d}t}(\|\Lambda u(t)\|_{L^{2}}^{2}+\|\Lambda\tau(t)\|_{L^{2}}^{2}) +\nu\|\Lambda^{\eta_{1}+1}u\|_{L^{2}}^{2}+\mu\|\Lambda^{\eta_{2}+1}\tau\|_{L^{2}}^{2} +\beta\|\Lambda\tau\|_{L^{2}}^{2}\\ & = &-\int_{{\Bbb R} ^{3}}\Lambda(u\cdot\nabla u)\cdot \Lambda u{\rm d}x-\int_{{\Bbb R} ^{3}}\Lambda(\nabla p)\cdot \Lambda u{\rm d}x+\int_{{\Bbb R} ^{3}}\Lambda[{\rm div}(\tau)]\cdot \Lambda u{\rm d}x\\ &&-\int_{{\Bbb R} ^{3}}\Lambda(u\cdot\nabla \tau): \Lambda\tau {\rm d}x+\int_{{\Bbb R} ^{3}}\Lambda(Du):\Lambda\tau {\rm d}x +\int_{{\Bbb R} ^{3}}\Lambda[Q(\nabla u, \tau)]: \Lambda\tau {\rm d}x\\ & = &:\sum\limits_{i = 1}^{6}J_{i}. \end{eqnarray}
(3.3)

利用 u 的散度自由条件且 \tau = \tau^{\top} ,通过分部积分,易验证

J_{2} = -\int_{{\Bbb R} ^{3}}\Lambda(\nabla p)\cdot \Lambda u{\rm d}x = 0,

J_{3}+J_{5} = \int_{{\Bbb R} ^{3}}\Lambda[{\rm div}(\tau)]\cdot \Lambda u{\rm d}x+\int_{{\Bbb R} ^{3}}\Lambda(Du):\Lambda\tau {\rm d}x = 0.

于是,在定理1.1的假设下,为了得到 (u, \tau) H^{1} -范数估计,需得到对 J_{1} , J_{4} J_{6} 的估计.为此,我们从以下两个方面进行考虑: (ⅰ) \eta_{1}>\frac{5}{2} , \eta_{2}>\frac{5}{2} ; (ⅱ) \frac{5}{4}\leq\eta_{1}\leq\frac{5}{2} , \frac{5}{4}\leq\eta_{2}\leq\frac{5}{2} .具体地有:

(ⅰ)对于 \eta_{1}>\frac{5}{2} , \eta_{2}>\frac{5}{2} ,我们可采用以下两个Gagliardo-Nirenberg型插值不等式,其证明类似于不等式(2.1)的证明.

\begin{eqnarray*} \|\Delta u\|_{L^{\infty}}\leq C\|u\|_{L^{2}}^{\frac{2\eta_{1}-5}{2(\eta_{1}+1)}}\|\Lambda^{\eta_{1}+1}u\|_{L^{2}}^{\frac{7}{2(\eta_{1}+1)}}, \end{eqnarray*}

\begin{eqnarray*} \|\Delta \tau\|_{L^{\infty}}\leq C\|\tau\|_{L^{2}}^{\frac{2\eta_{2}-5}{2(\eta_{2}+1)}}\|\Lambda^{\eta_{2}+1}\tau\|_{L^{2}}^{\frac{7}{2(\eta_{2}+1)}}. \end{eqnarray*}

再经过分部积分,并应用Hölder和Young不等式,结合(3.2)式, J_{1} , J_{4} J_{6} 可被估计为

\begin{eqnarray} |J_{1}|&\leq&\bigg|\int_{{\Bbb R} ^{3}}u\cdot\nabla u\cdot \Delta u{\rm d}x\bigg| \leq\|u\|_{L^{2}}\|\nabla u\|_{L^{2}}\|\Delta u\|_{L^{\infty}} \\ & \leq& C\|\Lambda^{\eta_{1}+1}u\|_{L^{2}}^{\frac{7}{2(\eta_{1}+1)}}\|\nabla u\|_{L^{2}} \leq \frac{\nu}{2}\|\Lambda^{\eta_{1}+1}u\|_{L^{2}}^{2}+C\|\nabla u\|_{L^{2}}^{\frac{4(\eta_{1}+1)}{4\eta_{1}-3}} \\ & \leq& \frac{\nu}{2}\|\Lambda^{\eta_{1}+1}u\|_{L^{2}}^{2}+C(\|\nabla u\|_{L^{2}}^{2}+1), \end{eqnarray}
(3.4)

\begin{eqnarray} |J_{4}|&\leq&\bigg|\int_{{\Bbb R} ^{3}}u\cdot\nabla \tau\cdot \Delta \tau{\rm d}x\bigg| \leq\|u\|_{L^{2}}\|\nabla \tau\|_{L^{2}}\|\Delta \tau\|_{L^{\infty}} \\ & \leq& C\|\Lambda^{\eta_{2}+1}\tau\|_{L^{2}}^{\frac{7}{2(\eta_{2}+1)}}\|\nabla \tau\|_{L^{2}} \leq \frac{\mu}{4}\|\Lambda^{\eta_{2}+1}\tau\|_{L^{2}}^{2}+C\|\nabla \tau\|_{L^{2}}^{\frac{4(\eta_{2}+1)}{4\eta_{2}-3}} \\ & \leq& \frac{\mu}{4}\|\Lambda^{\eta_{2}+1}\tau\|_{L^{2}}^{2}+C(\|\nabla \tau\|_{L^{2}}^{2}+1), \end{eqnarray}
(3.5)

\begin{eqnarray} |J_{6}|&\leq&\bigg|\int_{{\Bbb R} ^{3}}Q(\nabla u, \tau) \Delta \tau{\rm d}x\bigg| \leq\|\tau\|_{L^{2}}\|\nabla u\|_{L^{2}}\|\Delta \tau\|_{L^{\infty}} \\ & \leq& C\|\Lambda^{\eta_{2}+1}\tau\|_{L^{2}}^{\frac{7}{2(\eta_{2}+1)}}\|\nabla u\|_{L^{2}} \leq \frac{\mu}{4}\|\Lambda^{\eta_{2}+1}\tau\|_{L^{2}}^{2}+C\|\nabla u\|_{L^{2}}^{\frac{4(\eta_{2}+1)}{4\eta_{2}-3}} \\ & \leq &\frac{\mu}{4}\|\Lambda^{\eta_{2}+1}\tau\|_{L^{2}}^{2}+C(\|\nabla u\|_{L^{2}}^{2}+1), \end{eqnarray}
(3.6)

其中,由于 \eta_{1}>\frac{5}{2} , \eta_{2}>\frac{5}{2} ,我们用到了事实 \frac{4(\eta_{1}+1)}{4\eta_{1}-3}<2 , \frac{4(\eta_{2}+1)}{4\eta_{2}-3}<2 .

将估计式(3.4)–(3.6)代入(3.3)式,然后运用Gronwall不等式可得,对任意的 T>0 ,有

\begin{eqnarray*} &&\|\Lambda u(T)\|_{L^{2}}^{2}+\|\Lambda\tau(T)\|_{L^{2}}^{2} +\nu\int_{0}^{T}\|\Lambda^{\eta_{1}+1}u(t)\|_{L^{2}}^{2}{\rm d}t \nonumber\\ &&+\mu\int_{0}^{T}\|\Lambda^{\eta_{2} +1}\tau(t)\|_{L^{2}}^{2}{\rm d}t+2\beta\int_{0}^{T}\|\Lambda\tau(t)\|_{L^{2}}^{2}{\rm d}t\nonumber\\ &\leq& C(\|\Lambda u_{0}\|_{L^{2}}^{2} +\|\Lambda\tau_{0}\|_{L^{2}}^{2}+1)\exp\{CT\}<+\infty. \end{eqnarray*}

(ⅱ)对于 \frac{5}{4}\leq\eta_{1}\leq\frac{5}{2} , \frac{5}{4}\leq\eta_{2}\leq\frac{5}{2} ,我们需要用到以下四个Gagliardo-Nirenberg型三插值不等式

\begin{equation} \|\nabla u\|_{L^{3}}\leq C\|\nabla u\|_{L^{2}}^{\theta_{1}}\|\Lambda^{\eta_{1}} u\|_{L^{2}}^{\theta_{2}} \|\Lambda^{\eta_{1}+1} u\|_{L^{2}}^{\theta_{3}}, \end{equation}
(3.7)

其中 \theta_{1} = 1-\frac{5}{6\eta_{1}} , \theta_{2} = \frac{1}{3} , \theta_{3} = \frac{5}{6\eta_{1}}-\frac{1}{3} 满足 0<\theta_{1}, \theta_{2}, \theta_{3}<1 , \frac{3\theta_{1}}{2}+\frac{3\theta_{3}}{2} = 1 , \theta_{1}+\theta_{2}+\theta_{3} = 1 以及 \theta_{1}+\eta_{1}\theta_{2}+(\eta_{1}+1)\theta_{3} = \frac{3}{2} .

\begin{equation} \|\nabla \tau\|_{L^{3}}\leq C\|\nabla \tau\|_{L^{2}}^{\gamma_{1}}\|\Lambda^{\eta_{2}} \tau\|_{L^{2}}^{\gamma_{2}} \|\Lambda^{\eta_{2}+1} \tau\|_{L^{2}}^{\gamma_{3}}, \end{equation}
(3.8)

\begin{equation} \|\Delta \tau\|_{L^{\frac{3}{2}}}\leq C\|\nabla \tau\|_{L^{2}}^{\gamma_{1}}\|\Lambda^{\eta_{2}} \tau\|_{L^{2}}^{\gamma_{2}} \|\Lambda^{\eta_{2}+1} \tau\|_{L^{2}}^{\gamma_{3}}, \end{equation}
(3.9)

\begin{equation} \|\tau\|_{L^{\infty}}\leq C\|\nabla \tau\|_{L^{2}}^{\gamma_{1}}\|\Lambda^{\eta_{2}} \tau\|_{L^{2}}^{\gamma_{2}} \|\Lambda^{\eta_{2}+1} \tau\|_{L^{2}}^{\gamma_{3}}, \end{equation}
(3.10)

其中 \gamma_{1} = 1-\frac{5}{6\eta_{2}} , \gamma_{2} = \frac{1}{3} , \gamma_{3} = \frac{5}{6\eta_{2}}-\frac{1}{3} 满足 0<\gamma_{1}, \gamma_{2}, \gamma_{3}<1 , \frac{3\gamma_{1}}{2}+\frac{3\gamma_{3}}{2} = 1 以及 \gamma_{1}+\gamma_{2}+\gamma_{3} = 1 , \gamma_{1}+\eta_{2}\gamma_{2}+(\eta_{2}+1)\gamma_{3} = \frac{3}{2} .

因而,由于 \nabla\cdot u = 0 ,经过分部积分,再利用H \ddot{{\rm o}} lder不等式, Young不等式以及不等式(3.7)–(3.10),联合(3.2)式,我们将 J_{1} , J_{4} J_{6} 估计成

\begin{eqnarray} |J_{1}|&\leq& \bigg|\int_{{\Bbb R} ^{3}}\Lambda(u\cdot\nabla u)\cdot \Lambda u{\rm d}x\bigg| \leq\bigg|\int_{{\Bbb R} ^{3}}\partial_{i}u_{k}\partial_{k}u_{j} \partial_{i}u_{j}{\rm d}x\bigg|+\bigg|\int_{{\Bbb R} ^{3}}u_{k}\partial_{k}\partial_{i}u_{j} \partial_{i}u_{j}{\rm d}x\bigg|\\ &\leq&\|\nabla u\|_{L^{3}}^{3}\\ &\leq &C\|\nabla u\|_{L^{2}}^{3\theta_{1}}\|\Lambda^{\eta_{1}} u\|_{L^{2}}^{3\theta_{2}} \|\Lambda^{\eta_{1}+1} u\|_{L^{2}}^{3\theta_{3}}\\ &\leq&\frac{\nu}{8}\|\Lambda^{\eta_{1}+1} u\|_{L^{2}}^{2}+C\|\nabla u\|_{L^{2}}^{2}\|\Lambda^{\eta_{1}} u\|_{L^{2}}^{\frac{2\theta_{2}}{\theta_{1}}}, \end{eqnarray}
(3.11)

\begin{eqnarray} |J_{4}|&\leq&\bigg|\int_{{\Bbb R} ^{3}}\Lambda(u\cdot\nabla \tau): \Lambda\tau {\rm d}x\bigg| \leq\bigg|\int_{{\Bbb R} ^{3}}\partial_{i}u_{k}\partial_{k}\tau_{lj} \partial_{i}\tau_{lj} {\rm d}x\bigg|+\bigg|\int_{{\Bbb R} ^{3}}u_{k}\partial_{k}\partial_{i}\tau_{lj} \partial_{i}\tau_{lj}{\rm d}x\bigg|\\ &\leq&\|\nabla u\|_{L^{3}}\|\nabla \tau\|_{L^{3}}^{2}\\ &\leq& C\|\nabla u\|_{L^{3}}^{3}+C\|\nabla \tau\|_{L^{3}}^{3}\\ &\leq &C\|\nabla u\|_{L^{2}}^{3\theta_{1}}\|\Lambda^{\eta_{1}} u\|_{L^{2}}^{3\theta_{2}} \|\Lambda^{\eta_{1}+1} u\|_{L^{2}}^{3\theta_{3}}+C\|\nabla \tau\|_{L^{2}}^{3\gamma_{1}}\|\Lambda^{\eta_{2}} \tau\|_{L^{2}}^{3\gamma_{2}} \|\Lambda^{\eta_{2}+1} \tau\|_{L^{2}}^{3\gamma_{3}}\\ &\leq&\frac{\nu}{8}\|\Lambda^{\eta_{1}+1} u\|_{L^{2}}^{2}+\frac{\mu}{4}\|\Lambda^{\eta_{2}+1} \tau\|_{L^{2}}^{2} +C\|\nabla u\|_{L^{2}}^{2}\|\Lambda^{\eta_{1}} u\|_{L^{2}}^{\frac{2\theta_{2}}{\theta_{1}}} +C\|\nabla \tau\|_{L^{2}}^{2}\|\Lambda^{\eta_{2}} \tau\|_{L^{2}}^{\frac{2\gamma_{2}}{\gamma_{1}}}, \quad \end{eqnarray}
(3.12)

\begin{eqnarray} |J_{6}|& = &\bigg|\int_{{\Bbb R} ^{3}}\Lambda Q(\nabla u, \tau): \Lambda\tau {\rm d}x\bigg| \leq\frac{1}{2}\int_{{\Bbb R} ^{3}}\left|[(\nabla u-(\nabla u)^{\top})\tau-\tau(\nabla u-(\nabla u)^{\top})]: \Delta\tau\right| {\rm d}x\\ &\leq&2\|\Delta\tau\|_{L^{\frac{3}{2}}}\|\tau\|_{L^{\infty}}\|\nabla u\|_{L^{3}}\\ &\leq &C\|\Delta\tau\|_{L^{\frac{3}{2}}}^{3}+C\|\tau\|_{L^{\infty}}^{3}+C\|\nabla u\|_{L^{3}}^{3}\\ &\leq &C\|\nabla u\|_{L^{2}}^{3\theta_{1}}\|\Lambda^{\eta_{1}} u\|_{L^{2}}^{3\theta_{2}} \|\Lambda^{\eta_{1}+1} u\|_{L^{2}}^{3\theta_{3}}+C\|\nabla \tau\|_{L^{2}}^{3\gamma_{1}}\|\Lambda^{\eta_{2}} \tau\|_{L^{2}}^{3\gamma_{2}} \|\Lambda^{\eta_{2}+1} \tau\|_{L^{2}}^{3\gamma_{3}}\\ &\leq&\frac{\nu}{8}\|\Lambda^{\eta_{1}+1} u\|_{L^{2}}^{2}+\frac{\mu}{4}\|\Lambda^{\eta_{2}+1} \tau\|_{L^{2}}^{2}+C\|\nabla u\|_{L^{2}}^{2}\|\Lambda^{\eta_{1}} u\|_{L^{2}}^{\frac{2\theta_{2}}{\theta_{1}}} +C\|\nabla \tau\|_{L^{2}}^{2}\|\Lambda^{\eta_{2}} \tau\|_{L^{2}}^{\frac{2\gamma_{2}}{\gamma_{1}}}. \end{eqnarray}
(3.13)

将估计式(3.11)–(3.13)代入(3.3)式,可得

\begin{eqnarray*} &&\frac{{\rm d}}{{\rm d}t}(\|\Lambda u(t)\|_{L^{2}}^{2}+\|\Lambda\tau(t)\|_{L^{2}}^{2}) +\nu\|\Lambda^{\eta_{1}+1}u\|_{L^{2}}^{2}+\mu\|\Lambda^{\eta_{2}+1}\tau\|_{L^{2}}^{2} +\beta\|\Lambda\tau\|_{L^{2}}^{2}\nonumber\\ &\leq& C\|\nabla u\|_{L^{2}}^{2}\|\Lambda^{\eta_{1}} u\|_{L^{2}}^{\frac{2\theta_{2}}{\theta_{1}}} +C\|\nabla \tau\|_{L^{2}}^{2}\|\Lambda^{\eta_{2}} \tau\|_{L^{2}}^{\frac{2\gamma_{2}}{\gamma_{1}}}. \end{eqnarray*}

进一步地应用Gronwall不等式,我们得到:对任意的 T>0 , t \in [0, T] ,有

\begin{eqnarray*} &&\|\Lambda u(T)\|_{L^{2}}^{2}+\|\Lambda\tau(T)\|_{L^{2}}^{2} +\nu\int_{0}^{T}\|\Lambda^{\eta_{1}+1}u(t)\|_{L^{2}}^{2}{\rm d}t\\ &&+\mu\int_{0}^{T}\|\Lambda^{\eta_{2} +1}\tau(t)\|_{L^{2}}^{2}{\rm d}t+2\beta\int_{0}^{T}\|\Lambda\tau(t)\|_{L^{2}}^{2}{\rm d}t \nonumber\\ &\leq& C(\|\Lambda u_{0}\|_{L^{2}}^{2} +\|\Lambda\tau_{0}\|_{L^{2}}^{2})\exp \bigg\{\int_{0}^{T}(\|\Lambda^{\eta_{1}}u(t)\|_{L^{2}}^{\frac{2\theta_{2}}{\theta_{1}}} +\|\Lambda^{\eta_{2}}\tau(t)\|_{L^{2}}^{\frac{2\gamma_{2}}{\gamma_{1}}}){\rm d}t\bigg\}. \end{eqnarray*}

又由于 \frac{5}{4}\leq\eta_{1}\leq\frac{5}{2} , \frac{5}{4}\leq\eta_{2}\leq\frac{5}{2} ,可知 \frac{2\theta_{2}}{\theta_{1}}\leq2 , \frac{2\gamma_{2}}{\gamma_{1}}\leq2 ,结合(3.2)式,我们有

\int_{0}^{T}\|\Lambda^{\eta_{1}}u(t)\|_{L^{2}}^{\frac{2\theta_{2}}{\theta_{1}}}{\rm d}t <C(T, \|u_{0}\|_{L^{2}}, \|\tau_{0}\|_{L^{2}}),

\int_{0}^{T}\|\Lambda^{\eta_{2}}\tau(t)\|_{L^{2}}^{\frac{2\gamma_{2}}{\gamma_{1}}}{\rm d}t <C(T, \|u_{0}\|_{L^{2}}, \|\tau_{0}\|_{L^{2}}),

因此,我们推得

\begin{eqnarray*} &&\|\Lambda u(T)\|_{L^{2}}^{2}+\|\Lambda\tau(T)\|_{L^{2}}^{2} +\nu\int_{0}^{T}\|\Lambda^{\eta_{1}+1}u(t)\|_{L^{2}}^{2}{\rm d}t\\ &&+\mu\int_{0}^{T}\|\Lambda^{\eta_{2} +1}\tau(t)\|_{L^{2}}^{2}{\rm d}t +2\beta\int_{0}^{T}\|\Lambda\tau(t)\|_{L^{2}}^{2}{\rm d}t \\ &\leq &C(T, \|\Lambda u_{0}\|_{L^{2}}^{2} , \|\Lambda\tau_{0}\|_{L^{2}}^{2})<+\infty. \end{eqnarray*}

最后,由(ⅰ)和(ⅱ)的结论,联合 (u, \tau) 的基本能量估计(3.2)式,我们得到:若 \eta_{1}\geq\frac{5}{4} , \eta_{2}\geq\frac{5}{4} ,则

\begin{equation} u\in L^{\infty}([0, T];H^{1})\cap L^{2}([0, T];H^{\eta_{1}+1}), \quad\tau\in L^{\infty}([0, T];H^{1})\cap L^{2}([0, T];H^{\eta_{2}+1}). \end{equation}
(3.14)

3.3 (u, \tau) 的高阶导数估计

首先,我们给出 (u, \tau) H^2 -范数估计.在方程 (1.1)_1 和方程 (1.1)_2 的两端分别作用 \Lambda^2 ,并分别关于 \Lambda^2 u \Lambda^2 \tau 在全空间上作内积,对空间变量 x 进行分部积分,再将所得到的等式相加,然后利用H \ddot{{\rm o}} lder不等式,我们有

\begin{eqnarray} &&\frac{1}{2}\frac{{\rm d}}{{\rm d}t}(\|\Lambda^{2} u\|_{L^{2}}^{2}+\|\Lambda^{2}\tau\|_{L^{2}}^{2}) +\nu\|\Lambda^{\eta_{1}+2}u\|_{L^{2}}^{2}+\mu\|\Lambda^{\eta_{2} +2}\tau\|_{L^{2}}^{2}+\beta\|\Lambda^{2}\tau\|_{L^{2}}^{2}\\ &\leq&\|[\Lambda^{2}, u\cdot\nabla]u\|_{L^\frac{4}{3}}\|\Lambda^{2}u\|_{L^{4}} +\|[\Lambda^{2}, u\cdot\nabla]\tau\|_{L^\frac{4}{3}}\|\Lambda^{2}\tau\|_{L^{4}} +\|\Lambda^{2}(Q(\nabla u, \tau))\|_{L^\frac{4}{3}}\|\Lambda^{2}\tau\|_{L^{4}}\\ & = &:L_{1}+L_{2}+L_{3}. \end{eqnarray}
(3.15)

这里我们用到了以下事实

\int_{{\Bbb R} ^{3}}u\cdot\nabla\Lambda^{2} u\cdot \Lambda^{2}u{\rm d}x = 0, \quad\int_{{\Bbb R} ^{3}}u\cdot\nabla\Lambda^{2} \tau\cdot \Lambda^{2}\tau {\rm d}x = 0,

\int_{{\Bbb R} ^{3}}\Lambda^{2}(\nabla p)\cdot \Lambda^{2}u{\rm d}x = 0,

\int_{{\Bbb R} ^{3}}\Lambda^{2}[{\rm div}(\tau)]\cdot \Lambda^{2}u{\rm d}x+\int_{{\Bbb R} ^{3}}\Lambda^{2}(Du)\cdot \Lambda^{2}\tau {\rm d}x = 0.

由于 \eta_{1}\geq\frac{5}{4} , \eta_{2}\geq\frac{5}{4} ,通过引理2.1,我们可以得到下列不等式

\begin{equation} \|\Lambda^{2}u\|_{L^{4}}\leq C\|\Lambda^{2}u\|_{L^{2}}^{1-\frac{3}{4\eta_{1}}}\|\Lambda^{\eta_{1}+2}u\|_{L^{2}}^{\frac{3}{4\eta_{1}}}, \end{equation}
(3.16)

\begin{equation} \|\tau\|_{L^{4}}\leq C\|\tau\|_{L^{2}}^{\frac{1}{4}} \|\nabla\tau\|_{L^{2}}^{\frac{3}{4}}, \end{equation}
(3.17)

\begin{equation} \|\Lambda^{2}\tau\|_{L^{4}}\leq C\|\Lambda^{2}\tau\|_{L^{2}}^{1-\frac{3}{4\eta_{2}}}\|\Lambda^{\eta_{2}+2}\tau\|_{L^{2}}^{\frac{3}{4\eta_{2}}}. \end{equation}
(3.18)

下面我们分别对 L_{1} , L_{2} , L_{3} 进行估计.对于 L_{1} ,利用引理2.2中的交换子估计, Young不等式以及不等式(3.16),又由于 u\in L^{\infty}([0, T];H^{1}) ,我们推得

\begin{eqnarray*} L_{1}&\leq &C\|\nabla u\|_{L^{2}}\|\Lambda^{2} u\|_{L^{4}}^{2}\leq C\|\Lambda^{2}u\|_{L^{2}}^{2-\frac{3}{2\eta_{1}}}\|\Lambda^{\eta_{1}+2}u\|_{L^{2}}^{\frac{3}{2\eta_{1}}}\nonumber\\ &\leq&\frac{\nu}{8}\|\Lambda^{\eta_{1}+2}u\|_{L^{2}}^{2}+C\|\Lambda^{2}u\|_{L^{2}}^{2}. \end{eqnarray*}

同理,利用交换子估计, Young不等式以及不等式(3.16)和(3.18),又因为 u\in L^{\infty}([0, T];H^{1}) , \tau\in L^{\infty}([0, T];H^{1}) , L_{2} 可被估计为

\begin{eqnarray*} L_{2}&\leq& C\|\nabla u\|_{L^{2}}\|\Lambda^{2} \tau\|_{L^{4}}^{2}+C\|\nabla \tau\|_{L^{2}}\|\Lambda^{2} u\|_{L^{4}}\|\Lambda^{2} \tau\|_{L^{4}}\nonumber\\ &\leq &C\|\Lambda^{2}u\|_{L^{2}}^{2-\frac{3}{2\eta_{1}}}\|\Lambda^{\eta_{1}+2}u\|_{L^{2}}^{\frac{3}{2\eta_{1}}} +C\|\Lambda^{2}\tau\|_{L^{2}}^{2-\frac{3}{2\eta_{2}}} \|\Lambda^{\eta_{2}+2}\tau\|_{L^{2}}^{\frac{3}{2\eta_{2}}}\nonumber\\ &\leq&\frac{\nu}{8}\|\Lambda^{\eta_{1}+2}u\|_{L^{2}}^{2}+\frac{\mu}{4}\|\Lambda^{\eta_{2}+2}\tau\|_{L^{2}}^{2} +C\|\Lambda^{2}u\|_{L^{2}}^{2}+C\|\Lambda^{2}\tau\|_{L^{2}}^{2}. \end{eqnarray*}

对于 L_{3} ,由于 \eta_{1}\geq\frac{5}{4} ,我们有如下Gagliado-Nirenberg-Sobolev型插值不等式

\|\Lambda^{3} u\|_{L^{2}}\leq C\|\nabla u\|_{L^{2}}^{\frac{\eta_{1}-1}{\eta_{1}+1}}\|\Lambda^{\eta_{1}+2} u\|_{L^{2}}^{\frac{2}{\eta_{1}+1}},

进一步地,利用乘积的高阶导数估计, Hölder不等式, Young不等式以及不等式(3.16)–(3.18),结合(3.2)式和结论(3.14),我们得到

\begin{eqnarray*} L_{3}& = &\|\Lambda^{2}(Q(\nabla u, \tau))\|_{L^\frac{4}{3}}\|\Lambda^{2}\tau\|_{L^{4}}\nonumber\\ &\leq &C\|\Lambda^{3} u\|_{L^{2}}\|\tau\|_{L^{4}}\|\Lambda^{2} \tau\|_{L^{4}}+C\|\nabla u\|_{L^{2}}\|\Lambda^{2} \tau\|_{L^{4}}^{2}+C\|\nabla \tau\|_{L^{2}}\|\Lambda^{2} u\|_{L^{4}}\|\Lambda^{2} \tau\|_{L^{4}}\nonumber\\ &\leq &C(1+\|\Lambda^{\eta_{1}+2} u\|_{L^{2}})\|\tau\|_{L^{2}}^{\frac{1}{4}} \|\nabla\tau\|_{L^{2}}^{\frac{3}{4}}\|\Lambda^{2}\tau\|_{L^{4}}+C\|\Lambda^{2} \tau\|_{L^{4}}^{2}+C\|\Lambda^{2} u\|_{L^{4}}^{2}\nonumber\\ &\leq&\frac{\nu}{8}\|\Lambda^{\eta_{1}+2} u\|_{L^{2}}^{2} +C\|\Lambda^{2}u\|_{L^{2}}^{2-\frac{3}{2\eta_{1}}}\|\Lambda^{\eta_{1}+2}u\|_{L^{2}}^{\frac{3}{2\eta_{1}}} +C\|\Lambda^{2}\tau\|_{L^{2}}^{2-\frac{3}{2\eta_{2}}}\|\Lambda^{\eta_{2}+2}\tau\|_{L^{2}}^{\frac{3}{2\eta_{2}}}+C\nonumber\\ &\leq&\frac{\nu}{4}\|\Lambda^{\eta_{1}+2}u\|_{L^{2}}^{2}+\frac{\mu}{4}\|\Lambda^{\eta_{2}+2}\tau\|_{L^{2}}^{2} +C(1+\|\Lambda^{2}u\|_{L^{2}}^{2}+\|\Lambda^{2}\tau\|_{L^{2}}^{2}). \end{eqnarray*}

最后,将对 L_{1}, L_{2}, L_{3} 的估计结果代入不等式(3.15),然后应用Gronwall不等式,我们有

\begin{eqnarray*} &&\|\Lambda^{2} u(T)\|_{L^{2}}^{2}+\|\Lambda^{2}\tau(T)\|_{L^{2}}^{2}+\nu\int_{0}^{T}\|\Lambda^{\eta_{1} +2}u\|_{L^{2}}^{2}{\rm d}t\\ &&+\mu\int_{0}^{T}\|\Lambda^{\eta_{2} +2}\tau\|_{L^{2}}^{2}{\rm d}t +2\beta\int_{0}^{T}\|\Lambda^{2}\tau\|_{L^{2}}^{2}{\rm d}t\\ & \leq&(\|\Lambda^{2} u_{0}\|_{L^{2}}^{2}+\|\Lambda^{2}\tau_{0}\|_{L^{2}}^{2})\exp \{CT\}, \end{eqnarray*}

再联合(3.2)式,上述结果意味着

u\in L^{\infty}([0, T];H^{2})\cap L^{2}([0, T];H^{\eta_{1}+2}),

\tau\in L^{\infty}([0, T];H^{2})\cap L^{2}([0, T];H^{\eta_{2}+2}).

有了 \|(u, \tau)\|_{L^{2}} , \|(u, \tau)\|_{H^{1}} 以及 \|(u, \tau)\|_{H^{2}} 的估计之后, (u, \tau) 的更高阶导数估计可通过迭代的方法得到,且更为简便.具体地来说,在估计 \|(u, \tau)\|_{H^{s}}\ (s\geq3) 时,由于 \eta_{1}\geq\frac{5}{4} , \eta_{2}\geq\frac{5}{4} ,可利用以下Sobolev嵌入不等式

\|\tau\|_{L^{\infty}}\leq C\|\tau\|_{H^{s-1}}, \quad\|\nabla u\|_{L^{\infty}}\leq C\|u\|_{H^{s+\eta_{1}-1}},

\|\nabla \tau\|_{L^{\infty}}\leq C\|\tau\|_{H^{s+\eta_{2}-1}}, \quad\|\Lambda^{s+1} u\|_{L^{2}}\leq C\|u\|_{H^{{s+\eta_{1}}}},

再结合Gronwall不等式,利用已得到的低阶导数先验估计可得到:对任意的 T>0 , t\in[0, T] ,局部解 (u, \tau) H^{s} -范数一致有界,具体细节在此省略.至此我们就完成了对定理1.1的证明.

4 附录

为了该文证明的完整性,下面我们给出不等式(3.7)–(3.10)的证明.由于不等式(3.8)和(3.9)的证明分别和不等式(3.7)和(3.10)的证明相似,故在此只具体给出不等式(3.7)和不等式(3.10)的证明.具体地来说,利用Littlewood-Paley分解理论(详见文献[29]),可将 \nabla u 分解成

\nabla u = \sum\limits_{j = -1}^{\infty}\Delta_{j}\nabla u,

其中 \Delta_{j} 表示非齐次局部化傅里叶算子.接着应用Bernstein不等式(可参见文献[30])以及级数型的Young不等式,我们有

\begin{eqnarray} \|\nabla u\|_{L^{3}}&\leq&\sum\limits_{j = -1}^{N}\|\Delta_{j}\nabla u\|_{L^{3}}+\sum\limits_{j>N}\|\Delta_{j}\nabla u\|_{L^{3}}\\ &\leq&\sum\limits_{j = -1}^{N}2^{\frac{3j}{2}}\|\Delta_{j} u\|_{L^{2}}+\sum\limits_{j>N}2^{\frac{3j}{2}}\|\Delta_{j} u\|_{L^{2}}\\ &\leq&\bigg(\sum\limits_{j = -1}^{N}2^{j}\bigg)^{\frac{1}{2}}\bigg(\sum\limits_{j = -1}^{N}2^{2j}\|\Delta_{j} u\|_{L^{2}}^{2}\bigg)^{\frac{1}{2}} +\bigg(\sum\limits_{j>N}2^{2(\frac{3}{2}-\theta)j}\bigg)^{\frac{1}{2}} \bigg(\sum\limits_{j>N}2^{2\theta j}\|\Delta_{j} u\|_{L^{2}}^{2}\bigg)^{\frac{1}{2}}\\ &\leq & C2^{\frac{N}{2}}\|\nabla u\|_{L^{2}}+C2^{(\frac{3}{2}-\theta)N}\|\Lambda^{\theta}u\|_{L^{2}}, \end{eqnarray}
(4.1)

其中 \theta>\frac{3}{2} ,且 N 是一个待定的正实数.

N\doteq\frac{1}{\theta-1}\log_{2}(\frac{C\|\Lambda^{\theta}u\|_{L^{2}}}{\|\nabla u\|_{L^{2}}}) 使得

2^{\frac{N}{2}}\|\nabla u\|_{L^{2}} = C2^{(\frac{3}{2}-\theta)N}\|\Lambda^{\theta}u\|_{L^{2}},

将其代入不等式(4.1),可得

\begin{eqnarray} \|\nabla u\|_{L^{3}}\leq C\|\nabla u\|_{L^{2}}^{1-\frac{1}{2\theta-2}}\|\Lambda^{\theta}u\|_{L^{2}}^{\frac{1}{2\theta-2}}. \end{eqnarray}
(4.2)

运用相似的方法来估计 \|\Lambda^{\theta}u\|_{L^{2}} ,可以得到

\begin{eqnarray} \|\Lambda^{\theta}u\|_{L^{2}}\leq C\|\Lambda^{\eta_{1}}u\|_{L^{2}}^{\eta_{1}+1-\theta}\|\Lambda^{\eta_{1}+1}u\|_{L^{2}}^{\theta-\eta_{1}}, \end{eqnarray}
(4.3)

其中 \eta_{1}<\theta<\eta_{1}+1 ,于是联合不等式(4.2)和(4.3),我们推得

\|\nabla u\|_{L^{3}}\leq C\|\nabla u\|_{L^{2}}^{1-\frac{1}{2\theta-2}}\|\Lambda^{\eta_{1}}u\|_{L^{2}}^{\frac{\eta_{1}+1-\theta}{2\theta-2}} \|\Lambda^{\eta_{1}+1}u\|_{L^{2}}^{\frac{\theta-\eta_{1}}{2\theta-2}}.

由于 \frac{5}{4}\leq\eta_{1}\leq\frac{5}{2} ,可令 \theta = \frac{3\eta_{1}}{5}+1 使得

1-\frac{1}{2\theta-2} = \theta_{1} = 1-\frac{5}{6\eta_{1}}, \quad\frac{\eta_{1}+1-\theta}{2\theta-2} = \theta_{2} = \frac{1}{3}, \quad \frac{\theta-\eta_{1}}{2\theta-2} = \theta_{3} = \frac{5}{6\eta_{1}}-\frac{1}{3},

将其代入上述不等式就得到了不等式(3.7).

同理,应用Littlewood-Paley分解理论, Bernstein不等式以及级数型的Young不等式,我们可得

\begin{eqnarray*} \|\tau\|_{L^{\infty}}&\leq&\sum\limits_{j = -1}^{N}\|\Delta_{j}\tau\|_{L^{\infty}}+\sum\limits_{j>N}\|\Delta_{j}\tau\|_{L^{\infty}}\nonumber\\ &\leq&\sum\limits_{j = -1}^{N}2^{\frac{3j}{2}}\|\Delta_{j} \tau\|_{L^{2}}+\sum\limits_{j>N}2^{\frac{3j}{2}}\|\Delta_{j} \tau\|_{L^{2}}\nonumber\\ &\leq&\bigg(\sum\limits_{j = -1}^{N}2^{j}\bigg)^{\frac{1}{2}}\bigg(\sum\limits_{j = -1}^{N}2^{2j}\|\Delta_{j} \tau\|_{L^{2}}^{2}\bigg)^{\frac{1}{2}} +\bigg(\sum\limits_{j>N}2^{2(\frac{3}{2}-\gamma)j}\bigg)^{\frac{1}{2}} \bigg(\sum\limits_{j>N}2^{2\gamma j}\|\Delta_{j} \tau\|_{L^{2}}^{2}\bigg)^{\frac{1}{2}}\nonumber\\ &\leq & C2^{\frac{N}{2}}\|\nabla \tau\|_{L^{2}}+C2^{(\frac{3}{2}-\gamma)N}\|\Lambda^{\gamma}\tau\|_{L^{2}}, \end{eqnarray*}

其中 \gamma>\frac{3}{2} . N\doteq\frac{1}{\gamma-1}\log_{2}(\frac{C\|\Lambda^{\gamma}u\|_{L^{2}}}{\|\nabla u\|_{L^{2}}}) 使得

2^{\frac{N}{2}}\|\nabla \tau\|_{L^{2}} = C2^{(\frac{3}{2}-\gamma)N}\|\Lambda^{\gamma}\tau\|_{L^{2}},

则有

\|\tau\|_{L^{\infty}}\leq C\|\nabla \tau\|_{L^{2}}^{1-\frac{1}{2\gamma-2}}\|\Lambda^{\gamma}\tau\|_{L^{2}}^{\frac{1}{2\gamma-2}}.

类似的有

\|\Lambda^{\gamma}\tau\|_{L^{2}}\leq C\|\Lambda^{\eta_{2}}\tau\|_{L^{2}}^{\eta_{2}+1-\gamma}\|\Lambda^{\eta_{2}+1}\tau\|_{L^{2}}^{\gamma-\eta_{2}},

其中 \eta_{2}<\gamma<\eta_{2}+1 ,于是联合上述不等式,我们推得

\begin{eqnarray} \|\tau\|_{L^{\infty}}\leq C\|\nabla \tau\|_{L^{2}}^{1-\frac{1}{2\gamma-2}}\|\Lambda^{\eta_{2}}\tau\|_{L^{2}}^{\frac{\eta_{2}+1-\gamma}{2\gamma-2}} \|\Lambda^{\eta_{2}+1}\tau\|_{L^{2}}^{\frac{\gamma-\eta_{2}}{2\gamma-2}}. \end{eqnarray}
(4.4)

由于 \frac{5}{4}\leq\eta_{2}\leq\frac{5}{2} ,可令 \gamma = \frac{3\eta_{2}}{5}+1 ,于是有

1-\frac{1}{2\gamma-2} = \gamma_{1} = 1-\frac{5}{6\eta_{2}}, \quad\frac{\eta_{2}+1-\gamma}{2\gamma-2} = \gamma_{2} = \frac{1}{3}, \quad \frac{\gamma-\eta_{2}}{2\gamma-2} = \gamma_{3} = \frac{5}{6\eta_{2}}-\frac{1}{3},

将其代入不等式(4.4)就得到了不等式(3.10).

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